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==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 {{harvtxt|Carathéodory|1954}} will be followed. For triangles with one or two cusps, elementary arguments of {{harvtxt|Evans|1973}}, simplifying the approach of {{harvtxt|Hecke|1935}}, 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 {{harvtxt|Hardy|Wright|2008}} and {{harvtxt|Series|2015}}. 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 group]]s", described in detail in {{harvtxt|Mumford|Series|Wright|2015}}. Alternatively—by dividing the ideal triangle into six triangles with angles 0, {{pi}}/2 and {{pi}}/3—the tessellation by ideal triangles can be understood in terms of tessellations by triangles with one or two cusps.
===Triangles without cusps===
[[File:H2checkers 445.png|thumb|upright|Tessellation by triangles with angles {{pi}}/4, {{pi}}/4 and {{pi}}/5]]
[[File:H2checkers 357.png|thumb|upright|Tessellation by triangles with angles {{pi}}/3, {{pi}}/5 and {{pi}}/7]]
[[File:H2checkers 444.png|thumb|upright|Tessellation by equilateral triangles with angles {{pi}}/4]]
Suppose that the [[hyperbolic triangle]] Δ has angles {{pi}}/''a'', {{pi}}/''b'' and {{pi}}/''c'' with ''a'', ''b'', ''c'' integers greater than 1. The hyperbolic area of Δ equals {{pi}} – {{pi}}/''a'' – {{pi}}/''b'' – {{pi}}/''c'', so that
:<math>\frac1a + \frac1b + \frac1c < 1.</math>
The construction of a tessellation will first be carried out for the case when ''a'', ''b'' and ''c'' are greater than 2.<ref>{{harvnb|Carathéodory|1954|pages=177–181}}</ref>
The original triangle Δ gives a convex polygon ''P''<sub>1</sub> with 3 vertices. At each of the three vertices the triangle can be successively reflected through edges emanating from the vertices to produce 2''m'' copies of the triangle where the angle at the vertex is {{pi}}/''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 ''P''<sub>2</sub>. 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 {{pi}} and each side being the edge of a reflected triangle. In the case when an angle of Δ equals {{pi}}/3, a vertex of ''P''<sub>2</sub> will have an interior angle of {{pi}}, but this does not affect the convexity of ''P''<sub>2</sub>. Even in this degenerate case when an angle of {{pi}} arises, the two collinear edge are still considered as distinct for the purposes of the construction.
The construction of ''P''<sub>2</sub> 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 + (2''a'' – 3) + (2''b'' - 3) + (2''c'' - 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 + (2''a'' – 4) + (2''b'' - 4) + (2''c'' - 4) = 2(''a'' + ''b'' + ''c'') - 9 new vertices. By construction there is no overlapping. To see that ''P''<sub>2</sub> is convex, it suffices to see that the angle between sides meeting at a new vertex make an angle less than or equal to {{pi}}. 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{{pi}}/3 or {{pi}}, as required.
This process can be repeated for ''P''<sub>2</sub> to get ''P''<sub>3</sub> by first adding tiles to each edge of ''P''<sub>2</sub> and then filling in the tiles round each vertex of ''P''<sub>2</sub>. Then the process can be repeated from ''P''<sub>3</sub>, to get ''P''<sub>4</sub> and so on, successively producing ''P''<sub>''n''</sub> from ''P''<sub>''n'' – 1</sub>. 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 ''P''<sub>''n''</sub> from ''P''<sub>''n'' – 1</sub>, those attached to an edge of ''P''<sub>''n'' – 1</sub> 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 {{pi}} and hence that ''P''<sub>''n''</sub> is a convex polygon.{{efn|1=As in the case of ''P''<sub>2</sub>, if an angle of Δ equals {{pi}}/3, vertices where the interior angle is {{pi}} stay marked as vertices and colinear edges are not coallesced.}}
It therefore has to be verified that in constructing ''P''<sub>''n''</sub> from ''P''<sub>''n'' − 1</sub>:<ref>{{harvnb|Carathéodory|1954|pages=178−180}}</ref>
{{quote box|align=left|
(a) the new triangles do not overlap with ''P''<sub>''n'' − 1</sub> except as already described;
(b) the new triangles do not overlap with each other except as already described;
(c) the geodesic from any point in Δ to a vertex of the polygon ''P''<sub>''n'' – 1</sub> makes an angle ≤ 2{{pi}}/3 with each of the edges of the polygon at that vertex.}}
{{clear}}
To prove (a), note that by convexity, the polygon ''P''<sub>''n'' − 1</sub> is the intersection of the convex half-spaces defined by the full circular arcs defining its boundary. Thus at a given vertex of ''P''<sub>''n'' − 1</sub> there are two such circular arcs defining two sectors: one sector contains the interior of ''P''<sub>''n'' − 1</sub>, 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 point in the interior of Δ to the origin.
The proof of (c) proceeds by induction. Note that the radius joining the origin to a vertex of the polygon ''P''<sub>''n'' − 1</sub> makes an angle of less than 2{{pi}}/3 with each of the edges of the polygon at that vertex if exactly two triangles of ''P''<sub>''n'' − 1</sub> meet at the vertex, since each has an angle less than or equal to {{pi}}/3 at that vertex. To check this is true when three triangles of ''P''<sub>''n'' − 1</sub> meet at the vertex, ''C'' say, suppose that the middle triangle has its base on a side ''AB'' of ''P''<sub>''n'' − 2</sub>. By induction the radii ''OA'' and ''OB'' makes angles of less than or equal to 2{{pi}}/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 {{pi}}/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 {{pi}} (since ''OAB'' is a geodesic triangle), 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 {{math|''P''<sub>''n'' – 1</sub>}} meeting at ''C'' are less than or equal to {{pi}}/3 + {{pi}}/3 = 2{{pi}}/3, as claimed.
To prove (b), it must be checked how new triangles in ''P''<sub>''n''</sub> intersect.
First consider the tiles added to the edges of ''P''<sub>''n'' – 1</sub>. Adopting similar notation to (c), 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{{pi}}/3 with the edge ''AB'' and the reasoning in the proof of (c) 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 ''P''<sub>''n'' – 1</sub>. 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 ''P''<sub>''n'' – 1</sub>. Taking the vertex to be ''A'', three are two edges ''AB''<sub>1</sub> and ''AB''<sub>2</sub> of ''P''<sub>''n'' – 1</sub> that meet at ''A''. Let ''C''<sub>1</sub> and ''C''<sub>2</sub> 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 ''OB''<sub>1</sub> and ''OB''<sub>2</sub>. The polygon with vertices ''C''<sub>2</sub> ''O'', ''C''<sub>1</sub>, and then the vertices of the additional tiles has all its internal angles less than {{pi}} and hence is convex. It is therefore wholly contained in the sector defined by the radii ''OC''<sub>1</sub> and ''OC''<sub>2</sub>. Since the interiors of these sectors are all disjoint, this implies all the claims about how the added tiles intersect.
[[File:H2checkers 237.png|thumb|upright|Tessellation by triangles with angles {{pi}}/2, {{pi}}/3 and {{pi}}/7]]
[[File:H2checkers 245.png|thumb|upright|Tessellation by triangles with angles {{pi}}/2, {{pi}}/4 and {{pi}}/5]]
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 ''P''<sub>''n''</sub> and hence a polygon ''P''<sub>''n'' +1 </sub>. It therefore lies in a copy of the original triangle Δ as well as a copy of ''P''<sub>2</sub> entirely contained in ''P''<sub>''n'' +1 </sub>. The hyperbolic distance between Δ and the exterior of ''P''<sub>2</sub> is equal to ''r'' > 0. Thus the hyperbolic distance between ''z'' and points not coverered 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 ''P''<sub>''n''</sub>. 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
:<math>\frac1a + \frac1b + \frac1c < 1</math>.
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 {{pi}}/''c'', {{pi}}/''c'' and 2{{pi}}/''b''. If 2{{pi}}/''b'' ≤ {{pi}}/3, i.e. ''b'' is greater than 5, then all the angles of the doubled triangle are less than or equal to {{pi}}/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{{pi}}/''b'', only ''b''—and not 2''b''—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.<ref name="Cara1954">{{harvnb|Carathéodory|1954|pages=181–182}}</ref>
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 Δ<sub>1</sub>—with angles {{pi}}/2, {{pi}}/4 and {{pi}}/5 and hyperbolic area {{pi}}/20—needs to be considered (see below). {{harvtxt|Carathéodory|1954}} 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 {{pi}}/4. The previous method for constructing ''P''<sub>2</sub>, ''P''<sub>3</sub>, ... is modified by adding an extra triangle each time an angle 3{{pi}}/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.<ref name="Cara1954" />
On the other hand, the given configuration gives rise to an arithmetic triangle group. These were first studied in {{harvtxt|Fricke|Klein|1897}}. and have given rise to an extensive literature. In 1977 Takeuchi obtained a complete classification of arithmetic triangle groups (there are only finitely many) 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 Δ<sub>1</sub> contains the triangle group for the triangle Δ<sub>2</sub> with angles {{pi}}/4, {{pi}}/4 and {{pi}}/5 as a non-normal subgroup of index 6.<ref>See:
*{{harvnb|Takeuchi|1977a}}
*{{harvnb|Takeuchi|1977b}}
*{{harvnb|Weber|2005}}
</ref>
Doubling the triangles Δ<sub>1</sub> and Δ<sub>2</sub>, this implies that there should be a relation between 6 triangles Δ<sub>3</sub> with angles {{pi}}/2, {{pi}}/5 and {{pi}}/5 and hyperbolic area {{pi}}/10 and a triangle Δ<sub>4</sub> with angles {{pi}}/5, {{pi}}/5 and {{pi}}/10 and hyperbolic area 3{{pi}}/5. {{harvtxt|Threlfall|1932}} 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 Δ<sub>4</sub> can be tiled by 12 triangles of type Δ<sub>3</sub>. The tessellation by triangles of the type Δ<sub>4</sub> can be handled by the main method in this section; this therefore proves the existence of the tessellation by triangles of type Δ<sub>3</sub> and Δ<sub>1</sub>.<ref>See:
*{{harvnb|Threlfall|1932|pages=20–22}}, Figure 9
*{{harvnb|Weber|2005}}
</ref>
{{gallery| width=300px
|File:H2checkers 255.png|<small>Tessellation by triangles with angles {{pi}}/2, {{pi}}/5 and {{pi}}/5</small>
|File:H2-5-4-rhombic.svg|<small>Tessellation obtained by coalescing two triangles</small>
|File:Threlfall-245.jpeg|<small>Tiling with pentagons formed from 10 (2,5,5) triangles</small>
|File:Threlfall-245-a.jpeg|<small>Adjusting to tiling by triangles with angles {{pi}}/5, {{pi}}/10, {{pi}}/10</small>
|File:Threlfall-245-b.jpeg|<small>Tiling 2 (5,10,10) triangles with 12 (2,5,5) triangles</small>
}}
{{clear}}
===Triangles with one or two cusps===
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