On the Deformation of Thurston’s Circle Packings with Obtuse Intersection Angles

Abstract

We study Thurston’s circle packings with obtuse intersection angles on closed surfaces. By using combinatorial Ricci/Calabi flows and variational principle, we extend Thurston’s existence theorem for circle packings with non-obtuse intersection angles to those with obtuse intersection angles. As consequences, we generalize the existence and convergence results related to Chow-Luo’s combinatorial Ricci flows (J Differ Geom 63(1):97–129, 2018) and Ge’s combinatorial Calabi flows (Combinatorial Methods and Geometric Equations, Thesis (Ph.D.), Peking University, Beijing, 2012, Trans Am math Soc 370(2):1377–1391, 2018, Adv Math 333:528–533, 2018).

1 Introduction

1.1 Backgrounds

In the pioneering work of Thurston [32], he introduced circle packings (on triangulated surfaces with non-obtuse intersection angles) to construct hyperbolic metrics on a closed surface. The main idea is to take the triangulation as the nerve of a circle packing, from which the metric structure on the triangulated surface can be constructed via radii of circles and intersection angles between circles in the packing.

We recall Thurston’s construction first. Let M be a closed surface with a triangulation ${\mathcal {T}} = (V,E,F)$, where V, E, F denote the sets of vertices, edges and faces, respectively. A circle packing is a positive function on the vertices which defined as $r: V\rightarrow (0,+\infty ), v_{i}\mapsto r_{i}, i=1,\ldots ,N$, where $N = |V |$ is the number of vertices.. Fix a triangulated surface $(M, {\mathcal {T}})$. Each circle packing r endows a metric structure on $(M, {\mathcal {T}})$ as follows. Let ${\mathbb {E}}^2$ (${\mathbb {H}}^2$ resp.) denote the Euclidean plane (the hyperbolic plane resp.) with constant Gaussian curvature 0 (-1 resp.). Using the cosine law in ${\mathbb {E}}^2$ (${\mathbb {H}}^2$ resp.), one can equip each edge $\{ij\}\in E$ with the length

$$\begin{aligned} l_{ij}= & {} \sqrt{r^2_i+r^2_j+2r_ir_j\cos \Phi _{ij}} \end{aligned}$$

(1.1)

$$\begin{aligned} (l_{ij}= & {} \cosh ^{-1}(\cosh r_i \cosh r_j + \sinh r_i \sinh r_j \cos \Phi _{ij}) \;\;\text {resp.}). \end{aligned}$$

(1.2)

Thurston proved a three-circle configuration theorem, which reads as for each face $\{ijk\}\in F$, the three edge lengths $l_{ij}$, $l_{jk}$, $l_{ik}$ satisfy the triangle inequalities, see Lemma 13.7.2 in [32]. This makes each face in F isometric to a triangle in ${\mathbb {E}}^2$ (${\mathbb {H}}^2$ resp.). Furthermore, a triangulated surface $(M, {\mathcal {T}})$ could be constructed by gluing these Euclidean (hyperbolic resp.) triangles coherently, i.e. along common edges. The resulting surface has a flat (hyperbolic resp.) cone metric with cone points in V. Obviously, this metric has no singularity on $M-V$ (one should notice that there is no singularity in the interior of each edges). The singularities of this metric are recorded at each vertex i in V by the so called combinatorial Gaussian curvature (also called discrete Gaussian curvature) $K_i$, which equals to $2\pi$ minus the cone angle at i. Denote $\theta _{i}^{jk}$ by the inner angle at the vertex i in the triangle $\{ijk\}\in F$, then we can express the combinatorial Gaussian curvature at i as

$$\begin{aligned} K_{i} = 2\pi -\sum _{\{ijk\}\in F}\theta _{i}^{jk}, \end{aligned}$$

(1.3)

where the sum is taken over all triangles with i as one of its vertices. Similar to the smooth case, the following combinatorial version of Gauss-Bonnet formula holds true:

$$\begin{aligned} \sum _{i=1}^{N}K_{i}=2\pi \chi (M)-\lambda \text {Area}(M), \end{aligned}$$

(1.4)

where $\lambda =-1, 0$ correspond the two geometries, i.e., hyperbolic geometry ${\mathbb {H}}^{2}$ and Euclidean geometry ${\mathbb {E}}^{2}$. Consider the curvature map $K=K(r)$, where K varies as r varies. In Euclidean background geometry, we concerns a particular circle packing $r_{av}$ that determines a constant combinatorial curvature $K(r_{av})=(K_{av},\ldots ,K_{av})$ with

$$\begin{aligned} K_{av}=\frac{2\pi \chi (M)}{N} \end{aligned}$$

for the reason that the corresponding Euclidean cone metric on $(M, {\mathcal {T}})$ curves the same way around each vertex.

In hyperbolic background geometry, we concerns a particular circle packing $r_{ze}$ that determines the zero curvature $K(r_{ze})=0$ for the reason that the corresponding hyperbolic cone metric has no singularities, and hence is a complete hyperbolic metric on $(M, {\mathcal {T}})$. However, Thurston find that there are combinatorial and topological obstacles (see [32], 13.6) for the existence of such circle packings $r_{av}$ and $r_{ze}$. In fact, Thurston’s existence theorem characterized perfectly the image of the curvature map $K=K(r)$ by the information of the combinatorics ${\mathcal {T}}$ and the topology of M. Thurston also suggested an algorithm to find particular circle patterns with polynomial convergence rate. Inspired by Hamilton’s Ricci flow method, Chow-Luo [5] further introduced a combinatorial version of Ricci flow, which is the negative gradient flow of the Ricci potential. The flows can be used to deform the circle packing to a particular one with exponential convergence rate. Similarly, the thesis of Ge [8] (or see [9, 10]) introduced a combinatorial version of Calabi flow, which is the negative gradient flow of a more nature Calabi energy $\Vert K\Vert ^2=\sum _iK_i^2$. Since then, various discrete curvature flows were introduced and studied. We refer the the readers to Luo [30, 31], Guo [23], Glickenstein [21, 22], Ge-Xu [14, 16, 18], Ge-Jiang [11, 12], Lin-Zhang [28, 29].

1.2 Main Results

It is noticeable that the above work deal with circle packings with non-obtuse intersection angles, i.e. $\Phi \in [0,\frac{\pi }{2}]$. The purpose of this paper is to study combinatorial Ricci/Calabi flows under the condition $\Phi \in [0,\pi )$. See Fig. 1 for all possible arrangements of the circles.

Fig. 1

The position relations of two circles

The idea originated from Huang-Liu [26] and Zhou’s [36] pioneering observation:

(HLZ)::

For each triangle $\{ijk\}\in F$, either $\Phi _{ij}+\Phi _{jk}+\Phi _{ik}\le \pi$, or

$$\begin{aligned} \Phi _{ij}+\Phi _{jk}<\pi +\Phi _{ik}, \Phi _{ik}+\Phi _{jk}<\pi +\Phi _{ij}, \Phi _{ij}+\Phi _{ik}<\pi +\Phi _{jk}. \end{aligned}$$

Under the (HLZ) condition, by a complicated calculation, Zhou proved that Thurston’s three-circle configuration theorem (i.e. for each face $\{ijk\}\in F$, the three edge lengths $l_{ij}$, $l_{jk}$, $l_{ik}$ satisfy the triangle inequalities, see Lemma 13.7.2 in [32]) is still valid. Geometrically, the three circles at the vertices have a power center $O_{ijk}$, see Fig. 2. Assuming (HLZ) and without further assumptions on $\Phi$, the power center $O_{ijk}$ may lie outside the triangle $\triangle v_iv_jv_k$. This is the main obscure to extend Andreev-Thurston’s rigidity results and Chow-Luo and Ge’s combinatorial Ricci/Calaib flow results. To overcome this obscure, Zhou first introduced the following condition

(Z)::

Set $I_{st}=\cos \Phi _{st}$ for $st=ij, jk, ik$. For each triangle $\{ijk\}\in F$, there holds

$$\begin{aligned} I_{ij}+ I_{ik}I_{jk}\ge 0,\ I_{ik}+ I_{ij}I_{jk}\ge 0,\ I_{jk}+ I_{ij}I_{ik}\ge 0. \end{aligned}$$

As pointed by Zhou, (Z) implies (HLZ). Thus under (Z) condition, Thurston’s three-circle configuration theorem is valid, and the power center $O_{ijk}$ is inside the triangle $\triangle v_iv_jv_k$.

Fig. 2

A three-circle configuration

Fig. 3

Two adjacent triangles

Our first result says that Thurston’s existence theorem (i.e. the image of the curvature map $K=K(r)$ as a convex polytope whose boundary is characterized by the combinatorics of ${\mathcal {T}}$ and the topology of M) is the same as non-obtuse intersection angle case. Let $F_A$ be the sub-complex constituted of those t-simplex ($t=0,1,2$) that have at least one vertex in A, and Lk(A) is the set of pairs (e, v) of an edge e and a vertex v satisfying (i) $v\in A$; (ii) both the two end points of e are not in A; (iii) v and the two end points of e forms a triangle in F. We have

Theorem 1.1

Assume that $\Phi \in [0,\pi )$ satisfies (Z). In Euclidean background geometry, the image of the curvature map K consists of vectors $(K_1, K_2,\ldots , K_{|V|})$ satisfying

$$\begin{aligned} \sum _{i\in A}K_i>-\sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )+2\pi \chi (F_A) \end{aligned}$$

(1.5)

for any non-empty subset A of V, where the equality holds if and only if $A=V$. While in hyperbolic background geometry, the image of the curvature map K consists of vectors $(K_1, K_2,\ldots , K_{|V|})$ satisfying

$$\begin{aligned} \sum _{i\in A}K_i>-\sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )+2\pi \chi (F_A) \end{aligned}$$

(1.6)

for any non-empty subset A of V.

Remark 1

(1.5) was obtained by Ge-Jiang [13]. We restate here for completeness.

Thus the existence problem of a particular packing (such as $r_{av}$ and $r_{ze}$) transfers to the problem whether $K_{av}$ (zero resp.) is in the image of the curvature map K(r). Recall ${\mathcal {C}}(r)=\Vert K-K_{av}\Vert ^2$ (${\mathcal {C}}(r)=\Vert K\Vert ^2$ resp.) is the Euclidean (hyperbolic resp.) combinatorial Calabi energy, and a coordinate change $u_i=\ln r_i$ ($u_i=\ln \tanh \frac{r_i}{2}$ resp.) in Euclidean (hyperbolic resp.) background geometry. To find such a particular packing, we use Ge’s combinatorial Calabi flow [8,9,10]

$$\begin{aligned} u_i'(t)=-\frac{1}{2}\partial _{u_i}{\mathcal {C}}=\Delta K_i, \end{aligned}$$

(1.7)

for $i=1, 2, \ldots , N$, where ${\mathcal {C}}$ is considered as a function of $u=(u_1,\ldots ,u_N)$ in the expression $\partial _{u_i}{\mathcal {C}}$. Moreover, set

$$\begin{aligned} B_{ij}=\frac{\partial (\theta _i^{jk}+\theta _i^{jl})}{\partial u_j}, \end{aligned}$$

$\Delta$ is a discrete Laplacian and

$$\begin{aligned} \Delta K_i&=\sum _{j\thicksim i}B_{ij}(K_j-K_i)\quad \text {in}\ {\mathbb {E}}^{2}; \\ \Delta K_i&=\sum _{j\thicksim i}B_{ij}(K_j-K_i)-A_iK_i,\quad \text {in}\ {\mathbb {H}}^{2} \end{aligned}$$

with

$$\begin{aligned} A_i=\frac{\partial }{\partial u_i}\big (\sum _{\{ijk\}\in F}\text {Area}(\triangle v_iv_jv_k)\big ). \end{aligned}$$

We have

Theorem 1.2

Assume that $\Phi \in [0,\pi )$ satisfies (Z). In Euclidean background geometry, the solution r(t) to the combinatorial Calabi flow (1.7) exists for all the time $t\ge 0$, and the following properties $(E_1)$-$(E_3)$ are equivalent:

$(E_1)$:

r(t) converges as $t\rightarrow +\infty$.

$(E_2)$:

The vector $(K_{av},\ldots ,K_{av})$ belongs to the image of the curvature map.

$(E_3)$:

If A is a proper non-empty subset of V, then

$$\begin{aligned} 2\pi \chi (M)\frac{|A|}{|V|}>-\sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )+2\pi \chi (F_A). \end{aligned}$$

(1.8)

Moreover, if one of the above properties holds, then the combinatorial Calabi flow converges exponentially fast to a circle packing which produces an Euclidean cone metric on M with cone angles all equal to $2\pi -K_{av}$.

In hyperbolic background geometry, the results are more fruitful. Recall Chow-Luo’s hyperbolic combinatorial Ricci flow [5]

$$\begin{aligned} \frac{dr_i(t)}{dt}=-K_i\sinh r_i \end{aligned}$$

(1.9)

for $i=1, 2, \ldots , N$. We have

Theorem 1.3

Assume that $\Phi \in [0,\pi )$ satisfies (Z). In hyperbolic background geometry, the solutions to the combinatorial Ricci flow (1.9) exists for all the time $t\ge 0$, and the following properties $(H_1)$-$(H_6)$ are equivalent:

$(H_1)$:

The solution r(t) to the combinatorial Ricci flow (1.9) converges as $t\rightarrow +\infty$.

$(H_2)$:

If A is a proper non-empty subset of V, then

$$\begin{aligned} \sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )>2\pi \chi (F_A). \end{aligned}$$

(1.10)

$(H_3)$:

The genus $g>1$ and for any simple, null-homotopic closed path $e_1,e_2,\ldots ,e_s$, which is not the boundary of a triangle, there holds

$$\begin{aligned} \sum _{i=1}^s \Phi (e_i)<(s-2)\pi . \end{aligned}$$

(1.11)

$(H_4)$:

The origin $(0,\ldots ,0)$ belongs to the image of the curvature map.

$(H_5)$:

The solution r(t) to the combinatorial Calabi flow (1.7) converges as $t\rightarrow +\infty$.

$(H_6)$:

The image of the curvature map contains a point with non-positive coordinates.

Moreover, if one of the above properties holds, then the combinatorial Ricci/Calabi flow converges exponentially fast to a circle packing which produces a complete hyperbolic metric on M (with no cone points).

Remark 2

The results in the above theorem related to the combinatorial Ricci flows are essentially obtained by Ge-Hua-Zhou [19, 20]. In fact, they studied the combinatorial Ricci flow on surfaces of finite type. While our results are established on closed surfaces.

Our last result concerns the prescribed curvature problem. For any ${\bar{K}}=(\bar{K_1},\ldots ,\bar{K_N})$, the prescribed Calabi energy is $\bar{{\mathcal {C}}}(r)=\Vert K-{\bar{K}}\Vert ^2$. Consider the prescribed Ricci flow

$$\begin{aligned} \frac{du_i}{dt}=(\bar{K_i}-K_i) \end{aligned}$$

(1.12)

and the prescribed Calabi flow

$$\begin{aligned} u_i'(t)=-\frac{1}{2}\partial _{u_i}\bar{{\mathcal {C}}}=\Delta (K_i-\bar{K_i}) \end{aligned}$$

(1.13)

in Euclidean background geometry, we have

Theorem 1.4

Assume that $\Phi \in [0,\pi )$ satisfies (Z). The solutions r(t) to the prescribed Ricci flow (1.12) (Calabi flow (1.13) resp.) exists for all the time $t\ge 0$, and r(t) converges if and only if ${\bar{K}}$ belongs to the image of the curvature map. Moreover, r(t) converges exponentially fast to the unique (up to scaling in Euclidean background geometry) circle packing ${\bar{r}}$ with $K({\bar{r}})={\bar{K}}$.

This paper is organized as follows. In Sect. 2, we study the three-circle configurations and some useful lemmas. In Sect. 3, we give the proof of Theorem 1.2. In Sect. 4.1, we obtain an uniform estimate about the solutions to the combinatorial Calabi flow for hyperbolic background geometry. We give the proof of Theorem 1.3 in Sect. 4.2. In Appendix 1, we get the existence of the three-circle configurations for Euclidean background geometry; in Appendix 2, we obtain the image of K(r) for hyperbolic background geometry; in Appendix 3, we give a direct detailed proof of the uniform bounded from above to the $\frac{\partial \theta _i^{jk}}{\partial u_j}$.

2 Preliminaries: Three-Circle Configurations

To endow a metric structure on $(M,{\mathcal {T}})$ with the help of circle packings, we first need a three-circle configuration.

Lemma 2.1

Let $\Phi _{ij}$, $\Phi _{jk}$, $\Phi _{ik}\in [0,\pi )$ be three intersection angles satisfying (HLZ). For any three positive numbers $r_i$, $r_j$, $r_k$, there exits a configuration of three mutually intersecting closed disks in both Euclidean and hyperbolic geometry, unique up to congruence, having radii $r_i$, $r_j$, $r_k$ and meeting with exterior intersection angles $\Phi _{ij}$, $\Phi _{jk}$, $\Phi _{ik}\in [0,\pi )$.

Proof

In hyperbolic geometry, this was proved by Zhou, see Lemma 2.4 in [36]. In Euclidean geometry, we postpone its proof to Appendix 1.$\square$

Lemma 2.2

([36], Proposition 5.1) Let $\Phi _{ij}$, $\Phi _{jk}$, $\Phi _{ik}\in [0,\pi )$ be three intersection angles satisfying (Z), then they satisfy (HLZ). Consequently, the above three-circle configuration theorem is valid under the condition (Z).

Remark 3

Zhou’s proof of Proposition in [36] can be used to both Euclidean and hyperbolic background geometry.

Lemma 2.3

([33], Lemma 2.6) Let $\Phi _{ij}$, $\Phi _{jk}$, $\Phi _{ik}\in [0,\pi )$ be three intersection angles satisfying (Z). In Euclidean background geometry, the Jacobian matrix of functions $\theta _i^{jk}$, $\theta _j^{ik}$, $\theta _k^{ij}$ in terms of $u_i$, $u_j$, $u_k$ is symmetric and semi-negative definite with rank 2 and kernel $\{t(1,1,1)|t\in {\mathbb {R}}\}$. Moreover, $\frac{\partial \theta ^{jk}_i}{\partial u_i}<0$ and $\frac{\partial \theta ^{jk}_i}{\partial u_j}\ge 0$.

Lemma 2.4

([36], Lemma 5.5) Let $\Phi _{ij}$, $\Phi _{jk}$, $\Phi _{ik}\in [0,\pi )$ be three intersection angles satisfying (Z). In hyperbolic background geometry, the Jacobian matrix of functions $\theta _i^{jk}$, $\theta _j^{ik}$, $\theta _k^{ij}$ in terms of $u_i$, $u_j$, $u_k$ is symmetric and negative definite. Moreover, $\frac{\partial \theta _i^{jk}}{\partial u_i}<0$, $\frac{\partial \theta _i^{jk}}{\partial u_j}\ge 0$, $\frac{\partial (\theta _i^{jk}+\theta _j^{ik}+\theta _k^{ij})}{\partial u_i}<0$.

Lemma 2.5

([33], Corollaries 2.7 and 3.8) Let $\Phi _{ij}$, $\Phi _{jk}$, $\Phi _{ik}\in [0,\pi )$ be three intersection angles satisfying (Z). In Euclidean background geometry, the Jacobian matrix $\Delta =-\frac{\partial (K_1,\ldots , K_N)}{\partial (u_1,\ldots , u_N)}$ is symmetric and semi-negative definite with rank $N-1$ and kernel $\{t{{\textbf {1}}}|t\in {\mathbb {R}}\}$. In hyperbolic background geometry, the Jacobian matrix $\Delta =-\frac{\partial (K_1,\ldots , K_N)}{\partial (u_1,\ldots , u_N)}$ is symmetric and negative definite.

3 Euclidean Geometry Background

Suppose $\triangle v_iv_jv_k$ is a topological triangle in F. We use $l_{ij}, l_{jk},l_{ik}$ (defined as (1.1)) to denote the lengths of the edge $v_iv_j, v_jv_k,v_iv_k$, respectively. From Lemma 2.4 in [36], we can see that there is no restriction on the radii such that the lengths $l_{ij}, l_{jk},l_{ik}$ for $\triangle v_iv_jv_k\in F$ satisfy the triangle inequalities.

Proposition 3.1

([9], Proposition 4.1) Along the Combinatorial Calabi flow, the discrete Gauss curvature evolves according to

$$\begin{aligned} \frac{dK}{dt}=-L^2K, \end{aligned}$$

where $L=-\Delta$ and $\Delta$ defined as in Lemma 2.5.

Lemma 3.2

For any topological triangle $\triangle v_iv_jv_k\in F$ with fixed weights $\Phi _{ij},\Phi _{jk},\Phi _{ik}\in [0,\pi )$ as intersection angles satisfying (Z), then there exists a constant $C(\Phi )$ which only depends on the given $\Phi$ such that

$$\begin{aligned} 0<\frac{\partial \theta _i^{jk}}{\partial u_j}\le C(\Phi ). \end{aligned}$$

Fig. 4

The corresponding contact graph

Proof

Here we give a geometric proof by following He [25]. Under the (Z) condition, let O be the power center which is inside the triangle $\triangle v_iv_jv_k$, then $\theta _i^{Oj}\in [0,\frac{\pi }{2})$. See Fig. 4, let $h_{ij}$ be the length of the altitude from O onto side $v_iv_j$ and $h_{ik}$ be the length of the altitude from O onto side $v_iv_k$. Thus $\frac{\partial \theta _i^{jk}}{\partial u_j}=\frac{h_{ij}}{l_{ij}}$.

If $\theta _i^{Oj}$ is not close to $\frac{\pi }{2}$, say, $\theta _i^{Oj}\in [0,1)$, we have

$$\begin{aligned} \frac{\partial \theta _i^{ik}}{\partial u_j}=\frac{h_{ij}}{l_{ij}}<\tan \theta _i^{Oj}\le C(\Phi ), \end{aligned}$$

where $C(\Phi )$ is a positive constant which only dependent on the given $\Phi$.

If $\theta _i^{Oj}\in [1,\frac{\pi }{2})$, then $\theta _i^{jk}\ge \theta _i^{Oj}\ge 1$ is bounded from below. Since O lies in the convex hull of the union of circles $C_{i}$, $C_{j}$, we know $r_i<l_{ij}$ and $r_j<l_{ij}$. Thus $h_{ij}<2l_{ij}$. It follows that

$$\begin{aligned} \frac{\partial \theta _i^{ik}}{\partial u_j}=\frac{h_{ij}}{l_{ij}}<2. \end{aligned}$$

Which completes the proof. $\square$

By Lemma 2.5 in [33], we can see $B_{ij}=B_{ji}$. Directly from Lemma 3.2, we have

Lemma 3.3

For any two adjacent topological triangles $\triangle v_iv_jv_k,\ \triangle v_iv_jv_l\in F$ with fixed weights $\Phi _{ij},\Phi _{jk},\Phi _{ik},\Phi _{il},\Phi _{jl}\in [0,\pi )$ as intersection angles satisfying (Z), then there exists a constant $C(\Phi )$ which only depends on the given weight $\Phi$ such that

$$\begin{aligned} 0<B_{ij}\le C(\Phi ). \end{aligned}$$

3.1 Proof of Theorem 1.2

We first study the long time existence of the combinatorial Calabi flows (1.7) in Euclidean geometry background and get the following theorem.

Theorem 3.4

Given a triangulated surface $(M,{\mathcal {T}},\Phi )$ in ${\mathbb {E}}^{2}$ with weight $\Phi \in [0,\pi )$ satisfying (Z). For any initial circle packing metric $r(0)\in {\mathbb {R}}^{N}_{>0}$, the solution to the combinatorial Calabi flow (1.7) in ${\mathbb {E}}^{2}$ exists for all time $t\in [0, +\infty )$.

Proof

Let $d_{i}$ denote the degree at vertex $v_{i}$, which is the number of edges adjacent to $v_{i}$. Set $d=\max \{d_1,\ldots ,d_N\}$, then $(2-d)\pi< K_i < 2\pi$, and

$$\begin{aligned} |K_{j}-K_{i}|<d\pi ,\ \text {for all}\ i\in \{1, \ldots , N\}. \end{aligned}$$

Hence, by Lemma 3.3, the combinatorial Calabi flow equation $\sum _{j\thicksim i}B_{ij}(K_{j}-K_{i})$ are uniformly bounded by a positive constant $c_{1}= 2\pi dC(\Phi )$, which depends only on the triangulation and the fixed weight $\Phi$, where $C(\Phi )$ is a positive constant comes from Lemma 3.3. Then we have

$$\begin{aligned} c_{0}e^{-c_{1}t}\le r_{i}(t)\le c_{0}e^{c_{1}t}, \end{aligned}$$

where $c_{0} = c(r(0))$, which implies that the combinatorial Calabi flow (1.7) has a solution for all time $t\in [0, +\infty )$ for any $r(0)\in {\mathbb {R}}_{>0}^{N}.$ $\square$

Now, we give the proof of Theorem 1.2.

Proof

We first show “$(E_1)\Rightarrow (E_2)$”. From Theorem 3.4, we know the solution of the combinatorial Calabi flow (1.7) in ${\mathbb {E}}^{2}$ exists for all the time. Then we can denote $r(t),\ t\in [0,+\infty )$ as the solution of the combinatorial Calabi flow (1.7) in ${\mathbb {E}}^{2}$. We recall the definition of combinatorial Calabi energy in [9], that is

$$\begin{aligned} {\mathcal {C}}(r)=\Vert K-K(r_{av})\Vert ^2=\sum _{i=1}^N(K_i-K_{av})^2. \end{aligned}$$

(3.1)

In fact, the combinatorial Calabi flow (1.7) is the negative gradient flow of combinatorial Calabi energy, and the Calabi energy (3.1) is descending along this flow.

If $\{r(t)|t\in [0,+\infty )\}$ converges, i.e.,

$$\begin{aligned} r(+\infty )=\lim _{t\rightarrow +\infty }r(t)\in {\mathbb {R}}^N_{>0} \end{aligned}$$

exists, then both $K(+\infty )=\lim _{t\rightarrow +\infty }K(t)\in {\mathbb {R}}^N_{>0}$ and $L(+\infty )=\lim _{t\rightarrow +\infty }L(t)\in {\mathbb {R}}^N_{>0}$ exist. This leads to the existence of ${\mathcal {C}}(+\infty )$ and ${\mathcal {C}}'(+\infty )$. Combining with the fact that ${\mathcal {C}}(t)$ is uniformly bounded and using Lemma 2.5 and Proposition 3.1, we have

$$\begin{aligned} {\mathcal {C}}'(t)=2\sum _{i=1}^NK'_iK_i=2K^TK'=-2K^TL^2K\le 0, \end{aligned}$$

and then

$$\begin{aligned} {\mathcal {C}}'(+\infty )=-2K^T(+\infty )L^2(+\infty )K(+\infty )=0. \end{aligned}$$

Hence

$$\begin{aligned} K(+\infty )\in Ker(L^2)=Ker(L). \end{aligned}$$

By Lemma 2.3, we know $K(+\infty )$ is a constant and $r(+\infty )$ is a constant curvature metric.

Now we show “$(E_2)\Rightarrow (E_1)$”. Assume there exists a constant curvature circle packing metric $r_{av}$ which implies $K(r_{av})\in K({\mathbb {R}}^N_{>0})$. We claim $\{r(t)|t\in [0,+\infty )\}\subset \subset {\mathbb {R}}^N_{>0}$. Consider the combinatorial Ricci potential

$$\begin{aligned} F(u)=\int ^u_{u_{av}}\sum _{i=1}^N(K_i-K_{av})du_i,\ u\in {\mathbb {R}}^N, \end{aligned}$$

(3.2)

where $u_{av}=\ln r_{av}$. This type of line integral was first introduced by Verdière in [6]. By Lemma 2.5, we know

$$\begin{aligned} \frac{\partial K_i}{\partial u_j}=\frac{\partial K_j}{\partial u_i}, \end{aligned}$$

the smooth differential 1-form $\sum _{i=1}^N(K_i-K_{av})du_i$ is closed, and hence then (3.2) is well defined and is independent on the choice of piecewise smooth paths in ${\mathbb {R}}^N$ from $u_{av}$ to u. By Lemma B.2 in [9], we know that F(u) is strictly convex, $u_{av}$ is the unique critical point and

$$\begin{aligned} \lim _{\Vert u\Vert \rightarrow +\infty ,u\in {\mathbb {R}}^N}F(u)=+\infty . \end{aligned}$$

(3.3)

Set $\varphi (t)=F(u(t))$, then

$$\begin{aligned} \varphi '(t)=\sum _{i=1}^N(K_i-K_{av})\frac{du_i}{dt}=(K-K(r_{av}))^T(-LK)=-K^TLK\le 0, \end{aligned}$$

which means $\varphi (t)$ is descending as t increases. Combine with (3.3), we have $\{u(t)|t\in [0,+\infty )\}\subset \subset {\mathbb {R}}^N$, i.e., $\{r(t)|t\in [0,+\infty )\}\subset \subset {\mathbb {R}}^N_{>0}$. Hence r(t) converges.

Denote $\lambda _1$ as the minimum positive eigenvalue of $L=-\Delta$. Since the matrix L is semi-positive definite by Lemma 2.3, $\lambda ^2_1$ is the minimum positive eigenvalue of $L^2$. Then

$$\begin{aligned} K^TL^2K=(K-K(r_{av}))^TL^2(K-K(r_{av}))\ge \lambda ^2_1\Vert K-K(r_{av})\Vert ^2=\lambda ^2_1{\mathcal {C}}. \end{aligned}$$

Since r(t) converges, $\lambda ^2_1(t)$ has a uniform lower bound along the Combinatorial Calabi flow, i.e., $\lambda ^2_1(t)\ge \lambda >0$, where $\lambda$ is a positive constant. Hence

$$\begin{aligned} {\mathcal {C}}'(t)=-2K^TL^2K\le -2\lambda ^2_1(t){\mathcal {C}}(t)\le -2\lambda {\mathcal {C}}(t). \end{aligned}$$

So ${\mathcal {C}}(t)\le {\mathcal {C}}(0)e^{-2\lambda t}$ and using (3.1),

$$\begin{aligned} |K_{i}(t)-K_{av}|\le |K(t)-K(r_{av})|\le \sqrt{{\mathcal {C}}(t)}\le \sqrt{{\mathcal {C}}(0)}e^{-\lambda t}. \end{aligned}$$

Since $\{r(t)|t\in [0,+\infty )\}\subset \subset {\mathbb {R}}^N_{>0}$, $r_i$ is bounded along the combinatorial Calabi flow. By Lemma 3.3, we have

$$\begin{aligned} \left| \frac{dr_i}{dt}\right| =|r_i\Delta K_i|=|r_i\sum _{j\thicksim i}B_{ij}(K_j-K_i)|\le C(M,\Phi )e^{-\lambda t}, \end{aligned}$$

where $C(M,\Phi )$ is a positive constant and only dependent on M and $\Phi$. This implies that the solution converges with exponential rate.

In view of $\sum _{i\in V}K_i=2\pi \chi (M)$ and Theorem 1.1, “$(E_2)\Leftrightarrow (E_3)$” is obviously.

Moreover, from the step “$(E_1)\Rightarrow (E_2)$”, we can see that the combinatorial Calabi flow converges exponentially fast to a circle packing which produces an Euclidean cone metric on M with cone angles all equal to $2\pi -K_{av}$. $\square$

It is easy to see that we can also proof Theorem 1.4 by using the similar method of Theorems 3.4 and 1.2.

4 Hyperbolic Geometry Background

4.1 An Uniform Estimate

Suppose $\triangle v_iv_jv_k$ is a topological triangle in F. We use $l_{ij}, l_{jk},l_{ik}$ (defined as (1.2)) to denote the lengths of the edge $v_iv_j, v_jv_k,v_iv_k$, respectively. Zhou [36] obtained that there is no restriction on the radii such that the lengths $l_{ij}, l_{jk},l_{ik}$ for $\triangle v_iv_jv_k\in F$ satisfy the triangle inequalities.

If $\Phi _{ij},\Phi _{ik},\Phi _{jk}\in [0,\pi /2)$, the following result was first proved by Chow-Luo, Lemma 3.5 in [5], with a geometric argument. Moreover, Ge-Xu [17] Lemma 3.2, Ge-Jiang [13] Lemma 2.3 stated it by an analytic proof. Now we give a similar proof of Ge-Xu [17] Lemma 3.2 just for completeness.

Lemma 4.1

Let $\triangle v_iv_jv_k$ be a hyperbolic triangle which is patterned by three circles with fixed weighted $\Phi _{ij},\Phi _{ik},\Phi _{jk}\in [0,\pi )$ as intersection angles which satisfies (Z). Let $\theta _i^{jk}$ be the inner angle at $v_i$. Then for any $\epsilon >0$, there exists a number l so that when $r_i>l$, the inner angle $\theta _i^{jk}$ is smaller than $\epsilon$.

Proof

It is sufficient to prove that $\theta _i^{jk}\rightarrow 0$ uniformly as $r_i\rightarrow +\infty$. Set $a=\frac{\cosh (l_{ij}-l_{ik})}{\cosh (l_{ij}+l_{ik})}$ and $b=\frac{\cosh l_{jk}}{\cosh (l_{ij}+l_{ik})}$. By the hyperbolic cosine law, we have

$$\begin{aligned} \cos \theta _i^{jk}&=\frac{\cosh l_{ij}\cosh l_{ik}-\cosh l_{jk}}{\sinh l_{ij}\sinh l_{ik}}\nonumber \\&=\frac{\cosh (l_{ij}+l_{ik})+\cosh (l_{ij}-l_{ik})-2\cosh l_{jk}}{\cosh (l_{ij}+l_{ik})-\cosh (l_{ij}-l_{ik})}\nonumber \\&=\frac{1+a-2b}{1-a}. \end{aligned}$$

(4.1)

Noting that

$$\begin{aligned} 0<a<\frac{\cosh l_{ij}}{\cosh (l_{ij}+l_{ik})}<\frac{1}{\cosh l_{ik}}<\frac{1}{\cosh r_i}, \end{aligned}$$

we know $a\rightarrow 0$ uniformly as $r_i\rightarrow +\infty$. Now, we claim that $b\rightarrow 0$ uniformly as $r_i\rightarrow +\infty$.

Since $\cosh l_{jk}\le \cosh r_j\cosh r_k+\sinh r_j\sinh r_k=\cosh (r_j+r_k)$, we have

$$\begin{aligned} b\le \frac{\cosh (r_j+r_k)}{\cosh (l_{ij}+l_{ik})}. \end{aligned}$$

(4.2)

Set $c_{ij}=\min \{\cos \Phi _{ij},0\}$, then $-1<c_{ij}\le 0$. Since the triangulation ${\mathcal {T}}$ is a finite subdivision, we have

$$\begin{aligned} e^{l_{ij}}\ge \cosh l_{ij}&=\cosh r_i\cosh r_j+\sinh r_i\sinh r_j\cos \Phi _{ij}\\&=(1+\cos \Phi _{ij})\cosh r_i\cosh r_j-\cosh (r_i-r_j)\cos \Phi _{ij}\\&\ge (1+c_{ij})\cosh r_i\cosh r_j\\ {}&\ge C_1e^{ r_i+r_j}, \end{aligned}$$

where $C_1\in (0,1/4]$ is a constant number and only depended on the triangulation ${\mathcal {T}}$. That is $l_{ij}\ge r_i+r_j+\ln C_1$. Similarly, we have $l_{ik}\ge r_i+r_k+\ln C_2$, where $C_1\in (0,1/4]$ is also a constant number and only depended on the triangulation ${\mathcal {T}}$. Hence, we get

$$\begin{aligned} l_{ij}+l_{ik}-(r_j+r_k)\ge 2r_i+\ln C_1+\ln C_2 \rightarrow +\infty \end{aligned}$$

(4.3)

uniformly as $r_i\rightarrow +\infty$. Combine (4.2) and (4.3), we know $b\rightarrow 0$ uniformly as $r_i\rightarrow +\infty$. Hence, by (4.1), we complete the proof. $\square$

Lemma 4.2

Let $\triangle v_iv_jv_k$ be a hyperbolic triangle which is patterned by three circles with fixed weights $\Phi _{ij},\Phi _{jk},\Phi _{ik}\in [0,\pi )$ as intersection angles and satisfying (Z). There exists a constant $C>0$ which is only depending on the triangulation ${\mathcal {T}}$, such that if $r_i\ge C$, then

$$\begin{aligned} \frac{\partial }{\partial r_i}(2\text {Area}(\triangle v_iv_jv_k)+\theta _i^{jk})\ge 0. \end{aligned}$$

(4.4)

Fig. 5

Proof of Lemma 4.2

Proof

The proof is similar to the proof of Lemma 3.2 in [10]. Assume the triangle $\triangle v_iv_jv_k$ is embedded in ${\mathbb {H}}^2$, with $v_j, v_k$ and the corresponding radii $r_j, r_k$ fixed. Let ${\hat{v}}_i$ be the new vertex with a larger radius ${\hat{r}}_i>r_i$. Since ${\hat{r}}_i>r_i$, we have $l_{{\hat{v}}_iv_j} > l_{v_iv_j}$ and $l_{{\hat{v}}_iv_k} > l_{v_iv_k}$. We draw two triangles, $\triangle v_iv_jv_k$ and $\triangle {\hat{v}}_iv_jv_k$; with common edge $v_jv_k$ in the same half hyperbolic plane separated by the (extended) geodesic $v_jv_k$. By Lemma 2.4, for fixed $r_j$ and $r_k$ the angles $\theta _j^{ik}$ and $\theta _i^{jk}$ are increasing in $r_i$ which implies that the vertex $v_i$ lies in the interior of the triangle $\triangle v_iv_jv_k$, see Fig. 5. Denote ${\hat{\theta }}_i^{jk}$, ${\hat{\theta }}_j^{ik}$ and ${\hat{\theta }}_k^{ij}$ as three inner angles of the new triangle $\triangle {\hat{v}}_iv_jv_k$ respectively. To prove (4.4), it suffices to show that for any ${\hat{r}}_i>r_i$, sufficiently close to $r_i$,

$$\begin{aligned} 2\text {Area}(\triangle {\hat{v}}_iv_jv_k)-2\text {Area}(\triangle v_iv_jv_k)+{\hat{\theta }}_i^{jk}-\theta _i^{jk}\ge 0. \end{aligned}$$

(4.5)

Setting $x={\hat{\theta }}_j^{ik}-\theta _j^{ik}$ and $y={\hat{\theta }}_k^{ij}-\theta _k^{ij}$, we get

$$\begin{aligned}&2\text {Area}(\triangle {\hat{v}}_iv_jv_k)-2\text {Area}(\triangle v_iv_jv_k)+{\hat{\theta }}_i^{jk}-\theta _i^{jk} \nonumber \\ =&\text {Area}(\triangle {\hat{v}}_iv_iv_j)+\text {Area}(\triangle {\hat{v}}_iv_iv_k)-x-y\nonumber \\ =&[\text {Area}(\triangle {\hat{v}}_iv_iv_j)-x]+[\text {Area}(\triangle {\hat{v}}_iv_iv_k)-y]. \end{aligned}$$

(4.6)

Then it suffices to prove that

$$\begin{aligned} \text {Area}(\triangle {\hat{v}}_iv_iv_j)\ge x,\ \text {and}\ \text {Area}(\triangle {\hat{v}}_iv_iv_k)\ge y. \end{aligned}$$

By the symmetry, without loss of generality, we show that

$$\begin{aligned} \text {Area}(\triangle {\hat{v}}_iv_iv_j)\ge x. \end{aligned}$$

(4.7)

Let s be the point on the geodesic ${\hat{v}}_iv_j$ which attains the minimum distance from the vertex $v_i$ to a point on the geodesic ${\hat{v}}_iv_j$. Since $l_{{\hat{v}}_iv_j} > l_{v_iv_j}$, s is in the interior of the geodesic ${\hat{v}}_iv_j$, see Fig. 5. We assume that ${\hat{r}}_i$ is sufficiently close to $r_i$ such that $l_{{\hat{v}}_iv_i}\le 1$.

By the hyperbolic cosine law,

$$\begin{aligned} \cos x=\frac{\cosh l_{v_iv_j}\cosh l_{{\hat{v}}_iv_j}-\cosh l_{{\hat{v}}_iv_i}}{\sinh l_{v_iv_j}\sinh l_{{\hat{v}}_iv_j}}\rightarrow 1, \end{aligned}$$

uniformly as $r_i\rightarrow \infty$. Hence there is a universal constant $C_1$ such that if $r_i\ge C_1$, then $x\le \frac{\pi }{8}$. Set $\beta =\angle sv_iv_j$.

If $\beta <\frac{\pi }{4}$, then $\beta +x+x<\frac{\pi }{2}$. By the Gauss-Bonnet theorem in the hyperbolic case,

$$\begin{aligned} \beta +x+\frac{\pi }{2}=\pi -\text {Area}(\triangle sv_iv_j), \end{aligned}$$

we get $\text {Area}(\triangle {\hat{v}}_iv_iv_j)\ge \text {Area}(\triangle sv_iv_j)>x,$ which yields (4.7).

If $\beta \ge \frac{\pi }{4}$, then

$$\begin{aligned} \frac{\sinh l_{sj}}{\sinh l_{ij}}=\sin \beta \ge \frac{\sqrt{2}}{2}. \end{aligned}$$

Using the cosine law in the hyperbolic right triangle $\triangle sv_iv_j$, we have

$$\begin{aligned} \cos \beta =\sin x\cosh l_{sj},\ \cos x=\tanh l_{sj}/\tanh l_{ij},\ \sin \beta =\sinh l_{sj}/\sinh l_{ij}. \end{aligned}$$

This leads to

$$\begin{aligned} \sinh (\text {Area}(\triangle sv_iv_j))&=\cos (x+\beta )=\cos x\cos \beta -\sin x\sin \beta \nonumber \\ {}&=\frac{\tanh l_{sj}}{\tanh l_{ij}}\sin x\cosh l_{sj}-\sin x\frac{\sinh l_{sj}}{\sinh l_{ij}}\nonumber \\ {}&=\sin x(\cosh l_{ij}-1)\frac{\sinh l_{sj}}{\sinh l_{ij}}. \end{aligned}$$

(4.8)

Set $c_{ij}=\min \{\cos \Phi _{ij},0\}$, then $-1<c_{ij}\le 0$. So, we have

$$\begin{aligned} \cosh l_{ij}&=\cosh r_i\cosh r_j+\sinh r_i\sinh r_j\cos \Phi _{ij}\\&=(1+\cos \Phi _{ij})\cosh r_i\cosh r_j-\cosh (r_i-r_j)\cos \Phi _{ij}\\&\ge (1+c_{ij})\cosh r_i\cosh r_j. \end{aligned}$$

Hence, combine with (4.8), there exists a large enough universal constant $C_2$ which depending only on the triangulation ${\mathcal {T}}$, such that if $r_i\ge C_2$, then

$$\begin{aligned} \sin (\text {Area}(\triangle sv_iv_j))\ge \sin x. \end{aligned}$$

Noting that both x and $\text {Area}(\triangle sv_iv_j)$ are in $(0,\frac{\pi }{2})$, we obtain (4.7).

By setting $\max \{C_1,C_2\}$, combining all cases above, we complete the proof. $\square$

Lemma 4.3

There exists a universal number $C>0$, such that if $r_i>C$, then

$$\begin{aligned} A_i\ge \sum _{j\thicksim i}B_{ij}. \end{aligned}$$

Proof

By (4.4), we have

$$\begin{aligned}&A_i-\sum _{j\thicksim i}B_{ij}=\sum _{\{ijk\}\in F}\frac{\partial \text {Area}(\triangle v_iv_j v_k)}{\partial r_i}\sinh r_i-\sum _{j\thicksim i}\left( \frac{\partial \theta _i^{jk}}{\partial r_j}\sinh r_j+\frac{\partial \theta _i^{jl}}{\partial r_j}\sinh r_j\right) \\&\quad =\sum _{\{ijk\}\in F}\frac{\partial \text {Area}(\triangle v_iv_j v_k)}{\partial r_i}\sinh r_i-\sum _{\{ijk\}\in F}\left( \frac{\partial \theta _i^{jk}}{\partial r_j}\sinh r_j+\frac{\partial \theta _i^{kj}}{\partial r_k}\sinh r_k\right) \\&\quad =\sum _{\{ijk\}\in F}\frac{\partial \left( \text {Area}(\triangle v_iv_j v_k)-\theta _j^{ik}-\theta _k^{ij}\right) }{\partial r_i}\sinh r_i\\&\quad =\sum _{\{ijk\}\in F}\frac{\partial \left( 2\text {Area}(\triangle v_iv_j v_k)+\theta _i^{jk}\right) }{\partial r_i}\sinh r_i\ge 0. \end{aligned}$$

$\square$

Theorem 4.4

Let $(M,{\mathcal {T}})$ be a triangulated compact hyperbolic surface with an edge weight $\Phi : E\rightarrow [0,\pi )$ which satisfies (Z). Let r(t) be the unique solution to the combinatorial Calabi flow on a maximal time interval [0, T). Then all $r_i(t)$ are uniformly bounded above on [0, T).

Proof

By contradiction. Suppose it is not true, then there exists at least one vertex $i\in V$, such that

$$\begin{aligned} \limsup _{t\rightarrow T}r_i(t)=+\infty . \end{aligned}$$

(4.9)

For this vertex i, using Lemma 4.1, we can choose a large enough positive number l such that $r_i>l$, the inner angle $\theta _i$ is smaller than $\frac{\pi }{d_i}$, where $d_i$ is the degree of the vertex i. Then we have $K_i>\pi$.

Set $L=\max \{l,c,r_i(0)+1\}$, where c is given in Lemma 4.3. Now, we claim that for any $t\in (0,T)$ and if $r_i(t)>l$, then

$$\begin{aligned} \frac{dr_i}{dt}<0. \end{aligned}$$

(4.10)

Since

$$\begin{aligned} \frac{1}{\sinh r_i}\frac{d r_i}{d t}&=\Delta K_i=\sum _{j\thicksim i}B_{ij}(K_j-K_i)-A_iK_i<\sum _{j\thicksim i}B_{ij}(2\pi -K_i)-A_iK_i \\&=2\pi \sum _{j\thicksim i}B_{ij}-\left( \sum _{j\thicksim i}B_{ij}+A_i\right) K_i \le 2\pi \sum _{j\thicksim i}B_{ij}-\pi \left( \sum _{j\thicksim i}B_{ij}+A_i\right) \\&=\pi \left( \sum _{j\thicksim i}B_{ij}-A_i\right) \le 0. \end{aligned}$$

Hence we proved the claim.

By (4.10), we may choose $t_0\in (0,T)$ such that $r_i(t_0)>c$. Let $t_1\in [0,t_0]$ attain the maximum of $r_i(t)$ in $[0,t_0]$. By the definition of L, $t_1>0$. Hence

$$\begin{aligned} \frac{d r_i}{d t}(t_1)\ge 0, \end{aligned}$$

which contradicts to (4.10). This proves the theorem. $\square$

4.2 Proof of Theorem 1.3

Proof

“$(H_1)\Rightarrow (H_2)$”. Suppose the solution r(t) to the combinatorial Ricci flow (1.9) converges as $t\rightarrow +\infty$. Let $u^{*}$ be the corresponding u-coordinate of $r^{*}$, then u(t) converges to $u^{*}$.

$$\begin{aligned} u_i(n+1)-u_i(n)=u'_i(\xi _n)=-K_i(\xi _n)\rightarrow 0,\quad \text {as}\ n\rightarrow +\infty . \end{aligned}$$

As $u(t)\rightarrow u^{*}$, we have $K(t)\rightarrow K(u^{*})$. Thus $K_i(u^{*})=0$ for each vertex $i\in V$ and then $u^{*}$ has zero curvature. This implies that there exists a particular circle pattern with all its curvatures $K_i\le 0$. By Theorem 1.1, we know the image of the curvature map K consists of vectors $(K_1, K_2,\ldots , K_{|V|})$ satisfying

$$\begin{aligned} \sum _{i\in A}K_i>-\sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )+2\pi \chi (F_A) \end{aligned}$$

for any non-empty subset A of V, which implies (1.10).

“$(H_2)\Leftrightarrow (H_3)$”. First we prove $(H_2)\Rightarrow (H_3)$. We follow the way in [20] and [19]. Assume

$$\begin{aligned} -\sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )+2\pi \chi (F_A)<0 \end{aligned}$$

for each $A\subset V$, we need to prove $\sum _{i=1}^s \Phi (e_i)<(s-2)\pi$, whenever $e_1, e_2, \ldots , e_s$ form a simple, null-homotopic closed path which is not the boundary of a triangle. Given such a path, we take $A\subset V$ as the interior vertices that are bounded by the simple, null-homotopic closed path $e_1, e_2, \ldots , e_s$. Because the path is not a boundary of a triangle, A is nonempty. In addition, $\chi (F_A)=1$ since $F_A$ is contractible. Moreover, it is easy to see

$$\begin{aligned} \sum _{l=1}^s\big (\pi -\Phi (e_l)\big )=\sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )>2\pi , \end{aligned}$$

which implies what we need to prove.

Next we prove $(H_3)\Rightarrow (H_2)$. Assume $\sum _{i=1}^s \Phi (e_i)<(s-2)\pi$, or equivalently,

$$\begin{aligned} \sum _{i=1}^s \big (\pi -\Phi (e_i)\big )>2\pi \end{aligned}$$

whenever $e_1, e_2, \ldots , e_s$ form a simple, null-homotopic closed path which is not the boundary of a triangle. We need to prove

$$\begin{aligned} -\sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )+2\pi \chi (F_A)<0 \end{aligned}$$

(4.11)

for each $A\subset V$. Let $A\subset V, A\ne \varnothing$. If $A=V$, then the above inequality degenerates to $\chi (M)<0$, which is already guaranteed by the genus $g>1$. For $A\varsubsetneqq V$, we just need to prove $\sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )>2\pi \chi (F_A)$ on each connected component of $F_A$. Hence we may assume that $F_A$ is connected. In this case, it is easy to see $\chi (F_A)\le 1$. If $\chi (F_A)\le 0$, then (4.11) holds naturally. If $\chi (F_A)=1$, all triangles in $F_A$ (i.e. all triangles that has at least one boundary vertex in A) constitutes a simply-connected domain bounded by the edges e marked with a triangle f such that $(e,f)\in Lk(A)$. Denote all such edges as $e_1, e_2, \ldots , e_s$, then $e_1, e_2, \ldots , e_s$ form a simple, null-homotopic closed path which is not the boundary of a triangle. It follows that

$$\begin{aligned} \sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )\ge \sum _{i=1}^s \big (\pi -\Phi (e_i)\big )>2\pi =2\pi \chi (F_A), \end{aligned}$$

which completes the proof.

“$(H_2)\Rightarrow (H_4)$” is obvious. By Theorem 1.1, $(H_2)$ implies that $(0,\ldots ,0)$ belongs to the image of the curvature map.

“$(H_4)\Rightarrow (H_5)$”. Assume the origin $(0,\ldots ,0)$ belongs to the image of the curvature map, i.e., there exists a circle pattern $r^{*}\in {\mathbb {R}}_{>0}^N$ with zero curvature. Let r(t) be the unique solution to the combinatorial Calabi flow on a maximal time interval [0, T), we need to prove $T=+\infty$ and $r(t)\rightarrow r^{*}$ exponentially fast.

Let $u^{*}\in {\mathbb {R}}_{sps0}^N$ be the u-coordinate of $r^{*}$. Consider the combinatorial Ricci potential

$$\begin{aligned} F(u)\triangleq \int _{u^{*}}^u\sum _{i=1}^NK_idu_i,\quad u\in {\mathbb {R}}_{sps0}^N. \end{aligned}$$

(4.12)

By Lemma 2.5, we can see

$$\begin{aligned} \frac{\partial K_i}{\partial u_j}=\frac{\partial K_j}{\partial u_i}, \end{aligned}$$

the smooth differential 1-form $\sum _{i=1}^NK_idu_i$ is closed, and hence then (4.12) is well defined and is independent on the choice of piecewise smooth paths in ${\mathbb {R}}_{sps0}^N$ from $u^{*}$ to u. By Lemma B.1 in [16], there holds

$$\begin{aligned} \lim _{\Vert u\Vert \rightarrow +\infty ,u\in {\mathbb {R}}_{sps0}^N}F(u)=+\infty . \end{aligned}$$

(4.13)

By Lemma 2.5 and a direct calculation, we have

$$\begin{aligned} \frac{d}{dt}F(u(t))=\sum _{i}K_i\Delta K_i=K^T\Delta K\le 0, \end{aligned}$$

which implies that F(u(t)) is non-increasing along the Calabi flow. By (4.13), there is a positive constant $\delta$, depending only on the triangulation ${\mathcal {T}}$ and the initial circle pattern r(0). such that $u_i(t)\ge -\delta$ for all i and t. It follows that

$$\begin{aligned} r_i(t)\ge \ln \frac{1+e^{-\delta }}{1-e^{-\delta }}>0 \end{aligned}$$

(4.14)

for all i and t. By (4.14) and Theorem 4.4, we know that r(t) lies in a compact subset of ${\mathbb {R}}_{>0}^N$. Then, by Lemma 4.1 in [15], r(t) exists for all time and converges exponentially fast to $r^{*}$.

“$(H_5)\Rightarrow (H_6)$”. Suppose the solution r(t) to the combinatorial Calabi flow (1.7) converges as $t\rightarrow +\infty$. Let $u^{*}$ be the corresponding u-coordinate of $r^{*}$, then u(t) converges to $u^{*}$.

$$\begin{aligned} u_i(n+1)-u_i(n)=u'_i(\xi _n)=\Delta K_i(\xi _n)\rightarrow 0,\quad \text {as}\ n\rightarrow +\infty . \end{aligned}$$

As $u(t)\rightarrow u^{*}$, we have $K(t)\rightarrow K(u^{*})$. Thus $\Delta K_i(u^{*})=0$ for each vertex $i\in V$. From Lemma 2.5, we know the matrix $\Delta$ is negative definite, which implies $K_i(u^{*})=0$ for each $i\in V$ and then $u^{*}$ has zero curvature. This also shows the image of the curvature map contains at least one point with non-positive coordinates.

“$(H_6)\Rightarrow (H_1)$”. We deform the metric r(t) according to Ricci flow (1.9), beginning from a initial metric r(0) which is exactly the special metric with non-positive curvatures. Set $M(t) = \max \{ K_1(t), \ldots , K_{|V|}(t), 0\}$, $m(t)= \min \{K_1(t), \ldots , K_{|V|}(t), 0\}$. Using the maximum principle, Chow-Luo (Corollary 3.3, [5]) proved that M(t) is non-increasing while m(t) is non-decreasing. So $M(t)\le M(0)\le 0$, and hence all $K_i(t)\le 0$. Thus $\frac{dr_i}{dt}\ge 0$ and every $r_i(t)$ is increasing, which implies that all $r_i(t)$ are uniformly bounded below from a positive constant. By Corollary 3.6 in [5], all $r_i(t)$ are uniformly bounded from above. Thus the solution $\{r(t)\}$ lies in a compact region in ${\mathbb {R}}_N>0$. Using Proposition 3.7 in [5], we get $(H_1)$.

Moreover, from the step “$(H_4)\Rightarrow (H_5)$”, we can see if one of the above properties holds, then the combinatorial Ricci/Calabi flow converges exponentially fast to a circle packing which produces a complete hyperbolic metric on M (with no cone points) $\square$

Appendices

Appendix 1: Proof of Lemma 2.1

Proof

Assume the background geometry is Euclidean. Let $l_{ij}>0$ such that

$$\begin{aligned} l^2_{ij}=r^2_i+r^2_j+2r_ir_j\cos \Phi _{ij}. \end{aligned}$$

Define $l_{ik},\ l_{jk}$ similarly. The objective is to check that $l_{ij},\ l_{ik},\ l_{jk}$ satisfy the triangle inequalities. Namely,

$$\begin{aligned} l_{ij}+l_{ik}>l_{jk}, \end{aligned}$$

and

$$\begin{aligned} |l_{ij}-l_{ik}|<l_{jk}. \end{aligned}$$

Combining the above two relations, we have

$$\begin{aligned} (l^2_{ij}+l^2_{ik}-l^2_{jk})^2<4l^2_{ij}l^2_{ik}. \end{aligned}$$

(5.1)

To simplify the notations, we set $I_{st}=\cos \Phi _{st}$ for $st=ij, jk, ik$. And taking $I_{st}$ into (5.1), we need to prove that

$$\begin{aligned} r^2_ir^2_j(1-I^2_{ij})&+r^2_ir^2_k(1-I^2_{ik})+r^2_jr^2_k(1-I^2_{jk})+2r^2_ir_jr_k(I_{jk}+I_{ij}I_{ik})\nonumber \\&+2r_ir^2_jr_k(I_{ik}+I_{ij}I_{jk})+2r_ir_jr^2_k(I_{ij}+I_{ik}I_{jk})>0. \end{aligned}$$

(5.2)

Now there are two cases to distinguish.

If $\Phi _{ij}+\Phi _{jk}+\Phi _{ik}\le \pi$, then

$$\begin{aligned} I_{jk}+I_{ij}I_{ik}&=\cos \Phi _{jk}+\cos \Phi _{ij}\cos \Phi _{ik}\\&=\cos \Phi _{jk}+\cos (\Phi _{ij}+\Phi _{ik})+\sin \Phi _{ij}\sin \Phi _{ik}\\&\ge \cos \Phi _{jk}+\cos (\pi -\Phi _{jk})+\sin \Phi _{ij}\sin \Phi _{ik}\\&\ge 0. \end{aligned}$$

Similarly, $I_{ik}+I_{ij}I_{jk}\ge 0$ and $I_{ij}+I_{ik}I_{jk}\ge 0$. Note that $1-I^2_{ij}\ge 0$, $1-I^2_{ik}\ge 0$, $1-I^2_{jk}\ge 0$. Thus we deduce (5.2), by these six inequalities can not obtain “=” at the same time.

If $\Phi _{ij}+\Phi _{jk}+\Phi _{ik}>\pi$. Considering that

$$\begin{aligned} \Phi _{ij}+\Phi _{jk}<\pi +\Phi _{ik},\ \Phi _{ik}+\Phi _{jk}<\pi +\Phi _{ij},\ \Phi _{ij}+\Phi _{ik}<\pi +\Phi _{jk}, \end{aligned}$$

there is a spherical triangle with inner angles $\Phi _{ij},\ \Phi _{jk},\ \Phi _{ik}$. Let $\phi _{ij},\ \phi _{jk},\ \phi _{ik}$ denote the corresponding lengths of the three sides. By the second cosine law of spherical triangles,

$$\begin{aligned} \cos \phi _{jk}=\frac{\cos \Phi _{jk}+\cos \Phi _{ij}\cos \Phi _{ik}}{\sin \Phi _{ij}\sin \Phi _{ik}}. \end{aligned}$$

So

$$\begin{aligned} I_{jk}+I_{ij}I_{ik}=\cos \Phi _{jk}+\cos \Phi _{ij}\cos \Phi _{ik}=\cos \phi _{jk}\sin \Phi _{ij}\sin \Phi _{ik}, \end{aligned}$$

and

$$\begin{aligned} I_{ik}+I_{ij}I_{jk}=\cos \phi _{ik}\sin \Phi _{ij}\sin \Phi _{jk},\ I_{ij}+I_{ik}I_{jk}=\cos \phi _{ij}\sin \Phi _{ik}\sin \Phi _{jk}. \end{aligned}$$

Set $y_{st}=r_sr_t\sin \Phi _{st}$ for $st=ij,\ jk,\ ik$. Hence (5.2) is equivalent to

$$\begin{aligned} y^2_{ij}+y^2_{jk}+y^2_{ik}+2y_{ij}y_{jk}\cos \phi _{ik}+2y_{ij}y_{ik}\cos \phi _{jk}+2y_{ik}y_{jk}\cos \phi _{ij}>0. \end{aligned}$$

By the cosine law of spherical triangles, we obtain

$$\begin{aligned} \cos \phi _{ij}-\cos \phi _{ik}\cos \phi _{jk}=\cos \Phi _{ij}\sin \phi _{ik}\sin \phi _{jk}. \end{aligned}$$

It follows that

$$\begin{aligned}&y^2_{ij}+y^2_{jk}+y^2_{ik}+2y_{ij}y_{jk}\cos \phi _{ik}+2y_{ij}y_{ik}\cos \phi _{jk}+2y_{ik}y_{jk}\cos \phi _{ij}\\&\quad =(y_{ij}+\cos \phi _{ik}y_{jk}+\cos \phi _{jk}y_{ik})^2+y^2_{jk}\sin ^2\phi _{ik}+y^2_{ik}\sin ^2\phi _{jk}\\&\quad +2y_{jk}y_{ik}(\cos \phi _{ij}-\cos \phi _{ik}\cos \phi _{jk}) \\&\quad \ge y^2_{jk}\sin ^2\phi _{ik}+y^2_{ik}\sin ^2\phi _{jk}+2y_{jk}y_{ik}(\cos \phi _{ij}-\cos \phi _{ik}\cos \phi _{jk}) \\&\quad =(y_{jk}\sin \phi _{ik}+y_{ik}\sin \phi _{jk}\cos \Phi _{ij})^2+y^2_{ik}\sin ^2\phi _{jk}\sin ^2\Phi _{ij} \\&\quad \ge y^2_{ik}\sin ^2\phi _{jk}\sin ^2\Phi _{ij}=r^2_ir^2_j\sin ^2\phi _{jk}\sin ^2\Phi _{ij}\sin \Phi _{ik} \\&\quad >0. \end{aligned}$$

Thus the lemma is proved. $\square$

Remark 4

We refer a more geometric formulation of Lemma 2.1, see Ge-Jiang-Liu [7].

Appendix 2: Thurston’s Existence Theorem

Following Thurston’s formulation of Andreev’s theorem and Marden-Rodin’s original methods [1], we give the image of the curvature map $K=K(r)$. For Euclidean background geometry, the image of K(r) were already obtained by Ge-Jiang in [11,12,13]. For hyperbolic background geometry, we give a complete proof here for reader’s convenience. Firstly, by Lemma 2.5 in [36] and Lemma 2.2, we have

Lemma 6.1

Let $\Phi _{ij}$, $\Phi _{jk}$, $\Phi _{ik}\in [0,\pi )$ be three intersection angles satisfying (Z), then we have

$$\begin{aligned}&\lim _{r_i\rightarrow \infty }\theta _i^{jk}=0,\\&\lim _{(r_i,r_j,r_k)\rightarrow (0,a,b)}\theta _i^{jk}=\pi -\Phi _{jk},\\&\lim _{(r_i,r_j,r_k)\rightarrow (0,0,c)}(\theta _i^{jk}+\theta _j^{ik})=\pi ,\\&\lim _{(r_i,r_j,r_k)\rightarrow (0,0,0)}(\theta _i^{jk}+\theta _j^{ik}+\theta _k^{ij})=\pi . \end{aligned}$$

We use the above lemma to give the proof of Theorem 1.1.

Proof of Theorem 1.1 in hyperbolic background geometry

We can see that K(u) is injective as a function of u. Obviously, K(r) is also injective as a function of r. We next prove that the image of the curvature map K(r), $r\in {\mathbb {R}}_{>0}^{|V|}$ is

$$\begin{aligned} {\mathcal {Z}}=\bigcap _{A\subset V}\bigg \{K\in (-\infty , 2\pi )^{|V|}:\sum _{i\in A}K_i>-\sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )+2\pi \chi (F_A)\bigg \}. \end{aligned}$$

Obviously, all $K_i<2\pi$ by definition. For each vertex subset $A\subset V$, we consider all the triangles in F having a vertex in A. These triangles can be classified into three types $A_1, A_2$ and $A_3$. For each $i\in \{1, 2, 3\}$, a triangle is in $A_i$ if and only if it has exactly i vertices in A. Since $\frac{\partial \theta _i^{jk}}{\partial r_i}<0$ and the second limited of Lemma 6.1, we have $\theta _i^{jk}<\pi -\Phi _{jk}$. Noting that $\theta _i^{jk}+\theta _j^{ik}<\theta _i^{jk}+\theta _j^{ik}+\theta _k^{ij}<\pi$, it follows

$$\begin{aligned} \sum _{i\in A}K_i =&2\pi |A|-\sum _{i\in A}\sum _{\{ijk\}\in F}\theta _i^{jk}\\ =&2\pi |A|-\left( \sum _{i\in A, \{ijk\}\in A_1}\theta _i^{jk}+\sum _{i,j\in A, \{ijk\}\in A_2}(\theta _i^{jk}+\theta _j^{ik})\right. \\&\quad \left. +\sum _{\{ijk\}\in A_3}(\theta _i^{jk}+\theta _j^{ik}+\theta _k^{ij})\right) \\ >&2\pi |A|-\sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )-\pi |A_2|-\pi |A_3|\\ =&-\sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )+2\pi \chi (F_A). \end{aligned}$$

Then it follows that $K({\mathbb {R}}_{>0}^{|V|})$, the image of the curvature map, is contained in ${\mathcal {Z}}$. If we can further prove K(r) is proper map (that is, the preimage of every compact set in ${\mathcal {Z}}$ is compact in ${\mathbb {R}}_{>0}^{|V|}$), then by the invariance of domain theorem, K is a diffeomorphism from ${\mathbb {R}}_{>0}^{|V|}$ to ${\mathcal {Z}}$. We just need to prove, if there is a sequence $r^{(n)}$ tends to the boundary of ${\mathbb {R}}_{>0}^{|V|}$, then $K(r^{(n)})$ contains a subsequence that tends to the boundary of ${\mathcal {Z}}$. To see this, assume $r^{(n)}$ tends to the boundary of ${\mathbb {R}}_{>0}^{|V|}$, then there is a subsequence, which is still denoted as $r^{(n)}$ itself, there is a vertex subset $A\subset V$, so that $r^{(n)}_i\rightarrow 0$ for each $i\in A$, while $r^{(n)}_j\rightarrow c_j\in (0,+\infty ]$ for each $j\in V-A$.

In case $A=\emptyset$, which means that $r_i^{(n)}\rightarrow +\infty$ for all $i\in V$, all $\theta _i^{jk}(r^{(n)})\rightarrow 0$ by the first limit in Lemma 6.1, hence all curvatures $K_i(r^{(n)})\rightarrow 2\pi$. This implies that $K(r^{(n)})$ tends to the boundary of ${\mathcal {Z}}$.

In case $A\ne \emptyset$ and $A\ne V$, by Lemma 6.1, for $i\in A$ and $\{ijk\}\in A_1$, we have

$$\begin{aligned} \theta _i^{jk}(r^{(n)})\rightarrow \pi -\Phi _{jk}, \end{aligned}$$

for $i,j\in A$ and $\{ijk\}\in A_2$, we have

$$\begin{aligned} (\theta _i^{jk}+\theta _j^{ik})(r^{(n)})\rightarrow \pi , \end{aligned}$$

while for $\{ijk\}\in A_3$, we have

$$\begin{aligned} (\theta _i^{jk}+\theta _j^{ik}+\theta _k^{ij})(r^{(n)})\rightarrow \pi . \end{aligned}$$

Then it follows

$$\begin{aligned} \sum _{i\in A}K_i =&2\pi |A|-\left( \sum _{i\in A, \{ijk\}\in A_1}\theta _i^{jk}+\sum _{i,j\in A, \{ijk\}\in A_2}(\theta _i^{jk}+\theta _j^{ik})\right. \\&\quad \left. +\sum _{\{ijk\}\in A_3}(\theta _i^{jk}+\theta _j^{ik}+\theta _k^{ij})\right) \\ \rightarrow&2\pi |A|-\sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )-\pi |A_2|-\pi |A_3|\\ =&-\sum _{(e,v)\in Lk(A)}\big (\pi -\Phi (e)\big )+2\pi \chi (F_A). \end{aligned}$$

This implies that $K(r^{(n)})$ tends to the boundary of ${\mathcal {Z}}$.

In case $A=V$, by definition $A_1,\ A_2$, we have $A_1=\emptyset ,\ A_2=\emptyset$, and

$$\begin{aligned} \begin{aligned} \sum _{i\in V}K_i(r^{(n)})&=2\pi |V|-\sum _{\{ijk\}\in A_3}(\theta _i^{jk}+\theta _j^{ik}+\theta _k^{ij})\\&\rightarrow 2\pi |V|-\pi |F|=2\pi \chi (F_V) \end{aligned} \end{aligned}$$

(6.1)

by the fourth limited of Lemma 6.1. This implies that $K(r^{(n)})$ tends to the boundary of ${\mathcal {Z}}$. We finish the proof. $\square$