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LEMMA. With the notations i # j . Then (x,y) even for n m 2 if Indeed, of V and is nilpotent; i = 2 or V. 4(c). 5. 4(c). 9. LEMMA. i = 1 is invertible = (x,y) Passing to spaces we have (2). 4(c) Similarly = {Y2ZlY0 } ~ V1 is quasi-invertible. V is quasi- is quasi-inverti- is quasi-invertible. z = z 2 + z I + z 0 . Then = Q(Zl,Z2) (xi,Y i) z i = (xi)Yi ~ Rad V~l " For and hence V~ (i). Since every ideal of x. ~ Pad V~ , and let 1 1 yj ~ Vj . 2 (x2,Y2+Yl+Y0) i = 0 . 1, = B(zl,Y2+Y0)B(Zl,-y2-Y0) = in + z I 6 Rad V I V I 9 Rad V 0 .

Indeed, Q(x)Q(x -I - y) = Q(x)(Q(x -I) - Q(x-l,y) + Q(y)) x (resp. 1 = Id - Q(x)Q(x-l,y) + Q(x)Q(y). By JPI3 we have Q(x)Q(x-l,y) since - D(Q(x)x-l,y) = 2D(x,y) - D(x,y) = D(x,y) D(x,x -I) = D(x,x-l)Q(x)Q(x -I) = Q(x,Q(x)x-l)Q(x -I) = Q(x,x)Q(x -I) 2Q(x)Q(x -I) = 21d = = D(x,x-l)D(x,y) If (J,J) V = then formula by JPI. Similarly one proves the second formula. is the Jordan pair associated with a unital Jordan algebra (i) reads (2) B(x,y) = UxU(x-l- y) = U(x - y-l)u . 1. DEFINITION. simply write V z = xy Recall Instead (x,y) ~ V in the Jordan algebra In this case, and only if the following There exists b G J (ii) Ua is invertible; (iii) Ua is surjective; (iv) 1 V+ Y obtained (i - x) -I = i + z of conditions such that to the image of U .

This follows by a straightforward computation, using JP3 and JP26. 5. PROPOSITION. (x,Q(y)z) G V (Shifting principle) is quasi-invertible (a) Let x,z ~ V+ if and only if and y @ V- . Then (Q(y)x,z) 6 V ~ is quasi- invertible~ and in this case (i) (b) Q(y)(x Q(y)z) = (Q(y)x) z. Let (x,y) if and only if @nd (u,v) be in (B(u,v)x,y) V . Then (x,B(v,u)y) is quasi-invertible~ is quasi-invertible and in this case B(u,v)(x B(v'u)y) = (B(u,v)x) y. (2) Proof. (a) Let (x,Q(y)z) be quasi-invertible. 4 (a) is quasi-invertible (Q(y)x,z) that (Q(y)x,z) is quasi-invertible proved it follows that (b) elements it follows and that (i) holds.