As hyperbolic as they Comm

Nir Lazarovich is the speaker. He works at Technion in Israel. I am attending this seminar at the mathematics department at TIFR.

Introduction

$\Gamma$ a f.g. group is hyperbolic if it is $X = \mathrm{Cay}(\Gamma,S)$ is hyperbolic, where $\langle S\rangle = \Gamma$ is a finite generating set. The Cayley tree being hyperbolic is equivalent to \[\begin{equation} \exists \delta > 0, \ \forall x,y,z \in X, [x,z] \subseteq N_{\delta} ( [x,y] \cup [y,z]). \end{equation}\] Here $[x,y]$ is the shortest path between $x,y$ and $N_{\delta}(A)$ is the $\delta$-neighbourhood of a set $A$.

Motivation

Examples

Just to get out of the way, $\Gamma$ is always torsion free for the rest of this talk.

Some results

(Sela)

If $\Gamma$ is 1-ended hyperbolic, then $\Gamma$ is co-Hopfian: i.e. $\Gamma$ is not isomorphic to a proper subgroup of itself. ¹

(Sela)

If $\Gamma$ is normal in $G$, and both $\Gamma$ and $G$ are torsion-free hyperbolic, then $\Gamma$ is a free product of surface and free groups.

(Paulin)

If $\Gamma$ is hyperbolic and does not split over $\mathbb{Z}$ then $|\mathrm{Out}(\Gamma)|< \infty$.

A (abstract) commensurator of $\Gamma$ is an isomorphism $\varphi: H_{1} \xrightarrow[\simeq]{} H_{2}$ such that $H_{1} \subseteq \Gamma$, $| \Gamma : H_{i}| < \infty$. Here both $H_{1}, H_{2}$ are some subgroups of $\Gamma$.

(Whyte)

Can $1$-ended hyperbolic group $\Gamma$ have $H_{1} \simeq H_{2} < \Gamma$ of different indices?

(Stark-Woodhouse)

Yes, $\exists \Gamma$ 1-end hyperbolic such that \[\begin{equation} H_{1} \simeq H_{2} \leq \Gamma \text{ such that } [\Gamma: H_{1} ] < \infty \text{ and }[\Gamma : H_{2}] = \infty \end{equation}\]

$\Gamma$ is a finite index rigid if $\forall H_{1} , H_{2} \subseteq \Gamma$ of finite index if $H_{1} \simeq H_{2} \implies [\Gamma: H_{1}] = [\Gamma : H_{2}]$.

$\mathbb{Z}$ is not f.i.r because $\mathbb{Z} \simeq 2 \mathbb{Z}$.

(L., 2023)

If $\Gamma$ is hyperbolic non-elementary then $\Gamma$ is f.i.r.

(L.-Rahamain-Sisto, 25)

$\mathbb{Z} \simeq \Gamma$ is t.f. hyperbolic relative to type $F$ NRH proper subgroups, then $\Gamma$ is f.i.r.

$\Gamma \leq G$ then $\mathrm{Comm}_{G}(\Gamma)$ is defined as \[\begin{equation} \mathrm{Comm}_{G}(\Gamma) = \{ g\in G \mid [\Gamma : \Gamma \cap g \Gamma g^{-1}] < \infty, [g \Gamma g^{-1}, \Gamma \cap g \Gamma g^{-1} ] < \infty\}. \end{equation}\] We also define \[\begin{equation} \mathrm{N}_{G}(\Gamma) = \{ g\in G \mid g \Gamma g^{-1} = \Gamma \} \end{equation}\]

\[\begin{equation} \Gamma \text{ is normal in } G \Leftrightarrow N_{G}(\Gamma) = G \end{equation}\]

$\Gamma$ be almost normal subgroup in $G$ if $\mathrm{Comm}_{G}(\Gamma) = G$: aka commensurated subgroups.

There is an example of this given by \[\begin{equation} G = \langle a, t \mid t a^{p} t^{-1} = a^{q}\rangle. \end{equation}\] Here the subgroup $\Gamma = \langle a\rangle$ is almost normal.

If $G$ is finitely generated, one has \[\begin{equation} \Gamma \text{ almost normal in } G \Leftrightarrow \Gamma - \mathrm{Stab}_{G}(v) \text{ for an action } G \curvearrowright\underbrace{Y}_{\text{ hc. finite graph}.} \end{equation}\]

(L-Margolis-Mj)

If $\Gamma$ is almost normal in $G$ and both $\Gamma$ and $G$ are torsion-free hyperbolic, then $\Gamma$ is a free product of surface and free groups.

Commutator of a group

\[\begin{equation} \mathrm{Comm}(\Gamma) = \{ \phi : H_{1} \xrightarrow[\simeq]{} H_{2} \mid H_{i} \leq_{ \text{ f.i. }} \Gamma \} / \sim \end{equation}\] A priori $\mathrm{Comm}(\Gamma)$ is a groupoid but by passing to subgroup $H_{1} \cap H_{2}$ which are finite index in both $H_{1},H_{2}$ one can compose maps. Therefore this becomes a group.

For example, \[\begin{equation} \mathrm{Comm}(\mathbb{Z}) \simeq \mathbb{Q}^{\times}. \end{equation}\] To see this, observe that $q \mathbb{Z} \rightarrow p \mathbb{Z}$ is the map $x \mapsto \pm \tfrac{p}{q} x$.

\[\begin{equation} \underbrace{\pi_{1} ( \text{ closed hyperbolic $3$-manifolds })}_{\Gamma} \leq \mathrm{PSL}_{2}(\mathbb{C}) \end{equation}\]

\[\begin{equation} \mathrm{Comm}_{\mathrm{PSL}_{2}(\mathbb{C})}( \mathrm{PSL}_{2}(\mathbb{Z})) \geq \mathrm{PSL}_{2}(\mathbb{Q}) \end{equation}\]

The answer is negative. The example of $\mathrm{PSL}_{2}(\mathbb{Z})$ is supposed to illustrate this.

(Mostow rigidity):

For \[\begin{equation} \Gamma = \pi_{1}( \text{ closed hyperbolic 3-manifold}), \end{equation}\] one has \[\begin{equation} \mathrm{Comm}(\Gamma) = \mathrm{Comm}_{\mathrm{PSL}_{2}(\mathbb{C})} (\Gamma) \end{equation}\]

(Gromov)

$\Gamma$ rigid hyperbolic then $\exists$ locally compact topological $G$ s.t. $\Gamma \leq G$ is a discrete cocompact and $\mathrm{Comm}_{G}(\Gamma) = \mathrm{Comm}(\Gamma)$.

Nir adds: the condition of being 1-ended, under the assumption of torsion-free, is the same as freely-indecomposable.↩︎