MODULARITY OF AN ODD ICOSAHEDRAL REPRESENTATION

A. JEHANNE AND M. MÜLLER

1. Introduction

Consider an irreducible Galois representation

ρ : G_{ℚ} ⟶ GL_{2}(ℂ)

and the corresponding induced projective representation

\overline{ρ} : G_{ℚ} ⟶ PGL_{2}(ℂ),

where we denote by \overline{ℚ} an algebraic closure of ℚ and by G_{ℚ} =Gal(\overline{ℚ}/ℚ) the absolute Galois group over ℚ.

Artin has conjectured (for any nontrival irreducible representation ρ : G_{ℚ} ⟶ GL_{n}(ℂ)), that the analytic continuation of the L-series associated to the representation ρ is an entire function. It is known by results of Hecke, Langlands and Tunnel ([Lan], [Tun]) that ρ satisfies Artin’s conjecture, if Im(\overline{ρ}) is solvable. If Im(\overline{ρ}) is isomorphic to A_{5}, this conjecture is not yet proved in general, but it has been proved for particular cases by Buhler ([Buh]) in 1978, and by Kiming and Wang ([K-W]) in 1994. In 1999, Buzzard, Dickinson, Shepherd-Baron and Taylor proved the conjecture for infinitely many icosahedral representations ([B-D-SB-T]). Moreover, Buzzard and Stein gave another example for representations of conductor 1376.

In all these examples and theorems one has to assume, that the repesentation ρ is odd, that is, det(ρ)(c) = −1, where c denotes the complex conjugation. And for odd representations, Artin’s conjecture is equivalent to the fact that the L-series of this representation coincides with the L-series L(f, s) of a weight one modular form f (see [D-S]).

We give one more example for the Artin conjecture with det(ρ) = χ_{−43} and conductor 5203. With this method one can verify Artin’s conjecture for all odd representation with quadratic determinant and conductor lower than 10000. We prove in fact the modularity of this representation by using the computation of its L-function ([Jeh]) and a basis of the ℂ-vector space S_{2}(5203, 1) of cusp forms of weight 2, conductor 5203 and nebentype 1.

2. Modularity of ρ

We consider the L-function

L(ρ, s) = ∑_{n=1}^{∞} a_{n}n^{−s}

of our irreducible two-dimensional representation ρ. We want to prove that its Mellin transform

f(z) = ∑_{n=1}^{∞} a_{n}q^{n}

(where q = e^{2iπz}) is a newform of weight one and level equal to the conductor 𝔉(ρ) of ρ. We make use of the following proposition (see [Fre1], proposition 1.1).

Proposition 2.1. Assume that g(z) = ∑ b_{n}q^{n} belongs to the space M_{k}(N, χ) of modular forms of level N and nebentype χ, where N in an integer and χ : (ℤ/Nℤ)^{∗} → ℂ^{∗} a Dirichlet character. Let μ = N ∏_{p|N}(1 + \frac{1}{p}) (where p runs through the set of prime numbers dividing N). If b_{n} = 0 for 0 ≤ n ≤ μk/12, then g = 0.

Let N = 𝔉(ρ) be the conductor and let χ be the determinant of ρ. Via class field theory, we can consider χ, as a Dirichlet character (ℤ/Nℤ)^{∗} → ℂ^{∗}.

Proposition 2.2. Let θ be an element of M_{1}(N, χ) and μ = N ∏_{p|N}(1+ \frac{1}{p}). If there are two elements g_{1} and g_{2} in S_{2}(N, χ^{2}) such that

θf ≡ g_{1} mod q^{\frac{μ}{3} +1}

and

f^{2} ≡ g_{2} mod q^{\frac{μ}{3} +1},

then g_{1}/θ belongs to S_{1}(N, χ).

Proof. Since g_{1} belongs to S_{2}(N, χ) and θ to M_{1}(N, χ), it is enough to prove that the meromorphic function g_{1}/θ is holomorphic. Moreover we have g_{1}^{2} ≡ θ^{2}g_{2} mod q^{μ/3}. Furthermore g_{1}^{2} and θ^{2}g_{2} are both elements of S_{4}(N, χ^{4}), so we can apply proposition 2.1. This shows that g_{1}^{2} = θ^{2}g_{2} and therefore that (g_{1}/θ)^{2} is holomorphic. □

Now we consider a projective representation \overline{ρ} given by the polynomial P (X) = X^{5} + 5X^{2} − 6X + 27. It is the example 16 of [B-K]. We consider the lifting ρ of \overline{ρ} with conductor 5203 and determinant χ_{−43} (the character associated to ℚ(\sqrt{−43})) of [Jeh], example 6.

Theorem 2.3. The representation ρ is modular.

Proof. Let Q be the quadratic form Q(x, y) = x^{2} + xy + 11y^{2}; it is known that the associated theta series

Θ_{Q} = ∑_{(x,y)∈ℤ^{2}} q^{Q(x,y)} = 1 + 2 ∑_{n=1}^{∞} (∑_{d|n} χ_{−43}(d)) q^{n}

belongs to M_{1}(43, χ_{−43}). Let f be the Mellin’s transform of L(ρ, s). For N = 5203, the element μ/3 of proposition 2.2 is equal to 1936. We shall prove in section 3 that there are two elements g_{1} and g_{2} such that

Θ_{Q}f ≡ g_{1} mod q^{2000}

and

f^{2} ≡ g_{2} mod q^{2000}.

By proposition 2.2, we conclude that f is congruent to an element g = g_{1}/Θ_{Q} in S_{1}(5203, χ_{−43}) modulo q^{2000}. As the series f has an Euler product, it follows that g mod q^{1000} is an eigenvector for the Hecke operator T_{2} and the associated eigenvalue is equal to i\frac{1−\sqrt{5}}{2} and we obtain that

T_{2}g ≡ i\frac{1 − \sqrt{5}}{2}g mod q^{1000},

and then by proposition 2.1:

T_{2}g = i\frac{1 − \sqrt{5}}{2}g.

It proves that there exists an eigenform h in S_{1}(5203, χ_{−43}) with eigenvalue for T_{2} equal to i\frac{1−\sqrt{5}}{2}. By [D-S], theorem 4, we obtain a Galois representation

ρ_{h} : G_{ℚ} ⟶ GL_{2}(ℂ)

associated to h. This representation has conductor 5203 = 11^{2}.43 and determinant χ_{−43}. The trace i\frac{1−\sqrt{5}}{2} can be obtained neither by a tetrahedral representation (Im(\overline{ρ}_{h}) ≃ A_{4}) nor by an octahedral one (Im(\overline{ρ}_{h}) ≃ S_{4}). Assume that ρ_{h} is a dihedral representation. We know that there is a quadratic field ℚ(\sqrt{d}) and a character φ of the absolute Galois group of ℚ(\sqrt{d}) such that ρ_{h} is the induction of φ. The conductor 𝔉(ρ_{h}) is equal to |Δ_{d}|.N_{ℚ(\sqrt{d})/ℚ}(f_{φ}), where Δ_{d} is the discriminant of ℚ(\sqrt{d})/ℚ and f_{φ} the conductor of φ. Here, d ∈ {−11, −43, 473}.

In ℚ(\sqrt{−43}) and in ℚ(\sqrt{−11}), the prime number 2 is inert and then if d = −43 or −11, we cannot have Tr(Frob_{ρ_{h},2}) = i\frac{1−\sqrt{5}}{2} (where Frob_{ρ_{h},2} is the Frobenius element of ρ_{h} in 2). We denote by 𝔭 the ideal above 11 in the ring of integers of ℚ(\sqrt{473}). By [PARI], we compute that the ray class group of 𝔭 in ℚ(\sqrt{473}) has order 15. Since a matrix with trace i\frac{1−\sqrt{5}}{2} and determinant ±1 cannot have an order dividing 15, we conclude that if d = 473, we cannot have Tr(Fr_{ρ_{h},2}) = i\frac{1−\sqrt{5}}{2}. We have now proved that the representation ρ_{h} is icosahedral.

Let K_{h} be a quintic subfield of \overline{ℚ}^{Ker(ρ_{h})}. The only possibilities for the valuations of the discriminant d_{K_{h}} of K_{h} for prime numbers not dividing 60 are 2 and 4. Since the determinant of ρ_{h} is χ_{−43}, the only possibility for the valuation of d_{K_{h}} in 43 is 2. For the valuation of d_{K_{h}} in 11, we can have the valuations 4 (if the order of ρ̃_{h}(I_{5}) is equal to 5, where I_{5} is the inertia in 5), 2 or 0. We check then in table 1 of [B-K] that the only possibilities are d_{K_{h}} = 11^{2}.43^{2}, and then ρ_{h} is isomorphic to ρ or d_{K_{h}} = 11^{4}.43^{2}, that is too large for this table.

Now, we assume that d_{K_{h}} = 11^{4}.43^{2}. In this case, the order of \overline{ρ}_{h}(I_{5}) (where I_{5} is the inertia in 5) is equal to 5. Since ρ_{h} is modular, all the liftings ρ_{h} ⊗ χ (where χ runs though the Dirichlet characters) are modular. Moreover, there exists such a lifting ρ′ such that the order of ρ′(I_{5}) is 5 (see [Ser1], theorem 5 p. 228). Since ρ′ is modular, we can assume that ρ_{h} = ρ′. We can view ρ_{h} as a representation with image in GL_{2}(\overline{ℤ}), where \overline{ℤ} is the integral closure of ℤ in \overline{ℚ}. We choose a place v in \overline{ℤ} above 11 and we reduce ρ_{h} modulo v. We obtain a representation

ρ̃_{h} : G_{ℚ} ⟶ GL_{2}(\overline{𝔽}_{11}).

Consider the Eisenstein series E_{10} normalized such that its constant term is equal to 1. Then the coefficients of the q–expansion of E_{10} are 11-integers and E_{10} ≡ 1 mod 11. Moreover, the reduction S = ∑ b_{n}q^{n} of hE_{10} modulo v is a cusp form with coefficients in \overline{𝔽}_{11} of weight 11, level 5203 and determinant χ̃_{−43} (the reduction of χ_{−43} modulo v), that is an eigenform for the Hecke operators and such that

Tr(F_{ρ̃_{h},p}) = b_{p} and det(F_{ρ̃_{h},p}) = χ̃_{−43}(p)p^{10}

for all prime number p ∉ {11, 43} (see [D-S] p. 520 and 521). Then, ρ̃_{h} satisfies Serre’s conjecture (3.2.3_{?}) of [Ser2], that is, ρ̃_{h} arises from a cusp form with coefficients in \overline{𝔽}_{11}. The level of this cusp form is equal to 11^{2}.43. In this case, we know that we can “take off” the 11-part of this level. This representation arises from a cusp form with level 43 (see theorem (2.1) p. 643 of [Rib]). Finally, by theorem 4.5 p. 572 of [Edi], we obtain that ρ̃_{h} arises from a cusp form with the weight predicted by Serre. We conclude that ρ̃_{h} satisfies Serre’s conjecture (3.2.4_{?}), that is, with the level and weight predicted by Serre. Then this representation arises from a cusp form with coefficients in \overline{𝔽}_{11} of level 43, character χ̃_{−43} and weight k as in [Ser2], § 2. To compute this weight, we consider the restriction of ρ̃_{h} to the inertia group I_{11}. We denote by ϕ the cyclotomic character. Since ρ̃_{h}|_{I_{11}} has order 5 and has a trivial determinant, then ρ̃_{h}|_{I_{11}} is conjugated to

(\begin{matrix}ϕ^{4} & 0 \\ 0 & ϕ^{6} \end{matrix}) or (\begin{matrix}ϕ^{2} & 0 \\ 0 & ϕ^{8} \end{matrix}) ,

and we obtain k = 51 or 31, using the recipe of [Ser2]. By theorem 3.4 of [Edi], that we recall below (theorem 2.4), we see that we only have to deal with the spaces of cusp forms with coefficients in \overline{𝔽}_{11} of level 43, character χ̃_{−43} and weight 3 (resp. 7). We compute a basis for each of these spaces and check that no eigenform of these spaces corresponds to ρ̃_{h}. □

In order to write theorem 2.4, we have to introduce the derivation θ. If f = ∑ a_{n}q^{n} is a formal series, we define the series θf by setting θf = ∑ na_{n}q^{n}. If f is a modular form with coefficients in \overline{𝔽}_{p}, then θf is such a modular form too. Moreover, the application θ does neither change the level nor the character, but it increases the weight by p + 1 (see [E]).

Theorem 2.4 (Edixhoven). Let f be an eigenform in a space S̃_{k}(N, χ) of modular forms with coefficients in \overline{𝔽}_{p}. Then there exists integers n and k′ with 0 ≤ n ≤ p − 1, k′ ≤ p + 1, and an eigenform g in S̃_{k′}(N, χ) such that f and θ^{n}g have the same eigenvalues for all T̃_{ℓ} (ℓ ≠ p).

3. Existence of g_{1} and g_{2}

In this section we describe very briefly (since we use only standard techniques in computational number theory) how one can show, that there are two cusp forms g_{1}, g_{2} ∈ S_{2}(5203, 1) which satisfies

\begin{aligned}Θ_{Q}f & ≡ & g_{1} mod q^{2000} \\ f^{2} & ≡ & g_{2} mod q^{2000} \end{aligned} (1)

with

\begin{aligned}Θ_{Q}f & = & t_{1} + i.t_{2} + \sqrt{5}.t_{3} + i\sqrt{5}.t_{4} \\ f^{2} & = & f_{1} + i.f_{2} + \sqrt{5}.f_{3} + i\sqrt{5}.f_{4} \end{aligned}

with t_{i} = ∑_{n=1}^{∞} t_{i,n}q^{n}, f_{i} = ∑_{n=1}^{∞} f_{i,n}q^{n} and t_{i,n}, f_{i,n} ∈ ℚ for i ∈ {1, 2, 3, 4} and n ∈ ℕ as in section 2.

The method is as follows: We compute a basis {h_{1}, …, h_{g}} of S_{2}(5203, 1) with

h_{m} = ∑_{n=1}^{∞} h_{m,n}q^{n}, h_{m,n} ∈ ℤ, m = 1, …, g

(It is enough for us to compute the first 2000 coefficients of the q-expansion of the h_{m}). Here g = dim S_{2}(5203, 1) = 473 is the dimension of the space of cusp forms of level 5203. Such a basis exists due to results of [Shi], see also [D-D-T]. To compute the basis we use the modular symbol method (introduced by Birch) and the explicit knowledge of the action of the Hecke algebra on the modular symbols. For a description of this method see for example [Cre] or [Mer].

The rest is linear algebra. After computing such a basis we have to check, that there exists two cuspforms g_{1}, g_{2} which satisfies 1. Let

A = ( \begin{matrix}h_{m,n} \\ 2.f_{1,n} \\ … \\ 2.f_{4,n} \\ 2.t_{1,n} \\ … \\ 2.t_{4,n} \end{matrix} ) ∈ Mat^{481×2000}(ℤ)

be the matrix of by the coefficients of the basis {h_{1}, …, h_{g}} and the coefficients of the t_{i} and the f_{i}. Then one has to show that

rank(A) = dim S_{2}(5203, 1) = 473.

(We multiply f_{i}, t_{i} by 2 to get a q-expansion with integral coefficients for n ≤ 2000.) To compute the rank we reduce the matrix A modulo different prime numbers p and check that rank(A mod p) = 473. Hadamard’s inequality gives us a bound for the number of primes p that guarantee that rank(A) = 473 over ℚ. (We need 380 prime numbers p with p > 2 · 10^{9}. It is clear that these bound are not good, but the computation takes only 20 hours on a PII 450MHz Linux-PC, so we did not try to improve these bound.)

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