On Hopf Algebras over quantum subgroups

ON HOPF ALGEBRAS OVER QUANTUM SUBGROUPS

arXiv:1605.03995v1 [math.QA] 12 May 2016

´ ANDRES ´ GARCÍA AND JOAO ˜ MATHEUS JURY GIRALDI GASTON Abstract. Using the standard filtration associated to a generalized lifting method, we determine all finite-dimensional Hopf algebras over an algebraically closed field of characteristic zero whose coradical generates a Hopf subalgebra isomorphic to the smallest non-pointed Hopf algebra K of dimension 8 and the corresponding infinitesimal module is an indecomposable object in K K YD. As a byproduct we obtain new Hopf algebras of dimension 64.

Introduction Let k be an algebraically closed field of characteristic zero. The question of classification of all Hopf algebras over k of a given dimension up to isomorphism was posed by Kaplansky in 1975 [K]. Some progress has been made but, in general, it is a difficult question where there are no standard methods. One of the few general techniques is the so-called Lifting Method [AS2], under the assumption that the coradical is a subalgebra, i.e. the Hopf algebra has the Chevalley Property. More recently, Andruskiewitsch and Cuadra [AC] proposed to extend this technique by considering the subalgebra generated by the coradical and the related filtration. It turns out that this filtration is a Hopf algebra filtration, provided that the antipode is injective. We describe the lifting method briefly. Let H be a Hopf algebra. Recall that the coradical filtration {Hn }n≥0 of H is defined recursively by • H0 is the V coradical, • Hn = n+1 H0 = {h ∈ H : ∆(h) ∈ H ⊗ H0 + Hn ⊗ H}. This filtration corresponds to the filtration of H ∗ given by the powers of the Jacobson radical. It is always a coalgebra filtration and if H0 is a Hopf subalgebra, L then it is also an algebra filtration; in particular, its associated graded object gr H = n≥0 Hn /Hn−1 is a graded Hopf algebra, where H−1 = 0. Let π : gr H → H0 be the homogeneous projection. It turns out that gr H ≃ R#H0 as Hopf algebras, where R = (gr H)co π = {h ∈ H : (id ⊗π)∆(h) = h ⊗ 1} and # stands for the Radford-Majid biproduct or bosonization of R with H0 . R is not a usual Hopf algebra, but a graded connected Hopf algebra in the 0 category H H0 YD of Yetter-Drinfeld modules over H0 . It contains the algebra generated by the elements of degree one, called the Nichols algebra B(V ) of V ; here V = R1 is a braided vector space called the infinitesimal braiding. Assume we have a fixed cosemisimple Hopf algebra A. The lifting method then consists of the description of all graded connected braided Hopf algebras R ∈ A A YD, the determination of all possible deformations of the bosonization R#A and the proof that all Hopf algebras H with H0 = A satisfy that gr H ≃ R#A. In general, each step of the method constitute a difficult problem to solve. Through the use of the lifting method the complete classification, with non-trivial examples, of finite-dimensional pointed Hopf algebras H with H0 a group algebra kG, was obtained in the following cases • G a finite abelian group such that (|G|, 210) = 1 [AS3], • G = S3 , S4 the symmetric groups in 3 and 4 letters [AHS], [GGI]. • G = D4t the dihedral groups of order 8t, t ≥ 3 [FG]. Date: May 16, 2016. 2010 Mathematics Subject Classification: 16T05. Keywords: Nichols algebra; Hopf algebra; standard filtration. This work was partially supported by ANPCyT-Foncyt, CONICET, SeCyT (UNLP), CNPq (Brazil). 1

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G. A. GARCÍA, J. M. J. GIRALDI

• G any group admitting a principal realization over some affine racks, e.g. G ≃ Z5 ⋊ Zm [GIV]. Using different techniques, it was also shown that finite-dimensional pointed Hopf algebras over some finite simple non-abelian groups G are trivial, that is, isomorphic to group algebras: • alternating groups An with n ≥ 5 [AFGV1], • most simple sporadic groups [AFGV2], [FV], • infinite families of the projective special linear groups PSL(n, q) and the special linear groups SL(n, q) over finite fields Fq [ACG], [FGV]. The lifting method is also effective to study finite-dimensional copointed Hopf algebras, that is, Hopf algebras such that its coradical is isomorphic to a function algebra over a finite group G. In this case, the classification is known for • G = S3 [AV], • G any group admitting a principal realization over some affine racks [GIV]. The method also works for infinite-dimensional Hopf algebras of finite Gelfand-Kirillov dimension [AAH] or finite-dimensional Hopf algebras over finite fields. The main idea in [AC] is to replace the coradical filtration by a more general but adequate filtration: the standard filtration {H[n] }n≥0 , is defined recursively by • H[0] to be the subalgebra generated by H0 , called the Hopf coradical, V • H[n] = n+1 H[0] .

If the coradical H0 is a Hopf subalgebra, then H[0] = H0 and the coradical filtration coincides with the standard one. Let A be an arbitrary Hopf algebra. We will say that H is a Hopf algebra over A if H[0] ≃ A as Hopf algebras. Assume the antipode S of H is injective, then by [AC, Lemma 1.1] it holds that H[0] is a Hopf subalgebra of H, Hn ⊆ H[n] andL {H[n] }n≥0 is a Hopf algebra filtration of H. In particular, the graded algebra gr H = n≥0 H[n] /H[n−1] with H[−1] = 0 is a Hopf algebra associated with the standard filtration. If we denote as before, π : gr H → H[0] the homogeneous projection, it splits the inclusion of H[0] in gr H, the diagram R = (gr H)co π H

is a Hopf algebra in the category H[0] YD of Yetter-Drinfel’d modules over H[0] and gr H ≃ [0] R#H[0] as Hopf algebras. It turns out that R is also graded and connected. We call again the linear space R1 corresponding to degree one the infinitesimal braiding. This is summarized in the following theorem. Theorem. [AC, Theorem 1.3] Any Hopf algebra with injective antipode is a deformation of the bosonization of a Hopf algebra generated by a cosemisimple coalgebra by a connected graded Hopf algebra in the category of Yetter-Drinfeld modules over the latter. The procedure to describe explicitly any Hopf algebra as above defines a proposal for the classification of general Hopf algebras with injective antipode over a fixed Hopf subalgebra A which is generated by a cosemisimple coalgebra. The main steps are the following: (a) determine all Yetter-Drinfeld modules V in A A YD such that the Nichols algebra B(V ) is finite dimensional, (b) for such V , compute all Hopf algebras L such that gr L ≃ B(V )#A. We call L a lifting of B(V ) over A. (c) Prove that any finite-dimensional Hopf algebra over A is generated by the first term of the standard filtration. In this paper, we study this question in the case that A = K is the smallest Hopf algebra generated by a simple coalgebra of dimension 4. It is an 8-dimensional Hopf algebra which is the dual of a pointed Hopf algebra. As algebra, K is generated by the elements a, b, c, d


3

satisfying the relations ab = ξba,

ac = ξca,

0 = cb = bc,

ad = da,

ad = 1,

0 = b2 = c2 ,

cd = ξdc, a2 c = b,

bd = ξdb, a4 = 1.

The coalgebra structure and its antipode are determined by ∆(a) = a ⊗ a + b ⊗ c, ∆(c) = c ⊗ a + d ⊗ c, S(a) = d,

∆(b) = a ⊗ b + b ⊗ d,

ε(a) = 1,

ε(b) = 0,

ε(c) = 1,

ε(d) = 1

S(b) = ξb,

S(c) = −ξc,

S(d) = a.

∆(d) = c ⊗ b + d ⊗ d,

See Section 2 for more details. Notice that, by a result of Stefan [S¸, Theorem 1.5], K is a quantum subgroup of SLξ (2) since it is a quotient of the quantum group Oξ (SL2 ). For this reason, we call any Hopf algebra over K, a Hopf algebra over a quantum subgroup. In order to determine finite-dimensional Hopf algebras over K, we first compute the Drinfeld double D := D(Kcop ) of Kcop and describe the simple and indecomposable left D-modules, and the projective covers of the simple modules. In fact, we prove in Theorem 2.9 that there are 16 simple left D-modules pairwise non-isomorphic. Four one-dimensional given by characters and 12 two-dimensional. The former correspond to characters on Z4 and the latter are parametrized by the finite subset of Z4 × Z4 given by Λ = {(i, j) ∈ Z4 × Z4 | 2i 6= j}. We compute the separation diagram of D and show that D is of tame representation type. Using that the categories D M and K K YD are equivalent, we then translate the description above to simple and indecomposable modules in K K YD and prove in Theorem 4.5 that the Nichols algebra B(M ) is infinite-dimensional for any finite-dimensional non-simple indecomposable module M ∈ K K YD. In particular, if B(V ) is finite-dimensional, then V must be a semisimple module. Then, using the description of the braiding in K K YD, we obtain our first main result, see Section 3 for definitions. Theorem A. Let B(V ) be a finite-dimensional Nichols algebra over an indecomposable object in K K YD. Then V is simple and isomorphic either to kχ , kχ3 , V2,1 , V2,3 , V3,1 or V3,3 . V It turns out that B(kχℓ ) ≃ kχℓ is an exterior algebra for ℓ = 1, 3 with dim B(kχℓ ) = 2 and B(V ) is an 8-dimensional algebra for V = V2,1 , V2,3 , V3,1 and V3,3 . Since all objects in K K YD can be described as objects in the category of Yetter-Drinfeld modules over the pointed Hopf algebra K∗ = A′′4 , by [U] it follows that the associated braiding is triangular. These 8-dimensional examples are new examples of finite-dimensional Nichols algebras. They are isomorphic to quantum linear spaces as algebras, but not as coalgebras since the braiding differs; in our case, the bainding is not diagonal, see the Appendix. As the study of Nichols algebras over semisimple modules is a hard problem that demands different techniques to be applied, we focus on the description of Hopf algebras over K such that their infinitesimal braiding is simple. After proving in Theorem 5.1 that any such Hopf algebra is generated in degree one with respect to the standard filtration, we define two Hopf algebras A3,1 (µ) and A3,3 (µ) depending on a parameter µ ∈ k and prove our second main result, see Section 5 for definitions. Theorem B. Let H be a finite-dimensional Hopf algebra over K such that its infinitesimal braiding is an indecomposable module V in K K YD. Then V is simple and H is isomorphic either to V (i) kχℓ #K with ℓ = 1, 3; (ii) B(V2,1 )#K; (iii) B(V2,3 )#K; (iv) A3,1 (µ) for some µ ∈ k; (v) A3,3 (µ) for some µ ∈ k. V The Hopf algebras kχℓ #K with ℓ = 1, 3 have dimension 16 and are duals of pointed Hopf algebras. They have already appeared in [B]. The Hopf algebras B(V2,1 )#K and


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B(V2,3 )#K are dual of pointed Hopf algebras of dimension 64. The Hopf algebras A3,1 (µ) and A3,3 (µ) are non-pointed with non-pointed duals. As far as the authors knowledge, they constitute new examples of Hopf algebras of dimension 64. The paper is organized as follows. In Section 1 we recall some invariants associated to a Hopf algebra, define Yetter-Drinfeld modules, Nichols algebras and the Drinfeld double construction, and recall the relation between Hopf algebras with a projection and bosonizations. In Section 2 we describe the structure of K and give the presentation of the double D = D(Kcop ) by generators and relations. We also determine the simple left Dmodules, their proyective covers and some indecomposable left D-modules. We compute the Ext-Quiver of D and show that D is of tame representation type. Then, using the equivalence D M ≃ K K YD, we determine in Section 3 the corresponding objects of the latter and describe their braidings. The braided vector spaces corresponding to 2-dimensional simple modules are not diagonal. We give the proof of this fact in the Appendix. In Section 4 we show that if B(V ) is a finite-dimensional Nichols algebra in K K YD, then V is necessarily a semisimple object and prove Theorem A by describing first the Nichols algebra of the simple modules. Finally, in Section 5 we first show that a finite-dimensional Hopf algebra over K whose infinitesimal braiding is a simple module in K K YD is generated by the first term of the standard filtration, and then prove Theorem B.

Acknowledgements The authors thank M. I. Platzeck for the proof of Remark 2.18, L. Vendramin for providing the computation with GAP of the Nichols algebras over two-dimensional modules, and A. Garc´ıa Iglesias, N. Andruskiewitsch and I. Angiono for fruitfull discussions on the results of the paper.

1. Preliminaries 1.1. Conventions. We work over an algebraically closed field k of characteristic zero. Our references for Hopf algebra theory are [M], [R] and [Sw]. If H is a Hopf algebra over k then ∆, ε and S denote respectively the comultiplication, the counit and the antipode. Comultiplication and coactions are written using the Sweedler notation with summation sign suppressed, e.g., ∆(h) = h(1) ⊗ h(2) for h ∈ H. If H is a Hopf algebra in a braided monoidal category, we call it a braided Hopf algebra. Usual Hopf algebras are Hopf algebras in the category Vectk of k-vector spaces. We denote by H M the category of finite-dimensional left H-modules. The set G(H) = {h ∈ H \ {0} : ∆(h) = h ⊗ h} denotes the group of group-like elements. The coradical H0 of H is the sum of all simple subcoalgebras of H; in particular, kG(H) ⊆ H0 . If H has injective antipode, the algebra H[0] generated by H0 is a Hopf subalgebra which is called the Hopf coradical. For h, g ∈ G(H), the linear space of (h, g)primitives is: Ph,g (H) := {x ∈ H | ∆(x) = x ⊗ h + g ⊗ x}. If g = 1 = h, the linear space P(H) = P1,1 (H) is called the set of primitive elements. If M is a right H-comodule via δ(m) = m(0) ⊗ m(1) ∈ M ⊗ H for all m ∈ M , then the space of right coinvariants is M co δ = {x ∈ M | δ(x) = x ⊗ 1}. In particular, if π : H → L is a morphism of Hopf algebras, then H is a right L-comodule via (id ⊗π)∆ and H co π := H co Left coinvariants, written

(id ⊗π)∆

co π H

= {h ∈ H | (id ⊗π)∆(h) = h ⊗ 1}.

are defined analogously.


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1.2. Yetter-Drinfeld modules and Nichols algebras. Let H be a Hopf algebra. A left Yetter-Drinfeld module M over H is a left H-module (M, ·) and a left H-comodule (M, δ) with δ(m) = m(−1) ⊗ m(0) ∈ H ⊗ M for all m ∈ M , satisfying δ(h · m) = h(1) m(−1) S(h(3) ) ⊗ h(2) · m(0)

∀ m ∈ M, h ∈ H.

We denote by H H YD the category of left Yetter-Drinfeld modules over H. It is a braided monoidal category: for M, N ∈ H H YD, the braiding cM,N : M ⊗ N → N ⊗ M is given by (1)

cM,N (m ⊗ n) = m(−1) · n ⊗ m(0)

∀ m ∈ M, n ∈ N.

Definition 1.1. [AS2, Def. 2.1] Let H be a Hopf algebra and V ∈ H H YD. A braided L H N-graded Hopf algebra R = n≥0 R(n) ∈ H YD is called the Nichols algebra of V if

(i) (ii) (iii) In this

k ≃ R(0), V ≃ R(1) ∈ H H YD, R(1) = P(R) = {r ∈ R | ∆R (r) = r ⊗ 1 + 1 ⊗ r}. R is generated as an algebra by R(1). L case, R is denoted by B(V ) = n≥0 Bn (V ).

For any V ∈ H H YD there is a Nichols algebra B(V ) associated to it. It is the quotient of the tensor algebra T (V ) by the largest homogeneous two-sided ideal I satisfying: • I is generated by homogeneous elements of degree ≥ 2. • ∆(I) ⊆ I ⊗ T (V ) + T (V ) ⊗ I, i. e., it is also a coideal. In such a case, B(V ) = T (V )/I. See [AS2, Section 2.1] for details. Remark 1.2. An important observation is that the Nichols algebra B(V ) is completely determined, as algebra and coalgebra, by the braiding. Let V be a vector space and c ∈ End(V ⊗V ) be a solution of the braid equation (c⊗id)(id ⊗c)(c⊗id) = (id ⊗c)(c⊗id)(id ⊗c). Let AV , AC be the tensor algebra and the cotensor algebra, respectively. Both are braided bialgebras and there exists a unique bialgebra map S : AV → CV such that S|V = idV . The image Im S ⊆ CV is a braided bialgebra called the quantum symmetric algebra. If the braiding is rigid, then Im S = B(V ) is a Nichols algebra; in such a case, B(V ) is a braided Hopf algebra in a braided rigid category. See [AG] for details. If W ⊆ V is a subspace such that c(W ⊗ W ) ⊆ W ⊗ W , one may identify B(W ) with a subalgebra of B(V ); eventually belonging to different braided rigid categories. In particular, if dim B(W ) = ∞, then dim B(V ) = ∞. Thus, if V contains a non-zero element v such that c(v ⊗ v) = v ⊗ v, then dim(V ) = ∞. 1.3. Bosonization and Hopf algebras with a projection. Let H be a Hopf algebra and B a braided Hopf algebra in H H YD. The procedure to obtain a usual Hopf algebra from B and H is called the Majid-Radford biproduct or bosonization, and it is usually denoted by B#H. As vector spaces B#H = B ⊗ H, and the multiplication and comultiplication are given by the smash-product and smash-coproduct, respectively. That is, for all b, c ∈ B and g, h ∈ H, we have (b#g)(c#h) = b(g(1) · c)#g(2) h,

∆(b#g) = b(1) #(b(2) )(−1) g(1) ⊗ (b(2) )(0) #g(2) ,

where ∆B (b) = b(1) ⊗ b(2) denotes the comultiplication in B ∈ H H YD. If b ∈ B and h ∈ H, then we identify b = b#1 and h = 1#h; in particular we have bh = b#h and hb = h(1) · b#h(2) . Clearly, the map ι : H → B#H given by ι(h) = 1#h for all h ∈ H is an injective Hopf algebra map, and the map π : B#H → H given by π(b#h) = εB (b)h for all b ∈ B, h ∈ H is a surjective Hopf algebra map such that π ◦ ι = idH . Moreover, it holds that B = (B#H)co π . Conversely, let A be a Hopf algebra with bijective antipode and π : A → H a Hopf algebra epimorphism admitting a Hopf algebra section ι : H → A such that π ◦ ι = idH . Then B = Aco π is a braided Hopf algebra in H H YD and A ≃ B#H as Hopf algebras.


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1.4. The Drinfeld double. We breifly describe the structure of the Drinfeld double of a finite-dimensional Hopf algebra. Definition 1.3. [M, Def. 10.3.5] Let H be a finite dimensional Hopf algebra. The Drinfeld double is the Hopf algebra D(H), where D(H) = (H ∗ )cop ⊲⊳ H = H ∗ ⊗ H, as vector spaces, the product and the unit are given by (f ⊲⊳ h)(g ⊲⊳ k) = f (h1 ։ g2 ) ⊲⊳ (h2 և g1 )k,

1D(K) = ε ⊲⊳ 1,

where h ։ f = hf3 (S ∗ )−1 (f1 ), hif2 and h և f = hf, S −1 (h3 )h1 ih2 , and the coproduct and counity are ∆(f ⊲⊳ h) = f2 ⊲⊳ h1 ⊗ f1 ⊲⊳ h2 ε(f ⊲⊳ h) = f (1)ε(h). The following result establishes a categorical equivalence between H H YD and D(H cop ) M. This proposition will be central in Section 3, since we will study the simple and indecomposable left D(Kcop )-modules first and then translate the information to K K YD. Proposition 1.4. [M, Prop. 10.6.16] Let H be a finite-dimensional Hopf algebra. Then the Yetter-Drinfeld category H H YD can be identified with the category D(H cop ) M of left modules over the Drinfeld double D(H cop ). 2. The quantum subgroup K and its Drinfeld double D(Kcop ) In this section we describe the structure of the quantum subgroup K and present the Drinfeld double D = D(Kcop ) by generators and relations. We also determine the simple left D-modules, their proyective covers and some indecomposable left D-modules. We parametrize these objects by tuples of the characters of the commutative subalgebra generated by a, d and g. With this information we compute the Ext-Quiver of D and conclude that D is of tame representation type. Throughout the paper, we fix ξ a primitive 4-th root of 1. All pointed nonsemisimple Hopf algebras of dimension 8 were determined by [S¸]. Except for one case, given by A′′4 := khg, x | g 4 − 1 = x2 − g2 + 1 = gx + xg = 0i, with ∆(g) = g ⊗ g and ∆(x) = x ⊗ g + 1 ⊗ x, these pointed Hopf algebras have pointed duals. Moreover, it holds that K = (A′′4 )∗ , see [GV]. Up to isomorphism, K is the only Hopf algebra of dimension 8 which is neither semisimple nor pointed nor has the Chevalley property. The next proposition gives us a presentation of the Hopf algebra K and some useful relations that will be used in the sequel. The proof follows from [GV, Lemma 3.3]. Proposition 2.1. ab = ξba, ad = da,

(i) K is generated as an algebra by the elements a, b, c, d satisfying ac = ξca,

0 = cb = bc, 2

ad = 1,

2

0=b =c ,

cd = ξdc,

bd = ξdb,

2

a4 = 1.

a c = b,

(ii) A linear basis of K is given by {1, a, b, c, d, a2 , ab, ac}. (iii) The coalgebra structure is given by ∆(a) = a ⊗ a + b ⊗ c, ∆(d) = c ⊗ b + d ⊗ d,

∆(a2 ) = a2 ⊗ a2 ,

∆(b) = a ⊗ b + b ⊗ d, 2

∆(ab) = ab ⊗ 1 + a ⊗ ab, ε(a) = ε(d) = 1,

∆(c) = c ⊗ a + d ⊗ c,

∆(ac) = ac ⊗ a2 + 1 ⊗ ac, ε(b) = ε(c) = 0.

(iv) The antipode is given by S(a) = d,

S(a2 ) = a2 ,

S(b) = ξb,

S(ab) = −ac,

(v) The multiplication table is

S(c) = −ξc,

S(ac) = ab.

S(d) = a,


1 a b c d a2 ab ac ∗ (vi) K ≃ H4 ⊕ M (2, k) as

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1 a b c d a2 ab ac 1 a b c d a2 ab ac a a2 ab ac 1 d c b b −ξab 0 0 ξac −c 0 0 c −ξac 0 0 ξab −b 0 0 d 1 ac ab a2 a b c 2 a d c b a 1 ac ab ab −ξc 0 0 ξb −ac 0 0 ac −ξb 0 0 ξc −ab 0 0 coalgebras.

Remarks 2.2. (a) Denote by {1∗ , a∗ , b∗ , c∗ , d∗ , (a2 )∗ , (ab)∗ , (ac)∗ } the basis of K∗ dual to {1, a, b, c, d, a2 , ab, ac}. Using the multiplication table in Proposition 2.1 (v), it follows that ∆(1∗ ) = 1∗ ⊗ 1∗ + a∗ ⊗ d∗ + d∗ ⊗ a∗ + (a2 )∗ ⊗ (a2 )∗ ,

∆(a∗ ) = 1∗ ⊗ a∗ + a∗ ⊗ 1∗ + (a2 )∗ ⊗ d∗ + d∗ ⊗ (a2 )∗ ,

∆(d∗ ) = 1∗ ⊗ d∗ + d∗ ⊗ 1∗ + (a2 )∗ ⊗ a∗ + a∗ ⊗ (a2 )∗ ,

∆((a2 )∗ ) = 1∗ ⊗ (a2 )∗ + (a2 )∗ ⊗ 1∗ + a∗ ⊗ a∗ + d∗ ⊗ d∗ ,

∆(b∗ ) = 1∗ ⊗ b∗ + b∗ ⊗ 1∗ + a∗ ⊗ (ac)∗ − ξ(ac)∗ ⊗ a∗ +

+ (a2 )∗ ⊗ c∗ − c∗ ⊗ (a2 )∗ + ξ(ab)∗ ⊗ d∗ + d∗ ⊗ (ab)∗ ,

∆(c∗ ) = 1∗ ⊗ c∗ + c∗ ⊗ 1∗ − ξ(ab)∗ ⊗ a∗ + a∗ ⊗ (ab)∗ +

+ (a2 )∗ ⊗ b∗ − b∗ ⊗ (a2 )∗ + ξ(ac)∗ ⊗ d∗ + d∗ ⊗ (ac)∗ ,

∆((ab)∗ ) = 1∗ ⊗ (ab)∗ + (ab)∗ ⊗ 1∗ − ξb∗ ⊗ a∗ + a∗ ⊗ b∗ +

+ d∗ ⊗ c∗ + ξc∗ ⊗ d∗ − (ac)∗ ⊗ (a2 )∗ + (a2 )∗ ⊗ (ac)∗ ,

∆((ac)∗ ) = 1∗ ⊗ (ac)∗ + (ac)∗ ⊗ 1∗ − ξc∗ ⊗ a∗ + a∗ ⊗ c∗ +

+ d∗ ⊗ b∗ + ξb∗ ⊗ d∗ − (ab)∗ ⊗ (a2 )∗ + (a2 )∗ ⊗ (ab)∗ .

(b) Let α ∈ G(K∗ ) = Alg(K, k). As a4 = 1 and ad = 1, α(a) is a 4-th root of unity and α(d) = α(a)−1 . Further, since b2 = 0 = c2 , we have that α(b) = 0 = α(c). Thus G(K∗ ) = {αj = 1∗ + ξ −j a∗ + ξ j d∗ + (−1)j (a2 )∗ : 0 ≤ j ≤ 3}. Note that α0 = ε and αj1 = αj . In particular, (c) The multiplication table is 1∗ a∗ b∗ c∗ ∗ ∗ 1 1 0 0 0 ∗ a 0 a ∗ b∗ 0 ∗ b 0 0 0 a∗ ∗ c 0 c∗ d∗ 0 d∗ 0 0 0 c∗ 2 ∗ (a ) 0 0 0 0 (ab)∗ (ab)∗ 0 0 0 (ac)∗ 0 0 0 0

G(K∗ ) ≃ Z/4Z and αξ , α−ξ are generators. d∗ (a2 )∗ (ab)∗ (ac)∗ 0 0 0 (ac)∗ 0 0 0 0 b∗ 0 0 0 0 0 0 0 ∗ d 0 0 0 0 (a2 )∗ (ab)∗ 0 0 0 0 0 ∗ 0 (ac) 0 0

In order to compute the Drinfeld double D(Kcop ) of Kcop we need to describe the isomorphism K∗ ≃ A′′4 explicitly.

Lemma 2.3. The algebra map ϕ : A′′4 → K∗ given by ϕ(g) = α1 = 1∗ − ξa∗ + ξd∗ − (a2 )∗

is a Hopf algebra isomorphism.

and

ϕ(x) =

√ 2ξ(b∗ + c∗ + (ab)∗ + (ac)∗ ),

8


Proof. A direct computation shows that ϕ is a coalgebra map. Hence, the image of ϕ is a Hopf subalgebra of K∗ of dimension bigger than 4, since it contains the group algebra kG(K∗ ) and the image of the skew-primitive element x. Thus, by the Nichols-Zoeller theorem it follows that ϕ is suryective and whence an isomorphism. Remark 2.4. Let {gj , xg j }0≤j≤3 be a linear basis of A′′4 . By Remark 2.2 (c) and Lemma 2.3, it follows that ϕ(gj ) = αj = 1∗ + ξ −j a∗ + ξ j d∗ + (−1)j (a2 )∗ √ ϕ(xg j ) = 2ξ(ξ j b∗ + ξ −j c∗ + (ab)∗ + (−1)j (ac)∗ )

for all

0 ≤ j ≤ 3,

for all

0 ≤ j ≤ 3.

2.1. Desciption of D(Kcop ). Now we describe the Drinfeld double D(Kcop ) of Kcop . To make the notation lighter, from now on we denote D = D(Kcop ). Proposition 2.5. D(Kcop ) is the k-algebra generated by the elements a, b, c, d, x, g such that a, b, c, d satisfy the relations of Kcop , x, g satisfy the relations of (A′′4 )op cop and √ √ ax + ξxa = 2ξ(b + gc), bx − ξxb = 2ξ(a − gd), ag = ga, bg = −gb, cg = −gc, dg =√gd, √ cx + ξxc = 2ξ(d − ga), dx − ξxd = 2ξ(c + gb). Proof. Since (f ⊲⊳ 1)(g ⊲⊳ k) = f g ⊲⊳ k and (f ⊲⊳ h)(ε ⊲⊳ k) = f ⊲⊳ hk for all f, g ∈ (A′′4 )op cop and h, k ∈ Kcop , it is enough to describe the relations derived form products of the form (1A′′4 ⊲⊳ h)(y ⊲⊳ 1K ), where h ∈ Kcop and y ∈ (A′′4 )op cop are algebra generators. −1 (g) · Assume first that y = g. Since h ։ g = hScop op g, hig = hgS(g), hig = h1, hig = cop ε(h)g, for all h ∈ K , it follows that     a ։ g ⊲⊳ a և g   (1) (2) (2) (1) cop cop a          b    և g ։ g ⊲⊳ b (1) cop (2) (2) cop (1) b )(g ⊲⊳ 1K ) = (1A′′4 ⊲⊳ c(1) cop ։ g(2) ⊲⊳ c(2) cop և g(1)  c              d d (1) cop ։ g(2) ⊲⊳ d(2) cop և g(1)     a ։ g ⊲⊳ a և g + c ։ g ⊲⊳ b և g  aևg            b ։ g ⊲⊳ a և g + d ։ g ⊲⊳ b և g bևg = = g ⊲⊳ a ։ g ⊲⊳ c և g + c ։ g ⊲⊳ d և g  cևg            b ։ g ⊲⊳ c և g + d ։ g ⊲⊳ d և g dևg     −1 (a hg, Scop  (3) cop )a(1) cop ia(2) cop  hg, S(a(1) )a(3) ia(2)          hg, S −1 (b    hg, S(b(1) )b(3) ib(2) cop (3) cop )b(1) cop ib(2) cop = g ⊲⊳ = g ⊲⊳ −1 (c hg, Scop hg, S(c(1) )c(3) ic(2)    (3) cop )c(1) cop ic(2) cop           hg, S −1 (d  hg, S(d(1) )d(3) id(2) )d id cop (3) cop (1) cop (2) cop   hg, S(a)aia + hg, S(b)aic + hg, S(a)cib + hg, S(b)cid       hg, S(a)bia + hg, S(b)bic + hg, S(a)dib + hg, S(b)did = g ⊲⊳ hg, S(c)aia + hg, S(d)aic + hg, S(c)cib + hg, S(d)cid       hg, S(c)bia + hg, S(d)bic + hg, S(c)dib + hg, S(d)did   hg, daia + hg, ξbaic + hg, dcib + hg, ξbcid       hg, dbia + hg, ξbbic + hg, ddib + hg, ξbdid = g ⊲⊳ hg, −ξcaia + hg, aaic + hg, −ξccib + hg, acid       hg, −ξcbia + hg, abic + hg, −ξcdib + hg, adid     hg, 1ia + hg, abic + hg, abib + hg, 0id a             hg, acia + hg, 0ic + hg, a2 ib + hg, −acid −b = g ⊲⊳ = g ⊲⊳ , hg, −acia + hg, a2 ic + hg, 0ib + hg, acid  −c            d hg, 0ia + hg, abic + hg, abib + hg, 1id

which implies the relations ag = ga, bg = −gb, cg = −gc and dg = gd.


Assume now that y = x. Using the computations above, we have that      a  hx, 1ia + hx, abic + hx, abib + hx, 0id              √  b hx, acia + hx, 0ic + hx, a2 ib + hx, −acid ևx= = 2ξ 2 c  hx, −acia + hx, a ic + hx, 0ib + hx, acid               d hx, 0ia + hx, abic + hx, abib + hx, 1id

and

9

b+c a−d d−a b+c

    a a             b b ։ x = hS(x(1) )x(3) , ix c  c  (2)           d d       a  a  a                 b b b i1 ix + hS(1)x, ig + hS(1)g, = hS(x)g, c  c  c                 d d d       a  a  a                 b b b = h−xg 3 g, ig + hg, ix + hx, i1 c  c  c                 d d d       ξx a  a            √       2ξ(1 − g) b b √ . = hx, i(1 − g) + hg, ix = c  c  2ξ(1 − g)                 d d −ξx

Hence, it follows that    a ։ g ⊲⊳ a(1) և x + a(2) ։ x ⊲⊳ a(1) և 1 a          (2) b(2) ։ g ⊲⊳ b(1) և x + b(2) ։ x ⊲⊳ b(1) և 1 b (1A′′4 ⊲⊳ )(x ⊲⊳ 1K ) = c c ։ g ⊲⊳ c(1) և x + c(2) ։ x ⊲⊳ c(1) և 1          (2) d d(2) ։ g ⊲⊳ d(1) և x + d(2) ։ x ⊲⊳ d(1) և 1   ε(a(2) )g ⊲⊳ a(1) և x + a(2) ։ x ⊲⊳ a(1)       ε(b(2) )g ⊲⊳ b(1) և x + b(2) ։ x ⊲⊳ b(1) = ε(c(2) )g ⊲⊳ c(1) և x + c(2) ։ x ⊲⊳ c(1)       ε(d(2) )g ⊲⊳ d(1) և x + d(2) ։ x ⊲⊳ d(1)   g ⊲⊳ a և x + a ։ x ⊲⊳ a + c ։ x ⊲⊳ b       g ⊲⊳ b և x + b ։ x ⊲⊳ a + d ։ x ⊲⊳ b = g ⊲⊳ c և x + a ։ x ⊲⊳ c + c ։ x ⊲⊳ d       g ⊲⊳ d և x + b ։ x ⊲⊳ c + d ։ x ⊲⊳ d √ √   2ξg ⊲⊳ (b + c) +√ ξx ⊲⊳ a + 2ξ(1 − g) ⊲⊳ b     √   2ξg ⊲⊳ (a − d) + 2ξ(1 − g)√⊲⊳ a + −ξx ⊲⊳ b √ =  2ξg ⊲⊳ (d − a) +√ξx ⊲⊳ c + 2ξ(1 − g) ⊲⊳ d      √ 2ξg ⊲⊳ (b + c) + 2ξ(1 − g) ⊲⊳ c + −ξx ⊲⊳ d √ √   2ξg ⊲⊳ c + 2ξ ⊲⊳ b −ξx ⊲⊳ a +     √ √   ξx ⊲⊳ b − √2ξg ⊲⊳ d + √2ξ ⊲⊳ a = ,  −ξx ⊲⊳ c −√ 2ξg ⊲⊳ a +√ 2ξ ⊲⊳ d      ξx ⊲⊳ d + 2ξg ⊲⊳ b + 2ξ ⊲⊳ c

      

,

      

which gives us the other four relations of D(Kcop ).

2.2. Simple left D-modules. We begin by describing the one-dimensional D-modules. Lemma 2.6. There are four non-isomorphic one-dimensional left D-modules given by the characters χj , 0 ≤ j ≤ 3 where χj (a) = ξ j , χj (b) = 0, χj (c) = 0, χj (d) = ξ −j , χj (x) = 0, χj (g) = (−1)j .


10

Moreover, any one-dimensional D-module is isomorphic to kχj for some 0 ≤ j ≤ 3. Proof. Let λ : D → k be a character and write λ(a) = λ1 , λ(b) = λ2 , λ(c) = λ3 , λ(d) = λ4 , λ(x) = λ5 , λ(g) = λ6 . From the relations a4 = 1 = g 4 , it follows that λ(a) and λ(g) are 4-th roots of unity. Since ad = 1, we have that λ(d) = λ(a)−1 . As b2 = c2 = gx + xg = 0, we have that λ(b) = λ(c) √= λ(x) = 0 and then from √ g2 = 1 + x2 , it follows that λ(g)2 = 1. Since bx + ξxb = 2ξ(d − ga), it follows that 2ξ(λ(a)−1 − λ(g)λ(a)) = 0 which means that λ(g) = λ(a)−2 = λ(a)2 . Hence, λ is determined by its value in a and consequently must equal χj for some 0 ≤ j ≤ 3. It is immediate that these modules are pairwise non-isomorphic. Note that for all 0 ≤ j ≤ 3, χj coincides with the jth-power of the character χ = χ1 in the convolution algebra Homk (D, k). Thus, we will denote χj with j ∈ Z4 . Next we describe the simple D-modules of dimension two. For this, consider the finite subset of Z4 × Z4 given by Λ = {(i, j) ∈ Z4 × Z4 | 2i 6= j}. Clearly, |Λ| = 12. Lemma 2.7. For any pair (i, j) ∈ Λ, there exists a simple left D-module Vi,j of dimension 2. If we denote λ1 = ξ i and λ2 = ξ j , the action on a fixed basis is given by λ1 0 0 λ21 0 1 ρi,j (a) = , ρi,j (b) = , ρi,j (c) = , 0 −ξλ1 0 0 0 0 −1 λ1 0 λ2 0 ρi,j (d) = , ρi,j (g) = , 0 −λ2 0 ξλ−1 1   √ 2 ξ(λ1 + λ31 λ2 )  , 0 ρi,j (x) =  √ 2 0 2ξ(λ31 − λ1 λ2 )

Moreover, if V is a simple D-module of dimension 2, then V is isomorphic to Vi,j for some (i, j) ∈ Λ and Vi,j ≃ Vk,ℓ if and only if (i, j) = (k, ℓ). Proof. Let

a11 a12 b11 ρ(a) = , ρ(b) = a21 a22 b21 d11 d12 x11 ρ(d) = , ρ(x) = d21 d22 x21

b12 b22 x12 x22

, ρ(c) = , ρ(g) =

c11 c12 c21 c22 g11 g12 g21 g22

, ,

be a two-dimensional simple representation of D(Kcop ). As a4 = 1 = g 4 and ga = ag, ρ(a) and ρ(g) are simultaneously diagonalizable and we can assume that −1 λ1 0 λ1 0 λ3 0 ρ(a) = , ρ(d) = and ρ(g) = , 0 λ2 0 λ4 0 λ−1 2 where λ4i = 1 for 1 ≤ i ≤ 4. From ac = ξca we have that: λ1 c11 λ1 c12 λ1 c11 λ2 c12 =ξ λ2 c21 λ2 c22 λ1 c21 λ2 c22 what implies c11 = c22 = 0. Similarly, the relation gx = −xg implies x11 = x22 = 0. Since a2 c = b, we must have that 0 λ21 c12 ρ(b) = . 0 λ22 c21


11

Also note that from c2 = 0 we get c12 c21 = 0. Thus, by permuting the elements of the basis, we may assume that c21 = 0. Suppose c12 = 0. That is, 0 0 0 x12 ρ(b) = ρ(c) = and ρ(x) = . 0 0 x21 0

Clearly, √ these modules are simple if and only if x12 6= 0 and x21 6= 0. Since ax = −ξxa + 2ξ(b + gc), it follows that x12 (λ1 + ξλ2 ) = 0 and x21 (λ2 + ξλ1 ) = 0. Since x12 x21 6= 0, then λ1 + ξλ2 = λ2 + ξλ1 = 0, which implies that λ1 = λ2 = 0 a contradiction. Therefore, we must have that c12 6= 0. Clearly, we may also assume that c12 = 1. From the equality ac = ξca we get that λ2 = −ξλ1 with λ1√6= 0. Moreover, since cg = −gc, it follows that λ4 = −λ3 . Now, the relation ax + ξxa = 2ξ(b + gc) yields √ 0 λ1 x12 0 λ2 x12 0 λ21 + λ3 = −ξ + 2ξ , λ2 x21 0 λ1 x21 0 0 0 √ 2 which implies that x12 = ξ(λ1 + λ3 λ−1 1 ), since λ2 = −ξλ1 . This is the same information 2√ √ obtained from dx − ξxd = 2ξ(c + gb). Analogously, cx + ξxc = 2ξ(d − ga) yields −1 √ λ1 − λ1 λ3 0 x21 0 , = 2ξ 0 ξx21 0 ξ(λ−1 1 − λ1 λ3 ) √ which gives x21 = 2ξ(λ−1 1 −λ1 λ3 ). This is the same information obtained from bx−ξxb = √ 2ξ(a − gd). Note that, since g 2 = 1 + x2 , we must have that x12 x21 = λ23 − 1. In fact, √ √ 2 −1 −2 2 2 2 x12 x21 = ξ(λ1 + λ3 λ−1 1 ) 2ξ(λ1 − λ1 λ3 ) = −(1 − λ1 λ3 + λ3 λ1 − λ3 ) = λ3 − 1. 2 From the discussion above, the matrices defining the action on the simple module V are of the form λ1 0 0 λ21 0 1 λ3 0 ρ(a) = , ρ(b) = , ρ(c) = , ρ(g) = , 0 −ξλ1 0 0 0 0 0 −λ3   √ −1 2 −1 λ1 0 ξ(λ1 + λ3 λ1 )  , 0 , ρ(x) =  √ ρ(d) = 2 0 ξλ−1 −1 1 2ξ(λ1 − λ1 λ3 ) 0

with λ41 = 1 = λ43 . Moreover, a direct computation shows that V is simple if and only if λ3 6= λ21 . If we set λ1 = ξ i and λ3 = ξ j for some i, j ∈ Z4 , we have that 2i 6= j and consequently (i, j) ∈ Λ. Claim. Vi,j is isomorphic to Vk,ℓ if and only if (i, j) = (k, ℓ) in Z4 × Z4 . Let T : Vi,j → Vk,ℓ be an isomorphism of D-modules; in particular ρk,ℓ (t)T = T ρi,j (t) t11 t12 for any t ∈ D. Denote by [T ] = the matrix of T in the given bases. Using t21 t22 the action of c we have that t21 = 0 and t11 = t22 since t21 t22 0 1 t11 t12 t11 t12 0 1 0 t11 = = = . 0 0 0 0 t21 t22 t21 t22 0 0 0 t21

Moreover, acting by a we have k i i k ξ t11 ξ k t12 ξ 0 ξ t11 −ξ i+1 t12 ξ 0 [T ] = [T ] = = 0 −ξ i+1 t11 0 −ξ i+1 0 −ξ k+1 0 −ξ k+1 t11

which implies (ξ k − ξ i )t11 = 0 and (ξ k + ξ i+1 )t12 = 0. Since T is an isomorphism, this implies that ξ k = ξ i from which follows that t12 = 0 and consequently [T ] = t11 I. Finally, acting by g yields that ξ ℓ = ξ j and the claim follows. Remark 2.8. Let V be a left D-module. Since D is a Hopf algebra, V ∗ inherits a left D-module structure by the formula (h · f )(v) = f (S(h) · v) for all f ∈ V ∗ , v ∈ V and ∗ ≃V h ∈ D. A straightforward calculation yields Vi,j −i+1,j+2 for all (i, j) ∈ Z4 × Z4 .


12

Finally, we describe all simple left D-modules up to isomorphism. Theorem 2.9. There are 16 simple left D-modules pairwise non-isomorphic. Four onedimensional given by Lemma 2.6 and 12 two-dimensional given by Lemma 2.7. Proof. Assume there is a simple module of dimension d > 2 and let n denote the amount of simple d-dimensional modules pairwise non-isomorphic. By Lemmata 2.6 and 2.7 we have that 4.12 + 12.22 + nd2 = 52 + nd2 ≤ dim(D ∗ )0 < dim D ∗ = 64. Then nd2 < 12, which implies that d = 3 and n = 1. But in such a case, by [AN, Lemma 2.1] we must have that 4 = |G(D ∗ )| divides 32 = 9, a contradiction. 2.3. Projective covers of simple left D-modules. In this subsection we denote by b the set of isomorphism classes of simple left D-modules and by P (V ) the projective D cover of a simple D-module V . Projective covers are unique up to isomorphism and as left D-modules one has that M P (V )dim V . DD ≃ b V ∈D

For j ∈ Z4 , let kχj be the one-dimensional D-module associated to the character χj . Then b see [EG, Section 2]. In particular, by 1 < dim P (kε ) ≤ dim V dim P (V ), for all V ∈ D, item (ii) of the following lemma, it follows that dim P (kε ) ≤ 4. Lemma 2.10. (i) Vi,j ⊗ kχℓ ≃ Vi+ℓ,j+2ℓ and kχℓ ⊗ kχk ≃ kχk+ℓ for all (i, j) ∈ Λ, k, ℓ ∈ Z4 . (ii) P (Vi,j ) = Vi,j for all (i, j) ∈ Λ. (iii) P (kχℓ ) = P (kε ) ⊗ kχℓ and dim P (kχℓ ) = 4 for all ℓ ∈ Z4 .

Proof. (i) follows by a direct computation. (ii) Let (i, j) ∈ Λ and µ = χℓ be a character of D. Since Hom(P (Vi,j ) ⊗ kµ , Vi,j ⊗ kµ ) = Hom(P (Vi,j ), Vi,j ⊗ kµ ⊗ k∗µ ) = Hom(P (Vi,j ), Vi,j ) 6= 0, and P (Vi,j ) ⊗ kµ is projective, it follows that P (Vi,j )⊗kµ contains P (Vi,j ⊗kµ ) ≃ P (Vi+ℓ,j+2ℓ ). If dim P (Vi,j ) > Vi,j for some (i, j) ∈ Λ, then by [EO, Lemma 2.10] the socle of P (Vi,j ) is Vi,j , since D is unimodular. Thus, dim P (Vi,j ) ≥ 2 dim Vi,j and dim P (Vi−ℓ,j−2ℓ ) ≥ dim P (Vi,j ) ≥ 4. Thus, if we denote I = {(m, n) ∈ Λ : (m, n) 6= (i + k, j + 2k) for all 0 ≤ k ≤ 3} then dim D ≥

3 X j=0

dim P (kχj ) +

X

(m,n)∈I

2 dim P (Vm,n ) + 8 dim P (Vi,j ) ≥

3 X

dim P (kχj ) + 64,

j=0

a contradiction. (iii) Since P (kε ) is projective, P (kε )⊗kχℓ is also projective. As Hom(P (kε )⊗kχℓ , kχℓ ) = Hom(P (kε ), kχℓ ⊗ k∗χℓ ) = Hom(P (kε ), kε ) 6= 0, it follows that P (kε ) ⊗ kχℓ contains P (kχℓ ), whose dimension is at least dim P (kε ). Hence they must be equal. Since P (Vi,j ) = Vi,j for all (i, j) ∈ Λ, we have that 64 = 4 dim P (kε ) + 48, and whence dim P (kε ) = 4.

∗ = V Remark 2.11. From Lemma 2.10 (i) and Remark 2.8 follows that Vi,j i,j ⊗ kχ if i = 0, 2 and (i, j) ∈ Λ.

Let P = k{p1 , p2 , p3 , p4 } be the 4-dimensional D-module given by

(2)

· a b c d x g

p1 p2 p3 p1 −ξp2 ξp3 p3 ξp4 0 −p3 ξp4 0 p√ ξp −ξp 1 3 √2 p2 + 2p3 − 2p4 p4 p1 −p2 −p3

Lemma 2.12. P ≃ P (kε ) as D-modules.

p4 p4 0 0 p4 0 p4


13

Proof. A straightforward calculation shows that (2) endows P with a D-action. Moreover, consider the submodules given by kp4 and W = k{p2 , p3 , p4 }. Then kp4 and P/W are both isomorphic to the trivial D-module kε . Assume that P = M ⊕ N and let w = αp1 + βp2 + γp3 + δp4 ∈ M . If α 6= 0, then xb · w = αp4 and p4 ∈ M . Analogously, b · w = αp3 + ξβp4 which implies √ that p3 ∈ M . √ Thus, w′ = αp1 + βp2 ∈ M and consequently x · w′ = α(p2 + 2p3 ) − 2βp4 ∈ M which gives us p2 ∈ M and p1 ∈ M , implying that N = 0. If α = 0, then p1 ∈ N and since Dp1 = P , we have that N = P . Hence, P is indecomposable. Finally, we show that P ≃ P (kε ). Denote by ϕ : P → kε the epimorphism induced by the quotient P/W . Since P (kε ) is projective, there exists an epimorphism π such that the following diagram commutes P (kε ) π

P

|③

③

③ ϕ

③

③ / kε .

Thus, there exists an element w = αp1 + βp2 + γp3 + δp4 ∈ π(P (kε )) with α 6= 0. Following the argument in the paragraph above we have that π is surjective and whence an isomorphism. As a consequence of the results above we have the following theorem. Theorem 2.13. The D-modules Pℓ = P ⊗ kχℓ and Vi,j with i, j, ℓ ∈ Z4 and 2i 6= j are the projective covers of the simple D-modules and consequently DD

≃

3 X ℓ=0

Pℓ ⊕

X

2 Vi,j .

(i,j)∈Λ

For 0 ≤ ℓ ≤ 3, denote by {pi,ℓ = pi ⊗ 1}1≤i≤4 the linear basis of Pℓ , with pi,0 = pi . Using (2) and the following equalities, the D-module structure of Pℓ can be described explicitly.

(3)

a · (pi,ℓ ) = a · (pi ⊗ 1) = (a · pi ) ⊗ (a · 1) + (b · pi ) ⊗ (c · 1) = ξ ℓ (a · pi ) ⊗ 1 b · (pi,ℓ ) = b · (pi ⊗ 1) = (a · pi ) ⊗ (b · 1) + (b · pi ) ⊗ (d · 1) = ξ −ℓ (b · pi ) ⊗ 1 c · (pi,ℓ ) = c · (pi ⊗ 1) = (c · pi ) ⊗ (a · 1) + (d · pi ) ⊗ (c · 1) = ξ ℓ (c · pi ) ⊗ 1 d · (pi,ℓ ) = g · (pi ⊗ 1) = (c · pi ) ⊗ (b · 1) + (d · pi ) ⊗ (d · 1) = ξ −ℓ (d · pi ) ⊗ 1 x · (pi,ℓ ) = x · (pi ⊗ 1) = (x · pi ) ⊗ (g · 1) + (1 · pi ) ⊗ (x · 1) = (−1)ℓ (x · pi ) ⊗ 1 g · (pi,ℓ ) = g · (pi ⊗ 1) = (g · pi ) ⊗ (g · 1) = (−1)ℓ (g · pi ) ⊗ 1.

2.4. Some indecomposable D-modules. Let A be a finite-dimensional k-algebra and V1 , . . . , Vn a complete list of non-isomorphic simple left A-modules. The Ext-Quiver of A is the quiver ExtQ(A) with vertices 1, . . . , n and dim Ext1A (Vi , Vj ) arrows from the vertex i to the vertex j. Given a quiver Q with vertices 1, . . . , n, its separation diagram is the unoriented graph with vertices 1, . . . , n, 1′ , . . . , n′ and with an edge from i to j ′ for each arrow i → j in Q. The separation diagram of A is the separation diagram of its Ext-Quiver. It is well-known that a finite-dimensional algebra is of finite (tame) representation type if and only if its separation diagram is a disjoint union of finite (affine) Dynkin diagrams. In this section we compute the separation diagram of D and show that D is of tame representation type. In order to do so, we use the isomorphism of abelian groups between dim Ext1A (Vi , Vj ) and equivalence classes of extensions 0 → Vj → M → Vi → 0 of Vi by Vj ; here, the neutral element is given by the trivial extension. 2.4.1. 2-dimensional (non-simple) indecomposable modules. Let A be the subalgebra of D generated by a, d and g. Then A is an 8-dimensional commutative algebra given by A = kha, g : a4 = 1 = g 4 i. In particular, all simple modules are one-dimensional and the restriction to A of the characters of D induce characters on A.


14

Definition 2.14. Let 0 ≤ ℓ ≤ 3. Define Mℓ+ = k{m1 , m2 } to be the 2-dimensional D-module given by k m1 ≃ kχℓ and a · m2 = χℓ+1 (a) m2 = ξ ℓ+1 m2 ,

b · m2 = 0 = c · m2 ,

g · m2 = χℓ+1 (g) m2 = (−1)ℓ+1 m2 ,

x · m2 = m1 .

Then, Mℓ+ is an indecomposable module containing kχℓ and Mℓ+ /kχℓ = kχℓ+1 . Analogously, define Mℓ− = k{m1 , m2 } to be the 2-dimensional D-module given by k m1 ≃ kχℓ and √ √ 2 ℓ−1 2 ℓ−1 a · m2 = ξ m2 , b · m2 = ξ m1 , c · m2 = (−ξ)ℓ−1 m1 , 2 2 g · m2 = (−1)ℓ−1 m2 , x · m2 = m1 . Then, Mℓ− is an indecomposable module containing kχℓ and Mℓ− /kχℓ = kχℓ−1 .

Lemma 2.15. Let 0 ≤ ℓ ≤ 3. (i) Let M be a 2-dimensional indecomposable module containing kχℓ . Then M ≃ Mℓ+ or M ≃ Mℓ− . (ii) ( 1 if k = ℓ ± 1, 1 dim ExtD (kχk , kχℓ ) = 0 otherwise. Proof. (i) Assume M is a 2-dimensional indecomposable D-module containing kλ with λ = χℓ . We must have that M ≃ kλ ⊕ kµ as A-modules, with µ some character on D. That is, M has a linear basis {m1 , m2 } such that km1 ≃ kλ , and a · m2 = µ(a)m2 , d · m2 = µ(d)m2 and g · m2 = µ(g)m2 , with µ(a)2 = µ(g). In particular, M fits into an exact sequence of the form 0 → kλ → M → kµ → 0. Moreover, since M/kλ ≃ kµ , we must have that b · m2 = β m1 ,

a2 b

c · m2 = γ m1

λ(a)2 β

for some β, γ, α ∈ k. Since = c, then bg = −gb and xg = −gx we obtain the equalities β(λ(g) + µ(g)) = 0,

γ(λ(g) + µ(g)) = 0

and

x · m2 = α m1 ,

= γ. Using the relations cg = −gc, and

α(λ(g) + µ(g)) = 0.

Hence, if µ(g) 6= −λ(g) then β = 0, γ = 0 and α = 0 which implies that M ≃ kλ ⊕ kµ . √ Assume that µ(g) = −λ(g). From the relation bx − ξxb = 2ξ(a − gd) we deduce that λ(g) = λ(a)2 and µ(g) = µ(a)2 . Thus, we must have√that µ(a)2 = −λ(a)2 and consequently µ(a) = ±ξλ(a). Note that the relation cx + ξxc √ conclusion. √ = 2ξ(d − ga) yields the same On the other hand, the relations ax + ξxa = 2ξ(b + gc) and dx − ξxd = 2ξ(c + gb) give the equations √ √ α(λ(d) − ξµ(d)) = 2ξ2λ(a)2 β. α(λ(a) + ξµ(a)) = 2ξ2β,

Using that µ(a)2 = −λ(a)2 and λ(d) = λ(a)3 , it follows that the second equality is the first equality multiplied by λ(a)2 . If µ(a) = ξλ(a), then µ = χℓ+1 and λ(a) + ξµ(a) = 0, which implies that β = 0 = γ. In this case, M is an indecomposable module isomorphic to Mℓ+ . Denote this module by Mℓ+ (α). If µ(a)√= −ξλ(a), then µ = χℓ+3 = χℓ−1 and λ(a) + ξµ(a) = 2λ(a), which implies that αλ(a) = 2ξβ. In this case, M is isomorphic to Mℓ− and (i) follows. Denote this module by Mℓ− (α). (ii) By the preceding discussion, we have that dim Ext1D (kχk , kχℓ ) = 0 if k 6= ℓ ± 1. Let M be an extension of kχℓ by kχℓ±1 . Then, M = Mℓ± (α) for some α ∈ k× . Let α, α′ ∈ k× and assume Mℓ± (α) ≃ Mℓ± (α′ ) as extensions. If {m1 , m2 } and {m′1 , m′2 } denote


15

the linear basis of Mℓ± (α) and Mℓ± (α′ ), respectively, and ϕ the isomorphism, we must have that ϕ(m1 ) = m′1 and ϕ(m2 ) = γ m′1 + η m′2 with η 6= 0. Moreover, ϕ(x · m2 ) = αϕ(m1 ) = α m′1 must be equal to x · ϕ(m2 ) = η α′ m′1 . Hence, α = ηα′ implying that dim Ext1D (kχℓ±1 , kχℓ ) = 1. 2.4.2. 3-dimensional indecomposable modules. In this subsection we describe the 3-dimensional indecomposable modules as we did in the previous subsection for dimension 2. First, we prove that dim Ext1D (V, W ) = 0 for all simple modules V, W such that dim V dim W = 2. In particular, in the Ext-Quiver of D, the component of vertices corresponding to onedimensional modules is disconnected to the one corresponding to 2-dimensional modules. Lemma 2.16. (i) dim Ext1D (Vi,j , Vk,ℓ ) = 0 for all (i, j), (k, ℓ) ∈ Λ. 1 (ii) dim ExtD (Vi,j , kχℓ ) = 0 = dim Ext1D (kχℓ , Vi,j ) for all (i, j) ∈ Λ and ℓ ∈ Z4 . Proof. (i) follows inmediately since Vi,j is projective for all (i, j) ∈ Λ. (ii) By taking duals, we have that dim Ext1D (V, W ) = dim Ext1D (W ∗ , V ∗ ). Thus, it is enough to prove the assertion for V = Vi,j and W = kχℓ for (i, j) ∈ Λ and ℓ ∈ Z4 . Since Vi,j is projective for every (i, j) ∈ Λ by Lemma 2.10 (ii), the claim follows. Corollary 2.17. D is of tame representation type. Proof. Denote by i the vertex corresponding to the character χi for all 0 ≤ i ≤ 3. Lemma 2.15 implies that ExtQ(D) contains the quiver +

•K 0 k

+

•3 k

•K 1

•2 (1)

Thus, the separation diagram of D contains the quiver A3

•1′

⑤⑤ ⑤⑤ ⑤ ⑤ ⑤⑤

′

•0 ❇ •2

(1)

A3

•0 ⑥ ❆❆

❇❇ ❇❇ ❇❇ ❇❇

•3′

`

•1

⑥⑥ ⑥⑥ ⑥ ⑥ ⑥⑥

•2′

❆❆ ❆❆ ❆❆ ❆

•3

Moreover, by Lemma 2.16, ExtQ(D) consists of the quiver above and isolated points representing the simple modules Vi,j . Hence, D is of tame representation type. Remark 2.18. Notice that, since the simple representations Vi,j are projective for all (i, j) ∈ Λ, they cannot be contained in the socle or the top of any non-simple indecomposable module M . Hence, Soc(M ) and Top(M ) consist of direct sums of one-dimensional modules. Furthermore, if 0 ⊆ Soc(M ) ⊆ Soc2 (M ) ⊆ · · · ⊆ Socn (M ) = M denotes the socle series of M , then Soc(M/ Soci (M )) does not contain any simple projective module. Indeed, assume that Soc(M/ Soci (M )) = Soci+1 (M )/ Soci (M ) contains a simple projective module S, and denote by j : S → Soci+1 (M ))/ Soci (M ) the inclusion and by p : Soci+1 (M ) → Soci+1 (M ))/ Soci (M ) the projection. Since S is projective, there exists a morphism t : S → Soci+1 (M ) such that p ◦ t = j. Since S is simple, t is a monomorphism and whence Soci+1 (M ) contains a simple module t(S) isomorhic to S. But t(S) ⊆ Soc(Soci+1 (M )) ⊆ Soc(M ) ⊆ Soci (M ), which implies that j(S) = p(t(S)) = 0, a contradiction. In the following, we determine the 3-dimensional indecomposable modules.


16

Definition 2.19. Let Nℓ = k{n1 , n2 , n3 } be the 3-dimensional D-module given by k n1 ≃ kχℓ , k n2 ≃ kχℓ+2 and √ √ 2 ℓ+1 2 ℓ+1 a · n3 = ξ n3 , b · n3 = ξ n2 , c · n3 = − (−ξ)ℓ+1 n2 , 2 2 g · n3 = (−1)ℓ+1 n3 , x · n3 = n1 + n2 . Then, Nℓ is an indecomposable module with socle kχℓ ⊕ kχℓ+2 , Nℓ /(kχℓ ⊕ kχℓ+2 ) = kχℓ+1 , Nℓ /kχℓ = Mℓ+ and Nℓ /kχℓ+2 = Mℓ− . Lemma 2.20. Let N be a 3-dimensional indecomposable D-module. Then N ≃ Nℓ for some 0 ≤ ℓ ≤ 3. Proof. By Remark 2.18, Soc(N ) contains only one-dimensional modules. If Soc(N ) = kλ for some D-character λ, then N is injective by [ARS, Proposition II. 4.1 (d)]. But in such a case, N ∗ would be a projective module with an epimorphism to kλ−1 and whence dim N ∗ ≥ dim P (kλ−1 ) = 4, a contradiction. Hence, Soc(N ) = kλ ⊕ kµ for some Dcharacters λ, µ and N fits into an exact sequence (4)

0 → kλ ⊕ kµ → N → kτ → 0,

for some D-character τ . In particular, N ≃ kλ ⊕ kµ ⊕ kτ as A-modules. Let {n1 , n2 , n3 } be a linear basis of N such that kn1 ≃ kλ , kn2 ≃ kµ and a · n3 = τ (a)n3 , b · n3 = β1 n1 + β2 n2 ,

d · n3 = τ (d)n3 , c · n 3 = γ1 n 1 + γ2 n 2 ,

g · n3 = τ (g)n3 = τ (a)2 n3 , x · n3 = θ1 n1 + θ2 n2 .

2 As a2 b = c, we have that λ(a)2 β1 = γ1 and µ(a) √ β2 = γ2 . Further, using the relations cg = −gc, bg = −gb, xg = −gx and ax + ξxa = 2ξ(b + gc) we obtain the equalities

β1 (λ(g) + τ (g)) = 0,

β2 (µ(g) + τ (g)) = 0,

γ1 (λ(g) + τ (g)) = 0,

γ2 (µ(g) + τ (g)) = 0,

θ1 (λ(g) + τ (g)) = 0, √ θ1 (λ(a) + ξτ (a)) = 2 2ξβ1 ,

θ2 (µ(g) + τ (g)) = 0, √ θ2 (µ(a) + ξτ (a)) = 2 2ξβ2 .

If τ (g) 6= −λ(g) and τ (g) 6= −µ(g), then βi = γi = θi = 0 and consequently N ≃ kλ ⊕ kµ ⊕ kτ as D-modules, a contradiction. If τ (g) = −λ(g) but τ (g) 6= −µ(g), then β2 = γ2 = θ2 = 0 and this implies that N = L ⊕ kµ with L = k{n1 , n3 }. Analogously, N is decomposable if τ (g) = −µ(g) but τ (g) 6= −λ(g). Hence, −τ (g) = λ(g) = µ(g) and thus λ(a) = ±µ(a) = ±ξτ (a). The same reasoning shows that θ1 6= 0 6= θ2 since otherwise N would be decomposable. So, we may assume that θ1 = θ2 = 1. Moreover, λ = −µ = ξ 2 µ since otherwise N is also decomposable. Indeed, if λ = µ, let v = n1 + n2 . Then kv ≃ kλ and N ≃ kλ ⊕ k{v, n3 } as D-modules. Let λ = χℓ for some 0 ≤ ℓ ≤ 3. Then µ = χℓ+2 and τ = χℓ±3 = χℓ±1 . From the last equation above it follows that if τ (a) = χℓ+1 (a) = ξ ℓ+1 , then β1 = 0 and√ if τ (a) = √ 2 ℓ+1 ℓ−1 ℓ−1 ℓ+1 and γ2 = − 22 (−ξ)ℓ+1 . χ (a) = ξ , then β2 = 0. Assume τ = χ , then β2 = 2 ξ ℓ−1 In such a case, N ≃ Nℓ . If τ = χ , the same argument also shows that N ≃ Nℓ and the lemma is proved. 3. The category

K YD K

Using the equivalence D M ≃ K K YD we determine in this section the simple and some indecomposable objects of the latter and we describe their braidings. Note that by [AG, A ′′ ∗ Proposition 2.2.1], one has K K YD ≃ A YD with A = A4 = K .


17

3.1. Simple objects and projective covers in K K YD. Our intention is to describe the simple D-modules and their projective covers as left Yetter-Drinfeld modules over K. To do so, we simply need to describe the coaction of K. Proposition 3.1. Let kχj = kv be a one-dimensional D-module with j ∈ Z4 . Then k χj ∈ K K YD with its structure given by a · v = ξ j v,

b · v = c · v = 0,

d · v = ξ −j v,

δ(v) = a2j ⊗ v.

Proof. The action is given by the restriction of the action given in Lemma 2.6. Since kχj is one-dimensional, we must have that δ(v) = h ⊗ v with h ∈ G(K) = {1, a2 }. As f · v = hf, hiv for all f ∈ K∗ and hg, a2 i = −1, it follows that δ(v) = a2j ⊗ v. The following proposition gives the braiding of kχj for all j ∈ Z4 . Proposition 3.2. The braiding of the one-dimensional Yetter-Drinfeld module kχj is c(v ⊗ v) = (−1)j v ⊗ v. Proof. By formula (1) and Proposition 3.1 we have 2

c(v ⊗ v) = a2j · v ⊗ v = (ξ j )2j v ⊗ v = (−1)j v ⊗ v = (−1)j v ⊗ v. Proposition 3.3. Let Vi,j = k{e1 , e2 } be a two-dimensional simple D-module with (i, j) ∈ Λ. If we denote λ1 = ξ i and λ2 = ξ j , then Vi,j ∈ K K YD with its action given by a · e1 = λ1 e1 ,

b · e1 = 0,

c · e1 = 0,

b · e2 = λ21 e1 ,

a · e2 = −ξλ1 e2 ,

and its coaction by

δ(e1 ) = 1 ⊗ e1 − 2λ1 ac ⊗ e2 ,

δ(e1 ) = a2 ⊗ e1 + 2λ1 ab ⊗ e2 ,

c · e2 = e1 ,

d · e1 = λ−1 1 e1 ,

d · e2 = ξλ−1 1 e2 ,

δ(e2 ) = a2 ⊗ e2 ,

δ(e2 ) = 1 ⊗ e2 ,

for λ2 = 1, for λ2 = −1,

1 δ(e1 ) = d ⊗ e1 + (λ31 − ξλ1 )c ⊗ e2 , δ(e2 ) = a ⊗ e2 + (λ1 + ξλ31 )b ⊗ e1 , for λ2 = ξ, 2 1 δ(e1 ) = a ⊗ e1 + (λ31 + ξλ1 )b ⊗ e2 , δ(e2 ) = d ⊗ e2 + (λ1 − ξλ31 )c ⊗ e1 for λ2 = −ξ. 2 P8 i Proof. We just need to describe the coaction. Using that δ(v) = i=1 vi ⊗ v · v with i {vi }1≤i≤8 a basis of K and {v }1≤i≤8 its dual basis, and the isomorphism given in Lemma 2.3 have that 3 3 X X i ∗ i (xg i )∗ ⊗ xg i · e1 (g ) ⊗ g · e1 + δ(e1 ) = =

i=0 3 X

i=0 3 X

i=0 3 X

i=0

i=0

(gi )∗ ⊗ λi2 e1 +

i=0

(xgi )∗ ⊗ (−λ2 )i x21 e2 ,

3 3 X X i ∗ i (xg i )∗ ⊗ xg i · e2 (g ) ⊗ g · e2 + δ(e2 ) =

=

i=0

(gi )∗ ⊗ (−λ2 )i e2 +

3 X i=0

(xgi )∗ ⊗ λi2 x12 e1 ,

1 1 where (gi )∗ = (1 + ξ i a + (−ξ)i d + (−1)i a2 ), (xgi )∗ = √ ((−ξ)i b + ξ i c + ab + (−1)i ac) 4 4 2ξ √ √ 2 for all 0 ≤ i ≤ 3, x21 = 2ξ(λ31 − λ1 λ2 ), x12 = ξ(λ1 + λ31 λ2 ). Explicitly, if j = 0 and i ∈ 2

18


√

√ 2ξ(λ31 −λ1 ) = −2 2ξλ1 , x12 =

√

2 ξ(λ1 +λ31 ) = 0. 2 In such a case, δ(e1 ) = 1 ⊗ e1 − 2λ1 ac ⊗ e2 and δ(e2 ) = a2 ⊗ e2 . √ √ If j √ = 2 and i ∈ {0, 2}, then λ2 = −1, λ1 = ±1 and x21 = 2ξ(λ31 + λ1 ) = 2 2ξλ1 , 2 ξ(λ1 − λ31 ) = 0. In such a case, δ(e1 ) = a2 ⊗ e1 + 2λ1 ab ⊗ e2 and δ(e2 ) = 1 ⊗ e2 . x12 = 2 If j = 1 and i is arbitrary, then λ2 = ξ, and δ(e1 ) = d ⊗ e1 + (λ31 − ξλ1 )c ⊗ e2 and 1 δ(e2 ) = a ⊗ e2 + (λ1 + ξλ31 )b ⊗ e1 . 2 Finally, if If j = 3 and i is arbitrary, then λ2 = −ξ and δ(e1 ) = a ⊗ e1 + (λ31 + ξλ1 )b ⊗ e2 1 and δ(e2 ) = d ⊗ e2 + (λ1 − ξλ31 )c ⊗ e1 . 2 {1, 3} then λ2 = 1, λ1 = ±ξ and x21 =

Next we describe the braiding of the simple modules Vi,j in K K YD. To do so, we use a matrix-like notation to describe it in a compact way. Its proof follows by a direct computation using formula (1) and Proposition 3.3. Proposition 3.4. Denote as above λ1 = ξ i and λ2 = ξ j . The braiding of Vi,j ∈ K K YD is given by the following formulae: (i) If j = 0, i ∈ {1, 3}, then e1 e1 ⊗ e1 e2 ⊗ e1 + 2e1 ⊗ e2 c( ⊗ e1 e2 ) = . e2 −e1 ⊗ e2 e2 ⊗ e2 (ii) If j = 2, i ∈ {0, 2}, then e1 e1 ⊗ e1 −e2 ⊗ e1 + 2e1 ⊗ e2 c( ⊗ e1 e2 ) = . e2 e1 ⊗ e2 e2 ⊗ e2

(iii) If j = 1 and i is arbitrary, then ) ( 3 λ1 e1 ⊗ e1 ξλ31 e2 ⊗ e1 + (λ31 − ξλ1 )e1 ⊗ e2 e1 1 . c( ⊗ e1 e2 ) = e2 λ1 e1 ⊗ e2 −ξλ1 e2 ⊗ e2 + (λ31 + ξλ1 )e1 ⊗ e1 2 (iv) If j = 3 and i is arbitrary, then ) ( λ1 e1 ⊗ e1 −ξλ1 e2 ⊗ e1 + (λ1 + ξλ31 )e1 ⊗ e2 e1 1 . c( ⊗ e1 e2 ) = e2 λ31 e1 ⊗ e2 ξλ31 e2 ⊗ e2 + (λ1 − ξλ31 )e1 ⊗ e1 2

Remark 3.5. All the braidings given by Proposition 3.4 are not of diagonal type. See the Appendix for a proof. We end this section with the description of the projective covers of the one-dimensional modules kχj for j ∈ Z4 as objects in K K YD. Proposition 3.6. Let Pj be the projective cover of the one-dimensional D-module kχj , with j ∈ Z4 . Then Pj ∈ K K YD with its action given by (2) and (3) and its coaction by √ √ ξ 2 2 j δ(p1,j ) = (a ) ⊗ p1,j − (−a2 )j ac ⊗ (p2,j + 2p3,j ), 2 2 j+1 δ(p2,j ) = (a ) ⊗ p2,j + ξ(−a2 )j ab ⊗ p4,j , √ ξ 2 2 j+1 δ(p3,j ) = (a ) ⊗ p3,j − (−a2 )j ab ⊗ p4,j , 2 δ(p4,j ) = (a2 )j ⊗ p4,j . Proof. The action of K is given by the restriction of the action ofPD. To describe the 8 i coaction we use again that for all p ∈ Pj we have that δ(p) = i=1 vi ⊗ v · p with


19

{vi }1≤i≤8 a basis of K, and {v i }1≤i≤8 its dual basis. Using the isomorphism given in Lemma 2.3, it follows that δ(p1,j ) = =

3 3 X X (xgi )∗ ⊗ xg i · p1,j (gi )∗ ⊗ gi · p1,j + i=0 3 X i=0

i=0

i ∗

j i

(g ) ⊗ ((−1) ) p1,j +

3 X i=0

(xg i )∗ ⊗ ((−1)j+1 )i (−1)j (p2,j +

√

2p3,j )

√ √ ξ 2 (−a2 )j ac ⊗ (p2,j + 2p3,j ); = (a ) ⊗ p1,j − 2 3 3 X X (xgi )∗ ⊗ xg i · p2,j (gi )∗ ⊗ gi · p2,j + δ(p2,j ) = 2 j

i=0

i=0

3 3 X X √ (xg i )∗ ⊗ (− 2(−1)j ((−1)j )i )p4,j (gi )∗ ⊗ ((−1)j+1 )i p2,j + = i=0 2 j+1

= (a )

δ(p3,j ) = =

i=0

2 j

⊗ p2,j + ξ(−a ) ab ⊗ p4,j ;

3 3 X X (xgi )∗ ⊗ xg i · p3,j (gi )∗ ⊗ gi · p3,j + i=0 3 X i=0

i=0

i ∗

j+1 i

(g ) ⊗ ((−1)

) p3,j

3 X (xg i )∗ ⊗ (−1)j ((−1)j )i p4,j + i=0

√

ξ 2 (−a2 )j ab ⊗ p4,j ; 2 3 3 3 X X X (gi )∗ ⊗ ((−1)j )i p4,j (xgi )∗ ⊗ xg i · p4,j = (gi )∗ ⊗ gi · p4,j + δ(p4,j ) = = (a2 )j+1 ⊗ p3,j −

i=0 2 j

i=0

i=0

= (a ) ⊗ p4,j .

Finally, we describe the braiding for every Pj ∈ K K YD, j ∈ Z4 . The proof follows by a straightforward computation using (1) and the coaction given in Proposition 3.6. Proposition     c(p1,j ⊗        c(p2,j ⊗        c(p3,j ⊗        c(p4,j ⊗   

3.7. Let j ∈ Z4 . The braiding of Pj ∈ K K YD is given by the formulae:      jp p1,j  −p (−1)     3,j 1,j √         √ 2  (−1)j p4,j  p2,j p2,j )= ⊗ (p2,j + 2p3,j ), ⊗ p1,j + p3,j  p3,j 0    2            j p4,j 0 (−1) p4,j      p1,j p1,j  (−1)j+1 p3,j               p2,j (−1)j+1 p2,j −p4,j )= ⊗ p2,j + ⊗ p4,j , j+1 p3,j  (−1) p3,j  0              p4,j p4,j 0      p1,j (−1)j p3,j  p1,j     √          2 p4,j (−1)j+1 p2,j p2,j ⊗ p4,j , ⊗ p + )= 3,j 0 p3,j  (−1)j+1 p3,j    2            0 p4,j p4,j    j p1,j  (−1) p1,j         p2,j p2,j )= ⊗ p4,j . p3,j  p3,j         p4,j (−1)j p4,j

20


4. Nichols algebras in

K YD K

In this section we determine all finite-dimensional Nichols algebras of simple modules over K . They consist of exterior algebras of dimension 2 and 8-dimensional algebras with triangular braiding. Indeed, since all objects in K K YD can be described as objects in the category of Yetter-Drinfeld modules over the pointed Hopf algebra K∗ = A′′4 , by [U] it follows that the associated braiding is triangular. These 8-dimensional examples are new examples of finite-dimensional Nichols algebras. They are isomorphic to quantum linear spaces as algebras, but not as coalgebras since the braiding differs; in our case, the bainding is not diagonal. It remains an open question if they are twist equivalent and in such a case, in which category. We begin by studying the Nichols algebras of the one-dimensional simple modules and their projective covers. Lemma 4.1. Let j ∈ Z4 . The Nichols algebras B(kχj ) associated to kχj = kx are: Vk[x], if j = 0, 2, B(kχj ) = 2 k[x]/(x ) = kχj , if j = 1, 3.

Proof. Since by Proposition 3.2 we have that c = (−1)j τ , where τ represents the usual flip, the result follows immediately. Corollary 4.2. Let W = kχj1 ⊕ · · · ⊕ kχjt be a V direct sum of one-dimensional modules with js ∈ {1, 3} for all 1 ≤ s ≤ t. Then B(W ) = W ≃ B(kχj1 ) ⊗ · · · ⊗ B(kχjt ).

Proof. If jV s ∈ {1, 3} for all 1 ≤ s ≤ t, then the braiding on W ⊗ W is −τ and hence B(W ) = W . Let v ∈ kχjr , w ∈ kχjs , then c(v ⊗ w) = (a2 )jr w ⊗ v = (−1)jr js w ⊗ v. Hence, c2W = idW and the last assertion follows from [Gr, Theorem 2.2]. Lemma 4.3. Let j ∈ Z4 and Pj be the projective cover of kχj . Then dim B(Pj ) = ∞. Proof. In all cases, the braiding on Pj ⊗ Pj contains an eigenvector of eigenvalue 1 and the claim follows by Remark 1.2. Indeed, by Proposition 3.7 we have that c(p4,j ⊗ p4,j ) = p4,j ⊗ p4,j if j = 0, 2, and c(p3,j ⊗ p3,j ) = p3,j ⊗ p3,j if j = 1, 3. Before we describe the Nichols algebras associated to 2-dimensional simple modules, we analyze the Nichols algebras of non-simple indecomposable modules. It turns out that they are all infinite-dimensional. Remark 4.4. Let V ∈ K K YD be a finite-dimensional module such that dim B(V ) < ∞. Since taking the Nichols algebra defines a functor between the category of braided vector spaces and the category of braided Hopf algebras, see [AG], it follows that dim B(W ) < ∞ for all W ∈ Soc(V ) or W ∈ Top(V ). Furthermore, if 0 ⊆ Soc(V ) ⊆ Soc2 (V ) ⊆ · · · ⊆ Socn (V ) = V denotes the socle series of V , then dim B(V / Soci (V )), dim B(Soci (V )) and dim B(Soc(V / Soci (V ))) are all finite for all 1 ≤ i ≤ n. Theorem 4.5. Let M ∈ K K YD be a finite-dimensional non-simple indecomposable module. Then B(M ) is infinite-dimensional.

Proof. Assume first that dim M = 2. Then by Lemma 2.15 (i), we have that M ≃ Mℓ+ or M ≃ Mℓ− for some 0 ≤ ℓ ≤ 3. Since Soc(Mℓ+ ) = kχℓ , Top(Mℓ+ ) = kχℓ+1 and Soc(Mℓ− ) = kχℓ , Top(Mℓ− ) = kχℓ−1 , it follows that dim B(M ) = ∞ by Lemma 4.1 and Remark 4.4 because dim B(N ) is infinite for a simple module either in Soc(M ) or in Top(M ). Assume now that dim M = n ≥ 3. We prove the claim by induction on dim M . Suppose that dim B(N ) is infinite for all indecomposable module of dimension less than n. By ¯ be a simple module Remark 2.18, Soc(M ) consists of one-dimensional modules. Let N contained in Soc(M/ Soc(M )) and denote by N the corresponding submodule of M . Then, ¯ = 1 by Remark 2.18. If Soc(M ) = kλ , then N is an indecomposable module of dim N dimension 2. The previous paragraphs imply that dim B(N ) is infinite and consequently dim B(M ) is infinite. Assume Soc(M ) contains more than one simple module, and let


21

kλ ⊂ Soc(M ). If N/kλ is semisimple, then N contains an indecomposable module of dimension 2 and whence dim B(N/kλ ) is infinite. This implies again that dim B(N ) and dim B(M ) are both infinite. If N/kλ is not semisimple, then it contains an indecomposable module of dimension less than n. By induction, dim B(N/kλ ) is infinite and the theorem follows applying the arguments above. Remark 4.6. Let V ∈ K K YD such that dim B(V ) is finite. Then by Theorem 4.5, V is necessarily semisimple. In these notes we will analyse only the Nichols algebras over simple modules, since the case of semisimple modules demands much more work to be carried out. A first aproach could be done by studying the Yetter-Drinfeld submodules adn (V )(W ) of a given Nichols algebra B(V ⊕ W ) with V and W simple modules, see [HS] for details. A direct computation shows that dim B(V ⊕ W ) = ∞ for V = kχ and W = V3,1 , V3,3 ; and V = kχ3 and W = V2,1 , V2,3 . In fact, ad(kχ )(V3,1 ) ≃ V0,3 , ad(kχ )(V3,3 ) ≃ V0,1 , ad(kχ3 )(V2,1 ) ≃ V1,3 and ad(kχ3 )(V2,3 ) ≃ V1,1 . Now we proceed to analyze the Nichols algebras associated to two-dimensional simple modules. It turns out that 4 of these simple modules give rise to 8-dimensional Nichols algebras with triangular (and not diagonal) braiding. Lemma 4.7. Let Λ′ = Λ r {(2, 1), (3, 1), (2, 3), (3, 3)}. Then dim B(Vi,j ) = ∞ for all (i, j) ∈ Λ′ . Proof. In all cases, the braiding of Vi,j contains an eigenvector w ⊗ w with w ∈ Vi,j of eigenvalue 1, hence the lemma follows by Remark 1.2. Indeed, if (i, j) = (1, 1) or (1, 3), √ then w = e1 + 2ξe2 . Otherwise, take w = e1 . Next we describe the Nichols algebras associated to the pairs (2, 1), (3, 1), (2, 3) and (3, 3) in Λ. By Remark 2.8 we have that B(V2,1 )∗ ≃ B(V3,3 ) and B(V3,1 )∗ ≃ B(V2,3 ). It turns out that, as algebras B(V2,1 ) ≃ B(V2,3 ) and B(V3,1 ) ≃ B(V3,3 ) and all of them are isomorphic to algebras associated to quantum linear spaces. Remember that every graded Hopf algebra in K e duality [AG, K YD satisfies the Poincar´ LN i with N minimal such that RN 6= {0}, then Proposition 3.2.2], i.e. if R = R i=0 dim Ri = dim RN −i . Proposition 4.8. B(V2,1 ) = khx, y : x2 = 0, xy + ξyx = 0, y 4 = 0i; in particular dim B(V2,1 ) = 8. Proof. Write x = e1 , y = e2 for the generators of V2,1 . Then, the coaction given by 1 Proposition 3.3 is δ(x) = d ⊗ x + (−1 + ξ)c ⊗ y and δ(y) = a ⊗ y + (−1 − ξ)b ⊗ x. Using 2 the braiding given by Proposition 3.4, we get that ) ( −x ⊗ x −ξy ⊗ x + (ξ − 1)x ⊗ y x 1 , c( ⊗ x y )= y −x ⊗ y ξy ⊗ y − (1 + ξ)x ⊗ x 2

and one sees that the relations x2 = 0, xy + ξyx = 0 and y 4 = 0 must hold in B(V2,1 ). Indeed, the first two relations are easily checked since these elements are primitive. Let us focus in the last one; we show that it is also primitive. Since ∆(y 2 ) = (y ⊗ 1 + 1 ⊗ y)(y ⊗ 1 + 1 ⊗ y) 1 = y 2 ⊗ 1 + (1 + ξ)y ⊗ y − (1 + ξ)x ⊗ x + 1 ⊗ y 2 , 2 then 1 ∆(y 3 ) = (y ⊗ 1 + 1 ⊗ y)(y 2 ⊗ 1 + (1 + ξ)y ⊗ y − (1 + ξ)x ⊗ x + 1 ⊗ y 2 ) 2 1 1 3 2 = y ⊗ 1 + (1 + ξ)xy ⊗ x + ξy ⊗ y − (1 + ξ)x ⊗ xy + ξy ⊗ y 2 + 1 ⊗ y 3 2 2

22


1 because c(y ⊗ y 2 ) = (x2 − y 2 ) ⊗ y + (1 − ξ)(yx − xy) ⊗ x, and 2 1 ∆(y 4 ) = (y ⊗ 1 + 1 ⊗ y)(y 3 ⊗ 1 + (1 + ξ)xy ⊗ x + ξy 2 ⊗ y− 2 1 − (1 + ξ)x ⊗ xy + ξy ⊗ y 2 + 1 ⊗ y 3 ) 2 = y4 ⊗ 1 + 1 ⊗ y4, 1 because c(y ⊗ y 3 ) = ξ(−y 3 + x2 y + yx2 − xyx) ⊗ y + (1 + ξ)(y 2 x − x3 − yxy + xy 2 ) ⊗ x 2 1 2 and c(y ⊗ xy) = −ξxy ⊗ y + (1 + ξ)x ⊗ x. 2 Therefore, we have a graded braided Hopf algebra epimorphism π : T (V2,1 )/I ։ B(V2,1 ) where I is the ideal generated by the relations. Let R = T (V2,1 )/I, then R is a graded braided Hopf algebra with R5 = 0, R0 = k and R1 = V2,1 . Since R satisfies the Poincaré duality, we have that dim R4 = 1 and dim R3 = 2. Clearly, dim R2 = 2. As B5 (V2,1 ) = 0 and π is injective in degree 0 and 1, it follows that dim B4 (V2,1 ) = dim B0 (V2,1 ) = 1 and dim B3 (V2,1 ) = 2 = dim B1 (V2,1 ). Then π is injective in degree 4 and 3 also. In order to prove that π is injective, it remains to show that π is injective in degree 2. This is equivalent to check that the relations in degree 2 in the Nichols algebra are just x2 = 0, xy + ξyx = 0, which follows by a direct computation using the braiding. Proposition 4.9. B(V2,3 ) = khx, y : x2 = 0, xy − ξyx = 0, y 4 = 0i; in particular, dim B(V2,3 ) = 8. Proof. The proof follows the same lines of Proposition 4.8. We only show that the relations hold. Write x = e1 , y = e2 for the generators of V3,1 , then the coaction given by Proposition 1 3.3 is δ(x) = a ⊗ x + (−1 − ξ)b ⊗ y and δ(y) = d ⊗ y + (−1 + ξ)c ⊗ x. Using the braiding, 2 we see that ∆(x2 ) = x2 ⊗ 1 + 1 ⊗ x2 ,

∆(xy) = xy ⊗ 1 − ξx ⊗ y + ξy ⊗ x + 1 ⊗ xy,

∆(yx) = yx ⊗ 1 + y ⊗ x − x ⊗ y + 1 ⊗ yx, 1 ∆(y 2 ) = y 2 ⊗ 1 + (1 − ξ)y ⊗ y + (ξ − 1)x ⊗ x + 1 ⊗ y 2 , 2 which gives us the relations x2 = 0 and xy − ξyx = 0. Since c(y ⊗ y 2 ) = (x2 − y 2 ) ⊗ y + 1 (1 + ξ)(yx − xy) ⊗ x, we have that 2 1 1 ∆(y 3 ) = y 3 ⊗ 1 + (1 − ξ)xy ⊗ x − ξy 2 ⊗ y + (ξ − 1)x ⊗ xy − ξy ⊗ y 2 + 1 ⊗ y 3 , 2 2 and whence ∆(y 4 ) = ∆(y 3 )(y ⊗ 1 + 1 ⊗ y) 1 1 = (y 3 ⊗ 1 + (1 − ξ)xy ⊗ x − ξy 2 ⊗ y + (ξ − 1)x ⊗ xy − ξy ⊗ y 2 + 1 ⊗ y 3 ) 2 2 = y4 ⊗ 1 + 1 ⊗ y4, 1 since c(y ⊗ y 3 ) = −ξ(−y 3 + x2 y + yx2 − xyx) ⊗ y + (1 − ξ)(y 2 x − x3 − yxy + xy 2 ) ⊗ x 2 1 2 and c(y ⊗ xy) = ξxy ⊗ y + (1 − ξ)x ⊗ x. Hence, we have the relation y 4 = 0. 2 Proposition 4.10. B(V3,1 ) = khx, y : x2 − 2y 2 = 0, xy + yx = 0, y 4 = 0i; in particular dim B(V3,1 ) = 8.


23

Proof. The proof follows the same lines of Proposition 4.8. Again, we only show that the relations hold. Write x = e1 , y = e2 for the generators of V3,1 . First, note that the coaction 1 given by Proposition 3.3 is δ(x) = d ⊗ x + (ξ − 1)c ⊗ y and δ(y) = a ⊗ y + (−ξ − 1)b ⊗ x. 2 Then, using the action of K and the braiding given by Proposition 3.4, we get that ∆(x2 ) = x2 ⊗ 1 + (1 + ξ)x ⊗ x + 1 ⊗ x2 ,

∆(xy) = xy ⊗ 1 + ξx ⊗ y − y ⊗ x + 1 ⊗ xy, 1 ∆(yx) = yx ⊗ 1 + y ⊗ x − ξx ⊗ y + 1 ⊗ yx, ∆(y 2 ) = y 2 ⊗ 1 + (1 + ξ)x ⊗ x + 1 ⊗ y 2 , 2 2 2 which gives us the relations x − 2y = 0 and xy + yx = 0, since both elements are 1 primitive. Analogously, since c(y ⊗ y 2 ) = (y 2 − x2 ) ⊗ y − (1 + ξ)(xy + yx) ⊗ x, we have 2 that 1 1 ∆(y 3 ) = y 3 ⊗ 1 − (1 + ξ)xy ⊗ x − y 2 ⊗ y − (1 − ξ)x ⊗ xy + y ⊗ y 2 + 1 ⊗ y 3 , 2 2 and consequently ∆(y 4 ) = ∆(y 3 )(y ⊗ 1 + 1 ⊗ y) = 1 1 = (y 3 ⊗ 1 − (1 + ξ)xy ⊗ x − y 2 ⊗ y − (1 − ξ)x ⊗ xy + y ⊗ y 2 + 1 ⊗ y 3 )(y ⊗ 1 + 1 ⊗ y) 2 2 = y4 ⊗ 1 + 1 ⊗ y4 ,

1 since c(y ⊗ y 3 ) = (yx2 + xyx − y 3 + x2 y) ⊗ y + (1 + ξ)(yxy − x3 + xy 2 + y 2 x) ⊗ x and 2 1 c(y ⊗ xy) = ξxy ⊗ y + (1 − ξ)x2 ⊗ x. Thereby, we have the relation y 4 = 0. 2 Proposition 4.11. B(V3,3 ) = khx, y : x2 − 2y 2 = 0, xy + yx = 0, y 4 = 0i; in particular, dim B(V3,3 ) = 8. Proof. As in the proof of Proposition 4.10, we only show that the relations hold. Write x = e1 , y = e2 for the generators of V3,3 , then the coaction given by Proposition 3.3 is 1 δ(x) = a ⊗ x + (ξ + 1)b ⊗ y and δ(y) = d ⊗ y + (1 − ξ)c ⊗ x. Using the braiding given in 2 Proposition 3.4, we have ∆(x2 ) = x2 ⊗ 1 + (1 − ξ)x ⊗ x + 1 ⊗ x2 ,

∆(xy) = xy ⊗ 1 − ξx ⊗ y − y ⊗ x + 1 ⊗ xy, 1 ∆(yx) = yx ⊗ 1 + y ⊗ x + ξx ⊗ y + 1 ⊗ yx, ∆(y 2 ) = y 2 ⊗ 1 + (1 − ξ)x ⊗ x + 1 ⊗ y 2 , 2 2 2 which gives us the relations x − 2y = 0 and xy + yx = 0. Since c(y ⊗ y 2 ) = (y 2 − x2 ) ⊗ 1 y + (ξ − 1)(xy + yx) ⊗ x, it follows that 2 1 1 ∆(y 3 ) = y 3 ⊗ 1 + (ξ − 1)xy ⊗ x − y 2 ⊗ y − (1 + ξ)x ⊗ xy + y ⊗ y 2 + 1 ⊗ y 3 , 2 2 and consequently ∆(y 4 ) = ∆(y 3 )(y ⊗ 1 + 1 ⊗ y) 1 1 = (y 3 ⊗ 1 + (ξ − 1)xy ⊗ x − y 2 ⊗ y − (1 + ξ)x ⊗ xy + y ⊗ y 2 + 1 ⊗ y 3 )(y ⊗ 1 + 1 ⊗ y) 2 2 4 4 = y ⊗1+1⊗y ,

1 since c(y ⊗ y 3 ) = (yx2 + xyx − y 3 + x2 y) ⊗ y + (1 − ξ)(yxy − x3 + xy 2 + y 2 x) ⊗ x and 2 1 c(y ⊗xy) = −ξxy ⊗y + (1+ξ)x2 ⊗x. Thus, y 4 = 0 since it is also a primitive element. 2 We end this section with the characterization of the finite-dimensional Nichols algebras over indecomposable objects in K K YD.

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Proof of Theorem A. Let V be an indecomposable module such that B(V ) is finitedimensional. Then by Theorem 4.5, V is necessarily simple. The claim then follows by Lemmata 4.1 and 4.7, and Propositions 4.8, 4.9, 4.10 and 4.11. Clearly, Nichols algebras over distinct families are pairwise non-isomorphic, since they are generated by the set of primitive elements which are non-isomorphic as Yetter–Drinfeld modules. 5. Hopf algebras over K In this last section we determine all finite-dimensional Hopf algebras H such that H[0] = K K and the corresponding infinitesimal braiding is a simple module L in K YD. That is, the graded algebra with respect to the standard filtration is gr H = i≥0 H[i] /H[i−1] ≃ R#K 1 with R a connected graded braided Hopf algebra in K K YD and R isomorphic to a simple K object in K YD. We begin by proving that this type of Hopf algebra is generated in degree one, i.e. R ≃ B(V ) with V = R1 . Theorem 5.1. Let H be a finite-dimensional Hopf algebra over K such that its infinitesimal braiding is isomorphic to a simple object in K K YD. Then the diagram of H is a Nichols algebra, and consequently H is generated by the elements of degree one with respect to the standard filtration. Proof. As H is a Hopf algebra over K, we have that H[0] ≃ K, gr H = R#K, and by ∗ hypothesis R1 = V , with V a simple object in K K YD. Let S = R be the graded dual of the diagram R. By [AS1, Lemma 5.5], S is generated by W = S(1) and R is a Nichols algebra if and only if P (S) = S(1), that is, if S is also a Nichols algebra. Since B(W ) = T (W )/J, to show that P (S) = S(1), it is enough to prove that the relations that generate the ideal J hold in S. This can be done by a case by case computation. We perform here two cases, leaving the rest as exercise for the reader. V Assume W = kχℓ with 1 ≤ ℓ ≤ 3. Then by Lemma 4.1, ℓ = 1, 3 and B(W ) ≃ W = k[x]/(x2 ). If we denote r = x2 ∈ S, then r is primitive and c(r ⊗ r) = r ⊗ r, where c is the braiding in K K YD. Since S is finite-dimensional, it follows that r = 0 and there is a surjection B(W ) ։ S. In particular, P (S) = W = S(1) and S is a Nichols algebra. Assume dim W = 2, then by Theorem A we have that W ≃ V2,1 , V2,3 , V3,1 or V3,3 . We prove the statement only for W ≃ V2,1 . By Proposition 4.8, V2,1 is generated by the elements x, y and the ideal defining the Nichols algebra is generated by the elements x2 , xy + ξyx and y 4 . Moreover, by the proof of the same proposition, these elements are primitive. Thus, we need to prove that c(r ⊗ r) = r ⊗ r for r = x2 , xy + ξyx and y 4 . Using the coaction given by Proposition 3.3, we have that δ(x2 ) = d2 ⊗ x2 + (ξ − 1)dc⊗ (xy + ξyx) and whence c(x2 ⊗ x2 ) = (x2 )(−1) · x2 ⊗ (x2 )(0) = d2 · x2 ⊗ x2 + (ξ − 1)(dc) · x2 ⊗ (xy + ξyx) = (−1)2 x2 ⊗ x2 = x2 ⊗ x2 .

Analogously, since δ(xy) = 1 ⊗ xy + (ξ − 1)ca ⊗ y 2 and δ(yx) = 1 ⊗ yx + (ξ − 1)ac ⊗ y 2 , one has that δ(xy + ξyx) = 1 ⊗ (xy + ξyx). This implies that c((xy + ξyx) ⊗ (xy + ξyx)) = (xy + ξyx) ⊗ (xy + ξyx). Finally, we have that δ(y 2 ) = a2 ⊗ y 2 − 12 (1 + ξ)ba ⊗ (xy + ξyx) = a2 ⊗ y 2 . Thus, δ(y 4 ) = 1 ⊗ y 4 and consequently c(y 4 ⊗ y 4 ) = y 4 ⊗ y 4 . Recall that, if v ∈ V = R1 is a primitive element, then by the formula given by the bosonization we have that ∆(v#1) = v (1) #(v (2) )(−1) ⊗(v (2) )(0) #1 ∈ H[1] ⊗H[0] +H[0] ⊗H[1] . We denote v = v#1 for all v ∈ V , and k = 1#k for all k ∈ K. Next we show that the bosonizations of the Nichols algebras associated to the simple modules kχℓ with ℓ = 1, 3 and V2,1 , V2,3 do not admit deformations. Proposition 5.2. Let H be a finite-dimensional Hopf algebra over K such V that its infinitesimal braiding V is isomorphic either to kχℓ with ℓ = 1 or 3. Then H ≃ kχℓ #K.


25

L V Proof. By Theorem 5.1, we know that gr H = i≥0 H[i] /H[i−1] ≃ kχℓ #K with ℓ = 1 or 3. We prove that the relations also hold in V H. Since the relations are homogeneous, V it follows that H ≃ gr H ≃ kχℓ #K. Write kχℓ = k[x]/(x2 ). Then δ(x) = a2 ⊗ x and consequently ∆(x2 ) = x2 ⊗ 1 + 1 ⊗ x2 + (a2 · x + x) ⊗ x = x2 ⊗ 1 + 1 ⊗ x2 .

Since K does not contain primitive elements, it follows that the relation x2 = 0 must hold also in H. Proposition 5.3. Let H be a finite-dimensional Hopf algebra over K such that its infinitesimal braiding V is isomorphic either to V2,1 or V2,3 . Then H ≃ B(V )#K. L Proof. We know that gr H = i≥0 H[i] /H[i−1] ≃ B(V )#K with V isomorphic either to V2,1 or V2,3 . As before, we prove that the homogeneous relations also hold in H. Assume first that V ≃ V2,1 . Recall that B(V2,1 )#K is the algebra generated by x, y, a, b, c, d with x, y satisfying the the relations of B(V2,1 ), see Proposition 4.8, a, b, c, d satisfying the the relations of K, and all together satisfying the relations that give the commutativity: ax = −xa, ay = ξya + xc, bx = −xb, by = ξyb + xd, cx = −xc, cy = −ξyc + xa, dx = −xd, dy = −ξyd + xb.

(5) As

∆(x) = x ⊗ 1 + d ⊗ x + (ξ − 1)c ⊗ y

and

∆(y) = y ⊗ 1 + a ⊗ y −

ξ+1 b ⊗ x, 2

we have that ∆(xy + ξyx) = (xy + ξyx) ⊗ 1 + 1 ⊗ (xy + ξyx)

and

∆(y 4 ) = y 4 ⊗ 1 + 1 ⊗ y 4 .

Since K does not contain primitive elements, it follows that the relations xy + ξyx = 0 and y 4 = 0 hold in H. On the other hand, ∆(x2 ) = x2 ⊗ 1 + a2 ⊗ x2 + (ξ − 1)ab ⊗ (xy + ξyx) = x2 ⊗ 1 + a2 ⊗ x2 , which implies that x2 is a (1, a2 )-primitive element in H[1] . Since P1,a2 (H[1] ) = P1,a2 (K) = k{1 − a2 , ab}, we must have that x2 = µ1 (1 − a2 ) + µ2 ab

As ax2 = x2 a and bx2 = x2 b, but

for some µ1 , µ2 ∈ k.

a(µ1 (1 − a2 ) + µ2 ab) − (µ1 (1 − a2 ) + µ2 ab)a = µ2 (1 + ξ)c, b(µ1 (1 − a2 ) + µ2 ab) − (µ1 (1 − a2 ) + µ2 ab)b = 2µ1 c,

it follows that µ1 = µ2 = 0. Therefore, the relation x2 = 0 also holds in H and consequently, H ≃ gr H. Assume now V ≃ V2,3 and recall that B(V2,3 )#K = B(V2,1 )#K as algebras. Since ∆(x) = x ⊗ 1 + a ⊗ x − (1 + ξ)b ⊗ y

and

∆(y) = y ⊗ 1 + d ⊗ y +

ξ−1 c ⊗ x, 2

we have that ∆(xy − ξyx) = (xy − ξyx) ⊗ 1 + 1 ⊗ (xy − ξyx)

and

∆(y 4 ) = y 4 ⊗ 1 + 1 ⊗ y 4 .

Again, since K does not contain primitive elements, it follows that the relations xy −ξyx = 0 and y 4 = 0 hold in H. On the other hand, we have that ∆(x2 ) = x2 ⊗ 1 + a2 ⊗ x2 − (1 + ξ)ab ⊗ (xy − ξyx) = x2 ⊗ 1 + a2 ⊗ x2 .

In particular, x2 ∈ P1,a2 (H[1] ). Since P1,a2 (H[1] ) = P1,a2 (K) = k{1 − a2 , ab}, it follows that Since

ax2

=

x2 a

and

x2 = µ1 (1 − a2 ) + µ2 ab bx2

=

x2 b,

but

for some µ1 , µ2 ∈ k.

2

a(µ1 (1 − a ) + µ2 ab) − (µ1 (1 − a2 ) + µ2 ab)a = µ2 (1 + ξ)c b(µ1 (1 − a2 ) + µ2 ab) − (µ1 (1 − a2 ) + µ2 ab)b = 2µ1 c


26

we must have that µ1 = µ2 = 0 and therefore H ≃ gr H.

In the following we define two Hopf algebras A3,1 (µ) and A3,3 (µ) which are constructed by deforming the relations on the Nichols algebras B(V3,1 ) and B(V3,3 ) over K, respectively. We will show that they are indeed liftings of bosonizations. Definition 5.4. For µ ∈ k, denote by A3,1 (µ) the algebra generated by the elements x, y, a, b, c, d satisfying the relations ab = ξba, ad = da, ax = −ξxa, cx = ξxc,

ac = ξca,

0 = cb = bc,

ad = 1, ay = −ya − xc cy = −yc + xa,

x2 − 2y 2 = µ(1 − a2 ),

2

2

0=b =c , bx = −ξxb, dx = ξxd,

xy + yx = ξµac,

cd = ξdc,

bd = ξdb,

2

a4 = 1,

a c = b, by = −yb − xd dy = −yd + xb,

y 4 = −µy 2 (1 − a2 ) −

µ2 (1 − a2 ). 2

It is a Hopf algebra with its structure determined by ∆(x) = x ⊗ 1 + d ⊗ x + (ξ − 1)c ⊗ y ∆(a) = a ⊗ a + b ⊗ c, ∆(c) = c ⊗ a + d ⊗ c,

1 ∆(y) = y ⊗ 1 + a ⊗ y + (−ξ − 1)b ⊗ x, 2 ∆(b) = a ⊗ b + b ⊗ d, ∆(d) = c ⊗ b + d ⊗ d,

ε(a) = ε(d) = 1,

ε(b) = ε(c) = ε(x) = ε(y) = 0.

In particular, the antipode is given by S(x) = −ax − (1 + ξ)cy, S(b) = ξb,

1 S(y) = −dy + (ξ − 1)bx, 2 S(c) = −ξc,

S(a) = d, S(d) = a.

Remark 5.5. Clearly, A3,1 (0) ≃ B(V3,1 )#K. Also note that A3,1 (µ) is the quotient of the algebra T (V3,1 ) ⊗ K by the two-sided ideal generated by the elements given by the three last rows of equations. In particular, we have that ∆(ab) = ab ⊗ 1 + a2 ⊗ ab, S(ab) = −ac,

∆(ac) = ac ⊗ a2 + 1 ⊗ ac, S(ac) = ab,

∆(a2 ) = a2 ⊗ a2 , S(a2 ) = a2 .

Definition 5.6. For µ ∈ k, denote by A3,3 (µ) the Hopf algebra defined by A3,1 (µ) = A3,3 (µ) as algebra but as coalgebra ∆(x) = x ⊗ 1 + a ⊗ x + (ξ + 1)b ⊗ y, ε(x) = 0, 1 ∆(y) = y ⊗ 1 + d ⊗ y + (1 − ξ)c ⊗ x, ε(y) = 0, 2 and the same counit and comultiplication for the elements a, b, c, d in K. In particular, we have 1 S(x) = −dx − (ξ − 1)by, S(y) = −ay + (1 + ξ)cx. 2 Remark 5.7. As before, A3,3 (0) ≃ B(V3,3 )#K and A3,3 (µ) is the quotient of the algebra T (V3,3 ) ⊗ K by the two-sided ideal generated by the elements given by the three last rows of equations. In the next lemma we show that the algebras A3,1 (µ) and A3,3 (µ) are finite-dimensional Hopf algebras over K. Lemma 5.8. A3,1 (µ) and A3,3 (µ) are finite-dimensional for all µ ∈ k and (A3,1 (µ))[0] ≃ K ≃ (A3,3 (µ))[0] .


27

Proof. We prove the assertion for A3,1 (µ), being the proof for A3,3 (µ) completely analogous. Let J3,1 be the two-sided ideal generated by the elements given by the three last rows of equations. Then A3,1 (µ) = T (V3,1 ) ⊗ K/J3,1 . Note that T (V3,1 ) ⊗ K is a graded algebra with the gradation defined by the usual on T (V3,1 ) and all the elements in K to be of degree 0. Denote by A0 the subalgebra generated by the coalgebra C linearly spanned by a, b, c, d, then A0 is a Hopf subalgebra of A3,1 (µ) which is isomorphic to K. Indeed, consider the Hopf algebra map ϕ : K → A3,1 (µ) given by the composition K ֒→ T (V3,1 ) ⊗ K ։ T (V3,1 ) ⊗ K/J3,1 ; in particular, A0 ≃ ϕ(K). Since dim K = 8, to prove that ϕ(K) ≃ K it is enough to show that dim ϕ(K) > 4, since it is a divisor of 8 by the Nichols-Zoeller Theorem. This follows from the fact that ϕ(C) ≃ C as coalgebras, since with respect to the grading in T (V3,1 ) ⊗ K, the relations in J3,1 do not involve relations only in degree 0. Since the elements a, b, c, d are linearly independent in K, they are also l.i. in A3,1 (µ). If we set A1 = K{x, y}, A2 = A1 + K{xy, y 2 }, A3 = A2 + K{xy 2 , y 3 } and A4 = A3 + K{xy 3 }, we have that {An }0≤n≤3 is a coalgebra filtration of A3,1 (µ). In particular, (A3,1 (µ))0 ⊆ K and consequently (A3,3 (µ))[0] = K; that is, A3,1 (µ) is a Hopf algebra over K. Hence, A3,1 (µ) is a finite-dimensional Hopf algebra which is free over K. In particular, 8 divides dim A3,1 (µ). Besides, A3,1 is a K-module with a set of generators {1, x, y, xy, y 2 , xy 2 , y 3 , xy 3 }. Thus, dim A3,1 (µ) ≤ 8 dim K = 64. From the proof of the lemma above it follows that dim A3,1 (µ), dim A3,3 (µ) ≤ 8 dim K = 64. In the next lemma we show that the algebras A3,1 (µ) and A3,3 (µ) are liftings of B(V3,1 )#K and B(V3,3 )#K for all µ ∈ k, respectively. Lemma 5.9. gr A3,1 (µ) ≃ B(V3,1 )#K and gr A3,3 (µ) ≃ B(V3,3 )#K. Proof. To prove the lemma, it is enough to show that dim A3,1 (µ), dim A3,3 (µ) ≥ 64, since by the proof of Lemma 5.8 we have that gr A3,1 (µ) ≃ R3,1 #K and gr A3,3 (µ) ≃ R3,3 #K where R3,1 , R3,3 are K-modules linearly spanned by the set {1, x, y, xy, y 2 , xy 2 , y 3 , xy 3 }. We show that the set B = {xi y j ak bl : 0 ≤ i, l ≤ 1, 0 ≤ j, k ≤ 3} is linearly independent by using adequate representations. As A3,1 (µ) and A3,3 (µ) have the same algebra structure, we prove it only for A3,1 (µ). For λ a 4-th root of unity, consider the 8-dimensional representation Wλ of A3,1 (µ) given by the following matrices a 0 0 b12 ρ1 (a) = , ρ1 (b) = , 0 −ξa b21 0 0 x y11 0 ρ1 (x) = , ρ1 (y) = , id4 0 0 y22 where λ 0 0 0

a=

b21 = 

0 µ(λ3 −λ) 0 −λ 0 µ(λ−λ3 ) 0 −λ 0 0 0 λ

0 −λ3 0 µ(λ3 −λ) 0 0 0 0 0 0 0 λ3 0 0 0 0

0 0 0 y11 =  1 0 0

Assume (6)

010 001

!

!

,

b12 =

,

x=

 1 2 2 µ (λ −1)  2 ,

!

0 ξµ(λ3 −λ) 0 ξµ2 (λ−λ3 ) 0 0 0 0 , 0 2ξλ3 0 ξµ(λ−λ3 ) 0 0 0 0   µ(1−λ2 ) 0 µ2 (λ2 −1) 0 2 2 2 µ(1−λ ) 0 µ (λ −1)   0 , 2 0 µ(λ2 −1) 0 2 0 2 0 µ(λ −1)



 0 y22 =  −1

0 µ(λ2 −1) 0

X

0≤i,l≤1,0≤j,k≤3

0 0

µλ2

0

0 0 −1 0 0 −1

 1 2 2 µ (1−λ )  2 .

fi,j,k,l xi y j ak bl = 0.

0 µ 0

28


for some fi,j,k,l ∈ k. Applying this equation to the first vector of the canonical basis attached to the representation, we get that: fi,j,0,0 + λfi,j,1,0 + λ2 fi,j,2,0 + λ3 fi,j,3,0 = 0, for all 0 ≤ i ≤ 1, 0 ≤ j ≤ 3.

Since this equation must hold for any λ, it follows that fi,j,k,0 = 0 for all 0 ≤ i ≤ 1, and 0 ≤ j, k ≤ 3. To prove that the remaining coefficients are zero, we need another representation. For λ a 4-th root of unity, consider now the 16-dimensional representation Uλ given by   a 0 µ(d − a) 0 0 ξµ(c − b) 0 0  0 −a  0 µ(a − d) 0 0 0 0    0 0 −a 0 0 2ξc 0 −ξµ(b + c)     0 0  0 a 0 0 0 0 , ρ2 (a) =   0 −c  0 µc −ξa 0 ξµ(a − d) 0    0 0 0 0 0 ξa 0 ξµ(d − a)     0 0  0 c 0 0 ξa 0 0 0 0 0 0 0 0 −ξa   b 0 −µb 0 0 ξµ(d − a) 0 ξµ2 (a − d)  0 −b  0 µb 0 0 0 0    0 0 −b 0 0 2ξd 0 ξµ(a − d)     0 0  0 b 0 0 0 0  , ρ2 (b) =   0 −d 0 µ(d − a) −ξb 0 ξµb 0    0 0  0 0 0 ξb 0 −ξµb    0 0  0 d 0 0 ξb 0 0 0 0 0 0 0 0 −ξb 08 x ρ2 (x) = , id8 08   µ2 2 2 (a − id2 ) ξµac 0 −ξµ ab 0 0 0   0 2   0 0 0 ξµac 0 −ξµ2 ab   id2 0   µ(a2 − id2 ) 0 0 ξµac 0   0 id2 0   0 id2 0 0 0 0 ξµac   0 ρ2 (y) =  , 2 µ    0 0 0 0 0 µa2 0 (id2 − a2 )    2   0 0 0 0 −id2 0 0 0     0 0 0 0 0 −id2 0 µid2 0 0 0 0 0 0 −id2 0 where 3 λ 0 0 λ2 0 1 λ 0 a= , b= , c= , d= , and 0 −ξλ 0 0 0 0 0 ξλ3   µ(id2 − a2 ) 0 µ2 (a2 − id2 ) 0  0 µ(id2 − a2 ) 0 µ2 (a2 − id2 )  . x= 2   2id2 0 µ(a − id2 ) 0 2 0 2id2 0 µ(a − id2 ) Applying the residual equation of (6) to the second vector of the canonical basis attached to the new representation, we get that fi,j,0,1 + λfi,j,1,1 + λ2 fi,j,2,1 + λ3 fi,j,3,1 = 0, for all 0 ≤ i ≤ 1, 0 ≤ j ≤ 3

implying that fi,j,k,1 = 0 for all 0 ≤ i ≤ 1, 0 ≤ j, k ≤ 3. Therefore, B is a linearly independent set and A3,1 (µ) is a lifting of B(V3,1 )#K for any µ ∈ k. Proposition 5.10. Let H be a finite-dimensional Hopf algebra over K such that its infinitesimal braiding is isomorphic to V3,1 or V3,3 . Then H ≃ A3,1 (µ) or H ≃ A3,3 (µ) for some µ ∈ k, respectively.


29

Proof. Since H[0] ≃ K, we have that gr H = B(V )#K with V ≃ V3,1 or V ≃ V3,3 . Recall that B(V )#K is the algebra generated by x, y, a, b, c, d with x, y satisfying the the relations of B(V ), see Proposition 4.8, a, b, c, d satisfying the the relations of K, and all together satisfying the relations giving the commutativity: (7)

ax = −ξxa, ay = −ya − xc, bx = −ξxb, by = −yb − xd, cx = ξxc, cy = −yc + xa, dx = ξxd, dy = −yd + xb.

We prove the claim for V ≃ V3,1 . The proof for V ≃ V3,3 follows the same lines. In this ξ+1 b ⊗ x, case, we have that ∆(x) = x ⊗ 1 + d ⊗ x + (ξ − 1)c ⊗ y and ∆(y) = y ⊗ 1 + a ⊗ y − 2 which implies (8) (9)

∆(x2 − 2y 2 ) = (x2 − 2y 2 ) ⊗ 1 + a2 ⊗ (x2 − 2y 2 )

and

∆(xy + yx) = (xy + yx) ⊗ 1 + 1 ⊗ (xy + yx) − ξac ⊗ (x2 − 2y 2 ).

From the first equation, we get that the element x2 − 2y 2 ∈ P1,a2 ((V3,1 ⊕ k)#K) = P1,a2 (K) = k{1 − a2 , ab}. Then, there should exist µ1 , µ2 ∈ k such that x2 − 2y 2 = µ1 (1 − a2 ) + µ2 ab

in H.

However, a tedius calculation shows that (9) is possible only if µ2 = 0. In this case, ∆(xy + yx) = (xy + yx) ⊗ 1 + 1 ⊗ (xy + yx) − ξac ⊗ µ1 (1 − a2 )

= (xy + yx − µ1 ξac) ⊗ 1 + 1 ⊗ (xy + yx) + ξµ1 ac ⊗ a2

= (xy + yx − µ1 ξac) ⊗ 1 + 1 ⊗ (xy + yx − µ1 ξac) + ξµ1 ac ⊗ a2 + 1 ⊗ µ1 ξac

= (xy + yx − µ1 ξac) ⊗ 1 + 1 ⊗ (xy + yx − µ1 ξac) + ∆(ξµ1 ac),

which implies that xy + yx − µ1 ξac is a primitive element in H. Thus, we must have that xy + yx = ξµ1 ac. Finally, for y 4 we have ∆(y 4 ) = y 4 ⊗ 1 + 1 ⊗ y 4 − µ1 y 2 a2 ⊗ (1 − a2 ) + µ1 (1 − a2 ) ⊗ y 2 + µ1 xb ⊗ ya2 + µ1 xc ⊗ y 1+ξ 1−ξ 2 ξ−1 2 1+ξ + µ1 xa ⊗ xa2 + µ1 ac ⊗ ab + µ1 ab ⊗ ac − µ1 xd ⊗ x 2 2 2 2 1 + µ21 (1 − a2 ) ⊗ (1 − a2 ) − ξµ1 ab ⊗ xy − ξµ1 ac ⊗ xya2 2 1 which implies that y 4 = −µ1 y 2 (1 − a2 ) − µ21 (1 − a2 ). 2

We end the paper with the classification of finite-dimensional Hopf algebras over K such that their infinitesimal braiding is an indecomposable module in K K YD. Proof of Theorem B. Since H[0] ≃ K, by Theorem 5.1 we have that gr H = R#K and R ≃ B(V ) with V a simple object in K K YD. If V ≃ kχ , kχ3 , V2,1 or V2,3 , then H ≃ B(V )#K by Propositions 5.2 and 5.3. If V ≃ V3,1 or V ≃ V3,3 , then by Proposicion 5.10 it follows that H ≃ A3,1 (µ) or H ≃ A3,3 (µ) for some µ ∈ k, respectively. Conversely, it is clear that the algebras listed in items (i) and (iii) are liftings of Nichols algebras over K. The Hopf algebras A3,1 (µ) and A3,3 (µ) are liftings of Nichols algebras over K for all µ ∈ k by Lemma 5.9. Finally, two algebras from different families are not isomorphic as Hopf algebras since their infinitesimal braidings are not isomorphic as Yetter–Drinfeld modules


30

6. Appendix We are interested to give a full proof of Remark 3.5. We perform a case by case analysis. Let Vi,j ∈ K K YD with (i, j) ∈ Λ be a simple module. Throughout the section, we suppose the braiding is of diagonal type, i.e. there exists a basis {v1 , v2 } of Vi,j such that c(vi ⊗ vj ) = qij vj ⊗ vi , 1 ≤ i, j ≤ 2. We will see that in each case we arrive at a contradiction. If we write vi = αi1 e1 + αi2 e2 , i = 1, 2, we must have c(vi ⊗ vj ) = c((αi1 e1 + αi2 e2 ) ⊗ (αj1 e1 + αj2 e2 ))

= αi1 αj1 c(e1 ⊗ e1 ) + αi1 αj2 c(e1 ⊗ e2 ) + αi2 αj1 c(e2 ⊗ e1 ) + αi2 αj2 c(e2 ⊗ e2 ),

but on the other hand, c(vi ⊗ vj ) = qij vj ⊗ vi = qij (αj1 e1 + αj2 e2 ) ⊗ (αi1 e1 + αi2 e2 ) = qij αi1 αj1 e1 ⊗ e1 + qij αi2 αj1 e1 ⊗ e2 + qij αi1 αj2 e2 ⊗ e1 + qij αi2 αj2 e2 ⊗ e2 . • Case j = 0 and i ∈ {1, 3}: By Proposition 3.4, the braiding is given by e1 e1 ⊗ e1 e2 ⊗ e1 + 2e1 ⊗ e2 c( ⊗ e1 e2 ) = . e2 −e1 ⊗ e2 e2 ⊗ e2

Then we must have that (10)

q12 α11 α22 = α11 α22

(11)

q12 α12 α21 = 2α11 α22 − α12 α21

(12)

q21 α12 α21 = α12 α21

(13)

q21 α11 α22 = 2α12 α21 − α22 α11 . α11 α21 If we denote det := det 6= 0, from (10) and (11) we get that q12 det = − det. α12 α22 This implies that q12 = −1 and whence α11 α22 = 0. Similarly from (13) and (12) we get that q21 det = − det, q21 = −1 and α12 α21 = 0, which implies that det = 0, a contradiction. • Case j = 2 and i ∈ {0, 2}: By Proposition 3.4, the braiding is given by e1 e1 ⊗ e1 −e2 ⊗ e1 + 2e1 ⊗ e2 c( ⊗ e1 e2 ) = . e2 e1 ⊗ e2 e2 ⊗ e2

Then we have (14)

q12 α11 α22 = −α11 α22

(15)

q12 α12 α21 = 2α11 α22 + α12 α21

(16)

q21 α12 α21 = −α12 α21

(17)

q21 α11 α22 = 2α12 α21 + α22 α11 .

Computing (14)-(15)+(17)-(16) yields (q12 + q21 ) det = −2 det, which implies that q21 = −q12 − 2. Analogously, (14)+(17) yields α12 α21 = −α11 α22 . On the other hand, q11 α211 = α211 and q11 α212 = α212 , which implies that q11 = 1, since α12 6= 0 or α11 6= 0. As q11 α11 α12 = −α11 α12 , we get that α11 α12 = 0. Similarly, using the equations of q22 , it follows that α21 α22 = 0. Note that α11 α12 = 0 and α21 α22 = 0 imply that both columns of (αij )1≤i,j≤2 have zero elements. Therefore, this matrix is a diagonal matrix because it is a non-singular matrix. But α12 α21 = −α11 α22 , implying this matrix has determinant zero.

• Case j = 1 and i arbitrary: By Proposition 3.4, the braiding is given by ( 3 ) ξλ31 e2 ⊗ e1 + (λ31 − ξλ1 )e1 ⊗ e2 λ1 e1 ⊗ e1 e1 1 c( ⊗ e1 e2 ) = . e2 λ1 e1 ⊗ e2 −ξλ1 e2 ⊗ e2 + (λ31 + ξλ1 )e1 ⊗ e1 2


31

(i) Assume i = 2, then (18)

qij αi1 αj1 = −αi1 αj1 −

(19)

qij αi2 αj2 = ξαi2 αj2

(20)

qij αi1 αj2 = −ξαi1 αj2

(21)

1+ξ αi2 αj2 2

qij αi2 αj1 = (ξ − 1)αi1 αj2 − αi2 αj1 .

Since in this case the eigenvalues of e c := τ ◦ c are −1 and ±ξ, we have that q11 ∈ {−1, ±ξ}. Suppose q11 = −1. Then, by (19) we get −α212 = ξα212 and hence α12 = 0. This implies that α11 α22 = det 6= 0. But from (21) it follows that q12 α12 α21 = (ξ − 1)α11 α22 − α12 α21 , which implies that det = α11 α22 = 0, a contradiction. Assume now that q11 = −ξ. Then, by (19) we get −ξα212 = ξα212 which implies that α12 = 0 and α11 α22 6= 0. But, again by (21) we obtain q12 α12 α21 = (ξ1)α11 α22 − α12 α21 from which follows that α11 α22 = 0, a contradiction. Finally, suppose that q11 = ξ. Then, by (18) we have that ξα211 = −α211 − 1+ξ 2 α12 and consequently α212 = −2α211 . On the other hand, (20) yields that ξα11 α12 = 2 −ξα11 α12 and hence α11 α12 = 0, which implies that α11 = α12 = 0, a contradiction. (ii) Assume i = 3, then (22)

qij αi1 αj1 = ξαi1 αj1 +

(23)

qij αi2 αj2 = −αi2 αj2

(24) (25)

1+ξ αi2 αj2 2


qij αi2 αj1 = (ξ − 1)αi1 αj2 − ξαi2 αj1 .

In this case, the eigenvalues of e c are again −1 and ±ξ. Then q11 ∈ {−1, ±ξ}. Suppose first 1+ξ 2 that q11 = −1. Then, by (22) we have that −α211 = ξα211 + α12 which implies that 2 2 2 α12 = −2α11 . Also, (23) gives q12 α12 α22 = −α12 α22 and q21 α12 α22 = −α12 α22 . Thus, if α12 α22 6= 0 we would have that −1 = q11 = q12 = q21 , which is impossible. If α12 = 0, then α11 = 0, a contradiction. Hence, we should have that α22 = 0. But in such a case, (25) yields q21 α11 α22 = (ξ − 1)α12 α21 − ξα11 α22 from which follows that α12 α21 = 0, a contradiction. Suppose q11 = −ξ, then by (23) we have that −ξα212 = −α212 and whence 1+ξ 2 α12 = 0. Also, from (22) it follows that −ξα211 = ξα211 + α12 which implies α11 = 0, 2 a contradiction. Assume finally that q11 = ξ, then, by (23) we have that ξα212 = −α212 and then α12 = 0. Besides, by (25) we have that q12 α12 α21 = (ξ − 1)α11 α22 − ξα12 α21 which implies that α11 α12 = 0, a contradiction. (iii) Assume i = 0, then (26)

qij αi1 αj1 = αi1 αj1 +

(27)


1+ξ αi2 αj2 2

(28)


(29)

qij αi2 αj1 = (1 − ξ)αi1 αj2 + αi2 αj1 .

Here, the eigenvalues of e c are 1 and ±ξ. Then q11 ∈ {1, ±ξ}. Suppose q11 = 1, then by (27) we have that α212 = −ξα212 and α12 = 0. Also, by (29) we have that q12 α12 α21 = (1 − ξ)α11 α22 + α12 α21 from which follows that α11 α22 = 0, a contradiction. Suppose now 1+ξ 2 α12 and whence α212 = −2α211 . that q11 = −ξ. Then, by (26) we obtain −ξα211 = α211 + 2 Also, by (28) we have that −ξα11 α12 = ξα11 α12 and consequently α11 α12 = 0, which implies α11 = α12 = 0, a contradiction. Assume q11 = ξ, then by (27) we have ξα212 = −ξα212 , implying that α12 = 0. Also, (29) yields that q12 α12 α21 = (1 − ξ)α11 α22 + α12 α21 from which follows that α11 α22 = 0, a contradiction.


32

(iv) Assume i = 1, then 1+ξ αi2 αj2 2

(30)

qij αi1 αj1 = −ξαi1 αj1 −

(31)

qij αi2 αj2 = αi2 αj2

(32)


(33)

qij αi2 αj1 = (1 − ξ)αi1 αj2 + ξαi2 αj1 .

In this case, the eigenvalues of e c are 1 and ±ξ, then q11 ∈ {1, ±ξ}. Suppose q11 = 1, 1 +ξ 2 then (30) gives α211 = −ξα211 − α12 which implies that α212 = −2α211 . Also, 31 gives 2 q12 α12 α22 = α12 α22 and (31) gives q21 α12 α22 = α12 α22 . If α12 α22 6= 0, then 1 = q11 = q12 = q21 , which is impossible. If α12 = 0, then also α11 = 0, a contradiction. Thus, α22 = 0. In such a case, by (33) we have q21 α11 α22 = (1 − ξ)α12 α21 + ξα11 α22 which implies that α12 α21 = 0, a contradiction. Suppose now q11 = −ξ, then by (31) we have −ξα212 = α212 and α12 = 0. Also, (33) yields that q12 α12 α21 = (1 − ξ)α11 α22 + ξα12 α21 which implies that α11 α22 = 0, a contradiction. Assume finally that q11 = ξ, then by (31) we have that ξα212 = α212 and whence α12 = 0. Besides, (33) gives q12α12 α21 = (1 − ξ)α11 α22 + ξα12 α21 which implies that α11 α22 = 0, a contradiction. Finally, we prove the last case, also by a case-by-case argument. • Case j = 3 and i arbitrary: By Proposition 3.4, the braiding is given by ) ( λ1 e1 ⊗ e1 −ξλ1 e2 ⊗ e1 + (λ1 + ξλ31 )e1 ⊗ e2 e1 1 . c( ⊗ e1 e2 ) = e2 λ31 e1 ⊗ e2 ξλ31 e2 ⊗ e2 + (λ1 − ξλ31 )e1 ⊗ e1 2 (i) Assume i = 2, then (34)

qij αi1 αj1 = −αi1 αj1 +

(35)


ξ−1 αi2 αj2 2

(36)


(37)

qij αi2 αj1 = −(ξ + 1)αi1 αj2 − αi2 αj1 .

Since the eigenvalues of e c are −1 and ±ξ, we have that q11 ∈ {−1, ±ξ}. Suppose q11 = −1, then by (35) it follows that −α212 = −ξα212 which implies that α12 = 0. Also, (37) gives that q12 α12 α21 = −(ξ + 1)α11 α22 − α12 α21 from which follows that α11 α22 = 0, a contradiction. ξ−1 2 α12 and Suppose now that q11 = −ξ, then by (34) we obtain −ξα211 = −α211 + 2 2 2 therefore α12 = −2α11 . Also, by (36) we have that −ξα11 α12 = ξα11 α12 which implies that α11 α12 = 0 and hence α11 = α12 = 0, a contradiction. Suppose then that q11 = ξ. In this case, (35) yields ξα212 = −ξα212 which implies that α12 = 0. But (37) gives q12 α12 α21 = −(ξ + 1)α11 α22 − α12 α21 from which follows that α11 α22 = 0, a contradiction. (ii) Assume i = 3, then (38)

qij αi1 αj1 = −ξαi1 αj1 +

(39)


(40) (41)

1−ξ αi2 αj2 2


qij αi2 αj1 = −(1 + ξ)αi1 αj2 + ξαi2 αj1 .

Note that the eigenvalues of e c are −1 and ±ξ; then q11 ∈ {−1, ±ξ}. If q11 = −1, then by 1−ξ 2 2 α12 , which implies that α212 = −2α211 . Also, by (38) we have that −α11 = −ξα211 + 2 (39) we have that q12 α12 α22 = −α12 α22 q21 α12 α22 = −α12 α22 . If α12 α22 6= 0, we would have that q11 = q12 = q21 , a contradiction. If α12 = 0, then α11 = 0, also a contradiction. Thus, α22 = 0. In such a case, (41) yields q21 α11 α22 = −(1+ξ)α12 α21 +ξα22 α11 from which follows that α12 α21 = 0, a contradiction. If q11 = −ξ, then by (39) we have that −ξα212 =


33

−α212 giving α12 = 0. Also, (41) yields q12 α12 α21 = −(1+ξ)α11 α22 +ξα12 α21 which implies that α11 α22 = 0, a contradiction. If q11 = ξ, we have by (39) that ξα212 = −α212 which implies α12 = 0. Also, by (41) we have that q12 α12 α21 = −(1 + ξ)α11 α22 + ξα12 α21 which yields α11 α22 = 0, a contradiction. (iii) Assume i = 0, then (42)

qij αi1 αj1 = αi1 αj1 +

(43)


(44)


(45)

1−ξ αi2 αj2 2

qij αi2 αj1 = (1 + ξ)αi1 αj2 + αi2 αj1 .

Since the eigenvalues of e c are 1 and ±ξ, then q11 ∈ {1, ±ξ}. If q11 = 1, then by (43) we have that α12 = 0, and by (45) it follows that α11 α22 = 0, a contradiction. If q11 = −ξ, then by (43) we have that α12 = 0, and by (45) we have that α11 α22 = 0, a contradiction too. Finally, if q11 = ξ, then, by (42) we have that α212 = −2α211 and by (44) α11 α12 = 0. This implies that α11 = α12 = 0, a contradiction. (iv) Assume i = 1, then ξ−1 αi2 αj2 2

(46)

qij αi1 αj1 = ξαi1 αj1 +

(47)


(48)


(49)

qij αi2 αj1 = (1 + ξ)αi1 αj2 − ξαi2 αj1 .

Since the eigenvalues of e c are 1 and ±ξ, we have that q11 ∈ {1, ±ξ}. if q11 = 1, then by (46) 2 we have that α12 = −2α211 . Also, from (47) and (47) it follows that q12 α12 α22 = α12 α22 and q21 α12 α22 = α12 α22 . If α12 α22 6= 0, we would have that 1 = q11 = q12 = q21 , a contradiction. If α12 = 0, then α11 = 0, a contradiction too. Thus, we should have that α22 = 0. But in such a case, (49) yields that α21 α12 = 0, which is impossible. If q11 = −ξ, then (47) implies that α12 = 0 and (49) implies that α11 α22 = 0, a contradiction. Finally, if q11 = ξ, then (47) implies α12 = 0 and (49) implies α11 α22 = 0, which is impossible. References [AAH] N. Andruskiewitsch, I. Angiono and I. Heckenberger, Liftings of Jordan and super Jordan planes Preprint: arXiv:1512.09271. [ACG] N. Andruskiewitsch, G. Carnovale and G. A. Garc´ıa, Finite-dimensional pointed Hopf algebras over finite simple groups of Lie type III. Semisimple classes in PSLn (q), Preprint: arXiv:1506.06794. [AC] N. Andruskiewitsch and J. Cuadra, On the structure of (co-Frobenius) Hopf algebras, J. Noncommutative Geometry 7 (2013), Issue 1, pp. 83–104. ˜ a and L. Vendramin, Finite-dimensional pointed [AFGV1] N. Andruskiewitsch, F. Fantino, M. Gran Hopf algebras with alternating groups are trivial, Ann. Mat. Pura Appl. (4), 190 (2011), 225–245. [AFGV2] , Pointed Hopf algebras over the sporadic simple groups. J. Algebra 325 (2011), pp. 305– 320. ˜ a, Braided Hopf algebras over non-abelian finite groups, Col[AG] N. Andruskiewitsch and M. Gran loquium on Operator Algebras and Quantum Groups (Vaquer´ıas, 1997), Bol. Acad. Nac. Cienc. (C´ ordoba) 63 (1999), 45–78. [AHS] N. Andruskiewitsch, I. Heckenberger and H-J. Schneider, The Nichols algebra of a semisimple Yetter-Drinfeld module, Amer. J. Math. 132, no. 6, 2010, 1493–1547. [AN] Andruskiewitsch, N.; Natale, S. Counting arguments for Hopf algebras of low dimension. Tsukuba Math J. 25, n. 1, 187–201, 2001. [AS1] N. Andruskiewitsch and H.-J. Schneider, Finite quantum groups and Cartan matrices, Adv. Math. 154 (2000), 1–45. [AS2] , Pointed Hopf algebras, New directions in Hopf algebras, 1–68, Math. Sci. Res. Inst. Publ., 43, Cambridge Univ. Press, Cambridge, 2002. [AS3] , On the classification of finite-dimensional pointed Hopf algebras, Ann. Math. 171 (2010), No. 1, 375–417.

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´ tica, Facultad de Ciencias Exactas, Universidad NaG. A. G. : Departamento de Matema cional de La Plata. CONICET. Casilla de Correo 172, (1900) La Plata, Argentina. ´ tica, Universidade Federal do Rio Grande do Sul, Rio J.M.J.G. : Instituto de Matema Grande do Sul, Brazil E-mail address: [email protected] E-mail address: [email protected]