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

2.2. Singular quadratic Lie algebras
Recall that C is not unique (see Remark 2.2.20) and it depends on the choice of V . Let
a = X ⊥ 0 /CX0.
We denote by X the class of an element X ∈ X ⊥ 0 .
Proposition 2.2.23.
Keep the notation above. One has:
(1) the Lie algebra a is Abelian.
(2) define
B(X, Y ) = B(X,Y ), ∀ X,Y ∈ X ⊥ 0 .
Then B is a non-degenerate symmetric bilinear form on a.
(3) define
C(X) = C(X), ∀ X ∈ X ⊥ 0 .
Then C ∈ L (a) is a skew-symmetric map with rank( C) = rank(C) even and rank( C) ≥ 4.
(4) C does not depend on the choice of V . More precisely, if WI = Cα ⊕ φ(V ′ ) and C ′ is the
associated map to V ′ (see Remark 2.2.20), then C ′ = C.
(5) the Lie algebra g is reduced if and only if ker( C) ⊂ Im( C).
Proof.
(1) It follows from Proposition 2.2.22 (1).
(2) It is clear that B is well-defined. Now, since B(X0,Y0) = 1, B(X0,X0) = B(Y0,Y0) = 0, the
restriction of B to span{X0,Y0} is non-degenerate. So
g = span{X0,Y0} ⊥
⊕ span{X0,Y0} ⊥ ,
X ⊥ 0 = CX0 ⊕span{X0,Y0} ⊥ and X ⊥ 0 ⊥ = X ⊥ 0 ∩span{X0,Y0} = CX0. We conclude that B is
non degenerate.
(3) We have C(X ⊥ 0 ) = ad(Y0)(X ⊥ 0 ) ⊂ X ⊥ 0 since X ⊥ 0 is an ideal of g. Moreover, C(X0) = 0, so
C is well-defined. The image of C is contained in X ⊥ 0 and Im(C) ∩ CX0 = {0}, therefore
dim(Im(C)/CX0) = dim(Im( C)) = dim(Im(C)). Now it is enough to apply Proposition
2.2.22.
(4) By Remark 2.2.20, we have C ′ = C + α ⊗ X1 − β ⊗ X0. But α(X0) = 0, so C ′ = C.
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