Math 726: L-functions and modular forms Fall 2011

Lecture 25 : Applications of Rankin-Selberg

Instructor: Henri Darmon Notes written by: Jason K.C. Polak

Before continuing, recall that we stated the following facts:

Proposition 1. Let f be a normalized newform of weight k, level N, and character χ. Write

f = ∑ a_{n}q^{n}.

Then

1. The field K = ℚ(a_{n} : n ∈ ℕ) is a finite extension of ℚ and each a_{n} is an algebraic integer (i.e. a_{n} ∈ O_{K} for each n).

2. If σ : K_{f} ↪ ℂ is an embedding then σ(f) := ∑_{n=1}^{∞} σ(a_{n})q^{n} is a normalized newform of weight k, level N, and character σ ∘ χ.

3. S_{k}(Γ_{1}(N)) has a ℤ-basis.

We shall now prove Part 2. First we recall that we have already mentioned how to prove Part 1.

Proof. We do the case N = 1, the general case being just a teardrop more technical. First, we claim (exercise) that S_{k}(SL_{2}(ℤ)) is a free 𝕋_{ℂ} module of rank one where

𝕋_{ℂ} = ℂ[T_{n} : n ∈ ℤ],

Let f_{1}, …, f_{m} be normalized eigenforms such that

S_{k}(SL_{2}(ℤ)) ≅ ⊕_{i=1}^{m}ℂf_{i}.

For i = 1, …, m let φ_{i} : 𝕋_{ℚ} → ℂ be the associated homomorphism. In this case ∩ ker φ_{i} = {0} because φ_{i}(𝕋_{ℚ}) = K_{f_{i}} = ℚ(a_{n}(f_{i}) : n ∈ ℕ) so that the image is a field and hence ker φ_{i} is a maximal ideal of 𝕋_{ℚ}. Now choose a subset {i(1), …, i(ℓ)} ⊆ {1, …, m} such that ker φ_{i} ≠ ker φ_{j} whenever i ≠ j. Without loss of generality, we may assume that i(1) = 1 and so we have

∩_{j=1}^{ℓ} ker φ_{φ_{i(j)}},

Then by the Chinese remainder theorem,

𝕋_{ℚ} ≅ 𝕋_{ℚ}/ ∩ ker φ_{i(j)} ≅ ∏_{j} 𝕋_{ℚ}/ ker φ_{i(j)}

∏_{j} ≅ K_{f_{i(j)}}

and the first factor is K_{f}. Now

𝕋_{ℂ} ≅ 𝕋_{ℚ} ⊗_{ℚ} ℂ

≅ (K_{f} ⊗_{ℚ} ℂ) × (∏_{j=2}^{ℓ} K_{f_{i(j)}}) ⊗_{ℚ} ℂ

and K_{f} ⊗_{ℚ} ℂ ≅ ℂ^{r}. The isomorphism is

α ⊗ 1 ↦ (σ_{1}(α), …, σ_{r}(α))

where {σ_{1}, …, σ_{r}} are all the distinct embeddings into ℂ. For each i = 1, …, r, let e_{i} the standard ith basis vector in (K_{f} ⊗_{ℚ} ℂ) × (∏_{j=2}^{ℓ} K_{f_{i(j)}}) ⊗_{ℚ} ℂ. Then for n ∈ ℕ, we have

T_{n}(e_{i}) = σ_{i}(a_{n}(f))e_{i}.

Since S_{k}(SL_{2}(ℤ)) ≅ 𝕋_{ℂ} as 𝕋_{ℂ} modules, e_{i} in 𝕋_{ℂ} corresponds to some nonzero S_{k}(SL_{2}(ℤ)) and so

T_{n}(g) = σ_{1}(a_{n}(f))g

and thus

σ_{i}(f) = \frac{1}{a_{1}(g)}g

for all σ_{i} is a normalized Hecke eigenform.

1 The First Application of Rankin-Selberg

Now we return to Application I:

Theorem 1. For a suitable ℓ, the space M_{k}(SL_{2}(ℤ)) is spanned by the set {T_{n}E_{k−ℓ}E_{ℓ} : n ∈ ℕ}. In fact ℓ ≥ 4 and k − ℓ ≥ 4 will work.

Thus the products E_{k−ℓ}E_{ℓ} are in some sense as far away from eigenforms as possible.

Proof. Recall that M_{k} (SL_{2}(ℤ)) = ℂE_{k} ⊕ S_{k} (SL_{2} (ℤ)) as 𝕋_{ℂ} modules where

E_{k} = c_{k} + q + ∑_{n=2}^{∞} σ_{k−1}(n)q^{n}

where σ_{k−1}(n) ∑_{d|n} d^{k−1}. Also, a quick calculation shows that

E_{k−ℓ}E_{ℓ} = \frac{c_{k−ℓ}c_{ℓ}}{c_{k}} E_{k} + (E_{k−ℓ}E_{ℓ} − \frac{c_{k−ℓ}c_{ℓ}}{c_{k}} E_{k})

for all k and c_{k} ≠ 0. We claim (exercise) that the projection T_{E_{k}} : M_{k}(SL_{2}(ℤ)) → ℂE_{k} is in 𝕋_{ℂ} and so ℂE_{k} ⊆ 𝕋_{ℂ}E_{k−ℓ}E_{ℓ}. Now we want to show that S_{k} (SL_{2} (ℤ)) ⊆ 𝕋_{ℂ}(E_{k−ℓ}E_{ℓ}).

Suppose for the sake of contradiction that the orthogonal complement (𝕋_{ℂ}E_{k−ℓ}E_{ℓ})^{⊥} with respect to the Petersson inner product is nonzero. Hence we can choose a normalized Hecke eigenform in 𝕋_{ℂ}(E_{k−ℓ}E_{ℓ})^{⊥}.

Let f ∈ S_{k}(SL_{2}(ℤ)) and g ∈ M_{ℓ}(SL_{2}(ℤ)) be eigenforms and suppose that k > ℓ+2. Then

D(f, g, k − 1) = L(f ⊗ g, k − 1)ζ(2(k − 1) + 2 − k − ℓ)^{−1}
= ⟨E_{k−ℓ}g, f⟩_{k}.

Take g = E_{ℓ}. Then ⟨E_{k−ℓ}E_{ℓ}, f⟩ = L(f ⊗ g, k − 1)ζ(k − 1)^{−1} ≠ 0 and

L(f ⊗ g, s) = ∏_{p} L_{p}(f ⊗ g, s)

where^{1}

L_{p}(f ⊗ g, s) = [ ∏_{i,j=1}^{2}(1 − α_{p,i}β_{p,j}p^{−s}]^{−1}

and

g = E_{ℓ} = c_{ℓ} + q + ∑_{n=2}^{∞} σ_{ℓ−1}(n)q^{n}

= ∑_{n=0}^{∞} b_{n}q^{n} (b_{n} = σ_{ℓ−1}(n)).

Therefore, because β_{p,1} = 1 and β_{p,2} = p^{ℓ−1},

L_{p}(f ⊗ g, s) = (1 − α_{p,1}p^{−s})(1 − α_{p,2}p^{−s})(1 − α_{p,1}p^{ℓ−1−s})(1 − α_{p,2}p^{ℓ−1−s}),

1Recall that the α_{∗} and β_{∗} are the roots of the polynomials x^{2}−b_{p}x+p^{k−1} and x^{2}−a_{p}x+p^{ℓ−1} respectively.

and so

⟨E_{k−ℓ}E_{ℓ}, f⟩ = L(f ⊗ g, k − 1)ζ(k − ℓ)^{−1}
= L(f, k − 1)L(f, (k − 1) − ℓ + 1)ζ(k − 1)^{−1}.

Suppose k ≥ 12, k−ℓ ≥ k/2+2, and k−1 ≥ k/2+2. Since L(f, s) = ∏_{p}(1−a_{p}p^{−s}+p^{k−1−s})^{−1} for ℜ(s) ≥ k/2 + 2 so L(f, s) is nonzero for ℜ(s) ≥ k/2 + 2. Thus ⟨E_{k−ℓ}E_{ℓ}, f⟩_{k} ≠ 0 whenever k − ℓ ≥ k/2 + 2 and k − 1 ≥ k/2 + 2.