TH`ESE A NEW INVARIANT OF QUADRATIC LIE ALGEBRAS AND ...
TH`ESE A NEW INVARIANT OF QUADRATIC LIE ALGEBRAS AND ...
TH`ESE A NEW INVARIANT OF QUADRATIC LIE ALGEBRAS AND ...
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
If l = k then we stand at (k − 1)-step and<br />
2.2. Singular quadratic Lie algebras<br />
I = α1 ∧ ··· ∧ αk−1 ∧ ∑ ξ1,...,k−1, jk<br />
k≤ jk≤n<br />
α jk .<br />
Since αk ∧ I = 0 then ξJ = 0 if jk = k. Therefore, we obtain<br />
I = ξ α1 ∧ ··· ∧ αk.<br />
Lemma 2.2.5. WI = GI = {ιA(I) | A ∈ k−1 (V )}.<br />
Proof. Let X ∈ V such that ιX(I) = 0. If A ∈ k−1 (V ) then one has<br />
ιA(I)(X) = I(A ∧ X) = (−1) k−1 I(X ∧ A) = (−1) k−1 ιX(I)(A) = 0.<br />
So GI ⊂ WI. Let X ∈ G ⊥ I . If A ∈ k−1 (V ) then ιA(I)(X) = 0. That means ιX(I)(A) = 0, for<br />
all A ∈ k−1 (V ). Therefore ιX(I) = 0 and then X ∈ W ⊥ I . It implies that WI = GI.<br />
We turn now to the proof of the proposition.<br />
(1) Let α ∈ VI, we show that α ∈ WI. It means that if X ∈ V such that ιX(I) = 0 then<br />
α(X) = 0. Indeed, let X ∈ V such that ιX(I) = 0. Since α ∈ VI one has α ∧ I = 0. This<br />
implies that<br />
0 = ιX(α ∧ I) = ιX(α) ∧ I − α ∧ ιX(I) = α(X)I.<br />
Therefore, α(X) = 0 since I = 0.<br />
To prove dim(VI) ≤ k, we assume to the contrary, i.e. dim(VI) ≥ k+1. Let {α1,...,αk+1}<br />
be an independent system of vectors of VI. Apply Lemma 2.2.4 one has a non-zero<br />
complex ξ such that<br />
I = ξ α1 ··· ∧ αk.<br />
Since αk+1 ∧I = 0 we get ξ α1 ···∧αk = 0, i.e. ξ = 0. This is a contradiction. Therefore,<br />
dim(VI) ≤ k.<br />
To prove dim(WI) ≥ k, let {α1,...,αr} be a basis of VI and complete it to get a basis<br />
{α1,...,αn} of V ∗ . By Lemma 2.2.4, there exists a (k − r)-form β such that<br />
We can write β as follows:<br />
I = α1 ··· ∧ αr ∧ β.<br />
β = ∑ ξ jr+1,... jk<br />
r+1≤ jr+1