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

If l = k then we stand at (k − 1)-step and
2.2. Singular quadratic Lie algebras
I = α1 ∧ ··· ∧ αk−1 ∧ ∑ ξ1,...,k−1, jk
k≤ jk≤n
α jk .
Since αk ∧ I = 0 then ξJ = 0 if jk = k. Therefore, we obtain
I = ξ α1 ∧ ··· ∧ αk.
Lemma 2.2.5. WI = GI = {ιA(I) | A ∈ k−1 (V )}.
Proof. Let X ∈ V such that ιX(I) = 0. If A ∈ k−1 (V ) then one has
ιA(I)(X) = I(A ∧ X) = (−1) k−1 I(X ∧ A) = (−1) k−1 ιX(I)(A) = 0.
So GI ⊂ WI. Let X ∈ G ⊥ I . If A ∈ k−1 (V ) then ιA(I)(X) = 0. That means ιX(I)(A) = 0, for
all A ∈ k−1 (V ). Therefore ιX(I) = 0 and then X ∈ W ⊥ I . It implies that WI = GI.
We turn now to the proof of the proposition.
(1) Let α ∈ VI, we show that α ∈ WI. It means that if X ∈ V such that ιX(I) = 0 then
α(X) = 0. Indeed, let X ∈ V such that ιX(I) = 0. Since α ∈ VI one has α ∧ I = 0. This
implies that
0 = ιX(α ∧ I) = ιX(α) ∧ I − α ∧ ιX(I) = α(X)I.
Therefore, α(X) = 0 since I = 0.
To prove dim(VI) ≤ k, we assume to the contrary, i.e. dim(VI) ≥ k+1. Let {α1,...,αk+1}
be an independent system of vectors of VI. Apply Lemma 2.2.4 one has a non-zero
complex ξ such that
I = ξ α1 ··· ∧ αk.
Since αk+1 ∧I = 0 we get ξ α1 ···∧αk = 0, i.e. ξ = 0. This is a contradiction. Therefore,
dim(VI) ≤ k.
To prove dim(WI) ≥ k, let {α1,...,αr} be a basis of VI and complete it to get a basis
{α1,...,αn} of V ∗ . By Lemma 2.2.4, there exists a (k − r)-form β such that
We can write β as follows:
I = α1 ··· ∧ αr ∧ β.
β = ∑ ξ jr+1,... jk
r+1≤ jr+1