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

1.4 Invertible orbits
1.4. Invertible orbits
Definition 1.4.1. We say that the Iε-orbit OX is invertible if X is an invertible element in gε.
Keep the above notations. We say an element X ∈ V isotropic if B(X,X) = 0 and a subset
W ⊂ V totally isotropic if B(X,Y ) = 0 for all X,Y ∈ W. To classify invertible adjoint orbits in
gε, we need the following lemma:
Lemma 1.4.2. Let V be an even-dimensional vector space with a non-degenerate bilinear form
Bε. Assume that V = V+ ⊕V− where V± are totally isotropic vector subspaces.
(1) Let N ∈ L (V ) such that N(V±) ⊂ V±. We define maps N± by N+|V+ = N|V+ , N+|V− = 0,
N−|V− = N|V− and N−|V+ = 0. Then N ∈ gε(V ) if and only if N− = −N∗ + and, in this case,
N = N+ − N∗ +.
(2) Let U+ ∈ L (V ) such that U+ is invertible, U+(V+) = V+ and U+|V− = IdV− . We define
U ∈ L (V ) by U|V+ = U+|V+ and U|V− = U −1
∗ + |V− . Then U ∈ Iε(V ).
(3) Let N ′ ∈ gε(V ) such that N ′ satisfies the assumptions of (1). Define N± as in (1). More-
Proof.
over, we assume that there exists U+ ∈ L (V+), U+ invertible such that
|V+ .
N ′ +|V+ = U+ N+ U −1
+
We extend U+ to V by U+|V− = IdV− and define U ∈ Iε(V ) as in (2). Then
N ′ = U N U −1 .
(1) It is obvious that N = N+ + N−. Recall that N ∈ gε(V ) if and only if N ∗ = −N so
N ∗ + + N ∗ − = −N+ − N−. Since Bε(N ∗ +(V+),V ) = Bε(V+,N+(V )) = 0 then N ∗ +(V+) = 0.
Similarly, N ∗ −(V−) = 0. Hence, N− = −N ∗ +.
(2) We shows that Bε(U(X),U(Y )) = Bε(X,Y ), for all X,Y ∈ V . Indeed, let X = X+ +
X−,Y = Y+ +Y− ∈ V+ ⊕V−, one has
Bε(U(X+ + X−),U(Y+ +Y−)) = Bε(U+(X+) + U −1
+
= Bε(U+(X+), U −1
+
∗
(X−),U+(Y+) + U −1∗
(Y−))
∗
(Y−)) + Bε( U −1∗
(X−),U+(Y+))
= Bε(X+,Y−) + Bε(X−,Y+) = Bε(X,Y ).
(3) Since Bε(U −1 (V+),V+) = Bε(V+,U(V+)) = 0, one has U −1 (V+) = V+ and U −1 (V−) =
V−. Consequently, (U N U −1 )(V+) ⊂ V+ and (U N U −1 )(V−) ⊂ V−. Clearly, U N U −1 ∈
gε(V ). By (1), we only show that
This is obvious since U −1 |V+
= U−1
+ .
+
(U N U −1 )|V+ = N′ +
13
+