Is every Cantor set $C\subseteq\mathbb R^{\infty}$ the limit set of a Fuchsian group?

Question

Is every Cantor set $C\subseteq\mathbb R^{\infty}$ the limit set of a Fuchsian group?

Let's recall the definitions. A Fuchsian group $G$ is a discrete subgroup of $\operatorname{PSL}(2,\mathbb R)=\operatorname{SL}(2,\mathbb R)/\{\pm 1\}$. This acts on the upper half plane $\mathbb C^+=\{ z:\operatorname{Im}z>0\}$ by linear fractional transformations: $$ \begin{pmatrix}a & b\\ c& d\end{pmatrix} \cdot z = \frac{az+b}{cz+d} $$ The limit set $L$ can be defined as the collection of limits of the form $x=\lim A_n\cdot z\in\overline{\mathbb C^+}$, with distinct group elements $A_n\in G$ (and not necessarily distinct points $A_n\cdot z$, so for example a point with infinite stabilizer is in $L$).

Some basic observations about $L$ are: $L\subseteq\mathbb R^{\infty}$, $L$ is closed and invariant, and $L=\emptyset$, $L=\{ x\}$, $L=\{ x,y\}$, $L=\mathbb R^{\infty}$, or $L$ is a nowhere dense set with no isolated points.

My question is about the last case; in all other cases, it is easy to see that all possibilities occur. So, given such a $C\subseteq\mathbb R^{\infty}$, will there be a Fuchsian group $G$ with $L_G=C$?

I suspect this will be well known to experts; in this case, I would also appreciate pointers to the literature.

Edit: Moishe Kohan points out in the comments that the answer to the question as asked is no, due to "well known" (to those who know them well) extra restrictions on $L$, and suggests that there may be no good answer to the question of what sets $C$ exactly occur as limit sets.

Feel free of course to answer anyway (in the answer box) if you have something interesting to say.

Answer 1

Expanding on my comments, here are some obstructions coming from Hausdorff dimension and self-similarity.

1. One observation is that every nonelementary Kleinian group $\Gamma$ has positive critical exponent $\delta$. Furthermore, $\delta$ equals the Hausdorff dimension of the conical limit set of $\Gamma$, $\Lambda^c$, the subset of the limit set $\Lambda$ consisting of conical limit points. In particular, limit set cannot have zero Hausdorff dimension. Now, take a Cantor subset $C\subset \mathbb R$ which has zero Hausdorff dimension (see e.g. here for a construction). Then $C$ cannot be the limit set of a Fuchsian group.

2. Take two Schottky groups $\Gamma_1, \Gamma_2< PSL(2,\mathbb R)$ with disjoint limit sets $\Lambda_i, i=1,2$, that have different Hausdorff dimension. Then $C=\Lambda_1\cup \Lambda_2$ cannot be the limit set of a Fuchsian group (for instance, because of self-similarity of a limit set: It has the same "local" Hausdorff dimension everywhere).

With more thought, one can surely get other examples.

On the other hand, every nonempty compact subset $C\subset \mathbb R$ can be realized the Hausdorff-limit of a sequence of limit sets of Fuchsian groups $\Gamma_n$ (i.e. $\lim_{n\to\infty} d_{Haus}(C, \Lambda(\Gamma_n))=0$).

For proofs of basic facts about Hausdorff dimension of limit sets and critical exponents, see for instance

Nicholls, Peter J., The ergodic theory of discrete groups, London Mathematical Society Lecture Note Series, 143. Cambridge etc.: Cambridge University Press. xi, 221 p. (1989). ZBL0674.58001.

It is a bit dated, but, still the best textbook on this subject.