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An explicit Jacobian of dimension 3 with maximal Galois action

We gives an explicit genus 3 curve over Q such that the Galois action on the torsion points of its Jacobian is a large as possible. That such curves exist is a consequence of a theorem of D. Zureick-Brown and the author; however, those methods do not produce explicit examples. We shall apply the general strategies of Hall and Serre in their open image theorems. We also make use of Serre's conjecture to show that the modulo ell Galois actions are irreducible. While we computationally focus on a single curve, the methods of this paper can be applied to a large family of genus 3 curves.


Introduction
Consider a principally polarized abelian variety A of dimension g ≥ 1 defined over Q. Fix an algebraic closure Q of Q and define the absolute Galois group G Q := Gal(Q/Q). The Galois action on the torsion points of A(Q) can be expressed in terms of a Galois representation In [ZBZ15], Zureick-Brown and the author prove that for each integer g ≥ 3, there is a principally polarized abelian variety A/Q (in fact the Jacobian of a trigonal curve) such that ρ A (G Q ) = GSp 2g ( Z). For such an abelian variety, the Galois group acts on the torsion points in the most general way possible. Unfortunately, the methods of [ZBZ15] are not useful for constructing examples.
The goal of this paper is to give the first explicit A/Q for which the representation ρ A is surjective, i.e., the Galois group acts on the torsion points of A in the most general way possible.
When A/Q is an elliptic curve, the image of the representation ρ A is an important ingredient in several deep conjectures, for example the Lang-Trotter conjectures [LT76] and the Koblitz conjecture [Zyw11]. Our explicit example should be useful in providing numerical evidence for related higher dimension conjectures.
1.1. The example. Let C be the subscheme of P 2 Q defined by the quartic equation x 3 y − x 2 y 2 + x 2 z 2 + xy 3 − xyz 2 − xz 3 − y 4 + y 3 z − y 2 z 2 − yz 3 = 0. (1.1) The curve C is smooth and hence has genus 3. Let J be the Jacobian of the curve C; it is a principally polarized abelian variety of dimension 6 defined over Q. The Galois action on the torsion points of J is as large as possible.
Remark 1.2. Let A/Q be a principally polarized abelian variety of dimension g ≥ 1. In Proposition 2.5, we will show that if g ≤ 2 or if A is the Jacobian of a hyperelliptic curve, then ρ A is not surjective. This motivates why we have first considered the Jacobian of a smooth plane quartic.
Though we focus only on a specific curve, the methods will also apply to a large class of smooth plane quartics. Indeed, most of this paper can be viewed as describing how to make the criterion of C. Hall in [Hal11] effective. The largest difference from [Hal11] is that we use Serre's conjecture to prove that the modulo ℓ representations are irreducible; this is motivated by the work of Dieulefait [Die02] on abelian surfaces.
1.2. Overview. We now give a brief overview of the contents of this paper; none of the following will be needed later on.
For each prime ℓ, let J[ℓ] be the ℓ-torsion subgroup of J(Q). The natural G Q -action on J[ℓ] can be expressed by a representation ρ J,ℓ : G Q → GSp 6 (F ℓ ), see §2.2 for details. The constraint on the image of ρ J,ℓ arises from the Weil pairing.
We will show (Proposition 2.1) that ρ J is surjective if and only if ρ J,ℓ is surjective for all primes ℓ. So fix any odd prime ℓ (the prime ℓ = 2 can be dealt with separately).
We will see in §3 that the curve C, and hence also J, has good reduction at all primes away from the set S := {7, 11, 83}. Therefore, ρ J,ℓ is unramified at all primes p / ∈ S ∪ {ℓ}. The characteristic polynomial det(T I − ρ J,ℓ (Frob p )) ∈ F ℓ [T ] is the reduction modulo ℓ of a computable polynomial P p (T ) ∈ Z[T ] that does not depend on ℓ.
For p ∈ S with p = ℓ, we will show in §4 that ρ J,ℓ (I p ) is a cyclic group of order ℓ, where I p ⊆ G Q is an inertia subgroup for p. We will prove this by using the Picard-Lefschetz formula along with the fact that the only singularities for our model (1.1) modulo p are double ordinary points. If p ∈ {7, 11}, then ρ J,ℓ (I p ) will be generated by a transvection (an element with determinant 1 that fixes a codimension 1 subspace).
In §5, we shall give constraints on the semi-simplification of the representation ρ J,ℓ | I ℓ .
In §6, we will prove that the representation ρ J,ℓ is irreducible. The most involved case is when the composition factor of J[ℓ] (as an F ℓ [G Q ]-module) with smallest F ℓ -dimension has dimension 2; for this, we make use of Serre's conjecture.
In §7, we will prove that the representation ρ J,ℓ is primitive. More precisely, we show that there are no non-zero subspaces W 1 , . . . , W r of J[ℓ] such that J[ℓ] = W 1 ⊕ · · · ⊕ W r and such that the G Q -action permutes the spaces W 1 , . . . , W r .
Knowing that ρ J,ℓ is irreducible and primitive, and that ρ J,ℓ (G Q ) contains a transvection, we will be able to deduce that ρ J,ℓ is surjective.
Remark 1.3. Instead of using Serre's conjecture for irreducibility, one could use the explicit isogeny theorem of Gaudron and Rémond as done by Lombardo in [Lom15]. This gives an explicit ℓ 0 such that ρ J,ℓ is irreducible for all ℓ ≥ ℓ 0 ; unfortunately, ℓ 0 will be too large to feasibly check the irreducibility for primes ℓ < ℓ 0 . We finally remark that, independently, similar ideas as in this paper have been recently used to show that ρ A,ℓ is surjective for all ℓ > 2, where A is the Jacobian of an explicit genus 3 hyperelliptic curve over Q, cf. [ALS15].
2.1. Symplectic group background. For a commutative ring R, let M be a finitely generated free R-module equipped with a non-degenerate alternating bilinear form , : M × M → R. We define GSp(M ) to be the group of A ∈ Aut R (M ) such that for some mult(A) ∈ R × , we have Av, Aw = mult(A) v, w for all v, w ∈ M . The element mult(A) ∈ R × is called the multiplier of A and gives rise to a homomorphism mult : GSp(M ) → R × .
We call GSp(M ) the group of symplectic similitudes.
The rank of M over R is an even number, say 2g. There is an R-isomorphism between M and R 2g such that the pairing on M agrees with the pairing v, w = v t · J · w on R 2g , where we are viewing v and w as column vectors and J is the 2g × 2g matrix 0 Ig −Ig 0 . This gives an isomorphism between GSp(M ) and GSp 2g (R) := GSp(R 2g ). As before, we have a homomorphism mult : GSp 2g (R) → R × whose kernel, which we denote by Sp 2g (R), is called the symplectic group.

Galois representations.
For each integer n ≥ 1, let J[n] be the n-torsion subgroup of J(Q); it is a Z/nZ-module of rank 2g. There is a natural action of the Galois group G Q on J[n] that respects its group structure. The Weil pairing and the principal polarization of J give a nondegenerate and alternating pairing where µ n is the group of n-th roots of unity in Q. Let χ n : G Q → (Z/nZ) × be the modulo n cyclotomic character, i.e., σ(ζ) = ζ χn(σ) for all σ ∈ G Q and ζ ∈ µ n . The pairing e n satisfies e n (σ(v), σ(w)) = σ(e n (v, w)) = e n (v, w) χn(σ) for all v, w ∈ J[n] and σ ∈ G Q . The Galois action on J[n] can thus be expressed by a Galois representation ρ J,n : G Q → GSp(J[n], e n ) ∼ = GSp 2g (Z/nZ). Note that mult •ρ J,n = χ n . By combining over all n and choosing bases compatibly, we obtain a single Galois representation The character mult •ρ A : G Q → Z × is the cyclotomic character and is thus surjective. The following proposition, which will be proved in §2.4, will let us restrict our attention to the representations ρ J,ℓ .
Proposition 2.1. Let C/Q be a smooth projective and geometrically integral curve of genus g ≥ 3 and let J be its Jacobian. Then ρ J (G Q ) = GSp 2g ( Z) if and only if ρ J,ℓ (G Q ) ⊇ Sp 2g (F ℓ ) for all primes ℓ.
With the above proposition in mind, we now give a criteria for showing that a subgroup of GSp 2g (F ℓ ) contains Sp 2g (F ℓ ). First we need to introduces a few definitions.
Fix a representation G → Aut F ℓ (V ), where V is a finite dimensional F ℓ -vector space. We say that V is reducible (and irreducible otherwise) if there is a non-trivial proper subspace of V that is stable under the G-action. We say that V is imprimitive (and primitive otherwise) if there is an integer r ≥ 2 and non-zero subspaces W 1 , . . . , W r of V such that V = W 1 ⊕ · · · ⊕ W r and such that {σ(W 1 ), . . . , σ(W r )} = {W 1 , . . . , W r } for all σ ∈ G.
For A ∈ Aut F ℓ (V ), let V A=1 be the subspace of V consisting of the vectors that are fixed by A. We say that A is a transvection if V A=1 has codimension 1 in V and det(A) = 1.
Proposition 2.2. Fix an integer g ≥ 2 and an odd prime ℓ. Let G be a subgroup of GSp 2g (F ℓ ) with its natural action on V = F 2g ℓ . Suppose that G contains a transvection and that the action of G on V is irreducible and primitive. Then G ⊇ Sp 2g (F ℓ ).
Proof. Let R be the subgroup of G generated by transvections; it is a subgroup of G ∩ Sp 2g (F ℓ ). We have R = 1 since G contains a transvection by assumption. The group R is normal in G since the conjugate of a transvection is also a transvection.
Fix an irreducible R-submodule W of V . Using that R is normal in G, one can verify that σ(W ) is an R-module for all σ ∈ G. Let H be the group consisting of σ ∈ G for which σ(W ) = W . Using that V is an irreducible G-module, we find that that V = σ∈G/H σ(W ). Lemma 6 of [Hal11], which uses parts of [Hal08], says that we in fact have a direct sum V = ⊕ σ∈G/H σ(W ).
Therefore, V is the direct sum of the subspaces {σ(W ) : σ ∈ G/H} which are permuted by the natural action of G. Since G acts primitively on V by assumption, we deduce that W = V , i.e., V is an irreducible R-module.
The main theorem of Zalesskiȋ and Serežkin in [ZS76] shows that Sp 2g (F ℓ ) contains no proper subgroups that act irreducibly on V and are generated by transvections. Therefore, R = Sp 2g (F ℓ ). The lemma follows since R ⊆ G.
2.3. Compatibility. Take any prime p for which C/Q, and hence also J/Q, has good reduction. Let C p and J p be the reduction of C and J, respectively, modulo p. The abelian variety J p /F p agrees with the Jacobian of C p /F p .
Take any prime ℓ = p. Let be the Galois representation obtained by composing ρ J with the natural projection GSp 2g ( Z) → GSp 2g (Z ℓ ); it can also be obtained by taking the inverse limit of the representations ρ J,ℓ n . The representation ρ J,ℓ ∞ is unramified at p and we have det(T I − ρ J,ℓ ∞ (Frob p )) = P Jp (T ), for some polynomial P Jp (T ) ∈ Z[T ] that does not depend on the choice of ℓ. Here Frob p is an (arithmetic) Frobenius automorphism of p. Let π p : J p → J p be the Frobenius endomorphism of J p /F p . We may also characterize P Jp (T ) as the polynomial in Q[T ] for which P Jp (n) is the degree of the isogeny n − π p for every integer n.
We can also describe the polynomial P Jp (T ) in terms of the zeta function of C p . Recall that the zeta function of C p /F p is the formal power series  Proof. Fix g ≥ 3. Part (i) follows by observing that Sp 2g (Z) is its own commutator subgroup and that the reduction modulo n map Sp 2g (Z) → Sp 2g (Z/nZ) is surjective for all n ≥ 2, cf. [BMS67,§12].
Lemma 2.4. Fix an integer g ≥ 3 and let H be a closed subgroup of Sp 2g ( Z). Suppose that the reduction modulo ℓ map H → Sp 2g (F ℓ ) is surjective for all primes ℓ. Then H = Sp 2g ( Z).
Proof. For each integer n ≥ 2, let H(n) ⊆ Sp 2g (Z/nZ) be the image of H under the reduction modulo n map. We claim that H(n) = Sp 2g (Z/nZ) for all n ≥ 2. The lemma will follow directly from the claim since H is closed.
First suppose that n is a prime power, say n = ℓ e for some prime ℓ and integer e ≥ 1. One can show that there are no proper subgroups of Sp 2g (Z/ℓ e Z) whose image modulo ℓ is the full group Sp 2g (Z/ℓZ) (this for example follows from [Vas04, Theorem 2.2.5] with G = Sp 2g and k = F ℓ ). Since H(ℓ) = Sp 2g (Z/ℓZ) by hypothesis, we deduce that H(ℓ e ) = Sp 2g (Z/ℓ e Z). So the claim holds when n is a prime power. Now suppose that n ≥ 2 is not a prime power. By induction, we may assume that n = m 1 m 2 with m 1 , m 2 ≥ 2 relatively prime such that H(m 1 ) = Sp 2g (Z/m 1 Z) and H(m 2 ) = Sp 2g (Z/m 2 Z). We can thus view H(m) as a subgroup of Sp 2g (Z/m 1 Z) × Sp 2g (Z/m 1 Z) that projects surjectively on each of the two factors. If H(m) = Sp 2g (Z/mZ), then Goursat's lemma (cf. [Rib75, Lemma 3.2]) implies that Sp 2g (Z/m 1 Z) and Sp 2g (Z/m 2 Z) have a common simple group as a quotient; this is impossible by Lemma 2.3(ii). Therefore, H(n) = Sp 2g (Z/nZ). This completes our proof of the claim.
We now prove Proposition 2.1. First suppose that ρ J,ℓ (G Q ) ⊇ Sp 2g (F ℓ ) for all primes ℓ. Let H be the the commutator subgroup of ρ J (G Q ), i.e., the maximal closed normal subgroup of ρ J (G Q ) with abelian quotient. Observe that H is a closed subgroup of Sp 2g ( Z).
Take any prime ℓ.
Finally, the other direction of Proposition 2.1 is easy.
2.5. Further remarks. We now explain the claims from Remark 1.2; we will not use this later on.
Proposition 2.5. Let A/Q be a principally polarized abelian variety of dimension g ≥ 1. Suppose that g ∈ {1, 2} or that A is the Jacobian of a hyperelliptic curve. Then ρ A (G Q ) = GSp 2g ( Z).
where Q ab is the maximal abelian extension of Q. By the Kronecker-Weber theorem, we have Q ab = Q cyc and hence the commutator subgroup of GSp 2g ( Z) is equal to Sp 2g ( Z). In particular, the commutator subgroup of GSp 2g (Z/nZ) is Sp 2g (Z/nZ) for all n ≥ 2. However, when g ∈ {1, 2}, the group GSp 2g (Z/2Z) = Sp 2g (Z/2Z) is solvable and hence its commutator is a proper subgroup of Sp 2g (Z/2Z). Now suppose that A is the Jacobian of a hyperelliptic curve X/Q (of genus g ≥ 3). We have ρ A (G Q ) = GSp 2g ( Z) and hence ρ A,2 (G Q ) = GSp 2g (F 2 ) = Sp 2g (F 2 ). Let P 1 , . . . , P 2g+2 ∈ X(Q) be the Weierstrass points of X; they are the points fixed by the hyperelliptic involution. Let K be the smallest extension of Q for which all the points P 1 , . . . , P 2g+2 lie in X(K). The extension K/Q is Galois and the group Gal(K/Q) is isomorphic to a subgroup of S 2g+2 . One can show that the 2-torsion subgroup of A(Q) is generated by the points represented by the divisors P i − P j . Therefore, ρ A,2 (Gal(Q/K)) = {I} and hence Proceeding by induction on g, one can check that this inequality fails for all g ≥ 3. Therefore, ρ A is not surjective.
The case g = 1 of Proposition 2.5 was first observed by Serre [Ser72,Prop. 22]. Proposition 2.5 need not hold over a general number field when g ∈ {1, 2}. For example, Grecius [Gre10] found an explicit elliptic curve E/k, with k a cubic extension of Q, such that ρ E (Gal(k/k)) = GL 2 ( Z).

Good primes
Define the set of primes S := {7, 11, 83}; these are the primes for which C/Q, and hence J, has bad reduction, cf. Lemma 3.1.

3.1.
Singularities. Let C be the subscheme of P 2 Z defined by the equation (1.1). For each ring R, let C R be the scheme over Spec R obtained by base extending C to R. The curve C/Q is of course C Q . Let Q un p be the maximal unramified extension of Q p in Q p and let Z un p be its local ring. The residue field of Z un p is an algebraic closure F p of F p . Lemma 3.1.
(i) For any prime p / ∈ S, the curve C/Q has good reduction at p. Moreover, C Zp → Spec Z p is smooth and proper.
(ii) Take any prime p ∈ S. The morphism is smooth away from a finite set Σ of points that lie in the special fiber C Fp . The points in Σ are all ordinary double points of C Fp . We have |Σ| = 1 if p ∈ {7, 11} and |Σ| = 2 otherwise. For each P ∈ Σ, the completion of the local ring of C Z un p at P is isomorphic as a Z un palgebra to Z un p [[x, y]]/(xy + p). Proof. Let f (x, y, z) be the polynomial on the left hand side of (1.1). Define the polynomial g(x, y, z) := f (x − 69y − 1389z, y − 64z, z). Since f (x, y, z) = g(x + 69y + 5805z, y + 64, z), there is no harm in assuming instead that C is the subscheme of P 2 Z defined by g(x, y, z) = 0.
Let I be the the ideal of Z[x, y, z] generated by g and the partial derivatives g x , g y and g z . Let I ′ be the saturation of I (with respect to the irrelevant ideal of Z[x, y, z]). One can show, as in the following Magma code, that R<x,y,z>:=PolynomialRing(Integers(),3); f:=x^3*y-x^2*y^2+x^2*z^2+x*y^3-x*y*z^2-x*z^3-y^4+y^3*z-y^2*z^2-y*z^3; g:=Evaluate(f,[x-69*y-1389*z,y-64*z,z]); I:=ideal<R|[g,Derivative(g,x),Derivative(g,y),Derivative(g,z)]>; Saturation(I) eq ideal<R|[6391,x,83*y,y^2+11*y*z+616*z^2]>; That the primes divisors of 6391 are the elements of S is enough to show that C is smooth away from the fibers over the primes p ∈ S. Part (i) is an immediate consequence.
From our description of I ′ , we find that the only singular points of C are: • the point (0 : 0 : 1) in the fiber C F 7 , • the point (0 : 0 : 1) in the fiber C F 11 , • the points (0 : 32 : 1) and (0 : 40 : 1) in the fiber C F 83 . Take any singular point P of C. Take a prime p ∈ S and integer y 0 ∈ {0, 32, 40} such that P is the image of (0 : y 0 : 1) modulo p. Define the polynomial The completion of the local ring of is also non-degenerate, and a ≡ a 1 ≡ a 2 ≡ 0 (mod p) with a ≡ 0 (mod p 2 ).
Proposition 2.4 of [FK88,III] shows that we have an isomorphism of Z p -algebras More precisely, the proof of Proposition 2.4 of [FK88,III] shows that there are α 1 , α 2 ∈ pZ p and hence b has p-adic valuation 1 since a has p-adic valuation 1 and a 1 , a 2 , α 2 , α 2 ∈ pZ p . Since Z un p is strictly Henselian, there is a matrix A ∈ GL 2 (Z un p ) such that Q ′ (A 1,1 x+A 1,2 y, A 2,1 x+ A 2,2 y) = xy, cf. Proposition 2.2 of [FK88,III]. We deduce that the completion of the local ring of C Z un p is isomorphic as a Z un p -algebra to Z un p [[x, y]]/(xy + b). After replacing x by itself times an appropriate unit of Z un p , we may further assume that b = p.
3.2. Frobenius polynomials. Now take any prime p / ∈ S. Let C p be the curve in P 2 Fp defined by (1.1). By Lemma 3.1, we find that C p /F p is a smooth projective curve of genus 3. The abelian variety J thus has good reduction modulo p and its reduction J p /F p is equal to the Jacobian of C p .
Proof. Define G := ρ J,2 (G Q ); it is a subgroup of GSp 6 (F 2 ) = Sp 6 (F 2 ). Consider an odd prime p / ∈ S and let f p (x) ∈ F 2 [x] be the reduction of P p (x) modulo 2. Assume that f p (x) is separable and hence it is also the minimal polynomial of g p := ρ J,2 (Frob p ). Therefore, the order of g p is the smallest integer n p ≥ 1 for which f p (x) divides x n − 1 ∈ F 2 [x]. From §3, we find that f 23 (x) = x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 and f 73 (x) = (x 2 + x + 1)(x 4 + x 3 + x 2 + x + 1). One can then check that n 23 = 7 and n 73 = 15, and thus G contains elements of order 7 and 15.
A computation shows that Sp 6 (F 2 ) has no maximal subgroups with elements of order 7 and 15; therefore, G = Sp 6 (F 2 ). Moreover, any maximal subgroup of G that has order divisible by 7 · 15 is isomorphic to S 8 and hence has no element of order 15 (this can be easily deduced from the description of maximal subgroups of Sp 6 (F 2 ) in [CCN + 85, p.46]).

Inertia at bad primes
For a prime p ∈ S, let I p be an inertia subgroup of G Q at the prime p. The goal of this section is to prove the following using the Picard-Lefschetz formula.
Fix a prime p ∈ S. Set R = Z un p and let K = Q un p be its quotient field. Fix an algebraic closure K of K. With a choice of embedding Q ֒→ K, the restriction map gives an injective homomorphism G K := Gal(K/K) ֒→ G Q that we can view as an inclusion. The group G K is then conjugate to I p in G Q .
Fix a prime ℓ = p. So to prove Proposition 4.1, we need only consider the action of G K on J[ℓ] ⊆ J(K). Define the F ℓ -vector space V := H 1 (C K , F ℓ ); for background onétale cohomology, see [Mil80], [FK88] or [Del77]. There is a natural action of G K on V that we can express in terms of a representation ρ : G K → Aut(V ).
Let F ℓ (1) be the group of ℓ-th roots of unity in K and let F ℓ (−1) be the F ℓ -dual of F ℓ (1). Evaluation gives a natural isomorphism One knows that J[ℓ] is isomorphic to theétale cohomology group H 1 (C K , F ℓ (1)) as an F ℓ [G K ]module. The group G K acts trivially on F ℓ (1) since ℓ = p, so J[ℓ] and V are isomorphism F ℓ [G K ]modules. Therefore, ρ and ρ J,ℓ | G K are isomorphic representations. It thus suffices to prove that ρ(G K ) is a group of order ℓ and that it is generated by a transvection when p ∈ {7, 11}.
We have an alternating pairing where we are composing the cup product and trace map. The G K -action on V respects the pairing , , i.e., σ(v), σ(w) = σ( v, w ) = v, w for all σ ∈ G K and v, w ∈ V . The pairing , is non-degenerate (after taking Tate twists, we can identify this pairing with the Weil pairing on J[ℓ]).
The morphism C R → Spec R is a proper and flat morphism of relative dimension 1. From Lemma 3.1(ii), we know that C K is smooth and that C Fp is smooth away from a finite set Σ of ordinary double points.
The Picard-Lefschetz formula (see [SGA7-II, XV Théorème 3.4]) shows that there are non-zero and pairwise orthogonal vanishing cycles {δ x } x∈Σ in V such that for v ∈ V and σ ∈ G K , we have where ε x : G K → F ℓ (1) is a certain homomorphism and where we view ε x (σ) v, δ x as an element of F ℓ (1) ⊗ F ℓ F ℓ (−1) = F ℓ .
Let ε : G K → F ℓ (1) be the surjective homomorphism that satisfies σ( ℓ √ p) = ε(σ) ℓ √ p for all σ ∈ G K . From Lemma 3.1(ii), we find that the completion of the local ring of C R at a point x ∈ Σ is isomorphic as an R-algebra to R[[x, y]]/(xy + p). From [SGA7-II, XV §3.3], we find that ε x = ε (in general, you would need to raise ε to some power). Therefore, for v ∈ V and σ ∈ G K . The representation ρ thus factors through the order ℓ group Gal(K( ℓ √ p)/K).
Therefore, ρ(G K ) is a group of order 1 or ℓ. Since ε = 1 and v → v, δ x is non-trivial, we deduce from (4.1) that ρ(G K ) is a non-trivial group and hence is a cyclic of order ℓ. Now suppose that p ∈ {7, 11}. Fix any σ 0 ∈ G K with ρ(σ 0 ) = 1. It remains to prove that ρ(σ 0 ) is a transvection. Since ρ(σ 0 ) = 1 has order ℓ, it suffices to prove that σ 0 fixes an F ℓ -subspace of V of dimension dim F ℓ V − 1. Since p ∈ {7, 11}, we have |Σ| = 1 by Lemma 3.1(ii). We thus have where x is the unique element of Σ. Let W be the subspace of V consisting of v ∈ V for which v, δ x is trivial; it has F ℓ -dimension dim F ℓ V − 1 since the pairing is non-degenerate and δ x = 0. By (4.2), we deduce that σ 0 (v) = v for all v ∈ W . This completes the proof of Proposition 4.1.

Inertia at ℓ
Fix an odd prime ℓ / ∈ S and let I ℓ be any inertia subgroup of G Q = Gal(Q/Q) at the prime ℓ. In this section, we give some information on how I ℓ acts on J[ℓ].
We first need to recall some background on tame inertia groups and tame inertia weights, see §1 of [Ser72] for more details. Let P ⊆ I ℓ be the wild inertia subgroup of I ℓ ; it is the largest pro-ℓ subgroup of I ℓ . The quotient I t ℓ := I ℓ /P is the tame inertia group for the prime ℓ. For an integer d ≥ 1 relatively prime to ℓ, let µ d be the d-th roots of unity in Q. The map is a surjective homomorphism which factors through I t ℓ . Taking the inverse limit over all d relatively prime to ℓ (ordered by divisibility), we obtain an isomorphism By composing the homomorphism θ d with reduction modulo a place of Q lying over ℓ, we obtain a character I t ℓ → F × ℓ . For an integer m ≥ 1, setting d := ℓ m − 1 gives a surjective character φ : I t ℓ → F × ℓ m . The fundamental characters of level m are the m characters I t ℓ → F × ℓ m obtained by composing φ with the isomorphisms of F × ℓ m arising from field automorphisms of F ℓ m ; they are φ, φ ℓ , . . . , φ ℓ m−1 .
Let V be an irreducible F ℓ [I ℓ ]-module and set m := dim F ℓ (V ). There is then an isomorphism V ∼ = F ℓ m of F ℓ -vector spaces such that the induced character has image in F × ℓ m , where the first map gives the action of I ℓ on V (here we use scalar multiplication to identify F × ℓ m with a subgroup of Aut F ℓ (F ℓ m )). This representation arising from V thus factors through a character α : I t ℓ → F × ℓ m . Given a fundamental character φ : I t ℓ → F × ℓ m of level m, there are unique integers 0 ≤ e 1 , . . . , e m ≤ ℓ − 1 such that (5.1) α = φ e 1 +e 2 ℓ+···+emℓ m−1 .
The integers e 1 , . . . , e m are called the tame inertia weights of V . Let V be an F ℓ [I ℓ ]-module with V a finite dimensional F ℓ -vector space. Let V 1 , . . . , V r be the composition factors of V as an F ℓ [I ℓ ]-module. An integer is a tame inertia weight for V if it is a tame inertia weight for at least one of the V i .
We now consider the representations occurring in this paper. Proof. This follows from work of Raynaud, cf. [Ray74, Corollaire 3.4.4]. One could also deduce this from [Car08].
The following lemma gives some consequences of Proposition 5.1 that we will use later. Recall that χ ℓ : G Q → F × ℓ was defined in §2.2.
Lemma 5.2. Fix an odd prime ℓ / ∈ S and let V be any composition factor of the F ℓ [I ℓ ]-module J[ℓ]. Set m := dim F ℓ V and let ρ : I ℓ → Aut F ℓ (V ) be the representation describing the I ℓ -action.
The character det •ρ : is the norm map. Therefore, where we have used that N ℓ = N . We have 0 ≤ e 1 + . . . + e m ≤ m, so it suffices to prove that ..+ℓ m−1 which one can check is the (unique) fundamental character of level 1. Part (i) follows by noting that the fundamental character of level 1 is We now prove (ii). Set T := F × ℓ m and define the representation x → x e 1 +e 2 ℓ+···+emℓ m−1 . The representation ρ is isomorphic to the one obtained by composing the surjective character We can now use the rigidity of tori as described by Hall in [Hal11]. In the language of [Hal11, §2], the amplitude of β is max{e i } which in our case is 0 or 1. Lemma 3 of [Hal11] and our condition [T : S] < ℓ − 1 implies that β(T ) and β(S) have the same centralizer in Aut F ℓ (F ℓ m ).
Suppose that β| S : S → Aut F ℓ (F ℓ m ) is reducible. Since ℓ ∤ |S|, we have F ℓ m = W 1 ⊕ W 2 , where W 1 and W 2 are non-zero F ℓ -subspaces fixed under the action of S. Take A ∈ Aut F ℓ (F ℓ m ) such that A(w) = w for w ∈ W 1 and A(w) = −w for w ∈ W 2 . Since A commutes with β(S), we deduce that it also commutes with β(T ). For any B ∈ β(T ) and w ∈ W 1 , we have A(Bw) = B(Aw) = Bw. Therefore, B(W 1 ) ⊆ W 1 for all B ∈ β(T ) which contradicts that β is irreducible. Therefore, the representation β| S , and hence also ρ| H , is irreducible.

Irreducibility
For a prime ℓ, the F ℓ -vector space J[ℓ] has dimension 6 and comes with a natural G Q -action. The goal of this section is to prove the following: Suppose that there is an odd prime ℓ such that J[ℓ] is a reducible F ℓ [G Q ]-module; we will try to obtain a contradiction. We first exclude a few possibilities for ℓ.
We have computed P p (T ) for several small p, cf. §3. The polynomial P 17 (x) is irreducible modulo 3. The polynomial P 41 (x) is irreducible modulo 5. The polynomial P 2 (x) is irreducible modulo 7, 11 and 41. The polynomial P 19 (x) is irreducible modulo 83.
6.1. Determinants. Fix a finite dimensional F ℓ -vector space W with an action of G Q given by a representation ρ : G Q → Aut F ℓ (W ). Let W ∨ to be the dual vector space of W and let ρ * : G Q → Aut F ℓ (W ∨ ) be the contragredient representation, i.e., ρ * (σ) is the transpose of ρ(σ −1 ). Let W ∨ (1) be the vector space W ∨ where G Q acts via the representation χ ℓ · ρ * .
The following lemma constrains the possibilities for the characters det •ρ i : (iv) If V ∨ i (1) ∼ = V i , then d i is even and e i = d i /2. Proof. Fix an integer 1 ≤ i ≤ r. The semi-simplification of V i as an F ℓ [I ℓ ]-module is of the form W i,1 ⊕ · · · ⊕ W i,s , where W i,j is an irreducible F ℓ [I ℓ ]-module. By Lemma 5.2(i), the determinant of the action of I ℓ on W i,j is a character I ℓ → F × ℓ equal to χ e i,j ℓ | I ℓ for some integer 0 ≤ e i,j ≤ dim F ℓ W i,j . Therefore, where The representation ρ J,ℓ , and hence also α i , is unramified at all primes p / ∈ S ∪ {ℓ}. Since the order of F × ℓ is relatively prime to ℓ, Proposition 4.1 implies that α i is also unramified at the primes p ∈ S. The character α i is unramified at the prime ℓ by (6.1). We thus have α i = 1 since α i : G Q → F × ℓ is unramified at all primes and Q has no non-trivial extensions unramified at all primes. Therefore, det •ρ i = χ e i ℓ . This proves the existence of e i in (i); it remains to prove the uniqueness. Take any integer 0 ≤ f ≤ d i such that det •ρ i = χ f ℓ . We thus have χ f −e i ℓ = 1 and hence f − e i ≡ 0 (mod ℓ − 1) since χ ℓ has order ℓ − 1. We have d i ≤ 3 since r ≥ 2, so |f − e i | ≤ 3. Since |f − e i | ≤ 3 < ℓ − 1 and f − e i ≡ 0 (mod ℓ − 1), we must have f = e i .
We now prove part (iii). Fix an integer 1 ≤ i ≤ r. The representation (1) and its determinant is given by It remains to prove part (iv).
, then the computation above shows that det •ρ i is equal to both χ e i ℓ and χ d i −e i ℓ . Therefore, e i = d i − e i by the uniqueness in part (i) and hence d i = 2e i . 6.2. One-dimensional case. Suppose that d 1 = 1. The Galois action on V 1 is described by the character ρ 1 : So for any prime p / ∈ S ∪ {ℓ}, ρ 1 (Frob p ) is a root of P p (T ) ≡ det(T I − ρ J,ℓ (Frob p )) (mod ℓ). The character ρ 1 is 1 or χ ℓ by Lemma 6.3(i), so P p (1) ≡ 0 (mod ℓ) or P p (p) ≡ 0 (mod ℓ).
This completes our proof that the case d 1 = 1 does not occur.
After possibly renumbering the V i , we may assume by Lemma 6.4 that has determinant χ ℓ . The following lemma uses Serre's conjecture to relate ρ 1 to a newform of weight 2 and bounded level.
Lemma 6.5. There exists a newform f = q+ n≥2 a n (f )q n ∈ S 2 (Γ 0 (N )) with N dividing 7·11·83 = 6391 and a maximal ideal λ of the ring of integers of the number field Q(a n (f )) such that for all primes p / ∈ S ∪ ℓ.
Proof. The 2-dimensional representation ρ 1 is irreducible and is also odd since det ρ 1 = χ ℓ . Serre's conjecture [Ser87], proved by Khare and Wintenberger [KW09, KW09b], implies that the representation ρ 1 is isomorphic to one arising from some newform f . Moreover, the newform f = q + n≥2 a n (f )q n can be found in S k (Γ 1 (N )) with prescribed weight k and level N . Let K be the subfield of C generated by the Fourier coefficients of f ; it is a number field. Let ε : (Z/N Z) × → K × be the nebentypus of f . There is thus a maximal ideal λ of the ring of integers of K such that for all primes p ∤ N ℓ.
We now consider the level N . The representation ρ J,ℓ , and hence also ρ 1 , is unramified at all primes p / ∈ S∪{ℓ}. Take any p ∈ S; we have p = ℓ by Lemma 6.2. Let V Ip 1 be the subspace of V 1 fixed by I p . From [Ser87, §1.2], the Artin conductor of ρ 1 is N : there is no wild ramification since the cardinality of ρ J,ℓ (I p ) is not divisible by p by Lemma 4.1. Since ρ J,ℓ (I p ) is a group of order ℓ by Lemma 4.1, the group ρ 1 (I p ) has order 1 or ℓ. If ρ 1 (I p ) has order 1, then n p = 0. If ρ 1 (I p ) has order ℓ, then it is conjugate in Aut F ℓ (V 1 ) ∼ = GL 2 (F ℓ ) to the group generated by ( 1 1 0 1 ). In this last case, we have n p = 1. This completes the proof that N divides 7 · 11 · 83 = 6391.
Finally, it remains to show that f ∈ S 2 (Γ 1 (N )) actually lies in S 2 (Γ 0 (N )); equivalently, that the nebentypus ε is trivial. Let µ be the image of ε; it is a finite group of roots of unity in K.
Take any prime p / ∈ S ∪ {ℓ}. Let H p (x) be the characteristic polynomial of the Hecke operator T p acting on S 2 (Γ 0 (6391)); it is monic with integer coefficients. Take f and λ as in Lemma 6.5. Since p ∤ 6391, there is a cusp form f ′ ∈ S 2 (Γ 0 (6391)) such that T p (f ′ ) = a p (f )f ′ ; we can take f ′ to be an oldform if N properly divides 6391. Therefore, H p (a p (f )) = 0 and in particular H p (a p (f )) ≡ 0 (mod λ). Lemma 6.5 then implies that tr(ρ 1 (Frob p )) ∈ F ℓ is a root of H p (x).
A computation show that Q 2 (x) = x 3 + 3x 2 − 3 for Q 5 (x) = x 3 + 4x 2 − 5x − 23. For example, the following code gives Q 2 (x); one could also compute Q 2 (x) using approximations for the roots of P 2 (T ) in C and use that Q 2 (x) has integer coefficients. Let r p be the resultant of H p (x) and Q p (x); it is an integer. Since tr(ρ 1 (Frob p )) ∈ F ℓ is a common root of H p (x) and Q p (x), we deduce that ℓ divides r p .
The Magma code below shows that the greatest common divisor of r 2 and r 5 is 3 16 .
This shows that the case d 1 = 3 does not occur.

Primitivity
In this section, we prove the following: Proposition 7.1. The action of G Q on J[ℓ] is primitive for all odd primes ℓ.
The action of G Q on the set {W 1 , . . . , W r } can be expressed as a representation ϕ : G Q → S r , i.e., σ(W i ) = W ϕ(σ)i for 1 ≤ i ≤ r and σ ∈ G Q .
Lemma 7.3. The representation ϕ is unramified at all primes p.
Take any i ∈ {1, . . . , r}. The group H acts on W i so there is an irreducible H-submodule W i of W i . Define V i := σ∈I ℓ σ(W i ); it is an irreducible I ℓ -module. Lemma 5.2(ii) and [I ℓ : H] < ℓ − 1 implies that V i = W i . For any σ ∈ I ℓ , we have σ(W i ) = W i ⊆ W i . Since I ℓ permutes the spaces W 1 , . . . , W r , we deduce that σ(W i ) = W i for all σ ∈ I ℓ . Since i was arbitrary, we find that I ℓ acts on all the spaces W i and hence ϕ(I ℓ ) = 1.
Since Q has no non-trivial extensions unramified at all primes, Lemma 7.3 implies that ϕ = 1. Therefore, σ(W i ) = W i for all σ ∈ G Q and 1 ≤ i ≤ r. However, this implies that the action of G Q on J[ℓ] is reducible which contradicts Proposition 6.1. Therefore, the action of G Q on J[ℓ] is in fact primitive and this completes the proof of Proposition 7.1. 8. Proof of Theorem 1.1 Take any odd prime ℓ. The group ρ J,ℓ (G Q ) ⊆ GSp 6 (F ℓ ) contains a transvection by Proposition 4.1. By Propositions 6.1 and 7.1, the representation ρ J,ℓ is irreducible and primitive. By Proposition 2.2, we deduce that ρ J,ℓ (G Q ) ⊇ Sp 6 (F ℓ ). We also have ρ J,2 (G Q ) = GSp 6 (F 2 ) by Lemma 3.2.