2
$\begingroup$

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?
$\endgroup$
5
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

3
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.