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.