Some fundamental results on modular forms

Jordan Bell
May 2, 2007

1 Introduction

The purpose of this paper is to give complete proofs to several fundamental results about modular forms. Modular forms are complex functions with certain analytic properties, and that transform nicely under a certain group of transformations of the complex upper half plane. It turns out that modular forms can be used to study analytic number theory, by investigating the coefficients in series expansions of the modular forms. In fact, often using modular forms we can discover and prove things in number theory where a direct proof might not be obvious. To help the reader get a feel for this, we will give an example; all the terms used here are defined in the paper. An important class of modular forms, called Eisenstein series, have expansions that involve the divisor sum σr(n)=d|ndr. Another modular form, called the modular discriminant, has a series expansion that involves the Ramanujan tau function, τ(n). Modular forms comprise vector spaces whose dimensions we can explicitly determine. Using this information one can prove the congruence


We will develop all the results needed to prove this congruence and other similar results (cf. [9]).

We have tried to find the clearest proofs in the literature for the results about modular forms in this paper. The proofs use complex analysis and group theory, and do not assume advanced results.

In §2 we introduce a group action on the complex upper half plane, under which modular forms are well behaved.

Next in §3 we define modular forms and cusp forms, which are defined on the upper half plane. We show that modular forms have Fourier expansions. Indeed it is often convenient to study a modular form through its Fourier series. We also show that modular forms constitute complex vector spaces. We explicitly construct a class of modular forms, namely the Eisenstein series, in §4. However, these are not cusp forms. In §5 we construct a cusp form of weight 12, and prove estimates on the magnitude of the Fourier coefficients of cusp forms.

Finally in §6 we give a formula for the dimensions of spaces of modular forms. We also introduce a natural inner product on the spaces of cusp forms.

We will now introduce some notation that will be used throughout this paper. Let be the integers, be the complex numbers, be the real numbers, and 2=×, 2=×. For a commutative ring R with unity 1, let SL2(R) be the group of all 2×2 matrices over R with determinant 1. For integers m,n not both 0, let (m,n) denote their greatest common divisor.

2 Automorphisms of the complex upper half plane

Let ={z:(z)>0} be the complex upper half plane. Modular forms are defined on the upper half plane, and transform nicely under a certain group that acts on . We will present this group now, and prove several properties about it.

Γ=SL2() is called the modular group.

Lemma 1.

Γ has an action on H defined by γz=az+bcz+d for γ=(abcd)Γ and zH.


If z and γΓ, then

(γz) = (az+bcz+d)
= (az+bcz+dcz¯+dcz¯+d)
= (ad-bc)(z)|cz+d|2
= det(γ)(z)|cz+d|2
= (z)|cz+d|2
> 0.

Therefore γz.

Let γ=(abcd),β=(efgh)Γ and let z. Then

β(γz) = β(az+bcz+d)
= eaz+bcz+d+fgaz+bcz+d+h
= eaz+eb+fcz+fdcz+dgaz+gb+hcz+hdcz+d
= eaz+eb+fcz+fdgaz+gb+hcz+hd
= (ea+fc)z+eb+fd(gz+hc)z+gb+hd
= (ea+fceb+fdgz+hcgb+hd)z
= (βγ)z.

As well, for all z, (1001)z=1z+00z+1=z. Hence the identity element of Γ fixes all z. ∎

The proof of the following theorem follows [2, Chapter I, 5.7].

Theorem 2.

Γ is generated by A=(1101) and B=(0-110).


Let M=(abcd)Γ. We may assume that |c||d| since BM=(b-ad-c) is generated by A and B if and only if M is. If c=0, then ad=1, and so a=d=1 or a=d=-1. In the first case, M=Ab, and in the second case, M=AbB2.

We now assume that for n>1, and all |c|<n the element M is generated by A and B. Because B2M=(-a-b-c-d) is generated by A and B if and only if M is, we may assume that c>0. For c=n, let k be an integer such that 0ck+d<c=n. By the induction hypothesis,


is generated by A and B. Thus M is generated by A and B. ∎

Two points z0,z1 are said to be equivalent under Γ if there exists an MΓ such that Mz1=z2. A subset F of is said to be a fundamental domain for Γ if it is an open connected set (i.e. a domain) such that no two distinct points of F are equivalent, and every point of is equivalent to some point in F¯, the closure of F. In the following theorem we explicitly give a fundamental domain for Γ. Our proof follows [8, Chapter VII, Theorem 1].

Theorem 3.

F={z:|z|>1,|(z)|<12} is a fundamental domain for Γ.


Let zF and let g=(abcd)Γ such that gzF. We will show that g=±I and thus that gz=z. This implies that no two distinct points of F are equivalent. If (gz)<(z), then (g-1(gz))>(gz). But gzF, and g-1=±I if and only if g=±I; hence we may assume that (gz)(z). Then |cz+d|1, so, since (z)>1/2, c must either be -1,0 or 1. If c=0 then a=d=±1 and g is translation by b. But -12<(z)<12, so b must be =0, otherwise gzF. In this case indeed g=±I.

If c=1, then |x+iy+d|1 for z=x+iy. Then x2+2xd+d2+y21. As x2+y2>1, it follows that 2xd+d2<0. In this inequality d cannot be 0, so we can divide by d. If d1 then d<-2x, a contradiction since |x|<1/2, and if d-1 then d>-2x, again a contradiction since |x|<1/2. Hence, the case c=1 does not occur. If c=-1, we can replace g with g=-g (since gzF if and only if gzF), and the above argument shows that the case c=1 does not occur, and thus the case c=-1 does not occur. Therefore if zF and gΓ such that gzF, then g=±I.

Now, let z. Now, for a given C>0, there are only finitely many integers c,d such that |cz+d|<C. Hence there is a g=(abcd)Γ such that


is maximal. We may choose n to be an integer such that -12(Tngz)12. Put z=Tngz. If |z|<1 then Sz=-1/z would have an imaginary part strictly greater than (z)=(Tngz)=(gz), a contradiction. Thus |z|1, and zF¯. This proves that every point in is equivalent to some point in the closure of F. ∎

The fact that F={z:|z|>1,|(z)|<12} is a fundamental domain for Γ will be used in §6. The images of F under several elements of Γ are shown in Figure 1.

Figure 1: Images of the fundamental domain F under elements of Γ

If {(abcd),(-a-b-c-d)}SL2()/{I,-I} is identified with the mapping zaz+bcz+d,, then SL2()/{I,-I} is the automorphism group of the upper half plane (i.e., the group of all holomorphic bijections , cf. [5, Chapter VII, §3]). In the next section we define modular forms, which are “almost” invariant under automorphisms of the upper half plane.

3 Modular forms and cusp froms

In this section we define modular forms. We prove that they have Fourier expansions. Then we define cusp forms, which are an important class of modular forms, with 0 constant term in their Fourier expansions. We then show that modular forms and cusp forms constitute vector spaces, and prove a result about pointwise products of modular forms.

We define the factor of automorphy j(γ,z) by j(γ,z)=cz+d for γ=(abcd)Γ and z.

A holomorphic function f: is said to be holomorphic at infinity if lim(z)f(z) exists.

Definition 4.

For kZ, a modular form of weight k, with respect to Γ=SL2(Z), is a function f:HC that is holomorphic on H, holomorphic at infinity, and satisfies


for all γΓ,zH. The set of all modular forms of weight k is denoted by Mk(Γ).

Fourier series will be important in studying modular forms and cusp forms, so we first introduce these. We say that a function f: is ω-periodic if f(z+ω)=f(z) for all z.

Lemma 5.

If f:HC is holomorphic and 1-periodic, then f has an expansion

f(z)=n=-anqn,q=e2πiz,an, (1)

valid for all zH.


Let A={z:0<|z|<1} be the annulus obtained by removing the origin from the unit disc. We define F:A by F(q)=f(z) for all qA. Indeed, if q=e2πiz1=e2πiz2, then 2πiz1=2πiz2+2kπi for some integer k and z1=z2+k. Hence f(z1)=f(z2) by periodicity. Thus F is well defined.

For all q0A, there exists a holomorphic branch of the logarithm, say L0, defined in some neighborhood V0 of q0. Then for all qV0, F(q)=f(L0(q)2πi). Thus on V0A, F is the composition of holomorphic functions. Hence F is holomorphic at q0A. Therefore F is holomorphic on A.

Since F is holomorphic on the annulus A, it has a Laurent expansion

F(q)=n=-anqn,an, (2)

valid for all qA [5, Chapter V, Theorem 2.1], and


with q=e2πiz, as desired. ∎

The expansion (1) is called the Fourier expansion or q-expansion of the function f.

Since γ=(1101)Γ, for fMk(Γ), then f(γz)=f(z+1) for all z, i.e. f is 1-periodic. Thus by Lemma 5, modular forms have Fourier expansions. Let fMk(Γ) and let g(q)=f(z), q=e2πiz,z. There is an α such that limyf(x+iy)=α. Let ϵ>0 be given. Then there is a y0>0 such that for all y>y0, |f(x+iy)-α|<ϵ. Let δ=e-2πy0. Say q is such that |q|<δ. Now, q=e2πi(x+iy) for some x,y. Then |q|=e-2πy<δ=e-2πy0. As exp is an increasing function on , this implies y>y0. Hence |f(x+iy)-α|<ϵ, and so |g(q)-α|<ϵ. Thus limq0g(q)=α. Hence g is bounded in a neighborhood of q=0, so an=0 for all n<0 in its Laurent expansion (1). This means that an=0 for all n<0 in the Fourier expansion (1) of a modular form.

We recall [5, Chapter V, Theorem 2.1] that for 0<s<S<1, the Laurent series n=-anqn in (2) converges absolutely for q such that s<|q|<S. Thus in particular the Laurent series converges absolutely for q=e-2π. This shows that the Fourier series (1) converges absolutely for z=i. Now, let E={z:(z)1}; it is clear that |e2πiz| has a maximum value |e2πi2|=e-2π on E. We define Mn=|an|e-2nπ for n=0,1,2,. That the Fourier series (1) converges absolutely for z=i means that n=0Mn converges. But |ane2πinz|Mn for all zE. Therefore by the Weierstrass M-test, the Fourier series (1) converges uniformly on E. However, if a sequence fn converges uniformly on a set E and x is a limit point of E, then limtxlimnfn(t)=limnlimtxfn(t) [6, Theorem 7.11]. Thus,

limyf(z)=limyn=0ane2nπi(x+iy)=n=0anlimye2nπixe-2nπy=a0. (3)

This tells us that constant term in the Fourier series of a modular form is the limit of the function at i.

We will often need to consider modular forms with 0 constant term in their Fourier expansions. Thus we make the following definition.

Definition 6.

A cusp form of weight k is a modular form f of weight k whose Fourier expansion has a 0 constant term, i.e. fMk(Γ) and


The set of all cusp forms of weight k is denoted by Sk(Γ).

From (3), a modular form f is a cusp form if and only if lim(z)f(z)=0.

Suppose k is odd and fMk(Γ). Let γ=(-100-1)Γ. Then for all z,


so f is the zero function on . Thus for all odd k, Mk(Γ) contains only the zero function.

In the following theorem we show that the modular forms of weight k form a complex vector space, and that the cusp forms of weight k are a subspace.

Theorem 7.

For every integer, Mk(Γ) is a complex vector space and Sk(Γ) is a subspace of Mk(Γ).


Let f,gMk(Γ) and a. Since f and g are holomorphic on , af+g is holomorphic on , and since lim(z)f(z) and lim(z)g(z) exist, lim(z)(af+g)(z) exists. Therefore af+g is holomorphic at infinity. For γΓ and z,

(af+g)(γz) = af(γz)+g(γz)
= aj(γ,z)kf(z)+j(γ,z)kg(z)
= j(γ,z)k(af+g)(z),

hence af+g satisfies the automorphy condition. Thus af+gMk(Γ). This proves that Mk(Γ) is a complex vector space. If f,gSk(Γ), then lim(z)(af+g)(z)=alim(z)f(z)+lim(z)g(z)=a0+0=0, and thus af+gSk(Γ). Hence Sk(Γ) is a subspace of Mk(Γ). ∎

For modular forms f and g, we shall define their product fg by (fg)(z)=f(z)g(z). The next lemma shows that the product of a modular form of weight k and a modular form of weight l is a modular form of weight k+l. We will use this result later in §6.

Lemma 8.

If fMk(Γ) and gMl(Γ), then fgMk+l(Γ).


It is immediate that fg is holomorphic on and holomorphic at infinity. Let γΓ and z. Then

(fg)(γz) = f(γz)g(γz)
= j(γ,z)kf(z)j(γ,z)lg(z)
= j(γ,z)k+lf(z)g(z)
= j(γ,z)k+l(fg)(z),

showing that fg satisfies the automorphy condition. Therefore fg is a modular form of weight k+l. ∎

4 Eisenstein series

Now we will explicitly construct a class of modular forms of all even weights k>2, the Eisenstein series of weight k. In particular these will not be the zero function, thus giving us nontrivial examples of modular forms. Moreover, these are not cusp forms. In fact, we will show later in this section that for even k>2, any modular form of weight k is a linear combination of an Eisenstein series of weight k and a cusp form of weight k.

Definition 9.

The Eisenstein series Gk(τ) of weight k is defined by


for τH, where the primed summation means that summation is over all (c,d)Z2 such that (c,d)(0,0).

In the following theorem we prove that for all even k4, the Eisenstein series Gk is a nonzero modular form of weight k. This will be our first example of a modular form (aside from the zero function). The proof follows [7, Theorem 1, Chapter III].

Theorem 10.

For all even integers k4, the Eisenstein series Gk is a modular form of weight k.


We will use the fact [5, Chapter V, §1] that the series defining the Riemann zeta function ζ(s)=n=1n-s converges absolutely for (s)>1. First we show that the series Gk(τ) defines a holomorphic function . Let K be a compact subset of . Let S1={(x,y)2|x2+y2=1} be the unit circle in 2. Recall that S1 is a compact subset of 2. Thus the product K×S1 is a compact subset of ×2. Clearly, (τ,x,y)|xτ+y| is a continuous function K×S1. Hence it attains a minimum value μ. For all τK and m1,m2 with (m1,m2)(0,0),

|m1τ+m2|2 = |m1m12+m22τ+m2m12+m22|2(m12+m22)2

Thus for all τK, Gk(τ) is bounded above by the series




where the inner summation is over those m1,m2 such that |m1|=N and |m2|N, or |m2|=N and |m1|N. Fixing N, there are a most 22(2N+1)=8N+4 such pairs (m1,m2)2. Thus


We have shown that μ-kN=18N-k+1+4N-k is an upper bound for Gk(τ). Since k>2, the latter series converges, so by the Weierstrass M-test, Gk(τ) converges uniformly (and absolutely) on K.

This proves that Gk(τ) converges uniformly on compact subsets of . Therefore according to [5, Theorem 1.1, Chapter V], Gk is a holomorphic function on .

We now show that Gk satisfies the automorphy condition. Let τ and γ=(abcd)Γ. Then

Gk(γτ) = Gk(aτ+bcτ+d) (4)
= (m1aτ+bcτ+d+m2)-k
= (m1aτ+m1b+m2cτ+m2dcτ+d)-k
= (cτ+d)k((m1a+m2c)τ+m1b+m2d)-k.

Since ad-bc=1, (m1,m2)(m1a+m2c,m1b+m2d) is a bijection 22, by the Euclidean algorithm [3, §12.3]. Thus, after changing variables,

((m1a+m2c)τ+m1b+m2d)-k=(m1τ+m2)-k=Gk(τ). (5)

Combining (4) with (5),


Finally we show that Gk is holomorphic at infinity. For γ=(1101)Γ and τ, we have Gk(γτ)=Gk(1τ+10τ+1)=Gk(τ+1). But since Gk satisfies the automorphy condition, then Gk(γτ)=j(γ,z)kGk(τ)=(0τ+1)kGk(τ)=Gk(τ), and so Gk(τ+1)=Gk(τ). That is, Gk is 1-period. Since Gk is holomorphic on with period 1, by Lemma 5 it has a Fourier expansion


Now, since k is even and Gk(τ) converges absolutely,


For m1=1, the inner series is n=-(τ+n)-k. It follows from the Weierstrass M-test that the series n=--1(τ+n)-k and n=0(τ+n)-k converge uniformly on every compact subset of and hence define holomorphic functions on . Therefore n=-(τ+n)-k defines a holomorphic function on . Certainly n=-(τ+n)-k is 1-periodic. Thus it has a Fourier expansion


Then by Laurent’s formula [5, Chapter V, Theorem 2.1], for an arbitrary τ0,


Since k is even and k2, the series converges uniformly so we can interchange integration and summation

αν = n=-τ0τ0+1(τ+n)-ke-2πiντ𝑑τ (6)
= -+iy0+iy0τ-ke-2πiντ𝑑τ

where (τ0)=y0.

Put τ=x+iy0. Thus

|αν| e2πνy0-|x+iy0|-k𝑑τ
= e2πνy0-(x2+y02)-k/2𝑑τ
= e2πνy0y01-k-(x2+1)-k/2𝑑x.

This integral converges to some ck since k2. Thus


Since we can choose y0 to be arbitrarily large, it follows that

αν=0for all ν0.

For ν>0, we integrate τ-ke-2πiντ along the oriented paths in Figure 2, where y1<0 is arbitrary, and apply the residue formula [5, Theorem 1.2, Chapter VI] to obtain

Iτ-ke-2πiντ𝑑τ=II+III+IVτ-ke-2πiντ𝑑τ-Res(τ-ke-2πiντ;0). (7)
Figure 2: Paths of integration for τ-ke-2πiντ (only singularity is pole at origin)

On path II,

|τ-ke-2πiντ| = |T+iy|-k|e-2πiν(T+iy)|
= (T2+y2)-k/2e2πνy
= T-ke2πνy0.

That is, an upper bound for |τ-ke-2πiντ| on path II is T-ke2πνy0. The length of path II is y0-y1. Thus by [5, Theorem 2.3, Chapter III] we have the estimate


The limit of the righthand side of this inequality as T- is 0. Hence limT-IIτ-ke-2πiντ𝑑τ=0.

Similarly on path IV,

|τ-ke-2πiντ| = |T+iy|-k|e-2πiν(T+iy)|
= (T2+y2)-k/2e2πνy

The length of path IV is y0-y1, so we have the estimate


Hence limTIVτ-ke-2πiντ𝑑τ=0.

For path III,

|τ-ke-2πiντ| = |x+iy1|-k|e-2πiν(x+iy1)|
= |x+iy1|-ke2πνy1


|IIIτ-ke-2πiντ𝑑τ| |TT(x2+y12)-ke2πνy1dx
= e2πνy1TT(x2+y12)-k𝑑x.

However, -1(x2+y12)k𝑑x=π2k-113(2k-3)12(k-1)1y2k-1 [5, Chapter XV, §2, K7]. Thus the absolute value of the above integral is bounded above by e2πνy1π2k-113(2k-3)12(k-1)1y2k-1.

Since y1<0 is arbitrary, it must be that


Therefore (7) reduces to


Now, the Laurent series of τ-ke-2πiντ is

τ-ke-2πiντ = τ-kl=0(-2πiν)lτll!
= l=0(-2πiν)lτl-kl!.

Thus Res(τ-ke-2πiντ;0)=(-2πiν)k-1(k-1)!. Hence


By (6) we have


giving us the Fourier series when m1=1.

Then for m1>1,



Gk(z) = 2ζ(k)+2m1=1(2πi)k(k-1)!ν=1νk-1e2πim1νz (8)
= 2ζ(k)+2(2πi)k(k-1)!n=1σk-1(n)e2πinz,

where σr(n) is the sum of the rth powers of the positive divisors of n, e.g. σr(6)=1r+2r+3r+6r.

Therefore lim(z)Gk(z)=2ζ(k). Hence Gk is holomorphic at infinity, and the constant term in the Fourier expansion of Gk is 2ζ(k). Since ζ(k)=n=1n-k0, Gk is not the zero function. ∎

For even k4, in the following theorem we show that Mk(Γ) decomposes into an internal direct sum of the cusp forms Sk(Γ) and multiples of the Eisenstein series Gk.

Theorem 11.

For all even k4,


For fMk(Γ), f has a Fourier expansion


Let h=f-a02ζ(k)Gk. Then hSk(Γ), and f=h+a02ζ(k)Gk where a02ζ(k)GkGk. Furthermore, the only multiple of Gk with zero constant term in its Fourier expansion is the zero function. Hence the intersection of Sk(Γ) and Gk is the zero subspace of Mk(Γ). Thus Mk(Γ)=Sk(Γ)Gk. ∎

5 The Dedekind eta function

We would like to give an explicit example of a nonzero cusp form. We will construct this using the following function.

Definition 12.

The Dedekind eta function η:HC is defined by

η(z)=q1/24n=1(1-qn) (9)

for q=e2πiz.

We will now show that the Dedekind eta function is holomorphic on and has no zeros in . This will be done by showing that the infinite product defining η converges uniformly on all compact subsets of . We will prove the uniform convergence of this infinite product by results about the uniform convergence of series.

Let an be a sequence of complex numbers such that the series n=1log(1+an) converges. Here log(1+z)=-n=1znn for |z-1|<1 (the open unit disc with center 1+0i). For all N, exp(n=1Nlog(1+an))=n=1N(1+an). Since the exponential function is continuous, we have n=1(1+an)=exp(n=1log(1+an))0.

Let K be a compact subset of . Let z0K such that |e2πiz0| is maximum, and put q0=e2πiz0. Since |q0|<1, there is some m such that for all nm, |q0n|<12. Now, let zK, and put q=e2πiz. For all nm,

|log(1-qn)| |qn|+|qn|22+
= |q0n|11-|q0n|
< |q0n|11-12
= 2|q0n|.

But the series n=m|q0|n converges (since |q0|<1). Hence by the Weierstrass M-test, n=1log(1-e2πinz) converges uniformly on K.

Let hm be a sequence of continuous functions that converge uniformly on K to a (continuous) function h. Then there exists an N such that for all mN, |hm(z)-h(z)|<1 for all zK. Since h is continuous, h(K) is compact, hence it is contained in a closed disc, of radius r. Let D be the closed concentric disc with radius r+1. Then hm(K)D for all mN. Now let ϵ>0 be given. Since exp is continuous on the compact set D, it is uniformly continuous on D [6, Theorem 4.19]. Hence there exists a δ>0 such that if w1,w2D and |w1-w2|<δ, then |exp(w1)-exp(w2)|<ϵ. But since hmh uniformly on K, there exists an MN such that for all mM, |hm(z)-h(z)|<δ for all zK. So |exp(hm(z))-exp(h(z))|<ϵ for all mM,zK. Thus exphm converges uniformly on K to exph. Setting hm=n=1mlog(1-e2πinz), we find that the infinite product n=1(1-e2πinz) converges uniformly on K.

Since the infinite product n=1(1-e2πinz) converges uniformly on all compact subsets of , η(z)=eπiz/12n=1(1-e2πinz) is a holomorphic function [5, Chapter V, Theorem 1.1]. As η(z)=eπiz/12exp(n=1log(1-e2πinz)) for all z, η(z)0 for all z.

The following theorem shows how η transforms under B=(0-110), one of the two generators of the modular group Γ, by Theorem 2. After this theorem, we can then show how η transforms under the whole modular group Γ. Our proof of the following theorem follows [2, Theorem 2, Chapter VIII].

Theorem 13.

For all zH,


where is the principal branch of the square root, holomorphic on the plane with the ray {zC:(z)=0,(z)0} removed.


η(-1z) and ziη(z) are holomorphic functions of z on . If they are equal on a set with an accumulation point in then they are equal on all of by the identity theorem for holomorphic functions [5, Chapter III, Theorem 1.2]. It therefore suffices to prove the theorem for purely imaginary z=iy,y>0. Suppose now that z=iy,y>0.

Since η(z)0 for y>0, we can take logarithms in (9), and use the series expansion


to obtain

logη(z)=πiz12+m=1log(1-e2πimz)=πiz12-m=1k=11ke2πikmz. (10)

The double series converges absolutely, because |e2πikmz|=e-2πkmy,y>0.

Since (z)>0, we have (-1z)>0. We can therefore replace z with -1z in (10) to obtain

logη(-1z)=-πi12z-m=1k=11ke-2πikm/z. (11)



to prove to the theorem it suffices, by (10) and (11), to prove

-πiz12-πi12z+k=11k(e2πikz1-e2πikz-e-2πik/z1-e-2πik/z)=12logzi. (12)

Now, the nth partial sum of the infinite series in (12) can be written as

k=1n12k(2e2πikz1-e2πikz+1-2e-2πik/z1-e-2πik/z-1)=k=1ni2k(-i1+e2πikz1-e2πikz+i1+e-2πik/z1-e-2πik/z)=k=1ni2k(cotπkz+cotπkz)=k0k=-nni4k(cotπkz+cotπkz), (13)

where cots=ie2is+1e2is-1=cosssins is the cotangent function.

To prove the theorem, by (13) it suffices to prove that

-πi12(z+1z)+k0k=-nni4k(cotπkz+cotπkz)12logzi,as n+. (14)

This will be done by applying the residue theorem to a certain path integral, of a suitably chosen function.


φ(s)=cotscotsz,ν=π(n+12), (15)

Since sin(s)=eis-e-is2i has a simple zero at s=nπ and no other zeros, and cos(s)=eis+e-is2 has a simple zero at nπ+π/2 and no other zeros, n, the function cot(s) has a simple zero at s=nπi+π/2 and a simple pole at s=nπ, n, and no other zeros or poles. Therefore φ(νs)s is meromorphic, with simple poles at s=πkν and s=πkνz, where k is any nonzero integer, and a triple pole at s=0, and has no other poles.

The Laurent expansion of φ(νs)s about s=0 is found by noting that

cots = cosssins
= 1-s22+s(1-s26+)
= 1s(1-s22+)(1+s26+)
= 1s-s3+,

and so


Furthermore, cots has period π, because


Thus the Laurent series of cot(νs) about s=πkν is


and the Laurent series of cot(νsz) about s=πkνz is


The residues of φ(νs)s at s=0, s=πkν, and s=πkνz are respectively



Figure 3: Paths of integration of φ(νs)s, ν=π(n+12)

Let P denote the parallelogram with vertices at A(s=1), B(s=z=iy), C(s=1) and D(s=-z=-iy) oriented counterclockwise, shown in Figure 3. Then since ν=π(n+12), the residue theorem implies that


Here the initial term is the residue of φ(νs)s at the triple pole s=0, and the summation is over the residues of φ(νs)s at the simple poles s=πkν and s=πkνz respectively; the function φ(νs)s has no other poles inside P. Thus,


To prove (14) it therefore suffices to prove that

limnPφ(νs)8s𝑑s=12logzi,ν=π(n+12). (16)

Because cots has period π, and |cots|<M1 for sP with -π2(s)π2, |s|δ0>0 (since the only pole of cots with real part between -π/2 and π/2 is 0), it follows that

|φ(νs)s|<M,sP,ν=π(n+12),n=1,2,. (17)

Here M is independent of s and ν. As well, since


we have

cots{-i,as (s)+,+i,as (s)-. (18)

Let K be a compact subset of the upper half plane , and let ϵ>0. As ν, (νs) for sK. Thus by (18), cotνs-i as ν uniformly for sK (since as K is compact, there is an sK such that (s) is minimum). Similarly, cotνs+i as ν+ uniformly in every compact set in the lower half plane (s)<0, cotνsz-i as ν+ uniformly in every compact set in the left half plane (s)<0, and cotνsz+i as ν+ uniformly in every compact set in the right half plane (s)>0.

To prove (16), we shall show that

limnPφ(νs)s𝑑s=(1z-z-1+-1-z--z1)dss, (19)

where the integrals on the right hand side are along the respective paths AB, BC, CD, DA of the parallelogram P. We will then show that the sum of the integrals on the right hand side is equal to 4logzi, which would prove (16).

We first split the integral along the path AB into three parts, from A to A, A to B, and B to B, such that the distances |AA| and |BB| are equal to δ>0, to be chosen sufficiently small. From above, φ(νs)(+i)(-i)=1 uniformly as ν, for s in the compact set AB. Therefore, for any ϵ>0 there exists a ν0 such that for all νν0, |φ(νs)-1|<ϵ for sAB. But |s| has a minimum δ1>0 on AB, so |φ(νs)-1s|<ϵδ1 for sAB. By (17), |φ(νs)-1s|<M+1δ1 for all sAA,BB. Hence, given ϵ>0, there exist a δ>0 and a ν0 such that for all νν0, we have

|ABφ(νs)-1s𝑑s| |ABφ(νs)-1s𝑑s|+|AA|+|BB|
< |AB|ϵδ0+δ(M+1δ1)+δ(M+1δ1)
< ϵ.

We have just shown that

limnABφ(νs)s𝑑s=ABdss=1zdss. (20)

Similar results hold for the other three edges of P. Thus (19) follows, and

Figure 4: Simple curve L joining A and C

Let L be a simple curve in the upper half plane joining A and C and not passing through CB or AB, as in Figure 4. Since L is homotopic to the arc of the unit circle from -1 to 1, which is parameterized by s=eit,πt0, by [5, Chapter III, Theorem 5.1] we have


As 0 is not in the region contained by L and the line segments CB,AB, by the residue theorem we have

limnPφ(νs)s𝑑s = 41zdss+2Ldss
= 4logz-22πi2
= 4logz-4logi
= 4logzi.

This proves (16), and thus the theorem. ∎

The proof of the analog of the automorphy condition for η follows [2, Chapter VIII, Theorem 3].

Theorem 14.

Suppose a,d,b,c are integers such that ad-bc=1. Then for all zH,

η(az+bcz+d)=ωcz+dη(z) (21)

where ωC is a 24th root of unity that depends on a,b,c,d, but not on z.


First we note that for all z,

η(z±1)=e2π(z±1)24n=1(1-e2nπiz(z±1))=e±πi/12η(z). (22)

Let A=(1101),B=(0-1-10), and let M=(abcd)Γ. If M=A±1 or B=B-1 then

η(Mz)24=(cz+d)12η(z)24 (23)

holds because of (22) and Theorem 13 respectively. Now suppose (23) holds for some fixed M=(abcd)Γ. Then






by (22). Therefore so (23) holds for MA±1. Furthermore


by Theorem 13. Therefore (23) holds for MB=MB-1. Since Γ is generated by A and B, by Theorem 2, therefore (23) holds for all M=(abcd)Γ.

For M=(abcd)Γ, taking the 24th roots of both sides of (23) gives η(Mz)=ω(z)cz+dη(z) for some function γ that takes values in the 24th roots of unity. But η(Mz) and cz+dη(z) are continuous functions of z on , and cz+dη(z)0 for all z, so ω(z) is a continuous function on . Since ω(z) is a continuous function from the connected set to the discrete set of 24th roots of unity, it must be a constant 24th root of unity ω, which completes the proof. ∎

Now we can explicitly construct a cusp form. By taking the 24th powers of each side of (21) we find that η24 satisfies the automorphy condition η(γz)24=j(γ,z)12η(z) for all γΓ and z. This leads us to define the following function.

Definition 15.

The modular discriminant Δ(z) is defined by


for zH.

Corollary 16.

The modular discriminant Δ is a cusp form of weight 12.


That Δ is a holomorphic function and that Δ is holomorphic at infinity follows immediately from Δ=η24.

Suppose a,b,c,d such that ad-bc=1. Then for γΓ,z, Theorem 14 tells us that

Δ(γz) = η(γz)24
= (j(γ,z)η(z))24
= j(γ,z)12Δ(z).

This shows that Δ satisfies the automorphy condition. Therefore Δ is a modular form of weight 12.



showing that Δ is a cusp form of weight 12. ∎

Since Δ is a cusp form, it has a Fourier expansion Δ(z)=n=1anqn, q=e2πiz. The Ramanujan tau function is defined by τ(n)=an for all n, that is, the Ramanujan tau function is defined by the Fourier coefficients of the modular discriminant.

We give the following estimates for the magnitude of the Fourier coefficients of cusp forms. The following estimates will be employed in defining a certain inner product on Sk(Γ) in §6. Moreover, since the Fourier coefficients of cusp forms are related to number theoretic functions, estimates on their magnitude give us number theoretic information. The proof of the following theorem follows [8, Chapter VII, Theorem 5].

Theorem 17.

If fSk(Γ) with Fourier expansion


then an=O(nk/2) for all n1.


Indeed the series n=1anqn-1 defines a holomorphic function in a closed disc D about q=0 of radius r for any 0<r<1. Let such an r be fixed. Then n=1anqn-1 has a maximum value in the compact set D. Hence for z=x+iy, as y,


since for all sufficiently large y, e2πi(x+iy)D. Let φ(z)=|f(z)|yk/2 for z,y=(z). Then for all g=(abcd)Γ, (gz)=(z)|cz+d|2. Hence

φ(gz) = |f(gz)|(y|cz+d|2)k/2
= |(cz+d)kf(z)|yk/2|cz+d|k
= |cz+d|k|f(z)|yk/2|cz+d|k
= φ(z),

so φ is invariant under Γ. Since f is a cusp form, φ(z)0 as y. Since φ is continuous on F¯, the closure of the fundamental domain F, it is bounded on the compact set obtained by removing all zF¯ with (z)>y0 some y0>0. Therefore for some M>0, |φ(z)|M for all zF¯. The invariance of φ under Γ implies that |φ(z)|M for all z. Hence

|f(z)|My-k/2,z,y=(z). (24)

Let y>0 be fixed and let Cy be the circle about the origin with radius y, parametrized by q=e2πi(x+iy),0x1. Then using Cauchy’s formula [5, Chapter III, Theorem 7.1],


Consequently, by (24),


But this inequality is valid for all y>0. Letting y=1/n, we obtain |an|e2πMnk/2, proving the claim. ∎

6 The vector spaces of modular forms

In this section we will prove several results about the vector space Mk(Γ), and in particular the subspace Sk(Γ). We will explicitly determine the dimension of Mk(Γ). After this we will introduce an inner product on the subspace of cusp forms Sk(Γ).

For modular forms f,g where g has no zeros in , we shall define their quotient f/g by (f/g)(z)=f(z)/g(z). Let ordf be the order of the zero f at infinity.

Lemma 18.

If fMk(Γ),gMl(Γ) such that g has no zeros in H and ordfordg, then f/gMk-l(Γ).


Clearly f/g is holomorphic on . Since ordfordg, f/g is holomorphic at infinity.

Let γΓ, and let z. Then

(f/g)(γz) = f(γz)g(γz)
= j(γ,z)kf(z)j(γ,z)lg(z)
= j(γ,z)k-lf(z)g(z)
= j(γ,z)k-l(f/g)(z),

which shows that f/gMk-l(Γ). ∎

The following lemma shows that a modular form of weight 0 is constant. This lemma will be used in determining the dimension of Mk(Γ), for which we need to show that certain modular forms must be scalar multiples of other modular forms. Our proof of this lemma follows [4, Chapter IX, Notes on §9.11].

Lemma 19.

A modular form of weight 0 is constant.


Let f be a modular form of weight 0. Let F={τ:|τ|>1,|(τ)|<1/2}, a fundamental domain for Γ. Then for τ=x+iyF, limyf(τ)=α for some α. Since f-α is also a modular form of weight 0, we may suppose without loss of generality that α=0.

Because for τ=x+iy, limyf(τ)=0, there is a y0>0 such that for all τ=x+iy with y>y0, |f(τ)|<1. Then in the compact set {τ=x+iyF¯:yy0}, |f(τ)| is bounded. Hence |f(τ)| is bounded on F¯.

Let M=supτF¯|f(τ)|<. If M0, then there exists a y0>0 such that for all τ=x+iyF¯ with y>y0, |f(τ)|<M/2. Thus |f(τ)| attains its maximum in K={τ=x+iyF¯:yy0}. By the maximum modulus principle [5, Theorem 1.3, Chapter III], |f(τ)| attains its maximum at some τ0 on the boundary of K, so |f(τ0)|=M. If f is not constant then there must be a τ2 such that |f(τ2)|>|f(τ1)|. But since F is a fundamental domain for Γ, there exists a γΓ such that γτ2F¯. As f is a modular form of weight 0, M|f(γτ2)|=|f(τ2)|>M, a contradiction. Thus f is constant. ∎

It will be convenient to define the normalized Eisenstein series Ek by


where ζ(s)=n=1n-s is the Riemann zeta function. Certainly Ek is a modular form of weight k, since Gk is. Thus Ek has a Fourier series n=0anqn. We will use the fact that a0=1 and a1=(2πi)k(k-1)!ζ(k), which are immediate from the Fourier series (8) of Gk. In particular, since ζ(4)=π490 and ζ(6)=π6945 [8, Proposition 7, Chapter VII], the coefficient a1 of q in the Fourier expansions of E4 and E6 is respectively 240 and -504.

The proof of the following theorem follows [1, Proposition 1.3.3]. It will be used in the proof of the general formula for the dimension of Mk(Γ).

Theorem 20.

The space of cusp forms of weight 12 is one dimensional, spanned by the modular discriminant Δ. Moreover,

Δ=11728(E43-E62). (25)

Let fS12(Γ). Δ has no zeros in and ordf1=ordΔ. By Lemma 18, f/Δ is a modular form of weight 0. Thus f/Δ is a constant, so f=cΔ for some c.

We work out the first several terms of the Fourier expansion of 11728(E43-E62) to find


This implies that 11728(E43-E62) is a cusp form of weight 12, and by the above, 11728(G43-G62)=cΔ for some constant c. Thus comparing the Fourier coefficients of q, we see that c=1, completing the proof. ∎

The proof of the following fact follows the sketch [1, Exercise 1.3.3]. A consequence is that G4(ρ)=0 for ρ=e2πi/3, which we will use in our proof of the dimension formula for Mk(Γ).

Lemma 21.

Let ρ=e2πi/3. If 3 does not divide k, then f(ρ)=0 for any modular form of weight k.


γ=(11-10)Γ, and


Certainly then f(γρ)=f(ρ). On the other hand, as f is a modular form of weight k, f(γρ)=j(γ,ρ)kf(ρ). Since j(γ,ρ)=-ρ,


But k is not a multiple of 3, so it must be that f(ρ)=0. ∎

Now we can determine the dimension of Mk(Γ). Our proof of the following theorem follows [1, Proposition 1.3.4].

Theorem 22.

If k is an even nonnegative integer with k=12j+r for 0r10, then

dimM12j+r(Γ)={j+1,if r=0,4,6,8 or 10,j𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

We will first show that for k=4,6,8 or 10, Sk(Γ) is the zero space and thus that Mk(Γ) is generated by Gk. Let fSk(Γ), and suppose by contradiction that f is not the zero function. Now, for h=6(12-k), Gh(f/Δ)6 is a modular form of weight 0 by Lemma 18. Hence it is a constant c. So Gh=cΔ6/f6. This implies that Gh has no zeros in , because Δ has no zeros in and f is holomorphic in . For H=h/12, H=1,2,3 or 4. We consider ΔH/Gh. Since Gh has no zeros in and does not have a zero at infinity, it follows that ΔH/Gh is a modular form of weight 0. ΔH/Gh has a zero of order H at infinity but is not the zero function, a contradiction. This means that f must be the zero function. Hence for k=4,6,8 or 10, Mk(Γ) is spanned by the Eisenstein series Gk and thus is one dimensional.

We now show that M2(Γ) is the zero space. Suppose by contradiction that f is a nonzero element of M2(Γ). Then fG4M6(Γ). From the above, we know that M6(Γ) is generated by G6, hence fG4=cG6 for some c. Because f is not the zero function, c0. By Lemma 21, G4(ρ)=0 for ρ=e2πi/3. This implies that G6(ρ)=0, as c0. But then by (25), Δ(ρ)=0, which is a contradiction. Therefore f must be the zero function. So M2(Γ) is the zero space. Finally, M0(Γ) is of course one dimensional, as it spanned by any constant nonzero function .

For k12, we shall show that fΔf is a vector space isomorphism Mk-12(Γ)Sk(Γ). Clearly this mapping is linear over . If Δf is the zero function then f must be the zero function (since Δ has no zeros in ), so this mapping is injective. If fSk(Γ), then f/Δ is a modular form of weight k-12. However, f/Δ is sent by this mapping to f. This shows that this mapping is a surjection. Therefore the vector spaces Mk-12(Γ) and Sk(Γ) are isomorphic, and thus have the same dimension. ∎

We now introduce an inner product (,) on the space of cusp forms of a given weight k. We will show that indeed (,) is an inner product, that is, that it is linear in the first argument, conjugate symmetric, and positive definite.

Definition 23.

The Petersson inner product (,) on Sk(Γ) is defined, for f,gSk(Γ), by

(f,g)=Ff(z)g(z)¯ykdxdyy2,z=x+iy,x,y, (26)

where F={z:|z|>1,|(z)|<12}.

Theorem 24.

(,) is an inner product on Sk(Γ).


By (24), f,g=O(y-k/2), i.e., there exist C,D>0 such that |f(z)|<Cy-k/2,|g(z)|<Dy-k/2 for all sufficiently large y, z=x+iy. Then, since |g(z)|=|g(z)¯|,

|Ff(z)g(z)¯ykdxdyy2| F|f(z)||g(z)¯|ykdxdyy2
= CDFdxdyy2
= CD3/2dyy2
= CD3.

Therefore the integral (26) converges. We shall now show that (,) is an inner product on Sk(Γ).

That (αf1+f2,g)=α(f1,g)+α(f2,g) for all f1,f2,gSk(Γ),α, follows immediately from the fact that integration is linear.

Let f,gSk(Γ). Then

(f,g)¯ = Ff(z)g(z)¯ykdxdyy2¯
= Ff(z)¯g(z)ykdxdyy2
= (g,f),

showing that (,) is conjugate symmetric.

Let fSk(Γ). If f=0 on then (f,f)=F0𝑑x𝑑y=0. Otherwise, if f is not identically 0 on , then there is some z0 such that f(z0)0. Since F is a fundamental domain, there exist z0F¯,gΓ such that gz0=z0. Because f(gz0)=j(g,z0)kf(z0)0, then f(z0)0. Since f is continuous, there is some disc D of radius >0 about z0 on which f is nonzero. But indeed there is a disc D of radius >0 such that DDF. Hence,

(f,f) = Ff(z)f(z)¯ykdxdyy2
= F|f(z)|yk-2𝑑x𝑑y
> 0,

so (f,f)>0 if f0. This verifies that (,) is positive definite. Therefore (,) is an inner product on Sk(Γ). ∎


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