Schwarz triangle


In geometry, a Schwarz triangle, named after Hermann Schwarz, is a spherical triangle that can be used to tile a sphere, possibly overlapping, through reflections in its edges. They were classified in.
These can be defined more generally as tessellations of the sphere, the Euclidean plane, or the hyperbolic plane. Each Schwarz triangle on a sphere defines a finite group, while on the Euclidean or hyperbolic plane they define an infinite group.
A Schwarz triangle is represented by three rational numbers each representing the angle at a vertex. The value means the vertex angle is of the half-circle. "2" means a right triangle. When these are whole numbers, the triangle is called a Möbius triangle, and corresponds to a non-overlapping tiling, and the symmetry group is called a triangle group. In the sphere there are three Möbius triangles plus one one-parameter family; in the plane there are three Möbius triangles, while in hyperbolic space there is a three-parameter family of Möbius triangles, and no exceptional objects.

Solution space

A fundamental domain triangle, with vertex angles,, and, can exist in different spaces depending on the value of the sum of the reciprocals of these integers:
This is simply a way of saying that in Euclidean space the interior angles of a triangle sum to, while on a sphere they sum to an angle greater than, and on hyperbolic space they sum to less.

Graphical representation

A Schwarz triangle is represented graphically by a triangular graph. Each node represents an edge of the Schwarz triangle. Each edge is labeled by a rational value corresponding to the reflection order, being π/vertex angle.

Schwarz triangle on sphere
Schwarz triangle graph

Order-2 edges represent perpendicular mirrors that can be ignored in this diagram. The Coxeter-Dynkin diagram represents this triangular graph with order-2 edges hidden.
A Coxeter group can be used for a simpler notation, as for cyclic graphs, and = for, and = ×.

A list of Schwarz triangles

Möbius triangles for the sphere

Schwarz triangles with whole numbers, also called Möbius triangles, include one 1-parameter family and three exceptional cases:
  1. or – Dihedral symmetry,
  2. or – Tetrahedral symmetry,
  3. or – Octahedral symmetry,
  4. or – Icosahedral symmetry,

    Schwarz triangles for the sphere by density

The Schwarz triangles, grouped by density:
DensityDihedralTetrahedralOctahedralIcosahedral
d
1
2,
3
4
5
6,,
7
8
9
10,
11
13
14,
16
17
18,
19
21
22
23
26
27
29
32
34
38
42

Triangles for the Euclidean plane

Density 1:
  1. – 60-60-60,
  2. – 45-45-90,
  3. – 30-60-90,
Density 2:
  1. - 120-30-30 triangle
Density ∞:

  3. Triangles for the hyperbolic plane

Density 1:
  • ,, ...
  • ,, ...
  • ,, ...
  • ,, ...
  • ,, ...
  • ,, ...
  • ,, ...
  • ,, ...
  • ...
Density 2:
  • ,, ...
  • ,, ...
  • ,, ...
  • ,, ...
  • ...
Density 3:
  • ,, ...
Density 4:
  • ,,, ...
Density 6:
  • ,, ...
  • ,,...
Density 10:
The Schwarz triangle is the smallest hyperbolic Schwarz triangle, and as such is of particular interest. Its triangle group is the triangle group, which is the universal group for all Hurwitz groups – maximal groups of isometries of Riemann surfaces. All Hurwitz groups are quotients of the triangle group, and all Hurwitz surfaces are tiled by the Schwarz triangle. The smallest Hurwitz group is the simple group of order 168, the second smallest non-abelian simple group, which is isomorphic to PSL, and the associated Hurwitz surface is the Klein quartic.
The triangle tiles the Bolza surface, a highly symmetric surface of genus 2.
The triangles with one noninteger angle, listed above, were first classified by Anthony W. Knapp in. A list of triangles with multiple noninteger angles is given in.

Tessellation by Schwarz triangles

In this section tessellations of the hyperbolic upper half plane by Schwarz triangles will be discussed using elementary methods. For triangles without "cusps"—angles equal to zero or equivalently vertices on the real axis—the elementary approach of will be followed. For triangles with one or two cusps, elementary arguments of, simplifying the approach of, will be used: in the case of a Schwarz triangle with one angle zero and another a right angle, the orientation-preserving subgroup of the reflection group of the triangle is a Hecke group. For an ideal triangle in which all angles are zero, so that all vertices lie on the real axis, the existence of the tessellation will be established by relating it to the Farey series described in and. In this case the tessellation can be considered as that associated with three touching circles on the Riemann sphere, a limiting case of configurations associated with three disjoint non-nested circles and their reflection groups, the so-called "Schottky groups", described in detail in. Alternatively—by dividing the ideal triangle into six triangles with angles 0, /2 and /3—the tessellation by ideal triangles can be understood in terms of tessellations by triangles with one or two cusps.

Triangles without cusps

Suppose that the hyperbolic triangle Δ has angles /a, /b and /c with a, b, c integers greater than 1. The hyperbolic area of Δ equals − /a − /b − /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 /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 and each side being the edge of a reflected triangle. In the case when an angle of Δ equals /3, a vertex of P2 will have an interior angle of, but this does not affect the convexity of P2. Even in this degenerate case when an angle of arises, the two collinear edges 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 and then fill in the gaps at each vertex. This results in a total of 3 + + + = 2 − 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 + + + = 2 − 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. 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 2/3 or, 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 and hence that Pn is a convex polygon.
It therefore has to be verified that in constructing Pn from Pn − 1:
To prove, 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 is proved.
Before proving and, a Möbius transformation can be applied to map the upper half plane to the unit disk and a fixed point in the interior of Δ to the origin.
The proof of proceeds by induction. Note that the radius joining the origin to a vertex of the polygon Pn − 1 makes an angle of less than 2/3 with each of the edges of the polygon at that vertex if exactly two triangles of Pn − 1 meet at the vertex, since each has an angle less than or equal to /3 at that vertex. To check this is true when three triangles of Pn − 1 meet at the vertex, C say, suppose that the middle triangle has its base on a side AB of Pn − 2. By induction the radii OA and OB makes angles of less than or equal to 2/3 with the edge AB. In this case the region in the sector between the radii OA and OB outside the edge AB is convex as the intersection of three convex regions. By induction the angles at A and B are greater than or equal to /3. Thus the geodesics to C from A and B start off in the region; by convexity, the triangle ABC lies wholly inside the region. The quadrilateral OACB has all its angles less than , so is convex. Hence the radius OC lies inside the angle of the triangle ABC near C. Thus the angles between OC and the two edges of Pn − 1 meeting at C are less than or equal to /3 + /3 = 2/3, as claimed.
To prove, it must be checked how new triangles in Pn intersect.
First consider the tiles added to the edges of Pn − 1. Adopting similar notation to, let AB be the base of the tile and C the third vertex. Then the radii OA and OB make angles of less than or equal to 2/3 with the edge AB and the reasoning in the proof of applies to prove that the triangle ABC lies within the sector defined by the radii OA and OB. This is true for each edge of Pn − 1. Since the interiors of sectors defined by distinct edges are disjoint, new triangles of this type only intersect as claimed.
Next consider the additional tiles added for each vertex of Pn − 1. Taking the vertex to be A, three are two edges AB1 and AB2 of Pn − 1 that meet at A. Let C1 and C2 be the extra vertices of the tiles added to these edges. Now the additional tiles added at A lie in the sector defined by radii OB1 and OB2. The polygon with vertices C2 O, C1, and then the vertices of the additional tiles has all its internal angles less than and hence is convex. It is therefore wholly contained in the sector defined by the radii OC1 and OC2. Since the interiors of these sectors are all disjoint, this implies all the claims about how the added tiles intersect.
Finally it remains to prove that the tiling formed by the union of the triangles covers the whole of the upper half plane. Any point z covered by the tiling lies in a polygon Pn and hence a polygon Pn + 1. It therefore lies in a copy of the original triangle Δ as well as a copy of P2 entirely contained in Pn + 1. The hyperbolic distance between Δ and the exterior of P2 is equal to r > 0. Thus the hyperbolic distance between z and points not covered by the tiling is at least r. Since this applies to all points in the tiling, the set covered by the tiling is closed. On the other hand, the tiling is open since it coincides with the union of the interiors of the polygons Pn. By connectivity, the tessellation must cover the whole of the upper half plane.
To see how to handle the case when an angle of Δ is a right angle, note that the inequality
implies that if one of the angles is a right angle, say a = 2, then both b and c are greater than 2 and one of them, b say, must be greater than 3. In this case, reflecting the triangle across the side AB gives an isosceles hyperbolic triangle with angles /c, /c and 2/b. If 2/b ≤ /3, i.e. b is greater than 5, then all the angles of the doubled triangle are less than or equal to /3. In that case the construction of the tessellation above through increasing convex polygons adapts word for word to this case except that around the vertex with angle 2/b, only b—and not 2b—copies of the triangle are required to tile a neighborhood of the vertex. This is possible because the doubled triangle is isosceles. The tessellation for the doubled triangle yields that for the original triangle on cutting all the larger triangles in half.
It remains to treat the case when b equals 4 or 5. If b = 4, then c ≥ 5: in this case if c ≥ 6, then b and c can be switched and the argument above applies, leaving the case b = 4 and c = 5. If b = 5, then c ≥ 4. The case c ≥ 6 can be handled by swapping b and c, so that the only extra case is b = 5 and c = 5. This last isosceles triangle is the doubled version of the first exceptional triangle, so only that triangle Δ1—with angles /2, /4 and /5 and hyperbolic area /20—needs to be considered. handles this case by a general method which works for all right angled triangles for which the two other angles are less than or equal to /4. The previous method for constructing P2, P3,... is modified by adding an extra triangle each time an angle 3/2 arises at a vertex. The same reasoning applies to prove there is no overlapping and that the tiling covers the hyperbolic upper half plane.
On the other hand, the given configuration gives rise to an arithmetic triangle group. These were first studied in. and have given rise to an extensive literature. In 1977 Takeuchi obtained a complete classification of arithmetic triangle groups and determined when two of them are commensurable. The particular example is related to Bring's curve and the arithmetic theory implies that the triangle group for Δ1 contains the triangle group for the triangle Δ2 with angles /4, /4 and /5 as a non-normal subgroup of index 6.
Doubling the triangles Δ1 and Δ2, this implies that there should be a relation between 6 triangles Δ3 with angles /2, /5 and /5 and hyperbolic area /10 and a triangle Δ4 with angles /5, /5 and /10 and hyperbolic area 3/5. established such a relation directly by completely elementary geometric means, without reference to the arithmetic theory: indeed as illustrated in the fifth figure below, the quadrilateral obtained by reflecting across a side of a triangle of type Δ4 can be tiled by 12 triangles of type Δ3. The tessellation by triangles of the type Δ4 can be handled by the main method in this section; this therefore proves the existence of the tessellation by triangles of type Δ3 and Δ1.