3Dirichlet L-functions
III Modular Forms and L-functions
3 Dirichlet L-functions
We now move on to study a significant generalization of
ζ
-functions, namely
Dirichlet
L
-functions. While these are generalizations of the
ζ
-function, it turns
out the
ζ
function is a very particular kind of
L
-function. For example, most
L
-functions are actually analytic on all of
C
, except for (finite multiples of) the
ζ-function.
Recall that a Dirichlet series is a series of the form
∞
X
n=1
a
n
n
s
.
A Dirichlet
L
-function is a Dirichlet series whose coefficients come from Dirichlet
characters.
Definition
(Dirichlet characters)
.
Let
N ≥
1. A Dirichlet character mod
N
is
a character χ : (Z/NZ)
×
→ C
×
.
As before, we write
\
(Z/NZ)
×
for the group of characters.
Note that in the special case N = 1, we have
Z/NZ = {0 = 1} = (Z/NZ)
×
,
and so
\
(Z/NZ)
×
∼
=
{1}, and the only Dirichlet character is identically 1.
Not all characters are equal. Some are less exciting than others. Suppose
χ
is a character mod
N
, and we have some integer
d >
1. Then we have the
reduction mod N map
(Z/NdZ)
×
(Z/NZ)
×
,
and we can compose
χ
with this to get a character mod
Nd
. This is a rather
boring character, because the value of
x ∈
(
Z/NdZ
)
×
only depends on the value
of x mod N .
Definition
(Primitive character)
.
We say a character
χ ∈
\
(Z/nZ)
×
is primitive
if there is no M < N with M | N with χ
0
∈
\
(Z/MZ)
×
such that
χ = χ
0
◦ (reduction mod M).
Similarly we define
Definition
(Equivalent characters)
.
We say characters
χ
1
∈
\
(Z/N
1
Z)
×
and
χ
2
∈
\
(Z/N
2
Z)
×
are equivalent if for all
x ∈ Z
such that (
x, N
1
N
2
) = 1, we have
χ
1
(x mod N
1
) = χ
2
(x mod N
2
).
It is clear that if we produce a new character from an old one via reduction
mod Nd, then they are equivalent.
One can show the following:
Proposition.
If
χ ∈
\
(Z/NZ)
×
, then there exists a unique
M | N
and a primitive
χ
∗
∈
\
(Z/MZ)
×
that is equivalent to χ.
Definition
(Conductor)
.
The conductor of a character
χ
is the unique
M | N
such that there is a primitive χ
∗
∈
\
(Z/MZ)
×
that is equivalent to χ.
Example. Take
χ = χ
0
∈
\
(Z/NZ)
×
,
given by
χ
0
(
x
)
≡
1. If
N >
1, then
χ
0
is not primitive, and the associated
primitive character is the trivial character modulo
M
= 1. So the conductor is 1.
Using these Dirichlet characters, we can define Dirichlet L-series:
Definition
(Dirichlet
L
-series)
.
Let
χ ∈
\
(Z/NZ)
×
be a Dirichlet character. The
Dirichlet L-series of χ is
L(χ, s) =
X
n≥1
(n,N)=1
χ(n)n
−s
.
Since |χ(n)| = 1, we again know this is absolutely convergent for Re(s) > 1.
As
χ
(
mn
) =
χ
(
m
)
χ
(
n
) whenever (
mn, N
) = 1, we get the same Euler product
as we got for the ζ-function:
Proposition.
L(χ, s) =
Y
prime p-N
1
1 − χ(p)p
−s
.
The proof of convergence is again similar to the case of the ζ-function.
It follows that
Proposition.
Suppose
M | N
and
χ
M
∈
\
(Z/MZ)
×
and
χ
N
∈
\
(Z/NZ)
×
are
equivalent. Then
L(χ
M
, s) =
Y
p-M
p|N
1
1 − χ
M
(p)p
−s
L(χ
N
, s).
In particular,
L(χ
M
, s)
L(χ
N
, s)
=
Y
p-M
p|N
1
1 − χ
M
(p)p
−s
is analytic and non-zero for Re(s) > 0.
We’ll next show that
L
(
χ, s
) has a meromorphic continuation to
C
, and is
analytic unless χ = χ
0
.
Theorem.
(i) L
(
χ, s
) has a meromorphic continuation to
C
, which is analytic except for
at worst a simple pole at s = 1.
(ii)
If
χ 6
=
χ
0
(the trivial character), then
L
(
χ, s
) is analytic everywhere. On
the other hand, L(χ
0
, s) has a simple pole with residue
ϕ(N)
N
=
Y
p|N
1 −
1
p
,
where ϕ is the Euler function.
Proof. More generally, let φ : Z/NZ → C be any N-periodic function, and let
L(φ, s) =
∞
X
n=1
φ(n)n
−s
.
Then
(2π)
−s
Γ(s)L(φ, s) =
∞
X
n=1
φ(n)M(e
−2πny
, s) = M(f(y), s),
where
f(y) =
X
n≥1
φ(n)e
−2πny
.
We can then write
f(y) =
N
X
n=1
∞
X
r=0
φ(n)e
−2π(n+rN)y
=
N
X
n=1
φ(n)
e
−2πny
1 − e
−2πNy
=
N
X
n=1
φ(n)
e
2π(N −n)y
e
2πNy
− 1
.
As 0 ≤ N − n < N, this is O(e
−2πy
) as y → ∞. Copying for ζ(s), we write
M(f, s) =
Z
1
0
+
Z
∞
1
f(y)y
s
dy
y
≡ M
0
(s) + M
∞
(s).
The second term is analytic for all s ∈ C, and the first term can be written as
M
0
(s) =
N
X
n=1
φ(n)
Z
1
0
e
2π(N−n)y
e
2πNy
− 1
y
s
dy
y
.
Now for any L, we can write
e
2π(N−n)y
e
2πNy
− 1
=
1
2πNy
+
L−1
X
r=0
c
r,n
y
r
+ y
L
g
L,n
(y)
for some g
L,n
(y) ∈ C
∞
[0, 1]. Hence we have
M
0
(s) =
N
X
n=1
φ(n)
Z
1
0
1
2πNy
y
s
dy
y
+
Z
1
0
L−1
X
r=0
c
r,n
y
r+s−1
dy
!
+ G(s),
where G(s) is some function analytic for Re(s) > −L. So we see that
(2π)
−s
Γ(s)L(φ, s) =
N
X
n=1
φ(n)
1
2πN(s −1)
+
c
0,n
s
+ ··· +
c
L−1,n
s + L − 1
+ G(s).
As Γ(
s
) has poles at
s
= 0
, −
1
, ···
, this cancels with all the poles apart from the
one at s = 1.
The first part then follows from taking
φ(n) =
(
χ(n) (n, N) = 1
0 (n, N) ≥ 1
.
By reading off the formula, since Γ(1) = 1, we know
res
s=1
L(χ, s) =
1
N
N
X
n=1
φ(n).
If
χ 6
=
χ
0
, then this vanishes by the orthogonality of characters. Otherwise, it is
|(Z/NZ)
×
|/N = ϕ(N)/N .
Note that this is consistent with the result
L(χ
0
, s) =
Y
p|N
(1 − p
−s
)ζ(s).
So for a non-trivial character, our L-function doesn’t have a pole.
The next big theorem is that in fact
L
(
χ,
1) is non-zero. In number theory,
there are lots of theorems of this kind, about non-vanishing of
L
-functions at
different points.
Theorem. If χ 6= χ
0
, then L(χ, 1) 6= 0.
Proof. The trick is the consider all characters together. We let
ζ
N
(s) =
Y
χ∈
\
(Z/NZ)
×
L(χ, s) =
Y
p-N
Y
χ
(1 − χ(p)p
−s
)
−1
for
Re
(
s
)
>
1. Now we know
L
(
χ
0
, s
) has a pole at
s
= 1, and is analytic
everywhere else. So if any other
L
(
χ,
1) = 0, then
ζ
N
(
s
) is analytic on
Re
(
s
)
>
0.
We will show that this cannot be the case.
We begin by finding a nice formula for the product of (1
− χ
(
p
)
p
−s
)
−1
over
all characters.
Claim. If p - N, and T is any complex number, then
Y
χ∈
\
(Z/NZ)
×
(1 − χ(p)T ) = (1 −T
f
p
)
ϕ(N)/f
p
,
where f
p
is the order of p in (Z/nZ)
×
.
So
ζ
N
(s) =
Y
p-N
(1 − p
−f
p
s
)
−ϕ(N)/f
p
.
To see this, we write f = f
p
, and, for convenience, write
G = (Z/NZ)
×
H = hpi ⊆ G.
We note that
ˆ
G
naturally contains
[
G/H
=
{χ ∈
ˆ
G
:
χ
(
p
) = 1
}
as a subgroup.
Also, we know that
|
[
G/H| = |G/H| = ϕ(N)/f.
Also, the restriction map
ˆ
G
[
G/H
→
ˆ
H
is obviously injective, hence an isomorphism by counting orders. So
Y
χ∈
ˆ
G
(1−χ(p)T ) =
Y
χ∈
ˆ
H
(1−χ(p)T )
ϕ(N)/f
=
Y
ζ∈µ
f
(1−ζT )
ϕ(N)/f
= (1−T
f
)
ϕ(N)/f
.
We now notice that when we expand the product of
ζ
N
, at least formally, then we
get a Dirichlet series with non-negative coefficients. We now prove the following
peculiar property of such Dirichlet series:
Claim. Let
D(s) =
X
n≥1
a
n
n
−s
be a Dirichlet series with real
a
n
≥
0, and suppose this is absolutely convergent
for
Re
(
s
)
> σ >
0. Then if
D
(
s
) can be analytically continued to an analytic
function
˜
D on {Re(s) > 0}, then the series converges for all real s > 0.
Let
ρ > σ
. Then by the analytic continuation, we have a convergent Taylor
series on {|s − ρ| < ρ}
D(s) =
X
k≥0
1
k!
D
(k)
(ρ)(s − ρ)
k
.
Moreover, since
ρ > σ
, we can directly differentiate the Dirichlet series to obtain
the derivatives:
D
(k)
(ρ) =
X
n≥1
a
n
(−log n)
k
n
−ρ
.
So if 0 < x < ρ, then
D(x) =
X
k≥0
1
k!
(p − x)
k
X
n≥1
a
n
(log n)
k
n
−ρ
.
Now note that all terms in this sum are all non-negative. So the double series
has to converge absolutely as well, and thus we are free to rearrange the sum as
we wish. So we find
D(x) =
X
n≥1
a
n
n
−ρ
X
k≥0
1
k!
(ρ − x)
k
(log n)
l
=
X
n≥1
a
n
n
−ρ
e
(ρ−x) log n
=
X
n≥1
a
n
n
−ρ
n
ρ−x
=
X
n≥1
a
n
n
−x
,
as desired.
Now we are almost done, as
ζ
N
(s) = L(χ
0
, s)
Y
χ6=χ
0
L(χ, s).
We saw that
L
(
χ
0
, s
) has a simple pole at
s
= 1, and the other terms are all
holomorphic at
s
= 1. So if some
L
(
χ,
1) = 0, then
ζ
N
(
s
) is holomorphic for
Re
(
s
)
>
0 (and in fact everywhere). Since the Dirichlet series of
η
N
has
≥
0
coefficients, by the lemma, it suffices to find some point on
R
>0
where the
Dirichlet series for ζ
N
doesn’t converge.
We notice
ζ
N
(x) =
Y
p-N
(1 + p
−f
p
x
+ p
−2f
p
x
+ ···)
ϕ(N)/f
p
≥
X
p-N
p
−ϕ(N)x
.
It now suffices to show that
P
p
−1
=
∞
, and thus the series for
ζ
N
(
x
) is not
convergent for x =
1
ϕ(N)
.
Claim. We have
X
p prime
p
−x
∼ −log(x − 1)
as x → 1
+
. On the other hand, if χ 6= χ
0
is a Dirichlet character mod N, then
X
p-N
χ(p)p
−x
is bounded as x → 1
+
.
Of course (and crucially, as we will see), the second part is not needed for
the proof, but it is still nice to know.
To see this, we note that for any χ, we have
log L(χ, x) =
X
p-N
−log(1 − χ(p)p
−x
) =
X
p-N
X
r≥1
χ(p)
r
p
−rx
r
.
So
log L(χ, x) −
X
p-N
χ(p)p
−x
<
X
p-N
X
r≥2
p
−rx
=
X
p-N
p
−2x
1 − p
−x
≤
X
n≥1
n
−2
1/2
,
which is a (finite) constant for C < ∞. When χ = χ
0
, N = 1, then
log ζ(x) −
X
p
p
−x
is bounded as x → 1
+
. But we know
ζ(s) =
1
s − 1
+ O(s).
So we have
X
p
p
−x
∼ log(x − 1).
When
χ 6
=
χ
0
, then
L
(
χ,
1)
6
= 0, as we have just proved! So
log L
(
χ, x
) is
bounded as x → 1
+
. and so we are done.
Note that up to a finite number of factors in the Euler product (for
p | N
),
this
ζ
N
(
s
) equals to the Dedekind
ζ
-function of the number field
K
=
Q
(
n
√
1
),
given by
ζ
K
(s) =
X
ideals 06=I⊆O
K
1
(N(I))
s
.
We can now use what we’ve got to quickly prove Dirichlet’s theorem:
Theorem
(Dirichlet’s theorem on primes in arithmetic progressions)
.
Let
a ∈ Z
be such that (
a, N
) = 1. Then there exists infinitely many primes
p ≡ a
(mod N).
Proof. We want to show that the series
X
p prime
p≡a mod N
p
−x
is unbounded as
x →
1
+
, and in particular must be infinite. We note that for
(x, N) = 1, we have
X
χ∈
\
(Z/NZ)
×
χ(x) =
(
ϕ(N) x ≡ 1 (mod N)
0 otherwise
,
since the sum of roots of unity vanishes. We also know that
χ
is a character, so
χ(a)
−1
χ(p) = χ(a
−1
p). So we can write
X
p prime
p≡a mod N
p
−x
=
1
ϕ(N)
X
χ∈(Z/NZ)
×
χ(a)
−1
X
all p
χ(p)p
−x
,
Now if χ = χ
0
, then the sum is just
X
p-N
p
−x
∼ −log(x − 1)
as x → 1
+
. Moreover, all the other sums are bounded as x → 1
+
. So
X
p≡a mod N
p
−x
∼ −
1
ϕ(N)
log(x − 1).
So the whole sum must be unbounded as
x →
1
+
. So in particular, the sum
must be infinite.
This actually tells us something more. It says
P
p≡a mod N
p
−x
P
all p
p
−x
∼
1
ϕ(N)
.
as
x →
1
+
. So in some well-defined sense (namely analytic density),
1
ϕ(N)
of the
primes are ≡ a (mod N).
In fact, we can prove that
lim
X→∞
|{p ≤ X : p ≡ a mod N}|
|{p ≤ X}|
=
1
ϕ(N)
.
This theorem has many generalizations. In general, let
L/K
be a finite Galois
extension of number fields with Galois group
G
=
Gal
(
L/K
). Then for all
primes
p
of
K
which is unramified in
L
, we can define a Frobenius conjugacy
class [σ
p
] ⊆ G.
Theorem
(Cebotarev density theorem)
.
Cebotarev density theorem Let
L/K
be a Galois extension. Then for any conjugacy class
C ⊆ Gal
(
L/K
), there exists
infinitely many p with [σ
p
] = C.
If
L/K
=
Q
(
n
√
1
)
/Q
, then
G
∼
=
(
Z/NZ
)
×
, and
σ
p
is just the element of
G
given by
p
(
mod N
). So if we fix
a
(
mod N
)
∈ G
, then there are infinitely many
p
with
p ≡ a
(
mod N
). So we see the Cebotarev density theorem is indeed a
generalization of Dirichlet’s theorem.