亚洲免费av电影一区二区三区,日韩爱爱视频,51精品视频一区二区三区,91视频爱爱,日韩欧美在线播放视频,中文字幕少妇AV,亚洲电影中文字幕,久久久久亚洲av成人网址,久久综合视频网站,国产在线不卡免费播放

        ?

        Diagonal crossed product of multiplier Hopf algebras

        2020-07-20 09:20:48YanDongdongWangShuanhong

        Yan Dongdong Wang Shuanhong

        (School of Mathematics, Southeast University, Nanjing 211189, China)

        Abstract:Let A and B be two regular multiplier Hopf algebras. First, the notion of diagonal crossed product B#A of multiplier Hopf algebras is constructed using the bimodule algebra, which is a generalization of the diagonal crossed product in the sense of Hopf algebras. The result that the product in B#A is non-degenerate is given. Next, the definition of the comultiplication Δ# on B#A is introduced, which is composed of the multiplier ΔB(b) on B?B and the multiplier ΔA(a) on A?A, and the element Δ#(b?a) is a two-side multiplier of B#A?B#A, for any b∈B and a∈A. Then, a sufficient condition for B#A to be a regular multiplier Hopf algebra is described. In particular, Delvaux’s main theorem in the case of smash products is generalized. Finally, these integrals on a diagonal crossed product of multiplier Hopf algebras are considered.

        Key words:multiplier Hopf algebra; bimodule algebra; diagonal crossed product

        As is known, multiplier Hopf algebras are a generalization of Hopf algebras[1-2]. Differently from Hopf algebra, the underlying algebra is no longer assumed to be a unit, but the product of algebra is non-degenerate. Some studies of multiplier Hopf algebras and their applications can be found in Refs.[3-6]. In Ref.[7], the author considered the module algebra as a regular multiplier Hopf algebra, and the theory of smash products was generalized to multiplier Hopf algebras from Hopf algebras. The definition and properties of diagonal crossed products were introduced in Ref.[8]. In this article, we will consider the bimodule algebra as a regular multiplier Hopf algebra and generalize the theory of diagonal crossed products to the multiplier Hopf algebra case.

        1 Preliminaries

        Throughout this article, letkbe a fixed field of characteristic 0 (i.e., all algebraic systems are overk). In order to facilitate our computations, we always omit the summation symbolΣ.

        In the following, we recall some definitions.

        For an associative algebraA,Ahas a non-degenerate product with or without identity. We denote its multiplier algebra byM(A), andM(A) always contains a unit 1. In fact,M(A) can be characterized as the largest algebra with a unit, in whichAis regarded as an essential two-side ideal. Clearly,A=M(A) if and only ifAhas a unit (see the appendix in Ref.[2] for details), which is similar toM(A?A).

        A comultiplication onAis a homomorphismΔA:A→M(A?A) such thatΔA(a)(1?b) and (a?1)ΔA(b) belong toA?Afor alla,b∈A, andΔAis coassociative in the sense that

        (a?1?1)(ΔA?i)(ΔA(b)(1?c))=

        (i?ΔA)((a?1)ΔA(b))(1?1?c)

        (1)

        for alla,b,c∈A, whereiis the identity map.

        We say that (A,ΔA) is regular ifσΔAis again a comultiplication onAsuch that a pair (A,σΔA) is also a multiplier Hopf algebra, whereσis the flip.

        Remark11) The use of the Sweedler notation for regular multiplier Hopf algebra is discussed in Ref.[1]. Takea,b∈Aas an example and considerΔA(a)(1?b). If we chooseesuch thateb=b, whereeis called the local unit (see Ref.[1], Proposition 2.2), then,ΔA(a)(1?b)=(ΔA(a)(1?e))(1?b)=a1?a2b.

        Definition1[9]LetAbe a regular multiplier Hopf algebra. Then, algebraBis called anA-bimodule algebra if

        1)Bis a unital leftA-module and a unital rightA-module such that (a·b)·a′=a·(b·a′);

        2)a·bb′=(a1·b)(a2·b′);

        3)bb′·a=(b·a1)(b′·a2) for allb,b′∈Banda,a′∈A.

        2 Main Results

        In this section, we will give the construction of the diagonal crossed product in the sense of multiplier Hopf algebras. More information can be found in Refs.[3-4].

        Definition2LetAbe a regular multiplier Hopf algebra andBanA-bimodule algebra. Then, the diagonal crossed productB#Abuilt onB?Awith multiplication is given as

        (b?a)(b′?a′)=b(a1·b′·S-1(a3))?a2a′

        (2)

        for allb,b′∈Banda,a′∈A. Note that on the right side, each decomposition is well-covered.

        We will further investigate the algebraB#A.

        Lemma1LetAbe a cocommutative multiplier Hopf algebra andBis a regular multiplier Hopf algebra such thatBis anA-bimodule algebra. Then, the product inB#Ais non-degenerate.

        ProofSuppose thatbi#ai∈B#Aand that (bi#ai)(b#a)=0 for allb∈Banda∈A. Then, according to the definition of the product inB#Aand the non-degeneracy of the product inA, we have

        bi(ai1·b·S-1(ai3))#ai2=0

        ApplyingΔandS, multiplying withafrom the right and replacingbbya″b, we can obtain

        bi(ai1a″·b·S-1(ai4))?S(ai2)a?ai3=0

        for allb∈Banda,a′,a″∈A. Replacinga″ byS(ai2)a, we have

        bi(ai1S(ai2)a·b·S-1(ai4))?ai3=0

        That is

        bi(a·b·S-1(ai2))?ai1=0

        As thatAis cocommutative, we can obtain

        bi(a·b·S-1(ai1))?ai2=0

        ApplyingΔagain, multiplying witha′ from the left and replacingbbyba″, we obtain

        Hence,bi(a·b·a′)?ai=0. Note thatBhas a non-degenerate product and it is a unital leftA-module and a unital rightA-module, and we obtainbi?ai=0.

        On the other hand, suppose that (b#a)(bi#ai)=0 for allb∈Banda∈A. Similar to the proof of Lemma 5.6 in Ref.[1], we can obtainbi?ai=0.

        Moreover, we will construct a comultiplication for the diagonal crossed product such that it admits a structure of multiplier Hopf algebra. First, we give two maps as follows.

        LetAandBbe regular multiplier Hopf algebras such thatBis anA-bimodule algebra. The multiplication of diagonal crossed productB#Ais defined by two twist mapsR:A?B→B?Aviaa?ba1·b?a2andR′:B?A→B?Aviab?ab·S-1(a2)?a1. It can easily be checked thatRandR′ are two bijections.R-1:B?A→A?Bviab?aa2?S-1(a1)·bandR′-1:B?A→B?Aviab?ab·a2?a1.

        b1#a1?b2(a2·b′·S-1(a4))#a3a′

        Definition3For allb,b′,b″∈Banda,a′,a″∈A, we define

        Δ#(b#a)(b″#a″?b′#a′)=

        b1(a1·b″·S-1(a3))#a2a″?b2(a4·b′·S-1(a6))#a5a′

        and

        (b″#a″?b′#a′)Δ#(b#a)=

        Lemma2For anyb∈Banda∈A,Δ#(b#a) is a two-sided multiplier ofB#A?B#A.

        (b″#a″?1#1)(b1#a1?b2(a2·b′·S-1(a4))#a3a′)=

        Lemma3The comultiplicationΔ#is coassociative onB#A.

        For allb,b′,b″∈Banda,a′,a″∈A, we have

        b12#a12?b2(a2·b″·S-1(a4))#a3a″)

        and

        b22(a22·b″·S-1(a4))#a3a″)

        Before we proceed to give the main result, we need the following definition and lemma.

        Definition4LetAandBbe regular multiplier Hopf algebras andBis anA-bimodule algebra. Then,Bis anA-bimodule bialgebra if

        (3)

        ε#(a·b·a′)=εA(a)εB(b)εA(a′)

        (4)

        for allb,b′∈Banda,a′∈A. Observe that on the right side, all decompositions are well-covered.

        Lemma4AandBare the same as those in Definition 4. We denote the antipode ofB(A, resp.) bySB(SA, resp.). Then,

        a·SB(b)·a′=SB(a·b·a′)

        The proof is straightforward.

        Now, we can formulate the main result as follows.

        Theorem1LetAbe a cocommutative regular multiplier Hopf algebra andBis a regular multiplier Hopf algebra such thatBis anA-bimodule bialgebra. Then,Δ#is a comultiplication onB#Asuch that (B#A,Δ#) is a regular multiplier Hopf algebra.

        ProofAccording to Proposition 2.9 in Ref.[10], our proof is given as follows.

        First, we can easily check that the diagonal crossed productB#Ais an associative algebra, and we prove that it has a non-degenerate product by Lemma 1.

        The comultiplicationΔ#as a multiplier ofB#A?B#Ais coassociative by Lemma 3. We now show that Δ#:B#A→M(B#A?B#A) is a homomorphism. For allb,b′,b″∈Banda,a′,a″∈A, we have

        Δ#(b#a)(Δ#(b′#a′)(1#1?b″#a″))=

        Δ#(b(a1·b′·S-1(a3))#a2a′)(1#1?b″#a″)=

        Δ#((b#a)(b′#a′))(1#1?b″#a″)

        There is a counitε#defined as

        ε#(b#a)=εB(b)εA(a)

        for allb∈Banda∈A. It can easily be checked that these two equations (ε#?i)(Δ#(b#a)(1#1?b′#a′))=(b#a)(b′#a′) and (i?ε#)((b′#a′?1#1) Δ#(b#a))=(b′#a′)(b#a) are valid. From Eq.(4), it easily follows thatε#is a homomorphism.

        There is an invertible antipodeS#defined as

        S#(b#a)=R′oRo(SA?SB)oσ(b#a)=

        SA(a3)·SB(b)·a1#SA(a2)∈B#A

        For anyb,b′∈Banda,a′∈A, we obtain

        m#(S#?iB#iA)(Δ#(b#a)(1#1?b′#a′))=

        m#(S#?iB#iA)(b1#a1?b2(a2·b′·S-1(a4))#a3a′))=

        m#(S(a3)·S(b1)·a1#S(a2)?b2(a4·b′·S-1(a6))#a5a′)=(S(a5)·S(b1)·a1)(S(a4)·

        (b2(a6·b′·S-1(a8)))·a2)#S(a3)a7a′=

        S(a3)·(S(b1)b2(a4·b′·S-1(a6)))·a1#S(a2)a5a′=

        εB(b)b′·S-1(a4)·a1#S(a2)a3a′=εB(b)εA(a)b′#a′=

        εB#A(b#a)(b′#a′)

        Similarly, we can obtain

        m#(iB#iA?S#)((b′#a′?1#1)Δ#(b#a))=
        εB#A(b′#a′)(b#a)

        Using Lemma 4, we can easily prove thatS#is an anti-homomorphism.

        Finally, we can easily observe that the multipliersΔ#(b#a)(b′#a′?1#1) and (1#1?b′#a′)Δ#(b#a) are inB#A?B#A.

        The proof is complete.

        In Theorem 1, if the right action is trivial, we have the following corollary which is the main result of Ref.[7].

        Corollary1LetAbe a cocommutative multiplier Hopf algebra andBis a regular multiplier Hopf algebra such thatBis anA-bimodule bialgebra. Then,Δ#is a comultiplication onB#Asuch that the smash product (B#A,Δ#) is a regular multiplier Hopf algebra.

        In Ref.[10], for multiplier Hopf algebras, there is a natural notion of left and right invariance for linear functionals (called integrals in the theory of Hopf algebra). We now suppose thatAandBare multiplier Hopf algebras which are the same as those in Theorem 1. Furthermore, we suppose thatAandBhave invariant functions. In the following proposition, we will give these integrals on (B#A,Δ#).

        Proposition1AandBare the same as those in Theorem 1. LetφB(φA, resp.) be a left integral ofB(A, resp.). Then,φB?φAis the left integral ofB#A. The same statement yields for the right integral.

        亚洲最大中文字幕无码网站 | 日本看片一区二区三区| 亚洲av资源网站手机在线 | 99久久精品免费看国产情侣| 国产欧美亚洲精品第二区首页| 久久久亚洲成年中文字幕| 美女网站免费观看视频| 亚洲av综合久久九九| 亚洲色AV性色在线观看| 日本熟妇视频在线中出| 久久精品国产亚洲av网| 末发育娇小性色xxxx| 婷婷综合五月| 亚洲一码二码在线观看| 国产精品亚洲一区二区麻豆| 巨人精品福利官方导航| 999国产一区在线观看| 日韩美女av二区三区四区| 午夜福利视频一区二区二区| 国产精品亚洲а∨无码播放不卡| 国产免费破外女真实出血视频 | 国产人在线成免费视频| 久9热免费精品视频在线观看| 亚洲天堂一区二区三区| 风流老太婆大bbwbbwhd视频| 国产高潮国产高潮久久久| 欧美日韩中文字幕日韩欧美| 精品亚洲一区二区三洲| 热re99久久精品国99热| 国产美女遭强高潮网站| 国产精品一区二区三密桃| 国产自拍成人免费视频| 老少配老妇老熟女中文普通话 | 一区二区三区国产黄色| 高清不卡一区二区三区| 久久AV中文综合一区二区| 国产优质av一区二区三区| 日本熟妇美熟bbw| 国产av日韩a∨亚洲av电影| 一区二区三区精品偷拍av| 精品国产av一区二区三区四区|