Math 726: L-functions and modular forms Fall 2011

Lecture 23 : Applications of Rankin-Selberg

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

We shall describe the following two applications of Rankin-Selberg:

I For a suitable ℓ, the set {T_{n}E_{k−ℓ}E_{ℓ} : n ∈ ℕ} spans M_{k}(SL_{2}(ℤ)), and

II If f = ∑ a_{n}q^{n} is a newform of weight k, level N, and character χ then ∑ |a_{p}|^{2}p^{−s} converges for ℜ(s) > k, and

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

Before proving II we will first look at the simpler case of ∑_{p} p^{−s}.

Theorem 1. ∑_{p} p^{−s} = log (\frac{1}{s−k}) + O(1).

Proof. Recall

ζ(s) = ∏_{p}(1 − p^{−s})^{−1}, ℜ(s) > 1

and ζ(s) has a simple pole at s = 1. Taking logarithms and using the Taylor expansion for the logarithm gives

log ζ(s) = ∑_{p} − log(1 − p^{−s}) = ∑_{p} ∑_{n=1}^{∞} \frac{p^{−ms}}{m}

= ∑_{i=1}^{∞} g_{m}(s)

where g_{m}(s) = ∑_{p} \frac{p^{−ms}}{m} and the last equality is due to switching the order of summation. Since ζ(s) has a simple pole at s = 1 with a nonzero residue

lim_{s→1^{+}}(s − 1)ζ(s) ≠ 0,

taking logarithms gives

lim_{s→1^{+}}[log(s − 1) + log ζ(s)] = O(1).

Rearranging gives

lim_{s→1^{+}} log ζ(s) = log (\frac{1}{s − 1}) + O(1)

so as s → 1^{+},

g_{1}(s) + ∑_{m=2}^{∞} g_{m}(s) = log (\frac{1}{s − 1}) + O(1).

We claim that the summation ∑_{m=2}^{∞} g_{m}(s) converges for s = 1. Indeed,

∑_{m=2}^{∞} g_{m}(s) = ∑_{p} ∑_{m=2}^{∞} \frac{p^{−ms}}{m}

= ∑_{p} ∑_{m′=2}^{∞} (\frac{p^{−2m′s}}{2m} + \frac{p^{−2(m′+1)s}}{2m′ + 1} )

≤ ∑_{p} ∑_{m′=1}^{∞} \frac{p^{−2m′s}}{m′}

= log ζ(2s)

where the inequality comes from the inequality

\frac{p^{−2m′s}}{2m} + \frac{p^{−2(m′+1)s}}{2m′ + 1} ≤ 2p^{−2m′s}2m′.

Now we prove the more general theorem. We restate it for convenience.

Theorem 2. If f = ∑ a_{n}q^{n} is a newform of weight k, level N, and character χ then ∑_{p∤N} |a_{p}|^{2}p^{−s} converges for ℜ(s) > k, and

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

Note that in this Lecture, we show the convergence, and in Lecture 24 we will show the estimate.

Proof. We have already seen that

|a_{n}| ≤ cn^{k/2}

for some real c > 0. Thus

∑_{m=1}^{∞} |a_{n}|n^{−s}

converges for ℜ(s) > k/2 + 1. Let

L_{N}(f ⊗ \overline{f}, s) = ∏_{p∤N}L_{p}(f ⊗ \overline{f}, s)

where

L_{p}(f ⊗ \overline{f}, s) = ∏_{i,j=1}^{2}(1 − α_{p,i}\overline{α_{p,j}}p^{−s})^{−1}.

Here the α_{i,j} are the roots of x^{2} = a_{p}x + χ(p)p^{k−1} and \overline{f} = ∑ \overline{a_{n}}q^{n} ∈ S_{k}(N, \overline{χ}). Define

D_{N}(f, \overline{f}, s) = ∑_{(n,N)=1} \frac{|a_{n}|^{2}}{n^{s}} .

Then L_{N}(f ⊗ \overline{f}, s) = D_{N}(f, \overline{f}, s)ζ_{N}(2s − 2 − 2k) where

ζ_{N}(s) = ∏_{p∤N}(1 − p^{−s})^{−1} = ∏_{p} (∑_{n=0}^{∞} \frac{|a_{p^{r}}|^{2}}{p^{rs}} )

is the partial ζ function. Now let

D(f, \overline{f}, s) = ∑_{n∈ℕ} \frac{|a_{n}|^{2}}{n^{s}} .

Recall that for normalized Hecke eigenforms f, g ∈ S_{k}(SL_{2}(ℤ)),

L(f ⊗ g, s) = D(f, g, s)ζ(2s − 2 − 2k)

extends to a meromorphic function and has a unique simple pole at s = k if and only if ⟨\overline{f}, g⟩ ≠ 0. In particular, this means that D(f, \overline{f}, s)ζ(2s + 2 − 2k) extends to a meromorphic function with a unique simple pole at s = k because ⟨\overline{f}, \overline{f}⟩ ≠ 0.

Note that

H(k) = ∏_{p|N}(1 − p^{−2}) ∏_{p|N}(1 − |a_{p}|^{2}p^{−k}) ≠ 0

by our bound |a_{p}| < k/2. Since L_{N}(f ⊗ \overline{f}, s) has a simple pole, D(f, \overline{f}, s) extends to a meromorphic function holomorphic for ℜ(s) > k + 1, and hence via the Lemma below, a holomorphic function for ℜ(s) > k by looking at the product L_{N}(f ⊗ \overline{f}), s).

Lemma 1. Let Φ be a meromorphic function satisfying Φ(s) = ∑_{n=1}^{∞} \frac{a_{n}}{n^{s}} for ℜ(s) > σ ∈ ℝ for some fixed σ, c_{n} ≥ 0. If Φ is holomorphic at s = σ then there exists a δ > 0 such that Φ(s) = ∑_{n=1}^{∞} \frac{a_{n}}{n^{s}} for ℜ(s) > σ − δ.

Although we will not prove this lemma, we shall use this lemma to prove that

D(f, \overline{f}, s) = ∑_{n=1}^{∞} \frac{|a_{n}|^{2}}{n^{s}}

converges absolutely for ℜ(s) > k, where f = ∑ a_{n}q^{n} is our newform of weight k. Indeed, suppose that ∑ \frac{|a_{n}|^{2}}{n^{s}} diverges at s = σ for some σ ∈ ℝ with k < σ ≤ k + 1. Let

r_{0} = sup \{r ∈ ℝ : k < r ≤ k + 1, ∑ \frac{|a_{n}|^{2}}{n^{r}} diverges\} .

By definition of the supremum ∑ \frac{|a_{n}|^{2}}{n^{s}} converges for ℜ(s) > r_{0}. We already know that D(f, \overline{f}, s) is holomorphic at s = r_{0} because it is holomorphic except for a simple pole at s = k and by hypothesis r_{0} > k; and so by Lemma 1, there is a δ > 0 such that ∑ \frac{|a_{n}|^{2}}{n^{r}} converges for ℜ(s) > r_{0} − δ, and this is a contradiction!

Note, the Weil conjectures imply that |a_{p}| ≤ 2p^{\frac{k−1}{2}} for k ≥ 2.