Modularity of the Consani-Scholten Quintic

with an Appendix by José Burgos Gil and Ariel Pacetti

Luis Dieulefait ^{1}, Ariel Pacetti^{2} and Matthias Schütt^{3}^{,} ^{4}

Received: June 22, 2010 Revised: May 23, 2012

Communicated by Don Blasius

Abstract. We prove that the Consani-Scholten quintic, a Calabi-Yau threefold over ℚ, is Hilbert modular. For this, we refine several techniques known from the context of modular forms. Most notably, we extend the Faltings-Serre-Livné method to induced fourdimensional Galois representations over ℚ. We also need a Sturm bound for Hilbert modular forms; this is developed in an appendix by José Burgos Gil and the second author.

2000 Mathematics Subject Classification: Primary: 11F41; Secondary: 11F80, 11G40, 14G10, 14J32 Keywords and Phrases: Consani-Scholten quintic, Hilbert modular form, Faltings–Serre–Livné method, Sturm bound

1 Introduction

The modularity conjecture for Calabi-Yau threefolds defined over ℚ is a particular instance of the Langlands correspondence. Given a Calabi-Yau threefold X over ℚ we consider the compatible family of Galois representations ρ_{ℓ} of dimension, say, n, giving the action of G_{ℚ} = Gal(ℚ̄/ℚ) on H^{3}(Xℚ̄, ℚ_{ℓ}): the conjecture says that there should exist an automorphic form π of GL_{n} such that ℓ-adic Galois representations attached to π are isomorphic to the representations ρ_{ℓ}. This implies that the L-functions of π and ρ_{ℓ} agree, at least up to finitely many local factors. Observe that (according to Langlands functoriality) π should be cuspidal if and only if the representations ρ_{ℓ} are absolutely irreducible. The only case where the conjecture is known in general (among Calabi-Yau threefolds) is the rigid case, i.e., the case with n = 2. In this case modularity was established by the first author and Manoharmayum (cf. [DM03]) under some mild local conditions. These local assumptions are no longer required since it is known that modularity of all rigid Calabi-Yau threefolds defined over ℚ follows from Serre’s conjecture and the later has recently been proved (cf. [KW09b], [KW09a]). It was observed by Hulek and Verrill in [HV06] that the modularity result in [DM03] can be extended to show modularity of those Calabi-Yau threefolds such that the representations ρ_{ℓ} are (for every ℓ) reducible and have 2-dimensional irreducible components. In fact, using Serre’s conjecture, one can show that this is true even if reducibility occurs only after extending scalars, assuming that reducibility is a uniform property, i.e., independent of ℓ (this uniformity follows for instance from Tate’s conjecture).

1LD partially supported by MICINN grants MTM2009-07024 and MTM2012-33830 and by an ICREA Academia Research Prize.

2AP partially supported by CONICET PIP 2010-2012 and FonCyT BID-PICT 2010-0681.

3MS partially supported by DFG under grant Schu 2266/2-2 and by ERC through StG 279723 (SURFARI).

4JBG partially supported by grants MTM2009-14163-c02-01 and CSIC-2009501001.

Documenta Mathematica 17 (2012) 953–987

In this paper, we will prove modularity for a non-rigid Calabi-Yau threefold over ℚ such that the representations ρ_{ℓ} are absolutely irreducible. To our knowledge such an example has not been known before. We sketch the basic set-up: In [CS01], Consani and Scholten consider a quintic threefold X̃ which we will review in section 3. It has good reduction outside the set {2, 3, 5} and Hodge numbers:

h^{3,0} = 1 = h^{2,1}, h^{2,0} = 0 = h^{1,0} and h^{1,1} = 141.

In particular the third étale cohomology is four dimensional. If we fix a prime ℓ, the action of the Galois group Gal(ℚ̄/ℚ) on the third étale cohomology gives a 4-dimensional representation

ρ_{ℓ} : Gal(ℚ̄/ℚ) → GL(H^{3}(X̃ℚ̄, ℚ_{ℓ})) ≃ GL_{4}(ℚ_{ℓ}).

Let F = ℚ[\sqrt{5}] and 𝒪_{F} its ring of integers. In [CS01] it is shown that the restriction

ρ_{ℓ}|_{Gal(ℚ̄/F)} : Gal(ℚ̄/F) → Aut (H^{3}(X̃ℚ̄, ℚ_{ℓ}) ⊗ \overline{ℚ_{ℓ}}) ,

is the direct sum of two 2-dimensional representations (see Theorem 3.2 of [CS01]). More precisely, if λ is a prime of 𝒪_{F} over ℓ, then there exists a 2-dimensional representation

σ_{λ} : Gal(ℚ̄/F) → GL_{2}(𝒪_{λ}),

such that ρ_{ℓ}|_{Gal(ℚ̄/F)} is a direct sum of σ_{λ} and σ′_{λ} (the external conjugate of σ_{λ}) and ρ_{ℓ} = Ind_{F}^{ℚ} σ_{λ}.

In the same work, an holomorphic Hilbert newform 𝔣 on F of weight (2, 4) and level 𝔠_{𝔣} = (30) is constructed, whose L-series is conjectured to agree with that of σ. The aim of this work is to prove this modularity result. Let λ be a prime of F over a rational prime ℓ. Let 𝒪_{λ} denote the completion at λ of 𝒪_{F}. Since 𝔣 has F -rational eigenvalues, by the work of Taylor (see [Tay89]), there exists a two-dimensional continuous λ-adic Galois representation

σ_{𝔣,λ} : Gal(ℚ̄/F) → GL_{2}(𝒪_{λ}),

with the following properties: σ_{𝔣,λ} is unramified outside ℓ𝔠, and if 𝔭 is a prime of F not dividing ℓ𝔠, then

Tr σ_{𝔣,λ}(Frob 𝔭) = θ(T_{𝔭}),
det σ_{𝔣,λ}(Frob 𝔭) = θ(S_{𝔭})𝒩 𝔭.

Here T_{𝔭} denotes the 𝔭-th Hecke operator, S_{𝔭} denotes the diamond operator (given by the action of the matrix ( \begin{matrix}α & 0 \\ 0 & α \end{matrix}), for α = ∏_{𝔭} π_{𝔮}^{v_{𝔮}(𝔭)} and π_{𝔮} a local uniformizer) and θ(T) is the eigenvalue of the Hecke operator T on 𝔣. Let τ(𝔣) be the Hilbert modular form which is the external Galois conjugate of 𝔣, where τ is the order two element in Gal(F/ℚ). Observe that the λ-adic Galois representations attached to τ(𝔣) are obtained by applying τ to the traces of the images of Frobenius in the ones attached to 𝔣. Our result can be stated as:

Theorem 1.1. The representations σ_{λ} and σ_{ν(𝔣),λ} are isomorphic, where ν is either the identity or τ.

In particular the theorem implies that the L-series of σ_{λ} and σ_{ν(𝔣),λ} agree. This solves the conjecture from [CS01].

Remark 1.2. By known cases of automorphic base change (theta lift) and functoriality, Theorem 1.1 is known to imply that ρ_{ℓ} corresponds to a Siegel modular form of genus 2 and to a cuspidal automorphic form of GL_{4}.

Remark 1.3. Dimitrov has proved a Modularity Lifting Theorem that applies to Hilbert modular forms of non-parallel weight (cf. [Dim09]), and he and the first author checked that for ℓ = 7 the representation σ satisfies all the technical conditions of this theorem (cf. [DD06]). Thus it would be enough to prove residual modularity modulo 7 to deduce the modularity of σ from this result. We will follow, however, a different path.

We give an outline of the proof of Theorem 1.1. Since both Galois representations come in compatible families, it is enough to prove that they are isomorphic for a specific choice of primes λ over ℓ. We choose ℓ = 2 so as to apply a Faltings-Serre method of proving that two given 2-adic Galois representations are isomorphic (cf. [Liv87]). Actually, in [CS01] it is proven that σ_{λ} exists, but its trace at a prime is only determined up to conjugation by τ. Hence in Theorem 4.3 we give a version of the Faltings-Serre method that applies to reducible 4-dimensional representations of Gal(ℚ̄/F).

Theorem 4.3 implies that the 2-dimensional Galois representations σ_{λ}, σ_{ν(𝔣),λ} have isomorphic semisimplifications. Since 𝔣 is a cuspidal Hilbert eigenform, its λ-adic Galois representations is irreducible for all primes λ and Livné’s Theorem asserts in particular that the same is true for the representations attached to X̃. Thus we deduce Theorem 1.1. In the present situation, the problem is non-trivial notably because 2 is inert in 𝒪_{F}. Hence the residual representations lie a priori in GL_{2}(𝔽_{4}). Actually, they lie in SL_{2}(𝔽_{4}). This is clear for both representations: the representation σ_{𝔣,λ} has trivial nebentypus so the determinant image lies in 𝔽_{2}^{×} while the representation σ_{λ} at a prime ideal 𝔭 has real determinant of absolute value 𝒩 𝔭^{3}. The group SL_{2}(𝔽_{4}) is not a solvable group (it is in fact isomorphic to A_{5}, the alternating group in 5 elements). We will overcome this subtlety by showing that the images of the residual representations are 2-groups. For σ_{𝔣,2}, this will be achieved in section 2 by combining three techniques: the theory of congruences between Hilbert cuspforms, the explicit approach from [DGP10] and a Sturm bound for Hilbert modular forms that is developed by Burgos and the second author in the Appendix B. For σ_{2}, we will use the Lefschetz trace formula and automorphisms on the Calabi-Yau threefold X̃ (section 3). We collect the necessary data for a proof of Theorem 1.1 in section 4.

Acknowledgements: We would like to thank Lassina Démbelé for many suggestions concerning computing with Hilbert modular forms. Also we would like to thank José Burgos Gil for his contribution to the appendix with the proof of a Sturm bound. The computations of the a_{𝔭} where done using the Pari/GP system [PAR08]. We would like to thank Bill Allombert for implementing a routine in PARI that was not included in the original software for dealing with elements of small norm under a positive definite quadratic form. Special thanks are due to the referee for his comments and suggestions.

2 Computing the residual image of the Galois representation σ_{𝔣,2}

This section deals with the holomorphic Hilbert newform 𝔣 on F = ℚ(\sqrt{5}) of weight (2, 4) and conductor 𝔠_{𝔣} = (30) constructed in [CS01]. The aim of this section is to prove that the image of the residual 2-adic Galois representation σ̄_{𝔣,2} attached to 𝔣 has image a 2-group. This will enable us to apply methods for even trace Galois representations to σ_{𝔣,2}.

2.1 Properties of the Hilbert modular form 𝔣.

Let us recall some definitions of Hilbert modular forms. For 𝔠 ⊂ 𝒪_{F}, let Γ_{0}(𝔠) be the subgroup of SL_{2}(𝒪_{F}) whose second row and first column entry is divisible by 𝔠. Let ℌ denote the Poincare upper half plane. Then given k⃗ = (k_{1}, k_{2}), with k_{1} ≡ k_{2} (mod 2), a weight k⃗ Hilbert modular form of level 𝔠 is an holomorphic function f : ℌ^{2} ↦ ℂ such that for γ = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ_{0}(𝔠),

f(γ · z_{1}, τ(γ) · z_{2}) = f(z_{1}, z_{2})(cz_{1} + d)^{k_{1}}(τ(c)z_{2} + τ(d))^{k_{2}},

with the usual holomorphicity condition at the cusps. We denote such space Mk⃗(𝔠), and Sk⃗(𝔠) the subspace of cuspidal forms. We start studying some properties of the form 𝔣. In sections 6 and 7 of [CS01] such form was constructed using Eichler’s method on definite quaternion algebras and in Theorem 8.3 of loc. cit. it is proved that its coefficient field is exactly ℚ(\sqrt{5}). Since the primes 2 and 3 divide the level of the form 𝔣 to the first power, the automorphic representation attached to 𝔣 is Steinberg at both primes. To know its behaviour at the prime \sqrt{5}, we consider its twist by a suitable character.

Lemma 2.1. There exists a unique non-trivial quadratic Hecke character χ\sqrt{5} of 𝒪_{F} (of infinity type (sign, sign)), whose conductor is \sqrt{5}. The quadratic twist of 𝔣 by χ\sqrt{5} corresponds to a Hilbert newform of level 6\sqrt{5} and weight (2, 4) which we denote by 𝔣 ⊗ χ\sqrt{5}.

Proof. For any prime ideal 𝔭 of 𝒪_{F}, whose residue field has prime order p, we can consider the quotient map

𝒪_{F} ↠ 𝒪_{F} /𝔭 ≃ ℤ/pℤ,

to get a Dirichlet character χ_{𝔭} in 𝒪_{F}. To get the Hecke character we just need the infinity characters, but note that the infinite type (sign^{ϵ_{1}}, sign^{ϵ_{2}}) is uniquely defined by the conditions

χ_{𝔭}(−1) = (−1)^{ϵ_{1}+ϵ_{2}},
χ_{𝔭} (\frac{1 + \sqrt{5}}{2}) = (−1)^{ϵ_{2}}.

The uniqueness comes from the fact that the fundamental unit is not totally positive (which is equivalent to say that the class number and the narrow class number are the same). In our case, χ\sqrt{5} (\frac{1+\sqrt{5}}{2}) = −1, and χ\sqrt{5}(−1) = 1 so the first assertion follows. For the second statement of the Lemma, it is clear (as in the classical case) that the twist will have trivial character and level at most 30 (see for example [Shi78, Proposition 4.4]), so the proof goes by elimination. The data used was supplied by Lassina Démbelé, but is now available in the new versions of MAGMA for example. Here is a resume:

• The space S_{(2,4)}(30) is of dimension 74.

• Its subspace of newforms is of dimension 22.

• There are 16 eigenforms in the new subspace whose coefficient field is F and 2 eigenforms whose coefficient field has degree 3 over it.

• The space S_{(2,4)}(6\sqrt{5}) is of dimension 14.

• Its subspace of newforms is of dimension 4.

• There are 4 eigenforms in the new subspace whose coefficient field is F.

By computing the Hecke eigenvalue at the primes above 11 in the newspace of level 30, we find a unique form matching the eigenvalue of 𝔣 (in version 2.16 of Magma, the form is the first one to appear), so the form 𝔣 ⊗ χ\sqrt{5} does not lie in the newspace of level 30. Since the primes of norm 11 are generated by \frac{7+\sqrt{5}}{2}

and \frac{7−\sqrt{5}}{2}, and χ\sqrt{5} (\frac{7+\sqrt{5}}{2}) = 1, we search for a form with the same eigenvalue

in level 6\sqrt{5} and find a unique one (the third one), but none in level 6, hence 𝔣 ⊗ χ\sqrt{5} has level 6\sqrt{5}.

This implies the following.

Corollary 2.2. The form 𝔣 is also Steinberg at the prime \sqrt{5}.

Since the form 𝔣 and 𝔣 ⊗ \sqrt{5} are congruent modulo 2, we can work with the latter form which has smaller level.

2.2 Properties of the residual image of σ_{𝔣,2}

For reasons that will become clear later, we consider not the form 𝔣 but a form congruent to it modulo 2 and of parallel weight 2.

Lemma 2.3. If the residual image of σ_{𝔣,2} is irreducible, then there exists a Hilbert newform 𝔤 of level 12\sqrt{5} and parallel weight 2 which is congruent to 𝔣 modulo 2.

Proof. This follows from results on the determination of Serre’s weights, for example from [BDJ10], Corollary 2.12. In fact, item (2) of this corollary gives the sought for congruence, except for the fact that the value of the maximal power of 2 in the level of the Hilbert modular form of parallel weight 2, which is 4, follows from a close inspection of the proof of this corollary but is not given in the statement of item (2) (compare also with [BDJ10], Prop. 4.13 (a)). Since the value of the power of 2 is not specified in item (2), let us prove that this value is correct using item (3) of Corollary 2.12. The reader is warned that in doing this we are following a rather indirect path, because we are twisting by a character and then twisting back by its inverse, so from the conceptual point of view this is very unnatural: we just follow this path in order to show that the result we want follows immediately from a case covered by Corollary 2.12 as it is, but a direct and natural proof could be obtained by a detailed explanation of how this power of 2 in the level can be determined DIRECTLY using the techniques in [BDJ10]. This being said, we apply Corollary 2.12 (3) in loc. cit. and obtain a Hilbert modular form 𝔤′ of parallel weight 2 congruent modulo 2 to a twist by some character μ of 𝔣⊗χ\sqrt{5} such that the level at 2 of 𝔤′ is 2. The level at 3 and at \sqrt{5} of this form are the same as those of 𝔣⊗χ\sqrt{5}, but it can have some extra primes in the level, due to the fact that the global character μ whose local behaviour at 2 is prescribed in such a way that it changes the weight as desired, is only known to exist after allowing ramification at some auxiliary primes. From this congruence, we simply twist both sides by the inverse of μ and obtain what we wanted, the sought for form of parallel weight 2 is 𝔤 = 𝔤′ ⊗ μ^{−1}, which is easily seen to satisfy all the required properties: Notably it is congruent modulo 2 to 𝔣⊗χ\sqrt{5}, and its level only contains the primes 2, 3 and \sqrt{5} because the auxiliary primes introduced to the level while twisting by μ are removed after twisting by its inverse. In order to bound the level at 2 of 𝔤 recall that the level of 𝔤′ at 2 is 2 and apply the well-known formula for the behaviour of levels under twists. As final ingredient observe that the conductor at 2 of μ is just 2 because the order of this character is odd (it takes values on the multiplicative group of a finite field of characteristic 2) and 2 is inert in the quadratic field ℚ(\sqrt{5}) (thus a character of conductor 2 has order 3 while a character whose conductor at 2 is at least 4 has even order).

For proving that the image of the residual 2-adic Galois representation attached to 𝔤 has image a 2-group, eventually we will pursue a similar approach as in [DGP10]. We start by computing all subgroups of A_{5}.

Lemma 2.4. Any proper subgroup of A_{5} is isomorphic to one of the following:

{{1}, C_{2}, C_{3}, C_{2} × C_{2}, C_{5}, S_{3}, D_{5}, A_{4}}.

There is an easy classification of the orders of the elements in SL_{2}(𝔽_{4}) in terms of the traces.

Lemma 2.5. If M ∈ SL_{2}(𝔽_{4}), then

ord(M) = \begin{cases}1 & if M = id, \\ 2 & if Tr(M) = 0 and M ≠ id, \\ 3 & if Tr(M) = 1, \\ 5 & if Tr(M) ∉ 𝔽_{2}. \end{cases}

Recall how to derive a Fourier expansion at ∞ for a Hilbert modular form over F. Let τ denote the generator of Gal(F/ℚ). An element ν ∈ F is called totally positive if both ν > 0 and τ(ν) > 0. We denote this by ν ≫ 0. Since F has strict class number one, any Hilbert modular form G over F has a q-expansion

G(z_{1}, z_{2}) = ∑_{\begin{matrix}ξ∈ \frac{𝒪_{F}}{\sqrt{5}} \\ ξ≫0 \end{matrix}} exp(ξz_{1} + τ(ξ)z_{2}),  (1)

where exp(z) = e^{2πiz}. Recall that Aut(ℂ) acts on the space of Hilbert modular forms, just by acting on Fourier expansions. The following result is due to Shimura (see Proposition 1.2 of [Shi78]).

Proposition 2.6. Let σ ∈ Aut(ℂ), then σ(Mk⃗) = Ml⃗, where l⃗ = k⃗^{σ}.

In particular, for parallel weight k⃗, we have σ(Mk⃗) = Mk⃗ for all σ ∈ Aut(ℂ). This applies to the Hilbert modular form 𝔤 and thus indirectly to 𝔣:

Proposition 2.7. The residual image of σ_{𝔣,2} is not D_{5} nor A_{5}.

Proof. Suppose the residual image of σ_{𝔣,2} is D_{5} or A_{5}. In particular it is absolutely irreducible and Lemma 2.3 implies the existence of a parallel weight 2 form 𝔤 whose residual image (at 2) equals that of 𝔣. Let L be denote the coefficient field of 𝔤 and M its Galois closure. Due to the assumption on the coefficients of 𝔣 and the definition of 𝔤, we know that the residue field of (the ring of integers of) L modulo some prime ideal ˆ2 dividing 2 equals 𝔽_{4}. The groups D_{5} and A_{5} both have order 5 elements, so by Lemma 2.5 there exists a prime 𝔭_{0} such that the 𝔭_{0}-th Hecke eigenvalue a_{𝔭_{0}} of 𝔤 lies in 𝔽_{4}, but not in 𝔽_{2}. Consider the form 𝔤 + μ(𝔤), where μ is an element of Gal(M/ℚ) lying in the decomposition group of a prime above ˆ2 and such that it lifts a generator of Gal(𝔽_{4}/𝔽_{2}). It is a parallel weight 2 Hilbert modular form (by Proposition 2.6), whose 𝔭_{0}-th Fourier coefficient is non-zero modulo 2. We computed all Hecke eigenvalues with ideals generated by an element of trace smaller than 395 of 𝔣 and checked that they all lie in 2𝒪_{F} (a table for such eigenvalues can be found at [DPS]). In particular, all the Fourier coefficients of the Hilbert modular form 𝔤 + μ(𝔤) with trace smaller than 395 are zero modulo the ideal ˆ2, so the Sturm bound (Theorem B.21) implies that all Fourier coefficients of 𝔤 + μ(𝔤) are zero modulo ˆ2, which contradicts the fact that its 𝔭_{0}-th coefficient is not.

It remains to prove that the residual image at 2 cannot be any of the groups {C_{3}, C_{5}, S_{3}, A_{4}}. We recall some well known results from Class Field Theory:

Theorem 2.8. If L/F be an abelian Galois extension unramified outside the set of places {𝔭_{i}}_{i=1}^{n} then there exists a modulus 𝔪 = ∏_{i=1}^{n} 𝔭_{i}^{e(𝔭_{i})} such that Gal(L/F) corresponds to a subgroup of the ray class group Cl(𝒪_{F} , 𝔪).

A bound for e(𝔭) is given by the following result.

Proposition 2.9. Let L/F be an abelian Galois extension of prime degree p. Consider a modulus 𝔪 = ∏_{i=1}^{n} 𝔭_{i}^{e(𝔭_{i})} associated to the extension L/F by Theorem 2.8. If 𝔭 ramifies in L/F, then

\{ \begin{matrix}e(𝔭) = 1 & if 𝔭 ∤ p \\ 2 ≤ e(𝔭) ≤ ⌊\frac{pe(𝔭|p)}{p−1}⌋ + 1 & if 𝔭 | p, \end{matrix}

where 𝔭 is a prime above the rational prime p and e(𝔭|p) is the ramification index of 𝔭 in F/ℚ.

Proof. See [Coh00] Proposition 3.3.21 and Proposition 3.3.22.

Using these two results, we can compute for each possible Galois group all Galois extensions of F unramified outside {2, 3, 5}. In each extension, we find a prime 𝔭 where the Frobenius has non-zero trace (in 𝔽_{4}). If a_{𝔭} ≡ 0 (mod 2), for all such primes we are done:

Proposition 2.10. The residual representation σ̄_{𝔣,2} has image a 2-group.

Proof. Consider the different cases:

• The group A_{4} has C_{2} × C_{2} as a normal subgroup (with generators (12)(34) and (13)(24)). The quotient by this subgroup is a cyclic group of order 3, so the extension contains a cubic Galois subfield. Since there are no elements of order 6 in A_{4}, a non trivial element in this Galois group will have odd trace. Thus the case A_{4} and C_{3} can be discarded at the same time. In order to do so, we can take 𝔪 = 2 · 3^{2} · \sqrt{5} as the maximal modulus by Proposition 2.10. The ray class group Cl(𝒪_{F} , 𝔪) is isomorphic to C_{12} × C_{6} so there are 4 cubic extensions ramified at these primes. We consider the characters as additive characters (by taking logarithms), and denote by ψ_{1} and ψ_{2} two characters that generate the group of characters of order 3 (we take the fourth and the second power of the characters in the previous basis). Instead of computing a prime ideal where each character is non-zero, we compute two prime ideals 𝔭_{1} and 𝔭_{2} such that

⟨(ψ_{1}(𝔭_{1}), ψ_{2}(𝔭_{1})), (ψ_{1}(𝔭_{2}), ψ_{2}(𝔭_{2}))⟩ = ℤ/3ℤ × ℤ/3ℤ.

Then Proposition 5.4 of [DGP10] implies that any cubic character is non-trivial in one of these two ideals. The two ideals above the prime 11 have values (1, 0) and (1, 2), which is a basis for 𝔽_{3} × 𝔽_{3}. Since the modular form has even traces (in 𝔽_{2}) at both primes (see Table of [CS01]), we conclude that the residual representation cannot have image isomorphic to C_{3} nor A_{4}.

• To discard the C_{5} case, we take 𝔪 = 2 · 3 · (\sqrt{5})^{3}. The ray class group for this module is cyclic of order 10. The generator at primes above 11 has value 9, in particular the order 5 characters do not vanish at this primes. But the trace of Frobenius lies in 𝔽_{2} at these primes (as mentioned in the previous case), so the image cannot be cyclic of order 5.

• To discard the S_{3} case, we start by computing all the quadratic extensions of F ramified outside the set of primes {2, 3, 5}. The modulus in this case is 𝔪 = 2^{3} · 3 · \sqrt{5}. The ray class group is isomorphic to C_{4} × C_{4} × C_{2} × C_{2} × C_{2} so there are 31 such extensions. In Table 2.1 we put all the information of these extensions; the first column has an equation for each such extension, the second column its discriminant over ℚ, the third column the modulus considered (where by 𝔭_{2} (respectively 𝔭_{5}) we denote the unique prime ideal in the extension above the rational prime 2 (respectively 5)), the fourth column the ray class group and the last column the rational primes whose prime divisors in F generate the 𝔽_{3} vector space of cubic characters.

962 Luis Dieulefait, Ariel Pacetti and Matthias Schütt

Equation over F Disc. over ℚ Modulus  Ray Class Group  Rational Primes
x^{2} − 6\sqrt{5} −2^{6} · 3^{2} · 5^{3} 9 · 𝔭_{2} · 𝔭_{5}  C_{72} × C_{6} × C_{3} × C_{3}  \{7, 11, 13\}
x^{2} + 6\sqrt{5} −2^{6} · 3^{2} · 5^{3} 9 · 𝔭_{2} · 𝔭_{5}  C_{72} × C_{6} × C_{3} × C_{3}  \{7, 11, 13\}
x^{2} + \sqrt{5} + 1  −2^{6} · 5^{2} 9 · 𝔭_{2} · \sqrt{5}  C_{12} × C_{6} × C_{3}  \{7, 11, 13, 19\}
x^{2} − \sqrt{5} − 1  −2^{6} · 5^{2} 9 · 𝔭_{2} · \sqrt{5}  C_{12} × C_{6} × C_{3}  \{7, 11, 13, 19\}
x^{2} − 3\sqrt{5} − 3 −2^{6} · 3^{2} · 5^{2} 9 · 𝔭_{2} · \sqrt{5}  C_{36} × C_{6} × C_{3} × C_{3}  \{7, 11, 13, 17\}
x^{2} + 3\sqrt{5} + 3 −2^{6} · 3^{2} · 5^{2} 9 · 𝔭_{2} · \sqrt{5}  C_{36} × C_{6} × C_{3} × C_{3}  \{7, 11, 13, 17\}
x^{2} + 2\sqrt{5}  −2^{6} · 5^{3} 9 · 𝔭_{2} · 𝔭_{5}  C_{24} × C_{6} × C_{3}  \{7, 11, 13\}
x^{2} − 2\sqrt{5}  −2^{6} · 5^{3} 9 · 𝔭_{2} · 𝔭_{5}  C_{24} × C_{6} × C_{3}  \{7, 11, 13\}
x^{2} + \frac{3}{2} \sqrt{5} + \frac{3}{2} −2^{4} · 3^{2} · 5^{2} 9 · 𝔭_{2} · \sqrt{5}  C_{36} × C_{6} × C_{3}  \{7, 11, 13\}
x^{2} − \frac{3}{2} \sqrt{5} − \frac{3}{2} −2^{4} · 3^{2} · 5^{2} 9 · 𝔭_{2} · \sqrt{5}  C_{36} × C_{6} × C_{3}  \{7, 11, 13\}
x^{2} − \sqrt{5}  −2^{4} · 5^{3} 9 · 𝔭_{2} · 𝔭_{5}  C_{24} × C_{6} × C_{3}  \{7, 11, 13\}
x^{2} + \sqrt{5}  −2^{4} · 5^{3} 9 · 𝔭_{2} · 𝔭_{5}  C_{24} × C_{6} × C_{3}  \{7, 11, 13\}
x^{2} + 3\sqrt{5} −2^{4} · 3^{2} · 5^{3} 9 · 𝔭_{2} · 𝔭_{5}  C_{24} × C_{6} × C_{3} × C_{3}  \{7, 11, 13\}
x^{2} − 3\sqrt{5} −2^{4} · 3^{2} · 5^{3} 9 · 𝔭_{2} · 𝔭_{5}  C_{24} × C_{6} × C_{3} × C_{3}  \{7, 11, 13\}
x^{2} − \frac{1}{2} \sqrt{5} − \frac{1}{2}  −2^{4} · 5^{2} 9 · 𝔭_{2} · \sqrt{5}  C_{12} × C_{6} × C_{3} × C_{3}  \{7, 11, 13, 17\}
x^{2} + \frac{1}{2} \sqrt{5} + \frac{1}{2}  −2^{4} · 5^{2} 9 · 𝔭_{2} · \sqrt{5}  C_{12} × C_{6} × C_{3} × C_{3}  \{7, 11, 13, 17\}
x^{2} + \sqrt{5} + 5  2^{6} · 5^{3} 9 · 𝔭_{2} · 𝔭_{5}  C_{60} × C_{6} × C_{3} × C_{3}  \{7, 11, 13, 23\}
x^{2} − 6  2^{6} · 3^{2} · 5^{2} 9 · 𝔭_{2} · \sqrt{5}  C_{12} × C_{6} × C_{6}  \{7, 11, 13, 23\}
x^{2} + 2  2^{6} · 5^{2} 9 · 𝔭_{2} · \sqrt{5}  C_{24} × C_{12} × C_{3} × C_{3} × C_{3}  \{7, 11, 13, 23\}
x^{2} − 3\sqrt{5} − 15  2^{6} · 3^{2} · 5^{3} 9 · 𝔭_{2} · 𝔭_{5}  C_{12} × C_{6}  \{7, 11, 13\}
x^{2} + 6  2^{6} · 3^{2} · 5^{2} 9 · 𝔭_{2} · \sqrt{5}  C_{72} × C_{12} × C_{3} × C_{3} × C_{3}  \{7, 11, 13, 17\}
x^{2} − \sqrt{5} − 5  2^{6} · 5^{3} 9 · 𝔭_{2} · 𝔭_{5}  C_{36} × C_{18} × C_{3} × C_{3}  \{7, 11, 13\}
x^{2} + 3\sqrt{5} + 15  2^{6} · 3^{2} · 5^{3} 9 · 𝔭_{2} · 𝔭_{5}  C_{12} × C_{6} × C_{6}  \{7, 11, 13, 61\}
x^{2} − 2  2^{6} · 5^{2} 9 · 𝔭_{2} · \sqrt{5}  C_{36} × C_{18} × C_{3} × C_{3}  \{7, 11, 13\}
x^{2} + \frac{3}{2} \sqrt{5} + \frac{15}{2}  2^{4} · 3^{2} · 5^{3} 9 · 𝔭_{2} · 𝔭_{5}  C_{36} × C_{18} × C_{3} × C_{3}  \{7, 11, 13\}
x^{2} + 3  3^{2} · 5^{2}  18 · \sqrt{5} C_{36} × C_{3} × C_{3} × C_{3} × C_{3} × C_{3} \{7, 11, 13, 17, 19, 23\}
x^{2} + 1  2^{4} · 5^{2} 9 · 𝔭_{2} · \sqrt{5}  C_{24} × C_{12} × C_{3} × C_{3}  \{7, 11, 13, 17\}
x^{2} + \frac{1}{2} \sqrt{5} + \frac{5}{2}  5^{3}  18 · 𝔭_{5}  C_{60} × C_{3} × C_{3} × C_{3}  \{7, 11, 13\}
x^{2} − \frac{1}{2} \sqrt{5} − \frac{5}{2}  2^{4} · 5^{3} 9 · 𝔭_{2} · 𝔭_{5}  C_{12} × C_{6}  \{7, 11, 13\}
x^{2} − 3  2^{4} · 3^{2} · 5^{2} 9 · 𝔭_{2} · \sqrt{5}  C_{12} × C_{6} × C_{6} × C_{3}  \{7, 11, 13\}
x^{2} − \frac{3}{2} \sqrt{5} − \frac{15}{2}  3^{2} · 5^{3}  18 · 𝔭_{5}  C_{6} × C_{6}  \{7, 11, 13\}

Table 2.1: Quadratic extensions of F unramified outside {2, 3, 5}

Documenta Mathematica 17 (2012) 953–987

Note that the first 16 extensions are not Galois over ℚ and are listed so that they are isomorphic in pairs. The last 15 are indeed Galois over ℚ. This implies that we could consider one element of each pair for the first fields. Note that the primes at conjugate fields are the same, since the Frobenius at a prime in F of a field is the same as the Frobenius at the conjugate of the prime in the conjugate extension. The traces of Frobenius are even in all such primes (see Table of [CS01]), so we conclude that the image of the residual representation σ̄_{𝔣,2} cannot be isomorphic to S_{3} either.

3 Computing the residual image of the Galois representation σ_{2}

In this section we consider the Consani-Scholten quintic X̃ from [CS01]. Our goal is to prove that the ℓ-adic Galois representations ρ of H^{3}(X̃ℚ̄, ℚ_{ℓ}) have 4divisible trace. This will be used in 3.5 to deduce that modulo 2 the restricted 2-adic two-dimensional Galois representations have image in SL_{2}(𝔽_{2}), and in fact even trace, so that we can apply an adaptation of the Faltings-Serre-Livné method in order to prove Theorem 1.1.

3.1 Setup

Consider the Chebyshev polynomial

P (y, z) = (y^{5} + z^{5}) − 5yz(y^{2} + z^{2}) + 5yz(y + z) + 5(y^{2} + z^{2}) − 5(y + z).

Then we define an affine variety X in 𝔸^{4} by

X : P (x_{1}, x_{2}) = P (x_{4}, x_{5}).

Let X̄ ⊂ ℙ^{4} denote the projective closure of X. Then X̄ has 120 ordinary double points. Let X̃ denote a desingularisation obtained by blowing up X̄ at the singularities.

Remark 3.1. We might also consider a small resolution X̂, as many of the nodes lie on products of lines. Then we would have to check that X̂ is projective and can be defined over ℚ. This desingularisation would have the advantage of producing an honest Calabi-Yau threefold, but it does not affect the question of Hilbert modularity.

Consani and Scholten compute the Hodge diamond of X̃ as follows:

1

0  0
0  141  0
1  1  1  1
0  141  0
0  0

1 Hence the étale cohomology groups H^{3}(X̃ℚ̄, ℚ_{ℓ}) (ℓ prime) give rise to a compatible system of four-dimensional Galois representations {ρ_{ℓ}}. Since X̃ has good reduction outside {2, 3, 5}, ρ_{ℓ} is unramified outside {2, 3, 5, ℓ}. Let F = ℚ(\sqrt{5}) and fix some prime ℓ ∈ ℕ and a prime λ of F above ℓ. Then Consani and Scholten prove for the ℓ- resp. λ-adic representations:

Theorem 3.2 (Consani-Scholten [CS01]). The restriction ρ|_{Gal(ℚ̄/F)} is reducible as a representation into GL_{4}(F_{λ}): There is a Galois representation

σ : Gal(ℚ̄/F) → GL_{2}(F_{λ})

such that ρ = Ind_{F}^{ℚ} σ.

The theorem implies in particular that internal and external conjugation have the same effect on σ. Here we want to prove the following property:

Proposition 3.3. The Galois representation ρ has 4-divisible trace, and so has any restriction to a finite extension of ℚ. That is, ρ(Frob_{q}) ≡ 0 mod 4 for any odd prime power q.

In fact, for q ≡ 2, 3 mod 5, Consani-Scholten proved that ρ(Frob_{q}) has zero trace. Hence we would only have to consider the case q ≡ 1, 4 mod 5, although we will treat the problem in full generality. As a corollary, we will deduce that σ is even in 3.5 as required for the proof of Theorem 1.1.

3.2 Lefschetz fixed point formula

Choose a prime p ≠ ℓ of good reduction for X̃ and let q = p^{r}. Consider the geometric Frobenius endomorphism Frob_{q} on X̃/𝔽_{p}, raising coordinates to their q-th powers. Then the Lefschetz fixed point formula tells us that

#X̃(𝔽_{q}) = ∑_{i=0}^{6}(−1)^{i} trace Frob_{q}^{∗}(H^{i}(X̃ℚ̄, ℚ_{ℓ})).

In our situation, this simplifies as follows: h^{1} = h^{5} = 0; H^{2}(X̃) and H^{4}(X̃) are algebraic by virtue of the exponential sequence and Poincaré duality. Moreover Frob_{q}^{∗} factors through a permutation on H^{2}(X̃), i.e. all eigenvalues have the shape ζq where ζ is some root of unity. Denote the sum of these roots of unity by h_{q} (which is an integer in ℤ by the Weil conjectures). Finally geometric and algebraic Frobenius are compatible through ρ. Hence

t_{q} = trace ρ(Frob_{q}) = 1 + h_{q} q(1 + q) + q^{3} − #X̃(𝔽_{q}).  (2)

Prop. 3.3 claims that the left hand side is divisible by 4. If q ≡ −1 mod 4, this is a consequence of the following

Lemma 3.4. For any good prime p and q = p^{r}, #X̃(𝔽_{q}) ≡ 0 mod 4.

If q ≡ 1 mod 4, then we furthermore need the following

Lemma 3.5. For any good prime p and q = p^{r}, h_{q} is odd.

3.3 Proof of Lemma 3.4

To prove Lemma 3.4, we use the action of the dihedral group D_{4} on X̃ and the knowledge about the exceptional divisors from Consani-Scholten. Let ζ_{n} denote a primitive n-th root of unity. Then all the nodes are defined over ℚ(ζ_{15}). A detailed list can be found in [CS01]. Over the field of definition of the node, the exceptional divisor E is isomorphic to ℙ^{1} × ℙ^{1}. Hence #E(𝔽_{q}) = (q + 1)^{2} if the node is defined over 𝔽_{q}.

Lemma 3.6. For any good prime p and q = p^{r}, #X̃(𝔽_{q}) ≡ #X̄(𝔽_{q}) mod 32.

Proof: By [CS01], we have

#\{nodes over 𝔽_{q}\} = \begin{cases}0, & q ≡ 2, 7, 8, 13 mod 15, \\ 8, & q ≡ 14 mod 15, \\ 24, & q ≡ 4 mod 15, \\ 104, & q ≡ 11 mod 15, \\ 120, & q ≡ 1 mod 15. \end{cases}

Since the number of points on the exceptional divisor is the same for all nodes defined over 𝔽_{q}, the claim follows.

Lemma 3.7. For any good prime p and q = p^{r}, #X̄(𝔽_{q}) ≡ #X(𝔽_{q})−q mod 4.

Proof: The affine variety X is compactified by adding a smooth surface at ∞. In fact, this is the Fermat surface of degree five:

S = {x_{0}^{5} + x_{1}^{5} − x_{3}^{5} − x_{4}^{5} = 0} ⊂ ℙ^{3}.

Hence #X̄(𝔽_{q})−#X(𝔽_{q}) = #S(𝔽_{q}). Thus Lemma 3.7 amounts to the following

Lemma 3.8. For p ≠ 5 and q = p^{r}, #S(𝔽_{q}) ≡ 1 + q + q^{2} mod 4.

The proof of this lemma will be postponed to the end of this subsection. Lemma 3.7 follows.

To prove the corresponding statement about the affine variety X, we use the action of the dihedral group D_{4} generated by the involutions

(x_{1}, x_{2}, x_{3}, x_{4}) ↦ (x_{2}, x_{1}, x_{3}, x_{4}), (x_{1}, x_{2}, x_{3}, x_{4}) ↦ (x_{1}, x_{2}, x_{4}, x_{3})

and by the cyclic permutation

γ : (x_{1}, x_{2}, x_{3}, x_{4}) ↦ (x_{3}, x_{4}, x_{2}, x_{1}).  (3)

It follows that

#X(𝔽_{q}) ≡ #{x ∈ X(𝔽_{q}); #{y ∈ (D_{4} − orbit of x)} < 4} mod 4.

Here {x ∈ X(𝔽_{q}); #{y ∈ (D_{4} − orbit of x)} < 4} = {(x_{1}, x_{1}, x_{3}, x_{3}) ∈ X(𝔽_{q})}. We are led to consider the affine curve C in 𝔸^{2} defined by

C : P (y, y) = P (z, z).

Then the above subset of X(𝔽_{q}) is in bijection with C(𝔽_{q}), and we obtain

#X(𝔽_{q}) ≡ C(𝔽_{q}) mod 4.  (4)

Lemma 3.9. For any good prime p and q = p^{r}, #C(𝔽_{q}) ≡ q mod 4.

Proof of Lemma 3.9: C is reducible. The change of variables

u = \frac{y + z}{2} , v = \frac{y − z}{2}

allows us to write

P (y, y) − P (z, z) = = v(v^{4} + 5(2u^{2} − 4u + 1) v^{2} + 5(u^{2} − 3u + 1)(u^{2} − u − 1)) = v G(u, v).

Hence

#C(𝔽_{q}) = q + #(B(𝔽_{q}) ∩ {v ≠ 0})  (5)

where B is the affine curve in 𝔸^{2} given by G(u, v) = 0. Here B is endowed with involutions

(u, v) ↦ (u, −v), (u, v) ↦ (2 − u, v).

For the number of points, this implies

#(B(𝔽_{q}) ∩ {v ≠ 0}) ≡ #(B(𝔽_{q}) ∩ {u = 1, v ≠ 0}) mod 4
= #{v ∈ 𝔽_{q}; v^{4} − 5v^{2} + 5 = 0}.

The last polynomial factors as

4(v^{4} − 5v^{2} + 5) = (2v^{2} − 5 − \sqrt{5}) (2v^{2} − 5 + \sqrt{5}).

Since \frac{5+\sqrt{5}}{2} · \frac{5−\sqrt{5}}{2} = 4, a square, we deduce that the last equation has either zero or four solutions in 𝔽_{q}. In particular, (5) reduces to #C(𝔽_{q}) ≡ q mod 4, i.e. to the claim of Lemma 3.9.

Proof of Lemma 3.8: We shall again use the cyclic permutation γ from (3), but this time it operates on the homogeneous coordinates of ℙ^{3}. Hence

#S(𝔽_{q}) ≡ #(S ∩ Fix(σ^{2}))(𝔽_{q}) mod 4.  (6)

Here Fix(σ^{2}) = {[λ, μ, ±λ, ±μ]; [λ, μ] ∈ ℙ^{1}}.

One of these lines is contained in S, and it is easy to see that there are exactly (5, q − 1) further points of intersection unless p = 2. I.e.

#(S ∩ Fix(σ^{2}))(𝔽_{q}) = 1 + q + \begin{cases}0, & p = 2 \\ (5, q − 1), & p ≠ 2 \end{cases} ≡ 1 + q + q^{2} mod 4.

Lemma 3.8 follows from this congruence and (6).

Proof of Lemma 3.4: Lemma 3.9 and (4) imply that #X(𝔽_{q}) ≡ q mod 4. By Lemma 3.7 this gives #X̄(𝔽_{q}) ≡ 0 mod 4. The according statement for X̃ is obtained from Lemma 3.6.

3.4 Proof of Lemma 3.5

Lemma 3.5 states that the trace h_{q} of Frob_{q} on H^{2}(X̃ℚ̄, ℚ_{ℓ}(1)) is always odd. We shall first prove the following auxiliary result:

Lemma 3.10. The Galois group Gal(ℚ̄/ℚ(ζ_{15})) acts trivially on H^{2}(X̃ℚ̄, ℚ_{ℓ}(1)).

Proof: Denote the exceptional locus of the blow-up by E. Then E is defined over ℚ. The Leray spectral sequence for the desingularisation gives an exact sequence

0 → H^{2}(X̄ℚ̄, ℚ_{ℓ}(1)) → H^{2}(X̃ℚ̄, ℚ_{ℓ}(1)) → H^{2}(Eℚ̄, ℚ_{ℓ}(1)).  (7)

By construction, (7) is compatible with the Galois action. Here H^{2}(X̄ℚ̄, ℚ_{ℓ}(1)) is the same as for a general quintic hypersurface in ℙ^{4}. Hence it has dimension one and is generated by the class of a hyperplane section. In particular, Gal(ℚ̄/ℚ) acts trivially on H^{2}(X̄ℚ̄, ℚ_{ℓ}(1)). Recall that every component of E as well as both rulings on every component are defined over ℚ(ζ_{15}). Hence Gal(ℚ̄/ℚ(ζ_{15})) acts trivially on H^{2}(Eℚ̄, ℚ_{ℓ}(1)). By the Galois-equivariant exact sequence (7), the same holds for H^{2}(X̃ℚ̄, ℚ_{ℓ}(1)).

It follows from Lemma 3.10, that h_{q} = 141 if q ≡ 1 mod 15. The prove the parity for the other residue classes, we need two easy statements about sums of primitive roots of unity. They involve the Möbius function μ : ℕ → {−1, 0, 1}:

μ(n) = \begin{cases}0, & n not squarefree, \\ (−1)^{m}, & n squarefree with m prime divisors. \end{cases}

Lemma 3.11. Let n ∈ ℕ and ζ_{n} a primitive n-th root of unity. Then ζ_{n} has trace μ(n).

The lemma follows immediately from the factorisation of the cyclotomic polynomial x^{n} − 1 and the definition of μ(n).

Lemma 3.12. Let n ∈ ℕ and ζ_{n} a primitive n-th root of unity. Let m = 2^{s} · k with (k, n) = 1. Then

μ(n) = trace ζ_{n} = ∑_{j∈(ℤ/nℤ)^{∗}} ζ_{n}^{j} ≡ ∑_{j∈(ℤ/nℤ)^{∗}} ζ_{n}^{mj} mod 2.

Proof: If (m, n) = 1, then taking m-th powers permutes the primitive n-th roots of unity and both sums coincide. Hence it suffices to consider the case where m = 2^{s}(s > 0) and 2|n. If 4 ∤ n, then \{ζ_{n}^{mj}; j ∈ (ℤ/nℤ)^{∗}\} is the set of \frac{n}{2}-th primitive roots of unity. Hence

∑_{j∈(ℤ/nℤ)^{∗}} ζ_{n}^{mj} = μ (\frac{n}{2}) = −μ(n)

and the claim follows mod 2. If 4|n, then μ(n) = 0 and every element in {ζ_{n}^{mj}; j ∈ (ℤ/nℤ)^{∗}} appears with multiplicity (m, n). Hence

2 | (m, n) | ∑_{j∈(ℤ/nℤ)^{∗}} ζ_{n}^{mj},

and we obtain the claimed congruence.

Proof of Lemma 3.5: Let Ξ be the set of eigenvalues of Frob_{q} on H^{2}(X̃ℚ̄, ℚ_{ℓ}(1)) with multiplicities. Then

h_{q} = ∑_{ζ∈Ξ} ζ.

Recall that the Galois group Gal(ℚ̄/ℚ(ζ_{15})) acts trivially on H^{2}(X̃ℚ̄, ℚ_{ℓ}(1)). Since ℚ(ζ_{15})/ℚ is Galois of degree eight, we deduce that ζ^{8} = 1 for each ζ ∈ Ξ. In particular

∑_{ζ∈Ξ} ζ^{8} = 141.  (8)

In the present situation, h_{q} ∈ ℤ, i.e. h_{q} is a sum of traces of elements in Ξ. Hence we can apply Lemma 3.12 to deduce that h_{q} has the same parity as the sum in (8). That is, h_{q} is odd.

3.5 Evenness of σ

In this subsection we conclude our preparations for the proof of Theorem 1.1 by proving the following corollary of Proposition 3.3:

Corollary 3.13. The Galois representation σ is even.

Recall that σ induces the 4-dimensional Galois representation ρ over ℚ. The proof commences by spelling out the characteristic polynomial of ρ(Frob_{q}) for some odd prime power q:

μ_{q}(T) = T^{4} − t_{q}T^{3} + uT^{2} − q^{3}t_{q}T + q^{6} with u = (t_{q}^{2}−t_{q^{2}})/2 using the notation of (2). By Proposition 3.3, t_{q} ≡ t_{q^{2}} ≡ 0 mod 4, so u ≡ 0 mod 2. Now we turn to σ, the 2-dimensional Galois representation with values in F = ℚ(\sqrt{5}) inducing ρ. Let 𝔮 denote some power of a prime ideal in F with odd norm q ∈ ℤ. We write

s_{𝔮} = Tr σ(Frob_{𝔮})

and consider the case s_{𝔮} ∉ ℤ. Then s_{𝔮} = a + bω where ω solves v^{2} − v − 1 = 0 and a, b ∈ ℤ, b ≠ 0. By Theorem 3.2 we have

t_{q} = 2a + b.

Since t_{q} is even by Proposition 3.3, so is b, and we can write somewhat more intuitively s_{𝔮} = c + d\sqrt{5} with c, d ∈ ℤ, d ≠ 0. This already implies that the mod 2-reduction of σ has traces in 𝔽_{2}, so it will have image in SL(2, 𝔽_{2}). In the new notation, we obtain

t_{q} = 2c,

so by Proposition 3.3 the input c is even. But then, factoring μ_{q}(T) into quadratic factors over F corresponding to σ and its external conjugate, the coefficient of T^{2} reads

u = (c^{2} − 5d^{2}) + 2q^{3}.

Since we have already seen that u and c are even, we find that d is even, too. That is, σ has even trace at Frob_{𝔮}. The case s_{𝔮} ∈ ℤ is essentially the same argument, but even simpler.

4 Proof of the Main Theorem

There is version of the Faltings-Serre method in [Liv87] that allows to compare two-dimensional 2-adic Galois representations with even traces. Here we have to modify this approach slightly since the two-dimensional Galois representation σ_{2} is only determined up to conjugation of its coefficients in the quadratic field F. While the original result involved the notion of non-cubic test sets, in order to prove Theorem 1.1, we replace this notion by non-quartic sets:

Definition. A subset T of a finite dimensional vector space V is non-quartic if every homogeneous polynomial of degree 4 on V which vanishes on T, vanishes in the whole V.

The following lemma is useful to lower the cardinality of the test set T.

Lemma 4.1. Let V be a finite-dimensional vector space. Let T be a subset of V which contains 4 distinct hyperplanes through the origin and a point outside them. Then T \ {0} is non-quartic.

Proof. Let L_{1}, …, L_{4} denote linear homogeneous polynomials giving equations of the four hyperplanes. Let P (x_{1}, …, x_{n}) be a homogeneous quartic polynomial vanishing on all points of T. Division with remainder gives a representation

P (x_{1}, …, x_{n}) = L_{1}Q(x_{1}, x_{2}, …, x_{n}) + P_{2}(x_{1}, …, x_{n})

with L_{1} ∤ P_{2}. The crucial property here is the following: Since T contains the hyperplane {L_{1} = 0} and P vanishes on this hyperplane, P_{2} vanishes on all of V. (To see this, apply a linear transformation so that L_{1} = x_{1}; then P_{2} = P_{2}(x_{2}, …, x_{n}), and P_{2} vanishes on the hyperplane {x_{1} = 0} if and only if it vanishes on V). Since the hyperplanes are linearly independent, we can apply the same argument to the other three hyperplanes (starting with L_{1}Q instead of P). We obtain

P (x_{1}, …, x_{n}) = A · L_{1}L_{2}L_{3}L_{4} + P̃(x_{1}, …, x_{n}),

where A is a field constant and P̃ vanishes identically on V. Since T contains a point outside the union of the four hyperplanes, A must be zero. But then P = P̃ vanishes on all of V. Since this argument applies to any homogeneous quartic polynomial P, the test set T \ {0} is non-quartic.

Remark 4.2. Note that Lemma 4.1 does explicitly not require the hyperplanes to be linearly independent. It is immediate from the proof of Lemma 4.1 that the same argument works for test sets for homogeneous polynomials of degree n if we find n distinct hyperplanes through the origin and a point outside them.

We want to compare the two Galois representations, σ_{2} and σ_{𝔣,2}. It is crucial that in the present situation we know that the external Galois conjugate representation exists: this follows in the geometric example by construction, and in the modular example we can consider the 2-adic representation attached to the conjugate Hilbert modular forms τ(𝔣), where τ is a generator of the group Gal(F/ℚ). For any given ℓ-adic Galois representation ρ with field of coefficients F (i.e, the field generated by the traces of Frobenius elements is F) we will denote by ρ′ the external conjugate representation (if we know that such a representation exists). Since for the Calabi-Yau threefold X̃, we can only compute the traces of the 4-dimensional Galois representation σ_{2} ⊕ σ′_{2} of Gal(ℚ̄/F) (or actually of Gal(ℚ̄/ℚ)), we will need the following generalization of Theorem 4.3 in [Liv87] about Galois representations whose residual images are 2-groups:

Theorem 4.3. Let K be a global field, S a finite set of primes of K and E the unramified quadratic extension of ℚ_{2}. Denote by K_{S} the compositum of all quadratic extensions of K unramified outside S and by 𝒫_{2} the maximal prime ideal of 𝒪 := 𝒪_{E}. Suppose ρ_{1}, ρ_{2} : Gal(\overline{ℚ}/K) → GL_{2}(E) are continuous representations, unramified outside S, and with field of coefficients F, and assume also that their external Galois conjugates exist. We suppose that the following conditions are satisfied:

1. Tr(ρ_{1}) ≡ Tr(ρ_{2}) ≡ 0 (mod 𝒫_{2}) and det(ρ_{1}) ≡ det(ρ_{2}) ≡ 1 (mod 𝒫_{2}).

2. There exists a set T of primes of K, disjoint from S, for which

(i) The image of the set {Frob_{t}}_{t∈T} in Gal(K_{S}/K)\{0} is non-quartic.

(ii) Tr(ρ_{1}(Frob_{t})) + Tr(ρ′_{1}(Frob_{t})) = Tr(ρ_{2}(Frob_{t})) + Tr(ρ′_{2}(Frob_{t})) and det(ρ_{1}(Frob_{t})) = det(ρ_{2}(Frob_{t})) for all t ∈ T.

Then ρ_{1} ⊕ ρ′_{1} and ρ_{2} ⊕ ρ′_{2} have isomorphic semi-simplifications.

Proof. This is just a slight generalization of Proposition 4.7 and Theorem 4.3 in [Liv87], we reproduce most of the arguments for the reader convenience, adapted to our situation. Observe that due to assumption (1), and the fact that similar conditions hold also for ρ′_{1} and ρ′_{2}, the image of any of the representations ρ_{1}, ρ′_{1}, ρ_{2} and ρ′_{2} is a pro-2-group. Let G be the image of the product of the four representations. Then, G is a topologically finitely generated pro-2-group and the four representations can be thought of (and we will do so for the rest of this proof) as representations of G, each being obtained from a suitable projection. Set M_{2} to be the algebra of 2 by 2 matrices with coefficients in 𝒪. For g ∈ G, let ρ : G → M_{2} × M_{2} be the map:

ρ(g) = (ρ_{1}(g), ρ_{2}(g))

Keeping the notation in [Liv87], we call Σ the subset of G corresponding to T (the projection of the elements in {Frob_{t}}_{t∈T} to G). Let M be the ℤ_{2}-linear span of ρ(G). Since 𝒪 has rank 2 over ℤ_{2}, M is a subalgebra with unity of M_{2} × M_{2} which is free of rank at most 16 as a module over ℤ_{2}. We consider R = M/2M, which is an 𝔽_{2}-vector space of dimension at most 16. Denote the image of g ∈ G in R by ḡ. Set Γ = {ḡ | g ∈ G}. Then Γ ⊆ R^{×}. R is spanned as 𝔽_{2}-vector space by Γ and dim_{𝔽_{2}} R ≤ 16. We will show the following: Assertion: R is spanned over 𝔽_{2} by {σ̄ | σ ∈ Σ ∪ {1}}. In order to do so, following [Liv87], we have first to prove that

σ ∈ Σ ⇒ σ̄^{2} = ¯1 in Γ.  (9)

Let σ ∈ Σ. Set d = det ρ_{1}(σ) = det ρ_{2}(σ) (the last equality is due to assumption (2)(ii)) and t_{1} = Tr ρ_{1}(σ), t_{2} = Tr ρ_{2}(σ). By the Cayley-Hamilton theorem (for 2 by 2 matrices) we have:

ρ(σ)^{2} = (t_{1}ρ_{1}(g), t_{2}ρ_{2}(g)) − d(I, I)

in M_{2} × M_{2}, where I is the 2 by 2 identity matrix. Reducing modulo 𝒫_{2} this equality, since assumption (1) gives t_{1} ≡ t_{2} ≡ 0 (mod 𝒫_{2}) and d ≡ 1 (mod 𝒫_{2}) we obtain σ̄^{2} = ¯1 in Γ, and this is formula (9).

If we call G^{∗} the Frattini subgroup of G, and Γ^{∗} the one of Γ, since G/G^{∗} → Γ/Γ^{∗} is surjective, the image of Σ in Γ/Γ^{∗} is non-quadratic (we use assumption (2)(i) and the fact that non-quartic easily implies non-quadratic). Then, by Lemma 4.5 in [Liv87] we see that formula (9) implies that Γ^{∗} = 1 and hence Γ ≅ (ℤ/2ℤ)^{r}, for some r. In particular, Γ is commutative, and since Γ spans R, R is a commutative ring. Let r ∈ R be any element and write r = ∑_{γ∈Γ} k_{γ}γ, with k_{γ} ∈ 𝔽_{2}. Then r^{2} = (∑ k_{γ}^{2}) · ¯1. Hence r^{2} = 0 or else r is invertible. It follows that R is a local artinian algebra over 𝔽_{2} with maximal ideal

P = {r ∈ R | r^{2} = 0}

and R/P = 𝔽_{2}. Let us now show that P^{5} = 0. The proof of this fact goes as the one given in [Liv87], except that we consider now products of 5 elements x_{1}, ..., x_{5} in P (instead of 4 elements): assuming that their product is non-zero we derive a contradiction, as in [Liv87], by considering the 𝔽_{2}-algebra

R_{0} = 𝔽_{2}[T_{1}, ..., T_{5}]/(T_{1}^{2}, ..., T_{5}^{2})

and the injective map of 𝔽_{2}-algebras from R_{0} to R that sends T_{i} to x_{i}. The contradiction follows from the inequalities: dim_{𝔽_{2}} R_{0} = 32 > 16 ≥ dim_{𝔽_{2}} R. Using the fact that P^{5} = 0, the rest of the proof of the assertion follows as in [Liv87], changing “cubic polynomial” by “quartic polynomial”, and “noncubic” by “non-quartic”. The assertion being proved, we conclude as in [Liv87] from Nakayama’s lemma that the ℤ_{2}-span of {ρ(σ) | σ ∈ Σ ∪ {1}} is all of M (recall that M is the linear ℤ_{2}-span of ρ(G)). Here comes the only place where equality of the traces over Σ is used in Livné’s proof (cf. [Liv87]): he considers the map: α : M → 𝒪 defined by α(a, b) = Tr a−Tr b. Since this map is 𝒪-linear and α(I, I) = 0, assuming that α(ρ(σ)) = 0 for every σ ∈ Σ Livné concludes that α = 0, i.e., equality of the traces of ρ_{1} and ρ_{2}. We can argue in the same way but using the map: β : M → ℤ_{2} given by

β(a, b) = Tr a + (Tr a)^{ϕ} − Tr b − (Tr b)^{ϕ}

where ϕ is the order two element in Gal(E/ℚ_{2}). Observe that when applied to numbers in F, ϕ agrees with the order two element τ in Gal(F/ℚ).

Observe that if a = ρ_{1}(g), then (Tr a)^{ϕ} = (Tr ρ_{1}(g))^{ϕ} = Tr(ρ′_{1}(g)), and similarly for b = ρ_{2}(g). This map is ℤ_{2}-linear and satisfies β(I, I) = 0. Then, assumption (2) (ii) in the theorem implies that for elements of G in Σ (recall that these elements correspond to Frobenius elements for primes in T) the map β vanishes, thus we conclude (as Livné does for α) that β = 0, which is the equality of the traces of two 4-dimensional 2-adic Galois representations. Applying Brauer-Nesbitt we conclude that these 4-dimensional Galois representations have isomorphic semi-simplifications.

4.1 Proof of Theorem 1.1

We want to apply Theorem 4.3 to the 2-adic Galois representations σ_{2}, σ_{𝔣,2} over F = K. In Proposition 3.3 we proved that σ_{2} has even traces by geometric considerations, and in Section 2 we proved that the same is true for the representation σ_{𝔣,2}. Recall from the introduction that both residual representations σ̄_{2}, σ̄_{𝔣,2} have image in SL(𝔽_{4}). In particular their determinants are congruent modulo 2. Thus the first hypothesis of Theorem 4.3 is satisfied. The field K_{S} of Theorem 4.3 is the compositum of the thirty one field extensions computed in the proof of Proposition 2.10. The set of primes in K in the set

T = \{⟨61, 26−\sqrt{5}⟩, ⟨59, \sqrt{5}+8⟩, ⟨149, \sqrt{5}−68⟩, ⟨211, \sqrt{5}+65⟩, ⟨101, \sqrt{5}−45⟩,

⟨19, \sqrt{5} + 9⟩, ⟨229, \sqrt{5} − 66⟩, ⟨11, \sqrt{5} − 4⟩, ⟨11, \sqrt{5} + 4⟩, ⟨109, \sqrt{5} − 21⟩,

⟨19, \sqrt{5} − 9⟩, ⟨701, \sqrt{5} − 53⟩, ⟨211, \sqrt{5} − 65⟩, ⟨29, \sqrt{5} − 11⟩, ⟨59, \sqrt{5} − 8⟩,

⟨181, \sqrt{5} − 27⟩, ⟨239, \sqrt{5} + 31⟩, ⟨31, \sqrt{5} + 6⟩, ⟨79, \sqrt{5} − 20⟩, ⟨71, \sqrt{5} − 17⟩, 13,

⟨401, \sqrt{5} − 178⟩, ⟨449, \sqrt{5} − 118⟩, ⟨241, \sqrt{5} − 103⟩, ⟨89, \sqrt{5} − 19⟩, 7,

⟨79, \sqrt{5} + 20⟩, ⟨239, \sqrt{5} − 31⟩, ⟨41, \sqrt{5} − 13⟩, ⟨31, \sqrt{5} − 6⟩, ⟨71, \sqrt{5} + 17⟩\},

saturate the set Gal(K_{S}/K)\{0}. They are ordered in such a way that the primes (under class field theory) correspond to the extensions listed in Table 2.1 in the same order. Thus T saturates the set Gal(K_{S}/K)\{0}, but we can still eliminate some prime ideals of big norm by replacing T by a non-quartic test set by Lemma 4.1. To do so, we fix a standard basis of 𝔽_{2}^{5} corresponding to the following quadratic extensions of F:

x^{2} − 3\frac{\sqrt{5} − 5}{2}, x^{2} − 3, x^{2} + \frac{\sqrt{5} + 5}{2}, x^{2}2 − 3\sqrt{5}, x^{2} − 2.

In this basis, the elements corresponding to the primes above 701, 449 and 401 correspond to the elements (1, 1, 0, 0, 1), (0, 1, 1, 1, 0) and (1, 1, 0, 1, 0) respectively in 𝔽_{2}^{5}. We claim that the set T′ obtained from T by removing these three elements is a non-quartic set in 𝔽_{2}^{5}. To see this, we use Lemma 4.1 with the fact that T′ ∪ {0} contains the four hyperplanes

\begin{cases}x_{2} = 0 \\ x_{4} + x_{5} = 0 \\ x_{1} + x_{3} = 0 \\ x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 0 \end{cases}

and the extra point (0, 1, 1, 0, 1) (which corresponds to one of the primes above 59).

It was checked in [CS01] that the characteristic polynomials of the 4dimensional Galois representations σ_{2} ⊕σ′_{2} and σ_{𝔣,2} ⊕σ_{τ(𝔣),2} agree at the primes of K above rational primes smaller than 100. For the remaining set of primes above the set of rational primes

{101, 109, 149, 181, 211, 229, 239, 241},

the same was checked by us, see Appendix A and the table at [DPS]. By Theorem 4.3 we conclude that the two 4-dimensional Galois representations have indeed isomorphic semi-simplifications. In particular, any irreducible component of one of them must be isomorphic to some irreducible component of the other, thus since we know that σ_{𝔣,2} and σ_{τ(𝔣),2} are irreducible, one of them must be isomorphic to σ_{2} (and the other to σ′_{2}). This proves Theorem 1.1.

A Appendix: Counting points

In this appendix we indicate how to count the number of points on the Consani-Scholten quintic X̃ over finite fields. In particular, we give the traces of the Galois representations σ_{λ}, σ′_{λ} at the primes p > 100 needed to prove Theorem 1.1. For most part, we follow the approach from [CS01]. Recall that X̃ is given affinely by the symmetric equation in the Chebyshev polynomial P_{5}(y, z):

X = {P_{5}(x_{1}, x_{2}) = P_{5}(x_{4}, x_{5})} ⊂ 𝔸^{4}.

Thus we can count the number of points of X̃ over some finite field 𝔽_{q} as follows:

1. Compute the affine number of points #X(𝔽_{q}).

2. For the projective closure X̄ ⊂ ℙ^{4}, let S ⊂ ℙ^{3} be X̄ − X. Compute #S(𝔽_{q}).

3. Compute the contribution from the exceptional divisors in the resolution X̃ → X̄.

For (1), we can proceed by counting how often each value in 𝔽_{q} is attained by the Chebyshev polynomial P_{5}(y, z) over 𝔸^{2}. Due to the symmetry, #X(𝔽_{q}) is the sum of the squares of these numbers. For (2), note that S is the Fermat quintic surface given by the model

S = {x_{1}^{5} + x_{2}^{5} = x_{4}^{5} + x_{5}^{5}} ⊂ ℙ^{3}.

In [CS01] it was pointed out that for q ≢ 1 mod 5 one has #S(𝔽_{q}) = 1+q+q^{2}. Meanwhile for q ≡ 1 mod 5, we could either use the zeta function of S and its description in terms of Jacobi symbols due to A. Weil or proceed along the same lines as above, i.e. count points over 𝔸^{4} using symmetry and then take into account that we are actually working over ℙ^{3} (substract 1 and divide by q − 1). However either approach would a priori impose the same complexity as for computing #X(𝔽_{q}). Luckily counting {(y, z) ∈ 𝔸_{𝔽_{q}}^{2}; y^{5} + z^{5} = a} can be improved by noting that scalar multiplication acts as multiplication by fifth powers on the values. Hence #{(y, z) ∈ 𝔸_{𝔽_{q}}^{2}; y^{5} + z^{5} = 0} = (q − 1) · #{[y, z] ∈ ℙ_{𝔽_{q}}^{1}; y^{5} + z^{5} = 0} + 1 and for any a ≠ 0 with O(a) = a · (𝔽_{q}^{∗})^{5} denoting the a-orbit under multiplication by fifth powers in 𝔽_{q}^{∗}:

#{(y, z) ∈ 𝔸_{𝔽_{q}}^{2}; y^{5} + z^{5} = a} = 5 ∑_{o∈O(a)} #{[y, z] ∈ ℙ_{𝔽_{q}}^{1}; y^{5} + z^{5} = o}.

(Strictly speaking the sets on the right are ambiguous, but scalar multiples end up in the same orbit, so the contribution does not depend on the chosen representative [y, z] ∈ ℙ^{1}.) Essentially this simplification reduces the algorithm’s running time from q^{2} to q compared with computing #X(𝔽_{q}). Finally for (3) we recall from 3.3 that the 120 nodes are always defined over the extension of 𝔽_{q} containing the 15th roots of unity, and that their rulings are always defined over the same field. So if a node is defined over 𝔽_{q}, then its exceptional divisor contributes q^{2}+2q additional points. Thus we find #X̃(𝔽_{q}). This allows us to compute the trace

t_{q} = trace Frob_{q}^{∗}(H^{3}(X̃ℚ̄, ℚ_{ℓ}))

through the Lefschetz fixed point formula (2). Here we do not need to know h_{q} in advance since it is determined (for q > 20 and b_{3}(X̃) = 4) by the inequality

| t_{q} |≤ 4q^{3/2}.

We obtain the characteristic polynomial of Frob_{q}^{∗} on H^{3}(X̃ℚ̄, ℚ_{ℓ}):

μ_{q}(T) = T^{4} − t_{q}T^{3} + \frac{1}{2}(t_{q}^{2} − t_{q^{2}})T^{2} − t_{q}q^{3}T + q^{6}.

In the present situation, we know that L_{q}(T) will always split over ℚ(\sqrt{5}):

μ_{q}(T) = (T^{2} − α_{q}T + q^{3})(T^{2} − α_{q}^{σ}T + q^{3}), α_{q} ∈ ℚ(\sqrt{5}).

The traces α_{p}, α_{p}^{σ} appear as eigenvalues of the Hilbert modular form. In the following table, we collect one of the traces together with the numbers of points of X and S over 𝔽_{p} and 𝔽_{p^{2}} for all primes p > 100 needed to prove Theorem 1.1.

B Appendix: Sturm Bound (by José Burgos Gil and Ariel Pacetti)

The aim of this appendix is to show how a Sturm bound can be obtained for the modular form of level 6\sqrt{5} and parallel weight 2. We expect to extend the result to any real quadratic fields in a future work. Following the previous notation, F will denote the real quadratic field ℚ(\sqrt{5}).

\begin{matrix}p & #X(𝔽_{p}) & #X(𝔽_{p^{2}}) & #S(𝔽_{p}) & #S(𝔽_{p^{2}}) & α_{p} \\ 101 & 1222681 & 1063601210405 & 14655 & 104338955 & −598 − 476\sqrt{5} \\ 109 & 1338593 & 1679922873825 & 11991 & 141787855 & 890 + 468\sqrt{5} \\ 149 & 3395857 & 10952392903505 & 22351 & 494061055 & 150 − 344\sqrt{5} \\ 181 & 6562145 & 35183310464645 & 39455 & 1074841355 & −898 − 288\sqrt{5} \\ 211 & 10261235 & 88285583898085 & 49205 & 1984280555 & −1228 − 1616\sqrt{5} \\ 229 & 12214593 & 144270849112465 & 52671 & 2752837855 & −210 + 940\sqrt{5} \\ 239 & 13872967 & 186440164574105 & 57361 & 3265836055 & 3240 + 944\sqrt{5} \\ 241 & 15137985 & 195998061709305 & 65255 & 3375066455 & −4938 + 172\sqrt{5} \end{matrix}

B.1 Desingularization and a Hecke bound over ℂ

Let H(3) be the Hilbert modular surface obtained as the quotient of the product of two copies of the Poincare upper half plane modulo the action of the congruence group

Γ(3) := \{( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ∈ SL_{2}(𝒪_{F}) : α ≡ δ ≡ 1 (mod 3), β, γ ∈ 3𝒪_{F} \} .

The group Γ(3) has no fixed elliptic points for this action (see [vdG88] page 109); it has 10 non-equivalent cusps. Let \overline{H}(3) be the minimal compactification of H(3) obtained by adding one point for each cusp. The surface \overline{H}(3) is singular at all such points. Denoting by H̃(3) the minimal desingularization of \overline{H}(3), we get that the diagram of the desingularization of \overline{H}(3) at any cusp is the following (see [vdG88], page 193):

Denote by c_{i}, 1 ≤ i ≤ 10 the different cusps (where c_{1} is the cusp at infinity) and denote by S_{i}, 1 ≤ i ≤ 10 the exceptional divisor at the i-th cusp. The surface H̃(3) is of general type (by Theorem 3.4 of [vdG88]). We want a criterion to show that a Hilbert modular form whose Fourier expansion starts with many zeroes is actually the zero modular form. This is a generalization of the Sturm bound to Hilbert modular forms. To this end we need a nef (numerically eventually free) divisor. Let F_{1} be the curve defined in [vdG88] page 88. It has 30 disjoint connected components made of curves with self-intersection number −2 and it meets each connected component of the desingularization at the cusps in three points (see [vdG88], page 193).

Lemma B.1. The intersection numbers between the curves S_{i} and F_{1} are:

• S_{i} · S_{j} = \begin{cases}0 & if i ≠ j, \\ −4 & if i = j. \end{cases}

• F_{1} · F_{1} = −60.

• S_{i} · F_{1} = 12.

Consider the divisor

D′ := \frac{1}{5} (∑_{i=1}^{10} S_{i} + 2F_{1}) .

Lemma B.2. The divisor D′ is nef and it agrees with the canonical divisor. Its self-intersection number is given by

D′ · D′ = 8.

Proof. In example 7.5 of [vdG88] (page 179) it is proved that D′ equals the canonical divisor and its self-intersection number is computed. The fact that it is nef follows from the fact that H̃(3) is a minimal surface of general type.

Remark B.3. Since H̃(3) is a complex smooth projective surface, there is a unique canonical divisor (which equals the dualizing sheaf). Let M_{2k}(Γ(3)) denote the vector space of modular forms of parallel weight 2k for Γ(3). It is the space of global sections of the line bundle 𝒪(k(D′ + S)) = 𝒪(D′ + S)^{⊗k}:

M_{2k}(Γ(3)) = Γ(𝒪(k(D′ + S))),

where S = ∑_{i=1}^{10} S_{i}. Similarly, the space of cusp forms S_{2k}(Γ(3)) is given by the divisor k(D′ + S) − S. We want to add some vanishing conditions to M_{2k}(Γ(3)) such that the space of forms with these vanishing conditions is empty. Let a be a positive integer, and G be a Hilbert modular form. We say that G vanishes with order a at the cusp c_{i} if G is a section of 𝒪(k(D′ + S) − aS_{i}) ⊂ 𝒪(k(D′ + S)). Let G be a form which vanishes with order a at all the cusps and with order a + b at the infinity cusp, i.e. it belongs to the space given by the divisor

E = k(D′ + S) − aS − bS_{1}.

It follows from Lemma B.1 and Lemma B.2 that D′ · S_{i} = 4 for all 1 ≤ i ≤ 10, so

E · D′ = k(D′ · D′ + S · D′) − aS · D′ − bS_{1} · D′ = 48k − 40a − 4b.

If b + 10a > 12k, this intersection number is negative and, since D′ is nef, the space of global sections of 𝒪(E) is the zero space. This implies

Theorem B.4. If G is a Hilbert modular form of parallel weight 2k for Γ(3) which vanishes with order a at all cusps and with order a + b at the infinity cusp and b + 10a > 12k, then G is the zero form.

Corollary B.5. If G is a Hilbert modular form of weight (k_{1}, k_{2}), with k_{1} ≡ k_{2} (mod 2), for Γ(3) which vanishes with order a at all cusps and with order a + b at the infinity cusp and b + 10a > 3(k_{1} + k_{2}), then G is the zero form.

Proof. Just apply the previous Theorem to the form G(z_{1}, z_{2}) · G(z_{2}, z_{1}).

To relate the order of vanishing of a modular form at a cusp with the qexpansion we need to compute explicitly the first step of the desingularization of the cusp. In the case of the infinity cusp, this implies computing the local ring of the cusp, which is done in [vdG88] Chapter II, Section 2. The stabilizer of the infinity cusp for Γ(3) is given by

\{( \begin{matrix}ϵ & α \\ 0 & ϵ^{−1} \end{matrix} ) : α ∈ 3𝒪_{F} and ϵ ≡ 1 (mod 3)\} ,

i.e. it is of type (M, V) = (3𝒪_{F} , U_{F}^{8} ), where U_{F}^{8} = ⟨\frac{47}{2} + \frac{3}{2} \sqrt{5}⟩. The dual of M is given by M^{∨} = \frac{𝒪_{F}^{∨}}{3} = \frac{𝒪_{F}}{3\sqrt{5}}, so any Hilbert modular form for Γ(3) has a q-expansion at the infinity cusp of the form

∑_{\begin{matrix}ξ≫0 \\ ξ∈ \frac{𝒪_{F}}{3\sqrt{5}} \end{matrix}} a_{ξ} exp(ξz_{1} + τ(ξ)z_{2}).

Let M_{+} denote the elements of M which are totally positive, and consider the embedding of M_{+} in (ℝ_{+})^{2}, given by

μ ↦ (μ, τμ).

Denote by A_{k} = (A_{k}^{1}, A_{k}^{2}), k ∈ ℤ the vertices of the boundary of the convex hull of the image of M_{+}, ordered with the condition A_{k+1}^{1} < A_{k}^{1} for all k. Any pair (A_{k−1}, A_{k}) is a basis for M as ℤ-module (see [vdG88] Lemma 2.1). This determines an isomorphism

M\ℂ^{2} → ℂ^{×} × ℂ^{×},

which maps z = (z_{1}, z_{2}) to (u_{k−1}, u_{k}), where

exp(z_{j}) = u_{k−1}^{A_{k−1}^{j}}u_{k}^{A_{k}^{j}}, for j = 1, 2.  (10)

Let σ_{k} denote the cone spanned by A_{k−1} and A_{k}, i.e.

σ_{k} = {sA_{k−1} + tA_{k} : s, t ∈ ℝ_{+}}.

The desingularization of the infinity cusp is obtained by taking a copy of ℂ^{2} for each element σ_{k} and gluing them together in terms of the change of basis matrix (see [vdG88] page 31). Let ξ ∈ M^{∨} be a totally positive element. Then in the copy corresponding to σ_{k},

exp(Tr(ξz)) = exp(ξz_{1} + τ(ξ)z_{2}) = u_{k−1}^{ξA_{k−1}^{1}+τ(ξ)A_{k−1}^{2}} · u_{k}^{ξA_{k}^{1}+τ(ξ)A_{k}^{2}} =
= u_{k−1}^{Tr(ξA_{k−1})} · u_{k}^{Tr(ξA_{k})}.

Now, we denote by L_{k}, the component of the exceptional divisor S_{1}, over the infinity cusp, that corresponds to the ray through A_{k}. Observe that L_{k+4} = L_{k}. We have:

Proposition B.6. Let G be a Hilbert modular form with q-expansion at infinity

G(z_{1}, z_{2}) = ∑_{\begin{matrix}ξ≫0 \\ ξ∈ \frac{𝒪_{F}}{3\sqrt{5}} \end{matrix}} a_{ξ} exp(ξz_{1} + τ(ξ)z_{2}).

Let L_{k} be as above. Then ord_{L_{k}}(G) > K if and only if a_{ξ} = 0 for all ξ ∈ M^{∨}, ξ ≫ 0 with Tr(ξA_{k}) ≤ K.

Name  Point
A_{0}  3(1, 1)
A_{1}  3(1 + ω, 1 + τ(ω))
A_{2} 3(2 + 3ω, 2 + 3τ(ω))
A_{3} 3(5 + 8ω, 5 + 8τ(ω))

Table B.1: First boundary points

In Table B.1 it is shown a set of nonequivalent boundary points of the convex hull of M_{+}, where ω denotes the element \frac{1+\sqrt{5}}{2}. It is clear that they differ

by powers of ω^{2}, and since the matrix ( \begin{matrix}ω^{2} & 0 \\ 0 & ω^{−2} \end{matrix} ) ∈ Γ_{0}(3), a Hilbert modular form for Γ_{0}(3) will vanish with the same order in the four components L_{k}, k = 0, …, 3, of S_{1}. The vanishing condition corresponding to L_{0}, reads

ord_{L_{0}}(exp(ξz)) = 3 Tr(ξ).

In particular, a modular form vanishes at the cusp if and only if a_{0} = 0. The above discussion implies the following theorem for Γ_{0}(3).

Theorem B.7. Let G be a Hilbert modular form of parallel weight 2k for Γ_{0}(3) which vanishes with order a at all cusps and whose Fourier expansion at the infinity cusp is given by

G = ∑_{\begin{matrix}ξ≫0 \\ ξ∈ \frac{𝒪_{F}}{3\sqrt{5}} \end{matrix}} a_{ξ} exp(ξz_{1} + τ(ξ)z_{2}).

If a_{ξ} = 0 for all ξ with Tr(ξ) ≤ 4k − 3a then G is the zero form.

Proof. If a_{ξ} = 0 for all ξ with Tr(ξ) ≤ 4k − 3a, by Proposition B.6, G vanishes with order greater than 12k − 9a at the infinity cusp. Thus the result follows from Theorem B.4.

Corollary B.8. Let G be a Hilbert modular form of weight (k_{1}, k_{2}), with k_{1} ≡ k_{2} (mod 2), for Γ_{0}(3) which vanishes with order a at all cusps and whose Fourier expansion at the infinity cusp is given by

G = ∑_{\begin{matrix}ξ≫0 \\ ξ∈ \frac{𝒪_{F}}{3\sqrt{5}} \end{matrix}} a_{ξ} exp(ξz_{1} + τ(ξ)z_{2}).

If a_{ξ} = 0 for all ξ with Tr(ξ) ≤ (k_{1} + k_{2}) − 3a then G is the zero form.

B.2 Moduli interpretation and integral models

In order to make the computation of the previous section work over finite fields, we need to use the integral structure of the modular Hilbert surface and of the modular curve X(3). It comes from their moduli interpretation. Let us follow the notation of [BBGK07]. We fix ζ_{3} a third-root of unity and we denote by δ = (\sqrt{5})^{−1} the different of F. An abelian scheme A → S of relative dimension 2, together with a ring homomorphism

ι : 𝒪_{F} → End(A)

is called an abelian surface with multiplication by 𝒪_{F}, and is denoted by the pair (A, ι). This gives an 𝒪_{F} -multiplication in the dual abelian surface A^{∨}. An element μ ∈ Hom(A, A^{∨}) is called 𝒪_{F} -linear if μι(α) = ι(α)^{∨}μ for all α ∈ 𝒪_{F}. Denote by 𝒫(A) the sheaf for the étale topology on Sch /S defined by

𝒫(A)_{T} = {λ : A_{T} → A_{T}^{∨} : λ is symmetric and 𝒪_{F} -linear},

for all T → S. The subsheaf 𝒫(A)^{+} is the subsheaf of polarizations in 𝒫(A). The pair (A, ι) is said to satisfy the Deligne-Pappas condition, denoted by (DP), if the canonical morphism of sheaves

A ⊗_{𝒪_{F}} 𝒫(A) ↦ A^{∨}

is an isomorphism. In this case, 𝒫(A) is a locally constant sheaf of projective 𝒪_{F} -modules of rank 1.

Since the class number of 𝒪_{F} is one, we can restrict to consider only 𝒪_{F} - polarizations. An 𝒪_{F} -polarization on a pair (A, ι) is a morphism of 𝒪_{F} -modules ψ : 𝒪_{F} → 𝒫(A)_{S} taking 𝒪_{F}^{+} to 𝒫(A)^{+} such that the natural homomorphism

A ⊗_{𝒪_{F}} 𝒪_{F} → A^{∨}, a ⊗ α ↦ ψ(α)(a)

is an isomorphism. Suppose S is a scheme over Spec ℤ[1/3]. A level 3-structure on an abelian surface A over S with real multiplication by 𝒪_{F} is an 𝒪_{F} -linear isomorphism

φ : (𝒪_{F} /3)_{S}^{2} → A[3]

between the constant group scheme defined by (𝒪_{F} /3)^{2} and the 3-torsion of A.

Theorem B.9. The moduli problem “Abelian surfaces over S with real multiplication by 𝒪_{F} satisfying (DP) condition, 𝒪_{F} -polarization and level 3-structure” is represented by a regular algebraic scheme ℋ (3) which is flat and of relative dimension two over Spec ℤ[1/3, ξ_{3}]. Furthermore, it is smooth over Spec ℤ[1/15, ξ_{3}].

Proof. See [Gor02] Theorem 2.17, p. 57;Lemma 5.5, p. 99.

Remark B.10. The scheme ℋ (3) is not geometrically irreducible, it has #(𝒪_{F} /3)^{×} = 8 connected components over ℚ̄. In fact, the 8 components are defined over Spec ℤ[1/15, ξ_{3}]. This definition is not the same as the one given in [Rap78] (which is connected), it is a topological cover of degree 4 of it. Let S be a scheme over ℤ[1/3], then the abelian scheme A has a Weil pairing e_{3} : A[3] × A^{∨}[3] → μ_{3}, which satisfies e_{3}(αa, b) = e_{3}(a, αb). There exists an 𝒪_{F} -bilinear form e_{𝒪_{F}} : A[3] × A^{∨}[3] → (δ^{−1}/3δ^{−1})(1) such that e_{3} = Tr(e_{𝒪_{F}} ). Any element in λ ∈ 𝒫(A)_{T} defines a homomorphism between A and A^{∨} which is trivial on A[3] if and only if λ ∈ 3𝒫(A)_{T} for any morphism T → S. Since e_{𝒪_{F}} is non-degenerate, 𝒫(A) ⊗_{𝒪_{F}} Λ_{𝒪_{F}}^{2} A[3] = δ^{−1}/3δ^{−1}(1). Any element ϕ ∈ Isom(μ_{3}, ℤ/3) gives an isomorphism between δ^{−1}/3δ^{−1}(1) and δ^{−1}/3δ^{−1}. In [Rap78] the only level 3-structures considered are the φ such that in the following diagram

[figure: 𝒫(A) ⊗_{𝒪_{F}} Λ_{𝒪_{F}}^{2} A[3]; ∼; e_{𝒪_{F}} ⊗ψ; δ^{−1}/3δ^{−1}(1); δ^{−1} ⊗_{𝒪_{F}} Λ_{𝒪_{F}}^{2} (𝒪_{F} /3); δ^{−1}/3δ^{−1}]

the vertical dotted arrow is given by an element ϕ ∈ Isom(μ_{3}, ℤ/3). Since all such maps differ by (multiplication by) an element in (𝒪_{F} /3)^{×}, the two assertions follow.

Remark B.11. The group GL_{2}(𝒪_{F} /3) acts on ℋ (3), where an element M send a level 3-structure φ to the level 3-structure φ ∘ M. The subgroup SL_{2}(𝒪_{F} /3) acts on each connected component of ℋ (3) ⊗ ℚ̄, while the subgroup H_{F} = \{( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) such that α ∈ (𝒪_{F} /3)^{×}} acts transitively on the set of connected components.

Theorem B.12. There is a toroidal compactification h_{3} : ℋ̃(3) → ℤ[ζ_{3}, 1/3] of ℋ (3) that is smooth at infinity. The complement ℋ̃(3)\ℋ (3) is a relative divisor with normal crossings.

Proof. See [Cha90] Theorem 3.6, Theorem 4.3 and [Rap78] Theorem 5.1 and Theorem 6.7.

The set of complex points of ℋ (3) is equal to 8 copies of the surface H(3) considered in the previous section, while the set of complex points of ℋ̃(3) is equal to 8 copies of H̃(3). If we study the moduli problem for 1-dimensional abelian varieties (i.e. elliptic curves), we have the advantage that they are already principally polarized. As in the two-dimensional case, if S is a scheme over ℤ[1/3], a level 3-structure on an elliptic curve E over S is a ℤ-linear isomorphism

φ : (ℤ/3)_{S}^{2} → E[3]

between the constant group scheme defined by (ℤ/3)^{2} and the 3-torsion of E.

Theorem B.13. The moduli problem “elliptic curves over S with level 3structure” is represented by a smooth affine curve 𝒴 (3) over ℤ[1/3]. Furthermore, the category ℳ_{3}[1/3] of “generalized elliptic curves over S, that have smooth generic fibres, singular fibres whose Neron polygons have 3-sides and with a level 3-structure” is a projective smooth scheme 𝒳 (3) over ℤ[1/3].

Proof. See [DR73], Corollary 2.9.

Remark B.14. The group GL_{2}(ℤ/3) acts on 𝒳 (3), in the same way as in the two dimensional case, i.e. an element M ∈ GL_{2}(ℤ/3) sends the level 3-structure φ to the level 3-structure φ ∘ M. The subgroup SL_{2}(ℤ/3) acts on each connected component of 𝒳 (3)⊗ℚ̄ while the subgroup H_{ℚ} = \{( \begin{matrix}α & 0 \\ 0 & 1 \end{matrix} ) such that α ∈ (ℤ/3)^{×}} acts transitively in the set of connected components.

We want to study the inclusions of 𝒴 (3) into ℋ (3). If E is an elliptic curve over S, the abelian variety A_{E} = E ⊗_{ℤ}𝒪_{F} has a canonical 𝒪_{F} -action ι_{E} : 𝒪_{F} → End(A_{E}). Furthermore,

A_{E} ≃ E ×_{S} E,

where the isomorphism depends on chosing a basis for 𝒪_{F} as ℤ-module. The dual abelian variety A_{E}^{∨} is isomorphic to E ⊗_{ℤ} δ^{−1}. Furthermore, 𝒫(A_{E}) ≃ δ^{−1} ≃ 𝒪_{F} (i.e. A_{E} has a canonical principal polarization ψ_{E}), where the isomorphism preserves positivity and the Deligne-Pappas condition holds (see [BBGK07] Lemma 5.10). Let (E, φ) be an elliptic curve with level 3-structure.

Consider the natural inclusion SL_{2}(ℤ/3) ↪ SL_{2}(𝒪_{F} /3), and let 𝒩 = {N_{i}}_{i=1}^{30} be a set of representatives for the quotient set SL_{2}(𝒪_{F} /3)/SL_{2}(ℤ/3). Each element N of 𝒩 gives rise to an embedding of 𝒴 (3) into ℋ (3), which associates to the pair (E, φ) the element (A_{E}, ι_{E}, ψ_{E}, φ ∘ N). For shortness we abuse notation by writing φ ∘ N. The map φ extends by 𝒪_{F} -linearity to a map

φ̃: (ℤ/3)^{2} ⊗_{ℤ} 𝒪_{F} = (𝒪_{F} /3)^{2} → A_{E}[3],

and the element N acts in (𝒪_{F} /3)^{2}. By choosing a connected component of 𝒴 (3) ⊗ ℚ̄ and of ℋ ⊗ ℚ̄, the 30 embeddings obtained from the set 𝒩 (up to composition with an element of H_{F} if necessary) give us open dense subsets of the 30 connected components of the curve F_{1} from the previous section.

Theorem B.15. The closed inmersion 𝒴 (3) ↪ ℋ (3) of schemes over ℤ[ζ_{3}, 1/15] extends to a closed inmersion 𝒳 (3) ↪ ℋ̃(3).

Proof. Following the referee’s advice, we omit the proof of this Theorem.

Theorem B.15 has the following direct consequences. Let ℋ̃be one of the irreducible components of ℋ̃(3) over Spec ℤ[1/15, ξ_{3}]. The set of complex points of ℋ̃ agrees with the surface H̃(3) of section B.1. Let Z be any irreducible component of the divisor D′ of H̃(3) introduced in the same section. Then Z is defined over ℚ(ζ_{3}). Let 𝒵 = \overline{Z} be the Zariski closure of Z in ℋ̃.

Corollary B.16. For every prime 𝔭 of ℤ[1/15, ζ_{3}], the vertical cycle 𝒵_{𝔭} is irreducible.

Proof. If Z is a component of S this follows directly from Theorem B.12. If Z is a component of F_{1} this follows from Theorem B.15 and Theorem B.13.

Let 𝒟′ be the horizontal divisor of ℋ̃determined by D′.

Corollary B.17. For every prime 𝔭 of ℤ[1/15, ζ_{3}], the divisor 𝒟′_{𝔭} of the surface ℋ̃_{𝔭} over Spec ℤ[1/15, ζ_{3}]/𝔭 is nef.

Proof. Since the divisor 𝒟′_{𝔭} is effective, we only have to show that the intersection of 𝒟′_{𝔭} with any of its irreducible components is greater or equal to zero. By Corollary B.16, every irreducible component of 𝒟′_{𝔭} is the specialization of a irreducible component of D′. Thus the result follows from the fact that D′ is nef and that the intersection product is preserved by specialization.

Corollary B.18. Let G be a Hilbert modular form of weight (k_{1}, k_{2}), with k_{1} ≡ k_{2} (mod 2), for Γ_{0}(3) whose coefficients generate a finite field extension L of ℚ. Let 𝔭 be a prime ideal of 𝒪_{L.F [ξ_{3}]} not dividing 15. If G vanishes with order a at all cusps of ℋ̄_{𝔭} and if the Fourier coefficients of its q-expansion at the infinity cusp are algebraic integers satisfying a_{ξ} ≡ 0 (mod 𝔭) for all ξ with Tr(ξ) ≤ (k_{1} + k_{2}) − 3a then all a_{ξ} are divisible by 𝔭.

Proof. From the last Corollary, we have that the divisor 𝒟′_{𝔭} of the surface ℋ̃_{𝔭} over Spec ℤ[1/15, ζ_{3}]/𝔭 is nef, so we can apply the same argument as in the proof of Theorems B.4 and B.7. The result follows from the fact that 𝒟′_{𝔭} is the specialization of the divisor D consider in such theorems and the fact that intersection numbers are preserved by specialization.

Remark B.19. In practice, if one starts with a form whose coefficients vanish at all cusps of the complex Hilbert surface with order at least a (for example if it is a product of cusp forms, as will be the case in the next section) then it also vanishes at all the cusps of ℋ̄_{𝔭} with order at least a.

B.3 The case of the form 𝔤 + μ(𝔤) of level Γ_{0}(12\sqrt{5})

We want to apply the results of the last two sections to the Hilbert cusp form 𝔤 + μ(𝔤) of Section 2, which has weight (2, 2) and level Γ_{0}(12\sqrt{5}). Assuming that its q-expansion at the infinity cusp is zero for all elements of trace smaller than b̃ + 1 (hence the order of vanishing at the four lines L_{i}, i = 1, …, 3 is 3b̃ + 3), we want to determine which value of b̃ forces the form to be the zero form. We start with some general results.

Lemma B.20. Let 𝔭, 𝔮 be two distinct prime ideals of F relatively prime to 3. Then the index of Γ_{0}(3𝔭^{r}𝔮^{s}) in Γ_{0}(3) is

[Γ_{0}(3) : Γ_{0}(3𝔭^{r}𝔮^{s})] = 𝒩 𝔭^{r−1}(𝒩 𝔭 + 1)𝒩 𝔮^{s−1}(𝒩 𝔮 + 1).

In particular,

[Γ_{0}(3) : Γ_{0}(12\sqrt{5})] = 120.  (11)

Let 𝔥 be a modular form of weight (2, 2) for Γ_{0}(12\sqrt{5}), which vanishes with order 1 at all cusps, and with order b̃ + 1 at the infinity cusp. We consider its norm to Γ_{0}(3),

G := ∏_{γ∈ Γ_{0}(12\sqrt{5})\Γ_{0}(3)} 𝔥|_{2}[γ].

It is a parallel weight 2 · 120 Hilbert modular form for Γ_{0}(3). Since 𝔥 is a cusp form, looking at its q-expansion at the different cusps it is easy to see that G vanishes with order at least 16 at all cusps and with order at least 3b̃+96 at the infinity cusp. Then Theorem B.4 and Corollary B.18 (with a = 16, b = 3b̃ + 96 and k = 120) imply that if b̃ ≥ 395 then G is the zero form. So we get

Theorem B.21. Let 𝔥 be a Hilbert modular form of parallel weight (2, 2) for Γ_{0}(12\sqrt{5}). If its Fourier expansion is given by

𝔥 = ∑_{\begin{matrix}ξ≫0 \\ ξ∈𝒪_{F}^{∨} \end{matrix}} a_{ξ} exp(ξz_{1} + τ(ξ)z_{2}).

with a_{ξ} ≡ 0 (mod 2) for all ξ with Tr(ξ) ≤ 395, then a_{ξ} ≡ 0 (mod 2) for all ξ.

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Luis Dieulefait, Departament d’Àlgebra i Geometria Facultat de Matemàtiques Universitat de Barcelona Gran Via de les Corts Catalanes, 585 08007 Barcelona ldieulefait@ub.edu Ariel Pacetti Departamento de Matemática Universidad de Buenos Aires Pabellón I, Ciudad Universitaria.

C.P:1428 Buenos Aires, Argentina apacetti@dm.uba.ar

Matthias Schütt Institut für Algebraische Geometrie Leibniz Universität Hannover Welfengarten 1, 30167 Hannover, Germany schuett@math.uni-hannover.de José Burgos Gil ICMAT-CSIC Nicolas Cabrera 13-15 28049 Madrid, Spain burgos@icmat.es 988

Documenta Mathematica 17 (2012)