Schwarz triangle function

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 = (1-α-β-γ)/2, b = (1-α+β-γ)/2, c = 1-α, a' = a - c + 1 = (1+α-β-γ)/2, b' = b - c + 1 = (1+α+β-γ)/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, formula_2, and
formula_3. 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", formula_4, and the formula reduces to that given by the Schwarz–Christoffel transformation. In the special case of ideal triangles, where all the angles are zero, the triangle function yields the modular lambda function. 
This function was introduced by H. A. Schwarz as the inverse function of the conformal mapping uniformizing a Schwarz triangle. Applying successive hyperbolic reflections in its sides, such a triangle generates a tessellation of the upper half plane (or the unit disk after composition with the Cayley transform). The conformal mapping of the upper half plane onto the interior of the geodesic triangle generalizes the Schwarz–Christoffel transformation. By the Schwarz reflection principle, the discrete group generated by hyperbolic reflections in the sides of the triangle induces an action on the two dimensional space of solutions. On the orientation-preserving normal subgroup, this two dimensional representation corresponds to the monodromy of the ordinary differential equation and induces a group of Möbius transformations on quotients of solutions. Since the triangle function is the inverse function of such a quotient, it is therefore an automorphic function for this discrete group of Möbius transformations. This is a special case of a general method of Henri Poincaré that associates automorphic forms with ordinary differential equations with regular singular points. 
Hyperboloid and Klein models.
In this section two different models are given for hyperbolic geometry on the unit disk or equivalently the upper half plane.
The group "G" = SU(1,1) is formed of matrices
with
It is a subgroup of "G""c" = SL(2,C), the group of complex 2 × 2 matrices with determinant 1.
The group "G""c" acts by Möbius transformations on the extended complex plane. The subgroup "G" acts as automorphisms of the unit disk "D" and the subgroup "G"1 = SL(2,R) acts as automorphisms of the upper half plane. If
then
since the Möbius transformation corresponding "M" is the Cayley transform carrying the upper half plane onto the unit disk and the real line onto the unit circle.
The Lie algebra formula_9 of SU(1,1) consists of matrices
with "x" real. Note that "X"2 = (|"w"|2 – "x"2) "I" and
The hyperboloid formula_12 in formula_9 is defined by two conditions. The first is that det "X" = 1 or equivalently Tr "X"2 = –2. By definition this condition is preserved under conjugation by "G". Since "G" is connected it leaves the two components with "x" > 0 and "x" < 0 invariant. The second condition is that "x" > 0. For brevity, write "X" = ("x","w").
The group "G" acts transitively on "D" and formula_12 and the points 0 and (1,0) have stabiliser "K" consisting of matrices
with |ζ| = 1. Polar decomposition on "D" implies the Cartan decomposition "G" = "KAK" where "A" is the group of matrices
Both spaces can therefore be identified with the homogeneous space "G"/"K" and there is a "G"-equivariant map "f" of formula_12 onto "D" sending (1,0) to 0. To work out the formula for this map and its inverse it suffices to compute "g"(1,0) and
"g"(0) where "g" is as above. Thus "g"(0) = β/ and
so that
recovering the formula
Conversely if "z" = "iw"/("x" + 1), then |"z"|2 = ("x" – 1)/("x" + 1), giving the inverse formula
This correspondence extends to one between geometric properties of "D" and formula_12. Without entering into the correspondence of "G"-invariant Riemannian metrics, each geodesic circle in "D" corresponds to the intersection of 2-planes through the origin, given by equations Tr "XY" = 0, with formula_12. Indeed, this is obvious for rays arg "z" = θ through the origin in "D"—which correspond to the 2-planes arg "w" = θ—and follows in general by "G"-equivariance.
The Klein model is obtained by using the map "F"("x","w") = "w"/"x" as the correspondence between formula_12 and "D". Identifying this disk with (1,"v") with |"v"| < 1, intersections of 2-planes with formula_12 correspond to intersections of the same 2-planes with this disk and so give straight lines. The Poincaré-Klein map given by
thus gives a diffeomorphism from the unit disk onto itself such that Poincaré geodesic circles are carried into straight lines. This diffeomorphism does not preserve angles but preserves orientation and, like all diffeomorphisms, takes smooth curves through a point making an angle less than (measured anticlockwise) into a similar pair of curves. In the limiting case, when the angle is , the curves are tangent and this again is preserved under a diffeomorphism. The map "K" yields the "Klein model" of hyperbolic geometry. The map extends to a homeomorphism of the unit disk onto itself which is the identity on the unit circle. Thus by continuity the map "K" extends to the endpoints of geodesics, so carries the arc of the circle in the disc cutting the unit circle orthogonally at two given points on to the straight line segment joining those two points.
Convex polygons.
In this section the main results on convexity of hyperbolic polygons are deduced from the corresponding results for Euclidean polygons by considering the relation between Poincaré's disk model and the Klein model. A polygon in the unit disk or upper half plane is made up of a collection of a finite set of vertices joined by geodesics, such that none of the geodesics intersect. In the Klein model this corresponds to the same picture in the Euclidean model with straight lines between the vertices. In the Euclidean model the polygon has an interior and exterior (by an elementary version of the Jordan curve theorem), so, since this is preserved under homeomorphism, the same is true in the Poincaré picture.
As a consequence at each vertex there is a well-defined notion of interior angle.
In the Euclidean plane a polygon with all its angles less than is convex, i.e. the straight line joining interior points of the polygon also lies in the interior of the polygon. Since the Poincaré-Klein map preserves the property that angles are less than , a hyperbolic polygon with interior angles less than is carried onto a Euclidean polygon with the same property; the Euclidean polygon is therefore convex and hence, since hyperbolic geodesics are carried onto straight lines, so is the hyperbolic polygon. By a continuity argument, geodesics between points on the sides also lie in the closure of the polygon.
A similar convexity result holds for polygons which have some of their vertices on the boundary of the disk or the upper half plane. In fact each such polygon is an increasing union of polygons with angles less than . Indeed, take points on the edges at each ideal vertex tending to the two edges joining those points to the ideal point with the geodesic joining them. Since two interior points of the original polygon will lie in the interior of one of these smaller polygons, each of which is convex, the original polygon must also be convex.
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 "P"1 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 /"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"2. 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 angle of Δ equals 
/3, a vertex of "P"2 will have an interior angle of , but this does not affect the convexity of "P"2. Even in this degenerate case when an angle of arises, the two collinear edge are still considered as distinct for the purposes of the construction.
The construction of "P"2 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"2 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 "P"2 to get "P"3 by first adding tiles to each edge of "P"2 and then filling in the tiles round each vertex of "P"2. Then the process can be repeated from "P"3, to get "P"4 and so on, successively producing "P""n" from "P""n" – 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 "P""n" from "P""n" – 1, those attached to an edge of "P""n" – 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 "P""n" is a convex polygon.
It therefore has to be verified that in constructing "P""n" from "P""n" − 1:
To prove (a), note that by convexity, the polygon "P""n" − 1 is the intersection of the convex half-spaces defined by the full circular arcs defining its boundary. Thus at a given vertex of "P""n" − 1 there are two such circular arcs defining two sectors: one sector contains the interior of "P""n" − 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 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""n" − 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 "P""n" − 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 "P""n" − 1 meet at the vertex, "C" say, suppose that the middle triangle has its base on a side "AB" of "P""n" − 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 (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 meeting at "C" are less than or equal to /3 + /3 = 2/3, as claimed.
To prove (b), it must be checked how new triangles in "P""n" intersect.
First consider the tiles added to the edges of "P""n" – 1. 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 2/3 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""n" – 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 "P""n" – 1. Taking the vertex to be "A", three are two edges "AB"1 and "AB"2 of "P""n" – 1 that meet at "A". Let "C"1 and "C"2 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"1 and "OB"2. The polygon with vertices "C"2 "O", "C"1, 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 "OC"1 and "OC"2. 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 "P""n" and hence a polygon "P""n" +1 . It therefore lies in a copy of the original triangle Δ as well as a copy of "P"2 entirely contained in "P""n" +1 . The hyperbolic distance between Δ and the exterior of "P"2 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""n". 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 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.
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 (see below). 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 "P"2, "P"3, ... 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 (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 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.
Triangles with one or two cusps.
In the case of a Schwarz triangle with one or two cusps, the process of tiling becomes simpler; but it is easier to use a different method going back to Hecke to prove that these exhaust the hyperbolic upper half plane.
In the case of one cusp and non-zero angles /"a", /"b" with "a", "b" integers greater than one, the tiling can be envisaged in the unit disk with the vertex having angle /"a" at the origin. The tiling starts by adding 2"a" – 1 copies of the triangle at the origin by successive reflections. This results in a polygon "P"1 with 2"a" cusps and between each two 2"a" vertices each with an angle /"b". The polygon is therefore convex. For each non-ideal vertex of "P"1, the unique triangle with that vertex can be similar reflected around that vertex, thus adding 2"b" – 1 new triangles, 2"b" – 1 new ideal points and 2 "b" – 1 new vertices with angle /"a". The resulting polygon "P"2 is thus made up of 2"a"(2"b" – 1) cusps and the same number of vertices each with an angle of /"a", so is convex. The process can be continued in this way to obtain convex polygons "P"3, "P"4, and so on. The polygon "P""n" will have vertices having angles alternating between 0 and /"a" for "n" even and between 0 and /"b" for "n" odd. By construction the triangles only overlap at edges or vertices, so form a tiling.
The case where the triangle has two cusps and one non-zero angle /"a" can be reduced to the case of one cusp by observing that the trinale is the double of a triangle with one cusp and non-zero angles /"a" and /"b" with "b" = 2. The tiling then proceeds as before.
To prove that these give tessellations, it is more convenient to work in the upper half plane. Both cases can be treated simultaneously, since the case of two cusps is obtained by doubling a triangle with one cusp and non-zero angles /"a" and /2. So consider the geodesic triangle in the upper half plane with angles 0, /"a", /"b" with "a", "b" integers greater than one. The interior of such a triangle can be realised as the region "X" in the upper half plane lying outside the unit disk |"z"| ≤ 1 and between two lines parallel to the imaginary axis through points "u" and "v" on the unit circle. Let Γ be the triangle group generated by the three reflections in the sides of the triangle.
To prove that the successive reflections of the triangle cover the upper half plane, it suffices to show that for any "z" in the upper half plane there is a "g" in Γ such that "g"("z") lies in . This follows by an argument of , simplified from the theory of Hecke groups. Let λ = Re "a" and μ = Re "b" so that, without loss of generality, λ < 0 ≤ μ. The three reflections in the sides are given by
Thus "T" = "R"3∘"R"2 is translation by μ − λ. It follows that for any "z"1 in the upper half plane, there is an element "g"1 in the subgroup Γ1 of Γ generated by "T" such that "w"1 = "g"1("z"1) satisfies λ ≤ Re "w"1 ≤ μ, i.e. this strip is a fundamental domain for the translation group Γ1. If |"w"1| ≥ 1, then "w"1 lies in "X" and the result is proved. Otherwise let "z"2 = "R"1("w"1) and find "g"2
Γ1 such that "w"2 = "g"2("z"2) satisfies λ ≤ Re "w"2 ≤ μ. If |"w"2| ≥ 1 then the result is proved. Continuing in this way, either some "w""n" satisfies |"w""n"| ≥ 1, in which case the result is proved; or |"w""n"| < 1 for all "n". Now since "g""n" + 1 lies in Γ1 and |"w""n"| < 1,