# Starting letters of equivalent infinite geodesic paths of hyperbolic Coxeter groups

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.

• Beware that "hyperbolic Coxeter groups" usually denotes a proper subclass of the class of Gromov-hyperbolic Coxeter groups, so it would be useful if you are more specific. – YCor Aug 13 at 14:49
• I mean Gromov hyperbolic Coxeter groups. Thanks for the remark! – worldreporter14 Aug 13 at 16:13
• Did you try the infinite dihedral group? – Mark Sapir Aug 13 at 17:47
• I guess the infinite paths start at $1$, and $[\cdot]$ means the limit point, in which case there is no problem with $D_\infty$. – YCor Aug 13 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 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.

• Thank you for your great answer! Is it possible that at least in the right-angled, irreducible case my question has a positive answer? – worldreporter14 Aug 14 at 6:18
• The group $G$ given at the end above corresponds to the right-angled 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. – Jim Belk Aug 14 at 10:41
• (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, right-angled Coxeter group $W_\Gamma$ has the property you want if and only if it is a free product of finite groups. – Jim Belk Aug 14 at 10:49
• Thank you very much! – worldreporter14 Aug 14 at 13:02
• Note: In my previous comment, "path of length two as an isometrically embedded subgraph" should be "path of length two as an induced subgraph". – Jim Belk Aug 14 at 18:21