Rigidity of small Delaunay triangulations of the hyperbolic plane

It is well-known that a conformal diffeomorphism between two hyperbolic planes must be an isometry. In this paper we discretize the hyperbolic plane by small geodesic triangulations and prove an analogous discrete rigidity result.

In this paper we use the Poincaré disk model to represent the hyperbolic plane $\mathbb{H}^{2}$, and assume $\mathbb{H}^{2}=(D,g)$ where $D=\{z\in\mathbb{C}:|z|<1\}$ is the unit open disk in the complex plane and

$$g(z)=\frac{4|dz|^{2}}{(1-|z|^{2})^{2}}.$$

Given two points $z_{1},z_{2}$ in $D$, $d(z_{1},z_{2})$ denotes the hyperbolic distance between $z_{1},z_{2}$. Assume $T=(V,E,F)$ is an (infinite) simplicial topological triangulation of the hyperbolic plane $\mathbb{H}^{2}$, where $V$ is the set of vertices, $E$ is the set of edges and $F$ is the set of faces. Given a subcomplex $T_{0}=(V_{0},E_{0},F_{0})$ of $T$, denote $|T_{0}|$ as the underlying space of $T_{0}$. A mapping $\phi:|T_{0}|\rightarrow\mathbb{H}^{2}$ is called hyperbolic geodesic if

(a) $\phi$ maps each edge of $T_{0}$ to a hyperbolic geodesic arc in $\mathbb{H}^{2}$, and

(b) $\phi$ maps each triangle of $T_{0}$ to a hyperbolic geodesic triangle in $\mathbb{H}^{2}$.
A hyperbolic geodesic mapping $\phi$ of $T_{0}$ naturally induces an edge length $l=l(\phi)\in\mathbb{R}^{E_{0}}$ on $T_{0}$ by letting $l_{ij}=d(\phi(i),\phi(j))$.

Bobenko-Pinkall-Springborn [BPS15] introduced the following notion of hyperbolic discrete conformality, extending Luo’s notion of Euclidean discrete conformality in [Luo04].

A hyperbolic geodesic mapping $\phi$ is called Delaunay if $\phi(k^{\prime})$ is not in the open circumdisk of $\phi(\triangle ijk)$ for any pair of adjacent triangles $\triangle ijk$ and $\triangle ijk^{\prime}$ in $T_{0}$. The main result of the paper is the following.

The rigidity of triangulations of the Euclidean plane under Luo’s notion of discrete conformality was studied in [WGS15][DGM18][LSW20][Wu22]. In particular, [Wu22] relates the Euclidean discrete conformality with the hyperbolic conformality. We will use the tools developed in [Wu22] to prove our hyperbolic rigidity results.

It is elementary to verify that for all $x\in[0,1]$,

$$\frac{x}{2}\leq\sin x\leq x\leq\sinh x\leq e^{x}-1\leq 2x.$$

The following local maximum principle is indeed implied in [Wu22].

The hyperbolic discrete conformal change is related with the Euclidean discrete conformal change as follows.

Assume $l(\psi)=u*_{h}l(\phi)$. We will prove $u_{i}\geq 0$ for all $i\in V$, and then by symmetry $u_{i}\leq 0$ for all $i\in V$ and we are done. Let us prove by contradiction. Suppose $a\in V$ and $u_{a}<0$. Denote $z_{i}=\phi(i)$ and $z_{i}^{\prime}=\psi(i)$ for all $i\in V$. Without loss of generality, we may assume $z_{a}=z_{a}^{\prime}=0$.

Notice that $2d(z_{i}^{\prime},z_{j}^{\prime})=l_{ij}(\psi)\leq\epsilon/4$ for all $ij\in E$. By Lemma 2.2, there exists a map $\psi_{E}:|T|\rightarrow D$ such that

(a) $\psi_{E}(i)=z_{i}^{\prime}$ for all $i\in V$, and

(b) $\psi_{E}(\triangle ijk)$ is a Euclidean triangle in $D$ for all $\triangle ijk\in F$, and

(c) $\psi_{E}$ is an embedding restricted on the 1-ring neighborhood of $j$ for any $j\in V$, and

(d) all the inner angles in $\psi_{E}(\triangle ijk)$ are at least $\epsilon/2$ for all $\triangle ijk\in F$, and

(e) $\psi_{E}\circ\psi^{-1}$ is topologically positively oriented on $\psi(\triangle ijk)$ for all $\triangle ijk\in F$.

Pick $b\in(0,-u_{a})\cap(0,1/10000)$ and denote $\tilde{\psi}_{E}=e^{b}\psi_{E}$. Denote $z_{i}^{\prime\prime}=\tilde{\psi}_{E}(i)=e^{b}z_{i}^{\prime}$ for all $i\in V$. Given $z_{1},z_{2}\in\mathbb{C}$, denote $d(z_{1},z_{2})=\infty$ if $z_{1}\notin D$ or $z_{2}\notin D$. Let

$$E_{0}=\{ij\in E:d(z_{i}^{\prime\prime},z_{j}^{\prime\prime})\leq\frac{\epsilon}{16}\}$$

and

$$V_{0}=\{i\in V:i\in e\text{ for some }e\in E_{0}\}$$

and

$$F_{0}=\{\triangle ijk\in F:ij,jk,ik\in E_{0}\}.$$

Then $T_{0}=(V_{0},E_{0},F_{0})$ is a finite subcomplex of $T$, and it is elementary to verify that $a\in V_{0}$. By Lemma 2.2, there exists a hyperbolic Delaunay geodesic map $\tilde{\psi}$ from $T_{0}$ to $D$ such that

(a) $\tilde{\psi}(i)=z_{i}^{\prime\prime}$, and

(b) $\tilde{\psi}$ is an embedding restricted on the 1-ring neighborhood of $i$ if $ij\in E_{0}$ for all neighbors $j$ of $i$.

Denote $l=l(\phi)$ and $l^{\prime}=l(\tilde{\psi})$ and by Corollary 2.5 we have

$$l^{\prime}=\tilde{u}*_{h}l$$

on $T_{0}$ where

$$\tilde{u}_{i}=u_{i}+b+\log\frac{1-|z_{i}^{\prime}|^{2}}{1-|z_{i}^{\prime\prime}|^{2}}.$$

Then $\tilde{u}_{a}=u_{a}+b<0$. Let $i\in V_{0}$ be such that

$$u_{i}=\min_{j\in V_{0}}u_{j}<0.$$

Then by Lemma 2.3, there exists an edge $ik\in E$ not in $E_{0}$. Furthermore, there exists a triangle $\triangle ijk$ such that $ij\in E_{0}$ and $ik\notin E_{0}$.

Then

$$l^{\prime}_{ij}\leq\frac{\epsilon}{16},$$

and by Lemma 2.1

$$\frac{|z_{j}^{\prime\prime}-z_{i}^{\prime\prime}|}{1-|z_{j}^{\prime\prime}|}\leq\frac{\epsilon}{8},$$

and by the Euclidean sine law

$$\frac{|z_{k}^{\prime\prime}-z_{j}^{\prime\prime}|}{1-|z_{j}^{\prime\prime}|}\leq\frac{\epsilon}{8\sin(\epsilon/2)}\leq\frac{\epsilon}{8(\epsilon/2)/2}=\frac{1}{2},$$

and

$$1-|z_{k}^{\prime\prime}|\geq 1-|z_{j}^{\prime\prime}|-|z_{k}^{\prime\prime}-z_{j}^{\prime\prime}|\geq 1-|z_{j}^{\prime\prime}|-\frac{1}{2}(1-|z_{j}^{\prime\prime}|)=\frac{1}{2}(1-|z_{j}^{\prime\prime}|),$$

and by Lemma 2.1 and the Euclidean sine law

$$l^{\prime}_{jk}\leq\frac{2|z_{k}^{\prime\prime}-z_{j}^{\prime\prime}|}{\frac{1}{2}(1-|z_{j}^{\prime\prime}|)}\leq\frac{4|z_{j}^{\prime\prime}-z_{i}^{\prime\prime}|}{(1-|z_{j}^{\prime\prime}|)\sin(\epsilon/2)}\leq\frac{16}{\epsilon}\cdot\frac{|z_{j}^{\prime\prime}-z_{i}^{\prime\prime}|}{1-|z_{j}^{\prime\prime}|}\leq\frac{32}{\epsilon}l^{\prime}_{ij}\leq 2.$$

For the same reason $l^{\prime}_{ik}\leq 2$. On the other hand $l^{\prime}_{ik}\geq\epsilon/16$ since $ik\notin E_{0}$. Then we can derive a contracdiction using the hyperbolic sine law as the following.

$$1>e^{\tilde{u}_{i}}=\frac{(\sinh\frac{l_{ij}^{\prime}}{2}/\sinh\frac{l_{ij}}{2})\cdot(\sinh\frac{l_{ik}^{\prime}}{2}/\sinh\frac{l_{ik}}{2})}{\sinh\frac{l_{jk}^{\prime}}{2}/\sinh\frac{l_{jk}}{2}}=\frac{\sinh\frac{l_{ik}^{\prime}}{2}}{\sinh\frac{l_{ij}}{2}}\cdot\frac{\sinh\frac{l_{ij}^{\prime}}{2}}{\sinh\frac{l_{jk}^{\prime}}{2}}\cdot\frac{\sinh\frac{l_{jk}}{2}}{\sinh\frac{l_{ik}}{2}}$$

$$\geq\frac{1}{8}\cdot\frac{l_{ik}^{\prime}}{l_{ij}}\cdot\frac{l_{ij}^{\prime}}{l^{\prime}_{jk}}\cdot\frac{\sinh l_{jk}}{\sinh{l_{ik}}}\geq\frac{1}{8}\cdot\frac{\epsilon/16}{l_{ij}}\cdot\frac{\epsilon}{32}\cdot\sin\epsilon\geq\frac{\epsilon^{3}}{8192l_{ij}}\geq 1.$$