Sato–Tate distributions of twists of the Fermat and the Klein quartics

Francesc Fité; Elisa Lorenzo García; Andrew Sutherland

doi:10.1007/s40687-018-0162-0

Sato–Tate distributions of twists of the Fermat and the Klein quartics

Fité, Francesc; Lorenzo García, Elisa; Sutherland, Andrew 2018-10-15 00:00:00 drew@math.mit.edu https://math.mit.edu/~drew We determine the limiting distribution of the normalized Euler factors of an abelian Department of Mathematics, threefold A deﬁned over a number ﬁeld k when A is Q-isogenous to the cube of a CM Massachusetts Institute of Technology, 77 Massachusetts elliptic curve deﬁned over k. As an application, we classify the Sato–Tate distributions of Avenue, Cambridge, MA 02139, the Jacobians of twists of the Fermat and Klein quartics, obtaining 54 and 23, USA respectively, and 60 in total. We encounter a new phenomenon not visible in Full list of author information is available at the end of the article dimensions 1 or 2: the limiting distribution of the normalized Euler factors is not determined by the limiting distributions of their coeﬃcients. Contents 1 Introduction ........................................... 2 Equidistribution results for cubes of CM elliptic curves ................... 3 The Fermat and Klein quartics ................................. 3.1 Twists ............................................ 3.2 Moment sequences ..................................... 3.2.1 Independent coeﬃcient moment sequences ................... 3.2.2 Joint coeﬃcient moments ............................. 3.3 Sato–Tate groups ..................................... 3.4 Curve equations ...................................... 3.4.1 Constructing the Fermat twists .......................... 3.4.2 Constructing the Klein twists ........................... 3.5 Numerical computations ................................. 3.5.1 Naïve point-counting ............................... 3.5.2 An average polynomial-time algorithm ..................... 4Tables ............................................... References .............................................. 1 Introduction Let A be an abelian variety of dimension g ≥ 1 deﬁned over a number ﬁeld k. For a prime , let V (A):= Q ⊗ lim A[ ] be the (rational) -adic Tate module of A,and let ← − : G → Aut(V (A)) A k be the -adic representation arising from the action of the absolute Galois group G on V (A). Let p be a prime of k (a nonzero prime ideal of the ring of integers O )not lying © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. 0123456789().,–: volV 41 Page 2 of 40 Fité et al. Res Math Sci (2018) 5:41 above the rational prime .The L-polynomial of A at the prime p is deﬁned by L (A, T):= det(1 − (Frob )T; V (A) ) ∈ Z[T], p A p where Frob denotes a Frobenius element at p and I is the inertia subgroup at p;itdoes p p not depend on the choice of .Let S be a ﬁnite set of primes of k that includes all primes of bad reduction for A and all primes lying above .For p ∈ / S the polynomial L (A, T) has degree 2g and coincides with the numerator of the zeta function of the reduction of A modulo p.The L-function of A is deﬁned as the analytic continuation of the Euler product −s −1 L(A, s):= L (A, N(p) ) , where N(p):= [O : p] is the (absolute) norm of p.The normalized L-polynomial of A at p −1/2 is the monic polynomial L (A, T):= L (A, N(p) T) ∈ R[T]; its roots come in complex p p conjugate pairs and lie on the unit circle, as shown by Weil in [36]. As constructed by Serre in [26], the Sato–Tate group ST(A)isacompactrealLie sub- group of USp(2g), deﬁned up to conjugacy in GL (C), that comes equipped with a map 2g that assigns to each prime p ∈ / S a semisimple conjugacy class s(p)ofST(A) for which det(1 − s(p)T) = L (A, T). Let μ be the pushforward of the Haar measure of ST(A) to its set of conjugacy classes X,and let {s(p)} denote the sequence of conjugacy classes s(p)arrangedinanorder compatible with the partial ordering of primes p by norm. The generalized Sato–Tate conjecture predicts that: (ST) The sequence {s(p)} is equidistributed on X with respect to the measure μ. This conjecture has been proved for abelian varieties of dimension one (elliptic curves) over a totally real [15]orCMnumberﬁeld[1], and in several special cases for abelian varieties of higher dimension, including abelian varieties with potential CM [19]. 2g For each s ∈ X, we write det(1 − sT) =: a T , and deﬁne j=0 2g 2g I := − , , and I := I . j j j j j=1 For 0 ≤ j ≤ 2g we have a ∈ I and a = a for 0 ≤ j ≤ 2g, since the eigenvalues of any j j j 2g −j conjugacy class of USp(2g) come in complex conjugate pairs on the unit circle. Consider the maps : X −→I, := X −→ I −→ I , j j where is deﬁned by (s) = (a , ... ,a )and is the projection to the jth component. 1 g j Let μ (resp. μ ) denote the projection of the measure μ by the map (resp. ). We I I j will call μ the joint coeﬃcient measure and the set of measures {μ } ,the independent I I j coeﬃcient measures. The measures μ and μ are, respectively, determined by their moments I I n n M [μ ]:= a ··· a μ (a , ... ,a ), M [μ ]:= a μ (a ), (1.1) n ,...,n I g I 1 g n I I j 1 g 1 j j j I I for n , ... ,n ≥ 0and n ≥ 0. We denote by a (A)(p), or simply a (p), the jth coeﬃ- 1 g j j cient (s(p)) of the normalized L-polynomial, and by a(A)(p), or simply a(p), the g-tuple j Fité et al. Res Math Sci (2018) 5:41 Page 3 of 40 41 (s(p)) of coeﬃcients of the normalized L-polynomial. We can now consider the following successively weaker versions of the generalized Sato–Tate conjecture: (ST ) The sequence {a(p)} is equidistributed on I with respect to μ . p I (ST ) The sequences {a (p)} are equidistributed on I with respect to μ , for 1 ≤ j ≤ g. j p j I Let π(x) count the number of primes p ∈ / S for which N(p) ≤ x.Ifwedeﬁne 1 1 n n n 1 g M [a]:= lim a (p) ··· a (p) , M [a ]:= lim a (p) , n ,...,n 1 g n j j 1 g x→∞ x→∞ π(x) π(x) N(p)≤x N(p)≤x (1.2) then (ST )holds if andonlyifM [μ ] = M [a] for every n , ... ,n ≥ 0, while n ,...,n I n ,...,n 1 g 1 g 1 g (ST )holds if andonlyifM [μ ] = M [a ] for every n ≥ 0and 1 ≤ j ≤ g. n I n j Let A be an abelian variety deﬁned over a number ﬁeld k also of dimension g,and let X , μ , μ , μ be the data associated with A corresponding to X, μ, μ , μ , respectively. I I I I The following implications are immediate: ST(A) = ST(A ) ⇒ μ = μ ⇒{μ } ={μ } . (1.3) I I j j I j I The classiﬁcation of Sato–Tate groups of elliptic curves and abelian surfaces together with the explicit computation of their Haar measures implies that for g ≤ 2 the converses of the implications in (1.3)bothhold; see[10]. In this article, we show that for g = 3, there are cases in which the converse of the second implication of (1.3) fails to hold. Main result. In this article, we obtain the counterexamples alluded to in the previous paragraph by searching among abelian threefolds deﬁned over a number ﬁeld that are Q-isogenous to the cube of an elliptic curve with complex multiplication (CM). More precisely, we obtain a complete classiﬁcation of the Sato–Tate groups, the joint coeﬃcient measures, and the independent coeﬃcient measures of the Jacobians of twists of the Fermat and the Klein quartics (which are both Q-isogenous to the cube of a CM elliptic curve). The Fermat and Klein quartics are, respectively, given by the equations 0 4 4 4 0 3 3 3 ˜ ˜ C : x + y + z = 0, C : x y + y z + z x = 0, (1.4) 1 7 and they have the two largest automorphism groups among all genus 3 curves, of sizes 96 and 168, respectively. Our main result is summarized in the following theorem. Theorem 1 The following hold: (i) There are 54 distinct Sato–Tate groups of twists of the Fermat quartic. These give rise to 54 (resp. 48) distinct joint (resp. independent) coeﬃcient measures. (ii) There are 23 distinct Sato–Tate groups of twists of the Klein quartic. These give rise to 23 (resp. 22) distinct joint (resp. independent) coeﬃcient measures. (iii) There are 60 distinct Sato–Tate groups of twists of the Fermat or the Klein quartics. These give rise to 60 (resp. 54) distinct joint (resp. independent) coeﬃcient measures. Using Gassmann triples, one can construct examples (of large dimension) where the converse of the ﬁrst implication in (1.3) also fails to hold, but we will not pursue this here. 41 Page 4 of 40 Fité et al. Res Math Sci (2018) 5:41 One motivation for our work is a desire to extend the classiﬁcation of Sato–Tate groups that is known for dimensions g ≤ 2 to dimension 3. Of the 52 Sato–Tate groups that arise for abelian surfaces (see [10, Table 10] for a list), 32 can be realized as the Sato– Tate group of the Jacobian of a twist of one of the two genus 2 curves with the largest automorphism groups, as shown in [12]; these groups were the most diﬃcult to treat in [10] and notably include cases missing from the candidate list of trace distributions identiﬁed in [21, Table 13]. While the classiﬁcation of Sato–Tate groups in dimension 3 remains open, the 60 Sato–Tate groups identiﬁed in Theorem 1 and explicitly described in Sect. 3.3 are likely to include many of the most delicate cases and represent signiﬁcant progress toward this goal. Overview of the paper. This article can be viewed as a genus 3 analog of [12], where the 2 5 2 6 Sato–Tate groups of the Jacobians of twists of the curves y = x − x and y = x + 1 were computed. However, there are two important diﬀerences in the techniques we use here; these are highlighted in the paragraphs below that outline our approach. We also note [13], where the Sato–Tate groups of the Jacobians of certain twists of the genus 3 2 7 2 8 curves y = x − x and y = x + 1 are computed, and [2], where the Sato–Tate groups of 2 8 4 the Jacobians of twists of the curve y = x − 14x + 1 are determined. Like the Fermat and Klein quartics we consider here, these three curves represent extremal points in the moduli space of genus 3 curves, but they are all hyperelliptic, and their automorphism groups are smaller (of order 24, 32, 48, respectively). As noted above, the Sato–Tate conjecture is known for abelian varieties that are Q- isogenous to a product of CM abelian varieties [19, Cor. 15]. It follows that we can deter- mine the set of independent coeﬃcient measures {μ } by computing the sequences I j {M [a ]} , and similarly for μ and the sequences {M [a]} . Closed formu- n j j,n I n ,...,n n ,...,n 1 g 1 g las for these sequences are determined in Sect. 2 in the more general setting of abelian threefolds deﬁned over a number ﬁeld k that are Q-isogenous to the cube of an elliptic curve deﬁned over k; see Proposition 2.2 and Corollary 2.4. This analysis closely follows the techniques developed in [12,§3]. In Sect. 3, we specialize to the case of Jacobians of twists of the Fermat and Klein quar- tics. In Sect. 3.2, we obtain a complete list of possibilities for {M [a ]} : there are 48 in n j j,n the Fermat case, 22 in the Klein case, and 54 when combined; see Corollary 3.12.Wealso compute lower bounds on the number of possibilities for {M [a]} by com- n ,n ,n n ,n ,n 1 2 3 1 2 3 puting the number of possibilities for the ﬁrst several terms (up to a certain conveniently chosen bound) of this sequence. These lower bounds are 54 in the Fermat case, 23 in the Klein case, and 60 when combined; see Proposition 3.13. The ﬁrst main diﬀerence with [12] arises in Sect. 3.3, where we compute the Sato–Tate groups of the twists of the Fermat and Klein quartics using the results of [3]. Such an analysis would have been redundant in [12], since a complete classiﬁcation of Sato–Tate groups of abelian surfaces was already available from [10]. We show that there are at most 54 in the Fermat case, and at most 23 in the Klein case; see Corollaries 3.23 and 3.26. Combining the implications in (1.3) together with the lower bounds of Sect. 3.2 and upper bounds of Sect. 3.3 yields Theorem 1. The second main diﬀerence with [12] arises in Sect. 3.4, where we provide explicit equations of twists of the Fermat and Klein quartics that realize each of the possible Sato–Tate groups. Here, the computational search used in [12] is replaced by techniques Fité et al. Res Math Sci (2018) 5:41 Page 5 of 40 41 developed in [22,23] that involve the resolution of certain Galois embedding problems, and a moduli interpretation of certain twists X (7) of the Klein quartic as twists of the modular curve X(7), following [14]. In order to apply the latter approach, which also plays akey role in [25], we obtain a computationally eﬀective description of the minimal ﬁeld over which the automorphisms of X (7) are deﬁned (see Propositions 3.34 and 3.35), a result that may have other applications. Finally, in Sect. 3.5, we give an algorithm for the eﬃcient computation of the L- polynomials of twists of the Fermat and Klein quartics. This algorithm combines an average polynomial-time for computing Hasse–Witt matrices of smooth plane quartics [18] with a result speciﬁc to our setting that allows us to easily derive the full L-polynomial at p from the Frobenius trace using the splitting behavior of p in certain extensions; see Proposition 3.38. Our theoretical results do not depend on this algorithm, but it played a crucial role in our work by allowing us to check our computations and may be of inde- pendent interest. Notation. Throughout this paper, k denotes a number ﬁeld contained in a ﬁxed algebraic closure Q of Q. All the ﬁeld extensions of k, we consider are algebraic and assumed to lie in Q.Wedenoteby G the absolute Galois group Gal(Q/k). For an algebraic variety X deﬁned over k and a ﬁeld extension L/k, write X for the algebraic variety deﬁned over L obtained from X by base change from k to L. For abelian varieties A and B deﬁned over k, we write A ∼ B if there is an isogeny from A to B that is deﬁned over k.Weuse M to denote the transpose of a matrix M. We label the isomorphism class ID(H) = n, m of a ﬁnite group H according to the Small Groups Library [4], in which n is the order of H and m distinguishes the isomorphism class of H from all other isomorphism classes of groups of order n. 2 Equidistribution results for cubes of CM elliptic curves Let A be an abelian variety over k of dimension 3 such that A ∼ E , where E is an elliptic curve deﬁned over k with complex multiplication (CM) by an imaginary quadratic ﬁeld M.Let L/k be the minimal extension over which all the homomorphisms from E to A are deﬁned. We note that kM ⊆ L, and we have Hom(E ,A ) Hom(E ,A )and L L Q Q Q A ∼ E . Let σ and σ denote the two embeddings of M into Q.Consider Hom(E ,A ) ⊗ Q resp. End(A ) ⊗ Q , L L M,σ L M,σ where the tensor product is taken via the embedding σ : M → Q. Letting Gal(L/kM)act trivially on Q, it acquires the structure of a Q[Gal(L/kM)]-module of dimension 3 (resp. 9) over Q, and similarly for σ . Deﬁnition 2.1 Let θ := θ (E, A)(resp. θ (A)) denote the representation aﬀorded M,σ M,σ by the module Hom(E ,A ) ⊗ Q (resp. End(A ) ⊗ Q), and let us similarly deﬁne L L M,σ L M,σ θ := θ (E, A)and θ (A). Let θ := θ (E, A) (resp. θ (A)) denote the representation M,σ M,σ Q Q Q aﬀorded by the Q[Gal(L/k)]-module Hom(E ,A ) ⊗ Q (resp. End(A ) ⊗ Q). L L L For each τ ∈ Gal(L/kM), we write 2 3 det(1 − θ(τ)T) = 1 + a (θ)(τ)T + a (θ)(τ)T + a (θ)(τ)T , 1 2 3 so that a (θ) =− Tr θ and a (θ) =− det(θ). 1 3 41 Page 6 of 40 Fité et al. Res Math Sci (2018) 5:41 Fix a subextension F /kM of L/kM,and let S be the set of primes of F for which A or E F F has bad reduction. Note that by [27, Thm. 4.1] the set S contains the primes of F ramiﬁed in L.For z ∈ M, write |z| := σ (z) · σ (z). For p ∈ / S,there exists α(p) ∈ M, such that 1/2 |α(p)|= N(p) and σ (α(p)) + σ (α(p)) a (E )(p) =− . (2.1) 1 F 1/2 N(p) Proposition 2.2 Let A be an abelian variety of dimension 3 deﬁned over k with A ∼ E , where E is an elliptic curve deﬁned over k with CM by the imaginary quadratic ﬁeld M. Suppose that a (θ)(τ) is rational for every τ ∈ G := Gal(L/kM).Then, fori = 1, 2, 3, 2g 2g the sequence a (A ) is equidistributed on I = − , with respect to a measure i kM i i i that is continuous up to a ﬁnite number of points and therefore uniquely determined by its moments. For n ≥ 1,wehave M [a (A )] = M [a (A )] = 0 and: 2n−1 1 kM 2n−1 3 kM 1 2n 2n M [a (A )] = |a (θ)(τ)| , 2n 1 kM 1 τ ∈G |G| n n−i n n 2i 1 i 2 M [a (A )] = |a (θ)(τ)| |a (θ)(τ)| − 2 ·|a (θ)(τ)| , n 2 2 1 2 kM |G| τ ∈G i=0 i i n 2n 2i n−i 2i 1 2i−j M [a (A )] = (r (τ) − 3) 2n 3 1 kM |G| τ ∈G i=0 2i j=0 k=0 j n−i 2j+2n−2k 2n−2i k n−i−k ·r (τ) 4 (−1) . k j+n−k Here, r (τ) and r (τ) are the real and imaginary parts of a (θ)(τ)a (θ)(τ)a (θ)(τ),respec- 1 2 3 2 1 tively. Proof The proof follows the steps of [12, §3.3]. Deﬁne V (A) = V (A ) ⊗ Q , σ kM M⊗Q where the tensor product is taken relative to the map of Q -algebras M ⊗ Q → Q induced by σ ; similarly deﬁne V (A), V (E), and V (E). We then have isomorphisms of σ σ σ Q [G ]-modules kM V (A ) V (A) ⊕ V (A),V (E ) V (E) ⊕ V (E). σ σ σ σ kM kM It follows from Theorem 3.1 in [9] that V (A) θ (E, A) ⊗ V (E),V (A) θ (E, A) ⊗ V (E). σ M,σ σ σ M,σ σ We thus have an isomorphism of Q [G ]-modules kM V (A ) θ (E, A) ⊗ V (E) ⊕ θ (E, A) ⊗ V (E) . (2.2) M,σ σ M,σ σ kM For each prime p ∈ / S, let us deﬁne σ (α(p)) σ (α(p)) α (p):= , α (p):= , 1 1 1/2 1/2 N(p) N(p) where σ (α(p)), as in equation (2.1), gives the action of Frob on V (E). It follows from p σ (2.2) that a (A )(p) = a (p)α (p) + a (p)α (p), 1 1 1 1 1 kM 2 2 a (A )(p) = a (p)α (p) + a (θ)α (p) + a (p)a (p), 2 2 1 2 1 1 1 kM 3 3 a (A )(p) = a (p)α (p) + a (p)α (p) + a (p)a (p)α (p) + a (p)a (p)α (p), 3 kM 3 1 3 1 1 2 1 1 2 1 (2.3) Fité et al. Res Math Sci (2018) 5:41 Page 7 of 40 41 where to simplify notation we write a (p):= a (θ)(Frob )and a (p):= a (θ)(Frob ). Let i i p i i p r (p)and r (p) denote the real and imaginary parts of a (p)a (p)a (p), respectively. We 1 2 3 2 1 have a (p) = 1, since a (p) is a rational root of unity, and we can rewrite the above 3 3 expressions as a (A )(p) =|a (p)| z (p)α (p) + z (p)α (p) , ( ) 1 kM 1 1 1 1 1 a (A )(p) =|a (p)| z (p)α (p) + z (p)α (p) − 2|a (p)|+|a (p)| , ( ) 2 kM 2 2 1 2 1 2 1 a (A )(p) = a (p) (α (p) + α (p)) + (r (p) − 3) (α (p) + α (p)) 3 kM 3 1 1 1 1 1 ±r (p) 4 − (α (p) + α (p)) , 2 1 1 where 1/2 a (p) a (p) 1 2 z (p) = ,z (p) = ∈ U(1). 1 2 |a (p)| |a (p)| 1 2 Let α denote the sequence {α (p )} and, for each conjugacy class c of Gal(L/kM), 1 1 i i≥1 let α denote the subsequence of α obtained by restricting to primes p of ∈ / S such that 1,c 1 Frob = c. By the translation invariance of the Haar measure and [12, Prop. 3.6], for z ∈ U(1) and i ≥ 1wehave ⎨ i if i is even, i/2 M [zα + z α ] = M [α + α ] = (2.4) i 1,c 1,c i 1,c 1,c 0if i is odd. The formulas for M [a (A )], M [a (A )], M [a (A )] follow immediately from 2n 1 n 2 2n 3 kM kM kM (2.4) and the Chebotarev Density Theorem (see [12, Prop. 3.10] for a detailed explanation of a similar calculation). Remark 2.3 In the statement of the proposition, we included the hypothesis that a (θ) is rational, which is satisﬁed for Jacobians of twists of the Fermat and Klein curves (see Section 3.1), because this makes the formulas considerably simpler. This hypothesis is not strictly necessary; one can similarly derive a more general formula without it. Corollary 2.4 Let A be an abelian variety of dimension 3 deﬁned over k with A ∼ E , where E is an elliptic curve deﬁned over k with CM by the quadratic imaginary ﬁeld M, with k = kM. Suppose that a (θ)(τ) is rational for τ ∈ G := Gal(L/kM).Then, 2g 2g for i = 1, 2, 3, the sequence a (A ) is equidistributed on I = − , with respect i k i i i to a measure that is continuous up to a ﬁnite number of points and therefore uniquely determined by its moments. For n ≥ 1,wehave M [a (A )] = M [a (A )] = 0 and: 2n−1 1 k 2n−1 3 k 2n 1 2n M [a (A )] = |a (θ)(τ)| , 2n 1 1 2|G| τ ∈G n n−i 1 n n 2i i 2 M [a (A )] = |a (θ)(τ)| |a (θ)(τ)| − 2 ·|a (θ)(τ)| n 2 k 2 1 2 τ ∈G i=0 2|G| i i n n n + o(2)3 + o(4)(−1) + o(8) + o(12)2 , n 2n 2i n−i 2i 1 2i−j M [a (A )] = · (r (τ) − 3) 2n 3 1 2|G| τ ∈G i=0 2i j=0 k=0 j n−i 2j+2n−2k 2n−2i k n−i−k · r (τ) 4 (−1) . k j+n−k Here, o(n) denotes the number of elements in Gal(L/k) not in G of order n, and r (τ) and r (τ) are the real and imaginary parts of a (θ)(τ)a (θ)(τ)a (θ)(τ), respectively. 2 3 2 1 Proof For primes of k that split in kM,weinvokeProposition 2.2.Let N be such that τ = 1 for every τ ∈ Gal(L/k) \ G (the present proof shows a posteriori that one can take 41 Page 8 of 40 Fité et al. Res Math Sci (2018) 5:41 N = 24, but for the moment it is enough to know that such an N exists). For primes p of k that are inert in kM, we will restrict our analysis to those also satisfy: (a) p has absolute residue degree 1, that is, N(p) = p is prime. (b) p is of good reduction for both A and E. (c) Q( p) ∩ Q(ζ ) = Q, where ζ is a primitive 4Nth root of unity. 4N 4N These conditions exclude only a density zero set of primes and thus do not aﬀect the computation of moments. Now deﬁne D(T, τ):= det(1 − θ (E, A)(τ)T) for τ ∈ Gal(L/k) \ G,and let p be such that Frob = τ. In the course of the proof of [12, Cor. 3.12], it is shown that: (i) The polynomial L (A, T) divides the Rankin–Selberg polynomial L (E, θ (E, A),T). p p Q (ii) The roots of D(T, τ) are quotients of roots of L (A, T)and L (E, T). p p Since L (E, T) = 1 + T and the roots of D(T, τ)are Nth roots of unity, it follows from (i) that the roots of L (A, T) are 4Nth roots of unity. In particular, a (A)(p), a (A)(p), p 1 2 a (A)(p) ∈ Z[ζ ]. But (a) implies that 3 4N √ √ p · a (A)(p) ∈ Z,p · a (A)(p) ∈ Z,p p · a (A)(p) ∈ Z, (2.5) 1 2 3 which combined with (c) implies a (A)(p) = a (A)(p) = 0. This yields the desired moment 1 3 formulas for a (A )and a (A ), leaving only a (A )toconsider. 1 k 3 k 2 k From (2.5), we see that L(A, T) has rational coeﬃcients. Both L (E, T)and D(T, τ)have integer coeﬃcients, hence so does L (E, θ (E, A),T). Moreover, L (E, θ (E, A),T)isalso p Q p Q primitive, which by (i) and Gauss’ Lemma implies that L(A, T) has integer coeﬃcients. The Weil bounds then imply (see [20, Prop. 4], for example) that the polynomial L (A, T) has the form 2 4 6 1 + aT + aT + T =: P (T), (2.6) form some a ∈{−1, 0, 1, 2, 3}. To compute the moments of a (A ), it remains only to determine how often each value of a occurs as τ ranges over G.Wehave 2 2 2 P (T) = (1 − T) (1 + T) (1 + T ), −1 2 2 4 P (T) = (1 + T )(1 − T + T ), 2 4 P (T) = (1 + T )(1 + T ), (2.7) 2 2 2 P (T) = (1 − T + T )(1 + T + T )(1 + T ), 2 3 P (T) = (1 + T ) . The integer ord(τ) is even and then condition (ii) and the speciﬁc shape of the P (T) imply that the only possible orders of τ are 2, 4, 6, 8, 12. If ord(τ) = 2, then all the roots of L (E, θ (E, A),T) are of order 4. By (i), so are the roots of P (T), and (2.7)implies that p Q a a = 3. If ord(τ) = 4, then all the roots of L (E, θ (E, A),T) are of order dividing 4. Thus so are p Q the roots of P (T), which leaves the two possibilities a =−1or a = 3. But (ii) implies that the latter is not possible: if a = 3, then the roots of D(T, τ) would all be of order dividing 2 and this contradicts the fact that τ has order 4. Thus a =−1. Fité et al. Res Math Sci (2018) 5:41 Page 9 of 40 41 If ord(τ) = 6, then by (ii) we have a =−1, 1, 3 (otherwise the order of τ would not be divisible by 3). If a = 2, then again by (ii) the polynomial D(T, τ) would have a root of order at least 12, which is impossible for ord(τ) = 6. Thus a = 0. If ord(τ) = 8, then L (E, θ (E, A),T) has at least 8 roots of order 8 and thus P (T)has p Q a at least a root of order 8. Thus a = 1. If ord(τ) = 12, then by (ii) we have a =−1, 1, 3 (otherwise the order of τ would not be divisible by 3). If a = 0, then again by (ii) the polynomial D(T, τ) would only have roots of orders 1, 2, 3 or 6, which is incompatible for ord(τ) = 12. Thus, a = 2. 3 The Fermat and Klein quartics 0 0 ˜ ˜ The Fermat and the Klein quartics admit models over Q given by the equations C and C 1 7 of (1.4), respectively. The Jacobian of C is Q-isogenous to the cube of an elliptic curve deﬁned over Q (see Proposition 3.1), butthisisnot true for C , which leads us to choose 0 0 a diﬀerent model for the Klein quartic. Let us deﬁne C := C and 1 1 0 4 4 4 3 3 3 2 2 2 2 2 2 C : x + y + z + 6(xy + yz + zx ) − 3(x y + y z + z x ) + 3xyz(x + y + z) = 0. The model C is taken from [7, (1.22)], and its Jacobian is Q-isogenous to the cube of an elliptic curve deﬁned over Q, as we will prove below. One can explicitly verify that the curve C is Q-isomorphic to the Klein quartic by using (3.3) below to show that ID(Aut((C ) )) = 168, 42 . Q( −7) One similarly veriﬁes that C is Q-isomorphic to the Fermat quartic by using (3.2)below to show ID(Aut((C ) )) 96, 64 . 1 Q(i) 0 0 Let E and E be the elliptic curves over Q given by the equations 1 7 0 2 3 2 0 2 3 2 3 E : y z = x + xz,E : y z = x − 1715xz + 33614z , (3.1) 1 7 0 6 3 with Cremona labels 64a4 and 49a3, respectively. We note that j(E ) = 2 · 3 and 0 3 3 0 0 j(E ) =−3 · 5 , thus E has CM by the ring of integers of Q(i), and E has CM by the ring 7 1 7 of integers of Q( −7). For future reference, let us ﬁx some notation. The automorphisms s ([x : y : z]) = [z : x : y], (3.2) t ([x : y : z]) = [−y : x : z], u ([x : y : z]) = [ix : y : z] generate Aut((C ) ), whereas the automorphisms 1 Q s ([x : y : z]) = [y : z : x], ⎪ 7 t ([x : y : z]) = [−3x − 6y + 2z : −6x + 2y − 3z :2x − 3y − 6z], u ([x : y : z]) = [−2x + ay − z : ax − y + (1 − a)z : −x + (1 − a)y − (1 + a)z], (3.3) 41 Page 10 of 40 Fité et al. Res Math Sci (2018) 5:41 with −1 + −7 2 4 a := = ζ + ζ + ζ , 7 7 generate Aut((C ) ). 0 0 Proposition 3.1 For d = 1 or 7, the Jacobian of C is Q-isogenous to the cube of E . d d 0 0 Proof For d = 1, we have a nonconstant map ϕ : C → E , given by 1 1 3 2 3 3 ϕ ([x : y : z]) = [−x zy : x z : zxy ]. 0 0 Thus there exists an abelian surface B deﬁned over Q with Jac(C ) ∼ B × E . Suppose 1 1 0 0 that E was not a Q-factor of B. Then, the subgroup s ,t ⊆ Aut(C ), isomorphic to the 1 1 1 1 0 × symmetric group on 4 letters S , would inject into (End(Jac(C )) ⊗ C) . There are two r s options for this C-algebra: it is either GL (C) or GL (C) × GL (C) ,with r, s ∈ Z .In 1 2 1 ≥0 either case, we reach a contradiction with the fact that S has no faithful representations whose irreducible constituents have degrees at most 2. Thus, E is a Q-factor of B and 0 0 2 Jac(C ) ∼ (E ) × E, where E is an elliptic curve deﬁned over Q. Applying the previous 1 1 0 0 0 3 argument again shows E ∼ E ,soJac(C ) ∼ (E ) . 1 1 1 0 0 For d = 7, we have a nonconstant map ϕ : C → E , given by 7 7 ϕ ([x : y : z]) 2 2 2 2 2 = −7(x + y + z)(3x − y − 9z):2 · 7 (−x − 3xy − xz + 2z ):(x + y + z) , 0 0 thus Jac(C ) ∼ B × E for some abelian surface B deﬁned over Q.Since S is contained in 7 7 Aut((C ) ) PSL (F ), where M = Q( −7), we may reproduce the argument above to M 2 7 0 0 3 show that Jac(C ) ∼ (E ) . It follows that 7 7 M Jac(C ) ∼ E × E × E . 0 0 where E, E ,and E are either E or E ⊗ χ, where χ is the quadratic character of M.But 7 7 0 0 0 E ⊗ χ ∼ E ,since E has CM by M, and the result follows. 7 7 7 Remark 3.2 To simplify notation, for the remainder of this article d is either 1 or 7, and 0 0 0 0 we write C for C , E for E , M for Q( −d), and s, t, u for s , t , u . When d is not d d d d d speciﬁed, it means we are considering both values of d simultaneously. 3.1 Twists 2 0 0 Let C be a k-twist of C , a curve deﬁned over k that is Q-isomorphic to C .The set of k-twists of C ,upto k-isomorphism, is in one-to-one correspondence with 1 0 0 H (G , Aut(C )). Given an isomorphism φ : C → C , the 1-cocycle deﬁned by M Q σ −1 ξ(σ):= φ( φ) , for σ ∈ G , is a representative of the cohomology class corresponding to C. Let K /k (resp. L/k) denote the minimal extension over which all endomorphisms of ˜ ˜ Jac(C) (resp. all homomorphisms from Jac(C) to E ) are deﬁned. Let K /k (resp. L/k) Q Q denote the minimal extension over which all automorphisms of C (resp. all isomorphisms from C to C ) are deﬁned. 2 0 When we need not specify the number ﬁeld k over which C is deﬁned, we will simply say that C is a twist of C . Thus, by saying that C is a twist of C , we do not necessarily mean that C is deﬁned over Q. Fité et al. Res Math Sci (2018) 5:41 Page 11 of 40 41 Lemma 3.3 We have the following inclusions and equalities of ﬁelds: ˜ ˜ M ⊆ K = K ⊆ L = L. Proof The inclusion M ⊆ K follows from the fact that Tr(A ) ∈ M \ Q, where A is as in u u (3.12)and (3.13). From the proof of Proposition 3.1, we know that Jac(C) ∼ E , where E ˜ ˜ ˜ is an elliptic curve deﬁned over K with CM by M.Thisimplies K = KM and L = LM,as in the proof of [12,Lem.4.2]. 0 0 We now associate with C a ﬁnite group G that will play a key role in the rest of the article. 0 0 Deﬁnition 3.4 Let G := Aut(C ) Gal(M/Q), where Gal(M/Q) acts on Aut(C )in C M M the obvious way (coeﬃcient-wise action on rational maps). It is straightforward to verify that ID(G 0) = 192, 956 , ID(G 0) = 336, 208 . (3.4) C C 1 7 We remark that G 0 PGL(F ). As in [12, §4.2], we have a monomorphism of groups λ :Gal(L/k) = Gal(L/k) → G 0, λ (σ ) = (ξ(σ ), π(σ )), φ φ where π:Gal(L/k) → Gal(M/Q) is the natural projection (which by Lemma 3.3 is well 0 0 deﬁned). For each α ∈ Aut(C ), let α˜ denote its image by the embedding Aut(C ) → M M 0 3 End((E ) ). The 3-dimensional representation 0 0 0 θ 0 0:Aut(C ) → Aut (Hom(E , Jac(C )) ⊗ Q), M,σ E ,C M M M deﬁned by θ 0 0(α)(ψ):= α˜ ◦ ψ satisﬁes E ,C k 0 θ 0 0 ◦ Res λ θ (E , Jac(C)), (3.5) φ M,σ E ,C kM where Res λ denotes the restriction of λ from Gal(L/k)toGal(L/kM). φ φ kM Lemma 3.5 Let C be a twist of C .Then: 0 0 χ if C = C (see Table 3a), Tr θ 0 0 = E ,C ⎩ 0 χ if C = C (see Table 3b). Proof In the proof of Lemma 3.18, we will construct an explicit embedding 0 0 3 Aut(C ) → End((E ) ) ⊗ Q, α → α˜. M M 0 0 Fix the basis B ={id ×0 × 0, 0 × id ×0, 0 × 0 × id} for Hom(E , Jac(C )). In this basis, M M with the above embedding the representation θ 0 0 is given by E ,C −1 −1 −1 θ (s) = A , θ (t) = A , θ (u) = A , 0 0 0 0 0 0 E ,C s E ,C t E ,C u where A , A ,and A are as in (3.12)and (3.13). The lemma follows. s t u Remark 3.6 Observe that since det(θ ) is a rational character of Aut(C ), by (3.5)so 0 0 E ,C M is a (θ) = det θ (E , Jac(C)). Thus, Corollary 2.4 can be used to compute the moments 3 M,σ of a (Jac(C)) for i = 1, 2, 3. i 41 Page 12 of 40 Fité et al. Res Math Sci (2018) 5:41 Proposition 3.7 The ﬁelds K and L coincide. Proof Note that L/K is the minimal extension over which an isomorphism between E /n and E is deﬁned. It follows that L = K (γ ) for some γ ∈ K,with n = 4 for d = 1 and n = 2 for d = 7; see [28, Prop. X.5.4]. In either case, Gal(L/K)iscyclicoforder dividing 4 (note that Q(ζ ) ⊆ K ). Suppose that L = K,let ω denote the element in 0 ω 0 Gal(L/K ) of order 2, and write K = L . Fix an isomorphism ψ : E → E and an 1 L 3 3 isogeny ψ :(E 0) → Jac(C) 0.For i = 1, 2, 3, let ι : E 0 → (E 0) denote the natural 2 i K K K K injection to the ith factor. Then, {ψ ◦ ι ◦ ψ } constitute a basis of the Q[Gal(L/M)]- 2 i 1 i=1,2,3 0 ω ω ω module Hom(E , Jac(C) ) ⊗ Q.Since ψ =−ψ , ψ = ψ ,and ι = ι ,we L M,σ 1 1 2 2 i i have Trace θ (E , Jac(C))(ω) =−3. But this contradicts (3.5), because there is no α in M,σ Aut(C ) for which Trace θ (α) =−3. 0 0 M E ,C Remark 3.8 By Proposition 3.7, and the identities (2.3)and (3.5), the independent and joint coeﬃcient measures of Jac(C) depend only on the conjugacy class of λ (Gal(K /k)) in G 0.InProposition 3.22, we will see that this also applies to the Sato–Tate group of Jac(C). For this reason, henceforth, subgroups H ⊆ G 0 will be considered only up to conjugacy. Deﬁnition 3.9 Let G := Aut(C ) × 1 ⊆ G 0, and for subgroups H ⊆ G 0,let C C H := H ∩G .Wemay view H as a subgroup of Aut(C ) G whenever it is convenient 0 0 0 0 to do so. Noting that [G : G ] = 2, for any subgroup H of G there are two possibilities: 0 0 C C (c ) H ⊆ G , in which case [H : H ] = 1; 1 0 0 (c ) H G , in which case [H : H ] = 2. 2 0 0 Remark 3.10 We make the following observations regarding H ⊆ G 0 and cases (c )and (c ): (i) In Sect. 3.4, we will show that for each subgroup H ⊆ G , there is a twist C of C such that H = λ (Gal(K /k)). From the deﬁnition of λ ,wemustthenhave φ φ H = λ (Gal(K /kM)). The case (c ) corresponds to kM = k, and the case (c ) 0 φ 1 2 corresponds to k = kM. (ii) There are 83 subgroups H ⊆ G 0 up to conjugacy, of which 24 correspond to case (c ) and 59 correspond to case (c ). In Table 4 we list the subgroups in case (c ). 1 2 2 For any subgroup H in case (c ), there exists a subgroup H in case (c ) such that 1 2 H = H; thus the subgroups in case (c ) can be recovered from Table 4 by looking at the column for H . (iii) There are 23 subgroups H ⊆ G 0 up to conjugacy, of which 12 correspond to case (c ) and 11 correspond to case (c ). For all but 3 exceptional subgroups H in case 1 2 (c ), there exists a subgroup H in case (c ) such that H = H.InTable 5,welist 1 2 the subgroups in case (c ) as well as the 3 exceptional subgroups, which appear in rows #3, #8, and #12 of Table 5. As in (ii), the non-exceptional subgroups in case (c ) can be recovered from Table 5 by looking at the column for H , which for the 1 0 exceptional groups is equal to H. Note that this does not hold for the hyperelliptic curves considered in [12] where [L: K ] may be 1 or 2. Fité et al. Res Math Sci (2018) 5:41 Page 13 of 40 41 (iv) The subgroups H ⊆ G in Tables 4 and 5 are presented as follows. First, generators 0 0 of H ⊆ G Aut(C ) are given in terms of the generators s, t, u for Aut(C ) 0 0 M M listed in (3.2)and (3.3). For the 3 exceptional subgroups of Table 5 we necessarily have H = H , and for the others, H is identiﬁed by listing an element h ∈ Aut(C ) such that H = H ∪ H · (h, τ) ⊆ G , (3.6) 0 0 where τ is the generator of Gal(M/Q). 3.2 Moment sequences We continue with the notation of Sect. 3.1.If C is a k-twist of C ,wedeﬁne thejoint and independent coeﬃcient moment sequences M (C):={M [a(C)]} , M (C):={M [a (C)]} , joint n ,n ,n n ,n ,n indep n i j,n 1 2 3 1 2 3 where a(C)and a (C)denote a(Jac(C)) and a (Jac(C)), respectively, as deﬁned in Sect. 1; j j recall that these moment sequences are deﬁned by and uniquely determine corresponding measures μ and μ , respectively. I I Using Lemma 3.5 and (3.5), we can apply Corollary 2.4 to compute the moments M [a (C)] for any n, and as explained in Sect. 3.2.2,itiseasytocompute M [a(C)] n j n ,n ,n 1 2 3 for any particular values of n ,n ,n . Magma scripts [6] to perform these computations 1 2 3 are available at [11], which we note depend only on the pairs (H, H ) (or just H when 0 0 k = kM)listedinTables 4 and 5, and are otherwise independent of the choice of C. 3.2.1 Independent coeﬃcient moment sequences We now show that for any twist of the Fermat or Klein quartic, each of the independent coeﬃcient moment sequences (and hence the corresponding measures) is determined by the ﬁrst several moments. Proposition 3.11 Let C and C be k-twists of C . For each i = 1, 2, 3 there exists a positive integer N such that if M [a (C)] = M [a (C )] for 1 ≤ n ≤ N n i n i i then in fact M [a (C)] = M [a (C )] for all n ≥ 1. n i n i Moreover, one can take N = 6,N = 6,N = 10. 1 2 3 Proof For the sake of brevity, we assume k = kM (the case k = kM is analogous and easier). It follows from Corollary 2.4 that, for i = 1, 2, 3, the sequence {M [a (C)]} is n i n≥0 determined by |a (C)|, |a (C)|, a (C)a (C)¯a (C), and o¯(2), o¯(4), o¯(8) (note that G 0 and 1 2 3 2 1 G 0 contain no elements of order 12, so we ignore the o¯(12) term in the formula for M [a (C)]). We consider the Fermat and Klein cases separately. n 2 For the Fermat case, with the notation for conjugacy classes as in Table 3a, let x (resp. x , x , x , x ) denote the proportion of elements in Gal(L/k) lying in the conjugacy class 1a 2 3 4 5 (resp. 2a ∪ 2b ∪ 4c ∪ 4d,3a,4a ∪ 4b,8a ∪ 8b); note that by Lemma 3.5, we are interested in the representation with character χ listed in Table 3a, which motivates this partitioning of conjugacy classes. 41 Page 14 of 40 Fité et al. Res Math Sci (2018) 5:41 Let y (resp. y , y ) denote the proportion of elements in Gal(L/k) which do not lie in 1 2 3 Gal(L/kM) and have order 2 (resp. 4, 8). Applying Corollary 2.4, one ﬁnds that for n ≥ 1 we have 2n 2n 2n n n M [a (C)] = x · 9 + (x + x ) + x (−1) . (3.7) 2n 1 1 2 5 4 n n n Evaluating (3.7)at n = 1, 2, 3 yields an invertible linear system in x , x + x , x of 1 2 5 4 dimension 3. The moments M [a (C)] for n = 1, 2, 3 thus determine x , x + x , x and 2n 1 1 2 5 4 therefore determine all the M [a (C)]. 2n 1 For M [a (C)], one similarly obtains an invertible linear system in x , x + x , x , y , y , n 2 1 2 5 4 1 2 y of dimension 6, and it follows that the moments M [a (C)] for n ≤ 6 determine all the 3 n 2 M [a (C)]. n 2 For M [a (C)], one obtains an invertible linear system in x , x , x , x , x of dimension 2n 3 1 2 3 4 5 5, and it follows that the moments M [a (C)] for n ≤ 5 determine all the M [a (C)]. 2n 2 2n 3 In the Klein case one proceeds analogously. With the notation of Table 3b, let x (resp. x , x , x ) denote the proportion of elements in Gal(L/k) lying in the conjugacy class 1a 2 3 4 (resp. 2a ∪ 4a,3a,7a ∪ 7b), and let y , y , y be as in the Fermat case. Now x , x , x 1 2 3 1 2 4 determine M [a (C)]; x , x , x and y , y , y determine M [a (C)]; and x , x , x , x n 1 1 2 4 1 2 3 n 2 1 2 3 4 determine M [a (C)]. These proportions are, as before, determined by the ﬁrst several n 3 moments (never more than are needed in the Fermat case), and the result follows. We spare the reader the lengthy details. With Proposition 3.11 in hand we can completely determine the moment sequences M [a (C)] that arise among k-twists C of C by computing the moments M [a (C)] for n i n i 0 0 n ≤ N for the 59 pairs (H, H )listedinTable 4 in the case C = C , and for the 14 pairs i 0 (H, H )listedinTable 5 (as described in Remark 3.10). Note that each pair (H, H )with 0 0 H = H gives rise to two moments sequences M [a (C)] for each i, one with k = kM and 0 n i one with k = kM. After doing so, one ﬁnds that in fact Proposition 3.11 remains true with N = 4and N = 4. The value of N cannot be improved, but one also ﬁnds that the sequences 3 1 M [a (C)] and M [a (C)] together determine the sequence M [a (C)], and in fact just n 2 n 3 n 1 two well-chosen moments suﬃce. Corollary 3.12 There are 48 (resp. 22) independent coeﬃcient measures among twists of 0 0 C (resp. C ). In total, there are 54 independent coeﬃcient measures among twists C of 1 7 0 0 either C or C , each of which is uniquely distinguished by the moments M [a (C)] and 3 2 1 7 M [a (C)]. 4 3 The moments M [a (C)] and M [a (C)] correspond to the joint moments M [a(C)] 3 2 4 3 0,3,0 and M [a(C)] whose values are listed in Table 6 for each of the 60 distinct joint coef- 0,0,4 ﬁcient moment measures obtained in the next section (this includes all the independent coeﬃcient measures). 3.2.2 Joint coeﬃcient moments Instead of giving closed formulas for M [a(C)], analogous to those derived for n ,n ,n 1 2 3 M [a (C)] in Corollary 2.4, let us explain how to compute M [a(C)] for speciﬁc n j n ,n ,n 1 2 3 values of n ,n ,n (this will suﬃce for our purposes). Suppose k = kM (the other case is 1 2 3 Fité et al. Res Math Sci (2018) 5:41 Page 15 of 40 41 similar). By (2.3), we may naturally regard the quantity n n n 1 2 3 a (C) a (C) a (C) (3.8) 1 2 3 as an element of the formal polynomial ring Q[α , α¯ ]/(α α¯ − 1). The moment 1 1 1 1 M [a(C)] is simply the constant term of (3.8). For N ≥ 0 and each pair (H, H ) n ,n ,n 0 1 2 3 in Tables 4 and 5, one can then compute truncated joint moment sequences ≤N M (C):={M [a(C)] : n ,n ,n ≥ 0,n + n + n ≤ N }. n ,n ,n 1 2 3 1 2 3 joint 1 2 3 ≤4 By explicitly computing M (C) for all the pairs (H, H )listedinTables 4 and 5,we joint obtain the following proposition. Proposition 3.13 There are at least 54 (resp. 23) joint coeﬃcient measures (and hence Sato–Tate groups) of twists of the Fermat (resp. Klein) quartic, and at least 60 in total. These 60 joint coeﬃcient measures are listed in Table 6, in which each is uniquely distinguished by the three moments M [a(C)], M [a(C)],and M [a(C)]. 1,0,1 0,3,0 2,0,2 ≤N Computing M (C)with N = 5, 6, 7, 8, does not increase any of the lower bounds joint in Proposition 3.13, leading one to believe they are tight. We will prove this in the next section, but an aﬃrmative answer to the following question would make it easy to directly verify such a claim. Question 3.14 Recall the setting of the ﬁrst paragraph of Sect. 1. In particular, A is an abelian variety deﬁned over a number ﬁeld k of dimension g ≥ 1,and μ is the measure induced on I. By [10, Prop. 3.2] and [10,Rem.3.3], one expects a ﬁnite list of possibilities for the Sato–Tate group of A. One thus expects a ﬁnite number of possibilities for the sequence {M [μ ]} . In particular, one expects that there exists N ≥ 1, depending only n ,...,n I n ,...,n g 1 g 1 g on g, such that for any abelian variety A deﬁned over a number ﬁeld k ,if {M [μ ]} ={M [μ ]} (3.9) n ,...,n I n ,...,n n ,...,n n ,...,n 1 g 1 g 1 g I 1 g for all n + ··· + n ≤ N ,then (3.9) holds for all n , ... ,n ≥ 0. Is there an explicit and 1 g g 1 g eﬀectively computable upper bound for N ? Lacking an answer to Question, 3.14, in order to determine the exact number of distinct joint coeﬃcient measures, we take a diﬀerent approach. In the next section, we will classify the possible Sato–Tate groups of twists of the Fermat and Klein quartics. This classiﬁcation yields an upper bound that coincides with the lower bound of Proposition 3.13. Table 6 also lists z = [z ], z = [z ,z ,z ,z ,z ], and z = [z ], where z 1 1,0 2 2,−1 2,0 2,1 2,2 2,3 3 3,0 i,j denotes the density of the set of primes p for which a (C)(p) = j. For these, we record the following lemma. Lemma 3.15 Let C be a k-twist of C and let H := λ (Gal(K /kM)).Then, 0 φ ⎧ ⎧ o(3) ⎨ ⎨ if M ⊆ k, 0 if M ⊆ k, |H | z (Jac(C)) = z (Jac(C)) = 1 3 1 o(3) 1 ⎩ ⎩ + if M k, if M k, 2 2|H | 2 ⎨ 1 [0,o(3), 0, 0, 0] if M ⊆ k, |H | z (Jac(C)) = [¯o(4),o(3) + o¯(6), o¯(8), 0, o¯(2)] if M k. 2|H | 0 41 Page 16 of 40 Fité et al. Res Math Sci (2018) 5:41 Here, o¯(n) is as in Corollary 2.4 and o(n) denotes the number of elements of order n in H . Proof The formula for z is immediate from (2.3) and the study of the polynomial (2.6)in the proof of Corollary 2.4, together from the fact that τ ∈ Gal(L/k) satisﬁes a (θ)(τ) = 0 if and only if τ has order 3 (as can be seen from Table 3). The formula for z follows from a similar reasoning, once one observes that again a (θ)(τ) = 0ifand only if τ has order 3, and the discussion of the end of Corollary 2.4. Note also that, as G 0 contains no elements of order 12, we have o¯(12) = 0. For z , it suﬃces to note that a (τ) does not vanish. 3 3 3.3 Sato–Tate groups In this section, for any twist C of C , we explicitly construct ST(Jac(C)), which to simplify notation we denote by ST(C). The ﬁrst step is to compute a (non-canonical) embedding ι:Aut(C ) → USp(6) 1 0 1 0 (see [24] for a very similar approach). Let (E ) (resp. (C )) denote the M-vector M M 0 0 space of regular diﬀerentials of E (resp. C ). Deﬁne M M 0 1 0 ∗ −1 ι :Aut(C ) → Aut( (C )), ι (α) = (α ) , 1 1 M M ∗ 1 0 1 0 where α : (C ) → (C ) is the map induced by α. M M Remark 3.16 Let f (X, Y ) = 0 be an aﬃne model of the plane quartic C .Then, dX dX dX W := X ,W := Y ,W := , (3.10) 1 2 3 f f f Y Y Y ∂f 1 0 with f = , is a basis of the regular diﬀerentials (C ). If we denote by ω the regular Y i ∂Y M 0 0 3 diﬀerential of the ith copy of E in (E ) , then M M , ω , ω } (3.11) {ω 1 2 3 1 0 3 is a basis of the regular diﬀerentials (E ) . Consider the isomorphism 1 0 1 0 3 ι :End( (C )) → End( (E ) ) M M 1 0 1 0 3 induced by the isomorphism (C ) (E ) that sends W to ω . i i M M Fix an isomorphism [ ]: M → End(E ) ⊗ Q such that for any regular diﬀerential 1 0 ∗ ω ∈ (E ), one has [m] (ω) = mω for every m ∈ M (see [29, Chap. II, Prop. 1.1]) and then deﬁne 1 0 3 0 3 ι :End( (E ) ) → End((E ) ) ⊗ Q, ι ((m )) = ([m ]), 3 3 jk jk M M top where m ∈ M.Let f = i and let f = a.For d = 1, 7, let γ ∈ H ((E ) , Q) be such ij 1 7 d 1 d C top that {γ , [f ] γ } is a symplectic basis of H ((E ) , Q) with respect to the cup product, d d ∗ d 1 d C and use this basis to obtain an isomorphism top :End(H ((E ) , Q)) → GSp (Q). d 2 d C Then, deﬁne 0 3 ι :End((E ) ) → GSp (Q), ([m ]) → ( ([m ] )). 4 jk d jk ∗ M 6 d Fité et al. Res Math Sci (2018) 5:41 Page 17 of 40 41 Finally, deﬁne the matrices 10 0 −1 0 −2 I = ,J = ,K = . 2 2 2 01 10 1 −1 Remark 3.17 From now on, we ﬁx the following notation: denote by A the 3-diagonal embedding of a subset A of GL in GL . Throughout this section, we consider the general 2 6 symplectic group GSp /Q, the symplectic group Sp /Q, and the unitary symplectic group 6 6 USp(6) with respect to the symplectic form given by the 3-diagonal embedding (J ) . 2 3 Lemma 3.18 The map 0 ι 1 0 ι 1 0 3 ι 0 3 ι 1 2 3 4 ι: Aut(C ) → End( (C )) End( (E ) ) End((E ) ) ⊗ Q → GSp (Q) M M M M 6 0 0 is a monomorphism of groups that for C = C is explicitly given by ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 00 I 0 −I 0 −I 00 2 2 2 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ι(s ) = , ι(t ) = , ι(u ) = , ⎝ I 00⎠ ⎝ I 00⎠ ⎝ 0 −J 0 ⎠ 1 2 1 2 1 2 0 I 0 00 I 00 −J 2 2 2 0 0 T and for C = C is explicitly given by ι(s ) = ι(s ) and 7 1 ⎛ ⎞ ⎛ ⎞ −3I −6I 2I −2I − 4K 3I − K −I − 2K 2 2 2 2 2 2 2 2 2 1 1 ⎜ ⎟ ⎜ ⎟ ι(t ) = , ι(u ) = . ⎝ −6I 2I −3I ⎠ ⎝ 3I − K −I − 2K −2I + 3K ⎠ 7 2 2 2 7 2 2 2 2 2 2 7 7 2I −3I −6I −I − 2K −2I + 3K −4I − K 2 2 2 2 2 2 2 2 2 0 0 ∗ ∗ Proof We ﬁrst consider the case C = C .Inthe basisof (3.10), the elements s , t ,and 1 1 1 ∗ 1 0 u of End( (C )) are given by the matrices 1 M ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 010 01 0 −100 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ A = ,A = ,A = . (3.12) ⎝ 001⎠ ⎝ −100⎠ ⎝ 0 i 0⎠ s t u 1 1 1 100 00 1 00 i 1 0 3 Thus, in the basis of (3.11), the elements ι ι (s ), ι ι (t ), ι ι (u )ofEnd( (E ) )are 2 1 1 2 1 1 2 1 1 −1 −1 −1 given by the matrices A , A , A . It is then enough to check that in the basis {γ , [i] γ } 1 ∗ 1 s t u 1 1 1 we have (1) = I and (i) = J . 1 2 1 2 0 0 ∗ ∗ ∗ We now assume C = C .Inthe basisof (3.10), the matrices associated with s , t , u 7 7 7 7 are: ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 001 −3 −62 2 + 4a 4 + a 1 + 2a 1 1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ A = ,A = ,A = . ⎝ 100⎠ ⎝ −62 −3⎠ ⎝ 4 + a 1 + 2a −5 − 3a⎠ s t u 7 7 7 7 7 010 2 −3 −6 1 + 2a −5 − 3a −3 + a (3.13) 1 0 3 Thus, in the basis of (3.10), the elements ι ι (s ), ι ι (t ), ι ι (u )ofEnd( (E ) )are 2 1 7 2 1 7 2 1 7 M 7 −1 −1 −1 given by the matrices A , A , A . It is then enough to check that in the basis {γ , [a] γ } 7 ∗ 7 s u 7 7 7 we have (1) = I and (a) = K . But this is clear, since 7 2 7 2 [a] (γ ) = 0 · γ + 1 · ([a] γ ), ∗ 7 7 ∗ 7 [a] ([a] γ ) = [a ] γ = [−a − 2] γ =−2 · γ − 1 · ([a] γ ), ∗ ∗ 7 ∗ 7 ∗ 7 7 ∗ 7 and this completes the proof. 41 Page 18 of 40 Fité et al. Res Math Sci (2018) 5:41 Remark 3.19 Note that since Aut(C ) has ﬁnite order, the image of ι is contained in USp (Q). Remark 3.20 It is easy to check that the matrices ι(s ), ι(t ), ι(u ) (resp. ι(s ), ι(t ), ι(u )) 1 1 1 7 7 7 are symplectic with respect to J := (J ) . 2 3 The following theorem gives an explicit description of the Sato–Tate group of a twist of the Fermat or Klein quartic corresponding to a subgroup H of the group G := Aut(C ) × Gal(M/Q) 0 M associated with C (see Deﬁnition 3.4). Theorem 3.21 The following hold: (i) The monomorphism of Lemma 3.18 extends to a monomorphism ι: G → USp(6)/ −1 by deﬁning ⎧ ⎛ ⎞ ⎪ ii 1 0 0 ⎪ ⎝ ⎠ if C = C , i −i ⎛ ⎞ ι((1, τ)) := i −i 0 0 ⎝ ⎠ ⎪ if C = C , ⎪ 7 0 −i where τ denotes the non-trivial element of Gal(M/Q). 0 0 (ii) Let φ : C → C denote a k-twist of C and write H := λ (Gal(K /k)) ⊆ G .The Sato–Tate group of Jac(C) is given by ST(C) = ST(E , 1) · ι(H), where ! " cos(2πr)sin(2πr) ST(E , 1) = | r ∈ [0, 1] − sin(2πr)cos(2πr) 0 0 if C = C ,and ! " 1 4 √ √ cos(2πr) − sin(2πr) sin(2πr) 0 7 7 ST(E , 1) = | r ∈ [0, 1] 2 1 √ √ − sin(2πr)cos(2πr) + sin(2πr) 7 7 0 0 if C = C . Proof To prove (i) it is enough to note that ι((1, τ)) = 1 in USp(6)/ −1 and that ι((1, τ)) 0 0 acts by matrix conjugation on ι(Aut(C )) as τ acts by Galois conjugation on Aut(C ). If M M 0 0 0 0 C = C (resp. C = C ), the latter is equivalent to 1 7 −1 −1 ι((1, τ)) J ι((1, τ)) =−J , resp. ι((1, τ)) K ι((1, τ)) =−I − K , 2 2 2 2 2 which is straightforward to check. Fité et al. Res Math Sci (2018) 5:41 Page 19 of 40 41 For(ii),weconsideronlythe case kM = k,since thecase k = kM can be easily deduced from the case kM = k. Recall from [3]thatST(E ) is a maximal compact subgroup of the 0 0 0 algebraic Sato–Tate group AST(E ) ⊗ C attached to E . Recall that we have AST(E ) = 0 0 L(E , 1) ∪ L(E , τ), where for σ ∈ Gal(kM/k) one has 0 −1 σ 0 L(E , σ):={γ ∈ Sp | γ αγ = α for all α ∈ End(E ) ⊗ Q}. (3.14) 0 0 0 This induces a decomposition ST(E ) = ST(E , 1) ∪ ST(E , τ) that can be explicitly deter- mined. 0 0 For the case C = C ,wehave 0 T −1 L(E , 1)(C) ={A ∈ M (C)|A J A = J ,A J A = J } 2 2 2 2 2 ! " cb 2 2 = | c, b ∈ C,c + b = 1 . −bc Thus, a maximal compact subgroup of L(E , 1)(C)is ! " cos(2πr)sin(2πr) ST(E , 1) = | r ∈ [0, 1] . − sin(2πr)cos(2πr) Analogously, 0 T −1 L(E , τ)(C) ={A ∈ M (C)|A J A = J ,A J A =−J } 2 2 2 2 2 ! " ic ib 2 2 = | c, b ∈ C,c + b = 1 . ib −ic Thus, a maximal compact subgroup of L(E , τ)(C)is 1 ii 0 0 ST(E , τ) = ST(E , 1) · √ . (3.15) i −i There is a relation between the algebraic Sato–Tate groups AST(C)and AST(C ) attached to Jac(C)and Jac(C ), respectively, given by [12, Lemma 2.3]. If we put H := λ (Gal(K /kM)), this relation implies that 0 φ 0 0 −1 AST(C) = L(E , 1) · ι(H ) ∪ L(E , τ) · ι((1, τ) (H \ H )). 3 0 3 0 Then, (3.15)implies 0 0 ST(C) = ST(E , 1) ι(H ) ∪ ι(H \ H ) = ST(E , 1) · ι(H). (3.16) 3 0 0 3 Note that, even if ι(H) is only deﬁned as an element of USp(6)/ −1 , the product 0 0 ST(E , 1) · ι(H) is well deﬁned inside USp(6), provided that −1 ∈ ST(E , 1) . 3 3 0 0 For the case C = C ,wehave 0 T −1 L(E , 1)(C) ={A ∈ M (C)|A J A = J ,A K A = K } 2 2 2 2 2 ! " c − b 4b 2 2 = | c, b ∈ C,c + 7b = 1 . −2bc + b 41 Page 20 of 40 Fité et al. Res Math Sci (2018) 5:41 Thus, a maximal compact subgroup of L(E , 1)(C)is ! " 1 4 √ √ cos(2πr) − sin(2πr) sin(2πr) 7 7 ST(E , 1) = | r ∈ [0, 1] . 2 1 √ √ − sin(2πr)cos(2πr) + sin(2πr) 7 7 Analogously, 0 T −1 ST(E , τ) ={A ∈ M (C)|A J A = J ,A K A =−I − K } 2 2 2 2 2 2 ! " ic − ib 4ib 2 2 = | c, b ∈ C,c + 7b = 1 . ic 3ib + ib − ic 2 2 Thus, a maximal compact subgroup of L(E , τ)(C)is i −i 0 0 ST(E , τ) = ST(E , 1) · . 0 −i 0 0 We can now apply [12, Lemma 2.3] exactly as in the case C = C to complete the proof. The previous theorem describes the Sato–Tate group of a twist C of C . Now suppose that C and C are both twists of C . The next proposition gives an eﬀective criterion to determine when ST(C) and ST(C )coincide. Let H ⊆ G 0 (resp. H ) be attached to C (resp. C )asinRemark 3.10. Proposition 3.22 If H and H are conjugate in G 0,then ST(C) and ST(C ) coincide. Proof Since the Sato–Tate group is deﬁned only up to conjugacy, is suﬃces to exhibit −1 −1 A ∈ GL (C) such that A ST(C)A = ST(C ). Let g ∈ G 0 be such that H = g Hg.It is straightforward to check that ι(G 0) normalizes the group ST(E , 1) . In particular, by Theorem 3.21 (ii),wehave 0 −1 0 −1 ST(C ) = ST(E , 1) ι(H ) = ι(g) ST(E , 1) ι(H)ι(g) = ι(g) ST(C)ι(g). (3.17) 3 3 Corollary 3.23 There are at most 23 Sato–Tate groups of twists of the Klein quartic C . Proof There are 23 subgroups of G , up to conjugacy. In the Fermat case, ST(C) and ST(C ) may coincide when H and H are not conjugate in G . We thus require a sharper criterion. 0 0 0 0 Deﬁnition 3.24 Let C and C be twists of C and C , respectively (here C and C both 0 0 denote one of C or C , but possibly not the same curve in both cases), and let H and H 1 7 be the corresponding attached groups. We say that H and H are equivalent if there exists an isomorphism : H → H (3.18) such that (H ) = H and for every h ∈ H ,wehave 0 0 Tr(j(h)) = Tr(j((h))), (3.19) Fité et al. Res Math Sci (2018) 5:41 Page 21 of 40 41 where H and H are deﬁned as in Deﬁnition 3.9 and j denotes compositions ι ◦ ι of the 0 2 1 0 0 0 0 embeddings deﬁned in Lemma 3.18 for C and C (two diﬀerent maps j if C = C ). 0 0 Proposition 3.25 Let C and C be twists of C and C .IfH andH are equivalent, then ST(C) and ST(C ) coincide. 0 0 Proof Letusﬁrstassume that C = C .ByTheorem 3.21 (ii), we can consider the group isomorphism 0 0 : ST(C) = ST(E , 1) · ι(H) ST(C ) = ST(E , 1) · ι(H ) 3 3 deﬁned by sending an element of the form g = g ι(h)to g ι((h)). We aim to show that 0 0 ST(C) and ST(C ) are conjugate inside GL (C). This amounts to showing that ST(C)and ST(C ) are equivalent representations of the same abstract group, for which it suﬃces to prove the following claim: for every g ∈ ST(C), we have Tr(g) = Tr((g)). To prove the claim distinguish the cases: (a) h ∈ H , and (b) h ∈ H \ H . 0 0 Suppose we are in case (a). By (3.19), there exists A ∈ GL (M) such that we have −1 Aj(h)A = j((h)) for every h ∈ H . Moreover, if we let r denote the composition ι ◦ ι 0 4 3 of the embeddings deﬁned in Lemma 3.18,the fact that A hasentriesin M easily implies that r(A) centralizes ST(E , 1), and thus we have −1 −1 −1 (g) = g r(A)ι(h)r(A) = r(A)g ι(h)r(A) = r(A)gr(A) , 0 0 from which the claim follows. In case (b), we have that both g and (g) have trace 0, as follows for example from the proof of Corollary 2.4 and the Chebotarev Density Theorem. The claim follows immediately. 0 0 0 If C = C , then we may assume without loss of generality that C is a twist of C and C is a twist of C . Now consider the isomorphism 0 0 : ST(C) = ST(E , 1) · ι(H) ST(C ) = ST(E , 1) · ι(H ) 3 3 7 1 −1 deﬁned by sending an element of the form g = g ι(h)to Tg T ι((h)), where 0 0 T = √ √ . −1/ 74/ 7 0 −1 0 We now note that TST(E , 1)T = ST(E , 1), and the proof then proceeds exactly as 1 7 0 0 above; the hypothesis C = C implies that we have √ √ Tr(j(h)) = Tr(j((h))) ∈ Q( −1) ∩ Q( −7) = Q for every h ∈ H , and thus the matrix A from above can be taken in GL (Q). 0 3 Corollary 3.26 The following hold: (i) There are at most 54 distinct Sato–Tate groups of twists of the Fermat quartic. (ii) There are at most 60 distinct Sato–Tate groups of twists of the Fermat and Klein quartics. Proof Determining whether two subgroups H and H are equivalent is a ﬁnite problem. Using the computer algebra program [6], one can determine a set of representatives for equivalence classes of subgroups H that turn out to have size 54 in case (i), and of size 60 in case (ii). For the beneﬁt of the reader, here we give a direct proof of (ii), assuming (i). 41 Page 22 of 40 Fité et al. Res Math Sci (2018) 5:41 The 6 Sato–Tate groups of a twist of the Klein quartic that do not show up as the Sato– Tate group of a twist of the Fermat quartic are precisely those ruled out by the fact that H contains an element of order 7 (those in rows #10, #13, #14 of Table 5), since 7 does not divide #G 0. To show that the other 17 Sato–Tate groups of twists of the Klein quartic also arise for twists of the Fermat quartic, we proceed as follows. Let H ⊆ G 0 correspond to a twist C 0 0 of C such that H does not contain an element of order 7, and let H ⊆ C correspond 7 1 to a twist C of C .Inthiscase, from Table 3a, b, to ensure that H and H are equivalent it suﬃces to check that: (1) There exists an isomorphism : H → H such that (H ) = H ; (2) Tr(j(h)) = 1 for every h ∈ H such that ord(h) = 4. From Tables 4 and 5, it is trivial to check that for every H as above, one can always ﬁnd a subgroup H such that condition (1) is satisﬁed. Condition (2) is vacuous except for rows #6, #7, #11, and #12 of Table 5. In these cases, a subgroup H for which condition (2) is 3 3 also satisﬁed can be found by noting that both j(t )and j(t u t u ) have trace 1. More 1 1 1 1 1 precisely, one ﬁnds that the Sato–Tate groups corresponding to these cases coincide with the Sato–Tate groups of rows #13, #20, #34, and #55 of Table 4, respectively. Combining the lower and upper bounds proved in this section yields our main theorem, which we restate for convenience. Theorem 1 The following hold: (i) There are 54 distinct Sato–Tate groups of twists of the Fermat quartic. These give rise to 54 (resp. 48) distinct joint (resp. independent) coeﬃcient measures. (ii) There are 23 distinct Sato–Tate groups of twists of the Klein quartic. These give rise to 23 (resp. 22) distinct joint (resp. independent) coeﬃcient measures. (iii) There are 60 distinct Sato–Tate groups of twists of the Fermat or the Klein quartics. These give rise to 60 (resp. 54) distinct joint (resp. independent) coeﬃcient measures. Proof This follows immediately from Corollaries 3.12, 3.23, 3.26, and Proposition 3.13. Corollary 3.27 If C and C are twists of C corresponding to H and H , respectively, then ST(C) and ST(C ) coincide if and only if H and H are equivalent. Remark 3.28 One could have obtained the lower bounds of Proposition 3.13 by com- puting the joint coeﬃcient measures μ of the Sato–Tate groups explicitly described in Theorem 3.21. This is a lengthy but feasible task that we will not inﬂict on the reader. We note that this procedure also allows for case-by-case veriﬁcations of the equalities M [μ ] = M [a] and thus of the Sato–Tate conjecture in the cases considered. n ,n ,n I n ,n ,n 1 2 3 1 2 3 Remark 3.29 Let X denote the set of Sato–Tate groups of twists of C .Theorem 3.21 gives a map from the set of subgroups of G 0 to X that assigns to a subgroup H ⊆ G 0 d C the Sato–Tate group ST(E , 1) · ι(H). It also shows that X is endowed with a lattice 3 d structure compatible with this map and the lattice structure on the set of subgroups of G . Moreover, Proposition 3.22 says that this map factors via ε : C → X , d d d Fité et al. Res Math Sci (2018) 5:41 Page 23 of 40 41 where C denotes the lattice of subgroups of G up to conjugation. Parts (i) and (ii) of Theorem 1 imply that while the map ε is a lattice isomorphism, the map ε is far from 7 1 being injective. Corollary 3.27 can now be reformulated by saying that two subgroups H, H ∈ C lie in the same ﬁber of ε if and only if they are equivalent. 1 1 In virtue of the above remark, one might ask about conditions on twists C and C corresponding to distinct but equivalent groups H and H that ensure their Jacobians have the same Sato–Tate group. One such condition is that Jac(C)and Jac(C ) are isogenous (recall that the Sato–Tate group of an abelian variety is an isogeny invariant). The next proposition shows that, under the additional hypothesis that K and K coincide, the previous statement admits a converse. Proposition 3.30 Let C and C be k-twists of C . Suppose that the corresponding sub- groups H and H of G 0 are equivalent and that the corresponding ﬁelds K and K coincide. Then, Jac(C) and Jac(C ) are isogenous. Proof Let S be the set of primes of k which are of bad reduction for either Jac(C)orJac(C ) or lieover theﬁxedprime .Notethatby[27, Thm. 4.1] the set S contains the primes of k ramiﬁed in K or K . By Faltings’ Isogeny Theorem [8, Korollar 2], it suﬃces to show that for every p ∈ / S,wehave L (Jac(C),T) = L (Jac(C ),T). (3.20) p p If Frob ∈ / G , by the proof of Corollary 2.4, both polynomials of (3.20)havethe same p kM expression, which depends only on the order of (the projection of) Frob in Gal(K /kM). To obtain (3.20) for those p such that Frob ∈ G , we will show that V (Jac(C) ) p kM kM and V (Jac(C ) ) are isomorphic as Q [G ]-modules. Indeed, the fact that H and H kM kM are equivalent pairs implies that the restrictions from Aut(C )to H and H of the M 0 representations θ attached to C and C are equivalent. Together with (3.5), this shows 0 0 E ,C 0 0 that θ (E , Jac(C)) and θ (E , Jac(C )) are equivalent representations. The desired M,σ M,σ Q [G ]-module isomorphism follows now from (2.2). kM Remark 3.31 Let H and H be any two equivalent pairs attached to twists C and C of the same curve C . As one can read from Tables 4 and 5 (and as we will see in the next section), one can choose C and C so that K and K coincide. It follows from Proposition 3.30 that on Table 4 (resp. Table 5) two curves C and C satisfy ST(C) = ST(C )ifand only if Jac(C) ∼ Jac(C ). We conclude this section with an observation that is not directly relevant to our results but illustrates a curious phenomenon arising among the twists of the Fermat quartic in Table 4.Let 4 4 4 4 4 4 C :9x + 9y − 4z =0and C :9x − 4y + z = 0 5 8 be the curves listed in rows #5 and #8 of Table 4. As can be seen in Table 4, the groups ST(C ) and ST(C ) coincide, as do the respective ﬁelds K.Proposition 3.30 thus implies 5 8 that the Jacobians of C and C are isogenous, but in fact more is true. 5 8 Proposition 3.32 The curves C and C are not isomorphic (over Q), but their reductions 5 8 ˜ ˜ C and C modulo p are isomorphic (over F ) for every prime p > 3. 5 8 p 41 Page 24 of 40 Fité et al. Res Math Sci (2018) 5:41 Proof The twists C are C of C are not isomorphic because they arise from non- 5 8 ˜ ˜ conjugate subgroups H of G . For the reductions C and C , we ﬁrst consider the case 5 8 p ≡ 1 (mod 4). We claim that −4 is a fourth power modulo p; this follows from the factorization 4 2 2 x + 4 = (x − 2x + 2)(x + 2x + 2) in Q[x] 2 2 together with the fact that x − 2x + 2and x + 2x + 2 have discriminant −4. It follows 4 4 4 ˜ ˜ that C and C are both isomorphic to 9x + 9y + z = 0 (over F ). 5 8 p Suppose now that p ≡−1 (mod 4). Then, 9 is a fourth power modulo p since −3 3 4 2 2 x − 9 = (x − 3)(x + 3) and =− . p p 4 4 4 ˜ ˜ It follows that C and C are both isomorphic to x + y − 4z = 0 (over F ). 5 8 p 3.4 Curve equations In this section, we construct explicit twists of the Fermat and the Klein quartics realizing each of the subgroups H ⊆ G 0 described in Remark 3.10. Recall that each H has an associated subgroup H := H ∩ G of index at most 2 (see Deﬁnition 3.9), and there exists 0 0 a twist corresponding to H with k = kM if and only if H = H , where, as always, M 0 0 0 3 denotes the CM ﬁeld of E (the elliptic curve for which Jac(C ) ∼ (E ) ). Equations for these twists are listed in Tables 4 and 5 in Sect. 4. As explained in Remark 3.10, in the Fermat case every subgroup H ⊆ G 0 with [H : H ] = 1(case (c ) of Deﬁnition 3.9) arises as H for some subgroup H ⊆ G 0 for which [H : H ] = 2 0 C 0 (case (c ) of Deﬁnition 3.9), and a twist corresponding to H canthusbeobtainedasthe base change to kM of a twist corresponding to H . We thus only list twists for the 59 subgroups H in case (c ), since base changes of these twists to kM then address the 24 subgroups H in case (c ). In the Klein case, we list twists for the 11 subgroups H in case (c ) and also the 3 exceptional subgroups H in case (c ) that cannot be obtained as base 2 1 changes of twists corresponding to subgroups in case (c ); see Remark 3.10. Our twists are all deﬁned over base ﬁelds k of minimal possible degree, never exceeding 2. For the 3 exceptional subgroups H in the Klein case noted above, we must have k = kM, and we use k = M = Q( −7). In all but 5 of the remaining cases with [H : H ] = 2, we use k = Q. These 5 exceptions are all explained by Lemma 3.33 below (the second of the 4 pairs listed in Lemma 3.33 arises in both the Fermat and Klein cases, leading to 5 exceptions in total). In each of these 5 exceptions with [H : H ] = 2, the subgroup H also 0 0 arises as H for some subgroup H ⊆ G 0 with [H : H ] = 2 that is realized by a twist 0 C 0 with k = Q, allowing H to be realized over a quadratic ﬁeld as the base change to M of a twist deﬁned over Q. Lemma 3.33 Twists of the Fermat or Klein quartics corresponding to pairs (H, H ) with the following pairs of GAP identiﬁers cannot be deﬁned over a totally real ﬁeld: ( 4, 1 , 2, 1 ), ( 8, 1 , 4, 1 ), ( 8, 4 , 4, 1 ), ( 16, 6 , (8, 2 ). Proof If k is totally real, then complex conjugation acts trivially on k but not on kM, giving an involution in H = Gal(K /k) with non-trivial image in H /H = Gal(K /k)/Gal(K /kM). For the four pairs (H, H ) listed in the lemma, no such involution exists. 0 Fité et al. Res Math Sci (2018) 5:41 Page 25 of 40 41 In addition to listing equations and a ﬁeld of deﬁnition k for a twist C associated with each subgroup H,inTables 4 and 5 we also list the minimal ﬁeld K over which all the endomorphisms of Jac(C) are deﬁned, and we identify the conjugacy class of ST(C) and ST(C ), which depends only on H, not the particular choice of C.Asnoted in kM Remark 3.31, we have chosen twists C so that twists with the same Sato–Tate group have thesameﬁelds K and thus have isogenous Jacobians, by Proposition 3.30 (thereby demonstrating that the hypotheses of the proposition can always be satisﬁed). 3.4.1 Constructing the Fermat twists The twists of the Fermat curve over any number ﬁeld are parametrized in [23], and we specialize the parameters in Theorems 4.1, 4.2, 4.5 of [23] to obtain the desired examples. In every case, we are able to obtain equations with coeﬃcients in Q, but as explained above, we cannot always take k = Q; the exceptions can be found in rows #4, #13, #27, #33 of Table 4. The parameterizations in [23] also allow us to determine the ﬁeld L over which all the isomorphisms to (C ) are deﬁned, which by Lemma 3.3 and Proposition 3.7, this is the same as the ﬁeld K over which all the endomorphisms of Jac(C ) are deﬁned. 7 Q Specializing the parameters for each of the 59 cases with k = kM involves a lot of easy but tedious computations. The resulting equations are typically not particularly pleasing to the eye or easy to format in a table; in order to make them more presentable, we used the algorithm in [30] to simplify the equations. To give just one example, for the unique subgroup H with ID(H) = 24, 13 , the equation we obtain from specializing the parameterizations in [23]is 4 3 3 2 2 2 2 2 3 2 14x − 84x y + 392x z + 588x y − 2940x yz + 4998x z − 980xy + 9996xy z 2 3 4 3 2 2 − 28812xyz + 30184xz + 833y − 9604y z + 45276y z 3 4 − 90552yz + 69629z = 0, but the equation listed for this curve in row #48 of Table 4 is 4 3 3 2 2 2 2 3 2 4 3 2 2 4 3x +4x y+4x z +6x y +6x z +8xy +12xyz + 5y + 4y z + 12y z + z = 0. We used of the number ﬁeld functionality in [6]and [34] to minimize the presentation of the ﬁelds K listed in the tables (in particular, the function polredabs in PARI/GP). 3.4.2 Constructing the Klein twists Twists of the Klein curve over arbitrary number ﬁelds are parametrized in Theorems 6.1 and 6.8 of [23], following the method described in [22], which is based on the resolution of certain Galois embedding problems. However, in the most diﬃcult case, in which H = G 0 has order 336, this Galois embedding problem is computationally diﬃcult to resolve explicitly. This led us to pursue an alternative approach that exploits the moduli interpretation of twists of the Klein curve as twists of the modular curve X(7). As described in [14,§3] and[25, §4], associated with each elliptic curve E/Q is a twist X (7) of the Klein quartic deﬁned over Q that parameterizes isomorphism classes of 7-torsion Galois modules isomorphic to E[7], as we recall below. With this approach, we can easily treat the case H = G , and we often obtain twists with nicer equations. In one case, we also obtain a better ﬁeld of deﬁnition k, allowing us to achieve the minimal possible degree [k : Q] in every case. 41 Page 26 of 40 Fité et al. Res Math Sci (2018) 5:41 However, as noted in [25, §4.5], not every twist of the Klein curve can be written as X (7) for some elliptic curve E/Q, and there are several subgroups H ⊆ G for which the parameterizations in [23] yield a twist of the Klein quartic deﬁned over Q, but no twists of the form X (7) corresponding to H exist. We are thus forced to use a combination of the two approaches. For twists of the form X (7), we need to determine the minimal ﬁeld over which the endomorphisms of Jac(X (7)) are deﬁned; this is addressed by Propositions 3.34 and 3.35. Let E/Q be an elliptic curve and let E[7] denote the F [G ]-module of Q-valued points 7 Q of the kernel of the multiplication-by-7 map [7]: E → E. The Weil pairing gives a G - equivariant isomorphism E[7] μ , where μ denotes the F [G ]-module of 7th 7 7 7 Q roots of unity. Let Y (7) be the curve deﬁned over Q described in [14,§3] and[25, §4]. For any ﬁeld extension L/Q,the L-valued points of Y (7) parametrize isomorphism classes of pairs (E , φ), where E /L is an elliptic curve and φ : E[7] → E [7] is a symplec- tic isomorphism. By a symplectic isomorphism, we mean a G -equivariant isomorphism φ : E[7] → E [7] such that the diagram E[7] µ φ id E [7] µ , (3.21) ˜ ˜ commutes, where the horizontal arrows are Weil pairings. Two pairs (E , φ)and (E , φ) are isomorphic whenever there exists an isomorphism ε : E → E such that φ = ε ◦ φ. In [14] it is shown that X (7), the compactiﬁcation of Y (7), is a twist of C ,and an E E explicit model for X (7) is given by [14, Thm. 2.1], which states that if E has the Weierstrass 2 3 model y = x + ax + b with a, b ∈ Q, then 4 3 2 2 2 2 2 2 3 3 2 2 2 3 2 4 ax +7bx z+3x y −3a x z −6bxyz −5abxz +2y z+3ay z +2a yz −4b z = 0 (3.22) is a model for X (7) deﬁned over Q. We will use the moduli interpretation of X (7) to determine the minimal ﬁeld over which all of its automorphisms are deﬁned. Recall that the action of G on E[7] gives rise to a Galois representation : G → Aut(E[7]) GL (F ). E,7 Q 2 7 Let denote the composition π ◦ , where π:GL (F ) → PGL (F ) is the natural E,7 2 7 2 7 E,7 projection. Proposition 3.34 The following ﬁeld extensions of Q coincide: (i) The minimal extension over which all endomorphisms of Jac(X (7)) are deﬁned; (ii) The minimal extension over which all automorphisms of X (7) are deﬁned; ker E,7 (iii) The ﬁeld Q ; (iv) The minimal extension over which all 7-isogenies of E are deﬁned. In particular, if K is the ﬁeld determined by these equivalent conditions, then Gal(K /Q) Im( ) Im( )/(Im ∩ F ). E,7 E,7 E,7 7 Fité et al. Res Math Sci (2018) 5:41 Page 27 of 40 41 Proof The equivalence of (i) and (ii) follows from 3.3,since X (7)isatwistof C . Following [25], let Aut (E[7]) denote the group of symplectic automorphisms of E[7]. Given a ﬁeld extension F /Q, let us write E[7] for the F [G ]-module obtained from E[7] F 7 F by restriction from G to G . Note that E[7] (Z/7Z) . Under this isomorphism, for Q F any g ∈ Aut (E[7] ), diagram (3.21) becomes (Z/7Z) Z/7Z det(g) id (Z/7Z) Z/7Z , from which we deduce Aut (E[7] ) SL (F ). (3.23) ∧ 2 7 −1 Each g ∈ Aut (E[7] )actson Y (7) via (E , φ) → (E , φ ◦ g ). This action extends to ∧ E Q Q X (7), from which we obtain a homomorphism Aut (E[7] ) → Aut(X (7) ). (3.24) ∧ E Q Q This homomorphism is non-trivial, since elements of its kernel induce automorphisms of E ,but Aut(E ) is abelian and Aut (E[7]) SL (F ) is not, and it cannot be injective, ∧ 2 7 Q Q Q since the group on the right has cardinality 168 < #SL (F ) = 336. The only non-trivial 2 7 proper normal subgroup of SL (F )is ±1 , and thus (3.24) induces a G -equivariant 2 7 Q isomorphism Aut (E[7])/ ±1 → Aut(X (7) ). (3.25) ∧ E By transport of structure, we now endow SL (F )witha G -module structure that 2 7 Q turns (3.23) into a G -equivariant isomorphism ϕ:Aut (E[7] ) → SL (F ). Now deﬁne Q ∧ 2 7 : G → Aut(SL (F )) to be the representation associated with this G -module struc- Q 2 7 Q ture. Since the action of σ ∈ G on each g ∈ Aut (E[7] )isdeﬁnedby Q ∧ −1 σ σ σ ( g)(P) = (g( P)), −1 for P ∈ E[7], we have (σ )(ϕ(g)) = (σ ) · ϕ(g) · (σ ) . The G -action is trivial on E,7 E,7 Q ±1 and thus descends to PSL (F ). Let us write PSL (F ) for PSL (F ) endowed with 2 7 2 7 2 7 the G -action given by conjugation by . Then, (3.23)with (3.25)yielda G -equivariant Q E,7 Q isomorphism PSL (F ) Aut(X (7) ). 2 7 E Ker() This implies that the ﬁeld described in (ii) coincides with Q , and we now note that −1 ker() ={σ ∈ G | (σ ) · α · (σ ) = α, for all α ∈ PSL (F )} Q E,7 E,7 2 7 ={σ ∈ G | (σ ) ∈ F } Q E,7 = ker( ), E,7 ker() ker( ) E,7 thus Q = Q is the ﬁeld described in (iii). 41 Page 28 of 40 Fité et al. Res Math Sci (2018) 5:41 Now let P(E[7]) denote the projective space over E[7], consisting of its 8 linear F - subspaces, equivalently, its 8 cyclic subgroups of order 7. The G -action on P(E[7]) gives rise to the projective Galois representation : G → Aut(P(E[7])) PGL (F ). Q 2 7 E,7 The minimal ﬁeld extension K over which the G -action on P(E[7]) becomes trivial is precisely the minimal ﬁeld over which the cyclic subgroups of E[7] of order 7 all become Galois stable, equivalently, the minimal ﬁeld over which all the 7-isogenies of E are deﬁned. It follows that the ﬁxed ﬁeld of ker identiﬁed in (iii) is also the ﬁeld described in (iv). E,7 To explicitly determine the ﬁeld over which all the 7-isogenies of E are deﬁned, we rely on Proposition 3.35 below, in which (X, Y ) ∈ Z[X, Y ] denotes the classical modular polynomial; the equation for (X, Y ) is too large to print here, but it is available in [6] and can be found in the tables of modular polynomials listed in [33] that were computed via [5]; it is a symmetric in X and Y , and has degree 8 in both variables. The equation (X, Y ) = 0 is a canonical (singular) model for the modular curve Y (7) 7 0 that parameterizes 7-isogenies. If E and E are elliptic curves related by a 7-isogeny 1 2 then (j(E ),j(E )) = 0, and if j ,j ∈ F satisfy (j ,j ) = 0, then there exist elliptic 7 1 2 1 2 7 1 2 curves E and E with j(E ) = j and j(E ) = j that are related by a 7-isogeny. However, 1 2 1 1 2 2 this 7-isogeny need not be deﬁned over F! The following proposition characterizes the relationship between F and the minimal ﬁeld K over which all the 7-isogenies of E are deﬁned. Proposition 3.35 Let E be an elliptic curve over a number ﬁeld k with j(E) = 0, 1728.Let F be the splitting ﬁeld of (j(E),Y ) ∈ k[X], and let K be the minimal ﬁeld over which all the 7-isogenies of E are deﬁned. The ﬁelds K and F coincide. Proof Let S be the multiset of roots of (j(E),Y)in Q,viewedasa G -set in which the 7 k action of σ ∈ G preserves multiplicities: we have m(σ (r)) = m(r) for all σ ∈ G , where k k m(r) denotes the multiplicity of r in S.Let P(E[7]) be the G -set of cyclic subgroups P of E[7] of order 7. In characteristic zero every isogeny is separable, hence determined by its kernel up to composition with automorphisms; this yields a surjective morphism of −1 G -sets ϕ : P(E[7]) → S deﬁned by P → j(E/ P )with m(r) = #ϕ (r) for all r ∈ S (note #P(E[7]) = 8 = m(r)). The G -action on S factors through the G -action on k k r∈S P(E[7]), and we thus have group homomorphisms ¯ φ E,7 G −→ Aut(P(E[7]) −→ Aut(S), where φ:¯ (G ) → Aut(S)isdeﬁnedby φ(σ )(ϕ( P ):= ϕ(σ ( P )) for each σ ∈ ¯ (G ). E,7 E,7 k k ker ¯ ker(φ◦¯ ) E,7 E,7 We then have K = Q and F = Q ,so F ⊆ K . If E does not have complex multiplication, then m(r) = 1 for all r ∈ S, since otherwise over Q we would have two 7-isogenies α, β : E → E with distinct kernels, and then (ˆ α ◦ β) ∈ End(E ) is an endomorphism of degree 49 which is not ±7, contradicting End(E ) Z. It follows that ϕ and therefore φ is injective, so ker ¯ = ker(φ ◦ ¯ )and E,7 E,7 K = F. If E does have complex multiplication, then End(E ) is isomorphic to an order in an imaginary quadratic ﬁeld M. We now consider the isogeny graph whose vertices are j- invariants of elliptic curves E /FM with edges (j ,j ) present with multiplicity equal to 1 2 Fité et al. Res Math Sci (2018) 5:41 Page 29 of 40 41 the multiplicity of j as a root of (j ,Y ). Since j(E) = 0, 1728, the component of j(E ) 2 7 1 FM in this graph is an isogeny volcano, as deﬁned in [31]. In particular, there are at least 6 distinct edges (j(E ),j ) (edges with multiplicity greater than 1 can occur only at the FM 2 surface of an isogeny volcano and the subgraph on the surface is regular of degree at most 2). It follows that m(r) > 1 for at most one r ∈ S. The image of ¯ is isomorphic to a subgroup of PGL (F ), and this implies that if E,7 2 7 ¯ (σ ) ﬁxes more than 2 elements of P(E[7]) then σ ∈ ker ¯ . This necessarily applies E,7 E,7 whenever ¯ (σ)liesinker φ, since it must ﬁx 6 elements, thus ker ¯ = ker(φ ◦ ¯ ) E,7 E,7 E,7 and K = F. Corollary 3.36 Let E be an elliptic curve over a number ﬁeld k with j(E) = 0, 1728.The minimal ﬁeld K over which all the 7-isogenies of E are deﬁned depends only on j(E). Remark 3.37 The ﬁrst part of the proof of Proposition 3.35 also applies when j(E)is0 or 1728, thus we always have F ⊆ K . Equality does not hold in general, but a direct computationﬁndsthat[K :F] must divide 6 (resp. 2) when j(E) = 0 (resp. 1728), and this occurs when k = Q. 2 3 We now ﬁx E as the elliptic curve y = x + 6x + 7 with Cremona label 144b1.Note that is surjective; this can be seen in the entry for this curve in the L-functions and E,7 Modular Forms Database [35] and was determined by the algorithm in [32]. It follows that Gal(Q(E[7])/Q) GL (F ), and Proposition 3.34 implies that we then have H := 2 7 Gal(K /Q) PGL (F ). 2 7 For our chosen curve E,wehave a = 6and b = 7. Plugging these values into equation (3.22) and applying the algorithm of [30] to simplify the result yields the curve listed in entry #14 of Table 5 for H = G 0. To determine the ﬁeld K , we apply Proposition 3.35. Plugging j(E) = 48384 into (x, j(E)) and using PARI/GP to simplify the resulting poly- 8 7 4 nomial, we ﬁnd that K is the splitting ﬁeld of the polynomial x + 4x + 21x + 18x + 9. We applied the same procedure to obtain the equations for C and the polynomials deﬁning K that are listed in rows #4, #6, #9, #10 of Table 5 using the elliptic curves E with Cremona labels 2450ba1, 64a4, 784h1, 36a1, respectively, with appropriate adjustments for the cases with j(E) = 0, 1728 as indicated in Proposition 3.35. The curve in row #11 is a base change of the curve in row #6, and for the remaining 7 curves we used the parameterizations in [23]. 3.5 Numerical computations In the previous sections, we have described the explicit computation of several quantities 0 0 0 related to twists C of C = C , where C is our ﬁxed model over Q for the Fermat quartic d d 0 0 3 0 0 (d = 1) or the Klein quartic (d = 7), with Jac(C ) ∼ (E ) , where E = E is an elliptic curve over Q with CM by M = Q( −d)deﬁnedin (3.1). These include: • Explicit equations for twists C of C corresponding to subgroups H of G 0; • Deﬁning polynomials for the minimal ﬁeld K for which End(Jac(C) ) = End(Jac(C) ); • Independent and joint coeﬃcient moments of the Sato–Tate groups ST(Jac(C)). These computations are numerous and lengthy, leaving many opportunities for errors, both by human and by machine. We performed several numerical tests to verify our computations. 41 Page 30 of 40 Fité et al. Res Math Sci (2018) 5:41 3.5.1 Naïve point-counting A simple but eﬀective way to test the compatibility of a twist C/k and endomorphism ﬁeld K is to verify that for the ﬁrst several degree one primes p of k of good reduction for C that split completely in K , the reduction of Jac(C) modulo p is isogenous to the cube of the reduction of E modulo the prime p := N(p). By a theorem of Tate, it suﬃces to check that 0 3 L (Jac(C),T) = L (E ,T) . For this task, we used optimized brute force point-counting p p methods adapted from [20,§3].The L-polynomial L (Jac(C),T) is the numerator of the zeta function of C, a genus 3 curve, so it suﬃces to compute #C(F ), #C(F 2), #C(F 3), p p reducing the problem to counting points on smooth plane quartics and elliptic curves over ﬁnite ﬁelds. To count projective points (x : y : z) on a smooth plane quartic f (x, y, z) = 0 over a ﬁnite ﬁeld F , one ﬁrst counts aﬃne points (x : y : 1) by iterating over a ∈ F , computing the q q number r of distinct F -rational roots of g (x):= f (x, a, 1) via r = deg(gcd(x − x, g (x))), q a a and then determining the multiplicity of each rational root by determining the least n ≥ 1 (n) (n) for which gcd(g (x),g (x)) = 1, where g denotes the nth derivative of g ∈ F [x]; a a a a q q q note that to compute gcd(x − x, g (x)) one ﬁrst computes x mod g (x) using a square- a a and-multiply algorithm. Having counted aﬃne points (x : y : 1), one then counts the F -rational roots of f (x, 1, 0) and ﬁnally checks whether (1 : 0 : 0) is a point on the curve. To optimize this procedure, one ﬁrst seeks a linear transformation of f (x, y, z) that ensures g (x) = f (x, a, 1) has degree at most 3 for all a ∈ F ; for this, it suﬃces to translate a q a rational point to (1 : 0 : 0), which is always possible for q ≥ 37 (by the Weil bounds). 3 2+o(1) This yields an O(p (log p) )-time algorithm to compute L (Jac(C),T)thatisquite practical for p up to 2 , enough to ﬁnd several (possibly hundreds) of degree one primes p of k that split completely in K . 0 3 Having computed L (Jac(C),T), one compares this to L (E ,T) ; note that the poly- p p 0 2 0 nomial L (E ,T) = pT − a T + 1 is easily computed via a = p + 1 − #E (F ). If this p p p p comparison fails, then either C is not a twist of C , or not all of the endomorphisms of Jac(C) are deﬁned over K . The converse is of course false, but if this comparison succeeds for many degree-1 primes p it gives one a high degree of conﬁdence in the computations of C and K . Note that this test will succeed even when K is not minimal, but we also check that H Gal(K /k), which means that so long as C isatwist of C corresponding to the subgroup H ⊆ G , the ﬁeld K must be minimal. 3.5.2 An average polynomial-time algorithm In order to numerically test our computations of the Sato–Tate groups ST(C), and to verify our computation of the coeﬃcient moments, we also computed Sato–Tate statistics for all of our twists C/k of the Fermat and Klein quartics. This requires computing the L- polynomials L (Jac(C),T) at primes p of good reduction for C up to some bound N,and it suﬃces to consider only primes p of prime norm p = N(p), since nearly all the primes of norm less than N are degree-1 primes. In order to get statistics that are close to the values predicted by the Sato–Tate group one needs N to be fairly large. We used N = 2 , 3 2+o(1) which is far too large for the naive O(p (log p) )-time algorithm described above to be practical, even for a single prime p ≈ N, let alone all good p ≤ N. The objective of this test is not to prove anything, it is simply a mechanism for catching mistakes, of which we found several; most were our own, but some were due to minor errors in the literature, and at least one was caused by a defect in one of the computer algebra systems we used. Fité et al. Res Math Sci (2018) 5:41 Page 31 of 40 41 In [18], Harvey and Sutherland give an average polynomial-time algorithm to count points on smooth plane quartics over Q that allows one to compute L (Jac(C),T) mod p 3+o(1) for all good primes p ≤ N in time O(N(log N) ), which represents an average cost of 4+o(1) O((log p) ) per prime p ≤ N. This is achieved by computing the Hasse–Witt matrices of the reductions of C modulo p using a generalization of the approach given in [16,17] 1+o(1) for hyperelliptic curves. In [18], they also give an O( p(log p) )-time algorithm to compute L (Jac(C),T) mod p for a single good prime p, which allows one to handle reductions of smooth plane quartics C deﬁned over number ﬁelds at degree one primes; 3/2+o(1) this increases the total running time for p ≤ N to O(N ), which is still feasible with N = 2 . Having computed L (Jac(C),T) mod p, we need to lift this polynomial for (Z/pZ)[T]to Z[T], which is facilitated by Proposition 3.38 below. It follows from the Weil bounds that the linear coeﬃcient of L (Jac(C),T) is an integer of absolute value at most 6 p.For p > 144, the value of this integer is uniquely determined by its value modulo p, and for p < 144 we can apply the naive approach described above. This uniquely determines the value in the column labeled F (x)inTables 1 and 2 of Proposition 3.38, which then determines the values in columns F (x)and F (s), allowing the integer polynomial L (Jac(C),T) ∈ Z[T] 2 3 p to be completely determined. Proposition 3.38 For τ in Gal(L/k), let s = s(τ) and t = t(τ) denote the orders of τ and the projection of τ on Gal(kM/k), respectively. For C a twist of C , the following hold: 0 0 (i) The pair (s, t) is one of the 9 pairs listed on Table 1 if C = C ,and oneofthe 9 pairs 0 0 listed on Table 2 if C = C . (ii) For each pair (s, t),let F := F × F × F :[−2, 2] → [−6, 6] × [−1, 15] × [−20, 20] ⊆ R (s,t) 1 2 3 1 7 be the map deﬁned in Table 1 if C = C or Table 2 if C = C . For every prime p 0 0 0 0 unramiﬁed in K of good reduction for both Jac(C) and E ,wehave F (a (E )(p)) = a (Jac(C))(p),a (Jac(C))(p),a (Jac(C))(p) , (3.26) 1 1 2 3 (f (p),f (p)) L kM where f (p) (resp. f (p)) is the residue degree of p in L (resp. kM). kM Proof For every prime p unramiﬁed in K of good reduction for both Jac(C)and E , write 0 2 x := a (E )(p)and y :=± 4 − x . It follows from the proof of Proposition 2.2 that p 1 p a (Jac(C))(p) =− Re(a (p))x + Im(a (p))y , 1 1 p 1 p 2 2 a (Jac(C)(p)) = Re(a (p))(x − 2) − Im(a (p))x y +|a (p)| , 2 2 2 p p 1 a (Jac(C)(p)) =−a (p)(x − 3x ) − Re(a (p)a (p))x + Im(a (p)a (p))y , 3 3 p 1 2 p 1 2 p from which one can easily derive the assertion of the proposition. Tables 7 and 8 show Sato–Tate statistics for the Fermat and Klein twists C/k and their base changes C . In each row, we list moment statistics M , M , M for the kM 101 030 202 three moments M , M , M that uniquely determine the Sato–Tate group ST(C), by 101 030 202 41 Page 32 of 40 Fité et al. Res Math Sci (2018) 5:41 0 0 Table 1 For each possible pair (s, t), the corresponding values of F (x) if C = C .Inthe (s,t) table below, y denotes ± 4 − x (s, t) F (x) F (x) F (x) 1 2 3 2 3 (1, 1) 3x 3x + 3 x + 6x 2 3 −x −x + 3 x − 2x (2, 1) or 2 3 −x + 3 −x + 2x (3, 1) 0 0 x − 3x ⎪ 2 3 −x + 2y ⎪ −x + 7 − 2xy x − 6x + 4y 2 3 (4, 1) or x x − 1 x − 2x ⎪ 2 3 ⎩ x − 1 −x + 2x −x (8, 1) y −xy + 1 x − 4x (2, 2) 0 3 0 (4, 2) 0 −10 (6, 2) 0 0 0 (8, 2) 0 1 0 Table 2 For each possible pair (s, t), the corresponding values of F (x) if C = C .Inthe (s,t) table below, y denotes ± (4 − x )/7 (s, t) F (x) F (x) F (x) 1 2 3 2 3 (1, 1) 3x 3x + 3 x + 6x 2 3 (2, 1) −x −x + 3 x − 2x (3, 1) 0 0 x − 3x 2 3 (4, 1) xx − 1 x − 2x x 7 x 7 9 7 (7, 1) − + y − + 3 − xy x − x + y 2 2 2 2 2 2 (2, 2) 0 3 0 (4, 2) 0 −10 (6, 2) 0 0 0 (8, 2) 0 1 0 Proposition 3.13. These were computed by averaging over all good primes of degree one and norm p ≤ 2 . For comparison, we also list the actual value of each moment, computed using the method described in Sect. 3.2.2. In every case, the moment statistics agree with the corre- sponding moments of the Sato–Tate groups to within 1.5 percent, and in almost all cases, to within 0.5 percent. 4Tables In this ﬁnal section, we present tables of characters, curves, Sato–Tate distributions, and moment statistics referred to elsewhere in this article. Let us brieﬂy describe their contents. Table 3 lists characters of the automorphism groups of the Fermat and Klein quartics speciﬁed via conjugacy class representatives expressed using the generators s, t, u deﬁned in (3.2)and (3.3). Tables 4 and 5 list explicit curve equations for twists C of the Fermat and Klein quar- tics corresponding to subgroups H of G 0 := Aut(C ) Gal(M/Q), as described in C M Remark 3.10. The group H := H ∩ Aut(C ) is speciﬁed in terms of the generators s, t, u listed in (3.2)and (3.3). When kM = k we have H = H , and otherwise H is speciﬁed by listing an element h ∈ Aut(C ) for which H = H ∪ H · (h, τ), where Gal(M/Q) = τ ; 0 0 M Fité et al. Res Math Sci (2018) 5:41 Page 33 of 40 41 Table 3 Character tables of Aut(C ). See (3.2) and (3.3) for the generators s, t, u (A) Aut((C ) ) 96, 64 Class 1a 2a 2b 3a 4a 4b 4c 4d 8a 8b 2 2 3 3 Repr. 1 u uts u u tut t tu tu Order 1 22 3 4 4 4488 Size 1 3 12 32 3 3 6 12 12 12 χ 1 1 1 1 1 1 1111 χ 11 −1 1 111 −1 −1 −1 χ 22 0 −1 2 2 2000 χ 33 −10 −1 −1 −1 −11 1 χ 33 1 0 −1 −1 −11 −1 −1 χ 3 −11 0 −1 − 2i −1 + 2i 1 −1 i −i χ 3 −11 0 −1 + 2i −1 − 2i 1 −1 −ii χ 3 −1 −10 −1 + 2i −1 − 2i 11 i −i χ 3 −1 −10 −1 − 2i −1 + 2i 11 −ii χ 6 −20 0 2 2 −2 000 (B) Aut((C ) ) 168, 42 Class 1a 2a 3a 4a 7a 7b 2 3 2 3 Repr. 1 ts u tu tu uu Order 1 2 3 4 7 7 Size 1 21 56 42 24 24 χ 11 1 1 1 1 χ 3 −10 1 a a χ 3 −10 1 aa χ 62 0 0 −1 −1 χ 7 −11 −10 0 χ 80 −10 1 1 see (3.6). The isomorphism classes of H and H are speciﬁed by GAP identiﬁers ID(H) and ID(H ). The minimal ﬁeld K over which all endomorphisms of Jac(C ) are deﬁned is given as an explicit extension of Q, or as the splitting ﬁeld Gal(f (x)) of a monic f ∈ Z[x]. In the last 2 columns of Tables 4 and 5 we identify the Sato–Tate distributions of ST(C) and ST(C ) by their row numbers in Table 6. Among twists with the same Sato–Tate kM group ST(C) (which is uniquely identiﬁed by its distribution), we list curves with isogenous Jacobians, per Remark 3.31. In Table 6, we list the 60 Sato–Tate distributions that arise among twists of the Fermat and Klein quartics. Each component group is identiﬁed by its GAP ID, and we list the joint moments M ,M ,M suﬃcient to uniquely determine the Sato–Tate distribution, 101 030 202 along with the ﬁrst two non-trivial independent coeﬃcient moments for a ,a ,a .Wealso 1 2 3 list the proportion z of components on which the coeﬃcient a takes the ﬁxed integer i,j i value j; for i = 1,3welistonly z := z and z := z , and for i = 2 we list the vector 1 1,0 3 3,0 z := [z ,z ,z ,z ,z ]; see Lemma 3.15 for details. There are 6 pairs of Sato–Tate 2 2,−1 2,0 2,1 2,2 2,3 distributions whose independent coeﬃcient measures coincide; these pairs are identiﬁed by roman letters that appear in the last column. Tables 7 and 8 list moment statistics for twists of the Fermat and Klein quartics com- puted over good primes p ≤ 2 , along with the corresponding moment values. Twists with isogenous Jacobians necessarily have the same moment statistics, so we list only one twistineachisogeny class. 41 Page 34 of 40 Fité et al. Res Math Sci (2018) 5:41 Table 4 Twists of the Fermat quartic corresponding to subgroups H ⊆ G 0. See (3.2) for the deﬁnitions of s, t, u and (3.6) for the deﬁnition of h. We identify ST = ST(C) and ST = ST(C ) by row numbers in Table 6; here M = Q(i) kM kM #Gen(H ) h ID(H)ID(H )kK ST ST 0 0 k kM 1id id 2, 1 1, 1 QQ(i)31 0 4 4 4 C x + y + z 2id u t 2, 1 1, 1 QQ(i)31 4 2 2 4 4 x − 6x y + y − 2z 3id t utu 2, 1 1, 1 QQ(i)31 4 4 4 4x − y − z 2 4 2 4 t t 4, 1 2, 1 Q( −5) Gal(x −x −1) 5 2 4 3 3 4 4 12x + 40x y − 100xy − 75y − 2z 2 3 5 t t utu 4, 2 2, 1 QQ( 3,i)92 4 4 4 9x + 9y − 4z 2 2 6 t u t 4, 2 2, 1 QQ( 3,i)92 4 2 2 4 4 9x − 54x y + 9y − 2z 7 t id 4, 2 2, 1 QQ( 3,i)92 4 4 4 9x + y + z 8 t u 4, 2 2, 1 QQ( 3,i)92 4 4 4 9x − 4y + z 9 u t id 4, 2 2, 1 QQ( 3,i)92 4 2 2 4 4 2x + 36x y + 18y + z 2 3 10 utt utu 4, 2 2, 1 QQ( 3,i)92 4 2 2 4 4 9x + 18x y + y − 2z 2 3 11su t 6, 1 3, 1 Q Gal(x −3x −4) 11 4 4 3 2 2 2 2 2 2 4 2 2 3 4 x + 4x y + 12x y − 12x yz − 6x z + 36xyz + 6y − 36y z − 12yz + 9z 6 4 2 12 s id 6, 2 3, 1 Q Gal(x +5x +6x +1) 12 4 4 3 3 2 2 2 2 2 3 4 3 2 2 3 4 5x + 8x y − 4x z + 6x y + 12x z + 12xyz + 4xz + 2y + 4y z + 6y z + 4yz + 2z 3 3 8 4 13 t utu tu 8, 1 4, 1 Q( −5) Gal(x −2x −4) 14 6 4 3 2 2 4 4 x − 10x z + 30x z − 2y − 100z 3 8 4 14tt utu 8, 2 4, 1 Q Gal(x +15x +25) 16 6 4 3 2 2 3 4 4 3x − 4x y + 12x y + 4xy + 3y − 5z 3 3 2 8 4 15 t utu u t 8, 2 4, 1 Q Gal(x +15x +25) 16 6 4 3 3 4 4 12x + 40x y − 100xy − 75y + 10z 8 4 16 t id 8, 2 4, 1 Q Gal(x +15x +25) 16 6 4 3 2 2 3 4 4 3x + 4x y + 12x y − 4xy + 3y + 20z 2 2 3 3 4 2 17 u t, t t utu 8, 3 4, 2 Q Gal(x −6x +10) 19 8 4 3 2 2 3 4 4 11x + 12x y + 54x y − 12xy + 11y − 2z 2 2 2 4 2 18 t ,u u t 8, 3 4, 2 Q Gal(x −6x +10) 19 8 4 3 3 4 4 x + 5x y − 25xy − 25y + z 2 2 2 4 2 19 u t, t u 8, 3 4, 2 Q Gal(x −6x +10) 19 8 4 3 2 2 3 4 4 19x − 12x y + 6x y + 12xy + 19y + 2z 2 4 2 20tu 8, 3 4, 1 Q Gal(x −6x +12) 20 6 4 2 2 3 4 4 9x − 18x y − 12xy − 2y + 12z 3 3 4 2 21tt utu 8, 3 4, 1 Q Gal(x −6x +12) 20 6 4 2 2 3 4 4 9x − 18x y − 12xy − 2y − 3z 3 3 4 2 22 t utu u 8, 3 4, 1 Q Gal(x −6x +12) 20 6 4 4 4 9x + 3y − 4z 3 3 4 2 23 t utu id 8, 3 4, 1 Q Gal(x −6x +12) 20 6 4 4 4 9x + 3y + z 3 2 4 2 24 t utu u t 8, 3 4, 1 Q Gal(x −6x +12) 21 7 Fité et al. Res Math Sci (2018) 5:41 Page 35 of 40 41 Table 4 continued #Gen(H ) h ID(H)ID(H ) kK ST ST 0 0 k kM 4 2 2 4 4 x − 6x y + y − 6z 3 4 2 25 t utu u 8, 3 4, 1 Q Gal(x −6x +12) 21 7 4 4 4 3x − 4y + z 3 4 2 26 t utu id 8, 3 4, 1 Q Gal(x −6x +12) 21 7 4 4 4 3x + y + z 3 8 27 t utu t 8, 4 4, 1 Q( −2) Gal(x +9) 23 7 3 3 4 3x y − 3xy − 2z √ √ 2 2 28 t ,u id 8, 5 4, 2 QQ( 3, 5,i)24 8 4 4 4 9x + 25y + z √ √ 2 2 29 t ,u u 8, 5 4, 2 QQ( 3, 5,i)24 8 4 4 4 9x + 25y − 4z √ √ 2 2 3 30 u t, t t utu 8, 5 4, 2 QQ( 3, 5,i)24 8 4 2 2 4 4 x + 30x y + 25y − 18z √ √ 2 2 31 u t, t id 8, 5 4, 2 QQ( 3, 5,i)24 8 4 2 2 4 4 2x + 60x y + 50y + 9z 2 6 3 32 s, u t id 12, 4 6, 1 Q Gal(x +2x +2) 26 10 3 2 2 2 4 3 4x y − 3x z + 12xy z − 2y − 2yz 3 2 8 4 33 t utu, u ut 16, 6 8, 2 Q( −5) Gal(x −2x +5) 29 17 4 2 2 3 4 4 x − 30x y − 80xy − 55y − 2z 3 3 2 8 4 2 34 t utu ,utu 16, 7 8, 3 Q Gal(x −6x −8x −1) 31 18 4 2 2 3 4 4 x − 12x y − 32xy − 28y + z 2 2 8 4 35 tu tut tutu 16, 7 8, 1 Q Gal(x −8x −2) 32 15 3 3 4 2x y − xy − z 2 8 36 u tu u 16, 8 8, 1 Q Gal(x −2) 34 15 3 3 4 x y + 2xy + z 3 3 8 4 37 t utu ,t u 16, 8 8, 4 Q Gal(x −10x −100) 33 22 4 3 2 2 4 4 x + 10x y + 30x y − 100y − 10z 2 8 4 38 t, u utu 16, 11 8, 3 Q Gal(x −2x +9) 35 18 4 3 2 2 3 4 4 4x + 4x y + 6x y − 2xy + y − 2z 2 8 4 39 t, u id 16, 11 8, 3 Q Gal(x −2x +9) 35 18 4 3 2 2 3 4 4 4x + 4x y + 6x y − 2xy + y + 2z 3 3 2 8 4 40 t utu ,u t id 16, 11 8, 3 Q Gal(x −2x +9) 35 18 4 3 2 2 3 4 4 5x − 8x y + 12x y + 16xy + 20y + 2z 3 2 8 4 41 t utu, u id 16, 11 8, 2 Q Gal(x +5x +25) 36 17 4 4 4 9x + 5y + z 3 2 8 4 42 t utu, u u 16, 11 8, 2 Q Gal(x +5x +25) 36 17 4 4 4 9x + 5y − 4z 3 2 8 4 43 utu, t u 16, 11 8, 2 Q Gal(x +5x +25) 36 17 4 3 2 2 3 4 4 7x + 8x y + 6x y + 8xy + 7y + 10z 3 2 2 8 4 44 t utu, u u t 16, 13 8, 2 Q Gal(x −8x +25) 39 17 4 3 3 4 4 x + 5x y − 25xy − 25y + 2z 8 4 45 t, utu id 16, 13 8, 2 Q Gal(x −8x +25) 39 17 4 3 2 2 3 4 3 2 2 3 4 19x + 32x y + 21x y + 8xy + 3y + 2y z + 6y z + 8yz + 4z 3 3 8 4 46 t utu ,t id 16, 13 8, 4 Q Gal(x +12x +9) 37 22 4 3 2 2 3 4 4 x − 2x y + 6x y + 4xy + 4y + 3z 2 2 4 47 s, u u t 24, 12 12, 3 Q Gal(x −16x −24) 42 25 4 3 2 3 3 4 3 2 2 x − 3x z − 12x yz + 16xy − xz + 9y + 12y z + 6y z 2 6 4 2 48 s, u id 24, 13 12, 3 Q Gal(x −x −2x +1) 43 25 4 3 3 2 2 2 2 3 2 4 3 2 2 4 3x + 4x y + 4x z + 6x y + 6x z + 8xy + 12xyz + 5y + 4y z + 12y z + z 2 2 8 4 49 tu tut, u id 32, 7 16, 6 Q Gal(x −10x +20) 44 30 41 Page 36 of 40 Fité et al. Res Math Sci (2018) 5:41 Table 4 continued #Gen(H ) h ID(H)ID(H ) kK ST ST 0 0 k kM 4 3 2 2 4 4 4x − 8x y + 12x y + 2y + 5z 2 3 2 8 4 50 u, u tutu t 32, 11 16, 2 Q Gal(x −2x +5) 45 28 4 2 2 3 4 4 x − 30x y − 80xy − 55y − 2z 3 16 12 8 4 51 u, t ut id 32, 34 16, 2 Q Gal(x −4x +6x +20x +1) 47 28 4 4 4 2x + 3y + z 3 2 8 4 52 t utu, u ,t u 32, 43 16, 13 Q Gal(x +6x −9) 48 38 4 2 2 3 4 4 3x − 36x y − 96xy − 84y + 2z 2 2 8 4 53 tu tut, u u 32, 43 16, 6 Q Gal(x −10x +45) 49 30 4 3 3 4 4 9x + 36x y − 24xy − 4y − 10z 2 8 4 54 utu, u ,t id 32, 49 16, 13 Q Gal(x +8x +9) 50 38 4 3 2 2 3 4 4 x + x y + 24x y + 67xy + 79y + 2z 6 4 2 55 s, t id 48, 48 24, 12 Q Gal(x −x +5x +1) 53 41 4 3 2 3 2 3 4 3 2 2 3 3x + 2x z + 6x yz + 12xy − 30xy z + 2xz − 27y + 38y z + 18y z − 10yz 8 4 56 t, u id 64, 134 32, 11 Q Gal(x −4x −14) 54 46 4 2 2 3 4 4 x − 42x y − 168xy − 203y + z 2 12 4 57 s, u u t 96, 64 48, 3 Q Gal(x +6x +4) 55 52 3 2 2 2 2 3 2 4 3 3 4 6x z + 3x y + 27x z + 6xy + 18xyz + 4y + 2y z + 18yz − 36z 12 8 4 58 s, u id 96, 72 48, 3 Q Gal(x +x −2x −1) 57 52 3 3 2 2 3 2 2 4 3 4 x y − x z + 3x y + 28xy − 84xy z + 84xyz + 28y − 56y z + 98z 12 4 59 s, t, u id 192, 956 96, 64 Q Gal(x +48x +64) 59 56 4 3 3 2 2 2 3 3 4 2 2 44x + 120x y + 36x z + 60x yz + 9x z − 200xy + xz − 150y − 15y z Table 5 Twists of the Klein quartic corresponding to subgroups H ⊆ G 0. See (3.3) for the deﬁnitions of s, t, u and see (3.6) for the deﬁnition of h. We identify ST = ST(C) and √ √ ST = ST(C ) by row numbers in Table 6; here M = Q( −7) and a := (−1 + −7)/2 kM kM #Gen(H ) h ID(H)ID(H ) kK ST ST 0 0 k kM 1id id 2, 1 1, 1 QQ(a)31 4 4 4 3 3 3 2 2 2 2 2 2 C x + y + z + 6(xy + yz + zx ) − 3(x y + y z + z x ) + 3xyz(x + y + z) 2 t id 4, 2 2, 1 QQ(a, i)92 4 3 2 2 2 2 3 4 2 2 4 3x + 28x y + 105x y − 21x z + 196xy + 147y + 147y z − 49z √ √ 6 6 2 3 ustu ,sutu s – 4, 2 4, 2 Q(a) Q( 2, 3,a)8 8 4 2 2 2 2 4 2 2 4 x + 9ax y + 6ax z + 9y + 18ay z + 4z 3 2 4 st 6, 1 3, 1 Q Gal(x −x +2x −3) 11 4 4 3 3 2 2 2 2 3 2 2 4 3 2 2 3 x + 3x y − 9x z + 9x y − 6x z + 18xy + 3xy z − 3xyz + y + 4y z − 3y z + 7yz 5 s id 6, 2 3, 1 QQ(ζ)124 3 3 3 x y + xz + y z 2 3 2 5 2 8 7 4 6 u tu tu u tu 8, 1 4, 1 Q(i)Gal(x +2x −14x +16x +4) 14 6 4 2 2 2 2 3 2 2 3 x + 3x y − 3x z + 2y z + 3y z + 2yz 2 3 2 4 2 7 u tu tu id 8, 3 4, 1 Q Gal(x −4x −14) 20 6 4 3 2 2 2 2 3 2 4 2 2 4 12x − 80x y + 60x y − 24x z − 104xy + 24xyz + 83y + 36y z − 2z 6 2 8 su, tu – 12, 3 12, 3 Q(a)Gal(x −147x +343) 25 25 4 3 3 2 2 2 2 2 4 3x + (−18a + 12)x y + (12a + 4)x z + (−27a + 36)x y + (9a + 6)x z + 36xy z + 27y 3 2 2 3 4 + (54a − 36)y z + (−54a + 36)y z + (18a − 12)yz + (−3a + 2)z 3 2 9 s, t id 12, 4 6, 1 Q Gal(x −x +5x +1) · Q(a)26 10 3 2 2 2 3 4 7x z + 3x y − 6xyz + 2y z − 4z 7 3 2 10 u id 14, 1 7, 1 Q Gal(x +7x −7x +7x +1) 27 13 3 2 2 3 2 3 4 3 2 2 4 x y − 21x z + xy − 42xyz − 147xz + 2y + 21y z + 63y z − 196z 2 3 2 5 2 8 7 4 11 u tu tu ,u tu id 16, 7 8, 3 Q Gal(x +2x −14x +16x +4) 31 18 4 2 2 2 2 3 2 2 3 x + 3x y − 3x z + 2y z + 3y z + 2yz 6 2 2 4 3 2 12 sust, su s tu – 24, 12 24, 12 Q(a)Gal(x +2x +6x −6) · Q(a)41 41 4 3 2 2 2 2 3 2 (3a − 2)x + (30a − 20)x y + (90a − 60)x y + (9a + 6)x z + (150a − 100)xy + 60xy z 3 4 2 2 3 4 + (12a + 4)xz + (150a − 25)y + (−45a + 60)y z + (30a − 20)yz + 3z 13 u, s id 42, 1 21, 1 Q Gal(x −2) 51 40 3 3 3 2x y + xz + y z 8 7 4 14 t, u, s id 336, 208 168, 42 Q Gal(x +4x +21x +18x +9) 60 58 3 3 2 2 3 3 3 2 2 3 2x y − 2x z − 3x z − 2xy − 2xz − 4y z + 3y z − yz Fité et al. Res Math Sci (2018) 5:41 Page 37 of 40 41 Table 6 The 60 Sato–Tate distributions arising for Fermat and Klein twists #ID M M M M M M M M M z z z 101 030 202 200 400 010 020 002 004 1 2 3 1 1, 1 54 1215 4734 18 486 9 99 164 47148 0 0 0 0 0 0 0 2 2, 1 26 611 2374 10 246 5 51 84 23596 0 0 0 0 0 0 0 1 1 1 3 2, 1 27 621 2367 9 243 6 54 82 23574 /2 0000 /2 /2 2 2 4 3, 1 18 405 1578 6 162 3 33 56 15720 /3 0 /3 00 00 1 1 1 5 4, 1 13 305 1187 5 123 2 26 42 11798 /2 /2 000 0 /2 6 4, 1 14 309 1194 6 126 3 27 44 11820 0 0 0 0 0 0 0 a 7 4, 1 24 443 1614 10 198 5 43 68 14444 0 0 0 0 0 0 0 8 4, 2 12 309 1194 6 126 3 27 44 11820 0 0 0 0 0 0 0 a 1 1 1 9 4, 2 13 319 1187 5 123 4 30 42 11798 /2 0000 /2 /2 1 1 10 6, 1 8 206 796 4 84 2 18 30 7882 /3 0 /3 00 00 5 1 1 1 11 6, 1 9 216 789 3 81 3 21 28 7860 /6 0 /3 00 /2 /2 5 2 1 1 12 6, 2 9 207 789 3 81 2 18 28 7860 /6 0 /3 00 /6 /2 13 7, 1 12 201 732 6 90 3 21 32 6936 0 0 0 0 0 0 0 1 1 1 14 8, 1 7 155 597 3 63 2 14 22 5910 /2 00 /2 00 /2 15 8, 1 12 225 812 6 102 3 23 36 7236 0 0 0 0 0 0 0 1 1 1 1 16 8, 2 7 161 597 3 63 2 16 22 5910 /2 /4 000 /4 /2 b 17 8, 2 12 225 814 6 102 3 23 36 7244 0 0 0 0 0 0 0 18 8, 3 6 158 604 4 66 2 15 24 5932 0 0 0 0 0 0 0 c 1 1 1 1 19 8, 3 6 161 597 3 63 2 16 22 5910 /2 /4 000 /4 /2 b 1 1 1 20 8, 3 7 168 597 3 63 3 18 22 5910 /2 0000 /2 /2 d 1 1 1 21 8, 3 12 235 807 5 99 4 26 34 7222 /2 0000 /2 /2 22 8, 4 8 158 604 4 66 2 15 24 5932 0 0 0 0 0 0 0 c 1 1 1 23 8, 4 12 221 807 5 99 2 22 34 7222 /2 /2 000 0 /2 1 1 1 24 8, 5 6 168 597 3 63 3 18 22 5910 /2 0000 /2 /2 d 2 2 25 12, 3 4 103 398 2 42 1 9 16 3944 /3 0 /3 00 00 2 1 1 1 26 12, 4 4 112 398 2 42 2 12 15 3941 /3 0 /3 00 /3 /2 1 1 1 27 14, 1 6 114 366 3 45 3 15 16 3468 /2 0000 /2 /2 28 16, 2 12 183 624 6 90 3 21 32 4956 0 0 0 0 0 0 0 1 1 1 29 16, 6 6 113 407 3 51 2 12 18 3622 /2 00 /2 00 /2 30 16, 6 6 116 412 4 54 2 13 20 3636 0 0 0 0 0 0 0 1 1 1 1 31 16, 7 3 86 302 2 33 2 10 12 2966 /2 00 /4 0 /4 /2 e 1 1 1 32 16, 7 6 126 406 3 51 3 16 18 3618 /2 0000 /2 /2 1 1 1 1 33 16, 8 4 86 302 2 33 2 10 12 2966 /2 00 /4 0 /4 /2 e 1 1 1 1 34 16, 8 6 119 406 3 51 2 14 18 3618 /2 /4 000 /4 /2 1 1 3 1 35 16, 11 3 89 302 2 33 2 11 12 2966 /2 /8 000 /8 /2 f 1 1 1 36 16, 11 6 126 407 3 51 3 16 18 3622 /2 0000 /2 /2 1 1 3 1 37 16, 13 4 89 302 2 33 2 11 12 2966 /2 /8 000 /8 /2 f 38 16, 13 6 116 414 4 54 2 13 20 3644 0 0 0 0 0 0 0 1 1 1 1 39 16, 13 6 119 407 3 51 2 14 18 3622 /2 /4 000 /4 /2 2 2 40 21, 1 4 67 244 2 30 1 7 12 2316 /3 0 /3 00 00 1 1 41 24, 12 2 55 206 2 24 1 6 10 1994 /3 0 /3 00 00 5 1 1 1 1 42 24, 12 2 58 199 1 21 1 7 8 1972 /6 /4 /3 00 /4 /2 5 2 1 1 43 24, 13 2 56 199 1 21 1 6 8 1972 /6 0 /3 00 /6 /2 1 1 1 1 44 32, 7 3 65 206 2 27 2 9 10 1818 /2 00 /4 0 /4 /2 1 1 1 1 1 45 32, 11 6 95 312 3 45 2 12 16 2478 /2 /8 0 /4 0 /8 /2 46 32, 11 6 95 318 4 48 2 12 18 2496 0 0 0 0 0 0 0 1 1 1 47 32, 34 6 105 312 3 45 3 15 16 2478 /2 0000 /2 /2 1 1 1 1 48 32, 43 3 65 207 2 27 2 9 10 1822 /2 00 /4 0 /4 /2 1 1 3 1 49 32, 43 3 68 206 2 27 2 10 10 1818 /2 /8 000 /8 /2 41 Page 38 of 40 Fité et al. Res Math Sci (2018) 5:41 Table 6 continued #ID M M M M M M M M M z z z 101 030 202 200 400 010 020 002 004 1 2 3 1 1 3 1 50 32, 49 3 68 207 2 27 2 10 10 1822 /2 /8 00 0 /8 /2 5 2 1 1 51 42, 1 2 38 122 1 15 1 5 6 1158 /6 0 /3 00 /6 /2 2 2 52 48, 3 4 61 208 2 30 1 7 12 1656 /3 0 /3 00 0 0 2 1 1 5 1 53 48, 48 1 33 103 1 12 1 5 5 997 /3 /8 /3 00 /24 /2 1 1 1 5 1 54 64, 134 3 56 159 2 24 2 9 9 1248 /2 /16 0 /8 0 /16 /2 5 1 1 1 1 1 55 96, 64 2 34 104 1 15 1 5 6 828 /6 /8 /3 /4 0 /8 /2 1 1 56 96, 64 2 34 110 2 18 1 5 8 846 /3 0 /3 00 0 0 5 2 1 1 57 96, 72 2 35 104 1 15 1 5 6 828 /6 0 /3 00 /6 /2 1 1 58 168, 42 2 19 52 2 12 1 4 6 366 /3 0 /3 00 0 0 2 1 1 1 7 1 59 192, 956 121 55 1 9 1 44 423 /3 /16 /3 /8 0 /48 /2 2 1 1 1 1 60 336, 208 112 26 1 6 133 183 /3 0 /3 /4 0 /12 /2 Table 7 Sato–Tate statistics for Fermat twists over degree one primes p ≤ 2 #ST(C)ST(C ) k kM M M M M M M M M M M M M 101 101 030 030 202 202 101 101 030 030 202 202 126.99 27 620.72 621 2365.83 2367 53.99 54 1214.69 1215 4732.64 4734 412.98 13 304.55 305 1185.12 1187 25.98 26 610.36 611 2371.23 2374 512.99 13 318.75 319 1185.94 1187 25.99 26 610.61 611 2372.38 2374 11 8.99 9 215.63 216 787.39 789 17.98 18 404.33 405 1575.11 1578 12 9.00 9 206.88 207 788.45 789 18.00 18 404.84 405 1577.24 1578 13 6.98 7 154.56 155 595.23 597 13.97 14 308.25 309 1190.96 1194 14 6.99 7 160.84 161 596.37 597 13.99 14 308.75 309 1192.99 1194 17 5.99 6 160.80 161 596.20 597 11.99 12 308.67 309 1192.65 1194 20 6.99 7 167.74 168 595.89 597 13.98 14 308.54 309 1192.02 1194 24 11.99 12 234.80 235 806.10 807 23.99 24 442.70 443 1612.54 1614 27 11.99 12 220.64 221 805.56 807 23.98 24 442.43 443 1611.64 1614 28 5.99 6 167.77 168 596.03 597 11.98 12 308.60 309 1192.31 1194 32 4.00 4 111.98 112 397.87 398 8.00 8 205.99 206 795.91 796 33 6.00 6 112.89 113 406.55 407 12.00 12 224.88 225 813.43 814 34 3.00 3 85.98 86 301.92 302 6.00 6 157.99 158 603.96 604 35 5.99 6 125.77 126 405.04 406 11.98 12 224.57 225 810.25 812 36 5.99 6 118.74 119 404.95 406 11.98 12 224.52 225 810.06 812 37 4.00 4 85.99 86 301.98 302 8.00 8 158.00 158 604.09 604 38 2.99 3 88.81 89 301.24 302 5.99 6 157.65 158 602.60 604 41 5.99 6 125.74 126 405.90 407 11.98 12 224.53 225 811.97 814 44 6.00 6 118.88 119 406.47 407 11.99 12 224.80 225 813.12 814 46 3.99 4 88.73 89 300.88 302 7.98 8 157.50 158 601.89 604 47 2.00 2 57.88 58 198.48 199 3.99 4 102.77 103 397.05 398 48 1.99 2 55.81 56 198.20 199 3.99 4 102.64 103 396.49 398 49 2.99 3 64.85 65 205.42 206 5.99 6 115.72 116 410.92 412 50 6.00 6 94.92 95 311.67 312 12.00 12 182.87 183 623.48 624 51 5.99 6 104.72 105 310.80 312 11.98 12 182.47 183 621.72 624 52 3.00 3 64.94 65 206.72 207 6.00 6 115.89 116 413.52 414 53 2.99 3 67.84 68 205.30 206 5.99 6 115.71 116 410.68 412 54 3.00 3 67.88 68 206.50 207 5.99 6 115.78 116 413.08 414 Fité et al. Res Math Sci (2018) 5:41 Page 39 of 40 41 Table 7 continued #ST(C)ST(C ) k kM M M M M M M M M M M M M 101 101 030 030 202 202 101 101 030 030 202 202 55 1.00 1 32.93 33 102.73 103 1.99 2 54.86 55 205.50 206 56 2.99 3 55.77 56 158.06 159 5.98 6 94.56 95 316.20 318 57 1.99 2 33.87 34 103.46 104 3.99 4 60.76 61 206.96 208 58 2.00 2 34.93 35 103.71 104 4.00 4 60.87 61 207.45 208 59 1.00 1 20.99 21 54.95 55 2.00 2 33.98 34 109.92 110 Table 8 Sato–Tate statistics for Klein twists over degree one primes p ≤ 2 #ST(C)ST(C ) k kM M M M M M M M M M M M M 101 101 030 030 202 202 101 101 030 030 202 202 126.99 27 620.78 621 2366.04 2367 53.99 54 1214.67 1215 4732.50 4734 212.99 13 318.73 319 1185.85 1187 25.98 26 610.52 611 2371.92 2374 311.99 12 308.67 309 1192.59 1194 11.99 12 308.67 309 1192.59 1194 48.99 9 215.76 216 787.97 789 17.98 18 404.55 405 1576.08 1578 59.00 9 206.95 207 788.79 789 18.00 18 404.94 405 1577.71 1578 67.00 7 154.90 155 596.58 597 14.00 14 308.88 309 1193.45 1194 76.99 7 167.80 168 596.13 597 13.99 14 308.63 309 1192.36 1194 84.01 4 103.36 103 399.55 398 4.01 4 103.36 103 399.55 398 93.99 4 111.68 112 396.63 398 7.98 8 205.37 206 793.34 796 10 5.99 6 113.83 114 365.24 366 11.99 12 200.67 201 730.55 732 11 3.00 3 85.94 86 301.73 302 6.00 6 157.89 158 603.51 604 12 2.01 2 55.11 55 206.43 206 2.01 2 55.11 55 206.43 206 13 2.00 2 37.97 38 121.81 122 4.00 4 66.94 67 243.65 244 14 1.00 1 11.94 12 25.70 26 2.00 2 18.87 19 51.40 52 Author details 1 2 Institute for Advanced Study, Fuld Hall, 1 Einstein Drive, Princeton, NJ 08540, USA, Laboratoire IRMAR, Université de Rennes 1, Campus de Beaulieu, 35042 Rennes Cedex, France, Department of Mathematics, Massachusetts Institute of Technology, 77 Massachusetts Avenue, Cambridge, MA 02139, USA. Acknowledgements We thank Josep González for his help with Lemma 3.18, and we are grateful to the Banﬀ International Research Station for hosting a May 2017 workshop on Arithmetic Aspects of Explicit Moduli Problems where we worked on this article. Fité is grateful to the University of California at San Diego for hosting his visit in spring 2012, the period in which this project was conceived. Fité received ﬁnancial support from the German Research Council (CRC 701), the Excellence Program María de Maeztu MDM-2014-0445, and MTM2015-63829-P. Sutherland was supported by NSF Grants DMS-1115455 and DMS-1522526. This project has received funding from the European Research Council (ERC) under the European Unions Horizon 2020 research and innovation program (Grant Agreement No. 682152), and from the Simons Foundation (Grant #550033). Received: 25 May 2018 Accepted: 18 September 2018 References 1. 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Publisher: Springer Journals
Copyright: Copyright © 2018 by The Author(s)
Subject: Mathematics; Mathematics, general; Applications of Mathematics; Computational Mathematics and Numerical Analysis
eISSN: 2197-9847
DOI: 10.1007/s40687-018-0162-0
Publisher site: See Article on Publisher Site

Abstract

drew@math.mit.edu https://math.mit.edu/~drew We determine the limiting distribution of the normalized Euler factors of an abelian Department of Mathematics, threefold A deﬁned over a number ﬁeld k when A is Q-isogenous to the cube of a CM Massachusetts Institute of Technology, 77 Massachusetts elliptic curve deﬁned over k. As an application, we classify the Sato–Tate distributions of Avenue, Cambridge, MA 02139, the Jacobians of twists of the Fermat and Klein quartics, obtaining 54 and 23, USA respectively, and 60 in total. We encounter a new phenomenon not visible in Full list of author information is available at the end of the article dimensions 1 or 2: the limiting distribution of the normalized Euler factors is not determined by the limiting distributions of their coeﬃcients. Contents 1 Introduction ........................................... 2 Equidistribution results for cubes of CM elliptic curves ................... 3 The Fermat and Klein quartics ................................. 3.1 Twists ............................................ 3.2 Moment sequences ..................................... 3.2.1 Independent coeﬃcient moment sequences ................... 3.2.2 Joint coeﬃcient moments ............................. 3.3 Sato–Tate groups ..................................... 3.4 Curve equations ...................................... 3.4.1 Constructing the Fermat twists .......................... 3.4.2 Constructing the Klein twists ........................... 3.5 Numerical computations ................................. 3.5.1 Naïve point-counting ............................... 3.5.2 An average polynomial-time algorithm ..................... 4Tables ............................................... References .............................................. 1 Introduction Let A be an abelian variety of dimension g ≥ 1 deﬁned over a number ﬁeld k. For a prime , let V (A):= Q ⊗ lim A[ ] be the (rational) -adic Tate module of A,and let ← − : G → Aut(V (A)) A k be the -adic representation arising from the action of the absolute Galois group G on V (A). Let p be a prime of k (a nonzero prime ideal of the ring of integers O )not lying © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. 0123456789().,–: volV 41 Page 2 of 40 Fité et al. Res Math Sci (2018) 5:41 above the rational prime .The L-polynomial of A at the prime p is deﬁned by L (A, T):= det(1 − (Frob )T; V (A) ) ∈ Z[T], p A p where Frob denotes a Frobenius element at p and I is the inertia subgroup at p;itdoes p p not depend on the choice of .Let S be a ﬁnite set of primes of k that includes all primes of bad reduction for A and all primes lying above .For p ∈ / S the polynomial L (A, T) has degree 2g and coincides with the numerator of the zeta function of the reduction of A modulo p.The L-function of A is deﬁned as the analytic continuation of the Euler product −s −1 L(A, s):= L (A, N(p) ) , where N(p):= [O : p] is the (absolute) norm of p.The normalized L-polynomial of A at p −1/2 is the monic polynomial L (A, T):= L (A, N(p) T) ∈ R[T]; its roots come in complex p p conjugate pairs and lie on the unit circle, as shown by Weil in [36]. As constructed by Serre in [26], the Sato–Tate group ST(A)isacompactrealLie sub- group of USp(2g), deﬁned up to conjugacy in GL (C), that comes equipped with a map 2g that assigns to each prime p ∈ / S a semisimple conjugacy class s(p)ofST(A) for which det(1 − s(p)T) = L (A, T). Let μ be the pushforward of the Haar measure of ST(A) to its set of conjugacy classes X,and let {s(p)} denote the sequence of conjugacy classes s(p)arrangedinanorder compatible with the partial ordering of primes p by norm. The generalized Sato–Tate conjecture predicts that: (ST) The sequence {s(p)} is equidistributed on X with respect to the measure μ. This conjecture has been proved for abelian varieties of dimension one (elliptic curves) over a totally real [15]orCMnumberﬁeld[1], and in several special cases for abelian varieties of higher dimension, including abelian varieties with potential CM [19]. 2g For each s ∈ X, we write det(1 − sT) =: a T , and deﬁne j=0 2g 2g I := − , , and I := I . j j j j j=1 For 0 ≤ j ≤ 2g we have a ∈ I and a = a for 0 ≤ j ≤ 2g, since the eigenvalues of any j j j 2g −j conjugacy class of USp(2g) come in complex conjugate pairs on the unit circle. Consider the maps : X −→I, := X −→ I −→ I , j j where is deﬁned by (s) = (a , ... ,a )and is the projection to the jth component. 1 g j Let μ (resp. μ ) denote the projection of the measure μ by the map (resp. ). We I I j will call μ the joint coeﬃcient measure and the set of measures {μ } ,the independent I I j coeﬃcient measures. The measures μ and μ are, respectively, determined by their moments I I n n M [μ ]:= a ··· a μ (a , ... ,a ), M [μ ]:= a μ (a ), (1.1) n ,...,n I g I 1 g n I I j 1 g 1 j j j I I for n , ... ,n ≥ 0and n ≥ 0. We denote by a (A)(p), or simply a (p), the jth coeﬃ- 1 g j j cient (s(p)) of the normalized L-polynomial, and by a(A)(p), or simply a(p), the g-tuple j Fité et al. Res Math Sci (2018) 5:41 Page 3 of 40 41 (s(p)) of coeﬃcients of the normalized L-polynomial. We can now consider the following successively weaker versions of the generalized Sato–Tate conjecture: (ST ) The sequence {a(p)} is equidistributed on I with respect to μ . p I (ST ) The sequences {a (p)} are equidistributed on I with respect to μ , for 1 ≤ j ≤ g. j p j I Let π(x) count the number of primes p ∈ / S for which N(p) ≤ x.Ifwedeﬁne 1 1 n n n 1 g M [a]:= lim a (p) ··· a (p) , M [a ]:= lim a (p) , n ,...,n 1 g n j j 1 g x→∞ x→∞ π(x) π(x) N(p)≤x N(p)≤x (1.2) then (ST )holds if andonlyifM [μ ] = M [a] for every n , ... ,n ≥ 0, while n ,...,n I n ,...,n 1 g 1 g 1 g (ST )holds if andonlyifM [μ ] = M [a ] for every n ≥ 0and 1 ≤ j ≤ g. n I n j Let A be an abelian variety deﬁned over a number ﬁeld k also of dimension g,and let X , μ , μ , μ be the data associated with A corresponding to X, μ, μ , μ , respectively. I I I I The following implications are immediate: ST(A) = ST(A ) ⇒ μ = μ ⇒{μ } ={μ } . (1.3) I I j j I j I The classiﬁcation of Sato–Tate groups of elliptic curves and abelian surfaces together with the explicit computation of their Haar measures implies that for g ≤ 2 the converses of the implications in (1.3)bothhold; see[10]. In this article, we show that for g = 3, there are cases in which the converse of the second implication of (1.3) fails to hold. Main result. In this article, we obtain the counterexamples alluded to in the previous paragraph by searching among abelian threefolds deﬁned over a number ﬁeld that are Q-isogenous to the cube of an elliptic curve with complex multiplication (CM). More precisely, we obtain a complete classiﬁcation of the Sato–Tate groups, the joint coeﬃcient measures, and the independent coeﬃcient measures of the Jacobians of twists of the Fermat and the Klein quartics (which are both Q-isogenous to the cube of a CM elliptic curve). The Fermat and Klein quartics are, respectively, given by the equations 0 4 4 4 0 3 3 3 ˜ ˜ C : x + y + z = 0, C : x y + y z + z x = 0, (1.4) 1 7 and they have the two largest automorphism groups among all genus 3 curves, of sizes 96 and 168, respectively. Our main result is summarized in the following theorem. Theorem 1 The following hold: (i) There are 54 distinct Sato–Tate groups of twists of the Fermat quartic. These give rise to 54 (resp. 48) distinct joint (resp. independent) coeﬃcient measures. (ii) There are 23 distinct Sato–Tate groups of twists of the Klein quartic. These give rise to 23 (resp. 22) distinct joint (resp. independent) coeﬃcient measures. (iii) There are 60 distinct Sato–Tate groups of twists of the Fermat or the Klein quartics. These give rise to 60 (resp. 54) distinct joint (resp. independent) coeﬃcient measures. Using Gassmann triples, one can construct examples (of large dimension) where the converse of the ﬁrst implication in (1.3) also fails to hold, but we will not pursue this here. 41 Page 4 of 40 Fité et al. Res Math Sci (2018) 5:41 One motivation for our work is a desire to extend the classiﬁcation of Sato–Tate groups that is known for dimensions g ≤ 2 to dimension 3. Of the 52 Sato–Tate groups that arise for abelian surfaces (see [10, Table 10] for a list), 32 can be realized as the Sato– Tate group of the Jacobian of a twist of one of the two genus 2 curves with the largest automorphism groups, as shown in [12]; these groups were the most diﬃcult to treat in [10] and notably include cases missing from the candidate list of trace distributions identiﬁed in [21, Table 13]. While the classiﬁcation of Sato–Tate groups in dimension 3 remains open, the 60 Sato–Tate groups identiﬁed in Theorem 1 and explicitly described in Sect. 3.3 are likely to include many of the most delicate cases and represent signiﬁcant progress toward this goal. Overview of the paper. This article can be viewed as a genus 3 analog of [12], where the 2 5 2 6 Sato–Tate groups of the Jacobians of twists of the curves y = x − x and y = x + 1 were computed. However, there are two important diﬀerences in the techniques we use here; these are highlighted in the paragraphs below that outline our approach. We also note [13], where the Sato–Tate groups of the Jacobians of certain twists of the genus 3 2 7 2 8 curves y = x − x and y = x + 1 are computed, and [2], where the Sato–Tate groups of 2 8 4 the Jacobians of twists of the curve y = x − 14x + 1 are determined. Like the Fermat and Klein quartics we consider here, these three curves represent extremal points in the moduli space of genus 3 curves, but they are all hyperelliptic, and their automorphism groups are smaller (of order 24, 32, 48, respectively). As noted above, the Sato–Tate conjecture is known for abelian varieties that are Q- isogenous to a product of CM abelian varieties [19, Cor. 15]. It follows that we can deter- mine the set of independent coeﬃcient measures {μ } by computing the sequences I j {M [a ]} , and similarly for μ and the sequences {M [a]} . Closed formu- n j j,n I n ,...,n n ,...,n 1 g 1 g las for these sequences are determined in Sect. 2 in the more general setting of abelian threefolds deﬁned over a number ﬁeld k that are Q-isogenous to the cube of an elliptic curve deﬁned over k; see Proposition 2.2 and Corollary 2.4. This analysis closely follows the techniques developed in [12,§3]. In Sect. 3, we specialize to the case of Jacobians of twists of the Fermat and Klein quar- tics. In Sect. 3.2, we obtain a complete list of possibilities for {M [a ]} : there are 48 in n j j,n the Fermat case, 22 in the Klein case, and 54 when combined; see Corollary 3.12.Wealso compute lower bounds on the number of possibilities for {M [a]} by com- n ,n ,n n ,n ,n 1 2 3 1 2 3 puting the number of possibilities for the ﬁrst several terms (up to a certain conveniently chosen bound) of this sequence. These lower bounds are 54 in the Fermat case, 23 in the Klein case, and 60 when combined; see Proposition 3.13. The ﬁrst main diﬀerence with [12] arises in Sect. 3.3, where we compute the Sato–Tate groups of the twists of the Fermat and Klein quartics using the results of [3]. Such an analysis would have been redundant in [12], since a complete classiﬁcation of Sato–Tate groups of abelian surfaces was already available from [10]. We show that there are at most 54 in the Fermat case, and at most 23 in the Klein case; see Corollaries 3.23 and 3.26. Combining the implications in (1.3) together with the lower bounds of Sect. 3.2 and upper bounds of Sect. 3.3 yields Theorem 1. The second main diﬀerence with [12] arises in Sect. 3.4, where we provide explicit equations of twists of the Fermat and Klein quartics that realize each of the possible Sato–Tate groups. Here, the computational search used in [12] is replaced by techniques Fité et al. Res Math Sci (2018) 5:41 Page 5 of 40 41 developed in [22,23] that involve the resolution of certain Galois embedding problems, and a moduli interpretation of certain twists X (7) of the Klein quartic as twists of the modular curve X(7), following [14]. In order to apply the latter approach, which also plays akey role in [25], we obtain a computationally eﬀective description of the minimal ﬁeld over which the automorphisms of X (7) are deﬁned (see Propositions 3.34 and 3.35), a result that may have other applications. Finally, in Sect. 3.5, we give an algorithm for the eﬃcient computation of the L- polynomials of twists of the Fermat and Klein quartics. This algorithm combines an average polynomial-time for computing Hasse–Witt matrices of smooth plane quartics [18] with a result speciﬁc to our setting that allows us to easily derive the full L-polynomial at p from the Frobenius trace using the splitting behavior of p in certain extensions; see Proposition 3.38. Our theoretical results do not depend on this algorithm, but it played a crucial role in our work by allowing us to check our computations and may be of inde- pendent interest. Notation. Throughout this paper, k denotes a number ﬁeld contained in a ﬁxed algebraic closure Q of Q. All the ﬁeld extensions of k, we consider are algebraic and assumed to lie in Q.Wedenoteby G the absolute Galois group Gal(Q/k). For an algebraic variety X deﬁned over k and a ﬁeld extension L/k, write X for the algebraic variety deﬁned over L obtained from X by base change from k to L. For abelian varieties A and B deﬁned over k, we write A ∼ B if there is an isogeny from A to B that is deﬁned over k.Weuse M to denote the transpose of a matrix M. We label the isomorphism class ID(H) = n, m of a ﬁnite group H according to the Small Groups Library [4], in which n is the order of H and m distinguishes the isomorphism class of H from all other isomorphism classes of groups of order n. 2 Equidistribution results for cubes of CM elliptic curves Let A be an abelian variety over k of dimension 3 such that A ∼ E , where E is an elliptic curve deﬁned over k with complex multiplication (CM) by an imaginary quadratic ﬁeld M.Let L/k be the minimal extension over which all the homomorphisms from E to A are deﬁned. We note that kM ⊆ L, and we have Hom(E ,A ) Hom(E ,A )and L L Q Q Q A ∼ E . Let σ and σ denote the two embeddings of M into Q.Consider Hom(E ,A ) ⊗ Q resp. End(A ) ⊗ Q , L L M,σ L M,σ where the tensor product is taken via the embedding σ : M → Q. Letting Gal(L/kM)act trivially on Q, it acquires the structure of a Q[Gal(L/kM)]-module of dimension 3 (resp. 9) over Q, and similarly for σ . Deﬁnition 2.1 Let θ := θ (E, A)(resp. θ (A)) denote the representation aﬀorded M,σ M,σ by the module Hom(E ,A ) ⊗ Q (resp. End(A ) ⊗ Q), and let us similarly deﬁne L L M,σ L M,σ θ := θ (E, A)and θ (A). Let θ := θ (E, A) (resp. θ (A)) denote the representation M,σ M,σ Q Q Q aﬀorded by the Q[Gal(L/k)]-module Hom(E ,A ) ⊗ Q (resp. End(A ) ⊗ Q). L L L For each τ ∈ Gal(L/kM), we write 2 3 det(1 − θ(τ)T) = 1 + a (θ)(τ)T + a (θ)(τ)T + a (θ)(τ)T , 1 2 3 so that a (θ) =− Tr θ and a (θ) =− det(θ). 1 3 41 Page 6 of 40 Fité et al. Res Math Sci (2018) 5:41 Fix a subextension F /kM of L/kM,and let S be the set of primes of F for which A or E F F has bad reduction. Note that by [27, Thm. 4.1] the set S contains the primes of F ramiﬁed in L.For z ∈ M, write |z| := σ (z) · σ (z). For p ∈ / S,there exists α(p) ∈ M, such that 1/2 |α(p)|= N(p) and σ (α(p)) + σ (α(p)) a (E )(p) =− . (2.1) 1 F 1/2 N(p) Proposition 2.2 Let A be an abelian variety of dimension 3 deﬁned over k with A ∼ E , where E is an elliptic curve deﬁned over k with CM by the imaginary quadratic ﬁeld M. Suppose that a (θ)(τ) is rational for every τ ∈ G := Gal(L/kM).Then, fori = 1, 2, 3, 2g 2g the sequence a (A ) is equidistributed on I = − , with respect to a measure i kM i i i that is continuous up to a ﬁnite number of points and therefore uniquely determined by its moments. For n ≥ 1,wehave M [a (A )] = M [a (A )] = 0 and: 2n−1 1 kM 2n−1 3 kM 1 2n 2n M [a (A )] = |a (θ)(τ)| , 2n 1 kM 1 τ ∈G |G| n n−i n n 2i 1 i 2 M [a (A )] = |a (θ)(τ)| |a (θ)(τ)| − 2 ·|a (θ)(τ)| , n 2 2 1 2 kM |G| τ ∈G i=0 i i n 2n 2i n−i 2i 1 2i−j M [a (A )] = (r (τ) − 3) 2n 3 1 kM |G| τ ∈G i=0 2i j=0 k=0 j n−i 2j+2n−2k 2n−2i k n−i−k ·r (τ) 4 (−1) . k j+n−k Here, r (τ) and r (τ) are the real and imaginary parts of a (θ)(τ)a (θ)(τ)a (θ)(τ),respec- 1 2 3 2 1 tively. Proof The proof follows the steps of [12, §3.3]. Deﬁne V (A) = V (A ) ⊗ Q , σ kM M⊗Q where the tensor product is taken relative to the map of Q -algebras M ⊗ Q → Q induced by σ ; similarly deﬁne V (A), V (E), and V (E). We then have isomorphisms of σ σ σ Q [G ]-modules kM V (A ) V (A) ⊕ V (A),V (E ) V (E) ⊕ V (E). σ σ σ σ kM kM It follows from Theorem 3.1 in [9] that V (A) θ (E, A) ⊗ V (E),V (A) θ (E, A) ⊗ V (E). σ M,σ σ σ M,σ σ We thus have an isomorphism of Q [G ]-modules kM V (A ) θ (E, A) ⊗ V (E) ⊕ θ (E, A) ⊗ V (E) . (2.2) M,σ σ M,σ σ kM For each prime p ∈ / S, let us deﬁne σ (α(p)) σ (α(p)) α (p):= , α (p):= , 1 1 1/2 1/2 N(p) N(p) where σ (α(p)), as in equation (2.1), gives the action of Frob on V (E). It follows from p σ (2.2) that a (A )(p) = a (p)α (p) + a (p)α (p), 1 1 1 1 1 kM 2 2 a (A )(p) = a (p)α (p) + a (θ)α (p) + a (p)a (p), 2 2 1 2 1 1 1 kM 3 3 a (A )(p) = a (p)α (p) + a (p)α (p) + a (p)a (p)α (p) + a (p)a (p)α (p), 3 kM 3 1 3 1 1 2 1 1 2 1 (2.3) Fité et al. Res Math Sci (2018) 5:41 Page 7 of 40 41 where to simplify notation we write a (p):= a (θ)(Frob )and a (p):= a (θ)(Frob ). Let i i p i i p r (p)and r (p) denote the real and imaginary parts of a (p)a (p)a (p), respectively. We 1 2 3 2 1 have a (p) = 1, since a (p) is a rational root of unity, and we can rewrite the above 3 3 expressions as a (A )(p) =|a (p)| z (p)α (p) + z (p)α (p) , ( ) 1 kM 1 1 1 1 1 a (A )(p) =|a (p)| z (p)α (p) + z (p)α (p) − 2|a (p)|+|a (p)| , ( ) 2 kM 2 2 1 2 1 2 1 a (A )(p) = a (p) (α (p) + α (p)) + (r (p) − 3) (α (p) + α (p)) 3 kM 3 1 1 1 1 1 ±r (p) 4 − (α (p) + α (p)) , 2 1 1 where 1/2 a (p) a (p) 1 2 z (p) = ,z (p) = ∈ U(1). 1 2 |a (p)| |a (p)| 1 2 Let α denote the sequence {α (p )} and, for each conjugacy class c of Gal(L/kM), 1 1 i i≥1 let α denote the subsequence of α obtained by restricting to primes p of ∈ / S such that 1,c 1 Frob = c. By the translation invariance of the Haar measure and [12, Prop. 3.6], for z ∈ U(1) and i ≥ 1wehave ⎨ i if i is even, i/2 M [zα + z α ] = M [α + α ] = (2.4) i 1,c 1,c i 1,c 1,c 0if i is odd. The formulas for M [a (A )], M [a (A )], M [a (A )] follow immediately from 2n 1 n 2 2n 3 kM kM kM (2.4) and the Chebotarev Density Theorem (see [12, Prop. 3.10] for a detailed explanation of a similar calculation). Remark 2.3 In the statement of the proposition, we included the hypothesis that a (θ) is rational, which is satisﬁed for Jacobians of twists of the Fermat and Klein curves (see Section 3.1), because this makes the formulas considerably simpler. This hypothesis is not strictly necessary; one can similarly derive a more general formula without it. Corollary 2.4 Let A be an abelian variety of dimension 3 deﬁned over k with A ∼ E , where E is an elliptic curve deﬁned over k with CM by the quadratic imaginary ﬁeld M, with k = kM. Suppose that a (θ)(τ) is rational for τ ∈ G := Gal(L/kM).Then, 2g 2g for i = 1, 2, 3, the sequence a (A ) is equidistributed on I = − , with respect i k i i i to a measure that is continuous up to a ﬁnite number of points and therefore uniquely determined by its moments. For n ≥ 1,wehave M [a (A )] = M [a (A )] = 0 and: 2n−1 1 k 2n−1 3 k 2n 1 2n M [a (A )] = |a (θ)(τ)| , 2n 1 1 2|G| τ ∈G n n−i 1 n n 2i i 2 M [a (A )] = |a (θ)(τ)| |a (θ)(τ)| − 2 ·|a (θ)(τ)| n 2 k 2 1 2 τ ∈G i=0 2|G| i i n n n + o(2)3 + o(4)(−1) + o(8) + o(12)2 , n 2n 2i n−i 2i 1 2i−j M [a (A )] = · (r (τ) − 3) 2n 3 1 2|G| τ ∈G i=0 2i j=0 k=0 j n−i 2j+2n−2k 2n−2i k n−i−k · r (τ) 4 (−1) . k j+n−k Here, o(n) denotes the number of elements in Gal(L/k) not in G of order n, and r (τ) and r (τ) are the real and imaginary parts of a (θ)(τ)a (θ)(τ)a (θ)(τ), respectively. 2 3 2 1 Proof For primes of k that split in kM,weinvokeProposition 2.2.Let N be such that τ = 1 for every τ ∈ Gal(L/k) \ G (the present proof shows a posteriori that one can take 41 Page 8 of 40 Fité et al. Res Math Sci (2018) 5:41 N = 24, but for the moment it is enough to know that such an N exists). For primes p of k that are inert in kM, we will restrict our analysis to those also satisfy: (a) p has absolute residue degree 1, that is, N(p) = p is prime. (b) p is of good reduction for both A and E. (c) Q( p) ∩ Q(ζ ) = Q, where ζ is a primitive 4Nth root of unity. 4N 4N These conditions exclude only a density zero set of primes and thus do not aﬀect the computation of moments. Now deﬁne D(T, τ):= det(1 − θ (E, A)(τ)T) for τ ∈ Gal(L/k) \ G,and let p be such that Frob = τ. In the course of the proof of [12, Cor. 3.12], it is shown that: (i) The polynomial L (A, T) divides the Rankin–Selberg polynomial L (E, θ (E, A),T). p p Q (ii) The roots of D(T, τ) are quotients of roots of L (A, T)and L (E, T). p p Since L (E, T) = 1 + T and the roots of D(T, τ)are Nth roots of unity, it follows from (i) that the roots of L (A, T) are 4Nth roots of unity. In particular, a (A)(p), a (A)(p), p 1 2 a (A)(p) ∈ Z[ζ ]. But (a) implies that 3 4N √ √ p · a (A)(p) ∈ Z,p · a (A)(p) ∈ Z,p p · a (A)(p) ∈ Z, (2.5) 1 2 3 which combined with (c) implies a (A)(p) = a (A)(p) = 0. This yields the desired moment 1 3 formulas for a (A )and a (A ), leaving only a (A )toconsider. 1 k 3 k 2 k From (2.5), we see that L(A, T) has rational coeﬃcients. Both L (E, T)and D(T, τ)have integer coeﬃcients, hence so does L (E, θ (E, A),T). Moreover, L (E, θ (E, A),T)isalso p Q p Q primitive, which by (i) and Gauss’ Lemma implies that L(A, T) has integer coeﬃcients. The Weil bounds then imply (see [20, Prop. 4], for example) that the polynomial L (A, T) has the form 2 4 6 1 + aT + aT + T =: P (T), (2.6) form some a ∈{−1, 0, 1, 2, 3}. To compute the moments of a (A ), it remains only to determine how often each value of a occurs as τ ranges over G.Wehave 2 2 2 P (T) = (1 − T) (1 + T) (1 + T ), −1 2 2 4 P (T) = (1 + T )(1 − T + T ), 2 4 P (T) = (1 + T )(1 + T ), (2.7) 2 2 2 P (T) = (1 − T + T )(1 + T + T )(1 + T ), 2 3 P (T) = (1 + T ) . The integer ord(τ) is even and then condition (ii) and the speciﬁc shape of the P (T) imply that the only possible orders of τ are 2, 4, 6, 8, 12. If ord(τ) = 2, then all the roots of L (E, θ (E, A),T) are of order 4. By (i), so are the roots of P (T), and (2.7)implies that p Q a a = 3. If ord(τ) = 4, then all the roots of L (E, θ (E, A),T) are of order dividing 4. Thus so are p Q the roots of P (T), which leaves the two possibilities a =−1or a = 3. But (ii) implies that the latter is not possible: if a = 3, then the roots of D(T, τ) would all be of order dividing 2 and this contradicts the fact that τ has order 4. Thus a =−1. Fité et al. Res Math Sci (2018) 5:41 Page 9 of 40 41 If ord(τ) = 6, then by (ii) we have a =−1, 1, 3 (otherwise the order of τ would not be divisible by 3). If a = 2, then again by (ii) the polynomial D(T, τ) would have a root of order at least 12, which is impossible for ord(τ) = 6. Thus a = 0. If ord(τ) = 8, then L (E, θ (E, A),T) has at least 8 roots of order 8 and thus P (T)has p Q a at least a root of order 8. Thus a = 1. If ord(τ) = 12, then by (ii) we have a =−1, 1, 3 (otherwise the order of τ would not be divisible by 3). If a = 0, then again by (ii) the polynomial D(T, τ) would only have roots of orders 1, 2, 3 or 6, which is incompatible for ord(τ) = 12. Thus, a = 2. 3 The Fermat and Klein quartics 0 0 ˜ ˜ The Fermat and the Klein quartics admit models over Q given by the equations C and C 1 7 of (1.4), respectively. The Jacobian of C is Q-isogenous to the cube of an elliptic curve deﬁned over Q (see Proposition 3.1), butthisisnot true for C , which leads us to choose 0 0 a diﬀerent model for the Klein quartic. Let us deﬁne C := C and 1 1 0 4 4 4 3 3 3 2 2 2 2 2 2 C : x + y + z + 6(xy + yz + zx ) − 3(x y + y z + z x ) + 3xyz(x + y + z) = 0. The model C is taken from [7, (1.22)], and its Jacobian is Q-isogenous to the cube of an elliptic curve deﬁned over Q, as we will prove below. One can explicitly verify that the curve C is Q-isomorphic to the Klein quartic by using (3.3) below to show that ID(Aut((C ) )) = 168, 42 . Q( −7) One similarly veriﬁes that C is Q-isomorphic to the Fermat quartic by using (3.2)below to show ID(Aut((C ) )) 96, 64 . 1 Q(i) 0 0 Let E and E be the elliptic curves over Q given by the equations 1 7 0 2 3 2 0 2 3 2 3 E : y z = x + xz,E : y z = x − 1715xz + 33614z , (3.1) 1 7 0 6 3 with Cremona labels 64a4 and 49a3, respectively. We note that j(E ) = 2 · 3 and 0 3 3 0 0 j(E ) =−3 · 5 , thus E has CM by the ring of integers of Q(i), and E has CM by the ring 7 1 7 of integers of Q( −7). For future reference, let us ﬁx some notation. The automorphisms s ([x : y : z]) = [z : x : y], (3.2) t ([x : y : z]) = [−y : x : z], u ([x : y : z]) = [ix : y : z] generate Aut((C ) ), whereas the automorphisms 1 Q s ([x : y : z]) = [y : z : x], ⎪ 7 t ([x : y : z]) = [−3x − 6y + 2z : −6x + 2y − 3z :2x − 3y − 6z], u ([x : y : z]) = [−2x + ay − z : ax − y + (1 − a)z : −x + (1 − a)y − (1 + a)z], (3.3) 41 Page 10 of 40 Fité et al. Res Math Sci (2018) 5:41 with −1 + −7 2 4 a := = ζ + ζ + ζ , 7 7 generate Aut((C ) ). 0 0 Proposition 3.1 For d = 1 or 7, the Jacobian of C is Q-isogenous to the cube of E . d d 0 0 Proof For d = 1, we have a nonconstant map ϕ : C → E , given by 1 1 3 2 3 3 ϕ ([x : y : z]) = [−x zy : x z : zxy ]. 0 0 Thus there exists an abelian surface B deﬁned over Q with Jac(C ) ∼ B × E . Suppose 1 1 0 0 that E was not a Q-factor of B. Then, the subgroup s ,t ⊆ Aut(C ), isomorphic to the 1 1 1 1 0 × symmetric group on 4 letters S , would inject into (End(Jac(C )) ⊗ C) . There are two r s options for this C-algebra: it is either GL (C) or GL (C) × GL (C) ,with r, s ∈ Z .In 1 2 1 ≥0 either case, we reach a contradiction with the fact that S has no faithful representations whose irreducible constituents have degrees at most 2. Thus, E is a Q-factor of B and 0 0 2 Jac(C ) ∼ (E ) × E, where E is an elliptic curve deﬁned over Q. Applying the previous 1 1 0 0 0 3 argument again shows E ∼ E ,soJac(C ) ∼ (E ) . 1 1 1 0 0 For d = 7, we have a nonconstant map ϕ : C → E , given by 7 7 ϕ ([x : y : z]) 2 2 2 2 2 = −7(x + y + z)(3x − y − 9z):2 · 7 (−x − 3xy − xz + 2z ):(x + y + z) , 0 0 thus Jac(C ) ∼ B × E for some abelian surface B deﬁned over Q.Since S is contained in 7 7 Aut((C ) ) PSL (F ), where M = Q( −7), we may reproduce the argument above to M 2 7 0 0 3 show that Jac(C ) ∼ (E ) . It follows that 7 7 M Jac(C ) ∼ E × E × E . 0 0 where E, E ,and E are either E or E ⊗ χ, where χ is the quadratic character of M.But 7 7 0 0 0 E ⊗ χ ∼ E ,since E has CM by M, and the result follows. 7 7 7 Remark 3.2 To simplify notation, for the remainder of this article d is either 1 or 7, and 0 0 0 0 we write C for C , E for E , M for Q( −d), and s, t, u for s , t , u . When d is not d d d d d speciﬁed, it means we are considering both values of d simultaneously. 3.1 Twists 2 0 0 Let C be a k-twist of C , a curve deﬁned over k that is Q-isomorphic to C .The set of k-twists of C ,upto k-isomorphism, is in one-to-one correspondence with 1 0 0 H (G , Aut(C )). Given an isomorphism φ : C → C , the 1-cocycle deﬁned by M Q σ −1 ξ(σ):= φ( φ) , for σ ∈ G , is a representative of the cohomology class corresponding to C. Let K /k (resp. L/k) denote the minimal extension over which all endomorphisms of ˜ ˜ Jac(C) (resp. all homomorphisms from Jac(C) to E ) are deﬁned. Let K /k (resp. L/k) Q Q denote the minimal extension over which all automorphisms of C (resp. all isomorphisms from C to C ) are deﬁned. 2 0 When we need not specify the number ﬁeld k over which C is deﬁned, we will simply say that C is a twist of C . Thus, by saying that C is a twist of C , we do not necessarily mean that C is deﬁned over Q. Fité et al. Res Math Sci (2018) 5:41 Page 11 of 40 41 Lemma 3.3 We have the following inclusions and equalities of ﬁelds: ˜ ˜ M ⊆ K = K ⊆ L = L. Proof The inclusion M ⊆ K follows from the fact that Tr(A ) ∈ M \ Q, where A is as in u u (3.12)and (3.13). From the proof of Proposition 3.1, we know that Jac(C) ∼ E , where E ˜ ˜ ˜ is an elliptic curve deﬁned over K with CM by M.Thisimplies K = KM and L = LM,as in the proof of [12,Lem.4.2]. 0 0 We now associate with C a ﬁnite group G that will play a key role in the rest of the article. 0 0 Deﬁnition 3.4 Let G := Aut(C ) Gal(M/Q), where Gal(M/Q) acts on Aut(C )in C M M the obvious way (coeﬃcient-wise action on rational maps). It is straightforward to verify that ID(G 0) = 192, 956 , ID(G 0) = 336, 208 . (3.4) C C 1 7 We remark that G 0 PGL(F ). As in [12, §4.2], we have a monomorphism of groups λ :Gal(L/k) = Gal(L/k) → G 0, λ (σ ) = (ξ(σ ), π(σ )), φ φ where π:Gal(L/k) → Gal(M/Q) is the natural projection (which by Lemma 3.3 is well 0 0 deﬁned). For each α ∈ Aut(C ), let α˜ denote its image by the embedding Aut(C ) → M M 0 3 End((E ) ). The 3-dimensional representation 0 0 0 θ 0 0:Aut(C ) → Aut (Hom(E , Jac(C )) ⊗ Q), M,σ E ,C M M M deﬁned by θ 0 0(α)(ψ):= α˜ ◦ ψ satisﬁes E ,C k 0 θ 0 0 ◦ Res λ θ (E , Jac(C)), (3.5) φ M,σ E ,C kM where Res λ denotes the restriction of λ from Gal(L/k)toGal(L/kM). φ φ kM Lemma 3.5 Let C be a twist of C .Then: 0 0 χ if C = C (see Table 3a), Tr θ 0 0 = E ,C ⎩ 0 χ if C = C (see Table 3b). Proof In the proof of Lemma 3.18, we will construct an explicit embedding 0 0 3 Aut(C ) → End((E ) ) ⊗ Q, α → α˜. M M 0 0 Fix the basis B ={id ×0 × 0, 0 × id ×0, 0 × 0 × id} for Hom(E , Jac(C )). In this basis, M M with the above embedding the representation θ 0 0 is given by E ,C −1 −1 −1 θ (s) = A , θ (t) = A , θ (u) = A , 0 0 0 0 0 0 E ,C s E ,C t E ,C u where A , A ,and A are as in (3.12)and (3.13). The lemma follows. s t u Remark 3.6 Observe that since det(θ ) is a rational character of Aut(C ), by (3.5)so 0 0 E ,C M is a (θ) = det θ (E , Jac(C)). Thus, Corollary 2.4 can be used to compute the moments 3 M,σ of a (Jac(C)) for i = 1, 2, 3. i 41 Page 12 of 40 Fité et al. Res Math Sci (2018) 5:41 Proposition 3.7 The ﬁelds K and L coincide. Proof Note that L/K is the minimal extension over which an isomorphism between E /n and E is deﬁned. It follows that L = K (γ ) for some γ ∈ K,with n = 4 for d = 1 and n = 2 for d = 7; see [28, Prop. X.5.4]. In either case, Gal(L/K)iscyclicoforder dividing 4 (note that Q(ζ ) ⊆ K ). Suppose that L = K,let ω denote the element in 0 ω 0 Gal(L/K ) of order 2, and write K = L . Fix an isomorphism ψ : E → E and an 1 L 3 3 isogeny ψ :(E 0) → Jac(C) 0.For i = 1, 2, 3, let ι : E 0 → (E 0) denote the natural 2 i K K K K injection to the ith factor. Then, {ψ ◦ ι ◦ ψ } constitute a basis of the Q[Gal(L/M)]- 2 i 1 i=1,2,3 0 ω ω ω module Hom(E , Jac(C) ) ⊗ Q.Since ψ =−ψ , ψ = ψ ,and ι = ι ,we L M,σ 1 1 2 2 i i have Trace θ (E , Jac(C))(ω) =−3. But this contradicts (3.5), because there is no α in M,σ Aut(C ) for which Trace θ (α) =−3. 0 0 M E ,C Remark 3.8 By Proposition 3.7, and the identities (2.3)and (3.5), the independent and joint coeﬃcient measures of Jac(C) depend only on the conjugacy class of λ (Gal(K /k)) in G 0.InProposition 3.22, we will see that this also applies to the Sato–Tate group of Jac(C). For this reason, henceforth, subgroups H ⊆ G 0 will be considered only up to conjugacy. Deﬁnition 3.9 Let G := Aut(C ) × 1 ⊆ G 0, and for subgroups H ⊆ G 0,let C C H := H ∩G .Wemay view H as a subgroup of Aut(C ) G whenever it is convenient 0 0 0 0 to do so. Noting that [G : G ] = 2, for any subgroup H of G there are two possibilities: 0 0 C C (c ) H ⊆ G , in which case [H : H ] = 1; 1 0 0 (c ) H G , in which case [H : H ] = 2. 2 0 0 Remark 3.10 We make the following observations regarding H ⊆ G 0 and cases (c )and (c ): (i) In Sect. 3.4, we will show that for each subgroup H ⊆ G , there is a twist C of C such that H = λ (Gal(K /k)). From the deﬁnition of λ ,wemustthenhave φ φ H = λ (Gal(K /kM)). The case (c ) corresponds to kM = k, and the case (c ) 0 φ 1 2 corresponds to k = kM. (ii) There are 83 subgroups H ⊆ G 0 up to conjugacy, of which 24 correspond to case (c ) and 59 correspond to case (c ). In Table 4 we list the subgroups in case (c ). 1 2 2 For any subgroup H in case (c ), there exists a subgroup H in case (c ) such that 1 2 H = H; thus the subgroups in case (c ) can be recovered from Table 4 by looking at the column for H . (iii) There are 23 subgroups H ⊆ G 0 up to conjugacy, of which 12 correspond to case (c ) and 11 correspond to case (c ). For all but 3 exceptional subgroups H in case 1 2 (c ), there exists a subgroup H in case (c ) such that H = H.InTable 5,welist 1 2 the subgroups in case (c ) as well as the 3 exceptional subgroups, which appear in rows #3, #8, and #12 of Table 5. As in (ii), the non-exceptional subgroups in case (c ) can be recovered from Table 5 by looking at the column for H , which for the 1 0 exceptional groups is equal to H. Note that this does not hold for the hyperelliptic curves considered in [12] where [L: K ] may be 1 or 2. Fité et al. Res Math Sci (2018) 5:41 Page 13 of 40 41 (iv) The subgroups H ⊆ G in Tables 4 and 5 are presented as follows. First, generators 0 0 of H ⊆ G Aut(C ) are given in terms of the generators s, t, u for Aut(C ) 0 0 M M listed in (3.2)and (3.3). For the 3 exceptional subgroups of Table 5 we necessarily have H = H , and for the others, H is identiﬁed by listing an element h ∈ Aut(C ) such that H = H ∪ H · (h, τ) ⊆ G , (3.6) 0 0 where τ is the generator of Gal(M/Q). 3.2 Moment sequences We continue with the notation of Sect. 3.1.If C is a k-twist of C ,wedeﬁne thejoint and independent coeﬃcient moment sequences M (C):={M [a(C)]} , M (C):={M [a (C)]} , joint n ,n ,n n ,n ,n indep n i j,n 1 2 3 1 2 3 where a(C)and a (C)denote a(Jac(C)) and a (Jac(C)), respectively, as deﬁned in Sect. 1; j j recall that these moment sequences are deﬁned by and uniquely determine corresponding measures μ and μ , respectively. I I Using Lemma 3.5 and (3.5), we can apply Corollary 2.4 to compute the moments M [a (C)] for any n, and as explained in Sect. 3.2.2,itiseasytocompute M [a(C)] n j n ,n ,n 1 2 3 for any particular values of n ,n ,n . Magma scripts [6] to perform these computations 1 2 3 are available at [11], which we note depend only on the pairs (H, H ) (or just H when 0 0 k = kM)listedinTables 4 and 5, and are otherwise independent of the choice of C. 3.2.1 Independent coeﬃcient moment sequences We now show that for any twist of the Fermat or Klein quartic, each of the independent coeﬃcient moment sequences (and hence the corresponding measures) is determined by the ﬁrst several moments. Proposition 3.11 Let C and C be k-twists of C . For each i = 1, 2, 3 there exists a positive integer N such that if M [a (C)] = M [a (C )] for 1 ≤ n ≤ N n i n i i then in fact M [a (C)] = M [a (C )] for all n ≥ 1. n i n i Moreover, one can take N = 6,N = 6,N = 10. 1 2 3 Proof For the sake of brevity, we assume k = kM (the case k = kM is analogous and easier). It follows from Corollary 2.4 that, for i = 1, 2, 3, the sequence {M [a (C)]} is n i n≥0 determined by |a (C)|, |a (C)|, a (C)a (C)¯a (C), and o¯(2), o¯(4), o¯(8) (note that G 0 and 1 2 3 2 1 G 0 contain no elements of order 12, so we ignore the o¯(12) term in the formula for M [a (C)]). We consider the Fermat and Klein cases separately. n 2 For the Fermat case, with the notation for conjugacy classes as in Table 3a, let x (resp. x , x , x , x ) denote the proportion of elements in Gal(L/k) lying in the conjugacy class 1a 2 3 4 5 (resp. 2a ∪ 2b ∪ 4c ∪ 4d,3a,4a ∪ 4b,8a ∪ 8b); note that by Lemma 3.5, we are interested in the representation with character χ listed in Table 3a, which motivates this partitioning of conjugacy classes. 41 Page 14 of 40 Fité et al. Res Math Sci (2018) 5:41 Let y (resp. y , y ) denote the proportion of elements in Gal(L/k) which do not lie in 1 2 3 Gal(L/kM) and have order 2 (resp. 4, 8). Applying Corollary 2.4, one ﬁnds that for n ≥ 1 we have 2n 2n 2n n n M [a (C)] = x · 9 + (x + x ) + x (−1) . (3.7) 2n 1 1 2 5 4 n n n Evaluating (3.7)at n = 1, 2, 3 yields an invertible linear system in x , x + x , x of 1 2 5 4 dimension 3. The moments M [a (C)] for n = 1, 2, 3 thus determine x , x + x , x and 2n 1 1 2 5 4 therefore determine all the M [a (C)]. 2n 1 For M [a (C)], one similarly obtains an invertible linear system in x , x + x , x , y , y , n 2 1 2 5 4 1 2 y of dimension 6, and it follows that the moments M [a (C)] for n ≤ 6 determine all the 3 n 2 M [a (C)]. n 2 For M [a (C)], one obtains an invertible linear system in x , x , x , x , x of dimension 2n 3 1 2 3 4 5 5, and it follows that the moments M [a (C)] for n ≤ 5 determine all the M [a (C)]. 2n 2 2n 3 In the Klein case one proceeds analogously. With the notation of Table 3b, let x (resp. x , x , x ) denote the proportion of elements in Gal(L/k) lying in the conjugacy class 1a 2 3 4 (resp. 2a ∪ 4a,3a,7a ∪ 7b), and let y , y , y be as in the Fermat case. Now x , x , x 1 2 3 1 2 4 determine M [a (C)]; x , x , x and y , y , y determine M [a (C)]; and x , x , x , x n 1 1 2 4 1 2 3 n 2 1 2 3 4 determine M [a (C)]. These proportions are, as before, determined by the ﬁrst several n 3 moments (never more than are needed in the Fermat case), and the result follows. We spare the reader the lengthy details. With Proposition 3.11 in hand we can completely determine the moment sequences M [a (C)] that arise among k-twists C of C by computing the moments M [a (C)] for n i n i 0 0 n ≤ N for the 59 pairs (H, H )listedinTable 4 in the case C = C , and for the 14 pairs i 0 (H, H )listedinTable 5 (as described in Remark 3.10). Note that each pair (H, H )with 0 0 H = H gives rise to two moments sequences M [a (C)] for each i, one with k = kM and 0 n i one with k = kM. After doing so, one ﬁnds that in fact Proposition 3.11 remains true with N = 4and N = 4. The value of N cannot be improved, but one also ﬁnds that the sequences 3 1 M [a (C)] and M [a (C)] together determine the sequence M [a (C)], and in fact just n 2 n 3 n 1 two well-chosen moments suﬃce. Corollary 3.12 There are 48 (resp. 22) independent coeﬃcient measures among twists of 0 0 C (resp. C ). In total, there are 54 independent coeﬃcient measures among twists C of 1 7 0 0 either C or C , each of which is uniquely distinguished by the moments M [a (C)] and 3 2 1 7 M [a (C)]. 4 3 The moments M [a (C)] and M [a (C)] correspond to the joint moments M [a(C)] 3 2 4 3 0,3,0 and M [a(C)] whose values are listed in Table 6 for each of the 60 distinct joint coef- 0,0,4 ﬁcient moment measures obtained in the next section (this includes all the independent coeﬃcient measures). 3.2.2 Joint coeﬃcient moments Instead of giving closed formulas for M [a(C)], analogous to those derived for n ,n ,n 1 2 3 M [a (C)] in Corollary 2.4, let us explain how to compute M [a(C)] for speciﬁc n j n ,n ,n 1 2 3 values of n ,n ,n (this will suﬃce for our purposes). Suppose k = kM (the other case is 1 2 3 Fité et al. Res Math Sci (2018) 5:41 Page 15 of 40 41 similar). By (2.3), we may naturally regard the quantity n n n 1 2 3 a (C) a (C) a (C) (3.8) 1 2 3 as an element of the formal polynomial ring Q[α , α¯ ]/(α α¯ − 1). The moment 1 1 1 1 M [a(C)] is simply the constant term of (3.8). For N ≥ 0 and each pair (H, H ) n ,n ,n 0 1 2 3 in Tables 4 and 5, one can then compute truncated joint moment sequences ≤N M (C):={M [a(C)] : n ,n ,n ≥ 0,n + n + n ≤ N }. n ,n ,n 1 2 3 1 2 3 joint 1 2 3 ≤4 By explicitly computing M (C) for all the pairs (H, H )listedinTables 4 and 5,we joint obtain the following proposition. Proposition 3.13 There are at least 54 (resp. 23) joint coeﬃcient measures (and hence Sato–Tate groups) of twists of the Fermat (resp. Klein) quartic, and at least 60 in total. These 60 joint coeﬃcient measures are listed in Table 6, in which each is uniquely distinguished by the three moments M [a(C)], M [a(C)],and M [a(C)]. 1,0,1 0,3,0 2,0,2 ≤N Computing M (C)with N = 5, 6, 7, 8, does not increase any of the lower bounds joint in Proposition 3.13, leading one to believe they are tight. We will prove this in the next section, but an aﬃrmative answer to the following question would make it easy to directly verify such a claim. Question 3.14 Recall the setting of the ﬁrst paragraph of Sect. 1. In particular, A is an abelian variety deﬁned over a number ﬁeld k of dimension g ≥ 1,and μ is the measure induced on I. By [10, Prop. 3.2] and [10,Rem.3.3], one expects a ﬁnite list of possibilities for the Sato–Tate group of A. One thus expects a ﬁnite number of possibilities for the sequence {M [μ ]} . In particular, one expects that there exists N ≥ 1, depending only n ,...,n I n ,...,n g 1 g 1 g on g, such that for any abelian variety A deﬁned over a number ﬁeld k ,if {M [μ ]} ={M [μ ]} (3.9) n ,...,n I n ,...,n n ,...,n n ,...,n 1 g 1 g 1 g I 1 g for all n + ··· + n ≤ N ,then (3.9) holds for all n , ... ,n ≥ 0. Is there an explicit and 1 g g 1 g eﬀectively computable upper bound for N ? Lacking an answer to Question, 3.14, in order to determine the exact number of distinct joint coeﬃcient measures, we take a diﬀerent approach. In the next section, we will classify the possible Sato–Tate groups of twists of the Fermat and Klein quartics. This classiﬁcation yields an upper bound that coincides with the lower bound of Proposition 3.13. Table 6 also lists z = [z ], z = [z ,z ,z ,z ,z ], and z = [z ], where z 1 1,0 2 2,−1 2,0 2,1 2,2 2,3 3 3,0 i,j denotes the density of the set of primes p for which a (C)(p) = j. For these, we record the following lemma. Lemma 3.15 Let C be a k-twist of C and let H := λ (Gal(K /kM)).Then, 0 φ ⎧ ⎧ o(3) ⎨ ⎨ if M ⊆ k, 0 if M ⊆ k, |H | z (Jac(C)) = z (Jac(C)) = 1 3 1 o(3) 1 ⎩ ⎩ + if M k, if M k, 2 2|H | 2 ⎨ 1 [0,o(3), 0, 0, 0] if M ⊆ k, |H | z (Jac(C)) = [¯o(4),o(3) + o¯(6), o¯(8), 0, o¯(2)] if M k. 2|H | 0 41 Page 16 of 40 Fité et al. Res Math Sci (2018) 5:41 Here, o¯(n) is as in Corollary 2.4 and o(n) denotes the number of elements of order n in H . Proof The formula for z is immediate from (2.3) and the study of the polynomial (2.6)in the proof of Corollary 2.4, together from the fact that τ ∈ Gal(L/k) satisﬁes a (θ)(τ) = 0 if and only if τ has order 3 (as can be seen from Table 3). The formula for z follows from a similar reasoning, once one observes that again a (θ)(τ) = 0ifand only if τ has order 3, and the discussion of the end of Corollary 2.4. Note also that, as G 0 contains no elements of order 12, we have o¯(12) = 0. For z , it suﬃces to note that a (τ) does not vanish. 3 3 3.3 Sato–Tate groups In this section, for any twist C of C , we explicitly construct ST(Jac(C)), which to simplify notation we denote by ST(C). The ﬁrst step is to compute a (non-canonical) embedding ι:Aut(C ) → USp(6) 1 0 1 0 (see [24] for a very similar approach). Let (E ) (resp. (C )) denote the M-vector M M 0 0 space of regular diﬀerentials of E (resp. C ). Deﬁne M M 0 1 0 ∗ −1 ι :Aut(C ) → Aut( (C )), ι (α) = (α ) , 1 1 M M ∗ 1 0 1 0 where α : (C ) → (C ) is the map induced by α. M M Remark 3.16 Let f (X, Y ) = 0 be an aﬃne model of the plane quartic C .Then, dX dX dX W := X ,W := Y ,W := , (3.10) 1 2 3 f f f Y Y Y ∂f 1 0 with f = , is a basis of the regular diﬀerentials (C ). If we denote by ω the regular Y i ∂Y M 0 0 3 diﬀerential of the ith copy of E in (E ) , then M M , ω , ω } (3.11) {ω 1 2 3 1 0 3 is a basis of the regular diﬀerentials (E ) . Consider the isomorphism 1 0 1 0 3 ι :End( (C )) → End( (E ) ) M M 1 0 1 0 3 induced by the isomorphism (C ) (E ) that sends W to ω . i i M M Fix an isomorphism [ ]: M → End(E ) ⊗ Q such that for any regular diﬀerential 1 0 ∗ ω ∈ (E ), one has [m] (ω) = mω for every m ∈ M (see [29, Chap. II, Prop. 1.1]) and then deﬁne 1 0 3 0 3 ι :End( (E ) ) → End((E ) ) ⊗ Q, ι ((m )) = ([m ]), 3 3 jk jk M M top where m ∈ M.Let f = i and let f = a.For d = 1, 7, let γ ∈ H ((E ) , Q) be such ij 1 7 d 1 d C top that {γ , [f ] γ } is a symplectic basis of H ((E ) , Q) with respect to the cup product, d d ∗ d 1 d C and use this basis to obtain an isomorphism top :End(H ((E ) , Q)) → GSp (Q). d 2 d C Then, deﬁne 0 3 ι :End((E ) ) → GSp (Q), ([m ]) → ( ([m ] )). 4 jk d jk ∗ M 6 d Fité et al. Res Math Sci (2018) 5:41 Page 17 of 40 41 Finally, deﬁne the matrices 10 0 −1 0 −2 I = ,J = ,K = . 2 2 2 01 10 1 −1 Remark 3.17 From now on, we ﬁx the following notation: denote by A the 3-diagonal embedding of a subset A of GL in GL . Throughout this section, we consider the general 2 6 symplectic group GSp /Q, the symplectic group Sp /Q, and the unitary symplectic group 6 6 USp(6) with respect to the symplectic form given by the 3-diagonal embedding (J ) . 2 3 Lemma 3.18 The map 0 ι 1 0 ι 1 0 3 ι 0 3 ι 1 2 3 4 ι: Aut(C ) → End( (C )) End( (E ) ) End((E ) ) ⊗ Q → GSp (Q) M M M M 6 0 0 is a monomorphism of groups that for C = C is explicitly given by ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 00 I 0 −I 0 −I 00 2 2 2 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ι(s ) = , ι(t ) = , ι(u ) = , ⎝ I 00⎠ ⎝ I 00⎠ ⎝ 0 −J 0 ⎠ 1 2 1 2 1 2 0 I 0 00 I 00 −J 2 2 2 0 0 T and for C = C is explicitly given by ι(s ) = ι(s ) and 7 1 ⎛ ⎞ ⎛ ⎞ −3I −6I 2I −2I − 4K 3I − K −I − 2K 2 2 2 2 2 2 2 2 2 1 1 ⎜ ⎟ ⎜ ⎟ ι(t ) = , ι(u ) = . ⎝ −6I 2I −3I ⎠ ⎝ 3I − K −I − 2K −2I + 3K ⎠ 7 2 2 2 7 2 2 2 2 2 2 7 7 2I −3I −6I −I − 2K −2I + 3K −4I − K 2 2 2 2 2 2 2 2 2 0 0 ∗ ∗ Proof We ﬁrst consider the case C = C .Inthe basisof (3.10), the elements s , t ,and 1 1 1 ∗ 1 0 u of End( (C )) are given by the matrices 1 M ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 010 01 0 −100 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ A = ,A = ,A = . (3.12) ⎝ 001⎠ ⎝ −100⎠ ⎝ 0 i 0⎠ s t u 1 1 1 100 00 1 00 i 1 0 3 Thus, in the basis of (3.11), the elements ι ι (s ), ι ι (t ), ι ι (u )ofEnd( (E ) )are 2 1 1 2 1 1 2 1 1 −1 −1 −1 given by the matrices A , A , A . It is then enough to check that in the basis {γ , [i] γ } 1 ∗ 1 s t u 1 1 1 we have (1) = I and (i) = J . 1 2 1 2 0 0 ∗ ∗ ∗ We now assume C = C .Inthe basisof (3.10), the matrices associated with s , t , u 7 7 7 7 are: ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 001 −3 −62 2 + 4a 4 + a 1 + 2a 1 1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ A = ,A = ,A = . ⎝ 100⎠ ⎝ −62 −3⎠ ⎝ 4 + a 1 + 2a −5 − 3a⎠ s t u 7 7 7 7 7 010 2 −3 −6 1 + 2a −5 − 3a −3 + a (3.13) 1 0 3 Thus, in the basis of (3.10), the elements ι ι (s ), ι ι (t ), ι ι (u )ofEnd( (E ) )are 2 1 7 2 1 7 2 1 7 M 7 −1 −1 −1 given by the matrices A , A , A . It is then enough to check that in the basis {γ , [a] γ } 7 ∗ 7 s u 7 7 7 we have (1) = I and (a) = K . But this is clear, since 7 2 7 2 [a] (γ ) = 0 · γ + 1 · ([a] γ ), ∗ 7 7 ∗ 7 [a] ([a] γ ) = [a ] γ = [−a − 2] γ =−2 · γ − 1 · ([a] γ ), ∗ ∗ 7 ∗ 7 ∗ 7 7 ∗ 7 and this completes the proof. 41 Page 18 of 40 Fité et al. Res Math Sci (2018) 5:41 Remark 3.19 Note that since Aut(C ) has ﬁnite order, the image of ι is contained in USp (Q). Remark 3.20 It is easy to check that the matrices ι(s ), ι(t ), ι(u ) (resp. ι(s ), ι(t ), ι(u )) 1 1 1 7 7 7 are symplectic with respect to J := (J ) . 2 3 The following theorem gives an explicit description of the Sato–Tate group of a twist of the Fermat or Klein quartic corresponding to a subgroup H of the group G := Aut(C ) × Gal(M/Q) 0 M associated with C (see Deﬁnition 3.4). Theorem 3.21 The following hold: (i) The monomorphism of Lemma 3.18 extends to a monomorphism ι: G → USp(6)/ −1 by deﬁning ⎧ ⎛ ⎞ ⎪ ii 1 0 0 ⎪ ⎝ ⎠ if C = C , i −i ⎛ ⎞ ι((1, τ)) := i −i 0 0 ⎝ ⎠ ⎪ if C = C , ⎪ 7 0 −i where τ denotes the non-trivial element of Gal(M/Q). 0 0 (ii) Let φ : C → C denote a k-twist of C and write H := λ (Gal(K /k)) ⊆ G .The Sato–Tate group of Jac(C) is given by ST(C) = ST(E , 1) · ι(H), where ! " cos(2πr)sin(2πr) ST(E , 1) = | r ∈ [0, 1] − sin(2πr)cos(2πr) 0 0 if C = C ,and ! " 1 4 √ √ cos(2πr) − sin(2πr) sin(2πr) 0 7 7 ST(E , 1) = | r ∈ [0, 1] 2 1 √ √ − sin(2πr)cos(2πr) + sin(2πr) 7 7 0 0 if C = C . Proof To prove (i) it is enough to note that ι((1, τ)) = 1 in USp(6)/ −1 and that ι((1, τ)) 0 0 acts by matrix conjugation on ι(Aut(C )) as τ acts by Galois conjugation on Aut(C ). If M M 0 0 0 0 C = C (resp. C = C ), the latter is equivalent to 1 7 −1 −1 ι((1, τ)) J ι((1, τ)) =−J , resp. ι((1, τ)) K ι((1, τ)) =−I − K , 2 2 2 2 2 which is straightforward to check. Fité et al. Res Math Sci (2018) 5:41 Page 19 of 40 41 For(ii),weconsideronlythe case kM = k,since thecase k = kM can be easily deduced from the case kM = k. Recall from [3]thatST(E ) is a maximal compact subgroup of the 0 0 0 algebraic Sato–Tate group AST(E ) ⊗ C attached to E . Recall that we have AST(E ) = 0 0 L(E , 1) ∪ L(E , τ), where for σ ∈ Gal(kM/k) one has 0 −1 σ 0 L(E , σ):={γ ∈ Sp | γ αγ = α for all α ∈ End(E ) ⊗ Q}. (3.14) 0 0 0 This induces a decomposition ST(E ) = ST(E , 1) ∪ ST(E , τ) that can be explicitly deter- mined. 0 0 For the case C = C ,wehave 0 T −1 L(E , 1)(C) ={A ∈ M (C)|A J A = J ,A J A = J } 2 2 2 2 2 ! " cb 2 2 = | c, b ∈ C,c + b = 1 . −bc Thus, a maximal compact subgroup of L(E , 1)(C)is ! " cos(2πr)sin(2πr) ST(E , 1) = | r ∈ [0, 1] . − sin(2πr)cos(2πr) Analogously, 0 T −1 L(E , τ)(C) ={A ∈ M (C)|A J A = J ,A J A =−J } 2 2 2 2 2 ! " ic ib 2 2 = | c, b ∈ C,c + b = 1 . ib −ic Thus, a maximal compact subgroup of L(E , τ)(C)is 1 ii 0 0 ST(E , τ) = ST(E , 1) · √ . (3.15) i −i There is a relation between the algebraic Sato–Tate groups AST(C)and AST(C ) attached to Jac(C)and Jac(C ), respectively, given by [12, Lemma 2.3]. If we put H := λ (Gal(K /kM)), this relation implies that 0 φ 0 0 −1 AST(C) = L(E , 1) · ι(H ) ∪ L(E , τ) · ι((1, τ) (H \ H )). 3 0 3 0 Then, (3.15)implies 0 0 ST(C) = ST(E , 1) ι(H ) ∪ ι(H \ H ) = ST(E , 1) · ι(H). (3.16) 3 0 0 3 Note that, even if ι(H) is only deﬁned as an element of USp(6)/ −1 , the product 0 0 ST(E , 1) · ι(H) is well deﬁned inside USp(6), provided that −1 ∈ ST(E , 1) . 3 3 0 0 For the case C = C ,wehave 0 T −1 L(E , 1)(C) ={A ∈ M (C)|A J A = J ,A K A = K } 2 2 2 2 2 ! " c − b 4b 2 2 = | c, b ∈ C,c + 7b = 1 . −2bc + b 41 Page 20 of 40 Fité et al. Res Math Sci (2018) 5:41 Thus, a maximal compact subgroup of L(E , 1)(C)is ! " 1 4 √ √ cos(2πr) − sin(2πr) sin(2πr) 7 7 ST(E , 1) = | r ∈ [0, 1] . 2 1 √ √ − sin(2πr)cos(2πr) + sin(2πr) 7 7 Analogously, 0 T −1 ST(E , τ) ={A ∈ M (C)|A J A = J ,A K A =−I − K } 2 2 2 2 2 2 ! " ic − ib 4ib 2 2 = | c, b ∈ C,c + 7b = 1 . ic 3ib + ib − ic 2 2 Thus, a maximal compact subgroup of L(E , τ)(C)is i −i 0 0 ST(E , τ) = ST(E , 1) · . 0 −i 0 0 We can now apply [12, Lemma 2.3] exactly as in the case C = C to complete the proof. The previous theorem describes the Sato–Tate group of a twist C of C . Now suppose that C and C are both twists of C . The next proposition gives an eﬀective criterion to determine when ST(C) and ST(C )coincide. Let H ⊆ G 0 (resp. H ) be attached to C (resp. C )asinRemark 3.10. Proposition 3.22 If H and H are conjugate in G 0,then ST(C) and ST(C ) coincide. Proof Since the Sato–Tate group is deﬁned only up to conjugacy, is suﬃces to exhibit −1 −1 A ∈ GL (C) such that A ST(C)A = ST(C ). Let g ∈ G 0 be such that H = g Hg.It is straightforward to check that ι(G 0) normalizes the group ST(E , 1) . In particular, by Theorem 3.21 (ii),wehave 0 −1 0 −1 ST(C ) = ST(E , 1) ι(H ) = ι(g) ST(E , 1) ι(H)ι(g) = ι(g) ST(C)ι(g). (3.17) 3 3 Corollary 3.23 There are at most 23 Sato–Tate groups of twists of the Klein quartic C . Proof There are 23 subgroups of G , up to conjugacy. In the Fermat case, ST(C) and ST(C ) may coincide when H and H are not conjugate in G . We thus require a sharper criterion. 0 0 0 0 Deﬁnition 3.24 Let C and C be twists of C and C , respectively (here C and C both 0 0 denote one of C or C , but possibly not the same curve in both cases), and let H and H 1 7 be the corresponding attached groups. We say that H and H are equivalent if there exists an isomorphism : H → H (3.18) such that (H ) = H and for every h ∈ H ,wehave 0 0 Tr(j(h)) = Tr(j((h))), (3.19) Fité et al. Res Math Sci (2018) 5:41 Page 21 of 40 41 where H and H are deﬁned as in Deﬁnition 3.9 and j denotes compositions ι ◦ ι of the 0 2 1 0 0 0 0 embeddings deﬁned in Lemma 3.18 for C and C (two diﬀerent maps j if C = C ). 0 0 Proposition 3.25 Let C and C be twists of C and C .IfH andH are equivalent, then ST(C) and ST(C ) coincide. 0 0 Proof Letusﬁrstassume that C = C .ByTheorem 3.21 (ii), we can consider the group isomorphism 0 0 : ST(C) = ST(E , 1) · ι(H) ST(C ) = ST(E , 1) · ι(H ) 3 3 deﬁned by sending an element of the form g = g ι(h)to g ι((h)). We aim to show that 0 0 ST(C) and ST(C ) are conjugate inside GL (C). This amounts to showing that ST(C)and ST(C ) are equivalent representations of the same abstract group, for which it suﬃces to prove the following claim: for every g ∈ ST(C), we have Tr(g) = Tr((g)). To prove the claim distinguish the cases: (a) h ∈ H , and (b) h ∈ H \ H . 0 0 Suppose we are in case (a). By (3.19), there exists A ∈ GL (M) such that we have −1 Aj(h)A = j((h)) for every h ∈ H . Moreover, if we let r denote the composition ι ◦ ι 0 4 3 of the embeddings deﬁned in Lemma 3.18,the fact that A hasentriesin M easily implies that r(A) centralizes ST(E , 1), and thus we have −1 −1 −1 (g) = g r(A)ι(h)r(A) = r(A)g ι(h)r(A) = r(A)gr(A) , 0 0 from which the claim follows. In case (b), we have that both g and (g) have trace 0, as follows for example from the proof of Corollary 2.4 and the Chebotarev Density Theorem. The claim follows immediately. 0 0 0 If C = C , then we may assume without loss of generality that C is a twist of C and C is a twist of C . Now consider the isomorphism 0 0 : ST(C) = ST(E , 1) · ι(H) ST(C ) = ST(E , 1) · ι(H ) 3 3 7 1 −1 deﬁned by sending an element of the form g = g ι(h)to Tg T ι((h)), where 0 0 T = √ √ . −1/ 74/ 7 0 −1 0 We now note that TST(E , 1)T = ST(E , 1), and the proof then proceeds exactly as 1 7 0 0 above; the hypothesis C = C implies that we have √ √ Tr(j(h)) = Tr(j((h))) ∈ Q( −1) ∩ Q( −7) = Q for every h ∈ H , and thus the matrix A from above can be taken in GL (Q). 0 3 Corollary 3.26 The following hold: (i) There are at most 54 distinct Sato–Tate groups of twists of the Fermat quartic. (ii) There are at most 60 distinct Sato–Tate groups of twists of the Fermat and Klein quartics. Proof Determining whether two subgroups H and H are equivalent is a ﬁnite problem. Using the computer algebra program [6], one can determine a set of representatives for equivalence classes of subgroups H that turn out to have size 54 in case (i), and of size 60 in case (ii). For the beneﬁt of the reader, here we give a direct proof of (ii), assuming (i). 41 Page 22 of 40 Fité et al. Res Math Sci (2018) 5:41 The 6 Sato–Tate groups of a twist of the Klein quartic that do not show up as the Sato– Tate group of a twist of the Fermat quartic are precisely those ruled out by the fact that H contains an element of order 7 (those in rows #10, #13, #14 of Table 5), since 7 does not divide #G 0. To show that the other 17 Sato–Tate groups of twists of the Klein quartic also arise for twists of the Fermat quartic, we proceed as follows. Let H ⊆ G 0 correspond to a twist C 0 0 of C such that H does not contain an element of order 7, and let H ⊆ C correspond 7 1 to a twist C of C .Inthiscase, from Table 3a, b, to ensure that H and H are equivalent it suﬃces to check that: (1) There exists an isomorphism : H → H such that (H ) = H ; (2) Tr(j(h)) = 1 for every h ∈ H such that ord(h) = 4. From Tables 4 and 5, it is trivial to check that for every H as above, one can always ﬁnd a subgroup H such that condition (1) is satisﬁed. Condition (2) is vacuous except for rows #6, #7, #11, and #12 of Table 5. In these cases, a subgroup H for which condition (2) is 3 3 also satisﬁed can be found by noting that both j(t )and j(t u t u ) have trace 1. More 1 1 1 1 1 precisely, one ﬁnds that the Sato–Tate groups corresponding to these cases coincide with the Sato–Tate groups of rows #13, #20, #34, and #55 of Table 4, respectively. Combining the lower and upper bounds proved in this section yields our main theorem, which we restate for convenience. Theorem 1 The following hold: (i) There are 54 distinct Sato–Tate groups of twists of the Fermat quartic. These give rise to 54 (resp. 48) distinct joint (resp. independent) coeﬃcient measures. (ii) There are 23 distinct Sato–Tate groups of twists of the Klein quartic. These give rise to 23 (resp. 22) distinct joint (resp. independent) coeﬃcient measures. (iii) There are 60 distinct Sato–Tate groups of twists of the Fermat or the Klein quartics. These give rise to 60 (resp. 54) distinct joint (resp. independent) coeﬃcient measures. Proof This follows immediately from Corollaries 3.12, 3.23, 3.26, and Proposition 3.13. Corollary 3.27 If C and C are twists of C corresponding to H and H , respectively, then ST(C) and ST(C ) coincide if and only if H and H are equivalent. Remark 3.28 One could have obtained the lower bounds of Proposition 3.13 by com- puting the joint coeﬃcient measures μ of the Sato–Tate groups explicitly described in Theorem 3.21. This is a lengthy but feasible task that we will not inﬂict on the reader. We note that this procedure also allows for case-by-case veriﬁcations of the equalities M [μ ] = M [a] and thus of the Sato–Tate conjecture in the cases considered. n ,n ,n I n ,n ,n 1 2 3 1 2 3 Remark 3.29 Let X denote the set of Sato–Tate groups of twists of C .Theorem 3.21 gives a map from the set of subgroups of G 0 to X that assigns to a subgroup H ⊆ G 0 d C the Sato–Tate group ST(E , 1) · ι(H). It also shows that X is endowed with a lattice 3 d structure compatible with this map and the lattice structure on the set of subgroups of G . Moreover, Proposition 3.22 says that this map factors via ε : C → X , d d d Fité et al. Res Math Sci (2018) 5:41 Page 23 of 40 41 where C denotes the lattice of subgroups of G up to conjugation. Parts (i) and (ii) of Theorem 1 imply that while the map ε is a lattice isomorphism, the map ε is far from 7 1 being injective. Corollary 3.27 can now be reformulated by saying that two subgroups H, H ∈ C lie in the same ﬁber of ε if and only if they are equivalent. 1 1 In virtue of the above remark, one might ask about conditions on twists C and C corresponding to distinct but equivalent groups H and H that ensure their Jacobians have the same Sato–Tate group. One such condition is that Jac(C)and Jac(C ) are isogenous (recall that the Sato–Tate group of an abelian variety is an isogeny invariant). The next proposition shows that, under the additional hypothesis that K and K coincide, the previous statement admits a converse. Proposition 3.30 Let C and C be k-twists of C . Suppose that the corresponding sub- groups H and H of G 0 are equivalent and that the corresponding ﬁelds K and K coincide. Then, Jac(C) and Jac(C ) are isogenous. Proof Let S be the set of primes of k which are of bad reduction for either Jac(C)orJac(C ) or lieover theﬁxedprime .Notethatby[27, Thm. 4.1] the set S contains the primes of k ramiﬁed in K or K . By Faltings’ Isogeny Theorem [8, Korollar 2], it suﬃces to show that for every p ∈ / S,wehave L (Jac(C),T) = L (Jac(C ),T). (3.20) p p If Frob ∈ / G , by the proof of Corollary 2.4, both polynomials of (3.20)havethe same p kM expression, which depends only on the order of (the projection of) Frob in Gal(K /kM). To obtain (3.20) for those p such that Frob ∈ G , we will show that V (Jac(C) ) p kM kM and V (Jac(C ) ) are isomorphic as Q [G ]-modules. Indeed, the fact that H and H kM kM are equivalent pairs implies that the restrictions from Aut(C )to H and H of the M 0 representations θ attached to C and C are equivalent. Together with (3.5), this shows 0 0 E ,C 0 0 that θ (E , Jac(C)) and θ (E , Jac(C )) are equivalent representations. The desired M,σ M,σ Q [G ]-module isomorphism follows now from (2.2). kM Remark 3.31 Let H and H be any two equivalent pairs attached to twists C and C of the same curve C . As one can read from Tables 4 and 5 (and as we will see in the next section), one can choose C and C so that K and K coincide. It follows from Proposition 3.30 that on Table 4 (resp. Table 5) two curves C and C satisfy ST(C) = ST(C )ifand only if Jac(C) ∼ Jac(C ). We conclude this section with an observation that is not directly relevant to our results but illustrates a curious phenomenon arising among the twists of the Fermat quartic in Table 4.Let 4 4 4 4 4 4 C :9x + 9y − 4z =0and C :9x − 4y + z = 0 5 8 be the curves listed in rows #5 and #8 of Table 4. As can be seen in Table 4, the groups ST(C ) and ST(C ) coincide, as do the respective ﬁelds K.Proposition 3.30 thus implies 5 8 that the Jacobians of C and C are isogenous, but in fact more is true. 5 8 Proposition 3.32 The curves C and C are not isomorphic (over Q), but their reductions 5 8 ˜ ˜ C and C modulo p are isomorphic (over F ) for every prime p > 3. 5 8 p 41 Page 24 of 40 Fité et al. Res Math Sci (2018) 5:41 Proof The twists C are C of C are not isomorphic because they arise from non- 5 8 ˜ ˜ conjugate subgroups H of G . For the reductions C and C , we ﬁrst consider the case 5 8 p ≡ 1 (mod 4). We claim that −4 is a fourth power modulo p; this follows from the factorization 4 2 2 x + 4 = (x − 2x + 2)(x + 2x + 2) in Q[x] 2 2 together with the fact that x − 2x + 2and x + 2x + 2 have discriminant −4. It follows 4 4 4 ˜ ˜ that C and C are both isomorphic to 9x + 9y + z = 0 (over F ). 5 8 p Suppose now that p ≡−1 (mod 4). Then, 9 is a fourth power modulo p since −3 3 4 2 2 x − 9 = (x − 3)(x + 3) and =− . p p 4 4 4 ˜ ˜ It follows that C and C are both isomorphic to x + y − 4z = 0 (over F ). 5 8 p 3.4 Curve equations In this section, we construct explicit twists of the Fermat and the Klein quartics realizing each of the subgroups H ⊆ G 0 described in Remark 3.10. Recall that each H has an associated subgroup H := H ∩ G of index at most 2 (see Deﬁnition 3.9), and there exists 0 0 a twist corresponding to H with k = kM if and only if H = H , where, as always, M 0 0 0 3 denotes the CM ﬁeld of E (the elliptic curve for which Jac(C ) ∼ (E ) ). Equations for these twists are listed in Tables 4 and 5 in Sect. 4. As explained in Remark 3.10, in the Fermat case every subgroup H ⊆ G 0 with [H : H ] = 1(case (c ) of Deﬁnition 3.9) arises as H for some subgroup H ⊆ G 0 for which [H : H ] = 2 0 C 0 (case (c ) of Deﬁnition 3.9), and a twist corresponding to H canthusbeobtainedasthe base change to kM of a twist corresponding to H . We thus only list twists for the 59 subgroups H in case (c ), since base changes of these twists to kM then address the 24 subgroups H in case (c ). In the Klein case, we list twists for the 11 subgroups H in case (c ) and also the 3 exceptional subgroups H in case (c ) that cannot be obtained as base 2 1 changes of twists corresponding to subgroups in case (c ); see Remark 3.10. Our twists are all deﬁned over base ﬁelds k of minimal possible degree, never exceeding 2. For the 3 exceptional subgroups H in the Klein case noted above, we must have k = kM, and we use k = M = Q( −7). In all but 5 of the remaining cases with [H : H ] = 2, we use k = Q. These 5 exceptions are all explained by Lemma 3.33 below (the second of the 4 pairs listed in Lemma 3.33 arises in both the Fermat and Klein cases, leading to 5 exceptions in total). In each of these 5 exceptions with [H : H ] = 2, the subgroup H also 0 0 arises as H for some subgroup H ⊆ G 0 with [H : H ] = 2 that is realized by a twist 0 C 0 with k = Q, allowing H to be realized over a quadratic ﬁeld as the base change to M of a twist deﬁned over Q. Lemma 3.33 Twists of the Fermat or Klein quartics corresponding to pairs (H, H ) with the following pairs of GAP identiﬁers cannot be deﬁned over a totally real ﬁeld: ( 4, 1 , 2, 1 ), ( 8, 1 , 4, 1 ), ( 8, 4 , 4, 1 ), ( 16, 6 , (8, 2 ). Proof If k is totally real, then complex conjugation acts trivially on k but not on kM, giving an involution in H = Gal(K /k) with non-trivial image in H /H = Gal(K /k)/Gal(K /kM). For the four pairs (H, H ) listed in the lemma, no such involution exists. 0 Fité et al. Res Math Sci (2018) 5:41 Page 25 of 40 41 In addition to listing equations and a ﬁeld of deﬁnition k for a twist C associated with each subgroup H,inTables 4 and 5 we also list the minimal ﬁeld K over which all the endomorphisms of Jac(C) are deﬁned, and we identify the conjugacy class of ST(C) and ST(C ), which depends only on H, not the particular choice of C.Asnoted in kM Remark 3.31, we have chosen twists C so that twists with the same Sato–Tate group have thesameﬁelds K and thus have isogenous Jacobians, by Proposition 3.30 (thereby demonstrating that the hypotheses of the proposition can always be satisﬁed). 3.4.1 Constructing the Fermat twists The twists of the Fermat curve over any number ﬁeld are parametrized in [23], and we specialize the parameters in Theorems 4.1, 4.2, 4.5 of [23] to obtain the desired examples. In every case, we are able to obtain equations with coeﬃcients in Q, but as explained above, we cannot always take k = Q; the exceptions can be found in rows #4, #13, #27, #33 of Table 4. The parameterizations in [23] also allow us to determine the ﬁeld L over which all the isomorphisms to (C ) are deﬁned, which by Lemma 3.3 and Proposition 3.7, this is the same as the ﬁeld K over which all the endomorphisms of Jac(C ) are deﬁned. 7 Q Specializing the parameters for each of the 59 cases with k = kM involves a lot of easy but tedious computations. The resulting equations are typically not particularly pleasing to the eye or easy to format in a table; in order to make them more presentable, we used the algorithm in [30] to simplify the equations. To give just one example, for the unique subgroup H with ID(H) = 24, 13 , the equation we obtain from specializing the parameterizations in [23]is 4 3 3 2 2 2 2 2 3 2 14x − 84x y + 392x z + 588x y − 2940x yz + 4998x z − 980xy + 9996xy z 2 3 4 3 2 2 − 28812xyz + 30184xz + 833y − 9604y z + 45276y z 3 4 − 90552yz + 69629z = 0, but the equation listed for this curve in row #48 of Table 4 is 4 3 3 2 2 2 2 3 2 4 3 2 2 4 3x +4x y+4x z +6x y +6x z +8xy +12xyz + 5y + 4y z + 12y z + z = 0. We used of the number ﬁeld functionality in [6]and [34] to minimize the presentation of the ﬁelds K listed in the tables (in particular, the function polredabs in PARI/GP). 3.4.2 Constructing the Klein twists Twists of the Klein curve over arbitrary number ﬁelds are parametrized in Theorems 6.1 and 6.8 of [23], following the method described in [22], which is based on the resolution of certain Galois embedding problems. However, in the most diﬃcult case, in which H = G 0 has order 336, this Galois embedding problem is computationally diﬃcult to resolve explicitly. This led us to pursue an alternative approach that exploits the moduli interpretation of twists of the Klein curve as twists of the modular curve X(7). As described in [14,§3] and[25, §4], associated with each elliptic curve E/Q is a twist X (7) of the Klein quartic deﬁned over Q that parameterizes isomorphism classes of 7-torsion Galois modules isomorphic to E[7], as we recall below. With this approach, we can easily treat the case H = G , and we often obtain twists with nicer equations. In one case, we also obtain a better ﬁeld of deﬁnition k, allowing us to achieve the minimal possible degree [k : Q] in every case. 41 Page 26 of 40 Fité et al. Res Math Sci (2018) 5:41 However, as noted in [25, §4.5], not every twist of the Klein curve can be written as X (7) for some elliptic curve E/Q, and there are several subgroups H ⊆ G for which the parameterizations in [23] yield a twist of the Klein quartic deﬁned over Q, but no twists of the form X (7) corresponding to H exist. We are thus forced to use a combination of the two approaches. For twists of the form X (7), we need to determine the minimal ﬁeld over which the endomorphisms of Jac(X (7)) are deﬁned; this is addressed by Propositions 3.34 and 3.35. Let E/Q be an elliptic curve and let E[7] denote the F [G ]-module of Q-valued points 7 Q of the kernel of the multiplication-by-7 map [7]: E → E. The Weil pairing gives a G - equivariant isomorphism E[7] μ , where μ denotes the F [G ]-module of 7th 7 7 7 Q roots of unity. Let Y (7) be the curve deﬁned over Q described in [14,§3] and[25, §4]. For any ﬁeld extension L/Q,the L-valued points of Y (7) parametrize isomorphism classes of pairs (E , φ), where E /L is an elliptic curve and φ : E[7] → E [7] is a symplec- tic isomorphism. By a symplectic isomorphism, we mean a G -equivariant isomorphism φ : E[7] → E [7] such that the diagram E[7] µ φ id E [7] µ , (3.21) ˜ ˜ commutes, where the horizontal arrows are Weil pairings. Two pairs (E , φ)and (E , φ) are isomorphic whenever there exists an isomorphism ε : E → E such that φ = ε ◦ φ. In [14] it is shown that X (7), the compactiﬁcation of Y (7), is a twist of C ,and an E E explicit model for X (7) is given by [14, Thm. 2.1], which states that if E has the Weierstrass 2 3 model y = x + ax + b with a, b ∈ Q, then 4 3 2 2 2 2 2 2 3 3 2 2 2 3 2 4 ax +7bx z+3x y −3a x z −6bxyz −5abxz +2y z+3ay z +2a yz −4b z = 0 (3.22) is a model for X (7) deﬁned over Q. We will use the moduli interpretation of X (7) to determine the minimal ﬁeld over which all of its automorphisms are deﬁned. Recall that the action of G on E[7] gives rise to a Galois representation : G → Aut(E[7]) GL (F ). E,7 Q 2 7 Let denote the composition π ◦ , where π:GL (F ) → PGL (F ) is the natural E,7 2 7 2 7 E,7 projection. Proposition 3.34 The following ﬁeld extensions of Q coincide: (i) The minimal extension over which all endomorphisms of Jac(X (7)) are deﬁned; (ii) The minimal extension over which all automorphisms of X (7) are deﬁned; ker E,7 (iii) The ﬁeld Q ; (iv) The minimal extension over which all 7-isogenies of E are deﬁned. In particular, if K is the ﬁeld determined by these equivalent conditions, then Gal(K /Q) Im( ) Im( )/(Im ∩ F ). E,7 E,7 E,7 7 Fité et al. Res Math Sci (2018) 5:41 Page 27 of 40 41 Proof The equivalence of (i) and (ii) follows from 3.3,since X (7)isatwistof C . Following [25], let Aut (E[7]) denote the group of symplectic automorphisms of E[7]. Given a ﬁeld extension F /Q, let us write E[7] for the F [G ]-module obtained from E[7] F 7 F by restriction from G to G . Note that E[7] (Z/7Z) . Under this isomorphism, for Q F any g ∈ Aut (E[7] ), diagram (3.21) becomes (Z/7Z) Z/7Z det(g) id (Z/7Z) Z/7Z , from which we deduce Aut (E[7] ) SL (F ). (3.23) ∧ 2 7 −1 Each g ∈ Aut (E[7] )actson Y (7) via (E , φ) → (E , φ ◦ g ). This action extends to ∧ E Q Q X (7), from which we obtain a homomorphism Aut (E[7] ) → Aut(X (7) ). (3.24) ∧ E Q Q This homomorphism is non-trivial, since elements of its kernel induce automorphisms of E ,but Aut(E ) is abelian and Aut (E[7]) SL (F ) is not, and it cannot be injective, ∧ 2 7 Q Q Q since the group on the right has cardinality 168 < #SL (F ) = 336. The only non-trivial 2 7 proper normal subgroup of SL (F )is ±1 , and thus (3.24) induces a G -equivariant 2 7 Q isomorphism Aut (E[7])/ ±1 → Aut(X (7) ). (3.25) ∧ E By transport of structure, we now endow SL (F )witha G -module structure that 2 7 Q turns (3.23) into a G -equivariant isomorphism ϕ:Aut (E[7] ) → SL (F ). Now deﬁne Q ∧ 2 7 : G → Aut(SL (F )) to be the representation associated with this G -module struc- Q 2 7 Q ture. Since the action of σ ∈ G on each g ∈ Aut (E[7] )isdeﬁnedby Q ∧ −1 σ σ σ ( g)(P) = (g( P)), −1 for P ∈ E[7], we have (σ )(ϕ(g)) = (σ ) · ϕ(g) · (σ ) . The G -action is trivial on E,7 E,7 Q ±1 and thus descends to PSL (F ). Let us write PSL (F ) for PSL (F ) endowed with 2 7 2 7 2 7 the G -action given by conjugation by . Then, (3.23)with (3.25)yielda G -equivariant Q E,7 Q isomorphism PSL (F ) Aut(X (7) ). 2 7 E Ker() This implies that the ﬁeld described in (ii) coincides with Q , and we now note that −1 ker() ={σ ∈ G | (σ ) · α · (σ ) = α, for all α ∈ PSL (F )} Q E,7 E,7 2 7 ={σ ∈ G | (σ ) ∈ F } Q E,7 = ker( ), E,7 ker() ker( ) E,7 thus Q = Q is the ﬁeld described in (iii). 41 Page 28 of 40 Fité et al. Res Math Sci (2018) 5:41 Now let P(E[7]) denote the projective space over E[7], consisting of its 8 linear F - subspaces, equivalently, its 8 cyclic subgroups of order 7. The G -action on P(E[7]) gives rise to the projective Galois representation : G → Aut(P(E[7])) PGL (F ). Q 2 7 E,7 The minimal ﬁeld extension K over which the G -action on P(E[7]) becomes trivial is precisely the minimal ﬁeld over which the cyclic subgroups of E[7] of order 7 all become Galois stable, equivalently, the minimal ﬁeld over which all the 7-isogenies of E are deﬁned. It follows that the ﬁxed ﬁeld of ker identiﬁed in (iii) is also the ﬁeld described in (iv). E,7 To explicitly determine the ﬁeld over which all the 7-isogenies of E are deﬁned, we rely on Proposition 3.35 below, in which (X, Y ) ∈ Z[X, Y ] denotes the classical modular polynomial; the equation for (X, Y ) is too large to print here, but it is available in [6] and can be found in the tables of modular polynomials listed in [33] that were computed via [5]; it is a symmetric in X and Y , and has degree 8 in both variables. The equation (X, Y ) = 0 is a canonical (singular) model for the modular curve Y (7) 7 0 that parameterizes 7-isogenies. If E and E are elliptic curves related by a 7-isogeny 1 2 then (j(E ),j(E )) = 0, and if j ,j ∈ F satisfy (j ,j ) = 0, then there exist elliptic 7 1 2 1 2 7 1 2 curves E and E with j(E ) = j and j(E ) = j that are related by a 7-isogeny. However, 1 2 1 1 2 2 this 7-isogeny need not be deﬁned over F! The following proposition characterizes the relationship between F and the minimal ﬁeld K over which all the 7-isogenies of E are deﬁned. Proposition 3.35 Let E be an elliptic curve over a number ﬁeld k with j(E) = 0, 1728.Let F be the splitting ﬁeld of (j(E),Y ) ∈ k[X], and let K be the minimal ﬁeld over which all the 7-isogenies of E are deﬁned. The ﬁelds K and F coincide. Proof Let S be the multiset of roots of (j(E),Y)in Q,viewedasa G -set in which the 7 k action of σ ∈ G preserves multiplicities: we have m(σ (r)) = m(r) for all σ ∈ G , where k k m(r) denotes the multiplicity of r in S.Let P(E[7]) be the G -set of cyclic subgroups P of E[7] of order 7. In characteristic zero every isogeny is separable, hence determined by its kernel up to composition with automorphisms; this yields a surjective morphism of −1 G -sets ϕ : P(E[7]) → S deﬁned by P → j(E/ P )with m(r) = #ϕ (r) for all r ∈ S (note #P(E[7]) = 8 = m(r)). The G -action on S factors through the G -action on k k r∈S P(E[7]), and we thus have group homomorphisms ¯ φ E,7 G −→ Aut(P(E[7]) −→ Aut(S), where φ:¯ (G ) → Aut(S)isdeﬁnedby φ(σ )(ϕ( P ):= ϕ(σ ( P )) for each σ ∈ ¯ (G ). E,7 E,7 k k ker ¯ ker(φ◦¯ ) E,7 E,7 We then have K = Q and F = Q ,so F ⊆ K . If E does not have complex multiplication, then m(r) = 1 for all r ∈ S, since otherwise over Q we would have two 7-isogenies α, β : E → E with distinct kernels, and then (ˆ α ◦ β) ∈ End(E ) is an endomorphism of degree 49 which is not ±7, contradicting End(E ) Z. It follows that ϕ and therefore φ is injective, so ker ¯ = ker(φ ◦ ¯ )and E,7 E,7 K = F. If E does have complex multiplication, then End(E ) is isomorphic to an order in an imaginary quadratic ﬁeld M. We now consider the isogeny graph whose vertices are j- invariants of elliptic curves E /FM with edges (j ,j ) present with multiplicity equal to 1 2 Fité et al. Res Math Sci (2018) 5:41 Page 29 of 40 41 the multiplicity of j as a root of (j ,Y ). Since j(E) = 0, 1728, the component of j(E ) 2 7 1 FM in this graph is an isogeny volcano, as deﬁned in [31]. In particular, there are at least 6 distinct edges (j(E ),j ) (edges with multiplicity greater than 1 can occur only at the FM 2 surface of an isogeny volcano and the subgraph on the surface is regular of degree at most 2). It follows that m(r) > 1 for at most one r ∈ S. The image of ¯ is isomorphic to a subgroup of PGL (F ), and this implies that if E,7 2 7 ¯ (σ ) ﬁxes more than 2 elements of P(E[7]) then σ ∈ ker ¯ . This necessarily applies E,7 E,7 whenever ¯ (σ)liesinker φ, since it must ﬁx 6 elements, thus ker ¯ = ker(φ ◦ ¯ ) E,7 E,7 E,7 and K = F. Corollary 3.36 Let E be an elliptic curve over a number ﬁeld k with j(E) = 0, 1728.The minimal ﬁeld K over which all the 7-isogenies of E are deﬁned depends only on j(E). Remark 3.37 The ﬁrst part of the proof of Proposition 3.35 also applies when j(E)is0 or 1728, thus we always have F ⊆ K . Equality does not hold in general, but a direct computationﬁndsthat[K :F] must divide 6 (resp. 2) when j(E) = 0 (resp. 1728), and this occurs when k = Q. 2 3 We now ﬁx E as the elliptic curve y = x + 6x + 7 with Cremona label 144b1.Note that is surjective; this can be seen in the entry for this curve in the L-functions and E,7 Modular Forms Database [35] and was determined by the algorithm in [32]. It follows that Gal(Q(E[7])/Q) GL (F ), and Proposition 3.34 implies that we then have H := 2 7 Gal(K /Q) PGL (F ). 2 7 For our chosen curve E,wehave a = 6and b = 7. Plugging these values into equation (3.22) and applying the algorithm of [30] to simplify the result yields the curve listed in entry #14 of Table 5 for H = G 0. To determine the ﬁeld K , we apply Proposition 3.35. Plugging j(E) = 48384 into (x, j(E)) and using PARI/GP to simplify the resulting poly- 8 7 4 nomial, we ﬁnd that K is the splitting ﬁeld of the polynomial x + 4x + 21x + 18x + 9. We applied the same procedure to obtain the equations for C and the polynomials deﬁning K that are listed in rows #4, #6, #9, #10 of Table 5 using the elliptic curves E with Cremona labels 2450ba1, 64a4, 784h1, 36a1, respectively, with appropriate adjustments for the cases with j(E) = 0, 1728 as indicated in Proposition 3.35. The curve in row #11 is a base change of the curve in row #6, and for the remaining 7 curves we used the parameterizations in [23]. 3.5 Numerical computations In the previous sections, we have described the explicit computation of several quantities 0 0 0 related to twists C of C = C , where C is our ﬁxed model over Q for the Fermat quartic d d 0 0 3 0 0 (d = 1) or the Klein quartic (d = 7), with Jac(C ) ∼ (E ) , where E = E is an elliptic curve over Q with CM by M = Q( −d)deﬁnedin (3.1). These include: • Explicit equations for twists C of C corresponding to subgroups H of G 0; • Deﬁning polynomials for the minimal ﬁeld K for which End(Jac(C) ) = End(Jac(C) ); • Independent and joint coeﬃcient moments of the Sato–Tate groups ST(Jac(C)). These computations are numerous and lengthy, leaving many opportunities for errors, both by human and by machine. We performed several numerical tests to verify our computations. 41 Page 30 of 40 Fité et al. Res Math Sci (2018) 5:41 3.5.1 Naïve point-counting A simple but eﬀective way to test the compatibility of a twist C/k and endomorphism ﬁeld K is to verify that for the ﬁrst several degree one primes p of k of good reduction for C that split completely in K , the reduction of Jac(C) modulo p is isogenous to the cube of the reduction of E modulo the prime p := N(p). By a theorem of Tate, it suﬃces to check that 0 3 L (Jac(C),T) = L (E ,T) . For this task, we used optimized brute force point-counting p p methods adapted from [20,§3].The L-polynomial L (Jac(C),T) is the numerator of the zeta function of C, a genus 3 curve, so it suﬃces to compute #C(F ), #C(F 2), #C(F 3), p p reducing the problem to counting points on smooth plane quartics and elliptic curves over ﬁnite ﬁelds. To count projective points (x : y : z) on a smooth plane quartic f (x, y, z) = 0 over a ﬁnite ﬁeld F , one ﬁrst counts aﬃne points (x : y : 1) by iterating over a ∈ F , computing the q q number r of distinct F -rational roots of g (x):= f (x, a, 1) via r = deg(gcd(x − x, g (x))), q a a and then determining the multiplicity of each rational root by determining the least n ≥ 1 (n) (n) for which gcd(g (x),g (x)) = 1, where g denotes the nth derivative of g ∈ F [x]; a a a a q q q note that to compute gcd(x − x, g (x)) one ﬁrst computes x mod g (x) using a square- a a and-multiply algorithm. Having counted aﬃne points (x : y : 1), one then counts the F -rational roots of f (x, 1, 0) and ﬁnally checks whether (1 : 0 : 0) is a point on the curve. To optimize this procedure, one ﬁrst seeks a linear transformation of f (x, y, z) that ensures g (x) = f (x, a, 1) has degree at most 3 for all a ∈ F ; for this, it suﬃces to translate a q a rational point to (1 : 0 : 0), which is always possible for q ≥ 37 (by the Weil bounds). 3 2+o(1) This yields an O(p (log p) )-time algorithm to compute L (Jac(C),T)thatisquite practical for p up to 2 , enough to ﬁnd several (possibly hundreds) of degree one primes p of k that split completely in K . 0 3 Having computed L (Jac(C),T), one compares this to L (E ,T) ; note that the poly- p p 0 2 0 nomial L (E ,T) = pT − a T + 1 is easily computed via a = p + 1 − #E (F ). If this p p p p comparison fails, then either C is not a twist of C , or not all of the endomorphisms of Jac(C) are deﬁned over K . The converse is of course false, but if this comparison succeeds for many degree-1 primes p it gives one a high degree of conﬁdence in the computations of C and K . Note that this test will succeed even when K is not minimal, but we also check that H Gal(K /k), which means that so long as C isatwist of C corresponding to the subgroup H ⊆ G , the ﬁeld K must be minimal. 3.5.2 An average polynomial-time algorithm In order to numerically test our computations of the Sato–Tate groups ST(C), and to verify our computation of the coeﬃcient moments, we also computed Sato–Tate statistics for all of our twists C/k of the Fermat and Klein quartics. This requires computing the L- polynomials L (Jac(C),T) at primes p of good reduction for C up to some bound N,and it suﬃces to consider only primes p of prime norm p = N(p), since nearly all the primes of norm less than N are degree-1 primes. In order to get statistics that are close to the values predicted by the Sato–Tate group one needs N to be fairly large. We used N = 2 , 3 2+o(1) which is far too large for the naive O(p (log p) )-time algorithm described above to be practical, even for a single prime p ≈ N, let alone all good p ≤ N. The objective of this test is not to prove anything, it is simply a mechanism for catching mistakes, of which we found several; most were our own, but some were due to minor errors in the literature, and at least one was caused by a defect in one of the computer algebra systems we used. Fité et al. Res Math Sci (2018) 5:41 Page 31 of 40 41 In [18], Harvey and Sutherland give an average polynomial-time algorithm to count points on smooth plane quartics over Q that allows one to compute L (Jac(C),T) mod p 3+o(1) for all good primes p ≤ N in time O(N(log N) ), which represents an average cost of 4+o(1) O((log p) ) per prime p ≤ N. This is achieved by computing the Hasse–Witt matrices of the reductions of C modulo p using a generalization of the approach given in [16,17] 1+o(1) for hyperelliptic curves. In [18], they also give an O( p(log p) )-time algorithm to compute L (Jac(C),T) mod p for a single good prime p, which allows one to handle reductions of smooth plane quartics C deﬁned over number ﬁelds at degree one primes; 3/2+o(1) this increases the total running time for p ≤ N to O(N ), which is still feasible with N = 2 . Having computed L (Jac(C),T) mod p, we need to lift this polynomial for (Z/pZ)[T]to Z[T], which is facilitated by Proposition 3.38 below. It follows from the Weil bounds that the linear coeﬃcient of L (Jac(C),T) is an integer of absolute value at most 6 p.For p > 144, the value of this integer is uniquely determined by its value modulo p, and for p < 144 we can apply the naive approach described above. This uniquely determines the value in the column labeled F (x)inTables 1 and 2 of Proposition 3.38, which then determines the values in columns F (x)and F (s), allowing the integer polynomial L (Jac(C),T) ∈ Z[T] 2 3 p to be completely determined. Proposition 3.38 For τ in Gal(L/k), let s = s(τ) and t = t(τ) denote the orders of τ and the projection of τ on Gal(kM/k), respectively. For C a twist of C , the following hold: 0 0 (i) The pair (s, t) is one of the 9 pairs listed on Table 1 if C = C ,and oneofthe 9 pairs 0 0 listed on Table 2 if C = C . (ii) For each pair (s, t),let F := F × F × F :[−2, 2] → [−6, 6] × [−1, 15] × [−20, 20] ⊆ R (s,t) 1 2 3 1 7 be the map deﬁned in Table 1 if C = C or Table 2 if C = C . For every prime p 0 0 0 0 unramiﬁed in K of good reduction for both Jac(C) and E ,wehave F (a (E )(p)) = a (Jac(C))(p),a (Jac(C))(p),a (Jac(C))(p) , (3.26) 1 1 2 3 (f (p),f (p)) L kM where f (p) (resp. f (p)) is the residue degree of p in L (resp. kM). kM Proof For every prime p unramiﬁed in K of good reduction for both Jac(C)and E , write 0 2 x := a (E )(p)and y :=± 4 − x . It follows from the proof of Proposition 2.2 that p 1 p a (Jac(C))(p) =− Re(a (p))x + Im(a (p))y , 1 1 p 1 p 2 2 a (Jac(C)(p)) = Re(a (p))(x − 2) − Im(a (p))x y +|a (p)| , 2 2 2 p p 1 a (Jac(C)(p)) =−a (p)(x − 3x ) − Re(a (p)a (p))x + Im(a (p)a (p))y , 3 3 p 1 2 p 1 2 p from which one can easily derive the assertion of the proposition. Tables 7 and 8 show Sato–Tate statistics for the Fermat and Klein twists C/k and their base changes C . In each row, we list moment statistics M , M , M for the kM 101 030 202 three moments M , M , M that uniquely determine the Sato–Tate group ST(C), by 101 030 202 41 Page 32 of 40 Fité et al. Res Math Sci (2018) 5:41 0 0 Table 1 For each possible pair (s, t), the corresponding values of F (x) if C = C .Inthe (s,t) table below, y denotes ± 4 − x (s, t) F (x) F (x) F (x) 1 2 3 2 3 (1, 1) 3x 3x + 3 x + 6x 2 3 −x −x + 3 x − 2x (2, 1) or 2 3 −x + 3 −x + 2x (3, 1) 0 0 x − 3x ⎪ 2 3 −x + 2y ⎪ −x + 7 − 2xy x − 6x + 4y 2 3 (4, 1) or x x − 1 x − 2x ⎪ 2 3 ⎩ x − 1 −x + 2x −x (8, 1) y −xy + 1 x − 4x (2, 2) 0 3 0 (4, 2) 0 −10 (6, 2) 0 0 0 (8, 2) 0 1 0 Table 2 For each possible pair (s, t), the corresponding values of F (x) if C = C .Inthe (s,t) table below, y denotes ± (4 − x )/7 (s, t) F (x) F (x) F (x) 1 2 3 2 3 (1, 1) 3x 3x + 3 x + 6x 2 3 (2, 1) −x −x + 3 x − 2x (3, 1) 0 0 x − 3x 2 3 (4, 1) xx − 1 x − 2x x 7 x 7 9 7 (7, 1) − + y − + 3 − xy x − x + y 2 2 2 2 2 2 (2, 2) 0 3 0 (4, 2) 0 −10 (6, 2) 0 0 0 (8, 2) 0 1 0 Proposition 3.13. These were computed by averaging over all good primes of degree one and norm p ≤ 2 . For comparison, we also list the actual value of each moment, computed using the method described in Sect. 3.2.2. In every case, the moment statistics agree with the corre- sponding moments of the Sato–Tate groups to within 1.5 percent, and in almost all cases, to within 0.5 percent. 4Tables In this ﬁnal section, we present tables of characters, curves, Sato–Tate distributions, and moment statistics referred to elsewhere in this article. Let us brieﬂy describe their contents. Table 3 lists characters of the automorphism groups of the Fermat and Klein quartics speciﬁed via conjugacy class representatives expressed using the generators s, t, u deﬁned in (3.2)and (3.3). Tables 4 and 5 list explicit curve equations for twists C of the Fermat and Klein quar- tics corresponding to subgroups H of G 0 := Aut(C ) Gal(M/Q), as described in C M Remark 3.10. The group H := H ∩ Aut(C ) is speciﬁed in terms of the generators s, t, u listed in (3.2)and (3.3). When kM = k we have H = H , and otherwise H is speciﬁed by listing an element h ∈ Aut(C ) for which H = H ∪ H · (h, τ), where Gal(M/Q) = τ ; 0 0 M Fité et al. Res Math Sci (2018) 5:41 Page 33 of 40 41 Table 3 Character tables of Aut(C ). See (3.2) and (3.3) for the generators s, t, u (A) Aut((C ) ) 96, 64 Class 1a 2a 2b 3a 4a 4b 4c 4d 8a 8b 2 2 3 3 Repr. 1 u uts u u tut t tu tu Order 1 22 3 4 4 4488 Size 1 3 12 32 3 3 6 12 12 12 χ 1 1 1 1 1 1 1111 χ 11 −1 1 111 −1 −1 −1 χ 22 0 −1 2 2 2000 χ 33 −10 −1 −1 −1 −11 1 χ 33 1 0 −1 −1 −11 −1 −1 χ 3 −11 0 −1 − 2i −1 + 2i 1 −1 i −i χ 3 −11 0 −1 + 2i −1 − 2i 1 −1 −ii χ 3 −1 −10 −1 + 2i −1 − 2i 11 i −i χ 3 −1 −10 −1 − 2i −1 + 2i 11 −ii χ 6 −20 0 2 2 −2 000 (B) Aut((C ) ) 168, 42 Class 1a 2a 3a 4a 7a 7b 2 3 2 3 Repr. 1 ts u tu tu uu Order 1 2 3 4 7 7 Size 1 21 56 42 24 24 χ 11 1 1 1 1 χ 3 −10 1 a a χ 3 −10 1 aa χ 62 0 0 −1 −1 χ 7 −11 −10 0 χ 80 −10 1 1 see (3.6). The isomorphism classes of H and H are speciﬁed by GAP identiﬁers ID(H) and ID(H ). The minimal ﬁeld K over which all endomorphisms of Jac(C ) are deﬁned is given as an explicit extension of Q, or as the splitting ﬁeld Gal(f (x)) of a monic f ∈ Z[x]. In the last 2 columns of Tables 4 and 5 we identify the Sato–Tate distributions of ST(C) and ST(C ) by their row numbers in Table 6. Among twists with the same Sato–Tate kM group ST(C) (which is uniquely identiﬁed by its distribution), we list curves with isogenous Jacobians, per Remark 3.31. In Table 6, we list the 60 Sato–Tate distributions that arise among twists of the Fermat and Klein quartics. Each component group is identiﬁed by its GAP ID, and we list the joint moments M ,M ,M suﬃcient to uniquely determine the Sato–Tate distribution, 101 030 202 along with the ﬁrst two non-trivial independent coeﬃcient moments for a ,a ,a .Wealso 1 2 3 list the proportion z of components on which the coeﬃcient a takes the ﬁxed integer i,j i value j; for i = 1,3welistonly z := z and z := z , and for i = 2 we list the vector 1 1,0 3 3,0 z := [z ,z ,z ,z ,z ]; see Lemma 3.15 for details. There are 6 pairs of Sato–Tate 2 2,−1 2,0 2,1 2,2 2,3 distributions whose independent coeﬃcient measures coincide; these pairs are identiﬁed by roman letters that appear in the last column. Tables 7 and 8 list moment statistics for twists of the Fermat and Klein quartics com- puted over good primes p ≤ 2 , along with the corresponding moment values. Twists with isogenous Jacobians necessarily have the same moment statistics, so we list only one twistineachisogeny class. 41 Page 34 of 40 Fité et al. Res Math Sci (2018) 5:41 Table 4 Twists of the Fermat quartic corresponding to subgroups H ⊆ G 0. See (3.2) for the deﬁnitions of s, t, u and (3.6) for the deﬁnition of h. We identify ST = ST(C) and ST = ST(C ) by row numbers in Table 6; here M = Q(i) kM kM #Gen(H ) h ID(H)ID(H )kK ST ST 0 0 k kM 1id id 2, 1 1, 1 QQ(i)31 0 4 4 4 C x + y + z 2id u t 2, 1 1, 1 QQ(i)31 4 2 2 4 4 x − 6x y + y − 2z 3id t utu 2, 1 1, 1 QQ(i)31 4 4 4 4x − y − z 2 4 2 4 t t 4, 1 2, 1 Q( −5) Gal(x −x −1) 5 2 4 3 3 4 4 12x + 40x y − 100xy − 75y − 2z 2 3 5 t t utu 4, 2 2, 1 QQ( 3,i)92 4 4 4 9x + 9y − 4z 2 2 6 t u t 4, 2 2, 1 QQ( 3,i)92 4 2 2 4 4 9x − 54x y + 9y − 2z 7 t id 4, 2 2, 1 QQ( 3,i)92 4 4 4 9x + y + z 8 t u 4, 2 2, 1 QQ( 3,i)92 4 4 4 9x − 4y + z 9 u t id 4, 2 2, 1 QQ( 3,i)92 4 2 2 4 4 2x + 36x y + 18y + z 2 3 10 utt utu 4, 2 2, 1 QQ( 3,i)92 4 2 2 4 4 9x + 18x y + y − 2z 2 3 11su t 6, 1 3, 1 Q Gal(x −3x −4) 11 4 4 3 2 2 2 2 2 2 4 2 2 3 4 x + 4x y + 12x y − 12x yz − 6x z + 36xyz + 6y − 36y z − 12yz + 9z 6 4 2 12 s id 6, 2 3, 1 Q Gal(x +5x +6x +1) 12 4 4 3 3 2 2 2 2 2 3 4 3 2 2 3 4 5x + 8x y − 4x z + 6x y + 12x z + 12xyz + 4xz + 2y + 4y z + 6y z + 4yz + 2z 3 3 8 4 13 t utu tu 8, 1 4, 1 Q( −5) Gal(x −2x −4) 14 6 4 3 2 2 4 4 x − 10x z + 30x z − 2y − 100z 3 8 4 14tt utu 8, 2 4, 1 Q Gal(x +15x +25) 16 6 4 3 2 2 3 4 4 3x − 4x y + 12x y + 4xy + 3y − 5z 3 3 2 8 4 15 t utu u t 8, 2 4, 1 Q Gal(x +15x +25) 16 6 4 3 3 4 4 12x + 40x y − 100xy − 75y + 10z 8 4 16 t id 8, 2 4, 1 Q Gal(x +15x +25) 16 6 4 3 2 2 3 4 4 3x + 4x y + 12x y − 4xy + 3y + 20z 2 2 3 3 4 2 17 u t, t t utu 8, 3 4, 2 Q Gal(x −6x +10) 19 8 4 3 2 2 3 4 4 11x + 12x y + 54x y − 12xy + 11y − 2z 2 2 2 4 2 18 t ,u u t 8, 3 4, 2 Q Gal(x −6x +10) 19 8 4 3 3 4 4 x + 5x y − 25xy − 25y + z 2 2 2 4 2 19 u t, t u 8, 3 4, 2 Q Gal(x −6x +10) 19 8 4 3 2 2 3 4 4 19x − 12x y + 6x y + 12xy + 19y + 2z 2 4 2 20tu 8, 3 4, 1 Q Gal(x −6x +12) 20 6 4 2 2 3 4 4 9x − 18x y − 12xy − 2y + 12z 3 3 4 2 21tt utu 8, 3 4, 1 Q Gal(x −6x +12) 20 6 4 2 2 3 4 4 9x − 18x y − 12xy − 2y − 3z 3 3 4 2 22 t utu u 8, 3 4, 1 Q Gal(x −6x +12) 20 6 4 4 4 9x + 3y − 4z 3 3 4 2 23 t utu id 8, 3 4, 1 Q Gal(x −6x +12) 20 6 4 4 4 9x + 3y + z 3 2 4 2 24 t utu u t 8, 3 4, 1 Q Gal(x −6x +12) 21 7 Fité et al. Res Math Sci (2018) 5:41 Page 35 of 40 41 Table 4 continued #Gen(H ) h ID(H)ID(H ) kK ST ST 0 0 k kM 4 2 2 4 4 x − 6x y + y − 6z 3 4 2 25 t utu u 8, 3 4, 1 Q Gal(x −6x +12) 21 7 4 4 4 3x − 4y + z 3 4 2 26 t utu id 8, 3 4, 1 Q Gal(x −6x +12) 21 7 4 4 4 3x + y + z 3 8 27 t utu t 8, 4 4, 1 Q( −2) Gal(x +9) 23 7 3 3 4 3x y − 3xy − 2z √ √ 2 2 28 t ,u id 8, 5 4, 2 QQ( 3, 5,i)24 8 4 4 4 9x + 25y + z √ √ 2 2 29 t ,u u 8, 5 4, 2 QQ( 3, 5,i)24 8 4 4 4 9x + 25y − 4z √ √ 2 2 3 30 u t, t t utu 8, 5 4, 2 QQ( 3, 5,i)24 8 4 2 2 4 4 x + 30x y + 25y − 18z √ √ 2 2 31 u t, t id 8, 5 4, 2 QQ( 3, 5,i)24 8 4 2 2 4 4 2x + 60x y + 50y + 9z 2 6 3 32 s, u t id 12, 4 6, 1 Q Gal(x +2x +2) 26 10 3 2 2 2 4 3 4x y − 3x z + 12xy z − 2y − 2yz 3 2 8 4 33 t utu, u ut 16, 6 8, 2 Q( −5) Gal(x −2x +5) 29 17 4 2 2 3 4 4 x − 30x y − 80xy − 55y − 2z 3 3 2 8 4 2 34 t utu ,utu 16, 7 8, 3 Q Gal(x −6x −8x −1) 31 18 4 2 2 3 4 4 x − 12x y − 32xy − 28y + z 2 2 8 4 35 tu tut tutu 16, 7 8, 1 Q Gal(x −8x −2) 32 15 3 3 4 2x y − xy − z 2 8 36 u tu u 16, 8 8, 1 Q Gal(x −2) 34 15 3 3 4 x y + 2xy + z 3 3 8 4 37 t utu ,t u 16, 8 8, 4 Q Gal(x −10x −100) 33 22 4 3 2 2 4 4 x + 10x y + 30x y − 100y − 10z 2 8 4 38 t, u utu 16, 11 8, 3 Q Gal(x −2x +9) 35 18 4 3 2 2 3 4 4 4x + 4x y + 6x y − 2xy + y − 2z 2 8 4 39 t, u id 16, 11 8, 3 Q Gal(x −2x +9) 35 18 4 3 2 2 3 4 4 4x + 4x y + 6x y − 2xy + y + 2z 3 3 2 8 4 40 t utu ,u t id 16, 11 8, 3 Q Gal(x −2x +9) 35 18 4 3 2 2 3 4 4 5x − 8x y + 12x y + 16xy + 20y + 2z 3 2 8 4 41 t utu, u id 16, 11 8, 2 Q Gal(x +5x +25) 36 17 4 4 4 9x + 5y + z 3 2 8 4 42 t utu, u u 16, 11 8, 2 Q Gal(x +5x +25) 36 17 4 4 4 9x + 5y − 4z 3 2 8 4 43 utu, t u 16, 11 8, 2 Q Gal(x +5x +25) 36 17 4 3 2 2 3 4 4 7x + 8x y + 6x y + 8xy + 7y + 10z 3 2 2 8 4 44 t utu, u u t 16, 13 8, 2 Q Gal(x −8x +25) 39 17 4 3 3 4 4 x + 5x y − 25xy − 25y + 2z 8 4 45 t, utu id 16, 13 8, 2 Q Gal(x −8x +25) 39 17 4 3 2 2 3 4 3 2 2 3 4 19x + 32x y + 21x y + 8xy + 3y + 2y z + 6y z + 8yz + 4z 3 3 8 4 46 t utu ,t id 16, 13 8, 4 Q Gal(x +12x +9) 37 22 4 3 2 2 3 4 4 x − 2x y + 6x y + 4xy + 4y + 3z 2 2 4 47 s, u u t 24, 12 12, 3 Q Gal(x −16x −24) 42 25 4 3 2 3 3 4 3 2 2 x − 3x z − 12x yz + 16xy − xz + 9y + 12y z + 6y z 2 6 4 2 48 s, u id 24, 13 12, 3 Q Gal(x −x −2x +1) 43 25 4 3 3 2 2 2 2 3 2 4 3 2 2 4 3x + 4x y + 4x z + 6x y + 6x z + 8xy + 12xyz + 5y + 4y z + 12y z + z 2 2 8 4 49 tu tut, u id 32, 7 16, 6 Q Gal(x −10x +20) 44 30 41 Page 36 of 40 Fité et al. Res Math Sci (2018) 5:41 Table 4 continued #Gen(H ) h ID(H)ID(H ) kK ST ST 0 0 k kM 4 3 2 2 4 4 4x − 8x y + 12x y + 2y + 5z 2 3 2 8 4 50 u, u tutu t 32, 11 16, 2 Q Gal(x −2x +5) 45 28 4 2 2 3 4 4 x − 30x y − 80xy − 55y − 2z 3 16 12 8 4 51 u, t ut id 32, 34 16, 2 Q Gal(x −4x +6x +20x +1) 47 28 4 4 4 2x + 3y + z 3 2 8 4 52 t utu, u ,t u 32, 43 16, 13 Q Gal(x +6x −9) 48 38 4 2 2 3 4 4 3x − 36x y − 96xy − 84y + 2z 2 2 8 4 53 tu tut, u u 32, 43 16, 6 Q Gal(x −10x +45) 49 30 4 3 3 4 4 9x + 36x y − 24xy − 4y − 10z 2 8 4 54 utu, u ,t id 32, 49 16, 13 Q Gal(x +8x +9) 50 38 4 3 2 2 3 4 4 x + x y + 24x y + 67xy + 79y + 2z 6 4 2 55 s, t id 48, 48 24, 12 Q Gal(x −x +5x +1) 53 41 4 3 2 3 2 3 4 3 2 2 3 3x + 2x z + 6x yz + 12xy − 30xy z + 2xz − 27y + 38y z + 18y z − 10yz 8 4 56 t, u id 64, 134 32, 11 Q Gal(x −4x −14) 54 46 4 2 2 3 4 4 x − 42x y − 168xy − 203y + z 2 12 4 57 s, u u t 96, 64 48, 3 Q Gal(x +6x +4) 55 52 3 2 2 2 2 3 2 4 3 3 4 6x z + 3x y + 27x z + 6xy + 18xyz + 4y + 2y z + 18yz − 36z 12 8 4 58 s, u id 96, 72 48, 3 Q Gal(x +x −2x −1) 57 52 3 3 2 2 3 2 2 4 3 4 x y − x z + 3x y + 28xy − 84xy z + 84xyz + 28y − 56y z + 98z 12 4 59 s, t, u id 192, 956 96, 64 Q Gal(x +48x +64) 59 56 4 3 3 2 2 2 3 3 4 2 2 44x + 120x y + 36x z + 60x yz + 9x z − 200xy + xz − 150y − 15y z Table 5 Twists of the Klein quartic corresponding to subgroups H ⊆ G 0. See (3.3) for the deﬁnitions of s, t, u and see (3.6) for the deﬁnition of h. We identify ST = ST(C) and √ √ ST = ST(C ) by row numbers in Table 6; here M = Q( −7) and a := (−1 + −7)/2 kM kM #Gen(H ) h ID(H)ID(H ) kK ST ST 0 0 k kM 1id id 2, 1 1, 1 QQ(a)31 4 4 4 3 3 3 2 2 2 2 2 2 C x + y + z + 6(xy + yz + zx ) − 3(x y + y z + z x ) + 3xyz(x + y + z) 2 t id 4, 2 2, 1 QQ(a, i)92 4 3 2 2 2 2 3 4 2 2 4 3x + 28x y + 105x y − 21x z + 196xy + 147y + 147y z − 49z √ √ 6 6 2 3 ustu ,sutu s – 4, 2 4, 2 Q(a) Q( 2, 3,a)8 8 4 2 2 2 2 4 2 2 4 x + 9ax y + 6ax z + 9y + 18ay z + 4z 3 2 4 st 6, 1 3, 1 Q Gal(x −x +2x −3) 11 4 4 3 3 2 2 2 2 3 2 2 4 3 2 2 3 x + 3x y − 9x z + 9x y − 6x z + 18xy + 3xy z − 3xyz + y + 4y z − 3y z + 7yz 5 s id 6, 2 3, 1 QQ(ζ)124 3 3 3 x y + xz + y z 2 3 2 5 2 8 7 4 6 u tu tu u tu 8, 1 4, 1 Q(i)Gal(x +2x −14x +16x +4) 14 6 4 2 2 2 2 3 2 2 3 x + 3x y − 3x z + 2y z + 3y z + 2yz 2 3 2 4 2 7 u tu tu id 8, 3 4, 1 Q Gal(x −4x −14) 20 6 4 3 2 2 2 2 3 2 4 2 2 4 12x − 80x y + 60x y − 24x z − 104xy + 24xyz + 83y + 36y z − 2z 6 2 8 su, tu – 12, 3 12, 3 Q(a)Gal(x −147x +343) 25 25 4 3 3 2 2 2 2 2 4 3x + (−18a + 12)x y + (12a + 4)x z + (−27a + 36)x y + (9a + 6)x z + 36xy z + 27y 3 2 2 3 4 + (54a − 36)y z + (−54a + 36)y z + (18a − 12)yz + (−3a + 2)z 3 2 9 s, t id 12, 4 6, 1 Q Gal(x −x +5x +1) · Q(a)26 10 3 2 2 2 3 4 7x z + 3x y − 6xyz + 2y z − 4z 7 3 2 10 u id 14, 1 7, 1 Q Gal(x +7x −7x +7x +1) 27 13 3 2 2 3 2 3 4 3 2 2 4 x y − 21x z + xy − 42xyz − 147xz + 2y + 21y z + 63y z − 196z 2 3 2 5 2 8 7 4 11 u tu tu ,u tu id 16, 7 8, 3 Q Gal(x +2x −14x +16x +4) 31 18 4 2 2 2 2 3 2 2 3 x + 3x y − 3x z + 2y z + 3y z + 2yz 6 2 2 4 3 2 12 sust, su s tu – 24, 12 24, 12 Q(a)Gal(x +2x +6x −6) · Q(a)41 41 4 3 2 2 2 2 3 2 (3a − 2)x + (30a − 20)x y + (90a − 60)x y + (9a + 6)x z + (150a − 100)xy + 60xy z 3 4 2 2 3 4 + (12a + 4)xz + (150a − 25)y + (−45a + 60)y z + (30a − 20)yz + 3z 13 u, s id 42, 1 21, 1 Q Gal(x −2) 51 40 3 3 3 2x y + xz + y z 8 7 4 14 t, u, s id 336, 208 168, 42 Q Gal(x +4x +21x +18x +9) 60 58 3 3 2 2 3 3 3 2 2 3 2x y − 2x z − 3x z − 2xy − 2xz − 4y z + 3y z − yz Fité et al. Res Math Sci (2018) 5:41 Page 37 of 40 41 Table 6 The 60 Sato–Tate distributions arising for Fermat and Klein twists #ID M M M M M M M M M z z z 101 030 202 200 400 010 020 002 004 1 2 3 1 1, 1 54 1215 4734 18 486 9 99 164 47148 0 0 0 0 0 0 0 2 2, 1 26 611 2374 10 246 5 51 84 23596 0 0 0 0 0 0 0 1 1 1 3 2, 1 27 621 2367 9 243 6 54 82 23574 /2 0000 /2 /2 2 2 4 3, 1 18 405 1578 6 162 3 33 56 15720 /3 0 /3 00 00 1 1 1 5 4, 1 13 305 1187 5 123 2 26 42 11798 /2 /2 000 0 /2 6 4, 1 14 309 1194 6 126 3 27 44 11820 0 0 0 0 0 0 0 a 7 4, 1 24 443 1614 10 198 5 43 68 14444 0 0 0 0 0 0 0 8 4, 2 12 309 1194 6 126 3 27 44 11820 0 0 0 0 0 0 0 a 1 1 1 9 4, 2 13 319 1187 5 123 4 30 42 11798 /2 0000 /2 /2 1 1 10 6, 1 8 206 796 4 84 2 18 30 7882 /3 0 /3 00 00 5 1 1 1 11 6, 1 9 216 789 3 81 3 21 28 7860 /6 0 /3 00 /2 /2 5 2 1 1 12 6, 2 9 207 789 3 81 2 18 28 7860 /6 0 /3 00 /6 /2 13 7, 1 12 201 732 6 90 3 21 32 6936 0 0 0 0 0 0 0 1 1 1 14 8, 1 7 155 597 3 63 2 14 22 5910 /2 00 /2 00 /2 15 8, 1 12 225 812 6 102 3 23 36 7236 0 0 0 0 0 0 0 1 1 1 1 16 8, 2 7 161 597 3 63 2 16 22 5910 /2 /4 000 /4 /2 b 17 8, 2 12 225 814 6 102 3 23 36 7244 0 0 0 0 0 0 0 18 8, 3 6 158 604 4 66 2 15 24 5932 0 0 0 0 0 0 0 c 1 1 1 1 19 8, 3 6 161 597 3 63 2 16 22 5910 /2 /4 000 /4 /2 b 1 1 1 20 8, 3 7 168 597 3 63 3 18 22 5910 /2 0000 /2 /2 d 1 1 1 21 8, 3 12 235 807 5 99 4 26 34 7222 /2 0000 /2 /2 22 8, 4 8 158 604 4 66 2 15 24 5932 0 0 0 0 0 0 0 c 1 1 1 23 8, 4 12 221 807 5 99 2 22 34 7222 /2 /2 000 0 /2 1 1 1 24 8, 5 6 168 597 3 63 3 18 22 5910 /2 0000 /2 /2 d 2 2 25 12, 3 4 103 398 2 42 1 9 16 3944 /3 0 /3 00 00 2 1 1 1 26 12, 4 4 112 398 2 42 2 12 15 3941 /3 0 /3 00 /3 /2 1 1 1 27 14, 1 6 114 366 3 45 3 15 16 3468 /2 0000 /2 /2 28 16, 2 12 183 624 6 90 3 21 32 4956 0 0 0 0 0 0 0 1 1 1 29 16, 6 6 113 407 3 51 2 12 18 3622 /2 00 /2 00 /2 30 16, 6 6 116 412 4 54 2 13 20 3636 0 0 0 0 0 0 0 1 1 1 1 31 16, 7 3 86 302 2 33 2 10 12 2966 /2 00 /4 0 /4 /2 e 1 1 1 32 16, 7 6 126 406 3 51 3 16 18 3618 /2 0000 /2 /2 1 1 1 1 33 16, 8 4 86 302 2 33 2 10 12 2966 /2 00 /4 0 /4 /2 e 1 1 1 1 34 16, 8 6 119 406 3 51 2 14 18 3618 /2 /4 000 /4 /2 1 1 3 1 35 16, 11 3 89 302 2 33 2 11 12 2966 /2 /8 000 /8 /2 f 1 1 1 36 16, 11 6 126 407 3 51 3 16 18 3622 /2 0000 /2 /2 1 1 3 1 37 16, 13 4 89 302 2 33 2 11 12 2966 /2 /8 000 /8 /2 f 38 16, 13 6 116 414 4 54 2 13 20 3644 0 0 0 0 0 0 0 1 1 1 1 39 16, 13 6 119 407 3 51 2 14 18 3622 /2 /4 000 /4 /2 2 2 40 21, 1 4 67 244 2 30 1 7 12 2316 /3 0 /3 00 00 1 1 41 24, 12 2 55 206 2 24 1 6 10 1994 /3 0 /3 00 00 5 1 1 1 1 42 24, 12 2 58 199 1 21 1 7 8 1972 /6 /4 /3 00 /4 /2 5 2 1 1 43 24, 13 2 56 199 1 21 1 6 8 1972 /6 0 /3 00 /6 /2 1 1 1 1 44 32, 7 3 65 206 2 27 2 9 10 1818 /2 00 /4 0 /4 /2 1 1 1 1 1 45 32, 11 6 95 312 3 45 2 12 16 2478 /2 /8 0 /4 0 /8 /2 46 32, 11 6 95 318 4 48 2 12 18 2496 0 0 0 0 0 0 0 1 1 1 47 32, 34 6 105 312 3 45 3 15 16 2478 /2 0000 /2 /2 1 1 1 1 48 32, 43 3 65 207 2 27 2 9 10 1822 /2 00 /4 0 /4 /2 1 1 3 1 49 32, 43 3 68 206 2 27 2 10 10 1818 /2 /8 000 /8 /2 41 Page 38 of 40 Fité et al. Res Math Sci (2018) 5:41 Table 6 continued #ID M M M M M M M M M z z z 101 030 202 200 400 010 020 002 004 1 2 3 1 1 3 1 50 32, 49 3 68 207 2 27 2 10 10 1822 /2 /8 00 0 /8 /2 5 2 1 1 51 42, 1 2 38 122 1 15 1 5 6 1158 /6 0 /3 00 /6 /2 2 2 52 48, 3 4 61 208 2 30 1 7 12 1656 /3 0 /3 00 0 0 2 1 1 5 1 53 48, 48 1 33 103 1 12 1 5 5 997 /3 /8 /3 00 /24 /2 1 1 1 5 1 54 64, 134 3 56 159 2 24 2 9 9 1248 /2 /16 0 /8 0 /16 /2 5 1 1 1 1 1 55 96, 64 2 34 104 1 15 1 5 6 828 /6 /8 /3 /4 0 /8 /2 1 1 56 96, 64 2 34 110 2 18 1 5 8 846 /3 0 /3 00 0 0 5 2 1 1 57 96, 72 2 35 104 1 15 1 5 6 828 /6 0 /3 00 /6 /2 1 1 58 168, 42 2 19 52 2 12 1 4 6 366 /3 0 /3 00 0 0 2 1 1 1 7 1 59 192, 956 121 55 1 9 1 44 423 /3 /16 /3 /8 0 /48 /2 2 1 1 1 1 60 336, 208 112 26 1 6 133 183 /3 0 /3 /4 0 /12 /2 Table 7 Sato–Tate statistics for Fermat twists over degree one primes p ≤ 2 #ST(C)ST(C ) k kM M M M M M M M M M M M M 101 101 030 030 202 202 101 101 030 030 202 202 126.99 27 620.72 621 2365.83 2367 53.99 54 1214.69 1215 4732.64 4734 412.98 13 304.55 305 1185.12 1187 25.98 26 610.36 611 2371.23 2374 512.99 13 318.75 319 1185.94 1187 25.99 26 610.61 611 2372.38 2374 11 8.99 9 215.63 216 787.39 789 17.98 18 404.33 405 1575.11 1578 12 9.00 9 206.88 207 788.45 789 18.00 18 404.84 405 1577.24 1578 13 6.98 7 154.56 155 595.23 597 13.97 14 308.25 309 1190.96 1194 14 6.99 7 160.84 161 596.37 597 13.99 14 308.75 309 1192.99 1194 17 5.99 6 160.80 161 596.20 597 11.99 12 308.67 309 1192.65 1194 20 6.99 7 167.74 168 595.89 597 13.98 14 308.54 309 1192.02 1194 24 11.99 12 234.80 235 806.10 807 23.99 24 442.70 443 1612.54 1614 27 11.99 12 220.64 221 805.56 807 23.98 24 442.43 443 1611.64 1614 28 5.99 6 167.77 168 596.03 597 11.98 12 308.60 309 1192.31 1194 32 4.00 4 111.98 112 397.87 398 8.00 8 205.99 206 795.91 796 33 6.00 6 112.89 113 406.55 407 12.00 12 224.88 225 813.43 814 34 3.00 3 85.98 86 301.92 302 6.00 6 157.99 158 603.96 604 35 5.99 6 125.77 126 405.04 406 11.98 12 224.57 225 810.25 812 36 5.99 6 118.74 119 404.95 406 11.98 12 224.52 225 810.06 812 37 4.00 4 85.99 86 301.98 302 8.00 8 158.00 158 604.09 604 38 2.99 3 88.81 89 301.24 302 5.99 6 157.65 158 602.60 604 41 5.99 6 125.74 126 405.90 407 11.98 12 224.53 225 811.97 814 44 6.00 6 118.88 119 406.47 407 11.99 12 224.80 225 813.12 814 46 3.99 4 88.73 89 300.88 302 7.98 8 157.50 158 601.89 604 47 2.00 2 57.88 58 198.48 199 3.99 4 102.77 103 397.05 398 48 1.99 2 55.81 56 198.20 199 3.99 4 102.64 103 396.49 398 49 2.99 3 64.85 65 205.42 206 5.99 6 115.72 116 410.92 412 50 6.00 6 94.92 95 311.67 312 12.00 12 182.87 183 623.48 624 51 5.99 6 104.72 105 310.80 312 11.98 12 182.47 183 621.72 624 52 3.00 3 64.94 65 206.72 207 6.00 6 115.89 116 413.52 414 53 2.99 3 67.84 68 205.30 206 5.99 6 115.71 116 410.68 412 54 3.00 3 67.88 68 206.50 207 5.99 6 115.78 116 413.08 414 Fité et al. Res Math Sci (2018) 5:41 Page 39 of 40 41 Table 7 continued #ST(C)ST(C ) k kM M M M M M M M M M M M M 101 101 030 030 202 202 101 101 030 030 202 202 55 1.00 1 32.93 33 102.73 103 1.99 2 54.86 55 205.50 206 56 2.99 3 55.77 56 158.06 159 5.98 6 94.56 95 316.20 318 57 1.99 2 33.87 34 103.46 104 3.99 4 60.76 61 206.96 208 58 2.00 2 34.93 35 103.71 104 4.00 4 60.87 61 207.45 208 59 1.00 1 20.99 21 54.95 55 2.00 2 33.98 34 109.92 110 Table 8 Sato–Tate statistics for Klein twists over degree one primes p ≤ 2 #ST(C)ST(C ) k kM M M M M M M M M M M M M 101 101 030 030 202 202 101 101 030 030 202 202 126.99 27 620.78 621 2366.04 2367 53.99 54 1214.67 1215 4732.50 4734 212.99 13 318.73 319 1185.85 1187 25.98 26 610.52 611 2371.92 2374 311.99 12 308.67 309 1192.59 1194 11.99 12 308.67 309 1192.59 1194 48.99 9 215.76 216 787.97 789 17.98 18 404.55 405 1576.08 1578 59.00 9 206.95 207 788.79 789 18.00 18 404.94 405 1577.71 1578 67.00 7 154.90 155 596.58 597 14.00 14 308.88 309 1193.45 1194 76.99 7 167.80 168 596.13 597 13.99 14 308.63 309 1192.36 1194 84.01 4 103.36 103 399.55 398 4.01 4 103.36 103 399.55 398 93.99 4 111.68 112 396.63 398 7.98 8 205.37 206 793.34 796 10 5.99 6 113.83 114 365.24 366 11.99 12 200.67 201 730.55 732 11 3.00 3 85.94 86 301.73 302 6.00 6 157.89 158 603.51 604 12 2.01 2 55.11 55 206.43 206 2.01 2 55.11 55 206.43 206 13 2.00 2 37.97 38 121.81 122 4.00 4 66.94 67 243.65 244 14 1.00 1 11.94 12 25.70 26 2.00 2 18.87 19 51.40 52 Author details 1 2 Institute for Advanced Study, Fuld Hall, 1 Einstein Drive, Princeton, NJ 08540, USA, Laboratoire IRMAR, Université de Rennes 1, Campus de Beaulieu, 35042 Rennes Cedex, France, Department of Mathematics, Massachusetts Institute of Technology, 77 Massachusetts Avenue, Cambridge, MA 02139, USA. Acknowledgements We thank Josep González for his help with Lemma 3.18, and we are grateful to the Banﬀ International Research Station for hosting a May 2017 workshop on Arithmetic Aspects of Explicit Moduli Problems where we worked on this article. Fité is grateful to the University of California at San Diego for hosting his visit in spring 2012, the period in which this project was conceived. Fité received ﬁnancial support from the German Research Council (CRC 701), the Excellence Program María de Maeztu MDM-2014-0445, and MTM2015-63829-P. Sutherland was supported by NSF Grants DMS-1115455 and DMS-1522526. This project has received funding from the European Research Council (ERC) under the European Unions Horizon 2020 research and innovation program (Grant Agreement No. 682152), and from the Simons Foundation (Grant #550033). Received: 25 May 2018 Accepted: 18 September 2018 References 1. 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Sato–Tate distributions of twists of the Fermat and the Klein quartics

Sato–Tate distributions of twists of the Fermat and the Klein quartics

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Sato–Tate distributions of twists of the Fermat and the Klein quartics

Sato–Tate distributions of twists of the Fermat and the Klein quartics

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