Math 726: L-functions and modular forms Fall 2011

Lectures 32-33 : Values of zeta functions

Instructor: Henri Darmon Notes written by: Clement Gomez

Recall from last lectures

∀n ∈ ℕ, ζ(−n) ∈ ℚ

The proof is to consider the Mellin transform: L(f, s) := \frac{1}{Γ(s)} ∫_{0}^{∞} f(t)t^{s} \frac{dt}{t}. Then one can show that L(f, −n) = (−1)^{n}f^{(n)}(0). On the other hand, by a direct computation, ζ(s) = \frac{1}{s−1}L(f, s) where f(t) = \frac{t}{e^{t}−1} Since \frac{t}{e^{t}−1} has rational Taylor expansion, we conclude that ∀n ∈ ℕ, ζ(−n) ∈ ℚ.

Problem 1. Understand the p−adic properties of ζ(−n):

Idea: Express ζ(−n) as a p−adic Mellin transform of \frac{t}{e^{t}−1}.

Rudiments of p−adic integration theory:

Definition 1. A p-adic Banach space B is a ℚ_{p}-vector space, such that there exists a ℤ_{p}-submodule B_{0} with the following properties:

(i)B_{0} → lim_{←}B_{0}/p^{n}B_{0} is an isomorphism (ii)∀x ∈ B, ∃n ∈ ℤ such that p^{n}.x ∈ B_{0} We can define the valuation for any element of B, by v_{p}(x) := min{n ∈ ℤ|p^{n}.x ∈ B_{0}}. The norm is then defined by ||x|| := p^{−v_{p}(x)}.

Remark 1. The Banach space is complete relative to this norm and B_{0} = unit ball in B.

Example 2. B = \overline{ℚ_{p}} is NOT a Banach space.

Indeed, B_{0} = O\overline{Z_{p}} → lim_{←}B_{0}/p^{n}B_{0} is NOT an isomorphism \underline{Proof}: Considering elements of the form ∑_{i=1}^{∞} b_{i}p^{i} with deg(b_{i}) → ∞. Theses elements

are in lim_{←}B_{0}/p^{n}B_{0} but one can show that some of thoses elements can NOT have finite degree over Q_{p}.

Let’s give an example of this fact. Consider α := ∑_{k=0}^{∞} p^{\frac{1}{2^{k^{2}}}} .p^{k}.

By definition, α belongs to lim_{←}B_{0}/p^{n}B_{0}. Suppose that α has degree N over ℚ_{p}. Choose n such that \frac{2^{n^{2}}N}{2^{(n+1)^{2}}} < 1 then ∑_{k=n+1}^{∞} p^{\frac{1}{2^{k^{2}}}} .p^{k}

belongs to ℚ_{p}(p^{\frac{1}{2^{n^{2}}}} , x) which is an extension of degree ≤ 2^{n^{2}}.N.

So, v_{p}(∑_{k=n+1}^{∞} p^{\frac{1}{2^{k^{2}}}} .p^{k}).2^{n^{2}}.N ∈ ℤ. But on the other hand:

v_{p}( ∑_{k=n+1}^{∞} p^{\frac{1}{2^{k^{2}}}} .p^{k}).2^{n^{2}}.N = v_{p}(p^{\frac{1}{2^{(n+1)^{2}}}} .p^{n+1)}).2^{n^{2}}.N

= (n + 1).2^{n^{2}}.N + \frac{2^{n^{2}}.N}{2^{(n+1)^{2}}} is not an element of ℤ , a contradiction.□

Example 3. We can complete the first example to get a p−adic Banach space:

B = ℂ_{p} := ˆ \overline{ℚ_{p}}

B_{0} = O_{ℂ_{p}} = Ô\overline{ℚ_{p}}

Example 4. B = C(ℤ_{p}, ℚ_{p}) = {continuous ℚ_{p}-valued functions on ℤ_{p}} B_{0} = C(ℤ_{p}, ℤ_{p}) The property (ii) for a Banach space, follows from the compactness of ℤ_{p}. (f ∈ C(ℤ_{p}, ℚ_{p})

has bounded valuation)

Example 5. B = C(ℤ_{p}, ℂ_{p})

B_{0} = C(ℤ_{p}, O_{ℂ_{p}})

Example 6. Let I be an index set:

B = ℓ_{∞}(I, ℚ_{p}) := {(X_{i})_{i∈I} bounded ||X_{i}|| ≤ C, ∀i ∈ I} B_{0} = ℓ_{∞}(I, ℤ_{p})

Example 7. Let I be an index set:

B = ℓ_{1}(I, ℚ_{p}) := {(X_{i})_{i∈I}, (X_{i}) is ”summable” }

:= {(X_{i})|♯{i ∈ I|v_{p}(X_{i}) ≤ C} < ∞, ∀C > 0}.

B_{0} = ℓ_{1}(I, ℤ_{p}) In particular, if (X_{i}) belongs to ℓ_{1}(I, ℚ_{p}), then ∑_{i∈I} X_{i} ∈ ℚ_{p} makes sense.

Definition 8. A Banach basis of B is a family of elements (e_{i})_{i∈I} in B_{0}⧹pB_{0}, such that ∀x ∈ B there exists a unique (x_{i})_{i∈I} ∈ ℓ_{1}(I, ℚ_{p}) such that:

x = ∑_{i∈I} x_{i}e_{i}

Theorem 9. (e_{i})_{i∈I} is a Banach basis for B if and only if (\overline{e_{i}})_{i∈I} is a 𝔽_{p}-basis for B_{0}/p^{n}B_{0}

Sketch of the proof: (⇒) Let \overline{x} ∈ B_{0}/p^{n}B_{0} and choose x lifting \overline{x}. Then, x = ∑_{i∈I} x_{i}e_{i}. So, \overline{x} = ∑_{i∈I} \overline{x_{i}}.\overline{e_{i}} and

the sum is finite since (x_{i})_{i∈I} belongs to ℓ_{1}(I, ℤ_{p})

(⇐) Let φ: ℓ_{1}(I, ℚ_{p}) → B

(x_{i})_{i∈I} → ∑ x_{i}e_{i}

φ is injective: By multiplying (x_{i})_{i∈I} by v_{p}((x_{i})_{i∈I}) and by reducting modulo p gives: φ(x) = 0 ⇔ v_{p}(x) = ∞. φ is surjective: Let x ∈ B. Without a loss of generality, we can assume that x ∈ B_{0}. By hypothesis,

\overline{x} = ∑_{i} \overline{x_{1,i}} \overline{e_{i}}. Take x^{(0)} := ∑ x_{1,i}e_{i} (Recall that the sum is finite).

x − x^{(0)} ∈ pB_{0}. By recurrence, x = x^{(0)} + p.x^{(1)} + ... + p^{n}.x^{(n)} + ... with x^{(i)} linear combination of e_{i} with

coefficients in ℤ_{p}.

□

Corollary 1. Every Banach space has a Banach basis

\underline{Proof:} Consider B_{0}/pB_{0}. Let \overline{e}_{i∈I} be a basis for B_{0}/pB_{0}. Then, just take e_{i} lift for \overline{e_{i}}.□ Sometimes, we can easily find an explicit expression of the basis:

Example 10. B = ℓ_{1}(I, ℚ_{p}).

A basis is given by (e_{i})_{i∈I} such that e_{i} = δ_{i} (δ_{i}(i) = 1 , δ_{i}(j) = 0 if i ≠ j).

But, sometimes, the explicit expression is not that clear.

Example 11. B = ℓ_{∞}(I, ℚ_{p})

B_{0}/pB_{0} = functions(I, 𝔽_{p}) We know that there exists a basis. Its existence relies on the full strenght of the axiom

of choice but this basis is not countable.

Lemma 1. The functions \binom{x}{n} := \frac{x(x−1)...(x−n+1)}{n!} belong to B_{0}⧹pB_{0}

\underline{Proof}: (i) f_{n}(x) := \binom{x}{n} sends ℤ to ℤ. Since ℤ is dense in ℤ_{p}, it maps ℤ_{p} to ℤ_{p} ⇒ f_{n} ∈ B_{0}. (ii) f_{n}(n) = 1 ⇒ f_{n} ∈ B_{0}⧹pB_{0}.□

Theorem 12. (Mahler)

The functions (f_{n})_{n∈ℕ} are a Banach basis of C(ℤ_{p}, ℚ_{p})

\underline{Proof:} Define the ”discrete derivative” (δf)(x) := f(x + 1) − f(x) and the Mahler coefficients of f by a_{n}(f) := (δ^{n}f)(0)

The proof follows from the lemma

Lemma 2. If f belongs to C(ℤ_{p}, ℚ_{p}) then:

(i)(a_{n}(f))_{n≥0} belongs to ℓ_{1}(I, ℚ_{p}). (ii)f(x) = ∑_{0}^{∞} a_{n}(f)\binom{x}{n} , ∀x ∈ ℤ_{p} (iii)v_{p}(f) = v_{p}((a_{n}(f))_{n∈ℕ} , ∀f ∈ C(ℤ_{p}, ℚ_{p})

\underline{Proof of the lemma}:

(i) Let’s assume without a loss of generality that f ∈ C(ℤ_{p}, ℤ_{p})⧹pC(ℤ_{p}, ℤ_{p}). a_{n}(f) = (δ^{n}f)(0) v_{p}(a_{n}(f)) ≥ v_{p}(δ^{n}f) by definition. The sequence v_{p}(δ^{n}f) is clearly increasing, since δ preserves C(ℤ_{p}, ℤ_{p}) so it is enough to

show that v_{p}(δ^{n}f) is unbounded as n → ∞.

f is continuous on ℤ_{p} which is compact, hence f is uniformly continuous. Hence, for all M, ∃p^{k} satisfying: v_{p}(f(x + p^{k}) − f(x)) > M, ∀x ∈ ℤ_{p}. Let (sf)(x) := f(x + 1), δ = s − 1. (δ^{p^{k}}f) = (s − 1)^{p^{k}}(f) = ∑_{j=0}^{p^{k}}(−1)^{j}\binom{p^{k}}{j}s^{j}.f (δ^{p^{k}}f)(x) = f(x) − f(x + p^{k}) + ∑_{j=1}^{p^{k}−1}(−1)^{j}\binom{p^{k}}{j}f(x + j) which means: v_{p}(δ^{p^{k}}f) ≥ min(M, 1 + v_{p}(f)) So choosing M such that M ≥ 1 + v_{p}(f), we get v_{p}(δ^{p^{k}}f) ≥ 1 + v_{p}(f). Then taking δ^{p^{k}}f instead of f proves by recurrence that the sequence v_{p}(δ^{n}f) is un-

bounded.

(ii) Let f̃:= ∑_{0}^{∞} a_{n}(f)\binom{x}{n}. δ(\binom{x}{n}) = \binom{x+1}{n} − \binom{x}{n} = \frac{(x+1)...(x−n+2)}{n!} − \frac{x...(x−n+1)}{n!} = \frac{(x+1−(x−n+1))x...(x−n+2)}{n!} = \binom{x}{n−1} a_{n}(f̃) = (δ^{n}f̃)(0) = (δ^{n}. ∑_{0}^{∞} a_{n}(f)\binom{x}{n})(0) = a_{n}(f) The assignement f → (a_{n}(f))_{n∈ℕ} is injective. This is because, if δ^{n}(f)(0) = 0, ∀n ∈ ℕ

then f(j) = 0, ∀j ∈ ℕ.

(Indeed, f(0) = 0 so f(1) − f(0) = 0 ⇒ f(1) = 0, so f(2) − 2f(1) + f(0) = 0 ⇒ f(2) = 0 ...)

Hence f = 0 since ℕ is dense in ℤ_{p}.

Finally, by injectivity, f̃ = f.

(iii) Let f ∈ C(ℤ_{p}, ℚ_{p}), we know f(x) = ∑_{0}^{∞} a_{n}(f)\binom{x}{n}

Since v_{p}(δ^{n}f) → ∞, then ∃N such that v_{p}(f) = v_{p}(∑_{n=0}^{N} a_{n}(f)\binom{x}{n}).

But v_{p}(∑_{0}^{N} a_{n}(f)\binom{x}{n}) ≥ min(v_{p}(a_{n}(f)\binom{x}{n})) ≥ min(v_{p}(a_{n}(f)) since \binom{x}{n} (n ≥ 0) belongs to C(ℤ_{p}, ℤ_{p}).

So, v_{p}(f) ≥ v_{p}((a_{n}(f))_{n∈ℕ}).

Reciprocally, if p^{k}f ∈ B_{0}, then ∑_{n=0}^{∞} p^{k}a_{n}(f)\binom{x}{n} ∈ B_{0} and we want to show that v_{p}(p^{k}(a_{n}(f))_{n∈ℕ}) ≥ 0.

Let j = min_{n∈ℕ}{n|v_{p}(p^{k}(a_{n}(f)) < 0}. Then, v_{p}(p^{k}f(j)) ≥ 0 by hypothesis, but on the other hand v_{p}(∑_{0}^{∞} p^{k}a_{n}(f)\binom{j}{n}) = v_{p}(∑_{n=0}^{j−1} p^{k}a_{n}(f)\binom{j}{n} + p^{k}a_{j}(f)) = v_{p}(p^{k}a_{j}(f)) < 0 which is absurd. So {n|v_{p}(p^{k}(a_{n}(f)) < 0} is empty and v_{p}((a_{n}(f))_{n∈N}) ≥ v_{p}(f) which finishes the proof. □

In conclusion, we understand C(ℤ_{p}, ℚ_{p}) → ℓ_{1}(ℕ, ℚ_{p}). Now, we would like to understand its dual.

Dual spaces, measures, and integration:

For all this section ,let B be a Banach space.

Definition 13. A B-valued measure on ℤ_{p} is a continuous linear map from C(ℤ_{p}, ℚ_{p}) to B.

Notations: 1/ D(ℤ_{p}, ℚ_{p}) = space of ℚ_{p}-valued measures on ℤ_{p}.
D(ℤ_{p}, B) = space of B-valued measures on ℤ_{p}.
2/ If μ ∈ D(ℤ_{p}, ℚ_{p}), we write μ(f) = ∫_{ℤ_{p}} f(x)dμ(x)

Concrete description of D(ℤ_{p}, ℚ_{p})

Theorem 14. The map D(ℤ_{p}, ℚ_{p}) → ℓ_{∞}(ℕ, ℚ_{p})

μ → μ(\binom{x}{n})

is an isomorphism of p−adic Banach spaces, which identifies:

D(ℤ_{p}, ℤ_{p}) := {μ|μ(f) ∈ ℤ_{p}, ∀f ∈ C(ℤ_{p}, ℤ_{p})} with ℓ_{∞}(ℕ, ℤ_{p})

.

\underline{Sketch of the proof:} If μ ∈ D(ℤ_{p}, ℚ_{p}), μ(\binom{x}{n}) ∈ ℚ_{p} have to be bounded. (indeed \binom{x}{n} ∈ C(ℤ_{p}, ℤ_{p}), and C(ℤ_{p}, ℤ_{p}) is a compact so μ(C(ℤ_{p}, ℤ_{p}) is compact).

Conversely, given a sequence (b_{n})_{n∈ℕ} which is bounded, we can define:

μ_{b}(f) := ∑_{n∈ℕ}a_{n}(f)b_{n} ∈ ℚ_{p}

. □

Definition 15. Given μ ∈ D(ℤ_{p}, ℚ_{p}), the Amice transform of μ is the power serie;

A_{μ}(T) := ∑_{n=0}^{∞} μ\binom{x}{n}T^{n}

= ∑_{n=0}^{∞} ∫_{ℤ_{p}} \binom{x}{n}dμ(x)T^{n}

= ∫_{ℤ_{p}} ∑_{n=0}^{∞} \binom{x}{n}dμ(x)T^{n}

= ∫_{ℤ_{p}}(1 + T)^{x}dμ(x)

Remark 2. μ → A_{μ}(T) gives an isomorphism D(ℤ_{p}, ℚ_{p}) → ℚ_{p} ⊗ ℤ_{p}[[T]]

Let’s give some examples (or counterexamples) of measures

Example 16. Haar measure

We would like to have a measure invariant by translation (1) and with value 1 for the constant function 1 (2):

It should satisfy μ(x + a) = μ(x) by (1), but μ(x + a) = μ(x) + a by 2/. That’s absurd, so there exits NO Haar measure on ℤ_{p}.

Example 17. Dirac measure Let a ∈ ℤ_{p}. The dirac measure associated to a is defined by the evaluation at a: δ_{a}(f) := f(a) A_{δ_{a}}(T) = ∑_{n=0}^{∞} δ_{a}\binom{x}{n}T^{n} = ∑_{n=0}^{∞} \binom{a}{n}T^{n} = (1 + T)^{a}