By Dr. Ottmar Loos (auth.)
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11 this is a symmetric s_ystem of idempot@nts the and . 12. DEFINITION. if + however, as idem- pair). is primitive.
LEMMA. For a representation tities hold is any identity in of a Jordan pair V F* holds in A . the following iden- o's omitted; cf. 0). 7 (5) q(x)q(y)d(x,z) (6) q(Q(x)y,{xyz}) (7) q({xyz}) Proof. + q(Q(x)y,Q(z)y) We set Linearize y = z d(x,{yxz})q(x) which proves + (4). For + d(x,y)d(Q(x)z,y) - q(x)d(z,x)q(y) d(Q(x)z,y)q(x) + q(z)q(y)q(x) (5) we use This proves (i). + d(x,Q(y)Q(x)z) (3) follows + q(z,Q(x)y) from of (2) by the duality prin- (2) we have = q(x,{x,y,Q(x)z}) = q(x,Q(x){yxz}) = q(x)(d(y,x)q(y,z) = = d(Q(x)y,z)d(x,y) - q(Q(y)x,z)) d(x,y)d(x,z)d(x,y) + q(x)(q(y,z)d(x,y) - q(x)q(Q(y)x,z) - q(Q(y)x,z) - d(z,x)q(y)) .
Hence we can compare , R is a free k-module ~ SIT3V ~ i,j coefficients and at SiT j . E. E. E. i,j= 0 i 3 = (3). For Q(e a) . Similarly, = Ker(Q(e-O)) (6) let = Im(Q(e~)) R = k(e) Ker(Q(e-O))C . This proves be the algebra since (3). Now E~Q(e a) Ker(E~) (5) follows of dual numbers. If from x. x. = (l-r which proves {eae-~ means (c) (6). x. = x. {xyz} the coefficients of = E~(Q(x)y) at powers of Then by JP7, JP8, . This proves (3) and (7) and (5) we have i = 1 this (4). I " Then (i-2)x 2 + (i-l)x I + ix 0 = 0 .