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4.4. Symmetric Novikov algebras
Therefore, [[x,y]+,[x,x]+]+ = [x,[y,[x,x]+]+]+ if and only if x 2 (xy) + x 2 (yx) = x(yx 2 ) + x(x 2 y).
Remark that we have following identities:
x 2 (xy) = x 3 y − (x 2 ,x,y) = x 3 y − (x,x 2 ,y),
x 2 (yx) = (x 2 y)x − (x 2 ,y,x) = x 3 y − (y,x 2 ,x),
x(yx 2 ) = (xy)x 2 − (x,y,x 2 ) = x 3 y − (y,x,x 2 ),
x(x 2 y) = x 3 y − (x,x 2 ,y).
It means that we have only to check the formula (y,x 2 ,x) = (y,x,x 2 ). It is clear by the identities
(III) and (V). Then we can conclude that J(N) is a Jordan algebra.
Corollary 4.4.18. If (N,B) is a symmetric Novikov algebra satisfying (V) then (J(N),B) is a
pseudo-Euclidean Jordan algebra.
Proof. It is obvious since B([x,y]+,z) = B(xy + yx,z) = B(x,yz + zy) = B(x,[y,z]+), for all
x,y,z ∈ J(N).
Remark 4.4.19. Obviously, Jordan-Novikov algebras are power-associative but in general this
is not true for Novikov algebras. Indeed, if Novikov algebras were power-associative then they
would satisfy (V). That would imply they were Jordan-admissible. But, that is a contradiction
as shown in Example 4.4.16.
Lemma 4.4.20. Let N be a Novikov algebra then [x,yz]+ = [y,xz]+, for all x,y,z ∈ N.
Proof. By (III), for all x,y,z ∈ N one has (xy)z + y(xz) = x(yz) + (yx)z. Combined with (IV),
we obtain:
(xz)y + y(xz) = x(yz) + (yz)x.
That means [x,yz]+ = [y,xz]+, for all x,y,z ∈ N.
Proposition 4.4.21. Let (N,B) be a symmetric Novikov algebra then following identities:
(1) x[y,z] = [y,z]x = 0, and therefore [x,yz]+ = [x,zy]+,
(2) [x,y]+z = [x,z]+y,
(3) [x,yz]+ = [xy,z]+ = x[y,z]+ = [x,y]+z,
(4) x[y,z]+ = [y,z]+x,
hold for all x,y,z ∈ N.
Proof. Let x,y,z and t be elements in N.
(1) By Proposition 4.4.6 and Lemma 4.4.8, L [y,z] = 0 so one has (1).
(2) We have B([x,y]+z,t) = B(y,[x,zt]+) = B(y,[z,xt]+) = B([z,y]+x,t). Therefore, [x,y]+z =
[z,y]+x. Since the product [ , ]+ is commutative then [y,x]+z = [y,z]+x.
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