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