The purpose of this paper is to study the constructions between Bihom-alternative superalgebras and Bihom-Malcev superalgebras and Bihom-Jordan superalgebras. First, we explain in detail that every regular Bihom-alternative superalgebra could be Bihom-Malcev-admissible superalgebra or Bihom-Jordan-admissible superalgebra. Next, the bimodules and T∗θ-extensions of Bihom-alternative superalgebras are also discussed as properties of Bihom-alternative superalgebras.
Citation: Ying Hou, Liangyun Chen. Constructions of three kinds of Bihom-superalgebras[J]. Electronic Research Archive, 2021, 29(6): 3741-3760. doi: 10.3934/era.2021059
[1] | Ying Hou, Liangyun Chen . Constructions of three kinds of Bihom-superalgebras. Electronic Research Archive, 2021, 29(6): 3741-3760. doi: 10.3934/era.2021059 |
[2] | Ying Hou, Liangyun Chen, Keli Zheng . Super-bimodules and $ \mathcal{O} $-operators of Bihom-Jordan superalgebras. Electronic Research Archive, 2024, 32(10): 5717-5737. doi: 10.3934/era.2024264 |
[3] | Yusi Fan, Chenrui Yao, Liangyun Chen . Structure of sympathetic Lie superalgebras. Electronic Research Archive, 2021, 29(5): 2945-2957. doi: 10.3934/era.2021020 |
[4] | Hongyan Guo . Automorphism group and twisted modules of the twisted Heisenberg-Virasoro vertex operator algebra. Electronic Research Archive, 2021, 29(4): 2673-2685. doi: 10.3934/era.2021008 |
[5] | Maoji Ri, Shuibo Huang, Canyun Huang . Non-existence of solutions to some degenerate coercivity elliptic equations involving measures data. Electronic Research Archive, 2020, 28(1): 165-182. doi: 10.3934/era.2020011 |
[6] | Xuan Liu, Ting Zhang . $ H^2 $ blowup result for a Schrödinger equation with nonlinear source term. Electronic Research Archive, 2020, 28(2): 777-794. doi: 10.3934/era.2020039 |
[7] | Huafei Di, Yadong Shang, Jiali Yu . Existence and uniform decay estimates for the fourth order wave equation with nonlinear boundary damping and interior source. Electronic Research Archive, 2020, 28(1): 221-261. doi: 10.3934/era.2020015 |
[8] | Jens Lorenz, Wilberclay G. Melo, Suelen C. P. de Souza . Regularity criteria for weak solutions of the Magneto-micropolar equations. Electronic Research Archive, 2021, 29(1): 1625-1639. doi: 10.3934/era.2020083 |
[9] | Heesung Shin, Jiang Zeng . More bijections for Entringer and Arnold families. Electronic Research Archive, 2021, 29(2): 2167-2185. doi: 10.3934/era.2020111 |
[10] | Jianxing Du, Xifeng Su . On the existence of solutions for the Frenkel-Kontorova models on quasi-crystals. Electronic Research Archive, 2021, 29(6): 4177-4198. doi: 10.3934/era.2021078 |
The purpose of this paper is to study the constructions between Bihom-alternative superalgebras and Bihom-Malcev superalgebras and Bihom-Jordan superalgebras. First, we explain in detail that every regular Bihom-alternative superalgebra could be Bihom-Malcev-admissible superalgebra or Bihom-Jordan-admissible superalgebra. Next, the bimodules and T∗θ-extensions of Bihom-alternative superalgebras are also discussed as properties of Bihom-alternative superalgebras.
As for the non-associative algebras that have been gradually developed on the development path of associative rings and associative algebras, non-associative superalgebras are also important for superalgebras that delete the associativity of multiplication on the basis of associative superalgebras, and then satisfy certain identities. The basis of this paper is more typical non-associative superalgebra, namely alternative superalgebras, Malcev superalgebras and Jordan superalgebras.
As we know, Malcev superalgebra is a generalization of Lie superalgebra. Of course, it is not Lie superalgebra. It is a class of non-associative algebra that is very useful in the geometry of smooth rings. Therefore, it has also been widely concerned in recent years [5,6,21]. Also Jordan superalgebra has a very important position in quantum mechanics. For interested readers, please refer to the literature [2,16,18,22,20]. Closely related to Malcev superalgebras and Jordan superalgebras are alternative superalgebras, which is a superalgebra whose associator is a super-alternating function [25]. In particular, all associative superalgebras are alternative. We have known that alternative superalgebras are Malcev-admissible and Jordan-admissible. More related results on alternative, Malcev and Jordan (super)algebra can be found in [7,11,14].
In the past few years, many scholars have conducted indepth research on twisted algebras (also known as Hom-algebras or Bihom-algebras) excited by q-deformation of vector field algebras. We can find more related results in [8,9,10,12,13,15,19,23,24,26,27]. We noticed recently that quadratic Hom-Malcev superalgebras were studied in [17] and Hom-alternative, Hom-Malcev and Hom-Jordan superalgebras were studied in [1]. A Bihom-algebra is an algebra whose identity defining structure is twisted by two homomorphisms
The main purpose of this paper is to study Bihom-type generalizations of alternative superalgebras, Malcev (-admissible) superalgebras and Jordan (-admissible) superalgebras. Outline of the paper is as follows. In Section
Throughout the paper, all algebraic systems are supposed to be over a field
In this section, we introduce Bihom- alternative superalgebras, study their general properties and provide some construction results. A Bihom-superalgebra is a quadruple
Definition 2.1. Let
~asα,β=μ∘(μ⊗β−α⊗μ). | (1) |
For any
˜Jα,β(x,y,z)=↺x,y,z(−1)|x||z|μ(β2(x),μ(β(y),α(z))). | (2) |
In particular, when
Definition 2.2. [23] A Bihom-associative superalgebra is a quadruple
~asα,β(x,y,z)=0,forallx,y,z∈H(A).(Bihom−associativitycondition) |
Clearly, when
Definition 2.3. A left Bihom-alternative superalgebra (resp. right Bihom- alternative superalgebra) is a quadruple
~asα,β(β(x),α(y),z)+(−1)|x||y|~asα,β(β(y),α(x),z)=0, | (3) |
respectively, the right Bihom-alternative super-identity,
~asα,β(x,β(y),α(z))+(−1)|y||z|~asα,β(x,β(z),α(y))=0. | (4) |
A Bihom-alternative superalgebra is one which is both a left and right Bihom-alternative superalgebra.
Observe that when
Definition 2.4. Let
fμ=μ′(f⊗f)(weakmorphismcondition),fα=α′f,fβ=β′f. | (5) |
Definition 2.5. A Bihom-superalgebra
Proposition 1. Let
~asα,β(β2(x),αβ(y),α2(z))+(−1)|x||y|~asα,β(β2(y),αβ(x),α2(z))=0 | (6) |
and
~asα,β(β2(x),αβ(y),α2(z))+(−1)|y||z|~asα,β(β2(x),αβ(z),α2(y))=0. | (7) |
Proof. Straightforward by Definition 2.3.
Remark 1. Let
Lemma 2.6. Let
~asα,β(β2(x),αβ(y),α2(z))=−(−1)|z||x|+|z||y|+|x||y|~asα,β(β2(z),αβ(y),α2(x)), |
~asα,β(β2(x),αβ(y),α2(z))=(−1)|z||x|+|z||y|~asα,β(β2(z),αβ(x),α2(y)). |
Proof.
~asα,β(β2(x),βα(y),α2(z))=−(−1)|z||y|~asα,β(β2(x),βα(z),α2(y))=(−1)|z||y|+|z||x|~asα,β(β2(z),βα(x),α2(y))=−(−1)|z||y|+|z||x|+|x||y|~asα,β(β2(z),βα(y),α2(x)). |
Another equation is obvious.
Next, we provide a way to construct Bihom-alternative superalgebras starting from an alternative superalgebra and two commuting alternative superalgebra morphisms. This approach was applied to other algebras such as Bihom-Lie algebras, Bihom-Novikov algebras and so on in [9], [10].
Definition 2.7. Let
Aα,β=(A,μα,β=μ(α⊗β),α,β). |
Theorem 2.8. (i) Let
(ii) Suppose that
f:Aα,β⟶A′α′,β′ |
is a left Bihom-alternative superalgebra (resp. right Bihom-alternative superalgebra) morphism.
Proof. We show that
~asα,β(β(x),α(y),z)=(β(x)∘αβ(y))∘β(z)−αβ(x)∘(α(y)∘z)=(αβ(x)αβ(y))∘β(z)−αβ(x)∘(α2(y)αβ(z))=(α2β(x)α2β(y))β2(z)−α2β(x)(α2β(y)β2(z))=~as(α2β(x),α2β(y),β2(z))=−(−1)|x||y|~as(α2β(y),α2β(x),β2(z))=−(−1)|x||y|~asα,β(β(y),α(x),z). |
So conclusion
fμα,β=fμ(α⊗β)=μ′(f⊗f)(α⊗β)=μ′(α′⊗β′)(f⊗f)=μ′α′,β′(f⊗f). |
That is,
Remark 2. Similar to Theorem
On the contrary, we will give an approach to get an alternative superalgebra from Bihom-alternative superalgebra in the following proposition.
Proposition 2. Let
Proof. Denote
~as(x,y,z)+(−1)|x||y|~as(y,x,z)=0(leftalternativesuper−identity) |
and
~as(x,y,z)+(−1)|y||z|~as(x,z,y)=0.(rightalternativesuper−identity) |
Since
~as(x,y,z)+(−1)|x||y|~as(y,x,z)=(x⋄y)⋄z−x⋄(y⋄z)+(−1)|x||y|(y⋄x)⋄z−(−1)|x||y|y⋄(x⋄z)=(α−2(x)(αβ)−1(y))β−1(z)−α−1(x)((αβ)−1(y)β−2(z))+(−1)|x||y|(α−2(y)(αβ)−1(x))β−1(z)−(−1)|x||y|α−1(y)((αβ)−1(x)β−2(z))=~asα,β(α−2(x),(αβ)−1(y),β−2(z))+(−1)|x||y|~asα,β(α−2(y),(αβ)−1(x),β−2(z))=~asα,β(β(β−1α−2(x)),α(β−1α−2(y)),β−2(z))+(−1)|x||y|~asα,β(β(β−1α−2(y)),α(β−1α−2(x)),β−2(z))=0. |
Similarly, right alternative super-identity is easy to prove.
Corollary 1. Let
Proof. We write
In this section, we introduce Bihom-Malcev superalgebras (as a generalization of Bihom-Lie superalgebras) and provide a way to construct Bihom-Malcev superalgebras. Some general properties are studied. In the end, we show that Bihom-Malcev superalgebra could come into being from regular Bihom-alternative superalgebra via the Bihom-supercommutator bracket.
Definition 3.1. [23]
(i)αβ=βα,(ii)α([x,y])=[α(x),α(y)],β([x,y])=[β(x),β(y)],(iii)[β(x),α(y)]=−(−1)|x||y|[β(y),α(x)],(Bihom−skewsupersymmetry)(iv)˜Jcα,β(x,y,z)=(−1)|x||z|˜Jα,β(x,y,z).(Bihom−Jacobisuperidentity)(˜Jα,β(x,y,z)=↺x,y,z(−1)|x||z|[β2(x),[β(y),α(z)]]=0) |
Definition 3.2. A Bihom-Malcev superalgebra is a 4-tuple
(i)αβ=βα,(ii)α([x,y])=[α(x),α(y)],β([x,y])=[β(x),β(y)],(iii)[β(x),α(y)]=−(−1)|x||y|[β(y),α(x)],(Bihom−skewsupersymmetry)(iv)˜Jcα,β(αβ(x),αβ(y),[β(t),α(z)])+(−1)|x||y|+|t|(|x|+|y|)˜Jcα,β(αβ(t),αβ(y),[β(x),α(z)])=(−1)|t||z|[˜Jcα,β(β(x),β(y),β(z)),α2β2(t)]+(−1)|x|(|y|+|z|+|t|)+|t||y|[˜Jcα,β(β(t),β(y),β(z)),α2β2(x)].(Bihom−Malcevsuper−identity) |
If
2[t,˜J(x,y,z)]=˜J(t,x,[y,z])+(−1)|x|(|y|+|z|)˜J(t,y,[z,x])+(−1)|z|(|x|+|y|)˜J(t,z,[x,y]), |
for all
Theorem 3.3. Let
Proof. Step
[α(x),α(y)]′=[α2(x),αβ(y)]=α([α(x),β(y)])=α([x,y]′). |
Similarly, we can check that
Step
˜Jcα,β(αβ(x),αβ(y),[β(t),α(z)]′)=(−1)|x||t+z|˜Jα,β(αβ(x),αβ(y),[αβ(t),αβ(z)])=(−1)|x||t+z|αβ(↺x,y,[t,z](−1)|x||t+z|[β2(x),[β(y)[α(t),α(z)]]′]′)=(−1)|x||t+z|α2β3(↺x,y,[t,z](−1)|x||t+z|[x,[y,[t,z]]])=α2β3(˜J(x,y,[t,z])). |
On the other hand, we have
[˜Jcα,β(β(x),β(y),β(z)),α2β2(t)]′=[(−1)|x||z|(↺x,y,z(−1)|x||z|[β2(β(x)),[β(β(y)),α(β(z))]′]′),α2β2(t)]′=[(−1)|x||z|(↺x,y,z(−1)|x||z|αβ3[x,[y,z]]),α2β2(t)]′]=[(−1)|x||z|(↺x,y,z(−1)|x||z|α2β3[x,[y,z]]),α2β3(t)]]=α2β3[˜J(x,y,z),t]. |
So,
˜Jcα,β(αβ(x),αβ(y),[β(t),α(z)]′)+(−1)|x||y|+|t|(|x|+|y|)˜Jcα,β(αβ(t),αβ(y),[β(x),α(z)]′)=α2β3(˜J(x,y,[t,z])+(−1)|x||y|+|t|(|x|+|y|)˜J(t,y,[x,z]))=α2β3((−1)|t||z|[˜J(x,y,z),t]+(−1)|x|(|y|+|z|+|t|)+|t||y|[˜J(t,y,z),x])=(−1)|t||z|[˜Jcα,β(β(x),β(y),β(z)),α2β2(t)]′+(−1)|x|(|y|+|z|+|t|)+|t||y|[˜Jcα,β(β(t),β(y),β(z)),α2β2(x)]′. |
Thus,
Remark 3. Similar to Theorem
Lemma 3.4. Let
˜Jα,β∘(αβ)⊗3=αβ∘˜Jα,β. |
Proof. Let
˜Jα,β(αβ(x),αβ(y),αβ(z))=↺x,y,z(−1)|x||z|[β2(αβ(x)),[β(αβ(y)),α(αβ(z))]]=↺x,y,z(−1)|x||z|αβ[β2(x),[β(y),α(z)]]=αβ˜Jα,β(x,y,z). |
So the Lemma
Remark 4. Let
˜Jcα,β∘(αβ)⊗3=αβ∘˜Jcα,β. |
We recall the associative super-commutator in [23] in the following definition.
Definition 3.5. Let
A−=(A,[⋅,⋅]=μ∘(id−(α−1β⊗αβ−1)∘τ),α,β), |
where
[x,y]=μ(x,y)−(−1)|x||y|μ(α−1β(y),αβ−1(x)). |
We call a Bihom-superalgebra
Lemma 3.6. Let
˜Jcα,β=~asα,β(α−1β⊗β⊗α)∘(id+ξ+ξ2)∘(δ−id), |
where
Proof. For
O˜Jcα,β(x,y,z)=(−1)|x||z|(↺x,y,z(−1)|x||z|[β2(x),[β(y),α(z)]])=(−1)|x||z|(↺x,y,z(−1)|x||z|[β2(x),β(y)α(z)−(−1)|y||z|β(z)α(y)])=(−1)|x||z|(↺x,y,z(−1)|x||z|[β2(x),β(y)α(z)]−↺x,y,z(−1)|z|(|x|+|y|)[β2(x),β(z)α(y)])=(−1)|x||z|(↺x,y,z(−1)|x||z|β2(x)(β(y)α(z))−↺x,y,z(−1)|x||z|(α−1β2(x)β(y))αβ(z)−↺x,y,z(−1)|z|(|x|+|y|)β2(x)(β(z)α(y))+↺x,y,z(−1)|z|(|x|+|y|)(α−1β2(x)β(z))αβ(y))=−(−1)|x||z|(↺x,y,z(−1)|x||z|~asα,β(α−1β2(x),β(y),α(z)))+(−1)|x||z|(↺x,y,z(−1)|z|(|x|+|y|)~asα,β(α−1β2(x),β(z),α(y)))=~asα,β∘(α−1β2⊗β⊗α)∘(id+ξ+ξ2)∘(δ−id)(x,y,z). |
Proposition 3. Let
˜Jcα,β(x,y,z)=−6~asα,β(α−1β2(x),β(y),α(z)),forallx,y,z∈H(A), |
where the Bihom-associator is taken in
Proof. The result follows immediately from
Definition 3.7. Let
H(w,x,y,z)=~asα,β(β2(w)αβ(x),α2β(y),α3(z))−(−1)|w||x|+|w||y|+|w||z|~asα,β(β2(x)αβ(y),α2β(z),α3(w))+(−1)(|y|+|z|)(|w|+|x|)~asα,β(β2(y)αβ(z),α2β(w),α3(x)). |
Lemma 3.8. In a regular Bihom-alternative superalgebra
H(w,x,y,z)=α2β2(w)~asα,β(αβ(x),α2(y),α3β−1(z))+~asα,β(β2(w),αβ(x),α2(y))α3β(z) |
holds, for all
Proof. First, since
~asα,β(β2(x)αβ(y),α2β(z),α3(w))=~asα,β(β2(xαβ−1(y)),αβ(α(z)),α2(α(w)))=(−1)|w||x|+|w||y|+|w||z|~asα,β(αβ2(w),αβ(x)α2(y),α3(z)). |
In addition,
~asα,β(β2(y)αβ(z),α2β(w),α3(x))=~asα,β(β2(yαβ−1(z)),αβ(α(w)),α2(α(x)))=(−1)(|w|+|x|)(|y|+|z|)~asα,β(αβ2(w),α2β(x),α2(y)α3β−1(z)). |
So,
H(w,x,y,z)=~asα,β(β2(w)αβ(x),α2β(y),α3(z))−~asα,β(β2(w)αβ(x),α2β(y),α3(z))+~asα,β(β2(w)αβ(x),α2β(y),α3(z))=α2β2(w)~asα,β(αβ(x),α2(y),α3β−1(z))+~asα,β(β2(w),αβ(x),α2(y))α3β(z). |
The proof is finished.
Definition 3.9. Let
f(w,x,y,z)=~asα,β(β2(w)αβ(x),α2β(y),α3(z))−(−1)|w||x|+|w||y|+|w||z|~asα,β(β2(x),αβ(y),α2(z))α3β(w)−(−1)|w||x|α2β2(x)~asα,β(αβ(w),α2(y),α3β−1(z)), | (8) |
for all
Define another even muti-linear map
F=[⋅,⋅]∘((~asα,β∘(β2⊗αβ⊗α2))⊗α3β)∘(id−ξ+ξ2−ξ3), | (9) |
where
ξ(w⊗x⊗y⊗z)=(−1)|z||w|+|z||x|+|z||y|z⊗w⊗x⊗y, |
i.e., for all
F(w,x,y,z)=[~asα,β(β2(w),αβ(x),α2(y)),α3β(z)]−(−1)|z||w|+|z||x|+|z||y|[~asα,β(β2(z),αβ(w),α2(x)),α3β(y)]+(−1)(|y|+|z|)(|w|+|x|)[~asα,β(β2(y),αβ(z),α2(w)),α3β(x)]−(−1)|w||x|+|w||y|+|w||z|[~asα,β(β2(x),αβ(y),α2(z)),α3β(w)]. |
Lemma 3.10. Let
F=f∘(id−ρ+ρ2), |
where
F(w,x,y,z)=f(w,x,y,z)−(−1)|w||x|+|w||y|+|w||z|f(x,y,z,w)+(−1)(|y|+|z|)(|w|+|x|)f(y,z,w,x). |
Proof. For each
[~asα,β(β2(x),αβ(y),α2(z)),α3β(w)]=~asα,β(β2(x),αβ(y),α2(z))α3β(w)−(−1)|w||x|+|w||y|+|w||z|α2β2(w)~asα,β(αβ(x),α2(y),α3β−1(z)). |
Then, by
α2β2(w)~asα,β(αβ(x),α2(y),α3β−1(z))+~asα,β(β2(x),αβ(y),α2(z))α3β(z)=~asα,β(β2(w)αβ(x),α2β(y),α3(z))−(−1)|w||x|+|w||y|+|w||z|~asα,β(β2(x)αβ(y),α2β(z),α3(w))+(−1)(|y|+|z|)(|w|+|x|)~asα,β(β2(y)αβ(z),α2β(w),α3(x))=f(w,x,y,z)+(−1)|w||x|+|w||y|+|w||z|~asα,β(β2(x),αβ(y),α2(z))α3β(w)+(−1)|w||x|α2β2(x)~asα,β(αβ(w),α2(y),α3β−1(z))−(−1)|w||x|+|w||y|+|w||z|f(x,y,z,w)−(−1)(|w|+|x|)(|y|+|z|)~asα,β(β2(y),αβ(z),α2(w))α3β(x)−(−1)|w||x|+|w||y|+|w||z|+|x||y|α2β2(y)~asα,β(αβ(x),α2(z),α3β−1(w))+(−1)(|y|+|z|)(|w|+|x|)f(y,z,w,x)+(−1)|z||x|+|z||w|+|z||y|~asα,β(β2(z),αβ(w),α2(x))α3β(y)+(−1)(|y|+|z|)(|w|+|x|)α2β2(z)~asα,β(αβ(y),α2(w),α3β−1(x)). |
Moreover, since
~asα,β(αβ(y),α2(w),α3β−1(x))=αβ−1(~asα,β(β2(y),αβ(w),α2(x)))=(−1)|y||w|+|y||x|αβ−1(~asα,β(β2(w),αβ(x),α2(y)))=(−1)|y||w|+|y||x|~asα,β(αβ(w),α2(x),α3β−1(y)), |
Similarly,
~asα,β(αβ(x),α2(z),α3β−1(w))=(−1)|x||w|+|x||z|~asα,β(αβ(z),α2(w),α3β−1(x)), |
~asα,β(αβ(w),α2(y),α3β−1(z))=(−1)|w||y|+|w||z|~asα,β(αβ(y),α2(z),α3β−1(w)). |
Then
F(w,x,y,z)=f(w,x,y,z)−(−1)|w||x|+|w||y|+|w||z|f(x,y,z,w)+(−1)(|y|+|z|)(|w|+|x|)f(y,z,w,x), |
for all
Proposition 4. Let
Proof. The proof is similar to that Proposition 4.1 in Reference [1], it is easy to check it.
Next, we will give a relationship between Bihom-alternate superalgebra and Bihom-Malcev superalgebra, and use the above preparation knowledge to prove it.
Theorem 3.11. Every regular Bihom-alternative superalgebra is Bihom-Malcev-admissible.
Proof. Let
Step
Step
Step
˜Jcα,β(αβ(x),αβ(y),[β(t),α(z)])+(−1)|x||y|+|t|(|x|+|y|)˜Jcα,β(αβ(t),αβ(y),[β(x),α(z)])=(−1)|t||z|[˜Jcα,β(β(x),β(y),β(z)),α2β2(t)]+(−1)|x|(|y|+|z|+|t|)+|t||y|[˜Jcα,β(β(t),β(y),β(z)),α2β2(x)]. |
Note first that
In addition,
˜Jcα,β(αβ(x),αβ(y),[β(t),α(z)])+(−1)|x||y|+|t|(|x|+|y|)˜Jcα,β(αβ(t),αβ(y),[β(x),α(z)])=−6~asα,β(β3(x),αβ2(y),αβ(t)α2(z))+6(−1)|t||z|~asα,β(β3(x),αβ2(y),αβ(z)α2(t))−6(−1)|x||y|+|t|(|x|+|y|)~asα,β(β3(t),αβ2(y),αβ(x)α2(z))+6(−1)|x|(|y|+|z|)+|t|(|x|+|y|)~asα,β(β3(t),αβ2(y),αβ(z)α2(x))=−6(−1)(|t|+|z|)(|x|+|y|)~asα,β(α−1β3(t)β2(z),αβ2(x),α2β(y))+6(−1)|t||z|+(|t|+|z|)(|x|+|y|)~asα,β(α−1β3(z)β2(t),αβ2(x),α2β(y))−6(−1)|t||y|+|z||t|+|z||y|~asα,β(α−1β3(x)β2(z),αβ2(t),α2β(y))+6(−1)|x||z|+|t||y|+|z||t|+|z||y|~asα,β(α−1β3(z)β2(x),αβ2(t),α2β(y))=6(−1)(|t|+|z|)(|x|+|y|)α−1β{(−1)|t|(|x|+|y|+|z|)[~asα,β(β2(z),αβ(x),α2(y)),α3β(t)]−(−1)|z|(|x|+|y|)[~asα,β(β2(t),αβ(x),α2(y)),α3β(z)]}+6(−1)|t||y|+|z||t|+|z||y|α−1β{(−1)|x|(|z|+|t|+|y|)[~asα,β(β2(z),αβ(t),α2(y)),α3β(x)]−(−1)|z|(|t|+|y|)[~asα,β(β2(x),αβ(t),α2(y)),α3β(z)]}=(−1)|t||z|[˜Jcα,β(β(x),β(y),β(z)),α2β2(t)]+(−1)|x|(|y|+|z|+|t|)+|t||y|[˜Jcα,β(β(t),β(y),β(z)),α2β2(x)]. |
Thus, every regular Bihom-alternative superalgebra is Bihom-Malcev-admissible.
Proposition 5. Let
f=13F=~asα,β∘(([⋅,⋅]∘(β2⊗αβ))⊗α2β⊗α3)∘(id+ζ), | (10) |
where
ζ(w⊗x⊗y⊗z)=(−1)(|y|+|z|)(|w|+|x|)y⊗z⊗w⊗x. |
Proof. By
Use the definition of
2f(t,x,y,z)=f(t,x,y,z)−(−1)|x||t|f(x,t,y,z)=~asα,β(β2(t)αβ(x),α2β(y),α3(z)−(−1)|t|(|x|+|y|+|z|)~asα,β(β2(x),αβ(y),α2(z))α3β(t)−(−1)|t||x|α2β2(x)~asα,β(αβ(t),α2(y),α3β−1(z))−(−1)|x||t|~asα,β(β2(x)αβ(t),α2β(y),α3(z)+(−1)|x|(|y|+|z|)~asα,β(β2(t),αβ(y),α2(z))α3β(x)+α2β2(t)~asα,β(αβ(x),α2(y),α3β−1(z)), |
which implies that
~asα,β([β2(t),αβ(x)],α2β(y),α3(z))=2f(t,x,y,z)−[α2β2(t),~asα,β(αβ(x),α2(y),α3β−1(z))]+(−1)|t|(|x|+|y|+|z|)[α2β2(x),~asα,β(αβ(y),α2(z),α3β−1(t))]. | (11) |
Interchanging
(−1)(|t|+|x|)(|y|+|z|)~asα,β([β2(y),αβ(z)],α2β(t),α3(x))=(−1)|y|(|t|+|x|+|z|)[α2β2(z),~asα,β(αβ(t),α2(x),α3β−1(y))]−(−1)(|t|+|x|)(|y|+|z|)[α2β2(y),~asα,β(αβ(z),α2(t),α3β−1(x))]+2(−1)(|t|+|x|)(|y|+|z|)f(y,z,t,x)=(−1)|y|(|t|+|x|+|z|)[α2β2(z),~asα,β(αβ(t),α2(x),α3β−1(y))]−(−1)(|t|+|x|)(|y|+|z|)[α2β2(y),~asα,β(αβ(z),α2(t),α3β−1(x))]+2f(t,x,y,z), |
so
~asα,β([β2(t),αβ(x)],α2β(y),α3(z))+(−1)(|t|+|x|)(|y|+|z|)~asα,β([β2(y),αβ(z)],α2β(t),α3(z))=(4f−F)(t,x,y,z)=f(t,x,y,z)(sinceF=3f). |
Thus
For convenience and without confusion, we denote Bihom-alternative superalgebra
Definition 4.1. Let
αVL(x)=L(α(x))αV,αVR(x)=R(α(x))αV, | (12) |
βVL(x)=L(β(x))βV,βVR(x)=R(β(x))αV, | (13) |
L(β(x)∘α(y))βV(v)−L(αβ(x))L(α(y))v+(−1)|x||y|L(β(y)∘α(x))βV(v)−(−1)|x||y|L(αβ(y))L(α(x))v=0, | (14) |
(−1)|z||y|R(αβ(z))R(β(y))v−R(β(y)∘α(z))αV(v)+R(αβ(y))R(β(z))v−(−1)|y||z|R(β(z)α(y))αV(v)=0, | (15) |
(−1)|x||y|R(β(y))L(β(x))αV(v)−L(αβ(x))R(y)αV(v)+(−1)|x||y|R(β(y))R(α(x))βV(v)−R(α(x)∘y)αVβV(v)=0, | (16) |
(−1)|x||y|R(αβ(y))L(x)βV(v)−L(α(x))R(α(y))βV(v)+L(x∘β(y))αVβV(v)−L(α(x))L(β(y))αV(v)=0. | (17) |
Let
<L∗(x)u∗,v>=<u∗,L(x)v>,<R∗(x)u∗,v>=(−1)|u||x|<u∗,R(x)v>, |
α∗(x∗)(y)=x∗(α(y)),β∗(x∗)(y)=x∗(β(y)), |
α∗V(u∗)(v)=u∗(αV(v)),β∗V(u∗)(v)=u∗(βV(v)). |
Proposition 6. Let
βVL(β(x)∘α(y))(v)−(−1)|x||y|L(α(y))L(αβ(x))v+(−1)|x||y|βVL(β(y)∘α(x))(v)−L(α(x))L(αβ(y))v=0, | (18) |
R(β(y))R(αβ(z))v−αVR(β(y)∘α(z))(v)+(−1)|z||y|R(β(z))R(αβ(y))v−(−1)|y||z|αVR(β(z)α(y))(v)=0, | (19) |
αVL(β(x))R(β(y))(v)−(−1)|x||y|αVR(y)L(αβ(x))(v)+βVR(α(x))R(β(y))(v)−αVβVR(α(x)∘y)(v)=0, | (20) |
βVL(x)R(αβ(y))(v)−(−1)|x||y|βVR(α(y))L(α(x))(v)+αVβVL(x∘β(y))(v)−(−1)|x||y|αVL(β(y))L(α(x))(v)=0, | (21) |
Proof. Straightforward by calculation.
Theorem 4.2. Let
(x+u)⋆(y+v)=x∘y+θ(x,y)+L(x)v+(−1)|x||y|R(y)u, | (22) |
and
(α+αV)(x+u)=α(x)+αV(u),(β+βV)(x+u)=β(x)+βV(u), | (23) |
for any
θ(β(x1)∘α(x2),β(x3))+(−1)|x3|(|x1|+|x2|)R(β(x3))θ(β(x1),α(x2))−θ(αβ(x1),α(x2)∘x3)−L(αβ(x1))θ(α(x2),x3)+(−1)|x1||x2|θ(β(x2)∘α(x1),β(x3))+(−1)|x3|(|x1|+|x2|)+|x1||x2|R(β(x3))θ(β(x2),α(x1))−(−1)|x1||x2|θ(αβ(x2),α(x1)∘x3)−(−1)|x1||x2|L(αβ(x2))θ(α(x1),x3)=0, | (24) |
θ(x1∘β(x2),αβ(x3))+(−1)|x3|(|x1|+|x2|)R(αβ(x3))θ(x1,β(x2))−θ(α(x1),β(x2)∘α(x3))−L(α(x1))θ(β(x2),α(x3))+(−1)|x2||x3|θ(x1∘β(x3),αβ(x2))+(−1)|x1||x2|R(αβ(x2))θ(x1,β(x3))−(−1)|x2||x3|L(α(x1))θ(β(x3),α(x2))−(−1)|x2||x3|θ(α(x1),β(x3)∘α(x2))=0. | (25) |
Proof. First we prove the commutativity of maps,
(α+αV)(β+βV)(x+u)=(α+αV)(β(x)+βV(u))=αβ(x)+αVβV(u)=βα(x)+βVαV(u)=(β+βV)(α+αV)(x+u). |
By the fact that
~asα,β((β+βV)(x1+v1),(α+αV)(x2+v2),x3+v3)+(−1)|x1||x2|~asα,β((β+βV)(x2+v2),(α+αV)(x1+v1),x3+v3)=((β+βV)(x1+v1)⋆(α+αV)(x2+v2))⋆(β+βV)(x3+v3)−(α+αV)(β+βV)(x1+v1)⋆((α+αV)(x2+v2)⋆(x3+v3))+(−1)|x1||x2|((β+βV)(x2+v2)⋆(α+αV)(x1+v1))⋆(β+βV)(x3+v3)−(−1)|x1||x2|(α+αV)(β+βV)(x2+v2)⋆((α+αV)(x1+v1)⋆(x3+v3))=(β(x1)∘α(x2))∘β(x3)+θ(β(x1)∘α(x2),β(x3))+L(β(x1)∘α(x2))βV(v3)+(−1)|x3|(|x1|+|x2|)R(β(x3))θ(β(x1),α(x2))+(−1)|x3|(|x1|+|x2|)R(β(x3))L(β(x1))αV(v2)+(−1)|x3|(|x1|+|x2|)+|x1||x2|R(β(x3))R(α(x2))βV(v1)−αβ(x1)∘(α(x2)∘x3)−θ(αβ(x1),α(x2)∘x3)−L(αβ(x1))θ(α(x2),x3)−L(αβ(x1))L(α(x2))v3−(−1)|x2||x3|L(αβ(x1))R(x3)αV(v2)−(−1)|x1|(|x2|+|x3|)R(α(x2)∘x3)αβ(v1)+(−1)|x1||x2|(β(x2)∘α(x1))∘β(x3)+(−1)|x1||x2|θ(β(x2)∘α(x1),β(x3))+(−1)|x1||x2|L(β(x2)∘α(x1))βV(v3)+(−1)|x3|(|x1|+|x2|)+|x1||x2|R(β(x3))θ(β(x2),α(x1))+(−1)|x3|(|x1|+|x2|)+|x1||x2|R(β(x3))L(β(x2))αV(v1)+(−1)|x3|(|x1|+|x2|)R(β(x3))R(α(x1))βV(v2)−(−1)|x1||x2|αβ(x2)∘(α(x1)∘x3)−(−1)|x1||x2|θ(αβ(x2),α(x1)∘x3)−(−1)|x1||x2|L(αβ(x2))θ(α(x1),x3)−(−1)|x1||x2|L(αβ(x2))L(α(x1))v3−(−1)|x1|(|x2|+|x3|)L(αβ(x2))R(x3)αV(v1)−(−1)|x2||x3|R(α(x1)∘x3)αβ(v2). | (26) |
Since
(26)=~asα,β(β(x1),α(x2),x3)+(−1)|x1||x2|~asα,β(β(x2),α(x1),x3)+(a)+(−1)|x2||x3|(b)+(−1)|x1|(|x2|+|x3|)(c)+θ(β(x1)∘α(x2),β(x3))+(−1)|x3|(|x1|+|x2|)R(β(x3))θ(β(x1),α(x2))−θ(αβ(x1),α(x2)∘x3)−L(αβ(x1))θ(α(x2),x3)+(−1)|x1||x2|θ(β(x2)∘α(x1),β(x3))+(−1)|x3|(|x1|+|x2|)+|x1||x2|R(β(x3))θ(β(x2),α(x1))−(−1)|x1||x2|θ(αβ(x2),α(x1)∘x3)−(−1)|x1||x2|L(αβ(x2))θ(α(x1),x3)=θ(β(x1)∘α(x2),β(x3))+(−1)|x3|(|x1|+|x2|)R(β(x3))θ(β(x1),α(x2))−θ(αβ(x1),α(x2)∘x3)−L(αβ(x1))θ(α(x2),x3)+(−1)|x1||x2|θ(β(x2)∘α(x1),β(x3))+(−1)|x3|(|x1|+|x2|)+|x1||x2|R(β(x3))θ(β(x2),α(x1))−(−1)|x1||x2|θ(αβ(x2),α(x1)∘x3)−(−1)|x1||x2|L(αβ(x2))θ(α(x1),x3)=0. |
So the BiHom left super-alternativity is finished. By the same calculation, we also obtain
Theorem 4.3. Let
(x+a∗)⋆(y+b∗)=x∘y+θ(x,y)+(−1)|x||y|R∗(x)b∗+L∗(y)a∗, | (27) |
(α+α∗)(x+u)=α(x)+α∗(u),(β+β∗)(x+u)=β(x)+β∗(u), | (28) |
for any
θ(β(x1)∘α(x2),β(x3))+(−1)|x3|(|x1|+|x2|)L∗(β(x3))θ(β(x1),α(x2))−θ(αβ(x1),α(x2)∘x3)−R∗(αβ(x1))θ(α(x2),x3)+(−1)|x1||x2|θ(β(x2)∘α(x1),β(x3))+(−1)|x3|(|x1|+|x2|)+|x1||x2|L∗(β(x3))θ(β(x2),α(x1))−(−1)|x1||x2|θ(αβ(x2),α(x1)∘x3)−(−1)|x1||x2|R∗(αβ(x2))θ(α(x1),x3)=0, |
θ(x1∘β(x2),αβ(x3))+(−1)|x3|(|x1|+|x2|)L∗(αβ(x3))θ(x1,β(x2))−θ(α(x1),β(x2)∘α(x3))−R∗(α(x1))θ(β(x2),α(x3))+(−1)|x2||x3|θ(x1∘β(x3),αβ(x2))+(−1)|x1||x2|L∗(αβ(x2))θ(x1,β(x3))−(−1)|x2||x3|R∗(α(x1))θ(β(x3),α(x2))−(−1)|x2||x3|θ(α(x1),β(x3)∘α(x2))=0. |
For any
Proof. By Proposition
In Section 2, we introduced the construction from alternative superalgebra to Bihom-alternative superalgebra. As an application, next we will give the construction from bimodules of alternative superalgebras to bimodules of Bihom alternative superalgebra.
Proposition 7. Let
Proof. Let
˜L(β(x)∗α(y))βV(v)−˜L(αβ(x))˜L(α(y))v+(−1)|x||y|˜L(β(y)∗α(x))βV(v)−(−1)|x||y|˜L(αβ(y))˜L(α(x))v=˜L(αβ(x)∘αβ(y))βV(v)−˜L(αβ(x))L(α2(y))βV(v)+(−1)|x||y|˜L(αβ(y)∘αβ(x))βV(v)−(−1)|x||y|˜L(αβ(y))L(α2(x))βV(v)=L(α2β(x)∘α2β(y))β2V(v)−L(α2β(x))L(α2β(y))β2V(v)+(−1)|x||y|L(α2β(y)∘α2β(x))β2V(v)−(−1)|x||y|L(α2β(y))L(α2β(x))β2V(v)=0 |
and
(−1)|x||y|˜R(β(y))˜L(β(x))αV(v)−˜L(αβ(x))˜R(y)αV(v)+(−1)|x||y|˜R(β(y))˜R(α(x))βV(v)−˜R(α(x)∗y)αVβV(v)=(−1)|x||y|˜R(β(y))L(αβ(x))αVβV(v)−˜L(αβ(x))R(β(y))α2V(v)+(−1)|x||y|˜R(β(y))R(αβ(x))αVβV(v)−˜R(α2(x)∘β(y))αVβV(v)=(−1)|x||y|R(β2(y))L(α2β(x))α2VβV(v)−L(α2β(x))R(β2(y))α2VβV(v)+(−1)|x||y|R(β2(y))R(α2β(x))α2VβV(v)−R(α2β(x)∘β2(y))α2VβV(v)=0. |
Similarly, other identities can also be obtained.
In this section, We also give constructions for Bihom-Jordan and Bihom-Jordan admissible superalgebras. Additionally, we figure out that regular Bihom-alternative superalgebras are Bihom-Jordan admissible.
Definition 5.1. A BiHom superalgebra
(i)αβ=βα,(ii)μ(β(x),α(y))=(−1)|x||y|μ(β(y),α(x)),(Bihom−supercommutativitycondition)(iii)↺x,y,t(−1)|t|(|x|+|z|)~asα,β(μ(β2(x),αβ(y)),α2β(z),α3(t))=0.(Bihom−Jordansuper−identity) |
In particular, it is reduced to a Jordan superalgebra when
Definition 5.2. Let
x∗y=12(μ(x,y)+(−1)|x||y|μ(α−1β(y),αβ−1(x))). |
Note that product
β(x)∗α(y)=12(β(x)α(y)+(−1)|x||y|β(y)α(x))=(−1)|x||y|12(β(y)α(x)+β(x)α(y))=(−1)|x||y|β(y)∗α(x). |
Moreover, the plus Bihom-superalgebra
Proposition 8. Let
Proof. We need to prove that
x∗y=12(μ′(x,y)+(−1)|x||y|μ′(α−1β(y),αβ−1(x))). |
The Jordan superalgebra obtained from Jordan-admissible superalgebra
Also, it is easy to check that
~asα,β(β2(x)∗αβ(y),α2β(z),α3(t))=α3β2~as(x∘y,z,t). |
So,
↺x,y,t(−1)|t|(|x|+|z|)~asα,β(β2(x)∗αβ(y),α2β(z),α3(z))=α3β2↺x,y,t(−1)|t|(|x|+|z|)~as(x∘y,z,t)=0. |
Conclusion proof ends.
Theorem 5.3. Let
Proof. Firstly, note that
Step
μ′(β(x),α(y))=μ(αβ(x),αβ(y))=(−1)|x||y|μ(αβ(y),αβ(x))=(−1)|x||y|μ′(β(y),α(x)). |
Step
~asα,β(μ′(β2(x),αβ(y)),α2β(z),α3(t))=α3β2~as(μ(x,y),z,t). |
further, we have
↺x,y,t(−1)|t|(|x|+|z|)~asα,β(μ′(β2(x),αβ(y)),α2β(z),α3(t))=α3β2↺x,y,t(−1)|t|(|x|+|z|)~as(μ(x,y),z,t)=0. |
Thus,
Remark 5. Similar to Theorem
Theorem 5.4. Every regular BiHom-alternative superalgebra
Proof. As usual we write
↺x,y,t8(−1)|t|(|x|+|z|)~asα,β(β2(x)∗αβ(y),α2β(z),α3(z))=8(−1)|t|(|x|+|z|)(((β2(x)∗αβ(y))∗α2β(z))∗α3β(t)−(αβ2(x)∗α2β(y))∗(αβ(z)∗α3(t)))⏟ξ+8(−1)|x|(|y|+|z|)(((β2(y)∗αβ(t))∗α2β(z))∗α3β(x))−(αβ2(y)∗α2β(t))∗(αβ(z)∗α3(x)))⏟η+8(−1)|y|(|t|+|z|)(((β2(t)∗αβ(x))∗α2β(z))∗α3β(y)−(αβ2(t)∗α2β(x))∗(αβ(z)∗α3(y)))⏟γ, |
Further, we have the following via calculations:
ξ=(−1)|t|(|x|+|z|){((β2(x)αβ(y))α2β(z))α3β(t)+(−1)|t|(|x|+|y|+|z|)α2β2(t)((αβ(x)α2(y))α3(z))+(−1)|z|(|x|+|y|)(αβ2(z)(αβ(x)α2(y)))α3β(t)+(−1)|z|(|x|+|y|)+|t|(|x|+|y|+|z|)α2β2(t)(α2β(z)(α2(x)α3β−1(y)))+(−1)|x||y|((β2(y)αβ(x))α2β(z))α3β(t)+(−1)|t|(|x|+|y|+|z|)+|x||y|α2β2(t)((αβ(y)α2(x))α3(z))+(−1)|z|(|x|+|y|)+|x||y|(αβ2(z)(αβ(y)α2(x)))α3β(t)+(−1)|z|(|x|+|y|)+|t|(|x|+|y|+|z|)+|x||y|α2β2(t)(α2β(z)(α2(y)α3β−1(x)))−(αβ2(x)α2β(y))(α2β(z)α3(t))−(−1)(|z|+|t|)(|x|+|y|)(αβ2(z)α2β(t))(α2β(x)α3(y))−(−1)|z||t|(αβ2(x)α2β(y))(α2β(t)α3(z))−(−1)(|z|+|t|)(|x|+|y|)+|z||t|(αβ2(t)α2β(z))(α2β(x)α3(y))−(−1)|x||y|(αβ2(y)α2β(x))(α2β(z)α3(t))−(−1)(|z|+|t|)(|x|+|y|)+|x||y|(αβ2(z)α2β(t))(α2β(y)α3(x))−(−1)|x||y|+|z||t|(αβ2(y)α2β(x))(α2β(t)α3(z))−(−1)(|z|+|t|)(|x|+|y|)+|x||y|+|z||t|(αβ2(t)α2β(z))(α2β(y)α3(x))}, |
analogously, there are concrete equations of
↺x,y,t8(−1)|t|(|x|+|z|)~asα,β(β2(x)∗αβ(y),α2β(z),α3(z))=ξ+η+γ. |
So,
↺x,y,t8(−1)|t|(|x|+|z|)~asα,β(β2(x)∗αβ(y),α2β(z),α3(z))=ξ+η+γ={(−1)|t|(|x|+|z|)~asα,β(β2(x)αβ(y),α2β(z),α3(t))−(−1)|z|(|x|+|y|)+|t||y|~asα,β(αβ2(t),α2β(z),α2(x)α3β−1(y))+(−1)|x|(|y|+|z|)~asα,β(β2(y)αβ(t),α2β(z),α3(x))−(−1)|z|(|t|+|y|)+|t||x|~asα,β(αβ2(x),α2β(z),α2(y)α3β−1(t))+(−1)|y|(|t|+|z|)~asα,β(β2(t)αβ(x),α2β(z),α3(y))−(−1)|z|(|t|+|x|)+|y||x|~asα,β(αβ2(y),α2β(z),α2(t)α3β−1(x))+{(−1)|t|(|x|+|z|)+|x||y|~asα,β(β2(y)αβ(x),α2β(z),α3(t))−(−1)|z|(|x|+|y|)+|y|(|x|+|t|)~asα,β(αβ2(t),α2β(z),α2(y)α3β−1(x))+(−1)|x|(|y|+|z|)+|y||t|~asα,β(β2(t)αβ(y),α2β(z),α3(x))−(−1)|z|(|t|+|y|)+|t|(|x|+|y|)~asα,β(αβ2(x),α2β(z),α2(t)α3β−1(y))+(−1)|y|(|t|+|z|)+|t||x|~asα,β(β2(x)αβ(t),α2β(z),α3(y))−(−1)|z|(|t|+|x|)+|x|(|t|+|y|)~asα,β(αβ2(y),α2β(z),α2(x)α3β−1(t))} | (29) |
{+(−1)|t||y|α2β2(t)((αβ(x)α2(y))α3(z))+(−1)|t|(|x|+|z|)+|z|(|x|+|y|)(αβ2(z)(αβ(x)α2(y)))α3β(t)+(−1)|y|(|t|+|x|)α2β2(t)((αβ(y)α2(x))α3(z))+(−1)|t|(|x|+|z|)+|z|(|x|+|y|)+|x||y|(αβ2(z)(αβ(y)α2(x)))α3β(t)+(−1)|x||t|α2β2(x)((αβ(y)α2(t))α3(z))+(−1)|x|(|y|+|z|)+|z|(|t|+|y|)(αβ2(z)(αβ(y)α2(t)))α3β(x)+(−1)|t|(|x|+|y|)α2β2(x)((αβ(t)α2(y))α3(z))+(−1)|x|(|y|+|z|)+|z|(|t|+|y|)+|t||y|(αβ2(z)(αβ(t)α2(y)))α3β(x)+(−1)|y||x|α2β2(y)((αβ(t)α2(x))α3(z))+(−1)|y|(|t|+|z|)+|z|(|t|+|x|)(αβ2(z)(αβ(t)α2(x)))α3β(y)+(−1)|x|(|y|+|t|)α2β2(y)((αβ(x)α2(t))α3(z))+(−1)|y|(|t|+|z|)+|z|(|t|+|x|)+|t||x|(αβ2(z)(αβ(x)α2(t)))α3β(y)−(−1)(|z|+|t|)(|x|+|y|)(αβ2(z)α2β(t))(α2β(x)α3(y))−(−1)|z||t|(αβ2(x)α2β(y))(α2β(t)α3(z))−(−1)(|z|+|t|)(|x|+|y|)+|x||y|(αβ2(z)α2β(t))(α2β(y)α3(x))−(−1)|x||y|+|z||t|(αβ2(y)α2β(x))(α2β(t)α3(z))−(−1)(|z|+|x|)(|y|+|t|)(αβ2(z)α2β(x))(α2β(y)α3(t))−(−1)|z||x|(αβ2(y)α2β(t))(α2β(x)α3(z))−(−1)(|z|+|x|)(|y|+|t|)+|t||y|(αβ2(z)α2β(x))(α2β(t)α3(y))−(−1)|t||y|+|z||x|(αβ2(t)α2β(y))(α2β(x)α3(z))−(−1)(|z|+|y|)(|x|+|t|)(αβ2(z)α2β(y))(α2β(t)α3(x))−(−1)|z||y|(αβ2(t)α2β(x))(α2β(y)α3(z))−(−1)(|z|+|y|)(|x|+|t|)+|x||t|(αβ2(z)α2β(y))(α2β(x)α3(t))−(−1)|x||t|+|z||y|(αβ2(x)α2β(t))(α2β(y)α3(z))}. | (30) |
Since
↺x,y,t8(−1)|t|(|x|+|z|)~asα,β(β2(x)∗αβ(y),α2β(z),α3(z))=(30)=(−1)|t||y|[~asα,β(β(t),αβ(x),α2(y))+(−1)|x||y|~asα,β(β(t),αβ(y),α2(x)),α3β(z)]+(−1)|x||t|[~asα,β(β(x),αβ(y),α2(t))+(−1)|t||y|~asα,β(β(x),αβ(t),α2(y)),α3β(z)]+(−1)|y||x|[~asα,β(β(y),αβ(t),α2(x))+(−1)|x||t|~asα,β(β(y),αβ(x),α2(t)),α3β(z)]=0. |
Thus, it satisfies the Bihom-Jordan super-identity for
The authors would like to thank the referee for valuable comments and suggestions on this article.
1. | Ying Hou, Liangyun Chen, Keli Zheng, Super-bimodules and $ \mathcal{O} $-operators of Bihom-Jordan superalgebras, 2024, 32, 2688-1594, 5717, 10.3934/era.2024264 |