Math 726: L-functions and modular forms Fall 2011

Lecture 31 : Special values of L-functions.

Instructor: Henri Darmon Notes written by: Francesca Gala The goal of this lecture is to prove the following theorem.

Theorem 1. If n ∈ ℤ_{≥0}, then ζ(−n) ∈ ℚ and is non zero if and only if n is odd.

We are going to use the following integral representation of the Riemann zeta function for Re(s) > 1:

ζ(s) = \frac{1}{Γ(s)} ∫_{0}^{∞} f(t)t^{s−1} \frac{dt}{t},

where f(t) = \frac{t}{e^{t}−1}. Definition 2. If f : ℝ^{≥0} → ℂ is any C^{∞} function of rapid decay, let

L(f, s) := \frac{1}{Γ(s)} ∫_{0}^{∞} f(t)t^{s} \frac{dt}{t}

for Re(s) > 0. In this notation ζ(s) = \frac{1}{s−1}L(f, s − 1), where f(t) = \frac{t}{e^{t}−1}.

Proposition 1. If f : [0, ∞) → ℂ is any C^{∞} function of rapid decay, then L(f, s) has analytic continuation to all s ∈ ℂ and

L(f, −n) = (−1)^{n}f^{(n)}(0),

for n ∈ ℕ. Proof. Write f = f_{0} + f_{∞}, where

f_{0} = \{ \begin{matrix}f & on [0, 1] \\ 0 & on [2, ∞) \end{matrix}

and

f_{∞} = \{ \begin{matrix}0 & on [0, 1] \\ f & on [2, ∞) \end{matrix}

We have L(f, s) = L(f_{0}, s) + L(f_{∞}, s), but

h(s) = ∫_{0}^{∞} f_{∞}(s)t^{s} \frac{dt}{t}

converges for all s ∈ ℂ, so that L(f_{∞}, −n) = \frac{1}{Γ(−n)}h(−n) = 0, for all n ∈ ℕ, since Γ has simple poles at the negative integers and h is a holomorphic function at −n. On the other hand

Γ(s)L(f_{0}, s) = ∫_{0}^{∞} f_{0}(t)t^{s−1}dt = (1)

if we let u = f_{0}(t), with du = f′_{0}(t)du, and dv = t^{s−1}dt so that v = \frac{t^{s}}{s} we obtain for Re(s) > 0

(1) = \frac{f′_{0}(t)t^{s}}{s} |_{t=0}^{∞} − \frac{1}{s} ∫_{0}^{∞} f′_{0}(t)t^{s}dt = −\frac{1}{s} ∫_{0}^{∞} f′_{0}(t)t^{s+1} \frac{dt}{t}

⇒ L(f_{0}, s) = −\frac{1}{sΓ(s)} ∫_{0}^{∞} f′_{0}(t)t^{s+1} \frac{dt}{t} = −\frac{1}{Γ(s + 1)} ∫_{0}^{∞} f′_{0}(t)t^{s+1} \frac{dt}{t} = −L(f′_{0}, s + 1)

and since f′_{0} is also of rapid decay we can analytically continue the function L(f_{0}, s), which was a priori defined for Re(s) > 0, to all the values s ∈ ℂ for which Re(s) > −1. Moreover we immediately deduce that for Re(s) > −n, n ∈ ℕ:

L(f_{0}, s) = (−1)^{n}L(f_{o}^{(n)}, s + n).

Now we would like to determine explicitely the values of L at the negative integers.

L(f, −n) = L(f_{0}, −n) = (−1)^{n+1}L(f_{0}^{(n+1)}, 1) = \frac{(−1)^{n+1}}{Γ(1)} ∫_{0}^{∞} f_{o}^{(n+1)}dt =
= f_{o}^{(n+1)}|_{t=0}^{∞} = (−1)^{n}f_{o}^{(n)}(0) = (−1)^{n}f^{(n)}(0),

which is what we wanted.

Corollary 1. If n ∈ ℤ^{≥0}, then ζ(−n) ∈ ℚ. Moreover ζ(−n) = 0 if n is even.

Proof. If we let f(t) = \frac{t}{e^{t}−1} then

ζ(−n) = \frac{−1}{n + 1}L(f, −n − 1) =

= \frac{−1}{n + 1}(−1)^{n+1}f^{(n+1)}(0) = \frac{(−1)^{n}}{n + 1} f^{(n+1)}(0)

where

f = \frac{t}{e^{t} − 1} = \frac{t}{∑{∞ \atop n=1} \frac{t^{n}}{n!}} = ∑_{n=0}^{∞} B_{n} \frac{t^{n}}{n!}.

The coefficients B_{n} ∈ ℚ are called Bernoulli numbers. We have for example:

B_{0} = 1, B_{1} = \frac{−1}{2}, B_{2} = 16, B_{3} = 0, B_{4} = \frac{−1}{30}.

The formula for ζ(−n) in terms of B_{n} is then:

ζ(−n) = (−1)^{n} \frac{B_{n+1}}{n + 1} = \{ \begin{matrix}0 & if n is even \\ −\frac{B_{n+1}}{n+1} & if n is odd \end{matrix}

Let us show that B_{n} = 0 when n ≥ 3 is odd.

f(t) − f(−t) = \frac{t}{e^{t} − 1} − \frac{−t}{e^{−t} − 1} = \frac{t}{e^{t} − 1} − \frac{−te^{t}}{e^{t} − 1}

= \frac{t(1 − e^{t})}{e^{t} − 1} = t,

so that clearly B_{1} = −\frac{1}{2} and B_{k} = 0 for all odd k ≥ 3.

Kummer-Von Staudt congruences.

Theorem 3. Let p be a fixed prime and choose an integer a such that (a, p) = 1. If n_{1} and n_{2} are natural numbers such that n_{1} ≡ n_{2} mod (p − 1)p^{k} and n_{1}, n_{2} ≥ k then

(1 − a^{1+n_{1}})ζ(−n_{1}) ≡ ζ(−n_{2})(1 − a_{1+n_{2}}) mod p^{k+1}

This theorem suggests that we can modify the Riemann zeta function in such a way that it becomes an analytic function in some p-adic topology.

Theorem 4 (Reformulation, due to Kubota-Leopoldt). The function

\begin{matrix}ℤ^{≤0} & ⟶ & ℤ_{(p)} \\ n & ⟼ & (1 − a^{1−n})(1 − p^{−n})ζ(n) \end{matrix}

extends to a continuos function on ℤ/(p − 1)ℤ × ℤ_{p}. The resulting function on ℤ/(p − 1)ℤ × ℤ_{p} is called the Kubota-Leopoldt p-adic zeta function.

To understand ζ(−n) as a function of ℤ/(p − 1)ℤ × ℤ_{p} we will express ζ(−n) as a ‘p-adic Mellin transform’:

(1 − a^{1+n})(1 − p^{n})ζ(−n) = ∫_{ℤ_{p}^{×}} x^{n−1}dμ_{f}(x).  (1)

In the next lectures we will try to make sense of equality (1) and in particular we will explain how to define the measure dμ_{f}(x).

p-adic Functional analysis.

Definition 5. A p-adic Banach space is a vector space over ℚ_{p} equipped with ℤ_{p}-submodule B_{0} ⊂ B satisfying:

1. The natural map B_{0} ⟶ \underleftarrow{lim} B_{0}/p^{n}B_{0} is an isomorphism

2. For all r ∈ B, there exists p^{n} such that p^{n}r ∈ B_{0}.

Example 6. (1) If B is any finite dimensional ℚ_{p}-vector space then B is a p-adic Banach space. A basis for B over ℚ_{p} gives rise to an isomorphism

ϕ : B →^{∼} ℚ_{p}^{d}, then B_{0} := ϕ^{−1}(ℤ_{p}^{d}).

(2) The set of bounded ℚ_{p}-valued functions on X, where X is any set, is a p-adic Banach space. Then B_{0} is the set function from X to ℤ_{p}.

(3) The space of power series with bounded coefficients ℚ_{p} ⊗_{ℤ_{p}} ℤ_{p}[[t]] is a p-adic Banach space. (4)Non-example. ℚ_{p}[[t]] is not a p-adic Banach space.

Remark 1. If B is a p-adic Banach space, the datum of B_{0} induces a p-adic norm on B, namely for v ∈ B ∥v∥ = p^{−n}, where n = min{n|p^{n}v ∈ B_{0} \ pB_{0}}.