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

4.3. Pseudo-Euclidean 2-step nilpotent Jordan algebras
For a non-degenerate cyclic symmetric map θ of a, define a 3-form
I(x,y,z) = θ(x,y)z, ∀ x,y,z ∈ a.
Then I ∈ S 3 (a), the space of symmetric 3-forms on a. The non-degenerate condition of θ is
equivalent to ∂I
∂ p = 0, for all p ∈ a∗ .
Conversely, let a be a complex vector space and I ∈ S 3 (a) such that ∂I
∂ p = 0 for all p ∈
a∗ . Define θ : a × a → a∗ by θ(x,y) = I(x,y,.), for all x,y ∈ a then θ is symmetric and nondegenerate.
Moreover, since I is symmetric, then θ is cyclic and we obtain a reduced pseudo-
Euclidean 2-step nilpotent Jordan algebra T ∗(a)
defined by θ. Therefore, there is an one-to-one
θ
map from the set of all T ∗-extensions of a complex vector space a onto the subset {I ∈ S 3 (a) |
∂I
∂ p = 0, ∀ p ∈ a∗ }, such elements are also called non-degenerate.
Corollary 4.3.17. Let J1 and J2 be T ∗ -extensions of a with respect to I1 and I2 non-degenerate.
Then J and J ′ are i-isomorphic if and only if there exists an isomorphism A of a such that
I1(x,y,z) = I2(A(x),A(y),A(z)), ∀ x,y,z ∈ a.
In particular, J and J ′ are i-isomorphic if and only if there is an isomorphism t A on a ∗ which
induces the isomorphism on S 3 (a), also denoted by t A such that t A(I1) = I2. In this case, we
say that I1 and I2 are equivalent.
Example 4.3.18. Let a = Ce be a one-dimensional vector space then S 3 (a) = C(e ∗ ) 3 . By
Example 4.3.16, T ∗ -extensions of a by (e ∗ ) 3 and λ(e ∗ ) 3 , λ = 0, are i-isomorphic (also, these
3-forms are equivalent). Hence, there is only one i-isomorphic class of T ∗ -extensions of a, that
is J = Ce ⊕ C f with e 2 = f and B(e, f ) = 1, the other are zero.
Now, let a = Cx ⊕ Cy be 2-dimensional vector space then
S 3 (a) = {a1(x ∗ ) 3 + a2(x ∗ ) 2 y ∗ + a3x ∗ (y ∗ ) 2 + a4(y ∗ ) 3 , ai ∈ C}.
It is easy to prove that every bivariate homogeneous polynomial of degree 3 is reducible.
Therefore, by a suitable basis choice (certainly isomorphic), a non-degenerate element I ∈
S 3 (a) has the form I = ax ∗ y ∗ (bx ∗ + cy ∗ ), a,b = 0. Replace x ∗ by αx ∗ with α 2 = ab to get
the form of I λ = x ∗ y ∗ (x ∗ + λy ∗ ),λ ∈ C.
Next, we will show that I0 and I λ ,λ = 0 are not equivalent. Indeed, assume the contrary, i.e.
there is an isomorphism t A such that t A(I0) = I λ . We can write
Then
t A(x ∗ ) = a1x ∗ + b1y ∗ , t A(y ∗ ) = a2x ∗ + b2y ∗ , a1,a2,b1,b2 ∈ C.
t A(I0) = (a1x ∗ + b1y ∗ ) 2 (a2x ∗ + b2y ∗ ) = a 2 1a2(x ∗ ) 3 + (a 2 1b2 + 2a1a2b1)(x ∗ ) 2 y ∗ +
(2a1b1b2 + a2b 2 1)x ∗ (y ∗ ) 2 + b 2 1b2(y ∗ ) 3 .
Comparing the coefficients we will get a contradiction. Therefore, I0 and I λ , λ = 0 are not
equivalent.
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