7Lfunctions of eigenforms
III Modular Forms and Lfunctions
7 Lfunctions of eigenforms
Given any modular form, or in fact any function with a q expansion
f =
X
n≥1
a
n
q
n
∈ S
k
(Γ(1)),
we can form a Dirichlet series
L(f, s) =
X
n≥1
a
n
n
−s
.
Our goal of this chapter is to study the behaviour of this
L
function. There
are a few things we want to understand. First, we want to figure out when
this series converges. Next, we will come up with an Euler product formula
for the
L
series. Finally, we come up with an analytic continuation and then a
functional equation.
Afterwards, we will use such analytic methods to understand how
E
2
, which
we figured is not a modular form, transforms under Γ(1), and this in turns gives
us a product formula for ∆(z).
Notation.
We write
a
n

=
O
(
n
k/2
) if there exists
c ∈ R
such that for sufficiently
large n, we have a
n
 ≤ cn
k/2
. We will also write this as
a
n
 n
k/2
.
The latter notation might seem awful, but it is very convenient if we want to
write down a chain of such “inequalities”.
Proposition.
Let
f ∈ S
k
(Γ(1)). Then
L
(
f, s
) converges absolutely for
Re
(
s
)
>
k
2
+ 1.
To prove this, it is enough to show that
Lemma. If
f =
X
n≥1
a
n
q
n
∈ S
k
(Γ(1)),
then
a
n
 n
k/2
Proof.
Recall from the example sheet that if
f ∈ S
k
, then
y
k/2
f
is bounded on
the upper half plane. So
a
n
(f) =
1
2π
Z
q=r
q
−n
˜
f(q)
dq
q
for r ∈ (0, 1). Then for any y, we can write this as
Z
1
0
e
−2πin(x+iy)
f(x + iy)dx
≤ e
2πny
sup
0≤x≤1
f(x + iy) e
2πny
y
−k/2
.
We now pick y =
1
n
, and the result follows.
As before, we can write the
L
function as an Euler product. This time it
looks a bit more messy.
Proposition. Suppose f is a normalized eigenform. Then
L(f, s) =
Y
p prime
1
1 −a
p
p
−s
+ p
k−1−2s
.
This is a very important fact, and is one of the links between cusp forms and
algebraic number theory.
There are two parts of the proof — a formal manipulation, and then a
convergence proof. We will not do the convergence part, as it is exactly the same
as for ζ(s).
Proof. We look at
(1 −a
p
p
−s
+ p
k−1−2s
)(1 + a
p
p
−s
+ a
p
2
p
−2s
+ ···)
= 1 +
X
r≥2
(a
p
r
+ p
k−1
a
p
r−2
− a
p
a
r−1
p
)p
−rs
.
Since we have an eigenform, all of those coefficients are zero. So this is just 1.
Thus, we know
1 + a
p
p
−s
+ a
p
2
p
−2s
+ ··· =
1
1 −a
p
p
−s
+ p
k−1−2s
.
Also, we know that when (m, n) = 1, we have
a
mn
= a
m
a
n
,
and also a
1
= 1. So we can write
L(f, s) =
Y
p
(1 + a
p
p
−s
+ a
p
2
p
−2s
+ ···) =
Y
p
1
1 −a
p
p
−s
+ p
k−1−2s
.
We now obtain an analytic continuation and functional equation for our
L
functions. It is similar to what we did for the
ζ
function, but it is easier this
time, because we don’t have poles.
Theorem.
If
f ∈ S
k
then,
L
(
f, s
) is entire, i.e. has an analytic continuation to
all of C. Define
Λ(f, s) = (2π)
−s
Γ(s)L(f, s) = M(f(iy), s).
Then we have
Λ(f, s) = (−1)
k/2
Λ(f, k − s).
The proof follows from the following more general fact:
Theorem. Suppose we have a function
0 6= f(z) =
X
n≥1
a
n
q
n
,
with a
n
= O(n
R
) for some R, and there exists N > 0 such that
f 
k
0 −1
N 0
= cf
for some k ∈ Z
>0
and c ∈ C. Then the function
L(s) =
X
n≥1
a
n
n
−s
is entire. Moreover, c
2
= (−1)
k
, and if we set
Λ(s) = (2π)
−s
Γ(s)L(s), ε = c ·i
k
∈ {±1},
then
Λ(k − s) = εN
s−k/2
Λ(s).
Note that the condition is rather weak, because we don’t require
f
to even
be a modular form! If we in fact have
f ∈ S
k
, then we can take
N
= 1
, c
= 1,
and then we get the desired analytic continuation and functional equation for
L(f, s).
Proof. By definition, we have
cf(z) = f 
k
0 −1
N 0
= N
−k/2
z
−k
f
−
1
Nz
.
Applying the matrix once again gives
f 
k
0 −1
N 0

k
0 −1
N 0
= f 
k
−N 0
0 −N
= (−1)
k
f(z),
but this is equal to c
2
f(z). So we know
c
2
= (−1)
k
.
We now apply the Mellin transform. We assume Re(s) 0, and then we have
Λ(f, s) = M(f(iy), s) =
Z
∞
0
f(iy)y
s
dy
y
=
Z
∞
1/
√
N
+
Z
1/
√
N
0
!
f(iy)y
s
dy
y
.
By a change of variables, we have
Z
1/
√
N
0
f(iy)y
s
dy
y
=
Z
∞
1/
√
N
f
i
Ny
N
−s
y
−s
dy
y
=
Z
∞
1/
√
N
ci
k
N
k/2−s
f(iy)y
k−s
dy
y
.
So
Λ(f, s) =
Z
∞
1/
√
N
f(iy)(y
s
+ εN
k/2−s
y
k−s
)
dy
y
,
where
ε = i
k
c = ±1.
Since
f →
0 rapidly for
y → ∞
, this integral is an entire function of
s
, and
satisfies the functional equation
Λ(f, k − s) = εN
s−
k
2
Λ(f, s).
Sometimes, we absorb the power of N into Λ, and define a new function
Λ
∗
(f, s) = N
s/2
Λ(f, s) = εΛ
∗
(f, k − s).
However, we can’t get rid of the ε.
What we have established is a way to go from modular forms to
L
functions,
and we found that these
L
functions satisfy certain functional equations. Now is
it possible to go the other way round? Given any
L
function, does it come from
a modular form? This is known as the converse problem. One obvious necessary
condition is that it should satisfy the functional equation, but is this sufficient?
To further investigate this, we want to invert the Mellin transform.
Theorem
(Mellin inversion theorem)
.
Let
f
: (0
, ∞
)
→ C
be a
C
∞
function
such that
– for all N, n ≥ 0, the function y
N
f
(n)
(y) is bounded as y → ∞; and
–
there exists
k ∈ Z
such that for all
n ≥
0, we have
y
n+k
f
(n)
(
y
) bounded
as y → 0.
Let Φ(s) = M(f, s), analytic for Re(s) > k. Then for all σ > k, we have
f(y) =
1
2πi
Z
σ+i∞
σ−i∞
Φ(s)y
−s
ds.
Note that the conditions can be considerably weakened, but we don’t want
to do so much analysis.
Proof.
The idea is to reduce this to the inversion of the Fourier transform. Fix
a σ > k, and define
g(x) = e
2πσx
f(e
2πx
) ∈ C
∞
(R).
Then we find that for any
N, n ≥
0, the function
e
Nx
g
(n)
(
x
) is bounded as
x → +∞. On the other hand, as x → −∞, we have
g
(n)
(x)
n
X
j=0
e
2π(σ+j)x
f
(j)
(e
2πx
)
n
X
j=0
e
2π(σ+j)x
e
−2π(j+k)x
e
2π(σ−k)x
.
So we find that
g ∈ S
(
R
). This allows us to apply the Fourier inversion formula.
By definition, we have
ˆg(−t) =
Z
∞
−∞
e
2πσx
f(e
2πx
)e
2πixt
dx
=
1
2π
Z
∞
0
y
σ+it
f(y)
dy
y
=
1
2π
Φ(σ + it).
Applying Fourier inversion, we find
f(y) = y
−σ
g
log y
2π
= y
−σ
1
2π
Z
∞
−∞
e
−2πit(log y/2π)
Φ(σ + it) dt
=
1
2πi
Z
σ+i∞
σ−i∞
Φ(s)y
−s
ds.
We can now use this to prove a simple converse theorem.
Theorem. Let
L(s) =
X
n≥1
a
n
n
−s
be a Dirichlet series such that
a
n
=
O
(
n
R
) for some
R
. Suppose there is some
even k ≥ 4 such that
– L(s) can be analytically continued to {Re(s) >
k
2
− ε} for some ε > 0;
– L(s) is bounded in vertical strips {σ
0
≤ Re s ≤ σ
1
} for
k
2
≤ σ
0
< σ
1
.
– The function
Λ(s) = (2π)
−s
Γ(s)L(s)
satisfies
Λ(s) = (−1)
k/2
Λ(k − s)
for
k
2
− ε < Re s <
k
2
+ ε.
Then
f =
X
n≥1
a
n
q
n
∈ S
k
(Γ(1)).
Note that the functional equation allows us to continue the Dirichlet series
to the whole complex plane.
Proof.
Holomorphicity of
f
on
H
follows from the fact that
a
n
=
O
(
n
R
), and
since it is given by a
q
series, we have
f
(
z
+ 1) =
f
(
z
). So it remains to show
that
f
−
1
z
= z
k
f(z).
By analytic continuation, it is enough to show this for
f
i
y
= (iy)
k
f(iy).
Using the inverse Mellin transform (which does apply in this case, even if it
might not meet the conditions of the version we proved), we have
f(iy) =
1
2πi
Z
σ+i∞
σ−i∞
Λ(s)y
−s
ds
=
1
2πi
Z
k
2
+i∞
k
2
−i∞
Λ(s)y
−s
ds
=
(−1)
k/2
2πi
Z
k
2
+i∞
k
2
−i∞
Λ(k − s)y
−s
ds
=
(−1)
k/2
2πi
Z
k
2
+i∞
k
2
−i∞
Λ(s)y
s−k
ds
= (−1)
k/2
y
−k
f
i
y
.
Note that for the change of contour, we need
Z
σ±iT
k
2
±iT
Λ(s)y
−s
ds → 0
as
T → ∞
. To do so, we need the fact that Γ(
σ
+
iT
)
→
0 rapidly as
T → ±∞
uniformly for σ in any compact set, which indeed holds in this case.
This is a pretty result, but not really of much interest at this level. However,
it is a model for other proofs of more interesting things, which we unfortunately
would not go into.
Recall we previously defined the Eisenstein series E
2
, and found that
E
2
(z) = 1 − 24
X
n≥1
σ
1
(n)q
n
.
We know this is not a modular form, because there is no modular form of weight
2. However,
E
2
does satisfy
E
2
(
z
+ 1) =
E
(
z
), as it is given by a
q
expansion.
So we know that E
2
(−
1
z
) 6= z
2
E
2
(z). But what is it?
We let
f(y) =
1 −E
2
(iy)
24
=
X
n≥1
σ
1
(n)e
−2πny
.
Proposition. We have
M(f, s) = (2π)
−s
Γ(s)ζ(s)ζ(s − 1).
This is a really useful result, because we understand Γ and ζ well.
Proof. Doing the usual manipulations, it suffices to show that
X
σ
1
(m)m
−s
= ζ(s)ζ(s − 1).
We know if (m, n) = 1, then
σ
1
(mn) = σ
1
(m)σ
1
(n).
So we have
X
m≥1
σ
1
(m)m
−s
=
Y
p
(1 + (p + 1)p
−s
+ (p
2
+ p + 1)p
−2s
+ ···).
Also, we have
(1 −p
−s
)(1 + (p + 1)p
−s
+ (p
2
+ p + 1)p
−2s
+ ···)
= 1 + p
1−s
+ p
2−2s
+ ··· =
1
1 −p
1−s
.
Therefore we find
X
σ
1
(m)m
−s
= ζ(s)ζ(s − 1).
The idea is now to use the functional equation for
ζ
and the inverse Mellin
transform to obtain the transformation formula for
E
2
. This is the reverse of
what we did for genuine modular forms. This argument is due to Weil.
Recall that we defined
Γ
R
(s) = π
−s/2
Γ
s
2
, Z(s) = Γ
R
(s)ζ(s).
Then we found the functional equation
Z(s) = Z(1 − s).
Similarly, we defined
Γ
C
(s) = 2(2π)
−s
Γ(s) = Γ
R
(s)Γ
R
(s + 1),
where the last equality follows from the duplication formula. Then we know
(2π)
−s
Γ(s) = (2π)
−s
(s −1)Γ(s −1) =
s −1
4π
Γ
R
(s)Γ
R
(s −1).
This implies we have the functional equation
Proposition.
M(f, s) =
s −1
4π
Z(s)Z(s − 1) = −M(f, 2 − s).
This also tells us the function is holomorphic except for poles at
s
= 0
,
1
,
2,
which are all simple.
Theorem. We have
f(y) + y
−2
f
1
y
=
1
24
−
1
4π
y
−1
+
1
24
y
−2
.
Proof.
We will apply the Mellin inversion formula. To justify this application,
we need to make sure our
f
behaves sensibly ass
y →
0
, ∞
. We use the absurdly
terrible bound
σ
1
(m) ≤
X
1≤d≤m
d ≤ m
2
.
Then we get
f
(n)
(y)
X
m≥1
m
2+n
e
−2πmy
This is certainly very wellbehaved as
y → ∞
, and is
y
−N
for all
N
. As
y → 0, this is
1
(1 −e
2πy
)
n+3
y
−n−3
.
So f satisfies conditions of our Mellin inversion theorem with k = 3.
We pick any σ > 3. Then the inversion formula says
f(y) =
1
2πi
Z
σ+i∞
σ−i∞
M(f, s)y
−s
ds.
So we have
f
1
y
=
1
2πi
Z
σ+i∞
σ−i∞
−M(f, 2 −s)y
s
ds
=
−1
2πi
Z
2−σ+i∞
2−σ−i∞
M(f, s)y
2−s
ds
So we have
f(y) + y
−2
f
1
y
=
1
2πi
Z
σ+i∞
σ−i∞
−
Z
2+σ+i∞
2−σ−i∞
M(f, s)y
−s
ds.
This contour is pretty simple. It just looks like this:
×× ×
210
Using the fact that
M
(
f, s
) vanishes quickly as
Im
(
s
)
 → ∞
, this is just the
sum of residues
f(y) + y
−2
f
1
y
=
X
s
0
=0,1,2
res
s=s
0
M(f, s)y
−s
0
.
It remains to compute the residues. At s = 2, we have
res
s=2
M(f, s) =
1
4π
Z(2) res
s=1
Z(s) =
1
4π
·
π
6
· 1 =
1
24
.
By the functional equation, this implies
res
s=0
M(f, s) =
1
24
.
Now it remains to see what happens when s = 1. We have
res
s=1
M(f, s) =
1
4π
res
s=1
Z(s) res
s=0
Z(s) = −
1
4π
.
So we are done.
Corollary.
E
2
−
1
z
= z
2
E
2
(z) +
12z
2πi
.
Proof. We have
E
2
(iy) = 1 −24f (y)
= 1 −24y
−2
f
1
y
− 1 +
6
π
y
−1
+ y
−2
= y
−2
1 −24f
1
y
+
6
π
y
−1
= y
−2
E
−1
iy
+
6
π
y
−1
.
Then the result follows from setting
z
=
iy
, and then applying analytic con
tinuiation.
Corollary.
∆(z) = q
Y
m≥1
(1 −q
m
)
24
.
Proof.
Let
D
(
z
) be the righthandside. It suffices to show this is a modular
form, since
S
12
(Γ(1)) is onedimensional. It is clear that this is holomorphic on
H, and D(z + 1) = D(z). If we can show that
D 
12
0 −1
1 0
= D,
then we are done. In other words, we need to show that
D
−1
z
= z
12
D(z).
But we have
D
0
(z)
D(z)
= 2πi −24
X
m≥1
2πimq
1 −q
m
= 2πi
1 −24
X
m,d≥1
mq
md
= 2πiE
2
(z)
So we know
d
dz
log D
−
1
z
=
1
z
2
D
0
D
−
1
z
=
1
z
2
2πiE
2
−
1
z
=
D
0
D
(z) + 12
d
dz
log z.
So we know that
log D
−
1
z
= log D + 12 log z + c,
for some locally constant function c. So we have
D
−
1
z
= z
12
D(z) · C
for some other constant
C
. By trying
z
=
i
, we find that
C
= 1 (since
D
(
i
)
6
= 0
by the infinite product). So we are done.