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

2.2. Singular quadratic Lie algebras
Therefore, we can conclude that all solvable quadratic Lie algebras of type S3 are double
extensions of a quadratic vector space by a skew-symmetric map.
We remark that in the nilpotent Lie algebras classification, the Lie algebras g5 and g6 can be
identified respectively as g5,4 and g6,4, for more details the reader should refer to [Oom09] and
[Mag10].
2.2.5 Classification singular quadratic Lie algebras
Let (q,B) be a quadratic vector space. We recall the notations in Chapter 1 that O(q) is the
group of orthogonal maps and o(q) is its Lie algebra, i.e. the Lie algebra of skew-symmetric
maps. Also, recall that the adjoint action is the action of O(q) on o(q) by conjugation.
Theorem 2.2.30. Let (q,B) be a quadratic vector space. Let g = (CX1 ⊕ CY1) ⊥
⊕ q and g ′ =
(CX ′ 1 ⊕ CY ′ ⊥
1 ) ⊕ q be double extensions of q, by nonzero skew-symmetric maps C and C ′ respectively.
Then:
(1) there exists a Lie algebra isomorphism between g and g ′ if and only if there exist an
invertible map P ∈ L (q) and a non-zero λ ∈ C such that C ′ = λ PCP −1 and P ∗ PC = C
where P ∗ is the adjoint map of P with respect to B.
(2) there exists an i-isomorphism between g and g ′ if and only if C ′ is in the O(q)-adjoint
orbit through λC for some non-zero λ ∈ C.
Proof.
(1) Let A : g → g ′ be a Lie algebra isomorphism. We know by Proposition 2.2.28 that g and
g ′ are singular. Assume that g is of type S3. Then 3 = dim([g,g]) = dim([g ′ ,g ′ ]). So g ′
is also of type S3 ([PU07]). Therefore, g and g ′ are either both of type S1 or both of type
S3. Let us study these two cases.
(i) First, assume that g and g ′ are both of type S1. We start by proving that A(CX1 ⊕q) =
CX ′ 1 ⊕ q. If this is not the case, there is X ∈ q such that A(X) = βX ′ 1 + γY ′ 1 +Y with
Y ∈ q and γ = 0. Then
[A(X),CX ′ 1 ⊕ q] ′ = γC ′ (q) + [Y,q] ′ .
Since g ′ is of type S1, we have rank(C ′ ) ≥ 4 (see Proposition 2.2.28) and it follows
that dim([A(X),CX ′ 1 ⊕ q]′ ) ≥ 4. On the other hand, [A(X),CX ′ 1 ⊕ q]′ is contained in
A([X,g]) and dim([X,g]) ≤ 2, so we obtain a contradiction.
Next, we prove that A(X1) ∈ CX ′ 1 . By the definition of a double extension and B
non-degenerate, then there exist X, Y ∈ q such that X1 = [X,Y ]. Then A(X1) =
[A(X),A(Y )] ′ ∈ [CX ′ 1 ⊕ q,CX ′ 1 ⊕ q]′ = CX ′ 1 . Hence A(X1) = µX ′ 1 for some non-zero
µ ∈ C.
Now, write A|q = P + β ⊗ X ′ 1 with P : q → q and β ∈ q∗
. If X ∈ ker(P), then
A X − 1
µ β(X)X1
= 0, so X = 0 and therefore, P is invertible.
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