Algebra and Applications 1. Abdenacer Makhlouf
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Let Kan(n) = G(n) + G(n)v. The Grassmann superalgebra G(n) is embeddable in the associative commutative superalgebra A = F [t, ξ1,…, ξn] = F [t] ⊗F G(n).
For an arbitrary scalar α ∈ F, the Poisson bracket [ , ] extends to the Jordan bracket on A defined by [t, ξi] = 0, [ξi, ξj] = –δij, [ξi, 1] = 0, [t, 1] = αt. The Kantor double Kan(n) = G(n) + G(n)v embeds in the Kantor double Kan(A, [ , ]) = A + Av. The subspace V (α) = tG(n) + tG(n)v is an irreducible unital Jordan bimodule over K(n). The square of the multiplication operator by the element v acts on V (α) as the scalar multiplication by α.
The simple superalgebras Kan(n), n ≥ 2 are exceptional (see Martínez et al. (2001)). Therefore, they do not have non-zero one-sided Jordan bimodules.
THEOREM 1.9 (Stern (1995), Martínez and Zelmanov (2009), Solarte and Shestakov (2016)).– Every finite dimensional irreducible Jordan bimodule over Kan(n), n ≥ 2 is isomorphic to V (α) or V (α)op, α ∈ F.
In Solarte and Shestakov (2016), the theorem above was proved for algebras over a field of characteristic p > 2.
1.7.2(c) Jordan superalgebras of a superform. Let
Let Cl(m) be the Clifford algebra of the restriction of the form 〈 , 〉 to
be the simple Weyl algebra.
Then the tensor product S = Cl(m) ⊗F Wn is the universal associative enveloping superalgebra of the Jordan superalgebra J = V + F ∙ 1.
Since the algebra Wn, n ≥ 1 is infinite dimensional, it follows that the superalgebra J does not have non-zero finite dimensional one-sided Jordan bimodules unless n = 0.
Consider in the algebra S the chain of subspaces
where Sr = (0) for r < 0,
THEOREM 1.10 (Martin and Piard (1992)).–
1 1) For every r ≥ 1, Sr/Sr–2 is a unital irreducible Jordan bimodule over J.
2 2) Let V′= Fu ⊕ V, where |u| = 0. Extend the bilinear form 〈 , 〉 to V′ via 〈u, u〉 = 1, 〈u, V〉 = (0). Then for every even r ≥ 0, the quotient uSr /uSr–2 is a unital irreducible Jordan bimodule over J.
3 3) Every unital irreducible finite dimensional J-bimodule is isomorphic to Sr/Sr–2 or to uSr/uSr–2 for even r.
The classification of irreducible Jordan bimodules over M1+1(F)(+), D(t), K3, JP(2) is too technical for an Encyclopedia survey. For a detailed description of finite dimensional irreducible Jordan bimodules, (see Martínez and Zelmanov (2003), Martin and Piard (1992), Martínez and Zelmanov (2006), Martínez and Shestakov (2020)). We will make only some general comments.
1.7.2(d) The universal associative enveloping superalgebra of J = M1+1(F)(+) is infinite dimensional, and finite dimensional one-sided Jordan bimodules over J are not necessarily completely reducible.
There is a family of 4-dimensional unital Jordan J-bimodules V (α, β, γ), which are parameterized by scalars α, β, γ ∈ F. If γ2 – 1 – 4αβ ≠ 0, then the bimodule V (α, β, γ) is irreducible. If γ2 – 1 – 4αβ = 0, then it has a composition series with 2-dimensional irreducible factors.
Every irreducible finite dimensional unital Jordan J-bimodule is isomorphic to V (α, β, γ), γ2 – 1 – 4αβ ≠ 0, or to a factor of a composition series of V (α, β, γ), γ2 – 1 – 4αβ = 0 (see Martínez and Zelmanov (2009); Martínez and Shestakov (2020)).
1.7.2(e) Now let us discuss the superalgebras D(t) and K3. Recall that
Then D(0) = F ∙ 1 + K3, D(–1) ≅ M1+1(F)(+), D(1) is a Jordan superalgebra of a superform.
We will assume therefore that t ≠ –1, 1.
One-sided bimodules. The superalgebra K3 does not have any non-zero one-sided Jordan bimodules (it has non-zero one-sided bimodules if char F > 0). All finite dimensional one-sided Jordan bimodules over D(t), t ≠ –1, 1, are completely reducible. The superalgebra D(t) does not have non-zero one-sided bimodules unless
Unital bimodules. If J = D(t) and t cannot be represented as