# A Generalization forn-Cocycles

Authors: Beishang Ren; Shixun Lin
Journal: ISRN Algebra
 (2012)

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## Abstract

We will give generalized definitions called type IIn-cocycles and weak quasi-bialgebra and also show properties of type II n-cocycles and some results about weak quasi-bialgebras, for instance, construct a new structure of tensor product algebra over a module algebra on weak quasi-bialgebras.


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## Body

## 1. Introduction

We will introduce a new generalized definition called type IIn-cocycle; the namely, relax the invertible condition of associator in n-cocycle up to adding an associator satisfies all forms in definition of n-cocycle together with the original associator, and both need not to be invertible for each other; then we give examples to illustrate it clearly. Majid have shown many results about n-cocycle in [1], and we obtain some results including cohomologous concept through this new definition, main properties of type II n-cocycles, and its simple classification.It is well known that quasi-bialgebras and quasi-Hopf algebras play important roles in quantum group theory, and these concepts were introduced by Drinfel’d in [2] who relaxed the coassociative law of Δ up to conjugation. In this paper, we will show a new definition called weak quasi-bialgebra, a generalization of quasi-bialgebras, and there are simple examples to illustrate. Authors show results for weak quasi-bialgebras in place of quasi-bialgebras (cf. [1, 3]), including that there exists an algebra structure on A⊗H, a generalization of the algebra product in [3], where H is a weak quasi-bialgebra and A is a H-module algebra.We follow all the notation and conventions in [1], throughout the paper. In the following, k will be a fixed field throughout, and all algebras, coalgebras, vector spaces, and so forth are over k automatically unless specified. We recall definitions as follows.Definition 1.1.
For any bialgebra, if there is an invertible elementθ∈H⊗n such that
(1.1)∂θ=(Δ0θΔ2θ⋯Δsθ)(Δ1θΔ3θ⋯Δtθ)=1,
where integers s and t are max even number and max odd number, respectively, in {0,1,…,n}, we call θ an n-cocycle. If ϵi(θ)=1 (0⩽i⩽n), then the cocycle is counital. We define Δi=id⊗⋯⊗Δ⊗⋯⊗id, ϵi=id⊗⋯⊗ϵ⊗⋯⊗id(1⩽i⩽n),Δ0=1⊗θ, and Δn+1=θ⊗1.Definition 1.2.
LetH be a k-algebra with unit and homomorphisms Δ:H→H⊗H, ϵ:H→k. If there exists a counital 3-cocycle Φ∈H⊗3 rendering that, for all h∈H,
(1.2)(id⊗Δ)Δh=Φ(Δ⊗id)(Δh)Φ-1,(ϵ⊗id)Δh=1k⊗h,(id⊗ϵ)Δh=h⊗1k.
Then H is called a quasi-bialgebra together with coproduct Δ and counit ϵ, and call Φ an associator.
We will denote the tensor components ofΦ,ϕ with big and small letters, respectively, for instance,
(1.3)Φ=∑‍X1⊗X2⊗X3=∑‍Y1⊗Y2⊗Y3=∑‍Z1⊗Z2⊗Z3etc.,ϕ=∑‍x1⊗x2⊗x3=∑‍y1⊗y2⊗y3=∑‍z1⊗z2⊗z3etc.,
where xi is the ith factor.Definition 1.3.
LetH be a quasi-bialgebra and A a vector space. If A has a multiplication and the unit 1A obeying that
(1.4)a(bc)=∑‍(X1·a)[(X2·b)(X3·c)],h·(ab)=∑‍(h1·a)(h2·b),h·1A=ϵ(h)1A,
for any a,b,c∈A and h∈H, where ·:<H⊗A→A is the H-module structure map of A, then say A is a left H-module algebra.
In bialgebras, the composition of any two algebra homomorphismsΔi,Δj satisfies the equality ΔiΔj=Δj+1Δi for all i≤j. We will use this equation frequently.


## 2. Type II Cocycles and Weak Quasi-Bialgebras

Definition 2.1.
An associative algebraH with unit over a commutative ring R called a fake bialgebra, if there are two algebra homomorphisms Δ:H→H⊗H and ϵ:H→R.Definition 2.2.
LetH be a fake bialgebra, and elements Φ,ϕ∈H⊗n. Denote
(2.1)∂(Φ,ϕ)=(Δ0ΦΔ2Φ⋯ΔsΦ)(Δ1ϕΔ3ϕ⋯Δtϕ),χ(Φ,ϕ)=(ΔtΦΔt-2Φ⋯Δ1Φ)(ΔsϕΔs-2ϕ⋯Δ0ϕ),
where integers s and t in {0,1,…,n} are max even number and max odd number, respectively. If there exists an element ψ(Φ,ϕ)∈H⊗n+1 with ψ(Φ,ϕ)∂(Φ,ϕ)=χ(Φ,ϕ), which makes ∂(Φ,ϕ) and Φ, ϕ satisfy all equalities that one side of equalities is not a single item at least, similar to (1.1). Then call the pair (Φ,ϕ) a type II n-cocycle for H and denote it by (Φ,ϕ)nII. The cocycle (Φ,ϕ)nII is called counital, if both Φ and ϕ are counital.
There is a nature way to define type In-cocycle (Φ,ϕ)nI similarly. If we require the type II n-cocycle (Φ,ϕ)nII for H to satisfy all transformations of (1.1), but each side of formulas must have one item at least, then the type II n-cocycle (Φ,ϕ)nII is called type I n-cocycle and denoted by (Φ,ϕ)nI.
We write∂, ψ, and χ briefly for ∂(Φ,ϕ), ψ(Φ,ϕ), and χ(Φ,ϕ) without confusions, respectively. To clarify a new definition above, we give simple examples on a fake bialgebra H. In the following, we discuss type II n-cocycles (Φ,ϕ)nII only.Example 2.3.
Cocycle(Φ,ϕ)1II means both Φ and ϕ are in H, obeying that
(2.2)∂Δ1Φ=Δ0ΦΔ2Φ,Δ0ϕ∂=Δ2ΦΔ1ϕ.
And there is ψ∈H⊗2 such that
(2.3)Δ2ϕΔ0ϕ=Δ1ϕψ,Δ1ΦΔ2ϕ=ψΔ0Φ,
where ∂=Δ0ΦΔ2ΦΔ1ϕ.Example 2.4.
Cocycle(Φ,ϕ)3II and ∂=Δ0ΦΔ2ΦΔ4ΦΔ1ϕΔ3ϕ, ψ∈H⊗4, where Φ,ϕ∈H⊗3, satisfy that
(2.4)∂Δ3ΦΔ1Φ=Δ0ΦΔ2ΦΔ4Φ,∂Δ3ΦΔ1ΦΔ4ϕ=Δ0ΦΔ2Φ,∂Δ3Φ=Δ0ΦΔ2ΦΔ4ΦΔ1ϕ,(Δ0ϕ)∂Δ3ΦΔ1Φ=Δ2ΦΔ4Φ,etc.,Δ3ΦΔ1Φ=ψΔ0ΦΔ2ΦΔ4Φ,Δ3ΦΔ1ΦΔ4ϕ=ψΔ0ΦΔ2Φ,Δ3ΦΔ1ΦΔ4ϕΔ2ϕ=ψΔ0Φ,Δ1ΦΔ4ϕΔ2ϕ=(Δ3ϕ)ψΔ0Φ,etc.
Observing examples, we can see that∂ and Φ are replaced by ψ and ϕ, respectively, after moving to a corresponding place in the other side of equations, and vice versa. Obviously, n-cocycle θ must be a type II 2-cocycle (θ,θ-1)nII and ∂(θ,θ-1)=ψ(θ,θ-1)=1.Example 2.5.
LetA be an associative algebra with an idempotent q∈A over a field k. Define Δ:A→A⊗A by Δ(a)=a⊗a and ϵ:A→k by ϵ(a)=0k, for all a∈A. It is clear that A is a fake bialgebra. We set Φ=ϕ=q and ψ=q⊗q, then ∂=χ=q⊗q. It is easy to check that (Φ,ϕ) is a type II 1-cocycle.Proposition 2.6.
Let(Φ,ϕ)nII be a cocycle for a fake bialgebra H, and denote
(2.5)∂=(Δ0ΦΔ2Φ⋯ΔsΦ)(Δ1ϕΔ3ϕ⋯Δtϕ),χ=(ΔtΦΔt-2Φ⋯Δ1Φ)(ΔsϕΔs-2ϕ⋯Δ0ϕ),
where integers s and t in {0,1,…,n} are max even number and max odd number, respectively. Then one has the following.(1)
χ=ψΔ0(Φϕ)=Δt(Φϕ)ψ=(ΔtΦΔt-2Φ⋯Δ1Φ)(Δ1ϕΔ3ϕ⋯Δtϕ)=χ∂=χ2, and ∂=Δ0(Φϕ)∂=∂Δt(Φϕ)=(Δ0ΦΔ2Φ⋯ΔsΦ)(ΔsϕΔs-2ϕ⋯Δ0ϕ)=∂χ=∂2.(2)
If∂ is commutative with χ, then χ=∂. Especially, if either ∂ or χ is zero, then the other one is zero too. On the other hand, if either of elements ∂ and χ is not zero, then the rest elements in set {∂,χ,ψ} are not zero.(3)
IfΔ0(Φϕ)-1 is a left unit and Φ is not a right zero divisor, then ∂=χ=ψ=0.(4)
IfΦ has a right inverse ΦR-1, so do ∂, ψ, and χ. Similarly, if ϕ has a left inverse ϕL-1, so do ∂, ψ, and χ.Proof.
(
1
) We obtain that χ=Δt(Φϕ)ψ by
(2.6)(Δtϕ)ψ=Δt-2Φ⋯Δ3ΦΔ1ΦΔsϕ⋯Δ2ϕΔ0ϕ
and χ=ψΔ0(Φϕ) by
(2.7)ψΔ0Φ=ΔtΦ⋯Δ3ΦΔ1ΦΔsϕ⋯Δ4ϕΔ2ϕ,
since Δi is a homomorphism. And
(2.8)χ=ψ∂=(ψΔ0Φ⋯ΔsΦ)Δ1ϕ⋯Δtϕ=(ΔtΦ⋯Δ1Φ)(Δ1ϕ⋯Δtϕ).
Analogously, we have that
(2.9)Δ0(Φϕ)∂=Δ0Φ((Δ0ϕ)∂)=Δ0Φ(Δ2Φ⋯ΔsΦΔ1ϕ⋯Δtϕ)=∂,∂Δt(Φϕ)=(∂ΔtΦ)Δtϕ=(Δ0Φ⋯ΔsΦΔ1ϕ⋯Δt-2ϕ)Δtϕ=∂.
Then, we get easily that χ=ψ∂=ψΔ0(Φϕ)∂=χ∂ and
(2.10)∂χ=(∂ΔtΦΔt-2Φ⋯Δ1Φ)Δ1ϕ⋯Δt-2ϕΔtϕ=Δ0ΦΔ2Φ⋯ΔsΦΔ1ϕ⋯Δt-2ϕΔtϕ=∂.
Finally, there is the equality that
(2.11)∂=∂χ=(∂ΔtΦΔt-2Φ⋯Δ1Φ)(ΔsϕΔs-2ϕ⋯Δ0ϕ)=(Δ0ΦΔ2Φ⋯ΔsΦ)(ΔsϕΔs-2ϕ⋯Δ0ϕ).
We, last, compute that
(2.12)∂2=(∂χ)∂=∂(χ∂)=∂χ=∂,χ2=(χ∂)χ=χ(∂χ)=χ∂=χ.
Therefore ∂ and χ are idempotent.
(
2
) Obviously, we get this by statement (1). Let ∂≠0 and assume χ=0, from the previous part, that yields to ∂=0 contradicting ∂≠0. Therefore χ must be zero, and ψ=0 for the same reason.
(
3
) The equality (Δ0(Φϕ)-1)∂=0 suggests that ∂=0, and then χ=0. It is clear that Δ0Φ⋯ΔsΦ=0 because ∂ΔtΦ⋯Δ1Φ=Δ0Φ⋯ΔsΦ. In addition, ΔtΦ⋯Δ3ΦΔ1Φ=ψΔ0ΦΔ2Φ⋯ΔsΦ=0 implies that
(2.13)ψΔ0Φ=ΔtΦ⋯Δ3ΦΔ1ΦΔsϕ⋯Δ4ϕΔ2ϕ=0.
As a result, we have ψ=0 if Φ is not a right zero divisor.
(
4
) It is easy to obtain that ∂ΔtΦ⋯Δ3ΦΔ1ΦΔsΦR-1⋯Δ2ΦR-1Δ0ΦR-1=1 and ψΔ0Φ⋯ΔsΦΔ1ΦR-1⋯ΔtΦR-1=1 as ∂ΔtΦ⋯Δ1Φ=Δ0Φ⋯ΔsΦ and ψΔ0Φ⋯ΔsΦ=ΔtΦ⋯Δ1Φ, respectively. But then χ=ψ∂ and χ has a right inverse. Likewise, we can prove the rest part.Furthermore, if∂ or χ has a one-side inverse, it makes sense that ∂=χ=ψ=1 since both ∂ and χ are idempotent. We also have that Φϕ=1 which indicates Φ is a left unit and ϕ a right unit, by Δ0(Φϕ)∂=∂ if ∂=1. Hence ∂ and χ cannot be anything but the identity element if one of them is a one-side unit.Corollary 2.7.
The following statements are equivalent.(1)
Φ  has a right inverse.(2)
∂=χ=ψ=1.(3)
ϕ  has a left inverse.(4)
∂  is a one-side unit.(5)
χ  is a one-side unit.(6)
ψ  has a left inverse.Proof.
(Sketch of Proof).
Check by(1)⇒(2)⇒(3)⇒(4)⇒(5)⇒(6)⇒(1).Equality(Δ0(Φϕ)-1)∂=0 suggests that classification of ∂ is divided into three types. The first type ∂=0, and the second ∂=1 if Δ0(Φϕ)=1, that is, Φϕ=1. The last one is that ∂ is a right zero divisor.Example 2.8.
In algebraℤ6 over the integer ring ℤ, we define Δ:ℤ6→ℤ6⊗ℤ6 given by Δ3-=3-⊗3- and Δx-=x-⊗3- for any x-∈ℤ6-{3-}, and ϵ:ℤ6→ℤ by ϵ(y-)=y for all y-∈ℤ6 such that (ℤ6,Δ,ϵ) is a fake bialgebra. Set Φ=ϕ=3-∈ℤ6 such that ∂=3-⊗3-. The product of any two elements in the set {ΔiΦ,∂}(i=0,1,2) equals 3-⊗3-, obviously. We also set ψ=∂; then, it is easy to prove that (Φ,ϕ)1II is a cocycle and a right zero-divisor ∂.Proposition 2.9.
LetH be a fake bialgebra with counital law of ϵ. If Φ,ϕ∈H and ϵ(Φ)=1 (ϵ(ϕ)=1, resp.), then ∂(Φ,ϕ) (χ(Φ,ϕ), resp.) is counital if and only if Φϕ=1 (ϕΦ=1, resp.).Proof..
Since that∂(Φ,ϕ)=Δ0ΦΔ2ΦΔ1ϕ=(Φ⊗Φ)Δϕ, we have ϵi∂(Φ,ϕ)=ϵi(Φ⊗Φ)ϵiΔϕ=ϵ(Φ)Φϕ=Φϕ rendering that ϵi∂(Φ,ϕ)=1 if and only if Φϕ=1, where i=1,2.Proposition 2.10.
LetH be a fake bialgebra with coassociative law of Δ and there are elements Φ,ϕ in H. If ∂(Φ,ϕ) obeys ∂Δ1Φ=Δ0ΦΔ2Φ, then Δ3∂Δ1∂=Δ0∂Δ2∂. Especially, ∂ is a 2-cocycle if ∂ is invertible if H is a bialgebra.Proof..
To obtain the result, we observe that(2.14)Δ3∂Δ1∂=(∂⊗1)(Δ⊗id)∂=(∂⊗1)(Δ⊗id)((Φ⊗Φ)Δ1ϕ)=(∂⊗1)(ΔΦ⊗Φ)(Δ1Δ1ϕ)=(∂⊗1)(ΔΦ⊗Φ)(Δ2Δ1ϕ)=(∂ΔΦ⊗Φ)(Δ2Δ1ϕ)=(Φ⊗Φ⊗Φ)Δ2Δ1ϕ=(1⊗∂)(Φ⊗ΔΦ)Δ2Δ1ϕ=(1⊗∂)Δ2(Φ⊗Φ)Δ2Δ1ϕ=Δ0∂Δ2∂.We have known thatθγ=(γ⊗γ)θΔγ-1 is cohomologous to θ for a bialgebra H if θ is a counital 2-cocycle, which was mentioned by Majid in [1]. Let H be a bialgebra, Φ,ϕ∈H and cocycle (σ,δ)2II for H. Denote that σ(Φ,ϕ)=Δ0ΦΔ2ΦσδσΔ1ϕ and δ(Φ,ϕ)=Δ1ΦδσδΔ2ϕΔ0ϕ. Then we have the following.Proposition 2.11.
If equality∂(σ,δ)Δ2δΔ0δΔ3σ=Δ1δ holds and ∂(σ,δ) is a commutative element in set {Δ1σ,Δ3δ,Δ3σ}, and 1⊗Δ(ϕΦ)(=Δ(ϕΦ)⊗1) commutes with any element in set {Δ1σ,Δ1δ,Δ2σ,Δ2δ} as well, then
(2.15)Δ0σ(Φ,ϕ)Δ2σ(Φ,ϕ)=Δ3σ(Φ,ϕ)Δ1σ(Φ,ϕ).Proof.
A long equality showed that(2.16)Δ0σ(Φ,ϕ)Δ2σ(Φ,ϕ)=(1⊗(Φ⊗Φ)σδσΔϕ)(id⊗Δ)(σδσΔϕ)=(Φ⊗Φ⊗Φ)Δ0(σδσ)Δ0Δ1ϕΔ2Δ0ΦΔ2(σδσ)Δ2Δ1ϕ=(Φ⊗Φ⊗Φ)Δ0(σδσ)Δ2Δ0(ϕΦ)Δ2(σδσ)Δ1Δ1ϕ=(Φ⊗Φ⊗Φ)Δ0(σδσ)Δ2(σδσ)Δ1Δ2(ϕΦ)Δ1Δ1ϕ=(Φ⊗Φ⊗Φ)Δ0(σδ)Δ0σΔ2σΔ2δΔ2σΔ1Δ2(ϕΦ)Δ1Δ1ϕ=(Φ⊗Φ⊗Φ)Δ0(σδ)∂(σ,δ)Δ3σΔ1σΔ2δΔ2σΔ1Δ2(ϕΦ)Δ1Δ1ϕ=(Φ⊗Φ⊗Φ)∂(σ,δ)Δ3σΔ3δψ(σ,δ)Δ0σΔ2σΔ1Δ2(ϕΦ)Δ1Δ1ϕ=(Φ⊗Φ⊗Φ)Δ3(σδ)∂(σ,δ)ψ(σ,δ)∂(σ,δ)Δ3σΔ1σΔ1Δ2(ϕΦ)Δ1Δ1ϕ=(Φ⊗Φ⊗Φ)Δ3(σδ)∂(σ,δ)Δ3σΔ1σΔ2δΔ0δΔ3σΔ1σΔ1Δ2(ϕΦ)Δ1Δ1ϕ=(Φ⊗Φ⊗Φ)Δ3(σδσ)Δ1σ∂(σ,δ)Δ2δΔ0δΔ3σΔ1σΔ1Δ2(ϕΦ)Δ1Δ1ϕ=(Φ⊗Φ⊗Φ)Δ3(σδσ)Δ1(σδσ)Δ1Δ2(ϕΦ)Δ1Δ1ϕ=(Φ⊗Φ⊗Φ)(σδσ⊗1)(Δ(ϕΦ)⊗1)(Δ⊗id)(σδσΔϕ)=((Φ⊗Φ)σδσΔϕ⊗1)(Δ⊗id)((Φ⊗Φ)σδσΔϕ)=Δ3σ(Φ,ϕ)Δ1σ(Φ,ϕ).There exists a similar version forδ(Φ,ϕ), namely, the following preposition.Proposition 2.12.
If there is the equationΔ1δΔ3δψ(σ,δ)Δ0σ=Δ2δ and ∂(σ,δ) commutes with any element in set {Δ0σ,Δ0δ}, and 1⊗Δ(ϕΦ)(=Δ(ϕΦ)⊗1) is a commutative element in set {Δ0σ,Δ0δ,Δ3σ,Δ3δ} as well, then
(2.17)Δ1δ(Φ,ϕ)Δ3δ(Φ,ϕ)=Δ0δ(Φ,ϕ)Δ2δ(Φ,ϕ).Proposition 2.13.
LetH be a bialgebra and (Φ,ϕ)2II a counital cocycle for H, and define Δ(Φ,ϕ)(h)=ΦϕΦΔ(h)ϕΦϕ for all h∈H, then the algebra H with original ϵ and Δ(Φ,ϕ) consists a new coalgebra if χΔ3ΦΔ1Φ=Δ0ΦΔ2Φ. Moreover, Δ(Φ,ϕ) is an algebra map if (ϕΦ)3=1, then algebra H is a bialgebra with comultiplication Δ(Φ,ϕ).Proof.
It is clear that(ϵ⊗id)Δ(Φ,ϕ)(h)=(id⊗ϵ)Δ(Φ,ϕ)(h)=h. So we only need to show the coassociative law of Δ(Φ,ϕ). For all h∈H, we obtain
(2.18)(Δ(Φ,ϕ)⊗id)Δ(Φ,ϕ)(h)=(Δ(Φ,ϕ)⊗id)(ΦϕΦΔ(h)ϕΦϕ)=Δ3(ΦϕΦ)Δ1(ΦϕΦ)Δ1Δ1(h)Δ1(ϕΦϕ)Δ3(ϕΦϕ)=Δ3(Φϕ)ψΔ0ΦΔ2ΦΔ1ϕΔ1ΦΔ2Δ1(h)Δ1ϕΔ1ΦΔ2ϕΔ0ϕ∂Δ3(Φϕ)=χ∂Δ3ΦΔ1ΦΔ2Δ1(h)Δ1ϕΔ3ϕψ∂=χΔ3ΦΔ1ΦΔ2Δ1(h)Δ1ϕΔ3ϕχ=∂Δ3ΦΔ1ΦΔ2Δ1(h)Δ2ϕΔ0ϕ∂χ=∂ψΔ0ΦΔ2ΦΔ2Δ1(h)Δ2ϕΔ2ΦΔ1ϕΔ3ϕχ=Δ0(Φϕ)∂Δ3ΦΔ1ΦΔ2ΦΔ2Δ1(h)Δ2ϕΔ2ΦΔ1ϕΔ3ϕψΔ0(Φϕ)=Δ0(Φϕ)Δ0ΦΔ2ΦΔ2(ϕΦ)Δ2Δ1(h)Δ2(ϕΦ)Δ2ϕΔ0ϕΔ0(Φϕ)=Δ0(ΦϕΦ)Δ2(ΦϕΦ)Δ2Δ1(h)Δ2(ϕΦϕ)Δ0(ϕΦϕ)=Δ0(ΦϕΦ)(id⊗Δ)(ΦϕΦΔ(h)ϕΦϕ)Δ0(ϕΦϕ)=(id⊗Δ(Φ,ϕ))Δ(Φ,ϕ)(h).
Finally, for any g∈H,
(2.19)Δ(Φ,ϕ)(hg)=ΦϕΦΔ(hg)ϕΦϕ=ΦϕΦΔ(h)ϕΦϕΦϕΦΔ(g)ϕΦϕ=Δ(Φ,ϕ)(h)Δ(Φ,ϕ)(g).Definition 2.14.
Let(H,Δ,ϵ) be a fake bialgebra. If there exists a cocycle (Φ,ϕ)3II for H obeying that
(2.20)(id⊗Δ)Δ(h)Φ=Φ(Δ⊗id)Δ(h),ϕ(id⊗Δ)Δ(h)=(Δ⊗id)Δ(h)ϕ,(id⊗ϵ)Δ(h)=(ϵ⊗id)Δ(h)=h,
for all h∈H, then H is called a weak quasi-bialgebra.Example 2.15.
LetH be an associate algebra over field k, where the characteristic of k is not 2. And H is a 4-dimensional vector space with basis {1,i,j,ij} obeying that i2=i,j2=j, and ij=ji. We define homomorphisms Δ:H→H⊗H given by Δ(i)=i⊗j,Δ(j)=j⊗1 and ϵ:H→k given by ϵ(i)=ϵ(j)=0. Obviously, H is a fake bialgebra. Set Φ=j⊗j⊗j and ψ=j⊗1⊗1⊗j, then (Φ,Φ)3II is a cocycle with holding ∂=χ=j⊗j⊗j⊗j. It is routine to check (H,Δ,ϵ,(Φ,Φ)3II) is a weak quasi-bialgebra.We relax Definition1.3 by setting that H is a weak quasi-bialgebra so that we can define an algebra structure on A⊗H, if A is a left H-module algebra and H a weak quasi-bialgebra, given by
(2.21)(a#h)(b#g)=∑‍(y1X1x1·a)(y2X2x2h1·b)#y3X3x3h2g
for all a,b∈A,h,g∈H, while a#h is equal to a⊗h here.Theorem 2.16.
LetH be a weak quasi-bialgebra and A a left H-module algebra. Then A#H is an associative algebra under the multiplication mentioned above and 1A#1H is the unit.Proof..
For alla,b,andc∈A and h,g,andl∈H, we easily get that, by properties of ϵ,
(2.22)(1#1)(a#h)=(y1X1x1·1)(y2X2x2·a)#y3X3x3h=ϵ(y1X1x1)(y2X2x2·a)#y3X3x3h=a#h,(a#h)(1#1)=(y1X1x1·a)(y2X2x2h1·1)#y3X3x3h2=(y1X1x1·a)ϵ(y2X2x2h1)#y3X3x3h2=a#h.
Now we show the associative law:
(2.23)[(a#h)(b#g)](c#l)=[(y1X1x1·a)(y2X2x2h1·b)#y3X3x3h2g](c#l)=(w1Y1z1·(y1X1x1·a)(y2X2x2h1·b))×(w2Y2z2(y3X3x3h2g)1·c)#w3Y3z3(y3X3x3h2g)2l(2.24)=((w11Y11z11y1X1x1·a)(w12Y12z12y2X2x2h1·b))×(w2Y2z2y31X31x31h21g1·c)#w3Y3z3y32X32x32h22g2l.
But
(2.25)Δ1ϕΔ1ΦΔ1ϕΔ3ϕΔ3ΦΔ3ϕ=Δ1ϕΔ1ΦΔ4ϕΔ2ϕ(Δ0ϕ)∂Δ3ΦΔ3ϕ=Δ1ϕΔ1ΦΔ4ϕΔ2ϕ(Δ0ϕ)∂=Δ1ϕ(Δ3ϕ)ψ∂,
and then we obtain that (2.23) is
(2.26)((z11x1ψ1∂1·a)(z12x2ψ2∂2h1·b))(z2x31ψ3∂3h21g1·c)#z3x32ψ4∂4h22g2l=(Y1z11x1ψ1∂1·a)((Y2z12x2ψ2∂2h1·b)(Y3z2x31ψ3∂3h21g1·c))#z3x32ψ4∂4h22g2l.
On the other hand, the equation
(2.27)Δ4ΦΔ1ϕ(Δ3ϕ)ψ∂=Δ2ϕ(Δ0ϕ)∂ψ∂=Δ2ϕ(Δ0ϕ)∂χ=Δ2ϕ(Δ0ϕ)∂ψΔ0ΦΔ0ϕ=Δ2ϕΔ2ΦΔ4ΦΔ1ϕ(Δ3ϕ)ψΔ0ΦΔ0ϕ=Δ2ϕΔ2ΦΔ2ϕΔ0ϕΔ0ΦΔ0ϕ
makes (2.26) equal
(2.28)(w1Y1z1·a)((w21Y21z21y1X1x1h1·b)(w22Y22z22y2X2x2h21g1·c))#w3Y3z3y3X3x3h22g2l=(w1Y1z1·a)((w21Y21z21h11y1X1x1·b)(w22Y22z22h12y2X2x2g1·c))#w3Y3z3h2y3X3x3g2l=(w1Y1z1·a)(w2Y2z2h1·((y1X1x1·b)(y2X2x2g1·c)))#w3Y3z3h2y3X3x3g2l=(a#h)[(b#g)(c#l)].
Hence, [(a#h)(b#g)](c#l)=(a#h)[(b#g)(c#l)].Ifϕ is an inverse of Φ, then the multiplication becomes that
(2.29)(a#h)(b#g)=∑‍(x1·a)(x2h1·b)#x3h2g,
which is as exact as the one in [3].



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*Source: 596741-2012-12-30.xml*