Math 726: L-functions and modular forms Fall 2011

Lectures 34 : Values of zeta functions

Instructor: Henri Darmon Notes written by: Clement Gomez

Recall from last lecture

D(ℤ_{p}, ℚ_{p}) = space of ℚ_{p}-valued measures on ℤ_{p}. Definition 1. Given μ ∈ D(ℤ_{p}, ℚ_{p}), the Amice transform of μ is the power series;

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

Write \frac{1}{T} − \frac{a}{(1+T)^{a}−1} for some a ∈ ℤ_{p}^{∗} (for example we can choose a = 2 if p is odd) Lemma 1. \frac{1}{T} − \frac{a}{(1+T)^{a}−1} belongs to ℤ_{p}[[T]]

\underline{Proof:} \frac{1}{T} − \frac{a}{(1+T)^{a}−1} = \frac{1}{T} − _{\frac{a}{∑_{j=1}^{∞} \binom{a}{j}T^{j}}} = \frac{1}{T} (1 − _{\frac{1}{∑_{j=1}^{∞} a^{−1}\binom{a}{j}T^{j−1}}} ∈ \frac{1}{T} (T ℤ_{p}[[T]]) = ℤ_{p}[[T]] Using the isomorphism, D(ℤ_{p}, ℚ_{p}) → ℚ_{p} ⊗ ℤ_{p}[[T]], there is a measure μ_{a} such that:

A_{μ_{a}}(T) = \frac{1}{T} − \frac{a}{(1 + T)^{a} − 1}

This measure is a Zeta measure. (We’ll see after why it is called a Zeta measure)

Moments of a measure:

The nth-moment of a measure μ is : μ(x^{n}) = ∫_{ℤ_{p}} x^{n}dμ(x)

Theorem 3. ∫_{ℤ_{p}} x^{n}dμ_{a}(x) = (−1)^{n}(1 − a^{1+n})ζ(−n) ,∀n ≥ 0, n ∈ ℤ

Remark 2. If h(t) = ∑ λ_{n}(x)t^{n} is a power series with coefficients in C(ℤ_{p}, ℚ_{p}), we can define:

∫_{ℤ_{p}} h(t)dμ(x) = ∑_{n=0}^{∞}(∫_{ℤ_{p}} λ_{n}(x)μ(x))t^{n} ∈ ℚ_{p}[[T]]

Example 4. h(t) = (1 + t)^{x} = 1 + xt + \binom{x}{2}t^{2} + ... + \binom{x}{n}T^{n} + ... then:

∫_{ℤ_{p}} h(t)dμ(x) := ∑_{n=0}^{∞} μ\binom{x}{n}t^{n} = A_{μ}(t) ∈ ℚ_{p} ⊗ ℤ_{p}[[T]]

Example 5. h(t) = e^{tx} = 1 + xt + ... + \frac{x^{n}t^{n}}{n!} + ... then:

∫_{ℤ_{p}} h(t)dμ(x) := ∑_{n=0}^{∞} μ(x^{n})\frac{t^{n}}{n!} ∈ ℚ_{p}[[T]]

\underline{Proof of the theorem}:

∫_{ℤ_{p}} x^{n}dμ_{a}(x) = [(\frac{d}{dt})^{(n)} (∫_{ℤ_{p}} e^{tx}dμ_{a}(x))]_{|_{t=0}}

= [(\frac{d}{dt})^{(n)} (∫_{ℤ_{p}}(1 + T)^{x}dμ_{a}(x))]_{|_{t=0}} where T = e^{t} − 1

= [(\frac{d}{dt})^{(n)} (A_{μ_{a}}(T))]_{|_{t=0}}

= [(\frac{d}{dt})^{(n)} (A_{μ_{a}}(e^{t} − 1))]_{|_{t=0}}

= [(\frac{d}{dt})^{(n)} (\frac{1}{e^{t} − 1} − \frac{a}{e^{at} − 1})]_{|_{t=0}}

= f_{a}^{(n)}(0) where f_{a} = \frac{1}{e^{t} − 1} − \frac{a}{e^{at} − 1}

= (−1)^{n}L(f_{a}, −n) where L(f_{a}, s) := \frac{1}{Γ(s)} ∫_{0}^{∞} f_{a}(t)t^{s} \frac{dt}{t}

But, L(f_{a}, s) = \frac{1}{Γ(s)} ∫_{0}^{∞} [(e^{−t} + e^{−2t} + ...) − a (e^{−at} + e^{−2at} + ...)] t^{s} \frac{dt}{t} = ζ(s) − a^{1−s}ζ(s) by making the changes of variable u = nt and v = ant.

= (1 − a^{1−s})ζ(s)

Finally,

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

□

Corollary 1. If n_{1} ≡ n_{2} (mod (p − 1)p^{k}) and n_{1}, n_{2} ≥ k + 1, then:

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

\underline{Proof:} If n_{1} ≡ n_{2} (mod (p − 1)p^{k}) and n_{1}, n_{2} ≥ k + 1, then v_{p}(x^{n_{1}} − x^{n_{2}}) ≥ k + 1 because: x^{n_{1}} ≡ x^{n_{2}} (mod p^{k+1}) ∀x ∈ ℤ_{p}^{∗} using n_{1} ≡ n_{2} (mod (p − 1)p^{k}) and the fact that

(ℤ/p^{k+1}ℤ)^{∗} has order (p − 1)p^{k} x^{n_{1}} ≡ x^{n_{2}} ≡ 0 (mod p^{k+1}) ∀x ∈ pℤ_{p} using n_{1}, n_{2} ≥ k + 1. Hence, v_{p}(μ_{a}(x^{n_{1}})−μ_{a}(x^{n_{2}})) ≥ k+1 since μ_{a} sends C(ℤ_{p}, ℤ_{p}) to ℤ_{p}. (since \frac{1}{T} − \frac{a}{(1+T)^{a}−1} ∈

ℤ_{p}[[T]]) □.

But following the proof, n → (1−a^{1−n})ζ(−n) does NOT extend to a continuous function on ℤ_{p} × ℤ/pℤ (the problem comes from the extra hypothesis on n_{1} and n_{2}).

Question 6. How can we transform ∫_{ℤ_{p}} x^{n}dμ_{a}(x) in such a way that we could extend ζ to ℤ_{p}?

A related problem is the fact that we cannot define the nth-power of x for all n ∈ ℤ_{p} and all x ∈ ℤ_{p}.

For example, the definition x^{n} := ∑_{k=0}^{∞} \binom{n}{k}(x − 1)^{k} only works when x ∈ 1 + pℤ_{p}. Furthermore, one can give a good definition for x → x^{n} , ∀x ∈ ℤ_{p}^{∗} and ∀n ∈ ℤ_{p} In that sense, x → 1_{ℤ_{p}^{∗}}.x^{n} is a continuous function on ℤ_{p}. (1_{ℤ_{p}^{∗}} is the characteristic

function on ℤ_{p}^{∗} and is CONTINUOUS)

So we can take its μ_{a} measure:

∫_{ℤ_{p}} 1_{ℤ_{p}^{∗}}.x^{n}dμ_{a}(x) := ∫_{ℤ_{p}^{∗}} x^{n}dμ_{a}(x)

GOAL: Relate ∫_{ℤ_{p}^{∗}} x^{n}dμ_{a}(x) to ζ(−n).(simple factor)