Cohomology rings of finite-dimensional pointed Hopf algebras over abelian groups
Nicolás Andruskiewitsch, Iván Angiono, Julia Pevtsova, Sarah Witherspoon
aa r X i v : . [ m a t h . QA ] A p r COHOMOLOGY RINGS OF FINITE-DIMENSIONAL POINTED HOPFALGEBRAS OVER ABELIAN GROUPS, I & II
N. ANDRUSKIEWITSCH, I. ANGIONO, J. PEVTSOVA, S. WITHERSPOON
Abstract.
We show that the cohomology ring of a finite-dimensional complex pointedHopf algebra with an abelian group of group-like elements is finitely generated. The proofhas three parts. Part I itself splits further into two steps. First, we reduce the problem tofinite generation of cohomology for finite dimensional Nichols algebras of diagonal type.Second, using spectral sequence arguments and detailed analysis of cohomology via theAnick resolution we reduce the problem further to specific combinatorial properties ofNichols algebras.Parts II and III of the proof carry out verifications of these combinatorial propertiesusing case-by-case calculations through the classification of Heckenberger. Part II, thecomputational heart of the present paper, deals with the major parametric families,including Nichols algebras of Cartan and super types. It also develops all the theoreticalfoundations and provides detailed proofs necessary for the case-by-case analysis. PartIII, which will appear as a separate publication, addresses the remaining Nichols algebraswhich fall into smaller discrete families and require computer calculations as the numberof different possibilities to verify is considerable.As an application of the main theorem we deduce finite generation of cohomologyfor other classes of finite-dimensional Hopf algebras, including basic Hopf algebras withabelian groups of characters and finite quotients of quantum groups at roots of one.
Contents
1. Introduction 31.1. Antecedents 31.2. The main result and applications 31.3. Scheme of the proof of Theorem 1.2.1 41.4. Finite-generation of cohomology for Nichols algebras 51.5. Organization of the paper 61.6. Conventions 6Acknowledgements 7
Part I. From cohomology of Nichols algebras to cohomology of Hopfalgebras
72. Finite-dimensional Hopf algebras 72.1. Morita equivalence of Hopf algebras 72.2. The role of Nichols algebras 82.3. Nichols algebras of diagonal type 92.4. Realizations 10
Date : April 15, 2020.MSC 2020: 16E40;16T20.
Part II. Permanent cocycles for Nichols algebras of diagonal type A θ and A ( j | θ − j ), θ ≥ j ∈ I ⌊ θ +12 ⌋ B θ and B ( j | θ − j ), θ ≥ j ∈ I θ − B θ,j standard, j ∈ I θ − C θ D θ D ( j | θ − j ), θ ≥ j ∈ I θ − E θ F G Cartan 578.4. Type G standard 598.5. Type D (2 , α ) 608.6. Type F (4) 618.7. Type G (3) 679. Parametric modular types 729.1. Modular type wk (4) 729.2. Modular type br (2) 7510. Proofs of the Computational Lemmas 7610.1. 76References 114 OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 3 Introduction
Antecedents.
Let G be a finite group scheme over a field k , k [ G ] its coordinatealgebra and k [ G ] the dual of the latter; thus G k [ G ] gives an equivalence between thecategories of finite group schemes and of finite-dimensional cocommutative Hopf algebras.The fundamental result in representation theory of finite group schemes [FS, Theorem 1.1]could be phrased as follows. Let H be a finite-dimensional cocommutative Hopf algebra.Then(fgc-a) The cohomology ring H( H, k ) is finitely generated.(fgc-b) For any finitely generated H -module M , H( H, M ) is a finitely generated moduleover H( H, k ).This result was previously known for finite group algebras [G, V, Ev] and restrictedenveloping algebras [FP, AJ]. At the end of the introduction of [FS], the authors observethat the cohomology ring of a finite-dimensional commutative Hopf algebra is easily seento be finitely generated using the structure as in [Wa] and add:
We do not know whether it is reasonable to expect finite generation of thecohomology of an arbitrary finite-dimensional Hopf algebra.
Slowly, evidence confirming that this is indeed a reasonable question has emerged. In thepioneering paper [GK] the cohomology ring of Lusztig’s small quantum groups u q ( g ) (incharacteristic 0) under some restrictions on the parameters was identified as the coordinatering of the nilpotent cone of the Lie algebra g . The restrictions on the parameters wereweakened in [BNPP]. The finite generation of cohomology was established for the dualsof Lusztig’s small quantum groups (in characteristic 0) [Go], for Lusztig’s small quantumgroups in positive characteristic [Dru1], for finite supergroup schemes [Dru2], for finite-dimensional complex pointed Hopf algebras whose group of grouplike elements is abelianand has order coprime to 210 [MPSW], for some pointed Hopf algebras of dimension p [NWW, EOW] (in characteristic p > F K with the group algebra of S and its dual [SV], for Drinfeld doubles of someinfinitesimal group schemes [FN]. In all the cases above, the approach is based to a greateror lesser extent on the knowledge of the structure of the Hopf algebras under consideration.Finite tensor categories were introduced in [EO], where it was also conjectured thatfinite generation holds in this more general context. A systematic study of this questionwas started in [NP].1.2. The main result and applications.
In the present paper we work over an alge-braically closed field k of characteristic 0. For brevity we shall say that an augmentedalgebra H has finitely generated cohomology (abbreviated as fgc) when both (fgc-a) and(fgc-b) hold. Our main result is the following: Theorem 1.2.1.
Let H be a finite-dimensional pointed Hopf algebra whose group of group-like elements is abelian. Then H has finitely generated cohomology. The class of finite-dimensional pointed Hopf algebras is the best understood and thesubclass of those with abelian group of group-like elements is the only one whose classifica-tion is essentially complete. Theorem 1.2.1 goes beyond the situation treated in [MPSW]but uses the same approach to the classification of pointed Hopf algebras proposed in[AS1, AS2]. Let us mention the main differences between the setting of [MPSW], that
N. ANDRUSKIEWITSCH, I. ANGIONO, J. PEVTSOVA, S. WITHERSPOON invoked the classification result [AS3], and the present work. In the former, the associatedbraided vector space V (described below) was of Cartan type and the deformations of thedefining relations in the liftings took values in the group algebras. When the restrictionon the order of the group of group-like elements G ( H ) is dropped, V belongs to the listin the celebrated classification of [H2] but is not necessarily of Cartan type. Also thedefining relations of the Nichols algebras and their deformations are more involved, see[An1, An2, AAG, AnG, GaJ, He].We state two direct applications of Theorem 1.2.1 to give more answers to the questionof fgc for finite-dimensional Hopf algebras. We also observe that Theorem 3.1.7 providesanother class of Hopf algebras having finitely generated cohomology. Theorem 1.2.2.
Let H be a finite-dimensional basic Hopf algebra whose group of char-acters is abelian. Then H has finitely generated cohomology. Basic Hopf algebras with abelian group of characters are just the duals of the Hopfalgebras in Theorem 1.2.1; thus Theorem 1.2.2, that generalizes [Go], follows from Theorem1.2.1, Lemma 2.1.1, Corollary 3.2.3 and Theorem 5.0.6.
Theorem 1.2.3.
Let H be a finite-dimensional Hopf algebra that fits into an extension k → K → H → L → k , where K is semisimple and L is either pointed with abelian groupof group-like elements or else basic with abelian group of characters. Then H has fgc. Theorem 1.2.3 follows from Lemma 3.2.5 and one of the previous two theorems. Quo-tients of algebras of functions on quantum groups at roots of one (of various kinds) wereclassified in [AG, Ga, GaG]. In particular, these results provide families of Hopf algebras H that fit into an extension k → k | G | → H → L → k where G is a finite group and L is a finite-dimensional basic Hopf algebra with abelian group of characters; thus Theorem1.2.3 applies to them.1.3. Scheme of the proof of Theorem 1.2.1.
Let H be a finite-dimensional pointedHopf algebra with abelian group of group-like elements Γ := G ( H ) so that the coradicalof H is H ≃ k Γ. Let D ( H ) be the the Drinfeld double of H , let gr H be the gradedHopf algebra associated to the coradical filtration and let V be the infinitesimal braidingof H , see § H ≃ B ( V ) k Γ [An2]. Then the Nichols algebra B ( V )is finite-dimensional. We shall use V = Hom k ( V, k ) to denote the k -linear dual. The keypoint in the proof of Theorem 1.2.1 is the following. Theorem 1.3.1.
Let U be a braided vector space of diagonal type such that the Nicholsalgebra B ( U ) has finite dimension. Then B ( U ) has fgc. This Theorem being proved, the rest of the proof proceeds in the following steps:Theorem 1.3.1 + B ( V ), B ( V ) have fgc Theorem 3.1.6 + gr H , (gr H ) have fgc Theorem 5.0.6 (cid:11) (cid:19) H has fgc D ( H ) has fgc Theorem 3.2.1 k s D (gr H ) has fgc + Theorem 2.2.4 k s OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 5
Remark . Let H be a finite-dimensional Hopf algebra whose coradical H is a Hopfsubalgebra, beyond the setting of Theorem 1.2.1, for instance H pointed but with non-abelian group of group-like elements G ( H ). Let V be the infinitesimal braiding of H .Then gr H ≃ R H where R is a connected graded Hopf algebra in H H YD and B ( V ) isa graded Hopf subalgebra of R . To prove that H has fgc following the scheme above wewould need to address these problems:(i) Prove that B ( V ) and B ( V ) have fgc.(ii) Is R = B ( V )? (in all known examples in characteristic 0 the answer is positive). Ifnot, prove that R and R have fgc.(iii) Prove Lemma 3.1.4 for any semisimple Hopf algebra (and not just for group algebrasand their duals). Together with (ii) this would give that gr H , (gr H ) have fgc.(iv) Extend Theorem 5.0.6 to prove that D (gr H ) has fgc. Even in the pointed case, wewould need Lemma 3.1.4 for D ( k G ( H )) to prove this conjectural extension.(v) Is H a cocycle deformation of gr H or at least Morita equivalent to gr H as in § D ( H ), and a fortiori H , have fgc since Theorem 3.2.1 holds in general.We also notice that a large part of this approach could be used in positive characteristicunder approppriate assumptions, e.g. the coradical H needs to be a semisimple Hopfsubalgebra.1.4. Finite-generation of cohomology for Nichols algebras.
We next outline theproof of Theorem 1.3.1 referring to § Reduction to the connected case.
By Theorem 5.0.1, we conclude that Theorem 1.3.1holds for U if and only if it holds for U J for every connected component J ∈ X . We assume for the rest of this Subsection that the Dynkin diagram of U is connected. The Anick resolution.
The Nichols algebra B ( U ) has a convex PBW-basis, hencea suitable filtration. Its associated graded ring gr B ( U ) is a quantum linear space. Thecohomology ring of gr B ( U ) is well-known, but we provide a computation using the Anickresolution [Ani] specifically in order to relate it to permanent cycles in a suitable spectralsequence. See § B ( U ), wemay use a spectral sequence argument based on Evens Lemma 3.3.2 to reduce the finitegeneration of H( B ( U ) , k ) to the verification of the following statement. Condition 1.4.1.
For every positive root γ ∈ ∆ U + , there exists L γ ∈ N such that thecochain (cid:16) x L γ γ (cid:17) ∗ is a cocycle in the Anick resolution, that is, represents an element in H( B ( U ) , k ) . If Condition 1.4.1 holds, then Theorem 4.4.3 implies that B ( U ) has fgc. N. ANDRUSKIEWITSCH, I. ANGIONO, J. PEVTSOVA, S. WITHERSPOON
Reduction to Weyl-equivalence.
In practice, given U we shall prove that Condition1.4.1 holds for any braided vector space with the same Dynkin diagram as U , particularlyfor U . Let G be any finite abelian group such that U is realized in k G k G YD . By Theorem3.1.6 and Theorem 5.0.6 we see that D ( B ( U ) k G ) has fgc.We apply this last claim as follows: let U ′ be a braided vector space of diagonal typewhich is Weyl-equivalent to U . This implies that U ′ is realized as Yetter-Drinfeld moduleover G and there is an algebra isomorphism D ( B ( U ) k G ) ≃ D ( B ( U ′ ) k G ) . By Corollary 3.2.2 B ( U ′ ) has fgc.1.4.4. Verification of Condition 1.4.1.
We argue case-by-case using the list of [H2]; by thepreceding discussion we just need to consider one representative in each Weyl-equivalenceclass—and we could choose the most convenient for our purpose. We also argue recursivelyon dim U . All in all, we reduce the verification to claims on Nichols algebras of diagonaltype, see § Organization of the paper.
The proof of the main theorem consists of three parts.Only Parts I and II which lay out the fundamental theoretic foundations and developall the techniques necessary to complete the proof are published in this paper. A morecomputation heavy Part III will be published separately.We now describe in more details how the proof and the present paper are split. PartI starts with a recollection of facts on Hopf and Nichols algebras in Section 2. Section 3contains several preliminary results on cohomology including the passage from the coho-mology of B ( V ) to the cohomology of B ( V ) k Γ and versions of the Evens Lemma and theMay spectral sequence crucial for our arguments. Section 4 presents the Anick resolutionand the reduction to Condition 1.4.1. In the last Section 5 of this Part it is shown thatthe Drinfeld double of B ( V ) k Γ has fgc provided that B ( V ) has via considerations ofcohomology for twisted tensor products.Parts II and III are devoted to the proof of Condition 1.4.1. Section 6 presents thestrategy of the verification with proofs of technical Lemmas postponed to Section 10. PartII contains the verification of Condition 1.4.1 for finite-dimensional Nichols algebras ofdiagonal type belonging to families with continuous parameter. We proceed case by casein Sections 7, 8 and 9 corresponding respectively to classical (Cartan, standard and super)types, exceptional (Cartan, standard and super) types, and Nichols algebras with the sameroot systems as the modular Lie algebras wk (4) as br (2). The remaining Nichols algebrasof diagonal type are dealt with in Part III [AAPW] that will be posted separately on thearXiv.1.6. Conventions.
For ℓ < θ ∈ N , we set I ℓ,θ = { ℓ, ℓ + 1 , . . . , θ } , I θ = I ,θ . Let G N bethe group of roots of unity of order N in k and G ′ N the subset of primitive roots of order N ; G ∞ = S N ∈ N G N and G ′∞ = G ∞ − { } . If L ∈ N and q ∈ k × , then ( L ) q := P L − j =0 q j .All vector spaces, algebras and tensor products are over k . We use V to denote thelinear dual to a vector space V , V = Hom k ( V, k ).By abuse of notation, h a i : i ∈ I i denotes either the group, the subgroup or the vectorsubspace generated by all a i for i in an indexing set I , the meaning being clear from thecontext. Instead, the subalgebra generated by all a i for i ∈ I is denoted by k h a i : i ∈ I i . OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 7 If A is an associative augmented algebra and M is an A -module, then we setH n ( A, M ) = Ext nA ( k , M ) , H( A, M ) = ⊕ n ∈ N H n ( A, M ) . In particular H n ( A, k ) = Ext nA ( k , k ) ≃ Ext nA ⊗ A op ( A, k ). Thus H( A, k ) = ⊕ n ∈ N H n ( A, k ) isisomorphic to the Hochschild cohomology HH( A, k ) = ⊕ n ∈ N Ext nA ⊗ A op ( A, k ).Let P ∗ ( A ) be the normalized bar resolution of k in the category of left A -modules andlet Ω ∗ ( A ) = Hom A ( P ∗ ( A ) , k ), in particular Ω n ( A ) = Hom k ( A ( n )+ , k ).Let b Γ = Hom ab (Γ , k × ) be the character group of an abelian group Γ. Acknowledgements.
We thank H´ector Pe˜na Pollastri for developing a computer programused in Parts II and III. S. W. and J. P. thank Jon Carlson for bestowing his computerprowess and wisdom on them in the early stages of this project. N. A. thanks Cris Negronand Sonia Natale for very useful conversations. I. A. thanks Leandro Vendramin for jointwork on the program to compute root systems of Nichols algebras.The authors are extremely grateful to the hospitality and support of the AmericanInstitute for Mathematics (AIM) via their SQuaREs program. N. A. was supported bythe NSF under Grant No. DMS-1440140, while he was in residence at the MathematicalSciences Research Institute in Berkeley, California, in the Spring semester of 2020. J. P. wassupported by the NSF grants DMS-1501146 and DMS-1901854. S. W. was supported bythe NSF grants DMS-1401016 and DMS-1665286. The work of N. A. and I. A. was partiallysupported by CONICET and Secyt (UNC).
Part I. From cohomology of Nichols algebras to cohomology of Hopf algebras Finite-dimensional Hopf algebras
Morita equivalence of Hopf algebras.
In this subsection, no restrictions on thethe field k are needed. Let H be a finite-dimensional Hopf algebra. We refer to [R] forthe definitions of the Drinfeld double D ( H ) of H and of the (braided tensor) category HH YD of Yetter-Drinfeld modules over H . It is well-known that HH YD is the Drinfeld centerof the category of H -modules and that it is braided tensor equivalent to the category of D ( H )-modules.Let H ′ be another finite-dimensional Hopf algebra. Borrowing terminology from [Mu,ENO], we say that H and H ′ are Morita equivalent , denoted H ∼ Mor H ′ , if there is anisomorphism of quasitriangular Hopf algebras D ( H ) ≃ D ( H ′ ). This is not the same asMorita equivalent as algebras! Lemma 2.1.1. H is Morita equivalent to H ′ in the following cases: (a) H ′ ≃ H , the dual Hopf algebra. (b) H ′ ≃ H F is a twist of H [Dr, Re] , i.e. there exists F ∈ H ⊗ H invertible such that H F = H as algebra and has the comultiplication ∆ F = F ∆ F − . (c) H ′ ≃ H σ is a cocycle deformation of H [DoT] , i.e. there exists an invertible 2-cocycle σ : H ⊗ H → k such that H σ = H as coalgebra and has the multiplication x · σ y = σ ( x (1) ⊗ y (1) ) x (2) y (2) σ − ( x (3) ⊗ y (3) ) . N. ANDRUSKIEWITSCH, I. ANGIONO, J. PEVTSOVA, S. WITHERSPOON
Proof. (a) is folklore; (b) follows since the categories of H and H F -modules are tensorequivalent [Dr, p. 1422]. Finally (c) is a consequence of the preceding, as ( H σ ) ≃ ( H ) F where F = σ in H ⊗ H . (cid:3) The role of Nichols algebras.
The notion of a Nichols algebra originated in inde-pendent work of Nichols and Woronowicz. Here we give a brief account of its importancein the structure of finite-dimensional Hopf algebras and refer to the surveys [AS2, A] forthe precise definitions and details. Recall that k is algebraically closed of characteristic 0.The classification of finite-dimensional Hopf algebras can be organized in four classes.Let H be a Hopf algebra, H its coradical [R] and H [0] = k h H i its Hopf coradical, a Hopfsubalgebra of H [AC]. The classes are:(a) H = H , i.e. H is cosemisimple.(b) H = H [0] = H . (c) H = H [0] = H .(d) H = H [0] = H .Hopf algebras in class (a) are semisimple by a theorem of Larson and Radford. Albeitfamilies of examples and some classification results in low dimension are known, no sys-tematic approach to the classification is available. Similarly for class (b). Nichols algebrasare relevant to the study of classes (c) and (d).Let H be in class (c) i.e. H is a proper Hopf subalgebra. Let gr H be the graded Hopfalgebra associated to the coradical filtration of H ; thengr H ≃ R H (2.2.1)where R = ⊕ n ∈ N R n is a connected graded Hopf algebra in the braided monoidal category H H YD , called the diagram of H . We also say that H is a lifting of R , or of R H . Then R is coradically graded, hence its subalgebra generated by V := R is isomorphic to theNichols algebra B ( V ); see [AS2] for details. The braided vector space V is an importantinvariant of H called its infinitesimal braiding . It is expected that R ≃ B ( V ) , (2.2.2) ∃ σ : gr H ⊗ gr H → k such that (gr H ) σ ≃ H. (2.2.3)Assume that (2.2.2) and (2.2.3) hold. By (a conjectural generalization of) Theorem3.1.7 and Corollary 3.2.3, we see that the core of the question is to verify fgc for B ( V )and for the Drinfeld double D (gr H ).Assume that H is a group algebra k Γ, in which case H is called pointed, with Γ abelian(then V is of diagonal type). Then (2.2.2) holds by [An2, Theorem 2], see also [AS3,Theorem 5.5]; notice that the proof uses the classification in [H2] and the main resulton convex orders from [An1]. Moreover (2.2.3) also holds [AnG, Theorem 1.1], based onprevious studies of the lifting question and the explicit relations from [An2, Theorem 3.1],that again uses [H2, An1]. Summarizing, Theorem 2.2.4.
Let H be a finite-dimensional pointed Hopf algebra such that G ( H ) isabelian. Then H is a cocycle deformation of the bosonization of a Nichols algebra ofdiagonal type: H ≃ ( B ( V ) k G ( H )) σ . Hence H ∼ Mor gr H . (cid:3) We also point out that (2.2.2) and (2.2.3) have been verified in most known examplesin class (c) beyond pointed Hopf algebras with abelian group of group-like elements.
OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 9
Finally, let H be in class (d). Then one considers the graded Hopf algebra gr H associ-ated to the standard filtration of H [AC]; againgr H ≃ R H [0] (2.2.5)where R = ⊕ n ∈ N R n is a connected graded Hopf algebra in the braided monoidal category H [0] H [0] YD . But it is not known whether R is coradically graded, or its subalgebra R ′ generatedby V := R is isomorphic to the Nichols algebra B ( V ). Nevertheless B ( V ) is a quotientof R ′ but the present approach does not allow to reduce the question of finitely generatedcohomology for H to the analogous question for B ( V ).2.3. Nichols algebras of diagonal type.
Since finite-dimensional Nichols algebras ofdiagonal type are central in this paper, we present here the features more relevant forour goals and refer to [AA] for an exposition. The input is a matrix of non-zero scalars q = ( q ij ) i,j ∈ I where I = I θ , θ ∈ N . To this datum we attach a braided vector space ofdiagonal type V with a basis ( x i ) i ∈ I and braiding c q ∈ GL ( V ⊗ V ) given by c q ( x i ⊗ x j ) = q ij x j ⊗ x i , i, j ∈ I . The corresponding Nichols algebra is a graded connected algebra with strong propertiesdenoted here mostly as B q instead of B ( V ). For these Nichols algebras substantial infor-mation is available.2.3.1. Dynkin diagrams and positive roots.
We codify as usual the matrix q in a (gener-alized) Dynkin diagram D with vertices numbered by I and labelled with q ii , while twodifferent vertices i and j are joined by an edge only if e q ij := q ij q ji = 1 in which case theedge is labelled with e q ij : q ii ◦ i e q ij q jj ◦ j . (2.3.1)Two different matrices with the same Dynkin diagram are called twist-equivalent [AS2].The Nichols algebra B q has a very useful N I -grading given by the rule deg x i = α i , i ∈ I ,where ( α i ) i ∈ I is the canonical basis of Z θ . By [Kh, Theorem 2.2], B q has a PBW basis B = (cid:8) s e . . . s e t t : t ∈ N , s i ∈ S, s > · · · > s t , < e i < h ( s i ) (cid:9) . where S is an ordered set of N I -homogeneous elements and h : S N ∪ {∞} is a functioncalled the height . The following set does not depend on the choice of B :∆ q + := { deg s : s ∈ S } ⊂ N I . Occasionally we set ∆ V + = ∆ q + . The elements of ∆ q + are called the (positive) roots of B q .We assume from now on that dim B q < ∞ . Then ∆ q + is a finite set and the map S → ∆ q + , s deg s , is bijective. Also ∆ q + admits aconvex (total) order in the sense α, β, α + β ∈ ∆ q + , α < β = ⇒ α < α + β < β. See [An1]. The convex order is not unique; in the case-by-case analysis below we use thatof [AA] except when a more suitable choice is possible that we mention explicitly.A connected component of D is a subset J of I such that the matrix q J = ( q ij ) i,j ∈ J givesrise to a connected Dynkin subdiagram D J of D and is maximal with this property. Let X be the set of connected components of D . If J ∈ X , then we identify ∆ q J + with the subset∆ J + of ∆ q + of roots with support in J . We also denote by V J the subspace of V spannedby ( x j ) j ∈ J . By a result of Gra˜na, we have B q ≃ O J ∈X B q J , ∆ q + = a J ∈X ∆ J + . (2.3.2)Here ⊗ means the braided tensor product of algebras.2.3.2. Classification.
The classification of the matrices q such that dim B q < ∞ wasachieved in [H2] (the result is slightly more general). By the preceding discussion wemay assume that D is connected. As in [AA] we organize the classification in 4 types:(a) Cartan type.(b) Super type. (c) Modular type.(d) UFO type.The type refers to the connection with different parts of Lie theory, see loc. cit. Weshall check Condition 1.4.1 for each entry of the classification of [H2].2.3.3.
Root vectors.
For brevity, we set x ij = ad c x i ( x j ) , i = j ∈ I ;(2.3.3)more generally, the iterated braided commutators are x i i ··· i k := (ad c x i ) · · · (ad c x i k − ) ( x i k ) , i , i , · · · , i k ∈ I . (2.3.4)In particular, we will use repeatedly the following further abbreviation: x ( k l ) := x k ( k +1) ( k +2) ...l , k < l. (2.3.5)Using a fixed convex order, we define the root vector x α ∈ B q for every α ∈ ∆ q + asiterated braided commutators proceeding case-by-case, see [AA].For those q with dim B q < ∞ , the defining relations of B q were given in [An1, An2],again see [AA], but these are not needed in this paper. Instead, we use systematically[An1, Theorem 4.9]; that is, for α < β ∈ ∆ q + and a suitable defined q αβ ∈ k × , we have x α x β − q αβ x β x α ∈ X α<γ ≤ γ ≤···≤ γ t <β ∈ ∆ q + k x γ x γ . . . x γ t . Realizations.
Let Γ be a finite abelian group. A Yetter-Drinfeld module V over k Γis determined by families ( g i ) i ∈ I θ of elements of Γ and characters ( χ i ) i ∈ I θ in b Γ. Then V isa braided vector space of diagonal type with braiding matrix q = ( q ij ) i,j ∈ I with respect toa basis ( x i ) i ∈ I , i.e. c ( x i ⊗ x j ) = q ij x j ⊗ x i , i, j ∈ I , where q ij = χ j ( g i ). That is, the samebraided vector space V with braiding matrix q = ( q ij ) i,j ∈ I can be realized in many waysover many Γ. Even more, it can be realized over other Hopf algebras than group algebrasover abelian groups. To discuss the possible realizations we need the notion of a YD-pair. OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 11
Let H be a Hopf algebra. A pair ( g, χ ) ∈ G ( H ) × Hom alg ( H, k ) is a YD-pair for H if χ ( h ) g = χ ( h (2) ) h (1) g S ( h (3) ) , h ∈ H. (2.4.1)When this is the case, g ∈ Z ( G ( H )); also k χg = k with action and coaction given by χ and g respectively, is in HH YD . YD-pairs classify the 1-dimensional objects in HH YD . Note that,if dim H < ∞ , then ( g, χ ) is a YD-pair for H if and only if ( χ, g ) is a YD-pair for H . Definition 2.4.2.
Let V be a braided vector space of diagonal type with braiding matrix q = ( q ij ) i,j ∈ I . A principal realization of V over H is a family ( g i , χ i ) i ∈ I of YD-pairs suchthat χ j ( g i ) = q ij , i, j ∈ I , so that V ∈ HH YD via k x i ≃ k χ i g i , and the braiding c is thecategorical one from HH YD .Given a principal realization of V over H , we have Γ := h g , . . . , g θ i ≤ Z ( G ( H )); hencewe can also realize V as an object in k Γ k Γ YD . Example 2.4.3.
If Γ is a finite group, then the YD-pairs of H = k Γ are of the form( g, χ ) ∈ Z (Γ) × Hom grp (Γ , k × ). For example, the YD-pairs for GL n ( p ) are (diag( t ) , ϕ det n ),where t ∈ F × p and ϕ ∈ c F × p . Example 2.4.4.
Not all realization is principal: if g ∈ Z (Γ) and ρ ∈ Irrep Γ with dim ρ = d > ρ ( g ) = ζ id, then the simple Yetter-Drinfeld module M ( g, ρ ) [A, Example 24] isa braided vector space of diagonal type with braiding matrix ( q ij ) i,j ∈ I d where q ij = ζ forall i, j . Other examples arise from simple Yetter-Drinfeld modules M ( g, ρ ) such that theelements in the conjugacy class of g commute with each other.2.5. The Drinfeld double of a bosonization.
Recall that char k = 0. Let L be a Hopfalgebra whose coradical L is a Hopf subalgebra and let H be another Hopf subalgebra of L . Then H = L ∩ H by [R, 4.2.2] and this is a Hopf subalgebra of H . By [R, 4.4.11] wehave an injective map of graded Hopf algebras gr H ֒ → gr L . Let R and S be the diagramsof H and L respectively, see (2.2.1). Hence we have an injective map of graded braided Hopf algebras
R ֒ → S .Let L be a finite-dimensional Hopf algebra. The Drinfeld double of L , denoted by D ( L ),is a Hopf algebra whose underlying coalgebra is L ⊗ L . Let (cid:10) − , − (cid:11) : L ⊗ L → k denotethe evaluation map. The multiplication on D ( L ) is given by the following formula:( h ⊲⊳ f )( h ′ ⊲⊳ f ′ ) = (cid:10) f (1) , h ′ (1) (cid:11)(cid:10) f (3) , S ( h ′ (3) ) (cid:11) ( hh ′ (2) ⊲⊳ f ′ f (2) ) , f, f ′ ∈ L h, h ′ ∈ L , where h ⊲⊳ f := h ⊗ f in D ( L ) and f r = m ( f ⊗ r ) is the multiplication in L rather thanin L .Let K be a semisimple Hopf algebra, hence cosemisimple by the Larson-Radford the-orem. Let V ∈ KK YD with dim B ( V ) < ∞ and H = B ( V ) K . Then V ∈ K K YD appropriately and H ≃ B ( V ) K . Since D ( H ) ≃ H ⊗ H as coalgebras, thecoradical of H , respectively H , can be identified with K , respectively K . We identify D ( K ) with a Hopf subalgebra of D ( H ) in a natural way.The following result generalizes, with an analogous proof, Theorem 2.5 in [Be]. Proposition 2.5.1.
The Drinfeld double D ( H ) is a lifting of a Nichols algebra B ( W ) where W = V ⊕ V and V braided commutes with V . Proof.
First, the coradical D ( H ) of D ( H ) equals D ( K ); this follows from [R, 4.1.8]. Hencethe coradical filtration of D ( H ) is a Hopf algebra filtration and gr D ( H ) ≃ R D ( K ), where R = ⊕ n ≥ R n is the diagram of D ( H ). Let W = R . By the preceding paragraph appliedto L = D ( H ) and either H = H or H = H , we have morphisms of braided vectorspaces V ֒ → W and V ֒ → W ; we have V ⊕ V ֒ → W by comparing the D ( K )-comodulestructures. Recall that dim B ( V ) dim B ( V ) ≤ dim B ( V ⊕ V ) and the equality holds iff V and V braided commute [Gr, Theorem 2.2]. Thendim D ( H ) = dim B ( V ) dim K dim B ( V ) dim K ≤ dim B ( W ) dim D ( K ) ≤ dim R dim D ( K ) = dim D ( H ) , hence R = B ( W ) = B ( V ⊕ V ) and V and V braided commute. (cid:3) Assume next that K = k Γ where Γ is a finite abelian group; recall that b Γ is the groupof characters of Γ. Then k b Γ ∼ = ( k Γ) ,D ( k Γ) ∼ = k (Γ × b Γ) ∼ = k Γ ⊗ ( k Γ) . Let ( g i ) i ∈ I θ and ( χ i ) i ∈ I θ be (dual) generating families in Γ and b Γ respectively. Let V ∈ k Γ k Γ YD with a basis ( x i ) i ∈ I such that the action and coaction of Γ on x i are given by χ i and g i respectively, i ∈ I . Assume that B ( V ) has finite dimension and let H = B ( V ) k Γ. Let( y i ) i ∈ I be the basis of V dual to ( x i ) i ∈ I . Then H ≃ B ( V ) k b Γ where the actionand coaction of b Γ on y i are given by g i and χ − i respectively, i ∈ I . Also W = V ⊕ V canbe realized in k (Γ × b Γ) k (Γ × b Γ) YD extending these structures. See [Be, Theorem 2.5] for details. Let I ( V ) = ker( T ( V ) → B ( V )) be the ideal of defining relations of the Nichols algebra B ( V ).The following statement is well-known. Proposition 2.5.2. D ( H ) is isomorphic to the quotient of T ( W ) k (Γ × b Γ) by the idealgenerated by I ( V ) , I ( V ) and the relations x i y j − χ − j ( g i ) y j x i = δ i,j χ − i ( g i ) (1 − g i χ i ) , i, j ∈ I . (2.5.3) Outline of the proof.
By the preceding discussion there is a morphism of Hopf algebras T ( W ) k (Γ × b Γ) → D ( H ) whose kernel J contains I ( V ), I ( V ) and the relations (2.5.3),see the proof of [Be, Theorem 2.5]. The induced map T ( W ) k (Γ × b Γ) / J → D ( H ) isclearly surjective and preserves the coradical filtration. Since the associated graded mapis injective, the claim follows. (cid:3) Cohomology
Invariants.
Let H be a Hopf algebra and A an H -module algebra. The ring ofinvariants is the subalgebra A H = { x ∈ A : h · x = ε H ( h ) x ∀ h ∈ H } . OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 13
Let now K be a semisimple, hence finite-dimensional, Hopf algebra. Let t ∈ K be anormalized integral, that is kt = ε ( k ) t = tk for all k ∈ K and ε ( t ) = 1. Let A be a K -module algebra and let R : A → A be the Reynolds operator R ( x ) = t · x , x ∈ A . Then ◦ The Reynolds operator is a projector, R = R , and im R = A K . ◦ R is a morphism of K -modules. ◦ R is a morphism of R K -bimodules: R ( xyz ) = x R ( y ) z , for x, z ∈ A K and y ∈ A .The following result, a variation of a classical argument by Hilbert, is well-known. Lemma 3.1.1.
Let K be a semisimple Hopf algebra. Let A = ⊕ n ∈ N A n be a graded K -module algebra that is connected and (right) Noetherian. Let M be a finitely generated A K -module. Then A K is finitely generated and M K is a finitely generated A K -module.Proof. Let I = A ( A K ) + be the left ideal of A generated by the augmentation ideal of A K .Since A is Noetherian, I is finitely generated; we may assume that I = h f , . . . , f M i , where f i ∈ A K is homogeneous of degree d i . We claim that A K = k h f , . . . , f M i . For this we shallprove that any f ∈ A K homogeneous of degree d belongs to k h f , . . . , f M i . If d = 0, thisfollows by connectedness. If d >
0, then we may write f = P i a i f i with a i either 0 or elsehomogeneous of degree d − d i . Then f = R ( f ) = P i R ( a i ) f i and R ( a i ) ∈ k h f , . . . , f M i by the recursive hypothesis, so f ∈ k h f , . . . , f M i .For the module statement, note that the hypothesis of [Mo, Theorem 4.4.2] holds; namelythe map denoted there ˆ t is the Reynolds operator R . Hence A is a right Noetherian A K -module, and thus finitely generated over A K . Thus M is a Noetherian A K -module. Since M K is an A K -submodule, it is also Noetherian, therefore finitely generated over A K . (cid:3) If R is an H -module algebra, then H acts on Ω ∗ ( R ) via the comultiplication and theantipode, and a fortiori on H( R, k ). The following proposition is well-known. Proposition 3.1.2.
Let K be a semisimple Hopf algebra and let R be a finite-dimensional K -module algebra. Let M be an R K -module. Then H( R K, k ) ≃ H( R, k ) K , H( R K, M ) ≃ H( R, M ) K , and the action of H( R K, k ) on H( R K, M ) is precisely that induced by the isomorphismsand the action of H( R, k ) on H( R, M ) .Proof. This is well-known; the first isomorphism is for example [SV, Theorem 2.17]. Incase K is a group algebra, the relevant spectral sequence is the Lyndon-Hochschild-Serrespectral sequence [Ev, §§ K is semisimple. (cid:3) Now we pass to algebras in HH YD . An algebra A in HH YD is braided commutative if themultiplication m A satisfies m A = m A c A,A , that is xy = ( x ( − · y ) x (0) , x, y ∈ A. (3.1.3)If A is braided commutative, then A H is central in A . We elaborate on an idea of [MPSW];for this we do not need the commutativity of Γ. Lemma 3.1.4.
Let Γ be a finite group and let A be a braided commutative algebra eitherin k Γ k Γ YD or in k Γ k Γ YD . Assume that A is finitely generated (as an algebra). Then A isNoetherian. Proof.
Let N be the exponent of Γ. We deal first with k Γ k Γ YD . As an object in k Γ k Γ YD , A is Γ-graded: A = ⊕ g ∈ Γ A g . Thus, if A = k h f , . . . , f M i , then we may assume thateach f i belongs to A g i for some g i ∈ Γ. Then f Ni ∈ A g Ni = A e . Since A is braidedcommutative, f Ni f j = ( g Ni · f j ) f Ni = f j f Ni for all i, j . Then B = k h f N , . . . , f NM i is a centralsubalgebra of A and is Noetherian by Hilbert’s Basissatz. Now A is a finitely generated B -module, actually A = P ≤ a i ≤ N B f a . . . , f a M M . Thus A is a Noetherian B -module hencea Noetherian algebra. We deal next with k Γ k Γ YD . Since H = k Γ has a basis of idempotents δ g , g ∈ Γ, again A is Γ-graded: A = ⊕ g ∈ Γ A g where A g = δ g A . Thus, if A = k h f , . . . , f M i ,with each f i ∈ A g i for some g i ∈ Γ, then f Ni ∈ A g Ni = A e = δ e A . But δ e is the integral of k Γ , thus again f Ni ∈ A H is central. Then we proceed as previously. (cid:3) We wonder whether any finitely generated braided commutative algebra is Noetherian.We need the following result from [MPSW].
Proposition 3.1.5. [MPSW, Corollary 3.13]
Let H be a Hopf algebra and let R be abialgebra in HH YD . Assume that either H or R is finite-dimensional. Then the (opposite of )the Hochschild cohomology HH( R, k ) is a braided commutative graded algebra in HH YD . (cid:3) Actually [MPSW, Theorem 3.12] gives more: the claim is true if R is a bialgebra in anabelian braided monoidal category C where the needed hom-objects exist. Theorem 3.1.6.
Let Γ be a finite group and let R be a finite-dimensional Hopf algebra in k Γ k Γ YD . Let M be a finitely generated R k Γ -module. If R has fgc, then so does R k Γ .Proof. By Proposition 3.1.5, Lemma 3.1.4 and the hypothesis, H( R, k ) is Noetherian. ThenH( R, k ) Γ is finitely generated by Lemma 3.1.1. By Proposition 3.1.2, H( R k Γ , k ) ≃ H( R, k ) Γ is finitely generated. We next prove: If M is finitely generated, then H( R k Γ , M )is finitely generated as an H( R k Γ , k )-module. For this, we may induct on the length ofthe composition series of M , and so it suffices to prove it in case M is simple. Let R + denote the augmentation ideal of R . Note that R + M is an R k Γ-submodule of M andtherefore R + M = 0 (by Nakayama), that is, M | R is a trivial R -module. We concludethat H( R, M ) is finitely generated as an H( R, k )-module. By Lemma 3.1.1, it follows thatH( R, M ) K is finitely generated over H( R, k ) K . (cid:3) We are ready for one of our main results.
Theorem 3.1.7.
Let V be a braided vector space of diagonal type such that (a) the Nichols algebra B ( V ) is finite-dimensional, (b) V is realizable over a finite abelian group, (c) H( B ( V ) , k ) is finitely generated.Let K be a semisimple Hopf algebra and assume that V is realizable over K . Then B ( V ) K has fgc.Proof. The proof is the same as for the previous result. By (b), (c), Proposition 3.1.5and Lemma 3.1.4, H( B ( V ) , k ) is Noetherian. Then H( B ( V ) , k ) K is finitely generated byLemma 3.1.1. By Proposition 3.1.2 and (a), H( B ( V ) K, k ) ≃ H( B ( V ) , k ) K is finitelygenerated. The proof of the module statement is similar. (cid:3) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 15
Observation 3.1.8. If V admits a principal realization over K , then (b) holds. Noticethat (a) does not imply (b): take V of dimension 2 with braiding matrix q q q − q where q ∈ G ′ M , q / ∈ G ∞ , q ∈ G ′ N , N, M >
1. However we do not know if V beingrealizable over K semisimple implies (b).3.2. Subalgebras, extensions.Theorem 3.2.1.
Let R be an augmented subalgebra of a finite-dimensional augmentedalgebra A , over which A is projective as a right R -module under multiplication. If A hasfgc, then so does R .Proof. By the right module version of the Eckmann-Shapiro Lemma [Ben, Corollary 2.8.4],for each n , and any R -module M , there is an isomorphism of vector spaces,H n ( R, M ) ≃ Ext n A ( k , Hom R ( A , M )) = H n ( A , Hom R ( A , M )) , where Hom R ( A , M ) is the coinduced right A -module. (The action is given by ( f · a )( b ) = f ( ab ) for all a, b ∈ A , f ∈ Hom R ( A, M ). Then f · a is indeed a right R -module ho-momorphism.) These isomorphisms, one for each n , provide an isomorphism of H( A, k )-modules H( R, M ) ≃ H( A , Hom R ( A , M )). Now when M is a finite-dimensional R -module,Hom R ( A , M ) is finite-dimensional as a vector space. For M = k , a set of generators ofH( A , Hom R ( A , k )) as a module for H( A , k ), together with the restriction to R of a set ofgenerators of H( A , k ), generates H( R, k ) as a k -algebra. For an arbitrary finite-dimensionalmodule M , Hom R ( A , M ) is then a finite-dimensional module over H( A , k ) and, hence, overH( R, k ). (cid:3) If K is a Hopf subalgebra of a finite-dimensional Hopf algebra H , then H is free as aleft or right module over K with respect to multiplication by the Nichols–Z¨oller Theorem.Thus Theorem 3.2.1 applies to inclusions of Hopf algebras, in particular to the inclusionof a finite-dimensional Hopf algebra into its Drinfeld double, see [NP, Theorem 3.4]. Forfurther reference we state a useful application of Theorem 3.2.1. Corollary 3.2.2.
Let H be a finite-dimensional Hopf algebra and V ∈ HH YD such that dim B ( V ) < ∞ . If D ( B ( V ) H ) has fgc, then so does B ( V ) . These ideas apply in particular to Morita equivalence of Hopf algebras as in § Corollary 3.2.3. If H ∼ Mor H ′ and D ( H ) has fgc, then so does H ′ . Question 3.2.4.
Is the fgc property for Hopf algebras invariant under Morita equivalencein the sense of § Lemma 3.2.5.
Let k → K → H → L → k be an extension of finite-dimensional Hopfalgebras. If K is semisimple and L has finitely generated cohomology, then so does H . The proof makes use of a variation of the classical Hochschild-Serre spectral sequence.
Proof.
Let M be an L -module and N an H -module. By [CE, Chapter 16, Theorem 6.1]there exists a convergent spectral sequenceExt pL (cid:0) M, Ext qK ( k , N ) (cid:1) = ⇒ Ext p + qH ( M, N ) . In particular, if M ≃ k , thenExt pL (cid:0) k , Ext qK ( k , N ) (cid:1) = ⇒ Ext p + qH ( k , N ) . Now Ext K ( k , N ) ≃ N K and, since K is semisimple, Ext qK ( k , N ) = 0 when q >
0. If also M ≃ k , then E p,q := Ext pL (cid:0) k , Ext qK ( k , k ) (cid:1) = ⇒ Ext p + qH ( k , k ) . Now Ext K ( k , k ) ≃ k and, since K is semisimple, Ext qK ( k , k ) = 0 when q >
0. Hence E p,q ≃ E p,q ∞ , thus ⊕ n ∈ N Ext nH ( k , k ) is finitely generated, as it has a filtration whose associatedgraded algebra is ⊕ n ∈ N Ext nL ( k , k ). Similarly ⊕ n ∈ N Ext nH ( k , N ) is a finitely generatedmodule over ⊕ n ∈ N Ext nL ( k , N ) for any H -module N . (cid:3) Evens Lemma.
Let R = ⊕ n ∈ N R n be an N -graded ring with a decreasing algebrafiltration F n R , n ∈ N , compatible with the grading. We shall assume that F i R n = 0 for i sufficiently large. Then the associated graded ring E ( R ) = P i F i R/F i +1 R is N -graded.Similarly, the graded E ( R )-module associated to an N -graded R -module N with adecreasing module filtration is N -graded. Again, F i N j = 0 for i sufficiently large. Thefollowing proposition is [Ev, Section 2, Proposition 2.1]. Proposition 3.3.1.
Let R be a graded filtered ring and N a graded filtered R -module asabove. If E ( N ) is (left) Noetherian over E ( R ) , then N is Noetherian over R . (cid:3) The following result is a non-commutative version of [MPSW, Lemma 2.5], adapted inturn from [FS, Lemma 1.6] and inspired by early work of Evens.Let E p,q ⇒ E p + q ∞ be a multiplicative spectral sequence of bigraded k -algebras concen-trated in the half plane p + q ≥
0. Recall that x ∈ E p,qr is called a permanent cycle if d i ( x ) = 0 for all i ≥ r . More precisely, if i > r, d i is applied to the image of x in E i . Lemma 3.3.2. [Shr, Lemma 2.6](a)
Let C ∗ , ∗ be a bigraded k -algebra such that for each fixed q , C p,q = 0 for p sufficientlylarge. Assume that there exists a bigraded map of algebras φ : C ∗ , ∗ → E ∗ , ∗ such that (1) φ makes E ∗ , ∗ into a left Noetherian C ∗ , ∗ -module, and (2) the image of C ∗ , ∗ in E ∗ , ∗ consists of permanent cycles.Then E ∗∞ is a left Noetherian module over Tot( C ∗ , ∗ ). (b) Let e E p,q ⇒ e E p + q ∞ be a spectral sequence that is a bigraded module over the spectralsequence E ∗ , ∗ . Assume that e E ∗ , ∗ is a left Noetherian module over C ∗ , ∗ where C ∗ , ∗ acts on e E ∗ , ∗ via the map φ . Then e E ∗∞ is a finitely generated E ∗∞ -module. (cid:3) The May spectral sequence.
Let A be a Hopf algebra equipped with an increasingmultiplicative filtration A ⊂ A ⊂ A . . . ⊂ A . We fix a (non-canonical) vector spacesplitting A ≃ A ⊕ A + so that A/A + ≃ A . Let ( P, d ) = ( V n ⊗ A, d ) be a free resolutionof the trivial module k satisfying the following properties. Condition 3.4.1. (1) V n is a finite-dimensional vector space, the action of A on P n ison the last factor A . (2) P q is equipped with an increasing filtration . . . F i P n ⊂ F i +1 P n . . . . OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 17 (3)
For any x ∈ F i V n = F i ( V n ⊗
1) := F i P n ∩ ( V n ⊗ , we have d ( x ) ∈ F i V n − ⊗ A + F i − V n − ⊗ A + . Example 3.4.2.
We will apply this setting in at least two different situations.(i) When P q is the bar resolution and F q is the coradical filtration, see Theorem ?? . Inthis case A is the coradical.(ii) When P q is the Anick resolution of k for B ( V ), and the filtration is given by thePBW basis induced by the convex ordering of the roots, see Theorem 4.4.3. In thiscase A = k and we identify A + with the augmentation ideal.We set up a version of May spectral sequence analogous to the one in [GK, 5.5]. Wefollow the construction in May [May] but without assuming that the module M is filtered.Such a spectral sequence is also constructed in [BKN, §
9] using a non-canonical filtrationon M induced by the filtration on A . Theorem 3.4.3.
Let A be a filtered finite-dimensional Hopf algebra , ( P q , d ) be a projectiveresolution of the trivial module k , and assume that A and P q satisfy Condition 3.4.1. Let M be an A -module. Then there exists a converging cohomological spectral sequence e E ∗ = H ∗ (gr A, M A ) ⇒ H ∗ ( A, M ) equipped with a natural module structure over the multiplicative spectral sequence E ∗ = H ∗ (gr A, k ) ⇒ H ∗ ( A, k ) . The action of gr A on M A is via the projection gr A → A and then restricting the actionof A on M to the action of the subalgebra A ⊂ A .Proof. Let C q ( A, M ) := Hom A ( P q , M ) be the complex computing H ∗ ( A, M ). Let F i C n ( A, M ) := { f ∈ Hom A ( P n , A ) | f ↓ F i − P n = 0 } ⊂ C n ( A, M )be a decreasing filtration on C q ( A, M ) making it into a filtered complex.As V n ⊂ P n , we have an induced filtration (of vector spaces) on V q : F i V n = F i ( V n ⊗
1) = F i P n ∩ ( V n ⊗ A ( V n ⊗ A, M ) ≃ Hom k ( V n , M ), we make theidentifications: F i C n ( A, M ) F i +1 C n ( A, M ) = { f : V n → M | f ↓ F i − V n = 0 }{ f : V n → M | f ↓ F i V n = 0 }≃ Hom k ( V n /F i − V n , M )Hom k ( V n /F i V n , M ) ≃ Hom k (cid:18) F i V n F i − V n , M (cid:19) . Letting n be the total and i be the internal degree, we haveHom k (cid:18) F i V n F i − V n , M (cid:19) ≃ Hom gr A (cid:18) F i P n F i − P n , M tr (cid:19) ≃ H n,i (gr A, M tr ) . Let e E i,n − i ( M ) := F i C n ( A, M ) F i +1 C n ( A, M ) . Since A is finite-dimensional, the filtration is finiteand, hence, this defines the 0 page of the spectral sequence of a filtered complex C q ( A, M ) converging to H ( A, M ). We have identified the terms of the double complex e E ∗ ( M ) withthe complex C q (gr A, M A ) computing cohomology H ∗ (gr A, M A ). To identify e E ∗ ( M )and H (gr A, M A ) as complexes, it suffices to show that the differentials in e E ∗ ( M ) and e E ∗ ( M A ) are the same, that is, that the differential d M in the spectral sequence onlydepends on the A -module structure on M .Consider the differential d M : e E i,n − i ( M ) Hom k (cid:16) F i V n F i − V n , M (cid:17)e E i,n − − i ( M ) d M O O Hom k (cid:16) F i V n − F i − V n − , M (cid:17) . d M O O Let f ∈ Hom k (cid:18) F i V n − F i − V n − , M (cid:19) , ¯ x ∈ F i V n F i − V n , and let x ∈ F i V n be a representative of ¯ x . ByCondition 3.4.1(3) we can write d ( x ) = v ⊗ a + v ′ ⊗ a ′ with a ∈ A and v ′ ∈ F i − V n − .We now compute d M ( f )(¯ x ) = f ( d M ( x ))= f (¯ v ) a + f (¯ v ′ ) a ′∗ = f (¯ v ) a = f ( d M A ( x )) = d M A ( f )(¯ x )The equality (*) holds since ¯ v ′ = v ′ mod F i − V n − = 0.The statement about the action of E ( k ) acting on e E ( M ) follows from the constructionof the spectral sequence. (cid:3) The Anick resolution
The setup.
In this section we discuss the Anick graph and the construction of theAnick resolution [Ani, Fa, CoU]. Let V be a finite-dimensional vector space with a basis( x i ) i ∈ I , and I an ideal of T ( V ) such that ǫ ( I ) = 0, where ǫ : T ( V ) → k is the standardaugmentation map, ǫ ( x i ) = 0 for all i ∈ I . Thus the algebra A := T ( V ) / I has anaugmentation map ǫ : A → k .4.1.1. The tips.
Let X be the set of words on the letters ( x i ) i ∈ I (including the empty word1). Notice that X is a basis of T ( V ). Let x, y ∈ X . We say that x is a subword of y ifthere exist w, z ∈ X such that y = wxz . If w = 1, respectively z = 1, then we say that x is a prefix, respectively a suffix, of y .Let ℓ : X → N be the length function. The lex-length order < on X is defined asfollows: given v, w ∈ X , we say that v < w if either ℓ ( v ) < ℓ ( w ) or else ℓ ( v ) = ℓ ( w ) and v is less than w for the lexicographical order (induced by the numeration of the basis). Thisis a total order on X compatible with left and right multiplication. OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 19
Here is a way to give a set of generators of the ideal I . Given f ∈ I −
0, write f as alinear combination of elements of X ; let x f be the largest element of X (with respect to < ) with non-zero coefficient. Then x f is called the tip of f . Consider the set of all tips ofall elements in I −
0. A tip t is minimal if each subword of x is not a tip (Anick calls aminimal tip an obstruction ). Let T be the set of minimal tips of I . For each t ∈ T we pick ω t ∈ I such that t is the tip of ω t (which is not unique in general). Arguing recursively on < , it is possible to show that I = h ω t : t ∈ T i . For each w ∈ X we also denote by w its image in A = T ( V ) / I . By [Ani, Lemmas 1.1and 1.2], the set B = { w ∈ X : t is not a subword of w ∀ t ∈ T } (4.1.1)is a basis of A .4.1.2. The chains.
Let n ∈ N . We describe the n -chains which are words defined fromthe minimal tips; they will provide a basis of the n -th term of the Anick resolution of A .The unique 0-chain is the empty word 1. The 1-chains are the letters, i.e. the x i ’s. Let n ≥
1. An n -chain is a word w such that:(a) w admits a factorization w = uv such that u is an ( n − v doesnot contain any minimal tip as a subword (i.e., does not contain any tip);(b) for every suffix y = 1 of u as in (a), the word yv contains a minimal tip as a subword;(c) any other prefix w ′ of w does not satisfy (a) and (b) simultaneously.Let M ( n ) be the set of n -chains. We urge the reader to check that M (2) is the set of minimaltips—all requirements are needed.There exists an alternative way to express the definition of n -chains. A word w = x i · · · x i t , i j ∈ I , is an ( n + 1)-chain if there exist integers a j , b j , 1 ≤ j ≤ n , such that(1) 1 = a < a ≤ b < a ≤ b < · · · < a n ≤ b n − < b n = t ;(2) x i aj x i aj +1 . . . x i bj − x i bj ∈ T for all 1 ≤ j ≤ n ;(3) for all 1 ≤ m ≤ n , the words x i · · · x i s , s < b m , are not m -chains.By [Ani, Lemma 1.3] the integers a j , b j satisfying (1)–(3) are uniquely determined and • x i · · · x i bn − is the unique prefix which is an ( n − • x i bn − · · · x i t does not contain any element of T as a subword. Example 4.1.2.
We fix N ≥ V of dimension 1, x ∈ V − I = h x N i . Thus T = { x N } .Now M (0) = { } , M (1) = { x } , and we claim that M (2 k ) = { x Nk } , M (2 k + 1) = { x Nk +1 } , k ≥ . Moreover a i − = ( i − N + 1, a i = ( i − N + 2, b i − = iN , b i = iN + 1. We proceedby induction on k . If k = 2, then M (1) = { x N } since this is the unique minimal tip, and { x N +1 } is a 2-chain with a = 1, a = 2, b = N , b = N + 1; thus, each word x N + j , j > x N +1 is a prefix of x N + j . Assume that k ≥ k . To compute M (2 k + 2), we startwith the unique (2 k + 1)-chain x kN +1 and the integers a j , b j already determined: a k +1 should satisfy b k − = kN < a k +1 ≤ b k = kN + 1, hence a k +1 = kN + 1. Hence M (2 k + 2) = { x ( k +1) N } . For M (2 k + 3), we have that x ( k +1) N +1 is a (2 k + 3)-chain, hencethis is the unique (2 k + 3)-chain: indeed, if w = x s ∈ M (2 k + 3), then s > ( k + 1) N since w should contain the (2 k + 2)-chain x ( k +1) N as a prefix; but if s > ( k + 1) N , then w containsthe (2 k + 3)-chain x ( k +1) N +1 as a prefix so it cannot be a (2 k + 3)-chain.Let V ( n ) be the k -vector space with basis M ( n ). Then b M ( n ) := { u ⊗ w : u ∈ M ( n ) , w ∈ B } is a basis of V ( n ) ⊗ A . Given u ⊗ w, v ⊗ z ∈ b M ( n ), if uw = vz , then u = v , w = z ; indeed,if ℓ ( u ) ≤ ℓ ( v ), then the n -chain u is a prefix of the n -chain v and (3) implies that u = v .Hence the order on X induces an order on b M ( n ): u ⊗ w < v ⊗ z if uw < vz .4.1.3. The Anick graph.
We next introduce a graph which helps to compute the chains ofthe Anick resolution [CoU]. Let Γ be the graph whose set of vertices is given by the unionof { } , X and the set of all proper suffixes of the minimal tips. For the arrows, there existsone arrow from 1 → x for each x ∈ X , and one arrow u → v if the word uv contains aunique minimal tip such that it is a suffix of uv (possibly the word uv ).A basis of the free module of n -chains of the Anick resolution is given by paths of length n starting at 1. Thus: • There exists a unique 0-chain: 1. • The set X gives a basis of the 1-chains. • The set of minimal tips gives a basis of the 2-chains.Notice that vertices v not connected to 1 (that is, without a path from 1 to v ) do notcontribute new elements for the basis of chains, hence we may omit them and the relatedarrows. Example 4.1.3.
Let ζ ∈ G ′ , q ∈ k × . We want to determine the Anick graph of theNichols algebra B q of [AA, § q corresponds to q in loc. cit.) In termsof the PBW generators, B q is presented by generators x , x , x , x , x , and relations x = 0 , x x = ζ q x x , x x = − q x x + x ,x = 0 , x x = ζ q x x + x , x x = q x x + x x = 0 , x x = ζ q x x , x x = ζ q x x − q (1 + ζ ) x ,x = 0 , x x = ζq x x , x x = − q x x + ζ q x x ,x = 0 , x x = − q x x , x x = − q x x − q (1 + ζ ) x . Thus the set of obstructions is M (2) = { x , x , x , x , x , x x , x x , x x , x x ,x x , x x , x x , x x , x x , x x } and the Anick graph is OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 21 (cid:5) (cid:5) (cid:4) (cid:4) ✡✡✡✡✡✡✡✡✡✡✡✡✡✡✡✡✡✡✡✡✡✡✡✡ (cid:15) (cid:15) (cid:25) (cid:25) ✸✸✸✸✸✸✸✸✸✸✸✸✸✸✸✸✸✸✸✸✸✸✸✸ (cid:24) (cid:24) x (cid:7) (cid:7) / / & & ! ! (cid:31) (cid:31) x / / W W & & ! ! x / / (cid:7) (cid:7) % % x / / (cid:7) (cid:7) x W W x F F E E ☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛ < < ①①①①①①①①①①①①①①①①①①①①①①①①①①①①①①①① ♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣ ❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧ x F F E E ☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛☛ < < ③③③③③③③③③③③③③③③③③③③③③③③③③③③③③③ x F F F F ✌✌✌✌✌✌✌✌✌✌✌✌✌✌✌✌✌✌✌✌✌✌ There exists a vertex x with a loop on itself which we omit since this vertex is notconnected to 1. Using the graph we compute M (3) = { x , x x , x x , x x , x x , x x , x , x x , x x , x x ,x x , x x , x , x x , x x , x x , x x , x x , x , x x ,x x , x x , x x , x x , x , x x x , x x x , x x x , x x x ,x x x , x x x , x x x , x x x , x x x , x x x } . The resolution.
We consider the Anick resolution of the A -module k . In [Ani, The-orem 1.4] Anick introduced an A -free complex V ( n ) ⊗ A d n / / V ( n − ⊗ A d n − / / V (1) ⊗ A d / / A ǫ / / k / / k -linear maps s n : V ( n ) ⊗A → V ( n +1) ⊗A , n ∈ N , such that (4.1.4) is an A -resolutionof k and s q is a contracting homotopy: d n +1 s n + s n − d n = id V ( n ) ⊗A for all n ∈ N . (4.1.5)The maps d n , s n − are defined recursively; see for example [NWW, § n = 1, the map d is determined by d ( x ⊗
1) = x for all x ∈ M (1) , while for s we give the values on each w ∈ B : s ( w ) = x ⊗ z, w = xz, x ∈ M (1) , z ∈ B . Now assume that d , . . . , d n − , s , . . . , s n − were already defined and satisfy:(4.1.5) d i − d i = 0 , s i − s i − = 0 , for all 1 ≤ i ≤ n − . The morphisms of A -modules d n : V ( n ) ⊗ A → V ( n − ⊗ A are determined by d n ( u ⊗
1) = v ⊗ t − s n − d n − ( v ⊗ t ) , u ∈ M ( n ) , u = vt, v ∈ M ( n − , t ∈ B . Now we define s n − . From (4.1.5), V ( n − ⊗ A = ker d n − ⊕ im s n − . We start by defining( s n − ) | im s n − ≡
0, so s n − s n − = 0. Now we define ( s n − ) | ker d n − recursively on theorder of the leading term of each element of ker d n − , which we write in terms of the basis b M ( n − d n ( s n − ) | ker d n − = id ker d n − . Let K = X j ∈ I m a j u j ⊗ b j ∈ ker d n − , a j ∈ k × , u j ∈ M ( n − , b j ∈ B . We assume that u ⊗ b is bigger than u j ⊗ b j for all j >
1. We write u j = v j t j , where v j ∈ M ( n − t j ∈ B . Hence0 = d n − ( K ) = a v ⊗ t b + b K, b K := X j ≥ a j d n − ( u j ⊗ b j ) − s n − d n − ( a v ⊗ t b )Hence t b / ∈ B , otherwise v ⊗ t b is the biggest element of b M ( n −
1) in the previousexpression of d n − ( K ) with non-zero coefficient. Now we write b = w y , where w is theshortest prefix of b such that t w / ∈ B . Hence u w = v t w ∈ M ( n ). Hence we set s n − ( K ) := a u w ⊗ y + s n − X j ∈ I m a j u j ⊗ b j − d n ( a u w ⊗ y ) . (4.1.6)Below the differentials d n and the maps s n will be denoted simply by d and s .4.2. Application to Nichols algebras.
We consider now the Anick resolution of aNichols algebra B q of diagonal type associated to the presentation given by PBW gener-ators ( x β ) β ∈ ∆ q + . We fix a convex order on ∆ q + , which induces a total order on the letters: x β > x β > · · · > x β m . We set N β = ord q ββ .The defining relations of B q are x N β β = 0 , N β finite;(4.2.1) (cid:2) x β i , x β j (cid:3) c = X n i +1 ,...,n j − ∈ N c ( i,j ) n i +1 ,...,n j − x n j − β j − . . . x n i +1 β i +1 , i < j. (4.2.2)where c ( i,j ) n i +1 ,...,n j − ∈ k can be computed explicitly [An1, Lemma 4.5].We denote by x β , β ∈ ∆ q + , the set of letters for the tips and the chains, to distinguishthem from the generators x β of the Nichols algebra B q . The minimal tips are the following: x α x β , α > β ∈ β ∈ ∆ q + ; x N β β , β ∈ ∆ q + . (4.2.3) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 23
Hence the Anick graph looks locally as 1 z z ✈✈✈✈✈✈✈✈✈✈ $ $ ❍❍❍❍❍❍❍❍❍❍ x α (cid:4) (cid:4) / / x β (cid:5) (cid:5) x N α − α C C : : ✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉ x N β − β C C for α > β , where if N α = 2, then the loop between x α and x N α − α is understood to becollapsed to a loop from x α to itself, and similarly for x β . If N α >
3, then there existvertices x tα and x N α − tα for each 2 ≤ t ≤ N α −
2, and arrows between them. These verticesare not connected to 1 and are omitted.Now we describe the set of chains. For each δ ∈ ∆ q + we set f δ : N → N , f δ (2 k ) = N β k, f δ (2 k + 1) = N β k + 1 , k ∈ N . (4.2.4)The set of all n -chains, n ∈ N , is given by M ( n ) = n x f β ( n ) β x f β ( n ) β . . . x f βm ( n m ) β m : X n j = n o . (4.2.5)4.3. Quantum linear spaces.
These are the less complicated Nichols algebras of diagonaltype. Let q = ( q ij ) i,j ∈ I be as above and assume that q ij q ji = 1 for all i = j ∈ I . Let I ′ = { i ∈ I : q ii ∈ G ′∞ } and for i ∈ I ′ , set N i = ord q ii . Then B q is presented by generators y i , i ∈ I , with relations y i y j = q ij y j y i , i < j ∈ I , (4.3.1) y N i i = 0 , i ∈ I ′ . (4.3.2) Proposition 4.3.3.
The cohomology ring of B q is generated by η i for i ∈ I and ξ j , for j ∈ I ′ with relations ξ h ξ j = q N h N j hj ξ j ξ h , h < j ∈ I ′ ,η i ξ j = q N j ij ξ j η i , i ∈ I , j ∈ I ′ ,η i η k = − q ik η k η i , i < k ∈ I ,η i = 0 , i ∈ I , N i > ,η i ξ i = ξ i η i , i ∈ I ′ . (4.3.4) If N i = 2 , then η i = ξ i . If M is a finitely generated B q -module, then H( B q , M ) is finitelygenerated as a module over H( B q , k ) . Proof.
We will construct the Anick resolution K q of k as a B q -module. The followingnotation will be helpful. For each i , 1 ≤ i ≤ θ , let σ i , τ i : N → N be the functions definedby σ i ( a ) = , if a is odd N i − , if a is even , and τ i ( a ) = a X j =1 σ i ( j ) for a ≥ τ (0) = 0.We claim that the differential d of the Anick resolution is given as follows: d ( y τ ( a )1 · · · y τ θ ( a θ ) θ ⊗
1) = θ X i =1 Y ℓ
1. Identify these functions with thecorresponding elements in H ( S, k ) and H ( S, k ), respectively. We claim that ξ i , η i generateH( S, k ). We also denote by ξ i and η i the corresponding chain maps ξ i : K n → K n − and η i : K n → K n − given by ξ i ( y τ ( a )1 · · · y τ θ ( a θ ) θ ⊗
1) = Y ℓ>i q − N i τ ℓ ( a ℓ ) iℓ y τ ( a )1 · · · y τ i ( a i − i · · · y τ θ ( a θ ) θ ⊗ η i ( y τ ( a )1 · · · y τ θ ( a θ ) θ ⊗
1) = Y ℓi ( − a ℓ q − τ ℓ ( a ℓ ) iℓ y τ ( a )1 · · · y τ i ( a i − i · · · y τ θ ( a θ ) θ ⊗ y σ i ( a i ) − i . Note this implies that if a i is even and N i >
2, then η i ( y τ ( a )1 · · · y τ θ ( a θ ) θ ⊗
1) = 0, since y N i − i acts as 0 on k . Calculations show that these maps satisfy the following equations:(4.3.5) ξ i ξ j = q N i N j ij ξ j ξ i , η i ξ j = q N j ij ξ j η i , and η i η j = − q ij η j η i for all i < j , and η i ξ i = ξ i η i for all i , with one exception: If N i = 2, then η i = ξ i (sothat we may leave ξ i out of our choice of generators, or not, as is convenient), whileif N i >
2, then η i = 0. Due to these equations, any element in the subalgebra ofExt ∗ S ( k, k ) generated by the ξ i and η i may be written as a linear combination of ele-ments of the form ξ b · · · ξ b θ θ η c · · · η c θ θ with b i ≥ c i ∈ { , } . Such an element takes OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 25 y b N + c · · · y b θ N θ + c θ θ ⊗ S -basis elementsof K P (2 b i + c i ) to 0. Recall that the dimension of H n ( S, k ) is (cid:0) n + θ − θ − (cid:1) ; consequently, theelements ξ b · · · ξ b θ θ η c · · · η c θ θ form a k -basis for H( S, k ).For the last statement, we may induct on the length of a composition series of M , andit suffices to prove the statement in the case that M is a simple module. The generatorsof B q are all nilpotent, and so the only simple module is the trivial module k , for whichthe statement is clear. (cid:3) Cohomology of graded algebras with convex PBW basis.
Here we consider agraded connected algebra R = ⊕ n ∈ N R n with a finite PBW-basis B = B ( { } , X , <, h ); thatis X is a finite subset of R with r = | X | elements, < a total order on X (with a numeration X = { x , x , . . . } such that i < j if and only if x i < x j ) and a function h : X → N ∪ {∞} , x i N i ( h is called the height), such that B = (cid:8) x e r r x e r − r − . . . x e x e : 0 ≤ e i < N i (cid:9) is a k -basis of R . Let I = { , . . . , r } and I ′ = { i ∈ I : N i < ∞} . We assume that theelements of X are homogeneous: x j ∈ R d j , d j ∈ N . We setdeg b = ( e , . . . , e r , X j e j d j ) ∈ N r +10 , b = x e r r x e r − r − . . . x e x e ∈ B. Let (cid:22) be the lexicographical order, reading from the right, on N r +10 . We consider the N r +10 -filtration on R given by R f = h b ∈ B : deg b (cid:22) f i , f = ( f , . . . , f r +1 ) ∈ N r +10 . Inspired by [DCK], we also assume that the PBW-basis B is convex , i.e. ( R f ) f ∈ N r +10 is analgebra filtration. It can be shown that the PBW-basis B is convex if and only if(a) for every i, j ∈ I with i < j , there exists q ij ∈ k such that x i x j = q ij x j x i + X f ≺ deg x i +deg x j R f ;(4.4.1)(b) for every i ∈ I ′ , x N i i ∈ X f ≺ N i deg x i R f . (4.4.2)See [AAH]. We call this the PBW-filtration. Assume that in (4.4.1), q ij = 0 for all i < j .Then the associated graded algebra S := gr R is a quantum linear space, i.e. it is presentedby generators y i (the class of x i ) and relations (4.3.1), (4.3.2).The PBW basis gives rise to the Anick resolution computing H( R, k ) and H(
R, M ) for an R -module M as in § x ℓ i N i i are chains of cohomologicaldegree 2 ℓ i . In the next theorem we prove that if the dual cochains ( x ℓ i N i i ) ∗ are cocycles,then H ∗ ( R, k ) has fgc.
Theorem 4.4.3.
Let R be a graded connected algebra with a finite convex PBW-basissatisfying all of the assumptions above. Suppose that there exist positive integers ℓ i forany i with N i < ∞ , such that the cochains (cid:16) x ℓ i N i i (cid:17) ∗ are cocycles on the Anick resolution, that is, represent elements in H( R, k ) . Then H( R, k ) is finitely generated and H( R, M ) isfinitely generated as a module over H( R, k ) for any finitely generated R -module M .Proof. Observe that gr R is a quantum linear space. By Proposition 4.3.3, the cohomologyH(gr R, k ) is finitely generated over its subalgebra generated by ξ ℓ i i for all i ∈ I ′ , since itis generated by all ξ i , η i . Since the Anick resolution for R is compatible with the PBW-filtration on R , there exists an associated spectral sequence E convergent to H( R, k ) whose E -page is H(gr R, k ); see Theorem 3.4.3. Moreover, the cochains ( x ℓ i N i i ) ∗ are the imagesof ξ ℓ i i in the spectral sequence (see the proof of Proposition 4.3.3) and so by assumption,the ξ ℓ i i are permanent cycles. Thus the hypotheses of Lemma 3.3.2 are satisfied, and,hence, H( R, k ) is left Noetherian. (That is, gr H( R, k ) is Noetherian, from which it followsthat H( R, k ) is Noetherian.) Finite generation follows from Lemma 3.1.1 taking K to bea trivial algebra there.If M is a finitely generated R -module, then since the Condition 3.4.1 is satisfied, Theo-rem 3.4.3 implies that H( R, M ) is finitely generated as an H( R, k )-module. (cid:3) The following corollary is immediate since Nichols algebras of diagonal type have convex
P BW bases.
Corollary 4.4.4.
Let B q be a finite-dimensional Nichols algebra of diagonal type. If B q satisfies Condition 1.4.1, then it has fgc. (cid:3) Cohomology of the Drinfeld double
In this section we prove that if the bosonization of a Nichols algebra of diagonal typehas fgc then so does its Drinfeld double.We briefly recall the general definition of a twisting map τ for two algebras A and B :Let τ : B ⊗ A → A ⊗ B be a bijective k -linear map for which τ (1 B ⊗ a ) = a ⊗ B , τ ( b ⊗ A ) = 1 A ⊗ b for all a ∈ A and b ∈ B , and the following compositions of maps from B ⊗ B ⊗ A ⊗ A to A ⊗ B are equal: τ ◦ ( m B ⊗ m A ) = ( m A ⊗ m B )(1 ⊗ τ ⊗ τ ⊗ τ )(1 ⊗ τ ⊗ , where m A (respectively, m B ) denotes multiplication on A (respectively, on B ), and 1denotes an identity map. The twisted tensor product algebra A ⊗ τ B is A ⊗ B as a vectorspace, and its multiplication is the composition ( m A ⊗ m B )(1 ⊗ τ ⊗ A and B are Hopf algebras, we say that τ is a Hopf twisting if A ⊗ τ B is a Hopf algebrawith coalgebra structure being the usual tensor product of coalgebras (no twisting), and A , B are Hopf subalgebras. The augmentation map is ǫ A ⊗ ǫ B : A ⊗ τ B → k .We assume that the Hopf twisting τ is compatible with coradical filtrations, that is for C A ⊂ C A ⊂ · · · and C B ⊂ C B ⊂ · · · the coradical filtrations of A and B , we have τ ( C Bb ⊗ C Aa ) ⊂ X r + s ≤ a + b C Ar ⊗ C Bs . Then the associated graded space gr( A ⊗ τ B ) is again a Hopf algebra.We will need a special case of the twisting construction to apply it to Nichols algebras:Assume that A and B are graded by abelian groups Γ and Γ ′ . Let t : Γ × Γ ′ → k × be a OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 27 bicharacter (that is, it induces a homomorphism Γ ⊗ Z Γ ′ → k × of abelian groups). Define τ : B ⊗ A → A ⊗ B by τ ( b ⊗ a ) = t ( | a | , | b | ) a ⊗ b for all homogeneous a ∈ A , b ∈ B , where | a | ∈ Γ, | b | ∈ Γ ′ denote grading. In order to distinguish a twisted tensor product algebra A ⊗ τ B for which the twisting τ is defined by a bicharacter t in this way, we will write A ⊗ t B for this twisted tensor product algebra.Due to the following result of Bergh and Oppermann [BO, Theorem 3.7], the cohomologyof A ⊗ t B can be computed. Theorem 5.0.1.
Let A and B be augmented algebras graded by abelian groups Γ and Γ ′ .Let t be a bicharacter on Γ × Γ ′ . There is a twisting map ˆ t , induced by the bicharacter t ,for which H( A ⊗ t B, k ) ∼ = H( A, k ) ⊗ ˆ t H( B, k ) . Let A = R k Γ and B = ( R k Γ) with R = B ( V ). Let D = D ( A ) be the Drinfelddouble of A . Since A and B are subalgebras of D = D ( A ) and, as a vector space, D isisomorphic to A ⊗ B , there is an isomorphism of algebras, D ∼ = A ⊗ τ B, where A ⊗ τ B is a twisted tensor product algebra whose twisting map τ : B ⊗ A → A ⊗ B is defined to correspond to multiplication in D . The augmentation map on D ( A ) is ǫ D ( A ) = ǫ A ⊗ ǫ B ) . Recall that A and B are both coradically graded. With respect to the coradical filtrationon D , there is indeed an isomorphismgr D ∼ = A ⊗ t B for some bicharacter t on grading groups of A and B . The bicharacter t is defined bythe braiding c and the group action. See, for example, [Be]. Hence, as a consequence ofTheorem 5.0.1, H ∗ (gr D, k ) can be computed in terms of H ∗ ( A, k ) and H ∗ ( B, k ). Moreover,since gr D is a Hopf algebra, its cohomology is graded commutative so the bicharacter ˆ t takes values ± D has finitely generated cohomology we will establish that H ∗ ( D, k ) has“enough” cocycles and apply Evens Lemma (3.3.2).Let P q , Q q be bar resolutions of k as an A -module and as a B -module. Then as in [BO,SW] for left modules, we may form the twisted tensor product resolutions P q ⊗ τ Q q and P q ⊗ t Q q of k as a right A ⊗ τ B -module and a right A ⊗ t B -module, respectively. We recallhere briefly this construction, and translate to right modules: As a complex of vectorspaces, each of P q ⊗ τ Q q and P q ⊗ t Q q is simply P q ⊗ Q q , and it remains to define the A ⊗ τ B -and A ⊗ t B -module structures on each vector space P i ⊗ Q j . We will do this for A ⊗ τ B ,and A ⊗ t B is similar. For each j , define τ j : Q j ⊗ A → A ⊗ Q j by iterating τ . The rightmodule structure is defined by the following composition of maps:(5.0.2) P i ⊗ Q j ⊗ A ⊗ B ⊗ τ j ⊗ / / P i ⊗ A ⊗ Q j ⊗ B ρ Pi ⊗ ρ Qj / / P i ⊗ Q j , where ρ P i and ρ Q j denote the module structure maps. Lemma 5.0.3.
Let f ∈ Hom A ( P i , k ) be a cocycle. Then f extends to a cocycle representingan element in H ∗ ( D, k ) . A similar statement holds for H ∗ ( B, k ) .Proof. The first statement will follow from the construction of the resolution P q ⊗ τ Q q andthe definitions. The second statement involves switching the order of P q and Q q . Thisasymmetry in the proof is due to the asymmetry of choosing to work with right modulesinstead of left modules.Let f ∈ Hom A ( P i , k ) be a cocycle. We first claim that f ⊗ ǫ B , as a function on P i ⊗ τ Q = P i ⊗ τ B , is an A ⊗ τ B -module homomorphism.Consider(5.0.4) ( f ⊗ ǫ B )(( x ⊗ y ) · ( a ⊗ b ))where x ∈ P i , a ∈ A , and b, y ∈ B . Expression (5.0.4) can be evaluated by first applying1 ⊗ τ ⊗ x ⊗ y ⊗ a ⊗ b , then applying f ⊗ ǫ A ⊗ ǫ B ⊗ ǫ B to the result, since f is an A -module homomorphism and A acts trivially on k . We wish to show this is equal to(5.0.5) (( f ⊗ ǫ B )( x ⊗ y )) · ( a ⊗ b ) = ( f ⊗ ǫ B )( x ⊗ y ) ǫ D ( A ) ( a ⊗ b ) = f ( x ) ǫ B ( y ) ǫ A ( a ) ǫ B ( b ) . It suffices to show this for all y from a set of generators of B , for all a from a set ofgenerators of A , and for all x ∈ P i and b ∈ B . If either a or y is an element of the field k ,the expression (5.0.4) is equal to( f ⊗ ǫ B )(( x · a ) ⊗ ( y · b )) = f ( x · a ) ǫ B ( y · b ) = f ( x ) ǫ B ( y ) ǫ A ( a ) ǫ B ( b ) , as desired. Now assume that a and y are generators in the kernel of the augmentationmaps for A and B , respectively, so each is either a root vector or a difference of groupelements. The case where either is a difference of group elements is straightforward sinceapplying ǫ to either of the middle two factors will yield 0 after applying τ . Now assumethat a and y are both root vectors. Then by Proposition 2.5.2 we have that in D ( A ), ya = λay + κ (1 − gχ )where λ, κ are scalars, g ∈ Γ, χ ∈ b Γ. Then ( f ⊗ ǫ B )(( x ⊗ y ) · ( a ⊗ b )) =( f ⊗ ǫ A ⊗ ǫ B ⊗ ǫ B )( λx ⊗ a ⊗ y ⊗ b + κx ⊗ ⊗ ⊗ b − κx ⊗ g ⊗ χ ⊗ b ) = λf ( x ) ǫ A ( a ) ǫ B ( y ) ǫ B ( b ) + κ ( f ( x ) ǫ B ( b ) − f ( x ) ǫ A ( g ) ǫ B ( χ ) ǫ B ( b ) =0where the first term disappears since ǫ B ( y ) = 0 and the second two terms cancel out since ǫ A ( g ) = ǫ B ( χ ) = 1. We also have f ( x ) ǫ B ( y ) ǫ A ( a ) ǫ B ( b ) = 0 since ǫ B ( y ) = 0. Thereforethe expressions (5.0.4) and (5.0.5) are equal, as desired. It follows that f ⊗ ǫ B is an A ⊗ τ B -module homomorphism, that is, f ⊗ ǫ B ∈ Hom A ⊗ τ B ( P i ⊗ Q , k ).By hypothesis, 0 = d ∗ i +1 ( f ) = f d i +1 where d i +1 : P i +1 → P i is the differential. Letting d denote the differential on P q ⊗ τ Q q , d ∗ ( f ⊗ ǫ B ) = ( f ⊗ ǫ B )( d i +1 ⊗ − i ⊗ d ) = f d i +1 ⊗ ǫ B + ( − i f ⊗ ǫ B d = 0 . Therefore f ⊗ ǫ B is a cocycle representing an element of H( D, k ).Now let g ∈ Hom B ( Q j , k ) be a cocycle representing an element of H( B, k ). Note that A ⊗ τ B ∼ = B ⊗ τ − A . Let Q q ⊗ τ − P q be the twisted tensor product resolution corresponding OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 29 to this inverse twisting τ − . By the above arguments, g ⊗ ǫ is a cocycle representing anelement of H( A ⊗ τ B, k ). Since P q ⊗ τ Q q is quasi-isomorphic to Q q ⊗ τ − P q (in fact, acomparison map is given by iterating τ ), there is a cocycle g ′ defined on the resolution P q ⊗ τ Q q corresponding to g ⊗ ǫ on Q i ⊗ τ − P . Note that in general g ′ will not equal ǫ A ⊗ g ,due to the twisting. (cid:3) Theorem 5.0.6.
If the Nichols algebra R = B ( V ) and its dual R have fgc, then theDrinfeld double D = D ( B ( V ) k Γ) of the bosonization B ( V ) k Γ has fgc.Proof. Let A = R k Γ and B = A . By hypothesis, A and B have finitely generated co-homology, specifically, the cohomology H( A, k ) is a finite module over a finitely generatedcommutative subalgebra, and similarly for H( B, k ). Choose generators of these commuta-tive subalgebras and representative cocycles on P q and Q q ; we will use these in a spectralsequence argument in combination with Theorem 5.0.1.As a consequence of the Theorem 5.0.1, A ⊗ t B has finitely generated cohomology sinceboth A and B do. Of necessity, since A ⊗ t B is also a Hopf algebra, H(gr( A ⊗ τ B ) , k )is graded commutative, and so ˆ t will in the end only take values ±
1. We will next showthat A ⊗ τ B also has finitely generated cohomology. This relies on existence of neededcocycles. Let f ∈ Hom A ( P i , k ) be a cocycle representing an element of H( A, k ). Then byLemma 5.0.3, f extends to a cocycle representing an element of H( A ⊗ τ B, k ). A similarstatement holds for H( B, k ).Next note that the filtration on A ⊗ τ B induces a filtration on the resolution P q ⊗ τ Q q . Let E be the corresponding spectral sequence. Page E is H( A ⊗ t B, k ), which by Theorem 5.0.1is isomorphic to H( A, k ) ⊗ ˆ t H( B, k ). The cohomology H( A ⊗ τ B, k ) is the homology of thetotal complex of the bicomplex Hom A ⊗ τ B ( P q ⊗ τ Q q , k ) . By the above argument, a cocycle f ∈ Hom A ( P i , k ) representing an element of H i ( A, k )may be extended to a cocycle f ⊗ ǫ ∈ Hom A ⊗ τ B ( P i ⊗ τ Q , k ) representing an element ofH( A ⊗ τ B, k ). This is thus a permanent cocycle in the spectral sequence E . Moreover, itcorresponds to f ⊗ ǫ , this time representing an element of the E -page H( A, k ) ⊗ ˆ t H( B, k ).Similarly, a cocycle g ∈ Hom B ( Q j , k ) representing an element of H j ( B, k ) may be extendedto a cocycle g ′ ∈ Hom A ⊗ τ B ( P ⊗ τ Q j , k ) representing an element of H( A ⊗ τ B, k ). Thus weobtain, for each chosen generator of H(gr A ⊗ τ B, k ), a permanent cocycle in the spectralsequence. Applying the spectral sequence Lemma 3.3.2, since H( A ⊗ t B, k ) is a finite moduleover a finitely generated (commutative) subalgebra, H( A ⊗ τ B, k ) is finitely generated.(A commutative subalgebra can be found by taking high enough powers of the chosengenerators since the defining parameters and thus also the values of the bicharacter ˆ t areall roots of unity.)Now let M be a finitely generated A ⊗ τ B -module. Then H( A ⊗ τ B, M ) is a gradedmodule over H( A ⊗ τ B, k ). The coradical filtration on A ⊗ τ B induces a filtration on thebar resolution K q of k as A ⊗ τ B -module and thus on Hom A ⊗ τ B ( K q , M ). Let f E ∗ be thecorresponding spectral sequence. Arguing as in [Ja, § I.9.13], we get a spectral sequence e E ∗ M = H(gr A ⊗ τ B, k ) ⊗ M ⇒ H( A ⊗ τ B, M ) . which is a module over the spectral sequence f E ∗ . By Lemma 3.3.2, H( A ⊗ τ B, M ) is finitelygenerated over H( A ⊗ τ B, k ). (cid:3) Part
II.
Permanent cocycles for Nichols algebras of diagonal type
In this Part we deal with
Condition 1.4.1.
Let U be a braided vector space of diagonal type whose Nicholsalgebra is finite-dimensional. For every positive root γ ∈ ∆ U + , there exists L γ ∈ N suchthat the cochain (cid:16) x L γ γ (cid:17) ∗ is a cocycle on the Anick resolution, i.e. represents an element in H( B ( U ) , k ) . We shall prove that Condition 1.4.1 holds for one representative U of each Weyl-equivalence class in the classification of [H2]. By Theorem 4.4.3 this shows that B ( U )has fgc and as explained in § U ; in other words we often assume that the root γ has full support, i.e. supp γ = I .Towards this, we choose the representative U in the Weyl-equivalence class in such a waythat Condition 1.4.1 was already verified for any proper subdiagram.We discuss the strategy in Section 6, a summary been given in § wk (4) and br (2)). The remaining Nichols algebras of diagonal typein the classification are dealt with in Part III [AAPW].6. The strategy
The setup.
Let q be the braiding matrix of U and denote by B q the correspondingNichols algebra as before. For δ ∈ ∆ q + recall that N δ = ord q δδ . Recall that the set of n -chains, n ∈ N , is given by M ( n ) = n x f δ ( n ) δ x f δ ( n ) δ . . . x f δm ( n m ) δ m : X n j = n o , (4.2.5)where for δ ∈ ∆ q + we introduce f δ : N → N by f δ (2 k ) = N δ k, f δ (2 k + 1) = N δ k + 1 , k ∈ N . The starting point is the following straightforward observation.
Remark . Let γ ∈ ∆ q + and L ∈ N .(a) x Lγ is a chain if and only if L is of the form ℓN γ or ℓN γ + 1, for some ℓ ∈ N .(b) A cochain ( x Lγ ) ∗ is a cocycle if and only if for any chain c ∈ M ( n + 1) such that d ( c ⊗ ∈ V ( n ) ⊗ B q , when written as a linear combination of basis elements, theterm x Lγ ⊗ L = ℓN γ for ℓ ∈ N . OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 31
We reduce the set of chains c ∈ M ( n + 1) to be considered in (b) using degree and gradingconstraints. First, since the relations of B q are N I -homogeneous by definition we have: Lemma 6.1.2.
The differential of the Anick resolution preserves the N I -grading. (cid:3) Let c = x f δ ( n ) δ x f δ ( n ) δ . . . x f δm ( n m ) δ m ∈ M ( n + 1) such that d ( c ⊗
1) = . . . + λ x ℓN γ γ ⊗ . . . , λ = 0 , as a linear combination of basis elements. By Lemma 6.1.2 and since x ℓN γ γ is a 2 ℓ -chain,we have the following constraints: f δ ( n ) δ + · · · + f δ m ( n m ) δ m = ℓN γ γ, (6.1.3) n + · · · + n m = 2 ℓ + 1 . (6.1.4)Henceforth we refer to the conditions (6.1.3) and (6.1.4) on the chains as the N I -gradingand homological degree constraints . Writing the roots as linear combinations ofsimple roots, (6.1.3) and (6.1.4) boil down to a system of equations on { n , . . . , n m , ℓ } .Let γ ∈ ∆ q + and ℓ ∈ N . We summarize now the approaches to verify that ( x ℓN γ γ ) ∗ is acocycle using Remark 6.1.1 (b). Thus we need to consider the chains c ∈ M ( n + 1) in (b)up to degree and grading constraints. ◦ We introduce integers P γ , Q γ in § N γ > P γ , Q γ , then ( x N γ γ ) ∗ is a cocycle of degree2, by Lemma 6.2.4. Here ℓ = 1.We may assume that γ is not simple, otherwise it is covered by the previous discussion.Then there are β, δ ∈ ∆ q + such that β < γ < δ , γ = β + δ . If N γ is small, typically 2 or 3,then inequality ◦ Assume that N γ = 2 and condition (6.2.8) holds. Then ( x N γ γ ) ∗ is a 2-cocycle by a directcomputation, see Lemma 6.2.7. Again ℓ = 1. ◦ Condition (6.2.8) is about the relations between the root vectors x β , x γ , x δ . If it doesnot hold, then a finer analysis is needed. We summarize in Proposition 6.3.2 all possiblecases that we need to check in this setting when N γ = 2. ◦ Similarly we summarize in Proposition 6.3.23 all possible cases to check when N γ > Degree 2 cocycles.
We discuss two techniques to get generators of degree 2 incohomology from root vectors. First we introduce P γ and Q γ that under suitable conditionsimply the existence of the cocycles. Definition 6.2.1.
Let γ ∈ ∆ q + . We define P γ = max { P ∈ N : ∃ distinct δ , δ , δ ∈ ∆ q + such that δ + δ + δ = P γ } ,Q γ = max { Q ∈ N : ∃ distinct δ , δ ∈ ∆ q + such that N δ δ + δ = Qγ } . We set P γ = 0, respectively Q γ = 0 if no such relation exists. Example 6.2.2. If γ is simple, then P γ = Q γ = 0. Also, if γ is not simple, then P γ ≥ . (6.2.3)For, since γ is not simple, γ = β + δ for some distinct β, δ ∈ ∆ q + , hence 2 γ = γ + β + δ . Lemma 6.2.4.
Let γ ∈ ∆ q + . If N γ > P γ , Q γ , then ( x N γ γ ) ∗ is a cocycle of degree . Inparticular, if γ is simple, then ( x N γ γ ) ∗ is a cocycle of degree .Proof. To show that ( x N γ γ ) ∗ is a cocycle, we use Remark 6.1.1 (b). That is, we show thatthere is no chain c ∈ C such that x N γ γ ⊗ d ( c ⊗ ∈ C ⊗ A .The chains in C are of the form x N β +1 β , x N β β x δ , x β x N δ δ and x β x δ x η . We consider thesecases separately.For c = x N β +1 β , d ( c ) = x N β β ⊗ x β , which is not of the required form.Let c = x N β β x δ . If x N γ γ ⊗ d ( c ⊗ N β β + δ = N γ γ, which contradicts the assumption N γ > Q γ . The case x β x N δ δ is similar.Finally, if c = x β x δ x η , and x N γ γ ⊗ d ( c ⊗ β + δ + η = N γ γ which again contradicts the assumption N γ > P γ . (cid:3) Because of the previous Lemma we need to compute P γ and Q γ ; this is simplified viathe following result. Let W be the Weyl groupoid of the Nichols algebra B q , see [H1] or[AA]. Lemma 6.2.5.
Let δ , δ , δ ∈ ∆ q + . Then there exist w ∈ W and τ ∈ S θ such that w ( δ i ) ∈ ∆ p + ∩ (cid:0) Z γ τ (1) + Z γ τ (2) + Z γ τ (3) (cid:1) , i = 1 , , , for a suitable p .Proof. See [CuH, Theorem 2.3]. (cid:3)
Lemma 6.2.6.
Assume that q is of Cartan type and that γ is not simple. (a) In types A θ , D θ and E θ , we have P γ = 2 and Q γ = 1 . (b) In types B θ , C θ and F , we have P γ ≤ and Q γ = 2 . (c) In type G , P γ ≤ and Q γ ≤ . OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 33
Proof. (a) Type A : Let γ = γ ij with i < j . Suppose that there exists P ∈ N such that P γ ij = γ kℓ + γ mn + γ su with k ≤ m ≤ s (and the three roots in the right are different).Then the coefficient of γ k in the right hand side is at most 3, so P ≤
3. If P = 3, then k = m = s = i and ℓ , n , u are all different. If, say, u is the largest of them, then thecoefficient of γ u in the right hand side is 1, a contradiction. Thus P γ = 2 by (6.2.3). Next,suppose that there exists P, t ∈ N such that P γ ij = tγ kℓ + γ mn (and the two roots in theright are different). Arguing as before, we see that P ≤
1, and (6.2.3) applies. Types
D, E :this follows from Type A and Lemma 6.2.5.(b) Type B : Let γ , . . . , γ θ be the simple roots with γ θ the short root. Then the rootscome in three flavors: γ i + . . . + γ j , i < j < θ , γ i + . . . + γ θ , and γ i + . . . + γ j − +2 γ j + . . . +2 γ θ , i < j ≤ θ . Hence, the maximum P is 3 and P γ ≤
3; similarly, Q γ ≤
2. Note that P γ = 3and Q γ = 2 can occur, e.g.( γ θ − + γ θ − ) + ( γ θ − + γ θ − + γ θ ) + ( γ θ − + γ θ − + 2 γ θ ) = 3( γ θ − + γ θ − + γ θ ) . Type C : The coefficient of γ θ in γ is 0 or 1, but in the former, γ belongs to a sub-diagramof type A θ − that was already settled. Looking at the coefficient of γ θ in both sides of δ + δ + δ = P γ , we conclude that P ≤
3. Similarly, Q γ ≤
2. Note that P γ = 3 and Q γ = 2 can occur, e.g.(2 γ θ − + 2 γ θ − + γ θ ) + ( γ θ − + γ θ − + γ θ ) + γ θ = 3( γ θ − + γ θ − + γ θ ) . Type F : this follows from Types B and C and Lemma 6.2.5.(c) By inspection. (cid:3) By Lemma 6.2.4 we may assume γ is not simple. If N γ = 2, then the Lemma does notapply. The second technique provides a direct computation of the differential of a suitablechain in this setting. Let s = s n be as in (4.1.6). Lemma 6.2.7.
Let γ ∈ ∆ q + be such that N γ = 2 and the following conditions hold: (a) For all β, δ ∈ ∆ q + , β < δ , γ = β + δ , x β x γ = q βγ x γ x β , x γ x δ = q γδ x δ x γ , q ββ = q δδ . (6.2.8)(b) If γ , γ , γ ∈ ∆ q + are three different roots, γ i = γ , then γ + γ + γ = 2 γ . (c) If γ , γ ∈ ∆ q + , γ = γ , then N γ γ + γ = 2 γ .Then ( x γ ) ∗ is a cocycle of degree two.Proof. By Remark 6.1.1 we have to check that the coefficient of x γ ⊗ d ( c ⊗
1) is zerofor all 3-chains c of degree 2 γ . By (b) and (c) we have to deal with c = x β x γ x δ , where β, δ ∈ ∆ q + , β < δ , γ = β + δ : Here we use the convexity to deduce that β < γ < δ .Fix β, δ ∈ ∆ q + such that β < δ and γ = β + δ . By (4.2.2), x β x δ = q βδ x δ x β + b x γ + X β<ν ≤···≤ ν k <δ : P ν i = γ b ν ,...,ν k x ν k . . . x ν for some b x γ , b ν ,...,ν k ∈ k . Using the convexity again we see that if ν ≤ · · · ≤ ν k are suchthat P ν i = γ , then ν < γ < ν k .By definition of the differential on 2-chains and (a), d ( x β x δ ⊗
1) = x β ⊗ x δ − q βδ x δ ⊗ x β − bx γ ⊗ − X b ν ,...,ν k x ν k ⊗ x ν k − . . . x ν , d ( x β x γ ⊗
1) = x β ⊗ x γ − q βγ x γ ⊗ x β , d ( x γ x δ ⊗
1) = x γ ⊗ x δ − q γδ x δ ⊗ x γ . Using these computations and (a), d ( x β x γ x δ ⊗
1) = x β x γ ⊗ x δ − sd (cid:0) x β x γ ⊗ x δ (cid:1) = x β x γ ⊗ x δ − s (cid:0) x β ⊗ x γ x δ − q βγ x γ ⊗ x β x δ (cid:1) = x β x γ ⊗ x δ − s (cid:16) q γδ x β ⊗ x δ x γ − q βγ x γ ⊗ (cid:0) q βδ x δ x β + b x γ + X b ν ,...,ν k x ν k . . . x ν (cid:1)(cid:17) = x β x γ ⊗ x δ − q γδ x β x δ ⊗ x γ − s (cid:16) ( q γδ − q βγ ) bx γ ⊗ x γ − q βγ q βδ x γ ⊗ x δ x β + X b ν ,...,ν k ( q γδ x ν k ⊗ x ν k − . . . x ν x γ − q βγ x γ ⊗ x ν k . . . x ν ) + q βγ q βδ q γδ x δ ⊗ x γ x β (cid:17) = x β x γ ⊗ x δ − q γδ x β x δ ⊗ x γ + ( q βγ − q γδ ) bx γ ⊗ q βγ q βδ x γ x δ ⊗ x β + X b ν ,...,ν k q βγ x γ x ν k ⊗ x ν k − . . . x ν − s (cid:16) X b ν ,...,ν k x ν k ⊗ (cid:0) q γδ x ν k − . . . x ν x γ − q βγ q γν k x γ x ν k − . . . x ν (cid:1) − X b ν ,...,ν k q βγ s ( f x γ ,x νk x ν k − . . . x ν ) (cid:17) Here f x γ ,x νk = [ x γ , x ν k ] c − x γ x ν k + q γν k x ν k ⊗ x γ . We claim that d ( x β x γ x δ ⊗
1) = x β x γ ⊗ x δ − q γδ x β x δ ⊗ x γ + ( q βγ − q γδ ) bx γ ⊗ q βγ q βδ x γ x δ ⊗ x β + X b ν ,...,ν k q βγ x γ x ν k ⊗ x ν k − . . . x ν . According with the previous computation we should prove that s annihilates x ν k ⊗ (cid:0) q γδ x ν k − . . . x ν x γ − q βγ q γν k x γ x ν k − . . . x ν (cid:1) , s ( f x γ ,x νk x ν k − . . . x ν ) . (6.2.9)For the elements on the right of (6.2.9) we use that s = 0. For the elements on the leftof (6.2.9), due to the convexity of the PBW basis, x ν k − . . . x ν x γ and x γ x ν k − . . . x ν arelinear combinations of products x µ j . . . x µ , with ν ≤ µ ≤ · · · ≤ µ j ≤ ν k − and µ j ≤ γ .Hence µ j ≤ ν k , so they are linear combinations of x ν k ⊗ x µ j . . . x µ = s ( x ν k x µ j . . . x µ ) , and we use again that s = 0. Finally, using the claim and that q βγ − q γδ = q ββ q βδ − q βδ q δδ (6.2.8) = 0 , the coefficient of x γ ⊗ (cid:3) Higher degree cocycles.
We now assume that we are not in the situations of § c ∈ M (2 ℓ + 1) satisfying the degree and grading constraints(6.1.3) and (6.1.4) and verify that the condition in Remark 6.1.1 (b) is satisfied. As before γ ∈ ∆ q + is fixed.Let f δ : N → N be the function defined in (4.2.4) for δ ∈ ∆ + . OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 35 N γ = 2 . Here the constraints (6.1.3) and (6.1.4) take the form X δ ∈ ∆ + f δ ( n δ ) δ = Lγ, X δ ∈ ∆ + n δ = L + 1 . (6.3.1)In the following mega statement we collect all possible conditions that we may need toverify on γ to conclude that ( x Lγ ) ∗ is an L -cocycle. We explain the scheme of the proof upto the specific computation of differentials that is postponed to Section 10. Proposition 6.3.2.
Let L = 2 ℓ ∈ N even. Assume that each solution ( n δ ) δ ∈ ∆ + ∈ N ∆ + ofthe equations (6.3.1) is of one of the forms (A) , (B) , (C) , (D) , (E) , (F) , (G) , (H) , (I) or (J) . Then ( x Lγ ) ∗ is an L -cocycle. (A) n γ = L − , n α = n β = 1 and n ϕ = 0 for the other ϕ ∈ ∆ + ; where α, β ∈ ∆ + satisfy α < β, α + β = γ, (6.3.3) the corresponding PBW generators satisfy (10.1.8) ,and L satisfies ( L ) − qααqββ = 0 . (6.3.4)(B) n γ = L − , n α = n β = n δ = 1 and n ϕ = 0 for the other ϕ ∈ ∆ + . where α, β, δ, η ∈ ∆ + satisfy α < η < γ < β < δ, γ + η = α + β, η + δ = γ, (6.3.5) the corresponding PBW generators satisfy (10.1.11) ,and L satisfies c ( L ) αβγ := L X k =0 ( − e q αγ ) k ( k + 1) e q βγ = 0 . (6.3.6)(C) n γ = L − , n α = 2 , n β = n δ = 1 and n ϕ = 0 for the other ϕ ∈ ∆ + where α, β, δ, η, τ satisfy α < η < γ < τ < β < δ, γ + τ = α + β, η + δ = γ,N α = 2 α + τ = γ + η, η + β = 2 τ, (6.3.7) the corresponding PBW generators satisfy (10.1.16) ,and L satisfies ( L ) e q γα e q γβ + L − X j =1 c ( j ) ατγ = 0 . (6.3.8)(D) n γ = L − , n α = n β = n δ = 1 and n ϕ = 0 for the other ϕ ∈ ∆ + , where α, β, δ, η, τ ∈ ∆ + satisfy α < β < γ < τ < η < δ, α + δ = γ + η, β + δ = η + τ,γ + δ = 2 η + τ, η + β = γ. (6.3.9) the corresponding PBW generators satisfy (10.1.24) ,and L satisfies c ( L ) αβγ = 0 . (6.3.10)(E) n γ = L − , n α = n β = n δ = n η = 1 and n ϕ = 0 for the other ϕ ∈ ∆ + , where α, β, δ, τ, µ, η ∈ ∆ + satisfy α < β < δ < γ < τ < µ < η, α + µ = γ = β + τ, η + δ = γ + τ + µ. (6.3.11) the corresponding PBW generators satisfy (10.1.33) ,and L satisfies d ( L ) αβδγ := L − X k =0 e q kδγ ( k + 1) e q αγ ( k + 2) e q βγ = 0 . (6.3.12)(F) n γ = L − , n α = n β = n δ = 1 and n ϕ = 0 for the other ϕ ∈ ∆ + , where α, β, δ, η ∈ ∆ + satisfy α < η < γ < β < δ, γ + η = α + δ, η + β = γ, (6.3.13) the corresponding PBW generators satisfy (10.1.43) ,and L satisfies c ( L ) αδγ = 0 . (6.3.14)(G) n γ = L − , n α = n β = n δ = 1 and n ϕ = 0 for the other ϕ ∈ ∆ + , where α, β, δ, η ∈ ∆ + satisfy α < β < γ < η < δ, γ + η = β + δ, η + α = γ, (6.3.15) the corresponding PBW generators satisfy (10.1.47) ,and L satisfies c ( L ) − δαγ = 0 . (6.3.16)(H) n γ = L − , n α = n β = n δ = 1 and n ϕ = 0 for the other ϕ ∈ ∆ + , where α, β, δ, η, τ, µ, ν ∈ ∆ + satisfy α < τ < β < γ < µ < ν < η < δ, α + δ = η + τ, β + δ = ν + γ,β + η = µ + γ, α + ν = γ, (6.3.17) the corresponding PBW generators satisfy (10.1.53) ,and L satisfies c ( L ) − δαγ = 0 . (6.3.18)(I) n γ = L − , n α = n β = n δ = n η = 1 and n ϕ = 0 for the other ϕ ∈ ∆ + , where α, β, ν, µ, δ, η ∈ ∆ + satisfy α < β < ν < γ < µ < δ < η, β + δ = γ + ν, ν + η = µ + γ, α + µ = γ, (6.3.19) the corresponding PBW generators satisfy (10.1.61) ,and L satisfies d ( L ) α + β,δ,α,γ = 0 . (6.3.20)(J) n γ = L − , n α = n β = n δ = n η = 1 and n ϕ = 0 for the other ϕ ∈ ∆ + , where α, β, δ, η, ν, µ ∈ ∆ + satisfy α < β < δ < γ < µ < ν < η, β + η = γ + ν, δ + ν = µ + γ, α + µ = γ, (6.3.21) the corresponding PBW generators satisfy (10.1.61) ,and L satisfies d ( L ) β − ν αγ = 0 . (6.3.22) Proof. As N γ = 2, x nγ ⊗ n -chain for all n ∈ N ; hence x Lγ ⊗ L + 1)-chains of degree Lγ are one of the following forms: x α x L − γ x β , for a pair ( α, β )satisfying (6.3.3); x α x L − γ x β x δ , for a 4-tuple ( α, β, δ, η ) satisfying (6.3.5); x α x L − γ x β x δ , fora 5-tuple ( α, β, δ, η, τ ) satisfying (6.3.7), x α x β x L − γ x δ , for a 5-tuple ( α, β, δ, η, τ ) satisfying(6.3.9), x α x β x δ x L − γ x η , for a 6-tuple ( α, β, δ, τ, ϕ, η ) satisfying (6.3.11), x α x L − γ x β x δ , for OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 37 a 4-tuple ( α, β, δ, η ) satisfying (6.3.13), x α x β x L − γ x δ , for a 4-tuple ( α, β, δ, η ) satisfying(6.3.15), x α x β x L − γ x δ , for a 7-tuple ( α, β, δ, η, τ, µ, ν ) satisfying (6.3.17), x α x β x L − γ x δ x η , fora 6-tuple ( α, β, ν, µδ, η ) satisfying (6.3.19). ◦ Fix a pair ( α, β ) satisfying (6.3.3). To simplify the notation, call ζ := − q αγ q γβ = − q αα q ββ .We can apply Lemma 10.1.7 since conditions (10.1.8) hold by hypothesis. Assume firstthat L = 2 a + 1 is odd. Hence the coefficient of x Lγ ⊗ d ( x α x L − γ x β ⊗
1) is b q aγβ n ( − ζ −
1) ( a ) ( − ζ ) − ( − ζ ) a o = − b q L − γβ (cid:8) (1 + ζ ) ( a ) ζ + ζ a (cid:9) = − b q L − γβ ( L ) ζ . If L = 2 a is even, then the coefficient of x Lγ ⊗ d ( x α x L − γ x β ⊗
1) is b q a − γβ ( − ζ −
1) ( a ) ( − ζ ) = − b q L − γβ (1 + ζ ) ( a ) ζ = − b q L − γβ ( L ) ζ . By (A) such coefficient is zero in both cases. ◦ Fix a 4-tuple ( α, β, δ, η ) satisfying (6.3.5). We can apply Lemma 10.1.10 since conditions(10.1.11) hold by hypothesis. Hence the coefficient of x Lγ ⊗ d ( x α x L − γ x β x δ ⊗
1) iszero by (B). ◦ Fix a 5-tuple ( α, β, δ, η, τ ) satisfying (6.3.7). We can apply Lemma 10.1.15 since condi-tions (10.1.16) hold by hypothesis. Hence the coefficient of x Lγ ⊗ d ( x α x L − γ x β x δ ⊗ ◦ Fix a 5-tuple ( α, β, δ, η, τ ) satisfying (6.3.9). We can apply Lemma 10.1.23 since condi-tions (10.1.24) hold by hypothesis. Hence the coefficient of x Lγ ⊗ d ( x α x β x L − γ x δ ⊗ ◦ Fix a 6-tuple ( α, β, δ, τ, ϕ, η ) satisfying (6.3.11). We can apply Lemma 10.1.32 since con-ditions (10.1.33) hold by hypothesis. Hence the coefficient of x Lγ ⊗ d ( x α x β x δ x L − γ x η ⊗
1) is zero by (E). ◦ Fix a 4-tuple ( α, β, δ, η ) satisfying (6.3.13). We can apply Lemma 10.1.42 since (10.1.43)holds by hypothesis. Hence the coefficient of x Lγ ⊗ d ( x α x L − γ x β x δ ⊗
1) is zero by (F). ◦ Fix a 4-tuple ( α, β, δ, η ) satisfying (6.3.15). We apply Lemma 10.1.46 since (10.1.47)holds by hypothesis: The coefficient of x Lγ ⊗ d ( x α x β x L − γ x δ ⊗
1) is zero by (G). ◦ Fix a 7-tuple ( α, β, δ, η, τ, µ, ν ) satisfying (6.3.17). We can apply Lemma 10.1.52 since(10.1.53) holds by hypothesis: The coefficient of x Lγ ⊗ d ( x α x β x L − γ x δ ⊗
1) is zero by(H). ◦ Fix a 6-tuple ( α, β, ν, µ, δ, η ) satisfying (6.3.19). We apply Lemma 10.1.60 since (10.1.61)holds by hypothesis: The coefficient of x Lγ ⊗ d ( x α x β x L − γ x δ x η ⊗
1) is zero by (I). ◦ Fix a 6-tuple ( α, β, δ, η, ν, µ ) satisfying (6.3.21). We apply Lemma 10.1.68 since (10.1.69)holds by hypothesis: The coefficient of x Lγ ⊗ d ( x α x β x δ x L − γ x η ⊗
1) is zero by (J).Thus the coefficient of x Lγ ⊗ d ( c ) is zero for all c ∈ M ( L + 1) and Remark 6.1.1applies. (cid:3) N γ > . We carry out a similar analysis when the assumption is N γ > Proposition 6.3.23.
Let γ ∈ ∆ + be such that N γ > and for all pairs ( α, β ) ∈ ∆ suchthat α < β and α + β = ( N γ − γ, (6.3.24) the corresponding PBW generators satisfy (10.1.8) . Let f δ : N → N be the function de-fined in (4.2.4) for each δ ∈ ∆ + . Assume that ℓ ∈ N satisfies the following two conditions: (a) For each pair ( α, β ) ∈ ∆ satisfying (6.3.24) the scalars q αγ and q γβ satisfy (cid:0) q αγ q γβ − (cid:1) ( ℓ ) ( qαγqγβ ) Nγ = 0 . (6.3.25)(b) The solutions ( n δ ) δ ∈ ∆ + ∈ N ∆ + of the equations X δ ∈ ∆ + f δ ( n δ ) δ = ℓN γ γ, X δ ∈ ∆ + n δ = 2 ℓ + 1(6.3.26) are all of the form n γ = 2( ℓ −
1) + 1 , n α = n β = 1 , for any pair ( α, β ) satisfying (6.3.24) , and n δ = 0 for the remaining δ ∈ ∆ + .Then ( x ℓN γ γ ) ∗ is a ℓ -cocycle.Proof. Fix a pair of positive roots ( α, β ) satisfying (6.3.24). We can apply Lemma 10.1.7since conditions (10.1.8) hold by hypothesis and conclude that the coefficient of x LN γ γ ⊗ d ( x α x N γ ( L − γ x β ⊗
1) is zero by (a).By (b) all (2 L + 1)-chains of degree LN γ γ are of the form x α x N γ ( L − γ x β , for a pair( α, β ) satisfying (6.3.24). Thus the coefficient of x LN γ γ ⊗ d ( c ) is zero for all c ∈ M (2 L +1)and Remark 6.1.1 applies. (cid:3) Next we deal with the scalars c ( L ) αβγ and d ( L ) αβδγ . Given r, s, t ∈ k , let c ( L ) r,s := L − X k =0 r k ( k + 1) s , d ( L ) r,s,t := L − X k =0 r k ( k + 1) s ( k + 2) t . (6.3.27)Notice that c ( L ) αβγ = c ( L ) e q αγ , e q βγ and d ( L ) αδβγ = d ( L ) e q αγ , e q βγ , e q δγ . Lemma 6.3.28. (a)
Assume that ( L ) r = 0 . Then c ( L ) r,s = − s c ( L ) s,r (b) Assume that ( L ) r = 0 = ( L ) s and rs = 1 . Then c ( L ) r,s = 0 . (c) Assume that ( L ) r = 0 = ( L ) s for L ≥ and rs = 1 . Then d ( L ) r,s,s = 0 .Proof. For (a) we compute c ( L ) r,s = L − X k =0 r k (cid:16) k X j =0 s j (cid:17) = L − X i =0 s i (cid:16) L − X k = i r k (cid:17) = L − X i =1 s i (cid:16) ( L ) r − ( i ) r (cid:17) = − L − X i =1 s i ( i ) r = − L − X k =0 s k +1 ( k + 1) r = − s L − X k =0 s k ( k + 1) r = − s c ( L ) s,r . OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 39
Now (b) follows using (a). Indeed, we have that c ( L ) r,s = − s c ( L ) s,r = rs c ( L ) r,s ; as rs = 1 byhypothesis, we have that c ( L ) r,s = 0.Next we deal with (c). As rs = 1 and ( L ) r = ( L ) s = 0, we have that ( L ) rs = 0; thus c ( L ) rs,s = 0 by (b). Also, (cid:0) L +12 (cid:1) s = 0. Then we compute d ( L ) r,s,s = (2) s L − X k =0 r k (cid:18) k + 22 (cid:19) s = (2) s L − X k =0 r k (cid:16)(cid:18) k + 12 (cid:19) s + s k ( k + 1) s (cid:17) = (2) s L − X k =1 r k (cid:18) k + 12 (cid:19) s + (2) s c ( L ) rs,s = (2) s L − X j =0 r j +1 (cid:18) j + 22 (cid:19) s = (2) s r L − X j =0 r j (cid:18) j + 22 (cid:19) s = r d ( L ) r,s,s . As r = 1 we have that d ( L ) r,s,s = 0. (cid:3) For each δ ∈ ∆ + , let a δi ∈ N be the coordinate of α i in δ : that is, δ = P i ∈ I a δi α i .6.4. Summary of the algorithm.
The procedure is as follows: ◦ We fix one type in the classification. We choose a representative of the Weyl-equivalencewith the care that the proper subdiagrams were already treated. ◦ We fix γ ∈ ∆ q + . We assume that γ has full support; recall that Lemma 6.2.4 takes careof simple roots. ◦ We compute N γ , P γ , Q γ .Then we apply one of the following criteria:(I) If N γ > P γ , Q γ , then ( x N γ γ ) ∗ is a cocycle of degree 2.(II) If N γ = 2 and (6.2.8) holds, then ( x N γ γ ) ∗ is a 2-cocycle.(III) Assume that N γ = 2 but (6.2.8) does not hold. We define L γ = lcm (cid:16) { } ∪ (cid:8) ord( − q αα q ββ ) : γ = α + β, α, β ∈ ∆ q + (cid:9)(cid:17) . We find all families ( n δ ) δ ∈ ∆ q + of non-negative integers satisfying (6.3.1) with L = L γ .We check that any of these families ( n δ ) δ ∈ ∆ q + has one of the forms (A), . . . , or (J)in Proposition 6.3.2. Then ( x L γ γ ) ∗ is a cocycle.(IV) Assume that N γ > L γ = lcm (cid:16) { N γ } ∪ (cid:8) ord (cid:18)(cid:0) q αγ q γβ (cid:1) N γ (cid:19) : ( N γ − γ = α + β, α, β ∈ ∆ q + (cid:9)(cid:17) . We find all families ( n δ ) δ ∈ ∆ q + of non-negative integers satisfying (6.3.26) with L = L γ . We check that any of these families ( n δ ) δ ∈ ∆ q + has one of the forms (a), . . . inProposition 6.3.23. Then ( x L γ γ ) ∗ is a cocycle. Actually we distinguish two classes of types in the classification of finite-dimensionalNichols algebras of diagonal type. In the first the braiding matrices have continuousparameters and correspondingly the values of N γ might depend on these parameters. Thenarguments by hand are needed. These are the types treated in Sections 7, 8 and 9.The second class consists of the remaining types where the braiding matrices are so tosay discrete. For them we compute N γ , P γ , Q γ and the suitable families ( n δ ) δ ∈ ∆ q + usinga computer program developed by H´ector Pe˜na Pollastri towards these goals. We thencheck whether ( n δ ) δ ∈ ∆ q + has one of the forms (A), . . . , or (J), respectively (a), . . . by handusing the defining relations, or at least the convex order, of B q . The types in this classare those treated in Part III (to appear later).The implicit numeration of any generalized Dynkin diagram is from the left to the rightand from bottom to top; otherwise, the numeration appears below the vertices.7. Classical types
Types A θ and A ( j | θ − j ) , θ ≥ , j ∈ I ⌊ θ +12 ⌋ . Let q be a root of 1 of order N ≥
2. Inthis subsection, we deal with the Nichols algebra B q of standard diagonal type A θ , that isassociated to the Dynkin diagram q ◦ e q q ◦ e q q ◦ q θ − θ − ◦ e q θ − θ q θθ ◦ where the q ii ’s are either − q or q − and locally the edges are of the following forms: q ◦ q − q − , q − ◦ qq , − ◦ q − q , − ◦ qq − . For more information, see [AA, § § A θ and A ( j | θ − j ), θ ≥ j ∈ I ⌊ θ +12 ⌋ . That is, Proposition 7.1.1.
For every γ ∈ ∆ q + , there exists L γ ∈ N such that ( x L γ γ ) ∗ is a cocycle. We start by setting the notation. Let α ij = X k ∈ I i,j α k , i ≤ j ∈ I . (7.1.2)The set of positive roots is ∆ + = { α i j | i, j ∈ I , i ≤ j } ; this set is ordered lexicographicallyon the subindex ( i, j ). Let r = | ∆ + | = (cid:0) θ +12 (cid:1) ; we have a numeration ∆ + = { β i | i ∈ I r } sothat β k < β ℓ if k < ℓ .We set N ij := ord q α ij , that is N ij = N α if α = α ij . For simplicity we set N i = N ii forall i . Then N ij = ( |{ k ∈ I i,j : q kk = − | is odd, N |{ k ∈ I i,j : q kk = − | is even.(7.1.3)The root vectors are x α ii = x α i = x i , i ∈ I ,x α ij = x ( ij ) = [ x i , x α i +1 j ] c , i < j ∈ I , OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 41 see (2.3.5); we order them lexicographically: x < x (12) < · · · < x < · · · < x θ . Thus { x n θθ θ x n θ − θ ( θ − θ ) x n θ − θ − θ − . . . x n θ (1 θ ) . . . x n | ≤ n ij < N ij } is a PBW-basis of B q . The defining relations in terms of the PBW-generators are x ( ij ) x ( ik ) = q α ij α ik x ( ik ) x ( ij ) , i ≤ j < k ; x ( ik ) x ( jk ) = q α ik α jk x ( jk ) x ( ik ) , i < j ≤ k ; x ( ij ) x ( j +1 k ) = q α ij α j +1 k x ( j +1 k ) x ( ij ) + x ( ik ) , i ≤ j < k ; x ( ij ) x ( kℓ ) = q α ij α kℓ x ( kℓ ) x ( ij ) , i ≤ j < k − ≤ ℓ − x ( iℓ ) x ( jk ) = q α iℓ α jk x ( jk ) x ( iℓ ) , i < j ≤ k < ℓ ; x ( ij ) x ( kℓ ) = q α ij α kℓ x ( kℓ ) x ( ij ) + (1 − e q j,j +1 ) q α ij α kj x ( iℓ ) x ( kj ) , i < k ≤ j < ℓ ; x N ij ( ij ) = 0 , i ≤ j. The relations are homogeneous with respect to the N θ -grading, an observation that willbe useful later. As an example we draw the Cojocaru-Ufnarovski graph that encodes theAnick resolution for the Nichols algebra B q of type A in case N , N , N > u u ❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧ (cid:15) (cid:15) ) ) ❘❘❘❘❘❘❘❘❘❘❘❘❘❘❘ x (cid:3) (cid:3) / / x / / (cid:3) (cid:3) x (cid:3) (cid:3) x N − A A ; ; ✇✇✇✇✇✇✇✇✇✇✇✇✇✇✇ ❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦ x N − A A ; ; ✇✇✇✇✇✇✇✇✇✇✇✇✇✇✇✇ x N − A A Observe that in this case N = N = N and we are in Cartan type. The chains are then C = { x , x , x } ,C = { x N , x N , x N , x x , x x , x x } ,C = { x N +11 , x N +112 , x N +12 , x N x , x N x , x x N , x x N , x x N , x N x , x x x } ,C = { x N , x N , x N , x N x N , x N x x , x N x N , x N +11 x , x N +11 x , x x N +112 , x x N +12 , x x x N , x N +112 x , x N x N , x x N +12 } , and so on. In case N = 2, the loop between x and x N − is understood to be collapsedto a loop from x to itself, and similarly for the other root vectors.In order to apply Remark 6.1.1, we start by the following Claim. Claim 1.
Let α be a non-simple root with N α = 2. Let c ∈ C N +1 ( A ) a chain such that d ( c ) ∈ C N ( A ) has a term x Nα ⊗ c = x a β . . . x a r β r ⊗
1, where a j ∈ N for all j . If N β i >
2, then a i = 0 , X i ∈ I a i = N + 1 . (7.1.4) Indeed, we may safely assume that α = α θ = α + · · · + α θ for simplicity. Since d ishomogeneous, N α θ ∗ = P i ∈ I r a i β i . Assume that S := { i ∈ I r : N β i > , a i > } is non-empty. If i ∈ S , then a i ≥ N by looking at the Anick graph, hence a i = N by ∗ .Also, the supports of the β i with i ∈ S are disjoint. Let K = { k ∈ I : k / ∈ supp β i , i ∈ S } ;observe that K is non-empty because N α = 2, cf. (7.1.3). Now X i/ ∈ S a i β i by ∗ = N α − N X i ∈ S β i = N X k ∈ K α k . Pick k ∈ K and compute the coefficient of α k in the last expression; then X i/ ∈ S a i ≥ X i/ ∈ S ,k ∈ supp β i a i = N. Now the cohomological degree of x a β . . . x a r β r sums up a i for each i / ∈ S and 2 for each i ∈ S (one arrow in, one out). Thus N + 1 = X i/ ∈ S a i + 2 | S | ≥ N + 2 . This contradiction shows that S = ∅ and the claim is proved.The following result on root systems of type A should be well-known but we could notfind a reference. Lemma 7.1.5.
Let γ, γ , . . . , γ n +1 ∈ ∆ + with γ ≤ γ ≤ . . . ≤ γ n +1 . Assume that γ + . . . + γ n +1 = nγ. (7.1.6) Then γ = γ = . . . = γ n = γ, γ = γ + γ n +1 . Proof.
We may assume that γ = α + . . . + α θ for otherwise we reduce to a smaller θ . Sincewe have n +1 roots in the sum which contains each simple root with coefficient exactly n , wemust have that γ , . . . , γ n have α in the support but γ n +1 does not. Similarly, γ , . . . , γ n +1 must have α θ in their support but γ does not. Hence, the supports of γ , . . . , γ n containall simple roots. We conclude that γ = . . . = γ n = γ , thus γ + γ n +1 = γ by (7.1.6). (cid:3) Claim 2.
Let γ ∈ ∆ + and let c ∈ C n ( R ) be a chain of degree n + 1. Assume that(a) d ( c ⊗
1) = . . . + C x nγ ⊗ . . . for some C = 0,(b) The polynomial degree of c is n + 1.Then there exist α, β ∈ ∆ + c = x α x n − γ x β , γ = α + β. Proof.
By (b), we may write c = x γ . . . x γ n +1 ⊗
1, where γ ≤ γ ≤ . . . ≤ γ n +1 ∈ ∆ + .Since d is N θ -homogeneous, (7.1.6) holds by (a). Thus Lemma 7.1.5 applies. (cid:3) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 43
Claim 3.
Let α, β, γ ∈ ∆ + with γ = α + β . Let ℘ = − q αα q ββ . Assume that N γ = 2. Then d ( x α x j − β x γ ⊗
1) = x α x j − γ ⊗ x β − q γ,β x α x j − γ x β ⊗ x γ + ( − j q α,β q j − α,γ x j − γ x β ⊗ x α − q j − γ,β ( j ) ℘ x jγ ⊗ . Proof.
The defining relations say that x α x γ = q α,β x γ x α + x γ , x α x β = q α,γ x β x α , x β x γ = q γ,β x γ x β . Hence we are in the setting of Lemma 10.1.7. We consider two cases.First we assume that j = 2 a + 1 is odd. The only thing we need to prove is that thecoefficient of x jγ ⊗ q j − γ,β ((cid:18) q α,γ q γ,β − (cid:19) ( a ) ( qα,γqγ,β ) − (cid:18) q α,γ q γ,β (cid:19) a ) = q j − γ,β (( − ℘ − a ) ( − ℘ ) − ( − ℘ ) a )= − q j − γ,β (1 + ℘ )(1 + ℘ + . . . + ℘ a − + ℘ a ) = − q j − γ,β (2 a + 1) ℘ = − q j − γ,β ( j ) ℘ . Next we assume that j = 2 a + 2 is even. We do a similar calculation for the coefficientof x jγ ⊗ q j − γ,β (cid:18) q α,γ q γ,β − (cid:19) ( a + 1) (cid:18) qα,γqγ,β (cid:19) = q j − γ,β ( − ℘ − a + 1) ( − ℘ ) = − q j − γ,β (1 + ℘ )(1 + ℘ + . . . + ℘ a ) = − q j − γ,β (2 a + 2) ℘ = − q j − γ,β ( j ) ℘ . (cid:3) Proof of Proposition 7.1.1.
First, if N γ >
2, then ( x N γ γ ) ∗ is a cocycle in degree 2 byLemma 6.2.4, cf. Lemma 6.2.6 (a). Let γ be a non-simple root with N γ = 2.By Remark 6.1.1, it suffices to show that there is no c ∈ C N ( R ) such that d ( c ) contains x Nγ ⊗ c satisfiesthe hypothesis (b) of Claim 2. Hence c = x α x N − γ x β ⊗ α and β . But x Nγ ⊗ d ( x α x N − γ x β ⊗
1) with a non-zero coefficient for any α, β, γ ∈ ∆ + with γ = α + β . Indeed, ℘ = − q αα q ββ = q ± , see (7.1.3); taking j = N , the coefficient in Lemma3 is q j − γ,β ( j ) ℘ = 0. (cid:3) Types B θ and B ( j | θ − j ) , θ ≥ , j ∈ I θ − . Let q be a root of 1 of order N >
2. Inthis subsection, we deal with the Nichols algebras B q of diagonal types B θ or B ( j | θ − j ).In the first case, the Dynkin diagram is q ◦ q − q ◦ q − q ◦ q ◦ q − q ◦ while in the second we may assume that the corresponding diagram is q − ◦ q q − ◦ q − ◦ q − ◦ j q − q ◦ q ◦ q − q ◦ The set of positive roots in both cases is ∆ + = { α ik | i ≤ k ∈ I } ∪ { α iθ + α kθ | i < k ∈ I } . (7.2.1)We fix the following convex order: α < α < · · · < α θ < α θ + α θ < · · · < α θ + α θ (7.2.2) < α < α < · · · < α θ − < α θ − θ < α θ − θ + α θ < α θ . (7.2.3)We set N ik := ord q α ik , M ik := ord q α iθ + α kθ , that is N ik = N α if α = α ik , M ik = N α if α = α iθ + α kθ . Let M = ord q , P = ord( − q ). Then N ik = i ≤ j ≤ k < θ,M i ≤ k < j or j < i ≤ k < θ ; P i ≤ j < k = θ,N j < i < k = θ ; M ik = ( i ≤ j < kM i < k ≤ j or j < i < k. Here we set j = 0 if q is of Cartan type. For more information, see [AA, § § B θ and B ( j | θ − j ), θ ≥ j ∈ I θ − . That is, Proposition 7.2.4.
For every γ ∈ ∆ q + , there exists L γ ∈ N such that ( x L γ γ ) ∗ is a cocycle.Proof. Let γ be a positive non-simple root. Arguing recursively we may assume that γ hasfull support. Hence, either γ = α θ or else there exists k ∈ I ,θ such that γ = α θ + α kθ .First we consider γ = α θ . By direct computation, P γ = 3 and P ′ γ = 1. Hence, if N γ >
3, then Lemma 6.2.4 applies and ( x N γ γ ) ∗ is a 2-cocycle. Next we consider the case N γ = 3: that is, either N = 3 if the braiding is of Cartan type, or else N = 6 if not.All the pairs ( α, β ) as in (6.3.24) are of the form α = α k , β = α θ + α k +1 θ for some k ∈ I θ − . By direct computation, there exists b k ∈ k x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + b k x γ , so the root vectors satisfy (10.1.8). Also, q αγ q γβ = 1 for all of them, so we take L = 1. Nowwe look for solutions of (6.3.26). That is, P δ ∈ ∆ + f δ ( n δ ) δ = 3 γ , P δ ∈ ∆ + n δ = 3. We checkthat there is no solution with n δ = 3 neither with n δ = 2, δ ∈ ∆ + . Hence we are forcedto look for solutions with n γ t = 1 for three different roots γ t ∈ ∆ + (and 0 for the otherroots). We write γ t = P i ∈ I a ( t ) i α i . As P t a ( t ) i = 3 for all i ∈ I , we have a ( t )1 = 1 for all t ∈ I , and either a ( t ) θ = 1 for all t ∈ I , or a ( t ) θ = 2 for some t ∈ I . If a ( t ) θ = 1 for all t ∈ I ,then γ t = γ for all t , a contradiction. Hence we may assume a (1) θ = 0, a (2) θ = 2, a (3) θ = 1.Then γ = α k , γ = α θ + α k +1 θ , γ = γ , for some k ∈ I θ − . Thus Proposition 6.3.23applies and ( x γ ) ∗ is a 2-cocycle.Now we consider γ = α θ + α kθ , k ∈ I ,θ . By direct computation, P γ = 2 and P ′ γ = 1.Hence, if N γ >
2, then Lemma 6.2.4 applies and ( x N γ γ ) ∗ is a 2-cocycle. Next we considerthe case N γ = 2: that is, either N = 4 if the braiding is of Cartan type, or else k > j ifthe braiding is of type B ( j | θ − j ). Let α < β be a pair of positive roots as in (6.3.3). Wehave several possibilities: OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 45 ◦ α = α i − , β = α iθ + α kθ , i < k . By direct computation, there exists b ∈ k such that x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + b x γ , ◦ α = α i − , β = α kθ + α iθ , k < i ≤ θ . There exist b , b t ∈ k such that x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + b x γ + i − X t = k +1 b t x α kθ + α tθ x α t − . ◦ α = α θ , β = α kθ . There exist b , b t ∈ k such that x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + b x γ + θ − X t = k +1 b t x α θ + α tθ x α k t − . In all cases the root vectors satisfy (10.1.8), and − q αα q ββ = q ± , so we take L = M .Next we look for solutions of (6.3.1). In the Cartan case with N = 4, we have P δ ∈ ∆ + n δ = 3: by direct computation we discard the possibility that one n δ ≥
2, soexactly three of them are one. Arguing as in the case γ = α θ we check that all solu-tions are of form n γ = n α = n β = 1 for a pair ( α, β ) satisfying (6.3.3), and n ϕ = 0 forthe remaining ϕ ∈ ∆ + . Next we consider the case B ( j | θ − j ), k > j . We have that P δ ∈ ∆ + : 1 ∈ supp δ f δ ( n δ ) = M . Let η ∈ ∆ + such that 1 ∈ supp η and n η > • If η = α θ , then n η = 1: otherwise, f η ( n η ) ≥ P = 2 M > M = P δ ∈ ∆ + : 1 ∈ supp δ f δ ( n δ ), acontradiction. • If η = α i , i < j , then n η = 1. Suppose on the contrary that n η >
1. Then n η = 2, since f η ( n η ) ≤ M = N η . This implies that n δ = 0 for all δ = α i such that i ∈ supp δ .Let δ ∈ ∆ + such that a δi +1 = 2. Then n δ = 0 since i ∈ supp δ . Thus M = X δ ∈ ∆ + : a δi +1 =1 f δ ( n δ ) , X δ ∈ ∆ + : a δi +1 =1 n δ < M. Then n i +1 ℓ = 2 for some ℓ > i , which implies that n δ = 0 for all δ = α i +1 ℓ such that ℓ ∈ supp δ . Recursively, there exist i = 1 < i = i < i < · · · < i s = j − n δ = ( δ = α i r − i r , r ∈ I s , δ = α i r − i r , supp δ ∩ I j − = ∅ . On the other hand, if a δj = 1, then either N δ = 1, or δ = α iθ , i < j , in which case n δ ≤ N δ = P = 2 M . Hence n δ = f δ ( n δ ) for all δ such that a δj = 1. Using this fact, M = X δ ∈ ∆ + f δ ( n δ ) a δj = X δ ∈ ∆ + : a δj =1 n δ + 2 X δ ∈ ∆ + : a δj =2 f δ ( n δ ) ≤ X δ ∈ ∆ + −{ η } n δ + 2 X δ ∈ ∆ + : a δj =2 f δ ( n δ ) ≤ M − X δ ∈ ∆ + : a δj =2 f δ ( n δ ) . Thus n µ = 0 for some µ ∈ ∆ + such that a µj = 2; but j − ∈ supp µ , a contradiction. • If either η = α i or else η = α θ + α iθ , j ≤ i < θ , then N η = 2, so f η ( n η ) = n η . • If η = α θ + α iθ , i < j , then n η = 1. Otherwise, M = X δ ∈ ∆ + : j − ∈ supp δ f δ ( n δ ) a δj − ≥ f η ( n η ) ≥ M, and we get a contradiction.Hence f δ ( n δ ) = n δ for all δ such that 1 ∈ supp δ . Therefore, X δ ∈ ∆ + : 1 / ∈ supp δ n δ = M + 1 − X δ ∈ ∆ + : 1 ∈ supp δ n δ = M + 1 − X δ ∈ ∆ + : 1 ∈ supp δ f δ ( n δ ) = 1 . That is, there exists a unique root β ∈ ∆ + such that 1 / ∈ supp β and n β = 0; moreover n β = 1 for this root β . Thus f δ ( n δ ) = n δ for all δ and we may translate to solve (6.3.1)as follows: find γ t ∈ ∆ + , t ∈ I M +1 , not necessarily different such that P t γ t = M γ .Write γ t = P i ∈ I a ( t ) i α i . Hence we may assume that a ( t )1 = 1 for t ∈ I M , a ( M +1)1 = 0. As P t ∈ I M +1 a ( t ) k = 2 M and each a ( t ) k ≤
2, at least M − a ( t ) k ’s, either both are 1, or else one of them is 2 and the other is 0. Hence we may assumethat a ( t ) k = 1 for t ∈ I M − . This forces to have γ t = γ for t ∈ I M − , γ M + γ M +1 = γ .Hence all the hypotheses of Proposition 6.3.2 hold, and ( x L γ γ ) ∗ is a cocycle. (cid:3) Type B θ,j standard, j ∈ I θ − . Here ζ ∈ G ′ . In this subsection, we deal with Nicholsalgebras B q of standard type B θ,j . We assume that the corresponding diagram is − ζ ◦ − ζ − ζ ◦ − ζ ◦ − ζ − ◦ j − ζ − ζ ◦ − ζ ◦ − ζ ζ ◦ The set of positive roots is (7.2.1), and we fix the same convex order, see (7.2.2). For moreinformation, see [AA, § B θ,j standard: Proposition 7.3.1.
For every γ ∈ ∆ q + , there exists L γ ∈ N such that ( x L γ γ ) ∗ is a cocycle.Proof. Let γ be a positive non-simple root. Arguing recursively we may assume that γ hasfull support. Hence, either γ = α θ or else there exists k ∈ I ,θ such that γ = α θ + α kθ .First we consider γ = α θ . By direct computation, P γ = 3 and P ′ γ = 2, N γ = 3. All thepairs ( α, β ) as in (6.3.24) are of the form α = α k , β = α θ + α k +1 θ for some k ∈ I θ − . Bydirect computation, there exists b k ∈ k x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + b k x γ , so the root vectors satisfy (10.1.8). Also, q αγ q γβ = 1 for all of them, so we take L = 1. Nowwe look for solutions of (6.3.26). As in the proof of Proposition 8.6.1, there exists a pair( α, β ) as in (6.3.24) such that n γ = n α = n β = 1, and n δ = 0 for the remaining roots δ ∈ ∆ q + . Thus Proposition 6.3.23 applies and ( x γ ) ∗ is a 2-cocycle.Now we consider γ = α θ + α kθ , k ∈ I ,θ . By direct computation, P γ = 2 and P ′ γ = 1.Hence, if k ≤ j , then Lemma 6.2.4 applies since N γ = 6, so ( x γ ) ∗ is a 2-cocycle. Nextwe consider the case k > j , so N γ = 2. The pairs α < β as in (6.3.3) are the same as inProposition 8.6.1, the root vectors satisfy (10.1.8) and − q αα q ββ = − ζ ± , so we take L = 6. OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 47
Also, the same argument as in the case B ( j | θ − j ) in Proposition 8.6.1 shows that thereexists a pair ( α, β ) as in (6.3.24) such that n γ = 4, n α = n β = 1, and n δ = 0 for theremaining roots δ ∈ ∆ q + . Hence all the hypotheses of Proposition 6.3.2 hold, and ( x γ ) ∗ isa cocycle. (cid:3) Type C θ . Let q be a root of 1 of order N >
2. In this subsection, we deal with theNichols algebras B q of diagonal type C θ : the Dynkin diagram is q ◦ q − q ◦ q − q ◦ q ◦ q − q ◦ . The set of positive roots is ∆ + = { α ij | i ≤ j ∈ I } ∪ { α iθ + α j θ − | i ≤ j ∈ I θ − } . (7.4.1)We fix the following convex order: α < α < · · · < α θ − < α θ + α θ − < α θ + α θ − < · · · < α θ + α θ − < α θ < α < α < · · · < α θ − < α θ − θ < α θ − θ + α θ − < α θ . (7.4.2)Let M = ord q . By direct computation, N γ = ( M γ = α iθ + α iθ − or γ = α θ ; N γ = α iθ + α jθ − , i < j, or γ = α ij , ( i, j ) = ( θ, θ ) . For more information, see [AA, § C θ . More precisely, Proposition 7.4.3.
For every γ ∈ ∆ q + , ( x N γ γ ) ∗ is a -cocycle.Proof. Let γ be a positive non-simple root. Arguing recursively we may assume that γ hasfull support. Hence, either γ = α θ or else there exists j ∈ I θ − such that γ = α θ + α j θ − .First we consider γ = α θ . By direct computation, P γ = 3 and P ′ γ = 1. Hence, if N γ >
3, then Lemma 6.2.4 applies and ( x N γ γ ) ∗ is a 2-cocycle.Now assume that N γ = 3: that is, N = 3. The unique pair ( α, β ) as in (6.3.24) is α = α θ + α θ − , β = α θ . By direct computation, x α = [ x θ − , x θ ] c , x γ = x θ , so x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + q ( q − Y i ∈ I θ − q iθ x γ , and then the root vectors satisfy (10.1.8). Also, q αγ q γβ = 1 for all of them, so we take L = 1.Now we look for solutions of (6.3.26). That is, P δ ∈ ∆ + f δ ( n δ ) δ = 3 γ , P δ ∈ ∆ + n δ = 3. Wecheck that there is no solution with n δ = 3 neither with n δ = 2, δ ∈ ∆ + . Hence we look forsolutions with n γ t = 1, for three different roots γ t ∈ ∆ + (and n δ = 0 for the other roots).We write γ t = P i ∈ I a ( t ) i α i . As P t a ( t ) i = 3 for all i ∈ I , we have a ( t ) θ = 1 for all t ∈ I , andeither a ( t )1 = 1 for all t ∈ I , or a ( t )1 = 2 for some t ∈ I . The case a ( t ) θ = 1 for all t ∈ I gives a contradiction: either γ t = γ for all t , or else P t a ( t ) θ − >
3. Hence we may assume a (1)1 = 2, a (2)1 = 0, a (3)1 = 1. Then γ = α θ + α θ − , γ = α θ , γ = γ . Thus Proposition6.3.23 applies and ( x γ ) ∗ is a 2-cocycle. Now we consider γ = α θ + α jθ − , j ∈ I θ − . By direct computation, P γ = 2 and P ′ γ = 1.Therefore, if N γ >
2, then Lemma 6.2.4 applies and ( x N γ γ ) ∗ is a 2-cocycle. Hence we needto study the case N γ = 2: that is, N = 4 and γ = α θ + α θ − . Let α < β be a pair ofpositive roots as in (6.3.3). We have the following possibilities: ◦ α = α i − , β = α θ + α iθ , i ∈ I ,θ − . There exist b , b t ∈ k such that x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + b x γ + θ − X t = i +1 b t x α θ + α t θ − x α t − . ◦ α = α θ , β = α θ − . In this case, x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + x γ . In all cases the root vectors satisfy (10.1.8), and − q αα q ββ = −
1, so we take L = 2.Next we look for solutions of (6.3.1). That is, P δ ∈ ∆ + f δ ( n δ ) δ = 2 γ , P δ ∈ ∆ + n δ = 3. Let µ ∈ ∆ + be such that n µ = 0, θ ∈ supp µ . Notice that2 = X δ ∈ ∆ + : θ ∈ supp δ f δ ( n δ ) a δθ = X δ ∈ ∆ + : θ ∈ supp δ f δ ( n δ ) , so n µ ≤ f µ ( n µ ) ≤
2. Suppose that n µ = 2. Then µ = α iθ + α i θ − for some i ∈ I θ − since N µ = f µ (2) ≤
2, and there exists η = µ such that n η = 1, n δ = 0 if δ = µ, η . But then4 α i − = 2 γ − µ = X δ = µ f δ ( n δ ) δ = f η (1) η = η, a contradiction. Hence n µ = 1 for all µ ∈ ∆ + such that n µ = 0, θ ∈ supp µ . Thenthere exist three different roots γ i ∈ ∆ + such that n γ i = 1, and we may assume that θ ∈ supp γ ∩ supp γ , θ / ∈ supp γ . As γ = γ , we may assume γ = γ , so a γ = 1. Thisimplies that a γ = 1, so γ = α i for some i ∈ I θ − , and a γ = 2, so γ = γ .Hence all the hypotheses of Proposition 6.3.2 hold, and ( x N γ γ ) ∗ is a 2-cocycle. (cid:3) Type D θ . Let q ∈ G ′ N , n ≥
2. Let B q be a Nichols algebra of type D θ . That is, thegeneralized Dynkin diagram of B q has the form q ◦ q ◦ q − q ◦ q − q ◦ q ◦ q − q ◦ q − q − q ◦ The set of positive roots is ∆ q + = { α i j | i ≤ j ∈ I , ( i, j ) = ( θ − , θ ) }∪ { α i θ − + α θ | i ∈ I θ − } ∪ { α i θ + α j θ − | i < j ∈ I θ − } . (7.5.1)We fix the following convex order: α < α < · · · < α θ − < α θ − + α θ < α θ < α θ + α θ − < . . .< α θ + α θ − < α < α < · · · < α θ − < α θ − θ − < α θ − + α θ OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 49 < α θ − θ < α θ − < α θ . For more information, see [AA, § D θ . More precisely, Proposition 7.5.2.
For every γ ∈ ∆ q + , ( x N γ γ ) ∗ is a -cocycle.Proof. By Lemma 6.2.6, P α = 2 and Q α = 1 for all non-simple roots α . Hence, if N > x N γ γ ) ∗ is a 2-cocycle for all roots α .Next we assume N = 2, that is, q = −
1. We will apply Proposition 6.3.2. Let γ bea positive non-simple root. Arguing recursively we may assume that γ has full support.Hence, either γ = α θ or else there exists k ∈ I ,θ − such that γ = α θ + α k θ − . We lookfor pairs α < β ∈ ∆ + such that γ = α + β . Notice that q αα = q ββ = − − q αα q ββ in anycase so we may guess that L = 2.First we consider γ = α θ . All the pairs ( α, β ) as in (6.3.3) are of the form α = α j , β = α j +1 θ for some j ∈ I θ − . By direct computation, x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + x γ , so the root vectors satisfy (10.1.8). Next we look for solutions of (6.3.1). That is, 2 γ = δ + δ + δ , δ i ∈ ∆ + . We write δ j = P i ∈ I a ( j ) i α i . As a ( j ) i is 0 or 1 for i = 1 , θ − , θ , we mayfix a (1)1 = a (2)1 = 1, a (3)1 = 0 and see the possible pairs of roots such that a ( m ) θ − = a ( n ) θ − = 1,respectively a ( p ) θ = a ( r ) θ = 1. Suppose that no one of the δ i ’s has coefficient 1 for the threesimple roots simultaneously. Then we may assume a (1) θ − = a (3) θ − = 1, a (2) θ = a (3) θ = 1, so a ( j ) θ − > j ∈ I , a contradiction. Hence we assume a (1) θ − = a (1) θ = 1, so a (1) i ≥ i ∈ I . If either a (2) θ − = 1 or a (2) θ = 1, then a (2) θ − ≥
1, which implies a (1) θ − = 1 and so δ = γ . Otherwise a (3) θ − = 1 = a (3) θ , then a (3) θ − ≥
1, which implies again δ = γ .Finally, let γ = α θ + α k θ − . Let α < β be a pair of positive roots as in (6.3.3). Thenthe coefficient of α is one for just one of them (and zero for the other): it should be α ,since α < β . We have several possibilities: ◦ α = α j − , β = α jθ + α k θ − , j < k . By direct computation, x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + x γ , ◦ α = α j − , β = α kθ + α j θ − , k < j ≤ θ −
2. There exist b , b t ∈ k such that x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + b x γ + j − X t = k +2 b t x α kθ + α t θ − x α t − . ◦ α = α θ − , β = α kθ − + α θ . There exist b ∈ k such that x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + b x γ . ◦ α = α θ , β = α k θ − . There exist b , b t ∈ k such that x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + b x γ + j − X t = k +1 b t x α θ + α t θ − x α k t − . ◦ α = α θ + α j θ − , β = α k j − , k < j ≤ θ −
2. There exist b , b t ∈ k such that x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + b x γ + j − X t = k +1 b t x α θ + α t θ − x α k t − . In all cases the root vectors satisfy (10.1.8). Next we look for solutions of (6.3.1). Thatis, 2 γ = δ + δ + δ , δ i ∈ ∆ + . We write δ j = P i ∈ I a ( j ) i α i .When k = θ −
2, first consider the case δ = α θ − . Then a ( i ) j = 1 for i = 1 , θ − , θ and j = 2 ,
3, so δ , δ have full support. This implies that a ( i ) j = 1 for 2 ≤ i < θ − j = 2 ,
3, and we need that a ( θ − j = 1 for one of them; that is, either δ = γ or else δ = γ .If δ j = α θ − for all j ∈ I , then2 α θ = 2 s θ − ( γ ) = s θ − ( δ ) + s θ − ( δ ) + s θ − ( δ ) , s θ − ( δ j ) ∈ ∆ + . Applying the previous case, s θ − ( δ j ) = α θ for some j ∈ I , so δ j = γ .If k < θ −
2, then we argue recursively. Indeed, we first consider the case δ = α k andargue as in the case k = θ − δ = γ or else δ = γ . If δ j = α θ − forall j ∈ I , then2 α θ + α k +1 θ − = 2 s k ( γ ) = s k ( δ ) + s k ( δ ) + s k ( δ ) , s k ( δ j ) ∈ ∆ + . Hence s k ( δ j ) = α θ + α k +1 θ − for some j ∈ I , which means that δ j = γ for some j ∈ I .Hence all the hypotheses of Proposition 6.3.2 hold, and ( x γ ) ∗ is a 2-cocycle. (cid:3) Type D ( j | θ − j ) , θ ≥ , j ∈ I θ − . Let q be a root of 1 of order N >
2. In thissubsection, we deal with the Nichols algebras B q of type D ( j | θ − j ). We may assume thatthe corresponding diagram is q − ◦ q q − ◦ q − ◦ q − ◦ j q − q ◦ q ◦ q − q ◦ The set of positive roots is ∆ q + = { α ik | i ≤ k ∈ I } ∪ { α iθ + α k θ − | i < k ∈ I θ − }∪ { α iθ + α iθ − | i ∈ I j +1 ,θ − } . (7.6.1)Thus (7.6.1) is a subset of the set (7.4.1) of positive roots of type C θ : We fix the convexorder in ∆ q + obtained from (7.4.2). For more information, see [AA, § D ( j | θ − j ): Proposition 7.6.2.
For every γ ∈ ∆ q + , there exists L γ ∈ N such that ( x L γ γ ) ∗ is a cocycle.Proof. Let γ be a positive non-simple root. Arguing recursively we may assume that γ hasfull support. Hence, either γ = α θ or else there exists k ∈ I θ − such that γ = α θ + α k θ − .First we consider γ = α θ . By direct computation, N γ = 2, P γ = 3 and P ′ γ = 1. Let α < β be a pair of positive roots as in (6.3.3). Then there exists i ∈ I θ − such that α = α i , β = α i +1 θ . By direct computation, x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + x γ , OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 51 hence the root vectors satisfy (10.1.8), and − q αα q ββ ∈ { q − , q − } , so we take L = N .There exist 4-tuples ( α, β, δ, η ) ∈ ∆ as in (6.3.5): α = α i < η = α θ − < γ < β = α θ + α i +1 θ − < δ = α θ . The corresponding PBW generators satisfy (10.1.11); indeed, there exists b ∈ k such that[ x α , x β ] c = q αβ x β x α + b x γ x η , x η x δ = q ηδ x δ x η + x γ , and the other pairs of root vectors q -commute. Now c ( N ) αβγ = 0 by Lemma 6.3.28 (b) since e q αγ = e q βγ = q − .Next we look for solutions of (6.3.1). We claim that f η ( n η ) = n η for all η ∈ ∆ + . ◦ If N η = 2, then this holds by definition of f η . ◦ If η = α iθ + α k θ − , i < k ≤ θ −
1, then N η = N , so f η ( k ) ≥ N if k ≥
2. As2 f η ( n η ) ≤ X δ ∈ ∆ q + f η ( n η ) a δθ − = N, we have that n η ≤
1, so f η ( n η ) = n η . ◦ Let η = α ik , with i ≤ k < j . Then N η = N . Suppose that n η ≥
2: as f η ( s ) > N if s >
2, we may have n η = 2: moreover, n δ = 0 for all δ = α ik such that δ ∩ I i,k = ∅ since N = X δ ∈ ∆ q + f δ ( n δ ) a δt = N + X δ = α ik : t ∈ supp δ f δ ( n δ ) a δt for all t ∈ I i,k . Now if k + 1 < j , then n δ = 0 for all δ = α k +1 t , t ≥ k + 1: as P δ ∈ ∆ q + f δ ( n δ ) a δk +1 = N ,we have n α k +1 t = 2 for some k + 1 ≤ t < j . Thus we may assume k = j −
1. Let δ ∈ ∆ q + be such that j ∈ supp δ . Then n δ = 0 if j − ∈ supp δ , and N δ = 2 if j − / ∈ supp δ , so N = X δ ∈ ∆ q + : j ∈ supp δ f δ ( n δ ) a δj = X δ : j ∈ supp δ, j − / ∈ supp δ f α jt ( n α jt ) = X δ : j ∈ supp δ, j − / ∈ supp δ n α jt . This implies that P δ ∈ ∆ q + n δ ≥ N + 2, a contradiction. Then n η ≤
1, so f η ( n η ) = n η . ◦ Let η = α ik , j < i ≤ k . Then N η = N and an argument as in the previous case showsthat we have that n η ≤
1, so f η ( n η ) = n η . ◦ Similar situation holds for η = α iθ + α i θ − : N η = M but again n η ≤
1, so f η ( n η ) = n η .As the claim holds, we may rewrite the problem as follows: find γ i ∈ ∆ q + , i ∈ I N +1 ,such that P γ i = N γ . As a δ = 1 if 1 ∈ supp δ , a δθ = 1 if θ ∈ supp δ , there exist θ − , θ ∈ supp γ i : we may fix that 1 , θ ∈ supp γ i for i ≥
3. As N = P N +1 i =3 a γ i θ − and a γ i θ − ≥
1, there exists at most one i ≥ a γ i θ − = 2: • if a γ i θ − = 1 for all i ≥
3, then γ i = γ for all i ≥ γ + γ = γ . • if a γ θ − = 2, then γ i = γ for all i ≥ γ = α θ + α k θ − for some k ∈ I ,θ − . Hence γ + γ = α k − + α θ , so γ , γ are α k − , α θ .Hence all the hypotheses of Proposition 6.3.2 hold, and ( x L γ γ ) ∗ is a cocycle.Now we consider γ = α θ + α i θ − , i ∈ I ,θ − . By direct computation, P γ = 2 and P ′ γ = 1. Let i ≤ j . Then N γ = N >
2, so Lemma 6.2.4 applies and ( x N γ γ ) ∗ is a 2-cocycle.Next we assume that i > j , so N γ = 2. The pairs ( α, β ) as in (6.3.3) are the following: ◦ α = α θ + α k θ , β = α i k − , k ∈ I i +1 ,θ − . By direct computation, there exists b , b t ∈ k x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + b x γ + θ − X t = k +1 b t x α i t − x α θ + α t θ . ◦ α = α θ , β = α i θ − . By direct computation, there exists b ∈ k such that x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + b x γ . ◦ α = α k − , β = α k θ + α i θ − , k ∈ I j +1 ,i − . By direct computation, there exists b , b t ∈ k x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + b x γ + i − X t = k +1 b t x α tθ + α i θ x α t − . ◦ α = α i − , β = α iθ + α i θ − . By direct computation, there exists b ∈ k x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ , x α x β = q αβ x β x α + b x γ . Hence all the pairs of root vectors satisfy (10.1.8) and − q αα q ββ ∈ { q − , q − } , so we take L = N .Now we look for solutions of (6.3.1); i.e. P δ ∈ ∆ + f δ ( n δ ) δ = N γ , P δ ∈ ∆ + n δ = N + 1. Let η ∈ ∆ q + such that 1 ∈ supp η , N η > n η = 0. Suppose that n η ≥
2. Arguingas for the case γ = α θ , there exists t ∈ I j − such that η = α t and n η = 2. Thisimplies that n δ = 0 for all δ ∈ ∆ q + such that supp δ ∩ I t = ∅ . Recursively, there exist t = 0 < t = t < t < · · · < t s = j − n η = 2 if η = α i r − +1 i r , r ∈ I s ,and n δ = 0 for all δ ∈ ∆ q + such that supp δ ∩ I j − = ∅ . Now, if j ∈ supp δ , then either j − δ (so n δ = 0 by the previous argument) or a δj = 1, N δ = 2, so f δ ( n δ ) = n δ . Thus, N = X δ ∈ ∆ q + : j ∈ supp δ f δ ( n δ ) a δj = X δ ∈ ∆ q + : j ∈ supp δ n δ . But then N + 1 = X δ ∈ ∆ q + n δ ≥ n η + X δ ∈ ∆ q + : j ∈ supp δ n δ = N + 2 , a contradiction. Thus we have that f η ( n η ) = n η for all η ∈ ∆ q + such that 1 ∈ supp η sinceeither N η = 2 or else n η ≥
1. From here, N = X δ ∈ ∆ q + :1 ∈ supp δ f δ ( n δ ) a δj = X δ ∈ ∆ q + :1 ∈ supp δ n δ , As P δ ∈ ∆ q + n δ = N + 1, there exists a unique η ∈ ∆ + such that n η = 0 and 1 / ∈ supp η ;moreover, n η = 1. Again we may rewrite the problem as follows: find γ k ∈ ∆ q + , k ∈ I N +1 ,such that P k γ k = N γ . As a δ = 1 if 1 ∈ supp δ , a δθ = 1 if θ ∈ supp δ , there exist θ − , θ ∈ supp γ k : we may fix that 1 , θ ∈ supp γ k for k ≥
3. Also, a γ k i ≤ P t a γ k i = 2 N , so either a γ k i = 2 for exactly N of them and 0 for the remaining one a γ k i ,or else a γ k i = 2 for exactly N − a γ k i ’s. A detailedstudy case-by-case shows that γ i = γ for i ≥
3, and γ + γ = γ . Hence all the hypothesesof Proposition 6.3.2 hold, and ( x Nγ ) ∗ is a cocycle. (cid:3) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 53 Exceptional types
Type E θ . Let q ∈ G ′ N , n ≥
2. Let B q be a Nichols algebra of type E θ , 6 ≤ θ ≤ B q has the form q ◦ q ◦ q − q ◦ q ◦ q − q − q ◦ q − q ◦ θ Here ∆ q = ∆ is a root system of type E θ . We fix the following convex orders on the setsof positive roots: E :1 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , E : roots of support contained in I ordered as for E followed by1 , , , , , , , , , , , , , , , , , , , , , , , , , , E : roots of support contained in I ordered as for E followed by1 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . For more information, see [AA, § E θ . We need first the following result. Lemma 8.1.1.
Let β < δ ∈ ∆ q + be such that γ = β + δ ∈ ∆ q + . (a) If µ ≤ · · · ≤ µ k ∈ ∆ + satisfy P i µ i = β + γ , then either µ ≤ β or else µ k ≥ γ . (b) If µ ≤ · · · ≤ µ k ∈ ∆ + satisfy P i µ i = δ + γ , then either µ ≤ γ or else µ k ≥ δ . Proof.
Let ( · , · ) be the symmetric positive definite form on R θ such that ( ν, ν ) = 2 for all ν ∈ ∆ . Then − ≤ ( µ, µ ′ ) ≤ ν = ν ′ ∈ ∆ . As β + δ ∈ ∆ , ( β, δ ) = −
1; thus ( β, γ ) = 1.Next we prove (a): the proof of (b) is analogous. Let µ ≤ · · · ≤ µ k be such that P i µ i = β + γ . Note that k ≥
2, since β + γ = 2 β + δ / ∈ ∆ . Suppose on the contrary that β < µ ≤ · · · ≤ µ k < γ . Then k ≥
3, since ( µ i , β ) ≤ β + γ, β ) = X i ( µ i , β ) . Assume that k ≥
4. Then there exist j = ℓ such that ( µ j , µ ℓ ) = − β + γ, β + γ ) = X i ( µ i , µ i ) + X i = j ( µ i , µ j ) ≤ X i = j ( µ i , µ j ) . Thus µ j + µ ℓ ∈ ∆ + , µ j < µ j + µ ℓ < µ ℓ and we can replace the set { µ i } i ∈ I k by (cid:0) { µ i } i ∈ I k − { µ j , µ ℓ } (cid:1) ∪ { µ j + µ ℓ } . Hence, recursively, we may assume that k = 3. But using the computer we check that ν + ν + ν = β + γ for all the 3-uples β < ν ≤ ν ≤ ν < γ so we get a contradiction. (cid:3) Proposition 8.1.2.
For every γ ∈ ∆ q + , ( x N γ γ ) ∗ is a -cocycle.Proof. By Lemma 6.2.6, P γ = 2 and Q γ = 1 for all non-simple roots γ . Hence, if N > x N γ γ ) ∗ is a 2-cocycle for all roots γ .Next we assume N = 2, that is, q = −
1. We will apply Lemma 6.2.7. Let γ be a positivenon-simple root. For each pair β < δ ∈ ∆ q + such that γ = β + δ , we have that x β x γ = q βγ x γ x β , x γ x δ = q γδ x δ x γ by (4.2.2) and Lemma 8.1.1, and q ββ = q δδ = −
1. Hence (6.2.8) holds.Let γ , γ , γ ∈ ∆ q + be three different roots such that γ + γ + γ = 2 γ . By Lemma6.2.5 there exists w ∈ W such that the support of w ( γ i ) is of size ≤
3, and a fortiori γ too.As w ( γ ) + w ( γ ) + w ( γ ) = 2 w ( γ ) and these roots are contained in a subdiagram of type A or A × A , we conclude that γ i = γ for some i ∈ I . Using a similar argument we alsocheck that 2 γ + γ = 2 γ for all γ = γ ∈ ∆ q + . Hence all the hypothesis of Lemma 6.2.7hold, and ( x γ ) ∗ is a 2-cocycle. (cid:3) Type F . Let q be a root of 1 of order N > M = ord q . In this section, we dealwith a Nichols algebra B q of Cartan type F , that is associated to the Dynkin diagram q ◦ q − q ◦ q − q ◦ q − q ◦ For more information, see [AA, § { , , , , , , , , , } . The aim of this Section is to prove that Condition 1.4.1 holds for type F . More precisely, Proposition 8.2.1.
For every γ ∈ ∆ q + , ( x N γ γ ) ∗ is a -cocycle. OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 55
Proof.
By induction on the rank it is enough to consider γ with full support. By directcomputation,if γ ∈ { , , , , } : N γ = N, P γ = 3 , P ′ γ = 2;if γ ∈ { , , , , } : N γ = M, P γ = 2 , P ′ γ = 2 . Hence, if
N >
4, then N γ > P γ , Q γ for all γ with full support. Thus ( x N γ γ ) ∗ is a 2-cocyclefor all γ with full support by Lemma 6.2.4.Next we assume N = 4. If γ ∈ { , , , , } , then N γ = 4 >P γ , Q γ , so ( x N γ γ ) ∗ is a 2-cocycle by Lemma 6.2.4.Now we consider γ ∈ { , , , , } . Let α < β be apair of positive roots as in (6.3.3). We have the following possibilities: ◦ γ = 1 α = 3, β = 1
4. There exists b ∈ k such that [ x α , x β ] c = b x γ . ◦ γ = 1 α = 23, β = 1
4. There exist b , b ∈ k such that[ x α , x β ] c = b x γ + b x x . ◦ γ = 1 α = 2 β = 1
4. There exist b , b t ∈ k such that[ x α , x β ] c = b x γ + b x x + b x x . ◦ γ = 1 α = 123, β = 12
4. There exist b , b t ∈ k such that[ x α , x β ] c = b x γ + b x x + b x x + b x x . ◦ γ = 1 α = 12 β = 12
4. There exist b , b t ∈ k such that[ x α , x β ] c = b x γ + b x x + b x x + b x x + b x x . ◦ γ = 1 α = 1 β = 2
4. There exist b , b t ∈ k such that[ x α , x β ] c = b x γ + b x x + b x x + b x x + b x x + b x x . ◦ γ = 1 α = 2 β = 1
34. There exist b ∈ k such that [ x α , x β ] c = b x γ . ◦ γ = 1 α = 12 β = 12
34. There exist b , b ∈ k such that[ x α , x β ] c = b x γ + b x x . ◦ γ = 1 α = 1 β = 2
34. There exist b , b t ∈ k such that[ x α , x β ] c = b x γ + b x x + b x x . ◦ γ = 1 α = 2, β = 1
4. There exists b ∈ k such that [ x α , x β ] c = b x γ . ◦ γ = 1 α = 12, β = 12
4. There exist b , b t ∈ k such that[ x α , x β ] c = b x γ + b x x + b x x . ◦ γ = 1 α = 3, β = 1
34. There exists b ∈ k such that [ x α , x β ] c = b x γ . ◦ γ = 1 α = 123, β = 1234. There exist b , b ∈ k such that[ x α , x β ] c = b x γ + b x x . ◦ γ = 1 α = 1 β = 34. There exist b , b t ∈ k such that[ x α , x β ] c = b x γ + b x x + b x x . ◦ γ = 1 α = 1, β = 12
4. There exist b , b ∈ k such that[ x α , x β ] c = b x γ + b x x . ◦ γ = 1
34, ( α, β ) one of the pairs (12 , , , b ∈ k such that [ x α , x β ] c = b x γ . ◦ γ = 1 , α = 12 β = 12
34. There exists b ∈ k such that [ x α , x β ] c = b x γ . ◦ γ = 1 , α = 12 β = 1234. There exist b , b ∈ k such that[ x α , x β ] c = b x γ + b x x . ◦ γ = 1 , α = 1 β = 2
4. There exist b , b t ∈ k such that[ x α , x β ] c = b x γ + b x x + b x x . ◦ γ = 1 , α = 1 β = 2
34. There exist b , b t ∈ k such that[ x α , x β ] c = b x γ + b x x + b x x + b x x . ◦ γ = 1 , α = 1 β = 234. There exist b , b t ∈ k such that[ x α , x β ] c = b x γ + b x x + b x x + b x x + b x x . ◦ γ = 1 , α = 1 β = 34. There exist b , b t ∈ k such that[ x α , x β ] c = b x γ + b x x + b x x + b x x + b x x + b x x . ◦ γ = 1 , α = 1 β = 4. There exist b , b t ∈ k such that[ x α , x β ] c = b x γ + b x x + b x x + b x x + b x x + b x x + b x x . In all cases [ x α , x γ ] c = 0 = [ x γ , x β ] c , so the root vectors satisfy (10.1.8), and − q αα q ββ = − L = 2.Next we look for solutions of (6.3.1). That is, P δ ∈ ∆ q + f δ ( n δ ) δ = 2 γ , P δ ∈ ∆ q + n δ = 3.Notice that n δ = 3 for all δ ∈ ∆ q + : otherwise n η = 0 for all η = δ , so 2 γ = f δ (3) δ ,a contradiction. Hence, either 2 γ = f γ (2) γ + γ or else 2 γ = γ + γ + γ for some γ i = γ j ∈ ∆ q + . By Lemma 6.2.5 there exists w ∈ W such that γ ′ i = w ( γ i ) have supportin a rank 3 subdiagram, so γ ′ = w ( γ ) has the same support: this subdiagram is either ofCartan type B or C . Looking at the corresponding cases (see the proofs of Propositions8.6.1 and 7.4.3) the solutions for γ ′ are γ ′ = γ ′ , γ ′ + γ ′ = γ ′ . Hence, all solutions for γ areof form n γ = n α = n β = 1 for a pair ( α, β ) satisfying (6.3.3), and n ϕ = 0 for the remaining ϕ ∈ ∆ + . Hence all the hypotheses of Proposition 6.3.2 hold, and ( x L γ γ ) ∗ is a 2-cocycle. OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 57
Finally we consider N = 3. If γ ∈ { , , , , } , then N γ = 3 > P γ , Q γ , so ( x N γ γ ) ∗ is a 2-cocycle by Lemma 6.2.4.Now we consider γ ∈ { , , , , } . Let α < β be a pair ofpositive roots as in (6.3.24). We have the following possibilities: ◦ γ = 1 α = 1 β = 1
4. There exist b ∈ k such that [ x α , x β ] c = b x γ . ◦ γ = 1 α = 1 β = 1
34. There exist b , b ′ ∈ k such that[ x α , x β ] c = b x γ + b ′ x x . ◦ γ = 12 α = 1 β = 2
4. There exist b ∈ k such that [ x α , x β ] c = b x γ . ◦ γ = 12 α = 1 β = 2
34. There exist b , b ′ ∈ k such that[ x α , x β ] c = b x γ + b ′ x x . ◦ γ = 12 α = 1 β = 2
4. There exist b ∈ k such that [ x α , x β ] c = b x γ . ◦ γ = 12 α = 1 β = 34. There exist b , b ′ ∈ k such that[ x α , x β ] c = b x γ + b ′ x x . ◦ γ = 12 α = 1 β = 2
34. There exist b ∈ k such that [ x α , x β ] c = b x γ . ◦ γ = 12 α = 1 β = 4. There exist b , b ′ ∈ k such that[ x α , x β ] c = b x γ + b ′ x x . ◦ γ = 1234, α = 1 β = 34. There exist b ∈ k such that [ x α , x β ] c = b x γ . ◦ γ = 1234, α = 1 β = 4. There exist b , b ′ ∈ k such that[ x α , x β ] c = b x γ + b ′ x x . In all cases [ x α , x γ ] c = 0 = [ x γ , x β ] c , so the root vectors satisfy (10.1.8), and q αγ q γβ = 1, sowe take L = 1.Next we look for solutions of (6.3.1). That is, P δ ∈ ∆ q + f δ ( n δ ) δ = 3 γ , P δ ∈ ∆ q + n δ = 3.Notice that n δ = 3 for all δ ∈ ∆ q + : otherwise n η = 0 for all η = δ , so 3 γ = f δ (3) δ ,a contradiction. Hence, either 3 γ = f γ (2) γ + γ or else 3 γ = γ + γ + γ for some γ i = γ j ∈ ∆ q + . Again we use Lemma 6.2.5 to reduce to rank 3 subdiagrams and lookingat the proofs of Propositions 8.6.1 and 7.4.3 we conclude that all the solutions for γ are ofform n γ = n α = n β = 1 for a pair ( α, β ) satisfying (6.3.24), and n ϕ = 0 for the remaining ϕ ∈ ∆ + . Hence all the hypotheses of Proposition 6.3.23 hold, and ( x N γ γ ) ∗ is a 2-cocycle. (cid:3) Type G Cartan.
Let q be a root of 1 of order N >
3. Set M = ( N, N ; N/ , N. In this section, we deal with a Nichols algebra B q of Cartan type G , with Dynkin diagram q ◦ q − q ◦ . The set of positive roots is ∆ + = { , , , , , } . (8.3.1)We take as generators x , x , as well as x := (ad c x ) x , x := (ad c x ) x ,x := [ x , x ] c , x := (ad c x ) x . (8.3.2)We order these root vectors: x < x < x < x < x < x .The aim of this Section is to prove that Condition 1.4.1 holds for type G . More precisely, Proposition 8.3.3.
For every γ ∈ ∆ q + , ( x N γ γ ) ∗ is a -cocycle.Proof. ◦ For γ = 1
2, the case N > N = 2 (that is, N = 6). We will apply Proposition 6.3.2. The unique pair as in (6.3.3)is α = α , β = 2 α + α , since the following relations hold: x x = x + q q x x , x x = q q x x , x x = q q x x . In this case, − q αα q ββ = − L = 2. By direct computation, the unique solution of(6.3.1) is n = n = n = 1, and n δ = 0 for the remaining roots. Hence Proposition6.3.2 applies and ( x ) ∗ is a 2-cocycle. ◦ For γ = 1
2, the case N > N = 4. Wewill apply Proposition 6.3.23. The unique pair as in (6.3.24) is α = 3 α + α , β = 3 α +2 α ,since the following relations hold: x x = − ( q + 1) q x − q x x , x x = q q x x ,x x = q q x x . In this case, q αγ q γβ = 1 so we take L = 1. By direct computation, the unique solution of(6.3.26) is n = n = n = 1, and n δ = 0 for the remaining roots. Hence Proposition6.3.23 applies and ( x ) ∗ is a 2-cocycle. ◦ For γ = 1 , the case N > N = 2(that is, N = 6). We will apply Proposition 6.3.2. The pairs as in (6.3.3) are α = 2 α + α , β = α + α , and α = 3 α + α , β = α , since the following relations hold: x x = x + q q x x , x x = q x x ,x x = q x x ; x x = − (3) q q x − q x x − q q x x , x x = q x x ,x x = q x x . In both cases, − q αα q ββ = − L = 2. The solutions of (6.3.1) are • n = n = n = 1, and n δ = 0 for the remaining roots, or • n = n = n = 1, and n δ = 0 for the remaining roots.Hence Proposition 6.3.2 applies and ( x ) ∗ is a 2-cocycle. OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 59 ◦ For γ = 12, the case N > N = 4. Wewill apply Proposition 6.3.23. The unique pair as in (6.3.24) is α = 3 α + 2 α , β = α ,since the following relations hold: x x = − q ) q x + qq x x , x x = q q x x ,x x = q q x x . In this case, q αγ q γβ = 1 so we take L = 1. By direct computation, the unique solution of(6.3.26) is n = n = n = 1, and n δ = 0 for the remaining roots. Hence Proposition6.3.23 applies and ( x ) ∗ is a 2-cocycle. (cid:3) Table 1.
The roots with full support of G ; γ < γ = γ + γ < γ γ N γ , Cartan N γ , (8.4.1) b P γ Q γ γ γ L γ , Cartan L γ , (8.4.1) b1 M M N N M M N N Type G standard. Let ζ ∈ G ′ . In this section, we deal with a Nichols algebra B q of standard type G associated to any of the Dynkin diagramsa: ζ ◦ ζ ζ ◦ , b: ζ ◦ ζ − ◦ , c: ζ ◦ ζ − ◦ . (8.4.1)The set of positive roots is again (8.3.1). Thus we take as generators x , x , as well as(8.3.2) with the same order for these root vectors: x < x < x < x < x < x .For more information, see [AA, § G standard: Proposition 8.4.2.
For every γ ∈ ∆ q + , there exists L γ ∈ N such that ( x L γ γ ) ∗ is a cocycle.Proof. We just consider the diagram (8.4.1) b. ◦ For γ = 1
2, ( x ) ∗ is a 2-cocycle by Lemma 6.2.4. ◦ For γ = 1
2, we will use Proposition 6.3.23. The unique pair as in (6.3.24) is α = 3 α + α , β = 3 α + 2 α , since the following relations hold: x x = q x − q x x , x x = − q x x ,x x = ζ q x x . In this case, (cid:16) q αγ q γβ (cid:17) N γ = − L = 2. By direct computation, the unique solutionof (6.3.26) is n = 3, n = n = 1, and n δ = 0 for the remaining roots. HenceProposition 6.3.23 applies and ( x ) ∗ is a 4-cocycle. ◦ For γ = 1 , we will use Proposition 6.3.2. The pairs as in (6.3.3) are α = 2 α + α , β = α + α , and α = 3 α + α , β = α , since the following relations hold: x x = x + ζ q x x ,x x = ζ q x x ,x x = q x x ; x x = ζ (4) ζ q x − q x x − ζ (2) ζ q x x ,x x = ζ q x x ,x x = q x x . In both cases, − q αα q ββ = ζ so we take L = 8. The solutions of (6.3.1) are • n = 7, n = n = 1, and n δ = 0 for the remaining roots, or • n = 7, n = n = 1, and n δ = 0 for the remaining roots.Hence Proposition 6.3.2 applies and ( x ) ∗ is an 8-cocycle. ◦ For γ = 12, ( x ) ∗ is a 2-cocycle by Lemma 6.2.4. (cid:3) Type D (2 , α ) . Here q, r, s = 1, qrs = 1; N = ord q , M = ord r , P = ord s . In thissection, we deal with a Nichols algebra B q of super type D (2 , α ) with Dynkin diagram q ◦ q − − ◦ r − r ◦ . We fix the following convex order on the set of positive roots:1 < < < < < < . For more information, see [AA, § D (2 , α ): Proposition 8.5.1.
For every γ ∈ ∆ q + , there exists L γ ∈ N such that ( x L γ γ ) ∗ is a cocycle.Proof. By induction on the rank it is enough to consider γ with full support. We startwith γ = 123. The following relations between root vectors hold: x x = q q x x + x , x x = qq q x x , x x = − q q x x ; x x = q q x x + x , x x = − q q x x , x x = rq q x x . Hence for each pair ( γ , γ ) ∈ ∆ as in (6.3.3), the corresponding PBW generators satisfy(10.1.8). By direct computation, L = lcm(ord q, ord r ) satifies (A) and (B) of Proposition6.3.2 so ( x L γ γ ) ∗ is a cocycle.The pairs ( α, β ) of positive roots as in (6.3.3) are (1 ,
23) and (12 , x x = q q x x + x , x x = qq q x x , x x = − q q x x ; x x = q q x x + x , x x = − q q x x , x x = rq q x x . the root vectors satisfy (10.1.8). As − q αα q ββ = q , respectively = r , we may take L =lcm { N, M, P } . There exists a 4-tuples ( α, β, δ, η ) ∈ ∆ as in (6.3.5): α = 1 < η = 12 < γ = 123 < β = 12 < δ = 3 . OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 61
The corresponding PBW generators satisfy (10.1.11); indeed, x α x β = q αβ x β x α + qq q (1 − s ) x γ x η , x η x δ = q ηδ x δ x η + x γ , and the other pairs of root vectors q -commute. Now c ( N ) αβγ = 0 by Lemma 6.3.28 (b) since e q αγ = q , e q βγ = s . Next we look for solutions of (6.3.1). By direct computation we see thatthere exist three solutions: ◦ n = L − n = n = 1, n δ = 0 if δ = 123 , , ◦ n = L − n = n = 1, n δ = 0 if δ = 123 , , ◦ n = L − n = n = n = 1, n δ = 0 if δ = 123 , , , x L γ γ ) ∗ is a cocycle.Now we consider γ = 12
3. If N γ > P γ , then Lemma 6.2.4 applies and ( x N γ γ ) ∗ is a2-cocycle. Now assume that N γ = 2. The following relations between root vectors hold: x x = − q q x x + x ,x x = q q x x , x x = q q x x ; x x = − q q q x x − q x + q ( r − x x ,x x = − q q q x x , x x = − q q q x x . Hence for each pair ( α, β ) as in (6.3.3), the PBW generators satisfy (10.1.8). Now L = 2satifies the hypothesis of Proposition 6.3.2, so ( x γ ) ∗ is a 2-cocycle. (cid:3) Type F (4) . Let q be a root of 1 of order N >
3. Set M = ord q , P = ord q .In this subsection, we deal with the Nichols algebras B q of diagonal type F (4). We mayassume that the corresponding diagram is q ◦ q − q ◦ q − q ◦ − ◦ q − We fix the following convex order on ∆ q + :1 , , , , , , , , , , , , , , , , , . For more information, see [AA, § Proposition 8.6.1.
For every positive root γ , there exists a positive integer L γ such thatthe chain ( x L γ γ ) ∗ is a cocycle.Proof. Let γ be a positive non simple root. Arguing recursively we may assume that γ hasfull support. That is, γ ∈ { , , , , } .If γ = 12
4, then N γ = 2 = P γ . First we look for pairs α < β as in (6.3.3). We havethe following posibilities: ◦ ( α, β ) = (2 ,
4) or ( α, β ) = (23 , b ∈ k such that [ x α , x β ] c = b x γ . ◦ ( α, β ) = (12 , b , b ∈ k such that[ x α , x β ] c = b x γ + b x x . ◦ ( α, β ) = (123 , α, β ) = (12 , b , b ∈ k such that[ x α , x β ] c = b x γ + b x x . In all cases [ x α , x γ ] c = 0 = [ x γ , x β ] c , so the root vectors satisfy (10.1.8), and (cid:16) − q αα q ββ (cid:17) N =1. Hence we take L = N .Next we check that (12 , , , , , ) is a 6-uple ( α, β, δ, τ, ϕ, η ) sat-isfying (6.3.11). As e q αγ = q = e q βγ , e q δγ = q − , we have that d ( L ) αβδγ = d ( L ) q − ,q ,q = 0 byLemma 6.3.28 (c). Hence (E) holds in this case.Now we look for 4-uples ( α, β, δ, η ) satisfying (6.3.13). There are three possibilities:(2 , , , ), (2 , , , , , , e q αγ = q , e q βγ = q − . Hence c ( L ) αβγ = c ( L ) q ,q − = 0 by Lemma 6.3.28 (b), and (F) holds.We look for 4-uples ( α, β, δ, η ) satisfying (6.3.15). A possibility is (12 , , , e q αγ = q and e q δγ = q − ; thus c ( L ) − δαγ = c ( L ) q,q = 0 by Lemma 6.3.28 (b), and (G) holds.Now (12 , , , , , , ,
4) is a 7-uple ( α, β, δ, η, τ, µ, ν ) satisfying(6.3.17). As e q αγ = q , e q δγ = q − , we have that c ( L ) − δ,α,γ = c ( L ) q,q = 0 by Lemma 6.3.28 (a).Hence (H) holds in this case.Also, (2 , , , , , , , , , , α, β, ν, µ, δ, η ) satisfying (6.3.19). As e q αγ = e q βγ = q , and e q δγ = q − in both cases,we have that d ( L ) α + β,δ,α,γ = d ( L ) q ,q − ,q = 0 by Lemma 6.3.28 (b). Thus (I) holds in this case.Notice that γ = 2, γ = 12, γ = 12 , γ = γ = 12 satisfy P i ∈ I γ i = 4 γ .Hence, if N = 2, then n = n = n = 1, n γ = n = 2 is a solution of(6.3.1). The coefficient of x γ ⊗ d ( x x x x γ x ⊗
1) by Lemma 10.1.77.Finally we look for solutions of (6.3.1), i.e. P δ ∈ ∆ q + f δ ( n δ ) δ = N γ , P δ ∈ ∆ q + n δ = N + 1.Set η = 12 . Looking at the coefficient of α : N = X δ ∈ ∆ q + f δ ( n δ ) a δ = 2 f η ( n η ) + X δ = η, ∈ supp δ n δ ≥ f η ( n η ) . As N η = M , we have that n η ≤
3. Suppose that n η = 3: necessarily N = 3 M and X δ = η, , ∈ supp δ n δ = 3 M − n − f η (3) = M − n − . Looking at the coefficient of α : X δ :3 ∈ supp δ, / ∈ supp δ f δ ( n δ ) a δ = 6 M − X δ = η, , ∈ supp δ n δ a δ − f η (3) ≤ M − X δ = η, , ∈ supp δ n δ − f η (3) = 2 M + n − < M = N. By inspection, f δ (2) a δ = N δ a δ = N for all δ such that 3 ∈ supp δ, / ∈ supp δ , so n δ ≤ δ . This implies that f δ ( n δ ) = n δ for all δ = η such that 3 ∈ supp δ . Using thisfact, the coefficients of α and α give the following equalities:6 M = X δ :2 ∈ supp δ f δ ( n δ ) a δ = f ( n ) + f ( n ) + 2( M + 1) + X δ = η :2 , ∈ supp δ n δ a δ M = X δ :3 ∈ supp δ f δ ( n δ ) a δ = 3( M + 1) + X δ = η :3 ∈ supp δ n δ a δ From these two equalities: f ( n ) + f ( n ) = M + 1 + X δ = η :3 ∈ supp δ n δ a δ − X δ = η :2 , ∈ supp δ n δ a δ = M + 1 + n + n + n + n + n + n + n ≥ M + 1 . As n + n + n + n ≤ M − n − n , n , n ≤
1, we have that f ( n ) + f ( n ) ≤ M + 1 + 3 + M − n − ≥ M + 2 − n ≤ M − . But this is a contradiction since N = N = P . A similar argument holds if we supposethat n η = 2, so n η ≤ n δ = f δ ( n δ ) for all δ such that 4 ∈ supp δ , so we may translate the equationsto the following problem: Find γ i ∈ ∆ q + , i ∈ I N +1 , such that P i ∈ I N +1 γ i = N γ . As P a γ i = N and a δ ≤ δ ∈ ∆ q + , we may assume that a γ i = 1 for i ∈ I N , a γ N +1 = 0.As P a γ i = 2 N and a δ ≤ δ ∈ ∆ q + , there are two possible cases: either a γ i = 2for N of them, a γ i = 0 for the remaining root, or else a γ i = 2 for N − a γ i = 1for the remaining two roots. In any case N − a γ i = 2, and as a δ = 2 impliesthat a δ = 1, we may assume that a γ i = 2 for all i ∈ I N − , so a γ i ≥ i ∈ I N − . As P a γ i = 2 N , at most two of a γ i ’s are equal to 3. Therefore we have three cases:(a) a γ = a γ = 3, a γ i = 2 for i ∈ I ,N − . Hence a γ N = a γ N +1 = 0, which implies that a γ N = a γ N +1 = 0. As P a γ i = 2 N , at least one of them is equal to 2. With all theseconditions we find exactly two solutions: γ = 12 , γ = 12 , γ i = γ, i ∈ I ,N − , γ N = 12 , γ N +1 = 2; γ = γ = 12 , γ = 12 , γ i = γ, i ∈ I ,N − , γ N = 12 , γ N +1 = 2 . The last solution requires 2 = f ( n ) for some n ∈ N : the unique possibility is N = 2, n = 2.(b) a γ = 3, a γ i = 2 for i ∈ I ,N − . Hence either a γ N = 1, a γ N +1 = 0 or else a γ N = 0, a γ N +1 = 1. In the first case, γ N +1 = 2, so a N = 1. The solutions are: γ = 12 , γ i = γ, i ∈ I ,N − , γ N − = 12 , γ N = 1234 , γ N +1 = 2; γ = 12 , γ i = γ, i ∈ I ,N − , γ N = 123 , γ N +1 = 2; γ = 12 , γ i = γ, i ∈ I ,N − , γ N = 1234 , γ N +1 = 2 . Now we consider a γ N = 0, a γ N +1 = 1. Notice that a γ N , a γ N +1 ≤
1, so a γ N = a γ N +1 = 1.This implies that γ N = 12. We have three solutions: γ = 12 , γ i = γ, i ∈ I ,N − , γ N − = 12 , γ N = 12 , γ N +1 = 234; γ = 12 , γ i = γ, i ∈ I ,N − , γ N = 12 , γ N +1 = 23; γ = 12 , γ i = γ, i ∈ I ,N − , γ N = 12 , γ N +1 = 234 . (c) a γ i = 2 for all i ∈ I N − . In this case, exactly N of the a γ i ’s are 1, and the remainingone is 0. Hence either γ = 12 , γ i = γ, i ∈ I ,N − , γ N = 1234 , γ N +1 = 234 , or γ i = γ , for all i ∈ I N − , so γ N + γ N +1 = γ : the possible pairs ( γ N , γ N +1 ) are(123 , , , ), (12 , , x L γ γ ) ∗ is a cocycle.If γ = 12
4, then N γ = 2 = P γ . First we look for pairs α < β as in (6.3.3). We havethe following posibilities:(12 , , (123 , , (23 , , (3 , , (23 , , (123 , . In all cases [ x α , x γ ] c = 0 = [ x γ , x β ] c , so the root vectors satisfy (10.1.8), and (cid:16) − q αα q ββ (cid:17) N = 1.Hence we take L = N .Next we check that (123 , , , , , ) is a 6-uple ( α, β, δ, τ, ϕ, η ) sat-isfying (6.3.11). As e q αγ = q = e q βγ , e q δγ = q − , we have that d ( L ) αβδγ = d ( L ) q − ,q ,q = 0 byLemma 6.3.28 (c). Hence (E) holds in this case.Now we look for 4-uples ( α, β, δ, η ) satisfying (6.3.13). There are five possibilities:(123 , , , , (12 , , , , (123 , , , , (23 , , , , (12 , , , . Here e q αγ = q , e q βγ = q − . Hence c ( L ) αβγ = c ( L ) q ,q − = 0 by Lemma 6.3.28 (b), and (F) holds.We look for 4-uples ( α, β, δ, η ) satisfying (6.3.15). A possibility is (123 , , , e q αγ = q and e q δγ = q − ; thus c ( L ) − δαγ = c ( L ) q,q = 0 by Lemma 6.3.28 (b), and (G) holds.Now (23 , , , , ,
4) and (123 , , , , , α, β, ν, µ, δ, η ) satisfying (6.3.19). As e q αγ = e q βγ = q , and e q δγ = q − in bothcases, we have that d ( L ) α + β,δ,α,γ = d ( L ) q ,q − ,q = 0 by Lemma 6.3.28 (b). Thus (I) holds.Also γ = 12 , γ = 123 , γ = 23 , γ = γ = 12 satisfy P i ∈ I γ i = 4 γ . Hence,if N = 2, then n = n = n = 1, n γ = n = 2 is a solution of (6.3.1).The coefficient of x γ ⊗ d ( x x x x γ x ⊗
1) by Lemma 10.1.77.Finally we look for solutions of (6.3.1), i.e. P δ ∈ ∆ q + f δ ( n δ ) δ = N γ , P δ ∈ ∆ q + n δ = N + 1.Looking at the coefficient of α we get as in the previous case that n δ = f δ ( n δ ) for all δ such that 4 ∈ supp δ , so we may translate the equations to the following problem: Find γ i ∈ ∆ q + , i ∈ I N +1 , such that P i ∈ I N +1 γ i = N γ . As P a γ i = N and a δ ≤ δ ∈ ∆ q + ,we may assume that a γ i = 1 for i ∈ I N , a γ N +1 = 0. As P a γ i = 3 N and a δ ≤ δ ∈ ∆ q + , at least N − a γ i = 3 and we have three cases:(a) a γ = 0, a γ i = 3 for i ∈ I ,N +1 . Then a γ i = 1, a γ i = 2, a γ i ≥ i ∈ I ,N +1 but thereis no solution in this case.(b) a γ = 1, a γ = 2, a γ i = 3 for i ∈ I ,N +1 . Then a γ i = 1, a γ i = 2, a γ i ≥ i ≥
3, so a γ + a γ = 1 , a γ + a γ = 2 , a γ + a γ ≤ . If a γ = a γ = 0, then we obtain the following solutions γ = 123 , γ = 23 , γ = 12 , γ i = γ if i ∈ I ,N +1 ; γ = 123 , γ = 23 , γ = 12 , γ i = γ if i ∈ I ,N +1 ; OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 65 γ = 12 , γ = 3 , γ = 12 , γ i = γ if i ∈ I ,N +1 . Otherwise γ i = γ for all i ≥
3; that is, γ + γ = γ , and the possible pairs ( γ , γ ) are(12 , , (123 , , (23 , , (3 , , (23 , , (123 , . (c) a γ i = 2 for i ∈ I , a γ i = 3 for i ∈ I ,N +1 . Then a γ i = 1, a γ i = 2, a γ i ≥ i ≥
4, so a γ + a γ + a γ = 2 , a γ + a γ + a γ = 4 , a γ + a γ + a γ ≤ . If a γ = a γ = a γ = 0, then the unique solution is γ = 12 , γ = 123 , γ = 23 , γ = γ = 12 , γ i = γ if i ∈ I ,N +1 . For this solution we need N = 2, which implies that N = 6.If a γ = 1, a γ = a γ = 0, then the solutions are γ = 12 , γ = 123 , γ = 23 , γ = 12 , γ i = γ if i ∈ I ,N +1 ; γ = 123 , γ = 12 , γ = 23 , γ = 12 , γ i = γ if i ∈ I ,N +1 ; γ = 23 , γ = 123 , γ = 12 , γ = 12 , γ i = γ if i ∈ I ,N +1 . If a γ = a γ = 1, a γ = 0, then the unique solution is γ = 12 , γ = 23 , γ = 123 , γ i = γ if i ∈ I ,N +1 ; γ = 123 , γ = 12 , γ = 23 , γ i = γ if i ∈ I ,N +1 ; γ = 23 , γ = 123 , γ = 12 , γ i = γ if i ∈ I ,N +1 . Hence all the hypothesis of Proposition 6.3.2 hold, and ( x L γ γ ) ∗ is a cocycle.If γ = 123
4, then N γ = 2 = P γ . The pairs α < β as in (6.3.3) are (1 , , α, β ) = (123 , x α , x γ ] c = 0 = [ x γ , x β ] c , so the root vectors satisfy (10.1.8),and (cid:16) − q αα q ββ (cid:17) N = 1. Hence we take L = N .Now (123 , , ,
3) is a 4-uples ( α, β, δ, η ) satisfying (6.3.13). Here, e q αγ = q , e q βγ = q − , so c ( L ) αβγ = c ( L ) q ,q − = 0 by Lemma 6.3.28 (b), and (F) holds.Next we check that (1 , , ,
4) and (1 , , ,
4) are 4-uples ( α, β, δ, η )satisfying (6.3.15). As e q αγ = q and e q δγ = q − in both cases, c ( L ) − δαγ = c ( L ) q,q = 0 by Lemma6.3.28 (b), and (G) holds.Also, (1 , , , , ,
34) is a 6-uple satisfying (6.3.19). As e q αγ = e q βγ = q and e q δγ = q − , we have that d ( L ) α + β,δ,α,γ = d ( L ) q ,q − ,q = 0 by Lemma 6.3.28 (b). Thus (I)holds in this case.Finally we look for solutions of (6.3.1), i.e. P δ ∈ ∆ q + f δ ( n δ ) δ = N γ , P δ ∈ ∆ q + n δ = N + 1.Looking at the coefficient of α and arguing as in the case γ = 12
4, we find that n δ = f δ ( n δ ) for all δ , so we translate the equations to the following problem: Find γ i ∈ ∆ q + , i ∈ I N +1 , such that P i ∈ I N +1 γ i = N γ . As P a γ i = N and a δ ≤ δ ∈ ∆ q + , we may assume that a γ i = 1 for i ∈ I N , a γ N +1 = 0. Using a detailed study as the previous case wecheck that the solutions are γ = 12 , γ i = γ, i ∈ I ,N − , γ N − = 123 , γ N = 1 , γ N +1 = 34; γ = 12 , γ i = γ, i ∈ I ,N − , γ N = 1 , γ N +1 = 3; γ = 12 , γ i = γ, i ∈ I ,N − , γ N = 1 , γ N +1 = 34 .γ = 123 , γ i = γ, i ∈ I ,N − , γ N = 1234 , γ N +1 = 34 , or γ i = γ , for all i ∈ I N − , so γ N + γ N +1 = γ : the possible pairs ( γ N , γ N +1 ) are (123 , , , , x L γ γ ) ∗ is a cocycle.If γ = 1234, then N γ = 2 = P γ . The pairs α < β as in (6.3.3) are (1 , , , x α , x γ ] c = 0 = [ x γ , x β ] c , so the root vectors satisfy (10.1.8), and (cid:16) − q αα q ββ (cid:17) N = 1. Hence we take L = N .Next we check that (1 , , ,
4) and (12 , , ,
34) are 4-uples ( α, β, δ, η ) sat-isfying (6.3.15). As e q αγ = q and e q δγ = q − in both cases, c ( L ) − δαγ = c ( L ) q,q = 0 by Lemma6.3.28 (b), so (G) holds.Notice that γ = 1, γ = 12, γ = 12 , γ = 4 satisfy P i ∈ I γ i = 3 γ . The corre-sponding root vectors q -commute so the coefficient of x γ ⊗ d ( x x x x N − γ x ⊗ P δ ∈ ∆ q + f δ ( n δ ) δ = N γ , P δ ∈ ∆ q + n δ = N + 1.Looking at the coefficient of α and arguing as in the case γ = 12
4, we find that n δ = f δ ( n δ ) for all δ , so we translate the equations to the following problem: Find γ i ∈ ∆ q + , i ∈ I N +1 , such that P i ∈ I N +1 γ i = N γ . As P a γ i = N and a δ ≤ δ ∈ ∆ q + , we mayassume that a γ i = 1 for i ∈ I N , a γ N +1 = 0. Using a detailed study as the previous case wecheck that the solutions are γ = 12 , γ i = γ, i ∈ I ,N − , γ N − = 12 , γ N = 1 , γ N +1 = 4; γ = 12 , γ i = γ, i ∈ I ,N − , γ N = 1 , γ N +1 = 4 .γ = 12 , γ i = γ, i ∈ I ,N − , γ N = 123 , γ N +1 = 4 , or γ i = γ , for all i ∈ I N − , so γ N + γ N +1 = γ : the possible pairs ( γ N , γ N +1 ) are (123 , , , x L γ γ ) ∗ is a cocycle.Finally, if γ = 12 , then N γ = M , P γ = 2, P ′ γ = 1. If N = 6, then N γ > P γ = 2 , P ′ γ ,so ( x N γ γ ) ∗ is a 2-cocycle by Lemma 6.2.4. Next we assume N = 6; that is, N γ = 2. Let α < β be a pair of positive roots as in (6.3.3). We have the following posibilities: ◦ α = 23 β = 1234. There exists b ∈ k such that [ x α , x β ] c = b x γ . ◦ α = 123 β = 234. There exist b , b ∈ k such that[ x α , x β ] c = b x γ + b x x . OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 67 ◦ α = 12 β = 34. There exist b , b t ∈ k such that[ x α , x β ] c = b x γ + b x x + b x x . ◦ α = 12 β = 4. There exist b , b t ∈ k such that[ x α , x β ] c = b x γ + b x x + b x x + b x x . In all cases [ x α , x γ ] c = 0 = [ x γ , x β ] c , so the root vectors satisfy (10.1.8), and − q αα q ββ = − L = 2.Next we look for solutions of (6.3.1). That is, P δ ∈ ∆ q + f δ ( n δ ) δ = 2 γ , P δ ∈ ∆ q + n δ = 3.Suppose that n η = 3 for some η ∈ ∆ q + , then n δ = 0 for δ = η and 2 γ = ( N η + 1) η , acontradiction. Now suppose that n η = 2, n τ = 1 for η = τ : 2 γ = N η η + τ . As a η , a τ ≤ N η a η + a τ , we have that a η = 1, N η = 2, a τ = 0. As 4 = 2 a η + a τ and a η ≤ a τ ≤
1, we have that a η = 2, a τ = 0. Thus τ ∈ { , , } , but there is no solution for thesecases, a contradiction.Therefore, n η = n τ = n µ = 1 for three different roots η, τ, µ . As a η , a τ , a µ ≤ a η + a τ + a µ , we may assume a η = a τ = 1, a µ = 0. As a η , a τ ≤ a µ ≤ a η + a τ + a µ , either a η = a τ = 2, a µ = 0 or else a η = 2, a τ = a µ = 1. In the firstcase, a η , a τ ≤ a µ ≤ a η + a τ + a µ , so either a η = a τ = 3, a µ = 0 or else a η = 3, a τ = 2, a µ = 1; in both cases we are forced to get η = γ = τ + µ , and moreover weobtain only two possibilities, either τ = 12 µ = 34 or else τ = 12 µ = 4. In thesecond case, a η ≤ a τ , a µ ≤ a η + a τ + a µ , so a η = 2, a τ = a µ = 1, and againwe are forced to get η = γ = τ + µ , and moreover we obtain only two possibilities, either τ = 123 µ = 234 or else τ = 1234, µ = 23 x N γ γ ) ∗ is a 2-cocycle. (cid:3) Type G (3) . Let q be a root of 1 of order N >
3. In this section, we deal with aNichols algebra B q of super type G (3), associated to the Dynkin diagram − ◦ q − q ◦ q − q ◦ . For more information, see [AA, § { , , , , } . We fix the following convex order of ∆ q + :1 < < < < < < < < < < < < . It comes from Lyndon words once we fix the order of the letters 1 < < § G (3): Proposition 8.7.1.
For every γ ∈ ∆ q + , there exists L γ ∈ N such that ( x L γ γ ) ∗ is a cocycle.Proof. It is enough to prove the statement for γ with full support. ◦ For γ = 123, we apply Proposition 6.3.2. The pairs as in (6.3.3) are α = α , β = α + α ,and α = α + α , β = α since the following relations hold: x α x β = x γ + q αβ x β x α , x α x γ = q α x γ x α , x γ x β = q β x β x γ . As − q αα q ββ = q , respectively q , L should be a multiple of N .Let L = ord( − q ). Now (6.3.5) holds for α = 1, β = 12 δ = 3, η = 12, and the rootvectors satisfy (10.1.11); the scalars e q αγ = q − and e q βγ = q − satisfy c ( L ) αβγ = 0.Also, (6.3.7) holds for α = 1, β = 12 , δ = 3, η = 12, τ = 12
3, and the root vectorssatisfy (10.1.16); the scalars e q αγ = q − , e q βγ = q − , e q τγ = q − satisfy (6.3.8).Let ( n δ ) δ ∈ ∆ q + be a solution of (6.3.1). If n = 0, then L ( α + α ) = s ( Lγ ) = X δ ∈ ∆ q + f δ ( n δ ) s ( δ ) , and s ( δ ) ∈ ∆ q + if δ = α . As N δ = N s ( δ ) , we have that f δ = f s ( δ ) , so we have a systemas in (6.3.1) for α + α in place of γ and we may restrict the support to α , α . The newsystem has a unique solution, which gives place to the solution of the original system: • n = n = 1, n = L − n δ = 0 for all the other δ ∈ ∆ q + .Next we assume n = 0. Suppose that n >
1. Then f ( n ) ≥ N , so the coefficientof α in P δ ∈ ∆ q + f δ ( n δ ) δ is ≥ N , a contradiction. Hence n ≤
1, and then f ( n ) = n ≤
1. The coefficient of α in this sum is n + n + n + n + n + n + n = L. As the sum of all n δ ’s is L + 1 and n = 0, we have n = 1, n = n = n = n = n = 0. Now we look at the coefficients of α , α in the equality P δ ∈ ∆ q + f δ ( n δ ) δ = Lγ : L = n + n + 2 n + 3 n + 3 n + 4 n ,L = 1 + n + n + n + 2 n + 2 n . Thus n + n + 2 n + n + 2 n = 1, which implies that n = n = 0 andtwo of the three numbers n , n , n are zero (the remaining one being 1). Lookingat the three possibilities, we have three solutions: • n = n = 1, n = L − n δ = 0 for all the other δ ∈ ∆ q + ; • n = n = n = 1, n = L − n δ = 0 for all the other δ ∈ ∆ q + ; • n = 2, n = n = 1, n = L − n δ = 0 for all the other δ ∈ ∆ q + .Hence Proposition 6.3.2 applies and ( x L ) ∗ is a cocycle. ◦ For γ = 12
3, the case N γ > N γ = 3.We will work as in Proposition 6.3.23. The pairs as in (6.3.24) are (1 , ), (12 , ),(123 , x α x β = b αβ x γ + q αβ x β x α , x α x γ = q αγ x γ x α , x γ x β = q γβ x β x , b αβ ∈ k . As q αγ q γβ = 1 for the three cases, we take L = 1. We look for solutions of (6.3.26): X δ ∈ ∆ q + n δ = 3 , X δ ∈ ∆ q + f δ ( n δ ) δ = 3 γ. If n γ ≥
2, then f γ ( n γ ) ≥
3, a contradiction. Then n γ ≤
1, so f γ ( n γ ) = n γ . Looking atthe coefficient of α we get the equation:3 = n + n + n + n + n + n + n . (8.7.2) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 69
Hence n δ = 0 for δ = 2 , , , , ,
3. Looking at the coefficients of α and α ,6 = n + n + 2 n + 3 n + 3 n + 4 n , (8.7.3) 3 = n + n + n + 2 n + 2 n . (8.7.4)From (8.7.2) and (8.7.4), n + n = n + n . From (8.7.4), n + n ≤
1. If n = n = 0, then n = n = 0: the solution is n = n = n = 1. Nextwe assume n + n = 1 = n + n . If n = 1, then the solutions give pairs as in(6.3.24). Otherwise we have a unique solution: n = n = n = 1, n δ = 0 otherwise.Hence we have to compute d ( x x x ⊗ x x = − q q q q x x + b x x ,x x = − q q q q x x + b x x ,x x = − q q q q x x + b x + b x x ,x x = − q q q x x + b x , for some b j ∈ k . Using these relations, we get d ( x x x ⊗
1) = x x ⊗ x − s ( x ⊗ x x + q q x ⊗ x x )= x x ⊗ x − s (cid:0) − q q q q x ⊗ x x + b x ⊗ x x − q q q q x ⊗ x x + q q b x ⊗ x x (cid:1) = x x ⊗ x − b x x ⊗ x − s (cid:0) − q q q q x ⊗ x x − b q q q q x ⊗ x x + b b x ⊗ x + b b x ⊗ x x − q q q q x ⊗ x x + q q b x ⊗ x x (cid:1) = x x ⊗ x − b x x ⊗ x + q q q q x x ⊗ x − s (cid:0) q q q x ⊗ x x − b q q q q x ⊗ x x + b q q q q x ⊗ x x + b b x ⊗ x − qq q b b x ⊗ x x − q q q q x ⊗ x x + q q b x ⊗ x x (cid:1) = x x ⊗ x − b x x ⊗ x + q q q q x x ⊗ x − q q b x x ⊗ x + q q q q x x ⊗ x + ( q q b b − b b ) x ⊗ . We compute the scalars b j using the form of the Lyndon words and the q -Jacobi identity: b = q q q q (1 − q ) , b = q q q (1 − q ) , b = q q q q (1 + q ) , b = q q q (1 + q ) . Hence q q b b − b b = 0. Thus the coefficient of x ⊗ d ( c ) is zero for all 2-chains c , so ( x ) ∗ is a 2-cocycle. ◦ For γ = 12
3, we will apply Proposition 6.3.2. The pairs ( α, β ) as in (6.3.3) are (1 , , , x α x β = x γ + q αβ x β x α , x α x γ = q α x γ x α , x γ x β = q β x β x γ . As − q αα q ββ = q , respectively q , q , L should be a multiple of N . Let L = ord( − q ). Now (6.3.9) holds for α = 12, β = 12 δ = 2
3, and the root vectorssatisfy (10.1.11); the scalars e q αγ = q − , e q βγ = q − satisfy (6.3.10).Also, (6.3.7) holds for α = 12, β = 12 , δ = 2 η = 123, τ = 12
3, and the rootvectors satisfy (10.1.16); the scalars e q αγ = q − , e q βγ = q − , e q τγ = q − satisfy (6.3.8).Let ( n δ ) δ ∈ ∆ q + be a solution of (6.3.1). If n = 0, then N ( α + α + α ) = s ( N γ ) = X δ ∈ ∆ q + f δ ( n δ ) s ( δ ) , and s ( δ ) ∈ ∆ q + if δ = α . As N δ = N s ( δ ) , we have that f δ = f s ( δ ) , so we have a systemas in (6.3.1) for α + α + α in place of γ . The new system has four solutions, which givesplace to the following solutions of the original system: • n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + ; • n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + ; • n = n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + ; • n = 2, n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + .Next we assume n = 0. Suppose that n >
1. Then f ( n ) ≥ N , so the coefficientof α in P δ ∈ ∆ q + f δ ( n δ ) δ is ≥ N ; this forces to n δ = 0 for any δ = α since f δ ( n δ ) δ must have α , α with coefficient zero, and this gives a contradiction. Hence n = f ( n ) ≤ α in this sum is n + n + n + n + n + n + n = N. As the sum of all n δ ’s is N + 1 and n = 0, we have n = 1, n = n = n = n = n = 0. Now we look at the coefficients of α , α in the equality P δ ∈ ∆ q + f δ ( n δ ) δ = N γ :3 N = 1 + n + n + 2 n + 3 n + 3 n + 4 n ,N = n + n + n + 2 n + 2 n . Thus n + 2 n + n + 2 n + 4 n = 1, which implies that n = n = 0 andtwo of the three numbers n , n , n are zero (the remaining one being 1). Reducingthe three previous equations, we get n + n = N − n + 3 n ≥ N −
3, so wehave a unique solution: • n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + .Hence Proposition 6.3.2 applies and ( x L ) ∗ is a cocycle. ◦ For γ = 12 , we will apply Proposition 6.3.2. The pairs ( α, β ) as in (6.3.3) are (1 , ),(123 , , , x α x β = x γ + q αβ x β x α , x α x γ = q α x γ x α , x γ x β = q β x β x γ . As − q αα q ββ ∈ { q, q , q } , L should be a multiple of N .Let L = ord( − q ). Now (6.3.9) holds for α = 123, β = 12 δ = 2 , and the rootvectors satisfy (10.1.11); the scalars e q αγ = q − , e q βγ = q − satisfy (6.3.10).Also, (6.3.7) holds for α = 123, β = 12 , δ = 2 , η = 12, τ = 12
3, and the rootvectors satisfy (10.1.16); the scalars e q αγ = q − , e q βγ = q − , e q τγ = q − satisfy (6.3.8). OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 71
Let ( n δ ) δ ∈ ∆ q + be a solution of (6.3.1). If n = 0, then N ( α + 3 α + α ) = s ( N γ ) = X δ ∈ ∆ q + f δ ( n δ ) s ( δ ) , and s ( δ ) ∈ ∆ q + if δ = α . As N δ = N s ( δ ) , we have that f δ = f s ( δ ) , so we have a systemas in (6.3.1) for α + 3 α + α in place of γ . The new system has five solutions, whichgives place to the following solutions of the original system: • n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + ; • n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + ; • n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + ; • n = n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + ; • n = 2, n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + .Next we assume n = 0. An analogous analysis as for the root 12 • n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + .Hence Proposition 6.3.2 applies and ( x L ) ∗ is a cocycle. ◦ For γ = 12 , we apply Proposition 6.3.2 again. The pairs ( α, β ) as in (6.3.3) are(12 , ), (12 , , , , x α x β = x γ + q αβ x β x α , x α x γ = q α x γ x α , x γ x β = q β x β x γ . As − q αα q ββ ∈ { q, q , q } , L should be a multiple of N .Let L = ord( − q ). Now (6.3.9) holds for α = 12 β = 12 δ = 2 , and the rootvectors satisfy (10.1.11); the scalars e q αγ = q − , e q βγ = q − satisfy (6.3.10).Also, (6.3.7) holds for α = 12 β = 12 , δ = 2 , η = 1, τ = 12
3, and the rootvectors satisfy (10.1.16); the scalars e q αγ = q − , e q βγ = q − , e q τγ = q − satisfy (6.3.8).Let ( n δ ) δ ∈ ∆ q + be a solution of (6.3.1). If n = 0, then N ( α + 3 α + 2 α ) = s ( N γ ) = X δ ∈ ∆ q + f δ ( n δ ) s ( δ ) , and s ( δ ) ∈ ∆ q + if δ = α . As N δ = N s ( δ ) , we have that f δ = f s ( δ ) , so we have a systemas in (6.3.1) for α + 3 α + 2 α in place of γ . The new system has six solutions, whichgives place to the following solutions of the original system: • n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + ; • n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + ; • n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + ; • n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + ; • n = n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + ; • n = 2, n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + .Next we assume n = 0. An analogous analysis as for the root 12 • n = n = 1, n = N − n δ = 0 for all the other δ ∈ ∆ q + . Hence Proposition 6.3.2 applies and ( x L ) ∗ is a cocycle. (cid:3) Parametric modular types
Modular type wk (4) . Here θ = 4, q = ±
1. In this subsection, we deal with a Nicholsalgebra B q of diagonal type wk (4). We may assume that the corresponding diagram is q ◦ q − − ◦ − − ◦ − q − q − ◦ . (9.1.1)We fix the following convex order on ∆ q + :1 , , , , , , , , , , , , , , . For more information, see [AA, § M = ord( − q ): We may assume that N ≤ M .Note that N δ = N if δ ∈ { , , } ,M if δ ∈ { , , } , . We prove Condition 1.4.1 for type wk (4): Proposition 9.1.2.
For every γ ∈ ∆ q + , there exists L γ ∈ N such that ( x L γ γ ) ∗ is a cocycle.Proof. We may assume that γ has full support i.e. γ ∈ { , , , } .First we consider γ = 12
4. Here N γ = 2. The pairs α < β such that α + β = γ are:(3 , , (23 , , (123 , , (12 , , (2 , , (12 , . For all pairs, x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ . Also, there exist b i ∈ k such that[ x , x ] c = b x γ , [ x , x ] c = b x γ + b x x , [ x , x ] c = b x γ , [ x , x ] c = b x γ + b x x + b x x + b x x , [ x , x ] c = b x γ , [ x , x ] c = b x γ + b x x . Thus the root vectors satisfy (10.1.8), and − q αα q ββ ∈ {− , ± q ± } ; hence we take L = M .Next we check that (23 , , , , , ,
3) and (12 , , , α, β, δ, η ) satisfying (6.3.5). As e q αγ = − e q βγ = q , we have that c ( L ) αβγ = c ( L ) − ,q = 0 by Lemma 6.3.28 (b). Hence (B) holds.Finally we compute the solutions of (6.3.1). That is, X δ :1 ∈ supp δ f δ ( n δ ) = M, X δ :4 ∈ supp δ f δ ( n δ ) = M, (9.1.3) X δ :2 ∈ supp δ f δ ( n δ ) a δ = 2 M, X δ :3 ∈ supp δ f δ ( n δ ) a δ = 2 M. (9.1.4) X δ ∈ ∆ q + n δ = M + 1 . (9.1.5) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 73
Let ( n δ ) be a solution of (6.3.1). We claim that n δ ≤ N δ = 2. To prove it, first wenote that n , n , n ≤ n = 2, then n δ = 0 if 4 ∈ supp δ , δ = 4, so2 M = X δ ∈ ∆ q + f δ ( n δ ) a δ = f ( n ) + n + n + n ≤ f (2) + X δ =4 , n δ ≤ M + ( M −
1) = 2 M − , a contradiction. Hence n ≤
1. Next we suppose that n = 2. Then n δ = 0 if either1 ∈ supp δ or else 2 ∈ supp δ , δ = 12
3. From (9.1.5), n + n + n = M +1 − n = M − n ≤
1. Analogously, n ≤ n , n , n ≤
1. The proof is analogous to the cases n , n , n if N = M . Thus we assume that N < M : that is, M = 2 N , N odd. By (9.1.3), n , n , n ≤
4. We deal first with n . • Suppose that n = 4. Then n δ = 0 if supp δ ∩ { , , } 6 = ∅ by (9.1.3) and (9.1.4),and also n = 2 N . But from (9.1.5), n = 2 N −
3, a contradiction. • Suppose that n = 3. Then P δ =123 n δ = 2 N −
2. By (9.1.4), n = f ( n ) a ≤ N − . By the first equality of (9.1.4) and the previous computations3 N − X δ =123 f δ ( n δ ) a δ = 2 f ( n ) + 2 f ( n ) + 2 f ( n )+ X δ : a δ =1 n δ ≤ n + 2 + 2 + (2 N − − n ) ≤ N − . The equality holds if and only if n = N − n = n = 1, but in this casethe second equation of (9.1.4) does not hold. • Suppose that n = 2. Then P δ =123 n δ = 2 N −
1, and by (9.1.4), n ≤ N . A similar computation as for the previous case shows that the equality 3 N − P δ =123 f δ ( n δ ) a δ holds if and only if n = N , n = n = 1, but again thesecond equation of (9.1.4) does not hold.The same argument applies for n . Finally we check that n ≤
1. If n = 4, then n δ = 0for all δ = 4 such that 4 ∈ supp δ , so2 M = X δ :3 ∈ supp δ f δ ( n δ ) a δ = n + n + n + n ≤ X δ =4 n δ (9.1.5) = M − , a contradiction. Now suppose that 2 ≤ n ≤
3: by (9.1.3), n ≤ N and4 N = X δ =123 f δ ( n δ ) a δ ≤ n + 2 + 2 + (2 N − − n ) ≤ N + 2 , a contradiction. Hence n δ ≤ N δ = 2, so f δ ( n δ ) = n δ for all δ ∈ ∆ q + . Then we look for γ i ∈ ∆ q + , i ∈ I M +1 , such that P i ∈ I M +1 γ i = M γ . As a δ ≤ δ ∈ ∆ q + , there exist twopossibilities up to permutations of these roots: ◦ a γ i = 2 for I M , a γ M +1 = 0. As a δ ≤ δ ∈ ∆ q + , at least M − a γ i = 2, and we know that a γ M +1 ≤
1, so γ i = γ for i ∈ I M − up to permutation, and γ M + γ M +1 = γ . ◦ a γ i = 2 for I M − , a γ M = a γ M +1 = 1. Again at least M − a γ i = 2.If these roots are γ i , i ∈ I M − , then γ i = γ for i ∈ I M − up to permutation, and γ M + γ M +1 = γ . Otherwise we may assume that γ i = γ for i ∈ I M − , a γ M − = a γ M = 1, a γ M +1 = 2. We have three possibilities for ( γ M − , γ M , γ M +1 ):(123 , , , (23 , , , (123 , , . Hence all the hypotheses of Proposition 6.3.2 hold, and ( x Mγ ) ∗ is a cocycle.Next we consider γ = 1234. Here N γ = 2. The pairs α < β such that α + β = γ are (1 , , , x α x γ = q αγ x γ x α , x γ x β = q γβ x β x γ and thereexist b ∈ k such that [ x α , x β ] c = b x γ . Thus the root vectors satisfy (10.1.8) and − q αα q ββ ∈{− , ± q } ; hence we take L = M .Next we check that (12 , , , , , , , , , α, β, δ, η ) satisfying (6.3.15). As ( e q δγ , e q βγ ) are respectively ( − q − , − − , q ),( − q − , q ), we have that c ( L ) − δαγ = 0 by Lemma 6.3.28 (b). Hence (G) holds.Finally we check that (1 , , , , ,
34) is a 6-tuple ( α, β, δ, η, µ, ν ) satisfying(6.3.21). As e q αγ = q = e q βγ , e q νγ = −
1, we have that d ( L ) β − ναγ = 0 by Lemma 6.3.28 (c).Hence (J) holds.Next we look for solutions of (6.3.1). That is, (9.1.5) and X δ :1 ∈ supp δ f δ ( n δ ) = M, X δ :4 ∈ supp δ f δ ( n δ ) = M, (9.1.6) X δ :2 ∈ supp δ f δ ( n δ ) a δ = M, X δ :3 ∈ supp δ f δ ( n δ ) a δ = M. (9.1.7)Let ( n δ ) be a solution of (6.3.1). We claim that n δ ≤ N δ = 2. By (9.1.7), M ≥ f ( n ) , f ( n ) . As N = N = M we have that n , n ≤
1. Now suppose that n ≥
2. By(9.1.6) we have that n = 2. Then n δ = 0 for all δ = 4 such that 4 ∈ supp δ . By (9.1.7), M = X δ :3 ∈ supp δ f δ ( n δ ) a δ = n + n + n ≤ X δ =4 n δ = M − , a contradiction.We also have that n , n ≤ N = M , and n , n ≤ M = 2 N .Suppose that M = 2 N and n a = 2, a ∈ { , } . We have that n δ = 0 for all δ suchthat 3 ∈ supp δ , δ = 1 a
4. By (9.1.6), f ( n ) = M : that is, n = 2, a contradiction.Finally suppose that n ≥
2. By (9.1.6), M = X δ :4 ∈ supp δ f δ ( n δ ) = X δ :4 ∈ supp δ n δ ≤ X δ =1 n δ ≤ M − , OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 75 a contradiction. Hence n δ ≤ N δ = 2, so f δ ( n δ ) = n δ for all δ ∈ ∆ q + . Then we lookfor γ i ∈ ∆ q + , i ∈ I M +1 , such that P i ∈ I M +1 γ i = M γ . As a δ , a δ ≤ δ ∈ ∆ q + ,there exist exactly M roots such that a γ i = 1, respectively a γ i = 1. Thus there exist M − γ i such that a δ = a δ = 1, which implies that supp γ i = { , , , } for theseroots. We may assume that a γ i = a γ i = 1 for all i ∈ I M − and we have three possibilitiesup to permutations of the roots: either a γ M = a γ M = 1, a γ M +1 = a γ M +1 = 0, or else a γ M = a γ M +1 = 1, a γ M +1 = a γ M = 0. For the first case, a γ i , a γ i ≥ i ∈ I M and wehave a contradiction. Hence a γ M = a γ M +1 = 1, a γ M +1 = a γ M = 0. For i ∈ I M − we write γ i = 12 a i b i a i , b i ∈ I . We also write γ M = 12 a M b M , γ M +1 = 2 a M +1 b M +1
4. At mostone of the a i , respectively b i , is 2 for i ∈ I M − . We analyze each case. ◦ a i = 1 = b i = 1 for all i ∈ I M − . Hence γ i = γ for all i ∈ I M − and γ M + γ M +1 = γ . ◦ a i = 1 for all i ∈ I M − , b i = 1 for all i ∈ I M − , b M − = 2. Here b M = b M +1 = 0, so γ i = γ for all i ∈ I M − , γ M − = 123 , γ M = 12 , γ M +1 = 4 . ◦ a i = 1 for all i ∈ I M − , b i = 1 for all i ∈ I M − , a M − = 2. Here a M = a M +1 = 0, so γ i = γ for all i ∈ I M − , γ M − = 12 , γ M = 1 , γ M +1 = 34 . ◦ a i = b i = 1 for all i ∈ I M − , a M − = b M − = 2. Here a M = b M = a M +1 = b M +1 = 0, so γ i = γ for all i ∈ I M − , γ M − = 12 , γ M = 1 , γ M +1 = 4 . ◦ a i = 1 for all i ∈ I M − , b i = 1 for all i ∈ I M − − { M − } , a M − = b M − = 2. Here γ i = γ for all i ∈ I M − , γ M − = 123 , γ M − = 12 , γ M = 1 , γ M +1 = 4 . Hence all the hypotheses of Proposition 6.3.2 hold, and ( x Mγ ) ∗ is a cocycle.Finally we consider γ = 123 ,
34. By direct computation, P γ = 2, Q γ = 1, so P γ , Q γ < N γ . By Lemma 6.2.4 ( x N γ γ ) ∗ is a 2-cocycle. (cid:3) Modular type br (2) . Here θ = 2, ζ ∈ G , q / ∈ G . In this subsection, we dealwith a Nichols algebra B q of modular type br (2), that is associated to any of the Dynkindiagrams a ✤ / / ζ ◦ q − q ◦ , a ✤ / / ζ ◦ ζ q ζq − ◦ . (9.2.1)For more information, see [AA, § ζq − instead of q , we just discuss the latter. Essentially this is very similar tostandard B . The corresponding set of positive roots with full support is { α + α , α + α } . Let M = ord( ζq − ). We order the root vectors: x < x < x < x .We prove Condition 1.4.1 for type br (2). Proposition 9.2.2.
For every γ ∈ ∆ q + , ( x N γ γ ) ∗ is a -cocycle.Proof. As before we just consider non-simple roots, i.e. with full support.
Table 2.
The roots with full support of br (2); γ < γ , ( N γ − γ = γ + γ γ N γ P γ Q γ γ γ L γ M q ◦ For γ = 1
2, the case N > N = 2. Wewill apply Proposition 6.3.2. The unique pair as in (6.3.3) is α = α , β = α + α , sincethe following relations hold: x x = x + ζq x x , x x = ζ q x x , x x = ζ q x x . As − q αα q ββ = −
1, we take L = 2. The unique solution of (6.3.1) is n = n = n = 1, and n = 0. Hence Proposition 6.3.2 applies and ( x ) ∗ is a 2-cocycle. ◦ For γ = 12, we will apply Proposition 6.3.23. The unique pair as in (6.3.24) is α =2 α + α , β = α , since the following relations hold: x x = ( q − ζ ) q x + qq x x , x x = ζ q x x ,x x = qq x x . In this case, q αγ q γβ = ζ q − : as q = ζ , we take L = ord q . By direct computation, theunique solution of (6.3.26) is n = L − n = n = 1, and n = 0, so Proposition 6.3.23applies and ( x L ) ∗ is a 2 L -cocycle. (cid:3) Proofs of the Computational Lemmas γ ∈ ∆ + , let g γ : N → N be the function g γ ( n ) := f γ ( n ) − f γ ( n −
1) = ( , n odd ,N γ − n even . (10.1.1) Remark . Let β < β < β be positive roots such that the corresponding rootvectors q -commute: x β i x β j = q β i β j x β j x β i , for all i < j. For each n ∈ N , d ( x β x f β ( n ) β ⊗
1) = x β x f β ( n − β ⊗ x g β ( n ) β + ( − n q f β ( n ) β β x f β ( n ) β ⊗ x β , (10.1.3) d ( x f β ( n ) β x β ⊗
1) = x f β ( n ) β ⊗ x β − q g β ( n ) β β x f β ( n − β x β ⊗ x g β ( n ) β (10.1.4) d ( x β x f β ( n ) β x β ⊗
1) = x β x f β ( n ) β ⊗ x β − q g β ( n ) β β x β x f β ( n − β x β ⊗ x g β ( n ) β (10.1.5) − ( − n q β β q f β ( n ) β β x f β ( n ) β x β ⊗ x β ,d ( x f β ( n ) β x β x β ⊗
1) = x f β ( n ) β x β ⊗ x β − q β β x f β ( n ) β x β ⊗ x β (10.1.6) + q g β ( n ) β β q g β ( n ) β β x f β ( n − β x β x β ⊗ x g β ( n ) β . OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 77
In fact, the root vectors q -commute by hypothesis so we can compute the differentialsas in the proof of Proposition 4.3.3.The next results will allow us to identify some cocycles of degree higher than 2. Lemma 10.1.7.
Let n ∈ N . Let α < δ , . . . , δ n < γ < η , . . . , η n < β be positive roots.Assume that the relations among the corresponding root vectors take the form x α x β = q αβ x β x α + b x N γ − γ + n X j =1 b j x η j x δ j ,x α x γ = q αγ x γ x α , x δ j x γ = q δ j γ x γ x δ j ,x γ x β = q γβ x β x γ , x γ x η j = q γη j x η j x γ , (10.1.8) for some scalars b , b , . . . , b n . Then, for all a ≥ , d ( x α x aN γ γ x β ⊗
1) = x α x aN γ γ ⊗ x β − q N γ − γβ x α x N γ ( a − γ x β ⊗ x N γ − γ − q αβ q aN γ αγ x aN γ γ x β ⊗ x α − n X j =1 b j q aN γ αγ x aN γ γ x η j ⊗ x δ j + b q aN γ γβ ((cid:18) q αγ q γβ − (cid:19) ( a ) ( qαγqγβ ) Nγ − (cid:18) q αγ q γβ (cid:19) aN γ ) x aN γ +1 γ ⊗ x N γ − γ ; and for all a ≥ , d ( x α x aN γ +1 γ x β ⊗
1) = x α x aN γ +1 γ ⊗ x β − q γβ x α x aN γ γ x β ⊗ x γ + q αβ q aN γ +1 αγ x aN γ +1 γ x β ⊗ x α + n X j =1 b j q aN γ +1 αγ x aN γ +1 γ x η j ⊗ x δ j + b q aN γ +1 γβ (cid:18) q αγ q γβ − (cid:19) ( a + 1) (cid:18) qαγqγβ (cid:19) Nγ x N γ ( a +1) γ ⊗ . Notice that the first equality in (10.1.8) forces( N γ − γ = α + β = δ j + η j for all j ∈ I n . (10.1.9) Proof.
We need the following computation: d ( x α x β ⊗
1) = x α ⊗ x β − q αβ x β ⊗ x α − bx γ ⊗ x N γ − γ − n X j =1 b j x η j ⊗ x δ j The proof of the lemma is by induction on a . First we compute: d ( x α x γ x β ⊗
1) = x α x γ ⊗ x β − s ( d ( x α x γ ⊗ x β )= x α x γ ⊗ x β − s ( x α ⊗ x γ x β − q αγ x γ ⊗ x α x β )= x α x γ ⊗ x β − s (cid:16) q γβ x α ⊗ x β x γ − q αγ x γ ⊗ (cid:0) q αβ x β x α + b x N γ − γ + n X j =1 b j x η j x δ j (cid:1)(cid:17) = x α x γ ⊗ x β − q γβ x α x β ⊗ x γ + s (cid:16) q αγ q αβ (cid:0) x γ ⊗ x β − q γβ x β ⊗ x γ (cid:1) x α + b ( q αγ − q γβ ) x γ ⊗ x N γ − γ + n X j =1 b j (cid:0) q αγ x γ ⊗ x η j − q γβ q δ j γ x η j ⊗ x γ (cid:1) x δ j (cid:17) = x α x γ ⊗ x β − q γβ x α x β ⊗ x γ + n X j =1 b j q αγ x γ x η j ⊗ x δ j + q αγ q αβ x γ x β ⊗ x α + b ( q αγ − q γβ ) x N γ γ ⊗ s (cid:16) n X j =1 b j (cid:0) q αγ q γη j − q γβ q δ j γ (cid:1) s (1 ⊗ x η j x γ x δ j ) (cid:17) , which agrees with the second formula for a = 0 since s ◦ s = 0. Next we compute d ( x α x N γ γ x β ⊗
1) = x α x N γ γ ⊗ x β − s ( d ( x α x N γ γ ⊗ x β )= x α x N γ γ ⊗ x β − s (cid:16) q N γ − γβ x α x γ ⊗ x β x N γ − γ + q αβ q N γ αγ x N γ γ ⊗ x β x α + q N γ αγ bx N γ γ ⊗ x N γ − γ + n X j =1 b j q N γ αγ x N γ γ ⊗ x η j x δ j (cid:17) = x α x N γ γ ⊗ x β − q N γ − γβ x α x γ x β ⊗ x N γ − γ − b (cid:16) q N γ αγ − q N γ − γβ q αγ + q N γ γβ (cid:17) x N γ +1 γ ⊗ x N γ − γ − n X j =1 b j q N γ αγ x N γ γ x η j ⊗ x δ j − q αβ q N γ αγ x N γ γ x β ⊗ x α , and by (10.1.9), this agrees with the first formula in the lemma when a = 1. Now assumethe second formula given in the lemma holds when a is replaced by a −
1. Then d ( x α x aN γ γ x β ⊗
1) = x α x aN γ γ ⊗ x β − s ( d ( x α x aN γ γ ⊗ x β )= x α x aN γ γ ⊗ x β − s (cid:0) x α x N γ ( a − γ ⊗ x N γ − γ x β + q aN γ αγ x aN γ γ ⊗ x α x β (cid:1) = x α x aN γ γ ⊗ x β − s (cid:16) q N γ − γβ x α x N γ ( a − γ ⊗ x β x N γ − γ + q αβ q aN γ αγ x aN γ γ ⊗ x β x α + b q aN γ αγ x aN γ γ ⊗ x N γ − γ + n X j =1 b j q aN γ αγ x aN γ γ ⊗ x η j x δ j (cid:17) . Use the induction hypothesis to rewrite the term q N γ − γβ x α x N γ ( a − γ ⊗ x β x N γ − γ to obtain d ( x α x aN γ γ x β ⊗
1) = x α x aN γ γ ⊗ x β − s (cid:16) q N γ − γβ d ( x α x ( a − N γ +1 γ x β ⊗ x N γ − γ ) − q αβ q ( a − N γ +1 αγ q N γ − γβ x ( a − N γ +1 γ x β ⊗ x α x N γ − γ − b q aN γ γβ (cid:18) q αγ q γβ − (cid:19) ( a ) (cid:18) qαγqγβ (cid:19) Nγ x aN γ γ ⊗ x N γ − γ − n X j =1 b j q N γ − γβ q ( a − N γ +1 αγ x ( a − N γ +1 γ x η j ⊗ x δ j x N γ − γ + q αβ q aN γ αγ x aN γ γ ⊗ x β x α OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 79 + b q aN γ αγ x aN γ γ ⊗ x N γ − γ + n X j =1 b j q aN γ αγ x aN γ γ ⊗ x η j x δ j (cid:17) = x α x aN γ γ ⊗ x β − q N γ − γβ x α x ( a − N γ +1 γ x β ⊗ x N γ − γ + s (cid:16) q αβ q aN γ αγ q N γ − γβ x ( a − N γ +1 γ x β ⊗ x N γ − γ x α + b q aN γ γβ (cid:18) q αγ q γβ − (cid:19) ( a ) (cid:18) qαγqγβ (cid:19) Nγ − q aN γ αγ q aN γ γβ x aN γ γ ⊗ x N γ − γ − q αβ q aN γ αγ x aN γ γ ⊗ x β x α + n X j =1 b j q N γ − γβ q ( a − N γ +1 αγ q N γ − δ j γ x ( a − N γ +1 γ x η j ⊗ x N γ − γ x δ j − n X j =1 b j q aN γ αγ x aN γ γ ⊗ x η j x δ j (cid:17) . Now use the formula d ( x aN γ +1 γ ⊗
1) = x aN γ γ ⊗ x γ , (10.1.4) and (10.1.9) to rewrite the aboveexpression as x α x aN γ γ ⊗ x β − q N γ − γβ x α x N γ ( a − γ x β ⊗ x N γ − γ − q αβ q aN γ αγ x aN γ γ x β ⊗ x α + b q aN γ γβ { ( q αγ q γβ − a ) ( qαγqγβ ) Nγ − ( q αγ q γβ ) aN γ } x aN γ +1 γ ⊗ x N γ − γ − n X j =1 b j q aN γ αγ x aN γ γ x η j ⊗ x δ j . This agrees with the first claimed formula in the lemma.Now we use the first formula to obtain the second formula: d ( x α x aN γ +1 γ x β ⊗
1) = x α x aN γ +1 γ ⊗ x β − s ( d ( x α x aN γ +1 γ ⊗ x β )= x α x aN γ +1 γ ⊗ x β − s ( x α x aN γ γ ⊗ x γ x β − q aN γ +1 αγ x aN γ +1 γ ⊗ x α x β )= x α x aN γ +1 γ ⊗ x β − s (cid:16) q γβ x α x aN γ γ ⊗ x β x γ − q αβ q aN γ +1 αγ x aN γ +1 γ ⊗ x β x α − b q aN γ +1 αγ x aN γ +1 γ ⊗ x N γ − γ − n X j =1 b j q aN γ +1 αγ x aN γ +1 γ ⊗ x η j x δ j (cid:17) . By our induction hypothesis, we may use the first formula in the statement of the lemmato rewrite the term q γβ x α x aN γ γ ⊗ x β x γ , obtaining x α x aN γ +1 γ ⊗ x β − s (cid:16) q γβ d ( x α x aN γ γ x β ⊗ x γ ) + q αβ q aN γ αγ q γβ x aN γ γ x β ⊗ x α x γ − b q aN γ +1 γβ { ( q αγ q γβ − a ) ( qαγqγβ ) Nγ − ( q αγ q γβ ) aN γ } x aN γ +1 γ ⊗ x N γ − γ + n X j =1 b j q aN γ αγ q γβ x aN γ γ x η j ⊗ x δ j x γ − q αβ q aN γ +1 αγ x aN γ +1 γ ⊗ x β x α − b q aN γ +1 αγ x aN γ +1 γ ⊗ x N γ − γ − n X j =1 b j q aN γ +1 αγ x aN γ +1 γ ⊗ x η j x δ j (cid:17) = x α x aN γ +1 γ ⊗ x β − q γβ x α x aN γ γ x β ⊗ x γ − s (cid:16) q αβ q aN γ +1 αγ ( q γβ x aN γ γ x β ⊗ x γ x α − x aN γ +1 γ ⊗ x β x α ) − b q aN γ +1 γβ { ( q αγ q γβ − a ) ( qαγqγβ ) Nγ − ( q αγ q γβ ) aN γ + ( q αγ q γβ ) aN γ +1 } x aN γ +1 γ ⊗ x N γ − γ − n X j =1 b j q aN γ αγ (cid:0) q αγ x aN γ +1 γ ⊗ x η j − q γβ q δ j γ x aN γ γ x η j ⊗ x γ (cid:1) x δ j (cid:17) . Now we use the formula d ( x N γ ( a +1) γ ⊗
1) = x aN γ +1 γ ⊗ x N γ − γ , (10.1.4) and (10.1.9) to rewritethe above expression as x α x aN γ +1 γ ⊗ x β − q γβ x α x aN γ γ x β ⊗ x γ + q αβ q aN γ +1 αγ x aN γ +1 γ x β ⊗ x α + b q aN γ +1 γβ { ( q αγ q γβ − a + 1) ( qαγqγβ ) Nγ } x N γ ( a +1) γ ⊗ n X j =1 b j q aN γ +1 αγ x aN γ +1 γ x η j ⊗ x δ j , which agrees with the second claimed formula in the lemma. (cid:3) Lemma 10.1.10.
Let α < η < γ < β < δ be positive roots such that N γ = 2 and therelations among the corresponding root vectors take the form x α x β = q αβ x β x α + b x γ x η , x η x δ = q ηδ x δ x η + b x γ , (10.1.11) for some scalars b , b and the other pairs of root vectors q -commute. Then, for all n ≥ , d ( x α x nγ x β x δ ⊗
1) = x α x nγ x β ⊗ x δ − q βδ x α x nγ x δ ⊗ x β + q γβ q γδ x α x n − γ x β x δ ⊗ x γ + ( − q αγ ) n q αβ q αδ x nγ x β x δ ⊗ x α − q ηδ b ( − n − q nαγ ( n + 1) e q βγ x n +1 γ x δ ⊗ x η + b b c ( n ) αβγ ( − q γα ) − n x n +2 γ ⊗ . (10.1.12) where c ( n ) αβγ := n P k =0 ( − e q αγ ) k ( k + 1) e q βγ , n ∈ N . Notice that the equalities in (10.1.11) force γ + η = α + β, η + δ = γ. (10.1.13)Hence the following equality also holds: 2 γ = α + β + δ . Proof.
First we claim that d ( x α x nγ x β ⊗
1) = x α x nγ ⊗ x β − q γβ x α x n − γ x β ⊗ x γ − q αβ ( − q αγ ) n x nγ x β ⊗ x α − b ( − q αγ ) n ( n + 1) e q βγ x n +1 γ ⊗ x η (10.1.14)The proof is by induction on n . When n = 0, d ( x α x β ⊗
1) = x α ⊗ x β − s d ( x α ⊗ x β ) = x α ⊗ x β − q αβ x β ⊗ x α − b x γ ⊗ x η . Now assume that (10.1.14) holds for n . Let c n = ( − q αγ ) n ( n + 1) e q βγ . We compute: d ( x α x n +1 γ x β ⊗
1) = x α x n +1 γ ⊗ x β − sd ( x α x n +1 γ ⊗ x β )= x α x n +1 γ ⊗ x β − s (cid:0) q γβ x α x nγ ⊗ x β x γ + ( − q αγ ) n +1 x n +1 γ ⊗ ( q αβ x β x α + b x γ x η ) (cid:1) = x α x n +1 γ ⊗ x β − q γβ x α x nγ x β ⊗ x γ − s (cid:0) ( − q αγ ) n +1 q αβ x n +1 γ ⊗ x β x α + ( − q αγ ) n +1 b x n +1 γ ⊗ x γ x η + q γβ q αβ ( − q α γ ) n x nγ x β ⊗ x α x γ + q γβ b c n x n +1 γ ⊗ x η x γ (cid:1) = x α x n +1 γ ⊗ x β − q γβ x α x nγ x β ⊗ x γ − b (cid:0) q γβ q ηγ c n + ( − q αγ ) n +1 (cid:1) x n +2 γ ⊗ x η OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 81 − ( − q αγ ) n +1 q αβ sd (cid:0) x n +1 γ x β ⊗ x α (cid:1) . Now the inductive step follows using Remark 10.1.2 and q ηγ = q αγ q βγ q − γγ = − q αγ q βγ . Next we prove by induction on n that there exist e n ∈ k such that: d ( x α x nγ x β x δ ⊗
1) = x α x nγ x β ⊗ x δ − q βδ x α x nγ x δ ⊗ x β + q γβ q γδ x α x n − γ x β x δ ⊗ x γ + ( − q αγ ) n q αβ q αδ x nγ x β x δ ⊗ x α − q ηδ b c n − x n +1 γ x δ ⊗ x η + b b e n x n +2 γ ⊗ . The proof is again by induction. When n = 0, d ( x α x β x δ ⊗
1) = x α x β ⊗ x δ − s d ( x α x β ⊗ x δ )= x α x β ⊗ x δ − s ( q βδ x α ⊗ x δ x β − q αβ q αδ x β ⊗ x δ x α − b x γ ⊗ ( q ηδ x δ x η + b x γ ))= x α x β ⊗ x δ − q βδ x α x δ ⊗ x β − s (cid:0) − q αβ q αδ x β ⊗ x δ x α − q ηδ b x γ ⊗ x δ x η − b b x γ ⊗ x γ + q βδ q αδ x δ ⊗ ( q αβ x β x α + b x γ x η ) (cid:1) = x α x β ⊗ x δ − q βδ x α x δ ⊗ x β + b b x γ ⊗ q ηδ b x γ x δ ⊗ x η − s (cid:0) − q αβ q αδ x β ⊗ x δ x α + q βδ q αδ x δ ⊗ ( q αβ x β x α + b x γ x η ) − q γδ q ηδ b x δ ⊗ x γ x η (cid:1) = x α x β ⊗ x δ − q βδ x α x δ ⊗ x β + b b x γ ⊗ q ηδ b x γ x δ ⊗ x η + q αβ q αδ x β x δ ⊗ x α − b ( q βδ q αδ − q γδ q ηδ ) s (cid:0) x δ ⊗ x γ x η (cid:1) . Now assume that the formula holds for n . Using (10.1.14): d ( x α x n +1 γ x β x δ ⊗
1) = x α x n +1 γ x β ⊗ x δ − sd ( x α x n +1 γ x β ⊗ x δ )= x α x n +1 γ x β ⊗ x δ − s (cid:0) q βδ x α x n +1 γ ⊗ x δ x β − q γβ q γδ x α x nγ x β ⊗ x δ x γ − q αβ q αδ ( − q α γ ) n +1 x n +1 γ x β ⊗ x δ x α + b c n +1 x n +2 γ ⊗ ( q ηδ x δ x η + b x γ ) (cid:1) = x α x n +1 γ x β ⊗ x δ − q βδ x α x n +1 γ x δ ⊗ x β − s (cid:0) q βδ q γδ q γβ x α x nγ x δ ⊗ x β x γ + q βδ q αδ ( − q αγ ) n +1 x n +1 γ x δ ⊗ ( q αβ x β x α + b x γ x η ) − q γβ q γδ x α x nγ x β ⊗ x δ x γ − q αβ q αδ ( − q α γ ) n +1 x n +1 γ x β ⊗ x δ x α − b c n +1 x n +2 γ ⊗ ( q ηδ x δ x η + b x γ ) (cid:1) Next we use the inductive hypothesis, the relation x γ = 0, (10.1.5) and (10.1.6): d ( x α x n +1 γ x β x δ ⊗
1) = x α x n +1 γ x β ⊗ x δ − q βδ x α x n +1 γ x δ ⊗ x β + q γβ q γδ x α x nγ x β x δ ⊗ x γ − s (cid:0) − q γβ q γδ ( − q αγ ) n +1 q αβ q αδ x nγ x β x δ ⊗ x γ x α − q γβ q γδ q ηγ q ηδ b c n +1 x n +1 γ x δ ⊗ x γ x η + b b e n q γβ q γδ x n +2 γ ⊗ x γ + q βδ q αδ ( − q αγ ) n +1 x n +1 γ x δ ⊗ ( q αβ x β x α + b x γ x η ) − q αβ q αδ ( − q αγ ) n +1 x n +1 γ x β ⊗ x δ x α − b c n +1 x n +2 γ ⊗ ( q ηδ x δ x η + b x γ ) (cid:1) = x α x n +1 γ x β ⊗ x δ − q βδ x α x n +1 γ x δ ⊗ x β + q γβ q γδ x α x nγ x β x δ ⊗ x γ + q αβ q αδ ( − q αγ ) n +1 x n +1 γ x β x δ ⊗ x α − b b ( e n q γβ q γδ − c n +1 ) x n +3 γ ⊗ q ηδ b c n +1 x n +2 γ x δ ⊗ x η , and the inductive step follows. To finish the proof we have to compute e n . Note that e = 1 , e n +1 = − e n q γβ q γδ + c n +1 , for all n ≥ . By (10.1.13) and using that q γγ = − q γδ = q γγ q − γα q − γβ = q − γα q − γβ . Hence e n = c ( n ) αβγ ( − q γα ) − n for all n ≥ (cid:3) Lemma 10.1.15.
Let α < η < γ < τ < β < δ be positive roots such that N γ = N α = 2 and the relations among the corresponding root vectors take the form x α x β = q αβ x β x α + b x τ x γ , x η x δ = q ηδ x δ x η + b x γ ,x α x τ = q ατ x τ x α + b x γ x η , x η x β = q ηβ x β x η + b x τ , (10.1.16) for some scalars b j ∈ k and the other pairs of root vectors q -commute. Then, for all n ≥ , d ( x α x nγ x β x δ ⊗
1) = x α x nγ x β ⊗ x δ − q βδ x α x nγ x δ ⊗ x β + q γβ q γδ x α x n − γ x β x δ ⊗ x γ + ( − q αγ ) n q αβ q αδ x α x nγ x β x δ ⊗ x α + q nγβ q n +1 ηγ q ηδ b b x n +2 γ x δ ⊗ x η + q n +1 αγ q n +2 γα q γβ (cid:16) ( n + 1) e q γα e q γβ + n X j =1 c ( j ) ατγ (cid:17) b b b x n +3 γ ⊗ . (10.1.17)Notice that the equalities in (10.1.16) force γ + τ = α + β, η + δ = γ, α + τ = γ + η, η + β = 2 τ. (10.1.18)Thus the following equality also holds: 3 γ = 2 α + β + δ . Proof.
Let n ∈ N . A computation similar to (10.1.14) proves that d ( x α x nγ x τ ⊗
1) = x α x nγ ⊗ x τ − q γτ x α x n − γ x τ ⊗ x γ − q ατ ( − q αγ ) n x nγ x τ ⊗ x α − b ( − q αγ ) n ( n + 1) e q τγ x n +1 γ ⊗ x η , (10.1.19) d ( x α x nγ x β ⊗
1) = x α x nγ ⊗ x β − q γβ x α x n − γ x β ⊗ x γ − q αβ ( − q αγ ) n x nγ x β ⊗ x α − ( − q αγ ) n b x nγ x τ ⊗ x γ . (10.1.20)Now we compute more differentials: d ( x α x nγ x β ⊗
1) = x α x nγ ⊗ x β − q γβ x α x n − γ x β ⊗ x γ − q αβ ( − q αγ ) n x α x nγ x β ⊗ x α + q nγβ q n +1 ηγ b b x n +2 γ ⊗ x η − ( − q αγ ) n +1 b x α x nγ x τ ⊗ x γ , (10.1.21) d ( x α x nγ x β x δ ⊗
1) = x α x nγ x β ⊗ x δ − q βδ x α x nγ x δ ⊗ x β + q γβ q γδ x α x n − γ x β x δ ⊗ x γ + ( − q αγ ) n q αβ q αδ x nγ x β x δ ⊗ x α + ( − q αγ ) n q γδ b x nγ x τ x δ ⊗ x γ , (10.1.22)First we prove (10.1.21) by induction on n . For n = 0, d ( x α x β ⊗
1) = x α ⊗ x β − s (cid:0) q αβ x α ⊗ x β x α + b x α ⊗ x τ x γ (cid:1) = x α ⊗ x β − b x α x τ ⊗ x γ − q αβ x α x β ⊗ x α + s (cid:0) q ηγ b b x γ ⊗ x γ x η + ( q ατ q αγ + q αβ ) b x τ ⊗ x γ x α (cid:1) = x α ⊗ x β − b x α x τ ⊗ x γ − q αβ x α x β ⊗ x α + q ηγ b b x γ ⊗ x η . Assume that (10.1.21) holds for n . By inductive hypothesis, x γ = 0, (10.1.19) and(10.1.20): d ( x α x n +1 γ x β ⊗
1) = x α x n +1 γ ⊗ x β − s (cid:0) q γβ x α x nγ ⊗ x β x γ + ( − q αγ ) n +1 x α x n +1 γ ⊗ x α x β (cid:1) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 83 = x α x n +1 γ ⊗ x β − q γβ x α x nγ x β ⊗ x γ − s (cid:0) q γβ q αβ ( − q αγ ) n x α x nγ x β ⊗ x α x γ − q n +1 γβ q n +1 ηγ b b x n +2 γ ⊗ x η x γ + ( − q αγ ) n +1 q αβ x α x n +1 γ ⊗ x β x α + ( − q αγ ) n +1 b x α x n +1 γ ⊗ x τ x γ (cid:1) = x α x n +1 γ ⊗ x β − q γβ x α x nγ x β ⊗ x γ − ( − q αγ ) n +1 q αβ x α x n +1 γ x β ⊗ x α − s (cid:0) ( − q αγ ) n +1 b x α x n +1 γ ⊗ x τ x γ − q n +1 γβ q n +1 ηγ b b x n +2 γ ⊗ x η x γ + ( − q αγ ) n +2 q αβ b x n +1 γ x τ ⊗ x γ x α (cid:1) = x α x n +1 γ ⊗ x β − q γβ x α x nγ x β ⊗ x γ − ( − q αγ ) n +1 q αβ x α x n +1 γ x β ⊗ x α − ( − q αγ ) n +1 b x α x n +1 γ x τ ⊗ x γ + q n +1 γβ q n +2 ηγ b b s (cid:0) x n +2 γ ⊗ x γ x η ) (cid:1) , so (10.1.21) follows since s (cid:0) x n +2 γ ⊗ x γ x η ) = x n +3 γ ⊗ x η .Now we prove (10.1.22) by induction on n . For n = 0, d ( x α x β x δ ⊗
1) = x α x β ⊗ x δ − s (cid:0) q βδ x α ⊗ x δ x β − q αβ q αδ x β ⊗ x δ x α − q γδ b x τ ⊗ x δ x γ (cid:1) = x α x β ⊗ x δ − q βδ x α x δ ⊗ x β − s (cid:0) q βδ q αδ x δ ⊗ ( q αβ x β x α + b x τ x γ ) − q αβ q αδ x β ⊗ x δ x α − q γδ b x τ ⊗ x δ x γ (cid:1) = x α x β ⊗ x δ − q βδ x α x δ ⊗ x β + q γδ b x τ x δ ⊗ x γ + q αβ q αδ x β x δ ⊗ x α . Now assume that (10.1.22) holds for n . Using (10.1.20), Remark 10.1.2 three times, in-ductive hypothesis, x γ = 0 = x α , d ( x α x n +1 γ x β x δ ⊗
1) = x α x n +1 γ x β ⊗ x δ − s (cid:0) q βδ x α x n +1 γ ⊗ x δ x β − q γβ q γδ x α x nγ x β ⊗ x δ x γ − q αβ ( − q αγ ) n +1 q αδ x n +1 γ x β ⊗ x δ x α − ( − q αγ ) n +1 q γδ b x n +1 γ x τ ⊗ x δ x γ (cid:1) = x α x n +1 γ x β ⊗ x δ − q βδ x α x n +1 γ x δ ⊗ x β − s (cid:0) − q γβ q γδ x α x nγ x β ⊗ x δ x γ + q βδ q γδ q γβ x α x nγ x δ ⊗ x β x γ + q αδ ( − q αγ ) n +1 q βδ x n +1 γ x δ ⊗ ( q αβ x β x α + b x τ x γ ) − q αβ ( − q αγ ) n +1 q αδ x n +1 γ x β ⊗ x δ x α − ( − q αγ ) n +1 q γδ b x n +1 γ x τ ⊗ x δ x γ (cid:1) = x α x n +1 γ x β ⊗ x δ − q βδ x α x n +1 γ x δ ⊗ x β + q γβ q γδ x α x nγ x β x δ ⊗ x γ − s (cid:0) − q γβ q γδ ( − q αγ ) n +1 q αβ q αδ x nγ x β x δ ⊗ x γ x α + q αβ q αδ ( − q αγ ) n +1 q βδ x n +1 γ x δ ⊗ x β x α + q αδ ( − q αγ ) n +1 q βδ b x n +1 γ x δ ⊗ x τ x γ − q αβ ( − q αγ ) n +1 q αδ x n +1 γ x β ⊗ x δ x α − ( − q αγ ) n +1 q γδ b x n +1 γ x τ ⊗ x δ x γ (cid:1) = x α x n +1 γ x β ⊗ x δ − q βδ x α x n +1 γ x δ ⊗ x β + q γβ q γδ x α x nγ x β x δ ⊗ x γ + ( − q αγ ) n +1 q γδ b x n +1 γ x τ x δ ⊗ x γ − s (cid:0) − q γβ q γδ ( − q αγ ) n +1 q αβ q αδ x nγ x β x δ ⊗ x γ x α + q αβ q αδ ( − q αγ ) n +1 q βδ x n +1 γ x δ ⊗ x β x α − q αβ ( − q αγ ) n +1 q αδ x n +1 γ x β ⊗ x δ x α (cid:1) = x α x n +1 γ x β ⊗ x δ − q βδ x α x n +1 γ x δ ⊗ x β + q γβ q γδ x α x nγ x β x δ ⊗ x γ + ( − q αγ ) n +1 q γδ b x n +1 γ x τ x δ ⊗ x γ + q αβ ( − q αγ ) n +1 q αδ x n +1 γ x β x δ ⊗ x α . Finally we prove (10.1.17) by induction on n . Notice that root vectors corresponding to α < η < γ < τ < δ satisfy (10.1.11), so d ( x α x nγ x τ x δ ⊗
1) is given by (10.1.12). We claim that d ( x α x nγ x β x δ ⊗
1) = x α x nγ x β ⊗ x δ − q βδ x α x nγ x δ ⊗ x β + q γβ q γδ x α x n − γ x β x δ ⊗ x γ + ( − q αγ ) n q αβ q αδ x α x nγ x β x δ ⊗ x α + q nγβ q n +1 ηγ q ηδ b b x n +2 γ x δ ⊗ x η + c n b b b x n +3 γ ⊗ c n . For n = 0, d ( x α x β x δ ⊗
1) = x α x β ⊗ x δ − s (cid:0) q βδ x α ⊗ x δ x β − q γδ b x α x τ ⊗ x δ x γ − q αβ q αδ x α x β ⊗ x δ x α + q ηγ b b x γ ⊗ ( q ηδ x δ x η + b x γ ) (cid:1) = x α x β ⊗ x δ − q βδ x α x δ ⊗ x β − s (cid:0) − q γδ b x α x τ ⊗ x δ x γ − q αβ q αδ x α x β ⊗ x δ x α + q ηγ q ηδ b b x γ ⊗ x δ x η + q ηγ b b b x γ ⊗ x γ + q βδ q αδ x α x δ ⊗ ( q αβ x β x α + b x τ x γ ) (cid:1) = x α x β ⊗ x δ − q βδ x α x δ ⊗ x β + q γδ b x α x τ x δ ⊗ x γ − s (cid:0) − q αβ q αδ x α x β ⊗ x δ x α + q ατ q αδ q γδ b x τ x δ ⊗ x α x γ + q ηδ q γδ b b x γ x δ ⊗ x η x γ + q ηγ q ηδ b b x γ ⊗ x δ x η + ( q ηγ + q γδ ) b b b x γ ⊗ x γ + q βδ q αδ q αβ x α x δ ⊗ x β x α (cid:1) = x α x β ⊗ x δ − q βδ x α x δ ⊗ x β + q γδ b x α x τ x δ ⊗ x γ + q αβ q αδ x α x β x δ ⊗ x α + ( q ηγ + q γδ ) b b b x γ ⊗ − s (cid:0) ( q αβ + q ατ q αγ ) q αδ q γδ b x τ x δ ⊗ x γ x α + q ηδ q γδ b b x γ x δ ⊗ x η x γ + q ηγ q ηδ b b x γ ⊗ x δ x η (cid:1) = x α x β ⊗ x δ − q βδ x α x δ ⊗ x β + q γδ b x α x τ x δ ⊗ x γ + q αβ q αδ x α x β x δ ⊗ x α + q γδ (1 − e q − γδ ) b b b x γ ⊗ q ηγ q ηδ b b x γ x δ ⊗ x η . Now assume that (10.1.17) holds for n . Using (10.1.21), Remark 10.1.2, inductive hypoth-esis, x γ = 0 = x α , (10.1.12), (10.1.22), d ( x α x n +1 γ x β x δ ⊗
1) = x α x n +1 γ x β ⊗ x δ − s (cid:0) q βδ x α x n +1 γ ⊗ x δ x β − q γβ q γδ x α x nγ x β ⊗ x δ x γ − q αβ ( − q αγ ) n +1 q αδ x α x n +1 γ x β ⊗ x δ x α − ( − q αγ ) n +2 q γδ b x α x n +1 γ x τ ⊗ x δ x γ + q n +1 γβ q n +2 ηγ b b b x n +3 γ ⊗ x γ + q n +1 γβ q n +2 ηγ q ηδ b b x n +3 γ ⊗ x δ x η (cid:1) = x α x n +1 γ x β ⊗ x δ − q βδ x α x n +1 γ x δ ⊗ x β − s (cid:0) q γβ q γδ q βδ x α x nγ x δ ⊗ x β x γ + q αδ ( − q αγ ) n +1 q βδ x α x n +1 γ x δ ⊗ ( q αβ x β x α + b x τ x γ ) − q γβ q γδ x α x nγ x β ⊗ x δ x γ − q αβ ( − q αγ ) n +1 q αδ x α x n +1 γ x β ⊗ x δ x α − ( − q αγ ) n +2 q γδ b x α x n +1 γ x τ ⊗ x δ x γ + q n +1 γβ q n +2 ηγ b b b x n +3 γ ⊗ x γ + q n +1 γβ q n +2 ηγ q ηδ b b x n +3 γ ⊗ x δ x η (cid:1) = x α x n +1 γ x β ⊗ x δ − q βδ x α x n +1 γ x δ ⊗ x β + q γβ q γδ x α x nγ x β x δ ⊗ x γ − s (cid:0) − q γβ q γδ ( − q αγ ) n +1 q αβ q αδ x α x nγ x β x δ ⊗ x γ x α + q γβ q γδ c n b b b x n +3 γ ⊗ x γ + q γβ q γδ q ηγ q nγβ q n +1 ηγ q ηδ b b x n +2 γ x δ ⊗ x γ x η + q αβ q αδ ( − q αγ ) n +1 q βδ x α x n +1 γ x δ ⊗ x β x α + q αδ ( − q αγ ) n +1 q βδ b x α x n +1 γ x δ ⊗ x τ x γ − q αβ ( − q αγ ) n +1 q αδ x α x n +1 γ x β ⊗ x δ x α − ( − q αγ ) n +2 q γδ b x α x n +1 γ x τ ⊗ x δ x γ + q n +1 γβ q n +2 ηγ b b b x n +3 γ ⊗ x γ + q n +1 γβ q n +2 ηγ q ηδ b b x n +3 γ ⊗ x δ x η (cid:1) = x α x n +1 γ x β ⊗ x δ − q βδ x α x n +1 γ x δ ⊗ x β + q γβ q γδ x α x nγ x β x δ ⊗ x γ OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 85 + ( − q αγ ) n +2 q γδ b x α x n +1 γ x τ x δ ⊗ x γ − s (cid:0) − q γβ q γδ ( − q αγ ) n +1 q αβ q αδ x α x nγ x β x δ ⊗ x γ x α + q γβ q γδ c n b b b x n +3 γ ⊗ x γ + q γβ q γδ q ηγ q nγβ q n +1 ηγ q ηδ b b x n +2 γ x δ ⊗ x γ x η + q αβ q αδ ( − q αγ ) n +1 q βδ x α x n +1 γ x δ ⊗ x β x α − q αβ ( − q αγ ) n +1 q αδ x α x n +1 γ x β ⊗ x δ x α + q n +1 γβ q n +2 ηγ b b b x n +3 γ ⊗ x γ + q n +1 γβ q n +2 ηγ q ηδ b b x n +3 γ ⊗ x δ x η − q γδ q ηγ q ηδ q n +3 αγ ( n + 2) e q τγ b b x n +2 γ x δ ⊗ x γ x η − q γδ q n +4 αγ q ατ q αδ b x n +1 γ x τ x δ ⊗ x γ x α + ( − q αγ ) n +2 q γδ b b b ( − q γα ) − n − c ( n +1) ατγ x n +3 γ ⊗ x γ (cid:1) = x α x n +1 γ x β ⊗ x δ − q βδ x α x n +1 γ x δ ⊗ x β + q γβ q γδ x α x nγ x β x δ ⊗ x γ + ( − q αγ ) n +2 q γδ b x α x n +1 γ x τ x δ ⊗ x γ + q αβ ( − q αγ ) n +1 q αδ x α x n +1 γ x β x δ ⊗ x α − s (cid:0) q n +1 γβ q n +2 ηγ q ηδ b b x n +3 γ ⊗ x δ x η + ( q γβ q nγβ q n +1 ηγ q ηδ − q ηδ q n +3 αγ ( n + 2) e q τγ ) q γδ q ηγ b b x n +2 γ x δ ⊗ x γ x η + (cid:16) q n +1 γβ q n +2 ηγ + q γβ q γδ c n − q n +2 αγ q γδ q − n − γα c ( n +1) ατγ (cid:17) b b b x n +3 γ ⊗ x γ (cid:1) . Hence the claim follows using Remark 10.1.2 and that d ( x n +4 γ ⊗
1) = x n +3 γ ⊗ x γ . The scalars c n are defined recursively by the equation: c n +1 = q n +1 γβ q n +2 ηγ + q γβ q γδ c n − q n +2 αγ q γδ q − n − γα c ( n +1) ατγ . Thus (10.1.17) follows using (10.1.18) to express all the roots in terms of α , β , γ . (cid:3) Lemma 10.1.23.
Let α < β < γ < τ < η < δ be positive roots such that N γ = 2 and therelations among the corresponding root vectors take the form x α x δ = q αδ x δ x α + b x η x γ , x β x δ = q βδ x δ x β + b x η x τ ,x γ x δ = q γδ x δ x γ + b x η x τ , x α x τ = q ατ x τ x α + b x γ x β ,x β x η = q βη x η x β + b x γ , (10.1.24) for some scalars b j ∈ k and the other pairs of root vectors q -commute. Then, for all n ≥ , d ( x α x β x nγ x δ ⊗
1) = x α x β x nγ ⊗ x δ − q γδ x α x β x n − γ x δ ⊗ x γ − ( − q βγ ) n q βδ x α x nγ x δ ⊗ x β + q αβ ( − q αγ ) n q αδ x β x nγ x δ ⊗ x α − (cid:0) q n − γη ( n ) − qβγqγη b b + ( − q βγ ) n b (cid:1) x α x nγ x η ⊗ x τ + q αβ ( − q αγ ) n b b x β x nγ x η ⊗ x γ − b x α x β x n − γ x η ⊗ x η x τ − q αβ q nαγ q nγβ n X k =0 ( − e q αγ ) k ( k + 1) e q βγ ! b b x n +2 γ ⊗ . (10.1.25)Notice that the equalities in (10.1.24) force α + δ = γ + η, β + δ = η + τ, γ + δ = 2 η + τ, η + β = γ. (10.1.26)Thus the following equality also holds: 2 γ = α + β + δ . Proof.
A direct recursive computation on n ∈ N shows that d ( x nγ x δ ⊗
1) = x nγ ⊗ x δ − q γδ x n − γ x δ ⊗ x γ − b x n − γ x η ⊗ x η x τ . (10.1.27) The root vectors corresponding to β < γ < η satisfy (10.1.8), so by Lemma 10.1.7, d ( x β x nγ x η ⊗
1) = x β x nγ ⊗ x η − q γη x β x n − γ x η ⊗ x γ − q βη ( − q βγ ) n x nγ x η ⊗ x β + b q nγη ( n + 1) − qβγqγη x n +1 γ ⊗ . (10.1.28)We need more auxiliary results: d ( x α x β x nγ x η ⊗
1) = x α x β x nγ ⊗ x η − q γη x α x β x n − γ x η ⊗ x γ − ( − q βγ ) n q βη x α x nγ x η ⊗ x β + q αβ ( − q αγ ) n q αη x β x nγ x η ⊗ x α − q nγη ( n + 1) − qβγqγη b x α x n +1 γ ⊗ , (10.1.29) d ( x α x nγ x δ ⊗
1) = x α x nγ ⊗ x δ − q γδ x α x n − γ x δ ⊗ x γ − ( − q αγ ) n q αδ x nγ x δ ⊗ x α − b x α x n − γ x η ⊗ x η x τ − ( − q αγ ) n b x nγ x η ⊗ x γ , (10.1.30) d ( x β x nγ x δ ⊗
1) = x β x nγ ⊗ x δ − q γδ x β x n − γ x δ ⊗ x γ − ( − q βγ ) n q βδ x nγ x δ ⊗ x β − b x β x n − γ x η ⊗ x η x τ − (cid:0) ( − q βγ ) n b − b b q nγη ( n ) − qβγqγη (cid:1) x nγ x η ⊗ x τ . (10.1.31)We start with the proof of (10.1.29) by induction on n . For n = 0, d ( x α x β x η ⊗
1) = x α x β ⊗ x η − s (cid:16) q βη x α ⊗ x η x β + b x α ⊗ x γ − q αβ q αη x β ⊗ x η x α (cid:17) = x α x β ⊗ x η − q βη x α x η ⊗ x β + q αβ q αη x β x η ⊗ x α − b x α x γ ⊗ . Now assume that (10.1.29) holds for n . By Remark 10.1.2, inductive hypothesis and(10.1.28), d ( x α x β x n +1 γ x η ⊗
1) = x α x β x n +1 γ ⊗ x η − s (cid:16) q γη x α x β x nγ ⊗ x η x γ + ( − q βγ ) n +1 x α x n +1 γ ⊗ ( q βη x η x β + b x γ ) − q αβ ( − q αγ ) n +1 q αη x β x n +1 γ ⊗ x η x α (cid:17) = x α x β x n +1 γ ⊗ x η − q γη x α x β x nγ x η ⊗ x γ − s (cid:16) ( − q βγ ) n q βη q γη q βγ x α x nγ x η ⊗ x γ x β − q αβ ( − q αγ ) n q αη q γη q αγ x β x nγ x η ⊗ x γ x α − q αβ ( − q αγ ) n +1 q αη x β x n +1 γ ⊗ x η x α + ( − q βγ ) n +1 q βη x α x n +1 γ ⊗ x η x β + ( q n +1 γη ( n + 1) − qβγqγη + ( − q βγ ) n +1 ) b x α x n +1 γ ⊗ x γ (cid:17) = x α x β x n +1 γ ⊗ x η − q γη x α x β x nγ x η ⊗ x γ − q n +1 γη ( n + 2) − qβγqγη b x α x n +2 γ ⊗ − ( − q βγ ) n +1 q βη x α x n +1 γ x η ⊗ x β + q αβ ( − q αγ ) n +1 q αη x β x n +1 γ x η ⊗ x α . Next we prove (10.1.30) by induction on n . For n = 0, d ( x α x δ ⊗
1) = x α ⊗ x δ − q αδ x δ ⊗ x α − b x η ⊗ x γ . Now assume that (10.1.30) holds for n . Using Remark 10.1.2 three times, inductivehypothesis and (10.1.27): d ( x α x n +1 γ x δ ⊗
1) = x α x n +1 γ ⊗ x δ − s (cid:16) x α x nγ ⊗ ( q γδ x δ x γ + b x η x τ )+ ( − q αγ ) n +1 x n +1 γ ⊗ ( q αδ x δ x α + b x η x γ ) (cid:17) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 87 = x α x n +1 γ ⊗ x δ − b x α x nγ x η ⊗ x η x τ − q γδ x α x nγ x δ ⊗ x γ − s (cid:16) ( − q αγ ) n +1 b x n +1 γ ⊗ x η x γ + q αδ ( − q αγ ) n +1 x n +1 γ ⊗ x δ x α − ( − q αγ ) n +1 q αδ q γδ x nγ x δ ⊗ x γ x α + q αη ( − q αη ) n q ατ b x nγ x η ⊗ x η x τ x α + q αη ( − q αγ ) n b b x nγ x η ⊗ x η x γ x β (cid:17) = x α x n +1 γ ⊗ x δ − b x α x nγ x η ⊗ x η x τ − q γδ x α x nγ x δ ⊗ x γ − ( − q αγ ) n +1 b x n +1 γ x η ⊗ x γ − q αδ ( − q αγ ) n +1 x n +1 γ x δ ⊗ x α . Now we prove (10.1.31) by induction on n . For n = 0, d ( x β x δ ⊗
1) = x β ⊗ x δ − q βδ x δ ⊗ x β − b x η ⊗ x τ . We assume that (10.1.31) holds for n . Using Remark 10.1.2 three times, (10.1.28),inductive hypothesis and (10.1.27): d ( x β x n +1 γ x δ ⊗
1) = x β x n +1 γ ⊗ x δ − s (cid:16) q γδ x β x nγ ⊗ x δ x γ + b x β x nγ ⊗ x η x τ + q βδ ( − q βγ ) n +1 x n +1 γ ⊗ x δ x β + ( − q βγ ) n +1 b x n +1 γ ⊗ x η x τ (cid:17) = x β x n +1 γ ⊗ x δ − q γδ x β x nγ x δ ⊗ x γ − b x β x nγ x η ⊗ x η x τ − s (cid:16) d n q γδ x nγ x η ⊗ x τ x γ − q γδ ( − q βγ ) n +1 q βδ x nγ x δ ⊗ x γ x β + q βδ ( − q βγ ) n +1 x n +1 γ ⊗ x δ x β + q βη q βτ ( − q βγ ) n b x nγ x η ⊗ x η x τ x β + q βη ( − q βγ ) n q γτ b b x nγ x η ⊗ x τ x γ + (cid:0) ( − q βγ ) n +1 b − b b q nγη ( n + 1) − qβγqγη (cid:1) x n +1 γ ⊗ x η x τ (cid:17) = x β x n +1 γ ⊗ x δ − q γδ x β x nγ x δ ⊗ x γ − b x β x nγ x η ⊗ x η x τ − q βδ ( − q βγ ) n +1 x n +1 γ x δ ⊗ x β − (cid:0) ( − q βγ ) n +1 b − b b q n +1 γη ( n + 1) − qβγqγη (cid:1) x n +1 γ x η ⊗ x τ . Finally we prove (10.1.25). To do so, we prove that there exist c n , d n , e n ∈ k such that d ( x α x β x nγ x δ ⊗
1) = x α x β x nγ ⊗ x δ − q γδ x α x β x n − γ x δ ⊗ x γ − ( − q βγ ) n q βδ x α x nγ x δ ⊗ x β + q αβ ( − q αγ ) n q αδ x β x nγ x δ ⊗ x α − c n x α x nγ x η ⊗ x τ − b x α x β x n − γ x η ⊗ x η x τ + d n x nγ x η ⊗ x γ − e n b b x n +2 γ ⊗ . For n = 0, d ( x α x β x δ ⊗
1) = x α x β ⊗ x δ − s (cid:0) x α ⊗ ( q βδ x δ x β + b x η x τ ) − q αβ x β ⊗ x α x δ (cid:1) = x α x β ⊗ x δ − q βδ x α x δ ⊗ x β − b x α x η ⊗ x τ + s (cid:16) q αβ q αδ x β ⊗ x δ x α + q αβ b x β ⊗ x η x γ − ( q βδ b + q αη b b ) x η ⊗ x γ x β − q ατ q αη b x η ⊗ x τ x α − q αβ q αδ q βδ x δ ⊗ x β x α (cid:17) = x α x β ⊗ x δ − q βδ x α x δ ⊗ x β − b x α x η ⊗ x τ + q αβ q αδ x β x δ ⊗ x α + q αβ b x β x η ⊗ x γ + q αβ b b x γ ⊗ − s (cid:16) ( q βδ b + q αη b b − q βη q βγ q αβ b ) s ( x η x γ x β ) (cid:17) . We assume that (10.1.25) holds for n . Using Remark 10.1.2 twice, (10.1.29), inductivehypothesis, (10.1.30), (10.1.28), (10.1.31) d ( x α x β x n +1 γ x δ ⊗
1) = x α x β x n +1 γ ⊗ x δ − s (cid:16) q γδ x α x β x nγ ⊗ x δ x γ + b x α x β x nγ ⊗ x η x τ + ( − q βγ ) n +1 x α x n +1 γ ⊗ x β x δ − q αβ ( − q αγ ) n +1 x β x n +1 γ ⊗ x α x δ (cid:17) = x α x β x n +1 γ ⊗ x δ − q γδ x α x β x nγ x δ ⊗ x γ − b x α x β x nγ x η ⊗ x η x τ − s (cid:16) q γδ e n b b x n +2 γ ⊗ x γ + q γδ ( − q βγ ) n q βγ q βδ x α x nγ x δ ⊗ x γ x β + q γδ q αβ ( − q αγ ) n +1 q αδ x β x nγ x δ ⊗ x γ x α + ( − q βγ ) n +1 q βδ x α x n +1 γ ⊗ x δ x β + q γδ c n b x α x nγ x η ⊗ x τ x γ + ( − q βγ ) n q βη q βτ b x α x nγ x η ⊗ x η x τ x β + ( − q βγ ) n q βη q γτ b b x α x nγ x η ⊗ x τ x γ − q αβ ( − q αγ ) n q αη q ατ b x β x nγ x η ⊗ x η x τ x α − q αβ ( − q αγ ) n q αη b b x β x nγ x η ⊗ x η x γ x β + (cid:0) q nγη ( n + 1) − qβγqγη b b + ( − q βγ ) n +1 b (cid:1) x α x n +1 γ ⊗ x η x τ − q αβ ( − q αγ ) n +1 b x β x n +1 γ ⊗ x η x γ − q αβ ( − q αγ ) n +1 q αδ x β x n +1 γ ⊗ x δ x α (cid:17) = x α x β x n +1 γ ⊗ x δ − q γδ x α x β x nγ x δ ⊗ x γ − ( − q βγ ) n +1 q βδ x α x n +1 γ x δ ⊗ x β − b x α x β x nγ x η ⊗ x η x τ − (cid:0) q nγη ( n + 1) − qβγqγη b b + ( − q βγ ) n +1 b (cid:1) x α x n +1 γ x η ⊗ x τ − s (cid:16) − q αβ ( − q αγ ) n +1 b x β x n +1 γ ⊗ x η x γ − q αβ ( − q αγ ) n +1 q αδ x β x n +1 γ ⊗ x δ x α + (cid:0) q nγη ( n + 1) − qβγqγη b b + ( − q βγ ) n +1 b (cid:1) q αη ( − q αγ ) n +1 x n +1 γ x η ⊗ x α x τ + q γδ e n b b x n +2 γ ⊗ x γ + q αβ q n +1 αγ q αδ q n +1 βγ q βδ x n +1 γ x δ ⊗ x β x α + q γδ q αβ ( − q αγ ) n +1 q αδ x β x nγ x δ ⊗ x γ x α + ( − q βγ ) n +1 q βδ ( − q αγ ) n +1 b x n +1 γ x η ⊗ x γ x β − q αβ ( − q αγ ) n q αη q ατ b x β x nγ x η ⊗ x η x τ x α − q αβ ( − q αγ ) n q αη b b x β x nγ x η ⊗ x η x γ x β (cid:17) = x α x β x n +1 γ ⊗ x δ − q γδ x α x β x nγ x δ ⊗ x γ − ( − q βγ ) n +1 q βδ x α x n +1 γ x δ ⊗ x β − b x α x β x nγ x η ⊗ x η x τ − (cid:0) q nγη ( n + 1) − qβγqγη b b + ( − q βγ ) n +1 b (cid:1) x α x n +1 γ x η ⊗ x τ + q αβ ( − q αγ ) n +1 b x β x n +1 γ x η ⊗ x γ + q αβ ( − q αγ ) n +1 q αδ x β x n +1 γ x δ ⊗ x α − s (cid:16)(cid:0) q αβ ( − q αγ ) n +1 q n +1 γη ( n + 2) − qβγqγη + q γδ e n (cid:1) b b x n +2 γ ⊗ x γ + (cid:0) q nγη ( n + 1) − qβγqγη b b + ( − q βγ ) n +1 b (cid:1) q αη ( − q αγ ) n +1 b s ( x n +1 γ ⊗ x η x γ x β ) − q αβ ( − q αγ ) n q αη b b s ( x β x nγ ⊗ x η x γ x β ) (cid:17) . Hence the inductive step follows since s = 0 and s (cid:0) x n +2 γ ⊗ x γ (cid:1) = x n +3 γ ⊗
1; for the laststep we use the equalities s ( x n +1 γ ⊗ x η x γ x β ) = x n +1 γ x η ⊗ x γ x β , s ( x β x nγ ⊗ x η x γ x β ) = x β x nγ x η ⊗ x η x γ x β , OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 89 which follow since x γ = 0 and x γ x η = q γη x η x γ . The formula for c n , d n is explicit, whilefor the e n ’s we have the recursive expression: e = q αβ , e n +1 = q αβ q n +1 αγ q − n − γβ ( n + 2) e q βγ + q − γα q − γβ e n , n ≥ , where we use (10.1.26) to express δ and η in terms of α , β and γ . (cid:3) Lemma 10.1.32.
Let α < β < δ < γ < τ < ϕ < η be positive roots such that N γ = 2 andthe relations among the corresponding root vectors take the form x α x ϕ = q αϕ x ϕ x α + b x γ ,x β x τ = q βτ x τ x β + b x γ , x δ x η = q δη x η x δ + b x ϕ x τ x γ , (10.1.33) for some scalars b i , and the other pairs of root vectors q -commute. Then, for all n ≥ , d ( x α x β x δ x nγ x η ⊗
1) = x α x β x δ x nγ ⊗ x η − q γη x α x β x δ x n − γ x η ⊗ x γ − ( − q δγ ) n q δη x α x β x nγ x η ⊗ x δ + q βδ ( − q βγ ) n q βη x α x δ x nγ x η ⊗ x β − q αβ q αδ ( − q αγ ) n q αη x β x δ x nγ x η ⊗ x α − ( − q δγ ) n b x α x β x nγ x ϕ ⊗ x τ x γ − q αβ q γϕ ( n + 1) e q αγ ( − q δγ ) n b b x β x n +1 γ x τ ⊗ x γ − q αβ q γη ( − q δγ ) n d ( n ) αβδγ x n +3 γ ⊗ , (10.1.34) where d ( n ) αβδγ := n P k =0 e q kδγ ( k + 1) e q αγ ( k + 2) e q βγ , n ∈ N . Notice that the equalities in (10.1.33) force α + ϕ = γ = β + τ, η + δ = γ + τ + ϕ. (10.1.35)Hence the following equality also holds: 3 γ = α + β + δ + η . Proof.
We may apply Lemma 10.1.7 to the 3-tuples α < γ < ϕ and β < γ < τ to obtain d ( x α x nγ x ϕ ⊗
1) = x α x nγ ⊗ x ϕ − q γϕ x α x n − γ x ϕ ⊗ x γ − ( − q αγ ) n q αϕ x nγ x ϕ ⊗ x α + b q γϕ ( n + 1) e q αγ x n +1 γ ⊗ , (10.1.36) d ( x β x nγ x τ ⊗
1) = x β x nγ ⊗ x τ − q γτ x β x n − γ x τ ⊗ x γ − ( − q βγ ) n q βτ x nγ x τ ⊗ x β + b q γτ ( n + 1) e q βγ x n +1 γ ⊗ , (10.1.37)for all n ≥
0. The next step is to prove by induction on n the following equalities: d ( x α x β x nγ x ϕ ⊗
1) = x α x β x nγ ⊗ x ϕ − q γϕ x α x β x n − γ x ϕ ⊗ x γ − ( − q βγ ) n q βϕ x α x nγ x ϕ ⊗ x β + q αβ ( − q αγ ) n q αϕ x β x nγ x ϕ ⊗ x α + q αβ q nγϕ ( n + 1) e q αγ b x β x n +1 γ ⊗ , (10.1.38) d ( x δ x nγ x η ⊗
1) = x δ x nγ ⊗ x η − q γη x δ x n − γ x η ⊗ x γ − ( − q δγ ) n q δη x nγ x η ⊗ x δ − ( − q δγ ) n b x nγ x ϕ ⊗ x τ x γ , (10.1.39) d ( x α x δ x nγ x η ⊗
1) = x α x δ x nγ ⊗ x η − q γη x α x δ x n − γ x η ⊗ x γ − ( − q δγ ) n q δη x α x nγ x η ⊗ x δ + q αδ ( − q αγ ) n q αη x δ x nγ x η ⊗ x α − ( − q δγ ) n b x α x nγ x ϕ ⊗ x τ x γ − q γϕ ( n + 1) e q αγ ( − q δγ ) n b b x n +1 γ x τ ⊗ x γ , (10.1.40) d ( x β x δ x nγ x η ⊗
1) = x β x δ x nγ ⊗ x η − q γη x β x δ x n − γ x η ⊗ x γ − ( − q δγ ) n q δη x β x nγ x η ⊗ x δ + q βδ ( − q βγ ) n q βη x δ x nγ x η ⊗ x β − ( − q δγ ) n b x β x nγ x ϕ ⊗ x τ x γ . (10.1.41)We start with (10.1.38). For n = 0 we have d ( x α x β x ϕ ⊗
1) = x α x β ⊗ x ϕ − s (cid:0) q βϕ x α ⊗ x ϕ x β − q αβ x β ⊗ ( q αϕ x ϕ x α + b x γ ) (cid:1) = x α x β ⊗ x ϕ − q βϕ x α x ϕ ⊗ x β − s (cid:0) q βϕ q αϕ q αβ x ϕ ⊗ x β x α + q βϕ b x γ ⊗ x β − q αβ q αϕ x β ⊗ x ϕ x α − q αβ b x β ⊗ x γ (cid:1) = x α x β ⊗ x ϕ − q βϕ x α x ϕ ⊗ x β + q αβ q αϕ x β x ϕ ⊗ x α + q αβ b x β x γ ⊗ . We assume that (10.1.38) holds for n . Using Remark 10.1.2 three times, inductive hypoth-esis, (10.1.36), (10.1.33) and x γ = 0, we compute d ( x α x β x n +1 γ x ϕ ⊗
1) = x α x β x n +1 γ ⊗ x ϕ − s (cid:0) q γϕ x α x β x nγ ⊗ x ϕ x γ + ( − q βγ ) n +1 q βϕ x α x n +1 γ ⊗ x ϕ x β − q αβ ( − q αγ ) n +1 x β x n +1 γ ⊗ ( q αϕ x ϕ x α + b x γ ) (cid:1) = x α x β x n +1 γ ⊗ x ϕ − q γϕ x α x β x nγ x ϕ ⊗ x γ − ( − q βγ ) n +1 q βϕ x α x n +1 γ x ϕ ⊗ x β − s (cid:0) q αβ ( − q αγ ) n +1 q αϕ q γϕ x β x nγ x ϕ ⊗ x γ x α + q βϕ q αβ q n +1 αγ q n +1 βγ q αϕ x n +1 γ x ϕ ⊗ x β x α − q αβ ( − q αγ ) n +1 q αϕ x β x n +1 γ ⊗ x ϕ x α − ( − q βγ ) n +1 q βϕ b q γϕ ( n + 2) e q αγ x n +2 γ ⊗ x β − ( q γϕ ( n + 1) e q αγ + ( − q αγ ) n +1 ) q αβ b x β x n +1 γ ⊗ x γ (cid:1) = x α x β x n +1 γ ⊗ x ϕ − q γϕ x α x β x nγ x ϕ ⊗ x γ − ( − q βγ ) n +1 q βϕ x α x n +1 γ x ϕ ⊗ x β + q αβ q γϕ ( n + 2) e q αγ b x β x n +2 γ ⊗ q αβ ( − q αγ ) n +1 q αϕ x β x n +1 γ x ϕ ⊗ x α . Now we prove (10.1.39). For n = 0, d ( x δ x η ⊗
1) = x δ ⊗ x η − q δη x η ⊗ x δ − b x ϕ ⊗ x τ x γ . We assume that (10.1.39) holds for n . By Remark 10.1.2 and inductive hypothesis, d ( x δ x n +1 γ x η ⊗
1) = x δ x n +1 γ ⊗ x η − s (cid:0) q γη x δ x nγ ⊗ x η x γ + ( − q δγ ) n +1 x n +1 γ ⊗ x δ x η (cid:1) = x δ x n +1 γ ⊗ x η − q γη x δ x nγ x η ⊗ x γ − ( − q δγ ) n +1 b x n +1 γ x ϕ ⊗ x τ x γ − ( − q δγ ) n +1 q δη x n +1 γ x η ⊗ x δ . Next we prove (10.1.40). For n = 0, d ( x α x δ x η ⊗
1) = x α x δ ⊗ x η − s (cid:0) q δη x α ⊗ x η x δ + b x α ⊗ x ϕ x τ x γ − q αδ q αη x δ ⊗ x η x α (cid:1) = x α x δ ⊗ x η − b x α x ϕ ⊗ x τ x γ − q δη x α x η ⊗ x δ − s (cid:0) q δη q αη q αδ x η ⊗ x δ x α + q αϕ q ατ q αγ b x ϕ ⊗ x τ x γ x α + b b x γ ⊗ x τ x γ − q αδ q αη x δ ⊗ x η x α (cid:1) = x α x δ ⊗ x η − b x α x ϕ ⊗ x τ x γ − q δη x α x η ⊗ x δ + q αδ q αη x δ x η ⊗ x α − b b x γ x τ ⊗ x γ . We assume that (10.1.40) holds for n . Using Remark 10.1.2 three times, inductive hypoth-esis, (10.1.36), (10.1.39), we have d ( x α x δ x n +1 γ x η ⊗
1) = x α x δ x n +1 γ ⊗ x η − s (cid:0) q γη x α x δ x nγ ⊗ x η x γ + ( − q δγ ) n +1 q δη x α x n +1 γ ⊗ x η x δ + ( − q δγ ) n +1 b x α x n +1 γ ⊗ x ϕ x τ x γ + q αδ ( − q αγ ) n +1 q αη x δ x n +1 γ ⊗ x η x α (cid:1) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 91 = x α x δ x n +1 γ ⊗ x η − q γη x α x δ x nγ x η ⊗ x γ − ( − q δγ ) n +1 b x α x n +1 γ x ϕ ⊗ x τ x γ − s (cid:0) − ( − q δγ ) n +1 q δη q γη x α x nγ x η ⊗ x γ x δ + q αδ ( − q αγ ) n +1 q αη q γη x δ x nγ x η ⊗ x γ x α + ( − q δγ ) n +1 q δη x α x n +1 γ ⊗ x η x δ + q αδ ( − q αγ ) n +1 q αη x δ x n +1 γ ⊗ x η x α + q n +2 αγ q αϕ q ατ q n +1 δγ b x n +1 γ x ϕ ⊗ x τ x γ x α + q γϕ ( n + 2) e q αγ ( − q δγ ) n +1 b b x n +2 γ ⊗ x τ x γ (cid:1) = x α x δ x n +1 γ ⊗ x η − q γη x α x δ x nγ x η ⊗ x γ − ( − q δγ ) n +1 b x α x n +1 γ x ϕ ⊗ x τ x γ − ( − q δγ ) n +1 q δη x α x n +1 γ x η ⊗ x δ + q αδ ( − q αγ ) n +1 q αη x δ x n +1 γ x η ⊗ x α − q γϕ ( n + 2) e q αγ ( − q δγ ) n +1 b b s (cid:0) x n +2 γ ⊗ x τ x γ (cid:1) . As x γ = 0 and x γ x τ = q γτ x τ x γ , we have that s (cid:0) x n +2 γ ⊗ x τ x γ (cid:1) = x n +2 γ x τ ⊗ x γ ; hence theinductive step follows.Now we prove (10.1.41). For n = 0, d ( x β x δ x η ⊗
1) = x β x δ ⊗ x η − s (cid:0) q δη x β ⊗ x η x δ + b x β ⊗ x ϕ x τ x γ − q βδ q βη x δ ⊗ x η x β (cid:1) = x β x δ ⊗ x η − q δη x β x η ⊗ x δ − b x β x ϕ ⊗ x τ x γ − s (cid:0) q δη q βη q βδ x η ⊗ x δ x β + q βϕ b x ϕ ⊗ ( q βτ x τ x β + b x γ ) x γ − q βδ q βη x δ ⊗ x η x β (cid:1) = x β x δ ⊗ x η − q δη x β x η ⊗ x δ − b x β x ϕ ⊗ x τ x γ + q βδ q βη x δ x η ⊗ x β . We assume that (10.1.41) holds for n . Using Remark 10.1.2 three times, inductive hypoth-esis and (10.1.39), we have d ( x β x δ x n +1 γ x η ⊗
1) = x β x δ x n +1 γ ⊗ x η − s (cid:0) q γη x β x δ x nγ ⊗ x η x γ + ( − q δγ ) n +1 q δη x β x n +1 γ ⊗ x η x δ + ( − q δγ ) n +1 b x β x n +1 γ ⊗ x ϕ x τ x γ + q βδ ( − q βγ ) n +1 q βη x δ x n +1 γ ⊗ x η x β (cid:1) = x β x δ x n +1 γ ⊗ x η − q γη x β x δ x nγ x η ⊗ x γ − ( − q δγ ) n +1 b x β x n +1 γ x ϕ ⊗ x τ x γ − ( − q δγ ) n +1 q δη x β x n +1 γ x η ⊗ x δ − s (cid:0) q βδ ( − q βγ ) n +1 q βη q γη x δ x nγ x η ⊗ x γ x β + q βδ ( − q βγ ) n +1 q βη x δ x n +1 γ ⊗ x η x β + q n +1 δγ q n +1 βγ q βϕ b x n +1 γ x ϕ ⊗ ( q βτ x τ x β + b x γ ) x γ + q n +1 δγ q δη q n +1 βγ q βη q βδ x n +1 γ x η ⊗ x δ x β ) (cid:1) = x β x δ x n +1 γ ⊗ x η − q γη x β x δ x nγ x η ⊗ x γ − ( − q δγ ) n +1 b x β x n +1 γ x ϕ ⊗ x τ x γ − ( − q δγ ) n +1 q δη x β x n +1 γ x η ⊗ x δ + q βδ ( − q βγ ) n +1 q βη x δ x n +1 γ x η ⊗ x β . Finally, we prove (10.1.34) by induction on n . For n = 0, d ( x α x β x δ x η ⊗
1) = x α x β x δ ⊗ x η − s (cid:0) q δη x α x β ⊗ x η x δ + b x α x β ⊗ x ϕ x τ x γ − q βδ q βη x α x δ ⊗ x η x β + q αβ q αδ q αη x β x δ ⊗ x η x α (cid:1) = x α x β x δ ⊗ x η − b x α x β x ϕ ⊗ x τ x γ − q δη x α x β x η ⊗ x δ − s (cid:0) q βη q βδ q δη x α x η ⊗ x δ x β − q αβ q αη q αδ q δη x β x η ⊗ x δ x α + q βϕ q βτ q βγ b x α x ϕ ⊗ x τ x γ x β − q αβ q αϕ q ατ q αγ b x β x ϕ ⊗ x τ x γ x α − q αβ b b x β x γ ⊗ x τ x γ − q βδ q βη x α x δ ⊗ x η x β + q αβ q αδ q αη x β x δ ⊗ x η x α (cid:1) = x α x β x δ ⊗ x η − b x α x β x ϕ ⊗ x τ x γ − q δη x α x β x η ⊗ x δ + q βδ q βη x α x δ x η ⊗ x β − q αβ q αδ q αη x β x δ x η ⊗ x α + q αβ b b x β x γ x τ ⊗ x γ − q αβ q γτ [2] e q βγ b b b x γ ⊗ . We assume that (10.1.34) holds for n . Using Remark 10.1.2 twice, inductive hypothesis,(10.1.38), (10.1.40), (10.1.41), (10.1.37), we have d ( x α x β x δ x n +1 γ x η ⊗
1) = x α x β x δ x n +1 γ ⊗ x η − s (cid:0) q γη x α x β x δ x nγ ⊗ x η x γ + ( − q δγ ) n +1 q δη x α x β x n +1 γ ⊗ x η x δ + ( − q δγ ) n +1 b x α x β x n +1 γ ⊗ x ϕ x τ x γ − q βδ ( − q βγ ) n +1 q βη x α x δ x n +1 γ ⊗ x η x β + q αβ q αδ ( − q αγ ) n +1 q αη x β x δ x n +1 γ ⊗ x η x α (cid:1) = x α x β x δ x n +1 γ ⊗ x η − q γη x α x β x δ x nγ x η ⊗ x γ − ( − q δγ ) n +1 b x α x β x n +1 γ x ϕ ⊗ x τ x γ − s (cid:0) q γη ( − q δγ ) n q δγ q δη x α x β x nγ x η ⊗ x γ x δ + q n +2 βγ q βϕ q n +1 δγ b x α x n +1 γ x ϕ ⊗ q βτ x τ x γ x β − q γη q βδ ( − q βγ ) n q βγ q βη x α x δ x nγ x η ⊗ x γ x β − q γη q αβ q αδ ( − q αγ ) n +1 q αη x β x δ x nγ x η ⊗ x γ x α + ( − q δγ ) n +1 q δη x α x β x n +1 γ ⊗ x η x δ − q αβ q n +2 αγ q αϕ q ατ q n +1 δγ b x β x n +1 γ x ϕ ⊗ x τ x γ x α + q γη q αβ q − n − δγ d ( n ) αβδγ b b b x n +3 γ ⊗ x γ − q αβ q γϕ ( n + 2) e q αγ ( − q δγ ) n +1 b b x β x n +2 γ ⊗ x τ x γ − q βδ ( − q βγ ) n +1 q βη x α x δ x n +1 γ ⊗ x η x β + q αβ q αδ ( − q αγ ) n +1 q αη x β x δ x n +1 γ ⊗ x η x α (cid:1) = x α x β x δ x n +1 γ ⊗ x η − q γη x α x β x δ x nγ x η ⊗ x γ − ( − q δγ ) n +1 b x α x β x n +1 γ x ϕ ⊗ x τ x γ − ( − q δγ ) n +1 q δη x α x β x n +1 γ x η ⊗ x δ + q βδ ( − q βγ ) n +1 q βη x α x δ x n +1 γ x η ⊗ x β − s (cid:0) − q γη q αβ q αδ ( − q αγ ) n +1 q αη x β x δ x nγ x η ⊗ x γ x α − q αβ q n +2 αγ q αϕ q ατ q n +1 δγ b x β x n +1 γ x ϕ ⊗ x τ x γ x α − q γη q αβ q − n − δγ d ( n ) αβδγ b b b x n +3 γ ⊗ x γ − q αβ q γϕ ( n + 2) e q αγ ( − q δγ ) n +1 b b x β x n +2 γ ⊗ x τ x γ + q αβ q αδ ( − q αγ ) n +1 q αη x β x δ x n +1 γ ⊗ x η x α − q n +1 δγ q δη q αβ q n +1 αγ q αη q αδ x β x n +1 γ x η ⊗ x δ x α + q βδ q n +1 βγ q βη q αβ q αδ q n +1 αγ q αη x δ x n +1 γ x η ⊗ x β x α − q γϕ q n +1 δγ q βδ q n +1 βγ q βη ( n + 2) e q αγ b b x n +2 γ x τ ⊗ x γ x β (cid:1) = x α x β x δ x n +1 γ ⊗ x η − q γη x α x β x δ x nγ x η ⊗ x γ − ( − q δγ ) n +1 b x α x β x n +1 γ x ϕ ⊗ x τ x γ − ( − q δγ ) n +1 q δη x α x β x n +1 γ x η ⊗ x δ + q βδ ( − q βγ ) n +1 q βη x α x δ x n +1 γ x η ⊗ x β − q αβ q αδ ( − q αγ ) n +1 q αη x β x δ x n +1 γ x η ⊗ x α − s (cid:0) q αβ q γη q − n − δγ d ( n ) αβδγ b b b x n +3 γ ⊗ x γ − q αβ q γϕ ( n + 2) e q αγ ( − q δγ ) n +1 b b x β x n +2 γ ⊗ x τ x γ − q γϕ q n +1 δγ q βδ q n +1 βγ q βη ( n + 2) e q αγ b b x n +2 γ x τ ⊗ x γ x β (cid:1) = x α x β x δ x n +1 γ ⊗ x η − q γη x α x β x δ x nγ x η ⊗ x γ − ( − q δγ ) n +1 q δη x α x β x n +1 γ x η ⊗ x δ + q βδ ( − q βγ ) n +1 q βη x α x δ x n +1 γ x η ⊗ x β − q αβ q αδ ( − q αγ ) n +1 q αη x β x δ x n +1 γ x η ⊗ x α − ( − q δγ ) n +1 b x α x β x n +1 γ x ϕ ⊗ x τ x γ − q αβ q γϕ ( n + 2) e q αγ ( − q δγ ) n +1 b b x β x n +2 γ x τ ⊗ x γ − q αβ (cid:0) q γη ( − q δγ ) − n d ( n ) αβδγ − q γϕ ( n + 2) e q αγ ( − q δγ ) n +1 q γτ ( n + 3) e q βγ (cid:1) b b b s ( x n +3 γ ⊗ x γ ) . Thus the proof of the inductive step follows since q γδ q γδ = q γγ q γτ q γϕ = − q γτ q γϕ and s ( x n +3 γ ⊗ x γ ) = x n +4 γ ⊗ (cid:3) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 93
Lemma 10.1.42.
Let α < η < γ < β < δ be positive roots such that N γ = 2 and therelations among the corresponding root vectors take the form x α x δ = q αδ x δ x α + b x γ x η , x η x β = q ηβ x β x η + b x γ , (10.1.43) for some scalars b , b and the other pairs of root vectors q -commute. Then, for all n ≥ , d ( x α x nγ x β x δ ⊗
1) = x α x nγ x β ⊗ x δ − q βδ x α x nγ x δ ⊗ x β + q γβ q γδ x α x n − γ x β x δ ⊗ x γ + ( − q αγ ) n q αβ q αδ x nγ x β x δ ⊗ x α − b ( − n − q nαγ ( n + 1) e q δγ x n +1 γ x β ⊗ x η + b b c ( n ) αδγ ( − q γα ) − n x n +2 γ ⊗ . (10.1.44)Notice that the equalities in (10.1.43) force γ + η = α + δ, η + β = γ. (10.1.45)Hence the following equality also holds: 2 γ = α + β + δ . Proof.
The proof is analogous to Lemma 10.1.42. Indeed, first we apply (10.1.14) with δ in place of β to get d ( x α x nγ x δ ⊗
1) = x α x nγ ⊗ x δ − q γδ x α x n − γ x δ ⊗ x γ − q αδ ( − q αγ ) n x nγ x δ ⊗ x α − b ( − q αγ ) n ( n + 1) e q δγ x n +1 γ ⊗ x η Using this formula, (10.1.44) follows by induction on n . (cid:3) Lemma 10.1.46.
Let α < β < γ < η < δ be positive roots such that N γ = 2 and therelations among the corresponding root vectors take the form x β x δ = q βδ x δ x β + b x η x γ , x α x η = q αη x η x α + b x γ , (10.1.47) for some scalars b , b and the other pairs of root vectors q -commute. Then, for all n ≥ , d ( x α x β x nγ x δ ⊗
1) = x α x β x nγ ⊗ x δ − q γδ x α x β x n − γ x δ ⊗ x γ − ( − q βγ ) n q βδ x α x nγ x δ ⊗ x β + ( − q αγ ) n q αβ q αδ x β x nγ x δ ⊗ x α − ( − q βγ ) n b x α x nγ x η ⊗ x γ − q nγδ c ( n ) − δ,α,γ b b x n +2 γ ⊗ . (10.1.48)Notice that the equalities in (10.1.47) force γ + η = β + δ, η + α = γ. (10.1.49)Hence the following equality also holds: 2 γ = α + β + δ . Proof.
The following formula holds for all n ≥ d ( x β x nγ x δ ⊗
1) = x β x nγ ⊗ x δ − q γδ x β x n − γ x δ ⊗ x γ − ( − q βγ ) n q βδ x nγ x δ ⊗ x β − ( − q βγ ) n b x nγ x η ⊗ x γ . (10.1.50)The proof is analogous to (10.1.14), see also the proof of (10.1.30). Next, we apply Lemma10.1.7 for α < γ < η (no other intermediate roots) to get d ( x α x nγ x η ⊗
1) = x α x nγ ⊗ x η − q γη x α x n − γ x η ⊗ x γ − q αη ( − q αγ ) n x nγ x η ⊗ x α − ( − q αγ ) n ( n + 1) e q αγ b x n +1 γ ⊗ . (10.1.51) Now we prove (10.1.48) by induction on n . When n = 0, d ( x α x β x δ ⊗
1) = x α x β ⊗ x δ − s ( q βδ x α ⊗ x δ x β + b x α ⊗ x η x γ − q αβ q αδ x β ⊗ x δ x α )= x α x β ⊗ x δ − b x α x η ⊗ x γ − q βδ x α x δ ⊗ x β − s (cid:0) q αβ q αδ q βδ x δ ⊗ x β x α + q αγ q αη b x η ⊗ x γ x α + b b x γ ⊗ x γ − q αβ q αδ x β ⊗ x δ x α (cid:1) = x α x β ⊗ x δ − b x α x η ⊗ x γ − q βδ x α x δ ⊗ x β + q αβ q αδ x β x δ ⊗ x α − b b x γ ⊗ . Now assume that (10.1.48) holds for n . Using Remark 10.1.2, inductive hypothesis, therelation x γ = 0, (10.1.51), (10.1.50), d ( x α x β x n +1 γ x δ ⊗
1) = x α x β x n +1 γ ⊗ x δ − s (cid:0) q γδ x α x β x nγ ⊗ x δ x γ + ( − q βγ ) n +1 b x α x n +1 γ ⊗ x η x γ + ( − q βγ ) n +1 q βδ x α x n +1 γ ⊗ x δ x β − q αβ ( − q αγ ) n +1 q αδ x β x n +1 γ ⊗ x δ x α (cid:1) = x α x β x n +1 γ ⊗ x δ − q γδ x α x β x nγ x δ ⊗ x γ − ( − q βγ ) n +1 b x α x n +1 γ x η ⊗ x γ − ( − q βγ ) n +1 q βδ x α x n +1 γ x δ ⊗ x β − s (cid:0) − q αβ ( − q αγ ) n +1 q αδ x β x n +1 γ ⊗ x δ x α − q αγ ( − q αγ ) n q αβ q αδ q γδ x β x nγ x δ ⊗ x γ x α + ( q n +1 γδ c ( n ) − δ,α,γ + q n +1 βγ q n +1 αγ ( n + 2) e q αγ ) b b x n +2 γ ⊗ x γ + q n +1 βγ q αη q n +2 αγ b x n +1 γ x η ⊗ x γ x α − q n +1 βγ q βδ q αβ q n +1 αγ q αδ x n +1 γ x δ ⊗ x β x α (cid:1) = x α x β x n +1 γ ⊗ x δ − q γδ x α x β x nγ x δ ⊗ x γ − ( − q βγ ) n +1 b x α x n +1 γ x η ⊗ x γ − ( − q βγ ) n +1 q βδ x α x n +1 γ x δ ⊗ x β + q αβ ( − q αγ ) n +1 q αδ x β x n +1 γ x δ ⊗ x α − ( q n +1 γδ c ( n ) − δ,α,γ + q n +1) γγ q − n − δγ ( n + 2) e q αγ ) b b x n +3 γ ⊗ , and the inductive step follows. (cid:3) Lemma 10.1.52.
Let α < τ < β < γ < µ < ν < η < δ be positive roots such that N γ = 2 and the relations among the corresponding root vectors take the form x α x δ = q αδ x δ x α + b x η x τ , x β x δ = q βδ x δ x β + b x ν x γ ,x β x η = q βη x η x β + b x µ x γ , x α x ν = q αν x ν x α + b x γ , (10.1.53) for some scalars b i and the other pairs of root vectors q -commute. Then, for all n ≥ , d ( x α x β x nγ x δ ⊗
1) = x α x β x nγ ⊗ x δ − q γδ x α x β x n − γ x δ ⊗ x γ − ( − q βγ ) n q βδ x α x nγ x δ ⊗ x β + ( − q αγ ) n q αβ q αδ x β x nγ x δ ⊗ x α + q αβ ( − q αγ ) n b x β x nγ x η ⊗ x τ − ( − q βγ ) n b x α x nγ x ν ⊗ x γ − q nγδ c ( n ) − δ,α,γ b b x n +2 γ ⊗ . (10.1.54)Notice that the equalities in (10.1.53) force α + δ = η + τ, β + δ = ν + γ, β + η = µ + γ, α + ν = γ. (10.1.55)Hence the following equality also holds: 2 γ = α + β + δ . Proof.
We need some auxiliary computations. First we apply Lemma 10.1.7 to α < γ < ν : d ( x α x nγ x ν ⊗
1) = x α x nγ ⊗ x ν − q γν x α x n − γ x ν ⊗ x γ − q αν ( − q αγ ) n x nγ x ν ⊗ x α − ( − q αγ ) n ( n + 1) e q αγ b x n +1 γ ⊗ . (10.1.56) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 95
Next we claim that the following formulas hold for all n ≥ d ( x α x nγ x δ ⊗
1) = x α x nγ ⊗ x δ − q γδ x α x n − γ x δ ⊗ x γ − ( − q αγ ) n q αδ x nγ x δ ⊗ x α − ( − q αγ ) n b x nγ x η ⊗ x τ , (10.1.57) d ( x β x nγ x δ ⊗
1) = x β x nγ ⊗ x δ − q γδ x β x n − γ x δ ⊗ x γ − ( − q βγ ) n q βδ x nγ x δ ⊗ x β − ( − q βγ ) n b x nγ x ν ⊗ x γ , (10.1.58) d ( x β x nγ x η ⊗
1) = x β x nγ ⊗ x η − q γη x β x n − γ x η ⊗ x γ − ( − q βγ ) n q βη x nγ x η ⊗ x β − ( − q βγ ) n b x nγ x µ ⊗ x γ . (10.1.59)The proof of each equality is analogous to (10.1.14).Now we prove (10.1.54) by induction on n . When n = 0, d ( x α x β x δ ⊗
1) = x α x β ⊗ x δ − s (cid:0) q βδ x α ⊗ x δ x β + b x α ⊗ x ν x γ − q αβ x β ⊗ x α x δ (cid:1) = x α x β ⊗ x δ − b x α x ν ⊗ x γ − q βδ x α x δ ⊗ x β − s (cid:0) − q αβ q αδ x β ⊗ x δ x α − q αβ b x β ⊗ x η x τ + q αν b x ν ⊗ x α x γ + b b x γ ⊗ x γ + q αβ q αδ q βδ x δ ⊗ x β x α + q τβ q βδ b x η ⊗ x β x τ (cid:1) = x α x β ⊗ x δ − b x α x ν ⊗ x γ − q βδ x α x δ ⊗ x β + q αβ b x β x η ⊗ x τ + q αβ q αδ x β x δ ⊗ x α − s (cid:0) b b x γ ⊗ x γ − q αβ b b x µ ⊗ x γ x τ (cid:1) , which is (10.1.54) for n = 0 since s (cid:0) x γ ⊗ x γ (cid:1) = x γ ⊗ , s (cid:0) x µ ⊗ x γ x τ (cid:1) = s ◦ s (cid:0) x µ x γ x τ (cid:1) = 0 . Now assume that (10.1.48) holds for n . Using Remark 10.1.2, inductive hypothesis, therelation x γ = 0, (10.1.56), (10.1.57), (10.1.58), (10.1.59), d ( x α x β x n +1 γ x δ ⊗
1) = x α x β x n +1 γ ⊗ x δ − s (cid:0) q γδ x α x β x nγ ⊗ x δ x γ + ( − q βγ ) n +1 b x α x n +1 γ ⊗ x ν x γ + q βδ ( − q βγ ) n +1 x α x n +1 γ ⊗ x δ x β − q αβ ( − q αγ ) n +1 q αδ x β x n +1 γ ⊗ x δ x α − q αβ ( − q αγ ) n +1 b x β x n +1 γ ⊗ x η x τ (cid:1) = x α x β x n +1 γ ⊗ x δ − q γδ x α x β x nγ x δ ⊗ x γ − ( − q βγ ) n +1 b x α x n +1 γ x ν ⊗ x γ − s (cid:0) − q γδ ( − q βγ ) n +1 q βδ x α x nγ x δ ⊗ x γ x β + q γδ ( − q αγ ) n +1 q αβ q αδ x β x nγ x δ ⊗ x γ x α − q γδ q αβ ( − q αγ ) n b x β x nγ x η ⊗ x τ x γ + q γδ ( − q βγ ) n b x α x nγ x ν ⊗ x γ + ( q n +1 γδ c ( n ) − δ,α,γ + q n +1 αγ q n +1 βγ ( n + 2) e q αγ ) b b x n +2 γ ⊗ x γ + q βδ ( − q βγ ) n +1 x α x n +1 γ ⊗ x δ x β − q αβ ( − q αγ ) n +1 q αδ x β x n +1 γ ⊗ x δ x α − q αβ ( − q αγ ) n +1 b x β x n +1 γ ⊗ x η x τ + q αν q n +2 αγ q n +1 βγ b x n +1 γ x ν ⊗ x γ x α (cid:1) = x α x β x n +1 γ ⊗ x δ − q γδ x α x β x nγ x δ ⊗ x γ − ( − q βγ ) n +1 b x α x n +1 γ x ν ⊗ x γ − q βδ ( − q βγ ) n +1 x α x n +1 γ x δ ⊗ x β + q αβ ( − q αγ ) n +1 b x β x n +1 γ x η ⊗ x τ + q αβ ( − q αγ ) n +1 q αδ x β x n +1 γ x δ ⊗ x α − s (cid:0) − q αβ q n +1 αγ q n +1 βγ b b x n +1 γ x µ ⊗ x γ x τ + q n +1 γδ ( c ( n ) − δ,α,γ + q n +1 αγ q n +1 βγ q − n − γδ ( n + 2) e q αγ ) b b x n +2 γ ⊗ x γ (cid:1) . As α + β = 2 γ − δ , we have q n +1 αγ q n +1 βγ q − n − γδ = e q − n − γδ . Also, s ( x n +2 γ ⊗ x γ ) = x n +3 γ ⊗
1, so d ( x α x β x n +1 γ x δ ⊗
1) = x α x β x n +1 γ ⊗ x δ − q γδ x α x β x nγ x δ ⊗ x γ − ( − q βγ ) n +1 b x α x n +1 γ x ν ⊗ x γ − q βδ ( − q βγ ) n +1 x α x n +1 γ x δ ⊗ x β + q αβ ( − q αγ ) n +1 b x β x n +1 γ x η ⊗ x τ + q αβ ( − q αγ ) n +1 q αδ x β x n +1 γ x δ ⊗ x α − q n +1 γδ c ( n +1) − δ,α,γ b b x n +3 γ ⊗ q αβ q n +1 αγ q n +1 βγ b b s (cid:0) x n +1 γ x µ ⊗ x γ x τ (cid:1) . Next we claim that x n +1 γ x µ ⊗ x γ x τ = s ( x n +1 γ ⊗ x µ x γ x τ ). Indeed, d ( x n +1 γ ⊗ x µ x γ x τ ) = x nγ ⊗ x γ x µ x γ x τ = q γµ x nγ ⊗ x µ x γ x τ = 0 , so x n +1 γ ⊗ x µ x γ x τ ∈ ker d n , and we compute s (cid:0) x n +1 γ ⊗ x µ x γ x τ (cid:1) = x n +1 γ x µ ⊗ x γ x τ + s (cid:0) x n +1 γ ⊗ x µ x γ x τ − d ( x n +1 γ x µ ⊗ x γ x τ ) (cid:1) = x n +1 γ x µ ⊗ x γ x τ + s (cid:0) x n +1 γ ⊗ x µ x γ x τ − ( x n +1 γ ⊗ x µ x γ x τ − q γµ x nγ x µ ⊗ x γ x τ ) (cid:1) = x n +1 γ x µ ⊗ x γ x τ . From this claim, s (cid:0) x n +1 γ x µ ⊗ x γ x τ (cid:1) = 0, and the inductive step follows. (cid:3) Lemma 10.1.60.
Let α < β < ν < γ < µ < δ < η be positive roots such that N γ = 2 andthe relations among the corresponding root vectors take the form x β x δ = q βδ x δ x β + b x γ x ν , x ν x η = q νη x η x ν + b x µ x γ ,x α x µ = q αµ x µ x α + b x γ , (10.1.61) for some scalars b i , x γ q -commutes with the other root vectors and the following pairsof root vectors also q -commute: ( x α , x β ) , ( x α , x ν ) , ( x α , x δ ) , ( x α , x η ) , ( x β , x ν ) , ( x β , x η ) , ( x ν , x δ ) , ( x ν , x η ) , ( x µ , x δ ) , ( x µ , x η ) . Then, for all n ≥ , d ( x α x β x nγ x δ x η ⊗
1) = x α x β x nγ x δ ⊗ x η + ( − q βγ ) n ( n + 1) e q δγ b b x α x n +1 γ x µ ⊗ x γ − q δη x α x β x nγ x η ⊗ x δ + q νη ( − q βγ ) n ( n + 1) e q δγ b x α x n +1 γ x η ⊗ x ν + q γδ q γη x α x β x n − γ x δ x η ⊗ x γ + ( − q βγ ) n q βδ q βη x α x nγ x δ x η ⊗ x β − q αβ ( − q αγ ) n q αδ q αη x β x nγ x δ x η ⊗ x α − q αγ q nγδ q nγη d ( n ) α + β,δ,α,γ b b b x n +3 γ ⊗ . (10.1.62)Notice that the equalities in (10.1.61) force β + δ = γ + ν, ν + η = µ + γ, α + µ = γ. (10.1.63)Hence the following equality also holds: 3 γ = α + β + δ + η . Proof.
We need some auxiliary computations. By (10.1.14) d ( x β x nγ x δ ⊗
1) = x β x nγ ⊗ x δ − q γδ x β x n − γ x δ ⊗ x γ − q βδ ( − q βγ ) n x nγ x δ ⊗ x β − ( − q βγ ) n ( n + 1) e q δγ b x n +1 γ ⊗ x ν (10.1.64)Next we apply Lemma 10.1.7 to α < γ < µ : d ( x α x nγ x µ ⊗
1) = x α x nγ ⊗ x µ − q γµ x α x n − γ x µ ⊗ x γ − q αµ ( − q αγ ) n x nγ x µ ⊗ x α − ( − q αγ ) n ( n + 1) e q αγ b x n +1 γ ⊗ . (10.1.65) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 97
Now we prove the following equality by induction on n : d ( x α x β x nγ x δ ⊗
1) = x α x β x nγ ⊗ x δ − q γδ x α x β x n − γ x δ ⊗ x γ − ( − q βγ ) n q βδ x α x nγ x δ ⊗ x β + q αβ ( − q αγ ) n q αδ x β x nγ x δ ⊗ x α − ( − q βγ ) n ( n + 1) e q δγ b x α x n +1 γ ⊗ x ν . (10.1.66) d ( x β x nγ x δ x η ⊗
1) = x β x nγ x δ ⊗ x η − q δη x β x nγ x η ⊗ x δ + q γδ q γη x β x n − γ x δ x η ⊗ x γ + ( − q βγ ) n ( n + 1) e q δγ b b x n +1 γ x µ ⊗ x γ + ( − q βγ ) n q βδ q βη x nγ x δ x η ⊗ x β + ( − q βγ ) n q νη ( n + 1) e q δγ b x n +1 γ x η ⊗ x ν . (10.1.67)Indeed, for n = 0 we have: d ( x α x β x δ ⊗
1) = x α x β ⊗ x δ − s (cid:0) q βδ x α ⊗ x δ x β + b x α ⊗ x γ x ν − q αβ q αδ x β ⊗ x δ x α (cid:1) = x α x β ⊗ x δ − q βδ x α x δ ⊗ x β − b x α x γ ⊗ x ν − s (cid:0) − q αβ q αδ x β ⊗ x δ x α + q αγ q αν b x γ ⊗ x ν x α + q αβ q βδ q αδ x δ ⊗ x β x α (cid:1) = x α x β ⊗ x δ − q βδ x α x δ ⊗ x β − b x α x γ ⊗ x ν + q αβ q αδ x β x δ ⊗ x α . And for the other equality, d ( x β x δ x η ⊗
1) = x β x δ ⊗ x η − s (cid:0) q δη x β ⊗ x η x δ − q βδ q βη x δ ⊗ x η x β − b x γ ⊗ x ν x η (cid:1) = x β x δ ⊗ x η − q δη x β x η ⊗ x δ − s (cid:0) − q βδ q βη x δ ⊗ x η x β − q νη b x γ ⊗ x η x ν − b b x γ ⊗ x µ x γ + q βδ q βη q δη x η ⊗ x δ x β + q βη q δη b x η ⊗ x γ x ν (cid:1) = x β x δ ⊗ x η − q δη x β x η ⊗ x δ + b b x γ x µ ⊗ x γ + q νη b x γ x η ⊗ x ν + q βδ q βη x δ x η ⊗ x β . Now assume that (10.1.66) holds for n . Using Remark 10.1.2 repeatedly, inductivehypothesis, the relation x γ = 0 and (10.1.64), d ( x α x β x n +1 γ x δ ⊗
1) = x α x β x n +1 γ ⊗ x δ − s (cid:0) q γδ x α x β x nγ ⊗ x δ x γ + ( − q βγ ) n +1 b x α x n +1 γ ⊗ x γ x ν + ( − q βγ ) n +1 q βδ x α x n +1 γ ⊗ x δ x β − q αβ ( − q αγ ) n +1 q αδ x β x n +1 γ ⊗ x δ x α (cid:1) = x α x β x n +1 γ ⊗ x δ − q γδ x α x β x nγ x δ ⊗ x γ − s (cid:0) ( − q βγ ) n +1 q βδ x α x n +1 γ ⊗ x δ x β + ( − q βγ ) n +1 (1 + q γδ q νγ ( − q βγ ) − ( n + 1) e q δγ ) b x α x n +1 γ ⊗ x γ x ν − q αβ ( − q αγ ) n +1 q αδ x β x n +1 γ ⊗ x δ x α − ( − q βγ ) n +1 q βδ q γδ x α x nγ x δ ⊗ x γ x β + q αβ ( − q αγ ) n +1 q αδ q γδ x β x nγ x δ ⊗ x γ x α (cid:1) = x α x β x n +1 γ ⊗ x δ − q γδ x α x β x nγ x δ ⊗ x γ − ( − q βγ ) n +1 q βδ x α x n +1 γ x δ ⊗ x β − ( − q βγ ) n +1 ( n + 2) e q δγ b x α x n +2 γ ⊗ x ν − s (cid:0) − q αβ ( − q αγ ) n +1 q αδ x β x n +1 γ ⊗ x δ x α + q αβ ( − q αγ ) n +1 q αδ q γδ x β x nγ x δ ⊗ x γ x α + q αβ q n +1 αγ q αδ q n +1 βγ q βδ x n +1 γ x δ ⊗ x β x α + q n +2 αγ q n +1 βγ ( n + 2) e q δγ b x n +2 γ ⊗ x α x ν (cid:1) = x α x β x n +1 γ ⊗ x δ − q γδ x α x β x nγ x δ ⊗ x γ − ( − q βγ ) n +1 q βδ x α x n +1 γ x δ ⊗ x β − ( − q βγ ) n +1 ( n + 2) e q δγ b x α x n +2 γ ⊗ x ν + q αβ ( − q αγ ) n +1 q αδ x β x n +1 γ x δ ⊗ x α . Next we assume that (10.1.67) holds for n . Using (10.1.64), inductive hypothesis, Re-mark 10.1.2, the relation x γ = 0, d ( x β x n +1 γ x δ x η ⊗
1) = x β x n +1 γ x δ ⊗ x η − s (cid:0) q δη x β x n +1 γ ⊗ x η x δ − q γδ q γη x β x nγ x δ ⊗ x η x γ − ( − q βγ ) n +1 q βδ q βη x n +1 γ x δ ⊗ x η x β − ( − q βγ ) n +1 ( n + 2) e q δγ b x n +2 γ ⊗ x ν x η (cid:1) = x β x n +1 γ x δ ⊗ x η − q δη x β x n +1 γ x η ⊗ x δ + q γδ q γη x β x nγ x δ x η ⊗ x γ − s (cid:0) − ( − q βγ ) n +1 ( n + 2) e q δγ b b x n +2 γ ⊗ x µ x γ − ( − q βγ ) n +1 q νη ( n + 2) e q δγ b x n +2 γ ⊗ x η x ν − ( − q βγ ) n +1 q βδ q βη x n +1 γ x δ ⊗ x η x β + q νη q γη ( − q βγ ) n +1 (1 + e q γδ ( n + 1) e q δγ ) b x n +1 γ x η ⊗ x γ x ν + q δη ( − q βγ ) n +1 q βδ q βη x n +1 γ x η ⊗ x δ x β − q γδ q γη ( − q βγ ) n +1 q βδ q βη x nγ x δ x η ⊗ x γ x β (cid:1) = x β x n +1 γ x δ ⊗ x η − q δη x β x n +1 γ x η ⊗ x δ + q γδ q γη x β x nγ x δ x η ⊗ x γ + ( − q βγ ) n +1 ( n + 2) e q δγ b b x n +2 γ x µ ⊗ x γ + ( − q βγ ) n +1 q νη ( n + 2) e q δγ b x n +2 γ x η ⊗ x ν − s (cid:0) − ( − q βγ ) n +1 q βδ q βη x n +1 γ x δ ⊗ x η x β + q δη ( − q βγ ) n +1 q βδ q βη x n +1 γ x η ⊗ x δ x β − q γδ q γη ( − q βγ ) n +1 q βδ q βη x nγ x δ x η ⊗ x γ x β (cid:1) = x β x n +1 γ x δ ⊗ x η − q δη x β x n +1 γ x η ⊗ x δ + q γδ q γη x β x nγ x δ x η ⊗ x γ + ( − q βγ ) n +1 ( n + 2) e q δγ b b x n +2 γ x µ ⊗ x γ + ( − q βγ ) n +1 q νη ( n + 2) e q δγ b x n +2 γ x η ⊗ x ν + ( − q βγ ) n +1 q βδ q βη x n +1 γ x δ x η ⊗ x β . Finally we prove (10.1.62) by induction on n . When n = 0, d ( x α x β x δ x η ⊗
1) = x α x β x δ ⊗ x η − s (cid:0) q δη x α x β ⊗ x η x δ − q βη q βδ x α x δ ⊗ x η x β − q νη b x α x γ ⊗ x η x ν − b b x α x γ ⊗ x µ x γ + q αβ q αδ q αη x β x δ ⊗ x η x α (cid:1) = x α x β x δ ⊗ x η − q δη x α x β x η ⊗ x δ + q βη q βδ x α x δ x η ⊗ x β − s (cid:0) − q νη b x α x γ ⊗ x η x ν − b b x α x γ ⊗ x µ x γ + q βη q δη b x α x η ⊗ x γ x ν + q αβ q αδ q αη x β x δ ⊗ x η x α − q αβ q αη q αδ q δη x β x η ⊗ x δ x α + q αβ q αδ q αη q βη q βδ x δ x η ⊗ x β x α (cid:1) = x α x β x δ ⊗ x η − q δη x α x β x η ⊗ x δ + q βη q βδ x α x δ x η ⊗ x β + q νη b x α x γ x η ⊗ x ν + b b x α x γ x µ ⊗ x γ − s (cid:0) q αβ q αδ q αη x β x δ ⊗ x η x α + q αγ (2) e q αγ b b b x γ ⊗ x γ − q αβ q αη q αδ q δη x β x η ⊗ x δ x α + q αβ q αδ q αη q βη q βδ x δ x η ⊗ x β x α + q αγ q αη q αν q νη b x γ x η ⊗ x ν x α + q αµ q αγ b b x γ x µ ⊗ x γ x α (cid:1) = x α x β x δ ⊗ x η − q δη x α x β x η ⊗ x δ + q βη q βδ x α x δ x η ⊗ x β + q νη b x α x γ x η ⊗ x ν + b b x α x γ x µ ⊗ x γ − q αβ q αδ q αη x β x δ x η ⊗ x α − q αγ (2) e q αγ b b b x γ ⊗ , which is (10.1.62) for n = 0. Now assume that (10.1.62) holds for n . Using (10.1.66),Remark 10.1.2, inductive hypothesis, the relation x γ = 0, (10.1.65), (10.1.67), d ( x α x β x n +1 γ x δ x η ⊗
1) = x α x β x n +1 γ x δ ⊗ x η − s (cid:0) q δη x α x β x n +1 γ ⊗ x η x δ − ( − q βγ ) n +1 q βδ q βη x α x n +1 γ x δ ⊗ x η x β + q αβ ( − q αγ ) n +1 q αδ q αη x β x n +1 γ x δ ⊗ x η x α − q γδ q γη x α x β x nγ x δ ⊗ x η x γ − ( − q βγ ) n +1 ( n + 2) e q δγ b x α x n +2 γ ⊗ ( q νη x η x ν + b x µ x γ ) (cid:1) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 99 = x α x β x n +1 γ x δ ⊗ x η − q δη x α x β x n +1 γ x η ⊗ x δ − s (cid:0) − ( − q βγ ) n +1 q βδ q βη x α x n +1 γ x δ ⊗ x η x β + q αβ ( − q αγ ) n +1 q αδ q αη x β x n +1 γ x δ ⊗ x η x α − q γδ q γη x α x β x nγ x δ ⊗ x η x γ − q νη ( − q βγ ) n +1 ( n + 2) e q δγ b x α x n +2 γ ⊗ x η x ν + q γη q δη q γδ x α x β x nγ x η ⊗ x δ x γ − ( − q βγ ) n +1 ( n + 2) e q δγ b b x α x n +2 γ ⊗ x µ x γ − q αβ ( − q αγ ) n +1 q αδ q αη q δη x β x n +1 γ x η ⊗ x δ x α + ( − q βγ ) n +1 q βη q δη x α x n +1 γ x η ⊗ ( q βδ x δ x β + b x γ x ν ) (cid:1) = x α x β x n +1 γ x δ ⊗ x η − q δη x α x β x n +1 γ x η ⊗ x δ + q γδ q γη x α x β x nγ x δ x η ⊗ x γ − s (cid:0) q αβ ( − q αγ ) n +1 q αδ q αη x β x n +1 γ x δ ⊗ x η x α − ( − q βγ ) n +1 ( n + 2) e q δγ b b x α x n +2 γ ⊗ x µ x γ − ( − q βγ ) n +1 q βδ q βη x α x n +1 γ x δ ⊗ x η x β − q νη ( − q βγ ) n +1 ( n + 2) e q δγ b x α x n +2 γ ⊗ x η x ν + ( − q βγ ) n ( q νη ( n + 1) e q δγ q νγ q γδ q γη − q βγ q βη q δη ) b x α x n +1 γ x η ⊗ x γ x ν − q αβ ( − q αγ ) n +1 q αδ q αη q δη x β x n +1 γ x η ⊗ x δ x α + ( − q βγ ) n +1 q βδ q βη q δη x α x n +1 γ x η ⊗ x δ x β − ( − q βγ ) n +1 q βδ q βη q γδ q γη x α x nγ x δ x η ⊗ x γ x β − q αγ q n +1 γδ q n +1 γη d ( n ) α + β,δ,α,γ b b b x n +3 γ ⊗ x γ + q αβ ( − q αγ ) n +1 q αδ q αη q γδ q γη x β x nγ x δ x η ⊗ x γ x α (cid:1) = x α x β x n +1 γ x δ ⊗ x η − q δη x α x β x n +1 γ x η ⊗ x δ + q γδ q γη x α x β x nγ x δ x η ⊗ x γ + ( − q βγ ) n +1 ( n + 2) e q δγ b b x α x n +2 γ x µ ⊗ x γ + q νη ( − q βγ ) n +1 ( n + 2) e q δγ b x α x n +2 γ x η ⊗ x ν + ( − q βγ ) n +1 q βδ q βη x α x n +1 γ x δ x η ⊗ x β − s (cid:0) − q αβ ( − q αγ ) n +1 q αδ q αη q δη x β x n +1 γ x η ⊗ x δ x α + q αβ ( − q αγ ) n +1 q αδ q αη x β x n +1 γ x δ ⊗ x η x α + q αβ q n +1 αγ q αδ q αη q n +1 βγ q βδ q βη x n +1 γ x δ x η ⊗ x β x α + q αβ ( − q αγ ) n +1 q αδ q αη q γδ q γη x β x nγ x δ x η ⊗ x γ x α − q αγ q n +1 γδ q n +1 γη ( e q n +1 αγ e q n +1 βγ ( n + 2) e q δγ ( n + 3) e q αγ + d ( n ) α + β,δ,α,γ ) b b b x n +3 γ ⊗ x γ + q αµ q n +3 αγ q n +1 βγ ( n + 2) e q δγ b b x n +2 γ x µ ⊗ x γ x α + q αν q n +2 αγ q αη q νη q n +1 βγ ( n + 2) e q δγ b x n +2 γ x η ⊗ x ν x α (cid:1) = x α x β x n +1 γ x δ ⊗ x η − q δη x α x β x n +1 γ x η ⊗ x δ + q γδ q γη x α x β x nγ x δ x η ⊗ x γ + ( − q βγ ) n +1 ( n + 2) e q δγ b b x α x n +2 γ x µ ⊗ x γ + q νη ( − q βγ ) n +1 ( n + 2) e q δγ b x α x n +2 γ x η ⊗ x ν + ( − q βγ ) n +1 q βδ q βη x α x n +1 γ x δ x η ⊗ x β − q αβ ( − q αγ ) n +1 q αδ q αη x β x n +1 γ x δ x η ⊗ x α − q αγ q n +1 γδ q n +1 γη d ( n +1) α + β,δ,α,γ b b b s (cid:0) x n +3 γ ⊗ x γ (cid:1) , and the inductive step follows. (cid:3) Lemma 10.1.68.
Let α < β < δ < γ < µ < ν < η be positive roots such that N γ = 2 andthe relations among the corresponding root vectors take the form x β x η = q βη x η x β + b x ν x γ ,x δ x ν = q δν x ν x δ + b x µ x γ , x α x µ = q αµ x µ x α + b x γ , (10.1.69)
00 N. ANDRUSKIEWITSCH, I. ANGIONO, J. PEVTSOVA, S. WITHERSPOON for some scalars b i and all other pairs of root vectors q -commute except possibly ( x β , x µ ) .Then, for all n ≥ , d ( x α x β x δ x nγ x η ⊗
1) = x α x β x δ x nγ ⊗ x η − q γη x α x β x δ x n − γ x η ⊗ x γ − ( − q δγ ) n q δη x α x β x nγ x η ⊗ x δ + q βδ ( − q βγ ) n q βη x α x δ x nγ x η ⊗ x β + q βδ ( − q βγ ) n b x α x δ x nγ x ν ⊗ x γ − q αβ q αδ ( − q αγ ) n q αη x β x δ x nγ x η ⊗ x α + q βδ q nγη d ( n ) β − ν αγ b b b x n +3 γ ⊗ . (10.1.70)Notice that the equalities in (10.1.69) force β + η = γ + ν, δ + ν = γ + µ, α + µ = γ. (10.1.71)Hence the following equality also holds: 3 γ = α + β + δ + η . Proof.
We need some auxiliary computations. By (10.1.58), d ( x β x nγ x η ⊗
1) = x β x nγ ⊗ x η − q γη x β x n − γ x η ⊗ x γ − ( − q βγ ) n q βη x nγ x η ⊗ x β − ( − q βγ ) n b x nγ x ν ⊗ x γ , (10.1.72) d ( x δ x nγ x ν ⊗
1) = x δ x nγ ⊗ x ν − q γν x δ x n − γ x ν ⊗ x γ − ( − q δγ ) n q δν x nγ x ν ⊗ x δ − ( − q δγ ) n b x nγ x µ ⊗ x γ . (10.1.73)Now we prove the following equality by induction on n : d ( x α x β x nγ x η ⊗
1) = x α x β x nγ ⊗ x η − q γη x α x β x n − γ x η ⊗ x γ − ( − q βγ ) n b x α x nγ x ν ⊗ x γ − ( − q βγ ) n q βη x α x nγ x η ⊗ x β + q αβ ( − q αγ ) n q αη x β x nγ x η ⊗ x α . (10.1.74)Indeed for n = 0 we have: d ( x α x β x η ⊗
1) = x α x β ⊗ x η − s (cid:0) q βη x α ⊗ x η x β + b x α ⊗ x ν x γ − q αβ q αβ x β ⊗ x η x α (cid:1) = x α x β ⊗ x η − b x α x ν ⊗ x γ − q βη x α x η ⊗ x β + q αβ q αη x β x η ⊗ x α . Now assume that (10.1.74) holds for n . By Remark 10.1.2, inductive hypothesis and(10.1.72), d ( x α x β x n +1 γ x η ⊗
1) = x α x β x n +1 γ ⊗ x η − s (cid:0) q γη x α x β x nγ ⊗ x η x γ + ( − q βγ ) n +1 x α x n +1 γ ⊗ ( q βη x η x β + b x ν x γ ) − q αβ ( − q αγ ) n +1 q αη x β x n +1 γ ⊗ x η x α (cid:1) = x α x β x n +1 γ ⊗ x η − q γη x α x β x nγ x η ⊗ x γ − ( − q βγ ) n +1 b x α x n +1 γ x ν ⊗ x γ − ( − q βγ ) n +1 q βη x α x n +1 γ x η ⊗ x β + q αβ ( − q αγ ) n +1 q αη x β x n +1 γ x η ⊗ x α . Next we apply Lemma 10.1.46 to α < δ < γ < µ < ν to get: d ( x α x δ x nγ x ν ⊗
1) = x α x δ x nγ ⊗ x ν − q γν x α x δ x n − γ x ν ⊗ x γ − ( − q δγ ) n q δν x α x nγ x ν ⊗ x δ + ( − q αγ ) n q αδ q αν x δ x nγ x ν ⊗ x α − ( − q δγ ) n b x α x nγ x µ ⊗ x γ − q nγν c ( n ) − ν,α,γ b b x n +2 γ ⊗ . (10.1.75) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 101
Now we prove by induction on n that d ( x β x δ x nγ x η ⊗
1) = x β x δ x nγ ⊗ x η − q γη x β x δ x n − γ x η ⊗ x γ − ( − q δγ ) n q δη x β x nγ x η ⊗ x δ + ( − q βγ ) n q βδ q βη x δ x nγ x η ⊗ x β + q βδ ( − q βγ ) n b x δ x nγ x ν ⊗ x γ . (10.1.76)Indeed, for n = 0 we have d ( x β x δ x η ⊗
1) = x β x δ ⊗ x η − s (cid:0) q δη x β ⊗ x η x δ − q βδ q βη x δ ⊗ x η x β − q βδ b x δ ⊗ x ν x γ (cid:1) = x β x δ ⊗ x η − q δη x β x η ⊗ x δ + q βδ q βη x δ x η ⊗ x β + q βδ b x δ x ν ⊗ x γ . Now we assume that (10.1.76) holds for n . Using Remark 10.1.2, inductive hypothesis,(10.1.72), (10.1.73), d ( x β x δ x n +1 γ x η ⊗
1) = x β x δ x n +1 γ ⊗ x η − s (cid:0) q γη x β x δ x nγ ⊗ x η x γ + ( − q δγ ) n +1 q δη x β x n +1 γ ⊗ x η x δ − q βδ ( − q βγ ) n +1 q βη x δ x n +1 γ ⊗ x η x β − q βδ ( − q βγ ) n +1 b x δ x n +1 γ ⊗ x ν x γ (cid:1) = x β x δ x n +1 γ ⊗ x η − q γη x β x δ x nγ x η ⊗ x γ − ( − q δγ ) n +1 q δη x β x n +1 γ x η ⊗ x δ + q βδ ( − q βγ ) n +1 b x δ x n +1 γ x ν ⊗ x γ + q βδ ( − q βγ ) n +1 q βη x δ x n +1 γ x η ⊗ x β − q n +1 βγ q n +1 δγ q δη (1 − e q βδ ) b s (cid:0) x n +1 γ x ν ⊗ x γ x δ (cid:1) , and the inductive step follows since x n +1 γ x ν ⊗ x γ x δ = s ( x n +1 γ ⊗ x ν x γ x δ ).Finally we prove (10.1.70) by induction on n . When n = 0, d ( x α x β x δ x η ⊗
1) = x α x β x δ ⊗ x η − s (cid:0) q δη x α x β ⊗ x η x δ − q βδ q βη x α x δ ⊗ x η x β − q βδ b x α x δ ⊗ x ν x γ + q αβ q αδ q αη x β x δ ⊗ x η x α (cid:1) = x α x β x δ ⊗ x η − q δη x α x β x η ⊗ x δ + q βδ q βη x α x δ x η ⊗ x β + q βδ b x α x δ x ν ⊗ x γ − q αβ q αδ q αη x β x δ x η ⊗ x α + q βδ b b b x γ ⊗ . Now assume that (10.1.70) holds for n . Using Remark 10.1.2, inductive hypothesis,(10.1.74), (10.1.75) and (10.1.76) d ( x α x β x δ x n +1 γ x η ⊗
1) = x α x β x δ x n +1 γ ⊗ x η − s (cid:0) q γη x α x β x δ x nγ ⊗ x η x γ + ( − q δγ ) n +1 q δη x α x β x n +1 γ ⊗ x η x δ − q βδ ( − q βγ ) n +1 q βη x α x δ x n +1 γ ⊗ x η x β − q βδ ( − q βγ ) n +1 b x α x δ x n +1 γ ⊗ x ν x γ + q αβ q αδ ( − q αγ ) n +1 q αη x β x δ x n +1 γ ⊗ x η x α (cid:1) = x α x β x δ x n +1 γ ⊗ x η − q γη x α x β x δ x nγ x η ⊗ x γ − ( − q δγ ) n +1 q δη x α x β x n +1 γ x η ⊗ x δ + q βδ ( − q βγ ) n +1 b x α x δ x n +1 γ x ν ⊗ x γ + q βδ ( − q βγ ) n +1 q βη x α x δ x n +1 γ x η ⊗ x β − s (cid:0) q αβ q αδ ( − q αγ ) n +1 q αη x β x δ x n +1 γ ⊗ x η x α − q αβ q αδ q n +1 αγ q αη q n +1 δγ q δη x β x n +1 γ x η ⊗ x δ x α − q αβ q αδ ( − q αγ ) n +1 q αη q γη x β x δ x nγ x η ⊗ x γ x α + q βδ q n +1 βγ q n +2 αγ q αδ q αν b x δ x n +1 γ x ν ⊗ x γ x α − q βδ ( q n +1 γη d ( n ) β − ν αγ + q n +1 γν c ( n +1) − ν,α,γ ( − q βγ ) n +1 ) b b b x n +3 γ ⊗ x γ + q βδ q n +1 βγ q βη q αβ q αδ q n +1 αγ q αη x δ x n +1 γ x η ⊗ x β x α (cid:1) = x α x β x δ x n +1 γ ⊗ x η − q γη x α x β x δ x nγ x η ⊗ x γ − ( − q δγ ) n +1 q δη x α x β x n +1 γ x η ⊗ x δ + q βδ ( − q βγ ) n +1 b x α x δ x n +1 γ x ν ⊗ x γ + q βδ ( − q βγ ) n +1 q βη x α x δ x n +1 γ x η ⊗ x β
02 N. ANDRUSKIEWITSCH, I. ANGIONO, J. PEVTSOVA, S. WITHERSPOON − q αβ q αδ ( − q αγ ) n +1 q αη x β x δ x n +1 γ x η ⊗ x α + q βδ q n +1 γη d ( n +1) β − ν αγ b b b x n +4 γ ⊗ , and the inductive step follows. (cid:3) Lemma 10.1.77.
Let α < β < τ < δ < µ < ν < γ < κ < ι < η be positive roots such that N γ = N η = N κ = 2 and the relations among the corresponding root vectors take the form x α x δ = q αδ x δ x α + b x τ x β , x δ x η = q δη x η x δ + b x κ x γ x µ + b x γ x ν ,x τ x η = q τη x η x τ + b x κ x γ , x α x µ = q αµ x µ x α + b x τ x β ,x µ x η = q µη x η x µ + b x ι x γ , x ν x η = q νη x η x ν + b x ι x κ x γ ,x β x κ = q βκ x κ x β + b x γ , x α x ι = q αι x ι x α + b x γ , (10.1.78) for some scalars b j ∈ k and the other pairs of root vectors q -commute, except possibly ( x τ , x µ ) , ( x µ , x κ ) , ( x δ , x κ ) . Then d ( x α x β x δ x γ x η ⊗
1) = x α x β x δ x γ x η ⊗ x η + q γη x α x β x δ x γ x η ⊗ x γ + q δγ (3) e q γη b b x α x β x γ x ι ⊗ x κ x γ + q δγ q νη (3) e q γη b x α x β x γ x η ⊗ x ν + q δγ q µη q γη b x α x β x γ x κ x η ⊗ x γ x µ + q δγ q δη x α x β x γ x η ⊗ x δ + q δγ q γκ c (3) βηγ b b b x α x γ x ι ⊗ x γ + q δγ q γκ q µη c (3) βηγ b b x α x γ x η ⊗ x µ − q βδ q βγ q βη x α x δ x γ x η ⊗ x β + q αβ q βγ q βη q τγ q γη b b x β x τ x γ x κ x η ⊗ x γ x β − q αβ q βγ q βη q τγ q τη b x β x τ x γ x η ⊗ x τ x β + q αβ q αδ q αγ q αη x β x δ x γ x η ⊗ x α − q αβ q βτ q βγ q βη q − γβ q τγ c (3) βηγ b b b x τ x γ x η ⊗ x β − q δγ q γκ d (4) αβηγ b b b b x γ ⊗ . (10.1.79)Notice that the equalities in (10.1.16) forces α + δ = β + 2 τ, δ + η = κ + γ + µ, δ + η = γ + ν,τ + η = κ + γ, α + µ = τ + β, µ + η = ι + γ,ν + η = ι + κ + γ, β + κ = γ, α + ι = γ. (10.1.80)Thus the following equality also holds: 4 γ = α + β + δ + 2 η . Proof.
First we note that Lemma 10.1.7 applies for α < γ < ι , and Lemma 10.1.46 appliesfor β < τ < γ < κ < η . Hence the following formulas hold for all n ≥ d ( x α x nγ x ι ⊗
1) = x α x nγ ⊗ x ι − q γι x α x n − γ x ι ⊗ x γ − q αι ( − q αγ ) n x nγ x ι ⊗ x α − ( − q αγ ) n ( n + 1) e q αγ b x n +1 γ ⊗ , (10.1.81) d ( x τ x nγ x η ⊗
1) = x τ x nγ ⊗ x η − q γη x τ x n − γ x η ⊗ x γ − ( − q τγ ) n q τη x nγ x η ⊗ x τ − ( − q τγ ) n b x nγ x κ ⊗ x γ , (10.1.82) d ( x β x nγ x κ ⊗
1) = x β x nγ ⊗ x κ − q γκ x β x n − γ x κ ⊗ x γ − q βκ ( − q βγ ) n x nγ x κ ⊗ x β − ( − q βγ ) n ( n + 1) e q βγ b x n +1 γ ⊗ , (10.1.83) d ( x β x τ x nγ x η ⊗
1) = x β x τ x nγ ⊗ x η − q γη x β x τ x n − γ x η ⊗ x γ − ( − q τγ ) n q τη x β x nγ x η ⊗ x τ + ( − q βγ ) n q βτ q βη x τ x nγ x η ⊗ x β − ( − q τγ ) n b x β x nγ x κ ⊗ x γ − q nγη c ( n ) − η,β,γ b b x n +2 γ ⊗ . (10.1.84) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 103
We also need some auxiliar computations: d ( x α x β x δ ⊗
1) = x α x β ⊗ x δ − q βδ x α x δ ⊗ x β + q αβ b x β x τ ⊗ x τ x β + q αβ q αδ x β x δ ⊗ x α , (10.1.85) d ( x α x β x κ ⊗
1) = x α x β ⊗ x κ − b x α x γ ⊗ − q βκ x α x κ ⊗ x β + q αβ q ακ x β x κ ⊗ x α , (10.1.86) d ( x α x δ x γ ⊗
1) = x α x δ ⊗ x γ − q δγ x α x γ ⊗ x δ + q αδ q αγ x δ x γ ⊗ x α + q τγ q βγ b x τ x γ ⊗ x τ x β , (10.1.87) d ( x α x δ x η ⊗
1) = x α x δ ⊗ x η − q δη x α x η ⊗ x δ − b x α x κ ⊗ x γ x µ − b x α x γ ⊗ x ν + q αδ q αη x δ x η ⊗ x α + q βη b b x τ x κ ⊗ x γ x β + q βη q τη b x τ x η ⊗ x τ x β , (10.1.88) d ( x β x τ x κ ⊗
1) = x β x τ ⊗ x κ − q τκ x β x κ ⊗ x τ + q βτ b x τ x γ ⊗ q βτ q βκ x τ x κ ⊗ x β , (10.1.89) d ( x β x δ x η ⊗
1) = x β x δ ⊗ x η − q δη x β x η ⊗ x δ − b x β x κ ⊗ x γ x µ − b x β x γ ⊗ x ν + q βδ q βη x δ x η ⊗ x β − b b x γ ⊗ x µ , (10.1.90) d ( x β x κ ⊗
1) = x β x κ ⊗ x κ + b x γ x κ ⊗ q βκ x κ ⊗ x β , (10.1.91) d ( x β x κ x η ⊗
1) = x β x κ ⊗ x η − q κη x β x η ⊗ x κ + b x γ x η ⊗ q βκ q βη x κ x η ⊗ x β , (10.1.92) d ( x τ x κ x η ⊗
1) = x τ x κ ⊗ x η − q κη x τ x η ⊗ x κ + q τκ q τη x κ x η ⊗ x τ + q κτ (2) ηκ b x κ ⊗ x γ , (10.1.93) d ( x τ x η ⊗
1) = x τ x η ⊗ x η + q γη b x κ x η ⊗ x γ + q τη x η ⊗ x τ , (10.1.94) d ( x δ x γ x η ⊗
1) = x δ x γ ⊗ x η − q γη x δ x η ⊗ x γ + q δγ q δη x γ x η ⊗ x δ + q δγ b x γ x κ ⊗ x γ x µ + q δγ (2) e q γη b x γ ⊗ x ν . (10.1.95) d ( x δ x η ⊗
1) = x δ x η ⊗ x η + b b x γ x ι ⊗ x κ x γ + q νη b x γ x η ⊗ x ν + q γη q µη b x κ x η ⊗ x γ x µ + q δη x η ⊗ x δ . (10.1.96)The proof of these equalities is by direct computation.Next we compute differentials of some 4-chains, using the previous computations on3-chains and Remark 10.1.2: d ( x α x β x δ x γ ⊗
1) = x α x β x δ ⊗ x γ − q δγ x α x β x γ ⊗ x δ + q βδ q βγ x α x δ x γ ⊗ x β − q αβ q βγ q τγ b x β x τ x γ ⊗ x τ x β − q αβ q αδ q αγ x β x δ x γ ⊗ x α , (10.1.97) d ( x α x δ x γ ⊗
1) = x α x δ x γ ⊗ x γ + q δγ x α x γ ⊗ x δ − q αδ q αγ x δ x γ ⊗ x α − q τγ q βγ b x τ x γ ⊗ x τ x β , (10.1.98) d ( x α x β x δ x η ⊗
1) = x α x β x δ ⊗ x η − q δη x α x β x η ⊗ x δ − b x α x β x γ ⊗ x ν − b x α x β x κ ⊗ x γ x µ − b b x α x γ ⊗ x µ + q βδ q βη x α x δ x η ⊗ x β − q αβ q βη b b x β x τ x κ ⊗ x γ x β − q αβ q βη q τη b x β x τ x η ⊗ x τ x β − q αβ q αδ q αη x β x δ x η ⊗ x α + q βτ q αβ q βη b b b x τ x γ ⊗ x β , (10.1.99) d ( x α x β x κ x η ⊗
1) = x α x β x κ ⊗ x η − q κη x α x β x η ⊗ x κ + q βκ q βη x α x κ x η ⊗ x β + b x α x γ x η ⊗ − q αβ q ακ q αη x β x κ x η ⊗ x α , (10.1.100)
04 N. ANDRUSKIEWITSCH, I. ANGIONO, J. PEVTSOVA, S. WITHERSPOON d ( x α x β x γ x κ ⊗
1) = x α x β x γ ⊗ x κ − q γκ x α x β x κ ⊗ x γ + q βγ q βκ x α x γ x κ ⊗ x β + q − γβ (2) e q βγ b x α x γ ⊗ − q αβ q αγ q ακ x β x γ x κ ⊗ x α , (10.1.101) d ( x α x δ x γ x η ⊗
1) = x α x δ x γ ⊗ x η − q γη x α x δ x η ⊗ x γ + q δγ q δη x α x γ x η ⊗ x δ + q δγ b x α x γ x κ ⊗ x γ x µ − q αδ q αγ q αη x δ x γ x η ⊗ x α − q τγ q βγ q βη b b x τ x γ x κ ⊗ x γ x β + q δγ b x α x γ ⊗ x ν − q βγ q βη q τγ q τη b x τ x γ x η ⊗ x τ x β − q αγ q ακ q δγ b b x γ x κ ⊗ x τ x β , (10.1.102) d ( x α x δ x η ⊗
1) = x α x δ x η ⊗ x η + q νη b x α x γ x η ⊗ x ν + q δη x α x η ⊗ x δ + q µη q γη b x α x κ x η ⊗ x γ x µ − q βη q γη b b x τ x κ x η ⊗ x γ x β − q βη q τη b x τ x η ⊗ x τ x β − q αδ q αη x δ x η ⊗ x α , (10.1.103) d ( x β x τ x γ x κ ⊗
1) = x β x τ x γ ⊗ x κ − q γκ x β x τ x κ ⊗ x γ + q τγ q τκ x β x γ x κ ⊗ x τ − q βτ q − γβ (2) e q βγ b x τ x γ ⊗ − q βτ q βγ q βκ x τ x γ x κ ⊗ x β , (10.1.104) d ( x β x τ x κ x η ⊗
1) = x β x τ x κ ⊗ x η − q κη x β x τ x η ⊗ x κ + q τκ q τη x β x κ x η ⊗ x τ − q βτ b x τ x γ x η ⊗ − q γκ q − ηκ (2) e q κη b x β x κ ⊗ x γ − q βτ q βκ q βη x τ x κ x η ⊗ x β − q κη b b x γ x κ ⊗ , (10.1.105) d ( x β x τ x η ⊗
1) = x β x τ x η ⊗ x η + q τη x β x η ⊗ x τ − q βτ q βη x τ x η ⊗ x β + q γη b x β x κ x η ⊗ x γ + b b x γ x η ⊗ , (10.1.106) d ( x β x δ x γ x η ⊗
1) = x β x δ x γ ⊗ x η − q γη x β x δ x η ⊗ x γ + q δγ q δη x β x γ x η ⊗ x δ + q δγ b x β x γ x κ ⊗ x γ x µ + q δγ (2) e q γη b x β x γ ⊗ x ν + q βγ q δγ (2) e q γη b b x γ ⊗ x µ − q βδ q βγ q βη x δ x γ x η ⊗ x β , (10.1.107) d ( x β x δ x η ⊗
1) = x β x δ x η ⊗ x η + b b x β x γ x ι ⊗ x κ x γ + q νη b x β x γ x η ⊗ x ν + q µη q γη b x β x κ x η ⊗ x γ x µ + q δη x β x η ⊗ x δ − q βδ q βη x δ x η ⊗ x β + q µη b b x γ x η ⊗ x µ + b b b x γ x ι ⊗ x γ , (10.1.108) d ( x β x γ x κ ⊗
1) = x β x γ x κ ⊗ x κ + q γκ x β x κ ⊗ x γ − q βγ b (2) e q γκ x γ x κ ⊗ − q βγ q βκ x γ x κ ⊗ x β , (10.1.109) d ( x β x γ x κ x η ⊗
1) = x β x γ x κ ⊗ x η − q κη x β x γ x η ⊗ x κ + q γκ q γη x β x κ x η ⊗ x γ − q βγ (2) e q γκ b x γ x η ⊗ − q βγ q βκ q βη x γ x κ x η ⊗ x β , (10.1.110) d ( x τ x γ x κ x η ⊗
1) = x τ x γ x κ ⊗ x η − q κη x τ x γ x η ⊗ x κ + q γκ q γη x τ x κ x η ⊗ x γ − q τγ q τκ (2) e q κη b x γ x κ ⊗ x γ − q τγ q τκ q τη x γ x κ x η ⊗ x τ , (10.1.111) d ( x τ x γ x η ⊗
1) = x τ x γ x η ⊗ x η + q γη x τ x η ⊗ x γ − q τγ q γη b x γ x κ x η ⊗ x γ − q τγ q τη x γ x η ⊗ x τ , (10.1.112) d ( x δ x γ x η ⊗
1) = x δ x γ ⊗ x η − q γη x δ x γ x η ⊗ x γ − q δγ b x γ x κ ⊗ x γ x µ − q δγ (3) e q γη b x γ ⊗ x ν − q δγ q δη x γ x η ⊗ x δ , (10.1.113) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 105 d ( x δ x γ x η ⊗
1) = x δ x γ x η ⊗ x η + q γη x δ x η ⊗ x γ − q δγ (2) e q γη b b x γ x ι ⊗ x κ x γ − q δγ q νη (2) e q γη b x γ x η ⊗ x ν − q δγ q µη q γη b x γ x κ x η ⊗ x γ x µ − q δγ q δη x γ x η ⊗ x δ . (10.1.114)Next we compute differentials of some 5-chains: d ( x α x β x δ x γ ⊗
1) = x α x β x δ x γ ⊗ x γ + q δγ x α x β x γ ⊗ x δ − q βδ q βγ x α x δ x γ ⊗ x β + q αβ q βγ q τγ b x β x τ x γ ⊗ x τ x β + q αβ q αδ q αγ x β x δ x γ ⊗ x α , (10.1.115) d ( x α x β x δ x γ x η ⊗
1) = x α x β x δ x γ ⊗ x η + q δγ (2) e q γη b x α x β x γ ⊗ x ν − q γη x α x β x δ x η ⊗ x γ + q δγ q δη x α x β x γ x η ⊗ x δ + q δγ b x α x β x γ x κ ⊗ x γ x µ − q βδ q βγ q βη x α x δ x γ x η ⊗ x β + q αβ q βγ q βη q τγ q τη b x β x τ x γ x η ⊗ x τ x β − q δγ q − γβ c (2) βηγ b b x α x γ ⊗ x µ + q αβ q βγ q βη q τγ b b x β x τ x γ x κ ⊗ x γ x β + q αβ q αδ q αγ q αη x β x δ x γ x η ⊗ x α + q αβ q βτ q βη q − ηγ q − γβ c (2) βηγ b b b x τ x γ ⊗ x β , (10.1.116) d ( x α x β x δ x η ⊗
1) = x α x β x δ x η ⊗ x η + b b x α x β x γ x ι ⊗ x κ x γ + q δη x α x β x η ⊗ x δ + q νη b x α x β x γ x η ⊗ x ν + q µγ q µη q γη b x α x β x κ x η ⊗ x γ x µ − q βδ q βη x α x δ x η ⊗ x β + b b b x α x γ x ι ⊗ x γ − (4) e q αγ b b b b x γ ⊗ q µη b b x α x γ x η ⊗ x µ + q αβ q βη q τη b x β x τ x η ⊗ x τ x β + q αβ q βγ q βη q γη b b x β x τ x κ x η ⊗ x γ x β + q αβ q αδ q αη x β x δ x η ⊗ x α − q βτ q αβ q βη b b b x τ x γ x η ⊗ x β , (10.1.117) d ( x α x β x γ x κ ⊗
1) = x α x β x γ ⊗ x κ − q γκ x α x β x γ x κ ⊗ x γ − q βγ q βκ x α x γ x κ ⊗ x β − q − γβ (3) e q βγ b x α x γ ⊗ q αβ q αγ q ακ x β x γ x κ ⊗ x α , (10.1.118) d ( x α x β x γ x κ x η ⊗
1) = x α x β x γ x κ ⊗ x η − q κη x α x β x γ x η ⊗ x κ − q − γβ (2) e q βγ b x α x γ x η ⊗ q γκ q γη x α x β x κ x η ⊗ x γ − q βγ q βκ q βη x α x γ x κ x η ⊗ x β + q αβ q αγ q ακ q αη x β x γ x κ x η ⊗ x α , (10.1.119) d ( x α x δ x γ x η ⊗
1) = x α x δ x γ ⊗ x η − q γη x α x δ x γ x η ⊗ x γ − q δγ (2) e q γη b x α x γ ⊗ x ν − q δγ b x α x γ x κ ⊗ x γ x µ − q δγ q δη x α x γ x η ⊗ x δ + q αδ q αγ q αη x δ x γ x η ⊗ x α + q βη q τγ q βγ q τη b x τ x γ x η ⊗ x τ x β + q βη q τγ q βγ b b x τ x γ x κ ⊗ x γ x β , (10.1.120)
06 N. ANDRUSKIEWITSCH, I. ANGIONO, J. PEVTSOVA, S. WITHERSPOON d ( x α x δ x γ x η ⊗
1) = x α x δ x γ x η ⊗ x η − q δγ q νη (2) e q γη b x α x γ x η ⊗ x ν − q δγ (2) e q γη b b x α x γ x ι ⊗ x κ x γ + q δγ c (2) αηγ b b b x γ ⊗ x κ − q δγ q µη q γη b x α x γ x κ x η ⊗ x γ x µ + q γη x α x δ x η ⊗ x γ − q δγ q δη x α x γ x η ⊗ x δ + q βγ q βη q τγ q τη b x τ x γ x η ⊗ x τ x β + q τγ q βγ q βη q γη b b x τ x γ x κ x η ⊗ x γ x β + q αγ q ακ q βη q δγ b b b x γ x κ ⊗ x γ x β + q αδ q αγ q αη x δ x γ x η ⊗ x α + q αγ q ακ q βη q δγ q τη b b x γ x κ x η ⊗ x τ x β , (10.1.121) d ( x β x τ x γ x κ ⊗
1) = x β x τ x γ ⊗ x κ − q γκ x β x τ x γ x κ ⊗ x γ − q τγ q τκ x β x γ x κ ⊗ x τ + q βτ q − γβ (3) e q βγ b x τ x γ ⊗ q βτ q βγ q βκ x τ x γ x κ ⊗ x β , (10.1.122) d ( x β x τ x γ x κ x η ⊗
1) = x β x τ x γ x κ ⊗ x η − q κη x β x τ x γ x η ⊗ x κ − q τγ q τκ (2) e q κη b x β x γ x κ ⊗ x γ + q γκ q γη x β x τ x κ x η ⊗ x γ − q τγ q τκ q τη x β x γ x κ x η ⊗ x τ + q βτ q − γβ (2) e q βγ b x τ x γ x η ⊗ q βτ q βγ q βκ q βη x τ x γ x κ x η ⊗ x β − q κη q − ηγ (2) e q γη b b x γ x κ ⊗ , (10.1.123) d ( x β x τ x γ x η ⊗
1) = x β x τ x γ x η ⊗ x η + q γη x β x τ x η ⊗ x γ − q τγ q γη b x β x γ x κ x η ⊗ x γ − q τγ q τη x β x γ x η ⊗ x τ + q βτ q βγ q βη x τ x γ x η ⊗ x β + q − γη c (2) − β − ηγ b b x γ x η ⊗ , (10.1.124) d ( x β x δ x γ x η ⊗
1) = x β x δ x γ ⊗ x η − q γη x β x δ x γ x η ⊗ x γ − q δγ q δη x β x γ x η ⊗ x δ − q δγ b x β x γ x κ ⊗ x γ x µ − q δγ (3) e q γη b x β x γ ⊗ x ν + q βδ q βγ q βη x δ x γ x η ⊗ x β − q βγ q − γβ q δγ c (3) βη b b x γ ⊗ x µ , (10.1.125) d ( x β x δ x γ x η ⊗
1) = x β x δ x γ x η ⊗ x η − q δγ (2) e q γη b b x β x γ x ι ⊗ x κ x γ − q δγ (2) e q γη q νη b x β x γ x η ⊗ x ν − q δγ q µη q γη b x β x γ x κ x η ⊗ x γ x µ + q γη x β x δ x η ⊗ x γ − q δγ q δη x β x γ x η ⊗ x δ + q βδ q βγ q βη x δ x γ x η ⊗ x β − q βγ q δγ (2) e q γη b b b x γ x ι ⊗ x γ − q βγ q δγ q µη (2) e q γη b b x γ x η ⊗ x µ , (10.1.126) d ( x β x γ x κ ⊗
1) = x β x γ x κ ⊗ x κ + q γκ x β x γ x κ ⊗ x γ + q βγ (3) e q βγ b x γ x κ ⊗ q βκ q βγ x γ x κ ⊗ x β , (10.1.127) d ( x β x γ x κ x η ⊗
1) = x β x γ x κ ⊗ x η − q κη x β x γ x η ⊗ x κ + q γκ q γη x β x γ x κ x η ⊗ x γ + q βκ q βγ q βη x γ x κ x η ⊗ x β + q βγ (3) e q βγ b x γ x η ⊗ , (10.1.128) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 107 d ( x τ x γ x κ x η ⊗
1) = x τ x γ x κ ⊗ x η − q κη x τ x γ x η ⊗ x κ + q γκ q γη x τ x γ x κ x η ⊗ x γ + q τγ q τκ (2) e q κη b x γ x κ ⊗ x γ + q τγ q τκ q τη x γ x κ x η ⊗ x τ , (10.1.129) d ( x τ x γ x η ⊗
1) = x τ x γ x η ⊗ x η + q γη x τ x γ x η ⊗ x γ + q τγ q γη b x γ x κ x η ⊗ x γ + q τγ q τη x γ x η ⊗ x τ , (10.1.130) d ( x δ x γ x η ⊗
1) = x δ x γ x η ⊗ x η + q γη x δ x γ x η ⊗ x γ + q δγ (3) e q γη b b x γ x ι ⊗ x κ x γ + q δγ q νη (3) e q γη b x γ x η ⊗ x ν + q δγ q µη q γη b x γ x κ x η ⊗ x γ x µ + q δγ q δη x γ x η ⊗ x δ . (10.1.131)First we check (10.1.115). Using (10.1.97), Remark 10.1.2, (10.1.98), d ( x α x β x δ x γ ⊗
1) = x α x β x δ x γ ⊗ x γ − s (cid:0) − q δγ x α x β x γ ⊗ x γ x δ + q βδ q βγ x α x δ x γ ⊗ x γ x β − q αβ q βγ q τγ b x β x τ x γ ⊗ x γ x τ x β − q αβ q αδ q αγ x β x δ x γ ⊗ x γ x α (cid:1) = x α x β x δ x γ ⊗ x γ + q δγ x α x β x γ ⊗ x δ − q βδ q βγ x α x δ x γ ⊗ x β − s (cid:0) − q αβ q βγ q τγ b x β x τ x γ ⊗ x γ x τ x β − q αβ q αδ q αγ x β x δ x γ ⊗ x γ x α − q αβ q αγ q αδ q δγ x β x γ ⊗ x δ x α − q αβ q αγ q δγ b x β x γ ⊗ x τ x β + q βδ q βγ q τγ b x τ x γ ⊗ x τ x β + q αβ q αδ q αγ q βδ q βγ x δ x γ ⊗ x β x α (cid:1) = x α x β x δ x γ ⊗ x γ + q δγ x α x β x γ ⊗ x δ − q βδ q βγ x α x δ x γ ⊗ x β + q αβ q βγ q τγ b x β x τ x γ ⊗ x τ x β + q αβ q αδ q αγ x β x δ x γ ⊗ x α . For (10.1.116), we use (10.1.97), (10.1.99), Remark 10.1.2, (10.1.102), (10.1.104), (10.1.84),(10.1.107), x α x γ ⊗ x ν x β = s ( x α x γ ⊗ x γ x ν x β ), x γ ⊗ x µ x α = s ( x γ ⊗ x γ x µ x α ), d ( x α x β x δ x γ x η ⊗
1) = x α x β x δ x γ ⊗ x η − s (cid:0) q γη x α x β x δ ⊗ x η x γ − q δγ q δη x α x β x γ ⊗ x η x δ + q βδ q βγ q βη x α x δ x γ ⊗ x η x β − q δγ b x α x β x γ ⊗ x κ x γ x µ − q δγ b x α x β x γ ⊗ x γ x ν − q αβ q βγ q βη q τγ q τη b x β x τ x γ ⊗ x η x τ x β − q αβ q βγ q βη q τγ b b x β x τ x γ ⊗ x κ x γ x β − q αβ q αδ q αγ q αη x β x δ x γ ⊗ x η x α (cid:1) = x α x β x δ x γ ⊗ x η − q γη x α x β x δ x η ⊗ x γ + q δγ q δη x α x β x γ x η ⊗ x δ + q δγ b x α x β x γ x κ ⊗ x γ x µ + q δγ (2) e q γη b x α x β x γ ⊗ x ν − q βδ q βγ q βη x α x δ x γ x η ⊗ x β − q δγ q − γβ c (2) βηγ b b x α x γ ⊗ x µ + q αβ q βγ q βη q τγ q τη b x β x τ x γ x η ⊗ x τ x β + q αβ q βγ q βη q τγ b b x β x τ x γ x κ ⊗ x γ x β + q αβ q αδ q αγ q αη x β x δ x γ x η ⊗ x α + q αβ q βτ q βη q − ηγ q − γβ c (2) βηγ b b b x τ x γ ⊗ x β . For (10.1.117), we use (10.1.99), Remark 10.1.2, (10.1.100), (10.1.103), (10.1.81), (10.1.105),(10.1.106), (10.1.108) and (10.1.82): d ( x α x β x δ x η ⊗
1) = x α x β x δ x η ⊗ x η − s (cid:0) − q νη b x α x β x γ ⊗ x η x ν − b b x α x β x γ ⊗ x ι x κ x γ − q δη x α x β x η ⊗ x η x δ − q δη b x α x β x η ⊗ x κ x γ x µ − q δη b x α x β x η ⊗ x γ x ν − q γη q µη b x α x β x κ ⊗ x η x µ x γ − b b x α x γ ⊗ ( q µη x η x µ + b x ι x γ ) + q βδ q βη x α x δ x η ⊗ x η x β
08 N. ANDRUSKIEWITSCH, I. ANGIONO, J. PEVTSOVA, S. WITHERSPOON − q αβ q βη q γη b b x β x τ x κ ⊗ x η x β x γ − q αβ q βη q τη b x β x τ x η ⊗ ( q τη x η x τ + b x κ x γ ) x β − q αβ q αδ q αη x β x δ x η ⊗ x η x α + q βτ q αβ q βη b b b x τ x γ ⊗ x η x β (cid:1) = x α x β x δ x η ⊗ x η + b b x α x β x γ x ι ⊗ x κ x γ + q νη b x α x β x γ x η ⊗ x ν + q µγ q µη q γη b x α x β x κ x η ⊗ x γ x µ + q δη x α x β x η ⊗ x δ − q βδ q βη x α x δ x η ⊗ x β + b b b x α x γ x ι ⊗ x γ − (4) e q αγ b b b b x γ ⊗ q µη b b x α x γ x η ⊗ x µ + q αβ q βη q τη b x β x τ x η ⊗ x τ x β + q αβ q βγ q βη q γη b b x β x τ x κ x η ⊗ x γ x β + q αβ q αδ q αη x β x δ x η ⊗ x α − q βτ q αβ q βη b b b x τ x γ x η ⊗ x β . Next we check (10.1.118) using Remark 10.1.2, (10.1.101) and x γ ⊗ x α = s (cid:0) x γ ⊗ x γ x α (cid:1) : d ( x α x β x γ x κ ⊗
1) = x α x β x γ ⊗ x κ − s (cid:0) q γκ x α x β x γ ⊗ x κ x γ + q βγ q βκ x α x γ ⊗ x κ x β + q βγ b x α x γ ⊗ x γ − q αβ q αγ q ακ x β x γ ⊗ x κ x α (cid:1) = x α x β x γ ⊗ x κ − q γκ x α x β x γ x κ ⊗ x γ − q βγ q βκ x α x γ x κ ⊗ x β − q − γβ (3) e q βγ b x α x γ ⊗ q αβ q αγ q ακ x β x γ x κ ⊗ x α . For (10.1.119) we use (10.1.101), Remark 10.1.2, (10.1.100) and (10.1.110): d ( x α x β x γ x κ x η ⊗
1) = x α x β x γ x κ ⊗ x η − s (cid:0) q ηκ x α x β x γ ⊗ x η x κ − q γκ q γη x α x β x κ ⊗ x η x γ + q βγ q βκ q βη x α x γ x κ ⊗ x η x β + q − γβ (2) e q βγ b x α x γ ⊗ x η − q αβ q αγ q ακ x β x γ x κ ⊗ x α x η (cid:1) = x α x β x γ x κ ⊗ x η − q κη x α x β x γ x η ⊗ x κ − q − γβ (2) e q βγ b x α x γ x η ⊗ q γκ q γη x α x β x κ x η ⊗ x γ − q βγ q βκ q βη x α x γ x κ x η ⊗ x β + q αβ q αγ q ακ q αη x β x γ x κ x η ⊗ x α . To check (10.1.120) we use (10.1.98), (10.1.102), Remark 10.1.2, (10.1.82) and (10.1.113): d ( x α x δ x γ x η ⊗
1) = x α x δ x γ ⊗ x η − s (cid:0) q γη x α x δ x γ ⊗ x η x γ + q δγ q δη x α x γ ⊗ x η x δ + q δγ b x α x γ ⊗ x κ x γ x µ + q δγ b x α x γ ⊗ x γ x ν − q βη q τγ q βγ q τη b x τ x γ ⊗ x η x τ x β − q βη q τγ q βγ b b x τ x γ ⊗ x κ x γ x β − q αδ q αγ q αη x δ x γ ⊗ x η x α (cid:1) = x α x δ x γ ⊗ x η − q γη x α x δ x γ x η ⊗ x γ − q δγ (2) e q γη b x α x γ ⊗ x ν − q δγ b x α x γ x κ ⊗ x γ x µ − q δγ q δη x α x γ x η ⊗ x δ + q αδ q αγ q αη x δ x γ x η ⊗ x α + q βη q τγ q βγ q τη b x τ x γ x η ⊗ x τ x β + q βη q τγ q βγ b b x τ x γ x κ ⊗ x γ x β + q αδ q αγ q αη q δγ (3) e q γη b s (cid:0) x γ ⊗ x ν x α (cid:1) , and we use that x γ ⊗ x ν x α = s (cid:0) x γ ⊗ x γ x ν x α (cid:1) . To prove (10.1.121) we use (10.1.102),(10.1.103), (10.1.81), Remark 10.1.2, (10.1.111), (10.1.112) and (10.1.114): d ( x α x δ x γ x η ⊗
1) = x α x δ x γ x η ⊗ x η − s (cid:0) − q γη x α x δ x η ⊗ x η x γ + q δγ q δη x α x γ x η ⊗ x δ x η + q δγ b x α x γ x κ ⊗ x γ x µ x η − q αδ q αγ q αη x δ x γ x η ⊗ x α x η − q τγ q βγ q βη b b x τ x γ x κ ⊗ x γ x β x η + q δγ b x α x γ ⊗ x ν x η − q βγ q βη q τγ q τη b x τ x γ x η ⊗ x τ x β x η − q αγ q ακ q δγ b b x γ x κ ⊗ x τ x β x η (cid:1) = x α x δ x γ x η ⊗ x η + q γη x α x δ x η ⊗ x γ − q δγ q νη (2) e q γη b x α x γ x η ⊗ x ν − q δγ (2) e q γη b b x α x γ x ι ⊗ x κ x γ + q δγ c (2) αηγ b b b x γ ⊗ x κ − q δγ q δη x α x γ x η ⊗ x δ OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 109 − q δγ q µη q γη b x α x γ x κ x η ⊗ x γ x µ + q τγ q βγ q βη q γη b b x τ x γ x κ x η ⊗ x γ x β + q βγ q βη q τγ q τη b x τ x γ x η ⊗ x τ x β + q αγ q ακ q βη q δγ b b b x γ x κ ⊗ x γ x β + q αδ q αγ q αη x δ x γ x η ⊗ x α + q αγ q ακ q βη q δγ q τη b b x γ x κ x η ⊗ x τ x β . We compute (10.1.123) using (10.1.104), (10.1.84), (10.1.105), (10.1.109), (10.1.110),(10.1.82), (10.1.111) and Remark 10.1.2: d ( x β x τ x γ x κ x η ⊗
1) = x β x τ x γ x κ ⊗ x η − s (cid:0) q κη x β x τ x γ ⊗ x η x κ − q γκ q γη x β x τ x κ ⊗ x η x γ + q τγ q τκ q τη x β x γ x κ ⊗ x η x τ + q τγ q τκ b x β x γ x κ ⊗ x κ x γ − q βτ q − γβ (2) e q βγ b x τ x γ ⊗ x η − q βτ q βγ q βκ q βη x τ x γ x κ ⊗ x η x β (cid:1) = x β x τ x γ x κ ⊗ x η − q κη x β x τ x γ x η ⊗ x κ − q τγ q τκ (2) e q κη b x β x γ x κ ⊗ x γ + q γκ q γη x β x τ x κ x η ⊗ x γ − q τγ q τκ q τη x β x γ x κ x η ⊗ x τ + q βτ q − γβ (2) e q βγ b x τ x γ x η ⊗ q βτ q βγ q βκ q βη x τ x γ x κ x η ⊗ x β − q κη q − ηγ (2) e q γη b b x γ x κ ⊗ . Next we check (10.1.122) using Remark 10.1.2, (10.1.104) and (10.1.83): d ( x β x τ x γ x κ ⊗
1) = x β x τ x γ ⊗ x κ − s (cid:0) q γκ x β x τ x γ ⊗ x κ x γ + q τγ q τκ x β x γ ⊗ x κ x τ − q βτ q βγ q βκ x τ x γ ⊗ x κ x β − q βτ q βγ b x τ x γ ⊗ x γ (cid:1) = x β x τ x γ ⊗ x κ − q γκ x β x τ x γ x κ ⊗ x γ − q τγ q τκ x β x γ x κ ⊗ x τ + q βτ q − γβ (3) e q βγ b x τ x γ ⊗ q βτ q βγ q βκ x τ x γ x κ ⊗ x β . For (10.1.124) we use (10.1.84), (10.1.106), (10.1.110), Remark 10.1.2 and (10.1.112): d ( x β x τ x γ x η ⊗
1) = x β x τ x γ x η ⊗ x η − s (cid:0) − q γη x β x τ x η ⊗ x η x γ + q τγ q τη x β x γ x η ⊗ x η x τ + q τγ q τη b x β x γ x η ⊗ x κ x γ + q τγ q γη b x β x γ x κ ⊗ x η x γ − q βτ q βγ q βη x τ x γ x η ⊗ x η x β − q − ηγ (2) e q γη b b x γ ⊗ x η (cid:1) = x β x τ x γ x η ⊗ x η + q γη x β x τ x η ⊗ x γ − q τγ q γη b x β x γ x κ x η ⊗ x γ − q τγ q τη x β x γ x η ⊗ x τ + q βτ q βγ q βη x τ x γ x η ⊗ x β + q − γη c (2) − β − ηγ b b x γ x η ⊗ . The proof of (10.1.125) is similar, using Remark 10.1.2, (10.1.107), (10.1.83) and (10.1.113): d ( x β x δ x γ x η ⊗
1) = x β x δ x γ ⊗ x η − s (cid:0) q γη x β x δ x γ ⊗ x η x γ + q δγ q δη x β x γ ⊗ x η x δ + q δγ b x β x γ ⊗ x κ x γ x µ + q δγ b x β x γ ⊗ x γ x ν − q βδ q βγ q βη x δ x γ ⊗ x η x β (cid:1) = x β x δ x γ ⊗ x η − q γη x β x δ x γ x η ⊗ x γ − q δγ q δη x β x γ x η ⊗ x δ − q δγ b x β x γ x κ ⊗ x γ x µ − q δγ (3) e q γη b x β x γ ⊗ x ν + q βδ q βγ q βη x δ x γ x η ⊗ x β − q βγ q − γβ q δγ c (3) βη b b x γ ⊗ x µ . For (10.1.126) we use (10.1.107), (10.1.108), Remark 10.1.2, (10.1.110) and (10.1.114): d ( x β x δ x γ x η ⊗
1) = x β x δ x γ x η ⊗ x η − s (cid:0) − q γη x β x δ x η ⊗ x η x γ + q δγ q δη x β x γ x η ⊗ x η x δ + q δγ q δη b x β x γ x η ⊗ x κ x γ x µ + q δγ q δη b x β x γ x η ⊗ x γ x ν + q δγ q µη q γη b x β x γ x κ ⊗ x η x γ x µ + q δγ (2) e q γη q νη b x β x γ ⊗ x η x ν + q δγ (2) e q γη b b x β x γ ⊗ x ι x κ x γ + q βγ q δγ q µη (2) e q γη b b x γ ⊗ x η x µ
10 N. ANDRUSKIEWITSCH, I. ANGIONO, J. PEVTSOVA, S. WITHERSPOON + q βγ q δγ (2) e q γη b b b x γ ⊗ x ι x γ − q βδ q βγ q βη x δ x γ x η ⊗ x η x β (cid:1) = x β x δ x γ x η ⊗ x η + q γη x β x δ x η ⊗ x γ − q δγ (2) e q γη b b x β x γ x ι ⊗ x κ x γ − q δγ (2) e q γη q νη b x β x γ x η ⊗ x ν − q δγ q µη q γη b x β x γ x κ x η ⊗ x γ x µ − q δγ q δη x β x γ x η ⊗ x δ + q βδ q βγ q βη x δ x γ x η ⊗ x β − q βγ q δγ (2) e q γη b b b x γ x ι ⊗ x γ − q βγ q δγ q µη (2) e q γη b b x γ x η ⊗ x µ . We check (10.1.127) using (10.1.83), (10.1.109) and Remark 10.1.2: d ( x β x γ x κ ⊗
1) = x β x γ x κ ⊗ x κ − s (cid:0) − q γκ x β x γ x κ ⊗ x κ x γ − q βκ q βγ b x γ x κ ⊗ x γ − q βκ q βγ x γ x κ ⊗ x κ x β − q βγ (3) e q βγ b x γ ⊗ x κ (cid:1) = x β x γ x κ ⊗ x κ + q γκ x β x γ x κ ⊗ x γ + q βγ (3) e q βγ b x γ x κ ⊗ q βκ q βγ x γ x κ ⊗ x β . For (10.1.128) we use (10.1.83), Remark 10.1.2 and (10.1.110): d ( x β x γ x κ x η ⊗
1) = x β x γ x κ ⊗ x η − s (cid:0) q κη x β x γ ⊗ x η x κ − q γκ q γη x β x γ x κ ⊗ x η x γ − q βκ q βγ q βη x γ x κ ⊗ x η x β − q βγ (3) e q βγ b x γ ⊗ x η (cid:1) = x β x γ x κ ⊗ x η − q κη x β x γ x η ⊗ x κ + q γκ q γη x β x γ x κ x η ⊗ x γ + q βκ q βγ q βη x γ x κ x η ⊗ x β + q βγ (3) e q βγ b x γ x η ⊗ . Now we compute (10.1.129) using Remark 10.1.2, (10.1.82) and (10.1.111), d ( x τ x γ x κ x η ⊗
1) = x τ x γ x κ ⊗ x η − s (cid:0) q κη x τ x γ ⊗ x η x κ − q γη x τ x γ x κ ⊗ x η x γ − q τγ q τκ q τη x γ x κ ⊗ x η x τ − q τγ q τκ b x γ x κ ⊗ x κ x γ (cid:1) = x τ x γ x κ ⊗ x η − q κη x τ x γ x η ⊗ x κ + q γκ q γη x τ x γ x κ x η ⊗ x γ + q τγ q τκ (2) e q κη b x γ x κ ⊗ x γ + q τγ q τκ q τη x γ x κ x η ⊗ x τ . For (10.1.130) we use (10.1.82), (10.1.112) and Remark 10.1.2: d ( x τ x γ x η ⊗
1) = x τ x γ x η ⊗ x η − s (cid:0) − q γη x τ x γ x η ⊗ x η x γ − q τγ q τη b x γ x η ⊗ x κ x γ − q τγ q τη x γ x η ⊗ x η x τ − q τγ q γη b x γ x κ ⊗ x η x γ (cid:1) = x τ x γ x η ⊗ x η + q γη x τ x γ x η ⊗ x γ + q τγ q γη b x γ x κ x η ⊗ x γ + q τγ q τη x γ x η ⊗ x τ . Finally we compute (10.1.131) using (10.1.113), (10.1.114) and Remark 10.1.2: d ( x δ x γ x η ⊗
1) = x δ x γ x η ⊗ x η − s (cid:0) − q γη x δ x γ x η ⊗ x η x γ − q δγ q µη q γη b x γ x κ ⊗ x η x γ x µ − q δγ (3) e q γη b b x γ ⊗ x ι x κ x γ − q δγ q νη (3) e q γη b x γ ⊗ x η x ν − q δγ q δη b x γ x η ⊗ x γ x ν − q δγ q δη b x γ x η ⊗ x κ x γ x µ − q δγ q δη x γ x η ⊗ x η x δ (cid:1) = x δ x γ x η ⊗ x η + q γη x δ x γ x η ⊗ x γ + q δγ (3) e q γη b b x γ x ι ⊗ x κ x γ + q δγ q νη (3) e q γη b x γ x η ⊗ x ν + q δγ q µη q γη b x γ x κ x η ⊗ x γ x µ + q δγ q δη x γ x η ⊗ x δ . OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 111
Next we compute differentials of some 6-chains: d ( x α x β x δ x γ x η ⊗
1) = x α x β x δ x γ ⊗ x η − q γη x α x β x δ x γ x η ⊗ x γ − q δγ (3) e q γη b x α x β x γ ⊗ x ν − q δγ b x α x β x γ x κ ⊗ x γ x µ − q δγ q δη x α x β x γ x η ⊗ x δ + q βδ q βγ q βη x α x δ x γ x η ⊗ x β − q δγ q − γβ c (3) βηγ b b x α x γ ⊗ x µ − q αβ q βγ q βη q τγ b b x β x τ x γ x κ ⊗ x γ x β − q αβ q βγ q βη q τγ q τη b x β x τ x γ x η ⊗ x τ x β − q αβ q αδ q αγ q αη x β x δ x γ x η ⊗ x α + q αβ q βγ q βη q βτ q − γβ (3) e q βγ q τγ b b b x τ x γ ⊗ x β , (10.1.132) d ( x α x β x δ x γ x η ⊗
1) = x α x β x δ x γ x η ⊗ x η + q γη x α x β x δ x η ⊗ x γ − q δγ (2) e q γη q νη b x α x β x γ x η ⊗ x ν − q δγ (2) e q γη b b x α x β x γ x ι ⊗ x κ x γ − q δγ q γη q µη b x α x β x γ x κ x η ⊗ x γ x µ − q δγ q δη x α x β x γ x η ⊗ x δ + q αβ q βγ q βη q τγ q γη b b x β x τ x γ x κ x η ⊗ x γ x β + q δγ q − γβ c (2) βηγ b b b x α x γ x ι ⊗ x γ + q δγ q γκ d (3) αβηγ b b b b x γ ⊗ q δγ q − γβ q µη c (2) βηγ b b x α x γ x η ⊗ x µ + q βδ q βγ q βη x α x δ x γ x η ⊗ x β − q αβ q βγ q βη q τγ q τη b x β x τ x γ x η ⊗ x τ x β − q αβ q αδ q αγ q αη x β x δ x γ x η ⊗ x α − q αβ q βτ q βη q − ηγ q − γβ c (2) βηγ b b b x τ x γ x η ⊗ x β , (10.1.133) d ( x α x β x γ x κ x η ⊗
1) = x α x β x γ x κ ⊗ x η − q κη x α x β x γ x η ⊗ x κ + q γκ q γη x α x β x γ x κ x η ⊗ x γ + q − γβ (3) e q βγ b x α x γ x η ⊗ q βγ q βκ q βη x α x γ x κ x η ⊗ x β − q αβ q αγ q ακ q αη x β x γ x κ x η ⊗ x α , (10.1.134) d ( x α x δ x γ x η ⊗
1) = x α x δ x γ x η ⊗ x η + q γη x α x δ x γ x η ⊗ x γ + q δγ (2) e q γη b b x α x γ x ι ⊗ x κ x γ − q δγ c (3) αηγ b b b x γ ⊗ x κ + q δγ q νη (2) e q γη b x α x γ x η ⊗ x ν + q δγ q δη x α x γ x η ⊗ x δ + q δγ q µη q γη b x α x γ x κ x η ⊗ x γ x µ − q βη q τγ q βγ q γη b b x τ x γ x κ x η ⊗ x γ x β − q βη q τγ q βγ q τη b x τ x γ x η ⊗ x τ x β − q αδ q αγ q αη x δ x γ x η ⊗ x α , (10.1.135) d ( x β x τ x γ x κ x η ⊗
1) = x β x τ x γ x κ ⊗ x η − q κη x β x τ x γ x η ⊗ x κ + q γκ q γη x β x τ x γ x κ x η ⊗ x γ + q τγ q τκ (2) e q κη b x β x γ x κ ⊗ x γ + q τγ q τκ q τη x β x γ x κ x η ⊗ x τ − q βτ q − γβ (3) e q βγ b x τ x γ x η ⊗ − q βτ q βγ q βκ q βη x τ x γ x κ x η ⊗ x β − q γη q κη c (2) − η,β,γ b b x γ x κ ⊗ , (10.1.136) d ( x β x τ x γ x η ⊗
1) = x β x τ x γ x η ⊗ x η + q γη x β x τ x γ x η ⊗ x γ + q τγ q γη b x β x γ x κ x η ⊗ x γ + q τγ q τη x β x γ x η ⊗ x τ − q βτ q βγ q βη x τ x γ x η ⊗ x β + q γη c (2) − η,β,γ b b x γ ⊗ x η . (10.1.137)
12 N. ANDRUSKIEWITSCH, I. ANGIONO, J. PEVTSOVA, S. WITHERSPOON d ( x β x δ x γ x η ⊗
1) = x β x δ x γ x η ⊗ x η + q γη x β x δ x γ x η ⊗ x γ + q δγ (3) e q γη b b x β x γ x ι ⊗ x κ x γ + q δγ q νη (3) e q γη b x β x γ x η ⊗ x ν + q δγ q µη q γη b x β x γ x κ x η ⊗ x γ x µ + q δγ q δη x β x γ x η ⊗ x δ − q βδ q βγ q βη x δ x γ x η ⊗ x β − q βγ q γκ q δγ c (3) βη b b b x γ x ι ⊗ x γ − q βγ q γκ q δγ c (3) βη q µη b b x γ x η ⊗ x µ . (10.1.138)First we deal with (10.1.132): using (10.1.115), Remark 10.1.2, (10.1.116), (10.1.118),(10.1.120), (10.1.122), (10.1.84) and (10.1.125): d ( x α x β x δ x γ x η ⊗
1) = x α x β x δ x γ ⊗ x η − s (cid:0) q γη x α x β x δ x γ ⊗ x η x γ + q δγ x α x β x γ ⊗ q δη x η x δ + q δγ x α x β x γ ⊗ b x κ x γ x µ + q δγ x α x β x γ ⊗ b x γ x ν − q βδ q βγ q βη x α x δ x γ ⊗ x η x β + q αβ q βγ q βη q τγ b x β x τ x γ ⊗ ( q τη x η x τ + b x κ x γ ) x β + q αβ q αδ q αγ q αη x β x δ x γ ⊗ x η x α (cid:1) = x α x β x δ x γ ⊗ x η − q γη x α x β x δ x γ x η ⊗ x γ − q δγ (3) e q γη b x α x β x γ ⊗ x ν − q δγ b x α x β x γ x κ ⊗ x γ x µ − q δγ q δη x α x β x γ x η ⊗ x δ + q βδ q βγ q βη x α x δ x γ x η ⊗ x β − q δγ q − γβ c (3) βηγ b b x α x γ ⊗ x µ − q αβ q βγ q βη q τγ b b x β x τ x γ x κ ⊗ x γ x β − q αβ q βγ q βη q τγ q τη b x β x τ x γ x η ⊗ x τ x β − q αβ q αδ q αγ q αη x β x δ x γ x η ⊗ x α + q αβ q βγ q βη q βτ q − γβ (3) e q βγ q τγ b b b x τ x γ ⊗ x β . Next we compute (10.1.133): using (10.1.116), (10.1.117), Remark 10.1.2, (10.1.119),(10.1.121), (10.1.81), (10.1.123), (10.1.124), (10.1.126) and (10.1.82): d ( x α x β x δ x γ x η ⊗
1) = x α x β x δ x γ x η ⊗ x η − s (cid:0) q δγ (2) e q γη b x α x β x γ ⊗ ( q νη x η x ν + b x ι x κ x γ ) − q γη x α x β x δ x η ⊗ x η x γ + q δγ q δη x α x β x γ x η ⊗ ( q δη x η x δ + b x κ x γ x µ + b x γ x ν )+ q δγ q γη q µη b x α x β x γ x κ ⊗ x η x γ x µ − q βδ q βγ q βη x α x δ x γ x η ⊗ x η x β + q αβ q βγ q βη q τγ q τη b x β x τ x γ x η ⊗ ( q τη x η x τ + b x κ x γ ) x β − q δγ q − γβ c (2) βηγ b b x α x γ ⊗ x µ x η + q αβ q βγ q βη q τγ q γη b b x β x τ x γ x κ ⊗ x η x γ x β + q αβ q αδ q αγ q αη x β x δ x γ x η ⊗ x η x α + q αβ q βτ q βη q − ηγ q − γβ c (2) βηγ b b b x τ x γ ⊗ x η x β (cid:1) = x α x β x δ x γ x η ⊗ x η + q γη x α x β x δ x η ⊗ x γ − q δγ (2) e q γη b b x α x β x γ x ι ⊗ x κ x γ − q δγ (2) e q γη q νη b x α x β x γ x η ⊗ x ν − q δγ q γη q µη b x α x β x γ x κ x η ⊗ x γ x µ − q δγ q δη x α x β x γ x η ⊗ x δ + q αβ q βγ q βη q τγ q γη b b x β x τ x γ x κ x η ⊗ x γ x β + q δγ q − γβ c (2) βηγ b b b x α x γ x ι ⊗ x γ + q δγ q γκ d (3) αβηγ b b b b x γ ⊗ q δγ q − γβ q µη c (2) βηγ b b x α x γ x η ⊗ x µ + q βδ q βγ q βη x α x δ x γ x η ⊗ x β − q αβ q βγ q βη q τγ q τη b x β x τ x γ x η ⊗ x τ x β − q αβ q αδ q αγ q αη x β x δ x γ x η ⊗ x α − q αβ q βτ q βη q − ηγ q − γβ c (2) βηγ b b b x τ x γ x η ⊗ x β . For (10.1.134) we use (10.1.118), Remark 10.1.2, (10.1.119) and (10.1.128): d ( x α x β x γ x κ x η ⊗
1) = x α x β x γ x κ ⊗ x η − s (cid:0) q κη x α x β x γ ⊗ x η x κ − q γκ q γη x α x β x γ x κ ⊗ x η x γ − q βγ q βκ q βη x α x γ x κ ⊗ x η x β − q − γβ (3) e q βγ b x α x γ ⊗ x η + q αβ q αγ q ακ q αη x β x γ x κ ⊗ x η x α (cid:1) OHOMOLOGY RINGS OF FINITE-DIMENSIONAL HOPF ALGEBRAS 113 = x α x β x γ x κ ⊗ x η − q κη x α x β x γ x η ⊗ x κ + q γκ q γη x α x β x γ x κ x η ⊗ x γ + q − γβ (3) e q βγ b x α x γ x η ⊗ q βγ q βκ q βη x α x γ x κ x η ⊗ x β − q αβ q αγ q ακ q αη x β x γ x κ x η ⊗ x α . For (10.1.135) we use (10.1.120), (10.1.121), (10.1.81), (10.1.129), (10.1.130) and (10.1.131): d ( x α x δ x γ x η ⊗
1) = x α x δ x γ x η ⊗ x η − s (cid:0) − q γη x α x δ x γ x η ⊗ x η x γ − q δγ q νη (2) e q γη b x α x γ ⊗ x η x ν − q δγ (2) e q γη b b x α x γ ⊗ x ι x κ x γ − q δγ q µη q γη b x α x γ x κ ⊗ x η x γ x µ − q δγ q δη x α x γ x η ⊗ x η x δ − q δγ q δη b x α x γ x η ⊗ x κ x γ x µ − q δγ q δη b x α x γ x η ⊗ x γ x ν + q αδ q αγ q αη x δ x γ x η ⊗ x η x α + q βη q τγ q βγ q τη b x τ x γ x η ⊗ x η x τ x β + q βη q τγ q βγ q τη b b x τ x γ x η ⊗ x κ x γ x β + q βη q τγ q βγ q γη b b x τ x γ x κ ⊗ x η x γ x β (cid:1) = x α x δ x γ x η ⊗ x η + q γη x α x δ x γ x η ⊗ x γ + q δγ (2) e q γη b b x α x γ x ι ⊗ x κ x γ − q δγ c (3) αηγ b b b x γ ⊗ x κ + q δγ q νη (2) e q γη b x α x γ x η ⊗ x ν + q δγ q µη q γη b x α x γ x κ x η ⊗ x γ x µ + q δγ q δη x α x γ x η ⊗ x δ − q βη q τγ q βγ q γη b b x τ x γ x κ x η ⊗ x γ x β − q βη q τγ q βγ q τη b x τ x γ x η ⊗ x τ x β − q αδ q αγ q αη x δ x γ x η ⊗ x α . Next we compute (10.1.136) using (10.1.122), (10.1.84), (10.1.123), (10.1.127), (10.1.128),(10.1.82) and (10.1.129): d ( x β x τ x γ x κ x η ⊗
1) = x β x τ x γ x κ ⊗ x η − s (cid:0) q κη x β x τ x γ ⊗ x η x κ − q γκ q γη x β x τ x γ x κ ⊗ x η x γ − q τγ q τκ b x β x γ x κ ⊗ x κ x γ − q τγ q τκ q τη x β x γ x κ ⊗ x η x τ + q βτ q − γβ (3) e q βγ b x τ x γ ⊗ x η + q βτ q βγ q βκ q βη x τ x γ x κ ⊗ x η x β (cid:1) = x β x τ x γ x κ ⊗ x η − q κη x β x τ x γ x η ⊗ x κ + q γκ q γη x β x τ x γ x κ x η ⊗ x γ + q τγ q τκ (2) e q κη b x β x γ x κ ⊗ x γ + q τγ q τκ q τη x β x γ x κ x η ⊗ x τ − q βτ q − γβ (3) e q βγ b x τ x γ x η ⊗ − q βτ q βγ q βκ q βη x τ x γ x κ x η ⊗ x β − q γη q κη c (2) − η,β,γ b b x γ x κ ⊗ . For (10.1.137) we use (10.1.84), (10.1.124), (10.1.128), Remark 10.1.2 and (10.1.130): d ( x β x τ x γ x η ⊗
1) = x β x τ x γ x η ⊗ x η − s (cid:0) − q γη x β x τ x γ x η ⊗ x η x γ − q τγ q τη x β x γ x η ⊗ x η x τ − q τγ q γη b x β x γ x κ ⊗ x η x γ − q τγ q τη b x β x γ x η ⊗ x κ x γ + q βτ q βγ q βη x τ x γ x η ⊗ x η x β − q γη c (2) − η,β,γ b b x γ ⊗ x η (cid:1) = x β x τ x γ x η ⊗ x η + q γη x β x τ x γ x η ⊗ x γ + q τγ q γη b x β x γ x κ x η ⊗ x γ + q τγ q τη x β x γ x η ⊗ x τ − q βτ q βγ q βη x τ x γ x η ⊗ x β + q γη c (2) − η,β,γ b b x γ ⊗ x η . The proof of (10.1.138) is similar, using in this case (10.1.125), (10.1.126), Remark10.1.2, (10.1.128) and (10.1.131): d ( x β x δ x γ x η ⊗
1) = x β x δ x γ x η ⊗ x η − s (cid:0) − q γη x β x δ x γ x η ⊗ x η x γ − q δγ q δη b x β x γ x η ⊗ x γ x ν − q δγ q δη b x β x γ x η ⊗ x κ x γ x µ − q δγ q δη x β x γ x η ⊗ x η x δ − q δγ q µη q γη b x β x γ x κ ⊗ x η x γ x µ − q δγ (3) e q γη b b x β x γ ⊗ x ι x κ x γ − q δγ q νη (3) e q γη b x β x γ ⊗ x η x ν + q βδ q βγ q βη x δ x γ x η ⊗ x η x β − q βγ q − γβ q δγ c (3) βη b b b x γ ⊗ x ι x γ − q βγ q − γβ q δγ c (3) βη q µη b b x γ ⊗ x η x µ (cid:1)
14 N. ANDRUSKIEWITSCH, I. ANGIONO, J. PEVTSOVA, S. WITHERSPOON = x β x δ x γ x η ⊗ x η + q γη x β x δ x γ x η ⊗ x γ + q δγ (3) e q γη b b x β x γ x ι ⊗ x κ x γ + q δγ q νη (3) e q γη b x β x γ x η ⊗ x ν + q δγ q µη q γη b x β x γ x κ x η ⊗ x γ x µ + q δγ q δη x β x γ x η ⊗ x δ − q βδ q βγ q βη x δ x γ x η ⊗ x β − q βγ q γκ q δγ c (3) βη b b b x γ x ι ⊗ x γ − q βγ q γκ q δγ c (3) βη q µη b b x γ x η ⊗ x µ . Finally we compute (10.1.79). Using (10.1.132), (10.1.133), Remark 10.1.2, (10.1.134),(10.1.135), (10.1.136), (10.1.137), (10.1.138) and (10.1.82): d ( x α x β x δ x γ x η ⊗
1) = x α x β x δ x γ x η ⊗ x η − s (cid:0) − q γη x α x β x δ x γ x η ⊗ x η x γ − q δγ (3) e q γη b x α x β x γ ⊗ x ν x η − q δγ b x α x β x γ x κ ⊗ x γ x µ x η − q δγ q δη x α x β x γ x η ⊗ x δ x η + q βδ q βγ q βη x α x δ x γ x η ⊗ x β x η − q αβ q βγ q βη q τγ b b x β x τ x γ x κ ⊗ x γ x β x η − q δγ q − γβ c (3) βηγ b b x α x γ ⊗ x µ x η − q αβ q βγ q βη q τγ q τη b x β x τ x γ x η ⊗ x τ x β x η − q αβ q αδ q αγ q αη x β x δ x γ x η ⊗ x α x η + q αβ q βγ q βη q βτ q − γβ (3) e q βγ q τγ b b b x τ x γ ⊗ x β x η (cid:1) = x α x β x δ x γ x η ⊗ x η + q γη x α x β x δ x γ x η ⊗ x γ + q δγ (3) e q γη b b x α x β x γ x ι ⊗ x κ x γ + q δγ q νη (3) e q γη b x α x β x γ x η ⊗ x ν + q δγ q µη q γη b x α x β x γ x κ x η ⊗ x γ x µ + q δγ q δη x α x β x γ x η ⊗ x δ − q βδ q βγ q βη x α x δ x γ x η ⊗ x β + q δγ q γκ c (3) βηγ b b b x α x γ x ι ⊗ x γ + q δγ q γκ q µη c (3) βηγ b b x α x γ x η ⊗ x µ + q αβ q βγ q βη q τγ q γη b b x β x τ x γ x κ x η ⊗ x γ x β − q αβ q βγ q βη q τγ q τη b x β x τ x γ x η ⊗ x τ x β + q αβ q αδ q αγ q αη x β x δ x γ x η ⊗ x α − q αβ q βτ q βγ q βη q − γβ q τγ c (3) βηγ b b b x τ x γ x η ⊗ x β − q δγ q γκ d (4) αβηγ b b b b x γ ⊗ . This completes the proof. (cid:3)
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