ON THE NON-VANISHING OF L_{f}(s) AT THE

CENTER OF THE CRITICAL STRIP

J.B. Conrey D.W. Farmer

Abstract. We determine some conditions which imply the nonvanishing of L_{f} (k/2) for f a cusp form of weight k for the full modular group. These conditions are verified for weights k ≤ 500.

1. Introduction and statement of results

Let S_{k}(1) be the space of holomorphic cusp forms of weight k for the full modular group. This space has a basis of Hecke eigenforms

B = {f_{1}, …, f_{d}},

satisfying T_{n}f_{j} = a_{j}(n)f_{j} for all n, where T_{n} is the n^{th} Hecke operator. The f_{j}(z) have Fourier expansions

f_{j}(z) = ∑_{n=1}^{∞} a_{j}(n)e(nz)

where e(z) = exp(2πiz).

A Dirichlet series is associated to f_{j}(z) by

L_{f_{j}}(s) = ∑_{n=1}^{∞} \frac{a_{j}(n)}{n^{s}}.

In this paper we study the “critical value” L_{f_{j}}(k/2). Since L_{f_{j}}(s) has functional equation

(2π)^{−s}Γ(s)L_{f_{j}}(s) = (−1)^{k/2}(2π)^{k−s}Γ(k − s)L_{f_{j}}(k − s),

it follows that L_{f_{j}}(k/2) = 0 if k ≡ 2 mod 4. If k ≡ 0 mod 4, then L_{f_{j}}(k/2) = 0 implies that L_{f_{j}}(s) has a zero of order at least 2 at s = k/2. In this paper we make progress toward showing that L_{f_{j}}(k/2) ≠ 0 for k ≡ 0 mod 4.

Our motivation for this work came from a recent conjecture of Duke and Imamoglu [DI] concerning Siegel modular forms. They conjecture a lift of a modular form f of weight 2k − n to a Siegel modular form of degree n and weight k. The lift is a theta series provided L_{f}((2k − n)/2) ≠ 0.

Our main result is

Typeset by AMS-TEX

Research of the first author supported in part by a grant from the NSF.

Theorem 1. Suppose k ≡ 0 mod 4. If, for some n, the characteristic polynomial of T_{n} acting on S_{k}(1) is irreducible, then L_{f_{j}}(k/2) ≠ 0.

We have verified the irreducibility of the characteristic polynomial of T_{2} acting on S_{k}(1) for k ≤ 500, yielding

Theorem 2. L_{f_{j}}(k/2) ≠ 0 for k ≤ 500, k ≡ 0 mod 4.

Questions concerning the characteristic polynomials of Hecke operators have received considerable attention lately. For example, if N is large then the characteristic polynomial of T_{p} acting on S_{2}(Γ_{0}(N)) will have lots of linear factors corresponding to elliptic curves, but one can look for irreducible factors of a large degree (see [K]). At the other extreme, one expects the characteristic polynomial of T_{p} acting on S_{k}(Γ(1)) to be irreducible, so Buzzard [B] began a study of its Galois group. We applied the method of Buzzard to show

Theorem 3. For k ≤ 500, k ≡ 0 mod 4, with d = [k/12] prime, the Galois group of the characteristic polynomial of T_{2} acting on S_{k}(1) is the symmetric group S_{d}.

2. Background and Notation

Our methods involve the Hecke operators T_{n}. These can be defined as linear maps T_{n} : S_{k}(1) → S_{k}(1) given by

T_{n} ( ∑_{m=1}^{∞} a(m)e(mz)) = ∑_{m=1}^{∞} ( ∑_{d|(n,m)} d^{k−1}a (\frac{mn}{d^{2}})) e(mz)

We make extensive use of the fact that S_{k}(1) has a basis of functions with rational integer Fourier coefficients. Set B_{ℚ} = {F_{1}, …, F_{d}} where

\begin{aligned}F_{j} & = Δ^{j}E_{4}^{\frac{k}{4} −3j} \\ & = ∑_{m=j}^{∞} A_{j}(m)e(mz), A_{j}(m) ∈ ℤ, \end{aligned}

say, where Δ ∈ S_{12}(1) is the cusp form of weight 12, and E_{k} ∈ M_{k}(1) is the Eisenstein series of weight k for the full modular group. We denote by T_{n,k}(x) the characteristic polynomial of T_{n} acting on S_{k}(1). The matrix representing T_{n} in the basis B_{ℚ} has integer entries, so T_{n,k}(x) ∈ ℤ[x].

We require some information about the field 𝕂 obtained by adjoining to ℚ all of the Fourier coefficients a_{j}(n) of all of the f_{j}(z). This is the same as the splitting field of all of the T_{n,k}(x).

Lemma 1. 𝕂 = ℚ({a_{j}(n) : 1 ≤ j, n ≤ d}). In particular, 𝕂 is a finite Galois extension of ℚ.

Proof. Since B_{ℚ} is a basis, we can write

f_{j} = ∑_{m=1}^{d} α_{j,m}F_{m}  (2.1)

ON THE NON-VANISHING OF L_{f} (s) AT THE CENTER OF THE CRITICAL STRIP 3

The α_{j,m} can be found by solving the “upper triangular” system

a_{j}(n) = ∑_{m=1}^{d} α_{j,m}A_{m}(n), a ≤ j ≤ d.

Thus, the α_{j,m} are in the field ℚ({a_{j}(n) : 1 ≤ j, n ≤ d}). But then by (2.1) all of the Fourier coefficients of the f_{j} are in that field, so 𝕂 is as described.

The Galois group Gal(𝕂/ℚ) acts on functions with Fourier coefficients in 𝕂 by

σ ( ∑_{m=1}^{∞} a(m)e(mz)) = ( ∑_{m=1}^{∞} σ (a(m))e(mz)) .

Lemma 2. Gal(𝕂/ℚ) acts on B = {f_{1}, …, f_{d}}. The action is transitive if T_{n,k}(x) is irreducible for some n.

Proof. Gal(𝕂/ℚ) acts on B because the elements of Gal(𝕂/ℚ) commutes with each Hecke operator T_{n}, and the f_{j} are the simultaneous eigenfunctions of the T_{n}.

If T_{n,k}(x) is irreducible then Gal(𝕂/ℚ) acts transitively on {a_{j}(n) : 1 ≤ j ≤ d}, which is the set of n^{th} Fourier coefficients of the f_{j}. This forces the action to be transitive on B.

In the next section we relate the transitive action of Gal(𝕂/ℚ) on B to the nonvanishing of L_{f_{j}}(k/2). In the following section we describe our calculations which show that T_{2,k}(x) is irreducible for small k.

3. Proof of Theorem 1

Let

\begin{aligned}f(z) & = ∑_{m=1}^{∞} a(m)e(mz) \\ & = f_{j}(z) \end{aligned}

for some j. Let

σ_{ℓ}(n) = ∑_{d|n} d^{ℓ}

and set

F (s, ℓ) = ∑_{n=1}^{∞} \frac{a(n)σ_{ℓ−1}(n)}{n^{s}}.

Proposition 1. F (k − 1, k/2) = 0 if and only if L_{f}(k/2) = 0.

Proof. By standard methods,

F (s, ℓ) = \frac{L_{f}(s)L_{f}(s + 1 − ℓ)}{ζ(2s + 2 − k − ℓ)}.

In particular,

F (k − 1, k/2) = \frac{L_{f}(k − 1)L_{f}(k/2)}{ζ(k/2)}.

Finally, note that k − 1 is strictly within the plane of absolute convergence for the Euler product for L_{f}(s), so L_{f}(k − 1) ≠ 0.

The next Proposition uses the notation of the Petersson inner product ⟨·, ·⟩ on S_{k}(1).

Proposition 2. ⟨E_{k/2}^{2}, f⟩ = (4π)^{1−k}Γ(k)F (k−1, k/2). In particular, ⟨E_{k/2}^{2}, f⟩ = 0 if and only if L_{f}(k/2) = 0.

Proof. Let

I(s) = ∫_{0}^{∞} ∫_{0}^{1} \overline{f(z)}E_{ℓ}(z)y^{s−1} dx dy

where E_{ℓ}(z) is the holomorphic Eisenstein series of weight ℓ:

E_{ℓ}(z) = ∑_{γ∈Γ/Γ_{∞}} j(γ, z)^{ℓ} = ∑_{m=0}^{∞} σ_{ℓ−1}(m)e(mz)

for a suitable definition of σ_{ℓ−1}(0).

By substituting the Fourier series for E_{ℓ}(z) into the equation for I(s) and integrating term-by-term, we find that

I(s) = (4π)^{−s}Γ(s)F (s, ℓ − 1).

But by the Rankin-Selberg unfolding method, we also have

I(s) = ∫∫_{D} \overline{f(z)}E_{ℓ}(z) ∑_{γ} \overline{j(γ, z)}^{k}j(γ, z)^{ℓ}y(γz)^{s+1}dμ(z),

where D is a fundamental domain for the full modular group. The sum over γ equals y(z)^{k}E_{k−ℓ}(s+1−k, z), which involves the non-holomorphic Eisenstein series of weight k − ℓ. In the notation of the Petersson inner product, we have

(4π)^{−s}Γ(s)F (s, ℓ − 1) = ⟨E_{ℓ}(·)E_{k−ℓ}(s + 1 − k, ·), f(·)⟩.

If we take s = k − 1 here, and use the fact that

E_{k−ℓ}(0, z) = E_{k−ℓ}(z),

then, with ℓ = k/2, we obtain

(4π)^{1−k}Γ(k)F (k − 1, k/2) = ⟨E_{k/2}^{2}, f⟩,

as claimed.

Proof of Theorem 1. A Hecke eigenbasis for the space of modular forms of weight k is given by

{f_{1}, f_{2}, …, f_{d}, E_{k}}

where d = dim(S_{k}(1)). Since E_{k/2}^{2} is a modular form of weight k, it follows that there is a unique way to write

E_{k/2}^{2} = E_{k} + ∑_{j=1}^{d} c_{j}f_{j}  (3.1)

for some coefficients c_{j} ∈ 𝕂. In particular,

⟨E_{k/2}^{2}, f⟩ = c_{1}⟨f, f⟩,

so by Proposition 1, L_{f_{j}}(k/2) = 0 if and only if c_{j} = 0. We will show that c_{j} ≠ 0. Applying σ ∈ Gal(𝕂/ℚ) to (3.1) gives

E_{k/2}^{2} = E_{k} + ∑_{j=1}^{d} σ(c_{j})f\overline{σ},  (3.2)

where \overline{σ} is the permutation of \{1, 2, …, d\} induced by the action of σ on B = \{f_{1}, …, f_{d}\}. Since the expression (3.1) is unique, it follows that σ(c_{j}) = c\overline{σ}.

Suppose Gal(𝕂/ℚ) acts transitively on B, and c_{j}=0 for some j. Then by the previous observation we find that c_{j} = 0 for all j. This implies E_{k/2}^{2} = E_{k}, which is a contradiction if k > 4. So by Lemma 2, if T_{n,k}(x) is irreducible for some n, then c_{j} ≠ 0. This proves Theorem 1.

4. Calculation of the Hecke operators

In this section we describe some calculations involving the Hecke operators T_{n} acting on S_{k}(1). Our goal are to show that the the characteristic polynomial T_{n,k}(x) of T_{n} is irreducible and has Galois group S_{d}, where d = [k/12] = dim(S_{k}(1)).

Direct calculation of the polynomial T_{n,k}(x) is quite difficult for large k because of the huge size of the coefficients (T_{n,k}(x) is a polynomial of degree d, with roots of size d^{6}). To verify that T_{n,k}(x) is irreducible we calculate T_{n,k}(x) mod p for small p and check if the reduced polynomial is irreducible modp. Having established that the polynomial is irreducible, we use the following Lemma of Buzzard [B] to show that it has Galois group S_{d}:

Lemma 3. Suppose f(x) ∈ ℤ[x] is a monic irreducible polynomial of degree d, with d prime. If there is a prime q such that the mod q reduction f̄ ∈ 𝔽_{q}[x] satisfies f̄ = f_{0}f_{1} ⋯ f_{j} for f_{i} ∈ 𝔽_{q}[x] distinct irreducible polynomials with f_{0} having degree 2 and f_{i} having odd degree for i ≥ 1, then the Galois group of f is the full symmetric group S_{d}.

For fixed k, if one T_{n,k}(x) is irreducible then all of them are, and they all have the same Galois group. So we restrict our attention to T_{2,k}(x). The steps in our calculations are:

Step 1. Generate the basis B_{ℚ} = {F_{1}, …, F_{d}} by taking products of the weight 4 Eisenstein series E_{4} and the weight 12 cusp form Δ. We generate and store the first 200 terms in the Fourier expansion.

Step 2. Form the matrix M_{2} = A − xB, where A = (a_{j,l}) with a_{j,l} = A_{j}(2l) + 2^{k−1}A_{j}(l/2) and B = (A_{j}(l))

Note that M_{2} has integer entries, and the eigenvalues of T_{2} are the values of x for which M_{2} is singular. In particular, det(M_{2}) is the characteristic polynomial of T_{2}.

Step 3. Reduce M_{2} mod p, compute the characteristic polynimial of AB^{−1}, and factor that polynomial modp.