CHAPTER 5. LOCAL SUPERALGEBRA OF LIE SUPERALGEBRAS AT AD-NILPOTENT ELEMENTS

This chapter is part of an article that has been published in the journal Communications in Algebra and can be found in [30].

In this chapter we extend the ideas of local algebras of Jordan algebras to the super setting, and Jordan superstructures are attached to Lie superalgebras at ad-nilpotent homogeneous elements.

We also generalize in the section 6.3 the notion of subquotient to the Lie superalgebra. It comes attached to an abelian Lie inner ideal of a Lie superalgebra, and it is indeed a Jordan superpair. Moreover, in the particular case of an abelian inner ideal of the form [a, [a, L]], the subquotient agrees with the Jordan superobject obtained in the section 6.2.

The chapter is organized as follows. When a is even, we easily obtain a Jordan superalgebra by using the Grassmann envelope. But when we deal with an odd ad- nilpotent element a of index less than or equal to 4 we first define a triple product in [a, [a, L]], and then we double this triple and change a sign in one of the associated triple products to get a Jordan superpair. We introduce subquotients associated to abelian inner ideals of Lie superalgebras and show that they are Jordan superpairs.

Finally, we show that the Jordan superalgebras/superpairs obtained in the previous section agree with the subquotients associated to abelian inner ideals of the form [a, [a, L]].

In addition, we will assume that $\frac{1}{3}$ ϵ Φ.

5.0.1. Let M = M₀ ⊗ M₁ be a supermodule over Φ. Then the associative algebra End(M) is provided with the induced ℤ₂-grading End(M) = End(M)₀ ⊗ End(M)₁, in which

End(M)i = {f ϵ End(M) | f(M_j) ⊆ M_i+j}.

Let L = L₀ ⊗ L₁ be a Lie superalgebra over Φ then End(L) becomes an associative superalgebra and (End(L))⁻ with product [f, g] = fg − (−1)^|f||g|gf for homogeneous elements f, g ϵ End(L) becomes a Lie superalgebra. The set ad L of adjoint maps is a Lie superideal of (End(L))⁻, so if we denote by capital letters the adjoint maps associated to elements, i.e., A = ad_a, B = ad_b, etc., we have [A, B] = AB − (−1)|a||b|BA for homogeneous elements a,b ϵ L₀ ∪ L₁. This notation will be useful because it allows us to think in an associative way when we are doing calculations.

5.1. A Jordan superalgebra at an even homogeneous ad-nilpotent element

5.1.1. Let L = L₀ + L₁ be a Lie superalgebra, and let a ϵ L₀ such that ${\mathrm{ad}}_a^{3}L=0$. Such an element will be called Jordan element of L. In the Φ -module [a, [a, L]] we can define a new product

$$[a,[a,x\left]\right]\cdot [a,[a,y\left]\right]=\frac{1}{2}[a,[a,[x,[a,y\left]\right]\left]\right].$$

The (nonassociative) algebra ([a, [a, L]],·) is ℤ₂-graded with homogeneous parts [a, [a, L]]₀ = [a, [a, L₀]] and [a, [a, L]]₁ = [a, [a, L₁]]. The parity of an homogeneous element x̄ coincides with the parity of x as an element in the Lie superalgebra L, i.e., |x̄ | = |x| for every homogeneous element x ϵ L₀ ∪ L₁. In the next proposition we prove that this superalgebra is in fact a Jordan superalgebra.

Proposition 5.1.2. Let L = L₀ + L₁ be a Lie superalgebra and a ϵ L₀ be a Jordan element. Then ([a, [a, L]], •) is a Jordan superalgebra.

Proof. Let us check that the Grassmann envelope of [a, [a, L]] is a Jordan algebra with the induced product. Let us consider ã = a ⊗ 1 ϵ G(L), which is a Jordan element of the Lie algebra G(L). By Theorem [24, 2.4(ii)] and Remark [24, 2.44] we can consider the Jordan algebra [ã, [ã, G(L)]] of G(L) at ã with product

$$[\overline{a},[\overline{a},\tilde{x}\left]\right]\cdot [\tilde{a},[\overline{a},\tilde{y}\left]\right]=\frac{1}{2}[\overline{a},[\overline{a},[\overline{x},[\overline{a},\tilde{y}\left]\right]\left]\right]$$

for any x̃, ỹ ϵ G(L).

The map φ : [ã, [ã, G(L) → G([a, [a, L]]) given by φ([ã, [ã, x ⊗ ξ_i1 ξ_i2 ... ξi_k]]) = [a, [a, x]] ⊗ ξ_i1 ξ_i2 ... ξi_k is an isomorphism, so G([a, [a, L]]) is a Jordan algebra, giving that [a, [a, L]] is a Jordan superalgebra.

Remark 5.1.3. The induced triple product on [a, [a, L]] is given by

$$\{\overline{x},\overline{y},\overline{z}\}=(-1{)}^{\left|y\right|\left|z\right|}\frac{1}{4}{A}^{2}XZ{A}^2(y)$$

for homogeneous x̃,ỹ,z̃ ϵ [a, [a, L]]. Indeed,

$$\begin{array}{l}4\{\overline{x},\overline{y},\overline{z}\}=4\left(\overline{x}\cdot (\overline{y}\cdot \overline{z})+(-1{)}^{\left|y\right|\left|z\right|+\left|x\right|\left|z\right|}\overline{z}\cdot (\overline{x}\cdot \overline{y})-(-1{)}^{\left|x\right|\left|y\right|}\overline{y}\cdot (\overline{x}\cdot \overline{z})\right)=\\ =A^2[x,[a,[y,[a,z\left]\right]\left]\right]+(-1{)}^{\left|y\right|\left|z\right|+\left|x\right|\left|z\right|}{A}^2[z,[a,[x,[a,y]\left]\right]]-\\ -(-1{)}^{\left|x\right|\left|y\right|}{A}^2[y,[a,[x,[a,z]\left]\right]]=\\ =\overline{\left[\right[\mid x,a],[[\mid y,a],z\left]\right]}+(-1{)}^{\left|y\right|\left|z\right|+\left|x\right|\left|z\right|}\overline{[\mid z,a],[\mid x,a],y\left]\right]}-(-1{)}^{\left|x\right|\left|y\right|}\overline{\mid [y,a],[\mid x,a],z\left]\right]}=\\ \left.\left.=A^2\left[\right[\mid x,a],[y,a\left]\right],z\right]+(-1{)}^{\left|x\right|\left|y\right|+\left|y\right|\left|z\right|+\left|x\right|\left|z\right|}{A}^2[z,[a,[\mid y,a],x]]\right]=\\ =A^2\left[\right[x,[a,[y,a\left]\right]],z]=(-1{)}^{\left|z\right|\left|\right|x|+|y\left|\right|}{A}^2[z,[x,[a,[a,y]\left]\right]]=(-1{)}^{\left|z\right|\left|z\right|}{A}^{2}XZ{A}^2\left(y\right).\end{array}$$

Remark 5.1.4. An equivalent construction of the Jordan superalgebra [a, [a, L]] is the following: in L define a new product by x • y = ½[x, [a, y]] for any x, y ϵ L, and denote L^(a) the (nonassociative) ℤ₂-graded algebra (L, •), with L₀^(a) = L₀ and L₁^(a) = L₁. If we define Ker_l(a) := {x ϵ L| [a, [a, x]] = 0}, then Ker_l(a) is the kernel of the ℤ₂-graded algebra homomorphism φ : L^(a) → [a, [a, L]] given by φ(x) = [a, [a, x]], so L^(a)/Ker_L(a) and [a, [a, L]] are isomorphic as Jordan superalgebras.

5.2. Jordan superalgebras at odd homogeneous ad-nilpotent elements

Now we turn to odd ad-nilpotent elements. Notice that for every homogeneous element a ϵ L₁ we have ${\mathrm{ad}}_{[a,a]}=2{\mathrm{ad}}_a^2$. When dealing with ad-nilpotent elements of L₁ we will require ${\mathrm{ad}}_a^{4}L=0$. In this case the element b = [a, a] ϵ L₀ verifies ${\mathrm{ad}}_b^2={\mathrm{ad}}_{[a,a]}^2=4{\mathrm{ad}}_a^4=0.$

Remark 5.2.1. Given such an element a ϵ L₁ with ${\mathrm{ad}}_a^4=0$, if we consider the Φ -module [a, [a, L]] and we define the bilinear product as in 5.1.1 ([a, [a, x]] • [a, [a, y]] = ½[a, [a, [x, [a,y]]] for every x, y ϵ L) then [a, [a, L]] is ℤ₂-graded with [a, [a, L]]₀ = [a, [a, L₁]] and [a, [a, L₁]] = [a, [a, L₀]]. The parity of the homogeneous elements of [a, [a, L]] changes and |[a, [a, x]] | = |x| + 1 for any homogeneous element x ϵ L₀ U L₁. Moreover,

$$\overline{x}\cdot \overline{y}=-(-1{)}^{\left|\overline{x}\right|\left|\overline{y}\right|}\overline{y}\cdot \overline{x}$$

for homogeneous x̄ = [a, [a,x]], ȳ = [a, [a,y]] ϵ [a, [a, L]], i.e., ([a, [a, L]], •) is super-anticommutative. To avoid this situation and get a Jordan superstructure, we define a trilinear product on [a, [a, L].

5.2.2. For an element a ϵ L¹ with ${\mathrm{ad}}_a^4=0$, we consider the trilinear map { , , } on [a, [a, L]] defined by

$$\{\overline{x},\overline{y},\overline{z}\}:=\frac{1}{4}\left[\right[a,[a,x]],[y,[a,[a,z\left]\right]\left]\right]=\frac{1}{4}{A}^{2}XY{A}^2\left(z\right)$$

for every homogeneous x̄ = [a, [a,x]], ȳ = [a, [a,y]] and z̄ = [a, [a,z]] ϵ [a, [a,L]] (notice that [[a, [a,y]], [y [a, [a,z]]] = ¼[[[a, [,x],[y, [[a,a], z]]] = A²XYA²(z) because ad_[a,a]ad_[y]ad_[a,a] = 0). The Φ -module [a, [a, L]] is ℤ₂-graded with with respect to this trilinear product and [a, [a, L]]ⁱ = [a, [a, Lⁱ]], i ϵ {1,1}

We have that

$$\left[\right[a,[a,x]],[y,[a,[a,z\left]\right]\left]\right]=\left[\right[[a,[a,x\left]\right],y],[a,[a,z]\left]\right]$$

for every x, y, z ϵ L₀ ∪ L₁ because [[a, [a,x]], [a, [a,z]]] = ¼[[[a,a],x], [[a,a],z]] = 0 since [a, a] is an absolute zero divisor. This implies that the triple product is supersymmetric in the outer variables:

$$\{\overline{x},\overline{y},\overline{z}\}=(-1{)}^{\left|x\right|\left|y\right|+\left|x\right|\left|z\right|+|y\Vert |z|}\{\overline{z},\overline{y},\overline{x}\}. (5.2.2)$$

Moreover,

$$\{\overline{x},\overline{y},\overline{z}\}=(-1{)}^{\left|y\right|z\mid }\{\overline{x},\overline{z},\overline{y}\}. (5.2.3)$$

because $4\{\overline{x},\overline{y},\overline{z}\}=\left[\right[a,[a,x]],[y,[a,[a,z\left]\right]\left]\right]=(-1{)}^{\left|y\right|\left|z\right|+1+1}[[a,[a,x\left]\right],\left[\right[a,[a,z]],y]]=$ $\left.\left.\left.=\frac{1}{4}(-1{)}^{\left|s\right|\left|z\right|+1}[\mid [a,a],x],[[\mid a,a],z],y\right]\right]\left.=\frac{1}{4}(-1{)}^{\left|z\right|\left|z\right|}\right|[a,a],[x,[z,\left[\right[a,a],y]\left]\right]\right]=$ $=(-1{)}^{\left|y\right|z\mid }{A}^{2}XZ{A}^2(y)=4(-1{)}^{\left|y\right|\left|z\right|}\{\overline{x},\overline{z},\overline{y}\}.$. From equations (5.2.2) and (5.2.3) we get that the triple product defined in (5.2.1) is supercommutative on its three variables.

Lemma 5.2.3. For a homogeneous elementa ϵ L₁ with ${\mathrm{ad}}_a^4=0$, the trilinear map given in (5.2.1) satisfies

$$\begin{array}{r}\{\overline{x},\overline{y},\{\overline{z},\overline{u},\overline{v}\left\}\right\}=\left\{\right\{\overline{x},\overline{y},\overline{z}\},\overline{u},\overline{v}\}+(-1{)}^{\left|x\right|\left|y\right|+\left|x\right|\left|z\right|+\left|y\right|\left|z\right|}\{\overline{z},\{\overline{y},\overline{x},\overline{u}\},\overline{v}\}\\ +(-1{)}^{\left|x\right|\left|z\right|+\left|x\right|\left|u\right|+\left|y\right|\left|z\right|+\left|y\right|\left|u\right|}\{\overline{z},\overline{u},\{\overline{x},\overline{y},\overline{v}\}\}\end{array}$$

for every x, y, z, u, v ϵ L₀ ∪ L₁

Proof. For every x, y, z, u, v ϵ L₀ ∪ L₁,

$$\begin{array}{l}8\{\overline{x},\overline{y},\{\overline{z},\overline{u},\overline{v}\left\}\right\}=\left[\right[[a,[a,x\left]\right],y],[[\mid a,[a,z\left]\right],u],[a,[a,v]\left]\right]]=\\ =\left[\right[[\mid a,[a,x\left]\right],y],[[a,[a,z\left]\right],u\left]\right],[a,[a,v\left]\right]]+\\ \left.+(-1{)}^{\left(\right|x|+|y\left|\right)\left(\right|z|+|u\left|\right)}[\left[\right[a,[a,z]],u],\left[\right[\mid a,[a,x]],y],[a,[a,v\left]\right]]\right]=\\ =\left[\right[\left[\right[\mid a,[a,x]],y],[a,[a,z\left]\right]],u],[a,[a,v\left]\right]]+\\ \left.+(-1{)}^{\left(\right|\left|x\right|+\left|y\right|)\mid z]}[[\mid a,[a,z\left]\right],\left[\right[[a,[a,x\left]\right],y],u]],[a,[a,v]]\right]+\\ \left.+(-1{)}^{\left(\right|x|+|y\left|\right)\left(\right|z|+|u\left|\right)}[\left[\right[a,[a,z]],u],\left[\right[a,[a,x]],y],[a,[a,v\left]\right]]\right]=\end{array}$$

$$\begin{array}{l}\left.\left.=8\left\{\right\{\overline{x},\overline{y},\overline{z}\},\overline{u},\overline{v}\}+(-1{)}^{\left(\right|x|+|y\left|\right)\left|z\right|}[[\mid a,[a,z\left]\right],[\mid [a,[a,x]],y],u]\right],[a,[a,v\left]\right]\right]+\\ +(-1{)}^{\left|x\right|z|+|x\left|\right|u|+|y\left|z\right|+\left|z\right|\left|\right|u\mid }\mid 8\{\overline{z},\overline{u},\{\overline{x},\overline{y},\overline{v}\}\}\end{array}$$

Let us see that (-1)^(|x|+|y|)|z|[[[a, [a,z]], [[[a, [a,x]], y],u]],[a,[a,v]]] coincides with the second term on the right side of equality (*): from the definition of the triple product,

$$\left[\right[[a,[a,z\left]\right],\left[\right[[a,[a,x\left]\right],y],u]],[a,[a,v]\left]\right]=4\{\overline{z},\overline{\left[\right[a,[a,x]],y],u]},\overline{v}\}$$

and

$$\begin{array}{l}\left.\overline{\left.\left.\prod [a,[a,x\left]\right],y\right],u\right]}=[a,[a,\left[\right[[a,[a,x\left]\right],y],u]\left]\right]=(-1{)}^{1+\left|y\right|}[[a,[a,x\left]\right],[a,y]],[a,u]\right]+\\ +(-1{)}^{\left|y\right|}[[a,[a,x],[a,y\left]\right],[a,u]]+[\left[\right[a,[a,x]],y],[a,[a,u\left]\right]]=\\ =4\{\overline{x},\overline{y},\overline{u}\}=(-1{)}^{\left|x\right|\left|y\right|}4\{\overline{y},\overline{x},\overline{u}\}\end{array}$$

hence

$$\begin{array}{r}\left.\left.\left.(-1{)}^{\left(\right|x|+|y\left|\right)\left|z\right|}\mid [\mid a,[a,z]],[[\mid a,[a,x\left]\right],y],u\right]\right],[a,[a,v\left]\right]\right]=\\ =(-1{)}^{\left|x\right|\left|y\right|+\left|x\right|\left|z\right|+\left|y\right|\left|z\right|z\mid }8\{\overline{z},\{\overline{y},\overline{x},\overline{u}\},\overline{v}\}\end{array}$$

and we have shown (*).

5.2.4. A pair of ℤ₂-graded Φ -modules V = (V⁺, V⁻) is a (linear) Jordan superpair if there exist two trilinear maps { , , }^σ : V^σ × V ^−σ × V^σ × V^σ, σ = ±, both supersymmetric in the outer variables, and that satisfy (JSP15):

$$\begin{array}{r}{\left\{a,b,\{c,d,e{\}}^{\sigma }\right\}}^{\sigma }={\left\{\{a,b,c{\}}^{\sigma },d,e\right\}}^{\sigma }-(-1{)}^{\left|a\right|\left|b\right|+\left|a\right|\left|c\right|+\left|b\right|\mid d}{\left\{c,\{b,a,d{\}}^{-\sigma },e\right\}}^{\sigma }\\ +(-1{)}^{\left|a\right|\left|c\right|+\left|a\right|\left|d\right|+\left|b\right|c|+|b\left|\right|d\mid }{\left\{c,d,\{a,b,e{\}}^{\sigma }\right\}}^{\sigma },\sigma =\pm \end{array}$$

for homogeneous a, c, e ϵ V^σ and homogeneous b, d ϵ V^−σ, σ = ±.

We have just shown that when a ϵ L₁ has ad^4a = 0, [a, [a, L]] with the trilinear map {, ,} given in (5.2.1) is a (1,1)-Jordan supertriple in the sense of [47, §3], which are a particular case (∊, δ)-Freudenthal-Kantor supertriple systems, ∊ = ±1, ± = ±1 [47, §3]. We say that a ℤ₂-graded Φ -module M = M₀ + M₁ with a graded triple product { , , } : M × M × M → M is a (1, 1)-Jordan supertriple if

• {a,b,c} = (-1)^{|a||b|+|a||c|+|b||c|}{c,b,a} and

• {a, b, {c, d, e}} = {{a, b, c}, d, e} + (-1)^{|a||b|+|a||c|+|b|c|}{c, {b, a, d}, e} + (-1)^{|a||c|+|a||d|+|b|c|+|b|d|} {c,d, a, b, e,}}

for homogeneous elements a,b,c,d,e ϵ M. The second identity resembles (JSP15) but there is a change of sign in the second summand of its right side. Notice that every (1, 1)-Jordan supertriple M with triple product { , , } gives rise to a Jordan superpair V = (V⁺V⁻) = (M, M) with products {a, b, c}⁺ := {a, b, c} and {b, c, d}⁻ := -{b, c, d} for every a, c ϵ V⁺ and b, d ϵ V⁻. In our case we have shown that if we double [a, [a, L] and twist one of the triple products we have that ([a, [a, L], [a, [a, L]) is a Jordan superpair.

5.2.5. Another Jordan structure can be defined from an ad-nilpotent element a ϵ L₁: suppose that a ϵ L₁ has ${\mathrm{ad}}_a^6=0$. Then b = [a, a] ϵ L₀ is a Jordan element $({\mathrm{ad}}_b^3={\left(2{\mathrm{ad}}_a^2\right)}^3=0)$, and we can define a Jordan superalgebra on the Φ-module $[b,[b,L]]={\mathrm{ad}}_a^{4}L$ as in 5.1.2. The product is now given by

[b,|[b,x]]•[b,[b,y]] = ½[b,[b,[x,[b,y]]]]

or, equivalently,

$${\mathrm{ad}}_a^{4}x·{\mathrm{ad}}_a^{4}y=2{\mathrm{ad}}_a^4[x,{\mathrm{ad}}_a^{2}y].$$

5.3. Subquotients associated to abelian inner ideals of Lie superalgebras

5.3.1. Let L = L₀ + L₁ be a Lie superalgebra. We say that B = B₀ + B₁ ⊂ L is an inner ideal of L if [B, [B, L]] ⊂ B, and B is abelian if [B, B] = 0. Inner ideals can be easily produced from homogeneous ad-nilpotent elements.

Example 5.3.2. Let L = L₀ + L₁ a Lie superalgebra and let a ϵ L₀ with ${\mathrm{ad}}_a^3=0$ or a ϵ L₁ with ${\mathrm{ad}}_a^4=0$. Then

[a] := [a,[a,L]] (a) :=Φa + [a,[a,L]]

are inner ideals of L. Moreover, [a] is an abelian inner ideal.

Conversely, given an abelian inner ideal B = B₀ + B₁, any homogeneous b ϵ B₀ is a Jordan element and gives rise to the inner ideals [b] and (b) contained in B. If b ϵ B₁ then 0 = [b, b] implies 0 = ${\mathrm{ad}}_2^{\mathrm{b}}=0$ and (b) = Φb.

Proposition 5.3.3. Let L be a Lie superalgebra and B an abelian inner ideal of L. Let us consider KerB := {x ϵ L | [B, [B,x]] = 0}. Then (B,L/KerB) is a Jordan superpair with products:

$$\begin{array}{r}\{a,\overline{x},b\}:=[a,[x,b\left]\right]=\left[\right[a,x],b]\\ \{\overline{x},a,\overline{y}\}:=\overline{[x,[a,y\left]\right]}=\overline{[\left[\right[x,a],y]}\end{array}$$

for a, b ϵ B and x,y ϵ L (here x̄, ȳ, $\overline{[x,[a,y\left]\right]}$ and $\overline{\left[\right(x,a],y]}$ denote equivalence classes in the quotient L/KerB). This Jordan superpair is called the subquotient of L associated to B.

Proof. First notice that [a, [x, b]] = [[a, x], b] and $\overline{[x,[a,y\left]\right]}$ = $\overline{\left[\right(x,a],y]}$ for every a,b ϵ B and every x, y ϵ L because B is abelian and the definition of KerB.

The products are well defined: clearly {a, 0, b} = 0, and if we take homogeneous x̄, ȳ ϵ L/KerB with x̄ = 0̄ or ȳ = 0̄ then for homogeneous a, b, c ϵ L we have that

$$\begin{array}{c}[b,[c,[x,[a,y\left]\right]\left]\right]=[b,[[c,x],[a,y]\left]\right]+(-1{)}^{\left|c\right|\left|x\right|}[b,[x,[c,[a,y]\left]\right]]=\\ =(-1{)}^{\left|c\right|\left|x\right|}[[b,x],[c,[a,y\left]\right]]+(-1{)}^{\left|c\right|\left|\right|x|+|b\left|\right|x\mid }[x,[b,[c,[a,y\left]\right]\left]\right]=0\end{array}$$

Let us see that the triple products are supersymmetric in the outer variables:

$$\begin{array}{r}\{a,\overline{x},b\}\left.=[a,[x,b\left]\right]=(-1{)}^{1+\left|x\right|\left|b\right|}[a,[b,x]]=(-1{)}^{\left|b\right|\left|x\right|+\left|a\right|\left(\right|b|+|x\left|\right)}\left[\right[b,x],a]\right]=\\ =(-1{)}^{\left|b\right|\left|x\right|+\left|a\right|\left|b\right|+\left|a\right|\left|x\right|}[b,[x,a]]=(-1{)}^{\left|b\right|\left|x\right|+\left|a\right|\left|b\right|+\left|a\right|\left|x\right|}\{b,\overline{x},a\}\end{array}$$

$$\begin{array}{c}\{\overline{x},a,\overline{y}\}=\overline{[x,[a,y\left]\right]}=\overline{[\mid x,a],y]}+(-1{)}^{\left|x\right|\left|a\right|}\overline{[a,[x,y]}=\\ \left.\left.=(-1{)}^{\left|x\right|\left|y\right|+\left|x\right|\left|\right|a|+|y\left|\right|a\mid }\right]y,[a,x]\right]=(-1{)}^{\left|x\right|\left|y\right|+\left|x\right|\left|a\right|+\left|y\right|\left|a\right|}\{y,a,\overline{x}\}\end{array}$$

Let us prove (JSP15). For homogeneous a, b, c ϵ B and homogeneous x, y, z ϵ L,

$$\text{ - }\begin{array}{l}•\{a,\overline{x},\{b,\overline{y},c\left\}\right\}=\left[\right[a,x],[[b,y],c\left]\right]=\\ \left.=\left[\right[[\mid a,x],b],y],c]+(-1{)}^{\left(\right|a|+|x\left|\right)b}\left[\right[b,[\mid a,x],y\left]\right],c\right]+(-1{)}^{\left(\right|b|+|y\left|\right)\left(\right|a|+|x\left|\right)}[[b,y],\left[\right[a,x],c]]\\ =\left\{\right\{a,\overline{x},b\},\overline{y},c\}-(-1{)}^{\left|a\right|\left|\right|b|+|x\left|\right|b|+|a\left|\right|x\mid }\{b,\{\overline{x},a,\overline{y}\},c\}+\\ +(-1{)}^{\left(\right|b|+|y\left|\right)\left(\right|a|+|x\left|\right)}\{b,\overline{y},\{a,\overline{x},c\}\}.\end{array}$$

$$\text{ - }\begin{array}{c}•\{\overline{x},a,\{\overline{y},b,\overline{z}\left\}\right\}=\overline{[\mid x,a],\left[\right[y,b],z]]}=\\ =\overline{\left[\right[[x,a],y],b],z]}+(-1{)}^{\left|y\right|\left(\right|x|+|a\left|\right]}\overline{\left[\right[[y,[\mid x,a],b]],z]}+(-1{)}^{\left(\right|y|+|b\left|\right)(x+a)}\overline{[\mid y,b],[\mid x,a],z\left]\right]}\\ =\left\{\right\{\overline{x},a,\overline{y}\},b,\overline{z}\}-(-1{)}^{\left|y\right|\left|x\right|+\left|y\right|\left|a\right|+\left|a\right|\left|x\right|}\{\overline{y},\{a,\overline{x},b\},\overline{z}\}+\\ +(-1{)}^{\left(\right|y|+|b\left|\right)\left(\right|x|+|a\left|\right)}\{\overline{y},b,\{\overline{x},a,\overline{z}\}\}\end{array}$$

Therefore, (B, L/KerB) is a Jordan superpair.

Remark 5.3.4. Let a ϵ L₀ be a Jordan element or a ϵ L₁ with ${\mathrm{ad}}_a^4=0$. Then B = [a] = [a, [a, L]] is an abelian inner ideal and we can build the subquotient ([a], L/Ker[a]). In this particular case, for homogeneous x, y, z ϵ L the triple product

$$\begin{array}{c}\left\{\right[a,[a,x]],y+\mathrm{Ker}[a],[a,[a,z]\left]\right\}=\left[\right[a,[a,x]],[y,[a,[a,z\left]\right]\left]\right]=\\ =(-1{)}^{\left|y\right|\left|z\right|+1+\left|a\right|}{A}^{2}XZ{A}^2(y)\end{array}$$

coincides, up to a scalar, with the triple product we have already defined in [x], see Remark 5.1.3 when [a] is even and 5.2.2 when a is odd. In the following result we are going to prove that the Jordan superpair structures defined in this section and in the previous ones coincide.

Corollary 5.3.5. Let L be a Lie superalgebra, takea ϵ L₀ with ${\mathrm{ad}}_a^3=0$ ora ϵ L₁ with ${\mathrm{ad}}_a^4=0$, and let us consider the subquotient associated to the abelian inner ideal [a].

(a) Whena ϵ L₀, if we consider the Jordan superpair structure induced on ([a, [a, L]], [a, [a, L]]) by Remark 5.1.3, then the pair of maps

$$({\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2):\left(\right[a,[a,L]],[a,[a,L]\left]\right)\to \left(\right[a],L/\mathrm{Ker}[a\left]\right)$$

given by

$${\mathrm{\Psi }}_1=-\frac{1}{2}\mathrm{id}\text{and}{\mathrm{\Psi }}_2\left(\right[a,[a,x]\left]\right)=\frac{1}{2}x+\mathrm{Ker}\left[a\right]$$

is an isomorphism of Jordan superpairs.

(b) When a ϵ L₁, if we consider the Jordan superpair structure defined on ([a, [a, L]], [a, [a, L]]) by5.2.3, then the pair of maps

$$({\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2):\left(\right[a,[a,L]],[a,[a,L]\left]\right)\to \left(\right[a],L/\mathrm{Ker}[a\left]\right)$$

given by

$${\mathrm{\Psi }}_1=\frac{1}{2}\mathrm{id}\text{and}{\mathrm{\Psi }}_2\left(\right[a,[a,x]\left]\right)=\frac{1}{2}x+\mathrm{Ker}\left[a\right]$$

is an isomorphism of Jordan superpairs.

Proof. In both cases, the pair of maps given by

$$\begin{array}{r}{\mathrm{\Psi }}_1\left(\right[a,[a,x]\left]\right)=(-1{)}^{\left|a\right|+1}\frac{1}{2}[a,[a,x]]\in [a],\text{and}\\ {\mathrm{\Psi }}_2\left(\right[a,[a,x]\left]\right)=\frac{1}{2}x+\mathrm{Ker}\left[a\right]\in L/\mathrm{Ker}\left[a\right],\end{array}$$

for every x ϵ L, are well defined (if [a, [a, x]] = [a, [a,y]], then [a, [a,x–y]] = 0 implies x–y ϵ Ker[a]). They are clearly bijective. Let us see that they are Jordan superpair homomorphisms.

(a) Suppose that a ϵ L₀ and take homogeneous x, y, z ϵ L.

$$\begin{array}{c}{•}_{\mathrm{\Psi 1}}\left(\right\{[a,[a,x\left]\right],[a,[a,y\left]\right],[a,[a,z\left]\right]\left\}\right)={\mathrm{\Psi }}_1\left(\right(-1{)}^{|y\Vert z|}\frac{1}{4}{A}^{2}XZ{A}^2\left(y\right))=\\ =(-1{)}^{|y\Vert z|+1}\frac{1}{8}{A}^{2}XZ{A}^2(y)=\{-\frac{1}{2}[a,[a,x]],\frac{1}{2}y+\mathrm{Ker}[a],-\frac{1}{2}[a,[a,z]\left]\right\}=\\ =\left\{{\mathrm{\Psi }}_1\right([a,[a,x\left]\right]),{\mathrm{\Psi }}_2([a,[a,y\left]\right]),{\mathrm{\Psi }}_1([a,[a,z\left]\right]\left)\right\}.\end{array}$$

$$\begin{array}{c}{•}_{\mathrm{\Psi 2}}\left(\right\{[a,[a,x\left]\right],[a,[a,y\left]\right],[a,[a,z\left]\right]\left\}\right)={\mathrm{\Psi }}_2\left(\right(-1{)}^{\left|y\right|\left|z\right|}\frac{1}{4}{A}^{2}XZ{A}^2\left(y\right))=\\ =(-1{)}^{\left|y\right|\left|z\right|}\frac{1}{8}XZ{A}^2(y)+\mathrm{Ker}[a]=-\frac{1}{8}[x,\left[\right[a,[a,y]],z]]+\mathrm{Ker}[a]=\\ =\{\frac{1}{2}x+\mathrm{Ker}[a],-\frac{1}{2}[a,[a,y]],\frac{1}{2}z+\mathrm{Ker}[a\left]\right\}=\\ =\left\{{\mathrm{\Psi }}_2\right([a,[a,x\left]\right]),{\mathrm{\Psi }}_1([a,[a,y\left]\right]),{\mathrm{\Psi }}_2([a,[a,z\left]\right]\left)\right\}.\end{array}.$$

(b) Suppose that a ∊ L₁ and take homogeneous x, y, z ∊ L.

$$\begin{array}{c}{•}_{\mathrm{\Psi 1}}\left(\right\{[a,[a,x\left]\right],[a,[a,y\left]\right],[a,[a,z\left]\right]\left\}\right)={\mathrm{\Psi }}_1\left(\right(-1{)}^{\left|y\right|\left|z\right|}\frac{1}{4}{A}^{2}XZ{A}^2\left(y\right))=\\ =(-1{)}^{\left|y\right|\left|z\right|}\frac{1}{8}{A}^{2}XZ{A}^2(y)=\left\{\frac{1}{2}\right[a,[a,x]],\frac{1}{2}y+\mathrm{Ker}[a],\frac{1}{2}[a,[a,z]\left]\right\}=\\ =\left\{{\mathrm{\Psi }}_1\right([a,[a,x\left]\right]),{\mathrm{\Psi }}_2([a,[a,y\left]\right]),{\mathrm{\Psi }}_1([a,[a,z\left]\right]\left)\right\}.\end{array}.$$

$$\begin{array}{c}{•}_{\mathrm{\Psi 2}}\left(\right\{[a,[a,x\left]\right],[a,[a,y\left]\right],[a,[a,z\left]\right]\left\}\right)={\mathrm{\Psi }}_2(-\frac{1}{4}[\left[\right[a,[a,x]],y],[a,[a,z\left]\right]\left]\right)=\\ =-\frac{1}{4}{\mathrm{\Psi }}_2\left(\right(-1{)}^{\left|y\right|z\mid }{A}^{2}XZ{A}^2\left(y\right))=\frac{1}{8}(-1{)}^{1+\left|y\right|\left|z\right|}XZ{A}^2(y)+\mathrm{Ker}[a]=\\ =\frac{1}{8}[x,[[a,[a,y\left]\right],z\left]\right]+\mathrm{Ker}\left[a\right]=\{\frac{1}{2}x+\mathrm{Ker}[a],\frac{1}{2}[a,[a,y]],\frac{1}{2}z+\mathrm{Ker}[a\left]\right\}=\\ =\left\{{\mathrm{\Psi }}_2\right([a,[a,x\left]\right]),{\mathrm{\Psi }}_1([a,[a,y\left]\right]),{\mathrm{\Psi }}_2([a,[a,z\left]\right]\left)\right\}.\end{array}.$$

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