2Riemann ζ-function

III Modular Forms and L-functions

2 Riemann ζ-function
We ended the previous chapter by briefly mentioning Dirichlet series. The first
and simplest example one can write down is the Riemann ζ-function.
Definition (Riemann ζ-function). The Riemann ζ-function is defined by
ζ(s) =
X
n1
1
n
s
for Re(s) > 1.
This
ζ
-function is related to prime numbers by the following famous formula:
Proposition (Euler product formula). We have
ζ(s) =
Y
p prime
1
1 p
s
.
Proof.
Euler’s proof was purely formal, without worrying about convergence.
We simply note that
Y
p prime
1
1 p
s
=
Y
p
(1 + p
s
+ (p
2
)
s
+ ···) =
X
n1
n
s
,
where the last equality follows by unique factorization in
Z
. However, to prove
this properly, we need to be a bit more careful and make sure things converge.
Saying the infinite product
Q
p
convergence is the same as saying
P
p
s
converges, by basic analysis, which is okay since we know
ζ
(
s
) converges absolutely
when Re(s) > 1. Then we can look at the difference
ζ(s)
Y
pX
1
1 p
s
= ζ(s)
Y
pX
(1 + p
s
+ p
2s
+ ···)
=
Y
n∈N
X
n
s
,
where
N
X
is the set of all
n
1 such that at least one prime factor is
X
. In
particular, we know
ζ(s)
Y
pX
1
1 p
s
X
nX
|n
s
| 0
as X . So the result follows.
The Euler product formula is the beginning of the connection between the
ζ
-function and the distribution of primes. For example, as the product converges
for
Re
(
s
)
>
1, we know in particular that
ζ
(
s
)
6
= 0 for all
s
when
Re
(
s
)
>
1.
Whether or not
Re
(
s
) vanishes elsewhere is a less straightforward matter, and
this involves quite a lot of number theory.
We will, however, not care about primes in this course. Instead, we look at
some further analytic properties of the
ζ
function. To do so, we first write it as
a Mellin transform.
Theorem. If Re(s) > 1, then
(2π)
s
Γ(s)ζ(s) =
Z
0
y
s
e
2πy
1
dy
y
= M(f, s),
where
f(y) =
1
e
2πy
1
.
This is just a simple computation.
Proof. We can write
f(y) =
e
2πy
1 e
2πy
=
X
n1
e
2πny
for y > 0.
As y 0, we find
f(y)
1
2πy
.
So when Re(s) > 1, the Mellin transform converges, and equals
X
n1
M(e
2πny
, s) =
X
n1
(2πn)
s
M(e
y
, s) = (2π)
s
Γ(s)ζ(s).
Corollary. ζ
(
s
) has a meromorphic continuation to
C
with a simple pole at
s = 1 as its only singularity, and
res
s=1
ζ(s) = 1.
Proof. We can write
M(f, s) = M
0
+ M
=
Z
1
0
+
Z
1
y
s
e
2πy
1
dy
y
.
The second integral
M
is convergent for all
s C
, hence defines a holomorphic
function.
For any fixed N, we can expand
f(y) =
N1
X
n=1
c
n
y
n
+ y
N
g
N
(y)
for some g C
(R), as f has a simple pole at y = 0, and
c
1
=
1
2π
.
So for Re(s) > 1, we have
M
0
=
N1
X
n=1
c
n
Z
1
0
y
n+s1
dy +
Z
N
0
y
N+s1
g
N
(y) dy
=
N1
X
n=1
c
n
s + n
y
s+n
+
Z
1
0
g
N
(y)y
s+N1
dy.
We now notice that this formula makes sense for
Re
(
s
)
> N
. Thus we have
found a meromorphic continuation of
(2π)
s
Γ(s)ζ(s)
to
{Re
(
s
)
> N}
, with at worst simple poles at
s
= 1
N,
2
N, ··· ,
0
,
1. Also,
we know Γ(
s
) has a simple pole at
s
= 0
,
1
,
2
, ···
. So
ζ
(
s
) is analytic at
s = 0, 1, 2, ···. Since c
1
=
1
2π
and Γ(1) = 1, we get
res
s=1
ζ(s) = 1.
Now we note that by the Euler product formula, if there are only finitely
many primes, then ζ(s) is certainly analytic everywhere. So we deduce
Corollary. There are infinitely many primes.
Given a function
ζ
, it is natural to ask what values it takes. In particular,
we might ask what values it takes at integers. There are many theorems and
conjectures concerning the values at integers of
L
-functions (which are Dirichlet
series like the
ζ
-function). These properties relate to subtle number-theoretic
quantities. For example, the values of
ζ
(
s
) at negative integers are closely
related to the class numbers of the cyclotomic fields
Q
(
ζ
p
). These are also
related to early (partial) proofs of Fermat’s last theorem, and things like the
Birch–Swinnerton-Dyer conjecture on elliptic curves.
We shall take a tiny step by figuring out the values of
ζ
(
s
) at negative integers.
They are given by the Bernoulli numbers.
Definition
(Bernoulli numbers)
.
The Bernoulli numbers are defined by a gen-
erating function
X
n=0
B
n
t
n
n!
=
t
e
t
1
=
1 +
t
2!
+
t
2
3!
+ ···
1
.
Clearly, all Bernoulli numbers are rational. We can directly compute
B
0
= 1, B
1
=
1
2
, ··· .
Proposition. B
n
= 0 if n is odd and n 3.
Proof. Consider
f(t) =
t
e
t
1
+
t
2
=
X
n0,n6=1
B
n
t
n
n!
.
We find that
f(t) =
t
2
e
t
+ 1
e
t
1
= f(t).
So all the odd coefficients must vanish.
Corollary. We have
ζ(0) = B
1
=
1
2
, ζ(1 n) =
B
n
n
for
n >
1. In particular, for all
n
1 integer, we know
ζ
(1
n
)
Q
and vanishes
if n > 1 is odd.
Proof. We know
(2π)
s
Γ(s)ζ(s)
has a simple pole at s = 1 n, and the residue is c
n1
, where
1
e
2πy
1
=
X
n≥−1
c
n
y
n
.
So we know
c
n1
= (2π)
n1
B
n
n!
.
We also know that
res
s=1n
Γ(s) =
(1)
n1
(n 1)!
,
we get that
ζ(1 n) = (1)
n1
B
n
n
.
If
n
= 1, then this gives
1
2
. If
n
is odd but
>
1, then this vanishes. If
n
is even,
then this is
B
n
n
, as desired.
To end our discussion on the
ζ
-function, we shall prove a functional equation,
relating
ζ
(
s
) to
ζ
(1
s
). To do so, we relate the
ζ
-function to another Mellin
transform. We define
Θ(y) =
X
nZ
e
πn
2
y
= 1 + 2
X
n1
e
πn
2
y
.
This is convergent for for y > 0. So we can write
Θ(y) = ϑ(iy),
where
ϑ(z) =
X
nZ
e
πin
2
z
,
which is analytic for
|e
πiz
| <
1, i.e.
Im
(
z
)
>
0. This is Jacobi’s
ϑ
-function. This
function is also important in algebraic geometry, representation theory, and even
applied mathematics. But we will just use it for number theory. We note that
Θ(y) 1
as y , so we can’t take its Mellin transform. What we can do is
Proposition.
M
Θ(y) 1
2
,
s
2
= π
s/2
Γ
s
2
ζ(s).
The proof is again just do it.
Proof. The left hand side is
X
n1
M
e
πn
2
y
,
s
2
=
X
n1
(πn
2
)
s/2
M
e
y
,
s
2
= π
s/2
Γ
s
2
ζ(s).
To produce a functional equation for ζ, we first do it for Θ.
Theorem (Functional equation for Θ-function). If y > 0, then
Θ
1
y
= y
1/2
Θ(y), ()
where we take the positive square root. More generally, taking the branch of
which is positive real on the positive real axis, we have
ϑ
1
z
=
z
i
1/2
ϑ(z).
The proof is rather magical.
Proof. By analytic continuation, it suffices to prove (). Let
g
t
(x) = e
πtx
2
= g
1
(t
1/2
x).
In particular,
g
1
(x) = e
πx
2
.
Now recall that
ˆg
1
=
g
1
. Moreover, the Fourier transform of
f
(
αx
) is
1
α
ˆ
f
(
y
).
So
ˆg
t
(y) = t
1/2
ˆg
1
(t
1/2
y) = t
1/2
g
1
(t
1/2
y) = t
1/2
e
πy
2
/t
.
We now apply the Poisson summation formula:
Θ(t) =
X
nZ
e
πn
2
t
=
X
nZ
g
t
(n) =
X
nZ
ˆg
t
(n) = t
1/2
Θ(1/t).
Before we continue, we notice that most of the time, when we talk about the
Γ-function, there are factors of π floating around. So we can conveniently set
Notation.
Γ
R
(s) = π
s/2
Γ(s/2).
Γ
C
(s) = 2(2π)
s
Γ(s)
These are the real/complex Γ-factors.
We also define
Notation.
Z(s) Γ
R
(s)ζ(s) = π
s/2
Γ
s
2
ζ(s).
The theorem is then
Theorem (Functional equation for ζ-function).
Z(s) = Z(1 s).
Moreover, Z(s) is meromorphic, with only poles at s = 1 and 0.
Proof. For Re(s) > 1, we have
2Z(s) = M
Θ(y) 1,
s
2
=
Z
0
[Θ(y) 1]y
s/2
dy
y
=
Z
1
0
+
Z
1
[Θ(y) 1]y
s/2
dy
y
The idea is that using the functional equation for the Θ-function, we can relate
the
R
1
0
part and the
R
1
part. We have
Z
1
0
(Θ(y) 1)y
s/2
dy
y
=
Z
1
0
(Θ(y) y
1/2
)y
s/2
dy
y
+
Z
1
0
y
s1
2
y
1/2
dy
y
=
Z
1
0
(y
1/2
Θ(1/y) y
1/2
)
dy
y
+
2
s 1
2
s
.
In the first term, we change variables y 1/y, and get
=
Z
1
y
1/2
(Θ(y) 1)y
s/2
dy
y
+
2
s 1
2
s
.
So we find that
2Z(s) =
Z
1
(Θ(y) 1)(y
s/2
+ y
1s
2
)
dy
y
+
2
s 1
2
s
= 2Z(1 s).
Note that what we’ve done by separating out the
y
s1
2
y
s/2
term is that we
separated out the two poles of our function.
Later on, we will come across more
L
-functions, and we will prove functional
equations in the same way.
Note that we can write
Z(s) = Γ
R
(s)
Y
p primes
1
1 p
s
,
and the term Γ
R
(
s
) should be thought of as the Euler factor for
p
=
, i.e. the
Archimedean valuation on Q.