Conditions for graph embedded into hyperbolic space to be acyclic

Question

I am thinking about the following problem: Let $G$ be a graph embedded into hyperbolic space $\mathbb{H}^d$, i.e., the vertex set $V(G)$ is a subset of $\mathbb{H}^d$ and the edges are geodesic line segments connecting vertices.

Let us call such a graph an $(r, \alpha)$-graph when

• any two neighboring vertices have hyperbolic distance at least $r$ and
• whenever two edges meet in a common vertex, the angle between them is at least $\alpha$.

What I am looking for is the following statement: "For any $\alpha>0$ there exists an $r>0$ such that any $(r,\alpha)$-graph is acyclic".

This feels natural to me and I am relatively optimistic that it is true, however, I did not find it in the literature. The closest thing I found is Proposition 8 in this paper by Hansen and Müller. I believe that the proof given there could be modified to show the above statement, however I first wanted to check if it is already stated elsewhere.

So, does anyone know of a reference to this fact in the literature? (Or perhaps, do you know the statement to be false and have a counterexample?)

Any comments are greatly appreciated!

Answer 1

To expand my comment to an answer.

Theorem. Suppose that $f: [0,T]\to \mathbb H^d$ is a piecewise-geodesic map which is a concatenation of geodesic segments of length $\ge r$ which meet at angles $\ge \alpha$. Then, assuming that $\cosh(r/2)\sin(\alpha/2)\ge 1$, the map $f$ is 1-1.

This theorem implies the claim you are aiming for. The estimate in the theorem is sharp as one can see by considering regular polygons in the hyperbolic plane.

The proof of this theorem hinges on:

Lemma. Let $A_1A_2, A_2A_3$ be geodesic segments in $\mathbb H^d$ of length $\ge r$ meeting at the angle $\ge \alpha$. Let $H_1, H_2$ be hyperplanes which are perpendicular bisectors for these segments. Then, under the inequality $\cosh(r/2)\sin(\alpha/2)\ge 1$, these bisectors are disjoint. (They meet at infinity in the case of equalities.)

Proof of this lemma reduces to the 2-dimensional case (by considering the hyperbolic plane containing $A_1A_2, A_2A_3$) and hyperbolic trigonometry.

To apply this lemma, let $H_i^+$ denote the open hyperbolic half-space bounded by $H_i$ and containing $A_{i+1}, i=1, 2$. Then $H_2^+\subset H_1^+$. Given all this, to prove the theorem, you represent $f$ as a concatenation of segments $A_iA_{i+1}$, draw bisectors $H_i$ though these and observe, inductively, that $$ H_{i+1}\subset H_i, i=1,2,... $$ From this, it follows that $f(s)\ne f(t)$ for all $s<t$.

The details are straightforward and I skip these.