manuscripta math. 84, 73 - 87 (1994) manuscripta mathematica (~ Springer-Verlag 1994 Third Power Associative Composition Algebras Alberto Elduque* and Jos~ Maria P~rez** Classical composition algebras, with a unit element, are well-known and can be obtained by means of the Cayley-Dickson doubling process. If the condition on the existence of unit element is dropped, many new algebras arise. However, it is shown in this paper that if such a weak condition as the associativity of third powers of any element is imposed, only the known flexible composition algebras appear. 1991 Mathematics Subject Classification: 17 A 75 1. Introduction Let A be a nonassociative algebra with multiplication xy over a field F. Then, A is said to be a composition algebra if there is defined a nondegenerate symmetric bilinear form (x, y) on it satisfying the composition law: (xy, zy) = (x, y) (1.1) for ai1 x, y 6 A. Notice that we do not impose the existence of a unit element. The well-known generalized Hurwitz Theorem asserts that any unital com- position algebra over a field F of characteristic not two must have dimension 1,2,4 or 8 and is isomorphic to one of the algebras : F, F @ F, a quadratic field extension of F, a generalized quaternion algebra or a Cayley-Dickson (or octonion) algebra (see [22, Chap. 2]). Each of these algebras will be called a Hurwitz algebra over F. On the other hand, in any finite-dimensional composition algebra A, a new multiplication x. y can be defined so that (A, .) is a Hurwitz algebra relative to the same bilinear form ([7]). Therefore, the dimension of A is again 1, 2,4 or 8. * Partially supported by the DGICYT (PS 90-0129) and by the DGA (PCB-6/91) ** Supported by a grant from the 'Plan de Formaci6n det Personal Investigador' (DGICYT, Spain)
A l b e r t o E l d u q u e * a n d Jos~ M a r i a P~rez**
Classical composition algebras, with a unit element, are well-known and can be obtained by means of the Cayley-Dickson doubling process. If the condition on the existence of unit element is dropped, many new algebras arise. However, it is shown in this paper that if such a weak condition as the associativity of third powers of any element is imposed, only the known flexible composition algebras appear.
1991 Mathematics Subject Classification: 17 A 75
1. I n t r o d u c t i o n
Let A be a nonassociative algebra with multiplication xy over a field F. Then, A is said to be a composition algebra if there is defined a nondegenerate symmetric bilinear form (x, y) on it satisfying the composition law:
(xy, zy) = (x, y) (1.1)
for ai1 x, y 6 A. Notice that we do not impose the existence of a unit element. The well-known generalized Hurwitz Theorem asserts that any unital com-
position algebra over a field F of characteristic not two must have dimension 1,2,4 or 8 and is isomorphic to one of the algebras : F, F @ F, a quadratic field extension of F, a generalized quaternion algebra or a Cayley-Dickson (or octonion) algebra (see [22, Chap. 2]). Each of these algebras will be called a Hurwitz algebra over F.
On the other hand, in any finite-dimensional composition algebra A, a new multiplication x . y can be defined so that (A, .) is a Hurwitz algebra relative to the same bilinear form ([7]). Therefore, the dimension of A is again 1, 2,4 or 8.
* Partially supported by the DGICYT (PS 90-0129) and by the DGA (PCB-6/91) ** Supported by a grant from the 'Plan de Formaci6n det Personal Investigador' (DGICYT,
Spain)
74 ELDUQUE-PER.EZ
Other classes of composition algebras have appeared recently in the literature (see [1-3,10-18,21]). Obviously, the only composition algebra over the field F is F itself. Two-dimensional composition algebras have been classified in [18] as follows:
P r o p o s i t i o n 1.1 Let (A, , ) be a two-dimensional composition algebra over a field F of characteristic not two with corresponding symmetric bilinear fo rm ( , ) . Then, there is a new multiplication x �9 y on A such that A becomes a Hurwitz algebra with the same bilinear form and the multiplication * is given by one of the following:
i) x * y = x . y , ii) x * y = ~ . y ,
iii) x * y = x . y , iv) x * y = u . x . y, where (u, u) = 1,
where x ~ ~ denotes the standard involution of the Hurwitz algebra (A, .). m
Cases i) and iv) of the Proposition above are the ones in which the multipli- cation is commutative.
Composition algebras of dimension 4 have been classified in [21] over alge- braically closed fields of characteristic not two.
In what follows, the ground field F will be always assumed to be of charac- teristic not two.
If A is a Hurwitz algebra with multiplication xy and we consider the new multiplication given by x �9 y = ~ ([14,15,16]), we obtain a new composition algebra relative to the same bilinear form. The algebra (A, .) is termed the para- Hw'aritz algebra relative to A. The next result related to Proposition 1.1 will be useful. It extends [2, Proposition 4.4].
P r o p o s i t i o n 1.2 Let A be a two-dimensional composition algebra. Then, the following conditions are equivalent:
a) A is commutative. b) A is flexible; that is, ( xy )x = x ( yx ) for any x, y E A. c) The multiplication in A is given by cases i) or iv) of Proposition 1.1. d) A is either a Hurwitz algebra or a fo rm of a para-Hurwitz algebra.
Proof a)=~b) and d ) ~ a ) are immediate. Now, for the algebras in Proposition 1.1.ii), one checks that (x * y) * x = x �9 x �9 Y, while x * (y * x) -- (x, x)[/ and similarly for the algebras in Proposition 1.1.iii), so b) implies c).
Finally, to show c)=~d) we can assume that F is algebraically closed and have to show that any algebra as in Proposition 1.1.iv) is para-Hurwitz, actually, the para-Hurwitz algebra relative to A = F (9 F. For this, take v E A = F (9 F such that (v,v) = 1 and 03 = u and the new product x ~ y = v . x �9 y. (A,o) is a Hurwitz algebra with conjugation x ~ x J = 02~ (see [18, proof of Proposition 2]) and x . y = u . ffz . Tl = xg o y J. |
Another important family of known composition algebras is formed by those composition algebras with invariant bilinear form; that is,
(xy, z) = (z, y=) (1.2)
ELDUQUE-PEREZ 7S
for all x, y, z. Hurwitz algebras of dimension > 2 do not satisfy (1.2), but para-Hurwitz
algebras do satisfy it. We will make use of the following result (see [9, Lemma II.2.3] or [15]):
P r o p o s i t i o n 1.3 Let A be an algebra over F equipped with a symmetric bilinear form ( , ). I f ( , ) is nondegenerate and satisfies (1.1) and (1.2) then it also satisfies
(xy)x = z ( yx ) = (x, x )y (1.3)
for any x, y, z E A. Conversely, (1.3) implies (1.1) and (1.2). Moreover, any composition algebra
over F satisfying (1.3) is f inite-dimensional. |
The last assertion follows from the fact that (1.3) implies that the right or left multiplication by any element x with (x, x) ~ 0 is bijective and this is the condition for a new multiplication to exist on A so that A becomes a Hurwitz algebra ([7]).
Any finite-dimensional power associative composition algebra over a field of characteristic ~ 2, 3 was shown to be a Hurwitz algebra ([11]). This result has been extended by the second author in [17], where the same result is obtained with the weaker hypothesis of characteristic not two, the existence of at least 4 elements in F and the identities x~x = xx 2 (third power associativity) and x2x 2 = (x~x)x. Finite-dimensional flexible composition algebras have also been classified in [12,13,15,16,1,2,3]. This class includes, in characteristic ~ 2, 3, apart from the Hurwitz algebras and the forms of the para-Hurwitz algebras, forms of the algebra defined over sl(3, F) , the set of trace zero 3 x 3-matrices over the algebraic closure F of the ground field F, with multiplication given by
1 I x * y = #xy + ( 1 - # )yx - ~ t r ( zy ) 3,
where xy is the usual matrix multiplication, it is a root of the equation 3X(1 - X) = 1 and /3 is the identity matrix. The algebra thus obtained was called the algebra of pseudo-octonions ([10]) and denoted by Ps(F). Its forms are called Okubo algebras ([1,2]). There is a unique division Okubo algebra over the field of real numbers R, which is related to the Lie algebra su(3), and denoted by Ps(R).
Over I~, a related class of algebras are the absolute valued algebras. Finite- dimensional absolute valued algebras are precisely the composition algebras over R with positive definite form. E1-Mallah ([4,5,6]) showed that the only third power associative absolute valued algebras are the Hurwitz division algebras over ]R, their para-Hurwitz counterparts and Ps(R). In particular, any third power associative absolute valued algebra is flexible. The definiteness of the bilinear form is fundamental in E1-Mallah's proofs.
The main objective of this paper is to prove the following general result:
T h e o r e m A. A n y f ini te-dimensional third power associative composition alge- bra over a field of characteristic ~ 2, 3 is flexible. .
As a consequence of the methods developed to prove Theorem A, and using [17], we will obtain the next result, which extends Okubo's result and slightly the
76 ELDUQUE-PEREZ
second author's results for power-associative composition algebras mentioned above.
T h e o r e m B. Any finite-dimensional composition algebra over a field of char- acteristic not two satisfying the identities x2x = xx 2 and x2x 2 = (x2x)x is a Hurwitz algebra, m
Many of the arguments in what follows will make use of the Zariski topology on a finite dimensional vector space over an algebraically closed field (hence infinite). This will allow to extend to the whole vector space properties only checked on dense subsets. In particular, the following Lemma will be frequently used. We include the proof for completeness:
L e m m a 1.4 Let V be a finite-dimensional vector space over an infinite field F and let P l , . . . ,Pn : V ~ V be polynomial functions so that {pl (x) , . . .pn(x)} is a linearly dependent set for any x in a (Zariski) dense subset of V. Then, {pl(x) , . . .Pn(X)} is a linearly dependent set for any x in V.
Proof Let d imV = m and ( e l , . . . ,era} be a basis of V. Let T' be the dense subset in the statement of the Lemma. Consider the polynomial functions Pij :
m Y , F such that pi(x) = ~ j = l p i j ( x ) e j . Let A(x) be the matrix (pij(x)) for any x E V. If there exists an x0 E V such that (p l (x0) , . . . pn(xo) ) is linearly independent, then n < m and there is an n • n submatrix B(x) of A(x) such that det B(xo) ~ O. But A = {x E V : det B(x) r 0) is an open dense subset of V, so/~ M A ~ 0, a contradiction, m
A few words about notation. By alg(S) and F(S) we will denote the subalge- bra generated and subspace spanned by S, respectively. In case S = {Xl,. �9 xn}, we will just write a l g ( x l , . . . , xn) and F ( x l , . . . , xn). As usual, [x, y] denotes the commutator x y - yx of the elements x and y and x o y the symmetrized product xy + yx. Rx and Lx will denote the right and left multiplications by the elements x and x n will denote the right normalized n th power of x; that is, x 1 = x and X n + l ~ X X n for any n.
2. C o m m u t a t i v i t y o f a lg (x)
The aim of this section is to show that the subalgebra generated by any element of a third power associative composition algebra is commutative. Given a com- position algebra A with symmetric bilinear form (x, y), a linear map qo : A ~ A is said to be a similarity if (~(x), ~(y)) = a(x, y) for any x, y e A. The scalar a E F is called the similarity factor of ~. We start with some known results:
L e m m a 2.1 Let A be any finite-dimensional composition algebra. Then,
i) For any x, y E A, if xy = yx then
(x, x)y 2 + (y, y)x 2 = 2(x, y)xy . (2.1)
ii) For any x E A the following assertions are equivalent: a) Rx is bijective, b) L~ is bijective and c) (x, x) ~ 0.
iii) For any x e A with (x, x) ~ O, Lx and Rx are similarities of ( , ) with (x, x) as similarity factor.
ELDUQUE-PEREZ 77
Proof I t em i) appears in [2, L e m m a 2.4], i tem ii) is clear, while i tem iii) is well known ([20]) and follows directly from the l inearizations of (1.1):
Throughou t this section, A will denote a third power associative composi t ion a lgebra over a field F of characterist ic # 2. Under these circumstances, A p = A (~F -P is again a third power associative composi t ion algebra, where _P denotes the algebraic closure of F. Therefore, we will assume too tha t F is algebraically closed.
Subs t i tu t ing y = x 2 in (2.1) and using [x, x 2] = 0 we obtain
and Ix o y, z] + Ix o z, y] + [z o y, z] : 0 (2.4)
for any x , y , z E A. The set {x E A : (x, x) # 0} is an open dense subset of A (in the Zaxiski
topology) . Also, the set
S = {x e A : (x ,x) # 0 # (x, x2)}
is not empty, because if (x, x) # 0 = (x, x2), (x, x)x2x 2 + (x, x)2x 2 = 0 by (2.2), and wi th a = - ( x , x ) - l x 2, a 2 = a and (a,a) = 1 = (a, a2), so a E S. Therefore, S is again a dense subset of A tha t will be frequently used in wha t follows.
A subset X of A will be t e rmed commuta t ive in case [x, y] = 0 for any x , y ~ X . Let An(x) = F(x , x 2 , . . . , x n) (n > 1). Then:
L e m m a 2.2 Let x c S be such that An(x) is commutative and (x, x i) # 0 for any i = 1, . . . ,n. Then, A,~(x)An(x) C_ An+l(x).
Proof I f n = 1 it is clear. Assume the result is valid for n - 1, n > 2 and take x E S with ( z , x ~) # 0 for i = 1 , . . . , n and An(x) commuta t ive . Then An(x) = An-1 (x) @ Fx n so, by commuta t i v i t y of An (x) and induction hypothesis:
An(x)A,~(x) C_ An(x) + A,~-I(x)z n + F(xnz '~) �9
Since [x, x n] = 0, (2.1) implies
(X, x ) x n x n "4- ( x , x ) n x 2 = 2ix, x n ) x n+l ,
so xnx '~ E A,~+I(x). I f 2 < i < n - 1, [xi,x '~] = 0 so, again by (2.1),
(z, z)~znx ~ + (x, z ) ~ z ' z ~ = 2 ( z i, x~)z~xn = 2 ( x , z ) i - l ( z , z n - ~ + l ) z i z ~ ,
which shows t h a t An_l (x )x n C_ An+l(X). t
C o r o l l a r y 2.3 Let x ~ S be such that An(x) is commutative, (x ,x i) # 0 Vi = 1 , . . . ,n and An(x) = An+l(x). Then, alg(x) = An(x). ,
78 E L D U Q U E - P E R E Z
By third power associativity, A2(x) is commutative for any x E A. The same happens for A3(x):
L e m m a 2.4 A3(x) is commutative for any x E A.
Proof It must be checked that [x, x 3] = [x 2, x 3] = 0 for any x E A and to do so, it is enough to check it on a dense subset. Now, (2.3) with y = x 2 shows [x,x 3] = 0 for any x c A and (2.2) implies x 3 E F(x2, x2x 2} for any x E S, so t h a t [ x 2,x 3 ] = 0 f o r a n y z E S . t
The next result shows that either the symmetric bilinear form ( , ) of A satisfies a very specific restriction or alg<x) is determined for any x in the dense subset S and is nothing else but A3(x). This kind of situation will appear again.
P r o p o s i t i o n 2.5 If (x, x)(x, x 3) - (x, x2) 2 is not identically zero on A, then alg(z) = A3(x) for any x E S.
Proof Partially linearizing (2.2) (F is being assumed to be algebraically closed, hence infinite):
2(z, y)z2z 2 + (z, z ) ( z o y)z 2 + (z, z ) z2(z o y) + 4(z, y)(z, z ) z 2 + (z, z)2(z o y)
= 2(y, ~ ) ~ 3 + 2(~, z o y)x3 + 2(~, x 2) ( y ~ + ~(~ o y ) ) ,
which, with y = x 2 becomes, using the commutativity of A3(x) by Lemma 2.4:
and substituting (x, x)x2x 2 using (2.2), and simplifying by 2(x, x) for x E S,
(z ,~ )3z2 _ ( z , z ) (~ ,z2)x 3 = (x, z3)~ 4 - (x, ~ 2 ) ~ z 3 , (2.7)
which is valid for any x E S and, by density, for any x E A. Consider now the subset
Sl = {z e S : (z, z2) 2 - ( z , z ) ( z , x 3) # 0 } .
Then, either $1 is empty or S1 is an open dense subset of A. Moreover, for any x e S1, from (2.5) and (2.7) we conclude that x 4 and X2X 3 a r e in A3(x), from
ELDUQUE-PEREZ 79
(2.6) tha t x3x 3 E A3(x) too, and from (2.2) tha t x2x 2 E A3(x). Hence A3(x) is a suba lgebra for any x E $1. T h a t is, alg(x} = A3(x) for any x E $1.
Assume now tha t $1 is not empty. Then the sets {x, x 2, x 3, x2x2}, {x, x 2, x ~, x4}, {x, x 2, x 3, x2x 3 } and {x, x 2, x 3, x3x 3 } are linearly dependent for any x E $1. By L e m m a 1.4 they are linearly dependent for any x E A. Let 0 ~ x be any a rb i t r a ry element of A. If x, x 2 and x 3 are linearly independent , then by the above, I2X2,X4,X2X3,X3X 3 E A3(x) and alg(x) = A3(x). Otherwise, A2(x) --- A3(x) and by Corol lary 2.3, if x E S, alg(x) = A2(x) = Aa(x). .
C o r o l l a r y 2.6 If (x, x)(x, x 3) - (x, x2) 2 is not identically zero, then alg(x) is commutative for any x E A.
Proof By Proposi t ion 2.5 and L e m m a 2.4, alg(x) is commuta t ive for any x E S. This means tha t [u(x), v(x)] = 0 for any u(x) and v(x) monomials in x and for any x E S. By densi ty [u(x), v(x)] = 0 for any x E A and alg(x) is commuta t ive for a n y x E A . .
Now, assume tha t ( x , x ) ( x , x 3) = (x, z2) 2 for any x E A. Mult iplying (2.5) by (x, x2), (2.7) by (x, x) and subt rac t ing gives
L e m m a 2.7 The equality (~,x2) 2 = ( x , x ) ( x , x 3) is satisfied for any x E A if and only if so are
(z, z2) 2 = (x, z) 3 (2.9)
and
(z, x 3) = (z, z) 2 . (2.10)
Proof Obviously (2.9) and (2.10) imply (x, x2) 2 = ( x , x ) ( x , x 3) Vx E A. Con- versely, if (x, x2) 2 = ( x , x ) ( x , x 3) Vx E A and
s2 = {x e s : (x, z2) ~ # (x, z) 3} u {~ e s : (z, z 3) # (z, ~)2}
is not empty, then $2 is an open dense subset and for any x E $2 (2.8) implies t ha t x 2 and x 3 are l inearly dependent . Hence, by L e m m a 2.1.ii), x and x 2 are l inearly dependent too for any x E $2 and, by L e m m a 1.4, it is so for any z E A. Let us check tha t this implies t ha t d i m A = 1, which contradicts $2 :/= 0. Actually, if d im A > 1 we can choose x, y E A with (x, x) r 0 =/: (y, y) and (x ,y) = 0. Since x 2 = #x and y2 = uy for some # ,v , #2 = (x ,x) r 0 ~ (y ,y) -- u 2, so dividing by # and v we can assume # = 1 = u. Then ( x + y , x + y ) = 2 ~ O, so ( x + y ) 2 = a + ( z + y ) with (c~• 2 = 2. But , a + ( x + y ) + c ~ - ( x - y ) = ( x + y ) 2 + ( x - y ) 2 = 2x 2+2y~ = 2 x + 2 y , s o a + + a - = 2 = a + - a - and (~- = 0, a contradict ion. .
In order to show t h a t alg(x) is always commuta t ive also if (x, x ) ( x , x 3) - (x, x2) 2 is identically zero on A, we will work with As(x) instead of A3(x).
L e m m a 2.8 If (2.9) and (2.10) are satisfied for any x E A, then As(x) is commutative for any x E A.
80 ELDUQUE-PEREZ
Proof By Lemma 2.3, A3(x) is always commutative. On the other hand, it is enough, by continuity, to prove tha t As(x) is commutative for any x c S. So assume x E S. Now,
[x 4, x] = 0, by (2.3) with y = x 3.
Since (x , z 3) r 0 by (2.10) [x 3, x 4] = 0, by (2.1) with y = x 3,
[x2z 2, x a] = 0, by (2.2), [x2x 3, x 2] = 0, by (2.3) with x ~ x 2, y ~ x 3,
[xax3,x 2] = O, by (2.1) with x ~-* x2 ,y ~ x 3,
Ix 2, x 4] = 0, by (2.1) with y = x 3,
Up to now, we have checked tha t A4(x) is commutative for x C S. Also, by Lemma 2.2, Aa(x)A3(x) C_ Aa(x) for x e S. Moreover,
[x 5, z] = 0, by (2.3) with y = x 4,
[xax 4, x] = 0, by (2.1) with y = x 4,
Since (x2, x 4) = ( x , x ) ( x , x 3) ~ 0 ~ ( x , x ) 2 ( x , x 2) = (x3,x 4) for x E S
[x4x 2, x] = 0, by (2.1) with x ~-~ x 2, y ~ x 4,
[x4x 3, x] = 0, by (2.1) with x ~ x 3, y ~ x 4,
Ix 5,x 2] = 0, by (2.4) with y = x 4,z = x 2,
[x s, x 3] = 0, by (2.4) with y = x 4, z = x 3,
[z 5,z 4] = 0, by (2.3) w i t h z ~ z 4,y~-*z. m
P r o p o s i t i o n 2.9 I f (2.9) and (2.10) are satisfied for any x e A, then alg{x} = As(x) for any x e S.
Proof We first linearize (2.10) to obtain
(y, x 3) + (x, yx 2 + z ( x o y)) = 4(z, y)(z , z) . (2.11)
Now, take x E S
(z, x2x 2) : ( x , x ) (x , x~) , by (2.2) and (2.9), (2.12)
(x ,x 4) : ( z , z ) ( x , x2), by (2.11) with y : x 2 and (2.12), (2.13)
(x, z3x 3) = (x ,x)2(x , x2), by (2.1) with y : x 3 and (2.12), (2.14)
(x, x3x 2) = (x ,x ) 3, by (2.1) with x ~-* x2 ,y ~-* x 3, (2.12) and (2.14), (2.15)
(x ,x s) -- (x ,x) 3, by (2.11) with y = x 3, (2.10) and (2.15), (2.16) (x, zaz 4) : (x ,x)3(x , z2), by (2.1) with y : x 4, (2.13) and (2.16), (2.17)
(x, x2x 4) : (x ,x)2(x , x2), by (2.1) with x ~-~ x2 ,y ~ x 4,
(2.10),(2.12) and (2.17), (2.18) ( x , x ~) = (x , x )2 (x ,x2) , by (2.11) with y : z 4, (2.18) and (2.13), (2.19)
Now, let zl = (x 2, x)x 2 - (x, x)2x, z2 = z 3 - ( : r , x ) x , z 3 : z 4 - (x, :z)x 2, z4 --- x 5 - ( x , x )2x and zs = x 6 - (x ,x )2x 2. From (2.9), (2.10), (2.13), (2.16) and
ELDUQUE-PEREZ 8]
(2.19) it immediately follows that (z~,A6(x)) = O, so F(z l , z2 , z3,z4, zsl is a totally isotropic subspace and, since dim A _< 8 and ( , ) is nondegenerate, it follows that the z('s are linearly dependent. But Aa(x) = F(x , Zl, z2, z3, z4, Zst, so dimA6(x) < 5. Since (x, x i) ~ 0 for i = 1 , . . . , 5 by the formulae above since x �9 S, it follows from Corollary 2.3 that alg{x) = An(x) for some n _< 5, so alg(x) = A5(x) for any x �9 S. ,
Now, with the same argument as in Corollary 2.6, we conclude that under the hypothesis of Proposition 2.9, also alg(x / is commutative for any x �9 A. This and Corollary 2.6 are summarized in:
T h e o r e m 2.10 Let A be any f ini te-dimensional third power associative com- position algebra over an arbitrary field F of characteristic not two. Then alg(x) is commutative for any x �9 A. .
3. D e t e r m i n a t i o n o f a lg (x /
This section is devoted to improve our knowledge of alg(x / for x in a finite- dimensional third power associative composition algebra A. As in the previ- ous section, we assume, unless otherwise stated, that the ground feld F is al- gebraically closed and of characteristic not two, as always. It will be shown that alg(x} = As (x ) for any x c A. We still consider the dense subset S = {z �9 A: (x,x) • 0 r (z, x2)}.
For x �9 S, let L~ and R x be the adjoint endomorphisms of Lx and Rx relative to ( , ). That is, (xy, z) = (L~y, z) -- (y, L~z) and (yz, z) = (R~y, z) = (y, R~z) for any y , z �9 A. By Lemma 2.1.iii) L~Lx = ( x , x ) I = R~Rx or, equiv- alently, L~ = ( x , x ) L ; 1 and R~ = (x,x)R-~ 1. But by finite-dimensionality, L~ -1 �9 alg(Lx), R ; 1 �9 alg(Rx), so L ; l ( x ) and R ; l ( x ) are in alg(x) and so are L ; ( x ) and R ; ( x ) .
L e m m a 3.1 For any x �9 S, L~(x) = R*(x) .
Proof Since Lz is bijective (Lemma 2.1.iii)), there is an y E A such that xy = x, so (x, x )y = (x, x )L~ 1 (x) = L~(x) E alg(x/, which is commutative by Theorem 2.10. Hence x = xy = yx and ( x , x ) y = R~(x) too. .
P r o p o s i t i o n 3.2 In case ( x, x2) 2 - (x, x) 3 is identically zero in A, then alg(x) = A2(x) for any x C A.
P r o o f Linearizing (x, x2) 2 = (x, x) 3 we obtain
for any x �9 S, which shows A3(z) = A2(x). By Corollary 2.3, alg(x) = A2(x) Vz �9 S. Now, the same argument as in the end of the proof of Proposi t ion 2.5 shows alg(x) = A2(x) for any x �9 A. t
In case (x, x2) 2 - (x, x) 3 is not identically zero on A, Lemma 2.7 and Propo- sition 2.5 show tha t alg(z) = A3(x) for any x E S.
L e m m a 3 .3 I f x, y, z are commuting elements in the f ini te-dimensional com- position algebra A, then
(y , z ) x 2 - y) z - z ) x y + (x , x ) y z = o .
Proof See [2, Corollary 2.5]. a
Now take T = {x �9 S : (x, x2) 2 # (x, x)3}. It T is not empty, it is an open dense subset of A. Assume there is an x �9 T such tha t alg(x) = A3(x) # A2(x). Since x �9 T, ( , ) is nondegenerate on A2(x). On the other hand, in this ease dim alg(x) = 3, so alg(x) is not a composit ion algebra and ( , ) is degenerate on alg(x). Therefore, there is an element z �9 alg(x) such tha t {x, ( x , x ) x 2 - (x, x2)x, z ) is an or thogonal basis of alg(x), so (z, alg(x)) = 0. By Lemma 3.3 with y = ( x , x ) x 2 - (x, x2)x we obtain yz = O, so (y ,y) = 0 by Lemma 2.1.ii) and this is a contradict ion with ( , ) being nondegenerate on A2(x) . Therefore, alg{x) = A2(x) for any x �9 T and, as previously, this implies alg(x) = A2(x) for any x E A.
This and Proposi t ion 3.2 are summarized in the next result, which is proved by extending scalars:
P r o p o s i t i o n 3.4 Let A be any f ini te-dimensional third power associative composition algebra over an arbitrary field of characteristic not two. Then, alg(x) = F ( x , x 2) for any z �9 A. m
Assume again t ha t (x, x2) 2 - (x, x) 3 is not identically zero on A, so tha t T = {x �9 S : (x, x2) 2 # (x ,x) 3} is an open dense subset of A. Then, for any x �9 T, alg(x) = F ( x , x 2) and, since x �9 T, ( , ) is nondegenerate on alg(x} and d imalg(x) = 2. By Proposi t ion 1.2 and since F is assumed to be algebraically closed, either alg(x} is the Hurwitz algebra F @ F or its pa ra - Hurwitz counterpar t . In the last case, by (1.3) we have tha t x 3 = (x, x )x . On the other hand, if alg(x) = F @ F and e is its unit element, x 2 - 2 ( x , e ) x + ( x , x)e = O, so x 3 = 2(x, e)x 2 - (x, x )x and, since x and x 2 are linearly independent, x 3 (x, x )x . Thus, T = Th U Tph (disjoint union), with
Th = {X e T : x 3 ~ (x, 5g)x} ---~ {x �9 T : alg(x~ is Hurwitz}
Tph = {x �9 T : x 3 = (x, x ) x } = {x �9 T : alg(x I is para-Hurwi tz}
If Th is not empty, then it is a dense subset of A and alg(x / is associative for any x �9 Th. This implies easily tha t alg(x) is associative for any x �9 A, so A is
ELDUQUE-PEREZ 83
power-associative. Otherwise, T = Tph is dense, so x 3 = (x, z ) x for any x E A. We have arrived at the main result of this section:
T h e o r e m 3.5 Let A be any f ini te-dimensional third power associative compo- sition algebra over an arbitrary field of characteristic not two. Then alg(x) = F ( x , x 2) for any x E A and one of the following conditions holds:
a) (x, x2) 2 = (x,x) a for any x E A. b) z 3 = ( x , x ) x for any x E A. c) A is a strictly power associative algebra.
Proof Extend scalars to the algebraic closure and use the previous results of this section. .
4. P o w e r a s soc ia t ive c o m p o s i t i o n a lgebras
As mentioned in the introduction, Okubo ([11]) proved that finite-dimensional power associative composition algebras over a field F of characteristic -# 2, 3 are Hurwitz algebras. The second author has given a more direct proof which, besides, only requires characteristic # 2, that F contains at least four elements, third power associativity and the fourth power identity x2x 2 = (x2x)x. The proof consists first in showing that with these conditions the algebra is strictly power associative and then to show that any power associative composition algebra over an algebraically closed field of characteristic not two is Hurwitz. It will be shown in this section how to use Theorem 3.5 to prove the first result (strictly power associativity) without the restriction on the size of the field. This can be done by an ad hoc argument for the prime field of three elements or by a more general argument as it will be done here.
T h e o r e m B. Let A be a f ini te-dimensional composition algebra over a field of characteristic not two. Assume x2x = xx 2 and x2x 2 = (x2x)x for any x E A. Then A is a Hurwitz algebra.
Proof By using the results in [17] it is enough to prove that A is strictly power associative, and by Theorem 3.5 we are left with two cases: either (x, x2) 2 = (x, x) 3 or x 3 = (x, x ) x for any x E A. Substituting x2x 2 in (2.2) gives
(z,z)zaz + (z,z)2z = 2 ( z , x 2 ) z 3 ,
which, by Lemma 2.1.ii) gives
(X, X)X 3 + (X, X)2X : 2(X, X2)X 2 (4.1)
if (x, x) # 0. In case x 3 = (x, x ) x for any x E A, this implies
(x, x)2x = (z, x2) 2 (4.2)
for any x ~ A with (x, x) # 0. In the other case, (x, x2) 2 = (x, x) 3 for any x E A, (3.2) implies
(x, x2)x 3 + 2 ( x , x 2 ) ( x , x ) x - 3 ( x , x ) 2 x 2 = O,
84 ELDUQUE-PEREZ
which mult ipl ied by (x, x 2) gives
(Z,X)3X 3 + 2(Z,X)4Z -- 3(Z,X, )2(X, X2)X 2 : 0 ,
or (z , z ) z 3 + 2 ( z , z ) 2 z = 3(x, z 2 ) z 2 (4.3)
if (x, x) ~ 0. But (4.3) and (4.1) give (4.2) also in this case. Therefore , in bo th cases (4.2) is satisfied and x 2 E Fx for any x E A with
(x, x) ~ 0. The a rgument in the last pa r t of the proof of L e m m a 2.7 shows tha t d im A = 1 and A = F is a Hurwi tz algebra. .
5. P r o o f o f T h e o r e m A
In the last section, and unless otherwise s tated, A will be assumed to be a f ini te- dimensional third power associative composi t ion a lgebra over an algebraically closed field of character is t ic different from two and three.
T h e case c) of T h e o r e m 3.5 is deal t with in Section 4. Let us first consider the case a) in Theo rem 3.5. T h a t is, (X, X2) 2 = (X,X) 3 for any x E A. If d imalg(x} = 1 for any x E A with (x,x) ~ 0 then, as in the proof of L e m m a 2.7, d i m A = 1 and A is a Hurwi tz algebra. Therefore, assume there is an x E A with (x, x) ~ 0 and d im alg(x) = 2. Since F is algebraically closed, we can take a scalar mult iple a of x so t h a t (a, a) = 1 and changing a by - a if necessary we have (a, a) = 1 = (a, a2). Take z = a s - a . Then, (z, a) = 0 = (z, a s) and by (2.1) z 2 = 0. By (3.2), a 3 = - 2 a + 3 a 2, so if e = a - �89 we obtain e 2 = e , ( e , e ) = l a n d e z = z e = 2 z .
Let B = {x e A : (x ,e) = 0}. Then, (3.1) with x E B and y = e gives
(e, z~ ) ( z , z 2) + (z, e o z ) ( z , z 2) = o
and if (x, x) # 0, so is (x, x 2) and
(e, x 2) + (x, e o x) = 0 , (5.1)
for any x E B with (x, x) ~ 0. Since ( , ) is nondegenera te on B, by densi ty this is valid for any x E B.
Linearizing (1.1) gives (xy, xz) = (x, x)(y, z) and then (xy, uz) + (uy, xz) = 2(x, u)(y, z) which implies (ez, xe) + (x 2, e) = 0 for z E B, and by (5.1)
(e~, ~e) = (x, e o ~)
for any x E B, which, after l inearization, gives
(e~, ye) + (ey, ~e) = (~, e o y) + (y, e o ~)
for any x, y E B. In par t icular , with y = z:
(ex, 2z) + (2z, ze) = (x, 4z) + (z, e o x) ,
t ha t is, (z, e o z ) = 4 (x , z) .
But , (z, ex) = �89 ex) = l (z , x) = (z, xe), so (z, e o z) = (z, z) and we obta in 3(x, z) = 0. Thus , (x, z) = 0 for any x E B, a contradict ion.
ELDUQUE-PEREZ 85
Therefore, in case a) of Theorem 3.5, we obtain dim A = 1 and A is Hurwitz. Finally, let us consider the case b) of Theorem 3.5. T h a t is, x 3 = (x, x ) x
for any x E A. This condit ion appears also in the so-called pseudo-compos i t i on algebras studied in [8,19] and some of the a rguments t ha t follow car ry the same flavor as those in these references. We consider, as in Section 3, the dense subset T = {x c S : (x, x2) 2 ~ (x ,x) 3} and, since we are in case b) of Theorem 3.5, for any x c T, alg(x) is the pa r a -Hurwi t z a lgebra of dimension 2 relative to the unique Hurwi tz a lgebra of dimension 2 over F , which is F @ F. Identify- ing, as vector spaces, alg(x) = F �9 F and taking el = (1, 0), e2 = (0, 1), the mul t ip l icat ion of the p a r a - H u r w i t z a lgebra alg(z} is given by e 2 = e2, e22 = el and ele2 -- e2el = 0. The idempotents in A are easily shown to be the three e lements #el + ~t2e2, wi th # any cubed root of 1. In part icular , this a lgebra has a basis formed by idempoten ts and each of the three idempotents e verifies (e, e) = 1. This shows tha t any x E T is a linear combinat ion of idempotents e with (e, e) = 1.
Let e be such an idempotent : e 2 = e and (e, e) = 1. Linearizing x 3 = (x, x ) x writ ing x 3 = x2x gives
x2 y + (x o y ) x = ( x , x ) y + 2 (x ,y )x .
Subst i tu t ing x = e and y E B = {x
ey +
Now, linearizing x a = (x, x ) x with gives
ye + e(e o y) = y . (5.4)
Adding (5.3) and (5.4) and defining/~e(x) = e o x g ives / / r + / ~ - 2 I = 0 on B, so t h a t ( the character is t ic of F is not 3):
(5.2)
C A : (x,e) = 0}
(e o y )e = ~ . (5.3)
x 3 = x x 2 and subst i tu t ing in the same way
B = B - 2 �9 B1 ,
with Bi = { x ~ B : e o x = i x } , i = - 2 , 1 .
L e m m a 5.1 With the condit ions above:
i) B - 2 = { z ~ B : e x = z e = - x } . ii) (B-2 , B1) = 0.
iii) eBi C B i and B i e G Bi f o r any i = - 2 , 1.
Proo f i) Given x E B - 2 , e o x = - 2 x , so (e o x ) e = - 2 x e . Applying (5.3) leads to x - ex = - 2 x e , t h a t is, x - e o x = - 3 x e and xe = - x .
ii) For x e B - 2 , y E B1, ( x , y ) = (x, e o y ) = - ( e z , e y ) - ( z e , ye) = - 2 ( x , y ) , so (z , y) = 0.
iii) This is clear for i = - 2 by i tem i). I f x e B1 and y e B - 2 , (e, ex) = (e~ ,ex ) = ( e , x ) = 0 and (y, ez ) = - ( e y , ex) = - ( y , x ) = 0 by ii). Thus, (ex, F e + B - 2 ) = 0 and ex e B1. '
Now, let us prove t h a t (xe, y) = (x, ey) for any x, y E A = F e ~ B - 2 �9 B1. For this it is enough to take x , y E {e} U B - 2 U BI , and from L e m m a 5.1 it is clear if x or y equals e or if one of t hem belongs to B - 2 and the other to B1. For
86 ELDUQUE-PEREZ
x , y E B - 2 (xe, y) = - ( x , y ) = (x, ey) and for x , y e B1, (xe, y) -= (xe, y e + e y ) = (5, y) + (ze, ey) = (ez, ey) + (xe, ey) = (ex + xe, ey) = (x, ey).
Therefore, the equal i ty (xy, z) - (x, yz) is valid for a rb i t ra ry x, z E A and idempoten t y with (y, y) = 1. Since any d e m e n t y E T is a linear combina t ion of such idempoten ts and since T is dense in A, we conclude tha t (xy, z) = (x, yz) for any x, y, z E A. By Proposi t ion 1.3, this implies tha t A is flexible and the proof of Theorem A is complete.
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