Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 38 (1997), No. 1, 33-72.

Discriminant Algebras and Adjoints of Quadratic Forms

Ottmar Loos

Institut für Mathematik, Universität Innsbruck A-6020 Innsbruck, Austria

Introduction

The discriminant algebra of a nonsingular quadratic form q is usually defined as C^{C_{0}}, the centralizer of the even Clifford algebra in the full Clifford algebra. In [5], Haag introduced the discriminant algebra 𝔇(q) of a possibly singular quadratic form on a finitely generated free module by working with generic forms and then applying specialization. His method depends on a choice of basis, and although the result is up to isomorphism independent of that choice, an extra effort is required to extend the definition to non-free finitely generated projective modules [9, (4.8.11)].

In this paper, we present a definition of 𝔇(q) in a natural basis-free way which avoids the Clifford algebra entirely. We sketch this definition for the most interesting case of a quadratic module M of even rank 2n over a commutative ring k, and assume first that M is free and q = ∑ a_{ij}ξ_{i}ξ_{j} with a (in general not symmetric) 2n × 2n-matrix A = (a_{ij}). The construction is based on the elementary observation that there exists a unique polynomial qdet(A) in the a_{ij} with integral coefficients, homogeneous of degree 2n, such that

4 qdet(A) = det(A + A^{T}) − (−1)^{n} det(A − A^{T}).

We call this the quarter determinant of A. An explicit formula for qdet A is provided (1.1). Then the discriminant algebra is isomorphic to the free quadratic algebra with generators 1 and z and the relation

z^{2} − Pf(A − A^{T})z + (−1)^{n+1} qdet(A).1 = 0,

where Pf denotes the Pfaffian. This gives a convenient way of computing the discriminant algebra in the free case.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}1997 Heldermann Verlag

In general, let L be the invertible module ⋀^{2n} M and consider bilinear forms f, g on M representing q in the sense that q(x) = f(x, x) = g(x, x) for all x ∈ M. We introduce an Amitsur 1-cocycle κ_{fg} on such forms with values in the dual of L, defined in terms of Pfaffian powers of f − f^{T} and f − g (1.5). The module underlying 𝔇(q) is presented by generators 1 and s_{f}(u) (u ∈ L), k-linear in u, and satisfying s_{f}(u) − s_{g}(u) = κ_{fg}(u).1 for all bilinear forms f, g representing q. As a k-module, 𝔇(q) is isomorphic to k.1 ⊕ L but not in a canonical way. There is a linear form τ_{f} and a symmetric bilinear form γ_{f} on L such that

τ_{f}(u) = Pf (f(x_{i}, x_{j}) − f(x_{j}, x_{i})), γ_{f}(u, u) = (−1)^{n+1} qdet (f(x_{i}, x_{j})),

for decomposable u = x_{1} ∧…∧x_{2n} ∈ L. The algebra structure on 𝔇(q) is then determined by

s_{f}(u) · s_{f}(v) − τ_{f}(u)s_{f}(v) + γ_{f}(u, v)1 = 0 (u, v ∈ L).

The τ_{f} and γ_{f} interact correctly with the cocycle, ensuring that the multiplication is independent of the choice of f (2.3). As a byproduct, we obtain an explicit formula for the Dickson homomorphism (2.4). The functor 𝔇 from quadratic modules to graded quadratic algebras is multiplicative in the sense that 𝔇(q_{1} ⊥ q_{2}) ≅ 𝔇(q_{1})□𝔇(q_{2}) (2.11) for a suitably defined product □ which generalizes the well-known product in the separable case [6, 9].

In section 3 we show that there is a natural homomorphism ψ_{q}: 𝔇(q) → C^{C_{0}} which is an isomorphism if q is nonsingular or semiregular, depending on whether M has even or odd rank (3.14). This makes use of Bourbaki’s λ-operations between Clifford algebras [2], for which we derive an explicit formula (3.2).

In section 4 we introduce, for a quadratic module (M, q) of odd rank 2n + 1, a unital quadratic form q^{♯} of rank 2n + 2 on a module 𝔄(q) which is constructed from the secondto-the-highest exterior power of M by the same method as 𝔇(q) in the even rank case. We call this the adjoint of q since it is similar to the adjoint of a matrix; e.g., the discriminants of q and q^{♯} are related by δ_{q^{♯}} ≅ δ_{q}^{⊗2n} (4.4). In 4.6 we establish a natural homomorphism between the tensor product of q with its discriminant and the space of trace zero elements of 𝔄(q). In the opposite direction, there is a natural homomorphism from the □-product of a unital quadratic form Q of even rank with its discriminant algebra to the adjoint 𝔄(Q^{0}) where Q^{0} denotes the restriction of −Q to the trace zero elements (4.9).

Just as 𝔇(q) maps naturally into the Clifford algebra so does 𝔄(q), and its image is contained in a Jordan subalgebra J which we call the Jordan center since it is defined by suitable centralizer conditions (5.1). In case q is nonsingular of even rank or semiregular of odd rank, we determine the structure of J completely (5.4, 5.5).

Throughout, k denotes an arbitrary commutative base ring. Tensor symbols without subscripts are understood over k. We mostly follow the notation and terminology of [9]. A quadratic form on a k-module M is always assumed to be of the type q(x) = f(x, x) for some bilinear form f, in which case we say that q is represented by f and write q = [f]. A quadratic module (M, q) is a finitely generated projective k-module M with quadratic form q. To simplify statements, it is usually assumed that M has constant rank.

1. Exterior and Pfaffian powers of bilinear forms

1.1. The half- and the quarter-determinant. Let A and B be m × m-matrices with coefficients in k. Since det A is a homogeneous polynomial of degree m in the coefficients of A, there is an expansion

det(A + B) = ∑_{j=0}^{m} Δ_{j}(A, B)  (1)

where Δ_{j}(A, B) is homogeneous of bidegree (m − j, j) in (A, B). Explicitly,

Δ_{j}(A, B) = ∑_{|J|=j} det C_{J}(A, B)  (2)

where, for a subset J ⊂ {1, …, m}, the i-th column of the matrix C_{J}(A, B) equals the i-th column of B if i ∈ J and the i-th column of A otherwise. Clearly Δ_{0}(A, B) = det A and Δ_{j}(A, B) = Δ_{m−j}(B, A). Since the determinant is invariant under transposition we have Δ_{j}(A^{T} , B^{T}) = Δ_{j}(A, B). Combining these two formulas yields

Δ_{j}(A, A^{T}) = Δ_{m−j}(A, A^{T}).  (3)

For odd m = 2n + 1 we define the half-determinant

hdet A = ∑_{i=0}^{n} Δ_{i}(A, A^{T}).  (4)

This is a homogeneous polynomial of degree m with integer coefficients in the entries of A which satisfies

2 hdet A = det(A + A^{T}),  (5)

an easy consequence of (1) and (3). Note also that hdet A = P_{m}(a_{ii}, a_{ij} + a_{ji}), where P_{m} is the polynomial defined in [9, IV, (3.1.2)] or [11, (1.6.1)].

Now let m = 2n be even and define the quarter-determinant by

qdet A = ∑_{i<n, i≢n(2)} Δ_{i}(A, A^{T}).  (6)

This is again a homogeneous polynomial of degree m with integer coefficients in the entries of A, and it satisfies

4 qdet A = det(A + A^{T}) − (−1)^{n} det(A − A^{T}).  (7)

Indeed, by (1) and (3),

det(A + A^{T}) − (−1)^{n} det(A − A^{T}) = ∑_{i=0}^{2n} Δ_{i}(A, A^{T}) − (−1)^{n}Δ_{i}(A, −A^{T})

= ∑_{i=0}^{2n}(1 − (−1)^{n+i})Δ_{i}(A, A^{T})

= 2 ∑_{i≢n(2)} Δ_{i}(A, A^{T})

= 4 ∑_{i<n, i≢n(2)} Δ_{i}(A, A^{T}).

1.2. Formulas for the half- and quarter-determinant. The half-determinant of a 1 × 1-matrix (a) is obviously a, and qdet A = det A for a 2 × 2-matrix follows immediately from the definition. Next, let A_{i} ∈ Mat_{r_{i}}(k). Then for r_{i} = 2n_{i} even,

qdet ( \begin{matrix}A_{1} & 0 \\ 0 & A_{2} \end{matrix} ) = 4 qdet(A_{1}) qdet(A_{2}) + (−1)^{n_{2}} qdet(A_{1}) det(A_{2} − A_{2}^{T})
+ (−1)^{n_{1}} qdet(A_{2}) det(A_{1} − A_{1}^{T}),  (1)

while for r_{i} both odd,

qdet ( \begin{matrix}A_{1} & 0 \\ 0 & A_{2} \end{matrix} ) = hdet(A_{1}) hdet(A_{2}).  (2)

If r_{1} is odd and r_{2} is even, we have

hdet ( \begin{matrix}A_{1} & 0 \\ 0 & A_{2} \end{matrix} ) = hdet(A_{1}) det(A_{2} + A_{2}^{T}).  (3)

Since the half- and quarter-determinant is a polynomial with integral coefficients it suffices to prove these formulas for the torsion-free polynomial ring in the indeterminate entries of the A_{i} with integer coefficients. Then they follow by straightforward computations from 1.1.5 and 1.1.7. The details are left to the reader. By induction, we obtain

qdet diag(( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) , …, ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} )) = 0,  (4)
hdet diag(1, ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) , …, ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} )) = (−1)^{n}.  (5)

We also note

qdet ( \begin{matrix}0 & B \\ 0 & 0 \end{matrix} ) = 0  (6)

for a 2n × 2n-matrix subdivided into 4 blocks of size n × n, since

4 qdet ( \begin{matrix}0 & B \\ 0 & 0 \end{matrix} ) = det ( \begin{matrix}0 & B \\ B^{T} & 0 \end{matrix} ) − (−1)^{n} det ( \begin{matrix}0 & B \\ −B^{T} & 0 \end{matrix} )

= det ( \begin{matrix}0 & B \\ B^{T} & 0 \end{matrix} ) − (−1)^{n} det ( \begin{matrix}0 & B \\ B^{T} & 0 \end{matrix} ) det ( \begin{matrix}−1_{n} & 0 \\ 0 & 1_{n} \end{matrix} )

= (1 − (−1)^{2n}) det ( \begin{matrix}0 & B \\ B^{T} & 0 \end{matrix} ) = 0.

1.3. Exterior powers of bilinear forms. Let M be a k-module and f: M × M → k a bilinear form, with transposed form f^{T} (x, y) = f(y, x). For m-tuples x = (x_{1}, …, x_{m}) and y = (y_{1}, …, y_{m}) of elements of M let f(x, y) be the m × m-matrix with entries f(x_{i}, y_{j}). Then we have f(x, y)^{T} = f^{T} (y, x). Recall that the m-th exterior power of f is the unique bilinear form ⋀^{m} f on ⋀^{m} M satisfying

⋀^{m} f(x_{1} ∧ … ∧ x_{m}, y_{1} ∧ … ∧ y_{m}) = det f(x, y).

Let g be a second bilinear form. With Δ_{j} as in 1.1, we claim that the expression

Δ_{j}(f(x, y), g(x, y))  (1)

is a multilinear and alternating function of x and also of y. Indeed, let t be an indeterminate. Then h = f + tg is a bilinear form on M with values in k[t]. Therefore, det h(x, y) = det(f(x, y) + tg(x, y)) is a multilinear and alternating function of x and y, and so are the coefficients of t^{j} which are given by (1). By the universal property of the m-th exterior power, there exists a unique bilinear form ⋀^{(m−j,j)}(f, g) on ⋀^{m} M such that

⋀^{(m−j,j)}(f, g)(x_{1} ∧ … ∧ x_{m}, y_{1} ∧ … ∧ y_{m}) = Δ_{j}(f(x, y), g(x, y)),  (2)

for all x_{i}, y_{j} ∈ M. Now 1.1.1 yields the expansion

⋀^{m}(f + g) = ∑_{j=0}^{m} ⋀^{(m−j,j)}(f, g).  (3)

Clearly ⋀^{(m,0)}(f, g) = ⋀^{m}(f) and ⋀^{(m−j,j)}(f, g) is homogeneous of degree m − j in f and of degree j in g. From the corresponding properties of the Δ_{j} we obtain

⋀^{(m−j,j)}(f, g) = ⋀^{(j,m−j)}(g, f), ( ⋀^{(m−j,j)}(f, g))^{T} = ⋀^{(m−j,j)}(f^{T} , g^{T}),

which implies

( ⋀^{(m−j,j)}(f, f^{T}))^{T} = ⋀^{(j,m−j)}(f, f^{T}).  (4)

Define the bilinear form δ_{f} = δ_{f}^{(m)} on ⋀^{m} M by

δ_{f}^{(m)} = \{ \begin{matrix}(−1)^{n} ∑_{i=0}^{n} ⋀^{(2n+1−i,i)}(f, f^{T}) & if m = 2n + 1, \\ (−1)^{n} ⋀^{2n}(f + f^{T}) & if m = 2n. \end{matrix}  (5)

Then for decomposable u = x_{1} ∧ … ∧ x_{m} we have, by (2) and 1.1.4,

δ_{f}^{(m)}(u, u) = \{ \begin{matrix}(−1)^{n} hdet f(x, x) & if m = 2n + 1, \\ (−1)^{n} det (f(x, x) + f(x, x)^{T} ) & if m = 2n. \end{matrix}  (6)

Also, (3) and (4) imply

(−1)^{n} ⋀^{2n+1}(f + f^{T}) = δ_{f}^{(2n+1)} + (δ_{f}^{(2n+1)})^{T} .  (7)

Now let m = 2n be even and define the bilinear form γ_{f} = γ_{f}^{(2n)} on ⋀^{2n} M by

γ_{f}^{(2n)} = (−1)^{n+1} ∑_{i<n, i≢n(2)} ⋀^{(2n−i,i)}(f, f^{T}).  (8)

For decomposable u = x_{1} ∧ … ∧ x_{2n} we have

γ_{f}(u, u) = (−1)^{n+1} qdet f(x, x).  (9)

Moreover, (4) shows

γ_{f} + γ_{f}^{T} = (−1)^{n+1} ∑_{i≢n(2)} ⋀^{(2n−i,i)}(f, f^{T}),  (10)

and from (3) and (10) we obtain by a similar calculation as for 1.1.7,

δ_{f}^{(2n)} = ⋀^{2n}(f − f^{T}) − 2(γ_{f} + γ_{f}^{T}).  (11)

1.4. Pfaffian powers of alternating bilinear forms. Let a be an alternating bilinear form on M and x = (x_{1}, …, x_{2n}) ∈ M^{2n}. Then the matrix a(x, x) is alternating, and from the definition of the Pfaffian of an alternating matrix [1, §5.2] it follows easily that Pf(a(x, x)) is an alternating and multilinear function of x. Hence there exists a unique linear form π_{n}(a) on ⋀^{2n} M, called the n-th Pfaffian power of a, such that

π_{n}(a)(x_{1} ∧ … ∧ x_{2n}) = Pf(a(x, x)),  (1)

for all x ∈ M^{2n}. Since the square of the Pfaffian is the determinant, (1) implies

(π_{n}(a)(u))^{2} = (⋀^{2n} a)(u, u)  (2)

for all decomposable u = x_{1} ∧ … ∧ x_{2n}. If M is finitely generated projective of rank 2n then (2) holds for all u. Indeed, after localizing we may assume M is free and then every element of ⋀^{2n} M is decomposable.

The Pfaffian of an alternating 2n × 2n-matrix is a homogeneous polynomial of degree n in the matrix coefficients. Hence, π_{n}(a) is homogeneous of degree n in a. For the same reason, if A and A′ are alternating 2n × 2n-matrices, there is an expansion

Pf(A + A′) = ∑_{j=0}^{n} P_{j}(A, A′),  (3)

where P_{j}(A, A′) is homogeneous of bidegree (n − j, j) in (A, A′). Now let a and a′ be alternating bilinear forms on M. Similar arguments as in 1.3 show that there exist unique linear forms π_{n−j,j}(a, a′) on ⋀^{2n} M, homogeneous of bidegree (n − j, j), called the mixed Pfaffian powers of a and a′, such that

π_{n−j,j}(a, a′)(x_{1} ∧ … ∧ x_{2n}) = P_{j}(a(x, x), a′(x, x)),  (4)

for all x ∈ M^{2n}. Let t be an indeterminate and define the “difference quotient” Π_{n}(t, a, a′) by

π_{n}(a + ta′) = π_{n}(a) + tΠ_{n}(t, a, a′).  (5)

Then (3) and (4) show

Π_{n}(t, a, a′) = ∑_{j=1}^{n} t^{j−1}π_{n−j,j}(a, a′).  (6)

1.5. Definition. Let M be a k-module. Two bilinear forms f and g are said to be equivalent, f ∼ g, if they represent the same quadratic form [f] = [g], i.e., if f − g is alternating. In this case, define linear forms τ_{f} and κ_{fg} on any even exterior power ⋀^{2n} M by

τ_{f} = τ_{f}^{(n)} = π_{n}(f − f^{T}),  (1)
κ_{fg} = κ_{fg}^{(n)} = Π_{n}(−2, f − f^{T} , f − g).  (2)

We will prove below that κ_{fg} is an Amitsur cocycle on bilinear forms with respect to the equivalence relation ∼.

A similitude φ: (M̃, q̃) → (M, q) with multiplier μ between modules with quadratic forms is a linear map φ: M̃ → M such that q̃(x) = μq(φ(x)), for all x ∈ M̃. (This way of writing the multiplier μ on the “wrong” side allows for non-units as multipliers.) If q is represented by f then it is clear that q̃ is represented by

φ^{∗}f = μf ∘ (φ × φ).  (3)

We extend φ to a linear map Φ: ⋀^{m} M̃ → ⋀^{m} M by

Φ(u) = μ^{[ \frac{m}{2} ]} ⋀^{m} φ(u).  (4)

1.6. Proposition. Let M be a k-module with quadratic form q, represented by bilinear forms f, g, h.

(a) κ satisfies the relations

κ_{fg} + κ_{gh} = κ_{fh},  (1)
τ_{f} − τ_{g} = 2κ_{fg}.  (2)

(b) Let u = x_{1} ∧ … ∧ x_{m} ∈ ⋀^{m} M be decomposable. Then

γ_{g}(u, u) = γ_{f}(u, u) − κ_{fg}(u)τ_{f}(u) + κ_{fg}(u)^{2} (m = 2n),  (3)
δ_{g}^{(m)}(u, u) = δ_{f}^{(m)}(u, u).  (4)

If M is finitely generated projective of rank m then (3) and (4) hold for all u ∈ ⋀^{m} M.

(c) Let φ: (M̃, q̃) → (M, q) be a similitude with multiplier μ. Then with the notations of 1.5,

κ_{φ^{∗}f,φ^{∗}g} = κ_{fg} ∘ Φ,  (5)
τ_{φ^{∗}f} = τ_{f} ∘ Φ,  γ_{φ^{∗}f} = γ_{f} ∘ (Φ × Φ) (m = 2n),  (6)
δ_{φ∗f}^{(m)} = \{ \begin{matrix}δ_{f}^{(m)} ∘ (Φ × Φ) & (m = 2n), \\ μ δ_{f}^{(m)} ∘ (Φ × Φ) & (m = 2n + 1). \end{matrix}  (7)

Proof. (a) Let a, a′, a′′ be alternating bilinear forms on M. Then the polynomials Π_{n} of 1.4 satisfy

Π_{n}(t, a, a′) + Π_{n}(t, a + ta′, a′′) = Π_{n}(t, a, a′ + a′′).  (8)

Indeed, we replace a′ by a′ + a′′ in 1.4.5 and obtain

π_{n}(a) + tΠ_{n}(t, a, a′ + a′′) = π_{n}((a + ta′) + ta′′)
= π_{n}(a + ta′) + tΠ_{n}(t, a + ta′, a′′)
= π_{n}(a) + t(Π_{n}(t, a, a′) + Π_{n}(t, a + ta′, a′′))

which implies (8). Now let a = f − f^{T}, a′ = f − g, a′′ = g − h and t = −2. Then a − 2a′ = f − f^{T} − 2(f − g) = 2g − (f + f^{T}) = 2g − (g + g^{T}) = g − g^{T}, and therefore (8) implies (1) by definition of κ_{f,g}. Formula (2) follows in the same way from 1.4.5 and 1.5.1.

(b) Formulas (3) and (4) are polynomial identities with integer coefficients between the f_{ij} = f(x_{i}, x_{j}) and g_{ij} = g(x_{i}, x_{j}). Hence, it suffices to prove them when the f_{ij} and g_{ij} are indeterminates, the base ring is R = ℤ[f_{ij}, g_{ij}] and M = R^{m} is free with basis x_{1}, …, x_{n}, and then apply specialization. Let m be even. Then δ_{f}(u, u) = δ_{g}(u, u) by 1.3.5, and 1.3.11 and 1.4.2 yield 4γ_{f}(u, u) = τ_{f}(u)^{2} − δ_{f}(u, u). Hence by (2), 4(γ_{g}(u, u) − γ_{f}(u, u)) = τ_{g}(u)^{2} − τ_{f}(u)^{2} = (τ_{f}(u) − 2κ_{fg}(u))^{2} − τ_{f}(u)^{2} = 4(κ_{fg}(u)^{2} − κ_{fg}(u)τ_{f}(u)), and since R has no 2-torsion, we have (3). Similarly, 1.3.7 shows for m = 2n + 1 that 2δ_{f}(u, u) = (−1)^{n} ⋀^{2n+1}(f + f^{T})(u, u) = 2δ_{g}(u, u), which implies (4). To prove the last statement of (b), let M be finitely generated projective of rank m. After a faithfully flat base extension we may assume M free. Then every element of ⋀^{m} M is decomposable, so that (3) and (4) hold for all u ∈ ⋀^{m} M.

Finally, (5) – (7) are easily seen to hold for decomposable elements and therefore by linearity in general. This completes the proof.

1.7. Discriminants. Let M be a finitely generated projective module of rank r over k. Then L = ⋀^{r} M has rank 1 which implies ⋀^{2} L = 0. Hence there are no nonzero alternating bilinear forms on L. As a consequence, every bilinear form on L is symmetric, and bilinear and quadratic forms on L are in bijection, since every quadratic form on L is obtained from some bilinear form, unique modulo alternating bilinear forms, by setting the arguments equal.

Now let q = [f] be a quadratic form on M, with polar form b = f + f^{T}. The (signed) discriminant of q is the bilinear form δ_{q} = δ_{f}^{(r)} on L, which by the above remarks and 1.6.4 is symmetric and independent of the choice of f representing q. For even r = 2n we note the formula

δ_{q} = τ_{f}^{⟨2⟩} − 4γ_{f}  (1)

which follows from 1.3.11 and 1.4.2 for a = f − f^{T} due to the symmetry of γ_{f}. Here τ_{f}^{⟨2⟩} denotes the bilinear form (u, v) ↦ τ_{f}(u)τ_{f}(v). In case of odd rank r = 2n + 1, we have

2δ_{q} = (−1)^{n} ⋀^{r} b  (2)

by 1.3.7 and the symmetry of δ_{f}^{(r)}.

1.8. Proposition. Let M_{i} be finitely generated projective k-modules of rank r_{i} with quadratic forms q_{i} = [f_{i}] = [g_{i}], and let q = q_{1} ⊥ q_{2} on M = M_{1} ⊕ M_{2}, represented by f = f_{1} ⊥ f_{2} and g = g_{1} ⊥ g_{2}. Also, let L_{i} = ⋀^{r_{i}} M_{i} and identify L = ⋀^{r_{1}+r_{2}} M with L_{1} ⊗ L_{2} in the canonical way.

(a) Let r_{1} and r_{2} be even. Then

τ_{f} = τ_{f_{1}} ⊗ τ_{f_{2}}, γ_{f} = γ_{f_{1}} ⊗ τ_{f_{2}}^{⟨2⟩} + τ_{f_{1}}^{⟨2⟩} ⊗ γ_{f_{2}} − 4γ_{f_{1}} ⊗ γ_{f_{2}},  (1)
κ_{fg} = κ_{f_{1}g_{1}} ⊗ τ_{f_{2}} + τ_{f_{1}} ⊗ κ_{f_{2}g_{2}} − 2κ_{f_{1}g_{1}} ⊗ κ_{f_{2}g_{2}}.  (2)

(b) If r_{1} + r_{2} is odd,

δ_{q} = δ_{q_{1}} ⊗ δ_{q_{2}}.  (3)

(c) Let r_{1} and r_{2} be odd. Then

τ_{f} = 0, γ_{f} = δ_{q_{1}} ⊗ δ_{q_{2}},  (4)
κ_{fg} = 0.  (5)

Proof. By localization, we may assume M_{i} free. Suppose r_{i} = 2n_{i} is even and let n = n_{1} + n_{2}. Then the well-known formula

Pf ( \begin{matrix}A_{1} & 0 \\ 0 & A_{2} \end{matrix} ) = Pf(A_{1}) Pf(A_{2}),  (6)

for alternating matrices A_{i} of order 2n_{i} implies

π_{n}(a_{1} ⊥ a_{2}) = π_{n_{1}}(a_{1}) ⊗ π_{n_{2}}(a_{2})  (7)

for alternating bilinear forms a_{i} on M_{i}. Setting a_{i} = f_{i} − f_{i}^{T} yields the first formula of (1), while the second formula follows from 1.3.9 and 1.2.1. Now replace a_{i} by a_{i} + ta′_{i} in (7). Then 1.4.5 implies by a simple calculation

\begin{matrix}Π_{n}(t, a_{1} ⊥ a_{2}, & a′_{1} ⊥ a′_{2}) = π_{n_{1}}(a_{1}) ⊗ Π_{n_{2}}(t, a_{2}, a′_{2}) \\ + Π_{n_{1}}(t, a_{1}, a′_{1}) ⊗ π_{n_{2}}(a_{2}) + t · Π_{n_{1}}(t, a_{1}, a′_{1}) ⊗ Π_{n_{2}}(t, a_{2}, a′_{2}), \end{matrix}

and (2) follows by specializing a_{i} = f_{i} − f_{i}^{T}, a′_{i} = f_{i} − g_{i}, t = −2.

Formula (3) follows from 1.2.3. Now suppose r_{1} and r_{2} odd and let again A_{i} be alternating matrices of order r_{i}. Then

Pf ( \begin{matrix}A_{1} & 0 \\ 0 & A_{2} \end{matrix} ) = 0.  (8)

Indeed, it suffices to prove this when the base ring is the polynomial ring in the indeterminate coefficients of the A_{i} over the integers. Then det(A_{i}) = 0, and therefore Pf diag(A_{1}, A_{2})^{2} = det(A_{1}) det(A_{2}) = 0. Since the base ring is now an integral domain, we have (8). Let r_{1} + r_{2} = 2m. Then (8) implies π_{m}(a_{1} ⊥ a_{2}) = 0 for alternating bilinear forms a_{i} on M_{i}. Applied to a_{i} = f_{i} − f_{i}^{T} we obtain the first formula of (4). Also, (8) yields, by a similar argument as above, that Π_{m}(t, a_{1} ⊥ a_{2}, a′_{1} ⊥ a′_{2}) = 0. For a′_{i} = f_{i} − g_{i} and t = −2 we have (5). Finally, the second formula of (4) is a consequence of 1.2.2 and the definition of the discriminants in 1.7.

2. Discriminant algebras

2.1. Quadratic algebras. Recall [6, 9] that a quadratic (i.e., rank 2) algebra S over k has a unique standard involution σ such that all norms xσ(x) and traces x + σ(x) are scalar multiples of 1. Conversely, if S is a rank 2 module with a quadratic form N and a distinguished element 1 such that N(1) = 1, then S carries a unique structure of a quadratic algebra such that 1 is the unit element and N is the norm. It is given by

xy = F (x, 1)y + xF (1, y) − F (x, y)1  (1)

where F is any bilinear form representing N [10, 1.6]. Denote the polar form of N by N(x, y) and let T (x) = N(x, 1) be the trace form. Then x+σ(x) = T (x)1 and every x ∈ S satisfies x^{2} −T (x)x+N(x)1 = 0. A graded quadratic algebra is a quadratic algebra S with a specified ℤ/2ℤ-grading S = S^{0} ⊕ S^{1} such that S^{i}S^{j} ⊂ S^{i+j}. Then 1 ∈ S^{0} and S^{0} ⊥ S^{1} with respect to the polar form of N. Conversely, it is easy to see that these conditions suffice to make the quadratic algebra constructed from N by (1) graded. We say S is even or odd if S^{1} has (constant) rank 0 or 1, respectively.

2.2. Let M be a k-module with quadratic form q = [f] and L = ⋀^{2n} M. We define E_{2n}(q) as the k-module generated by symbols 1 and s_{f}(u), subject to the relations

s_{f}(u) is k-linear in u,  (1)
s_{f}(u) − s_{g}(u) = κ_{fg}(u)1,  (2)

for all u ∈ L and all bilinear forms f, g representing q. Clearly s_{f}: L → E_{2n}(q) is a module homomorphism, and there is a well defined homomorphism p: E_{2n}(q) → L such that p(1) = 0 and p(s_{f}(u)) = u. Moreover, the cocycle relation 1.6.1 guarantees that there is a well-defined linear form ρ_{f}: E_{2n}(q) → k such that

ρ_{f}(1) = 1, ρ_{f}(s_{g}(u)) = −κ_{fg}(u).  (3)

Hence, the sequence

[figure: 0; k; E_{2n}(q); p; ⋀^{2n} M; 0]  (4)

is exact and split, every f representing q giving rise to a splitting determined by s_{f} and ρ_{f}.

2.3. Theorem. Let (M, q) be a quadratic module M of rank 2n.

(a) There is a unique structure of a quadratic algebra on 𝔇(q) = E_{2n}(q) having 1 as unit element and satisfying

s_{f}(u) · s_{f}(v) − τ_{f}(u)s_{f}(v) + γ_{f}(u, v)1 = 0,  (1)

for all bilinear forms f representing q and all u, v ∈ ⋀^{2n} M. Norm and trace of 𝔇(q) are given by

N(λ1 + s_{f}(u)) = λ^{2} + λτ_{f}(u) + γ_{f}(u, u),  (2)
T (λ1 + s_{f}(u)) = 2λ + τ_{f}(u).  (3)

(b) 𝔇(q) depends functorially on q with respect to similitudes: If φ: (M̃, q̃) → (M, q) is a similitude of quadratic modules of rank 2n with multiplier μ, then there exists a unique algebra homomorphism 𝔇(φ): 𝔇(q̃) → 𝔇(q) such that, with the notations of 1.5, s_{φ^{∗}f}(u) ↦ s_{f}(Φ(u)) for all u ∈ ⋀^{2n} M̃.

(c) 𝔇(q) commutes with change of base ring: If k̃ is a commutative k-algebra then there is a natural isomorphism 𝔇(q) ⊗ k̃ ≅ 𝔇(q ⊗ k̃).

(d) The discriminants of q and N are related by the formula

δ_{q}(u, v) = δ_{N}(1 ∧ s_{f}(u), 1 ∧ s_{f}(v)).  (4)

Proof. (a) Fix a bilinear form f representing q. By 2.2, the map (λ, u) ↦ λ1 + s_{f}(u) is a module isomorphism between k × ⋀^{2n} M and 𝔇(q). Hence the right hand side of (2) defines a quadratic form N_{f} on 𝔇(q). If g ∼ f then by (2) and (3) of 1.6,

N_{f}(λ1 + s_{g}(u)) = N_{f}((λ − κ_{fg}(u))1 + s_{f}(u))
= λ^{2} − 2λκ_{fg}(u) + κ_{fg}(u)^{2} + (λ − κ_{fg}(u))τ_{f}(u) + γ_{f}(u, u)
= λ^{2} + λτ_{g}(u) + γ_{g}(u, u) = N_{g}(λ1 + s_{g}(u)),  (5)

whence N = N_{f} is independent of the choice of f. The bilinear form F defined by F (1, 1) = 1, F (s_{f}(u), 1) = τ_{f}(u), F (1, s_{f}(u)) = 0, F (s_{f}(u), s_{f}(v)) = γ_{f}(u, v) represents N. Now it follows from 2.1.1 that the quadratic algebra determined by N is given by (1), and (3) is an immediate consequence of (2).

(b) By 2.2 it is clear that there exists a unique module homomorphism h: 𝔇(q̃) → 𝔇(q) as indicated. We show that h is independent of the choice of f. Indeed, if g also represents q then by 1.6.5 and 2.2.2,

\begin{aligned}h(s_{φ^{∗}g}(u)) & = h(s_{φ^{∗}f}(u) − κ_{φ^{∗}f,φ^{∗}g}(u)1) = s_{f}(Φ(u)) − κ_{φ^{∗}f,φ^{∗}g}(u)1 \\ & = s_{f}(Φ(u)) − κ_{fg}(Φ(u))1 = s_{g}(Φ(u)), \end{aligned}

proving our assertion. To show that h is an algebra homomorphism, it suffices to verify that h preserves norms, which is an easy consequence of 1.6.6.

(c) This follows from the fact that the exterior powers and the linear and bilinear forms τ_{f}, γ_{f}, and κ_{fg} commute in the obvious sense with change of base ring.

(d) By (2) and the remarks at the beginning of 1.7 the polar form N(x, y) of N satisfies N(s_{f}(u), s_{f}(v)) = γ_{f}(u, v) + γ_{f}(v, u) = 2γ_{f}(u, v). Now we use 1.7.1:

\begin{aligned}δ_{N}(1 ∧ s_{f}(u), 1 ∧ s_{f}(v)) & = − det ( \begin{matrix}N(1, 1) & N(1, s_{f}(v)) \\ N(1, s_{f}(u)) & N(s_{f}(u), s_{f}(v)) \end{matrix} ) \\ & = − det ( \begin{matrix}2 & τ_{f}(v) \\ τ_{f}(u) & 2γ_{f}(u, v) \end{matrix} ) = δ_{q}(u, v). \end{aligned}

2.4. Corollary. Let (M, q) be a quadratic module of rank 2n.

(a) Let φ ∈ GO(q), the group of similitudes of q = [f], with multiplier μ ∈ k^{×}. Then with the notation of 1.5,

𝔇(φ)(s_{f}(u)) = κ_{f,φ^{∗}f}(u)1 + μ^{n} det(φ)s_{f}(u).  (1)

(b) Let q be nonsingular and suppose that q admits a representative f such that f − f^{T} is nonsingular. Then τ_{f}: ⋀^{2n} M → k is an isomorphism, and the Dickson homomorphism dick : GO(q) → Z_{2}(k) is given by the formula

dick(φ) = κ_{f,φ^{∗}f}(e)  (2)

where τ_{f}(e) = 1.

Proof. (a) We have Φ(u) = μ^{n} det(φ)u, for u ∈ ⋀^{2n} M. Hence, by 2.2.2 and 2.3(c),

\begin{aligned}𝔇(φ)(s_{f}(u)) & = 𝔇(φ)(κ_{f,φ^{∗}f}(u)1 + s_{φ^{∗}f}(u)) \\ & = κ_{f,φ^{∗}f}(u)1 + μ^{n} det(φ)s_{f}(u). \end{aligned}

(b) When M is free and f − f^{T} = \binom{0 1_{n}}{−1_{n} 0}, it is clear that τ_{f} = π_{n}(f − f^{T}) is an isomorphism between ⋀^{2n} M and k. The general case follows by descent and [9, I, §4]. By 2.3.4, q is nonsingular if and only if 𝔇(q) is separable. It is well-known that the automorphism group of a separable quadratic algebra with conjugation σ is isomorphic to Z_{2}(k), the group of idempotents of k with multiplication ε • ε′ = ε + ε′ − 2εε′, under the map ε ↦ α(ε) = (1 − ε)Id + εσ. The Dickson homomorphism is dick(φ) = α^{−1}(𝔇(φ)) (this is consistent with the usual definition in terms of the center of the even Clifford algebra by Theorem 3.14(c)). By 2.3.3, σ(s_{f}(u)) = τ_{f}(u)1 − s_{f}(u). Hence α(ε)(s_{f}(e)) = ετ_{f}(e)1 + (1 − 2ε)s_{f}(e). Now (2) follows from this and (1) by comparison.

We remark that representatives of the type required in (b) obviously exist for the split form on k^{2n} and also if k is local or 2 = 0 in k, but not in general since ⋀^{2n} M need not be free.

2.5. Definition. The discriminant algebra of a quadratic module (M, q) of even rank r = 2n is the even (trivially graded) quadratic algebra 𝔇(q) of Theorem 2.3. If M has odd rank r = 2n+1, the discriminant algebra is the odd quadratic algebra 𝔇(q) = 𝔇^{0}(q)⊕𝔇^{1}(q) defined by

𝔇^{0}(q) = k.1, 𝔇^{1}(q) = ⋀^{r} M, uv = δ_{q}(u, v)1  (1)

for u, v ∈ 𝔇^{1}(q), where δ_{q} is the discriminant of q as in 1.7. Hence norm, trace and standard involution are given by

N(u) = −δ_{q}(u, u), T (u) = 0, σ(u) = −u (u ∈ 𝔇^{1}(q)).  (2)

If M does not have constant rank, then k = k_{1} × … × k_{s} decomposes as a direct product of rings such that (M_{i}, q_{i}) = (M, q) ⊗ k_{i} has constant rank r_{i}. Then 𝔇(q) is the graded quadratic k-algebra 𝔇(q_{1}) × … × 𝔇(q_{s}). From 2.3(b) and 1.6(c) it follows that 𝔇(q) is a functor from quadratic modules of the same rank to quadratic graded algebras, and it is easy to see that it commutes with change of base ring.

2.6. Example. Let us compute 𝔇(q) for a hyperbolic module M = P ⊕ P^{∗} where P is finitely generated and projective of rank n, with the bilinear form f(x⊕α, y⊕β) = β(x), and the quadratic form q = [f]. We claim that γ_{f} = γ_{f}^{(2n)} = 0 and τ_{f} = τ_{f}^{(2n)}: L = ⋀^{2n} M → k is an isomorphism. Indeed, by localization we may assume P free, say with basis x_{1}, …, x_{n}. Let α_{i} = x_{n+i} (i = 1, …, n) be the dual basis of P^{∗} and x = (x_{1}, …, x_{2n}). Then

f(x, x) = ( \begin{matrix}0 & 1_{n} \\ 0 & 0 \end{matrix} ). Since u = x_{1} ∧ … ∧ x_{2n} is a basis of L, it follows from 1.2.6 and 1.3.9

that γ_{f}(u, u) = 0 and hence γ_{f} = 0, and by 1.4.1 and 1.5.1, τ_{f}(u) = Pf ( \begin{matrix}0 & 1_{n} \\ −1_{n} & 0 \end{matrix} ) = 1.

Thus if we identify k with L via τ_{f} and L with s_{f}(L), then 2.3.1 shows that 𝔇(q) ≅ k × k is a split quadratic algebra.

2.7. Free quadratic modules. Let M be free with basis e_{1}, …, e_{r} and let A = (f(e_{i}, e_{j})) be the matrix of a bilinear form f representing q. Then 𝔇(q) is isomorphic to the free quadratic algebra with generators 1 and z and the relation

z^{2} = \{ \begin{matrix}Pf(A − A^{T})z + (−1)^{n} qdet(A) & if r = 2n is even, \\ (−1)^{n} hdet(A) & if r = 2n + 1 is odd. \end{matrix}  (1)

Indeed, ⋀^{r} M is free with basis e = e_{1} ∧ … ∧ e_{r}. In the even rank case, set z = s_{f}(e). Then the assertion follows from 2.3.1 and the fact that τ_{f}(e) = Pf(A − A^{T}) and γ_{f}(e, e) = (−1)^{n+1} qdet(A) by 1.4.1, 1.5.1 and 1.3.9. For odd r, we let z = e and use 1.3.6 and 2.5.1 above.

We now show that 𝔇(q) is isomorphic to Haag’s discriminant algebra. Let R = ℤ[a_{i}, b_{ij}] where the a_{i} and b_{ij} (1 ≤ i < j ≤ r) are indeterminates. The generic quadratic form of rank r is q_{r} = ∑ a_{i}ξ_{i}^{2} + ∑_{i<j} b_{ij}ξ_{i}ξ_{j}. Define the matrix A = (a_{ij}) by

a_{ij} = \{ \begin{matrix}b_{ij} & if i < j, \\ a_{i} & if i = j, \\ 0 & if i > j. \end{matrix}

Haag’s discriminant algebra D(q_{r}) [5, Satz 1] is the free quadratic algebra R ⊕ Rz_{r} with the relation z_{r}^{2} = bz_{r} + c, where b = Pf(A − A^{T}) and b^{2} + 4c = (−1)^{n} det(A + A^{T}) if r = 2n, while b = 0 and c = (−1)^{n} hdet(A) in case r = 2n + 1. Let f be the bilinear form on R^{r} determined by f(e_{i}, e_{j}) = a_{ij}. Then f represents q_{r} and thus D(q_{2n+1}) ≅ 𝔇(q_{2n+1}) is clear from (1). In the even rank case, we have 4c = 4(−1)^{n} qdet(A) by 1.1.7, and since R has no 2-torsion, c = (−1)^{n} qdet(A), whence D(q_{2n}) ≅ 𝔇(q_{2n}) by (1). For an arbitrary free quadratic module, Haag defines D(q) by specialization. Since our discriminant algebra commutes with change of base ring, it follows again that the definitions are consistent.

2.8. The product of quadratic algebras. Let S_{i} be graded quadratic algebras with standard involutions σ_{i} and S_{1} ˆ⊗S_{2} their graded tensor product. If the algebras are separable, their product S_{1} □ S_{2} is the fixed point set of σ_{1} ˆ⊗σ_{2} [9, III, (6.2)]. The discriminant algebras of quadratic modules are in general not separable, so we need to extend this definition to the nonseparable case. Trivial examples show that in general the fixed point set of σ_{1} ˆ⊗σ_{2} is not even a rank 2 module; a different definition is therefore required. By decomposing the spectrum of the base ring, it suffices to consider the case where the odd part S_{i}^{1} has rank 0 or 1, i.e., where S_{i} is even or odd.

Let first both S_{1} and S_{2} be even, and let N_{i} and T_{i} be their norm and trace forms. Then the product S_{1} □ S_{2} is the even quadratic algebra determined as in 2.1 from the quadratic form N_{1} □ N_{2} constructed in [10, Theorem 2.4]. We quickly recall this construction and refer to [10, section 2] for details and proofs. Consider the exact sequences

[figure: 0; k; S_{i}; can; Ṡ_{i}; 0]  (1)

where Ṡ_{i} = S_{i}/k.1. A linear form α_{i} on S_{i} is called unital if α_{i}(1) = 1, i.e., if it determines a splitting of (1). We denote the corresponding section of can by s_{α_{i}}, so that Ker α_{i} = Im s_{α_{i}}. For unital linear forms α_{i} and β_{i} the formula c_{α_{i}β_{i}}(ẋ) = −α_{i}(x) + β_{i}(x) defines a linear form on Ṡ_{i}. Consider the linear and quadratic forms t_{α_{i}} and q_{α_{i}} on Ṡ_{i} given by

t_{α_{i}} = T_{i} ∘ s_{α_{i}}, q_{α_{i}} = N_{i} ∘ s_{α_{i}}.  (2)

For pairs α = (α_{1}, α_{2}) and β = (β_{1}, β_{2}) of unital linear forms, the linear form c_{αβ} on Ṡ_{1} ⊗ Ṡ_{2} defined by

c_{αβ} = c_{α_{1}β_{1}} ⊗ t_{α_{2}} + t_{α_{1}} ⊗ c_{α_{2}β_{2}} − 2c_{α_{1}β_{1}} ⊗ c_{α_{2}β_{2}}  (3)

satisfies the cocycle relation c_{αβ} + c_{βγ} = c_{αγ}. The module underlying S_{1} □ S_{2} is presented by generators 1 and u_{1} □_{α} u_{2} and the relations saying that u_{1} □_{α} u_{2} is bilinear in u_{1} and u_{2}, and

u_{1} □_{α} u_{2} − u_{1} □_{β} u_{2} = c_{αβ}(u_{1} ⊗ u_{2})1,  (4)

for all u_{i} ∈ Ṡ_{i} and all pairs of unital linear forms α, β. Similar to the situation in 2.2, there is a split exact sequence

[figure: 0; k; S_{1} □ S_{2}; p; Ṡ_{1} ⊗ Ṡ_{2}; 0]  (5)

where p(1) = 0 and p(u_{1} □_{α} u_{2}) = u_{1} ⊗ u_{2}, and any α yields a section s_{α} of p by u_{1} ⊗ u_{2} ↦ u_{1} □_{α} u_{2}. Now define linear and quadratic forms t_{α} and q_{α} on Ṡ_{1} ⊗ Ṡ_{2} by

t_{α} = t_{α_{1}} ⊗ t_{α_{2}}, q_{α} = q_{α_{1}} ⊗ t_{α_{2}}^{⟨2⟩} + t_{α_{1}}^{⟨2⟩} ⊗ q_{α_{2}} − 2q_{α_{1}} ⊗ q_{α_{2}}.  (6)

Then the norm form N = N_{1} □ N_{2} and its trace T are given by

N(λ1 + s_{α}(u)) = λ^{2} + λt_{α}(u) + q_{α}(u),
T (λ1 + s_{α}(u)) = 2λ + t_{α}(u).  (7)

In view of 2.1.1 this suffices to determine the algebra S_{1} □ S_{2}.

If S is any quadratic algebra, we denote by d_{S} the bilinear form on Ṡ = S/k.1 given by

d_{S}(ẋ, ẏ) = \{ \begin{matrix}T (x)T (y) − 2N(x, y) & if S is even, \\ ẋẏ & if S is odd. \end{matrix}  (8)

Here ẋ = x + k.1 ∈ Ṡ and we identify Ṡ canonically with S^{1} if S is odd. Note that in the even case d_{S} corresponds to the discriminant of N under the isomorphism Ṡ ≅ ⋀^{2} S, ẋ ↦ 1 ∧ x. On the other hand, an odd algebra S is entirely determined by d_{S} since its norm form is ⟨1⟩ ⊥ [−d_{S}].

Now we define the product of quadratic algebras S_{1} and S_{2} in case one of the factors is odd by S_{1} □ S_{2} = k.1 ⊕ (Ṡ_{1} ⊗ Ṡ_{2}), with the norm

N = N_{1} □ N_{2} = ⟨1⟩ ⊥ (−1)^{1+deg S_{1} deg S_{2}}[d_{S_{1}} ⊗ d_{S_{2}}],  (9)

and let the grading be even or odd depending on whether both or only one of the factors is odd. By 2.1.1, the multiplication in S_{1} □ S_{2} is

(ẋ_{1} ⊗ ẋ_{2})(ẏ_{1} ⊗ ẏ_{2}) = \{ \begin{matrix}(ẋ_{1}ẏ_{1})d_{S_{2}}(ẋ_{2}, ẏ_{2}) & if S_{1} is odd and S_{2} is even, \\ −(ẋ_{1}ẏ_{1})(ẋ_{2}ẏ_{2}) & if both S_{1} and S_{2} are odd. \end{matrix}  (10)

2.9. Proposition. Let S_{1} and S_{2} be graded quadratic algebras. There is a natural homomorphism η: S_{1} □ S_{2} → Fix(σ_{1} ˆ⊗ σ_{2}) ⊂ S_{1} ˆ⊗ S_{2} of unital graded algebras, given as follows.

(a) If S_{1} and S_{2} are both even then, with the notation of 2.8,

η(u_{1} □_{α} u_{2}) = s_{α_{1}}(u_{1}) ˆ⊗ t_{α_{2}}(u_{2}) + t_{α_{1}}(u_{1}) ˆ⊗ s_{α_{2}}(u_{2}) − 2s_{α_{1}}(u_{1}) ˆ⊗ s_{α_{2}}(u_{2}).  (1)

(b) If S_{1} is odd and S_{2} is even,

η(ẋ_{1} ⊗ ẋ_{2}) = ẋ_{1} ˆ⊗ (t_{α_{2}}(ẋ_{2}) − 2s_{α_{2}}(ẋ_{2})).  (2)

(c) If both S_{1} and S_{2} are odd,

η(ẋ_{1} ⊗ ẋ_{2}) = ẋ_{1} ˆ⊗ ẋ_{2}.  (3)

Proof. (a) By [10, Prop. 2.7(a)], there is a well-defined k-linear map η, independent of the choice of the α_{i}, satisfying η(1) = 1 ˆ⊗ 1 and (1). A tedious but straightforward computation shows that η is an algebra homomorphism. Noting that η(u_{1} □_{α} u_{2}) = (Id ˆ⊗ σ_{2} + σ_{1} ˆ⊗ Id)(s_{α_{1}}(u_{1}) ˆ⊗ s_{α_{2}}(u_{2})), we see that η takes values in the fixed point set of σ_{1} ˆ⊗ σ_{2}.

(b) If β_{2} is another unital linear form on S_{2} then by [10, 1.4], s_{α_{2}} − s_{β_{2}} = c_{α_{2}β_{2}} and t_{α_{2}} −t_{β_{2}} = 2c_{α_{2}β_{2}}, whence the right hand side of (2) is independent of the choice of α_{2} and therefore η is well-defined. A computation shows that η is an algebra homomorphism and similarly as before, η(ẋ_{1} ⊗ ẋ_{2}) = (Id ˆ⊗ σ_{2} + σ_{1} ˆ⊗ Id)(ẋ_{1} ˆ⊗ s_{α_{2}}(ẋ_{2})), so that η takes values in Fix(σ_{1} ˆ⊗ σ_{2}).

(c) This is immediate from the definitions.

We now show that the product □ is naturally associative and commutative and admits the even algebra E = k × k as neutral element, more precisely:

2.10. Proposition. There are natural isomorphisms

(S_{1} □ S_{2}) □ S_{3} ≅ S_{1} □ (S_{2} □ S_{3}),  (1)
S_{1} □ S_{2} ≅ S_{2} □ S_{1},  (2)
S □ E ≅ S ≅ E □ S,  (3)

satisfying the compatibility axioms of [8] and thus making the category of graded quadratic algebras a symmetric tensor category.

Proof. By the usual decomposition of the base ring, it suffices to assume that the S_{i} are either even or odd. When all S_{i} are even the proposition is a special case of [10, Theorem 2.6(a)], so we assume that at least one factor is odd. For odd S we have S□E = k.1⊕(Ṡ⊗Ė) with norm ⟨1⟩ ⊥ [d_{S} ⊗ d_{E}]. Let e = (1, 0) ∈ E. Then (Ė, d_{E}) ≅ (k, ⟨1⟩) by mapping ė ↦ 1 [10, 1.2], and this induces the first of the isomorphisms (3), the second being analogous. The symmetries (2) are defined by u_{1} □_{(α_{1},α_{2})} u_{2} ↦ u_{2} □_{(α_{2},α_{1})} u_{1} in case both S_{1} and S_{2} are even [10, 2.6.10], and by

u_{1} ⊗ u_{2} ↦ (−1)^{deg u_{1} deg u_{2}}u_{2} ⊗ u_{1}  (4)

when one factor is odd.

Next, we establish associativity. Setting d_{i} = d_{S_{i}}, we have the formula

d_{S_{i}□S_{j}} ≅ \{ \begin{matrix}d_{i} ⊗ d_{j} & if S_{i} or S_{j} is even, \\ −4d_{i} ⊗ d_{j} & if both S_{i} and S_{j} are odd. \end{matrix}  (5)

When both factors are even this is [10, Theorem 2.6(b)], otherwise it follows from 2.8.8 and 2.8.9. By listing all possibilities, one sees that the required isomorphism (1) is a consequence of (5) and the natural associativity of the tensor product of bilinear forms, (d_{1} ⊗ d_{2}) ⊗ d_{3} ≅ d_{1} ⊗ (d_{2} ⊗ d_{3}), except when two adjacent factors are odd and the third is even. Let, say, S_{1} and S_{2} be odd and S_{3} be even. By definition, S = S_{1} □ S_{2} is the even algebra determined by the quadratic form ⟨1⟩ ⊥ [d_{1} ⊗ d_{2}] on k.1 ⊕ (Ṡ_{1} ⊗ Ṡ_{2}). Let α_{12} be the unital linear form on S with kernel Ṡ_{1} ⊗ Ṡ_{2}. Then with the notations of 2.8, we have

t_{α_{12}} = 0, q_{α_{12}} = [d_{1} ⊗ d_{2}].  (6)

Now let α_{3} be any unital linear form on S_{3}, and set α = (α_{12}, α_{3}). Define φ: S □ S_{3} → S_{1} □ (S_{2} □ S_{3}) = k.1 ⊕ Ṡ_{1} ⊗ (Ṡ_{2} ⊗ Ṡ_{3}) by 1 ↦ 1 and (u_{1} ⊗ u_{2}) □_{α} u_{3} ↦ u_{1} ⊗ (u_{2} ⊗ u_{3}). Then φ is well-defined, independent of the choice of α_{3}. (Note that there is no ambiguity in the choice of α_{12} since by construction S_{1} □S_{2} comes equipped with the canonical unital linear form α_{12}). Indeed, if β_{3} is another unital linear form on S_{3} and β = (α_{12}, β_{3}) then by 2.8.3, c_{αβ} = 0 because t_{α_{12}} = c_{α_{12}α_{12}} = 0. Now 2.8.4 shows that φ is well-defined, and it is clearly an isomorphism of k-modules. It remains to show that it is a homomorphism of algebras, equivalently, that it preserves norms and traces. By 2.8.6 and 2.8.7 and (6) above, T ((u_{1} ⊗ u_{2}) □_{α} u_{3}) = t_{α_{12}}(u_{1} ⊗ u_{2})t_{α_{3}}(u_{3}) = 0, while T (u_{1} ⊗ (u_{2} ⊗ u_{3})) = 0 by 2.8.9. Similarly,

\begin{aligned}N((u_{1} ⊗ u_{2}) □_{α} u_{3}) & = q_{α_{12}}(u_{1} ⊗ u_{2})t_{α_{3}}(u_{3})^{2} − 4q_{α_{12}}(u_{1} ⊗ u_{2})q_{α_{3}}(u_{3}) \\ & = d_{1}(u_{1}, u_{1})d_{2}(u_{2}, u_{2})(t_{α_{3}}(u_{3})^{2} − 4q_{α_{3}}(u_{3})) \end{aligned}

while N(u_{1} ⊗ (u_{2} ⊗ u_{3})) = d_{1}(u_{1}, u_{1})d_{2}(u_{2}, u_{2})d_{3}(u_{3}, u_{3}). Since

d_{3}(u_{3}, u_{3}) = T_{3}(s_{α_{3}}(u_{3}))^{2} − 4N_{3}(s_{α_{3}}(u_{3})) = t_{α_{3}}(u_{3})^{2} − 4q_{α_{3}}(u_{3})

by 2.8.2, we have equality. This establishes the isomorphisms (1), and a tedious but straightforward verification shows that they satisfy the compatibility conditions of [8] defining a symmetric tensor category.

2.11. Theorem. The discriminant algebra functor 𝔇 is multiplicative, more precisely: There are natural isomorphisms

ϑ : 𝔇(q_{1}) □ 𝔇(q_{2}) → 𝔇(q_{1} ⊥ q_{2}),  (1)
ϑ_{0} : E → 𝔇(0),  (2)

which make 𝔇 a symmetric tensor functor from quadratic modules (with ⊥) to graded quadratic algebras (with □).

Proof. By the usual localization argument, it suffices to establish (1) when the (M_{i}, q_{i}) are quadratic modules of constant rank r_{i}. We use the notations of 1.8 and distinguish the following cases.

(a) r_{1} and r_{2} are even. Let α_{i} = ρ_{f_{i}}: 𝔇(q_{i}) → k be the unital linear forms defined in 2.2.3, set α = (α_{1}, α_{2}) and define ϑ by

ϑ(1) = 1, ϑ(u_{1} □_{α} u_{2}) = s_{f}(u_{1} ⊗ u_{2}) (u_{i} ∈ L_{i}).  (3)

Then ϑ is independent of the choice of representatives f_{i} of q_{i}. Indeed, let β_{i} and β be analogously constructed from g_{i}. Since s_{f_{i}} = s_{α_{i}}: L_{i} → 𝔇(q_{i}) and therefore

t_{α_{i}} = τ_{f_{i}}, q_{α_{i}} = [γ_{f_{i}}],  (4)

we have c_{α_{i}β_{i}}(u_{i}) = −α_{i}(s_{f_{i}}(u_{i})) + β_{i}(s_{f_{i}}(u_{i})) = κ_{f_{i}g_{i}}(u_{i}) by 2.8 and 2.2.3. Hence 2.8.3 and 1.8.2 imply κ_{fg} = c_{αβ}. Now 2.8.4 and 2.2.2 show that ϑ is independent of the choice of f_{i}, and it is clearly an isomorphism of k-modules. Further, (4), 1.8.1 and 2.8.6 show that it preserves norms and traces and is therefore an algebra isomorphism. Observe here that (q_{α_{1}} ⊗ q_{α_{2}})(u_{1} ⊗ u_{2}) = 2q_{α_{1}}(u_{1})q_{α_{2}}(u_{2}) by definition of the tensor product of quadratic forms and therefore 2q_{α_{1}} ⊗ q_{α_{2}} = 4[γ_{f_{1}} ⊗ γ_{f_{2}}].

(b) r_{1} + r_{2} is odd. First note that

d_{𝔇(q_{i})} = δ_{q_{i}}  (5)

which follows from 1.7.1 and 2.5.1. Now 1.8.3 shows that ϑ = Id is an isomorphism.

(c) r_{1} and r_{2} are odd. Then (5) shows 𝔇(q_{1}) □ 𝔇(q_{2}) = k.1 ⊕ (L_{1} ⊗ L_{2}) with norm ⟨1⟩ ⊥ [δ_{q_{1}} ⊗ δ_{q_{2}}]. Define ϑ by

ϑ(1) = 1, ϑ(u_{1} ⊗ u_{2}) = s_{f}(u_{1} ⊗ u_{2}).

Then 1.8.5 shows ϑ is independent of the choice of f_{i}, and by 1.8.4 it preserves norms and traces. In all three cases it is easy to see that ϑ depends functorially on q_{1} and q_{2}.

For M = {0}, q = 0 we have ⋀^{0} M = k and f = 0 is the unique bilinear form representing q. Define ϑ_{0} by ϑ_{0}((1, 0)) = s_{f}(1), and ϑ_{0}((1, 1)) = 1. Then ϑ_{0} is a module isomorphism. Since the Pfaffian and the quarter-determinant of an empty matrix are 1 and 0, respectively, we have τ_{f}(1) = 1 and γ_{f} = 0, whence ϑ_{0} is an algebra isomorphism.

To complete the proof, we must show that 𝔇 is compatible with the associativity, commutativity and unity constraints of 2.10. Let us check this for commutativity, leaving the other cases to the persistent reader. We must show that the diagram

𝔇(q_{1}) □ 𝔇(q_{2})  \overline{}^{ϑ} 𝔇(q_{1} ⊥ q_{2})
[figure: 𝔇(ω)]
𝔇(q_{2}) □ 𝔇(q_{1})  \overline{}_{ϑ} 𝔇(q_{2} ⊥ q_{1})

commutes, where the left vertical arrow is as in 2.10.2 and the right one is induced from the switch ω: M_{1} ⊕ M_{2} → M_{2} ⊕ M_{1}. Let x_{1}^{(i)}, …, … x_{r_{i}}^{(i)} ∈ M_{i} and u_{i} = x_{1}^{(i)} ∧ … ∧ x_{r_{i}}^{(i)} ∈ L_{i} = ⋀^{r_{i}} M_{i}. Let r = r_{1} + r_{2} and identify L_{1} ⊗ L_{2} with ⋀^{r}(M_{1} ⊕ M_{2}) via u_{1} ⊗ u_{2} ↦ x_{1}^{(1)} ∧ … ∧ x_{r_{1}}^{(1)} ∧ x_{1}^{(2)} ∧ … ∧ x_{r_{2}}^{(2)} and similarly L_{2} ⊗ L_{1} with ⋀^{r}(M_{2} ⊕ M_{1}) via u_{2} ⊗ u_{1} ↦ x_{1}^{(2)} ∧…∧x_{r_{2}}^{(2)} ∧x_{1}^{(1)} ∧…∧x_{r_{1}}^{(1)}. Then the map ⋀^{r} ω is given by u_{1} ⊗u_{2} ↦ (−1)^{r_{1}r_{2}}u_{2} ⊗u_{1}. Now it is easy to check the commutativity of the diagram from the definitions. Note that the sign occurring in 2.10.4 is essential for this.

3. Mapping the discriminant algebra into the Clifford algebra

3.1. Bourbaki’s λ-operation. Let M be a k-module, q a quadratic form on M and C(q) the Clifford algebra of q. Since we assume that q is represented by some bilinear form (cf. 1.5), the canonical map M → C(q) is injective and we may identify M with its image in C(q). Now let f be a bilinear form on M and consider the quadratic form q′ = q + [f]. By [1, §9.3, Prop. 3] there exists a unique isomorphism

λ_{f}: C(q′) → C(q)  (1)

of k-modules, compatible with the ℤ/2ℤ-gradings and the filtrations, such that

λ_{f}(1) = 1,  λ_{f}(xu) = xλ_{f}(u) + i_{x}^{f}(λ_{f}(u)),  (2)

for all x ∈ M, u ∈ C(q′). Here i_{x}^{f} denotes the graded derivation of odd degree of C(q) determined by

i_{x}^{f}(1) = 0,  i_{x}^{f}(y) = f(x, y),  (3)

for all x, y ∈ M [1, §9.2, Lemma 1]. We also recall that λ_{f} depends exponentially on f in the sense that

λ_{0} = Id,  λ_{f+g} = λ_{f} ∘ λ_{g} : C(q′′) → C(q),  (4)

for a second bilinear form g and q′′(x) = q′(x) + g(x, x). Furthermore, λ_{f} is compatible with orthogonal sums: If q′_{i} = q_{i} + [f_{i}] then the diagram

[figure: C(q′_{1}) ˆ⊗ C(q′_{2}); λ_{f_{1}} ˆ⊗ λ_{f_{2}}; C(q_{1}) ˆ⊗ C(q_{2}); C(q′_{1} ⊥ q′_{2}); λ_{f_{1}⊥f_{2}}; C(q_{1} ⊥ q_{2})]  (5)

is commutative, where the vertical arrows are the well-known isomorphisms [9, IV, (1.3.1)]. This is easily proved from the definitions.

We derive next an explicit formula for λ_{f}. Let I = {1, …, m}, let x_{1}, …, x_{m} ∈ M and let A = (a_{ij}) be the alternating m × m matrix satisfying a_{ij} = f(x_{i}, x_{j}) for i < j. For any subset J = {j_{1}, …, j_{r}} of I, j_{1} < … < j_{r}, let x_{J} = x_{j_{1}} … x_{j_{r}} and

A_{J} = (a_{ij})_{(i,j)∈J×J}.

Finally, if J and L are subsets of I we set ε(J, L) = 0 if J and L are not disjoint, and (−1)^{ν} otherwise, where ν is the number of pairs (j, l) ∈ J × L such that j > l.

3.2. Proposition. With the above notations, λ_{f} is given by the formula

λ_{f}(x_{I}) = ∑_{K⊂I, |K| even} ε(K, I−K) Pf(A_{K}) x_{I−K}.

The proof is by induction on the cardinality |I| of I, the case I = ∅ being trivial. For the induction step, suppose the formula holds for I = {1, …, m} and let I′ = {0} ∪ I. Also set w = λ_{f}(x_{I}) and write i_{x} = i_{x}^{f} for simplicity. Then

λ_{f}(x_{I′}) = λ_{f}(x_{0}x_{I}) = x_{0}w + i_{x_{0}}w
= ∑_{K⊂I, |K| even} ε(K, I−K) Pf(A_{K}) x_{0}x_{I−K} + i_{x_{0}}w.  (1)

Here ε(K, I−K) = ε(K, I′−K). Indeed, the ε(J, L) are the structure constants of the exterior algebra of the free ℤ-module with basis e_{0}, …, e_{m}, i.e., e_{J}∧e_{L} = ε(J, L)e_{J∪L} [2, III, §7.8, (20)]. Since |K| is even, e_{K} is central in this exterior algebra. Hence ε(K, I−K)e_{I′} = ε(K, I−K)e_{0} ∧ e_{I} = e_{0} ∧ e_{K} ∧ e_{I−K} = e_{K} ∧ e_{0} ∧ e_{I−K} = e_{K} ∧ e_{I′−K} = ε(K, I′−K)e_{I′}. Now (1) becomes

λ_{f}(x_{I′}) = ∑_{0∉K⊂I′, |K| even} ε(K, I′−K) Pf(A_{K}) x_{I′−K} + i_{x_{0}}w,

and it remains to show

i_{x_{0}}w = ∑_{0∈K⊂I′, |K| even} ε(K, I′−K) Pf(A_{K})x_{I′−K}.  (2)

From the definition of i_{x_{0}} it is easy to see that

i_{x_{0}}x_{J} = ∑_{j∈J} ε({j}, J−{j}) a_{0j} x_{J−{j}}.

Hence the induction hypothesis yields

i_{x_{0}}w = ∑_{K⊂I, |K| even} ε(K, I−K) Pf(A_{K}) ∑_{j∈I−K} ε({j}, I−(K ∪ {j})) a_{0j} x_{I−(K∪{j})}.

When K runs over all even subsets of I and j ∈ I−K then J = K ∪ {j} runs over all odd subsets of I. Hence by rearranging the double sum, we obtain

i_{x_{0}}w = ∑_{J⊂I, |J| odd} α_{J}x_{I−J},  (3)

where

α_{J} = ∑_{j∈J} ε(J−{j}, I−(J−{j}))ε({j}, I−J)a_{0j} Pf(A_{J−{j}}).  (4)

We claim that

ε(J−{j}, I−(J−{j}))ε({j}, I−J) = ε(J, I−J)ε({j}, J−{j}).  (5)

Indeed, interpreting the ε(J, L) as before, we have

e_{j} ∧ e_{J−{j}} ∧ e_{I−J} = ε({j}, J−{j})e_{J} ∧ e_{I−J} = ε(J, I−J)ε({j}, J−{j})e_{I}.

On the other hand, e_{J−{j}} is an element of even degree and therefore

\begin{aligned}e_{j} ∧ e_{J−\{j\}} ∧ e_{I−J} = e_{J−\{j\}} ∧ e_{j} ∧ e_{I−J} & = ε(\{j\}, I−J)e_{J−\{j\}} ∧ e_{I−(J−\{j\})} \\ & = ε(J−\{j\}, I−(J−\{j\}))ε(\{j\}, I−J)e_{I}, \end{aligned}

which proves (5). From (4) and (5) and the well-known expansion formula for the Pfaffian we obtain

α_{J} = ε(J, I−J) ∑_{j∈J} ε({j}, J−{j}) a_{0j} Pf(A_{J−{j}}) = ε(J, I−J) Pf(A_{{0}∪J}).  (6)

Let J′ = {0} ∪ J. Then I′−J′ = I−J and one sees easily that ε(J′, I′−J′) = ε(J, I−J). By substituting (6) in (3) we obtain (2) and the proof is complete.

3.3. Corollary. Let a be an alternating bilinear form on M. Then the map λ_{a}: ⋀ M → ⋀ M satisfies

λ_{a}(x_{I}) = ∑_{K⊂I, |K| even} ε(K, I−K)π_{|K|/2}(a)(x_{K}) x_{I−K}.

Indeed, since a is alternating, the matrix A introduced in 3.1 equals a(x, x), and Pf(A_{K}) = π_{|K|/2}(a)(x_{K}) by 1.4.1.

3.4. Let q be a quadratic form on a k-module M, represented by a bilinear form f, let t be an indeterminate and R = k[t, t^{−1}] the Laurent polynomial ring. We denote the scalar extensions of f and q to M ⊗ R by the same letters. Then f[t] = f − (1 + t)f^{T} = (f − f^{T}) − tf^{T} is a bilinear form on M ⊗ R, and

f{t} = t^{−1}f[t] = −f^{T} + t^{−1}(f − f^{T})

represents −q. Hence by 3.1, λ_{f{t}}: ⋀ M ⊗R → C(q)⊗R is an isomorphism of R-modules. Now define k-linear maps ζ_{2n}(t, f): ⋀^{2n} M → C_{0}(q) ⊗ R and ζ_{2n+1}(t, f): ⋀^{2n+1} M → C_{1}(q) ⊗ R by the formulas

τ_{f}(u)1 + tζ_{2n}(t, f)(u) = λ_{f{t}}(t^{n}u),  (1)
ζ_{2n+1}(t, f)(v) = λ_{f{t}}(t^{n}v),  (2)

where τ_{f} is as in 1.5.1. To derive explicit formulas, let x ∈ M^{m} and let A(t) = (a_{ij}(t)) be the alternating m × m-matrix such that a_{ij}(t) = f(x_{i}, x_{j}) − (1 + t)f(x_{j}, x_{i}) for i < j. Then A(0) = (f − f^{T})(x, x) and A(t) = A(0) + tȦ(0), where Ȧ(0) = (ȧ_{ij}(0)) is the alternating matrix satisfying ȧ_{ij}(0) = f(x_{i}, x_{j}) for i > j. With the notations and conventions introduced in 3.1 and 1.4, we then have 3.5. Lemma. Let I = {1, …, 2n} and I′ = {0, …, 2n}. Then

ζ_{2n}(t, f)(x_{1} ∧ … ∧x_{2n}) =
∑_{K⫋I, |K| even}t^{n− \frac{|K|}{2} −1} ε(K, I−K) Pf(A(t)_{K})x_{I−K}

+ ∑_{i=1}^{n} P_{i}(A(0), Ȧ(0))t^{i−1},  (1)
ζ_{2n+1}(t, f)(x_{0} ∧ … ∧x_{2n}) =
∑_{K⊂I′, |K| even}t^{n− \frac{|K|}{2}} ε(K, I′−K) Pf(A(t)_{K})x_{I′−K}.  (2)

In particular, ζ_{m}(t, f)(w) ∈ C(q) ⊗ k[t] is a polynomial in t, for all w ∈ ⋀^{m} M.

Proof. Since the Pfaffian of an alternating 2r × 2r-matrix is a homogeneous polynomial of degree r, we have Pf(t^{−1}A(t)_{K}) = t^{− \frac{|K|}{2}} Pf(A(t)_{K}). Proposition 3.2 implies

λ_{f\{t\}}(t^{n}x_{1}∧ … ∧ x_{2n}) = ∑_{K⊂I, |K| even} ε(K, I−K)t^{n− \frac{|K|}{2}} Pf(A(t)_{K})x_{I−K}

= t( ∑_{K⫋I, |K| even} t^{n− \frac{|K|}{2} −1} ε(K, I−K) Pf(A(t)_{K})x_{I−K}) + Pf(A(t)).

By 1.4,

Pf A(t) = Pf (A(0) + tȦ(0))

= Pf A(0) + t ∑_{i=1}^{n} P_{i}(A(0), Ȧ(0))t^{i−1}

= τ_{f}(x_{1} ∧ … ∧ x_{2n}) + t ∑_{i=1}^{n} P_{i}(A(0), Ȧ(0))t^{i−1}.

Now (1) follows from the definition of ζ_{2n}. The proof of (2) is similar, and the last statement is a consequence of (1) and (2).

Remark. It should be noted that the two terms on the right hand side of (1) are not alternating functions of x_{1}, …, x_{2n} but their sum is.

3.6. Definition. By the lemma, we can specialize t → −2 in ζ_{m}(t, f) and thus obtain a well-defined map

z_{f} = z_{f}^{(m)} = ζ_{m}(−2, f): ⋀^{m} M → C(q),

with values in C_{0}(q) or C_{1}(q), depending on whether m is even or odd. Note that a_{ij}(−2) = −a_{ji}(−2) = b(x_{i}, x_{j}) for i < j where b = f +f^{T} = f[−2] is the polar form of q. Hence 3.5.2 shows that z_{f}^{(m)} is independent of the choice of f representing q for m odd (but not so in the even case). Therefore, we will sometimes write z_{q} instead of z_{f}^{(2n+1)}: ⋀^{2n+1} M → C_{1}(q). In particular, the map z_{q}: M → C_{1}(q) is simply the canonical embedding of M into C(q), whereas z_{f}: ⋀^{2} M → C_{0}(q) is given by

z_{f}(x_{1} ∧ x_{2}) = x_{1}x_{2} − f(x_{2}, x_{1})1.  (1)

3.7. Proposition. Let M = M_{1} ⊕ M_{2} be an orthogonal sum of submodules and suppose the representative f = f_{1} ⊥ f_{2} of q decomposes accordingly. Let u_{i} ∈ L_{i} = ⋀^{m_{i}} M_{i} and identify L_{i} with a submodule of ⋀^{m_{i}} M. Then for m_{i} = 2n_{i} even,

z_{f}(u_{1} ∧ u_{2}) = z_{f}(u_{1})τ_{f}(u_{2}) + τ_{f}(u_{1})z_{f}(u_{2}) − 2z_{f}(u_{1})z_{f}(u_{2}).  (1)

If m_{1} is odd and m_{2} is even,

z_{f}(u_{1} ∧ u_{2}) = z_{f}(u_{1})(τ_{f}(u_{2})1 − 2z_{f}(u_{2})).  (2)

If m_{1} and m_{2} are both odd,

z_{f}(u_{1} ∧ u_{2}) = z_{f}(u_{1})z_{f}(u_{2}).  (3)

Proof. We only show (1) since the proof of the other formulas is similar. Clearly f{t} = f_{1}{t} ⊥ f_{2}{t} decomposes orthogonally. By 3.1.5 we have, identifying C(q) with C([f_{1}]) ˆ⊗ C([f_{2}]) and setting n = n_{1} + n_{2},

λ_{f{t}}(t^{n}u_{1} ∧ u_{2}) = λ_{f_{1}{t}}(t^{n_{1}}u_{1}) ˆ⊗ λ_{f_{2}{t}}(t^{n_{2}}u_{2}) = λ_{f{t}}(t^{n_{1}}u_{1})λ_{f{t}}(t^{n_{2}}u_{2}).

By substituting 3.4.1, expanding and comparing the coefficient of t, we obtain

\begin{aligned}ζ_{2n}(t, f)(u_{1} ∧ u_{2}) = τ_{f}(u_{1})ζ_{2n_{2}}(t, f)(u_{2}) & + ζ_{2n_{1}}(t, f)(u_{1})τ_{f}(u_{2}) \\ & + tζ_{2n_{1}}(t, f)(u_{1})ζ_{2n_{2}}(t, f)(u_{2}). \end{aligned}

Now (1) follows for t → −2.

3.8. Example. Let q = ξ_{1}ξ_{2} + … + ξ_{2n−1}ξ_{2n} be the standard form on k^{2n} with basis e_{1}, …, e_{2n}. Then f = f_{1} ⊥ … ⊥ f_{n} represents q where f_{i} is the bilinear form with matrix \binom{0 1}{0 0} on k^{2}. The elements c_{j} = e_{2j−1}e_{2j} are pairwise commuting idempotents of C(q) and 3.6.1. shows z_{f}(e_{2j−1}∧e_{2j}) = c_{j}. By 3.7.1 we have z_{f}(e_{1}∧…∧e_{4}) = c_{1}+c_{2}−2c_{1}c_{2} = c_{1}•c_{2}, the well-known product of commuting idempotents which is again an idempotent. Hence induction yields

z_{f}(e_{1} ∧ … ∧ e_{2n}) = c_{1} • … • c_{n} = c,  c^{2} = c.  (1)

Now let q′ = ξ_{0}^{2} ⊥ q on k^{2n+1} with basis e_{0}, …, e_{2n}, represented by f′ = ⟨1⟩ ⊥ f, and set e = e_{1} ∧ … ∧ e_{2n} and e′ = e_{0} ∧ e. By 1.5.1, τ_{f′}(e) = Pf diag(\binom{0 1}{−1 0}, …, \binom{0 1}{−1 0}) = 1, so that 3.7.2 implies

z_{f′}(e′) = z_{f′}(e_{0})(1 − 2z_{f′}(e)) = e_{0}(1 − 2c).  (2)

As e_{0}^{2} = 1 and e_{0} commutes with the c_{j} and hence with c, we obtain

z_{f′}(e′)^{2} = e_{0}^{2}(1 − 4c + 4c^{2}) = 1.  (3)

3.9. Proposition. Let q be a quadratic form on M, represented by bilinear forms f and g. Then for x ∈ M, u ∈ ⋀^{2n} M, v ∈ ⋀^{2n+1} M the following relations hold:

z_{f}(u) − z_{g}(u) = κ_{fg}(u)1,  (1)
z_{q}(x ∧ u) = τ_{f}(u)x − xz_{f}(u) − z_{f}(u)x,  (2)
2z_{f}(x ∧ v) = τ_{f}(x ∧ v)1 + [x, z_{q}(v)].  (3)

Proof. Let again R = k[t, t^{−1}] be the Laurent polynomial ring and define, for an alternating bilinear form a on M ⊗ R, the linear map η_{2n}(t, a): ⋀^{2n} M ⊗ R → ⋀ M ⊗ R by

π_{n}(a)(u)1 + tη_{2n}(t, a)(u) = λ_{a/t}(t^{n}u).  (4)

From 3.3 it follows easily that η_{2n}(t, a) is a polynomial in t whenever a is. If a′ is another alternating bilinear form then

λ_{a′} ∘ η_{2n}(t, a) = η_{2n}(t, a + ta′) + Π_{n}(t, a, a′)1.  (5)

For the proof, one replaces a by a+ta′ in (4) and uses 1.4.5 and 3.1.4. Now let in particular a = f − f^{T}. Then f{t} = −f^{T} + a/t and π_{n}(a) = τ_{f}. Hence by applying λ_{−f^{T}} to (4) we obtain

ζ_{2n}(t, f) = λ_{−f^{T}} ∘ η_{2n}(t, f − f^{T}).  (6)

Next, substitute t = −2 and a′ = f − g in (5). Since g − g^{T} = f − a′ − (f − a′)^{T} = a − 2a′, the result is, by 1.5.2,

λ_{a′} ∘ η_{2n}(−2, f − f^{T}) = η_{2n}(−2, g − g^{T}) + κ_{fg}1.  (7)

Finally, we apply λ_{−g^{T}} to (7) and use (6) and the relation −g^{T} + a′ = −f^{T}:

ζ_{2n}(−2, f) = λ_{−g^{T}} ∘ λ_{a′} ∘ η_{2n}(t, f − f^{T})
= λ_{−g^{T}} ∘ η_{2n}(−2, g − g^{T}) + κ_{fg}1
= ζ_{2n}(−2, g) + κ_{fg}1.

This proves (1).

Let b = f +f^{T} be the polar form of q and let w ∈ C(q) be a homogeneous element. Then it is easy to see that the graded derivation i_{x}^{b} of C(q) is just the inner graded derivation determined by x, i.e.,

i_{x}^{b}(w) = xw − (−1)^{deg w}wx.  (8)

In C(q) ⊗ R we have, since i_{x}^{f} is linear in f and annihilates 1,

ζ_{2n+1}(t, f)(x ∧ u) = λ_{f{t}}(t^{n}x ∧ u)
= (x + i_{x}^{f{t}})λ_{f{t}}(t^{n}u)
= (x + i_{x}^{f{t}})(τ_{f}(u)1 + tζ_{2n}(t, f)(u))
= τ_{f}(u)x + txζ_{2n}(t, f)(u) + i_{x}^{f[t]}ζ_{2n}(t, f)(u).  (9)

If we specialize t → −2 in this formula, observe f[−2] = f − f^{T} + 2f^{T} = b and use (8) we obtain (2). Similarly,

τ_{f}(x ∧ v)1 + tζ_{2n+2}(t, f)(x ∧ v) = λ_{f{t}}(t^{n+1}x ∧ v)
= (x + i_{x}^{f{t}})λ_{f{t}}(t.t^{n}v)
= (tx + i_{x}^{f[t]})ζ_{2n+1}(t, f)(v)  (10)

yields (3) for t → −2.

We remark that (9) and (10) give an inductive definition of ζ_{m}(t, f).

3.10. Corollary. (a) For any n, there is a well-defined homomorphism of k-modules ψ_{q}: E_{2n}(q) → C_{0}(q) such that ψ_{q}(1) = 1 and ψ_{q}(s_{f}(u)) = z_{f}(u) for all u ∈ ⋀^{2n} M.

(b) If M is finitely generated and projective of rank 2n and u ∈ ⋀^{2n} M then z_{f}(u) is central in C_{0}(q) and satisfies

xz_{f}(u) = (τ_{f}(u)1 − z_{f}(u))x,  (1)

for all x ∈ M.

(c) If M is finitely generated and projective of rank 2n + 1 and v ∈ ⋀^{2n+1} M then z_{q}(v) is central in C(q).

Indeed, (a) follows immediately from 3.9.1 and 2.2, while (b) is a consequence of 3.9.2 and implies, by a simple computation, that z_{f}(u) is central in C_{0}(q) since the latter is generated by all xy, x, y ∈ M. Similarly, (c) follows from 3.9.3.

3.11. We will need the behavior of ζ_{m}(t, f) under a similitude (φ, μ): (M̃, q̃) → (M, q), cf. 1.5. It is easy to see that (φ, μ) induces a k-linear map Φ = C(φ, μ): C(q̃) → C(q) of the Clifford algebras which generalizes 1.5.4 and is induced from the map x_{1} ⊗ … ⊗ x_{m} ↦ μ^{[ \frac{m}{2} ]}φ(x_{1}) ⊗ … ⊗ φ(x_{m}) of the tensor algebras [9, IV, (7.1.1)]. The restriction of Φ to the even Clifford algebras is an algebra homomorphism, and so is Φ provided μ = 1. In any case, Φ satisfies

Φ(yw) = μ^{deg w}φ(y)Φ(w)  (1)

for all y ∈ M̃ and homogeneous elements w ∈ C(q̃).

Let h be a bilinear form on M, let q′ = q + [h] and define h̃ = μh ∘ (φ × φ) and q̃′ = q̃ + [h̃]. Then (φ, μ) is a similitude between q̃′ and q′ as well, and the diagram

[figure: C(q̃′); Φ; C(q′); λh̃; λ_{h}; C(q̃); Φ; C(q)]  (2)

commutes. The proof is based on the inductive definition of λ_{h} and makes use of the formula

Φ(ih̃ ; yw) = μ^{deg w} i_{φ(y)}^{h}(Φ(w)).  (3)

The details are left to the reader.

3.12. Lemma. Let (φ, μ): (M̃, q̃) → (M, q) be a similitude, q = [f] and f̃ = μf ∘ (φ × φ). Then Φ ∘ ζ_{m}(t, f̃) = ζ_{m}(t, f) ∘ Φ, in particular, Φ ∘ zf̃ = z_{f} ∘ Φ.

Proof. We specialize 3.11.2 to the case h = f{t} and q′ = 0. Then h̃ = f̃{t}, and the diagram

⋀ M̃  \overline{}^{Φ} ⋀ M
[figure: λ_{f̃{t}}; λ_{f{t}}]
C(q̃)  \overline{}_{Φ} C(q)

commutes, with Φ acting on the exterior algebra by x_{1}∧…∧x_{m} ↦ μ^{[ \frac{m}{2} ]}φ(x_{1})∧…∧φ(x_{m}). Now the first formula follows easily from the definition of ζ_{m}(t, f), together with 1.6(c) in case m is even, and the second is obtained by specializing t → −2.

3.13. Proposition. Let q be a quadratic form represented by a bilinear form f on M. Then for decomposable multivectors u ∈ ⋀^{m} M the following relations hold in C(q):

z_{f}(u)^{2} − τ_{f}(u)z_{f}(u) + γ_{f}(u, u)1 = 0 (u ∈ ⋀^{2n} M),  (1)
z_{q}(u)^{2} = δ_{f}(u, u)1 (u ∈ ⋀^{2n+1} M).  (2)

Proof. We first reduce to the case where M = k^{m} is free of rank m and u is a basis of ⋀^{m} M. Indeed, if u = x_{1} ∧…∧x_{m}, let M̃ = k^{m} with standard basis e_{i}, and let φ: M̃ → M be the homomorphism mapping e_{i} to x_{i}. Set q̃ = q∘φ, f̃ = f ∘(φ×φ) and e = e_{1} ∧…∧e_{m}. Then φ is a similitude with multiplier 1, and Φ(e) = u. Therefore, if the asserted formulas hold for f̃, q̃ and e in place of f, q and u, they follow in general by applying Φ, because of Lemma 3.12 and (6) and (7) of 1.6.

Let M = k^{m} and consider the k-scheme B of all bilinear forms on M which, via the standard basis, may be identified with the affine scheme of m × m-matrices. As in [3] we consider schemes as set-valued functors on commutative k-algebras. First suppose m = 2n is even and define an open subscheme U ⊂ B by U(R) = {f ∈ B(R) : det(f + f^{T}) ∈ R^{×}}, for all k-algebras R. Since B is smooth with connected fibers and U(k) ≠ ∅, as for instance

( \begin{matrix}0 & 1_{n} \\ 0 & 0 \end{matrix} ) ∈ U(k), U is universally schematically dense in B [4, Exp. XVIII, sec. 1].

Let f ∈ B(R). Via the standard basis, we identify the R-module underlying C([f]) with the exterior algebra over R^{m}. Then the structure constants of C([f]) are polynomials in (the entries of) f and, therefore, so is the expression

p(f) = z_{f}(e)^{2} − τ_{f}(e)z_{f}(e) + γ_{f}(e, e)1.  (3)

Since all constructions are compatible with arbitrary base ring extension, (3) defines a morphism p from B to the affine k-scheme of the exterior algebra of k^{m} (which may be identified with affine 2^{m}-space). Formula (1) is equivalent to the vanishing of p on B, and since U is universally schematically dense, it suffices that p vanish on U.

Thus let now f ∈ U(R). Then [f] is nonsingular and therefore splits over some faithfully flat extension S of R [9, IV, (2.2.1)]. By faithfully flat descent, it suffices to prove p(f) ⊗ 1_{S} = 0. After extending the base ring from k to S we may assume k = S to simplify notation, and after a change of basis in k^{m}, which affects e by a unit and therefore p(f) by the square of that unit, we may assume that [f] is the standard form ξ_{1}ξ_{2} + … + ξ_{2n−1}ξ_{2n}. This means that f ∼ g = diag(\binom{0 1}{0 0}, …, \binom{0 1}{0 0}). A straightforward computation with 3.9.1 and 1.6(c) shows that p(f) = p(g) for f ∼ g. We may therefore even assume f = g. We compute τ_{f}(e) = Pf(f − f^{T}) = 1, and γ_{f}(e, e) = (−1)^{n+1} qdet(f) = 0 by 1.2.4 and 1.3.9. Now 3.8.1 shows p(f) = 0, and we have established (1).

Finally, let m = 2n+1 be odd. With B as above, consider the open subscheme U′ ⊂ B of all f′ for which hdet f′ is invertible. Again U′ is universally schematically dense. For f′ ∈ U′(R) the quadratic form [f′] is semiregular and therefore splits over a faithfully flat extension [9, IV, (3.2.1)]. Arguing as before, the validity of (2) needs to be shown only for u = e′ = e_{0} ∧ … ∧ e_{2n} and the standard form q′ = ξ_{0}^{2} + ξ_{1}ξ_{2} + … + ξ_{2n−1}ξ_{2n}, represented by f′ = diag(1, \binom{0 1}{0 0}, …, \binom{0 1}{0 0}). Since δ_{f′}(e′, e′) = 1 by 1.2.5 and 1.3.6 this follows from 3.8.3.

3.14. Theorem. Let (M, q) be a quadratic module of rank r.

(a) The map ψ_{q}: 𝔇(q) → C(q) defined as in 3.10(a) if r = 2n and by ψ_{q}(1) = 1 and ψ_{q}(u) = z_{q}(u) for all u ∈ ⋀^{r} M if r = 2n + 1, is a homomorphism of unital graded algebras.

(b) ψ: 𝔇 → C is a natural transformation of tensor functors: If φ: (M̃, q̃) → (M, q) is a similitude between quadratic modules of the same rank and the multiplier μ is 1 in case of odd rank, then the diagram

[figure: 𝔇(q̃); ψq̃; C(q̃); 𝔇(φ,μ); C(φ,μ); 𝔇(q); ψ_{q}; C(q)]  (1)

commutes. Moreover, if q = q_{1} ⊥ q_{2} is an orthogonal sum then the diagram

[figure: 𝔇(q_{1}) ˆ⊗ 𝔇(q_{2}); η; ψ_{q_{1}} ˆ⊗ ψ_{q_{2}}; 𝔇(q_{1}) □ 𝔇(q_{2})  C(q_{1}) ˆ⊗ C(q_{2}); ϑ; ≅; 𝔇(q_{1} ⊥ q_{2}); ψ_{q_{1}⊥q_{2}}; C(q_{1} ⊥ q_{2})]  (2)

is commutative, where η and ϑ are as in 2.9 and 2.11.

(c) ψ_{q} takes values in the center of C_{0}(q) in case of even rank, and in the center of C(q) in case of odd rank. If M has rank at most 2, or if q is nonsingular of even rank or semiregular of odd rank, then ψ_{q} is an isomorphism onto Z(C_{0}(q)) or Z(C(q)), respectively.

Proof. (a) Let r = 2n. By a localization argument, 3.13.1 holds for all u ∈ L = ⋀^{r} M. Now 2.3.1 and 3.13.1 show that the bilinear map ψ_{q}(s_{f}(u)s_{f}(v)) − ψ_{q}(s_{f}(u))ψ_{q}(s_{f}(v)) on the rank 1 module L is alternating and therefore vanishes. Hence ψ_{q} is a homomorphism. The case of odd rank follows similarly from 3.13.2 and 2.5.1.

(b) The commutativity of (1) follows easily from 3.12 and 2.3(b) in the even rank case, and 1.6.7 for odd rank. For (2) one uses 2.9, 2.11, and 3.7.

(c) The first statement is clear from 3.10. If M has rank 1 or 2 then ψ_{q} is an isomorphism onto C(q) or C_{0}(q), respectively. This is clear when the rank is 1. In the rank 2 case, C_{0}(q) = k.1 + MM, and hence 3.6.1 shows that ψ_{q}: 𝔇(q) → C_{0}(q) is surjective. Since both modules have rank 2 the assertion follows. Now assume q is nonsingular or semiregular. After a faithfully flat base ring extension, we may assume that q is the standard form. Then the assertion follows from 3.8 and [9, IV, (2.3), (3.2.4)].

4. Adjoints of quadratic forms of odd rank

4.1. Lemma. Let M be a finitely generated projective module of rank m+1 and p: ⋀^{m} M → N a quadratic map into some k-module N. If p vanishes on decomposable elements in all base extensions then p = 0. In particular, this applies to the formulas in 1.4.2, 1.6(b) and 3.13.

Proof. By localization we may assume M free, say with basis x_{0}, …, x_{m}. Then ⋀^{m} M is free with basis u_{i} = x_{0} ∧ … ∧ x̂_{i} ∧ … ∧ x_{m}. For i < j,

u_{i} + u_{j} = ((−1)^{i}x_{i} + (−1)^{j−1}x_{j}) ∧ x_{0} ∧ … ∧ x̂_{i} ∧ … ∧ x̂_{j} ∧ … ∧ x_{m}

is decomposable. Hence p(u_{i}) = p(u_{i} + u_{j}) = 0 which shows that p and its polar form vanishes on the basis u_{i}.

4.2. Definition. Let (M, q) be a quadratic module of rank 2n + 1 and E_{2n}(q) as in 2.2. Since ⋀^{2n} M has rank 2n + 1, 2.2.4 implies that E_{2n}(q) has rank 2n + 2. Except when n = 0, it is not possible to define an algebra structure on E_{2n}(q) as in 2.3.1. However, there is a well-defined quadratic form q^{♯} on E_{2n}(q), such that

q^{♯}(λ1 + s_{f}(u)) = λ^{2} + λτ_{f}(u) + γ_{f}(u, u)  (1)

for all λ ∈ k, u ∈ ⋀^{2n} M and f representing q. Indeed, the proof given in 2.3.5 carries over verbatim because it only uses 1.6(b) which continues to hold in the present situation due to 4.1. Since q^{♯}(1) = 1 and k.1 is a free direct summand of E_{2n}(q), the triple 𝔄(q) = (E_{2n}(q), q^{♯}, 1) is a unital quadratic form in the sense of [10], called the adjoint of q. By abuse of notation, we will sometimes write x ∈ 𝔄(q) instead of x ∈ E_{2n}(q).

Denote by B(x, y) the polar form of q^{♯}, by T (x) = B(1, x) the trace form and by σ(x) = x̄ = T (x)1 − x the conjugation. Clearly,

T (λ1 + s_{f}(u)) = 2λ + τ_{f}(u).  (2)

Also, σ is an orthogonal transformation of period two and by [7] 𝔄(q) becomes a quadratic Jordan algebra with unit element 1 and quadratic operators

U_{x}y = B(x, ȳ)x − q^{♯}(x)ȳ,  (3)

called the adjoint Jordan algebra of q.

Just as the discriminant algebra of a quadratic form of even rank, 𝔄(q) is functorial with respect to similitudes: If φ: (M̃, q̃) → (M, q) is a similitude with multiplier μ between quadratic modules of rank 2n + 1, then there is a unique morphism 𝔄(φ, μ) : 𝔄(q̃) → 𝔄(q) of unital quadratic forms and hence of Jordan algebras, given as in 2.3(b). The proof follows that of 2.3(b) and is therefore omitted. As in 2.3(c), it is easy to see that 𝔄(q) commutes with change of base ring.

If n = 0, i.e., M has rank 1, then ⋀^{2n} M = k and 𝔄(q) = k × k with q^{♯} the hyperbolic form, since τ_{f}(1) = 1 as the Pfaffian of an empty matrix, and γ_{f} = 0. Hence in this case, 𝔄(q) carries no information about q.

4.3. Example. Let us compute the adjoint of the standard form q = ξ_{0}^{2} + ∑ ξ_{2i−1}ξ_{2i} on M = k^{2n+1} with respect to the standard basis e_{0}, …, e_{2n}, represented by f = diag(1, \binom{0 1}{0 0}, …, \binom{0 1}{0 0}) as in 3.8. The elements v_{i} = e_{0} ∧ … ∧ ê_{i} ∧ … ∧ e_{2n} (i = 0, …, 2n) form a basis of ⋀^{2n} M. Direct computation yields

τ_{f}(v_{i}) = \{ \begin{matrix}1 & if i = 0 \\ 0 & if i > 0, \end{matrix}  (1)
γ_{f}(v_{i}, v_{i}) = 0 (i = 0, …, 2n),  (2)
γ_{f}(v_{i}, v_{j}) + γ_{f}(v_{j}, v_{i}) = \{ \begin{matrix}1 & if \{i, j\} = \{2l − 1, 2l\} \\ 0 & otherwise \end{matrix} (i ≠ j).  (3)

Let x_{−1} = 1 − s_{f}(v_{0}) and x_{i} = s_{f}(v_{i}), for i = 0, …, 2n. Then these formulas and 4.2.1 show that 𝔄(q) is the orthogonal sum of n+1 hyperbolic planes, spanned by the hyperbolic pairs (x_{2i−1}, x_{2i}), for i = 0, …, n, and 1 = x_{−1} + x_{0}.

Next, we determine the relation between the discriminants of q and its adjoint, which is more complicated than 2.3.5:

4.4. Proposition. Let (M, q) be a quadratic module of rank 2n + 1 with adjoint 𝔄(q) = (E_{2n}(q), q^{♯}, 1), and let L = ⋀^{2n+1} M. Then

δ_{q^{♯}} ≅ (δ_{q})^{⊗2n},  (1)

more precisely, there is a natural isomorphism φ: L^{⊗2n} ≅ ⋀^{2n+2} E_{2n}(q), under which (δ_{q})^{⊗2n} corresponds to δ_{q^{♯}}. Hence, if n > 0, q is semiregular if and only if q^{♯} is nonsingular.

Proof. There is a natural isomorphism r: W = ⋀^{2n} M → Hom(M, L) assigning to an element u ∈ W the map r_{u}: M → L, x ↦ x ∧ u, and Hom(M, L) ≅ L ⊗ M^{∗} since M is finitely generated projective. Moreover, L being of rank 1, we have ⋀^{2n+1}(L ⊗ M^{∗}) ≅ L^{⊗2n+1} ⊗ (⋀^{2n+1} M^{∗}) ≅ L^{⊗2n}, because ⋀^{2n+1} M^{∗} ≅ (⋀^{2n+1} M)^{∗} = L^{∗}. Altogether, this yields a natural isomorphism χ: ⋀^{2n+1} W → L^{⊗2n}.

Further, the map η: ⋀^{2n+1} W → ⋀^{2n+2} E_{2n}(q), w_{0} ∧ … ∧ w_{2n} ↦ 1 ∧ s_{f}(w_{0}) ∧ … ∧ s_{f}(w_{2n}), is an isomorphism independent of the choice of representative f of q, by 2.2.2. Let φ = η ∘ χ^{−1}. We must show (δ_{q})^{⊗2n} = δ_{q^{♯}} ∘ (φ × φ), equivalently,

(δ_{q})^{⊗2n} ∘ (χ × χ) = δ_{q^{♯}} ∘ (η × η).  (2)

By localization we may assume M free, say with basis e_{0}, …, e_{2n}. Let w_{i} = e_{0} ∧ … ∧ ê_{i} ∧ …∧e_{2n}, w = w_{0}∧…∧w_{2n} and e = e_{0}∧…∧e_{2n}. Then w and e are bases of ⋀^{2n+1} W and L, respectively, and we claim that χ(w) = (−1)^{n}e^{⊗2n}. Indeed, r_{w_{j}}(e_{i}) = e_{i} ∧ w_{j} = (−1)^{i}δ_{ij}e. Thus if e_{i}^{∗} is the basis dual to the e_{i} and e^{∗} = e_{0}^{∗} ∧ … ∧ e_{2n}^{∗} then e ⊗ e_{i}^{∗} corresponds to (−1)^{i}u_{i} under the isomorphism L ⊗ M^{∗} ≅ W, and therefore (e ⊗ e_{0}^{∗}) ∧ … ∧ (e ⊗ e_{2n}^{∗}) = e^{⊗2n+1} ⊗ e^{∗} = e^{⊗2n} corresponds to (−1)^{1+...+2n}w_{0} ∧ … ∧ w_{2n} = (−1)^{n}w. Hence (2) is equivalent to p(f) = 0 where

p(f) = δ_{q}(e, e)^{2n} − δ_{q^{♯}}(η(w), η(w)),  (3)

and of course q = [f]. As in the proof of 3.13 let B be the scheme of bilinear forms on M and U′ the open universally schematically dense subscheme of all f such that hdet f is invertible. Then (3) defines a morphism p: B → k_{a} since all constructions are compatible with arbitrary base ring extension. Also p(f) = p(g) for f ∼ g, because the right hand side of (3) depends only on [f]. By density, it suffices to show that p vanishes on U′. For f ∈ U′(R) the quadratic form [f] is semiregular and thus splits over a faithfully flat extension. Therefore, after a suitable base extension and a change of basis in M, which affects p(f) by a unit, we are reduced to proving p(f) = 0 in case f = diag(1, \binom{0 1}{0 0}, …, \binom{0 1}{0 0}) is the standard form. Then δ_{q}(e, e) = (−1)^{n} hdet f = 1 by 1.2.5. Defining x_{−1}, …, x_{2n} as in 4.3, we have η(w) = x_{−1} ∧ … ∧ x_{2n}, and the (x_{2i−1}, x_{2i}) are orthogonal hyperbolic pairs in 𝔄(q). Hence δ_{q^{♯}}(η(w), η(w)) = 1 and the proof is complete.

Remark. If q is semiregular then (L, δ_{q}) is a discriminant module in the sense that δ_{q}: L ⊗ L → k is an isomorphism, and therefore (L^{⊗2}, δ_{q}^{⊗2}) ≅ (k, ⟨1⟩). It follows that q^{♯} has discriminant ⟨1⟩ as well (and also trivial Arf invariant, see 5.6). Thus in contrast to 2.3(d), the discriminant of q cannot be recovered from 𝔄(q).

4.5. Lemma. Let q = [f] = [g] be a quadratic form on a k-module M and b = f + f^{T} = g + g^{T} the polar form of q. For any bilinear form h on M and x ∈ M let i_{x}^{h} be the graded derivation of degree −1 of ⋀ M given on M by i_{x}^{h}y = h(x, y). Then for all u ∈ ⋀^{2n+1} M,

τ_{f}(x ∧ u) = 2τ_{f}(i_{x}^{f}u) − τ_{f}(i_{x}^{b}u),  (1)
κ_{fg}(x ∧ u) = τ_{f}(i_{x}^{f}u) − τ_{g}(i_{x}^{g}u) − κ_{fg}(i_{x}^{b}u).  (2)

If u is decomposable then

γ_{f}(x ∧ u, x ∧ u) = q(x)δ_{q}(u, u) − τ_{f}(i_{x}^{f}u)^{2} + τ_{f}(i_{x}^{f}u)τ_{f}(x ∧ u) + γ_{f}(i_{x}^{b}u, i_{x}^{b}u). (3)

If M is finitely generated projective of rank ≤ 2n + 2 then (3) holds for all u ∈ ⋀^{2n+1} M.

Proof. Since (1) and (2) are linear in u it suffices to assume u = x_{2} ∧ … ∧ x_{2n+2} decomposable. Let x_{1} = x and x = (x_{1}, …, x_{2n+2}). If a is an alternating bilinear form on M, the expansion formula for Pf(a(x, x)) = π_{n+1}(a)(x ∧ u) implies

π_{n+1}(a)(x ∧ u) = π_{n}(a)(i_{x}^{a}u),  (4)

and setting a = f − f^{T} = 2f − b, we obtain (1) in view of 1.5.1.

Let a and a′ be alternating bilinear forms and t an indeterminate. A simple calculation with (4) and 1.4.5 shows

Π_{n+1}(t, a, a′)(x ∧ u) = π_{n}(a)(i_{x}^{a′}u) + Π_{n}(t, a, a′)(i_{x}^{a+ta′}u).

Now (2) follows by specializing a = f −f^{T}, a′ = f −g, t = −2 in this formula and observing a − 2a′ = 2g − b:

κ_{fg}(x ∧ u) = τ_{f}(i_{x}^{f−g}u) + κ_{fg}(i_{x}^{2g−b}u)
= τ_{f}(i_{x}^{f}u) + (2κ_{fg} − τ_{f})(i_{x}^{g}u) − κ_{fg}(i_{x}^{b}u)
= τ_{f}(i_{x}^{f}u) − τ_{g}(i_{x}^{g}u) − κ_{fg}(i_{x}^{b}u).

Here we used 1.6.2 for the last equality.

Formula (3) is a polynomial identity with integer coefficients in the f_{ij} = f(x_{i}, x_{j}). Hence it suffices to prove it when the f_{ij} are indeterminates and M is free with basis x_{1}, …, x_{2n+2} over the torsion-free ring ℤ[f_{ij}] and then apply specialization. By expanding the determinant det b(x, x) = ⋀^{2n+2} b(x∧u, x∧u) with respect to the first row and column, we obtain

⋀^{2n+2} b(x ∧ u, x ∧ u) = b(x, x) ⋀^{2n+1} b(u, u) − ⋀^{2n} b(i_{x}^{b}u, i_{x}^{b}u).  (5)

By 1.3.5, 1.3.11 and 1.4.2,

⋀^{2n+2} b(x ∧ u, x ∧ u) = (−1)^{n+1}δ_{f}(x ∧ u, x ∧ u)
= (−1)^{n+1}(τ_{f}(x ∧ u)^{2} − 4γ_{f}(x ∧ u, x ∧ u)).  (6)

Further, 1.3.7 implies

b(x, x) ⋀^{2n+1} b(u, u) = 4(−1)^{n}q(x)δ_{f}(u, u).  (7)

Clearly

v = i_{x}^{b}u = ∑_{j=2}^{2n+2}(−1)^{j}b(x_{1}, x_{j})x_{2} ∧ … ∧ x̂_{j} ∧ … ∧ x_{2n+2}

belongs to ⋀^{2n} N where N is the submodule of rank 2n+1 of M spanned by x_{2}, …, x_{2n+2}. By 1.4.2 and 4.1, ⋀^{2n}(f − f^{T})(v, v) = π_{n}(f − f^{T})(v)^{2} = τ_{f}(v)^{2}, so that, in analogy to (6),

⋀^{2n} b(v, v) = (−1)^{n}δ_{f}(v, v) = (−1)^{n}(τ_{f}(v)^{2} − 4γ_{f}(v, v)).  (8)

By (1), τ_{f}(v) = 2τ_{f}(i_{x}^{f}u) − τ_{f}(x ∧ u). By substituting this into (8) and then substituting (6), (7) and (8) into (5) we obtain (3) after canceling a factor 4.

Finally, suppose M has rank r ≤ 2n + 2. If r ≤ 2n we have u = 0 and there is nothing to prove. If r = 2n + 1 we may localize and thus assume M free. Then every u ∈ ⋀^{2n+1} M is decomposable. If r = 2n + 2, the assertion follows from Lemma 4.1, since the difference of the left and right hand side of (3) is, for fixed x, a quadratic form in u which vanishes on decomposable elements.

4.6. Theorem. Let (M, q) be a quadratic module of rank 2n+1 and L = ⋀^{2n+1} M with the discriminant form δ_{q}. Let 𝔄(q) = (E_{2n}(q), q^{♯}, 1) be the adjoint of q and E_{2n}(q)^{0} = Ker T the elements of trace 0, with the quadratic form q^{♯0} = −q^{♯}|E_{2n}(q)^{0}. Then there is a natural morphism of quadratic forms

φ : (M ⊗ L, q ⊗ δ_{q}) → 𝔄(q)^{0} = (E_{2n}(q)^{0}, q^{♯0})  (1)

such that, for all x ∈ M, u ∈ L and f representing q,

φ(x ⊗ u) = −τ_{f}(i_{x}^{f}u)1 + s_{f}(i_{x}^{b}u).  (2)

If q is semiregular then φ is an isomorphism.

Proof. Since x∧u ∈ ⋀^{2n+2} M = 0, 4.5.2 and 2.2.2 show that φ is well-defined, independent of the choice of f. An element v = λ1 + s_{f}(w) of E_{2n}(q)^{0} has 2λ + τ_{f}(w) = 0 by 4.2.2 and therefore q^{♯0}(v) = −λ^{2} − λτ_{f}(w) − γ_{f}(w, w) = λ^{2} − γ_{f}(w, w). Now T (φ(x ⊗ u)) = 0 follows from 4.5.1, and 4.5.3 yields q^{♯0}(φ(x ⊗ u)) = q(x)δ_{q}(u, u) = (q ⊗ δ_{q})(x ⊗ u). Since L is an invertible module, this implies by localization that φ is a morphism of quadratic forms. Finally, suppose that q is semiregular. By faithfully flat descent, we may assume q is the standard form as in 4.3. With the notations introduced there, e′ = e_{0} ∧ … ∧ e_{2n} is a basis of L, and a straightforward computation shows φ(e_{0} ⊗ e′) = x_{0} − x_{−1}, φ(e_{2i−1} ⊗ e′) = x_{2i}, φ(e_{2i} ⊗ e′) = −x_{2i−1}. Now the assertion follows from the description of 𝔄(q) given in 4.3.

A semiregular q of odd rank is determined by its adjoint and discriminant:

4.7. Corollary. Let (M, q) be a semiregular quadratic module of rank 2n + 1 and let dis(q) = (⋀^{2n+1} M, δ_{q}) be its discriminant module. Then (M, q) ≅ 𝔄(q)^{0} ⊗ dis(q).

Indeed, (M, q) ⊗ dis(q) ≅ 𝔄(q)^{0} by 4.6. Since q is semiregular, dis(q) ⊗ dis(q) ≅ ⟨1⟩ so the assertion follows by tensoring with dis(q).

As another corollary, we obtain an easy proof of the following result which was shown in [10, 6.7] by a cohomological argument.

4.8. Corollary. For a semiregular quadratic module (M, q) of rank 2n + 1 over k the following conditions are equivalent.

(i) There exists a nonsingular unital quadratic form 𝔛 = (X, Q, 1) of rank 2n + 2 such that (M, q) ≅ 𝔛^{0} = (X^{0}, Q^{0});

(ii) there exists a separable quadratic algebra R such that dis(R) ≅ dis(q); (iii) dis(q) ⊗ (k/4k) ≅ ⟨1⟩.

Proof. (i) ⟹ (ii): Let R = 𝔇(q). Then dis(q) ≅ dis(Q^{0}) ≅ dis(Q) (by [10, 3.10(b)]) ≅ dis(R) (by 2.3(d)).

(ii) ⟹ (i): (M, q) ≅ 𝔄(q)^{0} ⊗ dis(q) (by 4.7) ≅ 𝔄(q)^{0} ⊗ dis(R) ≅ (𝔄(q) □ R)^{0} (by [10, 2.12]).

(ii) ⟺ (iii): See [6, (12.6)].

4.9. Theorem. Let 𝔛 = (X, Q, 1) be a unital quadratic form of rank 2n+2 with surjective trace form T, and let (X^{0}, Q^{0}) be the quadratic module of rank 2n+1 defined by X^{0} = Ker T and Q^{0} = −Q|X^{0}. Then there is a natural morphism of unital quadratic forms

η : 𝔛 □ 𝔇(Q) → 𝔄(Q^{0}),  (1)

given as follows: Let α_{1} be any unital linear form on X, let F be a bilinear form representing Q, and set α_{2} = ρ_{F} as in 2.2 and α = (α_{1}, α_{2}). Also let B be the polar form of Q and F^{0} = −F |X^{0} × X^{0}. Then η(1) = 1 and

η(ẋ □_{α} w) = {τ_{F} (i_{x}^{F} i_{1}^{B}w) − α_{1}(x)τ_{F} (w)}1 + (−1)^{n+1}s_{F^{0}}(i_{x}^{B}i_{1}^{B}w),  (2)

for all ẋ ∈ X/k.1 and w ∈ ⋀^{2n+2} X ≅ 𝔇(Q)/k.1. If Q is nonsingular then η is an isomorphism.

Proof. Choose an element e ∈ X with T (e) = 1. Then X = k.e ⊕ X^{0}, so every w ∈ ⋀^{2n+2} X can be written uniquely

w = e ∧ u, u = i_{1}^{B}w ∈ ⋀^{2n+1} X^{0}.

Also, x − T (x)e ∈ X^{0} whence 0 = (x − T (x)e) ∧ u or

x ∧ u = T (x)w.  (3)

We begin by showing that the right hand side of (2) vanishes for x = 1 and therefore depends only on the class ẋ of x in X/k.1. This holds for the third term since (i_{1}^{B})^{2} = 0. Writing w = e ∧ u as above, let λ_{F} = τ_{F} (i_{1}^{F} u) − τ_{F} (w), and suppose G is a second bilinear form representing Q. Then by 4.5.2 and 1.6.2, λ_{F} − λ_{G} = κ_{F G}(1 ∧ u) + κ_{F G}(i_{1}^{B}u) − 2κ_{F G}(w) = κ_{F G}(1 ∧ u − 2w) = 0 since (3) shows, for x = 1, that 1 ∧ u = 2w. Thus λ_{F} = λ_{G} is independent of the choice of F. To prove λ_{F} = 0 we may assume k local. Then there are two cases: If 2 ∈ k^{×}, choose F = \frac{1}{2}B symmetric. Then i_{1}^{F} u = \frac{1}{2}i_{1}^{B}u = 0 and τ_{F} (w) = π_{2n+2}(F − F^{T})(w) = 0. If 2 is in the maximal ideal of k, let y = 1 − 2e ∈ X^{0}. Then by [10, 3.5], 1 − 4Q(e) = −Q(y) = Q^{0}(y) is a unit, so that Q^{0} is primitive. By [10, 3.6] we may assume, after a faithfully flat base extension, that X contains a hyperbolic pair (e^{+}, e^{−}) such that 1 = e^{+} + e^{−}. Decompose X = (k.e^{+} + k.e^{−}) ⊥ Y orthogonally and choose F such that F (e^{+}, e^{−}) = 1 and F (e^{−}, X) = F (e^{+}, e^{+} + Y) = F (Y, e^{±}) = 0. Also let e = e^{+}. Then X^{0} = k(e^{+} − e^{−}) ⊕ Y so that u = (e^{+} − e^{−}) ∧ v with v ∈ ⋀^{2n} Y. It follows i_{1}^{F} u = −v = i_{e}^{F} u = i_{e}^{B}u. Now 4.5.1 implies, for x = e, that λ_{F} = τ_{F} (i_{1}^{F} u) − 2τ_{F} (i_{e}^{F} u) + τ_{F} (i_{e}^{B}u) = 0.

Next, we show that η is well-defined, independent of the choice of F and α_{1}. Let η_{α} be the map defined by 1 ↦ 1 and the right hand side of (2). We must verify η_{β}(ẋ □_{α} w) = η_{α}(ẋ □_{α} w), where β = (β_{1}, β_{2}) for a second unital linear form β_{2} on X and β_{2} = ρ_{G} with Q = [G]. By definition of the product 𝔛 □ 𝔇(Q) of unital quadratic forms [10, 2.1], ẋ □_{α} w = ẋ □_{β} w + c_{αβ}(ẋ ⊗ w)1, where

c_{αβ}(ẋ ⊗ w) = c_{α_{1}β_{1}}(ẋ)τ_{F} (w) + t_{α_{F}} (ẋ)κ_{F G}(w) − 2c_{α_{1}β_{1}}(ẋ)κ_{F G}(w).  (4)

Here c_{α_{1}β_{1}}(ẋ) = −α_{1}(x) + β_{1}(x) and t_{α_{1}}(ẋ) = T (x − α_{1}(x)1) = T (x) − 2α_{1}(x), by [10, 1.4]. Substituting this into (4) and observing 1.6.2 and (3), we obtain

c_{αβ}(ẋ ⊗ w) = −α_{1}(x)τ_{F} (w) + β_{1}(x)τ_{G}(w) + κ_{F G}(x ∧ u).  (5)

Also, κ_{F^{0}G^{0}}(i_{x}^{B}u) = (−1)^{n}κ_{F G}(i_{x}^{B}u) since F^{0} and G^{0} are the restrictions of −F and −G to X^{0} and κ_{F G}^{(n)} is homogeneous of degree n in (F, G) by 1.5.2 and 1.4. Hence by 2.2,

(−1)^{n+1}s_{G^{0}}(i_{x}^{B}u) = (−1)^{n+1}s_{F^{0}}(i_{x}^{B}u) + κ_{F G}(i_{x}^{B}u)1,  (6)

and it follows

η_{β}(ẋ □_{α} w) = η_{β}(ẋ □_{β} w) + c_{αβ}(ẋ ⊗ w)1
= {τ_{G}(i_{x}^{G}u) − β_{1}(x)τ_{G}(w) + c_{αβ}(ẋ ⊗ w)}1 + (−1)^{n+1}s_{G^{0}}(i_{x}^{B}u)  (by (2))
= {τ_{G}(i_{x}^{G}u) − α_{1}(x)τ_{F} (w) + κ_{F G}(x ∧ u) + κ_{F G}(i_{x}^{B}u)}1 + (−1)^{n+1}s_{F^{0}}(i_{x}^{B}u)

(by (5) and (6))

= {τ_{F} (i_{x}^{F} u) − α_{1}(x)τ_{F} (w)}1 + (−1)^{n+1}s_{F^{0}}(i_{x}^{B}u)  (by 4.5.2)
= η_{α}(ẋ □_{α} w),

as desired.

Now we prove that η is a morphism of quadratic forms. By localization we may assume the rank 1 module ⋀^{2n+2} X free. Then every element of Ẋ □_{α} ⋀^{2n+2} X is of the form ẋ □_{α} w (instead of being a sum of such terms). Since η(1) = 1 it suffices to show that T (ẋ □_{α} w) = T (η(ẋ □_{α} w)) and the same for Q in place of T. By [10, 2.3.1] and (3), T (ẋ □_{α} w) = t_{α_{1}}(ẋ)t_{ρ_{F}} (w) = (T (x) − 2α_{1}(x))τ_{F} (w) = τ_{F} (x ∧ u) − 2α_{1}(x)τ_{F} (w), while T (η(ẋ □_{α} w)) = 2τ_{F} (i_{x}^{F} u) − 2α_{1}(x)τ_{F} (w) + (−1)^{n+1}τ_{F^{0}}(i_{x}^{B}u) = 2τ_{F} (i_{x}^{F} u) − τ_{F} (i_{x}^{B}u) − 2α_{1}(x)τ_{F} (w). Here we used τ_{F^{0}}(v) = τ_{−F} (v) = (−1)^{n}τ_{F} (v) for v ∈ ⋀^{2n} X^{0}, which follows from the fact that τ_{F} = τ_{F}^{(n)} is homogeneous of degree n in F. Now 4.5.1 shows T (ẋ □_{α} w) = T (η(ẋ □_{α} w)). For the quadratic forms we have, by [10, 2.3.2],

Q(ẋ □_{α} w) = q_{α_{1}}(ẋ)t_{ρ_{F}} (w)^{2} + {t_{α_{1}}(ẋ)^{2} − 4q_{α_{1}}(ẋ)}q_{ρ_{F}} (w)
= Q(x − α_{1}(x)1)τ_{F} (w)^{2} + {(T (x) − 2α_{1}(x))^{2} − 4Q(x − α_{1}(x)1)}γ_{F} (w, w)
= Q(x − α_{1}(x)1)τ_{F} (w)^{2} + {T (x)^{2} − 4Q(x)}γ_{F} (w, w)
= Q(x){τ_{F} (w)^{2} − 4γ_{F} (w, w)} + T (x)^{2}γ_{F} (w, w)
+ (α_{1}(x)^{2} − T (x)α_{1}(x))τ_{F} (w)^{2}
= Q(x)δ_{Q}(w, w) + γ_{F} (x ∧ u, x ∧ u)
+ α_{1}(x)^{2}τ_{F} (w)^{2} − α_{1}(x)τ_{F} (x ∧ u)τ_{F} (w).  (7)

Here we used (3) and 1.7.1 in the last step. On the other hand, γ_{F^{0}}(v, v) = γ_{−F} (v, v) = γ_{F} (v, v) for v ∈ ⋀^{2n} X^{0} since γ_{F} = γ_{F}^{(2n)} is homogeneous of degree 2n in F. Hence,

Q(η(ẋ □_{α} w)) = (τ_{F} (i_{x}^{F} u) − α_{1}(x)τ_{F} (w))^{2}

− (τ_{F} (i_{x}^{F} u) − α_{1}(x)τ_{F} (w))τ_{F} (i_{x}^{B}u) + γ_{F} (i_{x}^{B}u, i_{x}^{B}u)

= τ_{F} (i_{x}^{F} u)(τ_{F} (i_{x}^{F} u) − τ_{F} (i_{x}^{B}u)) − α_{1}(x)τ_{F} (w)(2τ_{F} (i_{x}^{F} u) − τ_{F} (i_{x}^{B}u))

+ α_{1}(x)^{2}τ_{F} (w)^{2} + γ_{F} (i_{x}^{B}u, i_{x}^{B}u) = τ_{F} (i_{x}^{F} u)(τ_{F} (x ∧ u) − τ_{F} (i_{x}^{F} u)) − α_{1}(x)τ_{f}(w)τ_{F} (x ∧ u)

+ α_{1}(x)^{2}τ_{F} (w)^{2} + γ_{F} (i_{x}^{B}u, i_{x}^{B}u)  (by 4.5.1)

= −Q(x)δ_{F} (u, u) + γ_{F} (x ∧ u, x ∧ u)

+ α_{1}(x)^{2}τ_{F} (w)^{2} − α_{1}(x)τ_{F} (x ∧ u)τ_{F} (w)  (by 4.5.3).

Comparing this with (7) we see that it remains to show δ_{Q}(w, w) = −δ_{F} (u, u). By [10, 3.10(b)], δ_{Q}(w, w) = δ_{Q^{0}}(u, u), and by 1.7, δ_{Q^{0}}(u, u) = δ_{F^{0}}(u, u) = δ_{−F} (u, u) = −δ_{F} (u, u) since δ_{F} = δ_{F}^{(2n+1)} is homogeneous of degree 2n + 1 in F by 1.3.5. This completes the proof that η is a morphism of quadratic forms.

Finally, we show η is an isomorphism if Q is nonsingular. By faithfully flat descent and [10, 3.7(a)] we may assume X free with basis e_{−1}, …, e_{2n}, Q the standard form represented by F = diag(\binom{0 1}{0 0}, …, \binom{0 1}{0 0}), and 1 = e_{−1} + e_{0}. Then X^{0} has basis e′_{0} = e_{−1} − e_{0} and e′_{i} = (−1)^{i}e_{i} (i = 1, …, 2n), and the matrix of F^{0} with respect to this basis is diag(1, \binom{0 1}{0 0}, …, \binom{0 1}{0 0}). Define v_{i} = e′_{0} ∧ … ∧ ê′_{i} ∧ … ∧ e′_{2n} and let x_{i} = s_{F^{0}}(v_{i}) for i ≥ 0 and x_{−1} = 1 − x_{0}. Then by 4.3, x_{−1}, …, x_{2n} is a hyperbolic basis for 𝔄(Q^{0}). Finally, let w = e_{−1} ∧ … ∧ e_{2n} and choose the unital linear form α_{1}(x) = F (1, x) on X. Then a computation shows that η(1) = 1, η(ė_{−1}□_{α}w) = 1−x_{0} = −η(ė_{0}□_{α}w), η(ė_{2j−1}□_{α}w) = x_{2j}, η(ė_{2j} □_{α} w) = x_{2j−1}. Hence η is an isomorphism and the proof is complete.

4.10. Corollary. A nonsingular unital quadratic form 𝔛 = (X, Q, 1) of even rank is naturally isomorphic to 𝔄(Q^{0}) □ 𝔇(Q).

Proof. Since 𝔇(Q) is separable, there is a natural isomorphism 𝔇(Q) □ 𝔇(Q) ≅ 𝔈 = k × k [10, 2.9], and 𝔈 acts like a unit with respect to the product □ [10, 2.6]. Hence □-multiplying 4.9.1 with 𝔇(Q) yields a natural isomorphism 𝔛 ≅ 𝔄(Q^{0}) □ 𝔇(Q).

5. The Jordan center of the Clifford algebra

5.1. Definition. Let C = C(q) = C_{0} ⊕ C_{1} be the Clifford algebra of a quadratic form q on M and Z = Z_{0} ⊕ Z_{1} its center. Define J = J(q) ⊂ C × C by

(u, v) ∈ J ⟺ ux − xv = xu − vx ∈ Z for all x ∈ M.  (1)

In view of the properties proved below, we call J the Jordan center of C. Clearly J is a Z-submodule of C × C containing 1_{J} = (1, 1), and there is a Z-linear map

χ: J → Hom_{k}(M, Z),  χ(u, v)(x) = ux − xv.  (2)

The kernel of χ is

K = Ker χ = {(u, v) ∈ C × C : ux = xv, vx = xu for all x ∈ M}.  (3)

It is easy to see that K is a subalgebra of C × C and J and K are stable under the exchange automorphism (u, v) ↦ \overline{(u, v)} = (v, u). Let u = u_{0} + u_{1}, v = v_{0} + v_{1} be the decomposition into homogeneous parts of (u, v) ∈ C × C. Since M ⊂ C_{1} it follows easily that (u, v) belongs to J or K if and only if this holds for (u_{i}, v_{i}). Hence J = J_{0} ⊕ J_{1} and K = K_{0} ⊕ K_{1} are ℤ/2ℤ-graded, and χ maps J_{i} to Hom_{k}(M, Z_{1−i}). We finally note that there is a homomorphism Δ: M → J_{1} sending y ∈ M to (y, −y) and making the diagram

[figure: M; b; Hom(M, k); Δ; i; J_{1}; χ; Hom(M, Z_{0})]  (4)

commutative, where b is given by the polar form of q and i is induced from k → Z_{0}. Indeed, yx − x(−y) = xy − (−y)x = b(x, y)1 for all x ∈ M.

5.2. Proposition. (a) If (u, v) ∈ K and w_{i} ∈ C_{i} then u and v commute with w_{0} and satisfy uw_{1} = w_{1}v and vw_{1} = w_{1}u; in particular, pr_{1}(K) ⊂ C^{C_{0}}. If q is primitive, pr_{1} : K → C^{C_{0}} is injective.

(b) If a = (u, v) ∈ J then Q(a) = uv = vu ∈ Z. The map Q: J → Z is a quadratic form on J with Q(1_{J}) = 1, polar form B(a, a′) = uv′ + u′v, trace form T (a) = B(a, 1_{J}) = u + v and conjugation σ(a) = T (a)1_{J} − a = ā.

(c) J is a (quadratic) Jordan subalgebra of C ×C and the Jordan algebra structure induced form C × C is the one determined by the quadratic form Q with base point 1_{J}, i.e.,

aba = B(a, b̄)a − Q(a)b̄ ∈ J  (1)

for all a, b ∈ J.

Proof. The first statement of (a) follows from 5.1.3. Let q be primitive and (0, v) ∈ K. Then q(M) generates k as an ideal which implies 1 ∈ MM, and vM = 0 by 5.1.3. Hence v = v1 = 0.

Let a = (u, v) ∈ J. By 5.1.1, (u+v)x = x(u+v) for all x ∈ M. This implies u+v = t ∈ Z and therefore uv = u(t − u) = (t − u)u = vu. Further, let c = ux − xv = xu − vx ∈ Z. Then uvx = u(xu − c) = (ux − c)u = xvu = xuv, and since x ∈ M was arbitrary, uv = vu ∈ Z. The remaining assertions of (b) are now clear. Finally, aā = (uv, vu) = Q(a)1_{J} = (vu, uv) = āa ∈ Z1_{J}. Hence B(a, b̄)1_{J} = (a + b̄)(ā + b) − aā − b̄b = ab + b̄ā which yields B(a, b̄)a − Q(a)b̄ = (ab + b̄ā)a − (aā)b̄ = aba.

5.3. Remark. By specializing b = 1_{J} in 5.2.1, we see that every a ∈ J satisfies the quadratic polynomial a^{2} −T (a)a+Q(a)1 = 0. Also, Q is multiplicative with respect to the Jordan product, Q(aba)1_{J} = abaāb̄ā = (aā)^{2}bb̄ = Q(a)^{2}Q(b)1_{J}. Since K_{0} is commutative by 5.2(a), we have Q(ab)1_{J} = abāb̄ = aābb̄ = Q(a)Q(b)1_{J} for a, b ∈ K_{0}. But K is in general not commutative and it is not clear whether Q is multiplicative on K. Also, I don’t know whether Q is in general a unital quadratic form in the sense of [10], i.e., whether Z.1_{J} is a free direct summand of the Z-module J.

5.4. Proposition. Let (M, q) be a quadratic module of rank 2n, 𝔇(q) its discriminant algebra and ψ_{q}: 𝔇(q) → C_{0} as in 3.14.

(a) The map a ↦ (ψ_{q}(a), ψ_{q}(ā)) is an algebra homomorphism Ψ_{q}: 𝔇(q) → K_{0}.

(b) Let q be nonsingular. Then Ψ_{q} is an isomorphism, K_{1} = 0, and J_{0} = K_{0}. Furthermore, Δ: M → J_{1} is an isomorphism and the quadratic form Q on J is isomorphic to N ⊥ (−q) where N is the norm form of 𝔇(q).

Proof. (a) Since conjugation a ↦ ā is an automorphism of 𝔇(q), 3.14(a) shows Ψ_{q} is an algebra homomorphism. By 3.10.1 it takes values in K_{0}.

(b) It is known [9, IV, §2, §4] that Z = Z_{0} = k.1 and C^{C_{0}} is the center of C_{0}. By 3.14(c), ψ_{q}: 𝔇(q) → C^{C_{0}} is an isomorphism. Since ψ_{q} = pr_{1} ∘Ψ_{q}, we have pr_{1} : K → C^{C_{0}} surjective. By 5.2(a), pr_{1} is injective so Ψ_{q} is an isomorphism. As the odd part of C^{C_{0}} is zero we obtain K_{1} = 0.

Via χ, J_{0}/K_{0} is isomorphic to a submodule of Hom(M, Z_{1}) = 0 whence J_{0} = K_{0}. Also, K_{1} = 0 implies χ: J_{1} → Hom(M, Z_{0}) = M^{∗} injective. Since q is nonsingular, b: M → M^{∗} is an isomorphism. Now 5.1.4 shows χ is surjective as well and Δ is an isomorphism. Finally, let a = (u_{0}, v_{0}) + (u_{1}, v_{1}) ∈ J_{0} ⊕ J_{1}. Then a_{0} = (u_{0}, v_{0}) ∈ K_{0} and v_{1} = −u_{1} ∈ M, and thus Q(a) = (u_{0} + u_{1})(v_{0} + v_{1}) = u_{0}v_{0} − u_{1}^{2} + u_{1}v_{0} − u_{0}u_{1} = N(a_{0}) − q(u_{1}) by 5.2(a). This completes the proof.

5.5. Theorem. Let (M, q) be a quadratic module of rank 2n + 1, 𝔄(q) its adjoint, and ψ_{q}: E_{2n}(q) → C_{0} as in 3.10(a).

(a) The k-linear map a ↦ (ψ_{q}(a), ψ_{q}(ā)) is a morphism Ψ_{q}: 𝔄(q) → J_{0} of quadratic forms, i.e.,

Q(Ψ_{q}(a)) = ψ_{q}(a)ψ_{q}(ā) = q^{♯}(a)1_{C} (a ∈ 𝔄(q)),  (1)

and a homomorphism of Jordan algebras.

(b) Let q be semiregular. Then K_{0} = k.1_{J}, Ψ_{q} is an isomorphism, and J ≅ J_{0} ⊗ Z is extended from J_{0}. If M has rank at least 3 then pr_{1} : J → C and ψ_{q}: 𝔄(q) → C_{0} are injective.

Proof. (a) By 3.9.2 and 3.10(c) we have z_{f}(u)x − x(τ_{f}(u)1 − z_{f}(u)) = xz_{f}(u) − (τ_{f}(u)1 − z_{f}(u))x = −z_{q}(x ∧ u) ∈ Z for all x ∈ M, u ∈ W = ⋀^{2n} M. Since 𝔄(q) = k.1 ⊕ s_{f}(W) and \overline{s_{f}(u)} = τ_{f}(u)1 − s_{f}(u) this implies that Ψ_{q} takes values in J_{0} and also shows that the diagram

[figure: 0; k; 𝔄(q); p; ⋀^{2n} M; 0; Ψ_{q}; η; 0; K_{0}; J_{0}; χ; Hom(M, Z_{1}); 0]  (2)

commutes, where η(u)(x) = −z_{q}(x ∧ u). Next, Ψ_{q} preserves quadratic forms since T (Ψ_{q}(s_{f}(u))) = z_{f}(u) + τ_{f}(u)1 − z_{f}(u) = T (s_{f}(u))1_{C} and Q(Ψ_{q}(s_{f}(u))) = z_{f}(u)(τ_{f}(u)1 − z_{f}(u)) = γ_{f}(u, u)1_{C} = q^{♯}(s_{f}(u))1_{C} for all u ∈ W, by 3.13.1, 4.1 and 4.2. From 4.2.3 and 5.2.1 it follows that Ψ_{q} is a Jordan homomorphism.

(b) By 3.14(c) the center of C is Z = k.1 ⊕ Z_{1} and z_{q}: L = ⋀^{2n+1} M → Z_{1} is an isomorphism. Since W ≅ Hom(M, L) via u ↦ (x ↦ x∧u), we see that η is an isomorphism. The top row of (2) is exact by 2.2.4. As both p and η are surjective so is χ, and thus the bottom row of (2) is exact. By 5.2(a) pr_{1} : K → C^{C_{0}} is injective. By [9, IV, (3.2.4)], C^{C_{0}} = Z = k.1 ⊕ Z_{1} whence K_{0} = k.1_{J}. Since k.1_{J} is a free direct summand of C × C, the map k → K_{0} in (2) is an isomorphism, and by the Five Lemma so is Ψ_{q}. That J is extended from J_{0} follows easily from C_{1} = C_{0}Z_{1} ≅ C_{0}⊗Z_{1} and the fact that multiplication induces an isomorphism Z_{1} ⊗ Z_{1} ≅ k [9, IV, (3.2.4)].

Finally, suppose M has rank ≥ 3. It suffices to show pr_{1} : J_{0} → C_{0} injective. Since Ψ_{q} is an isomorphism and ψ_{q} = pr_{1} ∘Ψ_{q} this is equivalent to ψ_{q} being injective. Thus assume ψ_{q}(λ1+s_{f}(u)) = λ1+z_{f}(u) = 0. Then by 3.9.2 and 3.10(c), z_{q}(x∧u) = (2λ+τ_{f}(u))x ∈ Z_{1} for all x ∈ M. This implies μ = 2λ + τ_{f}(u) = 0. Indeed, as M has rank ≥ 3, it contains a hyperbolic pair (e^{+}, e^{−}) after a faithfully flat base extension. Then 0 = e^{+}[e^{−}, μe^{+}] = μ(e^{+}e^{−}e^{+} −(e^{+})^{2}e^{−}) = μe^{+} shows μ = 0. Hence z_{q}(x∧u) = 0 for all x ∈ M, i.e., η(u) = 0 and therefore u = 0 and λ = 0, as desired.

5.6. Corollary. Let (M, q) be semiregular of odd rank 2n + 1 with adjoint form q^{♯}, and let A = C_{0}(q). Then C(q^{♯}) ≅ Mat_{2}(A) with the checkerboard grading. In particular, C_{0}(q^{♯}) ≅ A × A and 𝔇(q^{♯}) ≅ k × k, i.e., q^{♯} has trivial Arf invariant.

Proof. Define a map h: 𝔄(q) → Mat_{2}(A) by

h(a) = ( \begin{matrix}0 & ψ_{q}(a) \\ ψ_{q}(ā) & 0 \end{matrix} ) .

Then 5.5.1 shows h(a)^{2} = q^{♯}(a)1. Hence there exists a unique homomorphism h̃: C(q^{♯}) → Mat_{2}(A) extending h, and h̃ is clearly compatible with the gradings. Since q^{♯} is nonsingular by 4.4, C(q^{♯}) is an Azumaya algebra of rank 2^{2n+2} and so is Mat_{2}(A), for A is Azumaya of rank 2^{2n}. Therefore, h̃ is an isomorphism. The rest is immediate from 3.14(c).

5.7. Example. Let us work out ψ_{q} for the standard form q as in 4.3. By 3.8, the center of C(q) is k.1 ⊕ k.ε where ε = e_{0}(1 − 2c). With the notations of 4.3, the map z_{f}: ⋀^{2n} M → C_{0}(q) is given by

z_{f}(v_{0}) = c,  (1)
z_{f}(v_{2i−1}) = e_{2i}ε,  (2)
z_{f}(v_{2i}) = −e_{2i−1}ε.  (3)

Indeed, (1) is clear from 3.8.1. For the second, let u_{i} = ⋀_{j≠i}(e_{2j−1} ∧ e_{2j}). Then v_{2i−1} = e_{0} ∧ e_{2i} ∧ u_{i}, and by 3.7,

\begin{aligned}z_{f}(v_{2i−1}) & = z_{f}(e_{0})z_{f}(e_{2i} ∧ u_{i}) = e_{0}z_{f}(e_{2i})(τ_{f}(u_{i}).1 − 2z_{f}(u_{i})) \\ & = e_{0}e_{2i}(1 − 2c_{1} • … • ĉ_{i} • … • c_{n}). \end{aligned}

Now (1 − 2c_{i})^{2} = 1 and (1 − 2c_{i})(1 − 2c_{j}) = 1 − 2c_{i} • c_{j}, and furthermore e_{2i}(1 − 2c_{i}) = e_{2i} − 2e_{2i}e_{2i−1}e_{2i} = e_{2i} − 2e_{2i} = −e_{2i}. Hence z_{f}(v_{2i−1}) = e_{0}e_{2i}(1 − 2c_{i}) ∏_{j=1}^{n}(1 − 2c_{j}) = −e_{0}e_{2i}(1 − 2c) = e_{2i}ε. The proof of (3) is similar.

As an application, we obtain the following result, whose counterpart in the even rank case is well known [9, IV, (4.3.1)].

5.8. Corollary. Let (M, q) be semiregular of rank 2n + 1 ≥ 3, identify J_{0} with pr_{1}(J_{0}) = ψ_{q}(𝔄(q)) ⊂ A = C_{0}(q) as in 5.5(b) and let ρ be the standard involution of C(q), characterized by x^{ρ} = −x for x ∈ M.

(a) If n is even, J_{0} ⊂ S = {a + a^{ρ} : a ∈ A}, the set of traces of ρ in A. In particular, the restriction of ρ to J_{0} is the identity.

(b) If n is odd, the restriction of ρ to J_{0} corresponds to the conjugation of 𝔄(q) under the isomorphism ψ_{q}: 𝔄(q) → J_{0}.

Proof. By faithfully flat descent, we may assume that q is the standard form on k^{2n+1}. A direct computation with the formulas (1)–(3) of 4.3 shows that the conjugation in 𝔄(q) is given by

¯1 = 1, \overline{s_{f}(v_{0})} = 1 − s_{f}(v_{0}), \overline{s_{f}(v_{i})} = −s_{f}(v_{i}) for i > 0.

From e_{i}^{ρ} = −e_{i} follows c_{j}^{ρ} = (e_{2j−1}e_{2j})^{ρ} = e_{2j}e_{2j−1} = 1 − e_{2j−1}e_{2j} = 1 − c_{j}, and hence c_{1} • c_{2} = c_{1} + c_{2} − 2c_{1}c_{2} = c_{1}(1 − c_{2}) + c_{2}(1 − c_{1}) = c_{1}c_{2}^{ρ} + (c_{1}c_{2}^{ρ})^{ρ} is a trace of ρ. Moreover, if d_{1} and d_{2} are commuting idempotents which are traces then d_{1}d_{2} is fixed under ρ and therefore d_{1} • d_{2} = d_{1} + d_{2} − (d_{1}d_{2} + (d_{1}d_{2})^{ρ}) is a trace. By induction, z_{f}(v_{0}) = c = c_{1} • … • c_{n} is a trace for n even. If n is odd, let d = c_{2} • … • c_{n}. Then c = c_{1} • d and c^{ρ} = c_{1}^{ρ} • d^{ρ} = (1 − c_{1}) • d = 1 − c_{1} + d − 2(1 − c_{1})d = 1 − c. This implies

ε^{ρ} = (−e_{0})(1 − 2c^{ρ}) = (−1)^{n+1}ε,

and by (2) and (3) of 5.7,

z_{f}(v_{i})^{ρ} = (−1)^{n}z_{f}(v_{i}) (i > 0).

Now (b) follows from these formulas, and it remains to show that the e_{l}ε (l = 1, …, 2n) are traces when n is even. Suppose l = 2i or l = 2i − 1 and choose j ∈ {1, …, n} different from i, which is possible since n ≥ 2. Then e_{l} commutes with c_{j} and c_{j}^{ρ}. Hence e_{l}ε = 1.e_{l}ε = (c_{j} +c_{j}^{ρ})e_{l}ε = c_{j}e_{l}ε+(−ε)(−e_{l})c_{j}^{ρ} = c_{j}e_{l}ε+(c_{j}e_{l}ε)^{ρ} is a trace. This completes the proof.

5.9. Forms of low rank. For M of rank 3, a similar argument as in the proof of 3.14(c) shows that ψ_{q}: 𝔄(q) → C_{0} is an isomorphism (of modules and of Jordan algebras), even when q is not semi-regular. If q is semiregular of rank 5, it is known that the standard involution on C_{0} is of even symplectic type, and the space S of traces of ρ is of rank 6 [9, V, (5.4)]. This implies ψ_{q}: 𝔄(q) → S is an isomorphism. Indeed, since ψ_{q} is a homomorphism between finitely generated modules, it suffices that ψ_{q} ⊗ k/𝔪 be surjective for all maximal ideals 𝔪 of k. Since all constructions are compatible with change of base ring, this follows from injectivity of ψ_{q} and the equality of the ranks.

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Received February 26, 1996