Let $\left(W\text{, }S\right)$ be a Gromov hyperbolic Coxeter system and denote by $\partial W$ the corresponding Gromov boundary. For $z\in\partial W$ let $\alpha$, $\beta$ be infinite geodesic paths with $z=\left[\alpha\right]=\left[\beta\right]$ and assume that there exists $w\in W$ such that $\leftw^{1}\alpha_{n}\right=\left\alpha_{n}\right\leftw\right$ for all large enough $n$ (i.e. $\alpha_n$ starts with $w$). Here $\left\cdot\right$ denotes the word metric on $W$ regarding the generating set $S$. Is it true that then $\leftw^{1}\beta_{n}\right=\left\beta_{n}\right\leftw\right$ for all large enough $n$? My intuition for the Gromov boundary is not welldeveloped yet, so intuitively I would say yes. However, I doubt this is true for general hyperbolic groups.

1$\begingroup$ Beware that "hyperbolic Coxeter groups" usually denotes a proper subclass of the class of Gromovhyperbolic Coxeter groups, so it would be useful if you are more specific. $\endgroup$ – YCor Aug 13 '19 at 14:49

$\begingroup$ I mean Gromov hyperbolic Coxeter groups. Thanks for the remark! $\endgroup$ – worldreporter14 Aug 13 '19 at 16:13

1$\begingroup$ Did you try the infinite dihedral group? $\endgroup$ – user6976 Aug 13 '19 at 17:47

$\begingroup$ I guess the infinite paths start at $1$, and $[\cdot]$ means the limit point, in which case there is no problem with $D_\infty$. $\endgroup$ – YCor Aug 13 '19 at 19:19
This is false. The archetypical family of hyperbolic Coxeter groups are the hyperbolic triangle groups $$ T(l,m,n) = \langle a,b,c \mid a^2=b^2=c^2=(ab)^l=(bc)^m=(ca)^n=1\rangle $$ where $l,m,n\geq 2$ and $(1/l)+(1/m)+(1/n)<1$. Such a group acts as a group of symmetries of a tiling of the hyperbolic plane by congruent triangles with angles of $\pi/l$, $\pi/m$, and $\pi/n$. Here are some basic facts about these groups:
If we choose a base triangle $T_0$ (a.k.a. the "fundamental chamber"), then each triangle of the tiling can be expressed uniquely as $gT_0$ for some $g\in T(l,m,n)$.
As such, the (right) Cayley graph of $T(l,m,n)$ is precisely the dual graph to the triangular tiling, with the identity vertex lying in the triangle $T_0$.
The triangular tiling can be obtained by cutting the hyperbolic plane along a countable family of hyperbolic lines. A path of edges in the Cayley graph is a geodesic if and only if it crosses each line of the tiling at most one time.
The Gromov boundary $\partial T(l,m,n)$ is the circle $S^1$, which can naturally be identified with the boundary circle of the hyperbolic plane. An infinite geodesic path in the Cayley graph represents a point on the boundary if and only if it converges to that point in the closed unit disk.
Now, let $L$ be the hyperbolic line that separates $T_0$ from $aT_0$, let $z$ be an endpoint of $L$ on the boundary circle, and consider the following two geodesic paths in the Cayley graph:
The path $\alpha$ that goes from $T_0$ to $aT_0$, and then "follows along" $L$ in the direction of $z$.
The path $\beta$ that starts at $T_0$ and "follows along" $L$ in the direction of $z$.
To be precise, $\beta$ is the path in the Cayley graph corresponding to the sequence of triangles passed through by the hyperbolic ray from the identity vertex to $z$, and $\alpha$ consists of the dual edge from $T_0$ to $aT_0$ followed by the reflection of $\beta$ across $L$. Then $[\alpha]=[\beta]=z$, but $\alpha$ starts with $a$ and $\beta$ does not.
Edit: Part of the reason I chose the above example was to convey some intuition for hyperbolic Coxeter groups. However, there are simpler examples available. For example, consider the group $$ G = \langle a,b,c \mid a^2=b^2=c^2=(ab)^2=(ac)^2=1\rangle $$ This is the direct product of $\mathbb{Z}_2$ with the infinite dihedral group $\mathbb{Z}_2*\mathbb{Z}_2$, and its Cayley graph is a biinfinite "ladder" with $a$ edges as rungs and alternating $b$ and $c$ edges along the sides. This group is Gromov hyperbolic since it's virtually cyclic (making it an "elementary" hyperbolic group), and the Gromov boundary $\partial G$ is a twopoint set corresponding to the two ends of the ladder.
Now, if $\alpha$ is the geodesic path corresponding to the infinite word $a(bc)^\infty = abcbcbc\cdots$ and $\beta$ is the infinite path corresponding to $(bc)^\infty$, then $[\alpha]=[\beta]$ since the two paths go the same direction on the ladder, but $\alpha$ starts with $a$ and $\beta$ does not.

$\begingroup$ Thank you for your great answer! Is it possible that at least in the rightangled, irreducible case my question has a positive answer? $\endgroup$ – worldreporter14 Aug 14 '19 at 6:18

$\begingroup$ The group $G$ given at the end above corresponds to the rightangled Coxeter group $W_\Gamma$ for which $\Gamma$ is a path of length two. More generally, if $\Gamma$ is any graph which has a path of length two as an induced subgraph, then $W_\Gamma$ will have $G$ as an isometrically embedded subgroup, and the same phenomenon will occur. For example, $\mathbb{Z}_2* G$ is hyperbolic and irreducible (not a direct product  I'm assuming that's what you mean by "irreducible") , and its Cayley graph has the Cayley graph of $G$ as an isometrically embedded subgraph. $\endgroup$ – Jim Belk Aug 14 '19 at 10:41

$\begingroup$ (continued) Indeed, it is not hard to show that any graph $\Gamma$ which does not have a path of length two as an isometrically embedded subgraph must be a disjoint union of complete graphs. It follows that a Gromov hyperbolic, rightangled Coxeter group $W_\Gamma$ has the property you want if and only if it is a free product of finite groups. $\endgroup$ – Jim Belk Aug 14 '19 at 10:49


$\begingroup$ Note: In my previous comment, "path of length two as an isometrically embedded subgraph" should be "path of length two as an induced subgraph". $\endgroup$ – Jim Belk Aug 14 '19 at 18:21