5Modular forms of level 1

III Modular Forms and L-functions

5.2 The space of modular forms

In this section, we are going to find out all modular forms! For

k ∈ Z

, we write

M

k

=

M

k

(Γ(1)) for the set of modular forms of weight

k

(and level 1). We have

S

k

⊆ S

k

(Γ(1)) containing the cusp forms. These are

C

-vector spaces, and are

zero for odd k.

Moreover, from the definition, we have a natural product

M

k

· M

`

⊆ M

k+`

.

Likewise, we have

S

k

· M

`

⊆ S

k+`

.

We let

M

∗

=

M

k∈Z

M

k

, S

∗

=

M

k∈Z

S

k

.

Then M

∗

is a graded ring and S

∗

is a graded ideal. By definition, we have

S

k

= ker(a

0

: M

k

→ C).

To figure out what all the modular forms are, we use the following constraints

on the zeroes of a modular form:

Proposition.

Let

f

be a weak modular form (i.e. it can be meromorphic at

∞

)

of weight k and level 1. If f is not identically zero, then

X

z

0

∈D\{i,ρ}

ord

z

0

(f)

+

1

2

ord

i

(f) +

1

3

ord

ρ

f + ord

∞

(f) =

k

12

,

where ord

∞

f is the least r ∈ Z such that a

r

(f) 6= 0.

Note that if

γ ∈

Γ(1), then

j

(

γ, z

) =

cz

+

d

is never 0 for

z ∈ H

. So it follows

that ord

z

f = ord

γ(z)

f.

We will prove this using the argument principle.

Proof.

Note that the function

˜

f

(

q

) is non-zero for 0

< |q| < ε

for some small

ε

by the principle of isolated zeroes. Setting

ε = e

−2πR

,

we know f (z) 6= 0 if Im z ≥ R.

In particular, the number of zeroes of

f

in

D

is finite. We consider the

integral along the following contour, counterclockwise.

ρ

ρ

2

i

1

2

+ iR−

1

2

+ iR

C

0

C

We assume

f

has no zeroes along the contour. Otherwise, we need to go around

the poles, which is a rather standard complex analytic maneuver we will not go

through.

For ε sufficiently small, we have

Z

Γ

f

0

(z)

f(z)

dz = 2πi

X

z

0

∈D\{i,ρ}

ord

z

0

f

by the argument principle. Now the top integral is

Z

−

1

2

iR

1

2

+iR

f

0

f

dz = −

Z

|q|=ε

d

˜

f

dq

˜

f(q)

dq = −2πi ord

∞

f.

As

f

0

f

has at worst a simple pole at

z

=

i

, the residue is

ord

i

f

. Since we are

integrating along only half the circle, as ε → 0, we pick up

−πi res = −πi ord

i

f.

Similarly, we get −

2

3

πi ord

ρ

f coming from ρ and ρ

2

.

So it remains to integrate along the bottom circular arcs. Now note that

S : z 7→ −

1

z

maps C to C

0

with opposite orientation, and

df(Sz)

f(Sz)

= k

dz

z

+

df(z)

f(z)

as

f(Sz) = z

k

f(z).

So we have

Z

C

+

Z

C

0

f

0

f

dz =

Z

C

0

f

0

f

dz −

k

z

dz +

f

0

f

dz

− −k

Z

C

0

dz

z

→ k

Z

i

ρ

dz

z

=

πik

6

.

So taking the limit ε → 0 gives the right result.

Corollary. If k < 0, then M

k

= {0}.

Corollary. If k = 0, then M

0

= C, the constants, and S

0

= {0}.

Proof.

If

f ∈ M

0

, then

g

=

f − f

(1). If

f

is not constant, then

ord

i

g ≥

1, so

the LHS is > 0, but the RHS is = 0. So f ∈ C.

Of course, a

0

(f) = f. So S

0

= {0}.

Corollary.

dim M

k

≤ 1 +

k

12

.

In particular, they are finite dimensional.

Proof.

We let

f

0

, ··· , f

d

be

d

+ 1 elements of

M

k

, and we choose distinct points

z

1

, ··· , z

d

∈ D \ {i, ρ}. Then there exists λ

0

, ··· , λ

d

∈ C, not all 0, such that

f =

d

X

i=0

λ

i

f

i

vanishes at all these points. Now if

d >

k

12

, then LHS is

>

k

12

. So

f ≡

0. So (

f

i

)

are linearly dependent, i.e. dim M

k

< d + 1.

Corollary. M

2

=

{

0

}

and

M

k

=

CE

k

for 4

≤ k ≤

10 (

k

even). We also have

E

8

= E

2

4

and E

10

= E

4

E

6

.

Proof. Only M

2

= {0} requires proof. If 0 6= f ∈ M

2

, then this implies

a +

b

2

+

c

3

=

1

6

for integers a, b, c ≥ 0, which is not possible.

Alternatively, if

f ∈ M

2

, then

f

2

∈ M

4

and

f

3

∈ M

6

. This implies

E

3

4

=

E

2

6

,

which is not the case as we will soon see.

Note that we know

E

8

=

E

2

4

, and is not just a multiple of it, by checking the

leading coefficient (namely 1).

Corollary. The cusp form of weight 12 is

E

3

4

− E

2

6

= (1 + 240q + ···)

3

− (1 − 504q + ···)

2

= 1728q + ··· .

Note that 1728 = 12

3

.

Definition (∆ and τ ).

∆ =

E

3

4

− E

2

6

1728

=

X

n≥1

τ(n)q

n

∈ S

12

.

This function

τ

is very interesting, and is called Ramanujan’s

τ

-function. It

has nice arithmetic properties we’ll talk about soon.

The following is a crucial property of ∆:

Proposition. ∆(z) 6= 0 for all z ∈ H.

Proof. We have

X

z

0

6=i,ρ

ord

z

0

∆ +

1

2

ord

i

∆ +

1

3

ord

ρ

∆ + ord

∞

∆ =

k

12

= 1.

Since ord

ρ

∆ = 1, it follows that there can’t be any other zeroes.

It follows from this that

Proposition.

The map

f 7→

∆

f

is an isomorphism

M

k−12

(Γ(1))

→ S

k

(Γ(1))

for all k > 12.

Proof.

Since ∆

∈ S

12

, it follows that if

f ∈ M

k−1

, then ∆

f ∈ S

k

. So the map

is well-defined, and we certainly get an injection

M

k−12

→ S

k

. Now if

g ∈ S

k

,

since

ord

∞

∆ = 1

≤ ord

∞

g

and ∆

6

=

H

. So

g

∆

is a modular form of weight

k − 12.

Thus, we find that

Theorem.

(i) We have

dim M

k

(Γ(1)) =

0 k < 0 or k odd

k

12

k > 0, k ≡ 2 (mod 12)

1 +

k

12

otherwise

(ii) If k > 4 and even, then

M

k

= S

k

⊕ CE

k

.

(iii) Every element of M

k

is a polynomial in E

4

and E

6

.

(iv) Let

b =

(

0 k ≡ 0 (mod 4)

1 k ≡ 2 (mod 4)

.

Then

{h

j

= ∆

j

E

b

6

E

(k−12j−6b)/4

4

: 0 ≤ j < dim M

k

}.

is a basis for M

k

, and

{h

j

: 1 ≤ j < dim M

k

}

is a basis for S

k

.

Proof.

(ii) S

k

is the kernel of the homomorphism

M

k

→ C

sending

f 7→ a

0

(

f

). So the

complement of

S

k

has dimension at most 1, and we know

E

k

is an element

of it. So we are done.

(i)

For

k <

12, this agrees with what we have already proved. By the

proposition, we have

dim M

k−12

= dim S

k

.

So we are done by induction and (ii).

(iii)

This is true for

k <

12. If

k ≥

12 is even, then we can find

a, b ≥

0 with

4a + 6b = k. Then E

a

4

E

b

6

∈ M

k

, and is not a cusp form. So

M

k

= CE

a

4

E

b

6

⊕ ∆M

k−12

.

But ∆ is a polynomial in E

4

, E

6

, So we are done by induction on k.

(iv)

By (i), we know

k −

12

j −

6

k ≥

0 for

j < dim M

k

, and is a multiple of 4.

So

h

j

∈ M

k

. Next note that the

q

-expansion of

h

j

begins with

q

j

. So they

are all linearly independent.

So we have completely determined all modular forms, and this is the end of

the course.