Math 726: L-functions and modular forms Fall 2011

Lecture 26 : Artin Representations Attached to Forms of Weight One

Instructor: Henri Darmon Notes written by: Victoria de Quehen

Rankin-Selberg Method

Recall for a normalized eigenform

f = ∑ a_{n}q^{n} ∈ S_{k}(Γ_{0}(D), ε)

there is an estimate on the size of a_{p} given by

∑_{p prime} |a_{p}|^{2}p^{−s} ≤ log (\frac{1}{s − k} ) .

In particular we were interested in the case when k = 1.

Last week we showed:

Theorem 1. Let k = 1. For all η > 0, there is a finite set Y_{η} ⊂ ℂ such that a_{p} ∈ Y_{η} for all p outside a set of density less than η.

Our goal for this week is to prove:

Theorem 2 (Deligne-Serre). Given f ∈ S_{1}(Γ_{0}(D), ε) (where ε(−1) = −1), there exists ρ_{f} : G_{ℚ} → GL_{2}(ℂ) which is continuous (hence it has a finite image) with the property that

char(ρ_{f}(Frob_{p})) = x^{2} − a_{p}x + ε(p) for all p ∤ D.

Let K_{f} = ℚ(a_{1}, a_{2}, …) be a finite extension of ℚ.

Lemma 1. If the Deligne-Serre representation ρ_{f} exists, then it can be conjugated in such a way that it takes on values in GL_{2}(K_{f}) where (K_{f} ↪ ℂ).

Proof. Let C ∈ G_{ℂ} be a representative of complex conjugation. As the element C is of order two ρ_{f}(C) has eigenvalues 1 and/or −1. We note that det(ρ_{f}) : G_{ℚ} → ℂ^{∗} is ε when viewed as a Galois character. Furthermore, ε is an odd character, and so ε(−1) = −1. Thus we have:

det(ρ_{f})(C) = ε(−1) = −1.

Hence we may assume, by conjugating if necessary, that

ρ_{f}(C) = ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) .

For σ ∈ G_{ℚ}, we will write

ρ_{f}(σ) = ( \begin{matrix}a(σ) & b(σ) \\ c(σ) & d(σ) \end{matrix} )

where a, b, c, d : G_{ℚ} → ℂ such that (a(C), b(C), c(C), d(C)) = (1, 0, 0, −1). Thus

1. \underline{tr(σ) = a(σ) + d(σ) ∈ K_{f}} for all σ ∈ G_{ℚ}.

2. ( \begin{matrix}a(σ) & b(σ) \\ c(σ) & d(σ) \end{matrix} ) ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) = ( \begin{matrix}a(σ) & −b(σ) \\ c(σ) & −d(σ) \end{matrix} ) ∈ G_{ℚ}. Thus \underline{a(σ) − d(σ) ∈ K_{f}} for all σ ∈ G_{ℚ}.

3. The previous two points imply \underline{a(σ), d(σ) ∈ K_{f}} for all σ ∈ G_{ℚ}.

4. If σ, τ ∈ G_{ℚ}, then so is στ. Thus consider

( \begin{matrix}a(σ) & b(σ) \\ c(σ) & d(σ) \end{matrix} ) ( \begin{matrix}a(τ) & b(τ) \\ c(τ) & d(τ) \end{matrix} ) = ( \begin{matrix}a(σ)a(τ)+b(σ)c(τ) & ∗ \\ ∗ & ∗ \end{matrix} )

This implies a(σ)a(τ) + b(σ)c(τ) ∈ K_{f}. By point 3 this proves \underline{b(σ)c(τ) ∈ K_{f}} for all σ, τ ∈ G_{ℚ}.

5. Consider c : G_{ℚ} → ℂ.

\underline{Case 1: c = 0} Then ρ_{f} is reducible, hence by semisimplicity ρ_{f} = χ_{1} ⊕ χ_{2} where χ_{1}, χ_{2} are onedimensional. This proves

ρ_{f} = ( \begin{matrix}χ_{1} & 0 \\ 0 & χ_{2} \end{matrix} ) .

\underline{Case 2: c ≠ 0} There exists σ_{0} such that ρ_{f}(σ_{0}) = ( \begin{matrix}a(σ_{0}) & b(σ_{0}) \\ c(σ_{0}) & d(σ_{0}) \end{matrix} ) for c(σ_{0}) ≠ 0. We can conjugate by a

matrix of the form ( \begin{matrix}λ & 0 \\ 0 & λ^{−1} \end{matrix} ) so that ρ_{f}(σ_{0}) = ( \begin{matrix}a(σ_{0}) & b(σ_{0}) \\ 1 & d(σ_{0}) \end{matrix} ) . Therefore by point 4 we find \underline{b(σ) ∈ K_{f}} for all σ ∈ G_{ℚ}.

6. Consider c : G_{ℚ} → ℂ. We can assume that the function b : G_{ℚ} → ℂ is not identically 0, for if it were then ρ_{f} would be reducible. Choose σ such that b(σ) ≠ 0, then b(σ) ∈ K_{f}^{∗}. As b(σ)c(τ) ∈ K_{f} this implies \underline{c(τ) ∈ K_{f}} for all τ ∈ G_{ℚ}.

Now we will prove the Deligne-Serre Theorem.

Remark 1. If we had a Deligne-Serre representation ρ_{f}, we can conjugate it so that

ρ_{f} : G_{ℚ} → GL_{2}(K_{f})

By a similar argument (which will be part of the assignment) we can conjugate so that

ρ_{f} : G_{ℚ} → GL_{2}(O_{K_{f}}).

For each prime λ ⊲ O_{K_{f}} we would then have the reduced representation

ρ̄_{f,λ} : G_{ℚ} → GL_{2}(O_{K_{f}} /λ)

with char(ρ_{f,λ}(Frob_{ρ})) = x^{2} − a_{p}x + ε(p).

Strategy of Deligne-Serre Step I: Construct ρ_{f,λ} with this property for each λ ⊲ O_{K_{f}}. Step II: Try to lift these to a common representation ρ_{f}.

Theorem 3 (Deligne-Serre mod λ). For each λ ⊲ O_{K_{{}} where λ ∤ 6 there exists a representation

ρ̄_{f,λ} : G_{ℚ} → GL_{2}(O_{K_{{}}/λ)

satisfying char(ρ̄_{f,λ}(Frob_{p})) = x^{2} − a_{p}x + ε(p) for all p ∤ D Norm_{λ} .

Fix a prime λ ⊲ O_{K_{f}} . Let ℓ be the rational prime dividing Norm_{ℚ}^{K_{f}}(λ). Consider the Eisenstein series E_{ℓ−1} of weight ℓ − 1 where ℓ ≥ 5. Its Fourier expansion is given by

E_{ℓ−1}(q) = 1 − \frac{2(ℓ − 1)}{B_{ℓ−1}} ∑_{n=1}^{∞} σ_{ℓ−2}(n)q^{n}.

Proposition 1 (Clausen-Von Staudt). Let l be a prime. Then \frac{B_{k}}{2k} is integral at ℓ if and only if (ℓ − 1) ∤ k. More precisely if (ℓ − 1)|k then

ord_{ℓ} (\frac{B_{k}}{2^{k}} ) = −1 − ord_{ℓ}(k).

Corollary 1. The Eisenstein series E_{ℓ−1}(q) has Fourier coefficients in ℤ_{(ℓ)} (the localization of ℤ at ℓ) and

E_{ℓ−1}(q) ≡ 1 (mod ℓ).

This congruence, as well as the more general congruence

E_{ℓ^{j}(ℓ−1)}(q) ≡ 1 (mod ℓ^{j+1})

tells us that Eisenstein series are well behaved with respect to congruences. This is the basis for the theory of p-adic modular forms and modular forms mod p. Hence we can consider

F_{λ} = fE_{ℓ−1} ∈ S_{ℓ}(Γ_{0}(D), ε).

Properties of F_{λ}

1. The Fourier coefficients of F_{λ} belong to

O_{K,λ} = the localization of O_{K_{f}} at λ.

2. F_{λ} is not an eigenform, but it is an \underline{eigenform mod λ}, i.e.,

T_{N}(F_{λ}) − a_{n}F_{λ} ∈ λO_{K_{f} ,λ}[[q]].