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

3.6. Quasi-singular quadratic Lie superalgebras
By the non-degeneracy of B and J(X,Y ) = −J(Y,X), we obtain:
C(X) =
n
∑
i=1
Combined with B supersymmetric, one has:
B(X i 0 ,X)X i 1 and C(Y ) = −
n
∑
i=1
B(X i 1 ,Y )X i 0 .
−B(Y,C(X)) = B(C(X),Y ) = −B(C(Y ),X) = −B(X,C(Y )).
It shows that C is skew-supersymmetric. Finally, Im(C) ⊂ Z(g) since X i 0 ,X i ∈ Z(g), for all
1
i.
Proposition 3.6.4. Let X 1 = φ −1 (p) then for all X ∈ g 0, Y,Z ∈ g 1 one has:
(1) [X,Y ] = −B(C(X),Y )X 1 − B(X 1,Y )C(X),
(2) [Y,Z] = B(X 1,Y )C(Z) + B(X 1,Z)C(Y ),
(3) X 1 ∈ Z(g) and C(X 1) = 0.
Proof. Let X ∈ g 0, Y,Z ∈ g 1 then
B([X,Y ],Z) = J ∧ p(X,Y,Z) = −J(X,Y )p(Z) − J(X,Z)p(Y )
= −B(C(X),Y )B(X 1,Z) − B(C(X),Z)B(X 1,Y ).
By the non-degeneracy of B on g 1 × g 1, it shows that:
[X,Y ] = −B(C(X),Y )X 1 − B(X 1,Y )C(X).
Combined with B invariant and C skew-supersymmetric, one has:
[Y,Z] = B(X 1,Y )C(Z) + B(X 1,Z)C(Y ).
Since {p,I} = 0 then X1 ∈ Z(g). Moreover, {p, pi} = 0 imply B(X1,X i) = 0, for all i. It means
1
B(X1,Im(C)) = 0. And since B(C(X1),X) = B(X1,C(X)) = 0, for all X ∈ g then C(X1) = 0.
Let W be a complementary subspace of span{X 1 1 ,...,X n 1 ,X 1} in g 1 and Y 1 be an element in
W such that B(X 1,Y 1) = 1. Let X 0 = C(Y 1), q = (CX1 ⊕ CY 1) ⊥ and Bq = B|q×q then we have the
following corollary:
Corollary 3.6.5.
(1) [Y 1,Y 1] = 2X 0, [Y 1,X] = C(X) − B(X,X 0)X 1 and [X,Y ] = −B(C(X),Y )X 1, for all X,Y ∈
q ⊕ CX 1.
(2) [g,g] ⊂ Im(C)+CX 1 ⊂ Z(g) so g is 2-step nilpotent. If g is reduced then [g,g] = Im(C)+
CX 1 = Z(g).
(3) C 2 = 0.
Proof.
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