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Hexagonal Geometric Triangulations

It is well-known that the Euclidean plane has a standard 6-regular geodesic triangulation , and the unit sphere has a 5-regular geodesic triangulation, which is induced from the regular Dodecahedron, and the hyperbolic plane has an n-regular geodesic triangulation for any n>6. Here we constructed a 6-regular geodesic triangulation of the hyperbolic plane.


Introduction
Here we consider the problem on the existence of special classes of geometric triangulations on the Euclidean plane and the hyperbolic plane.
A triangulation T on a surface S is k-regular for some integer k > 0 if the degrees of all its vertices are k, and it is geometric if all the edges are embedded as geodesic arcs with respect to the metric. The most familiar geometric 6-regular triangulation is the standard hexagonal triangulation on the plane, and the 7-regular equilateral triangulation of the hyperbolic plane. The triangular groups [17] ∆(3, k) = {a, b, c|a 3 produce k-regular equilateral triangulations for any k > 6. We ask whether we can construct 6-regular geometric triangulations on the hyperbolic plane. Moreover, we can consider the general question as follows.
Question: Does there exist k-regular geometric triangulations on the Euclidean plane E 2 , the round 2-sphere S 2 and the hyperbolic plane H 2 for any k ≥ 3? Furthermore, can we construct such a geometric triangulation T with a uniform bound on the edge lengths of T ?
If k = 6, the number of vertices grows linearly with respect to the combinatorial distance from one point. If k > 6, the number of vertices grows exponentially. Intuitively, they correspond to the Euclidean and hyperbolic geometry. We show that we can construct 6-regular geometric triangulation on H 2 . The semi-regular tiling of the Euclidean and hyperbolic plane by polygons has been studied by Datta and Gupta [2].

Preliminary case: closed surfaces
The basic combinatorial formulas below of Euler characteristics of closed surfaces show that there is no k-regular triangulation on E 2 and H 2 if k = 3, 4, 5, and there is no k-regular triangulation on From the combinatorial formula we have a necessary condition for a closed orientable surface S to admit a k-regular triangulation T , namely (k − 6)|12(g − 1). It implies immediately that there is no k-regular triangulation on S 2 if k ≥ 6, and the only k-regular triangulation on tori is when k = 6. This condition turns out to be also sufficient for the existence of k-regular triangulation on S with genus g. Proof. The necessity is proved above. The sufficiency follows from the result by Jucovic and Trenkler [9]. First notice that if a k-regular triangulation T exists on S, then set V 0 , E 0 , and F 0 be the number of vertices, edges, and faces of T . Then V 0 = 6χ(S)/(6 − k), E 0 = 3kχ(S)/(6 − k), and F 0 = 2kχ(S)/(6 − k). By the Main Theorem in [9], if three positive numbers V , E, and F on S satisfies two conditions (2) kV = 3F = 2E, then a k-regular triangulation with V vertices, E edges and F faces exists on S. It is straightforward to check that V 0 , E 0 , and F 0 satisfies the two conditions. Proof. If such a triangulation T exists, take an arbitrary subcomplex T which is homeomorphic disk. A combinatorial formula for a triangulated disk is given by where V I and V B are the index sets for interior vertices and boundary vertices. Since T is k-regular for k < 6, if v i is a boundary vertex of T , then d i < 5 since T is part of T . Then the left side is no larger than 6. Hence the disk T can only contains at most 6 vertices. This leads to a contradiction because we can take T as large as we want.
For k = 3, 4, 5, we have k-regular geometric triangulations on S 2 , given by projecting regular tetrahedron, octahedron, and icosahedron to the unit sphere from the center. It is straightforward to construct a 6-regular triangulation on a torus using the standard hexagonal triangulation on E 2 .

General facts about regular geometric triangulations
The goal is to construct k-regular geometric triangulations on E 2 with k > 6, and 6-regular geometric triangulations on H 2 . We first point out that such triangulations contain skinny triangles. Lemma 3.0.1. Let T be a k-regular geometric triangulations on E 2 with k > 6, or a 6-regular geometric triangulations on H 2 . Then the infimum of the angles in T is zero.
Proof. If not, in the first case for E 2 we can construct a quasiconformal homeomorphism from E 2 to H 2 by sending a triangle in T to the corresponding triangle in the standard geometric triangulation of H 2 generated by reflecting equilateral triangles with inner angle 2π/k. Similarly, nn the second case we can construct a quasiconformal homeomorphism from H 2 to E 2 sending T to the standard equilateral hexagonal triangulation. This contradicts to the fact that there is no quasiconformal map from E 2 to H 2 .
In this sense, the "best" k-regular geometric triangulation we can expect is a geometric triangulation with a uniform low bound c > 0 and a uniform upper C on the lengths of edges.

k-Regular geometric triangulations on The Euclidean plane
It is known that one can construct k-regular geometric triangulation on E 2 for k > 6. The idea is to construct chains of circles with radius n and distribute the points on the circle evenly. The similar computation is given in [2] . We can show that these triangulations have the desired uniform bounds. Proof. The construction of a k-regular geometric triangulation is given above. Let C n be circles with radius r n = n centered at the origin of E 2 . Let a n be the number of vertices on C n on this construction. These vertices have combinatorial distance n from the vertex at the origin.
We have the following recursive relation: a n+1 = (k − 4)a n − a n−1 with a 0 = 0 and a 1 = k. Solve this series and we have the formula for a n with n ≥ 1: 4)/2) and β = 1/α. Thus, asymptotically a n grows as α n−1 to infinity. But the length of C n is given by l n = 2πn. So the arclength of each small arc goes to zero as n goes to infinity, hence the length between two consecutive vertices on C n goes to zero. By the construction above, if v is a vertex in C n , then the k − 4 or k − 3 vertices v i in C n−1 connecting to v forms an arc γ in C n+1 , and the ray starting from the origin passing through v intersects with γ. As the length of γ goes to zero, all the lengths of the edges connecting v to v i goes to r n+1 − r n = 1.
So as n → ∞, the lengths of edges approaches either 1 or 0. Hence we can take the maximal length of a compact part of T bounded by C n with n large enough so that its complement contains edges no longer than 2. Then we can find a uniform bound.
Using the same idea, we can construct a k-regular geometric triangulation of E 2 whose edges has a uniform lower bound. We can pick r n = 2 n and follow the construction above.

6-regular geometric triangulations on the Hyperbolic Plane
We now give a construction of the desired 6-regular geometric triangulations with uniform bound in the hyperbolic plane using the Klein disk model.
(1) Put one vertex v 0 in the origin, and shoot out two rays R 1 and R 2 along the positive x-axis and along the direction e iπ/3 . (2) Let r n = α log n with 0 < α < 1/2. Construct a sequence of points a n and b n on R 1 and R 2 with distance r n from v 0 in hyperbolic metric. Connect a n with a n+1 and b n with b n+1 using geodesics. Call these edges of type 0. (3) Connect a n with b n by circle arcs C n , called the n-th layer in this construction. Equally distribute n − 1 points on C n . Then each C n are divided to n arcs. Connect the consecutive vertices on C n to generate n edges. Call these edges of type 1. By symmetry, these edge have the same length.
(4) Connect points on C n and C n+1 based on the combinatorics of the hexagonal triangulation to generate 2(n + 1) edges. Call these edges of type 2. (5) Rotate the configuration by π/3 to generate the full hexagon geometric triangulation.
We can give explicit coordinates of the vertices on C n in this triangulation in the plane v k n = tanh r n e i kπ 3n , k = 0, 1, · · · , n.
And the vertex v k n is connected with vertices v k n+1 and v k+1 n+1 in this triangulation. Proof. The proof is based on induction on n. The base case n = 1 is trivial. Assume it is true for the triangulation generated up to the n-th layer. We will show that the triangles between C n and C n+1 lie in the ring R n bounded C n and C n+1 . There are two types of triangles: (1) Type 1: two vertices on C n and one vertex on C n+1 . The three vertices are v k n = tanh r n e i kπ 3n , v k+1 n = tanh r n e i (k+1)π 3n , and v k+1 n+1 = tanh r n+1 e i (k+1)π 3(n+1) .
Notice that k n < k + 1 n + 1 < k + 1 n , and r n ≤ r n+1 , (2) Type 2: one vertex on C n and two vertex on C n+1 . In this case, the three vertices are v k n = tanh r n e i kπ 3n , v k+1 n+1 = tanh r n+1 e i (k+1)π 3n , and v k n+1 = tanh r n+1 e i kπ
We want to show that the edges of type 2 lie in the ring bounded by two consecutive C n and C n+1 , so it does not intersect with previous layers. If r n increase too slowly, then it is possible that the edge determined by v k n+1 and v k+1 n+1 might intersect with v k n . We need to show that if r n = α log n with 0 < α < 1/2, this will not occur.
The critical figures for triangles of type 1 and type 2 are given below, where the angles at v k n or v k+1 n are right angles. In this case, we can show that for sufficiently large n > N , tanh r n < tanh r n+1 cos π 6(n + 1) , and tanh r n < tanh r n+1 cos π 6n , which is equivalent to sinh(r n+1 − r n ) > cosh r n sinh r n+1 (1 − cos π 6n ).
One can show that the left side has order O(1/n) and right side has order O(1/n 2−2α ), so the inequality holds for large n if 0 < α < 1/2. This means that for large n > N , the triangles of type 2 will not intersect with the previous layers. For 0 ≤ n ≤ N , we can treat it as a polygon in the Euclidean plane, and adjust the radius to avoid the intersections. Proof. The length of the edge of type 1 is bounded by the length of the arc of the circle connecting v k n and v k+1 n given by sinh |v k n v k+1 n | < π 3n sinh r n ∼ O( 1 n 1−α ) → 0. From the previous lemma, all the edges of type 0 and type 2 in the n-th layer lie in the ring R n . The length of the edges in R n with vertices on C n is bounded by the following situation The half length L of this edge is given by hyperbolic Pythagorean theorem L = 2 cosh −1 ( cosh r n+1 cosh r n ) ≤ 2 cosh −1 (2(1 + 1 n ) α ) < ∞.
Hence we can bound the edges of all the three types for large n > 0. The remaining part is naturally bounded by compactness.

Related work
The existence of certain types of geodesic triangulations is a fundamental question. Recently various theory upon geodesic triangle meshes were developed, where people discuss problems about rigidity, convergence, variational principles, discrete maps and discrete geometric structures. See [3] [13]. for example.