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Hodge Groups of certain Superelliptic Jacobians II

Let K be a field of characteristic zero, f(x) be a polynomial with coefficients in K and without multiple roots. We consider the superelliptic curve C_{f,q} defined by y^q=f(x), where q=p^r is a power of a prime p. We determine the Hodge group of the simple factors of the Jacobian of C_{f,q}.

Throughout this paper C is the field of complex numbers, K ⊆ C is a subfield of C, f (x) ∈ K[x] a polynomial without multiple roots and of degree n ≥ 4. Let p ∈ N be a prime that does not divide n and q = p r ∈ N an integral power of p. We write C f,q for the superelliptic K-curve y q = f (x), and J(C f,q ) for the Jacobian of C f,q . By definition, C f,q is the smooth projective model of the affine curve y q = f (x). The Jacobian J(C f,q ) is an abelian variety over K of dimension dimJ(C f,q ) = g(C f,q ) = (n − 1)(q − 1) 2 .
If q > p, the map C f,q → C f,q/p , (x, y) → (x, y p ) induces by Albanese fuctoriality a surjective K-map between the Jacobians J(C f,q ) → J(C f,q/p ). We write J (f,q) for the identity component of the kernel. If q = p, we set J (f,p) = J(C f,p ). It is follows easily that J (f,q) is an abelian variety over K of dimension (n − 1)ϕ(q)/2, where ϕ denotes the Euler ϕ-function. Moreover, J(C f,q ) is K-isogenous to the product r i=1 J (f,p i ) (See [15]). Since K ⊆ C, we may view J (f,q) as a complex abelian variety. We refer to [5], [10, Sect. 6.6.1 and 6.6.2] for the definition and basic properties of the Hodge group (aka special Mumford-Tate group). In [9], assuming that n > q and some other conditions on n, q and f (x), the authors showed that the (reductive Q-algebraic connected) Hodge group of J (f,q) coincides with the largest Q-algebraic subgroup of GL(H 1 (J (f,q) , Q)) that's "cut out" by the induced polarization from the canonical principal polarization of J(C f,q ) and the endomorphism ring of J (f,q) . Notice that when q = 2 (i.e., in the hyperelliptic case) this group was completely determined in [12] (when f (x) has "large" Galois group). In this paper, we study some additional properties of J (f,q) which will allow us to extend the result to the case n < q as well. This case is necessary in order to treat the infinite towers of superelliptic jacobians, which, in turn, are useful for the study of the ranks of Mordell-Weil groups in infinite towers of function fields (See [6]).
To state our main result, we make explicit the endomorphism ring and the polarization mentioned above. Let X be an abelian variety overK. We write End(X) for the ring of all itsK-endomorphisms and End 0 (X) for the endomorphism algebra End(X) ⊗ Z Q. In a series of papers [11,13,14,15], Yuri Zarhin discussed the structure of End 0 (J(C f,q )), assuming that n ≥ 5 and the Galois group Gal(f ) of f (x) over K is, at least, doubly transitive. Here Gal(f ) ⊆ S n is viewed as a permutation group on the roots of f (x). It is well known that f (x) is irreducible over K if and only if Gal(f ) acts transitively on the roots. For the sake of simplicity let's assume that K contains a primitive q-th root of unity ζ q . The curve C f,q : y q = f (x) admits the obvious periodic automorphism By an abuse of notation, we also write δ q for the induced automorphism of J(C f,q ). The subvariety J (f,q) is δ q -invariant and we have an embedding In particular, the q-th cyclotomic filed E := Q(ζ q ) is contained in End 0 (J (f,q) ). Zarhin showed ( [11,15,17]) that End(J (f,q) ) is isomorphic to Z[ζ q ] if either Gal(f ) coincides with the full symmetric group S n , n ≥ 4 and p ≥ 3, or Gal(f ) coincides with the alternating group A n (or S n ), and n ≥ 5. This result has also been extended to the case Gal(f ) = S n or A n , n ≥ 5 and p | n in [7].
The first rational homology group H 1 (J (f,q) , Q) carries a natural structure of E-vector space of dimension Notice that if q > 2, then E is a CM field with complex conjugation e →ē. Let be the maximal totally real subfield of E and let The canonical principal polarization on J(C f,q ) induces a polarization on J (f,q) , which gives rise to a nondegenerate E-sesquilinear Hermitian form ( [9]) We write U(H 1 (J (f,q) , Q), φ q ) for the unitary group of φ q of the Q(ζ q )-vector space H 1 (J (f,q) , Q), viewed as an Q-algebraic subgroup of GL(H 1 (J (f,q) , Q)) (via Weil's restriction of scalars from E + to Q ([5])). Since the Hodge group respects the polarization and commutes with endomorphisms of J (f,q) , If End 0 (J (f,q) ) = E, then U(H 1 (J (f,q) , Q), φ q ) is the largest connected reductive Q-algebraic subgroup of GL(H 1 (J (f,q) , Q)) that both respects the polarization and commutes with endomorphisms of J (f,q) . The following theorem is a natural extension of [9, Theorem 0.1].
Theorem 1. Suppose that n ≥ 4 and p is a prime that does not divide n. Let f (x) ∈ C[x] be a degree n polynomial without multiple roots. Let r be a positive integer and q = p r . Suppose that there exists a subfield K of C that contains all the coefficients of f (x). Let us assume that f (x) is irreducible over K and the Galois group Gal(f ) of f (x) over K is either S n or A n . Assume additionally that either n ≥ 5 or n = 4 and Gal(f ) = S 4 . Suppose that one of the following three conditions holds: (A) n = q + 1; (B) p is odd and n ≡ 1 mod q; (C) p = 2, n ≡ 1 mod q and n ≡ q − 1 mod 2q.
Corollary 2. Corollary 0.3, Theorem 4.2 and Theorem 4.3 of [9] all hold without the assumption that n > q.
Remark 3. We assume that n < q throughout the rest of the paper since the case n > q has already been treated in [9].
Remark 4. Since both Hdg(J (f,q) ) and U(H 1 (J (f,q) , Q), φ q ) are connected Qalgebraic groups, to prove Theorem 1, it suffices to show that It is known that Let hdg be the Q-Lie algebra of Hdg(J (f,q) ). It is a reductive Q-Lie subalgebra of End Q H 1 (J (f,q) , Q) , and thus splits into a direct sum hdg = c ⊕ hdg ss , of its center c and the semisimple part hdg ss = [hdg, hdg]. By [8,Theorem 1.3], if Gal(f ) = S n and n ≥ 4, or Gal(f ) = A n and n ≥ 5, the center c coincides with Theorem 1 follows if we show that The paper is organized as follows. In section 2 we study the Galois actions on certain vector spaces. In section 3 we recall some facts about the Hodge Lie algebra hdg. The proof of Theorem 1 is given at the end of section 3 except a key arithmetic lemma, which is proven in Section 4.

Galois Actions
Throughout this section, let E be a field that is a finite Galois extension of Q with Galois group G. Let V be a E-vector space of finite dimension. We write V Q for the underlying Q-vector space of V , and V C for the C-vector space Let's fix an embedding E ֒→ C. This allows us to identify each Galois automorphism σ : E → E with the embedding σ : E → E ⊂ C of E into C. It is well known that So every E C module W splits as a direct sum W = ⊕ σ∈G W σ , where In particular, Similarly, for each pair σ = τ , we write P σ,τ = P σ ⊕P τ : V C → V σ ⊕V τ for the projection map onto this pair of summands.
Let us define an action of Aut(C) on the set of projection Then for any element where all summations runs through σ ∈ G, and we used the fact that κ −1 (x σ ) belongs to V τ if and only if σ = κτ . Therefore, Clearly Aut(C) acts transitively on P. Since P σ,τ = P σ ⊕ P τ , we have similarly an action of Aut(C) on the set PP : The Aut(C)-orbit O σ,τ of each P σ,τ ∈ PP consists of all elements of the form P κσ,κτ with κ ∈ G.
Let R be a commutative ring with unity, and N be a free R-module of finite rank. We write Tr R : End R (N ) → R for the trace map, and for the R-Lie algebra of traceless endomorphisms of N . It is well-known that We will denote the projection map sl E (V ) ⊗ Q C → sl C (V σ ) again by P σ , and similarly for P σ,τ . Clearly, each For the rest of the section, we assume additionally that E is a CM-field. For any σ ∈ G, letσ : E → E be the complex conjugation of σ. In other words,σ is the composition E σ − → E → E, where the second arrow stands for the complex conjugation map e →ē.
Lemma 6. Let k be a semisimple Q-Lie subalgebra of sl E (V ), and k C := k ⊗ Q C be its complexification. Suppose that the following two conditions holds: Then Proof. Applying Lemma 5 with k in place of W and sl E (V ) in place of V , we see that Let us fix a CM-type Φ of E. By definition, Φ is a maximal subset of G = Hom(E, C) such that no two elements of Φ are complex conjugate to each other. Clearly, |Φ| = [E : Q]/2, and Let k ′ C be the projection of k C on ⊕ σ∈Φ sl C (V σ ). It follows that the projection k ′ C → sl C (V σ ) is surjective for all σ ∈ Φ, and k ′ C also projects surjectively onto sl C (V σ ) ⊕ sl C (V τ ) for all distinct pairs σ, τ ∈ Φ. Therefore, k ′ C = ⊕ σ∈Φ sl C (V σ ) by the Lemma on pp.790-791 of [4]. In particular, we get In the next section, we will show that our semisimple part of Hodge Lie algebra hdg ss = [hdg, hdg] satisfies (I) and (II) of Lemma 6 and thus prove our Main Theorem.

the hodge lie algebra
We keep all notation and assumptions of the previous sections. More specifically, ζ q is a primitive q-th root of unity, E = Q(ζ q ) and G = Gal(E/Q) = (Z/qZ) * , where each a ∈ (Z/qZ) * maps ζ q to ζ a q . In order to simplify the notation, we write X for the abelian variety J (f,q) , and V for its first rational homology group H 1 (X, Q). In addition, we assume that End 0 (X) = E.
Recall that E C = E ⊗ Q C. Let Lie(X) be the complex tangent space to the origin of X. By functoriality, E acts on Lie(X) and provides Lie(X) with a natural structure of E C -module. Therefore, Lie(X) splits into a direct sum where Lie(X) a := {x ∈ Lie(X) | (ζ q ⊗ 1)x = ζ a q x}. Let us put n a = dim C Lie(X) a . It is known that n a = [na/q] (see [15,16]), where [x] is the maximal integer that's less or equal to x, and we take the representative 1 ≤ a ≤ q − 1. Since V = H 1 (X, Q) carries a natural structure of E-vector space, the first complex homology group V C = H 1 (X, C) = H 1 (X, Q) ⊗ Q C carries a structure of E C -module, and therefore splits into a direct sum There is a canonical Hodge decomposition ([3, chapter 1], [1, pp. 52-53]) where H −1,0 (X) and H 0,−1 (X) are mutually "complex conjugate" dim(X)-dimensional complex vector spaces. This splitting is E-invariant, and H −1,0 (X) and Lie(X) are canonically isomorphic as E C -modules. In particular, It follows that the linear operator f 0 H : V a → V a is semisimple and its spectrum lies in the two-element set {−1/2, 1/2}. The multiplicity of eigenvalue −1/2 is n a = dim C H −1,0 (X) a , while the multiplicity of eigenvalue 1/2 is dim E V − n a . Clearly, the complex conjugate of a ∈ Gal(E/Q) = (Z/qZ) * is a = q − a. It is known ([1], [2]) that (2) n a + nā = dim E V.
This implies that the multiplicity of the eigenvalue 1/2 is nā.
The Hodge Lie algebra hdg of X is a reductive Q-Lie subalgebra of End Q (V ). Its natural representation in V is completely reducible and its centralizer in End Q (V ) coincides with End 0 (X) = E. Moreover, its complexification Recall that hdg = c ⊕ hdg ss , with c being the center of hdg and hdg ss = [hdg, hdg] the semisimple part. Let c C := c ⊗ Q C be the complexification of c and hdg ss C := hdg ss ⊗ Q C the complexification of hdg ss . Clearly, hdg ss ⊂ sl E (V ), and thus hdg ss . We write hdg ss a for the image of projection P a : hdg ss C → sl C (V a ). Clearly, each hdg ss a is a semisimple complex Lie subalgebra of sl C (V a ). Remark 8. Let us decompose f 0 H as f + f ′ with f ′ ∈ c C and f ∈ hdg ss C . By [9, Remark 3.2], the natural representation V a of hdg ss a is simple for all a ∈ G. It follows from Schur's Lemma that when restricted to each V a , f ′ coincides with multiplication by scalar c a ∈ C. Therefore, hdg ss C contains an operator (namely, f ) whose restriction on each V a is diagonalizable with at most two eigenvalues: −1/2 − c a of multiplicity n a and 1/2 − c a of multiplicity nā = dim E V − n a . Lemma 9. Let the assumptions be the same as in Theorem 1. There exists an a ∈ G = (Z/qZ) * such that hdg ss a = P a (hdg ss C ) coincides with sl C (V a ). Proof. The idea is to combine Remark 7, 8 together with Lemma 3.3 of [9]. This result is already contained in the proof of [9,Theorem 3.4], where we note that the assumption n > q in [9,Theorem 3.4] is not used for this particular step of the proof.
Notice that this is the place where assumptions (A)(B)(C) in Theorem 1 are used, since we need to make sure that there exists a ∈ G such that n a and dim E V are relative prime in order to apply Lemma 3.3 of [9].
Let h : (Z/qZ) * → R be the function such that for all 1 ≤ a ≤ q − 1 with gcd(a, q) = 1, , which is also easy to check directly from (3). The function h is non-increasing on the set of integers By Remark 3, we have 4 ≤ n < q. In particular, [n/q] = 0. On the other hand, let t be the maximal element of [1, q/2] Z . Then t = 1 and [nt/q] = 0. It follows that h is not a constant function.
Lemma 10. Let the assumption be the same as Theorem 1. Let (a, b) ∈ G 2 be a pair such that h(a) = h(b). Then P a,b (hdg ss h(b) if and only if n a = n b and n a = dim E V − n b . Let k ss = P a,b (hdg ss C ). By Lemma 9 and part (i) of Lemma 5, both projections k ss → sl C (V a ) and k ss → sl C (V b ) are surjective. By Remark 8, P a,b (f ) is a semisimple element of k ss ⊆ End C (V a ) ⊕ End C (V b ) such that P a,b (f ) acts on V a with (at most) 2 eigenvalues of multiplicities n a and dim E V − n a respectively, and similarly for b. Lemma 10 follows by setting d = 2 in [9, Lemma 3.6]. Last, we point out that the assumption that the multiplicities a i are positive in [9, Lemma 3.6] is not used in its proof, so the lemma applies to the case that n a or n b is zero, which may happen if n < q.
Proof of Theorem 1. As remarked at the end of Section 2, Theorem 1 follows if we show that the conditions (I) and (II) of Lemma 6 holds for k = hdg ss . Condition (I) holds by Lemma 9. To show that Condition (II) holds, by Lemma 10 it is enough to prove that for each (a, b) ∈ G 2 with a = b and a =b, there exists x ∈ G such that h(xa) = h(xb). Suppose that this is not the case, then there exists a pair (a, b) such that h(xa) = h(xb) for all x ∈ G. Without loss of generality, we may and will assume that b = 1 ∈ (Z/qZ) * , thus a = ±1. It follows that h(xa) = h(x) for all x ∈ (Z/qZ) * . Since h is not a constant function, such an a does not exists by Lemma 11 of next section. Contradiction.

Arithmetic Results
Throughout this section, G = (Z/qZ) * . For each a ∈ G, let θ a : G → G be the Lemma 11. Let h : (Z/qZ) * → R be an even function that's monotonic on [1, q/2] Z . If h • θ a = h for some a ∈ (Z/qZ) * and a = ±1, then h is a constant function.
Proof. We prove the Lemma in seven steps.
Step 1. Let ±a be the subgroup of (Z/qZ) * generated by a and −1. Clearly h • θ b = h for any b ∈ ±a since h • θ a = h and h is even. In particular, this holds true for the maximal element b max in the set in ±a ∩ [1, q/2] Z . If b max = 1, the group ±a is necessarily {±1}. Therefore, it is enough to prove that h being nonconstant implies that b max = 1. So with out lose of generality, we assume that a = b max throughout the rest of the proof. Notice that if a = 1, then 2a 2 > q, for otherwise it contradicts the maximality of a.
Every even function on (Z/qZ) * is constant if q is 2 or 4 so we assume that q = 2 r ≥ 8. The group (Z/2 r Z) * is isomorphic to Z/2Z × Z/2 r−2 Z, where the factor Z/2Z is generated by −1. Let us assume that ±a has order 2 s . Since ±a ⊇ ±1 , it follows that ±a ∼ = Z/2Z×Z/2 s−1 Z. In particular, if ±a = ±1 , then Z/2 s−1 Z is nontrivial, therefore ±a contains 3 elements of order two. But there are exactly 3 elements of order two in (Z/qZ) * : −1, 2 r−1 − 1, 2 r−1 + 1. Hence ±a contains all the above elements of order 2. So a = 2 r−1 − 1 since it is the largest element in [1, q/2] Z . Therefore, Since h is monotonic on [1, q/2] Z , the above equality implies that h is constant on [1, q/2] Z and therefore a constant function.
Step 3. Let p be an odd prime. Lemma 11 holds if either a is even, or a is odd and 3a ≥ q.
It is enough to prove that if a = 1, then First, let's assume that a = 2b is even. Then Next, assume that a is odd. Then Step 4. Lemma 11 holds if p = 3.
When p is odd, (Z/p r Z) * is cyclic of order ϕ(p r ) = (p − 1)p r−1 . For p = 3, In particular, if q ≥ 9, (Z/qZ) * contains a unique subgroup of order 3 which is generated by 3 r−1 + 1. If the order of ±a is coprime to 3, then ±a is necessarily {±1}, which leads to an contradiction. If the order of ±a is divisible by 3, then q ≥ 9 and ±a contains 3 r−1 + 1. By assumption on the maximality of a we must have a ≥ 3 r−1 + 1 and hence 3a > q.
Step 5. Assume that both p and a are odd, p = 3 and 3a < q. Lemma 11 holds if 7a ≥ q.
If the order of ±a is divisible by 5, then ±a contains the unique subgroup of order 5 in (Z/5 r Z) * . In particular, 2·5 r−1 +1 ∈ ±a . It follows that a > 2·5 r−1 +1 and therefore 3a > 5 r . The Lemma holds by Step 3.
Lemma 11 is proved by combining all the above steps.