Is every finitely generated classical Schottky group quasifuchsian?

Question

$\DeclareMathOperator\PSL{PSL}$(Classical, finitely generated) Schottky groups are groups generated by finitely many hyperbolic elements of $A_i\in \PSL(2,\mathbb{C}), $ $i<n$ such that the isometric circles of $\{A_i,A_i^{-1}\}_{i<n}$ are pairwise disjoint.

Quasi-Fuchsian groups are discrete subgroups $\Gamma$ of $\PSL(2, \mathbb{C})$ such that their limit set is contained in a Jordan curve, invariant under $\Gamma$.

Are there any examples of Schottky groups that are not quasi-Fuchsian?

It is a theorem of Denjoy-Riesz that any Cantor set on a plane is contained in a Jordan curve. Can the Jordan curve in this case be taken to be invariant under $\Gamma$?

Answer 1

Yes - every Schottky group is quasi-fuchsian. See Lemma 1 of Chuckrow's paper "On Schottky Groups with Applications to Kleinian Groups" published in Annals of Mathematics, 1968.

The argument there is nice. Start with a different, classical, Schottky group $\Gamma'$ where all of the circles are perpendicular to one, given round circle $C'$. So $\Gamma'$ is fuchsian (and thus quasi-fuchsian). Now use the fact (!) that the space of Schottky groups is connected - so we are given a quasi-conformal map conjugating $\Gamma'$ to the desired group $\Gamma$ and taking $C'$ to the desired Jordan curve $C$.

This proof is very slick, but does use some machinery. There is a more direct proof given by "connecting-the-dots". (However, in some sense, this proof is not really different.) For the sake of an easy life, let's restrict to the classical case. "Carefully" choose a pair of points in each isometric circle. (The points need to match up.) Since there are $2n$ circles, this gives us $4n$ points. Now "carefully" choose $4n$ (smooth) arcs meeting the isometric circles only in their endpoints, which are at the chosen points. (The arcs need to connect the points in a good cyclic order.) Now act on these arcs using the group and take the union. The result is the desired Jordan curve.