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

3.4. Quadratic Lie superalgebras with 2-dimensional even part
We keep the notations as in the previous sections. Then there exists an isotropic element
Y 0 ∈ g 0 such that B(X 0 ,Y 0 ) = 1 and one has the following proposition:
Proposition 3.4.4.
(1) The map C is skew-symmetric (with respect to B), that is, B(C(X),Y ) = −B(X,C(Y )) for
all X, Y ∈ g 1.
(2) [X,Y ] = B(C(X),Y )X 0, for all X,Y ∈ g 1 and C = ad(Y 0)|g 1 .
(3) Z(g) = ker(C) ⊕ CX 0 and [g,g] = Im(C) ⊕ CX 0. Therefore, g is reduced if and only if
ker(C) ⊂ Im(C).
(4) The Lie superalgebra g is solvable. Moreover, g is nilpotent if and only if C is nilpotent.
Proof.
(1) For all X, Y ∈ g 1, one has
(2) Let X ∈ g 0,Y,Z ∈ g 1 then
B(C(X),Y ) = Ω(X,Y ) = Ω(Y,X) = B(C(Y ),X) = −B(X,C(Y )).
B(X,[Y,Z]) = (α ⊗ Ω)(X,Y,Z) = α(X)Ω(Y,Z).
Since α(X) = B(X 0 ,X) and Ω(Y,Z) = B(C(Y ),Z) so one has
B(X,[Y,Z]) = B(X 0 ,X)B(C(Y ),Z).
The non-degeneracy of B implies [Y,Z] = B(C(Y ),Z)X 0. Set X = Y 0 then B(Y 0,[Y,Z]) =
B(C(Y ),Z). By the invariance of B, we obtain [Y 0,Y ] = C(Y ).
(3) It follows from the assertion (2).
(4) g is solvable since g 0 is solvable, or since [[g,g],[g,g]] ⊂ CX 0. If g is nilpotent then
C = ad(Y 0) is nilpotent obviously. Conversely, if C is nilpotent then it is easy to see that g
is nilpotent since (ad(X)) k (g) ⊂ CX 0 ⊕ Im(C k ) for all X ∈ g.
Remark 3.4.5. The choice of C is unique up to a non-zero scalar. Indeed, assume that I = α ′ ⊗Ω ′
and C ′ is the map associated to Ω ′ . Since Z(g)∩g 0 = CX 0 and φ −1 (α ′ ) ∈ Z(g) one has α ′ = λα
for some non-zero λ ∈ C. Therefore, α ⊗ (Ω − λΩ ′ ) = 0. It means that Ω = λΩ ′ and then we
get C = λC ′ .
77