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

4.2. Jordanian double extension of a quadratic vector space
Proof. Assume J and J ′ isomorphic by A. First, we will show that A(q ⊕ Cx1) = q ⊕ Cx ′ 1 .
Indeed, if A(x1) = y + βx ′ 1 + γy′ 1 where y ∈ q,β,γ ∈ C, then
0 = A(x 2 1) = A(x1)A(x1) = 2γC ′ (y) + (2βγ + B(C ′ (y),y))x ′ 1 + γ 2 y ′ 1.
Therefore, γ = 0. Similarly, if there exists x ∈ q such that A(x) = z + αx ′ 1 + δy′ 1 where z ∈
q,α,δ ∈ C. Then
B(C(x),x)A(x1) = A(x 2 ) = A(x)A(x) = 2δC ′ (z) + (2αδ + B(C ′ (z),z))x ′ 1 + δ 2 y ′ 1.
That implies δ = 0 and A(q ⊕ Cx1) = q ⊕ Cx ′ 1 .
The rest of the proof follows exactly the proof of Theorem 4.2.5, one has A(x1) = µx ′ 1 for
some non-zero µ ∈ C and there is an isomorphism of q such that A|q = P+β ⊗x ′ 1 where β ∈ q∗ .
Similarly as in the proof of Theorem 4.2.5, one also has P ∗ C ′ P = µC, where P ∗ is the adjoint
map of P with respect to B. Assume that A(y1) = λy ′ 1 +y+δx′ 1 . Since A(y1)A(y1) = A(y1), one
has λ = 1 and therefore C ′ = PCP −1 . Replace P by
1
(µ) 1 2
P to get P ∗ PC = C. However, since C
is invertible then P∗P = I. It means that P is an isometry of q.
Conversely, define A : J → J ′ by A(x1) = x ′ 1 , A(y1) = y ′ 1 and A(x) = P(x), for all x ∈ q then
A is an i-isomorphism.
An invertible symmetric endomorphism of q satisfying 3C − 2C2 = Id is diagonalizable by
an orthogonal basis of eigenvectors with eigenvalues 1 and 1 2 (see Appendix D). Therefore, we
have the following corollary:
Corollary 4.2.9. Let (q,B) be a quadratic vector space and let J = q ⊥
⊕ (Cx1 ⊕ Cy1) and
J ′ = q ⊥
⊕ (Cx ′ 1 ⊕ Cy′ 1
respectively. Then J and J ′ are isomorphic if and only if C and C ′ have the same spectrum.
) be diagonalizable double extensions of q, by invertible maps C and C′
Example 4.2.10. Let Cx be a one-dimensional Abelian algebra. Let J = Cx ⊥
⊕ (Cx1 ⊕Cy1) and
⊕ (Cx ′ 1 ⊕ Cy′ 1 ) be diagonalizable double extensions of Cx by C = Id and C = 1 2 Id. In
particular, the product on J and J ′ are defined by:
J ′ = Cx ⊥
y 2 1 = y1,y1x = x,y1x1 = x1,x 2 = x1;
(y ′ 1) 2 = y ′ 1,y ′ 1x = 1
2 x,y1x1 = x1,x 2 = 1
2 x1.
Then J and J ′ are not isomorphic. Moreover, J ′ has no unit element.
Remark 4.2.11. The i-isomorphic and isomorphic notions are not coinciding in general. For
example, the Jordan algebras J = Ce with e 2 = e, B(e,e) = 1 and J ′ = Ce ′ with (e ′ ) 2 = e ′ ,
B(e ′ ,e ′ ) = a = 1 are isomorphic but not i-isomorphic.
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