To appear in Mathematische Zeitschrift

On algebras satisfying x^{2}x^{2} = N(x)x

Alberto Elduque^{1}^{,}^{⋆}, Susumu Okubo^{2}^{,}^{⋆⋆}

1 Departamento de Matemáticas, Facultad de Ciencias, Universidad de Zaragoza, 50009 Zaragoza, Spain 2 Department of Physics and Astronomy, University of Rochester, Rochester, New York 14627, USA

Abstract. The commutative algebras satisfying the “adjoint identity”: x^{2}x^{2} = N(x)x, where N is a cubic form, are shown to be related to a class of generically algebraic Jordan algebras of degree at most 4 and to the pseudo-composition algebras. They are classified under a nondegeneracy condition.

As byproducts, the associativity of the norm of any pseudo-composition algebra is proven and the unital commutative and power-associative algebras of degree ≤ 3 are shown to be Jordan algebras.

AMS Subject Classifications: 17C50, 17A30

1 Introduction

The purpose of this paper is to study those commutative algebras A over a field F, endowed with a cubic form N (that is, an homogeneous polynomial mapping of degree three N : A → F) called the norm of A, and satisfying

x^{2}x^{2} = N(x)x  (1)

for any x ∈ A. Such algebras will be called admissible cubic algebras. The authors were lead to this equation through several examples in which the cubic form satisfied some extra requirements; but Eq.(1) also appears as the adjoint identity (x^{#})^{#} = N(x)x in the theory of Jordan algebras (see [McC2] and the references therein).

The admissible cubic algebras will be shown to be intimately related to a class of generically algebraic Jordan algebras of degree at most 4 (Theorem 5)

Correspondence to: A. Elduque ⋆ Supported by the Spanish DGES (Pb 97-1291-C03-03). ⋆⋆ Supported in part by U.S. Department of Energy Grant No. DE-F402-ER 40685.

on one side and to some pseudo-composition algebras, introduced in [MO], on the other side (Theorem 14).

Throughout the paper, the characteristic of the ground field F will always be assumed to be different from 2 and 3 and, unless otherwise stated, the term “algebra” will refer to an algebra over F. The paper is organized as follows. In Section 2 an associative symmetric bilinear form will be attached to the cubic norm of any admissible cubic algebra. This form will allow us to relate these algebras with a class of Jordan algebras of degree ≤ 4 in a very precise way in Section 3. Then Section 4 will classify our algebras assuming the bilinear form is nondegenerate. Finally Section 5 will explain the relationship between our algebras and the pseudo-composition algebras. Here it will be proven the associativity of the bilinear form of any pseudo-composition algebra (Theorem 12) and, as a consequence, the fact that any unital commutative and power– associative algebra of degree 3 is a Jordan algebra (Corollary 13), two results that have their own independent interest. The last Section 6 will provide some comments and connections with related work.

Acknowledgement. The authors are very much indebted to the anonymous referee, whose comments helped to improve and to put the original article in an adequate perspective. In particular, the definition of the class of quartic Jordan algebras given in Section 3 is due to the referee and the present form of Theorems 5 and 14 owes much to his/her suggestions.

2 Trace form

Let A be an admissible cubic algebra with norm N and let N(x, y, z) be the symmetric trilinear form with N(x) = N(x, x, x) for any x. First notice that given any scalar field extension K/F, the algebra K ⊗_{F} A still satisfies Eq.(1).

Use will be made of the Zariski topology on A (see [McC1] for its definition and main features on not necessarily finite dimensional spaces). All the references to density in what follows will refer to density in this topology, after a scalar extension to deal with infinite fields if necessary.

Substituting x by x^{2} in Eq.(1) we get

N(x)^{2}x^{2} = N(x^{2})x^{2}

for any x ∈ A, so that N(x^{2}) = N(x)^{2} for any x with x^{2} ≠ 0, and in particular for those x with N(x) ≠ 0. If N ≠ 0, the set of these elements is dense (remember, we first extend scalars if necessary to get an infinite field), so we conclude that

N(x^{2}) = N(x)^{2}  (2)

for any x ∈ A.

Linearizing Eq.(1) we obtain

4x^{2}(xy) = 3N(x, x, y)x + N(x)y  (3)

and substituting x by x^{2} in Eq.(3) and using Eqs.(1) and (2),

4N(x)x(x^{2}y) = 3N(x^{2}, x^{2}, y)x^{2} + N(x^{2})y  (4)

On algebras satisfying x^{2}x^{2} = N(x)x 3

for any x, y ∈ A. Under these circumstances:

Theorem 1. Let A be any admissible cubic algebra with norm N. Then there is an associative symmetric bilinear form ⟨ | ⟩ such that N(x) = ⟨x|x^{2}⟩ for any x ∈ A. If N ≠ 0 then this form is unique.

Proof . If N = 0 we can take ⟨x|y⟩ = 0 for any x, y and we are done. Hence, assume in what follows that N ≠ 0 and, by extending scalars if necessary, that the ground field F is infinite; therefore we may identify the polynomial functions on A or A × A with the polynomials defining them.

Let {v_{i} : i ∈ I} be a basis of A and let q_{i}, i ∈ I be the quadratic functions such that

x^{2} = ∑_{i∈I} q_{i}(x)v_{i}

for any x ∈ A. Then Eqs.(2) and (4) imply that N(x) divides N(x^{2}, x^{2}, y)q_{i}(x) for any i. Let us see first that N(x) divides N(x^{2}, x^{2}, y). Otherwise, there would be an irreducible factor f(x) (see [McC1, Theorem 1]) of N(x), of degree 1 or 2, such that f(x) divides q_{i}(x) for any i.

In case f(x) is linear, we get x^{2} = f(x)φ(x) for some linear map φ : A → A. Hence

N(x)x = x^{2}x^{2} = f(x)^{2}φ(x)^{2} = f(x)^{2}f(φ(x))φ^{2}(x)

for any x. Thus φ^{2}(x) ∈ F x for any x such that N(x) ≠ 0 and, by density, for any x. This implies the existence of a nonzero scalar α ∈ F such that φ^{2}(x) = αx for any x, and hence N(x) = αf(x)^{2}f(φ(x)). This in turn implies that

N(x, y, z) = \frac{α}{3}(f(x)f(y)f(φ(z)) + f(y)f(z)f(φ(x)) + f(z)f(x)f(φ(y)))

for any x, y, z ∈ A. Therefore

\begin{matrix}N(x^{2}, x^{2}, y) = f(x)^{2}N(φ(x), φ(x), y) \\ = \frac{αf(x)^{2}}{3} (f(φ(x))^{2}f(φ(y)) + 2αf(φ(x))f(y)f(x)) \end{matrix}

which is divisible by αf(x)^{2}f(φ(x)) = N(x).

If, on the contrary, f(x) is quadratic, then there is a fixed element a ∈ A such that x^{2} = f(x)a for any x ∈ A. Then

N(x)x = x^{2}x^{2} = f(x)^{2}f(a)a

for any x and we conclude that the dimension of A is 1. Then it is clear that A = F a, N(λa) = λ^{3}N(a) with N(a) ≠ 0 and N((λa)^{2}, (λa)^{2}, μa) = λ^{4}μN(a, a, a) = λμN(λa), as required.

In consequence, there is a bilinear form ⟨x|y⟩ such that

N(x^{2}, x^{2}, y) = N(x)⟨x|y⟩  (5)

for any x, y ∈ A. In particular, using Eq.(2),

N(x)^{2} = N(x^{2}) = N(x^{2}, x^{2}, x^{2}) = N(x)⟨x|x^{2}⟩

and hence N(x) = ⟨x|x^{2}⟩ for any x ∈ A.

Substituting x by x^{2} in Eq.(5) we obtain

N(x)^{2}N(x, x, y) = N(x^{2})⟨x^{2}|y⟩

so that

N(x, x, y) = ⟨x^{2}|y⟩

and, by linearization,

N(x, y, z) = ⟨xy|z⟩  (6)

for any x, y, z ∈ A.

The symmetry of N( , , ) implies

⟨xy|z⟩ = ⟨yz|x⟩  (7)

for any x, y, z ∈ A. Let us check now that ⟨ | ⟩ is symmetric. We have N(x) = ⟨x|x^{2}⟩, but also N(x) = ⟨x^{2}|x⟩ by Eq.(6). Linearizing ⟨x|x^{2}⟩ = ⟨x^{2}|x⟩ and using Eq.(7) we obtain

⟨y|x^{2}⟩ + 2⟨x|xy⟩ = 3⟨x^{2}|y⟩ ,

and substituting x by x^{2} once more

N(x)⟨y|x⟩ + 2⟨x^{2}|x^{2}y⟩ = 3N(x)⟨x|y⟩ .  (8)

But from Eq.(7) we know that ⟨x^{2}|x^{2}y⟩ = ⟨x(x^{2}y)|x⟩, and Eq.(4) may be rewritten now as

4x(x^{2}y) = 3⟨x|y⟩x^{2} + ⟨x|x^{2}⟩y .  (9)

Thus Eq.(8) becomes

N(x)⟨y|x⟩ + \frac{3}{2}⟨x|y⟩⟨x^{2}|x⟩ + \frac{1}{2}N(x)⟨y|x⟩ = 3N(x)⟨x|y⟩ ,

that is, by factoring out N(x) = ⟨x^{2}|x⟩, we check that ⟨y|x⟩ + \frac{3}{2}⟨x|y⟩ + \frac{1}{2}⟨y|x⟩ = 3⟨x|y⟩, or ⟨x|y⟩ = ⟨y|x⟩ for any x, y ∈ A, as required. Then, ⟨ | ⟩ is associative by Eq.(7).

The uniqueness of ⟨ | ⟩ follows from Eq.(5), because if N(x) = (x|x^{2}) for any x and some symmetric associative bilinear form ( |), then N(x, y, z) = (x|yz) so that

N(x)⟨x|y⟩ = N(x^{2}, x^{2}, y) = (x^{2}|x^{2}y) = (x^{2}x^{2}|y) = N(x)(x|y)

and ⟨x|y⟩ = (x|y) for any x, y ∈ A.

It only remains to check that if the ground field F is finite, the bilinear form ⟨ | ⟩, which has been defined on an infinite field extension K of F, takes values in F when applied to elements in A. Equation (3) becomes

4x^{2}(xy) = 3⟨x^{2}|y⟩x + ⟨x^{2}|x⟩y .  (10)

By Eq.(10), ⟨x^{2}|y⟩ ∈ F for any x, y ∈ A and, therefore, ⟨A^{2}|A⟩ ⊆ F. But if a is any element in A with N(a) ≠ 0, Eq.(1) implies that a ∈ A^{2} and Eq.(10), with x = a, that A^{2} = A. Hence ⟨A|A⟩ ⊆ F.

The bilinear form ⟨ | ⟩ referred to in Theorem 1 will be called the trace form of N. If N = 0 there may be several of them. The extreme case is the situation in which A^{2} = 0, so that any symmetric bilinear form on A is a trace form.

Notice that Eqs.(10) and (9) give

(x^{2}y)x − x^{2}(yx) = \frac{3}{4} (⟨x|y⟩x^{2} − ⟨x^{2}|y⟩x)  (11)

for any x, y in any admissible cubic algebra with nonzero norm. But Eq.(11) is not satisfied in general if the norm is trivial. For instance, let A = F u+F v +F w be the commutative algebra with u^{2} = v, uv = w = uw, the other products being zero. Then A^{2} = F v+F w and A^{2}A^{2} = 0, so that A is an admissible cubic algebra with zero norm. Then

(u^{2}u)u − u^{2}u^{2} = w ≠ \frac{3}{4} (⟨u|u⟩u^{2} − ⟨u^{2}|u⟩u)

no matter which bilinear form is used. Thus Eq.(11) is not satisfied.

3 Quartic Jordan algebras

Let us define a class of unital Jordan algebras which will be shown to be intimately related to the admissible cubic algebras. As a general rule, the multiplication in Jordan algebras will be denoted by a dot ·, while the multiplication in the admissible cubic algebras will be denoted, as before, by juxtaposition. Other symbols will be used when needed.

Definition. A unital Jordan algebra J over F is said to be a quartic Jordan algebra if it is endowed with a linear form t : J → F (the trace of J), a quadratic form s : J → F, a cubic form r : J → F and a quartic form n : J → F such that the following holds:

i) J satisfies the quartic equation

x^{·4} − t(x)x^{·3} + s(x)x^{·2} − r(x)x + n(x)1 = 0 .  (12)

ii) t(1) = 4. iii) The Newton formulae hold:

2s(x) = t(x)^{2} − t(x^{·2})
6r(x) = t(x)^{3} − 3t(x)t(x^{·2}) + 2t(x^{·3})
24n(x) = t(x)^{4} − 6t(x)^{2}t(x^{·2}) + 8t(x)t(x^{·3}) + 3t(x^{·2})^{2} − 6t(x^{·4}) .

For example, any generically algebraic Jordan algebra of generic degree 4 (see [McC1]) is a quartic Jordan algebra. Conversely, any quartic Jordan algebra is generically algebraic of generic degree ≤ 4. Because of our restriction on the characteristic, Eq.(12) is satisfied strictly (that is, it holds after scalar extensions). Hence by [McC1, Theorem 3 and its proof], the generic minimum polynomial m_{x}(λ) divides the polynomial

p_{x}(λ) = λ^{4} − t(x)λ^{3} + s(x)λ^{2} − r(x)λ + n(x)

in the polynomial ring P(J)[λ], where P(J) is the ring of polynomial functions on J.

The only quartic Jordan algebra of generic degree 1 is the ground field, while the quartic Jordan algebras of generic degree 4 are exactly the generically algebraic Jordan algebras of generic degree 4. Let us characterize the quartic Jordan algebras of generic degree 2 and 3.

Proposition 2. (Degree 2 quartic Jordan algebras) Let J be a quartic Jordan algebra with trace t, generic degree 2 and generic minimum polynomial

m_{x}(λ) = λ^{2} − τ(x)λ + σ(x)  (13)

for a linear form τ and a quadratic form σ. Then precisely one of the following holds:

i) t = 2τ. ii) There are primitive idempotents e_{1}, e_{2} = 1−e_{1} ∈ J with Peirce decomposition

J = F e_{1} ⊕ F e_{2} ⊕ J_{\frac{1}{2}} (J_{\frac{1}{2}} = \{x ∈ J : e_{1} · x = \frac{1}{2}x\}) such that J_{\frac{1}{2}} · J_{\frac{1}{2}} = 0 and for any α, β ∈ F and x ∈ J_{\frac{1}{2}}

t(αe_{1} + βe_{2} + x) = α + 3β .  (14)

Conversely, given a Jordan algebra J of a quadratic form (generic degree 2), with generic minimum polynomial as in Eq.(13), then J with the trace t = 2τ is a quartic Jordan algebra; and if J satisfies the conditions in ii), J with the trace in Eq.(14) is a quartic Jordan algebra too.

Proof . Let J be a quartic Jordan algebra with trace t and generic minimum polynomial m_{x}(λ) as in Eq.(13). Let J_{0} = {x ∈ J : t(x) = 0}, so that J = F 1 ⊕ J_{0}. In case τ(J_{0}) = 0, then t(α1 + x) = 4α = 2τ(α1 + x) for any α ∈ F and x ∈ J_{0}. We get i).

Otherwise, let a be any element in J_{0} with τ(a) = 1; then a^{·2} = a − σ(a)1, a^{·3} = (1 − σ(a))a − σ(a)1 and a^{·4} = (1 − 2σ(a))a + (σ(a)^{2} − σ(a))1. By the Newton formulae, since t(1) = 4 and t(a) = 0:

2s(a) = −t(a^{·2}) = 4σ(a),
6r(a) = 2t(a^{·3}) = −8σ(a),
24n(a) = 3t(a^{·2})^{2} − 6t(a^{·4})
= 48σ(a)^{2} − 24(σ(a)^{2} − σ(a)) = 24(σ(a)^{2} + σ(a)) ,

so that

0 = a^{·4} + 2σ(a)a^{·2} + \frac{4}{3}σ(a)a + (σ(a)^{2} + σ(a)) 1 = (1 + \frac{4}{3}σ(a)) a

and σ(a) = −\frac{3}{4}, a^{·2} = a + \frac{3}{4}. Let e_{1} = \frac{1}{4} + \frac{1}{2}a and e_{2} = 1 − e_{1} = \frac{3}{4} − \frac{1}{2}a. Then e_{1} and e_{2} are primitive orthogonal idempotents with t(e_{1}) = 1 and t(e_{2}) = 3. Besides, if V = {x ∈ J_{0} : τ(x) = 0}; then for any x ∈ V, τ(a + x) = 1, so that σ(a + x) = −\frac{3}{4} as above. Hence

(a + x)^{·2} = τ(a + x)(a + x) − σ(a + x)1 = a + x + \frac{3}{4}1 = a^{·2} + x,

so that 2a · x + x^{·2} = x for any x ∈ V. Thus, x^{·2} = 0 and a · x = \frac{1}{2}x for any x ∈ V, and hence e_{1} · x = \frac{1}{2}x for any x ∈ V and we obtain case ii).

For the converse, if J is a Jordan algebra of generic degree 2 and generic minimum polynomial m_{x}(λ) as in Eq.(13), then any x ∈ J is a root of

m_{x}(λ)^{2} = λ^{4} − 2τ(x)λ^{3} + (2σ(x) + τ(x)^{2}) λ^{2} − 2τ(x)σ(x)λ + σ(x)^{2}

and, since the Newton formulae are easily checked, J is a quartic Jordan algebra with

t(x) = 2τ(x), s(x) = 2σ(x) + τ(x)^{2}, r(x) = 2τ(x)σ(x), n(x) = σ(x)^{2}.

Finally, if J is such a Jordan algebra, but with two primitive idempotents e_{1}, e_{2} = 1 − e_{1} such that J = F e_{1} ⊕ F e_{2} ⊕ J_{\frac{1}{2}} and J_{\frac{1}{2}} · J_{\frac{1}{2}} = 0, we define t by Eq.(14). One has to check Eq.(12) for any z = αe_{1} + βe_{2} + x, for α, β ∈ F and x ∈ J_{\frac{1}{2}} . But the linear map φ from the subalgebra S = F e_{1} + F e_{2} + F x of J into the Jordan algebra Mat_{4}(F)^{+}, of 4 × 4 matrices under the symmetrized product u · v = \frac{1}{2}(uv + vu), given by

e_{1} ↦ ( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} ) , e_{2} ↦ ( \begin{matrix}0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} ) , x ↦ ( \begin{matrix}0 & ϵ & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} ) ,

with ϵ = 1 if x ≠ 0 and ϵ = 0 otherwise, is a one to one homomorphism of Jordan algebras which satisfies t(z) = tr(φ(z)) for any z ∈ S, where tr is the usual trace in Mat_{4}(F). The validity of Newton formulae in Mat_{4}(F)^{+} gives now its validity in J.

Proposition 3. (Degree 3 quartic Jordan algebras) Let J be a generically algebraic Jordan algebra of generic degree 3 and generic minimum polynomial

m_{x}(λ) = λ^{3} − τ(x)λ^{2} + σ(x)λ − ν(x)  (15)

for a linear form τ, quadratic form σ and cubic form ν. Then J is a quartic Jordan algebra if and only if there is a homomorphism of unital Jordan algebras J → F. Moreover, the forms t, s, r and n are given by

\begin{aligned}t(x) & = φ(x) + τ(x) \\ s(x) & = φ(x)τ(x) + σ(x) \\ r(x) & = φ(x)σ(x) + ν(x) \\ n(x) & = φ(x)ν(x) \end{aligned}  (16)

for a suitable homomorphism φ : J → F of unital Jordan algebras.

Proof . By applying [McC1, Theorem 2] (extending previously scalars if necessary) we obtain that J is a quartic Jordan algebra if and only if there are linear forms φ, t : J → F such that

λ^{4} − t(x)λ^{3} + s(x)λ^{2} − r(x)λ + n(x) = (λ − φ(x))m_{x}(λ)

for s, r and n given by the Newton formulae. The assertion in Eq.(16) is clear. But then, for any x ∈ J,

2s(x) = t(x)^{2} − t(x^{·2})
= (φ(x) + τ(x))^{2} − (φ(x^{·2}) + τ(x^{·2}))
= (φ(x)^{2} − φ(x^{·2})) + 2φ(x)τ(x) + (τ(x)^{2} − τ(x^{·2}))
= (φ(x)^{2} − φ(x^{·2})) + 2(φ(x)τ(x) + σ(x))
= (φ(x)^{2} − φ(x^{·2})) + 2s(x)

and hence φ(x^{·2}) = φ(x)^{2} and φ : J → F is a homomorphism. Besides 4 = t(1) = φ(1) + τ(1) = φ(1) + 3, so that φ(1) = 1.

And conversely, if φ : J → F is a homomorphism of unital Jordan algebras, then the Newton formulae for τ, σ and ν easily give the Newton formulae for t, s, r and n given by Eq.(16).

Remark . If J, φ and t are as in the previous Proposition, then J = F 1 ⊕ ker φ and for any x ∈ ker φ, ν(x)1 = x^{·3} − τ(x)x^{·2} + σ(x)x ∈ ker φ, so that ν(x) = 0. By [McC1, Theorem 2], φ divides ν, so there is a quadratic form ρ : J → F such that ν(x) = φ(x)ρ(x) for any x ∈ J. Then by [McC1, Theorem 4]

\begin{aligned}m_{x}(λ) & = ν(λ1 − x) = φ(λ1 − x)ρ(λ1 − x) \\ & = (λ − φ(x))(λ^{2} − μ(x)λ + ρ(x)) \end{aligned}

for a suitable linear map μ. In particular

τ(x) = φ(x) + μ(x), σ(x) = φ(x)μ(x) + ρ(x), ν(x) = φ(x)ρ(x) ,

and by Eq.(16)

\begin{aligned}t(x) & = 2φ(x) + μ(x) \\ s(x) & = φ(x^{·2}) + 2φ(x)μ(x) + ρ(x) \\ r(x) & = φ(x^{·2})μ(x) + 2φ(x)ρ(x) \\ n(x) & = φ(x^{·2})ρ(x) . \end{aligned}  (17)

Examples of Jordan algebras satisfying the conditions of Proposition 3 are the direct products F × Q, for Q a Jordan algebra of a quadratic form.

Lemma 4. Let J be a quartic Jordan algebra with trace t. Then the trace form (x, y) ↦ t(x · y) is associative.

Proof . This follows from [McC1, Theorem 4(ii)] if the generic degree of J is 4 and it is trivial if the degree is 1.

If the generic degree is 2, according to Proposition 2 we are left with two cases. In the first one, t is twice the generic trace of J, which is associative; otherwise J is of a very specific nature and the trace t is given by Eq.(14), which is immediately checked to be associative (J_{\frac{1}{2}} does not play any role here).

Finally, if the generic degree is 3, then t(x) = φ(x) + τ(x) for any x, where φ : J → F is a homomorphism, and hence trivially associative, and τ the generic trace, which is associative too.

So far, we have been dealing in an implicit way with two categories:

i) The category AC of admissible cubic algebras, whose objects are pairs (A, ⟨ | ⟩) of an admissible cubic algebra and a trace form of it, and whose morphisms ψ : (A_{1}, ⟨ | ⟩_{1}) → (A_{2}, ⟨ | ⟩_{2}) are the homomorphisms of algebras preserving the trace forms: ⟨ψ(x)|ψ(y)⟩_{2} = ⟨x|y⟩_{1} for any x, y ∈ A_{1}.

Notice that if ψ : A_{1} → A_{2} is a nonzero homomorphism of admissible cubic algebras with respective norms N_{1} and N_{2}, then for any x ∈ A_{1} with ψ(x) ≠ 0, N_{2}(ψ(x))ψ(x) = ψ(x)^{2}ψ(x)^{2} = ψ(x^{2}x^{2}) = N_{1}(x)ψ(x), so by density, ψ preserves norms. If N_{1} ≠ 0, then for any x ∈ A_{1} N_{1}(x)⟨ψ(x)|ψ(y)⟩_{2} = N_{2}(ψ(x))⟨ψ(x)|ψ(y)⟩_{2} = N_{2}(ψ(x)^{2}, ψ(x)^{2}, ψ(y)) = N_{1}(x^{2}, x^{2}, y) = N_{1}(x)⟨x|y⟩_{1} (see Eq.(5)), so that ψ preserves the trace forms too.

Use will be made of the full subcategory ÃC, whose objects are the admissible cubic algebras (A, ⟨ | ⟩) with nonzero norm (and hence with ⟨ | ⟩ uniquely determined by the norm because of Theorem 1).

ii) The category QJ of quartic Jordan algebras, whose objects are the pairs (J, t) for a quartic Jordan algebra J and its trace t (the same Jordan algebra can be endowed with two different traces making it a quartic Jordan algebra; this situation has already appeared in Proposition 2, and also in Proposition 3 when there are several homomorphisms J → F), and whose morphisms φ : (J_{1}, t_{1}) → (J_{2}, t_{2}) are the homomorphisms of unital Jordan algebras preserving traces, and therefore preserving the s’s, r’s and n’s too. For the quartic Jordan algebras of generic degree 4 this latter condition is superfluous.

Here the full subcategory Q̃J, whose objects are those quartic Jordan algebras (J, t) for which there are elements x ∈ J with t(x) = 0 and r(x) ≠ 0 (or t(x^{·3}) ≠ 0), which is equivalent to the fact that t does not divide the cubic form r (see [McC1, Theorem 2]).

It is easy to check that the generic degree 2 quartic Jordan algebras with t dividing r are the ones in Proposition 2.i). On the other hand, if (J, t) is a quartic Jordan algebra of degree 3 and t divides r, from Eq.(17) r(x) = φ(x)^{2}μ(x) + 2φ(x)ρ(x) = t(x)ρ(x) + μ(x)(φ(x)^{2} − ρ(x)) and it follows that either t divides μ or t divides φ^{2} − ρ. But if t divides μ, since t(1) = 4 and φ(1) = 1, t = 2μ, so μ = 2φ, t = 4φ and by the Newton formulae the quartic polynomial satisfied by J reduces to (λ − φ(x))^{4}. Also if t divides φ^{2} − ρ, then for any α ∈ F and x ∈ ker t, φ(x)^{2} = ρ(x), so

φ(α1 + x)^{2} − ρ(α1 + x) = α^{2} + 2αφ(x) + φ(x)^{2} − α^{2} − μ(x)α − ρ(x)

= α(2φ(x) − μ(x)) = 4αφ(x)

= t(α1 + x) (φ(α1 + x) − \frac{1}{4}t(α1 + x))

and therefore φ^{2} − ρ = t (φ − \frac{1}{4}t), so

ρ = φ^{2} + t (\frac{1}{4}t − φ) = (φ − \frac{1}{2}t)^{2} = (\frac{1}{2}μ)^{2} .

Let ψ = \frac{1}{2}μ so that ψ(1) = 1, then ρ = ψ^{2} and m_{x}(λ) = (λ − φ(x))(λ − ψ(x))^{2}. Besides, by the previous remark, τ = φ + 2ψ, σ = 2φψ + ψ^{2} and ν = φψ^{2}. The Newton formula 2σ(x) = τ(x^{2})−τ(x^{·2}) forces ψ(x^{·2}) = ψ(x)^{2}, so that ψ : J → F is a homomorphism of unital Jordan algebras too. Also, t = 2(φ + ψ) and r = 2φ^{2}ψ +2φψ^{2} = φψt by Eq.(17). Therefore the generic minimum polynomial of the generic degree 3 quartic Jordan algebras in which t divides r is of the form m_{x}(λ) = (λ − φ(x))(λ − ψ(x))^{2} with φ, ψ : J → F homomorphisms of unital Jordan algebras. If φ = ψ, then J = F 1 ⊕ N, with N(= ker φ) the nil radical of J.

Finally, it is clear that the Jordan algebras of generic degree 4 of the form J = F 1 ⊕ N with N the nilradical of J verify that t divides r; in this case the generic minimum polynomial is (λ − φ(x))^{4} = (λ − \frac{1}{4}t(x))^{4}. Assume now that J is a generic degree 4 quartic Jordan algebra with t dividing r and which is not the sum of F 1 and its nilradical. Then, by density, J contains an element x ∈ J with t(x) = 0, x^{·4} ≠ 0 and such that the subalgebra generated by x is four dimensional. Let J_{x} be such subalgebra. If s(x) = 0, then the minimum polynomial of x is λ^{4}−α for some 0 ≠ α ∈ F. But then t(x) = t(x^{·2}) = t(x^{·3}) = 0 since s(x) = 0 and J is not in Q̃J, and hence the element z = x + x^{·2} satisfies t(z) = 0 and

t(z^{·3}) = t(3α1 + 3αx + αx^{·2} + x^{·3}) = 12α ≠ 0 ,

a contradiction with t dividing r. Therefore s(x) ≠ 0 and the minimum polynomial of x is

λ^{4} + s(x)λ^{2} + n(x) = (λ^{2} + \frac{1}{2}s(x))^{2} + (n(x) − \frac{1}{4}s(x)^{2}) = (λ^{2} − α)^{2} − β

with 0 ≠ α = −\frac{1}{2}s(x) = \frac{1}{4}t(x^{·2}). But then t(x^{·2} − α1) = 0 and the element z = x + x^{·2} − α1 satisfies t(z) = 0 and

0 = t(z^{·3})
= t(x^{·3}) + t (3x^{·2} · (x^{·2} − α1)) + t (3x · (x^{·2} − α1)^{·2}) + t ((x^{·2} − α1)^{·3})

= 0 + 3t ((x^{·2} − α1)^{2}) + 3αt (x^{·2} − α1) + 3t(βx) + t (β(x^{·2} − α1)) = 12β .

Hence n(x) = \frac{1}{4}s(x)^{2} and (x^{·2} − \frac{1}{4}t(x^{·2})1)^{2} = 0. Therefore this is valid by density for any element of zero trace and for any z ∈ J,

(z − \frac{1}{4}t(z)1)^{·2} − \frac{1}{4}t ((z − \frac{1}{4}t(z)1)^{·2}) 1

squares to zero. Thus, the generic minimum polynomial is

(λ^{2} − \frac{1}{2}t(x)λ + σ(x))^{2}

for a suitable quadratic form σ. This shows that the reduced minimum polynomial of J (see [McC1, Section 5]) is a degree 2 polynomial.

In conclusion, the generic degree 4 quartic Jordan algebras in which t divides r are those generic degree 4 Jordan algebras whose reduced degree is 1 or 2; that is, the reduced minimum polynomial is either λ − \frac{1}{4}t(x) or λ^{2} − \frac{1}{2}t(x)λ + σ(x).

Examples of the latter algebras are the direct products J_{1} ×J_{2}, where J_{1} and J_{2} are Jordan algebras of trivial quadratic forms. An example which is not such a direct product is given by the Jordan algebra with a basis {e_{1}, e_{2}, v_{1}, v_{2}, w} and multiplication determined by

e_{i}^{·2} = e_{i}, e_{i} · v_{i} = v_{i}, e_{i} · w = \frac{1}{2}w (i = 1, 2)

and all other products equal to zero. Notice that the subalgebra H spanned by {e_{1}, e_{2}, v_{2}, w} is a generic degree 3 quartic Jordan algebra with generic minimum polynomial m_{x}(λ) = (λ − φ(x))(λ − ψ(x))^{2} for two different homomorphisms φ, ψ : J → F.

We are ready to relate the admissible cubic algebras with the quartic Jordan algebras. Let (J, t) be any quartic Jordan algebra and let J_{0} = {x ∈ J : t(x) = 0} be its subset of trace zero elements. Define a multiplication on J_{0} by

xy = x · y − \frac{1}{4}t(x · y)1 ,  (18)

that is, xy is the projection proj_{J_{0}}(x · y) of the product x · y onto J_{0} relative to the decomposition J = F 1 ⊕ J_{0}. Define too an associative symmetric bilinear form on J_{0} by means of

⟨x|y⟩ = \frac{1}{3}t(x · y)  (19)

for x, y ∈ J_{0}. Then:

Theorem 5. The assignment C : (J, t) ↦ (J_{0}, ⟨ | ⟩) gives, with the natural assignment for homomorphisms, a faithful and full functor from QJ to AC, which restricts to an equivalence between the subcategories Q̃J and ÃC.

Proof . Let (J, t) be a quartic Jordan algebra. For any x ∈ J_{0},

x^{·4} + s(x)x^{·2} − r(x)x + n(x)1 = 0

and s(x) = −\frac{1}{2}t(x^{·2}) and r(x) = \frac{1}{3}t(x^{·3}). From Eq.(18) we compute for any x ∈ J_{0}:

x^{2}x^{2} = proj_{J_{0}}(x^{2} · x^{2})

= proj_{J_{0}} ((x^{·2} − \frac{1}{4}t(x^{·2})1) · (x^{·2} − \frac{1}{4}t(x^{·2})1))

= proj_{J_{0}} (x^{·4} − \frac{1}{2}t(x^{·2})x^{·2}) = proj_{J_{0}} (x^{·4} + s(x)x^{·2})

= r(x)x = \frac{1}{3}t(x · x^{·2})x ,

and thus (J_{0}, \frac{1}{3}t( · )) is an admissible cubic algebra (recall that the trace form is associative by Lemma 4). If (J, t) is an object in Q̃J, then the restriction of r to J_{0} is nontrivial and hence the norm of this admissible cubic algebra is not zero, so it is an object in ÃC.

It is straightforward to check that the functor thus defined is faithful and full.

Now, if (A, ⟨ | ⟩) is an admissible cubic algebra with nonzero norm N(x) = ⟨x|x^{2}⟩, let J_{A} = F 1 ⊕ A with the commutative multiplication given by:

\{ \begin{matrix}1 · 1 = 1, 1 · x = x \\ x · y = xy + \frac{3}{4}⟨x|y⟩1 \end{matrix}  (20)

for x, y ∈ A. Denote by (X, Y, Z)^{·} = (X · Y) · Z − X · (Y · Z) the associator in J_{A}. Then for any X = α1 + x, Y = β1 + y ∈ J_{A}, with α, β ∈ F and x, y ∈ A:

((X · X) · Y) · X − (X · X) · (Y · X) = (X · X, Y, X)^{·} = (x^{2}, y, x)^{·}

since 1 associates with everything and (x, y, x)^{·} = 0 by commutativity. Hence, by associativity of ⟨ | ⟩,

((X · X) · Y) · X − (X · X) · (Y · X) = (x^{2}, y, x)^{·}

= (x^{2} · y) · x − x^{2} · (y · x)

= (x^{2}y) · x + \frac{3}{4}⟨x^{2}|y⟩x − x^{2} · (yx) − \frac{3}{4}⟨x|y⟩x^{2}

= (x^{2}y)x + \frac{3}{4}⟨x^{2}y|x⟩1 + \frac{3}{4}⟨x^{2}|y⟩x

− x^{2}(yx) − \frac{3}{4}⟨x^{2}|yx⟩1 − \frac{3}{4}⟨x|y⟩x^{2}

= (x^{2}y)x − x^{2}(yx) + \frac{3}{4}⟨x^{2}|y⟩x − \frac{3}{4}⟨x|y⟩x^{2} ,

which is zero by Eq.(11). Therefore, J_{A} is a unital Jordan algebra. Besides, with X as before,

X^{·2} = (α^{2} + \frac{3}{4}⟨x|x⟩) 1 + 2αx + x^{2} ,
X^{·3} = ((α^{2} + \frac{3}{4}⟨x|x⟩) 1 + 2αx + x^{2}) · (α1 + x)

= (α^{3} + \frac{9}{4}α⟨x|x⟩ + \frac{3}{4}⟨x|x^{2}⟩) 1 + (3α^{2} + \frac{3}{4}⟨x|x⟩) x + 3αx^{2} + x^{3} ,

X^{·4} = ((α^{2} + \frac{3}{4}⟨x|x⟩) 1 + 2αx + x^{2})^{·2}

= (α^{4} + \frac{9}{2}α^{2}⟨x|x⟩ + \frac{9}{16}⟨x|x⟩^{2} + 3α⟨x|x^{2}⟩ + \frac{3}{4}⟨x^{2}|x^{2}⟩) 1

+ (4α^{3} + 3α⟨x|x⟩ + ⟨x|x^{2}⟩) x + (6α^{2} + \frac{3}{2}⟨x|x⟩) x^{2} + 4αx^{3} .

Since N(x) = ⟨x|x^{2}⟩ is the norm of A, if we define

\begin{aligned}t(X) & = 4α \\ s(X) & = 6α^{2} − \frac{3}{2}⟨x|x⟩ \\ r(X) & = 4α^{3} − 3α⟨x|x⟩ + N(x) \\ n(X) & = α^{4} − \frac{3}{2}α^{2}⟨x|x⟩ + αN(x) + \frac{9}{16}⟨x|x⟩^{2} − \frac{3}{4}⟨x^{2}|x^{2}⟩ , \end{aligned}

then

X^{·4} − t(X)X^{·3} + s(X)X^{·2} − r(X)X + n(X)1 = 0

for any X = α1+x ∈ J_{A}. A direct computation now gives the Newton formulae, so that (J_{A}, t) is a quartic Jordan algebra. Moreover, by Eq.(20), t(x·y) = 3⟨x|y⟩ for any x, y ∈ A and hence the admissible cubic algebra ((J_{A})_{0}, \frac{1}{3}t( · )) is precisely (A, ⟨ | ⟩). This completes the proof.

Some remarks are in order:

1) We may have considered the category A, whose objects are the admissible cubic algebras and whose morphisms are those homomorphisms of algebras that preserve the norm (this latter condition being automatic for nonzero homomorphisms), but then the functor QJ → A which takes any quartic Jordan algebra (J, t) to the admissible cubic algebra defined by Eq.(18), fails to be full.

For example, take J = F [λ]/(λ^{4}) and Ĵ = F^{4}. Both J and Ĵ are generically algebraic, of generic degree 4, Jordan algebras and hence they are quartic Jordan algebras in a unique way. But while Ĵ is separable, J = F 1 ⊕ N, for a nilpotent ideal N. Let x be the class of λ in J. Then J_{0} is spanned by x, x^{2} and x^{3} (x^{4} = 0), while Ĵ_{0} = {(α, β, γ, δ) ∈ F^{4} : α + β + γ + δ = 0}. The linear map ψ : J_{0} → Ĵ_{0} given by ψ(x) = (1, 1, −1, −1) and ψ(x^{2}) = ψ(x^{3}) = 0 is a homomorphism of admissible cubic algebras since ψ((J_{0})^{2}) = 0 = ψ(J_{0})^{2}, because a^{2} = a^{·2} − \frac{1}{4}t(a^{·2})1 = 0 for a = (1, 1, −1, −1). However, ψ does not extend to a homomorphism φ : J → Ĵ of Jordan algebras, since otherwise

1 = a · a = φ(x) · φ(x) = φ(x · x) = ψ(x^{2}) = 0 .

Notice that the norm of the admissible cubic algebra J_{0} is trivial.

However, the restriction to Q̃J yields still an equivalence between Q̃J and the subcategory of admissible cubic algebras with nonzero norm.

2) Not all admissible cubic algebras are (isomorphic to) images under the functor QJ → AC of a quartic Jordan algebra. For instance, let A be the span of four basic elements a, b, c, d with the commutative multiplication given by a^{2} = b, ab = c, ac = d and all other products being zero. Hence (A^{2})^{2} = 0, so A is an admissible cubic algebra. Suppose there is a quartic Jordan algebra J such that J = F 1 ⊕ A and the product in A is the projection onto A of the product in J. Then a^{·2} = α1 + b, a^{·3} = β1 + αa + c and a^{·4} = γ1 + βa + αb + d for suitable scalars α, β, γ ∈ F. Hence 1, a, a^{·2}, a^{·3} and a^{·4} are linearly independent and hence J cannot be a quartic Jordan algebra.

3) In [MO], Meyberg and Osborn classified, under some restrictions, the commutative algebras A over fields of characteristic ≠ 2, 3 satisfying

(x ∗ x) ∗ x = (x|x)x  (21)

for any x ∈ A, where ( |) is a symmetric bilinear form. These are called pseudocomposition algebras. This condition (21) also appeared in [RW, 1.7] and in [GMW, Section 4]. Pseudo-composition algebras with associative bilinear form ( |) are closely related to Jordan algebras of generic degree three (this goes back to [S]). Let us sketch this connection here. Assume that Eq.(21) is satisfied and, moreover, that ( |) is associative. Linearizing Eq.(21) one gets

(x ∗ x) ∗ y + 2(x ∗ y) ∗ x = 2(x|y)x + (x|x)y.  (22)

Multiplying by x:

((x ∗ x) ∗ y) ∗ x + 2((y ∗ x) ∗ x) ∗ x = 2(x|y)x ∗ x + (x|x)y ∗ x,  (23)

while substituting y by y ∗ x in Eq.(22):

(x ∗ x) ∗ (y ∗ x) + 2((y ∗ x) ∗ x) ∗ x = 2(x|y ∗ x)x + (x|x)y ∗ x.  (24)

Subtracting Eq.(24) from Eq.(23) and using the associativity of ( |) give

((x ∗ x) ∗ y) ∗ x − (x ∗ x) ∗ (y ∗ x) = 2(x|y)x ∗ x − 2(x ∗ x|y)x.  (25)

Now consider the new algebra J = F 1 ⊕ A obtained by adjoining an identity element 1 and by defining

x · y = x ∗ y + 2(x|y)1  (26)

for any x, y ∈ A. Now as in the proof of Theorem 5 above, changing \frac{3}{4}⟨x|y⟩ by 2(x|y) after Eq.(20), the algebra J is shown to be a Jordan algebra. A direct computation shows that for any X = α1 + x ∈ J:

X^{·3} − 3αX^{·2} + 3 (α^{2} − (x|x)) X − (α^{3} − 3α(x|x) + 2(x|x ∗ x)) 1 = 0.

Therefore, J is a generically algebraic Jordan algebra of generic degree at most 3.

Moreover, if the degree is exactly 3, then the generic trace is given by

t(α1 + x) = 3α

so that A is just the algebra of trace zero elements of a Jordan algebra of degree 3 with the multiplication given by the projection of the multiplication in the Jordan algebra onto the set of trace zero elements. Besides, the bilinear form associated to the trace (the trace form) is given by

t(α1 + x, β1 + y) = t((α1 + x) · (β1 + y)) = 3αβ + 6(x|y)

for any α, β ∈ F and x, y ∈ A, which is nondegenerate if and only if so is ( |). In this case J is separable [McC1, Theorems 16 and 17] (for the finite dimensional case see alternatively [J, Theorem VI.5]).

If the generic degree of J is lower than 3, then there is a linear form φ such that

x ∗ x = φ(x)x

for any x ∈ A, so that (x|x)x = (x ∗ x) ∗ x = φ(x)^{2}x, and (x|x) = φ(x)^{2} for any x ∈ A. The product in A is then just given by

x ∗ y = \frac{1}{2} (φ(x)y + φ(y)x) .

As a consequence of our arguments in Section 5, it will be proved that the symmetric bilinear form associated to any pseudo-composition algebra is always associative, which allows to simplify some of the arguments in [MO].

4 Nondegenerate admissible cubic algebras

We start this section by describing the admissible cubic algebras obtained as C(J, t) (Theorem 5) for some interesting quartic Jordan algebras.

Proposition 6.

i) Let (J, t) be the quartic Jordan algebra where J = F 1 ⊕ V is the Jordan algebra of a quadratic form q : V → F (so v^{·2} = q(v)1 for any v ∈ V) and t(α1 + v) = 4α for any α ∈ F and v ∈ V (twice the generic trace of J if V ≠ 0). Then C(J, t) = (V, ⟨ | ⟩) where V^{2} = 0 and ⟨v|v⟩ = \frac{4}{3}q(v) for any v ∈ V. Conversely, given any trivial algebra A (A^{2} = 0) and symmetric bilinear form ⟨ | ⟩ defined on it, (A, ⟨ | ⟩) is the image under C of a suitable quartic Jordan algebra of generic degree ≤ 2.

ii) Let J = F e_{1} ⊕ F e_{2} ⊕ V be the Jordan algebra with multiplication given by

e_{i}^{·2} = e_{i}, e_{i} · v = \frac{1}{2}v, e_{1} · e_{2} = 0 = v^{·2}

for any v ∈ V; and let t : J → F be the linear form determined by t(V) = 0, t(e_{1}) = 1, t(e_{2}) = 3. Then (J, t) is a quartic Jordan algebra and C(J, t) = (A, ⟨ | ⟩), with A = F a ⊕ V, a = \frac{3e_{1}−e_{2}}{2} and

\begin{matrix}a^{2} = a, av = \frac{1}{2}v, v^{2} = 0, \\ ⟨a|a⟩ = 1, ⟨a|v⟩ = 0 = ⟨v|v⟩, \end{matrix}

for any v ∈ V. iii) Let J = J_{1} ⊕ J_{2} (direct sum of ideals) be the Jordan algebra where J_{i} =

F e_{i} ⊕ V_{i} is the Jordan algebra of a quadratic form q_{i} : V_{i} → F with identity e_{i}, and let t : J → F be the linear form determined by t(V_{1} + V_{2}) = 0, t(e_{1}) = t(e_{2}) = 2 (t is the generic trace in case V_{1} ≠ 0 ≠ V_{2}). Then (J, t) is a quartic Jordan algebra and C(J, t) = (A, ⟨ | ⟩) with A = F a ⊕ V_{1} ⊕ V_{2}, a = e_{1} − e_{2} and

\begin{matrix}(αa + v_{1} + v_{2})^{2} = \frac{1}{2} (q_{1}(v_{1}) + q_{2}(v_{2})) a + 2α(v_{1} − v_{2}), \\ ⟨αa + v_{1} + v_{2}|αa + v_{1} + v_{2}⟩ = \frac{2}{3} (2α^{2} + q_{1}(v_{1}) + q_{2}(v_{2})) , \end{matrix}

for any α ∈ F, v_{1} ∈ V_{1} and v_{2} ∈ V_{2}. iv) Let K/F be a quadratic field extension with Galois automorphism k ↦ k̄

and let J = K1 ⊕ V be the Jordan algebra (over K) of a quadratic form q : V → K. Let J^{F} be the algebra J, but viewed as an algebra over F. Then J^{F} is a quartic Jordan algebra with the trace given by t(k1 + v) = 2(k + k̄) (which is the generic trace of J^{F} if V ≠ 0) and C(J, t) = (A, ⟨ | ⟩) with A = K_{0} ⊕ V, K_{0} = {k ∈ K : k̄ = −k} and

\begin{matrix}(k + v)^{2} = \frac{1}{2} (q(v) + \overline{q(v)}) + 2kv, \\ ⟨k + v|k + v⟩ = \frac{2}{3} (2k^{2} + q(v) + \overline{q(v)}) , \end{matrix}

for any k ∈ K_{0} and v ∈ V.

v) Let (J, t) be the quartic Jordan algebra with J = F × H and H a generically algebraic Jordan algebra of degree 3 and generic minimum polynomial

λ^{3} − τ(x)λ^{2} + σ(x)λ − ν(x)

and where t((α, x)) = α+τ(x) (the generic trace) for any α ∈ F and x ∈ H. Then C(J, t) is isomorphic to the admissible cubic algebra defined on H by means of

\begin{matrix}x^{2}(= x^{#}) = x^{·2} − τ(x)x + σ(x)1_{H}, \\ ⟨x|x⟩ = \frac{1}{3}τ(x^{·2}). \end{matrix}

for any x ∈ H (here 1_{H} denotes the unit element in H).

Proof . All cases but v) are obtained with straightforward computations using Eqs.(18) and (19). The quartic Jordan algebras in i) and ii) are precisely the ones in Proposition 2 (generic degree 2), while case iii), with V_{1} = 0 ≠ V_{2}, fits into Proposition 3 (generic degree 3).

Let us prove case v). There J_{0} = F e ⊕ H_{0} with e = (\frac{3}{2}, −\frac{1}{2}) ∈ F × H and where H_{0} = {x ∈ H : τ(x) = 0} is identified with 0 × H_{0}. From Eq.(18) we conclude that:

e^{2} = e, ex = −\frac{1}{2}x,

x^{2} = x · x − \frac{1}{4}t(x · x)1 = −\frac{1}{6}τ(x · x)e + (x · x − \frac{1}{3}τ(x · x)1_{H}) ,

⟨e|e⟩ = 1, ⟨e|x⟩ = 0, ⟨x|x⟩ = \frac{1}{3}τ(x^{·2}),

for any x ∈ H_{0}. But 2σ(x) = τ(x)^{2} − τ(x^{·2}), so that

x^{#} = x^{·2} − τ(x)x + σ(x)1_{H} = x^{·2} − \frac{1}{2}τ(x^{·2})1_{H}

and

x^{2} = −\frac{1}{6}τ(x^{·2})e + (x^{#} + \frac{1}{6}τ(x^{·2})1_{H}) ∈ F e ⊕ H_{0}

for any x ∈ H_{0}. Therefore the linear map Φ : H → J_{0}, α1_{H} + x ↦ αe + x for α ∈ F and x ∈ H_{0}, satisfies

Φ ((α1_{H} + x)^{#}) = Φ (α^{2}1_{H} − αx + x^{#})

= Φ ((α^{2} − \frac{1}{6}τ(x^{·2}))1_{H} + (x^{#} + \frac{1}{6}τ(x^{·2})1_{H} − αx))

= (α^{2} − \frac{1}{6}τ(x^{·2}))e + (x^{#} + \frac{1}{6}τ(x^{·2})1_{H} − αx)

= α^{2}e − αx + x^{2} = (αe + x)^{2}

= Φ (α1_{H} + x)^{2} ,

and

\frac{1}{3}τ ((α1_{H} + x)^{·2}) = \frac{1}{3}τ (α^{2}1_{H} + x^{·2}) = α^{2} + \frac{1}{3}τ(x^{·2}) = ⟨αe + x|αe + x⟩,

for any x ∈ H_{0}, whence the result.

There is some overlapping in Proposition 6. Case iii) with V_{1} = V_{2} = 0 fits into case i) and the same happens with case iv) with V = 0. Also case v) with H = F × Q, Q being the Jordan algebra of a quadratic form, fits into case iii), because then J = F ×H ≅ (F ×F)×Q is the direct sum of two Jordan algebras of quadratic forms.

Only in case i) (including cases iii) and iv) with V_{1} = V_{2} = 0 or V = 0) the norm of the admissible cubic algebra is zero. The admissible cubic algebras that appear in iii) will be denoted by A((V_{1}, q_{1}), (V_{2}, q_{2})) and the ones in case iv) by A_{K}(V, q).

Given an admissible cubic algebra (A, ⟨ | ⟩), the radical of the associative symmetric bilinear form, R = {x ∈ A : ⟨x|A⟩ = 0}, is an ideal of A. If (A, ⟨ | ⟩) = C(J, t), R is the radical of the trace form (x, y) ↦ t(x · y) since ⟨x|y⟩ = \frac{1}{3}t(x · y) for any x, y ∈ J_{0}, and t(1) = 4 ≠ 0. R is then an ideal of J too and it is nil since x^{·4} = x^{·2} · x^{·2} = x^{2}x^{2} = ⟨x|x^{2}⟩x = 0 for any x ∈ R (because of Eq.(18), xy = x · y for any x, y ∈ R). Notice that since R is nil of bounded index 4, it is locally nilpotent, and even solvable if the characteristic is zero (see [Ze] and the references therein).

By factoring out R, we get an admissible cubic algebra with nondegenerate trace form.

Theorem 7. The functor C : QJ → AC restricts to an equivalence between the full subcategories QJ_{nd}, whose objects are those quartic Jordan algebras (J, t) with nondegenerate trace form (x, y) ↦ t(x · y), and AC_{nd}, whose objects are the admissible cubic algebras (A, ⟨ | ⟩) with nondegenerate trace form.

Proof . Since C(J, t) = (A, ⟨ | ⟩) with A = J_{0} = {x ∈ J_{0} : t(x) = 0}, ⟨x|y⟩ = \frac{1}{3}t(x · y) for any x, y ∈ A and since t(1) = 4 ≠ 0, it follows that if (J, t) is an object in QJ_{nd}, then C(J, t) is an object in AC_{nd}.

Now, let (A, ⟨ | ⟩) be an admissible cubic algebra with nondegenerate trace form. If the norm N(x) = ⟨x|x^{2}⟩ is not identically zero, then we already know (Theorem 5) that (J_{A} = F 1 ⊕ A, t), with t(α1 + x) = 4α for any α ∈ F and x ∈ A, and multiplication as in Eq.(20), is a quartic Jordan algebra, and since t(x · y) = 3⟨x|y⟩ for any x, y ∈ A, it is an object in QJ_{nd} whose image under C is (isomorphic to) (A, ⟨ | ⟩). Otherwise, if the norm N(x) = ⟨x|x^{2}⟩ is identically zero, then linearizing ⟨x|x^{2}⟩ = 0 and using the associativity of ⟨ | ⟩, we obtain ⟨y|x^{2}⟩ = 0 for any x, y ∈ A and, by nondegeneracy, A^{2} = 0. Then we apply Proposition 6.i) to get a (J, t) in QJ_{nd} such that C(J, t) = (A, ⟨ | ⟩), as required.

Corollary 8. The following list exhausts, up to isomorphism, all the admissible cubic algebras with nondegenerate trace form:

i) The field F with ⟨1|1⟩ = 1. ii) Any trivial algebra (A^{2} = 0) with any nondegenerate symmetric bilinear

form. iii) The algebras A((V_{1}, q_{1}), (V_{2}, q_{2})) with 0 ≠ dim_{F} V_{1} ≥ dim_{F} V_{2}, and q_{1} and

q_{2} (if V_{2} ≠ 0) nondegenerate quadratic forms. iv) The algebras A_{K}(V, q), where K is a quadratic field extension of F, V ≠ 0

and q : V → K is a nondegenerate quadratic form.

v) The finite dimensional simple Jordan algebras H of generic degree 3 and generic trace τ but with the new multiplication given by

x^{2} = x^{#} = x^{·2} − τ(x)x + \frac{1}{2} (τ(x)^{2} − τ(x^{·2})) 1

and trace form given by ⟨x|y⟩ = \frac{1}{3}τ(x · y) for any x, y ∈ H. vi) The algebras defined on the set of generic trace zero elements in a finite

dimensional simple Jordan algebra J of generic degree 4, generic trace t and whose center is not a quadratic field extension of F, with multiplication and trace form given by

xy = x · y − \frac{1}{4}t(x · y)1 and ⟨x|y⟩ = \frac{1}{3}t(x · y)

for any x, y ∈ J_{0}.

Moreover, algebras in different items are not isomorphic, algebras in item ii) are isomorphic (as algebras) if and only if they have the same dimension, algebras in item iii) or iv) are isomorphic if and only if the corresponding spaces with quadratic form are isometric, while algebras in items v) or vi) are isomorphic if and only if so are the corresponding Jordan algebras H or J.

Proof . By Theorem 7, one has to describe the quartic Jordan algebras (J, t) with nondegenerate trace form (x, y) ↦ t(x·y). By [McC1, Theorems 16 and 17], such a J is a direct sum of simple ideals which are either finite dimensional or of degree 2 over their centers. The different possibilities for J, with the exception of case vi), appear in Proposition 6, and the nondegeneracy of the trace form restricts case ii) of Proposition 6 by imposing V = 0 and also impose the nondegeneracy of the quadratic forms involved in other cases.

Notice that in case v) above, either H is a cubic field extension of F or a central simple degree 3 Jordan algebra. Albert algebras are the most distinguished examples. Also in case vi) either J is a quartic field extension of F or a central simple Jordan algebra of degree 4 (and hence special).

Also, the algebras defined on the set J_{0} of generic trace zero elements in finite dimensional central simple algebras J, with the multiplication given by the projection, xy = proj_{J_{0}}(x · y), in case vi) above, play a role in the determination of invariant affine connections on irreducible symmetric spaces (see [BDE]).

5 Relation with pseudo-composition algebras

In this section we will relate the admissible cubic algebras with nonzero norm with the pseudo-composition algebras, which satisfy Eq.(21).

We start by relating the nonzero idempotents in admissible cubic algebras with specific idempotents in quartic Jordan algebras. Notice that if e is a nonzero idempotent in the admissible cubic algebra (A, ⟨ | ⟩), then e = e^{2}e^{2} = N(e)e, so N(e) = 1. In particular the norm is nonzero and hence, by Theorem 5, there is a quartic Jordan algebra (J, t) with C(J, t) = (A, ⟨ | ⟩).

Also notice that if f is an idempotent of a quartic Jordan algebra (J, t), 0 ≠ f ≠ 1, and t(f) = α, the Newton formulae then give 2s(f) = α^{2} − α, 6r(f) = α^{3} − 3α^{2} + 2α, 24n(f) = α^{4} − 6α^{3} + 11α^{2} − 6α, so

(1 − α + \frac{α^{2} − α}{2} − \frac{α^{3} − 3α^{2} + 2α}{6}) f + \frac{α^{4} − 6α^{3} + 11α^{2} − 6α}{24} 1 = 0,

whence α^{3} − 6α^{2} + 11α − 6 = (α − 1)(α − 2)(α − 3) = 0. Hence t(f) = 1, 2 or 3. Now

Lemma 9. Let (J, t) be a quartic Jordan algebra with C(J, t) = (A, ⟨ | ⟩). Then the map

{idempotents of J of trace 1} ⟶ {nonzero idempotents of A}
f  ↦  2f − \frac{1}{2}

is a bijection.

Proof . If f^{·2} = f and t(f) = 1, then t (2f − \frac{1}{2}) = 0, so that e = 2f − \frac{1}{2} ∈ J_{0} = A and

e^{2} = e^{·2} − \frac{1}{4}t(e^{·2})1 = 2f + \frac{1}{4} − \frac{1}{4}t (2f + \frac{1}{4}) = 2f + \frac{1}{4} − \frac{3}{4} = e.

Conversely, if 0 ≠ e ∈ A is an idempotent and f = \frac{1}{2}e+ \frac{1}{4}, then t(f) = t (\frac{1}{4}) = 1, and since e^{·2} = e^{2} + \frac{3}{4}⟨e|e⟩1 and 1 = N(e) = ⟨e|e^{2}⟩ = ⟨e|e⟩, we obtain

f^{·2} = \frac{1}{4}e^{·2} + \frac{1}{4}e + \frac{1}{16}

= \frac{1}{4} (e^{2} + \frac{3}{4}) + \frac{1}{4}e + \frac{1}{16}

= \frac{1}{2}e + \frac{1}{4} = f.

Proposition 10. Let (J, t) be a quartic Jordan algebra in Q̃J. Then there is a field extension K/F of degree 1, 2 or 4 such that K ⊗_{F} J contains idempotents of trace 1.

Proof . Let A be the associated cubic algebra; it is enough to prove that A contains nonzero idempotents.

Assume first that there is a x ∈ A with N(x) ≠ 0 and such that x^{3} depends linearly on x and x^{2}. If x^{2} ∈ F x, then x^{2} = λx and N(x) = λ^{3} ≠ 0 since x^{2}x^{2} = N(x)x, thus e = \frac{1}{λ}x is a nonzero idempotent. Otherwise, x and x^{2} are linearly independent and there are scalars α, β ∈ F with x^{3} = αx + βx^{2}. From Eq.(9) we know that 4xx^{3} = 3⟨x|x⟩x^{2} + N(x)x, and since

4xx^{3} = 4x(αx + βx^{2}) = 4αx^{2} + 4β(αx + βx^{2}) = 4αβx + 4(α + β^{2})x^{2} ,

it follows that 4αβ = N(x) so, in particular, α ≠ 0 ≠ β. Let K be the extension of F obtained by adjoining a square root λ of 2α, with λ ≠ −2β (change λ by −λ if necessary). Then

(λx + x^{2})^{2} = λ^{2}x^{2} + 2λ(αx + βx^{2}) + N(x)x
= λ(λ + 2β)x^{2} + (2λα + 4αβ)x
= λ(λ + 2β)x^{2} + (λ^{3} + λ^{2}2β)x
= λ(λ + 2β)(λx + x^{2})

and e = \frac{1}{λ(λ+2β)}(λx + x^{2}) is a nonzero idempotent.

Now assume that x ∈ A satisfies N(x) ≠ 0 and x, x^{2}, x^{3} are linearly independent. Consider the unital subalgebra of J generated by x. It is four dimensional and, since t(x) = 0, isomorphic to the quotient of the polynomial algebra F [λ] modulo the ideal generated by a monic polynomial p(λ) = λ^{4} − ηλ^{2} − νλ − μ, where ν = r(x) = N(x) ≠ 0. Over the algebraic closure of F, p(λ) is neither of the form (λ − α)^{4} nor (λ − α)^{2}(λ − β)^{2}, because the coefficient of λ^{3} in p(λ) is 0, while the coefficient of λ is ≠ 0. Hence p(λ) has a root, say α, of multiplicity 1. Let K be the field obtained by adjoining this root α to F. Then p(λ) = (λ − α)q(λ) over K with q(α) ≠ 0 and

K ⊗_{F} F [λ]/(p(λ)) ≅ K[λ]/(p(λ)) ≅ K[λ]/(λ − α) × K[λ]/(q(λ)) ≅ K × B .

But in K × B, the element f = (1, 0) satisfies f^{·2} = f and its generic trace is 1. Notice that if p(λ) is irreducible, the extension K/F has degree 4, while if it is reducible, either it has a root of multiplicity 1 in F, so K = F, or α is a root of an irreducible factor of p(λ) of degree 2, so that K/F has degree 2.

In the presence of an idempotent, we can attach a pseudo-composition algebra to any admissible cubic algebra.

Let A be such an algebra and 0 ≠ e ∈ A with e^{2} = e; then Eq.(1) implies that N(e) = ⟨e|e^{2}⟩ = ⟨e|e⟩ = 1. Let V be the orthogonal complement to F e relative to ⟨ | ⟩. Equation (10) with x = e gives

4e(ex) = 3⟨e|x⟩e + x ,  (27)

so that V = {x ∈ A : (2e)((2e)x) = x}.

We define a commutative multiplication on V by means of

x ∗ x = 4(ex)(ex) − 4⟨e|(ex)(ex)⟩e  (28)

for any x ∈ V; that is, the projection of the square of 2ex onto V. Because of the associativity of ⟨ | ⟩,

4⟨e|(ex)(ex)⟩ = 4⟨e(ex)|ex⟩ = ⟨x|ex⟩ = ⟨e|x^{2}⟩ ,

and also Eq.(1) gives by linearization

4(xy)(xy) + 2x^{2}y^{2} = 3⟨y|x^{2}⟩y + 3⟨x|y^{2}⟩x .

Thus

4(ex)(ex) + 2ex^{2} = 3⟨e|x^{2}⟩e + 3⟨x|e⟩x

for any x ∈ A. Therefore Eq.(28) may be rewritten as

x ∗ x = −2ex^{2} + 2⟨e|x^{2}⟩e  (29)

for any x ∈ V, or

x ∗ y = −2e(xy) + 2⟨e|xy⟩e

for any x, y ∈ V.

Now, for any x ∈ V:

(x ∗ x) ∗ x = −2e((x ∗ x)x) + 2⟨e|(x ∗ x)x⟩e
= −2e ((−2ex^{2} + 2⟨e|x^{2}⟩e)x) + 2⟨e|(x ∗ x)x⟩e
= 4e((ex^{2})x) − 4⟨e|x^{2}⟩e(ex) + 2⟨e|(x ∗ x)x⟩e
= ⟨x|x^{2}⟩e − ⟨e|x^{2}⟩x + 2⟨e|(x ∗ x)x⟩e by Eqs.(9) and (27)
= −⟨e|x^{2}⟩x since (x ∗ x) ∗ x ∈ V.

Therefore, with (x|x) = −⟨e|x^{2}⟩ for any x ∈ V, we obtain

(x ∗ x) ∗ x = (x|x)x  (30)

for any x ∈ V, so that (V, ∗) is a pseudo-composition algebra.

A word of caution is needed here: we DO NOT assume the bilinear form ( |) to be nonzero, condition that is imposed in [MO]. Notice that if (x|x) = 0 for any x, then Eq.(25) shows that the algebra (V, ∗) is a Jordan algebra. Actually, Eq.(22) with y = x ∗ x shows that (x ∗ x)^{∗2} = 0 and, by linearization, that (x ∗ x) ∗ (x ∗ y) = 0, which by Eq.(25) shows that ((x ∗ x) ∗ y) ∗ x = 0 too. Also by Eq.(23) this implies that (L_{x}^{∗})^{3} = 0, where L_{x}^{∗} denotes the multiplication by x. These commutative algebras satisfying (x ∗ x) ∗ x = 0 for any x are a special case of the “(−1)-Jordan-Lie algebras” considered in [OK].

Also, for any x, y, z ∈ V, by associativity of ⟨ | ⟩, Eqs.(29) and (27):

(x ∗ y|z) = −⟨e(x ∗ y)|z⟩ = ⟨2e(e(xy)) − 2⟨e|xy⟩e|z⟩

= \frac{1}{2}⟨xy|z⟩ = \frac{1}{2}⟨x|yz⟩

= (x|y ∗ z) .

Let us consider the linear endomorphism φ of V given by φ(x) = −2ex. Equation (27) tells us that φ^{2} = 1. We also consider the linear endomorphism ϕ of A given by ϕ(e) = e and ϕ(x) = φ(x) for any x ∈ V, and its extension to the Jordan algebra (J_{A} = F 1 ⊕ A, ·) given in Theorem 5, denoted by Φ, obtained by means of Φ(1) = 1.

Proposition 11. φ is an automorphism of (V, ∗), ϕ is an automorphism of A and Φ is an automorphism of (J, ·).

Proof . For any x ∈ V, Eqs.(27) and (28) imply that

\begin{aligned}φ(x) ∗ φ(x) & = 4(ex) ∗ (ex) = 16(e(ex))(e(ex)) − 16⟨e|(e(ex))(e(ex))⟩e \\ & = x^{2} − ⟨e|x^{2}⟩e = x^{2} + (x|x)e . \end{aligned}

But from Eqs.(29) and (27) we deduce

φ(x ∗ x) = −2e (−2ex^{2} + 2⟨e|x^{2}⟩e)
= 4e(ex^{2}) − 4⟨e|x^{2}⟩e
= x^{2} − ⟨e|x^{2}⟩e = x^{2} + (x|x)e ,

and hence φ is an automorphism of (V, ∗).

Now from Eq.(30) it follows that φ is an orthogonal transformation of ( |). From e^{2} = e, ex = −\frac{1}{2}φ(x) and x^{2} = φ(x ∗ x) − (x|x)e, it readily follows that ϕ is an automorphism of A.

Since φ^{2} = 1, φ induces a ℤ_{2}-gradation of V: V = V_{0} ⊕ V_{1}, with V_{i} = {x ∈ V : φ(x) = (−1)^{i}x} (i = 0, 1), and since φ is an orthogonal transformation for ( |), V_{0} and V_{1} are orthogonal subspaces relative to ( |). Also, for any x, y ∈ V:

(x|y) = −⟨e|xy⟩ = −⟨ex|y⟩ = \frac{1}{2}⟨φ(x)|y⟩ ,

so that

⟨x|y⟩ = 2(φ(x)|y)  (31)

for any x, y ∈ V. From Eq.(31) it follows that ϕ is an orthogonal transformation relative to ⟨ | ⟩, and by the definition of the multiplication · on J, that Φ is an automorphism of (J, ·).

Moreover, for any α ∈ F and x ∈ V, since φ(x ∗ x) = x^{2} + (x|x)e, as shown in the proof of Proposition 11,

\begin{aligned}N(αe + x) & = ⟨αe + x|(αe + x)^{2}⟩ \\ & = ⟨αe + x|α^{2}e − αφ(x) + x^{2}⟩ \\ & = ⟨αe + x| (α^{2} − (x|x)) e − αφ(x) + φ(x ∗ x)⟩ \\ & = α (α^{2} − (x|x)) + ⟨x| − αφ(x) + φ(x ∗ x)⟩ \\ & = α (α^{2} − (x|x)) + 2 (φ(x)| − αφ(x) + φ(x ∗ x)) \\ & = α (α^{2} − (x|x)) + 2 (x| − αx + x ∗ x) \\ & = α^{3} − 3α(x|x) + 2(x|x ∗ x) . \end{aligned}  (32)

Conversely, let (V, ∗) be a ℤ_{2}-graded pseudo-composition algebra with grading automorphism φ. From Eq.(22) with y = x ∗ x it follows that

(x ∗ x) ∗ (x ∗ x) = 2(x|x ∗ x)x − (x|x)x ∗ x .  (33)

Then

(x ∗ x|x ∗ x)x ∗ x = ((x ∗ x) ∗ (x ∗ x)) ∗ (x ∗ x)
= 2(x|x ∗ x)x ∗ (x ∗ x) − (x|x)(x ∗ x) ∗ (x ∗ x)
= (x|x)^{2}x ∗ x ,

so that

(x ∗ x|x ∗ x) = (x|x)^{2}  (34)

for any x ∈ V (if x∗x = 0, then 0 = (x∗x)∗x = (x|x)x, so (x|x) = 0). Equation (34) has been proved in [RW, (1.11)].

We consider the new vector space A = F e ⊕ V, with commutative multiplication defined by

e^{2} = e , ex = −\frac{1}{2}φ(x) and x^{2} = φ(x ∗ x) − (x|x)e  (35)

for any x ∈ V. Then Eqs.(33), (34) and (35) imply for any α ∈ F and x ∈ V:

(αe + x)^{2} = (α^{2} − (x|x)) e − αφ(x) + φ(x ∗ x)
(αe + x)^{2}(αe + x)^{2} = ((α^{2} − (x|x)) e + φ(−αx + x ∗ x))^{2}
= [(α^{2} − (x|x))^{2} − (φ(−αx + x ∗ x)|φ(−αx + x ∗ x))] e
− (α^{2} − (x|x)) (−αx + x ∗ x) + (−αx + x ∗ x) ∗ (−αx + x ∗ x)
= [α^{4} − 2α^{2}(x|x) + (x|x)^{2} − α^{2}(x|x) + 2α(x|x ∗ x) − (x ∗ x|x ∗ x)] e
+ α^{3}x − α(x|x)x − α^{2}x ∗ x + (x|x)x ∗ x
+ α^{2}x ∗ x − 2αx ∗ (x ∗ x) + (x ∗ x) ∗ (x ∗ x)
= [α^{4} − 3α^{2}(x|x) + 2α(x|x ∗ x)] e + α^{3}x − 3α(x|x)x + 2(x|x ∗ x)x
= (α^{3} − 3α(x|x) + 2(x|x ∗ x)) (αe + x)

so that X^{2}X^{2} = N(X)X for any X ∈ A, with N(αe + x) = α^{3} − 3α(x|x) + 2(x|x ∗ x) (see Eq.(32)).

If we define the symmetric bilinear form ⟨ | ⟩ on A = F e ⊕ V by means of

⟨αe + x|βe + y⟩ = αβ + 2(φ(x)|y)

we obtain

⟨αe + x|(αe + x)^{2}⟩ = ⟨αe + x| (α^{2} − (x|x)) e + φ(−αx + x ∗ x)⟩
= α (α^{2} − (x|x)) + 2 (φ(x)|φ(−αx + x ∗ x))
= α^{3} − 3α(x|x) + 2(x|x ∗ x) = N(αe + x) .

If ⟨ | ⟩ were associative, this would be the unique form referred to in Theorem 1. But for any x, y, z ∈ V and α, β, γ ∈ F:

⟨(αe + x)|(βe + y)(γe + z)⟩

= ⟨αe + x| (βγ − (y|z)) e + φ(y ∗ z) − \frac{1}{2}βφ(z) − \frac{1}{2}γφ(y)⟩

= αβγ − α(y|z) + 2 (φ(x)|φ(y ∗ z) − \frac{1}{2}βφ(z) − \frac{1}{2}γφ(y))

= αβγ − [α(y|z) + β(x|z) + γ(x|y)] + 2(x|y ∗ z) ,

and ⟨ | ⟩ is associative in A if and only if so is ( |) in (V, ∗). Hence we need the following result, which has its own independent interest. It simplifies some of the arguments in [MO]. In particular it implies immediately that the radical of the form ( |), in any pseudo-composition algebra, is an ideal.

Theorem 12. The symmetric bilinear form of any pseudo-composition algebra is associative.

Proof . We can extend scalars and assume that the ground field is infinite. Let (V, ∗) be such an algebra, with (x ∗ x) ∗ x = (x|x)x. In case ( |) = 0 the result is trivial.

Otherwise, (V, ∗) is trivially ℤ_{2}-graded with V = V_{0} and V_{1} = 0 (that is, the grading automorphism φ is the identity). We construct as before the algebra A = F e ⊕ V with multiplication as in Eq.(35), which has cubic norm given by N(αe + x) = α^{3} − 3α(x|x) + 2(x|x ∗ x) for any α ∈ F and x ∈ V. Let ⟨ | ⟩ be the unique associative symmetric bilinear form associated to A by Theorem 1, so that N(αe + x) = ⟨αe + x|(αe + x)^{2}⟩ for any α ∈ F and x ∈ V. Then Eq.(10) with x = e and y ∈ V gives, since ey = −\frac{1}{2}y by Eq.(30) (remember that φ = 1 here):

y = 4e(ey) = 3⟨e|y⟩e + ⟨e|e⟩y

so that ⟨e|V ⟩ = 0 (and ⟨e|e⟩ = 1). Also, for any x ∈ V, Eq.(9) implies

\begin{aligned}4x(x^{2}x) & = 3⟨x|x⟩x^{2} + ⟨x|x^{2}⟩x \\ & = 3⟨x|x⟩(x ∗ x − (x|x)e) + ⟨x|x^{2}⟩x , \end{aligned}  (36)

so the projection of 4x(x^{2}x) over F e is −3⟨x|x⟩(x|x)e. But x^{2} = x ∗ x − (x|x)e and

x^{2}x = (x ∗ x − (x|x)e)x

= (x ∗ x) ∗ x − (x ∗ x|x)e + \frac{1}{2}(x|x)x

= \frac{3}{2}(x|x)x − (x ∗ x|x)e

and

\begin{aligned}4x(x^{2}x) & = 6(x|x)x^{2} − 4(x ∗ x|x)ex \\ & = 6(x|x)(x ∗ x − (x|x)e) + 2(x ∗ x|x)x . \end{aligned}  (37)

From Eqs.(36) and (37) we conclude that −3⟨x|x⟩(x|x) = −6(x|x)^{2} for any x ∈ V; so that (the field is infinite and ( |) ≠ 0) ⟨x|x⟩ = 2(x|x) for any x ∈ V. Now for any x, y, z ∈ V, using that ⟨ | ⟩ is associative with ⟨e|V ⟩ = 0 and Eq.(35), we obtain

(x ∗ y|z) = \frac{1}{2}⟨xy|z⟩ = \frac{1}{2}⟨x|yz⟩ = (x|y ∗ z) ,

as required.

As was remarked after Theorem 5, this result shows that a generically algebraic Jordan algebra of degree ≤ 3 can be attached to any pseudo-composition algebra.

Corollary 13. Any unital commutative strictly power-associative algebra which is generically algebraic of generic degree ≤ 3 is a Jordan algebra.

Proof . Let A be such a power associative algebra. In case the degree of A is ≤ 2 this is clear since (x^{2}, y, x) = (x^{2}y)x − x^{2}(yx) ∈ (F 1 + F x, y, x) = 0, by commutativity. Hence we will assume that the degree of A is exactly 3 and will extend scalars so as to have an infinite ground field. Any element x ∈ A satisfies the equation

x^{3} − τ(x)x^{2} + σ(x)x − ν(x)1 = 0

for a suitable linear form τ (the generic trace), quadratic form σ and cubic form ν. For any generic x ∈ A, its degree is 3 and τ(x), σ(x) and ν(x) are just the coefficients of the characteristic polynomial of the restriction of the right multiplication by x to the subalgebra generated by x. Therefore, σ(x) = \frac{1}{2}(τ(x)^{2} − τ(x^{2})). By density this is valid for any x. Let V be the set of trace zero elements in A and define the commutative multiplication on V by means of

x ∗ x = x^{2} − \frac{1}{3}τ(x^{2})1,

that is, the projection of x^{2} on V. Then for any x ∈ V

(x ∗ x) ∗ x = (x ∗ x)x − \frac{1}{3}τ((x ∗ x)x)1

= x^{3} − \frac{1}{3}τ(x^{2})x − \frac{1}{3}τ(x^{3})1

= \frac{1}{2}τ(x^{2})x + ν(x)1 − \frac{1}{3}τ(x^{2})x − \frac{1}{3}τ(x^{3})1

= \frac{1}{6}τ(x^{2})x + (ν(x) − \frac{1}{3}τ(x^{3})) 1

= \frac{1}{6}τ(x^{2})x since (x ∗ x) ∗ x ∈ V.

Therefore (V, ∗) is a pseudo-composition algebra, and by Theorem 12, its associated symmetric bilinear form: (x|y) = \frac{1}{6}τ(xy), is associative. Besides, for any x, y ∈ V,

x^{2} = x ∗ x + \frac{1}{3}τ(x^{2})x = x ∗ x + 2(x|x)1

and the arguments in remark 3) after Theorem 5 show that A is a Jordan algebra.

Remark . The word “strictly” in the statement above is only necessary for the field of 5 elements. If the ground field F is larger and A is just a unital commutative power-associative algebra over F in which x^{3} ∈ F 1 + F x + F x^{2} for any x ∈ A, then in ⋀^{4} A, 1 ∧ x ∧ x^{2} ∧ x^{3} = 0 and because the size of the field, its linearizations are also valid. Thus the algebra obtained by extending scalars to an infinite field K also satisfies that x^{3} ∈ K1 + Kx + Kx^{2} for any x ∈ K ⊗_{F} A. It also satisfies the associativity of the third and fourth powers, and hence it is power-associative too and its degree is ≤ 3.

Let us consider now the category AC_{e}, whose objects are the pairs (A, e), where A is an admissible cubic algebra and 0 ≠ e ∈ A a nonzero idempotent, and whose morphisms φ : (A_{1}, e_{1}) → (A_{2}, e_{2}) are the algebra homomorphisms with φ(e_{1}) = e_{2}. There is no need to worry about the trace form of the objects, since the existence of a nonzero idempotent forces the norm to be nonzero and hence there is a unique trace form associated (Theorem 1). Let us consider too the category PC_{2} of ℤ_{2}-graded pseudo-composition algebras with homomorphisms preserving the grading.

Given an object (A, e) in AC_{e}, let V be the orthogonal complement of F e relative to the trace form of A. Then (V = V¯0 ⊕ V¯1, ∗) is an object in PC_{2} by Proposition 11, where V¯0 = {x ∈ V : φ(x) = −2ex = x} and V¯1 = {x ∈ V : φ(x) = −2ex = −x}.

The next result summarizes the previous work done in this section.

Theorem 14. The assignment (A, e) ↦ (V = V¯0 ⊕V¯1, ∗) sets up an equivalence of categories between AC_{e} and PC_{2}.

Proof . Recall that any nonzero homomorphism of admissible cubic algebras which nonzero norm preserve the norms and the trace forms. Also any homomorphism of pseudo-composition algebras preserve the associated symmetric bilinear form, because of the identity (x ∗ x) ∗ x = (x|x)x. Now everything is straightforward.

6. Final remarks

1) Let (A, N, #, e) be an admissible cubic form with basepoint e ([McC2]) over our ground field F of characteristic ≠ 2, 3; that is, x^{#} is quadratic in x (the adjoint map), N is a cubic form, e ∈ A and they satisfy

i) (x^{#})^{#} = N(x)x (adjoint identity)
ii) T (x^{#}, y) = ∂_{y}N|_{x}
iii) e × y = T (y)e − y
iv) e^{#} = e
v) N(e) = 1

where T (x, y) = −∂_{x}∂_{y} log N|_{e}, T (y) = T (y, e) and x × y = (x + y)^{#} − x^{#} − y^{#} for x, y ∈ A.

The adjoint identity gives A the structure of an admissible cubic algebra, by defining the commutative multiplication by x^{2} = x^{#} for any x, and it has nonzero norm by v). Besides e = e^{#} = e^{2} and xy = \frac{1}{2}x × y for any x, y. Let ⟨ | ⟩ be the (unique) trace form of this admissible cubic algebra. Then for any x, y ∈ A, ii) implies

T (x^{2}, y) = T (x^{#}, y) = ∂_{y}N|_{x} = 3⟨x^{2}|y⟩

and, since x^{2}x^{2} = N(x)x, this proves that for any x, y ∈ A,

T (x, y) = 3⟨x|y⟩ .

With S(x) = T (x^{#}) and x^{·2} = x^{#} + T (x)x − S(x)e, A becomes a generically algebraic Jordan algebra with unit e and where

x^{·3} − T (x)x^{·2} + S(x)x − N(x)e = 0

for any x ∈ A (hence the generic degree is ≤ 3 and, conversely, any unital Jordan algebra of generic degree 3 fits into this construction). Besides, T, S and N verify the Newton formulae because

S(x) = T (x^{#}) = T (x^{·2} − T (x)x + S(x)e) = T (x^{·2}) − T (x)^{2} + 3S(x)

and since 2S(x) = 2T (x^{#}) = T (x)^{2} − T (x, x) too, by [McC2, (16)], T (x, x) = T (x^{·2}) for any x and

3N(x) = T (x^{#}, x)  by ii) above

= T (x^{·2} − T (x)x + S(x)e, x)

= T (x^{·3}) − T (x)T (x^{·2}) + S(x)T (x)

= \frac{1}{2} (2T (x^{·3}) − 3T (x)T (x^{·2}) + T (x)^{3}) .

Now the argument in the proof of Proposition 6.v) works here and shows that the admissible cubic algebra A is precisely C(J, t), where J = F ×A is the direct product of the Jordan algebras F and (A, ·), and where t((α, x)) = α + T (x) for any α ∈ F and x ∈ A.

Also, e is a nonzero idempotent of the admissible cubic algebra A and −2ex = −e × x = x for any x ∈ V = {z ∈ A : T (z) = 0} = {z ∈ A : ⟨e|z⟩ = 0}. Hence the automorphism φ on (V, ∗) given in Proposition 11 is just the identity (so we have a trivial ℤ_{2}-grading). Recall that for any x ∈ V,

x ∗ x = proj_{V} (2ex)^{2} = proj_{V} (−x)^{2}

= x^{2} − ⟨e|x^{2}⟩e = x^{2} − ⟨ex|x⟩e

= x^{2} + \frac{1}{2}⟨x|x⟩e

and (x ∗ x) ∗ x = (x|x)x with (x|x) = \frac{1}{2}⟨x|x⟩ by Eq.(31).

But from the arguments in Remark 3) after Theorem 5, A = F e⊕V becomes a Jordan algebra with the new commutative product determined by e ⋄ x = x for any x and

\begin{aligned}x ⋄ x & = x ∗ x + 2(x|x)e = x ∗ x + ⟨x|x⟩e \\ & = x^{2} + \frac{3}{2}⟨x|x⟩e \end{aligned}

for any x ∈ V. However, for x ∈ V,

x · x = x^{#} + T (x)x − S(x)e = x^{2} + \frac{1}{2}T (x, x)e = x^{2} + \frac{3}{2}⟨x|x⟩e = x ⋄ x,

so that the Jordan product ⋄ is just the previous Jordan product · on A.

2) The previous work gives a very interesting relationship between central simple Jordan algebras of degree 4 and central simple Jordan algebras of degree 3; which sheds some light on the relationship noticed in [AF, Section 5].

Given any quartic Jordan algebra (J, t) of generic degree 4 over the field F and an idempotent f ∈ J with generic trace t(f) = 1, we know from Theorem 5 that the set A = {x ∈ J : t(x) = 0}, with the multiplication given by xy = x · y − \frac{1}{4}t(x, y)1, is an admissible cubic algebra satisfying x^{2}x^{2} = \frac{1}{3}t(x, x · x)x, so ⟨x|y⟩ = \frac{1}{3}t(x, y) for any x, y ∈ A, where t(x, y) = t(x · y). The element e ∈ A given by e = 2f − \frac{1}{2} is a nonzero idempotent of the admissible cubic algebra A by Lemma 9. Now Theorem 14 gives a pseudo-composition algebra (V, ∗) with V = \{x ∈ A : ⟨x|e⟩ = 0\} and x ∗ x = ϕ(x^{2}) + \frac{1}{2}⟨ϕ(x)|x⟩e, where ϕ, given by ϕ(e) = e and ϕ(x) = −2ex for any x ∈ V, is an automorphism of A and φ = ϕ|_{V} an automorphism of (V, ∗). Now the arguments in the Remark 3) after Theorem 5 allows us to introduce a structure of Jordan algebra of degree ≤ 3 in A = F e ⊕ V by means of e ⋄ x = x for any x ∈ A and

x ⋄ x = x ∗ x + 2(x|x)e  see Eq.(26)

= x ∗ x + ⟨ϕ(x)|x⟩e by Eq.(31)

= (ϕ(x^{2}) + \frac{1}{2}⟨ϕ(x)|x⟩e) + ⟨ϕ(x)|x⟩e

= [(ϕ(x) · ϕ(x) − \frac{1}{4}t(ϕ(x), ϕ(x))1) + \frac{1}{6}t(ϕ(x), x)e] + \frac{1}{3}t(ϕ(x), x)e

for any x ∈ V. Notice that we do not add the last two summands in order to preserve the decomposition of x ⋄ x into the sum of an element of V and a scalar multiple of e.

Since Φ is an automorphism of J, the generic trace is invariant by Φ, so that we obtain

x ⋄ x = [(Φ(x · x) − \frac{1}{4}t(x, x)1) + \frac{1}{6}t(ϕ(x), x)e] + \frac{1}{3}t(ϕ(x), x)e (38)

for any x ∈ V.

The degree of the Jordan algebra (A, ⋄) may be strictly lower than 3, as the next example shows: A = F e + F a + F b with e^{2} = e, ea = −\frac{1}{2}a, eb = \frac{1}{2}b, a^{2} = b^{2} = ab = 0, ⟨e|e⟩ = 1, ⟨A|a⟩ = ⟨A|b⟩ = 0. It is straightforward to check that x^{2}x^{2} = ⟨x|x^{2}⟩x for any x ∈ A and that the element x = e + a + b satisfies that x^{2} = e−a+b and x^{3} = e+b, so that x, x^{2} and x^{3} are linearly independent. The associated Jordan algebra (J_{A}, ·), with J_{A} = F 1 ⊕ A, is the quotient of the polynomial algebra F [λ] by the ideal generated by λ^{4} − \frac{3}{2}λ^{2} − λ − \frac{3}{16} (the minimum polynomial of x in J), so that it is a Jordan algebra of degree 4. However, V = {z ∈ A : ⟨e|z⟩ = 0} = F a + F b and (V, ∗) is the zero algebra. By the definition above, the Jordan algebra (A, ⋄) satisfies e ⋄ z = z for any z ∈ A and z ⋄ z = 0 for any z ∈ V. Hence the degree of (A, ⋄) is 2.

However, for any separable Jordan algebra (J, ·) of degree 4, by extending scalars to the algebraic closure we may assume that J contains F^{4} as a subalgebra. We may also assume that our idempotent f equals (1, 0, 0, 0) ∈ F^{4} ⊆ J, so that e = (\frac{3}{2}, −\frac{1}{2}, −\frac{1}{2}, −\frac{1}{2}), and it is easy to find an element x ∈ V ∩ F^{4} such that x ∗ x ∉ F x (for instance, x = (0, 0, 1, −1)). Hence in this case, the degree of the Jordan algebra (A, ⋄) is exactly 3.

Things work very nicely for the central simple Jordan algebras of degree 4. Let (J, ·) be such an algebra. By extending scalars we may assume that J is the Jordan algebra H_{4}(B, ∗) of the symmetric elements in Mat_{4}(B) with respect to the involution ∗ given by (b_{ij})^{∗} = (\overline{b_{ji}}), where B is an associative composition algebra over F with standard involution x ↦ x̄; hence the dimension of B is 1, 2 or 4. Here the generic trace in J is the usual trace. We consider the idempotent f = diag(0, 0, 0, 1) with trace equal to 1, and the corresponding idempotent element e in A = H_{4}(B, ∗)_{0} = \{b ∈ H_{4}(B, ∗) : tr(b) = 0\}: e = 2f − \frac{1}{2} = diag(−\frac{1}{2}, −\frac{1}{2}, −\frac{1}{2}, \frac{3}{2}). Then we have the subspace

V = {b ∈ A : ⟨e|b⟩ = 0}
= {b ∈ H_{4}(B, ∗) : tr(b) = 0 = tr(e · b)}
= {b = (b_{ij}) ∈ H_{4}(B, ∗) : tr(b) = 0 and b_{44} = 0} .

Notice that with respect to the Peirce decomposition relative to f, we have that f · x = ix (i = 0, 1, \frac{1}{2}) if and only if (e + \frac{1}{2}) · x = 2ix, or e · x = (2i − \frac{1}{2})x. But for x ∈ V, e · x = ex (the product in A), so that e · x = ex = −\frac{1}{2}Φ(x) and, therefore, f · x = ix if and only if Φ(x) = φ(x) = (1 − 4i)x. Hence {x ∈ V : φ(x) = x} = V ∩ J_{0}(f) and \{x ∈ V : φ(x) = −x\} = V ∩ J_{\frac{1}{2}} (f). Since J = (F f ⊕ F e) ⊕ V = (F f ⊕ F (1 − f)) ⊕ V, we have J_{1}(f) = F f, J_{0}(f) = F (1 − f) ⊕ (V ∩ J_{0}(f)) and J_{\frac{1}{2}} (f) ⊆ V. Hence the automorphism Φ of J is just the automorphism which is the identity on J_{1}(f) + J_{0}(f) and minus the identity on J_{\frac{1}{2}} (f); that is, the Peirce reflection of the idempotent f.

Given the associative composition algebra B we can obtain a new composition algebra C = B × B by applying the Cayley-Dickson doubling process (see [Sc]), with multiplication given by

(b_{0}, b_{1})(c_{0}, c_{1}) = (b_{0}c_{0} + c_{1}\overline{b_{1}}, \overline{b_{0}}c_{1} + c_{0}b_{1})

and standard involution given by

\overline{(b_{0}, b_{1})} = (\overline{b_{0}}, −b_{1}) .

Now consider the Jordan algebra (H_{3}(C, ∗), ·), again with the standard involution: (x_{ij})^{∗} = (\overline{x_{ji}}), and the linear map

\begin{matrix}Ψ : (A, ⋄) & ⟶ (H_{3}(C, ∗), ·) \\ e & ↦ 1 \\ b = (b_{ij}) ∈ V & ↦ x = (x_{ij}) \end{matrix}

with

x_{ij} = (b_{ij}, ϵ_{ijk}b_{k4}) ∈ C

for 1 ≤ i, j ≤ 3 and ϵ_{ijk} the totally antisymmetric Levi-Civita symbol. That is, x_{ii} = (b_{ii}, 0) and x_{ij} = (b_{ij}, ±b_{k4}) for i ≠ j, where {i, j, k} = {1, 2, 3} and the ± sign is the signature of the permutation (i, j, k).

Theorem 15. Under the hypotheses above, Ψ is an isomorphism of Jordan algebras.

Proof . For any b = (b_{ij}) ∈ V, let x = Ψ(b). Then tr(x) = (b_{11}, 0) + (b_{22}, 0) + (b_{33}, 0) = 0 since tr(b) = 0 and b_{44} = 0. Hence Ψ gives a bijection of V with the set of trace zero elements in H_{3}(C, ∗) and, since Ψ(e) = 1, Ψ is a bijection. Again, with b ∈ V and x = Ψ(b), let us check the entries of x · x. By symmetry it is enough to deal with the (1,1) and (1,2) entries:

(x · x)_{11} = (b_{11}, 0)(b_{11}, 0) + (b_{12}, b_{34})(b_{21}, −b_{34}) + (b_{13}, −b_{24})(b_{31}, b_{24})

= (b_{11}^{2} + b_{12}b_{21} − b_{34}\overline{b_{34}} + b_{13}b_{31} − b_{24}\overline{b_{24}},

− \overline{b_{12}}b_{34} + b_{21}b_{34} + \overline{b_{13}}b_{24} − b_{31}b_{24})

= (b_{11}^{2} + b_{12}b_{21} + b_{13}b_{31} + b_{14}b_{41} − n(b_{14}) − n(b_{24}) − n(b_{34}) , 0)

= ((b · b)_{11} − ∑_{l=1}^{3} n(b_{l4}), 0)

= (Φ(b · b)_{11} − ∑_{l=1}^{3} n(b_{l4}), 0)

since b_{ij} = \overline{b_{ji}} and b_{ij}\overline{b_{ij}} = \overline{b_{ij}}b_{ij} = n(b_{ij}), the norm of b_{ij} in the composition algebra B. In the same way

(x · x)_{12} = (b_{11}, 0)(b_{12}, b_{34}) + (b_{12}, b_{34})(b_{22}, 0) + (b_{13}, −b_{24})(b_{32}, −b_{14})
= (b_{11}b_{12} + b_{12}b_{22} + b_{13}b_{32} + b_{14}\overline{b_{24}},
\overline{b_{11}}b_{34} + b_{22}b_{34} − \overline{b_{13}}b_{14} − b_{32}b_{24})
= ((b · b)_{12}, b_{11}b_{34} + b_{22}b_{34} − b_{31}b_{14} − b_{32}b_{24})
= ((b · b)_{12}, −b_{33}b_{34} − b_{31}b_{14} − b_{32}b_{24})
= ((b · b)_{12}, Φ(b · b)_{34})
= (Φ(b · b)_{12}, Φ(b · b)_{34})

since b_{ii} = \overline{b_{ii}} ∈ F and b_{11} + b_{22} + b_{33} = 0. On the other hand, by Eq.(38),

b ⋄ b = [Φ(b · b) − \frac{1}{4}t(b · b)1 + \frac{1}{6}t(ϕ(b), b)e] + \frac{1}{3}t(ϕ(b), b)e .

Hence, since Ψ(e) = 1,

Ψ(b ⋄ b)_{11}

= (Φ(b · b)_{11} − \frac{1}{4}t(b · b) + \frac{1}{6}t(ϕ(b), b) (−\frac{1}{2}) , 0) + (\frac{1}{3}t(ϕ(b), b), 0)

= (Φ(b · b)_{11} − \frac{1}{4}t(b · b) + \frac{1}{4}t(ϕ(b), b), 0)

= (Φ(b · b)_{11} − \frac{1}{4}t((b − ϕ(b)) · b), 0)

and for b = (b_{ij}),

b − ϕ(b) = 2 ( \begin{matrix}0 & 0 & 0 & b_{14} \\ 0 & 0 & 0 & b_{24} \\ 0 & 0 & 0 & b_{34} \\ b_{41} & b_{42} & b_{43} & 0 \end{matrix} ) ,

so that

\begin{aligned}t((b − ϕ(b)) · b) & = 2(b_{14}b_{41} + b_{24}b_{42} + b_{34}b_{43} + b_{41}b_{14} + b_{42}b_{24} + b_{43}b_{34}) \\ & = 4 (n(b_{14}) + n(b_{24}) + n(b_{34})) . \end{aligned}

Thus,

Ψ(b ⋄ b)_{11} = (Φ(b · b)_{11} − ∑_{l=1}^{3} n(b_{l4}), 0) .

With respect to the (1,2) entry, we get

Ψ(b ⋄ b)_{12} = (Φ(b · b)_{12}, Φ(b · b)_{34})

and we conclude that Ψ(b ⋄ b) = Ψ(b) · Ψ(b), as required.

In a similar vein, the algebra A = H_{4}(B, ∗)_{0} (where the multiplication is given by the projection of the Jordan product in H_{4}(B, ∗) onto A) is isomorphic to the algebra (H_{3}(C, ∗), ⋆), where

x ⋆ y = x̃ · ỹ − \frac{1}{4}(tr(x · y) + tr(x) tr(y))1

with x ↦ x̃ the automorphism of H_{3}(C, ∗) such that for x = (x_{ij}), x̃ = (x̃_{ij}), where (b_{0}, b_{1})˜ = (b_{0}, −b_{1}) (this is an involutive automorphism of the composition algebra C). The isomorphism is realized as:

\begin{matrix}Ψ̃ : A = H_{4}(B, ∗)_{0} & ⟶ (H_{3}(C, ∗), ⋆) \\ b & ↦ Ψ̃(b) = x = (x_{ij}) \end{matrix}

with x_{ij} = (b_{ij}, ϵ_{ijk}b_{k4}) as before, but this time for any b ∈ A and not only for the elements in V. Notice that for any b ∈ A and x = Ψ̃(b)

tr(x)^{2} = (b_{11} + b_{22} + b_{33})^{2} = (−b_{44})^{2} = b_{44}^{2}

and

\begin{aligned}tr(x · x) & = (b · b)_{11} + (b · b)_{22} + (b · b)_{33} − 3 ∑_{l=1}^{3} n(b_{l4}) \\ & = tr(b · b) − (b · b)_{44} − 3 ∑_{l=1}^{3} n(b_{l4}) \end{aligned}

(see the computation of (x · x)_{11} in the proof of Theorem 15). But

\begin{aligned}(b · b)_{44} & = b_{41}b_{14} + b_{42}b_{24} + b_{43}b_{34} + b_{44}^{2} \\ & = n(b_{14}) + n(b_{24}) + n(b_{34}) + b_{44}^{2} ; \end{aligned}

thus

tr(x · x) + tr(x)^{2} = tr(b · b) − 4 ∑_{l=1}^{3} n(b_{l4}) ,

and the same arguments as in the proof above now give

\begin{aligned}(Ψ̃(b) ⋆ Ψ̃(b))_{11} & = ((b · b)_{11} − ∑_{l=1}^{3} n(b_{l4}), 0) − (\frac{1}{4}(tr(b · b) − 4 ∑_{l=1}^{3} n(b_{l4})), 0) \\ & = ((b · b)_{11} − \frac{1}{4} tr(b · b)1, 0) = Ψ̃(b^{2})_{11} \end{aligned}

(recall that b^{2} is the projection of b · b onto A), and

(Ψ̃(b) ⋆ Ψ̃(b))_{12} = ((b · b)_{12}, Φ(b · b)_{34})˜
= ((b · b)_{12}, (b · b)_{34})
= ((b^{2})_{12}, (b^{2})_{34}) = Ψ̃(b^{2})_{12} .

Therefore Ψ̃(b) ⋆ Ψ̃(b) = Ψ̃(b^{2}) for any b ∈ A, as asserted.

3) The equation x^{2}x^{2} = ⟨x|x^{2}⟩x appears in [K, Equation (5)], where the bilinear form that appears is related to ⟨ | ⟩ by means of B(x, y) = \frac{3}{4}⟨x|y⟩. Then Equation (4) in [K] is just our Equation (9), so assuming B ≠ 0, this last equation can be dropped from the assumptions in [K, Theorem 1.2], since it is a consequence of x^{2}x^{2} = ⟨x|x^{2}⟩x. It may be interesting to relate the Jordan algebra (A, ⋄) that appears in the previous remark with the Jordan triple product constructed in [K].

4) Some of our algebras may have a signification in genetic. For the definitions here see, for example, the survey [Re] and the references therein.

Assume that the cubic form N(x) in Eq.(1) factorizes as N(x) = μρ(x)^{3} for some 0 ≠ μ ∈ F and nonzero linear form ρ. From Eq.(2) it follows that μρ(x^{2})^{3} = μ^{2}ρ(x)^{6}, so by unique factorization (after extending scalars if necessary), there is a nonzero scalar ν such that ρ(x^{2}) = νρ(x)^{2}. Then μν^{3}ρ(x)^{6} = μ^{2}ρ(x)^{6}, so μ = ν^{3}. Replace ρ(x) by ω(x) = νρ(x), so that N(x) = μρ(x)^{3} = (νρ(x))^{3} = ω(x)^{3} for any x, and besides, ω(x^{2}) = νρ(x^{2}) = ν^{2}ρ(x)^{2} = ω(x)^{2} for any x. Therefore we may assume that our commutative algebra A satisfies

x^{2}x^{2} = ω(x)^{3}x  (39)

for any x ∈ A, where ω : A → F is a nonzero homomorphism. Thus A is, in particular, a baric algebra. Equation (39) can be rewritten as

x^{[3]} = ω(x)^{3}x

for any x, where the principal powers x^{[i]} are defined recursively as x^{[1]} = x and x^{[i+1]} = x^{[i]}x^{[i]} for any x and any i > 0, which means that the elements x of our algebra with baric weight ω(x) equal to 1 oscillate with period two: x^{[i+2]} = x^{[i]} for any i, in contrast to Bernstein algebras, where x^{[i+1]} = x^{[i]} for i ≥ 2.

In this situation N(x, y, z) = ω(x)ω(y)ω(z) and

N(x^{2}, x^{2}, y) = ω(x^{2})ω(x^{2})ω(y) = ω(x)^{4}ω(y) = N(x)ω(x)ω(y)

so that the associative bilinear form given in Theorem 1 is given, according to Eq.(5), by

⟨x|y⟩ = ω(x)ω(y)

for any x, y ∈ A.

Moreover, Eq.(9) with y = x becomes here

4x^{4} = 3ω(x)^{2}x^{2} + ω(x)^{3}x ,

where x^{1} = x and x^{i+1} = x^{i}x for any i > 0. Hence the algebras satisfying Eq.(39) are train algebras.

Equation (11) implies that (x^{2}y)x = x^{2}(yx) for any x ∈ ker ω and any y ∈ A. In particular, the ideal R = ker ω is a locally nilpotent Jordan algebra, since it is nil by Eq.(39) of bounded index. Consider the chain of the principal powers of the ideal R = ker ω: R^{1} = R, R^{i+1} = R^{i}R for any i > 0; which are ideals of the Jordan algebra R. Extending scalars we may assume that there is an idempotent 0 ≠ e ∈ A; so that ω(e) = 1 and V = {x ∈ A : ⟨e|x⟩ = 0} = ker ω = R. By Proposition 11, we have an automorphism ϕ of A which preserves the ideal R and such that ϕ(x) = −2ex for any x ∈ R. Thus ϕ(R^{i}) ⊆ R^{i} for any i, that is to say, eR^{i} ⊆ R^{i}. Therefore R^{i} is an ideal of A for any i and, in consequence, A is a special train algebra.

Also, if e is an idempotent as before and V = ker ω = R, for any x ∈ V (x|x) = −⟨e|x^{2}⟩ = −ω(e)ω(x^{2}) = −ω(x)^{2} = 0, so that for any x ∈ V, Eq.(30) gives (x ∗ x) ∗ x = 0 for any x ∈ V. Now the comments after Eq.(30) apply.

A final comment here, the kernel of the weight of any Bernstein algebra satisfies x^{2}x^{2} = 0 for any x, which is our Eq.(1) with a zero cubic norm. These objects seem to be quite intractable. In particular there are examples of solvable nonnilpotent alternative or Jordan algebras, given by several authors (see [ZSSS, pp. 82, 84 and 127]), which satisfy (A^{2})^{2} = 0, so that they are admissible cubic algebras with trivial norm.

References

[AF] B.N. Allison and J.R. Faulkner: A Cayley-Dickson process for a class of structurable algebras. Trans. Amer. Math. Soc. 283 (1984), 185–210.

[BDE] P. Benito, C. Draper and A. Elduque: On some algebras related to simple Lie triple systems. to appear in J. Algebra.

[GMW] H. Gradl, K. Meyberg and S. Walcher: R-algebras with associative trace form. Nova J. Algebra Geom. 1 (1992), 397–417.

[J] N. Jacobson: Structure and representations of Jordan algebras. Amer. Math. Soc., Providence, Rhode Island 1968.

[K] N. Kamiya: A Structure Theory of Freudenthal-Kantor Triple Systems III. Mem. Fac. Sci. Shimane Univ. 23 (1989), 33–51.

[McC1] K. McCrimmon: Generically algebraic algebras. Trans. Amer. Math. Soc. 127 (1967), 527–551.

[McC2] K. McCrimmon: The Freudenthal-Springer-Tits constructions of exceptional Jordan algebras. Trans. Amer. Math. Soc. 139 (1969), 495–510.

[MO] K. Meyberg and J.M. Osborn: Pseudo-composition algebras. Math. Z. 214 (1993), 67–77.

[OK] S. Okubo and N. Kamiya: Jordan-Lie Super Algebra and Jordan-Lie Triple System. J. Algebra 198 (1997), 388–411.

[Re] M.L. Reed: Algebraic structure of genetic inheritance. Bull. Amer. Math. Soc. 34 (1997), no. 2, 107–130.

[RW] H. Röhrl and S. Walcher: Algebras of Complexity One. Algebras, Groups Geom. 5 (1988), 61–107.

[Sc] R.D. Schafer: An Introduction to Nonassociative Algebras. Academic Press, New York 1966.

[S] T.A. Springer: On a Class of Jordan Algebras. Indag. Math. 21 (1959), 259–264.

[Ze] E. Zelmanov: On the nilpotency of nil algebras. In: Proc. 5th National School in Algebra, Varna 1986. L.L. Avranov and K.B. Tchakerian, eds. Lecture Notes in Math. 1352, Springer-Verlag, Berlin-Heidelberg 1988.

[ZSSS] K.A. Zhevlakov, A.M. Slin’ko, I.P. Shestakov and A.I. Shirshov: Rings that are Nearly Associative. Academic Press, New York 1982.

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