Elliptic Modular Forms

Elliptic modular forms

Hironori Shiga

Version May 10, 2008

Contents

1 SL2(Z) and elliptic curves 2

1.1 SL2(Z) and the moduli of complex tori ...... 2 1.2 The Fundamental region and a system of generators ...... 3 1.3 The Weierstrass ℘ function ...... 5 1.4 Nonsingular cubics and the invariant j ...... 9 1.5 Elliptic modular function j(τ)...... 11

2 Modular forms for SL2(Z) 14 2.1 Cusps ...... 14 2.2 Concept of modular forms ...... 15 2.3 Eisenstein series ...... 16 2.4 Discriminant form ...... 19

2.5 Eisenstein series E2(z)...... 19 2.6 Algebra M(Γ)...... 21 2.7 The Dedekind η function ...... 24

3 Modular form for congruence subgroups 26 3.1 Geometry of congruence subgroups ...... 27 3.2 Principal congruence subgroup Γ(N)...... 27 3.3 Recalling the Riemann-Roch theorem ...... 30 3.4 Dimension formula for congruence subgroups ...... 30

4 Hecke operators and Hecke eigen forms 34 4.1 Preparatory consideration ...... 34 4.2 Hecke operator T (n)...... 35 4.3 Hecke eigen form ...... 37 4.4 Examples ...... 39

5 Theta functions 39

6 An example: how it works the theory applied to number theoretic problems 40

1 1 SL2(Z) and elliptic curves

1.1 SL2(Z) and the moduli of complex tori

Deﬁnition 1.1. For ω1, ω2 ∈ C − {0} with τ = ω2/ω1 ∈/ R, we deﬁne a lattice in C by

Λ = Λ(ω1, ω2) = {mω1 + nω2 : m, n ∈ Z}.

Proposition 1.1. We have

′ ′ ⇐⇒ ∃ ∈ ′ ′ Λ(ω1, ω2) = Λ(ω1, ω2) M GL(2, Z); (ω1, ω2) = (ω1, ω2)M.

Proposition 1.2. For τ, τ ′ ∈ H = {z ∈ C : Im (z) > 0} , we have ( ) a b aτ + b Λ(1, τ ′) = kΛ(1, τ), (k ∈ C − {0}) ⇐⇒ ∃M = ∈ SL (Z); τ ′ = . c d 2 cτ + d

Deﬁnition 1.2. We deﬁne a complex torus

T (ω1, ω2) = C/Λ(ω1, ω2),T (τ) = C/Λ(1, τ).

Theorem 1.1. ( ) a b aτ + b T (τ) ∼ T (τ ′) ⇐⇒ ∃M = ∈ SL (Z); τ ′ = . biholo c d 2 cτ + d [proof]. (⇐=) is apparaent from the above Proposition. We show (=⇒). Let π(resp. π′) be the canonical projection from C to T (τ)( resp.T (τ ′)). And let f : T (τ) → T (τ ′) be a biholomorphic map. We suppose f(O) = O′, with π(0) = O, π′(0) = O′. f ′ ˜ has a lifting f1 : C → T (τ ) in a natural way. It induces a analytic function f : U → C deﬁned in a neighborhood U of 0. And f˜ has an analytic continuation on the whole plane C as a possibly multivalued analytic function. Here, by the Monodromy theorem f˜ is single valued.

f˜ C - C

π π′ ? f ? T (τ) - T (τ ′)

Diagram 1：Lifting f˜ of f

If we consider f˜−1, by the Monodromy theorem again, we can show f˜ is injective. So f˜ : C → f˜(C) is a biholomorphic map. By the Riemann uniformization theorem, the image f˜(C) cannot be a proper subdomain. So f˜ : C → C is a bijective map, and it is biholomorphic. By the Weierstrass singularity theorem we can show that Aut(H) is a group of nontrivial linear functions. Hence we may put

f˜(z) = kz,

note that we supposed f(O) = O′. So we have Λ(1, τ ′) = kΛ(1, τ). q.e.d.

2 1.2 The Fundamental region and a system of generators

Notations:

PSL2(R) = SL2(R)/ ± I = Aut(H),

Γ := SL2(Z), Γ := SL2(Z)/ ± I, ⊂ − ∩ in general,( for a) subgroup G Γ, G := G/(< I > G), a b Γ(N) := { ∈ Γ) : a ≡ d ≡ 1 (mod N ), b ≡ c ≡ 0 (mod N )} ▹ Γ(N ∈ Z+). c d

Deﬁnition 1.3. Γ(N) is called the principal congruence subgroup of level N.

Remark 1.1. Note −I/∈ Γ(N) N > 2 . So

Γ(2) = Γ(2)/ ± I, Γ(N) = Γ(N)(N > 2).

Deﬁnition 1.4. ( ) a b Γ (N) := { ∈ Γ: c ≡ 0 (mod N )}, 0 c d ( ) a b Γ (N) := { ∈ Γ (N): a ≡ d ≡ 1 (mod N )}. 1 c d 0

Deﬁnition 1.5. Let G be a subgroup of Γ. G acts on H. Two points z1 and z2 are said to be G-equivalent, if we have z2 = g(z1) for some g ∈ G. A closed region in H is said to be a fundamental region of G, if (1) Every point z ∈ H is G-equivalent to a point in F . (2) Any two diﬀerent points in the interior of F are not G-equivalent.

Theorem 1.2. The closed region 1 F = {z ∈ H : |Re z| ≤ and |z| ≥ 1} (1.1) 2 is a fundamental region for Γ.

F i

e2 Πi3 eΠi3

0 Re - 1 1 2 2

Fundamental region for Γ

[proof]. (i) F contains all representatives. Set ( ) ( ) 1 1 0 −1 T = ,S = . 0 1 1 0

3 ( ) a b G := 〈S, T 〉 ⊂ Γ. For a ﬁxed z ∈ H and g = ∈ G we have c d

Im z Im g(z) = . |cz + d|2

2 2 {|cd + d| : c, d ∈ Z − {(0, 0)}, coprime} has the minimum, so Im g(z) has the maximum. Let g0 realizes k ± ··· k0 ∈ | | ≤ 1 the maximum, so do T g0 (k = 0, 1, ). And we can ﬁnd g = T g0 G such that g(z) 2 . We have g(z) ∈ F . In fact, if we have g(z) ∈/ F , it holds |g(z)| < 1. Take S(g(z)). We have

Im g(z) Im S(g(z)) = > Im g(z). |g(z)|2

This is contradicts the maximality. So we could ﬁnd a g ∈ G such that g(z) ∈ F . (ii) Equivalent points. ( ) a b Suppose z1, z2 are equivalent, and we have z2 = g(z1), g = ∈ Γ − {±I}. Assume Im z2 = √ c d Im z1 ≥ | | · 3 ≤ | | ≤ | | | | ≤ 2 Im z1 . So we have c c Im z1 (c Re z1 + d) + ic Im z1 = cz1 + d 1. We can |cz1+d| 2 see that it must hold |c| ≤ 1. By observing |cz1| ≤ 1/2 we have |d| ≤ 1. So we have only restricted possibilities: (a) c = 0, d = ±1 =⇒ g : az + b ± ⇒ − 1 (b) c = 1, d = 0 = g : a z ± ⇒ −1 2πi/3 (c) c = d = 1 = g : a + z+1 and z1 = e − ± ⇒ −1 πi/3 (d) c = d = 1 = g : a + z−1 and z1 = e . In any case z1 and z2 cannot stay inside F at the same time.

-1 -1 F2 F z + 1 z - 1 1

i i e2 Πi3 eΠi3 ¬ ®

- 1 1 - 1 1 2 0 2 2 0 2

Figures of case (c)(d)

q.e.d.

Proposition 1.3. Let z1, z2 ∈ ∂F . z1 and z2 are Γ-equivalent ⇐⇒

(1) z2 − z1 = ±1, |Re z1| = |Re z2| = 1/2 or 1 (2) z2 = − , |z1| = |z2| = 1. z1

We use the notation Gz = {g ∈ G : g(z) = z}, the isotropy group for z. Set ( ) ( ) 1 1 0 −1 T = ,S = . 0 1 1 0

4 Proposition 1.4. We have

(i) Γi = {±I, ±S} 2 2πi/3 (ii) Γω = {±I, ±ST, ±(ST ) }, ω = e 2 2πi/6 (iii) Γ−ω = {±I, ±T S, ±(TS) }, −ω = e

(iv) Γz = ±I, otherwise.

[proof]. We can solve

cz2 + (d − a)z − b = 0, ad − bc = 1 (a, b, c, d ∈ Z)

for z = i, ω, −ω2.

Corollary 1.1. Γ acts on H as a discrete group.

Theorem 1.3. We have Γ = 〈S, T. ± id〉.

[proof]. Take an element γ ∈ Γ. According to the part (i) of the proof of Theorem 1.2 we can ﬁnd g ∈ G =

〈S, T 〉 so that we have gγ(2i) ∈ F . Namely it holds gγ(2i) = 2i. So gγ ∈ I2i = {±id}. Hence we obtain the required conclusion. q.e.d.

1.3 The Weierstrass ℘ function

Deﬁnition 1.6. Let Λ = Λ(ω1, ω2) be a lattice in C. The Weierstrass ℘ function is deﬁned by ( ) 1 ∑ 1 1 ℘(z) = + − . (1.2) z2 (z − ω)2 ω2 ω∈Λ−{0}

Remark 1.2. For a lattice Λ,

′ ′ ∑ 1 ∑ ∑ ( means the sum ) ωk ω∈Λ−{0}

is absolute convergent for k ≥ 3, and is conditional convergent for k = 2.

Theorem 1.4. (i) ℘(z) is meromorphic on C and doubly periodic, i.e.

℘(z + ω) = ℘(z)(ω ∈ Λ).

(ii) ℘(z) has double poles at z = ω ∈ Λ, and is an even function of order 2. ′ (iii) ℘ is an odd function of order 3. It has zeros at half lattice points z = ω1/2, ω2/2, (ω1 + ω2)/2, and has a triple poles at z = ω ∈ Λ.

′ ′ [proof]. (i) We have ℘ (z + ωi) = ℘ (z)(i = 1, 2). So ℘(z + ωi) − ℘(z) = ci (c : constant). By putting z = −ωi/2 we get ci = 0. (ii)(iii) follows from the following general argument.

Lemma 1.1. (i) Let f be a doubly periodic meromorphic function (≠ 0) for a lattice Λ. f takes every

complex value α same times (counting multiplicities) in a period parallelogram Pa = {λω1 + µω2 + a : 0 <

λ < 1, 0 < µ < 1} provided f ≠ α, ∞ on ∂Pa.

5 (ii) Let f be a doubly periodic meromorphic function. a1, . . . , ar be its representatives of zeros, and

b1, . . . , br be the representatives of poles. Then we have

a1 + ... + ar − (b1 + ... + br) ∈ Λ.

[proof]. (i) By the argument principle ∫ ′ 1 f { ∈ } − { ∈ ∞} − dz = ♯ z Pa : f(z) = α ♯ z Pa : f(z) = . 2πi ∂Pa f α By the periodicity the left hand side is equal to 0. (ii) By the residue theorem ∫ 1 zf ′ ∑ zf ′ dz = Res( , z) = a1 + ··· ar − (b1 + ··· br). 2πi ∂Pa f f z∈Pa Again by the argument principle we see that the left hand side belongs to Λ.

Theorem 1.5. (i)

′ 2 ℘ (z) = 4(℘(z) − e1)(℘(z) − e2)(℘(z) − e3), (1.3)

ω1 ω1+ω2 ω2 where e1 = ℘( 2 ), e2 = ℘( 2 ), e3 = ℘( 2 ) , (ii)

′ 2 3 ℘ (z) = 4℘(z) − g2℘(z) − g3, (1.4) ∑ ∑ 1 1 where g2 = 60 ω∈Λ−{0} ω4 , g3 = 140 ω∈Λ−{0} ω6 .

[proof]. Set ( ) ∑ 1 1 ϕ(z) = ℘(z) − 1/z2 = − , (z − ω)2 ω2 ω∈Λ−{0} and let 2 2n ϕ(z) = a0 + a2z + ... + a2nz + ... be the power series expansion. Obviously ϕ(0) = 0, so a0 = 0. And we have ′ ′ ϕ′′(0) ∑ 1 ϕ′′′′(0) ∑ 1 a2 = = 3 4 , a4 = = 5 6 . 2! (mω1 + nω2) 4! (mω1 + nω2) Now we have 1 ℘(z) = + a z2 + a z4 + ··· z2 2 4 1 ℘′(z) = −2 + 2a z + 4a z3 + ··· . z3 2 4 So we have 1 3a ℘(z)3 = + 2 + 3a + ··· z6 z2 4 4 8a ℘′(z)2 = − 2 − 16a + ··· . z6 z2 4 Then it holds

′ 2 2 2 ℘ (z) − 4℘(z) + 20a2℘(z) = −28a4 + c1z + ··· .

The left hand side is holomorphic and doubly periodic on C. So (according to the theorem of Liouville)

it is a constant . Namely the right hand side is reduced to −28a4. Then we have the required equality. q.e.d.

6 ′ Remark 1.3. In a natural way ℘(z) (resp. ℘ (z)) becomes to be a function on the torus T (ω1, ω2).

Theorem 1.6. Let g2, g3 be the numbers given in Theorem 1.5. We set the corresponding algebraic curve

2 3 E = E(g2, g3): y = 4x − g2x − g3

in P 2(C) by adding a point at inﬁnity [X,Y,Z] = [0, 1, 0] with a homogeneous coordinate [X,Y,Z] such ′ that x = X/Z, y = Y/Z. The map Φ: T (ω1, ω2) → E deﬁned by z 7→ (x, y) = (℘(z), ℘ (z)) gives a

bijective correspondence between T (ω1, ω2) and E.

proof]. Denote the point at inﬁnity by P∞. Take an arbitrary point (x, y) ∈ E − {P∞}. From

℘(z) = x we obtain at most 2 possibilities z1, z2 with z2 = −z1. These two points are distinguished by ′ −1 the values of ℘ . So we can ﬁnd unique preimage Φ ((x, y)). Still we have limz→0 ℘(z) = ∞. So we

have limz→0 Φ(z) = P∞. q.e.d.

Theorem 1.7. (Addition formula for ℘(z)) We have ( ) 1 ℘′(z) − ℘′(u) 2 ℘(u + z) = −℘(u) − ℘(z) + . (1.5) 4 ℘(z) − ℘(u)

Corollary 1.2. ( ) 1 ℘′′(z) 2 ℘(2z) = −2℘(z) + . 4 ℘′(z)

[proof of the addition formula] We deﬁne ′ {( ) ( ) } ∏ z z 1 z 2 σ(z) = z 1 − exp[ + ] (1.6) ω ω 2 ω ω and ′ ( ) 1 ∑ 1 1 z ζ(z) = + + + . (1.7) z z − ω ω ω2 ω Proposition 1.5. σ(z) (i) is holomorphic on C, (ii) has a simple zero at z ∈ Λ, (iii) is an odd function. It holds σ′(z) d log σ(z) ζ(z) = = , σ(z) dz and σ′(z)2 − σ(z)σ′′(z) ℘(z) = −ζ′(z) = . σ(z)2 [proof]. We don’t make a precise argument about the absolute convergence of (1.6). Once it is done, the above properties are obvious.

Theorem 1.8. We have

ζ(z + ω1) − ζ(z) = η1 ··· a constant(quasi period),

ζ(z + ω2) − ζ(z) = η2 ··· a constant(quasi period),

η1ω2 − η2ω1 = 2πi ··· Legendre’s formula.

7 [proof]. We can obtain these properties by the Integral around a parallelogram.

Proposition 1.6. Set η = mη1 + nη2 for ω = mω1 + nω2. It holds { exp[η(z + ω )]σ(z)(ω ∈ 2Λ) σ(z + ω) = 2 − ω ∈ exp[η(z + 2 )]σ(z)(ω / 2Λ).

[proof]. By the relation d log σ(z + ω) d log σ(z) − = η dz dz − ω we have σ(z + ω) = C exp[ηz]σ(z). Determine the constant factor by putting z = 2 .

Proposition 1.7. Let f(z) be a doublly periodic meromorphic function (≠ 0). Let a1, . . . , ar and

b1, . . . , br be representatives of its zeros and poles with a1 + ··· + ar = b1 + ··· + br. Then we have σ(z − a ) ··· σ(z − a ) f(z) = c 1 r . σ(z − b1) ··· σ(z − br)

[proof]. Use the automorphic factor of σ to show the periodicity.

Proposition 1.8.

σ(z + u)σ(z − u) ℘(z) − ℘(u) = − . σ(z)2σ(u)2

[proof]. Use the preceeding proposition for f(z) = ℘(z) − ℘(u) and observe the Laurent expansion at z = 0.

Proposition 1.9.

1 ℘′(z) − ℘′(u) ζ(z + u) = ζ(z) + ζ(u) + . (1.8) 2 ℘(z) − ℘(u)

[proof]. Make two logarithmic derivatives of the preceeding formula. [proof of the addition formula] z derivative of (1.8): ( ) 1 ℘′′(z) ℘′(z)(℘′(z) − ℘′(u)) ζ′(z + u) = ζ′(z) + − . 2 ℘(z) − ℘(u) (℘(z) − ℘(u))2 u derivative of (1.8): ( ) 1 ℘′′(u) (−℘′(u))(℘′(z) − ℘′(u)) ζ′(z + u) = ζ′(u) + − − . 2 ℘(z) − ℘(u) (℘(z) − ℘(u))2

By ℘(z) = −ζ′(z) we have

1 ℘′′(z) − ℘′′(u) 1 (℘′(z) − ℘′(u))2 −2℘(z + u) = −℘(z) − ℘(u) + − . 2 ℘(z) − ℘(u) 2 (℘(z) − ℘(u))2

′′ 2 − 1 By using ℘ (z) = 6℘ (z) 2 g2 we have 1 ℘′′(z) − ℘′′(u) = 3(℘(z) + ℘(u)). 2 ℘(z) − ℘(u)

So we obtain the theorem. q.e.d.

8 1.4 Nonsingular cubics and the invariant j

Theorem 1.9. (Hurwitz’ formula) Let R,R0 be compact Riemann surfaces. Let π : R → R0 be a

nonconstant holomorphic map ( so necessarily a surjective ﬁnite map). p1, . . . , pr be the singular points

(or ramiﬁcation points) on R, namely dπ(pi) = 0, vi be its order of zero (ramiﬁcation index) . Set −1 n = ♯π π(p)(π(p) ≠ π(pi)). Then we have

∑r 2g(R) − 2 = n(2g0 − 2) + vi, i=1 where g(resp.g0) is the genus of R(resp.R0).

[proof]. Set χ (resp. χ0) be the Euler characteristic of R (resp. R0). Make a triangulation of R0 using π(pi) as its vertices. Obtain a triangulation of R by the lifting. If we don’t have any ramiﬁcation, by counting the numbers of triangles, sides and vertices down and up stairs, we have the relation of Euler characteristics

χ = nχ0.

When we have ramiﬁcations, we loose several vertices by the pinching at the ramiﬁcation. That is counted ∑ by i vi. q.e.d.

Example 1.1. (The Fermat curve) Set a projective algebraic curve

n n n Fn : X + Y = Z

n n X Y → 1 7→ with an affine representation x + y = 1 (x = Z , y = Z ). Set π : Fn P by (x, y) x. The ramification points come from x = 1, ζ, . . . , ζn−1, (ζn = 1). And at every ramification point ζi we have

vi = n − 1. So we have

2g − 2 = n(2g0 − 2) + n(n − 1), g0 = 0,

namely 1 g(F ) = (n − 1)(n − 2). n 2

Especially a cubic Fermat curve F3 is a Riemann surface of genus 1. Note that every nonsingular cubic is diﬀeomorphic to F3. That is because a cubic curve

3 3 3 2 2 2 2 2 2 a0X + a1Y + a2Z + a3X Y + a4Y Z + a5Z X + a6XY + a7YZ + a8ZX + a9XYZ = 0 (1.9)

is generically nonsingular. (So the family of nonsingular cubic curves makes a locally trivial topological ﬁbre space over the space of parameters.)

By the above argument, we have:

Proposition 1.10. Every nonsingular cubic curve is a Riemann surface of genus 1, that is homeomorphic to a complex torus.

Remark 1.4 (Genus Formula). Let C be an algebraic curve of degree n. Then we have 1 ∑ g(C˜) = (n − 1)(n − 2) − δ , 2 i i where C˜ is the nonsingular model of C, g(C˜) is its genus, δi is a contribution from every singular point. For example δ = 1 for an ordinary double point.

9 [Normal forms]

For an arbitrary nonsingular cubic (1.9), we can find a nice homogeneous coordinate with the conditions: [X,Y,Z] = [0, 1, 0] is a flex with the tangent Z = 0, namely [X,Y,Z] = [0, 1, 0] is the only intersection with Z = 0. (We used here the result that a nonsingular cubic has (9) flex points on it.)

Hence we get a1 = a3 = a6 = 0. So we obtain an aﬃne equation

′2 ′ ′ ′ ′3 ′ ′2 a4y + a7y + a9x y + a0x + a2 + a5x + a7x = 0. (1.10)

By an aﬃne transformation

x = αx′ + β, y = ay′ + bx′ + c (1.11)

of (1.10) we obtain a normal form

2 3 − − 3 − 2 ̸ E = E(g2, g3): y = 4x g2x g3, with g2 27g3 = 0, (1.12) 3 − 2 ̸ here g2, g3 are just some constants. g2 27g3 = 0 is the nonsingular condition. Remark 1.5. We note that the ambiguity of the transformation (1.11) contains only the proportionality

x˜ = t2x, y˜ = t3y (t ∈ C − {0}).

Deﬁnition 1.7. For an arbitrary given nonsingular cubic, we got an expression (1.12). We call it the Weierstrass normal form. Or we have

2 y = (x − e1)(x − e2)(x − e3), with (e1 − e2)(e2 − e3)(e3 − e1) ≠ 0 we call it the Legendre normal form.

Together with P∞ , the point at inﬁnity, the pair (E,P∞) is called an elliptic curve. Two elliptic curves ′ ′ ′ (E(g2, g3),P∞) and (E(g2, g3),P∞) are said to be projectively equivalent, when there is a projective linear ′ ′ ′ transformation f such that f(E(g2, g3)) = E(g2, g3) and f(P∞) = P∞. In some cases we will write only

E or E(g2, g3) for an elliptic curve. But always we are keeping in mind this ”marking” by P∞.

Deﬁnition 1.8. For an elliptic curve (E(g2, g3),P∞) we deﬁne the j invariant: g3 j(E(g , g )) = j = 123 2 , (1.13) 2 3 3 − 2 g2 27g3 3 − 2 and we call ∆ = g2 27g3 the discriminant. ′ ′ ′ Theorem 1.10. Two elliptic curves (E(g2, g3),P∞) and (E(g2, g3),P∞) are projectively equivalent if and only if we have ′ 4 ′ 6 g2 = t g2, g3 = t g3 for some t ∈ C∗. The projective linear transformation is given by { x′ = t2x y′ = t3y, and it is equivalent with the condition

′ ′ j(E(g2, g3)) = j(E(g2, g3)).

10 [proof] The ﬁrst ”if and only” statement is a direct consequence from Remark 1.5. The second ′3 3 ′2 2 condition for the j invariant is equivalent to the proportionality g2 : g2 = g3 : g3. So it follows our assertion. q.e.d.

Remark 1.6. The value λ has more geometric meaning. λ = (e1, e2, e3, ∞) is a cross ratio of the 4

tangents from a ﬂex. There is ambiguity of the order. If we take S3 inavariant (1 − λ + λ2)3 28 , λ2(1 − λ)2

that is the j invariant. We don’t speak about the invariant theory C[X,Y ]S3 .

2 3 Deﬁnition 1.9. For an elliptic curve E : y = 4x − g2x − g3, ∆ ≠ 0, we deﬁne an addition law

−P (x, y) = P (x, −y) and P1(x1, y1) + P2(x2, y2) = P3(x3, y3) by  ( )  − 2   1 y2 y1 y −y y x −y x −x1 − x2 + (P2 ≠ ±P1) − 2 1 − 1 2 2 1 ̸ ±  4 x2−x1  − − (P2 = P1) ( ) x2 x1 x2 x1 2− 2 2 x3 = 1 6x1 g2/2 , y3 = y + (x − x )(6x − g /2) (P = P ≠ P∞) −2x1 + (P1 = P2 ≠ P∞)  1 3 1 1 2 2 1  4 y1   y2 (P1 = P∞). x2 (P1 = P∞)

y y

HA+BL*

A B x x

A+B

Figure of A + B

Remark 1.7. Here we use the fact that every elliptic curve can be obtained as Φ(T (τ)) for some complex torus T (τ) with the marking by O, we shall prove it in the next subsection. So our addition law is just a translation of the group structure of T (τ). Hence it satisﬁes the associative law (P1 + P2) + P3 =

P1 + (P2 + P3) a priori. If we don’t use this transcription, it is not easy to get it.

1.5 Elliptic modular function j(τ)

Deﬁnition 1.10. For z ∈ H, we set ′ ∑ 1 g2(z) = 60 4 , (mz + n)   ′ ′ ∑ 1 ∑ ∑ g (z) = 140 ,  =  3 (mz + n)6 (m,n)∈Z2−{(0,0)} 3 2 ∆(z) = g2(z) − 27g3(z) .

The elliptic modular function g (z)3 j(z) := 123 2 (1.14) ∆(z) is deﬁned on H.

11 For a representaion Φ(T (τ)) = E(τ):

2 3 y = 4x − g2(τ)x − g3(τ),

we get an invariant j(τ). As we made an argument, we have

∼ ′ ′ ⇐⇒ ∼ ′ ⇐⇒ ∃ ′ ⇒ ′ (T (τ),O) biholo (T (τ ),O ) (1) τ PSL2(Z) τ (2) Λ(τ) = kΛ(τ ) = (3) j(τ) = j(τ ).

By the argument in the preceeding subsection we have

′ ′ ′ ′ ′ j(τ) = j(τ ) ⇐⇒(a) (E(τ),P∞) ∼projective equiv. (E(τ ),P∞) =⇒(b) (T (τ),O) ∼biholo (T (τ ),O ).

In fact (1) is assured by Theorem 1.1, (2) is obtained in Proposition 1.2 and (3) is deduced from the deﬁnition of j(τ). On the other hand (a) is shown in Theorem 1.10 and (b) is a consequence of Theorem . So we have

Theorem 1.11. 0) j(z) is holomorphic on H, ′ ′ ′ 1) (E(τ),P∞) ∼proj.equiv. (E(τ ),P∞) ⇐⇒ j(τ) = j(τ ), ′ ′ 2) j(z) = j(z ) ⇐⇒ z = g(z) ∃g ∈ PSL2(Z), 3) j(z) gives a biholomorphic equivalence between Γ\\ H and C ∪{∞} = P 1, where ˆ· means the one point compactiﬁcation. 4) We have the Fourier expansion: 1 j(z) = + c + c q + ··· , q = e2πiz. q 0 1

5) Every elliptic curve has a representative in a projective equivalence class given by (E(τ),P∞) coming from a marked complex torus (T (τ),O).

Remark 1.8. We know that j(τ) has a simple pole at q = 0 i.e. τ = i∞. But to get the exact value of the residue, we need some more argument. We shall show it in the later section.

Proposition 1.11. (i) We have j(ω) = 0, j(i) = 1728.

(ii) For a pure imaginary z, it holds j(z) ∈ R.

[proof]. (i) We have

′ ′ ′ ∑ 1 ∑ 1 1 ∑ 1 = = − . (mi + n)6 i6 (m + (−i)n)6 (mi + n)6 m,n m,n m,n

It means g3(i) = 0. So we have j(i) = 1728. Also we have

′ ′ ′ ′ ′ ∑ 1 ∑ 1 1 ∑ 1 1 ∑ 1 ∑ 1 = = = ω = ω . (mω + n)4 ω8 (m + nω2)4 ω8 (m + nω2)4 ((m − n) − nω)4 (mω + n)4 m,n m,n m,n m,n m,n

It means g2(ω) = 0. So we have j(ω) = 0 (ii) Set z = it (t > 0). It holds

′ ′ ′ ′ ∑ 1 ∑ 1 ∑ 1 ∑ 1 = = = . (m(it) + n)4 4 (−m(it) + n)4 (m(it) + n)4 m,n m,n (m(it) + n) m,n m,n

Hence g2(it) is always real valued. It is the same for g3(it). So j(z) takes a real value for every z = it. q.e.d.

12 Remark 1.9. There is no general formula for the Fourier coeﬃcients of j(τ). But they are always integers and have deep arithmetic meaning: 1 j(τ) = + 744 + 196884 q + 21493760 q2 + 864299970 q3 + ··· . q What does 196884 means ? ∑ − 1 n Setting J(z) = j(z) 744 = q + n≥1 c(n)q

Θ(Λ , z) J(z) = L − 24, ∆(z)

where ΛL means the Leech lattice, that is an even unimodular selfdual lattice with length 24 and minimum distance 2, and ∑ ∑ n Θ(ΛL, z) = exp[π〈α, α〉z] = |Λ2n|q (Λ2n = {α ∈ ΛL : 〈α, α〉 = 2n}),

α∈ΛL n≥0

where 〈α, α〉 is the inner product of ΛL. This is the starting point of the Monster theory. And the Leech lattice has many nice properties.

Example 1.2. (The oldest example of an elliptic curve found in ” Diophantus ”Number theory”.)

E : y2 = x3 − x + 9.

We have integer points (0, ±3) on E. Find other integer points as far as possible. (See also, E. Brawn and B. Myers,“Elliptic curves from Mordell to Diophantus and back”, Mathematics Monthly, August-September 2002,639-649)

-2 -1 1 2 3 4 5

-5

-10

Figure of the Diophantus elliptic curve

13 2 Modular forms for SL2(Z) 2.1 Cusps

We obtain the quotient space Γ \ H for Γ. How to make it complete ? We set

H = H ∪ {i∞} ∪ Q.

A point of H − H is called a cusp. Γ acts on the set of cusps transitively. In other words cusps belong to only one Γ equivalence class. For a subgroup Γ′ it acts on the cusps but in general not transitively. [Topology of H.] For a positive number C set

NC = {z ∈ H : Im z > C} ∪ {i∞}.

We define the( fundamental) system of open neighborhood of i∞ by {NC : C > 0}. Let a/c ∈ Q and take a b a matrix α = ∈ Γ. Then it holds α(i∞) = a . We define {αN : C > 0} to be the fundamental c d c C a ∈ system of open neighborhood of c . Together with a usual system of open neighborhood for z H, we define a topology of H = H ∪ {i∞} ∪ Q. This topology induces a one point compactification Γ\\ H = Γ \ (F ∪ {i∞}) of Γ \ H. Under the map q(z): z 7→ q := e2πiz

we have

H ∪ {i∞} → D = {|q| < 1}, (2.1)

−2πC and NC is mapped to a disc {|q| < e }. The topology of H ∪ {i∞} is the weakest topology that makes q(z) to be continuous. Let f(z) be a holomorphic function on H with a period f(z + 1) = f(z). Then we have a Fourier expansion (that is a Laurent series in q) ∑ ∑ 2πinz n f(z) = ane = anq . (2.2) n∈Z n∈Z We say f(z) is meromorphic (resp. holomorphic, 0 )at i∞, if it has only ﬁnite negative power terms (resp. no negative term, neither negative term nor constant term). If f(z) is holomorphic on H and has a periodicity

f(z + N) = f(z)

for a ﬁxed positive integer N, we consider a map

z 7→ e2πiz/N ,

and we may have an analogous argument.

Proposition 2.1. Let Γ′ be a index ﬁnite subgroup of Γ. Set [Γ: Γ′] = n. We have a coset decomposition

⊔n ′ Γ = αiΓ i=1 ∪ { } ′ n −1 with a complete set a1, . . . , an of coset representatives. Then F = i=1 αi F is a fundamental region of Γ′.

14 ′ ∼ Example 2.1. Γ = Γ(2). Because we have SL2(Z/2Z) = S3, we have [Γ, Γ(2)] = [Γ, Γ(2)] = 6. We have the following representatives of the cosets:

( )−1 ( )−1 1 0 1 1 α = I = , α = T −1 = 1 0 1 2 0 1 ( )−1 ( )−1 0 −1 1 −1 α = S−1 = , α = (TS)−1 = 3 1 0 4 1 0 ( )−1 ( )−1 0 −1 1 0 α = (ST )−1 = , α = (TST )−1 = . 5 1 1 6 1 1

Α-1 -1 2 F2 T F1 F2 F1 TF2 F2

- - 1 1 -1 TSF1 Α3 F2 Α4 F2 T SF2 SF1 SF2 -1 -1 2 Α5 F2 Α6 F2 STSF1 STF2 HTSL F1TSTF2

0 1 0 1

Fundamental region for Γ(2)

[proof of the proposition]. Take an arbitrary point z ∈ H. Find γ ∈ Γ; γ(z) ∈ F . We have unique ∈ ′ −1 ∈ ′ ∈ −1 αi; γ αiΓ . So we have gi = αi γ Γ , and gi(z) αi F . ∈ −1 ◦ ′ ∈ −1 ◦ ′ ∃ ∈ ′ − { } ∈ ◦ ′ ∈ ◦ Suppose z αi F , z αj F with z = gz, g Γ id . We have αiz F , αjz = αjgz F , and they are Γ equivalent. So αjg(z) = αi(z). Namely

−1 ∈ ◦ αjgαi (αiz) = αiz F .

−1 According to the triviality of the isometry subgroup (see Prop 1.4), we have αjgαi = id. So we have ′ ′ αiΓ = αjΓ . This is a contradiction. ∈ −1 ′ ∈ −1 For the case z ∂(αi F ) or z ∂(αj F ), we can shift a little bit them to some interior points all together. It is possible because g′ ∈ Γ′ is a homeomorphism that sends a small neighborfood of z to that of z′. So we may reduce the argument to the above case. q.e.d.

2.2 Concept of modular forms

Deﬁnition 2.1. Let f(z) be a holomorphic function on H. And let k be a non-negative integer. f(z) is said to be a modular form of weight k for Γ if the following conditions are satisﬁed: (i) f(z) is holomorphic at z = i∞, (ii) ( ) a b f(γ(z)) = (cz + d)kf(z) for ∀γ = ∈ Γ. (2.3) c d

The C vector space of modular forms of weight k for Γ is denoted by Mk(Γ).

15 In addition if we have (iii) f(z) is 0 at z = i∞, f(z) is said to be a cusp form of weight k for Γ. The C vector space of cusp forms of weight k for Γ is denoted by Sk(Γ).

Remark 2.1. Some times we use the term ”meromorphic modular form”. It means a meromorphic function f(z) on H that satisﬁes (i’) f(z) is meromorphic at z = i∞ together with the automorphic property (ii).

Remark 2.2. By putting γ = T,S we have

f((z + 1)) = f(z), (2.4) 1 f − = (−z)kf(z) (2.5) z for a modular form f(z) of weight k.

For a modular form for Γ with any weight k has a ”q-expansion”: ∑∞ n 2πiz f(z) = anq (q = e ). (2.6) n=0

Remark 2.3. (1) Putting γ = −I we have Mk(Γ) = 0 for an odd integer k. So we consider only the case k is even. (2) We have dγ(z) 1 = . (2.7) dz (cz + d)2 So we may rewrite the condition (2.3) by

f(γ(z))(dγ(z))k/2 = f(z)(dz)k/2. (2.8)

By this equality we know that if (2.3) holds for γ1 and γ2 then it holds for γ1γ2 also. Then two equalities (2.4) and (2.5) guarantee (2.3) for any γ ∈ Γ.

Remark 2.4. Observing (2.8) one may regard that a modular form is nothing but a multi diﬀerential form on Γ \ H. A modular form is very near to it. But they are not exactly the same thing . We should note that z does not necessarily give a local coordinate on it. We shall study this point in detail later.

2.3 Eisenstein series

Deﬁnition 2.2. Suppose k is an even integer with k ≥ 3. We deﬁne the Eisenstein series of weight k:

′ ∑ 1 G (z) = . (2.9) k (mz + n)k Lemma 2.1. The righthand side of (2.9) is absolutely uniform convergent on the compact set in H.

proof]. Suppose a positive real number δ(< 1). Let K ⊂ {Im z ≥ δ} ⊂ H be a compact set. ¯ ¯ ¯ ¯ ∑ ¯ ¯ ∑∞ ∑ ¯ ¯ ∑∞ ∑∞ ¯ 1 ¯ ¯ 1 ¯ 1 8 1 ¯ ¯ = ¯ ¯ ≤ 8L = . (mz + n)k (mz + n)k (Lδ)k δk Lk−1 (m,n)∈Z2−{(0,0)} L=1 max{|m|,|n|}=L L=1 L=1

So we obtain the conclusion.

16 z K

∆ ¯ 0 1

δ evaluates the distance

q.e.d.

Proposition 2.2. We have Gk(z) ∈ Mk(Γ).

[proof]. According to Lema 2.1, Gk(z) is holomorphic on H. And obviously it holds Gk(z+1) = Gk(z). Still we have ′ ′ 1 ∑ 1 ∑ 1 G (− ) = = (−z)k = (−z)kG (z). k z (−m/z + n)k (m − nz)k k It holds ∑∞ lim Gk(z) = 2 = 2ζ(k). z→i∞ n=1

It means Gk(z) is bounded at z = i∞, namely it is holomorphic there. q.e.d.

Proposition 2.3. (Fourier expansion of the Eisenstein series) Suppose k is an even integer greater than 2. We have

k ∑∞ ∑ 2(2πi) n k G (z) = 2ζ(k) + σ − (n)q with σ (n) = d , k (k − 1)! k 1 k n=1 d|n

( ) ∑∞ 2k n G (z) = 2ζ(k) 1 − σ − (n)q . (2.10) k B k 1 k n=1 Here we deﬁne the Bernoulli numbers by

∞ x ∑ xk = B . (2.11) ex − 1 k k! k=0 Several beginning numbers are given by 1 1 1 1 B = 1,B = − ,B = ,B = − ,B = , ··· . 0 1 2 2 6 4 30 6 42 Lemma 2.2. (Due to Euler)

1 (2πi)2k ζ(2k) = − B (k = 1, 2, 3,...). (2.12) 2 (2k)! 2k

− 1 ≥ [proof]. Note that B0 = 1,B1 = 2 ,B2k+1 = 0 (k 1). In fact we have the following argument. x 1 By putting ϕ(x) = ex−1 + 2 x, we have −x 1 xex 1 ϕ(−x) = − x = − x. e−x − 1 2 ex − 1 2

17 So it holds x xex ϕ(x) − ϕ(−x) = − + x = 0. ex − 1 ex − 1 It means that ϕ(x) is an even function. Hence it’s power series expansion does not contain any odd power ··· − 1 term. Namely B2k+1 = 0 for k = 1, 2, . We get B0 = 1,B1 = 2 by direct calculation. Now we have ∞ eπiz + e−πiz 2πiz ∑ 22kπ2kz2k πz cot πz = πzi · = πiz + = 1 + (−1)kB . eπiz − e−πiz e2πiz − 1 2k (2k)! k=1 On the other hand, from ( ) ∞ ( ) 2πi 1 ∑ 1 1 πi − = π cot πz = + + 1 − e2πiz z z + n z − n n=0̸ we have ( ) ∞ ( ) ∞ ∞ ( ) ∞ ∞ ∑ n2 ∑ ∑ z 2k ∑ ∑ 1 πz cot πz = 1 + 2 1 − = 1 − 2 = 1 − 2 z2k. n2 − z2 n n2k n=1 n=1 k=1 k=1 n=1 So by tem by term observation we obtain (2.12). q.e.d.

Lemma 2.3. For an integer k(≥ 2), we have

∞ ∑ 1 (2πi)k ∑ = nk−1e2πint (Im t > 0) (2.13) (t + n)k (k − 1)! n∈Z n=1 [proof]. We have

( ) ( ) ∞ 1 ∑ 1 1 2 ∑ ϕ(t) := + + = π cot πt = πi 1 + = πi − 2πi e2πint. t t + n t − n e2πit − 1 n=0̸ n=0 Making derivative we get ∞ ∑ 1 ∑ ϕ′(t) = − = −2πi 2πin · e2πint. (t + n)2 n∈Z n=0 This is the case k = 2. Taking higher derivatives we get equalities for all even k. q.e.d. [proof of the proposition] We have [ ] ∞ ∞ ∞ ∑ ∑ 1 ∑ (2πi)k ∑ G (z) = 2ζ(k) + 2 = 2ζ(k) + 2 nk−1e2πin(mz) k (mz + n)k (k − 1)! m=1 n∈Z m=1 n=1 [ ]   ∞ ∞ ∞ (2πi)k ∑ ∑ (2πi)k ∑ ∑ = 2ζ(k) + 2 nk−1e2πinmz = 2ζ(k) + 2  nk−1e2πiNz . (k − 1)! (k − 1)! m=1 n=1 N=1 n|N

So we obtain the required ﬁrst equality. The second formula follows from Lemma 2.2. q.e.d.

Deﬁnition 2.3. Let k be an even integer greater than 2. A normalized Eisenstein series is deﬁned by: ∑∞ ∑ 1 2k n 1 1 − − Ek(z) := Gk(z) = 1 σk 1(n)q = k . (2.14) 2ζ(k) Bk 2 (mz + n) n=1 m,n∈Z,(m,n)=1

18 We have  2 E4(z) = 1 + 240(q + 9q + ...),  (2.15)  2 E6(z) = 1 − 504(q + 33q + ...), − 1 1 here we used B4 = 60 ,B6 = 42 to get the equalities.

2.4 Discriminant form

2 3 For the torus C/Z + zZ and its realization as an elliptic curve y = 4x − g2(z)x − g3(z). We have

g2(z) = 60G4(z), g3(z) = 140G6(z).

Note that, due to Euler, π4 π6 ζ(4) = , ζ(6) = . 90 945 So we have 4 8 g (z) = π4E (z), g (z) = π6E (z). 2 3 4 3 27 6 These are modular forms of weight 4 and 6. The discriminant is expressed via Eisenstein series: (2π)12 ∆(z) = g (z)3 − 27g (z)2 = (E (z)3 − E (z)2). (2.16) 2 3 1728 4 6 By observing (2.14) we know that this is a cusp form of weight 12. By substituting (2.15) we get 43 g3(z) = π12(1 + ...), ∆(z) = (2π)12(q + ...). 2 33

3 43 12 1 So it holds 12 q · 3 π (1 + ...) 12 = 1. We can determine several other leading terms of 3 (2π) (q+...) |q=0 j(z), also. Hence we obtain the q expansion of j(z) as we stated in Theorem 1.11 (4) and mentioned in Remark 1.9: 1 j(z) = + 744 + 196884 · q + .... q

2.5 Eisenstein series E2(z)

Deﬁnition 2.4.  ∑ ∑  1 ∞ ∞ ′ 1 E2(z) = −∞ −∞ 2  2ζ(2) m= n= (mz+n)   ∑ ∑ 3 ∞ 1 = 1 + 2 ̸ −∞ 2 (2.17)  π m=0 n= (mz+n)   ∑ ∑  6 ∞ ∞ 1 = 1 + π2 m=1 n=−∞ (mz+n)2 . Here the sum is not absolutely convergent. So the order of the smmations is very important. According to (2.13) we have ∞ ∞ ∑ 1 ∑ = −4π2 nqn. (2.18) (z + n)2 n=−∞ n=1 Substitute mz for z, and take the summation for m. So we have ∑∞ ∑∞ md E2(z) = 1 − 24 dq . m=1 d=1

19 This double sum is absolutely convergent on |q| < 1. For example: for ∀K > 0, ∫ ∑K ∑K ∑ d|q|d ∞ xrx d|q|md ≤ ≤ dx ∼ − log(1 − r), r = |q| < 1. 1 − |q|d 1 − rx m≥1 d≥1 d 1

Hence we can rewrite ∑∞ n E2(z) = 1 − 24 σ1(n)q . (2.19) n=1

How it behaves under the invertion S(z) = −1/z ?

( ) ∞ ′′ ∞ ′ 1 1 1 ∑ ∑ 1 3 ∑ ∑ 1 E − = = 1 + . (2.20) z2 2 z 2ζ(2) (−m + nz)2 π2 (mz + n)2 m=∞ n n=−∞ m

(2.20) is not equal to ( 2.17), because our double sum is not absolutely convergent.

Proposition 2.4. ( ) 1 1 12 E − = E (z) + . (2.21) z2 2 z 2 2πiz

[proof]. Set 1 1 1 a (z) = = − m.n (mz + n − 1)(mz + n) mz + n − 1 mz + n and set ∞ ( ) 3 ∑ ∑ 1 E˜ (z) = 1 + − a (z) . 2 π2 (mz + n)2 m,n m=0̸ n=−∞ Note that 1 1 1 − a (z) = ∼ (mz + n)2 m,n (mz + n)2(mz + n − 1) (mz + n)3 is absolutely convergent. So we have

∞ ∞ ( ) 3 ∑ ∑ 1 3 ∑ ∑ 1 1 E˜ (z) = 1 + + − . 2 π2 (mz + n)2 π2 mz + n mz + n − 1 m=0̸ n=−∞ m=0̸ n=−∞

But the third term is equal to 0. Hence it holds

E˜2(z) = E2(z).

Because the double sum of E˜2(z) is absolutely convergent, we have

∞ ( ) ∞ 3 ∑ ∑ 1 1 1 3 ∑ ∑ E (z) = 1 + − a (z) = E (− ) − a (z). 2 π2 (mz + n)2 m,n z2 2 z π2 m,n n=−∞ m=0̸ n=−∞ m=0̸

So we are asked to evaluate the last term. We claim ¯ ¯ ¯ ¯ ∑∞ ¯ ∑ ¯ ¯ ¯ ¯ am,n(z)¯ ¯ ¯ n=−∞ m=0̸ is convergent. This is assured by the same way as the convergence ¯ ¯ ¯ ¯ ∑∞ ¯ ∑ ¯ ¯ 1 ¯ ¯ ¯ . ¯ (mz + n)2 ¯ n=−∞ m=0̸

20 And it is derived as the following: by using (2.18 ) it holds

∞ ∑ 1 1 ∑ 1 1 (2π)2 ∑ = = − − de−2πidn/z (n > 0). (−mz − n)2 z2 (−m − n/z)2 n2 z2 m=0̸ m=0̸ d=1

− 1 Set z = s + it (t > 0). So ¯ ¯ ¯ ¯ ¯ ∑ ¯ ∑∞ ¯ 1 ¯ 1 2 2 −2πdnt ¯ ¯ ≤ + 4π |s + it| de (n > 0). ¯ (mz + n)2 ¯ n2 m=0̸ d=1

And we have ¯ ¯ ¯ ¯ ∑∞ ¯ ∑ ¯ ∑∞ ∑∞ ∑∞ ¯ 1 ¯ 1 2 2 2 −2πdnt ¯ ¯ ≤ + |s + it| ζ(2) + 8π |s + it| de ¯ (mz + n)2 ¯ n2 n=−∞ m=0̸ n=−∞ n=1 d=1 ∑∞ ∑∞ ∑∞ 2 2 2 −2πdnt 2 2 2 −2πt N = (2 + |s + it| )ζ(2) + 8π |s + it| de = (2 + |s + it| )ζ(2) + 8π |s + it| σ1(N)(e ) . n=1 d=1 N=1 Hence we are allowed to write

∑∞ ∑ ∑N ∑ ∑ ∑N am,n(z) = lim am,n(z) = lim am,n(z) N→∞ N→∞ −∞ ̸ − ̸ ̸ − n= m=0 ( n= N+1)m=0 m=0 n= N+1 ∑ 1 1 = lim − . N→∞ mz − N mz + N m=0̸

Now we have ( ) ( ) [ ( ) ] ∑ 1 1 2 ∑ 1 1 2 πN z − = − = π cot − + . mz − N mz + N z m − N/z m + N/z z z N m=0̸ m=0̸

Here we have ( ) −2πiN/z 2 −πN 2π e + 1 −2πi lim π cot = lim i − = . N→∞ z z z N→∞ e 2πiN/z − 1 z

Hence 1 1 6i 1 1 12 E (z) = E (− ) + = E (− ) − . 2 z2 2 z πz z2 2 z 2πiz q.e.d.

2.6 Algebra M(Γ)

Theorem 2.1. For a nonzero meromorphic modular form f(z) of weight k, we have

1 1 ∑ k v∞(f) + v (f) + v (f) + v (f) = . (2.22) 2 i 3 ω a 12 a∈F −{i,ω}

Where va(f) means the order of f(z) at z = a.

[proof]. Let C be a closed arc indicated by the ﬁgure with possibly additional deviation around the zeros on ∂F .

21 A ¬ H C

B i G DE e2 Πi3C F eΠi3

O Re - 1 1 2 2

the path of integration C

According to the residue theorem ∫ 1 f ′(z) ∑ dz = v (f). (2.23) 2πi f(z) P C P ∈F −∂F ∑ The right hand side is nothing but the term a∈F −{iω} va(f). Let us evaluate the left hand side as the sum of piece-wise integration. (i) Because of the periodicity of f ∫ ∫ f ′(z) f ′(z) dz + dz = 0. AB f(z) GH f(z) ∑ ˜ n 2πiz ′ d ˜ dq (ii) Set f(q) = f(z) = anq (q = e ). Note that f (z) = dq f(q) dz . So ∫ ∫ 1 f ′(z) 1 df/dq˜ dz = − dq 2πi HA f(z) 2πi |q|=e−2πT f˜(q) Hence ∫ 1 f ′(z) dz = −v∞(f). 2πi HA f(z) (iii) ∫ ∫ 1 f ′(z) 1 f ′(z) 1 dz = dz → − vω(f) 2πi BC f(z) 2πi FG f(z) 6 as ε → 0. (iv) ∫ 1 f ′(z) 1 dz → − vi(f) 2πi DE f(z) 2 as ε → 0. (v) We claim ∫ ∫ 1 f ′(z) 1 f ′(z) k dz + dz → 2πi CD f(z) 2πi EF f(z) 12 as ε → 0. ( ) a b Lemma 2.4. For g = ∈ Γ with c ≠ 0 and a closed arc r on H we have c d ∫ ∫ ∫ f ′(z) f ′(z) dz dz − dz = −k (2.24) r f(z) g(r) f(z) r z + d/c

22 ( ) 0 −1 According to this lemma, setting g = , we have 1 0 ∫ ∫ 1 dz 1 1 1 lim = lim d log z = [arg z]z=i = − . → → z=ω ε 0 2πi CD z ε 0 2πi CD 2πi 12 So we have (v) and we complete the proof of the theorem. [proof of the lemma]. From f(g(z)) = (cz + d)kf(z) we have dg(z) f ′(g(z)) = (cz + d)kf ′(z) + kc(cz + d)k−1f(z). dz Then f ′(g(z)) f ′(z) cdz dg(z) = dz + k . f(g(z)) f(z) cz + d So ∫ ∫ f ′(z) f ′(g(z)) cdz dz − dg(z) = −k . C f(z) f(g(z)) C cz + d Hence we have the Lemma. q.e.d.

Theorem 2.2. We have the following.

(a) For k < 0, M0(Γ) = 0.

(b) M0(Γ) = C.

Sk(Γ) = C∆ k = 12  ∆Mk−12(Γ) k ≥ 14.

(f) Mk(Γ) = Sk(Γ) ⊕ CEk for k > 2.

[proof]. (a) The left hand side of (2.22) is a nongegative number. So k can not be negative. ∼ (b) When k = 0, f(z) becomes to be a holomorphic function on Γ \ H = P 1. The left hand side of (2.22) is zero only if all the terms are zero. That means f does not have zero. Hence f(z) must be a constant. 1 (c) 6 cannot be attained in the left hand side of (2.22). So M2(Γ) contains only 0. (d) For example the case k = 4 is realized only if νρ = 1 and other terms in LHS are all zero. So

f ∈ M4(Γ) is determined up to a constant multiple factor. According to Proposition 2.2 E4(z) ∈ M4(Γ).

So we have M4(Γ) = CE4(z). We can make a similar argument for other cases.

(e) For f ∈ S12(Γ) we have only possibility ν∞ = 1 and all other terms in LHS of (2.22) are zero. Hence

it is determined up to a constant multiple factor. In subsection 2.4 we discussed that ∆(z) ∈ S12(Γ).

Hence we have S12(Γ) = C∆(z). When k ≥ 14 we know that ∆Mk−12(Γ) ⊂ Sk(Γ). Take an arbitrary

f ∈ Sk(Γ). Then f/∆ is holomorphic on H ∪ {i∞}. Namely f ∈ ∆Mk−12(Γ).

(f) When k < 12 the assertion is reduced to (c). When k ≥ 12, take an element f ∈ Mk(Γ), and suppose

f(i∞) = c. Then f − cEk ∈ Sk(Γ). So f ∈ Sk ⊕ CEk. q.e.d.

⊕ i j Theorem 2.3. (1) Mk(Γ) = 4i+6j=kCE4E6, ⊕∞ ⊕ (2) M(Γ) = k=0Mk(Γ) = k:evenMk(Γ) = C[E4,E6]. Moreover E4 and E6 does not have any algebraic relation, namely M(Γ) is isomorphic to the polynomial algebra C[X,Y ].

23 [proof]. (1) The assertion is true for k ≤ 12. For k ≥ 14, we have a solution i, j such that k = 4i + 6j. i j Take E4E6. It belongs to Mk(Γ). So by the same argument as in the proof of the preceeding theorem we ⊕ i j have Mk = ∆Mk−12 CE4E6. By induction we obtain the required assertion. (2) The ﬁrst statement is a direct consequence of (1). Suppose that we have an algebraic relation between E4 and E6. There should exists a weighted homogeneous polynomial P (X,Y ) with weight 4 for X and weight 6 for Y such that P (E4,E6) = 0. Namely we have a homogeneous polynomial Q with 3 2 Q(E4 ,E6 ) = 0. That means j(z) satisﬁes an algebraic equation. According to Theorem 1.11 j(z) takes every complex value. Hence we have a contradiction. q.e.d.

2.7 The Dedekind η function

Definition 2.5. The Dedekind η function is defined by ∏∞ ( ) η(z) = e2πiz/24 1 − e2πinz z ∈ H. (2.25) n=1 ∑ | 2πinz| The absolute convergence of the infinte product follows from that of n≥1 e .

Proposition 2.5. ( ) 1 √ η − = −iz η(z), (2.26) z √ where we choose the branch so that we have Re −iz > 0.

[proof]. Put 1 h(z) = √ η(−1/z). −iz We claim that η′(z) h′(z) = . (2.27) η(z) h(z)

Then we have h(z) = cη(z) for some constant c. Observing h(i) = η(i) we obtain the required equality. So we concentrate ourselves to show (2.27).   η′ 2πi ∑ −2πine2πinz 2πi ∑ ne2πinz = + = 1 − 24  η 24 1 − e2πinz 24 1 − e2πinz n≥1 n≥1      2πi ∑ ∑ 2πi ∑ ∑ = 1 − 24 ne2πinz  e2πim(nz) = 1 − 24 ne2πim(nz) 24 24 n≥1 m≥0 n≥1 m≥1 ( ) 2πi ∑ = 1 − 24 σ (n)qn . 24 1 n

According to (2.19) we have

η′(z) 2πi = E (z). (2.28) η(z) 24 2

On the other hand, it holds

( ( ))′ h′(z) 1 1 1 η′(− 1 ) 1 = ( √ η − )/h(z) = z − . − 2 − 1 h(z) iz z z η( z ) 2z

24 Using ( 2.28) we have ( ( ) ) h′(z) 2πi 1 1 12 = E − − . h(z) 24 z2 2 z 2πiz

Due to the quasi automorphic property of E2 (Proposition 2.4), we obtain the required equality. q.e.d.

Proposition 2.6.

∞ 1 ∏ ∆(z) = q (1 − qn)24 = η(z)24, q = e2πiz. (2.29) (2π)12 n=1

3 − 2 ··· 12 ··· (Note that E4 E6 = 1728q + , ∆ = (2π) q + .)

Remark 2.5. (a) Note that we used the automorphic propertiy of E2(z) to get that of η(z). (b) For the q- expansion ∑∞ η(z)24 = τ(n)qn, n=1 τ(n) is called the Ramanujan τ function. There are many nice arithmetic properties for it. (i) τ(mn) = τ(m)τ(n) for (m, n) = 1 (We shall show it in the later section).

(ii) τ(n) ≡ σ11(n) ( mod 691). (iii) (Deligne 1973 from Weil conjecture)

11 |τ(p)| < 2p 2 (p : prime).

25 3 Modular form for congruence subgroups

In this section we use the following notation:

Γ = SL(2, Z),

for a positive( interger) N, a b Γ(N) = {g = ∈ Γ: g ≡ id (modN)}, c d ( ) ( ) a b ∗ ∗ Γ (N) = {g = ∈ Γ: g ≡ (modN)}, 0 c d 0 ∗ ( ) ( ) a b 1 ∗ Γ (N) = {g = ∈ Γ: g ≡ (modN)}, 1 c d 0 1 Γ′ : a congruence subgroup i.e. it holds Γ(N) ⊂ Γ′ ⊂ Γ, for some positive integer N, F (Γ′) : fundamental region for Γ′,

F = F (Γ) = F1 ∪ F2 : fundamental region for Γ (F1 = F ∩ {Re z ≥ 0},F2 = F ∩ {Re z ≤ 0}), µ = [Γ: Γ′], ′ πi/3 νρ : number of non multiple vertices of Γ(ρ) on F (Γ ), where, ρ = e , ′ νi : number of non multiple vertices of Γ(i) on F (Γ ), t : number of equivalence classes of cusps for Γ′, g : genus of Γ′\H.

Remark 3.1. Recall that in Proposition 2.1 we obtained

′ ∪µ −1 F (Γ ) = i=1αi F ⊔ µ ′ ∈ ∩ ′ for a cosets decomposition Γ = i=1 αiΓ . For a Γ(ρ) F (Γ ) we have the following two cases. First, the case a is a common vertex of αi(F1), αj(F1) and αk(F1) for diﬀerent three indices i, j, k. Second, the

case a is a vertex of unique triangle αi(F1). νρ is the number of αi(ρ)’s of the second case. We define νi by the same manner. 2 We find an example of non( multiple) vertex for Γ(i) in the configuration of F (Γ0(2)). Here (TS) F1 1 −1 is mapped to TSTF by g = ∈ Γ (2). Together with the matrix T , g determines the gluing of 1 2 −1 0 the edges of F (Γ0(2)).

+ F1 1 F2 F1 1+F2

SF2 TSF1 SF2 TSF1 2 2 HTSL F1 HTSL F1

TSTF1 g 0 1 0 1

1+i Fundamental region for Γ0(2) non multiple vertex 2 .

26 3.1 Geometry of congruence subgroups

Theorem 3.1. (genus formula) We have µ ν ν t g = 1 + − i − ρ − . (3.1) 12 4 3 2

[proof of the theorem]. Let us consider the triangulation of Γ \ H obtained by the Γ orbits of F1 and

F2. Set a0, a1, a2 be the number of vertices, sides and triangles in it. We have  a2 = 2µ,

a1 = 3µ,  − µ−νi µ νρ a0 = t + (νi + 2 ) + (νρ + 3 ). We are requested to show ν 2ν µ 2 − 2g = t + i + ρ − . 2 3 6 By the above counting we have the Euler characteristic of Γ \ H: µ − ν µ − ν 2 − 2g = a − a + a = (t + (ν + i ) + (ν + ρ )) − 3µ + 2µ. 0 1 2 i 2 ρ 3 So we obtain the assertion. q.e.d.

3.2 Principal congruence subgroup Γ(N)

Theorem 3.2. (1) ∼ Γ/Γ(N) = SL(2, Z/NZ).

(2) ( ) 1 ∏ 1 [Γ: Γ(N)] = µ(N) = N 3 1 − (N ≥ 3). 2 p2 p|N

[proof]. (1) We observe the reduction map

fN : SL2(Z) → SL2(Z/NZ).

So it holds

Γ(N) = ker(fN ). ∼ Because we have fN (Γ) = Γ/kerfN , the assertion follows from the surjectivity of fN . So we show that (fN is) surjective. Take an arbitrary element g ∈ SL2(Z/NZ). It is expressed by an integer matrix a b ∈ M2(Z) satisfying ad − bc ≡ 1 ( mod N). We can ﬁnd an interger t such that (a, b + Nt) = 1. c d ∏ (Put t = p|a,(p,b)=1 p (p : prime). Let us examine a common prime factor of a and b + Nt. We may consider only a prime factor p of a. If we don’t have p|b, it holds p|Nt so we don’t have p|(b + Nt). If p|b, from the condition ad − bc ≡ 1 ( mod N) and the fact that p is not contained in t, we know that p does not devide b + Nt.) We exchange b with b′ = b + Nt. So we have

ad − b′c = 1 + uN with some integer u. Because (a, b′) = 1, we can ﬁnd integers x, y so that we have −ax + b′y = u. Put

c′ = c + Ny, d′ = d + Nx.

27 ( ) a b′ Then we have ad′ − b′c′ = 1. Namely we found a representative ∈ SL (Z) of g. c′ d′ 2 (2) Suppose (N1,N2) = 1 for N1,N2 ∈ Z. We have ∼ SL2(Z/NZ) = SL2(Z/N1Z) × SL2(Z/N2Z). ( ) a b In fact. Take an element g = ∈ SL (Z/NZ), ad − bc ≡ 1 ( mod N), and observe the homomor- c d 2 phism ( g : SL) 2(Z/N( Z) →) SL2(Z/N1Z() × SL)2(Z/N2Z) a b a b a b . 7→ ( ( mod N ), ( mod N )) c d c d 1 c d 2 According to the Chinese Remainder Theorem it is a bijective homomorphism. So in this case, we obtain

µ(Γ(N)) = µ(Γ(N1)) · µ(Γ(N2)).

Hence it is enough to show 1 µ(Γ(pn)) = ♯SL (Z/pnZ) = p3n(1 − ) = p3n − p3n−2, p : prime. 2 p2 ( ) a b Let us count the possible choices of an element g = ∈ SL (Z/pnZ). c d 2 The case we have (a, p) = 1. For arbitrarily given b, y we have unique solution x for ax − by ≡ 1 ( mod pn). We have pn − pn−1 possibilities for a. Hence we have p3n − p3n−1 elements here. The case p|a. We have pn−1 possibilities for a. For ad − bc ≡ 1 ( mod pn) we need (b, p) = 1. If this is the case, for an arbitrary d we have unique solution c. For b we have pn − pn−1 possibilities, and pn choices for d. Then we have pn−1pn(pn − pn−1) = p3n−1 − p3n−2 elements here. So we get the assertion. q.e.d.

Proposition 3.1. We have µ(N) t = for Γ′ = Γ(N). N [proof]. Consider the reduction map

ϕ : Γ′ \ (H − H) → Γ \ (H − H).

It holds ( ) ( ) t = ♯ Γ′ \ (H − H) , ♯ Γ \ (H − H) = 1. Note that {( ) } {( ) } a k ′ a Nk Γ ∞ = : k = 0, ±1, ··· / ± I, Γ ∞ = : k = 0, ±1, ··· / ± I. i 0 1 i 0 1

We can regard H − H as a homogeneous space Γ/Γi∞, where the quotient means the set of right cosets. So we have ( ) ( ) ( ) ′ ′ ′ ′ ′ t = ♯ Γ \ (H − H) = ♯ Γ /(Γi∞ ∩ Γ ) \ Γ/Γi∞ = ♯ Γ /Γ i∞ \ Γ/Γi∞ ,

and ( ) ( ) ♯ Γ \ (H − H) = ♯ Γ/Γi∞ \ Γ/Γi∞ = 1. So we obtain ♯(Γ/Γ′) µ(N) t = ′ = . ♯(Γi∞/Γ i∞) N q.e.d.

28 name µ genus νi νρ t isomorphic Γ(2) 6 0 0 0 3 S3 Γ(3) 12 0 0 0 4 A4 : tetrahedral group Γ(4) 24 0 0 0 6 S4 : octahedral group Γ(5) 60 0 0 0 12 A5 : icosahedral group Γ(6) 72 1 0 0 12 Γ(7) 168 3 0 0 24 Klein’s max. aut. gr...... Table for Γ(N)

F1 1+F2

F2 F1 i Ρ i Ρ

1 1 0 1 2 0 3 1 2

Fundamental region for Γ(3) Fundamental region for Γ0(4) = Γ1(4)

F1 F2 +1 i Ρ

1 2 0 3 5 1

Fundamental region for Γ1(5)

By 2 −1 IF1,TSF1, (TS) F1,TSTF1,TF2,SF2,TSTF2,ST F2 ( ) ( ) ( ) ( ) 1 1 −2 1 1 1 1 0 we get a fundamental region of Γ (3) = 〈 , 〉 = 〈 , 〉. 1 0 1 −3 1 0 1 3 1

29 By

2 2 IF1,TSF1, (TS) F1,TSTF1,TST (TS)F1,TST (TS) F1, −1 −1 −1 2 TF2,SF2,TSTF2,ST F2,ST (ST )F2,ST (ST ) F2 ( ) ( ) ( ) ( ) 1 1 −3 1 1 1 1 0 we get a fundamental region of Γ (4) = 〈 , 〉 == 〈 , 〉. 1 0 1 −4 1 0 1 4 1

3.3 Recalling the Riemann-Roch theorem

In the next subsection we use the Riemann-Roch Theorem for a compact Riemann surface. Here we recall it. Let X be a compact Riemann surface of genus g. ∑ r Definition 3.1. Let P1,...,Pr be points on it, and let n1, . . . , nr be integers. The formal sum i=1 niPi is said to be a divisor on X. We define a divisor with coefficient in Q by the same manner. We say a divisor D is effective if all the coefficient are non negative, and we denote this by D ≥ 0. For D = ∑ ∑ r n P , r n is called the degree of D and is denoted by deg D. i=1 i i i=1 i ∑ For a meromorphic function f on X, by (f) we denote the divisor P nP (f). For a multi-differential ω on X, we define a divisor (ω) by the same way. Let K be the divisor of a meromorphic differential on X, that is called a canonical divisor. Any two canonical divisors are linearly equivalent, so we don’t distinguish such a difference. Theorem 3.3. Let X be a compact Riemann surface of genus g. Let D be a divisor on it. Then we have

dim H0(X, O(D)) − dim H1(X, O(D)) = 1 − g + deg D. (3.2) ∼ Here we note that H1(X, O(D)) = H0(X, O(K − D)) = H0(X, Ω(D)). Remark 3.2. H0(X, O(D)) is the vector space of meromorphic functions on X with (f) + D ≥ 0. H0(X, O(K − D)) is the vector space of meromorphic 1-forms with (ω) − D ≥ 0. H0(X, O(mK + D)) is the vector space of meromorphic multi-diﬀerentials of order m with (ω) + D ≥ 0. Remark 3.3. Note that we have H0(X, O(D)) = 0 for deg D < 0. It is an easy consequence from the fact deg(f) = 0 for any meromorphic function f on X. Example 3.1. By putting D = 0 we get

dim H0(X, O) = g.

By putting D = K and using the above equality, we get

deg K = 2g − 2.

3.4 Dimension formula for congruence subgroups

We ﬁx a cogruence subgroup Γ′ of Γ. For an even integer k we set

′ dk = dim Mk(Γ ), ′ ek = dim Sk(Γ ).

Theorem 3.4. (Dimension formula 1) For a congruence subgroup Γ′, we have:   dk = 0 (k < 0)  d0 ={ 1 g (t = 0) , d2 =  g + t − 1 (t ≥ 1)  ∑  s k 1 k dk = (k − 1)(g − 1) + [ (1 − )] + t j=1 2 nj 2

30 where [a] means the integer part of a, s = νi + νρ, and for a non multiple vertex Pj we set { ′ 2 Pj ∈ π (Γ(i)) nj = ′ . 3 Pj ∈ π (Γ(ρ))

[proof]. We denote the Riemann surface Γ′ \ H by X. We shall speak about the system of local coordinates later. Take a meromorphic modular form f(z) of weight k for Γ′. If we consider f(z)(dz)k/2, it seems to be something like a multidifferential form on X. But it is not the case. Because z does not necessarily give a local coordinate on X, it comes out a difference from an usual multidifferential form. We examine this difference in an exact way. Let π : H → Γ \ H, π′ : H → Γ′ \ H

be the natural projection. And set the natural covering map

ϕ : Γ′ \ H → Γ \ H.

According to Theorem 1.10 we know that j(z) gives a global coordinate of Γ \ H as P 1(C). (0) For any point p = π′(a) ∈ π′(H − Γ(i) ∪ Γ(ρ)), z gives a local coordinate at p. So we are safe. Namely, f(z)(dz)k/2 itself is locally a multidiﬀerential form there. Suppose p = π′(a) ∈ π′(Γ(i)∪Γ(ρ)) and a is a multivertex of F (Γ′). In this case π′ is locally biholomorphic, and z gives a local coordinate there. So we are safe here, also. (i) The case when p = π′(a) ∈ π′(Γ(i)) and a is a non-multiple vertex of F (Γ′). The map ϕ is locally biholomorphic there. So j(z) − j(π(a)) = j(z) − 1728 gives a local coordinate at p, recall j(i) = 1728. Note that j(z) gives a 2 : 1 map in a neighborhood of z = i. So j(z) − 1728 has a double zero at z = a. So we can take a local coordinate t at p such that t = (z − a)2. Hence we have 1 dz = t−1/2dt. 2 (ii) For a non-multiple vertex a ∈ Γ(ρ) ∩ F (Γ′), by the same consideration as above we get a local coordinate t = (z − a)3 at p = π′(a). Hence we have 1 dz = t−2/3dt. 3 (iii) Finally, the case when p = π′(a) with a ∈ Γ(i∞). We have t = e2πi/N for some positive integer N. Hence we have 2πi dz = t−1dt, N a ∈ recall( that) at a cusp r = c Q, always we observe the behavior of f(z) by that of f(α(z)) with a b α = ∈ Γ. c d According to this argument f(z) is a (holomorphic) modular form, if and only if the multi-diﬀerential form ω = f(z)(dz)k/2

on X has some poles those arising at p = Pj(j = 1, . . . , s) or p = Qℓ(ℓ = 1, . . . , t) with the condition   ∑s ∑t k  1  (ω) ≥ − · (1 − )Pj + Qℓ . (3.3) 2 nj j=1 ℓ=1

31 Remark 3.4. Note that we have the paraphrase of this argument for a cusp form. f(z) is a cusp form, if and only if the multi-diﬀerential form ω satisﬁes   ∑s ∑t ∑t k  1  (ω) ≥ − · (1 − )Pj + Qℓ + Qℓ. (3.4) 2 nj j=1 ℓ=1 ℓ=1 ∑ ∑ ∑ ∑ s 1 t s 1 t Set D = (1 − )Pj + Qℓ, and set D1 = (1 − )Pj,D2 = Qℓ. Via this j=1 nj ℓ=1 j=1 nj ℓ=1 correspondence between f and ω we have an isomorphism of vector spaces:

′ ∼ 0 k M (Γ ) = H (X, O( (K + D))). k 2 Naturally we have

∑s ∑t 0 k ∼ 0 k k 1 k H (X, O( (K + D))) = H (X, O( K + [ (1 − )]Pj + Qℓ)). 2 2 2 nj 2 j=1 ℓ=1 and ∑s ∑t 0 k 0 k k 1 k dim H (X, O( (K + D))) = dim H (X, O( K + [ (1 − )]Pj + Qℓ)). 2 2 2 nj 2 j=1 ℓ=1 By using Riemann-Roch theorem, let us count the right hand side. We use an abbreviation

k ∑s k 1 k ∑t k k k K + [ (1 − )]Pj + Qℓ = K + [ D1] + D2. 2 2 nj 2 2 2 2 j=1 ℓ=1 At ﬁrst we have deg(K + D) > 0.

2ν − νi ρ In fact, by R-R theorem deg K = 2g 2. And deg D = 2 + 3 + t. By the genus formula Theorem 3.1 2ν − νi ρ µ we have 2g 2 + 2 + 3 + t = 6 > 0. 0 O k k (a) For the case k < 0, we have H (X, ( 2 (K + D))) = 0 because deg 2 (K + D) < 0. 0 O k 0 O (b) For the case k = 0, we have H (X, ( 2 (K + D))) = H (X, ) = C. This is trivial. (c) Let us consider the case k ≥ 4. By the R-R theorem we have k k k k k k k k k dim H0(X, O( K + [ D ] + D ) = 1 − g + deg( K + [ D ] + D ) + dim H0(X, O((1 − )K − [ D ] − D ) 2 2 1 2 2 2 2 1 2 2 2 2 1 2 2 Note that we have k 1 k 1 k 2 k 2 −[ · ] = [(1 − ) · ], −[ · ] = [(1 − ) · ]. 2 2 2 2 2 3 2 3 It means that we have always k k k k k k k (1 − )K − [ D ] − D = (1 − )K + [(1 − )D ] + (1 − )D − D < (1 − )(K + D) < 0. 2 2 1 2 2 2 2 1 2 2 2 2 0 O − k − k − k − Hence dim H (X, ((1 2 )K [ 2 D1] 2 D2)) = 0. So, using deg K = 2g 2 again we have

k k k k ∑s k 1 k H0(X, O( (K + D))) = 1 − g + deg( K + [ D ] + D ) = (k − 1)(g − 1) + [ (1 − )] + t. 2 2 2 1 2 2 2 n 2 j=1 j

(d) For the case k = 2, the cohomologcal term on the right hand side becomes to be

H0(X, O(−[D1] − D2)) = H0(X, O(−D2)).

So we have { 0 t > 0 dim H0(X, O(−[D1] − D2)) = . 1 t = 0

32 By using deg K = 2g − 2, in case t > 0 we have (note that deg[D1] = 0)

′ dim M2(Γ ) = 1 − g + deg(K + [D1] + D2) = g − 1 + t.

′ In case t = 0 we have dim M2(Γ ) = g by the same way. Hence we obtain the assertion. q.e.d.

Theorem 3.5. (Dimension formula 2) For a congruence subgroup Γ′, we have:   ek ={ 0 (k < 0)   1 (t = 0) e0 =  0 (t > 0)  e2 = g  ek = dk − t (k ≥ 4).

[proof]. We can prove this formula by the same method just using the condition mentioned in Remark 3.4.

Remark 3.5. For the principal congruence subgroup Γ(N), we don’t have any nonmultiple vertex. So we get the genus g for by an easy calculation. But, for general cases we need to know the glueing of the ∂F (Γ′) by using the list of cosets representatives.

name µ genus t isomorphic dk(k ≥ 2) e2(= g) ek(= dk − t), k ≥ 4 Γ(2) 6 0 3 S3 k/2 + 1 0 k/2 − 2 Γ(3) 12 0 4 A4 k + 1 0 k − 3 Γ(4) 24 0 6 S4 2k + 1 0 2k − 5 Γ(5) 60 0 12 A5 5k + 1 0 5k − 11 Γ(6) 72 1 12 6k 1 6k − 12 Γ(7) 168 3 24 14k − 2 3 14k − 26 .. .. . Table for Γ(N)

name µ (νi, νρ, t) genus coinsidence d2 dk(k ≥ 4) e1(= g) ek(= dk − t) Γ1(2) 3 (1,0,2) 0 Γ0(2) k/2 + 1 0 Γ1(3) 4 (0,1,2) 0 Γ0(3) [2k/3] + 1 0 Γ1(4) 6 (0,0,3) 0 Γ0(4) k + 1 0 Γ1(5) 12 (0,0,4) 0 2k + 1 0 Γ1(6) 12 (0,0,4) 0 Γ0(6) 2k + 1 0 .. . . . Table for Γ1(N)

33 4 Hecke operators and Hecke eigen forms 4.1 Preparatory consideration

Deﬁnition 4.1. Let R be the set of all lattices in C, and let XR be the free Z module generated by R. We deﬁne an operator ∑ T (n)Λ = Λ′ [Λ:Λ′]=n

for Λ ∈ R and make a natural extension on XR. We call it a Hecke correspondence. Set

KαΛ = αΛ.

Remark 4.1. For a lattice Λ(ω1, ω2) in C, denote the volume of the parallelogram {z = kω1 + ℓω2 : 0 ≤ k ≤ 1, 0 ≤ ℓ ≤ 1} by vol(Λ). For a sublattice Λ′ ⊂ Λ it holds

[Λ : Λ′] = vol(Λ′)/vol(Λ).

When n = p : prime, we have p + 1 sublattices with [Λ : Λ′] = p. In fact, for Λ =< 1, τ > we have the choices (see Figure 4.1):

< 1, pτ >, < p, τ >, < p, 1 + τ >, . . . , < p, p − 1 + τ > .

Note that always it holds pΛ ⊂ Λ′.

pΤ

1 p

Figure 4.1 Sublattices with index p

Proposition 4.1. We have

(1) KαKβ = Kαβ,

(2) KαT (n) = T (n)Kα, (3) T (m)T (n) = T (mn) ((m, n) = 1), n n+1 n−1 (4) T (p )T (p) = T (p ) + pT (p )Kp (p : prime).

[proof]. We can obtain (1)(2) easily. ′′ ′′ ∼ (3) Suppose Λ ⊂ Λ, [Λ : Λ ] = mn. According to the isomorphism Z/mnZ = Z/mZ × Z/nZ we can ﬁnd a intermediate sublattice Λ′ such that Λ′′ ⊂ Λ′ ⊂ Λ with [Λ : Λ′] = n, [Λ′ :Λ′′] = m in a unique way. It shows the property. n n+1 n−1 (4) We consider T (p )T (p)Λ,T (p )Λ,T (p )KpΛ. They are elements in XR composed of sublattices of index pn+1. Set  ∑  n ′′ ′′ T (p )T (p)Λ =∑ a(Λ )Λ ∑ n+1 ′′ ′′ ′′ T (p )Λ = b(Λ )Λ = ′′ Λ  ∑ [Λ:Λ ]=pn+1  n−1 ′′ ′′ T (p )KpΛ = c(Λ )Λ ,

34 where Λ′′ be a sublattice with [Λ : Λ′′] = pn+1. We are requested to show a = b + pc(= 1 + pc). (i) The case Λ′′ is not a sublattice of pΛ. By deﬁnition we have c(Λ′′) = 0. On the other hand we have

a(Λ′′) = ♯{Λ′ : [Λ : Λ′] = p, Λ′′ ⊂ Λ′ ⊂ Λ}.

As we observed in Figure 4.1, this Λ′ necessarily contains pΛ. Note that Λ′/pΛ is a index p subgroup of ∼ ′′ Λ/pΛ = Z/pZ ×Z/pZ. It is determined by an element of order p in it. Here take an element λ ∈ Λ −pΛ. It is regarded as an element of Λ′/pΛ with order p. So it determines Λ′ in a unique way. It means a = 1. ′′ n−1 n−1 (ii) The case Λ is a sublattice of pΛ. By deﬁnition T (p )KpΛ = T (p )(pΛ), so c = 1. And we have b + pc = 1 + p. As we already saw, [Λ : Λ′] = p implies pΛ ⊂ Λ′. So especially Λ′′ ⊂ Λ′. Namely we have a = p + 1. By the above argument we obtained the required equality. q.e.d.

n Corollary 4.1. For prime number p, T (p ) is a polynomial of T (p) and Kp.

n Corollary 4.2. For prime number p, the algebra Z(T (p), Kp) is commutative and contains every T (p )

[proof of the corollaries]. These are direct consequence of (3)(4) in the proposition. q.e.d.

We consider a function F on XR satisfying the following condition:

F (αΛ) = α−kF (Λ) ∑ ∑ F ( Λi) = F (Λi) i i F (Λ(1, z))is a meromorphic function on H.

We call this function a meromorphic function of weight k deﬁned on XR. For such a function we deﬁne

K F (Λ) := F (K Λ) = F (αΛ), α α ∑ T (n)F (Λ) := F (T (n)Λ) = F (Λ′), [Λ:Λ′]=n and by the natural extension we deﬁne these operations for every element of XR. We have

−k Kα(T (n)F ) = T (n)(KαF ) = α T (n)F.

So we obtain a meromorphic function T (n)F of weight k again.

Proposition 4.2. For a meromorphic function F of weight k on XR we have { T (m)T (n)F = T (mn)F ((m, n) = 1), T (p)T (pn)F = T (pn+1)F + p1−kT (pn−1)F (p : prime)

[proof]. This is only a paraphrase of Proposition 4.1 .

4.2 Hecke operator T (n)

Lemma 4.1. Let Λ =< ω1, ω2 > be a lattice in C. Set ( ) a b S = { ∈ M (Z): ad = n, a ≥ 1, 0 ≤ b < d}. n 0 d 2 ∈ 〈 ′ ′ 〉 〈 〉 → { ′ ⊂ For σ Sn, we set Λσ = ω1, ω2 = aω1 + bω2, dω2 . Then the map σ : Sn Λ(n) = Λ Λ : [Λ : ′ Λ ] = n} deﬁned by σ(Λ) = Λσ is bijective.

35 [proof]. By Remark 4.1 we know [Λ, Λσ] = n. If we consider the right cosets Γσ1, Γσ2 for σ1, σ2 ∈ Sn, the injectivity follows. So let us suppose a sublattice Λ′ with [Λ : Λ′] = n. Set

′ ′ Y1 = Λ/(Λ + Zω2),Y2 = Zω2/(Λ ∩ Zω2).

These are cyclic groups generated by the mages of ω1, ω2, respectively. Set a and d be thier orders, respectively. By the exact sequence

′ 0 → Y2 → Λ/Λ → Y1 → 0 ∈ ′ ′ ∈ ′ we have ad = n. We have dω2 Λ . So, put ω2 = dω2. On the other hand we have aω1 Λ + Zω2. So ′ ∈ ′ ′ ′ we can ﬁnd ω1 = aω1 + bω2 Λ . Considering the indices, the lattice < ω1, ω2 > should coincides with Λ′. q.e.d.

Let F be a meromorphic function of weight k on XR.Then

f(z) := F (Λ(1, z)) is a meromorphic modular form of weight k for Γ, and vice versa. For it, note that −1 −1 −1 f( ) = F (Λ(1, )) = F ( )Λ(−z, 1)) = (−z)kF (Λ(1, z)) = (−z)kf(z). z z z Deﬁnition 4.2. For a meromorphic modular form f(z) of weight k we deﬁne

T (n)f(z) := nk−1T (n)F (Λ(1, z)).

So we have ∑ az + b T (n)f(z) = nk−1 d−kf( ). d a≥1,ad=n,0≤b

Proposition 4.3. Let f be a meromorphic modular form of weight k. (1) T (n)f is a meromorphic modular form of weight k also, (2) T (m)T (n)f = T (mn)f ((m, n) = 1), (3) T (p)T (pn)f = T (pn+1)f + pk−1T (pn−1)f.

[proof]. To make clear our argument, we denote T˜(n) the operator T (n) to the functions on XR. (1) is apparent. (2) Using Proposition 4.2 we have

LHS = T (m)(nk−1T˜(n)F (Λ(1, z)) = nk−1mk−1T˜(m)T˜(n)F (Λ(1, z)) = (mn)k−1T˜(mn)F (Λ(1, z)).

By deﬁnition (mn)k−1T˜(mn)F (Λ(1, z)) = T (mn)f(z). So we obtain the assertion. (3) We have

LHS = T (p)((pn)k−1T˜(pn)F (Λ(1, z)) = (pn)k−1pk−1T˜(p)T˜(pn)F (λ(1, z)) = (pn+1)k−1(T˜(pn+1)F (Λ(1, z)) + p1−kT˜(pn−1)F (Λ(1, z)) = (pn+1)k−1T˜(pn+1)F (Λ(1, z)) + (pn)k−1T˜(pn−1)F (Λ(1, z))

On the other hand

RHS = (pn+1)k+1T˜(pn+1)F (Λ(1, z)) + pk−1(pn−1)k−1T˜(pn−1)F (Λ(1, z)) = (pn+1)k+1T˜(pn+1)F (Λ(1, z)) + (pn)k−1T˜(pn−1)F (Λ(1, z)).

So we obtain the assertion. q.e.d.

36 Proposition 4.4. (Explicit description of T (n)) Let f be a meromorphic modular form of weight k. Suppose we have a ”q” expansion ∑ f(z) = c(m)qm. m∈Z

Then we have  ∑  m T (n)f(z) = m∈Z γ(m)q (4.1)  ∑  k−1 mn γ(m) = a|(m,n),a≥1 a c( a2 ).

[proof]. By deﬁnition we have ∑ ∑ T (n)f(z) = nk−1 d−k c(m)e2πim(az+b)/d. ad=n,a≥1,0≤b

Here we have { ∑ d d|m e2πibm/d = 0 otherwise. 0≤b

∑ ′ T (n)f(z) = nk−1 d−k+1c(m′d)qam . ad=n,a≥1,m′∈Z

Re arrange the sum following the power of q: ( ) ( ) ∑ ∑ n k−1 µd = qµ c d a µ∈Z a|(n,µ),a≥1

So we obtain the assertion. q.e.d.

Corollary 4.3.

γ(0) = σk−1(n)c(0), γ(1) = c(n).

Corollary 4.4. For n = p : prime, we have { c(pm)(p, m) = 1 γ(m) = k−1 m | c(pm) + p c( p ) p m.

Corollary 4.5. T (n) is a linear operator on Mk(Γ) and Sk(Γ).

4.3 Hecke eigen form

Deﬁnition 4.3. f ∈ Mk(Γ) is said to be a Hecke eigen form if it is a eigen form for every T (n)(n = 1, 2, ··· ). It is called a normalized Hecke eigen form if its Fourier expansion has the form f(z) = c(0) + q + c(2)q2 + ··· . ∑ n Theorem 4.1. Let f(z) = n≥0 c(n)q be amodular form of weight k (k > 0). Suppose f is an eigen function for every T (n), and put T (n)f = λ(n)f (n = 1, 2,...). Then (a) c(1) ≠ 0, (b) c(n) = λ(n)c(1) (n = 2, 3,...).

37 [proof]. Consider T (n)f = λ(n)c(0) + λ(n)c(1)q + ··· . By Corollary 4.3 to Proposition 4.4we have

T (n)f = σk−1(n)c(0) + c(n)q + ··· . Hence c(n) = λ(n)c(1) (n = 2, 3,...). If c(1) = 0, we have c(n) = 0 (n = 1, 2,...). It means that f is a constant. But a constant function is not a modular form of positive weight. So we obtain the assertion (a)(b) at the same time. q.e.d. Corollary 4.6. If f, g are eigen forms of weight k with same eigen values {λ(n)}, then diﬀers at most with a constant factor. ∑ n Corollary 4.7. Let f = n≥0 c(n)q be a normalized Hecke eigen form. Then we have (1) c(m)c(n) = c(mn) ((m, n) = 1) (2) c(p)c(pn) = c(pn+1) + pk−1c(pn−1) [proof]. Direct consequence of Proposition 4.3. q.e.d.

Set ∞ ∑ c(n) Φ (s) = . f ns n=1 This is convergent on the region {s ∈ C : Re (s) > k}. [Theorem according to Hecke] ∑ (1) For a cusp form f(z) = c(n)qn of weight k, we have

c(n) = O(nk).

(2) For a modular form f(z) that is not a cusp form, we have

c(n) = O(nk−1).

For example see the case f(z) = Ek(z).

[Convergence of Φf (s)] For some positive number A, it holds ¯ ¯ ¯ ¯ ¯ c(n) ¯ A A ¯ ¯ ≤ = . nk+ε nk+ε−(k−1) n1+ε

So Φf (s) converges on Re s > k. Corollary 4.8. For a normalized Hecke eigen form f(z), it holds ∏ 1 Φ (s) = . f 1 − c(p)p−s + pk−1 · p−2s p [proof]. By We have ∑ c(n) ∏ = (1 + c(p)p−s + ··· + c(pm)p−ms + ··· ). ns n≥1 p Putting k−1 2 Φf,p(T ) = 1 − c(p)T + p T , we may show ∞ ∑ 1 c(pn)T n = . Φ (T ) n=0 f,p

38 4.4 Examples

(a)

Proposition 4.5. The Eisenstein series Gk(z)(k ≥ 4, even) is a Hecke eigen form. The normalized eigen form is given by ( ) ( ) ∑∞ Bk Bk n − E (z) = − + σ − (n)q . 2k k 2k k 1 n=1

So T (n) has the eigen value λ(n) = σk−1(n), and the corresponding Dirichlet series is given by ζ(s)ζ(s − k + 1).

[proof]. Because of Proposition 4.3 we may consider only for the case n = p: prime. We can regard

Gk(z) as a function F (Λ(1, z)) on XR. Namely

′ ∑ 1 F (Λ) = F (Λ(1, z)) = G (Λ) = . k γk γ∈Λ

We have ∑ ∑ 1 T˜(p)F (Λ) = . γk [Λ:Λ′]=p γ∈Λ′ In case γ ∈ pΛ it belongs to all p + 1 sublattices with [Λ : Λ′] = p, and in case γ∈ / pΛ it belongs to only one index p sublattice. So ∑ 1 = F (Λ) + p = F (Λ) + pF (pΛ) = (1 + p1−k)F (Λ) γk γ∈pΛ

k−1 Because T (p)Gk(z) := p T˜(p)F (Λ(1, z)), we have

T (p)Gk(z) = σk−1(p)Gk(z).

For the Dirichlet series, we have     ∑∞ ∑ k−1 ∑ ∑ σ − (n) a 1 1 k 1 = =     = ζ(s)ζ(s − k + 1). ns asds ds as+1−k n=1 a,d≥1 d≥1 a≥1

5 Theta functions

Deﬁnition 5.1. Let a, b ∈ {0, 1}. The inﬁnte series [ ] ∑ a a a b ϑ (τ) = exp[πi(n + )2τ + 2πi(n + ) ], a, b ∈ {0, 1} b 2 2 2 n∈Z

converges on[H] as a function of τ, and it becomes to be holomorphic there. It is called a Jacobi theta constant. 1 Note that ϑ (τ) becomes to be constant zero, and others are non constant function. 1

Putq ˜ = exp[πiτ]. We can rewrite the deﬁnition in the form of Fourier series: [ ] ∑ [ ] ∑ [ ] ∑ 0 n2 0 n n2 1 (n+ 1 )2 ϑ (τ) = q˜ , ϑ (τ) = (−1) q˜ , ϑ (τ) = q˜ 2 . 0 1 0 n∈Z n∈Z n∈Z

39 Theorem 5.1. (Jacobi’s identity)

[ ]4 [ ]4 [ ]4 0 0 1 ϑ (τ) = ϑ (τ) + ϑ (τ). 0 1 0

The Jacobi theta constants have the following automorphic behavior:

Theorem 5.2. [ ] [ ] [ ] [ ] [ ] [ ] 0 0 0 0 1 πi 1 ϑ (τ + 1) = ϑ (τ), ϑ (τ + 1) = ϑ (τ), ϑ (τ + 1) = exp[ ]ϑ (τ), 0 1 1 0 0 4 0 [ ] [ ] 0 −1 −πi √ 0 ϑ ( ) = exp[ ] τϑ (τ), 0 τ 4 0 [ ] [ ] 0 −1 −πi √ 1 ϑ ( ) = exp[ ] τϑ (τ), 1 τ 4 0 [ ] [ ] 1 −1 −πi √ 0 √ ϑ ( ) = exp[ ] τϑ (τ)( τ ∈ H). 0 τ 4 1

We note here the above inversion formula is equivalent to the reﬂection formula of the Riemann zeta function via ”Mellin transformation”. By this theorem we have

[ ]4 a Proposition 5.1. ϑ (τ) ∈ M (Γ(2)), (a, b ∈ {0, 1}). b 2 For the Legendre normal form y2 = x(x − 1)(x − λ) of a complex torus T (τ) = C/(Z + Zτ), by the same argument as for j(τ) we have

Theorem 5.3. [ ]4 1 ϑ (τ) 0 λ(τ) = [ ]4 . 0 ϑ (τ) 0

According to this theorem we can obtain:

Theorem 5.4. [ ]4 [ ]4 0 1 M(Γ(2)) = C[ϑ (τ), ϑ (τ)]. 0 0

You can see precise descriptions of the modular function for some other congruence subgroups in [5] and [4](Klein).

6 An example: how it works the theory applied to number theoretic problems

As an example of application of the theory of modular form, here we show the following classical theorem due to Legendre and Jacobi:

Theorem 6.1. For a positive integer n, let a(n) be the number of integer solutions for

2 2 2 2 x1 + x2 + x3 + x4 = n.

40 We have { 8σ1(n), (n : odd) a(n) = r 24σ1(n0), (n = 2 n0, (2, n0) = 1). Especially every positive integer can be expressed as a sum of 4 squares.

The theorem itself is very classical, but the method is very suggestive. We have various applications with the same philosophy. You can see one enlarged application for the congruence number problem in [3] (Koblitz).

At ﬁrst we note: [Fact 1] [ ] 0 Set Θ(z) = ϑ (2z). Then we have 0 ∑∞ Θ(z)4 = 1 + a(n)qn (q = e2piiz). n=1

We shall work in the space M2(Γ0(4)) = M2(Γ1(4)). Recall [Fact 2] ( ) 1 0 (1) Γ (4) = 〈T, 〉 . (2) dim M (Γ (4)) = 2. 1 4 1 2 1

By using the automorphic property of the Dedekind η function we can show [Fact 3] (1) η8(4z) F (z) = ∈ M (Γ (4)). η4(2z) 2 1 (2) ∞ η8(4z) ∏ ∑ F (z) = = q (1 − q4n)4(1 + q2n)4 = σ (n)qn = q + 4q3 + 6q5 + ··· . η4(2z) 1 n=1 odd n>0 [ ] a By the automorphic property of ϑ (z) ( and the behavior at the cusps) we can show b [Fact 4] 4 Θ(z) ∈ M2(Γ1(4)).

[Hecke operator on Mk(Γ0(N))]

We can deﬁne Hecke operators acting on Mk(Γ0(N)) (k : even), and we have similar properties as

those acting on Mk(Γ). ∑ ∞ n ∈ Proposition 6.1. Let k be an even integer. Let p be a prime number. For f(z) = n=0 a(n)q 2πiz Mk(Γ0(N)) (q = e ), we have { ∞ ∑ a(pn)(p|N)or (p, n) = 1 T (p)f(z) = b(n)qn, b(n) = a(pn) + pk−1a( n ) otherwise n=0 p We obtain the same properties as Prop. 4.3,

Proposition 6.2. Let f be a meromorphic modular form of weight k for Γ0(N). (1) T (m)T (n)f ={T (mn)f ((m, n) = 1), T (p)T (pn)f − pk−1T (pn−1)f ((p, N) = 1), (2) T (pn+1)f = T (p)n+1f (p|N).

41 So we can use those two propositions as a deﬁnition of T (n) for Mk(Γ0(N)). As for Hecke eigen forms we have the paraphrase of Theorem 4.1 : ∑ n ∈ Theorem 6.2. Let f(z) = n≥0 c(n)q Mk(Γ0(N)) . Suppose f is an eigen form for every T (n), and put T (n)f = λ(n)f (n = 1, 2,...). Then (a) c(1) ≠ 0, (b) c(n) = λ(n)c(1) (n = 2, 3,...).

We consider a complementary operator on Mk(Γ0(N)): ( ) a b [γ] : f(z) 7→ (det γ)k/2(cz + d)−kf(γ(z)), γ = ∈ GL (Q). k c d 2

Especially we use ( ) −1 0 −1 [α ] : f(z) 7→ 4k/2(4z)−kf( ), α = . 4 k 4z 4 4 0 [Fact 5]

(1) [α4]2 is a linear transformation of the vector space (of dimension 2) M2(Γ0(4)) of order 2. 4 4 (2) We have [α4]2(Θ (z)) = −Θ (z), and F (z) is not an eigen form of [α4]2. So [α4]2 has diﬀerent eigen values +1 and −1. [Fact 6] (0) T (2)F (z) = 0,T (2)Θ4(z) = Θ4(z) + 16F (z),

(1) For an odd positive integer n,[α4]2T (n) = T (n)[α4]2 and T (2)T (n) = T (n)T (2).

(2) An eigen form for [α4]2 is an eigen form of every T (n) for every odd positive integer n, and for an eigen form for T (2), also. By Fact 6, we have three eigenforms F, Θ4 +16F, Θ4 of T (n) for every odd positive integer n. That means

T (n) acts on M2(Γ0(4)) as a constant map. [Conclusion] ∑ Apply Theorem 6.2 to F (z) = 1 + σ (n)qn. Then we know T (n) is a constant map λ(n)id = odd n>0 1 ∑ 4 ∞ n σ1(n)id. Again apply Theorem 6.2 to Θ (z) = 1 + n=1 a(n)q . Then we obtain

a(n) = a(1)λ(n) = 8σ1(n), n : odd.

To obtain the result for even n, we need some more detailed argument, but the method is quite the same.

[References]

1. D. Mumford, Tata Lectures on Theta I, (II, III,) Birkh¨auser.

2. J.P. Serre, Cours d’arithmetiques, PUF.

3. N. Koblitz, Introduction to Elliptic Curves and Modular Forms, Springer GTM.

4. F. Klein, Gesammelte Mathematische Abhandlungen, III, Springer 1923.

5. F. Klein and R. Fricke, Vorlesungen ¨uber die Theorie der Elliptischen Modulfunctionen, Zweiter Band, Especially ”F¨unfterAbschnitt, Viertes Kapitel”, Teubner, 1892.

6. Y. Kawada, Theory of modular functions of one variable, Seminary Note, Univ. Tokyo,1964(in Japanese)