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 ...
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Case 3: dim(V ) = 5<br />
Appendix C<br />
Proposition C.1. If I is an indecomposable 3-form on V then there exists a basis {α1,α2,α3,<br />
α4,α5} of V ∗ such that I has the form<br />
where a,b ∈ C.<br />
Proof. First, we prove the following lemma:<br />
I = α1 ∧ (aα2 ∧ α3 + bα4 ∧ α5),<br />
Lemma C.2. Let V1 be a 4-dimensional vector space and J be a 2-form on V1. Then there exists<br />
a basis {β1,β2,β3,β4} of V ∗ 1 such that<br />
where p,q ∈ C.<br />
Proof. Let {β1,β2,β3,β4} be a basis of V ∗ 1<br />
where a,b,c,d,e, f ∈ C.<br />
J = pβ1 ∧ β2 + qβ3 ∧ β4,<br />
then J has the form:<br />
J = β1 ∧ (aβ2 + bβ3 + cβ4) + β2 ∧ (dβ3 + eβ4) + f β3 ∧ β4,<br />
(1) If aβ2 + bβ3 + cβ4 = 0 and dβ3 + eβ4 = 0 then the result follows.<br />
(2) If aβ2 + bβ3 + cβ4 = 0 and dβ3 + eβ4 = 0 then we can assume d = 0. Replace with<br />
β ′ 3 = β3 + e d β4 and β ′ 2 = β2 − f<br />
d β4 then we have the result.<br />
(3) If aβ2 +bβ3 +cβ4 = 0 and dβ3 +eβ4 = 0 then we can assume that a = 0 because if a = 0<br />
then we return (2). Replace with β ′ 2 = β2 + b a β3 + c a β4 one has the result.<br />
(4) If f = 0 we can assume d = 0. Replace with β ′ 3 = β3 + e d β4 then we return (3).<br />
(5) If aβ2 + bβ3 + cβ4 = 0 and dβ3 + eβ4 = 0 then we can assume that d = 0. Replace with<br />
β ′ 3 = β3 + e d β4 then we return the case (4). Therefore, J has only the form<br />
where p,q ∈ C.<br />
J = pβ1 ∧ β2 + qβ3 ∧ β4,<br />
By the above Lemma we can choose {α1,α2,α3,α4,α5} a basis of V ∗ such that I has the<br />
form:<br />
I = α1 ∧ Ω + aα2 ∧ α3 ∧ α4 + bα2 ∧ α3 ∧ α5 + cα2 ∧ α4 ∧ α5 + dα3 ∧ α4 ∧ α5,<br />
where Ω = pα2 ∧ α3 + qα4 ∧ α5 and a,b,c,d, p,q ∈ C. Now, we need the next lemma:<br />
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