# Does the visual boundary of any one-ended Cayley graph contain at least three points?

Let $$\Gamma$$ be a Cayley graph of a finitely generated group. We can define the visual boundary of $$\Gamma$$ with respect to some base vertex $$b$$, denoted $$\partial \Gamma$$, as the set of geodesic rays based at $$b$$, modulo finite Hausdorff distance (since we have a transitive group action, the basepoint is not important). One can topologize this in the natural way, but that is not important here. We have the following straightforward question.

Suppose $$\Gamma$$ is a Cayley graph of a one-ended finitely generated group. Does $$\partial \Gamma$$ contain at least three points?

This is clearly known in the case of $$CAT(0)$$ and hyperbolic groups, but I'm interested in the most general case. It's not even immediately clear to me that $$\partial \Gamma$$ is non-empty, for example.

Any help or references would be appreciated. As with most questions about f.g. groups, I suspect this is either trivial or wrong.

• Every infinite locally finite connected vertex-transitive graph has a bi-infinite geodesic, so the visual boundary already has at least 2 points (so, it's not empty). Without the transitivity, there's a geodesic ray so it's not-empty (and of course in this case it can be 1-ended with a visual boundary reduced to a singleton).
– YCor
Jan 2 at 17:46

Yes. This holds for every vertex-transitive good graph except those with 0 or 2 ends, where I abbreviate "connected graph of finite valency" as "good graph".

First, if $$X$$ is a vertex-transitive infinite good graph, then it has a bi-infinite geodesic (find a geodesic segment of size $$2n$$, translative it so that its middle point is at a given basepoint, and use a compactness argument). This shows that the visual boundary has at least 2 elements. This is not enough, but the same idea will be used in the sequel.

So, there is a bi-infinite geodesic. If this geodesic is cobounded, the graph is quasi-isometric to $$\mathbf{Z}$$ and hence is 2-ended. So assuming otherwise, this geodesic is not cobounded. Hence for every $$n$$ there is a point at distance $$n$$ to the geodesic.

(Beware that we cannot assume that there are isometries preserving the fixed geodesic and translating it.) However, this shows that for every $$n$$, there exists, in the graph, a point $$v_0$$, three geodesic segments $$(w_k^i)_{0\le k\le n}$$, $$i=0,1,2$$, with $$w_0^i=v_0$$, such that the first two form a geodesic ($$d(w_k^0,w_\ell^1)=k+\ell$$ for all $$0\le k,\ell\le n$$), and $$d(w_k^i,w_\ell^2)\ge\ell$$ for all $$0\le k,\ell\le n$$ and $$i=0,1$$. Using transitivity and compactness, there exist three such infinite rays with the same two properties. Then they are pairwise at infinite Hausdorff distance and hence the visual boundary has at least 3 points.

• This is very slick, thank you. Am I right to observe that one can repeat this argument to gain infinitely many points in the boundary? The union of these first 3 geodesics cannot be cobounded (it has three ends), so I think we can apply identical logic ad infinitum. Jan 2 at 18:33
• @jpmacmanus probably, I haven't checked details. Or even better, if it can even show that the boundary has no isolated point (if it's compact), it would show that it has continuum cardinal.
– YCor
Jan 2 at 18:45
• I've looked a bit carefully, and don't even see how to get 4 points on the boundary. Once one has such a "tripod", since the number of ends is not 3, the tripod is not cobounded. But then it's not obvious how to pursue the argument (the new "branching" point can go to infinity and the compactness might thus yield another tripod instead of a quadripod).
– YCor
Jan 3 at 8:32
• that's a subtlety I hadn't noticed, good spot! Jan 3 at 10:31