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 ...
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
Appendix D<br />
In the last appendix, we will prove a result needed to Chapter 4 as follows:<br />
Lemma D.1. Let (V,B) be a quadratic vector space, C be an invertible endomorphism of V<br />
such that<br />
(1) B(C(x),y) = B(x,C(y)), for all x,y ∈ V ,<br />
(2) 3C − 2C 2 = Id.<br />
Then there is an orthogonal basis {e1,...,en} of B such that C is diagonalizable with eigenvalues<br />
1 and 1 2 .<br />
Proof. First, one has (2) equivalent to C(C − Id) = 1 2 (C − Id). Therefore, if x is a vector in V<br />
such that C(x)−x = 0 then C(x)−x is an eigenvector with respect to eigenvalue 1 2 . We prove the<br />
result by induction on dim(V ). If dim(V ) = 1, let {e} be an orthogonal basis of V and assume<br />
C(e) = λe for some λ ∈ C. Then by (2) one has λ = 1 or λ = 1 2 .<br />
Assume that the result is true for quadratic vector spaces of dimension n, n ≥ 1. Assume<br />
dim(V ) = n + 1. If C = Id then the result follows. If C = Id then there exists x ∈ V such that<br />
C(x) − x = 0. Let e1 = C(x) − x then C(e1) = 1 2e1. If B(e1,e1) = 0 then there is e2 ∈ V such that B(e2,e2) = 0, B(e1,e2) = 1 and<br />
V = span{e1,e2} ⊥<br />
⊕ V1 where V1 = span{e1,e2} ⊥ . Since 1 2 = B(C(e1),e2) = B(e1,C(e2)) one<br />
has C(e2) = 1 2e2 + y + βe1 with y ∈ V1,β ∈ C. Let f1 = C(e2) − e2 = − 1 2e2 + y + βe1 then<br />
C( f1) = 1 2 f1 and B(e1, f1) = − 1 2 . If B( f1, f1) = 0 then let e1 = f1. If B( f1, f1) = 0 then let<br />
e1 = e1 + f1. In the bold cases, we have B( e1, e1) = 0 and C( e1) = 1 2 e1.<br />
⊥<br />
Let V = C e1 ⊕ e1 ⊥<br />
then e1 ⊥ is non-degenerate, C maps e1 ⊥ into itself. Therefore the result follows by induction.<br />
139