2019
DOI: 10.1007/s00454-019-00156-0
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Semi-regular Tilings of the Hyperbolic Plane

Abstract: A uniform tiling of the hyperbolic plane is a tessellation by regular geodesic polygons with the property that each vertex has the same vertex-type, which is a cyclic tuple of integers that determine the number of sides of the polygons surrounding the vertex. We determine combinatorial criteria for the existence, and uniqueness, of a uniform tiling with a given vertex type, and pose some open questions.

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Cited by 10 publications
(8 citation statements)
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“…Since X has 18 squares, s 2 +s 3 +s 4 = 18. These give (s 2 , s 3 , s 4 ) = (6, 12, 0), (7, 10, 1), (8,8,2), (9,6,3), (10,4,4), (11,2,5) or (12,0,6).…”
Section: From Definition 37 and The (Geometric) Construction Of Trunc...mentioning
confidence: 99%
See 1 more Smart Citation
“…Since X has 18 squares, s 2 +s 3 +s 4 = 18. These give (s 2 , s 3 , s 4 ) = (6, 12, 0), (7, 10, 1), (8,8,2), (9,6,3), (10,4,4), (11,2,5) or (12,0,6).…”
Section: From Definition 37 and The (Geometric) Construction Of Trunc...mentioning
confidence: 99%
“…A semi-regular tiling of a surface S of constant curvature (eg., the round sphere, the Euclidean plane or the hyperbolic plane) is a semi-equivelar map on S in which each face is a regular polygon and each edge is a geodesic. It follows from the results in [4] that there exist semi-regular tilings of the hyperbolic plane of infinitely many different vertex-types. It is also shown that there exists a unique semi-regular tiling of the hyperbolic plane of vertextype [p q ] for each pair (p, q) of positive integers satisfying 1/p + 1/q < 1/2.…”
Section: Introductionmentioning
confidence: 99%
“…Call this tiling T . This [n, m, n, m] pattern about each vertex of the tiling satisfies condition (1) in [8]. Therefore there exists a semi-regular tiling of H 2 , T , by regular polygons with vertex type [n, m, n, m] at every vertex, which is equivalent to T [8, Lemma 2.5].…”
Section: Totally Geodesic Checkerboard Surfacesmentioning
confidence: 99%
“…For unit negative curvature this manifold is the hyperbolic plane H 2 and has the Poincaré disk representation (for a review see [52]). In this case, any 4-regular graph with 3 squares and a cycle of length larger than 4 per vertex can be promoted to the 1-skeleton of a tiling of a negative curvature Cartan-Hadamard manifold by assigning a fixed length to each edge (CH) [53]. In Fig.…”
Section: Cartan-hadamard Manifolds In 2dmentioning
confidence: 99%