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



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


2. 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:
$$
\lim\_{R\to 0} \frac{mes\_n(\Lambda \cap B(p,R))}{Vol(B(p,R))} = 1 
$$ 


