Elliptic Subcovers of a Curve of Genus 2 II. The Refined Humbert Invariant

Ernst Kani

1 Introduction

This paper is a continuation of the article[?] whose aim is to study of the set E(C) of all equivalence classes of elliptic subcovers of a given genus 2 curve C/K, where K is an arbitrary field. By what was explained in [?], there is a natural bijection between the set E(C) and the set of primitive representations of squares by a certain quadratic form q_{C} called the refined Humbert invariant; cf. [?], Theorem 20.

The main objective of this paper is to present a method for computing q_{C}. As in [?], this method depends on the knowledge of a presentation (E, E′, ψ) of degree N of C/K (which arises from a given elliptic subcover f : C → E of degree N). This triple consists of two elliptic curves E/K and E′/K and an isomorphism ψ : E_{1}[N] → E_{2}[N] which is an anti-isometry with respect to the e_{N}-pairings; cf. [?], Section 2.

As was shown [?], the discriminant of the quadratic form q_{C} can be determined from the N-presentation (E, E, ψ); cf. [?], Theorem 3. Moreover, in the case that rank(Hom(E, E′)) = 1, then q_{C} has extra structure which is encapsulated in the following definition.

Definition. Let N, m, d be positive integers. An integral binary quadratic form q is said to be of type (N, m, d) if it is positive-definite and satisfies the following properties:

(i) disc(q_{C}) = −16m^{2}d, and m|N with gcd(N/m, d) = 1.

(ii) q_{C} primitively represents N^{2};

(iii) q_{C}(X, Y) ≡ 0, 1 (mod 4), for all X, Y ∈ ℤ.

Such quadratic forms are studied in some detail in Section ??. They are important in the study of q_{C} due to the following result which is Theorem 2 of [?].

Theorem 1 Suppose that C/K is a curve which has a presentation (E, E′, ψ) of degree N with char(K) ∤ N. If Hom(E, E′) = ℤh, where d := deg(h) ≥ 1, then the refined Humbert invariant q_{C} is a binary quadratic form of type (N, m, d), where m = m_{ψ} is the isogeny defect of ψ which is defined by

m_{ψ} := min{m ≥ 1 : mψ = h_{|E[N]}, for some h ∈ Hom(E, E′)}.

In addition, q_{C} satisfies the condition (iv) q_{C}(X, Y) ≠ 1, for all X, Y ∈ ℤ.

Here we make this theorem more precise by giving an explicit formula for m_{ψ} and for q_{C}. More precisely, Proposition ?? and Remark ?? imply the following result.

Theorem 2 In the situation of Theorem ??, there exists a primitive matrix M of determinant −d such that M (mod N) is the matrix of ψ^{−1}h_{|E[N]} ∈ End(E[N]) with respect to some basis of E[N](\overline{K}). For any such matrix M = \binom{x y}{z w} we have

m_{ψ} = m := N/N̄, where N̄ := gcd(x − w, y, z, N),

and the refined Humbert invariant q_{C} is equivalent to the integral binary quadratic form

q_{M,N}(X, Y) := N^{2}X^{2} − 2m tr(M)XY + \frac{m^{2}}{N^{2}} (tr(M)^{2} + 4d)Y^{2}.

Now it turns out that every binary form of type (N, m, d) is equivalent to a form q_{M,N}, for a suitable primitive matrix M of determinant −d; cf. Theorem ??. By combining this result with Theorem ??, we obtain as a consequence that every such form of type (N, m, d) satisfying property (iv) is (equivalent to) the refined Humbert invariant q_{C} of some curve C/K.

Theorem 3 If K is an algebraically closed field of characteristic 0, then for each positive binary quadratic form q of type (N, m, d) which satisfies properties (iv) of Theorem ?? there exists a curve C/K such that its refined Humbert invariant q_{C} is equivalent to q.

This theorem is actually a special case of the more general existence results of Section ??; cf. Theorem ?? and its corollaries.

In the last section we apply the above theory to study curves C/K of genus 2 which have an elliptic subcover of degree 2. This is equivalent to the condition that C has an elliptic involution; cf. Proposition ??. Using the above results, we prove the following characterization of curves whose automorphism group contains a certain dihedral group D_{n} of order 2n.

Theorem 4 Let C/K be a curve of genus 2 where char(K) ≠ 2, 3. Then:

(a) Aut(C) contains a subgroup isomorphic to the dihedral group D_{4} (respectively, to D_{6}) if and only if the refined Humbert invariant q_{C} primitively represents the binary quadratic form q_{4}(X, Y) = 4X^{2} + 4Y^{2} (respectively, the form q_{6}(X, Y) = 4X^{2} + 4XY + 4Y^{2}).

(b) If the conditions of part (a) hold, then C/K has an elliptic subcover of degree N whenever N = 2N_{1}, where N_{1} = 1 of N_{1} is a product of primes p_{i} with p_{i} ≡ 1 (mod 4) (respectively, with p_{i} ≡ 1 (mod 3)).

(c) If the conditions of part (a) hold, and if also End(E) = ℤ, for some elliptic subcover f : C → E, then q_{C} is equivalent to q_{4} (respectively, to q_{6}). Then C/K has an elliptic subcover of degree N if and only if N = 2N_{1}, where N_{1} = 1 or N_{1} is a product of primes p_{i} with p_{i} ≡ 1 (mod 4) (respectively, with p_{i} ≡ 1 (mod 3)).

Moreover, if N is of this form, then C/K has precisely 2^{ω(N)+1} (respectively, 2^{ω(N)+1} · 3) such subcovers of degree N, where ω(N) denotes the number of distinct prime factors of N.

Some examples of curves C/K which satisfy the hypotheses of Theorem ??(c) are the Legendre curves

y^{2} = x(X^{2} − a)(x^{2} − a^{−1})

with a ∈ ℤ, a ≠ 0 ± 1, ±3 and K ⊃ ℚ(i), as is explained in Example ??. Thus, for N ≤ 150, these curves have an elliptic subcover of degree N if and only if N = 2, 10, 26, 34, 50, 58, 74, 82, 106, 122, 130, or 146. We also investigate the case a = 3 in some detail; cf. Proposition ??.

This paper is organized as follows. In Section ?? we study the basic properties of binary quadratic forms of type (N, m, d). For convenience of the reader, we recall in Section ?? the basic notation and results of [?], which is then used in Section ?? to prove Theorem ?? and other results. In Section ?? we prove the existence theorems and in the last section we apply Theorem ?? to study elliptic involutions and to compute some explicit examples; cf. Theorem ?? and Example ??.

Acknowledgment. This research was supported by a Discovery Grant from the Natural Sciences and Engineering Research Council of Canada (NSERC).

2 Quadratic forms of type (N, m, d)

In this section we analyze the binary quadratic forms which occur in the statement of Theorem ??, i.e., those binary forms q which are positive-definite and satisfy properties (i), (ii) and (iii) of the Introduction. Here, as usual, we use the abbreviation q = [a, b, c] to denote the binary quadratic form q(X, Y) = aX^{2} + bXY + cY^{2}. We begin with the following simple observation.

Proposition 5 Let q = [a, b, c] be an integral binary quadratic form. Then q satisfies property (iii) if and only there exist ā, c̄ ∈ {0, 1} such that

(1)  q(X, Y) ≡ (āX + c̄Y)^{2} (mod 4), for all X, Y ∈ ℤ.

Proof. If q satisfies (??), then clearly q(X, Y) ≡ 0, 1 (mod 4), for all X, Y ∈ ℤ, so condition (iii) holds.

Conversely, suppose that q = [a, b, c] satisfies condition (iii). Then a = q(1, 0) ≡ ā ≡ ā^{2} (mod 4) with ā ∈ {0, 1} and similarly c = q(0, 1) ≡ c̄ ≡ c̄^{2} with c̄ ∈ {0, 1}. Moreover, a ± b + c = q(1, ±1) ≡ 0, 1 (mod 4).

If ā = c̄ = 0, then the last two conditions imply that b ≡ 0 (mod 4), so (??) holds in this case. If ā = c̄ = 1, then we must have b ≡ 2 (mod 4), so again (??) holds. Finally, if ā = 1 and c̄ = 0 (or vice versa), then b ≡ 0 (mod 4), and so (??) holds in all cases. □

Corollary 6 If q satisfies property (iii), then 16|disc(q).

Proof. By the proposition, we have 4 cases to consider. If ā = c̄ = 0, then q ≡ [0, 0, 0] (mod 4), so q = [4a, 4b, 4c] with a, b, c ∈ ℤ and hence disc(q) = 16(b^{2} − 4ac). Next, if ā = 1 and c̄ = 0, then q = [1 + 4a, 4b, 4c] with a, b, c ∈ ℤ and so disc(q) = 16(b^{2}−a−ac). Similarly, if ā = 0 and c̄ = 1, then q = [4a, 4b, 1+4c] with a, b, c ∈ ℤ and so disc(q) = 16(b^{2} −c−ac). Finally, if ā = 1 and c̄ = 1, then q = [1+4a, 2+4b, 1+4c] with a, b, c ∈ ℤ and so disc(q) = (2+4b)^{2}−4(1+4a)(1+4c) = 16(b^{2}+b−a−c−4ac). □

We next observe that property (i) is a consequence of properties (ii) and (iii). More precisely:

Proposition 7 Let q be a positive binary quadratic form which primitively represents N^{2} for some N ≥ 1. If q satisfies property (iii) of Theorem ??, then there exist unique integers m, d ≥ 1 with m|N such that

(2)  disc(q) = −16m^{2}d, and gcd(N/m, d) = 1.

Proof. We first observe that m and d are uniquely determined by the condition (??) (and by N) because (??) implies that gcd((N/m)^{2}, disc(q)/(16m^{2})) = 1 and that hence gcd(N^{2}, disc(q)/16) = m^{2}.

To prove that there exist m and d satisfying (??), it suffices to verify that M := gcd(N^{2}, disc(q)/16) is a square, for then m = \sqrt{M} and d = −disc(q)/16m^{2} satisfy (??). (Note that disc(q)/16 ∈ ℤ by Corollary ?? and that disc(q) < 0 because q is positive-definite.)

For this, we observe that the first hypothesis implies that q is equivalent to the form q′ = [N^{2}, B, c], where b, c ∈ Z. Since 16|disc(q′) = disc(q) by Corollary ??, we see that B = 2b is even. Thus, M = gcd(N^{2}, (b^{2} − N^{2}c)/4).

If N is odd, then M = gcd(N^{2}, b^{2} − N^{2}c) = gcd(N, b)^{2} is a square. Thus, assume that N is even. Since q satisfies property (iii), we must have that B = 4b_{1} by Proposition ??. If 2|c, then c = 4c_{1} by Proposition ?? and so M = gcd(N^{2}, b_{1}^{2} − N^{2}c_{1}) = gcd(N, b_{1})^{2}. Thus, assume that c is odd, so c ≡ 1 (mod 4) by (iii).

Write N = 2^{r}N_{0} and b = 2^{s}b_{0}, where N_{0}b_{0} is odd. If r = s, then 2^{−2r}(b^{2} − N^{2}c) = b_{0}^{2} − N_{0}^{2}c ≡ 1 − 1 · 1 ≡ 0 (mod 4). Thus b^{2} − N^{2}c = 2^{2r+2}c_{0}, where c_{0} ∈ ℤ, and so M = 2^{2r} gcd(N_{0}^{2}, c_{0}) = 2^{2r} gcd(N_{0}, b_{0}^{2} − N_{0}^{2}c) = gcd(N, b)^{2}. If r < s, then b^{2} − N^{2}c = 2^{2r}d_{1}, where d_{1} = 2^{2s−2r}b_{0}^{2} − N_{0}^{2}c is odd, and so M = 2^{2r−2} gcd(4N_{0}^{2}, d_{1}) = 2^{2r−2} gcd(N_{0}^{2}, b_{0}^{2}) = \frac{1}{4} gcd(N, b)^{2}. Finally, if r > s, then b^{2} − N^{2}c = 2^{2s}d_{2}, where d_{2} = b_{0}^{2} − 2^{2r−2s}N_{0}^{2}c is odd, and so M = 2^{2s−2} gcd(2^{2r−2s−2}N_{0}^{2}, d_{2}) = 2^{2s−2} gcd(N_{0}^{2}, b_{0}^{2}) = \frac{1}{4} gcd(N, b)^{2}. Thus, M is a square in all cases, as claimed. □

We next study the existence of binary forms of type (N, m, d). For this, let

Q(N, m, d) := {[a, b, c] ∈ ℤ^{3} : q = [a, b, c] is a form of type (N, m, d)}.

Proposition 8 If N, m, d are positive integers with m|N and gcd(N/m, d) = 1, then

(3)  Q(N, m, d) ≠ ∅ ⇔ −d ≡ x^{2} (mod N/m), for some x ∈ ℤ.

As we shall see below, this result follows from the following more precise assertions.

Lemma 9 If q ∈ Q(N, m, d) is a quadratic form of type (N, m, d), then q is SL_{2}(ℤ)- equivalent to q′ = [N^{2}, 2mb, c], for some b, c ∈ ℤ where c = \frac{m^{2}}{N^{2}} (b^{2} + 4d). Moreover, if b_{1} ≡ b (mod \frac{N^{2}}{m}), and c_{1} = \frac{m^{2}}{N^{2}} (b_{1}^{2} + 4d), then [N^{2}, 2mb_{1}, c_{1}] is SL_{2}(ℤ)-equivalent to q′.

Proof. It follows from condition (ii) that q is SL_{2}(ℤ)-equivalent to q′ = [N^{2}, B, c], where B, c ∈ ℤ. Put N̄ = \frac{N}{m}. Since B^{2} − 4m^{2}N̄^{2}c = disc(q′) = −16dm^{2} by condition (i), we see that 4m^{2}|B^{2}, so B = 2mb, for some b ∈ ℤ, and hence q′ = [N^{2}, 2mb, c]. Moreover, from above we have that c = \frac{B^{2}+16dm^{2}}{4N^{2}} = \frac{m^{2}}{N^{2}} (b^{2} + 4d), as claimed. Finally, if b′ = b + s\frac{N^{2}}{m}, then the matrix M = \binom{1 s}{0 1} ∈ SL_{2}(ℤ) transforms q′ into [N^{2}, 2mb_{1}, c_{1}].

□

Lemma 10 In the situation of Proposition ??, suppose that k ∈ ℤ satisfies k^{2}d ≡ −1 (mod N̄), where N̄ = \frac{N}{m}. Then n := (k^{2}d+1)/N̄ ∈ ℤ, and for any r ∈ ℤ the form

q_{N,m,d,k,r} := [N^{2}, 2m(k(t − d) + rN̄^{2}), n^{2}t + 2k(t − d)r + r^{2}N̄^{2}], where t = d(N̄n + 3),

is a binary form of type (N, m, d), i.e., q_{N,m,d,k,r} ∈ Q(N, m, d).

Proof. It is clear from the hypothesis on k that n ∈ ℤ. Put q_{N̄,n,k} := [N̄^{2}, 2k(t−d), n^{2}t]. Since nN̄ − k^{2}d = 1, we have that (N̄, n, k) ∈ P (d) in the notation of [?], §5, so Theorem 13 of [?] shows that q_{N̄,n,k} is a binary quadratic form of type d in the sense of [?]. This means in particular that property (iii) holds for q_{N̄,n,k} and that disc(q_{N̄,n,k}) = −16d. Now since

(4)  q_{N,m,d,k,r}(X, Y) = q_{N̄,n,k}(mX + rY, Y),

we thus see that property (iii) also holds for q_{N,m,d,k,r}. Moreover, property (ii) clearly holds because q_{N,m,d,k,r}(1, 0) = N^{2}. Finally, since disc(q_{N̄,n,k}) = −16d and since q := q_{N,m,d,k,r} is the transform of q_{N̄,n,k} by the matrix M = \binom{m r}{0 1}, it follows that disc(q) = disc(q_{N̄,n,k}) det(M)^{2} = −16m^{2}d. Thus, property (i) holds for q, and so q has type (N, m, d). □

Proof of Proposition ??. If −d ≡ x^{2} (mod N̄), where N̄ = \frac{N}{m}, then ∃k ∈ ℤ such that kx ≡ 1 (mod N̄) because gcd(d, N̄) = 1, and so k^{2}d ≡ −1 (mod N̄). Thus Q(N, m, d) ≠ ∅ by Lemma ??.

Conversely, suppose that q ∈ Q(N, m, d). Since properties (i), (ii) and (iii) are stable under GL_{2}(ℤ)-equivalence of forms, so is the set Q(N, m, d). Thus, by Lemma ?? we may assume that q = [N^{2}, 2mb, c] ∈ Q(N, m, d), for some b, c ∈ ℤ with b^{2}+4d = cN̄^{2} ≡ 0 (mod N̄^{2}), where N̄ = \frac{N}{m}. If N̄ is odd, then there exists y ∈ ℤ such that 2y ≡ 1 (mod N̄), and so (by)^{2} ≡ −d (mod N̄). Thus, the right hand side of (??) holds with x = by. Next, suppose that N̄ is even, so also b = 2b_{1} is even. If c is even, then b_{1}^{2} + d = \frac{cN̄}{4} N̄ ≡ 0 (mod N̄), so we can take x = b_{1}. On the other hand, if c is odd, then (c + 1)N̄ ≡ 0 (mod 4) and then (b_{1} + \frac{N̄}{2} )^{2} + d = b_{1}N̄ + \frac{(c+1)N̄}{4} N̄ ≡ 0 (mod N̄), so we can take x = b_{1} + \frac{N̄}{2} . □

The above Lemma ?? constitutes the first construction principle for forms of type (N, m, d). We now give another construction of such forms by using integral 2 × 2 matrices.

Proposition 11 Let N and d be positive integers and let M = \binom{x y}{z w} ∈ M_{2}(ℤ) be an integral matrix such that det(M) ≡ −d (mod N). Assume that M is N-primitive, i.e., that gcd(x, y, z, w, N) = 1. Put

(5)  g = g_{M,N} := gcd(x − w, y, z, N) and m := N/g.

Then there is an integer k such that kx ≡ 1 (mod g). Fix such a k and consider

(6) q_{M,N,d,k} := [N^{2}, 2mT, (T^{2} + 4d)/g^{2}], where T := −tr(M) − dk^{3}(det(M) + d).

Then q_{M,N,d,k} is a form of type (N, m, d); in fact, we have that

(7)  q_{M,N,d,k} = q_{N,m,d,k,r},

where r is determined as follows. Put x̄ = (kx − 1)/g and I = \binom{1 0}{0 1}, so

(8) kM = I + gM_{1}, where M_{1} := x̄I + kM_{0} and M_{0} := (M − xI)/g,

are integral matrices. Then n := (k^{2}d + 1)/g and b := (x + kd)/g are integers, and (??) holds with

(9)  r := −kd det(M_{1}) − n(2b + tr(M_{0})).

Proof. Since gcd(x, g) = gcd(x, y, z, w, N) = 1, there is a k ∈ ℤ such that kx ≡ 1 (mod N̄). Thus x̄ = (kx − 1)/N̄ ∈ ℤ, and kM = kxI + gkM_{0} = I + g(x̄I + kM_{0}), which proves (??). Note that M_{0} ∈ M_{2}(ℤ) by the definition of g.

To show that q_{M,N,d,k} ∈ Q(N, m, d), it suffices in view of Lemma ?? to verify that k^{2}d ≡ −1 (mod g) and that (??) holds.

For this, put δ = det(M), so a := (δ + d)/N ∈ ℤ by our hypothesis on d. Moreover, put n_{1} := ak^{2}m−τ_{1} −gδ_{1} ∈ ℤ, where τ_{1} = tr(M_{1}) and δ_{1} = det(M_{1}). Since k^{2}δ = det(kM) = det(I +gM_{1}) = 1+τ_{1}g+δ_{1}g^{2}, we have gn_{1} = k^{2}(δ+d)−τ_{1}g−δ_{1}g^{2} = k^{2}d + 1 = gn, and so n = n_{1} ∈ ℤ. Moreover, b ∈ ℤ because by the definition of x̄ and n we have that g(xn − x̄kd) = xng − (kx − 1)kd = x(ng − k^{2}d) = kd = x + kd = bg, so b = xn − x̄kd ∈ ℤ.

To prove (??), note first that q_{N,m,d,k,r} is defined because k^{2}d = −1 + gn ≡ −1 (mod g). Thus, q_{N,m,d,k,d} = [N^{2}, 2mT_{1}, T_{2}], where T_{1} = k(t − d) + rg^{2} and T_{2} = n^{2}t + 2k(t − d)r + g^{2}r^{2} and t = d(ng + 3). Note that since q_{N,m,d,k,r} has type (N, m, d) by Lemma ??, we see that T_{2} = (T_{1}^{2} + 4d)/g^{2}; cf. Lemma ??. Thus, (??) follows once we have shown that T = T_{1}.

For this, put τ_{0} := tr(M_{0}). Then with r as in (??) we have

(10)  rg^{2} = −g(−k^{3}dam + kd(2x̄ − n) − 2nx − τ_{0})
= −k^{3}daN − kd(gn + 2) − tr(M).

Indeed, the first equality can be derived from (??) by using the fact that gδ_{1} = ak^{2}m − τ_{1} − n = −(−ak^{2}m + n + 2x̄ + kτ_{0}), which is a variant of the definition of n_{1} = n, together with the fact that gn − k^{2}d = 1, and the second follows from the first by using that gx̄ = kx − 1 and tr(M) = 2x + gτ_{0} and by using the identity gn − k^{2}d = 1 again.

Since k(t − d) = kd(gn + 2), it is clear from the definition of T that (??) shows that T_{1} = T. This proves (??). □

Corollary 12 If M ∈ M_{2}(ℤ) is an N-primitive matrix of determinant −d, and if m = N/g, where g = g_{M,N}, then q_{M,N} := [N^{2}, −2mtr(M), (tr(M)^{2} +4d)/g^{2}] is a form of type (N, m, d).

Proof. Clearly, M satisfies the hypotheses of Proposition ??. The extra hypothesis on M implies that T = −tr(M), so q_{M,N} = q_{M,N,d,k}, for any k as in Proposition ??. □

We note in passing that the quadratic form q_{M,N} only depends on the SL_{2}(ℤ)- conjugacy class of M.

Corollary 13 If M is an integral matrix and if M′ = P^{−1}MP, where P ∈ SL_{2}(ℤ), then for every N ≥ 1 we have g_{M′,N} = g_{M,N}. Thus, if M is N-primitive and if det(M) = −d, then q_{M′,N} = q_{M,N}.

Proof. Write M = \binom{x y}{z w}, and put g = g_{M,N}. Then as above M = xI + gM_{0}, where M_{0} = \binom{0 ȳ}{z̄ w̄}. Write M′ = \binom{x′ y′}{z′ w′}. Then M′ = xI + gM′_{0}, where M′_{0} = P^{−1}M_{0}P =

(\begin{matrix}x̄′ & ȳ′ \\ z̄′ & w̄′ \end{matrix}) ∈ M_{2}(ℤ). Thus x′ = x + gx̄′ and w′ = x + gw̄′, so w′ = x′ + gw̄′′ with w̄′′ =

w̄′ − x̄′. This means that M′ = x′I + g\binom{0 ȳ′}{z̄′ w̄′′}, and so g| gcd(x′ − w′, y′, z′, N) = g_{M′,N}. This shows that g_{M,N}|g_{M′,N}. But since M′ = P MP^{−1}, a similar argument shows that also g_{M′,N}|g_{M,N}, so g_{M,N} = g_{M′,N}. Moreover, if M is N-primitive, then so is M′ and hence q_{M′,N} = q_{M,N} because tr(M′) = tr(M) and det(M′) = det(M). □

We next observe that the equivalence class of the quadratic form q_{M,N,d,k} only depends on the congruence class mod N of the matrix M. More precisely:

Proposition 14 If M′ is an integral matrix with M′ ≡ M (mod N), where M is as in Proposition ??, and if k′ ≡ k (mod g), where g = g_{M,N} = g_{M′,N}, then the form q_{M,N,d,k} is SL_{2}(ℤ)-equivalent to q_{M′,N,d,k′}.

Proof. First note that since det(M) + d ≡ 0 (mod N), and k′ ≡ k (mod g), it follows that dk^{3}(det(M) + d) + tr(M) ≡ d(k′)^{3}(det(M) + d) + tr(M) (mod gN), and so q_{M,N,d,k} ∼ q_{M,N,d,k′} by Lemma ??, where ∼ denotes SL_{2}(ℤ)-equivalence.

Next, consider M′\binom{x′ y′}{z′ w′}. Since M′ ≡ M (mod N), it is clear from (??) that g_{M′,N} = g_{M,N} = g and that det(M′) ≡ det(M) ≡ −d (mod N). Moreover, since x′ ≡ x (mod N), we see that kx′ ≡ kx (mod N), so kx′ ≡ 1 (mod g) and hence q_{M′,N,d,k} is defined.

Moreover, we observe that M′_{0} := (M′ − x′I)/g ≡ M_{0} (mod m) and that x̄′ := (kx′ − 1)/g ≡ x̄ (mod m). Thus, M′_{1} := x̄′I + kM′_{0} ≡ M_{1} (mod m). Since b′ := (x′+kd)/g ≡ b (mod m), we see that r′ := −kd det(M′_{1})−n(2b′+tr(M′_{0})) ≡ r (mod m), and so T′_{1} := k(t − d) + r′g^{2} ≡ T_{1} (mod mg^{2}). Since mg^{2} = N^{2}/m, this implies by Lemma ?? that q_{N,m,d,k,r′} ∼ q_{N,m,d,k,r}, and so q_{M′,N,d,k} ∼ q_{M,N,d,k} by (??). Moreover, by replacing k by k′ in the above argument we see that q_{M′,N,d,k′} ∼ q_{M,N,d,k′}. Combining the above equivalences shows that q_{M,N,d,k} ∼ q_{M,N,d,k′} ∼ q_{M′,N,d,k′}, as claimed. □

We next show that every form of type (N, m, d) is equivalent to a form q_{M,N}, for a suitable integral matrix M of determinant −d.

Proposition 15 If q = [N^{2}, 2mb, c] has type (N, m, d), then there is a primitive matrix M of discriminant −d such that q = q_{M,N}.

Proof. Put N̄ = \frac{N}{m} and g = gcd(2, N̄). Since −4d = b^{2} − N̄c, we see that b_{1} := \frac{b}{g} ∈ ℤ. Moreover, since gcd(d, N̄) = 1 by property (i), we see that gcd(2d, N̄^{2}) = gcd(2, N̄^{2}) = g, so there exist integers α, β such that

(11)  2dα − βN̄^{2} = g.

Note that gcd(αd, g) = 1 because αd(\frac{2}{g}) − gβ(\frac{N̄}{g})^{2} = 1, so α ≡ 1 (mod 2) if g ≠ 1 and hence α^{2} ≡ 1 (mod g^{2}) (for any g).

By property (iii) we know that c = q(0, 1) ≡ C (mod 4), for some C ∈ {0, 1}, and so also C = C^{2} ≡ c (mod g^{2}). Since N̄ ≡ 2 ≡ 0 (mod g), we see that c_{1} := αβb_{1}^{2}g^{2} + 2gβ + β^{2}N̄^{2} − α^{2}cd ≡ −cd (mod g^{2}) and c_{2} := Cgb_{1}N̄ + C^{2}d ≡ cd (mod g^{2}). Thus the numbers

w_{1} := αb_{1} + CN̄/g, w̄ = βb_{1}N̄ + 2dC/g, and z̄ = (c_{1} + c_{2})/g^{2}

are all integers, as is x := w̄N̄ − w_{1}d. Put P = \binom{x N̄}{z̄N̄ w_{1}}. Then, after simplifying, det(P) = −(α^{2}b_{1}^{2}dg^{2}+N̄^{4}β^{2}−N̄^{2}α^{2}cd+2N̄^{2}βg)/g^{2}, and so, by using the relation (??) and the fact that c = (g^{2}b_{1}^{2} + 4d)/N̄^{2}, it follows that det(P) = 1. Thus P ∈ SL_{2}(ℤ), and hence M := \binom{1 0}{0 −d}P is a primitive matrix of discriminant −d.

We have tr(M) = x − w_{1}d = b_{1}(βN̄^{2} − 2αd) = −b_{1}g = −b by (??). Furthermore, since x + w_{1}d = w̄N̄, we see that g_{M,N} = gcd(w̄N̄, N̄, −dz̄N̄, N) = N̄, and so m = N/g_{M,N}. This shows that q = q_{M,N}, as claimed. □

We can summarize the above results in the following way.

Theorem 16 Let N, m, d be positive integers such that m|N and gcd(N̄, d) = 1, where N̄ = \frac{N}{m}, and let q be a binary quadratic form. If ∼ denotes SL_{2}(ℤ)-equivalence of binary quadratic forms, then the following conditions are equivalent.

(i) q ∈ Q(N, m, d).

(ii) q ∼ q_{N,m,d,k,r}, for some k, r ∈ ℤ with k^{2}d ≡ −1 (mod N̄).

(iii) q ∼ q_{M,N,d,k}, for some N-primitive matrix M = \binom{x y}{z w} with det(M) ≡ −d (mod N) and g_{M,N} = N̄ and some k ∈ ℤ such that kx ≡ 1 (mod N̄).

(iv) q ∼ q_{M,N}, for some primitive matrix M with g_{M,N} = N̄ and det(M) = −d.

Proof. (i) ⇒ (iv): Lemma ?? and Proposition ??.

(iv) ⇒ (iii): If M = \binom{x y}{z w} is as in (iv), then M also satisfies condition (iii). Since gcd(x, N̄) = gcd(x, g_{M,N}) = gcd(x, y, z, w, N) = 1, there is a k ∈ ℤ such that kx ≡ 1 (mod N̄), and we have q_{M,N,d,k} = q_{M,N}. Thus (iii) holds.

(iii) ⇒ (ii): Proposition ??.

(ii) ⇒ (i): By Lemma ??, q_{N,m,d,k,r} ∈ Q(N, m, d), and so q ∈ Q(N, m, d) because Q(N, m, d) is stable under SL_{2}(ℤ)-equivalence; cf. the proof of Proposition ??. □

Remark 17 Note that Lemma ?? shows that q_{N,m,d,k,r′} ∼ q_{N,m,d,k,r}, if r′ ≡ r (mod m). Thus, in property (ii) we can restrict to r’s satisfying the extra condition 0 ≤ r < m.

Corollary 18 Let N and d be positive integers. Then a binary quadratic form q has type (N, N, d) if and only if q is GL_{2}(ℤ)-equivalent to [N^{2}, 2Nr, r^{2} + 4d], for some integer r with 0 ≤ r ≤ N/2.

Proof. Since here N̄ = 1, we see that k = 0 satisfies k^{2}d ≡ −1 (mod N̄), and so by Theorem ?? we have that [N^{2}, 2Nr, r^{2} + 4d] = q_{N,N,d,0,r} ∈ Q(N, N, d), for all r ∈ ℤ.

Conversely, if q ∈ Q(N, N, d), then by Theorem ?? there is a matrix M such that q ∼ q_{M,N}. Since m = N, we have that q_{M,N} = [N^{2}, −2Nr, r^{2} + 4d] with r = tr(M). Moreover, by Remark ?? we see that q_{M,N} ∼ [N^{2}, 2Nr_{1}, r_{1}^{2}+4d], where r_{1} ≡ r (mod N) and |r_{1}| ≤ N/2. Since [N^{2}, −2Nr_{1}, r_{1}^{2}+4d] is GL_{2}(ℤ)-equivalent to [N^{2}, 2Nr_{1}, r_{1}^{2}+4d], we can replace r_{1} by −r_{1} if necessary to achieve that r_{1} ≥ 0. □

As an example, we can give a complete classification of the forms of type (N, m, d) in the case that N = 2.

Proposition 19 Let q be a binary quadratic form and let d ≥ 1. Then q has type (2, 2, d) if and only if q is GL_{2}(ℤ)-equivalent to one of the forms

(12)  [4, 0, 4d] or [4, 4, 4d + 1],

and q has type (2, 1, d) if and only if q is GL_{2}(ℤ)-equivalent to one of the forms

(13)  [4, 4s, d + s], where d ≡ 1 + 2s (mod 4) and s = 0 or 1.

Proof. The first assertion follows directly from Corollary ?? because here \frac{N}{2} = 1, so r = 0 or 1. Thus, assume now that m = 1.

It is clear that the forms (??) have discriminant −16d with gcd(2, d) = 1, so property (i) holds for them. Moreover, they satisfy property (ii) since q(1, 0) = 2^{2} when q = [4, ∗, ∗]. Finally, since all the forms are congruent to [0, 0, 1] or [0, 0, 0]] (mod 4), it is clear that they all satisfy property (iii). Thus, each is of type (2, 1, d).

Conversely, suppose that q ∈ Q(2, 1, d). By reduction theory, q is GL_{2}(ℤ)- equivalent to q′ = [a, b, c] with 0 ≤ b ≤ a ≤ c. If a < 4, then necessarily a = 1 because a ≡ 0, 1 (mod 4) by property (iii). Thus q′ is the principal form of discriminant −16d, so q′ = [1, 0, 4d]. But of these, the only form that primitively represents 4 is q′ = [1, 0, 4], which is equivalent to [4, 0, 1]. Thus we get the case d = 1 of (??).

Now suppose that a ≥ 4. Since a is the smallest value represented by q, we must have a = 4 by property (ii). Thus, 16|(disc(q)+16c) = b^{2}, so b = 0 or 4. If b = 0, then c = d ≥ 4. But since gcd(2, d) = 1 and c ≡ 0, 1 (mod 4), it follows that d ≡ 1 (mod 4). This gives a form of (??) with s = 0.

Now suppose that b = 4, so c = \frac{4^{2}+16d}{16} = d + 1 and gcd(d, 2) = 1. Thus c ≡ 0 (mod 4) by property (iii), and so d ≡ 3 (mod 4). This gives us the cases of (??) with s = 1. □

3 Basic definitions and previous results

In this section we review some of the notations, definitions and results which were presented in the first part[?].

Throughout, let K be an arbitrary field, and let (J, λ) be a principally polarized abelian surface over K. Thus, J/K is an abelian variety of dimension 2 and λ : J →^{∼} Ĵ is an isomorphism such that its base change λ\overline{K} : J\overline{K} → Ĵ\overline{K} to an algebraic closure \overline{K} of K is given by a theta-divisor θ ∈ NS(J\overline{K}), i.e., λ\overline{K} = ϕ_{θ} in the notation of [?].

Definition. If N ≥ 2, then an N-presentation over K is a triple (E, E′, ψ) where E/K and E′/K are elliptic curves and ψ : E[N] → E′[N] is a K-isomorphism of finite group schemes which is an anti-isometry with respect to the e_{N}-pairings.

If (J, λ) is a principally polarized abelian surface over K, then an N-presentation of (J, λ) over K is a 4-tuple (E, E′, ψ, π) such that (E, E′, ψ) is an N-presentation over K and π : E × E′ → J is an isogeny such that

(14)  Ker(π) = Graph(−ψ) and π̂ ∘ λ ∘ π = N(λ_{E} ⊗ λ_{E′}),

where λ_{E} : E → Ê is the canonical principal polarization of E/K and λ_{E}⊗λ_{E′} denotes the product polarization on E × E′, i.e., λ_{E} ⊗ λ_{E′} = ϕ_{θ_{E}+θ_{E′}}, where θ_{E} = pr_{E}^{∗}(0_{E}) and θ_{E′} = pr_{E′}^{∗}(0_{E′}), and pr_{E} : E × E′ → E and pr_{E′} : E × E′ → E′ are the projections.

Remark 20 (a) If (E, E′, ψ) is an N-presentation, and if π_{ψ} : A := E × E′ → A_{ψ} := A/Graph(−ψ) is the quotient map, then A_{ψ} has a unique principal polarization λ_{ψ} such that (E, E′, ψ, π_{ψ}) is an N-presentation of (A_{ψ}, λ_{ψ}); cf. [?], Proposition 10(b). We then say that (π_{ψ}, A_{ψ}, λ_{ψ}) is the quotient data associated to (E, E′ψ).

(b) If (J, λ) is principally polarized abelian surface over K with an elliptic subgroup E ≤ J of λ-degree N, then (J, λ) has an N-presentation (E, E′, ψ, π), for some elliptic curve E′/K and anti-isometry ψ and isogeny π : E × E′ → J; cf. [?], Proposition 10(a). Note that E′ = E_{λ}^{⊥} in the notation of [?], Proposition 10(a).

Moreover, if f : C → E is an elliptic subcover of degree N of a curve C/K of genus 2, then E = f^{∗}J_{E} is an elliptic subgroup of the Jacobian J_{C}/K of C/K, and E has λ_{C}-degree N, where λ_{C} denotes the canonical polarization of J_{C}, so f induces an N-presentation (E, E′, ψ_{f}, π_{f}) of (J_{C}, λ_{C}); cf. [?], Corollary 12.

We next recall the definition of the refined Humbert invariant q_{(J,λ)} of a principally polarized abelian surface (J, λ) over K. If K is algebraically closed, then we can use the Neron-Severi group NS(J) = Div(J)/ ≡ to define q_{(J,λ)}, but for a general base field K this group is no longer suitable, as was explained in [?], Section 3. Instead, one has to work with a slightly larger group NS′(J) which is defined as the saturation (or primitive hull) of NS(J) in the group NS(J\overline{K}), i.e.,

NS′(J) = \{D ∈ NS(J\overline{K}) : nD ∈ NS(J), for some n ≥ 1\}.

Note that the theta-divisor θ ∈ NS(J\overline{K}) of λ\overline{K} lies in NS′(J) (cf. proof of [?], Proposition 14), so the quotient group

NS(J, λ) := NS′(J)/ℤθ

is defined. Then the refined Humbert invariant is the positive-definite quadratic form q_{(J,λ)} : NS(J, λ) → ℤ defined by the formula

(15) q_{(J,λ)}(D + ℤθ) = q̃_{(J,λ)}(D) := (D.θ)^{2} − 2(D.D), for D ∈ NS′(J),

where ( ·) denotes the intersection pairing on the surface J\overline{K}.

In the case of a product surface A = E × E′, the group NS′(A) is just the usual Neron-Severi group NS(A). Moreover, if λ_{A} = λ_{E} ⊗ λ_{E′} is the product polarization on A, then q_{(A,λ_{A})} is given by formula (??) below.

Lemma 21 Let E/K and E′/K be elliptic curves over an arbitrary field K, and let A = E × E′ be the product surface. For a, b ∈ ℤ and h ∈ End(E, E′) put

D̃′(a, b, h) := (a − deg(h))θ_{E} + (b − 1)θ_{E′} + Γ_{−h} ∈ Div(A),

where Γ_{−h} denotes the graph of −h, and let D′(a, b, h) = cl(D̃′(a, b, h)) denote its class in NS(A). Then the rule (a, b, h) ↦ D′(a, b, h) ∈ NS(A) defines an isomorphism

D′_{E,E′} : H̃_{E,E′} := ℤ × ℤ × Hom(E, E′) →^{∼} NS′(A) = NS(A),

and if λ_{A} = λ_{E} ⊗ λ_{E′} denotes the product polarization, then

(16) q_{(A,λ_{A})}(D′(a, b, h) + ℤθ_{A}) = (a − b)^{2} + 4 deg(h) = (a − b)^{2} + 4q_{E,E′}(h),

where q_{E,E′} denotes the degree form on Hom(E, E′).

Proof. [?], Lemma 28. □

It is much more difficult to determine the refined Humbert invariant for an arbitrary principally polarized abelian surface (J, λ); indeed, this is precisely the aim of the present paper. A first step towards this was done in [?], which determined the discriminant (or determinant) of the quadratic form q_{(J,λ)} in terms of a presentation data (E, E′, ψ, π). This result depends on the isogeny defect m_{ψ} of ψ with respect to the group Hom(E, E′). For applications, it is useful to generalize this concept as follows; cf. [?], Section 4.

Definition. Let E/K and E′/K be elliptic curves, and let ψ ∈ Hom(E[N], E′[N]). Moreover, let H ≤ H_{E,E′} := Hom(E, E′) be a primitive subgroup of H_{E,E′}, i.e., H is a subgroup of H_{E,E′} such that H_{E,E′}/H is torsionfree. Then the isogeny defect of ψ with respect to H is

m_{ψ}(H) := min{m ≥ 1 : mψ = h_{|E[N]}, for some h ∈ H}.

As is mentioned in [?], Remark 21(b), m_{ψ}(H) can be expressed as an index of groups. Indeed, let

ρ_{N} : Hom(E, E′) → Hom(E[N], E′[N])

be the restriction map h ↦ h_{|E[N]}, and put H_{N} = ρ_{N}(H), which is a subgroup of Hom(E[N], E′[N]). Then by elementary group theory we have that

(17)  m_{ψ}(H) = [⟨ψ⟩ + H_{N} : H_{N}] = [⟨ψ⟩ : ⟨ψ⟩ ∩ H_{N}].

In particular, we see that m_{ψ}(H) | ord(ψ) | N, and that

(18)  kψ = ρ_{N}(h), for some k ∈ ℤ, h ∈ H ⇒ m_{ψ}(H) | k.

The main result which was used in the calculation of the discriminant of q_{(J,λ)} is the following index formula, which is Theorem 31 of [?].

Theorem 22 Let E/K and E′/K be elliptic curves and let ψ ∈ Hom(E[N], E′[N]). Moreover, let G_{ψ} = Graph(−ψ) be the graph of −ψ and let

π = π_{ψ} : A := E × E′ → A_{ψ} := (E × E′)/G_{ψ},

be the associated quotient isogeny. Assume that A_{ψ} has a principal polarization λ_{ψ}. In addition, let H ≤ Hom(E, E′) be a primitive subgroup, and put

H̃ := {D′(a, b, h)) : a, b ∈ ℤ, h ∈ H} ≤ NS′(A).

If r_{H} := rank(H) and if char(K) ∤ m_{ψ}(H), then

(19)  [H̃ : π_{ψ}^{∗}(NS′(A_{ψ})) ∩ H̃] = m_{ψ}(H)N^{r_{H}+2}.

Note that ψ does not have to be an anti-isometry in Theorem ??; indeed, if this is the case, then the existence of the principal polarization λ_{ψ} is automatic; cf. Remark ??(a). The generality of Theorem ?? is necessary to be able to apply it to the map ψ̄ of the following factorization lemma, which is Proposition 39 of [?].

Proposition 23 In the situation of Theorem ??, put m = m_{ψ}(H). Then there exists an h ∈ H such that mψ = mρ_{N}(h) and we have

(20)  ε_{N,m}^{E′} ∘ ψ̄ ∘ π_{N,m}^{E} = ψ − ρ_{N}(h),

for a unique ψ̄ ∈ Hom(E[m], E′[m]), where ε_{N,m}^{E′} : E′[m] ↪ E′[N] denotes the canonical inclusion and π_{N,m}^{E} : E[N] → E[m] denotes the map which is given by multiplication by N̄ = \frac{N}{m}.

Moreover, if ν = v_{N̄,h} := ( \begin{matrix}[N̄]_{E} & 0 \\ h & [1]_{E′} \end{matrix} ) ∈ End(A), then ν^{−1}Gψ̄ = G_{ψ}, and

hence there a unique isomorphism α_{ψ} : A_{ψ} →^{∼} Aψ̄ such that

(21)  α_{ψ} ∘ π_{ψ} = πψ̄ ∘ ν.

In addition, mψ̄(H) = m_{ψ}(H).

Another concept which was used in the calculation of the discriminant of q_{C} in [?] (and which is required here) was the notion of the trace tr(g) of an element g ∈ End(E[N]). Since tr(g) is closely linked to the transpose g^{t} of g, we first recall the definition of g^{t} and some of its properties. For this, assume henceforth in this section that char(K) ∤ N.

If g ∈ Hom(E[N], E′[N]), then by [?], Proposition 23, there exists a unique homomorphism g^{t} ∈ Hom(E′[N], E[N]) such that

(22)  e_{N}^{E′}(g(P), Q) = e_{N}^{E}(P, g^{t}(Q)), ∀P ∈ E[N](\overline{K}), Q ∈ E′[N](\overline{K}),

where e_{N}^{E} : E[N] × E[N] → μ_{N} is the usual e_{N}-pairing on E. Moreover, we have

(23)  (g^{t})^{t} = g and g^{t}g = [det(g)]_{E[N]},

for all g ∈ Hom(E[N], E′[N]), where det(g) ∈ ℤ/Nℤ is the unique number such that

(24)  e_{N}^{E′}(g(P), g(Q)) = e_{N}^{E}(P, Q)^{det(g)}, for all P, Q ∈ E[N](\overline{K}).

From this it follows (cf. [?], Corollary 24) that if we put tr(g) := det(1+g)−det(d)−1 for g ∈ End(E[N]), then

(25)  g^{t} + g = [tr(g)]_{E[N]}.

Note that if h ∈ Hom(E, E′) is a homomorphism of elliptic curves, then its transpose is defined as h^{t} := λ_{E′}^{−1}∘ĥ∘λ_{E} ∈ End(E′, E), where (as before) λ_{E} = ϕ_{0_{E}} : E →^{∼} Ê is the canonical principal polarization of E. This transpose is compatible with the above transpose in the sense that

(26) (h_{|E[N]})^{t} = (h^{t})_{|E′[N]}, and hence det(h_{|E[N]}) ≡ deg(h) (mod N);

cf. [?], Remark 25(a). In addition, the transpose is compatible with restriction to the subgroup E[M] for M|N. More precisely, if g_{M} ∈ Hom(E[M], E′[M]) is the unique map such that

(27)  π_{N,M}^{E′} ∘ g = g_{M} ∘ π_{N,M}^{E},

then by [?], Section 4, we have that

(28)  (g^{t})_{M} = (g_{M})^{t}, for all g ∈ Hom(E[N], E′[N]),

and from (??) it follows that

(29)  det(g_{M}) ≡ det(g) (mod M), for all g ∈ Hom(E[N], E′[N]).

Finally, we recall the following technical result which was used in the derivation of Theorem ?? and which is required in the proof of Theorem ?? below.

Proposition 24 In the situation of Theorem ??, assume also that char(K) ∤ N. If D = D′(a, b, f) ∈ NS′(A), then

(30) ND ∈ π_{ψ}^{∗}NS′(A_{ψ}) ⇔ p_{ψ}(D) := a + b det(ψ) − tr(f^{t}ψ) ≡ 0 (mod N).

Proof. [?], Proposition 37.

4 The computation of q_{C}

The purpose of this section is to compute the refined Humbert invariant q_{C} of C/K explicitly from the data given by a presentation (E, E′, ψ) of degree N in the case that rank Hom(E, E′) = 1.

Since essentially the same computation can also be done in a much more general context, we shall do it first in the general situation of a principal polarized abelian surface with an N-presentation (E, E′ψ, π), and then explain how to deduce the above case from it; cf. Corollary ??. Thus, we shall prove:

Theorem 25 Let (E, E′, ψ) be an N-presentation over K with associated quotient data (π_{ψ}, A_{ψ}, λ_{ψ}), and let d ≥ 1 be an integer. Then:

(a) There exists a cyclic isogeny h ∈ Hom(E, E′) such that d = deg(h) if and only if d is primitively represented by the degree form q_{E,E′}. If this is the case, put m = m_{ψ}(ℤh) and N̄ = \frac{N}{m}. Then there is an integer k ∈ ℤ such that the restriction ψN̄ of ψ to E[N̄] equals the restriction of kh to E[N̄], i.e.,

(31)  ψN̄ = ρN̄(kh).

(b) In the situation of part (a), fix a cyclic isogeny h with deg(h) = d and an integer k ∈ ℤ such that (??) holds, and put n := (k^{2}d + 1)/N̄. Then n ∈ ℤ and so the binary quadratic from q_{N̄,n,k} is defined as in the proof of Lemma ??. Moreover,

(32)  q̃_{(A,λ)}(XD_{1} + Y D_{2}) = N̄^{2}q_{N̄,n,k}(X, Y), for all X, Y ∈ ℤ,

where A = E × E′, λ is the product polarization on A, and D_{1} := D′(N̄^{2}, 0, 0) and D_{2} := D′(k(t − 2d), −kd, h) ∈ NS′(A) with t = d(k^{2}d + 4) = d(N̄n + 3).

(c) Assume henceforth that char(K) ∤ m. Then there is an integer r such that m(rD_{1} + D_{2}) ∈ π_{ψ}^{∗}(A_{ψ}), and r (mod m) is uniquely determined by this property. Moreover, if H = ℤh and θ = D′(1, 1, 0), then

(33)  H̃ ∩ π_{ψ}^{∗}NS′(A_{ψ}) = ℤNθ + ℤm^{2}D_{1} + ℤm(rD_{1} + D_{2}).

(d) There is a unique ψ̄ ∈ Hom(E[m], E′[m]) such that

(34)  ε_{N,m} ∘ ψ̄ ∘ π_{N,m} = ψ − ρ_{N}(kh),

and then the integer r of part (c) is given by the formula

(35)  r ≡ kd det(ψ̄) + ntr(h^{t}ψ̄) (mod m).

(e) Fix r as in part (c). Then there exist D̄_{1}, D̄_{2} ∈ NS(A_{ψ}, λ_{ψ}) such that ℤD̄_{1}+ℤD̄_{2} is a primitive submodule of NS(A_{ψ}, λ_{ψ}) of rank 2 and such that

(36)  q_{(A_{ψ},λ_{ψ})}(XD̄_{1} + Y D̄_{2}) = q_{N,m,d,k,r}(X, Y), for all X, Y ∈ ℤ,

where q_{N,m,d,k,r} is as in Lemma ??. In particular, q_{(A_{ψ},λ_{ψ})} primitively represents the binary quadratic form q_{N,m,d,k,r}.

Proof. (a) The first assertion is clear because h ∈ Hom(E, E′) is primitive if and only if h is a cyclic isogeny.

Now by the definition of m there exists a k′ ∈ ℤ such that mψ = ρ_{N}(k′h). Since m|N, we see that E[m] ≤ Ker(k′h), and so m|k′ because Ker(h) is cyclic. Thus k′ = mk, for some k ∈ ℤ, so m(ψ − ρ_{N}(kh)) = 0 and hence (??) holds by (??).

(b) Since deg(kh) = k^{2}d, we have by (??), (??) and (??) that k^{2}d ≡ det(ρN̄(kh)) ≡ det(ψN̄) ≡ −1 (mod N̄), the latter because ψ (and hence also ψN̄) is an anti-isometry. Thus N̄|(k^{2}d + 1), i.e., n ∈ ℤ.

To prove (??), write q̃ = q̃_{(A,λ)}. Then by (??) we see that q̃(D_{1}) = N̄^{4} and q̃(D_{2}) = k^{2}(t − d)^{2} + 4d = N̄^{2}n^{2}t, where the last equality follows by using the identity (6) of [?]. Similarly, since q̃(D_{1}+D_{2}) = (N̄^{2}+k(t−d))^{2}+4d = 2k(t−d)N̄^{2}+N̄^{4}+N̄^{2}n^{2}t by the same identity (6) again, it follows that (??) holds.

(c) To show that such an r exists, we shall use the results which were summarized in the previous section (and prove part (d) along the way).

First of all, since mψ = mρ_{N}(kh) (cf. the proof of part (a)), we have by Proposition ?? that (??) holds for a unique ψ̄. Let r ∈ ℤ be such that (??) holds. We now claim that m(rD_{1} + D_{2}) ∈ N_{ψ} := (π_{ψ})^{∗}NS′(A_{ψ}).

For this, recall from Proposition ?? that we have the isomorphism α : A_{ψ} →^{∼} Aψ̄ such that (??) holds, where ν = ν_{N̄,kh}. Since λ_{ψ} is a principal polarization on A_{ψ} (cf. Remark ??(a)), there is a unique principal polarization λψ̄ on Aψ̄ such that α̂ ∘ λψ̄ ∘ α = λ_{ψ}, so ψ̄ satisfies the hypothesis of Theorem ??, and hence we can apply Proposition ?? to ψ̄. Let pψ̄ be as in Proposition ??.

Put D′_{1} = D′(1, 0, 0) and D′_{2} = D′(0, −kd, nh), so rD′_{1} + D′_{2} = D′(r, −kd, nh). Then pψ̄(rD′_{1} + D′_{2}) ≡ r − (kd) det(ψ̄) − ntr(h^{t}ψ) ≡ 0 (mod m) by (??), so by Proposition ?? we have that m(rD′_{1} + D′_{2}) ∈ Nψ̄ := (πψ̄)^{∗}NS′(Aψ̄). Thus, since

(37)  ν^{∗}(D′_{1}) = D_{1} and ν^{∗}(D′_{2}) = D_{2}

by [?], formula (42), and the fact that N̄n − k^{2}d = 1, it follows that m(rD_{1} + D_{2}) ∈ ν^{∗}Nψ̄ = N_{ψ}, where the latter equality follows from (??). This proves the existence of r ∈ ℤ such that m(rD_{1} + D_{2}) ∈ N_{ψ}.

We next show that (??) holds. To see this, note first that if H̃_{1} denotes the module on the right hand side of (??), then clearly H̃_{1} ⊂ H̃. Next we verify that H̃_{1} ⊂ N_{ψ}.

For this, recall from above that m(rD_{1} + D_{2}) ∈ N_{ψ}. Moreover, since clearly pψ̄(mD′_{1}) ≡ 0 (mod m), we see that m^{2}D′_{1} ∈ Nψ̄ and so m^{2}D_{1} = ν^{∗}(m^{2}D′_{1}) ∈ N_{ψ}. Finally, we observe that Nθ ∈ N_{ψ}. Indeed, if θ_{ψ} ∈ NS′(A_{ψ}) is such that ϕ_{θ_{ψ}} = λ_{ψ}, then ϕ_{π_{ψ}^{∗}θ_{ψ}} = ϕ_{Nθ} because π̂_{ψ}λ_{ψ}π_{ψ} = Nλ = Nϕ_{θ} by (??), so we have

(38)  Nθ = π_{ψ}^{∗}θ_{ψ} ∈ N_{ψ}.

This shows the desired inclusion H̃_{1} ⊂ H̃ ∩ N_{ψ}.

Now since θ, D′_{1}, and D := D′(0, 0, h) form a basis of H̃, and since m^{2}D_{1} = N^{2}D′_{1} and m(rD_{1} + D_{2}) = aθ + bD_{1} + mD, for some a, b ∈ ℤ, we see that [H̃ : H̃_{1}] = N · N^{2} · m = mN^{3}. On the other hand, since H = ℤh is primitive in Hom(E, E′) because h is cyclic, it follows from Theorem ?? that [H̃ : H̃ ∩ N_{ψ}] = mN^{3}, and so we have that H̃_{1} = H̃ ∩ N_{ψ}. This proves (??).

We now show that r is uniquely determined by the given condition. Indeed, suppose that we have m(r′D_{1} + D_{2}) ∈ N_{ψ}, for some r′ ∈ ℤ. Then by (??) there exist a, b, c ∈ ℤ such that m(r′D_{1} + D_{2}) = aNθ + bm^{2}D_{1} + cm(rD_{1} + D_{2}). Since θ, D_{1} and D_{2} are linearly independent, it follows that c = 1 and a = 0, so mr′ = bm^{2} + mr, or r′ ≡ r (mod m), as claimed. Note also that the converse holds: if r′ = r + bm, b ∈ ℤ, then m(r′D_{1} + D_{2}) = m(rD_{1} + D_{2}) + m^{2}bD_{1} ∈ N_{ψ} by (??).

(d) This was established in the course of proving part (c).

(e) From (??) we see that there exist D′′_{1}, D′′_{2} ∈ NS′(A_{ψ}) such that

(39)  π_{ψ}^{∗}(D′′_{1}) = m^{2}D_{1} and π_{ψ}^{∗}(D′′_{2}) = m(rD_{1} + D_{2}).

Thus, if we put H̃_{2} := ℤθ_{ψ} + ℤD′′_{1} + ℤD′′_{2}, then π_{ψ}^{∗}(H̃_{2}) = H̃ ∩ N_{ψ} by (??), (??) and (??), and so H̃_{2} is a primitive submodule of NS′(A_{ψ}) because H̃ ∩ N_{ψ} is primitive in N_{ψ}. Thus, if D̄_{i} denotes the image of D′′_{i} in NS(A_{ψ}, λ_{ψ}) = NS′(A_{ψ})/ℤθ_{ψ}, then ℤD̄_{1} + ℤD̄_{2} = H̃_{2}/ℤθ_{ψ} is a primitive submodule of NS(A_{ψ}, λ_{ψ}).

Finally, to prove (??), we use (??), the projection formula (cf. [?], formula (14)) together with (??) and (??) to obtain that

N^{2}q̃_{(A_{ψ},λ_{ψ})}(XD′′_{1} + Y D′′_{2}) = q̃_{(A,λ)}(X(m^{2}D_{1}) + Y m(rD_{1} + D_{2}))
= m^{2}q̃_{(A,λ)}((mX + rY)D_{1} + Y D_{2})
= m^{2}N̄^{2}q_{N̄,n,k}(mX + rY, Y) = N^{2}q_{N,m,d,k,r}(X, Y),

and so (??) follows because q_{(A_{ψ},λ_{ψ})}(xD̄_{1} + yD̄_{2}) = q̃_{(A_{ψ},λ_{ψ})}(xD′′_{1} + yD′′_{2}) by definition.

□

We observe that the above theorem yields the following result which generalizes Theorem ??.

Corollary 26 In the situation of Theorem ?? assume in addition that char(K) ∤ N.

(a) If d ≥ 1 is an integer which is primitively represented by q_{E,E′}, then q_{(A_{ψ},λ_{ψ})} primitively represents a binary quadratic form of type (N, m, d), where m is some integer with m_{ψ}|m|N, where m_{ψ} is defined as in the Introduction.

(b) If Hom(E, E′) ≠ 0, then there exists an integer d ≥ 1 which is primitively represented by q_{E,E′} such that q_{(A_{ψ},λ_{ψ})} primitively represents a binary quadratic form of type (N, m_{ψ}, d).

Proof. (a) This clearly follows from Theorem ?? by taking m = m_{ψ}(ℤh) as in the theorem. Note that m|N by (??) and also that m_{ψ} = m_{ψ}(Hom(E, E′))|m by (??) because ℤh ≤ Hom(E, E′).

(b) By definition there exists h′ ∈ Hom(E, E′) such that m_{ψ}ψ = ρ_{N}(h′). We can write h′ = kh, where h is a cyclic isogeny. (If h′ = 0, then m_{ψ} = N and then we can take k = 0 and h any cyclic isogeny which exists because Hom(E, E′) ≠ 0.) Thus d = deg(h) is primitively represented by q_{E,E′} and m_{ψ} = m_{ψ}(ℤh). Thus, applying Theorem ?? to this d, h and k, we see that the last assertion follows. □

Corollary 27 Suppose that C/K has a presentation (E, E′, ψ) of degree N such that Hom(E, E′) = ℤh, where d = deg(h) ≥ 1. Put m = m_{ψ} and N̄ = \frac{N}{m}, and assume that char(K) ∤ m. Then there is an integer k such that (??) holds. Thus, if we choose the integer r according to the recipe of Theorem ??(d), then q_{C} is GL_{2}(ℤ)-equivalent to the binary quadratic form q_{N,m,d,k,r}.

Proof. The hypothesis implies that h is cyclic of degree d, so by Theorem ??(a) there is an integer k such that (??) holds. Since (J_{C}, λ_{C}) ≃ (A_{ψ}, λ_{ψ}), and since rank(NS′(J_{C})) = 3 by our hypothesis, it follows that the elements D̄_{1}, D̄_{2} of Theorem ??(e) form a basis of NS(A_{ψ}, λ_{ψ}) ≃ NS(J_{C}, λ_{C}), and so the last assertion of Theorem ??(e) shows that q_{C} and q_{(A_{ψ},λ_{ψ})} are both GL_{2}(ℤ)-equivalent to q_{N,m,d,k,r}. □

We next observe that the basic integers m, k, r which determine the quadratic form q_{N,m,d,k,r} can be computed directly from the matrix [ψ^{−1}h_{|E[N]}]_{B} of the endomorphism ψ^{−1}h_{|E[N]} ∈ End(E[N]) with respect to some basis B of E[N](\overline{K}).

More precisely, assume that char(K) ∤ N and that B = (P_{1}, P_{2}) is a basis of E[N](\overline{K}). Then there exist x, y, z, w ∈ ℤ such that

(40)  ψ^{−1}h(P_{1}) = xP_{1} + zP_{2} and ψ^{−1}h(P_{2}) = yP_{1} + wP_{2},

and so [ψ^{−1}ρ_{N}(h)]_{B} ≡ \binom{x y}{z w} (mod N) is the matrix of ψ^{−1}ρ_{N}(h) with respect to the basis B = (P_{1}, P_{2}) of E[N]. Note that x, y, z, w are uniquely determined (mod N) by h and ψ (and by the basis B).

Proposition 28 In the situation of Theorem ??, assume that char(K) ∤ N, and let h ∈ Hom(E, E′) be a cyclic isogeny of degree d. Let M = \binom{x y}{z w} ∈ M_{2}(ℤ) be an integral matrix such that M (mod N) is the matrix of ψ^{−1}ρ_{N}(h) with respect to some basis B = (P_{1}, P_{2}) of E[N]. Then M is N-primitive and det(M) ≡ −d (mod N), and

(41)  m := m_{ψ}(ℤh) = N/g_{M,N} where g_{M,N} = gcd(x − w, y, z, N).

Moreover, there exists an integer k such that kx ≡ 1 (mod \frac{N}{m}), and then (??) holds for this k. In addition, the quadratic form q_{N,m,d,k,r} of Theorem ??(e) is equivalent to the form q_{M,N,d,k} of Proposition ??, i.e., q_{N,m,d,k,r} ∼ q_{M,N,d,k}.

Proof. Write g := ψ^{−1}ρ_{N}(h) ∈ End(E[N]). Then det(g) = det(ψ)^{−1} det(ρ_{N}(h)) ≡ −d (mod N) by (??), and so det(M) ≡ det(g) ≡ −d (mod N). Moreover, we have that gcd(x, y, z, w, N) = 1 because E[n] ≰ Ker(h), ∀n > 1, since h is a cyclic isogeny. Thus, M is N-primitive.

To prove the other assertions, we first observe that it follows from the definitions that if n|N is any divisor, and if n̄ = \frac{N}{n}, then n̄B = (n̄P_{1}, n̄P_{2}) is a basis of E[n], and hence M (mod n) is the matrix of g_{n} ∈ End(E[n]) with respect to this basis.

Applying this to our situation, we thus see from Theorem ??(a) that ∃k′ ∈ ℤ such that k′M ≡ \binom{1 0}{0 1} (mod N̄), so k′x ≡ 1 ≡ k′w (mod N̄) and k′y ≡ k′z ≡ 0 (mod N̄). Since gcd(k′, N̄) = 1 by Theorem ??(b), it follows that x ≡ w (mod N̄) and y ≡ z ≡ 0 (mod N̄), which implies that N̄|N′ := g_{M,N}.

In order to show that N′ = N̄, note first that by the definition of N′ we have that gcd(x, N′) = gcd(x, y, z, w, N) = 1 and that M = \binom{x 0}{0 x} + N′M_{0}, for some integral matrix M_{0}. Thus, there exists x′ ∈ ℤ such that x′x ≡ 1 (mod N′), and so m′x′M ≡ \binom{m′ 0}{0 m′} (mod N), where m′ = N/N′. This means that m′x′g = [m′]_{E[N]}, or that ρ_{N}(m′x′h) = m′ψ. By (??) we thus have that m_{ψ} := m_{ψ}(ℤh)|m′, and so N′ = \frac{N}{m′} |\frac{N}{m_{ψ}} = N̄. This proves that N̄ = N′ and so (??) follows.

Since M is N-primitive and g_{M,N} = N̄ = \frac{N}{m}, there exists k ∈ ℤ such that kx ≡ 1 (mod N̄); cf. Proposition ??. Thus, by (??) we have that kM ≡ I (mod N̄), which is equivalent to (??).

To verify the last assertion, it suffices in view of Proposition ??, Remark ?? and (??) to verify that

(42) det(ψ̄) ≡ − det(M_{1}) (mod m) and tr(h^{t}ψ̄) ≡ −2b − tr(M_{0}) (mod m),

where ψ̄ is as in Theorem ??(d), M_{1} and M_{0} are as in (??) and b is as in (??).

Since by definition ψ̄(N̄P) = ψ(P)−h(P), for all P ∈ E[N], it follows immediately that ψ_{m}^{−1}ψ̄(N̄P) = P −ψ^{−1}h(P), and so we see that M_{1} (mod m) is the matrix of ψ_{m}^{−1}ψ̄ with respect to the basis N̄B = (N̄P_{1}, N̄P_{2}). Thus det(ψ_{m}^{−1}ψ̄) ≡ det(M_{1}) (mod m), and so det(ψ̄) ≡ det(M_{1}) det(ψ_{m}) ≡ − det(M_{1}) (mod m) by (??).

This proves the first formula of (??). To prove the second formula, it suffices to show that the matrix of ρ_{m}(h)^{t}ψ̄ with respect to the basis N̄B is given by the formula

(43)  [ρ_{m}(h)^{t}ψ̄]_{N̄B} ≡ −bI − M_{0}^{∗} (mod m),

where M_{0}^{∗} is the (classical) adjoint of M_{0}, because then the second formula of (??) follows by taking traces of both sides and observing that tr(M_{0}^{∗}) = tr(M_{0}).

In order to derive (??), we apply h^{t} to (??) to obtain that h^{t}ψ̄(N̄P) = h^{t}ψ(P) − kh^{t}h(P) = h^{t}ψ(P)−kdP, for all P ∈ E[N]. Thus ε_{N,ε}∘ρ_{m}(h)^{t}ψ̄∘π_{N,m} = h^{t}ψ−[kd]_{E[N]}. Now by (??) and by [?], formula (27), we have

ρ_{N}(h)^{t}ψ = ρ_{N}(h)^{t}ψ^{tt} = (ψ^{t}ρ_{N}(h))^{t} = −(ψ^{−1}ρ_{N}(h))^{t},

the latter because ψ^{t} = −ψ^{−1} by (??). Since [g^{t}]_{B} = ([g]_{B})^{∗}, for any g ∈ End(E[N]), we see that the matrix of −ρ_{N}(h)^{t}ψ is the adjoint M^{∗} of M = [ψ^{−1}ρ_{N}(h)]_{B}. Since M^{∗} = xI +N̄(M_{0})^{∗}, it follows that the matrix of ε_{N,m}∘ρ_{m}(h)^{t}ψ∘π_{N,m} = h^{t}ψ−[kd]_{E[N]} is −xI − N̄(M_{0})^{∗} − kdI = −N̄(bI + M_{0}^{∗}), and so (??) follows. This proves (??). □

Remark 29 It is useful to observe that we can choose the matrix M of Proposition ?? in such way that M is primitive and of determinant −d; cf. Lemma ?? below. If we do this, then the above quadratic form q_{M,N,d,k} equals the form q_{M,N} of Corollary ??, i.e., q_{M,N,d,k} = q_{M,N}.

Lemma 30 Let M′ = \binom{a_{1} a_{2}}{a_{3} a_{4}} be an integral matrix and let N and d be positive integers. If M′ is N-primitive, i.e., if gcd(a_{1}, a_{2}, a_{3}, a_{4}, N) = 1, and if det(M′) ≡ −d (mod N), then there exists a primitive integral matrix M of determinant −d such that M ≡ M′ (mod N).

Proof. Put g_{i} = gcd(a_{i}, N), so gcd(g_{1}, …, g_{4}) = 1. Then by Dirichlet’s theorem on primes in arithmetic progressions, there are distinct primes p_{i} ≡ \frac{a_{i}}{g_{i}} (mod \frac{N}{g_{i}}) with p_{i} ∤ g_{1} ⋯ g_{4}. Put b_{i} := p_{i}g_{i} and M̃ := \binom{b_{1} b_{2}}{b_{3} b_{4}}. Then M̃ ≡ M′ (mod N) and M̃ is primitive, i.e., gcd(b_{1}, …, b_{4}) = 1.

Note that δ := det(M̃) ≡ det(M′) ≡ −d (mod N). Thus, since M̃ is primitive, there is a matrix P ∈ SL_{2}(ℤ) such that M̃ = P \binom{1 b}{0 δ}, for some b ∈ ℤ, and so M := P \binom{1 b}{0 −d} is primitive with det(M) = −d and M ≡ M̃ ≡ M′ (mod N). □

The following result can be viewed as a partial converse of Theorem ??.

Theorem 31 Let (J, λ) be a principally polarized abelian surface over K and let N ≥ 2 be an integer. Suppose that NS(J, λ) contains a primitive submodule Ḡ of rank 2 such that the restriction qḠ of q_{(J,λ)} to Ḡ primitively represents N^{2}. Then there exist unique positive integers m, d with m|N and gcd(N/m, d) = 1 such that qḠ has type (N, m, d), and hence there exist integers k, r with k^{2}d ≡ −1 (mod \frac{N}{m}) such that qḠ ∼ q_{N,m,d,k,r}. Moreover, if char(K) ∤ m, then the following properties hold.

(a) There is an N-presentation (E, E′, ψ, π) of (J, λ) and a cyclic isogeny h ∈ Hom(E, E′) of degree d such that (??) holds. Thus m = m_{ψ}(H), where H = ℤh.

(b) If A = E × E′ and θ = D′(1, 1, 0), D_{1} = D′(N̄^{2}, 0, 0), and D_{2} = D′(k(t − 2d), −kd, h) ∈ NS′(A), where t = d(k^{2}d + 4), then the analogue of (??) holds, i.e.,

(44)  H̃ ∩ π^{∗}NS′(J) = ℤNθ + ℤm^{2}D_{1} + ℤm(rD_{1} + D_{2}).

In particular, the given r satisfies the condition of Theorem ??(c), and hence satisfies the congruence (??).

(c) If char(K) ∤ N, and if M is an integral matrix such that M (mod N) is the matrix of ψ^{−1} ∘ ρ_{N}(h) with respect to some basis B = (P_{1}, P_{2}) of E[N](\overline{K}), then qḠ ∼ q_{M,N,d,k}.

Proof. Recall that q_{(J,λ)} and also hence qḠ is positive definite. Since q_{(J,λ)}(D) ≡ (D.θ)^{2} (mod 4), ∀D ∈ NS(J, λ) (cf. the proof of Theorem 3 of [?]), we see that qḠ satisfies property (iii) of Theorem 1. Thus, since qḠ primitively represents N^{2} by hypothesis, the first assertion follows from Proposition ??, and the second follows from Theorem ??. To prove the other assertions, assume henceforth that char(K) ∤ m.

(a) Since qḠ ∼ q_{N,m,d,k,r}, there exist a basis (D̄_{1}, D̄_{2}) of Ḡ such that

(45)  q_{(J,λ)}(XD̄_{1} + Y D̄_{2}) = q_{N,m,d,k,r}(X, Y), for all X, Y ∈ ℤ.

Thus, D̄_{1} is a primitive element of NS(J, λ) and q_{(J,λ)}(D̄_{1}) = q_{N,d,m,k,r}(1, 0) = N^{2}, so by [?], Theorem 18, there is an elliptic subgroup E′ ≤ J such that D̄_{1} = cl′(E′) + ⟨θ̄⟩ and (E′.θ̄) = N, where θ̄ ∈ NS′(J) is such that λK̄ = ϕθ̄.

Let E = (E′)^{⊥}, where (E′)^{⊥} is as in [?], Proposition 10(a). Applying the latter to E ≤ J, and noting that E^{⊥} = E′, we have that (E, E′, ψ, π) is an N-presentation of (J, λ), for some ψ : E[N] → E′[N] and isogeny π : A := E × E′ → J. Furthermore, by [?], equation (6), we have that

(46)  π^{∗}E ≡ N^{2}θ_{E′} and π^{∗}E′ ≡ N^{2}θ_{E},

where θ_{E} = pr_{E}^{∗}0_{E} = D′(1, 0, 0) and θ_{E′} = pr_{E′}^{∗}0_{E′} = D′(0, 1, 0).

We next observe that there exists a D̃_{2} ∈ NS′(J) such that

(47)  D̄_{2} ≡ D̃_{2} + ⟨θ̄⟩ and (D̃_{2}.θ̄) = T_{3} := rN̄ + knd.

Indeed, by [?], Lemma 30, such a divisor class D̃_{2} exists if and only if T_{3} ≡ T_{2} (mod 2), where T_{2} := q_{(J,λ)}(D̄_{2}). To verify this congruence, note that by (??) and the definition of q_{N,m,d,k,r} we have that T_{2} = q_{(J,λ)}(D̄_{2}) = q_{N,m,d,k,r}(0, 1) = n^{2}t + 2k(t − d)r + N̄^{2}r^{2}, where n = (k^{2}d + 1)/N̄. Now since t = d(k^{2}d + 4) ≡ (kd)^{2} ≡ kd (mod 2), we see that nt − kd ≡ (n − 1)kd (mod 2), and hence n(nt − kd) ≡ 0 (mod 2). Thus T_{2}−T_{3} = N̄^{2}r^{2}+2k(t−d)r+n^{2}t−(rN̄ +knd) ≡ rN̄(rN̄ −1)+n(nt−kd) ≡ 0 (mod 2), and so there exists D̃_{2} ∈ NS′(J) satisfying (??).

Put G = ℤθ̄ + ℤD̃_{1} + ℤD̃_{2}, where D̃_{1} = E′, and let H̃′ be the primitive hull of π^{∗}G in NS′(A). Since Ḡ = G/⟨θ̄⟩ is primitive in NS′(J, λ) = NS′(J)/⟨θ̄⟩), we see that G is primitive in NS′(J). Thus π^{∗}G is primitive in π^{∗}NS′(J), and so it follows that H̃′ ∩ π^{∗}NS′(J) = π^{∗}G.

By (??) we see that θ_{E}, θ_{E′} ∈ H̃′. Since rank(H̃′) = rank(π^{∗}G) = rank(G) = 3, it follows that H̃′ = ℤθ_{E} + ℤθ_{E′} + ℤD′(0, 0, h), for some h ∈ End(E, E′). Note that h is necessarily cyclic because ℤh = D_{E,E′}^{−1}(H̃′)/D_{E,E′}^{−1}(⟨θ_{E}, θ_{E′}⟩) is primitive in Hom(E, E′). Thus H̃′ = H̃, where H = ℤh.

Put d′ = deg(h) and m′ = m_{ψ}(H). Then the proof of Theorem ??(e) (applied to (E, E′, ψ)) shows that the restriction of q_{(J,λ)} to Ḡ has type (N, m′, d′) because Ḡ = G/⟨θ̄⟩ and H̃ ∩ π^{∗}NS′(J) = π^{∗}G. Thus, this restriction has discriminant −16d′(m′)^{2}.

On the other hand, by (??) this restriction has discriminant −16dm^{2} = disc(q_{N,m,d,k,r}), so d′(m′)^{2} = dm^{2}. Now since gcd(N̄, d) = 1 by (??), we see that gcd(dm^{2}, N^{2}) = m^{2} and similarly gcd(d′(m′)^{2}, N^{2}) = (m′)^{2}. Thus m′ = m and d′ = d.

We next determine π^{∗}D̃_{2}. Since π^{∗}D̃_{2} ∈ H̃, we have π^{∗}D̃_{2} = D′(a, b, ch), for some a, b, c ∈ ℤ. Put Γ_{h}^{∗} := D′(0, 0, h). Then we have

(48)  π^{∗}θ̄ = Nθ, π^{∗}D̃_{1} = N^{2}θ_{E} and π^{∗}D̃_{2} = bθ + (a − b)θ_{E} + cΓ_{h}^{∗}.

Since (θ, θ_{E}, Γ_{h}^{∗}) is a basis of H and (π^{∗}θ̄, π^{∗}D̃_{1}, π^{∗}D̃_{2}) is one of π^{∗}G, we see from (??) that [H̃ : π^{∗}G] = |N · N^{2} · c|. On the other hand, since H̃ ∩ π^{∗}NS′(J) = π^{∗}G, we have by Theorem ?? that [H̃ : π^{∗}G] = mN^{3}, and so we see that |c| = m. Thus, by replacing h by −h if c < 0, we may assume that m = c. We now show:

(49) a = m(k(t − 2d) + N̄^{2}r), b = −mkd and so π^{∗}(D̃_{2}) = m(rD_{1} + D_{2}).

To prove this, write q = q̃_{(J,λ)}. Then by (??) we have that

(50) q(D + D′) − q(D) − q(D′) = 2(D.θ̄)(D′.θ̄) − 4(D.D′), ∀D, D′ ∈ NS′(J).

We apply this to D = D̃_{1} and D′ = D̃_{2}. Since (D̃_{1}.θ̄) = (E′.θ̄) = mN̄ and (D̃_{2}.θ̄) = T_{3} by (??) and since q(D̃_{1} + D̃_{2}) − q(D̃_{1}) − q(D̃_{2}) = 2mT_{1} by (??), we see that

(51)  4(D̃_{1}.D̃_{2}) = 2mN̄T_{3} − 2mT_{1} = −4mkd,

where the second equality holds because T_{1} = N̄^{2}r + k(t − d) = N̄^{2}r + kd(N̄n + 2) = N̄T_{3} + 2kd. From this and (??) we see that N^{2}b = N^{2}(θ_{E}.π^{∗}D̃_{2}) = (π^{∗}D̃_{1}.π^{∗}D̃_{2}) = N^{2}(D̃_{1}.D̃_{2}) = −N^{2}mkd, which proves the second equality of (??). Moreover, since N(a + b) = N(θ.π^{∗}D̃_{2}) = (π^{∗}θ̄.π^{∗}D̃_{2}) = N^{2}(θ̄.D̃_{2}) = N^{2}T_{3} by (??), we see that a = NT_{3} − b = m(N̄T_{3} + kd) = m(T_{1} − kd) = m(k(t − 2d) + N̄^{2}r). This proves the first equality of (??), and it is clear that the third equality follows from the first two.

Since m = m_{ψ}(ℤh), we have by Theorem ??(a) that ψN̄ = ρN̄(k′h) for some k′ ∈ ℤ, and then mψ = mk′h. Thus, applying [?], Corollary 40, to D = π^{∗}D̃_{2} = mD(a′, b′, h), where a′ = k(t − 2d) + N̄^{2}r and b′ = −kd, we see that ρN̄(h) = ρN̄(b′k′h) = ρN̄(−kdk′h). Since −k^{2}d ≡ 1 (mod N̄), we have that ρN̄(−k^{2}dh) = ρN̄(h) = ρN̄(−kdk′h), and so ρN̄(kh) = ρN̄(k′h) because gcd(kd, N̄) = 1. Thus ψN̄ = ρN̄(k′h) = ρN̄(kh), and hence (??) holds.

(b) From part (a) we know that H̃ ∩ π^{∗}NS′(J) = π^{∗}G, and that (π^{∗}θ, π^{∗}D̃_{1}, π^{∗}D̃_{2}) is a basis of π^{∗}G. Thus, (??) follows from (??) and (??). The last assertion follows from Theorem ??(c), (d) because m(rD_{1} + D_{2}) ∈ π^{∗}NS′(J) by (??).

(c) In view of part (b) we are in the situation of Theorem ?? and so qḠ ∼ q_{N,m,d,k,r} ∼ q_{M,N,d,k} by Proposition ??. □

5 The existence theorem

We now prove that for every binary quadratic form q of type (N, m, d) there exists a principally polarized abelian surface (A, λ) such that its refined Humbert invariant q_{(A,λ)} is equivalent to q. As a first step for this, we prove:

Theorem 32 (Existence Theorem) Let E/K and E′/K be two elliptic curves over an algebraically closed field K = \overline{K} with the property that the degree form q_{E,E′} primitively represents d ≥ 1. Then for any binary quadratic form q of type (N, m, d) with char(K) ∤ N there exists an anti-isometry ψ : E[N] → E′[N] such that the refined Humbert invariant q_{(A_{ψ},λ_{ψ})} of the abelian variety A_{ψ} = A/Graph(−ψ) primitively represents q.

Proof. By hypothesis, there is a cyclic isogeny h ∈ Hom(E, E′) with deg(h) = d, and by Theorem ?? there is a primitive matrix M of discriminant −d such that q is equivalent to q_{M,N}.

To find the desired anti-isometry ψ, fix an anti-isometry ψ′ : E[N] → E′[N] and a basis B = (P_{1}, P_{2}) of E[N](K). Then by Remark ?? there is primitive matrix M′ of discriminant −d such that [(ψ′)^{−1}ρ_{N}(h)]_{B} ≡ M′ (mod N).

Since M and M′ are both primitive of discriminant −d, there are matrices M_{1}, M_{2} ∈ SL_{2}(ℤ) such that M_{1}MM_{2} = M′. Put P = (M_{1}M_{2})^{−1} ∈ SL_{2}(ℤ) and let g ∈ End(E[N]) be such that [g]_{B} ≡ P (mod N). Then ψ := ψ′ ∘ g is an anti-isometry and [ψ^{−1}ρ_{N}(h)]_{B} = [g^{−1}(ψ′)^{−1}ρ_{N}(h)]_{B} ≡ P M′ ≡ M_{2}^{−1}MM_{2} (mod N). By Proposition ?? (and Remark ??) we see that q_{(A_{ψ},λ_{ψ}}) primitively represents q_{M_{2}^{−1}MM_{2},N}. But q_{M_{2}^{−1}MM_{2},N} = q_{M,N} by Corollary ?? and so the assertion follows. □

Corollary 33 Let E be an elliptic curve over an algebraically closed field K, and let q be a binary quadratic form of type (N, m, d) with char(K) ∤ N. Then there exists a principally polarized abelian surface (A, λ) with A isogenous to E × E such that q_{(A,λ)} primitively represents q. In particular, if End(E) = ℤ, then q_{(A,λ)} is equivalent to q.

Moreover, if q(X, Y) ≠ 1, for all X, Y ∈ ℤ, and if End(E) = ℤ, then there is a curve C/K such that q_{C} is equivalent to q.

Proof. It is well-known that there exists a cyclic subgroup scheme H ≤ E of order d. (Indeed, if char(K) ∤ d, then we can take H = ⟨P ⟩, where P ∈ E(K) is a point of order d. On the other hand, if d = p^{r}d_{0}, where p = char(K) > 0 and p ∤ d_{0}, then we can take H = F_{p^{r}}^{−1}(⟨P_{0}⟩), where F_{p^{r}} : E → E^{(p^{r})} denotes the Frobenius morphism and P_{0} ∈ E^{(p^{r})}(K) has order d_{0}.) Let h : E → E′ := E/H be the associated cyclic isogeny. Then q_{E,E′} primitively represents d, and so the first assertion follows directly from Theorem ?? because A_{ψ} ∼ E × E′ ∼ E × E.

Moreover, if End(A) = ℤ, then rank(NS(A_{ψ})) = rank(NS(E × E)) = 3, so q_{(A_{ψ},λ_{ψ})} is also a binary quadratic form and hence it follows that q_{(A_{ψ},λ_{ψ})} is equivalent to q.

The last assertion follows immediately from Proposition 6 of [?], which states that the given condition on q ∼ q_{(A_{ψ},λ_{ψ})} is equivalent to the condition that (A_{ψ}, λ_{ψ}) ≃ (J_{C}, λ_{C}) for some curve C/K of genus 2. □

Corollary 34 Let K be an algebraically closed field which is not the algebraic closure of a finite field, and let q be a binary quadratic form of type (N, m, d) with char(K) ∤ N. Then there exists a principally polarized abelian surface (A, λ) over K such that q_{(A,λ)} is equivalent to q. Moreover, if q(X, Y) ≠ 1, for all X, Y ∈ ℤ, then there is a curve C/K such that q_{C} is equivalent to q.

Proof. Since K is not the algebraic closure of a finite field, there exists an elliptic curve E/K with End(E) = ℤ. Thus the assertions follow from Corollary ??. □

6 Application to elliptic involutions

We now consider the special case that C/K is a curve of genus 2 which has a presentation (E, E′, ψ) of degree 2. As we shall see presently, this is equivalent to assumption that C/K has an elliptic involution, i.e., an automorphism σ ∈ Aut(C) of order 2 such that σ ≠ σ_{C}, where σ_{C} denotes the hyperelliptic involution of C. Let I(C) denote the set of elliptic involutions, and for σ ∈ Aut(C) let f_{σ} : C → C_{σ} := C/⟨σ⟩ denote the quotient cover.

Proposition 35 The rule σ ↦ f_{σ} defines a bijection between the set I(C) of elliptic involutions and the set E_{2}(C) of equivalence classes of elliptic subcovers of C/K, and hence the rule σ ↦ E_{σ} := f_{σ}^{∗}J_{C_{σ}} defines a bijection between I(C) and the set S_{2}(J_{C}, λ_{C}) of elliptic subgroups of J_{C} of λ_{C}-degree 2.

Proof. Let σ ∈ I(C). Since deg(f_{σ}) = ord(σ) = 2, the uniqueness of the hyperelliptic cover implies that g_{C_{σ}} > 0, and hence g_{C_{σ}} = 1 by Riemann-Hurwitz. By [?], Corollary 1.4, C_{σ} has a K-rational point and hence f_{σ} is an elliptic subcover of degree 2. Note that f_{σ} is uniquely defined by σ up to equivalence of covers. We thus obtain a map f_{C} : I(C) → E_{2}(C). This map is injective because if f_{σ} is equivalent to f_{σ′}, then ⟨σ⟩ = Gal(f_{σ}) = Gal(f_{σ′}) = ⟨σ′⟩, and hence σ = σ′. Moreover, this map is surjective because if f : C → E is an elliptic subcover of degree 2, then it is separable and hence Galois, so Gal(f) = ⟨σ⟩ with σ ∈ I(C), and then f_{σ} is equivalent to f. This proves the first statement, and the second statement follows from [?], Theorems 20 and 18. □

Proposition 36 Let σ ∈ I(C), and put σ′ = σσ_{C}. Then σ′ ∈ I(C) and there exists an isomorphism ψ_{σ} : E_{σ}[2] → E_{σ′}[2] and an isogeny π_{σ} : E_{σ} × E_{σ′} → J_{C} such that (E_{σ}, E_{σ′}, ψ_{σ}, π_{σ}) is a 2-presentation of C/K. Moreover, every 2-presentation of C/K is of this form.

Proof. Since σ_{C} lies in the centre of Aut(C), we see that σ and σ_{C} commute, and so it follows that σ′ ≠ σ_{C} has order 2, so σ′ ∈ I(C). We next show that (E_{σ})^{⊥} = E_{σ′} in the notation of [?], Proposition 10.

For this, consider f_{G} : C → C/G, where G := ⟨σ, σ_{C}⟩, and let f̄_{σ} : C_{σ} → C/G and f̄_{σ′} : C_{σ′} → C/G be the unique maps such that f̄_{σ} ∘ f_{σ} = f_{G} and f̄_{σ′} ∘ f_{σ′} = f_{G}. Note that G is a Klein 4-group, so f̄_{σ} and f̄_{σ′} also have degree 2. It thus follows (cf. [?], Lemma 3.3) that (f_{σ})_{∗}(f_{σ′})^{∗}(D) = (f̄_{σ})^{∗}(f̄_{σ′})_{∗}(D), for all D ∈ Div(C_{σ′}), and so (f_{σ})_{∗}(f_{σ})^{∗}(D) ∼ 0, for all D ∈ Div^{0}(C_{σ′}) because (f̄_{σ′})_{∗}(D) ∼ 0 as gC̄ = 0. It thus follows from [?], Remark 13, that E_{σ′} = f_{σ′}^{∗}J_{C_{σ′}} = (f_{σ}^{∗}J_{C_{σ}})^{⊥} = (E_{σ})^{⊥}, as claimed.

Now since E_{σ′} = (E_{σ})^{⊥}, we have f_{σ′} = (f_{σ})^{⊥} in the notation of [?], Corollary 12, and so by that corollary there exists an isomorphism ψ_{σ} : E_{σ}[2] → E_{σ′}[2] and an isogeny π_{σ} : E_{σ} × E_{σ′} → J_{C} such that (E_{σ}, E_{σ′}, ψ_{σ}, π_{σ}) is a 2-presentation of C/K.

Moreover, if (E, E′, σ, π) is a 2-presentation of C/K, then by Proposition 10(b) of [?] we know that E and E′ are isomorphic to elliptic subgroups Ē, Ē′ ∈ S_{2}(J_{C}, λ_{C}) with Ē′ = Ē^{⊥}, and so Ē = E_{σ} and Ē′ = E_{σσ_{C}} for some σ ∈ I(C) by Proposition ?? and by what was proved above. □

Corollary 37 Let C/K be curve of genus 2, where char(K) ≠ 2, and let d ≥ 1. Then the following conditions are equivalent.

(i) There exists σ ∈ I(C) and a cyclic isogeny h : E_{σ} → E_{σσ_{C}} of degree d.

(ii) q_{C} primitively represents a binary form of type (2, m, d), with m = 1 or m = 2.

Proof. (i) ⇒ (ii): By Proposition ??, there exist ψ and π such that (E_{σ}, E_{σσ_{C}}, ψ, π) is a 2-presentation of C/K. Since h ∈ Hom(E_{σ}, E_{σσ_{C}}) is cyclic of degree d, Theorem ?? shows that q_{C} primitively represents a binary form of type (2, m, d) with m|2.

(ii) ⇒ (i): By Theorem ?? we may assume that the binary form q represented by q_{C} is q = q_{2,m,d,k,r}, for some suitable k, r ∈ ℤ with k^{2}d ≡ −1 (mod \frac{2}{m}). Applying Theorem ?? shows that there exists an 2-presentation (E, E′, ψ, π) of C/K and a cyclic isogeny h ∈ Hom(E, E′) of degree d. By Proposition ?? we have that there exists an elliptic involution σ on C such E and E′ are as in (i), and so the assertion follows. □

Remark 38 In their paper, Accola and Previato[?] mention on p. 141-142 that “a condition for [two degree-2 elliptic subcovers] being isogenous to any degree does not appear to be known”. The above result can be viewed as a contribution towards filling in this knowledge gap. In particular, we can use it prove the following result.

Corollary 39 Let E/K be an elliptic curve with End(E) = ℤ, where K is an algebraically closed field of char(K) ≠ 2, and let d ≥ 1 be an integer. Then there exists a curve C/K with two elliptic subcovers f : C → E_{i} of degree 2 and a cyclic isogeny h : E_{1} → E_{2} of degree d such that E_{1} ≃ E.

Proof. Let q be a binary form of type (2, m, d), where m|2. Then q is equivalent to one of the forms listed in Proposition ??. Now all the forms listed there with the exception of the form [4, 0, 1], which has type (2, 1, 1), are reduced, so have 4 as their minimum, and hence satisfy condition (iv) of Theorem ??. (Thus, for d = 1 take m = 2.) By Corollary ?? we thus see that there exists a curve C/K with q_{C} equivalent to q, so C satisfies condition (ii) of Corollary ??, and hence the assertion follows from condition (i) of the corollary. □

We now study the case that C/K has several elliptic involutions in some detail. We first observe:

Proposition 40 Let σ_{1}, σ_{2} ∈ I(C) and put Ē_{σ_{1}} = E_{σ} + ℤθ_{C} ∈ NS(J_{C}, λ) and M̄ = M̄(σ_{1}, σ_{2}) = ℤĒ_{σ_{1}} + ℤĒ_{σ_{2}}. If σ_{2} ≠ σ_{1}, σ_{1}σ_{C}, then rankM̄ = 2 and

(52)  q_{C}(XĒ_{σ_{1}} + Y Ē_{σ_{2}}) = 4X^{2} + 4rXY + 4Y^{2}, for all X, Y ∈ ℤ,

where r = 2 − (E_{σ_{1}}.E_{σ_{2}}). Furthermore, |r| ≤ 1.

Proof. In view of (the proof of) Proposition ??, the hypothesis implies that E_{σ_{2}} ≠ E_{σ_{1}}, (E_{σ_{1}})^{⊥}, so Ē_{σ_{2}} ≠ ±Ē_{σ_{1}}. Thus, since Ē_{σ_{i}} is primitive in NS(J_{C}, λ_{C}) (cf. [?], Theorem 18), it follows that Ē_{σ_{1}} and Ē_{σ_{2}} are linearly independent, so rank(M̄) = 2.

Since q_{C}(Ē_{σ_{i}}) = 4 (cf. [?], Theorem 18) and since q_{C}(Ē_{σ_{1}} + Ē_{σ_{2}}) = 4^{2} − 2(E_{σ_{1}} + E_{σ_{2}})^{2} = 16 − 4(E_{σ_{1}}.E_{σ_{2}}) = 8 + 4r, we see that (??) holds.

Moreover, since E_{σ_{2}} ≠ E_{σ_{1}}^{⊥} = E_{σ_{1}σ_{C}}, we have that (E_{σ_{2}}.E_{σ_{1}}^{⊥}) ≥ 1. Thus, since E_{σ_{1}} + E_{σ_{1}}^{⊥} = 2θ_{C}, we have that 1 ≤ (E_{σ_{2}}.E_{σ_{1}}) = 4 − (E_{σ_{2}}.E_{σ_{1}}^{⊥}) ≤ 3, and so we see that |r| ≤ 1. □

Corollary 41 In the situation of Proposition ?? we have that M̄(σ_{1}, σ_{2}) is a primitive submodule of NS(J_{C}, λ_{C}). Thus, q_{C} primitively represents the form q, where q = [4, 0, 4], if r = 0, and q = [4, 4, 4], if r = ±1.

Proof. Let M̄′ be the primitive hull of M̄ = M̄(σ_{1}, σ_{2}) in NS(J_{C}, λ_{C}), and put c = [M̄′ : M̄]. Let q′ and q be the restriction of q_{C} to M̄′ and to M̄, respectively. By Theorem ?? we know that q′ has type (2, m, d), for some (unique) integers m, d, so disc(q′) = −16m^{2}d and hence disc(q) = −16m^{2}c^{2}d.

Now if r = ±1, then q ∼ [4, ±4, 4] by the proposition, so disc(q) = −16 · 3, and hence m^{2}c^{2}d = 3, which shows that m = c = 1 and d = 3. Thus, M̄ = M̄′ is primitive, and q ∼ [4, 4, 4]. On the other hand, if r ≠ ±1, then q ∼ [4, 0, 4] by the proposition, so disc(q) = −16 · 4, and hence m^{2}c^{2}d = 4. If c ≠ 1, then c = 2, m = 1 and d = 1, so q′ has type (2, 1, 1). Thus q ∼ [4, 0, 1] (cf. Proposition ??), which represents 1 and hence does not satisfy property (iv) of Theorem ??, contradiction. Thus, c = 1 and hence M̄ = M̄′ is primitive and q ∼ [4, 0, 4]. □

Note that the above results are valid for an arbitrary ground field K. In order to analyze the above situation further, it is useful to impose some mild restrictions on the characteristic of K.

Proposition 42 Let σ_{1}, σ_{2} ∈ I(C) be such that σ_{2} ≠ σ_{1}, σ_{1}σ_{C}, and put σ = σ_{1}σ_{2} and n := ord(σ). Then G := ⟨σ_{1}, σ_{2}⟩ ≃ D_{n} is a dihedral group of order 2n. Moreover, if char(K) ≠ 2, then n = 3, 4, or 6, and if n is even, then σ^{n/2} = σ_{C}.

Proof. The first assertion is clear by elementary group theory. To prove the other assertions, assume that char(K) ≠ 2. Moreover, we may assume without loss of generality that K is algebraically closed. Then the set W_{C} = Fix(σ_{C}) of Weierstrass points of C consists of 6 distinct points. Since each τ ∈ Aut(C) commutes with σ_{C}, we see that τ permutes the elements of W_{C}. We first show: Claim: τ ∈ Aut(C) acts trivially on W_{C} ⇔ τ = 1 or τ = σ_{C}. Indeed, if τ ≠ 1 acts trivially, then τ has at least 6 fixed points, so the different divisor Diff(f_{τ}) of f_{τ} : C → C_{τ} = C/⟨τ⟩ has degree at least 6(m − 1), where m = ord(τ). Thus, by Riemann-Hurwitz we have m(−2) + 6(m − 1) ≤ 2, so m ≤ 2. If m = 2 and τ ≠ σ_{C}, then τ has no fixed points on W_{C} because ⟨τ, σ_{C}⟩ is a Klein 4-group, so f_{σ} and f_{σ_{C}} have disjoint ramification. Since this is a contradiction, we must have τ = σ_{C}. This proves the claim because the other implication is trivial.

Let σ̄_{i} ∈ Sym(W_{C}) ≃ S_{6} be the permutation induced by σ_{i}, for i = 1, 2. By the claim and our hypotheses have that σ̄_{1} ≠ σ̄_{2}. Moreover, since σ_{i} does not have any fixed points on W_{C}, we see that σ̄_{i} is a (2, 2, 2)-cycle. Then σ̄ = σ̄_{1}σ̄_{2} is either a (2, 2)-cycle (if σ̄_{1}, σ̄_{2} have a common transposition) or a (3, 3)-cycle, as is easy to check. Thus ord(σ̄) = 2 or 3, and so σ has order 2, 3, 4 or 6. Now ord(σ) ≠ 2 because in that case σ ∈ I(C) and so σ̄ is a (2, 2, 2)-cycle and not a (2, 2)-cycle as above. This proves that n = 3, 4 or 6.

To prove the last assertion, note that if n is even, then the above shows that σ̄ is an (\frac{n}{2}, \frac{n}{2})-cycle, so σ̄^{n/2} = (1), and hence σ^{n/2} = σ_{C} by the claim. □

Proposition 43 Let σ_{1}, σ_{2}, σ be as in Proposition ??, and assume that char(K) ≠ 2. (a) If ord(σ) = 4, then (E_{σ_{1}}.E_{σ_{2}}) = 2.

(b) If 3|n := ord(σ) and char(K) ≠ 3, then (E_{σ_{1}}.E_{σ_{2}}) = \{ \begin{matrix}3 & if n = 3, \\ 1 & if n = 6. \end{matrix}

Proof. Again, we may assume without loss of generality that K is algebraically closed. Fix P_{0} ∈ C(K) and consider the map φ_{i} = φ_{σ_{i},P_{0}} : C(K) → J_{C}(K) defined by

φ_{i}(P) = cl(f_{σ_{i}}^{∗}(f_{σ_{i}})_{∗}(P − P_{0}) = cl(P − P_{0} + σ_{i}(P) − σ_{i}(P_{0})), for P ∈ C(K).

To analyze this map we first observe that if P ∈ C(K) and if σ′_{i} := σ_{C}σ_{i}, then

(53) h^{0}(P + σ_{i}(P)) > 1 ⇔ ∃P′ : P + σ_{i}(P) = P′ + σ_{C}(P′) ⇔ P ∈ Fix(σ′_{i}).

Indeed, h^{0}(P + σ_{i}(P)) > 1 if and only if P + σ_{i}(P) is a canonical divisor of C. Now every effective canonical divisor on C is of the form P +σ_{C}(P′), for P′ ∈ C(K), and so the first equivalence follows. To verify the second, note that if P +σ_{i}(P) = P′+σ_{C}(P′), then either P = P′ and σ_{i}(P) = σ_{C}(P′), so P = σ_{C}σ_{i}(P) ∈ Fix(σ′_{i}) or P = σ_{C}(P′) and σ_{i}(P) = P′, so P = σ_{i}σ_{C}(P) ∈ Fix(σ′_{i}). Conversely, if P ∈ Fix(σ′_{i}), then σ_{i}(P) = σ_{C}(P), so P + σ_{i}(P) = P + σ_{C}(P). This proves (??).

Using (??), we see easily that

(54)  φ_{i}(P_{1}) = φ_{i}(P_{2}) ⇔ P_{2} = P_{1} or P_{2} = σ_{i}(P_{1}).

Indeed, clearly φ_{i}(P_{1}) = φ_{i}(P_{2}) ⇔ cl(P_{1} + σ_{i}(P_{1})) = cl(P_{2} + σ_{i}(P_{2})). If h^{0}(P_{1} + σ_{i}(P_{1})) = 1, then this latter condition is equivalent to the equality P_{1} + σ_{i}(P_{1}) = P_{2} + σ_{i}(P_{2}), and so (??) follows in this case. If h^{0}(P_{1} + σ_{i}(P_{1})) > 1, then by (??) we see that the latter condition is equivalent to P_{1}, P_{2} ∈ Fix(σ′_{i}). Now if P ∈ Fix(σ′_{i}), then Fix(σ′_{i}) = {P, σ_{i}(P)} because |Fix(σ′_{i})| = 2 and σ_{i} and σ′_{i} commute and have disjoint fixed points. This proves (??) in this case as well.

Now suppose that P_{0} ∈ Fix(σ), i.e., that σ_{1}(P_{0}) = σ_{2}(P_{0}). Then we have:

(55)  φ_{1}(P) ∈ E_{σ_{1}}(K) ∩ E_{σ_{2}}(K) ⇔ P ∈ Fix(σ) ∪ Fix(σ′_{1}).

Indeed, since φ_{i}(C(K)) = E_{σ_{i}}(K), for i = 1, 2, we see that the condition of the left side of (??) is equivalent to the condition that φ_{1}(P) = φ_{2}(P′), for some P′ ∈ C(K). This in turn is equivalent to cl(P +σ_{1}(P)) = cl(P′ +σ_{2}(P′)) because σ_{1}(P_{0}) = σ_{2}(P_{0}). If h^{0}(P + σ_{1}(P)) = 1, then this is equivalent to P + σ_{1}(P) = P′ + σ_{2}(P′), for some P′ ∈ C(K), which in turn is equivalent to P ∈ Fix(σ). On the other hand, if h^{0}(P +σ_{1}(P)) > 1, then by (??) we have that P ∈ Fix(σ′_{1}). Conversely, if P ∈ Fix(σ′_{1}) then cl(P + σ_{1}(P)) = cl(P′ + σ_{C}(P′)) = cl(P′ + σ_{2}(P′) for P′ ∈ Fix(σ′_{2}), and so (??) follows.

(a) If ord(σ) = 4, then |Fix(σ)| = 2 (because Fix(σ) ⊂ Fix(σ^{2}) = W_{C} and σ̄ is a (2, 2)-cycle), so P_{0} ∈ Fix(σ) exists. Since Fix(σ) ⊂ W_{C} and Fix(σ′_{1}) are disjoint, we see that |φ_{1}(Fix(σ) ∪ Fix(σ′_{1}))| ≥ 2. Thus, (??) and (??) show that (E_{σ_{1}}.E_{σ_{2}}) ≥ 2. Moreover, since σ_{1}σ′_{2} = σσ_{C} also has order 4, we have by the same argument that (E_{σ_{1}}.E_{σ′_{2}}) ≥ 2. But since (E_{σ_{1}}.E_{σ_{2}}) + (E_{σ_{1}}.E_{σ′_{2}}) = 4 (cf. the proof of Proposition ??), the assertion follows.

(b) Now assume that char(K) ≠ 3 and that ord(σ) = 3. Since f_{σ} is tamely ramified, it follows from Riemann-Hurwitz that |Fix(σ)| = 1 or 4. Since Fix(σ) is σ_{C}-stable, the first case implies that Fix(σ) has a fixed point under σ_{C}, which contradicts the fact that σ̄ is a (3, 3)-cycle. Thus |Fix(σ)| = 4 (and g_{C_{σ}} = 0), so there exists P_{0} ∈ Fix(σ). Now Fix(σ) ∩ Fix(σ′_{1}) = ∅ because ⟨σ′_{1}⟩ is a maximal cyclic subgroup of ⟨σ, σ′_{1}⟩ ≃ D_{6}, so we obtain from (??) and (??) that (E_{σ_{1}}.E_{σ_{2}}) ≥ 3. Since (E_{σ_{1}}.E_{σ_{2}}) ≤ 3 by the proof of Proposition ??, this proves the assertion in this case.

Now suppose that ord(σ) = 6. Since σ^{3} = σ_{C} by Proposition ??, we see that σ_{1}σ′_{2} = σσ_{C} = σ^{4} has order 3, so by the above (applied to σ_{1}, σ′_{2}) we see that (E_{σ_{1}}.E_{σ′_{2}}) = 3, and hence (E_{σ_{1}}.E_{σ_{2}}) = 4 − (E_{σ_{1}}.E_{σ′_{2}}) = 1, as claimed. □

Remark 44 If K is algebraically closed and if char(K) ≠ 2, 3, then the above proof shows that

(56)  (E_{σ_{1}}.E_{σ_{2}}) = \frac{1}{2}|Fix(σ_{1}σ_{2})| + 1.

Indeed, if σ := σ_{1}σ_{2} has order 4 or 3, then this was implicitly shown in the above proof. If ord(σ) = 6, then (E_{σ_{1}}.E_{σ_{2}}) = 1 by Proposition ??(b), and we have Fix(σ) = ∅ (because Fix(σ) ⊂ Fix(σ^{3}) = W_{C} but σ̄ has type (3, 3) and hence has no fixed points on W_{C}), so (??) holds in this case as well. Thus (??) holds because there are no other cases by Proposition ??.

Proof of Theorem ??. (a) Suppose first that we have a subgroup G ≤ Aut(C) with G ≃ D_{n}, where n = 4 (respectively, n = 6). Then G = ⟨σ_{1}, ρ⟩, where ord(ρ) = n and ord(σ_{1}) = 2. Put σ_{2} := σ_{1}ρ. Then ρ = σ_{1}σ_{2} and ord(σ_{2}) = 2, so σ_{1} and σ_{2} do not commute and hence σ_{2} ≠ σ_{1}, σ_{1}σ_{C}. Thus, σ_{1} and σ_{2} satisfy the conditions of Corollary ??, and so q_{C} primitively represents q = [4, 4r, 4] with r = 2 − (E_{σ_{1}}, E_{σ_{2}}). By Proposition ?? we see that r = 0 if n = 4 and r = 1 if n = 6. This proves one direction of the assertion.

Conversely, suppose that q_{C} primitively represents q_{n}, where n = 4 (respectively, n = 6). This means that there exist D̄_{1}, D̄_{2} ∈ NS(J_{C}, λ_{C}) such that q_{C}(XD̄_{1}+Y D̄_{2}) = q_{n}(X, Y), ∀X, Y ∈ ℤ. Thus q_{C}(D̄_{i}) = 4, for i = 1, 2, and so by [?], Theorem 18, there exist elliptic subgroups E_{i} ≤ J_{C} of degree 2 such that E_{i} + ℤθ_{C} = D̄_{i}, for i = 1, 2. By Proposition ?? we see that E_{i} = E_{σ_{i}} with σ_{i} ∈ I(C). Moreover, since D̄_{2} ≠ ±D̄_{1}, it follows that σ_{1} and σ_{2} satisfy the hypotheses of Proposition ??, and so (E_{σ_{1}}.E_{σ_{2}}) = 2 − r, where r = 0 because q_{4} = [4, 0, 4] (respectively, r = 1 because q_{6} = [4, 4, 4]). By Propositions ?? and ?? we thus have that ord(σ_{1}σ_{2}) = n and that G := ⟨σ_{1}, σ_{2}⟩ ≃ D_{n}, as desired.

(b) Note first that if N^{2} is primitively represented by q_{n}, then it is also primitively represented by q_{C}, and so by [?], Theorem 20, each such representation gives rise to an elliptic subcover f : C → E of degree N.

Now N^{2} is primitively represented by q_{4} = [4, 0, 4] (respectively, by q_{6} = [4, 4, 4]) if and only if 2|N and N_{1}^{2} is primitively represented by [1, 0, 1], (respectively, by [1, 1, 1]), where N_{1} = \frac{N}{2} . But it is well-known that N_{1}^{2} is primitively represented by [1, 0, 1] (respectively, by [1, 1, 1]) if and only if N_{1}^{2} (and hence N_{1}) is a product of primes p_{i} ≡ 1 (mod 4) (respectively, p_{i} ≡ 1 (mod 3)); cf. Satz 161 of Landau[?].

(c) Since q_{C} represents q_{n}, which has type (2, m, d), for (m, d) = (2, 1) (respectively, for (m, d) = (1, 3)), we see by Proposition ?? and Corollary ?? that J_{C} ∼ E_{σ} ×E_{σ′} ∼ E_{σ}^{2}. Thus, if f : C → E is any elliptic subcover, then E is an isogeny factor of J_{C} and so E ∼ E_{σ}. Thus, the hypothesis implies that rank(NS(J_{C}, λ_{C})) = 2, and hence it follows that q_{C} is equivalent to q_{n}. Thus, by [?], Theorem 20, there is a bijection between the set E_{N} of equivalence classes of elliptic subcovers of degree N and the set of primitive representations of N^{2} by q_{n}. Thus, the statement about the existence of subcovers follows from what was said in the proof of part (b). The last assertion about the number of subcovers follows from the corresponding assertion about representations of q_{n}; cf. Satz 161 of Landau[?]. □

It is interesting to note the following alternate characterization of curves whose automorphism groups contain D_{4} as a subgroup.

Proposition 45 Let C/K be a genus 2 curve, where char(K) ≠ 2, 3. Then there exists G ≤ Aut(C) with G ≃ D_{4} if and only if there is an elliptic curve E/K and a K-rational ψ ∈ Aut(E[2]) of order 2 such that J_{C} ≃ (E × E)/Graph(ψ).

Proof. By the proof of Theorem ??(a), the first condition is equivalent to the existence of a 2-presentation (E_{σ}, E_{σ′}, ψ, π) of J_{C}/K and a primitive submodule M̄ = M̄(σ, σ′′) of NS(J_{C}, λ_{C}) such that the restriction of q_{C} is equivalent to [4, 0, 4]. Since the latter has type (2, 2, 1), this implies by Corollary ?? that there is an isomorphism h : E_{σ} →^{∼} E_{σ′}. We can thus replace E_{σ′} by E := E_{σ} and adjust ψ and π accordingly to obtain a 2-presentation of the form (E, E, ψ, π) of J_{C}/K with ψ ∈ Aut(E[2]) ≃ GL_{2}(2) ≃ S_{3}. Now since [4, 0, 4] = q_{2,2,1,0,0}, it follows from Theorem ??(d) that tr(ψ) = tr(1^{t}ψ) ≡ r ≡ 0 (mod 2). Now the elements of order 2 in GL_{2}(2) are conjugate to \binom{0 1}{1 0} and hence have trace 0, whereas the two elements of order 3 have trace 1. Now ψ ≠ 1 because otherwise m_{ψ} = 1, so we must have that ψ has order 2. This proves the one direction of the assertion, and the converse follows by reversing the steps. □

We now show that the classical Legendre curve family which was studied by Legendre in 1832 (cf. Krazer[?], p. 477) gives many examples of curves which satisfy the hypotheses of Proposition ?? and/or of Theorem ??(c).

Example 46 Let K be a field with char(K) ≠ 2 and let a ∈ K^{×}, a ≠ ±1. Then the plane curve

C_{a} : y^{2} = x(x^{2} − a)(x^{2} − a^{−1})

is a curve of genus 2 which is a twist of the classical Legendre curve

L_{κ} : y^{2} = x(1 − x^{2})(1 − κ^{2}x^{2})

because if a = b^{2}, then the transformation x = bX, y = Y transforms C_{a} to the curve Y^{2} = b(1 − X^{2})(1 − a^{2}X^{2}), which is a quadratic twist of L_{a}.

It is immediate that the rule σ : (x, y) ↦ (\frac{1}{x}, \frac{y}{x^{3}} ) defines an involution of C_{a}/K. Since σ ≠ σ_{C} : (x, y) ↦ (x, −y), it follows that σ is an elliptic involution.

Let F = K(x, y) denote the function field of C_{a}. Then the fixed field of F with respect to σ is K(u, v), where u = x + \frac{1}{x} and v = y(x + 1)/x^{2}. These are related by the cubic equation

E_{a} : v^{2} = (u + 2) (u^{2} − \frac{(a+1)^{2}}{a} ) ,

which defines an elliptic curve. Thus, the rule (x, y) ↦ (u, v) defines an elliptic subcover f : C_{a} → E_{a} of degree 2.

Now suppose that −1 = i^{2} is a square in K. Then the rule ρ : (x, y) ↦ (−x, iy) defines an automorphism of order 4 on C_{a}, and we have σρσ = ρ^{3} and ρ^{2} = σ_{C}. Thus, G = ⟨σ, ρ⟩ ≃ D_{4} is a dihedral subgroup of Aut(C_{a}), and so C_{a}/K satisfies the hypotheses of Theorem ??(a).

To find examples satisfying the hypotheses of Theorem ??(c), we need to impose further restrictions on a. For this, note that (by MAPLE) the j-invariant of E_{a} is

(57)  j(E_{a}) = \frac{64(a + 3)^{3}(3a + 1)^{3}}{(a + 1)^{2}(a − 1)^{4}} .

It is immediate that j(E_{a}) ∉ ℤ, if a ∈ ℤ (a ≠ 0, ±1), except for a = ±3. Indeed, if p is an odd prime which divides the denominator (a + 1)^{2}(a − 1)^{4} of j(E_{a}), then a ≡ ±1 (mod p) and then the numerator is not divisible by p because it is congruent to a power of 2 mod p. Thus, if j(E_{a}) is integral, then (a − 1)(a + 1) is (up to sign) a power if 2, and this can happen only for a = ±3. Thus, if a ≠ ±3, then End(E_{a} ⊗ K) = ℤ, for every extension field K of ℚ(i), and so these examples satisfy the hypotheses of Theorem ??(c).

Remark 47 Let C′_{a}/K denote the curve defined by y^{2} = (x^{3} − a)(x^{3} − a^{−1}), where a ∈ K^{×}, a ≠ ±1. If char(K) ≠ 2, 3, then C′_{a}/K is curve of genus 2 which has an elliptic involution defined by (x, y) ↦ (\frac{1}{x}, \frac{y}{x^{3}} ). Moreover, if K contains a primitive third root of unity ζ_{3}, then Aut(C′_{a}) contains a subgroup isomorphic to D_{6}, as is easy to see. Thus, a similar discussion as in Example ?? gives us examples for the D_{6} case of Theorem ??.

Finally, we consider the following example of a curve C/K for which the refined Humbert invariant is a ternary form.

Proposition 48 Let C/ℚ be the curve defined by y^{2} = x(x^{2} − 3)(x^{2} − \frac{1}{3}), and let K be any subfield of ℂ. Then:

(58) q_{C_{K}} ∼ \{ \begin{matrix}4X^{2} & if ℚ(i) ⊄ K, \\ 4X^{2} + 4Y^{2} & if ℚ(i) ⊂ K and ℚ(\sqrt{−3}) ⊄ K, \\ 4X^{2} + 4Y^{2} + 4Y Z + 4Z^{2} & if ℚ(i, \sqrt{−3}) ⊂ K. \end{matrix}

Thus, if K ⊅ ℚ(i), then C_{K}/K has precisely 2 elliptic subcovers, both of degree 2, and if ℚ(i) ⊂ K but ℚ(\sqrt{−3}) ⊄ K, then C_{K}/K has an elliptic subcover of degree N if and only if N = 2N_{1}, where N_{1} = 1 or N_{1} is a product of primes p_{i} with p_{i} ≡ 1 (mod 4). On the other hand, if ℚ(i, \sqrt{−3}) ⊂ K, then C_{K}/K has in addition an elliptic subcover of degree N whenever N = 2N_{2}, where N_{2} is a product of primes p_{i} with p_{i} ≡ 1 (mod 3).

Proof. By Example ?? we know that C/ℚ has an elliptic involution σ ∈ I(C), so by Proposition ?? we know that C/ℚ has a 2-presentation (E_{σ}, E_{σ′}, ψ, π). Thus, J_{C} ∼ E_{σ} × E_{σ′}, so by Lemma ?? we see that q_{C_{K}} is a quadratic form in 1 + r_{K} variables, where r_{K} = rank(Hom(E_{σ} ⊗ K, E_{σ′} ⊗ K)). Moreover, since Aut(C ⊗ ℚ(i)) contains a subgroup isomorphic to D_{4}, we see that E_{σ} ⊗ ℚ(i) ≃ E_{σ′} ⊗ ℚ(i); cf. Proposition ??. Since j(E_{σ}) = 0 by (??), we see that r_{K} = 1 in the second case of formula (??) and r_{K} = 2 in the third case. Moreover, r_{K} = 0 in the first case. (This can be verified either by verifying by a direct computation that E_{σ′} is a nontrivial quadratic twist of E_{σ} or by observing that the D_{4}-subgroup cannot be K-rational except when ℚ(i) ⊂ K and using Proposition ??.

In the first case we have q_{C_{K}} ∼ nX^{2}, for some n ≥ 1. Since C_{K} has an elliptic subcover of degree 2, we know that 4 is primitively represented by q_{C_{K}} (cf. [?], Theorem 20), and so n = 4 because n the only value which is primitively represented by nX^{2}. This proves (??) in the first case. Moreover, in the second case we are in the situation of Theorem ??(c), so (??) holds in this case as well.

Now suppose that ℚ(i, \sqrt{−3}) ⊂ K, so q_{C_{K}} is a ternary form. We first prove that m_{ψ_{K}} = 2. If not, then m_{ψ_{K}} = 1 because m_{ψ_{K}}|2, and then by Diem-Frey (cf. [?], Theorem 4) J_{C_{K}} ≃ (E_{σ} ×E_{σ′})_{K} ≃ E^{2}, where j(E) = 0. But this is impossible because this product surface cannot be a Jacobian by Hayashida-Nishi[?]; cf. [?], Remark 60. Thus, m_{ψ} = 2.

By Theorem ??(a) and Corollary ?? we know that NS(J_{C}, λ_{C}) has a primitive submodule M̄ such that the restriction of q_{C_{K}} to M̄ primitively represents [4, 0, 4] = q_{2,2,1,0,0}. Thus, by Theorem ?? we see that π^{∗}NS′(J_{C_{K}}) contains M := ℤ(2θ) + ℤ(4D_{1}) + ℤ(2D_{2}) as a primitive submodule, where D_{1} = D′(1, 0, 0) and D_{2} = D′(0, 0, 1_{E}). Thus, there exists D′_{3} ∈ NS(E^{2}) such that (2θ, 4D_{1}, 2D_{2}, D′_{3}) is a basis of π^{∗}NS′(J_{C_{K}}). Since m_{ψ} = 2, we have by [?], Corollary 38, that D′_{3} = 2D_{3}, for some D_{3} = D′(a, b, h). By replacing D′_{3} by D′_{2} − b(2θ), we may assume without loss of generality that b = 0. Moreover, since End(E) ≃ ℤ[ζ_{3}], there exists α ∈ End(E) of order 3 such that (1_{E}, α) is a basis of End(E). Thus, we can write h = c_{1} · 1_{E} + c_{2}α, and so, by replacing D′_{3} by D′_{3} − c_{1}(2D_{2}), we may assume that D′_{3} = 2D_{3}, where D_{3} = D′(a, 0, cα). It then follows from Proposition ?? that a ≡ ctr(α^{t}ψ) (mod 2). Now since the restriction of α to E[2] has order 3, and ψ has order 2, it follows that α^{t}ψ has order 2 and so tr(α^{t}ψ) = 0; cf. the proof of Proposition ??. Thus, a ≡ 0 (mod 2), and so by replacing D′_{3} by D′_{3} − 4\frac{a}{2}D_{1} we may assume that D′_{3} = 2cD_{3}, where D_{3} = D′(0, 0, α). Since θ, D_{1}, D_{2}, D_{3} is a basis of NS′(E^{2}), we see that [NS′(E^{2}) : π^{∗}NS′(J_{C_{K}})] = |2 · 4 · 2 · 2c| = 32|c|. But by Theorem ?? we know that this index equals m_{ψ}2^{2+2} = 2^{5}, so c = ±1, so we may take c = 1. Thus,

(59)  π^{∗}NS′(J_{C}) = ℤ(2θ) + ℤ(4D_{1}) + ℤ(2D_{2}) + ℤ(2D_{3})

where D_{3} = D′(0, 0, α), and so NS′(J_{C}) has a basis (θ_{C}, D̃_{1}, D̃_{2}, D̃_{3}) where π^{∗}D̃_{1} = 4D_{1}, π^{∗}D̃_{2} = 2D_{2}, and π^{∗}D̃_{3} = 2D_{3}.

Using the projection formula (cf. [?], Proposition 17) and (??), we thus obtain

q_{C}(XD̃_{1} + Y D̃_{2} + ZD̃_{3}) = \frac{1}{4}q_{(E^{2},θ)}(X(4D_{1}) + Y (2D_{2}) + Z(2D_{3}))
= \frac{1}{4}(16X^{2} + 4 deg(2(Y · 1_{E} + Z · α))),

where for X, Y, Z ∈ ℤ. Since deg(Y · 1_{E} + Z · α) = Y^{2} + Y Z + Z^{2}, we thus see that q_{C} ∼ X^{2} + Y^{2} + XY + Z^{2}, which proves the third case of (??).

The last assertion follows immediately from Theorem ??(a),(b) because the form 4Y^{2} + 4Y Z + 4Z^{2} is primitively represented by q_{C}. □

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