# Infinite geodesic rays leaving a K-quasiconvex subgroup stay K-close to it.

I am going through some basic properties of $\delta$-hyperbolic spaces and groups and I am having some difficulties proving precisey some things that are anyway intuitively clear to me.

Let $G$ be a $\delta$-hyperbolic group and $H \subset G$ a $K$-quasiconvex subgroup, i.e. geodesics joining points of $H$ lie in the $K$-neighbourhood of $H$. I want to prove a similar statement for infinite geodesics starting at some element of $H$ and converging at infinity to some limit point of $H$.

More precisely let $\partial G$ denote the Gromov boundary of $G$ and $\Lambda (H)$ the limit set of $H$, i.e. the subset of $\partial G$ of accumulation points for sequences from $H$. Now take a geodesic ray $\gamma : [0,+\infty [ \to G$ such that $\gamma (0)=h\in H$ and $\gamma \to \xi \in \Lambda (H)$ at infinity. I want to prove that $\gamma$ lies in the $K$-neighbourhood of $H$.

What I understand: I can approximate $\gamma$ with finite-length geodesics $\gamma_n$ leaving $h$ and coming back to $H$ at some $h_n \in H$, where $\{ h_n \}$ are a sequence in $H$ converging at infinity to $\xi$. By $K$-quasiconvexity, $\gamma_n$ lies in the $K$-neighbourhood of $H$ for every $n$, so this should remain true passing to the limit $n\to \infty$.

To make this hand-waving precise, we need a notion of convergence for geodesic rays in $\overline{G}=G \cup \partial G$ (which of course we have), but I would like to see a more elementary proof of this fact, maybe something that builds a convenient geodesic polygon and then uses nothing more than the definitions of hyperbolicity and quasiconvexity.

As stated, your result is in fact false. For example, take $G= \mathbb{Z} \times \mathbb{Z}_2$ and $H= \mathbb{Z} \times \{0\}$; with respect to $\{(1,0),(0,1)\}$, the Cayley graph of $G$ looks like a ladder $$\mathbb{R} \times \{0,1\} \cup \bigcup\limits_{n \in \mathbb{Z}} \{ n \} \times [0,1]$$ and $H$ is a $0$-quasiconvex (ie. convex) subgroup, and of course $G$ is hyperbolic (it is quasi-isometric to $\mathbb{Z}$). Now let $\gamma$ be the geodesic whose points are

$$(0,0), (0,1), (1,1), (2,1), (3,1), (4,1), \dots$$

Clearly, the ray $\gamma$ converges to a point of $\Lambda(H)= \partial G$, but it does not belong to the $0$-neighborhood of $H$. However, it belongs to the $1$-neighborhood, and indeed your result becomes true with a little modification:

Property: Let $G$ be a $\delta$-hyperbolic group, $H$ a $K$-quasiconvex subgroup and $\gamma$ a ray with $\gamma(0) \in H$ converging to some $\xi \in \Lambda(H)$. Then $\gamma$ belongs to the $(K+8\delta)$-neighborhood of $H$.

Sketch of proof. Let $(h_n) \subset H$ be a sequence converging to $\xi$, with $h_0= \gamma(0)$. Choose a sequence of geodesic segments $[h_0,h_1], [h_0,h_2], \dots$ and use Ascoli theorem to find a subsequence converging to some ray $r$. Now, $r$ is in the $K$-neighborhood of $H$; because $r$ and $\gamma$ converges to the same end, $\sup\limits_{t \geq 0} d(r(t),\gamma(t)) \leq 8 \delta$[1] and the conclusion follows. $\square$

The bound $K+8\delta$ is probably not optimal.

[1] I think it is a classical result. Unfortunately, the only reference I know is in French: P. de la Harpe et E. Ghys, Sur les groupes hyperboliques, chapitre 7, corollaire 3.