Chen Xiuli Li Fang Chen Jianlong

(1Department of Mathematics, Southeast University, Nanjing 211189, China)(2Department of Mathematics, Zhejiang University, Hangzhou 310028, China)

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Hom-dimodules and FRT theorem of Hom type

Chen Xiuli1Li Fang2Chen Jianlong1

(1Department of Mathematics, Southeast University, Nanjing 211189, China)(2Department of Mathematics, Zhejiang University, Hangzhou 310028, China)

In order to study the deformation of algebras, the notions of Hom-algebras are introduced. The Hom-algebra is a generalization of the classical associative algebra. First, the Hom-type generalization of dimodules, which is called the Hom-dimodule, is introduced, and its properties are discussed Moreover, the category of Hom-dimodules in connection with the Hom D-equationR12R23=R23R12forR∈Endk(M?M)and a Hom-moduleMis investigated. Some solutions of the Hom D-equation from Hom-dimodules over Hom-bialgebras are given, and the FRT-type theorem is constructed in the category of Hom-dimodules. The results generalize and improve the FRT-type theorem in the category of dimodules.

Hom-bialgebra; Hom-dimodule; Hom D-equation

Throughout this paper, all the modules are left modules without specification and all the spaces arek-spaces for a fixed fieldk.

A (long) dimodule over a bialgebraHis a vector spaceMwith a leftH-actionρMand a rightH-coactionρMso that the following compatibility condition holds: (hm)(0)?(hm)(1)=∑hm(0)?m(1)for allh∈Handm∈M. We denote this category withHLH, which is also a special case of the Doi-Hopf module category. This categoryHLHwas first defined by Long[1]for a commutative and cocommutativeHand was studied in connection with the construction of the Brauer group of anH-dimodule algebra. It is interesting to note that for a commutative and cocommutativeH,HYDH(the category of Yetter-Drinfel’d modules) is preciselyHLH[2]. Naturally, for an arbitraryH,HYDHandHLHare fundamentally different.HYDHplays a determinant role in describing the solutions of the quantum Yang-Baxter equation. It is natural to ask which equation will play a key role inHLH. In Ref.[3], considering thatHYDHis deeply involved in solving the quantum Yang-Baxter equation, the author studiedHLHin connection with the D-equation which is described presently. Given a vector spaceM, andR∈Endk(M?M),Ris said to be a solution of the D-equation ifR12R23=R23R12in Endk(M?M?M) whereR12=R?id, andR23=id?R.

The concept of a Hom-algebra was introduced by Larsson et al[4-5]. It is a special class in the deformation of algebras. Recently, the Hom-algebra has been studied by several authors. The further development of the Hom-algebra theory led Makhlouf et al.[6-8] consequently to Hom-associative algebra, Hom-coassociative coalgebra and Hom-bialgebra, and described many of the extending properties of classical associative algebras, coalgebras, bialgebras and Hopf algebra structures. In fact, the notion of Hom-associative algebras generalize associative algebras to a situation, where associativity law is twisted by a linear map. From Ref.[5], it is clear that the commutator bracket multiplication is defined using the multiplication in Hom-associative algebra and it leads naturally to Hom-Lie algebras. Following the patterns of Hom-Lie and Hom-associative algebras, one can define Hom-bialgebras as nonassociative and non-coassociative[9]. It is a generalization of the bialgebra in which the non(co)associativity is controlled by the twisted map. In this paper, we only investigate the case of Hom-associative algebras. Hom-versions of the Yang-Baxter equation was studied in Refs.[9-12]. Many classes of the solutions for the Hom-Yang-Baxter equation are constructed. Moreover, Hom-type generalizations of FRT quantum groups, including quantum matrices and related quantum groups, are obtained in Ref.[13]. The above motivates us to study the generalization of the D-equation.

In this paper, we mainly study Hom-versions of (Long) dimodules and D-equations of the Hom type, and investigate the category of Hom-dimodules in connection with the Hom D-equation.

1 Preliminaries

In this section, we recall the definition of a Hom-bialgebra and (comodules)modules over an Hom-algebra(coalgebra).

A Hom-module[14]is a pair (V,α) in whichVis a vector space andα:V→Vis a linear map. A morphism (V,α)→(V′,α′) of Hom-modules is a linear mapf:V→V′ such thatα′°f=f°α. The tensor product of the Hom-modules (V,αv)and (W,αw) consists of the vector spaceV?Wand the linear self-mapαV?αW.

Definition 1 A Hom-associative algebra[5]is a triple(A,μ,α)in which (A,α)is a Hom-module andμ:A?A→Ais a bilinear map such that 1)α°μ=μ° (α?α)(multiplicativity); and 2)μ° (α?μ)=μ° (μ?α)(Hom-associativity). In the following, we also writeμ(a?b)asab.

The Hom-associative algebra is said to be unital[7]if a homomorphism exists,η:k→Asatisfyingμ° (η?id)=idandμ° (id?η)=id.

A Hom-coassociative coalgebra[6-7]is a triple(C,Δ,α)in which (C,α) is a Hom-module andΔ:C→C?Cis a linear map such that 1)α?2°Δ=Δ°α(comultiplicativity); and 2) (α?Δ)°Δ=(Δ?α)°Δ(Hom-coassociativity).

A Hom-coassociative coalgebra is said to be counital[7]if there is a mapε:C→ksatisfying (ε?id)°Δ=idand (id?ε)°Δ=id.

Example 1(classical structures) 1) If (A,μ) is an associative algebra andα:A→Ais an algebra morphism, thenAα=(A,μα,α) is a Hom-associative algebra with the twisted multiplicationμα=α°μ. Indeed, the Hom-associativity axiomμα° (α?μα)=μα° (μα?α) is equal toα2when applied to the associativity axiom ofμ. Likewise, both sides of the multiplicativity axiomα°μα=μα?(α?α)are equal toα2°μ.

2) Dually, if (C,Δ) is a coassociative coalgebra andα:C→Cis a coalgebra morphism, thenCα=(C,Δα,α) is a Hom-coassociative coalgebra with the twisted comultiplicationΔα=Δ°α.

3) A bialgebra is an exact Hom-bialgebra withα=id. More generally, combining the previous two cases, if (A,μ,Δ) is a bialgebra andα:A→Ais a bialgebra morphism, thenAα=(A,μα,Δα,α) is a Hom-bialgebra.

Definition 2 1) Let (A,μA,αA) be a Hom-associative algebra, and (M,αM) be a Hom-module. A leftA-module structure onM[12]consists of a morphismρ:A?M→Mof Hom-modules such thatρ° (αA?ρ)=ρ° (μA?αM) andαM°ρM=ρM° (αH?αM) (*). We also writeρ(a?m) asamfora∈Aandm∈M. In this notation, (*) can be written asαA(a)(bm)=(ab)αM(M) andαM(ax)=αA(a)αM(x).

2) Dually, let (C,Δ,αC) be a Hom-coassociative coalgebra,AleftC-comodule structure onM[14]consists of a Hom-module (M,αM) together with a linear mapρM:M→C?Msuch that (Δ?αM)°ρM=(αC?ρM)°ρMand (αC?αM)°ρM=ρM°αM. Similarly, we can define a rightC-comodule.

IfMandNareA-modules, then a morphism ofA-modulesf:M→Nis a morphism of the Hom-modules such thatf°ρM=ρN° (idA?f). Similarly, morphisms ofC-comodule are defined in a clear way.

2 Hom-Dimodules

Definition 3 Let (H,μH,ΔH,αH) be a Hom-bialgebra, and (M,αM)be a Hom-module. A Hom-dimodule is a triple (M,ρM,ρM) such that 1) (M,ρM) is a leftH-module; 2) (M,ρM) is a rightH-comodule; 3)ρM°ρM=(ρM?αM)° (αH?ρM).

We may also write 3) as (hm)(0)?(hm)(1)=∑αH(h)m(0)?αH(m(1)).

The category of Hom-dimodules over a Hom-bialgebraHwithH-module morphisms andH-comodule morphisms are denoted byHHLH.

IfHis a bialgebra, i.e.,αH=idHandαM=idM, then the above definition of the Hom-dimodule coincides with the usual definition of the dimodule. The following result proves that dimodules deform into Hom-dimodules by endomorphisms.

Proposition 1 Let (H,μH,ΔH) be a bialgebra, and (M,ρM,ρM) be anH-dimodule. Assume thatαH:H→His a bialgebra morphism, andαM:M→Mis ak-linear map such that

αM°ρM=ρM° (αH?αM)

(1)

ρM°αM=(αH?αM)°ρM

(2)

1) (H,μα,Δα,αH) is a Hom-bialgebra, whereμα=αH°μHandΔα=ΔH°αH;

2) (M,ρα,M,ρα,M) is a Hom-dimodule over a Hom-bialgebraH, whereHis a Hom-bialgebra as 1).

Proof 1) It follows from 1) of Example 1.

2) First, by Definition 1 and Eq.(2), it is easy to prove thatρα,Mgives (M,αM) the structure of a leftH-module, i.e.,ρα,M° (αH?ρα,M)=ρα,M° (μH?αM).

Similarly, by Definition 2, we can prove thatρα,Mgives (M,αM) the structure of a rightH-comodule if and only if (ρα,M?αH)°ρα,M=(αM?ΔH)°ρα,M.

Finally, it is only needed to check thatρα,M°ρα,M=(ρα,M?αH)° (αH?ρα,M). In fact, we have

ρα,M°ρα,M=(ρM°αM)° (αM°ρM)=

(αM?αH)°ρM°ρM° (αH?αM)=

(αM?αH)° (ρM?idH)° (idH?ρM)° (αH?αM)=

(ρα,M?αH)° (αH?ρα,M)

where the second equality holds by Eqs.(1) and (2). BecauseMis aH-dimodule, the third equality holds.

Proposition 2 Let (H,μH,ΔH,αH)be a Hom-bialgebra, (M,αM) and (N,αN) be Hom-dimodules. ThenM?Nis a Hom-dimodule with the action and coaction, for allm∈M,n∈N,h∈H,

ρM?N(h?m?n)=αH(h(1))·m?αH(h(2))·n

ρM?N(m?n)=m(0)?n(0)?αH(m(1)n(1))

Proof We check thatρM?NandρM?Nis well defined. For anyh,l∈H,m∈Mandn∈N,

ρM?N° (αH?ρM?N)(h?l?m?n)=

ρM?N(αH(h)?αH(l(1))·m?αH(l(2))·n)=

αH((αH(h)(1))αH(l(1))·m)?αH(αH(h))(2))(αH(l(2))·n)=

(αH(h(1))αH(l(1))·αM(m)?(αH(h(2))αH(l(2)))αN(n)=

aH((hl)(1))αM(m)?αH((hl)(2))αN(n)=

ρM?N(hl?αM(m)?αN(n))=

pM?N° (μH?αM?N)(h?l?m?n)

ρM?N°ρM?N(h?m?n)=ρM?N(αH(h(1))·m?αH(h(2))·n)=

(αH(h(1))·m)(0)?(αH(h(2))·n)(0))?αH((αH(h(1))·m)(1)

αH(αH(m(1))αH(n(1)))=(ρM?N?αH)(αH(h)?m(0)?

n(0)?αH(m(1))αH(n(1)))=(ρM?N?αH)°

(αH?ρM?N)(h?m?n)

3 Solutions of Hom D-Equation

First we introduce the definition of the Hom D-equation. Moreover, we construct many solutions to the Hom D-equation by Hom-dimodules over the Hom-bialgebra.

The D-equation[3]is a special case of the Hom D-equation whenα=id.

Proposition 3 LetBbe a solution of the Hom D-equation for the Hom-module (M,αM).

1) Ifλ∈k, thenλBis also a solution of the Hom D-equation for (M,αM).

Proof 1) First,λB° (αM?αM)=λ(B° (αM?αM))=λ((αM?αM)°B)=(αM?αM)°λB, andλB?α=λ(B?α),α?λB=λ(α?B), so we have (λB)12(λB)23=(λB)23(λB)12.

Next, we will show that a Hom-bialgebra gives rise to many solutions of the Hom D-equation by its Hom-dimodules. The following shows thatHHLHplays a role in solving the Hom D-equation.

Proposition 4 Let (H,μH,ΔH,αH) be a Hom-bialgebra, and (M,αM) be a Hom-module. Assume that (M,ρα,M,ρα,M) is Hom-dimodule andαMis anH-module morphism. Then the map

B′:M?M→M?M,m?n|→αH(n(1))·m?αM(n(0))

(3)

is a solution of the Hom D-equation.

Proof First, it is easy to check thatB′° (αM?αM)(m?n)=(αM?αM)°B′(m?n).

Forl,m,n∈M, we have

B′12B′23(l?m?n)=(B′?αM)(αM?B′)(l?m?n)=

(B′?αM)(αM(l)?αH(n(1))·m?αM(n(0)))=

αH((αH(n(1))·m)(1))·αM(l)?αM((αH(n(1))·m)(0))?

B′23B′12(l?m?n)

The fourth equality holds because (M,αM) is a Hom-dimodule. SinceαMisH-module morphism, the fifth equality is true. By 2) of Definition 2, we haveαM(m(0))?αH(m(1))=(αM(m))(0)?(αM(m))(1), so the sixth equality holds by (3). Therefore,B′12B′23=B′23B′12, i.e.,B′ is a solution of the Hom D-equation.

Corollary 1 Let (H,μH,ΔH) be a bialgebra, andαH:H→Hbe a bialgebra morphism. Assume that (M,ρM,ρM) is a dimodule andαM:M→Mak-linear map such thatαM°ρM=ρM° (αH?αM), andρM°αM=(αM?αH)°ρM. IfαMisH-module morphism. Then the map (3) is a solution of the Hom D-equation.

In Corollary 1, ifαH=id:H→H, then the conditionsαM°ρM=ρM° (id?αM) andρM°αM=(αM?idH)°ρMmean thatαMisH-module morphism andH-comodule a morphism. So we have the following corollary.

Corollary 2 Let (H,μH,ΔH) be a bialgebra. Assume that (M,ρM,ρM) is a dimodule andαM:M→MisH-module morphism andH-comodule morphism. Then the map (3) is a solution of the Hom D-equation.

4 FRT Theorem of Hom Type

In Ref.[7], the authors give a FRT type theorem: ifMis finite dimensional, then any solutionRof the D-equation has the formR=R(M,ρM,ρM), where (M,ρM,ρM) is aD(R)-dimodule over a bialgebraD(R) andR(M,ρM,ρM)is the special mapR(M,ρM,ρM)(m?n):=∑n(1)m?n(0).

Next, we recall the details of the FRT type construction from Theorem 3.6 in Ref.[3].

(4)

for allu,v=1,2,…,n.

LetT(C)be an algebra generated by (Cij). Define the bialgebra

(5)

Lemma 1 LetD(R) be the bialgebra withR:M?M→M?Ma solution of theD-equation, and assume that (M,ρM,ρM) has the structure of the object inD(R)LD(R), whereR=R(M,ρM,ρM). Letλi∈kbe invertible scalars such that

(6)

In fact,

The third equality is right by (6), the fourth equality follows from (5), and the fifth equality also follows from (6).

Lemma 2 The linear mapαM:M→Mcan be defined asαM(mi)=λimifor alli. Then we have

αM°ρM=ρM° (α?αM)

(αM?α)°ρM=ρM°αM

(αM?αM)°R=R° (αM?αM)

whereρM:D(R)?M→Mis the leftD(R)-module structure map andρM:M→M?D(R) is the rightD(R)-comdule structure map.

Proof First, sinceΔis an algebra morphism, we have

This shows thatαis a bialgebra morphism. Next,

αM°ρM(Cju?mv)=αM(Cju·mv)=

and

ρM° (α?αM)(Cju?mv)=α(Cju)·αM(mv)=

soαM°ρM=ρM° (α?αM).

Since

(αM?α)°ρM=ρM°αMholds.

Finally,

So (αM?αM)°R=R° (αM?αM) is true.

Proof It follows with Proposition 1, Lemma 1 and Lemma 2.

Therefore, by Theorem 1, we obtain the Hom-bialgebraD(R)α=(D(R),μα=α°μ,Δα=Δ°α) such that (M,ρα,M=αM°ρM,ρα,M=αM°ρM) is a HomD(R)-dimodule.

Remark 2 Example 2 also gives an example of Hom-dimodules.

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Hom-dimodules與Hom型的FRT定理

陳秀麗1李 方2陳建龍1

(1東南大學(xué)數(shù)學(xué)系, 南京 211189)(2浙江大學(xué)數(shù)學(xué)系, 杭州 310028)

為了研究代數(shù)形變理論,引入了Hom-代數(shù)的概念．事實上Hom-代數(shù)是經(jīng)典結(jié)合代數(shù)的推廣．首先介紹了dimodule的Hom型推廣,即Hom-dimodule,并對其相關(guān)性質(zhì)進(jìn)行討論．進(jìn)一步研究了Hom-dimodule范疇與Hom D-方程R12R23=R23R12的關(guān)系,其中R∈Endk(M?M)且M為Hom模．針對Hom雙代數(shù)上的Hom-dimodule給出了Hom D-方程的一些解,并在Hom-dimodules范疇中構(gòu)造FRT-型定理．這些結(jié)果推廣并改進(jìn)了dimodule范疇中的FRT-型定理．

Hom雙代數(shù); Hom-dimodule; Hom D-方程

O153

s：The National Natural Science Foundation of China (No.11371089), the Specialized Research Fund for the Doctoral Program of Higher Education (No.20120092110020), the Postdoctoral Innovation Funds of Southeast University (No.3207013601).

：Chen Xiuli, Li Fang, Chen Jianlong．Hom-dimodules and FRT theorem of Hom type[J]．Journal of Southeast University (English Edition),2014,30(3):391-395．

10.3969/j.issn.1003-7985.2014.03.025

10.3969/j.issn.1003-7985.2014.03.025

Received 2013-03-28．

Biographies：Chen Xiuli (1980—), female, doctor; Chen Jianlong (corresponding author), male, doctor, professor, jlchen@seu.edu.cn.