## Notes of Peter Haissinsky’s lecture

Hyperbolic groups with planar ideal boundaries

I will prove the following theorem.

Theorem 1 Let ${G}$ be a word hyperbolic group whose boundary is homeomorphic to a proper subset of the 2-sphere. Then the following are equivalent.

1. ${G}$ is virually a convex cocompact Kleinian group.
2. ${\mathrm{dim}_{AR}(\partial G)<2}$.
3. ${G}$ acts cellularly and geometrically on a ${CAT(0)}$ cube complex.

This allows us to give new proofs of the following results.

Corollary 2 (Bonk-Kleiner) If ${\partial G}$ is homeomorphic to the Sierpinski carpet, then ${G}$ is virtually Kleinian if and only if ${\mathrm{dim}_{AR}(\partial G)<2}$.

(note that no proof of this longstanding announcement has appeared yet).

Corollary 3 (Markovic) If ${\partial G}$ is homeomorphic to the 2-sphere and ${G}$ acts cellularly and geometrically on a ${CAT(0)}$ cube complex, then ${G}$ is virtually a uniform lattice of ${PSL(2,\mathbb{C})}$.

1. Tools

1.1. Kleinian groups

Kleinian means discrete group of isometries of ${\mathbb{H}^3}$. Such a group ${G}$ acts conformally on the 2-sphere, with a limit set ${\Lambda_G}$ and an ordinary set ${\Omega_G}$ where the action is properly discontinuous. If ${G}$ is convex-cocompact, ${G\setminus(\mathbb{H}^3 \cup \Omega_G)}$ is a compact 3-manifold with boundary.

I will need Thurston’s hyperbolization theorem in the following form.

Theorem 4 (Thurston) If ${M}$ is a compact 3-manifold with non empty boundary and word hyperbolic fundamental group, then ${M=M_G}$ for some Kleinian group ${G}$.

1.2. Conformal gauge

Gromov, Coornaert, Bowditch. The ideal boundary ${\partial G}$ of a word hyperbolic group ${G}$ carries a quasi-symmetry (in fact, quasi-Möbius) class of metrics whch are Ahlfors-regular

Definition 5 (Bourdon-Pajot) The Ahlfors-regular conformal dimension ${\mathrm{dim}_{AR}(\partial G)}$ is the infimum of Hausdorff dimensions of Ahlfors-regular metrics in the gauge.

(1)${\Rightarrow}$(2) in our Theorem is due to D. Sullivan.

(1)${\Rightarrow}$(3) in our Theorem is due to N. Bergeron and D. Wise.

2. Sketch of proof of (2)${\Rightarrow}$(1)

2.1. Planar actions

First treat a special case : planar actions. Assume that the given homeomorphism ${\partial G \rightarrow \Lambda\subset S^2}$ has the following extra property: for any connected component ${\Omega}$ of the complement of ${\Lambda}$, and any ${g\in G}$, there exists an other component ${\Omega'}$ such that ${g(\partial\Omega)=\partial\Omega'}$.

The heart of the proof is the following

Proposition 6 Let ${G}$ be one-ended with ${\mathrm{dim}_{AR}(\partial G)<2}$. Assume ${G}$ admits a planar action. Then the action is conjugate to that of a Kleinian group.

2.2. JSJ decomposition

Theorem 7 (Bowditch) Let ${G}$ be a word hyperbolic group. Then ${G}$ acts on a tree ${T}$ with finite quotient and no edge inversions. Furthermore,

1. Edge stabilizers are virtually cyclic.
2. Vertex stabilizers belong to one of the following classes
1. Virtually cyclic,
2. Virtually free and preserving a canonical cyclic order on adjacent edges.
3. Rigid, i.e. quasi-convex, non elementary, and not in former classes.

The keypoint of reduction from general case to special case (planar action) is the following

Proposition 8 Let ${G}$ be one ended and let ${\partial G\rightarrow\Lambda}$ be a planar embedding. Then the action of any rigid vertex subgroup of the JSJ decomposition of ${G}$ on ${\Lambda}$ extends to a convergence action on the whole 2-sphere.

This provides us with a compact 3-manifold with boundary for each rigid vertex.

The next steps (getting further pieces for non rigid vertices and for edges, gluing them together) require to improve slightly the JSJ decomposition. One needs

1. Elementary vertex groups are cyclic and fix components of the complement of their limit sets.
2. Surface (with boundary) vertex groups act freely.
3. Rigid vertex groups are torsion free.

Proposition 9 If all rigid vertex groups have conformal dimension ${<2}$, then there exists a finite index subgroup ${H\subset G}$ with a regular JSJ decomposition.

The proof relies on Agol and Wise’s hierarchical decompositions.

NB: The proof gives a slightly stronger result in case ${G}$ has no 2-torsion: It suffices to assume that Sierpinsky carpet subgroups have conformal dimension ${<2}$.

3. Rigid vertex groups are planar

3.1. Special case : rigid groups

When ${G}$ itself is rigid, i.e. it has a trivial JSJ decomposition, ${\partial G}$ has no cut points, so it is a Sierpinsky carpet. The planar action assumption is automatically satisfied. Indeed, components of the complement are disks bounded by peripheral circles embedded in the carpet. Any self-homeomorphism of the carpet permutes peripheral circles, and thus permutes components of the complement.

3.2. General case

Being a rigid vertex subgroup in the JSJ decomposition of a group is a weaker assumption. Components of the complement of ${\Lambda}$ are simply connected, the Riemann mapping onto them extends continuously to the boundary, but the extension may be non injective, in case ${\Lambda}$ has cut points. One must get rid of them.

Each time two cut points arise on the boundary of the same component, join them with an arc and pinch it. Show that the resulting space is again a 2-sphere, that components of the complement of the limit set are disks. By Gabai’s theorem, the action extends to these disks.