Derivations and automorphisms of Jordan algebras in characteristic two

Pablo Alberca Bjerregaard^{a,1} Ottmar Loos^{b}

Cándido Martín González^{c}

aDepartamento de Matemática Aplicada, University of Málaga, 29012 Málaga, Spain

bInstitut für Mathematik, University of Innsbruck, 6020 Innsbruck, Austria

cDepartamento de Álgebra, Geometría y Topología, University of Málaga, 29080 Málaga, Spain

Abstract

A Jordan algebra J over a field k of characteristic 2 becomes a 2-Lie algebra L(J) with Lie product [x, y] = x∘y and squaring x^{[2]} = x^{2}. We determine the precise ideal structure of L(J) in case J is simple finite-dimensional and k is algebraically closed. We also decide which of these algebras have smooth automorphism groups. Finally, we study the derivation algebra of a reduced Albert algebra J = H_{3}(𝒪, k) and show that Der J has a unique proper nonzero ideal V_{J}, isomorphic to L(J)/k · 1_{J}, with quotient Der J/V_{J} independent of 𝒪. On the group level, this gives rise to a special isogeny between the automorphism group of J and that of the split Albert algebra, whose kernel is the infinitesimal group determined by V_{J}.

Introduction

We study finite-dimensional simple quadratic Jordan algebras J over fields of characteristic 2. This situation is of particular interest because it presents phenomena without counterpart in the linear theory:

(i) The bilinear trace form may be degenerate, giving rise to a proper outer ideal Def(J), the defect of J.

(ii) The algebra may be traceless, i.e., its linear trace form may be zero.

1 The first and third authors are supported by the Spanish DGICYT with project number BFM 2001-1886, by the Junta de Andalucía projects: FQM-336, and ‘Estudio analítico-algebraico de Sistemas Triples y de Pares en diferentes clases de estructuras no asociativas’.

Preprint submitted to Elsevier Science 26 October 2004

(iii) J may not have capacity in the sense of [7].

(iv) Squaring and circle product make the vector space J into a restricted Lie algebra L(J).

(v) The left multiplications V_{J} = {V_{x} : x ∈ J} form an ideal in the derivation algebra Der(J), isomorphic to L(J) modulo its centre.

(vi) The automorphism group of J is not necessarily smooth.

Our main concern is to explore (iv) – (vi), but this also necessitates a study of (i) – (iii).

Here is a more detailed description of the contents. In §1 we introduce the trace forms, the defect, and the notions of rank and primitive rank. We also classify (Prop. 1.9) the orbits of the automorphism group of a simple Jordan pair, and as an application give the classification, in a form suitable for our purposes, of the simple finite-dimensional Jordan algebras over an algebraically closed field of characteristic 2.

§2 contains a detailed study of the Lie algebra L(J). We compute the derived series and show that L(J) is solvable if and only if J has primitive rank ≤ 2 (Cor. 2.9). The ideal structure of L(J) is completely determined in Th. 2.11, and simplicity of subquotients of L(J) is studied in Cor. 2.12. We also show that some of the Lie algebras L(J) occur as Lie algebras of classical algebraic groups.

The question of smoothness of the automorphism group Aut(J) is discussed in §3. Our method is based on the result of [14] that the structure group of a separable Jordan algebra is always smooth. This simplifies and completes work of Springer [23] who studied smoothness of Aut(J) in his framework of J-structures which excludes a priori traceless algebras.

The last section (§4) is devoted to the case where J = H_{3}(𝒪, k) is a reduced Albert algebra. We show (Th. 4.9) that Der(J)/V_{J} is a simple 26-dimensional Lie algebra whose isomorphism class does not depend on 𝒪, and that V_{J} is the unique proper nonzero ideal of Der(J). Finally, using the result of Schafer-Tomber [22] that Der(J)/V_{J} ≅ V_{J^{s}}, where J^{s} is the split Albert algebra, we describe in Th. 4.14 a homomorphism β: Aut(J) → Aut(J^{s}) with infinitesimal kernel, which gives a concrete realization of the special isogeny between an isotropic and a split group of type F_{4}.

Acknowledgements. We wish to thank A. Elduque and H. P. Petersson for their valuable remarks. During the final preparation of this paper, the second-named author was a guest of the Department of Algebra, Geometry and Topology of the University of Málaga. The hospitality of this institution is gratefully acknowledged.

1. Preliminaries

1.1. Diagonalizability and defect in Jordan pairs. Let 𝔙 = (𝔙^{+}, 𝔙^{−}) be a finite-dimensional semisimple Jordan pair over an algebraically closed field k, see [13] for a general reference. We review some properties of the defect and the rank function from [15], specialized to the present situation.

Every x ∈ 𝔙^{σ} (σ ∈ {+, −}) is von Neumann regular [13, Th. 10.17] and can therefore be embedded into a (Jordan pair) idempotent e = (e^{+}, e^{−}) as x = e^{σ} [13, 5.2]. The rank of x, denoted rk(x), is the capacity of the Peirce space 𝔙_{2}(e) [15, Prop. 3]. In particular, x has rank one if and only if 𝔙_{2}(e) is a division pair. Since k is algebraically closed, x is of rank one if and only if Q_{x}𝔙^{−σ} = k · x, i.e., x is reduced in the sense of [16]. Also, 𝔙 itself is reduced in this sense, i.e., it is spanned by its rank one elements. By [16, §1], there exists a well-defined bilinear form f: 𝔙^{+} × 𝔙^{−} → k, the Faulkner form, with the property that

f(x, v)x = Q_{x}v, f(z, y)y = Q_{y}z  (1)

for all rank one elements x ∈ 𝔙^{+}, y ∈ 𝔙^{−}, and arbitrary v ∈ 𝔙^{−}, z ∈ 𝔙^{+}.

An element x ∈ 𝔙^{σ} is called diagonalizable if there exist orthogonal division idempotents d_{1}, …, d_{t} such that x = d_{1}^{σ} + ⋯ + d_{t}^{σ}, and it is called defective if Q_{y}x = 0 for all rank one elements y ∈ 𝔙^{−σ}. Then 0 is the only element which is both diagonalizable and defective, and if 𝔙 is simple, every element is either diagonalizable or defective [15, Cor. 1 of Th. 1]. (For 𝔙 semisimple but not simple, there may be “mixed” elements which are neither diagonalizable nor defective). Letting Def(𝔙^{σ}) denote the set of defective elements of 𝔙^{σ}, the defect Def(𝔙) = (Def(𝔙^{+}), Def(𝔙^{−})) is either zero or a proper outer ideal of 𝔙, which itself has defect zero: Def(Def(𝔙)) = 0. Also, the defect can only be nonzero if k has characteristic two [15, Th. 2].

We next relate the defect to the generic trace form m_{1}: 𝔙^{+} × 𝔙^{−} → k [13, §16].

1.2. Lemma. Let 𝔙 be a semisimple finite-dimensional Jordan pair over an algebraically closed field k.

(a) The Faulkner form agrees with the generic trace: f = m_{1}.

(b) The defect is the kernel of m_{1} in the sense that

x ∈ Def(𝔙^{+}) ⟺ m_{1}(x, 𝔙^{−}) = 0, y ∈ Def(𝔙^{−}) ⟺ m_{1}(𝔙^{+}, y) = 0.

Proof. (a) Consider a frame F = (d_{1}, …, d_{r}) of division idempotents of 𝔙, and let ℭ = ∑_{i=1}^{r} 𝔙_{ii} be the Cartan subpair defined by F. Since both f and m_{1} are invariant under Aut(𝔙), and the orbit of ℭ^{+} × ℭ^{−} under Aut(𝔙) is Zariski-dense in 𝔙^{+}×𝔙^{−} [13, 15.15], it suffices to show that f and m_{1} agree on ℭ^{+}×ℭ^{−}. We have m_{1}(∑_{i=1}^{r} λ_{i}d_{i}^{+}, ∑_{i=1}^{r} μ_{i}d_{i}^{−}) = ∑_{i=1}^{r} λ_{i}μ_{i} by [13, 16.15]. On the other hand, f(d_{i}^{+}, d_{j}^{−}) = 0 for i ≠ j by [16, 1.7(b)], and f(d_{i}^{+}, d_{i}^{−}) = 1 since the d_{i} are division idempotents, by 1.1.1. Hence also f(∑_{i=1}^{r} λ_{i}d_{i}^{+}, ∑_{i=1}^{r} μ_{i}d_{i}^{−}) = ∑_{i=1}^{r} λ_{i}μ_{i}.

(b) This follows from (a) and [16, 1.9(a)].

1.3. Jordan algebras. We recall from [13, §1] the correspondence between isotopy classes of unital Jordan algebras on the one hand and isomorphism classes of Jordan pairs containing invertible elements on the other: If v ∈ (𝔙^{−})^{×} is invertible, then the vector space 𝔙^{+} becomes a unital quadratic Jordan algebra, denoted 𝔙_{v}^{+}, with unit element 1 = v^{−1}, quadratic operators U_{x} = Q_{x}Q_{v}, squaring x^{2} = Q_{x}v and circle product x∘y = {xvy}. Conversely, if J is a unital Jordan algebra then (J, J) is a Jordan pair containing invertible elements, with quadratic operators Q_{x} = U_{x}. This correspondence is easily seen to preserve simplicity.

Let J be a finite-dimensional Jordan algebra over a field k, and let tr(x) = t(x) and Tr(x, y) = t(x, y) be the linear and bilinear trace forms as in [10]. By [10, (15)], we have

tr(x) = Tr(x, 1)  (1)

for all x ∈ J, and from [10, (14)] it is evident that Tr is a symmetric bilinear form on J.

Now let J = 𝔙_{v}^{+} be obtained from a Jordan pair 𝔙 and an invertible element v ∈ 𝔙^{−} as described above. Then tr and Tr are related to the generic trace form m_{1} of 𝔙 by

tr(x) = m_{1}(x, v),  Tr(x, y) = m_{1}(x, Q_{v}y).  (2)

Indeed, by [13, 16.3(ii)], the generic minimum polynomial of the Jordan pair 𝔙̃ := (J, J) is given by

m̃(τ, x, y) = ∑_{i=0}^{d}(−1)^{i}m̃_{i}(x, y)τ^{d−i} = N(x)N(τx^{−1} − y),  (3)

for all y ∈ J and all invertible x ∈ J^{×}. Here N is the generic norm of J and τ an indeterminate. By [13, Prop. 1.11], (Id, Q_{v}): 𝔙̃ → 𝔙 is an isomorphism of Jordan pairs. Hence the coefficients of the respective generic minimum polynomials are related by

m̃_{i}(x, y) = m_{i}(x, Q_{v}y).

From (3) we obtain by comparing coefficients at τ^{d−1} that

m̃_{1}(x, y) = N(x) · ∂_{y}N|_{x^{−1}}.

By [18, Th. 3,(17)], we have ∂_{y}N|_{x^{−1}} = Tr((x^{−1})^{♯}, y), and by standard properties of the adjoint [10, Sec. 3], (x^{−1})^{♯} = (x^{♯})^{−1} = (N(x)x^{−1})^{−1} = N(x)^{−1}x. Hence, m̃_{1}(x, y) = N(x)Tr(N(x)^{−1}x, y) = Tr(x, y) holds for all invertible x and thus by density for all x ∈ J. (Since the generic minimum polynomial is compatible with base field extension, we may assume k algebraically closed, so density arguments involving the Zariski topology are justified) . This proves the second formula of (2), and the first follows by specializing y = v^{−1} = 1_{J}. Now [13, 16.8.2] yields the identity

Tr(x, y ∘ z) = Tr(x ∘ y, z).  (4)

Also, if c is any (algebra) idempotent of J and A = J_{2}(c) = U_{c}J its Peirce-2-space, then the bilinear trace and the trace of A are just the restrictions of those of J to A:

Tr_{A} = Tr|A × A, tr_{A} = tr|A.  (5)

Indeed, this follows from the corresponding property of the Faulkner form [16, 1.7] and Lemma 1.2(a), applied to 𝔙̃ = (J, J) and the Jordan pair idempotent e = (c, c) of 𝔙̃ whose Peirce-2 space is Ṽ_{2}(e) = (J_{2}(c), J_{2}(c)).

We define the defect of J to be Def(J) := Def(𝔙^{+}). From the properties of Def(𝔙) and Lemma 1.2 it follows easily that Def(J) is an outer ideal of J, connected with Tr via

Def(J) = {x ∈ J : Tr(x, J) = 0}.  (6)

Thus Def(J) = 0 if and only if Tr is a nondegenerate bilinear form on J. The algebra J will be called traceless if tr = 0, equivalently, if v ∈ Def(𝔙^{−}) or 1_{J} ∈ Def(J). Clearly, if J has zero defect then its trace will be nonzero, but not conversely.

1.4. Jordan algebras of quadratic forms with base point. Let k be an algebraically closed field of characteristic 2. As an example, we discuss the Jordan algebras obtained from a Jordan pair of a nondegenerate quadratic form q on k^{n}, where q is said to be nondegenerate if q(x) = b(x, y) = 0 for all y ∈ k^{n} implies x = 0. Here b(x, y) = q(x + y) − q(x) − q(y) is the bilinear form associated with q. We may then assume that q is in standard form, given by q(x) = ∑_{i=1}^{m} x_{2i−1}x_{2i} if n = 2m, and x_{0}^{2} + ∑_{i=1}^{m} x_{2i−1}x_{2i} if n = 2m + 1, where x = ∑ x_{i}ε_{i} with respect to the standard basis (ε_{i}) of k^{n}. We always assume n ≥ 3 because the case n = 2 yields a non-simple Jordan pair, and the case n = 1 is without interest. The quadratic operators of the Jordan pair 𝔙 = (k^{n}, k^{n}) determined by q are

Q_{x}y = b(x, y)x − q(x)y,

and b = m_{1} is the generic trace form of 𝔙. Thus by Lemma 1.2, Def(𝔙) = 0 if n is even, while Def(𝔙^{σ}) = k · ε_{0} if n is odd.

An element v is invertible in the Jordan pair sense if and only if q(v) ≠ 0, and it suffices to consider the case q(v) = 1. The isotope with respect to v is then the Jordan algebra J = Jor(k^{n}, q, v) of the quadratic form with base point (k^{n}, q, v), with underlying vector space k^{n}, unit element v, and U-operators given by

U_{x}y = b(x, ȳ)x − q(x)ȳ, where ȳ = b(y, v)v − y.  (1)

Hence the squaring and the circle product are

x^{2} = b(x, v)x − q(x)v, x ∘ y = b(x, v)y + b(y, v)x − b(x, y)v.  (2)

The trace form tr of J is tr(x) = b(x, v). A typical non-defective invertible element is v = ε_{1} + ε_{2}, while in the odd-dimensional case, v = ε_{0} is the only defective element with q(v) = 1. The traceless Jordan algebra Jor(k^{2m+1}, q, ε_{0}), then has ȳ = y, and hence squaring and circle product are given by

x^{2} = q(x)ε_{0}, x ∘ y = b(x, y)ε_{0}.  (3)

1.5. Lemma. Let J be a simple finite-dimensional Jordan algebra of rank r over an algebraically closed field k of characteristic 2. Then J has nonzero trace if and only if J has capacity in the sense of [9, Ch. 6], i.e., there exists an orthogonal system (c_{1}, …, c_{r}) of division idempotents of the algebra J such that 1 = c_{1} + ⋯ + c_{r}. For the Peirce decomposition J = ⨁_{1≤i≤j≤r} J_{ij} of J with respect to such a system we have

J_{ii} = k · c_{i} and J_{ij} ≠ 0 for i ≠ j.  (1)

The defect of J is

Def(J) = ⨁_{i<j} J_{ij}^{♮},  (2)

where

J_{ij}^{♮} = {x ∈ J_{ij} : Tr(x, J_{ij}) = 0}.  (3)

Proof. Let J = 𝔙_{v}^{+} where 𝔙 is a simple Jordan pair. Then v ∉ Def(𝔙^{−}) implies, by 1.1, that v is diagonalizable in 𝔙. Hence there exists a frame F = (d_{1}, …, d_{r}) of division idempotents of 𝔙 such that v = d_{1}^{−} +⋯+d_{r}^{−}, and then c_{i} = d_{i}^{+} are the required algebra idempotents. The properties (1) follow from well-known corresponding ones for the Peirce decomposition of 𝔙 with respect to F. Conversely, if J contains an algebra division idempotent c then (c, c) is a Jordan pair division idempotent and (1 − c) ⊥ c, so 1 = Tr(c, c) = Tr(c, 1) = tr(c) and thus tr ≠ 0. The formula for the defect follows easily from 1.3.6, because the Peirce spaces are orthogonal with respect to Tr and Tr(c_{i}, c_{i}) = 1.

We will call such a system of algebra division idempotents a frame of the Jordan algebra J. From the conjugacy of frames in the Jordan pair 𝔙 [13, 17.1] it follows easily that any two frames of J are conjugate by an automorphism of J.

In contrast, a traceless J cannot contain any division idempotent in the algebra sense. As a substitute for the Peirce decomposition above, we have the following result. By the rank of an element of J we mean its Jordan pair rank when considered as an element of 𝔙^{+}. The rank of J is defined as the rank of 1_{J}.

1.6. Lemma. Let J be a simple finite-dimensional traceless Jordan algebra of rank r over an algebraically closed field k of characteristic 2.

(a) Any algebra idempotent c of J has even rank and belongs to the defect of J. In particular, r = rk(1_{J}) = 2s is even. Moreover, c is primitive if and only if it has rank 2.

(b) There exist orthogonal systems c_{1}, …, c_{s} of primitive algebra idempotents c_{i} with c_{1} + ⋯ + c_{s} = 1. Any two such systems are conjugate under an automorphism of J. In the Peirce decomposition J = ⨁_{1≤i≤j≤s} J_{ij} with respect to such a system, J_{ii} is the Jordan algebra of a traceless nondegenerate quadratic form with base point of odd dimension ≥ 3, and J_{ij} ≠ 0 for i ≠ j.

(c) The defect of J is given by

Def(J) = ⨁_{i=1}^{s} k · c_{i} ⊕ ⨁_{i<j} J_{ij}.  (1)

Proof. (a) As in 1.3 we write J = 𝔙_{v}^{+} where 𝔙 is a simple Jordan pair, and v is a defective invertible element. We put 𝔚 = Def(𝔙). An idempotent c of the algebra J satisfies c = c^{2} = Q_{c}v ∈ 𝔚^{+} = Def(J) because 𝔚 is an outer ideal. Hence also rk(c) = 2 rk_{𝔚}(c) (by [15, Lemma 6(b)]) is even. It follows that rk(c) = 2 implies c is primitive, else c = c′+c′′ could be decomposed as the sum of two orthogonal algebra idempotents, which would have rk(c′) = rk(c′′) = 1. Conversely, let c be a primitive idempotent of J. Then e′ = (c, Q_{v}c) and e′′ = (1 − c, Q_{v}(1 − c)) are orthogonal Jordan pair idempotents in 𝔚 with e′+e′′ = (1, v). Since 𝔚 has defect zero, e′ = d_{1}+⋯+d_{t} is the orthogonal sum of division idempotents d_{i} ∈ 𝔚, so rk(d_{i}) = 2. Also, d_{i} ∈ W_{2}(e′) = W_{0}(e′′), so by the Peirce rules, (d_{i}^{+})^{2} = Q(d_{i}^{+})v = Q(d_{i}^{+})e^{′−} = Q(d_{i}^{+})d_{i}^{−} = d_{i}^{+} is an algebra idempotent. Similarly, one shows that the d_{i}^{+} are orthogonal. Since c = d_{1}^{+} + ⋯ + d_{t}^{+}, it follows from primitivity of c that t = 1, so c has rank 2.

(b) As 𝔚 has defect zero, v is diagonalizable in 𝔚, so we can write v = d_{1}^{−} + ⋯ + d_{s}^{−} where (d_{1}, …, d_{s}) is a frame of division idempotents in 𝔚. In particular, the d_{i} have rank 1 in 𝔚 and therefore rank 2 in 𝔙. Hence the c_{i} := d_{i}^{+} are orthogonal algebra idempotents of J with sum 1_{J}, and they are primitive because they have rank two. Now J_{ii} is the isotope of the Jordan pair 𝔙_{2}(d_{i}) with respect to d_{i}^{−}, and the latter is simple of rank two and has d_{i} in its defect. By [15, p. 260, Example (a)], 𝔙_{2}(d_{i}) is the Jordan pair of a nondegenerate defective quadratic form, so the structure of J_{ii} follows from 1.4. It is a general fact that Peirce-2-spaces inherit simplicity. This implies J_{ij} ≠ 0, else J_{2}(c_{i} + c_{j}) = J_{ii} ⊕ J_{jj} would not be simple.

By the proof of (a), there is a natural bijection between frames (d_{1}, …, d_{s}) of 𝔚 with ∑ d_{i}^{−} = v and systems of primitive orthogonal idempotents of J with sum 1_{J}. Now the conjugacy statement follows easily from conjugacy of frames in 𝔚.

(c) By (a), all c_{i} belong to the defect, and hence also all J_{ij} = c_{i} ∘ J_{ij}, because the defect is an outer ideal. Thus we have the inclusion from right to left in (1). On the other hand, J_{ii} ∩ Def(J) is the defect of J_{ii} [15, Prop. 2(c)], and this is k · c_{i} by 1.4.

1.7. The primitive rank. Let J be a simple Jordan algebra over an algebraically closed field of characteristic 2. By Lemmas 1.5 and 1.6, it makes sense to define the primitive rank prk(J) of J by

prk(J) = \{ \begin{matrix}rk(J) & if J has nonzero trace \\ \frac{1}{2} rk(J) & if J is traceless \end{matrix} \} .

The primitive rank can also be characterized as the maximal length of a system of orthogonal primitive algebra idempotents of J.

1.8. Classification. Using the correspondence between Jordan algebras and Jordan pairs (cf. 1.3), the classification of simple unital Jordan algebras is equivalent to

(i) the classification of simple Jordan pairs 𝔙 containing invertible elements,

(ii) the determination of the orbits of Aut(𝔙) on the set (𝔙^{−})^{×} of invertible elements of 𝔙^{−}.

The classification of simple finite-dimensional Jordan algebras over algebraically closed fields of characteristic ≠ 2 is well known, see, e.g., [8]. In the characteristic two case, the classification could be extracted from the much more general results of [19], but there seems to be no explicit handy reference (the one given in [7] is incomplete). However, under our rather restrictive assumptions, it seems simpler to use the procedure outlined above. Step (i) is well-known [13, §17], so it remains to carry out step (ii). It is actually not difficult to determine all orbits of Aut(𝔙) on 𝔙^{±} which will be important in §3.

1.9. Proposition. Let 𝔙 be a simple finite-dimensional Jordan pair of rank r over an algebraically closed field k and let σ ∈ {±}. Then the automorphism group Aut(𝔙) and the inner automorphism group Inn(𝔙) have the same orbits on 𝔙^{σ}, and these orbits are described as follows:

(a) If Def(𝔙) = 0 then x and x̃ belong to the same orbit if and only if rk(x) = rk(x̃). Hence there are r+1 orbits, corresponding to the possible values 0, …, r of the rank function.

(b) If Def(𝔙) ≠ 0 then x and x̃ belong to the same orbit if and only if rk(x) = rk(x̃), and both x and x̃ are diagonalizable or both are defective. Since the rank of a defective element is always even, there are r + 1 + [r/2] orbits.

Proof. The rank function and the defect are clearly invariant under automorphisms, so the conditions listed in (a) and (b) are certainly necessary for x and x̃ to belong to the same orbit. To prove that they are sufficient, let first x, x̃ ∈ 𝔙^{σ} both be diagonalizable of the same rank t. Then x = d_{1}^{σ}+⋯+d_{t}^{σ} and x̃ = d̃_{1}^{σ} + ⋯ + d̃_{t}^{σ} can be embedded into frames (d_{1}, …, d_{r}) and (d̃_{1}, …, d̃_{r}) of division idempotents of 𝔙. By [13, 17.1], there exists an inner automorphism φ = (φ_{+}, φ_{−}) of 𝔙 such that φ(d_{i}) = d̃_{i}, whence x = φ_{σ}(x̃). As the rank function takes all values between 0 and r, there are r + 1 orbits in case 𝔙 has defect zero.

Next, let x and x̃ both be nonzero and defective of the same rank, and put 𝔚 := Def(𝔙). Then 𝔚 is simple and has Def(𝔚) = 0 [15, Th. 2], and the rank function of 𝔚 is given by rk_{W} (x) = (1/2) rk(x) [15, Lemma 6(b)]. In particular, x and x̃ have even rank. Now x and x̃ are diagonalizable and of the same rank in 𝔚, so they are, by what we proved above, conjugate under some ψ ∈ Inn(𝔚). Here ψ is a finite product of inner automorphisms β(u, v), where (u, v) ∈ 𝔚 is quasi-invertible in 𝔚. Since quasi-invertibility in 𝔚 implies quasi-invertibility in 𝔙, it follows that ψ extends to an inner automorphism φ of 𝔙. The formula for the number of orbits is then immediate.

1.10. Corollary. Let 𝔙 be a simple finite-dimensional Jordan pair containing invertible elements over an algebraically closed field k.

(a) If 𝔙 contains no defective invertible elements then Aut(𝔙) acts transitively on (𝔙^{σ})^{×}.

(b) If 𝔙 contains defective invertible elements then Aut(𝔙) has two orbits on (𝔙^{σ})^{×}. In this case, k has characteristic 2 and the rank of 𝔙 is even.

1.11. Classification, continued. Let k be an algebraically closed field of characteristic 2. We now carry out the classification of simple finite-dimensional Jordan algebras outlined in 1.8. Of the list of simple Jordan pairs [13, §17], precisely the following contain invertible elements: I_{n}, II_{2m}, III_{n}, IV_{n}, VI. From the computation of the generic trace form in [13, §17], we see that m_{1} is degenerate, and hence, by Lemma 1.2, the defect of 𝔙 is nonzero, only in the cases III_{n} and IV_{2m+1}.

(a) The isotopes of the types I_{n}, III_{n} and VI with respect to the unit matrix yield the Jordan algebras of hermitian matrices with diagonal coefficients in k over k ⊕ k, k, and the split octonions 𝒪, respectively. It is well known that the

isotope of II_{2r} with respect to the element v = ( \begin{matrix}0 & 1_{r} \\ 1_{r} & 0 \end{matrix} ) is the Jordan algebra

H_{r}(𝒬, k) of hermitian matrices over the split quaternion algebra 𝒬 = Mat_{2}(k)

with respect to the involution ( \begin{matrix}α & β \\ γ & δ \end{matrix} )^{∗} = ( \begin{matrix}δ & β \\ γ & α \end{matrix} ), with diagonal coefficients in k.

(b) The defect of the Jordan pair III_{n} (n × n symmetric matrices) is the Jordan pair II_{n} of alternating n × n-matrices, which contains invertible

elements if and only if n = 2p is even. In particular, v = ( \begin{matrix}0 & 1_{p} \\ 1_{p} & 0 \end{matrix} ) is such an

element, and the isotope with respect to v is easily seen to be isomorphic to H_{p}(𝒬, 𝒬_{0}), hermitian matrices over 𝒬 as above, with diagonal coefficients in the three-dimensional fixed point set 𝒬_{0} of ∗.

(c) The case of Jordan pairs of type IV_{n}, that is, Jordan pairs of a nondegenerate quadratic form, was done in 1.4.

We collect these results in the following table, which also lists the spaces J ∘ J, determined in 2.7 below.

𝔙  v  J  rk prk tr Def(J) J ∘ J
\begin{matrix}I_{r} \\ (r ≥ 1) \end{matrix} 1_{r}  H_{r}(k ⊕ k, k)  r r ≠ 0  0  Ker tr
\begin{matrix}II_{2r} \\ (r ≥ 1) \end{matrix} \binom{0 1_{r}}{1_{r} 0}  H_{r}(𝒬, k)  r r ≠ 0  0  Ker tr
\begin{matrix}III_{r} \\ (r ≥ 1) \end{matrix} 1_{r}  H_{r}(k)  r r ≠ 0 Alt_{r}(k) Alt_{r}(k)
\begin{matrix}III_{2p} \\ (p ≥ 1) \end{matrix} \binom{0 1_{p}}{1_{p} 0}  H_{p}(𝒬, 𝒬_{0})  2p p 0 H_{p}(𝒬, k) H_{p}(𝒬, k)
\begin{matrix}IV_{2m} \\ (m ≥ 2) \end{matrix} ε_{1} + ε_{2} Jor(k^{2m}, q, ε_{1} + ε_{2}) 2 2 ≠ 0  0  Ker tr
\begin{matrix}IV_{2m+1} \\ (m ≥ 1) \end{matrix} ε_{1} + ε_{2} Jor(k^{2m+1}, q, ε_{1} + ε_{2}) 2 2 ≠ 0 k · ε_{0} Ker tr
\begin{matrix}IV_{2m+1} \\ (m ≥ 1) \end{matrix} ε_{0}  Jor(k^{2m+1}, q, ε_{0}) 2 1 0 k · ε_{0}  k · ε_{0}
VI  1_{3}  H_{3}(𝒪, k)  3 3 ≠ 0  0  Ker tr

There are the following isomorphisms in low ranks:

H_{1}(k) ≅ H_{1}(k ⊕ k, k) ≅ H_{1}(𝒬, k) ≅ k,
H_{1}(𝒬, 𝒬_{0}) ≅ Jor(k^{3}, q, ε_{0}), H_{2}(k) ≅ Jor(k^{3}, q, ε_{1} + ε_{2}),
H_{2}(k ⊕ k, k) ≅ Jor(k^{4}, q, ε_{1} + ε_{2}), H_{2}(𝒬, k) ≅ Jor(k^{6}, q, ε_{1} + ε_{2}).

Finally, we note that H_{r}(k ⊕ k, k) is isomorphic to the full matrix algebra Mat_{r}(k), considered as Jordan algebra.

2. The Lie algebra L associated with J 2.1. The structure Lie algebra and the derivation algebra. Let J be a unital quadratic Jordan algebra over a commutative ring R. Recall [8, p. 437] that the structure Lie algebra Strl(J) of J is the set of all linear maps A: J → J such that there exists a linear map A^{♯} from J to itself with the property that AU_{x} + U_{x}A^{♯} = U_{x,A(x)} for all x ∈ J. This A^{♯} is uniquely determined and given by A^{♯} = U_{1,A(1)} − A; in fact, the map A ↦ A^{ϑ} := −A^{♯} is an automorphism of period 2 of Strl(J). Thus

A ∈ Strl(J) ⟺ U_{x,A(x)} = [A, U_{x}] + U_{x}U_{1,A(1)} for all x ∈ J. (1)

The structure Lie algebra is isomorphic to the derivation algebra Der(𝔙) of the associated Jordan pair 𝔙 = (J, J) via A ↦ (A, A^{ϑ}). For x, y, z ∈ J let as usual {x, y, z} = V_{x,y}z = U_{x,z}y. Then it follows from the defining identities of J that V_{x,y} ∈ Strl(J), with V_{x,y}^{♯} = V_{y,x}.

A derivation of J is an endomorphism Δ satisfying Δ(1) = 0 and [Δ, U_{x}] = U_{Δ(x),x}, for all x ∈ J; equivalently, Δ ∈ Der(J) if and only if Δ ∈ Strl(J) and Δ(1) = 0.

Let us specialize now to the case where 2 = 0 in R. Then each left multiplication V_{x} = V_{x,1} : y ↦ x ∘ y is in Der(J) since x ∘ 1 = 2x = 0. Moreover, Der(J) is a restricted subalgebra of the restricted Lie algebra End(J) (with Lie product given by the commutator) because it is easy to see that for any Δ ∈ Der(J), its square Δ^{2} is again in Der(J). This is of course just a special case of the general fact that the Lie algebra of a linear algebraic group in characteristic p is closed under p-th powers and hence a p-Lie algebra. In particular, Strl(J) is also a 2-Lie algebra.

Recall the following well-known fact:

2.2. Theorem [7, Theorem 4, p. 1.28]. Let J be a quadratic Jordan algebra over a ring R with 2R = 0. Then the R-module J is a 2-Lie algebra, denoted by L = L(J), with commutator [a, b] := a ∘ b and squaring a^{[2]} := a^{2}.

Thus the adjoint representation of L is given by ad(x)y = x ∘ y. From the definitions it follows easily that every derivation of J is also a derivation of the restricted 2-Lie algebra L because [Δ, ad(x)] = ad(Δ(x)) = V_{Δ(x)} and Δ(x^{2}) = x ∘ Δ(x) = [x, Δ(x)]. This also shows that V_{J} is an ideal of the Lie algebra Der(J), and it is even an ideal in the sense of restricted Lie algebras, i.e., a 2-ideal, since the identity QJ20 of [7] says V_{x}^{2} − V_{x^{2}} = 2U_{x} = 0.

2.3. Lemma. Let J be a finite-dimensional Jordan algebra over a field k of characteristic 2 and let tr be its generic trace. Then tr(x)^{2} = tr(x^{2}), in particular, Ker tr is closed under squares.

Proof. This is a consequence of Newton’s identities and probably well known. Since there seems to be no reference in the literature covering the present situation, we provide a proof. Let 𝔙 = (J, J) be the Jordan pair associated to J and let N(x, y) = 1 − m_{1}(x, y) + m_{2}(x, y) ∓ ⋯ be its generic norm [13, 16.9]. Also, let R = k(ε) ⊗ k(δ) be the tensor product of two copies of the dual numbers. For the moment, we don’t assume that k has characteristic 2. Since (ε + δ)^{2} = 2εδ and (ε + δ)^{i} = 0 for i ≥ 3 and the m_{i} are homogeneous of degree i in x and in y, we have in 𝔙 ⊗ R,

N(x, (ε + δ)y) = 1 − (ε + δ)m_{1}(x, y) + 2εδm_{2}(x, y).  (1)

On the other hand, (x, εy) is quasi-invertible with quasi-inverse x^{εy} = x+εU_{x}y, so by [13, Th. 16.11],

N(x, (ε + δ)y) = N(x, εy) · N(x + εU_{x}y, δy)
= (1 − εm_{1}(x, y)) · (1 − δm_{1}(x + εU_{x}y, y))
= 1 − (ε + δ)m_{1}(x, y) + εδ(m_{1}(x, y)^{2} − m_{1}(U_{x}y, y)). (2)

Comparing coefficients at εδ and putting y = 1_{J} yields, because m_{1}(x, 1_{J}) = tr(x) by 1.3.1 and 1.3.2, that tr(x)^{2} − tr(x^{2}) = 2m_{2}(x, 1_{J}), and this vanishes in characteristic 2.

Our next aim is to determine the ideal structure of L in the finite-dimens-ional simple case. Recall that a simple algebra over a field k is said to be absolutely simple if it remains simple under any base field extension.

2.4. Lemma. Let J be a simple Jordan algebra of primitive rank p over an algebraically closed field k of characteristic 2 and and let c_{1}, …, c_{p} be a complete orthogonal system of primitive idempotents, with associated Peirce decomposition J = ⨁_{1≤i≤j≤p} J_{ij}.

(a) We have

J_{ii}^{2} = k · c_{i} and J_{ij}^{2} = k · (c_{i} + c_{j}) for i ≠ j.  (1)

(b) Let a_{ii} ∈ J_{ii}. Then

a_{ii} ∘ J_{ii} = \{ \begin{matrix}0 & if a_{ii} ∈ k · c_{i} \\ k · c_{i} & if a_{ii} ∉ k · c_{i} \end{matrix} \} ,  (2)
J_{ii} ∘ J_{ii} = \{ \begin{matrix}0 & if dim J_{ii} = 1 \\ k · c_{i} & if dim J_{ii} > 1 \end{matrix} \} .  (3)

(c) Suppose J has nonzero trace. Then for i ≠ j and a_{ij} ∈ J_{ij},

a_{ij} ∘ J_{ij} = \{ \begin{matrix}0 & if a_{ij} ∈ J_{ij}^{♮} \\ k · (c_{i} + c_{j}) & otherwise \end{matrix} \}  (4)

where J_{ij}^{♮} is defined in 1.5.3. Also,

dim J_{ij}^{♮} = \{ \begin{matrix}0 & if dim J_{ij} is even \\ 1 & if dim J_{ij} is odd \end{matrix} \}  (5)

and hence

J_{ij} ∘ J_{ij} = \{ \begin{matrix}0 & if dim J_{ij} = 1 \\ k · (c_{i} + c_{j}) & if dim J_{ij} > 1 \end{matrix} \} .  (6)

Proof. J_{ii} is either k · c_{i} or the Jordan algebra of a traceless nondegenerate quadratic form q with base point c_{i} of dimension ≥ 3. Now the first formula of (1), formula (2) and hence (3) follow easily from 1.4.3.

Let A := k · c_{i} ⊕ J_{ij} ⊕ k · c_{j}. If J has nonzero trace, the c_{i} are division idempotents so J_{ii} = k · c_{i} and therefore A = J_{2}(c_{i} + c_{j}). If, on the other hand, J is traceless and K := Def(J), then the c_{i} are division idempotents of K and 1.6.1 shows that A = K_{2}(c_{i} + c_{j}). Thus in any case, A is a subalgebra which is simple of rank 2 and has nonzero trace. By the classification, A ≅ Jor(k^{n}, q, ε_{1} + ε_{2}) is the Jordan algebra of a nondegenerate quadratic form q with nonzero trace of dimension n ≥ 3. Thus q is given by q(λc_{i} + μc_{j} + x_{ij}) = λμ + q(x_{ij}) and q|J_{ij} is again nondegenerate. Since J_{ij} is orthogonal to c_{i} + c_{j} = 1_{A}, formula 1.4.2 shows x^{2} = q(x)(c_{i} + c_{j}) for x ∈ J_{ij}. This proves the second formula of (1).

Now let J have nonzero trace. By 1.4, the bilinear trace of A is Tr_{A} = b, the bilinear form associated to q, and by 1.3.5 this is also the restriction of the bilinear trace Tr of J to A. Thus we have x ∘ y = Tr(x, y)(c_{i} + c_{j}) for all x, y ∈ J_{ij} and therefore (4). In characteristic 2, a nondegenerate quadratic form has its associated bilinear form equal to zero if and only if it is onedimensional. Hence we have (5) and (6).

2.5. Proposition. Let J be a finite-dimensional absolutely simple Jordan algebra over a field k of characteristic 2. Then V_{x} = 0 for x ∈ J if and only if x ∈ k · 1, so the centre of the associated Lie algebra L = L(J) is k · 1.

Proof. We have V_{1} = 2Id = 0 since k has characteristic two. For the converse, it is no restriction, after extending scalars, to assume k algebraically closed. We use the Peirce decompositions given in Lemmas 1.5 and 1.6. Choose an orthogonal system c_{1}, …, c_{p} of primitive idempotents of J such that c_{1} +⋯+ c_{p} = 1, decompose x = ∑_{1≤i≤j≤p} x_{ij} relative to (c_{1}, …, c_{p}), and put x_{ij} = x_{ji} for convenience. Then for all l = 1, …, p, using the Peirce rules and 2 = 0 in k, we have

0 = V_{x}c_{l} = x_{ll} ∘ c_{l} + ∑_{i<j} x_{ij} ∘ c_{l} = 2x_{ll} + ∑_{i≠l} x_{il} ∘ c_{l} = ∑_{i≠l} x_{il}.

This shows that all off-diagonal x_{il} vanish, so x = ∑_{i=1}^{p} x_{ii}. Furthermore, 0 = x ∘ J_{ii} = x_{ii} ∘ J_{ii} implies x_{ii} = λ_{i}c_{i} is a scalar multiple of c_{i}, by 2.4.2.

Now let i ≠ j. Since all off-diagonal Peirce spaces J_{ij} are nonzero, we may choose 0 ≠ z_{ij} ∈ J_{ij} and then obtain 0 = x ∘ z_{ij} = λ_{i}c_{i} ∘ z_{ij} + λ_{j}c_{j} ∘ z_{ij} = (λ_{i} + λ_{j})z_{ij}, whence λ_{i}+λ_{j} = 0 or λ_{i} = λ_{j}. It follows that x = λ(c_{1}+⋯+c_{p}) = λ·1, as asserted.

2.6. Corollary. We keep the assumptions of 2.5, and assume that Tr is nondegenerate. Then [L, L] = Ker tr is a 2-ideal of codimension one.

Proof. For a subspace X of J let X^{⊥} = {y ∈ J : Tr(X, y) = 0}. Then y ∈ [L, L]^{⊥} ⟺ 0 = Tr(J ∘ J, y) = Tr(J, J ∘ y) (by 1.3.4) ⟺ J ∘ y = 0 (by nondegeneracy of Tr) ⟺ y ∈ k · 1 (by 2.5). Again by nondegeneracy of Tr, it follows that [L, L] = [L, L]^{⊥⊥} = (k · 1)^{⊥} = Ker tr, and Ker tr is closed under squares by Lemma 2.3.

2.7. Proposition. Let J be a finite-dimensional simple Jordan algebra of rank r ≥ 2 over an algebraically closed field k of characteristic 2, and L the associated Lie algebra.

(a) If J is traceless we have [L, L] = L^{[2]} = Def(J).

(b) If J has nonzero trace, let c_{1}, …, c_{r} be a frame of division idempotents of J, and let J = ⨁_{i} k ·c_{i} ⊕⨁_{i≤j} J_{ij} the associated Peirce decomposition. Then

[L, L] = \{ \begin{matrix}Def(J) = ⨁_{i<j} J_{ij} & if dim J_{12} = 1 \\ Ker tr = \{∑ λ_{i}c_{i} : ∑ λ_{i} = 0\} ⊕ ⨁_{i<j} J_{ij} & if dim J_{12} > 1 \end{matrix} \}

and L^{[2]} = L.

Proof. (a) We have 1 ∈ Def(J), and hence also J ∘ J = {J, 1, J} ⊂ Def(J), because the defect is an outer ideal. To prove the reverse inclusion, we choose a frame c_{1}, …, c_{p} of primitive idempotents and use formula 1.6.1 for the defect: J_{ij} = c_{i}∘J_{ij} ⊂ J ∘J for i ≠ j, and J_{ii} has dimension ≥ 3 by 1.6(b), so c_{i} ∈ J ∘J by 2.4.3. Finally, L^{[2]} = J^{2} = ∑_{i} J_{ii}^{2} + ∑_{i<j} J_{ij}^{2} + J ∘ J, so L^{[2]} = J ∘ J follows from Lemma 2.4(a).

(b) Always [L, L] ⊂ Ker tr, because

tr(x ∘ y) = Tr(x ∘ y, 1) = Tr(x, y ∘ 1) = 2Tr(x, y) = 0.

As before, J_{ij} = c_{i}∘J_{ij} ⊂ J∘J for i ≠ j. The only way elements in (J∘J)∩∑ J_{ii} can arise is from J_{ij} ∘ J_{ij} for i ≠ j, because J_{ii} ∘ J_{jj} = 0, and J_{ij} ∘ J_{jl} ⊂ J_{il} for different i, j, l. From conjugacy of frames, it follows that dim J_{12} = dim J_{ij} for all i ≠ j. Hence there are the following two cases:

(i) dim J_{ij} = 1 for all i ≠ j. Then J_{ij} ∘ J_{ij} = 0 by 2.4.6, so J ∘ J = ⨁_{i<j} J_{ij} = Def(J), by 1.5.2 and 2.4(c).

(ii) dim J_{ij} > 1 for all i ≠ j. Then J ∘ J contains all J_{ij} as well as all c_{i} + c_{j}, for i ≠ j, by 2.4.6. Now an element x = ∑_{i=1}^{n} λ_{i}c_{i} has tr(x) = ∑_{i=1}^{n} λ_{i} because tr(c_{i}) = 1, and the formula x = tr(x)c_{1} + ∑_{i=2}^{n} λ_{i}(c_{i} + c_{1}) shows that every element of trace zero belongs to J ∘ J. The final assertion is clear from the fact that all c_{i} = c_{i}^{2} ∈ L^{[2]}.

It is now easy to determine the derived series of the Lie algebra L. Recall that this is defined inductively by L^{(0)} = L, L^{(n+1)} = [L^{(n)}, L^{(n)}]. We also use the notation L′, L′′ etc. instead of L^{(1)}, L^{(2)} etc.

2.8. Corollary. Let J be a finite-dimensional simple Jordan algebra of rank r ≥ 2 over an algebraically closed field k of characteristic 2, with associated Lie algebra L.

(a) If J has nonzero trace and r ≥ 3 then

L⫌L′ = L′′ ≠ 0,

where L′ = [L, L] is described in 2.7(b).

(b) Let J be traceless and r = 2s ≥ 6, and put K = Def(J). Then K is a simple Jordan algebra of rank s with nondegenerate trace form Tr_{K} and hence tr_{K} ≠ 0, and the derived series of L is

L⫌L′ = K⫌L′′ = Ker(tr_{K}) = L′′′ ≠ 0.

(c) If J is traceless of rank 2, i.e., J = Jor(k^{2m+1}, q, ε_{0}) is the Jordan algebra of a traceless quadratic form with base point of dimension 2m + 1 ≥ 3, then

L⫌L′ = k · 1_{J}⫌L′′ = 0.

(d) If J has rank 2, dimension n ≥ 4 and nonzero trace, i.e., J = Jor(k^{n}, q, ε_{1} + ε_{2}) is the Jordan algebra of a quadratic form with base point and nonzero trace, then

L⫌L′ = Ker tr⫌L′′ = k · 1⫌L′′′ = 0.

(e) If J = Jor(k^{3}, q, ε_{1} + ε_{2}) ≅ H_{2}(k) is the Jordan algebra of a 3dimensional quadratic form with base point and nonzero trace, then

L⫌L′ = k · ε_{0}⫌L′′ = 0.

(f) Let J be traceless and r = 4, and let K = Def(J). Then K = Jor(k^{6}, q, ε_{1} + ε_{2}) is as in case (d), and the derived series is

L⫌L′ = K⫌L′′ = Ker(tr_{K})⫌L′′′ = k · 1⫌L^{(4)} = 0.

Proof. (a) By 2.7(b), J_{ij} ⊂ [L, L] for i ≠ j. Since r ≥ 3, there exists l ≠ i, j, and therefore J_{ij} = J_{il}∘J_{lj} ⊂ L′′. If dim J_{12} = 1 then by 2.7(b), L′ = ⨁_{i<j} J_{ij} ⊂ L′′. If dim J_{12} > 1 then again by 2.7(b), [L, L] = ⨁_{i<j} J_{ij} + ∑_{i<j} k · (c_{i} + c_{j}). Here c_{i} + c_{j} ∈ J_{ij} ∘ J_{ij} ⊂ L′′, so we again have L′′ = L′.

(b) Here L′ = K by 2.7(a), and K has rank s ≥ 3 and nondegenerate trace form, so the assertion follows from what we proved in (a) (applied to K) and Cor. 2.6.

(c), (d), (e) This follows by easy computations in the Jordan algebra of a quadratic form with base point. The details are left to the reader.

(f) By 2.7(a), K = Def(J) is a rank two algebra with nondegenerate trace form Tr_{K}, hence of the form K = Jor(k^{2m}, q, ε_{1} + ε_{2}), and m ≥ 2. These algebras have been dealt with in case (d), so the form of the derived series in this case follows from (d). In fact, by the classification in 1.11 we have J = H_{2}(𝒬, 𝒬_{0}) and K = H_{2}(𝒬, k) ≅ Jor(k^{6}, q, ε_{1} + ε_{2}).

2.9. Corollary. Under the assumptions of 2.8, L is a solvable Lie algebra if and only if J has primitive rank ≤ 2.

We can now determine all ideals of the Lie algebra L, which we also call Lie ideals to distinguish them from the Jordan ideals. A Lie ideal which is closed under the squaring map will be called a 2-ideal. Of course, 0, L, and the centre Z(L) = k · 1 (by 2.5) are always Lie ideals and even 2-ideals, so we will concentrate on the non-central proper ideals. Also the defect, being an outer ideal of J, is a Lie ideal (but not necessarily a 2-ideal, as the example J = H_{r}(k), Def(J) = Alt_{r}(k) shows). Note that in any Lie algebra L, an arbitrary subspace containing the derived algebra L′ = [L, L] is always an ideal.

2.10. Lemma. Let J be a simple finite-dimensional Jordan algebra of rank r and primitive rank p over an algebraically closed field k of characteristic 2. Let c_{1}, …, c_{p} be an orthogonal system of primitive idempotents of J, with associated Peirce decomposition J = ∑_{1≤i≤j≤p} J_{ij}. Also let 𝔞 be a Lie ideal of J.

(a) We have

𝔞 = 𝔞 ∩ ( ∑_{i=1}^{p} J_{ii}) ⊕ ⨁_{i<j}(𝔞 ∩ J_{ij}).  (1)

(b) Suppose that p ≥ 2. Then 𝔞 is central if and only if 𝔞 ∩ J_{ij} = 0 for all i ≠ j.

(c) Suppose that r ≥ 3. Then a non-central Lie ideal 𝔞 contains all J_{ij}, i ≠ j. If 𝔞 is a 2-ideal then it also contains all c_{i} + c_{j}, i ≠ j.

Proof. (a) Decompose an arbitrary element a ∈ 𝔞 into its Peirce components: a = ∑ a_{ij}. Then c_{i} ∘ (c_{j} ∘ a) = a_{ij} for i ≠ j by the Peirce rules and because k has characteristic 2, so a_{ij} ∈ 𝔞, and we have (1).

(b) If 𝔞 ⊂ k · 1 is central then clearly 𝔞 ∩ J_{ij} = 0. Conversely, let 𝔞 ∩ J_{ij} = 0 for all i ≠ j, so 𝔞 ⊂ ⨁_{i=1}^{p} J_{ii} by (1). Decompose an element a ∈ 𝔞 accordingly as a = ∑ a_{ii}. If a_{ii} ∉ k ·c_{i} for some i then by 2.4.2, a∘J_{ii} = a_{ii} ∘J_{ii} = k ·c_{i} ⊂ 𝔞, and hence c_{i} ∘ J_{ij} = J_{ij} ⊂ 𝔞, contradiction. Thus we have a_{ii} = λ_{i}c_{i} ∈ k · c_{i} for all i. Now a ∘ J_{ij} = (λ_{i} + λ_{j})J_{ij} ⊂ 𝔞 ∩ J_{ij} = 0 for i ≠ j shows λ_{i} + λ_{j} = 0, so all λ_{i} are equal, and therefore a ∈ k · 1.

(c) Let 𝔞 be a non-central Lie ideal. First, let J have nonzero trace. Then p = r ≥ 3 and J_{ii} = k·c_{i} is one-dimensional. By (b) we have, say, 𝔞∩J_{12} ≠ 0. If J_{12} has dimension 1, this already implies J_{12} ⊂ 𝔞. Otherwise, as a consequence of the classification (r ≥ 3 is essential here), J_{12} has even dimension ≥ 2, and it follows from 2.4(c) that a_{12} ∘ J_{12} = k · (c_{1} + c_{2}) for any nonzero a_{12} ∈ 𝔞 ∩ J_{12}, so we conclude that c_{1} +c_{2} ∈ 𝔞. Since r ≥ 3, (c_{1} +c_{2})∘J_{13} = c_{1} ∘J_{13} = J_{13} ⊂ 𝔞. Now the well-known relations

J_{ij} = J_{il} ∘ J_{lj}  (2)

for three distinct indices i, j, l imply that 𝔞 contains all J_{ij}, i ≠ j.

Next, assume J traceless. Since the rank of a traceless algebra is even, we then have r = 2p ≥ 4. By the classification 1.11, J = H_{p}(𝒬, 𝒬_{0}) is the algebra of hermitian matrices over the split quaternion algebra 𝒬 = Mat_{2}(k) with diagonal coefficients in the fixed point space 𝒬_{0} = \{\binom{αβ}{γα} : α, β, γ ∈ k\} of the involution ∗. Denote the usual matrix units by E_{ij} and put a[ii] := aE_{ii} and b[ij] := bE_{ij} + b^{∗}E_{ji}, for i ≠ j, where a ∈ 𝒬_{0}, b ∈ 𝒬. Then c_{i} = E_{ii} (i = 1, …, p) is a frame of primitive idempotents of J, with Peirce spaces J_{ii} = 𝒬_{0}[ii] and J_{ij} = 𝒬[ij].

Let again 𝔞 be a non-central Lie ideal. Then by (b) we have, say, M := 𝔞 ∩ J_{12} ≠ 0. Since 𝔞 is a Lie ideal, it follows from the Peirce rules that J_{ii}∘M ⊂ M. A computation shows a[11] ∘ b[12] = (ab)[12] and a[22] ∘ b[12] = (ba)[12], for all a ∈ 𝒬_{0}, b ∈ 𝒬. Hence M = B[12] where B ⊂ 𝒬 is a subspace with the property that 𝒬_{0}B + B𝒬_{0} ⊂ B. It is an easy exercise to show that B is then a two-sided ideal of the associative algebra 𝒬. As 𝒬 is simple and M ≠ 0, it follows that M = J_{12} ⊂ 𝔞. Now J_{ij} ⊂ 𝔞 for all i ≠ j follows again from (2). Finally, the last statement is immediate from the fact that J_{ij}^{2} = k · (c_{i} + c_{j}) by 2.4.1.

2.11. Theorem. Let J be a simple finite-dimensional Jordan algebra of rank r ≥ 2 and primitive rank p over an algebraically closed field k of characteristic 2, and let L = L(J) be the associated 2-Lie algebra with underlying vector space J, Lie product [x, y] = x∘y and squaring x^{[2]} = x^{2}. We choose a complete orthogonal system c_{1}, …, c_{p} of primitive idempotents of J, with associated Peirce decomposition J = ∑_{1≤i≤j≤p} J_{ij}.

(a) If J has nonzero trace and dim J_{ij} = 1, i.e., J ≅ H_{r}(k), then the proper non-central Lie ideals of L are precisely the subspaces 𝔞 satisfying L⫌𝔞 ⊃ L′ ≅ Alt_{r}(k), while Ker tr is the only proper non-central 2-ideal.

(b) If J has nonzero trace, r ≥ 3 and dim J_{ij} > 1 for i ≠ j, i.e., J ≅ H_{r}(𝒞, k) where 𝒞 is a composition algebra of dimension ≥ 2 (and r = 3 in case 𝒞 = 𝒪) then L′ = Ker tr is the only non-central proper Lie ideal of L, and it is a 2-ideal.

(c) If J is traceless of rank r = 2p ≥ 4, i.e., J ≅ H_{p}(𝒬, 𝒬_{0}) then the proper non-central Lie ideals of L are L′′ and all subspaces 𝔞 with L⫌𝔞 ⊃ L′. Here L′ = K := Def(J) ≅ H_{p}(𝒬, k) and L′′ = Ker tr_{K}. All Lie ideals are 2-ideals.

(d) Let J have nonzero trace, rank 2 and dimension n ≥ 4, so that J = Jor(k^{n}, q, ε_{1} + ε_{2}) is the Jordan algebra of a quadratic form with base point and nonzero trace. Then for n even, the proper non-central Lie ideals of L are precisely the subspaces 𝔞 with k · 1⫋𝔞 ⊂ L′ = Ker tr, and in addition, for n odd, the 1-dimensional ideal Def(J) = k · ε_{0}. All Lie ideals are 2-ideals.

(e) If J is traceless of rank 2, i.e., J = Jor(k^{2m+1}, q, ε_{0}) is the Jordan algebra of a traceless quadratic form with base point, then L′ = Z(L) = k · 1 by 2.8(c), and the proper non-central ideals of L are precisely the subspaces L⫌𝔞⫌k · 1. All Lie ideals are 2-ideals.

Proof. (a) We leave the case r = 2, which consists of simple computations with 2 × 2-matrices, to the reader, and assume r ≥ 3. By Lemma 2.10(c) and 2.7(b), we have 𝔞 ⊃ L′ = ⨁_{i<j} J_{ij}, and by the remark made after Cor. 2.9, these spaces are indeed Lie ideals. If 𝔞 is a 2-ideal then it contains all c_{i} + c_{j} by Lemma 2.4(a), and therefore 𝔞 = Ker tr. Conversely, Ker tr is closed under squares by 2.3 and hence a 2-ideal.

(b) By Lemma 2.10(c), 𝔞 contains all J_{ij} (i ≠ j), and c_{i} + c_{j} ∈ J_{ij} ∘ J_{ij} ⊂ 𝔞 by 2.4.6, whence 𝔞 = Ker tr.

(c) By 1.6.1, the defect is K = ∑ k · c_{i} ⊕ ⨁_{i<j} J_{ij}. By (b) and (f) of 2.8, we have L′ = K and L′′ = Ker(tr_{K}) = {∑ λ_{i}c_{i} : ∑ λ_{i} = 0} + ⨁_{i<j} J_{ij}. All J_{ij} ⊂ 𝔞 for i ≠ j, by Lemma 2.10(c). Furthermore, the classification shows that dim J_{ij} = 4. Now 2.4.6, applied to K, which has nonzero trace and Peirce spaces K_{ij} = J_{ij}, yields c_{i} + c_{j} ∈ 𝔞 for all i ≠ j. This shows L′′ ⊂ 𝔞. Now assume L′′⫋𝔞. As 𝔞 contains all J_{ij} for i ≠ j, there must be an element of the form a = ∑ a_{ii} ∈ 𝔞 for which either a_{ii} ∉ k · c_{i} for some i, or for which all a_{ii} = λ_{i}c_{i} but ∑ λ_{i} ≠ 0. In the first case, a ∘ J_{ii} = a_{ii} ∘ J_{ii} = k · c_{i} ⊂ 𝔞 by 2.4.2, so K = k · c_{i} + K′ ⊂ 𝔞. In the second case, a ∈ K and tr_{K}(a) = ∑ λ_{i} ≠ 0, so again K = k · a + K′ ⊂ 𝔞. Since L′′ is closed under squares by Lemma 2.3 and L^{[2]} = K by Prop. 2.7(a), all these ideals are 2-ideals.

(d) A frame of division idempotents of J is c_{i} = ε_{i} (i = 1, 2), with Peirce spaces J_{ii} = k · c_{i} and J_{12} = ∑_{j≠1,2} k · ε_{j}, of dimension n − 2 ≥ 2. The trace is given by tr(c_{i}) = 1 and tr(J_{12}) = 0. Hence by 2.7(b), L′ = Ker tr = k · 1 + J_{12}, so that a subspace 𝔞 with k · 1⫋𝔞 ⊂ L′ has the form 𝔞 = k · 1 ⊕ M, for an arbitrary nonzero subspace M of J_{12}. From 1.4.2 it follows easily that such 𝔞 are non-central proper ideals of L. Also, in the odd-dimensional case, it is clear that Def(J) = k · ε_{0}, which is an outer ideal of J, is an ideal of L. Conversely, let 𝔞 be a non-central proper ideal of L. Then by Lemma 2.10, 𝔞 = (𝔞 ∩ (k · c_{1} + k · c_{2})) ⊕ M, where M := 𝔞 ∩ J_{12} ≠ 0. We claim that 𝔞∩(k ·c_{1} +k ·c_{2}) ⊂ k ·1. Indeed, suppose 𝔞 contains an element a = λ_{1}c_{1} +λ_{2}c_{2} with λ_{1} ≠ λ_{2}. Then a ∘ J_{12} = (λ_{1} + λ_{2})J_{12} = J_{12} ⊂ 𝔞. As dim J_{12} > 1, the restriction to J_{12} of the bilinear form b associated to q is nonzero, so by 1.4.2, J_{12} ∘ J_{12} = k · 1 ⊂ 𝔞, which implies c_{1}, c_{2} ∈ 𝔞, and thus 𝔞 = L would not be a proper ideal. Now there are two possibilities:

Case 1: 𝔞 ∩ (k · c_{1} + k · c_{2}) = k · 1. Then 𝔞 = k · 1 ⊕ M has the required form.

Case 2: 𝔞∩(k·c_{1}+k·c_{2}) = 0. Then M∘J_{12} = b(M, J_{12})·1 ⊂ 𝔞∩(k·c_{1}+k·c_{2}) = 0. From b(M, c_{i}) = 0 we see that M is contained in the kernel of b. As M ≠ 0, it follows that n is odd and 𝔞 = M = k · ε_{0}, the defect of J.

Finally, part (e) (where r = 2) is evident from 2.8(c). The fact that all Lie ideals are 2-ideals follows easily from the formulas for x^{2} in 1.4.

We now discuss simplicity of subquotients of L. Since L is solvable for p = prk(J) ≤ 2, we assume p ≥ 3.

2.12. Corollary. Let J be a simple finite-dimensional Jordan algebra over an algebraically closed field k of characteristic 2, of rank r and of primitive rank p ≥ 3.

(a) Let J have nonzero trace, so r = p ≥ 3.

(i) If dim J_{12} = 1 (and thus J ≅ H_{r}(k)) then L′ = Alt_{r}(k) is simple for r ≠ 4, and isomorphic to Alt_{3}(k) × Alt_{3}(k) for r = 4.

(ii) If dim J_{12} > 1 then L′/Z(L) ∩ L′ is simple.

(b) Let J be traceless, so p = r/2 ≥ 3 and J = H_{p}(𝒬, 𝒬_{0}) by 1.11. Then L′′/Z(L) ∩ L′′ is simple.

Proof. (a) The well-known proof about the structure of the Lie algebra Alt_{r}(k) works in any characteristic and yields (i).

Next, consider case (ii). By 2.7(b), L′ = Ker tr contains all c_{i} + c_{j}. Now let 𝔞 ⊂ L′ be an ideal of L′ with 𝔞 ⊄ k ·1. By 2.4.6, it suffices to show that J_{ij} ⊂ 𝔞 for all i ≠ j. Decompose an element a ∈ 𝔞 \ k · 1 as a = ∑ λ_{i}c_{i} + ∑_{i<j} a_{ij} where tr(a) = ∑ λ_{i} = 0. If all a_{ij} = 0 then not all λ_{i} can be equal; say, λ_{1} ≠ λ_{2}. Hence a ∘ J_{12} = (λ_{1} + λ_{2})J_{12} = J_{12} ⊂ 𝔞.

If, say, a_{12} ≠ 0 then by 2.4(c), c_{1} + c_{2} ∈ 𝔞, hence (c_{1} + c_{2}) ∘ J_{13} = J_{13} ⊂ 𝔞. Now 2.10.2 implies that 𝔞 contains all J_{ij}.

(b) Here K = Def(J) = L′ has nonzero trace and rk(K) = p ≥ 3. Also dim K_{12} = dim J_{12} = 4, so the assertion follows from case (ii) of (a) applied to K.

2.13. Corollary. Let J = H_{3}(𝒪, k) where 𝒪 is an octonion algebra and k has characteristic 2 but is not necessarily algebraically closed. Then V_{J} is a simple 2-ideal of dimension 26 in Der(J).

Proof. As noted after Th. 2.2, V_{J} is always a 2-ideal in Der(J), so it remains to show that V_{J} is simple. Since J is absolutely simple we may assume k algebraically closed. Let L = L(J) as in 2.2. We have Z(L) = k · 1_{J} by Prop. 2.5, and Ker(tr) = L′ by Prop. 2.7(b). It follows that Z(L) ∩ L′ = {0}, so V_{J} ≅ J/k · 1_{J} ≅ L′ = L′/Z(L) ∩ L′ is simple by case (ii) of the previous Corollary.

For later use, we now show that the Lie algebras L(H_{n}(𝒬, k)) are isomorphic to Lie algebras of orthogonal groups of even rank, and that their trace forms correspond to the spinor trace. We begin by introducing this concept, the infinitesimal version of Bass’ spinor norm [3].

2.14. The spinor trace. Let k be a an arbitrary field and let q be a nondegenerate quadratic form on a k-vector space X of dimension 2n. We denote by Cliff(q) the Clifford algebra of q, by Cliff_{0}(q) its even part, by O(q) the orthogonal group, and by

Spin(q) = {u ∈ Cliff_{0}(q)^{×} : uXu^{−1} = X and uu^{∗} = 1}  (1)

the spin group, cf. [11, Ch. IV]. Here ∗ denotes the main involution of Cliff(q), characterized by x^{∗} = x for all x ∈ X, where we identify X with its canonical image in Cliff(q). By “varying the base ring” we obtain the spin group and orthogonal group schemes of q, i.e.,

Spin(q)(R) = Spin(q ⊗ R) for all R ∈ k-alg,

and similarly for O(q). The Lie algebras of Spin(q) and O(q) are then

𝔰𝔭𝔦𝔫(q) = {v ∈ Cliff_{0}(q) : [v, X] ⊂ X and v + v^{∗} = 0},  (2)
𝔬(q) = {A ∈ End(X) : b(x, A(x)) = 0 for all x ∈ X},  (3)

where b is the bilinear form associated with q. Let χ: Spin(q) → O(q) be the vector representation, and χ̇ := Lie(χ): 𝔰𝔭𝔦𝔫(q) → 𝔬(q) the corresponding Lie algebra homomorphism. It is well known that Cliff(q) is an Azumaya algebra. From this, one sees easily that Ker(χ) = μ_{2} · 1 where μ_{2} denotes the group of second roots of unity. Hence, if k has characteristic ≠ 2, we have Lie(μ_{2}) = 0 and χ̇: 𝔰𝔭𝔦𝔫(q) ⟶^{≅} 𝔬(q) is injective, while Ker(χ̇) = Lie(Ker(χ)) = k · 1 if k has characteristic 2

We now consider the question of surjectivity of χ̇. Let k(ε) be the algebra of dual numbers. Every A ∈ 𝔬(q) defines an element g = Id+εA ∈ O(q)(k(ε)). Using the universal property of the Clifford algebra, one sees that g induces an automorphism α of Cliff(q) ⊗ k(ε) such that α(x) = g(x), for all x ∈ X. Since the Clifford algebra is an Azumaya algebra and k(ε) is a local ring, α is inner [12, IV, Cor. 1.3], given by conjugation with an element u, and it is easily seen that we may assume u in the form u = 1 + εv for some v ∈ Cliff_{0}(q). Then we have [v, x] = A(x) for all x ∈ X, but v + v^{∗} ≠ 0 in general, so v is not necessarily in 𝔰𝔭𝔦𝔫(q). However, v + v^{∗} ∈ k · 1, because [v + v^{∗}, x] = [v, x] + [v^{∗}, x^{∗}] = [v, x] + [x, v]^{∗} = [v, x] + [x, v] = 0 for all x ∈ X, so v + v^{∗} = λ · 1 is central in Cliff(q). If k has characteristic ≠ 2, then v′ := \frac{1}{2}(v −v^{∗}) = v −λ/2 is in 𝔰𝔭𝔦𝔫(q) and satisfies χ̇(v′) = A, so χ̇ is surjective and therefore an isomorphism. Now let k have characteristic 2. Then there is a linear form trs: 𝔬(q) → k, called the spinor trace, given by trs(A) · 1 = v + v^{∗}. This is well-defined, because if also [w, x] = A(x) for all x ∈ X then w−v = γ·1 for γ ∈ k, so w + w^{∗} = v + v^{∗} + 2γ · 1 = v + v^{∗}. By construction, it is clear that A belongs to the image of χ̇ if and only if trs(A) = 0. Also, it is easily seen that trs is in fact a Lie algebra homomorphism, so we have the exact sequence

0 ⟶ k · 1 ⟶ 𝔰𝔭𝔦𝔫(q) ⟶^{χ̇} 𝔬(q) ⟶^{trs} k ⟶ 0

of Lie algebras. We put

𝔬′(q) := χ̇(𝔰𝔭𝔦𝔫(q)) = Ker(trs).

2.15. Proposition. Let q be a nondegenerate quadratic form on a vector space X of dimension 2n over a field k of characteristic 2. Let 𝒬 = Mat_{2}(k) be the split quaternion algebra, and L = L(J) the Lie algebra associated to J = H_{n}(𝒬, k).

(a) There is an isomorphism φ: 𝔬(q) → L of 2-Lie algebras such that Id_{X} ↦ 1_{J} and the spinor trace on 𝔬(q) corresponds to the trace form of J.

(b) If n = 2m + 1 ≥ 3 then 𝔬(q) = k · Id_{X} ⊕ 𝔬′(q) (direct sum of ideals) and 𝔬′(q) is a simple Lie algebra, while for n = 2m ≥ 4, we have Id_{X} ∈ 𝔬′(q) and 𝔬′(q)/k · Id_{X} is simple.

Proof. (a) Since k has characteristic 2, the bilinear form b associated with q is symplectic. Hence there exists a basis e_{1}, …, e_{2n} of X such that b(e_{2i−1}, e_{2i}) = 1 for i = 1, …, n, while b(e_{i}, e_{j}) = 0 otherwise. Using the characterization 2.14.3 of an element A ∈ 𝔬(q), it is an easy exercise to check that the matrix φ(A) of A with respect to e_{1}, …, e_{2n} belongs to J and this yields the desired isomorphism of 2-Lie algebras.

We identify 𝔬(q) and L and show that tr_{J} = trs. Since trs is a Lie algebra homomorphism, it vanishes on [L, L]. By 1.11, J has nondegenerate bilinear trace form Tr so [L, L] = Ker(tr_{J}) by Cor. 2.6. Thus tr_{J} and trs have the same kernel, and it remains to prove that tr_{J} and trs take the same nonzero value on one element, say, on the matrix unit A = E_{11} ∈ H_{n}(𝒬, k). Now A(e_{1}) = e_{1}, A(e_{2}) = e_{2} while A(e_{i}) = 0 otherwise. Consider the element v = e_{1}e_{2} ∈ Cliff_{0}(q). We claim that A(x) = [v, x] for all x ∈ X (where the products are to be taken in Cliff(q)). Indeed, b_{q}(e_{1}, e_{2}) = 1 implies e_{1}e_{2} + e_{2}e_{1} = 1 in Cliff(q), hence ve_{1}−e_{1}v = e_{1}e_{2}e_{1}−e_{1}e_{1}e_{2} = −e_{2}e_{1}^{2}+e_{1}−e_{1}^{2}e_{2} = −q(e_{1})e_{2}+e_{1}−q(e_{1})e_{2} = e_{1}, and similarly ve_{2} − e_{2}v = e_{2}, while ve_{i} − e_{i}v = 0 for i > 2 follows from e_{1}e_{i} + e_{i}e_{1} = e_{2}e_{i} + e_{i}e_{2} = 0, because b_{q}(e_{1}, e_{i}) = b_{q}(e_{2}, e_{i}) = 0. Furthermore, v + v^{∗} = e_{1}e_{2} + e_{2}e_{1} = 1, so trs(E_{11}) = 1 = tr_{J}(E_{11}), as required.

(b) This follows from Prop. 2.5 and Cor. 2.12(a), part (ii).

We leave it to the reader to prove in a similar way the following result.

2.16. Proposition. (a) The Lie algebra 𝔰p_{2n}(k) of the symplectic group over a field k of characteristic 2 is isomorphic as a 2-Lie algebra to L(H_{n}(𝒬, 𝒬_{0})).

(b) L(H_{n}(k)) is the Lie algebra of the automorphism group of the standard bilinear form h(x, y) = ∑_{i=1}^{n} x_{i}y_{i} on k^{n}.

3. Smoothness of the automorphism group

3.1. The structure group. In this section, we discuss the question of smoothness of the automorphism group scheme of a separable finite-dimensional Jordan algebra. This is closely related to the structure of the orbit of the unit element under the structure group. Recall that the structure group Str(J) of a Jordan algebra J (over a commutative ring R) is the set of g ∈ GL(J) for which there exists g^{♯} ∈ GL(J) such that U_{g(x)} = gU_{x}g^{♯} for all x ∈ J. Such a g^{♯} is uniquely determined by g; in fact, g^{♯} = g^{−1}U_{g(1)}, and the map ϑ: g ↦ g^{ϑ} := (g^{♯})^{−1} is an automorphism of period two of Str(J). Also, Str(J) is isomorphic to the automorphism group of the Jordan pair (J, J) associated to J under the map g ↦ (g, g^{ϑ}). The automorphism group Aut(J) is just the isotropy group of the unit element 1_{J} in Str(J).

We establish some notation and terminology for algebraic groups. Let k be a field. Following [5], we will always embed algebraic k-groups into the category of group functors on the category k-alg of (commutative associative unital) k-algebras. For a k-group functor G and R ∈ k-alg, we denote by G(R) the associated (abstract) group. The Lie algebra of G is denoted by Lie(G).

For a finite-dimensional Jordan algebra J over k we have the group functors

Str(J)(R) := Str(J ⊗ R), Aut(J)(R) := Aut(J ⊗ R) (R ∈ k-alg),

which are affine algebraic k-groups. By abuse of language, these will also be referred to simply as the structure group and the automorphism group. Their Lie algebras are then just Strl(J) and Der(J), respectively.

Example. Let J = Jor(k^{n}, q, 1) be the Jordan algebra of a nondegenerate quadratic form q with base point 1 = 1_{J} as in 1.4, and let GO(q) be the general orthogonal group of q, i.e., g ∈ GO(q)(R) if and only if g ∈ GL_{n}(R) and there exists λ ∈ R^{×} such that q(g(x)) = λq(x) for all x ∈ R^{n}. Note that λ = λ(g) = q(g(1)) is uniquely determined by g. The Lie algebra of GO(q) consists of all A ∈ End(k^{n}) for which there exists μ ∈ k such that b(x, A(x)) = μq(x), for all x ∈ k^{n}. It is easily seen that GO(q) ⊂ Str(J), with g^{♯} = λ(g)g^{−1}, and consequently, Lie(GO(q)) ⊂ Strl(J).

The bilinear trace Tr is invariant under the structure group and the structure Lie algebra in the following sense:

3.2. Lemma. Let J be a finite-dimensional Jordan algebra over a field k and put G := Str(J) and 𝔤 := Strl(J). Then

Tr(g(x), y) = Tr(x, g^{♯}(y)) for all g ∈ G(R), x, y ∈ J ⊗ R, R ∈ k-alg, (1)
Tr(A(x), y) = Tr(x, A^{♯}(y)) for all A ∈ 𝔤, x, y ∈ J.  (2)

Consequently, the defect is stable under G and 𝔤, i.e.,

g(Def(J) ⊗ R) ⊂ Def(J) ⊗ R  for all g ∈ G(R), R ∈ k-alg, (3)
A(Def(J)) ⊂ Def(J)  for all A ∈ 𝔤.  (4)

Proof. Formula (1) is a consequence of [13, Prop. 16.7] and the fact that Tr(x, y) = m_{1}(x, y) where m_{1} is the generic trace of the Jordan pair (J, J) associated with J, cf. 1.3. Formula (2) then follows by letting R = k(ε) (dual numbers) and g = Id + εA, and (3) and (4) are immediate from (1) and (2).

Recall that a Jordan algebra (or pair) over a field is said to be separable if any base field extension has trivial lower radical [16, 3.5]. The basic fact on Str(J) is

3.3. Theorem [14, Cor. 6.6]. The structure group Str(J) of a finite-dimen-sional separable Jordan algebra is a reductive, hence in particular smooth, algebraic k-group.

In contrast, the automorphism group of J is in general not smooth. However, this can only happen in characteristic 2.

3.4. Group actions. We recall some notions for actions of algebraic groups. Let k be a field with algebraic closure k̄, let G be a smooth algebraic k-group acting on a smooth algebraic k-scheme X on the left, let x ∈ X(k) be a krational point of X, and denote by H = Cent_{G}(x) the stabilizer of x in G. Also let π: G → X be the orbit map sending g ∈ G(R) to g · x_{R} for all R ∈ k-alg (where x_{R} ∈ X(R) is the image of x under the map X(k) → X(R) induced from k → R).

The orbit of x under G is the image sheaf Im(π) (in the flat topology) of π, cf. [5, III, §1, 2.3]. Denote by G ˜/H the sheaf (in the flat topology) associated to the functor R ↦ G(R)/H(R), cf. [5, III, §3, 1.4]. Then π induces a canonical isomorphism G ˜/H ≅ Im(π) by [5, III, §3, 1.6].

For an algebraic k-scheme Y and a k-rational point y ∈ Y(k), let T_{y}(Y) be the Zariski tangent space of Y at y. Finally, let e ∈ G(k) be the unit element of G(k), and let 𝔤 = Lie(G) = T_{e}(G) and 𝔥 = Lie(H) be the respective Lie algebras. Then we have 𝔥 = Ker(d_{e}π), where d_{e}π: 𝔤 → T_{x}(X) denotes the differential of π at the unit element of G(k).

3.5. Lemma. In the situation of 3.4, assume furthermore that G acts transitively on X in the sense that π: G(k̄) → X(k̄) is surjective.

(a) π induces an isomorphism G ˜/H ⟶^{≅} X, and

dim G = dim H + dim X.  (1)

(b) The following conditions are equivalent:

(i) H is smooth, (ii) π: G → X is smooth, (iii) d_{e}π: 𝔤 → T_{x}(X) is surjective, (iv) dim H = dim 𝔥.

Proof. (a) The first statement follows from [5, III, §3, Prop. 2.1], and the dimension formula follows from [5, III, §3, 5.5(a)].

(b) (i) ⟹ (ii): By [5, III, §3, Cor. 2.6] the canonical morphism G → G ˜/H is smooth, and hence so is the composite π: G → G ˜/H ⟶^{≅} X (note that the subgroups H and H′ of loc. cit. are {e} and H in our situation).

(ii) ⟹ (iii): This follows from [5, I, §4, Cor. 4.14, Remark 4.15].

(iii) ⟹ (iv): We have dim 𝔥 = dim Ker(d_{e}π) = dim 𝔤 − dim Im(d_{e}π) = dim G − dim X = dim H (by (1)).

(iv) ⟹ (i): See [5, II, §5, Th. 2.1(vi)].

3.6. Let J be a separable finite-dimensional Jordan algebra over a field k. We denote by J the affine scheme defined by the vector space J, and by J^{×} the open dense subscheme of invertible elements of J; thus

J(R) = J ⊗ R and J^{×}(R) = (J ⊗ R)^{×} for all R ∈ k-alg.  (1)

For the rest of this section, we will always let

G := Str(J),

which is smooth by Th. 3.3, act on a suitably chosen G-stable subscheme X of J containing the unit element x = 1 of J, so that

H := Cent_{G}(1) = Aut(J).

We denote the Lie algebras of G and H by

𝔤 = Lie(G) = Strl(J) and 𝔥 = Lie(H) = Der(J).

The orbit map π is just evaluation of an element g ∈ G(R) at 1, and likewise, d_{e}π: 𝔤 → T_{x}(X) is simply evaluation A ↦ A(1) of an element A in the structure Lie algebra at the unit element of J.

Clearly, J and J^{×} are smooth affine schemes; in fact, a defining function for J^{×} is the generic norm of J. The defect of J gives rise to functors J_{def} and J_{def}^{×} in analogy to (1) by

J_{def}(R) = Def(J) ⊗ R for all R ∈ k-alg, and J_{def}^{×} = J_{def} ∩ J^{×}. (2)

Again, J_{def} and J_{def}^{×} are smooth and affine, and J_{def}^{×} is open in J_{def}. We will also need the subfunctor of non-defective elements J_{nd} of J, defined by

J_{nd}(R) = {x ∈ J(R) : there exists y ∈ J(R) such that Tr(x, y) ∈ R^{×}}, (3)

for all R ∈ k-alg. This is an open (hence smooth) but in general not affine subscheme of J. To see this, choose a vector space basis v_{1}, …, v_{n} of J and define functions f_{i} on J by f_{i}(x) := Tr(x, v_{i}). Then J_{nd} is the open subscheme of J defined by the f_{i} in the sense of [5, I, §1, 3.7], i.e., x ∈ J_{nd}(R) if and only if R is generated as an ideal by f_{1}(x), …, f_{n}(x). We finally put

J_{nd}^{×} := J_{nd} ∩ J^{×}.  (4)

From the fact that elements of the structure group preserve invertibility and the bilinear trace form Tr by Lemma 3.2, it follows that G acts on each of the schemes J^{×}, J_{def}^{×} and J_{nd}^{×}.

3.7. Proposition. If k has characteristic ≠ 2 then the orbit of 1 under G is J^{×}, and H is smooth.

Proof. Let X = J^{×}. By Prop. 1.9, G(k̄) acts transitively on X(k̄) = (J ⊗k̄)^{×} so the first assertion follows from Lemma 3.5(a). Also, T_{1}(X) is canonically identified with the vector space J because X is open in J. For any given a ∈ J, we have A = \frac{1}{2}V_{a} ∈ 𝔤, and d_{e}π(A) = A(1) = \frac{1}{2}(a ∘ 1) = a. Hence H is smooth by Lemma 3.5(b).

3.8. To decide smoothness of the automorphism group in characteristic 2 requires a more detailed discussion. We remark that Springer [23], in his framework of J-structures, has also studied this problem. However, his definition of J-structure is rather restrictive and in characteristic 2 rules out, a priori, the case where the orbit of 1 under the structure group is not open. For an absolutely simple Jordan algebra, this happens precisely when the algebra is traceless. Our approach includes these cases and it is simpler than Springer’s since it uses the result 3.3, not available to him. — Until further notice, we assume that

J is simple and k is an algebraically closed field of characteristic 2.

We will use repeatedly the fact that V_{J,J} ⊂ 𝔤 (cf. 2.1) and hence

J ∘ J = V_{J,J}(1) ⊂ Im(d_{e}π).  (1)

3.9. Lemma. If J has nondegenerate bilinear trace form Tr, i.e., Def(J) = 0, then the orbit of 1 under G is J^{×}, and H is smooth.

Proof. We apply Lemma 3.5 in case X = J^{×}. By Prop. 1.9, G(k) acts transitively on X(k), so the first statement follows from Lemma 3.5(a). Also, we have T_{1}(X) = J and by Cor. 2.6, [L, L] = J ∘ J = Ker(tr) has codimension one in J. Hence by 3.8.1, it suffices, for d_{e}π to be surjective, to find an element of trace one in the image of d_{e}π. We remark also that Id_{J} always belongs to Strl(J). Now we distinguish the following cases.

Case 1: r := rk(J) is odd. Here tr(1) = r · 1_{k} = 1_{k} because k has characteristic 2. Hence 1 = Id(1) is an element of trace one in the image of d_{e}π.

When the rank is even we use the classification.

Case 2: r = 2. Then J = Jor(k^{2n}, q, ε_{1} + ε_{2}) is the Jordan algebra of an even-dimensional quadratic form with base point 1 = ε_{1} + ε_{2}, n ≥ 2. Define A ∈ End(J) by A(ε_{i}) = ε_{i} if i is odd and A(ε_{i}) = 0 if i is even. One checks easily that A belongs to the Lie algebra of the general orthogonal group of q, in fact, b(x, A(x)) = q(x), and hence A belongs to the structure Lie algebra by the example in 3.1. Now A(ε_{1}) = ε_{1}, A(ε_{2}) = 0, and therefore A(1) = ε_{1} which has trace one.

Case 3: r even and ≥ 4. Then J = H_{r}(𝒞, k) is the Jordan algebra of hermitian matrices with scalar diagonal coefficients over a composition algebra 𝒞. The assumption r ≥ 4 eliminates the case where 𝒞 is an octonion algebra, and the assumption Def(J) = 0 eliminates the case 𝒞 = k. Thus either 𝒞 = k ⊕ k or 𝒞 = Mat_{2}(k), the split quaternions. In both cases, 1_{𝒞} = ε_{1} + ε_{2} is the sum of two primitive orthogonal idempotents which are interchanged by the involution of 𝒞. One checks easily that every a ∈ Mat_{r}(𝒞) defines an element A_{a} in the structure Lie algebra by A_{a}(x) = ax + xa^{∗}, so all A_{a}(1) = a + a^{∗} ∈ Im(d_{e}π). Now let in particular a = ε_{1}e_{11} where the e_{ij} are the usual matrix units. Then a + a^{∗} = e_{11} is the desired element of trace one in the image of d_{e}π.

3.10. Lemma. If J is traceless, i.e., 1 ∈ Def(J), then the orbit of 1 under G is J_{def}^{×} and H is smooth.

Proof. Let X = J_{def}^{×} as in 3.6.2. Then 1 ∈ X(k) and the tangent space of X at 1 is just the vector space Def(J). By Prop. 1.9, G(k) acts transitively on X(k), so the orbit of 1 under G is X by Lemma 3.5(a). By Prop. 2.7(a), J ∘ J = [L, L] = Def(J), so H is smooth by 3.8.1 and Lemma 3.5(b).

3.11. Lemma. Let J have tr ≠ 0 and Def(J) ≠ 0, and let r := rk(J). Then the orbit of 1 under G is J_{nd}^{×} and dim 𝔥 − dim H = r − 1 ≥ 1; in particular, H is not smooth.

Proof. Let X = J_{nd}^{×} as in 3.6.4. Then 1 ∈ X(k), and by Prop. 1.9, G(k) acts transitively on X(k) so the orbit of 1 is X. Also, T_{1}(X) = T_{1}(J) = J because X is open in J. We determine the image of the evaluation map d_{e}π: 𝔤 → J, using the classification. There are two cases:

Case 1: J = H_{r}(k), symmetric matrices over k, r ≥ 2, with unit element 1 = 1_{r} the r × r unit matrix, and Def(J) = Alt_{r}(k), the alternating r × rmatrices. We claim that

Im(d_{e}π) = k · 1 ⊕ Alt_{r}(k).  (1)

For the inclusion from right to left, let e_{ij} be the usual matrix units. Then c_{i} = e_{ii} (i = 1, …, r) is a frame of division idempotents of J and the Peirce space J_{ij} = k · (e_{ij} + e_{ji}) (for i < j) is 1-dimensional. Hence by the first case of Prop. 2.7(b), Alt_{r}(k) = J ∘ J is contained in Im(d_{e}π), and so is 1. To prove the inclusion from left to right, let A ∈ Strl(J). Since now U_{x}y = xyx (matrix product), formula 2.1.1 says

xyA(x) + A(x)yx = A(xyx) − xA(y)x + x(A(1)y + yA(1))x,  (2)

for all x, y ∈ J. Write A(1) = ∑_{l,m} α_{lm}e_{lm} and A(x) = ∑_{l,m} β_{lm}e_{lm} as linear combinations of the matrix units. Now put x = y = e_{ij} + e_{ji} (where i ≠ j) in (2), and multiply the resulting equation with e_{ii} on the left and with e_{ji} on the right. An elementary matrix calculation then yields the relation

β_{ij} + β_{ij} = β_{ij} − β_{ji} + α_{jj} + α_{ii}.

Since A(x) ∈ H_{r}(k) is symmetric we have β_{ij} = β_{ji}. Hence 2 = 0 in k implies α_{ii} = α_{jj}, so all diagonal coefficients of A(1) are equal, proving the inclusion from left to right in (1).

Case 2: J = Jor(k^{2m+1}, q, ε_{1} + ε_{2}) with m ≥ 1. In fact, we can assume m ≥ 2 because Jor(k^{3}, q, ε_{1} + ε_{2}) ≅ H_{2}(k). Then ε_{1} and ε_{2} form a frame of division idempotents of J whose Peirce space J_{12} has dimension 2m − 1 ≥ 3. We claim that

Im(d_{e}π) = Ker(tr).  (3)

Indeed, the inclusion from right to left holds because Ker(tr) = J ∘ J by the second case of Prop. 2.7(b), and 3.8.1.

Let us prove the inclusion from left to right. From 1.4.1 we obtain ε̄_{0} = ε_{0} and U_{x}ε_{0} = q(x)ε_{0} whence U_{x,z}ε_{0} = b(x, z)ε_{0}, for all x, z ∈ J. In particular, U_{ε_{0}}ε_{0} = ε_{0} (because q(ε_{0}) = 1) and U_{ε_{0},z}ε_{0} = 0 for all z ∈ J. Now let A ∈ Strl(J). By 3.2.4, A(ε_{0}) ∈ k · ε_{0}, which implies [A, U_{ε_{0}}]ε_{0} = 0. Hence 2.1.1 yields

\begin{aligned}0 = U_{ε_{0},A(ε_{0})}ε_{0} & = [A, U_{ε_{0}}]ε_{0} + U_{ε_{0}}U_{1,A(1)}ε_{0} \\ & = 0 + U_{ε_{0}}b(1, A(1))ε_{0} = b(1, A(1))ε_{0} = tr(A(1))ε_{0}, \end{aligned}

which proves the inclusion from left to right in (3).

Observe that the results of the previous three lemmas also hold for an arbitrary base field k, provided J is absolutely simple. This follows by passing to the algebraic closure of k because the structure group commutes with base change, and smoothness of G and G ⊗ k̄ are equivalent [5, I, §4, 4.1]. We collect our results in the following theorem.

3.12. Theorem. Let J be an absolutely simple finite-dimensional Jordan algebra of rank r over a field k of arbitrary characteristic. Then Aut(J) is not smooth if and only if k has characteristic 2 and J has both nonzero trace and nonzero defect. In this case, dim Der(J) − dim Aut(J) = r − 1.

4. The exceptional case

4.1. Definitions and notations. Let R be a commutative ring and let either 𝒞 = R or let 𝒞 be a composition algebra of constant rank r ≥ 2 over R in the sense of [21], with norm form q and associated bilinear form b_{q}. Thus 𝒞 is finitely generated and projective of rank r = 1, 2, 4, 8 as an R-module. If r = 1 and 2 is not a unit in R then b_{q} is singular, while in the other cases, it is nonsingular. To cover the case r = 1 as well, we will let B = b_{q} if r ≥ 2, and put B(a, b) = ab in case r = 1, i.e., 𝒞 = R. As usual, ā denotes the involution of 𝒞. For an endomorphism h ∈ End(𝒞) we define h̄ by h̄(a) = \overline{h(ā)}. We also introduce the trilinear form ⟨a, b, c⟩ = B(ab, c̄) on 𝒞. Then ⟨ , , ⟩ behaves as follows under permutation of its arguments:

⟨a_{σ(1)}, a_{σ(2)}, a_{σ(3)}⟩ = \{ \begin{matrix}⟨a_{1}, a_{2}, a_{3}⟩ & if σ is even \\ ⟨ā_{1}, ā_{2}, ā_{3}⟩ & if σ is odd \end{matrix} \} .  (1)

This follows from well-known formulas for b_{q} (and is of course trivial in case r = 1).

Consider the Jordan algebra J = H_{3}(𝒞, R) of hermitian matrices over 𝒞 with scalar diagonal coefficients which by [17] is generically algebraic of degree 3. Denoting the usual matrix units by e_{ij}, we put e_{i} = [ii] = e_{ii} and a[ij] = ā[ji] = ae_{ij} + āe_{ji} for i ≠ j and a ∈ 𝒞. There is an action of the symmetric group 𝔖_{3} on J by automorphisms given by

σ(e_{i}) = e_{σ(i)}, σ(a[ij]) = a[σ(i), σ(j)].

This is easily verified. Let

E := R · e_{1} ⊕ R · e_{2} ⊕ R · e_{3} ≅ R^{3}

be the subalgebra of diagonal matrices in J. We will denote the subgroup of all g ∈ Aut(J) fixing E pointwise by M = M(J). Then 𝔖_{3} acts on M on the right via g^{σ} = σ^{−1} ∘ g ∘ σ. Finally, we denote by P the set of all pairs (i, j) in {1, 2, 3}^{2} with i ≠ j. Clearly, 𝔖_{3} acts simply transitively on P by σ(i, j) = (σ(i), σ(j)).

4.2. Lemma. With the above notations, let g ∈ M. For every pair (i, j) ∈ P there exists a unique g_{ij} ∈ O(q), the orthogonal group of q, such that

g(a[ij]) = g_{ij}(a)[ij],  (1)

for all a ∈ 𝒞. The g_{ij} satisfy the relations

\overline{g_{ij}} = g_{ji},  (2)
g_{ij}(a)g_{jk}(b) = g_{ik}(ab),  (3)
⟨g_{ij}(a), g_{jk}(b), g_{ki}(c)⟩ = ⟨a, b, c⟩,  (4)

for all a, b, c ∈ 𝒞 and all pairwise different i, j, k. The map g ↦ (g_{ij}) is an isomorphism η from M to the subgroup S = S(𝒞) ⊂ O(q)^{P} of all tuples (g_{ij})_{(i,j)∈P} satisfying (2) and (3), or, equivalently, (2) and (4). This isomorphism is 𝔖_{3}-equivariant with respect to the natural right action of 𝔖_{3} on S, i.e., (g^{σ})_{ij} = g_{σ(i),σ(j)}.

Proof. An element g ∈ M preserves the Peirce spaces J_{ij} = {a[ij] : a ∈ 𝒞}, so there exist unique g_{ij} ∈ GL(𝒞) satisfying (1). Since g preserves squares and a[ij]^{2} = q(a)([ii] + [jj]), we have g_{ij} ∈ O(q), and (2) follows from the fact that a[ij] = ā[ji]. Finally, (3) is a consequence of the formula a[ij] ∘ b[jk] = ab[ik]. Conversely, given (g_{ij}) ∈ S, define g ∈ GL(J) by g(e_{i}) = e_{i} and (1). Then a simple computation shows that g preserves squares and traces in J. As J is generically algebraic of degree 3, we have g ∈ Aut(J) by [17, Th. 5.1]. The 𝔖_{3}-equivariance is clear from the definitions. The equivalence of (3) and (4) for (g_{ij}) ∈ O(q)^{P} satisfying (2) follows easily from the definition of ⟨ , , ⟩ and nondegeneracy of B.

Remark. In [6], A. Elduque introduces the notion of related triples (φ_{0}, φ_{1}, φ_{2}) ∈ O(q)^{3} by requiring φ_{1}(ab) = φ_{0}(a)φ_{2}(b) for all a, b ∈ 𝒞. Any element (g_{ij}) of S(𝒞) gives rise to related triples, e.g., (g_{12}, g_{13}, g_{23}) or (g_{31}, g_{32}, g_{12}). The approach via related triples seems less natural because it involves a particular choice of indices. Also, the action of the symmetric group becomes somewhat cumbersome to describe, and the trilinear form considered in [6, p. 52] does not satisfy the equivariance property 4.1.1.

4.3. Theorem. Let 𝒪 be an octonion algebra over a ring R with norm form q, Clifford algebra Cliff(q) and even part Cliff_{0}(q). Let Spin(q) ⊂ Cliff_{0}(q)^{×} be the spin group and χ: Spin(q) → O(q) its vector representation.

(a) The maps φ, ψ: 𝒪 → End(𝒪 ⊕ 𝒪), a ↦ ( \begin{matrix}0 & lā \\ l_{a} & 0 \end{matrix} ) and a ↦ ( \begin{matrix}0 & rā \\ r_{a} & 0 \end{matrix} ),

where l_{a} and r_{a} denotes left and right multiplication in 𝒪, induce isomorphisms

Φ, Ψ: Cliff(q) ⟶^{≅} End(𝒪 ⊕ 𝒪)

which restrict to isomorphisms

Φ_{0}, Ψ_{0}: Cliff_{0}(q) ⟶^{≅} ( \begin{matrix}End(𝒪) & 0 \\ 0 & End(𝒪) \end{matrix} ) .

(b) For (i, j) ∈ P, define homomorphisms ϱ_{ij}: Spin(q) → GL(𝒪) by

ϱ_{12}(u) = χ(u),  Φ_{0}(u) = ( \begin{matrix}ϱ_{23}(u) & 0 \\ 0 & ϱ_{13}(u) \end{matrix} ) ,

ϱ_{21}(u) = \overline{ϱ_{12}(u)},  Ψ_{0}(u) = ( \begin{matrix}ϱ_{31}(u) & 0 \\ 0 & ϱ_{32}(u) \end{matrix} ) ,

for all u ∈ Spin(q). Then ϱ = (ϱ_{ij})_{(i,j)∈P}: Spin(q) → S(𝒪), u ↦ (ϱ_{ij}(u))_{(i,j)∈P}, is an isomorphism of groups.

(c) The centre of Spin(q) contains μ_{2}(R) × μ_{2}(R), where μ_{2}(R) = {λ ∈ R : λ^{2} = 1}. More precisely, we have: For λ_{1}, λ_{2} ∈ μ_{2}(R) there exists a unique element u in the centre of Spin(q) such that ϱ_{23}(u) = ϱ_{32}(u) = λ_{1}Id_{𝒪}, ϱ_{13}(u) = ϱ_{31}(u) = λ_{2}Id_{𝒪}, and ϱ_{12}(u) = ϱ_{21}(u) = λ_{1}λ_{2}Id_{𝒪}.

Proof. (a) and (b) are proved in [6, Th. 1.1] for the case of a base field, but the proof applies with slight modifications also in case of a base ring. We therefore omit the details. For part (c), define g_{ij} ∈ O(q) by g_{23} = g_{32} = λ_{1}Id, g_{13} = g_{31} = λ_{2}Id and g_{12} = g_{21} = λ_{1}λ_{2}Id. Then it is easy to verify that g = (g_{ij}) is a central element of S(𝒪), so the assertion follows from (b).

4.4. Corollary. Let J = H_{3}(𝒪, k) be a reduced Albert algebra over a field k. Define group functors M ⊂ H = Aut(J) and S ⊂ O(q)^{P} by M(R) = M(J ⊗R) and S(R) = S(𝒞⊗R) for all R ∈ k-alg, and let Spin(q) be the spin group as in 2.14.1. Then M, S and Spin(q) are smooth (in fact, semisimple) algebraic group schemes of dimension 28 over k, and the maps η of Lemma 4.2 and ϱ of Th. 4.3 induce isomorphisms

η: M ⟶^{≅} S and ϱ: Spin(q) ⟶^{≅} S.  (1)

Hence the Lie algebra 𝔪 of M is isomorphic to the subalgebra 𝔰 = Lie(S) of 𝔬(q)^{P} consisting of all (A_{ij}) satisfying \overline{A_{ij}} = A_{ji} and A_{il}(ab) = A_{ij}(a)b + aA_{jl}(b), for all a, b ∈ 𝒪 and all i, j, l ≠, and also to the Lie algebra 𝔰𝔭𝔦𝔫(q) of Spin(q), and all three are of dimension 28. If k has characteristic p > 0 then these isomorphisms are isomorphisms of restricted Lie algebras.

Proof. Since the maps η and ϱ of 4.2 and 4.3 are compatible with base ring extension, we have the asserted isomorphisms of group functors. The rest follows from well-known facts about the spin group of a nondegenerate quadratic form.

4.5. The scheme of frames. With J as above, let F ⊂ J^{3} be the functor of frames of J, i.e., for every R ∈ k-alg, F(R) is the set of complete systems c⃗ = (c_{1}, c_{2}, c_{3}) of orthogonal idempotents of J ⊗ R whose Peirce 2-spaces are R-modules of rank 1. We claim that F is an affine algebraic k-scheme. Indeed, it is easily seen that the conditions c_{i}^{2} = c_{i}, U_{c_{i}}c_{j} = δ_{ij}c_{i} and c_{2} + c_{2} + c_{3} = 1, which express the fact that c⃗ is a complete system of orthogonal idempotents, define a closed subscheme F′ of J^{3}. Now F ⊂ F′ is singled out by the conditions that the c_{i} be in addition rank 1 idempotents. Since the Peirce 2-spaces of the c_{i} are direct summands of J ⊗ R and hence finitely generated and projective R-modules, their rank functions are locally constant on Spec(R). From this, it easily seen that F is an open and closed subscheme of F′; in particular, it is affine algebraic.

4.6. Proposition. (a) F is smooth of dimension 24.

(b) H acts transitively on F and the stabilizer of e⃗ = (e_{1}, e_{2}, e_{3}) ∈ F(k) is M. Hence the dimension of H and of 𝔥 is 52.

Proof. (a) Since F is open in F′, it suffices to show that the latter is smooth. By [5, I, §4, Cor. 4.6], we must show that the map F′(R) → F′(R/I) induced from the canonical map R → R/I is surjective, for every ideal I of square zero of R. The kernel of the map J ⊗ R → J ⊗ (R/I) is a nilideal, so the assertion follows from the well-known lifting of finite orthogonal systems of idempotents through nil ideals. Now let c⃗ = (c_{1}, c_{2}, c_{3}) ∈ F(k). Then v⃗ = (v_{1}, v_{2}, v_{3}) ∈ Tc⃗(F) if and only if c⃗ + εv⃗ ∈ F(k(ε)), where k(ε) denotes the dual numbers. It is an easy exercise to show that this is equivalent to the conditions v_{i} = x_{ij} + x_{il} where {i, j, l} = {1, 2, 3} and x_{ij} = −x_{ji} ∈ J_{ij}, the Peirce spaces of J with respect to c⃗. Hence Tc⃗(F) as a vector space is isomorphic to J_{12} ⊕ J_{23} ⊕ J_{13}, of dimension 24.

(b) Obviously, H acts on F via g·c⃗ = (g(c_{1}), g(c_{2}), g(c_{3})), and the stabilizer of e⃗ is just M. By conjugacy of frames (cf. the remark after Lemma 1.5), H(k̄) acts transitively on F(k̄), and H is smooth by Th. 3.12. Thus we are in the situation of Lemma 3.5 and conclude that dim 𝔥 = dim H = dim M+dim F = 28 + 24 = 52.

4.7. From now on, we will always assume that

k is a field of characteristic 2 and J = H_{3}(𝒪, k)

is a reduced Albert algebra over k. As before, we let 𝔥 = Der(J) denote the Lie algebra of H = Aut(J). By Cor. 2.13, V_{J} is a simple 26-dimensional ideal of 𝔥. Our aim is to determine the structure of V_{J} as well as that of the quotient algebra 𝔥/V_{J}. Since tr(e_{i}) = 1, we have tr(1_{J}) = 3 = 1 ≠ 0 and therefore J = k · 1_{J} ⊕ J_{0} where J_{0} = Ker(tr) ≅ V_{J} as a 2-Lie algebra.

The following remark will be useful: Let L and L′ be p-Lie algebras over a ring R with pR = 0 and let f: L → L′ be an isomorphism of Lie algebras. If L (and hence L′) has trivial centre then f is an isomorphism of restricted Lie algebras. Indeed, this follows easily from the formula ad(x^{[p]}) = (ad x)^{p} and the fact that the adjoint representations of L and L′ are faithful.

4.8. Lemma. We have 𝔥 = V_{J} + 𝔪 and V_{J} ∩ 𝔪 = V_{E} is a 2-dimensional central ideal of 𝔪.

Proof. Let D ∈ 𝔥 be a derivation and e an idempotent of J. Then e = e^{2} implies

D(e) = e ∘ D(e).  (1)

Now put D′ := D + V_{D(e_{1})} and D′′ := D′ + V_{D′(e_{2})}. Then

D = V_{D(e_{1})} + V_{D′(e_{2})} + D′′

holds because 2 = 0 in k, and we claim that D′′ ∈ 𝔪. Indeed, we have first D′(e_{1}) = D(e_{1}) + e_{1} ∘ D(e_{1}) = 0 by (1). Moreover, e_{1} ∘ e_{2} = 0 implies, because D′ is a derivation, 0 = D′(e_{1}) ∘ e_{2} + e_{1} ∘ D′(e_{2}) = e_{1} ∘ D′(e_{2}). Hence, D′′(e_{1}) = D′(e_{1}) + D′(e_{2}) ∘ e_{1} = 0, and also D′′(e_{2}) = D′(e_{2}) + D′(e_{2}) ∘ e_{2} = 0, again by (1). This shows D′′ ∈ 𝔪. That V_{J} ∩𝔪 = V_{E} is an easy consequence of the Peirce decomposition of J with respect to the e_{i}. For D ∈ 𝔪 and x ∈ E, we have [D, V_{x}] = V_{D(x)} = 0, so V_{E} is central in 𝔪.

4.9. Theorem. Let 𝔷 be the centre of Cliff_{0}(q), which by Th. 4.3 is isomorphic to k · Id_{𝒪} ⊕ k · Id_{𝒪} under the isomorphism Φ_{0}.

(a) 𝔷 ⊂ 𝔰𝔭𝔦𝔫(q) is the centre of 𝔰𝔭𝔦𝔫(q), V_{E} is the centre of 𝔪, and there are isomorphisms of 2-Lie algebras

𝔥/V_{J} ⟶^{φ_{1}} 𝔪/V_{E} ⟶^{φ_{2}} 𝔰𝔭𝔦𝔫(q)/𝔷 ⟶^{φ_{3}} 𝔬′(q)/k · Id_{𝒪} ⟶^{φ_{4}} H_{4}(𝒬^{s}, k)_{0}/k · 1_{4}, (1)

where H_{4}(𝒬^{s}, k)_{0} denotes the subspace of trace 0 elements in the Jordan algebra of 4 × 4 hermitian matrices with scalar diagonal entries over the split quaternion algebra 𝒬^{s} = Mat_{2}(k). In particular, the quotient 𝔥/V_{J} is a simple Lie algebra, which is up to isomorphism independent of the Cayley algebra 𝒪.

(b) V_{J} is the unique proper nonzero ideal of 𝔥.

Proof. (a) The isomorphism φ_{1} is clear from Lemma 4.8. We next establish φ_{3}. By Th. 4.3(c), Spin(q) contains μ_{2} ×μ_{2} as a central subgroup, and the Lie algebra of this is 𝔷 because k has characteristic 2 and therefore Lie(μ_{2}) = k. Thus 𝔷 is a central subalgebra of 𝔰𝔭𝔦𝔫(q). Also, k·1 = Ker(χ̇) ⊂ 𝔷. Hence χ̇(𝔷) is a central ideal of dimension 1 in χ̇(𝔰𝔭𝔦𝔫(q)) = 𝔬′(q), cf. 2.14. By Prop. 2.15(b), 𝔬′′(q) := 𝔬′(q)/k · Id_{𝒪} is a simple Lie algebra. Hence can(χ̇(𝔷)) = 0 in 𝔬′′(q) where can: 𝔬′(q) → 𝔬′′(q) denotes the canonical map, so we have χ̇(𝔷) = k·Id_{𝒪}, and therefore 𝔰𝔭𝔦𝔫(q)/𝔷 ≅ 𝔬′′(q). This establishes φ_{3}, and also shows that 𝔷 equals the centre of 𝔰𝔭𝔦𝔫(q). By Lemma 4.8, V_{E} is a 2-dimensional central ideal of 𝔪. By Cor. 4.4, we have a Lie algebra isomorphism ψ = Lie(ϱ^{−1} ∘ η): 𝔪 → 𝔰𝔭𝔦𝔫(q), which of course preserves centres, so V_{E} equals the centre of 𝔪. Now ψ induces the isomorphism φ_{2}. Finally, φ_{4} follows from Prop. 2.15(a).

(b) Let 𝔞 be a proper nonzero ideal of 𝔥 and can: 𝔥 → 𝔥/V_{J} the canonical map. Then can(𝔞) ⊂ 𝔥/V_{J} is either zero or 𝔥/V_{J}, so either 𝔞 ⊂ V_{J} or 𝔞 + V_{J} = 𝔥. The first case yields 𝔞 = V_{J} because V_{J} is simple, so we must derive a contradiction from the second case. By simplicity of V_{J} we have 𝔞 ∩ V_{J} = 0, so 𝔥 = 𝔞 ⊕ V_{J} (direct sum of ideals) and therefore dim 𝔞 = 26. We also have 𝔪 ∩ 𝔞 ≠ 0, otherwise dim 𝔥 ≥ dim 𝔪 + dim 𝔞 = 28 + 26 which is impossible. Choose 0 ≠ D ∈ 𝔪 ∩ 𝔞. Because D annihilates the e_{i}, there exist i ≠ j and x ∈ J_{ij} such that D(x) ≠ 0, and D(x) ∈ J_{ij}, because the elements of 𝔪 stabilize the Peirce spaces. This implies e_{i} ∘ D(x) = D(x) ≠ 0. On the other hand, [D, V_{x}] = V_{D(x)} ∈ [𝔞, V_{J}] = 0, because 𝔥 = 𝔞 ⊕ V_{J} is a direct sum of ideals, and therefore also V_{D(x)}(e_{i}) = e_{i} ∘ D(x) = 0, contradiction.

Remarks. (i) The isomorphism (1) was also pointed out to us by A. Elduque with a different proof.

(ii) We remind the reader that the above theorem deals only with reduced Albert algebras. When J is a division algebra, it is clear, by passing to the algebraic closure, that 𝔥/V_{J} is still a simple Lie algebra, but it is not clear whether it is isomorphic to H_{4}(𝒬, k)_{0}/k ·1_{4}, independently of J. However, part (b) continues to hold in case of a division algebra, as is seen by passing to the algebraic closure of k.

4.10. Schafer’s isomorphism. Let 𝒪^{s} be a split Cayley algebra and 𝒬 a quaternion algebra over k. In [22], Schafer and Tomber prove that there is a Lie algebra isomorphism Σ: H_{3}(𝒪^{s}, k)_{0} → H_{4}(𝒬, k)_{0}/k · 1_{4}, where the subscript 0 indicates the spaces of trace zero elements. We now give a more convenient description of this isomorphism.

Let 𝒪^{s} = 𝒬 ⊕ 𝒬 be the Cayley-Dickson double of 𝒬, with multiplication, involution and norm given by

(u, v)·(z, w) = (uz+wv̄, ūw+zv), \overline{(u, v)} = (ū, −v), q(u, v) = uū−vv̄. (1)

There is a natural embedding of H_{3}(𝒬, k)_{0} into H_{4}(𝒬, k)_{0} by adding a row and column of zeros. We extend this to a linear map f: H_{3}(𝒪^{s}, k) → H_{4}(𝒬, k)_{0} by

f([ii]) := [ii] + [44], f((u, v)[ij]) := u[ij] + v[k4],  (2)

where {i, j, k} = {1, 2, 3}. Note that this is well defined: We have a[ij] = ā[ji] for a ∈ 𝒪^{s}, but because of (1) and 2 = 0 also f(a[ij]) = f(ā[ji]). Now it is easy to compute that f behaves as follows with respect to squaring and circle products, where a = (u, v) ∈ 𝒪^{s} and i, j, k ∈ {1, 2, 3}:

f([ii]^{2}) = f([ii])^{2},  (3)
f(a[ij]^{2}) = f(a[ij])^{2} + vv̄ · 1_{4},  (4)

f(a[ij] ∘ b[jk]) = f(a[ij]) ∘ f(b[jk]) ({i, j, k} = {1, 2, 3}),  (5)
f([ii] ∘ [jj]) = f([ii]) ∘ f([jj]),  (6)
f([ii] ∘ a[jk]) = f([ii]) ∘ f(a[jk]).  (7)

Let x = ∑_{i=1}^{3} ξ_{i}[ii]+∑_{1≤i<j≤3} a_{ij}[ij] ∈ H_{3}(𝒪^{s}, k), where a_{ij} = (u_{ij}, v_{ij}) ∈ 𝒬⊕𝒬. Then (3) – (7) imply

f(x^{2}) − f(x)^{2} = ( ∑_{1≤i<j≤3} v_{ij}\overline{v_{ij}}) · 1_{4}.

Now it is clear that

Σ: H_{3}(𝒪^{s}, k)_{0} ⟶^{f} H_{4}(𝒬, k)_{0} ⟶^{can} H_{4}(𝒬, k)_{0}/k · 1_{4}

is a vector space isomorphism preserving the squaring, and hence is an isomorphism of restricted Lie algebras.

4.11. Corollary. Let 𝒪 and 𝒪^{s} be, respectively, an octonion algebra and a split octonion algebra, and let J = H_{3}(𝒪, k) and J^{s} = H_{3}(𝒪^{s}, k) be the corresponding reduced and split Albert algebras over k. We identify the 2-Lie algebra J_{0} = Ker(tr) with V_{J} under the map x ↦ V_{x} as in 4.7, and similarly for J^{s}.

(a) There is an isomorphism of restricted Lie algebras

ψ := Σ^{−1} ∘ φ : 𝔥/V_{J} ⟶^{φ} H_{4}(𝒬, k)_{0}/k · 1_{4} ⟶^{Σ^{−1}} J_{0}^{s},  (1)

obtained by composing the isomorphism φ := φ_{4} ∘ φ_{3} ∘ φ_{2} ∘ φ_{1} of Th. 4.9(a) with the inverse of Schafer’s isomorphism Σ.

(b) If 𝒪 is an octonion division algebra then 𝔥/V_{J} is not isomorphic to V_{J} as a Lie algebra.

Proof. (a) This is evident from 4.10 and Th. 4.9(a).

(b) Assume that 𝔥/V_{J} ≅ V_{J}. Then J_{0} ≅ V_{J} ≅ 𝔥/V_{J} ≅ J_{0}^{s} (by (1)) as Lie algebras. Since J_{0} and J_{0}^{s} are, by Cor. 2.13, simple Lie algebras, they have trivial centres. Hence J ≅ J^{s} follows from [17, Cor. 5.3(b)]. Now the Albert-Jacobson Theorem [20] implies that 𝒪 ≅ 𝒪^{s} is split.

4.12. Remark. The results of Th. 4.9 and Cor. 4.11(a) can also be obtained by computational algebra techniques contained in the first author’s PHD thesis, see [1]. This computational approach allows a detailed description (involving root space decomposition relative to a Cartan subalgebra) of the ideal V_{J} and also of the quotient algebra Der(J)/V_{J} for a general octonion algebra. The split case is contained in [1]. The main routines of [1] where published in the more handy reference [2].

4.13. The homomorphism β: H → H^{s}. We keep the notations and assumptions of Corollary 4.11. The isomorphism ψ induces an isomorphism between Aut(𝔥/V_{J}) and Aut(J_{0}^{s}), where Aut refers to the algebraic group of Lie algebra automorphisms. In view of the remark made in 4.7 and our previous results, this is also the group of automorphisms in the restricted sense. By [17, Cor. 5.4] we have a further isomorphism

Aut(J_{0}^{s}) ≅ H^{s} := Aut(J^{s}),  (1)

which assigns to an automorphism g_{0} of J_{0}^{s} ⊗ R the R-linear extension to J^{s} ⊗R = R·1⊕(J_{0}^{s} ⊗R) fixing the unit element, for all R ∈ k-alg. Composing these isomorphisms, we obtain an isomorphism

ϑ: Aut(𝔥/V_{J}) ⟶^{≅} H^{s}.  (2)

The group H acts by 2-Lie algebra automorphisms on its own Lie algebra 𝔥 via the adjoint representation Ad, and V_{J} is stable under Ad H. Hence we have an induced homomorphism α: H → Aut(𝔥/V_{J}), and by composing with ϑ we obtain a homomorphism

β = ϑ ∘ α: H ⟶^{α} Aut(𝔥/V_{J}) ⟶^{ϑ} H^{s}.

We introduce the notations

I := Ker(β), β̇ := Lie(β) and 𝔦 := Lie(I) = Ker(β̇),

and recall that an algebraic group G over a field k is called infinitesimal if G(K) = {1} for every field K ∈ k-alg [5, II, §4, 7.1].

4.14. Theorem. (a) For D ∈ 𝔥 = Der(J) we have

β̇(D) = V_{ψ(can(D))} ∈ 𝔥^{s},  (1)

where can: 𝔥 → 𝔥/V_{J} is the canonical map, ψ is as in 4.11.1, and 𝔥^{s} = Lie(H^{s}) = Der(J^{s}). Hence Ker(β̇) = V_{J} and Im(β̇) = V_{J^{s}}.

(b) I is an infinitesimal group with Lie algebra 𝔦 = V_{J}.

(c) β is faithfully flat, so the sequence

1 ⟶ I ⟶ H ⟶^{β} H^{s} ⟶ 1  (2)

is exact in the flat topology.

Proof. (a) It is a standard fact that Lie(Ad) = ad. Hence for D, D′ ∈ 𝔥 we have, putting α̇ = Lie(α),

α̇(D)(D′ + V_{J}) = [D, D′] + V_{J} = [D + V_{J}, D′ + V_{J}],

i.e., α̇(D) = ad_{𝔥/V_{J}} (can(D)). Since ψ is an isomorphism of Lie algebras, this implies ψ ∘ α̇(D) ∘ ψ^{−1} = ad_{J_{0}^{s}}(ψ(can(D)) ∈ Der(J_{0}^{s}). From the description of the isomorphism 4.13.1 it follows at once that the corresponding isomorphism Der(J_{0}^{s}) ≅ Der(J^{s}) on the Lie algebra level is just the k-linear extension Δ̃ of Δ ∈ Der(J_{0}^{s}) to J^{s} satisfying Δ̃(1_{J^{s}}) = 0. Now ad_{J_{0}^{s}}(x) · y = x ∘ y = V_{x}(y) for all x, y ∈ J_{0}^{s} and V_{x}(1) = 2x = 0, so the extension of ad_{J_{0}^{s}}(x) to J^{s} is V_{x}. For x = ψ(can(D)) we obtain formula (1). The statements about kernel and image of β̇ follow from the isomorphism ψ of 4.11.1 and the fact that J^{s} = k · 1 ⊕ J_{0}^{s} with J_{0}^{s} ≅ V_{J^{s}} under x ↦ V_{x}, cf. 4.7.

(b) For I to be infinitesimal, it suffices that I(K) = {1} for an algebraically closed field K. It is known [23, 14.20–14.25] that H is a group of type F_{4}, in particular, it is an almost simple connected algebraic k-group with trivial centre (in the sense of group schemes). Hence by [24], the group H(K) of Krational points is a simple abstract group. It follows that the homomorphism β_{K}: H(K) → H^{s}(K) is either constant or injective. The first alternative would imply, since H is smooth by Th. 3.12, that Lie(β) = 0 which is not the case. Hence β_{K} is injective and therefore I(K) = Ker(β_{K}) = {1}.

(c) Let H′ := H ˜/I be the quotient sheaf and π: H → H′ the canonical homomorphism. By [5, III, §3, 5.6, 2.7, 2.6], H′ is a smooth affine group scheme and π is faithfully flat. Moreover, by [5, III, §3, 1.6], β factors as β = ι∘π where ι: H′ → H^{s} is a monomorphism. By Lemma 4.8 and smoothness of H and H^{s}, both groups have dimension 52, and dim H′ = dim H − dim I (by [5, III, §3, 5.5(a)]) = dim H, because I as an infinitesimal group has dimension zero. Now Lie(ι): Lie(H′) → Lie(H^{s}) is injective and H′ and H^{s} are smooth of the same dimension, so Lie(ι) is bijective. It follows that ι is an open embedding [5, II, §5, 5.5(b)]. But H^{s} is connected, so ι is an isomorphism. This completes the proof.

Remark. By [4, Lemma 3.7, Cor. 3.11], β is the special isogeny between an isotropic group of type F_{4} and the split group of type F_{4}.

References

[1] P. Alberca, Sobre los teoremas de Cartan-Jacobson y Chevalley-Schafer, Ph. D. Thesis, Publications of the University of Málaga, 2000.

[2] P. Alberca and C. Martín, Computational approach to the simplicity of f_{4}(O_{s}, −) in the characteristic two case, J. Comput. Appl. Math. 158 (2003), 1–10.

[3] H. Bass, Clifford algebras and spinor norms over a commutative ring, Amer. J. Math. 96 (1974), 156–206.

[4] A. Borel and J. Tits, Homomorphismes “abstraits” de groupes algébriques simples, Ann. of Math. 97 (1973), 499–571.

[5] M. Demazure and P. Gabriel, Groupes algébriques, vol. 1, Masson, Paris, 1970.

[6] A. Elduque, On triality and automorphisms and derivations of composition algebras, Linear Algebra Appl. 314 (2000), 49–74.

[7] N. Jacobson, Lectures on quadratic Jordan algebras, Tata Institute of Fundamental Research, Bombay, 1969, Tata Institute of Fundamental Research Lectures on Mathematics, No. 45.

[8] , Structure and Representations of Jordan Algebras, Amer. Math. Soc. Colloq. Publ., vol. 39, Amer. Math. Soc., 1969.

[9] , Structure theory of Jordan algebras, University of Arkansas Lecture Notes in Mathematics, vol. 5, University of Arkansas, Fayetteville, Ark., 1981.

[10] N. Jacobson and J. Katz, Generically algebraic quadratic Jordan algebras, Scr. Math. 29 (1973), 215–227.

[11] M. A. Knus, Quadratic and Hermitian forms over rings, Grundlehren der Mathematischen Wissenschaften, vol. 294, Springer-Verlag, Berlin, 1991.

[12] M.-A. Knus and M. Ojanguren, Théorie de la descente et algèbres d’Azumaya, Lecture Notes in Math., vol. 389, Springer-Verlag, 1974.

[13] O. Loos, Jordan pairs, Lecture Notes in Math., vol. 460, Springer-Verlag, 1975.

[14] , On algebraic groups defined by Jordan pairs, Nagoya Math. J. 74 (1979), 23–66.

[15] , Diagonalization in Jordan pairs, J. Algebra 143 (1991), 252–268.

[16] , Trace and norm for reduced elements of Jordan pairs, Comm. Algebra 25 (1997), 3011–3042.

[17] , Generically algebraic Jordan algebras over commutative rings, Preprint, 2004.

[18] K. McCrimmon, The generic norm of an isotope of a Jordan algebra, Scripta Math. 29 (1973), 229–236.

[19] K. McCrimmon and E. Zel’manov, The structure of strongly prime quadratic Jordan algebras, Adv. in Math. 69 (1988), 133–221.

[20] H. P. Petersson, The isotopy problem for Jordan matrix algebras, Trans. Amer. Math. Soc. 244 (1978), 185–197.

[21] , Composition algebras over algebraic curves of genus zero, Trans. Amer. Math. Soc. 337 (1993), 473–491.

[22] R. D. Schafer and M. L. Tomber, On a simple Lie algebra of characteristic 2, Mem. Amer. Math. Soc. 14 (1955), 11–14.

[23] T. A. Springer, Jordan algebras and algebraic groups, Springer, New York-Heidelberg, 1973.

[24] J. Tits, Algebraic and abstract simple groups, Ann. of Math. 80 (1964), 313–329.