Jacobians isomorphic to a product of two elliptic curves and ternary quadratic

forms

Ernst Kani

1 Introduction

Let E_{1} and E_{2} be two elliptic curves over an algebraically closed field K. The purpose of this paper is to study the question of whether or not the product surface E_{1} × E_{2} can be the Jacobian of a (smooth, irreducible) curve C of genus 2. By properties of the Jacobian, this question is equivalent to the question of whether or not there is such a curve C on E_{1} × E_{2}.

This question was first investigated in 1965 by Hayashida and Nishi[7], [6] who obtained partial results. Later Ibukiyama, Katsura and Oort[8] settled the case that E_{1} and E_{2} are supersingular (see Theorem 5 below).

In studying the moduli spaces of genus 2 curves C whose Jacobians are isomorphic to a product of two elliptic curves, the following result was obtained in [11]:

Theorem 1 Suppose that Hom(E_{1}, E_{2}) = ℤh ≠ 0, and put d = deg(h). Then there is no genus 2 curve on E_{1} × E_{2} if and only if d = 1 or if d is an even idoneal number which is not divisible by 8. This is the case for the following 21 values of d,

d = 1, 2, 4, 6, 10, 12, 18, 22, 28, 30, 42, 58, 60, 70, 78, 102, 130, 190, 210, 330, 462, (1)

and for at most one more value d = d^{∗} > 462. If the Euler/Gauss Conjecture (or if the Generalized Riemann Hypothesis) is true, then no such extra d^{∗} exists.

This, therefore, answers the above question in the case that E_{1} and E_{2} are isogenous elliptic curves without complex multiplication, if we leave aside the difficult number-theoretic question of whether or not an extra idoneal number d^{∗} exists. (A discussion of this question and its history can be found in [12].)

In this paper we answer the above question in the remaining case that E_{1} and E_{2} are isogenous elliptic curves with complex multiplication. (Note that if E_{1} and E_{2} are not isogenous, then this question is uninteresting; cf. Remark 9.) Before stating the result, we make the important observation (cf. Corollary 8) that the existence of genus 2 curves on E_{1} × E_{2} depends only on the nature of the quadratic form q_{E_{1},E_{2}} on Hom(E_{1}, E_{2}) which is defined by

q_{E_{1},E_{2}}(f) := deg(f) for f ∈ Hom(E_{1}, E_{2}).

Note that by fixing a basis of Hom(E_{1}, E_{2}) we obtain an explicit binary quadratic form and hence (by considering all bases) a GL_{2}(ℤ)-equivalence class of forms.

Theorem 2 Let E_{1} ∼ E_{2} be two isogenous CM elliptic curves over K. Then there is no genus 2 curve on E_{1} × E_{2} if and only if q_{E_{1},E_{2}} is equivalent to one of the 15 forms f(x, y) = ax^{2} + bxy + cy^{2} whose coefficients (a, b, c) are in the following list:

L = {k(1, 1, 1) : k = 1, 2, 4, 6, 10} ∪ {k(1, 0, 1) : k = 1, 2, 6} ∪ {(1, 1, 2), (1, 1, 4)}
∪ {2(1, 1, c) : c = 3, 9} ∪ {2(1, 0, c) : c = 2, 5} ∪ {2(2, 0, 3)}.

In particular, there are only finitely many pairs (E_{1}, E_{2}) of isomorphism classes of CM-curves such that E_{1} × E_{2} has no genus 2 curve, and these can be determined explicitly; cf. Remark 60. However, since different pairs may give rise to isomorphic product surfaces, it is more useful to count isomorphism classes of CM abelian product surfaces (i.e. abelian surfaces which are isogenous to E ×E, where E is a CM elliptic curve). By using the results of [13], we obtain:

Corollary 3 Up to isomorphism, there are at most 15 isomorphism classes of CM abelian product surfaces over K which do not contain a genus 2 curve. Moreover, if char(K) = 0, then there are precisely 15 such surfaces.

These 15 product surfaces are given explicitly in Corollary 59 and Remark 60. Moreover, Remark 58 discusses what happens when char(K) ≠ 0.

Note that if we restrict attention to those CM-curves E_{i} for which End(E_{i}) is a maximal order, then q_{E_{1},E_{2}} is a primitive form (cf. Remark 38) and so Theorem 2 shows that there are only 4 such exceptional forms and 4 exceptional surfaces. Thus, the above result includes the result of Hayashida and Nishi[7], who considered only the maximal order case.

This result can also be turned into an existence theorem which is valid over an arbitrary ground field K_{0}, as follows.

Corollary 4 Let A/K_{0} be an abelian surface such that A is isogenous to E × E, where E/K_{0} is a CM elliptic curve with rank(End_{K_{0}}(E)) = 2, and let Δ(A) be the discriminant of the intersection pairing on the Néron-Severi group NS(A). Then there exist elliptic curves E_{1} and E_{2}/K_{0} such that A ≃ E_{1} × E_{2}. Moreover, if −Δ(A) is not equal to one of the 15 discriminants of the forms in L, i.e. if

Δ(A) ∉ {3, 4, 7, 12, 15, 16, 32, 44, 48, 80, 96, 108, 140, 144, 300}, (2)

then there is a genus 2 curve C/K_{0} on A and hence J_{C} ≃ A ≃ E_{1} × E_{2}.

Note that the above corollary is a special case of a more precise result (Theorem 61) which is proven in section 7.

For the sake of completeness we also mention the following analogue of Theorems 1 and 2 in the supersingular case which was proven (but not explictly stated) by Ibukiyama, Katsura and Oort [8].

Theorem 5 (Ibukiyama/Katsura/Oort) If E_{1} and E_{2} are two supersingular curves over K, then there is no genus 2 curve on E_{1} × E_{2} if and only if char(K) = 2 or 3.

The first main ingredient of the proof of Theorem 2 is the refined Humbert invariant q_{θ} which was defined in [11] (see also [10]). As is explained in §2, this allows us to translate the problem of finding genus 2 curves on E_{1} × E_{2} into a problem about the classification of quadratic forms with certain properties.

This classification problem is also of independent interest, for it can be viewed as a generalization of the problem of classifying idoneal numbers, as is explained in more detail in [12]. Indeed, one of the key steps of this result is the classification (given in §5) of (special) idoneal-valued binary quadratic forms which are introduced here in §3. (Roughly speaking, a special idoneal-valued form is a quadratic form whose small values are all special idoneal numbers, i.e. those idoneal numbers which appear in Theorem 1.)

In addition, it turns out that there is close connection between this problem and the class-number one problem for forms in r ≥ 3 variables. This problem, which was studied in a series of papers by Watson[20], is to classify all positive definite quadratic forms q whose class number c(q) := #gen(q) = 1; i.e. to determine those forms q whose genus gen(q) consists only of a single equivalence class.

As the following result shows, these two classification problems are also connected to other classification problems, and this connection is established via the θ-construction, which is an abstract version of the refined Humbert invariant q_{θ} in the context of quadratic forms. This construction, which associates to a pair (Q, θ) (where Q is an arbitrary quadratic form in r variables and θ ∈ ℤ^{r} satisfies Q(θ) = 1) a certain equivalence class Q_{θ} of forms in r − 1-variables, is studied in detail in §4, and leads to the following classification theorem of quadratic forms.

Theorem 6 Let q(x, y) = ax^{2} + bxy + cy^{2} be a positive-definite binary quadratic form such that either b is odd or that q(x, y) ≢ 3 (mod 4), for all x, y ∈ ℤ. If f_{q}(x, y, z) = z^{2}+4q(x, y) and Q(x, y, z, w) = xy−q(z, w), then the following conditions are equivalent:

(i) c(f_{q}) = 1;

(ii) 1 ∈ f′(ℤ^{3}), for all f′ ∈ gen(f_{q});

(iii) 1 ∈ Q_{θ}(ℤ^{3}), for all θ ∈ ℤ^{4} with Q(θ) = 1;

(iv) q is a special idoneal-valued form;

(v) q is equivalent to one of 15 forms of the list L of Theorem 2.

The proof this theorem, which uses the mass formula for ternary quadratic forms (cf. §6) and the analytic results of Weinberger[22] (cf. §5), occupies most of this paper (§4-7). As was mentioned above (and is explained in detail in §3), Theorem 2 follows immediately from it once one has established certain properties of the refined Humbert invariant which are presented in §2.

This research was partially supported by a Discovery Grant from the Natural Sciences and Engineering Research Council of Canada (NSERC), and also by the Graduiertenkolleg of the Institute of Experimental Mathematics (IEM) of the University of Duisburg/Essen. I would like to express my appreciation to Gerd Frey and to the IEM for their hospitality, and to thank him for his interest and helpful comments on this paper.

2 Curves on product surfaces

As was mentioned in the introduction, the first step in proving Theorem 2 is to translate the problem of finding genus 2 curves on the product surface E_{1} × E_{2} into a problem about quadratic forms. For this, we recall from [11] the following facts about such curves.

Let A be an abelian surface over an algebraically closed field K, and let P^{irr}(A) denote the set of smooth, irreducible genus 2 curves C on A. By the adjunction formula on A, the self-intersection number of C is C^{2} = 2, so

P^{irr}(A) ⊂ P(A) := {θ ∈ Div(A) : θ ≥ 0, θ^{2} = 2},

where Div(A) denotes the group of divisors on A. Here the notation P^{irr}(A) reflects the fact (due to Weil[21]) that if θ ∈ P(A), then θ ∈ P(A)^{irr} if and only if θ is an irreducible curve. (Thus, if θ is irreducible, then it is also smooth.)

To decide whether or not θ ∈ P(A) is irreducible, we shall use the refined Humbert invariant q_{θ} defined in [11] (which by [10] is closely related to the classical Humbert invariant defined by Humbert). This is the quadratic form defined by the formula

q_{θ}(D) = (D.θ)^{2} − 2D^{2}, D ∈ Div(A), (3)

where (.) denote intersection numbers. It is clear that q_{θ} can be viewed as a quadratic form on the Néron-Severi group of A, i.e. on the quotient group NS(A) = Div(A)/≡, where the equivalence relation D_{1} ≡ D_{2} (numerical equivalence) means that (D_{1}.D) = (D_{2}.D), ∀D ∈ Div(A). Moreover, a short computation (using the fact that θ^{2} = 2) shows that q_{θ} is actually a quadratic form on NS(A, θ) := NS(A)/ℤθ, and the Hodge Index Theorem shows that q_{θ} is positive definite (on NS(A, θ)).

One of the key properties of q_{θ} is the following irreducibility criterion (cf. [11], Proposition 6): if θ ∈ P(A), then

(4) θ is irreducible ⇔ q_{θ}(D) ≠ 1, for all D ∈ Div(A).

The above criterion translates the existence of genus 2 curves on A into a problem that only involves the integral quadratic form q_{A} defined by the intersection pairing on NS(A) ≃ ℤ^{ρ}. To make this precise, put

q_{A}(D) = \frac{1}{2}(D.D), for D ∈ NS(A),

which is an integral quadratic form by the Riemann-Roch theorem ([15], p. 150). Then the refined Humbert invariant q_{θ} = (q_{A})_{θ} associated to θ ∈ P(A) is given by

(q_{A})_{θ}(D) = β_{q_{A}}(D, θ)^{2} − 4q_{A}(D), for D ∈ NS(A), (5)

where β_{q_{A}} denotes the bilinear form associated to the quadratic form q_{A} (cf. §3).

Proposition 7 There is a genus 2 curve on the abelian surface A if and only if there is a θ ∈ NS(A) with q_{A}(θ) = 1 and (q_{A})_{θ}(D) ≠ 1, for all D ∈ NS(A).

Proof. If such a curve C exists, then its class θ = cl(C) in NS(A) satisfies q_{A}(θ) = \frac{1}{2}C^{2} = 1 and (q_{A})_{θ}(D) ≠ 1, for all D ∈ NS(A) by (4). Conversely, if θ ∈ NS(A) satisfies these properties, then by [10], Corollary 2.4, there is an effective curve C on A such that its class cl(C) equals either θ or −θ. Thus, C ∈ P(A). Since (q_{A})_{−θ} = (q_{A})_{θ}, it thus follows from (4) that C ∈ P(A)^{irr}.

So far, the above results are true for an arbitrary abelian surface A. If we now specialize to the case that A = E_{1} × E_{2} is a product surface, then we can relate q_{A} to the binary quadratic form q_{E_{1},E_{2}} defined by the degree map on Hom(E_{1}, E_{2}), i.e. by

q_{E_{1},E_{2}}(f) = deg(f), if f ∈ Hom(E_{1}, E_{2}) ≃ ℤ^{r},

where r = rank(Hom(E_{1}, E_{2})). (Note that this defines a GL_{r}(ℤ)-equivalence class of quadratic forms in r variables.)

Now by Proposition 22 of [11] we have an isomorphism NS(A) ≃ ℤ^{2}⊕Hom(E_{1}, E_{2}) and via this identification we have

q_{E_{1}×E_{2}}(x, y, f) = xy − q_{E_{1},E_{2}}(f), for x, y ∈ ℤ, f ∈ Hom(E_{1}, E_{2}). (6)

In other words, q_{E_{1}×E_{2}} ∼ (xy) ⊥ (−q_{E_{1},E_{2}}), where xy denotes the quadratic form defined by the hyperbolic plane (and ∼ denotes equivalence of quadratic forms). We thus obtain:

Corollary 8 Let E_{1} and E_{2} be two elliptic curves over K, and let q ∼ q_{E_{1},E_{2}} be a quadratic form in r variables. Put Q = xy ⊥ (−q). Then there is no curve of genus 2 on E_{1} × E_{2} if and only if for every θ ∈ ℤ^{r+2} with Q(θ) = 1 there exists an x = x_{θ} ∈ ℤ^{r+2} such that Q_{θ}(x) = 1.

Remark 9 Note that the above corollary applies to arbitrary elliptic curves E_{1}, E_{2} (including the supersingular case). It also applies to the case that E_{1} and E_{2} are not isogenous, i.e. r = 0. Then we have that Q(x, y) = xy, so the only solution of Q(θ) = 1 is θ = ±(1, 1) and then Q_{θ}(x, y) = (x+y)^{2}. Since Q_{θ}(1, 0) = 1, we conclude that in the non-isogenous case there is never a curve of genus 2 on E_{1} × E_{2}.

3 The main result

By Corollary 8, the geometric problem of finding genus 2 curves on the product surface A = E_{1} × E_{2} has been translated into a purely arithmetic problem involving the quadratic form q_{A}. In this section we formulate this problem in the context of quadratic forms and present the main result which connects the original problem with other interesting problems about quadratic forms.

For this, let Q(x_{1}, …, x_{r}) = ∑_{i≤j} a_{ij}x_{i}x_{j} be an integral quadratic form in r variables, and let

P(Q) = {x ∈ ℤ^{r} : Q(x) = 1}.

denote the set of representations of 1 by Q. For each θ ∈ P(Q), let Q_{θ} be defined by

Q_{θ}(x) = β_{Q}(x, θ)^{2} − 4Q(x), (7)

where β_{Q} is the bilinear form associated to Q. Since this is naturally a quadratic form on the quotient ℤ^{r}/ℤθ (cf. §4), we thus obtain (after choosing a basis of ℤ^{r}/ℤθ ≃ ℤ^{r−1}), an equivalence class Q̄_{θ} of quadratic forms in r − 1 variables. We call this construction of Q̄_{θ} the θ-construction; it will be studied in more detail in the next section.

The discussion in the previous section (cf. Corollary 8) leads us to consider the following problem.

Problem 10 Classify the positive definite quadratic forms q in r variables such that Q := xy ⊥ (−q) satisfies the following property:

Q_{θ} → 1, for all θ ∈ P(Q). (8)

Here, the symbol Q_{θ} → 1 means that Q_{θ} represents 1, i.e. that P(Q_{θ}) ≠ ∅.

Remark 11 If θ ∈ P(Q)^{ev} := {θ ∈ P(Q) : β_{Q}(x, θ) ≡ 0 (mod 2), ∀x ∈ ℤ^{r}}, then (7) shows that Q_{θ}(x) ≡ 0 (mod 4), for all x ∈ ℤ^{r}, so clearly Q_{θ} cannot represent 1. Thus, if P(Q)^{ev} ≠ ∅ or, equivalently, if P(Q) ≠ P(Q)^{odd} := P(Q) \ P(Q)^{ev}, then Q cannot satisfy (8).

In the case that r = 1, i.e. that q(x) = nx^{2}, for some n ≥ 1, Problem 10 was (implicitly) solved in [11]. In this case it was found that (8) holds if and only if n is an idoneal number (in the sense of Euler) satisfying certain extra conditions. Thus, by using the results of Euler, Grube[5] and Weinberger[22] on the classification of idoneal numbers, one obtains Theorem 1.

The key result in the above case was the fact (cf. [11], Proposition 15) that

{Q̄_{θ} : θ ∈ P(Q)^{odd}} = gen(x^{2} + 4ny^{2}); (9) here gen(f) denotes the genus of the (binary) quadratic form f. This yields a direct connection to the idoneal numbers of Euler because we have the well-known relation (due to Grube; cf. [12]) that

(10) n ≥ 1 is idoneal ⇔ #gen(x^{2} + ny^{2}) = 1.

A (partial) generalization of (9) is proved for all r ≥ 1 in the next section; cf. Theorem 20. For this, we require the (well-known) generalization of the notion of a genus for binary forms to forms in an arbitrary number of variables: if q is a quadratic form in r variables, then its genus gen(q) is the set of equivalence classes of quadratic forms in r variables which are genus-equivalent (or semi-equivalent) to q; cf. Watson[18], p. 72. (Here, as in [18], equivalence means GL_{r}(ℤ)-equivalence.)

Following Watson[20], we call

c(q) = #gen(q)

the class number of the form q. In view of (10), the following problem, which was studied by Watson in a series of papers in the years 1963–1978 (cf. [20] and the references therein), can be viewed as a generalization of the classification problem of idoneal numbers (cf. [12]):

Problem 12 (Watson) Find all the equivalence classes of positive definite primitive forms q with class number 1, i.e. with c(q) = 1.

As we shall see below in Theorem 13, Watson’s problem is closely related to Problem 10, but this fact is far from obvious.

Another problem that turns out to be is closely related to the above problems is the problem of classifying all (special) idoneal-valued quadratic forms which are defined as follows.

Definition. Let q(x_{1}, …, x_{r}) be a positive definite quadratic form in r variables. We say that q is an idoneal-valued form if its only (proper) values ≤ |Δ(q)| are idoneal numbers, i.e. if

(11) q → n ≤ |Δ(q)| ⇒ n is an idoneal number;

here, the symbol q → n means that q properly represents n, i.e. that there exist x_{1}, …, x_{r} ∈ ℤ with gcd(x_{1}, …, x_{r}) = 1 such that q(x_{1}, …, x_{r}) = n, and Δ(q) denotes the discriminant of q; cf. [18], p. 2.

In addition, we say that q is a special idoneal-valued form if

(12) q → n ≤ |Δ(q)| ⇒ 4n is an idoneal number,

and if in addition we have that q ↛ n, for any n ≡ 3 (mod 4) when Δ(q) ≢ 1 (mod 4), and that q ↛ n, for any n ≡ 3 (mod 4) with n < |Δ(q)| when Δ(q) ≡ 1 (mod 4).

At first sight it might seem unlikely that (special) idoneal-valued quadratic forms in 2 variables exist at all. However, there are some, and they are classified in the following theorem which can be viewed as the main result of this paper:

Theorem 13 Let q(x, y) = ax^{2} + bxy + cy^{2} be a positive definite binary quadratic form, and let f_{q}(x, y, z) = z^{2} + 4q(x, y) and Q(x, y, z, w) = xy − q(z, w). Assume that P(Q) = P(Q)^{odd} or, equivalently, that either Δ := b^{2} − 4ac ≡ 1 (mod 4) or that q ↛ n, for any n ≡ 3 (mod 4). Then the following conditions are equivalent:

(i) c(f_{q}) = 1;

(ii) f′ → 1, for all f′ ∈ gen(f_{q});

(iii) Q_{θ} → 1, for all θ ∈ P(Q);

(iv) q is a special idoneal-valued form;

(v) q is equivalent to one of 15 forms whose coefficients (a, b, c) are in the following list:

L = {k(1, 1, 1) : k = 1, 2, 4, 6, 10} ∪ {k(1, 0, 1) : k = 1, 2, 6} ∪ {(1, 1, 2), (1, 1, 4)}
∪ {2(1, 1, c) : c = 3, 9} ∪ {2(1, 0, c) : c = 2, 5} ∪ {2(2, 0, 3)}.

The proof of this theorem (which is a restatement of Theorem 6) is quite long and will occupy the rest of this paper; cf. §7, where all parts are put together. Note that this theorem immediately implies Theorem 2 of the introduction, as we now show.

Proof of Theorem 2. By Corollary 8, there is no genus 2 curve on E_{1} × E_{2} if and only if q = q_{E_{1},E_{2}} satisfies condition (iii) of Theorem 13. Since E_{1} ∼ E_{2} has CM, it follows that rank(Hom(E_{1}, E_{2})) = 2, and so q is a positive definite binary quadratic form. Thus, the assertion of Theorem 2 follows from the equivalence (iii) ⇔ (v) of Theorem 13, together with Remark 11.

Note that the notion of a special idoneal-valued quadratic form can be used to unify the CM and non-CM cases (Theorems 1 and 2) as follows:

Corollary 14 Let E_{1} ∼ E_{2} be two isogenous elliptic curves over K, and assume that E_{1} is not supersingular. Then there is no genus 2 curve on E_{1} × E_{2} if and only if q_{E_{1},E_{2}} is a special idoneal-valued form.

Proof. If E_{1} has CM, i.e. if r = rank(Hom(E_{1}, E_{2})) = 2, then by the above proof of Theorem 2, this follows from the equivalence (iii) ⇔ (iv) of Theorem 13.

If E_{1} does not have CM, i.e. if End(E_{1}) = ℤ, then Hom(E_{1}, E_{2}) = ℤh, for some h and q_{E_{1},E_{2}}(xh) = nx^{2}, where n = deg(h). Since Δ(q_{E_{1},E_{2}}) = n, and n = q(1 · h) is the only value which is primitively represented by n, it is clear that q_{E_{1},E_{2}} is a special idoneal-valued form if and only if n ≢ 3 (mod 4) and 4n is not idoneal. Since this is equivalent to the condition that either n = 1 or that n is even but n ≢ 0 (mod 8) (cf. [11], Corollary 34 or [12], Theorem 36) we see from Theorem 1 that this condition is equivalent to the condition that E_{1} × E_{2} does not contain any genus 2 curve.

4 The θ-construction

In this section we study the θ-construction (cf. §3) in the general context of quadratic R-modules. Here R can be an arbitrary commutative ring (in which 2 is not a zerodivisor), but in the applications we only need the cases that R = ℤ, ℤ_{p} or ℝ.

Let (X, q) be a quadratic R-module. Thus, X is an R-module and q is a quadratic form in the sense of Milnor/Husemoller[14], p. 110, i.e. q : X → R is a function which satisfies (i) q(rx) = r^{2}q(x), for all r ∈ R and (ii) the map β_{q} : X × X → R defined by

β_{q}(x, y) = q(x + y) − q(x) − q(y), for all x, y ∈ X, (13)

is an R-bilinear map. If this is the case, then we call β_{q} the bilinear map associated to q. Note that β_{q}(x, x) = 2q(x).

Let θ ∈ P(q) = P(X, q) := {x ∈ X : q(x) = 1}, and put

q_{θ}(x) = β_{q}(x, θ)^{2} − 4q(x), for x ∈ X. (14)

Clearly, q_{θ} is again a quadratic form on X with associated bilinear form

β_{q_{θ}} = 2β_{θ}, where β_{θ}(x, y) = β_{q}(x, θ)β_{q}(y, θ) − 2β_{q}(x, y), ∀x, y ∈ R. (15)

Indeed, clearly β_{θ} is symmetric and R-bilinear and we have

q_{θ}(x) = β_{θ}(x, x), (16)

and so the formula β_{q_{θ}} = 2β_{θ} follows immediately.

Since β_{q}(θ, θ) = 2q(θ) = 2, it follows from the definition (15) of β_{θ} that

β_{θ}(x, rθ) = 0, for all r ∈ R, (17)

and so β_{θ} defines a bilinear form β̄_{θ} on the quotient module X̄_{θ} = X/Rθ. In particular, q_{θ} induces a quadratic form q̄_{θ} on X_{θ} and we have by (15) that βq̄_{θ} = 2β̄_{θ}. Thus, (X̄_{θ}, q̄_{θ}) is quadratic R-module, and if π_{θ} : X → X_{θ} = X/Rθ denotes the quotient map, then we have q_{θ}(x) = q̄_{θ}(π_{θ}(x)), for all x ∈ X.

Remark 15 (a) The above construction clearly generalizes the θ-construction presented in §3. To see this, note first that if q : ℤ^{r} → ℤ is an integral quadratic form in r variables, then (ℤ^{r}, q) is a quadratic ℤ-module, and if q′ : ℤ^{r} → ℤ is another quadratic form, then (ℤ^{r}, q) ≃ (ℤ^{r}, q′) ⇔^{def} ∃α ∈ Aut(ℤ^{r}) = GL_{r}(ℤ) such that q′ ∘ α = q ⇔ q ∼ q′, where ∼ denotes GL_{r}(ℤ)-equivalence of forms. Thus, the isomorphism classes of quadratic modules (X, q) with X ≃ ℤ^{r} can be identified with the equivalence classes of quadratic forms in r variables.

From this we see that if q : ℤ^{r} → ℤ is an integral quadratic form in r variables, and if θ ∈ P(q) = P(ℤ^{r}, q), then (ℤ^{r}/ℤθ, q̄_{θ}) defines an equivalence class q̄_{θ} of quadratic forms in r − 1 variables because the condition q(θ) = 1 implies that the quotient ℤ^{r}/ℤθ is torsionfree and hence is free of rank r − 1.

(b) For later use we observe that the θ-construction is compatible with basechange, i.e. if φ : R → R′ is a ring homomorphism, then for any θ ∈ P(X, q) we have an isomorphism

(\overline{(X_{R′})}_{θ⊗1}, \overline{(q_{R′})}_{θ⊗1}) →^{∼} (X̄_{θ}, q̄_{θ}) ⊗ R := ((X̄_{θ}) ⊗_{R} R′, q̄_{θ} ⊗ id_{R′}), (18)

where X_{R′} = X ⊗_{R} R′ and q_{R′} = q_{R} ⊗ R′ : X ⊗_{R} R′ → R ⊗_{R} R′ = R′. (Indeed, by the right-exactness of the tensor product, the canonical map π_{θ} ⊗id_{R′} : X_{R′} = X ⊗_{R} R′ → X̄_{θ} ⊗_{R} R′ induces an isomorphism X_{R′}/R′(θ ⊗ 1) ≃ X̄_{θ} ⊗_{R} R′ because R′(θ ⊗ 1) is the image of (Rθ) ⊗_{R} R′ in X_{R′}, and from this (18) follows immediately.)

As was explained in the previous section, we are interested in studying the set

Θ(X, q) = {(X̄_{θ}, q̄_{θ}) : θ ∈ P(q)}/≃

of isomorphism classes of quadratic modules resulting from the θ-construction. A first result in this direction is given by:

Proposition 16 If α ∈ Aut(X, q) := {α ∈ Aut_{R}(X) : q ∘ α = q}, then

q_{α(θ)} ∘ α = q_{θ}, for all θ ∈ P (q), (19)

and hence α induces an isometry ᾱ : (X̄_{θ}, q̄_{θ}) →^{∼} (X̄_{α(θ)}, q̄_{α(θ)}).

Proof. Clearly, the group Aut(X, q) acts (on the right) on P (q). If α ∈ Aut(X, q), then by (13) we see that β_{q} ∘ (α × α) = β_{q}, and so for x ∈ X we have q_{α(θ)}(α(x)) = β_{q}(α(x), α(θ))^{2} − 4q(α(x)) = β_{q}(x, θ)^{2} − 4q(x) = q_{θ}(x), which proves (19).

Moreover, since α(Ker(π_{θ})) = α(Rθ) = Rα(θ) = Ker(π_{α(θ)}), there is a unique R-module isomorphism ᾱ : X_{θ} →^{∼} X_{α(θ)} such that π_{α(θ)} ∘ α = ᾱ ∘ π_{θ}, and by (19) we see that ᾱ is an isometry.

Remark 17 The above result shows that the map θ ↦ (X̄_{θ}, q̄_{θ}) induces a surjection

P(X, q)/Aut(X, q) → Θ(X, q).

Now if [R : 2R] ≤ 2, then one can show that this is in fact a bijection; indeed, one can show more precisely that if θ_{1}, θ_{2} ∈ P(X, q), then any isometry ᾱ : (X̄_{θ_{1}}, q̄_{θ_{1}}) →^{∼} (X̄_{θ_{2}}, q̄_{θ_{2}}) can be lifted to an isometry α ∈ Aut(X, q) with α(θ_{1}) = θ_{2}. Since this is not needed below (and is somewhat tedious to prove), the proof of this fact will not be given here.

By combining the above proposition with a basic fact about isometries of quadratic modules over local rings, we obtain the following important result.

Proposition 18 If R is a local ring with 2 ∈ R^{×}, then Aut(X, q) acts transitively on P(X, q) and hence we have (X̄_{θ_{1}}, q̄_{θ_{1}}) ≃ (X̄_{θ_{2}}, q̄_{θ_{2}}), for all θ_{1}, θ_{2} ∈ P(X, q).

Proof. By [14], Lemma (4.2), we know that for every θ_{1}, θ_{2} ∈ P(X, q) there is a reflection α ∈ Aut(X, q) such that α(θ_{1}) = θ_{2}. Thus, Aut(X, q) acts transitively on P(X, q), and so the last assertion follows from this and Proposition 16.

The above result implies in particular that when p is an odd prime, then the q̄_{θ}’s are all p-adically equivalent (notation: ∼_{p}) in the sense of Watson[18], p. 50, and/or Jones[9], p. 82.

Corollary 19 If q : ℤ^{r} → ℤ is a quadratic form in r variables, and p is an odd prime or p = ∞, then q̄_{θ_{1}} ∼_{p} q̄_{θ_{2}}, for all θ_{1}, θ_{2} ∈ P(q).

Proof. If p is an odd prime, then 2 ∈ ℤ_{p}^{×}, and so by Proposition 18 we have that (\overline{(X_{p})}_{θ_{1}}, \overline{(q_{p})}_{θ_{1}}) ≃ (\overline{(X_{p})}_{θ_{2}}, \overline{(q_{p})}_{θ_{2}}), where q_{p} denotes the canonical extension of q to X_{p} = ℤ_{p}^{r} = X ⊗ ℤ_{p} (with X = ℤ^{r}). By base-change (cf. Remark 15(b)), this means that (X̄_{θ_{1}}, q̄_{θ_{1}}) ⊗ ℤ_{p} ≃ (X̄_{θ_{2}}, q̄_{θ_{2}}) ⊗ ℤ_{p}, which by definition means that q̄_{θ_{1}} ∼_{p} q̄_{th_{2}}.

If p = ∞, then by replacing ℤ_{p} by ℝ, an analogous argument yields the result.

If R is a (local) ring for which 2 ∉ R^{×}, then the results of Proposition 18 are in general not true. One reason for this is that there may exist proper Aut(X, q)- invariant subsets P(X, q)^{ev} and P(X, q)′ of P(X, q), which would therefore prevent Aut(X, q) from acting transitively on P(X, q). These sets are defined as follows:

P(X, q)^{ev} = {θ ∈ P(X, q) : β_{q}(x, θ) ∈ 2R, for all x ∈ X}
P(X, q)′ = {θ ∈ P(X, q) : q_{θ}(x) ∈ 1 + 8R, for some x ∈ X}

Thus, if 2 ∉ R^{×}, then P(X, q)′ ⊂ P(X, q)^{odd} := P(X, q) \ P(X, q)^{ev}. It is immediate that these sets are Aut(X, q)-invariant, and hence give rise to sets

Θ(X, q)^{ev}, Θ(X, q)^{odd} and Θ(X, q)′

by replacing in the definition of Θ(X, q) the set P(X, q) by P(X, q)^{ev}, P(X, q)^{odd}, and P(X, q)′, respectively. We shall prove:

Theorem 20 For any integral quadratic form q : X = ℤ^{r} → ℤ in r variables with Δ(q) ≠ 0, the set Θ(X, q)′ lies in a single genus, i.e. for any θ ∈ P(X, q)′ we have

gen(q̄_{θ}) ⊃ {q̄_{θ_{1}} : θ_{1} ∈ P(ℤ^{r}, q)′}/∼ . (20)

Moreover, if there exists θ_{0} ∈ P(X, q) with P(X̄_{θ_{0}}, q̄_{θ_{0}}) ≠ ∅, then P(X, q)′ = P(X, q)^{odd} and hence Θ(X, q)^{odd} = Θ(X, q)′ lies in a single genus.

As a first step towards proving this, we note the following:

Lemma 21 If R = ℤ_{2}, then P(X, q)′ = {θ ∈ P(X, q) : 1 ∈ q_{θ}(X)}. (21) Proof. Let θ ∈ P(X, q)′. Then there exists x ∈ X such that μ := q_{θ}(x) ∈ 1 + 8ℤ_{2}. Then μ is the square of a 2-adic unit, so μ = μ_{1}^{2}, for some μ_{1} ∈ ℤ_{2}^{×}. Then x_{1} := μ_{1}^{−1}x satisfies q_{θ}(x_{1}) = μ_{1}^{−2}q_{θ}(x) = 1, so 1 ∈ q_{θ}(X). This proves that P(X, q)′ is contained in the right hand side of (21), and so (21) follows since the other inclusion is trivial.

By the above lemma we see that the key to proving Theorem 20 is to analyze the condition that 1 ∈ q_{θ}(X) or, equivalently, that P(X̄_{θ}, q̄_{θ}) ≠ ∅. It turns out that this condition is closely related to the existence of a hyperbolic plane in X. Definition. A hyperbolic plane in a quadratic R-module (X, q) is an R-submodule H = Rx_{1} + Rx_{2} of X such that q(r_{1}x_{1} + r_{2}x_{2}) = r_{1}r_{2}, for all r_{1}, r_{2} ∈ R. (22) Any basis {x_{1}, x_{2}} of H for which (22) holds is called a hyperbolic basis of H. Remark 22 If H = Rx_{1} + R_{x} is a hyperbolic plane in (X, q), then β_{q}(r_{1}x_{1} + r_{2}x_{2}, r′_{1}x_{1} + r′_{2}x_{2}) = r_{1}r′_{2} + r_{2}r′_{1}, for all r_{1}, r_{2}, r′_{1}, r′_{2} ∈ R, (23) and so we see that q_{|H} is non-degenerate, and that x_{1} and x_{2} are R-linearly independent. Thus, H = Rx_{1} ⊕ Rx_{2} ≃ R^{2}. Moreover, we have that

X = H ⊕ H^{⊥}, where H^{⊥} = {x ∈ X : β_{q}(x, h) = 0, ∀h ∈ H}

because if x ∈ X, then

(24) x = x_{H}+x^{⊥}, where x_{H} = β_{q}(x, x_{2})x_{1}+β_{q}(x, x_{1})x_{2} ∈ H and x^{⊥} = x−x_{H} ∈ H^{⊥}. The following result classifies the structure of the spaces (X̄_{θ}, q̄_{θ}) when P(X̄_{θ}, q̄_{θ}) ≠

∅. For this we shall use the following (usual) notation: if a ∈ R, then ⟨a⟩_{R} denotes the (rank 1) quadratic R-module (Rx, q_{a}), where q_{a}(rx) = r^{2}a, for all r ∈ R. Proposition 23 Suppose that [R : 2R] ≤ 2. If θ ∈ P(X, q), then the following conditions are equivalent:

(i) P(X̄_{θ}, q̄_{θ}) ≠ ∅, i.e. ∃x ∈ X such that q_{θ}(x) = 1;

(ii) there is a hyperbolic plane H in (X, q) with θ ∈ H;

(ii′) there is a hyperbolic plane H in (X, q) such that θ = x_{1}+x_{2} for some hyperbolic basis {x_{1}, x_{2}} of H;

(iii) there is an R-submodule X′ of X such that (X̄_{θ}, q̄_{θ}) ≃ ⟨1⟩_{R} ⊕ (X′, −4q_{|X′}); (25)

(iii′) there is a hyperbolic plane H in (X, q) containing θ such that (25) holds for X′ = H^{⊥}.

Proof. Since the implications (iii′) ⇒ (iii) ⇒ (i) and (ii′) ⇒ (ii) are trivial, it is enough to verify that (ii) ⇒ (i) ⇒ (ii′) ⇒ (iii′).

(ii) ⇒ (i): Write H = Rx_{1} + Rx_{2} and θ = rx_{1} + sx_{2} with r, s ∈ R. Put x = rx_{1}. Since rs = q(θ) = 1, we have by (23) that β_{q}(x, θ) = rs = 1, and so q_{θ}(x) = β_{q}(x, θ)^{2} − 4q(x) = 1^{2} − 0 = 1.

(i) ⇒ (ii′): We first note that since q_{θ}(x) = β_{q}(x, θ)^{2} − 4q(x) = 1, there is an r ∈ R such that β_{q}(x, θ) = 1 + 2r. Indeed, if R = 2R, then this is trivial. Otherwise we have by hypothesis that R = 2R ˙∪(1 + 2R), and then β_{q}(x, θ) ∈ 1 + 2R because if β_{q}(x, θ) ∈ 2R, then 1 = β_{q}(x, θ)^{2} − 4q(x) ∈ 2R, contradiction. Thus, in both cases β_{q}(x, θ) = 1 + 2r for some r ∈ R.

Put x_{1} = x − rθ and x_{2} = θ − x_{1}. Then θ = x_{1} + x_{2}, so it is enough to show that H = Rx_{1} + Rx_{2} is a hyperbolic plane in (X, q). For this we first observe that β_{q}(x_{1}, θ) = β_{q}(x, θ)−rβ_{q}(θ, θ) = 1+2r−r(2) = 1, and so q(x_{1}) = 0 because by (17) we have q_{θ}(x_{1}) = q_{θ}(x) = 1 and so 4q(x_{1}) = β_{1}(x_{1}, θ)^{2}−q_{θ}(x_{1}) = 1^{2}−1 = 0. Next we note that β_{q}(x_{1}, x_{2}) = β_{q}(x_{1}, θ) − β_{q}(x_{1}, x_{1}) = 1 − 2(0) = 1. Finally, β_{q}(x_{2}, x_{2}) = 0 because β_{q}(x_{2}, θ) = β_{q}(θ, θ) − β_{q}(x_{1}, θ) = 2 − 1 = 1 and so β_{q}(x_{2}, x_{2}) = β_{q}(θ, x_{2}) − β_{q}(x_{1}, x_{2}) = 1 − 1 = 0. From these identities and the R-bilinearity of β_{q} it is clear that (22) holds, and so H = Rx_{1} + Rx_{2} is a hyperbolic plane with hyperbolic basis {x_{1}, x_{2}}.

(ii′) ⇒ (iii′): Let x̄_{1} = π_{θ}(x_{1}) ∈ X̄_{θ}. Then q̄_{θ}(x̄_{1}) = q_{θ}(x_{1}) = 1, the latter by the proof of (ii) ⇒ (i). Then we have

X̄_{θ} = Rx̄_{1} ⊕ (Rx̄_{1})^{⊥} (26)

because if x̄ ∈ X̄_{θ}, then x̄ = β_{θ}(x̄, x̄_{1})x̄_{1} + x̄′, where x̄′ = x − β_{θ}(x̄, x̄_{1})x̄_{1} ∈ (Rx̄_{1})^{⊥}. (Note that β̄_{θ}(x̄′, x̄_{1}) = β̄_{θ}(x̄, x̄_{1}) − β̄_{θ}(x̄, x̄_{1})β̄_{θ}(x̄_{1}, x̄_{1}) = 0 because by (16) we have that β̄_{θ}(x̄_{1}, x̄_{1}) = q_{θ}(x_{1}) = 1 .)

Thus, (25) follows once we have shown that the restriction π′ = (π_{θ})_{|H^{⊥}} of π_{θ} to H^{⊥} induces an isometry

π′ : (H^{⊥}, −4q_{|H^{⊥}}) →^{∼} ((Rx̄_{1})^{⊥}, (q̄_{θ})_{|(Rx_{1})^{⊥}}). (27)

For this, note first that π′ is clearly injective because Ker(π_{θ}) = Rθ ⊂ H and so Ker(π′) = Ker(π_{θ}) ∩ H^{⊥} ⊂ H ∩ H^{⊥} = 0.

Next we observe that π′(H^{⊥}) = (Rx̄_{1})^{⊥}. Indeed, if x ∈ H^{⊥}, then β_{q}(x, x_{i}) = 0, for i = 1, 2, and so β_{q}(x, θ) = 0, and hence β_{θ}(x, x_{1}) = 0 − 2β_{q}(x_{1}, x_{1}) = 0, which means that π′(x) ∈ (Rx̄_{1})^{⊥}. Conversely, let x̄ ∈ (Rx̄_{1})^{⊥}; thus, β̄_{θ}(x̄, x̄_{1}) = 0. Let x̃ ∈ X be such that π_{θ}(x̃) = x̄ and put r := β_{q}(x̃, x_{1}) and x := x̃ − rθ. Then π_{θ}(x) = π_{θ}(x̃) = x̄. Moreover, since β_{q}(θ, x_{i}) = 1, we have β_{q}(x, x_{1}) = β_{q}(x̃, x_{1})−rβ_{q}(θ, x_{1}) = r−r(1) = 0. Thus 0 = β̄_{θ}(x̄, x̄_{1}) = β_{θ}(x, x_{1}) = β_{q}(x, θ)β_{q}(x_{1}, θ) − 2β_{q}(x_{1}, x_{1}) = β_{q}(x, θ) · 1, i.e. β_{q}(x, θ) = 0 and hence also β_{q}(x, x_{2}) = β_{q}(x, θ − x_{1}) = 0 − 0 = 0. This means that x ∈ H^{⊥}, and so x̄ = π′(x) ∈ π′(H^{⊥}).

We thus have that π′ : H^{⊥} →^{∼} (Rx̄_{1})^{⊥} is an isomorphism of R-modules. Now if x ∈ H^{⊥}, then β_{q}(x, θ) = 0, and hence q̄_{θ}(π′(x)) = q_{θ}(x) = β(x, θ)^{2} − 4q(x) = −4q(x). This shows that π′ defines an isometry (27) and hence (iii′) holds.

For later reference we observe the following consequence of the above results.

Corollary 24 Let q : X = ℤ^{r} → ℤ be an integral quadratic form such that (X, q) contains a hyperbolic plane H. If q′ = −q_{|H^{⊥}} is positive definite, then q_{θ} is positive definite, for all θ ∈ P(X, q).

Proof. Let {x_{1}, x_{2}} be a hyperbolic basis of H and let θ_{1} = x_{1}+x_{2}. Then θ_{1} ∈ P(X, q) and by Proposition 23 we know that q_{θ_{1}} ∼ x^{2} + 4q′ is positive definite. Now if θ ∈ P(X, q), then q_{θ} ∼_{∞} q_{θ_{1}} by Proposition 18 (with R = ℝ), and so q_{θ} is also positive-definite.

For further applications of the above proposition, we shall use the following fundamental Cancellation Theorem in the theory of quadratic modules:

Proposition 25 Suppose that R = ℤ_{p} (or that R = ℝ) and that q : X = R^{r} → R is a quadratic form with Δ(q) ≠ 0. If H_{1} and H_{2} are two hyperbolic planes in (X, q), then their orthogonal complements are isometrically isomorphic.

Proof. If p is odd (or if R = ℝ), then this follows form the general Cancellation Theorem (O’Meara[16], Theorem 92:3) because H_{1} and H_{2} are isomorphic. (Here we do not require that H_{1} and H_{2} are hyperbolic planes.) If p = 2, then this is Theorem 93:14 of O’Meara[16], p. 256.

Corollary 26 If q_{1} and q_{2} are two integral quadratic forms in r variables with Δ(q_{i}) ≠ 0 such that xy ⊥ q_{1} ∼ xy ⊥ q_{2}, then q_{1} and q_{2} are genus-equivalent.

Proof. By Proposition 25 we know that q_{1} ∼_{p} q_{2} are p-adically equivalent for all primes p (including p = ∞), so q_{1} and q_{2} are genus-equivalent.

Remark 27 The converse of this corollary is also true, as is mentioned in Conway and Sloane[4], p. 378 (without proof). Thus, q_{1} and q_{2} are genus-equivalent if and only if xy ⊥ q_{1} ∼ xy ⊥ q_{2}. (To prove the converse, note that the class number c(xy ⊥ q_{1}) = 1, which follows from Theorem 1(iii) of [19] by observing that xy and hence xy ⊥ q_{1} is universal.)

From this, together with Proposition 23, we can conclude that the quadratic modules in Θ(X, q)′ are all 2-adically isomorphic (and hence lie in the same genus).

Corollary 28 If q : X = ℤ_{2}^{r} → ℤ_{2} is a 2-adic quadratic form with Δ(q) ≠ 0, then (X̄_{θ_{1}}, q̄_{θ_{1}}) ≃ (X̄_{θ_{2}}, q̄_{θ_{2}}), for all θ_{1}, θ_{2} ∈ P(X, q)′.

Proof. By Lemma 21 and Proposition 23 we know that there exist hyperbolic planes H_{i} in (X, q) such that (X̄_{θ_{i}}, q̄_{θ_{i}}) ≃ ⟨1⟩_{R} ⊕ (H_{i}^{⊥}, −4q_{|H_{i}^{⊥}}), for i = 1, 2. By the Cancellation Theorem (Proposition 25) we know that (H_{1}^{⊥}, q_{|H_{1}^{⊥}}) ≃ (H_{2}^{⊥}, q_{|H_{2}^{⊥}}), so also (H_{1}^{⊥}, −4q_{|H_{1}^{⊥}}) ≃ (H_{2}^{⊥}, −4q_{|H_{2}^{⊥}}), and hence (X̄_{θ_{1}}, q̄_{θ_{1}}) ≃ (X̄_{θ_{2}}, q̄_{θ_{2}}).

Proof of Theorem 20. Let θ_{1}, θ_{2} ∈ P(ℤ^{r}, q)′. Then by Corollaries 19 and 28 we have that q̄_{θ_{1}} ∼_{p} q̄_{θ_{2}}, for all primes p (including p = ∞), and so q̄_{θ_{1}} and q̄_{θ_{2}} are genus equivalent. This proves (20).

Now suppose that ∃θ_{0} ∈ P(X, q) such that 1 ∈ q̄_{θ_{0}}. Then by Proposition 23 we know that there exists a hyperbolic plane H = ℤx_{1} + ℤx_{2} in (X, q).

Let θ ∈ P(X, q)^{odd}. Then we can write θ = n_{1}x_{1} + n_{2}x_{2} + y with n_{1}, n_{2} ∈ ℤ and y ∈ H^{⊥}. Note that for any x = m_{1}x_{1} + m_{2}x_{2} + x^{⊥} ∈ X we have by (23) that

β_{q}(x, θ) = n_{1}m_{2} + n_{2}m_{1} + β_{q}(x^{⊥}, y). (28)

To show that θ ∈ P(X, q)′, we distinguish two cases. Case 1: n_{1} ≡ 1 (mod 2) or n_{2} ≡ 1 (mod 2). Assume first that n_{1} ≡ 1 (mod 2). Then by (28) we have β_{q}(x_{1}, θ) = n_{1}, so q_{θ}(x_{2}) = β_{q}(x_{1}, θ)^{2} − 4q(x_{1}) = n_{1}^{2} − 0 = n_{1}^{2} ≡ 1 (mod 8). Thus θ ∈ P(X, q)′. Similarly, if n_{2} ≡ 1 (mod 2), then θ ∈ P(X, q)′. Case 2: n_{1} ≡ n_{2} ≡ 0 (mod 2). By hypothesis, ∃x_{0} ∈ X such that β_{q}(x_{0}, θ) ≡ 1 (mod 2). By (24) we have x_{0} = (x_{0})_{H} + x_{0}^{⊥} with (x_{0})_{H} ∈ H and x^{⊥} ∈ H^{⊥}, and by (28) we see that b := β_{q}(x_{0}^{⊥}, θ) ≡ β_{q}(x_{0}, θ) ≡ 1 (mod 2). Thus, if we put x′ = x_{1} − q(x_{0}^{⊥})x_{2} ∈ H and x = x′ + x_{0}^{⊥}, and m_{i} = β_{q}(x, x_{i}), then β_{q}(x, θ) = n_{1}m_{1} + n_{2}m_{2} + b ≡ b ≡ 1 (mod 2). Moreover, since q(x′) = 1(−q(x_{0}^{⊥})) = −q(x_{0}^{⊥}), we have that q(x) = q(x′) + q(x_{0}^{⊥}) = −q(x_{0}^{⊥}) + q(x_{0}^{⊥}) = 0. Thus q_{θ}(x) = β_{q}(x, θ)^{2} − 4q(x) = b^{2} − 0 = b^{2} ≡ 1 (mod 8), and so θ ∈ P(X, q)′.

From the above two cases we see that P(X, q)^{odd} ⊂ P(X, q)′. Since the opposite inclusion is trivial, we thus have that P(X, q)^{odd} = P(X, q)′, and the theorem follows.

In view of what was proved above, the last part of Theorem 20 can also be stated in the following way.

Corollary 29 If q : X = ℤ^{r} → ℤ is an integral quadratic form with Δ(q) ≠ 0, and if (X, q) contains a hyperbolic plane H, then Θ(X, q)^{odd} ⊂ gen(1 ⊥ (4q′)), where q′ = −q_{|H^{⊥}}.

Proof. Let x_{1}, x_{2} be a hyperbolic basis of H and put θ_{0} = x_{1} + x_{2}. Then θ_{0} ∈ P(X, q), and by Proposition 23 we know that q̄_{θ_{0}} ∼ 1 ⊥ (4q′), and so the corollary follows from Theorem 20.

In the case that r = 3, a much stronger result than Corollary 29 was proved in [11], §5. Indeed, by using the composition theory of binary quadratic forms (and a precise classification of the binary quadratic forms q̄_{θ}), it was shown there that in fact equality holds in (20). While the argument there does not carry over to arbitrary r, it does yield the following useful general result (Proposition 30) which shows that equality holds in (20) in the case that r = 3.

In order to state this result, we first introduce the following (well-known) concept:

Definition. If (X, q) and (X′, q′) are two quadratic R-modules, then we say that (X, q) represents (X′, q′) if there exists an injective R-module homomorphism φ : X′ ↪ X such that q′ = q ∘ φ. If φ can be chosen such that in addition X/φ(X′) is R-torsionfree, then we say that (X, q) primitively represents (X′, q′) and write (X, q) → (X′, q′) or q → q′.

Proposition 30 Let q : X = ℤ^{r} → ℤ be an integral quadratic form which primitively represents xy − dz^{2} (i.e. q → xy − dz^{2}), where d ≥ 1. If q_{0} ∈ gen(x^{2} + 4dy^{2}) or if q_{0} ∈ gen(4x^{2} + 4xy + (d − 1)y^{2}) and d ≡ 3 (mod 4), then there is a θ ∈ P(X, q) such that q̄_{θ} → q_{0}.

Proof. Since q → xy−dz^{2}, there is a submodule X′ of X such that X/X′ is torsionfree and such that (X′, q_{|X′}) ≃ (ℤ^{3}, xy − dz^{2}). Thus, X′ has a basis {x_{1}, x_{2}, x_{3}} such that H = ℤx_{1} + ℤx_{2} is a hyperbolic plane and x_{3} ∈ H^{⊥} satisfies q(x_{3}) = −d.

By Proposition 15 of [11], there exists s = (n_{1}, n_{2}, k) ∈ ℤ^{3} with n_{1}n_{2}−dk^{2} = 1 such that q_{0} ∼ q_{s} := n_{2}^{2}x^{2}−2k(t−d)xy+n_{1}^{2}ty^{2}, where t = d(n_{1}n_{2}+3). (Note that although the case that q_{0} ∼ x^{2} + 4y^{2} was excluded in that result, we can include it here by taking s = (1, 1, 0).) Put θ = n_{1}x_{1} + n_{2}x_{2} + kx_{3} ∈ X′. Then θ ∈ P(X′, q′) ⊂ P(X, q), and the proof of Proposition 28 of [11] shows that q_{θ} → q_{s} ∼ q_{0}.

For convenience of the reader, we recall the proof here. Put y = kdx_{2} + n_{1}x_{3}. Then x_{2} = n_{1}θ −ky −n_{1}^{2}x_{1} and x_{3} = n_{2}y −kdθ +n_{1}kdx_{1}, so {x_{1}, θ, y} is another basis of X′. Thus, if x̄_{1} := π_{θ}(x) and ȳ:= π_{θ}(y), then {x̄_{1}, ȳ} is a basis of X̄′ := π_{θ}(X′).

Next we observe that β_{q}(mx_{1} + ny, θ) = n_{2}m − n_{1}kdn and q(mx_{1} + ny) = kdmn − n_{1}^{2}dn^{2}, and so q_{θ}(mx + ny) = β_{q}(mx + ny, θ)^{2} − 4q(mx + ny) = n_{2}^{2}m^{2} − 2kd(n_{1}n_{2} + 2)mn + n_{1}^{2}d(k^{2}d + 4)n^{2} = q_{s}(m, n); here we used the fact that k^{2}d + 4 = n_{1}n_{2} + 3 by equation (5) of [11]. From this we see that (q̄_{θ})_{|X̄′} ∼ q_{s} ∼ q_{0}, and so q̄_{θ} → q_{0} because X̄/X̄′ ≃ X/X′ is torsionfree.

Remark 31 The above proof of Proposition 30 actually yields the following more precise statement: if X′ is a submodule of X such that X/X′ is torsionfree and q_{|X′} ∼ xy − dz^{2}, then ∃θ ∈ X′ ∩ P(X, q) such that (ℤ^{2}, q_{0}) ≃ (X̄′, (q̄_{θ})_{|X̄′}), where X̄′ = π_{θ}(X′).

In addition, we note that if we write θ = θ_{H} + kx_{3} with θ_{H} ∈ H = ℤx_{1} + ℤx_{2}, then it follows from Lemma 16 of [11] that q_{0} ∈ gen(x^{2} + 4dy^{2}) ⇔ θ_{H} ∉ 2H.

We will use the above result to show that (under suitable hypotheses) there exists a θ ∈ P(X, q) such that q_{θ} ↛ 1. We begin by characterizing the trivial case that θ ∈ P(X, q)^{ev}; recall that by Remark 11 we know that for such a θ we have q_{θ} ↛ 1.

Proposition 32 Let q : X = ℤ^{r} → ℤ be an integral quadratic form, and suppose that (X, q) contains a hyperbolic plane H with complement X′ = H^{⊥}. Write q′ = −q_{|X′}. Then P(X, q)^{ev} ≠ ∅ if and only if

(29) ∃x′ ∈ X′ with q′(x′) ≡ 3 (mod 4) and β_{q′}(x′, x′′) ≡ 0 (mod 2), ∀x′′ ∈ X′.

Proof. Let {x_{1}, x_{2}} be a hyperbolic basis of H, and let θ ∈ P(X, q). Then θ = n_{1}x_{1} + n_{2}x_{2} + x′ with n_{1}, n_{2} ∈ ℤ and x′ ∈ X′ = H^{⊥}. Moreover, by definition and (23) we have that

(30) θ ∈ P(X, q)^{ev} ⇔ n_{1} ≡ n_{2} ≡ 0 (mod 2) and β_{q′}(x′, x′′) ≡ 0 (mod 2).

Thus, if θ ∈ P(X, q)^{ev}, then n_{1} ≡ n_{2} ≡ 0 (mod 2) and so 1 = q(θ) = n_{1}n_{2} − q′(x′) ≡ −q′(x′) (mod 4), which shows that (29) holds.

Conversely, suppose that x′ ∈ X′ satisfies condition (29). Then −q′(x′) = 1 + 4m with m ∈ ℤ, and so θ := 2x_{1} + 2mx_{2} + x′ satisfies q(θ) = 1, and from (30) we see that θ ∈ P(X, q)^{ev}.

Remark 33 It is clear that if q′(x) = mx^{2}, then q′ satisfies condition (29) if and only if m ≡ 3 (mod 4).

Similarly, if q′(x, y) = ax^{2} + bxy + cy^{2}, then q′ satisfies condition (29) if and only if Δ(q′) = b^{2} − 4ac ≡ 0 (mod 4) and q′(r, s) ≡ 3 (mod 4) for some r, s ∈ ℤ.

Indeed, if this latter condition holds, then b ≡ 0 (mod 2) and then (29) holds for x′ = (r, s) ∈ ℤ^{2} = X′ because β_{q′}((r, s), (m, n)) = 2arm + bsm + brn + 2csn ≡ 0 (mod 2), for all m, n ∈ ℤ. Conversely, suppose that (29) holds for x′ = (r, s) ∈ ℤ^{2} = X′. Since β_{q′}((r, s), (m, n)) = 2arm + bsm + brn + 2csn ≡ 0 (mod 2), for all m, n ∈ ℤ, we see that either b ≡ 0 (mod 2) or r ≡ s ≡ 0 (mod 2). However, in the latter case q′(x′) ≡ 0 (mod 4), contradiction. Thus b ≡ 0 (mod 2) and hence Δ(q′) ≡ 0 (mod 4).

We now turn to investigate when q_{θ} ↛ 1 in the case that θ ∈ P(X, q)^{odd}. The following result suffices for our purposes.

Proposition 34 Let q : X = ℤ^{r} → ℤ be an integral quadratic form, and suppose that (X, q) contains a hyperbolic plane H with complement X′ = H^{⊥}. Suppose that q′ := −q_{|X′} is a positive-definite binary quadratic form, and that θ ∈ P(X, q) has the form θ = θ_{H} + kθ^{⊥}, where θ_{H} ∈ H, k ∈ ℤ and θ^{⊥} ∈ X′ is primitive. If a := q′(θ^{⊥}) ≤ |Δ(q′)|, then

q_{θ} → 1 ⇔ q_{θ}(y) = 1, for some y ∈ H + ℤθ^{⊥}, (31)

except in the case that a = |Δ(q′)| ≡ 3 (mod 4), θ_{H} ∈ 2H and q′ ∼ ax^{2} ± axy + \frac{a+1}{4}y^{2}.

The proof of this will be based on the following elementary identity.

Lemma 35 Suppose (X, q) contains a hyperbolic plane H. Let θ ∈ P(X, q), and write θ = θ_{H} + kθ^{⊥} where θ_{H} ∈ H, θ^{⊥} ∈ H^{⊥} and k ∈ ℤ. If a := q(θ^{⊥}), then

4a^{2}q_{θ}(x + y) = q_{θ}(2ax + β_{q}(y, θ^{⊥})θ^{⊥}) + 4ad(y), (32)

for all x ∈ H + ℤθ^{⊥} and y ∈ H^{⊥}, where d(y) = β_{q}(y, θ^{⊥})^{2} − 4q(θ^{⊥})q(y).

Proof. Put H_{θ} = H + ℤθ^{⊥}. We first observe that

q_{θ}(x + z) = q_{θ}(x) − 4q(z), if x ∈ H_{θ}, z ∈ (H_{θ})^{⊥} (33)

because q_{θ}(x+z) = β_{q}(x+z, θ)^{2}−4q(x+z) = β_{q}(x, θ)^{2}−4q(x)−4q(z) = q_{θ}(x)−4q(z).

Now let x ∈ H_{θ} and y ∈ H^{⊥}, and put z = 2ay − bθ^{⊥} ∈ H^{⊥}, where b = β_{q}(y, θ^{⊥}). Then β_{q}(z, θ^{⊥}) = 0, and so z ∈ (H_{θ})^{⊥}. Moreover, q(z) = −ad(y) because 4a^{2}q(y) = q(2ay) = q(z +bθ^{⊥}) = q(z)+q(bθ^{⊥}) = q(z)+ab^{2}. Thus, if we put x̃ = 2ax+bθ^{⊥} ∈ H_{θ}, then x̃ + z = 2a(x + y) and so by (33) we obtain 4a^{2}q_{θ}(x + y) = q_{θ}(x̃ + z) = q_{θ}(x̃) − 4q(z) = q_{θ}(x̃) + 4ad(y), which proves (32).

Proof of Proposition 34. Since θ^{⊥} is primitive in X′, there exists x′ ∈ X′ such that {θ^{⊥}, x′} is a basis of X′. Put b = −β_{q′}(θ^{⊥}, x′) = β_{q}(θ^{⊥}, x′) and c = q′(x′); then Δ(q′) = (−b)^{2} −4ac = d(x′) (in the notation of Lemma 35). Note that we can assume without loss of generality that |b| ≤ a by replacing x′ by x′ + nx, for a suitable n ∈ ℤ (because β_{q′}(θ^{⊥}, x′ + nx) = 2na − b).

Suppose q_{θ}(x) = 1 for some x ∈ X. Write x = x_{H} +mθ^{⊥} +nx′ with m, n ∈ ℤ, and put x̃ = x_{H} +mθ^{⊥} ∈ H_{θ} = H +ℤθ^{⊥}. Since d(nx′) = n^{2}d(x′) = n^{2}Δ(q′) = −n^{2}|Δ(q′)|, it follows from (32) (applied to (x̃, nx′, −a) in place of (x, y, a)) that

4a^{2} = 4a^{2}q_{θ}(x) = q_{θ}(−2ax̃ + nbθ^{⊥}) + 4an^{2}|Δ(q′)| ≥ 4an^{2}|Δ(q′)|, (34)

where the last inequality follows from fact that q_{θ} is positive definite (cf. Corollary 24). We thus see that if a < |Δ(q′)|, then n^{2} < 1 and so n = 0 and x = x̃ ∈ H_{θ}. This proves (31) in this case.

If a = |Δ(q′)| and n ≠ 0, then (34) shows that n = ±1 and q_{θ}(−2ax̃ + nbθ^{⊥}) = 0, so −2ax̃ + nbθ^{⊥} = tθ, for some t ∈ ℤ because q̄_{θ} is positive definite. To analyze this equation, let {x_{1}, x_{2}} be a hyperbolic basis of H and write x_{H} = m_{1}x_{1} + m_{2}x_{2} and θ = n_{1}x_{1} + n_{2}x_{2} + kθ^{⊥}. Then the identity −2ax̃ + nbθ^{⊥} = tθ is equivalent to the three equations

−2am_{1} = tn_{1}, −2am_{2} = tn_{2}, −2am + nb = tk.

Since q(θ) = n_{1}n_{2} − k^{2}a = 1, we see from the first two equations that n_{i}|2m_{i}, for i = 1, 2, and so a|t. Thus, the third equation yields that a|nb = ±b. Since |b| ≤ a, this is only possible if b = 0 or b = ±a. Now b ≠ 0 because otherwise a = |Δ(q′)| = 4ac, contradiction. Thus b = ±a, so c = \frac{|Δ(q′)|+b^{2}}{4a} = \frac{a+a^{2}}{4a} = \frac{1+a}{4}, and hence a ≡ 3 (mod 4) and q′ ∼ ax^{2} ± axy + \frac{1+a}{4}y^{2}. Moreover, from the third equation we now see that tk and hence t is odd, so the first two equations show that n_{1} and n_{2} are even. This means that θ_{H} ∈ 2H, and so we are in the exceptional case, contradiction. Thus n = ±1 is not possible, and hence n = 0. This proves (31).

Note that the above results yield the implication (iii) ⇒ (iv) of Theorem 13:

Corollary 36 Suppose q = xy ⊥ (−q′), where q′ is a positive definite binary quadratic form. If q′ is not a special idoneal-valued form, then there exists θ ∈ P(ℤ^{4}, q) such that q_{θ} ↛ 1.

Proof. By hypothesis, X = ℤ^{4} = H ⊕ X′, where H is a hyperbolic plane, X′ = H^{⊥} and q_{|X′} = −q′. Since q′ is not a special idoneal-valued form, we have the following three possibilities: Case 1: Δ(q′) ≡ 0 (mod 4) and q′ → n ≡ 3 (mod 4). From Proposition 32 together with Remark 33 it follows that in this case ∃θ ∈ P(X, q)^{ev}, and so q_{θ} ↛ 1 by Remark 11. Case 2: q′ → n ≡ 3 (mod 4) and n < |Δ(q′)|. Let x′ ∈ X′ be such that q′(x′) = n, and put X′′ = H + ℤx′. Note that X/X′′ ≃ X′/ℤx′ is torsionfree since x′ is primitive (by hypothesis). Moreover, since q_{|X′′} ∼ xy−nz^{2}, we see that q → xy−nz^{2}. Let q_{0} := 4x^{2}+4xy+(n+1)y^{2}. Then by Proposition 30 and Remark 31 we have that ∃θ ∈ X′′ ∩ P(X, q) such that (X̄′′, (q̄_{θ})_{X′′}) ≃ (ℤ^{2}, q_{0}), where X̄′′ = π_{θ}(X′′). Since q_{0}(x) ≡ 0 (mod 4) for all x ∈ ℤ^{2}, we see that q_{θ}(x′′) ≠ 1, for all x′′ ∈ X′′, and so by Proposition 34 it follows that q_{θ} ↛ 1. Case 3: q → n, n ≤ |Δ(q′)| and 4n is not idoneal. Since 4n is not idoneal, we have that c(x^{2} + 4ny^{2}) > 1, so ∃q_{0} ∈ gen(x^{2} + 4ny^{2}) with q_{0} ↛ 1. Then by the same argument (and notation) as in Case 2, there exists θ ∈ X′′ ∩ P(X, q) such that (X̄′′, (q̄_{θ})_{X′′}) ≃ (ℤ^{2}, q_{0}). Thus, q_{θ}(x′′) ≠ 1, ∀x′′ ∈ X′′. Since θ_{H} ∉ 2H by Remark 31, it follows from Proposition 34 that q_{θ} ↛ 1.

5 Idoneal-valued quadratic forms

By the previous Corollary 36 we see that the hypothesis of Problem 10 (for r = 2) naturally leads to the condition that q is a special idoneal-valued binary quadratic form, or, more briefly, a special form.

We shall now classify the special forms q(x, y) = ax^{2} + bxy + cy^{2} and begin by classifying those whose content cont(q) = gcd(a, b, c) is odd.

Proposition 37 Let q be a special form with cont(q) ≡ 1 (mod 2). Then q is equivalent to one of the following four forms f(x, y) = ax^{2} + bxy + cy whose coefficients (a, b, c) are in the list L_{1} = {(1, 1, 1), (1, 0, 1), (1, 1, 2), (1, 1, 3)}.

Remark 38 The above Proposition 37 is all that is needed to deduce (a generalization of) of the existence theorem of Hayashida and Nishi[7]. Indeed, if we assume that End(E_{i}) ≃ 𝔒_{F} is a maximal order of an imaginary quadratic field F of discriminant −d, then by [13], Corollary 42, we know that q = q_{E_{1},E_{2}} is a form of discriminant −d and content cont(q) = 1. Thus, if E_{1} × E_{2} does not contain a curve of genus 2, then by Corollary 8 and Corollary 36 we know that q is special, and so by Proposition 37 we have that d = 3, 4, 7, 15 and that E_{1} ≃ E_{2} (and that the same is result true if the conductors of E_{i} are odd). This, therefore, proves the existence part of the theorem in [7], p. 14.

Note that the above proof is quite different from that of [7] because here we consider the problem of whether the (positive definite) ternary forms Q_{θ} represent 1 whereas Hayashida and Nishi consider the problem of determining whether specific numbers are represented by certain (positive definite) quaternary forms; cf. [7], p.

10. Note that if cont(q_{E_{1},E_{2}}) > 1, then it is questionable whether their criterion (for the existence of genus 2 curves) still holds, and so their method cannot be adapted readily to the more general case.

In order to prove Proposition 37 and other related results, it is useful to introduce the following notation.

Notation. If q(x, y) = ax^{2} + bxy + cy^{2} is a positive definite binary quadratic form, then we write q = [a, b, c]. Moreover, we let R(q) = {n ∈ ℤ : q → n} denote the set of values which are primitively represented by q. Furthermore, we let

R′(q) = {n ∈ R(q) : n < |Δ(q)|} and R^{∗}(q) = {n ∈ R(q) : n ≤ |Δ(q)|}.

Lemma 39 If q = [a, b, c] is reduced, then a, c ∈ R′(q) and a ± b + c ∈ R^{∗}(q). Moreover, a ± b + c ∈ R′(q) except when a = b = c = 1.

Proof. Put d = |Δ(q)|. Since q is reduced, we have a ≤ \sqrt{d/3} < d and c ≤ \frac{1}{4}(\sqrt{d/3} + d) ≤ \frac{d}{2} < d (cf. e.g. [3], p. 184), so a = q(1, 0) ∈ R′(q) and c = q(0, 1) ∈ R′(q). Furthermore, q(1, ±1) = a±b+c ≤ 2a+c ≤ 2\sqrt{d/3}+ \frac{1}{4} \sqrt{d/3}+ \frac{1}{4}d = \frac{9}{4} \sqrt{d/3}+ \frac{1}{4}d ≤ d, and equality holds only if \sqrt{d/3} = d/3, i.e. d = 3. Thus, a±b+c ∈ R′(q) except when d = 3 (and then q = [1, 1, 1] because this is the only reduced form with Δ(q) = −3).

We next recall some elementary properties of idoneal numbers. For this, it is useful to introduce the following notation.

Notation. Let I = {n ≥ 1 : c(x^{2} + ny^{2}) = 1} denote the set of idoneal numbers, and put

S̃ = {n : 4n ∈ I} and S = {n : 4n ∈ I, n ≢ 3 (mod 4)}.

We call S the set of special idoneal numbers.

Remark 40 If q is a positive definite binary quadratic form, then

(35) q is a special idoneal-valued form ⇔ R′(q) ⊂ S and R^{∗}(q) ⊂ S̃.

Indeed, if q is special, then the first condition of the definition is equivalent to the inclusion R^{∗}(q) ⊂ S̃. Moreover, since q ↛ n ≡ 3 (mod 4), if n < |Δ(q)|, we also have R′(q) ⊂ S. Conversely, if R′(q) ⊂ S and R^{∗}(q) ⊂ S̃, then it is clear that q is special if Δ(q) ≡ 1 (mod 4). Thus, assume Δ(q) ≡ 0 (mod 4), and (without loss of generality) that q = [a, b, c] is reduced. Then a, c, a ± b + c ∈ R′(q) ⊂ S (by Lemma 39) and 2|b. Thus, either 2|cont(q) or (a, b, c) ≡ (0, 0, 1), (1, 0, 0), (1, 0, 1), (1, 2, 1), (1, 2, 2), (2, 2, 1) (mod 4). In all these cases we see that q does not represent any number n ≡ 3 (mod 4), and so q is special.

The following properties of idoneal numbers are essentially due to Euler and Grube[5].

Lemma 41 (a) If n ∈ S and n > 1, then n ≡ 2, 4, 6 (mod 8). Moreover, if S_{kn} := {n ∈ S : n ≤ 10^{5}} denotes the set of “known” special idoneal numbers, then

S_{kn} = {1, 2, 4, 6, 10, 12, 18, 22, 28, 30, 42, 58, 60, 70, 78, 102, 130, 190, 210, 330, 462}.

(b) If n ∈ I and n ≡ 3 (mod 4), then n = 3, 7, 15. In particular, S̃ = S ∪ {3, 7, 15}.

(c) If n ∈ S and n > 60, then n is squarefree. Thus

S_{4} := {n ∈ S : 4|n} = {4, 12, 28, 60}.

Proof. (a) This follows immediately from Euler’s results; cf. Theorem 1 and Corollary 10 of [12]. (See also [11].)

(b) The first assertion is due to Grube; cf. [12], Corollary 8. The second assertion follows from this and Euler’s result ([12], Corollary 10(b)).

(c) Since n ≢ 0 (mod 8) by (a), it follows from Grube’s Theorem ([12], Theorem 12) that n is squarefree when n > 60. If n ∈ S_{4}, then n ≡ 4 (mod 8) (because n ≢ 0 (mod 8)), so n ∈ S_{4} ⇔ n ∈ {4, 12, 28, 60} by Grube (cf. [12], Theorem 12(c)).

Corollary 42 The form q is not special if any one of the following conditions hold: (i) ∃n ∈ R′(q) with n ≡ 1 (mod 2) and n > 1; (ii) q = [a, b, c] is reduced and one of a, b, c is an odd number ≠ 1; (iii) 1 ∈ R(q), Δ(q) ≡ 0 (mod 4), and Δ(q) ≠ −4; (iv) 1 ∈ R(q), Δ(q) ≡ 1 (mod 4) but Δ(q) ≢ 1 (mod 8) and Δ(q) ≠ −3.

Proof. (i) Since n ∉ S by Lemma 41(a), we have R′(q) ⊄ S, and hence q cannot be special (cf. (35)).

(ii) Since a, c ∈ R′(q) by Lemma 39, it follows from (i) that q cannot be special if a ≠ 1 is odd or if c ≠ 1 is odd. Now suppose that b ≠ 1 is odd. Then c ≥ a ≥ |b| ≥ 3, so if a or c are odd, then we are done. If not, then a + |b| + c ∈ R′(q) is odd, and so q is not special by (i).

(iii) If 1 ∈ R(q) and Δ(q) = −4d, then q ∼ q_{1} := [1, 0, d]. If d > 1 and odd, then q_{1} and hence q are not special by (ii). If d is even, then 1 + 0 + d ∈ R′(q) is odd, and so q_{1} and q are not special by (i).

(iv) If 1 ∈ R(q) and Δ(q) = −4d + 1, then q ∼ q_{1} := [1, 1, d]. By hypothesis, d > 1 is odd, so q_{1} and q are not special by (ii).

Proof of Proposition 37. Let q ∼ q_{1} := [a, b, c], where q_{1} is reduced. Since cont(q) = cont(q_{1}) is odd, at least one of a, b, c must be odd. If a > 1, then q is not special because if a or c ≥ a is odd, then we are done by Corollary 42(ii), and if a and c are even, then a + |b| + c ∈ R′(q) is odd, so we are done by Corollary 42(i). Thus, assume that a = 1. If Δ(q) ≡ 0 (mod 4), then q ∼ [1, 0, 1] by Corollary 42(iii), and if Δ(q) ≡ 1 (mod 4) but Δ(q) ≢ 1 (mod 8) then q ∼ [1, 1, 1] by Corollary 42(iv).

Thus, assume that Δ(q) ≡ 1 (mod 8). Then q ∼ q_{1} = [1, 1, d] with d ≡ 0 (mod 2). Thus 4d − 1 = q_{1}(1, −2) ∈ R^{∗}(q) ⊂ S̃, and so 4d − 1 ∈ {3, 7, 15} by Lemma 41. Thus d = 1, 2, or 4, and the assertion follows.

In view of Proposition 37, this leaves us to classify the special forms whose content is even. If, in fact, 4|cont(q), then the answer is simple.

Proposition 43 If q is special and 4|cont(q), then q ∼ [4, 4, 4].

The proof of this is based on the following fact which gives a useful lower bound on the number of elements in R′(q).

Lemma 44 Let q = [a, b, c] be reduced and b ≥ 0. If n ≥ 1 satisfies the inequality

n(n + 1) < c(3 − 1/a) + c − b, (36)

then q(k, ±1) ∈ R′(q), for 0 ≤ k ≤ n. Thus

(37) #R′(q) ≥ n + 1 and even #R′(q) ≥ 2n + 1, if a > b > 0.

Proof. For 0 ≤ k ≤ n we have that

q(k, ±1) = ak^{2} ± bk + c ≤ an^{2} + an + c = an(n + 1) + c.

Combining this inequality with (36) yields

q(k, ±1) ≤ an(n + 1) + c < a(c(3 − 1/a) + c − b) + c = 4ac − ab ≤ 4ac − b^{2} = |Δ(q)|,

and so q(k, ±1) ∈ R′(q), as claimed.

Since the values q(k, 1) are distinct for 0 ≤ k ≤ n, the first inequality of (37) is clear. The second follows from the fact that the function q(±k, 1) = q(k, ±1) takes on 2n + 1 distinct values for 0 ≤ k ≤ n because for x, y ∈ ℤ we have that

(38) q(x, 1) = q(y, 1) ⇔ x = y, provided that a > b > 0.

Indeed, suppose that ax^{2} + bx + c = ay^{2} + by + c. Then ax^{2} − by^{2} = by − bx, so if x ≠ y, then −a(x + y) = b. Thus a|b which is impossible since 0 < b < a.

Proof of Proposition 43. By reduction theory there exists a reduced form q_{1} := [a, b, c] with b ≥ 0 such that q ∼ q_{1} (GL_{2}(ℤ)-equivalence). Since 4|cont(q_{1}), we have that c ≥ 4. If c = 4, then also a = 4 and b = 0 or 4. Thus q_{1} = [4, 0, 4] or ]4, 4, 4]. However, since 4 + 0 + 4 = 8 ∉ S_{4} by Lemma 41(c), the form [4, 0, 4] is not special, so only q_{1} = [4, 4, 4] is possible if c = 4.

Now assume that c > 4. Since c ∈ S_{4}, we have by Lemma 41(c) that c ≥ 12. Since 4(4 + 1) = 20 < 12(\frac{11}{4}) ≤ c(3 − 1/a) + c − b (because a ≥ 4 and c ≥ b), it follows from Lemma 44 that q(k, 1) ∈ R′(q_{1}) for 0 ≤ k ≤ 4. If q_{1} were special, then these would give 5 distinct numbers in S_{4}, which contradicts the fact that #S_{4} = 4 by Lemma 41(c). Thus q_{1} is not special for c > 4.

The above results reduce the classification problem to the case that cont(q) ≡ 2 (mod 4). Here we first prove:

Proposition 45 If q = [a, b, c] is a reduced special form with cont(q) ≡ 2 (mod 4) and 4|c, then q = [2, 0, 4].

Proof. As in the proof of Proposition 43, we may assume that b ≥ 0.

Suppose first that c = 4. If a = 2, then b = 0 or 2. But q = [2, 2, 4] is not special since q(1, 1) = 8 ∉ S_{4}, so we must have q = [2, 0, 4]. If a > 2, then a = 4 and b = 2. But q = [4, 2, 4] is not special because q(2, 1) = 24 < |Δ(q)| = 60 but 24 ∉ S_{4}.

Thus, assume c > 4. Then c ≥ 12 because c ∈ S_{4}. If a = 2, then q = [2, 0, c] or q = [2, 2, c]. In the first case q(2, 1) = 8 + c < 8c = |Δ(q)|, but 8 + c ∉ S_{4} (when c ∈ S_{4} and c ≥ 12), so [2, 0, 4] is not special. Similarly, if q = [2, 2, c], then q(2, −1) = c + 4 < 8c − 4 = |Δ(q)|, but c + 4 ∉ S_{4} (when c ∈ S_{4} and c ≥ 12), and so [2, 2, c] is not special.

We thus have that a > 2, so a ≥ 4. Then, as in the proof of Proposition 43, we have that q(k, ±1) ∈ R′(q) for 0 ≤ k ≤ 4. Thus, if q is special, then q(4, 1) ∈ S_{4} (because 4|q(4, 1)). But q(4, 1) ≥ 4^{2}a + 4b + c > 4^{2}4 = 64 > 60, so q(4, 1) ∉ S_{4}, contradiction. Thus q is not special, and so the only possible case is q = [2, 0, 4].

Proposition 46 If q = [a, b, c] is a reduced special form with cont(q) ≡ 2 (mod 4) and 4|a, then q = [4, 0, 6].

Here we shall use the following technical fact.

Lemma 47 If q = [a, b, c] is reduced, and if a ≥ 2 and b ≥ 0, then

(39) q(−1, 2) ∈ R′(q), and also q(1, 2) ∈ R′(q), if c > b.

Proof. Since a ≥ 2, it follows that we have

(40) (b − 1)^{2} < (4a − 5)c + 1, and even (b + 1)^{2} < (4a − 5)c + 1, if c > b.

Indeed, if b = 0, then this is clear, so assume that b ≥ 1. Then 3a − 6 ≥ 3 · 2 − 6 = 0, so b + 1 ≤ a + 1 ≤ 4a − 5. Thus 0 ≤ b − 1 < b + 1 ≤ 4a − 5 and 0 ≤ b − 1 < c, so (b − 1)^{2} < (4a − 5)c < (4a − 5)c + 1, which proves the first part of (40). Note that if c > b, then b + 1 ≤ c, so a similar argument shows that the second part of (40) holds.

From the first part of (40) we obtain that q(−1, 2) = a − 2b + 4c = (b − 1)^{2} + 4c + a − b^{2} + 1 < (4a − 5)c + 1 + a − b^{2} − 1 = 4ac − b^{2} − (c − a) ≤ 4ac − b^{2}. Thus, q(−1, 2) < 4ac − b^{2} = |Δ(q)|, and so q(−1, 2) ∈ R′(q). Similarly, if c > b, then a similar argument (using the second part of (40)) shows that q(1, 2) < |Δ(q)|, so (39) follows.

Proof of Proposition 46. Note first that b < c for else q = [a, a, a] and then cont(q) = a ≡ 0 (mod 4), contradiction. Thus, since a ≥ 4, we have by (39) that q(1, ±2) = a ± 2b + 4c ∈ S_{4} (and a ∈ S_{4}).

If a ≥ 12, then q(1, 2) = a + 2b + 4c ≥ 5a ≥ 60, so q(1, 2) ∉ S_{4} unless a = c = 12 and b = 0. But q = [12, 0, 12] is not special since q(1, 1) = 24 ∉ S_{4}.

Thus, a < 12, and hence a = 4 because a ∈ S_{4}. Then b = 0, 2, or 4. Suppose first that b = 4. Since q(±1, 2) ∈ S_{4} and since q(1, 2) − q(−1, 2) = 4b = 16, it follows that q(−1, 2) = 12 because (12, 28) is the only pair of numbers in S_{4} whose difference is 16. Thus c = 4 and so 4|cont(q), contradiction. Next, suppose that b = 2. Here we have that q(−1, 2) = 4 because (4, 12) is the only pair of numbers in S_{4} whose difference is 2b = 8, and so c = 1, contradiction.

Thus, only b = 0 is possible. Then q(1, 2) = 4 + 4c ∈ S_{4}, so 4c ∈ {8, 25, 56} or c ∈ {2, 6, 14}. Now c = 2 < a is impossible. Moreover, since c = 14 ∉ S (cf. Lemma 41(a)), we must have that c = 6, so q = [4, 0, 6], as claimed.

Proposition 48 Let q = [a, b, c] be a reduced special form with cont(q) ≡ 2 (mod 4). If q ≢ [2, 2, 2] (mod 4), then q ∈ L_{0} := {[2, 0, 2], [2, 0, 4], [4, 0, 6], [6, 0, 6], [2, 0, 10]}.

Proof. If 4|c or 4|a, then q = [2, 0, 4] ∈ L_{0} or q = [4, 0, 6] ∈ L_{0} by Propositions 45 and 46, respectively. Thus, assume that a ≡ c ≡ 2 (mod 4). Then b ≡ 0 (mod 4), for otherwise q ≡ [2, 2, 2] (mod 4). Thus, if k is odd, then q(k, 1) ≡ a + c ≡ 0 (mod 4). As before, we may assume that b ≥ 0.

Suppose first that c < b + 4. Then we must have that a = c = b + 2, so we can write q = [4k + 2, 4k, 4k + 2], for some integer k ≥ 0. Since q(1, ±1) ∈ R′(d) by Lemma 39, we have that 12k + 4 = q(1, 1) ∈ S_{4} and 4k + 4 = q(1, −1) ∈ S_{4}. This is only possible for k = 0 or k = 2. If k = 2, then we obtain that q = [10, 8, 10]. But since 124 = q(3, 1) ∈ R′(q) (because 124 < |Δ(q)| = 336) yet 124 ∉ S_{4}, we see that [10, 8, 10] cannot be special, and so we must have k = 0, and hence q = [2, 0, 2] ∈ L_{0}.

Now suppose that c ≥ b + 4. Then in fact c ≥ b + 6 because c ≢ b (mod 4). Since 3(3 + 1) = 12 < 6(\frac{5}{2}) ≤ c(3 − \frac{1}{a}) + c − b, we see from (36) that q(k, ±1) ∈ R′(q) for 0 ≤ k ≤ 3. Now if b ≠ 0, then 0 < b < a (because b ≢ a (mod 4)), and so it follows from (38) that q(1, ±1) and q(3, ±1) are 4 distinct numbers S_{4}. But since they are ≥ c ≥ 6, this is impossible.

Thus, b = 0. Now 10a ≤ 9a + c = q(3, 1) ∈ S_{4}, so 10a ≤ 60 or a ≤ 6. If a = 6, then the inequalities 60 = 10a ≤ 9a + c ≤ 60 force that c = a = 6, and so q = [6, 0, 6] ∈ L_{0}. Thus, assume that a < 6. Then a = 2 because a ≡ 2 (mod 4). Since 18 + c = q(3, 1) ∈ S_{4}, only the cases c = 28 − 18 = 10 and c = 60 − 18 = 42 are possible. But since 2 + 42 = 44 ∉ S_{4}, it follows that [2, 0, 42] is not special, and so only q = [2, 0, 10] ∈ L_{0} is possible. This proves the assertion.

By the above results, the only case left to investigate is the case that q ≡ [2, 2, 2] (mod 4). The analysis of this case, however, seems to require much deeper results about idoneal numbers than the previous cases. Indeed, we shall use here the following basic fact which will be deduced from the (unconditional!) results of Weinberger[22]:

Theorem 49 If q is a special idoneal-valued form, then |Δ(q)| ≤ 10^{6} and hence R′(q) ⊂ S_{kn}.

Proof. We first note that if there exists d^{∗} ∈ S \ S_{kn}, then by Lemma 41(c) we know that d^{∗} is squarefree and so −4d^{∗} is a fundamental discriminant. By Weinberger[22] we therefore know that d^{∗} is unique and that 4d^{∗} > 10^{11}. We thus have that

either S = S_{kn} or S = S_{kn} ∪ {d^{∗}} with d^{∗} > 10^{10}.

Thus, if q = [a, b, c] is reduced and special, then by Lemma 39 we have that a, c, a + |b| + c ∈ S. But if c ∈ S \ S_{kn} = {d^{∗}}, then a + |b| + c > c = d^{∗}, so a + b + c ∉ S, contradiction. Thus, c ∈ S_{kn}, and hence a ≤ c ≤ 462. Thus |Δ(q)| ≤ 4 · 462^{2} < 10^{6} < d^{∗}, and hence R′(q) ⊂ S_{kn}.

Using this result, we can now prove the following:

Proposition 50 If q = [a, 2, c] is a reduced special form with a ≡ c ≡ 0 (mod 2), then q ∈ L_{2} := {[2, 2, 2], [2, 2, 6], [2, 2, 18]}.

Proof. By Lemma 39 and Theorem 49 we have that a, c, a + c ± 2 ∈ S_{kn}, and so a + c − 2 ∈ T_{4} := {n ∈ S_{kn} : n + 4 ∈ S_{kn}}. Since S_{kn} is known explicitly (cf. Lemma 41), one checks easily that

T_{4} =^{def} {n ∈ S_{kn} : n + 4 ∈ S_{kn}} = {2, 6, 18}. (41)

Thus, a + c = 4, 8 or 20. If a + c = 4, then 2a ≤ a + c = 4 and so a ≤ 2. Thus a = c = 2 and hence q = [2, 2, 2] ∈ L_{2}.

Next, if a + c = 8, then 2a ≤ a + c = 8, so a ≤ 4, i.e. a = 2 or 4. If a = 4, then q = [4, 2, 4] which is not special by Proposition 45. Thus, a = 2 and q = [2, 2, 6] ∈ L_{2}.

We are thus left with the case that a + c = 20, so 2a ≤ a + c = 20 or a ≤ 10. The cases a = 4 and a = 8 are not possible by Proposition 46. Moreover, the case a = 6 is also not possible because then c = 14, but c ∉ S_{kn}. In addition, a = 10 is out because for q = [10, 2, 10] we have that 54 = q(1, 2) ∈ R′(q) by (39) but 54 ∉ S_{kn}. Thus, only q = [2, 2, 18] ∈ L_{2} is possible, and the assertion follows.

The last remaining case to be studied is the following.

Proposition 51 If q = [a, b, c] is a reduced special form with cont(q) ≡ 0 (mod 2) and b > 2, then q ∈ L_{3} := {[k, k, k] : k = 4, 6, 10}.

Proof. First note that if 4|cont(q), then q = [4, 4, 4] ∈ L_{3} by Proposition 43, so we can assume that cont(q) ≡ 2 (mod 4). It thus follows from Proposition 48 that a ≡ b ≡ 2 (mod 4), and hence b ≥ 6 because b > 2. Thus also c ≥ a ≥ 6. In addition, we know from Theorem 49 that R′(q) ⊂ S_{kn}.

Suppose first that c > 10. Then c ≥ 18 because c ∈ S_{kn} and c ≡ 2 (mod 4). Since a ≥ 6 we have 42 < 51 = 18(3 − \frac{1}{6}) ≤ c(3 − \frac{1}{a}) + (c − b), so we see that (36) holds with n = 6 and hence q(6, 1) ∈ R′(q) by Lemma 44. Then a ≤ 10, for otherwise a ≥ 18 (because a ∈ S_{kn} and a ≡ 2 (mod 4)) and then q(6, 1) ≥ 36(18) + 6(6) + 18 = 702 > 462, so q(6, 1) ∉ S_{kn} and hence q is not special. If a = 10, then b = 6 or 10. In the first case q(6, 1) = 396 + c ∈ S_{kn}. Since q(6, 1) > 330, it follows that q(6, 1) = 462, so c = 462 − 396 = 66. But then c ∉ S_{kn}, contradiction. Similarly, for b = 10 we have q(6, 1) = 420 + c ∈ S_{kn}, so c = 462 − 420 = 42 and hence q = [10, 10, 42]. But q(1, 1) = 62 ∉ S_{kn}, contradiction.

We thus have that a < 10. Then a = b = 6 and so q(6, 1) = 252 + c ≥ 270. Thus, only q(6, 1) = 330 or 462 are possible. In the first case we get c = 330−252 = 78, but q = [6, 6, 78] is not special since q(1, 1) = 90 ∉ S_{kn}. Similarly, c = 462 − 252 = 210 is impossible since then q(1, 1) = 222 ∉ S_{kn}.

We are thus left with the case that c ≤ 10. If c = 6, then q = [6, 6, 6] ∈ L_{3}, so assume that c = 10. If b ≥ 10, then q = [10, 10, 10] ∈ L_{3}, so assume b = 6. Thus a = 6 or 10, so either q = [6, 6, 10] or q = [10, 6, 10]. But neither of these is special because in the latter case q(1, 1) = 26 ∉ S_{kn}, and in the first case q(3, 1) = 82 ∉ S_{kn} but 82 ∈ R′(q) because 82 < 204 = |Δ(q)|. This proves the assertion.

Combining Propositions 37, 43, 48, 50 and 51, we obtain:

Theorem 52 If q is a special idoneal-valued binary form, then q ∼ [a, b, c], where [a, b, c] ∈ L := L_{0} ∪ L_{1} ∪ L_{2} ∪ L_{3}.

Proof. As in the proof of Proposition 43, there is a reduced form q_{1} = [a, b, c] with b ≥ 0 such that q ∼ q_{1}. If cont(q_{1}) is odd, then q_{1} ∈ L_{1} by Proposition 37. Thus, assume that cont(q_{1}) is even. If b > 2, then q_{1} ∈ L_{3} by Proposition 51, and if b = 2, then q_{1} ∈∈ L_{2} by Proposition 50. If b = 0, then by Proposition 43 we must have that cont(q_{1}) ≡ 2 (mod 4), and so Proposition 48 shows that q_{1} ∈ L_{0}. Thus, in all cases q_{1} ∈ L, as claimed.

6 The mass formula for ternary forms

In the previous section we saw that if q(x, y) is a special idoneal-valued binary quadratic form, then q is equivalent to one of the 15 forms in L; cf. Theorem 52. This does not imply, however, that the forms in L are actually special idoneal-valued forms. In order to verify that this is indeed the case, we shall show first that the class number c(f_{q}) of the ternary form f_{q}(x, y, z) = x^{2} + 4q(y, z) is equal to 1, and then use the theory developed in §4 to deduce from this that q is special; cf. Theorem 13, which is proved in the next section.

In order to show that c(f_{q}) = 1, we shall use the mass formula for positive definite ternary forms which is due to Eisenstein, Smith and Brandt; cf. Brandt[1]. This formula gives a simple expression for the mass M(f) of a ternary form, which is defined as

M(f) = ∑_{f′∈gen(f)} \frac{1}{w(f′)}, where w(f′) = |Aut^{+}(f′)|;

cf. Smith[17], p. 483. In order to state the result, it is useful to introduce the abbreviation

ψ(n, Δ) := n ∏_{p|n} (1 + (\frac{Δ}{p}) \frac{1}{p});

here Δ ≡ 0 or 1 (mod 4), and (\frac{Δ}{p}) denotes the Kronecker-Legendre symbol. Note that this function generalizes the usual Dedekind function ψ(n) = ψ(n, 1). If ϕ(n) denotes the Euler ϕ-function, then the mass formula of Eisenstein/Smith/Brandt for the form f_{q} is as follows.

Proposition 53 Let q′ be a primitive positive definite binary quadratic form of discriminant Δ′ = −d′, and let t ≥ 1 be an integer. Put q = tq′ and δ = (d′, 2t). Then the mass of f_{q}(x, y, z) = x^{2} + 4q(y, z) is given by

M(f_{q}) = \frac{1}{24g(−16t)g(Δ′)δ^{2}} ϕ(δ)ψ(δ)ψ(t, Δ′)ψ(d′, −16tr′), (42)

where g(d) denotes the number of genera of binary quadratic forms of discriminant d, and r′ ∈ R(q′) is any number represented by q′ with (r′, d′) = 1.

Proof. Let f be any primitive positive definite ternary form with primitive positive definite adjoint F, and let I_{1} and I_{2} be their invariants as defined on p. 316 of [1]. Let r_{f} (respectively, r_{F}) be a number represented by f with (r_{f}, I_{1}) = 1 (respectively, by F with (r_{F} , I_{2}) = 1). Then the mass formula on p. 316 of Brandt[1] is

M(f) = \frac{κI}{6 · 2^{ν}} ∏_{p|(I_{1},I_{2})} (1 − \frac{1}{p^{2}} ) ∏_{p|I} (1 + (\frac{I_{2}r_{f}}{p} ) \frac{1}{p}) (1 + (\frac{I_{1}r_{F}}{p} ) \frac{1}{p}) , (43)

where I = I_{1}I_{2}/16, κ ∈ \{1, \frac{1}{2}, \frac{1}{3}\} is as defined on p. 317 of [1] and 2^{ν} denotes the total number of characters attached to f and to F. Thus, 2^{ν} = 2^{ν_{1}}2^{ν_{2}}, where 2^{ν_{k}} = [(ℤ/I_{k}ℤ)^{×} : ((ℤ/I_{k}ℤ)^{×})^{2}], for k = 1, 2. Note that 2^{ν_{k}} equals the total number of factorizations of I_{k} into prime discriminants, and so this definition of 2^{ν_{k}} agrees with that of Brandt[2], p. 339. In addition, we observe that by Gauss’s genus theory we have 2^{ν_{k}} = 2g(I_{k}).

We now apply this formula to f = f_{q}. For this, we need to determine F, I_{1}, I_{2}, etc. To calculate F, recall that if A(f) is the symmetric matrix associated to f (so f(x⃗) = \frac{1}{2}x^{t}A(f)x⃗ as on p. 2 of [18]), then by [18], p. 25, the adjoint form adj(f) of f is given by the formula

A(adj(f)) = (−2)adj(A(f)) = (−2) det(A(f))A(f)^{−1},

and we have adj(f) = I_{1}F, where F is positive definite and primitive and I_{1} ∈ ℤ.

Applying this to f = f_{q}, and writing q(x, y) = ax^{2} + bxy + cy^{2} = t(a′x^{2} + b′xy + c′y^{2}) = tq′(x, y) and d = −Δ(q), we have

A(f) = ( \begin{matrix}2 & 0 & 0 \\ 0 & 8a & 4b \\ 0 & 4b & 8c \end{matrix} ) and hence adj(A(f)) = (−2) ( \begin{matrix}16d & 0 & 0 \\ 0 & 8c & −4b \\ 0 & −4b & 8a \end{matrix} ) ,

because det(A(f)) = 32(4ac − b^{2}) = 32d. Thus adj(f_{q})(x, y, z) = (−2)2(8dx^{2} + 4cy^{2} − 4byz + 4az^{2}) = −16tF (x, y, z), where F (x, y, z) := d′tx^{2} + c′y^{2} − b′yz + a′z^{2}. Since F is clearly primitive and positive definite (because q′ is), we see that the primitive adjoint of f_{q} is

F (x, y, z) = d′tx^{2} + c′y^{2} − b′yz + a′z^{2} ∼ d′tx^{2} + q′(y, z). (44)

In addition, we see that I_{1} = −16t (because adj(f_{q}) = −16tF = I_{1}F). Moreover, since f_{q} has discriminant Δ(f_{q}) = −\frac{1}{2} det(A(f_{q})) = −16d, it follows from the last line of [1], §5, that I_{2} = 16Δ(f_{q})/I_{1}^{2} = −16^{2}d/(16t)^{2} = −d/t^{2} = Δ′, and so

I_{1} = −16t and I_{2} = Δ′.

(Note that above we used a slightly different definition of F, I_{1} and I_{2} than that given in [1]. However, this gives the same form and invariants because one can check that the above form F (x, y, z) and the invariants I_{1} and I_{2} do indeed satisfy the two formulae in the middle of p. 316 of [1].)

Since 16|I_{1} we have κ = 1 by [1], p. 317. Moreover, since f_{q}(1, 0, 0) = 1 and F (0, y, −x) = q′(x, y), we see that we can take r_{f} = 1 and r_{F} = r′, where r′ = r_{q′} is any number represented by q′ with (d′, r′) = 1.

Since the prime divisors of δ = (d′, 2t) are the same as those of (I_{1}, I_{2}) = (16t, d′), we thus see that the mass formula (43) for f_{q} becomes

M(f) = \frac{td′}{6 · 2^{ν}} ∏_{p|δ} (1 − \frac{1}{p^{2}} ) ∏_{p|td′} (1 + (\frac{Δ′}{p} ) \frac{1}{p}) (1 + (\frac{−16tr′}{p} ) \frac{1}{p})

= \frac{1}{6δ^{2} · 2^{ν}} ϕ(δ)ψ(δ)ψ(t, Δ′)ψ(d′, −16tr′),

where the latter equality uses the fact that (1 + (\frac{Δ′}{p} ) \frac{1}{p}) = (1 + (\frac{−16tr′}{p} ) \frac{1}{p}) = 1 when p|(t, d′). From this, equation (42) follows immediately because 2^{ν} = 4g(I_{1})g(I_{2}) = 4g(−16t)g(Δ′), as was mentioned above.

We now want to analyze the weight (or mass) w(f) = |Aut^{+}(f)| of the form f = f_{q}. For this we shall prove the following general result.

Proposition 54 If q_{1} and q_{2} are two positive definite binary quadratic forms, then we have a bijection

(45) Isom(f_{q_{1}}, f_{q_{2}}) ≃ {±1} × Isom(q_{1}, q_{2}),

where Isom(·, ·) denotes the set of isometries between two forms. In particular, for any positive binary form q we have c(f_{q}) ≥ c(q) and wt(f_{q}) = |Aut(q)|.

Proof. Let e_{1}, e_{2}, e_{3} be the standard basis of M := ℤ^{3}, and put M_{1} = ℤe_{1} and M_{2} = ℤe_{2}+ℤe_{3}. Then f_{q_{i}}(xe_{1}+ye_{3}+ze_{3}) = x^{2}+4q_{i}(y, z), so (M_{1})_{q_{i}}^{⊥} = M_{2} and (f_{q_{i}})_{|M_{2}} = 4q_{i}, for i = 1, 2. Thus, if φ ∈ Isom(q_{1}, q_{2}) = {φ ∈ Aut(M_{2}) = GL_{2}(ℤ) : q_{1} = q_{2} ∘ φ}, and if ε = ±1, then εid_{M_{1}} ⊕ φ ∈ Isom(f_{q_{1}}, f_{q_{2}}) = {φ ∈ Aut(M) : f_{q_{1}} = f_{q_{2}} ∘ φ}, and so the rule (ε, φ) ↦ εid_{M_{1}} ⊕ φ defines an injection {±} × Isom(q_{1}, q_{2}) → Isom(f_{q_{1}}, f_{q_{2}}).

To show that this map is also surjective, note first that for m ∈ M we have that f_{q_{i}}(m) = 1 ⇔ m = ±e_{1} because 4q_{i}(y, z) ≥ 4 for (y, z) ≠ (0, 0). Thus, if φ ∈ Isom(f_{q_{1}}, f_{q_{2}}), then f_{q_{2}}(φ(e_{1})) = f_{q_{1}}(e_{1}) = 1, and so φ(e_{1}) = ±e_{1}. It thus follows that φ(M_{2}) = φ((M_{1})_{q_{1}}^{⊥}) = (M_{1})_{q_{2}}^{⊥} = M_{2}, so φ_{|M_{2}} ∈ Isom(q_{1}, q_{2}), and hence φ = ±id_{M_{1}} ⊕ φ_{|M_{2}}. Thus, the above map is surjective and hence bijective.

From (45) we see that the natural map gen(q) → gen(f_{q}) (given by q_{1} ↦ f_{q_{1}}, for q_{1} ∈ gen(q)) is injective, and so c(q) = #gen(q) ≤ #gen(f_{q}) = c(f_{q}).

To verify the last assertion, it is enough to show that |Aut^{+}(f_{q})| = |Aut(q)| (because wt(f_{q}) = |Aut^{+}(f_{q})|), and this follows immediately from (45) together with the observation that if α ∈ Aut(q), then there is a unique choice of sign ε ∈ {±1} such that det(εid_{M_{1}} ⊕ α) = 1, i.e. such that εid_{M_{1}} ⊕ α ∈ Aut^{+}(f_{q}).

Corollary 55 Let q′ be a primitive positive definite binary quadratic form of discriminant Δ′ and put q = tq′, where t ≥ 1. Then

(46) wt(f_{q}) = \{ \begin{matrix}12 & if Δ′ = −3, \\ 8 & if Δ′ = −4, \\ 4 & if Δ′ < −4 and q′ is ambiguous, \\ 2 & if Δ′ < −4 and q′ is not ambiguous. \end{matrix}

Proof. By Proposition 54 we have wt(f_{q}) = |Aut(q)| = |Aut(q′)|. Now it is well-known that |Aut(q′)| = ε|Aut^{+}(q′)|, where ε = 2 if q is ambiguous, and ε = 1 otherwise (cf. e.g. Jones[9], p. 147). Since |Aut^{+}(q′)| = 6 and 4, for Δ = −3 and −4, respectively, and |Aut^{+}(q′)| = 2, for Δ < −4, the formula (46) follows.

We are now ready to prove the main result of this section.

Proposition 56 For every q ∈ L we have M(f_{q}) = 1/wt(f_{q}), and so c(f_{q}) = 1.

Proof. We shall apply the mass formula (42) to the form q = tq′ ∈ L. Since each form q′ except q′ = [2, 0, 3] represents 1, we can take r′ = 1 for all except for the last, where we can take r′ = 5. To evaluate the right hand side of (42), note that for any discriminant Δ we have that

g(Δ) = 2^{ω(Δ)−1+ε(Δ)}, (47)

where ω(Δ) denotes the number of distinct prime divisors of Δ and ε(Δ) = 1 if Δ ≡ 0 (mod 32), ε = −1 if Δ ≡ 4 (mod 16) and ε(Δ) = 0 otherwise; cf. e.g. [12], §2.2.

Now if we let ν^{∗} be defined by 2^{ν^{∗}} = g(−16t)g(Δ′) and put ϕψ = ϕ(δ)ψ(δ), ψ_{1} = ψ(t, Δ′) and ψ_{2} = ψ(d′, −16tr′), then the quantities appearing on the right hand side of (42) for q ∈ L are given in the Table 1 below.

Since all the forms q′ here are ambiguous, we see from Table 1 and (46) that M(f_{q}) = 1/wt(f_{q}), for all q ∈ L. Now it is clear from the definition that

M(f_{q}) ≥ 1/wt(f_{q}) and M(f_{q}) = 1/wt(f_{q}) ⇔ #gen(f_{q}) = 1,

and so we obtain from this that c(f_{q}) = #gen(f_{q}) = 1, for all q ∈ L.

Remark 57 It is perhaps worthwhile to mention that the forms q which are equivalent to those in L are not the only binary forms q for which c(f_{q}) = 1 (so the “converse” of Proposition 56 is false). Indeed, the forms q ∈ {[1, 0, 2], [1, 0, 3], [1, 0, 6], [3, 0, 3]} also satisfy c(f_{q}) = 1. But each of these represents an integer n ≡ 3 (mod 4) (with n < |Δ(q)|), so they cannot be special idoneal-valued forms. Note that such forms are excluded from consideration in the hypothesis of Theorem 13, and hence this shows that this extra hypothesis is necessary for the validity of Theorem 13.

\begin{matrix}q & d′ & t & δ & ν^{∗} & ϕψ & ψ_{1} & ψ_{2} & M(f_{q})^{−1} \\ [1, 1, 1] & 3 & 1 & 1 & 0 & 1 & 1 & 2 & 12 \\ [2, 2, 2] & 3 & 2 & 1 & 1 & 1 & 1 & 4 & 12 \\ [4, 4, 4] & 3 & 4 & 1 & 1 & 1 & 2 & 2 & 12 \\ [6, 6, 6] & 3 & 6 & 3 & 2 & 8 & 3 & 3 & 12 \\ [10, 10, 10] & 3 & 10 & 1 & 2 & 1 & 4 & 2 & 12 \\ [1, 0, 1] & 4 & 1 & 2 & 0 & 3 & 1 & 4 & 8 \\ [2, 0, 2] & 4 & 2 & 4 & 1 & 12 & 2 & 4 & 8 \\ [6, 0, 6] & 4 & 6 & 4 & 2 & 12 & 4 & 4 & 8 \\ [1, 1, 2] & 7 & 1 & 1 & 0 & 1 & 1 & 6 & 4 \\ [1, 1, 4] & 15 & 1 & 1 & 1 & 1 & 1 & 12 & 4 \\ [2, 2, 6] & 11 & 2 & 1 & 1 & 1 & 1 & 12 & 4 \\ [2, 2, 18] & 35 & 2 & 1 & 2 & 1 & 1 & 24 & 4 \\ [2, 0, 4] & 8 & 2 & 4 & 1 & 12 & 2 & 8 & 4 \\ [2, 0, 10] & 20 & 2 & 4 & 2 & 12 & 2 & 16 & 4 \\ [4, 0, 6] & 12 & 2 & 4 & 1 & 12 & 2 & 8 & 4 \end{matrix}

Table 1: Evaluating the mass formula for q ∈ L.

7 Proofs of the main results

We are now ready to put the above results together to prove the main theorems of this paper. We begin with the main theorem on quadratic forms.

Proof of Theorems 6 and 13. (i) ⇒ (ii): Trivial, since f_{q} → 1.

(ii) ⇒ (iii): Here we use the hypothesis that P(Q) = P(Q)^{odd}. (Note that by Remark 33 this hypothesis is equivalent to the condition that either Δ ≡ 1 (mod 4) or that q ↛ n, for any n ≡ 3 (mod 4)). Thus, if θ ∈ P(Q) = P(Q)^{odd}, then by Corollary 29 we have that Q_{θ} ∈ gen(f_{q}) and so it follows from (ii) that Q_{θ} → 1.

(iii) ⇒ (iv): Corollary 36.

(iv) ⇒ (v): Theorem 52.

(v) ⇒ (i): Proposition 56.

As was already explained after the statement of Theorem 13, this theorem, together with Corollary 8, immediately implies Theorem 2 of the introduction.

We next prove Corollary 3, which requires some results from [13].

Proof of Corollary 3. Let A_{K}^{ex} denote the set of isomorphism classes of the CM abelian product surfaces A/K which do not contain a genus 2 curve. We will construct an injective map Φ_{K} : A_{K}^{ex} → L and show that it is surjective when char(K) = 0.

The map Φ_{K} : A_{K}^{ex} → L is defined by the relation

Φ_{K}(A) = q ⇔ q_{A} ∼ xy ⊥ (−q). (48)

To see that this relation defines a function, we show first that q ∈ L is uniquely determined by this condition. For this we shall use the fact that

c(q) = 1, for all q ∈ L, (49)

which follows from the fact that if q ∈ L, then q = tq′, where q′ is primitive and Δ(q′) ∈ {−3, −4, −7, −8, −11, −15, −20, −24, −35}, which are all idoneal discriminants. (Alternately, we can deduce (49) by noting that c(f_{q}) = 1 by Proposition 56, and that hence also c(q) = 1 by Proposition 54.) Thus, if xy ⊥ (−q_{1}) ∼ xy ⊥ (−q_{2}), then by Corollary 26 we know that gen(q_{1}) = gen(q_{2}). But by (49) it then follows that q_{1} ∼ q_{2} and that hence q_{1} = q_{2} because the forms in L are distinct reduced forms. Thus, q is uniquely determined by (48).

Now let A ∈ A_{K}^{ex}. Since (by definition) A ∼ E × E, for some CM elliptic curve E/K, we have by [13], Theorem 2, that A ≃ E_{1} × E_{2}, for some elliptic curves E_{i}/K, and so by Theorem 2 we know that q_{E_{1},E_{2}} ∼ q, for some q ∈ L. Then by (6) we have that q_{A} ∼ xy ⊥ (−q_{E_{1},E_{2}}) ∼ xy ⊥ (−q), and hence Φ_{K}(A) = q. This shows that (48) defines a map Φ_{K} : A_{K}^{ex} → L.

To show that Φ_{K} is injective and/or bijective, we examine the fibre Φ_{K}^{−1}(q), for q ∈ L. For this we first note that if q ∈ L, then

(50) q ∈ Im(Φ_{K}) ⇔ there exists a CM elliptic curve E/K with Δ_{E} = Δ(q).

Indeed, suppose that q ∈ Im(Φ_{K}). Then by the above discussion we know that q ∼ q_{E_{1},E_{2}}, for some CM elliptic curves E_{1} ∼ E_{2}. By [13], (79), we know that Δ(q) = Δ(q_{E_{1},E_{2}}) = −lcm(Δ_{E_{1}}, Δ_{E_{1}}), and by [13], Proposition 29, it follows that there is an elliptic curve E/K with Δ_{E} = −lcm(Δ_{E_{1}}, Δ_{E_{1}}) = Δ(q).

Conversely, if E/K is a CM elliptic curve with Δ_{E} = Δ(q), then for any pair (E_{1}, E_{2}) of isogenous CM elliptic curves with q_{E_{1},E_{2}} ∼ q, it follows from [13], Corollary 42 and [13], Proposition 36 that E_{i} ∼ E. Thus, Φ_{K}^{−1}(q) = {A ∼ E × E : q_{A} ∼ xy ⊥ (−q)}/ ≃, which by [13], Corollary 70, has cardinality N_{q} = h(Δ′)/g(Δ′) ≥ 1, where Δ′ = Δ(q)/cont(q)^{2}. Thus q ∈ Im(Φ_{K}), and so (50) follows. Moreover, since it follows from (49) that N_{q} = h(Δ′)/g(Δ′) = 1, we see that Φ_{K} is injective, and so #A_{K}^{ex} ≤ #L = 15, as asserted. In addition, if char(K) = 0, then for each discriminant Δ < 0 there is a CM elliptic curve E′/K with Δ_{E′} = Δ (cf. [13], Proposition 37(a)), and so Φ_{K} is surjective. Thus, there are precisely 15 such non-isomorphic surfaces.

Remark 58 The above proof also determines the number Ex_{K} := #A_{K}^{ex} of exceptional surfaces when char(K) = p ≠ 0. Indeed, since a CM elliptic curve E/K with Δ_{E} = Δ exists if and only if (\frac{Δ}{p}) = 1 (cf. [13], Proposition 37(b)), it follows from (50) that Ex_{K} = #(L_{p}), where L_{p} = \{q ∈ L : (\frac{Δ(q)}{p}) = 1\}. From this we see that Ex_{K} is determined by suitable congruence conditions on p = char(K). In particular, there are infinitely many characteristics p for which no exceptional surfaces exist, i.e. Ex_{K} = 0 (the smallest is p = 479) and there are infinitely many for which all 15 surfaces exist, i.e. Ex_{K} = 15 (the smallest is p = 2689). Similarly, since #L_{2} = 2, #L_{3} = 4 and #L_{5} = 5, we see that there are precisely 2, 4, and 5 exceptional surfaces when char(K) = 2, 3, and 5, respectively.

By using the theory of [13], the exceptional surfaces can be written down quite explicitly. Here we use the method due to Deuring (see the reference in [13]) which associates to each ideal I of End(E) a finite subgroup scheme H(I) of E.

Corollary 59 Let q ∈ L and suppose that there exists an elliptic curve E/K with Δ_{E} = Δ := Δ(q). Fix an isomorphism End(E) ≃ R_{Δ} := ℤ + ℤ\frac{Δ+\sqrt{Δ}}{2}, and put c = cont(q). Then the unique exceptional surface A_{q}/K with q_{A_{q}} ∼ xy ⊥ (−q) is given by

(51) A_{q} = E × E/H(I_{q}), where I_{q} = \{ \begin{matrix}cR_{Δ/c^{2}} & if Δ ≠ −96 \\ 4ℤ + \sqrt{−24}ℤ & if Δ = −96. \end{matrix}

Proof. In view of the bijection of [13], Theorem 67 (see also [13], Remark 68), together with equations (101) and (78) of [13], it is enough to show that q ∼ [R(I) : R]q_{I_{q}}^{+}, where q_{I_{q}}^{+} denotes the equivalence class of (primitive) forms associated to the lattice (ideal) I_{q} of R = R_{Δ}. Now if Δ(q) ≠ −96, then q ∼ c1_{Δ′}, where Δ′ = Δ/c^{2}, and so the desired equivalence is clear in this case. If Δ(q) = −96, then q ∼ [4, 0, 6], and it is easy to see that here q_{I_{q}}^{+} ∼ [2, 0, 3] and so the result is true here as well.

Remark 60 If K = ℂ, then it follows from Corollary 59 (and [13], Proposition 33) that the exceptional surfaces A_{q} with q ∈ L have a very simple analytic description:

(52) A_{q} ≃ ℂ/R_{Δ} × ℂ/L_{q}, where L_{q} = \{ \begin{matrix}R_{Δ/c^{2}} & if Δ ≠ −96, \\ 2ℤ + \sqrt{−6}ℤ & if Δ = −96. \end{matrix}

Here, as before, Δ = Δ(q), c = cont(q) and R_{Δ} = ℤ + ℤ\frac{Δ+\sqrt{Δ}}{2}.

From (52) we can determine explictly the j-invariants of the pair (E_{1}, E_{2}) which define A_{q} by calculating the (known) values j_{1} := j(E_{1}) = j(R_{Δ}) and j_{2} = j(E_{2}) = j(L_{q}) of the j-function. Thus, there are precisely 5 cases for which (j_{1}, j_{2}) ∈ ℚ^{2}; these are given in the following table:

\begin{matrix}−Δ(q) & 3 & 4 & 7 & 12 & 16 \\ (j_{1}, j_{2}) & (0, 0) & (2^{6}3^{3}, 2^{6}3^{3}) & (−3^{3}5^{3}, −3^{3}5^{3}) & (2^{4}3^{2}5^{3}, 0) & (2^{3}3^{3}11^{3}, 2^{6}3^{2}) \end{matrix}

The values of other pairs can also be computed, but lead to complicated algebraic numbers of degree ≤ max{h(Δ(q)) : q ∈ L} = 6.

One can also determine the pairs (E_{1}, E_{2}) of all elliptic curves such that q_{E_{1},E_{2}} ∼ q ∈ L by using Corollary 42 of [13]. A somewhat tedious count shows that there are 46 such pairs.

We now turn to the proof of Corollary 4. This will we be deduced from the following more precise statement.

Theorem 61 If A/K_{0} is an abelian surface such that A is isogenous to E × E, where E/K_{0} is a CM elliptic curve with rank(End_{K_{0}}(E)) = 2, then there exist elliptic curves E_{1} and E_{2}/K_{0} such that A ≃ E_{1} × E_{2}. Moreover, the following conditions are equivalent:

(i) There is a curve C/K_{0} such that J_{C} ≃ A ≃ E_{1} × E_{2}; (i′) There is a genus 2 curve on A ≃ E_{1} × E_{2}; (ii) q_{A} is not equivalent to xy ⊥ (−q), for any q ∈ L; (iii) q_{E_{1},E_{2}} is not equivalent to any q ∈ L.

Proof. The existence of E_{1} and E_{2} with A ≃ E_{1} × E_{2} follows directly from [13], Theorem 2. It thus remains to show that the conditions (i) – (iii) are equivalent.

(i) ⇒ (iii): Let K = \overline{K}_{0} be the algebraic closure of K_{0} and write \overline{E}_{i} = E_{i} ⊗ K. Since E_{1} ∼ E_{2} ∼ E, the hypothesis implies that Hom(E_{1}, E_{2}) = Hom(\overline{E}_{1}, \overline{E}_{2}) (cf. [13], Lemma 14(b)), and so q_{E_{1},E_{2}} ∼ q_{\overline{E}_{1},\overline{E}_{2}}. Since \overline{C} = C ⊗ K is a genus 2 curve on J\overline{C} ≃ \overline{E}_{1} × \overline{E}_{2}, it follows from Theorem 2 that q_{E_{1},E_{2}} cannot be equivalent to any q ∈ L.

(iii) ⇒ (ii): Put \overline{A} = A ⊗ K = \overline{E}_{1} × \overline{E}_{2}. Then by (53) below we have that NS(A) = NS(\overline{A}), and so by (6) (applied to \overline{A}) we have that q_{A} ∼ q\overline{A} ∼ xy ⊥ (−q_{\overline{E}_{1},\overline{E}_{2}}) ∼ xy ⊥ (−q_{E_{1},E_{2}}). Now suppose that (iii) is false, i.e. that q_{A} ∼ xy ⊥ (−q), for some q ∈ L. Then by Corollary 26 we know that q_{E_{1},E_{2}} ∈ gen(q). But for q ∈ L we have c(q) = 1 (cf. proof of Corollary 3), and so q_{E_{1},E_{2}} ∼ q, which is contrary to the hypothesis (ii).

(ii) ⇒ (i′): If (ii) holds, then it is clear that also (iii) holds because if q_{E_{1},E_{2}} ∼ q, then q_{A} ∼ xy ⊥ (−q_{E_{1},E_{2}}) ∼ xy ⊥ (−q), contradiction. Thus, by Theorem 2 (applied to \overline{A} ≃ \overline{E}_{1} × \overline{E}_{2}), we see that there exists a genus 2 curve \overline{C} on \overline{A}. By Lemma 62 below we then also have a genus 2 curve on A, and so (i′) holds.

(i′) ⇒ (i): This follows from Weil[21], Satz 2.

In the above proof we used the following fact.

Lemma 62 Let E_{1} and E_{2} be elliptic curves over a field K_{0}, and suppose that Hom(E_{1}, E_{2}) = Hom(\overline{E}_{1}, \overline{E}_{2}), where \overline{E}_{i} = E_{i} ⊗ K and K is the algebraic closure of K_{0}. If \overline{C} is genus 2 curve on \overline{A} := \overline{E}_{1} × \overline{E}_{2}, then there exists a genus 2 curve C on A = E_{1} × E_{2} such that C ⊗ K is a translate of \overline{C}.

Proof. We first show that

(53) \overline{D} ∈ Div(\overline{A}) ⇒ ∃D ∈ Div(A) such that D ⊗ K ≡ \overline{D},

where D ⊗K denotes the pullback of the divisor D under base change, and ≡ denotes numerical equivalence (as in §2). Indeed, let θ_{i} = pr_{i}^{∗}(0_{E_{i}}) ∈ Div(A) and \overline{θ}_{i} = θ_{i} ⊗ K = pr_{i}^{∗}(0_{\overline{E}_{i}}) ∈ Div(\overline{A}), for i = 1, 2. Then by the fundamental identification NS(\overline{A}) ≃ ℤ^{2} ⊕ Hom(\overline{E}_{1}, \overline{E}_{2}) (cf. [11], Proposition 22), there exist integers a, b ∈ ℤ and \overline{h} ∈ Hom(\overline{E}_{1}, \overline{E}_{2}) such that \overline{D} ≡ a\overline{θ}_{1} + b\overline{θ}_{2} + Γ\overline{h}, where Γ\overline{h} denotes the graph of \overline{h}. By hypothesis, there is an h ∈ Hom(E_{1}, E_{2}) such that h ⊗ K = \overline{h}. Thus, if we put D = aθ_{1} + b\overline{θ}_{2} + Γ_{h} ∈ Div(A), then D ⊗ K = a\overline{θ}_{1} + b\overline{θ}_{2} + Γ\overline{h}, and so (53) follows.

Applying this to \overline{D} = \overline{C}, we thus see that there exists D ∈ Div(A) such that D⊗K ≡ \overline{C}. Since \overline{C}^{2} = 2 by the adjunction formula, we conclude from the Riemann-Roch theorem on \overline{A} (cf. Mumford[15], p. 150) that \overline{C} defines a principal polarization and that hence D ⊗ K ∼ \overline{C}_{x}, for some x ∈ \overline{A}(K). Thus dim_{K_{0}} H^{0}(A, O(D)) = dim_{K} H^{0}(\overline{A}, O(D⊗K)) = 1 (the latter by Riemann-Roch and the Vanishing Theorem; cf. [15], p. 150), and hence D ∼ C, for a unique effective divisor C ≥ 0, C ∈ Div(A). By uniqueness we then have C ⊗ K = \overline{C}_{x}, which proves the lemma.

Proof of Corollary 4. Since q_{A} ∼ xy ⊥ (−q_{E_{1},E_{2}}), it follows that Δ(q_{A}) = −Δ(q_{E_{1},E_{2}}), and so it is clear that Corollary 4 follows from Theorem 61 because the discriminants of the forms in L are the negative of the numbers listed in (2).

Proof of Theorem 5. As is mentioned on p. 145 of [8], it follows from a result of Deligne that E_{1} × E_{2} ≃ E_{0} × E_{0}, where E_{0}/K is a fixed supersingular curve, so all such product abelian surfaces are isomorphic. Thus, using Weil[21], Satz 2, again, there is a genus 2 curve C on E_{1} × E_{2} if and only if there is a genus 2 curve C with J_{C} ≃ E′_{1} × E′_{2}, for some supersingular curves E′_{1} and E′_{2}, and so the assertion follows immediately from the mass formula of [8], p. 149.

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