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

4.2. Jordanian double extension of a quadratic vector space
4.2.1 Nilpotent double extensions
Consider J1 = q as an Abelian algebra, J2 = Cy1 the nilpotent one-dimensional Jordan
algebra, π(y1) = C and identify J∗ 2 with Cx1. Then by Definition 4.1.17, J = J2 ⊕ J1 ⊕ J∗ 2 is a
pseudo-Euclidean Jordan algebra with a bilinear form B given by B = Bq + Bt. In this case, C
obviously satisfies the condition C3 = 0.
An immediate corollary of the definition is:
Corollary 4.2.2. If J = q ⊥
⊕ (Cx1 ⊕ Cy1) is the nilpotent double extension of q by C then
y1x = C(x),xy = B(C(x),y)x1 and y 2 1 = x1J = 0, ∀ x ∈ q.
As a consequence, J 2 = Im(C) ⊕ Cx1 and Ann(J) = ker(C) ⊕ Cx1.
Remark 4.2.3. In this case, J is k-step nilpotent, k ≤ 3 since R k x(J) ⊂ Im(C k ) ⊕ Cx1.
Definition 4.2.4. Let (J,B) and (J ′ ,B ′ ) be pseudo-Euclidean Jordan algebras, if there exists a
Jordan algebra isomorphism A : J → J ′ such that it is also an isometry then we say that J, J ′ are
i-isomorphic and A is an i-isomorphism.
Theorem 4.2.5. Let (q,B) be a quadratic vector space. Let J = q ⊥
⊕ (Cx1 ⊕ Cy1) and J ′ =
q ⊥
⊕ (Cx ′ 1 ⊕ Cy′ 1 ) be nilpotent double extensions of q, by symmetric maps C and C′ respectively.
Then:
(1) there exists a Jordan algebra isomorphism A : J → J ′ such that A(q ⊕ Cx1) = q ⊕ Cx ′ 1
if and only if there exist an invertible map P ∈ End(q) and a non-zero λ ∈ C such that
λC ′ = PCP −1 and P ∗ PC = C where P ∗ is the adjoint map of P with respect to B.
(2) there exists an i-isomorphism A : J → J ′ such that A(q ⊕ Cx1) = q ⊕ Cx ′ 1 if and only if
there exists a non-zero λ ∈ C such that C and λC ′ are conjugate by an isometry P ∈ O(q).
Proof.
(1) Assume that A : J → J ′ is an isomorphism such that A(q ⊕ Cx1) = q ⊕ Cx ′ 1 . By Corollary
4.2.2 and B non-degenerate, there exist x,y ∈ q ⊕ Cx1 such that xy = x1. Therefore
A(x1) = A(x)A(y) ∈ (q ⊕ Cx ′ 1 )(q ⊕ Cx′ 1 ) = Cx′ 1 . That means A(x1) = µx ′ 1 for some nonzero
µ ∈ C. Write A|q = P + β ⊗ x ′ 1 with P ∈ End(q) and β ∈ q∗
. If x ∈ ker(P) then
A x − 1
µ β(x)x1
= 0, so x = 0 and therefore, P is invertible. For all x,y ∈ q, one has
µB(C(x),y)x ′ 1 = A(xy) = A(x)A(y) = B(C ′ (P(x)),P(y))x ′ 1.
So we obtain P∗C ′ P = µC. Assume that A(y1) = y + δx ′ 1 + λy′ 1 , with y ∈ q. For all x ∈ q,
one has
P(C(x)) + β(C(x))x ′ 1 = A(y1x) = A(y1)A(x) = λC ′ (P(x)) + B(C ′ (y),P(x))x ′ 1.
Therefore, λC ′ = PCP−1 . Combined with P∗C ′ P = µC to get P∗PC = λ µC. Replace P
by
1
P to obtain λC ′ = PCP−1 and P∗PC = C.
(µλ) 1 2
105