Consider the surface group $\Gamma=\langle a,b,c,d\mid [a,b][c,d]=1\rangle$: it is a Gromov hyperbolic group; its Gromov boundary $\partial\Gamma$ is homeomorphic to $S^1$ (the unit circle). I would like to define a family of subsets of $\partial\Gamma$ as follows: fix $x,y\in\Gamma$. Then $$ U(x,y):=\left\{\xi\in\partial\Gamma\mid\text{ there exists a geodesic }g\text{ in }\Gamma\text{ starting from }x,\text{ passing through }y\text{ and s.t. }g(\infty)=\xi\right\}. $$

  1. Does this definition make sense? Is $U(x,y)$ trivial?
  2. Is $U(x,y)$ open?
  3. Is $U(x,y)$ connected?
  • $\begingroup$ For it to make sense, you need to fix a metric on $\Gamma$. Maybe you have in mind the word metric w.r.t. some finite generating subset (e.g., the given one), and probably the definition depends on this choice. $\endgroup$
    – YCor
    Jul 1, 2016 at 8:53
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    $\begingroup$ With the word metric w.r.t. this choice of generators, the Cayley graph is a plane tiling by octogons with vertices of valency 8. Geodesic rays are injective paths that do not follow more than 4 consecutive times a given 8-gon. It rather seems then that $U(x,y)$ is a segment for all $x,y$. (In more generality, the set of geodesic rays from $x$ thru $y$ seems compact, so I'd always expect $U(x,y)$ to be compact.) $\endgroup$
    – YCor
    Jul 1, 2016 at 8:57
  • $\begingroup$ The slightly coarser notion of the shadow of an element $y$ of a hyperbolic group is commonly used. (One may as well take $x=1$.) See, for instance, arXiv:1111.0029v2 . $\endgroup$
    – HJRW
    Jul 1, 2016 at 10:25
  • $\begingroup$ @YCor: My fault, I forgot to state: I clearly consider the word metric on $\Gamma$ w.r.t. the generating subset $\left\{a,b,c,d,a^{-1},b^{-1},c^{-1},d^{-1}\right\}$. I knew about the plane tiling and the fact that geodesic rays do not follow more than 4 consecutive times a given octagon. I have the same feeling that $U(x,y)$ is a segment, but I still can't show this... $\endgroup$
    – EM90
    Jul 1, 2016 at 10:26
  • $\begingroup$ @HJRW: I know the notion of shadow, I am not completely sure of the relation between the definition of $U(x,y)$ and that of shadow; still, I think that a shadow may be a (finite) union of $U(x,y)$ for our $\Gamma$... $\endgroup$
    – EM90
    Jul 1, 2016 at 10:28

1 Answer 1


You definition is closely related to the notion of the "cone type" introduced by Jim Cannon. As Yves noted, $U(x,y)$ is compact. It is also connected.

  1. Compactness part is immediate from the Arzela-Ascoli theorem: Take a sequence of rays $r_i: [0,\infty)\to X$ (where $X$ is the Cayley graph; here it does not matter for what hyperbolic group and what generating set) such that $r_i(0)=x$, $r_i(n)=y$ for all $i$ and a fixed $n$. Then the sequence $r_i$ subconverges to a geodesic ray $r$ from $x$ passing through $y$, such that $r(\infty)= \lim_i r_i(\infty)$. Hence, $U(x,y)$ is compact.

  2. To prove connectedness you need to use the fact that the Cayley graph $X$ in your case is planar, more precisely, is a subset of the hyperbolic plane $H^2$, so that the ideal boundary of $X$ is the boundary circle $S^1$ of $H^2$. In what follows, all the metric notions are with respect to the Cayley graph $X$. I will assume that $x\ne y$. Let $\xi_1, \xi_2$ be distinct points in $U(x,y)$. Consider geodesic rays $r_1, r_2$ from $x$ passing through $y$ and asymptotic to $\xi_1, \xi_2$. There exists the smallest $T\ge n=d(x,y)$ such that for all $t\ge T$, $r_1(t)\ne r_2(t)$. Thus, the union $A=r_1([T,\infty))\cup r_2([T,\infty))$ is a topological line in $X\subset H^2$. This line splits $H^2$ in two components, one of then, call it $C$, does not contain $x$. The ideal boundary $\alpha$ of $C$ equals one of the two arcs of $S^1$ with the end-points $\xi_1, \xi_2$. I claim that all points $\eta\in \alpha$ belong to $U(x,y)$. Indeed, consider a geodesic ray $r$ from $x$ asymptotic to $\eta$. This ray has to cross $A$ at some point $z=r(s)$. Say, $z\in r_1([0,\infty))$. Now, replace the portion $r[0,s])$ with $r_1([0,s])$. I will leave you to verify that the new ray $r_3$ is a geodesic ray from $x$ which passes through $y$. Hence, $\eta\in U(x,y)$. qed

  • $\begingroup$ Thank you very much, Misha! Your argument on connectedness is really interesting. Actually, my definition of the set $U(x,y)$ is somehow intended to get a kind of "projection" or "shadow" of a cone type on the boundary, thus it is indeed related to that notion. $\endgroup$
    – EM90
    Jul 3, 2016 at 14:04

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