Tesselation of hyperbolic plane are hyperbolic

Question

Let’s take a tesselation $T$ of the hyperbolic plane (not necessarily regular), my intuition tells me that clearly $T$ should be hyperbolic itself (in the sense of Gromov or using $\delta$-slim triangles), but I haven’t find a proof of that. I think it can be done by showing that $T$ is quasi-isometric to the hyperbolic plane, but I am not sure how to show this, or if there is another proof of this. Also I haven’t find much references about how the curvature is “preserved” after a triangulation. Can you please help me find a reference about this? Thanks.

Answer 1

Because you are allowing irregular tesselations whose edges have no geometric relation to the geometry of $\mathbb H^2$, it seems unreasonable to expect that a tesselation of $\mathbb H^2$ would inherit a large scale metric property of $\mathbb H^2$ such as Gromov hyperbolicity.

Keeping that in mind, and remembering also that $\mathbb H^2$ is actually homeomorphic to $\mathbb E^2$, it is then easy to construct an explicit counterexample.

To start with, pick a homeomorphism $f : \mathbb E^2 \to \mathbb H^2$.

Next, consider the Euclidean tesselation with vertex set $\mathcal V = \mathbb Z \times \mathbb Z$ and edge set $$\mathcal E = \bigl(\{[m-1,m] \mid m \in \mathbb Z\} \times \mathbb Z\bigr) \cup \bigl(\mathbb Z \times \{[n-1,n] \mid n \in \mathbb Z\bigr) $$

Now transport that Euclidean tesselation from $\mathbb E^2$ to $\mathbb H^2$ using the homeomorphism $f$. You get a tesselation of $\mathbb H^2$ with vertices $$f(\mathcal V) = \{f(v) \mid v \in \mathcal V\}, \quad f(\mathcal E) = \{f(e) \mid e \in \mathcal E\} $$ From this construction, the tesselation of $\mathbb H^2$ with vertex set $f(\mathcal V)$ and edge set $f(\mathcal E)$ is isometric to the tesselation of $\mathbb E^2$ with vertex set $\mathcal V$ and edge set $\mathcal E$. The latter tesselation is not Gromov hyperbolic, and so neither is the former.

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Here's an even more explicit construction, which produces an example whose edges are actually $\mathbb H^2$ geodesics. Start with the upper half plane model of $\mathbb H^2$ and the homeomorphism $f : \mathbb E^2 \to \mathbb H^2$ given by $$f(x,y) = (x,e^y) $$ Now transport the $\mathbb E^2$ tesselation to $\mathbb H^2$ as described above. This homeomorphism already takes vertical edges $m \times [n-1,n] \subset \mathbb E^2$ to geodesic segments in $\mathbb H^2$, because the vertical segment $m \times [e^{n-1},e^n]$ is already geodesic. However, the homeomorphism does not yet take a horizontal edge $[m-1,m] \times n \subset \mathbb E^2$, to a geodesic segment in $\mathbb H^2$, because $[m-1,m] \times e^n$ is a horocyclic segment in $\mathbb H^2$. To repair that, you simply replace that $\mathbb H^2$ horocyclic segment by the geodesic segment with the same endpoints $(m-1,e^n)$ and $(m,e^n)$, which is the segment of the unique Euclidean semicircle that passes through those two points and hits the horizontal axis $\mathbb R \times \{0\}$ at right angles.