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 $\left|w^{-1}\alpha\_{n}\right|=\left|\alpha\_{n}\right|-\left|w\right|$ 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 $\left|w^{-1}\beta\_{n}\right|=\left|\beta\_{n}\right|-\left|w\right|$ for all large enough $n$? My intuition for the Gromov boundary is not well-developed yet, so intuitively I would say yes. However, I doubt this is true for general hyperbolic groups.



Starting letters of equivalent infinite geodesic paths of hyperbolic Coxeter groups
This is false. The archetypical family of hyperbolic Coxeter groups are the [hyperbolic triangle groups](https://en.wikipedia.org/wiki/Triangle_group#The_hyperbolic_case)
$$
T(l,m,n) = \langle a,b,c \mid a^2=b^2=c^2=(ab)^l=(bc)^m=(ca)^n=1\rangle
$$
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 bi-infinite "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 two-point 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.


