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1.6.4. Non-unital simple Jordan superalgebras
ОглавлениеFinally, let us consider non-unital simple Jordan superalgebras. As we have seen, K3 the three-dimensional Kaplansky superalgebra and K9 the nine-dimensional degenerate Kac superalgebra are examples of such superalgebras.
EXAMPLE 1.23.– Let Z be a unital associative commutative algebra, D : Z → Z a derivation. Assume that Z is D-simple and that the only constants are elements α1, α ∈ F.
Let us consider in Z the bracket { , } given by:
The above bracket is a Jordan bracket, so the Kantor double V(Z, D) = Z + Zv = KJ(Z, { , }) is a simple unital Jordan superalgebra.
Now we will change the product in V (Z, D), modifying only the action of the even part on the odd part and preserving the product of two even (respectively, two odd) elements. Denote with juxtaposition the product on V (Z, D). Let a, b ∈ Z. We define the new product ∙ by:
In this way, we get another Jordan superalgebra V1/2(Z, D) that is simple but not unital.
It was proved in Zelmanov (2000) that:
THEOREM 1.4.– Let J be a finite dimensional simple central non-unital Jordan superalgebra over a field F. Then J is isomorphic to one of the superalgebras on the list:
1 i) the Kaplansky superalgebra K3 (example 1.12);
2 ii) the field F has characteristic 3 and J is the degenerate Kac superalgebra (example 1.15);
3 iii) a superalgebra V1/2(Z, D) (example 1.23).
DEFINITION 1.14.– Let A be a Jordan superalgebra and let N be its radical, that is, the largest solvable ideal of A. The superalgebra A is said to be semisimple if N = (0).
EXAMPLE 1.24.– Let B be a simple non-unital Jordan superalgebra and let H(B) = B + F1 be its unital hull. Then H(B) is a semisimple Jordan superalgebra that is not simple.
THEOREM 1.5 (Zelmanov (2000)).– Let J be a finite dimensional Jordan superalgebra. Then J is semisimple if and only if
where J(1),…, J(t) are simple Jordan superalgebras and for every i = 1,…, s, the superalgebras Jij are simple non-unital Jordan superalgebras over the field extension Ki of F.