Remark A.6. Appendix A (1) Caution: O(n)-adjoint orbits are generally not the same as SO(n)-adjoint orbits. (2) Lemma A.5 (1) is a particular case of a general and classical result on semisimple Lie algebras: any semisimple element of a semisimple Lie algebra belongs to a Cartan subalgebra and all Cartan subalgebras are conjugate under the adjoint action [Sam80]. Here, gε are semisimple Lie algebras and the corresponding adjoint groups are SO(n) and Sp(2p). 132
Appendix B Here we prove: Lemma B.1. Let (g,B) be a non-Abelian 5-dimensional quadratic Lie algebra. Then g is a singular quadratic Lie algebra. Proof. • We assume g is not solvable and then we write g = s ⊕ r where s is semisimple and r is the radical of g [Bou71]. Then s sl(2) and B|s×s = λκ where κ is the Killing form. If λ = 0, consider Ψ : s → r ∗ defined by Ψ(S)(R) = B(S,R), for all S ∈ s, R ∈ r. Then Ψ is one-to-one and Ψ(ad(X)(S)) = ad ∗ (X)(ψ(S)), for all X, S ∈ s. So Ψ must be a homomorphism from the representation (s,ad|s) of s into the representation (r ∗ ,ad ∗ |s), so Ψ = 0, a contradiction. So λ = 0. Then B|s×s is non-degenerate. Therefore g = s ⊥ ⊕ s⊥ and ad(s)| s⊥ is an orthogonal 2-dimensional representation of s. Hence, ad(s)| s⊥ = 0 and [s,s⊥ ] = 0. We have B(X,[Y,Z]) = B([X,Y ],Z) = 0, for all X ∈ s, Y ∈ s⊥ , Z ∈ g. It follows that s⊥ is an ideal of g and therefore a quadratic 2-dimensional Lie algebra. So s⊥ is Abelian. Finally, g = s ⊥ ⊕ s⊥ with s⊥ a central ideal of g, so dup(g) = dup(s) = 3. • We assume that g is solvable and we write g = l ⊥ ⊕ z with z a central ideal of g (Proposition 2.1.5). Then dim(l) ≥ 3. If dim(l) = 3 or 4, then it is proved in Proposition 2.2.15 that l is singular, so g is singular. So we can assume that g is reduced, i.e. Z(g) ⊂ [g,g]. It results that dim(Z(g)) = 1 or 2 (Proposition 2.1.5 (3) and Remark 2.2.10). – If dim(Z(g)) = 1, Z(g) = CX0. Then dim([g,g]) = 4 and [g,g] = X ⊥ 0 . We can choose Y0 such that B(X0,Y0) = 1 and B(Y0,Y0) = 0. Let q = (CX0 ⊕ CY0) ⊥ . Then g = (CX0 ⊕ CY0) ⊥ ⊕ q. If X, X ′ ∈ q, then B(X0,[X,X ′ ]) = B([X0,X],X ′ ) = 0, so [X,X ′ ] ∈ X ⊥ 0 . Write [X,X ′ ] = λ(X,X ′ )X0 + [X,X ′ ]q with [X,X ′ ]q ∈ q. Remark that [X,[X ′ ,X ′′ ]] = λ(X,[X ′ ,X ′′ ]q)X0 + [X,[X ′ ,X ′′ ]q]q, for all X, X ′ , X ′′ ∈ q. So [·,·]q satisfies the Jacobi identity. Moreover B([X,X ′ ],X ′′ ) = −B(X ′ ,[X,X ′′ ]q). But also B([X,X ′ ],X ′′ ) = B([X,X ′ ]q,X ′′ ). So (q,[·,·]q,B|q×q) is a 3-dimensional quadratic Lie algebra. If q is an Abelian Lie algebra, then [X,X ′ ] ∈ CX0, for all X, X ′ ∈ q. Write B(Y0,[X,X ′ ]) = B([Y0,X],X ′ ) to obtain [X,X ′ ] = B(ad(Y0)(X),X ′ )X0, for all X, X ′ ∈ q. Since 133
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Remerciements Je remercie l’Ambas
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Contents 3.4.2 Quadratic dimension
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Introduction structures can be dete
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Notations 0 Ik where J = ∈ gl(2
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1.1. Definitions and gε(V ) = {C
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1.2. Nilpotent orbits Lie algebra g
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Consider Then dim(g) = 6 ∑ i=3 2.
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(2) g g ′ if and only if g i g
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