Integer-Valued Quadratic Forms

and Quadratic Diophantine Equations

Goro Shimura

Received: September 12, 2006

Communicated by Don Blasius

Abstract. We investigate several topics on a quadratic form Φ over an algebraic number field including the following three: (A) an equation ξΦ ·^{t}ξ = Ψ for another form Ψ of a smaller size; (B) classification of Φ over the ring of algebraic integers; (C) ternary forms. In (A) we show that the “class” of such a ξ determines a “class” in the orthogonal group of a form Θ such that Φ ≈ Ψ ⊕ Θ. Such was done in [S3] when Ψ is a scalar. We will treat the case of nonscalar Ψ, and prove a class number formula and a mass formula, both of new types. In [S5] we classified all genera of Z-valued Φ. We generalize this to the case of an arbitrary number field, which is topic (B). Topic (C) concerns some explicit forms of the formulas in (A) when Φ is of size 3 and Ψ is a scalar.

2000 Mathematics Subject Classification: 11E12 (primary), 11D09, 11E41 (secondary)

Introduction

A quadratic Diophantine equation in the title means an equation of the form ξΦ · ^{t}ξ = Ψ, where Φ and Ψ are symmetric matrices of size n and m, and ξ is an (m × n)-matrix. We assume that n > m, det(Φ) det(Ψ) ≠ 0, and all these matrices have entries in an algebraic number field F. The purpose of this paper is to present various new ideas and new results on such an equation. In the simplest case m = 1, we take a vector space V of dimension n over F, take also a nondegenerate symmetric F -bilinear form φ : V × V → F, and put φ[x] = φ(x, x) for x ∈ V. Then the equation can be written φ[x] = q with q ∈ F, ≠ 0. In our recent book [S3] we formulated a new arithmetic framework of such an equation. In the present paper we attempt to extend the theory so that equations of the type ξΦ ·^{t}ξ = Ψ can be treated along the same line of ideas, and give essential improvements on the results in [S3], as well as some new results.

Let us first recall the basic ideas of [S3] on this topic. Let 𝔤 denote the ring of algebraic integers in F. For a 𝔤-lattice L in V and a fractional ideal 𝔟 in F we put

L[q, 𝔟] = { x ∈ V | φ[x] = q, φ(x, L) = 𝔟 },
Γ (L) = { γ ∈ SO^{φ}(V) | Lγ = L},
SO^{φ}(V) = { α ∈ SL(V, F) | φ[xα] = φ[x] for every x ∈ V }.

We call L integral if φ[x] ∈ 𝔤 for every x ∈ L, and call an integral lattice maximal if it is the only integral lattice containing itself.

Given h ∈ L[q, 𝔟], put G = SO^{φ}(V), W = (F h)^{⊥}, that is,

W = {x ∈ V | φ(x, h) = 0},

and H = SO^{φ}(W), where we use φ also for its restriction to W. Then in [S3] we proved, for a maximal L, that

(1a) There is a bijection of ⨆_{i∈I} {L_{i}[q, 𝔟]/Γ (L_{i})} onto H\H_{A}/(H_{A} ∩ C), and consequently

(1b)  ∑_{i∈I} #{L_{i}[q, 𝔟]/Γ (L_{i})} = #{H\H_{A}/(H_{A} ∩ C)}.

Here the subscript A means adelization, C = {x ∈ G_{A} | Lx = L}, and {L_{i}}_{i∈I} is a complete set of representatives for the classes in the genus of L with respect to G. If F = Q, 𝔤 = 𝔟 = Z, L = Z^{3}, and φ is the sum of three squares, then the result is a reformulation of the result of Gauss that connects the number of primitive representations of an integer q as sums of three squares with the class number of primitive binary forms of discriminant −q. This result can be formulated as a statement about #L[q, Z] for such φ and L as Gauss did, and also as another statement concerning #{L[q, 𝔟]/Γ (L)} as in (1b), which Gauss did not present in a clear-cut form. Though Eisenstein and Minkowski investigated #L[q, Z] when φ is the sum of five squares, neither (1a) nor (1b) appears in their work. It seems that the idea of L[q, 𝔟]/Γ (L) was taken up for the first time in [S3].

Now the major portion of this paper consists of several types of new results, which are the fruits of the ideas developed from (1a) and (1b). More explicitly, they can be described as follows.

I. The main theorems about #{L[q, 𝔟]/Γ (L)} in [S3], formula (1b) in particular, were proved under the following condition: if n is odd, then det(φ)𝔤 is a square ideal. We will show that this condition is unnecessary.

II. The same types of problems about the equation ξΦ·^{t}ξ = Ψ for m > 1 were discussed in [S3] to some extent, but mainly restricted to the case m = n−1. In Theorem 2.2 we will prove more general results for an arbitrary m formulated from a new viewpoint. Namely, we present the principle that the class of a solution ξ determines a class in the orthogonal group in dimension n − m, which is complementary to that of Ψ. We then obtain generalizations of (1a) and (1b).

III. In [S3] we proved a certain mass formula which connects the “mass” of the set L[q, 𝔟] with the mass of H with respect to a subgroup of H_{A}. If F is totally real and φ is totally definite, this can be given in the form

(2)  ∑_{i∈I} #L_{i}[q, 𝔟]/#Γ (L_{i}) = ∑_{ε∈E}[H ∩ εCε^{−1} : 1]^{−1},

where E is a subset of H_{A} such that H_{A} = ⨆_{ε∈E} Hε(H_{A} ∩ C). The right-hand side may be called the mass of H with respect to H_{A} ∩ C. In Theorem 3.2 we will generalize this to the case of ξΦ · ^{t}ξ = Ψ with m ≥ 1 and definite or indefinite φ. Our formulas are different from any of the mass formulas of Siegel. See the remark at the end of Section 3 for more on this point.

IV. For n = 3, the right-hand side of (1b) can be written [H_{A} : H(H_{A}∩C)]. In [S3] we gave an explicit formula for this index under the condition mentioned in I. In Theorem 5.7 of the present paper we prove the result without that condition. In general it is difficult to determine when L[q, 𝔟] ≠ ∅. We will investigate this problem for a ternary form.

Every ternary space is isomorphic to a space of type (B^{∘}, β) obtained from a quaternion algebra B over F as follows. Denote by ι the main involution of B and put B^{∘} = {x ∈ B | x^{ι} = −x} and β[x] = dxx^{ι} for x ∈ B^{∘} with d ∈ F^{×}. Then we will determine for a maximal L exactly when L[q, Z] ≠ ∅ and give an explicit formula for #L[q, Z] for (B^{∘}, β) over Q under the following conditions: (i) B is definite and the genus of maximal lattices in B^{∘} consists of a single class; (ii) the discriminant e of B is a prime number; (iii) d is one of the following four types: d = 1, d = e, d is a prime ≠ e, d/e is a prime ≠ e (Theorems 6.6 and 6.7). In fact there are exactly 30 positive definite ternary forms over Q satisfying these three conditions, including of course the case of the sum of three squares. If we drop conditions (ii) and (iii), then there are exactly 64 ternary quadratic spaces over Q of type (i). Though our methods are applicable to those 64 spaces, we impose the last two conditions in order to avoid complicated analysis.

Indeed, though the results about L[q, Z] are not so complicated, they are not of the type one can easily guess, even under all three conditions. Also, to make transparent statements, it is better to consider an equation ξ·λΦ^{−1}·^{t}ξ = s = λq with a suitable λ, where Φ is the matrix representing φ with respect to a Zbasis of L. Take, for example, a ternary form 2x^{2} + 3y^{2} − yz + z^{2}. Then we consider the equation ξ · λΦ^{−1} ·^{t}ξ = s for s ∈ Z with λ = 22, which can be written

(3) 11x^{2} + 8(y^{2} + yz + 3z^{2}) = s.

We can show that there is a bijection of L[s/22, Z] for this φ onto the set of solutions (x, y, z) of (3) such that xZ + yZ + zZ = Z. Moreover, such a solution exists if and only if s = r^{2}m with a squarefree positive integer m such that m − 3 ∈ 8Z and an odd positive integer r such that 11|r if 11 remains prime in K, where K = Q(\sqrt{−m}). Thus #L[s/22, Z] equals the number of such solutions of (3), and can be given as

(4) \frac{2^{3−μ_{0}−μ_{1}}c}{w} · r ∏_{p|r} \{1 − (\frac{−m}{p})\}.

Here μ_{0} = 0 if 2 is unramified in K and μ_{0} = 1 otherwise; μ_{1} = 0 if 11|r or 11 ∤ m, and μ_{1} = 1 otherwise; c is the class number of K; w is the number of roots of unity in K; ∏_{p|r} is the product over all prime factors p of r. We can state similar results in the 30 cases mentioned above, and the same can be said, in principle, even in the 64 cases too, though we will not do so in the present paper. The condition that the class number is 1 is necessary, as the left-hand side of (2) has more than one term otherwise.

A much smaller portion of this paper, Section 4, is devoted to the classification of Φ over 𝔤. In [S5] we classified the genera of matrices that represent reduced Z-valued quadratic forms. Here a quadratic form is called reduced if it cannot be represented nontrivially over Z by another Z-valued quadratic form. This is different from Eisenstein’s terminology for ternary forms. We will treat the same type of problem over the ring of algebraic integers of an arbitrary algebraic number field. For this we first have to define the genus of a symmetric matrix in a proper way, so that every genus of maximal lattices can be included. The formulation requires new concepts, and the classification has some interesting features. It should be noted that the obvious definition of a genus employed by Siegel applies only to a special case.

As a final remark we mention the article [S6], in which the reader will find a historical perspective of this topic that we do not include here. For example,

[S6] contains a more detailed account of the work of Gauss and his predecessors, Lagrange and Legendre, and also comparisons of our formulas with Siegel’s mass formulas. Therefore, [S6] is complementary to the present paper in that sense.

1. Basic symbols and a crucial local result

1.1. Throughout the paper we denote by V a finite-dimensional vector space over a field F and by φ a nondegenerate F -bilinear symmetric form V ×V → F; we put then φ[x] = φ(x, x) for x ∈ V, n = dim(V), and

O^{φ}(V) = {α ∈ GL(V) | φ[xα] = φ[x] for every x ∈ V },
SO^{φ}(V) = O^{φ}(V) ∩ SL(V).

For every subspace U of V on which φ is nondegenerate, we denote the restriction of φ to U also by φ, and use the symbols O^{φ}(U) and SO^{φ}(U). We denote by A(V) the Clifford algebra of (V, φ), and define the canonical automorphism α ↦ α′ and the canonical involution α ↦ α^{∗} of A(V) by the condition −x′ = x^{∗} = x for every x ∈ V. We then put

A^{+}(V) = A^{+}(V, φ) = {α ∈ A(V) | α′ = α},

G^{+}(V) = {α ∈ A^{+}(V)^{×} | α^{−1}V α = V }.

We can define a homomorphism ν : G^{+}(V) → F^{×} by ν(α) = αα^{∗} and also a surjective homomorphism τ : G^{+}(V) → SO^{φ}(V) by xτ(α) = α^{−1}xα for α ∈ G^{+}(V) and x ∈ V. Then Ker(τ) = F^{×}. We denote by δ(φ) the coset of F^{×}/F^{×2} represented by (−1)^{n(n−1)/2} det(φ), where F^{×2} = {a^{2} | a ∈ F^{×}}.

We note here an easy fact [S3, Lemma 1.5 (ii)]:

(1.1) {k ∈ V | φ[k] = q} = h·SO^{φ}(V) if n > 1, h ∈ V, and φ[h] = q ∈ F^{×}.

1.2. We now consider symbols F and 𝔤 in the following two cases: (i) F is an algebraic number field of finite degree and 𝔤 is its maximal order; (ii) F and 𝔤 are the completions of those in Case (i) at a nonarchimedean prime. We call a field of type (i) a global field, and that of type (ii) a local field. In this paper, we employ the terms global and local fields only in these senses. If F is a local field, we denote by 𝔭 the maximal ideal of 𝔤. In both local and global cases, by a 𝔤-lattice (or simply a lattice) in a finite-dimensional vector space V over F, we mean a finitely generated 𝔤-module in V that spans V over F. When F is a global field, we denote by a and h the sets of archimedean primes and nonarchimedean primes of F respectively, and put v = a ∪ h. For each v ∈ v we denote by F_{v} the v-completion of F. Given an algebraic group G defined over F, we define an algebraic group G_{v} over F_{v} for each v ∈ v and the adelization G_{A} as usual, and view G and G_{v} as subgroups of G_{A}. We then denote by G_{a} and G_{h} the archimedean and nonarchimedean factors of G_{A}, respectively. In particular, F_{A}^{×} is the idele group of F. For v ∈ v and x ∈ G_{A} we denote by x_{v} the v-component of x.

Given a 𝔤-lattice L in V and another 𝔤-lattice M contained in L in both local and global cases, we can find a finite set {𝔞} of integral ideals 𝔞 such that L/M as a 𝔤-module is isomorphic to ⨁_{𝔞∈{𝔞}} 𝔤/𝔞. We then put [L/M] = ∏_{𝔞∈{𝔞}} 𝔞. For x ∈ G_{A} with G acting on V in the global case, we denote by Lx the 𝔤-lattice in V such that (Lx)_{v} = L_{v}x_{v} for every v ∈ h. We call the set of all such Lx the G-genus of L, and call the set of Lα for all α ∈ G the G-class of L.

In our later treatment we will often use a quaternion algebra over a local or global field F. Whenever we deal with such an algebra B, we always denote by ι the main involution of B; we then put

(1.2) B^{∘} = { x ∈ B | x^{ι} = −x },

Tr_{B/F} (x) = x + x^{ι}, and N_{B/F} (x) = xx^{ι} for x ∈ B.

Given (V, φ) as in §1.1 over a local or global F, we call a 𝔤-lattice L in V integral if φ[x] ∈ 𝔤 for every x ∈ L, and call such an L maximal if L is the only integral lattice containing L. For an integral lattice L in V we denote by A(L) the subring of A(V) generated by 𝔤 and L, and put A^{+}(L) = A(L) ∩ A^{+}(V). For a 𝔤-lattice Λ in V, an element q in F^{×}, and a fractional ideal 𝔟 in F, we put

(1.3a)  Λ̃ = { y ∈ V | 2φ(y, Λ) ⊂ 𝔤 },

(1.3b) Λ[q, 𝔟] = { x ∈ V | φ[x] = q, φ(x, Λ) = 𝔟 }, (1.3c) D(Λ) = {γ ∈ O^{φ}(V) | Λγ = Λ}, C(Λ) = D(Λ) ∩ SO^{φ}(V) (F local).

Theorem 1.3. If F is local, L is a maximal lattice in V, and n > 2, then #{L[q, 𝔟]/C(L)} ≤ 1.

This was given in [S3, Theorem 10.5] under the condition that if n is odd, then δ(φ) ∩ 𝔤^{×} ≠ ∅; see [S3, (8.1)]. Our theorem says that this condition is unnecesary. It is sufficient to prove the case 𝔟 = 2^{−1}𝔤, since cL[q, 𝔟] = L(c^{2}q, c𝔟] for every c ∈ F^{×}. We devote §§1.4 and 1.5 to the proof. To avoid possible misunderstandings, we note that a result which looks similar to the above theorem was stated in [E1, Satz 10.4]. This result of Eichler has no relevance to our theory, as it does not consider the set φ(x, L), an essential ingredient of our theorem. Besides, we have #{L[q, 𝔟]/C(L)} = 2 for certain nonmaximal L; see [S7, Theorem 4.2 and (4.3)].

1.4. Given (V, φ) over a local field F and a maximal lattice L in V, by Lemma 6.5 of [S3] we can find decompositions

(1.4a) V = Z ⊕ U, U = ∑_{i=1}^{r}(F e_{i} + F f_{i}), φ(Z, U) = 0,

(1.4b) L = M ⊕ R, R = ∑_{i=1}^{r}(𝔤e_{i} + 𝔤f_{i}), M = {z ∈ Z | φ[z] ∈ 𝔤},

(1.4c) 2φ(e_{i}, f_{j}) = δ_{ij}, φ(e_{i}, e_{j}) = φ(f_{i}, f_{j}) = 0, φ[z] ≠ 0 for 0 ≠ z ∈ Z.

We put t = dim(Z), and call it the core dimension of (V, φ). To prove Theorem 1.3, we assume hereafter until the end of §1.6 that n is odd and δ(φ) contains a prime element of F. Then t is 1 or 3. We denote by π any fixed prime element of F. We first note a few auxiliary facts:

(1.5) {x ∈ 𝔤 | x − 1 ∈ 4𝔭^{m}} = {a^{2} | a − 1 ∈ 2𝔭^{m}} if 0 < m ∈ Z. (1.6) If k ∈ U and 2φ(k, R) = 𝔤, then kα = e_{1} + sf_{1} with some α ∈ D(R)

and s ∈ 𝔤; α can be taken from C(R) if r > 1. (1.7) If h ∈ L[q, 2^{−1}𝔤], ∉ Z, then there exists an element α of C(L) such that

hα = π^{m}(e_{1} + sf_{1}) + z with 0 ≤ m ∈ Z, s ∈ 𝔤, and z ∈ M.

The first of these is [S3, Lemma 5.5 (i)]; (1.6) is proven in [S3, §10.7, (A1), (A2)]. Notice that the proof there is valid even when φ[k] = 0. Now let h = k + w ∈ L[q, 2^{−1}𝔤] with k ∈ U and w ∈ Z. If h ∉ Z, then k ≠ 0, and we can put 2φ(k, R) = 𝔭^{m} with 0 ≤ m ∈ Z. By (1.6), there exists an element α_{0} ∈ D(R) such that π^{−m}kα_{0} = e_{1} + sf_{1} with s ∈ 𝔤. Since D(M) contains an element of determinant −1, we can extend α_{0} to an element α of C(L) such that Mα = M. This proves (1.7).

1.5. By virtue of (1.7), it is sufficient to prove Theorem 1.3 when r ≤ 1. If r = 0, then L = M and SO^{φ}(V) = C(L) by [S3, Lemma 6.4]. Therefore, if n = t = 3 and φ[z] = φ[w] with z, w ∈ V, then by (1.1), z = wα with α ∈ C(L), which gives the desired fact. For this reason we hereafter assume that r = 1, and write e and f for e_{1} and f_{1}. We represent the elements of O^{φ}(V) by (n × n)-matrices with respect to {e, g_{1}, , …, g_{t}, f}, where {g_{1}, , …, g_{t}} is a 𝔤-basis of M.

(A) Case t = 1. In this case M = 𝔤g with an element g such that φ[g] is a prime element of F. Thus we can take φ[g] as π.

(a1) Suppose L[q, 2^{−1}𝔤] ∩ Z ≠ ∅; let w ∈ L[q, 2^{−1}𝔤] ∩ Z and h ∈ L[q, 2^{−1}𝔤]. Then we can put w = [0 c 0] with c ∈ F such that 2c𝔭 = 𝔤. By (1.1), h = wα with α ∈ SO^{φ}(V). Now SO^{φ}(V) = P C(L) with the subgroup P of SO^{φ}(V) consisting of the upper triangular elements; see [S3, Theorem 6.13 (ii)]. Put α = βγ with β ∈ P and γ ∈ C(L). The second row of β is of the form [0 1 j] with j ∈ F. Thus hγ^{−1} = wβ = [0 c cj], and so cj ∈ 𝔤. Since 2c𝔭 = 𝔤, we can find an element p of 𝔤 such that 2cπp = cj. Take the matrix

(1.8) η = [ \begin{matrix}1 & −p & −πp^{2} \\ 0 & 1 & 2πp \\ 0 & 0 & 1 \end{matrix} ] .

Then η ∈ C(L) and wη = wβ = hγ^{−1}, which gives the desired fact.

(a2) Thus we assume that L[q, 2^{−1}𝔤] ∩ Z = ∅. Let h ∈ L[q, 2^{−1}𝔤]. By (1.7) we can put h = π^{m}(e + sf) + cg with 0 ≤ m ∈ Z and s, c ∈ 𝔤. Then q = π^{2m}s + πc^{2} and 𝔭^{m} + 2c𝔭 = 𝔤. Put d = 2πc. Suppose d ∈ 𝔤^{×}. Then by (1.5), we can put d^{2} + 4π^{2m+1}s = d_{1}^{2} with d_{1} ∈ 𝔤^{×}. We easily see that (2π)^{−1}d_{1}g ∈ L[q, 2^{−1}𝔤], a contradiction, as g ∈ Z. Thus d ∈ 𝔭, so that m = 0 and h = e + sf + (2π)^{−1}dg, and q = s + (4π)^{−1}d^{2}. Suppose we have another element h_{1} = e+s_{1}f +(2π)^{−1}d_{1}g with s_{1} ∈ 𝔤 and d_{1} ∈ 𝔭 such that φ[h_{1}] = q. Then d_{1}^{2} − d^{2} = 4π(s − s_{1}) ∈ 4𝔭, and so d_{1} − d ∈ 2𝔭 or d_{1} + d ∈ 2𝔭. Since −2d ∈ 2𝔭, we have d_{1} − d ∈ 2𝔭 in both cases. Let α be the element of (1.8) with p = (d_{1} − d)/(2π). Then α ∈ C(L) and h = h_{1}α ∈ h_{1}C(L). This completes the proof when t = 1.

(B) Case t = 3. As shown in [S3, §§7.3 and 7.7 (III)], there is a division quaternion algebra B over F with which we can put Z = B^{∘} and 2φ(x, y) = dTr_{B/F} (xy^{ι}) for x, y ∈ Z, where d is an element of F^{×} that represents the determinant of the restriction of φ to Z. Thus we can take a prime element π of F as d; then M = {x ∈ Z | πxx^{ι} ∈ 𝔤}. Let K be an unramified quadratic extension of F, 𝔯 the maximal order of K, and ρ the generator of Gal(K/F). We can put B = K + Kη with an element η such that η^{2} = π and xη = ηx^{ρ} for every x ∈ K. Take u ∈ 𝔯 so that 𝔯 = 𝔤[u] and put σ = u − u^{ρ}. Then M = 𝔤σ + 𝔯η^{−1} and M̃ = (2𝔭)^{−1}σ + 𝔯η^{−1}. Since r = 1, φ is represented by a matrix

[ \begin{matrix}0 & 0 & 2^{−1} \\ 0 & ζ & 0 \\ 2^{−1} & 0 & 0 \end{matrix} ] ,

where ζ is an element of 𝔤_{3}^{3} that represents the restriction of φ to Z. Define λ = (λ_{ij}) ∈ 𝔤_{3}^{3} by λ_{ii} = ζ_{ii}, λ_{ij} = 2ζ_{ij} if i < j and λ_{ij} = 0 if i > j. Then λ +^{t}λ = 2ζ. Let P be the subgroup of SO^{φ}(V) consisting of the elements that send F f onto itself, Then SO^{φ}(V) = P C(L); see [S3, Theorem 6.13 (ii)].

(b1) Suppose L[q, 2^{−1}𝔤] ∩ Z ≠ ∅. Let h ∈ L[q, 2^{−1}𝔤] and w ∈ L[q, 2^{−1}𝔤] ∩ Z. Identify w with a row vector w = [0 y 0], where y = (y_{i})_{i=1}^{3} with y_{i} ∈ 𝔤. By (1.1), h = wα with α ∈ SO^{φ}(V). Put α = βγ with β ∈ P and γ ∈ C(L). The middle three rows of β can be written in the form [0 ε j] with ε ∈ SO^{φ}(Z) and a column vector j. Replacing γ by diag[1, ε, 1]γ, we may assume that ε = 1. Then hγ^{−1} = wβ = [0 y yj] and yj ∈ 𝔤. Since w ∈ L[q, 2^{−1}𝔤], we see that 2yζ is primitive, and so we can find p ∈ 𝔤_{3}^{1} such that −2yζ ·^{t}p = yj. Let

(1.9) ψ = [ \begin{matrix}1 & p & −pλ ·^{t}p \\ 0 & 1 & −2ζ ·^{t}p \\ 0 & 0 & 1 \end{matrix} ] .

Then ψ ∈ C(L) and wψ = wβ = hγ^{−1}, and so h ∈ wC(L) as expected.

(b2) Let h ∈ L[q, 2^{−1}𝔤], ∉ Z. By (1.7) we may assume that h = π^{m}(e + sf) + z with 0 ≤ m ∈ Z, s ∈ 𝔤, and z ∈ Z. Put z = (2π)^{−1}cσ + xη^{−1} with c ∈ F and x ∈ K. Then 𝔤 = 𝔭^{m} + c𝔤 + Tr_{K/F} (x𝔯), q = π^{2m}s − πz^{2}, and πz^{2} = (4π)^{−1}c^{2}σ^{2} + xx^{ρ}. Suppose m > 0; then c ∈ 𝔤^{×} or x ∈ 𝔯^{×}. If c ∈ 𝔤^{×}, then by (1.5) we can put c^{2} − 4π^{2m+1}σ^{−2}s = b^{2} with b ∈ 𝔤^{×}. Put w = (2π)^{−1}bσ + xη^{−1}. Then we see that w ∈ L[q, 2^{−1}𝔤] ∩ Z. Next suppose x ∈ 𝔯^{×}; then xx^{ρ}−π^{2m}s ∈ 𝔤^{×}. Since N_{K/F} (𝔯^{×}) = 𝔤^{×}, we can find x_{1} ∈ 𝔯^{×} such that x_{1}x_{1}^{ρ} = xx^{ρ} −π^{2m}s. Put x = (2π)^{−1}cσ +x_{1}η^{−1}. Then x ∈ L[q, 2^{−1}𝔤]∩Z. Therefore, if m > 0, then the problem can be reduced to case (b1), which has been settled.

(b3) Thus if L[q, 2^{−1}𝔤]∩Z = ∅ and h ∈ L[q, 2^{−1}𝔤], then we may assume that h = e+sf +z with s ∈ 𝔤 and z ∈ Z. Take another element h_{1} = e+s_{1}f +z_{1} ∈ L[q, 2^{−1}𝔤] with s_{1} ∈ 𝔤 and z_{1} ∈ Z. We have z = (2π)^{−1}cσ + xη^{−1} and z_{1} = (2π)^{−1}c_{1}σ+x_{1}η^{−1} with c, c_{1} ∈ 𝔤 and x, x_{1} ∈ 𝔯. Since s−πz^{2} = s_{1} −πz_{1}^{2}, we have (4π)^{−1}(c^{2} − c_{1}^{2}) ∈ 𝔤, so that c^{2} − c_{1}^{2} ∈ 4𝔭. Then c − c_{1} ∈ 2𝔭 or c + c_{1} ∈ 2𝔭. Now the map xσ + yη^{−1} ↦ −xσ + y^{ρ}η^{−1} is an element of C(M). Therefore, replacing z_{1} by its image under this map if necessary, we may assume that c − c_{1} ∈ 2𝔭, which implies that z − z_{1} ∈ M. Define ψ by (1.9) by taking p to be the vector that represents z − z_{1}. Then h_{1}ψ = h, which gives the desired fact.

1.6. We can now remove condition (8.1) from [S3, Lemma 10.8], which concerns the case where t = 1 and r > 0. To be precise, that lemma can be restated, without condition (8.1), as follows:

Let L = 𝔤e_{1}+𝔤f_{1}+𝔤g with g such that 𝔭 ⊂ φ[g]𝔤 ⊂ 𝔤, and let h ∈ L[q, 2^{−1}𝔤] with q ∈ F^{×}.Then there exists an element γ of C(L) such that hγ = cg with c satisfying 2cφ[g]𝔤 = 𝔤 or hγ = ae_{1} + f_{1} + cg with a ∈ 𝔤 and c ∈ 2^{−1}𝔤.

This follows from the discussions of (a1) and (a2) combined with (1.7).

1.7. Still asssuming F to be local, take the symbols as in (1.4a, b, c) with even or odd n, put C = C(L) for simplicity, and define subgroups T and J of G^{+}(V) by (1.10) T = { α ∈ G^{+}(V) | τ(α) ∈ C},

(1.11) J = { γ ∈ T | ν(γ) ∈ 𝔤^{×} }.

In [S3], Proposition 8.8, Theorems 8.6, and 8.9 we discussed the structure of A(L), A^{+}(L), ν(J), and the connection of τ(J) with C under the condition that δ(φ) ∩ 𝔤^{×} ≠ ∅ if n is odd. Let us now prove the results in the cases in which the condition is not satisfied.

Theorem 1.8. Suppose F is local, 1 < n − 1 ∈ 2Z, and δ(φ) contains a prime element. Then the following assertions hold.

(i) If t = 1, then there exist an order 𝔒 in M_{2}(F) of discriminant 𝔭 (see §5.1 for the definition) and an isomorphism θ of A^{+}(V) onto M_{s}(M_{2}(F)) that maps A^{+}(L) onto M_{s}(𝔒), where s = 2^{r−1}.

(ii) If t = 3, then there exist a division quaternion algebra B over F and an isomorphism ξ : B^{∘} → Z such that φ[xξ] = dxx^{ι} for every x ∈ B^{∘} with a prime element d of F^{×} independent of x, where B^{∘} is defined by (1.2). Moreover there exists an isomorphism θ of A^{+}(V) onto M_{s}(B) that maps A^{+}(L) onto M_{s}(𝔒), where s = 2^{r} and 𝔒 is the unique maximal order in B.

(iii) In both cases t = 1 and t = 3 we have ν(J) = 𝔤^{×}, C = τ(T), [C : τ(J)] = 2, and T = F^{×}(J ∪ Jη) with an element η such that η ∉ F^{×}J, η^{2} ∈ F^{×}, Jη = ηJ, and ν(η)𝔤 = 𝔭.

This will be proved after the proof of Lemma 5.4.

Theorem 1.9. In the setting of Theorem 1.3 with even or odd n > 2, let t be the core dimension of (V, φ) and let σ : SO^{φ}(V) → F^{×}/F^{×2} be the spinor norm map of [S3, (3.7)]. Then σ(C) = 𝔤^{×}F^{×2} in the following three cases: (i) t = 0; (ii) t = 1 and δ(φ)∩𝔤^{×} ≠ ∅; (iii) t = 2 and L̃ = L. We have σ(C) = F^{×} in all the remaining cases.

Proof. That σ(C) = 𝔤^{×}F^{×2} in the three cases specified above is shown by Proposition 8.8 (iii) and Theorem 8.9 (i) of [S3]. Therefore we assume that those cases do not apply. If δ(φ) ∩ 𝔤^{×} ≠ ∅ or n is even, then Theorem 8.9 (ii) of [S3] shows that σ(C) = F^{×}. If n is odd and δ(φ) ∩ 𝔤^{×} = ∅, then from (iii) of Theorem 1.8 we obtain σ(C) = ν(T) = F^{×}, which completes the proof.

2. Global quadratic Diophantine equations

2.1. Let us now turn to the global case. The main theorems of [S3, Sections 11 through 13] were given under a condition stated in [S3, (9.2)]: if n is odd, then δ(φ), at every nonarchimedean prime, contains a local unit. By virtue of Theorem 1.3 we can now remove this condition from [S3, Theorem 11.6] and some others, which we will indicate in Remark 2.4 (3), (5), (6), and (7).

Throughout this section we assume that F is an algebraic number field. Given a quadratic space (V, φ) over F, we can define, for each v ∈ v, the v-localization (V_{v}, φ_{v}) of (V, φ) as a quadratic space over F_{v} by putting V_{v} = V ⊗_{F} F_{v} and extending φ to an F_{v}-valued quadratic form φ_{v} on V_{v} in a natural way.

In [S3, Section 13], we treated the equation φ[h] = q not only for a scalar q, but also for a symmetric matrix q of size n − 1. We now consider a more general case by taking a new approach. Given ^{t}q = q ∈ GL_{m}(F) and ^{t}φ = φ ∈ GL_{n}(F), we consider the solutions h ∈ F_{n}^{m} of the equation hφ ·^{t}h = q. Here and throughout this and the next sections we assume that n > 2 and n > m > 0. More intrinsically, take (V, φ) as before and take also (X, q) with a nondegenerate quadratic form q on a vector space X over F of dimension m. We put

(2.1) 𝒱 = Hom(X, V),

and consider h ∈ 𝒱 such that φ[xh] = q[x] for every x ∈ X. Since q is nondegenerate, h must be injective. To simplify our notation, for every k ∈ 𝒱 we denote by φ[k] the quadratic form on X defined by φ[k][x] = φ[xk] for every x ∈ X. Then our problem concerns the solutions h ∈ 𝒱 of the equation φ[h] = q for a fixed q. If m = 1 and X = F, then q ∈ F^{×}, and an element h of V defines an element of 𝒱 that sends c to ch for c ∈ F, and 𝒱 consists of all such maps. Thus we can put 𝒱 = V if m = 1 and the problem about φ[h] = q with q ∈ F^{×} is the one-dimensional special case. For m ≥ 1, if h ∈ 𝒱 and det(φ[h]) ≠ 0, then

(2.2) {k ∈ 𝒱 | φ[k] = φ[h]} = h · SO^{φ}(V).

This is a generalization of (1.1) and follows easily from the Witt theorem; see [S3, Lemma 1.5 (i)].

For a fixed h ∈ 𝒱 such that det (φ[h]) ≠ 0, put W = (Xh)^{⊥}, G = SO^{φ}(V), and H = SO^{φ}(W). We identify H with the subgroup of G consisting of the elements that are the identity map on Xh; thus

(2.3) H = {α ∈ G | hα = h}.

For ξ ∈ G_{A} the symbol hξ is meaningful as an element of 𝒱 _{A}, and so for a subset Ξ of G_{A} the symbol hΞ is meaningful as a subset of 𝒱 _{A}.

Theorem 2.2. Let D = D_{0}G_{a} with an open compact subgroup D_{0} of G_{h}. Then the following assertions hold.

(i) For y ∈ G_{A} we have H_{A} ∩ GyD ≠ ∅ if and only if 𝒱 ∩ hDy^{−1} ≠ ∅.

(ii) Fixing y ∈ G_{A}, for every ε ∈ H_{A} ∩ GyD take α ∈ G so that ε ∈ αyD. Then the map ε ↦ hα gives a bijection of H\(H_{A} ∩ GyD)/(H_{A} ∩ D) onto (𝒱 ∩ hDy^{−1})/Γ^{y}, where Γ^{y} = G ∩ yDy^{−1}.

(iii) Take Y ⊂ G_{A} so that G_{A} = ⨆_{y∈Y} GyD. Then

(2.4) #{H\H_{A}/(H_{A} ∩ D)} = ∑_{y∈Y} #{(𝒱 ∩ hDy^{−1})/Γ^{y}}.

(iv) In particular, suppose m = 1 and n > 2. With a fixed maximal lattice L in V put C = {ξ ∈ G_{A} | Lξ = L}, q = φ[h], and 𝔟 = φ(h, L). Then

(2.5) V ∩ hCy^{−1} = (Ly^{−1})[q, 𝔟] for every y ∈ G_{A}.

Note: Since H\H_{A}/(H_{A} ∩ D) is a finite set, from (ii) we see that (𝒱 ∩ hDy^{−1})/Γ^{y} is a finite set.

Proof. Let y, ε, and α be as in (ii); then clearly hα ∈ 𝒱 ∩ hDy^{−1}. If ηεζ ∈ βyD with η ∈ H, ζ ∈ H_{A} ∩ D, and β ∈ G, then β^{−1}ηα ∈ G ∩ yDy^{−1} = Γ^{y}, and hence hα = hηα ∈ hβΓ^{y}. Thus our map is well defined. Next let k ∈ 𝒱 ∩ hDy^{−1}. Then k = hδy^{−1} with δ ∈ D, and moreover, by (2.2), k = hξ with ξ ∈ G. Then h = hξyδ^{−1}, so that ξyδ^{−1} ∈ H_{A} by (2.3). Thus ξyδ^{−1} ∈ H_{A} ∩ GyD. This shows that k is the image of an element of H_{A} ∩ GyD. To prove that the map is injective, suppose ε ∈ αyD ∩ H_{A} and δ ∈ βyD ∩ H_{A} with α, β ∈ G, and hα = hβσ with σ ∈ Γ^{y}. Put ω = βσα^{−1}. Then hω = h, so that ω ∈ H. Since σ ∈ yDy^{−1}, we have βyD = βσyD = ωαyD, and hence δ ∈ βyD ∩H_{A} = ωαyD ∩H_{A} = ω(αyD ∩H_{A}) = ω(εD ∩H_{A}) = ωε(D ∩H_{A}) ⊂ Hε(D ∩ H_{A}). This proves the injectivity, and completes the proof of (ii). At the same time we obtain (i).

Now H_{A} = ⨆_{y∈Y} (H_{A} ∩ GyD), and so (iii) follows immediately from (ii). As for (iv), clearly V ∩ hC ⊂ L[q, 𝔟]. Conversely, every element of L[q, 𝔟] belongs to hC by virtue of Theorem 1.3. Thus

(2.6) V ∩ hC = L[q, 𝔟].

Let k ∈ V ∩ hCy^{−1} with y ∈ G_{A}; put M = Ly^{−1}. Then φ[k] = q, φ(k, M) = φ(h, L) = 𝔟, and kyCy^{−1} = hCy^{−1}. Taking k, M, and yCy^{−1} in place of h, L and C in (2.6), we obtain V ∩ hCy^{−1} = V ∩ kyCy^{−1} = M[q, 𝔟] = (Ly^{−1})[q, 𝔟]. This proves (2.5) when V ∩ hCy^{−1} ≠ ∅. To prove the remaining case, suppose ℓ ∈ (Ly^{−1})[q, 𝔟]; then φ(ℓy_{v}, L_{v}) = 𝔟_{v} = φ(h, L)_{v} for every v ∈ h. By Theorem 1.3, ℓy ∈ hC, and hence ℓ ∈ hCy^{−1}. This shows that if (Ly^{−1})[q, 𝔟] ≠ ∅, then V ∩hCy^{−1} ≠ ∅, and hence (2.5) holds for every y ∈ G_{A}. This completes the proof.

If k and ℓ are two elements of 𝒱 ∩hDy^{−1}, then k = ℓx_{v} with x_{v} ∈ y_{v}D_{v}y_{v}^{−1} for every v ∈ h. Therefore we are tempted to say that k and ℓ belong to the same genus. Then each orbit of (𝒱 ∩ hDy^{−1})/Γ^{y} may be called a class. Thus (ii) of Theorem 2.2 connects such classes of elements of 𝒱 with the classes of H with respect to H_{A} ∩ D.

Combining (2.4) with (2.5), we obtain, in the setting of (iv),

(2.7) #{H\H_{A}/(H_{A} ∩ C)} = ∑_{y∈Y} #{(Ly^{−1})[q, 𝔟]/Γ^{y}}.

This was stated in [S3, (11.7)] under the condition mentioned at the beginning of this section, which we can now remove.

The above theorem concerns SO^{φ}((Xh)^{⊥}). Let us now show that we can formulate a result with respect to SO^{q}(X) instead, when m = n − 1. The notation being as above, put Y = Xh and J = SO^{φ}(Y). We identify J with {α ∈ G | α = id. on W }. For every δ ∈ J there is a unique element δ′ of SO^{q}(X) such that δ′h = hδ, and δ ↦ δ′ gives an isomorphism of J onto SO^{q}(X). In this section we employ the symbol δ′ always in this sense. We note a simple fact:

(2.8) If m = n − 1 and hξ = hη for ξ, η ∈ G_{A}, then ξ = η.

Indeed, for each v ∈ v we have hξ_{v}η_{v}^{−1} = h, and so ξ_{v}η_{v}^{−1} is the identity map on X_{v}h. Since m = n − 1, it must be the identity map on the whole V_{v}. Thus ξ_{v}η_{v}^{−1} = 1 for every v ∈ v, which proves (2.8).

Theorem 2.3. Let D = D_{0}G_{a} as in Theorem 2.2 and let E = E_{0}J_{a} with an open compact subgroup E_{0} of J_{h}. Suppose m = n − 1 and J_{A} ∩ D ⊂ E. Then for every z ∈ J_{A} and y ∈ G_{A} there exists a bijection

(2.9) J\(JzE ∩ GyD)/(J_{A} ∩ D) ⟶ Δ′\(𝒱 ∩ hzEDy^{−1})/Γ,

where Γ = yDy^{−1} ∩ G and Δ′ = {δ′ | δ ∈ Δ}, Δ = zEz^{−1} ∩ J.

Proof. Given σ ∈ JzE∩GyD, take α ∈ G and β ∈ J so that σ ∈ βzE∩αyD, and put k = hβ^{−1}α. Then k ∈ 𝒱 ∩hzEDy^{−1}. If σ ∈ β_{1}zE∩α_{1}yD with α_{1} ∈ G and β_{1} ∈ J, then β_{1}^{−1}β ∈ Δ and α_{1}^{−1}α ∈ Γ, and so σ → hβ^{−1}α is a welldefined map as in (2.9). To show that it is surjective, take k ∈ 𝒱 ∩ hzEDy^{−1}; then φ[k] = q, and so k = hα with α ∈ G by (2.2). We have also k = hzεζy^{−1} with ε ∈ E and ζ ∈ D. By (2.8), zεζy^{−1} = α, and so zε = αyζ^{−1} ∈ zE ∩αyD. This shows that k is the image of zε, which proves the surjectivity. To prove the injectivity, let σ ∈ βzE ∩ αyD and σ_{1} ∈ β_{1}zE ∩ α_{1}yD with α, α_{1} ∈ G and β, β_{1} ∈ E. Suppose hβ_{1}^{−1}α_{1} = δ′hβ^{−1}αγ with δ ∈ Δ and γ ∈ Γ. Then β_{1}^{−1}α_{1} = δβ^{−1}αγ by (2.8). Put λ = αγα_{1}^{−1}. Then λ = βδ^{−1}β_{1}^{−1} ∈ J. Since γ ∈ yDy^{−1}, we have σ ∈ αyD = αγyD = λα_{1}yD = λσ_{1}D ⊂ Jσ_{1}D. Also, since σ, σ_{1} ∈ J_{A}, we have σ ∈ Jσ_{1}(J_{A} ∩ D). This proves the injectivity, and completes the proof.

Remark 2.4. (1) We can take E = J_{A} ∩ D in the above theorem. Then (2.9) takes a simpler form

(2.10) J\(JzE ∩ GyD)/E ⟶ Δ′\(𝒱 ∩ hzDy^{−1})/Γ,

(2) Let G_{A} = ⨆_{y∈Y} GyD as in Theorem 2.2 and let J_{A} = ⨆_{z∈Z} JzE with a finite subset Z of J_{A}. Then J_{A} = ⨆_{z,y}(JzE ∩ GyD). For each (z, y) such that JzE ∩ GyD ≠ ∅ pick ξ ∈ JzE ∩ GyD and denote by Ξ the set of all such ξ. Then J_{A} = ⨆_{ξ∈Ξ}(JξE ∩ GξD), and we obtain

(2.11) #{J\J_{A}/(J_{A} ∩ D)} = ∑_{ξ∈Ξ} #{Δ′_{ξ}\(𝒱 ∩ hξEDξ^{−1})/Γ^{ξ}},

where Γ^{ξ} = ξDξ^{−1} ∩ G and Δ′_{ξ} = {δ′ | δ ∈ J ∩ ξEξ^{−1}}.

(3) For ℓ ∈ V let us denote by φ(h, ℓ) the element of Hom(X, F) defined by xφ(h, ℓ) = φ(xh, ℓ) for x ∈ X; then for a subset S of V let us put φ(h, S) = {φ(h, s) | s ∈ S}. Now, fixing a maximal lattice L in V, put Λ = L̃, 𝔅 = φ(h, Λ), and E = {ε ∈ J_{A} | ε′𝔅 = 𝔅}. Since 𝔅 is a 𝔤-lattice in Hom(X, F), we see that E is a subgroup of J_{A} of the type considered in Theorem 2.3. Then we can prove

(2.12) 𝒱 ∩ hzEDy^{−1} = {k ∈ 𝒱 | φ[k] = q, φ(k, Λy^{−1}) = z′𝔅}.

This is similar to (2.5), and proved in the proof of [S3, Theorem 13.10]. It should be noted that condition (9.2) imposed in that theorem and also (8.1) in [S3 Theorem 13.8] are unnecessary for the reason explained at the beginning of this section.

(4) Suppose m = n − 1 in the setting of Theorem 2.2; then H = {1}. Therefore H_{A} ∩ GyD ≠ ∅ only when GyD = GD. Then (2.4) gives #{(𝒱 ∩ hD)/(G ∩ D)} = 1, but this is an immediate consequence of (2.2).

(5) Let us note some more statements in [S3] from which condition (9.2) of [S3] can be removed. First of all, that is the case with Theorem 9.26 of [S3]. In fact, we can state an improved version of that theorem as follows. Using the notation employed in the theorem, let ℑ′_{φ} be the subgroup of ℑ generated by ℑ_{0} and the prime ideals for which σ(C_{v}) = F_{v}^{×}. (Such prime ideals are determined by Theorem 1.9.) Then #{SO^{φ}\SO_{A}^{φ}/C) = [ℑ : ℑ′_{φ}] for C of [S3, (9.15)], which equals C of Theorem 2.2 (iv). The last group index is 1 if F = Q, and so the genus of maximal lattices consists of a single class. Consequently we can state a clear-cut result as follows. Let 𝔖_{n} denote the set of symmetric matrices of GL_{n}(Q) that represent Z-valued quadratic forms on Z^{n}, and 𝔖_{n}^{0} the subset of 𝔖_{n} consisting of the reduced elements of 𝔖_{n} in the sense of [S5], that is, the set of Φ ∈ 𝔖_{n} which cannot be represented nontrivially over Z by another element of 𝔖_{n}. For Φ ∈ 𝔖_{n} put σ(Φ) = p − q where Φ as a real matrix has p positive and q negative eigenvalues. Now the number of genera of Φ ∈ 𝔖_{n}^{0} with given σ(Φ) and det(2Φ) was determined in [S5, Theorem 6.6]. If n ≥ 3 and σ(Φ) ≠ n, then the number of genera of Φ given there equals the number of classes, as each genus of maximal lattices consists of a single class as explained above.

(6) Next, (9.2) is unnecessary in Theorem 12.1 of [S3]. What we need, in addition to Theorem 1.3, is the following fact: If v ∈ h, φ[h]𝔤_{v} = φ(h, L_{v})^{2}, then L_{v} ∩W_{v} is maximal and C ∩H_{v} = {α ∈ H_{v} | (L_{v} ∩W_{v})α = L_{v} ∩W_{v}}. Here the notation is the same as in the proof of Theorem 12.1. The statement is valid irrespective of the nature of L_{v} and t_{v}. Indeed, replacing h by an element of F^{×}h, we may assume that φ[h]𝔤_{v} = φ(h, L_{v}) = 𝔤_{v}. Then, by Lemma 10.2 (i) of [S3], we have L_{v} = 𝔤_{v}h ⊕ (L_{v} ∩ W_{v}), from which we immediately obtain the expected facts on L_{v} ∩ W_{v}. Consequently, 𝔤_{v}^{×} ⊂ σ(C ∩ H_{v}) by Theorem 1.9, as n − 1 > 2, and the original proof of Theorem 12.1 is valid without (9.2).

(7) In [S4] we proved a result of type (2.7) in terms of G^{+}(V), but imposed the condition that if n is odd, then det(φ)𝔤 is a square in the ideal group of F; see Theorem 1.6 (iv) of [S4]. We can remove this condition by virtue of Theorem 1.3.

3. New mass formulas

3.1. We first recall the symbols 𝔪(G, D) and ν(Γ) introduced in [S1] and [S2]. Here G = SO^{φ}(V) and D is an open subgroup of G_{A} containing G_{a} and such that D∩G_{h} is compact. Fixing such a D, we put Γ^{a} = G∩aDa^{−1} for every a ∈ G_{A}. Let C_{a} be a maximal compact subgroup of G_{a} and let 𝒵 = G_{a}/C_{a}.

Then 𝒵 is a symmetric space on which G acts through its projection into G_{a}. Taking a complete set of representatives ℬ for G\G_{A}/D, we put

(3.1) 𝔪(G, D) = 𝔪(D) = ∑_{a∈ℬ} ν(Γ^{a}).

Here ν(Γ) is a quantity defined by

(3.2)  ν(Γ) = \{ \begin{matrix}[Γ : 1]^{−1}  if G_{a} is compact, \\ vol(Γ \𝒵 )/#(Γ ∩ \{±1\}) otherwise. \end{matrix}

We fix a Haar measure on G_{a}, which determines a unique G_{a}-invariant measure on 𝒵 by a well known principle. (In fact, 𝔪(D) and ν(Γ) are the measure of D ∩ G_{a} and that of Γ \G_{a}; see [S6, Theorem 9].) We easily see that 𝔪(D) does not depend on the choice of ℬ . We call 𝔪(G, D) the mass of G relative to D. If D′ is a subgroup of G_{A} of the same type as D and D′ ⊂ D, then, as proven in [S1, Lemma 24.2 (1)],

(3.3) 𝔪(G, D′) = [D : D′]𝔪(G, D).

Next, in the setting of §2.1 we consider a subset S of 𝒱 of the form S = ⨆_{ζ∈Z} hζΓ, where h is as in (2.2), Z is a finite subset of G, and Γ = G ∩ D with D as above. Then we put

(3.4) 𝔪(S) = ∑_{ζ∈Z} ν(Δ_{ζ})/ν(Γ), Δ_{ζ} = H ∩ ζΓ ζ^{−1},

and call 𝔪(S) the mass of S. Here, to define ν(Δ_{ζ}), we need to fix a measure on H_{a}. Thus 𝔪(S) depends on the choice of measures on G_{a} and H_{a}, but 𝔪(S) is independent of the choice of Z and Γ. (Let 𝔜 = {x ∈ 𝒱 _{a} | φ[x] = q}. Since 𝔜 can be identified with H_{a}\G_{a}, once a measure on G_{a} is fixed, a measure on 𝔜 determines a measure on H_{a}, and vice versa. The replacement of h by an element of hG changes the group H, but that does not change 𝔪(S) if we start with a fixed measure on 𝔜. Notice also that an identification of 𝒱 _{a} with a Euclidean space determines a G_{a}-invariant measure on 𝔜.) Since the left-hand side of (2.4) is finite, we see that (𝒱 ∩ hDy^{−1})/Γ^{y} is a finite set for every y ∈ G_{A}. Thus we can define 𝔪(𝒱 ∩ hDy^{−1}) for every y ∈ G_{A}.

If G_{a} is compact, we naturally take the measures of G_{a} and H_{a} to be 1. Then 𝔪(S) can be defined in a unique way. We easily see that S = ⨆_{ζ∈Z} ⨆_{γ∈Δ′_{ζ}\Γ} hζγ, where Δ′_{ζ} = ζ^{−1}Δ_{ζ}ζ, and so #S = ∑_{ζ∈Z}[Γ : Δ′_{ζ}] = 𝔪(S) if G_{a} is compact. Thus we obtain

(3.5) 𝔪(S) = #(S) if G_{a} is compact.

Theorem 3.2. The notation being the same as in Theorem 2.2, we have

(3.6) 𝔪(H, H_{A} ∩ D) = ∑_{y∈Y} ν(Γ^{y})𝔪(𝒱 ∩ hDy^{−1}).

In particular if m = 1 and the notation is as in (iv) of Theorem 2.2, then

(3.7) 𝔪(H, H_{A} ∩ C) = ∑_{y∈Y} ν(Γ^{y})𝔪((Ly^{−1})[q, 𝔟]).

Proof. Let E_{y} = H\(H_{A} ∩ GyD)/(H_{A} ∩ D). For each ε ∈ E_{y} take ζ_{ε} ∈ G so that ε ∈ ζ_{ε}yD. By Theorem 2.2 (ii) we have 𝒱 ∩ hDy^{−1} = ⨆_{ε∈E_{y}} hζ_{ε}Γ^{y} and H ∩ ε(H_{A} ∩ D)ε^{−1} = H ∩ H_{A} ∩ εDε^{−1} = H ∩ ζ_{ε}yDy^{−1}ζ_{ε}^{−1} = H ∩ ζ_{ε}Γ^{y}ζ_{ε}^{−1}, so that

ν(Γ^{y})𝔪(𝒱 ∩ hDy^{−1}) = ∑_{ε∈E_{y}} ν(H ∩ ζ_{ε}Γ^{y}ζ_{ε}^{−1}) = ∑_{ε∈E_{y}} ν(H ∩ ε(H_{A} ∩ D)ε^{−1}).

Since ⨆_{y∈Y} E_{y} gives H\H_{A}/(H_{A} ∩ D), we obtain

𝔪(H, H_{A} ∩ D) = ∑_{y∈Y} ∑_{ε∈E_{y}} ν(H ∩ ε(H_{A} ∩ D)ε^{−1}) = ∑_{y∈Y} ν(Γ^{y})𝔪(𝒱 ∩ hDy^{−1}).

This proves (3.6), which combined with (2.5) gives (3.7).

Formula (3.7) was given in [S3, (13.17b)] under the condition mentioned at the beginning of Section 2. That condition can be removed by the above theorem.

It should be noted that (3.6) and (3.7) are different from any of the mass formulas of Siegel. Indeed, the left-hand side of (3.6) concerns an orthogonal group in dimension n − m, and the right-hand side concerns the space 𝒱 , whereas both sides of Siegel’s formulas are defined with respect to matrices of the same size. See also [S3, p. 137] for a comment on the connection with the work of Eisenstein and Minkowski on the sums of five squares.

4. Classification and genera of quadratic forms in terms of matrices

4.1. Traditionally the genus and class of a quadratic form over Q were defined in terms of matrices, but it is easy to see that they are equivalent to those defined in terms of lattices. In the general case, however, there is a standard definition in terms of lattices, but the definition in terms of matrices is nontrivial. Also, we treated in [S5] the classification of quadratic forms over a number field, but gave explicit results in terms of matrices only when Q is the base field. Let us now discuss how we can handle such problems over an arbitrary number field, as the generalization is far from obvious and requires some new concepts. Before proceeding, we recall two basic facts: Given (V, φ) over a global F and a lattice L in V, the O^{φ}(V)-genus of L is the same as the SO^{φ}(V)-genus of L; all the maximal lattices in V form a single O^{φ}(V)-genus; see [S3, Lemmas 6.8 and 6.9].

We take a global field F, and denote by F_{n}^{1} the vector space of all ndimensional row vectors with components in F, and by 𝔤_{n}^{1} the set of elements of F_{n}^{1} with components in 𝔤. Define subgroups E and E_{ξ} of GL_{n}(F)_{A} by

(4.1) E = GL_{n}(F)_{a} ∏_{v∈h} GL_{n}(𝔤_{v}), E_{ξ} = ξ^{−1}Eξ (ξ ∈ GL_{n}(F)_{A}).

Put L_{0} = 𝔤_{n}^{1}. An arbitrary 𝔤-lattice L in F_{n}^{1} can be given as L = L_{0}ξ with ξ ∈ GL_{n}(F)_{A}; then E_{ξ} = {y ∈ GL_{n}(F)_{A} | Ly = L}. It is well known that the map x ↦ det(x)𝔤 gives a bijection of E\GL_{n}(F)_{A}/GL_{n}(F) onto the ideal class group of F; see [S1, Lemma 8.14], for example. Thus the ideal class of det(ξ)𝔤 is determined by the GL_{n}(F)-class of L_{0}ξ, and vice versa. Consequently, L is isomorphic as a 𝔤-module to the direct sum of 𝔤_{n−1}^{1} and det(ξ)𝔤.

To define the generalization of Z-valued quadratic forms, we denote by S_{n} the set of all symmetric elements of GL_{n}(F), fix an element ξ of GL_{n}(F)_{A}, and denote by S_{n}(ξ) the set of all T ∈ S_{n} such that xT · ^{t}x ∈ 𝔤 for every x ∈ L_{0}ξ. We call such a T reduced (relative to ξ) if the following condition is satisfied:

(4.2) T ∈ S_{n}(ζ^{−1}ξ) with ζ ∈ GL_{n}(F)_{h} ∩ ∏_{v∈h} M_{n}(𝔤_{v}) ⟹ ζ ∈ E.

We denote by S_{n}^{0}(ξ) the set of all reduced elements of S_{n}(ξ). These depend essentially on EξGL_{n}(F), as will be seen in Proposition 4.3 (i) below.

4.2. To define the genus and class of an element of S_{n}, put

(4.3)  Δ_{ξ} = E_{ξ} ∩ GL_{n}(F),  Δ_{ξ}^{1} = E_{ξ} ∩ SL_{n}(F).

We say that two elements Φ and Ψ of S_{n}(ξ) belong to the same genus (relative to ξ) if there exists an element ε of E_{ξ} such that εΦ ·^{t}ε = Ψ; they are said to belong to the same O-class (resp. SO-class) if αΦ ·^{t}α = Ψ for some α ∈ Δ_{ξ} (resp. α ∈ Δ_{ξ}^{1}). These depend on the choice of L = L_{0}ξ, or rather, on the choice of the ideal class of det(ξ)𝔤. There is no reason to think that ξ = 1 is the most natural choice.

Given Φ ∈ S_{n}, put V = F_{n}^{1} and φ[x] = xΦ ·^{t}x for x ∈ V. Then we obtain a quadratic space (V, φ), which we denote by [Φ]; we put then O(Φ) = O^{φ}(V) and SO(Φ) = SO^{φ}(V). Now put L = L_{0}ξ with ξ ∈ GL_{n}(F)_{A}. Clearly L is integral if Φ ∈ S_{n}(ξ), in which case L is maximal if and only if Φ ∈ S_{n}^{0}(ξ).

Proposition 4.3. (i) If η ∈ Eξα with ξ, η ∈ GL_{n}(F)_{A} and α ∈ GL_{n}(F), then S_{n}(ξ) = αS_{n}(η) ·^{t}α and S_{n}^{0}(ξ) = αS_{n}^{0}(η) ·^{t}α.

(ii) For Φ, Ψ ∈ S_{n}^{0}(ξ), ξ ∈ GL_{n}(F)_{A}, the spaces [Φ] and [Ψ] are isomorphic if and only if they belong to the same genus.

(iii) Let X be a complete set of representatives for E\GL_{n}(F)_{A}/GL_{n}(F) and for each ξ ∈ GL_{n}(F)_{A} let Y_{ξ} be a complete set of representatives for the genera of the elements of S_{n}^{0}(ξ). Then the spaces [Φ] obtained from Φ ∈ Y_{ξ} for all ξ ∈ X exhaust all isomorphism classes of n-dimensional quadratic spaces over F without overlapping.

Proof. Assertion (i) can be verified in a straightforward way. Let Φ and Ψ be elements of S_{n}^{0}(ξ) belonging to the same genus. Then there exists an element ε ∈ E_{ξ} such that εΦ ·^{t}ε = Ψ, and the Hasse principle guarantees an element α of GL_{n}(F) such that Ψ = αΦ ·^{t}α. Thus [Ψ] is isomorphic to [Φ]. Conversely, suppose [Φ] and [Ψ] are isomorphic for Φ, Ψ ∈ S_{n}^{0}(ξ). Then Φ = βΨ ·^{t}β for some β ∈ GL_{n}(F). Now L_{0}ξ is maximal in both [Φ] and [Ψ], and L_{0}ξβ is maximal in [Ψ]. Thus L_{0}ξβ = L_{0}ξγ with γ ∈ O(Ψ)_{A}. Put ζ = βγ^{−1}. Then ζ ∈ E_{ξ}, and ζΨ · ^{t}ζ = Φ. Therefore Ψ belongs to the genus of Φ. This proves (ii). As for (iii), clearly every n-dimensional quadratic space is isomorphic to [Ψ] for some Ψ ∈ S_{n}. Putting V = F_{n}^{1}, pick a maximal lattice L in V and put L = L_{0}η with η ∈ GL_{n}(F)_{A}. Then η ∈ EξGL_{n}(F) for some ξ ∈ X, and by (i) we can replace η by ξ. Take a member Φ of Y_{ξ} belonging to the genus of Ψ. Then by (ii), [Ψ] is isomorphic to [Φ]. Thus every n-dimensional quadratic space is isomorphic to [Φ] for some Φ ∈ Y_{ξ} with ξ ∈ X. To prove that there is no overlapping, take Φ_{i} ∈ Y_{ξ_{i}} with ξ_{i} ∈ X, i = 1, 2, and suppose that [Φ_{1}] is isomorphic to [Φ_{2}]. Then Φ_{1} = αΦ_{2} ·^{t}α with α ∈ GL_{n}(F). Now L_{0}ξ_{1} is maximal in [Φ_{1}], and so L_{0}ξ_{1}α is maximal in [Φ_{2}]. Therefore L_{0}ξ_{1}α = L_{0}ξ_{2}γ with γ ∈ O(Φ_{2})_{A}. Then ξ_{1}αγ^{−1}ξ_{2}^{−1} ∈ E, and so det(ξ_{1}ξ_{2}^{−1}) ∈ F^{×} det(E), which implies that ξ_{1} = ξ_{2}, as ξ_{i} ∈ X. Thus both Φ_{1} and Φ_{2} belong to Y_{ξ_{1}}. By (ii) they must belong to the same genus. This completes the proof.

4.4. Take Φ ∈ S_{n}^{0}(ξ) and suppose another member Ψ ofS_{n}^{0}(ξ) belongs to the genus of Φ. Put L = L_{0}ξ. Then Ψ = εΦ · ^{t}ε = αΦ · ^{t}α with ε ∈ E_{ξ} and α ∈ GL_{n}(F), as observed in the above proof. Clearly ε^{−1}α ∈ O(Φ)_{A}. Since Lα = Lε^{−1}α, we see that Lα belongs to the genus of L. We associate the O(Φ)-class of Lα to the O-class of Ψ. We easily see that the class of Lα is determined by the class of Ψ. Moreover, it gives a bijection of the set of O-classes in the genus of Φ onto the set of O(Φ)-classes in the genus of L_{0}ξ. Indeed, let M = Lσ with σ ∈ SO(Φ)_{A}. Then det(σ)𝔤 = 𝔤, and so σ = τβ with τ ∈ E_{ξ} and β ∈ GL_{n}(F). Put Ψ = τ^{−1}Φ ·^{t}τ^{−1}. Then Ψ = βΦ ·^{t}β and M = Lβ, which corresponds to Ψ. This proves the surjectivity. The injectivity can be easily verified too.

For Φ ∈ S_{n}^{0}(ξ) we define the SO-genus of Φ to be the set of all Ψ in the genus of Φ such that det(Ψ) = det(Φ). For Ψ = εΦ · ^{t}ε = αΦ · ^{t}α as above, suppose det(Ψ) = det(Φ); then det(α)^{2} = 1. Since −1 ∈ det (O(Φ)), changing α for αγ with a suitable γ ∈ O(Φ), we may assume that det(α) = 1. We then associate the SO(Φ)-class of Lα to Ψ We can easily verify that the set of all SO-classes in the SO-genus of Φ contained in S_{n}^{0}(ξ) are in one-to-one correspondence with the set of SO(Φ)-classes in the genus of L.

In general, if Ψ = εΦ·^{t}ε = αΦ·^{t}α as above, then det(α)^{2} = det(ε)^{2} ∈ det(E_{ξ}), and so det(α) ∈ 𝔤^{×}. Therefore, if F = Q, then det(Ψ) = det(Φ), and the SOgenus of Φ coincides with the genus of Φ.

4.5. Define (V, φ) by V = F_{n}^{1} and φ[x] = xΦ ·^{t}x as above with any Φ ∈ S_{n}. Put L = L_{0}ξ with ξ ∈ GL_{n}(F)_{A}. We easily see that L̃ = L_{0}(2Φ ·^{t}ξ)^{−1}, and so

(4.4) [L̃/L] = det(2Φξ^{2})𝔤 if L = L_{0}ξ.

In order to state our main results, we need a basic fact on an algebraic extension K of F. Let 𝔯 be the maximal order of K, and 𝔡 the different of K relative to F; let 𝔡_{0} = N_{K/F} (𝔡), m = [K : F], and d(K/F) = det [(Tr_{K/F} (e_{i}e_{j}))_{i,j=1}^{m}] with an F -basis {e_{i}}_{i=1}^{m} of K. Strictly speaking, d(K/F) should be viewed as a coset in F^{×}/F^{×2} represented by that determinant, but we denote any number in that coset by d(K/F). By the characteristic ideal class for the extension K/F we understand the ideal class 𝔨 in F determined by the property that 𝔯 is isomorphic as a 𝔤-module to 𝔤_{m−1}^{1} ⊕ 𝔵 with an ideal 𝔵 belonging to 𝔨. Then we have (4.5) The characteristic ideal class for K/F contains an ideal 𝔵 such that 𝔡_{0} =

d(K/F)𝔵^{2}.

To prove this, take V = K, φ(x, y) = Tr_{K/F} (xy) for x, y ∈ K, and L = 𝔯 in (4.4). Then L̃ = (2𝔡)^{−1}, and (4.4) shows that 𝔡_{0} = d(K/F)𝔵^{2} with 𝔵 = det(ξ)𝔤, which proves (4.5). This fact was noted, in substance, by Artin in 1949; that 𝔡 defines a square ideal class in K is due to Hecke. We can easily generalize (4.5) to the case of a maximal order 𝔬 in a simple algebra over F. The ideal class which is an obvious analogue of 𝔨 is independent of the choice of 𝔬.

We also need a few more symbols. We denote by r the set of all real primes in a. For T ∈ S_{n} and v ∈ r we put s_{v}(T) = p_{v}−q_{v} if T as a real symmetric matrix in GL_{n}(F_{v}) = GL_{v}(R) has p_{v} positive and q_{v} negative eigenvalues. Given a quadratic space (V, φ), we denote by Q(φ) the characteristic quaternion algebra of (V, φ) in the sense of [S5, §3.1]; if φ is obtained from Φ ∈ S_{n} as above, we put δ(Φ) = δ(φ) and Q(Φ) = Q(φ). In [S5, Theorem 4.4] we showed that the isomorphism class of (V, φ) is determined by {n, δ(φ), Q(φ), {s_{v}(φ)}_{v∈r}}. We will now state this fact in terms of the matrices Φ in S_{n}^{0}(ξ).

Theorem 4.6 (The case of even n). Let the symbols n, {σ_{v}}_{v∈r}, δ, K_{0}, 𝔢_{0}, 𝔢_{1}, and ξ be given as follows: 4 ≤ n ∈ 2Z, σ_{v} ∈ 2Z, |σ_{v}| ≤ n; δ ∈ F^{×}, K_{0} = F (\sqrt{δ}); 𝔢_{0} and 𝔢_{1} are squarefree integral ideals in F; ξ ∈ GL_{n}(F)_{A}. Let r be the number of prime factors of 𝔢_{0}𝔢_{1}, and 𝔡 the different of K_{0} relative to F; put 𝔡_{0} = 𝔡^{2} ∩F. Suppose that 𝔢_{0} divides 𝔡_{0}, 𝔢_{1} is prime to 𝔡_{0}, (−1)^{σ_{v}/2}δ > 0 at each v ∈ r, det(ξ)𝔢_{1}^{−1} belongs to the characteristic ideal class for K_{0}/F, and

(4.6) r + #{v ∈ r | σ_{v} ≡ 4 or 6 (mod 8)} ∈ 2Z.

Then there exists an element Φ of S_{n}^{0}(ξ) such that

(4.7) δ ∈ δ(Φ), det(2Φξ^{2})𝔤 = 𝔡_{0}𝔢_{1}^{2}, s_{v}(Φ) = σ_{v} for every v ∈ r,

and Q(Φ) is ramified at v ∈ h if and only if v|𝔢_{0}𝔢_{1}. Moreover, every element of S_{n}^{0}(ξ) is of this type, and its genus is determined by ({σ_{v}}_{v∈r}, δ, 𝔢_{0}, 𝔢_{1}). These statements are true even for n = 2 under the following additional condition on 𝔢_{1} : if 𝔢_{1} ≠ 𝔤, then K_{0} ≠ F and every prime factor of 𝔢_{1} remains prime in K_{0}.

Proof. Given the symbols as in our theorem, let B be the quaternion algebra over F ramified exactly at the prime factos of 𝔢_{0}𝔢_{1} and at those v ∈ r for which σ_{v} ≡ 4 or 6 (mod 8). Such a B exists because of (4.6). By the main theorem of classification [S5, Theorem 4.4] there exists a quadratic space (V, φ) over F such that B = Q(φ), δ ∈ δ(φ), and σ_{v} = s_{v}(φ) for every v ∈ r. By Proposition 4.3 (ii) we can take (V, φ) = [Ψ] with Ψ ∈ S_{n}^{0}(η), η ∈ GL_{n}(F)_{A}. Put L = L_{0}η. By [S5, Theorem 6.2 (ii)] we have [L̃/L] = 𝔡_{0}𝔢_{1}^{2}, which combined with (4.4) shows that det(2Ψη^{2})𝔤 = 𝔡_{0}𝔢_{1}^{2}. By our assumption on ξ and (4.5) we have 𝔡_{0}𝔢_{1}^{2} = c^{2}δ det(ξ)^{2}𝔤 with c ∈ F^{×}. Thus det(2Ψη^{2})𝔤 = c^{2}δ det(ξ)^{2}𝔤. Since (−1)^{n/2}δ = b^{2} det(2Ψ) with b ∈ F^{×}, we see that det(η)𝔤 = bc det(ξ)𝔤, which implies that η ∈ Eξα with α ∈ GL_{n}(F). Then by Proposition 4.3 (i) we can replace η by the given ξ, and we obtain the first part of our theorem.

Next, given Φ ∈ S_{n}^{0}(ξ), put L = L_{0}ξ. Let 𝔢 be the product of all the prime ideals in F ramified in Q(Φ) and let K_{0} = F (\sqrt{δ}) with δ ∈ δ(Φ). By [S5, (4.2b)], Q(Φ) is ramified at v ∈ r if and only if s_{v}(Φ) ≡ 4 or 6 (mod 8). Put 𝔢 = 𝔢_{0}𝔢_{1}, where 𝔢_{0} is the product of the prime factors of 𝔢 ramified in K_{0}. Then in [S5, Theorem 6.2 (ii)] we showed that [L̃/L] = 𝔡_{0}𝔢_{1}^{2}, where 𝔡_{0} is defined for this K_{0} as in our theorem. Combining this with (4.4) we obtain det(2Φξ^{2})𝔤 = 𝔡_{0}𝔢_{1}^{2}. We have 𝔡_{0} = δ𝔵^{2} with an ideal 𝔵 as in (4.5), and so det(ξ)𝔤 = 𝔵𝔢_{1}, if we take δ to be (−1)^{n/2} det(2Φ). This proves the second part of our theorem. The last part concerning the case n = 2 follows from [S5, Theorem 4.4, (4.4b)].

Theorem 4.7 (The case of odd n). Let the symbols n, {σ_{v}}_{v∈r}, δ, K_{0}, 𝔢, and ξ be given as follows: 0 < n − 1 ∈ 2Z, σ_{v} − 1 ∈ 2Z, |σ_{v}| ≤ n; δ ∈ F^{×}, K_{0} = F (\sqrt{δ}); 𝔢 is a squarefree integral ideal in F; ξ ∈ GL_{n}(F)_{A}. Let r be the number of prime factors of 𝔢; let δ𝔤 = 𝔞𝔟^{2} with a squarefree integral ideal 𝔞 and a fractional ideal 𝔟 in F; put 𝔢 = 𝔢_{0}𝔢_{1} with 𝔢_{1} = 𝔞 + 𝔢. Suppose (−1)^{(σ_{v}−1)/2}δ > 0 at each v ∈ r, det(ξ)𝔟 belongs to the ideal class of 𝔢_{0}, and

(4.8) r + #{v ∈ r | σ_{v} ≡ ±3 (mod 8)} ∈ 2Z.

Then there exists an element Φ of S_{n}^{0}(ξ) such that

(4.9) δ ∈ δ(Φ), det(2Φξ^{2})𝔤 = 2𝔞𝔢_{0}^{2}, s_{v}(Φ) = σ_{v} for every v ∈ r,

and Q(Φ) is ramified at v ∈ h if and only if v|𝔢. Moreover, every element of S_{n}^{0}(ξ) is of this type, and its genus is determined by ({σ_{v}}_{v∈r}, δ, 𝔢).

Proof. In [S5, Theorem 6.2 (iii)] we showed that [L̃/L] = 2𝔞^{−1}𝔢^{2} ∩ 2𝔞 for a maximal lattice L if δ(φ)𝔤 = 𝔞𝔟^{2} as above. We easily see that 2𝔞^{−1}𝔢^{2} ∩ 2𝔞 = 2𝔞𝔢_{0}^{2}. Therefore the proof can be given in exactly the same fashion as for Theorem 4.6.

Remark. (1) Let 𝔵 be the ideal belonging to the characteristic ideal class for K_{0}/F such that 𝔡_{0} = δ𝔵^{2}. Suppose Φ has the last two properties of (4.7) and det(ξ)𝔢_{1}^{−1} belongs to the class of 𝔵. Then b^{2} det(2Φ)𝔤 = δ𝔤 with b ∈ F^{×}, and so (−1)^{n/2} det(2Φ) = b^{−2}δc with c ∈ 𝔤^{×}. Since s_{v}(Φ) = σ_{v} for every v ∈ r, we see that c > 0 at every v ∈ r. Suppose an element of 𝔤^{×} positive at every v ∈ r is always a square. Then we obtain δ ∈ δ(Φ). Thus the first property of Φ in (4.7) follows from the assumption on det(ξ)𝔤 and the last two properties of (4.7) under that condition on 𝔤^{×}. The same comment applies to (4.9).

(2) The last two statements of (1) apply to the case F = Q. Therefore the above two theorems for F = Q are the same as [S5, Theorems 6.4 and 6.5]. The new features in the general case are that we need ξ ∈ GL_{n}(F)_{A} and we have to impose a condition on the ideal class of det(ξ)𝔤. In other words, the coset EξGL_{n}(F) is determined by {σ_{v}}_{v∈r}, δ, and the characteristic quaternion algebra.

(3) A 𝔤-valued symmetric form is represented by an element of S_{n} with entries in 𝔤. This is different from the notion of a 𝔤-valued quadratic form, and so the classification of 𝔤-valued symmetric forms is different from that of 𝔤-valued quadratic forms. We refer the reader to [S7] for the classification of 𝔤valued symmetric forms and its connection with the classification of quadratic forms.

5. Ternary forms

5.1. Before proceeding further, let us recall several basic facts on quaternion algebras. Let B be a quaternion algebra over a global field F, and 𝔒 an order in B containing 𝔤. For ξ ∈ B_{A}^{×} the principle of §1.2 about the lattice Lx enables us to define 𝔒ξ as a 𝔤-lattice in B. We call a 𝔤-lattice in B of this form a proper left 𝔒-ideal. Thus the B^{×}-genus of 𝔒 consists of all proper left 𝔒-ideals, and the genus is stable under right multiplication by the elements of B^{×}, and so a B^{×}-class of proper left 𝔒-ideals is meaningful. Define a subgroup U of B_{A}^{×} by U = B_{a}^{×} ∏_{v∈h} 𝔒_{v}^{×}. Then #(U\B_{A}^{×}/B^{×}) gives the number of B^{×}- classes in this genus, which we call the class number of 𝔒. Next, for x ∈ B_{A}^{×} we denote by x𝔒x^{−1} the order 𝔒′ in B such that 𝔒′_{v} = x_{v}𝔒_{v}x_{v}^{−1} for every v ∈ h. By the type number of 𝔒 we mean #S for a minimal set S of such orders 𝔒′ with the property that every order of type x𝔒x^{−1} can be transformed to a member of S by an inner automorphism of B.

Given an order 𝔒 in B over a local or global F, put 𝔒̃ = {x ∈ B | Tr_{B/F} (x𝔒) ⊂ 𝔤}. It can be shown that [𝔒̃/𝔒] is a square of an integral ideal 𝔱. We call 𝔱 the discriminant of 𝔒. In the local case, if B is a division algebra, then 𝔒 is maximal if and only if the disriminant is the prime ideal of 𝔤. In the global case, if 𝔒 is maximal, then the discriminant of 𝔒 is the product of all the prime ideals where B is ramified. Thus we call it the discriminant of B. If F = Q, then writing the discriminant of 𝔒 or of B in the form tZ with a positive integer t, we call t the discriminant of 𝔒 or of B. Let us quote here a result due to Eichler [E2, Satz 3]:

(5.1) If F is local, B is isomorphic to M_{2}(F), and the discriminant of an order 𝔒 in B is the prime ideal 𝔭 of 𝔤, then there is an isomorphism of B onto M_{2}(F) that maps 𝔒 onto

(5.1a) \{ [ \begin{matrix}a & b \\ c & d \end{matrix} ] ∈ M_{2}(𝔤) | c ∈ 𝔭\}.

If 𝔒 is the order of (5.1a), then we can easily verify that

(5.1b) 𝔒̃ = 𝔒η^{−1} = \{ [ \begin{matrix}a & b \\ c & d \end{matrix} ] | a, c, d ∈ 𝔤, b ∈ 𝔭^{−1}\}, η = [ \begin{matrix}0 & −1 \\ π & 0 \end{matrix} ] ,

where π is a prime element of F.

Let 𝔒 be an order in B in the global case, and let 𝔢 be the discriminant of B; suppose the discriminant of 𝔒 is squarefree. Then the discriminant of 𝔒 is of the form 𝔞_{0}𝔢 with a squarefree integral ideal 𝔞_{0} prime to 𝔢. For each v|𝔞_{0} the order 𝔒_{v} can be transformed to an order of type (5.1a). We will be considering such an order 𝔒 in the following treatment.

5.2. Let us now consider (V, φ) with n = 3 over a local or global F. As shown in [S3, §7.3], we can find a quaternion algebra B over F with which we can put V = B^{∘}ξ, A(V) = B + Bξ, A^{+}(V) = B, φ[xξ] = dxx^{ι} and 2φ(xξ, yξ) = dTr_{B/F} (xy^{ι}) for x, y ∈ B^{∘}, where B^{∘} is as in (1.2) and ξ is an element such that ξ^{2} = −d ∈ δ(φ); F + F ξ is the center of A(V); G^{+}(V) = B^{×}; ν(α) = N_{B/F} (α) and xξτ(α) = α^{−1}xαξ for x ∈ B^{∘} and α ∈ B^{×}.

Given h ∈ V such that φ[h] ≠ 0, take k ∈ B^{∘} so that h = kξ. Put W = (F h)^{⊥} and K = F + F k. It can easily be seen that K = {α ∈ B | αk = kα}, and K is either a quadratic extension of F, or isomorphic to F × F. In either case we can find an element ω of B such that

(5.2) B = K + ωK, ω^{2} ∈ F^{×}, ωk = −kω.

Then B^{∘} = F k + ωK and Tr_{B/F} (kωK) = 0, and so W = ωKξ = F ωξ + F ωkξ, and φ[ωxξ] = −dω^{2}xx^{ρ} for x ∈ K, where ρ is the nontrivial automorphism of K over F. Since −dk^{2} = φ[h], we have K = F × F if −dφ[h] is a square in F; otherwise K = F (κ^{1/2}) with κ = −dφ[h]. We easily see that

(5.3)  A^{+}(W) = K,  G^{+}(W) = K^{×}.

The group SO^{φ}(W) can be identified with {b ∈ K^{×} | bb^{ρ} = 1}, and the map τ : K^{×} → SO^{φ}(W) is given by τ(a) = a/a^{ρ} for a ∈ K^{×}. To be precise, ωxξτ(a) = ω(a/a^{ρ})xξ for x ∈ K.

Clearly the isomorphism class of (V, φ) is determined by B and dF^{×2}, and vice versa. If F is a totally real number field and φ is positive definite at every v ∈ a, then B is totally definite and d is totally positive.

Assuming F to be global, put d𝔤 = 𝔞𝔵^{2} with a squarefree integral ideal 𝔞 and a fractional ideal 𝔵; let 𝔢 be the product of all the prime ideals of F ramified in B. In general the pair (𝔢, 𝔞) does not necessarily determine (V, φ), but it does if F = Q and φ is positive definite.

Lemma 5.3. Suppose F is global; let (V, φ), B, 𝔞, and 𝔢 be as above. Put 𝔢_{1} = 𝔞 + 𝔢, 𝔢 = 𝔢_{0}𝔢_{1}, and 𝔞 = 𝔞_{0}𝔢_{1}. Let L be an integral lattice in V. Then the following assertions hold:

(i) L is maximal if and only if [L̃/L] = 2𝔞𝔢_{0}^{2}.

(ii) If L is maximal, then there is a unique order of B ( = A^{+}(V)) of discriminant 𝔞_{0}𝔢 containing A^{+}(L).

Proof. Let Λ be a maximal lattice containing L. Since [L̃/L] = [Λ̃/Λ][Λ/L]^{2}, the “if”-part of (i) follows from the “only if”-part. Clearly the problem can be reduced to the local case. In the local case, we may assume that d𝔤 equals 𝔤 or the prime ideal 𝔭 of 𝔤. Let t and M be as in §1.4. Then [L̃/L] = [M̃/M]. If t = 1, we can easily find [M̃/M]. Suppose t = 3; if d𝔤 = 𝔭, then the explicit forms of M and M̃ given in §1.5, (B) show that [M̃/M] = 2𝔭. If d𝔤 = 𝔤. we have [M̃/M] = 2𝔭^{2} as shown in [S3, (7.9)]. This proves (i).

Next, since all the maximal lattices form a genus, it is sufficient to prove (ii) for a special choice of L. We take an order 𝔒 in B of discriminant 𝔞𝔢_{0}, and take the 𝔤-lattice 𝔐 in B such that 𝔐_{v} = 𝔒̃_{v} if v|𝔞 and 𝔐_{v} = 𝔒_{v} if v ∤𝔞. We then put

(5.4) L = 𝔵^{−1}(𝔐 ∩ B^{∘})ξ,

Our task is to show that L is maximal and to prove (ii) for this L. The problems can be reduced to the local case. For simplicity we fix v ∈ h and denote F_{v} and 𝔤_{v} by F and 𝔤, suppressing the subscript v. We can take 𝔵 = 𝔤, and d𝔤 to be 𝔤 or 𝔭. If B = M_{2}(F) and d𝔤 = 𝔤, then 𝔒 = 𝔒̃ = M_{2}(𝔤). If B = M_{2}(F) and d𝔤 = 𝔭, then 𝔒 and 𝔒̃ can be given by (5.1a, b). Thus, for either type of d𝔤 we have

(5.5a) L = \{ [ \begin{matrix}a & b \\ c & −a \end{matrix} ] | a, c ∈ 𝔤, b ∈ d^{−1}𝔤\} · ξ,

(5.5b)  L̃ = \{ [ \begin{matrix}a & b \\ c & −a \end{matrix} ] | a ∈ (2d)^{−1}𝔤, c ∈ 𝔤, b ∈ d^{−1}𝔤\} · ξ,

so that [L̃/L] = 2d𝔤, which together with (i) shows that L is maximal. If v|𝔢, then there is a unique maximal lattice in V given in the form {x ∈ V | φ[x] ∈ 𝔤}. It can easily be seen that this coincides with (5.4). Now, in the local case we easily see that A^{+}(L) = 𝔒 except when B is a division algebra and d𝔤 = 𝔤, in which case B has a unique order of prime discriminant. This completes the proof.

Lemma 5.4. In the setting of Lemma 5.3 let L be a maximal lattice in V and let 𝔒 be the order in B of discriminat 𝔞_{0}𝔢 established in Lemma 5.3 (ii). Put

(5.6) C = {x ∈ SO^{φ}(V)_{A} | Lx = L},

(5.7) T = {y ∈ B_{A}^{×} | y𝔒 = 𝔒y}, T_{v} = B_{v}^{×} ∩ T.

Then C = τ(T); moreover, for v ∈ h we have T_{v} = B_{v}^{×} if v|𝔢, T_{v} = F_{v}^{×}𝔒_{v}^{×} if v ∤𝔞_{0}𝔢, and T_{v} = F_{v}^{×}(𝔒_{v}^{×}∪𝔒_{v}^{×}η_{v}) if v|𝔞_{0}, where η_{v} is η of (5.1b). Furthermore,

(5.8) {γ ∈ SO^{φ}(V) | Lγ = L} = τ({α ∈ B^{×} | α𝔒 = 𝔒α}).

Proof. The statements about T_{v} are well known if 𝔒_{v} is maximal, that is, if v ∤𝔞_{0}. If v|𝔞_{0}, then 𝔒_{v} is of type (5.1a), and the desired fact follows from [E2, Satz 5]. To prove τ(T) = C, we first recall the equality SO^{φ}(V) = τ(G^{+}(V)), which holds over any field. Thus it is sufficient to show that τ(T_{v}) = C_{v} for every v ∈ h, where C_{v} = C ∩SO^{φ}(V)_{v}. This is trivial if v|𝔢, as C_{v} = SO^{φ}(V)_{v} and T_{v} = B_{v}^{×} then. For v ∈ h put D_{v} = {α ∈ B_{v}^{×} | α^{−1}L_{v}α = L_{v}}. Then τ(D_{v}) = C_{v}. We have seen in the proof of Lemma 5.3 that A^{+}(L_{v}) = 𝔒_{v} if v ∤ 𝔢. Therefore if α ∈ D_{v} for such a v, then α^{−1}𝔒_{v}α = 𝔒_{v}, and so α ∈ T_{v}. Conversely, if α ∈ T_{v}, then α𝔒_{v} = 𝔒_{v}α and we easily see that α𝔒̃_{v} = 𝔒̃_{v}α.

We may assume that L is of the form (5.4). Since 𝔐_{v} is 𝔒_{v} or 𝔒̃_{v}, we have α^{−1}L_{v}α = L_{v}, and so α ∈ D_{v}. Thus T_{v} = D_{v}. Since τ(D_{v}) = C_{v}, we obtain τ(T) = C. Since Ker(τ) = F_{A}^{×} ⊂ T, if τ(α) ∈ C with α ∈ B^{×}, then α ∈ T ∩B^{×}, from which we obtain (5.8).

Proof of Theorem 1.8. Since τ(G^{+}(V)) = SO^{φ}(V), the equality τ(T) = C stated in (iii) is clear from the definition of T. To prove the remaining statements, we first consider the case n = 3. Then C of §1.7 and Theorem 1.8 is C_{v} of Lemma 5.4 with a divisor v of 𝔞, and A^{+}(L) = 𝔒_{v} as noted at the end of the proof of Lemma 5.3. Moreover, by Lemma 5.4, we have F_{v}^{×} ⊂ T_{v} and τ(T_{v}) = C_{v}, and so T_{v} coincides with T of our theorem. Thus T = B_{v}^{×} if v|𝔢 and T = F_{v}^{×}(𝔒_{v}^{×} ∪ 𝔒_{v}^{×}η_{v}) if v|𝔞_{0}. The last expression for T is valid even for v|𝔢, if we take η_{v} to be an element of B_{v} such that η_{v}^{2} is a prime element of F_{v}. Clearly J = 𝔒_{v}^{×} and ν(J) = 𝔤_{v}^{×}; also, all the other statements of our theorem can easily be verified.

Next suppose n > 3; using the symbols of (1.4a, b, c), put W = Z and N = M if t = 3; put W = Z + F e_{r} + F f_{r} and N = M + 𝔤e_{r} + 𝔤f_{r} if t = 1. Then V = W +∑_{i=1}^{u}(F e_{i} +F f_{i}) and L = N +∑_{i=1}^{u}(𝔤e_{i} +𝔤f_{i}), where u = r if t = 3 and u = r − 1 if t = 1. Observe that N contains an element g such that φ[g] ∈ 𝔤^{×}. Therefore, as shown in [S3, §8.2], there exists an isomorphism θ of A^{+}(V) onto M_{2}(A^{+}(W)) such that θ(A^{+}(L)) = M_{s}(A^{+}(N)), where s = 2^{u}. Thus (i) and (ii) of our theorem follow from what we proved in the case n = 3; namely, A^{+}(L) can be identified with M_{s}(𝔒) with an order 𝔒 as stated. As for (iii) in the general case, clearly ν(J) = 𝔤^{×}. We already found an element η of G^{+}(W) such that Nτ(η) = N, η^{2} ∈ F^{×}, and ν(η)𝔤 = 𝔭; also ηJ = Jη as can easily be seen. Put T_{0} = F^{×}(J ∪ Jη). Then T_{0} is a subgroup of G^{+}(V), τ(T_{0}) ⊂ C, and ν(T_{0}) = F^{×}. Let α ∈ T. Take β ∈ T_{0} so that ν(α) = ν(β). Then ν(β^{−1}α) = 1, and so β^{−1}α ∈ J. Thus α ∈ βJ ⊂ T_{0}, and we obtain T = T_{0} and C = τ(T_{0}) = τ(J) ∪ τ(Jη). If τ(η) ∈ τ(J), then η ∈ F^{×}J, which is impossible, as ν(η)𝔤 = 𝔭. Thus τ(η) ∉ τ(J), and so [C : τ(J)] = 2. This completes the proof.

5.5. We now return to (V, φ) over a global F of an odd dimension n > 1, and fix a maximal lattice L in V. For each v ∈ h we take an element ε_{v} ∈ δ(φ_{v}) that is either a unit or a prime element of F_{v}. Then ε_{v}𝔤_{v}^{×2} is determined by φ_{v}, where 𝔤_{v}^{×2} = {a^{2} | a ∈ 𝔤_{v}^{×}}. Given h ∈ V such that φ[h] ≠ 0, take an element β_{v} ∈ F_{v} such that φ(h, L_{v}) = β_{v}𝔤_{v}, and put r_{v}(h) = ε_{v}^{−1}φ[h]β_{v}^{−2}. Then r_{v}(h) determines a coset of F_{v}^{×}/𝔤_{v}^{×2}, which depends only on φ_{v}, L, and F_{v}^{×}hC(L_{v}). Strictly speaking, r_{v}(h) should be defined as a coset, but for simplicity we view it as an element of F_{v}^{×}, and write r_{v}(h) ∈ X or r_{v}(h) ∈ X for any subset X of F_{v} stable under multiplication by the elements of 𝔤_{v}^{×2}. For example, we can take as X the set

(5.9) 𝔈_{v} = {u^{2} + 4w | u, w ∈ 𝔤_{v}}.

Then the condition r_{v}(h) ∈ 𝔈_{v} is meaningful. Obviously 𝔈_{v} = 𝔤_{v} if v ∤ 2. We can also define r_{v}(h) for h ∈ V_{v}.

The following lemma concerns the local case, that is, v is fixed, and so we write r(h) and 𝔈 for r_{v}(h) and 𝔈_{v}.

Lemma 5.6. Let (V, φ), B, ξ, d, and B^{∘} be as in §5.2 with a local field F and B = M_{2}(F); suppose d is a prime element of F; let 𝔒 be the order of (5.1a) and L be as in (5.5a); put C = {α ∈ SO^{φ}(V) | Lα = L}. Fixing h = kξ as in §5.2, put K = F [k], 𝔣 = K ∩ 𝔒, 𝔶 = 𝔒̃ ∩ K, W = (F h)^{⊥}, and

(5.10) J = {α ∈ SO^{φ}(W) | (L ∩ W)α = L ∩ W }.

Then the following assertions hold:

(i) 𝔣 is the order of K whose discriminant is r(h)𝔭^{2}.

(ii) There exists an element ω of B^{×} such that (5.2) is satisfied, τ(ω) ∈ C, W = ωKξ; L ∩ W = ω𝔣ξ if r(h) ∉ 𝔭^{−1}, and L ∩ W = ω𝔶ξ if r(h) ∈ 𝔭^{−1}.

(iii) If r(h) ∉ 𝔭^{−1}, then K ≅ F × F, 𝔣 is the maximal order of K, L ∩ W is a maximal lattice in W, and J = SO^{φ}(W) ∩ C = τ(𝔣^{×}).

(iv) If r(h) ∈ 𝔈, then 𝔶 is the order in K whose discriminant is r(h)𝔤, 𝔣 is not maximal, J = τ({a ∈ K^{×} | a/a^{ρ} ∈ 𝔶^{×}}), and SO^{φ}(W) ∩ C = τ(𝔣^{×}).

(v) If r(h) ∈ 𝔭^{−1} and r(h) ∉ 𝔈, then K is ramified over F, 𝔣 is the maximal order of K, J = τ({a ∈ K^{×} | a/a^{ρ} ∈ 𝔣^{×}}), and SO^{φ}(W) ∩ C = {x ∈ K | xx^{ρ} = 1}, which has τ(𝔣^{×}) as a subgroup of index 2.

(vi) Cases (iii), (iv), and (v) cover all possibilities for h, and mutually exclusive.

Proof. To prove our assertions, we can replace h by any element of F^{×}hC. Indeed, if we replace h by chτ(α) with c ∈ F^{×} and α in the set T_{v} of (5.7), then K, 𝔣, and other symbols are replaced by their images under the inner asutomorphism x ↦ α^{−1}xα of A(V). Thus we may assume that 2φ(h, L) = 𝔤.

We can take [ \begin{matrix}0 & −d^{−1} \\ 0 & 0 \end{matrix} ] ξ, [ \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ] ξ, and [ \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ] ξ as the elements e_{1}, f_{1},

and g of §1.6. Therefore the result there guaratees an element γ ∈ C such that hγ = jξ with j of the following two types: (1) j = diag[c, −c], 2c𝔭 = 𝔤; (2)

j = [ \begin{matrix}c & b \\ 1 & −c \end{matrix} ] with b ∈ 𝔭^{−1} and c ∈ 2^{−1}𝔤. Thus we may assume that k = j

with such a j. Take ω = [ \begin{matrix}0 & d^{−1} \\ −1 & 0 \end{matrix} ] for type (1) and ω = [ \begin{matrix}−1 & 2c \\ 0 & 1 \end{matrix} ] for type

(2). Then τ(ω) ∈ C and ω^{−1}kω = −k; also W = ωKξ and L∩W = (𝔒̃∩ωK)ξ. We now treat our problems according to the type of j.

Type (1). Put ℓ = diag[2c, 0]. Then K = F + F ℓ, as ℓ = j + c. Clearly K consists of the diagonal matrices, and 𝔣 is its maximal order. Since 2c𝔭 = 𝔤, we have 𝔣 = 𝔤[dℓ]. Now, 𝔒̃ ∩ ωK = ω(𝔒 ∩ K) = ω𝔣, so that L ∩ W = ω𝔣ξ, which is a maximal lattice in W. It can easily be seen that J = SO^{φ}(W) ∩ C = τ(𝔣^{×}). Also, r(h)𝔤 = 4c^{2}𝔤 = 𝔭^{−2}, so that r(h) ∉ 𝔭^{−1}, and r(h)𝔭^{2} = 𝔤, which is the discriminant of 𝔣. Since r(h) ∈ 𝔭^{−1} for type (2) as can easily be seen, we have type (1) if and only if r(h) ∉ 𝔭^{−1}.

Type (2). For j of type (2), we have r(h) ∈ 𝔭^{−1}. We easily see that r(h) ∈ 𝔈 if b ∈ 𝔤. Conversely suppose r(h) ∈ 𝔈; then we have −4φ[h] = d(u^{2} + 4w)

with u, w ∈ 𝔤. Put h′ = j′ξ with j′ = [ \begin{matrix}u/2 & w \\ 1 & −u/2 \end{matrix} ] . Then φ[h′] = φ[h] and

2φ(h′, L) = 𝔤 = 2φ(h, L), so that h′ ∈ hC by Theorem 1.3. Thus if r(h) ∈ 𝔈, then we can put h = jξ with j of the above form such that b ∈ 𝔤. Without

assuming b ∈ 𝔤, put ℓ = [ \begin{matrix}2c & b \\ 1 & 0 \end{matrix} ] . Then ℓ = c + j and K = F + F ℓ. For

y, z ∈ F we have y + zℓ = [ \begin{matrix}y + 2cz & bz \\ z & y \end{matrix} ] . This belongs to 𝔒 if and only if

y ∈ 𝔤 and z ∈ 𝔭. Thus 𝔣 = 𝔤[dℓ]. This has discriminant 4(b + c^{2})𝔭^{2}, which equals r(h)𝔭^{2}. Since ω^{2} = 1 and ω ∈ 𝔒^{×}, we have L ∩ W = ω(𝔒̃ ∩ K)ξ. We see that y + zℓ ∈ 𝔒̃ if and only if y ∈ 𝔤 and z ∈ 𝔤. Thus 𝔶 = 𝔤 + 𝔤ℓ and L ∩ W = ω𝔶ξ. For x ∈ K and a ∈ K^{×} = G^{+}(W), we have ωxξτ(a) = ωexξ, where e = a/a^{ρ}. Thus (L ∩ W)τ(a) = L ∩ W if and only if (a/a^{ρ})𝔶 = 𝔶. Also τ(a) ∈ C if and only if τ(a) = τ(s) with s ∈ 𝔒^{×} ∪ 𝔒^{×}η^{−1}, as can be seen from Lemma 5.4. If s ∈ 𝔒^{×}, then a ∈ F^{×}𝔒^{×} ∩ K^{×} = F^{×}𝔣^{×}, so that τ(a) ∈ τ(𝔣^{×}). If s ∈ 𝔒^{×}η^{−1}, then s ∈ 𝔒^{×}η^{−1} ∩ K ⊂ 𝔶.

Suppose now r(h) ∈ 𝔈; then we may assume that b ∈ 𝔤. Since ℓ^{2} = 2cℓ + b, we see that 𝔶 is an order, and so J is as in (iv). If s ∈ 𝔒^{×}η^{−1}, then dss^{ι} ∈ 𝔤^{×}, which is impossible as 𝔶 is an order. Thus SO^{φ}(W) ∩ C = τ(𝔣^{×}). Since 𝔣 = 𝔤[dℓ] ≠ 𝔤[ℓ] = 𝔶, 𝔣 is not maximal and the discriminant of 𝔶 is r(h)𝔤. This proves (iv).

Next suppose r(h) ∉ 𝔈; then b ∉ 𝔤 and b𝔭 = 𝔤. We easily see that ℓ𝔣 = 𝔶, and so e𝔶 = 𝔶 if and only if e ∈ 𝔣^{×}. Thus J is as in (v). Put π = b^{−1} and σ = πℓ. Then K = F [σ]. Since σ^{2} = 2cπσ+π, K is ramified over F and its maximal order is 𝔤[σ], which coincides with 𝔣. Now SO^{φ}(W) = {x ∈ K^{×} | xx^{ρ} = 1}, which has τ(𝔣^{×}) as a subgroup of index 2 by a general principle [S3, Lemma 5.6 (iii)]. Since ℓ ∈ 𝔒^{×}η^{−1}, we have τ(ℓ) ∈ SO^{φ}(W) ∩ C. If τ(ℓ) ∈ τ(𝔣^{×}), then ℓ ∈ F^{×}𝔣^{×}, which is impossible. Thus τ(𝔣^{×}) ⫋ SO^{φ}(W) ∩ C ⊂ SO^{φ}(W), which proves (v). Finally (vi) is clear from the above discussion. This completes the proof, as (i) and (ii) have been proved.

If we apply (2.7) to the present setting, then H = {b ∈ K^{×} | bb^{ρ} = 1} with K = F + F k as noted in §5.2. Thus H is commutative and the left-hand side of (2.7) becomes [H_{A} : H(H_{A} ∩ C)]. We can determine this index explicitly as follows.

Theorem 5.7. In the setting of §5.2 with a global F, let L be a maximal lattice in V, and 𝔢 the product of all the prime ideals in F ramified in B; put d𝔤 = 𝔞𝔵^{2} with a squarefree integral ideal 𝔞 and a fractional ideal 𝔵. Let 𝔒 be the order in B of discriminant 𝔞 ∩ 𝔢 containing A^{+}(L). (See Lemma 5.3 (ii).) Given h = kξ ∈ V with k ∈ B^{∘} such that φ[h] ≠ 0, put K = F + F k; denote by 𝔯 the maximal order of K and by 𝔡 the different of K relative to F. Let 𝔞^{∗} be the product of the prime factors v of 𝔞 such that v ∤ 𝔢, r_{v}(h) ∈ 𝔭_{v}^{−1}, and r_{v}(h) ∉ 𝔈_{v}, where 𝔭_{v} is the local prime ideal at v. Then the order 𝔣 in K given by 𝔣 = K ∩𝔒 has conductor 𝔠, which can be determined by the condition that 𝔠_{v} = 𝔤_{v} if v|𝔞^{∗}𝔢 and 𝔠_{v}^{2}N_{K/F} (𝔡)_{v} = 𝔞_{v}φ[h]φ(h, L)_{v}^{−2} if v ∤ 𝔢. Moreover, put H = SO^{φ}(W) and identify H with {α ∈ K^{×} | αα^{ι} = 1}; define C by (5.6). Suppose K is a field; then

(5.11) [H_{A} : H(H_{A} ∩ C)] = (c_{K}/c_{F}) · 2^{1−μ−ν}[𝔤^{×} : N_{K/F} (𝔯^{×})]

· [U : U′]^{−1}N(𝔠) ∏_{𝔭|𝔠} {1 − [K/F, 𝔭]N(𝔭)^{−1}}.

Here c_{K} resp. c_{F} is the class number of K resp. F; μ is the number of prime ideals dividing 𝔞^{∗}𝔢 and ramified in K; ν is the number of v ∈ a ramified in K; U = {x ∈ 𝔯^{×} | xx^{ρ} = 1} and U′ = { x ∈ U | x − 1 ∈ 𝔠_{v}𝔡_{v} for every v ∤𝔞^{∗}𝔢 };

𝔭 runs over all prime factors of 𝔠; [K/F, 𝔭] denotes 1, −1, or 0 according as 𝔭 splits in K, remains prime in K, or is ramified in K.

Proof. If d ∈ 𝔤^{×}, this is [S3, Theorem 12.3]. Our proof here is a modified version of the proof there. For v|𝔢 we have 𝔯_{v} ⊂ 𝔒_{v}, so that 𝔣_{v} = 𝔯_{v} and 𝔠_{v} = 𝔤_{v}. Next suppose v ∤𝔢; then we can put B_{v} = M_{2}(F_{v}), and so [S3, Lemma 11.11] is applicable if v ∤𝔞; by (i) of that lemma, 𝔠_{v}^{2}N_{K/F} (𝔡)_{v} = φ[h]φ(h, L)_{v}^{−2}. If v|𝔞, then we use (i) of Lemma 5.6. If v|𝔞^{∗}, then 𝔠_{v} = 𝔤_{v} by (v) of Lamma 5.6. Thus we obtain our assertion concerning 𝔠.

To prove (5.11), suppose K is a field. Then we have

[H_{A} : HE] = (c_{K}/c_{F}) · 2^{1−κ}[𝔤^{×} : N_{K/F} (𝔯^{×})],

where E = H_{a} ∏_{v∈h} E_{v} with E_{v} = 𝔯_{v}^{×} ∩ H_{v} and κ is the number of v ∈ v ramified in K. Indeed, [H_{A} : HE] equals the number of classes in the genus of 𝔤-maximal lattices in W, and in [S3, (9.16)] we noted that it is the right-hand side of the above equality. Put D = H_{A} ∩C and D_{v} = D ∩H_{v}. If v|𝔢, then φ_{v} is anisotropic, so that C_{v} = G_{v}; thus D_{v} = H_{v} = E_{v} if v|𝔢. If v ∤𝔞^{∗}𝔢, then, by [S3, Lemma 11.11 (iv)] and Lemma 5.6 (iii), (iv), we have D_{v} = τ(𝔣_{v}^{×}). If v|𝔞^{∗}, then D_{v} = E_{v} by Lemma 5.6 (v). Thus U ∩ D = {x ∈ U | x ∈ τ(𝔣_{v}^{×}) for every v ∤𝔞^{∗}𝔢}. Applying [S3, Lemma 11.10 (iii)] to τ(𝔣_{v}^{×}), we obtain U ∩ D = U′. Now we have U = E ∩ H,

[HE : HD]=[E : E∩HD]=[E : UD]=[E : D]/[UD : D]=[E : D]/[U : U ∩D], [E : D] = ∏_{v∤𝔞^{∗}𝔢}[E_{v} : D_{v}] = ∏_{v∤𝔞^{∗}𝔢} [E_{v} : τ(𝔯_{v}^{×})][τ(𝔯_{v}^{×}) : τ(𝔣_{v}^{×})].

By [S3, Lemma 5.6 (iii)], [E_{v} : τ(𝔯_{v}^{×})] = 2 if v is ramified in K, and = 1 otherwise. The index [τ(𝔯_{v}^{×}) : τ(𝔣_{v}^{×})] is given by [S3, Lemma 11.10 (i), (iv)]. Thus we obtain

[E : D] = 2^{b}N(𝔠) ∏_{𝔭|𝔠} {1 − [K/F, 𝔭]N(𝔭)^{−1}},

where b is the number of the primes v ∤𝔞^{∗}𝔢 ramified in K. Now [H_{A} : HD] = [H_{A} : HE][HE : HD]. Combining all these, we obtain (5.11).

Corollary 5.8. The notation and assumption being as in Theorem 5.7, let c(𝔣) denote the class number of 𝔣 in the sense of [S3, §12.5] and let U_{𝔣} = U ∩𝔣^{×}. Then U′ ⊂ U_{𝔣} and (5.12) [H_{A} : H(H_{A}∩ C)] = (c(𝔣)/c_{F} ) · 2^{1−μ−ν}[𝔤^{×} : N_{K/F} (𝔣^{×})][U_{𝔣} : U′]^{−1}.

Proof. By [S3, Lemma 11.10 (ii) ] we have U_{𝔣} = { x ∈ U | x^{ρ} −x ∈ 𝔠𝔡 }. Since 𝔠 is prime to 𝔞^{∗}𝔢, for x ∈ U we have x^{ρ} − x ∈ 𝔠𝔡 if and only if x^{ρ} − x ∈ 𝔠_{v}𝔡_{v} for every v ∤ 𝔞^{∗}𝔢. We have also x^{ρ} − x = (1 − x)(1 + x^{ρ}), and hence U′ ⊂ U_{𝔣}. Now we recall a well known formula (see [S3, (12.3)])

(5.13) c(𝔣) = c_{K} · [𝔯^{×} : 𝔣^{×}]^{−1}N(𝔠) ∏_{𝔭|𝔠} {1 − [K/F, 𝔭]N(𝔭)^{−1}}.

As shown in [S3, p. 117, line 7], we have

(5.14) [𝔯^{×} : 𝔣^{×}] = [U : U_{𝔣}][N_{K/F} (𝔯^{×}) : N_{K/F} (𝔣^{×})].

Combining (5.11), (5.13), and (5.14) together, we obtain (5.12).

Theorem 5.9. In the setting of Lemma 5.4 and Theorem 5.7, the class number of the genus of maximal lattices in the ternary space (V, φ) equals the type number of 𝔒.

Proof. We easily see that the type number of 𝔒 equals #{B^{×}\B_{A}^{×}/T } with T of (5.7). By Lemma 5.4, τ(T) = C and clearly F_{A}^{×} ⊂ T, and so τ gives a bijection of B^{×}\B_{A}^{×}/T onto SO^{φ}(V)\SO^{φ}(V)_{A}/C. This proves our theorem.

This theorem should not be confused with the results in [Pe] (Satz 9 and its corollary), which concern the classes with respect to the group of similitudes.

Theorem 5.10. In the setting of Theorem 5.7, suppose the genus of maximal lattices consists of a single class (which is the case if 𝔒 has type number 1). Put Γ (L) = {γ ∈ SO^{φ}(V) | Lγ = L}. Then

(5.15a) #{L[q, 𝔟]/Γ (L)} = (c(𝔣)/c_{F} ) · 2^{1−μ−ν}[𝔤^{×} : N_{K/F} (𝔣^{×})][U_{𝔣} : U′]^{−1}.

Moreover, if B is totally definite, then

(5.15b) #L[q, 𝔟]/#Γ (L) = (c(𝔣)/c_{F} ) · 2^{1−μ−ν}[𝔤^{×} : N_{K/F} (𝔣^{×})]#(U_{𝔣})^{−1}.

Proof. The left-hand side of (2.7) in the present case is [H_{A} : H(H_{A} ∩ C)], as H is commutative; the right-hand side consists of a single term. Thus (2.7) combined with (5.12) gives (5.15a). Since H ∩ C = U ∩ D = U′, we have 𝔪(H, H_{A} ∩ C) = [H_{A} : H(H_{A} ∩ C)]#(U′)^{−1}, which together with (3.7) and (5.15a) proves (5.15b).

Lemma 5.11. Let B be a quaternion algebra over a global field F, and 𝔢 the product of all the prime ideals in F ramified in B; let 𝔞_{0} be a squarefree integral ideal prime to 𝔢. Further let K be a quadratic extension of F contained in B and 𝔣 an order in K containing 𝔤 that has conductor 𝔠. Then there exists an order 𝔒 in B of discriminant 𝔞_{0}𝔢 such that 𝔒 ∩ K = 𝔣 if and only if 𝔠 + 𝔢 = 𝔤 and every prime factor of 𝔞_{0} not dividing 𝔠 does not remain prime in K.

This is due to Eichler [E2, Satz 6].

Lemma 5.12. Let B, 𝔢, 𝔞_{0}, K, 𝔣 and 𝔠 be as in Lemma 5.11; let 𝔒 be an order in B of discriminant 𝔞_{0}𝔢. Suppose that 𝔒 has type number 1, 𝔠 + 𝔢 = 𝔤 and every prime factor of 𝔞_{0} not dividing 𝔠 does not remain prime in K. Then there exists an element α of B^{×} such that α^{−1}𝔒α ∩ K = 𝔣.

Proof. Since 𝔒 has type number 1, every order in B of discriminant 𝔞_{0}𝔢 is of the form α^{−1}𝔒α with α ∈ B^{×}. Therefore our assertion follows from Lemma 5.11.

6. Positive definite ternary forms over Q

6.1. Every positive definite ternary quadratic space (V, φ) over Q is obtained by taking B to be a definite quaternion algebra over Q and d to be a squarefree positive integer in the setting of §5.2. If we represent φ with respect to a Zbasis of an integral Z-lattice L in V, then we obtain a Z-valued ternary form. If L is maximal, then the form is reduced in the sense that it cannot be represented nontrivially by another Z-valued ternary form, and vice versa. This definition of a reduced form is different from Eisenstein’s terminology for ternary forms.

Define 𝔞 and 𝔢 as in §5.2. Then 𝔞 = dZ and 𝔢 = eZ with a squarefree positive integer e. We have 𝔢∩𝔞 = d_{0}eZ with a positive divisor d_{0} of d prime to e. Thus 𝔒 of Theorem 5.7 is an order of discriminant d_{0}e. The pair (e, d) determines the isomorphism class of (V, φ) and hence the genus of a reduced Zvalued ternary form. If Φ is a matrix which represents a ternary form belonging to that genus, then det(2Φ) = 2d_{0}^{2}e^{2}/d; this follows from Lemma 5.3 (i). As to the general theory of reduced forms and det(2Φ) for an arbitrary n, the reader is referred to [S5]. Taking a maximal lattice L in V, put C = {SO^{φ}(V)_{A} | Lα = L}. Then we have

Lemma 6.2. 𝔪(SO^{φ}(V), C) = \frac{1}{12} ∏_{p|e} \frac{p − 1}{2} ∏_{p|d_{0}} \frac{p + 1}{2}.

Proof. Define a quadratic form β on B^{∘} by β[x] = xx^{ι} for x ∈ B^{∘}; put G = SO^{β}(B^{∘}). Then x ↦ xξ for x ∈ B^{∘} gives an isomorphism of (B^{∘}, dβ) onto (V, φ), and also an isomorphism of G onto SO^{φ}(V). Moreover, for α ∈ B^{×} we have α^{−1}xαξ = α^{−1}xξα. Thus the symbol τ(α) is consistent. We can identify C with the subgroup {γ ∈ G_{A} | (𝔐 ∩ B^{∘})γ = (𝔐 ∩ B^{∘})} of G_{A}, where 𝔐 is as in (5.4). Let 𝔒_{0} be a maximal order in B containing 𝔒; put M = 𝔒_{0} ∩ B^{∘} and D = {γ ∈ G_{A} | Mγ = M}. Then M is a maximal lattice with respect to β, and from [S2, Theorem 5.8] we obtain

(6.1) 𝔪(G, D) = \frac{1}{12} ∏_{p|e} \frac{p − 1}{2}.

By (3.3) we have [C : C ∩ D]𝔪(G, C) = 𝔪(G, C ∩ D) = [D : C ∩ D]𝔪(G, D). Clearly D_{p} = C_{p} if p ∤ d_{0}. Suppose p|d_{0}; then we can put B_{p} = M_{2}(Q_{p}), (𝔒_{0})_{p} = M_{2}(Z_{p}), D_{p} = τ(GL_{2}(Z_{p})), and 𝔒_{p} is of the type (5.1a). From Lemma 5.4 we obtain [C_{p} : C_{p}∩D_{p}] = 2 and [D_{p} : C_{p}∩D_{p}] = p+1.

Thus 𝔪(G, C) = 𝔪(G, D) ∏_{p|d_{0}}{(p + 1)/2}. Combining this with (6.1), we obtain our lemma.

6.3. Now Eichler gave a formula for the class number of 𝔒 and also a formula for the type number of 𝔒 in [E2, (64)]. However, his formula for the type number is not completely correct, and correct formulas were given by Peters in [Pe] and by Pizer in [Pi]. There is a table for the type number of 𝔒 for d_{0}e ≤ 30 in [Pe, p. 360]; a larger table for d_{0}e ≤ 210 is given at the end of [Pi]; in these papers, e and d_{0} are denoted by q_{1} and q_{2}. From these tables we see that the type number of 𝔒 for d_{0}e ≤ 210 is 1 exactly when (e, d_{0}) is one of the following 20 pairs:

(2, 1), (2, 3), (2, 5), (2, 7), (2, 11), (2, 23), (2, 15), (3, 1), (3, 2), (3, 5), (3, 11), (5, 1), (5, 2), (7, 1), (7, 3), (13, 1), (30, 1), (42, 1), (70, 1), (78, 1).

Now d is d_{0} times a factor of e. Therefore if e has exactly t prime factors, then there are 2^{t} choices for (e, d) with the same (e, d_{0}). Consequently there are exactly 64 choices for (e, d) obtained from the above 20 pairs of (e, d_{0}), and each (e, d) determines (V, φ); det(2Φ) = 2d_{0}^{2}e^{2}/d as noted in §6.1. By Theorem 5.9, #{SO^{φ}(V)\SO^{φ}(V)_{A}/C} = 1 in all these cases. We have actually

Theorem 6.4. The spaces (V, φ) obtained from these 64 pairs (e, d) exhaust all positive definite ternary quadratic spaces over Q for which the genus of maximal lattices has class number 1. In other words, there are exactly 64 genera of positive definite, Z-valued, and reduced ternary forms consisting of a single class.

Proof. Our task is to show that the class number is not 1 for d_{0}e > 210. If the inverse of the right-hand side of the equality of Lemma 6.2 is not an integer, then the class number cannot be 1. For d_{0}e > 210, the inverse is an integer exactly when (e, d_{0}) is one of the following six: (2 · 7 · 17, 1), (2 · 5 · 13, 3), (2 · 3 · 17, 5), (2 · 3 · 13, 7), (2 · 3 · 5, 23), (2 · 3 · 5, 11). The verification is easy, as the mass ≤ 1 for relatively few cases of (e, d_{0}). Among those six, the mass is 1/2 for the last one; in the other five cases the mass is 1. Now, by the formula in [Pe, p. 361] the type number of 𝔒 (which is the class number in question by virtue of Theorem 5.9) is given by 2^{−κ} ∑_{t} Sp{P^{∗}(t)}, where κ is the number of prime factors of d_{0}e, t runs over all positive divisors of d_{0}e, and Sp{P^{∗}(t)} is given on the same page. From the formulas there we can easily verify that Sp{P^{∗}(1)} = 2^{κ} and Sp{P^{∗}(d_{0}e)} > 0 in the first five cases, and hence the type number is greater than 1. As for the last case (2 · 3 · 5, 11), computing Sp{P^{∗}(t)} for all t|d_{0}e, we find that the type number is 2. Thus we obtain our theorem.

Remark. In the setting of §6.1, if det(2Φ)/2 is squarefree, then Φ must be reduced and det(2Φ)/2 = d = d_{0}e with squarefree d_{0} and e. If the genus of Φ has class number 1, then in view of the above theorem such a Φ is obtained from one of the pairs listed in §6.3 by taking d = d_{0}e. Thus there are exactly 20 genera of Z-valued positive definite ternary forms with class number 1 such that det(2Φ)/2 is squarefree. This result was obtained by Watson [W], and we derived it from a stronger result, Theorem 6.4.

6.5. We are going to specialize Theorem 5.7 and (5.15b) to these cases, and give explicit formulas for #L[q, Z]. For simplicity, we consider only the cases in which e is a prime, and d_{0} is 1 or a prime; thus we consider only 15 pairs (e, d_{0}), and consequently 30 pairs (e, d) with d = d_{0} or d = d_{0}e. In this setting we have

(6.2) #{Γ (L)} = 48/[(d_{0} + 1)(e − 1)].

This is because #{Γ (L)} equals the inverse of the quantity of Lemma 6.2, as the class number is 1.

To study the nature of q for which L[q, Z] ≠ ∅, we consider the localization of B at e. Let 𝔯 denote the maximal order in the unramified quadratic extension of Q_{e}. We can put B_{e}^{∘} = Q_{e}σ + Q_{e}η, 𝔒_{e} = 𝔯 + 𝔯η, and 𝔒̃_{e} = 𝔯 + 𝔯η^{−1} with σ and η as in (B) of §1.5; we can take η^{2} = e, though this is often unnecessary. We have L = (𝔐 ∩ B^{∘})ξ, and so

(6.3a) L_{e} = (Z_{e}σ + 𝔯η)ξ and L̃_{e} = (2^{−1}Z_{e}σ + 𝔯η^{−1})ξ if e∤d,

(6.3b) L_{e} = (Z_{e}σ + 𝔯η^{−1})ξ and L̃_{e} = ((2e)^{−1}Z_{e}σ + 𝔯η^{−1})ξ if e|d.

Here is an easy fact: given an element η_{0} of B_{e}^{×} such that η_{0}^{2} is a prime element, we can find an element α of B_{e}^{×} such that (6.3a, b) are true with (α^{−1}σα, α^{−1}𝔯α, η_{0}) in place of (σ, 𝔯, η). Indeed, since η_{0}^{2}/η^{2} ∈ Z_{e}^{×}, we have η_{0}^{2}/η^{2} = aa^{ι} with a ∈ 𝔯^{×}. Then η_{0}^{2} = (aη)^{2}, and hence there exists an element α of B_{e}^{×} such that α^{−1}aηα = η_{0}. Since α^{−1}L_{e}α = L_{e} and α^{−1}L̃_{e}α = L̃_{e}, we obtain the desired result.

Theorem 6.6. Given 0 < q ∈ Q in the setting of §6.5, put K_{0} = Q(\sqrt{−dq}) and denote by δ the discriminant of K_{0}. Then the following assertions hold:

(i) L[q, Z] ≠ ∅ only if q is as follows:

(6.4a) d_{0}eq = r^{2}m when e ≠ 2 or e|d, (6.4b) d_{0}q = r^{2}m when e = 2 and e∤d.

Here m is a squarefree positive integer such that e does not split in K_{0}, and also that m − 3 ∈ 8Z if e = 2 and 2|d; r is a positive integer prime to e.

(ii) Moreover, put a^{∗} = d_{0} in the following two cases: (A) d_{0} ≥ 2, d_{0}|m, and d_{0} ∤ r; (B) d_{0} = 2, r − 2 ∈ 2Z, and 2 is ramified in K_{0}; put a^{∗} = 1 if neither (A) nor (B) applies. Let ℭ be the set of all positive integers c prime to a^{∗}e such that d_{0}|c if d_{0} is a prime that remains prime in K_{0}. (Thus ℭ depends on d, e, and q.) Then r/2 ∈ ℭ if 4|δ and e ≠ 2; r ∈ ℭ otherwise.

(iii) Conversely, given r and m satisfying all these conditions, determine q by (6.4a, b). Then L[q, Z] ≠ ∅.

Notice that K_{0} = Q(\sqrt{−m}) if e|d or e = 2, and K_{0} = Q(\sqrt{−em}) otherwise. Also, ℭ = ∅ if a^{∗} is a prime that remains prime in K_{0}.

Proof. Suppose h = kξ ∈ L[q, Z] with k ∈ B^{∘}; put K = F [k] as in §5.2 and 𝔣 = K ∩ 𝔒; let c be the conductor of 𝔣. Then K ≅ K_{0} and h ∈ 2L̃. Then from (6.3a, b) we see that eq ∈ Z_{e} if e ≠ 2 or e|d, and q ∈ Z_{e} otherwise. The set L̃_{p} for p ≠ e can be given by (5.5b), and so dq ∈ Z_{p}. Thus d_{0}eq ∈ Z if e ≠ 2 or e|d; d_{0}q ∈ Z otherwise. Take 0 < r ∈ Z and a squarefree positive integer m as in (6.4a, b). Define 𝔞^{∗} as in Theorem 5.7 and put 𝔞^{∗} = a^{∗}Z with 0 < a^{∗} ∈ Z. Clearly a^{∗} = 1 if d_{0} = 1. Suppose d_{0} is a prime number. To make our exposition easier, denote this prime by s; then s ≠ e and r_{s}(h) = −d^{−1}q. We can easily verify that a^{∗} = s exactly in Cases (A) and (B) stated in (ii) above. By Theorem 5.7 and Lemma 5.11, we see that c ∈ ℭ. Notice that s is ramified in K_{0} if a^{∗} = s.

(1) We first consider the case e|d. Then d = d_{0}e and dq = d_{0}eq and so K ≅ Q(\sqrt{−m}). Suppose e|r; put ℓ = e^{−1}k. Then d^{2}ℓℓ^{ι} = de^{−2}q = (r/e)^{2}m ∈ Z, and hence dℓ ∈ 𝔒_{e} ∩ B^{∘}. Thus ℓ ∈ e^{−1}𝔒_{e} ∩ B^{∘} = e^{−1}Z_{e}σ + η^{−1}𝔯, and so ℓξ ∈ 2L̃_{e} by (6.3b). Therefore h = eℓξ ∈ 2eL̃_{e}, which implies that φ(h, L)_{e} ⊂ eZ_{e}, a contradiction. Thus e∤r.

(2) Next suppose e ∤ d; then d = d_{0}. Assuming e|r, put ℓ = e^{−1}k. Then d^{2}ℓℓ^{ι} = de^{−2}q ∈ e^{−1}Z_{e}, and hence ℓ ∈ 𝔒̃_{e} ∩ B^{∘} = Z_{e}σ + η^{−1}𝔯. Thus ℓξ ∈ 2L̃_{e} by (6.3a), which leads to a contradiction for the same reason as in Case (1). Therefore e∤r in this case too. This reasoning is valid even when e = 2.

(3) Since K_{0} must be embeddable in B, the prime e cannot split in K_{0}. Suppose e = 2, e|d, and e|m. Since kk^{ι} = d^{−2}r^{2}m, we have k ∈ η^{−1}𝔒_{e}^{×} ∩B^{∘}, so that k = aσ +η^{−1}b with a ∈ Z_{e} and b ∈ 𝔯. Since ηk ∈ 𝔒_{e}^{×}, we have b ∈ 𝔯^{×}. By (6.3b), 2φ(h, L)_{e} = 2Tr_{B/F} (k(Z_{e}σ + η^{−1}𝔯)) = 4aZ_{e} + Tr_{K/Q}(b𝔯) = Z_{e}, a contradiction. Therefore m must be odd if e = 2 and e|d. In this case, k ∈ 2^{−1}𝔒_{e}^{×} ∩ B^{∘}, so that 2k = xσ + ηy with x ∈ Z_{e}^{×} and y ∈ 𝔯. Thus 2Z_{e} = 2φ(h, L)_{e} = 2Tr_{B/F} (k(Z_{e}σ+η^{−1}𝔯)) = 2Z_{e}+Tr_{K/Q}(y𝔯), so that y ∈ 2𝔯. Now r^{2}m = dq = d^{2}kk^{ι} = −d_{0}^{2}(x^{2}σ^{2} + 2yy^{ι}). We can take 𝔯 ⊂ Q_{2}(\sqrt{5} ) and σ = \sqrt{5}. Therefore we see that m − 3 ∈ 8Z. Thus we obtain the condition on m as stated in our theorem.

(4) By Theorem 5.7, c is prime to a^{∗}e and c^{2}δZ_{p} = d_{0}qZ_{p} = r^{2}mZ_{p} for every p ≠ e. We have seen that r is prime to e. First suppose 4|δ and e ≠ 2. Then we see that δZ_{p} = 4mZ_{p} for p ≠ e, and so 4c^{2}Z_{p} = r^{2}Z_{p} for p ≠ e. Since both 2c and r are prime to e, we obtain 2c = r. Similarly we easily find that c = r if 4 ∤ δ or e = 2. Thus we obtain (ii). This completes the proof of the “if”-part.

(5) Conversely, suppose q is given with r and m as in our theorem; put c = r/2 or c = r according as r/2 ∈ ℭ or r ∈ ℭ. Let 𝔬 be the order in K_{0} whose conductor is c. By Lemma 5.12, our conditions on r and m guarantees an injection θ of K_{0} into B such that θ(𝔬) = θ(K_{0}) ∩ 𝔒. Our task is to find an element h such that φ[h] = q and φ(h, L) = Z. Put μ = θ(\sqrt{−m}) if K_{0} = Q(\sqrt{−m}).

(5a) First we consider the case e|d and |δ| = m. Then d = d_{0}e, r = c, and K_{0} = Q(\sqrt{−m}). Put h = kξ with k = d^{−1}rμ. Then φ[h] = q and by Theorem 5.7, dqφ(h, L)_{p}^{−2} = c^{2}δZ_{p} = r^{2}mZ_{p} for every p ≠ e, so that φ(h, L)_{p} = Z_{p} for every p ≠ e. Now 2φ(h, L)_{e} = dTr_{B/F} (k(Z_{e}σ + η^{−1}𝔯)) = Tr_{B/F} (μ(Z_{e}σ + η^{−1}𝔯)) by (6.3b). If e|m, then we can take μ as η, changing 𝔯 by an inner automorphism, as noted after (6.3b). (Since |δ| = m, we have e ≠ 2.) If e ∤ m, then we can take 𝔯 ⊂ θ(K_{0})_{e} and σ = μ. In either case we easily see that 2φ(h, L)_{e} = 2Z_{e}. Thus φ(h, L) = Z, and L[q, Z] ≠ ∅ as expected.

(5b) Next suppose e|d and |δ| = 4m; then r = 2c if e ≠ 2 and r = c if e = 2. We have K_{0} = Q(\sqrt{−m}) and we again take h = kξ with k = d^{−1}rμ. Then φ[h] = q and φ(h, L)_{p} = Z_{p} for every p ≠ e for the same reason as in (5a). If e ≠ 2, the same argument as in (5a) shows that φ(h, L)_{e} = Z_{e}, which leads to the desired result. Thus suppose e = 2. Then m−3 ∈ 8Z as stipulated in our theorem. We can take σ = \sqrt{5} as we did in (3). Put μ = aσ + ηb with a ∈ Z_{e} and b ∈ 𝔯. Then a ∈ Z_{e}^{×} and −m = a^{2}σ^{2}+2bb^{ρ}. Since m+σ^{2} ∈ 8Z_{e}, we see that b ∈ 2𝔯. Thus 2φ(h, L)_{e} = Tr_{B/F} (μ(Z_{e}σ+η^{−1}𝔯)) = 2Z_{e}+Tr_{K/Q}(b𝔯) = 2Z_{e}, which gives the expected result.

(5c) Suppose e ∤ d and e = 2. Then d = d_{0} and K_{0} = Q(\sqrt{−m}); we take h = kξ with k = d^{−1}rμ. We find that φ[h] = q and φ(h, L)_{p} = Z_{p} for every p ≠ 2 in the same manner as in (5a). Now 2φ(h, L)_{2} = Tr_{B/F} (μ(Z_{2}σ + η𝔯)) by (6.3a). If 2|m, then taking μ as η, we obtain 2φ(h, L)_{2} = 2Z_{2} as expected. Suppose 2 ∤ m; then μ = aσ + ηb with a ∈ Z_{2}^{×} and b ∈ 𝔯, and so Tr_{B/F} (μ(Z_{2}σ + η𝔯)) = 2Z_{2}, which gives the desired result.

(5d) Finally suppose e ∤ d and e ≠ 2. Then d = d_{0} and K_{0} = Q(\sqrt{−em}); we take h = kξ with k = θ(d^{−1}e^{−1}r\sqrt{−em}); then φ[h] = q. For p ≠ e, δZ_{p} equals mZ_{p} or 4mZ_{p}, and r = c or r = 2c accordingly. Then we easily see that φ(h, L)_{p} = Z_{p} for every p ≠ e in the same manner as in (5a). Now 2φ(h, L)_{e} = Tr_{B/F} (e^{−1}θ(\sqrt{−em})(Z_{2}σ + η𝔯)). If e|m, then we can put θ(\sqrt{−em}) = e(aσ + ηb) with a ∈ Z_{e}^{×} and b ∈ 𝔯, and obtain φ(h, L)_{e} = Z_{e}. If e∤m. then taking η = θ(\sqrt{−em}), we obtain the desired result. This completes the proof.

Theorem 6.7. If L[q, Z] ≠ ∅ in the setting of Theorem 6.6, then

#L[q, Z] = \frac{2^{1−μ} · 48 · c(K_{0})}{(d_{0} + 1)(e − 1)w} · c ∏_{p|c} \{1 − [K_{0}/Q, p]p^{−1}\}.

Here μ is the number of prime factors of a^{∗}e ramified in K_{0}; c(K_{0}) is the class number of K_{0}; w is the number of roots of unity in K; c = r/2 if 4|δ and e ≠ 2; c = r otherwise; [K_{0}/Q, p] is defined as in Theorem 5.7.

Proof. Specialize (5.15b) to the present case. Then [U : 1] = w and #Γ (L) is given by (6.2); c(𝔣) can be connected to c(K_{0}) by (5.13) and (5.14). Thus we obtain our formula.

6.8. Before discussing examples, let us insert here a remark applicable to (V, φ) over Q with an arbitrary n. If F = Q and 𝔤 = Z, it is natural to consider L[q, Z], but it is not always best to formulate the result with respect to a Z-basis of L. To see this, first take the standard basis {e_{i}}_{i=1}^{n} of Z^{n} = L, and define a matrix Φ by Φ = [φ(e_{i}, e_{j})]_{i,j=1}^{n}; put f_{i} = e_{i}Φ^{−1}. Then 2L̃ = ∑_{i=1}^{n} Zf_{i}. Let h ∈ L[q, Z]. Then h ∈ 2L̃, and so h = ∑_{i=1}^{n} a_{i}f_{i} with a_{i} ∈ Z. We easily see that φ(h, L) = Z if and only if ∑_{i=1}^{n} a_{i}Z = Z, and also that Φ^{−1} = [φ(f_{i}, f_{j})]_{i,j=1}^{n}. This means that

(6.5) L[q, Z] = { ∑_{i=1}^{n} a_{i}f_{i} | aΦ^{−1} ·^{t}a = q, ∑_{i=1}^{n} a_{i}Z = Z}, a = (a_{i})_{i=1}^{n}.

Therefore if we follow the traditional definition of primitivity, we have to use the matrix Φ^{−1} instead of Φ. Recall that Γ (L) = Γ (L̃). Thus Γ (L) can be represented, with respect to the basis {f_{i}}_{i=1}^{n}, by the group

(6.6) Γ′ = {γ ∈ SL_{n}(Z) | γΦ^{−1} ·^{t}γ = Φ^{−1}}.

Consequently, L[q, Z]/Γ (L) corrresponds to the vectors a such that aΦ^{−1}·^{t}a = q and ∑_{i=1}^{n} a_{i}Z = Z, considered modulo Γ′. We note here an easy fact:

(6.7) | det(2Φ)| = [L̃ : L].

6.9. Let us now illustrate Theorem 6.6 by considering five examples and formulating the results in terms of the matrix Φ^{−1} of §6.8. For (e, d) = (2, 1) we obtain the result on sums of three squares, which Gauss treated; we do not include this in our examples, as it is easy and too special. Indeed, in this case we have Φ = Φ^{−1}, and so the result concerns the primitive solutions of ∑_{i=1}^{3} x_{i}^{2} = q. But in all other cases, Φ ≠ Φ^{−1}, and the results have more interesting features. From Lemma 5.3 (i) and (6.7) we obtain det(2Φ) = 2d_{0}^{2}e^{2}/d. In each case we state only the condition for L[q, Z] ≠ ∅. We will dispense with the statement about #L[q, Z], as it is merely a specialization of Theorem 6.7.

(1) We first take e = d = 3 in Theorem 6.6. Then

(6.8) 2Φ = diag[2, [ \begin{matrix}2 & −1 \\ −1 & 2 \end{matrix} ] ], 3Φ^{−1} = diag[3, [ \begin{matrix}4 & 2 \\ 2 & 4 \end{matrix} ] ].

These forms of matrices can be obtained by analyzing L of (5.4) in the present case. Alternatively, since det(2Φ) = 6 for Φ of (6.8), we can conclude that Φ is the matrix representing φ by the principle explained in [S5]. Theorem 6.6 (or the form of Φ^{−1}) in the present case shows that L[q, Z] ≠ ∅ only when 3q ∈ Z. Thus with s = 3q ∈ Z, the principle of §6.8 says that L[s/3, Z] corresponds to the set of vectors (x, y, z) such that

(6.9) 3x^{2} + 4(y^{2} + yz + z^{2}) = s and xZ + yZ + zZ = Z.

Theorem 6.6 specialized to this case means that given 0 < s ∈ Q, we can find (x, y, z) satisfying (6.9) if and only if s = r^{2}m with a squarefree positive integer m such that m + 1 ∉ 3Z and a positive integer r prime to 3 such that 2|r if m + 1 ∉ 4Z; K_{0} = Q(\sqrt{−m}).

(2) Next we take e = 3 and d = 1. Then

(6.10) 2Φ = diag[6, [ \begin{matrix}2 & −1 \\ −1 & 2 \end{matrix} ] ], 3Φ^{−1} = diag[1, [ \begin{matrix}4 & 2 \\ 2 & 4 \end{matrix} ] ].

Therefore L[s/3, Z] corresponds to the set of vectors (x, y, z) such that (6.11) x^{2} + 4(y^{2} + yz + z^{2}) = s and xZ + yZ + zZ = Z.

Such vectors exist if and only if s = r^{2}m with a squarefree positive integer m such that 3 does not split in K_{0} = Q(\sqrt{−3m}) and a positive integer r prime to 3 such that 2|r if 4 divides the discriminant of K_{0}.

In both cases (1) and (2) the group Γ (L) has order 12. Represent (x, y, z) ∈ Z^{3} by (x, β) with β = y + zζ, where ζ is a primitive cubic root of unity. Then Γ′ of (6.6) consists of the maps (x, β) ↦ (x, εβ) and (x, β) ↦ (−x, ε\overline{β}), where ε is a sixth root of unity. Therefore a set of representatives for L[q, Z]/Γ (L) can be found numerically. For instance, take s = m = 79 in (6.9); then we easily find that #{L[q, Z]/Γ (L)} = 5, which combined with (5.15a) confirms that Q(\sqrt{−79} ) has class number 5.

(3) Our third example concerns the case e = d = 2. We have

(6.12) 2Φ = [ \begin{matrix}2 & −1 & 0 \\ −1 & 2 & −1 \\ 0 & −1 & 2 \end{matrix} ] , 2Φ^{−1} = [ \begin{matrix}3 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 3 \end{matrix} ] .

Then L[s/2, Z] corresponds to the set of vectors x = (x_{i})_{i=1}^{3} ∈ Z^{3} such that

(6.13) x · 2Φ^{−1} ·^{t}x = s and ∑_{i=1}^{3} x_{i}Z = Z.

Such vectors x exist if and only if s = r^{2}m with an odd integer r and a squarefree positive integer m such that m − 3 ∈ 8Z. In this case, K_{0} = Q(\sqrt{−m}).

(4) Take e = d = 7. We have then

(6.14) 2Φ = diag[2, [ \begin{matrix}4 & −1 \\ −1 & 2 \end{matrix} ] ], 7Φ^{−1} = diag[7, [ \begin{matrix}4 & 2 \\ 2 & 8 \end{matrix} ] ],

and L[s/7, Z] with s ∈ Z corresponds to the set of vectors (x, y, z) such that

(6.15) 7x^{2} + 4(y^{2} + yz + 2z^{2}) = s and xZ + yZ + zZ = Z.

Such vectors exist if and only if s = r^{2}m with a positive integer r prime to 7 and a squarefree positive integer m such that 7 does not split in K_{0} = Q(\sqrt{−m}); 2|r if m + 1 ∉ 4Z.

(5) Let us finally take e = 2 and d = 22. We have then

(6.16) 2Φ = diag[4, [ \begin{matrix}6 & −1 \\ −1 & 2 \end{matrix} ] ], 22Φ^{−1} = diag[11, [ \begin{matrix}8 & 4 \\ 4 & 24 \end{matrix} ] ],

and the result about L[s/22, Z] is what we stated in the introduction.

References

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[S2] G. Shimura, An exact mass formula for orthogonal groups, Duke Math. J. 97 (1999), 1–66 (=Collected Papers, IV, 509–574).

[S3] G. Shimura, Arithmetic and analytic theories of quadratic forms and Clifford groups, Mathematical Surveys and Monographs, vol. 109, Amer. Math. Soc. 2004.

[S4] G. Shimura, Quadratic Diophantine equations and orders in quaternion algebras, Amer. J. Math. 128 (2006), 481–518.

[S5] G. Shimura, Classification, construction, and similitudes of quadratic forms, to appear in Amer. J. Math. 128 (2006).

[S6] G. Shimura, Quadratic Diophantine equations, the class number, and the mass formula, Bulletin, Amer. Math. Soc. 43 (2006), 285–304.

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[W] G. L. Watson, One-class genera of positive ternary quadratic forms, Mathematika 19 (1972), 96–104.

Goro Shimura

Department of Mathematics

Princeton University

Princeton, New Jersey 08544-1000 U.S.A. goro@Math.Princeton.EDU 368

Documenta Mathematica 11 (2006)