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

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