Asian Journal of Algebra

Year: 2017 | Volume: 10 | Issue: 1 | Page No.: 1-9
DOI: 10.3923/aja.2017.1.9

Flexible Lie-admissible Superalgebras of Vector Type

G. Lakshmi Devi and K. Jayalakshmi

Abstract: Background: First examples of simple nonassociative superalgebras were constructed by Shestakov in (1991 and 1992). Since then many researchers showed interest towards the study of superalgebras and superalgebras of vector type. Materials and Methods: Multiplication in M is uniquely defined by a fixed finite set of derivations and by elements of A. The types of derivations used in this article to obtain the results are the near derivation δ_x,y : a ↦ (a, x, y) the derivation D : a ↦ (x, a, x) and the derivation D_ij : a ↦ (x_i ,a, x_j) Results: The flexible Lie-admissible superalgebra F_FLSA[φ; x] over a 2, 3-torsion free field Φ on one odd generator e is isomorphic to the twisted superalgebra B₀ (Φ[Γ], D, γ₀) with the free generator . In a 2, 3-torsion free flexible Lie-admissible superalgebras of vector type F, the even part A is differentiably simple, associative and commutative algebra and the odd part M is a finitely generated associative and commutative A-bimodule. Conclusion: A connection between the integral domains, the finitely generated projective modules over them, the derivations of an integral domain and the flexible Lie-admissible superalgebras of vector type has been established. If A is an integral domain and M = Ax₁+…+Ax_n be a finitely generated projective A-module of rank 1, then F (A, Δ, Γ) is a flexible Lie-admissible superalgebra with even part A and odd part M provided that the mapping M = Ax_i+…+Ax_n is a nonzero derivation of A into the A-module (M⊗_A M)*, Δ = {D_ij |i, j = 1,…, n} is a set of derivations of A where D_ij (a) = ā (x⊗x_j).

Keywords: Superalgebra, super algebra of vector type, flexible Lie-admissible superalgebra, differentiably simple algebra and grassmann envelope

INTRODUCTION

Okubo and Myung¹ gave the classification of simple flexible Lie-admissible algebra A such that A‾ is reductive and the Levi-factor of A‾ is simple. Later using the results of Okubo and Myung¹ and Benkart and Osborn² determined flexible Lie-admissible algebras A by module approach for which the radical A‾ is abelian. Kamiya and Okubo³ constructed some class of Lie superalgebras from non-super flexible Lie-admissible algebras. They concentrated especially on the case of any associative algebra giving rise to Lie superalgebras.

In Shestakov⁴, the concept of a (-1, 1)-superalgebra of vector type was introduced and it was proved that a Jordan superalgebra of vector type may be obtained as the (super) symmetrized algebra A⁽⁺⁾ of a (-1, 1)-superalgebra of vector type A. Furthermore, the (-1, 1)-superalgebras A_VF: = A(V, Φ, τ, λ) and the Jordan superalgebras J_VF: = J(V, Φ, τ, λ) of vector fields on a line associated with an additive homomorphism τ: V→Φ of abelian groups and partial map λ: V→V were introduced in Shestakov⁴. It was proved that if V ≠ 0 and the map τ is injective, then the superalgebras A_VF and J_VF are prime. These superalgebras were then used to construct prime degenerate (-1, 1) and Jordan algebras.

Simple (-1, 1)-superalgebras were obtained from the associative-commutative algebras which are differentiably simple with respect to a derivation. The 2, 3-torsion free simple (-1, 1)-superalgebras were described in Shestakov⁵. The (-1, 1)-superalgebras obtained in Shestakov⁵ are also called the (-1, 1)-superalgebras of vector type. Zhelyabin⁶ and Zhelyabin and Shestakov⁷ described the product in M by some fixed finite sets of derivations and the elements of A for (-1, 1)-superalgebras.

In terms of the Lie algebra of derivations of an integral domain, some necessary and sufficient conditions were found by Zhelyabin⁸, which allowed him to construct the Jordan superalgebras of vector type whose even part is the initial integral domain. Zhelyabin⁹ studied analogous questions for the (-1, 1)-superalgebras. In particular, he gave some new examples of simple and prime (-1, 1)-superalgebras of vector type whose odd part is generated as a module by two elements in the case of simple superalgebras and by an arbitrary number of elements in the case of prime superalgebras. Some properties of the universal envelopings of the simple Jordan superalgebras of vector type were described as well. In particular, he proved that the even part of the universal enveloping contains two central orthogonal idempotents such that their sum is equal to unity.

Zhelyabin⁹ described simple nonassociative 2, 3-torsion free (-1, 1)-superalgebras. If B has a positive characteristic, which is the only possibility in the finite dimensional case, then the even part A is local and B is isomorphic to B (Γ, D, γ). He posed an open question that whether M is always generated by one element as an A-bimodule. He also noticed that if there exists a simple (-1, 1)-superalgebra B which does not satisfy this condition and is therefore not isomorphic to B (Γ, D, γ) the attached superalgebra B⁺ will infact exemplify a new simple Jordan superalgebra.

In the present article using the constraints of Shestakov⁵ and Zhelyabin⁹, a description of a simple 2, 3-torsion free flexible Lie-admissible superalgebra is obtained. It is proved that the flexible Lie-admissible superalgebra F_FLSA [φ; x] over a 2, 3-torsion free field Φ on one odd generator e is isomorphic to the twisted superalgebra B₀ (Φ [Γ], D, γ₀) with free generator . The even part A in such a superalgebra F is necessarily an associative and commutative algebra and the odd part M is finitely generated associative and commutative A-bimodule over A, which is a projective module of rank 1. Multiplication in M is defined by a fixed set of derivations and by elements of A. If the bimodule M is one-generated, i.e., M = Am for a certain m∈M, then F is isomorphic to a twisted superalgebra of vector type B (Γ, D, γ).

MATERIALS AND METHODS

A superalgebra R = R₀⊕R₁ is a Z₂-graded algebra. A typical example of a superalgebra is the associative Grassmann algebra G = G₀⊕G₁ on a countable set of generators with a natural Z₂-grading. If is an arbitrary homogeneous variety of algebras then R = R₀⊕R₁ is called an if its Grassmann envelope G (R) = G₀⊗R₀⊕G₁⊗R₁ belongs to . In particular R = R₀⊕R₁ is a flexible Lie-admissible superalgebra if and only if it satisfies the (super) identities:

where, x, y, z∈R₀∪R₁ and |r|∈{0, 1} is the parity index of an homogeneous element r: |r| = i for r∈R_i; [x, y]_s = xy-(-1)^|x||y| yx is the supercommutator of homogeneous elements x, y.

Let flexible Lie-admissible superalgebra satisfy the following two identities:

In a flexible Lie-admissible algebra A, the center defined by U (A) = {u∈A|[u, x] = 0, for all x∈A} is the commutative center. Throughout this study A is an associative bimodule of U (A).

Any algebra satisfies Teichmuller and semi Jacobi identities:

For r, s∈A and x, y∈M, Eq. 1 and (*) yield:

One can easily check that Eq. 1 and (*) gives rise to:

Anderson and Outcult¹⁰ have derived an important identity for r, s, x, y∈A∪M, (r, [x, y], s) = 0 for antiflexible rings. Applying the same argument for flexible Lie-admissible superalgebras, one can obtain:

One can easily check that flexible ring with (*) satisfies:

Let u∈U(A), a, x, y, z∈A. The following identities are valid in flexible Lie-admissible algebras. From Eq. 4 and 5:

From Eq. 5 and 1:

From Eq. 5:

From Eq. 6 and 8:

RESULTS

Free flexible Lie-admissible superalgebra F_FLSA [φ; x]: Let Φ[Γ] be the algebra of polynomials on a countable set of variables Γ = {γ₀,…, γ_n,…} and let D be the derivation of Φ[Γ] defined by the condition D(γ_i) = γ_i+1, i = 0, 1,… consider the twisted superalgebra of vector type B(Φ[Γ], D, γ₀). Let Φ₀[Γ] be a subalgebra of polynomials without constant terms, then the subspace is a subsuperalgebra of B(Φ[Γ], D, γ₀) and is denoted by B₀(Φ[Γ], D, γ₀). It can be checked that it is generated by the odd element .

Preliminary lemmas
Lemma 1: e²∈U (F), where, U (F) is the commutative center.

Proof: Identity Eq. 2 implies:

3(e, e, e) = 0, (e², e, e)-(e, e², e)+(e, e, e²) = 0

that is:

Identity Eq. 3 in the form of generators are:

(e³, e, e) = -(e², e², e)+(e², e, e²) = e²(e, e, e)+(e², e, e)e,
(e², e², e)-(e, e³, e)+(e, e, e³) = e(e, e², e)+(e, e, e²)e,
(e², e, e²)-(e, e², e²)+(e, e, e³) = e(e, e, e²)+(e, e, e)e²

which from Eq. 5 implies:

But the Jacobi identity Eq. 2 implies:

Hence by Eq. 15 one can see that 2(e², e², e)+2(e², e, e)e+ (e, e³, e) = 0 and eventually using Eq. 17, 16, 14, 5 and 18, one obtains:

In particular:

[e³, e²] = 0 = (e, e², e²)

Let p(x) be a monomial of degree n>3 in x. And suppose that [q(x), x²] = 0 for every monomial q(x) of degree less than n and flexible Lie-admissible superalgebra A satisfies the identity [[r, s], D(A)] = 0 where, D(A) is the associator ideal of A. Then in F, [[x, x]_s, D(F)] = [2x², D(F)] = 0 holds. Now, [p(x), x²] = [x²(x² q(x)), x²], where, deg. q(x)≥0. Replacing x² by r and q(x) by q, applying Eq. 4, induction assumption, (*) and Eq. 5 yields:

[p(x), x²] = [r(rq), r]
= r[rq, r]+[r, r]rq+(r, rq, r)-(r, r, rq)+(r, r, rq)
= (r, r, rq) = (r, r, q)r

The above arguments show that (r, r, q) = [rq, r] = 0, which proves the lemma.

Lemma 2: Let F be a flexible Lie-admissible superalgebra. Then [[F, F]_s, F]_s = 0.

Proof: It is sufficient to show that p(x)∈U(F) for every monomial p of even degree. This can be done by using induction on degree of p. One can check that the square of an ideal is an ideal in a flexible Lie-admissible algebra satisfying (*), it may be assumed that p = qx or p = xq, where, q is a monomial of odd degree. On the same lines again assume that q = tx or q = xt for some even t∈F. By lemma 1, x²∈U(F). By theorem 13.10 in Zhevlakov et al.¹¹, x²∈U(F) implies that:

Applying this to q, we have (q∘x)∈U(F). So it suffices to prove that (tx)x and x(xt)∈U(F). By induction t∈U(F), thus tx², x²t∈U(F) and hence Eq. 8 gives:

So, (tx)x∈U(F) as (tx)x = tx²+(t, x, x).

Applying the same argument one obtains x(xt) = x²t-(x, x, t) = x²t+(t, x, x)∈U(F).

Theorem 1: A 2, 3-torsion free flexible Lie-admissible superalgebra F_FLSA[φ; x] over a field Φ on one odd generator e is isomorphic to the superalgebra B₀ (Φ[Γ], D, γ₀) with the free generator .

Proof: Let A = F₀, then F₁ = M¹x and F = A⊕M¹x, where by assumption A is an associative and commutative Φ-algebra and M¹x is a commutative and associative A-module generated by x. If then D is a derivation of A. But and since it is sufficient to prove that is sub superalgebra of F. Since A₀⊆U(F), for all r, s∈A, it can be seen that r⋅sx = sx⋅r = (rs)x. Now the product of the elements from M¹x can be calculated:

Hence, is a subsuperalgebra of F and , A₀ = A = F₀. Also Eq. 21 shows that the superalgebra F is a homomorphic image of the superalgebra B₀ (M¹, D, x²) under the homomorphism Consider the homomorphism Obviously, this is a homomorphism of differential algebras and it can be extended to a homomorphism of superalgebras : B₀(Φ[Γ], D, γ₀)→B₀(M¹, D, x²). The composition map π∘ is onto map from B₀ (Φ[Γ], D, γ₀) to F. As F is a free superalgebra generated by x, this map is an isomorphism.

Lemma 3: Let F = A+M be a flexible Lie-admissible superalgebra. Then for any x, y∈M the mapping is a near derivation of A.

Proof: Equation 5 and 7 yield:

δ_{x, y}(rs) = (rs, x, y) = r(s, x, y)+(r, x, y)s+(r, x∘y, s)
= r⋅δ_{x, y}(s)+δ_{x, y}(r)⋅s+(r(x∘y))s-r((s∘y)s)
= r⋅(δ_{x, y}-L_x∘y)(s)+(δ_{x, y}+R_x∘y)(r)⋅s

Hence, the lemma is proved.

Lemma 4: If F is simple then A is Δ-simple, that is A does not contain proper Δ-invariant ideals.

Proof: Assume that I ≠ 0 is an ideal of A such that (I, M, M)⊆I. Since F is simple, = F where, is the ideal generated by I in F. Hence, M = IM+MI. The M+M² is a graded ideal of B; therefore M² = A. To prove that M² = (IM+MI)²⊆I, from which I = A consider:

IM⋅M⊆I⋅MM+(I, M, M)⊆I

Let x, y∈M and i, j∈I; xi∘jy ≡ [x, i]∘jy = -jy∘[x, i] ≡ 0 by (**). The super-semi Jacobi identity Eq. 4 along with (**) gives:

Thus, M²⊆I and I = A.

Remark 1: Suppose that F = A+M is a flexible Lie-admissible superalgebra with its even part A not containing non zero Δ-invariant nil ideals. Then (A, B, A) = [A, B] = 0.

Theorem 2: Let F = A+M be a flexible Lie-admissible superalgebra. Then (A, B, A) = [A, B] = 0 there exists x₁, x₂,…, x_n∈M such that M = Ax₁+Ax₂+…+Ax_n and the product in M is defined by rx_i⋅sx_j = γ_ij⋅rs+2D_ij (r)s+D_ij (s)r, i, j = 1,.., n, where, γ_ij∈A, D_ij = D_ji∈DerA satisfy the conditions:

for any i, j, k = 1,…, n, r∈A.

Proof: First, to prove that A satisfies the assumptions of the above remark, by lemma 4, A does not contain proper Δ-ideals and so it is sufficient to show that A itself is not nil. Since, A satisfies the identity (**), if A were nil, it would be locally nilpotent by Eq. 4 which is impossible by Shestakov¹², lemma 5. Thus A is associative and commutative and M is an associative and commutative A-bimodule.

Next to show that (A, M, M) = A, for any x, y∈M and r∈A, (*) and (**) gives:

(xr)y = (rx)y = (r, x, y)+r(xy) = (r, x, y)+(xy)r = (r, x, y)+2x(ry)-(xr)y

which yields:

Assume (A, M, M) = 0; then (M, A, M) = 0 by Eq. 25. This yields A⊆Z₀, where Z = Z(F) is the center of F. Since F is simple, Z₀ is a field and F may be treated as a superalgebra over it. Let x, y∈M and xy ≠ 0. Then, by Eq. 1:

(xy)z = x(yz)+(zy)x-z(yx), for any z∈M

Which implies that If M = Z₀⋅x then (F, F, F)⊆Z₀⋅ (x, x, x) = Z₀⋅[x², x] = 0, a contradiction. Now, let M = Z₀⋅ x+Z₀⋅y, x² = α, y² = β, xy = 0, yx = η. Hence, α = 0, η = 0 since (x, x, y) = (y, x, x,). Similarly, (x, y, y) = (y, y, x) implies that β = 0, η = 0. Hence, α = β = γ = 0 and M² = 0, a contradiction. Thus (A, M, M) is a nonzero Δ-subspace of A. From super-linearized Eq. 6, it follows that (A, M, M) is an ideal of A and hence (A, M, M) = A.

It is known that for any x, y∈M, the mapping is a derivation of A. Thus A is differentially simple and by Posner¹³ it contains unity 1. For any derivation δ∈Δ, δ(1) = 0. Moreover, (x, y, 1) = (1, y, x) = 0 for any x, y∈M. Hence, 1∈Z(F) and so 1 is unity in F. Since 1∈(A, M, M), there exist x_i, y_i∈M and r_i∈A, i = 1,…, n, such that:

For any r∈A and x, y, z∈M, by super Eq. 6 and the associativity of the A-bimodule M, one has:

Substituting r = r_i, x = x_i, y = y_i and then summing the resulting equations over i, by Eq. 26 one can obtain:

Thus M = Ax₁+…+Ax_n and hence F = A+Ax₁+…+Ax_n. Let D_ij be the mapping By Eq. 25, is a derivation of A. For any r, s∈A, super Eq. 5 and 25 yields:

rx_i⋅sx_j = (rx_i⋅s)x_j-(rx_i, s, x_j)
                   = (rs⋅x_i)x_j-r(x_i, s, x_j)+(r, s, x_j)x_i
                    = rs⋅x_ix_j+2 (x_i, rs, x_j)-r (x_i, s, x_j)
           = rs⋅x_ix_j+2D_ij (rs)-rD_ij (s)
          = γ_ij⋅rs+2D_ij (r)s+rD_ij (s)

where, γ_ij = x_ix_j∈A.

Equalities 23 and 24 are consequences of Eq. 1 and (*).

Remark 2: If there exists x∈M such that (A, x, M) = A or A is a local algebra, then B is isomorphic to a superalgebra B (Γ, D, γ), where Γ = Γ₀ is a commutative and associative D-simple algebra with 0 ≠ D∈Der Γ and γ∈Γ.

Remark 3: If A is a local algebra with a maximal ideal J, the ideal cannot contain (A, M, M) since otherwise it would be a nonzero Δ-invariant ideal in A. Hence, (A, M, M)⊈J and there exists x∈M such that (A, x, M)⊈J. By super Eq. 5 and (*), for any a, b∈A and y∈M we have (a, x, y)b = (a, b, y)x+(a, x, by)+(a, b, xy) = (a, x, by). Consequently, (A, x, M) is an ideal in A and (A, x, M) = A.

Flexible Lie-admissible superalgebra F(A, Δ, Γ): As shown in theorem 2, given x, y∈M, the mapping, defined by the rule is a derivation. For all r, s∈A and x, y∈M:

sD_{x, y} (r) = D_{sx, y} (r) = D_{x, syy} (r), D_{x, y} (r) D_{u, v} (r) = D_{u, y} (r) D_{x, v} (r)

Since F is a flexible superalgebra, D_{x, y} = D_{y, x}. Also, D_ij = D_{xi xj} for i, j = 1,…, n.

From D_ij (r)x_k = D_ik (r)x_j one can deduce that:

D_ij (r)x_k⋅x_l = D_ik (r)x_j⋅x_l

and:

x_l⋅D_ij (r)x_k = x_l⋅D_ik (r)x_j for every r∈A

Let M* = Hom_A(M, A). Given r, define by setting Similarly the linear mapping given by is a derivation of A into the A-module (M⊗_AM)* Zhelyabin⁸.

In what follows, all rings are considered as algebras over a field of characteristic 0. Now, let A be an associative-commutative ring with unity and without zero divisors (i.e., an integral domain) and let M be a finitely generated projective A-module of rank 1. Let be a nonzero mapping. Assume that:

for all r, s∈A. By the definition of ‾, each pair x, y∈M gives a derivation Then Zhelyabin⁸:

for r∈A and x, y, z∈M.

Let x₁,…, x_n be some generators of the A-module M, i,e., M = Ax_n+…+Ax_n. Put Then D_ij = D_ji.

Lemma 5: D_ij (r) D_kl = D_kl (r) D_ij and D_ij (r) D_kl = D_kj (r) D_il for every r∈A.

Proof: Let r, s∈A. Then:

Similarly:

D_ij (r) D_kl = D_kj (r)D_il

Fix some elements γ_ij of A, where i, j = 1,…, n.

Lemma 6: Assume that:

for all i, j, k, l = 1,…, n. Then Eq. 23 holds, i.e., γ_ij x_k-γ_ik x_j = (γ_jk-γ_kj)x_i for all i, j, k, = 1,…, n.

Proof: From identities Eq. 29 and 30 it can be seen that:

(γ_kl-2γ_lk) D_ij = (γ_jl-2γ_lj) D_ik

This implies:

for every r∈A. Therefore:

By the conditions on A and M Zhelyabin⁸:

(γ_kl-2γ_lk)x_j = (γ_jl-2γ_lj)x_k

By Eq. 30 and 28:

2D_klD_ij = -γ_kl D_ij+γ_jl D_ik+2D_jl D_ik = -γ_kl D_ij+γ_jl D_ik+2D_lj D_ki
= -γ_kl D_ij+γ_jl D_ik-γ_lj D_ki+γ_ij D_kl+2D_ij D_kl

Then:

2[D_kl, D_ij] = -γ_klD_ij+(γ_jl-γ_lj)D_ik+γ_ijD_kl

for all i, j, k, l = 1, …, n. Hence:

2[D_lk, D_ij] = -γ_lkD_ij+(γ_jk-γ_kj)D_li+γ_ijD_lk

Thus:

2[D_kl, D_ij]-2[D_lk, D_ij] = (γ_lk-γ_kl) D_ij+(γ_jl-γ_lj)D_ki+(γ_kj-γ_jk)D_li = 0

Similarly:

(γ_lk-γ_kl)x_j+(γ_jl-γ_lj)x_k+(γ_kj-γ_jk)x_l = 0

Then by the above:

-γ_lkx_j+γ_ljx_k+(γ_kj-γ_jk)x_l = 0

Hence:

γ_ljx_k-γ_lkx_j = (γ_jk-γ_kj)x_l

Therefore:

γ_ljx_k-γ_ikx_j = (γ_jk-γ_kj)x_i

Lemma 7: Assume that:

γ_lkD_ij+D_lkD_ij = γ_ljD_ik+D_ljD_ik, γ_ijx_k-γ_ikx_j = (γ_jk-γ_kj)x_i

for all i, j, k, l = 1,…, n. Then Eq. 30 holds, i.e.,

γ_klD_ij+2D_klD_ij = γ_jlD_ik+2D_jlD_ik

for all i, j, k = 1,…, n.

Proof: By Eq. 28 and 29:

γ_lkD_ij+D_lkD_ij = γ_ljD_ik+D_ljD_ik, γ_jiD_kl+D_jiD_kl = γ_jlD_ik+D_jlD_ik

Therefore by Eq. 28:

[D_ij, D_kl] = γ_lkD_ij-γ_jiD_kl+(γ_jl-γ_lj) D_ik

Then with l = j:

[D_ij, D_kj] = γ_jkD_ij-γ_jiD_kj

By the hypothesis of the lemma:

By proposition Eq. 5 of Zhelyabin⁸:

(γ_ij-γ_ji)D_lk+D_ij D_lk = (γ_lj-γ_jl) D_ik+D_lj D_ik

Since γ_jiD_kl+D_jiD_kl = γ_jlD_ik+D_jlD_ik, by summing up the last two equalities and by replacing i by k and j by l, it is obvious that:

γ_kl D_ij+2D_kl D_ij = γ_jl D_ik+2D_jl D_ik

Similarly the following lemma may be proved.

Lemma 8: Assume that:

for all i, j, k, l = 1,…, n. Then Eq. 29 holds, i.e., for all i, j, k = 1,…, n:

γ_lk D_ij+D_lk D_ij = γ_lj D_ik+D_lj D_ik

Lemma 9: Assume that:

for all i, j, k, l = 1,…, n. Then Eq. 29 and 30 hold.

Proof: For every r∈A., from Eq. 31 and 32:

Then:

Hence:

By Eq. 31 and corollary of Zhelyabin⁸:

Therefore:

Let G (l.k.i, j) = D_lk D_ij-D_lj D_ik. Now for every r∈A, lemma 5 gives:

Since, A is an integral domain:

Hence, γ_lkD_ij+D_lkD_ij = γ_ljD_ik+D_jD_ik. Analogously, Eq. 30 may be proved.

Corollary 1: Equation 29 and 30 are equivalent to Eq. 31 and 32.

Proof: By lemma 9 it is sufficient to show that Eq. 29 and 30 imply Eq. 31 and 32.

Assume Eq. 29 and 30. Then lemma 6 implies:

As it was shown in lemma 7:

So, Eq. 31 is valid.

As it was shown in lemma 6:

Now replacing j by k and l by j:

This implies:

i.e., Eq. 32 is valid.

Theorem 3: Let A be an integral domain. Let M = Ax₁+…+Ax_n be a finitely generated projective A-module of rank 1. Assume that the mapping is a nonzero derivation of A into the A-module (M⊗_AM)* and Δ = {D_ij|i, j = 1,…, n} is a set of derivations of A, where D_ij (a) =r̄(x_i⊗x_j). Fix the subset Γ = {γ_ij∈A, i, j = 1,…, n} of A. Consider the vector space F(A, Δ, Γ) = A⊕M and equip it with the structure of a Z₂-graded algebra putting:

where, r, s∈A and rs is the product of r and s in A. Assume that the derivations D_ij and the elements of Γ satisfy Eq. 31 and 32 then F(A, Δ, Γ) is a flexible, Lie-admissible Superalgebra with even part A and odd part M.

Proof: First to show that the product in F(A, Δ, Γ) is correctly defined, let r₁x₁+…+r_nx_n = 0. Then for every fixed k = 1,…, n.

Then by lemma 9, Eq. 29 holds. So:

On the other hand, by Eq. 30:

Hence:

Therefore:

Since A is an integral domain:

Hence:

Therefore, the product in F(A, Δ, Γ) is correctly defined.

Hence F(A, Δ, Γ) is a flexible Lie-admissible superalgebra.

DISCUSSION

Kamiya and Okubo³ constructed some class of Lie superalgebras from nonsuper flexible Lie-admissible algebras without resorting to (-1, -1)-freudenthal-kantor tiple systems. Their motivation is a base for constructing the flexible Lie-admissible Grassmann algebra and hence the free unital flexible Lie-admissible superalgebra with one odd generator is isomorphic to the flexible Lie-admissible superalgebra of vector type B(Φ(Γ), D, γ₀) where, D(γ_i) = γ_i+1 for i = 0, 1,… which was analogously proved for (-1, 1) superalgebras of vector type in Zhelyabin⁹.

If the superalgebra is not a superalgebra of nondegenerate bilinear superform then it’s even part A is differentiably simple algebra and the odd part M is a finitely generated projective A-module of rank 1. Yuan¹⁴ restricted his studies on derivations of differentiably simple associative commutative algebra of n-torsion free (n>0). Whereas in this study remarks 2 and 3 to theorem 2 state that every simple n-torsion free (n>3) nonassociative flexible Lie-admissible superalgebra is local algebra and hence isomorphic to the superalgebra B(Γ, D, γ). Pchelintsev and Shestakov¹⁵ proved that (-1, 1) monster constructed by Pchelintsev¹⁶ generates the same variety of algebras as the Grassmann (-1, 1) algebra. Following Pchelintsev¹⁶ flexible Lie-admissible monster may be constructed.

CONCLUSION

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