page_title: Schwarz triangle function
section_title: Triangles without cusps
hierarchical_section_title: Schwarz triangle function / Tessellation by Schwarz triangles / Triangles without cusps
caption_attribution_description: Tiling of the hyperbolic plane by triangles: π/3, π/5, π/7. Generated by Python code at User:Tamfang/programs. Dual:
context_page_description: In complex analysis, the Schwarz triangle function or Schwarz s-function is a function that conformally maps the upper half plane to a triangle in the upper half plane having lines or circular arcs for edges. Let πα, πβ, and πγ be the interior angles at the vertices of the triangle. If any of α, β, and γ are greater than zero, then the Schwarz triangle function can be given in terms of hypergeometric functions as:
  where a = /2, b = /2, c = 1-α, a' = a - c + 1 = /2, b' = b - c + 1 = /2, and c' = 2 - c = 1+α. This mapping has singular points at z=0, 1, and ∞, corresponding to the vertices of the triangle with angles πα, πγ, and πβ respectively. At these singular points, and
  . This formula can be derived using the Schwarzian derivative.
  This function can be used to map the upper half-plane to a spherical triangle on the Riemann sphere if α + β + γ > 1, or a hyperbolic triangle on the Poincaré disk if α + β + γ < 1. When α + β + γ = 1, then the triangle is a Euclidean triangle with straight edges: a=0, and the formula reduces to that given by the Schwarz–Christoffel transformation.
context_section_description: Suppose that the hyperbolic triangle Δ has angles π/a, π/b and π/c with a, b, c integers greater than 1. The hyperbolic area of Δ equals π – π/a – π/b – π/c, so that
  The construction of a tessellation will first be carried out for the case when a, b and c are greater than 2.
  The original triangle Δ gives a convex polygon P₁ with 3 vertices. At each of the three vertices the triangle can be successively reflected through edges emanating from the vertices to produce 2m copies of the triangle where the angle at the vertex is π/m. The triangles do not overlap except at the edges, half of them have their orientation reversed and they fit together to tile a neighborhood of the point.  The union of these new triangles together with the original triangle form a connected shape P₂. It is made up of triangles which only intersect in edges or vertices, forms a convex polygon with all angles less than or equal to π and each side being the edge of a reflected triangle. In the case when angle of Δ equals 
  π/3,  a vertex of P₂ will have an interior angle of π, but this does not affect the convexity of P₂. Even in this degenerate case when an angle of π arises, the two collinear edge are still considered as distinct for the purposes of the construction.
  The construction of P₂ can be understood more clearly by noting that some triangles or tiles are added twice, the three which have a side in common with the original triangle. The rest have only a vertex in common. A more systematic way of performing the tiling is first to add a tile to each side (the reflection of the triangle in that edge) and then fill in the gaps at each vertex. This results in a total of 3 + (2a – 3) + (2b - 3) + (2c - 3) = 2(a + b + c) - 6 new triangles. The new vertices are of two types. Those which are vertices of the triangles attached to sides of the original triangle, which are connected to 2 vertices of Δ. Each of these lie in three new triangles which intersect at that vertex.  The remainder are connected to a unique vertex of Δ and belong to two new triangles which have a common edge. Thus there are 3 + (2a – 4) + (2b - 4) + (2c - 4) = 2(a + b + c) - 9 new vertices. By construction there is no overlapping. To see that P₂ is convex, it suffices to see that the angle between sides meeting at a new vertex make an angle less than or equal to π. But the new vertices lies in two or three new triangles, which meet at that vertex, so the angle at that vertex is no greater than 2π/3 or π, as required.
  This process can be repeated for P₂ to get P₃ by first adding tiles to each edge of P₂ and then filling in the tiles round each vertex of P₂. Then the process can be repeated from P₃, to get P₄ and so on, successively producing Pₙ from Pₙ – 1. It can be checked inductively that these are all convex polygons, with non-overlapping tiles.
  Indeed, as in the first step of the process there are two types of tile in building Pₙ from Pₙ – 1, those attached to an edge of Pₙ – 1 and those attached to a single vertex. Similarly there are two types of vertex, one in which two new tiles meet and those in which three tiles meet. So provided that no tiles overlap, the previous argument shows that angles at vertices are no greater than π and hence that Pₙ is a convex polygon.
  It therefore has to be verified that in constructing Pₙ from Pₙ − 1:
  To prove (a), note that by convexity, the polygon Pₙ − 1 is the intersection of the convex half-spaces defined by the full circular arcs defining its boundary. Thus at a given vertex of Pₙ − 1 there are two such circular arcs defining two sectors: one sector contains the interior of Pₙ − 1, the other contains the interiors of the new triangles added around the given vertex. This can be visualized by using a Möbius transformation to map the upper half plane to the unit disk and the vertex to the origin; the interior of the polygon and each of the new triangles lie in different sectors of the unit disk. Thus (a) is proved.
  Before proving (c) and (b), a Möbius transformation can be applied to map the upper half plane to the unit disk and a fixed po
caption: Tessellation by triangles with angles π/3, π/5 and π/7
exif: {}