Math 726: L-functions and modular forms Fall 2011

Lecture 35 : The operators φ and ψ on measures.

Instructor: Henri Darmon Notes written by: Francesc Castella

Recall from last week that we obtained the integral representation

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

for the values of the Riemann zeta function ζ at the negative integers. Here a was taken to be a p-adic unit, and μ_{a} is the so-called “zeta-measure”, characterised by the property that its Amice transform is given by

A_{μ_{a}}(T) := ∫_{ℤ_{p}}(1 + T)^{x}dμ_{a}(x) = \frac{1}{T} − \frac{a}{(1 + T)^{a} − 1} ∈ ℤ_{p}[[T]].

Now we consider the natural diagonal embedding

ℤ_{≥0} ↪ ℤ/(p − 1)ℤ × ℤ_{p} : n ↦ ([n]_{p−1}, n).  (2)

Lemma 1. Let x be a p-adic integer. If p does not divide x, the function on the non-negative integers

n ↦ x^{n}

extends to a continuous (in fact, analytic) function on ℤ/(p − 1)ℤ × ℤ_{p}.

Proof. We have to show that there exists a continuous function on ℤ/(p − 1)ℤ × ℤ_{p} which, when restricted to ℤ_{≥0} via the embedding (2), it agrees with the function n ↦ x^{n}. Fix an integer n_{o} ∈ {0, 1, …, p − 1}, and let

S_{n_{o}} := {n ∈ ℤ_{≥0} : n ≡ n_{o} (mod p − 1)} ⊂ ℤ_{≥0},

which we view as sitting inside ℤ/(p − 1)ℤ × ℤ_{p} via the natural embedding (2), thus yielding a dense subset of {[n_{o}]_{p−1}} × ℤ_{p}.

Since p ∤ x, x ∈ ℤ_{p} is a p-adic unit, and we can write it uniquely as a product

x = ω(x) · ⟨x⟩

with ω(x) ∈ μ_{p−1}(ℤ_{p}^{×}) (the Teichmüller lift of x modulo p), and ⟨x⟩ := \frac{x}{ω(x)} a 1-unit, say ⟨x⟩ = 1 + pt with t ∈ ℤ_{p}. Now for any n ∈ S_{n_{o}}, we can write

x^{n} = ω(x)^{n} · ⟨x⟩^{n} = ω(x)^{n_{o}} · (1 + pt)^{n}

= ω(x)^{n_{o}} · exp(nρ), (ρ := log(1 + pt) = pt − \frac{p^{2}t^{2}}{2} + ⋯ ∈ pℤ_{p})

= ω(x)^{n_{o}} ∑_{j=0}^{∞} \frac{ρ^{j}}{j!} n^{j},

which is analytic in n. Since p divides ρ, v_{p}(\frac{ρ^{j}}{j!} ) → ∞ as j → ∞, and the result follows.

On the other hand, note that if p divides x, then the function n ↦ x^{n} does not extend to a continuous function of a p-adic variable. Hence the naive approach to defining a continuous function ζ_{p,a}(s) of s ∈ ℤ_{p} analogous to the Riemann zeta function via the tentative definition

ζ_{p,a}(s) =^{??} ∫_{ℤ_{p}} x^{−s}dμ_{a}(x)

(as one may be tempted to set after the integral representation (1)) does not work, since when p|x we can not give a sensible definition of x^{−s} for an arbitrary value of s ∈ ℤ_{p}. Nevertheless, there is a way to amend this, suggested by the previous Lemma. Namely to restrict the domain of integration to ℤ_{p}^{×}, and so to consider the p-adic Mellin transform

ζ_{p,a}(s) := ∫_{ℤ_{p}^{×}} x^{−s}dμ_{a}(x) := ∫_{ℤ_{p}} 1_{ℤ_{p}^{×}}x^{−s}dμ_{a}(x)

where 1_{ℤ_{p}^{×}} denotes the characteristic function of ℤ_{p}^{×}. We will see that this modification to the naive approach does yield a well-behaved function ζ_{p,a}(s) on ℤ_{p}, with the difference that the resulting interpolation of the values ζ(−n) will not be exactly in the form that they appear in (1), but affected by a corresponding value of the Euler factor at p.

Remark 1. One should compare this phenomenon with the need to “complete” the classical Riemann zeta function with the Gamma-factors —or the Euler factor at ∞, if one wishes— in order for the result to admit an integral representation as a Mellin transform, from which the nice analytic properties are deduced.

Since ℤ_{p} = ℤ_{p}^{×} ⊔ pℤ_{p}, we start by the following.

Lemma 2. If μ is a measure on ℤ_{p}, then

A_{1_{pℤp}·μ}(T) = \frac{1}{p} ∑_{ξ∈μ_{p}} A_{μ}(ξ(1 + T) − 1).

Proof. Indeed, for any x ∈ ℤ_{p} we have

1_{pℤ_{p}}(x) = \frac{1}{p} ∑_{ξ∈μ_{p}} ξ^{x}  (3)

(since the expression on the right is 1 or 0 depending on whether x is divisible by p or not), and since we can write

A_{ξ^{x}·μ}(T) = ∫_{ℤ_{p}}(1 + T)^{x}d(ξ^{x} · μ)(x) = ∫_{ℤ_{p}} ξ^{x}(1 + T)^{x}dμ(x)
= ∫_{ℤ_{p}}{1 + [ξ(1 + T) − 1]}^{x}dμ(x)
= A_{μ}(ξ(1 + T) − 1),

the result follows from (3).

Now we introduce two important operators on D(ℤ_{p}, ℂ_{p}), φ and ψ, whose action on a measure μ is given by the formulae

φ : ∫_{ℤ_{p}} f(x)d(φμ)(x) := ∫_{ℤ_{p}} f(px)dμ(x)
ψ : ∫_{ℤ_{p}} f(x)d(ψμ)(x) := ∫_{pℤ_{p}} f(\frac{x}{p})dμ(x) = ∫_{ℤ_{p}} 1_{pℤ_{p}}(x)f(\frac{x}{p})dμ(x).

Lemma 3. The following relations hold:

1. ψφ = id.

2. φψ = res_{pℤ_{p}}. That is, ψ is a left inverse of φ, and φψ is multiplication by the characteristic function 1_{pℤ_{p}}. Proof. It follows immediately from the definitions:

1.

∫_{ℤ_{p}} f(x)d(ψφμ)(x) = ∫_{ℤ_{p}} 1_{pℤ_{p}}(x)f(\frac{x}{p})d(φμ)(x)  (by definition of ψ)
= ∫_{ℤ_{p}} 1_{pℤ_{p}}(px)f(x)dμ(x)  (by definition of φ)
= ∫_{ℤ_{p}} f(x)dμ(x).

∫_{ℤ_{p}} f(x)d(φψμ)(x) = ∫_{ℤ_{p}} f(px)d(ψμ)(x)  (by definition of φ)
= ∫_{pℤ_{p}} f(x)dμ(x)  (by definition of ψ)
= ∫_{ℤ_{p}} 1_{pℤ_{p}}(x)f(x)dμ(x).

The following two lemmas describe the action of the operators φ and ψ on measures in terms of their Amice transform.

Lemma 4.

A_{φμ}(T) = A_{μ}((1 + T)^{p} − 1).

Proof. Indeed,

A_{φμ}(T) = ∫_{ℤ_{p}}(1 + T)^{x}d(φμ)(x) = ∫_{ℤ_{p}}(1 + T)^{px}dμ(x)
= ∫_{ℤ_{p}}{1 + [(1 + T)^{p} − 1]}^{x}dμ(x)
= A_{μ}((1 + T)^{p} − 1).

Lemma 5. Suppose that A_{μ}(T) can be written as

A_{μ}(T) = ∑_{n=0}^{∞} λ_{n}(1 + T)^{n} with λ_{n} ∈ ℂ_{p}.

Then

A_{ψμ}(T) = ∑_{n=0}^{∞} λ_{pn}(1 + T)^{n}.

In particular, if λ_{pn} = λ_{n} for all n ≥ 0, then A_{ψμ} = A_{μ}, and hence ψμ = μ.

Proof. Since φψμ = 1_{pℤ_{p}} · μ by Lemma 3, we have

A_{φψμ}(T) = A_{1_{pℤp}·μ}(T) = \frac{1}{p} ∑_{ξ∈μ_{p}} A_{μ}(ξ(1 + T) − 1)  (by Lemma 2)

= \frac{1}{p} ∑_{ξ∈μ_{p}} ∑_{n=0}^{∞} λ_{n}(ξ^{n}(1 + T)^{n})

= ∑_{n=0}^{∞}(\frac{1}{p} ∑_{ξ∈μ_{p}} ξ^{n})λ_{n}(1 + T)^{n}

= ∑_{p|n} λ_{n}(1 + T)^{n}

= ∑_{n=0}^{∞} λ_{pn}(1 + T)^{pn}.  (4)

On the other hand, by Lemma 4 applied to the measure ψμ:

A_{φψμ}(T) = A_{ψμ}((1 + T)^{p} − 1).  (5)

Comparing (4) and (5), we obtain

A_{ψμ}((1 + T)^{p} − 1) = ∑_{n=0}^{∞} λ_{pn}(1 + T)^{pn},

and hence after the change of variables T′ = (1 + T)^{p} − 1 ⟺ (1 + T)^{p} = 1 + T′:

A_{ψμ}(T′) = ∑_{n=0}^{∞} λ_{pn}(1 + T′)^{n}.  (6)

This shows the first claim. Now if λ_{pn} = λ_{n} for all n ≥ 0, then this last equation becomes (setting T = T′)

A_{ψμ}(T) = ∑_{n=0}^{∞} λ_{n}(1 + T)^{n};

but the expression on the right is just A_{μ}(T) and so A_{ψμ}(T) = A_{μ}(T), which implies ψμ = μ by injectivity of the Amice transform.

Now armed with all this, we go back to the zeta measure μ_{a}.

Corollary 1.

ψμ_{a} = μ_{a}.

Proof. We assume that our a is a positive integer prime to n. Then from the expansion

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

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

= a[(1 + T)^{a} + (1 + T)^{2a} + (1 + T)^{3a} + ⋯] − [(1 + T) + (1 + T)^{2} + ⋯]

we see that the coefficient λ_{n} of (1 + T)^{n} in A_{μ_{a}}(T) is given by

λ_{n} = \{ \begin{matrix}a − 1 & if a|n \\ −1 & otherwise. \end{matrix}  (7)

Since a is prime to n, we have a|n ⟺ a|pn, and hence (7) implies λ_{pn} = λ_{n}, for all n ≥ 0. Thus the fact that ψμ_{a} = μ_{a} follows from the last claim in Lemma 5.

Now we can finally compute the moments of the restriction to ℤ_{p}^{×} of the measure μ_{a}.

Theorem 1. For every n ∈ ℤ_{≥0},

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

Proof. Using ℤ_{p}^{×} = ℤ_{p} ∖ pℤ_{p}, we compute

∫_{ℤ_{p}^{×}} x^{n}dμ_{a}(x) = ∫_{ℤ_{p}} x^{n}dμ_{a}(x) − ∫_{pℤ_{p}} x^{n}dμ_{a}(x)
= ∫_{ℤ_{p}} x^{n}dμ_{a}(x) − ∫_{ℤ_{p}} d(φψμ_{a})(x)  (by Lemma 3)
= ∫_{ℤ_{p}} x^{n}dμ_{a}(x) − ∫_{ℤ_{p}}(px)^{n}d(ψμ_{a})(x)  (by definition of φ)
= ∫_{ℤ_{p}} x^{n}dμ_{a}(x) − p^{n} ∫_{ℤ_{p}} x^{n}dμ_{a}(x)  (by Corollary 1)
= (1 − p^{n}) ∫_{ℤ_{p}} x^{n}dμ_{a}(x).

The result thus follows from the formula (1) for the n-th moments of μ_{a}.