Hausdorff Dimensions of Limit set of subgroups of SL(2,Z)

Question

In a recent paper by Bourgain, Sarnak, Gamburd [1] talks about subgroups of $SL(2,\mathbb{Z})$.

Let $\Lambda$ be a finitely generated non-elementary subgroup of $SL(2,\mathbb{Z})$ with Hausdorff dimension $\delta(\Lambda) > \frac{1}{2}$ ...

Are these necessarily Fuchsian groups?

I would like to know what it means for such a group to have a Hausdorff dimension and whether any examples which have known $\delta$ (larger or smaller than $ \frac{1}{2}$). Guessing here we identify these $2 \times 2$ matrices as fractional linear transformations in $PSL(2, \mathbb{Z})$ acting on $\mathbb{H}$.

With some searching I found that $SL(2, \mathbb{Z})$ is a Fuchsian group itself with a limit set of $\mathbb{R}$. This seems to be related to the fact that all real numbers have continued fraction expansions.

Apparently there are subgroups $\Lambda \subset SL(2, \mathbb{Z}) $ where the limit set is smaller than $\mathbb{R}$ - in fact a Cantor set. What are some known examples where this is the case?

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Here are notes handling the case where the limit set has $0,1,2$ points. Referring is to Svetlana Katok's book for the others. The notes claim:

$$ z \mapsto 4z, z \mapsto \frac{7z-6}{3z-2} $$

is a Fuchsian group with a cantor set as a limit set. Now I am worried these are not subgroups of $SL(2,\mathbb{Z})$, but $PSL(2,\mathbb{Z})$.

Are there any limit sets with Hausdorff dimension known to be on either side of $\frac{1}{2}$?

Answer 1

Added later: One key word you might pursue is Patterson-Sullivan theory, which among other things equates the Hausdorff dimension of the limit set of a geometrically finite group action on hyperbolic space and the so-called "exponent of convergence" for that group action.

Original answer: Here is an answer to the question about known examples of Fuchsian subgroups of $SL(2,\mathbb{Z})$ whose limit set is a Cantor set, using a simple basic construction.

The action of $SL(2,\mathbb{Z})$ preserves the Farey tiling by ideal triangles. Dual to the Farey tiling is a trivalent tree $T$ which is invariant under the action. Elements $M \in SL(2,\mathbb{Z})$ with $|Trace(M)| \ge 2$ have an invariant line $L$ (meaning a bi-infinite edge path) in this tree. If $|Trace(M)| =2$ then $M$ acts "parabolically" and $L$ is a "quasi-horocycle" in the hyperbolic plane, having both ends converging to the same rational point at infinity. And if $|Trace(M)| > 2$ then $M$ is "loxodromic" and $L$ is a "quasi-geodesic" in the hyperbolic plane, having ends converging to distinct quadratic irrationalities (i.e. numbers with periodic continued fraction expansions).

Given two loxodromic transformations $M_1,M_2$ with distinct invariant lines $L_1,L_2$, the intersection $L_1 \cap L_2$ must be a finite segment $\alpha$, or a point, or the empty set. If the intersection is a finite segment $\alpha$, and if for each $i$ there exists a subsegment $D_i \subset L_i$ which is a fundamental domain for the action of $M_i$ whose interior contain $\alpha$, then $M_1,M_2$ freely generate a rank $2$ Fuchsian group (if this was not already true for $M_1,M_2$, it will become true after replacing $M_1,M_2$ by sufficiently high powers). All the groups obtained from this construction are "convex cocompact" Fuchsian groups, meaning that they act cocompactly on the convex hull of their limit set. Also, if the intersection $L_1 \cap L_2$ is a point or the empty eset then $M_1,M_2$ automatically generate a convex cocompact Fuchsian group.

From convex cocompactness of a Fuchsian group which is free of rank $2$, it automatically follows that the limit set is homeomorphic to the end space aka Gromov boundary of the rank $2$ free group, which is a Cantor set.

Answer 2

One may obtain subgroups whose limit sets are arbitrarily close to $0$ or to $1$.

Consider a torsion-free subgroup $\Lambda < PSL_2(\mathbb{Z})$ of genus $>0$, then $b_1(\Lambda) >0$. So there is a homomorphism $\Lambda\to \mathbb{Z}$ (let's say which is trivial on parabolic subgroups), and let $\Lambda'$ be the kernel of this homomorphism. The space $\mathbb{H}^2/\Lambda'$ has limit set all of $S^1=\partial_\infty\mathbb{H}^2$, in particular with Hausdorff dimension $1$. Moreover, $\mathbb{Z}$ acts on $\mathbb{H}^2/\Lambda'$ in such a way that there is a finite-area fundamental domain $D$ for this action with compact geodesic boundary. Now, take a union $D_n= D\cup 1(D)\cup \cdots \cup n(D)$. The claim is that $\pi_1(D_n)<\Lambda'$ has Hausdorff dimension of the limit set approaching $1$. One way to see this is via a geometric convergence argument. One may also see that $D_n$ has Cheeger constant approaching zero. Hence the smallest eigenvalue of the Laplacian $\lambda_n$ approaches zero by Cheeger's inequality. But $\lambda_n=\delta_n(1-\delta_n)$, so $\delta_n\to 1$ (since it is an increasing sequence).

On the other hand, the construction described in Mosher's answer will give subgroups with Hausdorff dimension $\to 0$.