AND AUTOMORPHIC FORMS ON SHIMURA CURVES
FANG-TING TU AND YIFAN YANG
ABSTRACT. In this paper, we will obtain new algebraic transformations of the 2F1 -hypergeometric functions. The main novelty in our approach is the interpretation of iden-tities among2F1-hypergeometric functions as identities among automorphic forms on dif-ferent Shimura curves.
1. INTRODUCTION
For a real number a and a nonnegative integer n, let (a)n=
(
1, if n = 0,
a(a + 1) . . . (a + n − 1), if n ≥ 1,
be the Pochhammer symbol. Recall that, for real numbers a, b, c with c 6= 0, −1, −2, . . ., the2F1-hypergeometric function is defined by
2F1(a, b; c; z) = ∞ X n=0 (a)n(b)n (c)nn! zn
for z ∈ C with |z| < 1. The hypergeometric function is a solution of the differential equation
θ(θ + c − 1)F − z(θ + a)(θ + b)F = 0, θ = z d dz.
Hypergeometric functions arise naturally in many branches of mathematics. For exam-ple, the periods
Z ∞
1
dx
px(x − 1)(x − λ)
of the Legendre family of elliptic curves Eλ: y2= x(x − 1)(x − λ) can be expressed as 2F1 1 2, 1 2; 1; λ . Also, it is well-known that
E4(τ ) =2F1 1 12, 5 12; 1; 1728 j(τ ) 4
where E4(τ ) is the Eisenstein series of weight 4 on SL(2, Z) and j(τ ) is the elliptic
j-function.
Date: December 6, 2011.
2000 Mathematics Subject Classification. Primary 11F12 secondary 11G18, 33C05.
The authors were partially supported by Grant 99-2115-M-009-011-MY3 of the National Science Council, Taiwan (R.O.C.).
1
In this paper, we are concerned with algebraic transformations of hypergeometric func-tions, that is, identities of the form
(1) 2F1(a, b; c; z) = R(z)2F1(a0, b0; c0; S(z))
with suitable parameters a, b, c, a0, b0, c0 and algebraic functions R(z) and S(z). If w = R(z) is of degree m over the field C(z) or if z is of degree m over the field C(w), we say the algebraic transformation has degree m. Two examples of algebraic transformations of degree 1 are given by
2F1(a, b; c; z) = (1−z)c−a−b2F1(c−a, c−b; c; z) = (1−z)−a2F1
a, c − b; c; z z − 1
. These identities can be easily proved using the well-known result in the classical analy-sis that a Fuchsian differential equation with precisely 3 singularities at 0, 1, and ∞ is completely determined by the local exponents at these three points.
Beyond transformations of degree 1, one of the simplest examples is Kummer’s qua-dratic transformation 2F1 2a, 2b; a + b +1 2; z =2F1 a, b; a + b + 1 2; 4z(1 − z) ,
valid for any real numbers a, b with a + b + 1/2 6= 0, −1, −2, . . .. In [3], Goursat gave more than 100 algebraic transformations of degrees 2, 3, 4, 6. One such example is
2F1 a, a +1 3; 1 2; z(9 − 8z)2 (4z − 3)3 =1 + z 3 3a 2F1 3a, a +1 6; 1 2; z
of degree 3. (See Entry (96) on Page 132 of [3].) More recently, Vid¯unas [10] gave dozens of new algebraic transformations of degrees 6, 8, 9, 10, 12. For example, he showed that if we set β = ±√−2, (2) S(z) = 4z(z − 1)(8βz + 7 − 4β) 8 (2048βz3− 3072βz2− 3264z2+ 912βz + 3264z + 56β − 17)3, and R(z) = 1 +16 9 (4 − 17β)z − 64 243(167 − 136β)z 2+2048 6561(112 − 17β)z 3 −1/16 , then 2F1 5 24, 13 24; 7 8; z = R(z)2F1 1 48, 17 48; 7 8; S(z) ,
which is a transformation of degree 10. (See (32) of [10].) Vid¯unas’ examples usually involve Gr¨obner-basis computation. This is perhaps one of the reasons why Goursat could not find these transformations.
In this paper, we will present several new algebraic transformations. For example, one of our favorite identities is
2F1 1 20, 1 4; 4 5; 64z(1 − z − z2)5 (1 − z2)(1 + 4z − z2)5 = (1 − z2)1/20(1 + 4z − z2)1/42F1 3 10, 2 5; 9 10; z 2 . (3)
The main novelty in our approach is the interpretation of2F1-hypergeometric functions
as automorphic forms on Shimura curves. Then proving identities such as the one above amounts to showing two certain automorphic forms on two Shimura curves are equal. This point of view is especially useful in determining the function R(z) in (1). We will review
the basic definitions regarding Shimura curves and their automorphic forms in the next section.
2. PRELIMINARIES
The materials in this section are mostly taken from [11].
Quaternion algebras. A quaternion algebra B over a field K is a central simple algebra of dimension 4 over K. (Here central means that the center of the algebra is K and simple means that B has no proper nontrivial two-sided ideals.) If the characteristic of K is not 2, then one can show that there are elements i and j in B and a, b ∈ K×such that
i2= a, j2= b, ij = −ji,
and B = K + Ki + Kj + Kij. In this case, we denote this algebra bya,bK. For example, we have M (2, K) ' 1,1K and −1,−1
R is the set of Hamilton’s quaternions. Moreover,
for α = a0+ a1i + a2j + a3ij ∈ B, we set α = a0− a1i − a2j − a3ij. Then the reduced
tracetr (α) is defined to be α + α = 2a0∈ K and the reduced norm n(α) is defined to be
αα = a20− a21a − a22b + a23ab ∈ K.
If K = C, then up to isomorphisms, there is only one quaternion algebra over C, which is M (2, C). If K = R or a non-Archimedean local field, then up to isomorphism, there are only two quaternion algebras. One is M (2, K) and the other is a division algebra.
Now assume that K is a number field. Let v be a place of K and Kvbe the completion
of K with respect to v. If the localization B ⊗KKvis isomorphic to M (2, Kv), we say
B splits at v. If B ⊗KKvis isomorphic to a division algebra, we say B ramifies at v. It is
known that the number of ramified places is finite and in fact an even integer. The product of finite ramified places is called the discriminant of the quaternion algebra.
Still assume that K is a number field. Let R be its ring of integers. An order in B is a finite generated R-module that is also a ring with unity containing a basis for B. An order is maximal if it is not properly contained in another order. An Eichler order is the intersection of two maximal orders. For example, if B = M (2, Q), then M (2, Z) is a maximal order and Z Z
N Z Z = M (2, Z) ∩ ( 1 0
0 N) M (2, Z) (1 00 N) −1
is an Eichler order. Shimura curves. To define a Shimura curve, we assume that K is a totally real number field and take a quaternion algebra B over K that splits at exactly one infinite place, that is,
B ⊗QR ' M (2, R) × H[K:Q]−1,
where H is Hamilton’s quaternion algebra −1,−1R . Then, up to conjugation, there is a unique embedding ι of B into M (2, R). Note that we have n(α) = det ι(α) for all α ∈ B. Let Ø be an order and Ø∗1 = {α ∈ Ø : n(α) = 1} be the norm-one group of Ø. Then the image Γ(Ø) of Ø∗1under the embedding ι is a discrete subgroup of SL(2, R). Let Γ(Ø) act on the upper half-plane H in the usual manner
a b c d
: τ 7−→aτ + b cτ + d.
Then the quotient space Γ(Ø)\H is called the Shimura curve associated to Ø. For example, if B = M (2, Q) and Ø = M (2, Z), then Γ(Ø) = SL(2, Z) and Γ(Ø)\H is just the usual modular curve Y0(1). Thus, Shimura curves are generalizations of classical modular curves
and they are moduli spaces of certain abelian varieties with quaternionic multiplication [5]. In a broader setting, if Γ is any discrete subgroup of SL(2, R) commensurable with Γ(Ø), then the quotient space Γ\H will also be called a Shimura curve.
An element γ of Γ(Ø) is parabolic, elliptic, or hyperbolic, according to whether |tr (γ)| = 2, |tr (γ)| < 2, or |tr (γ)| > 2. The fixed point of a parabolic element is called a cusp. This can appear only when B = M (2, Q). The fixed point τ of an elliptic element in H is called an elliptic point of order n, where n is the order of the isotropy subgroup of τ in Γ(Ø)/±1. Note that if B 6= M (2, Q), then the quotient space Γ(Ø)\H is a compact Riemann surface, which we denote by X(Ø). If B = M (2, Q), we compactify the Riemann surface Γ(Ø)\H by adding cusps and the resulting compact surface will also be denoted by X(Ø). Now suppose that the compact Riemann surface X(Ø) has genus g. Then a classical result says that there exist hyperbolic elements A1, . . . , Ag, B1, . . . , Bg, and elliptic or
parabolic elements C1, . . . , Crthat generate Γ(Ø)/ ± 1 with the single relation
[A1, B1] . . . [Ag, Bg]C1. . . Cr= Id,
where [Ai, Bi] = AiBiA−1i B −1
i is the commutator of Ai and Bi. (See [4, Chapter 4].)
We let (g; e1, . . . , er) be the signature of X(Ø).
Triangle groups. Suppose that a Shimura curve X(Ø) has signature (0; e1, e2, e3). Then
we say Γ(Ø) is an arithmetic triangle group. The complete lists of all arithmetic triangle groups and their commensurability classes were determined by Takeuchi [7, 8].
If we cut each fundamental domain of an arithmetic triangle group Γ(Ø) into 2 halves in a suitable way, then the fundamental domains give a tessellation of the upper half-plane H by congruent triangles with internal angles π/e1, π/e2, and π/e3. The
follow-ing figure shows the tessellation of the unit disc, which is conformally equivalent to H through τ → (τ − i)/(τ + i), by fundamental half-domains of the arithmetic triangle group (0; 2, 3, 7).
Here each triangle represents a fundamental half-domain. Any combination of a grey trian-gle with a neighboring white triantrian-gle will be a fundamental domain for the triantrian-gle group (0; 2, 3, 7). The triangle group (0; 2, 3, 7) and its associated Shimura curve have been stud-ied in details in [2].
In general, for any discrete subgroup Γ of SL(R) such that Γ\H has finite volume, we can define its signature in the same way. If the signature is (0; e1, e2, e3), then we say Γ
is a (hyperbolic) triangle group. (There are also notions of parabolic and elliptic triangle groups, corresponding to tessellation of C and P1(C), respectively.)
Automorphic forms on Shimura curves. The definition of an automorphic form on Shimura curves is the same as that of a modular form on classical modular curves.
For simplicity, we assume that B 6= M (2, Q) so that we do not need to consider cusps. Then an automorphic form of weight k on Γ(Ø) is a holomorphic function f : H → C
such that (4) f aτ + b cτ + d = (cτ + d)kf (τ ) for all a b
c d ∈ Γ(Ø) and all τ ∈ H. The space of automorphic forms of weight k on Γ(Ø)
will be denoted by Sk(Ø). Also, if a meromorphic function f : H → C satisfies (4) with
k = 0, we say f is an automorphic function. If X(Ø) has genus 0, we call an automorphic function a Hauptmodul if it generates the field of automorphic functions on Γ(Ø).
Using the Riemann-Roch formula, one can calculate the dimension of Sk(Ø).
Proposition 1 ([6, Theorem 2.23]). Assume that B 6= M (2, Q). Suppose that the Shimura curveX(Ø) associated to an order Ø in B has signature (g; e1, . . . , er). Then for even
integersk, we have dim Sk(Ø) = 0, ifk < 0, 1, ifk = 0, g, ifk = 2, (k − 1)(g − 1) + r X j=1 k 2 1 − 1 ej , ifk ≥ 4.
The dimension formula for the case B = M (2, Q) is slightly different.
In the case B = M (2, Q), there are many methods to construct modular forms, such as Eisenstein series, theta series, the Dedekind eta function, and etc. In practice, most explicit methods for modular curves rely on the Fourier expansions of modular forms and modular functions, i.e., the expansions with respect to the local parameter at the cusp ∞. However, in the case B 6= M (2, Q), because of the lack of cusps on Shimura curves, very few explicit methods are available for Shimura curves. One of the few methods is given by the second author of the present paper.
One of the key ideas in [12] is the following characterization of Sk(Ø). Here we assume
that the quaternion algebra is not M (2, Q).
Proposition 2 ([12, Theorem 4, Propositions 1 and 6]). Assume that a Shimura curve X has genus zero with elliptic pointsτ1, . . . , τrof ordere1, . . . , er, respectively. Lett(τ ) be
a Hauptmodul ofX and set ai= t(τi), i = 1, . . . , r. For a positive even integer k ≥ 4, let
dk = dim Sk(Ø) = 1 − k + r X j=1 k 2 1 − 1 ej .
Then a basis for the space of automorphic forms of weightk on X is t0(τ )k/2t(τ )j
r
Y
i=1,ai6=∞
(t(τ ) − aj)−bk(1−1/e1)/2c, j = 0, . . . , dk− 1.
Moreover, the functionst0(τ )1/2 andτ t0(τ )1/2, as functions oft, satisfy the differential equation f00+ Q(t)f = 0, where Q(t) = 1 4 r X j=1,aj6=∞ 1 − 1/e2 j (t − aj)2 + r X j=1,aj6=∞ Bj t − aj
for some constantsBj. Moreover, ifaj6= ∞ for all j, then the constants Bjsatisfy r X j=1 Bj = r X j=1 ajBj+ 1 4(1 − 1/e 2 j) = r X j=1 a2jBj+ 1 2aj(1 − 1/e 2 j) = 0. Also, ifar= ∞, then Bjsatisfy
r−1 X j=1 Bj = 0, r−1 X j=1 ajBj+ 1 4(1 − 1/e 2 j) = 1 4(1 − 1/e 2 r).
Remark 3. In [12], the differential equation f00+ Q(t)f = 0 is called the Schwarzian differential equationassociated to t because Q(t) is related to the Schwarzian derivative by the relation 2Q(t)t0(τ )2+ {t, τ } = 0, where {t, τ } = t 000(τ ) t0(τ ) − 3 2 t00(τ ) t0(τ ) 2
is the Schwarzian derivative. In general, in literature [1], if f is a thrice-differentiable function of z, then
D(f, z) := − {f, z} 2f0(z)2
is called the automorphic derivative associated to f and z. In the case f is an automorphic function on a Shimura curve, then D(f, τ ) is also an automorphic function. In particular, if t is a Hauptmodul on a Shimura curve of genus 0, then Q(t) = D(t, τ ) is a rational function of t.
The upshot of this result is that it is often possible to determine the differential equation without explicitly constructing a Hauptmodul. For example, if Γ(Ø) is a triangle group with signature (0; e1, e2, e3), then there always exists a (unique) Hauptmodul t with a1=
0, a2= 1, and a3= ∞. Then the relations between Bjuniquely determine the differential
equation. In general, one can usually use coverings between Shimura curves of genus 0 to determine the differential equation. This is done by the first author [9] of the present paper for many Shimura curves of genus 0 associated to orders in quaternion algebras over Q. Once the differential equation is determined, one can express automorphic forms in terms of t-series and then study properties of automorphic forms using these t-series. For example, in [12] the second author devised a method to compute Hecke operators on these t-series.
In the case of triangle groups, since the number of singularities of the differential equa-tion is 3, the differential equaequa-tion is essentially a hypergeometric differential equaequa-tion. Proposition 4 ([12, Theorem 9]). Assume that a Shimura curve X has signature (0; e1, e2, e3).
Lett(τ ) be the Hauptmodul of X with values 0, 1, and ∞ at the elliptic points of order e1,e2, ande3, respectively. Letk ≥ 4 be an even integer. Then a basis for the space of
automorphic forms of weightk on X is given by t{k(1−1/e1)/2}(1 − t){k(1−1/e2)/2}tj
2F1(a, b; c; t) + Ct1/e12F1(a0, b0, c0; t)
k , j = 0, . . . , bk(1 − 1/e1)/2c + bk(1 − 1/e2)/2c + bk(1 − 1/e3)/2c − k, for some constant
C, where for a rational number x, we let {x} denote the fractional part of x, a = 1 2 1 − 1 e1 − 1 e2 − 1 e3 , b = a + 1 e3 , c = 1 − 1 e1
and a0= a + 1 e1 , b0= b + 1 e1 , c0= c + 2 e1 .
For general Shimura curves of genus 0, the following properties of automorphic deriva-tives are very useful in determining the Schwarzian differential equation associated to a Hauptmodul.
Proposition 5. Automorphic derivatives have the following properties. (1) D((az + b)/(cz + d), z) = 0 for all a b
c d ∈ GL(2, C).
(2) D(g ◦ f, z) = D(g, f (z)) + D(f, z)/(dg/df )2.
Proposition 6. Let z(τ ) be a Hauptmodul for a Shimura curve X(Ø) of genus 0. Let R(x) ∈ C(x) be the rational function such that the automorphic derivative Q(z) = D(z, τ ) is equal to R(z). Assume that γ is an element of SL(2, R) normalizing the norm-one group of Ø and let σ be the automorphism of X(Ø) induced by γ. If σ : z 7→ (az + b)/(cz + d), then R(x) satisfies
(ad − bc)2 (cx + d)4R ax + b cx + d = R(x).
Proof. We shall compute D(z(γτ ), τ ) in two ways. By Proposition 5, we have D(z(γτ ), τ ) = D az(τ ) + b cz(τ ) + d, z(τ ) + D(z(τ ), τ ) (dz(γτ )/dz(τ ))2 = 0 + (cz + d)4R(z) (ad − bc)2 .
On the other hand, by the same proposition, we also have D(z(γτ ), τ ) = D(z(γτ ), γτ ) + D(γτ, τ ) (dz(γτ )/dγτ )2 = R(z(γτ )) = R az + b cz + d .
Comparing the two expressions, we get the formula.
Algebraic transformations of hypergeometric functions. Consider the following situa-tion. Suppose that Γ1 < Γ2are two arithmetic triangle groups with Hauptmoduls z1and
z2, respectively. Since any automorphic function on Γ2 is also an automorphic function
on Γ1, we have z2 = S(z1) for some S(x) ∈ C(x). Likewise, if f1and f2are two
auto-morphic forms of the same weight k on Γ1and Γ2, respectively, then the ratio f1/f2is an
automorphic function on Γ1and hence is equal to R(z1) for some R(x) ∈ C(x). In view
of Proposition 4, after taking the kth roots of the two sides of f1/f2= R(z1), we obtain an
algebraic transformation of hypergeometric function. This explains the existence of Kum-mer’s, Goursat’s and Vid¯unas’ transformations. (Of course, the triangle groups appearing in their transformations may not be arithmetic, but the argument above is still valid.)
More generally, if Γ1and Γ2are two commensurable arithmetic triangle groups such
that the Shimura curve associated to Γ = Γ1∩ Γ2has genus 0. Let z be a Hauptmodul on
Γ. Then each of z1and z2is a rational function of z. Similarly, the ratio f1/f2is also a
rational function of z. Again, Proposition 4 yields an algebraic transformation of the form
2F1(a1, b1; c1; S1(z)) = R(z)2F1(a2, b2; c2; S2(z))
for some rational functions S1(z) and S2(z) and some algebraic function R(z). This is the
theory behind (3) and other algebraic transformations given in the paper.
Definition 7. Let S(z) ∈ C(z) be a rational function. If the finite covering P1(C) → P1(C) defined by S : z → S(z) is ramified at most at three points 0, 1, and ∞, then S is called a Belyi function.
In practice, the Belyi functions S1(z) and S2(z) can be determined by the
ramifica-tion data of the coverings of Shimura curves. The funcramifica-tion R(z) can be determined by Propositions 2 and 4.
We now obtain algebraic transformations of hypergeometric functions using the above idea. Note that according to [8], arithmetic triangle groups fall in 19 commensurability classes. The first class in his list corresponds to classical modular curves. In this case, it is easier to use classical modular forms to derive identities. We will not discuss this case here. Identities arising from Classes II, V, and XII are special cases of a family of identities, and so are identities from Classes IV, VIII, XI, XIII, XV, and XVII. These cases will be treated in a later section. Here we first consider Class III in Section 3 and Class VI in Section 4. (There are no identities from Classes IX and XIX since these classes consist of a single group. Also, identities from Classes VII, XIV, XVI, and XVIII are just Kummer’s quadratic transformations.)
3. ALGEBRAIC TRANSFORMATIONS ASSOCIATED TOCLASSIII
According to [8], Takeuchi’s Class III of commensurable arithmetic triangle groups has the following subgroup diagram. Here because all groups involved have genus zero, we omit the genus information in the signatures of the groups.
(2, 6, 8) (2, 3, 8) (4, 6, 6) (3, 8, 8) (3, 3, 4) (2, 4, 8) (4, 4, 4) (2, 8, 8) (4, 8, 8) 2 H H H H H H 2 10 2 H H H H H H 3 3 2 2 2
The main goal in this section is to prove an algebraic transformation associated to the pair of triangle groups (4, 6, 6) and (4, 4, 4).
Theorem 1. Let α be a root of x2+ 3 = 0 and β a root of x2+ 2 = 0. We have
(1 + z)1/8(1 − 3z)1/8 (1 + αz)5/4 2F1 5 24, 3 8; 3 4; 12αz(1 − z2)(1 − 9z2) (1 + αz)6 = 1 (1 + (4 + 2β)z − (1 + 2β)z2)1/2 2F1 1 8, 3 8; 3 4; R(z) , (5) and (1 − z)1/4(1 + z)5/8(1 − 3z)1/4(1 + 3z)5/8 (1 + αz)11/4 2F1 11 24, 5 8; 5 4; 12αz(1 − z2)(1 − 9z2) (1 + αz)6 = (1 + (−7 + 4β)z 2/3) (1 + (4 + 2β)z − (1 + 2β)z2)3/2 2F1 3 8, 5 8; 5 4; R(z) (6)
where
R(z) = − 4(1 + β)
4z(1 + (−7 + 4β)z2/3)4
(1 + z)(1 − 3z)(1 + (4 + 2β)z − (1 + 2β)z2)4.
We first determine the signatures of the intersections. Lemma 8. We have (2, 6, 8) (2, 3, 8) Γ1= (4, 6, 6) Γ2= (3, 8, 8) Γ3= (3, 3, 4) (2, 4, 8) Γ5= (3, 4, 3, 4) Γ4= (4, 4, 4) (2, 8, 8) Γ6= (46) (4, 8, 8) 2 H H H H HH 2 10 2 H H H H HH 3∗ H H H H HH 2 2 10 3 2 2 3 10 2
Moreover, the group of signature (46) is a normal subgroup of the group of signature
(3, 4, 3, 4). (Here (46) is a shorthand for (4, 4, 4, 4, 4, 4).)
Proof. Let Γ2= (3, 8, 8) and Γ02be its commutator subgroup. From the group presentation
Γ2' hγ1, γ2: γ13, γ28, (γ1γ2)8i
for Γ2, we know that Γ2/Γ02 is cyclic of order 8. Thus, Γ2 has exactly one subgroup
of index 2, which must be the common intersection of the groups (4, 6, 6), (3, 8, 8) and (3, 3, 4). The signature of this subgroup can be easily determined by observing that a covering of degree 2 from a Shimura curve to the Shimura curve associated to (3, 8, 8) can only ramify at the two elliptic points of order 8. We find that the signature must be (3, 4, 3, 4).
We next observe that the commutator subgroup Γ03of the group Γ3= (3, 3, 4) is cyclic
of order 3. Thus, Γ03is a normal subgroup of index 3 in Γ3. This Γ03must be the same as
the group of signature (4, 4, 4). If Γ036= (4, 4, 4), then Γ0
3∩ (4, 4, 4) is a normal subgroup
of (4, 4, 4) of index 3, but the group (4, 4, 4) cannot have a normal subgroup of index 3. We next determine the signature of the intersection of Γ4= (4, 4, 4) and Γ5= (3, 4, 3, 4).
Let Xj denote the Shimura curve associated to the group Γj. Since Γ4 is a normal
subgroup of Γ3of index 3, the intersection Γ6of Γ4and Γ5is a normal subgroup of index
3 in Γ5, which implies that the two elliptic points of order 4 of X5must split completely
on X6. In view of the Riemann-Hurwitz formula, the two elliptic points of order 3 of X5
must be totally ramified. We conclude that Γ6has signature (46).
In fact, the subgroup relations mentioned above can be visualized by the following figures.
Here the small triangles are (2, 3, 8)-triangles. Let G be the group of all symmetries of the tessellation of the hyperbolic plane by the (4, 4, 4)-triangles and G0be the subgroup
gen-erated by the reflections across the edges of (4, 4, 4)-triangles. Then G/G0is isomorphic
to D3. The (3, 3, 4)-triangle group corresponds to the cyclic subgroup of order 3 in G/G0,
while the group (2, 3, 8) corresponds the whole group G/G0. Similarly, if we piece 12
copies of (2, 6, 8)-triangles around the vertex of inner angle π/4, we get a regular hexagon with inner angles π/4. Let H be the group of all symmetries of the tessellation by this regular hexagon and H0be the subgroup generated by the reflections across the edges of
hexagons. Then H/H0 is isomorphic to D6. The unique cyclic subgroup of order 3 in
H/H0corresponds to the group (3, 4, 3, 4). See the figures below.
(The groups (2, 6, 8), (4, 6, 6), and (3, 8, 8) correspond to the whole H/H0, the cyclic
subgroup of order 6 of H/H0, and one of the D3-subgroups, respectively.)
Now let Γ1= (4, 6, 6), Γ2 = (3, 8, 8), Γ3= (3, 3, 4), Γ4= (4, 4, 4), Γ5 = (3, 4, 3, 4),
and Γ6 = (46). Let Xj = X(Γj), j = 1, . . . , 6, be the corresponding Shimura curves.
Label the elliptic points on X1by P4, P6, and P60, those on X2by Q3, Q8, and Q08, those
on X3by R3, R03, and R4, those on X4by S4, S40, S400, and those on X5by T3, T30, T4, and
T40 (with the subscripts carrying the obvious meaning) such that the ramification data are given by
T3 T30 T4 T40 T3 T30 T4 T40
P6 P60 P4 Q3 Q8 Q08
T3 T30 T4 T40
R3 R03 R4
1 3 3 3 1 3 3 3 1 1 4 4
Label the elliptic points of X6 by U1, . . . , U6such that the rotation around the center
of the (46)-polygon by the angle π/3 permutes the six points cyclically. From the figures
above, we know that if we label the points such that U1lies above T4, then the ramification
data for X6→ X5are
U1 U3 U5 U2 U4 U6 U0 U00
T4 T40 T3 T30
1 1 1 1 1 1 3 3
where U0and U00 are the centers of the (46)-polygons. (The reader is reminded that each
(46)-polygon represents only half of the fundamental domain for the Shimura curve X 6.
Referring to the figure in the proof of the lemma above, a fundamental domain consists of a grey (46)-polygon and a neighboring white (46)-polygon.)
Lemma 9. The two elliptic points of X6at the two ends of a diagonal of a(46)-polygon
lie above the same elliptic point ofX4. That is, labeling the elliptic points ofX4suitably,
we have
U1 U4 U2 U5 U3 U6
S4 S40 S004
1 1 4 4 1 1 4 4 1 1 4 4
Moreover, if we choose Hauptmodulszj(τ ) for Xj,j = 1, . . . , 6, by requiring
z1(P4) = 0, z1(P6) = 1, z1(P60) = ∞, z2(Q8) = 0, z2(Q3) = 1, z2(Q08) = ∞, z3(R4) = 0, z3(R3) = 1, z3(R30) = ∞, z4(S4) = 0, z4(S40) = 1, z4(S400) = ∞, z5(T4) = 0, z5(T3) = 1, z5(T40) = ∞, z6(U1) = 0, z6(U3) = 1, z6(U4) = ∞, then we have z1= 4z5 (1 + z5)2 , z2= z52, z3= 3(ζ − ζ2)z 4(1 − z4) (1 + ζz4)3 , z5= 3(ζ − ζ2)z 6(1 − z62) 1 − 9z2 6 , z3= (28 + 16β)z5(1 + (−17 + 56β)z52/81)4 (1 + z5)(1 + (13 + 8β)z5/3 − (25 + 32β)z25/9 + (17 − 56β)z 3 5/81)3 ,
and z4= − 4(1 + β)4z 6(1 + (−7 + 4β)z62/3)4 (1 + z6)(1 − 3z6)(1 + (4 + 2β)z6− (1 + 2β)z26)4 , whereζ is a 3rd root of unity and β is a root of x2+ 2 = 0.
Proof. The ramification data for the covering X5 → X2and the assumption z2(Q3) =
z5(T3) = 1 imply that z2= z52and
z(T30) = −1.
The relation between z1and z5is easy to determine. We find z1= 4z5/(1 + z5)2.
To determine the relation between z3 and z4, we recall from Lemma 8 that Γ4 is a
normal subgroup Γ3. Any element of Γ3 not in Γ4 induces an automorphism of order
3 on X4. Such an automorphism must permute the three elliptic points S4, S40, and S400
cyclically. In term of the Hauptmodul z4, such an automorphism is either
σ : z47−→
−1 z4− 1
or its square. Moreover, the fixed points of such an automorphism are the ramified points in the covering X4→ X3. That is, if we let S0and S00 be the points lying above R3and R03
respectively, then z4(S0), z4(S00) ∈ {−ζ, −ζ
2}, where ζ is a primitive 3rd root of unity.
Then from the ramification data, we easily deduce that z3= (ζ −ζ2)z4(1−z24)/(1+ζz4)3.
To determine the relation between z5and z6, we argue similarly as above. The
tessel-lation of the hyperbolic plane by Γ6 has a D6-symmetry, in addition to the symmetries
arising from the reflections across the edges of the (46)-polygons. This provides many useful informations. For example, if we let τ be the reflection across the diagonal joining U1and U4, then τ induces an involution on X6, which, in terms of z6, is given by
τ : z67−→ −z6,
which implies that
z6(P5) = −1.
Furthermore, let ρ denote the rotation by angle π/3 around the center of the hexagon. Then ρ : z67−→
cz6+ 1
−cz6+ c
for some zero constant c since ρ maps 1 to ∞ and ∞ to −1. In light of ρ2 : 0 → 1, we
conclude that c = 3 and
z6(U2) = 1/3, z6(U6) = −1/3.
It follows that z5= Az6(1 − z62)/(1 − 9z62) for some A. This constant A has the property
that Ax(1 − x2) − (1 − 9x2) has repeated roots. We find A = ±3√−3. The choice of the
sign must be synchronized with the choice of the third root of unity in the relation between z4and z5. This will be done later.
We now come to the more complicated part of the lemma. Let π : X6 → X4be the
covering of the Shimura curves. Let γ be an element of Γ5not in Γ6. Then γ normalizes
both Γ4and Γ6and induces automorphisms ρ1and ρ2on X4and X6, respectively. We may
assume that ρ2 = ρ2, where ρ permutes U1, . . . , U6cyclically, as defined in the previous
paragraph. It is easy to check that π ◦ ρ1 = ρ2◦ π. Thus, π(U1), π(U3), and π(U5) are
three different elliptic points on X4. We label them by S4, S40, and S400, respectively. Let
V1, V2be the two ramified points lying above S4. Now there are three possibilities
We will show that the correct one is {U1, U4, V1, V2}. Let Vj0 = ρ2(Vj) and Vj00 = ρ 2 2(Vj) for j = 1, 2. If π−1(S4) = {U1, U2, V1, V2}, then we have z4= Bz6(1 − 3z6)(z6− z6(V1))4(z6− z6(V2))4 (1 + z6)(1 + 3z6)(z6− z6(V100))4(z6− z6(V200))4
for some constant B. The values of z6(V1) and etc. must satisfy
Bx(1 − 3x)(1 − x/z6(V1))4(1 − x/z6(V2))4 − (1 + x)(1 + 3x)(1 − x/z6(V100)) 4(1 − x/z 6(V200)) 4 = C(1 − x)(1 − x/z6(V10)) 4(1 − x/z 6(V20)) 4 (7)
for some constant C. Now if we let p1(x) = 1+ax+bx2= (1−x/z6(V1))(1−x/z6(V2)),
then (1−x/z6(V10))(1−x/z6(V20)) and (1−x/z6(V100)(1−x/z6(V200)) are scalar multiples
of p2(x) = (1 + 3x)2p1 x − 1 3x + 1 = (1 − a + b) + (6 − 2a − 2b)x + (9 + 3a + b)x2, p3(x) = (1 − 3x)2p1 x + 1 1 − 3x = (1 + a + b) + (−6 − 2a + 2b)z + (9 − 3a + b)x2, respectively. Substituting these into (7) and equating the coefficients in the two sides, we find A = B = 0, a = −2, b = −3, but obviously this is invalid. This means that π−1(S4) 6= {U1, U2, V1, V2}. Likewise, π−1(S4) 6= {U1, U6, V1, V2}. Thus, we must
have π−1(S4) = {U1, U4, V1, V2}. Now equating the coefficients in the two sides of
Bx(1 + ax + bx2)4− (1 − x)(1 + 3x)p2(x)4= C(1 + x)(1 − 3x)p3(x)4
and excluding the invalid solutions, we get the claimed relation between z4 and z6. The
relation between z3and z5 can be determined by the known relation between z3and z4,
that between z4 and z6, and that between z5 and z6. This process also determines the
choices of the third roots of unity in the relation between z3and z4 and that between z5
and z6. We omit the details.
Lemma 10. The automorphic derivative Q(z6) = D(z6, τ ) is equal to
15 64 1 z2 6 + 1 (1 − z6)2 + 1 (1 + z6)2 + 1 (z6− 1/3)2 + 1 (z6+ 1/3)2 + 45 128 1 1 − z6 + 1 1 + z6 + 3 1 − 3z6 + 3 1 + 3z6 . (8)
Proof. By Proposition 2, the rational function R(x) such that automorphic Q(z6) = D(z6, τ )
is equal to R(z6) is equal to R(x) = 15 64 1 x2 + 1 (1 − x)2 + 1 (1 + x)2 + 1 (x − 1/3)2+ 1 (x + 1/3)2 +B1 x + B2 x − 1+ B3 x + 1+ B4 x − 1/3+ B5 x + 1/3 for some constants Bjsatisfying
(9) B1+ B2+ B3+ B4+ B5= 0, B2− B3+ 1 3B4− 1 3B5+ 15 16= 0.
Now the normalizer of Γ6in SL(2, R) contains at least the group of signature (2, 6, 8). The
factor group, in terms of the Hauptmodul z6, is generated by σ : z67→ (3z6+1)/(−3z6+3)
and τ : z67→ −z6. By Proposition 6, R(x) satisfies
R(−x) = R(x), 144 (−3x + 3)4R 3x + 1 −3x + 3 = R(x). Combining these informations with (9), we find
B1= 0, B2= B4= −
45
128, B3= B5= 45 128.
This gives us the formula.
We now prove the theorem.
Proof of Theorem 1. By Proposition 1, we have dim S6(Γ1) = 1 − 6 + 6 2 1 −1 4 + 2 6 2 1 −1 6 = 1, dim S6(Γ4) = 1 − 6 + 3 6 2 1 −1 4 = 1, dim S6(Γ6) = 1 − 6 + 6 6 2 1 −1 4 = 7.
By Proposition 4, the one-dimensional spaces S6(Γ1) and S6(Γ4) are spanned by
(10) F1= z 1/4 1 (1 − z1)1/2 2F1 5 24, 3 8; 3 4; z1 + C1z 1/4 1 2F1 11 24, 5 8; 5 4; z1 6 and (11) F2= z 1/4 4 (1 − z4)1/4 2F1 1 8, 3 8; 3 4; z4 + C2z 1/4 4 2F1 3 8, 5 8; 5 4; z4 6
for some complex numbers C1and C2, respectively. Furthermore, by Proposition 2, if we
let f1= z 3/8 6 1 −15 7 z 2 6− 111 14 z 4 6− 2045 46 z 6 6− 11355195 39928 z 8 6− 77997477 39928 z 10 6 − · · · f2= z 5/8 6 1 −5 3z 2 6− 245 34z 4 6− 7269 170 z 6 6− 115223 408 z 8 6− 55230121 27880 − · · ·
be a basis for the solution space of the Schwarzian differential equation d2f /dz62+Q(z6)f =
0, where Q(z6) is the rational function in (8), then a basis for S8(Γ6) is
{z6jg : j = 0, . . . , 6}, g =
(f1+ C3f2)8
z2
6(1 − z62)2(1 − 9z62)2
. Now from Lemma 9, we have
z1= 12αz6(1 − z62)(1 − 9z62) (1 + αz6)6 and z4= − 4(1 + β)4z 6(1 + (−7 + 4β)z62/3)4 (1 + z6)(1 − 3z6)(1 + (4 + 2β)z6− (1 + 2β)z26)4 ,
where α is a root of x2+ 3 = 0 and β is a root of x2+ 2 = 0. Substituting these into (10) and (11) and comparing the coefficients, we find
F1= c1(1 + 3z62) 3g and F2= c2 1 +−7 + 4β 3 z 2 6 1 + (4 + 2β)z6− (1 + 2β)z26 × 1 − (4 + 2β)z6− (1 + 2β)z62 g
for some constants c1 and c2. Taking the sixth roots of F1 and F2 and simplifying, we
obtain the identities claimed in the theorem.
4. ALGEBRAIC TRANSFORMATIONS ASSOCIATED TOCLASSVI According to Appendix A, the subgroup diagram for Takeuchi’s Class VI is
(2, 4, 5) (2, 4, 10) (2, 5, 5) (4, 4, 5) (2, 10, 10) (2, 2, 5, 5) (5, 10, 10) (5, 5, 5, 5) 2 H H H H HH 6 2 2 H H H H HH 6 2 2 2 2 2
Let Γ1 = (2, 5, 5), Γ2 = (5, 10, 10), Γ3 = (5, 5, 5, 5), and X1, X2, X3 be the Shimura
curves associated to these three groups. (The reader is reminded that the subgroup diagram should be read as “there are arithmetic Fuchsian subgroups of SL(2, R) such that their subgroup relations are given by the diagram”.) The subgroups relations Γ3< Γ1, Γ2admit
Coxeter decompositions as the following figures show.
Here the small triangles are (2, 4, 5)-triangles. Associated to this triplet of groups is the following identities. Theorem 2. We have 2F1 1 20, 1 4; 4 5; 64z(1 − z − z2)5 (1 − z2)(1 + 4z − z2)5 = (1 − z2)1/20(1 + 4z − z2)1/42F1 3 10, 2 5; 9 10; z 2 . (12)
and (1 − z − z2)2F1 1 4, 9 20; 6 5; 64z(1 − z − z2)5 (1 − z2)(1 + 4z − z2)5 = (1 − z2)1/4(1 + 4z − z2)5/42F1 2 5, 1 2; 11 10; z 2 . (13)
Proof. Label the elliptic points of Xjby P2, P5, P50 for X1, Q5, Q10, Q010for X2, and Ri,
i = 1, . . . , 4, for X3such that the ramifications data are given by
R1 R3 S1 S3 R2 R4 S2 S4 R2 R4 R1 R3
P5 P50 Q5 Q10 Q010
1 1 5 5 1 1 5 5 1 1 2 2
Here the numbers next to the lines are the ramification indices. We have omitted P2from
the diagram. There are 6 points lying above P2. Each has ramficiation index 2. Choose
Hauptmoduls zjfor Xjby requiring
z1(P5) = 0, z1(P2) = 1, z1(P50) = ∞, z2(Q10) = 0, z2(Q5) = 1, z2(Q010) = ∞
and
z3(R1) = 0, z3(R2) = 1, z3(R3) = ∞.
The relation between z2and z3is easy to figure out. We have
(14) z2= z23,
which implies that z3(R4) = −1. To determine the relation between z1and z3, we observe
that the tessellation of the hyperbolic plane by the (5, 5, 5, 5)-polygons has extra symme-tries by rotation by 90 degree around the center of any (5, 5, 5, 5)-polygon. In terms of groups, this means that Γ3 has a supergroup Γ normalizing Γ3 such that Γ/Γ3 is cyclic
of order 4. (In fact, Γ is the (4, 4, 5)-triangle group in the subgroup diagram.) Therefore, the automorphism group of X3has an element σ of order 4 that permutes R1, R2, R3, R4
cyclically. In terms of the Hauptmodul, we have σ : z37−→
z3+ 1
z3− 1
. Thus, if the value of z3at S1is a, then we have
z3(S1) = a, z3(S2) = a − 1 a + 1, z3(S3) = − 1 a, z3(S4) = − a + 1 a − 1. Therefore, the relation between z1and z3is
z1=
Bz3(z3− a)5(z3+ 1/a)5
(1 − z2
3)(z3− (a − 1)/(a + 1))5(z3+ (a + 1)/(a − 1))5
for some constant B. Moreover, the automorphism σ of X3rotates 4 of the six points lying
above P2cyclically and fixes the other two. (The reader is reminded that each (5, 5, 5,
5)-polygon represents only half of the fundamental domain for Γ3. The two fixed of σ are the
centers of the (5, 5, 5, 5)-polygons.) In terms of the Hauptmodul z3, this means that the
values of z3at the two fixed points of σ are ±i and if the value of z3at one of the other 4
points above P2is b, then the values at the other 3 points are −1/b, (b − 1)/(b + 1), and
−(b + 1)/(b − 1). Thus, we have z1− 1 = C(1 + z2 3)2(z3− b)2(z3+ 1/b)2(z3− (b − 1)/(b + 1))2(z3+ (b + 1)/(b − 1))2 (1 − z2 3)(z3− (a − 1)/(a + 1))5(z3+ (a + 1)/(a − 1))5
for some constant C. Comparing the two sides, we find a = 0, ±1, ±i, a2+ a − 1 = 0, or a2− a − 1 = 0. The first five solutions are invalid. The other two solutions give
(15) z1= 64z3(1 − z3− z32)5 (1 − z2 3)(1 + 4z3− z23)5 or (16) z1= − 64z3(1 + z3− z32)5 (1 − z2 3)(1 − 4z3− z32)5 .
Both are valid because of the following reason. Notice that Γ2normalizes Γ3. If we take an
element γ of Γ2not in Γ3, then γ−1Γ1γ is again a triangle of signature (2, 5, 5) containing
the same Γ3. If the relation between the Hauptmoduls of Γ1and Γ3is (15), then the relation
between the Hauptmoduls of γ−1Γ1γ and Γ3will be (16).
Having determining the relations among Hauptmoduls, we can composite identity (36) in [10] with Kummer’s quadratic transformation several times to get the identities in the theorem. However, the procedure is very tedious. Here we provide a better proof using the theory of automorphic forms on Shimura curves.
By Proposition 1, we have dim S8(Γ1) = 1 − 8 + 8 2 1 − 1 2 + 2 8 2 1 − 1 5 = 1, dim S8(Γ2) = 1 − 8 + 8 2 1 − 1 5 + 2 8 2 1 − 1 10 = 2, dim S8(Γ3) = 1 − 8 + 4 8 2 1 − 1 5 = 5.
By Proposition 4, the one-dimensional space S8(Γ1) is spanned by
(17) F1= z 1/5 1 2F1 1 20, 1 4; 4 5; z1 + C1z 1/5 1 2F1 1 4, 9 20; 6 5; z1 8
for some constant C1, and the function
(18) F2= z 3/5 2 (1 − z2)1/5 2F1 3 10, 2 5; 9 10; z2 + C2z 1/10 2 2F1 2 5; 1 2; 11 10; z2 8
is contained in S8(Γ2) for some constant C2. To get a basis for S8(Γ3), we need to work
out the Schwarzian differential equation associated to z3. It is actually easy in this case.
By Proposition 2, the function z30(τ ), as a function of z3, satisfies
d2 dz2 3 f + Q(z3)f = 0, where (19) Q(z3) = 6 25 1 z2 3 + 1 (1 − z3)2 + 1 (1 + z3)2 +B1 z3 + B2 z3− 1 + B3 z3+ 1
for some complex numbers satisfying
(20) B1+ B2+ B3= 0, B2− B3+
12 25 = 0.
To determine the values of Bj, we use the automorphism of X3coming from the normal
subgroup relation Γ3C Γ1. Let γ be an element of Γ2not in Γ3. We know that
Now by Proposition 5, we have D(−z3(τ ), τ ) = D(z3(γτ ), τ ) = D(z3(γτ ), γτ ) + D(γτ, τ )/(dγτ /dτ )2= Q(z3(γτ )) = 6 25 1 z2 3 + 1 (1 − z3)2 + 1 (1 + z3)2 + B1 −z3 + B2 −z3− 1 + B3 −z3+ 1 (21)
On the other hand, we also have, by the same proposition,
(22) D(−z3, τ ) = D(−z3, z3) + D(z3, τ )/(−1)2= Q(z3)
Comparing (19), (21), and (22), we find B1= 0 and B2 = −B3. Together with (20), this
gives us B1= 0, B2= − 6 25, B3= 6 25 and Q(z3) = 6 25 1 z2 3 + 1 (1 − z3)2 + 1 (1 + z3)2 + 1 1 − z3 + 1 1 + z3 . Now a basis for the solution space of the Schwarzian differential equation d2f /dz2
3 + Q(z3)f = 0 is given by f1= z 2/5 3 1 − 4 15z 2 3− 52 475z 4 3− 13436 206625z 6 3− 46348 1033125z 8 3− 2024924 60265625z 10 3 − · · · f2= z 3/5 3 1 −12 55z 2 3− 28 275z 4 3− 2708 42625z 6 3− 393636 8738125z 8 3− 7503908 218453125z 10 3 − · · · . By Proposition 2, g, z3g, z23g, z 3 3g, z 4 3g, g = (f1+ C3f2)8 z3 3(1 − z3)3(1 + z3)3 ,
form a basis for S8(Γ3) for some constant C3. That is, after substituting (15) and (14)
into (17) and (18), respectively, we have F1 = h1(z3)g and F2 = h2(z3)g for some
polynomials h1(x) and h2(x) of degree ≤ 4. Indeed, by comparing the coefficients, we
find
F1= 26/5(1 − z3− z32)(1 + 4z3− z32)g, F2= z3g.
(The computation becomes easier if we take the 8th roots of the functions first.) Simplify-ing the relation z3F1= 26/5(1 − z3− z23)(1 + 4z3− z32)F2, we get the two identities in
the theorem. This completes the proof.
5. ALGEBRAIC TRANSFORMATIONS ASSOCIATED TO OTHER CLASSES
5.1. Classes II, V, and XII. The subgroup diagrams of Class II, V, and XII are all of the form (2, 4, 2n) (2, 2n, 2n) (4, 4, n) (n, 2n, 2n) (2, n, 2, n) (n, n, n, n) 2 H H H H 2 2 H H H H 2 2 H H H H 2 2
The subgroup relation (2, 2n, 2n) ∩ (4, 4, n) = (2, n, 2, n) is a special case of (2, 2m, 2n) (m, 2n, 2n) (n, 2m, 2m) (m, n, m, n) 2 H H HH 2 H H HH 2 2
which arises from the Coxeter decompositions of a quadrilateral polygon that is symmetric with respect to both the diagonals as shown below
2 @ @@ 2 @ @@ 2 2
Associated to this family of subgroup relations is the following identity.
Theorem 3. For real numbers a and b such that neither b + 3/4 nor 2b + 1/2 is a nonpos-itive integer, we have
(1 + z)2a+2b2F1 a + b, a +1 4; b + 3 4; z 2 =2F1 a + b, b +1 4; 2b + 1 2; 4z (1 + z)2 in a neighborhood ofz = 0.
This identity can be easily proved using Kummer’s quadratic transformation. Alterna-tively, one can verify that both sides are solutions of the differential equation
2z(1−z)(1+z)2F00−(1+z)((3−4b)z2+8(a+b)z−4b−1)F0−(a+b)(1+4b)(1−z)F = 0.
5.2. Classes IV, VIII, XI, XIII, XV, XVII. The subgroups diagrams of Classes IV, VIII, XI, XIII, XV, and XVIII are either of the form
(2, 3, 12n) (3, 3, 6n) (3, 4n, 12n) (2, 6n, 12n) (3, 3, 2n, 6n) (6n, 6n, 6n) (2n, 4n, 6n, 12n) (3n, 12n, 12n) (2n3, 6n3) (3n2, 6n2) (n, 3n, 4n2, 12n2) (1; n2, 2n2, 3n2, 6n2) 2 4 H H H H H HH 3 4 H H H H H HH H H H H H HH 4 H H H H H HH 2 H H H H H H 3 4 H H H H H H 2 2 HH H H H H 2 4 H H H H H H 2 4 2
or sub-diagram of it with Class XI having one extra node. There are two families of essen-tially new identities associated to these classes. One corresponds to the pair of (3, 3, 6n) and (3, 4n, 12n). (Theorem 4 below.) One corresponds to the pair of (3, 4n, 12n) and (2, 6n, 12n). (Theorem 5 below.)
Theorem 4. For a real number a such that neither 3a + 1 nor 2a + 1 is a nonpositive integer, we have (1 + z)a+1/6(1 − z/3)3a+1/22F1 2a +1 3, a + 1 3; 3a + 1; z 2 =2F1 a +1 6, a + 1 2; 2a + 1; 16z3 (1 + z)(3 − z)3 in a neighborhood ofz = 0.
Theorem 5. For a real number a such that neither 6a + 1 nor 4a + 1 is a nonpositive integer, we have (1 − z)9a+3/42F1 4a +1 3, 2a + 1 3; 6a + 1; − 27z2(1 − z) 1 − 9z = (1 − 9z)a+1/122F1 3a +1 4, a + 1 4; 4a + 1; − 64z3 (1 − z)3(1 − 9z) in a neighborhood ofz = 0.
In principle, these two identities can be deduced from Kummer’s and Goursat’s transfor-mations, once the related Belyi functions are determined. Here we briefly indicate how one can prove the theorems in the cases where the parameters correspond to discrete Fuchsian groups using theory of automorphic forms.
Proof of Theorem 4 in the cases of Shimura curves. For the pair of (3, 3, 6n) and (3, 4n, 12n), the subgroup relations admit Coxeter decompositions, as shown in the figures
Here the parameter n in the figures is 1 and the smaller triangles are (2, 3, 12)-triangles. Let Γ1 = (3, 3, 6n), Γ2 = (3, 4n, 12n), Γ3 = Γ1∩ Γ2, and let Xi, i = 1, . . . , 3 be
the associated Shimura curves. Denote by P3, P30, and P6n the elliptic points of
or-ders 3, 3, and 6n on X1, by Q3, Q4n, and Q12n the elliptic points of orders 3, 4n, and
12n on X2, and by R3, R03, R2n, and R6n the elliptic points of order 3, 3, 2n, and 6n
on X3. The points are labelled in a way such that the ramification data are given by
R3 S1 R03 S2 R2n R6n R3 R30 R2n R6n
P3 P30 P6n Q3 Q4n Q12n
1 3 1 3 3 1 1 1 2 2
Choose Hauptmoduls zjon Xj, j = 1, 2, 3, by requiring
z1(P6n) = 0, z1(P3) = 1, z1(P30) = ∞, z2(Q4n) = 0, z2(Q3) = 1, z2(Q12n) = ∞
and
z3(R2n) = 0, z3(R3) = 1, z3(R6n) = ∞.
It is easy to see from the ramification informations that
(23) z2= z23,
which implies that z3(R03) = −1. For z1, we have
z1=
Az3 3
(1 + z3)(1 − az3)3
for some complex numbers A and a, where 1/a is the value of z3at S1. These two numbers
satisfy (24) 1 − Az 3 3 (1 + z3)(1 − az3)3 = 1 − z1= (1 − z3)(1 − bz3)3 (1 + z3)(1 − az3)3 ,
where 1/b is the value of z3at S2. Now observe that Γ3is a normal subgroup of Γ2. Thus,
an element of Γ2not in Γ3induces an automorphism on X3. In terms of the Hauptmodul
z3, it is easy to see that this automorphism sends z3to −z3. Since this automorphism maps
S1 to S2, we find b = −a. Then comparing the two sides of (24), we get A = 16/27,
a = 1/3, and
(25) z1=
16z33 (1 + z3)(3 − z3)3
. Now by Proposition 1, we have
dim S6(Γ1) = dim S6(Γ2) = 1, dim S6(Γ3) =
(
2, if n = 1, 3, if n ≥ 2. From now on, we assume that n ≥ 2.
By Proposition 4, the one-dimensional spaces S6(Γ1) and S6(Γ2) are spanned by F1= z 1−1/2n 1 2F1 1 6 − 1 12n, 1 2 − 1 12n; 1 − 1 6n; z1 + C1z 1/6n 1 2F1 1 6 + 1 12n, 1 2+ 1 12n; 1 + 1 6n; z1 !6 (26) and F2= z 1−3/4n 2 2F1 1 3− 1 6n, 1 3− 1 12n; 1 − 1 4n; z2 + C2z 1/4n 2 2F1 1 3 + 1 12n, 1 3 + 1 6n; 1 + 1 4n; z2 !6 , (27)
respectively, for some constants C1and C2. Also, if we let f1= z
1/2−1/4n
3 (1 + c1z + · · · )
and f2 = z
1/2+1/4n
3 (1 + d1z + · · · ) be a basis of the solution space of the Schwarzian
differential equation d2f /dz2
3 + Q(z3)f = 0 associated to z3, then by Proposition 2,
S6(Γ3) is spanned by g, z3g, and z23g, where
g = (f1+ C3f2)
6
z2
3(1 − z3)2(1 + z3)2
for some constant C3. Now we substitute (25) and (23) into (26) and (23), respectively.
We find F1= a1z 3−3/2n 3 + · · · , F2= z 2−3/2n 3 + · · · ,
where a1= (16/27)1−1/2n, and thus
F1= a1z32g, F2= (z3+ a2z32)g
for some constant a2. That is, a1zF2/F1 = 1 + a2z3. We then take the 6th roots of the
two sides and compare the coefficients of z3/2−1/4n, we find that a
2is actually 0. After simplifying, we arrive at (1 + z)1/6−1/12n(1 − z/3)1/2−1/4n2F1 1 3 − 1 6n, 1 3 − 1 12n; 1 − 1 4n; z 2 =2F1 1 6 − 1 12n, 1 2− 1 12n; 1 − 1 6n; 16z3 (1 + z)(3 − z)3 .
This proves Theorem 4 in the case the parameters correspond to arithmetic triangle groups. Proof of Theorem 5 in the cases of Shimura curves. The subgroups (3, 4n, 12n), (2, 6n, 12n) and their intersection admit Coxeter decompositions as the figures below show.
Denote the groups (3, 4n, 12n), (2, 6n, 12n), and (2n, 4n, 6n, 12n) by Γ1, Γ2, and Γ3,
respectively. Label the elliptic points of (3, 4n, 12n) by P3, P4n, and P12n, those of
(2, 6n, 12n) by Q2, Q6n, and Q12n, and those of (2n, 4n, 6n, 12n) by R2n, R4n, R6n,
and R12n. The ramifications are shown as follows.
R2n R4n R6n R12n R2n R6n R4n R12n
P4n P12n P3 Q6n Q12n Q2
2 1 2 1 3 3 1 3 1 2 2
Choose Hauptmoduls zjfor Γj, j = 1, . . . , 3, by requiring that
z1(P4n) = 0, z1(P3) = 1, z1(P12n) = ∞,
z2(Q6n) = 0, z2(Q2) = 1, z2(Q12n) = ∞,
z3(R2n) = 0, z3(R4n) = 1, z3(R6n) = ∞.
It is easy to work out the relation between z1and z3and that between z2and z3. They are
(28) z1= 27z23(1 − z3) 1 − 9z3 , z2= − 64z33 (1 − z3)3(1 − 9z3) .
Here 1/9 is the value of z3at R12n. We then follow the same arguments as before to obtain
the claimed identities. We omit the details.
APPENDIXA. LIST OF ARITHMETIC TRIANGLE GROUPS
In this section, we determine the signatures of the intersections of commensurable tri-angle groups.
According to [7, 8], there are totally 85 arithmetic triangle groups, falling in 19 different commensurability classes. Here we give the subgroup diagrams. Note that since most groups here have genus 0, we omit the genus information from the signature, unless the group has a positive genus. Also, to save space, the notation (g; en1
1 , . . . , enrr) means that
the Shimura curve has ni elliptic points order ei. Furthermore, for convenience, we will
often call the groups by their signatures, even though this raises some ambiguity.
Remark 11. There is some ambiguity when we say “the intersections of commensurable triangle groups” because there may be more than one orders whose norm-one groups have the same signature and the intersections of these groups with another group may have different signatures. For example, in the case B = M (2, Q), the subgroups Γ0(2) and
Γ0
(2) of SL(2, Z) have the same signature (0; 2, ∞, ∞) and the group Γ0(4) has signature
(0; ∞, ∞, ∞). The intersection of Γ0(2) and Γ0(4) is just Γ0(4), but the intersection of
Γ0(2) and Γ0(4) has signature (0; ∞, ∞, ∞, ∞). Thus, the subgroup diagrams described
here should be read as “there are arithmetic groups whose subgroup relations are given by the subgroup diagrams”.
Since it is not easy to describe explicitly the orders associated to arithmetic triangle groups, here we use group theory and properties of discrete subgroups of SL(2, R) to determine the signatures. We will work out the case of Class IV in [8] and omit the proof of the others.
According to [8], Class IV of arithmetic triangle groups has the following subgroup diagram. (2, 3, 12) (3, 3, 6) (3, 4, 12) (2, 6, 12) (6, 6, 6) (3, 12, 12) 2 4 H H H H H H 3 H H H H H H 3 2 H H H H H H 2
Here the numbers next to the lines are the indices. Set
Γ1= (2, 3, 12), Γ2= (3, 3, 6), Γ3= (3, 4, 12),
Γ4= (2, 6, 12), Γ5= (6, 6, 6), Γ6= (3, 12, 12),
and let Xi, i = 1, . . . , 6, denote the respective Shimura curves. To determine Γ2∩ Γ3, we
observe that Γ2is a normal subgroup of Γ1of index 2 and Γ1= Γ2Γ3. Thus, Γ2∩ Γ3is a
normal subgroup of Γ3of index 2. Now the elliptic point of order 3 on X3must split into
two points in X(Γ2∩ Γ3) because 2 - 3. Then from the Riemann-Hurwitz formula, we see
that the elliptic points of order 4 and 12 must be ramified. That is, the curve X(Γ2∩ Γ3)
must have signature (2, 3, 3, 6). In fact, this can also be seen from the following figures.
Here the smaller triangles are (2, 3, 12)-triangles. The figures show that the triangle group (2, 3, 12) contains two subgroups of signatures (3, 3, 6) and (3, 4, 12), respectively, whose intersection has signature (2, 3, 3, 6). (In fact, the theoretical argument above shows that for any pair of subgroups of Γ1 with signatures (3, 3, 6) and (3, 4, 12), respectively, the
intersection must have signature (2, 3, 3, 6).) Likewise, the figures
show that there are two subgroups of Γ1 of signatures (2, 6, 12) and (3, 4, 12) such that
there intersection has signature (2, 4, 6, 12). We have the following subgroup diagram. (2, 3, 12) (3, 3, 6) (3, 4, 12) (2, 6, 12) (2, 3, 3, 6) (6, 6, 6) (2, 4, 6, 12) (3, 12, 12) 2 4 H H H H H H 3 4 H H H H H H H H H H H H H H H H H H 2 4
Let Γ7 = (2, 3, 3, 6) and Γ8 = (2, 4, 6, 12) and X7and X8be their associated Shimura
curves. Again, because Γ5 is a normal subgroup of Γ4 of index 2 and Γ5Γ8 = Γ4, the
intersection of Γ5and Γ8is a subgroup of index 2 of Γ8. Now the group (2, 4, 6, 12) has
many subgroups of index 2. (The structure of the quotient group of (2, 4, 6, 12) over its commutator subgroup is C2× C4× C6.) To determine which of them is contained is the
group (6, 6, 6), we use the following properties.
(1) If p is an elliptic point of order e on X8, then its preimage in the covering X(Γ5∩
Γ8) → X8 consists of either a single elliptic point of order e/2 or two elliptic
points of order e.
(2) The total branch number of any finite covering of compact Riemann surface is always even.
(3) The volume of X(Γ5∩ Γ8) is twice of that of X8. Thus, if (g; e1, . . . , er) is the
signature of X(Γ5∩ Γ8), then we must have
2g − 2 + r X i=1 1 − 1 ej = 2 2 −1 2 − 1 4− 1 6 − 1 12 = 2.
From these informations, we find that possible signatures of a subgroup of index 2 of (2, 4, 6, 12) are
(1; 2, 3, 6), (0; 2, 62, 122), (0; 3, 42, 122), (0; 42, 63) (0; 23, 3, 122), (0; 23, 63), (0; 22, 3, 42, 6). (29)
Likewise, an elliptic point of order 6 on X5can
(1) splits into 4 elliptic points of order 6, or (2) splits into 2 elliptic points of order 3, or
(3) splits into 1 elliptic point of order 3 and 2 elliptic point of order 6, or (4) splits into 1 elliptic point of order 2 and 1 elliptic point of order 6,
in the covering X(Γ5∩ Γ8) → X5of degree 4. Also, the total branch number of X(Γ5∩
Γ8) → X5must be a positive even integer and the volume of X(Γ ∩ Γ8) is 2. We find the
possible signatures of a subgroup of index 4 of Γ5are
(30) (0; 23, 63), (0; 22, 32, 62), (0; 2, 34, 6), (0; 36).
From (29) and (30), we conclude that the signature of Γ5∩ Γ8must be (0; 23, 63). This
By the same argument, we can also show that the intersection of Γ6and Γ8must have
signature (0; 3, 42, 122) and the intersection of Γ
5and Γ6has signature (0; 3, 3, 6, 6). The
subgroup diagram becomes
(2, 3, 12) (3, 3, 6) (3, 4, 12) (2, 6, 12) (2, 3, 3, 6) (6, 6, 6) (2, 4, 6, 12) (3, 12, 12) (23, 63) (32, 62) (3, 42, 122) 2 4 H H H H HH 3 4 H H H H HH H H H H HH H H H H HH 2 4 H H H H H H 3 4 H H H H H H 2 H H H H H H 4
Finally, we can show that the only possible signatures of subgroups of index 2 in (23, 63)
are
(0; 26, 32, 62), (0; 24, 3, 64), (0; 22, 66), (1; 24, 33), (1; 22, 32, 62), (1; 3, 64), (2; 33), while the only possible signatures of subgroups of index 2 in (3, 42, 122) are
(0; 32, 44, 62), (0; 2, 32, 42, 6, 122), (0; 22, 32, 124), (1; 22, 32, 62).
From these, we see that the common intersection of (23, 63), (3, 42, 122), and (32, 62) has
signature (1; 22, 32, 62). This completes the proof of the case of Class IV.
Now we give the subgroup diagrams for arithmetic triangle groups. Class II. (2, 4, 6) (2, 6, 6) (3, 4, 4) (3, 6, 6) (2, 2, 3, 3) (3, 3, 3, 3) 2 H H H 2 2 H H H 2 2 H H H 2 2
Class III. (2, 6, 8) (2, 3, 8) (4, 6, 6) (3, 8, 8) (3, 3, 4) (2, 4, 8) (22, 43) (3, 4, 3, 4) (4, 4, 4) (2, 8, 8) (46) (2, 4, 2, 4) (4, 8, 8) ? (4, 4, 4, 4) ? 2 H H H H HH 2 10 2 H H H H HH 3∗ 3∗ H H H H HH 2 2 10 3∗ 2 2 H H H H HH 2 3∗ 10 2 2 2 2 10 2 2 2 10 Class IV. (2, 3, 12) (3, 3, 6) (3, 4, 12) (2, 6, 12) (3, 3, 2, 6) (6, 6, 6) (2, 4, 6, 12) (3, 12, 12) (23, 63) (32, 62) (3, 42, 122) (1; 2, 2, 3, 3, 6, 6) 2 4 H H H H HH 3 4 H H H H HH H H H H HH 4 H H H H HH 2 H H H H H H 3∗ 4 H H H H H H 2 2 H H H H H H 2 4 H H H H H H 2 4 2
Class V. (2, 4, 12) (2, 12, 12) (4, 4, 6) (6, 12, 12) (2, 2, 6, 6) (6, 6, 6, 6) 2 H H H 2 2 H H H 2 2 H H H 2 2 Class VI. (2, 4, 5) (2, 4, 10) (2, 5, 5) (4, 4, 5) (2, 10, 10) (2, 2, 5, 5) (5, 10, 10) (5, 5, 5, 5) 2 H H H H HH 6 2 2 H H H H HH 6∗ 2 2 2 2 2 Class VII. (2, 5, 6) (3, 5, 5) 2 Class VIII. (2, 3, 10) (3, 3, 5) (2, 5, 10) (5, 5, 5) 2 Q Q QQ 3 Q Q QQ 3 2 Class IX. (3, 4, 6)
Class X. (2, 4, 7) (2, 3, 7) (2, 3, 14) (2, 7, 7) (3, 3, 7) (2, 7, 14) (1; 7, 7) (7, 7, 7) (1; 76) Q Q QQ 2 9 Q Q QQ 8 2 Q Q QQ 3 Q Q QQ 8 9 Q Q QQ 3 2 Q Q Q Q 3 9 Class XI. (2, 3, 9) (2, 3, 18) (3, 3, 9) (3, 6, 18) (2, 9, 18) (3, 3, 3, 9) (9, 9, 9) (3, 6, 9, 18) (33, 93) 4 2 4 H H H H HH 3 4 H H H H HH H H H H HH 4 H H H H H H 3∗ 4 2 Class XII. (2, 4, 18) (2, 18, 18) (4, 4, 9) (9, 18, 18) (2, 2, 9, 9) (9, 9, 9, 9) 2 H H H 2 2 H H H 2 2 H H H 2 2 Class XIII. (2, 3, 16) (2, 8, 16) (3, 3, 8) (4, 16, 16) (8, 8, 8) (4, 4, 8, 8) 3 H H H 2 2 H H H 2 3 H H H 2 2
Class XIV. (2, 5, 20) (5, 5, 10) 2 Class XV. (2, 3, 24) (3, 3, 12) (3, 8, 24) (2, 12, 24) (3, 3, 4, 12) (12, 12, 12) (4, 8, 12, 24) (6, 24, 24) (43, 123) (62, 122) (2, 6, 82, 242) (1; 22, 42, 62, 122) 2 4 H H H H HH 3 4 H H H H HH H H H H HH 4 H H H H HH 2 H H H H HH 3∗ 4 H H H H HH 2 2 HH H H HH 2 4 H H H H HH 2 4 2 Class XVI. (2, 5, 30) (5, 5, 15) 2 Class XVII. (2, 3, 30) (3, 3, 15) (3, 10, 30) (2, 15, 30) (3, 3, 5, 15) (15, 15, 15) (5, 10, 15, 30) (53, 153) 2 4 H H H H HH 3 4 H H H H HH H H H H HH 4 H H H H H H 3∗ 4 2
Class XVIII. (2, 5, 8) (4, 5, 5) 2 Class XIX. (2, 3, 11) REFERENCES
[1] Pilar Bayer and Artur Travesa. Uniformizing functions for certain Shimura curves, in the case D = 6. Acta Arith., 126(4):315–339, 2007.
[2] Noam D. Elkies. Shimura curves for level-3 subgroups of the (2, 3, 7) triangle group, and some other examples. In Algorithmic number theory, volume 4076 of Lecture Notes in Comput. Sci., pages 302–316. Springer, Berlin, 2006.
[3] ´Edouard Goursat. Sur l’´equation diff´erentielle lin´eaire, qui admet pour int´egrale la s´erie hyperg´eom´etrique. Ann. Sci. ´Ecole Norm. Sup., 10:3–142, 1881.
[4] Svetlana Katok. Fuchsian groups. Chicago Lectures in Mathematics. University of Chicago Press, Chicago, IL, 1992.
[5] Goro Shimura. Construction of class fields and zeta functions of algebraic curves. Ann. of Math. (2), 85:58– 159, 1967.
[6] Goro Shimura. Introduction to the arithmetic theory of automorphic functions, volume 11 of Publications of the Mathematical Society of Japan. Princeton University Press, Princeton, NJ, 1994. Reprint of the 1971 original, Kano Memorial Lectures, 1.
[7] Kisao Takeuchi. Arithmetic triangle groups. J. Math. Soc. Japan, 29(1):91–106, 1977.
[8] Kisao Takeuchi. Commensurability classes of arithmetic triangle groups. J. Fac. Sci. Univ. Tokyo Sect. IA Math., 24(1):201–212, 1977.
[9] Fang-Ting Tu. Schwarzian differential equations associated to Shimura curves of genus zero. preprint, 2011. [10] Raimundas Vid¯unas. Transformations of some Gauss hypergeometric functions. J. Comput. Appl. Math.,
178(1-2):473–487, 2005.
[11] Marie-France Vign´eras. Arithm´etique des alg`ebres de quaternions, volume 800 of Lecture Notes in Mathe-matics. Springer, Berlin, 1980.
[12] Yifan Yang. Schwarzian differential equations and Hecke eigenforms on Shimura curves. arxiv:1110.6284, 2011.
DEPARTMENT OFAPPLIEDMATHEMATICS, NATIONALCHIAOTUNGUNIVERSITY, 1001 TAHSUEH ROAD, HSINCHU, TAIWAN30010, ROC
E-mail address: ft12.am95g@nctu.edu.tw
DEPARTMENT OFAPPLIEDMATHEMATICS, NATIONALCHIAOTUNGUNIVERSITY ANDNATIONALCEN -TER FORTHEORETICALSCIENCES, HSINCHU300, TAIWAN
