TH`ESE A NEW INVARIANT OF QUADRATIC LIE ALGEBRAS AND ...

(ii) For all x,y,z,t ∈ N, one has
4.4. Symmetric Novikov algebras
B((x,y,z),t) = B((xy)z − x(yz),t) = B(x,y(zt) − (yz)t) = −B(x,(y,z,t)).
Therefore, As(N) = (N,N,N) ⊥ . To prove N(N) = As(N), we fix x ∈ As(N) and let
y,z,t ∈ N. Since (x,z,t) = 0 and (III) one has (z,x,t) = 0. Moreover, B((y,z,x),t) =
−B(y,(z,x,t)) and B non-degenerate imply (y,z,x) = 0, for all y,z ∈ N. Hence,
N(N) = As(N).
(iii) Let x ∈ LAnn(N) then B(xN,N) = B(x,N 2 ) = 0. It means that x ∈ (N 2 ) ⊥ . Conversely,
if x ∈ (N 2 ) ⊥ then since B(xN,N) = B(x,N 2 ) = 0 one has x ∈ LAnn(N).
It implies that LAnn(N) = (N 2 ) ⊥ . Similarly, we obtain LAnn(N) = RAnn(N) =
Ann(N) = (N 2 ) ⊥ .
Proposition 4.4.7. Let N be a Novikov algebra then
(1) Z(N) is a commutative subalgebra.
(2) As(N), N(N) are ideals.
Proof.
(1) Let x,y ∈ Z(N) then (xy)z = (xz)y = (zx)y = z(xy)+(z,x,y) = z(xy), for all z ∈ N. Therefore,
xy ∈ Z(N) and then Z(N) is a subalgebra of N. Certainly, Z(N) is commutative.
(2) Let x ∈ As(N),y,z,t ∈ N. By the equality
(xy,z,t) = ((xy)z)t − (xy)(zt) = ((xz)t)y − (x(zt))y = (x,z,t)y = 0,
one has xy ∈ As(N). Moreover,
(yx,z,t) = ((yx)z)t − (yx)(zt) = (y(xz))t − y(x(zt))
= (y,xz,t) + y((xz)t) − y(x(zt)) = y(x,z,t) = 0
since xz ∈ As(N). Therefore As(N) is an ideal of N.
Similarly, let x ∈ N(N),y,z,t ∈ N one has:
and
(y,z,xt) = (yz)(xt) − y(z(xt)) = ((yz)x)t − (yz,x,t) − y((zx)t − (z,x,t))
= ((yz)x)t − (y(zx))t + (y,zx,t) = (y,z,x)t = 0
(y,z,tx) = (yz)(tx) − y(z(tx)) = ((yz)t)x − (yz,t,x) − y((zt)x − (z,t,x))
Then N(N) is also an ideal of N.
= ((yz)x)t − y((zx)t) = (y,z,x)t + (y,zx,t) = 0.
118