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

2.3. Quadratic dimension of quadratic Lie algebras
2.3.2 Quadratic dimension of reduced singular quadratic Lie algebras
and the invariance of dup number
Proposition 2.3.6. Let g be a reduced singular quadratic Lie algebra and D ∈ L (g) be a
symmetric map. Then:
(1) D is a centromorphism if and only if there exist µ ∈ C and a symmetric map Z : g → Z(g)
such that Z| [g,g] = 0 and D = µ Id+Z. Moreover D is invertible if and only if µ = 0.
(2)
Proof.
dq(g) = 1 +
dim(Z(g))(1 + dim(Z(g))
.
2
(1) If g = o(3), with B = λκ and κ the Killing form, the two results are obvious. So, we
examine the case where g is solvable, and then g can be realized as a double extension:
g = (CX1 ⊕ CY1) ⊥
⊕ q, with corresponding bilinear form B on q, C = ad(Y1), C = C|q ∈
o(q).
Let D be an invertible centromorphism. One has D ◦ ad(X) = ad(X) ◦ D, for all X ∈ g
and that implies DC = CD. Using formula (1) of Lemma 2.2.27 and CD = DC, from
[D(X),Y1] = [X,D(Y1)], we find D(C(X)) = B(D(X1),Y1)C(X). Let µ = B(D(X1),Y1).
Since D is invertible, one has µ = 0 and C(D− µ Id) = 0. Since ker(C) = CX1 ⊕ker(C)⊕
CY1 = Z(g) ⊕ CY1, there exists a map Z : g → Z(g) and ϕ ∈ g ∗ such that D − µ Id =
Z + ϕ ⊗Y1. But D maps [g,g] into itself, so ϕ| [g,g] = 0. One has [g,g] = CX1 ⊕ Im(C).
If X ∈ Im(C), let X = C(Y ). Then D(X) = D(C(Y )) = µC(Y ), so D(X) = µX. For
Y1, D([Y1,X]) = DC(X) = µC(X) for all X ∈ g. But also, D([Y1,X]) = [D(Y1),X] =
µC(X) + ϕ(Y1)C(X), hence ϕ(Y1) = 0.
Assume we have shown that D(X1) = µX1. Then if X ∈ q, B(D(X1),X) = µB(X1,X) = 0.
Moreover, B(D(X1),X) = B(X1,D(X)), so ϕ(X) = 0. Thus, to prove (1), we must prove
that D(X1) = µX1. We decompose q respectively to C as in Appendix A. Let l = ker(C).
Then:
q = (l ⊕ l ′ ) ⊥
⊕ (u ⊕ u ′ )
⊥
and C is an isomorphism from l ′ ⊕ (u ⊕ u ′ ) onto l ⊥
⊕ (u ⊕ u ′ ). Or
q = (l + l ′ ) ⊥
⊕ CT ⊥
⊕ (u ⊕ u ′ )
⊥
and C is an isomorphism from l ′ ⊕ CT ⊥
⊕ (u ⊕ u ′ ) onto l ⊥
⊕ CT ⊥
⊕ (u ⊕ u ′ ).
If u⊕u ′ = {0}, there exist X ′ , Y ′ ∈ u⊕u ′ such that B(X ′ ,Y ′ ⊥
) = −1 and X, Y ∈ l ′ ⊕ (u⊕u ′ )
⊥
(resp. l ′ ⊕ CT ⊥
⊕ (u⊕u ′ )) such that X ′ = C(X), Y ′ = C(Y ). It follows that [C(X),Y ] = X1
and then D(X1) = [DC(X),Y ] = µ[C(X),Y ] = µX1.
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