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

Appendix A
Moreover Im(C) = L ⊥
⊕ Cv ⊥
⊕ (U ⊕U ′ ⊥
) and C : L ′ ⊕ Cv ⊥
⊕ (U ⊕U ′ ) → L ⊥
⊕ Cv ⊥
⊕ (U ⊕
U ′ ) is a bijection.
(3) If ε = 1, in both cases, rank(C) is even.
Proof. Since (ker(C)) ⊥ = Im(C), L is isotropic.
(1) If dim(V ) is even, there exist maximal isotropic subspaces W1 and W2 such that V =
W1 ⊕W2 [Bou59] and L ⊂ W1. Let U be a complementary subspace of L in W1, W1 =
L ⊕U and {U1,...,Us} a basis of U. Consider the isomorphism Ψ : W2 → W ∗ 1 defined by
Ψ(w2)(w1) = Bε(w2,w1), for all w1 ∈ W1, w2 ∈ W2. Define L ′ i = ψ−1 (L∗ i ), 1 ≤ i ≤ r, L′ =
) =
span{L ′ 1 ,...,L′ r}, U ′ j = ψ−1 (U ∗ j ), 1 ≤ j ≤ s, U′ = span{U ′ 1 ,...,U′ s}. Then Bε(Li,L ′ j
δi j, 1 ≤ i, j ≤ r, L and L ′ are isotropic, Bε(Ui,U ′ j ) = δi j, for all 1 ≤ i, j ≤ s, U and U ′ are
isotropic and
V = (L ⊕ L ′ ) ⊥
⊕ (U ⊕U ′ ).
Since Im(C) = L⊥ , we have Im(C) = L ⊥
⊕ (U ⊕U ′ ⊥
). Finally, if v ∈ L ′ ⊕ (U ⊕U ′ ) and
⊥
C(v) = 0, then v ∈ L. So v = 0. Therefore C is one to one from L ′ ⊕ (U ⊕ U ′ ) into
L ⊥
⊕ (U ⊕U ′ ) and since the dimensions are the same, C is a bijection.
(2) There exist maximal isotropic subspaces W1 and W2 such that V = (W1 ⊕W2) ⊥
⊕ Cv, with
v ∈ V such that Bε(v,v) = 1 and L ⊂ W1 [Bou59]. Then the proof is essentially the same
as in (1).
⊥
(3) Assume that dim(V ) is even. Define a bilinear form ∆ on L ′ ⊕ (U ⊕U ′ ) by ∆(v1,v2) =
⊥
Bε(v1,C(v2)), for all v1, v2 ∈ L ′ ⊕ (U ⊕U ′ ). Since C ∈ o(V ), ∆ is skew-symmetric. Let
⊥
v1 ∈ L ′ ⊕ (U ⊕U ′ ⊥
) such that ∆(v1,v2) = 0, for all v2 ∈ L ′ ⊕ (U ⊕U ′ ). Then Bε(v1,w) = 0,
for all w ∈ L ⊥
⊕ (U ⊕U ′ ). It follows that Bε(v1,w) = 0, for all w ∈ V , so v1 = 0 and ∆ is
⊥
non-degenerate. So dim(L ′ ⊕ (U ⊕U ′ ) is even. Therefore dim(L ′ ) = dim(L) is even and
rank(C) is even. If V is odd-dimensional, the proof is completely similar.
Corollary A.3. If C ∈ o(V ), then rank(C) is even.
Proof. By Lemma A.1, Im(C) = Im(CW ) and rank(CW ) is even by the preceding Lemma.
For instance, if C ∈ o(V ) and C is invertible, then dim(V ) must be even. But this can also
be proved directly: when C is invertible, then the skew-symmetric form ∆C on V defined by
∆C(v1,v2) = Bε(v1,C(v2)), for all v1, v2 ∈ V , is clearly non-degenerate.
When C is semisimple (i.e. diagonalizable), we have V = ker(C) ⊥
⊕ Im(C) and C| Im(C) is
invertible. So semisimple elements are completely described by:
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