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

Appendix A
Lemma A.4. Assume that C is semisimple and invertible. Then there is a basis {e1,...,ep,
f1,..., fp} of V such that Bε(ei,ej) = Bε( fi, f j) = 0, Bε(ei, f j) = δi j, 1 ≤ i, j ≤ p. For 1 ≤ i ≤ p,
there exist non-zero λi ∈ C such that C(ei) = λiei and C( fi) = −λi fi.
Moreover, if Λ denotes the spectrum of C, then λ ∈ Λ if and only if −λ ∈ Λ, λ and −λ have
the same multiplicity.
Proof. We prove the result by induction on dim(V ). Assume that dim(V ) = 2. Let {e1,e2}
be an eigenvector basis of V corresponding to eigenvalues λ1 and λ2. We have Bε(C(v),v ′ ) =
−Bε(v,C(v ′ )) and C is invertible, so Bε(e1,e1) = Bε(e2,e2) = 0, Bε(e1,e2) = 0 and λ2 = −λ1.
1
Let f1 =
Bε(e1,e2) e2, then the basis {e1, f1} is a convenient basis.
Assume that the result is true for vector spaces of dimension n with n ≤ 2(p − 1). Assume
dim(V ) = 2p. Let {e1,...,e2p} be an eigenvector basis with corresponding eigenvalues
λ1,...,λ2p. As before, Bε(ei,ei) = 0, 1 ≤ i ≤ 2p, so there exists j such that Bε(e1,ej) =
1
0. Then λ j = −λ1. Let f1 =
Bε(e1,ej) e j. Then Bε| span{e1, f1} is non-degenerate, so V =
span{e1, f1} ⊥
⊕ V1 where V1 = span{e1, f1} ⊥ . But C maps V1 into itself, so we can apply the
induction assumption and the result follows.
As a consequence, we have this classical result, used in Chapter 1:
Lemma A.5.
(1) Let C be a semisimple element of o(n). Then C belongs to the SO(n)-adjoint orbit
of an element of the standard Cartan subalgebra of o(n) (i.e., an element with matrix
diag 2p (λ1,...,λp,−λ1,...,−λp) if n = 2p and diag 2p+1 (λ1,...,λp,0,−λ1,...,−λp) if
n = 2p + 1 in a canonical basis of C n ).
(2) Let C be a semisimple element of sp(2p). Then C belongs to the Sp(2p)-adjoint orbit
of an element of the standard Cartan subalgebra of sp(2p) (i.e., an element with matrix
diag 2p (λ1,...,λp,−λ1,...,−λp)).
(3) Let C and C ′ be semisimple elements of gε. Then C and C ′ are in the same Iε-adjoint orbit
if and only if they have the same spectrum, with same multiplicities.
Proof.
(1) We have C n = ker(C) ⊥
⊕ Im(C) and rank(C) is even. So dim(ker(C)) is even if n = 2p
and odd, if n = 2p + 1. Then apply Lemma A.4 to C| Im(C) to obtain the result.
(2) The proof is similar to (1) with n even.
(3) If C and C ′ have the same spectrum and their eigenvalues having same multiplicities, they
are Iε-conjugate to the same element of the standard Cartan subalgebra.
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