Lectures on Quadratic Jordan Algebras

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Lectures On Quadratic Jordan Algebras

By N. Jacobson

Tata Institute of Fundamental Research, Bombay 1969 Lectures on Quadratic Jordan Algebras

By N. Jacobson

No part of this book may be reproduced in any from by print, microﬁlm or any other means without written permission from the Tata Institute of Fundamental Research, Colaba, Bombay 5

Tata Institure of Fundamental Reasearch, Bombay 1969 Preface

Until recently the structure theory of Jordan algebras dealt exclusively with finite dimensional algebras over fields of characteristic , 2. In 1966 the present author succeeded in developing a structure theory for Jordan algebras of characteristic , 2 which was an analogue of the Wedderburn-Artin structure theory of semi-simple associate rings with minimum condition on left or right ideals. In this the role of the left or right ideals of the associative thory was played by quadratic ideals (new called inner ideals) which were defined as subspaces Z of the Jor- = 2 dan algebra J such that J Ub ⊆ Z where Ub 2Rb − mRb2 , Ra the multiplication by b in J . The operator Ub (P(b) in the notation of Braun and Kocher), which in an associative algebra is x → bxb, was introduced into abstract Jordan algebras by the author in 1955 and it has playedan increasingly important role in the theory and lits applications. It has been fairly clear for some time that an extension of the structure theory which was to encompass the characteristic two case would have to be “quadratic” in character, that is, would have to be based on the 1 + composition yUx rather than the usual x, y (which is 2 (xy yx) in associative algebras). The first indication of this appeared already in 1947 in a paper of Kaplansky’s which extended a result of Ancocheas’s on Jor- dan homomorphisms (then called semi-homomorphisms) of associative algebras to the characteristic two case by redefining Jordan homomorphisms using the product xyx in place of xy + yx. a completely satisfactory extension of the author’s structure theory which include characteristic two or more precisely algebras over an arbitrary commutative ring has been given by McCrimmon in [5] and [6],

iii iv 0. Preface

McCrimmon’s theory begins with a simple and beautiful axiomatiza- tion of the composition YUx. In addition to the quadratic character of the mapping xøUx of into its algebra of endomorphisms and the existence of unit 1 such that U1 = 1 one has to assume only the so-called = “fundamental formula” UxUyUx UyUx , one additional indentity, and the linearizations of these. Instead of assuming the linearizations it is equivalent and neater to assume that the two identities carry over on ex- ffi ffi Φ 1 tension of the coe cient ring. If the coe cient ring contains 2 then the notion of a quadratic Jordan algebra is equivalent to the classical notion of a (linear) Jordan algebra there is a canonical way of passing from the operator U to the usual multiplication R and back. Based on these fundations one can carry over the fundamental notions (inverses, isotopy, powers) of the linear theory to the quadratic case and extend the Artin like structure theory to quadratic Jordan algebras. In particular, one obtains for the first time a satisfactory Jordan structure theory for finite dimensional algebras over a field of characteristic two. In these lectures we shall detailded and self-contained exposition of McCrimmon’s structure theiry including his recently developed theory of radicals and absolute zero divisors which constitute an important addition even to the classical linear theory. In our treatment we restrict attention to algebras with unit. This effects a sybstantial simplication. However, it should be noted that McCrimmon has also given an axiom- atization for quadratic Jordan algebras withour unit and has developted the structure theory also for these. Perhaps the reader should be warned at the outsetthat he may find two (hopefully no more) parts of the exposition somewhat heavynamely, the derivation of the long list of identities in §1.3 and the proof of Osborn’s thorem on algebras of capacity two. The first of these could have been avoided by proving a general theorem in identities due to Macdonald. However, time did not permit this. The simplification of the proof of Osborn’s theorem remains an open problem. We shall see at the end of our exposition that this difficulty evaporates in the important special case of finite dimensional quadratic Jordan algebras over an algebraically closed field.

Nathan Jacobson Contents

Preface iii

0 Artinian Semi-simple Rings with Involution 1 1 Determination of the semi-simple artinian rings...... 1

1 Basic Concepts 11 1 Special Jordan and quadratic Jordan algebras ...... 11 2 Definition of Jordan and quadratic Jordan algebras . . . 13 3 Basic identities ...... 21 Φ 1 4 Category isomorphism for ∋ 2 . Characteristic two case. 29 5 Inner and outer ideals. Difference algebras...... 30 6 Special universal envelopes ...... 33 7 Quadratic Jordan algebras of quadratic forms...... 34 8 The exceptional quadratic Jordan algebra ...... 38 9 Quadratic Jordan algebras defined by certian cubic forms. 45 10 Inverses ...... 51 11 Isotopes ...... 54

2 Pierce Decompositions. Standard Quadratic... 59 1 Idempotents. Pierce decompositions ...... 59 2 Standard quadratic Jordan matrix algebras ...... 66 3 Connectedness and strong connectedness of...... 71 4 Strong coordinatization theorem ...... 76

v vi Contents

3 Structure Theory 83 1 The radical of a quadratic Jordan algebra ...... 84 2 Properties of the radical ...... 86 3 Absolute zero divisors...... 88 4 Minimal inner ideals ...... 90 5 Axioms for the structure theory...... 93 6 Capacity...... 97 7 First structure theorem ...... 100 8 A theorem on alternative algebras with involution. . . . . 102 9 Simple quadratic Jordan algebras of capacity two . . . . 108 10 Second structure theorem...... 121 Chapter 0 Artinian Semi-simple Rings with Involution

In this chapter we shall determine the Artinian semi-simple rings with 1 involutions. The results are all well known. The ﬁnal formulation in terms of matrix rings (in §2) will be particularly useful in the sequel. The results which we shall state without proof can be found in any standard text on associative ring theory.

1 Determination of the semi-simple artinian rings with involution.

Throughout these notes associative rings and algebras will be assumed to be unital, that is to contain a unit 1 such that a1 = a = 1a for all a in the ring. We recall that such a ring is called right artinian if it satisﬁes one of the following equivalent conditions: Minimum Condition: Any non-vacuous set of right ideals of the ring contains a minimal element. Desending chain condition: There exist no properly descending in- ﬁnite chain of right ideals I1 ⊃ I2 ⊃ I3 ··· A right artinian ring is called semi-simple if it contains no non-zero nilpotent (two-sided) ideal. An ideal N is nilpotent if there exists a positive integer N such that every product of Nzi in N is 0. Equivalently, if

1 2 0. Artinian Semi-simple Rings with Involution

Z L is deﬁned to be the ideal generated by all be, b ∈ Z , c ∈ L and Z m for m = 1, 2,... is deﬁned by Z 1 = Z , Z k = Z k−1Z then NN = 0. 2 We recall the fundamental Wedderburn-Artin structure theorems on semi-simple (right) artinian rings.

I. If a is semi-simple artinian (, 0) then a = a1 ⊕ a2 ⊕···⊕ a3 where ai is an ideal which regarded as a ring is simple artinian. (A ring a is simple if a , 0 and 0 and a are the only ideals in a.) Conversely, if a has the indicated structure then it is semi-simple artinian.

II. A ring a is simple artinian if and only if a is isomorphic to a complete ring ∆n of n × n matrices over a division ring ∆. This is equivalent to isomorphism to the ring End∆V of linear transformations of an n dimensional (left) vector space V over a division ring ∆.

It is easily seen that the simple components ai in the ﬁrst structure theorem are uniquely determined. In the second structure theorem, n and the isomorphism class of ∆ are determined by a. This follows from the following basic isomorphism theorem.

III. Let Vi, i = 1, 2, be a vector space over a division ring ∆i and

let ρ be an isomorphism of End∆1 V1 onto End∆2 V2. Then there exists a semi-linear isomorphism S of V1 onto V2 with associated isomorphism s of ∆1 onto ∆2 such that

σ −1 A = S AS, A ∈ EndV1. (1)

We now consider semi-simple artinian rings with involution. First, we give the basic deﬁnitions, which we formulate more generally 3 for rings which need not be associative. As in the associative case, we assume the rings are unital. Also homorphisms are assumed to map 1 into 1 and subrings contain 1.

Deﬁnition 1. A ring with involution is a pair (a, J) where a in a ring (with 1) and J is an involution (= anti-automorphism such that J2 = 1) in a.A homomorphism σ of (a, J) into a second ring with involution (Z , K) in a homomorphism of a into Z (sending 1. Determination of the semi-simple artinian rings... 3

1 into 1) such that Jσ = σk.A subring of (a, J) is a subring Z (containing 1) of a which is stable under J. An ideal of (a, J) is an ideal of a which is J-stable. ( , ) is simple if a , 0 and a and 0 are the only ideals of (a, J).

Let (a, J) be simple and assume Z is an ideal , 0 in a. Then Z + Z J is an ideal in (a, J). Hence Z + Z J = a. Also Z ∩ Z J is an ideal in (a, J) so Z ∩ Z = 0. Thus a = Z ⊕ Z . If L is an ideal in Z then L + LJ is an ideal in (a, J). It follows that Z is simple. This shows that if (a, J) is simple then either a is simple or a = Z ⊕ Z J where Z is simple. An associative ring with involution (a, J) is artinian semi-simple if a is artinian semi-simple. It follows from the ﬁrst Wedderburn- Artin structure theorem that such a ring with involution is a direct sum of ideals which are artinian simple rings with involution and conversely. An artinian simple ring with involution is of one of the J following types:a = Z ⊕ Z where Z ∆n, ∆ a division ring or a ∆n (or End V where V is n dimensional vector space over a 4 division ring ∆). We now consider the latter in greater detail. Thus consider End V where V is an n-dimensional vector space over ∆. Assume End V has an involution J. Let V∗ be the right vector space of linear functions on V. We denote the elements of V∗ as x∗, y∗ etc. And write the value of x∗ at y by hy, x∗i. This gives a bilinear pairing of the left vector space V/∆ with the right vector space V∗/∆ in the sense that ∗ ∗ ∗ hy1 + y2, x i = hy1, z i + hy2, x i ∗ + ∗ = ∗ + ∗ hy1 x1 x2i hy, x1i hy, z2i (2) hαy, x∗i = αhy, x∗i, hy, x∗αi = hy,∗ iα, α ∈ ∆ Also the pairing is non-degenerate in the sense that if hy, x∗i = 0 for all x∗ ∈ V∗ then y = 0 and if hy, x∗i = 0 for all y ∈ V then x∗ = 0. Let ∆◦ be the opposite ring of ∆ : ∆◦ is the same additive group as ∆ and has the multiplication α ◦ β = αβ. Then if we put αx∗ = x∗α, V∗ becomes a(left) vector space over ∆◦ and the last ∗ ∗ equation in (2) becomes hy,αx i = hy1, x iα. 4 0. Artinian Semi-simple Rings with Involution

Let A ∈ End V. Then we have a uniquely determined linear trans- formation A∗ in End V∗ satisfying

∗ ∗ ∗ ∗ ∗ hyA, x i = hy1, x A i, y ∈ V, x ∈ V (3)

This is called the transpose of A. If we use the usual functional 5 notation x∗(y) for hy, x∗i then x∗A∗(y) = x∗ is the resultant of A followed by x∗(yA) or x∗A∗. It follows directly that

(A + B)∗ = A∗ + B∗, (AB)∗ = B∗A∗ (4)

and A → A∗ is bijective from End V to End V∗. Thus A → A∗ is an anti-isomorphism of End V onto End V∗. Now suppose End V has an involution J. Since A → AJ and A → A∗ are anti-isomorphisms, AJ → A∗ is an isomorphism of End V onto End V∗ (considered as left vector space over ∆◦). Hence by the isomorphism theorem III we have a semi-linear isomorphism (S, s) (S with associated division ring isomorphism s) of V/∆ onto V∗/∆◦ such that

A∗ = S −1AJS, A ∈ EndV (5)

Now put g(x, y) = hx, yS i, x, y ∈ V. (6)

Then g is additive in both factors, g(αx, y) = αg(x, y) and

g(x,αy) = hx, (αy)S i = hx,αs(ys)i = hx, (yS )αsi = hx, yS iαs

The mapping α → αs is an isomorphism of ∆ onto ∆◦; hence an anti-automorphism in ∆. The conditions just noted for g are that g is a sesquilinear form on V/∆ relative to the anti-automorphism s 6 in ∆. Since the pairing h , i is non-degenerate it follows that g is a non-degenerate form: g(x, V) = 0 implies x = 0 and g(V, x) = 0 implies x = 0. We have

g(x, yAJ ) = g(x, yS A∗S −1) = hx, yS A∗i 1. Determination of the semi-simple artinian rings... 5

= hxA, yS i = g(xA, y).

Hence AJ is the uniquely determined adjoint of A relative to g in the sense that g(x, yAJ ) = g(xA, y), x, y ∈ V. So far we have not used the involutorial character J2 = 1 of J. We note ﬁrst that the sesquilinear character of g implies that if v ∈ V then x → g(v, x)s−1 is a linear function on V. Hence there exists a v′ ∈ V such that −1 g(v, x)s = g(x, v′), x ∈ V. (7) Next we consider the linear mapping x → g(x, u)v in V where u, v ∈ V. Since g(g(x, u)v, y) = g(x, u)g(v, y) −1 = g(x, g(v, y)s u),

− it is clear that if we put A : x → g(x, u)v then AJ is y → gA(v, y)s 1 u = g(y, v′)u. Since AJ2 = A this gives g(x, u′)v′ = g(x, u)v for all x, u, v. This implies that v′ = γv, γ , 0, in ∆, (independent of v) so by (7), we have g(x, y)s = δg(y, x), δ , 0 in ∆. (8)

2 By (8) we have g(x, y)s = δg(x, y)δs. We can choose x, y so that 7 g(x, y) = 1. Then we get δs = δ−1. If δ = −1 we have g(x, y)s = −g(y, x). Then g(x, y)s2 = g(x, y) and g(αx, y)s2 = αs2 g(x, y) = g(αx, y) = αg(x, y). Then αs2 = α. So s is an involution in ∆ and g is a semi-degenerate skew hermitian form relative to this involution. If δ , −1 then we put ρ = δ + 1 and h(x, y) = g(x, y)ρ. Then h is sesquilinear relative to the anti-automorphism t : α → ρ−1αsρ and h(x, y)t = (g(x, y)ρ)t = ρtg(x, y)t = ρtρ−1g(x, y)sρ = ρtρ−1δg(y, x)ρ = ρtρ−1δh(y, x). Also ρtρ−1 = ρ−1ρs so ρtρ−1δ = (1 + δ)−1(1 + δs)δ = 1 since δsδ = 1. Hence h(x, y)t = h(y, x). Then αt2 = α so t is an involution in ∆ and h is a non-degenerate hermitian form relative to this involution. Clearly h(xA, y) = h(x, yAJ ) so A → AJ is the adjoint mapping determined by h. We have now proved. 6 0. Artinian Semi-simple Rings with Involution

IV. Let V be a ﬁnite dimensional vector space over a division ring ∆ and assume that End V has an involution J. Then ∆ has an involution j : δ → δ and there exists a non-degenerate hermitian or skew-hermitian form h on V such that J is the adjoint mapping determined by h. The converse of IV is trivial. Given a non-degernate hermitian or skew-hermitian form h on V/∆ then the adjoint mapping relative to h is an involution in End V. We recall next how such forms are constructed. Let (∆, j) be a division ring with involution and let V 8 be an n-dimensional vector space over ∆. Let (v1, v2,..., vn) be a base for V/∆ and H = (ηi j) ∈ ∆n be hermintian or skew hermitian, t = = that is, H ±H where H (ηi j) and the t denotes the transpose. If x = ξivi, y = ηivi then we deﬁne P P n = h(x, y) ξiηi jη j. (9) iX, j=1

Then direct verification shown that h is a hermitian or skew hermitian form accordings as H is hermitian or skew hermitian. More- over, h is non-degernate if and only if H is invertible in ∆n. Since it is clear that there exist hermitian invertible matrices for any involution j and any n (e.g the matrix 1) it follows that End V has an involution if and only if ∆ has an involution. We remark that there exist ∆ which have no involutions. For example, any finite dimensional central division algebra over the rationals Q of dimensionality > 4 has no involution. We remark also that if ∆ = Φ is field then j = 1 is an involution. The construction we gave yields all hermitian and skew hermitian forms on V/∆. Suppose h is a hermitian or skew hermitian form on V/∆ and, as before, (v1, v2,..., vn) is a base for V/∆. Then the matrix H = (h(vi, v j)) of h relative to the given base is hermitian or skew hermitian and if = ξivi, y = ηivi then = h(x, y) ξih(vi1 v j)η j as before. P P

Let h haveP the matrix H = (ηi j) relative to the base (v1, v2,..., vn) and assume h is non-degenerate or, equivalently, H is invertible. 1. Determination of the semi-simple artinian rings... 7

Let A ∈ End V and write viA = αi jv j, so (α) = (αi j) in the 9 J matrix of A relative to this base. LetP A be the adjoint of A rel- J ative to h and write viA = βi jv j. It is clear that the deﬁn- J ing conditions: h(xA, y) = h(Px, yA ) are equivalent to the con- J ditions h(viA, v j) = h(vi, v jA ), i, j = 1,..., n. Using the matrices (α) and (β) there n2 conditions give the matrix condition t t (α)H = H(β) , H = (h(vi jv j)), where (β) is the transposed of the = = = matrix (β) (βi j). For, h(viA, v j) h(ikvk, v j) h( αikvk, v j) and J = = = h(v j, v jA ) h(vi, β jkvk) h(vi, vk)β jk and Pαikh(vk, v j) k k =P P P h(vi, vk)β jk, i, j 1,..., n are equivalent to the matrix condition k weP noted. Then the matrix of AJ is

(β) = H(α)t H−1. (10)

t −1 Now the mapping K : (α) → H(α) H is an involution in ∆n. Also A → (α) is an isomorphism of End V onto ∆n and since this J k maps A → (α) it is an isomorphism of (End V, J) onto (∆n, K). We note next that unless j = 1 then we may normalize h to be hermitian. Then suppose j , 1 and h is skew hermitian. Choose j so that j , j and put ρ = j − j , 0. Then h′ = hρ is sesquilinear relative to the involution α → β−1αρ and ρ−1h′(x, y)ρ = ρ−1h(x, y)ρρ = −(−h(y, x)ρ) = h′(y, x). Hence h′ is hermitian. Clearly the adjoint mappings determined by h and h′ are identical. If j = 1 then ∆ = Φ is commutative and again h is hermitian unless the characteristic is , 2 and h(x, y) = −h(y, x). Hence the two two cases we need 10 consider are 1) h is hermitian, 2) ∆ = Φ a field of characteristic , 2, j = 1, h, skew symmetric. It is easily seen that in the first case unless ∆ , Φ a field of characteristic 2, j = 1 and h( , ) ≡ 0 then there exists a base (u1, u2,..., un) such that the matrix (h(ui, u j)) is a diagonal matrix γ = diag{γ1, γ2,...,γn} where γi = γi , 0. This is proved on pp. 152-157 and pp. 170-171 of Jacobson’s Lectures in Abstract Algebra, Vol. II. The foregoing argument shows that (End V, J) is t −1 isomorphic to (∆n, K) where K is the involution (α) → γ(α) γ in 8 0. Artinian Semi-simple Rings with Involution

∆n. If we take into account to the case omitted in 1) and 2) we see that it remains to assume that ∆ = Φ a ﬁeld, j = 1 and h(x, x) ≡ 0. In this case n = 2r is even and as is well known, there exists a base (u1, u2,..., un) such that the matrix (h(ui, u j)) is

0 1 s = diag{Q, Q,..., Q} where Q = . (11) −1 0!

−1 Then (End V, J) is isomorphic to (Φn, K) where K is (α) → S (α)tS . 2. Standard and canonical involutions in matrix rings. Let

(O, j) be an associative ring with involution and let On be the ring of n × n matrices with entries in O. As usual, we denote by ei j the matrix whose (i, j) entry is 1 and other entries are 0 and we identify O with the set of scalar matrices d = diag{d,..., d}, d ∈ O. Then dei j = ei jd and every element of On can be written in one and only

one way as di j ei j , di j ∈ O. Also we have the multiplication table P ei jekl = δ jkeil (12)

11 and n eii = 1. (13) X1

j t Write d = d and consider the mapping Jl : D = (di j) → D . As is well-known and redily veriﬁed, J1 is an involution in On. We shall call this the standard involution (associated with j) in On. More generally, let C = diag{c1, c2,..., cn} be a diagonal matrix with = −1 = −1 −1 −1 inverteble diagonal element ci ci, C diag{c1 , c2 ,..., cn }. The the mapping

t −1 J1 −1 JC : D → CD C = CD C (14)

is an involution. We shall call such an involution a canonical involution (associated with j). We shall now show that we have the following matrix form of the determination of simple artinian rings with involution given in §1. 1. Determination of the semi-simple artinian rings... 9

V. Any artinian simple ring with involution is isomorphic to a matrix ring with canonical involution (On, Jc) where O is of one of the following types:

1. O = ∆⊕∆◦, ∆ a division ring, j the exchange involution (a, b) → (b, a). 2. O = ∆ a division ring, j an involution in ∆.

3. ∆ = Φ2 the ring of 2 × 2 matrices over a ﬁeld Φ the involution 0 1 X → Q−1XtQ where Q = −1 0.!

The first possibility we noted for an artinian simple ring with in- 12 ∆ ∆J = ∆ ∆◦ volution is n ⊕ n are ideals Let O ⊕ and consider the matrix ring On with the standard involution Jl determined by the involution j : (a, b) → (b, a) in O. Let (ai j), (bi j) ∈ ∆n and con- + J ∆ ∆J sider the element (ai j) (vi j) of n ⊕ n. We map this into the element of On whose (i, j) entry is (ai j, b ji). Then direct verifica- ∆ ∆J tion shows that this mapping is an isomorphism of ( n ⊕ n, J) onto (On, J1). Thus these artinian simple rings with involution are matrix rings with standard involution with coefficient rings of the form 1 above. It remains to consider the simple artinian rings with involution (a, J) such that a is simple. The considerations of the last part of §1 show that such an (a, J) is isomorphic either to a (∆n, Jc) where (∆, J) is a division ring with involution and Jc is a corresponding canonical involution or to (Φ2r, JS ) where JS is the involution 0 1 A → S −1AtS, S = diag{Q, Q,..., Q}, Q = . −1 0! The first possibility is the case 2 listed above. Suppose we have the second possibility. We consider the standard isomorphism of Φ2r onto (Φ2)r which maps a 2r × 2r matrix onto the corresponding r × r matrix of 2 × 2 blocks. It is easy to check that this is an isomorphism of (Φ2r, JS ) onto ((Φ2)r, Jl) where J1 is the standard −1 tl involution based on the involution X → Q X Q in Φ2. Thus we have the case 3. 10 0. Artinian Semi-simple Rings with Involution

13 We shall show later (Theorem of Herstein-Kleinfold-Osborn- McCrimmon, Chap.III) that the three possibilities for the coefficient ring (O, J) we noted have a uniform characterization as the simple rings with involution whose non-zero symmetric elements are invertible. It is easy to check that the rings with involution listed in 1, 2, 3, have these properties. The converse of V is clear by re-tracing the steps. We remark finally that the considerations of matrix rings with involutions can be generalized to the case in which the coefficient ring is not necessarily associative. If (O, J) is such a ring then t the matrix ring On has the standard involution D → D . Also the notion of canonical involution Jc can be generalized. Here it is required that the entries ci of the diagonal matrix C are the nucleous of O, that is, there associate with all pairs of elelments a, b of O((a, b)c = a(bc), (ac)b = a(cb), (ca)b = c(ab)). Chapter 1 Basic Concepts

In this chapter we give the basic deﬁnitions and general results for 14 quadratic Jordan algebras. These algebras are Φ-modules for a commutative ringΦ equipped with a multiplicative composition which is linear in one of the variables and quadratic in the other. If the base ring Φ 1 contains L then the notion of aquadratic Jordan algebra is equivalent to the usual notion of a (linear) Jordan algebra (see §4). The results of this chapter parallel those of Chapter I and a part of Chapter II of the author’s book [4].

1 Special Jordan and quadratic Jordan algebras

It will be convenient from now on to deal with algebras over a (unital) commutative ring Φ. An associative algebra a over Φ is a left (unital) Φ-module together with a product xy which is Φ-bilinear and associative. The results of chapter 0 carry over without change to algebras. We remark that rings are just algebras over Φ= Z the ring of integers. Let (a, J) be an associative algebra with involution and let H (a, J) denote the subset of symmetric elements (aJ = a) of. It is clear that H (a, J) is a Φ-submodule. What other closure proportions does H (a, J) have? Clearly if a ∈ H (a, J) and n = 1, 2, 3,... then an ∈ H = H (a, J). In particular, a2 ∈ H and hence ab + ba = (a + b)2 − a2 − b2 ∈ H if a, b ∈ H . We note also that if, a, b ∈ H then aba ∈ H . We now observe that this last fact implies all the others since H isaΦ-module

11 12 1. Basic Concepts

and contains 1 ∈ a. For, let J be any Φ-submodule of a containing 1 15 and aba for every a, b ∈ J . Then J contains abc + cba = (a + c) − aba − cba, a, b, c in J . Hence contains ab + ba = ab1 + 1ba. Also J contains a2 = a1a, a3 = aaa and an = aan−2a, n ≥ 4. In view of this it is natural to consider aba as the primary composition in H besides the module composition and the property that H contains 1. There is one serious drawback in using the composition aba, namely, this is quadratic in a. It is considerably easier to deal with bilinear Φ 1 compositions. We now note that if contains an element 2 such that 1 + 1 = 2 2 1 (necessarily unique)then we can replace aba by the bilinear = 1 + Φ product a · b 2 (ab ba). More precisely, let J be a -submodule of the associative algebra a such that 1 ∈ J . Then J is closed under a.b if and only if it closed under aba. We have see that if J is closed under + = 1 + aba then it is closed under ab ba, hence, under a · b 2 (ab ba). Conversely, if J is closed under a · b then it is closed under aba since

1 2(a, b), a = (ba2 + a2b) + aba 2 so 2(b, a) · a − b, a2 = aba. (1) These observations lead us to define (tentatively) a special quadratic Jordan algebra J as a Φ-submodule of an associative algebra a/Φ, Φ a commutative associative ring (with 1) containing 1 and aba for a, b ∈ J . We call J a special (linear) Jordan algebra if Φ contains 1 2 . In this case the closure conditions are equivalent to : 1 ∈ J and = 1 + a · b 2 (ab ba) ∈ J if a, b ∈ J . We have seen that if a, J) is an associative algebra with involution then H (a, J) the set of J-symmetric 16 elements is a special qudratic Jordan algebra. Of course, a itself is a special quadratic Jordan algebra. We now give another important example as follows. Let V be a vector space over a field Φ, Q a quadratic form on V and C(V, Q) the corresponding Clifford algebra. Thus if T(V) is the tensor algebra Φ⊕V ⊕(V ⊗V)⊕...⊕V(i) ..., V(i) = V ⊗V ⊗...⊗V (i times) with the usual multiplication then C(V, Q) = T(V)/k where k is the ideal in T = T(V) generated by the elements x⊗ x−Q(x), x ∈ V. It is known that 2. Definition of Jordan and quadratic Jordan algebras 13 the mapping α + x → α + x + k of Φ ⊕ V into C = C(V, Q) is injective. Hence we may identify Φ ⊕ V with the corresponding subspace of C. Then C is generated by Φ⊕V and we have the relation x2 = Q(x), x ∈ V, in C. We claim that J ≡ Φ+ V is a special quadratic Jordan algebra in C. Let a = α + x, b = β + y,α,β ∈ V. Then aba = (α + x)(β + y)(α + x) = α2β + 2αβx + α2y + α(xy + yx) + βx2 + xyx. Now x2 = Q(x) gives xy+yx = (x+y)2 − x2y2 = Q(x, y) where A(x, y) = Q(x+y)−Q(x)−Q(y) is the symmetric bilinear form associated with Q. Hence xyx = −yx2 + Q(x, y)x and aba = (α2β + αQ(x, y) + βQ(x)) + (2αβ + Q(x, y))x + (α2 − Q(x))y (2) ∈ J =Φ+ V. Since J ⊕ 1 and is a subspace of C/Φ it is clear that J is a special quadratic Jordan algebra.

2 Deﬁnition of Jordan and quadratic Jordan algebras

These notions arise in studying the properties of the compositions a, b = 17 1 + Φ 2 (ab ba) and aba is an associative algebra, over where in the ﬁrst Φ 1 = 2 = 2 = case ∋ 2 . We note that a · b b · a and if a a.a then (a , b) · a 1 2 + 2 + 2 + 2 = 1 2 + 2 + 3 + 3 2 = 4 [(a b ba )a a(a b ba )] 4 (a ba aba ba a6 b), a ·(b·a) 1 2 + + + 2 = 1 3 + 3 + 2 + 2 4 a (ab ba) (ab ba)a 4 (a b ba a ba aba ). These observations and the fac, which can be veriﬁed by experimentation, that other simple identities on a · b are consequences of a · b = b · a and (a2 · b) · a = a2 · (b · a) lead to the following

Definition 1′. An algebra J over Φ is called a (unital linear) Jordan Φ 1 = algebra if 1) contains 2 , 2)J contains an element 1 such that a · 1 a = 1 · a, a ∈ J , 3) the product a · b, satisfies a · b = b · a, (a2 · b) · a = a2 · (b · a) where a2 = a · a. It is clear that if J is a special Jordan algebra then J is a Jordan = 1 + Φ 1 algebra with a · b 2 (ab ba). If a is an associative algebra over ∋ 2 then a defines the Jordan algebra a+ whose underlying Φ-module is a = 1 + and whose multiplication composition is a · b 2 (ab ba). 14 1. Basic Concepts

If J is Jordan we denote the mapping x → x · a by Ra. This is a Φ-endomorphism of J . We can formulate the Jordan conditions on a · b in terms of Ra and this will give our preferred deﬁnition of a Jordan algebra as follows:

Deﬁnition 1. A(unital linear)Jordan algebra over a commutative ring Φ 1 (with 1)containing 2 is a triple (J , R, 1) such that J is a (unital) left Φ-module, R is a mapping of J into End J (the associative Φ-algebra of endomorphisms of J ) such that

18 J1 R : a → Ra is a Φ-homomorphism.

J2 R1 = 1 = J3 RaRaRa RaRa Ra

J4 If La is defined by xLa = aRx then La = Ra. Definitions 1 and 1′ are equivalent: If J is Jordan in the sense of ′ Defintion 1 then we define Ra as x → x · a and obtain Jl − J4 of definition 1. Conversely, if J is Jordan in the sense of definition ′ 1 then we define a · b = aRb. Then the conditions of Definition 1 hold. Moreover, the passage from the bilinear composition a · b to the mapping R is the inverse of that from R to a · b. Let J be Jordan in the sense of the second definition and write = = 2 = = = a·b aRb bRa, a a·a. Then [RaRa2 ] 0 where [AB] AB−BA Φ = for the -endomorphisms A, B. Let (a) [Ra, Ra2 ] and consider the identity

0 = f (a + b + c) − f (a + b) − f (b + c) − f (a + c) + f (a) + f (b) + f (c).

This gives

[RaRb·c] + [RaRc·b] + [RbRa·c] + [RbRc·a]

+ [RcRa·b] + [RcRb·a] = 0,

as is readily checked. Since a·b = b·a we get 2[RaRb·c]+2[RbRa·c]+ = Φ 1 19 2[RcRa·b] 0. Since contains 2 we obtain 2. Deﬁnition of Jordan and quadratic Jordan algebras 15

J5 [RaRb·c] + [RbRa·c] + [RcRa·b] = 0.

Let ρ be a commutative associative algebra over Φ (= commutative associative ring extension of Φ). If m is a Φ-module we write mρ = ρ ⊗Φ m regarded as (left unital)ρ-module in the usual way. We have the Φ -homomorphism ν : x → 1 ⊗ x of m into mρ as Φ-module. In the cases in which this is injective we shall identify x and 1 ⊗ x and m and ν its image 1 ⊗ m(= m ). In any case 1 ⊗ m generates mρ as ρ-module. If n is a second Φ-module and η is a homorphism of m into n then there exists a unique homorphism ηρ of mρ into nρ such that

η m / n ν ν (3) mρ / nρ ηρ is commutative. It follows that if γ End m andγ ˜ denotes the resultant of ν: End m →(End m)ρ and the canonical mapping of (End m)ρ into End mρ. Then we have a unique homomorphism η of mρ into End mρ such that η e m / End m

ν ν (4) / mρ / End emρ η is commutative. e 20 Φ 1 Φ Now supppose ∋ 2 and (J , R, 1) is a Jordan algebra over . Let R be homomorphism of Jρ into End Jρ determined as in (4) by R and = pute 1 1 ⊗ 1. If we use the deﬁnition of J J5, and the fact that 1 ⊗ J generates J it is straight forward to check that (J , R, 1) is a Jordan e ρ ρ algebra. e e We formulate next the notion of a (unital) quadratic Jordan algebra. This is arrived at by considering the properties of the product aba in an associative algebra or, equivalently, the mapping Ua : x → a × a. Note that Ua ∈ End a where a the given associate algebra. Also U : a → Ua is quadratic is in the sense of the following 16 1. Basic Concepts

Definition 2. Let m and n be left (unital) Φ-modules, Φ-modules, Φ an arbitrary (unital) commutative associative ring. Then a mapping Q : a → Q(a) (or Qa) of n into m is called quadratic if 1)Q(αa) = 21 α2Q(a),α ∈ Φ, a ∈ m, 2)Q(a, b) ≡ Q(a + b) − Q(a) − Q(b) is Φ -bilinear from m to n. The kernel of Q is the set of z such that Q(z) = 0 = Q(a, z), a ∈ m. The associated Φ-bilinear mapping Q(a, b) is symmetric: Q(a, b) = Q(b, a). The kernel ker Q is a submodule. If Q and Q′ are quadratic mappings of m into n then so is Q + Q′ and βQ, β ∈ Φ. Hence the set of these mappings is a Φ-module. The resultant of a quadratic mapping and a Φ -homomorphism and of a Φ -homomorphism and a quadratic mapping is a quadratic mapping. If Q is a quadratic mapping of m into n and R is contained in ker Q then Q(a + R) = Q(a) defines a quadratic mapping of m = m/R into n. If Q and Q′ are quadratic mappings and ′ ′ Q(ai) = Q (ai), Q(ai, a j) = Q (ai, a j) for all ai, a j on a set of generators ′ {ai} then Q = Q . In particular, if Q(ai) = 0, A(ai, a j) = 0 then Q = 0. Let F be a free left module with base {xi|i ∈ I} and let i → bi, {i, j} → bi j be mappings of the index set I and of the set I2 of distinct unordered paris of elements i1 j, i1 jI into n. = = 2 + If x ∈ F and x ξi xi (finite sum) then we define Q(x) ξi bi ξiξ jbi j. Then it is easyP to check that Q is a quadratic mappingP of F Xi< j into n. It F is free with base {xi} and ρ is a commutative associative algebra over Φ then Fρ is free with base {1 ⊗ xi}. It follows the remark just made that the following lemma holds for m = F free:

22 Lemma . Let Q be a quadratic mapping of m into n where these are left modules over Φ and let ρ be an associative commutative ring extension of Φ. Then exists a unique quadratic mappings Qρ of mρ into nρ such that the following diagram is commutative

Q m / n ν ν (5) mρ / n Qρ 2. Deﬁnition of Jordan and quadratic Jordan algebras 17

η Proof. Let F −→ m → 0 be an exact sequence of modules where F is free and put K = ker η the kernel of η. Then we have the corresponding homomorphism ηρ of Fρ onto mρ (as in (3)) and, as is well-known, ker ηρ = ρ(1 ⊗ K) the ρ -submodule generated by 1 ⊗ K = {1 ⊗ k|k ∈ K}. ηρ We have the isomorphism x + ρ(1 ⊗ K) → x of F /ρ(1 ⊗ k) onto mρ. We deﬁne Qη of F to η by Q(x) = Q(xη), x ∈ F . Since this is the resultant of η and Q it is aquadratice mapping.e Also ker Qη ⊇ K. Since η η Fρ is ρ -free Q determines the quadratic mapping Qρ of F into η so η that (5) is commutative for m = F . We have ker Qρ ⊇ ρ(1 ⊗ K). Hence η x + ρ(1 ⊗ K) → Qρ(x) is a quadratic mapping of Fρ/ρ(1 ⊗ K) into nρ. Using the isomorphism of Fρ/ρ(1 ⊗ K) and mρ this can be transferred e e ηρ to the quadratic mapping Qρ : x → Qρ(x) of mρ into n. If x ∈ F then η ηρ η η η 1 ⊗ x = (1 ⊗ x) → Qρ(1 ⊗ x) = 1 ⊗ Q (x) = 1 ⊗ Q(x ). Hence Qρ satisﬁes the commutativity in (5). The uniqueness of Qρ is clear since 23 1 ⊗ m generates mρ. Let n = End m. Then it is immediate from the lemma that if Q is a quadratic mapping of m into End m then there exists a unique quadratic mapping Q of mρ into End mρ such that commutativiy holds in: e Q m / Endm

ν ν (6) / mρ / Endemρ Q e wherenu ˜ is A → 1 ⊗ A and (ρ × A) = ρ ⊗ xA. We are now ready to define a quadratic Jordan algebra. We have two objectives in mind: first, to give simple axioms which will be adequate for studying the composition axa = xUa in associative algebras and sec- 2 ond, to characterize the mapping Ua ≡ 2Ra −Ra2 in Jordan algebras. We = 2 recall that in an associative algebra xUa x(2Ra − Ra2 ). The following definiion is due to McCrimmon [5].

Deﬁnition 3. A (unital) quadratic Jordan algebra over a commutative associative ring Φ (with 1) is a triple (J , U, 1) where J is a (unital) 18 1. Basic Concepts

left Φ -module, 1 a distinguished element of J and U is a mapping of J into End J such that 24

QJ1 U is quadratic

QJ2 U1 = 1

QJ3 UaUbUa = UbUa

QJ4 If Ua,b = Ua+b − Ua − Ub and Va,b is deﬁned by xVa,b = aUx,b then UbVa,b = Vb,aUb.

QJ5 If ρ is any commuatative associative algebra over Φ and U is the quadratic mapping of Jρ into End Jρ as in (6), then U esatisﬁes QJ3 and 5. e

It is clear from QJ5 that (Jubρ, U, 1), 1 = 1⊗1 is a quadratic Jordan = algebra over ρ. We remark also that QJ4 states that bUa,xUb aUb,xUb ·. e e e = Since the left side is symmetric in a and so is the right. Hence aUb,xUb

xUb,aUb ·. This gives the following addendum to QJ4: = = ′ UbVa,b Vb,aUb UaUb,b · QJ4

Let a be an associative algebra over Φ and deﬁne Ua to be x → axa. Then Ua ∈ End a and QJ1 and QJ2 are evidently satisﬁed since Ua,b is the mapping x → axb + bxa. We have xUaUbUa = a(b(axa)b)a and = = xUbUa xUaba (aba)x(aba) so QJ3 holds by the associative law. Now xVa.b = aUx,b = xab + bax. Hence xUbVa,b = b(xba + abx)b = 25 bxbab + babxb and Vb,aUb = b(xba + abx)b = bxbab + babxb. Thus QJ4 holds. Now QJ5 is clear since aρ is an associative algebra and the

mapping U of aρ is a → Ua where xUa = a˜ x˜ a˜. We denote (a, U, 1) by a. e e e If (J , U, 1) is ae quadratic Jordan algebra, a subalgebra B of J is a Φ-submodule containing 1 and every aUb, a, b ∈ B, a homomorphism η of J into a second quadratic Jordan algebra is a module homomor- η = η = η η phism such that 1 1, (aUb) a Ub. Monomorphism, isomorphism, 2. Definition of Jordan and quadratic Jordan algebras 19 automorphism are defined in the obvious way. A quadratic Jordan algebra J will be called special if there exists a monomorphism of J into algebra a(q), a associative. It is immediate that this is essentially the same definition we gave before. Φ 1 If a is an associative algebra over ∋ 2 then we can form the Jordan + + = 1 + algebra a and the quadratic Jordan algebra a(q). For a , xRa 2 (a a) (q) 2 and for a , xUa = axa. The relation (1) : 2(x, a), a − xa = axa shows that = 2 Ua 2Ra − Ra2 (7) is the formula expressing U in terms of R. Conversely, we can express R in terms of U by noting that Ua,b = Ua+b − Ua − Ub is x → axb + bxa so Va ≡ Ua,1 = U1,a is x → ax + xa. Hencec 1 R = V , V = U = U . (8) a 2 a a a,1 1,a Φ 1 Now let (J , R, 1) be any Jordan algebra (over ∋ 2 ) and define U 26 by (7). Then we claim that (J , U, 1) is a quadratic Jordan algebra. It is = 2 = = clear that a → Ua 2Ra −Ra2 is quadratic in a and R1 1 gives U1 1. Also, since Jρ is Jordan for any commutative associative algebra ρ over Φ, it is enough to prove that QJ3 and QJ4 hold. We need to recall some basic identities, namely,

J6 RaRbRc + RcRbRa + R(a·c)·b = Ra·bRc + Rb·cRa + Rc·aRc = J7 [Rc[RaRb]] Rc[RaRb]. The ﬁrst of these is obtained by writing J5 in element form: (d·a), (b· c) − (d · (b · c))a+etc., interchanging b and d and re-interpreting this as operation identity. This gives RaRbRc + RcRbRa + R(a·c)·b = RaRb·c + RbRa·c + RcRa·b. This and J5 give J6. To obtain J7 we interchange a and b in J6 and subtract the resulting relations frm J6. Special cases of J5 and J6 are ′ + = J5 [Ra2 Rb] 2[Ra,bRa] 0 ′ 2 + 2 + = + J6 RaRb RbRa R(a,b),b Ra2 Rb 2Ra,bRa. 20 1. Basic Concepts

using (7) we obtain Ua,b = 2(RaRb + RbRa) − 2Ra,b. Then xVa,b = aUx,b gives Va,b = 2(RaRb − RbRa + Ra·b) (9)

27 We shall now prove QJ4, which is equivalent to:

2 + (2Ra−Ra2 )(Ra·b RbRa − RaRb)− + 2 = (Ra·b RaRb − RbRa)(2Ra − Ra2 ) 0

The left hand side of this after a little juggling becomes

2 + 2 + + + 2[RaRb RbRa, Ra] 2Ra[Ra Ra·b] 2[RaRa·b]Ra + + + −[Ra2 Rb]Ra Ra[RbRa2 ] [RaRa2 Ra2 Ra, Rb] + [Ra·ba2 ] = 2 + 2 + + 2[RaRb RbRa, Ra] [RaRa2 Ra2 RaRb] + ′ [Ra·bRa2 ] · (byJ5 ) = + + 2[Ra2 Rb 2Ra·bRa − R(a·b)·a′ Ra] 2[RaRa2 , Rb] ′ ′ + 2[R(a·b)·a, Ra](J6 and J5 with b → a · b) = + = ′ 4[Ra·bRa]Ra 2[Ra2 Rb]Ra 0(J5 ).

Hence QJ4 holds. 28 For the proof of QJ3 we begin with the following identity

J8. [Va,bVc,d]Va,bVc,d − VaVd,c , b (cf. the author’s book [2], (5) on p.325). To derive this we note that J7 shows that [RaRb] is a derivation in J . For any derivation D we = + = have directly : [Va,bD] VaD,b Va,bD. Also [Va,bRc] Va,bRc − VaRc,b follows directly from J5 and J7. Then J9, is a consequence of these two relations. We note next that the left hand side of J8 is skew in the pairs (a, b), (c, d). Hence we have the consequence = J9 Va,bVc,d − VaVd,c,b VcVb,a , d − Vc,dVa,b .

We now use the formula xVa,b = aUx,b deﬁning Va,b to write J8 and J9 in the following equivalent forms: 3. Basic identities 21

′ = J8 UaUc,b,d − Uc,dVa,b Ub,dVa,c − Vd,aU − c, b ′ = J9 Vc,dVa,b − VdUa,c,b Va,cUb,d − Ua,dVc,d. ′ Taking d = aUb, c = b in J8 gives = 2UaUb Ub,aUb Va,b − VaUb,aUb (10)

Replacing a → b, c → b,b → a, d → a in J9′ gives 29 = 2 VaUb,a Vb,a − 2UbUb (11) = If we substitute this in the last term of (10) we get 2UaUb Ub,aUb 2 + ′ Va,b − Vb,aUb 2UbUaUb. Since QJ4 has the consequence QJ4 : = = 1 UbVa,b Vb,aUb UaUb , b (as above) the foregoing reduces to QJ3

3 Basic identities

In this section we shall derive a long but of identities which will be adequate for the subsequent considerations. No attempt has been made to reduce the set to a minimal one. On the contrary we have tried to list almost every identity which will occur in the sequel. Let (J , U, 1) be a quadratic Jordan algebra over Φ. We write aba = bUa, abc = bUa,c so b → aba is the Φ -endomorphism Ua for ﬁxed a and a → aba is a quadratic mapping of J into itself for ﬁxed b. 2 2 2 2 We put a = 1Ua, a ◦ b = (a + b) − a − b = 1Ua,b = aV1,b and 2 Va = Ua,1 = U1,a. We have a ◦ b = b ◦ a, a ◦ a = 2a , Ua,a = 2Ua, V1 = V1,1 = 2. Taking b = 1 in QJ4 gives Va,1 = V1,a so 1Ua, = aU1,x. Then a ◦ x = aVx. Since a ◦ x = x ◦ a we have xVa = aVx : Also 30 xV1,a = 1Ua,x = a ◦ x = xVa so Va = V1,a = Va,1. We shall now apply a process of linearization to deduce consequences of QJ3 and QJ4. This method consists of applying QJ3 and QJ4 to Jρ = Φ[λ] the polynomial algebra over Φ in the indeterminate λ. Since ρ is Φ -free

1The proof we have given of QJ4 was communicated to us by McCrommon, that of QJ3 by Meyberg. The ﬁrst direct proof of QJ3 was given by Macdonald. Subsequently he gave a general theorem on identities from which QJ3 and QJ4 are immediate consequences. See Macdonald [1] and the author’s book [2] pp.40-48). 22 1. Basic Concepts

the canonical mapping of J into Jρ is injective so we may identify

J with its image 1 ⊗ J in Jρ and regard U as the unique extension of U to a quadratic mapping of Jρ into EndeJρ. We write U for U. The elements of Jρ can be written in one and only one way in the form 2 n e ao + λa1 + λ a2 + ··· + λ an, ai ∈ J , and the endomorphism of Jρ can be written in one and only way as Ao + λA1 + ··· . where Ai is the endomorphism in Jρ which extends the endomorphism Ai of J . Now ∈ + + = let a, b, c J and consider the identity Ua λcUbUa λc UbUa+λc which 2 holds in Jρ by QJ5. Comparing coeﬃcients of λ and λ we obtain + = = QJ6 UaUbUa,c Ua,cUbUa UbUa UbUa,bUa,c + + = + QJ7 UaUbUc UcUbUa Ua,cUbUa,c UbUa,bUc Ub,Ua,c .

The same method applied to the variable a in QJ6 and b in QJ4 gives

UaUbUc,d + Ua,cUbUa,d + Uc,dUbUa + Ua,dUbUa,c

= + QJ8 UbUa,c,bUa,d UbUa,bUc,d

QJ9 Vb,aUb,c + Vc,aUb = Ub,cVa,b + UbVa,c 31 We remark that comparison of the coefficients of the other powers of λ in the foregoing identities yields identitites which we have already displayed. Also, if the method is applied to a variable in which the identity is quadratic, say Q(a) = 0 (e.g. QJ3 and the variable a) then we obtain in this way the bilinerizaation Q(a, b) = Q(a + b) − Q(a) − Q(b) = 0. We shall usually not display these bilinearizations.2 . We shall now show that if (J , U, 1) satisifies QJ1−4, 6−9 then QJ5 holds, so (J , U, 1) is a quadratic Jordan algebra. Hence QJ 1 − 4, 6 − 9 constitute an intrinsic set of conditions defining quadratic Jordan algebras. Let ρ be any commutative associative algebra over Φ and consider Jρ where it is assumed that (J , U, 1) satisfies QJ1−4, 6−9. If a ∈ J we put a′ = 1 ⊗ a. Then QJ1 − 4, 6 − 9 hold in J ρ for all choices of

2An exception to this rule in QJ8 which is the bilinerization of QJ7 with respecet to c 3. Basic identities 23 the arguments in J ′ = 1 ⊗ J . Also the bilinearizations of these conditions hold for all values of the argument in J ′. Since QJ7 − QJ9 are either Φ-linear (= Φ. endomorphisms) or Φ -quadratic in their arguments these hold for all choices of the arguments in Jρ. Similarly, QJ6 ′ holds for all a ∈ J and all b, c in Jρ. The validity of QJ8 in Jρ now implies that if QJ6 holds for the arguments (a = a1, b, c) and (a2, b, c) in Jρ then it holds for (a = a1 + ρa2, b, c) for any ρ ∈ ρ. It follows from this that QJ6 holds in Jρ. A similar argument using 32 QJ9 shows that QJ4 holds in Jρ. Similarly, using QJ6 and QJ7 in Jρ one sees that if QJ3 holds for b in Jρ and a = a1, a = a2 in Jρ then it holds for b and a = a1 + ρa2. It follows that QJ3 holds in Jρ. We have therefore proved.

Theorem 1. Let J be a left Φ -module, U a mapping of J into End J satisfying QJ1 − QJ4, QJ6 − QJ9. Then QJ5 holds so J is a quadratic Jordan algebra.

The same argument implies the following result

Theorem 2. Let J be a left Φ -module, U a quadratic mapping of J into End J such that U1 = 1 and QJ3, 4, 6−9 and all their bilinearizations hold for all choices of the arguments in a set of generators of the Φ -module J . Then J is a quadratic Jordan algebra.

It is easy to prove by a Vandermonde determinant argument that if Φ is a ﬁeld of cardinality |Φ|≥ 4 then QJ6−9 follows from QJ3, 4 without the intervention of QJ5. Hence in this case QJ1 − 4 are a deﬁning set of conditions for a quadratic Jordan algebra over Φ. If we put b = 1 in QJ3, 6 and 7 we obtain respectively

2 = Ua Ua2 QJ 10 + = UaUa,c Ua,cUa Ua2,aoc QJ 11 + = + + 2 Ua2,c2 Uaoc UaUc UcUa Ua,c QJ 12

If we replace b by b + 1 in QJ3, 6, 7 and use the foregoing we obtain 33

= 2 UaVbUa UbUa,a QJ13 24 1. Basic Concepts

+ = + 2 UaVbUa,c Ua,cVbUa UbUa,aoc Ua ,bUa,c QJ14

2 + 2 + Ua ,bUc UbUa,c UbUa,c,aoc QJ15 = UaVbUc + UcVbUa + Ua,cVbUa,c

Putting c = 1 in QJ6, QJ7 and QJ9 gives

+ = UaUbVa VaUbUa UbUa,boa QJ16 + = + + UbUa,b Uboa UaUb UbUa VaUbVa (QJ17) UbVa + VbVa,b = Vb,aVb + VaUb. QJ18

If we put a = 1 in QJ6 we get

UbVc + VcUb = Ub,boc. QJ19

Putting c = 1 in QJ12 and replacing b by b+1 in QJ17 give respectively:

= 2 2Ua Va − Va2 QJ20 + + = = + VbUa Ua2,b 2Ua,aob UaVb VbUa VaVbVa. QJ21

If we apply the two sides of QJ3 to 1 we obtain

2 2 a UbUa = (bUa) QJ22

34 which for b = 1 is 2 2 2 a Ua = (a ) QJ23 Next we put a = 1 in QJ4′ to obtain

= = VbUb UbVb Ub,b2 QJ24

Putting b = 1 in QJ9 gives

VaVc + Vc,a = VcVa + Va,c QJ25

or (x ◦ a) ◦ c + cUa, = (x ◦ c) ◦ a + aUc, QJ 25’ 3. Basic identities 25 putting x = c we obtain

(c ◦ a) ◦ c + cUa,c = (c ◦ c) ◦ a + aUc,c 2 = 2c ◦ a + 2aUc which can be simpliﬁed by QJ20 to give

{acc} = c2 ◦ a. QJ26

It is useful to list also the bilinearization of this:

{abc} + {bac} = (a ◦ b) ◦ c QJ 27 which has the operator form 35

′ Vb,c = VaVc − Ub,c QJ 27 If we operate with the two sides of QJ17 on 1 we obtain

2 2 2 (a ◦ b) = a Ub + b Ua + 2aUb ◦ a − bUa ◦ b. Also, if we replace a by a + λb in QJ24 and compare coeﬃcients of λ we obtain + = + Ua,bVa UbVb Ua,a◦b Ub,a2 VaUa,b + VbUa. QJ28 Applying the ﬁrst and last of these to b gives

2 bUa ◦ b = −{abb} ◦ a + {ab ◦ ab} + 2b Ua = 2 2 + 2 + −b Va 2b Ua {aa ◦ bb} QJ 26 = −b2 ◦ a2 + {aa ◦ bb} QJ 20 which is symmetric in a and b. Hence we have

bUa ◦ b = aUb ◦ a QJ29 Using this and the foregoing formula for (a ◦ b)2 we obtain

2 2 2 (a ◦ b) = a Ub + b Ua + aUb ◦ a QJ 30 26 1. Basic Concepts

2 2 = a Ub + b Ua + bUa ◦ b

We wish to prove next 36 = VaUb,a Vb,bUa QJ 31

In element form this is {caUba} = {cbbUa}. Using QJ27 this is equivalent to (c ◦ aUb) ◦ a −{aUbca} = (c ◦ b) ◦ bUa −{bcbUa} which is equivalent to = VaUb Va − UaUb,a VbVbUa − Ub,bUa QJ 32

If we interchange a and b in QJ17 and subtract we obtain VaUbVa − = VbUaVb UbUa,b − UaUb,a which implies that QJ32 is equivalent to = ﬃ VaUbVa − VbUaVb VbVbUa − VaUb Va. Hence it su cs to prove + = + (VaUb VaUb)Va Vb(VbUa UaVb) QJ33

We note next that bilinearization of QJ29 relative to b gives bUa ◦ c + cUa ◦ b = aUb,c ◦ a = cVa,b ◦ a. hence + = VbUa UaVb Va,bVa QJ 34

Using this on the right hand side of QJ33 gives VbVa,bVa also, by = + = + = QJ34, VaUb Vb,aVb −UbVa so VaUb VaUb Vb,aVb −UbVa VaUb VbVa,b(QJ18). Hence the left hand side of QJ33 reduces to Vb, Va,bVa 37 also . This proves QJ33 and with it QJ32 and 31. ◦ 1 2 We shall now deﬁne the powers of a by a = 1, a = a, a = 1Ua, n = n−2 = n as before, and a a Ua, n 2. Then QJ3 implies that Uan Ua. Also by induction on n we have

n (am)n = am (QJ 35)

We shall now prove am ◦ an = 2am+n (QJ 36) by induction on m + n. This clear if n + m ≦ 2. Moreover, we may assume m ≦ n. We now note that QJ36 will follow if we can show m n m m−2 that UaVan = Van Ua. for then a ◦ a = b Van = a UaVan (since n−2 n+n−2 m+n m ≧ 2)= a Van Ua = 2a Ua = 2a . To prove the required 3. Basic identities 27

operation commutativity we shall show that Van is in the subalgebra a of End J generated by the commuting operations Ua, Va(QJ24). More genrally, we shall prove that Uam,an and Vam,an ∈ a. Since Vam,an = ′ Vam Van − Uam,an (QJ27 ) = Uam,1Uan,1 − Uam,an it suﬃces to show this for Uam,an . We use induction on m + n. The result is clear for m + n ≦ 2 by QJ20 and Ua,a = 2Ua so we assume m ≧ n, m ≧ 2. If n ≧ 2, Uan,an = UaUan−2,an−2 Ua by QJ3, so the result holds by induction in this case if ′ = m = m−2 = m−2 n 1, Ua ,a Ua Ua,a Va,a Ua by QJ4 , so the result is valid = m = m = m−2 = m−2 − in this case. Finally, if n 0, Ua ,1 Va Va Ua Va,a Va UaVam−2 (QJ34). Hence the result holds in this case also. This completes 38 the proof that Uam,an , Vam,an ∈ a and consequently of QJ36. We shall now prove a general theorem on operator identities involving the operators Uam,an , Vam,an

Theorem 3. If f (λ1, λ2,...) is a polynomial in indeterminates λ1, λ2,... Φ with coeﬃcients in such that f (Uan1 ,Uan2 ,..., Uann,amn ,...,Van1 ,am1 , ...) = 0 is an identity for all special quadratic Jordan algebras then this is an identity for all quadratic Jordan algebras.

Proof. If X is an arbitrary non-vacuous set then there exists a free quadratic Jordan algebra F(X) over Φ (freely) generated by X whose characteristic property is that F(X) contains X and every mapping X → J of X into a quadratic Jordan algebra (J , U, 1) has a unique extension to a homomorphism of F(X) into (J , U, 1).3 Let X contain more than one element one of which is denoted as x. It is clear from the universal property of F(X) that of f (Un1, Un2,...,...) = 0 holds in = F(X) then f (Uan1 , Uan2 ,...) 0 holds in every quadratic Jordan algebra. Hence it suﬃces to prove f (Un1, Un2,...,...) = 0. Let Y be a set of the same cardinality as X and suppose x → y is a bijective mapping of X onto Y. Let Φ{Y} be the free associative algebra (with 1) gener- 39 (q) ated by Y and let Fs(Y) be the subalgebra of φ{y} generated by Y. We have a homomorphism of F(X) onto Fs(Y) such that x → y. If

3This is a special case of a general result proved in Cohn, Universal Algebra, pp.116- 121 and p.170. A simple construction of free Jordan algebras and more generally of (linear) algebras deﬁned by identities is given in Jacobson [2], pp.23-31. It is not diﬃ- cult to modify this so that it applies to quadratic Jordan algebras. 28 1. Basic Concepts

J is a quadratic Jordan algebra then we denote the subalgebra of End J generated by the Ua, a ∈ J , as Env U(J ). It is clear that Env U(J ) contains all Ua,b and all Va,b. Moreover it is easily seen that if a → aη is a homomorphism of (J , U, 1) onto a second quadratic Jor- dan algebra (J ′, U′, 1′) then there exists a (unique) homomorphism of ′ ′ Env U(J ) onto Env U(J ) such that Ua → Uaη , a ∈ J . Then also ′ ′ Ua,b → Uaη,bη and Va,b → Vaη,bη . In particular we have such a homoor- phism of Env U(F(X)) onto Env U(Fs(Y)). Let X and Y respectively denote the subalgebra of Env U(F(X)) and Env U(Fs(Y)) generated by all Uxn , Uxn,xm , Vxn,xm and Uyn , Uyn,ym , Vyn,ym . Then the restriction of our homomorphism of Env U(F(X)) onto Env U(F(Y)) to X is a homomorphism of X onto Y such that Uxn → Uyn , Uxn,xm → Uyn,ym , Vxn,xm → = Vyn,ym . Since F(y) is special, f (Uyn1 , Uyn2 ,...,...) 0 holds. It will = follow that f (Uxn1 , Uxn2 ,...) 0 holds in F(X) if we can show that the homomorphism of X onto Y is an isomorphism. We have seen that X is generated by Ux and Vx and Y is generated by Uy and Vy. Since Ux → Uy and Vx → Vy the isomorphism will follow by showing that Uy (q) and Vy are algebraically independent over Φ. Now in Φ{Y} we have Uy = yRyL, Vy = yR = yL where aR is b → ba and aL is a b → ab and 40 yL and yR commute and are algebraically independent over Φ since if , k l = l k l k = z ∈ Y, z y, then zkRyL y zy and the elements y zy , l, k 0, 1, 2,... are Φ -independent. Now Vy = yR + yL and Uy = yRyL are the “elementary symmetric” functions of yR and yL. The usual proof of the algebraic independence of the elementary symmetric function (e.g. Ja- cobson, Lectures in Abstract Algebra, p.108) carries over to show that (q) Uy and Vy are algebraically independent operators in Φ{Y} . It follows that they are algebraically independent operators also in Fs(Y). This completes the proof of the theorem. We now give two important instances of Theorem 3 which we shall need. Let f (λ) ∈ Φ[λ]. an indeterminate and let f (a) be deﬁned in the obvious way if a ∈ J a quadratic Jordan algebra. Suppose J is a subalgebra of a(q), a associative. Then we claim that U f (a)Ug(a) = (q) U( f g)(a) holds in a , hence in J . For, xu( f g)(a) = f (a)g(a)x f (a)g(a) = g(a) f (a)x f (a)g(a) = xU f (a)Ug(a). It follows from Theorem 3 that

U f (a)Ug(a) = U( f g)(a) QJ 37 Φ 1 4. Category isomorphism for ∋ 2 . Characteristic two case. 29 in any quadratic Jordan algebra. Another application of the theorem is the proof of Van,an = Vam+n QJ 38 m n This follows since in a(q), a associative, we have xVam,an = xa a + n m m+n m+n a a x = xa + a x = xVam+n . Similarly, one proves

Van Uan = Uan,am+n (QJ 39) and 41 = Van Van−1 Va − Van−2 Ua, n ≥ 2 QJ 40 The list of identities we have given will be adequate for the results which will be developed in this monograph. Other aspects of the theory require additional identities. Nearly all of these are consequences of the analogues for quadratic Jordan algebras of Macdomalli theorem. This result, which states that the extension of Theorem 3 to subalgebras with two generators is valid, has been proved by McCrimmon in [7].

Φ 1 4 Category isomorphism for ∋ 2. Characteristic two case.

Φ 1 We shall show first that if ∋ 2 then the two notions of Jordan algebra and quadratic Jordan algebra are equivalent. Let CJ(CQJ) denote the category whose objects are Jordan algebras (quadratic Jordan algebras) over Φ with morphisms as homomorphisms. We have the following Cat- egory Isomorphism Theorem. Let (J , R, 1) be a Jordan algebra over a Φ 1 = 2 commutative ring containing 2 . Define U by Ua 2Ra − Ra2 . Then (J , U, 1) is a quadratic Jordan algebra. Let (J , U, 1) be a quadratic Φ = 1 = Jordan algebra over and define R by Ra 2 Va, Va Ua,1. Then (J , R, 1) is a Jordan algebra. The two constructions are inverses. More- over, a mapping η of J is a homomorphism of (J , R, 1) if and only if it is a homomorphism of (J , U, 1). Hence (J , R, 1) → (J , U, 1), η → η is an isomorphism of the category CJ onto CQJ. 42 Φ 1 = 2 Proof. Let (J , R, 1) be unital jordan over ∋ 2 and Ua 2Ra − Ra2 . Then we have shown in §2 that (J , U, 1) is a quadratic Jordan algebra. 30 1. Basic Concepts

= + = = 1 = We have Ua,b 2(RaRb RbRa−Ra,b) so Va Ua,1 2Ra and 2 Va Ra. Φ 1 = Next let (J , U, 1) be a quadratic Jordan algebra over ∋ 2 and let Ra 1 = = 2 = = 2 Va. By QJ24, 20, [Va, Va2 ] 0 so [RaRa2 ] 0. Moreover, a 1Ua 1 = = = = 2 (a ◦ a) aRa. Also aVb bVa gives aRb bRa and we have R1 1 and a → Ra is a Φ -homomorphism of J into End J . Hence (J , R, 1) = 1 2 1 = 2 is a linear Jordan algebra. By QJ20, UA 2 Va − 2 Va2 2Ra − Ra2 . This proves the assertions on the passage from (J , R, 1) to (J , U, 1) and back. The rest is clear. We consider next the opposite extreme of the foregoing, namely, that in which 2Φ = 0 or, equivalently, 2(1) = 1 + 1 = 0. Let (J , U, 1) be a quadratic Jordan algebra over Φ. We claim that J is a 2 -Lie algebra (= restricted Lie algebra of characteristic two) if we deﬁne [ab] = a ◦ b and a[2] = a2. We have [aa] = a ◦ a = 2a2 = 0 and [[ab]c] + [[bc]a] + [[ca]b] = (a ◦ b) ◦ c = (b ◦ c) ◦ a + (c ◦ a) ◦ b = {abc} + {bac} + {bca} + {cba} + {cab} + {acb}(QJ27) = 2{abc} + 2{bca} + 2{cab} = 0. Also + 2 = 2 + 2 + 2 = = 2 (a b) a b [a, b] and ba [[ba]a] since Va2 Va by QJ20. Hence the axioms for a 2-Lie algebra hold (Jacobson, Lie Algebras P. 6). This proves

Theorem 4. Let (J , U, 1) be a quadratic Jordan algebra over Φ such that 2Φ = 0. Then J is a 2 lie algebra relative to [ab] = a ◦ b and a[2] = a2.

5 Inner and outer ideals. Diﬀerence algebras.

43 Deﬁnition 4. Let (J , U, 1) be a quadratic Jordan algebra. A subset L of J is called an inner (outer) ideals if L is a sub-module and bab = aUb(aba = bUa) ∈ L for all a ∈ J , b ∈ L L is an ideal if it is both an inner and an outer ideal.

The condition can be written symbolically as J UL ⊆ L for an ideal, L UJ ⊆ L for an outer ideal. If L is an inner ideal and a ∈ = = = J then LUa is an inner ideal since for c bUa, J Uc J UbUa J UaUbUa ⊆ J UbUa ⊆ L Ua. In particular, J Ua is an inner ideal called the principal inner ideal determined by a. This need not contain a. The inner ideal generated by a is Φa + J Ua. For this contains 5. Inner and outer ideals. Diﬀerence algebras. 31 a, is contained in every inner ideal containing a and is an inner ideal, since a typical element of Φa + J Ua is αa + bUa,α ∈ Φ, b ∈ J , and = 2 + + = ′ Uαa+bUa α Ua αUa,bUa UaUbUa. Since Ua,bUa Va,bUa by QJ4 Φ + we see that J Uαa+bua ⊆ J Ua, so a J Ua is an inner ideal. The outer ideal generated by a is the smallest submodule of J contianing a and stable under all Ub, b ∈ J . The principal inner ideal detrermined by 1 is J U1 = J . On the other hand, as we shall see, the outer ideal generated by 1 need not be J . We shall call this the cone of J . = = If L is an outer ideal then {a1ba2} bUa1,a2 bUa1+a2 − bUa1 − = bUa2 ∈ L for b ∈ L , ai ∈ J . In particular, b ◦ a bua,1 ∈ L , b ∈ L , a ∈ L . By QJ27 it follows that {ba1a2} ∈ L , b ∈ L , ai ∈ J . Φ 1 If contains 2 then L is an outer ideal if and only if it is an ideal and if and only if L is an ideal in (J , R, 1) where (J , R, 1) is the Jordan 44 algebra corresponding to (J , U, 1) in the usual way. For, if L is an = 1 outer ideal then b · a 2 b ◦ a ∈ L , a ∈ J , b ∈ L . On the other hand, if L is an ideal of (J , R, 1) then bRa = aRb ∈ L and this implies that bUa and aUb ∈ L . It is clear that the intersection of inner (outer) ideals is an inner (outer) ideal and the sum of outer ideals is an outer ideal. It is easily checked that the sum of an inner ideal and an ideal is an inner ideal. Let L be an ideal in J , U, 1), bi ∈ L , ai ∈ J . Then we have seen = that a1Ua2,b2 {a2a1b2}∈ L . Hence

+ = + + + (a1 b1)Ua2+b2 (a1 b1)(Ua2 Ua2,b2 Ub2 )

≡ a1Ua2 ( mod L )

It follows that if we deﬁne in J = J /L = {a + L |a ∈ J }, a1 = Ua2 a1Ua2 then this is single valued. It is immediate that (J , U, 1) is a quadratic Jordan algebra and we have the canonical homomorphism a → a of (J , U, 1) onto (J , U1). Conversely, if η is a homomorphism of (J , U, 1) then L = ker η is an ideal and we have the isomorphism η : a → aη of (J = J /L , U, 1) onto (J η, U, 1). This fundamental theorem has its well-known consequences.

Examples. (1) Let J = H (Zn) the quadratic Jordan algebra (over the ring of integers Z) of n × n integral symmetric matrices. Let 32 1. Basic Concepts

E = diag {1,..., 1, 0,..., 0}. Then E is an idempotent (E2 = E) and the principal inner ideals J UE = EH E is the set of 45 matrices of the form A 0 , 0 0!

A ∈ Zr. Next let B = (bi j) ∈ H (Zn) have even diagonal elements and let A ∈ H (Zn). Then (i, i) -entry of C = ABA is

cii = ai jb jkaki = ai jb jkaik Xj,k Xj,k = 2 + ai jb j j 2 ai jb jkaik Xj Xj

Hence C has even diagonal elements. Thus the set L of integral symmetric matrices with even diagonal elements is an outer ideal in H (Zn). If m ∈ Z the set mH (Zn) of integral symmetric matrices whose entries are divisible by m is an ideal in H (Zn).

(2) Let ρ = Φ(λ) the ﬁled of rational expressions in an indeterminate λ over a ﬁeld Φ of characteristic two. let H (ρn) be the set of n × n symmetric matrices with entries in ρ. This is a quadratic Jordan algebra over Φ (with ABA as usual). Let L be the subset of matrices with diagonal entries in Φ(λ2). Then L is an outer ideal containing 1. It is easy that this is the cone of H (ρn).

(3) Let Φ be a ﬁeld of characteristic two, a = Φ[λ], λ an indeterminate. Consider a(q) and the subspace L = Φλ2 + Φλi. i≧4 P 46 It is readily checked that L is an ideal in a(q). Let J = a(q)/L 2 and put λ = λ + L . Then λ = λ + L . Then λ = 0 but 3 λ , 0 in J . If J is a special Jordan algebra, say, J is a subalgebra of L(q), L, associative, then the Jordan power Xn of X ∈ J coincides with the associative power Xn in L since n n−2 n−2 X = X UX = XX X. Hence it is clear that in a special Jor- + dan algebra Xn = 0 implies Xn 1 = 0. It follows that J = a(q)/L is not special. 6. Special universal envelopes 33

We have deﬁned ker U = {z|Uz = 0 = Uz,a, a ∈ J }. This is an ideal since it is a submodule and if a ∈ J , z ∈ ker U, then = = = aUz 0, and UzUa UaUzUa 0. Also we can show that for = = b ∈ J , UzUa,b 0. To see this we note that Uz,a 0, a ∈ J ′ implies Vz = Uz,1 = 0. Then Vz,a = 0 = Va,z by QJ27 . By QJ9, we have {b{zba}c} + b{zca}b = {{bzc}ab} + {(bzb)ac} which + = + gives (using a as operand): Vb,zUb,c Vc, Ub Ub,bVz,c Uc,zUb . = = This implies Uc,zUb 0 or Ub,zUa 0. Hence zUa ∈ker U. The argument we have used show that every { } and −U− with one of the arguments z ∈ ker U is 0, with the exception of zUa. In particular 2z = z · 1 = 0 which show that ker U = 0 if J has no two torsion. We call J nondegenerate if ker U = 0.

6 Special universal envelopes

A homomorphism of (J , U, 1) into a(q) where a is associative is called an associative specialization of J into a.A special universal envelops for J is a pair (S (J ),σu) where S (J ) is an associative algebra and σu is an associative specilalization of into S (J ) such that if σ is an- associative specialisation of J into an associative algebra a then there 47 exists a unique homomorphism η of s(J ) into a such that

σ J u / S (J ) w ww ww (12) ww η {ww a is commutative. To construct an (S (J ),σu) let T(J ) be the tensor algebra deﬁned by the Φ -module J : T(J ) = Φ ⊕ (J ⊕ (J ⊗ J ) ⊕ ...... where all these tensor products are taken over Φ. Multiplication in T(J ) is deﬁned by (x1 ⊗ ... ⊗ xr)(xr+1 ⊗ ... ⊗ xs) = x1 ⊗···⊗ xs, xi ∈ J , and the rule that the unit element Φ of 1Φ is unit for T(J ). Then T(J ) is an associative algebra over Φ. Let k be the ideal in T(J ) generated by the elements 1 − 1Φ(1 ∈ J ), aba − a ⊗ b ⊗ a, a, b ∈ J . Put S (J ) = T(J )/K and aσu = a + K, a ∈ J . Then it is readily seen 34 1. Basic Concepts

4 that ((J ),σu) is a special universal envelope for (J , U, 1). . It is clear ′ π′ that we have an involution π of T(J ) such that (x1 ⊗ x2 ... ⊗ xr) = π′ xr ⊗ xr−1 ⊗ ... ⊗ x1, xi ∈ J . Since (aba − a ⊗ b ⊗ a) = aba − a ⊗ b × a ′ it is clear that Kπ ⊆ K. Hence π′ induces an involution π in S (J )/K. We have aσu π = aσu , a ∈ J , and since the aσu generate S (J ) it is clear that π is the only involution satisfying aσuπ = aσuπ. We shall call π the main involution of S (J ). If ξ is a homomorphism of (J , U, 1) into ′ ′ ′ (J , U , 1 ) then we have a unique homomorphism ξu of S (J ) into S (J ′) such that ξ J / J ′

σu σu (13) S (J ) / S (J ′) ξ

48 It is immediate that (J , U, 1) in special if and only if the mapping σu of J into S (J ) is injective. In this case it is convenient to identify J with its image in S (J ) and so regard J as a subset of S (J ),σu as the injection mapping. Then J is a sub-algebra of the quadratic (q) Jordan algebra S (J ) and the universal property of σu states that any homomorphism of J into an a(q), a associative, has a unique extension to a homomorphism of S (J ) into a.

7 Quadratic Jordan algebras of quadratic forms with base points.

We consider a class of quadratic Jordan algebras (J , U, 1) over a ﬁeld Φ satisfying the following conditions:

1. There exists a linear function T and a quadratic form Q on J (to Φ) such that

X2 − T(X)X + Q(X) = 0 (14)

4cf. Jacobson [2],pp. 65-72, for the corresponding discussion for Jordan algebras 1 Φ 1 K = k+1 This condition is superﬂuous if contains 2 since in this case X · X X so (14) implies (15). 7. Quadratic Jordan algebras of quadratic forms... 35

X3 − T(X)X2 + Q(X)X = 01 (15)

2. The same conditions hold for Jρ where ρ is any extension ﬁeld of Φ and T and Q for Jρ are the extenstions of these functions 49 on J to a linear function and a quadratic form on Jρ respectively. (We assume J imbedded in Jρ and write, U, 1 for the U-operator and unit in Jρ).

3. J , Φ Taking ρ = Φ(λ), an indeterminate and replacing x by x + y in (14) we obtain x ◦ y = T(x)y + T(y)x − Q(x, y) (16) where Q(x, y) is the symmeteric bilinear form associated with the quadratic form Q(Q(x, y) = Q(x + y) − Q(x) − Q(y)). Similarly, (15) and 3 X = xUx give

2 2 yUx = T(x)x ◦ y + T(y)x − Q(x)y − Q(x, y)x − x ◦ y (17)

Putting y = 1 in (16) gives 2x = T(x)1 + T(1)x − Q(x, 1)1. If we take x < Φ1 T(1) = 2 (18) Then x = 1 in (14) gives Q(1) = 1 (19) Also using the formulas for X2 and x ◦ y, (17) becomes

yUx = Q(x)y + T(y)T(x)x − Q(x, y)x − T(y)Q(x)1 (20)

We can write this in a somewhat more compact form by introducing x = T(x)1 − x. Then (20) becomes 50

yUx = Q(y, x)x − Q(x)y 20’

Conversely, suppose we are given a quadratic form Q on a vector space J with a base point 1 such that Q(1) = 1. Deﬁne T(x) = ′ Q(x, 1), x = T(x)1 − x and Ux by (20 ) (or (20)). Then one can verify by direct calculation that (J , U, 1) is a quadratic Jordan algebra 36 1. Basic Concepts

satisfying condition 1 and 2. We proceed to prove more that this by showing that (J , U, 1) is a special quadratic Jordan algebra satisfying 1 and 2. For this purpose we introduce the Clifford algebra of Q with base point 1 which is defined as follows. Let T(J ) be the tensor algebra over J and let L be the ideal in T(J ) generated by 1Φ − 1 and all x ⊗ x − T(x)x + Q(x)1 where 1, x ∈ J and T(x) = Q(x, 1). Then we define the clifford algebra C(J , Q, 1) of the quadratic form Q will base point (such that Q(1) = 1) to be T(J )/L . If x ∈ J we put xσu = x + L . Then we have (xσu )2 − T(x)xσu + Q(x)1 = 0. This implies that xσu yσy + yσu xσu = T(x)yσu + T(y)xσu − Q(x, y)1 = 0. Since σ σ σ σ y u Uxσu = x u y u x u we obtain

σu σu σu σu 1 y uxσu = Q(x)y + T(y)T(x)x − Q(x, y)x − T(y)Q(x) . (21) Also we have 1σu = 1. This and (21) show that J σu = {xσu |x ∈ J } (q) is a subalgebra of C(J , Q, 1) . We shall that σu is injective. Then it 51 will follow from (20) and (21) that (J , U, 1) is a quadratic Jordan algebra and σu is an associative specialization of J in C = C(J , Q, 1). It is clear from the definition of C(J , Q, 1) that if x → xσ is a linear mapping of J into an associative algebra a suchthat 1σ = 1 and (xσ)2 − T(x)x + Q(x)1 = 0 then there exists a unique homomorphism of C(J , Q, 1) into a such that xσu → xσ, x ∈ J . We consider first the case in which (Q, 1) is pure in the sense that J = Φ1 + V where V is a subspace such that T(v) = 0, v ∈ V. If the characteristic is , 2 then T(1) = Q(1, 1) = 2 , 0 and J = Φ1⊕(Φ1)⊥(1Φ1)⊥ the orthogonal complement of Φ1 relative to Q(x, y)). Then T(v) = Q(1, v) = 0 for v ∈ (Φ1)⊥ and (Q, 1) is pure. If the characteristic is two then T(1) = 0 so (Q, 1) is pure if and only if T ≡ 0. In this case V can be taken to be any subspace such that J = Φ1 ⊕ V. Now let C(V, −Q) be the Clifford algebra of V relative to the restriction of −Q to V. The canonical mapping of Φ1 + V into C(V, −Q) is injective so we can identify J = Φ1 ⊕ V with the corresponding subset of C(V, −Q). Let x = α1 + v,α ∈ Φ, v ∈ V. Then T(x) = 2α, Q(x) = α2 + Q(v) and in C(V, −Q), x2 = α21 + 2αv + v2 = (α2 − Q(v))1 + 2αv. Hence x2 − T(x) + Q(x) = 0. It follows from the universal property of C(J , Q, q) that we have a homomorphism of C(J , Q, 1) into C(V, −Q) σ such that x v → x. Clearly this implies that σu is injective. 7. Quadratic Jordan algebras of quadratic forms... 37

Suppose next that (Q, 1) isnot pure so the characteristic is two and T , 0. We can choose d so that T(d) = 1 and write J = Φd ⊕ W 52 where W is the hyperplane in J defined by T(x) = 0. Then 1 ∈ W and W = Φ1 ⊕ V, V a subspace. We again consider C(V, Q)(Q = −Q) since char= 2) and we identify W = Φ1 + V with the corresponding subset of C(V, Q). Let D′ be the derivation in the tensor algebra T(V) such that vD′ = v + Q(v, d)1. Since this maps v ⊗ v + Q(V) into 0 it maps the ideal K defining C(V, Q) into itself. Hence this induces a derivartion D in C(V, Q) such that vD = v + Q(v, d)1. Now put L = C(V, Q) and let L [t, D] be the algebra of differential polynomials in an indeterminate t with coefficients in L such that

ct + tc = cD, c ∈ L (22) since the characteristic is two, D2 is a derivation. Since vD2 = vD, v ∈ V, and V generates L , D2 = D. Also ct2 + t2c = cD2 so c(t2 + t) = (t2 + t)c, c ∈ L . Since t2 + t commutes with t also, it is clear that this polynomial is in the center of L [t, D]. Hence also g(t) = t2 + t + Q(d)1 is in the center. Let (g(t)) be the ideal in L [t, D] generated by g(t) and put O = L [t, D]/(g(t)). It is clear from the division algorithm (which is applicable to g(t) since its leading coeﬃcient is 1) that every element of L [t, D] is congrument modulo (g(t)) to an element of the form co +c1t, ci ∈ L . Also co +c1t ≡ 0( mod g(t)) implies co = c1 = 0. Hence we can identify O with the set of elements of the form co+c1t, ci ∈ L , and we have the realtions vt + tv = v + Q(v, d), t2 + t + Q(d) = 0. We have the injective linear mapping x = α1 + v + βd → y = α1 + v + 53 βt, ,αβ ∈ Φ, v ∈ V, of J = Φ1+V +Φd into O. Moreover, T(x) = β, Q(x) = α2 + Q(v) + β2Q(d) + βQ(v, d) + αβ and

y2 = α2 + Q(v) + β2t2 + β(vt + tv) = α2 + Q(v) + β2(t + Q(d)1) + β(v + Q(v, d)) = T(x)y + Q(x)1 = T(x)y − Q(x)′.

Hence by the universal property of C(J , Q, 1) we have a homomorphism of C(J , Q, 1) into O such that (α1+v+βd)σu → y = α1+v+βt. Clearly this implies that σu is injective. 38 1. Basic Concepts

We have now proved that (J , U, 1) is a special quadratic Joradan algebra and σu is an associative specialization of J into, C(J , Q, 1). We now take y = 1 in (20) to obtain x2 = Q(x)1+2T(x)x−T(x)x−2Q(x)1 (Since T(x) = Q(x, 1), T(1) = Q(1, 1) = 2) = T(x)x − Q(x). Since J is special we have x3 − T(x)x2 + Q(x)x = 0 in J . If ρ is an extension ﬁeld of Φ then it is clear that the extension of U to a quadratic mapping of Jρ into End Jρ is given by (20) where Q and T are the extensions of Q and T to a quadratic form a linear function on Jρ. It follows as in J that we have x2 − T(x)x + Q(x)1 = 0 = x3 − T(x)x2 + Q(x)x also Jρ. Thus conditions 1 and 2 hold. Now let σ be an associative specialization of J into a. Since σ is a homomorphism of J into a(q) we have (xk)σ = (xσ)k, k = 0, 1, 2,... Since x2 − T(x)x + Q(x)1 = 0 in J we have (xσ)2 − T(x)xσ + Q(x)1 = 54 0. By the universal property of C(J , Q, 1) we have a unique homomorphism of C(J , Q, 1) into a such that xσu → xσ. It follows that (C(J , Q, 1),σu) is a special universal envelope for (J , U, 1). We shall call (J , U, 1) the (quadratic Jordan) algebra of the form Q with base point 1. If we to indicate Q and 1 then we use the notation Jord (Q, 1) for this (J , U, 1).

8 The exceptional quadratic Jordan algebra

H (O3), O an Octonion algebra. A quadratic Jordan algebra which is not special will be called exceptional. We have already given one example of this sort, example (3) of §5. We shall now give the most important examples of exceptional quadratic Jordan algebra. These are based on Octonion algebras. We proceed to define these for an arbitrary basic field. Let Φ be a field and let ρ = Φ[u] be the algebra over Φ with base (1, u) over Φ where 1 is unit and u2 − y + ρ (ρ = ρ1), (23) ρ ∈ Φ, 4ρ , −1. This is a commutative associative algebra which has the involution x = α + βu → x = α + β(1 − u),α,β ∈ Φ (24) 8. The exceptional quadratic Jordan algebra 39

Next we deﬁne a quaternion algebra over Φ which as Φ -module is a direct sum of two copies of Φ[u], so its elements are pairs (a, b), a, b ∈ Φ[u] with the usual vector space structure. We deﬁne a product in O = 55 Φ[u] ⊕ Φ[u] by

(a, b)(c, d) = (ac + σdb, da + bc) (25) where a, b, c, d ∈ Φ[u] and σ is a ﬁxed non-zero element of Φ. Then O is an associative algebra with 1 = (1, 0) and σ/ has the standard involution.

x(a, b) → x = (a, −b). (26)

Finally, let O = σ/ ⊕ σ/ as vector space over Φ and deﬁne a product in O by (25) where σ replaced by τ , 0 in Φ and the elements are now in O. The resulting algebra O is called an Octonion algebra over Φ. It has the standard involution (26). These algebras are not associative but are alternative in the sense that they satisfy the following weakening of the associative law called the alternative laws:

x2y = x(xy), yx2 = (yx)x (27)

In O we have x + x = t(x) where t is a linear funtion and xx = n(x) = xx where n(x) is quadratic form on O (values in Φ). t and n are called respectively the trace and norm. We write v = (0, 1) in O. Then u and v generate O and we have the basic rules: vu = uv = (1 − u)v, u2 = u + ρ, v2 = σ. Similarly, we put w = (0, 1) in O and we have wu = uw, wv = vw = −vw, w2 = τu, v, w generate O and every element of O can be written in one and only one 56 was as a+bw, a, b ∈ O. Suppose the base ﬁeld Φ is algebraically closed. Then Φ[u] is a direct sum of two copies of Φ since the polynomial λ2 − λ − ρ is a product of distinct linear factors. Then Φ[u] = Φ[e] where e2 = e, e = 1 − e. Thus in this case we may take ρ = 0. Also replacing v and w by mutliples of these elements we may suppose v2 = 1, w2 = 1. Then (1, u, v, uv, w, uw, vw, (uv)w) is a base for O whose multiplication table has coeﬃcients which are 0, ±1. For arbitrary Φ we shall say that O is a split Octonion algebra if ρ = 0, σ = 0 = τ = 1, or equivalently the base (1, u, v,...) has the multiplication table just indicated. 40 1. Basic Concepts

Now suppose Φ is of characteristic , 2, O an octonion algebra over Φ. Let O3 be the set of 3 × 3 matrices with entries in O. This is an algebra over O with the usual vector space compositions and matrix multiplication. We have the standard involution in this algebra: A → t A where A = (ai j) for A = (ai j). Let H (O3) be the Φ -subspace of t matrices satisfying A = A. This is closed under the bilinear product = 1 + A · B 2 (AB BA) and it is well-known that (H (O3), R, 1) is a Jordan algebra if XRA = X·A (See Jacobson’s book [2], p.21). We now consider = 2 the quadratic Jordan algebra (H (O3), U, 1) where Ua 2Ra − Ra2 and we wish to analize the U operator in this algebra. For this we introduce the following notation.

α[ii] = αeii, α ∈ Φ ′ a[i j] = aei j + aei j, a ∈O, i , j (29 )

57 Here the ei j are the usual matrix units ei j has 1 in the (i, j) -position 0’s elsewhere. We have a[ ji] = a[i j] (27′)

and if Hii = {α[ii],α ∈ Φ}, Hi j = {a[i j]|i , j, a ∈ O}, then

H = (O3) = H11 ⊕ H22 ⊕ H33 ⊕ H12 ⊕ H23 ⊕ H13 (28)

The Hii are one dimensional and the Hi j, i , j, are eight dimensional so dim H = 27. Any AUB, A, B ∈ H is a sum of elements xUy where x, y are in the spaces Hi j and xUy,z where x, y, z are in the Hi j and y and z are not in the same subspace. It is easily checked that the non-zero xUy, xUy,z of the type just indicated are the following:

(i) β[ii]Uα[ii] = αβα[ii]

(ii) α[ii]Ua[i j] = aαa[ j j]

(iii) b[i j]Ua[i j] = aba[i j] (It is easily seen that (ax)a = a(xa) in any alternative algebra. Hence this is abbreviated to axa.) (iv) {α[ii]a[i j]b[ ji]} = (αab + α(ab))[ii] 8. The exceptional quadratic Jordan algebra 41

(v) {α[ii]β[ii]a[i j]} = αβa[i j]

(vi) {α[ii]a[i j]β[ j j]} = αβa[i j]

(vii) {α[ii]a[i j]b[ jk]} = αab[ik]

(viii) {a[i j]α[ j j]b[ jk]} = αab[ik] 58

(ix) {a[i j]b[ ji]c[ik]} = a(bc)[ik]

(x) {a[i j]b[ jk]c[ki]} = (a(bc) + a(bc))[ii].

Now let Φo be a subring of φ containing 1. Then (H , U, 1) can be regarded as a quadratic Jordan algebra over Φ0. The foregoing formulas show that if Oo is a subalgebra of (O/Φo, j), that is, a subalgebra of O/Φo stable under j, then the subset Ho of H of matrices having entries in Oo is a subalgebra of (H , U, 1). This is clear since Ho is the set of sums of α[ii], a[i j] where α, a ∈ Oo. It is clear also that if K is an ideal in (Oo, j) then the set Z of matrices with entries in K is an ideal in (Ho, U, 1). Hence we have the quadratic Jordan algebra (Ho/Z , U, 1). If Φ has characteristic , 2 then a = a in O if and only if a = α ∈ Φ. This is not the case for characteristic two accordingly, in this case we let t H (O3) denote the set of 3×3 matrices with entries in O such that A = A and the diagonal entries are in Φ. (For Φ of characteristic , 2 the latter 3 condition is implied by the former.) Then H = H (O3) = Hi j i≤ j=1 P where Hi j is as before. Then it is easily seen that there is a unique quadratic mapping of H into End H such that the formulas (i)-(x) hold and all other xUy, xUy,z are 0 where x, y, z are in the subspaces Hi j, y and z not in the same subspace. For any characteristic we have

Theorem 5. (H , U, 1) is a quadratic Jordan algebra. 59 Proof. The case in which the characteristic , 2 has been settled before, so we assume the characteristic is 2. Assume ﬁrst that Φ= Z2 the ﬁeld of two elements and O is the split octonion algebra over Φ. Let O′ be the split octonion algebra over the rationals Q, Oo the Z -subalgebra of (O′ j) of integral linear combinations of the base (1, u, v,...) Then we 42 1. Basic Concepts

Z have the -quadratic Jordan algebra (H (Oo3), U, 1)/2H (O03 ) which can be regarded as a Z2 -quadratic Jordan algebra. Moreover, it is clear that (H (O3), U, 1) (over Z2) is isomorphic to this. Hence H (O3, U, 1) is a quadratic Jordan algebra over Z2. Now let Φ be arbitrary of characteristic two, O an arbitrary octonion algebra. To prove (H (O3), U, 1) is Jordan it is enough to show that the conditions QJ3, 4, 6 − 9 hold for the U -operator (Theorem 1). These hold if and only if they hold for (H (O3)Ω, U, 1) where Ω is the algebraic closure of Φ. Also we may identify H (O3)Ω with H (OΩ)3. Hence it suffices to assume Φ algebraically closed. Then O is split. Now it is clear from the definition of a split algebra that if Oo is the split octonion algebra over Z2 then O = OoΦ = Φ ⊗Z2 O0 so (H (O3), U1) = (H (Oo3)Φ, U, 1). Since we have just seen that the latter is a quadratic Jordan algebra it follows that (H (O3), U, 1) is a quadratic Jordan algebra. We have seen in Theorem 3 that a quadratic Jordan algebra over Φ with 2Φ= 0, is a 2-Lie algebra relative to [a, b] = a◦b = (a+b)2 −a2 −b2 [2] 2 60 and a = a . In particular, this holds for (H (O3), U, 1) where O is an octonion algebra over a field of characteristic two. We now note that 2 in this case A = 1UA is the same as the square of the matrix A ∈ H as defined in O3. To see this it is sufficient to show that 1Ua[i j] = (aei j + a · e ji)(aei j + ae ji) = n(a)(eii + e j j) = n(a)[ii] + n(a)[ j j], 1Uα[ii] = 2 (αeii)(αeii) = α [ii], {x1y} = xy + yx if x, y are in different spaces Hi j. By the defining formulas 1Ua[i j] = 1[ii]Ua[i j] + 1[ ji]Ua[i j] = n(a)[ j j] + 2 n(a)[ii] (by (ii)), 1Uα[ii] = 1[ii]Uα[ii] = 1[ii]Uα[ii] = α [ii] (by (i)). By (v),{α[ii]1c[i j]} = {α[ii]1[ii]c[i j]} = αc[i j] = αcei j +αce ji. On the other hand, αeii(ceii + ce ji) + (cei j + ce ji)(αeii) = αcei j + αce ji. By(Viii). {a[i j]1c[ jk]} = ac[ik]. Also a[i j]c[ jk] + c[ jk]a[i j] = (aei j + ae ji)(ce jk + cek j) + (ce jk + cek j)(aei j + ae ji) = aceik + caeki = ac[ik]. The remaining {x1y} and xy + yx are 0. Hence we have proved our assertion and we have the following consequence of Theorems 4 and 5:

Corollary . Let O be an octonion algebra over a ﬁeld of characteristic two, H (O3) the set of 3 × 3 hermitian matrices in O3 with diagonal entries in Φ. Then H (O3) is a 2-Lie algebra relative to AB = AB + BA and A[2] = A2. Theorem 5 has an important generalization in which the octonion 8. The exceptional quadratic Jordan algebra 43 algebra is replaced by an alternative algebra with involution (O, j) such that all norms dd = dd j, d ∈ O, are in the nucleus. We recall that the nucleus of a non-associative algebra is the set of elements α such that [α, x, y] = (αx)y − α(xy) = 0, [x,α, y] = 0, [x, y,α] = 0 for all x, y in 61 the algebra. In an alternative algebra the associator [x, y, z] ≡ (xy)z − x(yz) is an alternating function in the sense that [x, y, z] is unchanged under even permutation of the arguments and changes sign under odd permutation. Hence α ∈ N(O) the nucleus of the alternative algebra O if and only if [α, x, y] = 0, x, y ∈O. Now suppose (O, j) is an alternative algebra satisfying the condition that norms are in the nucleous N(O). Let No be the Φ -submodule of N(O) generated by the norms. If x, y ∈O then (x + y)(x + y) − xx − yy = xy + yx ∈ No. In particular, t(x) = + Φ 1 x x ∈ No. It follows that if contains 2 then H (O, j) ⊆ N(O) so the condition in this case is that the symmetric elements of O are contained in the nucleus. Again suppose O arbitrary and (O, j) satisﬁes the norm condition. Then we have the following results (McCrimmon):

1) xNo x ⊆ No, x ∈O 2) xNx ⊆ N 3) If N′ = N ∩ H (O, j)then xN′ x ⊆ N′ Proof. 1. We shall use (xa)x = x(ax) which we write as xax. Also we shall need Moufang’s identity: (ax)(ya) = a(xy)a which holds in any alternative algebra. It is enough to prove x(yy)x ∈ No, x, y ∈ O. We have x(yy) = x(y(t(y) − y)) = x(yt(y)) − xy2 = (xy)t(y) − (xy)t(y) − (xy)y = (xy)y. Hence x(yy)x = (x(yy))(t(x) − x) = (x(yy))t(x) − (x(yy))x = ((xy)y)t(x) − (xy)(yx) (by Moufang)= (xy)(yt(x) − yx) = (xy)(yx) = (xy)(xy) ∈ No.

2. (2) We use α[x, y, z] = [αx, y, z] = [xα, y, z] = [x, y, z]α for 62 x, y, z ∈ O,α ∈ N, and (xyx)z = x(y(xz)) (see the author’s book [2], pp. 18-19). We have to show that [xαx, y, z] = 0 if α ∈ N, x, y, z ∈ O. Since xx ∈ N this will follow by showing that [xαx, y, z] = [xx, y, z]α. For this we have the following calculation: [xαx, y, z] = [xαt(x), y, z] − [xαx, y, z] 44 1. Basic Concepts

= t(x)[x, y, z]α + (x(α(x(yz))) − (x(α(xy)))z = t(x)[x, y, z]α − (xα)[x, y, z] + (xα)((xy)z) − [xα, xy, z] = (xα)((xy)z) = g(x, y, z)α

where g(x, y, z) = t(x)[x, y, z] − x[x, y, z] − [x, xy, z]. Taking α = 1 we have g(x, y, z) = [xx, y, z]. Hence [xαx, yz, z] = [xx, y, z]α = 0.

3. is an immediate consequence of this. Now consider the algebra O3 of 3 × 3 matrices with entries in O. By H (O3) we shall now understand the set of hermitian matrices of t ′ O3(A = A)with diagonal entries in N = N ∩ H ( , j). If O is associative then H (O3) is just the set of hermitian matrices. In any case the ′ elements of H (O3) are sums of elements α[ii], α ∈ N , and a[i j], a ∈ ′ O, i , j. Since No ⊆ N and all traces are in No it is clear from (1) − (3) that the right hand sides of (i) − (x) are contained in H (O3). Hence we can deﬁne a unique quadratic mapping of H (O3) into End H (O3) such that (i) − (x) hold and the remaining xUy, xUy,z = 0 for x, y of the form 63 α[ii] or a[i j]. It has been proved by McCrimmon that H (O3, U, 1) is a quadratic Jordan algebra. The algebras H (O3) with O not associative are exceptional. In fact, we have the following stronger result:

Theorem 6. If (H (O3), U, 1) is a homomorphic image of a special quadratic Jordan algebra then O is associative.

Proof. The proof we sketch is due to Glennie and is given in detail on p.49 of the author’s book [2]. One can show that the following identity holds in every a(q), a associative:

xzx ◦{y(zy2z)x}− yzy ◦{x(zy2z)y} = x(z{x(yzy)y}z)x − y(z{y(xzx)x}z)y. (29)

On the other hand, if one takes

x = 1[12], y = 1[23], z = a[21] + b[13] + c[32] 9. Quadratic Jordan algebras deﬁned by certian cubic forms. 45 then one can see that the (1, 3) entry in the matrix on the left side of (29) is a(bc) − (ab)c while the (1, 3) -entry on the right hand side is 0. Hence if (29) is to hold in H (O3) then a(bc) = (ab)c for all a, b, c ∈O so O is associative. Clearly this identity holds if (H (O3), U, 1) is a homomorphic image of a special quadratic Jordan algebra.

9 Quadratic Jordan algebras deﬁned by certian cubic forms.

In this section we assume the base ring Φ is an infinite field. We shall give another definition of the quadratic Jordan structure on H (O3), O an octonion algebra over Φ. As before, H (O3) denotes the set of 3 × 64 3 hermitian matrices with entries in O, diagonal entries in Φ. If a = 3 3 αi[ii] + ai[ jk], where (i, j, k) is a cyclic permutation of (1, 2, 3) and 1 i=1 theP notationsP are as in §8, then we define a “determinant” by

3 N(a) = det a = α1α2α3 − αin(ai) + t((a1a2)a3) (30) X1

Here n(a) = aa, t(a) = a + a in O. It is known that t((a1a2)a3) = t(a1(a2a3)) so we write this as t(a1a2a3). Also it is known that t(a1a2a3) is unchanged under cyclic permutation of the arguments. If f is a rational mapping of H into a second ﬁnite dimensional space then we let ∆b a f denote the directional derivative of f at a in the direction b (see the author’s book, pp, 215-221). In particular, if f is a polynomial function + + ∆b 2 ∆b then we have f (a λb) ≡ f (a) ( a f )λ(modλ ) and a f is determined by this condition. Since N is polynomial mapping which is homogeneous of degree three we have + = + ∆b + ∆a 2 + 3 N(a λb) N(a) ( aN)λ ( bN)λ N(b)λ (31)

By (30) we have for a = αi[ii] + ai[ jk], b = βi[ii] bi[ jk] that P P P P ∆b = + aN βiα jαk − βin(ai) − αit(ai, bi) t(bia j, ai) (32) Xi Xi Xi Xi 46 1. Basic Concepts

= ∆a∆b = ∆a ∆b ∆a ∆b 65 We deﬁne T(a, b) − 1 logN ( 1N)( 1N) − 1( N) so T is a symmetric bilinear form in a and b. By (32), we have

T(a, b) = αi βi − βi(α j + αk) − t(ai, bi) X X Xi X

= αiβi − t(ai, bi) (33) X X Since t(a, b) is non-degenerate on O, T(a, b) is non-degenerate on H . If we deﬁne the “adjoint matrix”, a♯ by

♯ a (α jαk − n(ai)[ii] + (a jak − αiai)[ jk] (34) Xi Xi it is easy to check that ∆b = ♯ aN T(a , b) (35) A straight forward verifiction using Moufang’s identity shows that we have a♯♯ = N(a)a. (36) It is clear from the definition of N that N(1) = 1. We now define T(a) = T(a, 1) = T(a, 1♯) since 1♯ = 1 by (34). Then (33) gives ♯ ♯ ♯ T(a) = αi. We define a × b = (a + b) − a − b . We have T(a, b) = ∆a ∆b ∆a ∆b = ∆a ∆b ( 1N)( P1N) − 1( N) T(a)T(b) − 1( N) (by (35)). Since N is cubic form (=homogeneous polynomial function of degree three we have ∆a ∆b = ∆c ∆a ∆b b i ( N) x( ( ( N))) is independent of x and is symmetric in ∆a ∆b = ∆a ∆a = ∆a ♯ = a, b, c. Hence 1( N) b( N) bT(a ) (by (35)) T(a × b). Hence we have T(a × b) = T(a)T(b) − T(a, b) (37)

66 We deﬁne the characteristic polynomial fa(λ)N(λ1 − a). By (31) and the deﬁnition of T(a) we have

3 2 fa(λ) = N(λ1 − a) = λ − T(a)λ + S (a)λ − N(a) (38) where S (a) = T(a♯). Direct veriﬁcation, using (34) and the foregoing deﬁnitions shows that

a♯ = a2 − T(a)a + S (a)1 (39) 9. Quadratic Jordan algebras deﬁned by certian cubic forms. 47 where a2 is the usual matrix square. Since S (a) = T(a♯) and T is linear and satisﬁes T(1) = 3 this gives 2T(a♯) = T(a)2 − T(a2) (40)

We now suppose that H = H (O3) is endowed with the quadratic Jor- 2 dan structure given in §8. Then a = 1Ua is the usual square of a and 3 a = aUa. We shall now establish the following formula for the U- operator in H : ♯ bUa = T(a, b)a − a × b (41) We shall establish this using the foregoing formulas and the Hamil- ton-Cayley type theorem that 3 2 fa(a) = a − T(a)a + S (a)a − N(a)1 = 0, (42) which we prove first. Suppose first that the characteristic is , 2. Then 67 3 = 1 2 + 2 a aUa 2 (aa a a). Then one can verify (42) by direct calculation (see the author’s book [2], p.232). Next assume char. = 2. Then we shall establish (42) by a reduction mod 2 argument similar to that used in §8. We note first that we may assume the base field is algebraically closed. Then O is split and has a canonical base (u1, u2,..., u8) with multiplication table in Z2 as in§8. We obtain a corresponding canon- ial base (v1,..., v27) for O/Φ where vi = i[ii], i = 1, 2, 3 and v j, for j > 3, has the form uk[12], uk[13] or uk[23], k = 1, 2,..., 8. Now let ξ1,ξ2,...,ξ27 be indeterminates and consider the “generic” element 27 x = ξ jv j in Hρ, ρ = Φ(ξ),ξ = (ξ1,...,ξ27). By specialization it suf- 1 ficesP to prove (42) for a = x. Let ρ = Z2(ξ), Ho = ρ v j, so Ho is o o quadratic Jordan algebra over ρ and x ∈ H . The functions S, T, N on o o P Ho are the restriction of the corresponding ones on Ho are the restriction of the corresponding ones on H . Hence it suffices to prove the result for x in Ho. This follows by applying a Z -homomorphism of an ′ = ′ ′ ′ = Z ′ algebra H ρ v j where ρ (ξ) and the v j are obtained from a ′ canonical base ofP the split octonion algebra O /Q as in §8. We now begin with (42). A linearization of this relation by replacing a by a + λb and taking the coefficient of λ gives 2 2 ♯ ♯ bUa = −a ◦ b + T(a)(a ◦ b) + T(b)a − T(a × b)a − T(a )b + T(a , b)1 48 1. Basic Concepts

= −(a♯ + T(a)a − T(a)a − T(a♯)1) ◦ b + T(b)(a♯+ T(a)a − T(a♯)1) + T(a)a × b − T(a♯)b − (T(a)T(b) − T(a, b))a + T(a♯, b) ((39), (37)) = −a♯ ◦ b + T(a♯)b + T(b)a♯ − (T(b)T(a♯) − T(a♯, b)) + T(a, b)a = T(a, b)a − a♯ × b

68 using (37) and a × b = a ◦ bT(a)b − T(b)a + T(a × b)1 which is the linearization of (39). Hence (41) holds. We now assume we have a finite dimensional vector space J over an infinite field Φ equipped with a cubic form N, a point 1 satisfying N(1) = 1, such that:

= ∆a∆b = ∆a ∆b ∆a ∆b (i) T(a, b) − 1 log N ( 1N)( 1N) − 1( N) is a non-degenerate symmetric bilinear form in a and b.

♯ ♯ = ∆b ♯♯ = (ii) If a is deﬁned by T(a , b) aN then a N(a)a. We deﬁne

♯ (iii) bUa = T(a, b)a − a × b

where a × b = (a + b)♯ − a♯ − b♯. Then we have

Theorem 7. (J , U, 1) is a quadratic Jordan algebra.

Proof. We can linearize (ii) to obtain

a × (a × b) = N(a)b + T(a♯, b)a (43) a♯ × b♯ + (a × b)♯ = T(a♯, b)b + T(b♯, a)a (44)

= ∆′ ∆b ∆′ ∆b = ∆b ∆b we have T(1, b) ( 1N)( 1N)− 1( N) 3N(1) 1 N −2 1N (by Eu- = ∆b = ♯ ler’s theorem on homogeneous polynomial functions) 1N T(1 , b). 69 Hence 1♯ = 1 (45) 9. Quadratic Jordan algebras deﬁned by certian cubic forms. 49

∆c ∆a∆b by the non-degeneracy of T. Since N is a cubic form x( N) is independent of x and symmetric in a, b, c. By(ii) we have T(a × c, b) = ∆a∆b = ∆c ∆a∆b c N x( N). Hence T(a × c, b) is symmetric in a, b, c and in = = ∆a∆b = ∆a ∆b particular we have T(a, b × 1) T(a × b, 1) 1 N ( 1N)( 1N) − T(a, b) = T(a)T(b) − T(a, b) where T(a) = T(a, 1♯) = T(a, 1). This and the non-degeneracy of T imply

b × 1 = T(b)1 − b (46)

By (43) and the symmetry of T(a × c, b) we have T(b, (c × a♯) × a) = T(b × a, c × a♯) = T(b × a) × a♯, C = T(N(a)b, c) + T(T(a♯, b)a, c) = T(b, N(a)c) + T(b, T(a, c)a♯). Hence

(c × a♯) × a = N(a)c + T(a, c)a♯ (47)

♯ Since T(bUa, c) = T(a, b)T(a, c) − T(a × b, c) is symmetric in b and c we have T(bUa, c) = T(b, cUa) (48) ♯ ♯ ♯ Next we note that T(bUa) , c) = T((T(a, b)a − a × b) , c) = T(T(a, b)2 a♯ +(a♯ ×b)♯ −T(a, b)(a♯ ×b)×a, c) = T9(a, b)2T(a♯, c)−T(N)(a)a× b♯−N(a)T(a, b)b−T(a♯, b♯)a♯, c−T(a, b)T(N(a)b+T(a, b)a♯ , c) (by (44), ♯ ♯ ♯ ♯ ♯ (47) and (ii))= T(a , b )T(a , c) − N(a)T(a × b , c) = T(b Ua♯, c). Hence

♯ ♯ (bUa) = b Ua♯ (49)

70 Linearization of this relative to b gives

= bUa × cUa (b × c)Ua♯ (50) = We can now provce QJ3. For this we consider T(xUbUa,y) T(T ♯ ♯ (bUa, x)bUa −(bUa) ×x, y). Since T(bUa,x) = T(b, xUa) and T((bUa) × = ♯ = ♯ = ♯ = ♯ x, y) T(b Ua♯ ×x, y) T(b Ua♯ , x×y) T(b , (x×y)Ua♯ ) T(b , xUa× ♯ ♯ yUa) (by (50))= T(b × xUa, yUa) = T((b × xUa)Ua, y) the foregoing = ♯ = relation becomes T(xUbUa , y) T(T(b, xUa)bUa − (b × xUa)Ua, y) T(xUaUbUa, y). Hence QJ3 holds. To prove QJ4 we note that the 50 1. Basic Concepts

deﬁnition of Ua and Va,b give xVa,b = T(x, a)b + T(a, b)x − (x × b) × a. Hence

xVa,bUa − xUaVb,a = T(x, a)bUa − ((x × b) × a)Ua

− T(xUa, b)a + (xUa × a) × b Using the symmetry of T(x × y, z) and (48) we obtain

T(T(x, a)bUa − ((x × b) × a)Ua − T(xUa, b)a+

(xUa × a) × b, y) = T(b, T(x, a)yUa−

T(b, ((yUa × a) × x) − T(b, T(a, y)xUa + T(b, y × (xUa × a)). It suffices to show this is 0 and this will follows by showing that 71 T(x, a)yUa −(yUa ×a)× x is symmetric in x and y. We have T(x, a)yUa − ♯ ♯ (yUa×a)×x = T(x, a)T(y, a)a−T(x, a)a ×y−T(a, y)(a×a)×x+((a ×y)× x = T(x, a)T(y, a)a−T(x, a)a♯ ×y−2T(a, y)a♯ ×x+N(a)(y×x)+T(a, y)a♯ × x = T(x, a)T(y, a)a−T(x, a)a♯ ×y−T(a, y)a♯×x+N(a)(y×x). Since this is symmetric in x and y we have QJ4. Also we have xU1 = T(x)1−1× x = x by (46). To prove QJ5 we observe that if ρ is an extension field of Φ then QJ3 and QJ4 are valid for Jρ since the hypothesis made on N carry over to the polynomial extension of N to Jρ. In particular, there hold if ρ = Φ(λ). Then the argument in §3 shows that QJ6 − 9 hold in J . Hence J is a quadratic Jordan algebra by Theorem 1. A cubic form N and element 1 with N91) = 1 satisfying (i)-(iii) will be called admissible. We shall now give another imporatant example of an admissible (N, 1) which is due to Tits (see the author’s book [2], pp.412-422). Let a be a central simple associative algebra of degree three (so dim a = q) and let n be the generic (= reduced) norm on a, t the generic trace. Let J = a ⊕ a ⊕ a a direct sum of three copies of a. We write the elements of J as triples x = (a0, a1, a2), ai ∈ a. Let µ be a non-zero element of Φ and define −1 N(x) = n(ao) + µn(a1) + µ n(a2) − t(aoa1a2) (51) If we put 1 = (1, 0, 0) we have N(1) = 1. It is not difficult to verify that (N, 1) is admissible. Hence J with the U-operator defined by (iii) is a 72 quadratic Jordan algebra. It can be shown that these algebras are of the form H (O3) and hence are exceptional also. 10. Inverses 51

10 Inverses

If a, b are elements of an associative algebra such that aba = a, ab2a = 1 then a is invertiable by the second condition, so ab = 1 = ba by the ﬁrst. Thus a is invertible with inverse b = a−1. Conversely, if a is invertible then aa−1a = a and aa−2a = 1. This motivates the following:

Deﬁnition 5. An element a of a quadratic Jordan algebra (J , U, 1) is invertible if there exists ab in J such that aba = a, ab2a = 1. Then b is called an inverse of a. The foregoing remark shows that if J = a(q), a associative then a is invertible in J with inverse b if and only if ab = 1 = ba in a. If σ is a homomorphism of J into J ′ and a is invertible with inverse b then aσ is invertible in J ′ with inverse bσ. In particular if J ′ = a(q) then aσbσ = 1 = bσaσ.

We have the following

Theorem on Inverses 1. The following conditions are equivalent: (i) a is invertible, (ii) Ua is invertible in End J , (iii) 1 ∈ J Ua(2) If = −1 = −1 a is invertible the inverse b is unique and b aUa . Also Ub Ua . and if we put b = a−1 then a−1 is invertible and (a−1)−1 = a. (3) We have −1 = 2 −1 = = −1 = −1 a ◦ a 2, a ◦ a 2a, V VaUa Ua Va. (4) aba is invertibel if and only if a and b are invertible, in which case (aba)−1 = a−1b−1a−1. 73 2 = = = 2 = 2 Proof. (1) If b Ua 1 then 1 U1 Ub Ua UaUb Ua so Ua is invertible. Then (i)⇒.(ii). Evidently(ii) ⇒ (iii). Now assume (iii). Then there exists a c such that 1 = cUa. Then 1 = UaUcUa so Ua is invertible. Then there exists ab such that bUa = a. Hence UbUbUa = Ua and since Ua is invertible, UaUb = 1 = UbUa. 2 Then b Ua = 1UbUa = 1 so a is invertible with b as inverse. Thus (iii) ⇒ (i).

= −1 (2) If a is invertible with b as inverse then bUa a and since Ua = −1 = = exists, b aUa is unique. Also UaUbUa Ua gives UaUb = −1 1UbUa so Ua−1 Ua . Also Ub -invertible implies that b is in- −1 = = vertible and its inverse is bUb bUa a. This completes the 52 1. Basic Concepts

proof of (2). = = −1 = (3) By QJ24 we have UaVa Ua,a2 VaUa. Hence (a ◦ a)Ua −1 = −1 = = 2 = −1 a VaUa a UaVa aVa 2a 2Ua. Since Ua exists −1 = −1 2 = −1 = this gives a ◦ a 2. By QJ20 we have a ◦ a a Va2 −1 2 = = = a (Va −2Ua) 4a−2a 2a. Also, by QJ13 and 24, UaVa−1 Ua −1 U −1 = U 2 = UaVa = VaUa. Hence V −1 = U Va = a Ua,a2 a,a a a −1 = VaUa (4). The first assertion is clear since Uaba UaUbUa. −1 = −1 = −1 Also if a and b are invertible then (aba) (aba)Uaba bUaUa −1 −1 = −1 −1 = −1 = −1 −1 −1 Ub Ua b Ua b Ua−1 a b a . Remark by M.B. Rege. There are two other conditions for a to be invertible with b as inverse which can be added to those given in (1): (iv) aba = a and ba2b = 1, (v) aba = a and b is the only element 74 of J satisfying this condition. These are well-known for associative algebras. The associative case of (iv) applied to Ua and Ub given (iv) in the Jordan case. (v) is an immediate consequence. If n is a positive integer then we define a−n = (a−1)n. Then it is easy to extend QJ32, 33 to all integral powers. It is easy to see also that for arbitrary integral m, n, Uam,an , Vam,an are contained in the (commutative) −1 subalgebra of End J generated by Ua, Va and Ua . A quadratic Jordan algebra J is called a division algebra if 1 , 0 in J and every non-zero element of J is invertible. If is an associative division algebra then a(q) is a quadratic Jordan division algebra. Also if (a, J) is an associative division algebra with involution then H (a, J) is a quadratic Jordan division algebra since if 0 , h ∈ H then (h−1)J = (hJ )−1 = h−1 ∈ H . If Q is a quadratic form with basic point 1 on a vector space J then we have seen that the quadratic Jordan algebra J =Jord (Q, 1) is special and can be identified with a subalgebra of C(J , Q, 1)(q) where C(J , Q, 1) is the Clifford algebra of Q with base point 1. In C we have the equation x2 − T(x)xQ(x) = 0, x ∈ J . Hence xx = Q(x)1 = xx for x = T(x)1 − x. This shows that x is invertible in C if and only if Q(x) , 0 in which case x−1 = Q(x)−1 x. It follows that x is invertible in J if and only if Q(x) , 0. Hence J =Jord(Q, 1) is a division algebra if and onky if Q is unisotropic in the sense that Q(x) , 0 if x , 0 in J . 75 The existence of exceptional Jordan division algebras was first es- 10. Inverses 53 tablished by Albert. Examples of these can be obtained by using Tits construction of algebras defined by cubic forms as in §9. In fact, it can be seen that if the algebra a used in Tits construction is a division algebra and µ is not a generic norm in a then the Tits’ algebra defined by a and µ is a division algebra.

An element a ∈ J is called a zero divisor if Ua is not injective, equivalently, there exists ab , 0 in J such that bUa = 0. Clearly, if a is invertible then it is not a zero divisor. An element z is called an absolute zero divisor if Uz = 0 and J is called strongly non-degenerate if J contains no absolute zero divisors , 0. This condition is stronger than the condition that J is non-degenerate which was defined by: ker U = 0, where ker U = {z|Uz = 0 = uz,a, a ∈ J }. Let Φ be a field. An element a ∈ (J /Φ, U, 1) is called algebraic if the subalgebra Φ[a] generated by a is finite dimensional. Clearly Φ[a] is the Φ-subspace spanned by the powers am, m = 0, 1, 2,..., and we have the homomorphism of Φ[λ](q), λ an indeterminate, onto Φ[a] such that λ → a. Let ka be the kernal of this homomorphism. If the characteristic is , 2 then the ideals of Φ[λ](q) are the same as those of Φ[λ]+, which is the Jordan algebra associated with Φ[λ](q) by the category isomorphism. = 1 + = Φ Φ + = Φ Since ab 2 (ab ba) a · b in [λ] we have [λ] [λ] as algebras. Hence the ideals of Φ[λ](q) are ideals of the assoicative algebra Φ[λ] if char Φ , 2. If char Φ = 2, Example (3) of §5 shows that 76 there exist ideals of Φ[λ](q) which are not ideals of Φ[λ]. Let K be an ideal , 0 in Φ[λ](q), f (λ) , 0 an element of K. Then g(λ) f ()2 = 2 g(λ)U f (λ) ∈ K. Hence K contains the ideal ( f (λ) ) of Φ[λ]. The sum of all such ideals is an ideal (m(λ)) of Φ[λ]. We may assume m(λ) monic. In particular, if a is an algebraic element of J then Ka contains a unique ideal (ma(λ)) of Φ[λ] maximal in Ka where ma(λ) is monic. We shall call ma(λ) the minimim polynomial of the algebraic element a. If ma(0) = 0 so ma(λ) = λh(λ) then h(λ) < (ma(λ)) so these exists ag(λ) ∈ Φ[λ] such that h(λ)g(λ) ∈ Ka. Hence h(a)g(a) , 0 and h(a)g(a)Ua = 0. Thus ma(0) = 0 implies that a is a zero divisor. On the other hand, suppose there exists a polynomial f (λ) such that f (a) = 0 and f (0) , 0. Then we have a relation 1 = g(a) where g(0) = 0. Then 1 = g(a)2 and 2 2 g(λ) = λ h(λ). Then 1 = h(a)Ua and a is invertible by the Theorem 54 1. Basic Concepts

on Inverses. Hence an algebraic element a is either a zero divisior or is invertible according as ma(0) = 0 or ma(0) , 0. It is easily seen also that if a is algebraic then Φ[a] is a quadratic Jordan division algebra if and only if Ka = (ma(λ)) where ma(λ) is irreducible. We have it to the reader to prove this. If J is strongly non-degenerate then Ka = (ma(λ)) for every algebraic element a. For, if g(λ) ∈ Ka and f (λ) ∈ Φ[λ] then U( f g)(a)Ug(a) = 0(QJ34). Hence ( f, g)(a) = 0 and f (λ)g(λ) ∈a. Thus Ka is an ideal of Φ[λ] and Ka = (ma(λ)) by deﬁnition of (ma(λ)).

11 Isotopes

77 This is an important notion in the Jordan theory which, like inverses, has an associative back ground. Let a be an associative algebra, c an invertible element a. Then we can define a new algebra a(c) which is the same Φ-module as a and which has the product xcy = xcy. We have (xcy)c = xcycz and xc(ycz) = (c) −1 = −1 = −1 = −1 = xcycz so a is associative. Also xcc xcc x and cc x c cx −1 x so c is unit for a(c). The mapping cR : x → xc is an isomorphism (c) of a onto a since (xcy)cR = xcyc = (xcR)(ycR). An element u is invertible in a if and only if, it is invertiable in a(c) since uv = 1 = vu −1 −1 −1 −1 −1 is equivalent to uc(c vc ) = c = (c vc )cu. If d is invertible in a (or a(c)) then we can form the algebra (a(c))(d). The product here is (c) (d) (cdc) xc,dy = xcdcy = xcdcy. Hence (a ) = a . In particular, if we − taked d = c−2 then we see that (a(c))c 2 = a. Finally, we consider the quadratic Jordan algebras a(q) and a(c)(q). The U-operator in the first is (c) Ua : x → axa and in the second it is Ua : x → ac xca = acxca. Hence (c) we have Ua = UcUa. The considerations lead to the definition and basic properties of isotopy for quadratic Jordan algebras. Let (J , U, 1) be a quadratic Jordan algebra c an invertible element of J . We define a mapping U(c) of J into End J by (c) Ua = UcUa (52)

78 and we put 1(c) = c−1. Evedently U(c) is a quadratic mapping we have 11. Isotopes 55

(c) = = U1(c) UcUc−1 1. Hence the axioms QJ1 and QJ2 for quadratic Jordan algebras hold. Also (c) = = U (c) UcUaUcUb UcUbUcUaUcUb aUb = (b) (c) (c) Ub Ua Ub (c) (c) = (c) Hence QJ3 holds. Next we deﬁne Va,b by xVa,b aUx,b where (c) (c) (c) (c) = = + = = Ua,b Ua+b − Ua − Ub UcUa b − UcUa UcUb UcUa,b. Then (c) = = (c) = xVa,b aUcUx,b xVaUc,b. Thus Va,b VaUcb. Now we have (c) (c) = (c) (c) = xVb,aUb bUx,aUb bUcUx,aUcUb = bUxUc,aUc Ub(bilinearization ofQJ3) = xUcUb,aUc Ub = xUcUbVaUc,b (QJ4) = (c) (c) xUb Va,b Hence QJ4 holds for (J , U(c), 1(c)). It is clear also that these properties carry over to Jρ for ρ any commutative associative algebra over Φ. Hence (J , U(c), 1(c)) is a quadratic Jordan algebra.

Deﬁnition 6. If c is an invertible element of (J , U, 1) then the quadratic 79 (c) (c) (c) (c) (c) −1 Jordan algebra J = (J , U , 1 ) where Ua = UcUa, 1 = c is called the c-isotope of (J , U, 1).

(c) It is clear from the formula Ua = UcUa and the fact that a is invertible in J if and only if Ua is invertible in End J that a is invertible in J if and only if it is invertible in the isotope J (c). If d is another invertible element then we can form the d-isotope (J (c))(d) of J (c). Its U operator is U(c)(d) where (c)(d) = (c) (c) = = Ua Ud Ua UcUdUcUa UcdcUa (cdc) = Ua Also we recall that cdc is invertible and 1(c)(d) = (cdc)−1 since (c) −1 = −1 = −1 = −1 −1 −1 = −1 d(Ud ) d(UcUd) dUd Uc−1 c d c (cdc) . Hence 56 1. Basic Concepts

(J (c))(d) = J (cdc). In this sense we have transitivity of the construction of isotopes. Also since J (1) = J , J is its own isotope. Finally, −2 −2 we have (J (c))(c ) = J (cc c) = J (1) = J so J is the c−2 -isotope of the c -isotope (J , c). In this sence the construction is symmetric. Unlike the situation in the associative case isotopic quadratic Jor- dan algebras need not be isomorphic. An important instance of isotopy whihc gives examples of isotopic, non-isomorphic algebras is obtained 80 as follows. Let (a, J) be an associative algebra with involution, h an invertible element of H (a, J). Then the mapping K : x → h−1 xJh is also an involution in a. We claim that the quadratic Jordan algebra H (a, K) is isomorphic to the h-isotope of H (a, J). Let x ∈ H (a, J) then xhR = xh ∈ H (a, K) since (xh)K = h−1(xh)J h = h−1(hx)h = xh. It follows that hR is a Φ -isomorphism of H (a, J) onto H (a, K). Moreover, if (h) = = = x, y ∈ H (a, J) then xUy hR (xUhUy)hR yhxhyh (xhR)UyhR . (h) Hence hR is an isomorphism of the quadratic Jordan algebra H (a, J) onto H (a, K). It is easy to give examples such that H (a, J) and H (a, K) are not isomorphic. This gives examples of H (a, J) which is not isomorphic to (h) the isotope to the isotope H (a, J) For example, let a = R2 the algebra of 2 × 2 matrices over the reals R, J the standard involution in R2. If R = 2 = 2 , a ∈ H ( 2) say a (ai j) then tr a ai j 0. Hence a is not nilpotent. −1 Let h =diag{1, −1} and consider theP involution K : x → h xh in R2. 2 1 −1 1 1 1 0 1 −1 Then = ∈ H (R2, K). Also = 0. 1 −1! 1 1! 0 −1! 1 −1! Hence H (R2, K) contains non-zero nilpotent elements. Thus H (R2) is not isomorphic to H (R2, K) and the latter is isomorphic to the isotope (h) H (R2) . It is convenient (as in the foregoing discussion) to extend the notion of isotopy to apply to different algebras and to define isotopic mappings. Accordingly we give 81 Definition 7. Let (J , U, 1), (J ′, U′, 1) be quadratic Jordan algebras. A mapping η of J into J ′ is called an isotopy if η is an isomorphism ′ of (J , U, 1) onto an isotope J ′(c ) of J ′. If such a mapping exists then (J , U, 1) and (J ′, U′, 1′) are called isotopic (or isotopes) 11. Isotopes 57

Using η = 1 we see that the isotope J (c) and J are isotopic in the sense of the present deﬁnition. Also it is clear that isomorphic algebras η = ′ −1 η = η ′ ′ = are isotopic. The deﬁnition gives 1 (c ) and (xUa) x Uaη (c ) η ′ ′ = ′ ′ x Uc′ Uaη or Uaη ηUc′ Uaη . Hence we have ′ = ∗ Uaη η Uaη (53)

∗ ′−1 −1 ′ −1 ′ where η = U ′ η U η is a module isomorphism of onto c (c′)−1 J J . Conversely, let η be a module isomorphism of J onto J ′ such that there exists a module isomorphism η∗ of J ′ onto J satisfying (53). Then (53) implies that a is invertible in J if and only if aη is ′ ′ = ∗ ∗ = ′ −1 = ′−1 −1 invertible on J . Also U1η η η so η U1η η Uc′ η where ′ η −1 η ′ ′ ′ c = (1 ) . Then U = ηU ′ U = ηU ′ and η is an isomorphism a c aη a(c ) ′ ′ ′ of (J , U, 1) onto (J ′, U′(c ), 1′(c ), 1′(c )) so η is an isotopy of J onto J ′. Hence we have shown that a module isomorphism η of J onto J ′ is an isotopy if and only if there exists a module isomorphism η∗ of J ′ onto J satisfying (53). It is cleat r that the isotopy η is an isomorphism of J onto J ′ if and only if η∗ = η−1 and 1η = 1′. The ∗ = ′ −1 latter condition implies the former since we have η U1η . Hence an 82 isotopy η is an isomorphism if and only if 1η = 1′. ′ −1 If η is an isotopy of J onto J and (53) holds then Ua′ η = ∗ −1 ′ −1 ′ (η ) Ua′ η−1 which shows that η is an isotopy of J onto J . If J ′ ′′ ′′ ′ ′ ′′ is an isotopy of onto and U = ζU ′ ζ, a ∈ , then U = J J a′ζ a J aηζ ∗ ∗ ′′ ∗ ∗ ∗ ζ η Uaηζ. Hence ηζ is an isotopy of J onto J and (ηζ) = ζ η . It is clear from this that isotopy is an equivalence relation. Since η−1 is an isotopy it is an isomorphism of J ′ onto an isotope of J . Hence η is also an isomorphism of an isotope of J onto J ′ (as well as of J on an isotope of J ′). The set of isotopies of J onto J is a group of transformations of J . Following Koecher, we call this the structure group of J and we denote it as Str J . Clearly Str J contians the group of automorphisms Aut J as a subgroup. Moreover, Aut J is the subgroup of η StrJ of such that 1 = 1. If c is invertible then Uc is a module iso- = morphism of J onto J and UaUc UcUaUc. Hence (53) holds for ∗ η = Uc, η = Uc so Uc ∈ Str J . It is clear from the foregoing discussion −2 −1 that Uc is an isomorphism of (J , U, 1) onto the c = (1Uc) isotope 58 1. Basic Concepts

(c−2) 2 2 (J , U , c ). In particular, if c = 1 then Uc is an automorphism of (J , U, 1). The subgroup of Str J generated by the Uc, c invertible is called the inner structure group. We denote this as Instr. J . If η ∈ Str −1 −1 −1 J , Uaη = U1ηη Uaη so η Uaη = (U1η) Uaη which implies that Instr J is a normal subgroup of Str J . It follows also that Aut J ∩ Instr 83 J is normal in Str J . We call this the group of inner automorphisms. ′ ∗ −1 We have seen that if η is an isotopy of J on J then η = (U1η ) −1 −1 ′ −1 ′ η . since (U1η ) ∈ Str J and η is an isotopy of J onto J . We see that η∗ is an isotopy of J ′ onto J . In particular, if η ∈Str J ∗ ∗ = ∗ ∗ ∗ = then η ∈ Str J . We have (ηζ) ζ η and Uc Uc for invertible ∗ = ∗ −1 −1 = −1 = ∗ c. Hence η (η ) U1η ηU1η U1n η Thus η → η is an anti- automorphism of Str J such that η∗∗ = η and this stabilzes Instr J .If −1 ∗ −1 η is an automorphism then Uaη = η Uaη so η = η . Let c be invertible and consider the isotope (kmathscrJ, c). Let = ∗ η (c) = = ∗ = η ∈Str J , so Uaη η Ua, a ∈ J . Then Uaη UcUaη Ucη Uaη ∗ −1 = ∗ −1 (c) (c) (Ucη Uc )UcUaη (Ucη Uc )Ua η. Hence η ∈Str J . By symmetry, Str J (c) =StrJ . Similarly, one sees that Instr J (c) =Instr J . If Z is an inner(outer) ideal in J then Z is an inner (outer) ideal (c) (c) = of the isotope J . For, if Z is inner and b ∈ Z then J Ub (c) J UcUb = J Ub ⊆ Z and if Z is outer and a ∈ J then Z Ua = ′ Z UcUa ⊆ Z . Since an isotopy of J into J maps an isotope of J onto J ′ it is clear that if Z is an inner (outer) ideal in J then Z η is an inner (outer) ideal of J ′. Chapter 2 Pierce Decompositions. Standard Quadratic Jordan Matrix Algebras

In this chapter we shall develop two of the main tools for the structure 84 theory: Peirce decomposition and the strong coordinatization Theorem. The corresponding discussion in the linear case is given in the author’s book [2], Chapter III.

1 Idempotents. Pierce decompositions

An element e of J = (J , U, 1) is idempotent if e2 = e. Then e3 = 2 2 2 2 n n−2 eUe = e Ue = (e ) (QJ23) = e = e. Then e = e Ue = e for n = = n ≧ 1, by induction. Also Ue Uen Ue. The idempotents e and f are said to be orthogonal (e⊥ f ) if e ◦ f = fUe = eU f = 0. If we apply QJ12 with a = e, c = f to 1 we see that e ◦ f = 0 and fUe = 0 imply eU f = 0. Hence e and f are orthogonal if e ◦ f = 0 and either eU f = 0 ar fUe = 0. If e and f are orthogonal then e + f is idempotent since (e + f )2 = e2 + e ◦ f + f 2 = e + f . If e, f, g are orthogonal idempotents (that is, e⊥ f, g and f ⊥g) then (e + f ) ◦ g = (e + f )Ug = 0 so (e + f )⊥g. It follows that if e, f, g, h are orthogonal idempotents then e + f and g + h are orthogonal idempotents. If e is idempotent then f = 1 − e

59 60 2. Pierce Decompositions. Standard Quadratic...

is idempotent since f 2 = (1 − e)2 = 1 = −e ◦ 1 + e = 1 − e. Also e ◦ f = e ◦ 1 − e ◦ e = 2e − 2e = 0 and fU◦ = (1 − e)Ue = e − e = 0 so e and f are orthogonal. 85 We recall that an endomorphism E of a module is called a projection if E is idempotent: E2 = E, and the projections E and F are orthogonal if EF = 0 = FE. If E is a projection and X is an endomorphism satisfying the Jordan conditions: EXE = EX + XE = 0 then it clear that EX = 0 = XE. We now prove

Lemma 1. If e and f are orthogonal idempotents in J then Ue,U f and Ue, f are orthogonal projections. If e, f, g are orthogonal idempotents then the projections Ue,Ue, f and U f,g are orthogonal. If e, f, g, h are orthogonal idempotents then Ue, f and Ug,h are orthogonal. 2 = 2 = Proof. We have seen that Ue Ue2 , so Ue Ue is a projection. We = = = = have UeUe, f Ue Ue2.eo f (QJ11) 0 and UeUe, f Ue UeUe, fUe (QJ3) 0. Hence UeUe, f = 0 = Ue, f Ue (1) + = 2 + = + + Also UeU f U f Ue Ue U f U f Ue2 −Ue, f UeUe, f UeUe,eU f = = = = 2 = UeUe, f (QJ7) 0 by (1), eU f 0 and eUe, f ee f e ◦ f 0. Since = = UeU f Ue U fUe 0 we have

UeU f = 0 = U f Ue. (2)

Now e + f is idempotent so Ue+ f = Uc + Ue, f + U f is idempotent. + 2 + = + + By (1) and (2) this gives Ue Ue, f U f Ue Ue, f U f . Hence 2 = Ue, f Ue, f (3)

86 Since e⊥ f, g, e⊥ f + g, so UeU f +g = 0 = U f +gUe. By (2) this gives

UeU f,g = 0 = U f,gUe. (4) = = + we have Ue+ f U f,gUe+ f U fUe+ f , gUe+ f 0 since g and e f are orthogonal. Since Ue+ f = Ue +U f +Ue, f this, (1) and (4) gives Ue, f U f,g = + + + = + Ue, f 0. Also Ue, f U f,g U f,gUe, f U f Ue,g Ue,gU f Ue◦ f f og U f 2,eog. (taking b = 1, a = f, c = e, d = g in QJ8)= 0. Hence

Ue, f U f,g = 0 = U f,gUe, f . (5) 1. Idempotents. Pierce decompositions 61

Finally, e⊥g, h and f ⊥g, h so e + f ⊥g, h. Hence, by (4), Ue, f Ug,h = (Ue+ f − Ue − U f )Ug,h = 0 and Ug,hUe, f = 0. A set of orthogonal idempotents {ei|i = 1, 2,..., n} will be called supplementery if ei = 1. Then this gives

P n = = + 1 U1 Uei Uei,e j . (6) X1 X1< j + The foregoing lemma shown that the n(n 1)/2 operators Uei , Uei,e j with distinct subscripts are orthogonal projections. Since they are supplementary in End J in the sense that their sum is 1 we have

= = = J Ji j, Jii J Uei , Ji j J Uei,e j , i < j (7) Xi≤ j M which we call the pierce decomposition of J relative to the ei. We shall = call Ji j the pierce (i, j)-component of J relative to the ei. Jii Uei is 87 an inner ideal called the Pierce inner ideal determined by the idempotent ei.

We shall now derive a list of formulas for the products ai jUbkl where ai jǫJi j, bklǫJkl. For this purpose we require

Lemma 2. If ai jǫJi j then = Uei Uaii Uei Uaii (8) = + + , Uaij Uei Uaij Ue j Ue j Uaij Uei Ueie j , Uaij Uei,e j , i j (9) = + + + Vaii Uei Uaii,ei Uei (Uei VaiiUe j Ue j Vaii Uei Xj,i

Uei,e j Vaii Uei,e j ) (10) = + , Uaii,cij Uei Uaii,cij Uei,e j Uei,e j Uaiicij Uei i j (11) = Proof. The ﬁrst is clear from QJ3 and ai jUei aii. The second follows by taking a = ei, b = ai j, c = e j in QJ7. For (1) we have = = + = Vaii U1,aii Uei,aii Ue jaii . Then Uei,aii Uei Uei,aii Uei by QJ3 j,i = +P + and Ue j,aii Uei Vaii Ue j Ue j Vaii Uei Uei,e j Vaii Uei,e j follows by putting a = ei, b = aii, c = e j in QJ15. Hence (10) holds. To obtain (11) 62 2. Pierce Decompositions. Standard Quadratic...

we belinearrize QJ6 relative to b to obtain UaUb.dUa,c + Ua,cUb,dUa = + = = = = UbUa,dua,c UdUa,bUa,c and put a ei, b ai j, c e j, d ci j in this.

88 To formulate the results on the products ai jUbkl of elements in Pierce components in a compact from we consider triples of unordered pairs of induced taken from {1, 2,..., n} : (pq, rs, uv). In any pair pq we allow p = q and we assume pq = qp. Also we identify (pq, rs, uv) = (uv, rs, pq). We shall call such a triple connected if it can be written as

(pq, qr, rs)

It is easily seen that the only triples which are not connected are those of one of the following two forms:

(pq, rs, −) with {p, q}∩{r, s} = (pq, qr, qs) with r , p, q, s.

We can now state the important

Pierce decomposition theorm. Let {ei|i = 1, 2,..., n} be a supplementary set of orthogonal idempotents, J = Ji j the corresponding Pierce decomposition of J . Let apqǫJpq etc.P Then for any connected triple (pq, qr, rs) we have

{apqbqrcrs}ǫJps PD 1 = = bqrUapq apqbqrapqǫJPS if pq rs PD 2 If (pq, rs, uv) is not connected then = = {apqbrscuv} 0 and brsUapq apqbrsapq PD 3 = 0 for pq = uv

89 Also

{apqbqrcrs} = (apq ◦ bqr) ◦ crs if(qr, pq, rs) is not connected PD 4 = , {apqbqrcrp} ((apq ◦ bqr) ◦ crp)Uep ifp r. 1. Idempotents. Pierce decompositions 63

If p , q then = = apqVep apq, apqVapp Ubpp apqVbpp Vapp Vbpp . PD 5

(In other words, if Vapp denotes the restriction of Vapp to Jpq then app → Vapp is a homomorphism of the quadratic Jordan algebra (q) (Jpp, U, ep) into (End Jpq) . Finally, we have = 2 , eqUapq apqUep , p q. PD 6

Proof. We prove ﬁrst PD1 − 3. The formulas in this set of {apqbrscuv} = with pq ur are obtained by bilinearization of brsUapq . Hence we may drop {apqbrscuv} for pq = uv. Then the only formula in PD1 − 3 involving just one index that we have to prove is biiUaii ǫJ . This is clear from (8). Next we consider the formulas in PD 1−3 which involve two distinct induces i, j. These are: 90

biiUaij ǫJ j j (12)

bi jUaij ǫJi j (13) {ai jb jicii}ǫJii (14)

{aiibiici j}ǫJi j (15)

{aiibi jc j j}ǫJi j (16) = = = = b jiUaii 0, bi jUaii 0, {aiib j jci j} 0, {aiib j jc j j} 0 (17)

(12) and (13) follow from (9), and the first two equations in (17) follow from (8). (14) and (15) and the third part of (17) follow from (11). To prove (16) and the last part of (17) we note first that Jii ◦Ji j ⊆ Ji j and Jii ◦J j j = 0 if i , j. The first of these is an immediate consequence of (10). Also (10) implies that Jii ◦J j j ⊆ Jii. By symmetry, Jii ◦J j j ⊆ J j j and since Jii ∩ J j j = 0 we have Jii ◦ J j j = 0. By QJ27, we have {aiibi jc j j} = −{bi jaiic j j} + (aii ◦ bi j) ◦ c j j = (aii ◦ bi j) ◦ c j j (third of (17)) ǫJi j. Also {aiib j jc j j} = −{b j jaiic j j} + (aii ◦ b j j) ◦ c j j = 0, by the first of (17) and Jii ◦ J j j = 0. Next we consider PD1 − 2 for three distinct indices i, j, k. The formulas we have to establish are

{ai jbi jc jk}ǫJik (18) 64 2. Pierce Decompositions. Standard Quadratic...

{ai jb j jc jk}ǫJik (19)

{ai jb jicik}ǫJik (20)

{ai jb jkcki}ǫJii (21)

91 To prove these we make the following observation. Let S and T be non-vacaous disjoint subsets of the index set {1, 2,..., n} and put es = ei, eT = e j. It follows easily, as before, that eS and eT are or- iǫS jǫT P P = + thogonal idempotents. Also J UeS J U ei ⊆ Uei J Uei ei iǫS i,i′′ǫS P P P and since the Uei and Uei,ei′ are orthogonal projections with sum UeS , J = ′ = ′ + Uei J Uei UeS and J Uei,e J Uei,e UeS . Hence J Uei i i iǫS i,i′ǫS P= P+ J Uei,ei ⊆ J UeS and we have the equality J UeS J Uei iǫS ′ = ′ = P J Uei,e Jii . Similary, we have UeS ,eT Ji j. We i,i′ǫS i i,i′ǫS iǫS P P jPǫT now consider the supplementary set of orthogonal idempotents {e j + , = + ek, el, l j, k}. Since aiiǫJ Uei , ai j, bi j, cik, ckiǫJ Uei,es , es e j ek and

b j j, c jk, b jkǫJ UeS (14)-(16) imply that the left hand sides of (18)-(21) are contained in Ji j + Jik, Ji j + Jik, Ji j + Jik, Jii respectively. Similarly, if we use the set of orthogonal idempotents {ei +e j, el, l , i, j} we see that the left hand sides of (18)-(20) are contained in Jik + Jik. Since (Ji j + Jik) ∩ (Jik + J jk) = Jik we obtain (18)-(20). Next 92 we consider the case of four distinct induces i, j, k, l. The only connected triple here is (i j, jk, kl). If we use the set of orthogonal idempotents {ei + e j + ek, em, m , i, j, k} as just indicated we obtain that {ai jb jkckl}ǫJil + J jl + Jkl. Similarly, using {e j + ek + el, em, m , j, k, l} we get that {ai jb jkckl}ǫJi j + Jk j + Jil. Taking the intersection of the right hand sides gives {ai jb jkckl}∈ Jil. Since a connected triple cannot have more than four distinct induces this concludes the proof of PD1 and PD2. We consider next the triples which are not connected. The ﬁrst possibility is (pq, rs, −) with {p, q}∩{r, s} = φ. Choose a subset S of the index set so that p, qǫS, r, s < S and put eS = ei, eT = e j. iǫS j

~~plied to the set of orthogonal idempotents {eS , eT } since apqǫJ UeS and , brsǫJ UeT . Finally suppose we have (pq, qr, qs) where r p, q, s. In 1. Idempotents. Pierce decompositions 65~~

this case we obtain {apqbqrcqr} = 0 by applying the second part of (17) ′ = to the two orthogonal idempotents er and er 1 − er. This proves PD3. To prove PD4 we note that the hypothesis that (qr, pq, rs) is not connected and PD3 imply that {bqrapqcrs} = 0. The ﬁrst part of PD4 follows from this and QJ27. For the second part of PD4 we note that {apqbqrcrp}ǫJpp by PD1 so {apqbqrcrp} = {apqbqrcrp}. By QJ27, {apqbqrcrp} = −{bqrapqcrp} + (apq ◦ bqr) ◦ crp. Since {bqrapqcrp}ǫJrr , and r p, applying Uep to the two sides of the foregoing equations , = = + gives PD4. If p q we have apqVep apqU1,ep apqUeq , ep = + + = apqUel, ep apqUep , eq apqUem,ep 2apqUep apq since apq l,q m,q,p P P since apqǫJpq and the Uei , Uei,e j are orthogonal projections. This is 93 the ﬁrst part of PD5. The second part follows directly from QJ21 and = = 2 = = apqUapp 0 apqUapp,bpp . To obtian PD6 we use apqUep 1Uapq Uep + + epUapq Uep eqUapq Uep eqUapq Uep (by PD 3). Since epUapq ∈ Jpp and qqUapq ∈ Jpp this reduced to eqUapq , which proves PD6. The formaulas PD4 4 imply some uesful associatively formulas for ◦. Suppose we have a connected triple (pq, qr, rs) such that (qr, pr, rs) and (pq, rs, qr) are note connected. Then we can apply PD also to {cpsbqrapq} to obtain.

~~(apq ◦ bqr) ◦ crs = apq ◦ (bqr ◦ crs) (22) If (qr.pr.rs) and (pq, rs, qr) are not connected. Special cases of this are~~

~~(aii ◦ ai j) ◦ a j j = aii ◦ (ai j ◦ a j j), i , j (23)~~

~~(ai j ◦ a j j) ◦ a jk = ai j ◦ (a j j ◦ a jk), i, j, k , (24)~~

~~(ai j ◦ a jk) ◦ akl = ai j ◦ (a jk ◦ akl, i, j, k, l , . (25) Similarly we have the following consequece of the second part of PD4:~~

~~= ◦ , ((apq ◦ bqr) ◦ crp)Uep (apq ◦ (bqr crp))Uep , p q, r (26)~~

We note also that the PD theorem permits us to deduce the following 94 formulas for the squaring composition and its bilinearization: 2 + Φ apqǫJpp Jqq, apq ◦ aqrǫJprif{p, q} {qr} (27) 66 2. Pierce Decompositions. Standard Quadratic...

~~apq ◦ ars = 0 if {p, q}∩{r, s} = ∅.~~

~~We leave this to the reader to check. It is easily veriﬁed also that~~

~~= = , Ji j{x|xVei x xVe j }, i j (28)~~

~~2 Standard quadratic Jordan matrix algebras~~

Let (O, j) be a (unital) non-associative algebra with involution, Oo a Φ- submodule of H (O, j) containing all the norms aa(a = a j).aǫO. Then Oo contains every ab + ba, a, bǫO hence all the tracis a + a. It follows Φ 1 = that, if contains 2 then Oo H (O, j).On the other hand, as the example of an octonian algebra with standard involution of a ﬁeld of characteristic two shows, we may have Oo ⊂ H (O, j). We consider the algebra On of n×n matrices with entries in O and the standard involution t J1 : A → A in On. Let H (On, Oo) be the set of matrices with entries t in O satisfying A = A and having diagonal entries in Oo. We use the notation we introduced in considering H (O3) ≡ H (O3Φ) and write

~~α[ii] = αeii, αǫOo (29)~~

~~a[i j] = aei j + aei j, aǫO, i , j.~~

~~95 Then it is clear that H (On, Oo) is the set of sums of the matrices α[ii] and a[ii]. Let Hii(Oo) = {α[ii]|αǫOo}, Hi j = {a[i j]|aǫO} for i , j. Then we have~~

~~H (On, o) = Hii(Oo) + Hi j (30) Xi Xi< j~~

~~2 and the sum is direct. Let A denote the usual square of the matrix AǫOn and put A ◦ B = AB + BA. Then we have the following formulas:~~

~~α[ii]2 = α2[ii] M1 a[i j]2 = aa[ii] + aa[ j j], i , j M2 2. Standard quadratic Jordan matrix algebras 67~~

~~α[ii] ◦ a[i j] = αa[i j], i , j M3 a[i j] ◦ b[ jk] = ab[ik], i, j, k , M4 α[ii] ◦ β[ j j] = 0,α[ii] ◦ a[ jk] = 0 a[i j]◦b[kl] = 0, i, j, k, l , M5~~

It is clear that these formulas together with a[i j] = a[ ji] determine 2 2 A for AǫH (On, Oo) and they show that A ǫH (On, Oo) if AǫH (On, Oo). Hence also A ◦ BǫH (On, Oo) if A, BǫH (On, Oo). Let VA in H (On, Oo) denote the endomorphism X → X ◦ A, and suppose from now on that n ≥ 3. Suppose we have the following identity in H (On, Oo):

~~[VA, VB◦C] + [VBVA◦C] + [VCVA◦B] = 0 (31)~~

~~If we write [A, B, C]o for the associator (A ◦ B) ◦C − A ◦ (B ◦C) then 96 (31) is the same as~~

~~′ [A, D, B ◦ C]o + [B, D, A ◦ C]o + [C, DA ◦ B]o = 0 (31 )~~

Assume ﬁrst n ≥ 4 and take A = a[i j], B = b[ jk], C = c[kl], D = 1[ll] where i, j, k, l ,. This gives [a, b, c][il] = 0 where [a, b, c] = (ab)c − a(bc) the associator in O. Hence [a, b, c] = 0, a, b, cǫO, so O must be associative if n ≥ 4 and (31) holds. Next let n = 3. Let αǫOo and take A = α[ii], B = b[ jk], C = l[kk], D = d[i j] in (25’) where i, j, k ,. This gives [α, d, b][ik] = 0 so

~~[α, a, b] = 0, αǫOo, a, bǫO (32)~~

Assume next that in addition to (25) we have the identity + = [VAVA◦B] [VBVA2 ] 0 (33) or 2 [A, D.A ◦ B]0 + [B, D, A ]o = 0 (33’) Taking A = a[i j], D = 1[kk], B = b[ jk], i, j, k ,, we obtain [a, a, b] [ jk] = 0 so [a, a, b] = 0, a, bǫO. Since Oo contians all the traces a + a we have by (33), [a + a, a, b] = 0 and, by the result just proved, 68 2. Pierce Decompositions. Standard Quadratic...

[a, a, b] = 0. Applying the involution we obtain [b, a, a] = 0. Thus O must be alternative. Then (33) implies that Oo ⊆ N(O) the nucleus of O. In particular, we see that we have the conditions on O we noted in 97 §1.8, namely, O is alternative and all norms xxǫN(O). We saw also that if these conditions hold then the Φ -module No spanned by the ′ norms and N = N(O) ∩ H (O, j) have the property that xNo x ⊆ No, ′ xN x ≡ No, xǫO. We note also that the two conditions (31) and (33) = are consequences of [VAVA2 ] 0 and the hypothesis that this carries over Oρ where ρ is commutative associative algebra over the base ring Φ. Our results give the following

Lemma . Let (O, j) be an algebra with involution over Φ, Oo a Φ- submodule of O containing all xx, xǫO. Let H (On, Oo) be the Φ- t module of matrices AǫOn such that A = A and the diagonal elements 2 of A are in Oo. IfAǫH (On, Oo) let A be usual square of A, A ◦ B = AB + BA. Assume n ≧ 3.Then the identities (31) and (33) in H (On, Oo) imply that O is associative if n ≧ 4 and O is alternative and Oo ⊆ N(O) if n = 3. We can now prove the following

Theorem 1. Let (O, j) be an algebra with involution, Oo a Φ -sub module of H (O. j) containing all the norms xx, xǫO. Let H (On, Oo) be t the set of n × n matrices with entries in O such that A = A and the diagonal elements are in Oo. Assume n ≥ 3. Then there exists at most one quadratic Jordan structure on H (On, Oo) satisfying the fol- 2 lowing conditions: 1UA = A the usual matrix square, the elements ei = 1[ii] = eii, i = 1, 2,..., n are a supplementary set a orthogonal 98 idempotents in H (On, Oo), the submodule Hi j = {a[i j]|aǫO}, i , j is the pierce (i, j) -module and Hii(Oo) = {α[i, i]|αǫOo} is the pierce (i, i)-module relative to the set {ei}. Necessary condition for the existence of such a structure are: O associative if n > 3, O alternative with Oo ⊆ NO if n = 3 and xOo x ⊆ O, xǫO. Proof. Suppose we have a quadratic mapping U so that (J , U, 1) is quadratic Jordan and the given conditions hold. Then we shall establish the following formulas for the U operator, in which i, j, k, l ,, α, βǫOo, a, b, ǫO: 2. Standard quadratic Jordan matrix algebras 69

~~QM 1 β[ii]Uα[ii] = (α(β)α[ii]~~

~~QM 2 α[ii]U[i j] = a(αa)[ j j]~~

~~QM 3 b[i j]Ua[i j] = a(ba)[i j]~~

~~QM 4 {α[ii]a[i j]b[ ji]} = ((αa)b + (αa)b)[ii]~~

~~QM 5 {α[ii]β[ii]a[ii]} = α(βa)[i j]~~

~~QM 6 {α[ii]a[i j]β[ j j]} = α(aβ)[i j]~~

~~QM 7 {α[ii]a[i j]b[ jk]} = α(ab)[ik]~~

~~QM 8 {a[i j]α[ j j]b[ jk]} = a(αb)[ik]~~

~~QM 9 {a[i j]b[ ji]c[ik]} = a(bc)[ik]~~

~~QM 10 {a[i j]b[ jk]c[ki]} = (a(bc) + a(bc))[ii]~~

~~QM 11 {a[i j]b[ jk]c[kl] = a(bc)[il].~~

The formulas QM4 − QM11 are immediate consequences of PD4 99 and M1−M4. To prove QM1 we note that β[ii]Uα[ii]ǫH (Oo) so this has = = = the form γ[ii], γǫOo. Then γ[i j] β[ii]Uα[ii]o1[i j] 1[i j]Vβ[ii]Uα[ii] 1[i j]Vα[ii]Vβ[ii](PD6) = (αβ)α[i j](M3) Hence QM1 holds. For QM2 we recall the identity

~~UbVa = Vb,aVb + VaVb − VbVa,b (QJ 18)~~

Let k , i, j. Then α[ii]Ua[i j] ◦ 1[ jk] = α[ii]Ua[i j]V1[ jk] = α[ii] Va[i j],1[ jk], Va[i j] + α[ii]V1[ jk]Ua[i j] − α[ii]Va[i j]V1[ jk],a[i j] = a(αa)[ jk] by PD4 and M1 − 4. Since α[ii]Ua[i j]ǫJ j j this proves QM2. To prove QM3 we again use QJ18 to write b[i j]Ua[i j] ◦1[ jk] = b[i j]Ua[i j]V1[ jk] = b[i j]Va[i j],1[ jk]Va[i j] since the other two terms given by QJ18 are 0 be PD3. Also the ﬁrst term is {1[ jk]b[i j]a[i j]} ◦ a[i j] = ((1[ jk] ◦ b[i j]) ◦ a[i j]) ◦ a[i j] = a(ba)[ik]. Hence QM3 holds. The PD theorem and the = argument just used shows that U is unique. Since the identity (VAVA2 ) 70 2. Pierce Decompositions. Standard Quadratic...

0 holds in H (On, Oo) and in extensions obtained by extending the ring Φ, the Lemma implies that is associative if n > 3 and alternative with O0 ⊆ N(O) if n = 3. It is clear from QM2 that xOo x ⊆ Oo. This completes the proof. The conditions for n > 3 given in this theorem are clearly sufficient since in this case On is associative and H (On, Oo) is a subalgebra of (q) On . Moreover, it is easy to see that the square 1UA in H (On, Oo) co- 100 incides with the usual A2 (cf. the proof of the Corollary to Theorem 1.5 (§1.8)) and the conditions on the ei1[ii] hold. The unique quadratic Jor- dan structure given in this case is that defined by BUA = ABA. We now consider the case n = 3. Suppose O is an alternative algebra such that all norms xxǫN(O). Let No be the submodule generated by all norms, N′ = H (O, j) ∩ N(O). Then we have shown in §1.8 that ′ ′ xNo x ⊆ No and xN x ⊆ N . Hence we can take these as choices for the submodule Oo. It is clear also that any Oo satisfying the condi- ′ tions of the theorem satisfies N ⊇ Oo ⊇ No. It can be verified by a ′ rather lengthy fairly direct calculation that H (O3, N ) with the usual 1 and the U operator defined by QM1 − 11 is a quadratic Jordan algebra. We omit the proof of this (due to McMrimmon). In the associative ′ ′ case N = H (O, j) ∩ N(O) = H (O, j) and H (On, N ) = H (On) the complete set of hermitian matrices with entries in O. Accordingly, we shall now define a standard quadratic Jordan matrix algebra to be any algebra of the form H (On), n = 1, 2, 3,..., to be any algebra of the form H (On), n = 1, 2, 3,..., where (O, j) is an associative algebra with ′ involution or an algebra H (O3, N ) where (O, j) is alternative with involution such that all norms xxǫN(O) and N′ = H (O, j) ∩ N(O). For ′ the sake of uniformity we abbreviate H (O3, N ) = H (O3). It is easily seen that if n ≧ 3 and No is the submodule of H (O, j) spanned by the norms then H (On, No) is the come of H (On).

101 Theorem 2. Let H (On), n ≧ 3, be a standard quadratic Jordan matrix algebra. A subset J of H (On) is an outer ideal containing 1 if and ′ only if J = H (On, Oo) where Oo is a Φ-submodule of N = H (O, j)∩ N(O) such that 1ǫOo and xOo x ⊆ Oo, xǫO. J = H (On, Oo) is simple if and only if (O, j) is simple. ′ Proof. Let Oo be a submodule of N containing 1 and every xαx, 3. Connectedness and strong connectedness of... 71

xǫO, αǫOo. Then Oo contains all the norms xx and all the traces x + x. It is therefore clear from QM1 − 11 (especially QM1, 2, 4, 10) that H (On, Oo) is an outer ideal. Since 1ǫOo, H (On, Oo) contains 1 = n 1[ii]. Conversely, let J be an outer ideal of H (On) containing 1. 1 P = = = Then J contains ei 1Uei , ei 1[ii] and every a[i j] {eieia[i j]}, aǫO, i , j. Also, if β[ii]ǫJ then β[ j j] = β[ii]U1[i j]ǫJ for j , i. = = If bǫJ and b bi j, bi jǫHi j, then bii bUei ǫJ . It is clear from i≦ j P ′ these results that J = H (On, Oo) where Oo is a submodule of N containing 1. Since α[ii]Ua[i j] = aαa[ j j]ǫJ if αǫOo, aǫO, i , j, it is clear that aOoa ⊆ Oo, aǫO. Let Z be an ideal in (O, j) and let k be the subset of J = H (On, Oo) of matrices all of whose entries are in Z . Then inspection of QM1 − QM11 shows that k is an ideal of J . Hence simplicity of J implies simplicity of (O, j). Conversely, suppose (O, j) is simple and let k be an ideal , 0 in J . If bǫk and = b bi j, bi jǫHi j, then operating on b with Uei or Uei,e j shows that ev- 102 i≦ j P ery bi jǫk. Let Z = {bǫO|b[12]ǫk}. We now use the formulas M1 − M5 for the squaring operator, which are consequences of QM1 − QM11. Since k is an ideal it follows from M4 that Z = {b|b[i j]ǫk, i , j} and Z is an ideal of (O, j). Also, by M3, if β[ii]ǫk then βǫZ . It is clear from these results that Z , 0 so Z = O, hence a[i j]ǫk for all aǫO, i , j. Now let αǫOo. Then, by QM2, α[ j j] = α[ii]U1[i j]ǫk if i , j. Hence k = J and J is simple. It is clear from the ﬁrst part of Theorem 2 that H (On.No), No the submodule generated by the norms is the core of H (On) (=outer ideal generated by:1).

~~3 Connectedness and strong connectedness of orthogonal idem-potents~~

In this section we shall give some lemmas on orthogonal idem-potents which will be used to prove the Strong Coordinatization Theorem (in the next section) and will play a role in the structure theory of chapter III. 72 2. Pierce Decompositions. Standard Quadratic...

Deﬁnition 1. If e1 and e2 are orhtogonal idempotents in J then e1 and e2 are connected (strongly connected) in J if the Pierce submod- = ule J12 J Ue1,e2 contains an element u12 which is invertible in = + 2 = + J Ue, e e1 e2 (satisﬁes u12 e1 e2). Then we say also that e1 and e2 are connected(strongly connected) by U12. 2 = 2 = Note that u 1 implies Uu 1 so u is invertible. Thus strong connectedness implies connectedness. Moreover, if e1 and e2 are strongly = 103 connected by u12 then u12 is its own inverse in J Ue. For, u12Uu12 = = = 2 u12Ue1,e2 Uu12 e1Vu12,e2 Uu12 e1Uu12 Ve2,U12 (QJ4) u12Ue2 Ve2,u12 = = { } = 2 ◦ = (PD6) e2Ve2,u12 e2e2u12 e2 u12 u12.

If U12 connects e1 and e2 then J11Uu12 ⊆ J22, J22Uu12 ⊆ J11, ′ J12Uu12 ⊆ J12 by PD2. Hence if U is the inverse of the restriction ′ = = ′ of Uu12 to Ue then J12U J12. Hence the inverse u21 u12U of u12 in J Ue is contained in J12. If f = 1 − e then f and e are orthogonal idempotents and since eiǫJ Ue, {e1, e2, f } is a supplementary set of orthogonal idempotents in J . It is clear also from the PD formulas that J Ue + J U f is a subalgebra of J , that Ue and U1 are ideals in this subalgebra and J Ue + J U f = J Ue ⊕ J U f . If follows that if xǫJ Ue, yǫJ U f then x + y is invertible in J Ue + J U f if and only if x is invertible in J Ue and y is invertible in J U f . In this case, the deﬁnition of invertibility shows that x + y is invertible in J . Also if x′ ′ and y respectively are the inverse of x and y in J Ue and J U f then ′ ′ x + y is the invese of x + y in J Ue + J U f and so in J . In particu- = lar, if u12ǫJ12 J Ue1,e2 is invertible in J Ue with inverse u21ǫJ12 then c12 = f + u12 is invertible in J with inverse c21 = f + u21. We have the Pierce decomposition J = J Ue ⊕ J Ue, f ⊕ J U f relative to {e, f } and we have the following usefull formulas for the action of = + + Uc12 Uu12 u f u f,u12 on these submodules:

xUc12 = xUu12, xǫJ Ue(PD3) 104 = yUC12 y ◦ u12, yǫJ Ue, f (PD3, 4, 5) (34) = zUc12 z, zǫJ U f (PD3) 2 = = 2 = If U12 e so u21 u12 then c12 1 and Uc12 is an automorphism of J such that U 2 = 1 (see §1.11). c12 3. Connectedness and strong connectedness of... 73

Lemma 1. Let e1, e2, e3 be pairwise orthogonal idempotents in J such that e1 and e2 are connected (strongly connected) by u12 and e2 and e3 are connected (strongly connected) by u23. Then e1 and e3 are connected (strongly connected) by u13 = u12 ◦ u23 In the strongly connected case c12 = u12 + 1 − e1 − e2, c23 = u23 + 1 − e2 − e3, c13 = u13 + 1 − e1 − e3 2 = 2 = satisfy ci j 1,Ucij is an automorphism such that Ucij 1 and we have

~~= = c13 c12Uc23 c23Uc12 (35) = = Uc13 Uc23 Uc12 Uc23 Uc12 Uc23 Uc12 so~~

Proof. Put e4 = 1 − e1 − e2 − e3 so {ei|i = 1, 2, 3, 4} is a supplementary set of orthogonal idempotents. Let J = Ji j be the corresponding = + + = + + Pierce decomposition. Then c12 u12 e3 Pe4, c23 u23 e1 e4. Since u12 is invertible in J11 + J12 + J22 with inverse u21, c12 is invertible in J with inverse c21 = u21 + e3 + e4. Similarly, c23 is invertible with = + + = inverse c32 u32 e1 e4. By the Theorem on inverse c13 c23Uc12 is 105 = invertible with inverse c31 c32Uc21 . We have

= = + + c13 c23Uc12 u23uc12 e1Uc12 e4Uc12 = + 2 + + u13 u12 e1Uu12 e4 (by(34)) = + 2 + u13 u12Ue2 e4(PD6). = + + 2 2 Similarly, c31 u31 e1Uu21 e4. Since u12Ue2 and u21Ue2 ǫJ22 it follows that u13 is invertible in J11 + J13 + J31 with inverse u31. = 2 = + Hence e1 and e3 are connected by u13 u12 ◦ u23. If u12 e1 e2 2 = + 2 = 2 = and u23 e2 e3 then c12 1 and c23 1. Then Uc12 , Uc23 are au- = 2 = tomorphisms with square 1. Also c13 c23Uc12 satisﬁes c13 1 so 2 = Uc13 is an automorphism and Uc13 1. The formula for c13 now be- = + + = comes c13 u23 ◦ u12 e2 e4. A similar calculation gives c12Uc23 + + = u12 ◦ u23 c2 c4. Hence c12Uc23 c23Uc12 so (35) and its consequence = Uc23 Uc12 Uc23 Uc12 Uc23 Uc12 hold. The following lemma is of technical importance since it permits the reduction of considerations on connected idempotents to strongly connected idempotents. 74 2. Pierce Decompositions. Standard Quadratic...

Lemma 2. Let {eii = 1,..., n} be a supplementary set of orthogonal idempotents in J , J = Ji j the corresponding Pierce decompostion. Assume e1 and e j, j >P1, are connected by u1 j with inverse u ji in J11 + J1 j + J j j. Then = 2 = 2 u j ui jUe j and v j u j1Ue j , j > 1 (36)

n n 106 are inverses in J j j and if we put u1 = e1 = v1 then u = ui, v = vi 1 1 are inverses in J . The set {ui} is a supplementry setP of orthogonalP (v) idempotents in the v-isotope J = J and u j is strongly connected by u1 j to u1 in J . The Pierce submodule Ji j of J relative to the ui coincides with . Moreover, = are algebras and and Ji j J11 J11 f f J j j , j > 1, are isotpic. Also, if j > 1, x ǫ , x ǫ , x V = J j j f11 J11 1 j J1 j 1 j x11 x V where V is the V-operator in . 1fj x11 J 2 2 + f + Proof.e It is cleare that u1 j, u j1ǫJ11 J j j and these are inverses in J11 = 2 = 2 J j j. It follows that u j u1 jUe j and v1 u jlUe j are inverses in J j j n n and u = ui, v = vi(u1 = e1 = v1) are inverses in J . Now consider 1 1 P = P (v) (v) = = the isotope J J with unit element u. We have Uui UvUui + j Uvi Uui Uv j,vk Uui . It is clear from the PD. Theorem (PD1 − 3) jf

~~Similarly, replacing ei by ei + e j, ui by ui + u j, vi by vi + v j, i , 1. (v) = + We obtain Uui+u j uei e j . This and (37) imply~~

~~(v) = , Uui,u j Uei,e j , i j (38)~~

n (v) (v) (v) 107 = = = = Now uUui uk Uei ui, uUui,u j ukUei,e j 0 and uiUU 1 ! k j = ,P P uiUe j 0 if i j. These shows that the ui are orthogonal idempotents 3. Connectedness and strong connectedness of... 75 in the isotope J = J (v). Since their sum is u they are supplementary. Then (37) and (38) show that = , = for the correspond- f Jii Jii Ji j Ji j ing Pierce submodules. Since u ǫJ , u ǫJ . More over, uU(v) = f1 j i j fj 1 j u1 j = = + = + = 2 + 2 uUvUulj vUulj (e1 v j)Uu1 j e1Uu1 j v1Uu1 j u1 jUe j u j1Ue j Uu1 j = + = + (PD 6 and ((36))) u j e1Uu jl Uulj u j e1. Hence u1 j strongly connects e1 and ui in J . Let xi, yiǫJii = Jii, xǫJl j = Jl j, j > 1. Then x U(v) = x U U = x U U . If i = 1, v = e so x U(v) = x U . Thus i yi i v yi fi vi yi 1f 1 1 y1f i y1 Jii and Jii are isotopic J11 and J11 are identical as algebras. Finally, n = (v) ======xVx1 uUfx,x1 uUvUx,x1 vUx,x1f {xvx1} {xe1 x1} {x ei x1} 1 P xVex1 . This completes the proof.

Lemma . Let {ei| = i = 1, 2,..., n} be a supplementary set os orthogonal idempotents in J such that e1 is strongly connected to e j, j > 1, by u1 j. = + = Put c1 j u1 j 1 − e1 − e j as above and U(1 j) Uc1 j .Then there exists a unique isomorphism π → Uπ of the symmetric group S n into Aut J ′ ′ such that (1 j) → U(i j). Moreover, eiUπ = eiπ and if iπ = iπ , jπ = jπ then Uπ = Uπ′ on Ji j (i = j allowed).

Proof. We have seen that U(1 j) is an automorphism of period two. Now it is known that the symmetric group S n is generated by the transpositions (1 j) and that the deﬁning realations for there is (1 j)2 = 1, ((1 j) (1k)3 = 1, ((1 j)(1k)(1 j)1l))2 = 1, j, k, l ,. By lemma 1, we have 3 U(1 j)U(1k)U(1 j) = U(1k)U(1 j)U(1k). Hence (U(1 j)U(1k)) = U(1 j)U(1k) 108 U(1 j)U(1k)U(1 j)U(1k) = U(1 j)U(1k)U(l j)U(1 j)U(1k)U(1 j) = 1. Also, if j, k, , = l , then U(1 j)U(1k)U(1 j) U(1l)U(1 j) U(1 j)U(1k)U(1 j)U(1l) Uc1 Uc Uc l 1 j 1 j = = Uc1 Uc1 Uc jk Uc1l where c jk clkUc1 j . The form of c jk derived in 1 1 = = Lemma 1 and (34) imply that c1lUc jk c1l. Hence Uc1l Uc jk Uc1l 1. These relations imply that we have a unique monpmorphism of S n into = = = Aut J such that (1 j) → U(1 j) Uc1 j . By (34) e1U(1 j) e1Uc1 j = 2 = = e1UU1 j u1 jUe j (PD6) e j. Hence eiUπ eiπ for πǫS n. Now sup- ′ ′ ′ ′ ′′ ′ −1 pose π and π satisfy eiπ = eiπ , e jπ = e jπ . Put π = π π . Then ′′ ′′ ′′ eiπ = ei, e jπ = e j so π is a product of transpositions which ﬁx i and j. If (kl) is such a transposition then U(kl) = U(1l)U(1k)U(1l). By (34) this ′′ ′′ acts as identity on Ji j. Hence π is 1 on Ji j and π = π on Ji j. 76 2. Pierce Decompositions. Standard Quadratic...

~~4 Strong coordinatization theorem~~

We shall now obtain an important characterization of the quadratic Jor- dan algebras H (On, Oo), n ≧ 3, or equivalently, in view of Theorem 2, of the outer ideals containing 1 in standard quadratic Jordan algebras H (On), n ≧ 3. We shall call a triple (O, j, Oo) a coordinate algebra, if (O, j) is an alternative algebra with involution and Oo is a Φ -submodule ′ of N (O) = H (O, j) ∩ N(O) suchthat 1ǫOo and xOo x ⊆ Oo, xǫO. We call (O, j, Oo) associative if O is associative. We shall show that J = H (On, Oo), n ≧ 3, is characterised by the following two conditions: (1) J contains n ≧ 3 supplementary strongly connected orthog- 109 onal idempotents, (2) J is non-degenerate in the sense that ker U = 0. Consider an H (On, Oo), n ≧ 3. Let ei = 1[ii], u1 j = 1[1 j], j > 1 (no- n tation as in §2). Then the ei are orthogonal idempotents and ei = 1. 1 Also u1 j is in the Pierce (i, j) component of H (On, Oo) relativeP to the ei 2 = + and u1 j e1 e j so e1 and e j are strongly connected by u1 j. Hence (1) holds. To prove (2) we recall that ker U is an ideal (§1.5) so if = = = , z zi j[i j]ǫ ker U then zii[ii] zUei ǫ ker U and zi j[i j] zUei,e j , i j, 1≦ j ker UP. We recall also that all products involving an element of ker U are 0 except those of the form zUa, z ∈ ker U. Hence zi j[i j] = 0 and zii[i j] = zii[ii] ◦ 1[i j] = 0. Then zii = 0 and z = 0. Thus H (On, Oo) is non-degenerate. We shall now prove that the conditions (1) and (2) are suﬃcient for a quadratic Jordan algebra to be isomorphic to an algebra H (On.Oo), n ≧ 3. We have the following Strong coordinatization Theorm. Let J be a quadratic Jordan algebra satisfying:(1)J is non-degenerate, (2) J contains n ≧ 3 supplementary strongly connected orthogonal idempotents. Then there exists a coordinate algebra (O, jOo) which is associative if n ≧ 4 such that J is isomorphic to H (On, Oo). More precisely, let {ei|i = 1,..., n} be a supplementary set of strongly connected orthogonal idempotents and let e1 be strongly connected to e j, j > 1, by u1 j. Then there an iso- η = = morphism η of J onto H (On, Oo) such that ei 1[ii], i 1,..., n,, η = = u1 j 1[1 j], j 2,..., n. 4. Strong coordinatization theorem 77

Proof. Put ci j = ui j + 1 − e1 − e j. By lemma 3 of §3 we have an iso- 110 = = morphism π → Uπ of S n into Aut J such that U(i j) Uc1 j , eiUπ eiπ. Then for the Pierce module Jpq (relative to the ei) we have JpqUπ = ′ ′ ′ JpqUpπ,qπ. Also if π,π ǫS n satisfy pπ = pπ , qπ = qπ then Uπ and Uπ′ have the same restrictions to Jpq. This implies that if pπ = p and = = qπ q then Uπ is the identity on Jpq. ByLemma1of §3, c jk c1 jUc1k , , = = = + for i, j, k satisﬁes c jk ck j c1kUc1 j u jk 1 − e j − ek where = ◦ 2 = + = = u jk u1 j u1k also u jk e j ek, U( jk) U(1k)U(1 j)U(1k) Uc jk . By (34), , = = 2 = if i, j, k then xi jU( jk) xi j ◦u jk. Hence (xi j ◦u jk)◦u jk xi jU( jk) xi j. In particular, if 1, j, k , then u jk ◦ u1k = (u1 j ◦ ulk) ◦ ulk = u1 j. We note also that −1 = Uπ Uei Uπ Ueiπ (39) = = = = since for π (1 j) we have U(1 j)Uei U(1 j) Uc1 j Uei Ue1 j Uei Uc1 j = −1 ′ = −1 −1 ′ = Uei U1 j Uei(1 j) and Uππ′ Uei Uππ Uπ′ Uπ Uei UπUπ . Let O J12 and deﬁne for x, y xy = xU(23) ◦ yU(13) (40)

~~Since xU(23)ǫJ13 and yU(13)ǫJ23, xyǫO = J12. Also the product is Φ -bilinear. Deﬁne for xǫO.~~

~~j : x → x = xU(12) (41) 2 = Φ Since U(12) maps J12 into itself and U(12) 1, j is a -isomorphism 111 2 of J12 such that j = 1. Also if x, yǫO,~~

~~xy = (xU(23) ◦ yU(13))U(12)~~

~~= xU(23)U(12) ◦ yU(13)U(12)~~

~~= yU(12)U(23) ◦ xU(12)U(13) = yx U xu12 = xU(23) ◦ 12 (13)~~

~~= xU(23) ◦ U23 = 2 xU23 = x = 3 = u12 u12 u12 (see§3) 78 2. Pierce Decompositions. Standard Quadratic...~~

These relations imply that u12 acts as the unit element of O relative to its product and j is an involution in O. 2 We now deﬁne n “coordinate mappings” ηpq, p, q = 1, 2,..., n of J into O as follows: = , = = ηi j Uei,e j Uπ if i j, iπ 1, jπ 2 (42) = = ηii Uei UπVc12 if iπ 1 (43)

112 It is clear from the preliminary remark that this is independent of the choice of π. Also ηpq = 1 on all the Pierce submodules except Jpq. If i , j, Uπ is a Φ isomorphism of Ji j onto O = J12. Hence ηi j is a Φ = -isomorphism of Ji j onto O. Since JiiUei Uπ J11 it is clear also Φ ≡ η11 = that ηii is a homomorphism of Jii onto Oo J11 J11Vc12 . We now prove for xǫJ xηpq = xηpq . (44) ′ , ηij = = ′ = If i j, we have x xUei,e j UπU(12) xUe j,ei Uπ , where iπ ′ 2, jπ = 1. Hence xηij = xη ji . To prove xηii = xηii we require = = Vc12 Uc12 Vc12 Uc12 Vc12 (45) = 2 = We have Vc12 Uc12 U 2 UC12 Vc12 by QJ24. Since c 1 and c12,c12 12 = ηii = = = ηii U1,c12 Vc12 we have (45). Now x xUei UVc12 xUei UVc12 X . Deﬁne the mapping η of J onto H (On, Oo) by

~~x = xηpq [pq]. (46) Xp≦q~~

This is a Φ-homomorphism of J onto H (On, Oo) since, if xǫJpq, ηpq η x = x [pq]. It is clear that Jpq = Hpq where these are deﬁned η = = = = as usual. We have ei eiUei UπVc12 [ii] e1Vc12 [ii] e1Vu12 [ii] = = η = η = U12[ii] 1[ii](U12 1 in O). Hence 1 1 also. Also u12 n12[12] , η = = = = 113 and if 1, 2, j then ui j u1 jU(2 j)[i j] (U1 j (u1 j ◦ u2 j)[1 j] u12[1 j] = 1[1 j] (since u1 j ◦ u2 j = u12 was shown in the ﬁrst paragraph). 2 We now consider J relative to the squaring operation (a = 1Ua) and we shall prove for i, j, k ,, xi jǫJi j etc: η = η η (xi j ◦ y jk) xi j ◦ y jk (47) 4. Strong coordinatization theorem 79

~~η = η η (xii ◦ yi j) xii ◦ yi j (48) 2 η = η 2 (xii) (xii) (49) 2 η = η 2 (xi j) (xi j) (50)~~

~~For (47) let π be a permutation such that iπ = 1, jπ = 3, kπ = 2 and put xi jUπ = xǫJ13, y jkUπ = yǫJ23. Then~~

~~η ◦ η = ◦ xi j y jk xU(23)[i j] yU(13)[ jk]~~

~~(xU(23))(yU(13))[ik] = 2 2 (xU(23) ◦ yU(13))[ik] = (x ◦ y)[ik]~~

~~= (xi j ◦ y jk)Uπ[ik] η = (xi j ◦ y jk) .~~

For (48) let π be a permutation such that iπ = 1, jπ = 2. Put xiiUπ = 114 xǫJ11, yi jUπ = yǫJ12. Then = xii ◦ yi j xVc12 [ii] ◦ y[i j] = ((xVc12 )y)[i j] = (xVc12 U(23) ◦ yU(13))[i j]. = = Since xǫJ11, xU(23) x (see ﬁrst paragraph) and xVc12 x ◦ u12. = = = = Hence xVc12 U(23) (x ◦ u12)U(23) x ◦ u12U(23) x ◦ (u12 ◦ u23) = = = x ◦ u13 xVc13 xVc13 Uc13 (cf. (45)) (x ◦ u13)Uc13 . Then xVc12 U23 ◦ yU(13) = ((x ◦ u13) ◦ y)U(13) = ((x ◦ y) ◦ u13)U13 = x ◦ y = (xii ◦ yii)Uπ. η Hence xii ◦ yii j = (xii ◦ yi j)Uπ[i j] = (xii ◦ yi j) . For (49) choose π so that iπ = 1 and put x = xiiUπǫJ11. Then

~~2 = 2 = 2 (xi j) (xVc12 [ii]) (xVc12 ) [ii] = (xVc12 U(23) ◦ Vc12 U(13)[ii] = (xVc13 ◦ xVc12 )U(13)[ii] (proof of(48)) = ((x ◦ u13) ◦ (x ◦ u12))U(13)[ii]~~

~~= ((x ◦ (x ◦ U12)) ◦ u13)U(13)[ii] 80 2. Pierce Decompositions. Standard Quadratic...~~

~~= 2 2 (x ◦ u12)U(13)[ii] 2 = (x ◦ u12)[ii] = 2 xiiUπVc12 [ii] = 2 η (xii) .~~

~~115 For (50) let π satisfy iπ = 1, jπ = 2 and put x = xi jUπǫJ12. Then~~

~~2 2 2 (xi j) = xi jUπ[i j] = (x[i j]) = xx[ii] + xx[ j j].~~

Now xx = xU(23) ◦ xU(12)U(13) = xU(23) ◦ xU(13)U(23) = (x ◦ = = 2 = 2 = xU(13))U(23) (x◦(x◦u13))U(23) (x ◦u13)U(23) (x Ue1 ◦u13)U(23) 2 ◦ = 2 = 2 = 2 x Ue1 u12 x Ue1 Vc12 xi jUπUe1 Vc12 . Similary, xx xi jUπUe2 Vc12 . Hence 2 = 2 + 2 (xi j) (xi jUπUe1 Ue12 )[ii] (xi jUUe2 Vc12 )[ j j]. 2 = 2 + 2 2 η = 2 On the other hand, xi j xi jUei xi jUe j so (xi j) (xi jUei UπVc12 ) + 2 2 = 2 [ii] (xi jUe j UπU(12)Vc12 Vc12 )[ j j] and x1Uei UπVc12 xi j Uπ Ue1 Vc12 , 2 = 2 = 2 xi jUe j Uπ U(12)Vc12 xi jUπUe2 Uc12 Vc12 xi jUπUe2 Vc12 . Hence (50) holds. It is clear from these formulas that (x2)η = (xη)2 for xǫJ and H (On, Oo) is closed under squaring. We now introduce a 0-operator in H (On, Oo) by QM1 − 11 and the formulas giving 0. Clearly AUBǫOn but it is not immediately clear that AUgǫH (On, Oo). We claim that η η η 116 xU ) = x Uyη , x, y, ǫJ . It is sufficient to prove this and {xyz} = {xηyηzη} for x, y, z in Pierce components. We note first that since η maps η = η η = η Jpq into Hpq we have (xUei ) x U1[ii], (xUei,e j ) x U1[ii],1[ j j]. From this it follows that we can carry over the proof of theorem 1. For η the relation corresponding to QM4 we have {xiiyz} = (((xii ◦ yi j) ◦ η = ◦ ◦ { η η η } = ◦ ◦ zi j)Uei ) ((xii zi j) zi j)U1[ii] and xiiyi jzi j ((xii yi j) zi j)U1[ii] by η = η η η the formulas in H (On, Oo). Hence {xiiyi jzi j} {xiiyi jzi j} holds. The formulas corresponding to QM5 − 11 are obtained in a similar manner. η = , For QM2 we note that u jk 1[ jk], j k and V1[ jk] is injective on Hi j if , ffi η ◦ η = η η ◦ η i, j, k . Hence it su ces to prove (xiiUyij ) u jk xiiUyij u jk. Now η ◦ η = η = η (xiiUyij ) u jk (xiiUyij Vu jk ) (xiiVyij,u jk Vyij ) as in the proof of QM2 4. Strong coordinatization theorem 81

= { } ◦ η = { η η η } ◦ η = η η ◦ η in Theorem 1) ( xiiyi ju jk yi j) xiiyi ju jk yi j xiiUyij u jk by the formulas in H (On, Oo). A similar argument applies to QM1, 3. η η η Hence (xUy) = x Uyη which with J = H (On, Oo) implies that AUBǫH (On.Oo) for A, BǫH (On.Oo). It follows that (H (On, Oo), U, 1) is Jordan and η is a homomorphism. It now follows from Theorem 1 that O is associative if n ≧ 4 and alternative with Oo ⊆ N(O) if n = 3. Also xOo x ⊆ Oo if xǫO. It remains to prove that η is an isomorphism. Let K = kerη. Since K = is an ideal, K = Ki jK = K ∩ Ji j also since ηi j is injective if = = ffi i , j we have Ki j 0,P if i , j. Hence K Kii and it su ces to show = = Kii 0, i 1,..., n. Let zǫKii and considerP the products {xyz}, x, y in 117 Pierce modules Jpq. Such a product is 0 since Kpq = K ∩ Jpq = 0 Unless x, yJi j, i , j or x, y, ǫJii. In the first case we note that x ◦ z = 0 since Ki j = 0 so {xyz} = 0 by PD4. In the second case we write = , x wUuij (ui j as above), wǫJ j j where j i. This can be done since = = = = J j jUuij J j jU(i j) Jii. Then {xyz} wUuij Vy,z 0 by QJ9, the PD relations and Ki j = 0. Our result implies that {xyz} = 0 for all x, yǫJ so Ux,z = 0 for all x. We show that Uz = 0. For this it suffices to show = = , that xUz 0 if xǫJii or wUuij uz 0 for wǫJ j j, i j. This follows from QJ17 and the PD relations. We have now shown that Kii ⊆ ker U. Hence Kii = 0 and the proof is complete.

Remarks . The hypothesis that J is non-degenerate is used only at the last stage of the proof. If this is dropped the argument shows that K = ker η ⊆ ker U. The converse inequality holds since (ker U)η is η contained in the radical of U in J = H (On, Oo). Since H (On, Oo) is non-degenerate we have (radU)η = 0 so ker U ≡ K and K = ker U in any case. This gives a characterization of the algebras satisfying the ﬁrst hypothesis of the S.C.T. IF J has no two torsion ker u = 0 so in this case we can drop the second hypothesis and obtain the conclusion of S.C.T. The S.C.T can be strenghtened to give a coordinatization Theorem in which the second hypothesis is replaced by the weaker one that the ei are connected. In this case we get an isomorphism onto a “canonical matrix algebra” (cf. Jacobson [3], p.137).

~~Chapter 3 Structure Theory~~

In this chapter we shall develop the structure theory of quadratic Jor- 118 dan algebras which is analogous to and is intimately connected with the structure with the structure theory of semi-simple Artinion associative rings. We consider first the theory of the radical of a quadratic Jordan algebra which was given recently by McCrimmon in [5]. McCrimmon’s definition of the radical is analogous to that of the Jacobson radical in the associative case and is an important new notion even for Jordan algebras. We shall determine the structure of the quadratic Jordan algebras which are semi-simple (that is have 0 radical) and satisfy the minimum condition for principal inner ideals. Such an algebra is a direct sum of simple ones satisfying the same minimum condition. The simple quadratic Jordan algebras satisfying the minimum condition are either division algebras, outer ideals containing 1 in algebras H (a, J) where (a, J) is simple Artinian with involution, outer ideals containing 1 in quadratic Jordan algebras of quadratic forms with base points (§1.7), or certain isotopes H (O3, Jc) of algebras H (O3), O an actonion algebra over a field (§1.8, 1.9). The only algebras in this list which are of capacity ≧ 4 (definition in §6) are outer ideals cintaining 1 in H (a, J), (a, J) simple Artinian with involution. In this sense the general case in the classification of simple quadratic Jordan algebras satisfying the minimum condition is constituted by the algebras defined by the (a, J) we 119 determined in Chapter0.

~~83 84 3. Structure Theory~~

~~These results reduce in the case of (linear) Jordan algebras to those given in Chapter IV of the author’s book [2].~~

~~1 The radical of a quadratic Jordan algebra~~

~~As in the associative ring theory the notation of the radical will be based on the Quasi-invertiblity which we deﬁne as follows:~~

Deﬁnition 1. An element z of a quadratic Jordan algebra J is called quasi-invertible if 1 − z is invertible. If the inverse of 1 − z is denoted as 1 − w then w is called the quasi-inverse of z. The condition that z be quasi-invertible with quasi-inverse w are

~~2 (1 − w)U1−z = 1 − z, (1 − w) U1−z = 1. (1)~~

~~2 Since U1−z = 1+Uz−Vz the conditions are 1+z −2z−w−wUz+w◦z = 2 2 2 2 1 − z, 1 + z − 2z − 2w − 2wUz + 2w ◦ z + w + w Uz − w ◦ z = 1. These reduce to~~

~~2 w + z − z − w ◦ z + wUz = 0 (2) 2 2 2 2 2w − z + 2wUz − w Uz + 2z − 2w ◦ z − w + w ◦ z = 0. (3)~~

~~The quasi-inverse of z is~~

~~= 2 −1 w (z − z)U1−z (4) −1 = −1 = 2 −1 2 −1 = 2 −1 since (1−z) (1−z)U1−z (1−z) U1−z−(z −z)U1−z 1−(z −z)U1−z.~~

~~120 Also (1 − z) ◦ (1 − w) = 2 which gives~~

~~z ◦ w = 2(z + w) (5)~~

~~Then 1 − z = (1 − w)U1−z = 1 − w + (1 − w)Uz − (1 − w) ◦ z = 1 − w + (1 − w)Uz − 2z + 2z + 2w = 1 + w + (1 − w)Uz. Thus~~

~~w + z + (1 − w)Uz = 0 (6)~~

~~An immediate consequence of this is 1. The radical of a quadratic Jordan algebra 85~~

~~Lemma 1. If Z is an inner ideal and zǫZ is quasi-invertible then the quasi-inverse w of z is in Z. We prove next Lemma 2. If zn is quasi-invertible then z is quasi-invertible.~~

n n−1 Proof. We have 1 − λ = (1 − λ)(1 + λ + ... + λ ). Hence U1−zn = n−1 n U1−zUy = UyU1−z where y = 1 + z + ... + z (QJ37). Since z is quasi-invertible, U1−zn is invertible. Hence 1 − z is invertible and z is quasi-invertible.

Remarks. The argument shows also that −(z + z2 + ... + zn−1) is quasi- invertibel. The converse of lemma 2 is false since −1 is quasi-invertible in any (linear) Jordan algebra. On the other hand, 1 = (−1)2 is not quasi-invertible. An immediate corollary of Lemma 2 is: Any nilpontent element is quasi-invertible. By a nilpotent z we mean an element such that zn = 0 for some n. Then zn = 0 for all m > 2n. Since no idempotent , 1 is 121 invertibel and c idempotent implies 1 − e idempotent it is clear that no idempotent , 0 is quasi-invertible.

Deﬁnition 2. An ideal (inner ideal, outer ideal) Z is calledquasi - invertible if very zǫZ is quasi-invertible Z is called nil if every zǫZ is nilpotent. The foregoing result shows that if Z is nil then Z is Quasi-invertible.

~~Lemma 3. If Z is a quasi-invertible ideal and uǫJ is invertible then u − z is invertible for every z.~~

2 −1 Proof. u − z is invertible if and only if (u − z) Uu is invertible. We have 2 −1 = 2 + 2 −1 = = 2 −1 (u − z) Uu (u − u ◦ z z )Uu 1 − w where w (u ◦ z − z )Uu ǫZ. Then w is quasi-invertible and u − z is invertible.

Lemma 4. If Z and L are quasi-invertible ideals then Z + c is a quasi- invertible ideal. Proof. Let xǫZ, yǫc. Then 1−(x+y) = (1− x)−y is invcertible by lemma 3 since 1 − x is invertible and yǫc. Hence x + y is quasi-invertible. Thus every element of Z + c is quasi-invertible. 86 3. Structure Theory

Lemma 5. If z is quasi-invertible in J and η is a homomorphism of J into J ′ then zη is quasi-invertible in J ′. If Z is a quasi-invertible ideal and z = z + Z is quasi invertible in J = J /Z then z is quasi invertible in J .

Proof. The ﬁrst statement is clear since invertible elements are mapped into invertible elements by a homomorphism. To prove the second result consider 1 − z where z where z is quasi-invertible in J = J /Z. We 2 122 have a wǫJ such that (1 − w) U1−z = 1 − y, yǫZ. Since Z is quasi- invertible, 1 − y is invertible. Hence 1 − z is invertible (Theorem on inverses). Thus z is quasi-invertible. We are now in position to prove our ﬁrst main result.

~~Theorem 1. There exists a unique maximal quasi-invertible ideal R in J .R contains every quasi-invertible ideal and J = J /R contains no quasi-invertible ideal , 0.~~

Proof. Let {Zα} be the collection of quasi-invertible ideals of J and put R = ∪Zα. If x, yǫR, xǫZα, yǫZβ for some α, β. Bylemma4, Zα+Zβ = Zγ. Hence x + yǫR and x + y is quasi-invertible. It follows that R is a quasi- invertible ideal. Clearly R contains every quasi-invertible ideal so R is the unique maximal quasi-invertible ideal. Now let Z be a quasi- invertible ideal of J = J /R. Then Z = Z/R where Z is an ideal of J containing R. Let zǫZ. Then z = z + R is quasi invertible in J . Hence by Lemma 5, z is quasi-invertible in J . Hence Z is a quasi- invertible ideal of J . Then Z ⊆ R, Z = R and Z = Z/R = 0. The ideal R is called the (Jacobson)radical of J and will be denoted also as rad J ·J is called semi-simple if rad J = 0. The second statement of Theorem 1 is J = J / rad J is semi-simple. Since nil ideals are quasi-invertible rad J contains every nil ideal.

~~2 Properties of the radical~~

~~We show ﬁrst that rad J is independent of the base ring Φ. This is a consequence of 2. Properties of the radical 87~~

123 Theorem 2. Let J be a quadratic Jordan algebra over Φ, R = rad J and γ the radical of J regarded as a quadratic Jordan algebra over Z. Then R = γ. Proof. It is clear that R ⊆ γ. The reverse inequality will follow if we can show that Φγ the Φ-submodule generated by γ is a quasi-invertible ideal. The elements of Φγ have the form αizi,αiǫΦ, ziǫγ. If aǫJ then = Φ = 2 + ( αizi)Ua αi(ziUa)ǫ γ since ziUaǫγP. Also aU αizi αi aUzi i P P P P αiα j(aUzi,z j ) since aUzi , aUzi,z j ǫγ for any aǫJ . Next we show that i< j P 3 z = αizi is quasi-invertible. The result just proved show that z = zUzǫγP. Hence this is quasi-invertible and z is quasi-invertible by lemma 2. Hence Φγ is a quasi-invertible ideal, which completes the proof.

Definition 3. An element aǫJ is called regular if aǫJ UaJ is called regular if every aǫJ is regular. If Φ is a field and a is an algebraic element of J then Φ[a] is fi- Φ Φ Φ = 2 nite dimensional (§1.10). Then [a] [a]Ua ⊇ [a]Ua2 (Ua2 Ua) ⊇ Φ Φ = Φ = [a]Ua3 ... Hence we have an n such that [a]Uan [a]Uan+1 ... 2n Φ = Φ 2n Then a ǫ [a]Uan [a]Ua2n . Hence a is regular. Thus if a is algebraic (Φ a field) then there exists a power of a which is regular.

~~Theorem 3. (1) rad J contains no non-zero regular elements. (2) If Φ is a ﬁeld and zǫ rad J is algebraic then z is nilpotent.~~

Proof. (i) Let zǫ rad J be regular, so z = xUz for some xǫJ . Suppose 124 ﬁrst that x is invertible. Since zUxǫ rad J , x−zUx is invertible by lemma = + 3. Hence Ux−zUx Ux − Ux,zUx UxUzUx is invertible in End J . Now = + zUx−zUx zUx − zUx,zUx zUxUzUx ′ = zUx − zVx,zUx + zUxUzUx (QJ4 )

~~= zUx − xUz,zUx + zUxUzUx~~

= zUx − 2zUx + xUzUx = = Since xU2 z this is 0 and since Ux−zUx is invertible, z 0. Now = = = let x be arbitrary. Since z xUz xUxUz xUzUxUz. We may 2 2 replace x by xUzUx and thus assume x ∈ rad J . Now z (xUz) = 88 3. Structure Theory

2 2 z UxUz(QJ22) = yUz where y = z Uxǫ rad J . Then u = 1 + x − y is 2 2 invertible by lemma 3. Also uUz = z + z − z = z. Hence z = 0 since u is invertible by the ﬁrst case. (2) If zǫ rad J is algebraic then we have seen that z2n is regular for some n. Since z2nǫ rad J , (1) implied that z2n = 0. Hence z is nilpotent.

~~Theorem 4. If u is an invertible element of J then rad J (u) = rad J .~~

Proof. Since rad J is an ideal in J it is ideal in the isotope J (u). If z ∈ rad J then u−1 − z is invertible by lemma 3. Then u−1 − z is invertible in J (u). Since u−1 is the unit of J (u) this states that z is quasi-invertible in J (u). Thus rad J is a quasi-invertible ideal of 125 J (u). Hence rad J ⊆ rad J (u). By symmetry rad J = rad J (u).

~~3 Absolute zero divisors.~~

We recall that z is an absolute zero divisor (§1.10) if Uz = 0. We shall now show that such a z generates a nil ideal of J . For the proof we shall need some information on the ideal (inner ideal, outer ideal) generated by a non-vacuous subset S of J . Clearly the outer ideal generated by S is the smallest submodule containing S which is stable under all Ux, xǫJ . This is the set of Φ-linear combinations of the elements of the

form sUx1 Ux2 ··· Uxk s ∈ S, xi ∈ J . We have seen also that the linear ideal generated by a single element s is Φs + J Us (§1.5). Beyond this we have no information on the inner ideal generated by a subset. We now prove

~~Lemma 6. If Z is an inner ideal then the outer ideal c generated by Z is an ideal.~~

Proof. The elements of c are sums of elments of the form bUx1Ux2 ... Uxk, bǫZ, xiǫJ . We have to show that c is an inner ideal. For this ﬃ it su ces to prove that if a, xiǫJ , bǫZ then aUbUx1 ... Uxk ǫc and for c, dǫc, aUc,d ǫc. Since aUc,d = {cad} the second is clear since cǫc and c is an outer ideal (§1.5). For the ﬁrst we use

~~aU = aU ... U U U ... U . bUx1 ...Uxk x x1 b x1 x 3. Absolute zero divisors. 89~~

Since Z is an inner ideal aU ··· U U ǫZ. Hence aU ǫc. x x1 b bUx1 ···Ux Let z be an absolute zero divisor. Then the inner ideal generated by z is Φz. Hence the last result shows that the ideal generated by z is the set Φ of linear combinations of elments of the form zUx1 Ux2 ··· Uxk , xiǫJ . 126 Since the ideal generated by a set of elements is the sum of the ideals generated by the individual elements we see that the ideal M = zerJ generated by all the absolute zero divisors is the set of Φ -linear combinations of elements zUx1 Ux2 ··· Ux, z an absolute zero divisor, xiǫJ . Φ Since for αǫ ,αzUx1 Ux2 ··· Uxk is an absolute zero divisor we have the following

Lemma 7. The ideal M = zerJ generated by the absolute zero divisors is the set of sums z1 + z2 + ··· + zk where the zi are absolute zero divisors We prove next Lemma 8. If z is an absolute zero divisor and y is nilpotent then x = y+z is nilpotent. Proof. We show ﬁrst that for n = 0, 1, 2,...

n n x = y + zMn−1 (7) where = = = + M−1 0, Mo 1, Mn+1 Vyn+1 Mn−1Uy, n ≧ 0. (8) = = + We note that by (8) we have M1 Vy, M2 Vy2 Uy and in general = + + + + M2k Vy2k Vy2k−2 Uy Vy2k+1 Uy2 ··· Uyn , ≥ 1 (9) = + + + M2k+1 Vy2k+1 Vy2k−1 Uy ··· VyUyk , k ≧ 1

n+2 n Now (7) is clear if n = 0, 1. Assume it for n. Then x = x Ux = 127 n n n x (Uy + Uy,z) (since Uz = 0)= y Uy + zMn−1Uy + y Uy,z + zMn−1Uy,z. n = n+2 n = n = = Now y Uy y and y Uy,z {zy y} zVyn,y zVyn+1 (QJ38). Hence n+2 = n+2 + + + x y z(Mn−1Uy Vyn+1 ) zMn−1Uy,z

~~Thus we shall have (7) by induction if we can show that zMn−1Uy,z = = 0. By (9), this will follow if we can show that zVyi Uy j Uy,z 0 for 90 3. Structure Theory~~

= i > 0, j ≧ 0 and zUy j Uy,z 0 for j ≧ 0. For the second of these we use zU j U = yzU j z = yV = yV j j (QJ31) = 0 since y y,z y zUy j,z y ,y Uz Uz = 0. For the ﬁrst we use the bilinearization of QJ31 relative to = + = = + a: VbUa,c,b VaUb,c VcUb,a in zVyi Uy j Uy,z yzVyi Uy j z yzUy j,yi j z (QJ39) = yV = y(V j i+ j + V i+ j i ) = 0. This proves (7). zUy j,yi, j,z y Uz,y y Uz,y n Since y is nilpotent it is clear from (9) that y = 0 and Mn = 0 for suﬃciently large n. Hence xn = 0 by (7). Repeated application of Lemma 8 shows that if the zi are absolute zero sivisors then z1 + z2 + ··· + zk is nilpotent. Hence, by lemma 7, we have

~~Theorem 5. The ideal ker J generated by the absolute zero divisors is a nil ideal.~~

~~It is clear from this that ker J ⊆ rad J . It clear also that a simple quadratic Jordan algebra contains no absolute zero divisors, 0.~~

~~4 Minimal inner ideals~~

128 An inner ideal Z in J is called minimal (maximal) if Z , 0(Z , J ) there exists no inner ideal c in J such that Z ⊃ c ⊃ 0(Z ⊂ c ⊂ J ). If Z is a minimal inner ideal then J Ub = 0 or J Ub = Z for every bǫZ since J ub is an inner ideal contained in Z. Similarly, either ZUb = 0 or ZUb = Z, bǫZ. We shall now prove a key result on minimal inner ideals which will serve as the starting point of the structure theory. As a preliminary to the proof we note the following

~~Lemma . Let a, bǫJ satisfy aUb = b. Then E = UaUb and F = UbUa are idempotent elements of End J and if d = bUa then b and d are related in the sense that dUb = b and bUd = d.~~

Proof. Since aUb = b, UbUaUb = Ub. Then UaUbUaUb = UaUb and UbUaUbUa = UbUa so E and F are idempotents. If d = bUa then dUa = bUaUb = aUaUaUb = aUb = aUb = b and bUd = aUbUaUbUa = aUbUa = bUa = d. We shall now prove the following 4. Minimal inner ideals 91

Theorem on Minimal Inner Ideals. Any minimal inner ideal Z of J is of one of the following types: IZ = Φz where z is a non-zero absolute 2 zero divisor, II Z = J Ub for every b , 0 in Z but ZUb = 0 and b = 0 2 for every b ∈ Z, III Z is a Pierce inner ideal J Ue, e = e, such that (Z, U, e) is a division algebra. Moreover, if J contains no idempotent , 0, 1 and contains a minimal inner ideal Z of type II then 2J = 0 and for every b , 0 in Z there exists an element dǫJ such that

~~2 2 (i) dUb = b, bUd = d, b = 0 = d , b◦d = 1, O = J Ud is a minimal 129 inner ideal of type II.~~

(ii) c = b + d satisﬁes c2 = 1, c−1 = c and in the isotope J (c), b and d are supplementary strongly connected orthogonal idempo- (c) (c) = tents such that the Pierce inner ideals J Ub Z, J Ud O are minimal of type III.

Proof. Suppose ﬁrst that Z contains an absolute zero divisor z , 0. Then Φz is a non-zero inner ideal contained in Z so Z = Φz, by the minimality of Z. From now on we assume that Z contains no absolute zero divisor , 0. Then Z = J Ub for every b , 0 in Z. Also ZUb = 0 or ZUb = Z. Suppose Z contains ab , 0 such that ZUb = 0 and let yǫZ. Then there exists an aǫJ such that aUb = y. Then ZUy = ZUbUaUb = 0. Thus either ZUb = 0 for every bǫZ or ZUb = Z for every b , 0 in Z. In the ﬁrst = 2 = = 2 = case J Ubi J Ub J UbUb ⊂ ZUb 0, bǫZ and since b 1UbǫZ and Z contains no absolute zero divisors , 0 we have b2 = 0, bǫZ. Thus Z is of type II. Now assume ZUb = Z for every b , 0 in Z. Let b be such an element and let aǫZ satisfy aUb = b. By the lemma, b and d = bUa are related. Also dǫZ and E = UbUd and F = UdUb are idempotent operators. We have J E = J UbUd = Z and J F = Z so the restrictions E and F of E and F to Z are Z the identiy on Z. Put e = 2 = 2 , = 2 = d UbǫZ, f b UdǫZ. Then e 0 since J Ue J UbUdUb Z and , 2 = 2 2 = 2 2 = 2 = 2 = similarly f 0. Also e (d Ub) b UdUb b UdF b Ud f . 2 = 2 = 2 2 = 2 = 2 2 = Similarly, f e. Then e ( f ) f U f (QJ23) f UdUbUd 2 2 f FE = f = e. Then e is a non-zero idempotent in Z, Z = J Ue and this is a division algebra since ZUb = Z for every non-zero b hence Z is 130 type III. 92 3. Structure Theory

Now suppose J contains no idempotents , 0, 1 and contains the minimal inner ideal Z of type II. Let b , 0 in Z and let aǫJ satisfy 2 aUb = b. We claim that c = a Ub = 0. Otherwise c is a non-zero element of Z and there exist bo, coǫJ suchthat boUc = b, coUc = c. Put = 2 = 2 = 2 = 2 = e coUbUa. Then e (coUbUa) a UcOUb Ua a UbUco UbUa = = = cUco UbUa cUco UcUbo UcUa cUbo UcUa (since cUco Uc c by the = = = = = Lemma) coUcUbo UcUa coUboUc Ua coUbUa e. Since eUaUb 2 = 2 = 2 = = , coUbUaUb coUbUa Ub coUa Ub coUc c 0. More over, if 2 e = 1 then 0 = b Ua = 1UbUa = eUbUa = coUbUaUbUa = coUbUa (Lemma)= e. Hence e is an idempotent , 0, 1 contrary to hypothesis. 2 Thus we have shown that a Ub = 0 for every aǫJ such that aUb = b. 2 2 Put d = bUa. Then bUd = d, dUb = b and d Ub = 0. Then d = 2 2 2 2 2 2 (bUd) = d UbUd = 0. By QJ30, (b◦d) = b Ud +d Ub +d Ub +bUd ◦ b = b ◦ d. By QJ17, bUb◦d = −bUd,b + bUdUb + bUbUd + bVaUbVd = 2 −b ◦ d + b + 0 + dVbUbVd = b (since dVbUb = dUbVb = b ◦ b = 0). Hence b ◦ d is an idempotent , 0 and so b ◦ d = 1. We have now established all the relations on b, d in (i). Now put c = b + d. Then 2 = 2 + + 2 = 2 = c b b ◦ d d 1. Hence Uc is an automorphism such that Uc 1 and bUc = bUb+bUb,d+bUd = d. Thus Uc maps the minimal inner ideal Z = J Ub onto the minimal inner ideal O = J Ud. Next we consider 2 2 (d + 1)Ub = dUb + b = b. As before, this implies (d + 1) Ub = 0 so 0 = 2dUb = 2b. Then 4Z = 4J Ub = J U2b = 0. Since 2Z is 131 an inner ideal contained in Z, which is minimal, this implies 2Z = 0. Applying the automorphism Uc shows that 2O = 2ZUc = 0. Also 2b = 0 implies 2J Ub,d = U2b,d = 0. We now have 2J = 2J Uc = 2J Ub + 2J Ub,d + 2J Ud = 0. (c) −1 Next we consider the isotope J . We have c = cUc = (b + d)Ub+d = bUb + bUb,d + bUd + dUbdUb,d + dUd = b + d = c. Hence c (c) (c) = = = + = is the unit of J . Since cUb cUcUb cUb (b d)Ub b, b is idempotent in J (c). Hence d = c − b is an idempotent orthogonal to b (c) (c) = = = = (c) in J . We have cU1 cUcU1 c and 1 b ◦ d 1Ub,dǫJ Ub,d. (c) (c) = Hence b and d are strongly connected by 1 in J . Finally, J Ub = = (c) = J UcUb J Ub Z and J Ud O , so Z and O are the Pierce inner ideals determined by the idempotents b and d in J (c). Since Z and O are minimal inner ideals of J they are minimal inner ideals of 5. Axioms for the structure theory. 93

J (c). Clearly they are of type III in J . All the possibilities indicated in the theorem can occur. To see this Φ(q) consider 2 the special quadratic Jordan algebra of 2 × 2 matrices over a ﬁeld Φ. Then Φe11 is a minimal inner ideal of type III and Φe12 is a = Φ(q) = Φ + Φ minimal iner ideal of type II in J 2 . Next let J 1 e12 . Φ(q) This is a subalgebra of 2 and e12 is an absolute zero divisor in J . Hence Φe12 is a minimal inner ideal of type I. Finaly, assume Φ has characteristic two. Then J = Φ1 + Φe12 + Φe21 is a subalgebra of Φ(q) = + + Φ 2 = 2 + 2 since x α1 βe12 γe21,α,β,γǫ , then x (α βγ)1 so x2 and xy + yxǫΦ1 for x, yǫJ . Then xyx + yx2ǫΦx and xyxǫΦx + Φy. The formula for x2 shows that J contains no idempotents , 0, 1. Also 132 Z = Φe12, O = Φe21 are minimal inner ideals of type II in J and b = e12, d = e21 satisfy (i).

~~5 Axioms for the structure theory.~~

We shall determine the structure of the quadratic Jordan algebras which satisfy the following two conditions: (1) strong non-degeneracy (=non- existance of absolute zero divisors , 0), (2) the descending chain condition (D C C) for principal inner ideals.The latter is equivalent to the minimum condition for principal inner ideals. We have called a quadratic Jordan algebra J regular if for every aǫJ there exists xǫJ such that xUa = a (§1.10, Deﬁnition 3 of §2). Clearly this implies strong non- degeneracy.

~~Lemma 1. If J is strongly non-degenerate and satisﬁes the DCC for principal inner ideals then every non-zero inner ideal Z of J contains a minimal inner ideal of J .~~

Proof. If b , 0 is in Z then J Ub is a principal inner ideal contained in Z and J Ub , 0 by the strong non-degeneracy. By the minimum condition (= D C C) for principal inner ideals contained in contains a minimal element K. We claim that K is a minimal inner ideal of J . Otherwise, we have an inner ideal Z such that K ⊃ Z ⊃ 0. The argument used for Z shows that Z contains a non-zero principal inner ideal J Uc and K ⊃ J Uc contrary to the choice of K. 94 3. Structure Theory

~~As a ﬁrst application of this result we note that the foregoing conditions (1) and (2) are equivalent to : (1) and (2′) J is semi-simple. For, we have~~

~~133 Theorem 6. If a quadratic Jordan algebra J satisﬁes the DCC for principal inner ideals then J is semi-simple if and only if J is strongly non-degenerate.~~

Proof. If z is an absolute zero divisor then the ideal generated by z is nil and so is contained in rad J . Hence is J is semi-simple, so rad J = 0, then J contains no absolute zero divisors , 0. Conversely, suppose radJ , 0. Then we claim that J contains non-zero absolute xero divisors. Otherwise, we can apply lemma 1 to conclude that rad J contains a minimal inner ideal K of J . K is not of type III since rad J contains no non-zero idempotents. Also K is not of type II since in this case the Theorem on Minimal Inner Ideals shows that every element of K is regular. Hence K is os type I and J contains an absolute zero divisor , 0, contrary to hypothesis. It is immediate that if J satisﬁes the DCC for principal inner ideals, or is strongly non-degenerate, or is regular then the same condition holds for every isotope J (c). The same is true of the quadratic Jordan algebra (J Ue, U, e) if e is an idempotent in J . This follows from

Lemma 2. Let e be an idempotent in J . Then any inner (principal inner) ideal of (J Ue, U, e) is an inner(principal inner) ideal of J and any absolute zero divisor of J Ue is an absolute zero divisor of J . Moreover, if J is regular then J Ue is regular.

Proof. If Z is an inner ideal of J Ue and b ∈ Z then b = bUe. Hence = = = J Ub J UbUe (J Ue)UbUe ⊆ ZUe Z. Hence Z is an inner ideal of J . If Z is principal in J Ue, Z = J UeUb, bǫJ Ue. Then 134 b = bUe so Z = J UeUeUbUe = J UeUbUe = J Ub is a principal inner ideal of J . Let zǫJ Ue be an absolute zero-divisor in J Ue. Then z = zUe. Hence if xǫJ then xUz = (xUe)UzUe = 0. Thus z is an absolute zero divisor in J . FInally, suppose J is regular and let a = aUeǫJ Ue. Then a = xUa for some xǫJ . Hence a = xUa = 5. Axioms for the structure theory. 95

~~xUeUaUe = (γUe)UeUaUe = x(Ue)Ua. Since xUeǫJ Ue this shows that J Ue is regular.~~

~~We shall now give the principal examples of quadratic Jordan algebras which are strongly non-degenerate and satisfy the DCC for principal inner ideals.~~

Examples. 1. If a is asemi-simple right Artinian algebra then it is well-known that a is left Artinian and every right (left) ideal of a has a complementary right (left) ideal. Let aǫa. Then a a is a left ideal so there exists a left ideal J such that a = aa ⊕ J. Then 1 = e + e′ where eǫaa, e′ǫJ. It follows that aa = ae and e2 = e. (q) (q) Then ae = a and e = xa. Hence axa = a so aǫa Ua. Thus a is regular and consequently strongly non-degenerate. We note that aaa = aa ∩ aa. Clearly aaa ⊆ aa ∩ aaa. On the other hand, if x = au = va then, by regularity, x = xyx, yǫa so x = auyvaǫaaa. Hence aa ∩ aa ⊆ aaa. If aaa ⊇ bab then b = bzb = awaǫaaa. Then aa ⊇ ba and aa ⊇ ab. Since a satisﬁes the descending chain condition on both left and right ideals it is now clear that a(q) satisﬁes the descending chain condition for principal inner ideals.

2. Let (a, J) be an associative algebra with involution. Suppose a(q) 135 is regular. Then H (a, J) is regular. For if hǫH (a, J) there exists an aǫa such that hah = a. Then haJ h = h so h = ha(haJ h) = h(ahaJ )h and ahaJ ǫH . Hence H is regular. We note next that if a(q) is regular and satisiﬁes the descending chain condition for principle inner ideals then H satisﬁes these conditions. We have

~~seen that H is regular. Now suppose H Ub1 ⊇ H Ub2 ⊇ H~~

~~Ub3 ... where biǫH . Then bi+1 ∈ H Ubi+1 by regularity so bi+1 ∈ = H Ubi and so bi+1 bihibi, hi ∈ H . Then aUbi Ubi Ubi ⊆ abi.~~

Hence aUb1 ⊇ aUb2 ⊇ ... is a descending chain of principal inner = = ideals of a(q). Hence we have an m such that aUbm aUbm+1 ... = By regularity, biǫaUbi so if n ≧ m, bn bn+1an+1bn+1, an+1ǫa. j = + + = Then bn bn 1an+1bn 1. Also, by regularity of H , bn bnknbn, j = + + + + + = + + knǫH . Then bn (bn 1an 1bn 1)kn(bn 1an+1bn 1) bn 1ln 1 96 3. Structure Theory

j + + = + + + ∈ bn 1 where ln 1 an 1bn 1knbn 1an+1 H . Hence bnǫH Ubn+1 = = and H Ubn ⊆ H Ubn+1 . Thus H Ubm H Ubm+1 ... and H satisﬁes the DCC for principal inner ideals. It is clear from (1) and the foregoing results that if (a, J) is semi- simple Artinian with involution then H (a, J) is regular and satis- ﬁes the DCCfor principal inner ideals.

3. If J is a quadratic Jordan algebra over Γ and Φ is a subring Γ then J /Φ and J /Γ have the same principal inner ideals. Hence J /Φ has DCC on these if and only if this holds for J /Γ. It 136 is clear also that J /Φ is regular if and only if J /Γ is, and is strongly non-degenerate if and only if J /Γ is. Now let Γ be a field and let J = Jord (Q, 1), Q a quadratic form on J /Γ with vase point 1 (cf. §1.7). We have the formulas yUx = Q(x, y)x − Q(x)y, x = T(x)1 − x, T(x) = Q(x, 1) in J . If Q(x) , 0 then x is invertible and J Ux = J (§1.7). If Q(x) = 0 the Q(x) = 0 the formula for Ux shows that J UX ⊆ Γx. This implies that J /Γ, hence J /Φ, satisfies the DCC for principal inner ideals. If xǫJ satisfies Q(x) = 0, Q(x, y) = 0, yǫJ , then Ux = 0. On the other hand, suppose Q is non-degenerate. Then for xǫJ either Q(x) , 0 there exists a y such that Q(x, y) , 0. In either case xǫJ Ux. Hence J = Jord (Q, 1) has non-zero absolute zero divisors or is regular according as Q is degenerate or not. We remark that the formula for U shows that if K is a subspace such that Q(k) = 0, kǫK, then K is an inner ideal. This can be used to construct examples of algebras which are regular with DCC on principal inner ideal but not all inner ideals. 4. Let O be an octonion algebra over a field Γ, Φ a subring of Γ. We consider H (O3) as quadratic Jordan algebra over Φ(cf. §§1.8, 1.9).since H (O3) is a finite dimensional vector space over Γ, H /Γ satisfies the DCC for principal inner ideals. Hence H /Φ satisfies this condition. We proceed to show that H is strong non- degenerate. Let ei = 1[ii], fi = 1 − ei (notations as in §1.7). Then = = + + Φ H U f3 H11 ⊕H12 ⊕H22 {α[11] β[22] a[12]|α, βǫ , aǫO}. If xα[11]+β[22]+a[12] the Hamilton-Cayley theorem in H (O3) 6. Capacity 97

shows that x3 − T(x)x2 + S (x)x = 0 (since N(x) = 0, see §1.9). Here T(x) = α+β and S (x) = T(x♯) = αβ−n(a) by direct calcula- 137 tion. Also, direct calculation using the usual matrix square shows 2 + = = that x − T(x)x S (x) f3 0. Hence (H U f3 , U, f3) Jord (S, f3) (see §1.7). Since the symmetric form n(a, b) of the norm form n(a) of O is non-degenerate the same is trur of the symmetric bilinear form of S (x) = αβ − n(a). Hence S (x) is non-degenerate so = H U f3 Jord (S, f3) is strongly non-degenerate. By symmetry, = H U fi is strongly non-degenerate for i 1, 2 also. Now let ǫH

be an absolute zero divisor in H . Then zU fi is an absolute zero = = = divisor in H U fi so zU fi 0, i 1, 2, 3. Clearly this implies z 0 so H is strongly non-degenerate. 5. It is not diﬃcult to shows by an argument similar to that used in 2 that if J is regular then any ideal Z and J contianing 1 is regular and if J is regular and satisﬁes the minimum condion then the same is true of Z. We leave the proofs to the reader.

~~6 Capacity~~

An Idempotent eǫJ is called primitive If e , 0 and e is the only non- ′ zero idempotent of J Ue. If e is not primitive and e is an idempotent ′ ′′ ′ ′′ , 0, e in J Ue then e = e + e where e and e are orthogonal idempotents , 0. Conversely if e = e′ + e′′ where e′ and e′′ are orthogonal ′ ′′ idempotents then e , e ǫJ Ue (cf. §2.1) so e is not primitive. Hence e is primitive if and only if it is impossible to write e = e′ + e′′ where e′ and e′′ are non-zero orthogonal idempotents. An idempotent e is called completely primitive if (J Ue, U, e) is a division algebra. Since a division algebra contains no idempotents , 0, 1 it is clear that if e is completely primitive then e is primitive. Lemma 1. If J , 0 satisﬁes the DCC for Pierce inner ideals then 138 J contains a (ﬁnite) supplementary set of orthogonal primitive idempotents. Proof. Consider the set of non-zero Pierce inner ideals of J . By the

~~DCC on these there exists a minimal element J Ue1 in the set. Clearly 98 3. Structure Theory~~

~~e1 is primitive. If e1 = 1 we are done. Otherwise, put f1 = 1 − e1 so , f1 0 and consider J U f1 . Since f1 is an idempotent the hypothesis~~

carries over to J U f1 . Hence J U f1 contains a primitive idempotent e2 and this is orthogonal to e1. If 1 = e1 + e2 we are done. Otherwise, put f2 = 1 − e1 − e2 and apply the argument to obtain a primitive idempotent = + , e3 in J U f2 . Also J U f1 ⊇ J U f2 since f1 e2 f2, e2 0. Now e3 is orthogonal to e1 and e2 so if 1 = e1 +e2 +e3 we are done. Otherwise, we = repeat the argument with f3 1−e1−e2−e3. Then J U f2 ⊃ U f2 ⊃ U f3 ⊃ ... Since the DCC holds for Pierce inner ideals this process terminates with a supplementary set of orthogonal primitive idempotents.

Deﬁnition 4. A quadratic Jordan algebra J is said to have a capacity if it contains a supplementary set of orthogonal completely primitive idempotents. Then the minimum number of elements in such a set is called the capacity of J .

~~Theorem 7. If J is strongly non-degenerate and satisﬁes the DCC for principal inner ideals then J has an isotope J (c) which has a capacity. If J has no two torsion then J itself has a capacity.~~

~~139 Proof. Let {ei} be a supplementary set of orthogonal primitive idempotents in J (Lemma 1). Suppose for some i, {ei} is not completely~~

~~primitive. Since the hypothesis carry over to J Uei , Uei contains a min-~~

~~imal inner ideal Z (Lemmas 1, 2 of §5). Since J Uei is not a division~~

~~algebra Z ⊂ J Uei Now Z is not of type I by the strong non-degeneracy~~

and it is not of type III since Z ⊂ J Uei and ei is primitive. Hence Z is , of tyoe II. Also since J Uei contains no idempotent 0, ei the Theorem = on Minimal Inner Ideals implies that 2J Uei 0 and J Uei contains 2 = −1 = an element ci such that ci ei, ci ci (in J Uei ) and in the isotope (ci) = + (J Uei ) , ci bi di where bi, di are orthogonal idempotents such that the corresponding pierce inner ideals are minimal of type III. Let c1 = e j if e j is completely primitive; otherwise let c j be as just indicated. Put (c) c = c j. Then c is invertible and it is clear that J has a capacity. ItP is clear from the deﬁnition that J has capacity 1 is and only if J is a division algebra. We consider next the algebras of capacity two and we shall prove the following usefull lemma for these 6. Capacity 99

Lemma 2. Let J have capacity two, so 1 = e1 + e2 where the ei are orthogonal completely primitive idempotents, J = J11 ⊕ J12 ⊕ J22 2 the corresponding Pierce decomposition. If xǫJ12 either x = 0 or x invertible. The set of absolute zero divisors of J is the set of xǫJ12 2 such that x = 0 and x ◦ y = 0, yǫJ12 and this set is an ideal. Either e1 and e2 are connected or every element of J12 is an absolute zero 140 divisor. J is simple if and only if J12 , 0 and J is strongly non- degenerate. If J is simple e1 and e2 are connected and every outer ideal containing 1 in J is simple of capacity two.

2 Proof. If xǫJ12, x = x1 + x2, xiǫJii. Since Jii is a division algebra, 2 either xi = 0 or xi is invertible in Jii. Clearly if x , 0 and x2 , 0 then x 2 and hence x is invertible. Suppose x1 = 0 so x = x2ǫJ22. Then since = = = 2 = 2 = e1◦x x, and VxVx2 Vx2 Vx we have x2◦x x ◦(e1◦x) (x ◦e1)◦x = (x2 ◦ e1) ◦ x 0. By PD 5, if a2ǫJ22 the mapping a2 → Va2 the restriction of Va2 to J12 is a homomorphism of (J22, U, e2) into (End (q) J12) . Since J22 is a division algebra this is a monomorphism and (q) the image is a division subalgebra of (End J12) . We recall also that (q) invertibility in (End J12) is equivalent to invertibility in End J12. = = = Since we had xVx2 0 it now follows that either x2 0 or x 0. In 2 2 either case x = x2 = 0. Thus x1 = 0 implies x = 0 and, by symmetry, 2 2 x2 = 0 implies x = 0. It is now clear that either x = 0 or x is invertible. 2 Let xǫJ12 satisfy x = 0, x ◦ y = 0 for all yǫJ12. Let aǫJ11. Then 2 2 aUxǫJ22 and (aUx) = x UaUx = 0. Since J22 is a division algebra this implies that aUx = 0. Similarly bUx = 0 if bǫJ22. By QJ17, = = + + = + Ux Ux◦e1 UxUe1 Ue1 Ux Vs xUe1 Vx − Ue1 Vx − Ue1Ux,e1 UxUe1 + = Ue1 Ux VxUe1 Vx since e1Ux 0 by the PD theorem. If yǫJ12 we have = = = yUe1 0 y12Vx. Hence yUx yUxUe1 ǫJ11. By symmetry yUxǫJ22 so yUx = 0 Thus Ux = 0 and x is an absolute zero divisor. Conversely 141 suppose x is an absolute zero divisor. Then xUei is an absolute zero = = divisor in the division algebra Jii so xUei 0. Then x xUe1,e2 ǫJ12. 2 2 ALso x = 1Ux = 0 and if yǫJ12 then y ◦ xǫJ11 + J22 and (y ◦ x) = 2 2 y Ux + x Uy + yUx ◦ y(QJ30) = 0. As before, this implies that y ◦ x = 0. Hence the set of absolute zero divisors coincides with the set of xǫJ12 2 such that x = 0 and x ◦ y = 0, yǫJ12. To see that this set is an ideal it is enough to prove that it is closed under addition. This is immediate. 100 3. Structure Theory

2 Suppose e1 and e2 are not connected. Then x = 0 for all xǫJ12. 2 2 2 Then x ◦ y = (x + y) − x − y = 0. for all x, yǫJ12. Then the preceding result shows that every xǫJ12 is an absolute zero divisor. Now suppose J12 , 0 and J is strongly non-degenerate. Let , = Z be an ideal 0 in J . We have Z ZUe1 ⊕ ZUe2 ⊕ ZUe1,e2 and = = Zii ≡ ZUei Z ∩ Jii, Z12 ≡ ZUe1,e2 Z ∩ J12. Since Jii is a division algebra and Zii is an ideal in Jii either Zii = 0 or Zii = Jii. Since Z , 0, Z22 , 0. If Z11 , 0 so Z11 = J11 then Z ⊇ e1 ◦ J12 = J12. Similarly, if Z22 , 0 then Z ⊇ J12. Next suppose Z12 , 0 and let x , 0 in 2 Z12. Since x is not an absolute zero divisor either x , 0 or there exists 2 ayǫJ12 such that x◦y , 0. In either case, since x and x◦yǫJ11 +J22 we obtain either Z11 , 0 or Z22 , 0. Then, as before, Z12 = J12. Thus Z ⊇ J12. Since J12 , 0 and J is strongly non-degenerate J12 contains as invertible element. Then Z contains an invertible element and so Z = J . Hence J is simple. If J12 = 0 then J = J11 ⊕ J22 142 and the Jii are ideals. Hence in this case J is not simple. Also if J contains absolute zero divisors, 0 then the set of these is an ideals and J is not simple. Hence simplicity of J implies J12 , 0 and J is strongly non-degenerate. If J is simple J12 contains an invertible element. Then e1 and e2 are connected. If Z is an outer ideal containing 1 then contains the ei and J12 = J12 ◦ ei (of. the proof of Theorem 2.2) Clearly, this and the previous results imply that Z is simple of capacity two.

~~7 First structure theorem~~

The results of the last section have put us into position to prove rather quickly the First structure Theorem. Let J be a strongly non-degenerate quadratic Jordan algebra satisfying the DCC for principal inner ideals (equivalently, by Theorem 6 J is semi-simple with DCC for principal inner ideals). Then J is a direct sum of ideals which are simple quadratic Jordan algebras satisfying the DCC on principal inner ideals. Con- versely, if J = J1 ⊕ ... ⊕ Js where the Ji are ideals which are simple quadratic Jordan algebras with DCC on principal inner ideals 7. First structure theorem 101 then J is strongly non-degenerate with DCC on prinicipal inner ideals.

Proof. By Theorem 7, J has an isotope J˜ = J (c) whose unit c−1 = c is a sum of completely primitive orthogonal idempotenets ei. Let J J˜i j be the corresponding Pierce decompostion. It is clear that J˜ Pand hence every Pierce inner ideal of J is strongly non-degenerate. If c = e1 so J˜ = J11 then J is a division algebra and the result is clear. Hence assume the number of ei is > 1. Let i , j and consider 143 = + ˜ + ˜ + ˜ the Pierce inner ideal J Uei+e j Jii Jii Ji j J j j. By lemma 2 of the proceding section, either J˜i j = 0 or ei and e j are connected and J˜ii + J˜i j + J˜j j is simple. Since connectedness of orthogonal idempotents is a transitive relation (§2.3) we may decompose the set of indices i into non over-lapping subsets I1, I2,..., Is such that if i, jǫIk, i , j, the ei and e j are connected but if iǫIk and jǫIl, k , l, then ei = = ˜ = ˜ (1) and e j are not connected so Ji j 0. Put 1k ei, Jk J U1 . iǫI k Pk J˜k = J˜1 ⊕ ... ⊕ J˜s and the Pierce relations show that J˜k is an ideal. We claim that Jk is simple. We may suppose Ik = {1, 2,..., m}, m > 1. m Then J˜k = J˜i j and ei and e j are connected if i , jǫ{1,..., m}. i≧ j=1 P Also J˜ii + J˜i j + J˜j j is simple. Let Z be a non-zero ideal in J˜k. Then, as before, Z = Zi j where Zi j = J˜i j ∩Z. Since J˜ii +J˜i j +J˜j j is sim- = = = ple either Z containsP this or Zii Zi j Z j j 0. Clearly Z , 0 implies that for some i , j we have Zii, Z j j or Zi j , 0. Then Z ⊇ J˜ii, J˜j j and consequently Z ⊇ J˜ll, J˜lr for all l, rǫ{1,..., m}. Then Z = J˜k and J˜k is simple. Since J is an isotope of J˜ we have J = J1 ⊕ ... ⊕ Js where Ji = J˜i as module, id an ideal of J which is a simple algebra (since any ideal of Ji is an ideal of J because of the direct decomposition) Since Ji is a Pierce inner ideals of J it satisﬁes the DCC for principal inner ideals. Conversely, suppose J = J1 ⊕J2 ⊕...⊕Js where J is an ideal and is a simple quadratic Jordan algebra with unit 1i, satisfying the DCC 144 in principal inner ideals. Since the absolute zero divisors generate a nil ideal Ji is strongly non-degenerate. If z is an absolute zero divisor in = = J in J then z zi, zi zU1i , and zi is an absolute zero divisor of = = = Ji. Hence zi 0, iP 1, 2,..., s and z 0. Thus J is strongly non- 102 3. Structure Theory

= = degenerate. Let aǫJ and write a ai, ai aU1i , then it is immediate = = = = that J Ua J Uai JiUai .P Also if b bi, bi bU1i then = J Ua ⊇ J UbPif and onlyP if J Uai ⊇ J Ubi , i P1, 2. It follows from this that the minimum condition for principal inner ideals carries over from the Ji to J . = + + It is easy to show that if Z is an ideal of J then Z Ji1 Ji2 + ... Jik for some subset {i1,..., ik} of the index set. Clearly this implies that the decompostion J = J1 ⊕ J2 ⊕ ... ⊕ Js into simple ideals is unique. We shall call the Ji the simple components of J .

~~8 A theorem on alternative algebras with involution.~~

Our next task is to determine the simple quadratic Jordan algebras which satisfy the DCC for principal inner ideals. By passing to an isotope we may assume J has a capacity. If the capacity is 1, J is a division algebra. We shall have nothing further to say about this. The case of capacity two will be treated in the next section by a rather lengthy direct analysis. The determination for capacity 3 will be based on the Strong coordinatization Theorem supplemented by information by information on the coordinate algebra. Both for this and for the study of the capacity two case we shall need to determine the coordinate algebra (O, j, Oo) 145 (§2.4 for the deﬁnition) such that (O, j) is simple and the non-zero elements of Oo are invertible. We now consider this problem. We deﬁne an absolute zero divisor in an altenative algebra O to be an element z such that zaz = 0 for all aǫO.

Deﬁnition 5. An alternative algebra with involution (O, j) is called a composition algebra if 1 O has no absolute zero divisors , 0 and 2) for any xǫO, Q(x) = xxǫΦ1. A complete determination of these algebras over a ﬁeld is given in the following.

Theorem 8. Let (O, j) be a composition algebra over a field Φ. Then (O, j) is of one of the following types: I a purely is separable field P/Φ 8. A theorem on alternative algebras with involution. 103 of exponent one and charateristic two, j = 1. II (O, j) = (Φ, 1). III (O, j) a quadratic algebra with standard involtion (§1.8).IV. (O, j) = (a, j), a a quaternion, j the standard involution. V. (O, j) an octonion algebra with standard involution. Proof. We have xx = Q(x)ǫΦ, from which it is immediate that Q is a quadratic form on O/Φ whose associated bilinear form satisfies Q(x, y)1 = xy + yx. Hence

~~T(x) ≡ x + x = Q(x, 1)1ǫΦ1. (10)~~

~~Also [x, x, y] = [x; x + x, y] − [x, x, y] = 0 for all yǫO and Q(x)x = xQ(x) = x(xx) = (xx)x = Q(x)x. Hence~~

~~Q(x) = Q(x)Q(x, y) = Q(x, y) (11)~~

Next we note that Q(xz, y)−Q(x, yz) = (xz)y+y(z x)− x(zy)−(yz)x = 146 [x, z, y] − [y, z, x] = −[x, y, z] − [y, z, x] = [x, y, z] − [x, y, z] = 0 (by x + xǫN(O) and the alternating character of [x, y, z]. Hence Q(xz, y) = Q(x, yz) and Q(xz, y) = Q(zx, y) = Q(z, yx) = Q(z, xy)). Thus

~~Q(xz, y) = Q(x, yz) = Q(z, xy). (12)~~

~~We have x(xy) = Q(x)y = (yx)x so by bilinearization we have~~

~~x(zy) + z(xy) = Q(x, z)y = (yz)x + (yx)z (1)~~

We suppose ﬁrst that Q(x, y) is generate which means that we have a non-zero z such that Q(x, z) = 0 for all x. Then xz + zx = 0 and z + z = 0 so xz = zx. Also z2 = −zz = −Q(z)1. Hence zxz = −Q(z)x, xǫO. If Q(z) = 0, z is an absolute zero divisor contrary to hypothesis. Hence Q(z) , 0 and x = αzxz, α = −Q(z)−1, xǫO. Then xy = yx = αz(y x)z = α(zy(xz) = α(yz)(zx). In particular, xz = αz2(zx) = zx and consequently x = αzxz = αz2 x = x. Thus j = 1 and consequently xy = y x given xy = yx so O is commutative. Also z + z = 0 gives 2z = 0 and x = αzxz gives 2x = 0. Hence 2O = 0. This implies that O has no 3 torsion. Then

~~3[x, y, z] = [x, y, z] + [y, z, x] + [y, z, x] + [z, x, y] 104 3. Structure Theory~~

~~= (xy)z − x(yz) + (yz)x − y(zx) + (zx)y − z(xy) = [xy, z] + [yz, x] + [zx, y] = 0~~

147 and commutativity imply that O is associative. Since x2 = xx = Q(x)1 and xyx = Q(x)y it is clear that Q(x) , 0 if x , 0 so O is a purely inseparable extension ﬁeld of exponent one over Φ. Now assume Q(x, y) is non-degenerate. If O = Φ1 we have type II. Hence assume O ⊃ Φ1. If xǫO we have x2 − T(x)x + xx = x2 − (x + x)x + xx = 0 so x2 − Q(x, 1)x + Q(x)1 = 0. If Φ has characteristic two then Q(1) = 1 and Q(1, 1) = 2 = 0. Hence we can choose a uǫO such that Q(1, u) = 1. Then u2u + ρ1 and 4ρ + 1 = 1 , 0 so Φ[u] is a quadratic algebra. If Φ has characteristic , 2 then Q(1, 1) , 0 and we = = , = + 1 can choose qǫO such that Q(1, q) 0 and Q(q) β 0. Put u q 2 1. = = 1 + 2 = + = 1 Then T(u) 1 and Q(u) 4 β, u u ρ1, ρ −β − 4 . Since 4ρ + 1 = −4β , 0, Φ[u] is a quadartic algebra. Hence in both cases we obtian a quadratic subalgebra Φ[q] which is a subalgebra of (O, j) since u = 1 − u. Thus the induced involution is the standard one in Φ[q]. It is clear also that Φ[u] is non-isotropic as a subspace relative to Q(x, y). Now let Z be any ﬁnite deminsional non-isotropic subalgebra of (O, j) and assume Z ⊂ O. As is well-known, O = Z ⊕ Z⊥ and Z⊥ , 0 is non- isotropic. Hence there exists a vǫZ⊥ such that Q(v) = −σ , 0. Since 1ǫZ, Q(1, v) = 0 so v = −v and v2 = σ1. If aǫZ then Q(a, v) = av + va = −av + va = 0 so av = va, aǫZ (14)

If a, bǫZ, Q(av, b) = Q(v, ab)(by (12)) = 0. Hence Z v = {xv|xǫZ } 148 ⊆ Z ⊥ and L = Z + Z v = Z ⊕ Z v has dimensionality= 2dimZ . Also Q(av, bv) = Q((av)v, b) = Q(a(vv), b) = −σQ(a, b). It follows that x → xv is a linear isomorphism of Z onto Z v and Z v and L are non-isotropic. By (13) with z = v we obtain

~~a(bv) = (ba)v, (av)b = (ab)v, a, b, ǫZ . (15)~~

~~Also (av)(bv) = v(ab)v (Moufang identity)= (ba)v2. Hence~~

~~(av)(bv) = σba, a, bǫZ . (16) 8. A theorem on alternative algebras with involution. 105~~

We have a + bv = a − vb = a − bv. We apply these considerations to the quadratic subalgebra Φ[u]. If O = Φ[u] we have case III. Otherwise, we take Z = Φ[u] and obtian the quaternion algebra a = Φ[u] + Φ[u]v. If O = a we have caseIV. Otherwise, we take Z = a and repeat the argument. Then L = O is an octonion algebra. We now claim that O = O so we have case V. Otherwise, we can apply the construction to Z = O and obtain L = Z + Z v such that (15) and (16) hold. Put x = a + bv, y = dv, a, b, dǫO. Then we have [x, x, y] = 0 and x(xy) = (a − bv)(σdb + (da)v). Since (xy)y is a multiple of y = dv this implies that a(db) = (ad)b. Since this holds for all a, b, dǫO we see that O we see that O must be associative. Since it is readily veriﬁed that it is not we have a contradiction. This completes the proof that only I − V can occur. It is readily seen that the algebras with involution I − V are composition algebras. We prove next 149

Theorem 9 (Herstein-Kleinfeld-Osborn-McCrimmon). Let (O, j, Oo) be a coordinate algebra (over any Φ) such that (O, j) is simple and every non-zero element of Oo is invertible in Oo. Then we have one of the following alternatives:

~~I. O = ∆ ⊕ ∆◦, ∆ an associative division algebra j the exchange involution, Oo = H (O, j).~~

~~II. an associative division algebra with involution.~~

~~III. a split quaternion algebra Γ2 over its center Γ which is a ﬁeld over Φ, standard involution, Oo =Γ.~~

~~IV. an algebra of octonions over its center Γ which is a ﬁeld over Φ standard involution, Oo =Γ.~~

Proof. We recall that the hypothesis that (O, jOo) is a coordinate algebra means that (O, j) is an alternative algebra with involution, Oo is a Φ submodule of O contianed in H (O, j) ∩ N((O)) and containing 1 and every xdx, dǫOo, xǫO. Hence Oo contains all the norms xx and all the traces x + x. Then [x, x, y] = 0, x, yǫO. We recall also the following 106 3. Structure Theory

~~realtion in any alternativfe algebra~~

~~n[x, y, z] = [nx, y, z] = [xn, y, z] = [x, y, z]n (17)~~

for nǫN(O), x, y, zǫO (see the author’s book pp. 18-19). Suppose first that O is not simple. Then O = ∆ ⊕ ∆(∆ = ∆ j) where ∆ is an ideal. The elements of H (O, j) are the elements a + a, aǫ∆. Hence H (O, j) = Oo. Since these are in N(O) every aǫN(O) so ∆ and 150 ∆ ⊆ N(O). Then O = N(O) is associative. Also if a , 0 is in ∆ then a + a is invertible which implies a is invertible. Hence ∆ is a division algebra and we have case I. From now on we assume O simple. Then its center C(O)(defined as the subset of N(O) of elements which commute with every xǫO) is a field over Φ (see the author’s book p.207). It follows that Γ = H (O, j) ∩ C(O) is a field over Φ. We can regard O as an algebra over Γ when we wish to do so. We note first that the following conditions on aǫO are equvalent: (i)aa , 0, (ii) as has a right inverse (iii) aa , 0, (iv) a has a left inverse, Asssume (i). Then aa is invertible in N(O) so we have ab such that (aa)b = 1. Then a(ab) = 1 and a has a right inverse. Hecne (i) ⇒ (ii). Next assume aa = 0. Then 0 =(a, a)b = a(ab) and ab , 1.Hence (ii) → (iii). By symmetry, (iii) → (iv) → and (iv) → (i). Let zǫz, aǫO. Then (az)(az) = (az)(z a) = (az)(z(a + a)) − (az)(za) = ((az)z)(a + a) − a(zz)a = 0. Hence azǫz. Also z = z so zaǫz . Moreover, z is closed under multiplication by elements of Φ and 1 < z. Hence if z is closed under addition it is an ideal , O, of (O, j) and so z = 0. Then every non-zero element of O has a left and a right inverse in O. Suppose z = 0. If O is associative (O, j) is an associative algebra with involution and we have case II. Next assume O = N(O). We claim that in this case N(O) = C(O). By (17), if nǫN(O), xǫO, [nx] = nx − xnǫN(O) and n commutes with all associators Direct verification shows that if x, yǫO, nǫN(O) then [xy, n] = [xn]y + x[yn] where 151 [ab] = ab − ba and x[x, y, z] = [x2, y, z] − [x, xy, z]. The last implies that 0 = [x[x, y, z], n] = [xn][x,, y, z]. Hence we have

~~[xn][x, y, z] = 0, nǫN(O), x, y, z, ǫO (18) 8. A theorem on alternative algebras with involution. 107~~

~~Bilinearization of this gives~~

~~[xn][w, y, z] + [wn][x, y, z] = 0 (19)~~

Suppose x < N(O). Then we can choose y, z such that [x, y, z] , 0 so this has a right inverse. Since [x, n]ǫN(O) this and (18) imply [xn] = 0. If xǫN( ), (19) gives [x, n][w, y, z] = 0. Since N(O) , O we can choose [w, y, z] , 0 and again conclude [xn] = 0. Hence [xn] = 0 for all x and N(O = C(O) if z = 0 and O is not associative. In this case xxǫC(O) ∩ H (O, j) =Γ. Also we have no absolute zero divisors since O is a division algebra. Treating (O, j) as a algebra over Γ we have a composition algebra. Since O is not associative we have the octonion case and we shall have case IV if we can show that Oo = Γ. Since N(O) ⊆ Γ for an actonion algebra over a ﬁeld, Oo ⊆ Γ. To prove the opposite inequality it is enough to show that every element of Γ is a 2 trace. Now O contains a quadratic algebra Φ[u] in whixh u u + ρ1 and u = 1 − u. Thus u + u = 1 and if γǫΓ then γ = γu + γu is a trace. It remains to consider the situation in which z is not closed under addition. Then we have z1, z2ǫz such that z1 +z2 = u is invertible. Hence −1 e1 + e2 = 1, ei = ziu ǫz and e1 + e2 = 1. Also eiei = 0 and e1 = e1(e1 + e2) = e1e2 = (e1 + e2)e2 = e2, Then e2e1 = 0, e1 + e2 = 1 152 so the ei are orthongonal idempotents and e1 = e2, e2 = e1. Let O = O11 ⊕ O12 ⊕ O21 ⊕ O22 be the corresponding Pierce decomposition (see shafer [1] pp. 35-37 and the author’s [2’ pp. (165-166). Since O is simple O12 + O21 , 0 and Since O12O21 + O12 + O21 + O21O12 is an ideal, O12O21 = O11, O21O12 = O22. Also since e1 = e2, e2 = e1 we have O11 = O22, O22 = O11, O12 = O12, O21 = O21. Let x = x11 + x12 + x21 + x22 where xi jǫOi j. Then the Pierce relations give

~~x11 x11 = x22 x22 = x11 x21 = x22 x12 = x12 x22 = x21 x11 = 0.~~

Also x12 = e1 xe2ǫz since eiǫz so x12 x12 = 0. Similarly, x21 x21 = 0. If yǫO12, y + yǫO12 ∪ Oo ⊆ z ∩ Oo = 0. Similarly, if yǫO21, y + y = 0. Hence x11 x12 + x12 x11 = 0 = x22 x21 + x21 x22. Combining we see that

~~xx = x11 x22 + x22 x11 + x12 x21 + x21 x12 = y + y 108 3. Structure Theory~~

where y = x11 x22+x12 x21ǫO11. We show next that if yǫO11 then y = yǫΓ. Since y + yǫOo ⊆ N(O) it is enough to show that y + y commutes with every xǫO. We have seen that if xi jǫOi j then xi j = −xi j if i , j so xii xi j = −xi j xii = xi j xii. Hence if yǫO11 then (y + y)xi j = yxi j + yxi j = xi jy + xi jy = xi j(y + y). Also

~~(y + y)(x12 x21) = ((y + y)x12)x21 = (x12(y + y))x21~~

~~= x12((y + y)x21) = x12(x12(x21(y + y))~~

~~= x12 x21(y + y).~~

153 Similarly, [y+y, x21 x12] = 0 so y+y commutes with every x and y+yǫΓ. Thus we have xxǫΓ. We claim that if Q(x, y) = xy + yxǫΓ then this is non-degenerate. The formulas (12) show that the set of z such that Q(z, x) = 0 for all xǫO is an ideal of (O, j). Hence if this is not 0 it contains 1. But Q(1, e1) = e1 + e1 = 1. Hence Q(x, y) is non-degenerate. Then the proof of Theorem 7 shows that we have one of cases II-V of that theorem, one sees easily that the only possibilities allowed here are (O, j) is split quaternion or split octoion over Γ. As before, we have Oo =Γ in the octonion case and we are in case IV. In the split quaternion case, O = Γ2, the argument used before shows that Oo ⊇ Γ. If the characteristic is , 2Γ = H (O, j) , hence, Oo = Γ. If the characteristic is two then it is easily seen that we have a base of matrix units ei j such that e11 = e22, e22 = c11, e12 = e12, e21 = e21. If aǫOo, aǫH (O, j) so a = α1 + βe12 + γe21,α,β,γǫΓ. Since e12ae12 = γe21ǫO0 and the non-zero elements of Oo are invertible, γ = 0. Similarly β = 0 so again Oo =Γ. Thus we have case III.

~~9 Simple quadratic Jordan algebras of capacity two~~

Let J be of capacity two, so 1 = e1 + e2 where the ei are completely primitive orthogonal idempotents, J = J11 ⊕ J12 ⊕ J22 the corresponding Pierce decomposition. Then Jii is a division algebra. Put = m J12. If xiǫJii then νi : xi → V xi the restricition of Vxi to m is a (q) homomorphism of Jii into (End m) . Since Jii is a division algebra, 9. Simple quadratic Jordan algebras of capacity two 109

νi is a monomorphism. Hence Jii is special so this can be indentiﬁed 154 (q) with a subalgebra of S (Jii) where S (Jii) is the special universal envelope of Jii (see §1.6). If π is the main involution of S (Jii) then Jii ⊆ H (S (Jii),π) and s(Jii) is generated by Jii. The homomorphism νi has a unique extension to a homomorphism of S (Jii). The latter permits us to consider m as a right s(Jii) module in the natural = = way. If mǫm and xiǫJii then the deﬁnitions give mxi mVxi m ◦ xi and if xi, yi,..., ziǫJii then

~~m(xiyi ... zi) = (... ((m ◦ xi) ◦ yi) ◦ ... ◦ zi) (20)~~

~~Also by the associativity conseqences of the PD theorem (m ◦ x1) ◦ x2 = (m ◦ x2) ◦ x1 from which follows~~

~~(ma1)a2 = (ma2)a1, aiǫS (Jii). (21)~~

We recall that if mǫm, either m2 = 0 or m is invertible (Lemma 2 2 of §6). Suppose m = 0. Then xiUm = 0 for xiǫJIii (proof of Lemma 2 = 2 + 2 + = 2 §6). Then (xi ◦ m) xi Um m Uxi xiUm ◦ xi(QJ30) 0. This implies that if m is invertible and x , 0 then mxi is invertible. Otherwise 2 2 2 (mx2) = 0 and (mxiyi) = (mxi ◦ y) = 0 for all yiǫJii. If we choose 2 yi to be the inverse of xi and Jii we obtain the contradiction m = 0. If = + 2 = m is invertible and, as in Lemma 2 of §2.3. we put u c1 m Ue2 , ·v + −2 ˜ = (v) = e1 m Ue2 then we have seen that in the isotope J J , u1 = 2 e1 and u2 m Ue2 are supplementary orthogonal idempotents which are strongly connected by m. The Pierce submodule Ji j relative to the ui coincides with Ji j. Moreover, J11 = J˜11 as quadratic Jordan 155 = = = ˜ = algebras, and for mǫm J12 m J12 and x1ǫJ11 we have mVx1 mV x1 . Hence the S (J11) module structure on m is unchanged in passing from J to J˜. Also J22 and J˜22 are isotopic so J˜22 is a division algebra and u1 and u2 are completely primitive in J˜. Clearly, J is of capactiy two. Since the isotope J˜ is determined by the choice of the invertible element m it will be convenient to denote this as Jm. We shall now assume J simple and we shall prove the following structure theorem which is due to Osborn [11] in the linear case and to McCrommon in the quadratic case. 110 3. Structure Theory

Theorem 10. Let J be a simple quadratic Jordan algebra of capacity two. Then either J is isomorphic to an outer ideal ∋ 1 of a quadratic Joradan algebra of a non-degenerate quadratic form on a vector space over a field P/Φ or J is isomorphic to an outer ideal ∋ 1 of an algebra H (O2, JH) where (O, J) is either an associative division algebra with j involution or O = ∆ ⊕ ∆ , ∆ an associative division algebra and JH is −1 −1 the involution X → H X H,HǫH (O2). We have seen in §1.11 that H (O2, JH) is isomorphic to the H- isotope of H (O2). Now consider Jord (Q, 1) the quadratic Jordan algebra of a quadartic form Q with base point 1 on a vector space over a field P. Let u be an invertible elements so Q(u) , 0. Then Q′ = Q(u)Q is a quadratic form which has the base point u−1 = Q(u)−1u since Q(u)Q(u−1) = Q(Q(u)−1u) = Q(Q(u, 1)1 − u) = Q(u)−1Q(u) = 1. Now 156 consider Jord (Q(u)Q, u−1). Put x′ = Q(u)Q(x, u−1)u−1 − x and let U′ denote the U-operator in this algebra. A straight forward calculation ′ = = shows that xUa Q(a, xUu)a − Q(a)xUu xUvUa. It follows that Jord (Q(u)Q, u−1) is identical with the u-isotope of Jord (Q, 1). These remarks show that to prove Theorem 10 it suffices to show that there exists an isotope of J which is isomorphic to an outer ideal containing 1 in a Jord (Q, 1) with non-degenerate Q(over a field) or to an outer ideal containing 1 in an H(O2). By passing to an isotope we may assume at the start that 1 = e1 + e2 where the ei are orthogonal completely primitive and are strongly connected by an element uǫJ12. The proof will be divided into a series of lemmas. An important point in the argument will be that except for trivial cases m = J12 is spanned by invertible elements. This fact is contained in

Lemma 1. Suppose J11 , {0, ±e1} (that is, J11 , Z2 or Z3). Let m, nǫγm, m invertible. Then there exist x1 , 0, y1 , 0 in J11 such that x1 ◦ m + y1 ◦ n is invertible. Any element of m is a sum of invertible elements. Proof. Since xǫm is either invertible or x2 = 0, if the result is false, then 2 x1 ◦ m + y1 ◦ n) = 0 for all x1 , 0, y1 , 0 in J11. By QJ30 and the PD theorem the component in J22 of this element is 2 + 2 = x1Um y1Un{x1 ◦ m, e1, y1 ◦ m} 0 (22) 9. Simple quadratic Jordan algebras of capacity two 111

~~Take (x1, y1) = (x1, e1), (e1, y1), (e1, e1), (x1, y1) and add the ﬁrst two equations thus obtained to the negative of the last two. This gives~~

~~{z1 ◦ me1w1 ◦ n}, z1 = x1 − e1, w1 = y1 − e1 (23)~~

~~157 Using this and (22) we obtain~~

~~2 + 2 = z1Um w1Un 0 if z1, w1 , 0, −e1 in J11. (24)~~

2 + 2 = 2 2 = In particular, w1Um w1Un 0, so, by (24) (z1 − w1)Um 0 and 2 = 2 , , since m is invertible, z1 w1 is z1, w1 0, −e1. Let z1 0, ±e1 so , 2 = 2 = 2 + = z1 − e1 0, −e1 and so z1 (z1 − e1) z1 − 2z1 e1. Hence 2z1 e1. Also since −z1 , 0, ±e1 we have also −2z1 = e1. Then 4z1 = 0 and since J11 is a quadratic Jordan division algebra, 2z1 = 0. This gives e1 = 0 contrary to J11 , 0. Hence the ﬁrst statement holds. For the second we note that m contains an invertible element m and if n is any element of m then there exist x1, y1 , 0 in J11 such that p = x1 ◦ m + y1 ◦ n is invertible. Then if z1 is the inverse of y1 in J11, n = z1 ◦ p − z1 ◦ (x1 ◦ m) and z1 ◦ p and −z1 ◦ (x1 ◦ m) are invertible elements of m. We are assuming that e1 and e2 are strongly corrected by uǫm. Then η = Uu is an automorphism of period two in J , η maps m onto itself and exchange J11 and J22. Hence η deﬁnes an isomorphism of J11 onto J22. This extends uniquely to an isomorphism η of (S (J11),π) 3 η 3 onto (S (J22),π). We have u = u(§2.3). Hence u = uUu = u = u. We shall now derive a number of results in which u plays a distinguished roles. These will be applied later to any invertible mǫm by passing to (v) η η η the isotope Jm(= J as above). Let xǫJ . Then (x◦n) = x ◦u ◦u. η = = = = = Also x ◦ u xUuVu xUu,u2 xUu,1 xVu x ◦ u. Thus we have

~~(x ◦ u)η = xη ◦ u = x ◦ u, xǫJ (25)~~

~~We prove next 158~~

~~+ η = = + η = Lemma 2. Let mǫm. Then m m ux1, x1 (u◦m)Ue1 . Ifm m 0 then u ◦ m = 0. 112 3. Structure Theory~~

= = = Proof. Let m, nǫm. Then mUn mUe1,e2 Un e jVm,ei Un −e jVn,ei + + = + 2 + Um,n e jUnVei , m e1Um,nVei , n (by QJ9) −{nnm} {n Uei eim} {(m◦ = 2 + 2 n)Uei ein} (PD 6 and its bilinearization). Hence mUn −n ◦ m n Uei ◦ + + = + + m (m ◦ n)Uei ◦ m (replacing ei by ei e j). Since 1 Uei Ue j Uei,e j this gives

~~= 2 + mUn −n Ue j ◦ m (m ◦ n)Uei ◦ m, n, ∈ m (26)~~

= η = = = + = Taking n u we get m mUn −nµn −m ux1, x1 (m◦u)Ue1 . η This is the ﬁrst statment of the lemma. If m + m = 0 we have x1 = = η = = (u ◦ m)Ue1 0. Applying η gives (u ◦ m )Ue2 0 (u ◦ m)Ue2 , by (25). + = = Since u ◦ mǫJ11 J22 the realtions (u ◦ m)Uei 0, i 1, 2, imply u ◦ m = 0.

~~π Lemma 3. If mǫm satisﬁes mx1 = mx1 for all x1ǫJ11 then ma = η ma , aǫS (J11).~~

Proof. Since J11 generates S (J11) it suﬃces to prove the conclusion for a = x1, x2 ... xk, xi ∈ J11. We use induction on k. As- π η π η = + = sume m(x1, x2 ... xk) m(x1 ... xk) . Then m(x1 ... xk 1) mxk+1 π = π n = η η = (x1 ... xk) m(x1 ... xk) xk+1 (by (21)) m(x1 ... xk) xn+1 m η η + = 159 (x1 ... xk 1) which proves the inductive step. We have ux1 ux1, x1ǫ J11, by (25). Hence Lemma 3 and (25) give

~~π η η ua = ua = (ua) , aǫS (J11). (27)~~

~~Now suppose ua = 0 for an aǫS (J11). Then for bǫS (J11), uba = ubηπa = ubπηa = uabπη (by (21))= 0. Hence we have~~

~~Lemma 4. If ua = 0 for aǫS (J11) then uba = 0 for all bǫS (J11). We prove next~~

π Lemma 5. Let nǫn = uS (J11), aǫS (J1). Then n(a + a ) = nx1, x1 = ◦ π = = η (u ua)Ue1 . Also if y1ǫJ11 then n(a y, a) nz1 where z1 y1UuaJ11. π + = η + = = Proof. We have ua ua (ua) (ua) (by (5)) ux1, x1 (u ◦ ua)Ue1 π π (by Lemma 1). Hence u(a +a−x1) = 0 so, by Lemma 4, n(a +a−x1) = 0 for all nǫuS (J11). This proves the ﬁrst statment. To prove the second 9. Simple quadratic Jordan algebras of capacity two 113

ﬃ π = η π + π = it su ces to show that n(a y1a) n(y1Uua) and n(a y1b b y1a) η n(y Uua,ub) for a = t1 ... z1, y1, t1,..., z1ǫJ11, bǫS (J11, bǫ(J11). For m η = η = η mǫ , we have y1Umt1 y1Um◦t1 y1UmUt1 by QJ17 and the PD the- = η = η = orem. Iteration of this gives y1Uua y1Uut1 ... z1 y1Uuut1 ... Uz1 = = π y1Ut1 ... Uz1 z1 ... t1y1t1 ... z1 (in A(J11)) a y1a. Next we use the ﬁrst statement of the lemma to obtain

π + π = π n(a uy1b b y1a) n((u ◦ ua y1b)Ue1 , nǫn (28) m = = If m, nǫ , y1ǫJ11 then {my1n} ((m ◦ y1) ◦ n)Ue2 (PD theorem) = ((m ◦ y1) ◦ (n ◦ e1)Ue2 {m ◦ y1e1n}. Since {my1n} is symmetric in m and n we have {m ◦ y1e1, n} = {me1n ◦ y1}. If we take a = t1 ... z1, 160 t1,..., z1ǫJ11 then we can iterate this to obtain

~~π {my1ae1n} = {m, y1, na } = {my1e1na} (29)~~

Since J11 generates S (J11) this holds for all ǫS (J11). In particular, we have π π π {u.e1ub y1a} = {ua y1ub } (30) π = π + π = π Now (u ◦ ua y1b)Ue1 {ue1ua y1b}Ue1 {ue2uay1b}Ue1 {ue2ua = π = π η = y1b}Ue1 (PD theorem) {ue2ua y1b} {ue1ub y1a} (by (27)) π η {ua y1ub} (by (30))= {uay1ub} (by (27)). Going back to (28) we obtain π + π = η = η n(a y1b b y1a) n{uay1ub} ny1Uua,ub as required. This completes the proof. We obtain next an important corollary of Lemma 5 nam- ley

~~Lemma 6. Let n be as in Lemma 5 and let ua be invertible, aǫS (J11). Then there exists abǫS (11). Such that nab = n = nba, n.~~

Proof. The hypothesis implies that Uua is a invertible. Hence z1 = , −1 e2Uua 0 in J11. Then z1 exits in S (J11). Applying the second π part of Lemma 5 to y1 = e1 shows that na a = nz1 holds for all nǫn. −1 = = −1 η = π Hence replacing n by nz1 gives nba n for b z1 a. Also (ua) ua π is invertible so w1 = e2Uuaπ invertible in S (J11) and naa = nw1, nǫn. 161 −1 = = π −1 Multiplying by w1 on the right gives nac n, c a w1 . It now follows that nb = na and nab = n = nba, nǫn. 114 3. Structure Theory

Lemma 7. If x2ǫJ22 and mǫm is invertible then mx2 = mx1m1, x1 = −1 = 2 x2Um , m1 m Ue2 . = = −1 = = Proof. We have x2 x1Um, where x1 x2Um . Then mx2 x2 ◦ m = = 2 = 2 = = x1UmVm x1Um,m2 {mx1m } {mxx1m Ue1 } {mx1m1} (m ◦ x1) ◦ m1 = mx1m1. We prove next

~~= η = ∗ = = η Lemma 8. If mo {mǫm|m −m}, m {mǫm|mx1 mx1, x1ǫJ11} ∗ then mo ⊆ m ∪ n, n = u ∈ s(J11).~~

∗ Proof. Let mǫmo and assume mǫm . Then we have an x1ǫJ11 such that = η , η = = n mx1 − mx1 0 and we have to show that mǫn. Since m −m, n + η = = mx1 (mx1) uy1, y1 (u◦mx1)Ue1 , byLemma2. Now m is invertible 2 2 since otherwise, m = 0 and hence xUm = 0 and x Um = 0 for x = x1 − η + 2 = 2 = = x1ǫJ11 J22. Then n (m◦ x) 0 by QJ 30. However, n uy1 and Since y1ǫJ11 is , 0 u is invertible, n is invertible. This contradiction = − η = − η −1 2 proves m invertible. We have n mx1 mx1 mx1 m(x1Um )(m Ue1 ) = = − η −1 2 (by Lemma 7) a x1 (x1Um )(m Ue1 ǫ(J11). Thus

~~n = uy1 = ma, y1 , 0 in J11, aǫsu(J11) (31)~~

We now apply lemma 6 to m(replacing u) in the isotope Jm. Since ma = n is invertible in J , hence in Jm, and since the S (J11) module 162 structure on m is unchanged in passing from J to Jm it follows from Lemma 6 that there exists ab ∈ S (J11) such that mab = m. Then m = nb = uy1bǫn as requried.

As before, Let ν1 be the monomorphism x1 → V x1 of J11 into (q) (End m . Also let ν1 denote the (unique) extension of this to a homo- ν ν morphism of s(J11) into End m and let E1 = S (J11) 1 . Then S (J11) 1

~~is the algebra of endomorphisms generated by the V x1,x1 ǫJ11. We shall now prove the following important result on E1.~~

Lemma 9. The involution π in S (J11) induces an involution π in E1. If ν1 we identity J11 with its image J11 in E1 then J11 ⊆ H (E1π), J11 π contians 1 and every a x1a, x1ǫJ11, aǫE1. Also (E1,π) is simple and the nonzero elements of J11(⊆E1) are invertible. 9. Simple quadratic Jordan algebras of capacity two 115

Proof. If J11 = {0, ±e1} then E1 = Z2 or Z3 and the result is clear. From now on we assume J11 , {0, ±e1} so lemma 1 is applicable. In particular, m is spanned by invertible elements. To show that π in- π duces an involution in E1 we have to show that (ker ν1) ⊆ ker ν1 and for this it suffices to show that if kǫ ker ν1 and mǫm is invertible then π mk = 0. Let Jm be the isotope of J defined by m as before. By (27) applied to m (in place of u) we have mkπ = (mk)η = 0. Hence the first statement is proved. It is clear that J11 ⊆ H (E1,π) and 1ǫJ11. Let x1ǫJ11, aǫS (J11), m an invertible element of m. Then, by Lemma π 5, there exists an element ymǫJ11 such that xa x1a = xym for all x in mS (J11). Let n be a second invertible element of and let ynǫJ11 satisfy π xa x1a = xyn, xǫnS (J11). As in Lemma let u1 , 0, v1 , 0 be elements of J11 such that p = mu1 + nv1 is invertible and let ypǫJ11 satisfy π xa x1a = xyp, xǫ pS (J11). Suppose ym , yn. Then d1 = ym − yn , 0 −1 −1 π = is invertible in J11 with inverse d1 . Then nv1d1 (a x1a − yn) 0 so 163 −1 π = −1 π = −1 = pd1 (a x1a − yn) mu1d1 (a x1a − yn) mu1d1 (ym − yn) mu1. π Hence mǫ pS (J11) and, similarly, nǫ pS (J11). Then ma x1a = myp π and na y1a = nyp. This implies yp = ym = yn contradicting ym , yn. π Hence there exists an element y1ǫJ11 such that ma x1a = my1 for all invertible mǫm. Since m is spanned by invertible elements this gives the second statement of the Lemma. We note next that the non-zero elements of J11 are invertible in E1 since J11 is a division subalgebra of (q) E1 . π π It remains to show that (E1,π) is simple. Let aǫE1 then a a, aa ǫ J11 and the non-zero element of J11 are invertible; the proof of the theorem of Herstein-Kleinfeld. Osbon-McCrimmon shows that either aaπ = 0 = aπa or a is invertible. Let u be an element strongly connecting π = η = = e1 and e2, as before. By Lemma 5, we have ua a ue1Uua ue2Uua 2 π = 2 = 2 + u((ua) Ue1 ). Hence a a 0 implies (ua) Ue1 0. Since (ua) ǫJ11 2 J22 this implies (ua) not invertible. Then ua is not invertible and 2 π π (ua) = 0. Thus we see that if aǫE1 then either a a and aa are invertible on (ua)2 = 0. π π Now let Z be a proper ideal of (E1,π) and let zǫZ. Then z+z , z zǫZ∩ π π J11. Since Z contains no invertible elements, we have z + z = 0 = zz . 2 2 η Hence z = 0 and (uz) = 0. By Lemma 2, if mǫm, m + m = ux1, x1 = 116 3. Structure Theory

= = Φ (u◦m)Ue1 . Then f (m) x1 (u◦m)Ue1 deﬁnes a -homomorphism of m into J11. Since η is an automorphism of J mapping m onto m and 164 J11 onto J22 it is clear that η deﬁnes an isomorphism of E1 onto the ν2 subalgebra E2 of End m generated by J22 extending the isomorphism η η η of J11 onto J22. Moreover, we have (ma) = m a , mǫm, aǫE1. Iter- ation of the result of Lemma 7 shows that if m is an invertible element η of m then for any aǫE1 there exists abǫE1 such that ma = mb. Then u f (ma) = ma+(ma)η = ma+mηaη = m(a−aη)+(m+mη)+(m+mη)aη = m(a − b) + u f (m)aη = m(a − b) + uaη f (m) = m(a − b) + uaπ f (m) (by (27)). Hence m(a − b) = u( f (ma) − aπ f (m)) (32)

In particular, taking a = zǫZ we obtain w so that mzη = mw and π π m(z−w) = ur, r = f (mz)+z f (m)ǫE1. Since wzǫZ we have w z+z w = 0. Also mw2 = mzηw = mwzη = m(zη)2 = 0 since z2 = 0. Hence w is not invertible and consequently wπw = 0. Then (z−w)π(z−w) = zπz−zπw− wπz + wπw = 0. This relation and the second part of Lemma 5 applied to the isotope Jn imply that m(z−w) is not invertible in this isotope. Hence m(z−w) is not invertible in J and consequently (ur)2 = (m(z−w))2 = 0. Then a reversal of the argument shows that rπr = 0. Then r is not invertible and since r = f (mz) + z f (m) and z is in Z which is a nil ideal, f (mz) is not invertible. Since f (mz)ǫJ11 it follows that f (mz) = 0. Hence we have mz + (mz)η = 0. By the second part of Lemma 2, this = = = implies mz ◦ u 0. Then 0 (mz ◦ u)Ue2 {ue1mz} (by linearization 2 = = π = = of x Ue2 e1Ux, xǫm) {uz e1m} (by (29)) −{uze1m} −(uz ◦ m)Ue2 . η = If we replace m by m in this and apply η we obtian (uz ◦ m)Ue1 0. Hence we have proved that uz◦m = 0 for all invertible m. It follows that 165 this holds for all mǫm and since (uz)2 = 0, Lemma 2 of §6, shows that this is an absolute zero divisor. Since J is simple we have uz = 0. On passing to the isotpe m we can replace u by any invertible mǫm. Then mz = 0 for all invertible mǫm so z = 0. Hence Z = 0 and (E1,π) is simple. Lemma 9 shows that (E1,π, J11) is an associative coordinate algebra satisfying the hypotheses of the Herstein-Kleinfeld-Osborn- McCrimmon theorem. Also in the present case J11 generates E1. This excludes case III given in that theorem so we have only the possibilities 9. Simple quadratic Jordan algebras of capacity two 117

I and II given in the theorem. It is convenient to separate the cases in which E1 is a division algebra into the subcases: π non-trivial and π = 1 in which case E1 is ﬁeld. Accordingly, the list of possibilities for (E1,π, J11) we

~~π I E1 = ∆ ⊕ ∆ , ∆ an assoiative division algebra, H (E1,π) = J11.~~

~~II E1 an associative division algebra, π , 1~~

~~III E1 a ﬁeld, π = 1.~~

~~Lemma 10. If E1 is of thye I or II then m = uE1(= uS (J11)) and if E1 = ∗ = = η is of type III then m m {mǫm|mX1 mX1, X1ǫJ11} (as in Lemma 7).~~

Proof. We show ﬁrst that in types I and II any mǫm such that mη = −m as contained in uE1. By Lemma 7, it is enough to show this for for m η = − η = π = η = with m m and mx1 mx1, x111 . Then, by Lemma 3, ma ma ma, aǫS (J11). Now in types I and II there exists an invertible element 166 π a in E1 such that a = +b − b . In the case I we choose b invertible in ∆ π π π π then b is invertible in ∆ and a = b − b , is invertible in Ei = ∆ ⊕ ∆ . in case II we choose an element b is the division algebra E1 such that bπ , b. This can be done since π , 1. Then a = b − bπ , 0 is invertible. η π η Now let m be as indicated (m = −m, ma = ma , aǫS (J11)). Then π η η η η ma = mb − mb = mb − mb = mb + m b = ma + (ma) = ux1, x1ǫJ11. −1 Then m = ux1a ∈ uS (J11). η Suppose we have type I and let mǫm. Then m + m = ux1, x1ǫJ11. π η π Since the type is I, x1 = a + a , aǫS (J11). Then m + m = ua + ua = ua + (ua)η (by (27)). Hence (m − ua)η = mη − (ua)η = ua − m. Then m − uaǫuS (J11) and mǫUS (J11) = uE1. Suppose we have type II. Then J11 , {0, ±e1} so m is spanned by invertible elements. Hence it suﬃces to show that if mǫm is invertible η then mǫuE1. By Lemma 2, m + m = u f (m) where f (m)ǫJ11. By η Lemma 6, if aǫE1 there exists a bǫE1 such that ma = mb. By (31), π m(a − b) = u( f (ma) − a f (m)) where f (m), f (ma)ǫJ11. If a − b is 118 3. Structure Theory

invertible for some a this implies mǫuE1. Otherwise, since E1u a division algebra, a = b for all a. Then f (ma) = aπ f (m) and applying II, f (ma)π = f (ma) = f (m)a. Hence aπ f (m) = f (m)a. In particular, x1 f (m) = f (m)x1, x1ǫJ11 and since J11 generates E1, f (m) is in the π center of E1. Then (a − 1) f (m) = 0. Since we can choose a so that π π a − a is inverible, f (m) = 0. Then m + m = 0. and mǫuE1 by the result proved before. 167 Now suppose we have type III. The case J11 = {0, ±e1} is trivial so we may assume m is spanned by invertible elements. It suﬃces to = η show that if mǫm is invertible then mx1 mx1, x1ǫJ11. As in the last η case, m + m = u f (m) and if aǫE1 then there exists bǫE1 such that m(a − b) = u( f (ma) − a f (m)) (since π = 1). If a − b , 0, m = uc, cǫE1 = = = η = η = η and mx1 ucx1 ux1c (by commutativity of E1) ux1c ucx1 mx1 (by (27) and (21)). Hence the result holds in this case. It remains to conisder the case in which a = b for all a. Then f (ma) = a f (m) = η η f (m)a, aǫE1. Then mx1 + (mx1) = u f (mx1) = u f (m)x1 = (m + m )x1. η = η = η Then (mx1) m x1 and mx1 mx1, x1ǫJ11 as required. We can now complete the

Proof of Theorem 10. Suppose ﬁrst (E1,πJ11) is of type I or II. Since = η = m uE1 and J11 J22 any element of J can be written in the form + η + = x1 y1 ua, x1, y1ǫJ11, aǫE1. Also a is unique since ua 0 implies ma = (uE1)a = 0 (Lemma 4). Hence the mapaping + η + + + ζ : x1 y1 ua → x1[11] y1[22] a[21] (33)

is a module isomorphism of J onto H ((E1)2, J11). It is clear that ′ = the mapping η : X → 1[12]X1[12] XU1 [12] is an automorphism in η ζ ζ η′ H ((E1)2, J11) and by inspection we have (X ) = (x ) . We shall now show that ζ is an algebra isomorphism. Because of the properties of the Pierce decomposition, the relation between η and η′ and the quadratic Jordan matrix algebra properties QN1 − QM6 this will follow if we can 168 establish the following formulas: ζ = (i) (x1Uy1 ) Y1 x1y1[11]

~~ζ π (ii) (x1Uua) = (a x1a)[11] 9. Simple quadratic Jordan algebras of capacity two 119~~

~~ζ π (iii) ((ua)Uub) = ba b[21]~~

~~π ζ π (iv) {x1ua ub} = (X1ab + 1(X1ab) )[11]~~

~~ζ (v) {x1y1ua} = (ay1 x1)[21]~~

~~η ξ = (vi) {y1uax1} y1ax1[21]. = Since x1Uy1 y1 x1y1 in E1, (i) is clear. For (ii) we use Lemmas 5 π = η and 10 to obtain a x1a x1Uua. For (iii) we have~~

= 2 + (ua)Uab −(ub) Ue1 ◦ ua (ua◦)ubUe2 ◦ ub (by(26)) = −e1Uub ◦ ua + e1Uua,ub ◦ ub (PD) η η π η = + π π −e1Uub ◦ ua ((e1Uua ,ub ) ◦ ub ) = −uabπb + (u(bπabπ + bπbaπ))η = −uabπb + ubaπb + uabπb = ubaπb.

π = π = This implies (iii). For (iv), we use {x1ua ub} ((x1◦ua )◦ub))Ue1 π η π π ◦ = π = + (ua x1 ub)Ue1 e1Uua x1,ub x1ab b a x1. This gives (iv). For (v) = = η = we have {x1y1ua} (ua◦y1)◦ x1 uay1 x1. (vi) follows from {y1uax1} η = η = (ua ◦ y1) ◦ x1 uy1ax1 uy1ax1 This completes the proof of the first = η part. Now suppose we have type III. Then mx1 mx1, mǫm, x1ǫJ11. 169 η Also E1 is a field and π = 1. Hence, by Lemma 3, ma = ma , aǫE1. We 2 consider the mapping Q : m → −m Ue1 of m into E1. We claim that Q is quadratic mapping of m as E1 module into E1. If mǫm, x1ǫJ11 then 2 = ◦ 2 = 2 = 2 = 2 2 (mx1) Ue1 (m x1) Ue1 m Ux1 Ue1 m Ue1 Ux1 x1(m Ue1 )x1 since E1 is commutative. It follows that for a = x1 ... z1, x,... z1ǫJ11, 2 = 2 2 we have (ma) Ue1 (m Ue1 )a . We show next that if m, nǫm and = = y1ǫJ11 then (mx1 ◦ n)Ue1 ((mon)Ue1 x1. Since this is clear for n 0 ffi = and both sides are in E1 it su ces to show that (mx1 ◦ n)Ue1 ◦ n = (((m ◦ n)Ue1 )x1) ◦ n. Since E1 is commutative (((m ◦ n)Ue1 )x1) ◦ n = = = (n ◦ x1) ◦ ((m ◦ n)Ue1 ) {nx1(m ◦ n)Ue1 } (PD theorem) {nx1m ◦ n} η x1Un,m◦n = x1(UnVm +VmUn) (QJ19) = m(x1Un) +(mx1)Un. By QJ33, = = = we have (mx1)Un (mx1)ve1 UnVe1 (mx1)VnUe1 Vn − (mx1)Ve1Un 120 3. Structure Theory

~~(mx1)VnUe Vn − (mx1)V 2 (PD 6)= ((mx1 ◦ n)Ue ) ◦ n − (mx1)V 2 . 1 n Ue2 1 n Ue2 This and the following relation give~~

~~(((m ◦ n)Ue1 )x1) ◦ n − (mx1 ◦ n)Ue1 ) ◦ n (34) η = m(x1U − n) − (mx1)V 2 n Ue2~~

The left hand side is a multiple (in E1) of n and the right hand side is a multiple of m. Hence if m and n are E1 independent then we obtian = (((m ◦ n)Ue1 )x1) ◦ n (((mx1)Ue1 )x1) ◦ n which gives the required relation. Now suppose m = na, aǫE1. Then again (34) will yield the result η 170 provided we can prove that n(x1Un) = (nx1)V 2 . This follows since n Ue2 η = = = = 2 = 2 = n(x1Un) n(x1Un) x1UnVn x1Un,n2 {nx1n } {nx1n Ue2 } 2 (n ◦ x1) ◦ n Ue = (nx1)V 2 . 2 n Ue2 = 2 = We have now proved that for Q(m) −m Ue1 we have Q(ma) 2 a Q(m) for all a = x1 ... z1, x1,..., z1ǫJ11 and Q(mx1, n) = x1Q(m, n). The latter implies that Q(ma, n) = aQ(m, n) m, nǫm, aǫE1. This and 2 the ﬁrst result imply that Q(m) = a Q(m) for all aǫE1. Then Q is 2 = 2 η a quadratic mapping. We note next that m Ue2 (m Ue1 ) . Since 2 = 2 η ﬃ 2 = 2 m(m Ue1 ) m(m Ue1) it su ces to show that m(m Ue1 ) m(m Ue2 ). 2 = 2 = 2 = 2 = = We have m(m Ue1 ) m Ue1 ◦m {m Ue1 e1m} {m e1m} e1Um,m2 = 2 = 2 = 2 = 2 e1UmVm m Ue2 Vm m ◦ m Ue2 mm Ue2 mm Ue2 . Thus 2 = 2 + 2 = 2 + 2 η m m Ue1 m Ue2 m Ue1 (m Ue1 ) . Since the elements mǫm such that m2 = 0 and m ◦ n = 0, nǫn, are absolute zero divisors it now follows that Q is non-degenerate.

We now introduce K = f1ǫ1 ⊕ f2ǫ1 ⊕ ma direct sum of m and two one dimensional (right) vector spaces over E1 and extend Q to K by deﬁning Q( f1a + f2b + m) = ab + Q(m), a, b ∈ E1. Then Q is a non-degenerate quadratic form on K and Q( f ) = 1 for f + f1 + f2. Hence we can form ′ Jord (Q, f ). It is immediate that J ≡ f1J11 + f2J11 + m is an outer ideal containing f in K = Jord (Q, f ). We now deﬁne the mapping ζ of = + η + + + J onto K by x1 y1 m → f1 x1 f1y1 m. It is easy to check that this is a monomorphism. 10. Second structure theorem. 121

~~10 Second structure theorem.~~

Let J be a simple quadratic Jordan algebra satisfying the minimum condition (for principal inner ideals). Then J contains no absolute zero divisors , 0 since these generate a nil ideal (Theorem 5). By The- orem 7. J has an isotope J˜ which has a capacity. If the capacity 171 is one then J˜, hence J , is a division algebra. If the capacity is two then the structure of J is given by Theorem 10. This implies that J itself has the form given in Theorem 10. Now assume the capacity of n J˜ is n ≧ 3 and let 1˜ = fi be decomposition of the unit 1˜ of J˜ 1 P into orthogonal completely primitive idempotents. Since J˜ is simple the proof of the First structure Theorem shows that every f j, j > 1, is connected to f1. Since we can replace J˜ by an isotope, by Lemma 2 of §2.3, we may assume the connectedness is strong. Then we can apply the strong coordinatization Theorem to conclude that J˜ is isomorphic to an algebra H (On, Oo) with the coordinate algebra (O, j, Oo). By Theorem 2.2, H (On, Oo) is an outer ideal in H (On) and the simplicity of H (On, Oo) implies that (O, j) is simple. The Pierce inner ideal determined by the idempotent 1[11] in H = H (On, Oo) is the set of elements α[11], αǫO0 ⊆ N(O). Since this Pierce inner ideal is a division algebra it follows that every non-zero element of Oo is invertible in N(O). Hence (O, j, Oo) satisﬁes the hypothesis of the Herstein- Kleinfeld-Osborn-McCrimmon theorem. Hence (O, j, Oo) has one of the types I − V given in the H − K − O − M theorem. If the type is I-IV then the consideration of chapter 0 show that On with its standard involution J1 is a simple Aritinian algebra with involution. Since H (On, Oo) is an outer ideal containing 1 in H (On) it follows that H is isomorphic to an outer ideal containing 1 in an H (a, J), (a, J) simple Artinian with involution. Then J also has this form. The remaining type of coordinate algebra allowed in the H − K −O− M theorem is an octoion algebra 172 with standard involution over a ﬁeld Γ with Γ = O. In this case we must have n = 3. Thus if we take into account the previous results we see that J is of one of the following types. 1) a division algebra, 2) an outer ideal containing 1 in a Jord (Q, 1) where (Q, 1) is a non-degenerate quadratic form with base point on a vector space, 3)an outer ideal con- 122 3. Structure Theory

taining 1 in an H (a.J) where (a, J) is simple Artinian with involution, 4)an isotope of an algebra H (O3) where O is an octonian algebra over a ﬁeld Γ/Φ with standard involution. We now consider the last possibility in greater detail. Let c γi[ii], Γ γi , 0 in . Since c is an invertible element of N(O3) it determinesP −1 t an involution Jc : X → c X c in O3. Let H (O3, Jx) denote the set of matrices in O3 which are symmetric under Jc and have diagonal elements in Γ. If αǫΓ we put α{ii} = α[ii] = αeii and if aǫO we put = + −1 , a{i j} aei j γ j γiae ji, i j. Then α[ii], a{i j}ǫH (O3, Jc) and every element of H (O3, Jc) is a sum of α[ii] and a{i j}. If XǫH (O3) −1 t then XcǫH (O3 Jc) since c (Xc) c = Xc. In view of the situation for algebras H (a, J), a associative, it is natural to introduce a quadratic Jordan structure in H (O3, Jc) so that bijective mapping X → Xc of (c) H (O3) onto H (O3, Jc) becomes an isomorphism of quadratic Jor- dan algebras. We shall call H (O3, Jc) endowed with this structure a cononical quadratic Jordan matrix algrbra. It is easy to check that 173 the elements ei = 1{ii} are orthogonal idempotents in H (O3, Jc) and ei is the unit of H (O3, Jc). The pierce spaces relative to these are = Γ = PH (O3, Jc)ii {α{ii}|ǫ , H (O3, Jc)i j {a{i j}|aǫO}, i , j. It is easy to verify that the formulas for the U-operator for elements in these submodules are identical with (i)-(x) of §1.8 with the exceptions that (ii) and (iii) become = −1 ′ α{ii}Ua{i j} γ j γiaγa{i j} (ii) = −1 ′ b{i j}Ua{i j} γ j γiaba{i j} (iii)

It is clear from these formulas that if ρ , 0 in Γ then H (O3, Jρc) = H (O3, Jc). We note next that if aiǫO, n(ui) , 0, δi = n(ui)γi and d =diag{γ1, γ2, γ3} then there exists an isomorphism of H (O3, Jc) onto H (O3, Jd) fixing the eii. First, one can verify directly that if uǫO, n(u) , 0, then the Γ -linear mapping of H (O3, Jc) onto H (O3, Jd), d = ′ diag{γ1, n(u)γ2, n(u)γ3} such that eii → eii, a{12} → au{12} , a{23} → −1 ′ ′ ′ = + −1 = u au{23} , a{13} → au{13} where a{i j} aei j δ j δiae ji, δ1 γ1, δi = n(u)γi, i = 2, 3, is an isomorphism. (Because of the Pierce relations it is sufficent to verify QM(ii)], (iii)′ and M4, §2.2). Similarly, one can define isomorphism of H (O3, Jc) onto H (O3, Jd), d =diag{n(u) 10. Second structure theorem. 123

γ1, γ2, n(u)γ3} or diag {n(u), γ1n(u)γ2, γ3}. Combining these and taking = account the fact that H (O3, Jρc ) H (O3, Jc) we obtain an isomorphism of H (O3, Jc) onto H (O3,d ) ﬁxing the eii, d as above. We shall now prove the following

Lemma . Let J be a quadratic Jordan algebra which is an isotope of 174 ′ ′ Γ H (O3), O octonian over a ﬁeld . Then J is isomorphic to a canonica Jordan matrix algebra H (O3, Jc) where O is an octonian algebra.

Proof. It is easily seen that if (O′, j) is an octonion algebra with standard ′ ′ involution then (O , j) is simple. Hence, by Theorem 2.2, H (O3) and ′ every isotope J of H (O3) is simple. Let e1, e2,..., ek be a supplementary set of primitive orthogonal idempotents in J (Lemma 1 of §6). We show ﬁrst that k = 3 and the ei are completely primitive. Since J is not a division algebra 1 is not completely primitive. Hence if 1 is primitive then the Theorem on Minimal Inner Ideals shows that an isotope of J has capacity two. Since this is simple it follows. from Theorem 10 that this algebra is special. Since an isotope of a special quadratic Jordan al- ′ gebra is special this implies that H (O3) is special. Since this is not the case (§1.8), 1 is not primitive, so k > 1. If k > 3 or k = 3 and one the e0 is not completely primitive then the Minimal Inner Ideal Theorem implies that J has an isotope containing l > 3 supplementary orthogonal completely primitive idempotents. Then the foregoing results show that ′ this isotope, hence H (O3) is special. Since this is ruled out we see tjhat k = 2 or 3 and if k = 3 then the ei are completely primitive. It remains to exclude the possibility k = 2. In this case the arguments just used show that we may assume e1 completely primituve, e2 not. By the MII theorem and Lemma 2 of §2.3 we have an isotope J˜ = J (v) where = + ˜ + v e1 v2, v2ǫJ Ue2 such that the unit of J is e1 u2, u2ǫJ Ue2 and u2 is a sum of two completley primitive strongly connected orthogonal 175 ˜ = = ˜ idempotents in J . Then J Uu2 J Ue1+v2 Uu2 J Ue2 and J Uu2 ˜ is an isotope of J Ue2 . Since J is exceptional the foregoing results ˜ ′′ ′′ show that we can identify J with an algebra H (O3 ) where O is an octonion algebra. Moreover, we can identify e1 with 1[11]. Then, as ˜ ˜ we saw in §5, J Uu2 is the quadratic Jordan algebra of a quadratic form S with base point such that the associated symmetric bilinear form is 124 3. Structure Theory

~~˜ ˜ non-degenerate. Since J Ue2 is an isotope of J Uu2 it is the quadratic Jordan algebra of a quadratic form Q with base point such that Q(x, y)~~

is non-degenerate. Moreover, J Ue2 is not a division algebra since e2 , is not completely primitive. Hence we can choose x 0 in J Ue2 such that Q(x) = 0. Then x2 = T(x)x, T(x) = Q(x, 1), and if T(x) , 0, −1 e = T(x) x is an idempotent , 0, e2, contrary to the primitivity of e2. = Hence T(x) 0. Since Q is non-degenerate there exists a y in J Ue2 such that Q(x, y) = 1 and we may assume also that Q(y) = 0. Then, as for x, we have T(y) = 0. Since Q(a, b) is non-degenerate there exists a w such that T(w) = Q(w, 1) = 1. Then z = w−Q(x, w)y−Q(y, w)x sarisﬁes Q(x, z) = 0 = Q(y, z), T(z) = 1. Put e = z + x − Q(z)y. Then T(e) = 1 and Q(e) = 0. Hence e is an idempotent , 0, e2. This contradiction proves our assertion on the idempotents. Now let e1, e2, e3 he supplementary completely primitive orthogonal idempotents in J . These are connected so we have an isotope J = (v) = + + J where v e1 v2 v3, viǫJ Uei and the unit u is a sum of three strongly connected primitive orthogonal idempotents. 176 As before, we can identify J˜ with an H (O3). O an octonian algebra over a ﬁeld, e1 with 1[11]. Then J is the isotope of H (O3) determined by an element of the form e1 + e2 + e3, eiǫH (O3)U1[ii]. Then ei = γi[ii]. Then J is isomorphic to H (O3, Jc). The foregoing lemma and previous results prove the direct part of the

Second structure Theorem. Let J be a simple quadratic Jordan algebra satisfying DCC for principal inner ideals. Then J is of one of the follwing types: 1) a quadratic Jordan division algebra, 2) an outer ideal containing 1 in a quadratic Jordan algebra of a non-degenerate quadratic form with base point over a field Γ/Φ, 3 ) an outer ideal containing in H (a, J) where (a, J) is simple associative Artinian O with involution, 4) a canonical Jordan matrix algebra H (O3, Jc) where is an octonion algebra over a field Γ/Φ and c =diag{1, γ2, γ3}, γi , 0 in Γ. Conversely, any algebra of one of the types 1)-4) satisfies the DCC for principal inner ideals and all of these are simple with the exception of certian algebras of type 2) which are direct sums of two division algebras isomorphic to outer ideals of Ω(q). 10. Second structure theorem. 125

We consider the exceptional case indicated in the foregoing statement. Let J =Jord(Q, 1) where Q is a non-degenerate quadratic form on J /Γ with base point 1. We have seen in §5 that J satisfies the DCC for principal inner ideals and J is regular, hence, strongly non- degenerate. Hence J = J1 ⊕ J2 ⊕ ... ⊕ Js where Ji is an ideal and is simple with unit 1. If e is an idempotent , 0, 1 in J then Q(e) = 0 177 since e is not invertible, and T(e) = 1 since e2 − T(e)e + Q(e) = 0. Then the formula yUx = Q(x, y)x − Q(x)y in shows that J Ue = Ωe. = = Ω = In particular Ji J U1i 1i. If s > 1 we put u 11 − 12. Then 2 2 T(u) = T(11) − T(12) = 0 so u + Q(u) = 0. But u = 11 + 12. Hence Q(u) = −1 and 11 + 12 = 1 = 1i, which implies that s = 2. Thus = =Ω Ω either J Jord(Q, 1), Q non-degenerate,P is simple or J 11 ⊕ 12. Suppose J is simple and not a division algebra. Suppose first that J contians an idempotent e , 0, 1. Then J Ue = Ωe so e is completely primitive. The same is true of 1 − e. Hence J is of capcity two. Next suppose J contains no idempotent , 0, 1 (and is simple and not a division algebra). Then the Theorem on Minimal Inner Ideals shows that there exists an isotope of J which is of capacity two. Thus we have the following possibilities for J =Jord (Q, 1), Q non-degenerate I J =Ω11 ⊕ Ω12, II J is a division algebra III J is simple and has an isotope of capacity two. Now let K be an outer ideal in J containing 1. In case I it is immediate that K =Ω111 ⊕Ω212 where Ωi is an outer ideal containing 1i in Ω so Ωi is a division algebra. In case II K is a division algebra. In case III K is simple by Lemma 2 of §6. If J is of types 1), 3) or 4) then we have seen in §5 that satisfies the DCC for pricipal inner ideals. Also in there cases it follows from Theorem 2.2 and lemma 2 of §6 that J is simple. This completes the proof of the second statement of the second structure Theorem. We now consider a special case of this theorem, namely, that in 178 which J is finite dimensional over algebraically closed field Φ. The only finite dimensional quadratic Jordan division algebra over Φ is Φ itself (see §1.10). It is clear also that the field Γ in the statement of the theorem is finite dimensional over Φ, so Γ = Φ. The simple algebra with involution over Φ are: Φn ⊕ Φn with exchange involution, Φn with −1 t standard involution, Φ2m with the involution J : X → S X S where 126 3. Structure Theory

0 1 S =diag{Q, Q ..., Q},Q = . In all cases it is easy to check that −1 0! any outer ideals of H )(a, J) containing 1 in H (a, J) coincides with H (a, J). The same is true of Jord (Q, 1) for a non-degenerate Q. There is only one algebra of octonion’s O over Φ (the split one). Since the norm form for this represents every ρ , 0 in Φ it is clear that there is only one exceptional simple quadratic Jordan algebra over Φ, namely, H (O3). The determination of the simple quadratic Jordan algebras of capacity two, which was so arduous in the general case, can be done quickly for finite dimensional algebras over an algebraically closed field. In this = Φ Φ case J e1 ⊕ e2 ⊕ m where the ei are supplementary orthogonal 2 idempotents and m = J12. If mǫm, m = µe1 + νe2,µ,νǫΦ. As be- = = Γ = 2 = = = fore, µm µei ◦ m { e1e1m} {m e1m} e1Um.m2 e1UmVm 2 = = 2 = = m Ue2 Vm νm. Hence µ ν and m µ1 −Q(m)1 where Q is a quadratic form on m. We extend this to J by defining Q(αe1 + βe2 + = + = + + m) αβ Q(m). It is easy to check that if x αe1 βe2 m and 179 T(x) = Q(x, 1) then x2 = T(x)x+ Q(x)1 = 0 and x3 −T(x)x2 + Q(x) = 0. Hence J =Jord (Q, 1). Since J is simple, Q is non-degenerate. There remains the problem of isomorphism of simple quadratic Jor- dan algebras with DCC for principal inner ideals. This can be discussed as in the linear case considered on pp 183 − 187 and 378 − 381 of Ja- cobson’s book [2]. Bibliography

[1] Cohn, P.M. Universal Algebra, Harper and Row, New York, 1965. [2] Jacobson, N.Structure theory for a class of Jordan algebras, Proc. Nat. Acad. Sci. U.S.A 55(1966), 243-251. [3] : Structure and Reprsebtations of Jordan Algebras, A.M.S. colloq. Publ. Vol. XXXIX, 1968. [4] Jacobson N. and McCrimmon, R. Quadratic Jordan algebras of quadratic forms, forthcoming. [5] McCrimmon, K.A general theory of Jordan rings, Proc. Nat. Acad. Sci. U.S.A. 56(1966), 1072-1079. [6] : A general theory of Jordan rings, unpublished article giving the proofs of results announced in [5]. [7] : Macdonald’s theorems for Jordan algebras, to appear. [8] : The Freudenthal-Springer-Tits constructions of exceptional Jor- dan algebras, Trans. Amer. Math. Soc. 139 (1969), 495-510. [9] : The Freudenthal-Springer-Tits constructions revisited, to appear in Trans. [10] Amer. Math. Soc.The radical of a Jordan algebra, Proc. Nat. Acad. Sci. U.S.A. 62 (1969), 671-678. [11] Osborn. J. M.Jordan algebras of capacity two, Proc. Nat. Acad. Sci. U.S.A. 57 (1967), 582-589.

~~127 128 BIBLIOGRAPHY~~

~~[12] Schafer, R. D.An introduction to non-associative algebras, Aca- demic Press, New york, 1966.~~