JACOBIAN VARIETIES of HURWITZ CURVES with AUTOMORPHISM GROUP PSL(2, Q)

JACOBIAN VARIETIES OF HURWITZ CURVES WITH AUTOMORPHISM GROUP PSL(2, q)

ALLISON FISCHER, MOUCHEN LIU, AND JENNIFER PAULHUS

Abstract. Curves with automorphism groups of sizes the upper bound for the particular curve’s genus are called Hurwitz curves. In many cases, the automorphism group of these curves is the projective special linear group PSL(2, q). We present a decomposition of the Jacobian varieties for all curves of this type.

1. Introduction Let X be a compact Riemann surface of genus g (henceforth called a “curve”), and G its automorphism group with identity element denoted idG. A result of Wedderburn gives the following decomposition of the group ring QG, ∼ M QG = Mni (∆i) i where Mni (∆i) denotes ni × ni matrices with coeﬃcients in a division ring ∆i. It is possible to decompose the Jacobian variety, JX, of the curve X into abelian varieties as

M ni (1) JX ∼ ((ei(JX)) i where ei are certain idempotents in End(JX) ⊗Z Q. The dimension of the factors 1 ei(JX ) as abelian varieties is 2 hχ, ϕii where hχ, ϕii denotes ith inner product of χ, a special character we define below, with ϕi, the ith irreducible Q-character, labeled according to the Wedderburn decomposition. More details about this decomposition may be found in [9]. To define χ we need to consider the covering from X to its quotient Y = X/G, a curve with genus denoted gY . Let g1, . . . , gs be the monodromy of this covering, a set of elements g1, . . . , gs ∈ G such that g1 ··· gs = idG and the set of all gi generates G. For any subgroup H of G, define the characters χH to be the trivial character of H induced to G, and 1G to be the trivial character of G. In particular, we consider cyclic subgroups generated by one element of the monodromy which we write as hgii. Note that χhidGi is the character associated to the regular representation. Then define s X (2) χ = 2 · 1G + 2(gY − 1)χhidGi + (χhidGi − χhgj i). j=1 Hence to determine the dimension of factors of JX using this technique, we must know the automorphism group of X, the Q-character table for that particular group, and the monodromy of the covering X → Y .

Date: June 17, 2014. 1 2 ALLISON FISCHER, MOUCHEN LIU, AND JENNIFER PAULHUS

The upper bound on the size of the automorphism group of a curve of genus g > 1 is given by 84(g − 1). Curves whose automorphism groups attain this bound are called Hurwitz curves and the groups themselves are called Hurwitz groups. Additionally, the monodromy for these coverings is always of type (2, 3, 7), meaning it consists of an element of order 2, an element of order 3, and an element of order 7, denoted in this paper as g2, g3, and g7, respectively. (Equivalently, a Hurwitz group is a ﬁnite, nontrivial quotient of the (2, 3, 7)-triangle group.) In these cases (2) may be simpliﬁed to give the character χ as

(3) χ = 2 · 1G + χhidGi − χhg2i − χhg3i − χhg7i. Let PSL(2, q) denote the projective special linear group with coefficients in the finite field of order q. MacBeath determines precisely for which q the group PSL(2, q) is a Hurwitz group. Theorem 1. [8] The group PSL(2, q) is a Hurwitz group if and only if 1. q = 7 or 2. q is a prime and congruent to ±1 mod 7 or 3. q = p3 for a prime p ≡ ±2 or ±3 mod 7. Note that both Case 2 and Case 3 imply that q ≡ ±1 mod 7. Case1 corresponds to a curve of genus 3, and the Jacobian is known to decompose as JX ∼ E3 where E is an elliptic curve [7]. In Case 2 when q = 13 (and g = 14), the technique above may be used to show that JX ∼ E14, again for E some elliptic curve. Case 3 includes the special case where q = 8. This corresponds to a genus 7 curve sometimes called the MacBeath curve. It has long been known that JX ∼ E7 CITE. For odd q, PSL(2, q) has a well understood and relatively small character table. Additionally, the monodromy of the coverings is not hard to find and, in fact, (3) only requires knowledge of the monodromy up to conjugation so we only need to determine the conjugacy classes of the monodromy. It turns out that, as we show below in Proposition 1 for almost all q satisfying Theorem 1, PSL(2, q) has only one conjugacy class each of elements of order 2 and 3, and three conjugacy classes of elements of order 7. This then allows us to compute (2) in all such examples and prove very general results about the Jacobian decomposition of curves with these groups as automorphism groups. The exceptions to this are either discussed above or at the end of the paper. We begin in Section 2 by reviewing known results about PSL(2, q). In Section 3 we compute the special character χ, and in Section 4 we compute the inner products needed to determine the dimension of the factors. Finally we put the pieces together and present the Jacobian decomposition in Section 5.

Using a diﬀerent set of idempotents in QG and the fact that PSL(2, q) has a partition, (a set of subsets of G whose pairwise intersection is the identity and whose union is the whole group), Ernst Kani and Michael Rosen describe a decomposition of a power of the Jacobian variety of curves with such automorphisms [5, Example 2]. The factors are themselves Jacobians of quotients of the curve by p-Sylow subgroups of G or Cartan subgroups of G. Also, there are Hurwitz curves with diﬀerent automorphism groups besides those listed in Theorem 1. See [1] for additional information about Hurwitz curves. Future work may include studying other families of Hurwitz groups. JACOBIAN VARIETIES OF HURWITZ CURVES 3

2. Properties of PSL(2, q) Here we collect the relevant information about the group PSL(2, q). More details may be found in [6]. We follow the notation in that book. For the rest of the paper, assume q is odd and satisfies Case 2 or Case 3 in Theorem 1 and q > 27. All cases not covered by this are discussed above, except for q = 27 which we discuss in q(q+1)(q−1) Section 6. First, the size of PSL(2, q) is 2 . We start by defining several special elements of SL(2, q). Let α be a generator of the group of units of the finite field with q2 elements, let β = αq+1 and define b as q−1 the element of SL(2, q) determined by the map x → α x for x ∈ Fq2 . Additionally let β 0 1 0 1 0 a = , c = , and d = . 0 β−1 1 1 β 1 The image of the elements a, b, c, and d in the quotient PSL(2, q) are denoted ¯ ¯ q−1 ¯ q+1 asa ¯, b,c ¯, and d. The elementa ¯ has order 2 while the element b has order 2 and the elementsc ¯ and d¯ each have order q.

2.1. Conjugacy Classes. To determine the monodromy of the covering, we need to understand the conjugacy classes of elements of order 2, 3, and 7. The representatives of the conjugacy classes of PSL(2, q) are 1,¯ c ¯, d¯,a ¯n, and ¯bm q−1 q−3 q+1 where 1 ≤ n, m ≤ 4 if q ≡ 1 mod 4 while 1 ≤ n ≤ 4 and 1 ≤ m ≤ 4 if q ≡ −1 mod 4. We will write the conjugacy class of an element h ∈ G as [h]. Conjugacy classes with a representativea ¯n have size q(q + 1), and conjugacy classes with a representative ¯bm have size q(q − 1), with the exception of the conjugacy class containing elements of order 2 which has order half that [6]. It turns out, as we will see in Section 3, that χ as deﬁned in (3) is 0 outside of the the conjugacy classes of elements of order 1, 2, 3, and 7. So it will be suﬃcient to just study these conjugacy classes of PSL(2, q) since any other conjugacy class will not contribute to our goal of computing the inner product of χ with the irreducible Q-characters. First we determine how many such conjugacy classes there are. Proposition 1. When G =PSL(2, q) for the q we consider, then G has three dis- tinct conjugacy classes of elements of order 7, and one each of elements of order 2 and 3. Proof. When q ≡ ±1 mod 4 and q > 27, then by considering orders of elements in the conjugacy classes, as described above, the elements of order 7 can only lie in conjugacy classes represented by some power ofa ¯ or ¯b. (For q = 7 this need not be true asc ¯ and d¯ both have order q = 7.) Recall for any group G, the order of gk for any g ∈ G and positive integer k is k o(g) ¯ o(g ) = gcd(k,o(g)) . Thus, 7 must divide the order ofa ¯ or the order of b but not q+1 q−1 both, else it divides 2 − 2 = 1. Thus the conjugacy class(es) of order 7 are either represented by some power(s) ofa ¯ or some power(s) of ¯b. First consider the case where q ≡ 1 mod 4. Suppose that the conjugacy classes of order 7 are represented by powers ofa ¯ (this means that q ≡ 1 mod 7). The number q−1 o(¯a) 2 of conjugacy classes will be the number of i such that 7 = gcd(o(¯a),i) = q−1 , gcd( 2 ,i) q−1 q−1 where 1 ≤ i ≤ 4 . Since 7 divides the order ofa ¯, we let o(¯a) = 2 = 7j for some positive integer j. We can rephrase our question to ask how many possible 7j integers i are there that satisfy gcd(7j, i) = j and 1 ≤ i ≤ 2 . Obviously, there are 4 ALLISON FISCHER, MOUCHEN LIU, AND JENNIFER PAULHUS

q−1 q−1 3(q−1) always three of them since q > 13: j = 14 , 2j = 7 , and 3j = 14 . Hence q−1 q−1 the elements of order 7 are in the conjugacy classes represented bya ¯ 14 ,a ¯ 7 , and 3(q−1) a¯ 14 . A similar argument works if these classes are represented by powers of ¯b (or q ≡ −1 mod 7). The elements of order 7 are in the conjugacy classes represented q+1 q+1 3(q+1) by ¯b 14 , ¯b 7 , and ¯b 14 . Now, when q ≡ −1 mod 4, the argument is identical except the bounds on i q−3 q−1 change to 1 ≤ i ≤ 4 if q ≡ 1 mod 7 and 1 ≤ i ≤ 4 if q ≡ −1 mod 7. This does not change the rest of the argument and so there are three conjugacy classes q−1 q−1 3(q−1) of elements of order 7, again deﬁned asa ¯ 14 ,a ¯ 7 , anda ¯ 14 if q ≡ 1 mod 7 or q+1 q+1 3(q+1) ¯b 14 , ¯b 7 , and ¯b 14 if q ≡ −1 mod 7. The argument for elements of order 2 or 3 follows similarly. When q ≡ 1 mod 4, q−1 the elements of order 2 are in the conjugacy class [¯a 4 ], when q ≡ −1 mod 4, the q+1 elements of order 2 are in the conjugacy class [¯b 4 ]. For 3 the conjugacy class is q−1 q+1 [¯a 6 ] if q ≡ 1 mod 3 and [¯b 6 ] if q ≡ −1 mod 3. (If q = 27 there are two conjugacy classes of elements of order 3. See Section 6 for this special case.) 2.2. Character Tables. Let ε be a primitive (q − 1)-th root of unity and let 2kn −2kn δ be a primitive (q + 1)-th root of unity where εkn = ε + ε and δtm = −(δ2tm + δ−2tm). When q ≡ 1 mod 4, the character table of G =PSL(2, q) is [6, Theorem 8.9]

[1]¯ [¯an] [¯bm] [¯c] [d¯]

1G 1 1 1 1 1 λ q 1 −1 0 0 µk q + 1 εkn 0 1 1 θ q − 1 0 δ −1 −1 t tm √ √ χ q+1 (−1)n 0 1+ q 1− q 1 2 2√ 2√ q+1 n 1− q 1+ q χ2 2 (−1) 0 2 2

q−1 q−5 for 1 ≤ m, n, t ≤ 4 and 1 ≤ k ≤ 4 .

When q ≡ −1 mod 4, the character table of G =PSL(2, q) is then [6, Theorem 8.11] [1]¯ [¯an] [¯bm] [¯c] [d¯]

1G 1 1 1 1 1 λ q 1 −1 0 0 µk q + 1 εkn 0 1 1 θ q − 1 0 δ −1 −1 t tm √ √ γ q−1 0 (−1)m+1 −1+ −q −1− −q 1 2 2√ 2√ q−1 m+1 −1− −q −1+ −q γ2 2 0 (−1) 2 2 q−3 q+1 where 1 ≤ n, k, t ≤ 4 and 1 ≤ m ≤ 4 .

2.3. Irreducible Q-characters. The character tables above give the irreducible C-characters of PSL(2, q) but we need Q-characters to compute the dimension of the factors of the Jacobian decompositions. Since all irreducible C-characters of JACOBIAN VARIETIES OF HURWITZ CURVES 5

PSL(2, q) have Schur index 1 [4], it is suﬃcient to ﬁnd the Galois conjugates of all C-characters which are not already Q-characters themselves. The characters 1G and λ are already Q-characters, and it is clear that χ1 + χ2 and γ1 +γ2 are Q-characters as their non-integer entries are Galois conjugates. This leaves the µk and θt characters.

q−1 Theorem 2. (a) Let r be a divisor of 2 and deﬁne the set

( q−5 q−1 {µi | 1 ≤ i ≤ 4 and gcd(i, 2 ) = r} if q ≡ 1 mod 4, Sr = q−3 q−1 {µi | 1 ≤ i ≤ 4 and gcd(i, 2 ) = r} if q ≡ −1 mod 4.

The sum of the characters in each Sr is a irreducible Q-character of PSL(2, q). q+1 (b) Let s be a divisor of 2 and deﬁne the set

( q−1 q−1 {θi | 1 ≤ i ≤ 4 and gcd(i, 2 ) = s} if q ≡ 1 mod 4, Ts = q−3 q−1 {θi | 1 ≤ i ≤ 4 and gcd(i, 2 ) = s} if q ≡ −1 mod 4.

The sum of the characters in each Ts is an irreducible Q-character of PSL(2, q). Proof. We prove (a) below. The argument for (b) is almost identical. Since the n only non-rational values of the µi characters are their values on the [¯a ], we only need to consider the values on these conjugacy classes. For simplicity of notation, 2 q−1 we deﬁne ρ to be ε , so ρ is a primitive 2 -th root of unity. Then the values of n the µi on the conjugacy classes [¯a ] are given in Table 1.

n Table 1. Values of µk on conjugacy classes of elementsa ¯

q−1 [¯a] [¯a2] ... [¯a 4 ] q−1 q−1 −1 2 −2 − µ1 ρ + ρ ρ + ρ ... ρ 4 + ρ 4 q−1 2 −2 4 −4 − q−1 µ2 ρ + ρ ρ + ρ ... ρ 2 + ρ 2 ...... q−5 q−5 q−5 q−5 (q−5)(q−1) (q−5)(q−1) − − − µ q−5 ρ 4 + ρ 4 ρ 2 + ρ 2 ... ρ 16 + ρ 16 4

q−1 Fix a particular µk with gcd(k, 2 ) = r. It is enough to ﬁnd the Galois conjugates of µk([¯a]) since the values of µk on the other conjugacy classes with representatives powers ofq ¯ are powers of this entry (as seen in Table 1). Hence the Galois orbit is completely determined by µk[¯a]. k −k k q−1 Now µk([¯a]) = ρ +ρ where ρ is a primitive 2r -th root of unity. The Galois q−1 conjugates of this will be sums of the other primitive 2r -th roots of unity. By q−2 i 2 a simple order argument, we determine that ρ is a primitive q−1 -th root of gcd( 2 ,i) q−1 unity. So the other primitive 2r -th roots of unity appear for exactly those µi q−1 such that gcd(i, 2 ) = r. Hence the irreducible Q-character associated with µk q−1 q−1 will be the sum of µk with the other characters µi such that (i, 2 ) = (k, 2 ). 6 ALLISON FISCHER, MOUCHEN LIU, AND JENNIFER PAULHUS

Example. We demonstrate this previous theorem with an example. Consider q−1 q+1 q−5 q−1 q = 29 ≡ 1 mod 4. Here 2 = 14, 2 = 15, 4 = 6, and 4 = 7. Hence there are 6 µk characters and 7 θt characters. q−1 The only divisors of 2 less than 6 are 1 and 2. Hence we have two sets:

S1 = {µi | gcd(i, 14) = 1} = {µ1, µ3, µ5} and S2 = {θi | gcd(i, 14) = 2} = {µ2, µ4, µ6}.

q+1 The divisors of 2 less than 7 are 1, 3, and 5 so here we have three sets:

T1 = {θi | gcd(i, 15) = 1} = {θ1, θ2, θ4, θ7} and T3 = {θi | gcd(i, 15) = 3} = {θ3, θ6} and T5 = {θi | gcd(i, 15) = 5} = {θ5}. So there are two irreducible Q-characters of degree q + 1 (µ1 + µ3 + µ5 and µ2 + µ4 + µ6) and three irreducible Q-characters of degree q − 1 (θ1 + θ2 + θ4 + θ7, θ3 + θ6, and θ5).

But then what are the values of the irreducible Q-characters which are sums of the elements in Sr? To determine the values of these Q-characters, we need to q−1 know the size Sr. This is half the number of i such that gcd(i, 2 ) = r, or half q−1 q−1 φ( 2r ) the number of i such that gcd(i, 2r ) = 1. We get 2 where φ(x) is the Euler q+1 φ( 2s ) phi function. Similarly, for Ts we get 2 . For our computations, we will only need the values of the characters on conjugacy classes of order 1, 2, 3, or 7, as it turns out that χ is 0 outside these conjugacy classes. This means the inner product we use to compute the dimension of the factors of the Jacobian will not be impacted by the values outside of these conjugacy classes. Again, see Section 3 and (6). Let us ﬁx one particular k and consider its values. Each case boils down to whether elements of that order are powers ofa ¯ or ¯b. When q ≡ 1 mod 4, the conjugacy class of elements of order 2 is represented by the power ofa ¯ and when q ≡ −1 mod 4, the conjugacy class of elements of order 2 is represented by the power of ¯b. Similarly, when q ≡ 1 mod 3 (or q ≡ 1 mod 7) , the conjugacy class of elements of order 3 (or 7) is represented by the power ofa ¯ and is represented by a power of ¯b when q ≡ −1 mod 3 (or q ≡ −1 mod 7). We will later distinguish cases modulo 84 which will determine precisely which of 2, 3, and 7 reside in conjugacy classes of powers ofa ¯ and which reside in conjugacy classes of powers of ¯b. q−1 q+1 For these cases, let r = gcd(k, 2 ) and let s = gcd(t, 2 ) Conjugacy class of elements of order 2. As we saw above, the conjugacy class of elements of order 2 is represented by a power of eithera ¯ or ¯b, depending on q−1 whether q ≡ ±1 mod 4. In the ﬁrst case, it is [¯a 4 ]. Consider the value of one µk q−1 q−1 k − k on this conjugacy class, ε q−1 = ε 2 + ε 2 . Since ε is a primitive (q − 1)th 4 k q−1 root of unity, then ε 2 is a primitive second root of unity, which is −1. Thus, k −k ε q−1 = (−1) + (−1) . When k is odd, this value is −2, and when k is even, 4 k JACOBIAN VARIETIES OF HURWITZ CURVES 7 this value is 2. Combining this value with the number of µi added together to get a particular irreducible Q-character (this is the size of Sr computed above) yields q − 1 (−1)kφ . 2r

All the Q-characters which as sums of the θt are 0 on this class.

When q ≡ −1 mod 4, the conjugacy class is represented by a power of ¯b and so the Q-characters which are sums of the µk are 0 on that class while a similar argument as above gives that each θt will be 2 when t is odd and −2 when t is even. Then the irreducible Q-character of which θt is a sum will be this value multiplied by the number of θt added together to get one irreducible Q-character (again see the computation for the size of Ts above). This gives q + 1 (−1)t+1φ . 2s Conjugacy class of elements of order 3. As was discussed in the proof of Proposition q−1 q+1 1, the conjugacy class of order 3 is represented bya ¯ 6 or ¯b 6 . Then, consider q−1 q−1 k − k the value of µk, ε q−1 = ε 3 + ε 3 . Since ε is a primitive (q − 1)-th root of 6 k q−1 k −k unity, then ε 3 is a third root of unity, call it ω. Thus, ε q−1 = ω + ω . When 4 k 2 3 | k, this is 2 and when 3 k, this is ε q−1 = ω + ω = −1. Putting this together - 4 k gives when q ≡ 1 mod 3 the irreducible Q-characters which are sums of θt’s are zero but the irreducible Q character of which µk is a sum evaluates to ( q−1 φ 2r if k ≡ 0 mod 3, q−1 φ( 2r ) − 2 otherwise. A similar argument may be used when q ≡ −1 mod 3 (or the elements of order ¯ q+1 3 are in the conjugacy class represented by b 6 to get that the µk are zero but irreducible Q-character of which θt is a sum evaluates to ( q+1 −φ 2s if t ≡ 0 mod 3, q+1 φ( 2s ) 2 otherwise. Conjugacy classes of elements of order 7. From Proposition 1 we know that the q−1 q−1 3(q−1) three conjugacy classes of order 7 are represented bya ¯ 14 ,a ¯ 7 , anda ¯ 14 or q+1 q+1 3(q+1) ¯b 14 , ¯b 7 , and ¯b 14 . If q ≡ 1 mod 7 (equivalently the conjugacy classes of elements of order 7 are k −k represented by a power ofa ¯) then µk = ζ + ζ where ζ is a primitive 7th root of unity. If 7 | k, then the value is 2. If 7 - k, the value will be −1. Note that on q−1 3(q−1) the conjugacy class represented bya ¯ 7 ora ¯ 14 the values will be the same. So together we get when q ≡ 1 mod 7 the θt are zero but the irreducible Q character of which µk is a sum evaluates to ( q−1 φ 2r if k ≡ 0 mod 7, q−1 φ( 2r ) − 2 otherwise. 8 ALLISON FISCHER, MOUCHEN LIU, AND JENNIFER PAULHUS

And when q ≡ −1 mod 7 the µk are zero but the irreducible Q-character of which θt is a sum evaluates to

( q+1 −φ 2s if t ≡ 0 mod 7, q+1 φ( 2s ) 2 otherwise.

3. Computation of the Hurwitz Character

Recall from (3) that in order to compute χ, we need to determine χhidGi, χhg2i,

χhg3i and χhg7i. Let H be a subgroup of G. By the deﬁnition of χH , the induced character of the trivial character of H is

( 1 X o −1 o 1 if g ∈ H, (4) χH (g) = χ (xgx ), where χ (g) = H 0 if g∈ / H. x∈G To compute these characters, we need two lemmas which are immediate conse- quences of basic algebra facts.

Lemma 1. Let G be a group and g ∈ G, not the identity. There is no x ∈ G so −1 that xgx = idG.

Lemma 2. Let G be a group and g, h ∈ G with g not the identity. The number of x ∈ G such that xgx−1 = h is the size of the centralizer of h if g ∈ [h] and 0 otherwise.

The second Lemma follows from the Orbit-Stabilizer theorem considering the group action of conjugation.

Returning to the characters, note that χhidGi is just the regular representation, ( |G| if g = idG, χhidGi(g) = 0 if g 6= idG.

Next, consider χhg2i. We know

1 X o −1 (5) χhg2i(g) = 2 · χ (xgx ). x∈G

−1 We need to ﬁgure out the number of x ∈ G for all g ∈ G such that xgx = idG −1 or g2, since hg2i = {idG, g2}. The case of xgx = idG is covered by Lemma

1. It is the size of G when g is the identity and 0 otherwise. For χhg2i(g) when g 6= idG, we calculate the size of the centralizer of g2 by Lemma 2. Recall that under the action of conjugacy, orbits are conjugacy classes. By the Orbit-Stabilizer |G| q(q+1) Theorem, |CG(g2)| = . For g2 of order 2, |[g2]| = when q ≡ 1 mod 4 |[g2]| 2 q(q−1) and |[g2]| = 2 when q ≡ −1 mod 4 hence |CG(g2)| = q − 1 if q ≡ 1 mod 4 and |CG(g2)| = q + 1 if q ≡ −1 mod 4. Therefore, plugging these values into (5) gives JACOBIAN VARIETIES OF HURWITZ CURVES 9

 |G| if g = id ,  2 G  q−1  2 if g ∈ [g2] and q ≡ 1 mod 4, χhg2i(g) = q+1  2 if g ∈ [g2] and q ≡ −1 mod 4,  0 otherwise.

Now, we calculate χhg3i. As before, we need to ﬁnd the number of x ∈ G for any −1 2 g ∈ G that xgx ∈ hg3i = {1, g3, g3}, and the formula in this case is

1 X o −1 χhg3i(g) = 3 · χ (xgx ). x∈G

|G| When g = idG, then χhg3i(idG) = 3 . Otherwise by Lemma 2 and the fact that 2 2|CG(g3)| q+1 g3 ∈ [g3], χhg3i(g) = 3 . Since from Section 2.1 |[g3]| = q(q − 1) if 3| 2 and q−1 q+1 q+1 q−1 |[g3]| = q(q + 1) if 3| 2 we know |CG(g3)| = 2 if 3| 2 and |CG(g3)| = 2 if q−1 3| 2 . Thus,  |G| if g = id ,  3 G  q−1  3 if g ∈ [g2] and q ≡ 1 mod 3, χhg3i(g) = q+1  3 if g ∈ [g2] and q ≡ 2 mod 3,  0 otherwise.

|G| For χhg7i as with the proof of the other characters, χhg7i(idG) = 7 . To compute the value on other elements, observe that for any g of order 7, g and g−1 are in the same conjugacy class [6, Corollary 8.3] but g, g2, and g3 are all in dis- tinct conjugacy classes. Combining Lemma 2 with this information gives us that 2|CG(g7)| χhg7i(g) = 7 . We know the sizes of the conjugacy classes by Section 2.1.

Putting all this information together we get that the value of χhg7i(g) is

 |G| if g = id ,  7 G  q−1  7 if g ∈ [g7] and q ≡ 1 mod 7, χhg7i(g) = q+1  7 if g ∈ [g7] and q ≡ −1 mod 7,  0 otherwise. Note that this description shows that the values of χ are invariant under the three conjugacy classes of elements of order 7. This means we do not have to ﬁnd in which conjugacy class of elements of order 7 the monodromy exists in order to compute (3).

We want to use χ to calculate inner products with irreducible Q-characters to ﬁnd the dimension of the factors in the Jacobian variety decomposition. To simplify later 0 0 calculations, we rewrite χ as χ = 2·1G +χ , where χ = χh1Gi −χhg2i −χhg3i −χhg7i. Then, the inner product of χ and an irreducible Q-character ϕi will be hχ, ϕii = 0 2 · h1G, ϕii + hχ , ϕii. But since ϕi and 1G are orthogonal when ϕi 6= 1G, hχ, ϕii is 0 simply hχ , ϕii in all cases except for the trivial character. Below is a table for the value of χ0 on the conjugacy classes of elements of order 1, 2, 3, or 7, computed by combining all the data in this section. Additionally, χ0(g) = 0 if g is not in one of these conjugacy classes. 10 ALLISON FISCHER, MOUCHEN LIU, AND JENNIFER PAULHUS

Table 2. Values of χ0 on conjugacy classes of elements of order 1, 2, 3, and 7

q Value for elements of order q Value for elements of order mod84 1 2 3 7 mod84 1 2 3 7 |G| q−1 q−1 q−1 |G| q+1 q+1 q+1 1 42 − 2 − 3 − 7 −1 42 − 2 − 3 − 7 |G| q−1 q−1 q+1 |G| q+1 q+1 q−1 13 42 − 2 − 3 − 7 −13 42 − 2 − 3 − 7 |G| q−1 q+1 q−1 |G| q+1 q−1 q+1 29 42 − 2 − 3 − 7 −29 42 − 2 − 3 − 7 |G| q+1 q−1 q−1 |G| q−1 q+1 q+1 43 42 − 2 − 3 − 7 −43 42 − 2 − 3 − 7

4. Inner Product Computations Our next goal is to use our computation of χ0 in Section 3 and the irreducible Q- 0 0 characters in Section 2.3 to compute the inner products hχ , ϕii. Consider hχ , ϕii, where ϕi is an irreducible Q-character of PSL(2, q). The formula of the inner 0 1 P 0 −1 −1 product is hχ , ϕii = |G| g∈G χ (g)ϕi(g ). Since g and g are in the same conjugacy class and χ0 is 0 for all elements that are not order 1, 2, 3, or 7 , we have the formula

0 1 0 0 0 0 hχ , ϕii = |G| χ (idG)ϕi(idG) + |[g2]|χ (g2)ϕi(g2) + |[g3]|χ (g3)ϕi(g3) + 3|[g7]|χ (g7)ϕi(g7) .

0 |G| 0 2|G| In section 3 we saw that |[g2]| · χ (g2) = − 2 , |[g3]| · χ (g3) = − 3 and 3 · |[g7]| · 0 6|G| χ (g7) = − 7 . The formula of the inner product reduces to

0 ϕi(1) ϕi(g2) 2ϕi(g3) 6ϕi(g7) (6) hχ , ϕii = 42 − 2 − 3 − 7 .

Since the values of the irreducible Q-characters are based on whether the conjugacy classes of elements of order 2, 3 and 7 are represented bya ¯ or ¯b (which depend on what q is modulo 3, 4, or 7, the values of these characters, and the subsequent inner products, will depend on what q is modulo 3 · 4 · 7 = 84.

4.1. Trivial Character.

Theorem 3. hχ, 1Gi = 0

0 Proof. By the calculation of χ, hχ, 1Gi = 2h1G, 1Gi + hχ , 1Gi and h1G, 1Gi = 1. 0 Consider hχ , 1Gi. We use (6) to get

0 1G(1) 1G(g2) 2·1G(g3) 6·1G(g7) 1 1 2 6 hχ , 1Gi = 42 − 2 − 3 − 7 = 42 − 2 − 3 − 7 = −2.

Thus, hχ, 1Gi = 2 − 2 = 0.

All other irreducible Q-characters have degree greater than 1. Hence by (1), where the ni correspond to the degree of the ith irreducible Q-character, the decomposition of JX must have more than one factor.

Corollary 1. No Hurwitz curve with automorphism group PSL(2, q) has a simple Jacobian variety. JACOBIAN VARIETIES OF HURWITZ CURVES 11

4.2. Character of degree q. Recall λ is the character of degree q. We again apply (6). Since the value of λ is either 1 or −1 depending on whether the element is in ¯ 0 q−u the conjugacy classes represented bya ¯ or b, we get that hχ , λi = 42 where u is given in Table 3 and the positive u values correspond to positive q mod 84 and the negative u values correspond to negative q mod 84 values.

Table 3. Values for hχ0, λi

q mod 84 Value of u q mod 84 Value of u ±1 ±85 ±29 ±29 ±13 ±13 ±43 ±43

q±1 4.3. Characters of degree 2 . When q ≡ 1 mod 4, this irreducible Q-character n is χ1 + χ2 and evaluates to q + 1 on the identity, 2(−1) on the conjugacy classes [¯an], and 0 on the conjugacy classes [¯bm]. Furthermore, the conjugacy class of elements of order 2 will always be in the set of conjugacy classes [¯an]. We use (6) again which becomes. q + 1 (χ + χ )(g ) 2(χ + χ )(g ) 6(χ + χ )(g ) hχ0, χ + χ i = − 1 2 2 − 1 2 3 − 1 2 7 . 1 2 42 2 3 7 Determining these values depends on whether q ≡ ±1 mod 3 and whether q ≡ ±1 mod 7 (as always, this distinguishes the cases where the elements of order 3 or order 7 are in conjugacy classes represented by powers ofa ¯ or ¯b). But additionally we need to determine if n are even or odd to determine the sign of χ1 + χ2. Recall q−1 q−1 n is given by 6 for elements of order 3 and 14 for elements of order 7. This requires us to consider values modulo 3 · 4 · 7 · 2 = 168. Similar arguments will give us the values for γ1 + γ2 when q ≡ −1 mod 4. In all q−v cases, the inner product is given by 42 where v is given in Table 4. In the table, the positive value of q mod 168 corresponds to the positive v value and the negative value of q mod 168 corresponds to the negative v value.

0 0 Table 4. Values for hχ , χ1 + χ2i or hχ , γ1 + γ2i

q mod 168 Values of v q mod 168 Values of v ±1 ±169 ±43 mod 168 ±43 ±13 ±13 ±85 mod 168 ±85 ±29 ±29 ±97 mod 168 ±97 ±41 ±41 ±113 mod 168 ±113

4.4. Characters of degree q ± 1. The computations for the inner products of χ0 with sums of µk or θt happens similarly. Once more, using (6) we recall the values of these Q-characters on the conjugacy classes of order 1, 2, 3, and 7. In Section 2.3 we determined the values of these characters, which all depended on whether the conjugacy classes were powers ofa ¯ or ¯b. To describe the value in all cases, we need two other values. For each possible r or s value, deﬁne f to be the number of 12 ALLISON FISCHER, MOUCHEN LIU, AND JENNIFER PAULHUS

2, 3, and 7 which divide r or s. Also deﬁne z to be the least residue of q modulo 84. Then inner product with sums of the µk in Sr will be φ( q−1 ) q − w 2r · 2 84 and the inner product with sums of the θt in Ts will be φ( q+1 ) q − w 2s · 2 84 where w is given in Table 5.

Table 5. Values of w for the inner products of χ0 with characters of degree q ± 1

q ≡ 1 mod 4 q ≡ −1 mod 4

µk z + (f − 1) · 84 z − f · 84 θt z + f · 84 z − (f − 1) · 84

Example. Continuing from the example in Section 2.3, let q = 29, so z also is 29. When r = 1 or s = 1or 5 then f = 0 and when r = 2 or s = 3 we have f = 1. In this case, since q = Z, if f = 1, the value of the inner product on the corresponding µk will be zero and if f = 0 the value on the inner product of the corresponding θt will be zero. This just leaves two non-zero values to compute,

φ 28 29 + 55 6 hχ0, µ + µ + µ i = 2 · = · 1 = 3 1 3 5 2 84 2 and φ 30 29 + 55 4 hχ0, θ + θ i = 6 · = · 1 = 2. 3 6 2 84 2

5. Decomposition of Jacobian Varieties As described in the introduction, Jacobian varieties may be factored into the direct product of abelian varieties as in (1). Collecting the information in the previous section for the dimension of the factors, provides the following result.

Theorem 4. Let X be a Hurwitz curve with full automorphism group PSL(2, q) where q is odd, q > 27, and q satisﬁes one of Case 2 or Case 3 in Theorem 1. y Let Ax denote the product of y copies of an abelian variety of dimension x. Let u, v, and w be as given in Table 3, Table 4, and Table 5, respectively.

When q ≡ 1 mod 4 then the Jacobian variety of X is isogenous to

q+1 q 2 Y q+1 Y q−1 A q−u × A q−v × A q−1 × A q+1 . φ ·(q−w) φ ·(q−w) 84 84 ( 2d ) ( 2d ) q−1 168 q+1 168 d| 2 d| 2 q−5 q−1 d< 4 d< 4 JACOBIAN VARIETIES OF HURWITZ CURVES 13

When q ≡ −1 mod 4, then the Jacobian variety of X is isogenous to q−1 q 2 Y q+1 Y q−1 A q−u × A q−v × A q−1 × A q+1 . φ ·(q−w) φ ·(q−w) 84 84 ( 2d ) ( 2d ) q−1 168 q+1 168 d| 2 d| 2 q−3 q−3 d< 4 d< 4 6. Special Case In the special case when q = 27 = 33, there are still three conjugacy classes of elements of order 7 and one of elements of order 2, however there are now two conjugacy classes of elements of order 3. When we apply the decomposition technique to this special case we ﬁnd 13 26 27 JX ∼ E1 × A3 × E2 where the Ei are elliptic curves and A3 is a dimension 3 abelian variety. These factors correspond to non-zero inner products of χ with the character γ1 + γ2, a sum of θt, and λ, respectively.

Acknowledgments The authors wish to thank Grinnell College for their generous summer funding through the Mentored Advanced Project program for undergraduate students.

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Department of Mathematics and Statistics, Grinnell College, Grinnell, IA 50112 E-mail address: [email protected]