5Modular forms of level 1

III Modular Forms and L-functions

5.1 Basic definitions
We can now define a modular form. Recall that we have SL
2
(Z) = Γ(1).
Definition
(Modular form of level 1)
.
A holomorphic function
f
:
H C
is a
modular form of weight k Z and level 1 if
(i) For any
γ =
a b
c d
Γ(1),
we have
f(γ(z)) = (cz + d)
k
f(z).
(ii) f is holomorphic at (to be defined precisely later).
What can we deduce about modular forms from these properties? If we take
γ = I, then we get
f(z) = (1)
k
f(z).
So if
k
is odd, then
f
0. So they only exist for even weights. If we have even
weights, then it suffices to consider
¯
Γ = hS, T i. Since
f(z) 7→ (cz + d)
k
f(γ(z))
is a group action of Γ(1) on functions on
H
, it suffices to check that
f
is invariant
under the generators S and T . Thus, (i) is equivalent to
f(z + 1) = f(z), f(1/z) = z
k
f(z).
How do we interpret (ii)? We know
f
is
Z
-periodic. If we write
q
=
e
2πiz
, then
we have
z H
iff 0
< |q| <
1, and moreover, if two different
z
give the same
q
,
then the values of
f
on the two
z
agree. In other words,
f
(
z
) only depends on
q
,
and thus there exists a holomorphic function
˜
f(q) on {0 < |q| < 1} such that
˜
f(e
2πiz
) = f(z).
Explicitly, we can write
˜
f(q) = f
1
2πi
log q
.
By definition,
˜
f
is a holomorphic function on a punctured disk. So we have a
Laurent expansion
˜
f(q) =
X
n=−∞
a
n
(f)q
n
,
called the Fourier expansion or
q
-expansion of
f
. We say
f
is meromorphic
(resp. holomorphic) at if
˜
f is meromorphic (resp. holomorphic) at q = 0.
In other words, it is meromorphic at
if
a
n
(
f
) = 0 for
n
sufficiently negative,
and holomorphic if
a
n
(
f
) = 0 for all
n
0. The latter just says
f
(
z
) is bounded
as Im(z) .
The following definition is also convenient:
Definition
(Cusp form)
.
A modular form
f
is a cusp form if the constant term
a
0
(f) is 0.
We will later see that “almost all” modular forms are cusp forms.
In this case, we have
˜
f =
X
n1
a
n
(f)q
n
.
From now on, we will drop the ˜, which should not cause confusion.
Definition
(Weak modular form)
.
A weak modular form is a holomorphic form
on H satisfying (i) which is meromorphic at .
We will use these occasionally.
The transformation rule for modular forms seem rather strong. So, are there
actually modular forms? It turns out that there are quite a lot of modular forms,
and remarkably, there is a relatively easy way of listing out all the modular
forms.
The main class (and in fact, as we will later see, a generating class) of modular
forms is due to Eisenstein. This has its origin in the theory of elliptic functions,
but we will not go into that.
Definition (Eisenstein series). Let k 4 be even. We define
G
k
(z) =
X
m,nZ
(m,n)6=(0,0)
1
(mz + n)
k
=
X
0
(m,n)Z
2
1
(mz + n)
k
.
Here the
P
0
denotes that we are omitting 0, and in general, it means we
don’t sum over things we obviously don’t want to sum over.
When we just write down this series, it is not clear that it is a modular form,
or even that it converges. This is given by the following theorem:
Theorem. G
k
is a modular form of weight
k
and level 1. Moreover, its
q
-
expansion is
G
k
(z) = 2ζ(k)
1
2k
B
k
X
n1
σ
k1
(n)q
n
, (1)
where
σ
r
(n) =
X
1d|n
d
r
.
Convergence of the series follows from the following more general result. Note
that since z 6∈ R, we know {1, z} is an R-basis for C.
Proposition. Let (e
1
, ··· , e
d
) be some basis for R
d
. Then if r R, the series
X
0
mZ
d
km
1
e
1
+ ··· + m
d
e
d
k
r
converges iff r > d.
Proof. The function
(x
i
) R
d
7→
X
i=1
x
i
e
i
is a norm on
R
d
. As any 2 norms on
R
d
are equivalent, we know this is equivalent
to the sup norm k · k
. So the series converges iff the corresponding series
0
X
mZ
d
kmk
r
converges. But if 1
N Z
, then the number of
m Z
d
such that
kmk
=
N
is (2N + 1)
d
(2N 1)
d
2
d
dN
d1
. So the series converges iff
X
N1
N
r
N
d1
converges, which is true iff r > d.
Proof of theorem.
Then convergence of the Eisenstein series by applying this
to
R
2
=
C
. So the series is absolutely convergent. Therefore we can simply
compute
G
k
(z + 1) =
0
X
m,n
1
(mz + (m + n))
k
= G
k
(z).
Also we can compute
G
k
1
z
=
0
X
m,n
=
z
k
(m + nz)
k
= z
k
G
k
(z).
So
G
k
satisfies the invariance property. To show that
G
k
is holomorphic, and
holomorphic at infinity, we’ll derive the q-expansion (1).
Lemma.
X
n=
1
(n + w)
k
=
(2πi)
k
(k 1)!
X
d=1
d
k1
e
2πidw
for any w H and k 2.
There are (at least) two ways to prove this. One of this is to use the series
for the cotangent, but here we will use Poisson summation.
Proof. Let
f(x) =
1
(x + w)
k
.
We compute
ˆ
f(y) =
Z
−∞
e
2πixy
(x + w)
k
dx.
We replace this with a contour integral. We see that this has a pole at
w
. If
y > 0, then we close the contour downwards, and we have
ˆ
f(y) = 2πi Res
z=w
e
2πiyz
(z + w)
k
= 2πi
(2πiy)
k1
(k 1)!
e
2πiyw
.
If
y
0, then we close in the upper half plane, and since there is no pole, we
have
ˆ
f(y) = 0. So we have
X
n=−∞
1
(n + w)
k
=
X
nZ
f(n) =
X
dZ
ˆ
f(d) =
(2πi)
k
(k 1)!
X
d1
d
k1
e
2πidw
by Poisson summation formula.
Note that when we proved the Poisson summation formula, we required
f
to decrease very rapidly at infinity, and our
f
does not satisfy that condition.
However, we can go back and check that the proof still works in this case.
Now we get back to the Eisenstein series. Note that since
k
is even, we can
drop certain annoying signs. We have
G
k
(z) = 2
X
n1
1
n
k
+ 2
X
m1
X
nZ
1
(n + mz)
k
= 2ζ(k) + 2
X
m1
(2πi)
k
(k 1)!
X
d1
d
k1
q
dm
.
= 2ζ(k) + 2
(2πi)
k
(k 1)!
X
n1
σ
k1
(n)q
n
.
Then the result follows from the fact that
ζ(k) =
1
2
(2πi)
k
B
k
k!
.
So we see that G
k
is holomorphic in H, and is also holomorphic at .
It is convenient to introduce a normalized Eisenstein series
Definition (Normalized Eisenstein series). We define
E
k
(z) = (2ζ(k))
1
G
k
(z)
= 1
2k
B
k
X
n1
σ
k1
(n)q
n
=
1
2
X
(m,n)=1
m,nZ
1
(mz + n)
k
.
The last line follows by taking out any common factor of
m, n
in the series
defining G
k
.
Thus, to figure out the (normalized) Eisenstein series, we only need to know
the Bernoulli numbers.
Example. We have
B
2
=
1
6
, B
4
=
1
30
, B
6
=
1
42
, B
8
=
1
30
B
10
=
5
66
, B
12
=
631
2730
, B
14
=
7
6
.
Using these, we find
E
4
= 1 + 240
X
σ
3
(n)q
n
E
6
= 1 504
X
σ
5
(n)q
n
E
8
= 1 + 480
X
σ
7
(n)q
n
E
10
= 1 264
X
σ
9
(n)q
n
E
12
= 1 +
65520
691
X
σ
11
(n)q
n
E
14
= 1 24
X
σ
13
(n)q
n
.
We notice that there is a simple pattern for k 14, except for k = 12.
For more general analysis of modular forms, it is convenient to consider the
following notation:
Definition (Slash operator). Let
a b
c d
= γ GL
2
(R)
+
, z H,
and f : H C any function. We write
j(γ, z) = cz + d.
We define the slash operator to be
(f |
k
γ)(z) = (det γ)
k/2
j(γ, z)
k
f(γ(z)).
Note that some people leave out the
det γ
k/2
factor, but if we have it, then
whenever γ = Ia, then
f |
k
γ = sgn(a)
k
f,
which is annoying. In this notation, then condition (i) for
f
to be a modular
form is just
f |
k
γ = f
for all γ SL
2
(Z).
To prove things about our j operator, it is convenient to note that
γ
z
1
= j(γ, z)
γ(z)
1
. ()
Proposition.
(i) j(γδ, z) = j(γ, δ(z))j(δ, z) (in fancy language, we say j is a 1-cocycle).
(ii) j(γ
1
, z) = j(γ, γ
1
(z))
1
.
(iii) γ
:
ϕ 7→ f |
k
γ
is a (right) action of
G
=
GL
2
(
R
)
+
on functions on
H
. In
other words,
f |
k
γ |
k
δ = f |
k
(γδ).
Note that this implies that if if Γ GL
2
(R)
+
and Γ = hγ
1
, ··· , γ
m
i then
f |
k
γ = f f |
k
γ
i
= f for all i = 1, ··· , m.
The proof is just a computation.
Proof.
(i) We have
j(γδ, z)
γδ(z)
1
= γδ
z
1
= j(δ, z)γ
δ(z)
1
= j(δ, z)j(γ, δ(z))
z
1
(ii) Take δ = γ
1
.
(iii) We have
((f |
k
γ)|
k
δ)(z) = (det δ)
k/2
j(δ, z)
k
(f |
k
γ)(δ(z))
= (det δ)
k/2
j(δ, z)
k
(det γ)
k/2
j(γ, δ(z))
k
f(γδ(z))
= (det γδ)
k/2
j(γδ, z)
k
f(γδ(z))
= (f |
k
γδ)(z).
Back to the Eisenstein series.
G
k
arise naturally in elliptic functions, which
are coefficients in the series expansion of Weierstrass function.
There is another group-theoretic interpretation, which generalizes in many
ways. Consider
Γ(1)
=
±
1 n
0 1
: n Z
Γ(1) = SL
2
(Z),
which is the stabilizer of . If
δ = ±
1 n
0 1
Γ(1)
,
then we have
j(δγ, z) = j(δ, γ(z))j(γ, z) = ±j(γ, z).
So j(γ, z)
2
depends only on the coset Γ(1)
γ. We can also check that if
γ =
a b
c d
, γ =
a
0
b
0
c
0
d
0
Γ(1),
then Γ(1)
γ = Γ(1)
γ
0
iff (c, d) = ±(c
0
, d
0
).
Moreover, gcd(c, d) = 1 iff there exists a, b such that
a b
c d
= 1.
We therefore have
E
k
(z) =
X
γΓ(1)
\Γ(1)
j(γ, z)
k
,
where we sum over (any) coset representatives of Γ(1)
.
We can generalize this in two ways. We can either replace
j
with some other
appropriate function, or change the groups.