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Vietnam Journal of Mathematics 34:2 (2006) 189–207 The Quantum Double of a Dual Andruskiewitsch-Schneider Algebra Is a Tame Algebra * Meihua Shi Dept. of Math. Zhejiang Education Institute Hangzhou, Zhejiang 310012, China Received July 5, 2005 Revised December 22, 2005 Abstract. In this paper, we study the representation theory of the quantum double D(Γ n,d ). We give the structure of projective modules of D(Γ n,d ) at first. By this, we give the Ext-quiver (with relations) of D(Γ n,d ) and show that D(Γ n,d ) is a tame algebra. 2000 Mathematics Subject Classification: 16W30 Keywords: Representation Theory, Quantum double Tame Algebra 1. Introduction In this paper, k is an algebraically closed field of characteristic 0 and an algebra is a finite dimensional associative k-algebra with identity element. Although the quantum doubles of finite dimension Hopf algebras are impor- tant, not very much is known about their representations in general. A complete list of simple modules of the quantum doubles of Taft algebras is given by Chen in [2]. He also gives all indecomposable modules for the quantum double of a special Taft algebra in [3]. From this, we can deduce immediately that the quantum double of this special Taft algebra is tame. The authors of [7] study ∗ Project(No.10371107) supported by the Natural Science Foundation of China. 190 Meihua Shi the representation theory of the quantum doubles of the duals of the general- ized Taft algebras in detail. They describe all simple modules, indecomposable modules, quivers with relations and AR-quivers of the quantum doubles of the duals of the generalized Taft algebras explicitly and show that these quantum doubles are tame. The structures of basic Hopf algebras of finite representation type are gotten in [11]. In fact, the authors of [11] show that basic Hopf algebras of finite representation type and monomial Hopf algebras (see [4]) are the same. But for the structure of tame basic Hopf algebras, we know little. In [10], the author gives the structure theorem for tame basic Hopf algebras in the graded case. In order to study tame basic Hopf algebras or generally tame Hopf algebras, we need more examples of tame Hopf algebras. The Andruskiewitsch-Schneider algebra is a kind of generalization of gener- alized Taft algebra and of course Taft algebra. Therefore, it is natural to ask the following question: whether is the quantum double of dual Andruskiewitsch- Schneider algebra a tame algebra? In this paper, we give an affirmative answer. As a consequence, we give some new examples of tame Hopf algebra. Our method is direct. Explicitly, we firstly give the structure of projective modules of the Drinfel Double of a dual Andruskiewitsch-Schneider algebra by direct computations. Then using this, we get its Ext-quiver with relations which will help us to get the desired conclusion. 2. Main Results In this section, we will study the Drinfeld Double D(Γ n,d ), which is a general- ization of [7], of (A(n, d, μ, q)) ∗cop . Our main result is to give the structure of projective modules of D(Γ n,d ). By this, we give the Ext-quiver (with relations) of D (Γ n,d )andshowthatD(Γ n,d ) is a tame algebra. This section relays heavily on [7] and we refer the reader to this paper. The algebra Γ n,d := kZ n /J d with d|n is described by quiver and relations. The quiver is a cycle, with n vertices e 0 , ,e n−1 . We shall denote by γ m i the path of length m starting at the vertex e i . The relations are all paths of length d  2. We give the Hopf structure on Γ n,d . We fix a primitive d-th root of unity q The Quantum Double of a Dual Andruskiewitsch-Schneider 191 and a μ ∈ k. Δ(e t )=  j+l=t e j ⊗ e l + α 0 t − β 0 t , Δ(γ 1 t )=  j+l=t e j ⊗ γ 1 l + q l γ 1 j ⊗ e l + α 1 t − β 1 t ε(e t )=δ t0 ,ε(γ 1 t )=0,S(e t )=e −t ,S(γ 1 t )=−q t+1 γ 1 −t−1 where α s t = μ d−1  l=s+1  i+j= t q jl (s)! q l! q (d − l + s)! q γ l i ⊗ γ d+s−l j , β s t = d−1  l=s+1  i+j+ d=t q jl (s)! q l! q (d − l + s)! q γ l i ⊗ γ d+s−l j . Proposition 2.1. With above comultiplication, counit and antipode, Γ n,d is a Hopf algebra. Proof. We only prove that Δ is an algebra morphism. The other axioms of Hopf algebras can be proved easily from this. In order to do it, it is enough to prove that, for s, t ∈{0, ··· ,n− 1}, Δ(e s )Δ(e t )=Δ(δ st e t ), Δ(γ 1 s e t )=Δ(γ 1 s )Δ(e t ), Δ(e t γ 1 s )=Δ(e t )Δ(γ 1 s ). We have Δ(e s )Δ(e t )=   j+l=s e j ⊗ e l + α 0 s − β 0 s   j+l=t e j ⊗ e l + α 0 t − β 0 t  =   j+l=s e j ⊗ e l   j+l=t e j ⊗ e l  +   j+l=s e j ⊗ e l  α 0 t −   j+l=s e j ⊗ e l  β 0 t + α 0 s   j+l=t e j ⊗ e l  − β 0 s   j+l=t e j ⊗ e l  + r where r = α 0 s α 0 t −α 0 s β 0 t −β 0 s α 0 t +β 0 s β 0 t and clearly r ∈ J d ⊗kZ n +kZ n ⊗J d .Thus r =0. Notethatinα 0 t = μ  d−1 l=1  i+j= t q jl l! q (d−l)! q γ l i ⊗ γ d−l j every component, say γ l i ⊗ γ d−l j , the end point of γ l i is e i+l and that of γ d−l j is e j+d−l .Thus (e m ⊗ e n )(γ l i ⊗ γ d−l j ) =0 implies m + n = i + l + j + d − l = t + d. Similarly, (γ l i ⊗ γ d−l j )(e m ⊗ e n ) =0 implies m + n = t. Therefore, if s = t + p,   j+l=s e j ⊗ e l  α 0 t =0,β 0 s   j+l=t e j ⊗ e l  =0 and if s = t   j+l=s e j ⊗ e l  β 0 t =0,α 0 s   j+l=t e j ⊗ e l  =0 192 Meihua Shi Thus if s = t + p and s = t,Δ(e s )Δ(e t )=0. If s = t + p, Δ(e s )Δ(e t )=   j+l=s e j ⊗ e l  α 0 t − β 0 s   j+l=t e j ⊗ e l  = α 0 t − β 0 s = μ d−1  l=1  i+j= t q jl l! q (d − l)! q γ l i ⊗ γ d−l j − μ d−1  l=1  i+j+ d=t+d q jl l! q (d − l)! q γ l i ⊗ γ d−l j =0. If s = t,Δ(e s )Δ(e t )=  j+l=s e j ⊗e l −(  j+l=s e j ⊗e l )β 0 t +α 0 t (  j+l=t e j ⊗e l )=  j+l=s e j ⊗ e l − β 0 t + α 0 t =Δ(e t ). Thus,inaword,Δ(e s )Δ(e t )=Δ(δ st e t ). Next, let us show that Δ(γ 1 s e t )=Δ(γ 1 s )Δ(e t ). Δ(γ 1 s )Δ(e t )=   j+l=s (e j ⊗ γ 1 l + q l γ 1 j ⊗ e l )+α 1 s − β 1 s   j+l=t e j ⊗ e l + α 0 t − β 0 t  =   j+l=s e j ⊗ γ 1 l + q l γ 1 j ⊗ e l   j+l=t e j ⊗ e l  +   j+l=s e j ⊗ γ 1 l + q l γ 1 j ⊗ e l  α 0 t −   j+l=s e j ⊗ γ 1 l + q l γ 1 j ⊗ e l  β 0 t + α 1 s   j+l=t e j ⊗ e l  − β 1 s   j+l=t e j ⊗ e l  . Similar to the computation of Δ(e s )Δ(e t )=Δ(δ st e t ), if s = t,   j+l=s e j ⊗ γ 1 l + q l γ 1 j ⊗ e l  β 0 t =0,α 1 s v(  j+l=t e j ⊗ e l  =0 and if s = t + p,   j+l=s e j ⊗ γ 1 l + q l γ 1 j ⊗ e l  α 0 t =0,β 1 s   j+l=t e j ⊗ e l  =0. Thus if s = t and s = t + p,Δ(γ 1 s )Δ(e t )=0. If s = t + p, Δ(γ 1 s )Δ(e t )=   j+l=s e j ⊗ γ 1 l + q l γ 1 j ⊗ e l  α 0 t − β 1 s (  j+l=t e j ⊗ e l ) The Quantum Double of a Dual Andruskiewitsch-Schneider 193 =  j+l=s (e j ⊗ γ 1 l + q l γ 1 j ⊗ e l )  μ d−1  l=1  i+j= t q jl l! q (d − l)! q γ l i ⊗ γ d−l j  − β 1 s = μ d−1  l=1  i+j= t q jl l! q (d − l)! q (γ l i ⊗ γ d−l+1 j + q j+d−l γ l+1 i ⊗ γ d−l j ) − β 1 s = μ d−1  l=2  i+j= t  q jl l! q (d − l)! q + q j(l−1) q j+d−l+1 1 (l − 1)! q (d − l +1)! q  γ l i ⊗ γ d−l+1 j − β 1 s = μ d−1  l=2  i+j= t q jl l! q (d − l +1)! q γ l i ⊗ γ d−l+1 j − β 1 s =0. If s = t, Δ(γ 1 s )Δ(e t )=   j+l=s e j ⊗ γ 1 l + q l γ 1 j ⊗ e l  −   j+l=s e j ⊗ γ 1 l + q l γ 1 j ⊗ e l  β 0 t + α 1 t   j+l=t e j ⊗ e l  =   j+l=s e j ⊗ γ 1 l + q l γ 1 j ⊗ e l  − β 1 t + α 1 t =Δ(γ 1 t ) where the second equality can be gotten by a similar computation in the case of s = t + p. Therefore, we have Δ(γ 1 s e t )=Δ(γ 1 s )Δ(e t ). The equality Δ(e t γ 1 s )=Δ(e t )Δ (γ 1 s ) can be gotten similarly.  By [11], we know that the most typical examples of basic Hopf algebras of finite representation type are Taft algebras and the dual of A(n, d, μ, q), which as an associative algebra is generated by two elements g and x with relations g n =1,x d = μ(1 − g d ),xg= qgx with comultiplication Δ, counit ε, and antipode S given by Δ(g)=g ⊗ g, Δ(x)=1⊗ x + x ⊗ g ε(g)=1,ε(x)=0 S(g)=g −1 ,S(x)=−xg −1 . We call this Hopf algebra the Andruskiewitsch-Schneider algebra. If μ =0,then it is the so-called generalized Taft algebra (see [8]). If μ =0andd = n,then clearly it is the usual Taft algebra. 194 Meihua Shi Lemma 2.2. As a Hopf algebra, (Γ n,d ) ∗cop ∼ = A(n, d, μ, q) by  γ 0 1 → G,  γ 1 0 → X and Δ(γ m i )=   s+t=i,v+l=m  m v  q q vt γ v s ⊗ γ l t  + α m i − β m i . Proof. It is a direct computation.  We always denote  s+t=i,v+l=m  m v  q q vt γ v s ⊗ γ l t by M m i . Lemma 2.3. (id ⊗ Δ)M m l =  m 1 +m 2 +m 3 =m,l 1 +l 2 +l 3 =l q m 1 (l 2 +l 3 )+m 2 l 3 m! q (m 1 )! q (m 2 )! q (m 3 )! q γ m 1 l 1 ⊗ γ m 2 l 2 ⊗ γ m 3 l 3 , (id ⊗ Δ)α m l = μ  m 1 +m 2 +m 3 =d+m,l 1 +l 2 +l 3 =l q m 1 (l 2 +l 3 )+m 2 l 3 m! q (m 1 )! q (m 2 )! q (m 3 )! q γ m 1 l 1 ⊗ γ m 2 l 2 ⊗ γ m 3 l 3 , (id ⊗ Δ)β m l = μ  m 1 +m 2 +m 3 =d+m,l 1 +l 2 +l 3 +d=l q m 1 (l 2 +l 3 )+m 2 l 3 m! q (m 1 )! q (m 2 )! q (m 3 )! q γ m 1 l 1 ⊗ γ m 2 l 2 ⊗ γ m 3 l 3 . Proposition 2.4. The Drinfeld double D(Γ n,d ) is described as follows: as a coalgebra, it is (Γ n,d ) ∗cop ⊗ Γ n,d . We write the basis elements G i X j γ m l ,with i, l ∈{0, 1, ,n− 1}, 0 ≤ j, m ≤ d − 1. The following relations determined the algebra structure completely: G n =1,X d = μ(1 − G d ),GX= q −1 XG the product of elements γ m l is the usual product of paths, and γ m l G = q −m Gγ m l (∗1) in particular e l G = Ge l since e l = γ 0 l by the definition of γ m l ,and γ m l X =  q −m Xγ m l+1 − q −m (m) q γ m−1 l+1 + q l+1−m (m) q Gγ m−1 l+1 if m  1 Xγ 0 l+1 − μ (d−1)! q (γ d−1 l+1 − γ d−1 l+1−d )+ μq l+1 (d−1)! q G(γ d−1 l+1 − γ d−1 l+1−d ) if m =0. (∗2) Proof. We only prove equality (∗1), (∗2). For equality (∗1), by the definition of Drinfeld double, γ m l G =(1⊗ γ m l )(  γ 0 1 ⊗ 1) =  m 1 +m 2 +m 3 =m,l 1 +l 2 +l 3 =l C m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 0 1 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 (I) + μ  m 1 +m 2 +m 3 =d+m,l 1 +l 2 +l 3 =l C m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 0 1 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 (II) − μ  m 1 +m 2 +m 3 =d+m,l 1 +l 2 +l 3 +d=l C m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 0 1 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 (III) The Quantum Double of a Dual Andruskiewitsch-Schneider 195 where C m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3 = q m 1 (l 2 +l 3 )+m 2 l 3 m! q (m 1 )! q (m 2 )! q (m 3 )! q . By observation, we can find the following results. For term (I),  γ 0 1 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) = 0 only if l 1 =1,l 3 = n − 1,l 2 = l, m 1 = 0,m 3 =0,m 2 = m.InthiscaseC m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3 = q −m .Thus(I)=q −m  γ 0 1 ⊗ γ m l = q −m Gγ m l . In a similar way, we can find (II)=0and(III)=0. Thus(∗1) is proved. For equality (∗2), γ m l X =(1⊗ γ m l )(  γ 1 0 ⊗ 1) =  m 1 +m 2 +m 3 =m,l 1 +l 2 +l 3 =l C m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 (I) + μ  m 1 +m 2 +m 3 =d+m,l 1 +l 2 +l 3 =l C m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 (II) − μ  m 1 +m 2 +m 3 =d+m,l 1 +l 2 +l 3 +d=l C m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 (III) For term (I), it is easy to find that there are only three cases satisfying  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) =0. Theyare (1): l 1 =0,l 2 = l +1,l 3 = n − 1,m 1 =0,m 2 = m − 1,m 3 =1 (2): l 1 =0,l 2 = l +1,l 3 = n − 1,m 1 =0,m 2 = m, m 3 =0 (3): l 1 =0,l 2 = l +1,l 3 = n − 1,m 1 =1,m 2 = m − 1,m 3 =0. For case (1), it is straightforward to prove that C m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 = −q −m (m) q 1 ⊗ γ m−1 l+1 . For case (2), we have C m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 = −q −m  γ 1 0 ⊗ γ m l+1 = −q −m Xγ m l+1 . For case (3), we have C m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 = − q l+1−m (m) q  γ 0 1 ⊗ γ m−1 l+1 = −q l+1−m (m) q Gγ m−1 l+1 . For term (II), there are also three possible cases such that  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) = 0. They are (1): l 1 =0,l 2 = l +1,l 3 = n − 1,m 1 =0,m 2 = d + m − 1,m 3 =1 (2): l 1 =0,l 2 = l +1,l 3 = n − 1,m 1 =0,m 2 = d + m, m 3 =0 (3): l 1 =0,l 2 = l +1,l 3 = n − 1,m 1 =1,m 2 = d + m − 1,m 3 =0. Thus if m  1, we have that γ m 2 l 2 ∈ J d which is zero by the definition of Γ n,d .Thus  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 = 0 only if m =0. 196 Meihua Shi Assume m = 0. For case (1), μC m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 = −μ (d − 1)! q γ d−1 l+1 . For case (2), μC m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 =0. For case (3), μC m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 = μq l+1 (d − 1)! q Gγ d−1 l+1 . For term (III), there are also three cases which we need to consider. (1): l 1 =0,l 2 = l +1− d, l 3 = n − 1,m 1 =0,m 2 = d + m − 1,m 3 =1 (2): l 1 =0,l 2 = l +1− d, l 3 = n − 1,m 1 =0,m 2 = d + m, m 3 =0 (3): l 1 =0,l 2 = l +1− d, l 3 = n − 1,m 1 =1,m 2 = d + m − 1,m 3 =0. If m  1, we also have term (III) = 0. Assume m = 0. For case (1), μC m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 = −μ (d − 1)! q γ d−1 l+1−d . For case (2), μC m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 =0. For case (3), μC m 1 ,m 2 ,m 3 l 1 ,l 2 ,l 3  γ 1 0 (S −1 (γ m 3 l 3 )?γ m 1 l 1 ) ⊗ γ m 2 l 2 = μq l+1 (d − 1)! q Gγ d−1 l+1−d .  In order to study the structure of projective modules of D(Γ n,d ), we first decompose D(Γ n,d ) into a direct sum of algebras Γ 0 , ··· , Γ n−1 and study each of these algebras. Our method is from [7]. This method were used by several authors, see [13, 14]. Proposition 2.5. The elements E u := 1 n  i,j∈Z n q −i(u+j) G i e j ,foru ∈ Z n ,are central orthogonal idempotents, and  u∈Z n E u =1. Therefore, D(Γ n,d ) ∼ =  u∈Z n D(Γ n,d )E u . Proof. We only prove that E u X = XE u , the others are easy. E u X = 1 n  i,j∈Z n q −i(u+j) G i e j X = 1 n  i,j∈Z n q −i(u+j) G i  Xe j+1 − μ (d − 1)! q (γ d−1 j+1 − γ d−1 j+1−d ) + μq j+1 (d − 1)! q G(γ d−1 j+1 − γ d−1 j+1−d )  . The Quantum Double of a Dual Andruskiewitsch-Schneider 197 Note that  j∈Z n q −i(u+j) μ (d − 1)! q (γ d−1 j+1 − γ d−1 j+1−d ) =  j∈Z n q −i(u+j) μ (d − 1)! q γ d−1 j+1 −  j∈Z n q −i(u+j) μ (d − 1)! q γ d−1 j+1−d =  j∈Z n q −i(u+j) μ (d − 1)! q γ d−1 j+1 −  l∈Z n q −i(u+l+d) μ (d − 1)! q γ d−1 l+1 = μ (d − 1)! q  j∈Z n (q −i(u+j) − q −i(u+j+d) )γ d−1 j+1 =0. Similarly,  j∈Z n q −i(u+j) μq j+1 (d − 1)! q G(γ d−1 j+1 − γ d−1 j+1−d )=0. Thus E u X = 1 n  i,j∈Z n q −i(u+j) G i Xe j+1 = 1 n  i,j∈Z n q −i(u+j) q −i XG i e j+1 = X 1 n  i,j∈Z n q −i(u+j+1) G i e j+1 = XE u .  Define Γ u := D(Γ n,d )E u . The above propositions tell us that we need to study Γ u . We now define some idempotents inside Γ u , which are not central, but we will use them to describe a basis for Γ u . Proposition 2.6. Set E u,j =  n d −1 v=0 e j+vd E u ,forj ∈ Z d .ThenE u,j E u,l = δ jl E u,j and  d−1 j=0 E u,j = E u . We also have E u,j = E u,j  if and only if j ≡ j  (mod d). Moreover, the following relations hold within Γ u : GE u,j = q u+j E u,j = E u,j G, XE u,j = E u,j−1 X γ m l E u,j =  E u,j+m γ m l if l ≡ j (mod d) 0 otherwise. Proof. We only prove that XE u,j = E u,j−1 X. The proof of other equalities is the same with that of Proposition 2.7 in [7]. 198 Meihua Shi E u,j−1 X = n d −1  v=0 (e j−1+vd X)E u = n d −1  v=0  Xe j+vd − μ (d − 1)! q  γ d−1 j+vd − γ d−1 j+(v−1)d  + μq j+vd (d − 1)! q G(γ d−1 j+vd − γ d−1 j+(v−1)d )  E u = n d −1  v=0 Xe j+vd E u = XE u,j .  WecannowdescribeabasisforΓ u and a grading on Γ u , as follows: Γ u =  d−1 s=1−d (Γ u ) s with (Γ u ) s =span{X t γ m j E u,j |j ∈ Z n , 0 ≤ m, t ≤ d−1,m−t = s}. This is a sum of eigenspaces for G:ifyE u,j is an element in (Γ u ) s ,wehavethat G · yE u,j = q s+j+u yE u,j ,yE u,j · G = q j+u yE u,j . Now set F u,j := γ d−1 j E u,j for j ∈ Z n .Ifj is an element in Z n ,weshall denote its representative modulo d in {1, ,d} by <j>and its representative modulo d in {0, ,d− 1} by <j> − . Proposition 2.7. The module Γ u F u,j has the following form: where H u,j := X <2j+u−1>−1 F u,j ,  F u,j := X <2j+u−1> − F u,j and  H u,j := X d−1 F u,j . In this diagram, ↓ represents the action of X and ↑ represents the action of the suitable arrow up to a nonzero scalar; the basis vectors are eigenvectors for the action of G. Note that when 2j + u − 1 ≡ 0(mod d), the single arrow does not occur, the module is simple, and we have H u,j =  H u,j = X d−1 F u,j and  F u,j = F u,j . In order to prove this proposition, we require the following lemma: [...]... in [7] An algebra A is said to be of finite representation type provided there are finitely many non-isomorphic indecomposable A- modules A is of tame type or A is a tame algebra if A is not of finite representation type, whereas for any dimension d > 0, there are finite number of A- k[T ]-bimodules Mi which are free as right k[T ]-modules such that all but a finite number of indecomposable A- modules of dimension... Snashall, and R Taillefer, Representation Theory of the Drinfeld double of a family of Hopf algebras, J Pure and Applied Algebra 204 (2006) 413–454 8 H-L Huang, H-X Chen, and P Zhang, Generalized Taft algebras, Alg Collo 11 (2004) 313–320 9 H Krause, Stable Equivalnece Preserves Representation Type, Comment Math Helv 72 (1997) 266–284 10 Gongxiang Liu, On the structure of tame graded basic Hopf algebras,... 2.19 is not a The Quantum Double of a Dual Andruskiewitsch-Schneider 207 union of finite Dynkin diagrams Indeed, it has two components A 2n −1 (Eud clidean diagram) Let J denote the Jacobson radical of D(Γn,d ) Since the separated quiver of D(Γn,d ) is not a union of finite Dynkin diagrams and the quivers of D(Γn,d ) and D(Γn,d )/J 2 are identical, Theorem 2.6 in Chapter X of [1] implies D(Γn,d )/J 2 is. .. Finite-dimensional representation of a quantum double, J Algebra 251(2002) 751–789 4 Xiao-Wu Chen, Hua-Lin Huang, Yu Ye, and Pu Zhang, Monomial Hopf algerbas, J Algerba 275 (2004) 212–232 5 C Cibils, A Quiver quantum group, Comm Math Phys 157 (1993) 459–477 6 K Erdmann, Blocks of Tame Representation Type and Related Algebras, Lecture Notes Math Vol 1428, Springer–Verlag, Berlin, 1990 7 K Erdmann, E L.Green, N Snashall,... dimension d are isomorphic to Mi ⊗k[T ] k[T ]/(T − λ) for λ ∈ k The following conclusion is our main aim Theorem 2.20 D(Γn,d ) is a tame algebra Proof By Theorem 2.19, we know that D(Γn,d ) is a special biserial algebra (for definition, see [6]) and thus it is tame or of finite representation type (see II.3.1 of [6]) Given a quiver Γ, we associate with Γ the following quiver Γs called the separated quiver of. .. )/J 2 is not of finite representation type Thus D(Γn,d ) is not of finite representation type and D(Γn,d ) is tame Acknowledgment The author is grateful to the referee for his/her valuable comments References 1 M Auslander and I Reiten, Representation Theory of Artin Algebras, Cambridge University Press, 1995 2 H-X Chen, Irreducible representations of a class of quanrum doubles, J Algebra 225 (2000)... , n} are the vertices of Γ, then the vertices of Γs i j are {1, , n, 1 , , n } For each arrow · −→ · in Γ, we have by definition an i j arrow · −→ · in Γs It is known that for a finite quiver Q, path algebra kQ is of finite representation type if and only if the underlying graph Q of Q is one of finite Dynkin diagrams: An , Dn , E6 , E7 , E8 Clearly, the separated quiver of the quiver drawn in... algebras, J Algebra, (in press) 11 G X Liu and F Li, Pointed Hopf algebras of finite Corepresentation type and their classifications, Proc A. 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Proposition 2.11 Assume that < 2j + u − 1 >= d Then module Γu Ku,j has the following structure: Meihua Shi 202 − where Du,j := X d– 2j+u–1 –1 Ku,j , Ku,j = X d– 2j–u+1 and Du,j := X d−1 Ku,j As before, ↓ denotes the action of X and ↑ the action of suitable arrow up to a nonzero scalar also denotes the action of suitable arrow up to a nonzero scalar To prove this, we need some preliminaries Lemma 2.12 For... case Using this relation, we see that Hu,j X d− −1 is a nonzero multiple of The Quantum Double of a Dual Andruskiewitsch-Schneider d−− −1 205 − (q 2j−1+u+t − 1)X d−1 γj+d−− −1 Eu,−j−u t=1 which is nonzero We now can decompose Γu into a sum of indecomposable modules Note that if < 2j + u − 1 >= d, the module Γu Ku,j has dimension 2d while if < 2j + u − 1 >= d, it has . Representation Theory, Quantum double Tame Algebra 1. Introduction In this paper, k is an algebraically closed field of characteristic 0 and an algebra is a finite dimensional associative k -algebra with. whether is the quantum double of dual Andruskiewitsch- Schneider algebra a tame algebra? In this paper, we give an a rmative answer. As a consequence, we give some new examples of tame Hopf algebra. Our. Vietnam Journal of Mathematics 34:2 (2006) 189–207 The Quantum Double of a Dual Andruskiewitsch-Schneider Algebra Is a Tame Algebra * Meihua Shi Dept. of Math. Zhejiang Education Institute Hangzhou,

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