Is the canonical map from isometry group of a Gromov hyperbolic space to homeomorphisms of its Gromov boundary injective?

Question

Suppose X is a proper Gromov hyperbolic space and $\partial X$ is its Gromov boundary. It is well-known that there is a canonical group homomorphism $\Phi$ from the isometry group of X to the group of homeomorphisms of $\partial X$.

My question is whether this map $\Phi$ is injective.

What about if we do not assume X is proper? Will the map $\Phi$ be injective?

Thank you in advance !

Answer 1

Let $X$ be the planar graph

• whose vertices are the points $(i,j)$ for $i \in \mathbb{Z}$ and $j \in \{-1,0,1\}$;
• whose edges connect $(i,0)$ with $(i+1,0)$ for every $i \in \mathbb{Z}$ and $(i,\pm 1)$ with $(i,0)$ for every $i \in \mathbb{Z}$.

In other words, $X$ is a bi-infinite line with top and bottom neighbours added to every vertex.

Of course, $X$ is a locally finite hyperbolic graph (it's a tree). It is not difficult to verify that its isometry group is the unrestricted wreath product $$\mathbb{Z}/2\mathbb{Z}\ \mathrm{wr} \ \mathbb{D}_\infty = \left( \prod\limits_{i \in \mathbb{Z}} \mathbb{Z}/2\mathbb{Z} \rtimes \mathbb{Z} \right) \rtimes \mathbb{Z}/2\mathbb{Z}.$$ Each $\mathbb{Z}/2\mathbb{Z}$ from the product corresponds to an isometry that swaps the top and bottom neighbours $(i,1)$ and $(i,-1)$ for some $i \in \mathbb{Z}$ but fixes all the other vertices. The $\mathbb{Z}$ corresponds to the obvious translation. And the right $\mathbb{Z}/2\mathbb{Z}$ corresponds to a left-right reflection.

The boundary $\partial X$ of $X$ contains only two points, and the action of $\mathrm{Isom}(X)$ on $\partial X$ corresponds to the projection onto the right factor $\mathbb{Z}/2\mathbb{Z}$. Consequently, the kernel of the action is the unrestricted wreath product $$\mathbb{Z}/2\mathbb{Z} \ \mathrm{wr} \ \mathbb{Z}:= \prod\limits_{i \in \mathbb{Z}} \mathbb{Z}/2\mathbb{Z} \rtimes \mathbb{Z},$$ which is uncountably infinite.

The key point is that hyperbolicity has no control on small scale geometry. By using the same trick of adding top and bottom neighbours, you can start from your favourite hyperbolic graph $Y$ and create a new hyperbolic graph $Y^+$ (quasi-isometric to $Y$) whose isometry group contains an uncountably infinite subgroup $\prod \mathbb{Z}/2\mathbb{Z}$ in the kernel of its action on $\partial Y^+$.

Answer 2

Let $G$ be the isometry group of a quasi-geodesic Gromov-hyperbolic space $X$. If $X$ is empty there's not much so say, so assume otherwise. Say that $B\subset G$ is bounded if $Bx$ is bounded for some/every $x\in X$. When $X$ is proper, this just means that $B$ has compact closure.

Let $W$ be the kernel of the $G$-action on the Gromov boundary $\partial X$.

From Gromov's classification, an isometric action on a Gromov-hyperbolic falls into exactly one of the following (a) bounded orbits (b) horocyclic: unbounded, a unique fixed point on the boundary, no loxodromic (c) axial: a fixed pair on the boundary and some loxodromic, no other finite orbit on the boundary (d) focal: a unique fixed point on the boundary, some loxodromic (e) general type two loxodromic with disjoint ends.

We apply this to the $W$-action. (d), (e) are excluded since not all points can be fixed. (c) (axial) is possible only if the boundary is reduced to a pair (b) (horocyclic) is possible only if the boundary is a singleton. (a) is possible.

A) If the $G$-action is axial and the boundary is a pair, the $G$-action is axial as well. Let $E$ be a $G$-orbit: it is a quasi-geodesic. Suppose by contradiction that $E$ is not cobounded in $G$. Then since the $G$-action on $E$ is cobounded, for some $C$ and all $n$ there exists $x_n$ with $d(x_n,x_0)\le n$ and $d(x_n,E)\ge n-C$. Using hyperbolicity, one can show that the sequence $(x_n)$ is not equivalent to any one representing one of the ends of $E$. So there's a third point on the boundary, contradiction. Then $E$ is cobounded. In this case, the kernel $W$ is $G$ or has index 2 in $G$ according to whether $G$ switches the two boundary points.

Hence, if the $W$-action is axial, we are in (A) as above.

If the $W$-action is horocyclic, there's a single boundary point. Hence the $G$-action is horocyclic as well.

Otherwise the $W$-action is bounded, so $W$ is indeed bounded.

A bounded subgroup $H$ need not act trivially on the boundary. However, this holds if the $G$-action is cobounded and $H$ is normal in $G$.

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In summary, we have exactly one of the following:

1. $W$ is bounded (and if $G$ acts coboundedly, it is the largest bounded normal subgroup)
2. $G$ acts horocyclically, and $\partial X$ is a singleton (example: $X$ horodisc in hyperbolic plane) — in this case $G$ can't act coboundedly
3. $G$ acts axially coboundedly on $X$ (so $X$ is quasi-isometric to $\mathbf{Z}$); $W$ has index 1 or 2 on $G$ according to whether $G$ fixes the two boundary points.