Drinfeld twists for monoidal Hom-bialgebras
aa r X i v : . [ m a t h . R A ] O c t Drinfeld twists for monoidal Hom-bialgebras
Xiaohui Zhang ∗ and Xiaofan Zhao † Department of Mathematics, Southeast UniversityJiangsu Nanjing 210096, P. R. CHINA
Abstract.
The aim of this paper is to define and study Drinfeld twists for monoidalHom-bialgebras. We show that a new Hom-bialgebra could be constructed bychanging the coproduct of a monoidal Hom-bialgebra via a Drinfeld twist, and thisconstruction preserves R -matrixes if there exist one. Moreover, their representationcategories are monoidal isomorphic. Mathematics Subject Classification(2000).
Keywords:
Drinfeld twist; monoidal Hom-bialgebra; R -matrix; Hom-bialgebra In 2006, Hartwig, Larsson and Silvestrov introduced the Hom-Lie algebras when theyconcerned about the q -deformations of Witt and Virasoro algebras (see [13]). In a Hom-Liealgebra, the Jacobi identity is replaced by the so called Hom-Jacobi identity via a homo-morphism. Hom-associative algebras, the corresponding structure of associative algebras,were introduced by Makhlouf and Silvestrov in [4]. The associativity of a Hom-algebra istwisted by an endomorphism (here we call it the Hom-structure map). The generalized no-tions, Hom-bialgebras, Hom-Hopf algebras were developed in [3], [5], [6]. Further researchon various Hom-Lie structures and Hom-type algebras by many schlors could be found in[7], [9], [16], [17]. Quasitriangular Hom-bialgebras were considered by Yau ([8], [9]), whichprovided a solution of the quantum Hom-Yang-Baxter euqation, a twisted version of thequantum Yang-Baxter equation ([10], [11]).The notions of Hom-categories and monoidal Hom-Hopf algebras were introduced byCaenepeel and Goyvaerts ([18]) in order to provide a categorical approach to Hom-typealgebras. In a Hom-category, the associativity and unit constraints are twisted by theHom-structure maps. A (co)monoid in the Hom-category is a Hom-(co)algebra, and abimonoid in the Hom-category is a monoidal Hom-bialgebra. Further research on monoidalHom-bialgebras can be found in [22], [23], and [14].Moreover, through a direct computation (see Example 2.1 and Example 2.2 for details),we can get that there is a one to one correspondence between the collection the monoidal ∗ E-mail: [email protected] † Corresponding author, E-mail: [email protected] k , and the collection of the unital Hom-bialgebraover k which Hom-structure map is a bijection. Does there another way to get a Hom-bialgebra through a given monoidal Hom-bialgebra? Is there any relation between theirrepresentation categories? This is the motivation of the present article. In order toinvestigate these questions, we introduce the definition of Drinfeld twists for monoidalHom-bialgebras, and construct a new Hom-bialgebra via a Drinfeld twist.A Drinfeld twist for a Hopf algebra H is an invertible element σ ∈ H ⊗ H , satisfyingthe 2-cocycle condition ( σ ⊗ ⊗ id )( σ ) = (1 ⊗ σ )( id ⊗ ∆)( σ ) . Note that our definition of a Drinfeld twist is inverse with Drinfeld’s (see [20] and [12]).In our paper, we always assume that σ is normalized, i.e. ( ε ⊗ id )( σ ) = ( id ⊗ ε )( σ ) = 1 H .The twisting elements or twists were first introduced by Drinfeld [20] on quasi-Hopfalgebras, in order to twist the coproduct without changing its product. They have becomean important tool in the classification of finite-dimensional Hopf algebras ([12]). Thetwisting elements for a Hom-bialgebra have been discussed in [2].The paper is organized as follows. In Section 2 we recall some notions of monoidalHom-type algebras and Hom-type algebras. In section 3, we describe the category of rep-resentations of a monoidal Hom-bialgebra which is more generalized than Caenepeel’s, andgive the definition of quasitriangular monoidal Hom-bialgebras. In section 4, we introducethe notion of the Drinfeld twists for a monoidal Hom-bialgebra, and construct a Hom-bialgebra by changing the coproduct via a Drinfeld twist. If the monoidal Hom-bialgebrais quasitriangular, then the Hom-bialgebra which we obtained is also quasitriangular. Fur-thermore, we show that their representation categories are monoidal isomorphic. Throughout the paper, we let k be a fixed commutative ring and char ( k ) = 0. Allalgebras are supposed to be over k . For the comultiplication ∆ of a k -module C , we usethe Sweedler-Heyneman’s notation: ∆( c ) = c ⊗ c , for any c ∈ C . τ means the flip map τ ( a ⊗ b ) = b ⊗ a . When we say a ”Hom-algebra” ora ”Hom-coalgebra”, we mean the unital Hom-algebra and counital Hom-coalgebra. In this section, we will review several definitions and notations related to monoidalHom-bialgebras (see [18]).Let C be a category. We introduce a new category f H ( C ) as follows: the objects arecouples ( M, α M ), with M ∈ C and α M ∈ Aut C ( M ). A morphism f : ( M, α M ) → ( N, α N )is a morphism f : M → N in C such that α N ◦ f = f ◦ α M .Specially, let M k denote the category of k -modules. H ( M k ) will be called the Hom-category associated to M k . If ( M, α M ) ∈ M k , then α M : M → M is obviously an2somorphism in H ( M k ). It is easy to show that f H ( M k ) = ( H ( M k ) , ⊗ , ( k, id ) , e a, e l, e r ))is a monoidal category by Proposition 1.1 in [18]: • the tensor product of ( M, α M ) and ( N, α N ) in f H ( M k ) is given by the formula( M, α M ) ⊗ ( N, α N ) = ( M ⊗ N, α M ⊗ α N ). • for any m ∈ M , n ∈ N , p ∈ P , the associativity is given by the formulas e a M,N,P (( m ⊗ n ) ⊗ p ) = α M ( m ) ⊗ ( n ⊗ α − P ( p )) , • for any m ∈ M , the unit constraints are given by the formulas e l M ( x ⊗ m ) = e r M ( m ⊗ x ) = xα M ( m ) . A monoidal Hom-algebra over k is an object ( A, α ) ∈ f H ( M k ) together with a k -linearmap m A : A ⊗ A → A and an element 1 A ∈ A such that for all a, b, c ∈ A , (1) α ( ab ) = α ( a ) α ( b );(2) α ( a )( bc ) = ( ab ) α ( c );(3) α (1 A ) = 1 A ;(4) 1 A a = a A = α ( a ) . A morphism f : A → B of monoidal Hom-algebras is a linear map such that α B ◦ f = f ◦ α A , f (1 A ) = 1 B and µ B ◦ ( f ⊗ f ) = f ◦ µ A .A monoidal Hom-coalgebra over k is an object ( C, γ ) ∈ f H ( M k ) together with k -linearmaps ∆ : C → C ⊗ C, ∆( c ) = c ⊗ c and α : C → C such that for all c ∈ C , (1) ∆( γ ( c )) = γ ( c ) γ ( c );(2) γ − ( c ) ⊗ ∆( c ) = ∆( c ) ⊗ γ − ( c );(3) ε ◦ γ = ε ;(4) ε ( c ) c = c ε ( c ) = γ − ( c ) . A morphism f : C → D of monoidal Hom-coalgebras is a linear map such that γ D ◦ f = f ◦ γ C , ε C = ε D ◦ f and ∆ D ◦ f = ( f ⊗ f ) ◦ ∆ C .A monoidal Hom-bialgebra H = ( H, α, m, η, ∆ , ε ) is a bimonoid in f H ( M k ). Thismeans that ( H, α, m, η ) is a monoidal Hom-algebra, (
H, α, ∆ , ε ) is a monoidal Hom-coalgebra and ∆, ε are morphisms of monoidal Hom-algebras preserving unit. Example 2.1.
Suppose ( A, m, η, ∆ , ε ) is a bialgebra over k endowed with a bialgebraisomorphism α : A → A . Then ( A, α, α ◦ m, η, ∆ ◦ α − , ε ) is a monoidal Hom-bialgebraover k . We denote this monoidal Hom-bialgebra by α A .Conversely, if ( H, α, m, η, ∆ , ε ) is a monoidal Hom-bialgebra, then ( H, α − ◦ m, η, ∆ ◦ α, ε ) is a bialgebra over k . We write this bialgebra for α H .Thus we immediately get a bijective map A → α A between the collection of all bialgebrasover k endowed with an invertible endomorphism on it and the collection of all monoidalHom-bialgebras over k . A monoidal Hom-Hopf algebra over k is a monoidal Hom-bialgebra ( H, α ) togetherwith a linear map S : H → H in f H ( M k ) (called the antipode ) such that S ∗ id = id ∗ S = ηε, S ◦ α = α ◦ S. a, b ∈ H , S satisfies S ( ab ) = S ( b ) S ( a ) , S (1 H ) = 1 H , ∆( S ( a )) = S ( a ) ⊗ S ( a ) , ε ◦ S = ε. Note that the map A → α A given in Example 2.1 is also a bijection between the collec-tion of k -Hopf algebra ( A, m, η, ∆ , ε, S ) which endowed with an invertible endomorphismand the collection of k -monoidal Hom-Hopf algebra α A = ( A, α, α ◦ m, η, ∆ ◦ α − , ε, S ). In this section, we will review several definitions and notations related to Hom-bialgebras.Recall from [4] that a
Hom-algebra over k is a quadruple ( A, µ, η, α ), in which A isa k -module, α : A → A , µ : A ⊗ A → A and η : k → A are linear maps, with notation µ ( a ⊗ b ) = ab and η (1 k ) = 1 A , satisfying the following conditions, for all a, b, c ∈ A : (1) α ( ab ) = α ( a ) α ( b );(2) α ( a )( bc ) = ( ab ) α ( c );(3) α (1 A ) = 1 A ;(4) 1 A a = a A = α ( a ) . Note that a monoidal Hom-algebra is also a Hom-algebra. Conversely, a Hom-algebrais a monoidal Hom-algebra if its Hom-structure map is invertible.
A morphism f : A → B of Hom-algebras is a linear map such that α B ◦ f = f ◦ α A , f (1 A ) = 1 B and µ B ◦ ( f ⊗ f ) = f ◦ µ A .Recall from [3] that a Hom-coalgebra over k is a quadruple ( C, ∆ , ε, α ), in which C isa k -module, α : C → C , ∆ : C → C ⊗ C and ε : C → k are linear maps, with notation∆( c ) = c ⊗ c , satisfying the following conditions for all c ∈ C : (1) ∆( α ( c )) = α ( c ) ⊗ α ( c );(2) α ( c ) ⊗ ∆( c ) = ∆( c ) ⊗ α ( c );(3) ε ◦ α = ε ;(4) ε ( c ) c = c ε ( c ) = α ( c ) . A morphism f : C → D of Hom-coalgebras is a linear map such that α D ◦ f = f ◦ α C , ε C = ε D ◦ f and ∆ D ◦ f = ( f ⊗ f ) ◦ ∆ C .Recall from [6] that a Hom-bialgebra B = ( B, α, m, η, ∆ , ε ) over k is a sextuple ( B, µ, η ,∆, ε , α ), in which ( B, α, m, η ) is a Hom-algebra, (
B, α, ∆ , ε ) is a Hom-coalgebra, and ∆, ε are morphisms of Hom-algebras preserving unit. Example 2.2.
Suppose ( A, m, η, ∆ , ε ) is a bialgebra over k endowed with a bialgebra map α : A → A . Then ( A, α, α ◦ m, η, ∆ ◦ α, ε ) is a Hom-bialgebra over k . We denote thisHom-bialgebra by A α .Conversely, if ( B, α, m, η, ∆ , ε ) is a Hom-bialgebra and α is invertible, then ( B, α − ◦ m, η, ∆ ◦ α − , ε ) is a bialgebra over k . We write this bialgebra for B α .Thus we immediately get a bijective map A → A α between the collection of all bialgebrasover k endowed with an invertible endomorphism on it, and the collection of all Hom-bialgebras with invertible Hom-structure maps. efinition 2.3. A Hom-Hopf algebra over k is a Hom-bialgebra ( H, α ) together with a k -linear map S : H → H (called the antipode ) such that S ∗ id = id ∗ S = ηε, S ◦ α = α ◦ S. Lemma 2.4.
Let ( H, α ) be a Hom-Hopf algebra. If α is invertible, then for all a, b ∈ H , S satisfies S ( ab ) = S ( b ) S ( a ) , S (1 H ) = 1 H , ∆( S ( a )) = S ( a ) ⊗ S ( a ) , ε ◦ S = ε. Proof.
We will only prove the first statement. We compute as follows. S ( ab ) = S ( α − ( a ) α − ( b )) ε ( b ) ε ( a )= S ( α − ( a ) α − ( b )) ε ( b )( α − ( a ) S ( α − ( a )))= S ( α − ( a ) α − ( b ))(( α − ( a )( α − ( b ) S ( α − ( b )))) S ( α − ( a )))= S ( α − ( a ) α − ( b ))(( α − ( a ) α − ( b ))( S ( α − ( b )) S ( α − ( a ))))= ( S ( α − ( a ) α − ( b ))( α − ( a ) α − ( b )))( S ( α − ( b )) S ( α − ( a )))= ε ( a ) ε ( b ) S ( α − ( b )) S ( α − ( a )) = S ( b ) S ( a ) . ✷ Note that the map A → A α given in Example 2.2 is also a bijection between the col-lection of all k -Hopf algebras endowed with an invertible endomorphism and the collectionof all k -Hom-Hopf algebras with invertible Hom-structure maps. Definition 2.5.
Let ( B, α ) be a Hom-bialgebra. If there exists an invertible element R ∈ B ⊗ B , such that the following conditions hold: ( q
1) ( α ⊗ α ) R = R ;( q R ∆( x ) = ∆ op ( x ) R ;( q R (1)1 ⊗ R (1)2 ⊗ α ( R (2) ) = α ( r (1) ) ⊗ α ( R (1) ) ⊗ r (2) R (2) ;( q α ( R (1) ) ⊗ R (2)1 ⊗ R (2)2 = r (1) R (1) ⊗ α ( R (2) ) ⊗ α ( r (2) ) , for any x ∈ B , where r = R = R (1) ⊗ R (2) = r (1) ⊗ r (2) , then R is called an R -matrix of B , ( B, α, R ) is called a quasitriangular Hom-bialgebra . Note that the above definition of quasitriangular Hom-bialgebras is slightly differentfrom Yau’s (see [8]). In order to make sure the representation category of a quasitriangularHom-bialgebra is braided, we need R is invertible and Eq.(q1) is hold.Let ( B, α ) be a Hom-algebra. A left (
B, α ) -Hom-module is a triple ( M, α M , θ M ), where M is a k -module, α M : M → M and θ M : B ⊗ M → M are linear maps with notation θ M ( b ⊗ m ) = b · m , satisfying the following conditions, for all b, b ′ ∈ B , m ∈ M (1) α ( b · m ) = α ( b ) · α M ( m );(2) α ( b ) · ( b ′ · m ) = ( bb ′ ) · α M ( m );(3) 1 B · m = α M ( m ) . A morphism f : M → N of B -modules is a k -linear map such that α N ◦ f = f ◦ α M and θ N ◦ ( id A ⊗ f ) = f ◦ θ M . 5 The representations of monoidal Hom-bialgebras
Let (
H, α ) be a monoidal Hom-bialgebra. A category
Rep i,j ( H ) is defined as followsfor any fixed i, j ∈ Z (the domain of integrals): • Rep i,j ( H ) is the category of left Hom-modules of the monoidal Hom-algebra H andthe morphisms of H -modules; • the tensor product M ⊗ N for ( M, α M ) , ( N, α N ) ∈ Rep i,j ( H ) is obtained by ( M ⊗ N, α M ⊗ α N ) with the action of H given by h · ( m ⊗ n ) = α i ( h ) · m ⊗ α j ( h ) · n, where i, j ∈ Z , h ∈ H , m ∈ M , n ∈ N ; • the tensor product of two arrows f, g ∈ Rep i,j ( H ) is given by the tensor product of k -linear morphisms, i.e. the forgetful functor from Rep i,j ( H ) to the category of k -modulesis faithful; • for any λ ∈ k , ( k, id k ) is an object in Rep i,j ( H ) with the action h · λ = ε ( h ) λ. Lemma 3.1. k is the unit object of the tensor product ⊗ in Rep i,j ( H ) . Proof.
Firstly, it is easy to check that ( k, id k ) and ( M ⊗ N, α M ⊗ α N ) under the H -action defined above are objects in Rep i,j ( H ).Secondly, for any M ∈ Rep i,j ( H ), define a k -linear map l M : k ⊗ M → M, λ ⊗ m λα − j +1 M ( m ) , λ ∈ k, m ∈ M. Obviously l is natural, and we have α M ◦ l M = l M ◦ ( id k ⊗ α M ). Furthermore, l M is H -linear, actually, l M ( h · ( λ ⊗ m )) = l M ( ε ( h ) λ ⊗ α j ( h ) · m )= λα − j +1 M ( α j − ( h ) · m )= λh · α − j +1 M ( m ) = h · l M ( λ ⊗ m ) . The inverse of l is given by l − M : M → k ⊗ M, m k ⊗ α j − M ( m ) , m ∈ M. Similarly, we define the k -linear maps r M : M ⊗ k → M, m ⊗ λ λα − i +1 M ( m ) , λ ∈ k, m ∈ M, and r − M : M → M ⊗ k, m α i − M ( m ) ⊗ k , m ∈ M. It is easy to check that r is a natural isomorphism with the inverse r − .This completes the proof. ✷ Theorem 3.2.
Rep i,j ( H ) is a monoidal category. roof. Firstly, for any
M, N, P, Q ∈ Rep i,j ( H ), define an associativity constraint by a M,N,P (( m ⊗ n ) ⊗ p ) = α − i +1 M ( m ) ⊗ ( n ⊗ α j − P ( p )) , m ∈ M, n ∈ N, p ∈ P. Obviously that a is natural and satisfies a M,N,P ◦ ( α M ⊗ ( α N ⊗ α P )) = (( α M ⊗ α N ) ⊗ α P ) ◦ a M,N,P . For any h ∈ H , we obtain a M,N,P ( h · (( m ⊗ n ) ⊗ p ))= α − i +1 M ( α i ( h ) · m ) ⊗ ( α i + j ( h ) · n ⊗ α j − P ( α j ( h ) · p ))= α i ( h ) · α − i +1 M ( m ) ⊗ ( α i + j ( h ) · n ⊗ α j ( h ) · α j − P ( p ))= h · ( α − i +1 M ( m ) ⊗ ( n ⊗ α j − P ( p )))= h · ( a M,N,P (( m ⊗ n ) ⊗ p )) , thus a M,N,P is H -linear. Since a is invertible, a is a natural isomorphism in Rep i,j ( H ).Secondly, one can see that a satisfies the Pentagon Axiom. Actually,(( id M ⊗ a N,P,Q ) ◦ a M,N ⊗ P,Q ◦ ( a M,N,P ⊗ id Q ))((( m ⊗ n ) ⊗ p ) ⊗ q )= ( id M ⊗ a N,P,Q )( α − i +2 M ( m ) ⊗ (( n ⊗ α j − P ( p )) ⊗ α j − Q ( q )))= α − i +2 M ( m ) ⊗ ( α − i +1 N ( n ) ⊗ ( α j − P ( p ) ⊗ α j − Q ( q )))= a M,N,P ⊗ Q (( α − i +1 M ( m ) ⊗ α − i +1 N ( n )) ⊗ ( p ⊗ α j − Q ( q )))= ( a M ⊗ N,P,Q ◦ a M,N,P ⊗ Q )((( m ⊗ n ) ⊗ p ) ⊗ q ) . At last, it is also a direct check to prove that a, l, r satisfy the Triangle Axiom. Thiscompletes the proof. ✷ Definition 3.3.
Let ( H, α ) be a monoidal Hom-bialgebra. If there exists an invertibleelement R = R (1) ⊗ R (2) ∈ H ⊗ H , such that the following conditions hold: ( Q
1) ( α ⊗ α ) R = R ;( Q R ∆( h ) = ∆ op ( h ) R ;( Q R (1)1 ⊗ R (1)2 ⊗ R (2) = r (1) ⊗ R (1) ⊗ r (2) R (2) ;( Q R (1) ⊗ R (2)1 ⊗ R (2)2 = r (1) R (1) ⊗ R (2) ⊗ r (2) , where h ∈ H , r = R = R (1) ⊗ R (2) = r (1) ⊗ r (2) , then R is called an R -matrix of H . ( H, α, R ) is called a quasitriangular monoidal Hom-bialgebra . Example 3.4.
Let ( A, R ) be a quasitriangular bialgebra over k and α : A → A be aninvertible bialgebra homomorphism. If R satisfies ( α ⊗ α )( R ) = R , then ( α A, α, R ) inExample 2.1 is a quasitriangular monoidal Hom-bialgebra.Conversely, if ( H, α, R ) is a quasitriangular monoidal Hom-bialgebra, then ( α H, R ) inExample 2.1 is a quasitriangular bialgebra over k . Proposition 3.5.
Let ( H, α, R ) be a quasitriangular monoidal Hom-bialgebra. Then R satisfies the quantum Hom-Yang-Baxter equations ( R R ) R = R ( R R ) , R ( R R ) = ( R R ) R , where R = R ⊗ H , R = 1 H ⊗ R , R = ( τ ⊗ id ) R . roof. Straightforward. ✷ Theorem 3.6.
Let ( H, α ) be a monoidal Hom-bialgebra. For the fixed elements R, S ∈ H ⊗ H , define maps c M,N : M ⊗ M → N ⊗ M, m ⊗ n α i ( R (2) ) · α i − j − N ( n ) ⊗ α j ( R (1) ) · α j − i − M ( m ) , and c ′ M,N : N ⊗ M → M ⊗ M, n ⊗ m α i ( S (1) ) · α i − j − M ( m ) ⊗ α j ( S (2) ) · α j − i − N ( n ) , for any ( M, α M ) , ( N, α N ) ∈ Rep i,j ( H ) , then c is a braiding in Rep i,j ( H ) with the inverse c ′ if and only if R is an R -matrix with the inverse S . Proof.
It is easy to check that c and c ′ are natural in Rep i,j ( H ).(1). If R satisfies ( α ⊗ α ) R = R , then we immediately get that c M,N ◦ α M,N = α N ⊗ M ◦ c M,N . If R also satisfies R ∆( h ) = ∆ op ( h ) R for any h ∈ H , then for any m ∈ M , n ∈ N , we have c M,N ( h · ( m ⊗ n )) = α i ( R (2) ) · α i − j − N ( α j ( h ) · n ) ⊗ α j ( R (1) ) · α j − i − M ( α i ( h ) · m )= ( α i − ( R (2) ) α i − ( h )) · a i − jN ( n ) ⊗ ( α j − ( R (1) ) α j − ( h )) · a j − iM ( m )= α i ( h ) · ( α i R (2) ) · a i − j − N ( n ) ⊗ α j ( h ) · ( α j R (1) ) · a j − i − M ( m )= h · ( c M,N ( m ⊗ n )) , thus c M,N ∈ M or ( Rep i,j ( H )).Conversely, if c M,N ∈ M or ( Rep i,j ( H )), then c is H -linear and satisfies c M,N ◦ α M ⊗ N = α N ⊗ M ◦ c M,N . Take M = N = H and m = n = 1 H , then we directly get Eq.(Q1) andEq.(Q2) for any h ∈ H .(2). If S is the inverse of R , thus S satisfies ( α ⊗ α )( S ) = S . Then for any m ∈ M , n ∈ N , we obtain c ′ M,N ( c M,N ( m ⊗ n ))= α i ( S (1) ) · α i − j − M ( α j ( R (1) ) · α j − i − M ( m )) ⊗ α j ( S (2) ) · α j − i − N ( α i ( R (2) ) · α i − j − N ( n ))= α i − ( S (1) R (1) ) · α − M ( m ) ⊗ α j − ( S (2) R (2) ) · α − N ( n )= m ⊗ n. Similarly, we have c M,N ◦ c ′ M,N = id N ⊗ M , hence c ′ is the inverse of c .Conversely, if c ′ is the inverse of c , take M = N = H and m = n = 1 H , then weimmediately get RS = SR = 1 H ⊗ H . 83). Assume that R satisfies Eq.(Q4), thus we obtain( a N,P,M ◦ c M,N ⊗ P ◦ a M,N,P )(( m ⊗ n ) ⊗ p )= a N,P,M (( α i ( R (2)1 ) · α i − j − N ( n ) ⊗ α i + j ( R (2)2 ) · α i − P ( p )) ⊗ α j ( R (1) ) · α j − iM ( m ))= α i +1 ( R (2) ) · α − jN ( n ) ⊗ ( α i + j ( r (2) ) · α i − P ( p ) ⊗ α j − ( r (1) R (1) ) · α j − i − M ( m )))= α ( R (2) ) · α − jN ( n ) ⊗ ( α i ( r (2) ) · α i − P ( p ) ⊗ α j ( r (1) ) · ( α j − i − ( R (1) ) · α j − i − M ( m )))= ( id N ⊗ ( c M,P ))( α ( R (2) ) · α − jN ( n ) ⊗ ( α j ( R (1) ) · α j − i − M ( m ) ⊗ α j − P ( p )))= (( id N ⊗ c M,P ) ◦ a N,M,P ◦ ( c M,N ⊗ id P ))(( m ⊗ n ) ⊗ p ) , for any M, N, P ∈ Rep i,j ( H ).Similarly, if R satisfies Eq.(Q3), then c satisfies a − P,M,N ◦ c M ⊗ N,P ◦ a − M,N,P = ( c M,P ⊗ id N ) ◦ a − M,P,N ◦ ( id M ⊗ c N,P ) . Conversely, if c is a braiding, then take M = N = P = H and m = n = p = 1 H , it isa direct computation to get Eq.(Q3) and Eq.(Q4).Combining (1)-(3), the conclusion holds. ✷ Theorem 3.7.
Let ( H, α ) be a monoidal Hom-bialgebra. For any i, j, i ′ , j ′ ∈ Z , Rep i,j ( H ) and Rep i ′ ,j ′ ( H ) are monoidal isomorphic. Moreover, if H is a quasitriangular monoidalHom-bialgebra, then Rep i,j ( H ) and Rep i ′ ,j ′ ( H ) are isomorphic as braided categories. Proof.
Define a functor F = ( F, F , F ) : ( Rep i,j ( H ) ⊗ , k, a, l, r ) → ( Rep i ′ j ′ ( H ) , ⊗ ′ , k, a ′ , l ′ , r ′ )by F ( M ) := M, F ( f ) := f, F = id k ,F ( M, N ) : F ( M ) ⊗ ′ F ( N ) → F ( M ⊗ N ) , m ⊗ ′ n α i − i ′ M ( m ) ⊗ α j − j ′ N ( n ) , where M, N ∈ Rep i,j ( H ), f : M → N ∈ M or ( Rep i,j ( H )), and ⊗ ′ , a ′ , l ′ , r ′ mean thecorresponding structures in Rep i ′ ,j ′ ( H ).Obviously F is natural and compatible with the Hom structure map.Firstly, since F ( M, N ) is invertible, and F ( M, N )( h · ( m ⊗ ′ n )) = F ( M, N )( α i ′ ( h ) · m ⊗ ′ α j ′ ( h ) · n )= h · ( α i − i ′ M ⊗ α j − j ′ N ( n )) = h · F ( M, N )( m ⊗ ′ n ) ,F is a natural isomorphism in Rep i ′ ,j ′ ( H ).Secondly, we have( F ( M, N ⊗ P ) ◦ ( id F ( M ) ⊗ ′ F ( N, P )) ◦ ( a F ( M ) ,F ( N ) ,F ( P ) ))(( m ⊗ ′ n ) ⊗ ′ p )= ( F ( M, N ⊗ P ) ◦ ( id F ( M ) ⊗ ′ F ( N, P )))( α − i ′ +1 M ( m ) ⊗ ′ ( n ⊗ ′ α j ′ − P ( p )))= α i − i ′ +1 M ( m ) ⊗ ( α i + j − i ′ − j ′ N ( n ) ⊗ α j − j ′ − P ( p ))= F ( a M,N,P )( α i − i ′ M ( m ) ⊗ α i + j − i ′ − j ′ N ( n )) ⊗ α j − j ′ P ( p )= ( F ( a M,N,P ) ◦ F ( M ⊗ N, P ) ◦ ( F ( M.N ) ⊗ ′ id P ))(( m ⊗ ′ n ) ⊗ ′ p ) .
9t last, it is easy to get that F ( l M ) ◦ F ( k, M ) ◦ ( F ⊗ ′ id M ) = l ′ F ( M ) , and F ( r M ) ◦ F ( M, k ) ◦ ( id M ⊗ ′ F ) = r ′ F ( M ) , hence F = ( F, F , F ) is a monoidal functor.Obviously F is invertible, thus the conclusion holds.Furthermore, if H is a quasitriangular monoidal Hom-bialgebra with the R -matrix R ∈ H ⊗ H , for any m ∈ M , n ∈ N , we compute( F ( N, M ) ◦ c ′ F ( M ) ,F ( N ) )( m ⊗ ′ n )= F ( N, M )( α i ′ ( R (2) ) · α i ′ − j ′ − N ( n ) ⊗ α j ′ ( R (1) ) · α j ′ − i ′ − M ( m ))= α i ( R (2) ) · α i − j ′ − N ( n ) ⊗ α j ( R (1) ) · α j − i ′ − M ( m )= F ( c M,N )( α i − i ′ M ( m ) ⊗ α j − j ′ N ( n )) = ( F ( c M,N ) ◦ F ( M, N ))( m ⊗ ′ n ) , which implies F is a braided monoidal functor. ✷ Let (
H, α, m, η, ∆ , ε ) be a monoidal Hom-bialgebra. Definition 4.1. A Drinfeld twist for H is an invertible element σ ∈ H ⊗ H such that ( T
1) ( α ⊗ α ) σ = σ ;( T
2) ( ε ⊗ id H )( σ ) = ( id H ⊗ ε )( σ ) = 1 H ;( T σ (1) ⊗ ¯ σ (1) σ (2)1 ⊗ ¯ σ (2) σ (2)2 = ¯ σ (1) σ (1)1 ⊗ ¯ σ (2) σ (1)2 ⊗ σ (2) , where σ = σ (1) ⊗ σ (2) = ¯ σ (1) ⊗ ¯ σ (2) . Note that if α = id , then our definition of Drinfeld twists for a bialgebra is inversewith Drinfeld’s (see [20] and [12]). Notation.
We write σ − = ̺ = ̺ (1) ⊗ ̺ (2) = ¯ ̺ (1) ⊗ ¯ ̺ (2) = · · · ∈ H ⊗ H , and σ = σ (2) ⊗ σ (1) . Lemma 4.2. ̺ satisfies ( α ⊗ α )( ̺ ) = ̺ ; (4.1) ̺ satisfies the 2-cocycle condition ̺ (1) ⊗ ̺ (2)1 ¯ ̺ (1) ⊗ ̺ (2)2 ¯ ̺ (2) = ̺ (1)1 ¯ ̺ (1) ⊗ ̺ (1)2 ¯ ̺ (2) ⊗ ̺ (2) . (4.2) Proof. α ⊗ α )( ̺ ) = 1 H ̺ (1) ⊗ H ̺ (2) = ( α − (¯ ̺ (1) ) α − ( σ (1) )) ̺ (1) ⊗ ( α − (¯ ̺ (2) ) α − ( σ (2) )) ̺ (2) T = α − (¯ ̺ (1) )( α − ( σ (1) ) α − ( ̺ (1) )) ⊗ α − (¯ ̺ (2) )( α − ( σ (2) ) α − ( ̺ (2) ))= α − (¯ ̺ (1) )1 H ⊗ α − (¯ ̺ (2) )1 H = ̺. ̺ (1) ⊗ ̺ (2)1 ¯ ̺ (1) ⊗ ̺ (2)2 ¯ ̺ (2) on the left to Eq.(T3). Since( ̺ (1) ⊗ ̺ (2)1 ¯ ̺ (1) ⊗ ̺ (2)2 ¯ ̺ (2) )( σ (1) ⊗ ¯ σ (1) σ (2)1 ⊗ ¯ σ (2) σ (2)2 )= ̺ (1) σ (1) ⊗ α ( ̺ (2)1 )(( α − (¯ ̺ (1) ) α − (¯ σ (1) )) σ (2)1 ) ⊗ α ( ̺ (2)2 )(( α − (¯ ̺ (2) ) α − (¯ σ (2) )) σ (2)2 )= ̺ (1) σ (1) ⊗ ∆( α ( ̺ (2) σ (2) )) = 1 H ⊗ H ⊗ H , we have 1 H ⊗ H ⊗ H = ( ̺ (1) ⊗ ̺ (2)1 ¯ ̺ (1) ⊗ ̺ (2)2 ¯ ̺ (2) )(¯ σ (1) σ (1)1 ⊗ ¯ σ (2) σ (1)2 ⊗ σ (2) ) . (4.3)Secondly, we multiply by ˜ ̺ (1)1 ˙ ̺ (1) ⊗ ˜ ̺ (1)2 ˙ ̺ (2) ⊗ ˜ ̺ (2) on the right to Eq.(4.3). We compute(( ̺ (1) ⊗ ̺ (2)1 ¯ ̺ (1) ⊗ ̺ (2)2 ¯ ̺ (2) )(¯ σ (1) σ (1)1 ⊗ ¯ σ (2) σ (1)2 ⊗ σ (2) ))(˜ ̺ (1)1 ˙ ̺ (1) ⊗ ˜ ̺ (1)2 ˙ ̺ (2) ⊗ ˜ ̺ (2) )= ( α ( ̺ (1) ) ⊗ α ( ̺ (2)1 ¯ ̺ (1) ) ⊗ α ( ̺ (2)2 ¯ ̺ (2) ))((¯ σ (1) σ (1)1 ⊗ ¯ σ (2) σ (1)2 ⊗ σ (2) )( α − (˜ ̺ (1)1 ˙ ̺ (1) ) ⊗ α − (˜ ̺ (1)2 ˙ ̺ (2) ) ⊗ α − (˜ ̺ (2) )))= ( ̺ (1) ⊗ ̺ (2)1 ¯ ̺ (1) ) ⊗ ̺ (2)2 ¯ ̺ (2) )(¯ σ (1) ˙ ̺ (1) ⊗ ¯ σ (2) ˙ ̺ (2) ⊗ H ) (since Eq.(T1) and (4.1))= ̺ (1) ⊗ ̺ (2)1 ¯ ̺ (1) ⊗ ̺ (2)2 ¯ ̺ (2) , thus Eq.(4.2) holds. ✷ Example 4.3.
Let A be a bialgebra over k , σ ∈ A ⊗ A be the usual normalized Drinfeldtwist for A , α : A → A be an invertible bialgebra homomorphism. If σ satisfies ( α ⊗ α )( σ ) = σ , then σ is a Drinfeld twist for α A .Conversely, if ( H, α ) is a monoidal Hom-bialgebra endowed with a Drinfeld twist σ ,then σ is a normalized Drinfeld twist for k -bialgebra α H . The following property is a generalization of (Theorem 1.3, [1]).
Proposition 4.4. ( A, α A , m A , η A ) is an algebra in Rep , ( H ) (the H -Hom-modulealgebra ), define a new multiplication by a ◦ b := ( ̺ (1) · a )( ̺ (2) · b ) , for any a, b ∈ A , then ( A, α A , ◦ , A ) is a monoidal Hom-algebra over k .2).If ( C, α C , ∆ C , ε C ) is a coalgebra in Rep , ( H ) (the H -Hom-module coalgebra ), de-fine a new comultiplication by b ∆ C ( c ) := c (1) ⊗ c (2) = σ (1) · c ⊗ σ (2) · c , for any c ∈ C , then ( C, b ∆ C , ε C ) is a coassociative coalgebra over k . roof. A is a left H -Hom-module algebra iff ( A, α A ) is botha monoidal Hom-algebra over k and a left H -Hom module, such that (cid:26) (1) h · ( ab ) = ( h · a )( h · b );(2) h · A = ε ( h )1 A , for any a, b ∈ A and h ∈ H . Thus we have( a ◦ b ) ◦ α A ( c ) = (( ̺ (1)1 · (¯ ̺ (1) · a ))( ̺ (1)2 · (¯ ̺ (2) · b )))( ̺ (2) · α A ( c )) (4 . = (( α − ( ̺ (1) ) · α A ( a ))( α − ( ̺ (2)1 ¯ ̺ (1) ) · α A ( b )))(( ̺ (2)2 ¯ ̺ (2) ) · α A ( c ))= ( ̺ (1) · α A ( a ))(( ̺ (2)1 · (¯ ̺ (1) · b ))( ̺ (2)2 · (¯ ̺ (2) · c )))= a A ( a ) ◦ ( b ◦ c ) , where a, b, c ∈ A , and 1 A ◦ a = ( ̺ (1) · A )( ̺ (2) · a )= 1 A ( ε ( ̺ (1) ) ̺ (2) · a ) = a A ( a )= a ◦ A . So (
A, α A , ◦ , A ) is a monoidal Hom-algebra over k .2). Note that C is a left H -Hom-module coalgebra if and only if ( C, α C ) is both amonoidal Hom-coalgebra over k and a left H -Hom-module, such that (cid:26) (1) ∆ C ( h · c ) = h · c ⊗ h · c ;(2) ε C ( h · c ) = ε ( h ) ε C ( c ) , for any c ∈ C . Thus we have b ∆ C ( c (1) ) ⊗ c (2) = ¯ σ (1) · ( σ (1)1 · c ) ⊗ ¯ σ (2) · ( σ (1)2 · c ) ⊗ σ (2) · c T = σ (1) · α C ( c ) ⊗ (¯ σ (1) σ (2)1 ) · α C ( c ) ⊗ (¯ σ (2) σ (2)2 ) · c = σ (1) · c ⊗ ( σ (2) · c ) (1) ⊗ ( σ (2) · c ) (2) = c (1) ⊗ b ∆ C ( c (2) ) , and ε C ( c (1) ) c (2) = ε ( σ (1) ) σ (2) · ( ε C ( c ) c )= c = c (1) ε C ( c (2) ) . Hence ( C, b ∆ C , ε C ) is a coassociative coalgebra over k . ✷ For any x ∈ H , define a new comultiplication ∆ σ on H by∆ σ ( x ) = x [1] ⊗ x [2] = ( σ ∆( x )) ̺. (4.4)Note that it is easy to get ∆ σ ( x ) = ( σ ∆( x )) ̺ = σ (∆( x ) ̺ ). Lemma 4.5. ∆ σ is a Hom-algebra map preserving unit. Proof.
For one thing, obviously ∆ σ preserves unit and satisfies ∆ σ ( α ( x )) = ( α ⊗ α )∆ σ ( x ) for any x ∈ H . 12or another, for any x, y ∈ H , we have x [1] y [1] ⊗ x [1] y [1] = (( σ (1) x ) ̺ (1) )((¯ σ (1) y )¯ ̺ (1) ) ⊗ (( σ (2) x ) ̺ (2) )((¯ σ (2) y )¯ ̺ (2) )= (( σ (1) x ) α − ( ̺ (1) ¯ σ (1) )) α ( y ¯ ̺ (1) ) ⊗ (( σ (2) x ) α − ( ̺ (2) ¯ σ (2) )) α ( y ¯ ̺ (2) )= ( α ( σ (1) )( x y )) α (¯ ̺ (1) ) ⊗ ( α ( σ (2) )( x y )) α (¯ ̺ (2) )= ( xy ) [1] ⊗ ( xy ) [2] . (since Eq.(T1) and (4.1))Thus the conclusion holds. ✷ Theorem 4.6. H σ = ( H, α, m, η, ∆ σ , ε ) is a Hom-bialgebra. Proof.
Firstly, for any x ∈ H , we have x [1] ε ( x [2] ) = ( σ (1) x ) ̺ (1) ε ( σ (2) ) ε ( x ) ε ( ̺ (2) )= (1 H α − ( x ))1 H = α ( x )= ε ( σ (1) ) ε ( x ) ε ( ̺ (1) )( σ (2) x ) ̺ (2) = ε ( x [1] ) x [2] . Secondly, we have∆ σ ( x [1] ) ⊗ α ( x [2] )= (¯ σ (1) (( σ (1)1 x ) ̺ (1)1 ))¯ ̺ (1) ⊗ (¯ σ (2) (( σ (1)2 x ) ̺ (1)2 ))¯ ̺ (2) ⊗ α (( σ (2) x ) ̺ (2) )= (( α − (¯ σ (1) ) α ( σ (1)1 )) α ( x ))( α ( ̺ (1)1 ) α − (¯ ̺ (1) )) ⊗ (( α − (¯ σ (2) ) α ( σ (1)2 )) α ( x ))( α ( ̺ (1)2 ) α − (¯ ̺ (2) )) ⊗ α (( σ (2) x ) ̺ (2) )= ((¯ σ (1) σ (1)1 ) α ( x ))( ̺ (1)1 ¯ ̺ (1) ) ⊗ ((¯ σ (2) σ (1)2 ) α ( x ))( ̺ (1)2 ¯ ̺ (2) ) ⊗ ( σ (2) α ( x )) ̺ (2)( T , (4 . = ( σ (1) α ( x )) ̺ (1) ⊗ ((¯ σ (1) σ (2)1 ) α ( x ))( ̺ (2)1 ¯ ̺ (1) ) ⊗ ((¯ σ (2) σ (2)2 ) α ( x ))( ̺ (2)2 ¯ ̺ (2) )= α (( σ (1) x ) ̺ (1) ) ⊗ (( α − (¯ σ (1) ) α ( σ (2)1 )) α ( x ))( α ( ̺ (2)1 ) α − (¯ ̺ (1) )) ⊗ (( α − (¯ σ (2) ) α ( σ (2)2 )) α ( x ))( α ( ̺ (2)2 ) α − (¯ ̺ (2) ))= α (( σ (1) x ) ̺ (1) ) ⊗ (¯ σ (1) (( σ (2)1 x ) ̺ (2)1 ))¯ ̺ (1) ⊗ (¯ σ (2) (( σ (2)2 x ) ̺ (2)2 ))¯ ̺ (2) = α ( x [1] ) ⊗ ∆ σ ( x [2] ) . Finally, we have∆ σ ( α ( x )) = ( σ (1) α ( x )) ̺ (1) ⊗ ( σ (2) α ( x )) ̺ (2) = α (( σ (1) x ) ̺ (1) ) ⊗ α (( σ (2) x ) ̺ (2) ) , (since Eq.(T1) and (4.1))which implies ( H, α, ∆ σ , ε ) is a Hom-coalgebra.Since ( H, α, m, η ) is already a Hom-algebra, by Lemma 4.5, the conclusion holds. ✷ For any given monoidal Hom-bialgebra (
H, α ) endowed with a Drinfeld twist σ ∈ H ⊗ H , we already know that α H is a bialgebra and σ is a Drinfeld twist on it. Thus( α H ) σ is a new bialgebra which coproduct is also given by Eq.(4.4). By Example 2.2,((( α H ) σ ) α , α ) is a Hom-bialgebra. 13 heorem 4.7. (( α H ) σ ) α = H σ as Hom-bialgebras. Furthermore, ( α H ) α = H σ if andonly if σ = 1 H ⊗ H . Proof.
We only need to show ∆ (( α H ) σ ) α = ∆ H σ .In fact, denote the multiplication m α H ( x ⊗ y ) = α − ( x ) α − ( y ) by x ∗ y for any x, y ∈ H .We have ∆ (( α H ) σ ) α ( x ) = ∆ ( α H ) σ ( α ( x ))= ( σ ∗ ∆ α H ( α ( x ))) ∗ ̺ = ( σ ∗ ∆( α ( x ))) ∗ ̺ = ( σ ∆( x )) ̺. Thus the conclusion holds. ✷ Theorem 4.8. If ( H, α, S ) is a monoidal Hom-Hopf algebra, then H σ is a Hom-Hopfalgebra. Proof.
Firstly, since(1 H ⊗ ¯ ̺ (1) ⊗ ¯ ̺ (2) )(( σ (1) ⊗ ¯ σ (1) σ (2)1 ⊗ ¯ σ (2) σ (2)2 )( α ( ̺ (1)1 ) ⊗ α ( ̺ (1)2 ) ⊗ ̺ (2) ))= (1 H ⊗ ¯ ̺ (1) ⊗ ¯ ̺ (2) )((¯ σ (1) σ (1)1 ⊗ ¯ σ (2) σ (1)2 ⊗ σ (2) )( α ( ̺ (1)1 ) ⊗ α ( ̺ (1)2 ) ⊗ ̺ (2) )) , we have α ( σ (1) ) ⊗ ̺ (1) σ (2) ⊗ α ( ̺ (2) ) = σ (1) α ( ̺ (1)1 ) ⊗ α ( σ (2)1 ) α ( ̺ (1)2 ) ⊗ α ( σ (2)2 ) ̺ (2) . (4.5)Define S σ : H → H by S σ ( x ) = ( σ (1) ( S ( α − ( σ (2) ))( S ( α − ( x )) S ( α − ( ̺ (1) ))))) ̺ (2) , for any x ∈ H , where p, q ∈ Z . Obviously S ◦ α = α ◦ S .Since S σ ( x [1] ) x [2] = (( σ (1) ( S ( α − ( σ (2) ))(( S ( α − ( ̺ (1) ))( S ( α − ( x )) S ( α − ( σ (1) )))) S ( α − ( ̺ (1) ))))) ̺ (2) )(( σ (2) x ) ̺ (2) )= (( σ (1) ( S ( α − ( σ (2) ))(( S ( α − ( ̺ (1) )) S ( α − ( x ))) S ( α − ( ̺ (1) ) α − ( σ (1) )))))( α − ( ̺ (2) ) σ (2) ))( α ( x ) ̺ (2) )= ( α ( σ (1) )( S ( σ (2) )(( S ( α − ( ̺ (1) )) S ( α − ( x ))))))( α ( x ) ̺ (2) )= (( σ (1) ( S ( α − ( σ (2) )) S ( α − ( ̺ (1) )))) S ( α ( x )))( α ( x ) ̺ (2) )= (( σ (1) ( S ( α − ( σ (2) )) S ( α − ( ̺ (1) ))))( S ( x ) x )) α ( ̺ (2) )= ( α ( σ (1) )( S ( σ (2) ) S ( α − ( ̺ (1) )))) α ( ̺ (2) ) ε ( x )= ( α ( σ (1) )( S ( σ (2) ) S ( ̺ (1) ))) α ( ̺ (2) ) ε ( x )= ( α ( σ (1) ) S ( ̺ (1) σ (2) )) α ( ̺ (2) ) ε ( x ) (4 . = (( σ (1) α ( ̺ (1)1 ))( S ( α ( ̺ (1)2 )) S ( α ( σ (2)1 ))))( α ( σ (2)2 ) α ( ̺ (2) )) ε ( x )= (( σ (1) ( ̺ (1)1 S ( α ( ̺ (1)2 )))) S ( α ( σ (2)1 )))( α ( σ (2)2 ) α ( ̺ (2) )) ε ( x )14 α ( σ (1) )( S ( α ( σ (2)1 )) α ( σ (2)2 )) ε ( x )= α ( σ (1) ) ε ( σ (2) ) ε ( x ) = 1 H ε ( x ) , which implies S σ is the antipode of H σ . ✷ Proposition 4.9. If ( H, α ) is a quasitiangular monoidal Hom-bialgebra with the R -matrix R = R (1) ⊗ R (1) , then H σ is a quasitriangular Hom-bialgebra. Proof.
Define R σ = ( σ R ) ̺ ∈ H ⊗ H , we will prove R σ is an R -matrix in H σ .Firstly, we will check Eq.(q2). For any x ∈ H , we have∆ σop ( x ) R σ = (( σ (2) x ) ̺ (2) )((¯ σ (2) R (1) )¯ ̺ (1) ) ⊗ (( σ (1) x ) ̺ (1) )((¯ σ (1) R (2) )¯ ̺ (2) )= ( σ (2) ( x (( ̺ (2) ¯ σ (2) ) R (1) )))¯ ̺ (1) ⊗ ( σ (1) ( x (( ̺ (1) ¯ σ (1) ) R (2) )))¯ ̺ (2)( Q = ( σ (2) ( R (1) x ))¯ ̺ (1) ⊗ ( σ (1) ( R (2) x ))¯ ̺ (2) = (( σ (2) R (1) ) ̺ (1) )((¯ σ (1) x )¯ ̺ (1) ) ⊗ (( σ (1) R (2) ) ̺ (2) )((¯ σ (2) x )¯ ̺ (2) )= R σ ∆ σ ( x ) . Secondly, to verify Eq.(q3), we compute∆ σ ( R σ (1) ) ⊗ α ( R σ (2) )= (¯ σ (1) (( σ (2)1 R (1)1 ) ̺ (1)1 ))¯ ̺ (1) ⊗ (¯ σ (2) (( σ (2)2 R (1)2 ) ̺ (1)2 ))¯ ̺ (2) ⊗ α (( σ (1) R (2) ) ̺ (2) ) ( Q = ((¯ σ (1) σ (2)1 ) α ( R (1) ))( ̺ (1)1 ¯ ̺ (1) ) ⊗ ((¯ σ (2) σ (2)2 ) α ( r (1) ))( ̺ (1)2 ¯ ̺ (2) ) ⊗ ( σ (1) ( α ( R (2) r (2) ))) ̺ (2)( T , (4 . = ( α (¯ σ (2) )( σ (1)2 R (1) )) ̺ (1) ⊗ ( σ (2) ( r (1) α − ( ̺ (2)1 ))) α (¯ ̺ (1) ) ⊗ ( α (¯ σ (1) )( σ (1)1 R (2) ))(( r (2) α − ( ̺ (2)2 ))¯ ̺ (2) ) ( Q = (¯ σ (2) ( R (1) σ (1)1 )) α ( ̺ (1) ) ⊗ ( σ (2) ( ̺ (2)2 r (1) )) α (¯ ̺ (1) ) ⊗ (¯ σ (1) ( R (2) σ (1)2 ))(( ̺ (2)1 r (2) )¯ ̺ (2) )= α (( α − (¯ σ (2) ) R (1) )( σ (1)1 α − ( ̺ (1) ))) ⊗ ( σ (2) α ( ̺ (2)2 ))( α ( r (1) )¯ ̺ (1) ) ⊗ (( α − (¯ σ (1) ) R (2) )( σ (1)2 ̺ (2)1 ))( α ( r (2) )¯ ̺ (2) )= α (( α − (¯ σ (2) ) R (1) )(( ˙ ̺ (1) ˙ σ (1) ) α − ( σ (1)1 α − ( ̺ (1) )))) ⊗ ( σ (2) α ( ̺ (2)2 ))( α ( r (1) )¯ ̺ (1) ) ⊗ (( α − (¯ σ (1) ) R (2) )(( ˙ ̺ (2) ˙ σ (2) ) α − ( σ (1)2 ̺ (2)1 )))( α ( r (2) )¯ ̺ (2) ) ( T = α (( α − (¯ σ (2) ) R (1) )(( ˙ ̺ (1) α − ( σ (1) )) α − ( ̺ (1) ))) ⊗ (( ˙ σ (2) σ (2)2 ) α ( ̺ (2)2 ))( α ( r (1) )¯ ̺ (1) ) ⊗ (( α − (¯ σ (1) ) R (2) )(( ˙ ̺ (2) α − ( ˙ σ (1) σ (2)1 )) ̺ (2)1 ))( α ( r (2) )¯ ̺ (2) )= α ((¯ σ (2) R (1) ) α ( ˙ ̺ (1) )) ⊗ α ( ˙ σ (2) )( r (1) ¯ ̺ (1) ) ⊗ ((¯ σ (1) R (2) )( ˙ ̺ (2) ˙ σ (1) ))( r (2) ¯ ̺ (2) )= ( α (¯ σ (2) ) α ( R (1) )) α ( ˙ ̺ (1) ) ⊗ ( α ( ˙ σ (2) ) r (1) ) α (¯ ̺ (1) ) ⊗ α − ((( α (¯ σ (1) ) α ( R (2) )) α ( ˙ ̺ (2) ))(( α ( ˙ σ (1) ) r (2) ) α (¯ ̺ (2) )))= α ( R σ (1) ) ⊗ α ( r σ (1) ) ⊗ R σ (2) r σ (2) . Eq.(q4) could be obtained in a similar way.15t last, since ( α ⊗ α )( R σ ) = R σ , ( H σ , α, R σ ) is a quasitriangular Hom-bialgebra. ✷ Recall from [21] that if (
B, α B ) is a Hom-bialgebra over k and α B is invertible, then therepresentation category Rep i,j ( B ) of B is a monoidal category with the following structure: • the objects are left B -Hom-modules and the morphism f : ( M, α M ) → ( N, α N ) is B -linear and satisfies f ◦ α M = α N ◦ f ; • the tensor product of ( M, α M ) and ( N, α N ) is ( M ⊗ N, α M ⊗ α N ), with the B -actiongiven by b · ( m ⊗ n ) = α iB ( b ) · m ⊗ α jB ( b ) · n, b ∈ B, m ∈ M, n ∈ N ; • the unit object is ( k, id k ) with the B -action b · λ = ε ( b ) λ , where b ∈ B , λ ∈ k ; • the associativity constraint a is given by a M,N,P (( m ⊗ n ) ⊗ p ) = α − i − M ( m ) ⊗ ( n ⊗ α j +1 P ( p )); • the left unit constraint l is given by l M ( λ ⊗ m ) = λα − j − M ( m ) , for any m ∈ M and λ ∈ k ; • the right unit constraint r is given by r M ( m ⊗ λ ) = λα − i − M ( m ) , for any m ∈ M and λ ∈ k .Furthermore, if ( B, α B ) is a quasitriangular Hom-bialgebra with the R -matrix R = R (1) ⊗ R (2) , then Rep i,j ( B ) is a braided category with the braiding c M,N ( m ⊗ n ) = α iB ( R (2) ) · α i − j − N ( n ) ⊗ α jB ( R (1) ) · α j − i − M ( m ) , for any m ∈ M , n ∈ N . Theorem 4.10.
Rep i +3 ,j +3 ( H ) and Rep i,j ( H σ ) are isomorphic as monoidal categories.Moreover, if H is a quasitriangular monoidal Hom-bialgebra, then Rep i +3 ,j +3 ( H ) and Rep i,j ( H σ ) are braided isomorphic. Proof.
Define a functor G = ( G, G , G ) : ( Rep i +3 ,j +3 ( H ) , ⊗ , k, a, l, r ) → ( Rep i,j ( H σ ) , ¯ ⊗ , k, ¯ a, ¯ l, ¯ r )by G ( M ) := M, G ( f ) := f, G = id k , and G ( M, N ) : G ( M ) ¯ ⊗ G ( N ) → G ( M ⊗ N ) , m ¯ ⊗ n α i ( ̺ (1) ) · m ⊗ α j ( ̺ (2) ) · n, where the H σ -Hom-module structure of M is given by h ⇀ m = h · m, h ∈ H σ , m ∈ M, m ∈ M , n ∈ N , and f : M → N ∈ M or ( Rep i +3 ,j +3 ( H )). Obviously G is naturaland compatible with the Hom structure map.Firstly, since G ( M, N ) is invertible, and G ( M, N )( h ⇀ ( m ¯ ⊗ n ))= G ( M, N )( α i ( h [1] ) ⇀ m ¯ ⊗ α j ( h [2] ) ⇀ n )= α i +3 ( ̺ (1) ) · ( α i (( σ (1) h )¯ ̺ (1) ) · m ) ⊗ α j +3 ( ̺ (2) ) · ( α j (( σ (2) h )¯ ̺ (2) ) · n )= (( α i +1 ( ̺ (1) ) α i +1 ( σ (1) ))( α i +1 ( h ) α i (¯ ̺ (1) ))) · α M ( m ) ⊗ (( α j +1 ( ̺ (2) ) α j +1 ( σ (2) ))( α j +1 ( h ) α j (¯ ̺ (2) ))) · α N ( n )= α i +3 ( h ) · ( α i +1 (¯ ̺ (1) ) · m ) ⊗ α j +3 ( h ) · ( α j +1 (¯ ̺ (2) ) · n )= h · ( α i (¯ ̺ (1) ) · m ⊗ α j (¯ ̺ (2) ) · n ))= h ⇀ ( G ( M, N )( m ¯ ⊗ n )) ,G is a natural isomorphism in Rep i,j ( H σ ).Secondly, we have( G ( M, N ⊗ P ) ◦ ( id G ( M ) ¯ ⊗ G ( N, P )) ◦ (¯ a G ( M ) ,G ( N ) ,G ( P ) ))(( m ¯ ⊗ n ) ¯ ⊗ p )= ( G ( M, N ⊗ P ) ◦ ( id G ( M ) ¯ ⊗ G ( N, P )))( α − i − M ( m ) ¯ ⊗ ( n ¯ ⊗ α j +1 P ( p )))= α i (¯ ̺ (1) ) · α − i − M ( m ) ⊗ α j (¯ ̺ (2) ) · ( α i ( ̺ (1) ) · n ⊗ α j ( ̺ (2) ) · α j +1 P ( p ))= α i (¯ ̺ (1) ) · α − i − M ( m ) ⊗ (( α i + j +2 (¯ ̺ (2)1 ) α i ( ̺ (1) )) · α N ( n ) ⊗ ( α j +2 (¯ ̺ (2)2 ) α j ( ̺ (2) )) · α j +2 P ( p ))= α i − j − (¯ ̺ (1) ) · α − i − M ( m ) ⊗ (( α i (¯ ̺ (2)1 ) α i ( ̺ (1) )) · α N ( n ) ⊗ ( α j (¯ ̺ (2)2 ) α j ( ̺ (2) )) · α j +2 P ( p )) (4 . = α i − j − (¯ ̺ (1)1 ̺ (1) ) · α − i − M ( m ) ⊗ ( α i (¯ ̺ (1)2 ̺ (2) ) · α N ( n ) ⊗ α j (¯ ̺ (2) ) · α j +2 P ( p ))= ( α i (¯ ̺ (1)1 ) α − ( ̺ (1) )) · α − i − M ( m ) ⊗ (( α i + j +2 (¯ ̺ (1)2 ) α j ( ̺ (2) )) · α N ( n ) ⊗ ( a j +2 (¯ ̺ (2) )) · α j +2 P ( p ))= G ( a M,N,P )((( α i +2 (¯ ̺ (1)1 ) α i ( ̺ (1) )) · α M ( m ) ⊗ ( α i + j +2 (¯ ̺ (1)2 ) α j ( ̺ (2) )) · α N ( n )) ⊗ a j (¯ ̺ (2) ) · p )= G ( a M,N,P )( α i (¯ ̺ (1) ) · ( α i ( ̺ (1) ) · m ⊗ α j ( ̺ (2) ) · n ) ⊗ a j (¯ ̺ (2) ) · p )= ( G ( a M,N,P ) ◦ G ( M ⊗ N, P ) ◦ ( G ( M, N ) ¯ ⊗ id P ))(( m ¯ ⊗ n ) ¯ ⊗ p ) . At last, it is easy to get G ( l M ) ◦ G ( k, M ) ◦ ( G ¯ ⊗ id M ) = ¯ l G ( M ) , and G ( r M ) ◦ G ( M, k ) ◦ ( id M ¯ ⊗ G ) = ¯ r G ( M ) , hence G = ( G, G , G ) is a monoidal functor.Obviously G is invertible, thus Rep i +3 ,j +3 ( H ) and Rep i,j ( H σ ) are monoidal isomorphic.Furthermore, if R ∈ H ⊗ H is an R -matrix, from Proposition 4.9, ( H σ , α, R σ ) is aquasitriangular Hom-bialgebra. Suppose that the braiding in Rep i +3 ,j +3 ( H ) is c and thebraiding in Rep i,j ( H σ ) is ¯ c , then for any m ∈ M , n ∈ N , we have( G ( N, M ) ◦ ¯ c G ( M ) ,G ( N ) )( m ¯ ⊗ n )= G ( N, M )( α i ( σ (1) ( R (2) ̺ (2) )) ⇀ α i − j − N ( n ) ¯ ⊗ α j ( σ (2) ( R (1) ̺ (1) )) ⇀ α j − i − M ( m ))= α i ((¯ ̺ (1) σ (1) )( R (2) ̺ (2) )) · α i − jN ( n ) ⊗ α j ((¯ ̺ (2) σ (2) )( R (1) ̺ (1) )) · α j − iM ( m )= G ( c M,N )( α i ( ̺ (1) ) · m ⊗ α j ( ̺ (2) ) · n ) = ( G ( c M,N ) ◦ G ( M, N ))( m ⊗ ′ n ) , G is a braided monoidal functor. ✷ ACKNOWLEDGEMENT
The work was partially supported by the Fundamental Research Funds for the CentralUniversities (NO. 3207013906), and the NSF of China (NO. 11371088), and the NSF ofJiangsu Province (NO. BK2012736).
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