On the local-global principle for isogenies of abelian surfaces

Let $\ell$ be a prime number. We classify the subgroups $G$ of $\operatorname{Sp}_4(\mathbb{F}_\ell)$ and $\operatorname{GSp}_4(\mathbb{F}_\ell)$ that act irreducibly on $\mathbb{F}_\ell^4$, but such that every element of $G$ fixes an $\mathbb{F}_\ell$-vector subspace of dimension 1. We use this classification to prove that the local-global principle for isogenies of degree $\ell$ between abelian surfaces over number fields holds in many cases -- in particular, whenever the abelian surface has non-trivial endomorphisms and $\ell$ is large enough with respect to the field of definition. Finally, we prove that there exist arbitrarily large primes $\ell$ for which some abelian surface $A/\mathbb{Q}$ fails the local-global principle for isogenies of degree $\ell$.


Introduction
Let K be a number field and A be an abelian variety over K. For all primes v of K we denote by F v the residue field at v, and -if A has good reduction at v -we write A v for the reduction of A modulo v. If A/K has some kind of global level structure (say, a K-rational isogeny or a K-rational torsion point), then so do all the reductions A v . Localglobal principles ask about the converse: if A v has some level structure for (almost) all v, is the same true for A/K? A question of this form was first raised by Katz [Kat81], who considered the property |E(K) tors | ≡ 0 (mod m) when E is an elliptic curves and m is a fixed positive integer (if m = ℓ is prime, this is equivalent to asking that E(K) contains a non-trivial ℓ-torsion point). He showed that this property does not satisfy the local-global principle, but also proved [Kat81, Theorem 2] that, if |E(K v ) tors | ≡ 0 (mod m) for almost all v, then E is isogenous over K to an elliptic curve E ′ with |E ′ (K) tors | ≡ 0 (mod m).
Seen in this light, the local-global principle for the existence of isogenies is perhaps more natural, because the existence of isogenies is itself an isogeny invariant. In this paper, we consider in particular the local-global problem for (prime-degree) isogenies of abelian surfaces. The analogous question for abelian varieties of dimension one, namely elliptic curves, has received much attention in recent years [Sut12,Ann14,Vog20], and is Thus, the situation for abelian surfaces is strikingly different from that of elliptic curves, for which [Ann14] provides a uniform bound for every fixed number field. In addition to showing that no such uniform bound exists in the case of abelian surfaces, Proposition 6.28 is significant also for another reason, namely, it helps explain where the difficulty lies in proving Conjecture 2.2. Indeed, the latter is a statement about Galois representations, and in order to prove it one should in particular show that -for ℓ large enough -the mod-ℓ Galois representation attached to a non-CM abelian surface A/K is non-isomorphic to the Galois representation attached to certain CM abelian surfaces. This is a notoriously difficult problem, so we suspect that a full solution to Conjecture 2.2 is out of reach at present.
Computer calculations. While writing this paper, we have often relied on the computer algebra software MAGMA to double-check our results. However, our proofs are independent of computer calculations, except for the precise list of groups given in Table  1 and for the proof of Theorem A.1 in the Appendix. All the MAGMA scripts to verify these results are available online [LV22]. The same repository also contains tables of the maximal Hasse subgroups of Sp 4 (F ℓ ) for ℓ < 100. These tables are obtained by a direct computation independent from the results in this paper, and agree in all cases with Table  3.1.

Notation
Throughout the paper, K denotes a number field and A an abelian surface over K. We write G K for the absolute Galois group of K, and denote by G ℓ the image of the natural Galois representation ρ ℓ : where we will usually fix an F ℓ -basis of A[ℓ] and therefore identify Aut(A[ℓ]) with GL 4 (F ℓ ). We let χ ℓ : G K → F × ℓ denote the mod-ℓ cyclotomic character. Let k be a field and n be a positive integer. For a subgroup G of GL n (k), we denote by PG the image of G under the canonical projection GL n (k) → PGL n (k). Given a matrix M ∈ GL n (k), we write M −T for the inverse of the transpose of M. As it is well-known, this is also the transpose of the inverse of M.
We say that a matrix M ∈ GL 4 (F ℓ ) is block-diagonal if it is of the form M = x 0 0 y with x, y ∈ GL 2 (F ℓ ). If M is block-diagonal and x and y are scalar multiples of the identity, then we say that M is block-scalar. Moreover, we say that M is block-anti-diagonal if it is of the form M = 0 x y 0 with x, y ∈ GL 2 (F ℓ ).
Definition 1.5. For a choice of symplectic form on F 4 ℓ , represented by a matrix J, we set GSp 4 (F ℓ ) = M ∈ GL 4 (F ℓ ) | ∃k ∈ F × ℓ such that M T JM = kJ . Given M ∈ GSp 4 (F ℓ ), there is a unique k ∈ F × ℓ such that M T JM = kJ: we call it the multiplier of M, and denote it by λ(M). The map M → λ(M) is a group homomorphism, whose kernel is denoted Sp 4 (F ℓ ).
We will use several choices of symplectic forms. The two main ones correspond to the matrices  (2)

Structure of the paper
In Section 2 we collect some preliminary observations about counterexamples to the localglobal principle for isogenies between abelian surfaces and formulate a conjecture about the boundedness of counterexamples for a given number field. We also briefly review some well-known facts about GL 2 (F ℓ ) and its subgroups. In Section 3 we classify the maximal Hasse subgroups of Sp 4 (F ℓ ), and in Section 4 we study the Hasse subgroups H of GSp 4 (F ℓ ) with the property that H ∩ Sp 4 (F ℓ ) acts reducibly. Combining these results, in Section 5 we obtain a classification of the maximal Hasse subgroups of GSp 4 (F ℓ ). Finally, Section 6 contains our main arithmetical results about abelian surfaces: we give sufficient conditions (in terms of the field of definition of the endomorphisms of A) that ensure that (A, ℓ) is not a strong counterexample, and provide an infinite family of counterexamples (A/Q, ℓ) with ℓ unbounded.

Endomorphism rings and algebraic monodromy groups
Let A be an abelian surface over a number field K. By the classification of the geometric endomorphism algebras of abelian surfaces, one of the following holds: 1. A is geometrically irreducible: (a) Trivial endomorphisms: End K (A) = Z.
(b) Real multiplication: End K (A) ⊗ Z Q is a real quadratic field.
(c) Quaternion multiplication: End K (A)⊗ Z Q is a non-split quaternion algebra over Q.
(d) Complex multiplication: End K (A) ⊗ Z Q is a quartic CM field.

2.
A is geometrically reducible: (e) A K is isogenous to the product of two non-isogenous elliptic curves E 1 , E 2 . This gives rise to three sub-cases, according to whether none, one, or both of E 1 , E 2 have CM.
(f) A K is isogenous to the square of an elliptic curve without CM.
(g) A K is isogenous to the square of an elliptic curve with CM.
We now describe certain predictions on strong counterexamples (A/K, ℓ) that follow from well-established conjectures on Galois representations. Denote by T ℓ A = lim ← −n A[ℓ n ] the ℓ-adic Tate module of A, and by G ℓ the ℓ-adic monodromy group of A, namely, the Zariski closure inside GL T ℓ (A)⊗Q ℓ of the image of the ℓ-adic Galois representation Gal K/K . By general conjectures on Galois representations, one expects |G ℓ | to differ at most by a fixed multiplicative constant from [G ℓ : G 0 ℓ ]ℓ dim G 0 ℓ , where in cases (a)-(g) above dim G 0 ℓ is 11, 7, 4, 3, 7 or 5 or 3, 4, 2 respectively, see [FKRS12]. More precisely, G ℓ is by definition a subgroup of G ℓ (F ℓ ), which for ℓ > 2 is a group of order [G ℓ : G 0 ℓ ] · |G 0 ℓ (F ℓ )|, and one knows that asymptotically |G 0 ℓ (F ℓ )| ∼ ℓ dim G 0 ℓ , see [HR12, Proposition 2.2]. In particular, we see that the ratio |G ℓ | [G ℓ : G 0 ℓ ] · ℓ dim G 0 ℓ is bounded above by a universal constant; it is also bounded away from zero because the Mumford-Tate conjecture holds for abelian surfaces. One may then conjecture that, for a fixed number field K, there exists a uniform lower bound c(K) such that for every abelian surface A/K and every prime ℓ we have Remark 2.1. This conjecture does not seem to appear in print in this form. However, at least in the case of abelian surfaces, the results of [Lom16a,Lom16b,Lom17] imply that the existence of c(K) would follow from the uniform boundedness of the degrees of minimal isogenies for abelian varieties of a fixed dimension over a number field of fixed degree. This latter statement has been conjectured by many authors, and is closely related to many other well-known uniformity conjectures, see [Ré18].
On the other hand, if (A/K, ℓ) is a strong counterexample to the local-global principle for cyclic isogenies of abelian surfaces, Lemma 1.2 and Theorem 5.5 show that |G ℓ | is bounded above by an absolute constant f times ℓ 3 : if we assume that (3) holds, we obtain which is only possible if ℓ is 'small' (that is, bounded above by a constant depending only on K) or dim G 0 ℓ ≤ 3. In turn, this latter inequality is satisfied only in cases (d), (e) and (g), and we show in Theorem 6.23 and Lemma 6.24 that -for a fixed number field Kcounterexamples in cases (d) and (e) arise only for finitely many primes ℓ (in fact, case (e) gives no counterexamples at all). This suggests the following conjecture: Conjecture 2.2. For every number field K there is a constant b = b(K) such that, for all primes ℓ > b(K) and for all strong counterexamples (A, ℓ) to the local-global principle for isogenies of prime degree between abelian surfaces, A is geometrically isogenous to the square of an elliptic curve with complex multiplication.
We make some progress on this conjecture in Theorem 6.1, and show in Proposition 6.28 that the case of A being geometrically isogenous to the square of a CM elliptic curve does need to be excluded if we aim for a uniform bound on ℓ. We remark explicitly that, while we make significant headway on this conjecture for all cases when End K (A) = Z, our methods do not allow us to say much for generic surfaces (that is, those with End K (A) = Z). It should be pointed out that even finding examples of violations of the local-global principle for isogenies of generic abelian surfaces seems very hard, and the examples in [Ban21] are all non-generic.

Invariance under isogeny
We now show that the property of being a strong counterexample is an isogeny invariant.
Lemma 2.3. Let (A/K, ℓ) be a strong counterexample to the local-global principle for isogenies of abelian surfaces. Let B/K be an abelian surface that is K-isogenous to A. There exists an isogeny φ : A → B with ℓ ∤ deg φ.
Corollary 2.4. Let K be a number field and A/K be an abelian surface. Suppose that (A, ℓ) is a strong counterexample and that B/K is an abelian variety K-isogenous to A: then (B, ℓ) is also a strong counterexample.
Proof. By Lemma 2.3, there exists an isogeny ϕ : A → B of degree not divisible by ℓ. It induces an isomorphism A[ℓ] ∼ = B[ℓ] of G K -modules. Since the property of being a strong counterexample depends only on the image of the mod-ℓ Galois representation (Lemma 1.2), the claim follows.
In particular, we obtain that, when (A, ℓ) is a strong counterexample, G ℓ preserves a non-trivial symplectic form, even if A is not principally polarised: Corollary 2.5. Suppose that (A/K, ℓ) is a strong counterexample. The image G ℓ of the mod-ℓ Galois representation is contained in GSp 4 (F ℓ ) with respect to a suitable symplectic form on A[ℓ].
Proof. As is well-known, the dual abelian surface A ∨ is isogenous to A over K. By Lemma 2.3, there exists a K-isogeny ϕ : A → A ∨ of degree prime to ℓ. Via ϕ, the Weil pairing A[ℓ]×A ∨ [ℓ] → µ ℓ induces the desired non-degenerate, Galois-invariant, antisymmetric form

Group theory
We briefly review some basic group theory we will need in the rest of the paper. We begin with a rather standard definition and a simple lemma: Definition 2.6. Let I and J be arbitrary groups. We say that G < I × J is a sub-direct product of I and J if G projects surjectively onto both I and J.
Lemma 2.7. The following hold: 1. An element g ∈ GL 2 (F ℓ ) has an F ℓ -rational eigenvalue if and only if both its eigenvalues are F ℓ -rational.
2. An element g ∈ GL n (F ℓ ) has an F ℓ -rational eigenvalue if and only if 1 is an eigenvalue of g ℓ−1 .
3. Let g ∈ GL n (F ℓ ) have order prime to ℓ. The eigenvalues of g are all F ℓ -rational if and only if g ℓ−1 = Id. This applies in particular to all elements of any subgroup We will have to make extensive use of the classification of the maximal subgroups of GL 2 (F ℓ ), so we briefly recall it here. The result is classical and goes back to Dickson; see also [Ser72,§2].
Theorem 2.8. Let ℓ ≥ 2 be a prime and let G be a maximal proper subgroup of GL 2 (F ℓ ).
One of the following holds: 2. Borel: up to conjugacy, G is contained in the subgroup of upper-triangular matrices.
3. Normaliser of Split Cartan: G is conjugate to the group a b , a b : 5. Exceptional: G contains the scalars, and PG is isomorphic to A 4 , S 4 or A 5 .
Variants of the same classification also hold for SL 2 (F ℓ ) and PGL 2 (F ℓ ), see Tables 8.1 and 8.2 of [BHRD13] for a modern reference. In particular, the exceptional maximal subgroups G of SL 2 (F ℓ ) are as follows: according to whether they have projective image A 4 , S 4 or A 5 , they are isomorphic respectively to SL 2 (F 3 ), S 4 or SL 2 (F 5 ), where S 4 , the group with GAP identifier (48, 28), is a Schur double cover of the symmetric group S 4 .
We will be especially interested the maximal subgroup of SL 2 (F ℓ ) given by the intersection of the normaliser of a split Cartan subgroup of GL 2 (F ℓ ) with SL 2 (F ℓ ). This is a generalised quaternion group, which we now describe in more detail. The generalised quaternion group Q 4n of order 4n is generated by an element of order 2n, that we will denote by r, and by an element of order 4, that we will denote by s and we will call a symmetry, subject to the relations s 2 = r n and s −1 rs = r −1 . Up to conjugacy, there is a unique maximal subgroup of SL 2 (F ℓ ) isomorphic to Q 2(ℓ−1) . A representative of the conjugacy class is generated by the matrices with δ a generator of F × ℓ . We will denote this specific subgroup of SL 2 (F ℓ ), which is the normaliser of a split Cartan subgroup of SL 2 (F ℓ ), by N(C s ). When considering the group Q 4n , we denote by Z/(2n)Z the subgroup generated by r. This subgroup is unique if n = 2. If j | 2n, we then denote by Z/jZ the unique subgroup of Z/(2n)Z < Q 4n of order j.
3 Hasse subgroups of Sp 4 (F ℓ ) Let us formally define the group-theoretic objects we are interested in: if every g ∈ G possesses an F ℓ -rational eigenvalue. We further say that G is Hasse if it has property (E) and acts irreducibly on F n ℓ .
Our objective in this section is to classify the maximal Hasse subgroups of Sp 4 (F ℓ ). The result is as follows.
Theorem 3.2. Let G be a subgroup of Sp 4 (F ℓ ). If G is Hasse, then ℓ ≡ 1 (mod 4) and up to conjugacy it is contained in one of the following groups: 1. An extension of degree 2 of the normaliser of a split Cartan subgroup of GL 2 (F ℓ ).
For a full description, see Equation (4).
3. An extension of degree 2 of an extension of the cyclic group of order (ℓ − 1)/2 by a finite group of order at most 240.
In Table 1 we give an exhaustive list containing all maximal Hasse subgroups of Sp 4 (F ℓ ). More precisely, the table lists Hasse subgroups that are maximal within a given maximal subgroup of Sp 4 (F ℓ ). We do not make any statement about possible containments between (conjugates of) subgroups that are contained in maximal subgroups of different types (first column). The only exception to this is in Remark 3.19, where we show that (a conjugate of) the group in the first line of the table is always contained in the groups of the fifth or sixth line.
Remark 3.3. In order to obtain the list of groups given in Table 1 we made extensive use of the computer algebra software MAGMA. However, note that we prove Theorem 3.2 as stated, without the explicit list of finite groups that may arise in case (4), without relying on any computer calculations. We use the detailed classification of the maximal Hasse subgroups of Sp 4 (F ℓ ) only in order to prove a fine point of the classification of the Hasse subgroups of GSp 4 (F ℓ ), see Theorem A.1.
Remark 3.4. Primes ℓ ≤ 7 cannot be handled directly by our methods, both because the technique of Section 3.2, which we use to analyse certain small groups H, requires the assumption ℓ ∤ |H|, and because the classification of the maximal subgroups of Sp 4 (F ℓ ) is slightly different for small ℓ. A direct computation reveals that Sp 4 (F ℓ ) and GSp 4 (F ℓ ) contain no Hasse subgroups at all for ℓ = 2, 3. Moreover, one can check that Theorems 3.2, 5.5, and 4.6 all hold for ℓ ≤ 7. Hence, from now on, we will tacitly assume that ℓ > 7.

Preliminary lemmas
Lemma 3.5. Let G < GL 2 (F ℓ ) be a Hasse subgroup such that every matrix in G is diagonal or anti-diagonal. Let M ∈ GL 2 (F ℓ ) be a matrix that normalises G and such that MM −T is diagonal or anti-diagonal. At least one of the following holds: • M is diagonal or anti-diagonal. There exists g ∈ G such that gM is diagonal.
• PG ∼ = Z/2Z × Z/2Z and there exists g ∈ G such that gM is symmetric. This case is only possible if ℓ ≡ 1 (mod 4).
x y z w . Note that G contains a diagonal matrix D = a 0 0 d with a = d, because otherwise PG would have order ≤ 2 and G would not act irreducibly.
Let D = a 0 0 d ∈ G be a diagonal matrix that is not a multiple of the identity. If MDM −1 is diagonal, then by direct computation we have xy = zw = 0, so M is diagonal or anti-diagonal. By irreducibility, G contains an anti-diagonal matrix g; if M is antidiagonal, gM is diagonal, and we are done. Otherwise, we may suppose that MDM −1 is anti-diagonal for all diagonal matrices Finally, we prove that ℓ is congruent to 1 modulo 4. Let g ∈ G be an anti-diagonal matrix, with characteristic polynomial t 2 + det(g). The condition that g has rational eigenvalues implies that − det(g) is a square. The matrix D 0 g is anti-diagonal, and the condition that − det(D 0 g) = (−a 2 )(− det g) is a square implies that −1 is a square modulo ℓ, so ℓ ≡ 1 (mod 4).
Lemma 3.6. Let G < SL 2 (F ℓ ) be a Hasse subgroup of N(C s ) and let M ∈ GL 2 (F ℓ ) normalise G. One of the following holds: • M is diagonal or anti-diagonal. There exists g ∈ G such that gM is diagonal; Proof. If |G| > 8, the subgroup of diagonal matrices is characteristic in G, hence M normalises it. This forces M to be diagonal or anti-diagonal; the conclusion follows easily.
Remark 3.7. Let A ∈ GL 4 (F ℓ ) be a block-anti-diagonal matrix of the form 0 g 1 g 2 0 with g 1 , g 2 ∈ GL 2 (F ℓ ). The eigenvalues of A are given by ± √ λ 1 , ± √ λ 2 , where λ 1 , λ 2 are the eigenvalues of g 1 g 2 . In particular, A admits an F ℓ -rational eigenvalue if and only if one of the eigenvalues of g 1 g 2 is a square in F × ℓ . If det(g 1 g 2 ) = λ 1 λ 2 is a square in F × ℓ , then A has an F ℓ -rational eigenvalue if and only if all of its eigenvalues are F ℓ -rational.
We now briefly describe the general strategy of proof of Theorem 3.2, which is inspired by [Cul12], even though the details are significantly different. The idea is to recursively explore the lattice of subgroups of Sp 4 (F ℓ ), starting with the maximal ones and considering smaller and smaller subgroups as needed. More precisely, given a subgroup G < Sp 4 (F ℓ ), one of the following holds: 1. G is Hasse, in which case we add it to the list of Hasse subgroups of Sp 4 (F ℓ ); 2. G acts reducibly, in which case it contains no Hasse subgroups; 3. G acts irreducibly, but it contains elements without any F ℓ -rational eigenvalues. We then consider each maximal subgroup of G, and iterate the same analysis.
At the top level, we start with G = Sp 4 (F ℓ ) itself, which contains elements without F ℓrational eigenvalues. Thus, we need to consider the maximal proper subgroups of Sp 4 (F ℓ ), which are as in Table 3.1 (see [BHRD13] for the notion of Aschbacher type of a maximal subgroup and Tables 8.12 and 8.13 of op. cit. for the classification). We exclude from our list the groups of type C 1 , since these act reducibly by definition. The cases corresponding to each of these maximal subgroups will be considered in turn in Sections 3.4 to 3.7. It is useful to point out at the outset that most groups H in this list have the property that all maximal subgroups of Sp 4 (F ℓ ) isomorphic to H are conjugate inside Sp 4 (F ℓ ), so that -for our purposes -we may work with a single, fixed maximal subgroup in the given isomorphism class. More precisely, this property holds for all the groups but 2 1+4 − .O − 4 (2) and 2.S 6 , for which two conjugacy classes exist (these groups will be handled using the methods of Section 3.2 and cause no difficulties).

Handling the 'small' groups
In this section we describe a computational technique to classify the Hasse subgroups of Sp 4 (F ℓ ) that are isomorphic to a subgroup of a fixed abstract group G, as ℓ varies among the primes that do not divide |G|. The technique is based on basic representation theory, so we only give a sketch, but we point out that we have implemented the algorithm resulting from the arguments in this section as a MAGMA script. Since there is nothing specific about Sp 4 (F ℓ ), we actually consider more generally subgroups of arbitrary matrix groups over finite fields.
Notice first that since ℓ ∤ |G| all representations of G in characteristic ℓ are semi-simple (Maschke's theorem) and come by reduction from representations defined in characteristic 0, so that we have at our disposal all the usual machinery of characters and representations theory in characteristic 0. In particular, for a fixed k ≥ 1 we can describe all representations G ֒→ GL k (F ℓ e ) (and even G ֒→ Sp k (F ℓ e )): 1. we construct all k-dimensional representations of G by looking at complex characters; 2. by [Ser98, Theorem 24, p. 109], the representation corresponding to each complex character can be realised over the number field K := Q(ζ |G| ). The prime ℓ is unramified in this field, so by reducing modulo a place p of K of characteristic ℓ we obtain a corresponding representation defined over a finite extension of F ℓ ; 3. we may also determine the minimal extension of F ℓ over which a given representation is defined: by [Ser98, Corollaire on p. 108], since the Brauer group of any finite field vanishes, a representation ρ over F ℓ is defined over the finite field F ℓ e if and only if F ℓ e contains the field generated by the image of the character of ρ (which we obtain by reducing the corresponding complex character modulo the place p); 4. finally, when the dimension k is even, in order to test whether a given representation V has image in Sp k (F ℓ e ) (that is, whether V admits an invariant alternating bilinear form), it suffices to test whether Λ 2 V * contains a copy of the trivial representation. This can also be understood in terms of characters: the character of V determines the character of Λ 2 V * , and in order to check whether Λ 2 V contains a copy of the trivial representation we simply need to take the scalar product of this character with the trivial character. An obvious variant of this procedure, using Sym 2 V * , can be used to test whether a representation is orthogonal.
Suppose now that we wish to know for which primes ℓ (not dividing |G|) there exist • an embedding ρ : G ֒→ Sp k (F ℓ ) • a subgroup H of G such that ρ(H) is a Hasse subgroup. The inclusion ρ gives in particular a symplectic representation of G on a k-dimensional space, which comes by reduction from a faithful representation ρ : G ֒→ GL k (K). Since we can list all irreducible k-dimensional representations of G, we may assume that the representation ρ is fixed. We may then proceed as follows: 1. for each subgroup H of G, we restrict ρ to H; 2. we decompose ρ| H as a direct sum of representations of H, using character theory; 3. for each sub-representation W of ρ| H we test whether W is defined over F ℓ . Notice that this amounts to testing whether ℓ splits completely in the sub-field of K generated by the traces of the character of ρ| H . Since the field K is cyclotomic, by class field theory (or even just the Kronecker-Weber theorem) this amounts to some congruence conditions on ℓ. If no non-trivial sub-representation W of ρ| H is defined over F ℓ , then ρ| H is irreducible over F ℓ ; 4. for each h ∈ H we compute the characteristic polynomial of ρ(h). Its roots are all roots of unity, of orders (say) n 1 , . . . , n k . The condition that ρ(h) has an F ℓ -rational eigenvalue again translates into a congruence condition: ℓ must be congruent to 1 modulo at least one of the integers n 1 , . . . , n k .
The output of this algorithm is a collection of pairs (H, congruence conditions on ℓ): the Hasse subgroups of ρ(G) < Sp k (F ℓ ) are precisely the ρ(H) for which the corresponding congruence conditions on ℓ are met. Notice that each subgroup H of G will correspond to different conditions in general, and for some subgroups the conditions will correspond to the empty set of prime numbers. Naturally we can also list the maximal Hasse subgroups by checking for inclusions between the various subgroups. We shall use this procedure repeatedly to handle cases when the relevant subgroups of Sp 4 (F ℓ ) to be studied have order independent of the prime ℓ.

Further input from representation theory
Let G be a finite group and let ℓ be a prime such that ℓ ∤ |G|. As explained in the previous section, there is a bijective correspondence between irreducible representations of G over F ℓ and over C.
Proposition 3.8. Let G, ℓ be as above, let G 0 be a subgroup of G of index 2, and let ρ : G → GL n (F ℓ ) be a representation. Suppose that, for every g ∈ G, all eigenvalues of ρ(g) are F ℓ -rational. Then the following hold: 1. ρ is irreducible if and only if it is absolutely irreducible.
2. Let χ be the character of the complex representation lifting ρ. Then ρ is irreducible if and only if χ, χ G = 1, where ·, · G is the usual scalar product on characters.
3. Suppose that the restriction of ρ to G 0 decomposes as the direct sum of two isomorphic representations over F ℓ . Then ρ is reducible.
Proof. 1. One implication is trivial. For the other, let χ be the character of the complex representation lifting ρ, and let χ 1 be an irreducible character appearing as a summand of χ. For every g ∈ G, the reduction modulo ℓ of χ 1 (g) is a sum of eigenvalues of g, hence is F ℓ -rational. By [Ser98, Corollaire on p. 108], the representation ρ 1 with character (the reduction modulo ℓ of) χ 1 is defined over F ℓ and is a subrepresentation of ρ.
2. Follows combining (1), the correspondence between representations over C and F ℓ , and the well-known fact that a complex representation is irreducible if and only if its character has norm 1 with respect to the natural scalar product.
3. Let χ be as above.
3.4 G of type C 2 : G < GL 2 (F ℓ ).2 In this section we prove: Proposition 3.9. Let G < Sp 4 (F ℓ ) be a Hasse group contained in a group isomorphic to GL 2 (F ℓ ).2. Then, one of the following holds: • ℓ ≡ 1 (mod 4) and G is contained (up to conjugacy) in G ′ , the group described in Equation (4).
• G is contained in one of the groups of Proposition 3.16.
The group GL 2 (F ℓ ).2 sits in the exact sequence 1 / / GL 2 (F ℓ ) i / / GL 2 (F ℓ ).2 π / / S 2 / / 0 and up to conjugacy in Sp 4 (F ℓ ), considered as the group of isometries of the symplectic form given in (1), we have see the beginning of [Cul12, Section 3.1]. Let G < GL 2 (F ℓ ).2 be a Hasse subgroup and let G 0 := G ∩ ker π: every element of G 0 can be written as A 0 0 A −T . We can then identify Since there are elements of GL 2 (F ℓ ) that do not have any rational eigenvalues, G 0 is a proper subgroup of GL 2 (F ℓ ). By Theorem 2.3.1, G 0 contains SL 2 (F ℓ ) or is contained in the normaliser of a Cartan subgroup, in a Borel subgroup, or in groups that have projective image A 4 , S 4 , or A 5 . Observe that there are elements of SL 2 (F ℓ ) without a rational eigenvalue: it follows that G 0 does not contain SL 2 (F ℓ ), hence it is a subgroup of one of the groups above.

Case G 0 in the normaliser of a split Cartan subgroup
In a suitable basis, the normaliser NC s of a split Cartan can be written as G is Hasse and then contains a block-anti-diagonal matrix 0 The possible matrices M are described in Lemma 3.5. If we are in the second case of Lemma 3.5, then PG 0 ∼ = Z 2 × Z 2 and ℓ ≡ 1 (mod 4). It follows that G 0 is exceptional, and we will study this case in Section 3.4.4.
If we are in the first case of Lemma 3.5, then M is diagonal or anti-diagonal. Put Since G is Hasse, it must contain matrices of all four types above (for otherwise it would stabilise a 2-dimensional subspace). In particular, the set G\G 0 is non-empty and contains an element of the form 0 . A matrix of this form has characteristic polynomial (t 2 + 1) 2 , so it has a rational eigenvalue if and only if −1 is a square modulo ℓ. Hence, in order for every element of G to have a rational eigenvalue, we need ℓ ≡ 1 (mod 4), which we assume from now on. As above, G 0 contains at least one element of the form B(i 0 , j 0 ). The matrix B(i, j) has a rational eigenvalue if and only if δ i+j is a square, hence i 0 + j 0 is even. Since A(i, j)B(i 0 , j 0 ) = B(i + i 0 , j + j 0 ) is also an element of G, we must also have i + i 0 + j + j 0 ≡ 0 (mod 2), so i + j must be even and Moreover, G contains an element of the form 0 , so it has a rational eigenvalue if and only if i − j ≡ 0 (mod 2) (recall that −1 is a square modulo ℓ ≡ 1 (mod 4)). We conclude that On the other hand, if ℓ ≡ 1 (mod 4) one checks immediately that the group G ′ is a (necessarily maximal) Hasse subgroup.

Case G 0 in the normaliser of a non-split Cartan subgroup
Up to conjugacy, the normaliser NC ns of a non-split Cartan is It follows that G 0 is contained in the group generated by the scalar matrices and by M. In particular, G 0 fixes the eigenspaces of M, so G 0 is contained in a Borel subgroup, which we treat next.

Case G 0 in a Borel subgroup
Let v be a line in F 4 ℓ fixed by G 0 . Let g ∈ G \ G 0 and consider the two-dimensional subspace V = v, gv : one checks immediately that V is G-invariant, hence G does not act irreducibly.
Proof. One checks that HH 1 = H 1 H, hence that HH 1 is a group. We check that HH 1 is Hasse. By assumption every h ∈ H has at least one F ℓ -rational eigenvalue. If h is blockdiagonal, then it is easy to see that any element of the form hh 1 for h 1 ∈ H 1 has at least we get again −BB −T , which by assumption has an eigenvalue that is a square in F × ℓ , so hh 1 has at least one F ℓ -rational eigenvalue, as desired. Finally, since H acts irreducibly on F 4 ℓ , then a fortiori so does HH 1 , hence HH 1 is Hasse as claimed.
Corollary 3.11. Every subgroup of GL 2 (F ℓ ).2, maximal among Hasse subgroups, contains the group H 1 of the previous lemma.
Corollary 3.12. Let ℓ > 3 be a prime and let H be a subgroup of GL 2 (F ℓ ).2 that contains ℓ stable under the action of H. We will show that either . To see this, simply notice that W is stable under the action of H 1 , hence in particular under the action of which -for λ = ±1 (and there is such an element in F × ℓ , since ℓ > 3) -is the projector on V 1 ; one reasons similarly for the projection on V 2 . The subspace W ∩ V 1 is stable under the action of H 0 , so by assumption it is either trivial or all of V 1 (and the same applies to W ∩ V 2 ). Finally, since H contains an element that exchanges V 1 with V 2 , the subspaces W ∩ V 1 and W ∩ V 2 are either both trivial or both 2-dimensional. In the two cases, one 2 is a Hasse subgroup, then H 0 = H ∩ ker π is a Hasse subgroup of GL 2 (F ℓ ): the condition on rational eigenvalues is satisfied, and if F 2 ℓ were reducible under the action of H 0 , then H 0 would be contained in a Borel subgroup, which contradicts the arguments of Section 3.4.3.
By [Sut12, Lemma 1] we see that if PH 0 is not contained in PSL 2 (F ℓ ), then PH 0 cannot be an exceptional group, so we fall back into the cases of the previous sections. Hence we may assume that PH 0 is contained in PSL 2 (F ℓ ). By [Ann14, Lemma 3.5] we then obtain that ℓ is 1 modulo 4 and PH 0 is isomorphic to one among A 4 , S 4 , A 5 . Notice that GL 2 (F ℓ ) := {g ∈ GL 2 (F ℓ ) det(g) ∈ F ×2 ℓ } coincides with the subgroup of GL 2 (F ℓ ) generated by SL 2 (F ℓ ) and the scalar matrices. We record what we have just shown as a lemma: Lemma 3.14. Let ℓ ≡ 1 (mod 4) be a prime. Let H 0 be a subgroup of GL 2 (F ℓ ), contained in GL 2 (F ℓ ) and containing F × ℓ Id. 1. Suppose that H 0 has projective image isomorphic to S 4 or A 5 . Then the normaliser N of H 0 in GL 2 (F ℓ ).2 satisfies [N : H 0 ] = 2, and an element of the non-trivial coset is Suppose that H 0 has projective image isomorphic to A 4 . Then the normaliser N of H 0 in GL 2 (F ℓ ).2 satisfies [N : H 0 ] = 4, and representatives of the three non-trivial cosets are given by 3. With notation as in (2), assume that PH 0 ∼ = A 4 is a maximal subgroup of PSL 2 (F ℓ ).
Proof. We begin by noticing the following matrix identity: for every A ∈ GL 2 (F ℓ ) one has The fact that J ′ is in N follows from a simple calculation using the above matrix identity. 3. Observe that det(σ) is not a square in F × ℓ , for otherwise P H 0 , σ would be a proper overgroup of PH 0 in PSL 2 (F ℓ ). Let A 0 0 A −T be an element in H 0 and notice that

The group PGL
By Remark 3.7, in order to check if this matrix has F ℓ -rational eigenvalues, we need to test whether the matrix −J 2 σ −T A −T J 2 σA has an eigenvalue that is a square in F × ℓ . Using the matrix identity at the beginning of the proof, we need to understand whether 1 det(σA) (σA) 2 admits an eigenvalue in F ×2 ℓ . We may choose A in such a way that σA represents a transposition in S 4 . Notice that det(A) is a square (since this is true for all elements in H 0 ). From the choice of A it follows that (σA) 2 = Id, so the eigenvalues of 1 det(σA) (σA) 2 are 1 det(σA) , which is not a square (since det(A) ∈ F ×2 ℓ but det σ ∈ F ×2 ℓ ).
Corollary 3.15. Let ℓ ≡ 1 (mod 4) be a prime. Let H 0 be a subgroup of GL 2 (F ℓ ), contained in GL 2 (F ℓ ) and containing F × ℓ Id. Suppose that H 0 is Hasse. 1. Suppose that one of the following holds: Then H := H 0 , J ′ is Hasse and is the unique maximal Hasse subgroup Writing A = λB with det(B) = 1 and using the matrix identity in the proof of Lemma 3.14 one checks easily that (AJ 2 )(−A −T J 2 ) = B 2 . Since by assumption A (and hence also B) has an F ℓ -rational eigenvalue, this matrix has an F ℓ -rational eigenvalue that is a square. Combining this observation with Corollary 3.12 we see that H is Hasse.
In the cases PH 0 ∼ = S 4 or A 5 , it follows immediately from the previous lemma that either G = H 0 (which, however, is not Hasse, since H 0 obviously stabilizes two 2-dimensional subspaces) or G = N = H, as claimed. 2. Consider the normaliser N of H 0 in GL 2 (F ℓ ).2. By Lemma 3.14 we know that N = is a square in F × ℓ , because by assumption PH 0 extends to a subgroup of PSL 2 (F ℓ ) isomorphic to S 4 . Note that this happens only if ℓ ≡ ±1 (mod 8), and since ℓ ≡ 1 (mod 4) we obtain ℓ ≡ 1 (mod 8). Reasoning as in the proof of part (3) of Lemma 3.14 we see easily that N is Hasse (notice that the elements of S 4 \ A 4 have order dividing 4, so their lifts to SL 2 (F ℓ ) have order dividing 8; it follows that the elements of the coset H 0 σ have F ℓ -rational eigenvalues since ℓ ≡ 1 (mod 8)). If G is a group with G 0 = H 0 , then H 0 is normal in G and hence G < N. By maximality of G we should have G = N, but N 0 = H 0 , as desired.
Combining the previous lemmas we obtain: The maximal Hasse subgroups G of G ′ with PG 0 isomorphic to A 4 , S 4 or A 5 are as follows.

Group
Condition Proof. Let G be a Hasse subgroup of G ′ and such that PG 0 is isomorphic to A 4 , S 4 or A 5 . If G is maximal with such properties, then by Corollary 3.11 we know that it contains the group H 1 . By Lemma 3.13, we have ℓ ≡ 1 (mod 4) and PG 0 is contained in PSL 2 (F ℓ ), so G 0 is contained in GL 2 (F ℓ ) and contains F × ℓ Id. The hypotheses imply that G 0 has elements of order 3, so the condition that every element of G 0 has F ℓ -rational eigenvalues implies ℓ ≡ 1 (mod 12). Consider the following cases: 1. if ℓ ≡ 1 (mod 5), then by [BHRD13, Table 8.2] the group SL 2 (F ℓ ) contains a maximal subgroup isomorphic to SL 2 (F 5 ) with projective image A 5 . This group satisfies the assumptions of Corollary 3.15, so we get a maximal subgroup isomorphic to ).2. From the previous discussion it is clear that the conditions ℓ ≡ 1 (mod 60) are necessary and sufficient in order for this subgroup to be Hasse. Moreover, in this case we do not get any Hasse maximal subgroup X such that PX 0 ∼ = A 4 : this is proven exactly as in part (2) of Corollary 3.15, using the fact that in this case PX 0 extends to a subgroup isomorphic to A 5 .
• G is contained in one of the groups described in Section 3.5.5.
3.5.1 Case G 0 < G Since SL 2 (F ℓ ) contains matrices without a rational eigenvalue, G 0 cannot be all of G. Hence G 0 is contained in a group of the form {(g, ±g, 1) | g ∈ M} for a certain proper maximal subgroup M of SL 2 (F ℓ ). In particular, G 0 is a subgroup of M × M with M a maximal subgroup of SL 2 (F ℓ ), so this case is included in one of the cases below.

Case M Borel
Recall that G 0 = G ∩ ker π. The group G 0 fixes a line v , and G does not act irreducibly by the same argument as in Section 3.4.3, so G is not Hasse.
. In both cases, G 0 fixes a line and G does not act irreducibly, contradiction. The case ℓ ≡ 1 (mod 4) is similar: one proves that q 1 or q 2 has order that divides 4, hence G 0 < Z/4Z × Z/4Z, and this subgroup fixes a line. So, G is not Hasse.

Case
Recall the description of the group Q 4n from Section 2.3.1. Assume first ℓ ≡ 3 (mod 4).
Observe that G 0 cannot contain an element (s 1 , s 2 , 1) with s 1 , s 2 / ∈ Z/(ℓ − 1)Z since such an element does not have a rational eigenvalue as ord(s 1 ) = ord(s 2 ) = 4 ∤ ℓ − 1. Therefore, Proceeding as in the previous case we conclude that G does not act irreducibly.
Assume now that ℓ ≡ 1 (mod 4). We start by showing that the exponent of G divides ℓ − 1. The elements of G 0 have order dividing ℓ − 1. Let g ∈ G \ G 0 . Its characteristic polynomial is of the form x 4 + bx 2 + 1, hence its eigenvalues are of the form ±λ ±1 . If one such eigenvalue is rational, then they all are, and it follows as desired that the order of g divides ℓ − 1. Let H be the set of subgroups of SL 2 (F ℓ ) ≀ S 2 with exponent that divides ℓ − 1, that act irreducibly, and such that the intersection with ker π is contained in Q 2(ℓ−1) × Q 2(ℓ−1) . Observe that (up to conjugacy) G is contained in a maximal element of H with respect to inclusion. We want to classify these maximal elements. Let H be a maximal element of H, let H 0 = H ∩ ker π and (z, w, −1) ∈ H \ H 0 .
Assume that each of z and w is diagonal or anti-diagonal. Let H ′ be the subgroup of Q 2(ℓ−1) ≀ S 2 defined by H ′ := {(x, y, 1) | x, y ∈ Q 2(ℓ−1) and xy ∈ Z/((ℓ − 1)/2)Z}. (5) The group H ′ is normalised by H, so H, H ′ = HH ′ . One can easily show that, given Otherwise, assume that at least one between z and w is neither diagonal nor antidiagonal. By Lemma 3.6 we have In conclusion, the maximal groups in H are isomorphic to ( We study this case in Section 3.5.5. If G is contained in a maximal group H of H that contains H ′ , then ℓ ≡ 1 (mod 4).
Remark 3.18. Let H be a maximal Hasse subgroup that contains H ′ . Note that H ′ is normal in Q 2(ℓ−1) ≀ S 2 and (Q 2(ℓ−1) ≀ S 2 )/H ′ ∼ = (Z/2Z) 3 . So, H corresponds to a subgroup H of (Q 2(ℓ−1) ≀S 2 )/H ′ ∼ = (Z/2Z) 3 . Let X −1 be the subset of (Q 2(ℓ−1) ≀S 2 )/H ′ given by the classes of elements of the form (x, y, −1). Since H is Hasse, H contains an element in X −1 . If ℓ ≡ 5 (mod 8), the only class in X −1 that can belong to H is the class of (Id, Id, −1)H ′ , since the other classes contain elements without a rational eigenvalue. Hence, H = H ′ , (Id, Id, −1) and |H| = 2(ℓ − 1) 2 . If ℓ ≡ 1 (mod 8), three of the four classes in X −1 have the property that every element in the class has a rational eigenvalue. By maximality we obtain that H is generated by two of these three classes, hence that it has order 4. It follows that there are 3 possible choices of H, each leading to a maximal subgroup H of order 4(ℓ − 1) 2 .
Remark 3.19. Let G ′ be the maximal Hasse subgroup described in Equation (4), that is, the group listed in the first line of Table 1 Table 1 is always contained (up to conjugay) in the groups of the fifth or sixth line of the table.

Case M ∼ = E
All these cases can be treated using the algorithm of Section 3.2. The results are listed in Table 1 and correspond to (part of) Proposition 2 in [Cul12].
3.6 G of type C 3 Proposition 3.20. Let G be a Hasse subgroup of Sp 4 (F ℓ ) contained in a group isomorphic to GU 2 (F ℓ ).2 or SL 2 (F ℓ 2 ).2. Then, G is contained in one of the groups described in Proposition 3.26.
We start by describing explicitly the two (conjugacy classes of) maximal subgroups of Sp 4 (F ℓ ) of type C 3 . Table 8.12 in [BHRD13] shows that all the maximal subgroups of type C 3 that are abstractly isomorphic form a single conjugacy class, so it suffices to study a specific subgroup of each type.
A subgroup G of type C 3 consists of all transformations in Sp 4 (F ℓ ) that act either F ℓ 2 -linearly or F ℓ 2 -anti-linearly for a given F ℓ 2 -vector space structure on F 4 ℓ . In order to construct such groups we start with the vector space V 2 = F 2 ℓ 2 , whose basis vectors we denote by e 1 = 1 0 , e 2 = 0 1 . We denote by σ the non-trivial element of Gal(F ℓ 2 /F ℓ ) and equip V 2 with one of the following forms: 1. the symplectic form characterised by e 1 , e 2 = 1; 2. the Hermitian form characterised by e 1 , e 1 H = e 2 , e 2 H = 0, e 1 , e 2 H = √ d.
Remark 3.21. Recall that a Hermitian form on We fix once and for all d ∈ F × ℓ a non-square; in case ℓ is congruent to 3 modulo 4, we take d = −1. Setting e 3 = √ de 1 and e 4 = √ de 2 , we obtain that e 1 , e 2 , e 3 , e 4 is an F ℓ -basis of V 2 . We will represent F ℓ -linear transformations of V 2 in the basis e 1 , . . . , e 4 . In particular, we let denote the matrix giving the natural action of σ on V 2 . We are now ready to describe the maximal subgroups of Sp 4 (F ℓ ) of type C 3 .
acts on F 4 ℓ (with respect to our coordinates) via and it is easy to check that the condition det(g) = 1 implies that ι(g) preserves the symplectic form with matrix Notice that this is the F ℓ -bilinear form obtained as tr F ℓ 2 /F ℓ ( ·, · ). The subgroup ι(SL 2 (F ℓ 2 )) of Sp 4 (F ℓ ) is normalised by τ , and we write SL 2 (F ℓ 2 ).2 for the group generated by ι(SL 2 (F ℓ 2 )) and by τ . This subgroup preserves the bilinear form just described. From now on, we shall identify SL 2 (F ℓ 2 ) with its image via ι. For a subgroup G of SL 2 (F ℓ 2 ).2, we denote by G 0 the intersection of G with SL 2 (F ℓ 2 ).
The subgroup GU 2 (F ℓ ).2. Let GU 2 (F ℓ ) ⊆ GL 2 (F ℓ 2 ) be the isometry group of ·, · H , that is, the subgroup of GL 2 (F ℓ 2 ) consisting of those g that satisfy Lemma 3.23. Let µ ∈ F × ℓ 2 be an element of norm −1 and let H be the group One checks that all the elements given in the statement preserve ·, · H , hence that they are in GU 2 (F ℓ ). On the other hand, by [BHRD13, Theorem 1.6.22] we have which concludes the proof.
is anti-symmetric and invariant under the action of GU 2 (F ℓ ) by definition of this group. We consider Sp 4 (F ℓ ) and GSp 4 (F ℓ ) as the groups of transformations that preserve (resp. preserve up to scalars) this symplectic form. We denote by GU 2 (F ℓ ).2 the subgroup of Sp 4 (F ℓ ) generated by ι(GU 2 (F ℓ )) and τ (this latter element normalises ι(GU 2 (F ℓ ))). For a subgroup G of GU 2 (F ℓ ).2, we denote by G 0 the intersection of G with ι(GU 2 (F ℓ )).
. It is clear that G 0 contains elements that do not have F ℓ -rational eigenvalues, so G cannot be Hasse.
2. G 0 is contained in a Borel subgroup. Using the fact that all the eigenvalues of the elements of G 0 are rational (Remark 3.22), we see that the group G 0 ⊆ Sp 4 (F ℓ ) stabilises a 1-dimensional subspace V of F 4 ℓ . If G = G 0 , let g be an element of G \ G 0 : then g normalises G 0 (since [G : G 0 ] = 2) and the subspace W = V + gV , of dimension at most 2, is stable under the action of G. Thus G cannot be Hasse.
3. G 0 is contained in Q 2(ℓ 2 +1) . An element g ∈ G 0 has one F ℓ -rational eigenvalue if and only if both its eigenvalues are F ℓ -rational (their product is 1), if and only if g ℓ−1 = Id. This implies that the order of every g ∈ G 0 divides (ℓ 2 + 1, ℓ − 1) = 2, so G 0 is either Z/2Z or (Z/2Z) 2 . In both cases, G 0 stabilizes a line in (F ℓ 2 ) 2 and we are reduced to the previous case. The conclusion is that G cannot be Hasse.
4. G 0 is contained in Q 2(ℓ 2 −1) . Reasoning as in the previous case, we obtain that G 0 is contained in Q 2(ℓ−1) , which -up to conjugacy -is a subgroup of SL 2 (F ℓ ).
We prove more generally that G 0 cannot be (conjugate to) a subgroup of SL 2 (F ℓ ). Indeed, if this is the case, ι(G 0 ) stabilizes the non-trivial subspaces e 1 , e 2 F ℓ and e 3 , e 4 F ℓ of F 4 ℓ . Moreover, it acts on both subspaces with the same character. Proposition 3.8 (3), which we can apply by Remark 3.22, implies that G cannot be Hasse.
5. G 0 is isomorphic to a subgroup of SL 2 (F 3 ), S 4 , or SL 2 (F 5 ). In the cases SL 2 (F 3 ) and S 4 , the subgroup G 0 is conjugate to a subgroup of SL 2 (F ℓ ), and by what we proved in the previous case we obtain that G cannot be Hasse. In the case SL 2 (F 5 ), either G 0 is again conjugate to a subgroup of SL 2 (F ℓ ), or ℓ ≡ ±3 (mod 10), see [BHRD13, Table 8.2]. However, in the latter case no element of G 0 of order 5 can have F ℓ -rational eigenvalues, so 5 ∤ |G 0 |. Any such G 0 is conjugate to a subgroup of S 4 , so we obtain a contradiction as above.
6. G 0 is contained in SL 2 (F ℓ ).2. Let G 00 be the intersection of G 0 with SL 2 (F ℓ ). If the order of G 00 is not divisible by ℓ, then ℓ ∤ |G 0 | and G 0 is contained in a subgroup maximal among those of order not divisible by ℓ, which are covered by the previous points. On the other hand, if ℓ | |G 00 |, then by the classification of the subgroups of SL 2 (F ℓ ) we know that either G 00 = SL 2 (F ℓ ) or G 00 is contained in a Borel subgroup. In the former case, G 00 contains elements that do not have F ℓ -rational eigenvalues, which is impossible since G is assumed to be Hasse. In the latter case, G 00 is normal inside G 0 , of index at most 2. Since G 00 fixes precisely one line w in F 2 ℓ (any element of order ℓ in SL 2 (F ℓ ) has this property, and we know that ℓ | |G 00 |), by normality we obtain that G 0 also fixes that line (let g be a representative of the possible non-trivial coset of G 00 inside G 0 . Then g w is G 00 -stable, hence it must coincide with w ). This implies that G stabilizes a non-trivial subspace, contradiction. The conclusion is that G cannot be Hasse, unless it is already covered by one of the previous cases. But since no Hasse subgroup existed for any of the previous cases, putting everything together we have established: Proposition 3.24. The maximal subgroups of Sp 4 (F ℓ ) isomorphic to SL 2 (F ℓ 2 ).2 contain no Hasse subgroups.
3.6.2 Subgroups of GU 2 (F ℓ ).2 Let G be a maximal Hasse subgroup of GU 2 (F ℓ ).2. We consider G 0 as subgroup of GU 2 (F ℓ ), hence of F × ℓ 2 · GL 2 (F ℓ ). We will show below that the group G fixes a non-trivial subspace of F 4 ℓ (of dimension at most 2) whenever G 0 fixes a line in F 2 ℓ 2 . Therefore, if G is a maximal Hasse subgroup of GU 2 (F ℓ ).2, then all the elements in G 0 have F ℓ -rational eigenvalues and G 0 does not stabilize any line in F 2 ℓ 2 . We now distinguish cases according to the structure of PG 0 , relying on the classification of the maximal subgroups of PGL 2 (F ℓ ) = P GU 2 (F ℓ ), see §2.3.1.

Assume
. It is easy to show that the only subgroup of SL 2 (F ℓ ) that projects onto PSL 2 (F ℓ ) is SL 2 (F ℓ ) itself. But this would imply that (G 0 ) ′ (hence also G 0 ) contains SL 2 (F ℓ ), contradicting the fact that every element of G 0 has F ℓ -rational eigenvalues.
2. PG 0 is contained in a Borel subgroup. Then (up to conjugating by a matrix in GL 2 (F ℓ )) all matrices in G 0 are of the form λ µ 1 ⋆ 0 µ 2 with µ 1 , µ 2 ∈ F × ℓ and λ ∈ F × ℓ 2 . Such a matrix admits a rational eigenvalue if and only if λ is in fact in F × ℓ . This implies that G 0 is contained in GL 2 (F ℓ ), so it stabilises an F ℓ -line v . As [G : G 0 ] ≤ 2, this implies that G stabilises a subspace of dimension at most 2, contradiction.
3. PG 0 is contained in the normaliser of a split Cartan subgroup. Up to conjugacy, G 0 is then contained in A matrix of the form λ α 0 0 β has F ℓ -rational eigenvalues if and only if λα or λβ are in F ℓ ; since α, β are in F × ℓ , this implies that λ is also in F × ℓ . On the other hand, consider a matrix of the form λ 0 α β 0 . The condition of rational eigenvalues translates to the fact that λ 2 αβ is in F ×2 ℓ . Since α, β are in F × ℓ , this implies that λ is either in Notice that the set of matrices of the form λ 0 α β 0 is a coset for the subgroup , all of whose elements have F ℓ -rational coefficients.
This shows that either all elements λ 0 α β 0 have λ ∈ F × ℓ (case 1), or they all have In case (2), applying ι we see that G 0 acts on F 4 ℓ via the matrices From this description we see that V 1 = e 1 , e 4 and V 2 = e 2 , e 3 are stable under the action of G 0 , and that the characters of G 0 on V 1 and V 2 are equal. By Proposition 3.8, we conclude that G does not act irreducibly, contradiction. Note that, in order to apply Proposition 3.8, we need that all eigenvalues of every matrix of G 0 are rational.
All the eigenvalues of the diagonal matrices are rational. The matrices ι λ 0 α β 0 have eigenvalues ± λ 2 αβ (with multiplicity 2), that are F ℓ -rational since, as we noted before, λ 2 αβ is a square. In case (1) the proof is similar, but simpler.
4. PG 0 is contained in the normaliser N of a non-split Cartan subgroup C, which is the maximal cyclic subgroup of N.
Suppose first that PG 0 is contained in C. This implies in particular that PG 0 is cyclic, say generated by the projective image of g ∈ G 0 . Since the kernel of G 0 → PG 0 consists of scalars that lie in F × ℓ 2 and have both F ℓ -rational eigenvalues and norm equal to 1, we see that this kernel is contained in {± Id} (and in fact, by maximality of G, equal to it). This implies that G is generated by ι(g), ι(− Id), and any element h in G \ G 0 (assuming G = G 0 ). Notice that h 2 ∈ G 0 and that by assumption g ∈ GU 2 (F ℓ ) has at least one F ℓ -rational eigenvalue, so ι(g) possesses that same eigenvalue. Letting v ∈ F 4 ℓ denote a corresponding eigenvector, one checks easily that v, hv F ℓ is a non-trivial subspace of F 4 ℓ stable under the action of G, contradiction. Suppose now that PG 0 meets N \ C, the non-trivial coset of the cyclic group C inside the dihedral group N. Recall from §2.3.1 that -up to conjugacy in PGL 2 (F ℓ ) -elements Any lift of such a matrix is of the form λ α dβ −β −α , with characteristic polynomial t 2 − λ 2 (−α 2 + dβ 2 ), hence eigenvalues ±λ −α 2 + dβ 2 . Since −α 2 + dβ 2 is in F ℓ , we see that λ is either in F × ℓ or in F × ℓ √ d. Now consider two elements of G 0 that project to classes lying in N \ C. The group G 0 contains their product: The eigenvalues of this matrix are In particular, there can be an F ℓ -rational eigenvalue only if we have α 1 α 2 − dβ 1 β 2 = 0 or α 1 β 2 − α 2 β 1 = 0.
Suppose now that for at least one element of PG 0 ∩ (N \ C) we have β 1 = 0 (otherwise, consists of at most one element, the projective class of 1 0 0 −1 , hence |PG 0 | = 2. We will rule out below the possibility that |PG 0 | | 4). Then the equations (8) imply the equality which -at the level of projective classes -means Since α 2 dβ 2 −β 2 −α 2 is an arbitrary element in PG 0 ∩(N \C), this shows that PG 0 ∩(N \C) consists of at most 2 elements, so PG 0 has cardinality at most 4 and all elements of order at most 2. It follows that PG 0 is isomorphic to a subgroup of (Z/2Z) 2 . Since any subgroup of PGL 2 (F ℓ ) isomorphic to (Z/2Z) 2 acts on P(F 2 ℓ ) with a fixed point, this implies that (up to conjugacy in PGL 2 (F ℓ )) the group PG 0 is contained in a Borel subgroup, contradicting what we already proved.
5. PG 0 is contained in an exceptional subgroup isomorphic to A 4 , S 4 or A 5 . As observed above, the kernel of the projection map G 0 → PG 0 is {±1}, so G 0 is a central extension of degree 2 of a subgroup of one among A 4 , S 4 , and A 5 . In fact, one checks easily that if PG 0 is a proper subgroup of A 4 , or a proper subgroup of S 4 distinct from A 4 , or a proper subgroup of A 5 distinct from A 4 , then PG 0 falls in one of the previous cases, so we may assume PG 0 ∈ {A 4 , S 4 , A 5 }. (c) ℓ ≡ 2 (mod 3).
Next suppose that PG 0 ∼ = S 4 . Then reasoning as above we obtain that (G 0 ) ′ is a subgroup of SL 2 (F ℓ ) having projective image the exceptional subgroup A 4 , so (G 0 ) ′ ∼ = SL 2 (F 3 ). Since elements in (G 0 ) ′ < SL 2 (F ℓ ) have one F ℓ -rational eigenvalue if and only if they have all their eigenvalues in F ℓ , and since SL 2 (F 3 ) contains elements of order 3 and 4, we obtain ℓ ≡ 1 (mod 12). Take an element g in PG 0 that under the isomorphism PG 0 ∼ = S 4 corresponds to a transposition. The element g has exactly two lifts ±g in GL 2 (F ℓ ) with order 4. Since 4 | ℓ−1, the elements ±g have all their eigenvalues in F ℓ . It follows that no multiple λg with λ ∈ F ℓ 2 \ F ℓ has any F ℓ -rational eigenvalues, hence the elements of G 0 that project to g must be precisely ±g ∈ GL 2 (F ℓ ). Since transpositions generate S 4 , it follows that all elements of G 0 are contained in GL 2 (F ℓ ). Reasoning as in the case of SL 2 (F ℓ ).2, this gives a contradiction to the fact that G acts irreducibly on F 4 ℓ . Having excluded the possibilities PG 0 ∼ = S 4 , A 5 , this concludes the proof of (a). Suppose now that PG 0 ∼ = A 4 , hence P ((G 0 ) ′ ) ∼ = (Z/2Z) 2 . It is easy to see that (G 0 ) ′ contains elements of order 4: otherwise, the 2-Sylow subgroup would only have elements of order 2 and would therefore be commutative. Since elements of order 2 are diagonalisable, and they all commute, all matrices in the 2-Sylow of (G 0 ) ′ would be simultaneously diagonalisable in GL 2 (F ℓ 2 ); but there are only 4 diagonal elements of order at most 2 in GL 2 (F ℓ 2 ), while the 2-Sylow of (G 0 ) ′ has order 8. Reasoning as above we then obtain that ℓ ≡ 1 (mod 4), that is, (b). Finally, suppose by contradiction ℓ ≡ 1 (mod 3). Any element g of PG 0 has a lift g in GL 2 (F ℓ ), and such an element has order dividing 6 or 4. Since ℓ ≡ 1 (mod 12), the element g has both its eigenvalues in F × ℓ , so no multiple of g by a scalar in F ℓ 2 \ F ℓ has any F ℓ -rational eigenvalues. It follows that the elements of G 0 whose projective image is g are precisely ±g, hence that G 0 ⊆ GL 2 (F ℓ ). Reasoning as above, this gives a contradiction to the fact that G acts irreducibly on F 4 ℓ .
The above analysis shows that |G| = 48, G contains a subgroup G 0 isomorphic to SL 2 (F 3 ), and ℓ ≡ 2 (mod 3). The problem can now be handled by the methods of Section 3.2, and the result is as follows.
Proposition 3.26. Let G ′ be a maximal subgroup of Sp 4 (F ℓ ) isomorphic to the group GU 2 (F ℓ ).2. The maximal Hasse subgroups G of G ′ are as follows: 3.7 G of type C 6 and S These cases can be handled by the algorithm in Section 3.2. For groups of class S, one also needs to contend with certain subgroups of SL 2 (F ℓ ) whose order depends on ℓ, but these can be excluded using the arguments in [Cul12, Proposition 4]. The results are listed in Table 1 and correspond to Propositions 3 and 4 and Lemmas 2 and 3 of [Cul12].
4 Hasse subgroups of GSp 4 (F ℓ ) that become reducible upon intersection with Sp 4 (F ℓ ) Let G be a Hasse subgroup of GSp 4 (F ℓ ).
Definition 4.1. Let G be a subgroup of GL n (F ℓ ). The saturation G sat of G is the subgroup of GL n (F ℓ ) generated by G and by F × ℓ · Id. We say that G is saturated if G = G sat .
The following lemma is obvious: Lemma 4.2. Let G be a subgroup of GL n (F ℓ ).
1. The groups G and G sat (acting on F n ℓ ) have the same invariant subspaces. In particular, G acts irreducibly if and only if G sat does.
2. G has property (E) if and only if G sat does.

G is Hasse if and only if G sat is.
We note the following formal consequence of the above: Remark 4.4. Let G be a saturated subgroup of GSp 4 (F ℓ ) and let G 1 := G ∩ Sp 4 (F ℓ ). Then (G 1 ) sat coincides with the subgroup of G consisting of elements having square multiplier, which has index at most 2 in G.

Lemma 4.5. Let G be a maximal Hasse subgroup of GSp
, then G = (G 1 ) sat and so G 1 acts irreducibly, contradiction. So, there is δ ∈ F × ℓ \ (F × ℓ ) 2 in the image of λ(G). Hence, λ(G) = F × ℓ . Given a Hasse subgroup G of GSp 4 (F ℓ ) there are two possibilities: either G 1 = G ∩ Sp 4 (F ℓ ) is irreducible, in which case it is one of the groups described in Theorem 3.2, or G 1 is reducible, and is then described by the following result.
• ℓ ≡ 3 (mod 4) and G is conjugate to (C (ℓ−1)/2 .H).2, where H is a subgroup of N GL 2 (F ℓ ) (C s ) of index 2. Under the action of G 1 , the module F 4 ℓ decomposes as the direct sum of two totally isotropic subspaces of dimension 2.
We split the proof into several lemmas. Theorem 4.6 follows from Lemmas 4.12 and 4.13 below, which also give a more explicit description of the groups in question.
Remark 4.7. In the third case of the Theorem, one can prove that PG has order dividing 2 9 · 3 2 · 5 2 .
Remark 4.8. Let G be a maximal Hasse subgroup such that G 1 acts reducibly and corresponds to one of the groups of the first two cases of the theorem. In both cases, G has a subgroup of index 2 that decomposes the module F 4 ℓ as the direct sum of two non-singular subspaces of dimension 2. In the same way, G has a subgroup of index 2 that decomposes F 4 ℓ as the direct sum of two isotropic subspaces of dimension 2. This follows easily from the description of the groups given in Lemma 4.12 and 4.13. In both cases, the base change that exchanges the two non-singular spaces with the two isotropic spaces is the same as in Remark 3.19. The main difference between the two cases is that, when ℓ ≡ 1 (mod 4), then G (that has index 2) decomposes F 4 ℓ in two non-singular subspaces, and, when ℓ ≡ 3 (mod 4), then G decomposes F 4 ℓ in two isotropic subspaces.
Lemma 4.9. Let G be a maximal Hasse subgroup of GSp 4 (F ℓ ). Suppose that G 1 acts reducibly: then there exist two subspaces V 1 , V 2 of F 4 ℓ , both of dimension 2 and irreducible under the action of G 1 , such that F 4 ℓ ∼ = V 1 ⊕ V 2 and with the property that for every g ∈ G \ G one has g(V i ) = V 3−i for i = 1, 2. Finally, either the restriction of the symplectic form to both V 1 and V 2 is trivial, or the restriction of the symplectic form to both V 1 and V 2 is non-degenerate.
Proof. By Corollary 4.3 we know that G is saturated. By Lemma 4.2 (1) we know that G 1 and (G 1 ) sat = G have the same invariant subspaces, so it suffices to prove the result with G 1 replaced by G . Since [G : G ] ≤ 2, it follows from Clifford's theorem that the irreducible G-module F 4 ℓ either stays irreducible upon restriction to G or splits as the direct sum of two irreducible sub-modules of the same dimension. As the first possibility is ruled out by the assumption of the lemma, the first claim follows. As G acts irreducibly, there is an element in G \ G that exchanges V 1 and V 2 (hence the same holds for every element in G \ G ). Let ω be the anti-symmetric bilinear form we consider on F 4 ℓ . The radical of ω| V i is a G -submodule of the irreducible module V i , hence (for each i = 1, 2) it is either trivial or all of V i . Since any element of G \ G exchanges V 1 with V 2 , the same case must happen for both representations V i .
Lemma 4.10. Let G be a maximal Hasse subgroup of GSp 4 (F ℓ ) such that G 1 acts reducibly. Write F 4 ℓ = V 1 ⊕ V 2 as in the previous lemma.
This group preserves the symplectic form of Equation (2).
2. If V 1 , V 2 are both totally isotropic, then up to conjugacy in GL 4 (F ℓ ) the group G is contained in This group preserves the symplectic form of Equation (1).
The following hold: (a) for h = g 1 0 0 g 2 ∈ G ns or h = 0 g 1 g 2 0 ∈ G ns we have λ(h) = det(g 1 ) = det(g 2 ); (c) given a subgroup G of g 1 0 0 g 2 , 0 g 1 g 2 0 g 1 , g 2 ∈ GL 2 (F ℓ ) , denote by G 0 the subgroup of G consisting of block-diagonal matrices. If G is as in the statement of the lemma, all matrices h ∈ G 0 satisfy λ(h) ∈ F ×2 ℓ , and all matrices h ∈ G \ G 0 satisfy λ(h) ∈ F × ℓ \ F ×2 ℓ . Proof. Let e 1 , . . . , e 4 be the standard basis of F 4 ℓ . Up to conjugacy, we may assume that the invariant subspaces are e 1 , e 2 , and e 3 , e 4 . The claim is then easy to check by direct computation, taking into account the fact that every h ∈ G either stabilizes both V 1 , V 2 or exchanges them. Part (c) follows from the fact that, by Lemma 4.9, (G 1 ) sat = G is precisely the subgroup of matrices that send each V i into itself.
Lemma 4.11. Let I be a subgroup of Q 2(ℓ−1) not contained in the subgroup generated by r (see §2.3.1). Let G < I × I be a sub-direct product of I by itself. The group G contains an element of the form (s 1 , s 2 ) with s 1 and s 2 symmetries of Q 2(ℓ−1) .
Proof. As I is not contained in r , the group G contains two elements of the form g 1 = (s ′ 1 , q 1 ) and g 2 = (q 2 , s ′ 2 ), where s ′ 1 , s ′ 2 are symmetries. One of the elements g 1 , g 2 , g 1 g 2 satisfies the conclusion of the lemma.
Recall that we defined G 1 = G ∩ Sp 4 (F ℓ ). We now set G 1 0 = G 0 ∩ G 1 , where G 0 is as in Lemma 4.10.
Lemma 4.12. Let G be a maximal Hasse subgroup of GSp 4 (F ℓ ). Suppose that G 1 acts reducibly on F 4 ℓ and that we are in case 1 of Lemma 4.10. Then, ℓ ≡ 1 (mod 4). Moreover, one of the following holds: 1. G 1 is conjugate to a subgroup of Q 2(ℓ−1) × Q 2(ℓ−1) . The matrices with multiplier a square are block-diagonal with blocks diagonal or anti-diagonal. The matrices with multiplier not a square are block-anti-diagonal with blocks diagonal or anti-diagonal.
2. PG has order smaller than 2 7 · 3 2 · 5 2 In case (1), one of the following holds: (i) G\G 0 contains a block-anti-diagonal matrix M = 0 x y 0 with both x and y diagonal.
(ii) The fourth power of any block-anti-diagonal matrix is a scalar.
Proof. By definition we have G 1 0 < SL 2 (F ℓ ) × SL 2 (F ℓ ). Let I (resp. J) be the projection of G 1 0 on the first (resp. second) factor SL 2 (F ℓ ), and let δ be a fixed generator of F × ℓ . Since G acts irreducibly on F 4 ℓ , it contains an element of the form M = 0 x y 0 with x, y ∈ GL 2 (F ℓ ) and, by Lemma 4.5, we have det x = det y / ∈ F ×2 ℓ . Multiplying the matrix M by a rational constant (recall that G contains all matrices λ Id for λ ∈ F × ℓ ), we can assume det x = det y = δ. The group G 0 has index 2 in G, so it is normal in it, and M belongs to N G (G 0 ). The map ϕ x : J → I defined as ϕ x (j) = xjx −1 induces an isomorphism I → J.
We now proceed as in Section 3.5. The group G 1 0 cannot be a sub-direct product of SL 2 (F ℓ ) by itself, hence G 1 0 < I × J with I ∼ = J proper subgroups of SL 2 (F ℓ ).
• If I is contained in a Borel subgroup, then G 1 0 fixes a line and G does not act irreducibly on F 4 ℓ , contradiction. Note that (G 1 0 ) sat = G 0 by part (3) of Lemma 4.10.
• If I is contained in Q 2(ℓ+1) , then imposing that all of its elements have an F ℓ -rational eigenvalue yields that G does not act irreducibly, unless |I| ≤ 8, in which case |PG| is smaller than 2 7 · 3 2 · 5 2 . This follows from arguments very similar to those in Section 3.5.
Assume that PG is greater than 2 7 · 3 2 · 5 2 . Thanks to Lemma 3.6, x and y are diagonal or anti-diagonal.
Note that I is not cyclic since otherwise G would not act irreducibly. By Lemma 4.11, G contains a matrix of the form (s 1 , s 2 ). If the blocks x and y of M are both anti-diagonal, then multiplying M by (s 1 , s 2 ) we find that G contains a block-anti-diagonal matrix with x and y diagonal. Thus, the following are equivalent: (a) G contains no block-anti-diagonal matrix 0 x ′ y ′ 0 with x ′ and y ′ both diagonal; or vice-versa.
Assume that property (i) in the statement of the lemma does not hold. Then (a) is true, anti-diagonal matrix in Q 2(ℓ−1) , so its square is scalar. We conclude that A 4 is a scalar, that is, (ii) holds.
Lemma 4.13. Let G be a maximal Hasse subgroup of GSp 4 (F ℓ ) and suppose that we are in case (2) of Lemma 4.10. Then we have ℓ ≡ 3 (mod 4) and up to conjugacy in GSp 4 (F ℓ ) the group G is given by where H is a subgroup of index 2 of N GL 2 (F ℓ ) (C s ). In particular, G has order (ℓ − 1) 3 .
Proof. Observe that the group G 1 0 is of the form with H a subgroup of GL 2 (F ℓ ). Proceeding as in the case GL 2 (F ℓ ).2, we can easily show that H < N(C s ) or H is exceptional. Note that the diagonal matrix g 0 0 g −T has an F ℓ -rational eigenvalue if and only if g ∈ GL 2 (F ℓ ) does.
We consider first the case when H is exceptional. We will show that no Hasse subgroups arise in this case. If ℓ ≡ 3 (mod 4), then H cannot contain any elements of order 4, because such elements would not have F ℓ -rational eigenvalues. It is easy to check that a subgroup of GL 2 (F ℓ ) of exceptional type and without elements of order 4 has cyclic projective image, hence it acts reducibly on F 2 ℓ , contradiction. Suppose now that ℓ ≡ 1 (mod 4). Arguing as in Corollary 3.11, we may assume that H contains all the scalars. By the assumption that we are in case (2) of Lemma 4.10 and the surjectivity of the symplectic multiplier (Lemma 4.5) we know that, for every µ ∈ F × ℓ \ F ×2 ℓ , there exists in G an element of the form M := 0 x −µx −T 0 that normalises G . This , which is in the subgroup GL 2 (F ℓ ).2 of Sp 4 (F ℓ ), normalises H sat . Notice that v is not in G (its multiplier is 1, but v is not block-diagonal).
We have described the normaliser of a group like H sat inside GL 2 (F ℓ ).2 in Lemma 3.14.
With notation as in that lemma, this allows us to conclude that x = gJ 2 or x = gJ 2 σ −T .
Multiplying M by g 0 0 g −T −1 ∈ H, we obtain an element of G of the form where the second case can only arise if PH is isomorphic to A 4 and PA 4 is not maximal in PSL 2 (F ℓ ) (see Lemma 3.14 (3)). In particular, H sat is normalised by an element σ ∈ SL 2 (F ℓ ) with Pσ representing a transposition in P( H, σ ) ∼ = S 4 . Note that σ 2 = − Id.
If G contains an element of the form u ′ (which is automatic if PH ∼ = A 4 ) we get a contradiction: it is clear that u ′ does not have F ℓ -rational eigenvalues, since the product of the off-diagonal blocks is −µJ 2 2 = µ, whose eigenvalues are not squares in F × ℓ (see Remark 3.7). If instead G contains an element of the form u ′′ (hence in particular PH ∼ = A 4 ), then similarly −µJ 2 σ −T J 2 σ = −µ σ 2 det σ = µ, contradiction. Hence H cannot be an exceptional subgroup.
So we may assume that H is a subgroup of N(C s ) and H = H sat . In particular, the condition that every element of H has an F ℓ -rational eigenvalue gives is not diagonal or anti-diagonal, then we are in the second case of Lemma 3.5 and ℓ ≡ 1 (mod 4). In this case, up to multiplying M by an element of G 1 0 , we can then assume that x is symmetric, which implies M 2 = −µ. Therefore, M ℓ−1 = (−µ) (ℓ−1)/2 = − Id since µ is not a square, which is absurd since M must have a rational eigenvalue. Otherwise, if we are in the first case of Lemma 3.5, up to multiplying M by an element of G 1 0 we can assume Observe that M 2 = (−µ) and M ℓ−1 = (−µ) (ℓ−1)/2 . If −1 is a square mod ℓ, then M ℓ−1 = − Id and M does not have a rational eigenvalue, contradiction. Therefore, −1 must not be a square, that is, ℓ ≡ 3 (mod 4), and we can take µ = −1. One checks that 0 If we show that G ′ is Hasse, then necessarily G = F × ℓ · G ′ since G is maximal. The fact that G ′ acts irreducibly follows from the character formula, similarly to the case GL 2 (F ℓ ).2. The fact that every matrix has a rational eigenvalue follows from the fact that every matrix has order that divides ℓ − 1.

Hasse subgroups of GSp 4 (F ℓ )
The goal of this section is to describe all maximal Hasse subgroups of GSp 4 (F ℓ ) having surjective multiplier.
Definition 5.1. Let G 1 be a Hasse subgroup of Sp 4 (F ℓ ). If G 1 is not contained in one of the groups of the first three cases of Theorem 3.2, then we say that G 1 is exceptional.
Lemma 5.2. Let G be a subgroup of GSp 4 (F ℓ ) containing the scalar multiples of Id and such that λ(G) = F × ℓ . Let G 1 = G ∩ Sp 4 (F ℓ ). The index [PG : PG 1 ] is at most 2.
Proof. The kernel of the projection π : G → PG has order |F × ℓ | = ℓ − 1, while G 1 → PG 1 has kernel of order k ≤ 2 (the only scalar matrices in Sp 4 (F ℓ ) are ± Id). On the other hand, |G|/|G 1 | = |λ(G)| = ℓ − 1. It follows that [PG : Lemma 5.3. Let G be a maximal Hasse subgroup of GSp 4 (F ℓ ) with λ(G) = F × ℓ such that G 1 = G ∩ Sp 4 (F ℓ ) acts irreducibly. One of the following holds: • G 1 is of class C 2 . In particular, as in Section 3.5 we can choose a basis of F 4 ℓ with respect to which all elements in G are either block-diagonal or block-anti-diagonal.
Proof. As G 1 acts irreducibly, it is a Hasse subgroup of Sp 4 (F ℓ ). The assumption λ(G) = F × ℓ implies that PG is not contained in P Sp 4 (F ℓ ). Thus there exists a maximal subgroup M of P GSp 4 (F ℓ ) with M = P Sp 4 (F ℓ ) and M containing PG. We let M be the inverse image of M in GSp 4 (F ℓ ). The maximal subgroups M of P GSp 4 (F ℓ ) are classified in [BHRD13, Tables 8.12 and 8.13].
1. Suppose first that M is of Aschbacher type C i for some i = 2, or lies in class S. Then by definition G 1 is contained in a maximal subgroup of Sp 4 (F ℓ ) of the same Aschbacher type, or is of class S. By Theorem 3.2, Table 1, and Definition 5.1, G 1 is exceptional.
2. Suppose instead that M is of Aschbacher type C 2 . By definition, M (hence also G) preserves a decomposition of F 4 ℓ as the direct sum of two 2-dimensional subspaces: thus, in a suitable basis, all matrices in M are either block-diagonal or block-anti-diagonal. Note that by Theorem 3.2 we know that G 1 is contained in a maximal subgroup isomorphic to SL 2 (F ℓ ) ≀ S 2 and the present choice of basis is compatible with that of Section 3.5.
Lemma 5.4. Let G be a Hasse subgroup of GSp 4 (F ℓ ) such that λ(G) = F × ℓ . If G 1 is not exceptional, then it acts reducibly.
Proof. Suppose by contradiction that G 1 acts irreducibly. Up to conjugacy, G 1 is contained in a maximal Hasse subgroup of one of the first three types listed in Theorem 3.2. In particular, we have ℓ ≡ 1 (mod 4). By Lemma 5.3, we can assume that every matrix in G is block-diagonal or block-anti-diagonal. We will find a contradiction by showing that G contains a matrix without rational eigenvalues. Note that we can assume that G contains all the scalars.
1. Assume G 1 < (Q 2(ℓ−1) × Q 2(ℓ−1) ).C 2 . As we did in Section 3.5, we write elements of (Q 2(ℓ−1) × Q 2(ℓ−1) ).C 2 as triples (g, h, ±1). As above, G 1 contains a block-anti-diagonal matrix. Let M ∈ G be an operator with λ(M) = δ. Multiplying if necessary M by a block-anti-diagonal matrix in G 1 , we can assume that M is block-diagonal. So, nor anti-diagonal, then G 1 < (Q 8 × Q 8 ).C 2 thanks to Lemma 3.6. In this case G 1 is exceptional, contradiction. So, we can assume that M 1 and M 2 are diagonal or antidiagonal. By Lemma 4.11, we can assume that M ′ = (s 1 , s 2 , 1) is in G 1 . So, without loss of generality, we can assume that M 1 is diagonal. If M 2 is diagonal, then M ′ M does not have a rational eigenvalue and G is not Hasse. If M 2 is anti-diagonal, then M 2 = δ(r a , ±1, 1) with a odd. Let M 3 ∈ G 1 \ G 1 0 . As we showed in Section 3.5.4, M 3 = (q 1 , q 2 , −1) with q 1 , q 2 ∈ Q 2(ℓ−1) . There are three possible cases: r d , −1). Under the assumption that M 3 has a rational eigenvalue, the order of M 3 divides ℓ − 1 and c + d is even. So, M 2 M 3 = δ(r a+c , ±r d , −1) does not have a rational eigenvalue since a + c + d is odd. Hence, G is not Hasse.
• M 3 = (s 3 , s 4 , −1) with s 3 and s 4 symmetries. Then, M ′ 3 = M ′ M 3 is of the form (r c , r d , −1). So, one between M ′ 3 and M 2 M ′ 3 does not have a rational eigenvalue, as we proved in the previous case.
• M 3 = (q 1 , q 2 , −1) with q 1 q 2 a symmetry. Multiplying by M, we see that G contains an element of the form N = 0 N 1 N 2 0 with det(N 1 ) = det(N 2 ) = δ and N 1 and N 2 both diagonal. Since N has a rational eigenvalue, we have (N 1 N 2 ) (ℓ−1)/2 = 1. In this case, M 2 N does not have a rational eigenvalue, contradiction.
2. Assume that G 1 < (N GL 2 (F ℓ )(Cs) .2). From Remark 3.19, G 1 is contained in a maximal group of the previous case and so the lemma holds from the previous one.
3. Assume G 1 < (C (ℓ−1)/2 .E).2 with E exceptional. We know that G 1 0 has projective image A 4 , A 5 , or S 4 . Assume G 1 0 has projective image A 5 or S 4 . Proceeding as above, we obtain that G contains an element of the form N = 0 x −δx −T 0 . Observe that x normalises G 1 0 , so, as we pointed out in the proof of Lemma 3.14, x is in G 1 0 (when we see it as a subgroup of GL 2 (F ℓ )). So, M = Assume now that G 1 0 has projective image A 4 . The normaliser of G 1 0 in GL 2 (F ℓ ), that we denote with G ′ , has projective image contained in S 4 . Since G 1 acts irreducibly, it contains a matrix of the form M 2 = 0 y −y −T 0 , and since λ(G) = F × ℓ the group G contains a matrix of the form M 1 = 0 x −δx −T 0 (notice that, up to multiplication by M 2 , we can assume that M 1 is block-anti-diagonal). Since x normalises G 1 0 , it belongs to G ′ . If x ∈ G 1 0 , we conclude as in the case projective image A 5 or S 4 . Otherwise, we may assume that PG ′ = S 4 and that x is an element of G ′ \ G 1 0 . Since [G ′ : G 1 0 ] = 2 all elements in G ′ \ G 1 0 appear as x for some choice of M 1 (simply multiply by a suitable element in Passing to the projective quotient, this induces an automorphism ϕ of order ≤ 2 of PG ′ ∼ = S 4 . All automorphisms of S 4 are inner, so ϕ is conjugation by some element w ∈ S 4 of order ≤ 2. In particular, ϕ(w) = w, so if x ∈ G ′ \G 1 0 lifts w we have x −T = ±x and xx −T = ± Id. Now for this x we have M 2 1 = ±δ Id, hence M ℓ−1 1 = − Id and M 1 does not have any rational eigenvalues, contradiction.
Remark 5.7. With more work in the style of Section 3, one could probably improve the bound on the order of |PG| in the third case of the theorem, and also classify the groups of the form PG that arise from the Hasse subgroups of GSp 4 (F ℓ ). We have decided not to pursue this, since the qualitative form of the result given above will be enough for our applications Remark 5.8. The assumption λ(G) = F × ℓ is less restrictive than it may seem: indeed, by Corollary 4.3 we know that for every maximal Hasse subgroup G of GSp 4 (F ℓ ) the multiplier group λ(G) contains λ(F × ℓ Id) = F ×2 ℓ . The assumption λ(G) = F × ℓ is then equivalent to the requirement that G contains an element whose multiplier is not a square. If this is not the case, then G is simply the saturation of G 1 , which is a Hasse subgroup of Sp 4 (F ℓ ). These cases are therefore already covered by Theorem 3.2. 6 Strong counterexamples 6.1 Statement of the main result Theorem 6.1. Let A be an abelian surface defined over a number field K. There exists a constant C 1 , depending only on K, such that the following hold for all primes ℓ > C 1 .
• If End K (A) is an order O in a real quadratic field, there exists an extension K ′ /K, of degree at most 2, such that End is not a strong counterexample. In particular, if all the endomorphisms of A are defined over K, then (A, ℓ) is not a strong counterexample.
• If A K is isogenous to the square of an elliptic curve E without CM, then there exists an extension K ′ /K of degree at most 3 such that A K ′ is either isogenous to the product of two elliptic curves or satisfies that End K ′ (A) ⊗ Q is a quadratic field. If [K ′ : K] = 1 or 3, then (A, ℓ) is not a strong counterexample. If [K ′ : K] = 2 and ℓ is unramified in K ′ , then (A, ℓ) is not a strong counterexample.
• If End K (A) is an order in a (nonsplit) quaternion algebra and End K (A) is an order in a quaternion algebra or an order in a quadratic field, then (A, ℓ) is not a strong counterexample. If End K (A) = Z, then there is a field extension K ′ /K of degree 2 such that End K ′ (A) is an order in a quadratic field. If ℓ is unramified in K ′ , then (A, ℓ) is not a strong counterexample.
• If End K (A) is an order in a CM field, then (A, ℓ) is not a strong counterexample.
Strong counterexamples (A, ℓ) for which A is geometrically isogenous to the square of an elliptic with CM are not bounded in the same sense as in the above theorem. Indeed, as we will show in Proposition 6.28, we can find infinitely many ℓ such that there exists an abelian surface defined over Q and geometrically isogenous to the square of an elliptic curve with CM such that (A, ℓ) is a strong counterexample.
We will also obtain the following consequence: Corollary 6.2. Let A be an abelian surface defined over a number field K. Assume that End K (A) = Z. There exists a constant C 1 , depending only on K, such that (A, ℓ) is not a strong counterexample for ℓ > C 1 .
We will make the following assumptions on ℓ: • ℓ is unramified in K.

Lower bounds on the image of Galois
We shall need the following result, proven in [SZ05]: Theorem 6.3. Let A be an abelian surface over a number field K, and let v be a place of K. Let L be a minimal extension of K over which A acquires semi-stable reduction at a place w above v. Suppose that the residue characteristic of v is at least 7: then the ramification index e(w|v) is bounded by 12.
From now on, we will always assume that ℓ ≥ 7, so that the previous theorem applies. • V has a structure of F ℓ n -vector space; • the action of I t v on V is given by a character ψ : I t v → F × ℓ n ; • ψ = ϕ e 1 1 . . . ϕ en n , where ϕ 1 , . . . , ϕ n are the fundamental characters of I t v of level n; • for every i = 1, . . . , n the inequality 0 ≤ e i ≤ e holds.
Remark 6.5. Raynaud's theorem is usually stated for places of good reduction. However, as shown in [LV14,Lemma 4.9], the extension to the semi-stable case follows easily upon applying results of Grothendieck [Gro71].
Theorem 6.6. Let A/K be an abelian surface over a number field K. Given a finite group G, we write exp(G) = mcm{ord(g) : g ∈ G}.
2. Without the assumption of semi-stable reduction, we have exp(PG ℓ ) ≥ 1 12 Proof. We first show that the first statement implies the second. Let L/K be a minimal extension of K over which A acquires semi-stable reduction at some place of characteristic ℓ. Since ℓ > 5, by Theorem 6.3 we have [L : K] ≤ 12 (hence [L : Q] ≤ 12[K : Q]), and since clearly exp (Pρ ℓ (G K )) ≥ exp (Pρ ℓ (G L )) the claim follows from part (1) applied to A/L. We now prove part 1. Consider the action of an inertia group I v at v on A[ℓ]. If the wild inertia subgroup (which is pro-ℓ) acts non-trivially, then G ℓ contains an element of order ℓ, and since ker(G ℓ → PG ℓ ) has order prime to ℓ we see that PG ℓ contains an element of order ℓ, so that exp(PG ℓ ) ≥ ℓ and we are done. We may therefore assume that the wild inertia subgroup acts trivially, hence that the action of I v on A[ℓ] factors through I t v , the tame inertia quotient. Recall that this is a pro-cyclic group, hence all its finite homomorphic images are cyclic.
The representation ρ ℓ induces, by restriction to I v and then passage to the quotient I t v , a group homomorphism (which we still denote by ρ ℓ ) from I t v to G ℓ . By composing with the projection G ℓ → PG ℓ , we obtain a map φ : I t v → PG ℓ , and it suffices to show that the image of this map has order at least ℓ−1 [K:Q] . Indeed, the image of this map is cyclic, , we now want to study the kernel of φ. Furthermore, since [K : Q] ≥ e(v|ℓ), it suffices to show the theorem with [K : Q] replaced by the ramification index e := e(v|ℓ).
If σ ∈ I t v lies in the kernel of φ, then ρ ℓ (σ) is a scalar matrix. Notice that A[ℓ] is a semisimple I t v -module, because ρ ℓ (I v ) has no elements of order ℓ. Write A[ℓ] ∼ = W i , where W i is irreducible and of dimension l i . By Theorem 6.4, the eigenvalues of ρ ℓ (σ)| W i are given by the conjugates of ψ i = ϕ a i l i , where ϕ l i is a fundamental character of level l i and if we write a i = a i,0 + a i,1 ℓ + · · · + a i,l i −1 ℓ l i −1 we have 0 ≤ a i,j ≤ e. Moreover, if i > 1 then we cannot have a i,0 = . . . = a i,l i −1 (otherwise, ψ i = χ a i,0 ℓ would take values in F × ℓ and W i would not be irreducible, see also [Lom16b, Proposition 3.15]). We distinguish several cases: 1. At least one l i is 2 or more. Without loss of generality, assume that l 1 ≥ 2, and let ϕ b = ϕ a 1 l 1 be a character giving one of the eigenvalues of the action of inertia. Write for simplicity ϕ := ϕ l 1 and b := Notice that ϕ(σ) b and ϕ(σ) ℓb are both eigenvalues of ρ ℓ (σ), so if ρ ℓ (σ) is a scalar we must have ϕ(σ) b(ℓ−1) = 1. Since I t v is a pro-cyclic group, the subgroup H = {σ ∈ I t v : ϕ(σ) b = ϕ(σ) bℓ } is also pro-cyclic, and its index in I t v is Now b 0 + b 1 ℓ + · · · + b l−1 ℓ l−1 , 1 + ℓ + · · · + ℓ l−1 is equal to is non-zero since we already remarked that the b i cannot all be equal. It follows that the denominator of (10) is at most e(1 + ℓ + · · · + ℓ l−2 ) = e ℓ l−1 −1 ℓ−1 , and therefore |( − 1). It follows in particular that Pρ ℓ (I v ) has order at least ℓ(ℓ−1) e > ℓ−1 e . 2. All l i are equal to 1, at least one character ψ i is trivial, and at least one character ψ j is non-trivial. Write ψ j = χ b ℓ with b > 0. For every σ ∈ I t v the endomorphism ρ ℓ (σ) admits 1 as an eigenvalue, and therefore ker φ is contained in {σ ∈ I t v : χ b ℓ (σ) = 1}, which has index (ℓ − 1, b) in I. Since b ≤ e, the claim follows.
3. All l i are equal to 1, and there are two indices i, j such that a i = a j . Write b 1 = a i and b 2 = a j . We have ker φ ⊆ {σ ∈ I v : χ ℓ (σ) b 1 −b 2 = 1}, which again has index at least 4. All l i are equal to 1 and all the a i are equal to each other. We show that this case cannot arise for ℓ > 2[K : Q]. All the characters ϕ a i l i are equal to χ b ℓ for some b with 0 ≤ b ≤ e. Then for every σ ∈ I t v we have χ ℓ (σ) = λ(ρ ℓ (σ)) = χ 2b ℓ (σ), whence ℓ − 1 | 2b − 1 ≤ 2e − 1 ≤ 2[K : Q] − 1, contradicting our assumption ℓ > 2[K : Q].

Preliminary lemmas
For simplicity of notation, from now on we write ρ instead of ρ ℓ . We choose a place v of K of characteristic ℓ and let I v < Gal K/K be a corresponding inertia group.
Lemma 6.8. Let A be an abelian surface defined over a number field K. Let ℓ > C 1 be a prime and let G = ρ(Gal(K/K)). Assume that (A, ℓ) is a strong counterexample, so that G is Hasse. The order of PG is at least 2 7 · 3 2 · 5 2 . Up to conjugacy, G contains only block-diagonal and block-anti-diagonal matrices, with blocks that are diagonal or antidiagonal. The matrices whose multiplier is a square are block-diagonal, and the matrices whose multiplier is not a square are block-anti-diagonal. Moreover, 1 (mod 4), then G is contained in a group as in Lemma 4.12, case (1).
• If ℓ ≡ 3 (mod 4), then G is contained in the group described in Lemma 4.13.
Every element of G has four rational eigenvalues and λ(G) = F × ℓ . Finally, G contains a matrix M of the form 0 x y 0 such that the following all hold: x and y are either both diagonal or both anti-diagonal, λ(M) generates F × ℓ , and M 4 is not a scalar.
Proof. Since ℓ is unramified in K by the assumption ℓ > C 1 , the multiplier of G is χ ℓ (Gal K/K ) = F × ℓ . As (A, ℓ) is a strong counterexample, it follows that up to conjugacy G is contained in one of the groups described in Theorem 5.5. By Theorem 6.6, the order of PG is greater than 2 7 · 3 2 · 5 2 since ℓ > C 1 . So, if ℓ ≡ 3 (mod 4), then G is necessarily contained in the group described in Lemma 4.13. If ℓ ≡ 1 (mod 4), then G is contained in a group as in Lemma 4.12, case (1). From these explicit descriptions the first part of the lemma follows easily.
Let M = ρ(σ) be an element in ρ(I v ) such that λ(M) generates F × ℓ . Such an element exists because ℓ is unramified in K (since ℓ > C 1 ). By Corollary 6.7, M 4e is not a scalar, hence M 4 is not a scalar. Since the multiplier of M is not a square, M is a block-antidiagonal matrix of the form 0 x y 0 . By what we already proved, x and y are diagonal or anti-diagonal. We just need to show that it is impossible for x to be diagonal and y anti-diagonal (or viceversa). If this were the case, by direct computation M 4 would be a scalar, contradiction.
Lemma 6.9. Let G be as in Lemma 6.8 and let M be as in the conclusion of that lemma. The matrix M has four different eigenvalues.
Given a ring R, we will denote by Nilrad(R) the ideal of nilpotent elements.
Lemma 6.10. Let R = End K (A) be an order in a field. If ℓ is ramified in R ⊗ Q or it divides the conductor of R, then Nilrad(R ⊗ F ℓ ) is non-trivial and Gal(K/K)-invariant.
Proof. The assumptions imply that R ⊗ F ℓ is not a product of fields. The ring R ⊗ F ℓ is finite, hence Artinian. Every Artinian ring can be written as a product of Artinian local rings. Hence, R ⊗F ℓ is isomorphic to A i , where at least one of the A i is not a field, hence contains a non-trivial non-invertible element. Since Nilrad(R ⊗ F ℓ ) = i Nilrad (A i ), the claim follows from the well-known fact that a finite local Artinian ring A with a non-zero non-invertible element has non-trivial nilradical. Finally, Nilrad(R ⊗ F ℓ ) is Gal(K/K)invariant because the condition x n = 0 clearly is.
Lemma 6.11. Any group G as in Lemma 6.8 contains at most 4(ℓ − 1) 2 diagonal matrices having at most 3 distinct eigenvalues.
Proof. Assume ℓ ≡ 3 (mod 4), so that G is contained in the group described in Lemma 4.13. Then, the eigenvalues of a diagonal matrix are µδ ±i , µδ ±j with µ ∈ F × ℓ , i + j even, and δ a generator of F × ℓ . If a 4 × 4 matrix has at most three different eigenvalues, then two of them are equal.
If δ i = δ j , then we have ℓ − 1 choices for i, one choice for j and (ℓ − 1)/2 choices for µ (up to sign). So, there are (ℓ − 1) 2 /2 matrices such that δ i = δ j . The same holds for every other pair of eigenvalues. Since there are 6 pairs to consider, there are at most 3(ℓ − 1) 2 diagonal matrices with at most three different eigenvalues.
If instead ℓ ≡ 1 (mod 4), then G is in particular contained in a group as in Lemma 4.12, case (1) . Then, the eigenvalues of a diagonal matrix are µδ ±a , µδ ±b . Reasoning as above we see that there are at most (ℓ − 1) 2 /2 matrices such that δ ±a = δ ±b . Moreover, we have at most (ℓ − 1) 2 matrices such that δ a = δ −a , and at most (ℓ − 1) 2 matrices such that δ b = δ −b . In conclusion, there are at most 4(ℓ − 1) 2 matrices with at most three different eigenvalues.
Lemma 6.12. Let ρ : G → GL(V ) be a 4-dimensional representation of a group G. Assume that V splits as V = V 1 ⊕ V 2 , where V 1 and V 2 are two-dimensional G-invariant subspaces. Suppose that there is λ = 0, 1 and an element g of G such that ρ(g)(v 1 ) = v 1 for all v 1 ∈ V 1 and ρ(g)(v 2 ) = λv 2 for all v 2 ∈ V 2 . Then at least one of the following holds: 1. V 1 and V 2 are the only G-invariant subspaces of dimension 2; 2. there exists a G-invariant subspace of dimension 1.
Proof. The assumptions imply that g commutes with every h ∈ G: the restrictions of g, h to V 1 , V 2 commute since g| V i is a scalar. Notice that V 1 , V 2 are the eigenspaces of g. Since g is in the center, every element of G preserves the eigenspaces of g, hence every G-invariant subspace W splits as (W ∩ V 1 ) ⊕ (W ∩ V 2 ), which easily implies the statement. Proof. First note that if M is a 2 × 2 diagonal matrix and N is a 2 × 2 diagonal or antidiagonal matrix, then NMN −1 is diagonal. From this it follows easily that D is normal in G. Assume ℓ ≡ 3 (mod 4). In this case, D has index 4, with cosets represented by with A (resp. B) diagonal (resp. anti-diagonal). Note that every one of these cosets must appear since G acts irreducibly. Therefore, G/D has order 4 and every element has order that divides 2, so G/D ∼ = (Z/2Z) 2 .
Assume now ℓ ≡ 1 (mod 4). As we showed at the end of the proof of Lemma 4.12, there are eight possible cosets, namely r 1 0 0 r 2 , 0 xr 1 yr 2 0 , s 1 0 0 s 2 , 0 xs 1 ys 2 0 r 1 0 0 s 2 , 0 xr 1 ys 2 0 , s 1 0 0 r 2 , 0 xs 1 yr 2 0 with r i diagonal, s i anti-diagonal, and x and y diagonal. As we observed in Lemmas 4.12 and 6.8, G must contain elements from each of the first 4 cosets since it acts irreducibly. From this it follows easily that either G/D has order 4, in which case G/D ∼ = (Z/2Z) 2 , or it has order 8, and is then isomorphic to D 4 .
Lemma 6.14. Let G be a group as in Lemma 6.8 and let G ′ be a subgroup of index 2 of G such that G ′ acts reducibly on A[ℓ]. Let D < G be the subgroup of diagonal matrices of G and D ′ < G ′ be the subgroup of diagonal matrices of G ′ . Assume that G ′ contains a block-anti-diagonal matrix whose square is not a scalar. Then, [G ′ : D ′ ] = 2.
Proof. We assume that [G • Assume that H ′ acts reducibly on V 1 and V 2 . We have V 1 = V 1,1 ⊕ V 1,2 , with each of the two 1-dimensional subspaces invariant under the action of H ′ . Denote by H ′ 1 the projection of H ′ to GL(V 1 ) ∼ = GL 2 (F ℓ ). All elements in H ′ 1 are simultaneously diagonalisable by the assumption that H ′ acts reducibly on V 1 , hence in particular H ′ 1 is commutative. Since anti-diagonal matrices commute if and only if they differ by a scalar, every diagonal matrix in H ′ 1 is a scalar. The same holds for V 2 , so the diagonal matrices in H ′ (hence also in G ′ ) are block-scalar.
Suppose instead that ℓ ≡ 3 (mod 4). Let M ∈ G ′ be a block-anti-diagonal matrix. Using Equation (9) one can easily check that, if M 2 is block-scalar, then it is a scalar. So, the square of every block-anti-diagonal matrix in G ′ is a scalar. This contradicts the hypothesis.
Assume now ℓ ≡ 1 (mod 4). Note that, since χ 1 = χ 2 , the group H ′ contains no matrices of the form r a s 1 where s 1 is a symmetry in Q 2(ℓ−1) , unless H ′ is a sub-direct product of Q 8 × Q 8 . In this case, |G| = 4|H ′ | ≤ 2 8 , contradicting Lemma 6.8. Therefore 1 is a scalar, say M 2 1 = λ, then λ 2 = det x det y. But det x det y = (det x) 2 / ∈ F ×4 ℓ , while λ 2 is a fourth power since λ is an eigenvalue of xy, which is a square by Remark 3.7. So, M 2 1 cannot be a scalar. Hence, M 2 1 = diag(a, b, a, b) with a = b. Similarly, G contains a matrix M 2 = 0 x 2 y 2 0 with x 2 and y 2 anti-diagonal and diag(a, b, a, b)v and diag(a, b, b, a)v contains at least one of the basis vectors e i . We assume e 1 ∈ W , the other cases being identical. Multiplying e 1 by a block-diagonal but non-diagonal matrix in G ′ we have that e 2 ∈ W . So, W = e 1 , e 2 . Multiplying e 1 by an anti-block-diagonal we have that e 3 ∈ W or e 4 ∈ W , contradiction.
Lemma 6.15. Let K be a number field and let (A, ℓ) be a strong counterexample with ℓ > C 1 . Assume that there exists a degree-2 extension K ′ of K such that ρ(Gal(K/K ′ )) acts reducibly. Assume that ℓ is unramified in K ′ . The following hold: • There exist precisely two ρ(Gal(K/K ′ ))-invariant subspaces V 1 and V 2 of dimension 2.
• Let v K ′ be a place of K ′ and let L be a minimal extension of K ′ over which A acquires semi-stable reduction at a place above v K ′ . Let v L be a place of L above v K ′ and e = e(v L | v K ′ ) ≤ 12 be its ramification index. Choose σ in an inertia group corresponding to v K ′ with the property that χ ℓ (σ) generates F × ℓ and let M = ρ(σ) ∈ Gal(K/K ′ ). Up to exchanging V 1 and V 2 , we have M 2e |V 1 = Id and M 2e |V 2 = χ ℓ (σ 2e ).
2. Assume that End K (A) contains an order O in the (not necessarily real) quadratic field L = Q( √ d). If ℓ > C 1 , then (A, ℓ) is not a strong counterexample.
Proof. We begin with the proof of part (1). Let c be the conductor • If ℓ divides c or is ramified in O ℓ , then by Lemma 6.10 we have that Nilrad(O ℓ ) ⊂ O ℓ is nontrivial and Galois-stable, hence so is the subspace Nilrad is not a strong counterexample.
be an eigenvector with eigenvalue λ. Observe that λ ∈ F ℓ by Lemma 6.8 and that j(M) · j(v) = λj(v), so each eigenvalue of M is also an eigenvalue of j(M). Thus, M has at most two different eigenvalues.
Assume that (A, ℓ) is a strong counterexample. Up to conjugacy we may then assume that G is as in Lemma 6.8. Let M be the element of G whose existence is assured by that result: by Lemma 6.9, M has four different eigenvalues, contradiction.
For part (2), in the first two cases we immediately get nontrivial Galois-invariant subspaces defined over K, while the third case is handled exactly as above.

Squares of elliptic curves
We will need the following lemma, that is contained in [FKRS12,Proposition 4.7]: Lemma 6.17. Let K be a number field and let A/K be an abelian surface such that A K is isogenous to the square of an elliptic curve E without CM. There exists an extension K ′ /K of degree at most 3 such that A K ′ is either isogenous to the product of two elliptic curves or satisfies that End K ′ (A) ⊗ Q is a quadratic field. Moreover, this quadratic field can be taken to be either real or equal to Q(ζ n ) with n ∈ {3, 4, 6}.
Lemma 6.18. In the setting of the previous lemma, suppose that R = End K ′ (A) is an order in a quadratic field. Let ℓ > 2 be a prime that does not divide the conductor of R and splits in R ⊗ Q. The action of R ⊗ F ℓ ∼ = F 2 ℓ decomposes A[ℓ] as the direct sum of two 2-dimensional sub-modules W 1 , W 2 , corresponding to the non-trivial idempotents of F 2 ℓ . The determinant of the action of Gal K ′ /K ′ on each of W 1 , W 2 is the product of the cyclotomic character with a character of order dividing 4 or 6.
Similarly, if ℓ divides the conductor of R or ramifies in R ⊗ Q, let x be a non-trivial nilpotent element in R ⊗ F ℓ . Let V be the kernel of the action of x on A[ℓ]. Then V is a 2dimensional subspace with the following property: for all σ ∈ Gal K ′ /K ′ , the determinant of ρ(σ | V ) is χ ℓ (σ)ε(σ) for some character ε of order dividing 4 or 6.
is a real quadratic field, this follows (in a stronger form) from [Rib76, Lemma 4.5.1], see also the comments on page 784 of [Rib76]. For the general case, note that W 1 , W 2 are the reduction modulo ℓ of Z ℓ -sub-modules W 1 , W 2 (each of rank 2) of T ℓ (A), coming from the decomposition R ⊗ Z ℓ ∼ = Z 2 ℓ , so it suffices to prove that the determinant of the action of σ ∈ Gal K/K on W i is given by the product of the ℓ-adic cyclotomic character and a character of order dividing 4 or 6. Since T ℓ (A) embeds into T ℓ (A) ⊗ Z ℓ Q ℓ =: V ℓ (A), it suffices to work with the latter. Let W 1 , W 2 be the subspaces of V ℓ (A) corresponding to W 1 , W 2 .
Let L be the minimal (Galois) extension of K over which all the endomorphisms of A are defined. By [FKRS12, Theorem 3.4 and Table 8], the degree [L : K] divides 8 or 12, and [L : K ′ ] divides 4 or 6 (indeed, if [L : K] = 12 or 8, then K ′ /K is a nontrivial extension). There exists an L-isogeny A → E 2 , which induces an isomorphism ψ : V ℓ (A) → V ℓ (E 2 ) = V ℓ (E) 2 . We will use ψ to identify W 1 , W 2 to subspaces of V ℓ (E 2 ) that we still denote by the same symbol. Note that ψ is equivariant for the action of the absolute Galois group of L.
The hypothesis that ℓ splits in Q( β)x + λ 12 y = 0. Now observe that β cannot be a rational number (since d is not a square in Q), so λ 11 − β is non-zero. This shows that W 1 = ker(M − β) is the graph of the (Gal L/L -equivariant) map so the determinant of the action of Gal(L/L) on W 1 is the same as the determinant of the action on V ℓ (E), namely, the cyclotomic character. A similar argument applies to W 2 , and shows that for i = 1, 2 one has det (σ | W i ) = χ ℓ (σ) for all σ ∈ Gal(L/L). Finally, consider the character ε i (σ) = det (σ | W i ) · χ ℓ (σ) −1 , defined on all of Gal K ′ /K ′ . By the above, ε is trivial on Gal(L/L), so its image has order dividing [Gal(K ′ /K ′ ) : Gal(K ′ /L)] = [L : As already observed, this quantity divides 4 or 6, which proves the lemma. The second half of the statement is proved in the same way.
Theorem 6.19. Let K be a number field and let A/K be an abelian surface such that A K is isogenous to the square of an elliptic curve E without CM. Let K ′ be as in Lemma 6.17. If ℓ is unramified in K ′ and ℓ > C 1 , then (A, ℓ) is not a strong counterexample.
Assume now that R = End K ′ (A) is an order in a quadratic field. If ℓ ramifies in R or divides its conductor, Lemma 6.10 implies that A[ℓ] is reducible under the action of Gal(K/K ′ ). If [K ′ : K] is equal 1 or 3, then we conclude as above by Clifford's theorem. If [K ′ : K] = 2, then we are in the hypotheses of Lemma 6.15. Reasoning as in the proof of Theorem 6.16, but replacing [Rib76, Lemma 4.5.1] with Lemma 6.18, we find that there are a 2-dimensional subspace V of A[ℓ], an element M 2e = ρ(σ 2e ), and an element ζ ∈ F × ℓ of order dividing 12 such that det ρ(σ 2e ) V = ζδ 2e = 1 or δ 4e .
Raising to the 12th power, this implies δ 24e = 1, which contradicts the fact that 0 < 24e ≤ 24 · 12 < ℓ − 1. The same argument applies if ℓ does not divide the conductor of R and splits in R ⊗ Q. Finally, if ℓ is inert, the proof is identical to the proof of Theorem 6.16 in the inert case.

Quaternion algebra
Theorem 6.20. Let A be an abelian surface over a number field K. Assume that End K (A) is an order in a quaternion algebra and that ℓ > C 1 . If End K (A) is an order in a quaternion algebra or an order in a quadratic field, then (A, ℓ) is not a strong counterexample. If End K (A) = Z, then there is a field extension K ′ /K of degree 2 such that End K ′ (A) is an order in a quadratic field. If ℓ is unramified in K ′ , then (A, ℓ) is not a strong counterexample.
Proof. Assume by contradiction that (A, ℓ) is a strong counterexample. Let R = End K (A), R = End K (A), and write R ℓ = R ⊗ F ℓ , R ℓ = R ⊗ F ℓ . If R = Z we are done by Theorem 6.16. Assume instead that R = Z. Table 8 in [FKRS12] then shows that the Sato-Tate group of A/K must be of type J(E n ) for some n ∈ {2, 3, 4, 6}. In this case, there exists a quadratic extension K ′ /K such that the Sato-Tate group of A over K ′ is of type E n , and from [FKRS12, Table 8] we see that End K ′ (A) ⊗ Q is an (imaginary) quadratic number field. Let R ′ = End K ′ (A) and R ′ ℓ = R ′ ⊗ F ℓ . If the Jacobson radical J := rad(R ℓ ) of R ℓ is non-trivial, then J is a Galois-invariant ideal in R ℓ , hence A[ℓ][J] := {x ∈ A[ℓ] : jx = 0 ∀j ∈ J} is a non-trivial, Galois-invariant subspace of A[ℓ] defined over K. This cannot happen since we are assuming that (A, ℓ) is a strong counterexample, hence we may assume that J = (0). The condition J = (0) implies that R ℓ is semisimple, that is, it is a direct product of simple algebras. However, a simple algebra of dimension at most 3 is commutative, and the product of commutative algebras is commutative, so R ℓ cannot be a non-trivial product. Therefore, R ℓ is simple. As the Brauer group of finite fields is trivial, this implies that R ℓ is a matrix algebra over some finite field F ℓ k . Combined with dim F ℓ R ℓ = 4, this yields R ℓ ∼ = Mat 2 (F ℓ ). There are three cases: • If ℓ divides the conductor of R ′ ℓ or is ramified in R ′ ℓ ⊗ Q, let x ∈ R ′ ℓ be a non-trivial nilpotent element (which exists by Lemma 6.10). Let σ ∈ Gal(K/K) and note that σ(x) ∈ R ′ ℓ . Indeed, for all τ ∈ Gal(K/K ′ ), we have τ (σ(x)) = σ(x) since σ −1 τ σ ∈ Gal(K/K ′ ) and x is defined over K ′ . So, σ(x) is a nilpotent element in R ′ ℓ , which implies σ(x) = b σ x for some b σ ∈ F × ℓ (notice that the nilpotent elements in R ℓ form a proper F ℓ -subspace of R ′ ℓ , that has dimension 2). This shows that the ideal (x) is stable under Gal K/K , hence ker(x) ⊆ A[ℓ] is a nonzero proper subspace of A[ℓ] defined over K, contradiction.
• If R ′ ℓ ∼ = F ℓ 2 , we proceed as in the proof of Theorem 6.16. A[ℓ] acquires the structure of an F ℓ 2 -vector space of dimension 2 and Gal(K/K ′ ) acts F ℓ 2 -linearly on it. So, each matrix in ρ(Gal(K/K ′ )) has at most two rational eigenvalues. Choose M ∈ G ′ such that λ(M) generates F × ℓ . Proceeding as in the proof of Lemma 6.9, we show that M 2 is a scalar since it has at most two rational eigenvalues. This contradicts Corollary 6.7.
Proof. Basic linear algebra.
Lemma 6.22. Let A be an abelian surface over a number field K. Assume that End K (A) = R is an order in a quartic CM field. Assume that ℓ is not ramified in R ⊗ Q and does not divide the conductor of R. If ℓ > C 1 , then (A, ℓ) is not a strong counterexample.
Proof. Let K ′ be a cyclic extension of K such that End K ′ (A) = R with [K ′ : K] | 4 (see [FKRS12,§4.3]) and let ρ( ℓ } be the group of F ℓ -rational points of the Mumford Tate group of A, where σ denotes the automorphism of (R ⊗ F ℓ ) × induced by complex conjugation on R. Theorem 1.3 (2) in [Lom17] gives where C K is a constant that depends only on K. In the notation of [Lom17], we have [K : E * ] ≤ [K : Q], µ * ≤ 12 (a field of degree 4 cannot contain more than 12 roots of unity), and |F | = 1, as noticed in [Lom17,§6.4]. Thus we may take C K = 12[K : Q]. By our assumptions on ℓ, the ring R ⊗ F ℓ is a product of fields.
Theorem 6.23. Let A be an abelian surface defined over a number field K. Assume that End K (A) = R is an order in a CM field. If ℓ > C 1 , then (A, ℓ) is not a strong counterexample.
Proof. If ℓ is unramified in End K (A) and does not divide the conductor of this order, we conclude using Lemma 6.22. Otherwise, we use Lemma 6.10.

Proof of the main results
We can now easily conclude the proof of our main results.
Proof of Theorem 6.1. If End K (A) is an order in a real quadratic field, the claim follows from Theorem 6.16. If A K is isogenous to the square of an elliptic curve without CM, we conclude using Theorem 6.19. If End K (A) is an order in a quaternion algebra, we apply Theorem 6.20. Finally, if End K (A) is an order in a quartic CM field, the conclusion follows from Theorem 6.23.
Lemma 6.24. Let A be an abelian surface over a number field K. If End K (A) ⊗ Q ⊇ Q 2 , then (A, ℓ) is not a strong counterexample.
Proof. The assumption End K (A) ⊗ Q ⊇ Q 2 implies that A is isogenous (over K) to the product of two elliptic curves E 1 and E 2 . By Corollary 2.4, this implies that (E 1 × E 2 , ℓ) is a strong counterexample, but this is obviously a contradiction since ( Proof of Corollary 6.2. If End K (A) is larger than Z, then it contains an order in a quadratic field or in Q 2 (see §2.1, and notice that a quartic CM field contains a real quadratic field). The claim follows from Theorem 6.16 and Lemma 6.24.

Squares of CM elliptic curves
The goal of this section is to construct infinitely many strong counterexamples (A/Q, ℓ) with A geometrically isogenous to the square of a CM elliptic curve and ℓ unbounded. Such examples will be obtained as twists of E 2 , where E/Q is the elliptic curve with Weierstrass equation y 2 = x 3 + x. We begin by finding suitable Galois extensions of Q with Galois group D 8 (the dihedral group with 16 elements), which we will then use to construct our twists. The following is a special case of [Kim90, Theorems 5 and 6].
Theorem 6.25. Let F be a field of characteristic different from 2. Let a and b in F be such that the following hold: • a, b, and ab are not squares in F ; • b = a − 1; • the equation X 2 − aY 2 − 2Z 2 − 2abV 2 = 0 has a solution in F with (X, Y ) = (0, 0).
There exists q ∈ F * such that the Galois extension F ( √ a, √ b, 2q(a + √ a))/F has Galois group D 4 and can be embedded in a D 8 -extension, cyclic over F ( √ b).
Proposition 6.28. Let ℓ > 5 be a prime with ℓ ≡ 5 (mod 8). There exists an abelian surface A, defined over Q and geometrically isogenous to the square of a CM elliptic curve, such that (A, ℓ) is a strong counterexample.
Proof. Let E be the elliptic curve y 2 = x 3 + x. The prime ℓ (which is in particular congruent to 1 modulo 4) splits in Z where i is one of the two primitive fourth roots of unity in F × ℓ . By [Lom17, Theorem 1.3], the image G ℓ of the mod-ℓ Galois representation attached to E/Q is the normaliser of a split Cartan subgroup of GL 2 (F ℓ ). In particular, in the basis above G ℓ is given by the set  {A(a, b), B(a, b) The subgroup ρ E,ℓ Gal Q(i)/Q(i) is given by those Galois automorphisms that commute with the action of [i], that is, {A(a, b) : a, b ∈ F × ℓ }. In other words, ρ E (σ) is of the form A(a, b) for suitable a, b if σ(i) = i, and it is of the form B(a, b) otherwise. Moreover, in the two cases one has χ ℓ (σ) = det ρ E,ℓ (σ) = ±ab; since −1 is a square modulo ℓ, the quantity ab ∈ F × ℓ is a square if and only if χ ℓ (σ) is a square, if and only if σ fixes √ ℓ. We now construct the desired abelian surface A as a twist of E 2 . Let L be as in Lemma 6.26 and identify End(E 2 Q ) with Mat 2×2 (End(E Q )). We define a cocycle c : Gal(Q/Q) → Aut(E 2 Q ) ⊂ End(E 2 Q ) as the composition of the canonical projection Gal Q/Q → Gal(L/Q) ∼ = r, s r 8 = s 2 = 1, srs = r −1 with the unique cocycle of Gal(L/Q) mapping r to 0 Id [i] 0 and s to 0 Id Id 0 . One checks easily that these conditions do in fact define a cocycle. Let now A denote the twist of E 2 by the class of c in H 1 (Gal Q/Q , Aut(E 2 Q )), so that for σ ∈ Gal(Q/Q) we have ρ A,ℓ (σ) = c(σ)ρ E 2 ,ℓ (σ).
We now show that (A, ℓ) is a strong counterexample. We start by checking that ρ A,ℓ (σ) admits at least one F ℓ -rational eigenvalue for every σ ∈ Gal Q/Q , distinguishing cases according to the image σ |L of σ in Gal(L/Q). Recall that we denote by N = Here ab is not a square in F × ℓ , because by construction r (hence also σ) does not fix √ ℓ. Thus ρ A,ℓ (σ) has the rational eigenvalue √ iab: note that iab is a square since i and ab are not (here we use ℓ ≡ 5 (mod 8) to deduce that i is not a square modulo ℓ). We may reason similarly for all other cases. If σ |L = s, then σ(i) = i, so with ab ∈ F ×2 ℓ , so that ρ A,ℓ (σ) has the rational eigenvalues ± √ ab. If σ |L = 1, then ρ A,ℓ (σ) is represented by a diagonal matrix, hence admits F ℓ -rational eigenvalues.
Remark 6.29. With more work, the construction given in the proof can be adapted to y 2 = x 3 + 1, and probably to all elliptic curves over Q with potential CM (in each case, one would get a different congruence condition on the prime ℓ).
Remark 6.30. A variant of the same construction can be used to obtain weak counterexamples over many number fields K. Let E/Q be a CM elliptic curve, with CM by an order in the quadratic imaginary field F , and let E K denote the base-change of E to K. Suppose that K does not contain F . For ℓ sufficiently large and split in F , the image of ρ E K ,ℓ is the full normaliser of a split Cartan subgroup of GL 2 (F ℓ ). Let L = Q( √ ℓ * ) be the quadratic subfield of Q(ζ ℓ ) and let A = Res KL/K (E L ), where Res denotes the Weil restriction of scalars. Using the fact that the mod-ℓ Galois representations attached to the abelian surface A/K are given by Ind G K G KL (ρ E,ℓ ), one checks easily that (A, ℓ) is a weak counterexample to the local-global principle for isogenies.
6.10 The semistable case for K = Q To finish our discussion of strong counterexamples, we show the following non-existence result for semistable counterexamples over the rational numbers: Theorem 6.31. Let A/Q be a semistable abelian surface and let ℓ = 5 be a prime. The pair (A/Q, ℓ) is not a strong counterexample to the local-global principle for isogenies.
The idea is that such a counterexample would lead to the existence of an everywhere unramified number field. The proof relies on the following theorem: