# Can the limit set of an infinitely generated Schottky group have positive area?

Dear Mathoverflow Community,

Suppose that $\Omega$ is a domain in the Riemann Sphere $\widehat{\mathbb{C}}$ with $\infty \in \Omega$, and assume that every connected component of $\partial \Omega$ is a round circle.

Let $\Gamma(\Omega)$ denote the Schottky group of $\Omega$, that is, the free group of Möbius and anti-Möbius transformations generated by the family of reflections across the boundary circles.

Question : Can the limit set $\widehat{\mathbb{C}} \setminus \bigcup_{T \in \Gamma(\Omega)} T(\Omega)$ have positive area?

Remark : If there are only finitely many boundary circles, then the limit set is a Cantor set and it is not difficult to prove that it must have zero area. We can therefore assume that there are infinitely many boundary circles.

Thank you

• Which definition of "area" are you using? Lebesgue measure and (2-dimensional) Hausdorff measure are the natural choices. These can differ for finitely generated Kleinian groups. – Ian Agol May 24 '17 at 22:26
• "Area" certainly refers to Lebesgue measure. Hausdorff measure is natural to consider but it would be awkward to call it "area". – YCor May 25 '17 at 10:14
• I indeed was refer to Lebesgue measure, but aren't they proportional? – Malik Younsi May 25 '17 at 20:06
• @Karim: Okay, you're right, I was confused. There are limit sets of finitely generated Kleinian groups with Hausdorff dimension 2, but Lebesgue/Hausdorff measure 0. So I suppose a separate question is whether the Hausdorff dimension can be 2? – Ian Agol May 26 '17 at 16:31
• @IanAgol No problem! As for dimension 2, I think that Theorem 5.3 in the paper by Startmann and Urbanski cited in Igor's answer gives such an example. – Malik Younsi May 26 '17 at 17:54

## 3 Answers

Two relevant references:

1. W. Abikoff, Some remarks on Kleinian groups. 1971 Advances in the Theory of Riemann Surfaces (Proc. Conf., Stony Brook, N.Y., 1969) pp. 1–5. Ann. of Math. Studies, No. 66. Princeton Univ. Press, Princeton, N.J.

Among other things, he constructs an infinitely generated free Kleinian subgroup $\Gamma< PSL(2,C)$ whose limit set is a Jordan curve of positive planar measure.

It is worth looking more closely at his construction to see if it can be made using fundamental domains with a single accumulation point of boundary faces.

Remark 1. I looked: Abikoff's argument is a variation on the construction in Remark 2 and, hence, is useless for your purposes.

1. K. Matsuzaki, The Hausdorff dimension of the limit sets of infinitely generated Kleinian groups. Math. Proc. Cambridge Philos. Soc. 128 (2000), no. 1, 123–139.

In Theorem 2' he proves that if $\Gamma< PSL(2,R)$ is a discrete subgroup such that the convex core $C$ of $H^2/\Gamma$ consists of infinitely many boundary loops $c_i$ which satisfy $$\sum_{i} \ell(c_i)^{1/2}<\infty$$ then the limit set of $\Gamma$ (in $S^1$) has positive linear measure. (Here $\ell$ denotes the length of a curve) Using this it is easy to construct an example of an infinite rank Schottky subgroup of $PSL(2,R)$ whose fundamental domain has unique accumulation point (of pairwise disjoint boundary arcs) and such that the limit set has positive linear measure. This is a 1-dimensional version of an example you are asking for.

In order to construct an example, note that there exists a convex infinitely-sided right-angles polygon $P\subset H^2$ whose boundary is connected, whose ideal boundary is a single point, such that $\partial P$ is a concatenation of alternating odd and even edges $e_i$, $I\in {\mathbb N}$, such that the sequences of hyperbolic lengths $$(\ell(e_2i))_{i>0}, (\ell(e_2i))_{i<0}$$ converge to zero as fast as you wish. Thus, you can assume that $$\sum_{i\in {\mathbb Z}} \ell(e_{2i})^{1/2}<\infty.$$ Now, take the subgroup of isometries of $H^2$ generated by reflections in the odd-numbered edges of $P$. Let $\Gamma$ denote the orientation preserving index 2 subgroup of this reflection group. The convex core $C$ of $H^2/\Gamma$ is isometric to the surface obtained by doubling $P$ across its odd-numbered edges. Hence, $\Gamma$ satisfies Matsuzaki's condition.

Remark 2. There is a variety of inequivalent definitions of Schottky groups in the literature. An easy and well-known construction (which some people call "Schottky" but you do not!) is the following:

Start with a compact nowhere dense subset $K\subset R^2$ of positive measure (say the product of two thick Cantor sets in $R$). Let $D_i\subset R^2$ be a collection of pairwise disjoint closed disks disjoint from $K$, such that the closure of $$\bigcup_{i} D_i$$ contains $K$. Now, take the group $\Gamma_K< Mob(S^2)$ generated by inversions in the boundary circles of the disks $D_i$. Its limit set will contain $K$ and, hence, will have positive measure. By taking $K$ totally disconnected one obtains examples with totally disconnected limit sets.

I think this is what Rich Schwartz had/has in mind. Furthermore, the examples of Stratmann and Urbanski mentioned by Igor are variations on this construction (there is a minor and inessential difference that instead of reflections they use pairwise matchups wings of "isometric spheres"): Their $K$, while it may have zero measure, is never a singleton, it always has positive Hausdorff dimension. (You have to dig through the proof of Theorem 5.3 of their paper in order to understand this, their $W$ is my $K$.)

This is all fine and well, but you want the union of disks to accumulate at a single point in $S^2$. Here is the essential difference between the two classes of groups: The above example $\Gamma_K$ will have dissipative action on the limit set (i.e. there is a measurable wondering domain of positive measure, namely, $K$). On the other hand, you are asking for a Schottky group such that the action on the limit set is conservative, i.e. admits no measurable wondering domains (this is not immediate, but follows from Sullivan's work; the key thing is that the closure of the $H^3$-fundamental domain in your case intersects the limit set in a subset of measure zero, since it is a singleton).

Edit. Here is a conjectural construction (in arbitrary dimension), but making it work requires doing some computations and I do not have time for this.

Let $\{D_i\}_{i\in {\mathbb N }}$ be a collection of pairwise disjoint closed round disks in the $n$-dimensional sphere which accumulate to a single point $p\in S^n$. Suppose that this collection has the following property: Given $R>0$ let $B(p,R)\subset S^n$ denote the $R$-ball centered at $p$ and set $$F_R:= B(p,R) - \bigcup_{i\in {\mathbb N }} D_i.$$ Assume that $$\lim_{R\to 0} \frac{Vol(F_R)}{R^n}= 0.$$ Conjecture. Then the discrete group $\Gamma$ of Moebius transformations generated by inversions in the spheres $\partial D_i$ has limit set $\Lambda$ of positive $n$-dimensional measure and, moreover, the point $p$ is a density point of the limit set: $$\lim_{R\to 0} \frac{mes_n(\Lambda \cap B(p,R))}{Vol(B(p,R))} = 1$$

• Thank you very much for the interesting references, I'll look at them closely, especially Abikoff's construction. – Malik Younsi May 26 '17 at 18:02
• Thanks for the remarks, that is very helpful. I knew what I had in mind for a Schottky group was different than some definitions in the literature, but your answer makes the distinction much more clear. I do want the union of the closed disks to be compact. – Malik Younsi May 26 '17 at 20:04
• You mention that Abikoff's construction is useless for my purposes. I don't know if it is the construction you had in mind, but I found one construction in the book Kleinian groups and Uniformization in examples and problems, Example 36, based on Osgood's construction of a positive area Jordan curve as the closure of countably many linear segments. Packing pairwise tangent disks on those segments leads to a Schottky group whose limit set has positive area. Do you think that construction could be modified to give what I want (maybe shrinking the disks a bit to make them disjoint)? – Malik Younsi May 26 '17 at 20:08
• @Kalim: It is the same construction as in Abikoff's paper. I do not believe you can modify it to suit your needs. The main thing is that the positive area comes from the boundary of a fundamental domain. This is not what you want. – Misha May 26 '17 at 20:44
• It thus seems that the construction of such a domain remains open. I accepted your answer since it is very detailed and contains interesting information on relevant related constructions. Thank you! – Malik Younsi May 30 '17 at 18:14

Yes. This is due Z.-X. He, and in greater generality to Rich Schwartz, see

Schwartz, Richard, The limit sets of some infinitely generated Schottky groups, Trans. Am. Math. Soc. 335, No.2, 865-875 (1993). ZBL0815.30033.

For a packing with positive area, the following is given by this same Rich Schwartz (private communication):

Suppose you look at the Schottky group generated by inversions in disks of the Apollonian gasket. Call this union of disks $D_0.$ Note that $D_0$ has full measure in the sphere. Let $D_1$ be the unions, taken over all disks of $D_0,$ of the images of $D_0$ under inversion in the disks of $D_0.$ Then $D_1$ also has full measure in the sphere and $D_1 \subset D_0.$ Next define $D_2 \subset D_1,$ etc. The nested intersection $D_0 \cap D_1 \cap D_2\dots$ again has full measure in the sphere and I think that every point in this intersection is a point of the limit set.

• In the paper, he writes "We remark that it is well known that one can produce (infinitely generated) Schottky groups whose limit sets have positive volume". Of course, this doesn't help with finding a reference. But, can't you build an example by hand, just make denser and denser packings with tiny radii as you approach some point of the boundary? – Lasse Rempe-Gillen May 24 '17 at 11:17
• @LasseRempe-Gillen Yes it does, one can write to Rich and ask :) – Igor Rivin May 24 '17 at 14:53
• @LasseRempe-Gillen See the edit - I did what I had suggested. – Igor Rivin May 24 '17 at 21:04
• @Kalim It should not matter whether the discs are disjoint or not in my construction. Just make them a little bit smaller than in the packing, but closer and closer to it as you go to the boundary. – Lasse Rempe-Gillen May 26 '17 at 14:50
• @Kalim: It is due to Abikoff. – Misha May 26 '17 at 18:43

Almost surely such examples are constructed by Stratmann and Urbanski:

Stratmann, Bernd O.; Urba\'nski, Mariusz, Pseudo-Markov systems and infinitely generated Schottky groups, Am. J. Math. 129, No. 4, 1019-1062 (2007). ZBL1121.37014.

I say "almost surely" because they only compute the Hausdorff dimension (which can be as high as $2$ in your setting), not the Hausdorff measure.

• Interesting reference, thanks! However, it is not clear to me that the authors' condition of "rapid decay" can produce limit sets of positive area. I need to think more about this, but if I had to guess, I would say that their examples always give zero area. – Malik Younsi May 23 '17 at 17:18