## Notes of Matias Carrasco’s lecture

Conformal dimension and canonical splittings of hyperbolic groups

Can one characterize hyperbolic groups whose conformal dimension equals one ? For the definition of conformal dimension, see Haissinsky’s talk.

1. The Kleinian dimension

For Kleinian groups, an other dimension-like invariant arises, the infimum ${D(G)}$ of Hausdorf dimensions of limit sets of fundamental groups of Kleinian manifolds homeomorphic to ${M_G}$.

By definition, ${\mathrm{dim}_{AR}(G)\leq D(G)}$. Groups with ${D(G)=1}$ have a nice characterization.

Theorem 1 (Canary-Minsky-Taylor) ${D(G)=1}$ if and only if ${M}$ is a generalized book of ${I}$-bundles, i.e. there exists a collection ${A}$ of essential annuli in ${M}$ such that all closures of connected components of their complement are either

1. Solid tori.
2. ${I}$-bundles.

In particular, there is no rigid vertex in the JSJ decomposition of ${G}$.

Question. ${\mathrm{dim}_{AR}(G)=D(G)}$ ?

We shall see that the answer is no.

2. Splittings

In general, one understands ideal boundaries of groups which admit quasiconvex splittings: ${\partial G}$ contains one copy of the ideal boundary of each vertex group ${\partial G_v}$, ${v}$ a vertex of the splitting tree ${T}$, whose size decays as ${v}$ tends to infinity, and is compactified by adjoining ${\partial T}$. We shall first split along finite groups, and then along virtually cyclic groups.

Theorem 2 (Stability under Dunwoody-Stalling splitting) Recall that a Dunwoody-Stalling splitting is a splitting over finite groups with one-ended vertex groups. In this case,

$\displaystyle \begin{array}{rcl} \mathrm{dim}_{AR}(G)=\max\{\mathrm{dim}_{AR}(G_v)\,;\,|G_v|=\infty\}. \end{array}$

Theorem 3 (Well spread local cut points condition) Let ${G}$ be one-ended. Assume

$\displaystyle \begin{array}{rcl} (WS)\quad &&\forall \delta>0,~\exists P_{\delta}\subset\partial G \textrm{ finite subset such that }\\ &&\sup\{\mathrm{diam}(A)\,;\,A\textrm{ connected component of }\partial G\setminus P_{\delta}\}\leq\delta. \end{array}$

Then

$\displaystyle \begin{array}{rcl} \mathrm{dim}_{AR}(G)=1. \end{array}$

Recall Bowditch’s JSJ decomposition for hyperbolic groups (see Haissinsky’s talk).

Corollary 4 If all vertex groups in the JSJ decomposition are virtually free, then ${\partial G}$ satisfies condition (WS). Therefore ${\mathrm{dim}_{AR}(G)=1}$.

Example 1 (of a group with conformal dimension 1 and ${D(G)>1}$) Remove a disk to a 2-torus, take a product with an interval. Then glue to it a solid torus and two ${I}$-bundles. This satisfies (WS) but it is not a generalized ${I}$-bundle.

Remark 1 If there exists a rigid vertex groups whose ideal boundary is a circle, then ${\partial G}$ does not satisfy (WS).

It would desirable to fully understand the following family of examples.

Example 2 Glue two surfaces along filling geodesics.

3. A global picture

Above results should be compared to

Theorem 5 (McKay) If the JSJ decomposition of ${G}$ is trivial (no local cut points), then ${\mathrm{dim}_{AR}(\partial G)>1}$.

McKay’s theorem rules out Sierpinsky curves or Menger curves from the list of possible 1-conformal dimensional ideal boundaries of groups.

Start from a group ${G}$ with ${\mathrm{dim}_{AR}(\partial G)=1}$. Assume that ${G}$ has no 2-torsion. DS split it, then JSJ split vertex groups, then DS vertex groups again… In the end, only Fuchsian cocompact groups and finite groups may arise. This is why Example 2 is so important.

4. Proof of Theorem 3

The argument is inspired by J. Tyson and J.-M. Wu’s treatment of the Sierpinski gasket.

Thanks to (WS), and by self-similarity, there is a uniform upper bound on the number of points needed to disconnect an ${R}$-ball into sets of diameter ${\leq R/2}$, for all ${R}$. Fix a large ${N}$. Split a ball of radius ${1}$ into concentric annuli of width ${2^{-1},\ldots,2^{-N}}$. Each of them contains a bounded number of cut points sufficient to disconnect it from the next layers. Consider the family ${\Gamma}$ of curves joining the center to the complement of the ball. Any curve of ${\Gamma}$ must pass through at least ${N}$ such cut-points. So make a conformal change of metric where the weight is ${1/N}$ at cut-points and 0 elsewhere. This will give length ${\geq 1}$ to all curves in ${\Gamma}$. On the other hand, the ${p}$-volume is

$\displaystyle \begin{array}{rcl} \frac{1}{N^p}O(N) \end{array}$

which tends to ${0}$ as ${N}$ tends to infinity. This shows that the ${p}$-modulus of ${\Gamma}$ vanishes for all ${p>1}$, and thus, that conformal dimension equals 1.