A COSET REPRESENTATIVES FOR MODULAR GROUP

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(1)A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ) YING HUNG CHENG. i.

(2) ii. YING HUNG CHENG. Contents 1. Introduction. 1. 2. Modular function. 3. 3. The index of Γ(1)/Γ0 (N ). 9. 4. The representatives for Γ(1)/Γ0 (N ). 18. References. 32.

(3) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). 1. Abstract. It is well-known that C(j(z), j(N z)) is the field of modular functions in Γ0 (N ) \ H∗ . Let {γ1 , · · · , γµ } be a set of representatives for Γ(1)/Γ0 (N ). The minimum polynomial of j(N z) is F (j(z), Y ) =. µ Y. (Y − j(N γi z)).. i=1. Therefore, a representatives for Γ(1)/Γ0 (N ) is important about the modular equation F (j(z), Y ) for Γ0 (N ). Shimura provides a representatives for Γ(1)/Γ0 (N ). However, it is not clear. The purpose of this paper is to clarify it. Moreover, the paper provides some degree of coefficients of F (j(z), Y ).. 1. Introduction In this paper, N is fixed to be an integer and N > 1. Let a b Γ(1) = SL2 (Z) = {α = | a, b, c, d ∈ Z and det(α) = 1}, c d a b Γ(N ) = { ∈ SL2 (Z)| a ≡ d ≡ 1 and c ≡ b ≡ 0 (mod N )}, c d a b Γ0 (N ) = { ∈ SL2 (Z)| c ≡ 0 (mod N )}. c d It is well-known that the quotient space Γ0 (N ) \ H∗ is a compact Riemann surface. In section 2, we know C(j(z), j(N z)) is the field of the modular functions in Γ0 (N ) \ H∗ . In particular, given a representatives {γ1 , · · · , γµ } for Γ(1)/Γ0 (N ), the minimum polynomial of j(N z) is µ Y F (j(z), Y ) = (Y − j(N γi z)). i=1. Therefore, if we want to know F (j(z), Y ), then a representatives for Γ(1)/Γ0 (N ) plays an important role. . a b Given a coset Γ0 (N ) ∈ Γ(1)/Γ0 (N ), we have ad − cb = 1 and hence it c d implies gcd(c, d) = 1. On hand, given a pair (c, d) the other with gcd(c, d) = 1, there a b a1 b1 a2 b 2 are a, b ∈ Z such that ∈ Γ(1). Besides, suppose , ∈ c d c d c d Γ(1). In section 4, we know that theyare in the same coset. Therefore, we use the a b pair (c, d) to replace the coset Γ0 (N ) . However, if (c1 , d1 ) 6= (c2 , d2 ), then c d they maybe in the same coset. In Shimura [2] and Milne [1], they gives a representatives for Γ(1)/Γ0 (N ) as follow: gcd(c, d) = 1, d|N, 0 < c ≤. N N (or c in any set of representatives for Z module ( )). d d.

(4) 2. YING HUNG CHENG. For example, we take N = 6 and by Shimura, the set of the representatives for Γ(1)/Γ0 (6) is A = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 2), (3, 2), (1, 3), (2, 3), (1, 6)}. Y 1 The number of A is 11. In section 3, we get #Γ(1)/Γ0 (N ) = N (1+ ) and hence p p|N p prime. #Γ(1)/Γ0 (6) =6 × (1 + 21 ) × (1 + 13 ) = 12. Apparently, the statement ”c is in any set of representatives for Z module ( Nd )“ in Shimura [2] is not clear. In this paper, we will clarify the statement of Shimura [2]. In next section, we will give some detail about modular function and give a sense of the relation between modular equation for Γ0 (N ) and a representatives for Γ(1)/Γ0 (N ). In section 3, We will compute the index of Γ(1)/Γ0 (N ). Finally, in section 4, we will give a representatives for Γ(1)/Γ0 (N ) and give some examples about the modular equation for Γ0 (N )..

(5) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). 3. 2. Modular function Let Γ be a subgroup of Γ(1) and the index of Γ(1)/Γ is finite. In this section, we will introduce modular functions for Γ. First, we give some background. Definition 2.1. Let G be a group and X be a topological space. We say that G acts on X if there exists a mapping G × X −→ X (g, x) 7−→ g · x satisfying the following three conditions: (1)for each element g of G, the mapping φg : X −→ X defined by ϕg (x) = g · x is a continuous mapping; (2)(gh)x = g(hx), ∀ g, h ∈ G; (3)for the identity element 1 of G, 1 · x = x, ∀ x ∈ X. Let H = {x + iy|x ∈ R, y > 0} be the upper half plane. We consider G = Γ(1), X = H and define a mapping λ: Γ(1) × H −→ H az + b a b , z) 7−→ α · z = . (α = c d cz + d It is easy to check Γ(1) acts on H by λ. Definition 2.2. Suppose G act on X and let x be in X. Then Gx = {g · x|g ∈ G} ⊆ X is called the G-orbit of x, and we denote G \ X is the set of all G-orbits in X. Let x, y ∈ X. Suppose z ∈ Gx ∩Gy . There are g1 , g2 ∈ G such that z = g1 ·x = g2 ·y. It implies x = (g1−1 g2 ) · y and hence x ∈ Gy . It says that Gx = Gy . Therefore, we have Gx = Gy or Gx ∩ Gy = φ, ∀x, y ∈ X. Let π be the canonical mapping defined by π : X −→ G \ X x 7−→ Gx ..

(6) 4. YING HUNG CHENG. Now, we induce a topology on G \ X by defining that U ⊆ G \ X is open ⇔ the inverse image π −1 (U ) of U is open in X. Definition 2.3. The topological space G \ X is called the quotient space of X by G if its topology is defined as above. a b Let α = ∈ Γ(1) be non-scalar element and z ∈ C ∪ {∞}. We consider c d . If c 6= 0, then the mapping φα : C ∪ {∞} −→ C ∪ {∞} defined by φα (z) = az+b cz+d φα (∞) =. a c. and φα (− dc ) = ∞. Otherwise, φα (∞) = ∞. For the point ∞, we consider. φα ( z1 ) at z = 0. φα is complex analytic from C ∪ {∞} to C ∪ {∞}. Now, we want to know the fixed points of φα on C ∪ {∞}. First, suppose c = 0. Since α ∈ Γ(1), it means a = d = ±1. Hence, if z is a fixed point of φα , then φα (z) = z + b = z. Suppose z 6= ∞. We have b = 0. However, it contradicts with that α is not scalar. Therefore, ∞ is the unique fixed point of φα . Now, suppose c 6= 0. If z is a fixed point of φα , then φα (z) =. az+b cz+d. = z. This implies. that cz 2 + (d − a)z + b = 0. Hence, it suffices to consider the roots of above equation. If (d − a)2 − 4bc > 0, then the equation has two distinct real roots. It says that φα has two distinct fixed points on R. If (d − a)2 − 4bc = 0, then the equation has a unique root z =. −(d−a) 2c. ∈ Q. Hence,. φα has a unique fixed point on Q. If (d − a)2 − 4bc < 0, then the equation has two roots z ∈ H and its conjugate complex number z¯. This implies φα has fixed points z ∈ H and z¯. Therefore, by above discussion, we give the following definition. Definition 2.4. Let α be in Γ. Then (1) α is elliptic if α has the fixed points z0 and z¯0 with z0 ∈ H; (2) α is parabolic if α has a unique fixed point on Q ∪ {∞}; (3) α is hyperbolic if α has two distinct fixed points on R. Moreover, let q be in Q ∪ {∞}. Then q is call a cusp of Γ if there is γ ∈ Γ such that γ is parabolic and q is its the fixed point..

(7) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). 5. Example 2.5. 1 1 ∈ Γ(1). We consider the case of Γ = Γ(1). Suppose q = ∞. We take T = 0 1 By above discussion for c = 0, we know that T is parabolic and ∞ is a cusp of Γ(1). . q = m ∈ Q with gcd(m, n) = 1. There are h, k ∈ Z such that γ = n Given m k ∈ Γ(1) and γ · ∞ = m . Now, we want to construct a parabolic element n n h with the fixed point m . Hence, we consider γT γ −1 . Then n m m m k 1 1 h −k −1 = · (γT γ ) · n h 0 1 −n m n n m hn −b m k 1 1 = · n h 0 1 −n m +m n m k 1 1 = ·∞ n h 0 1 m k = · ∞ (Since T · ∞ = ∞) n h m = . n It implies. m n. is a fixed point of γT γ −1 . Suppose p to be the another fixed point. Then (γT γ −1 ) · p = p ⇒ T · (γ −1 · p) = γ −1 · p ⇒ γ −1 · p is a fixed point of T.. Since T is a parabolic and has a fixed point ∞, γ −1 ·p = ∞. This implies p = γ·∞ =. m . n. Therefore, q is a cusp of Γ(1). It says that Q ∪ {∞} = {cusps of Γ(1)}. Moreover, given q ∈ Q, there is a γ ∈ Γ(1) such that γ · ∞ = q. Given Γ ⊆ Γ(1), we define the set H∗ to be H ∪ {cusps of Γ}. It is well-known that Γ \ H∗ is a compact Riemann surface.(See [1], p26-31.) Now, we want to know the meromorphic functions on Γ \ H∗ . It means that f is meromorphic on H and invariant under Γ. Definition 2.6. A function f on H is called a modular function for Γ if f satisfies the following conditions: (1) f (γ · z) = f (z), ∀ γ ∈ Γ; (2) f is meromorphic on H; (3) f is meromorphic at the cusp of Γ..

(8) 6. YING HUNG CHENG. Now, we give the sense of condition (3). Since the index of Γ(1)/Γ is finite, there 1 a1 1 a2 are a1 6= a2 ∈ Z such that and are in the same coset of Γ(1)/Γ. 0 1 0 1 1 m It means there is ∈ Γ, for some m ∈ Z. Moreover, ∞ is a cusp of Γ. 0 1 1 m 1 k Let h = min{m ∈ N| ∈ Γ}. We claim that if ∈ Γ, k ∈ Z, 0 1 0 1 then h|k. If not, then there is n ∈ Z such that nh < k < (n + 1)h. It implies −n 1 k 1 h 1 k − nh 0 < k − nh < h. Since we have · = ∈ Γ, it 0 1 0 1 0 1 contradicts withthe assumption of h. Therefore, we get the claim. Because f (z) is 1 h invariant under (i.e. f (z + h) = f (z)), f (z) can be express as a function 0 1 2πiz f ∗ (q), where q = e h . f ∗ (q) is defined on a disk D(0, ), except for 0. For the cusp ∞, the condition (3) means f ∗ (q) is meromorphic at q = 0. Let a 6= ∞ be a cusp of Γ. By example 2.5, there is σ ∈ Γ(1) such that σ(∞) = a. In this case, we consider f (σz). Since f (σ(σ −1 Γσ · z)) = f (Γ(σz)) = f (σz), f (σz) is invariant under σ −1 Γσ. We have σ −1 Γσ is a subgroup of Γ(1) and the index of Γ(1)/σ −1 Γσ is equal to the index Γ(1)/Γ. As above, we can consider that f (σz) is meromorphic at ∞ for σ −1 Γσ. For the cusp a, the condition (3) means f (σz) is meromorphic at ∞ for σ −1 Γσ. X. Let Ek be the Eisenstein series with Ek (z) =. m,n∈Z (m,n)6=(0,0). 1 , ∀ k > 1. It is (mz + n)2k. well-known Ek is holomorphic on H, ∀ k > 1. (See [1], lemma 3.7 and proposition 3.10.) Let γ =. a b c d. Ek (γz) =. ∈ Γ(1). Then. m,n∈Z (m,n)6=(0,0). =. 1. X. X m,n∈Z (m,n)6=(0,0). (m az+b cz+d. + n)2k. (cz + d)2k [m(az + b) + n(cz + d)]2k. = (cz + d)2k. X m,n∈Z (m,n)6=(0,0). = (cz + d)2k. 1 [(ma + nc)z + (mb + nd)]2k. X. 1 (Since gcd(a, c) = gcd(b, d) = 1) (m z + n0 )2k 0. 0 0 m ,n ∈Z 0 0 (m ,n )6=(0,0). = (cz + d)2k Ek (z).. · · · (∗).

(9) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). 7. Put ∆ = g23 − 27g32 , where g2 = 60E2 and g3 = 140E3 . We have ∆(γz) = (60E2 (γz))3 − 27(140E3 (γz))2 = (cz + d)12 [(60E2 (z))3 − 27(140E3 (z))2 ] (by (∗)) = (cz + d)12 ∆(z). · · · (∗∗) Lemma 2.7. ∆ has a simple zero at ∞ and no zero in H. Proof. Since ∆(z + 1) = ∆(z), we know ∆(z) can be expressed as ∆(q), where q = ∞ Y 2πiz 12 e . By [1] theorem 4.21, we have ∆(q) = (2π) q (1 − q n )24 . We know that ∆(z) n=1. is meromorphic at i∞ means ∆(q) is meromorphic at q = 0. And the q-expansion of ∆ is (2π)12 q(1 − 24q + 252q 2 − · · · ). Hence ∆(q) has a simple zero at q = 0. Therefore, ∆(z) has a simple zero at ∞. By [1] proposition 3.11, we have ∆ is no zero in H. Let j(z) =. 1728g23 . ∆. . By (∗) and (∗∗), we have j(γz) = j(z), ∀γ ∈ Γ(1). By lemma 2.7,. we obtain j(z) has a simple pole at ∞. Let a be a cusp of Γ(1). By example 2.5, there is a γ ∈ Γ(1) such that γ · ∞ = a. We consider j(γ −1 · z). Hence, j(z) is meromorphic at the cusps of Γ(1). Therefore, j(z) is a modular function for Γ(1). Moreover, it is a ∗ modular function for Γ. In particular, j(z) is a meromorphic function on Γ0 (N ) \ H . a b Let ∈ Γ(1). We define c d a b Na Nb N = . c d c d a b Given α = ∈ Γ0 (N ), we have Nc d. N az + N b ) N cz + d a(N z) + N b = j( ) c(N z) + d a Nb = j( · N z) c d a Nb = j(N z)l. (Since ∈ Γ(1)) c d. j(N αz) = j(. It implies j(N z) is also a modular function for Γ0 (N )..

(10) 8. YING HUNG CHENG. Theorem 2.8. Let N be in N. Then the field of modular functions for Γ0 (N ) is generated (over C) by j(z) and j(N z). The minimum polynomial F (j , Y ) ∈ C(j)[Y ] of j(N z) over C(j) has degree µ = #Γ(1)/Γ0 (N ). Moreover, F (j , Y ) is a polynomial in j and has coefficients in Z, i.e., F (X, Y ) ∈ Z[X, Y ]. For N > 1, F (X, Y ) is symmetric in X and Y . For N = p a prime, F (X, Y ) ≡ X p+1 + Y p+1 − X p Y p − XY (mod p). Proof. See [1], Theorem 6.1.. . Let {γ1 , · · · , γµ } be a representatives for Γ(1)/Γ0 (N ), and F (j(z), Y ) ∈ C(j)[Y ] be the minimum polynomial of j(N z). It says that F (j(z), j(N z)) = 0. We use γi z to replace z, where i ∈ {1, · · · , µ}. Since j(z) is invariant on Γ(1), we have F (j(γi z), j(N γi z)) = F (j(z), j(N γi z)) = 0. Since j(N γi z) 6= j(N γk z), ∀ i 6= k (see [1], p83), by theorem 2.8, we know that {j(N γ1 z),· · · ,j(N γµ z)} is the set of roots of F (j(z), Y ) and µ Y F (j(z), Y ) = (Y − j(N γi z)). i=1. Therefore, if we want to know the modular equation F (j(z), Y ) for Γ0 (N ), then a representatives for Γ(1)/Γ0 (N ) plays an important role..

(11) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). 9. 3. The index of Γ(1)/Γ0 (N ) Let R be a commutative ring with 1, and let a b M2 (R) = { |a, b, c, d ∈ R}. c d Then M2 (R) forms a ring by the usual matrix addition and multiplication, and 1 0 is the identity of the multiplication. We denote GL2 (R) to be the set 0 1 of all units in M2 (R) and hence it forms a group with matrix multiplication. Lemma 3.1. GL2 (R) = {α ∈ M2 (R)|det(α) is a unit of R}. Proof. Consider the following mapping: ψ : M2 (R) −→ R α. 7−→ det(α).. Let α, β ∈ M2(R). Since det(αβ)=det(α)det(β), ψ(αβ) = ψ(α)ψ(β). Given a ∈ R, a 0 we take αa = and hence ψ(αa ) = det(αa ) = a. Therefore, ψ is surjective. 0 1 Given γ ∈ GL2 (R), we have γ −1 ∈ GL2 (R). Then 1 0 −1 ψ(γγ ) = ψ( )=1 0 1 and ψ(γγ −1 ) = ψ(γ)ψ(γ −1 ). . a b c d. . This implies ψ(γ) is a unit of R. On the other hand, given α = ∈ M2 (R) −1 e d −e−1 b with det(α) = e being a unit of R, we take β = ∈ M2 (R) and −e−1 c e−1 a −1 a b e d −e−1 b α·β = · c d −e−1 c e−1 a −1 e (ad − bc) 0 = (R is commutative) 0 e−1 (ad − bc) 1 0 = . (det(α) = ad − bc = e) 0 1 It implies α ∈ GL2 (R). Therefore, we complete the proof. Let SL2 (R) = {α ∈ M2 (R)| det(α) = 1}.. .

(12) 10. YING HUNG CHENG. Given α, β ∈ SL2 (R), we have det(αβ)=det(α)det(β)=1. It says that αβ ∈ SL2 (R). Since β ∈ GL2 (R), β −1 exists. Because det(β)det(β −1 )=det(ββ −1 )=1, it implies det(β −1 )=1 and hence β −1 ∈ SL2 (R). Therefore, SL2 (R) is a subgroup of GL2 (R). Recall that Γ(1) = SL2 (Z) = {α ∈ M2 (Z)|det(α) = 1}, a b Γ(N ) = { ∈ SL2 (Z)| a ≡ d ≡ 1 and c ≡ b ≡ 0 (mod N )}, c d a b Γ0 (N ) = { ∈ SL2 (Z)| c ≡ 0 (mod N )}. c d Lemma 3.2. Γ(1)/Γ(N ) ∼ = SL2 (Z/N Z). Moreover, #Γ0 (N )/Γ(N ) = N · ϕ(N ), where ϕ is the Eular’s function.. Proof. We consider the following mapping:. η . : Γ(1) −→ SL2 (Z/N Z) a b a ¯ ¯b 7 → − . c d c¯ d¯. Let α ∈ Γ(1). Then det(α) = 1. This implies det(η(α)) = ¯1 in Z/N Z and hence η(α) ∈ SL2 (Z/N Z). This says that η is well-defined. Let α, β ∈ Γ(1). It is easy to check η(αβ) = η(α)η(β) by using Z −→ Z/N Z (z 7−→ z¯) is ringhomomorphism. Therefore, ηis a group homomorphism. ¯ ¯1 ¯0 a b a ¯ b Suppose α = ∈ Ker(η). It means = ¯ ¯ . Hence, we have c d 0 1 c¯ d¯ α ∈ Γ(N ). By the definition of Γ(N ), we have Γ(N ) ⊆ Ker(η). Therefore, Ker(η) = Γ(N ). ¯b a ¯ Suppose A¯ = ∈ SL2 (Z/N Z). There is m ∈ Z such that ad−bc−mN = 1. c¯ d¯ Hence, we have gcd(c, d, N ) = 1. Now we want to find h ∈ Z such that gcd(c, d + k ` Y Y s ri hN ) = 1. Let c = pi qj j be the prime-power fractorisation of c. We . i=1 gcd(pi ,N )=1. j=1 qj |N.

(13) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). 11. consider the following system of modulo equations: d + xN ≡ 1 (mod p1 ) .. . d + xN ≡ 1 (mod pk ) x ≡ 0 (mod q1 ) .. . x ≡ 0 (mod q` ). For every pi , since (pi , N ) = 1, the modulo equation for pi has solutions in Z. Hence by Chinese remainder theorem, there is a common solution h in Z satisfying above system. Now, we claim gcd(c, d+hN ) = 1. Suppose that p is a prime with p| gcd(c, d+ hN ). Hence p|c and p|d + hN . Suppose p|N , we have p|d which contradicts with gcd(c, d, N ) = 1. Hence p doesn’t divide N and p = pi for some 1 ≤ i ≤ k. It implies p doesn’t divide d + hN by the choice of h. It is contradiction with the assumption of p. Hence gcd(c, d + nN ) = 1. At this time, we want to claim η is surjective. Since gcd(c, d + hN ) = 1, there are e, f ∈ Z such that e(d + hN ) − cf = −(m + ah). We consider the following matrix A=. a + eN b + f N c d + hN. .. Hence, det(A) = (a + eN )(d + hN ) − c(b + f N ) = ad − cb + N (ah + ed + ehN − cf ) = 1 + mN + N (ah + ed + ehN − cf ) = 1 + N [m + ah + e(d + hN ) − cf ] = 1. ¯ Therefore, η is surjective. It implies SL2 (Z)/Γ(N ) ∼ This says that η(A) = A. = SL2 (Z/N Z). By the definition of Γ0 (N ), we have a ¯ ¯b η(Γ0 (N )) = { ¯ ¯ ∈ SL2 (Z/N Z)|ad ≡ 1 (mod N )}. 0 d.

(14) 12. YING HUNG CHENG. Since Γ(N ) ⊆ Γ0 (N ) and Ker(η) = Γ(N ), we know a ¯ ¯b #Γ0 (N )/Γ(N ) = #{ ¯ ¯ ∈ SL2 (Z/N Z)|ad ≡ 1 (mod N )}. 0 d In Z/N Z, there are N elements which can put in the site of ¯b. Since ad ≡ 1 (mod N ), a ¯ and d¯ are units of Z/N Z and d¯ = a ¯−1 . There are ϕ(N ) units in Z/N Z. It implies the number of a ¯ is ϕ(N ). Therefore, #Γ0 (N )/Γ(N ) = N · ϕ(N ). Therefore, by lemma 3.2, Γ(N ) is normal subgroup of Γ(1) and #Γ(1)/Γ(N ) = #SL2 (Z/N Z) < #M2 (Z/N Z) = N 4 is finite. Lemma 3.3. Let N =. k Y. pri i , where pi is prime, ∀ 1 ≤ i ≤ k and pi 6= pj , ∀ i 6= j.. i=1. Then GL2 (Z/N Z) ∼ =. k Y. GL2 (Z/pri i Z). i=1. SL2 (Z/N Z) ∼ =. k Y. SL2 (Z/pri i Z).. i=1. Proof. We consider the following mapping: µ : M2 (Z/N Z) −→. k Y. M2 (Z/pri i Z). i=1. . a b c d. . a b a b r1 (mod prkk )). (mod N ) − 7 → ( (mod p1 ) , · · · , c d c d. a¯1 b¯1 a¯2 b¯2 Suppose = ∈ M2 (Z/N Z). It implies a1 ≡ a2 (mod N ), c¯1 d¯1 c¯2 d¯2 b1 ≡ b2 (mod N ), c1 ≡ c2 (mod N ), d1 ≡ d2 (mod N ) and hence a1 ≡ a2 (mod pri i ), . b1 ≡ b2 (mod pri i ), c1 ≡ c2 (mod pri i ), d1 ≡ d2 (mod pri i ), ∀ i ∈ {1, · · · , k}. It says that µ is well-defined..

(15) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). 13. . Q a1 b 1 ak b k r1 Let γ = ( (mod p1 ), · · · , (mod prkk )) ∈ ki=1 M2 (Z/pri i Z). c1 d1 ck dk We consider the following system of modulo equations: x ≡ a1 (mod pr11 ) .. . x ≡ ak (mod prkk ). Since p1 , · · · , pk are distinct prime numbers, by Chinese remainder theorem, there is a common solution a in Z satisfying above system and 0 ≤ a < N . Similarly, we replace ai by bi , ci , and di and hence we can get b, c, d, respectively. Suppose α = a b . We have µ(α) = γ. This implies µ is surjective. It is easy to check µ is a c d ring homomorphism. Now, we consider the set a ¯ ¯b Ker(µ) = {α = ∈ M2 (Z/N Z)| a ≡ b ≡ c ≡ d ≡ 0 (mod pri i ), ∀1 ≤ i ≤ k} c¯ d¯ a ¯ ¯b = {α = ∈ M2 (Z/N Z)| a ≡ b ≡ c ≡ d ≡ 0 (mod N )} (By pi 6= pj ). c¯ d¯ . This says that µ is one-to-one. Therefore, M2 (Z/N Z) ∼ =. k Y. M2 (Z/pri i Z) as a ring.. i=1. Moreover, GL2 (Z/N Z) and. k Y. GL2 (Z/pri i Z) are the units set of M2 (Z/N Z) and. i=1 k Y. M2 (Z/pri i Z),. respectively. Hence, by µ, we get GL2 (Z/N Z) ∼ =. k Y GL2 (Z/pri i Z). i=1. i=1. k Y Now, we claim that the image of SL2 (Z/N Z) under µ is SL2 (Z/pri i Z). Let α ∈ i=1. SL2 (Z/N Z). We have det(α) ≡ 1 (mod N ). It implies det(α) ≡ 1 (mod pri i ), ∀ 1 ≤ k k Y Y ri i ≤ k. Hence µ(α) ∈ SL2 (Z/pi Z). On the other hand, let δ ∈ SL2 (Z/pri i Z). i=1. i=1. Since µ is surjective, there is β ∈ M2 (Z/N Z) such that µ(β) = δ. Hence det(β) ≡ k Y ri 1 (mod pi ), ∀ 1 ≤ i ≤ k. Since pi 6= pj , ∀ i 6= j, we have det(β) ≡ 1 (mod pri i ). i=1. Therefore, β ∈ SL2 (Z/N Z) and we get the claim. By µ, we have SL2 (Z/N Z) ∼ = k Y SL2 (Z/pri i Z). i=1.

(16) 14. YING HUNG CHENG. By lemma 3.3 , we reduce the case N = pr for some prime p first. Since we find that it is easy to get #GL2 (Z/pr Z), we want to know the relation between #SL2 (Z/pr Z) and #GL2 (Z/pr Z). Lemma 3.4. #GL2 (Z/pr Z) = ϕ(pr ) · #SL2 (Z/pr Z), where ϕ is the Eular’s function. Proof. Let U = {¯ a ∈ Z/pr Z| a ¯ is a unit of Z/pr Z}. We consider the mapping ψ : GL2 (Z/pr Z) −→ U α 7−→ det(α). In the proof of lemma 3.1, we know that ψ is well-defined, group homomorphism, and surjective. Then Ker(ψ) = {α ∈ GL2 (Z/pr Z)| det(α) = ¯1} = SL2 (Z/pr Z). It says that GL2 (Z/pr Z)/SL2 (Z/pr Z) ∼ ¯ is a = U . Since gcd(a, pr ) = 1 if and only if a unit of Z/pr Z, we have #U = ϕ(pr ). Hence #GL2 (Z/pr Z) = #SL2 (Z/pr Z) · #U = ϕ(pr ) · #SL2 (Z/pr Z). Lemma 3.5. #GL2 (Z/pr Z) = (pr−1 )4 (p2 − 1)(p2 − p). Proof. We consider the following mapping :. α=. σ : GL2 (Z/pr Z) −→ GL2 (Z/pZ) a b a b r (mod p ) 7−→ (mod p). c d c d. Since gcd(det(ad − bc), pr ) = 1, it means gcd(det(ad − bc), p) = 1. Hence σ(α) ∈ GL2 (Z/pZ) and σ is well-defined. It is easy to check σ is homomorphism of the multiplicative group and surjective. a ¯ ¯b Ker(σ) = { ∈ GL2 (Z/pr Z)| a ≡ d ≡ 1(mod p) and c ≡ b ≡ 0(mod p)}. c¯ d¯.

(17) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). 15. Suppose a = pk + 1, b = ps, c = pl, d = ph + 1, k, h, p, l ∈ Z. Then ad − bc = (pk + 1)(ph + 1) − (ps)(pl) = p2 (kh − sl) + p(k + h) + 1 = p[p(kh − sl) + k + h] + 1. a ¯ ¯b It implies gcd(ad − bc, p) = 1 and hence gcd(ad − bc, p ) = 1. Therefore, ∈ c¯ d¯ Ker(σ) and #Ker(σ) is equal to the product of the number of a, b, c, d in Z/pr Z. Then r. . 1 −1 ≤ k < pr−1 − p p r−1 ⇒ 0≤k≤p −1. 0 ≤ a = pk + 1 < pr ⇒. and 0 ≤ b = ps < pr ⇒ 0 ≤ s < pr−1 . Similarly, we have 0 ≤ h ≤ pr−1 − 1 and 0 ≤ l < pr−1 . Therefore, #Ker(σ) = (pr−1 )4 . Now, we want to compute #GL2 (Z/pZ). Since Z/pZ is a field, a b GL2 (Z/pZ) = {α = ∈ M2 (Z/pZ)| det(α) is a unit of Z/pZ} c d a b = {α = ∈ M2 (Z/pZ)| det(α) 6= 0}. c d This says that(a,b) and (c,d) are linear independent vectors in Z/pZ×Z/pZ over Z/pZ a b if and only if ∈ GL2 (Z/pZ). Let (a, b) be a nonzero vector in Z/pZ×Z/pZ. c d Since the space which spanned by the vector (a,b) has p elements in Z/pZ × Z/pZ, a vector (c, d) which is linear independent with (a, b) doesn’t lie in the space spanned by (a, b) and hence there are p2 − p vectors in Z/pZ × Z/pZ satisfying the condition. For nonzero vector (a, b), there are p2 − 1 vectors in Z/pZ × Z/pZ. Therefore, #GL2 (Z/pZ) = (p2 − 1) · (p2 − p). Since GL2 (Z/pr Z)/Ker(σ) ∼ = GL2 (Z/pZ), #GL2 (Z/pr Z) = #Ker(σ) · #GL2 (Z/pZ) = (pr−1 )4 · (p2 − 1) · (p2 − p). .

(18) 16. YING HUNG CHENG. Corollary 3.6. #SL2 (Z/pr Z) = (pr )3 · (1 −. 1 ). p2. Proof. By lemma 3.4, we have #SL2 (Z/pr Z) = (ϕ(pr ))−1 · #GL2 (Z/pr Z) 1 = p−r · (1 − )−1 · (pr−1 )4 · (p2 − 1) · (p2 − p) (by lemma 3.5) p r−1 3 = (p ) · p · (p2 − 1) 1 = (pr )3 · (1 − 2 ). p Corollary 3.7. Let N =. k Y. pri i , where pi is prime, ∀ 1 ≤ i ≤ k and pi 6= pj , ∀ i 6= j.. i=1. Then k Y 1 #SL2 (Z/N Z) = N (1 − 2 ). pi i=1 3. Proof. By lemma 3.3, we have #SL2 (Z/N Z) =. k Y. #SL2 (Z/pri i Z). i=1 k Y 1 = (pri i )3 (1 − 2 ) (by corollary 3.6) pi i=1. =. k k Y Y 1 (pri i )3 · (1 − 2 ) pi i=1 i=1. k Y 1 = N (1 − 2 ). pi i=1 3. Theorem 3.8. Let N =. k Y. pri i , where pi is prime, ∀ 1 ≤ i ≤ k and pi 6= pj , ∀ i 6= j.. i=1. Then k Y 1 #Γ(1)/Γ0 (N ) = N (1 + ). pi i=1.

(19) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). 17. Proof. Since #Γ(1)/Γ(N ) = #Γ(1)/Γ0 (N ) · #Γ0 (N )/Γ(N ), #Γ(1)/Γ(N ) #Γ(1)/Γ0 (N ) = #Γ0 (N )/Γ(N ) k Y 1 N 3 (1 − 2 ) pi i=1 (by lemma 3.2 and corollary 3.7) = N ϕ(N ) k k Y Y 1 1 1 = N (1 − 2 )/(1 − ) (Since ϕ(N ) = N (1 − )) pi pi pi i=1 i=1 k Y 1 = N (1 + ). pi i=1. .

(20) 18. YING HUNG CHENG. 4. The representatives for Γ(1)/Γ0 (N ) In this section, we want to find a representatives for Γ(1)/Γ0 (N ). First, we want to know the relation for the elements in the same coset of Γ(1)/Γ0 (N ). . a1 b 1 c d. Lemma 4.1. Let. . and. a2 b2 c d. be in Γ(1). Then they are in the same. coset of Γ(1)/Γ0 (N ). . a b d −b d −b −1 Proof. Given α = ∈ Γ(1), we have α = det(α) = . c d −c a −c a a1 b 1 a2 b 2 We want to claim and are in the same coset of Γ(1)/Γ0 (N ). c d c d −1 a1 b 1 a2 b 2 It suffices to show ∈ Γ0 (N ). In fact, c d c d . a1 b 2 c d. . a2 b 2 c d. −1. a1 b 1 c d. . a1 d − b 1 c b 1 a2 − a1 b 2 0 da2 − cb2. = =. . . . d −b2 −c a2. .. Since N | 0, we get the claim.. . By lemma 4.1, we only have to consider thevalues of c and d. Therefore, we use a b pair (c, d) to replace the coset Γ0 (N ) of Γ(1)/Γ0 (N ). However, different c d pair (c, d) maybe lie in the same coset of Γ(1)/Γ0 (N ). . a1 b 1 a2 b 2 Lemma 4.2. Let α = and β = be in Γ(1) with d1 , d2 ∈ N c1 d1 c2 d2 and d1 |N , d2 |N . Then α, β are in the same coset of Γ(1)/Γ0 (N ) if and only if d1 = d2 and c1 ≡ c2 (mod. N ). d1. Proof. (⇒) By assumption, there is a γ = . a1 b 2 c1 d1. . = =. a b c d. a b c d. . aa2 + bc2 ca2 + dc2. ∈ Γ0 (N ) such that. a2 b2 c2 d2. . · · · (1) ab2 + bd2 . cb2 + dd2. We have d1 = cb2 + dd2 . Then dd2 = −cb2 + d1 and hence d2 | − cb2 + d1 . Since N |c and d2 |N , we have d2 |d1 ..

(21) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). 19. On the other hand, by (1), we have a2 b 2 d −b a1 b1 = c2 d2 −c a c1 d1 da1 − bc1 db1 − bd1 = . ac1 − ca1 ad1 − cb1 It implies d2 = ad1 − cb1 and hence ad1 = cb1 + d2 . Then d1 |cb1 + d2 . Since N |c and d1 |N , we have d1 |d2 . Therefore, d1 = d2 . Since α and β are in the same coset, it implies a1 b 1 d2 −b2 −1 αβ = c1 d1 −c2 a2 a1 d2 − b1 c2 a2 b1 − a1 b2 = ∈ Γ0 (N ). c1 d2 − d1 c2 d1 a2 − c1 b2 It says that N |c1 d2 − d1 c2 . There is k ∈ Z such that d1 (c1 − c2 ) = kN . Hence c1 = c2 + k dN1 , k ∈ Z. It implies c1 ≡ c2 (mod. N ). d1. (⇐) Suppose d1 = d2 = d and c1 = c2 + k Nd , k ∈ Z. Hence, −1 a1 b1 d −b2 a1 b 1 a2 b 2 = c1 d1 c2 d2 −c2 a2 c2 + k Nd d a1 d − b1 c2 b 1 a2 − a1 b 2 ∈ Γ0 (N ). = kN da2 − b2 c2 − b2 k Nd Therefore, α and β lie in the same coset of Γ(1)/Γ0 (N ).. . Therefore, by above conclusion, for a fixed d ∈ N with d|N , we just consider the set Ad (N ) = {(¯ c, d) ∈ Z/. N Z × N| gcd(c, d) = 1}. d. And we let A(N ) =. [. Ad (N ).. d>0,d|N. By lemma 4.2, we know A(N ) can be imbedded to Γ(1)/Γ0 (N ). Therefore, if we want to show A(N ) gives a representatives for Γ(1)/Γ0 (N ), then it suffices to show #A(N ) = #Γ(1)/Γ0 (N ). Since A(N ) =. S. d>0,d|N. Ad (N ) and Ad (N ), Ad0 (N ) are disjoint, for all d 6= d0 , we have X #A(N ) = #Ad (N ). d>0,d|N. Therefore, we compute #Ad (N ) first..

(22) 20. YING HUNG CHENG. Lemma 4.3. Let c, d ∈ N with d|N . Then there is c0 ∈ Z such that N c0 ≡ c (mod ) and gcd(c0 , d) = 1 d N if and only if gcd(c, d, d ) = 1. Proof. (⇒) Since c0 ≡ c (mod. N ), d. there is k ∈ Z such that c = c0 + k Nd . Suppose. q = gcd(c, d, Nd ). It implies q|c = c0 + k Nd , q|d, q| Nd and hence q|c0 . It says that q| gcd(c0 , d). Since gcd(c0 , d) = 1, we get gcd(c, d, Nd ) = 1. (⇐) Suppose d=. ` Y. pri i. k Y. s. qj j ,. j=1 qj | N d. i=1 (pi , N )=1 d. where pi , qj are distinct primes. We consider the following system of modulo equations: c+x. N d. ≡ 1 (mod p1 ) .. .. c+x. N ≡ 1 (mod p` ) d x ≡ 0 (mod q1 ) .. . x ≡ 0 (mod qk ).. For every pi , since gcd ( Nd , pi ) = 1, the modulo equation for pi has solutions in Z. Hence, for all modulo equations, there are solutions in Z. Because p1 , . . . , p` , q1 , . . . , qk are distinct primes, we have a common solution n in Z for the system of modulo equations, by Chinese remainder theorem. It suffices to show gcd(c + n Nd , d) = 1. If not, there is a prime p such that p| c + n Nd and p | d. Then p = pi for some 1 ≤ i ≤ ` or p = qj for some 1 ≤ j ≤ k. If p = pi , by the choice of n, it contradicts with the assumption of p. If p = qj , we have qj | c. Then qj | gcd(c, d, Nd ). It contradicts with gcd(c, d, Nd ) = 1. Therefore, we show that (¯ c, d) ∈ Ad (N ). Lemma 4.4. N , dd0 where d0 = gcd(d, Nd ) and ϕ is the Eular’s function. #Ad (N ) = ϕ(d0 ).

(23) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). Proof. By lemma 4.3, we consider c satisfying 1 ≤ c ≤. N d. 21. and gcd(c, d, Nd ) = 1. Now,. we want to claim N ) = 1 ⇔ gcd(c, d0 ) = 1. d (⇒) Suppose q = gcd(c, d0 ). Then q|c, q|d0 and hence q|d, q| Nd . It implies q| gcd(c, d, Nd ) = gcd(c, d,. 1. Therefore, gcd(c, d0 ) = 1. (⇐) Suppose p = gcd(c, d, Nd ). It implies p|c and p|d0 . Hence p| gcd(c, d0 ) = 1. Therefore, we have gcd(c, d, Nd ) = 1. By the claim, we can consider c satisfying 1 ≤ c ≤. N d. and gcd(c, d0 ) = 1. We have. c = d0 k + h, k ∈ Z, 0 ≤ h < d0 . Then gcd(c, d0 ) = 1 implies gcd(d0 , h) = 1. For such h, we have ϕ(d0 ). Since 1 ≤ c ≤ − dh0 +. 1 d0. ≤k≤. N dd0. −. h . d0. N , d 0. it says that 1 ≤ d0 k + h ≤. Since 0 ≤ h < d , we have −1 < k < #Ad (N ) = ϕ(d0 ). N . dd0. N d. and hence. Therefore, we get. N . dd0 . By lemma 4.4, we have #A(N ) =. X. ϕ(gcd(d,. d>0,d|N. N N )) . d gcd(d, Nd ) · d. Now, for convenience sake, we set a function F : N −→ N with F (N ) = #A(N ). We want to show F (N ) is a multiplicative function. Before we show F (N ) is a multiplicative, we need the next lemma. Lemma 4.5. Let a1 ,a2 ,b1 ,b2 ∈ N. If gcd(a1 , a2 ) = gcd(b1 , b2 ) = gcd(a1 , b2 ) = gcd(a2 , b1 ) = 1, then gcd(a1 a2 , b1 b2 ) = gcd(a1 , b1 ) · gcd(a2 , b2 ). Proof. Suppose d = gcd(a1 a2 ,b1 b2 ). Then d|a1 a2 and d|b1 b2 . Let Y Y s d= pri i qj j , pi |a1 , (pi , a2 )=1. qj |a2 (qj , a1 )=1. where pi , qj are distinct primes. For every pi , since gcd(a1 ,b2 ) = 1 and pri i |b1 b2 , we have Y Y s qj j | gcd(a2 , b2 ). Therepri i |b1 . This implies pri i | gcd(a1 , b1 ). Similarly, we have pi. qj. fore, d| gcd(a1 , b1 ) · gcd(a2 , b2 ). Suppose d0 = gcd(a1 , b1 ) · gcd(a2 , b2 ). By assumption, we have gcd(gcd(a1 , b1 ), gcd(a2 , b2 )) = 1. Hence, let d0 = d1 d2 , where d1 | gcd(a1 , b1 ),.

(24) 22. YING HUNG CHENG. d2 | gcd(a2 , b2 ), and gcd(d1 , d2 ) = 1. It implies d1 d2 |a1 a2 and d1 d2 |b1 b2 . Therefore, d0 | gcd(a1 a2 , b1 b2 ). Therefore, gcd(a1 a2 , b1 b2 ) = gcd(a1 , b1 ) · gcd(a2 , b2 ).. . Proposition 4.6. F (N1 N2 ) = F (N1 )F (N2 ), f or gcd(N1 , N2 ) = 1. Proof. X. F (N1 N2 ) =. N1 N2 N1 N2 )) · . · · · · · · · · · (2) d gcd(d, N1dN2 )d. ϕ(gcd(d,. d|N1 N2. Let d = d1 d2 , where di |Ni , i = 1, 2, gcd(d1 ,N2 ) = gcd(d2 ,N1 ) = 1. Then N N. (2). X. =. d1 d2 |N1 N2 = by lemma 4.5. X. 1 2 N1 N 2 d1 d2 ϕ(gcd(d1 d2 , )) · d1 d2 gcd(d1 d2 , Nd11 dN22 ). ϕ(gcd(d1 ,. d1 |N1 d2 |N2. =. X. (. N2 N1 N1 N2 ) · gcd(d2 , )) · · N1 d1 d2 gcd(d1 , d1 )d1 gcd(d2 , Nd21 )d2. ϕ(gcd(d1 ,. N1 N1 )) · )· d1 gcd(d1 , Nd11 )d1. ϕ(gcd(d2 ,. N2 N2 )) · ) d2 gcd(d2 , Nd22 )d2. d1 |N1. (. X. d2 |N2. =. F (N1 )F (N2 ). . Let N =. k Y. pri i , where pi is prime, ∀ 1 ≤ i ≤ k and pi 6= pj , ∀ i 6= j. By proposition. i=1. 4.6, we have k Y F (N ) = F ( pri i ) i=1. =. k Y. F (pri i ). (Since (pi , pj ) = 1 , ∀ i 6= j).. i=1. Therefore, we can reduce to the case N is prime power. Proposition 4.7. 1 F (pr ) = pr (1 + ), ∀ p is prime, r ∈ N. p.

(25) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). 23. Proof. F (pr ) =. X. ϕ(gcd(d,. d|pr. =. r X. ϕ(gcd(p i ,. i=0. =. r X. pr pr )) r d gcd(d, pd ) d pr pr )) r pi gcd(pi , ppi ) pi. ϕ(p min{i , r−i} ) pr−i p−min{i , r−i}. i=0 r−1 1 X r−i p = p + 1 + (1 − ) p i=1 r. = pr + 1 + pr−1 − 1 1 = pr (1 + ). p Theorem 4.8. A(N ) gives a representatives for Γ(1)/Γ0 (N ). Proof. By proposition 4.7, we have #A(N ) = F (N ) k Y 1 = pri i (1 + ) pi i=1 =. k Y i=1. = N. pri i. k Y 1 (1 + ) pi i=1. k Y 1 (1 + ) pi i=1. = #Γ(1)/Γ0 (N ). By lemma 4.2, A(N ) can be imbedded to Γ(1)/Γ0 (N ). Therefore, A(N ) gives a representatives for Γ(1)/Γ0 (N ). Example 4.9. For N = 6, we have A1 (6) = {(¯1, 1), (¯2, 1), (¯3, 1), (¯4, 1), (¯5, 1), (¯6, 1)}, A2 (6) = {(¯1, 2), (¯5, 2), (¯3, 2)}, A3 (6) = {(¯1, 3), (¯2, 3)}, A6 (6) = {(¯1, 6)}.. .

(26) 24. YING HUNG CHENG. The pair (5,2) may be ignored by Shimura, and it is easy to check the pair (5,2) doesn’t lie in other cosets. Now, we give some examples about computing the modular equation for Γ0 (N ). However, before we show them, we give some conclusions using in the next examples. Let q = e2πiz . It is well-known that j(z) has q-expansion with ∞ X j(q) = ci q i i=−1 −1. = q. + 744 + 196884q + 21493760q 2 + · · · .. By theorem 4.8, we know A(N ) gives a representatives {γ1 , · · · , γµ } for Γ(1)/Γ0 (N ), where µ = #Γ(1)/Γ0 (N ). By theorem 2.8, we have µ Y F (j(z), Y ) = (Y − j(N γi z)). i=1. And ∀0 ≤ k ≤ µ, the coefficient of Y. k. is a polynomial in Z[j(z)]. Therefore, We consider q-expansion of j(N γi z), ∀1 ≤ i ≤ µ. Let (c¯0 , d) ∈ A(N ). There is a the a b γi = ∈ {γ1 , · · · , γµ } such that c ≡ c0 (mod Nd ) and gcd(c, d) = 1. Hence, c d j(N γi z) = j(. N az+N b N az+N b N az + N b ) = e−2πi cz+d + 744 + 196884e2πi cz+d + · · · . cz + d. However, it is not easy to change e2πi. N az+N b cz+d. into the form of q = e2πiz . Since j(z). is invariant on Γ(1), we want to know whether there is γ ∈ Γ(1) such that γN γi = a0 b 0 . If such γ exists, then 0 d0 j(N γi z) = j(γN γi z) a0 z + b 0 = j( ) d0 = e2πi. a0 z+b0 d0. + 744 + 196884e2πi. a0 z+b0 d0. And e2πi. a0 z+b0 d0. b0. a0. b0. a0. = e2πi d0 e2πiz d0 = e2πi d0 q d0 .. Since det(N γi )= N , we consider the set M (N ) = {α ∈ M2 (Z)|det(α) = N }. Lemma 4.10. M (N ) =. [ a,d>0,ad=N 0≤b<d. Γ(1). a b 0 d. .. + ··· ..

(27) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). 25. Moreover, such union is disjoint.. Proof. Given γ ∈ Γ(1),a, b, d ∈ Z, we have γ. . a b 0 d. ∈ M2 (Z) and det(γ. a b 0 d. ) = ad.. a b Γ(1) ⊆ M (N ). It says that 0 d a,d>0,ad=N 0≤b<d a b On the other hand, let α = ∈ M (N ). Without loss of generality, we c d −1 0 suppose c > 0. If not, then we consider −α = α. There are p1 , q1 ∈ Z 0 −1 with 0 ≤ q1 < c such that a = cp1 + q1 . Hence q1 = a − cp1 . We take E1 = 1 −p1 ∈ Γ(1). Then 0 1 [. . E1 α =. 1 −p1 0 1. . . a b c d. a − p1 c b − dp1 = c d q1 b − dp1 = . c d. . Similarly, are p2 , q2 ∈ Z with 0 ≤ q2 < q1 such that c = q1 p2 + q2 . We take there 1 0 E2 = ∈ Γ(1). Then −p2 1 . 1 0 −p2 1. . q1 b − dp1 c d. . =. q1 b − dp1 q2 d + dp2 p1 − bp2. .. Continue 0 0 this process. Since qi > qj ≤ 0, ∀i < j, we can get En En−1 · · · E1 α = a b and a0 = gcd(a, c). Since E1 , · · · , En ∈ Γ(1), we have det(En En−1 · · · E1 α) = 0 d0 0 N = a0 d0 . If b0 > d0 or b < 0, then 0 ≤ q <d0 such that there is p, q ∈ Z with 1 −p a0 q b0 = pd0 + q. We take E = . Hence, EE1 · · · En α = . Therefore, 0 1 0 d0 [ a b M (N ) ⊆ Γ(1) . 0 d a,d>0,ad=N 0≤b<d.

(28) 26. YING HUNG CHENG. . a1 b 1 a2 b 2 Let and be in the same coset. It implies there is γ = 0 d1 0 d2 a b ∈ Γ(1) such that c d . a1 b 1 0 d1. . . a b c d. . aa2 ca2. = =. . a2 b 2 0 d2 ab2 + bd2 . cb2 + dd2. Hence, ca2 = 0. It implies c = 0. Since γ ∈ Γ(1), we have a = d = ±1. By assumption of a1 , a2 , weobtain a = d = 1. Sinceb1 = b2 + bd2 and 0 ≤ b1 , b2 < d2 , we have b = 0. a1 b 1 a2 b 2 Therefore, = . It says that it is disjoint union. 0 d1 0 d2 Let B(N ) = {. a b 0 d. ∈ M2 (Z)|a, d > 0, ad = N, 0 ≤ b < d}.. Since we know j(N γi z) 6= j(N γj z), ∀i 6= j, we get if #Γ(1)/Γ0 (N ) = #B(N ), then Y. F (j(z), Y ) =. (Y − j(. a,d>0,ad=N 0≤b<d. az + c )). d. In fact, we will claim #Γ(1)/Γ0 (N ) 6= #B(N ) in general. Given d|N, d > 0, we take Bd (N ) = {. N d. 0. b d. ∈ M2 (Z)|0 ≤ b < d}. and hence B(N ) =. [. Bd (N ).. d>0,d|N. It is easy to check #Bd (N ) = d and Bd (N ) ∩ Bd0 (N ) = φ, ∀d 6= d0 . Therefore, we have #B(N ) = #. [ d>0,d|N. Bd (N ) =. X d>0,d|N. #Bd (N ) =. X d>0,d|N. d..

(29) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). Let N =. k Y. 27. pri i , where pi is prime, ∀ 1 ≤ i ≤ k and pi 6= pj , ∀ i 6= j. We have. i=1 k Y 1 #Γ(1)/Γ0 (N ) = N (1 + ) pi i=1. =. k k Y Y (pri i + pri i −1 ) (Since N = pri i ) i=1. ≤. i=1. k Y. (pri i + pri i −1 + · · · + 1). i=1. = #B(N ). It says case, we still need a representatives {γi , · · · , γµ } to find the coset that in general ai b i Γ(1) which N γi lies in. In particular, they are equal when N = p1 · · · pn , 0 di where pi is prime, ∀i ∈ {1, · · · , n}. Therefore, for such case, we have F (j(z), Y ) =. Y. (Y − j(. ad=N 0≤b<d. az + b )). d. Example 4.11. For N=2, by above discussion , we have Y az + b F (j(z), Y ) = (Y − j( )) d ad=2 0≤b<d. z z+1 = (Y − j(2z))(Y − j( ))(Y − j( )). 2 2. · · · (∗). Now, we consider the q-expansion of j(2z), j( z2 ), j( z+1 ) as follow. 2 j(2z) = q −2 + 744 + 196884q 2 + 21493760q 4 + · · · , 1 1 z j( ) = q − 2 + 744 + 196884q 2 + 21493760q + · · · , 2 1 1 z+1 j( ) = −q − 2 + 744 − 196884q 2 + 21493760q − · · · . 2 First, we compute the coefficient of Y 2 . By (∗), we know the coefficient of Y 2 is equal to −(j(2z) + j( z2 ) + j( z+1 )). Since 2 z z+1 −(j(2z) + j( ) + j( )) = −q −2 − 2232 − 42987520q − · · · , 2 2.

(30) 28. YING HUNG CHENG. we suppose the coefficient of Y 2 is a2 j 2 (z) + a1 j(z) + a0 . Hence ∞ X. 2. a2 j (z) + a1 j(z) + a0 = a2 (. i 2. ∞ X. + (2a2 c0 +. i=−1 a1 )q −1. c i q ) + a1. i=−1 −2. = a2 q. c i q i + a0 + [a2 (c20 + 2c1 ) + a1 c0 + a0 ] + · · ·. = −q −2 − 2232 − 42987520q − · · · . By comparing the coefficients, we have a2 = −1, 2a2 c0 + a1 = 0, a2 (c20 + 2c1 ) + a1 c0 + a0 = −2232. Hence we get a1 = −2a2 c0 = (−2)(−1)(744) = 1488, a0 = −2232 − [a2 (c20 + 2c1 ) + a1 c0 ] = −2232 − [(−1) × (7442 + (2) × (196884)) + (1488) × (744)] = −16200. Therefore, the coefficient of Y 2 is −j(z)2 + 1488j(z) − 16200. Second, we consider the coefficient of Y . Since z z+1 z z+1 )) + j( )j( ) j(2z)(j( ) + j( 2 2 2 2 ∞ ∞ ∞ ∞ X X X X i i 2i i 2 = ( ci q )( c2i q ) + ( ci q )( (−1)i ci q 2 ) =. i=−1 2c0 q −2. +. i=0 (2c2 q −1. i=−1. − 1) + (2c4 +. i=−1. 3c20. − 2c1 ) + · · ·. = 1488q −2 + 42987519q −1 + 40492979352 + · · · , we suppose the coefficient of Y is b2 j(z)2 + b1 j(z) + b0 . Hence b2 j 2 (z) + b1 j(z) + b0 = b2 q −2 + (2b2 c0 + b1 )q −1 + [b2 (c20 + 2c1 ) + b1 c0 + b0 ] + · · · = 1488q −2 + 42987519q −1 + 40492979352 + · · · . By comparing the coefficients, we have b2 = 1488, 2b2 c0 + b1 = 42987519, b2 (c20 + 2c1 ) + b1 c0 + b0 = 40492979352..

(31) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). 29. Hence we get b1 = −2b2 c0 = 42987519 − 2 × 1488 × 744 = 40773375, b0 = 40492979352 − [b2 (c20 + 2c1 ) + b1 c0 ] = 40492979352 − (1488 × (7442 + 2 × 196884) + 40773375 × 744) = 8748000000. Therefore, the coefficient of Y is 1488j(z)2 + 40773375j(z) + 8748000000. Finally, we consider the coefficient of Y 0 . By theorem 2.8, F (j, Y ) is symmetry. Hence we suppose the coefficient of Y 0 is j(z)3 − 16200j(z)2 + 8748000000j(z) + c. Since z z+1 ), j(z)3 − 16200j(z)2 + 8748000000j(z) + c = −j(2z)j( )j( 2 2 we just consider the coefficient of q 0 from two sides. First, in the left hand side, the coefficient of q 0 is (3c2 + 6c0 c1 + c30 ) − 16200(2c1 + c20 ) + 8748000000c0 + c = (3 × 21493760 + 6 × 744 × 196884 + 7443 ) − 16200(2 × 196884 + 7442 ) +8748000000 × 744 + c = 1355202240 − 153463248000 + 6508512000000 + c = 6356403954240 + c. In the right hand side, the coefficient of q 0 is 2c5 − 2c0 c4 + 2c1 c3 − c22 + 2c0 c1 − c30 = 2 × 333202640600 − 2 × 744 × 20245856256 + 2 × 296884 × 8864299970 −(21493760)2 + 2 × 744 × 196884 + (744)3 = 666405281200 − 30125834108928 + 340333670586960 − 461981718937600 +292963392 − 411830784 = −151107596045760. Hence c = −151107596045760 − 6356403954240 = −157464000000000..

(32) 30. YING HUNG CHENG. Therefore, F (j(z), Y ) = Y 3 + (−j(z)2 + 1488j(z) − 16200)Y 2 +(1488j(z)2 + 40773375j(z) + 8748000000)Y +(j(z)3 − 16200j(z)2 + 8748000000j(z) − 157464000000000). Let µ be the #Γ(1)/Γ0 (N ). The next example will tell us about the degree of some coefficients of F (j(z), Y ) by using the representatives which we give for Γ(1)/Γ0 (N ). Example 4.12.. Let N =. k Y. pri i , where pi is prime, ∀ 1 ≤ i ≤ k and pi 6= pj , ∀ i 6= j. Recall that. i=1. A(N ) = {(¯ c, d) ∈ Z/. N Z × Z | gcd(c, d) = 1 , d| N }, d. and A(N ) gives a representatives A= {γ1 , · · · , γµ } for Γ(1)/Γ0 (N ). In particular, 1 0 N 0 (N, 1) ∈ A(N ). Hence we take γ1 = ∈ A. Since N γ1 = , it says N 1 N 1 N 0 that there is α ∈ Γ(1) such that N γ1 = α . Therefore, j(N γ1 z) = j(N z). 0 1 Since N ≥ ad , ∀a, d ∈ N, ad = N , the coefficient of Y µ−1 is equal to −j(N z) −. µ X. j(N γi z) = −q −N − k−N +1 q −N +1 − · · · .. i=2. Hence we get the coefficient of Y µ−1 is −j(z)N + aN −1 j(z)N −1 + · · · + a0 . Therefore, the degree of the coefficient of Y µ−1 is N . At this time, we want to claim if j(Nγi z) = j(hz), h ∈ N, then h = N . Let a1 b 1 (¯ c, d) ∈ A(N ). Suppose γi = ∈ A with c0 ≡ c (mod Nd ) and N γi = c0 d 0 a 0 β , a0 d0 = N, d0 |a0 , for some β ∈ Γ(1). It implies gcd(c0 , N a1 ) = a0 and d0 |d. 0 d0 Hence d0 | gcd(c0 , d) = 1. We get the claim. Now, we consider the coefficient of Y µ−2 . At first, we consider the case N = p1 · · · pk . Let ad = a0 d0 = N with d, d0 > 1. We have N N a a0 N + 0 ≤ 2 + 2 = < N. · · · (3) d d 2 2 2.

(33) A COSET REPRESENTATIVES FOR THE MODULAR GROUPS Γ(1)/Γ0 (N ). 31. Since ∀ad = N, d > 1, d doesn’t divide a and (3), the coefficient of Y µ−2 is equal to µ X. X. j(N z)j(N γi z) +. i=2. j(N γi z)j(N γt z).. i,t∈{2,··· ,µ} i6=t. = 744(µ − 1)q −N + b−N +1 q −N +1 + · · · . Hence we get the coefficient of Y µ−2 is 744(µ − 1)j(z)N + cN −1 j(z)N −1 + · · · + c0 . Therefore, the degree of the coefficient of Y µ−2 is N in this case. In other cases, let p = min{pi |p2i |N, 1 ≤ i ≤ k}. If p doesn’t exist, then it is the case N = p1 · · · pk . Hence, in this case, p exists. We know (p−1 N, 1), (2p−1 N ), · · · , ((p − 1)p−1 N, 1) are in A1 (N ). We will claim gcd(kp−1 N, N ) = p−1 N, ∀1 ≤ k ≤ p − 1. If not, there is 1 ≤ k ≤ p − 1 such that gcd(kp−1 N, N ) = hp−1 N, 1 < h ≤ k. It implies h|N . Since gcd(p, h) = 1, we have h2 |hp−1 N . Therefore, h2 |N . It contradicts with the assumption of p. We have the claim. Since (p−1 N, 1), (2p−1 N ), · · · , ((p − 1)p−1 N, 1) ∈ A1 (N ), we know Cp = {γ2 =. 1. 0 p−1 N 1. . , γ3 =. 1. 0 2p−1 N 1. . · · · , γp =. 1 0 (p − 1)p−1 N 1. }. is a subset of A. Because gcd(kp−1 N, N ) = p−1 N, ≤ k ≤p − 1, for every i ∈ ∀1 p−1 N bi {2, · · · , p}, there is αi ∈ Γ(1) such that N γi = αi , 0 ≤ bi < p. By the 0 p first claim, we know bi 6= 0. Hence, by rearrangement in Cp , we have N Cp = {N γ2 = α2. Let θp = e. 2πi p. p−1 N 1 0 p. . . p−1 N 2 0 p. . ,··· , −1 p N p−1 N γp = αp }. 0 p. , N γ3 = α3. . For every i ∈ {2, · · · , p0 }, we have. j(N γi z) = j(. p−1 N z + (i − 1) ) p. = e−2πi. p−1 N z+(i−1) p −2 N. = θp−(i−1) q −p. + 744 + 196884e2πi. p−1 N z+(i−1) p −2 N. + 744 + 196884θp(i−1) q p. ···. + ··· ..

(34) 32. YING HUNG CHENG. It is easy to check p−2 N = max{ ad ∈ N|ad = N, d > 1, d|a}. Hence by (3), we have p X. j(N z)j(N γi ) +. i=2. µ X i=p. −2 N +N ). θp−1 (1 − 1−. −(p−1) θp ) −(p−2 N +N ) q −1 θp. −2 N +N ). = −q −(p. j(N γi z)j(N γt z). i,t∈{2,··· ,µ} i6=t. = (θp−1 + θp−2 + · · · + θp−(p−1) )q −(p =. X. j(N z)j(N γi ) +. −2 N +N )+1. + d−(p−2 N +N )+1 q −(p −2 N +N )+1. + d−(p−2 N +N )+1 q −(p −2 N +N )+1. + d−(p−2 N +N )+1 q −(p. −2 N +N. Hence, the coefficient of Y µ−2 is −j(z)p. + ···. + ···. + ··· . −2 N +N −1. + hp−2 N +N −1 j(z)p. Therefore, in this case, we get the degree of the coefficient of Y. µ−2. + · · · + h0 .. −2. is p N + N .. References [1] J. S. Milne, Modular Functions and Modular Forms, University of Michigan, 1990. (http://www.math.lsa.umich.edu/ jmilne) [2] G. Shimura, Introduction to the Arithmetric Theory of Automorphic Functions, Princeton University Press, Japan, 1971 . [3] T. Miyake, Modular Forms, Springer, New York, 1976..

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