REPRESENTATIONS BY SPINOR GENERA

OF TERNARY QUADRATIC FORMS

Rainer Schulze-Pillot and Fei Xu

§1. Introduction

It is our basic question to study the following Diophantine equations

(1.1)  ^{t}XAX = B

over the ring of integers ℤ, where A and B are non-degenerate and symmetric matrices of size m × m and n × n over ℤ respectively, and A is indefinite with m ≥ 3. It is a necessary condition for solubility of equation (1.1) that it is solvable over ℤ_{p} for all primes p and the real numbers ℝ. This necessary condition is already sufficient if m − n ≥ 3 [Kn1,Hs]. However the equation (1.1) is no longer a purely local problem when m − n ≤ 2. By the Hasse principle, the necessary condition implies there is a rational solution of (1.1). In the previous papers [CX] and [X1], one of us has given conditions that allow to decide for m − n ≤ 2 whether the equation (1.1) is solvable over ℤ by looking at a given rational solution whose denominator is prime to the determinant of A. Can one also determine the solubility of (1.1) if the denominator of the rational solution is not prime to the determinant of A? In this note, we try to give such a condition.

Notation and terminology are standard if not explained, or adopted from [CX] and [X1]. Let V be a quadratic space over a number field F with a non-degenerate symmetric bilinear form ⟨x, y⟩, Q(x) = ⟨x, x⟩ be the quadratic map on V and SO(V) be the special orthogonal group of V. A lattice in V means a finitely generated 𝔬_{F} module in V such that it generates a non-degenerate quadratic subspace of V. A full lattice means a lattice which generates the whole space. For a full lattice L, L^{♯} denotes the dual lattice of L. For two lattices K and L in V, ⟨K, L⟩ denotes the fractional ideal generated by ⟨x, y⟩ for x ∈ K and y ∈ L. We use τ_{z} for the reflection if Q(z) ≠ 0. We also denote 𝔫(L) and 𝔰(L) as norm and scale of a lattice L in the sense of [O] respectively.

For any prime 𝔭 of F, V_{𝔭} (resp. F_{𝔭}, etc.) denotes the local completion of V (resp. F, etc.). Let 𝔬_{F} be the ring of integers of F. If 𝔭 is a finite prime, the group of units of 𝔬_{F_{𝔭}} is denoted by 𝔲_{𝔭}, and π_{𝔭} is a uniformizer of F_{𝔭}. We use θ_{𝔭} to denote the spinor norm map of SO(V_{𝔭}). For a lattice K_{𝔭} and a full lattice L_{𝔭} in V_{𝔭}, let

X(L_{𝔭}/K_{𝔭}) = {σ ∈ SO(V_{𝔭}) : K_{𝔭} ⊆ σL_{𝔭}}.

It is clear that X(L_{𝔭}/K_{𝔭}) is not empty if and only if K_{𝔭} is represented by L_{𝔭}. Then there is σ ∈ X(L_{𝔭}/K_{𝔭}) such that K_{𝔭} ⊆ σL_{𝔭}. By [HSX, Thm.2.1], one has that

Typeset by AMS-TEX

θ_{𝔭}(X(σL_{𝔭}/K_{𝔭}) is a group. It can be easily verified that this group is independent of choice of σ ∈ X(L_{𝔭}/K_{𝔭}). Then one can define

θ_{𝔭}(L_{𝔭}, K_{𝔭}) := θ_{𝔭}(X(σL_{𝔭}/K_{𝔭})).

The invariant θ_{𝔭}(L_{𝔭}, K_{𝔭}) was introduced and computed in [SP1] for ternary forms with codimension two over non-dyadic and 2-adic local fields. For such case over general dyadic fields, this invariant is computed in [X]. For general high dimensional cases, the invariant is computed in [HSX] and [HSX1] over non-dyadic and 2-adic local fields respectively. Recently, Beli has announced a computation of θ_{𝔭}(L_{𝔭}, K_{𝔭}) over general dyadic local fields.

By the strong approximation theorem for spin groups, the solubility of (1.1) is reduced to determine in which coset of the quotient group F_{𝔭}^{×}/θ_{𝔭}(L_{𝔭}, K_{𝔭}) the set θ_{𝔭}(X(L_{𝔭}/K_{𝔭})) is for all 𝔭. The results of [CX,X1] show that this coset depends only on [K_{𝔭} : L_{𝔭} ∩K_{𝔭}] if 𝔭 does not divide the discriminant of L_{𝔭}. A question raised in [X1] is to find some invariants derived from L_{𝔭} and K_{𝔭} directly to describe to which coset θ_{𝔭}(X(L_{𝔭}/K_{𝔭})) belongs if 𝔭 divides the discriminant of L_{𝔭}.

As a first step, we show that (under a suitable primitivity condition in case that rank(K_{𝔭}) ≥ 2) the coset in question depends only on the image of (a suitable multiple of) K_{𝔭} in the discriminant group L_{𝔭}^{#}/L_{𝔭} of L_{𝔭}.

We focus then on the most interesting case that rank(L) = 3 and rank(K) = 1. The group θ_{𝔭}(L_{𝔭}, K_{p}) is then (as always in the case that the codimension of K in L is 2) a subgroup of index ≤ 2 of F_{𝔭}^{×}, and we consider the case that this subgroup is the group of units 𝔲_{𝔭} of F_{𝔭}. For this unramified situation we obtain a complete answer in the special case F = ℚ (or more generally for F/ℚ such that 2 splits completely in F). In fact it turns out that the coset in question then depends only on ⟨K_{𝔭}, L_{𝔭}⟩ and (in the case of 𝔭 = (2)) information about the primitivity of (a suitable multiple of) K_{𝔭} in the discriminant group. It appears that the case of dyadic F_{𝔭} ≠ ℚ_{2} is more complicated in the sense that one needs more information about the position of K_{𝔭} in the discriminant group in that case.

The case that θ_{𝔭}(L_{𝔭}, K_{𝔭}) is the norm group of a ramified quadratic extension of F_{𝔭} is apparently also quite a bit more complicated. We have no general result in that case but give some examples of the type of calculation arising in one case of the example below and in the final section dealing with the question of the existence of regular forms in an indefinite genus.

As an example for the usefulness of our (admittedly rather technical) results for getting concrete answers about the solubility of equation (1.1), we prove the following

Example 1.2. Let m, n and k be positive integers. The Diophantine equation

m^{2}x^{2} + n^{2k}y^{2} − nz^{2} = 1

is solvable over ℤ if and only if (m, n) = 1 and

n ≡ \{ \begin{matrix}1, 3, 7 mod 8 & if 2 | m and 4 ∤ m \\ 1, 7 mod 8 & if 4 | m. \end{matrix}

We always assume that X(L_{𝔭}/K_{𝔭}) is not empty for all primes 𝔭. If 𝔫(K_{𝔭}) ⊂ 𝔫(L_{𝔭}), there is a full lattice L′_{𝔭} ⊂ L_{𝔭} such that

𝔫(K_{𝔭}) ⊆ 𝔫(L′_{𝔭}) ⊂ 𝔫(L_{𝔭})), X(L′_{𝔭}/K_{𝔭}) = X(L_{𝔭}/K_{𝔭})

and

θ_{𝔭}(L′_{𝔭}, K_{𝔭}) = θ_{𝔭}(L_{𝔭}, K_{𝔭}).

Therefore we also assume that 𝔫(K_{𝔭}) = 𝔫(L_{𝔭}) by the above standard argument.

§2. An application of Witt’s theorem

We recall that a nonzero submodule N of a module M over a ring R is called primitive if it is a direct summand of M.

Lemma 2.1. Let R be a discrete valuation ring with field of fractions F and maximal ideal (π). Let M be a finitely generated free R-module with quadratic form q : M → R and associated symmetric bilinear form ⟨x, y⟩ = q(x + y) − q(x) − q(y) and let N_{1}, N_{2} be two R-submodules of F M that are isometric with respect to the extension of q to F M.

Assume that there is j ∈ ℤ such that π^{j}N_{ν} is a primitive submodule of M^{#} = {y ∈ F M | ⟨y, M⟩ ⊆ R} for ν = 1, 2. Assume moreover that there is an isometry τ : N_{1} → N_{2} such that τ(x) − x ∈ M for all x ∈ π^{j}N_{1}.

Then there is σ ∈ O(M, q) with σ(N_{1}) = N_{2}.

Proof. For x ∈ M and a submodule N ⊆ M^{#} let β_{N}(x) ∈ N^{∗} be the linear form on N given by β_{N}(y) := ⟨x, y⟩. It is then easily seen that a submodule N of M^{#} is a primitive submodule of M^{#} if and only if one has β_{N}(M) = N^{∗}. A submodule of M satisfying this last property is called sharply primitive with respect to the symmetric bilinear form ⟨ , ⟩ in [Kn3, Definition 2.17]. If in addition to our assumptions π^{j}N_{ν} ⊆ M is true for ν = 1, 2, our assertion is therefore just Folgerung 4.4 in [Kn3]. An inspection of the proof given there shows that it remains valid in the present situation. □

Proposition 2.2. Let L_{𝔭} be as in the introduction with \frac{1}{2}⟨x, x⟩ ∈ 𝔬_{F_{𝔭}} for all x ∈ L_{𝔭}, let K_{𝔭}^{(1)}, K_{𝔭}^{(2)} be isometric sublattices of F_{𝔭}L_{𝔭}. Assume that there are an isometry τ : K_{𝔭}^{(1)} → K_{𝔭}^{(2)} and j ∈ ℤ such that 𝔭^{j}K_{𝔭}^{(1)} and 𝔭^{j}K_{𝔭}^{(2)} are primitive sublattices of L_{𝔭}^{#} and such that τ(x) − x ∈ L_{𝔭} for all x ∈ 𝔭^{j}K_{𝔭}^{(1)}.

Then one has θ_{𝔭}(X(L_{𝔭}/K_{𝔭}^{(1)})) = θ_{𝔭}(X(L_{𝔭}/K_{𝔭}^{(2)})).

In particular, for one dimensional K_{𝔭} = 𝔬_{F_{𝔭}}x as above with fixed ⟨x, x⟩ ∈ 2𝔬_{F_{𝔭}} the set θ_{𝔭}(X(L_{𝔭}/K_{𝔭})) depends only on ⟨K_{𝔭}, L_{𝔭}⟩ and on the class of x modulo (⟨K_{𝔭}, L_{𝔭}⟩)^{−1}L_{𝔭}.

Proof. Since θ_{𝔭}(X(L_{𝔭}/K_{𝔭})) is (if θ_{𝔭}(L_{𝔭}, K_{𝔭}) is non-empty) a coset of θ_{𝔭}(L_{𝔭}, K_{𝔭}) ⊇ θ_{𝔭}(SO(L_{𝔭})) this is an immediate consequence of Lemma 2.1. □

Remark. If 𝔭 does not divide the discriminant of L and K = 𝔬_{F} x is one dimensional, Proposition 2.2 implies the well known fact that θ_{𝔭}(X(L_{𝔭}/K_{𝔭})) depends only on ⟨x, x⟩ and ⟨K_{𝔭}, L_{𝔭}⟩ or equivalently on [K_{𝔭} : K_{𝔭} ∩ L_{𝔭}] (see [CX]). In the cases of 𝔭 dividing the discriminant of L discussed below we see that at least in the dyadic case it is sometimes necessary to use (part of) the additional information about the position of K_{𝔭} in the local discriminant group L_{𝔭}^{#}/L_{𝔭}.

§3. Nondyadic primes

In this section, we assume that 𝔭 is non-dyadic.

Proposition 3.1. Suppose θ_{𝔭}(L_{𝔭}, K_{𝔭}) = 𝔲_{𝔭} and 𝔫(K_{𝔭}) = 𝔫(L_{𝔭}) = 𝔬_{F_{𝔭}}. Write ⟨K_{𝔭}, L_{𝔭}⟩ = 𝔭^{k}. Then

θ_{𝔭}(X(L_{𝔭}/K_{𝔭})) = \{ \begin{matrix}θ_{𝔭}(L_{𝔭}, K_{𝔭}) & if k ≥ 0 \\ π^{k}θ_{𝔭}(L_{𝔭}, K_{𝔭}) & if k < 0. \end{matrix}

Proof. Without loss of generality, one can write

L = 𝔬_{F_{𝔭}}x ⊥ 𝔬_{F_{𝔭}}y ⊥ 𝔬_{F_{𝔭}}z with Q(x) = 1, Q(y) = ε_{1}π^{r_{1}} and Q(z) = ε_{2}π^{r_{2}}

and K = 𝔬_{F_{𝔭}}w with Q(w) = 1. The condition θ_{𝔭}(L_{𝔭}, K_{𝔭}) = 𝔲_{𝔭} implies that 0 ≤ r_{1} ≤ r_{2} and that both r_{1} and r_{2} are even. Let

w = αx + βy + γz with α, β and γ in F_{𝔭}.

i) k ≥ 0. Since θ_{𝔭}(L_{𝔭}, K_{𝔭}) = 𝔲_{𝔭}, one has ord(Q(v)) is even for any vector v in F_{𝔭}y ⊥ F_{𝔭}z and F_{𝔭}y ⊥ F_{𝔭}z is anisotropic. It is clear that

τ_{w−x}τ_{x} ∈ X(L_{𝔭}/K_{𝔭}) and Q(w − x) = 2(1 − α).

Since K_{𝔭} ⊂ L_{𝔭}^{♯}, one has ord(α) ≥ 0 and

1 − α^{2} = Q(βy + γz)

Then

ord(1 − α) = 0 or ord(1 − α) = ord(Q(βy + γz))

which is even. Therefore θ_{𝔭}(X(L_{𝔭}/K_{𝔭})) = θ_{𝔭}(L_{𝔭}, K_{𝔭}).

ii) k < 0. One can assume that F_{𝔭}y ⊥ F_{𝔭}z is anisotropic. It is clear that

k = min{ord(α), r_{1} + ord(β), r_{2} + ord(γ)}.

Suppose ord(α) > min{r_{1} + ord(β), r_{2} + ord(γ)}. Then one of ord(β), ord(γ) is negative and smaller than ord(α). This implies

min{ord(Q(βy)), ord(Q(γz))} < ord(1 − Q(αx)) = ord(Q(βy + γz)),

which is impossible since F_{𝔭}y + F_{𝔭}z is anisotropic. Therefore

ord(α) ≤ min{r_{1} + ord(β), r_{2} + ord(γ)} and ord(α) = k.

Since τ_{w−x}τ_{x} ∈ X(L_{𝔭}/K_{𝔭}) and

ord(Q(w − x)) = ord(2(1 − α)) = ord(α) = k,

the proof is complete. □

§4. Over ℤ_{2}

It becomes very complicated to study such a question over dyadic local fields. The simple result as in the previous section is no longer true over general dyadic local fields even if 2 is unramified. Therefore we restrict ourselves to F_{𝔭} = ℚ_{2} and 𝔬_{F_{𝔭}} = ℤ_{2}. As pointed out in previous sections, one only needs to consider 𝔫(K_{𝔭}) = 𝔫(L_{𝔭}) and θ_{𝔭}(L_{𝔭}, K_{𝔭}) = 𝔲_{𝔭}. According to the classification in [X], we only need to study Case II, III and IV in [X]. For convenience, we list them as follows by [X, Theorem 2.1], taking into account the additional restrictions imposed in these cases by the condition θ_{2}(L_{2}, K_{2}) = 𝔲_{2}.

Case II: L_{2} = ℤ_{2}x ⊥ ℤ_{2}y ⊥ ℤ_{2}z with

Q(x) = 1, Q(y) = 2^{r_{1}}γ and Q(z) = 2^{r_{2}}δ

where 0 < r_{1} < r_{2}, both r_{1} and r_{2} are even and γδ = −5.

Then K_{2} = ℤ_{2}w with Q(w) = 1.

Case III: L_{2} = ℤ_{2}x ⊥ (ℤ_{2}y + ℤ_{2}z) with

Q(x) = 1, Q(y) = 2^{r_{1}+r_{2}}γ, Q(z) = −2^{r_{2}−r_{1}+2}γ^{−1}, and ⟨y, z⟩ = 2^{r_{2}}

where r_{2} ≥ r_{1} ≥ 0, r_{2} > 0 and r_{1} ≡ r_{2} mod 2.

Then K_{2} = ℤ_{2}w with Q(w) = 1.

Case IV: If 𝔰(L_{2}) = 𝔫(L_{2}), L_{2} = ℤ_{2}x ⊥ ℤ_{2}y ⊥ ℤ_{2}z with

Q(x) = 1, Q(y) = γ and Q(z) = 2^{r_{2}}δ

where r_{2} ≥ 2 and is even, γδ = −5 and γ = −1 or −5.

Then K_{2} = ℤ_{2}w with Q(w) = 1.

If 2𝔰(L_{2}) = 𝔫(L_{2}), L_{2} = (ℤ_{2}x + ℤ_{2}y) ⊥ ℤ_{2}z with

Q(x) = 2, Q(y) = 2Γ, ⟨x, y⟩ = 1 and Q(z) = δ2^{r_{2}}

where r_{2} ≥ 3 and is odd, δ = 1 or 5 and

Γ = \{ \begin{matrix}−1 & if δ = 1 \\ 0 & if δ = 5. \end{matrix}

Then K_{2} = ℤ_{2}w with Q(w) = 2.

§5. Over ℤ_{2}, Rank of the first Jordan component equals one

In this section, we study the case that the first Jordan component of L_{2} is of rank one which corresponds to the cases II and III in §4.

Proposition 5.1. Suppose θ_{2}(L_{2}, K_{2}) = 𝔲_{2}, 𝔰(L_{2}) = ℤ_{2} and the first Jordan component of L_{2} is of rank one . Let ⟨K_{2}, L_{2}⟩ = 2^{k}ℤ_{2}.

If k > 0, then θ_{2}(X(L_{2}/K_{2})) = 2θ_{2}(L_{2}, K_{2}).

If k = 0, then

θ_{2}(X(L_{2}/K_{2})) = \{ \begin{matrix}2θ_{2}(L_{2}, K_{2}) & if (K_{2} + L_{2})/L_{2} is primitive in L_{2}^{♯}/L_{2}, \\ θ_{2}(L_{2}, K_{2}) & otherwise. \end{matrix}

If k < 0, then θ_{2}(X(L_{2}/K_{2})) = 2^{k+1}θ_{2}(L_{2}, K_{2}).

Proof. It is clear that one only needs to consider Case II and III in §4. We first consider Case II. Write w = ax + by + cz where a, b and c are in ℚ_{2}. Then

(5.2)  1 = a^{2} + b^{2}Q(y) + c^{2}Q(z)

and

(5.3)  min{ord(a), r_{1} + ord(b), r_{2} + ord(c)} = k

It is clear that τ_{w−x}τ_{x} ∈ X(L_{2}/K_{2}) and Q(w − x) = 2(1 − a).

For k > 0, the result follows from the same argument as in Proposition 3.1 i).

For k = 0, we have the following two cases.

If (K_{2} + L_{2})/L_{2} is primitive in L_{2}^{♯}/L_{2}, then ord(b) = −r_{1} or ord(c) = −r_{2}. This implies

ord(Q(by + cz)) ≤ −r_{1} + 2 ≤ 0,

since ℚ_{2}y + ℚ_{2}z is anisotropic. By (5.2) one has ord(Q(by + cz)) = 0, and a^{2} + Q(by + cz) = 1 implies ord(a) > 0 and hence (1 − a) ∈ ℤ_{2}^{×} (the fact that we are working over ℤ_{2} will be used in a similar way several times in this section).

Otherwise, ord(a) = 0 by (5.3). The result follows clearly for ord(1 − a) = 1. We can assume that ord(1 − a) > 1. Then ord(1 + a) = 1, and ord(1 − a^{2}) = ord(1 − a) + ord(1 + a) = ord(2(1 − a)) is even since ℚ_{2}y + ℚ_{2}z is anisotropic, and the result follows from (5.2).

For k < 0, we claim that ord(a) = k.

If ord(a) > r_{1} + ord(b), then

ord(a) ≥ 3 + ord(b) and ord(a^{2}) ≥ 4 + ord(Q(by)).

By (5.3)

min{ord(Q(by)), ord(Q(cz))} < 0.

Then

r_{1} + 2ord(b) = ord(Q(by)) = ord(Q(cz)) = r_{2} + 2ord(c)

by (5.2), and hence ord(b) > ord(c) and r_{2} + ord(c) > r_{1} + ord(b). Therefore we have

ord(b) + r_{1} = k < 0 and ord(Q(by)) = 2k − r_{1} ≤ −4.

Since ℚ_{2}y ⊥ ℚ_{2}z is anisotropic, one has

ord(Q(by) + Q(cz)) ≤ 2 + ord(Q(by)).

A contradiction is derived from (5.2). This implies that ord(a) ≤ r_{1} + ord(b) < r_{2} + ord(c) and the claim follows.

Case III follows from similar arguments as above and the domination principle in [R].

The proof is complete. □

Remark 5.4. The case that (K_{2} + L_{2})/L_{2} is primitive in L_{2}^{♯}/L_{2} can only happen for case II with r_{1} = 2 and Case III with r_{2} ≤ 2.

§6. Over ℤ_{2}, Rank of the first Jordan component equals two

In this section, we study the case that the first Jordan component of L_{2} is of rank two which corresponds to case IV in §4.

Proposition 6.1. Suppose θ_{2}(L_{2}, K_{2}) = 𝔲_{2}, 𝔰(L_{2}) = ℤ_{2} and the first Jordan component of L_{2} is of rank two. Let ⟨K_{2}, L_{2}⟩ = 2^{k}ℤ_{2}.

If k > 0, then θ_{2}(X(L_{2}/K_{2})) = 2θ_{2}(L_{2}, K_{2}).

If k = 0, then θ_{2}(X(L_{2}/K_{2})) = θ_{2}(L_{2}, K_{2}).

If k < 0 and 𝔫(L_{2}) = ℤ_{2}, then K_{2} is a sublattice of 2^{k−r}L and

θ_{2}(X(L_{2}/K_{2})) = \{ \begin{matrix}2^{k}θ_{2}(L_{2}, K_{2}) & if K_{2} is primitive in 2^{k−r}L \\ 2^{k+1}θ_{2}(L_{2}, K_{2}) & otherwise. \end{matrix}

where r = [\frac{1}{2}(ord(𝔫(L_{2}^{♯}))].

If k < 0 and 𝔫(L_{2}) = 2ℤ_{2}, then θ_{2}(X(L_{2}/K_{2})) = 2^{μ}θ_{2}(L_{2}, K_{2}) where

μ = \{ \begin{matrix}k & when the first Jordan component of L_{2} is hyperbolic \\ k + 1 & otherwise. \end{matrix}

Proof. It is clear that we only need to consider Case IV in §3. We first consider the case 𝔫(L_{2}) = ℤ_{2}. Write w = ax + by + cz. Then

(6.2)  min{ord(a), ord(b), r_{2} + ord(c)} = k

and

(6.3)  1 = a^{2} + γb^{2} + δ2^{r_{2}}c^{2}.

One also has τ_{w−x}τ_{x} ∈ X(L_{2}/K_{2}) and Q(w − x) = 2(1 − a).

For k > 0, the arguments in Proposition 3.1 i) are still valid.

For k = 0, we claim that ord(a) = 0. Otherwise

ord(b) = 0 and ord(δ2^{r_{2}}c^{2}) ≥ 2

by (6.2) and (6.3) (r_{2} +2ord(c) = 0 is impossible since then a^{2} +γb^{2} +δ2^{r_{2}}c^{2} would be even). A contradiction is derived by considering (6.3) modulo 4. Then the result follows from the same arguments as in Proposition 5.1 for k = 0 in the case ord(a) = 0.

For k < 0, (6.3) and (6.2) imply r_{2} + 2ord(c) ≥ min{0, 2ord(a), 2ord(b)} ≥ 2k, and hence (with r = \frac{r_{2}}{2} )

ord(δ2^{r_{2}}c^{2}) ≥ 2k and ord(c) ≥ k − r.

Therefore K is a sublattice of 2^{k−r}L and

min{ord(a), ord(b)} = k.

If K is primitive in 2^{k−r}L, then ord(c) = k − r and hence

ord(δ2^{r_{2}}c^{2}) = 2k = min{ord(a^{2}), ord(γb^{2})}.

ord(a) = k would (since δ = 1 or 5) imply ord(a^{2} + δ2^{r_{2}}c^{2}) = 2k + 1 ≠ ord(γb^{2}), and (6.3) leads to a contradiction for ord(b) = k as well as for ord(b) > k. We have therefore ord(a) > k, ord(b) = k and ord(γb^{2} + δ2^{r_{2}}c^{2}) = 2 + 2k since γδ = −5. Equation (6.3) implies then ord(a) > 0 for k = −1 and ord(a) = k + 1 for k < −1, and hence ord(Q(w − x)) = 1 + ord(1 − a) is congruent to k modulo 2 in both cases.

If K is not primitive in 2^{k−r}L we have ord(c) > k −r, and (6.3) implies ord(a) = ord(b) = k and hence

ord(Q(w − x)) = 1 + ord(1 − a) = k + 1.

Now we consider the case 𝔫(L_{2}) = 2ℤ_{2}. Write w = ax + by + cz. Then

(6.4)  min{ord(a), ord(b), ord(c) + r_{2}} = k

and

(6.5)  2 = 2a^{2} + 2ab + 2Γb^{2} + δ2^{r_{2}}c^{2}.

It is clear that τ_{x−w}τ_{x} ∈ X(L_{2}/K_{2}) and Q(w − x) = 2(2(1 − a) − b).

For k > 0, one has ord(a) > 0 and ord(b) > 0. We claim that ord(b) ≥ 2.

If δ = 1, one has

ord(a^{2} + ab − b^{2}) = min{2ord(a), 2ord(b)} ≥ 3

by (6.5). Therefore ord(b) ≥ 2.

If δ = 5, then ord(a^{2} + ab) = 2 by (6.5). This also implies that ord(b) ≥ 2, and we get ord(Q(x) + ord(Q(w − x)) = 3 as asserted in this case.

For k = 0, ord(b) = 0 implies ord(Q(w − x)) = 1 = ord(Q(x)) and hence θ_{2}(X(L_{2})/K_{2}) = θ(L_{2}, K_{2}). We can therefore assume ord(b) ≥ 1 which implies ord(a) = 0. Then

ord(δ2^{r_{2}}c^{2}) ≥ 3 and ord(b) ≥ 2

by (6.5). In fact, one can further assume that ord(2(1 − a) − b) > 2. Then

ord(2(1 − (a + 2^{−1}b))) = ord(1 − (a + 2^{−1}b)^{2}) = ord((2^{−1}b)^{2}(4Γ − 1) + δ2^{r_{2}−1}c^{2})

by (6.5), and this order is always even.

For k < 0, (6.5) and (6.4) imply ord(δ2^{r_{2}−1}c^{2}) ≥ min{2ord(a), 2ord(b)} ≥ 2k and hence

ord(δ2^{r_{2}−1}c^{2}) ≥ 2k and ord(c) ≥ k − \frac{1}{2}(r_{2} − 1)

by (6.4) and (6.5).

If the first Jordan component of L_{2} is not a hyperbolic plane, then

δ = 1 and ord(c) = k − \frac{1}{2}(r_{2} − 1)

by (6.4), (6.5) and the domination principle in [R]. We have

τ_{2^{− \frac{1}{2} (r_{2}−1)}z−w}τ_{x} ∈ X(L_{2}/K_{2}),

and

ord(Q(2^{− \frac{1}{2} (r_{2}−1)}z − w)) = ord(4(1 − 2^{\frac{1}{2} (r_{2}−1)}c)) = k + 2.

If the first Jordan component is a hyperbolic plane, then δ = 5 and Γ = 0.

If ord(c) = k − \frac{1}{2}(r_{2} − 1), one has ord(a) = ord(a + b) = k and

ord(a^{2} − δ2^{r_{2}−1}c^{2}) = 2k + 2.

by (6.4) and (6.5). Since

a(2a + b) = a^{2} + a(a + b) = a^{2} − δ2^{r_{2}−1}c^{2} + 1

by (6.5), one has

ord(2a + b) = \{ \begin{matrix}> 1 & if k = −1 \\ k + 2 & if k < −1. \end{matrix}

Therefore ord(Q(w − x)) = k + 3.

Otherwise ord(c) > k − \frac{1}{2}(r_{2} − 1). By (6.4) and (6.5), we have

ord(a) ≠ ord(a + b) and min{ord(a), ord(a + b)} = k.

Therefore ord(Q(w − x)) = k + 1. The proof is complete. □

§7. Proof of Example 1.2

Before we give a proof of Example 1.2, we recall some basic facts (see [Kn1, SP1, HSX]):

To a lattice K in F L of codimension 2 that is represented locally everywhere by L one associates the quadratic extension E = F (\sqrt{d}) of F where d is the discriminant of F K^{⊥} and puts N_{𝔭}(E) := N_{F_{𝔭}}^{E_{𝔓}}(E_{𝔓}^{×}) for any place 𝔭 of F with a place 𝔓 in E dividing 𝔭. The lattice K is then represented by all classes in the genus of L if and only if either d is a square or there is some 𝔭 such that θ_{𝔭}(L_{𝔭}, K_{𝔭}) ≠ N_{𝔭}(E) (notice that θ_{𝔭}(L_{𝔭}, K_{𝔭}) ⊇ N_{𝔭}(E) is true automatically). If this is not the case, K is represented by precisely half the classes in the genus of L; such K are called exceptional for gen(L).

If K is exceptional for gen(L) and σ = (σ_{𝔭})_{𝔭} is in the adelic special orthogonal group of F L such that K is represented by σL, then K is represented by L if and only if

(θ_{𝔭}(σ_{𝔭}))_{𝔭} ⊆ F^{×} · ∏_{𝔭} N_{𝔭}(E).

We have the following proposition which explains that the solubility of (1.1) for the case of codimension two depends only on the local computation of θ_{𝔭}(X(L_{𝔭}/K_{𝔭})). For any two lattices K and L, let

P (L, K) = {𝔭 prime : θ_{𝔭}(X(L_{𝔭}/K_{𝔭})) ≠ θ_{𝔭}(L_{𝔭}, K_{𝔭})}.

It is clear that | P (L, K) | is finite.

Proposition 7.1. If K is an exceptional lattice for gen(L) with codimension two, then K is represented by L if and only if | P (L, K) | is even.

Proof. As above let σ = (σ_{𝔭})_{𝔭} be in adelic special orthogonal group of F L such that K ⊆ σL. By Hilbert’s reciprocity law

(θ_{𝔭}(σ_{𝔭}))_{𝔭} ⊆ F^{×} · ∏_{𝔭} N_{𝔭}(E).

is true if and only if θ_{𝔭}(σ_{𝔭}) ⊈ N_{𝔭}(E) holds for an even number of places 𝔭, i. e., if and only if |P (L, K)| is even. □

Proof of Example 1.2. Let L = ℤe_{1} + ℤe_{2} + ℤe_{3} be a quadratic lattice such that

Q(xe_{1} + ye_{2} + ze_{3}) = m^{2}x^{2} + n^{2k}y^{2} − nz^{2}

and K = ℤ(m^{−1}e_{1}).

It is easily seen that 1 is represented by gen(L) if and only if (m, n) = 1. We have then K_{p} ⊆ L_{p} for all p ∤ m, and it is clear that ⟨L, K⟩ = mℤ.

We can conclude from Proposition 3.1 that θ_{𝔭}(X(L_{p}/K_{p})) = θ_{𝔭}(L_{p}, K_{p}) for all odd p | m, hence 1 is certainly represented by L if m is odd. If m is even (which we assume from now on) and 4 ∤ m we see from [SP1,X] that K is not exceptional if n is congruent to 3 or 7 modulo 8. In all other cases of even m, the lattice K is exceptional, and in these remaining cases 1 is represented by L if and only if θ_{2}(X(L_{2}/K_{2})) = θ_{2}(L_{2}, K_{2}) by Proposition 7.1.

If n ≡ 1 mod 8 there is nothing to check at p = 2 and the representability of 1 by L follows. If n ≡ 5 mod 8 Proposition 6.1 applies and yields θ_{2}(X(L_{2}/K_{2})) = 2θ_{2}(L_{2}, K_{2}), thus P (L, K) = {2} and 1 is not represented by L.

It remains to consider the case that m is even and ℚ_{2}(\sqrt{n})/ℚ_{2} is ramified, i.e., n ≡ 3 mod 8 or n ≡ 7 mod 8; by the argument above we need to discuss this only for 4 | m.

It is clear that

τ_{\frac{1}{n^{k}} e_{2}− \frac{1}{m} e_{1}}τ_{e_{2}} ∈ X(L_{2}/K_{2})

and

Q(\frac{1}{n^{k}} e_{2} − \frac{1}{m}e_{1}) = 2.

We have 2 ∉ θ_{2}(L_{2}, K_{2}) = N_{2}(ℚ(\sqrt{n})) for n ≡ 3 mod 8 and 2 ∈ θ_{2}(L_{2}, K_{2}) = N_{2}(ℚ(\sqrt{n})) for n ≡ 7 mod 8, which finishes the proof. □

§8. Regularity of indefinite ternary forms

In this section, we discuss the regularity of indefinite ternary forms in the sense of Dickson [Di].

Definition 8.1. A quadratic form is called regular if it represents all integers which are locally represented.

It is well known that there are only finitely many regular positive definite ternary quadratic forms by the results of Watson. Jagy, Kaplansky and Schiemann [JaKaSc] try to give a complete list of such positive definite ternary forms. For indefinite ternary forms, it is easy to see that there are infinitely many genera consisting only of one class and hence infinitely many regular forms. In [BH] an example was given of genera with arbitrary large class number such that all forms in the genus are regular. We consider here some examples of other types of behaviour. Let L be an indefinite lattice of rank 3.

Definition 8.2. ([BH]) A set of exceptional square classes {c_{1}, ⋯, c_{n}} for gen(L) is called independent if {−c_{1}det(L), ⋯, −c_{n}det(L)} are linearly independent vectors in the 𝔽_{2} vector space ℚ^{×}/(ℚ^{×})^{2} (or equivalently, if the fields ℚ(\sqrt{−c_{i}det(L)}) are linearly disjoint). It is called complete if in addition 2^{n} is the number of spinor genera in gen(L).

The following proposition is already known in [BH]. Here we give a proof which also provides a method of explicit construction.

Proposition 8.3. If the set of all exceptional square classes for gen(L) is independent, then there is a regular form in gen(L).

Proof. Let {c_{1}, ⋯, c_{n}} be all exceptional square classes for gen(L) and

Gal(ℚ(\sqrt{−c_{i}det(L)})/ℚ) = \{1, g_{i}\}

for 1 ≤ i ≤ n. Define

h_{i} = \{ \begin{matrix}1 & if L represents c_{i}, \\ g_{i} & otherwise. \end{matrix}

Since {−c_{1}det(L), ⋯, −c_{n}det(L)} are linearly independent, by Kummer theory and Cebotarev density Theorem, there is a prime l ∤ 2det(L) such that

l is unramified in Gal(ℚ(\sqrt{−c_{i}det(L)})/ℚ)

and the Frobenius of l is h_{i} for 1 ≤ i ≤ n. Then the l-neighbor of L in the sense of Kneser [Kn2,SP2] is in gen(L) and regular. □

Corollary 8.4. If gen(L) has only two exceptional square classes, then there is a regular form in gen(L).

Proof. It is clear that any two exceptional square classes are independent. □

Example 8.5. Let p, q and r be primes such that p ≡ 1 mod 8, q ≡ r ≡ 3 mod 8 and (\frac{q}{p}) = (\frac{r}{p}) = 1. Let

f(x, y, z) = x^{2} − pqry^{2} + p^{3}q^{2}r^{2}z^{2}

Then 1 and p are the only two exceptional square classes and f is regular.

Proof. It is clear that 1 and p are represented by gen(f) and are also exceptional integers for gen(f). In fact, they are the only exceptional square classes. Indeed, let a be an exceptional integer for gen(f). By [CX, (3.2)] at ∞, one has that a is positive. Write a = a_{0}b^{2} where both a_{0} and b are positive integers and a_{0} is square-free. Besides 1 and p, the other possible values of a_{0} are q, r, pq, pr and pqr.

Since q and r are symmetric, one can assume (\frac{r}{q}) = 1 and (\frac{q}{r}) = −1. By a simple computation of Hasse symbols, one has that a is not represented by f over ℚ_{q} if a_{0} = q, pq or pqr. For a_{0} = r or pr, the relative spinor norm is ℚ_{r}^{×} at r but r is not split in the corresponding quadratic field ℚ(\sqrt{q}) or ℚ(\sqrt{pq}) respectively. Then a is not exceptional.

In order to prove that f is regular, one only needs to show that p is represented by f. Equivalently, one only needs to show that the following equation

px^{2} − qry^{2} + p^{2}q^{2}r^{2}z^{2} = 1

is solvable over ℤ. Let L = ℤe_{1} ⊥ ℤe_{2} ⊥ ℤe_{3} be a quadratic lattice such that

Q(xe_{1} + ye_{2} + ze_{3}) = px^{2} − qry^{2} + p^{2}q^{2}r^{2}z^{2}

and K = ℤ\frac{1}{pqr} e_{3}. It is clear that the primes in P (L, K) in §7 are from p, q and r.

At p, one has

τ_{\sqrt{−qr}^{−1}e_{2}−(pqr)^{−1}e_{3}}τ_{e_{2}} ∈ X(L_{p}/K_{p})

and

Q(\sqrt{−qr}^{−1}e_{2} − (pqr)^{−1}e_{3}) = 2 ∈ (Q_{p}^{×})^{2}

Therefore p ∉ P (L, K).

At q, one has

τ_{\sqrt{p}^{−1}e_{1}−(pqr)^{−1}e_{3}}τ_{e_{1}} ∈ X(L_{q}/K_{q})

and

Q(\sqrt{p}^{−1}e_{1} − (pqr)^{−1}e_{3}) = 2 ∉ (ℤ_{q}^{×})^{2}

Therefore q ∈ P (L, K). Similarly one has r ∈ P (L, K). The result follows from Prop.7.1. □

If there are more than two exceptional square classes, the set of all exceptional square classes is not necessarily independent and there may be no regular forms in the given genus.

Remark 8.6. It should be pointed out that the term “regular spinor genus” in [BH] denotes a spinor genus which represents a maximal independent set of exceptional square classes but not necessarily all exceptional square classes.

In fact, we’ll construct a genus in the following steps with only three exceptional square classes which are not independent and with arbitrarily large class number but such that there is no regular form in the genus.

Step 1): Suppose p_{1}, ⋯, p_{n} are odd primes with p_{i} ≡ 1 mod 8 for 1 ≤ i ≤ n and (\frac{p_{i}}{p_{j}}) = 1 for i ≠ j. Let

f(x, y, z) = x^{2} − (p_{1} ⋯ p_{n})y^{2} + (p_{1} ⋯ p_{n})^{2}z^{2}.

It is clear that the class number of f is 2^{n} and there are 2^{n} − 1 exceptional square classes for gen(f) which are the classes of 1 and of the p_{i_{1}} ⋯ p_{i_{s}} for i_{1} < ⋯ < i_{s} with 1 ≤ s ≤ n − 1. The form f is equivalent to

f′(x, y, z) = (p_{1} ⋯ p_{k})^{2}x^{2} − (p_{1} ⋯ p_{n})y^{2} + (p_{k+1} ⋯ p_{n})^{2}z^{2}.

To see equivalence of these forms, one can write L = ℤe_{1} ⊥ ℤe_{2} ⊥ ℤe_{3} such that

Q(xe_{1} + ye_{2} + ze_{3}) = (p_{1} ⋯ p_{k})^{2}x^{2} − (p_{1} ⋯ p_{n})y^{2} + (p_{k+1} ⋯ p_{n})^{2}z^{2}.

Then we only need to show that L is equivalent to L′ = ℤ(p_{k}^{−1}e_{1}) ⊥ ℤe_{2} ⊥ ℤ(p_{k}e_{3}). It is clear that L_{p} = L′_{p} if p ≠ p_{k}. At p = p_{k}, one can check that

τ_{\frac{1}{p_{1}···p_{k}} e_{1}− \frac{1}{p_{k+1}···pn} e_{3}}(L_{p_{k}}) = L′_{p_{k}}

and

Q(\frac{1}{p_{1} ⋯ p_{k}} e_{1} − \frac{1}{p_{k+1} ⋯ p_{n}} e_{3}) = 2 ∈ (ℚ_{p_{k}}^{×})^{2}.

Step 2): To decide if f represents (p_{1} ⋯ p_{k}) is equivalent to determine if the following Diophantine equation

(p_{1} ⋯ p_{k})x^{2} − (p_{k+1} ⋯ p_{n})y^{2} + (p_{1} ⋯ p_{k})(p_{k+1} ⋯ p_{n})^{2}z^{2} = 1

is solvable over ℤ. Let L = ℤe_{1} ⊥ ℤe_{2} ⊥ ℤe_{3} such that

Q(xe_{1} + ye_{2} + ze_{3}) = (p_{1} ⋯ p_{k})x^{2} − (p_{k+1} ⋯ p_{n})y^{2} + (p_{1} ⋯ p_{k})(p_{k+1} ⋯ p_{n})^{2}z^{2}.

Write

p_{1} ⋯ p_{k} = m^{2} + n^{2}

and

K = ℤ(\frac{m}{p_{1} ⋯ p_{k}} e_{1} + \frac{n}{p_{1} ⋯ p_{n}} e_{3}).

Since p_{i} with 1 ≤ i ≤ k splits in ℚ(\sqrt{p_{k+1} ⋯ p_{n}}), P (L, K) in §7 contains only primes from {p_{k+1}, …, p_{n}}.

Step 3): For any p ∈ {p_{k+1}, …, p_{n}}, (p_{1} ⋯ p_{k}) is a square in ℚ_{p}. It is clear that

τ_{(\sqrt{p_{1}···p_{k}})^{−1}e_{1}−u}τ_{e_{1}} ∈ X(L_{p}/K_{p})

where u = \frac{m}{p_{1}···p_{k}} e_{1} + \frac{n}{p_{1}···p_{n}} e_{3}. Therefore p ∈ P (L, K) if and only if

Q((\sqrt{p_{1} ⋯ p_{k}})^{−1}e_{1} − u) = 2(1 − m(\sqrt{p_{1} ⋯ p_{k}})^{−1})

is a non-square unit or p times a non-square unit up to a square in ℚ_{p}.

Step 4): Let n = 2 and p_{1} = 17 and p_{2} = 89 (this is the first pair of primes satisfying the conditions above).

To see if f represents 17, one can write 17 = 1^{2} + 4^{2} and have

1 − \sqrt{17}^{−1} ≡ 1 − 33 ≡ −2^{5} mod 89.

Therefore P (L, K) is empty and f represents 17.

To see if f represents 89, one can write 89 = 8^{2} + 5^{2} and have

1 − 8(\sqrt{89})^{−1} ≡ 1 − 8 × 2^{−1} ≡ −3 mod 17.

Therefore P (L, K) = {17} and f does not represent 89.

By the Cebotarev density theorem, for any given class in gen(f) there are primes l ∤ 2 · 17 · 89 such that the l-neighbor of f is equivalent to this given class.

However, there are no primes l ∤ 2 · 17 · 89 such that Frobenius of l is trivial at Gal(ℚ(\sqrt{89})/ℚ) and Gal(ℚ(\sqrt{17 · 89})/ℚ) but non-trivial at Gal(ℚ(\sqrt{17})/ℚ). Hence no form in the genus of f can represent all of 1, 17, 89, i. e., there are no regular forms in gen(f).

Step 5): Based on the above construction, one can have

g(x, y, z) = x^{2} − 17 · 89 · (q_{1} ⋯ q_{m})^{2}y^{2} + (17 · 89)^{2}(q_{1} ⋯ q_{m})^{4}z^{2}

where q_{i} are odd primes with q_{i} ≡ 1 mod 8 and different from 17 and 89, and (\frac{17}{q_{i}}) = (\frac{89}{q_{i}}) = 1 for 1 ≤ i ≤ m and (\frac{q_{i}}{q_{j}}) = 1 for i ≠ j.

If there were an exceptional integer a that is divisible by one of the q_{i} it had to be divisible by q^{3} in order to be locally representable at q_{i}. But then it is seen from [SP1] that a can not be exceptional. All the exceptional squares for gen(g) are therefore 1, 17 and 89. By the same argument as above, g represents 1 and 17 but not 89 and there are no regular forms in gen(g). At the same time, the class number in gen(g) can be arbitarily large. □

Acknowledgement. The first author’s travel to the conference in Talca was supported by DFG. The second author is supported by Alexander von Humboldt Foundation in Universität des Saarlandes, the Hundred Talent Program of Chinese Academy of Sciences and Morningside Center of Mathematics.

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FR 6.1 Mathematik, Universität des Saarlandes, Postfach 151150, D-66041 Saarbrücken, Germany.

   E-mail address: schulzep@math.uni-sb.de

Academy of Mathematics and System Sciences, CAS, Beijing 100080, P.R.China.

   E-mail address: xufei@math.ac.cn