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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Appendix B<br />
dim(q) = 3 and ad(Y0)|q is skew-symmetric, there exists Q0 ∈ q such that ad(Y0)(Q0) =<br />
0. It follows that Q0 ∈ Z(g) and that is a contradiction since dim(Z(g)) = 1.<br />
Therefore (q,[·,·]q) sl(2). Consider<br />
0 → CX0 → X ⊥ 0 → q → 0.<br />
Then there is a section σ : q → X ⊥ 0 such that σ([X,X ′ ]q) = [σ(X),σ(X ′ )], for all X,<br />
X ′ ∈ q [Bou71]. Then σ(q) is a Lie subalgebra of g, isomorphic to sl(2) and that is<br />
a contradiction since g is solvable.<br />
– If dim(Z(g)) = 2, then we choose a non-zero X0 ∈ Z(g) and Y0 ∈ g such that<br />
B(X0,Y0) = 1 and B(Y0,Y0) = 0. Let q = (CX0 ⊕CY0) ⊥ . Then g = (CX0 ⊕CY0) ⊥<br />
⊕ q<br />
and as in the preceding case, [X,X ′ ] ∈ X ⊥ 0 , for all X, X ′ ∈ q. Write [X,X ′ ] =<br />
λ(X,X ′ )X0 +[X,X ′ ]q with [X,X ′ ]q ∈ q. Same arguments as in the preceding case allow<br />
us to conclude that [·,·]q satisfies the Jacobi identity and that B|q×q is invariant.<br />
So (q,[·,·]q,B|q×q) is a 3-dimensional quadratic Lie algebra.<br />
If q sl(2), then apply the same reasoning as in the preceding case to obtain a<br />
contradiction with g solvable.<br />
If q is an Abelian Lie algebra, then [X,X ′ ] ∈ CX0, for all X, X ′ ∈ q. Again, as in the<br />
preceding case, [X,X ′ ] = B(ad(Y0)(X),X ′ )X0, for all X, X ′ ∈ q. Then it is easy to<br />
check that g is the double extension of the quadratic vector space q by C = ad(Y0)|q.<br />
By Proposition 2.2.28, g is singular.<br />
Remark B.2. Let us give a list of all non-Abelian 5-dimensional quadratic Lie algebras:<br />
• g i o(3) ⊥<br />
⊕ C 2 with C 2 central, o(3) equipped with bilinear form λκ, λ ∈ C, λ = 0 and<br />
κ the Killing form. We have dup(g) = 3.<br />
• g i ⊥<br />
g4 ⊕ C with C central, g4 the double extension of C2 <br />
1 0<br />
by , g is solvable,<br />
0 −1<br />
non-nilpotent and dup(g) = 3.<br />
• g i g5, the double extension of C3 ⎛<br />
0<br />
by ⎝0<br />
1<br />
0<br />
⎞<br />
0<br />
−1⎠,<br />
g is nilpotent and dup(g) = 3.<br />
0 0 0<br />
⊥<br />
See Proposition 2.2.29 for the definition of g4 and g5. Remark that g4 ⊕ C is actually the<br />
double extension of C3 ⎛ ⎞<br />
1 0 0<br />
by ⎝0<br />
0 0 ⎠<br />
0 0 −1<br />
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