Modular Forms and Dirichlet Series

ANDREW OGG

Preface

These are the official notes for a course given at Berkeley during the fall and winter quarters of 1967–68 on Hecke’s theory of modular forms and Dirichlet series. The reader who is conversant with Hecke’s Werke will find nothing new here, except I have taken the liberty of including a recent paper of Weil, which stimulated my interest in this field. The prerequisites for reading these notes are the theory of analytic functions of one complex variable and some number theory. Unattributed theorems are generally due to Hecke. A. P. Ogg Berkeley, California March, 1968

TEX edition. This is a re-issue of Ogg’s book [8] published in 1969, typeset with TEX. In particular, the numbering system (theorems, propositions) diﬀers from [8]. Marked text in [8] is emphasized here and included in the index. Detected typos in [8] have been corrected; they are listed at the end.

c 2018 TEX version Berndt E. Schwerdtfeger, v1.0, 29th August 2018

Contents

Preface 1

TEX edition 1 Introduction2 1. Dirichlet series with functional equation4 2. Hecke operators for the full modular group 23 3. The Petersson inner product 29 4. Congruence subgroups of the modular group 34 5. A theorem of Weil 50 6. Quadratic forms 56 Corrected typos in Ogg’s book 66 References 66 Index 68

2010 Mathematics Subject Classiﬁcation. Primary 11F11; Secondary 11F25, 11F66. Key words and phrases. modular forms, Dirichlet series. 1 2 ANDREW OGG

Introduction

The simplest and most famous series is the Riemann zeta-function ζ(s), defined for Re(s) > 1 by ∞ X Y ζ(s) = n−s = (1 − p−s)−1, n=1 p the product being over all primes p; the equality of the two expressions is just an analytic statement of the fundamental theorem of arithmetic. Riemann [10, VII.] proved in 1859 that ζ(s) has an analytic continuation to the whole s-plane except for a simple pole of residue 1 at s = 1 and satisfies the functional equation: s Z(s) = π−s/2Γ( )ζ(s) 2 is invariant under s 7→ 1 − s. In fact, this functional equation almost character- izes ζ(s), for Hamburger [4] showed in 1921 that any Dirichlet series satisfying this functional equation and suitable regularity conditions is necessarily a constant multiple of ζ(s). However, the situation did not become clear until greatly generalized by Hecke in his paper “Über die Bestimmung Dirichletscher Reihen durch ihre Funktionalgleichung”, published in 1936 [5, 33.]. Let us sketch that proof of the functional equation for ζ(s) which leads naturally to Hecke’s generalization. Starting from Z ∞ Γ(s) = ts−1e−tdt (Re (s) > 0) 0 we find ∞ X Z ∞ π−sΓ(s)ζ(2s) = (πn2)−sts−1e−tdt (Re (s) > 1) n=1 0 ∞ Z ∞ X 2 = ts−1e−πn tdt n=1 0 Z ∞ 1 = ts−1(ϑ(it) − )dt 0 2 where ∞ 1 X 2 ϑ(τ) = e+πin τ (Im τ > 0) 2 −∞ ∞ 1 X 2 = + eπin τ 2 n=1 is the basic theta-function. Now ϑ(τ) is holomorphic on the upper half plane, and satisfies ϑ(τ + 2) = ϑ(τ) τ ϑ(−1/τ) = ( )1/2ϑ(τ), i where the square root is defined on Re (z) > 0 to be real on the real axis. These 1 two equations say that ϑ(τ) is a modular form of dimension − 2 for the group G(2) generated by τ 7→ τ + 2, τ 7→ −1/τ, and ϑ is up to a constant multiple the only solution of these equations. (These facts have been known for ages, and will be MODULAR FORMS AND DIRICHLET SERIES 3 proved later in these notes.) The functional equation for ζ(s) is now a consequence of that for ϑ(τ):

Z ∞ s Z 1 −s s−1 1 1 t 1 s−1 π Γ(s)ζ(2s) = t (ϑ(it) − )dt − |0 + t ϑ(it)dt 1 2 2 s 0 Z ∞ 1 1 Z ∞ i = ts−1(ϑ(it) − )dt − + t−s−1ϑ( )dt 1 2 2s 1 t Z ∞ 1 1 1 = (ts−1 + t1/2−s−1)(ϑ(it) − )dt − − , 1 2 2s 1 − 2s 1 1 visibly invariant under s 7→ 2 −s; furthermore, the integral is entire, since ϑ(it)− 2 = O(e−ct), for some c > 0. On the other hand, by Mellin inversion we have 1 1 Z ϑ(ix) − = x−s(π−sΓ(s)ζ(2s))ds 2 2πi Re (s)=c for suﬃciently large c > 0, and by similar reasoning (carried out in detail in a more general situation) the functional equation for ϑ(τ) can be derived from that for ζ(s); this is Hecke’s proof that ζ(s) is determined by its functional equation. The above proof generalizes directly, as follows. Given a sequence of complex c numbers a0, a1, a2, . . . , an = O(n ) for some c > 0, and given λ > 0, k > 0, C = ±1, form ∞ ∞ X 2π X ϕ(s) = a n−s Φ(s) = ( )−sΓ(s)ϕ(s) f(τ) = a e2πinτ/λ n λ n n=1 n=0 (the O-condition ensures that ϕ(s) converges somewhere, and f(s) is holomorphic in the upper half plane.) Theorem. The following two conditions are equivalent:

a0 Ca0 (A) Φ(s) + s + k−s is entire and bounded in every vertical strip (henceforth abbreviated to EBV ) and satisﬁes Φ(k − s) = CΦ(s); τ k (B) f(−1/τ) = C( i ) f(τ).

τ τ k k log i (( i ) = e , where log is real on the real axis.)

1 Note for ϕ(s) = ζ(2s), f(τ) = ϑ(τ), we have C = 1, λ = 2, k = 2 . Generally, let G(λ) be the group of substitutions of the upper half plane generated by τ 7→ τ + λ, τ 7→ −1/τ.A modular form of dimension −k and multiplier C for G(λ) is a holomorphic function f(τ) on the upper half plane satisfying

(1) f(τ + λ) = f(τ) τ k (2) f(−1/τ) = C( i ) f(τ) (3) the expansion of f(τ) in a Laurent series in e2πiτ/λ (from (1)) has no neg- P∞ 2πinτ/λ ative terms: f(τ) = n=0 ane , i.e. f is “holomorphic at ∞”.

We denote the space of such f by M(λ, k, C). We also denote by M0(λ, k, C) the subspace of those f which satisfy the additional condition that the Fourier c coeﬃcients an satisfy an = O(n ) for some c > 0. The theorem then says there is a one-one correspondence between the elements of M0(λ, k, C) and Dirichlet series satisfying (A); note that ϕ(s) is regular at s = k if and only if a0 = 0, i.e. f(τ) “vanishes at ∞”. We say ϕ(s) has signature (λ, k, C) if (A) holds. 4 ANDREW OGG

Remark. If ϕ(s) = ζ(K, s) is the zeta-function of an algebraic number ﬁeld K, its functional equation is that s |d|s/2((2π)−sΓ(s))r2 (π−s/2Γ( ))r1 ϕ(s) 2 is invariant under s 7→ 1 − s, where d is the discriminant of K and r1 resp. r2 is the number of real resp. complex primes of K. Note this falls within the scope of the theorem only when there is only one Γ-function, i.e. K is rational or imaginary quadratic. If K is imaginary quadratic, then ϕ(s) has signature (λ, k, C) = (p|d|, 1, 1); it turns out that ϕ(s) is determined by its signature when d = −3, −4 but not for d < −4.

The other part of Hecke’s theory concerns the question of whether ϕ(s) has an Q −s Euler product, i.e. ϕ(s) = p ϕp(s), where ϕp(s) is a power series in p . Suppose for concreteness that λ = 1, so G(λ) = Γ is the modular group. It turns out that M(1, k, C) = M0(1, k, C), and this space is 0 unless k is an even integer ≥ 4, and C = ik, and the only possible Euler product for ϕ(s), of signature (1, k, ik), is ∞ X −s Y −s k−1−2s −1 ϕ(s) = ann = (1 − app + p ) , n=1 p and ϕ(s) has this Euler product if and only if the associated modular form f is an eigenfunction for a certain ring of operators on M(1, k, ik), the Hecke operators. The question of the existence of Euler products is of course fundamental for number theory, since in practice the numbers an will be the number of solutions of some number-theoretic problem and the knowledge of an Euler product reduces knowledge of all the an to knowledge of the ap for primes p.

1. Dirichlet series with functional equation

c All that we need about Dirichlet series is that if an = O(n ), then ϕ(s) = P∞ −s n=1 ann converges absolutely and uniformly in Re (s) ≥ c + 1 + ε, since it is P∞ −1−ε dominated term-by-term uniformly by n=1 n < ∞, and hence ϕ(s) deﬁnes a holomorphic function in (at least) the half plane Re (s) > c + 1. Conversely, if ϕ(s) σ0 −s converges at s0 = σ0 + it, we see an = O(n ) since the general term ann tends P∞ −s c to 0. Thus ϕ(s) = n=1 ann converges somewhere if and only if an = O(n ). As to the gamma-function, we need:

R ∞ s−1 −t (a) Γ(s) = 0 t e dt, for Re (s) > 0 1 √ (b) Γ(s + 1) = sΓ(s), Γ(1) = 1, Γ( 2 ) = π. (c) Γ(s) is never 0, and it is entire except for simple poles at s = −n of residue (−1)n n! , n = 0, 1, 2,... . (d) Stirling’s formula√ (cf., e.g., Ahlfors [1, chap 5, 2.5, (38), p.204]) s− 1 −s+µ(s) (i) Γ(s) = 2πs 2 e , where µ(s) → 0 as |s| → ∞, uniformly in a half plane√ σ ≥ σ0 > 0, where s = σ + it, σ− 1 − π |t| (ii) Γ(s) ∼ 2πt 2 e 2 , as t → ∞, uniformly in σ1 ≤ σ ≤ σ2. (This follows from (di) when σ1 > 0, and then in general from (b).) −x 1 R −s (e) Mellin inversion formula. e = 2πi σ=c>0 x Γ(s)ds, for x > 0, the integral taken upwards on a vertical line. In fact, by the calculus of residues, the right side is ∞ ∞ X X (−x)n Res x−sΓ(s) = = e−x s=−n n! n=0 n=0 MODULAR FORMS AND DIRICHLET SERIES 5

This being said, let us begin the basic theorem. Given a0, a1, a2,... with an = c P∞ −s P∞ 2πinτ/λ O(n ), we form ϕ(s) = n=1 ann , f(τ) = n=0 ane for some ﬁxed λ > 0. We have already noted that the growth condition on the an means that ϕ(s) converges somewhere; for f(τ) it means that f(x + iy) grows only as a power of y as y → 0, a condition on the growth of f(τ) as τ approaches the real axis, the boundary of the upper half plane. More precisely: P∞ 2πinτ/λ Proposition 1.1. Given f(τ) = n=0 ane , with the series converging in the upper half plan.

c −c−1 (a) If an = O(n ), then f(x + iy) = O(y ) as y → 0, uniformly in all real x. −c c (b) If f(x + iy) = O(y ) as y → 0, uniformly in x, then an = O(n ). √ x− 1 −x Proof. By Stirling’s formula Γ(x) ∼ 2πx 2 e , we see that −c − 1 (c + 1) ··· (c + n) Γ(c + n + 1) (−1)n = = ∼ (const.)nc, n n! Γ(c + 1)Γ(n + 1) c so if an = O(n ), then f(x + iy) is dominated term-by-term by ∞ X −c − 1 (−1)n e−2πyn/λ = (1 − e−2πy/λ)−c−1 = O(y−c−1). n n=0 Conversely, if |f(x + iy)| ≤ By−c, then Z 1 i −2πin(x+ i )/λ c 2π/λ |an| = | f(x + )e n | ≤ Bn e 0 n

For future convenience, we state our theorem for two functions instead of one. Thus we are given two sequences a0, a1, a2,... and b0, b1, b2,... of complex num- c bers, an, bn = O(n ) for some c > 0, and λ > 0, k > 0, C 6= 0.(C need not be real.) We form ∞ ∞ X 2π X ϕ(s) = a n−s Φ(s) = ( )−sΓ(s)ϕ(s) f(τ) = a e2πinτ/λ n λ n n=1 n=0 ∞ ∞ X 2π X ψ(s) = b n−s Ψ(s) = ( )−sΓ(s)ψ(s) g(τ) = b e2πinτ/λ n λ n n=1 n=0 (ϕ, ψ are analytic in some right half plane; f, g are analytic in the upper half plane, with the boundary growth condition of Proposition 1.1.) Theorem 1.2. The following two conditions are equivalent:

a0 Cb0 (A) Φ(s) + s + k−s is EBV and Φ(s) = CΨ(k − s); τ −k (B) f(τ) = C( i ) g(−1/τ).

Proof. ∞ X Z ∞ 2πn Φ(s) = a ( )−sts−1e−tdt n λ n=1 0 ∞ Z ∞ X s−1 −2πnτ/λ = ant e dt n=1 0 Z ∞ s−1 = t (f(it) − a0)dt, 0 6 ANDREW OGG for Re (s) suﬃciently large, the interchange of integral and summation being jus- tiﬁed by absolute convergence. The integral is improper at both ends, but since −ct R ∞ s−1 f(it) − a0 = O(e ) as t → ∞ for some c > 0, we see 0 t (f(it) − a0)dt converges uniformly on vertical strips, and so is EBV. Now assume (B). Then Z 1 s Z ∞ s−1 t 1 1−s i dt t (f(it) − a0)dt = −a0 |0 + t f( ) 2 = 0 s 1 t t Z ∞ −a0 k−s−1 Cb0 = + C t (g(it) − b0)dt − s 1 k − s Thus Z ∞ a0 Cb0 s−1 k−s−1 Φ(s) + + = t (f(it) − a0) + t C(g(it) − b0) dt s k − s 1 is EBV, with Φ(s) = CΨ(k − s), which is (A). −x 1 R −s Similarly, by Mellin inversion, or directly from e = 2πi Re(s)=c>0 x Γ(s)ds, we have Z 1 −s f(ix) − a0 = x Φ(s)ds, 2πi σ=c for x > 0, where σ = Re (s), and c is chosen large enough to be in the domain of absolute convergence of ϕ(s). Assuming now (A), we can push the line of integration −k to the left, past 0, picking up residues of Cb0x at s = k and −a0 at s = 0. Then Z −k 1 −s f(ix) − Cb0x = x Φ(s)ds = 2πi σ=c<0 C Z = x−sΨ(k − s)ds = 2πi σ=c<0 C Z = x−(k−s)Ψ(s)ds = 2πi σ=c>k i = Cx−k(g( ) − b ) or x 0 i f(ix) = Cx−kg( ), x which is (B)

Theorem 1.2 was a great step forward, 75 years after the functional equation for the zeta-function, for it reduces a question about Dirichlet series to one about modular forms, which are easier to work with. Taking for some time hereafter an = bn, f = g, ϕ = ψ, etc., and hence C = ±1, we now have the problem of ﬁnding the ϕ(s) of signature (λ, k, C), i.e. the f(τ) ∈ M(λ, k, C). We consider the domain B(λ): Re (τ) ≤ λ/2, |τ| ≥ 1, which will turn out to be a fundamental domain for G(λ) in certain cases, and has diﬀerent topological character as λ > 2, λ = 2, or λ < 2. Proposition 1.3. Every τ in the upper half plane is a translate under G(λ) of a point in B(λ).

∗ Proof. Let G (λ) be the group generated by the three reﬂections T1, T2, T3:

2 (1) T1 = reflection in unit circle, i.e. T1(τ) = τ/|τ| (2) T2 = reflection in y-axis, i.e. T2(τ) = −τ (3) T3 = reflection in x = −λ/2, i.e. T3(τ) = −(τ + λ). MODULAR FORMS AND DIRICHLET SERIES 7

2 ∗ Then Tj = 1, T1T2(τ) = −1/τ, T1T3(τ) = −1/τ + λ, T2T3(τ) = τ + λ, so G (λ) ⊃ ∗ G(λ), and in fact G(λ) consists of all words of even length in T1,T2,T3. Let B (λ): |τ| ≥ 1, −λ/2 ≤ Re (τ) ≤ 0 be the left half of B(λ); the proposition is equivalent with showing that the G∗(λ)-translates of B∗(λ) cover the upper half plane. Given τ = x + iy, we can assume −λ/2 ≤ x ≤ 0; if |τ| ≥ 1 we are done, so 0 2 0 2 assume |τ| < 1. Then τ = T1(τ) = τ/|τ| is higher, y = y/|τ| , and so we have only to verify that we arrive in B∗(λ) after finitely many steps. If λ < 2 or λ > 2, we can cover the arc at the bottom of B∗(λ), i.e. the unit circle, and the upper half plane, by reflecting in the sides of B∗(λ), and hence assume |τ| ≤ c < 1, and 0 aτ+b the result is clear. If λ = 2, then an element of G(2) is a substitution τ = cτ+d a b with ∈ SL(2, Z), y0 = y/|cτ + d|2; if τ 0 is higher than τ, then |cτ + d|2 < 1, c d and there are only finitely many possibilities for c, d, given τ, since c, d ∈ Z. If τ 0 is 0 λ 0 G(2)-equivalent to τ and maximally high, and |Re (τ )| ≤ 2 , then τ ∈ B(λ).

We now dispose of the least interesting case λ > 2, where M0(λ, k, C) has infinite dimension for every value of k and C. Roughly speaking, the great number of solutions is because we can solve our problem in the upper half plane with arbitrary singularities in the lower half plane, since B(λ) extends into the lower half plane. Given λ > 2, let z = g(τ) map the interior of B∗(λ) one-one conformally on the upper half plane, so that g defines a homeomorphism of B∗(λ) onto the closed upper half plane, normalized by ∞, i, −i 7→ 1, 0, ∞. For the existence of g(τ), one can appeal to a strong form of the Riemann mapping theorem (cf. [11, 14.8 Theorem]), or a generalized Schwarz-Christoffel transformation. By the reflection principle, and Proposition 1.3, we extend g to a function defined on the upper half plane and invariant under G(λ). The only corner of B∗(λ) in the upper half plane π is at τ = i, where there is an angle of 2 , so the extended g is analytic on the upper half plane, single-valued by the monodromy theorem. g is bounded on the upper half plane since g is G(λ)-invariant and g(−i) = ∞. It is clear that g(τ) is one-one near points τ not equivalent to i or ∞. At ∞ we have the Fourier expansion ∞ X 2πinτ/λ g(τ) = 1 + ane n=1 2πiτ/λ with a1 6= 0 since g is one-one near ∞ as a function of e ; g(τ) has a double zero at τ = i and a double pole at τ = −i. To show dim M0(λ, k, C) = ∞ it n suffices to show M0(λ, k, C) 6= 0, since if f 6= 0 is one such function, so is fg for n = 0, 1, 2,... , and they are linearly independent since gn has a pole of order 2n at τ = −i.

Thus we only have to construct one non-zero function f ∈ M0(λ, k, C). We have already noted g is locally one-one except at points equivalent to i or ∞, so g0(τ) 6= 0 except at thoses points. Since g(τ) has a double zero at i, we can deﬁne a function pg(τ) in the upper half plane, since it is simply connected; since 2πiτ/λ p P∞ 2πinτ/λ g(τ) = 1 + a1e + ... , a1 6= 0, we have g(τ) = 1 + bne near ∞, p p p n=1 p so g(τ + λ) = g(τ). Since g(−1/τ) = g(τ), we get g(−1/τ) = ±√ g(τ); the p τ−i g(−1/τ) minus sign is correct, for g(τ) = ·g1(τ), where g1(i) 6= 0, so ±1 = √ = τ+i g(τ) − g1(1/τ) = −1, substituting τ = i. g1(τ) 0 Now let h(τ) = √ g (τ) , analytic and never 0 in the upper half plane and g(τ)(g(τ)−1) at ∞ (g0(τ) has simple zeros at i and at ∞, canceling zeros of the denominator; the 8 ANDREW OGG zero at ∞ is measured in z = e2πiτ/λ.) Then h(τ + λ) = h(τ), and g(−1/τ) = g(τ) 0 0 1 2 τ 2 gives g (τ) = g (−1/τ) · τ 2 , so h(−1/τ) = −τ h(τ) = ( i ) h(τ).

Thus h ∈ M(λ, 2, 1). Now let k > 0. Since h is never 0, we can deﬁne h(τ)k/2 = k log h(τ) k/2 e 2 , analytic in the upper half plane and at ∞; we have h(τ + λ) = k/2 k/2 τ k k/2 ε1h(τ) , h(−1/τ) = ε2( i ) h(τ) for some constants ε1, ε2; evaluating at k/2 τ = ∞, we see ε1 = 1, and at τ = i, we see ε2 = 1. Thus h ∈ M(λ, k, 1). Finally, to obtain the O-condition, consider f(τ) = h(τ)k/2 · (g(τ) − 1)n, where n is an integer > k/2; then f ∈ M(λ, k, 1), and to show that f satisﬁes the

g0(x+iy) O-condition, it suﬃces to show that √ = O(y−1), which is true since g(x+iy)

yg0(x+iy) √ is invariant under G(λ) and bounded in the intersection of B(λ) and the g(x+iy) upper half plane (vanishes at ∞), and so bounded in the upper half plane. Thus √ f ∈ M0(λ, k, 1), and f g ∈ M0(λ, k, −1), which proves:

Theorem 1.4. If λ > 2, then M0(λ, k, C) has inﬁnite dimension for every k > 0, C = ±1.

We consider next the case λ < 2. This time we work in the upper half plane only, so let us change the notation so B(λ), B∗(λ) are the intersections of the previous domains with the upper half plane. Let τ0 be the lower left corner of B(λ), so |τ0| = 1, Re (τ0) = −λ/2, and τ0 is a ﬁxed point of τ 7→ −1/τ + λ, which is in G(λ). λ 1 Let πα be the angle of B(λ) at τ0, i.e. cos πα = 2 , 0 < α < 2 . i is a ﬁxed point of τ 7→ −1/τ; the two halves of the bottom of B(λ) are equivalent under τ 7→ −1/τ, and the two vertical sides under τ 7→ τ + λ.

Let f ∈ M(λ, k, C), f 6= 0. Let N be the number of zeros of f in B(λ), counting multiplicities, except at τ0, i, ∞, with appropriate identiﬁcations, e.g. a zero on a side of B(λ) should be counted on only one of the two sides. Let n0, ni, n∞ be the 2πiτ/λ order of zero of f at τ0, i, ∞, the zero at ∞ measured in z = e .

Lemma 1.5.

ni k 1 (a) N + n∞ + 2 + n0α = 2 ( 2 − α) k 1 (b) dim M(λ, k, C) ≤ 1 + [ 2 ( 2 − α)].

Proof. Let C be a contour enclosing the zeros of f in the interior of B(λ); the λ bottom of C follows the unit circle from τ0 to τ0 + λ, the right side follows x = 2 λ λ to 2 + iT , the top follows y = T to −λ/2 + iT , and the left side follows x = − 2 back down to τ0, except that we must detour around small circular arcs to avoid any zeros on the boundary of B(λ). Then

1 Z N = d log f(τ) 2πi C

ni n0α n0α and the integrals over the arcs about i, τ0, τ0+λ approaches − 2 , − 2 , − 2 , while the integrals over the top approaches −n∞, and the two integrals on the vertical k 1 sides cancel. Thus we want the integral on the bottom to be 2 ( 2 − α). Now MODULAR FORMS AND DIRICHLET SERIES 9

τ k dτ f(−1/τ) = C( i ) f(τ), so d log f(−1/τ) = k τ + d log f(τ), so this last integral is 1 Z τ0+λ d log f(τ) − d log f(−1/τ) 2πi i 1 Z τ0+λ −kdτ = 2πi i τ −k = arg τ|τ0+λ 2π i k π = ( − πα), 2π 2 k 1 as desired. Finally, if m > 1 + [ 2 ( 2 − α)] > n∞, and f1, . . . , fm ∈ M(λ, k, C), then a suitable non-trivial linear combination of f1, . . . , fm has a zero of order ≥ m − 1 > n∞ at ∞, and so vanishes, by (a); this proves (b). √ Remark. The zeta function of Q( −3) is 1 X0 ϕ(s) = (n2 + nm + m2)−s 6 n,m∈Z √ √ 1 and is known to have signature ( 3, 1, 1). Here α = 6 , and dim M( 3, 1, 1) ≤ 1 by the above. Hence ϕ(s) is determined by its functional equation.

At this point we digress brieﬂy for some general considerations. A substitution a b L(τ) = aτ+b , where ∈ SL(2, R) is elliptic if it has two non-real ﬁxed cτ+d c d 2 points τ1, τ 1, where Im τ1 > 0. Since cτ1 + (d − a)τ1 − b = 0, L is elliptic ⇐⇒ (d − a)2 < −4bc ⇐⇒ (d + a)2 < 4 ⇐⇒ the eigenvalues of L, i.e. the roots of x2 − (a + d)x + 1 = 0, are non-real. We now prove a very useful result:

Proposition 1.6. Let L be elliptic, with ﬁxed point τ1 in the upper half plane. Suppose there is a non-zero holomorphic function f(τ) on Im τ > 0 such that

(a) f(τ + λ) = f(τ) cτ+d k (b) f(−1/τ) = ε( i ) f(τ) for some constants λ > 0, k > 0, and ε. Then L is periodic, i.e. its eigenvalues are roots of 1.

Proof. In the variable t = τ−τ1 , L is a (complex) linear fractional transformation τ−τ¯1 ﬁxing 0 and ∞ , i.e. L(t) = ρ · t, and we want ρ to be a root of 1.

k Let g(τ) = (τ − τ 1) , for Im (τ) > 0. Then aτ + b aτ + bk g(L(τ)) = − 1 cτ + d cτ 1 + d cτ + d−k = g(τ)η i

k cτ1+d for some constant η; evaluating at τ1 = L(τ1), we see η = i . Now let P∞ n h(τ) = f(τ)g(τ). Then h(L(τ)) = εηh(τ), and writing h(τ) = n=0 cnt , we P∞ n n P n have n=0 cnρ t = εη cnt . Now cn 6= 0 for two distinct values of n, since n f(τ + λ) = f(τ), and ρ = εη when cn 6= 0; hence ρ is a root of 1. Furthermore, if k n1 cτ1+d n1 is the order of zero of f at τ1, we have the formula ρ = εη = ε i . 10 ANDREW OGG

As an incidental result, if 0 6= f ∈ M(λ, k, C), then applying this last formula ni to L(τ) = −1/τ, τ1 = i, we have clearly ρ = −1, so (−1) = ε = C.

Proposition 1.7. For arbitrary λ, if 0 6= f ∈ M(λ, k, C), and ni is the order of zero of f at i, then C = (−1)ni .

Returning to our development of the case λ < 2: Lemma 1.8. If M(λ, k, C) 6= 0, then α and k are rational.

−1 2 Proof. τ0 is a ﬁxed point of L(τ) = τ+λ , whose eigenvalues are the roots of x − πiα λx + 1, i.e. ρ, ρ, where ρ = τ0 + λ = e . Then α is rational, by Proposition 1.6, as is then k, by Lemma 1.5. 1 Example. λ = 1 (modular group), i.e. α = 3 . If 0 6= f ∈ M(1, k, C), then ni n0 k 1 ni 1 2 + 3 ≡ 2 ( 6 ) (mod 1), and then 2 ≡ 4 (mod 1). Hence k is an even integer, and ni k/2 k 1 k/2 k C = (−1) = (−1) . Also 12 ≥ 3 , so k ≥ 4, and dim M(1, k, (−1) ) ≤ 1+ 12 . 1 Lemma 1.9. If M(λ, k, C) 6= 0, then α = q , where q is an integer ≥ 3 (and hence π λ ≥ 2 cos 3 = 1.)

Proof. Assume α = p/q, where p, q are relatively prime and p ≥ 2; we will then ﬁnd an element of G(λ) which is elliptic but not periodic, contrary to Proposition 1.6. n 1 λ 0 −1 In fact, one computes that = L has trace s = 2 sin(n+1)pπ/q ; 0 1 1 λ sin pπ/q choosing n so (n + 1)p ≡ 1 (mod q), we have |s| < 2, so L is elliptic. But s has conjugates s0 with |s0| > 2, so the eigenvalues of L are imaginary but not roots of 1, which is the desired contradiction. 1 Remark. We shall see there are forms when α = q ; G(λ) is discrete exactly in that π −1 πi/q 0 πp case. Note that λ = 2 cos q = ζ + ζ , ζ = e is conjugate to λ = 2 cos q = ζp + ζ−p for (p, q) = 1, so G(λ) and G(λ0) are isomorphic as abstract groups, but only G(λ) is of interest for modular forms.

So far we have proved that if 0 6= f ∈ M(λ, k, C), λ < 2, then

π (1) λ = 2 cos q , q ∈ Z, q ≥ 3, ni n0 k(q−2) (2) N + n∞ + 2 + q = 4q (3) C = (−1)ni π Lemma 1.10. Given λ = 2 cos q . Then there exist f0.fi, f∞ ∈ M(λ, k, C), for suitable k and C in each case, with simple zeros at τ0, i, ∞, respectively, and no other zeros. Thus:

4 (a) for f0, C = +1 and k = q−2 , the smallest possible value for k; 2q (b) for fi, C = −1, k = q−2 ; 4q (c) for f∞, C = +1, k = q−2 .

Proof. As in the case λ > 2, by the mapping theorem there exists a homeomorphism ∗ λ g of B (λ): − 2 ≤ Re (τ) ≤ 0, |τ| ≥ 1, Im τ > 0 onto the closed upper half plane, mapping the interior of B∗(λ) conformally onto the open upper half plane, normalized by τ0, i, ∞ 7→ 0, 1, ∞. We continue g(τ) to the upper half plane by repeated reﬂections in the sides of B∗(λ), obtaining an analytic function g(τ) on the upper half plane by Proposition 1.3, the simple connectivity of the upper half MODULAR FORMS AND DIRICHLET SERIES 11

∗ π π plane, and the fact that the angles at the corners of B (λ) are q at τ0 and 2 at i, an integral fraction of π, which makes the extended g analytic at the corners. Thus g(τ) is analytic for Im τ > 0, invariant under G(λ), and one-one on the interior of B(λ). The opposite sides of B(λ) are equivalent under τ 7→ τ + λ and τ 7→ −1/τ. Thus we have made G\(λ)\H = G(λ)\(H ∪ {∞}), where H : Im τ > 0, into a ∼ Riemann surface of genus zero, in fact so g : G\(λ)\H −→ Cb = Riemann sphere. B(λ) is a fundamental domain for G(λ). (If λ = 1, g = J is called the elliptic modular invariant.)

Since g is one-one on G(λ)\H, we see g has a zero of order q at τ0, takes the value P∞ n 2πiτ/λ 1 doubly at i, and has a simple pole at ∞, i.e. g(τ) = n=−1 anz , z = e , 0 −2 dz −a−1 2πiz a−1 6= 0. Note g (τ) = −a−1z dτ + ··· = z2 · λ + ... also has a simple pole at ∞. We also have g0(τ + λ) = g0(λ), g0(−1/τ) = τ 2g0(τ). By comparing zeros and poles at τ0, i, ∞, we check the desired f0, fi, f∞ are: (g0)2 1/(q−2) f = 0 g(g − 1) (g0)2q 1/(q−2) f = ∞ g2q−2(g − 1)q (g0)q 1/(q−2) f = i gq−1(g − 1)

It is now easy to compute the dimension of M(λ, k, C). Let f ∈ M(λ, k, +1), k(q−2) f 6= 0. Then, in the above notation, ni is even and m = 4 is an integer, i.e. 4 k = mk0, k0 = q−2 (the minimal value of k, corresponding to f0.) Then for a m m unique constant α0, f − α0f0 vanishes at ∞. Since f0 ∈ M(λ, k, +1) also, we m have f − α0f0 = f∞ · f1, where f1 ∈ M(λ, k − qk0, +1) (if m > q; f1 is constant if m ≤ q). Continuing, we see that [m/q] X m−qν ν f = αν f0 f∞ ν=0 is a polynomial in f0, f∞, in a unique way. Thus dim M(λ, mk0, +1) = 1 + [m/q]. 2q Similarly, if 0 6= f ∈ M(λ, k, −1), then ni is odd and f/fi ∈ M(λ, k − q−2 , +1). 2q 2q−4 Thus k − q−2 = (m − 1)k0, where m is an integer ≥ 1, k = mk0 + q−2 = mk0 + 2; 2 dim M(λ, k, −1) = dim M(λ, (m − 1)k0, +1) = 1 + [(m − 1)/q]. Finally, fi ∈ 4q 2 q M(λ, q−2 , +1), so fi = αf0 + βf∞, with α, β 6= 0 (look at the zeros). Thus f∞, hence any f ∈ M(λ, k, ±1), is a polynomial in f0, fi. Summing up: Theorem 1.11. Given 0 < λ < 2. Then M(λ, k, C) = 0 except in the case π 4m λ = 2 cos q , where q is an integer ≥ 3, and k = q−2 + 1 − C; in this case " # m + C−1 dim M(λ, k, C) = 1 + 2 q

Of course, from the point of view of Theorem 1.2, Theorem 1.11 gives the an- swer to the wrong question. Let us call f ∈ M(λ, k, C) a cusp form (of dimension −k and multiplier C for G(λ)) if it vanishes at ∞, and let S(λ, k, C) be the space of all such cusp forms. Thus S(λ, k, C) ⊂ M(λ, k, C), and dim M(λ, k, C) − dim S(λ, k, C) ≤ 1. 12 ANDREW OGG

π 4m Theorem 1.12. Given λ = 2 cos q , k = q−2 + 1 − C as in Theorem 1.11. Then:

−k/2 (a) S(λ, k, C) ⊂ M0(λ, k, C); in fact f(x + iy) = O(y ) if f ∈ S(λ, k, C) h C−1 i m+ 2 (b) dim S(λ, k, C) = q , the dimension of the space of Dirichlet series of signature (λ, k, C) which are regular at s = k. (c) dim M0(λ, k, C) = δ(λ)+dim S(λ, k, C), where δ(λ) = 0 or 1 is independent of k and C.

Proof. (b) follows from the proof of Theorem 1.11, since we did construct modular forms not vanishing at ∞, so dim M(λ, k, C) = 1 + dim S(λ, k, C). For (a), F (x + iy) = yk/2|f(x+iy)| is bounded on B(λ) (vanishes at ∞) and G(λ)-invariant and so bounded. (c) is equivalent with the following statement: if there exists one function 0 0 f ∈ M0(λ, k, C) with f(∞) 6= 0, then every h ∈ M(λ, k ,C ) also satisﬁes the O- condition. In fact, since k, k0 are rational, we can choose positive integers n, m and a constant α so that αf n + hm ∈ S(λ, k00, +1) and hence satisﬁes the O-condition; then h must also satisfy the O-condition.

The question of whether δ(λ) = 0 or 1 appears to be open in general. For λ = 1, the functions not vanishing at ∞ are provided explicitly by the Eisenstein series, as follows. X0 Lemma 1.13. If k > 2, then |nτ + m|−k converges in Im τ > 0, uni- n,m∈Z formly on compact subsets. (The prime on the summation symbol means the term for (m, n) = (0, 0) is omitted.)

Proof. For τ in a compact set, there exists B > 0 with |xτ + y| ≥ B(|x| + |y|) for all real x, y. Since there are only 4r pairs (n, m) with |n| + |m| = r, our series is dominated term-by-term by ∞ X 4r = 4ζ(k − 1), rk r=1 which is ﬁnite for k > 2.

Hence for k = 4, 6, 8,..., the Eisenstein series 0 X −k Gk(τ) = (nτ + m) n,m∈Z is holomorphic on Im τ > 0, and satisﬁes aτ + b a b G ( ) = (cτ + d)kG (τ) for ∈ SL(2, Z). k cτ + d k c d

(The g2 and g3 of Weierstrass theory are g2 = 60G4, g3 = 140G6.) The Fourier expansion is: Proposition 1.14. ∞ 2(2πi)k X G (τ) = 2ζ(k) + σ (n)zn, k (k − 1)! k−1 n=1 2πiτ P ν where z = e , σν (n) = d . d|n d>0 MODULAR FORMS AND DIRICHLET SERIES 13

P∞ P −k Proof. Clearly Gk(τ) = 2ζ(k) + n=1 m∈Z(m + nτ) . Now

π2 X = (m + τ)−2, sin2 πτ m∈Z for the diﬀerence is entire, of period 1, even, and → 0 as Im τ → ∞. Thus

∞ X (2πi)2e2πiτ (2πi)2z X (m + τ)−2 = = = (2πi)2 nzn. (1 − e2πiτ )2 (1 − z)2 m∈Z n=1

dz Diﬀerentiating with respect to τ ( dτ = 2πiz):

∞ X (−2πi)k X (m + τ)−k = nk−1zn. (k − 1)! m∈Z n=1 Thus ∞ X X (2πi)k X (m + nτ)−k = νk−1znν (k − 1)! n=1 m∈Z n,ν=1

k/2 Thus Gk has signature (1, k, (−1) ) for k = 4, 6,... , and satisﬁes the O- condition, so δ(1) = 1 in the notation of Theorem 1.12. The corresponding Dirichlet series is ∞ X −s ϕk(s) = ζ(s)ζ(s + 1 − k) = σk−1(n)n , n=1 which satisﬁes the functional equation

k/2 Φk(k − s) = (−1) Φk(s)

−s where Φk(s) = (2π) Γ(s)ϕk(s), which of course can also be derived from the functional equation for ζ(s), which we have not yet proved. Note

k/2 −k (−1) a0 = Ress=k Φk(s) = (2π) Γ(k), in agreement with the above, assuming known that Ress=1 ζ(s) = 1.

k/2 Actually, since dim M(1, k, (−1) ) = 1 for k = 4, 6, we see that G4,G6 are the 2πi/3 f0, fi of lemma 1.10 (up to a constant multiple) and in particular τ0 = e is the only zero of G4 in B(1). It follows that the modular group Γ = SL(2, Z)/ ± I, the group of all linear fractional transformations from SL(2, Z), is generated by τ 7→ τ + 1, τ 7→ −1/τ, since G4 is a modular form for Γ but has only one zero in the fundamental domain for the subgroup.

The normalized Eisenstein series are the series ∞ Gk(τ) X E (τ) = = 1 + (−1)k/2A σ (n)zn k 2ζ(k) k k−1 n=1

(2π)k for k = 4, 6,... , where Ak = Γ(k)ζ(k) is a positive real number. Actually Ak is th rational (Ak = 2k/B2k, B2k = k Bernoulli number), as follows. Starting from 14 ANDREW OGG

Q∞ s2 sin s = s n=1(1 − n2π2 ), we get d s cot s = s log sin s ds ∞ X 1 = 1 − 2s2 n2π2(1 − s2/n2π2) n=1 ∞ ∞ ν X X s2 = 1 − 2 n2π2 n=1 ν=1 ∞ X ζ(2ν) = 1 − 2 s2ν π2ν ν=1 which shows Ak is rational since s cot s is a power series in s with rational coefficients. One finds A4 = 240, A6 = 504. Recalling that the first cusp form occurs when k = 12 (since the formula of k/2 k k Theorem 1.12 gives dim S(1, k, (−1) ) = [ 12 ] if k 6≡ 2 (mod 12), [ 12 ] − 1 if k ≡ 2 3 2 E4(τ) −E6(τ) (mod 12)) let us define ∆(τ) = 1728 . Then ∆ ∈ S(1, 12, 1). Writing ∞ ∞ X 1 X ∆(τ) = a zn = (1 + 240 σ (n)zn)3 n 1728 3 n=1 n=1 ∞ X n 2 −(1 − 504 σ5(n)z ) n=1 one finds a1 = (3 · 240 + 2 · 504)/1728 = 1, and all an ∈ Q. Actually, the an are integers; to see this, we need, in an obvious notation, (1 + 240U)3 ≡ (1 − 504V )2 (mod 123), i.e. 3 · 240U ≡ −2 · 504V (mod 123), for which it suffices that U ≡ V (mod 12), i.e. σ3(n) ≡ σ5(n) (mod 12), which is true since d3 ≡ d5 (mod 12). P∞ n Thus ∆ ∈ S(1, 12, 1), ∆(τ) = n=1 anz , where an ∈ Z, a1 = 1. By the k formula for the number of zeros, we see ∆(τ) 6= 0 for Im τ > 0 ( 12 = n∞ = 1); −12 3 2 in the Weierstrass theory of elliptic functions ∆(τ) = (2π) (g2 − 27g3) is the discriminant. Now we take the quotient of two forms of dimension −12 to get the elliptic modular invariant 3 ∞ E4(τ) 1 X j(τ) = = + b zn, ∆(τ) z n n=0 0 where bn ∈ Z, a crucial fact in arithmetic applications; j(τ) = j(τ ) if and only if τ 0 is equivalent to τ under the modular group. Remark. ∆ has the product expansion ∞ Y ∆(τ) = z (1 − zn)24; n=1 cf. Siegel [14] for a short proof. Another proof follows Theorem 1.18 in these notes.

Thus we have shown M(λ, k, C) = M0(λ, k, C) in the case λ = 1 by explicit construction (the Eisenstein series) of forms not vanishing at ∞ which do satisfy the O√-condition.√ We can use this result for λ = 1 to prove the same thing for λ = 2, 3, as follows. Let f ∈ M(1, k, C). If ` = 1, 2, 3,... , then g(τ) = MODULAR FORMS AND DIRICHLET SERIES 15 √ √ √ √ −k/2 f( `τ) + ` f(τ/ `) ∈ M( `, k, C).√ Taking√ ` = 2, 3, so ` < 2, we conclude that M(λ, k, C) = M0(λ, k, C) for λ = 2, 3. Finally we consider the case λ = 2; note G(2) is a subgroup of G(1) = Γ, and B(2) contains three copies of B(1). Since every point of H is G(2)-equivalent to a point of B(2), by Proposition 1.3, and B(1) is a fundamental domain for G(1), we see that (Γ : G(2)) ≤ 3. Defining Γ(2), the principal congruence subgroup of Γ of level 2, by f 0 → Γ(2) −→ Γ−→SL(2, Z/2Z) → 0 where f is reduction modulo 2, one checks f is onto and so (Γ : Γ(2)) = 6. Since clearly (G(2)Γ(2) : Γ(2)) = 2, we conclude G(2) ⊃ Γ(2) and G(2) has index 3 in Γ, and hence that B(2) is a fundamental domain for G(2). Now B(2) has two cusps (points where it meets the boundary of the upper half plane), ∞ and −1; +1 is equivalent to −1 under τ 7→ τ + 2 and so is not counted. Note B(2) has an angle of 0 at each cusp. We make G\(2)\H = H∪{∞, −1} modulo G(2) into a Riemann surface (of genus 0) by assigning local parameters t as follows: (1) t = τ at points not equivalent to i, ∞, −1 τ−i 2 (2) t = τ+i at i (3) t = eπiτ at ∞ (4) t = e−2πi/(τ+1) at −1 The reason for these choices is as follows. Except at the three corners i, ∞, −1, a neighborhood of τ contains no equivalent point, whence (1). At those three points, one computes the stability groups in G(2), which has order 2 (generated by τ 7→ −1/τ) at i, an elliptic fixed point, and infinite cyclic (generated by the least translation τ 7→ τ + 2) at ∞, a parabolic fixed point, or cusp, whence (2), (3). Now −1 is a fixed point of τ 7→ −1/(τ + 2) (in G(2)), and τ 0 = −1/(τ + 1) throws −1 to ∞, changes τ 7→ −1/(τ + 2) into

0 −1 −τ − 2 0 τ 7→ −1 = = τ − 1 τ+2 + 1 τ + 1 0 thus t = e2πiτ = e−2πi/(τ+1) is the appropriate local variable at −1. (Generally speaking, to treat questions of analyticity at any rational cusp, the procedure is to send it to ∞ by a linear fractional transformation and proceed as before.) We now investigate the meaning of the O-condition:

Lemma 1.15. If f ∈ M0(2, k, C), then f is quasi-regular at τ = −1, in the variable t = e−2πi/(τ+1), in the sense that τ + 1 f(τ)( )k = tnh(t) i where h(t) is holomorphic and 6= 0 at t = 0, and n ≥ 0; tn = e−2πin/(τ+1).(n is fractional in general, and is called the order of zero of f(τ) at τ = −1.)

Proof. Let τ 0 = −1/(τ + 1), as above, so τ 7→ −1/(τ + 2) is τ 0 7→ τ 0 − 1. Now τ+2 k f(τ) i = Cf(−1/(τ + 2)), so 0 0 τ + 1 k τ −k τ − 1 −k −1 f(τ) = f(τ) = ε f( ) i i i τ + 2 for some constant ε of absolute value 1. Write ε = e2πiρ, where ρ is real, in general irrational. Then 0 τ −k 0 f(τ) e−2πiρτ i 16 ANDREW OGG is invariant under τ 7→ −1/(τ +2) (i.e. τ 0 7→ τ 0 −1) and so has a Laurent expansion P∞ 2πinτ 0 −∞ ane : ∞ τ + 1 k X 0 f(τ) = a e2πi(n+ρ)τ i n −∞

We now show that an = 0 if n + ρ < 0. We have

0 Z τ0+1 τ + 1k −2πi(n+ρ)τ 0 0 an = f(τ) e dτ 0 i τ0

0 τ+1 k τ 0 −k Take τ = u + ib, 1 ≤ u ≤ 2, b large. The term i = i can be ignored; if f(x + iy) = O(y−c), then τ 0 + 1 f(τ) = f− = O(bc), τ 0 2πb(n+ρ) c so an = O(e b ) for large b, so an = 0 for n + ρ < 0.

Let us denote M1(2, k, C) the subspace of M(2, k, C) consisting of those f which are quasi-regular at τ = −1; thus M0(2, k, C) ⊂ M1(2, k, C).

Remark (1). If k is an integer, f ∈ M1(2, k, C), then the order of zero n−1 of f at −1 satisﬁes e2πin−1 = Cik.

k/2 k/2 Remark (2). Let f ∈ M(1, k, (−1) ) ⊂ M0(2, k, (−1) ). Then n−1 is an integer, τ+1 k by Remark (1). In fact, f(τ) i = Cf(−1/(τ + 1)), so n−1 = n∞, the order of zero at ∞ as previously deﬁned. This is as it should be, since the cusps −1 and ∞ are equivalent under Γ = G(1). Note that if f ∈ M1(2, k, C) and k = p/q is rational, then h = f 12q/∆p is a quotient of elements of M(2, 12p, +1) and is then a meromorphic function on G\(2)\H. (By remark (1), the numerator and denominator are regular, not just quasi-regular, at the cusps.) Thus h has as many zeros as poles on the Riemann surface. Now ∆ has one zero on B(1), hence 3 on B(2), so f has 3p k 12q = 4 zeros on B(2), measured in local parameters on the Riemann surface. This is actually true generally:

Lemma 1.16. The number of zeros of f ∈ M1(2, k, C) in B(2) is n k N + n + i + n = ∞ 2 −1 4

(ni/2 would be called ni in local variables; the others are adjusted. This lemma can be considered as a limiting case, as α → 0, of the similar formula for λ < 2, see lemma 1.5)

Proof. This is proved exactly as the analogous result for λ < 2; we only have to 1 R check that 2πi γ d log f(τ) tends to −n−1/2 as a little arc γ about τ = −1 in B(2) R τ+1 k shrinks to zero. Now γ d log( i ) → 0, so we want 1 Z τ + 1 −n d log f(τ)( )k → −1 , 2πi γ i 2 0 −1 which follows from the substitution τ = τ+1 , which carries Re (τ) = −1 on 0 0 1 Re (τ ) = 0 and |τ| = 1 on Re (τ ) = − 2 . MODULAR FORMS AND DIRICHLET SERIES 17

We also have

C = (−1)ni k dim M (2, k, +1) ≤ 1 + 1 4 k − 2 dim M (2, k, −1) ≤ 1 + 1 4 1 as before. Hence ζ(2s), if known to have signature (2, 2 , 1), is determined by its functional equation, as is the zeta-function of Q(i), 1 X0 ϕ(s) = (n2 + m2)−s, 4 n,m∈Z which has signature (2, 1, 1).

1 P∞ πin2τ 1 Lemma 1.17. The theta-function ϑ(τ) = 2 n=−∞ e belongs to M0(2, 2 , 1), 1 and its only zero in B(2) is one of order 8 at τ = −1.

Proof. ϑ(τ) is clearly holomorphic in Im τ > 0, satisfies ϑ(τ + 2) = ϑ(τ), is holomorphic at ∞, and satisfies the O-condition. We want now to show ϑ(−1/τ) = 1/2 P∞ (τ/i) ϑ(τ). For this we apply the Poisson summation formula: if n=−∞ f(x+n) converges absolutely, uniformly on compact subsets, to a continuously differentiable function F (x), where x is a real variable, then

∞ X 2πinx F (x) = ane n=−∞ is represented by its Fourier series; Z 1 Z ∞ −2πint −2πint an = F (t)e dt = f(t)e dt. 0 −∞

2 Applying this to f(x) = eπiτx , where τ is a parameter with Im τ > 0:

∞ X 2 ϑ(τ, x) = eπiτ(n+x) n=−∞ ∞ Z ∞ X 2 2 2 = e2πinx eπi(τu −2nu+n /τ−n /τ)du n=−∞ −∞ ∞ Z ∞ X 2 2 = e2πinx−πin /τ eπiτ(u−n/τ) du. n=−∞ −∞

We claim the integral is (τ/i)−1/2; it suﬃces to take τ = iy, y > 0: ∞ ∞ Z 2 Z 2 e−πy(u+in/y) du = e−πyn du, −∞ −∞ by Cauchy’s theorem, ∞ Z 2 = (πy)−1/2 e−u du = y−1/2. −∞

1/2 P∞ −πin2/τ+2πinx 1/2 Thus ϑ(τ, x)(τ/i) = −∞ e . Taking x = 0, we get ϑ(τ)(τ/i) = 1/2 ϑ(−1/τ). To ﬁnd n−1 = order of zero of ϑ(τ)((τ + 1)/i) , measured in t = 18 ANDREW OGG e−2πi/(τ+1): 1/2 τ + 1 τ + 1 X 2 ϑ(τ)( )1/2 = eπin (τ+1)+πin i i −1 1 = ϑ( , ) τ + 1 2 X 2 = e−πi(n +n+1/4)/(τ+1) = e−πi/4(τ+1)h(t) = t1/8h(t), where h(0) 6= 0. This proves the lemma 1.17

k We can now prove dim M0(2, k, 1) ≥ 1 + [ 4 ], whence M0(2, k, 1) = M1(2, k, 1), k of dimension 1 + [ 4 ]. Since ϑ(τ) 6= 0 for Im τ > 0, and ϑ(τ) is a power series in z = eπiτ with non-zero constant term, we can deﬁne log ϑ(τ) in Im τ > 0, still a power series in z. We then deﬁne, for k > 0, ϑ2k(τ) = e2k log ϑ(τ), satisfying ϑ2k(τ + 2) = ϑ2k(τ), holomorphic at ∞, and satisfying the O-condition; also τ k ϑ2k(−1/τ) = ϑ2k(τ) · ε i 2k for some constant ε; substituting τ = i we see ε = 1. Thus ϑ ∈ M0(2, k, 1), with its only zero at τ = −1.

The Eisenstein series E4(τ) ∈ M(1, 4, 1) ⊂ M0(2, 4, 1), with its only zero at 2πi/3 2k τ0 = e . Now if f ∈ M1(2, k, 1), then f − α0ϑ vanishes at τ0 for a unique 2k constant α0, so f − α0ϑ = E4 · f1, where f1 ∈ M(2, k − 4, 1).(f1 is constant if k = 4, 0 if k < 4.) Continuing, any such f is uniquely of the form X 2k−4i i αiϑ E4 i≤k/4 k which proves M1(2, k, 1) = M0(2, k, 1), of dimension 1 + [ 4 ]. 8 For C = −1, choose α 6= 0 so that g = ϑ −αE4 vanishes at i. Clearly g does not have a zero at τ = −1, so the formula 1 = k = N+n + ni shows that g has a double 4 √ ∞ 2 zero at τ = i and no other zeros. Then h = g ∈ M(2, 2, −1). If f ∈ M1(2, k, −1), ni f 6= 0, then −1 = (−1) , so f(i) = 0, f = h · f1, where f1 ∈ M1(2, k − 2, −1). k−2 Thus dim M1(2, k, −1) = dim M0(2, k, −1) = 1 + [ 4 ]. Note E4 is a polynomial P 2k−2i i in ϑ and h, so any f ∈ M1(2, k, C) is of form f = αiϑ h . We have proved:

Theorem 1.18. M0(2, k, C) = M1(2, k, C), and k + C − 1 dim M (2, k, C) = 1 + 0 4 Remark. ∞ r X πiτ(n2+···+n2) X πiντ (2ϑ(τ)) = e 1 r = 1 + ar(ν)e , where ar(ν)

n1,...,nr ∈Z ν=1 r r is the number of ways of writing ν as the sum of r squares; (2ϑ(τ)) ∈ M(2, 2 , 1). r P∞ πinτ If one knows an explicit basis for M(2, 2 , 1), one can write 1 + n=1 ar(n)e in terms of this basis (which involves only knowing the ﬁrst few coeﬃcients) and thus get an explicit formula for ar(n) for all n. For example:

P∞ −s P∞ −s P −4 (1) n=1 a2(n)n = 4ζ(Q(i), s) = 4 n=1 n d|n d . Thus a2(n) = P −4 4 d|n d , as is well known. MODULAR FORMS AND DIRICHLET SERIES 19

P∞ −s −s s 2−s (2) n=1 a4(n)n = C · 2 ζ(s)ζ(s − 1)(2 − 2 ) P∞ −s −s s 4−s (3) n=1 a8(n)n = 2 ζ(s)ζ(s − 3)(C1 + C2(2 + 2 )) P∞ −s −s s 6−s (4) n=1 a12(n)n = C12 ζ(s)ζ(s − 5)(2 − 2 ) + C2ϕ(s) where ϕ(s) is associated to p∆(τ).

(The first three formulas are classical.) The general principle has vast applicability— if you know a basis of a space of forms, then you get arithmetic identities. For 2 example, for the modular group Γ, we have E4 = E8, E10 = E4E6, etc. Remark. Now that we have the functional equation for ζ(s), we can give a very Q∞ n 24 natural proof, due to Weil, of the product expansion ∆(τ) = z n=1(1 − z ) , where z = e2πiτ . Taking this product as the definition of ∆(τ), it suffices to show ∆(−1/τ) = τ 12∆(τ), since clearly ∆(τ + 1) = ∆(τ) and so ∆(τ) is the unique cusp form of dimension −12 for Γ. Extracting the 24th root, we have Dedekind’s function ∞ Y η(τ) = eπiτ/24 (1 − zn) = ∆1/24(τ) n=1

τ 1/2 πiτ and it suﬃces to prove η(−1/τ) = i η(τ). Now let f(τ) = 12 − log η(τ) = P∞ −1 nm n,m=1 m z , to which we associate as usual the Dirichlet series ∞ X ϕ(s) = m−1(mn)−s = ζ(s)ζ(s + 1). m,n=1 Thus 1 Z τ −s f(τ) = Φ(s)ds 2πi σ=σ1 i −s where σ1 > 1, and Φ(s) = (2π) Γ(s)ϑ(s). The functional equation for ζ(s) (and identities for the Γ-function) give the functional equation Φ(s) = Φ(−s). Fur- π thermore, Φ(s) is entire except for simple poles at s = ±1 of residue ± 12 and a 1 double pole at s = 0 with Φ(s) + 2s2 regular at 0; excluding a neighborhood of the poles, Φ(s) is bounded on vertical strips. The method of Theorem 1.2 is therefore τ −s applicable–shifting the line of integration to the left of −1, noting that i Φ(s)ds πi 1 τ −πτ has residues 12τ , 2 log i , 12i at s = 1, 0, −1, and applying Φ(s) = Φ(−s), we get πi 1 τ πτ f(τ) = + log − + f(−1/τ) 12τ 2 i 12i 1 τ i.e. log η(−1/τ) = log η(τ) + 2 log i .

Quite similarly, we can derive the product expansion for the theta-function: ∞ X 2 Y ϑ(τ) = zn = (1 − z2n)(1 + z2n−1)2, n∈Z n=1 where z = e2πiτ ; as above, it suﬃces to take the product as a deﬁnition of ϑ(τ) and prove τ 1/2 ϑ(−1/τ) = ϑ(τ) i Here ∞ X −zmn 2(−1)m−1zm(2n−1) log ϑ(τ) = ( + ) m m n,m=1 20 ANDREW OGG is associated to the Dirichlet series ∞ X −(2mn)−s 2(−1)m−1m−s(2n − 1)−s ϕ(s) = + ) m m n,m=1 = 2−sζ(s)ζ(s + 1)(−5 + 2(2s + 2−s)).

1 R τ −2 −s Thus log ϑ(τ) = Γ(s)Φ(s)ds, where σ1 > 1 and Φ(s) = π Γ(s)ϕ(s). 2πi σ=σ1 i Again Φ(s) = Φ(−s); since −5 + 2(2s + 2−s) vanishes at s = ±1, we have this time 1 1 τ Φ(s) − 2s2 is entire. By the same method as before, we get log ϑ(τ) + 2 log i = log ϑ(−1/τ). This completes our general program of determining all solutions to the functional equation of Theorem 1.2, except for the ambiguity on the O-condition of Theorem 1.2. Before going on to the theory of the Euler product, we close this chapter with a theorem on the zeros of a Dirichlet series with functional equation; this theorem, due to Hardy and Hecke, is a good illustration of the technique of passing back and forth between Dirchlet series and associated Fourier series, à la Theorem 1.2. First we need: Deﬁnition 1.1. A function f(s), regular in some domain, has order c if f(s) = c+ε O(e|s| ) as |s| → ∞ in that domain, for all ε > 0. Theorem 1.19 (Phragmen-Lindelöf). Let f(s) be regular and of ﬁnite order in the strip σ1 ≤ σ ≤ σ2, 1 ≤ t, where s = σ + it, and suppose

kj f(σj + it) = O(t ) for j = 1, 2. k(σ) Then f(σ + it) = O(t ), uniformly in σ1 ≤ σ ≤ σ2, where k(σ) is the linear function with k(σj) = kj, j = 1, 2.

tc Proof. Say |f(σ + it)| ≤ Ae . Let us consider ﬁrst the case k1 = k2 = 0, so |f(s)| ≤ B on σ = σ1, σ = σ2, and t = 1. Let m be an interger > c, m ≡ 2 (mod 4). Then Re (sm) = −tm + p(σ, t), where p(σ, t) is a polynomial in σ and t2 of degree < m in t, so |p(σ, t)| ≤ atm−2. Note Re (sm) ≤ −tm + atm−2 ≤ D (some constant).

εsm εD For any ε > 0, let g(s) = e f(s). Then |g(s)| ≤ Be on σ = σ1, σ = σ2, and c m m−2 t = 1, and |g(s)| ≤ Aet +ε(−t +at ) ≤ B for t ≥ T (ε), Hence |g(s)| ≤ BeεD for m all s, by the maximum principle, so |f(s)| ≤ Beε(D+|s| ). Letting ε → 0 (for ﬁxed s), we get |f(s)| ≤ B.

k(s) log s For the general case, let h(s) = e i . Now s Re (k(s) log ) = k(σ) log |s| − at arg(t − iσ) = k(σ) log t + O(1) i k(σ) O(1) f(s) so h(s) = t e . Then h(s) satisﬁes the conditions of the special case, so f(s) = O(h(s)) = O(tk(σ))

Corollary 1.20. Let f(s) be regular and of ﬁnite order in σ1 ≤ σ ≤ σ2, 1 ≤ t, and suppose f(σ + it) = O(tc(σ)) for some constant c(σ), for each σ ∈ [σ1, σ2]. Let µ(σ) be the growth function of f, σ1 ≤ σ ≤ σ2, i.e. µ(σ) is the inﬁmum of the numbers c(σ) so that f(σ + it) = O(tc(σ)). Then µ(σ) is a convex function. MODULAR FORMS AND DIRICHLET SERIES 21

P∞ −s Proposition 1.21. Given ϕ(s) = n=1 ann , a Dirichlet series converging somewhere, and λ > 0, k > 0, C = ±1. Then ϕ(s) has signature (λ, k, C), i.e. satisﬁes the condition (A) of Theorem 1.2, if and only if ϕ(s) satisﬁes the following condition:

(A’) (k − s)ϕ(s) is entire of ﬁnite order, and satiﬁes the functional equation 2π −s Φ(s) = CΦ(k − s), where Φ(s) = ( λ ) Γ(s)ϕ(s).

Proof. Let ϕ(s) have signature (λ, k, C). Now Γ(s) has order 1 in σ ≥ σ0 > 0, 1 sin πs by Stirling’s formula, as does then Γ(s) = π Γ(1 − s) in σ ≤ −σ0 < 0. By 2π −s absolute convergence, ϕ(s) is bounded in σ ≥ σ1 > k, so Φ(s) = ( λ ) Γ(s)ϕ(s) has order 1 there and also in σ ≤ k − σ1 by the functional equation, as does 2π s Φ(s) ϕ(s) = ( λ ) Γ(s) . Only the strip k − σ1 ≤ σ ≤ σ1 remains; ϕ(s) is of order 1 there a0 Ca0 since Φ(s) + s + k−s is bounded. Thus if ϕ(s) satisﬁes (A), then it has order 1. Conversely, if ϕ(s) satisﬁes (A’), then ϕ(σ + it) = O(1) for σ ≥ σ1, so also Φ(σ + it) = O(1) for σ ≥ σ1 and σ ≤ k − σ1 (by the functional equation). Hence Φ(σ + it) = O(1) in any strip a ≤ σ ≤ b, 1 ≤ t, by the Phragmen-Lindelöf theorem, which proves (A). Remark. You can state Proposition 1.21 for two functions, as in Theorem 1.2.(A’) is actually the condition in Hecke’s papers.

P∞ −s Theorem 1.22. Let ϕ(s) = n=1 ann be a non-zero real Dirichlet series (i.e. P∞ 2πinτ/λ an ∈ R) of signature (λ, k, C) and f(τ) = n=0 ane the associated modular form. Fix σ0, 0 < σ0 < k, and let

R(t) = Re Φ(σ0 + it),

I(t) = Im Φ(σ0 + it),

2π −2 iy −β where Φ(s) = ( λ ) Γ(s)ϕ(s). Suppose f(e ) = O(y ) as y → 0 through positive 1 values, for some constant β, 0 ≤ β < σ0 + 2 . Then either R(t) or I(t) changes k k sign inﬁnitely many times. (In particular, if σ0 = 2 , Φ( 2 + it) is real, if C = 1, or purely imaginary, if C = −1, and so ϕ(s) has inﬁnitely many zeros on the middle k k+1 line σ = 2 , provided β < 2 .)

Proof. Since Φ(s) is real on the real axis, 1 R(t) = Φ(σ + it) + Φ(σ + it) 2 0 0 1 = Φ(σ + it) + Φ(σ − it) 2 0 0 1 = Φ(σ + it) + CΦ(k − σ + it); 2 0 0 similarly, 1 I(t) = Φ(σ + it) − CΦ(k − σ + it). 2i 0 0 Note that replacing σ0 by k − σ0 does not aﬀect the statement to be proved, so we may assume k/2 ≤ σ0 < k.

For any σ1, 0 < σ1 < k, we have as usual: τ −k 1 Z τ −s f(τ) − a0 − Ca0 = Φ(s)ds. i 2πi σ=σ1 i 22 ANDREW OGG

iy τ i(y−π/2) Put τ = e , y > 0, so i = e . The right side above is then 1 Z ∞ e−iσ1(y−π/2) Z ∞ −i(σ1+it)(y−π/2) t(y−π/2) e Φ(σ1 + it)dt = e Φ(σ1 + it)dt 2π −∞ 2π −∞ Hence Z ∞ t(y−π/2) −β e Φ(σ1 + it)dt = O(y ) −∞ Since Z ∞ t(y−π/2) e Φ(σ1 + it)dt = O(1), 0 π say for 0 < y ≤ 4 , we get (change t → −t, and conjugate): Z ∞ −t(y−π/2) −β e Φ(σ1 + it)dt = O(y ). 0 Hence G(t) = R(t) or I(t) satisﬁes Z ∞ e−t(y−π/2)G(t)dt = O(y−β). 0 Now assume G(t) has constant sign for t >> 0. Then the integral converges absolutely: Z ∞ e−t(y−π/2)|G(t)|dt = O(y−β). 0 If the theorem is false, this is true for both R(t) and I(t) and hence for

Z σ0+iT Then |sαϕ(s)||ds| = O(T β), so σ0+i

Z σ0+iT |mssαϕ(s)||ds| = O(T β) σ0+i for any natural number m, and so a fortiori

Z σ0+iT mssαϕ(s)ds = O(T β) σ0+i where β < α + 1, which leads to a contradiction, via Phragmen-Lindilöf, as follows.

Choose m so am 6= 0 and let Z s Z(s) = zαmzϕ(z)dz σ0+i for σ > 0, t ≥ 1. Since ϕ(z) is of finite order (of order 1, actually), we see Z(s) β has finite order, Z(σ0 + it) = O(t ). We will now show the growth function µ(σ) MODULAR FORMS AND DIRICHLET SERIES 23 for Z(s) is α + 1 for σ sufficiently large (in the domain of absolute convergence of ϕ(s)), contrary to the convexity of µ(σ). In fact, for large σ: Z σ+it X Z σ+it m Z(σ + it) = a zαdz + a zα( )zdz m n n σ+i n6=m σ+i (σ + it)α+1 = a + O(tα) + O(1) m α + 1 a = m tα+1 + O(tα) α + 1 since Z σ+it α m z σ+it Z σ+it α−1 m z α m z z ( n ) αz ( n ) z ( ) dz = m − m dz σ+i n log n σ+i σ+i log n = O(tαn−σ). Remark. Looking at other lines in the critical strip besides the middle line was suggested by Berlowitz [2].

Examples. (1) If λ < 2, and a0 = 0, i.e. f ∈ S(λ, k, C), then one can take k β = 2 by Theorem 1.12. Hence the theorem applies for any σ0. (2) ϕ(s) = ζ(s)ζ(s+1−k), for k = 4, 6,... corresponds to the Eisenstein series k −1 Gk(τ). Now Gk(τ)(τ − 1) = Gk( τ−1 ) 6= 0, ∞ as τ → 1. Thus we can only 1 take β = k; the theorem applies for k − 2 < σ0 < k.

2. Hecke operators for the full modular group

Let k be a fixed positive integer (unlike the first chapter, where k was only a b required to be real.) If L = is a real matrix with positive determinant, and c d f(τ) is a holomorphic function on the upper half plane, let f|L be the holomorphic function on the upper half plane defined by aτ + b f|L(τ) = (ad − bc)k/2(cτ + d)−kf( ); cτ + d a 0 note f| = f for a > 0. Passing to the corresponding homgeneous function 0 a

−k F (ω1, ω2) = ω2 f(ω1/ω2)

(on the space of two variables (ω1, ω2) with Im (ω1/ω2) > 0) we get ω1 −k aω1 + bω2 (F ◦ L) = (cω1 + dω2) f ω2 cω1 + dω2 −k/2 −k = (ad − bc) ω2 (f|L)(ω1/ω2). Thus f ↔ F induces f|L ↔ (ad − bc)k/2f ◦ L = |L|f ◦ L which proves the rule (f|L)|M = f|(LM) f will be a modular form of dimension −k for a subgroup G of GL+(2, R) if, besides certain regularity conditions, f|L = f for all L ∈ G; note that it suﬃces to check this for a set of generators of G. For the homogeneous modular group SL(2, Z), 24 ANDREW OGG

1 1 0 −1 −1 0 , , will do (the modular group is Γ = SL(2, Z)/ ± I, 0 1 1 0 0 −1 generated by τ 7→ τ + 1, τ 7→ −1/τ); the regularity condition is just that f(τ) is holomorphic at ∞. We let M(Γ, k) be the space of all such, where we now take k to k/2 k/2 be an even integer; M(Γ, k) = M(1, k, (−1) ) = M0(1, k, (−1) ) in the earlier notation. Similarly, S(Γ, k) is the space of f ∈ M(Γ, k) vanishing at ∞, the cusp forms for Γ of dimension −k. We know

M(Γ, k) = CEk ⊕ S(Γ, k). We now give a “geometric” definition of the Hecke operators. Let X be the set of lattices L in C; each L has a basis ω1, ω2 with Im (ω1/ω2) > 0, unique up to the action of SL(2, Z). Let D be the divisor group of X, i.e. the free abelian group on P X, i.e. the set of formal finite sums nL · L = A, with nL ∈ Z; A is positive L∈X P if all nL ≥ 0, written A ≥ 0; the degree of A is deg A = L nL. An abstract correspondence on X of degree n is a “one-to-n mapping” of X into itself, i.e. a homomorphism D → D which carries each positive divisor of degree 1 to a positive divisor of degree n. For n = 1, 2, 3,... we define three types of correspondences:

(1) T (n) associates to a lattice L all sublattices L0 of index n. (2) T (1, n) associates to a lattice L all primitive sublattices of index n, i.e. such that L/L0 is cyclic of order n. (3) T (n, n) associates nL to L. Lemma 2.1. Any T (a, a) commutes with any T (n) and any T (n, n).

Proof. Reading from left to right, say, T (a, a)T (n) associates to a lattice L all L0 of index n in aL; note (L : L0) = a2n. Since L0 ⊂ aL, L0 = aL00 for a unique L00 of index n in L. P nm Theorem 2.2. T (n)T (m) = d|n,m dT (d, d)T ( d2 ); in particular, T (n) is multiplicative, i.e. T (n)T (m) = T (nm) if (n, m) = 1. Hence the T (n) generate a commutative ring of correspondences on X. Remark. The correspondences were known to Hurwitz, but he did not know their commutativity.

Proof. The ﬁrst and key step is the special case n = p is prime, m = ps, so we want: T (p)T (ps) = T (ps+1) + pT (p, p)T (ps−1). 0 0 The left side takes a lattice L to the p + 1 sublattices L1,...,Lp+1 of L of index 0 00 s 0 00 s+1 p, and then takes each Li to the L of index p in Li. Thus L has index p in L, and each L00 certainly occurs; what duplication takes place? If L00 comes from 0 0 Li 6= Lj, then 00 0 0 L ⊂ Li ∩ Lj = pL 00 0 and then L is contained in all (p + 1) of the Li; thus we get the right side. By induction, left to the reader, we get X T (pr)T (ps) = pvT (pv, pv)T (pr+s−2v). v≤r,s Finally, if (n, m) = 1, then T (n)T (m) = T (nm), for if L00 has index nm in L, then 0 00 0 L ⊃ L ⊃ L for a unique L of index n in L; this proves the theorem. MODULAR FORMS AND DIRICHLET SERIES 25

Remark (C.T.C. Wall). A facetious proof that T (n)T (m) = T (m)T (n) for arbitrary m, n is to observe that given L00 of index nm, on the one side we wish to know the number of subgroups of order n of L/L00 and on the other the number of factor groups of order n, and these two numbers are the same, by Pontrjagin duality.

Now let F (ω1, ω2), for Im (ω1/ω2) > 0, be a homogeneous function of dimension −k which is invariant under SL(2, Z). Then F (ω1, ω2) = F (L) may be regarded as a function of the lattice L = Zω1 +Zω2. Deﬁne an operator T (n) on such functions by X F · T (n)(L) = nk−1 F (L0) L0 where L0 runs over the sublattices of index n in L. (The factor nk−1 turns out to be convenient.) Similarly, F · T (d, d)(L) = (d2)k−1F (dL) = dk−2F (L) so the identity of Theorem 2.2 becomes X nm T (n)T (m) = dk−1T ( ) d2 d|n,m operating on F . Another way to do all this is as follows. Let M(n) be the set of all 2 × 2 integer matrices of determinant n, and M ∗(n) the primitive ones. Lemma 2.3. 1 0 [ a b M ∗(n) = Γ0 Γ0 = Γ0 , 0 n 0 d ad=n d>0 b mod d (a,b,d)=1 where Γ0 = SL(2, Z), and the union is disjoint; the index (number of cosets) is Y 1 (M ∗(n):Γ0) = ψ(n) = n (1 + ). p p|n Similarly, [ a 0 [ a b M(n) = Γ0 Γ0 = Γ0 0 d 0 d ad=n ad=n d>0 b mod d a|d d>0

Proof. The decompositions follow from the fact that left multiplications from Γ0 are row operations, right multiplications are column operations. For the index formula, ∗ S 0 note ﬁrst that the index is multiplicative, for if (n, m) = 1, M (n) = Γ αi, ∗ S 0 M (m) = Γ βj. then 1 0 [ M ∗(mn) = Γ0 Γ0 = M ∗(n)M ∗(m) = Γ0α β ; 0 mn i j the union is disjoint, for if −1 0 (αiβj)(αi0 βj0 ) ∈ Γ , 26 ANDREW OGG

−1 1 1 0 then βjβj0 is a matrix with coeﬃcients in n Z and m Z, hence in Z, so j = j , 1 b i = i0. For n = pr, we get pr representatives , pr−1 − pr−2 representatives 0 pr p b pr−1 b pr 0 ,..., p − 1 representatives , and , so 0 pr−1 0 p 0 1 1 ψ(pr) = pr + pr−1 = pr(1 + ). p

Now given a lattice L, and a matrix α of determinant n, αL is a sublattice of index n, primitive if and only if α is a primitive matrix; furthermore, αL depends only on the coset Γ0α, and we get a one-one correspondence between cosets Γ0α −k ω1 and sublattices αL of L. Hence if f ∈ M(Γ, k), and F (ω1, ω2) = ω f( ) is the 2 ω2 corresponding homogeneous function, we deﬁne the nth Hecke operator by

k −1 X f|T (n) = n 2 f|α α S 0 where M(n) = α Γ α, which is the inhomogeneous function corresponding to F · T (n). Then f|T (n)|A = f|T (n) for all A ∈ Γ0, so f|T (n) is still formally a modular form for Γ, of dimension −k; we will verify shortly that f|T (n) is holomorphic at ∞. We still have the same identities: X nm T (n)T (m) = dk−1T ( ) d2 d|n,m on M(Γ, k). Proposition 2.4. Let f ∈ M(Γ, k), say ∞ X 2πiντ f(τ) = aν e . ν=0 Then ∞ X 2πiντ f|T (n)(τ) = aν (n)e , ν=0 where X k−1 aν (n) = d a nν , a0(n) = σk−1(n) · a0. d2 d|ν,n Hence f|T (n) ∈ M(Γ, k), and is a cusp form if f is.

Proof. ∞ X X X nk−1 f|T (n)(τ) = a e2πiaντ/d e2πibν/d ν dk ad=n ν=0 b mod d ∞ X X 2πiaντ k−1 = aνde a ad=n ν=0 ∞ X 2πiµτ X k−1 = e a a µn a2 µ=0 a|µ,n

MODULAR FORMS AND DIRICHLET SERIES 27

P 2πibν/d (We used the fact that the character sum b mod d e is d if d|ν and 0 otherwise.) P∞ −s We now form the formal Dirichlet series D(s) = n=1 T (n)n , whose coef- ﬁcients T (n) are operators on M(Γ, k), a ﬁnite dimensional vector space. The identities among the Hecke operators are most neatly expressed as: Theorem 2.5. ∞ X Y D(s) = T (n)n−s = (I − T (p)p−s + pk−1−2sI)−1 n=1 p where I is the identity matrix.

Proof. In view of the multiplicativity, we have only to check that ∞ X I = T (pν )p−νs(I − T (p)p−s + pk−1−2sI), ν=0 which follows from the previous identities (and vice versa). P∞ 2πiντ Corollary 2.6. Let f ∈ M(Γ, k), f(τ) = ν=0 aν e , and suppose f is an eigenfunction for the Hecke algebra, say f|T (n) = cn · f. Then cn · a1 = an, so normalizing to have a1 = 1, the eigenvalue cn of T (n) is the same as Fourier coeﬃcient an, and the associated Dirichlet series has the Euler product ∞ X −s Y −s k−1−2s −1 ϕ(s) = ann = (1 − app + p ) n=1 p

Proof. By Proposition 2.4, cna1 = a1(n) = an; the rest is clear.

We now prove the converse of the corollary in a strong form–the only possible Euler product for ϕ(s) is when f(τ) is an eigenfunction for the T (n). First we prove a useful preliminary result: P∞ 2πinτ Proposition 2.7. Let f ∈ M(Γ, k), f(τ) = n=0 ane . (1) If f|α ∈ M(Γ, k) for some α ∈ M ∗(n), n > 1, then f = 0. (2) Let p be a ﬁxed prime. (a) If am = 0 for all m with p - m, then f = 0. (b) If apn = 0 for all n, then f = 0.

1 0 1 0 Proof. (1) We may as well take α = , since α ∈ Γ0 Γ0. Then 0 n 0 n 1 0 1 1 1 0 1 0 1 1 n 0 f| = f| , so f = f| = 0 n 0 1 0 n 0 n 0 1 0 1 n 1 n 1 1 0 1 0 f| . Now = L M, where L, M ∈ Γ0, so f| = 0 n 0 n 0 n2 0 n2 τ −k f, i.e. f(τ) = f( n2 ) · n . But then f = 0 (look at the Fourier series). p 1 (2) If a = 0 for p m, then f(τ + 1 ) = f(τ), i.e. f = f| , so f = 0. If m - p 0 p k −1 p 0 all a = 0, then f|T (p) = p 2 f| ∈ M(Γ, k), so f = 0. pn 0 1

28 ANDREW OGG

P∞ −s Now let ϕ(s) = n=1 ann 6= 0 correspond to f ∈ M(Γ, k). We say ϕ(s) has an Euler product relative to the prime p if ∞ X −s X ν −νs ϕ(s) = ( amm )( c(p )p ) p-m ν=0 ν i.e. ampν = amc(p ) for p - m, ν ≥ 0. Note c(1) = 1, by (2a) of Proposition 2.7. Theorem 2.8. ϕ(s) has an Euler product relative to p if and only if f is an eigenfunction for T (p), say f|T (p) = c · f. If so, then the p-factor is necessarily ∞ X c(pν )p−νs = (1 − cp−s + pk−1−2s)−1 ν=0

ν Proof. Assume ﬁrst ϕ(s) has an Euler product relative to p, i.e. ampν = amc(p ) for p - m, ν ≥ 0. Now 1 X τ + ` f|T (p)(τ) = pk−1f(pτ) + f( ) p p ` mod p k−1 X n = p f(pτ) + apnz so k−1 X n p f|T (p)(τ) − c(p)f(τ) = p f(pτ) + (apn − c(p)an)z = a power series in z so is 0, by Proposition 2.7. Conversely, suppose f|T (p) = c · f. Then, for any n ( anp if p - n c · an = an(p) = k−1 anp + p an/p if p|n by Proposition 2.4. Hence ϕ(s)(1 − cp−s + pk−1−2s) = X −s k−1 X 2 −s X −s k−1 X 2 −s ann + p an(np ) − apn(pn) − p an(np ) n n n n X −s = amm p-m

Thus ϕ(s) has an Euler product if and only if the associated f(τ) is an eigenfunction for the Hecke operators. For example, we know the Eisenstein series Gk corresponds to (a constant times) ϕ(s) = ζ(s)ζ(s + 1 − k), which has an Euler product, so Gk is an eigenfunction. Thus the decomposition

M(Γ, k) = C · Gk ⊕ S(Γ, k) is preserved by all Hecke operators, and to show M(Γ, k) has a basis of eigenvectors it suﬃces to show S(Γ, k) does, which is the goal of the next chapter. There we will show S(Γ, k) has an inner product in which the T (n) are Hermitian operators, and so can be simultaneously diagonalized by standard linear algebra. S(Γ, 12) is one-dimensional, generated by ∞ Y n 24 X n ∆(τ) = z (1 − z ) = anz , n=1 n=1 MODULAR FORMS AND DIRICHLET SERIES 29

2πiτ P −s z = e ; hence the corresponding Dirichlet series ϕ(s) = n=1 ann has the Euler product Y −s 11−2s −1 ϕ(s) = (1 − app + p ) , p ﬁrst proved by Mordell [7] in 1917. Ramanujan’s conjecture says that the polynomial 11 2 11/2 1−apt+p t has conjugate roots, i.e. |ap| ≤ 2p . Petersson’s conjecture is that k−1 any eigenvalue ap of T (p) on S(Γ, k) satisﬁes |ap| ≤ 2p 2 , for any prime p.

As a ﬁnal remark, the Eisenstein series Gk is characterized as the eigenfunction not vanishing at ∞. In fact, if f|T (n) = λn · f, for all n, where f is not a cusp form, we write f = a · Gk + g, where g is a cusp form; then (λn − σk−1(n))f = g|T (n) − σk−1(n) · g is a cusp form, so λn = σk−1(n), f = constant times Gk.

3. The Petersson inner product

Although we are mainly interested in the full modular group Γ, it is convenient at this point to develop some machinery relative to a subgroup G of Γ of finite index µ. In particular, we want to make G[\H into a Riemann surface, so that the natural map G[\H → Γd\H is holomorphic; there is very little to do, since the fundamental domain D(G) will be a union of µ copies of the fundamental domain D(Γ). Any element of Γ carries ∞ onto ∞ or a rational number; in this way we get σ, say, inequivalent (under G) points P1 = ∞,P2,...,Pσ, the cusps of G. Let U(τ) = τ + 1, so the stability group Γ(P1) of P1 in Γ is infinite cyclic, generated by e1 U. If e1 is the least positive integer with U ∈ G, then the stability group of P1 e1 2πiτ/e1 in G is infinite cyclic, generated by U . We take z1 = e as local parameter at P1 (it is one-one on G\H near P1). At one of the other cusps Pj = LjP1, where −1 Lj ∈ Γ, the stability group Γ(Pj) is infinite cyclic on Uj = LjULj , G(Pj) is cyclic e −1 j 2πiLj τ/ej on Uj (ej > 0), and zj = e is the appropriate local parameter. Then

σ ej −1 [ [ i D(G) = Uj LjD(Γ) j=1 i=0 is a fundamental domain for G, and we have made G[\H = G\(H ∪ {P1,...,Pσ}) into a Riemann surface, with G[\H −→ Γd\H holomorphic (and ej the ramification index of Pj over ∞, i.e. the number of sheets in the covering which stick together there), except we still have to treat the elliptic fixed points. But this is trivial, for 2πi/3 H −→ Γ\H ramifies only over i and τ0 = e , with stability groups of order 2 and 3, so if P is an elliptic fixed point, we take the same local parameter for G as for Γ. Note if P 7→ i or τ0, then P is an elliptic fixed point if and only if P is unramified over i or τ0. Hence if µi resp. µ0 is the number of elliptic fixed points µ−µi µ−µ0 of order 2 resp. 3, then 2 resp. 3 are the number of ramified points over i resp. τ0. Now the Riemann-Hurwitz formula for the genus p(G) of G[\H is (since Γd\H has genus 0): X 2p(G) − 2 = µ(−2) + (eP − 1) P where P ∈ G[\H, and eP is the ramification index of P over Γd\H. By the above,

X µ − µi X µ − µ0 X (e − 1) = , (e − 1) = 2( ), (e − 1) = µ − σ P 2 P 3 P P 7→i P 7→τ0 P 7→∞ 30 ANDREW OGG

Thus µ − µ µ − µ µ − σ p(G) = 1 − µ + i + 0 + 4 3 2 µ µ µ σ = 1 + − i − 0 − , 12 4 3 2 proving:

Proposition 3.1. If G is a subgroup of Γ of ﬁnite index µ, the genus of G[\H is µ µ µ σ p(G) = 1 + − i − 0 − , 12 4 3 2 where µi resp. µ0 is the number of elliptic ﬁxed points of order 2 resp. 3, and σ is the number of cusps.

To define modular forms, it is convenient to use homogeneous notation. Let G0 be s subgroup of Γ0 = SL(2, Z), of finite index, and G the group of linear fractional −1 0 transformations defined by G0; thus G = G0/ ± I if −I = ∈ G0, and 0 −1 otherwise G = G0. We have already defined a b aτ + b f| (τ) = (ad − bc)k/2(cτ + d)−kf( ); c d cτ + d A modular form for G0 is a holomorphic function f(τ) on H such that

(1) f|L = f for all L ∈ G0, (2) f is holomorphic at the cusps.

P∞ n The meaning of (2) is as follows. At the cusp ∞, we can write f(τ) = n=−∞ anz1 2πiτ/e1 as a Laurent series in z1 = e , and the condition is that an = 0 for n < 0; 0 a0 is called the value of f at ∞. At a cusp Pj = Lj(∞), where Lj ∈ Γ , we throw −1 Pj to ∞ by Lj and proceed as before. More precisely, f|Lj satisﬁes condition −1 0 (1) for the conjugate group Lj G Lj, and condition (2) at Pj is just that f|Lj is holomorphic at ∞; the value of f|Lj at ∞ will be called the value of f at Pj. We denote by M(G0, k) the space of all such f. Actually, if −I ∈ G0 then k must be even (if M(G0, k) 6= 0) and there is no danger in confusing G and G0, so let us write M(G, k) = M(G0, k). f ∈ M(G, k) is a cusp form if f vanishes at all cusps; S(G, k) denotes the space of cusp forms.

If f ∈ M(G, k), then f is not a function on G[\H, but we can still speak of the order of zero of f at P ∈ G[\H, measured in local parameters on the Riemann surface. Proposition 3.2. Let f ∈ M(G, k), f 6= 0. Then the total number of zeros of f, µk µk counting multiplicities, is 12 . Hence dim M(G, k) ≤ 1 + [ 12 ].

12 12 k f [ Proof. f , ∆ ∈ M(G, 12k), so h = ∆k is a meromorphic function on G\H, with kµ poles and hence an equal number of zeros. Remark. This is not a good bound. For k ≥ 2, an exact formula for dim M(G, k) follows from the Riemann-Roch theorem; cf. Gunning [3, Chap. II, § 8, Th. 1]. a b Remark. Take k = 2. Since for ∈ SL(2, R), d( aτ+b ) = dτ , we see c d cτ+d (cτ+d)2 f satisﬁes condition (1) if and only if f(τ)dτ is G-invariant, i.e. can be regarded as a diﬀerential on G[\H. If f ∈ M(G, 2), then f(τ)dτ is certainly holomorphic MODULAR FORMS AND DIRICHLET SERIES 31

P∞ n 2πiτ/e dz on G[\H except at the cusps. At ∞, f(τ) = n=0 anz , z = e ; then z = 2πidτ e dz e , so f(τ)dτ = ( 2πi )a0 z + ... , is holomorphic at ∞ if and only if f vanishes at ∞. A similar consideration holds at the other cusps, so we see that there is an isomorphism of S(G, 2) onto the space of holomorphic diﬀerentials (and hence dim S(G, 2) = p(G).)

The holomorphic diﬀerentials on a compact Riemann surface X have a natural inner product Z (ω, ω0) = ω ∧ ω0 X The Petersson inner product is the natural generalization of this inner product to S(G, k) for arbitrary k. Given holomorphic functions f and g on the upper half plane, consider the double diﬀerential i δ(f, g) = f(τ)g(τ)(Im τ)k−2dτ ∧ dτ = f(x + iy)g(x + iy)yk−2dx ∧ dy 2 a b Proposition 3.3. For any real matrix L = of positive determinant, we c d have δ(f|L, g|L) = δ(f, g) ◦ L

aτ+b Proof. The right side means replace τ by L(τ) = cτ+d throughout and so depends only on the linear fractional transformation L(τ), as is also the case with the left side; hence we can assume det L = 1. Then f(L(τ)) = (cτ + d)kf|L(τ) dL(τ) = (cτ + d)−2dτ Im L(τ) = |cτ + d|−2Im τ and the proposition follows.

In view of Proposition 3.3, if f, g ∈ M(G, k), then Z (f, g) = (f, g)G = δ(f, g) D(G) is well deﬁned, i.e. independent of the choice of fundamental domain D(G), provided the integral converges. Lemma 3.4. (f, g) exists if f or g is a cusp form.

Proof. Clearly the integral converges if we exclude a neighborhood of the cusps. At the cusp ∞, f(τ)g(τ) = O(e−cy), for some c > 0, so the integral is dominated by R ∞ e−cyyk−2dy < ∞. The other cusps go the same way, using Proposition 3.3. y1

The inner product satisﬁes the rules:

(1) (f, g) is linear in f, conjugate linear in g (2) (g, f) = (f, g) (3) (f, f) ≥ 0, and (f, f) = 0 only if f = 0 (4) If H is a subgroup of G of ﬁnite index ν, then (f, g)H = ν(f, g)G (DH contains ν copies of DG). 32 ANDREW OGG

Thus S(G, k) is a finite-dimensional Hilbert space. We wish to show the Hecke operators T (n) are Hermitian on S(G, k), i.e. (f|T (n), g) = (f, g|T (n)). Since the T (n) are polynomials in the T (p) with real coefficients, it suffices to treat the case n = p is prime. Lemma 3.5. There exists a set {α} of common representatives for the left and right cosets of M(p) modulo Γ0 = SL(2, Z), i.e. [ [ M(p) = Γ0α = αΓ0.

Proof. Since M(p) = Γ0αΓ0 is a single double coset, every right coset meets every left coset, which proves the lemma. a b a b 1 0 Lemma 3.6. Let Γ0(n) = ∈ Γ0 : ≡ (mod n) , and c d c d 0 1 Γ(n) the corresponding group of linear fractional transformations. Let γ ∈ M(n). Then Γ(mn) ⊂ γ−1Γ(m)γ.

a b d −c Proof. Say γ = , γ−1 = 1 , α ∈ Γ(mn). Then γαnγ−1 ≡ c d n −b a a b d −c n 0 = (mod mn). c d −b a 0 n

0 −1 a b d −c a b Let = = (ad−bc) . Now M(p) = S Γ0α = S αΓ0, c d −b a c d 0 0 0 k −1 k −1 0 so M(p) = M(p) = S Γ α . Then T (p) = p 2 P α = p 2 P α , so to show (f|T (p), g)Γ = (f, g|T (p))Γ, for f, g ∈ S(Γ, k), or equivalently (f|T (p), g)Γ(p) = 0 (f, g|T (p))Γ(p) (by rule (4) above), it suﬃces to prove (f|α, g)Γ(p) = (f, g|α )Γ(p), since by Lemma 3.6, f|α, g|α0 ∈ S(Γ(p), k). Let D(p) be a fundamental domain for Γ(p), so α−1D(p) is one for α−1Γ(p)α; then f|α and g are forms for both Γ(p) and α−1Γ(p)α, which have the same index. Hence:

Proof. The eigenvalues are real by the theorem. Now any element of S(Γ, k) is a polynomial in E4 and E6, so S(Γ, k) has a basis g1, . . . , gr of elements gi with rational Fourier coeﬃcients. Hence T (n) is represented by a rational matrix, so its characteristic polynomial r Y (j) (x − λn ) = det(xI − T (n)) j=1 (1) has rational coeﬃcients; hence λn = λn is algebraic and all its conjugates are real.

P∞ n 2πiτ If f(τ) = n=1 anz , z = e , is an eigenfunctions for all T (n), then we know a1 6= 0, and if we normalize to have a1 = 1, then the Fourier coeﬃcients are the eigenvalues, i.e. f|T (n) = an · f for all n (Corollary to Theorem 2.5). Proposition 3.9. Let f, g be normalized eigenfunctions. Then either f = g or (f, g) = 0.

P∞ n P∞ n Proof. Say f(τ) = n=−∞ anz , g(τ) = n=−∞ bnz , a1 = b1 = 1. If (f, g) 6= 0, then for all n, ¯ an(f, g) = (f|T (n), g) = (f, g|T (n)) = bn(f, g) ¯ so an = bn = bn. Hence f = g.

It follows that if f1, . . . , fr are a maximal set of normalized eigenfunctions, then they are linearly independent and r ≤ dim S(Γ, k). That actually there exists a basis of eigenfunctions follows from Theorem 3.7 and linear algebra: Lemma 3.10. Let R be a commutative ring of Hermitian operators on a ﬁnite dimensional Hilbert space V . Then V has an orthogonal basis f1, . . . , fn of eigenvectors of R.

Proof. Let S1,...,Sm span R (ﬁnite-dimensional, being a matrix ring). Assume V 6= 0. We show ﬁrst V contains one eigenvector f1 of S1,...,Sm, by induction on m. Let λ1 be an eigenvalue of S1, and V1 = {f ∈ V : S1f = λ1·f} the corresponding eigenspace. Then SjV1 ⊂ V1 since SjS1 = S1Sj, and so V1 contains an eigenvector ⊥ f1 of S2,...,Sm by the induction hypothesis. Then V = (Cf1) ⊕ (Cf1) , where ⊥ (Cf1) = {g ∈ V :(f1, g) = 0} is invariant under S1,...,Sm since the Sj are Hermitian, and then has an orthogonal basis f2, . . . , fn, by induction on n.

Hence S(Γ, k) has an orthogonal basis f1, . . . , fr of normalized eigenfunctions of the Hecke operators. The set {f1, . . . , fr} is the set of all normalized eigenfunctions, by Proposition 3.9. The ring R of Hecke operators is a ring of diagonal r × r matrices, and so has rank ≤ r; actually the rank is r, since the r × ∞ matrix of Fourier coeﬃcients (eigenvalues) of f1, . . . , fr has rank r. We have proved:

Theorem 3.11. S(Γ, k) has a basis f1, . . . , fr of normalized eigenfunctions of the Hecke operators R, and {f1, . . . , fr} is the set of all normalized eigenfunctions of R. R has maximal rank r = dim S(Γ, k).

In terms of Dirichlet series, there exist exactly r = dim S(Γ, k) normalized Dirich- k/2 let series ϕ1(s), . . . , ϕr(s) with Euler product, which have signature (1, k, (−1) ) and are regular at s = k. They have real coeﬃcients and so have inﬁnitely many zeros on the line σ = k/2, by Theorem 1.22. 34 ANDREW OGG

If T (n1),...,T (nr) are a basis of R, their eigenvalues lie in a totally real number ﬁeld F ; then F contains all eigenvalues of all T (n). For example, if dim S(Γ, k) = 2, i.e. 24 ≤ k ≤ 48, k 6= 26, 36, then R is√ spanned by I = T (1) and some T (n), diagonalized over a real ﬁeld F (k) = Q( d), d > 0. When k = 24, Hecke found d = 144169, a prime.

4. Congruence subgroups of the modular group

If N is an integer ≥ 1, the homogeneous principal congruence subgroup of level N of Γ0 = SL(2, Z) is a b a b 1 0 Γ0(N) = ∈ Γ0 : ≡ (mod N) . c d c d 0 1 Proposition 4.1. We have an exact sequence 0 → Γ0(N) −→ Γ0 −→ SL(2, Z/NZ) → 0 0 0 3 Q 1 Hence (Γ :Γ (N)) = N (1 − p2 ). p|N

Proof. We have only to show SL(2, Z) → SL(2, Z/NZ) is onto, since the kernel is Γ0(N) by deﬁnition, i.e. given a, b, c, d ∈ Z with ad − bc ≡ 1 (mod N), we have to adjust a, . . . , d by a multiple of N to get determinant 1. Now ad − bc = 1 + kN, so (c, d, N) = 1, so (c, d + `N) = 1 for some ` and we can assume (c, d) = 1. Thus we

a + eN b + fN 1−ad+bc want = 1, i.e. 1 = ad−bc+(ed−fc)N, i.e. ed−fc = = c d N −k; e and f can be so chosen since (c, d) = 1. The index formula is the same as showing that GL(2, Z/NZ) = Aut((Z/nZ)2) 3 Q 1 has order ϕ(N)N (1 − p2 ). The question is multiplicative, so assume N = p|N pr is a prime power. The order is then the number of ways choosing a basis of 2 1 Z/NZ ⊕ Z/NZ; if a, b is a ﬁxed basis, there are N (1 − p2 ) choices for σ(a) (σ is an automorphism), and, given σ(a), N · ϕ(N) choices for σ(b), for this is the order of the union of the cosets of {σ(a)} which are primitive in Z/NZ. The principal congruence subgroup of Γ of level N is the group of linear fractional transformations determined by Γ0(N). Γ(N) = Γ0(N) if N > 2. Γ(N) is a normal subgroup of Γ of index  1 if N = 1  (Γ : Γ(N)) = 6 if N = 2 1 3 Q 1  2 N (1 − p2 ) otherwise.  p|N A congruence subgroup of level N is an intermediate group G, Γ(N) ⊂ G ⊂ Γ. A modular form of level N and dimension −k is an element of M(Γ(N), k). Modular forms of higher level arise from modular forms of level 1 as follows. If f ∈ M(Γ, k), and L ∈ M(N), then f|L is a form for the group −1 GL = Γ ∩ L ΓL, N 0 and G contains Γ(N) (Lemma 3.6). In particular, L = gives L 0 1 a b Γ (N) = G  = ∈ Γ: N | c 0 N 0 c d  0 1 MODULAR FORMS AND DIRICHLET SERIES 35

N 0 Any other primitive L of determinant N is in the double coset ΓL Γ, L = , 0 0 0 1 say L = AL0B, where A, B ∈ Γ, giving −1 −1 GL = B GL0 B = B Γ0(N)B, 1 0 a conjugate subgroup. In particular L = gives 0 N a b Γ0(N) = ∈ Γ: N | b . c d All these conjugate subgroups have isomorphic Riemann surfaces; we will give particular attention to Γ0(N). Note that the function ﬁeld of X0(N) = Γ0\(N)\H is C(j(τ), j(Nτ)), where j(τ) is the elliptic modular invariant. ∗ S N 0 Remark. Write M (N) = ΓL0Γ = ΓL (disjoint), where L0 = . Γ L 0 1 permutes the cosets ΓL transitively, and the stability group of ΓL1 is GL1 , so the index is Y 1 µ(N) = (Γ : Γ (N)) = (Γ : G ) = ψ(N) = N (1 + ) 0 L1 p p|N a formula from section2, Lemma 2.3 (it is the degree of T (1,N).)

Proof. This is the formula of Proposition 3.1; we have to check that the number of elliptic ﬁxed points of order 2 rep. 3 is µi resp. µ0 as claimed, and that the number of cusps is σ. a b Now ΓL Γ = S ΓL, where we can take the L’s to be L = , d > 0, 0 L 0 d ad = N, b mod d, (a, b, d) = 1, i.e. (b, t) = 1 where t = (a, d). Thus there are ϕ(t) d · t b’s for each d, and hence X d µ(N) = ϕ(t). t d|N

2πi/3 Let G = GL0 = Γ0(N); let P = MP0, where P0 = i, τ0 = e , or ∞, and M ∈ Γ. The stability group G(P ) is determined as follows. Let A ∈ G. Then −1 A ∈ G(P ) ⇐⇒ AP = P ⇐⇒ M AMP0 = P0. Now write L0M = BL, B ∈ Γ. 36 ANDREW OGG

−1 −1 Then A ∈ G(P ) ⇐⇒ A ∈ MΓ(P0)M ⇐⇒ A = MEM , where E ∈ Γ(P0), −1 −1 −1 −1 and E ∈ M GM, i.e. E ∈ M L0 ΓL0M = L ΓL. For P = ∞, the ramification index e is the least positive integer such that Γ −1 1 e a b 1 e a b 1 ea contains L L−1 = = d , i.e. ea ≡ 0 0 1 0 d 0 1 0 d 0 1 (mod d), i.e. e ≡ 0 (mod d ). Thus e = d . We noted above there are d ϕ(t) = eϕ(t) t t tP values of b for each d, so ϕ(t) cusps corresponding to d, whence σ = d|N ϕ(t) = P N d|N ϕ((d, d )). 0 −1 For P = i, since generates Γ(P ), we see P is an elliptic fixed point if 1 0 0 −1 2 2 a b 0 −1 a b b − b +a and only if γ contains = a N , i.e. a = 1, 0 d 1 0 0 d d b a a 2 2 b + 1 ≡ 0 (mod N). Thus µi is the number of solutions (mod N) of x + 1 ≡ 0 −4 (mod N), which is given by the formula stated, where p is the Legendre symbol. −1 a b 0 −1 a b For P = τ , the question is whether Γ contains = 0 0 0 d 1 1 0 d b b2−ab+a2 a − N 2 d b , i.e. a = 1, b − b + 1 ≡ (mod N); this gives the formula for a 1 − a µ0. Remark. If G is a subgroup of the modular group, of finite index, then the definition of a modular form for G can be stated as

(1) f|L = f for L ∈ G (2) for all A ∈ Γ, f|A is holomorphic at ∞ (resp. vanishes at ∞ for cusp P∞ 2πinτ/N forms), i.e. f|A(τ) = n=0 ane (resp. also a0 = 0), for some positive integer N. If we replace G by a smaller group, then f is a form for that group, for condition (2) is independent of what group we regard f as being a form for. In particular, if G is a congruence subgroup, f can be regarded as a form of various levels, and the question of whether f is a cusp form is independent of the level.

We now define the Hecke operators in a reasonably general setting. Let G be a subgroup of Γ0 = SL(2, Z) of finite index and ∆ ⊂ GL+(2, R) a set of real matrices of positive determinant, closed under multiplication, and such that for each α ∈ ∆, the double coset (α) = GαG contains only finitely many right and left cosets with respect to G. (Chapter2 treated the case G = Γ0, ∆ = integer matrices of positive determinant.) Let R = R(G, ∆) be the free Z-module (or C-module) on the double cosets (α) = GαG, for α ∈ ∆. R is a ring under X γ (α) · (β) = Cα,β · (γ) (γ) S S γ where if (α) = Gαi, (β) = Gβj (disjoint), then Cα,β is the number of pairs γ (i, j) with Gαiβj = Gγ; Cα,β depends only on the double cosets (α), (β), (γ). We leave the verification of the associative law to the reader. Now take G = Γ0 = SL(2, Z), ∆ = integer matrices of positive determinant. A 0 0 S 0 double coset (α) = Γ αΓ = Γ αi operates on the group D of divisors (i.e. the i P free abelian group on the set X of lattices) by L 7→ αi(L). Thus the mapping T : R(Γ0, ∆) −→ R(X) = End(D) MODULAR FORMS AND DIRICHLET SERIES 37 of R into the ring of correspondences which is linear (by definition) and multiplicative: X X X T ((α) · (β))L = αiβjL 0 0 (γ) k Γ αiβj =Γ γk X = αiβjL = T (α)(T (β) · L) i,j S 0 S 0 S 0 (where (α) = Γ αi, (β) = Γ βj, (γ) = Γ γk.) For fixed L, the left cosets Γ0α correspond one-one to sublattices αL of L, so T : R −→ R(X) is injective. a 0 Furthermore, any double coset (α) may be written (α) = Γ0 Γ0, where 0 d a 0 P d > 0, a | d, and we write T (a, d) = T , and T (n) = ad=n T (a, d). 0 d a|d Then R(Γ0, ∆) is the ring of Hecke operators defined in Chapter2, and we have the basic identity X nm T (n)T (m) = dT (d, d)T ( ) d2 d|n,m In general, we get a representation of R(G, ∆) on V = M(G, k) or S(G, k) by

k −1 X f|T (α) = |α| 2 f|αi S ((α) = Gαi). Now let G = Γ0(N). Let ∆0(N) be the set of integer matrices of positive determ- 1 0 inant n such that (n, N) = 1, and ∆(N) the set of α ∈ ∆0(N) with α ≡ 0 n (mod N). We have a natural map ϕ(N): R(Γ0(N), ∆(N)) −→ R(Γ0, ∆0(N)) by Γ0(N)αΓ0(N) 7→ Γ0αΓ0. Lemma 4.3. ϕ(N) is an isomorphism.

Proof. We know the set of primitive matrices of determinant m is 1 0 [ a b M ∗ = Γ0 Γ0 = Γ0 m 0 m 0 d ad=m,d>0 b mod d (a,b,d)=1 ∗ We claim that the set Mm(N) of primitive matrices of determinant m and ≡ 1 0 (mod N) (for (m, N) = 1) is similarly 0 m 1 0 [ a bN M ∗ (N) = Γ0(N) Γ0(N) = Γ0(N)R m 0 m a 0 d ad=m,d>0 b mod d (a,b,d)=1 a−1 0 where R ∈ Γ0 satisﬁes R ≡ (mod N).(R is not well deﬁned, but the a a 0 a a ∗ coset Γ(N) · Ra is.) It is clear (from the decomposition of Mm) we have distinct ∗ cosets adding up to Mm(N), and so only have to check there is only a single double coset mod Γ0(N). Thus we have to solve α β a bN 1 0 ∈ Γ0(N) γ δ 0 d 0 m 38 ANDREW OGG with α β α β a−1 0 ∈ Γ0, ≡ (mod N). γ δ γ δ 0 a Now α β a bN ∗ ∗ = , γ δ 0 d aγ dδ + bNγ so take γ = dN, δ + bN 2 = ax ≡ a (mod N) with (γ, δ) = 1, i.e. x ≡ 1 (mod N) and (dN, ax − bN 2) = 1; we can choose x, hence γ, δ, since (a, b, d) = 1. Then α β r sN ∗ 0 rδ − sγN = 1, and = ≡ will do. The lemma now γ δ γ δ 0 a follows.

Let T N (a, d), T N (n) be the elements of R(Γ0(N), ∆(N)) mapped on T (a, d), T (n) by ϕ(N). The basic identity in R0(N) = R(Γ0(N), ∆(N)) is then X nm T N (n)T N (m) = dT N (d, d)T N ( ) d2 d|n,m for (nm, N) = 1. N 2 k −1 d 0 However, when we operate on M(Γ(N), k), T (d, d) operates as (d ) 2 R = d 0 d d 0 1 0 dk−2R , since R ≡ (mod N). Thus, on M(Γ(N), k), the basic d d 0 d 0 d2 identity is X nm T N (n)T N (m) = dk−1R T N ( ) d d2 d|n,m for (nm, N) = 1. This suggests we should make the following decomposition of V = M(Γ(N), k) × or S(Γ(N), k). d 7→ Rd is a representation of the abelian group (Z/NZ) on V , and so the irreducible subspaces are one-dimensional. If f|Rd = ε(d) · f for all d ∈ (Z/NZ)×, then ε(d) is a character of (Z/NZ)×. Thus M V = V (ε), ε × summed over all characters of (Z/NZ) , where Rd operates as ε(d) on V (ε). N Lemma 4.4. Rn and T (m) commute, (nm, N) = 1. Hence V (ε) is invariant under T N (m).

k N 2 −1 P Proof. T (m) = m L L, where the set of all matrices of determinant m, 1 0 ≡ (mod N) is 0 m [ 0 Mm(N) = Γ (N) · L −1 S 0 −1 N −1 N But Mm(N) = Rn Mm(N)Rn = Γ (N)Rn LRn, so T (m) = Rn T (m)Rn.

Thus T N (n) operates on V (ε), with basic identity X nm T N (n)T N (m) = dk−1ε(d)T N ( ) d2 d|n,m MODULAR FORMS AND DIRICHLET SERIES 39

N P N −s for (nm, N) = 1, or equivalently: the Dirichlet series D (s) = (n,N)=1 T (n)n (whose coeﬃcients are operators on V (ε)) has the Euler product Y DN (s) = (1 − T N (p)p−s + ε(p)pk−1−2s)−1.

p-N Thus the Hecke operators T N (n) for (n, N) = 1 behave much as in level 1, except for the introduction of the characters ε(n). We will prove later they are ε-Hermitian, i.e. (f|T N (n), g) = ε(n)(f, g|T N (n)) for f, h ∈ S(Γ(N), k)(ε). 1 1 In order to deﬁne T N (n) for n | N we must split V up further. Let U = . 0 1 If f ∈ V , then f|U N = f and ∞ X n 2πiτ/N f(τ) = anz (z = e ). n=0

Let t | N. Let us say f has divisor t if an 6= 0 =⇒ (n, N) = t. The set of all f ∈ V of divisor t is then a subspace V (t). Now ν ∈ Z/NZ operates on V by f 7→ f|U ν ; since the group is abelian we can L again decompose V according to characters, say V = χ Vχ, where f ∈ Vχ =⇒ ν N f|U = χ(ν) · f, for each character χ of Z/NZ. If the exact period of χ is t , i.e. N 2πin/N f|U = ζ · f, ζ = primitive t -th root of 1, then an 6= 0 =⇒ ζ = e , and then (n, N = t); thus Vχ ⊂ V (t). This proves that M V = V (t) t|N P N Note f ∈ V (t) ⇐⇒ f = fi, where fi|U = ζi · fi, ζi = primitive t -th root of 1. N Lemma 4.5. V (t) is invariant under Rn and T (n), for (n, N) = 1.

n n Proof. RnU ≡ URn (mod N); if f|U = ζ · f then f|RnU = ζf|Rn, which shows 2 Rn carries V (t) into itself, since n generates Z/NZ. Similarly, if L ∈ Mn(N), so 1 0 L ≡ (mod N), then U −1LU n ∈ M (N); hence T N (n) = U −1T N (n)U n, 0 n n N and T (n) carries V (t) into itself.

L Thus V = ε,t V (ε, t), where any element f of V (ε, t) has divisor t and f|Rn = ε(n) · f, (n, N) = 1. Note that if V = M(Γ(N), k), then V (1, k) = M(Γ0(N), k). We now deﬁne Hecke operators T t(p) = T N,t(p) for p | N on V (t); as the notation suggests, they will be diﬀerent for the various t. 1 b N Lemma 4.6. On V (t), t depends only on b mod m. 0 m

Proof. If f ∈ V (t), then f|U N/t = f, so 1 b N 1 N 1 b N 1 (b + m) N f| t = f| t t = f| t . 0 m 0 1 0 m 0 m

40 ANDREW OGG

Now if m is such that every prime factor of m divides N, we deﬁne N t k −1 X 1 b T (m) = m 2 t 0 m b mod m on V (t). We will check shortly that f|T t(m) ∈ V ; to show the divisor is still t, we look at the Fourier expansion:

X 2πinτ· t f(τ) = ane N N (n, t )=1 (change of notation). Then ! 1 X τ + b N f|T t(m)(τ) = f t m m b mod m X 1 X = a e2πinτt/Nm · e2πinb/m n m N b mod m (n, t )=1 X 2πinτt/Nm = anme . N (n, t )=1 t Thus T (m) has the eﬀect of replacing an by anm, whence

(1) f|T t(m) still has divisor t, (2) the T t(m) commute with each other, t N (3) T (m) = 0 if (m, t ) 6= 1

To check that f|T t(m) is still a form for Γ(N), we can assume m = p is prime, α 0 and p N , by the remarks above. Let A ∈ Γ0, A ≡ (mod N). One - t 0 δ −1 1 b N 1 b N α β0 computes that A0 = t A t ∈ Γ0, A0 ≡ (mod N), β0 ≡ 0 p 0 m 0 δ N t t 0 (mod t ). This proves f|T (p) ∈ V (t) if f ∈ V (t), and that T (p) commutes with t t Rn for (n, N) = 1; hence T (p) operates on V (t, ε), and hence also T (m). Finally, the T t(m) (where p | m =⇒ p | N) and the T (n) = T N (n) (for (n, N) = 1) commute on V (t). To see this, we can take n = p to be prime. Now k −1 p 0 T (p) is the sum of the operator p 2 R and the one which replaces the p 0 1 t Fourier coeﬃcient an by anp; this second operator certainly commutes with T (m), by the remarks above, and we just proved that Rp does, so it remains to prove that p 0 the T t(m) and commute. But 0 1

−1 p 0 1 b p 0 1 bp = 0 1 0 m 0 1 0 m which proves it. Finally, we deﬁne T t(nm) = T (n)T t(m) on V (t), where (n, N) = 1 and p | m =⇒ p | N. If we agree that ε(n) = 0 for (n, N) 6= 1 by the usual convention, then we have deﬁned Hecke operators T t(n) on V (t, ε) for all n ≥ 1, satisfying X nm T t(n)T t(m) = ε(d)dk−1T t( ) d2 d|n,m MODULAR FORMS AND DIRICHLET SERIES 41

(for all n, m ≥ 1). Equivalently, we have an identity of formal Dirichlet series ∞ X Y T t(n)n−s = (1 − T t(p)p−s + ε(p)pk−1−2s)−1; n=1 p t N note ε(p) = 0 if p | N and T (p) = 0 if p | t . P n Proposition 4.7. If f ∈ V (t, ε) has Fourier expansion f(τ) = n anz , z = e2πiτ·t/N , then t X ν f|T (n)(τ) = aν (n)z , ν P k−1 where aν (n) = ε(d)d a νn . d|ν,n d2

Proof. We can assume (n, N) = 1, by the remarks above; in that case the proof is the same as that of Proposition 2.4 in Chapter2. Theorem 4.8. Let T (n) = T t(n) be the Hecke operator on M(Γ(N), k, t, ε), for n ≥ 1.

(1) ∞ X Y T (n)n−s = (1 − T (p)p−s + ε(p)pk−1−2s)−1 n=1 p

(2) If f is an eigenfunction for all the T (n), normalized to have a1 = 1, then the associated Dirichlet series has the Euler product ∞ X −s Y −s k−1−2s −1 ann = (1 − app + ε(p)p ) . n=1 p

Next we generalize Theorem 2.8 on the uniqueness of the p-factor in the Euler product (for p - N). We need ﬁrst the corresponding generalization of Proposition 2.7: Theorem 4.9. Let f, f|α ∈ M(Γ(N), k), where α is a primitive integer matrix of determinant n, with n > 1, (n, N) = 1. Then f = 0.

1 0 1 0 1 0 1 N Proof. Since α ∈ Γ0 Γ0, assume α = . Since f| = 0 n 0 n 0 n 0 1 1 0 f| , 0 n 1 0 1 N n 0 n N f = f| = f| . 0 n 0 1 0 1 0 n n N n N The powers of are not all primitive, so replace it by V = A , 0 n 0 n 1 ∗ 0 N where A ∈ Γ(N), A ≡ (mod n). Then V ≡ (mod n), V ` ≡ 1 ∗ 0 N 0 N ` (mod n), and all powers V ` are primitive. Since V ` is primitive of 0 N ` 1 0 n 0 determinant n2`, write V ` = B C, where B,C ∈ Γ0. Now V ≡ 0 n2` 0 n (mod N), so if we choose ` so that n` ≡ 1 (mod N), we have BC ∈ Γ(N), so f|B = f|C−1. Finally, 1 0 f = f|V ` = f|B C, 0 n2` 42 ANDREW OGG so g = f|B ∈ M(Γ(N), k) satisﬁes g(τ) = g(τ/n2`) · n−k`, so g = 0, f = 0.

P 2πinτ/N Corollary 4.10. Let p be a prime, p - N; let f ∈ M(Γ(N), k), f(τ) = ane .

(1) If am = 0 for all m with p - m, then f = 0. (2) If apn = 0 for all n, then f = 0.

p N Proof. (1): Then f(τ) = f(τ + N ); taking α = , we get f = 0. p 0 p (2): Then k −1 p 0 f|T (p) = p 2 f|R p 0 1 p 0 has level N; taking α = R , we get f = 0. p 0 1

P −s We recall that ϕ(s) = ann has an Euler product relative to p if ∞ X −s X ν −νs ϕ(s) = ( amm )( c(p )p ) p-m ν=0 ν i.e. ampν = amc(p ) for p - ν. If ϕ(s) 6= 0, and p - N, then c(1) = 1, by (1) of the corollary. P 2πinτ/N Theorem 4.11. Let ϕ(s) 6= 0 be the Dirichlet series associated to f(τ) = ane ∈ M(Γ(N), k). Let p - N. Then ϕ(s) has an Euler product relative to p if and only N if f is an eigenfunction for Rp and T (p) = T (p), say f|Rp = ε · f, f|T (p) = c · f. If so, the p-factor is necessarily of the form ∞ X c(pν )p−νs = (1 − cp−s + εpk−1−2s)−1 ν=0

ν Proof. For “only if”, assume apν m = amc(p ) for p - m, ν ≥ 0. Then f|T (p) − c(p) · f k −1 p 0 X 2πinτ/N = p 2 f|R + (a − c(p)a )e p 0 1 pn n n = power series in e2πipτ/N = 0, by (1) in the corollary. Similarly, k −1 p 0 X n p 2 f|R = f|T (p) − a z p 0 1 pn X n = (c(p)an − apn)z n X pn = (c(p)apn − ap2n)z , n MODULAR FORMS AND DIRICHLET SERIES 43 where z = e2πiτ/N . Then k−1 X n p f|Rp = (c(p)apn − ap2n)z n = (c(p)2 − c(p2))f + power series in zp = (c(p)2 − c(p2))f

k−1 Conversely, suppose f|T (p) = c · f, f|Rp = ε · f. Then c · f(τ) = p εf(pτ) + P n apnz , so ( apn if p - n c · an = k−1 apn + p εan/p if p | n Thus ϕ(s)(1 − cp−s + εpk−1−2s) X −s X k−1 2 −s X −s k−1 X 2 −2 = ann + ε anp (np ) − apn(pn) − εp an(np )

X −s = amm p-m Theorem 4.12 (Petersson). Let f, g be cusp forms for Γ(N), of character ε. Then (f|T (n), g) = ε(n)(f, g|T (n)), for (n, N) = 1, where T (n) = T N (n).

Proof. It suﬃces to prove this for n a prime power, and then for n = p, for if we have the result for n = p, p2, . . . , pν , then from T (pν )T (p) = T (pν+1) + ε(p)pk−1T (pν−1), we get (f|T (pν+1), g) = (f|T (pν )T (p), g) − ε(p)pk−1(f|T (pν−1), g) = ε(pν+1)((f, g|T (pν )T (p)) − ε(p)pk−1(f, g|T (pν−1))) = ε(pν+1)(f, g|T (pν+1)) using the fact that (f, g) is conjugate-linear in g. Thus we assume n = p is prime, p - N. 1 0 Now the set of integer matrices of determinant p and ≡ (mod N) is a 0 p single double coset 1 0 M ∗(N) = Γ0(N) Γ0(N), p 0 p which shows (as in the case N = 1, Theorem 3.7) that every left coset meets every right coset, and so there exists a set {α} of left and right representatives:

∗ [ 0 [ 0 Mp (N) = Γ (N)α = αΓ (N). 0 a b d −b Letting = , we see that c d −c a

[ 0 ∗ ∗ 0 [ 0 0 Γ (N)α = Mp (N) = Rp(Mp (N)) = Γ (N)Rpα . Hence k −1 X k −1 X 0 T (p) = p 2 α = p 2 Rpα , 44 ANDREW OGG

Note the eigenvalues λn of T (n) ((n, N) = 1) are not necessarily real this time; the rule is that

λn = ε(n)λn − 1 i.e. λnε(n) 2 is real. In particular, λn is real if ε(n) = 1, purely imaginary if ε(n) = −1. It follows as before that the space S(N, k, t, ε) of cusp forms of dimension −k divisor t, and character ε for γ(N) has an orthogonal basis f1, . . . , fr of eigenfunctions for all T (n) with (n, N) = 1. If f is any eigenfunction, and (f, fj) 6= 0, say, then

λn(f, fj) = (f|T (n), fj)

= ε(n)(f, fj|T (n))

(j) (j) = ε(n)λn (f, fj) = λn (f, fj) and so f and a suitable constant times fj have the same Fourier coefficients an for N (n, N) = 1. This is all you can say in general; however, if p | t =⇒ p | t , then T (n) = 0 for (n, N) 6= 1, and f is a constant times fj. We have the following general estimate on the Fourier coefficients of cusp forms: Proposition 4.13. Let G be a subgroup of Γ of finite index, and f ∈ S(G, k). Then f(τ) = O(y−k/2) as y → 0, uniformly in x, and hence the Fourier coefficients k/2 of f satisfy an = O(n ).

S Proof. Write Γ = L GL (disjoint). Then X h(τ) = (Im τ)k |f(L(τ))|2 L is invariant under Γ, and bounded on the fundamental domain D of Γ since it −k −k/2 k/2 vanishes at ∞. Thus h(τ) = O(y ), so f(τ) = O(y ). Then an = O(n ) by Proposition 1.1. MODULAR FORMS AND DIRICHLET SERIES 45

Now if f ∈ S(Γ(N), k, ε) is an eigenfunction for T (p), where p - N, then the p-factor of the associated Dirichlet series is

−s k−1−2s −1 (1 − app + ε(p)p ) .

2 Let η satisfy ηη = 1, η = ε(p), so apη is real, Petersson’s conjecture states that

k−1 2 1 − apηt + p t ,

k−1 which has real coeﬃcients, has conjugate roots, i.e. |ap| ≤ 2p 2 , which would of course be much stronger than the general estimate of Proposition 4.13. Thus we see that the theory of the Hecke operators T (n) in level N, for cusp forms, parallels that in level 1, at least for (n, N) = 1. To treat the non-cusp forms, we again need the explicit construction of Eisenstein series. Let k ≥ 3 and c, d ∈ Z, and consider the Eisenstein series 0 X −k Gk(τ; c, d; N) = (mτ + n) ; m≡c (mod N) n≡d (mod N) since k ≥ 3, this is an absolutely converging double series, for Im (τ) > 0. (There are also Eisenstein series for k = 1, 2; a modiﬁcation to ensure convergence is 0 necessary—cf. Hecke’s paper [5, No. 24].) If L ∈ Γ = SL(2, Z), then Gk|L has the term for (m, n) replaced by that for (m, n)L:

Gk(τ;(c, d); N)|L = Gk(τ;(c, d)L; N).

Since clearly Gk(τ; c, d; N) depends only on c and d modulo N, we see it is a form for Γ(N), provided it is holomorphic at the cusps. To show this, we determine the Fourier expansion:

P∞ λ 2πiτ/N Proposition 4.14. Gk(τ; c, d; N) = λ=0 aλz , z = e , where ( 0 if c 6≡ 0 (mod N) a = 0 P −k n≡d (N) n if c ≡ 0 (mod N) and for λ ≥ 1, (−2πi)k X a = (sgn ν)νk−1e2πiνd/N λ N kΓ(k) mν=λ m≡c (N)

Proof. As in the proof of Proposition 1.14, we start from ∞ X (−2πi)k X (m + τ)−k = nk−1e2πinτ . Γ(k) m n=1 Clearly

X X −k Gk(τ) − a0 = (mτ + nN + d) m≡c n m6=0 −k k ∞ −∞ N (−2πi) X X k−1 2πiν mτ+d X X k−1 2πiν mτ+d = ν e N + (−ν) e N , Γ(k) m≡c ν=1 m≡c ν=−1 m>0 m<0 and this proves the proposition. 46 ANDREW OGG

Thus Gk(τ; c, d; N) ∈ M(Γ(N), k).

Gk(τ; c, d; N) is called primitive if (c, d, N) = 1; if (c, d, N) = t > 1, then c d N G (τ; c, d; N) = t−kG (τ; , ; ) k k t t t N is a primitive Eisenstein series of level t . Let E(N) = E(N, k) be the space generated by all primitive Eisenstein series of level N. Now N fundamental domains for Γ meet at ∞; since Γ(N) is a normal subgroup of Γ, the same is true at any other cusp. The number of cusps is then  1 if N = 1 (Γ : Γ(N))  σ(N) = = 3 if N = 2 N  1 2 Q 1 2 N p|N (1 − p2 ) if N > 2 We have an obvious map E(N) −→ Cσ(N) by evaluating at the cusps, and we want to show this is an isomorphisms. Now the 2 Q 1 number of primitive pairs (c, d) mod N is N p|N (1 − p2 ), and clearly k Gk(τ; c, d; N) = (−1) Gk(τ; −c, −d; N), so dim E(N) ≤ σ(N) and so it suﬃces to prove the map is onto. For this, it is convenient to consider the restricted Eisenstein series ∗ X −k Gk(τ; c, d; N) = (mτ + n) ; m≡c (N) n≡d (N) (m,n)=1 again a form for Γ(N), with ∗ ∗ Gk(τ; c, d; N)|L = Gk(τ;(c, d)|L; N) for L ∈ Γ0. To connect the two kinds of Eisenstein series, we use the Möbius function µ(n), the multiplicative function of positive integers n with µ(1) = 1, µ(p) = −1 for p prime, and µ(n) = 0 if n has a square factor. The Möbius function satisﬁes ( X 1 if n = 1 µ(d) = 0 if n > 1 d|n (d>0) Thus 0 ∗ X −k X Gk(τ; c, d; N) = (mτ + n) µ(a) m≡c a|(n,m) n≡d ∞ X X0 = µ(a)a−k (mτ + n)−k. a=1 ma≡c na≡d ∗ Now assume (c, d, N) = 1 (otherwise Gk(τ; c, d; N) = 0). Then in the sum above we can assume (a, N) = 1. For such an a, choose a0 with aa0 ≡ 1 (mod N). Then ∗ −k 0 0 Gk(τ; c, d; N) = µ(a)a Gk(τ; a c, a d; N) X = ct · Gk(τ; ct, dt; N), (t,N)=1 t mod N P −k ∗ where ct = at≡1 (N) µ(a)a . Thus Gk(τ; c, d; N) ∈ E(N). a>0 MODULAR FORMS AND DIRICHLET SERIES 47

∗ th The value of Gk(τ; c, d; N) at ∞, i.e. its 0 Fourier coeﬃcient, is visibly ( X −k 1 if (c, d) ≡ (0, 1) (mod N) a0 = n = 0 otherwise. m=0≡c n≡d (m,n)=1 α β Since G∗(τ; 0, 1; N)| = G∗(τ; γ, δ; N), we see G∗(τ; 0, 1; N) takes the value k γ δ k k ∗ 1 at the cusp ∞ and 0 at the other cusps; similarly, Gk(τ; c, d; N) takes the value 1 at the cusp −d/c and 0 at all other cusps. This proves: Proposition 4.15. The map E(N) → Cσ(N) is an isomorphism, and E(N, k) is generated by the restricted Eisenstein series. Proposition 4.16. If N 0 | N, then E(N 0, k) ⊂ E(N, k). Hence E(N) is the space of all Eisenstein series of level N, primitive or not.

Proof. ∗ 0 0 0 X ∗ Gk(τ; c , d ; N ) = Gk(τ; c, d; N) c≡c0 (N) d≡d0 (N) c,d mod N which proves the ﬁrst statement, in view of Proposition 4.15; we have already N observed that an imprimitive series of level N is a primitive series of lower level t , which proves the second statement. Proposition 4.17. If L is a primitive integer matrix of determinant m ≥ 1, and f ∈ E(N), then f|L ∈ E(mN). Also, E(mN, k) ∩ M(Γ(N), k) = E(N, k).

Proof. We have proved this for m = 1, i.e. L ∈ Γ0, so we may as well take 1 0 L = ; then G (τ; c, d; N)|L = Pm G (τ; c + νN, dm; Nm). 0 m k ν=1 k For the second statement, note that we have as a result of Proposition 4.15 a direct sum decomposition M(Γ(N), k) = E(N, k) ⊕ S(N, k). If f ∈ E(mN, k) ∩ M(Γ(N), k), write f = E+g, where E ∈ E(N, k), g ∈ S(N, k). Then g ∈ E(mN, k), by Proposition 4.16, and g is a cusp form, so g = 0 by Proposition 4.15.

In view of these propositions, let us call any element of E(N, k) an Eisenstein series.

Since E(N, k) is invariant under all modular transformations, in particular by Rn 1 1 for (n, N) = 1 and U = , we can decompose E(N, k) according to divisors 0 1 t of N and characters ε of (Z/NZ)×, getting M(Γ(N), k, ε, t) = E(N, k, ε, t) ⊕ S(N, k, ε, t). Furthermore, this decomposition is respected by all Hecke operators T (n) = T t(n), n ≥ 1, by Proposition 4.17. A way to construct modular forms (of higher level) from given ones, using characters, is given by the following theorem; this technique is also emphasized in the following chapter. Given an integer m ≥ 1, a character modulo m is a character χ on (Z/mZ)×; we extend χ to a function of all positive integers n by the usual convention that χ(n) = 0 if (n, m) > 1. Note that we do not in general require χ to be primitive (not deﬁned modulo a proper divisor of m); in particular, even the identity character χ = 1 modulo m satisﬁes the convention χ(n) = 0 for (n, m) > 1. 48 ANDREW OGG

P nt 2πiτ/N Theorem 4.18. Let f(τ) = N anz ∈ M(Γ(N), k, ε, t), z = e , and (n, t )=1 N let χ be a character modulo m · t , where (n, N) = 1. Let X nt fχ(τ) = χ(n)anz n 2 2 2 Then fχ ∈ M(Γ(m N), k, εχ , tm ), and fχ is a cusp form (resp. Eisenstein series) if f is.

N Proof. Let M = m · t , and consider the operator 1 X m y L = χ(x)e−2πixy/M ; χ n 0 m x,y mod M 2 nt 2πinτt/N then f|Lχ ∈ M(Γ(m N), k). Applying Lχ to anz = ane contributes a factor of 1 X χ(x)e2πi(n−x)y/M , M x,y mod M which is χ(n), since the sum over y is 0 for x 6= n. Thus fχ = f|Lχ is a form of level m2N and divisor m2t, and a cusp form (resp. Eisenstein series) if f is; it remains n−1 0 to check that the character is εχ2. Let R ∈ Γ0, R ≡ (mod m2N). n n 0 n m y m yn2 Then R ≡ R (mod N), so 0 m n n 0 m 2 ε(n) X 2 m yn f |R = χ(xn2)e−2πixn y/M f| χ n M 0 m x,y mod M 2 = (εχ )(n)fχ

For example, we know the Eisenstein series of level 1 is associated to the Dirichlet series ∞ X −s ζ(s)ζ(s + 1 − k) = σk−1(n)n ; n=1 if χ is a character modulo m, then ∞ X −s X −s−1+k X −s L(χ, s)L(χ, s + 1 − k) = χ(n1)n1 χ(n2)n2 = χ(n)σk−1(n)n n1 n1 n=1 is associated to an Eisenstein series of level m2, divisor m2 and character χ2, and is an eigenfunction for the T (n), (n, m) = 1. More generally: Theorem 4.19. The space of Dirichlet series associated to E(N, k) is generated by the series of form −s (t1t2) L(χ1, s)L(χ2, s) N k where t1, t2 | N, χj is a character module , (χ1χ2)(−1) = (−1) , L(χ, s) = tj P∞ −s χ1,χ2 n=1 χ(n)n . The corresponding Eisenstein series E has character ε = χ1χ2, divisor t = (t1t2,N), and is an eigenfunction of the T (n), (n, N) = 1.

Proof. By Proposition 4.14, the Dirichlet series ζc,d(s) associated to Gk(τ; c, d; N) is (a constant times) ∞ X −s X k−1 2πiνd/N ζc,d(s) = n sgn (ν)ν e . n=1 mν=n m≡c (N) MODULAR FORMS AND DIRICHLET SERIES 49

c,d 1 P −2πiad/N c,d Let ζ (s) = N a mod N e ζc,a(s); the ζ (s) generate the same space. The coeﬃcient of n−s in ζc,d(s) is 1 X X e−2πiad/N sgn (ν)νk−1e2πiaν/N N a mod N mν=n m≡c (N) X = sgn (ν)νk−1 mν=n m≡c ν≡d X X = sgn (ν)νk−1 + (−1)k sgn (ν)νk−1 mν=n mν=n m≡c m≡−c ν≡d,ν>0 ν≡−d,ν>0 and so ζc,d(s) = ψc,d(s) + (−1)kψ−c,−d(s), where X X ψc,d(s) = m−s nk−1−s. m≡c (N) n≡d (N)

c,d Now ﬁx t1, t2 | N. Then the space generated by the ζ (s) with (c, n) = t1 and (d, N) = t2 is the same as that generated by the

χ1,χ2 X b1t1,b2t2 L (s) = χ1(b1)χ2(b2)ζ (s), N bi mod ti

N where χi is a character modulo . Now ti

X b1t1,b2t2 χ1(b1)χ2(b2)ψ (s) N bi mod ti X −s k−1−s X −s X k−1−s = χ1(b1)χ2(b2)t1 t2 m n N N N bi mod m≡b1 ( ) n≡b2 ( ) ti t1 t2 k−1 −s = t2 L(χ1, s)L(χ2, s)(t1t2) hence

χ1,χ2 k−1 k −s L (s) = t2 (1 + (−1) (χ1χ2)(−1))(t1t2) L(χ1, s)L(χ2, s).

This proves the theorem.

Corollary 4.20. M(Γ(N), k, ε, t) has a basis of eigenfunctions for all T (n) with (n, N) = 1.

Proof. In view of the decomposition

M(Γ(N), k, ε, t) = E(N, k, ε, t) ⊕ S(N, k, ε, t), and the fact we have already diagonalized the T (n) on the cusp forms, we only have to diagonalize the T (n) on the Eisenstein series; this is done by Theorem 4.19, in view of Theorem 4.11.

That one cannot in general diagonalize the T (p) for p | N is shown by the 3 2 following example. Let q be an odd prime, N = q , t1 = t2 = q , and χ1, χ2 k characters modulo q with (χ1χ2)(−1) = (−1) . (They exist.) The theorem gives 50 ANDREW OGG an Eisenstein series

f(τ) = Eχ1,χ2 (τ) X 2πinq4τ/q3 = a0 + ane q-n X 2πinqτ = a0 + ane , q-n with divisor t = N = q3. Since T (q) replaces the nth Fourier coeﬃcient by the (nq)th, we have X 2πinτ h(τ) = f|T (q)(τ) = a0 + ane , q-n 0 1 h(τ)|T (q) = 0. Thus T (q) has matrix on the space spanned by f, h, and 0 0 so T (q) is not diagonalizable. Remark. We have shown that any form f of level N has an associated Dirichlet const. series, i.e. an = O(n ); by Theorem 4.19 if f is an Eisenstein series and by Proposition 4.13 if f is a cusp form.

5. A theorem of Weil

From now on we deal only with forms f of level N and maximal divisor t = N, i.e. f(τ + 1) = f(τ). If ε is a character modulo N, let M(N, k, ε) = M(Γ(N), k, ε, N), and similarly (at least for k ≥ 3) S(N, k, ε), E(N, k, ε). Thus if f ∈ M(N, k, ε), and a b a b ∈ Γ0 (N), i.e. N|c, then f| = ε(d) · f; f = 0 unless ε(−1) = (−1)k, c d 0 c d −1 0 since f| = (−1)kf. Note M(Γ (N), k) = M(N, k, 1). 0 −1 0

0 −1 a b d −c/N Let H = . Note that H H−1 = ∈ Γ0 (N) N N 0 N c d N −bN a 0 a b if ∈ Γ0 (N), and so: c d 0

(1) Γ∗(N) = Γ0(N) ∪ Γ0(N) · HN is a group (of substitutions of the upper half plane), containing Γ0(N) as a (normal) subgroup of index 2 (for N > 1). ∼ (2) f 7→ f|HN deﬁnes an isomorphism M(N, k, ε) −→ M(N, k, ε¯). (3) In particular, if ε =ε ¯ is real (i.e. its values are ±1), then f 7→ f|HN is an −1 0 automorphism of M(N, k, ε), and f|H2 = f| = (−1)kf. N 0 −1

We can diagonalize this representation of the group of order 2: M(N, k, ε) = M+(N, k, ε) ⊕ M−(N, k, ε)

± k + − + where f ∈ M (N, k, ε) =⇒ f|HN = ±i f.(f = f + f , where f = f + −k − −k ± i f|HN , f = f − i f|HN ). If f ∈ M (N, k, ε), we will say f has a functional equation, or that f is a form for the extended group Γ∗(N), of character ε and multiplier C = ±1. The T (n), for (n, N) = 1, operate on M±(N, k, ε), as follows from:

Lemma 5.1. HN T (n) = ε(n)T (n)HN , on M(N, k, ε) MODULAR FORMS AND DIRICHLET SERIES 51

Proof. T (n) = nk/2−1 P L, where S Γ0L is all matrices of determinant n, and 1 ∗ L ≡ (mod N). Now S Γ0H LH−1 is still disjoint, as one checks, and 0 n N N n ∗ H LH−1 ≡ (mod N), so R H T (n)H−1 = T (n). N N 0 1 N N N

The theorem of Weil [15] we are about to prove is in the spirit of Hecke’s basic Theorem 1.2. When N = 1, Γ = Γ∗(1) is generated by two elements 0 −1 1 1 H = , U = , so modular forms for Γ are deﬁned by two func- N 1 0 0 1 tional equations (plus regularity); periodicity (functional equation for U) gives an associated Dirichlet series ϕ(s), and the functional equation for H1 gives a functional equation for ϕ(s), and vice versa. Weil’s result is to characterize forms for Γ∗(N), which has in general more than two generators, by functional equations for many associated Dirichlet series. Besides Theorem 1.2, we need the notion of Gauss sums. Let χ be a character of conductor m, i.e. χ is a primitive character modulo m, i.e. χ is a character modulo m and not modulo any proper divisor of m. For n ∈ Z/mZ, we have the Gauss sum X 2πixn/m gχ(n) = χ(x)e x mod m

Let gχ = gχ(1). Proposition 5.2.

(1) gχ(n) =√χ ¯(n)gχ (2) |gχ| = m.

Proof. If (n, m) = 1, then

X 2πinx/m χ(n)gχ(n) = χ(nx)e = gχ, x mod m as desired. Now suppose (n, m) = t > 1; we want to show gχ(n) = 0. Write n0t = n, m0t = m. Then

X 2πin0x/m0 gχ(n) = χ(x)e

x mod m0t

X 2πin0x0/m0 X = e χ(x0y),

x0 mod m0 y≡1 (mod m0) (x0,m)=1 y mod m and this is 0, since P χ(y) = 0, because χ is a non-trivial character on the group of y ≡ 1 (mod m0), since χ is not a character modulo m0. For (2), 2 X X 2πix(1−y)/m |gχ| = gχgχ = χ(x) χ(xy)e (x,m)=1 y X = χ(y)e2πix(1−y)/m (x,m)=1

P 2πix(1−y)/m 2πix/m If (x, m) > 1, then y χ(xy)e = e gχ(x) = 0. Thus 2 X 2πix(1−y)/m |gχ| = χ(y)e = m, x,y since the sum over x is 0 for y 6= 1, m for y = 1. 52 ANDREW OGG

P∞ 2πinτ Now if f(τ) = n=0 ane is a Fourier series, and χ a character of conductor m, we deﬁne, as in Theorem 4.18, ∞ X 2πinτ fχ(τ) = anχ(n)e . n=0 By the proof of Theorem 4.18, this is also 1 X y f (τ) = χ(x)e−2πixy/mf(τ + ) χ m m x,y mod m 1 X y = g (−y)f(τ + ) m χ m y mod m gχ X y = χ(−y)(τ + ) m m y mod m

Now let a0, a1, a2,... and b0, b1, b2,... be two sequences of complex numbers, σ an, bn = O(n ) for some σ > 0, and form ∞ ∞ X 2πinτ X −s −s f(τ) = ane ϕ(s) = ann Φ(s) = (2π) Γ(s)ϕ(s) n=0 n=1 ∞ ∞ X 2πinτ X −s −s g(τ) = bne ψ(s) = bnn Ψ(s) = (2π) Γ(s)ψ(s) n=0 n=1 Let C 6= 0, A > 0, k > 0; recall that EBV means entire and bounded in every vertical strip. Lemma 5.3. Equivalent are:

k −s/2 a0 Cb0 2 −s (A1) Φ(s) + A s + k−s is EBV, and Φ(s) = CA Ψ(k − s) k/2 Aτ −k −1 (B1) f(τ) = CA i g Aτ .

2 τ Proof. This is a reformulation of Theorem 1.2, with A = λ , f1(τ) = f( λ ), Φ1(s) = s λ Φ(s), etc.

a b Now assume k is a positive integer, and so we have the notation f| c d 0 −1 for k. Letting H = ,(B1) reads f = Cikg|H . The group ring A A 0 A + P P C[GL (2, R)] = R operates on f by f| ciLi = cif|Li, and

Ωf = {ω ∈ R : f|ω = 0} is a right ideal in R.

σ We now change the notation. Given a0, a1, a2,... , an = O(n ), and a character χ of conductor m, deﬁne ∞ ∞ X X 2π −s f (τ) = χ(n)a e2πinτ ,L (s) = χ(n)a n−s, Λ (s) = Γ(s)L (s). χ n χ n χ m χ n=0 n=1 1 x For m = 1, we write simply f = f , L = L , Λ = Λ . Letting α(x) = for 1 1 1 0 1 x ∈ R, we have gχ X y f = χ(−y)f|α( ). χ m m y mod m Let N be a ﬁxed positive integer, and C = ±1. MODULAR FORMS AND DIRICHLET SERIES 53

Lemma 5.4. Equivalent are:

k −s/2 a0 Ca0 2 −s (A2) Λ(s) + N s + k−s is EBV, and Λ(s) = CN Λ(k − s) k (B2) 1 ≡ Ci HN (mod Ωf ).

Proof. This is immediate from lemma 5.3, with f = g etc., A = N.

Lemma 5.5. Equivalent are (for m > 1, Cχ 6= 0):

k −s (A3) Λχ(s) is EBV, and Λχ(s) = CχN 2 Λχ(k − s) P y k P y (B3) gχ y mod m χ(y)α( m ) ≡ Cχi gχ χ(y)α( m )Hm2N (mod Ωf ).

2 Proof. Again immediate from lemma 5.3, with A = m N.

Remark. If the an are real, then Λχ = Λχ, and |cχ| = 1 (if (A3) holds.)

Now suppose (m, N) = 1. Then for each a with (a, m) = 1, there exists a b with abN ≡ −1 (mod m), and so m −b γ(a, b) = −Na n 0 belongs to Γ0(N), for some n. One computes a b m 0 α( )H 2 = H γ(a, b)α( ) ; m Nm N m 0 m m −b since b runs over (b, m) = 1, b mod m as a does, and writing γ(b) = = −Na n γ(a, b), we see that (B3) is equivalent with k X Cχi gχ b (B3’) χ(b) 1 − HN γ(b) α( ) ≡ 0 (mod Ωf ). gχχ(−N) m b mod m

(−abN ≡ 1 (m), so χ(x) = χ(−bN).) k Now let f ∈ M(N, k, ε), with functional equation f = Ci f|HN (C = ±1); here P∞ 2πinτ σ ε is a real character modulo N. Then f(τ) = n=0 ane with an = O(n ) by the preceding chapter. Thus (B2) holds by hypothesis, so (A2) is true; also Cgχχ(−N)ε(m) γ(b) ≡ ε(n) = ε(m) (mod Ωf ). Thus, taking Cχ = , we see (B3’) gχ holds, and hence (A3). This proves:

k Theorem 5.6. Let f ∈ M(N, k, ε) with functional equation f = Ci f|HN , C = ±1 (so ε is real). Then Λ satisﬁes (A2), and Λχ satisﬁes (A3), for every character χ whose conductor m is relatively prime to N, the values of Cχ being taken as

Cχ = Cε(m)χ(−N)gχ/gχ

We now turn to the converse, in a strong form: i.e. assuming the functional equation for Lχ for “suﬃciently many” χ’s. Let M = {4, 3, 5, 7, 11,... }; any non- identity character χ modulo m ∈ M is primitive.

0 Lemma 5.7. Let m ∈ M, with (m, N) = 1; let Cm 6= 0. Then equivalent are:

0 Cmχ(−N)gχ (A4) for every primitive χ modulo m, (A3) holds with Cχ = k i gχ 0 0 (B4) λ(b) ≡ λ(b ) (mod Ωf ), whenever (b, m) = (b , m) = 1, where 0 b λ(b) = (1 − CmHN γ(b))α( ); P m (C4) b mod m χ(b)λ(b) ≡ 0 (mod Ωf ), for every primitive χ modulo m. 54 ANDREW OGG

Proof. That (A4) is equivalent with (C4) follows from Lemma 5.5. Clearly (B4) implies (C4); conversely, given (C4), we have X 0 ≡ χ(b0) − χ(b00)χ(b)λ(b) = ζ(m)λ(b0) − λ(b00) χ,b where the sum is over all primitive χ, hence over all χ.

The last key lemma is: m −b Lemma 5.8. Let γ = ∈ Γ0 (N), with m, n ∈ M. Assume (A4) holds −aN n 0 0 0 k Cik for m and n, with CmCn = (−1) , and assume (A2) holds. Then f|γ = 0 f. Cm

m b Proof. Let γ0 = , so γ0 = γ(−b), γ = γ(b) in the notation above. Now aN n k (A2), hence (B2), holds, so 1 ≡ Ci HN (mod Ωf ), and (A4), hence (B4), holds; 0 b 0 −b 0 −k taking b, b = b, −b, we have (1 − ξγ)α( m ) ≡ (1 − ξγ )α( m ) where ξ = CmCi . Thus 2b (1) 1 − ξγ0 ≡ (1 − ξγ)α( ) (mod Ω ). m f n b n −b Now γ−1 = , γ0−1 = , so we get similarly (m replaced by Na m −Na m n):

2b (2) (1 − ξ−1γ−1) ≡ (1 − ξ−1γ0−1)α( ) (mod Ω ) m f 0 −k −1 since CnCi = ξ . Thus 1 − ξγ0 = −(1 − ξ−1γ0−1)ξγ0 −2b ≡ −ξ(1 − ξ−1γ−1)α( )γ0 by (2) n −2b ≡ (1 − ξγ)γ−1α( )γ0 n 2b ≡ (1 − ξγ)α( ), by (1). m

Thus (1 − ξγ)(1 − µ) ≡ 0 (mod Ωf ), where −2b 2b −2b −1 0 1 m µ = γ α( )γ α( ) = 2Na 4 . n m n mn − 3 2 4 The eigenvalues of µ are the root of x − ( mn − 2)x + 1, and are imaginary but not roots of 1; thus µ is elliptic of inﬁnite order. Hence, by Proposition 1.6, f|(1−ξγ) = 0.

Theorem 5.9. Let M0 be a subset of M meeting every primitive arithmetic pro- gression a+nb, (a, b) = 1, with (m, N) = 1 for m ∈ M0. Let ε be a character modulo N, and C = ±1. Suppose (A2) is satisfied, and (A3) is satisfied for every character 0 χ of conductor m ∈ M , with Cχ = Cε(m)χ(−N)gχ/gχ. Then f ∈ M(N, k, ε), k and f satisfies the functional equation f = Ci f|HN . If L(s) converges absolutely at s = k − δ for some δ > 0, then f is a cusp form. MODULAR FORMS AND DIRICHLET SERIES 55

a b Proof. (A2) holds, hence (B2), so f = Cikf|H . Let γ = ∈ Γ0 (N). We N Nc d 0 a 0 1 −c are to show f|γ = ε(d) · f. If b = 0, then γ = = H H−1 and Nc d N 0 1 N f = f|γ, as desired. Assume now b 6= 0. Then (a, Nb) = (d, Nb) = 1, so m = a + Nbs n = d + Nbt

0 0 k 0 for some m, n ∈ M and s, t ∈ Z. Now (A4) holds, with Cm = Ci ε(m), Cn = 1 0 1 0 m b Cikε(n), so C0 C0 = (−1)k. Let γ0 = γ = ∈ Γ0 (N); m n Nt 1 Ns 1 ∗ n 0 then f|γ0 = ε(m)−1f = ε(n)f = ε(d)f, by Lemma 5.8. Hence 1 0 1 0 f|γ = f| γ0 = f. −Nt 1 −Ns 1

Thus f is formally a modular form for Γ0(N), with character ε, and with the desired functional equation; it remains to verify the regularity conditions at the cusps. Lemma 5.10. Let f(τ) be holomorphic in the upper half plane, satisfying

(a) f|L = f for L ∈ Γ(N) (b) f is holomorphic at ∞, i.e. at 0 in the variable z = e2πiτ/N (c) f(x + iy) = O(y−σ) as y → 0, uniformly in x.

Then f(τ) is a modular form of level N, and a cusp form if σ < k, and f(∞) = 0.

Proof. Let τ0 = L(∞) be a rational cusp, where L ∈ Γ. By (a), f|L has a Laurent expansion in z = e2πiτ/N : ∞ X n f|L(τ) = anz n=−∞ and we want to show an = 0 for n < 0 (and also a0 = 0 if σ < k.) Now

Z τ0+1 −n an = f|L(τ)z dτ τ0 a b Let L = ; note c 6= 0; let τ = u + ib, where b is a large constant, u ≤ u ≤ c d 0 Im τ 1 σ u0 + 1. Then x + iy = L(τ), where y = |cτ+d|2 = O( b ). Hence f(L(τ)) = O(b ), σ−k σ−k 2πnb/N f|L(τ) = O(b ), an = O(b e ) as b → ∞; thus an = 0 for n < 0, and a0 = 0 if σ < k. Remark. It is reasonable that the condition (c), uniform in all x, should imply regularity at the cusps; we know the converse only for congruence subgroups, via the Eisenstein series.

c Returning to the proof of Theorem 5.9, we are given an = O(n ), and so by Proposition 1.1, f(x + iy) = O(y−c−1), and f is then a modular form, by the lemma. Finally, suppose L(s) converges absolutely at s = σ, σ < k. Then L(s) converges absolutely in the half plane Re (s) > σ and so a0 = 0, by (A2). By the lemma, it suﬃces to show that f(x + iy) = O(y−σ) as y → 0. Now n ∞ X σ X −σ σ sn = |aν | ≤ n |aν |ν = O(n ). ν=1 ν=1 56 ANDREW OGG

Hence X −2πny |f(x + iy)| ≤ |an|e

2πy X −2πny ≤ (1 − e ) sne = O(y)O(y−σ−1) = O(y−σ),

(Cf. proof of Proposition 1.1.)

Weil’s theorem leads to a very interesting conjecture on the zeta-function of ζ(s)ζ(s−1) an elliptic curve E defined over Q. This zeta-function can be written L(s) , Q where L(s) = p Lp(s), the local factor Lp(s)being defined as follows. If E has non-degenerate reduction at p (which will be the case for all but a finite number −s 1−2s −1 of the p), then Lp(s) = (1 − app + p ) , where 1 + p − ap is the number of points on the reduced curve with coordinates in the field Fp of p elements. If the reduction of E at p is singular, we set Lp(s) = 1 if the singular point is a cusp, and −s −1 Lp(s) = (1−app ) if the singular point is a node, where ap = +1 if the tangents at the double point are rational over Fp, and ap = −1 otherwise. One can define Q np an integer N = p p , the conductor of E, with np = 0 if E is non-degenerate at p, np = 1 if the reduction has a node, and np ≥ 2 if the reduction has a cusp (and np = 2 for p 6= 2, 3). The Hasse-Weil conjecture is that the hypotheses of Theorem 5.9 hold for L(s), with N the conductor of E, k = 2, and ε = 1. It would then follow that L(s) is associated to a form f of dimension −2 for Γ0(N), with a functional 1/2 equation. Actually, one knows |ap| ≤ 2p ; this is the Riemann hypothesis for elliptic curves, proved by Hasse in 1934. Hence L(s) converges for Re (s) > 3/2, so f will be a cusp form. Note the agreement of the Riemann hypothesis and the Petersson conjecture. A good introduction to algebra-geometric aspects of this subject is Shimura [13] (and many other papers of Shimura.)

6. Quadratic forms

The most fruitful method of constructing modular forms of higher level is to form theta-series of positive integral quadratic forms. The basic references are Hecke’s “Analytische Arithmetik der positiven quadratischen Formen” [5, No. 41], and Schoeneberg [12]; we will give only a very small part of the theory here. Let r 1 1 X Q(x) = xtAx = a x x , xt = (x , . . . , x ), x ∈ R,A = r × r matrix 2 2 ij i j 1 r j i,j=1 be a positive deﬁnite integral quadratic form, i.e. Q(x) > 0 for x 6= 0, and aij = aji is an integer, aii is an even integer, i.e. A = (aij) is an integral symmetric matrix. The theta-function associated to Q is ∞ X 2πiQ(n)τ X 2πiντ ϑ(τ; Q) = e = 1 + aQ(ν)e n ν=1 t r where n = (n1, . . . , nr) ∈ Z , and aQ(ν) is the number of integral solutions of Q(x) = ν. We will show eventually that ϑ(τ; Q) is a modular form of dimension −k Pr 2 of a certain level N, in the case where r = 2k is even. Note Q(x) ≥ C i=1 xi , for some C > 0, since Q is positive deﬁnite, and so ϑ(τ; Q) is dominated term-by-term by-term X −2πyC P n2 X −2πyCm2 r e i = ( e ) , n∈Zr m∈Z MODULAR FORMS AND DIRICHLET SERIES 57 where y = Im τ, and so ϑ(τ; Q) is a holomorphic function on the upper half plane. A is the matrix of Q, its determinant D is the determinant of Q, and (if r = 2k is even) ∆ = (−1)kD is the discriminant of Q.

Lemma 6.1. Let A = (aij) be an integral symmetric matrix (aii even). If r is odd, then D = det A is even; if r is even, then ∆ = (−1)r/2D is ≡ 0, 1 (mod 4). P Proof. D = σ sgn (σ)a1σ1 a2σ2 ··· arσr , the sum ranging over all permutations σ of r letters. Since A is symmetric, the terms for σ and σ−1 are the same, and so cancel modulo 2 if σ 6= σ−1, i.e. σ2 6= 1. If r is odd, any σ with σ2 = 1 ﬁxes some letter i, and since aii is even, we get D ≡ 0 (mod 2). We leave the second statement, which we do not need, to the reader.

−1 Let A be integral symmetric, with r = 2k even. Then DA = (Aij) is the cofactor mytrix, which is still integral symmetric, by the lemma. Thus DA−1 is 1 t −1 again the matrix of an integral positive definite quadratic form 2 x DA x. The least positive integer N with NA−1 = A∗ integral is the level (Stufe) of Q, and the corresponding quadratic form 1 Q∗(x) = xtNA−1x 2 ∗ 1 ∗ is the adjoint form to Q; note N | D, and Q (x) is primitive; i.e. ν Q (x) is 1 ∗ not integral for ν > 1, i.e. the greatest common divisor of the coefficients 2 aii ∗ ∗∗ ∗ ∗∗ ∗ ∗−1 N ∗ 1 and aij is 1. The adjoint Q of Q has matrix A = N A = N A = β A, 1 ∗ where β = gcd( 2 aii, aij). In particular, if Q is primitive, then N = N , i.e. Q and its adjoint Q∗ have the same level, and Q∗∗ = Q. In general, N ∗ | N, and D∗ = N 2kD−1, so D | N 2k. Thus: Proposition 6.2. The determinant D and level N of Q satisfy N | D | N 2k; hence N and D have the same prime factors, and for a given level N and number of variables 2k, there are only finitely many corresponding discriminants ∆.

The basic result, due to Schoeneberg [12], which we prove eventually, is that ∆ ϑ(τ; Q) ∈ M(N, k, ε), where ε(n) = n (Jacobi symbol) for n > 0, ε(−n) = (−1)kε(n); the two uses of the word level then agrees. We also need a modified theta-function, using spherical functions. Let A be a symmetric positive definite real matrix of degree r, defining a quadratic form xtAx. By linear change of variables y = Bx, we diagonalize the quadratic form: r X 2 t t −1 t −1 yi = x Ax = y (B ) AB y i=1 i.e. I = (B−1)tAB−1, or A = BtB.(B is a real matrix.) Now a function f(x) is a spherical function with respect to the quadratic form 2 2 t P ∂ f P ∂ f ∂xj ∂xk −1 ∗ −1 x Ax if 2 = 0, i.e. 0 = i,j,k . Letting A = (aij), B = (bij), ∂yi ∂xj ∂xk ∂yi ∂yi ∂x we see P j ∂xk = P b∗ b∗ = a∗ , since A−1 = B−1(B−1)t, so f(x) is a spherical i ∂yi ∂yi i ji ki jk 2 function relative to Q if and only if P a∗ ∂ f = 0. ij ∂xj ∂xk There is an inner product on functions f(x), x ∈ Rr, by Z (f, g)A = f(x)g(x)dx1 . . . dxr xtAx≤1 1 Z = f(x)g(x)dy1 . . . dyr |B| yty≤1 58 ANDREW OGG

Theorem 6.3. Let f(x) be a homogeneous polynomial of degree ν in x1, . . . , xr, with complex coeﬃcients. Then the following statements are equivalent:

(1) f(x) is a spherical function with respect to xtAx (2) f(x) is orthogonal (in the above inner product) to all homogeneous polynomials of degree < ν. (3) f is a linear sum of functions of the form (ζtAx)ν , where ζ ∈ Cr, ζtAζ = 0.

Proof. Translating (3) into variables y, if we let η = Bζ, then ζtAζ = ηtη, and ζtAx = ηty; thus we can assume without loss of generality that A = I is the identity. If (3) holds, then so does (1), since 2 2 X ∂ f X ∂ X ν X 2 ν−2 X 2 2 = 2 ( ζjxj) = ν(ν − 1)( ζi )( ) = 0 if ζi = 0. ∂xi ∂xi

P ∂f In general, f is homogeneous of degree ν, and so · xi = νf; hence the ∂xi divergence theorem gives: Z Z X ∂f Z 1) ν fω = ( · x )ω = ∆fdx ∂x i ∂K ∂K i i K

P 2 P 2 P ∂2f P where K : xi ≤ 1, ∂K : xi = 1, ∆f = 2 and ω = xiωi, ωi = ∂xi i−1 (−1) dx1 ... dxci . . . dxr, dx = dxiωi = dx1 . . . dxr. ∆ satisﬁes X ∂f ∂g 2) ∆(fg) = f∆g + g∆f + 2 . ∂xi ∂xi By Stokes’ theorem, Z Z X Z X ∂ Z fω = fxiωi = (fxi)dx = (ν + r) fdx. ∂K ∂K K ∂xi K Thus: Z Z 3) ∆f = ν(ν + r) f. K K We now prove (1) =⇒ (2) by induction on ν. Note that ∆f = 0 =⇒ ∆( ∂f ) = 0. ∂xi Assuming ∆f = 0, and deg(g) < deg(f): Z Z fg = ( ) ∆(fg), by 3) K K Z = ( ) f∆g, by 2) and induction K Z = ( ) f∆2g = ··· = 0. K

Finally, to show (2) =⇒ (3), let f be orthogonal to all g of lower degree, and orthogonal to all (ζtx)ν , where ζtζ = 0; we are to show f = 0. Let g(x) = (ζtx)ν ; then g and all of its partial derivatives satisfy (3), hence (1) and (2). Then Z Z Z X ∂f ∂g 0 = fg = ( ) ∆(fg) = ( ) K K K ∂xi ∂xi Z X ∂ν f ∂ν g = ··· = ( ) . K ∂xi1 . . . ∂xiν ∂xi1 . . . ∂xiν MODULAR FORMS AND DIRICHLET SERIES 59

P ∂f P ∂ν f ∂ν g Now iteration of νf = ·xi gives ν!f = , ν!ζi1 . . . ζiν = , ∂xi ∂xi1 ...∂xiν ∂xi1 ...∂xiν so the above gives simply that f(ζ) = 0 when ζtζ = 0, and hence f(x) is divisible t P 2 by δ(x) = x x = xi , say f(x) = δ(x)g(x). Then, from the equation 1) and 3): Z Z gg = ( ) ggω K ∂K Z = ( ) δggω since δ = 1 on ∂K ∂K Z = ( ) fg = 0. K Thus g = 0, so f = 0.

Corollary 6.4. The space Hν of spherical functions which are homogeneous polynomials of degree ν has dimension r − 1 + ν r − 3 + ν − r − 1 r − 1

Proof. Let Pν be the homogeneous polynomials of degree ν. Then Hν = {f ∈ Pν : f ⊥ P µ for µ < ν}. Now Pµ ⊥ Pν if µ+ν is odd, so given f ∈ Pν , f ∈ Hν ⇐⇒ f ⊥ Pν−2,Pν−4,... . On the other hand, (f, g) = ( )(f, δg), so f ∈ Hν ⇐⇒ f ⊥ Pν−2; r−1+ν r−1+ν−2 thus dim Hν = dim Pν − dim Pν−2 = r−1 − r−1 .

1 t Given an integral positive definite quadratic form Q(x) = 2 x Ax, and P (x) a spherical function of order ν with respect to Q, we have a theta-series X ϑ(τ; Q, P ) = P (n)e2πiQ(n)τ n∈Zr which we will prove is a modular form with character ε for Γ0(N), of dimension −(k +ν) (if r = 2k is even), and a cusp form if ν > 0. The introduction of the P (x) is somewhat like passing from zeta-functions to L-series by introducing characters. Proposition 6.5. Given a positive definite symmetric real matrix A of degree r, 1 t Q(x) = 2 x Ax, define X ϑ(τ, x) = e2πiQ(n+x)τ n∈Zr for a parameter x ∈ Rr. Then

τ r/2 1 X t t −1 ϑ(τ, x) = √ e2πin x−(πi/τ)n A n. i D n

Proof. (Cf. the same result for r = 1 in Chapter1.) ϑ(τ, x) = periodic function of x = its Fourier series X 2πimtx = ame m∈Zr where Z 1 Z 1 −2πimtx am = ··· ϑ(τ, x)e dx1 . . . dxr 0 0 Z 1 Z 1 πiτxtAx−2πimtx = ··· e dx1 . . . dxr. −∞ −∞ 60 ANDREW OGG

Completing the square, τ(x − τ −1A−1m)tA(x − τ −1A−1m) = τxtAx − 2mtx + τ −1mtA−1m. −πiτ −1mtA−1m Hence am = e bm, where Z πiτ(x−τ −1A−1m)tA(x−τ −1A−1m) bm = e dx Rr

Z t = eπiτx Axdx, by Cauchy’s theorem Rr 1 Z t = √ eπiτy ydy, D Rr where y = Bx, dy = |B|dx, A = BtB ∞ r 1 Z 2 = √ e2πiτu du D −∞ 1 τ −r/2 = √ D i Corollary 6.6. If Q(x) is integral in r = 2k variables, then

τ k −1 ϑ(τ, Q) = D−1/2ϑ ,Q∗. i Nτ

∗ −1 Proof. Set x = 0, and note Q has matrix NA .

We now generalize the above corollary to ϑ(τ; Q, P ) = P P (n)e2πiQ(n)τ , where P (x) is a spherical function relative to Q(x). From now on, Q(x) will always be an integral positive definite quadratic form in an even number 2k of variables. We take first a typical spherical function P (x) = (ζtAx)ν , where Q(ζ) = 0. The transformed theta-function will involve P ∗(x) = P (A−1x) = (ηtA−1x)ν , where η = Aζ satisfies ηtA−1η = ζtAζ = 0, i.e. Q∗(η) = 0; thus P ∗(x) is a spherical function relative to Q∗(x). P 2πiQ(n+x)τ P ∂ We start with ϑ(τ, x) = e and apply L = ζi , ν times. Note n ∂xi LQ(x) = ζtAx, L2Q(x) = ζtAζ = 0. Thus: X (1) Lν ϑ = (2πiτ)ν (ζtA(xn + x))ν e2πiQ(n+x)τ n But by the transformation formula (Proposition 6.5), we have also τ −k X t −1 t ϑ = D−1/2 e(−πi/τ)n A n+2πin x i n and hence τ −k X t −1 t (2) Lν ϑ = D−1/2(2πi)ν (ζtn)ν e(−πi/τ)n A n+2πin x i n and since P ∗(x) = P (A−1x) = (ζtx)ν , we have proved for the spherical function and hence (by Theorem 6.3) for any spherical function P (x), by comparing (1) and (2): Theorem 6.7 (Schoeneberg). If P (x) is a spherical function for Q(x), and P ∗(x) = P (A−1x) the adjoint function (a spherical function for Q∗(x)), then: k X i X t −1 t P (n + x)e2πiQ(n+x)τ = √ P ∗(x)e(−πi/τ)n A n+2πin x k+ν n Dτ n MODULAR FORMS AND DIRICHLET SERIES 61

This suggests our ‘k’ will be k + ν, so we set a b k+ν −(k+ν) aτ + b f| (τ) = (ad − bc) 2 (cτ + d) f( ); c d cτ + d 0 −1 setting x = 0 above, and H = as usual, we have N N 0 ik −1 ϑ(τ; Q, P ) = √ ϑ( ; Q∗,P ∗) Dτ k+ν Nτ k k+ν i N 2 ∗ ∗ = √ ϑ(τ; Q ,P )|HN D Thus: Corollary 6.8. k k+ν i N 2 ∗ ∗ ϑ(τ; Q, P ) = √ ϑ(τ; Q ,P )|HN D

h t Taking x = N , where h = (h1, . . . , h2k) is integral, the formula of Theorem 6.7 reads: X 2 (3) ϑ(τ; Q, P, h) := N −ν P (n)e2πiQ(n)τ/N n≡h (N) k i X t −1 t = √ P ∗(n)e(−πi/τ)n A n+2πin h/N . k+ν Dτ n On the right, substitute m = NA−1n. Then m is integral, and Am ≡ 0 (mod N); on the other hand, if m is integral, and Am ≡ 0 (mod N), then n = N −1Am is integral. Also, P ∗(n) = P (A−1n) = N −ν P (m). The right side of (3) is thus:

k i X 2 t 2 (4) √ P (m)e(−2πi/τ)Q(m)/N +2πim Ah/N k+ν ν Dτ N m Am≡0 (N)

t 2 Now suppose Ah ≡ 0 (mod N). Then e2πim Ah/N depends only on m modulo N, so (4) becomes:

k i X t 2 −1 (5) √ e2πig Ah/N ϑ ; Q, P, g. k+ν τ Dτ g mod N Ag≡0 (N) This proves: Corollary 6.9. For an integral vector h with Ah ≡ 0 (mod N), deﬁne X 2 ϑ(τ; Q, P, h) = N −ν P (n)e2πiQ(n)τ/N . n≡h (N) (Note ϑ(τ; Q, P, 0) = ϑ(τ; Q, P ).) Then

k i X 2πigtAh/N 2 ϑ(τ; Q, P, h) = √ e ϑ(τ; Q, P, g)|H1 D g mod N Ag≡0 (N) We also have obviously that

2 ϑ(τ + 1; Q, P, h) = e2πiQ(h)/N ϑ(τ; Q, P, h) 62 ANDREW OGG

Hence the vector space generated by the ϑ(τ; Q, P, h) is operated on by the full modular group Γ. These series are clearly regular at ∞, and vanishing at ∞ if ν > 0. Hence we only need to check their invariance under Γ(N) to know they are modular forms of level N (cusp forms if ν > 0).

Remark. Suppose N = 1, i.e. D = 1. Then ϑ(τ + 1; Q) = ϑ(τ; Q) and ϑ(τ; Q) = τ −k −1 k ( i ) ϑ( τ ; Q). By Chapter1, we know i = 1, i.e. 4 | k; the number of variables is thus divisible by 8. We have ϑ(τ, Q) is a modular form of dimension −k and level 1. An example with r = 8 is

8 8 1 X 1 X Q(x) = x2 + ( x )2 − x x − x x , 2 i 2 i 1 2 2 8 i=1 i=1

P∞ 2πiντ with ϑ(τ, Q) = 1 + ν=1 aQ(ν)e . Since dim M(Γ, 4) = 1, we have necessarily

∞ X 2πiντ ϑ(τ, Q) = E4(τ) = 1 + 240 σ3(ν)e , ν=1

so Q has representation numbers aQ(ν) = 240σ3(ν). Similarly, if Q has 16 variables, then aQ(ν) = 480σ7(ν). For k = 12, Siegel has proved there exists two forms Q1,Q2 in 24 variables with discriminant 1 and diﬀerent theta-series; then

∞ X 2πiντ ϑ(τ; Q1) − ϑ(τ; Q2) = (aQ1 (ν) − aQ2 (ν))e ν=1 is a non-zero cusp form of dimension −12, i.e. c∆(τ), c 6= 0. Ramanujan’s conjecture can be thought of an assertion about the diﬀerences aQ1 (ν) − aQ2 (ν). In general, if Q(x) is a form of discriminant 1 in 2k variables (4 | k), then

ϑ(τ, Q) = Ek(τ) + g(τ) where g(τ) is a cusp form. In terms of Fourier coeﬃcients, we have

aQ(ν) = Akσk−1(ν) + bν

k/2 where bν = O(ν ), by Proposition 4.13, so the theory of modular forms gives asymptotic results about the representation numbers aQ(ν).

Returning to our general development, besides the rules above, we also have, for any natural number c:

X (6) ϑ(τ; Q, P, h) = ϑ(cτ; cQ, P, g) g≡h (N) g mod cN MODULAR FORMS AND DIRICHLET SERIES 63

a b Let ∈ Γ0, with c > 0. Then c aτ+b = a − 1 , so: c d cτ+d cτ+d

a b X 1 ϑ(τ; Q, P, h)| = (cτ + d)−(k+ν) ϑ(a − ; cQ, P, g) c d cτ + d g≡h (N) g mod cN 2 X e2πiaQ(g)/cN −1 = ϑ( ; cQ, P, g) (cτ + d)k+ν cτ + d g≡h (N) g mod cN k k+ν i (−1) X 2 = √ e2πiaQ(g)/cN × ck D g≡h (N) g mod cN X t 2 e2πi` Ag/cN ϑ(cτ + d; cQ, P, `) ` mod cN A`≡0 (N) (The last by Corollary 6.9; the determinant of cQ is c2kD.) Thus:

a b i−k−2ν X (7) ϑ(τ; Q, P, h)| = √ ζ(h, `)ϑ(cτ; cQ, P, `) c d k c D ` mod cN A`≡0 (N) where X t 2 (8) ζ(h, `) = e2πi(aQ(g)+` Ag+dQ(`))/cN g≡h (N) g mod cN a b for ∈ Γ0, c > 0. One computes: c d

t 2 (9) ζ(h, `) = e−2πi(h A`+dQ(`))b/N ζ(h + d`, 0). This shows that ζ(h, `) depends only on ` modulo N; hence, using (6), we rewrite (7) as

a b i−k−2ν X (10) ϑ(τ; Q, P, h)| = √ ζ(h, `)ϑ(τ; Q, P, `) c d k c D ` mod N A`≡0 (N) In particular, if d ≡ 0 (mod N), then (10) becomes a b ζ(h, 0) X t 2 (11) ϑ(τ; Q, P, h)| = √ e−2πih A`·b/N ϑ(τ; Q, P, `) c d k+2ν k i c D ` mod N A`≡0 (N) 0 −1 and applying H = , we get 1 1 0 b −a ζ(h, 0) X t 2 (12) ϑ(τ; Q, P, h)| = e2πi` A(g−bh)/N ϑ(τ; Q, P, g) d −c (−1)kckD `,g

P 2πi`tA(g−bh)/N 2 Now ` mod N e is a character sum on a ﬁnite group with D ele- A`≡0 (N) ments, so this sum is D if g ≡ bh (mod N) and 0 otherwise. Thus (12) becomes: b −a ζ(h, 0) (13) ϑ(τ; Q, P, h)| = ϑ(τ; Q, P, bh) d −c (−c)k 64 ANDREW OGG

a b for c > 0 and N | D. This then holds also for c < 0, since if we replace c d −a −b by , this gives a factor of (−1)k+ν on both sides of (13), since ϑ(−h) = −c −d (−1)ν ϑ(h). Changing the notation, a b ϕ(h) (14) ϑ(τ; Q, P, h)| = ϑ(τ; Q, P, ah), c d dk a b for ∈ Γ0 (N), where (cf. (8)) c d 0 X 2 (15) ϕ(h) = e2πibQ(g)/dN g≡h (N) g mod dN

In this sum, we can write g = adh + Ng1, g1 mod d, and (15) becomes 2 X (16) ϕ(h) = e2πiabQ(h)/N e2πibQ(g1)/d

g1 mod d Thus, (14) becomes a b 2 (17) ϑ(τ; Q, P, h)| = e2πiabQ(h)/N ϑ(τ; Q, P, ah)Φ(b, d) c d a b where Φ(b, d) = d−k P e2πibQ(g)/d, for ∈ Γ0 (N). g mod d c d 0 In order to investigate the Φ(b, d) further, let us take P = 1, h = 0, so we know 1 n ϑ(τ; Q, P, h) = ϑ(τ; Q) 6= 0. Applying to both sides of (17), we find 0 1 Φ(b, d) = Φ(b + na, d + nc) is in the field of (d + nc)th roots of 1 for all n ≥ 1 and hence Φ(b, d) is rational. Applying the automorphism e2πib/d 7→ e2πi/d, we find a bc0 a b Φ(b, d) = Φ(1, d) = ε(d). Finally, since ∈ Γ0 (N) if ∈ Γ0 (N), N d 0 Nc0 d 0 we see ε(d) depends only on d modulo N. We have proved: Theorem 6.10 (Schoeneberg). We have a b 2 ϑ(τ; Q, P, h)| = e2πiabQ(h)/N ε(d)ϑ(τ; Q, P, ah) c d a b for ∈ Γ0 (N), where ε is a real character modulo N, satisfying ε(−1) = c d 0 a b (−1)k. In particular, taking ∈ Γ0(N), c d a b ϑ(τ; Q, P, h)| = ϑ(τ; Q, P, h), c d and hence ϑ(τ; Q, P, h) is a modular form of dimension −(k + ν), of level N, and a cusp form if ν > 0.

It remains to determine ε(d) for d > 0 (we know ε(−1) = (−1)k). We have X ε(d) = d−k e2πiQ(g)/d, g mod d and ε(d) depends only on d modulo N. Let p be an odd prime with p ≡ d (mod N) P 2 and so that Q(x) can be diagonalized modulo p with integers, say Q(x) ≡ j ajxj MODULAR FORMS AND DIRICHLET SERIES 65

(mod p), where aj ∈ Z. Then 2k p −k X X 2πi P a g2/p ε(d) = ε(p) = p e j j

j=1 gj =1 2k −k Y X 2πia g2/p = p e j j

j=1 gj mod p 2k Y X zj = p−k 1 + e2πiaj zj /p p j=1 zj mod p

zj zj 2 where p = χ(zj) is the Legendre symbol, i.e. p = 0, 1, −1 as p | zj, x ≡ zj Q2k (mod p) is solvable, or otherwise. Note that D ≡ j=1(2aj) (mod p); hence p - aj (otherwise p | D, p | N, p | d, contrary to (d, N) = 1). Thus P e2πiaj zj /p = zj mod p 0, and the above becomes: 2k −k Y X 2πiaj zj /p ε(d) = ε(p) = p χ(zj)e

j=1 zj mod p 2k −k Y = p gχ(aj) (Gauss sums) j=1 2k −k Y = p (χ(aj)gχ), by Proposition 5.2 j=1 −k 2k = p gχ χ(D), Q2k 2 2 since D ≡ j=1(2aj) (mod p). Now p = |gχ| = gχgχ = χ(−1)gχ, so we have finally that ε(d) = ε(p) = χ((−1)kD) = χ(∆). ∆ Thus we have proved that ε(d) = p for any prime p which is sufficiently large and satisfies p ≡ d (mod N). Hence ∆ is a discriminant, we necessarily have ∆ ≡ 0, 1 ∆ (mod 4), and ε(d) = d (Jacobi symbol). (Here we are appealing to basic facts 0 ∆ about quadratic number fields; if ∆ ≡ 2, 3 (mod 4), then the conductor N of x is 4 or 8 times some of the odd primes dividing D, whence N 0 does not divide N, since N | D. Cf., e.g., Hecke’s book [6, Chapter VII.]) Thus: ∆ Theorem 6.11. ∆ ≡ 0, 1 (mod 4), and ε(d) = d for d > 0. Remark. As sketched earlier for the case N = 1, one gets asymptotic results on the representation numbers aQ(ν) by writing ϑ(τ, Q) as an Eisenstein series plus a cusp form. Finally, the theory of the T (n) gives a way of deriving knowledge of the representation numbers aQ(n) from those for primes. Starting from ∞ X 2πinτ f1(τ) = ϑ(τ, Q) = aQ(n)e , n=0 one gets a basis f1, . . . , fr for the least space V (Q) containing ϑ(τ, Q) and closed under all T (n), (n, N) = 1, by taking fj = f1|T (nj) for suitable integers 1 = n1 < n2 < ··· < nr. Note each fj has integral Fourier coefficients; hence we can diagonalize the T (n) over a certain number field K. Thus certain K-linear sums g1, . . . , gr of f1, . . . , fr will be eigenfunctions for the T (n), (n, N) = 1, and hence their Fourier coefficients an, (n, N) = 1, are known once the ap, p - N, are known 66 ANDREW OGG

(for g1, . . . , gr). Furthermore, one can determine f2, . . . , fr, g1, . . . , gr from Q by a ﬁnite process. For details and examples, we refer again to Hecke’s “Analytische Arithmetik der positiven quadratischen Formen” [5, No. 41].

Corrected typos in Ogg’s book

Typos in Ogg’s book [8] are listed with reference to the chapter, page, line (from the top > 0, from the bottom < 0). They have been corrected in this TEX edition. Typos occurring on several lines repeatedly have only been listed once (like τ 0 = −1/τ + 1 instead of τ 0 = −1/(τ + 1)) Chap. Page Line replace this by this Intro xv −5 c = ik C = ik I 27 +3 (m + τ)−k (m + nτ)−k I 29 −7 k 6≡= 2 (mod 12) k 6≡ 2 (mod 12) I 32 −3 t = e−2πi/τ+1 t = e−2πi/(τ+1) I 37 −2 (n2 + m2)−5 (n2 + m2)−s II 16 +5 Peterson’s Petersson’s III 2 −6 P1,...,P0 P1,...,Pσ III 13 −4, −3 f f1 IV 3 −9 M(N) M(N) IV 13 +5 ineducible irreducible IV 16 +1 if f|U = ζ · U f|U = ζ · f N N IV 17 −2 (m, t ) 6= 0 (m, t ) 6= 1 V 9 +5 abN ≡ −1 (mod M) abN ≡ −1 (mod m) V 13 −8 Cχ = Cε(m)χ(−N)gχ|gχ Cχ = Cε(m)χ(−N)gχ/gχ 0 k 0 k V 14 +7 Cn = Ci ε(m) Cn = Ci ε(n) V 15 +6 x0 τ0 VI 2 −8 c C VI 3 +4 σ of n letters σ of r letters VI 8 −7 ∂fi ∂gi ∂f ∂g ∂xi ∂xi ∂xi ∂xi VI 17 −3 Ramamyan Ramanujan VI 19 −5 A`|0 (N) A` ≡ 0 (N)

References [1] Lars V. Ahlfors, Complex Analysis, 3rd ed., McGraw-Hill, New York, 1979. [2] B. Berlowitz, Extensions of a theorem of Hardy, Acta Arith. 14 (1968), 203–207. [3] Robert C. Gunning, Lectures on Modular Forms, Annals of Mathematics Studies, vol. 48, Princeton University Press, Princeton, 1962. [4] H. Hamburger, Über die Riemannsche Funktionalgleichung der ζ-Funktion, Math. Zeitschr. 10, 11, 13 (1921, 1921, 1922), 240–254, 224–245, 283–311. [5] Erich Hecke, Mathematische Werke, Vandenhoeck & Ruprecht, Göttingen, 1959. 3. Auﬂage 1983. [6] , Vorlesungen über die Theorie der Algebraischen Zahlen, 2nd ed., AMS Chelsea Pub- lishing, New York, 1923. re-issue 1970. [7] Louis Joel Mordell, On Mr. Ramanujan’s Empirical Expansions of Modular Functions, Proc. Cambridge Phil. Soc. 19 (1917), 117–124. [8] Andrew P. Ogg, Modular Forms and Dirichlet Series, Mathematics lecture note series, Ben- jamin, 1969. [9] Hans Petersson, Konstruktion der sämtlichen Lösungen einer Riemannschen Funktion- algleichung durch Dirichlet-Reihen mit Eulerscher Produktentwicklung, Math. Ann. 116, 117 (1939, 1940/41), 401–412, 39–64, 277–300. [10] Bernhard Riemann, Gesammelte Mathematische Werke, 2. Auﬂ., Teubner, Leipzig, 1892. [11] Walter Rudin, Real and complex analysis, 3rd ed., McGraw-Hill, 1987. MODULAR FORMS AND DIRICHLET SERIES 67

[12] Bruno Schoeneberg, Das Verhalten von mehrfachen Thetareihen bei Modulsubstitutionen, Math. Ann. 116 (1939), 511–523. [13] Goro Shimura, The zeta-function of an algebraic variety and automorphic functions, Arith- metical Algebraic Geometry, Proc. Conf. at Purdue Univ., Harper and Row, New York, 1965, pp. 6–31. [14] Carl Ludwig Siegel, A simple proof of η(−1/τ) = η(τ)pτ/i, Mathematika 1 (1954), 4–4. [15] André Weil, Über die Bestimmung Dirichletscher Reihen durch Funktionalgleichungen, Math. Ann. 168 (1967), 149–156. Index A homogeneous principal congruence adjoint form subgroup ...... 34 of quadratic form Q...... 57 I C integral symmetric matrix ...... 56 character . . 27, 38, 39, 43, 44, 47, 50, 52–55, J 59, 63, 64 Jacobi symbol ...... 57, 65 primitive ...... 51 conductor ...... 51–54, 56, 65 L congruence subgroup ...... 34 Legendre symbol ...... 36, 65 conjecture level of quadratic form ...... 57 Hasse-Weil...... 56 Petersson...... 29, 45, 56 M Ramanujan ...... 29, 62 Möbius function...... 46 correspondence ...... 24, 37 matrix A cusp . . . . 15, 16, 29–31, 35, 36, 45–47, 55, 56 of quadratic form Q...... 57 cusp form . . . . . 11, 14, 19, 24, 26, 29–31, 36, Mellin inversion formula ...... 4 43–45, 47–50, 54–56, 59, 62, 64, 65 modular form ...... 23, 30 of dimension −k ...... 3 D of level N and dimension −k ...... 34 Dedekind’s η-function ...... 19 modular group ...... 4, 13, 24 degree of divisor ...... 24 determinant D O of quadratic form Q...... 57 O-condition ...... 3 discriminant ∆ ...... 65 order of function ...... 20 of quadratic form Q...... 57 order of zero ...... 15, 30 divisor ...... 39 P divisor group...... 24 parabolic ﬁxed point ...... 15 E Petersson conjecture ...... 29, 45, 56 EBV. . .see entire, bounded in vertical strip Petersson inner product ...... 31 Eisenstein series . . 12, 14, 18, 28, 45, 47–50, Poisson summation formula ...... 17 55, 65 positive divisor...... 24 primitive character ...... 47, 51 Gk ...... 12, 23 normalized...... 13 primitive quadratic form ...... 57 primitive ...... 46 primitive sublattice ...... 24 restricted ...... 46, 47 principal congruence subgroup elliptic ﬁxed point...... 15 of level N ...... 34 elliptic modular invariant ...... 11, 14, 35 of level 2...... 15 elliptic substitution ...... 9 Q entire, bounded in vertical strip...... 3, 52 quasi-regular ...... 15 ε-Hermitian ...... 39 Euler product ...... 4, 20, 27–29, 33, 39, 41 R relative to p ...... 28, 42 Ramanujan conjecture ...... 29, 62 Riemann hypothesis...... 56 F Riemann zeta-function ...... 2 functional equation2–4,6,9, 13, 17, 19–21, Riemann-Hurwitz formula ...... 29 50, 51, 53–56 fundamental domain . .6, 11, 13, 15, 29, 31, S 32, 44, 46 signature (λ, k, C) ...... 3 spherical function...... 57–60 G Stirling’s formula...... 4 Gauss sum ...... 51, 65 Stufe...... 57 genus ...... 29 growth function ...... 20 T theta-function ...... 2, 17, 19, 56, 57, 60 H theta-series ...... 56, 59, 62 Hasse-Weil conjecture ...... 56 Hecke operator .4, 24, 26–28, 32, 33, 36, 37, V 39, 41, 45, 47 value holomorphic at ∞ ...... 3, 24 of f at ∞ ...... 30 homogeneous modular group ...... 23 vanishes at ∞ ...... 3 68