Finite GK-Dimensional pre-Nichols algebras of super and standard type
aa r X i v : . [ m a t h . QA ] S e p FINITE GK-DIMENSIONAL PRE-NICHOLS ALGEBRAS OF SUPERAND STANDARD TYPE
IV ´AN ANGIONO, EMILIANO CAMPAGNOLO, AND GUILLERMO SANMARCO
Abstract.
We prove that finite GK-dimensional pre-Nichols algebras of super and stan-dard type are quotients of the corresponding distinguished pre-Nichols algebras, exceptwhen the braiding matrix is of type super A and the dimension of the braided vector spaceis three. For these two exceptions we explicitly construct substitutes as braided centralextensions of the corresponding pre-Nichols algebras by a polynomial ring in one variable.Via bosonization this gives new examples of finite GK-dimensional Hopf algebras. Introduction
Fix an algebraically closed field k of characteristic zero. We abbreviate GKdim A forthe Gelfand-Kirillov dimension of an associative algebra A .The study of Hopf algebras with finite GKdim can be approached from two differentperspectives. A first possibility consists on looking for particular features of the underlyingalgebra structure. For instance noetherian Hopf algebras were intensively studied [BG,G] and many authors gave classification results under this or similar hypothesis, see forexample [BZ, L, GZ, WZZ, G].A second approach comes from the underlying coalgebra structure. Following the LiftingMethod [AS], the first invariant to be considered is the coradical (the sum of all simplesubcoalgebras). An intensively studied family is that of pointed Hopf algebras — thosewhose coradical coincide with the group algebra of the so called group-like elements. If H is a pointed Hopf algebra with group of group-like elements Γ then the coradical filtrationgives raise to a graded Hopf algebra gr H endowed with maps k Γ ֒ → gr H ։ k Γ composingto the identity of k Γ. Thus gr H decomposes as the bosonization gr H ≃ R k Γ, where R (the so called diagram of H ) is a coradically graded Hopf algebra in k Γ k Γ YD such that R = k
1. The degree 1 component V := R ∈ k Γ k Γ YD is the infinitesimal braiding of H ,the subalgebra B ( V ) generated by V is the Nichols algebra of V and R is a post-Nicholsalgebra of V , see § V over Γ such that GKdim B ( V ) < ∞ ,(II) for each V as in (I), determine all finite GKdim post-Nichols algebras R of V ;(III) for all such R , obtain all possible H such that gr H ≃ R k Γ.Assume from now on that Γ is abelian and finitely generated. Each object V ∈ k Γ k Γ YD either splits as direct sum of one-dimensional submodules or it contains an indecomposable Key words and phrases.
Hopf algebras, Nichols algebras, Gelfand-Kirillov dimension.MSC2020: 16T05, 16T20, 17B37, 17B62.The work of the three authors was partially supported by CONICET and Secyt (UNC). submodule of dimension >
1. In the former scenario we say that V is of diagonal type ; thebraiding is described by a matrix q ∈ ( k × ) θ × θ , see § B q . Regarding (I), based on the evidence from [AAH2], the same authors proposed: Conjecture 1.1. [AAH1, 1.5] If GKdim B q < ∞ , then the root system ∆ q is finite. Throughout this paper we will assume the validity of this statement : it means that q belongs to the classification given in [H2]. About non-diagonal braided vector spaces, awide classification result was performed in [AAH1].For (II) the situation is quite different from the finite-dimensional context, where the generation in degree-one problem has a positive answer [An2]: it means that the uniquefinite dimensional post-Nichols algebra of each q is B q itself. This is not longer truewhen we move to finite GKdim. For example, if dim B q < ∞ , the so-called distinguished pre-Nichols algebra e B q is a finite GKdim pre-Nichols algebra different from B q , see § § q as in (I), determine all pre-Nichols algebras B of V with GKdim B < ∞ .Assume that q is as in (I). Pre-Nichols algebras of a fixed q form a poset with minimalelement T ( V ) and maximal one B q . Those with finite GKdim determine a sub-poset anda first question is if there exist minimal elements (called eminent pre-Nichols algebras).If dim B q < ∞ , then it is natural to ask whether the distinguished pre-Nichols algebra iseminent, as it was formulated in [An4].Following [AA], the list given in [H2] containing all matrices q with connected Dynkindiagram and finite generalized root system can be organized as follows: ⋄ Standard type: the Cartan matrix is constant in the Weyl equivalence class. This familyincludes Cartan type [AS], but also braidings with Cartan matrices B θ and G . ⋄ Super type: the Weyl groupoid coincides with the one of a finite-dimensional contragre-dient Lie superalgebras in characteristic 0. ⋄ (Super) modular type: related to finite-dimensional contragredient Lie (super)algebrasin positive characteristic. ⋄ UFO type: a list of twelve Weyl equivalence classes of braiding matrices.Eminent pre-Nichols algebras of Cartan type were considered in [ASa]. Here we giveanother step towards the classification of pointed Hopf algebras with finite GKdim andabelian group of group-likes: we answer the question above for some exceptional casesof Cartan type G , braidings of standard type (but not Cartan) and the whole family ofsuper type. Super modular and UFO type will be considered in a forthcoming paper.As in [ASa] we find that the distinguished pre-Nichols algebra is eminent for almostall braidings considered here and determine the eminent pre-Nichols algebras in theseexceptional cases. Coming back to (II), this means that in almost all the cases, post-Nichols algebras with finite GKdim are subalgebras of the Lusztig algebras introduced in[AAR], containing the corresponding Nichols algebras.More explicitly, the main result of this paper can be summarized as follows. Theorem 1.2. (1) The distinguished pre-Nichols algebra e B q is eminent when q is: • of Cartan type G with parameter q ∈ G ′ ∪ G ′ . RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 3 • of super type, except A ( q | J ) , q ∈ G ∞ , J = { } or J = { , , } . • of standard type B θ and G .(2) Suppose that q is of type A ( q |{ } ) , q ∈ G ∞ . Then b B q = T ( V ) / h x , x , x , x i is an eminent pre-Nichols algebra of q , and GKdim b B q = 3 .(3) Suppose that q is of type A ( q |{ , , } ) , q ∈ G ∞ . Then b B q = T ( V ) / h x , x , x , x , [ x , x ] c i is an eminent pre-Nichols algebra of q , and GKdim b B q = 3 . The structure of the paper is the following. In § § G is contained in §
4, where first we determinea minimal presentation when the parameter is a root of unity of order 4 ,
6. The proof forsuper type braidings is given in § b B q contains a central Hopf subalgebra b Z q that fits in an extension of braided Hopf algebras k → b Z q ֒ → b B q ։ e B q → k . Finally, § G type. Notation.
For a positive integer θ we set I θ = { , . . . , θ } ; if θ is understood we shall writesimply I . The canonical basis of Z I is denoted { α i : i ∈ I } .Let N ∈ N . The subgroup of k × of N -th roots of unity is denoted by G N , and G ′ N denotes the subset of those of order N . We also set G ∞ = S N ∈ N G N .We always consider Hopf algebras over k and assume they have bijective antipode. Thesubspace of primitive elements of a (braided) Hopf algebra H is P ( H ). The category of(left) Yetter-Drinfeld modules over H is denoted by HH YD . The subalgebra generated bya subset X ⊆ H is denoted by k h X i .We refer to [R] for any unexplained notion on Hopf algebras and to [KL] for the definitionand basic properties of Gelfand-Kirillov dimension of associative algebras.2. Preliminaries
Nichols algebras.
A braided vector space is a pair (
V, c ), where V is a vector spaceand the braiding c ∈ GL( V ⊗ V ) satisfies the Yang-Baxter equation:( c ⊗ id)(id ⊗ c )( c ⊗ id) = (id ⊗ c )( c ⊗ id)(id ⊗ c ) . (2.1)The tensor algebra T ( V ) is a braided Hopf algebra in the sense of [T]; a pre-Nicholsalgebra of V is a quotient of T ( V ) by a braided graded Hopf ideal contained in ⊕ n > V ⊗ n . ANGIONO, CAMPAGNOLO, AND SANMARCO
Dually, the tensor coalgebra T c ( V ) is a braided Hopf algebra; a post-Nichols algebra of V is a graded subalgebra of T c ( V ) containing V .By the universal properties of T ( V ) and T c ( V ), there exists a (unique) graded Hopfalgebra map Ω : T ( V ) → T c ( V ) such that Ω | V = id V . The Nichols algebra of V isΩ( T ( V )); thus, there exists a braided graded Hopf ideal J ( V ) contained in ⊕ n > V ⊗ n suchthat B ( V ) = T ( V ) / J ( V ). Moreover, J ( V ) is the maximal Hopf ideal among those.Let H be a Hopf algebra and V ∈ HH YD . As HH YD is a braided tensor category, thebraiding c : V ⊗ V → V ⊗ V of HH YD makes ( V, c ) a braided vector space. Hence we mayconsider B ( V ), which is a Hopf algebra in HH YD .2.2. Braidings of diagonal type.
Let q = ( q ij ) ≤ i,j ≤ θ ∈ ( k × ) I × I . Consider the pair( V, c q ), where V has basis ( x i ) i ∈ I and c q ∈ GL( V ⊗ V ) is determined by: c q ( x i ⊗ x j ) = q ij x j ⊗ x i , i, j ∈ I . (2.2)Then c q satisfies (2.1). The braided vector space ( V, c q ) and the associated Nichols algebra B q := B ( V ) are said of diagonal type . The generalized Dynkin diagram of q is a graphwith labelled vertices and edges, defined as follows: • the set of vertices is I ; the vertex i is labelled with q ii . • two vertices i = j ∈ I are connected by an edge if and only if e q ij := q ij q ji = 1. If so, theedge is labelled with e q ij .We also denote by q the Z -bilinear form Z I × Z I → k × associated to the matrix q , thatis q ( α j , α k ) := q jk , j, k ∈ I . If α, β ∈ Z I and i ∈ I , then we set q αβ = q ( α, β ) , q α = q ( α, α ) , N α = ord q α , N i = ord q α i = N α i . (2.3)The braided vector space ( V, c q ) admits a realization over kZ θ kZ θ YD , where • the coaction is given by δ ( x i ) = α i ⊗ x i , i ∈ I ; • the action is defined by β · x i = q ( β, α i ) x i , i ∈ I , β ∈ Z θ .In this context, T ( V ) is a Hopf algebra in kZ θ kZ θ YD , J ( V ) is a Yetter-Drinfeld submoduleof T ( V ) and B q is a Hopf algebra in kZ θ kZ θ YD . In particular, T ( V ) is a Z I -graded braidedHopf algebra, with grading deg x i = α i , i ∈ I , and B q inherits the graduation.Following [Kh], any Z I -graded pre-Nichols algebra B (and particularly B q ) has a PBWbasis with homogeneous PBW generators. This means that there exists a subset ∅ 6 = S ⊂ B of homogeneous elements (the PBW-generators ) provided with a total order < , and afunction h : S → N ∪ {∞} (the height ) such that the following set is a k -basis of B : B ( S, <, h ) := (cid:8) s e . . . s e t t : t ∈ N , s i ∈ S, s > · · · > s t , < e i < h ( s i ) (cid:9) . Adjoint action and braided bracket.
Any braided Hopf algebra R admits a leftadjoint representation ad c : R → End R ,(ad c x ) y = m ( m ⊗ S )(id ⊗ c )(∆ ⊗ id)( x ⊗ y ) , x, y ∈ R. Also, the braided bracket [ · , · ] c : R ⊗ R → R is the map given by[ x, y ] c = m (id − c )( x ⊗ y ) , x, y ∈ R. Notice that (ad c x ) y = [ x, y ] c if x ∈ P ( R ). RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 5
We are interested in Z I -graded pre-Nichols R algebras of V . In this case the braidedcommutator satisfies [ u, vw ] c = [ u, v ] c w + q αβ v [ u, w ] c , (2.4) [ uv, w ] c = q βγ [ u, w ] c v + u [ v, w ] c , (2.5) (cid:2) [ u, v ] c , w (cid:3) c = (cid:2) u, [ v, w ] c (cid:3) c − q αβ v [ u, w ] c + q βγ [ u, w ] c v, (2.6)for all homogeneous elements u ∈ R α , v ∈ R β , w ∈ R γ . Given i , · · · , i k ∈ I , j ≥
2, we set x i ··· i k := (ad c x i ) x i ··· i k = x i x i ··· i k − q i i · · · q i i k x i ··· i k x i . Weyl groupoids and Cartan roots.
Next we briefly recall the notions of Weylgroupoid and generalized root systems [H1, HY]. We assume here that GKdim B q < ∞ .Let C q = ( c q ij ) i,j ∈ I ∈ Z I × I be the generalized Cartan matrix defined by c q ii := 2 and c q ij := − min { n ∈ N : ( n + 1) q ii (1 − q nii q ij q ji ) = 0 } , i = j. (2.7)When q is fixed we simply write ( c ij ). Let i ∈ I . The reflection s q i ∈ GL ( Z I ) is given by s q i ( α j ) := α j − c q ij α i , j ∈ I . (2.8)Next we introduce the matrix ρ i ( q ), given by( ρ i ( q )) jk := q ( s q i ( α j ) , s q i ( α k )) = q jk q − c q ij ik q − c q ik ji q c q ij c q ik ii , j, k ∈ I , (2.9)and ρ i ( V ) is the braided vector space of diagonal type with matrix ρ i ( q ). Finally we set X := { ρ j . . . ρ j n ( q ) : j , . . . , j n ∈ I , n ∈ N } . This set is called the Weyl-equivalence class of q .For each p ∈ X the set ∆ p + of positive roots consists of the Z I -degrees of the generatorsof a PBW-basis of B p , counted with multiplicities. Let ∆ p := ∆ p + ∪ − ∆ p + . The generalizedroot system of q is the fibration ∆ = (∆ p ) p ∈X over X .The reflections s p i , i ∈ I and p ∈ X , generate a subgroupoid of X × GL ( Z θ ) × X calledthe Weyl groupoid W of q . The groupoid W acts on ( C p ) p ∈X and on the generalized rootsystem ( ∆ p ) p ∈X , generalizing the classical Weyl group.Next we assume that ∆ q + is finite. Let ω q ∈ W be an element of maximal length and ω q = σ q i σ i · · · σ i ℓ be a reduced expression. Then β k := s q i · · · s i k − ( α i k ) , k ∈ I ℓ (2.10)are pairwise different; moreover ∆ q + = { β k : k ∈ I ℓ } [CuH], so | ∆ q + | = ℓ . In this setting,there exist PBW generators x β of degree β such that the following set is a basis of B q : { x n β · · · x n ℓ β ℓ : 0 ≤ n k < N β k } . ANGIONO, CAMPAGNOLO, AND SANMARCO
Distinguished and eminent pre-Nichols algebras.
We start with the definitionof Cartan roots [An2]. An element i ∈ I is a Cartan vertex of q if q ij q ji = q c q ij ii , for all j = i. (2.11)The set of Cartan roots of q is the orbit of Cartan vertices under the action of W ; explicitly, O q = { s q i s i . . . s i k ( α i ) ∈ ∆ q : i ∈ I is a Cartan vertex of ρ i k . . . ρ i ρ i ( q ) } . Now we assume that dim B q < ∞ and recall the definition of the distinguished pre-Nichols algebra introduced in [An2, An4]. The presentation of B q given in [An2, Theorem3.1] includes a long list of relations in two, three and four letters x i , and powers of rootvectors x N β β for β ∈ O q + . Let I ( V ) denote the ideal of T ( V ) generated by all the defining re-lations of B q except x N β β , β ∈ O q + , and adding the quantum Serre relations (ad c x i ) − c ij ( x j )when i = j are such that q c ij ii = q ij q ji = q ii . This happens to be a Hopf ideal. The distin-guished pre-Nichols algebra is the braided Hopf algebra e B q := T ( V ) / I ( V ).When q is of Cartan type, e B q is the positive part of a multiparametric version of thede Concini-Procesi quantum group U q ( g ) (under restrictions on the order of q ).We recall some properties of e B q :(A) Set O q + = O q ∩ N θ . Define e N β := N β if β / ∈ O q + , and e N β := ∞ if β ∈ O q + (see (2.3)).Let x β denote the canonical preimage in e B q of the PBW generator of degree β for B q . The set { x n β · · · x n ℓ β ℓ : 0 ≤ n k < e N β k } is a basis of e B q .(B) Let Z q be the subalgebra generated by x N β β , β ∈ O q + . Then Z q is a q -polynomial ringin variables x N β β , β ∈ O q + , which is also a Hopf subalgebra.(C) We have that GKdim e B q = | O q + | .Let q be such that GKdim B q < ∞ . The set of all pre-Nichols algebras of q is a poset Pre ( V ) with T ( V ) minimal and B q maximal. Let Pre
GKd ( V ) be the subposet of Pre ( V )of all finite GK-dimensional pre-Nichols algebras. We say that a pre-Nichols algebra b B is eminent if it is a minimum in Pre
GKd ( V ); that is, for any B ∈ Pre
GKd ( V ),there is anepimorphism of braided Hopf algebras which is id V when we restrict to degree one. Remark 2.1.
Let B be a pre-Nichols algebra of q . Fix J ⊆ I and q ′ = ( q j,k ) j,k ∈ J , V ′ thebraided vector subspace of V with basis x j , j ∈ J . The subalgebra B ′ of B generated by x j , j ∈ J , is a pre-Nichols algebra of q ′ , and GKdim B ′ ≤ GKdim B .In particular, if B ∈ Pre
GKd ( V ), then B ′ ∈ Pre
GKd ( V ′ ). Remark 2.2.
Since e B q ∈ Pre
GKd ( V ), in order to prove that e B q is eminent it sufficesto show that each defining relation of e B q (of the presentation fixed above) holds in any B ∈ Pre
GKd ( V ).2.6. Extensions of graded braided Hopf algebras.
Let H be a Hopf algebra withbijective antipode. Recall that a sequence of morphisms of Hopf algebras in HH YD k → A ι → C π → B → k (2.12)is an extension of braided Hopf algebras (cf. [AN, § RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 7 (a) ι is injective,(b) π is surjective, (c) ker π = Cι ( A + ) and(d) A = C co π , or equivalently A = co π C .For simplicity, we shall write A ι ֒ → C π ։ B instead of (2.12).Assume further that C is connected (i.e. the coradical of C is k ), so for any extension A ι ֒ → C π ։ B , also B and A are connected. In this context we recall the following result: Proposition 2.3. [A+, 3.6]
Let C ∈ HH YD be a connected Hopf algebra. The map { right coideal subalgebras of C } → { quotient left C − module coalgebras } , A C/CA + , is bijective, with inverse B C co B .If A is a right coideal subalgebras of C and B = C/CA + , then there exists a left B -colinear and right A -linear isomorphism B ⊗ A ≃ −→ C . (cid:3) Hence in order to get extensions of a given connected C , it is enough to consider either • a surjective Hopf algebra morphism C π ։ B and set A = C co π , or • a normal Hopf subalgebra A of C , ι the inclusion and set B = CA + .In either case there exists a left B -colinear and right A -linear isomorphism B ⊗ A ≃ C .An example of extensions of connected braided Hopf algebras is k → Z q ֒ → e B q ։ B q → k ,where e B q and Z q are defined as in the previous subsection.It is more suitable to our purposes to consider N θ -graded versions of these objects.For each α = ( a , · · · , a θ ) ∈ Z θ , set t α = t a · · · t a θ θ . Given a N θ -graded object U withfinite-dimensional homogeneous components, the Hilbert (or Poincar´e) series is H U = X α ∈ N θ dim U α t α ∈ N [[ t , . . . , t θ ]] . For example, from the definition of ∆ q and the PBW bases of B q , e B q , we have that H B q = Y α ∈ ∆ q + − t N α α − t α , H e B q = Y α ∈ O q + − t α Y α ∈ ∆ q + − O q + − t N α α − t α . If U ′ is another N θ -graded object, then we say that H U ≤ H U ′ ifdim U α ≤ dim U ′ α for all α ∈ N θ . Next we relate extensions of (braided) connected Hopf algebras and Hilbert series.
Lemma 2.4.
Assume there is a degree-preserving extension A ι ֒ → C π ։ B of N θ -gradedconnected Hopf algebras in HH YD with finite-dimensional homogeneous components. Then H C = H A H B .Proof. Consider the Hopf algebra b H = kZ θ ⊗ H . Then every N θ -graded object U ∈ HH YD is canonically a Yetter-Drinfeld module over b H , where • the action of H on U is extended to an action of b H , where Z θ acts trivially; • for each β ∈ N θ and u ∈ U β , the coaction on u is given by δ ( u ) = β ⊗ u − ⊗ u ∈ N θ ⊗ H ⊗ U ⊂ b H ⊗ U. ANGIONO, CAMPAGNOLO, AND SANMARCO
Since ι and π preserve the degrees, A ι ֒ → C π ։ B is an extension in the category of b H -Yetter-Drinfeld modules and [A+, Proposition 3.6 (d)] applies: there is an b H -colinear,isomorphism B ⊗ A ≃ −→ C , which implies that H C = H A H B . (cid:3) Tool box for pre Nichols algebras of diagonal type.
We collect here some tech-nical results which help to conclude that a given pre-Nichols algebra has infinite GKdim.
Remark 2.5. If W is of diagonal type and has Dynkin diagram q ◦ q − q ◦ where q ∈ k × is a root of unity of order larger than 2, then GKdim B ( W ) = ∞ .Indeed, one can see that this diagram is of Cartan type with non-finite associated Cartanmatrix, so the rank-two result of [AAH2] applies. Lemma 2.6. [AAH1, Proposition 4.16]
Let W be a braided vector space of diagonal typewith diagram ◦ p q ◦ , p = 1 . Then GKdim B ( W ) = ∞ . Next we state a well-known result: for a proof we refer to [ASa, Lemma 2.8].
Lemma 2.7.
Let R be a graded braided Hopf algebra and W a braided subspace of P ( R ) .Then GKdim B ( W ) ≤ GKdim R . (cid:3) As we assume Conjecture 1.1, it is useful to describe the shape of some Dynkin diagramswith finite root systems.
Lemma 2.8. [H2, Lemma 9 (ii)]
Assume q = ( q ij ) i,j ∈ I has connected Dynkin diagramand finite root system. Then e q e q e q = 1 and ( e q + 1)( e q + 1)( e q + 1) = 0 .Moreover, if q = − , then q e q = q e q = 1 . Lemma 2.9. [H2, Lemma 23]
Let q be such that ∆ q is finite. Then the Dynkin diagramof q does not contain cycles of length larger than . Finally we summarize the main result of [ASa] on pre-Nichols algebras of Cartan type:
Theorem 2.10.
Let q be of Cartan type with connected Cartan matrix X and parameter q ∈ G ′ N , N ≥ . If ( X, N ) is different from ( A θ , , ( D θ , , ( A , , ( G , , ( G , , thenthe distinguished pre-Nichols algebra B q is eminent. (cid:3) Defining relations and finite GK dimensional pre-Nichols algebras
Fix θ ≥ q = ( q ij ) i,j ∈ I θ with connected Dynkin diagram such that GKdim B q < ∞ .By Lemma 2.6, q ii = 1 for all i ∈ I θ . Let V = ( V, c q ) be the associated braided vectorspace with basis ( x i ) i ∈ I θ , and B a pre-Nichols algebra of q such that GKdim B < ∞ .We determine sufficient conditions under which some defining relations from the pre-sentation of Nichols algebras in [An2, Theorem 3.1] are annihilated in B .For a given relation x u the strategy to show that x u = 0 in B is the following:(a) We suppose that x u = 0, and either we check that x u ∈ P ( B ) or we assume this fact.(b) By Lemma 2.7, the braiding matrix q ′ of V ⊕ k x u ⊂ P ( B ) satisfies GKdim B q ′ < ∞ .(c) We compute a subdiagram of q ′ and prove that GKdim B q ′ = ∞ using results in § RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 9
We start with the relations x N i i when i is not of Cartan type. After that we take care ofquantum Serre relations (ad c x i ) − c ij x j , which are primitive in T ( V ) by [H3, (4.45)], seealso [AS, Lemma A.1]; first we give sufficient conditions for them to vanish when c ij = 0,and later we do the same when c ij <
0. Finally we consider other relations in [An2,Theorem 3.1] which appear for braidings of super or standard type.3.1.
Powers of non Cartan vertices.Lemma 3.1.
Let i ∈ I θ be a non-Cartan vertex such that q ii ∈ G ′ N . Then x Ni = 0 .Proof. As i is not Cartan, there exists j = i such that e q ij / ∈ { q − nii : n ∈ N } = G N .Note that x Ni is a primitive element of B since q ii ∈ G N . Suppose that x Ni = 0. Then P ( B ) contains k x Ni ⊕ k x j , which has Dynkin diagram ◦ e q Nij q jj ◦ . As e q Nij = 1, we getGKdim B q ′ = ∞ from Lemma 2.7, a contradiction. Hence x Ni = 0. (cid:3) Remark 3.2.
Let i, j ∈ I θ be such that q ii ∈ G ′ N , e q ij / ∈ G N . Then (ad c x i ) N x j = 0.This follows by Lemma 3.1 and [H3, (4.45)].3.2. Quantum Serre relations, c ij = 0 .Lemma 3.3. Let i, j ∈ I θ be such that e q ij = 1 and either q ii = 1 or q jj = 1 . Then x ij = 0 .Proof. Recall that q ii , q jj = 1. Hence the statement follows by [ASa, Proposition 3.2]. (cid:3) Lemma 3.4.
Let i, j ∈ I θ be such that q ii q jj = 1 , e q ij = 1 and there exists ℓ ∈ I θ − { i, j } such that e q iℓ e q jℓ = 1 . Then x ij = 0 .Proof. Suppose that x u := x ij = 0. The subdiagram of q ′ with vertices u and l is q ℓℓ ◦ e q iℓ e q jℓ ◦ , so GKdim B q ′ = ∞ by Lemma 2.6, a contradiction. (cid:3) Quantum Serre relations, c ij < , q − c ij ii = 1 . Here the condition q − c ij ii = 1assures that the relation (ad c x i ) − c ij x j = 0 appears in the essentially minimal presentation[An2, Theorem 3.1]. Following the strategy in [An2, Proposition 4.1], we prove: Lemma 3.5.
Let i, j ∈ I θ be such that c ij < , q − c ij ii = 1 and one of the following hold: (a) q − c ij ii = 1 , or (b) q c ij (1 − c ij ) ii q jj = 1 .Then (ad c x i ) − c ij x j = 0 in B .Proof. We set q = q ii , m = − c ij . Suppose that x u := (ad c x i ) m +1 x j = 0. By definition of m , q m e q ij = 1, hence the Dynkin diagram of k x j ⊕ k x i ⊕ k x u is q m +1 q jj ◦ q jj ◦ q − m q − m ( m +1) q jj ♥♥♥♥♥♥♥♥♥♥♥♥♥ q ◦ . q m +2 PPPPPPPPPPPPP
Next we use the hypothesis to split the proof in three cases. (1) q m +2 ii = 1, q − m ( m +1) ii q jj = 1. By Lemma 2.8,1 = e q ij e q ju e q iu = q − m q m +2 q − m ( m +1) q jj = q − m ( m +1) q jj = q − m ( m +1) q jj . Moreover, at least one of the vertices has label −
1. We study each case: notice that q = − < m < ord( q ii ) − • q jj = −
1. By Lemma 2.8, m = 1 and 1 = ( q m +1 q jj )( q − m ( m +1) q jj ) = ( − q )( q − ) = −
1, a contradiction. • q m +1 q jj = −
1. By Lemma 2.8 we also have that 1 = qq m +2 = q m +3 and1 = ( q jj )( q − m ( m +1) q jj ) = q − m ( m +3)+2 m q jj = q m q jj . Hence − − = ( q m +1 q jj ) = ( q m +3 q − m ) = q m +3 = 1, a contradiction.(2) q m +2 = 1. By hypothesis e q ju = q − m ( m +1) q jj = 1, and we have the diagram q ◦ i q q jj ◦ j q − q jj q − q jj ◦ u . We see that this diagram does not belong to [H2, Table 2] since • q jj = q (otherwise we have a vertex with label 1 connected with another vertex); • the extreme vertices have labels q ii , q uu = − q ii q uu = q jj (thelabel of the middle vertex) and the product of the two edges is e q ij e q ju = q jj ; • if q ii = e q − ij and q jj = −
1, then q ii = 1 and q uu e q ju = − q uu = e q − ju ).(3) q − m ( m +1) q jj = 1; that is, q jj = ± q m ( m +1)2 . By hypothesis, q m +2 = 1. Hence thediagram of q ′ has the form vq ( m +1)( m +2)2 ◦ u q m +2 q ◦ i q − m vq m ( m +1)2 ◦ j , v ∈ {− , } . We check that this diagram does not belong to [H2, Table 2] since • the vertex in the middle has label q ii = − • the product of the two edges is e q ij e q iu = q ii (the square of the vertex in the middle), • the vertex on the left has label q uu = e q ij e q iu q ii .All the possibilities lead to a contradiction, so x u = 0. (cid:3) Next we analyze what happens when q does not fulfill the hypothesis of Lemma 3.5. Lemma 3.6.
Let i, j ∈ I θ be such that • c ij < , • ord q ii = 2 − c ij , • q c ij (1 − c ij ) ii q jj = 1 .Then q ii = q jj = e q − ij ∈ G ′ and c ij = − .Proof. By hypothesis, 1 = q c ij (1 − c ij ) ii q jj = q − c ij ii q jj = q − ii q jj . Hence q ii = ± q jj . Wediscard the case q jj = − q ii using Remark 2.5 since the subdiagram with vertices i, j is q ii ◦ q cijii − q ii ◦ . Therefore q ii = q jj and the Dynkin diagram is q ii ◦ q cijii q ii ◦ . Thus the RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 11 braiding matrix is of Cartan type with Cartan matrix (cid:16) c ij c ij (cid:17) . As GKdim B q < ∞ wehave that c ij = − q ii = q jj = e q − ij ∈ G ′ . (cid:3) Hence the quantum Serre relations hold when c ij < q − c ij ii = 1 and the Dynkindiagram spanned by i, j is not of the form ζ ◦ ζ ζ ◦ , ζ ∈ G ′ . We get now a sufficientcondition for the validity of the quantum Serre relations for this exceptional diagram,assuming the existence of another vertex connected either with i or j . Lemma 3.7.
Let i, j ∈ I θ be such that q ii = q jj = e q − ij ∈ G ′ , and there exists k ∈ I θ − { i, j } such that e q ik e q jk = 1 . Then (ad c x i ) x j = 0 .Proof. Suppose that x u := (ad c x i ) x j = 0. As e q uk = e q ik e q jk = 1 and q uu = 1, thesubdiagram of q ′ with vertices u and k is q kk ◦ e q ik e q jk ◦ . So GKdim B q ′ = ∞ by Lemma2.6, a contradiction. (cid:3) Remark 3.8.
The hypothesis on k ∈ I θ −{ i, j } of Lemma 3.7 is fulfilled for example when:(a) either e q ik = ± e q jk = 1, or (b) e q jk = 1, e q ik = 1.3.4. Quantum Serre relations, c ij < , q − c ij ii = 1 . When q − c ij ii = 1 and i is not aCartan vertex, the relations (ad c x i ) − c ij x j hold in the Nichols algebra but are not minimalrelations since they follow from x − c ij i = 0. Anyway (ad c x i ) − c ij x j is primitive in the tensoralgebra and belongs to the defining ideal of the distinguished pre-Nichols algebra.First we work with conditions depending only on i , j , and later we involve a third vertex. Lemma 3.9.
Let i, j ∈ I θ such that q − c ij ii = 1 and e q ij = q ii . If either q jj = − or c ij ≤ − , then (ad c x i ) − c ij x j = 0 .Proof. Suppose that x u := (ad c x i ) − c ij x j = 0. As e q ju = q jj , e q ui = e q ij and q uu = q − c ij ii q jj = q jj , the Dynkin diagram of W = k x i ⊕ k x j ⊕ k x u is q jj ◦ uq jj ◦ j q ii q jj ♥♥♥♥♥♥♥♥♥ q ii ◦ i . q ii PPPPPPPPP
Assume first that q jj = −
1, so e q ij , e q iu , e q uj = 1. If q ii = −
1, then the three vertices havelabels = −
1, so GKdim B ( W ) = ∞ by Lemma 2.8. If q ii = −
1, then the diagram doesnot belong to [H2, Table 2].It remains to consider the case q jj = −
1. By hypothesis ord q ii ≥ W is − ◦ j q ii q ii ◦ i q ii − ◦ u , which does not belong to [H2, Table 2]. In any case we get a contradiction, so x u = 0. (cid:3) Lemma 3.10.
Let i, j ∈ I θ such that q ii = e q ij = − , and assume there exists k = i, j suchthat e q jk , e q ik e q jk = 1 . Then (ad c x i ) x j = 0 .Proof. Notice that c ij = −
1. Suppose that x u := (ad c x i ) x j = 0. By direct computation, q uk = e q ik e q jk = 1, e q iu = q ii = −
1, so the Dynkin diagram of q ′ contains a 4-cycle, acontradiction with Lemma 2.9. Thus x u = 0. (cid:3) The relation [ x ijk , x j ] c . Next subsections include results about the validity of otherrelations. First we consider the relation [ x ijk , x j ] c , which holds in the Nichols algebra when e q ij = e q − jk , e q ik = 1, q jj = −
1. We begin the subsection stating a formula of the coproductof [ x ijk , x j ] c which becomes important to determine if this element is primitive. Lemma 3.11. If i, j, k ∈ I θ satisfy q jj = − , e q ij = e q − jk , e q ik = 1 , then the followingformula holds in T ( V ) : ∆ (cid:0) [ x ijk , x j ] c (cid:1) = [ x ijk , x j ] c ⊗ ⊗ [ x ijk , x j ] c + (1 − e q ij ) e q ij q kj x i ⊗ x jjk + (1 − e q jk ) q kj [ x ij , x j ] c ⊗ x k − (1 − e q jk ) q ij q kj x j ⊗ x jik + 2(1 − e q jk ) q ij q kj x j ⊗ x ik . Proof.
This is an straightforward computation using the conditions on q . (cid:3) Lemma 3.12.
Let i, j, k ∈ I θ be such that q jj = − , e q ik = 1 and e q ij = e q − jk = ± . Then (a) x jik = 0 . (b) [ x ijk , x j ] c ∈ P ( B ) .Proof. (a) Notice that x jik ∈ P ( T ( V )) by [ASa, Lemma 2.7], because x ik is primitive and e q ( α j , α i + α k ) = 1. Also, q ( α i + α k + α j , α i + α k + α j ) = − q ii q kk , e q ( α i + α k + α j , α t ) = q jt q tj t ∈ { i, j, k } . Let q = e q jk . If x jik = 0 in B , then P ( B ) contains the 4-dimensional subspace k x i ⊕ k x j ⊕ k x k ⊕ k x jik , which has Dynkin diagram − q ii q kk ◦ jikq − ♠♠♠♠♠♠♠♠♠♠♠ q ◗◗◗◗◗◗◗◗◗◗◗ q ii ◦ i q − − ◦ j q q kk ◦ k . Now Lemma 2.9 implies that this Dynkin diagram is not arithmetic. Assuming the validityof Conjecture 1.1, Lemma 2.7 says that GKdim B = ∞ , a contradiction.(b) By Lemma 3.11 and (a), it is enough to prove that x j = x jji = x jjk = [ x ij , x j ] c = 0.For, we observe that j ∈ I θ is not a Cartan vertex since e q ij = ±
1, so x j = 0 by Lemma3.1: this relation implies that x jji = x jjk = [ x ij , x j ] c = 0. (cid:3) Lemma 3.13.
Let i, j, k ∈ I θ be such that q jj = − , e q ik = 1 , and e q ij = e q − jk = ± . Ifeither q ii = − or q kk = − , then [ x ijk , x j ] c = 0 .Proof. By Lemma 3.12, x u := [ x ijk , x j ] c ∈ P ( B ). Suppose that x u = 0. Set q = e q jk = e q − ij .It is enough to consider the case q ii = − q uu = − q kk , e q ui = q − , e q uj = 1 , e q uk = q q kk . RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 13
Hence the Dynkin diagram of the subspace W = k x i ⊕ k x j ⊕ k x k ⊕ k x u ⊂ P ( B ) is − q kk ◦ u − ◦ i q − q − ♥♥♥♥♥♥♥♥♥♥♥♥ − ◦ j q q kk ◦ k . q q kk ◗◗◗◗◗◗◗◗◗◗◗ If q q kk = 1, then the diagram of W contains a 4-cycle, so from Lemma 2.9 (andassuming Conjecture 1.1), we get GKdim B ( W ) = ∞ , a contradiction.If q q kk = 1, that is q kk = ± q − , then the previous Dynkin diagram becomes − q kk ◦ u q − − ◦ i q − − ◦ j q q kk ◦ k . We see that this diagram does not belong to [H2, Table 3] since • the vertices in the middle are − • the labels of the extreme vertices are opposite and both different of − x u = [ x ijk , x j ] c = 0. (cid:3) Lemma 3.14.
Let i, j, k ∈ I θ be such that q jj = − , e q ik = 1 and e q ij = e q − jk = ± . If q ii q kk = 1 and there exists ℓ ∈ I θ such that one of the following conditions holds (a) e q iℓ = 1 = e q jℓ = e q kℓ , (b) e q jℓ = 1 = e q iℓ = e q kℓ , (c) e q kℓ = 1 = e q jℓ = e q iℓ ,then [ x ijk , x j ] c = 0 .Proof. By Lemma 3.12, x u := [ x ijk , x j ] c ∈ P ( B ). Suppose that x u = 0. Set q = e q jk = e q − ij .As q uu = q ii q kk = 1 and e q uℓ = e q iℓ e q jℓ e q kℓ = 1, the subdiagram of q ′ with vertices u and ℓ is ◦ e q uℓ q ℓℓ ◦ . Thus GKdim B q ′ = ∞ by Lemma 2.6 and we get a contradiction. (cid:3) Other relations.
Next we consider other relations listed in [An2, Theorem 3.1] thatappear in the defining ideal of Nichols algebras of super type.
Lemma 3.15.
Let i, j ∈ I θ be such that q ii e q ij ∈ G ′ ∪ G ′ , q jj = − , and either q ii ∈ G ′ or c ij ≤ − . If [ x iij , x ij ] c ∈ P ( B ) , then [ x iij , x ij ] c = 0 .Proof. We follow the proof of [An2, Lemma 4.3 (i)]. Suppose that x u := [ x iij , x ij ] c = 0.Since q uu = q ii e q ij = q ii , e q ui = e q ij q jj = e q ij , e q uj = q ii e q ij = e q − ij , the Dynkin diagram of k x j ⊕ k x i ⊕ k x u is q ii ◦ u − ◦ j e q ij e q ij ♦♦♦♦♦♦♦♦♦♦♦♦ q ii ◦ i . e q − ij ❖❖❖❖❖❖❖❖❖❖❖❖ If q ii ∈ G ′ , then GKdim B q ′ = ∞ by Lemma 2.6, since either e q ij = 1 or e q ij = 1.If c ij ≤ − loc. cit. with some c rs ≤ − e q ij = 1, e q ij = q − ii = −
1. In anycase we get a contradiction so x u = 0. (cid:3) Lemma 3.16.
Let i, j, k ∈ I θ be such that q ii = ± e q ij ∈ G ′ , e q ik = 1 and one of the followingconditions on q jj , e q ij , e q jk hold: (a) q jj = − , e q ij = e q − jk , or (b) q − jj = e q ij = e q jk .If [ x iijk , x ij ] c ∈ P ( B ) , then [ x iijk , x ij ] c = 0 .Proof. Suppose that x u := [ x iijk , x ij ] c = 0. Set ξ = q ii . By direct computation, e q uk = e q jk q kk , e q uj = q jj e q ij e q jk , e q ui = ξ = 1 , q uu = q jj q kk e q jk . Assume first that (a) holds. The Dynkin diagram of k x i ⊕ k x j ⊕ k x k ⊕ k x u is ξ − q kk ◦ uξ rrrrrrrrrrrrrrrr ξ − q kk ▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼ ξ ◦ i ± ξ − ◦ j ± ξ − ξ − q kk ◦ k , so x i , x j , x u determine a triangle: by Lemma 2.8, 1 = q ii e q ij = ± ξ , a contradiction.Assume now that (b) holds. The subdiagram of q ′ spanned by the vertices i, j, k, u is ξq kk ◦ uξ ♠♠♠♠♠♠♠♠♠♠♠♠♠♠ ξ q kk ❘❘❘❘❘❘❘❘❘❘❘❘❘❘ ξ ◦ i ± ξ ± ξ − ◦ j ± ξ q kk ◦ k . If e q uk = ξ q kk = 1, then the diagram above is a 4-cycle, a contradiction to Lemma 2.9. If q kk = ξ − , then q uu = 1 and e q iu = 1, a contradiction with Lemma 2.6. Finally we assume q kk = − ξ − . The diagram above becomes − ◦ u ξ ξ ◦ i ± ξ ± ξ − ◦ j ± ξ − ξ − ◦ k which does not belong to [H2, Table 3] since • there is a unique vertex with label = −
1, and this vertex is an extreme; • q uu = − q ii = e q − ui ∈ G ′ and q kk ∈ G ′ .In any case we get a contradiction, so x u = 0. (cid:3) Lemma 3.17.
Let i, j ∈ I θ be such that q ii = e q ij = q jj = − . (a) The following formula holds in T ( V ) : ∆( x ij ) = x ij ⊗ ⊗ x ij + 2 q ji x i ⊗ x j − q ji [ x i , x ij ] c ⊗ x j − q ji x i ⊗ [ x ij , x j ] c . (b) If there exists k ∈ I θ − { i, j } such that e q ik e q jk = 1 , then x ij = 0 . RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 15
Proof.
For (a) we note that ∆( x ij ) = x ij ⊗ ⊗ x ij + 2 x i ⊗ x j . Then∆( x ij ) = x ij ⊗ ⊗ x ij + 2 q ji x i ⊗ x j + (1 + q ii e q ij q jj ) x ij ⊗ x ij − q ji x i x ij ⊗ x j + 2 x ij x i ⊗ x j + 2 x i ⊗ ⊗ x j x ij − q ji x i ⊗ x ij x j = x ij ⊗ ⊗ x ij + 2 q ji x i ⊗ x j − q ji [ x i , x ij ] c ⊗ x j − q ji x i ⊗ [ x ij , x j ] c . For (b), we suppose that x u := x ij = 0. We check first that x ij ∈ P ( B ). By hypothesis,either e q ik = 1 or e q jk = 1: we may assume that e q jk = 1 so j is a not a Cartan vertex.By Lemma 3.1, x j = 0, and this relation implies that [ x ij , x j ] c = 0. If e q ik = 1, then thesame result implies that x i = 0, thus [ x i , x ij ] c = 0. Otherwise, e q ik e q jk = e q jk = 1 and thenLemma 3.10 says that [ x i , x ij ] c = 0. Hence x ij ∈ P ( B ).By direct computation, q uu = 1, e q uk = e q ik e q jk = 1. Then the Dynkin diagram of k x k + k x u is q kk ◦ e q ik e q jk ◦ , and by Lemma 2.6 we get a contradiction. Thus x ij = 0. (cid:3) Lemma 3.18.
Let i, j, k ∈ I θ be such that q jj = e q − ij = e q jk ∈ G ′ , e q ik = 1 , and either q ii = − or else q kk = − . Then [[ x ijk , x j ] c , x j ] c = 0 .Proof. First we prove that x u = [[ x ijk , x j ] c , x j ] c ∈ P ( B ). We have that: • x ik = 0 in B applying Lemma 3.3; • x jji = 0 in B applying Lemma 3.5 if q ii = q jj , and Lemma 3.7 otherwise; and • x jjjk = 0 in B by Lemma 3.9 (here − c jk = 2).By direct computation, x u = [[ x ijk , x j ] c , x j ] c = x i x j x k x j − q jk x i x k x j − q ij q ik x j x k x i x j + q jk q ij q ik x k x j x i x j − q ij q jj q kj (1 + q jj ) x j x i x j x k x j + q ij q jj e q jk (1 + q jj ) x j x i x k x j + q ij q jj q kj q ik (1 + q jj ) x j x k x i x j − q ij q ik q jj e q jk (1 + q jj ) x j x k x j x i x j + q ij q kj x j x i x j x k − q jk q ij q kj x j x i x k x j − q ik q ij q kj x j x k x i + q jk q ik q ij q kj x j x k x j x i . Using explicit formulas for the comultiplication of each summand we get∆( x u ) = x u ⊗ ⊗ x u + 3 q ij q kj (1 − e q jk ) x j ⊗ x ik + q kj (1 − e q jk )( x i x j − q ij x j x i ) ⊗ x k + ( e q ij − q kj x i ⊗ ( x j x k − q jk x k x j ) . Notice that x jjji = (ad c x j ) x jji = 0, so x i x j = q ij x j x i . Similarly, x j x k = q jk x k x j . Usingthese relations and x ik = 0, we get that x u ∈ P ( B ).Suppose that x u = 0. We have that e q uk = q kk , e q uj = 1 , e q ui = q ii , q uu = q ii q kk . Hence the subdiagram of q ′ spanned by the vertices i, j, k, u is q ii q kk ◦ uq ii ♦♦♦♦♦♦♦♦♦♦♦♦ q kk ❖❖❖❖❖❖❖❖❖❖❖❖ q ii ◦ i q − jj q jj ◦ j q − jj q kk ◦ k • If q kk = −
1, then q ii = − i , u is as inRemark 2.5, then GKdim B q ′ = ∞ . • If q ii = −
1, then q kk = − B q ′ = ∞ by Remark 2.5. • If q ii , q kk = −
1, the subdiagram with vertices i, j, k, u is a 4-cycle, so GKdim B q ′ = ∞ by Lemma 2.9.In any case we get a contradiction, thus x u = 0. (cid:3) Lemma 3.19.
Let i, j, k ∈ I θ be such that e q ik = 1 , q jj = q ii = − , e q ij = e q − jk = 1 , andeither q kk = − or e q ij = 1 . If [[ x ij , x ijk ] c , x j ] c ∈ P ( B ) , then [[ x ij , x ijk ] c , x j ] c = 0 .Proof. Suppose that x u := [[ x ij , x ijk ] c , x j ] c = 0. We have that e q uk = e q jk q kk , e q uj = 1 , e q ui = e q ij , q uu = − q kk . Then the Dynkin diagram of q ′ contains that of k x i ⊕ k x j ⊕ k x k ⊕ k x u , which is − q kk ◦ u e q ij ♣♣♣♣♣♣♣♣♣♣♣♣ e q ij q kk ❖❖❖❖❖❖❖❖❖❖❖❖ − ◦ i e q ij − ◦ j e q − ij q kk ◦ k . • If q kk = −
1, then e q ij = 1 by hypothesis and the subdiagram with vertices i, u is as inLemma 2.6, thus GKdim B q ′ = ∞ . • If e q ij = 1, then q kk = 1 and the subdiagram with vertices i, u is as in Remark 2.5, henceGKdim B q ′ = ∞ . • If q kk , e q ij = 1, then the Dynkin diagram above is a 4-cycle, so GKdim B q ′ = ∞ byLemma 2.9.In any case we get a contradiction, thus x u = 0. (cid:3) Lemma 3.20.
Let i, j, k ∈ I θ be such that q ii = e q ij = − , q jj = e q − jk = − , e q ik = 1 andeither q kk = q jj or else q jj / ∈ G ′ . Then [ x ij , x ijk ] c = 0 .Proof. Suppose that x u := [ x ij , x ijk ] c = 0. First we prove that x u ∈ P ( B ). It is enoughto prove that all defining relations of B q of degree lower that x u hold in B : • If i is not a Cartan vertex, then x i = 0 by Lemma 3.1. • If i is a Cartan vertex, then x iij = 0 in B by Lemma 3.10: indeed, e q jk = e q ik e q jk = 1. • By hypothesis, q kk = q jj or q jj / ∈ G ′ , therefore x jjk = 0 in B by Lemma 3.5. • If q kk = −
1, then x ik = 0 by Lemma 3.3. If q kk = −
1, then x ik = 0 by Lemma 3.4 since e q ij e q jk = − e q jk = 1.By direct computation, e q uk = q − jj q kk , e q uj = q jj = 1 , e q ui = 1 , q uu = q jj q kk . RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 17
Hence the subdiagram of q ′ spanned by the vertices i, j, k, u is q jj q kk ◦ uq jj q − jj q kk ▲▲▲▲▲▲▲▲▲▲▲▲▲▲▲▲ − ◦ i − q jj ◦ j q − jj q kk ◦ k . If q kk = − q jj , then the subdiagram with vertices j, k does not belong in [H2, Table 1], so q kk = − q jj . We split the study in the following three cases: • q jj = 1, q kk = q jj : the subdiagram with vertices j, k, u is a 3-cycle. By Lemma 2.8,1 = e q jk e q uk e q uj = q kk , so q kk = −
1, and 1 = e q uk q uu = −
1, a contradiction. • q jj = 1, q kk = q jj . The subdiagram spanned by the vertices i, j, k is − ◦ i − q jj ◦ j q − jj q kk ◦ k . Looking at [H2, Table 2], the unique possibility is q kk = −
1. This says that the Dynkindiagram of k x k + k x u is − ◦ q − jj − q jj ◦ , which does not belong at [H2, Table 1]. • q jj = 1, q kk = q jj : that is, q kk = q jj . The subdiagram spanned by i, j, k, u is: q jj ◦ uq jj − ◦ i − q jj ◦ j q − jj q jj ◦ k . As q jj / ∈ G ∪ G , this diagram does not belong to [H2, Table 3].In any case we get a contradiction, thus x u = 0. (cid:3) Lemma 3.21.
Let i, j, k, ℓ ∈ I θ be such that q kk = − , q jj e q ij = q jj e q jk = 1 , e q ik = e q il = e q jℓ =1 and e q jk = e q − kℓ = q ℓℓ . If [[[ x ijkℓ , x k ] c , x j ] c , x k ] c ∈ P ( B ) , then [[[ x ijkℓ , x k ] c , x j ] c , x k ] c = 0 .Proof. Suppose that x u := [[[ x ijkℓ , x k ] c , x j ] c , x k ] c = 0. We set q = e q ij and p = q ii . We have q uu = − p, e q uℓ = q − = 1 , e q uk = 1 , e q uj = q , e q ui = p q , Hence the Dynkin diagram of q ′ contains the following one: − p ◦ uq ✂✂✂✂✂✂✂✂✂ p q ❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧❧ q − ❘❘❘❘❘❘❘❘❘❘❘❘❘❘❘❘❘❘❘❘❘❘❘❘❘ p ◦ i q q − ◦ j q − ◦ k q − q ◦ ℓ . If either q / ∈ G ′ or p = ± q − , then GKdim B = ∞ by Lemma 2.9. Next we assume that q ∈ G ′ , p = ± q − . Then the diagram with vertices j, k, ℓ, u is q − ◦ j q − ◦ k − − ◦ l − ± q ◦ u . This diagram does not belong to [H2, Table 3] since: • the labels of the vertices of degree one are roots of order 4; • each vertex of degree two and the edge between them are labelled with − • the label of one of extreme vertices is not the inverse of the adjacent edge.As we are assuming Conjecture 1.1 we get x u = 0. (cid:3) Lemma 3.22.
Let i, j, k ∈ I θ be such that e q ij , e q ik , e q jk = 1 , e q ij e q ik e q jk = 1 . Then x ijk = q ij (1 − e q jk ) x j x ik − − e q jk q kj (1 − e q ik ) [ x ik , x j ] c . Proof.
Suppose that x u := x ijk − q ij (1 − q ) x j x ik + q jk (1 + q − )[ x ik , x j ] c = 0. By directcomputation, x u ∈ P ( B ). The subdiagram of q ′ with vertices i, j, k, u is q ii ◦ e q ij sssssssssssssssssssssssssssssss q ii e q − jk e q ik ❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑❑ q ii q jj q kk ◦ q jj e q − ik ❣❣❣❣❣❣❣❣❣❣❣❣❣❣❣❣❣❣❣❣❣❣❣❣❣ q kk e q − ij ❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳ q jj ◦ e q jk q kk ◦ . By Lemma 2.8 there exists ℓ ∈ { i, j, k } such that q ℓℓ = −
1. If there exist ℓ = ℓ ∈ { i, j, k } such that q ℓ ℓ = q ℓ ℓ = −
1, then e q uℓ , e q uℓ = 1. Thus the previous diagram contains a4-cycle, so GKdim B = ∞ by Lemma 2.9.Next we assume that there exists a unique ℓ ∈ { i, j, k } such that q ℓℓ = −
1; relabelingthe vertices, we can assume that q ii = − = q jj , q kk , and also q jj e q − ik = 1 = q kk e q − ij (otherwise the diagram still contains a 4-cycle since q ui = e q − jk = 1). By Lemma 2.8, q jj e q ij = 1 = q kk e q ik . Set q = e q jk . Notice that1 = q jj e q − ik = e q − ij e q − ik = e q − ij q, q kk e q − ij = e q − ik e q − ij = e q − ik q, so e q ij = e q ik = q , and therefore q ∈ G ′ . Now the previous diagram becomes q − ◦ q − q − ◦ q − − ◦ q ✐✐✐✐✐✐✐✐✐✐✐ q ❯❯❯❯❯❯❯❯❯❯❯ q − ◦ , which does not belong to [H2, Table 3]. This contradiction shows that x u = 0. (cid:3) RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 19
Lemma 3.23.
Let i, j, k, ℓ ∈ I θ and q ∈ k − ( G ∪ G ) be such that q ℓℓ = e q − kℓ = q kk = e q − jk = q , q jj = − , q ii = e q − ij = q − , e q ik = e q iℓ = e q jℓ = 1 . If [[[ x ijk , x j ] c , [ x ijkℓ , x j ] c ] c , x jk ] c ∈ P ( B ) , then [[[ x ijk , x j ] c , [ x ijkℓ , x j ] c ] c , x jk ] c = 0 .Proof. Suppose that x u := [[[ x ijk , x j ] c , [ x ijkℓ , x j ] c ] c , x jk ] c = 0. By direct computation, e q ui = q ii e q ij = q − = 1 , e q uℓ = q ℓℓ e q kℓ = q − = 1 . Then the subdiagram with vertices i, j, k, ℓ, u contains a 5-cycle. Therefore GKdim B = ∞ by Lemma 2.9, a contradiction. (cid:3) Lemma 3.24.
Let i, j, k, ℓ ∈ I θ and q ∈ k − ( G ∪ G ) be such that q ii = e q − ij = q , q kk = − , q − ℓℓ = e q kℓ = q , e q − jk = q jj = q, e q ik = e q iℓ = e q jℓ = 1 . If [[ x ijkℓ , x j ] c , x k ] c − q jk ( q − q )[[ x ijkℓ , x k ] c , x j ] c ∈ P ( B ) , then [[ x ijkℓ , x j ] c , x k ] c − q jk ( q − q )[[ x ijkℓ , x k ] c , x j ] c = 0 . Proof.
Suppose that x u := [[ x ijkℓ , x j ] c , x k ] c − q jk ( q − q )[[ x ijkℓ , x k ] c , x j ] c = 0. We have that e q ui = e q uj = e q uℓ = 1 and e q uk = q uu = q . Then the subdiagram with vertices i, j, k, ℓ, u is q ◦ uq ◦ i q − q ◦ j q − − ◦ k q q q − ◦ ℓ . If q / ∈ G ′ , then this diagram has a unique vertex labelled with − q ∈ G ′ , then − c uk = 5. In any case case this diagram does not belong to [H2, Table 3]and we get a contradiction. Thus x u = 0. (cid:3) Lemma 3.25.
Let i, j, k ∈ I θ such that q jj = e q ij = e q jk ∈ G ′ and e q ik = 1 . If either q ii = − or else q kk = − , then [[[ x ijk , x j ] c , x j ] c , x j ] c = 0 .Proof. First we check that x u := [[[ x ijk , x j ] c , x j ] c , x j ] c ∈ P ( B ). Using previous results, • x ik = 0 in B by Lemma 3.3. • x jji = 0 in B : this follows either by Lemma 3.5 if q ii = q jj or by Lemma 3.7 if q ii = q jj . • x jjjjk = 0 in B applying Lemma 3.9: here − c jk = 3.Let q := q jj ∈ G ′ , z = [ x ijk , x j ] c , y = [[ x ijk , x j ] c , x j ] c . Using the relations x ik = 0 and x ij x j = qq ij x j x ij , we compute recursively the following coproducts:∆( x jk ) = x jk ⊗ − e q jk ) x j ⊗ x k + 1 ⊗ x jk , ∆( x ijk ) = x ijk ⊗ − q ) x ij ⊗ x k + (1 − q − ) x i ⊗ x jk + 1 ⊗ x ijk , ∆( z ) = z ⊗ ⊗ z + 2 x ijk ⊗ x j + ( q − q kj x ij ⊗ x jk + (1 − q ) x ij ⊗ (2 x k x j − q kj qx jk ) + (1 + q ) x i ⊗ (2 x jk x j − q kj x jjk ) , ∆( y ) = y ⊗ ⊗ y + 2 qz ⊗ x j + 2(1 + q ) x ijk ⊗ x j + q kj x ij ⊗ (cid:0) ( q − q kj x jjk − qx jk x j − qq jk x k x j (cid:1) + (1 + q ) x i ⊗ (2(1 + q ) x jk x j − qq kj x jjk x j + qq kj x jjjk ) , ∆( x u ) = x u ⊗ ⊗ x u + ( q − q kj x i ⊗ x jjjjk . Since x jjjjk = 0 in B , it follows that x u ∈ P ( B ). Suppose that x u = 0. As e q ui = q ii , e q uk = q kk , e q uj = 1, q uu = q ii q kk , the Dynkin subdiagram of q ′ with vertices i, j, k, u is q ii q kk ◦ uq jj ♦♦♦♦♦♦♦♦♦♦♦♦ q kk PPPPPPPPPPPP q ii ◦ i q − q ◦ j q − q kk ◦ k . If q ii = − = q kk , then the diagram above is a 4-cycle so GKdim B = ∞ by Lemma 2.9.Otherwise, either q ii = − q kk = −
1, and the other vertex ℓ has label q ℓℓ = −
1. Thesubdiagram corresponding to ℓ, u is q ℓℓ ◦ q ℓℓ − q ℓℓ ◦ , so GKdim B = ∞ by Remark 2.5. Inboth cases we get a contradiction, so x u = 0. (cid:3) Lemma 3.26.
Let i, j, k ∈ I θ be such that q ii = q jj = − , e q ik = 1 , e q ij = q , q − kk = q − = e q jk for some q ∈ k − ( G ∪ G ) . If x u := [[ x ij , [ x ij , x ijk ] c ] c , x j ] c ∈ P ( B ) , then x u = 0 .Proof. Suppose on the contrary that x u = 0. As e q ui = q , e q uj = 1, e q uk = q − , q uu = − q ,the Dynkin subdiagram with vertices i, j, k, u is − q ◦ uq qqqqqqqqqqqq q − ▼▼▼▼▼▼▼▼▼▼▼▼ − ◦ i q − ◦ j q − q ◦ k . If q ∈ G ′ , then the subdiagram with vertices i, u is − ◦ q ◦ , so GKdim B = ∞ byLemma 2.6 . Otherwise the subdiagram with vertices k, u is q ◦ q − − q ◦ where q = ± B = ∞ . In any case we get a contradiction, so x u = 0. (cid:3) Lemma 3.27.
Let i, j, k ∈ I θ be such that q ii = − q − , e q ik = 1 , q jj = − , e q ij = q , e q jk = q − = q − kk , for some q ∈ k − ( G ∪ G ) .If x u := [ x i , [ x ijk , x j ] c ] c − q ij q kj q [ x ij , x ijk ] c + ( q − − q − ) q ij q ik x ijk x ij ∈ P ( B ) , then x u = 0 .Proof. Suppose on the contrary that x u = 0. As e q ui = e q uk = 1, e q uj = q uu = q , thesubdiagram with vertices i, j, k, u is q ◦ uq − q − ◦ i q − ◦ j q − q ◦ k . As either − c uj = 5 if q ∈ G ′ , or the unique vertex with label − (cid:3) RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 21 Eminent pre-Nichols algebras of Cartan type G Consider a braiding q of Cartan type G , so the Dynkin diagram is q ◦ q − q ◦ , where q is a root of unity of order N >
3. If N = 4 ,
6, then the distinguished pre-Nichols algebraof q is eminent by [ASa, Lemma 4.13]. Here we extend the result for N = 4 ,
6. In order todo that, we give first a minimal presentation of these Nichols algebras.4.1.
Minimal presentation of B q when N = 4 or . Let x := [ x , x ] c . By[An2, Theorem 3.1], the algebra B q is presented by generators x , x and relations x , x , x N α α , α ∈ ∆ + , [ x , x ] c , [ x , x ] c , [ x , x ] c , [ x , x ] c − q q − qq + 1 x . (4.1)Compare this presentation with [An1, Theorem 5.25] and [AA]. Lemma 4.1.
Assume that q is of Cartan type G and N = 4 . Then B q is minimallypresented by generators x , x and relations x , [ x , x ] c , x N α α , α ∈ ∆ + . Proof.
Let B be the algebra generated by x , x subject to these relations. Since q ∈ G ′ ,we have x = 0, hence x = 0 in B . We verify with GAP that relations [ x , x ] c ,[ x , x ] c − q ( q − q )(2) − q x and [ x , x ] c also vanish in B . Hence B = B q .By a degree argument, to verify the minimality of this presentation it is enough to provethat [ x , x ] c does not vanish in the algebra presented by the relations x , x , x ,which is checked using GAP . (cid:3) Lemma 4.2.
Assume that q is of Cartan type G and N = 6 . Then B q is minimallypresented by generators x , x and relations x , x , [ x , x ] c , x N α α , α ∈ ∆ + . Proof.
We argue as in Lemma 4.1. (cid:3)
Eminent pre-Nichols algebras of Cartan type G . The goal now is to provethat the distinguished pre-Nichols algebra is eminent. By [ASa, Lemma 4.13] we alreadyknow that relations x = 0 and x = 0 hold in any finite GKdim pre-Nichols algebra. Lemma 4.3.
Let B be a pre-Nichols algebra of q . The following hold: (a) If x = 0 = x hold in B , then the following relations also hold: [ x , x ] c = 0(4.2) [ x , x ] c = q q ( q − x (4.3) [ x , x ] c = q q ( q − q − x + q q ( q − x x (4.4) [ x , x ] c = [ x , x ] c = q ( q − − q − x (4.5) [ x , x ] c = q q (3) q x x (4.6) Proof.
Follow from rutinary computations. (cid:3)
Assume now N = 4. By Lemma 4.1, the distinguished pre-Nichols algebra of q is e B q = T ( V ) / h x , x , [ x , x ] c i ; Lemma 4.4.
Assume N = 4 and let B be a pre-Nichols algebra of q . The following hold: (a) If x = 0 = x hold in B , then [ x , x ] c is primitive in B . (b) If GKdim( B ) < ∞ , then [ x , x ] c = 0 hold in B . Hence e B q is eminent.Proof. (a) Using [H3, Lemma 4.23] and that the coproduct is braided multiplicative,∆([ x , x ] c ) = x x ⊗ X u =0 (cid:18) u (cid:19) q u Y t =1 (1 − q − − t ) x x u ⊗ (ad c x ) − u x + X r =0 (cid:18) r (cid:19) q r Y s =1 (1 − q − s ) q − r )11 q − r q q x r x ⊗ (ad c x ) − r x + X ≤ r ≤ ≤ u ≤ (cid:18) r (cid:19) q (cid:18) u (cid:19) q Y ≤ s ≤ r ≤ t ≤ u ( q − u − q − s )( q − q − − t ) q u x r + u ⊗ (ad c x ) − r x (ad c x ) − u x − q q x x ⊗ − q q
12 3 X r =0 (cid:18) r (cid:19) q r Y s =1 (1 − q − s ) x x r ⊗ (ad c x ) − r x − q q
12 2 X u =0 (cid:18) u (cid:19) q u Y t =1 (1 − q − − t ) q − u )11 q − u q q x u x ⊗ (ad c x ) − u x − X ≤ r ≤ ≤ u ≤ (cid:18) r (cid:19) q (cid:18) u (cid:19) q Y ≤ s ≤ r ≤ t ≤ u ( q − q − s )( q − r − q − − t ) q r − x r + u ⊗ (ad c x ) − u x (ad c x ) − r x . Notice that the terms on the homogeneous components B (5 , ⊗ k and k ⊗ B (5 , are[ x , x ] c ⊗ ⊗ [ x , x ] c , respectively. Next we show that the other homoge-neous components vanish.Component in B (3 , ⊗ B (2 , :( x + (1 − q − ) q ( x x − q q x x ) − q e q x ) ⊗ x = 0 . Component in B (4 , ⊗ B (1 , :((2) q (1 − q − ) x x + (3) q (1 − q − )(1 − q − ) q q x x − q q (3) q (1 − q − )(1 − q − ) x x − q q (2) q (1 − q − ) q q x x ) ⊗ x =(2) q (1 − q − )( x x x − q q x x + q q x x − q x x − q q x x + q x x x ) ⊗ x = 0 . Component in B (5 , ⊗ B (0 , :(1 − q − )(1 − q − )( x x − q q x x + (1 − q − ) q q [ x , x ] c ) ⊗ x =(1 − q − )(1 − q − )(1 − q + (1 − q − ) q ) x x ⊗ x = 0 , where we are using that [ x , x ] c = 0 and (4.6). RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 23
Component in B (2 , ⊗ B (3 , : ( q q q q − q q ) x ⊗ x = 0.Component in B (1 , ⊗ B (4 , : using (4.5),(1 − q − ) x ⊗ ((1 − q − )(3) q x + (2) q q q [ x , x ] c )=(1 − q − ) x ⊗ ((3) q (1 − q − ) + (2) q q ) x = 0 . Component in B (2 , ⊗ B (3 , : by definition of x and (4.4) we get,(1 − q − ) qx ⊗ ( q (2) q q x + (2) q ( x x − (2) q q x x + (1 − q ) qq [ x , x ] c )=(1 − q − ) qx ⊗ (( q q (2) q + (1 − q ) q q ( q − q − x + ((2) q + q (1 − q )( q − x x − (2) q q x x )=(1 − q − ) q (2) q x ⊗ ( q x − qx x − q x x ) = 0 . Component in B (3 , ⊗ B (2 , : using (4.3),(1 − q − )(1 − q − )(1 − q ) x ⊗ ( − q (1 + q ) x x + q (1 + q ) x x )+ (1 − q − )(1 − q − ) (2) q qx ⊗ ( q q x − q q x ) = 0 . Component in B (4 , ⊗ B (1 , : by (4.2),(1 − q − )(1 − q − ) (1 − q − ) x ⊗ ((2) q q x x − (2) q q q x x + (3) q q q x x − (3) q q x x )= − (1 − q − )(1 − q − ) (1 − q − ) q x ⊗ [ x , x ] c = 0 . Component in B (5 , ⊗ B (0 , : (1 − q − )(1 − q − ) (1 − q − ) x ⊗ ( q − q q q ) x = 0.(b) Assume [ x , x ] c = 0. By (a) we get k x ⊕ k x ⊕ k [ x , x ] c ⊂ P ( B ), wherethe braiding has Dynkin diagram q ◦ q − q ◦ q − q ◦ , which is of affine Cartan type D (3)4 . By [AAH2, Theorem 1.2 (a)] the Nichols algebra withthis braiding has infinite GKdim, and by [ASa, Lemma 2.7] it follows GKdim B = ∞ . (cid:3) Assume now N = 6. By Lemma 4.1, the distinguished pre-Nichols algebra of q is e B q = T ( V ) / h x , x , [ x , x ] c i ; Lemma 4.5.
Assume N = 6 and let B be a pre-Nichols algebra of q . The following hold: (a) If x = 0 = x hold in B , then [ x , x ] c is primitive in B . (b) If GKdim( B ) < ∞ , then [ x , x ] c = 0 hold in B . Hence e B q is eminent.Proof. (a) We compute ∆( x ) in the tensor algebra and get∆( x ) = x ⊗ ⊗ x + 2 q x ⊗ x + 2 q q x ⊗ [ x , x ] c + 2[ x , x ] c ⊗ x + 2(1 − q − ) q q [ x , x ] c ⊗ x ++ 4(1 − q − )(1 − q − ) q x ⊗ x + 2( q − x ⊗ [ x , x ] c . Using this and ∆( x ) = x ⊗ ⊗ x + 2 x ⊗ x , we compute ∆([ x , x ] c )in B . The components of degrees B (4 , ⊗ k and k ⊗ B (4 , are [ x , x ] c ⊗ ⊗ [ x , x ] c , respectively. The other homogeneous components vanish, most of themdue to commutations between powers of x and powers of x , which follow from (4.2).(b) If [ x , x ] c = 0 in B , then by (a) we get k x ⊕ k x ⊕ k [ x , x ] c ⊂ P ( B ),where the braiding has Dynkin diagram q ◦ q − q ◦ q − q ◦ , which is of indefinite Cartan type. By Lemma 2.7 it follows GKdim B = ∞ . (cid:3) Theorem 4.6. If q is of Cartan type G , then the distinguished pre-Nichols algebra e B q iseminent.Proof. If N = 4 , N = 4 , (cid:3) Eminent pre-Nichols algebras of super type
In this Section we describe case-by-case all eminent pre-Nichols of braiding of supertype. We prove that the associated distinguished pre-Nichols algebra is eminent exceptfor two exceptional cases in type A; for these exceptions we construct eminent pre-Nicholsalgebras and show that they are central extension of the distinguished one.5.1.
Type A.
Here q is a braiding of type A θ ( q | J ), where q ∈ k × is a root of unity oforder N > ∅ 6 = J ⊆ I θ [AA, § q is q ◦ e q q ◦ q θ − θ − ◦ e q θ − θ q θθ ◦ where q ii , e q ij satisfy the following conditions:(A) q θθ e q θ − θ = q ;(B) if i ∈ J , then q ii = − e q i − i = e q − ii +1 ;(C) if i / ∈ J , then e q i − i = q − ii = e q i i +1 (as long as i ± ∈ I θ ).Given q and J , these conditions determine the diagram. One can deduce that: • q ii = q ± = ± i ∈ I − J , and e q ii +1 = q ± = ± i < θ ; • if i ∈ J but i − , i + 1 / ∈ J , then q i − i − q i +1 i +1 = 1.These two remarks will allow us to apply Lemmas 3.13 and 3.14.The distinguished pre-Nichols algebra is presented by generators ( x i ) i ∈ I and relations x ij = 0 , i < j − x ii ( i ± = 0 , q ii = − x i = 0 , q ii = −
1; [ x ( i − i +1) , x i ] c = 0 , q ii = − . (5.1)Let B be a pre-Nichols algebra of q such that GKdim B < ∞ . Next we prove that eachrelation in (5.1) must hold in B , except for two pairs ( θ, J ). Lemma 5.1.
Assume that q is not of type A ( q |{ , , } ) . If i, j ∈ I θ , i < j − , then x ij = 0 in B . RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 25
Proof.
Suppose on the contrary that x ij = 0 in B . By Lemma 3.3, q ii = q jj = −
1, thatis i, j ∈ J . Also, i = 1 by Lemma 3.4: otherwise i − ∈ I and e q ii − e q ji − = e q ii − = 1.Analogously j = θ , and i = j −
2: if i < j −
2, then e q ij − e q jj − = e q jj − = 1.Hence i = 1, j = 3 = θ and i, j ∈ J . Thus 2 / ∈ J since q is not of type A ( q |{ , , } ).But then e q e q = q = 1, a contradiction with Lemma 3.4. Therefore x ij = 0. (cid:3) Lemma 5.2. If i ∈ I θ − J , then x ii ( i ± = 0 in B .Proof. Let i ∈ I θ − be such that i J , and set j = i + 1. Suppose that x iij = 0. ByLemmas 3.5 and 3.6, q ii = q jj = e q − ij ∈ G ′ . If i >
1, then e q ( i − i = ± e q ( i − j = 1, and weget a contradiction with Lemma 3.7. If j < θ , then e q j ( j +1) = ± e q i ( j +1) = 1, and again weget a contradiction with Lemma 3.7. Otherwise i = 1, θ = 2 and J = ∅ , a contradiction.Therefore x iij = 0. The proof of x ii ( i − = 0 for i ∈ I ,θ − J is analogous. (cid:3) Lemma 5.3. If q is not of type A ( q |{ } ) and i ∈ I ,θ − ∩ J , then [ x ( i − ,i +1) , x i ] c = 0 .Proof. Let i ∈ I ,θ − ∩ J . If either i − / ∈ J or i + 1 / ∈ J , then [ x ( i − ,i +1) , x i ] c = 0 by Lemma3.13. If 2 < i (respectively i + 1 < θ ), then ℓ = i − ℓ = i + 2) satisfies(a) (respectively (b)) of Lemma 3.14, hence [ x ( i − ,i +1) , x i ] c = 0. Otherwise i = 2, θ = 3, J = { } , so q is of type A ( q |{ } ). (cid:3) Theorem 5.4.
Let q be of type A θ ( q | J ) , where the pair ( θ, J ) is not one of the following: (I) θ = 3 , J = { } . (II) θ = 3 , J = { , , } .Then the distinguished pre-Nichols algebra e B q is eminent.Proof. By Lemmas 3.1, 5.1, 5.2 and 5.3 the defining relations of e B q hold in any finiteGKdim pre-Nichols algebra B of q , hence e B q projects onto B . As GKdim e B q < ∞ , itfollows that e B q is eminent. (cid:3) Next we find eminent pre-Nichols algebras for braidings of type (I) and (II).5.1.1.
Type A ( q |{ } ) . In this case the distinguished pre-Nichols algebra is e B q = T ( V ) / h x , x , x , x , [ x , x ] c i . We consider the following algebra: b B q = T ( V ) / h x , x , x , x i , (5.2)which is a braided Hopf algebra since x , x , x , and x are primitive in T ( V ). Alsothe projections from T ( V ) induce a surjective map of braided Hopf algebras π : b B q ։ e B q .Next we prove that the pre-Nichols algebra b B q is eminent. Proposition 5.5. If q of type A ( q |{ } ) , then b B q as in (5.2) an eminent pre-Nicholsalgebra of q .Let x u = [ x , x ] c . The set B = (cid:8) x a x b x c x du x e x f x g : b, c, e, f ∈ { , } , a, d, g ∈ N (cid:9) (5.3) is a basis of b B q , so GKdim b B q = 3 . Proof.
We split the proof in steps. Let B be a finite GKdim pre-Nichols algebra of q . Step 1.
The projection T ( V ) ։ B induces a projection b B q ։ B of braided Hopf algebras.Indeed the defining relations of b B q annihilate in B by Lemmas 3.1, 3.3 and 3.5.To see that b B q is eminent, it remains to prove that GKdim b B q < ∞ . For, we will checkthat B is a basis of b B q and that the later is a braided central extension of e B q . Step 2.
The following relations hold in b B q : x = 0 , x = 0 , x = 0 . (5.4)Using the relations x = 0 and x = 0 we compute(1 + q ) x = (1 + q )( x x x x − q x x x + q x x x x )= q q x x x − q (1 + q ) x x x + q x x x = 0As q = −
1, we get x = 0. Analogously x = 0.Using (2.6) and the defining relations we check that the following also hold: x = x x − q q x x , (5.5) x x = q q x x , x x = q x x , (5.6) x x = x ( x x − q q x x ) = q q q x x (5.7)Using now that ad c x is a skew-derivation and that (ad c x ) x = 0 by (5.7),0 = (ad c x ) ( x ) = (ad c x )( x x + q q x x ) = (1 + q ) q q x . As q = −
1, we have that x = 0. Step 3. b B q is spanned by B .Let I be the subspace spanned by B . It suffices to prove that I is a left ideal of b B q ,which in turn follows from the following statement: x i I ⊂ I for all i ∈ I .We note that x I ⊂ I by definition. Next we check that x x = q q x x , so x I ⊂ I . Also, x x = x + q x x , x x = − q x x , so x I ⊂ I .It remains to check that x I ⊂ I . Using the defining relations we check that x x = − q x x , x x = q ( e q − x x − q q q x x − q x u ,x x = − q q x x , x x = − q q x x , x x = − q q x x ,x x u = q q q x u x , x u x = q q q x x u , x u x = q q x x u .x x u = q q q x u x , x x u = − q q q x u x , x u x = q q q x x u , Using these relations, (5.4), (5.5), (5.6), (5.7), and the definition of the PBW generatorswe see that I is stable by left multiplication by x , x and x u , and so by x .Next we check that b B q fits into an exact sequence of braided Hopf algebras, in order toprove that B is linearly independent. Let Z be the subalgebra generated by x u = [ x , x ] c . Step 4. Z is a skew-central Hopf subalgebra isomorphic to a polynomial algebra in onevariable. RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 27 As x u is primitive and c ( x u ⊗ x u ) = x u ⊗ x u , Z is a Hopf subalgebra of b B q isomorphic toa polynomial algebra in one variable. It is central since x i x u = q i q i q i x u x i for all i ∈ I . Step 5.
There is a degree-preserving extension of braided Hopf algebras Z ֒ → b B q ։ e B q .Let Z ′ = b B co π q . As Z is normal (since it is central) and x u ∈ P ( e B q ) ∩ ker π , we havethat Z ⊆ Z ′ . By Lemma 2.4 and the known PBW basis of e B q we have that H b B q = H e B q H Z ′ ≥ H e B q H Z = (1 + t t )(1 + t )(1 + t t t )(1 + t t )(1 − t )(1 − t ) · − t t t . On the other hand, as b B q is spanned by B , H b B q ≤ (1 + t t )(1 + t )(1 + t t t )(1 + t t )(1 − t )(1 − t )(1 − t t t ) . Thence the two series above are equal, i.e. H b B q = (1 + t t )(1 + t )(1 + t t t )(1 + t t )(1 − t )(1 − t )(1 − t t t ) . (5.8)so Z = Z ′ , and the claim follows. Step 6. B is a basis of b B q and GKdim b B q = 3.By (5.8) B is a basis of b B q . Also, the decomposition above of the Hilbert series saysthat GKdim b B q = GKdim Z + GKdim e B q = 3. (cid:3) Remark 5.6.
Let Z q denote the subalgebra of b B q generated by x u , x N and x N . Onecan verify that this is subalgebra is skew-central (more precisely, it is annihilated by thebraided adjoint action of b B q ), and that it fits in a degree-preserving extension of braidedHopf algebras Z q ֒ → b B q ։ B q .5.1.2. Type A ( q |{ , , } ) . In this case the distinguished pre-Nichols algebra is given by e B q = T ( V ) / h x , x , x , x , [ x , x ] c i . Remark 5.7.
Let B be a pre-Nichols algebra of q such that GKdim B < ∞ . The relations x = x = x = 0 , x = 0 , [ x , x ] c = 0hold in B by Lemmas 3.1, 3.12 (a) and 3.13. Remark 5.8.
Let b B q denote the following quotient of T ( V ): b B q = T ( V ) / h x , x , x , x , [ x , x ] c i . Note that the defining ideal is actually a Hopf ideal, see the proof of Lemma 3.12, so b B q is a pre-Nichols algebra of q ; next we show that it is eminent. Proposition 5.9.
The pre-Nichols algebra b B q is eminent, with GKdim b B q = 3 and basis B = (cid:8) x a x b x c x d x e x f x g : a, c, e, g ∈ { , } , b, d, f ∈ N (cid:9) . (5.9) Proof.
By Remark 5.7, every finite GKdim pre-Nichols algebra of q is covered by b B q . Therest of the proof is carried out in several parts. Step 1.
The following relations hold in b B q : x = 0; x iij = 0 , i = j ;(5.10) [ x , x ] c = 0 , [ x , x ] c = 0 , [ x , x ] c = 0 , x = 0 . (5.11)For (5.10), the relations x iij = 0 for i = j follow from the condition q ii = − x i = 0. Similarly, x = x x − q q (1 + q ) x x x + q q q x x = 0 . For (5.11), notice that x = 0 = x imply that [ x , x ] c = 0; this last equality togetherwith [ x , x ] c = 0 give[ x , x ] c = [ x , [ x , x ] c ] c − q q x x + q q x x = q q (1 − e q ) x x . Since x = 0 = x , it follows [ x , [ x , x ] c ] c = 0. Then[ x , x ] c = [[ x , x ] c , x ] c + q q q x [ x , x ] c − q [ x , x ] c x = q q q [ x , [ x , x ] c ] c = 0 . Next, use [ x , x ] c = 0 = x to get[ x , x ] c = [ x , [ x , x ] c ] c − q q x x + q q q q x x = 0 . Finally, use x x = − ( q q ) − x x and x x = q q e q x x to compute x = x ( x x − q q x x ) = − e q ( x x − q q x x ) x = − e q x . Since N = ord e q >
2, this implies x = 0. Step 2.
The set B linearly spans b B q .It is enough to show that the linear span I of B is a left ideal. The inclusion x L ⊂ L isclear, and x L ⊂ L follows from the commutation x = 0, cf. (5.10). In order to verifythat x L ⊂ L , we argue inductively on b ≥ x x b = ( q q ) b − ( b ) e q x b − x + ( q q ) b x b x ,x x x b = ( q q q q ) b x b x + q b − q b ( b ) e q x x b − x + q b q b +113 x x b x .x x b x = − q b q b − q ( b ) e q x b − x x + ( q q ) b x b x + ( q q ) b q x b x x ,x x x b x = q b +112 ( q q q ) b q x b x x − q b q b q ( b ) e q x x b − x x + q b q b +113 x x b x + ( q q ) b +1 x x b x x . These equalities prove that x B ∈ I .The following is an auxiliary tool to compute the Hilbert series b B q . Step 3.
The subalgebra Z of b B q generated by x is a skew-central Hopf subalgebraisomorphic to a polynomial algebra in one variable.The generator of Z is annihilated by the braided adjoint action of x i , i ∈ I . In fact,(ad c x i ) x = 0 hold in b B q either by definition for i = 2 or by (5.10) if i = 1 ,
3. As x is primitive, Z is a Hopf subalgebra. Since x is a non-zero primitive element with q ( α + α , α + α ) = 1, it generates a polynomial algebra. RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 29
Step 4.
There is a degree-preserving extension of braided Hopf algebras Z ֒ → b B q ։ e B q .Let π : b B q ։ e B q denote the canonical projection, and put Z ′ = b B co π q . Notice that Z ⊂ Z ′ because x projects to zero, and Lemma 2.4 gives H b B q = H e B q H Z ′ ≥ H e B q H Z = (1 + t )(1 + t )(1 + t t t )(1 + t )(1 − t t )(1 − t t ) · − t t . On the other hand, since B linearly spans b B q we get the opposite inequality, so H b B q = (1 + t )(1 + t )(1 + t t t )(1 + t )(1 − t t )(1 − t t )(1 − t t ) , (5.12)and this warranties Z = Z ′ . The desired extension follows. Step 5.
The set B is a basis of b B q , which has Gelfand-Kirillov dimension 3.Now this follows as in the proof of Proposition 5.5. (cid:3) Remark 5.10.
Let Z q denote the subalgebra of b B q generated by x , x N and x N . Onecan verify that this subalgebra is skew-central (more precisely, it is annihilated by thebraided adjoint action of b B q ), and that here is a degree-preserving extension of braidedHopf algebras Z q ֒ → b B q ։ B q .5.2. Type B.
Here q is of type B θ ( q | J ), where q ∈ G ∞ has order N = 2 , ∅ 6 = J ⊆ I θ .Up to relabeling we may assume that θ / ∈ J (see [AA, § A θ − ( q | J ) q − q ◦ . If N = 3, then the pre-Nichols algebra e B q has the following presentation. x θθθθ − = 0; x ij = 0 , i < j − x iii ± = 0 , q ii = − , i ∈ I θ − ;[ x ( i − i +1) , x i ] c = 0 , q ii = − x i = 0 , q ii = − . (5.13)And if N = 3, then e B q is defined by the following relations: x ij = 0 , i < j − x iii ± = 0 , q ii = − , i ∈ I θ − ; x i = 0 , q ii = −
1; [ x ( i − i +1) , x i ] c = 0 , q ii = − x θθθθ − = 0; [ x θθ ( θ − , x θ ( θ − ] c = 0 , q θ − θ − = − x θθ ( θ − θ − , x θ ( θ − ] c = 0 . (5.14)Let B denote a pre-Nichols algebra of q with GKdim B < ∞ . We will check that eachrelation in (5.13), respectively (5.14), must hold in B . Lemma 5.11.
Let i, j ∈ I θ , i < j − . Then x ij = 0 in B .Proof. We fix first j = θ . Then x iθ = 0 by Lemma 3.3. Assume now that j < θ . If q ii = − q jj = −
1, then x ij = 0 by Lemma 3.3. If q ii = q jj = −
1, then x ij = 0 by Lemma 3.4,since e q jj +1 = 1 = e q ij +1 . (cid:3) Lemma 5.12. If i ∈ I θ − − J , then x iii ± = 0 in B . Proof.
We check first that x θ − θ − θ = 0 if i = θ − / ∈ J . If q θ − θ − = 1, i.e. q ∈ G ′ , then q c θ − θ (1 − c θ − θ ) θ − θ − q θθ = q − q = q = 1 . Hence either q − c θ − θ θ − θ − = 1 or q c θ − θ (1 − c θ − θ ) θ − θ − q θθ = 1. By Lemma 3.5, x θ − θ − θ = 0.Now we take i ∈ I θ − − J . If either q / ∈ G ′ or e q i i +1 = −
1, then x iii +1 = 0 by Lemma 3.5.Otherwise q ii = q i +1 i +1 = e q − ii +1 ∈ G ′ . As e q i +1 i +2 = 1 and e q ii +2 = 1, Lemma 3.7 applies toprove that x iii +1 = 0. The proof for i ∈ I ,θ − − J , x iii − = 0 follows analogously. (cid:3) Lemma 5.13. x θθθθ − = 0 in B .Proof. Here c θθ − = −
2. If N = 3, then x θθθθ − = 0 by Lemma 3.9. Now we assume N = 3. Notice that q − c θθ − θθ = q , q c θθ − (1 − c θθ − ) θθ q θ − θ − = ( q − ( − = q − , si θ − ∈ J ,q − q = q − , si θ − / ∈ J . Hence either q − c θθ − θθ or q c θθ − (1 − c θθ − ) θθ q θ − θ − . By Lemma 3.5, x θθθθ − = 0. (cid:3) Lemma 5.14. If i ∈ J , then [ x ( i − ,i +1) , x i ] c = 0 in B .Proof. If either q i − i − = − q i +1 i +1 = −
1, then x u = 0 by Lemma 3.13. Hence weassume that q i − i − , q i +1 i +1 = −
1. If i < θ −
1, then i + 2 ∈ I θ , q i − i − q i +1 i +1 = 1, e q i +1 i +2 = 1, e q i − i +2 = e q ii +2 = 1 and e q i − i = e q − ii +1 = ±
1, so x u = 0 by Lemma 3.14Finally we consider i = θ − θ − / ∈ J . Suppose that x u = 0. As q uu = q − , e q u ( θ − = 1, e q u ( θ − = 1, e q θu = q − = 1, the diagram of q ′ contains the following subdiagram: q − ◦ θ − q − ◦ θ − q − q ◦ θ q − q − ◦ u . This diagram does not belong to [H2, Table 3] since the two extremal vertices and one inthe middle have label = −
1, and these three labels are pairwise different, a contradictionwith Conjecture 1.1. Hence x u = 0 in any case. (cid:3) Lemma 5.15. If N = 3 and θ − ∈ J , then [ x θθ ( θ − , x θ ( θ − ] c = 0 in B .Proof. By Lemmas 3.1, 5.12, 5.13 and [An1, Lemma 5.9 (a)], [ x θθ ( θ − , x θ ( θ − ] c ∈ P ( B ).Now Lemma 3.15 applies and we get [ x θθ ( θ − , x θ ( θ − ] c = 0. (cid:3) Lemma 5.16. If N = 3 , then [ x θθ ( θ − θ − , x θ ( θ − ] c = 0 in B .Proof. By Lemmas 3.1, 5.12, 5.13 and [An1, Lemma 5.9 (b)], [ x θθ ( θ − θ − , x θ ( θ − ] c ∈P ( B ). Now Lemma 3.16 applies, so [ x θθ ( θ − θ − , x θ ( θ − ] c = 0. (cid:3) Theorem 5.17.
Let q be of type B θ ( q | J ) . Then the distinguished pre-Nichols algebra e B q is eminent.Proof. If N = 3, then the statement follows by Lemmas 3.1, 5.12, 5.13 and 5.14. If N = 3,then we apply the same results together with Lemmas 5.15 and 5.16. (cid:3) RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 31
Type D.
Here q is of type D θ ( q | J ), where θ ≥ q ∈ k × is a root of unity of order N >
2, and ∅ 6 = J ⊆ I θ , see [AA, § A θ − ( q ; J ) q − q ◦ , (5.15) − ◦ q − q ❅❅❅❅❅❅❅❅ A θ − ( q ; J ∩ I θ − ) q − − ◦ , θ − ∈ J , (5.16) q − ◦ q A θ − ( q − ; J ∩ I θ − ) q q − ◦ , θ − / ∈ J . (5.17) Theorem 5.18.
Let q be of type D θ ( q | J ) . Then the distinguished pre-Nichols algebra e B q is eminent. We give a proof for each diagram above in Propositions 5.24, 5.29 and 5.33.5.3.1.
The diagram (5.15) . The presentation of the distinguished pre-Nichols algebra e B q depends on N and q θ − θ − as we describe below: • q θ − θ − = − N >
4: the set of defining relations is x ij = 0 , i < j − x iii ± = 0 , i ∈ I θ − − J ; x θ − θ − θ − = 0; x θ − θ − θ − θ = 0; x θθθ − = 0; [ x ( i − i +1) , x i ] c = 0 , i ∈ I ,θ − ∩ J ; x i = 0 , i ∈ J ;(5.18) • q θ − θ − = − N = 4: the set of defining relations is[ x ( θ − θ ) , x θ − θ ] c = 0; x iii ± = 0 , i ∈ I θ − − J ; x θ − θ − θ − = 0; x ij = 0 , i < j − x θ − θ − θ − θ = 0; [ x ( i − i +1) , x i ] c = 0 , i ∈ I ,θ − ∩ J ; x i = 0 , i ∈ J ; x θθθ − = 0 , . (5.19) • q θ − θ − = − N = 3: the set of defining relations is[[ x ( θ − θ ) , x θ − ] c , x θ − ] c = 0; x iii ± = 0 , i ∈ I θ − − J ; x θ − θ − θ − = 0; x ij = 0 , i < j − x θθθ − = 0; [ x ( i − i +1) , x i ] c = 0 , i ∈ I ,θ − ∩ J ; x i = 0 , i ∈ J ; x θ − θ − θ − θ = 0 , (5.20) • q θ − θ − = − N = 4: the set of defining relations is[[ x θ − θ − , x ( θ − θ ) ] c , x θ − ] c = 0 , θ − ∈ J ;[[[ x ( θ − θ ) , x θ − ] c , x θ − ] c , x θ − ] c = 0 , θ − / ∈ J ;[ x ( i − i +1) , x i ] c = 0 , i ∈ I ,θ − ∩ J ; x ij = 0 , i < j − x iii ± = 0 , i ∈ I θ − − J ; x θθθ − = 0; x i = 0 , i ∈ J ; . (5.21) • q θ − θ − = − N = 4: the set of defining relations is[[[ x ( θ − θ ) , x θ − ] c , x θ − ] c , x θ − ] c = 0 , θ − / ∈ J ;[[ x θ − θ − , x ( θ − θ ) ] c , x θ − ] c = 0 , θ − ∈ J ; x iii ± = 0 , i ∈ I θ − − J ; x ij = 0 , i < j − x ( i − i +1) , x i ] c = 0 , i ∈ I ,θ − ∩ J ; x i = 0 , i ∈ J ; x θθθ − = 0; x θ − θ = 0 . (5.22)Fix a finite GKdim pre-Nichols algebra B of q . We check that each relation in (5.18),(5.19), (5.20), (5.21) and (5.22) holds in B . Lemma 5.19. If i, j ∈ I θ , i < j − , then x ij = 0 in B .Proof. Let j = θ . Notice that q ii ∈ { q ± , − } . Hence, • if either N = 4 or q ii = −
1, then x iθ = 0 by Lemma 3.3; • if N = 4, q ii = −
1, then q ii q θθ = 1, e q ii +1 e q i +1 θ = ± q ± = 1. Thus x iθ = 0 by Lemma 3.4.Assume now that j < θ . If q ii = − q jj = 1, then x ij = 0 by Lemma 3.3. If q ii = q jj = −
1, then x ij = 0 by Lemma 3.4, since e q jj +1 = 1 and e q ij +1 = 1. (cid:3) Lemma 5.20.
Te relation x θθθ − = 0 holds in B .Proof. Recall that c θθ − = −
1. If N = 4, the desired relation follows from Lemma 3.5since q − c θθ − θθ = q = 1 and q c θθ − (1 − c θθ − ) θθ q θ − θ − = q − q θ − θ − ∈ { q − , q − } .In the case N = 4 we can directly apply Lemma 3.10, since q − c θθ − θθ = 1, q θθ = − e q θ − θ = − e q θ − θ − = q ± = 1, and e q θθ − e q θ − θ − = e q θ − θ − = 1. (cid:3) Lemma 5.21. If q θ − θ − = − then x θ − θ − θ − θ = 0 in B .Proof. Now c θ − θ = −
2. When N = 3 the relation follows from Lemma 3.5, since q − c θ − θ θ − θ − = q = 1 and q c θ − θ (1 − c θ − θ ) θ − θ − q θθ = q − q = q − = 1.If N = 3, we are under the hypothesis of Lemma 3.9 (indeed, q − c θ − θ θ − θ − = q = 1, q θ − θ − = q = q − = e q θ − θ , and q θθ = − x θ − θ − θ − θ = 0. (cid:3) Lemma 5.22. (a) If i ∈ I θ − − J , then x iii +1 = 0 in B . (b) If i ∈ I θ − − J , then x iii − = 0 en B .Proof. (a) Notice that c i i +1 = − q ii = 1. If x iii +1 = 0 in B then Lemma 3.5 impliesthat q ii = 1 and q − ii q i +1 i +1 = 1. So Lemma 3.6 assures that q ii = q i +1 i +1 = e q − ii +1 ∈ G ′ .This last equality together with e q ii +2 e q i +1 i +2 = e q i +1 i +2 = 1 are exactly the hypothesis ofLemma 3.7, which states that x iii +1 = 0, a contradiction. RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 33 (b) Note that c i i − = − q ii = 1. If x iii − = 0 in B then Lemma 3.5 implies that q ii = 1 and q − ii q i − i − = 1. So Lemma 3.6 assures that q ii = q i − i − = e q − ii − ∈ G ′ .This and e q ii +1 e q i − i +1 = e q ii +1 = 1 allow us to apply Lemma 3.7 to get x iii − = 0, acontradiction. (cid:3) Lemma 5.23.
Given i ∈ I ,θ − ∩ J we have [ x ( i − ,i +1) , x i ] c = 0 in B .Proof. If [ x ( i − ,i +1) , x i ] c = 0, Lemma 3.13 implies i − i + 1 / ∈ J . Now condition (C) gives q i − i − = e q − i − i = e q ii +1 = q − i +1 i +1 . Note also that e q i +1 i +2 = 1 = e q ii +2 = e q i − i +2 so Lemma3.14(c) assures [ x ( i − ,i +1) , x i ] c = 0, a contradiction. (cid:3) Proposition 5.24. If q has Dynkin diagram (5.15), then e B q is eminent.Proof. Fix a pre-Nichols algebra B with finite GKdim. From Lemmas 3.1,5.19, 5.20, 5.21,5.22 and 5.23 we know that the quantum Serre relations as well as x i = 0 (for i ∈ J ) and[ x ( j − ,j +1) , x j ] c = 0 (for j ∈ I ,θ − ∩ J ) hold in B . Other previous results apply: • If N = 4 and q θ − θ − = −
1, then e q θ − θ − e q θθ − = q = 1. So x θ − θ = 0 by Lemma 3.17. • If N = 4 and q θ − θ − = −
1, then i = θ , j = θ − k = θ − x θθ − , x θθ − θ − ] c = 0. Thus [ x ( θ − ,θ ) , x θ − θ ] c = 0. • If N = 3 and q θ − θ − = −
1, we can apply Lemma 3.18 to i = θ − j = θ − k = θ and get [[ x ( θ − ,θ ) , x θ − ] c , x θ − ] c = 0.It only remain to verify that the elements[[[ x ( θ − ,θ ) , x θ − ] c , x θ − ] c , x θ − ] c , [[ x θ − θ − , x ( θ − ,θ ) ] c , x θ − ] c , vanish in B when q fulfills the corresponding constraints. In each case the defining relationsof e B q with smaller degrees certainly hold in B , so these elements are primitive in B . Hencethey annihilate in B thanks to Lemmas 3.21 and 3.19, respectively. (cid:3) The diagram (5.16) . The presentation of e B q depends on q as follows: • q θ − θ − = − N = 4: the set of defining relations is x θ − θ − θ = 0; x iii ± = 0 , i ∈ I θ − − J ; x ij = 0 , i < j − , θ −
2; [ x ( i − i +1) , x i ] c = 0 , i ∈ I θ − ∩ J ; x i = 0 , i ∈ J ; x ( θ − θ ) = q θ − θ − (1 − q ) x θ − x θ − θ − q θ − θ (1 + q − )[ x θ − θ , x θ − ] c . (5.23) • q θ − θ − = − N = 4: the set of defining relations is x iii ± = 0 , i ∈ I θ − − J ; [ x ( i − i +1) , x i ] c = 0 , i ∈ I θ − ∩ J ; x θ − θ − θ = 0; x ij = 0 , i < j − , θ − x i = 0 , i ∈ J ; x θθθ − = 0; x θ − θ − θ = 0; x ( θ − θ ) = q θ − θ − (1 − q ) x θ − x θ − θ − q θ − θ (1 + q − )[ x θ − θ , x θ − ] c . (5.24) • q θ − θ − = − N = 4: the set of defining relations is[ x θ − θ − θ , x θ − ] c = 0; x iii ± = 0 , i ∈ I θ − ∩ J ; x ij = 0 , i < j − , θ −
2; [ x ( i − i +1) , x i ] c = 0 , i ∈ I θ − ∩ J ; x i = 0 , i ∈ J ; x ( θ − θ ) = q θ − θ − (1 − q ) x θ − x θ − θ − q θ − θ (1 + q − )[ x θ − θ , x θ − ] c . (5.25) • q θ − θ − = − N = 4: the set of defining relations is x iii ± = 0 , i ∈ I θ − − J ; x ij = 0 , i < j − , θ − x θ − θ − θ , x θ − ] c = 0; [ x ( i − i +1) , x i ] c = 0 , i ∈ I θ − ∩ J ; x i = 0 , i ∈ J ; x θθθ − = 0; x θ − θ − θ = 0; x ( θ − θ ) = 2 q θ − θ − x θ − x θ − θ − q θ − θ (1 + q − )[ x θ − θ , x θ − ] c . (5.26)Next we verify that all relations (5.23), (5.24), (5.25) and (5.26) hold in any finite GKdimpre-Nichols algebra B of q . Lemma 5.25.
For i, j ∈ I θ with i < j − , θ − , we have x ij = 0 in B .Proof. The argument depends on the value of q ii q jj as follows: • If q ii = − q jj = −
1, then x ij = 0 by Lemma 3.3. • If q ii = q jj = −
1, by Lemma 3.4 it is enough to find l = i, j such that e q il e q jl = 1. Now – for j = θ , note that e q iθ − e q θθ − = q = 1; – if j = θ − e q iθ e q θ − θ = q = 1; – when j < θ − e q ij +1 e q jj +1 = e q jj +1 = 1.So x ij = 0 in any case. (cid:3) Lemma 5.26. (a) If i ∈ I θ − − J , then x iii ± = 0 in B . (b) If θ − / ∈ J , then x θ − θ − θ = 0 in B .Proof. (a) Note that c i i ± = − q ii = 1. Assume x iii +1 = 0, so Lemma 3.5 gives q ii = 1 and q − ii q i +1 i +1 = 1. Hence q ii = q i +1 i +1 = e q − ii +1 ∈ G ′ by Lemma 3.6. Using thisand e q ii +2 e q i +1 i +2 = e q i +1 i +2 = 1, we apply Lemma 3.7 and get x iii +1 = 0, a contradiction.The same argument leads to x iii − = 0, using in the last step that e q ii +1 e q i − i +1 = e q ii +1 = 1.(b) Now c θ − θ = − q θ − θ − = 1. Since q − θ − θ − q θθ = q − θ − θ − = 1, Lemma 3.5 applies. (cid:3) Lemma 5.27. If N = 4 , then x θθθ − = 0 and x θ − θ − θ = 0 in B .Proof. Note that if { i, j } = { θ − , θ } , then Lemma 3.10 applies. Indeed, c ij = − q ii = 1and q ii = e q ij = −
1; moreover k = θ − e q jk = q − = 1 and e q ik e q jk = q − = 1. (cid:3) Lemma 5.28. (a) If i ∈ I ,θ − ∩ J , then [ x ( i − ,i +1) , x i ] c = 0 in B . (b) If θ − ∈ J , then [ x θ − θ − θ , x θ − ] c = [ x ( θ − θ − , x θ − ] c = 0 in B .Proof. (a) Let i ∈ I θ − ∩ J . If either i − ∈ J or i + 1 ∈ J , then [ x ( i − ,i +1) , x i ] c = 0 byLemma 3.13. Assume now i − , i + 1 / ∈ J , so e q i +1 i +2 = 1 = e q ii +2 = e q i − i +2 , and Lemma3.14 (c) gives [ x ( i − ,i +1) , x i ] c = 0. RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 35 (b) If k ∈ { θ − , θ } , then i = θ − j = θ − k fulfill the hypothesis of Lemma 3.13, since q θ − θ − = − e q θ − k = 1, e q θ − θ − = e q − θ − k = q = ± q kk = − (cid:3) Proposition 5.29. If q has Dynkin diagram (5.16) , then e B q is eminent.Proof. All defining relations of e B q hold in any finite GKdim pre-Nichols algebra of q byLemmas 3.1, 3.22, 5.25, 5.26, 5.27 and 5.28. (cid:3) The diagram (5.17) . The following is a presentation of e B q : • q θ − θ − = − N = 4: x iii ± = 0 , i ∈ I θ − − J ; x θ − θ − θ = 0; x θ − θ − θ − = 0; x ij = 0 , i < j − , θ − x θ − θ = 0; x θθθ − = 0;[ x ( i − i +1) , x i ] c = 0 , i ∈ I θ − ∩ J ; x i = 0 , i ∈ J . (5.27) • q θ − θ − = − N = 4: x θθθ − = 0; x iii ± = 0 , i ∈ I θ − − J ; x θ − θ − θ = 0; [ x ( i − i +1) , x i ] c = 0 , i ∈ I θ − ∩ J ; x θ − θ − θ − = 0; x ij = 0 , i < j − , θ − x θ − θ = 0; x i = 0 , i ∈ J . (5.28) • q θ − θ − = − N = 4: x iii ± = 0 , i ∈ I θ − − J ; x θθθ − = 0; x θ − θ − θ − = 0;[ x ( i − i +1) , x i ] c = 0 , i ∈ I θ − ∩ J ; x θ − θ = 0; [ x θ − θ − θ , x θ − ] c = 0; x ij = 0 , i < j − , θ − x i = 0 , i ∈ J . (5.29) • q θ − θ − = − N = 4: x θ − θ = 0; x ij = 0 , i < j − , θ − x θ − θ − θ , x θ − ] c = 0; [ x ( i − i +1) , x i ] c = 0 , i ∈ I θ − ∩ J ; x θθθ − = 0; x iii ± = 0 , i ∈ I θ − − J ; x θ − θ − θ − = 0; x i = 0 , i ∈ J (5.30)Given a finite GKdim pre-Nichols algebra B of q , we show that the relations in (5.27),(5.28), (5.29) and (5.30) hold in B . Lemma 5.30.
For i, j ∈ I θ with i < j − , θ − we have x ij = 0 in B ; also x θ − θ = 0 .Proof. The relation x θ − θ = 0 follows from Lemma 3.3, since q θθ = 1.Consider now i + 1 < j ≤ θ −
1. Then the vertices i, j belong to the subdiagram spannedby I θ − , which is of type A θ − ( q − | J ). Since θ − / ∈ J , this last diagram is not of type A ( p |{ , , } ), so the relation x ij = 0 follows from Lemma 5.1.For the case i + 1 < j = θ , we may apply the same argument but taking care of thesubdiagram spanned by I θ − ∪ { θ } instead of I θ , and using that θ / ∈ J . (cid:3) Lemma 5.31. (a) If i ∈ I θ − − J , then x iii +1 = 0 in B . (b) If i ∈ I ,θ − − J , then x iii − = 0 in B . (c) If θ − / ∈ J , then x θ − θ − θ = 0 in B . (d) The relation x θθθ − = 0 holds in B .Proof. Note that the relations considered are of the form x iij for some i, j with c ij = − q ii = 1 or q − ii q i +1 i +1 = 1, we have x iii +1 = 0 by Lemma 3.5. In the othercase, Lemma 3.6 gives q ii = q i +1 i +1 = e q − ii +1 ∈ G ′ . By Lemma 3.7, the relation x iii +1 willfollow if we find k = i, i + 1 such that e q ik e q i +1 k = 1. This is easily achieved: • if i < θ −
2, then e q ii +2 e q i +1 i +2 = e q i +1 i +2 = q ± = 1; • if i = θ −
2, then e q θ − θ e q θ − θ = e q θ − θ = q − = 1.(b) Follows similarly.(c)(d) The subdiagram spanned by { θ − , θ − , θ } is of type A ( q − | J ′ ) with J ′ ⊂ { θ − } .Hence both claims follow from Lemma 5.2. (cid:3) Lemma 5.32. (a) If i ∈ I ,θ − ∩ J , then [ x ( i − ,i +1) , x i ] c = 0 in B . (b) If θ − ∈ J , then [ x ( θ − θ − , x θ − ] c = 0 and [ x θ − θ − θ , x θ − ] c = 0 in B .Proof. (a) Lemma 3.13 assures the desired relation if either i − ∈ J or i + 1 ∈ J . Ifthat is not the case, it follows that q i − i − q i +1 i +1 = 1; also, since i < θ −
2, we have e q i +1 i +2 = 1 = e q ii +2 = e q i − i +2 . Thus Lemma 3.14 (c) gives [ x ( i − ,i +1) , x i ] c = 0.(b) By Lemma 3.13, we only need to consider the case θ − / ∈ J . Since also θ − / ∈ J we have q i − i − q i − i − = 1. Now Lemma 3.14 (b) gives [ x ( θ − θ − , x θ − ] c = 0, as e q θ − θ = q − = 1 = e q θ − θ = e q θ − θ . A similar argument shows that [ x θ − θ − θ , x θ − ] c = 0. (cid:3) Proposition 5.33. If q has Dynkin diagram (5.17), then e B q is eminent.Proof. Follows directly from Lemmas 3.1, 5.30, 5.31, 5.32. (cid:3)
Type D (2 , α ) . The possible Dynkin diagrams of q are: q ◦ q − − ◦ r − r ◦ , (5.31) − ◦ s qqqqqqq r ▼▼▼▼▼▼▼ − ◦ q − ◦ , (5.32)where q, r, s ∈ k − { } are such that qrs = 1. It is required that, if q = −
1, then r, s = − q = −
1, either r = 1 or s = 1. See [AA, § Theorem 5.34.
Let q of type D (2 , α ) . Then the distinguished pre-Nichols algebra e B q is eminent. The diagram (5.31) . If q, r, s = −
1, then e B q is presented by x = 0; x = 0; x = 0; x = 0 . (5.33)If q = − r, s = −
1, then e B q is presented by the relations x = 0; x = 0; x = 0; x = 0; x = 0 . (5.34) Proposition 5.35. If q has Dynkin diagram (5.31) , then e B q is eminent. RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 37
Proof.
Let B a be pre-Nichols algebra of q such that GKdim B < ∞ .5.33 Assume q, r, s = −
1. By Lemma 3.1, x = 0 . As either q = − q = −
1, therelation x = 0 follows from Lemma 3.3. For x = 0 we apply Lemma 3.5 (b) since c = − q = q = 1 and q − q = q − = 1. Similarly, using r ± = 1 we have x = 0.5.34 Consider now q = −
1. Again x = 0 by Lemma 3.1. For x , note that either q = − q = −
1, so Lemma 3.3 applies. The argument given in the previous paragraph for x = 0 also works here. For x = 0 we apply Lemma 3.10 since c = − q = q = 1, e q = − e q = r − = 1 and e q e q = e q = 1. Finally x = 0 follows from Lemma 3.17(b) since q = e q = q = − e q e q = r − = 1. (cid:3) The diagram (5.32) . Up to relabeling, we can assume that e q , e q = −
1. In thiscase the defining relations of the distinguished pre-Nichols algebra are x = 0; x = 0; x = 0; x = 0 si e q = − x (13) − − sq (1 − r ) [ x , x ] c − q (1 − s ) x x = 0 . (5.35) Proposition 5.36. If q has Dynkin diagram (5.32) , then e B q is eminent.Proof. It follows by Lemmas 3.1, 3.17 and 3.22. (cid:3)
Type super F (4) . Fix q of type F (4) [AA, § q ◦ q − q ◦ q − q ◦ − ◦ q − q ◦ q − q ◦ q − − ◦ − ◦ qq ◦ q − q − ✾✾✾✾✾✾✾✾ q ◦ q − − ◦ q − ◦ − ◦ q q − ✝✝✝✝✝✝✝ − ◦ q − ◦ q ◦ q − q ◦ q − q ◦ q − − ◦ q q − ◦ , q ◦ q − q ◦ q − − ◦ q q − ◦ , (5.36)where q ∈ G ′ N , with N ≥
4. Treating each of these diagrams separately, we prove:
Theorem 5.37.
Let q of type F (4) . Then the distinguished pre-Nichols algebra is eminent. The diagram (5.36 a) . When
N > e B q are: x = 0; x = 0; x = 0; x = 0; x = 0; x = 0; x = 0; x = 0; x = 0 . (5.37)For N = 4, the distinguished pre-Nichols algebra has the following presentation: x = 0; x = 0; x = 0; x = 0; [ x (13) , x ] c = 0; x = 0; x = 0; x = 0; x = 0; x = 0; [ x , x (24) ] c = 0 . (5.38) Proposition 5.38. If q has diagram (5.36 a), then e B q is eminent. Proof.
Let B be a finite GKdim pre-Nichols algebra of q . • The subdiagram spanned by { , , } is of Cartan type B , hence all the defining relationsinvolving only x , x and x hold in B by Theorem 2.10. • Similarly, the subdiagram spanned by { , , } is of type D ( q |{ } ) (5.15) hence all thedefining relations supported in x , x , x hold in B by Proposition 5.24.Finally x = 0 holds in B by Lemma 3.3 since q = q = − (cid:3) The diagram (5.36 b) . When
N >
4, a set of defining relations of the e B q is x = 0; x = 0; x = 0; x = 0; [[ x , x ] c , x ] c = 0; x = 0; x = 0; x = 0; x = 0 . (5.39)In the case N = 4, a presentation of e B q is x = 0; x = 0; x = 0; x = 0; [ x (13) , x ] c = 0; x = 0; x = 0; x = 0; x = 0; x = 0; [[ x , x ] c , x ] c = 0 . (5.40) Proposition 5.39. If q has diagram (5.36 b), then e B q is eminent.Proof. Let B be a pre-Nichols algebra of q with GKdim B < ∞ . • On one hand { , , } span a subdiagram of the form q ◦ q − A ( q ; { , } ) hence allthe relations in 5.39 and 5.40 supported in x , x , x hold in B by Proposition 5.24. • If N > { , , } is of type A ( q |{ } ), hence all the definingrelations involving only x , x and x hold in B by Theorem 5.4. • If N = 4 the vertices { , , } span a subdiagram of Cartan type A at −
1, so x = x = 0 by [ASa, Lemma 5.3]. We get x = 0 from Lemma 3.4, since q q = 1and e q e q = e q = 1. The element x must vanish in B since it is primitive and theDynkin diagram of V ⊕ k x is − ◦ − ◦ − − ◦ − − ❑❑❑❑❑❑❑❑❑❑ − ◦ − ◦ , q which does not belong to [H2, Table 4] because q = −
1. Finally turn to x u = [ x (13) , x ] c .From Lemma 3.11 and the facts x = x = [ x , x ] c = 0, it follows that x u ∈ P ( B ).But the Dynking diagram of k x u + k x is ◦ q − ◦ , so Lemma 2.6 gives [ x (13) , x ] c = 0.Finally x = 0 holds in B since either q = 1 (for N >
4) so Lemma 3.3 applies, orelse q q = 1, e q e q = 1 (for N = 4) and Lemma 3.4 applies here. (cid:3) The diagram (5.36 c) . A set of defining relations of e B q when N > x = 0; x = 0; x = 0; x (24) − q q [ x , x ] c − q (1 − q − ) x x = 0; x = 0; x = 0; x = 0; x = 0; [ x (13) , x ] c = 0 . (5.41) RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 39
A presentation when N = 4 is x = 0; x = 0; x = 0; [ x (13) , x ] c = 0; x = 0; x = 0; x = 0; x = 0; x = 0; x (24) − q q [ x , x ] c − q (1 − q − ) x x = 0 . (5.42) Proposition 5.40. If q has diagram (5.36 c), then e B q is eminent.Proof. Let B be a pre-Nichols algebra of q , GKdim B < ∞ . • The vertices { , , } determine a subdiagram of type (5.16) hence all the defining rela-tions in (5.41) and (5.42) involving only x , x and x hold in B by Proposition 5.29. • If N > { , , } determine a subdiagram of type A ( q |{ , } ), hence therelations in 5.41 involving only x , x and x hold in B by Theorem 5.4. • If N = 4 we check case-by-case that all the defining relations 5.42 supported in 1 , , B . Since q q = 1 and e q e q = 1, Lemma 3.4 gives x = 0, while x = x = 0hold by Lemma 3.1. Also x = 0 in B by Lemma 3.10 with k = 3, and x = 0 byLemma 3.17 with k = 4. Now focus on x u = [ x (13) , x ] c which is primitive in B byLemma 3.11 since [ x , x ] c = x = x = 0. The Dynking diagram of k x u + k x is ◦ q − q ◦ , so Lemma 2.6 gives [ x (13) , x ] c = 0.Finally x = 0 holds in B by Lemma 3.3 since q = − (cid:3) The diagram (5.36 d) . A presentation of e B q when N > x = 0; x = 0; x = 0; [ x , x ] c = 0; x = 0; x = 0; x = 0; [[ x , x ] c , x ] c = 0; x (24) + q − q − q [ x , x ] c − q (1 − q − ) x x = 0 . (5.43)When N = 4 a set of defining relations of e B q is x = 0; x = 0; x = 0; [ x , x ] c = 0; x = 0; x = 0; x = 0; x = 0; [[ x , x ] c , x ] c = 0; x (24) + q − q − q [ x , x ] c − q (1 − q − ) x x = 0 . (5.44) Proposition 5.41. If q has Dynkin diagram (5.36 d), then e B q is eminent.Proof. Let B be a pre-Nichols algebra of q with GKdim B < ∞ . • The vertices { , , } span a subdiagram of type (5.15), hence the relations in (5.43) and(5.44) supported in x , x , x hold in B by Proposition 5.24. • The subdiagram determined by { , , } is of type (5.32), hence the relations in (5.43)and (5.44) supported in x , x , x hold in B by Proposition 5.36. • If N >
4, the relations in (5.43) involving only x , x and x hold in B by Theorem 5.4,since { , , } determine a subdiagram of type A ( q |{ , } ). • If N = 4, we check that each relation in (5.44) with support contained in x , x , x holdin B . Note that x = 0 by Lemma 3.4 since q q = 1 and e q e q = q − = 1, while x = x = 0 by Lemma 3.1. Also, x = 0 by [ASa, Lemma 5.3], and x = 0 byLemma 3.17 with k = 3. Let x u = [ x , x ] c which is primitive in B by Lemma 3.11since [ x , x ] c = x = x = 0. The Dynking diagram of k x u + k x is ◦ q − − ◦ , soLemma 2.6 gives [ x , x ] c = 0.Hence the relations in (5.43) and (5.44) hold in B . (cid:3) The diagram (5.36 e) . A presentation of e B q when N = 4 , x = 0; x = 0; x = 0; x = 0; x = 0; x = 0; [[[ x , x ] c , [ x , x ] c ] c , x ] c = 0; x = 0; x = 0;(5.45)A presentation of e B q when N = 6 is x = 0; x = 0; x = 0; x = 0; x = 0; x = 0; [[[ x , x ] c , [ x , x ] c ] c , x ] c = 0; x = 0; x = 0;(5.46)A set of defining relations of e B q when N = 4 is x = 0; x = 0; x = 0 : x = 0; [ x (13) , x ] c = 0; x = 0; x = 0; x = 0; x = 0; x = 0; [[[ x , x ] c , [ x , x ] c ] c , x ] c = 0 . (5.47) Proposition 5.42. If q has Dynkin diagram (5.36 e), then e B q is eminent.Proof. Let B be a finite GKdim pre-Nichols algebra of q . • The subdiagram spanned by { , , } is of type (5.31), hence those relations in (5.45),(5.46) and (5.47) that do not involve x hold in B by Proposition 5.35. • If N > { , , } determine a subdiagram of type A ( q |{ } ), hence therelations (5.45), (5.46) that do not involve x hold in B by Theorem 5.4. • If N = 4 we check that each relation in (5.46) involving only x , x and x hold in B .Note that x = 0 by Lemma 3.4 since q q = 1 and e q e q = q = 1, while x = 0hold by Lemma 3.1. Also x = x = 0 by [ASa, Lemma 5.3] and x = 0 by Lemma3.17 with k = 4. If x = 0 we get V ⊕ k x ⊂ P ( B ) with Dynkin diagram − ◦ − ◦ − − ◦ − − ❏❏❏❏❏❏❏❏❏❏ − ◦ q ◦ , q − which is not in [H2, Table 4]. We get a contradiction with Conjecture 1.1, so x = 0.Now [ x (13) , x ] c is primitive in B by Lemma 3.11. Moreover, the diagram of k [ x (13) , x ] c + k x is ◦ q q − ◦ , so Lemma 2.6 gives [ x (13) , x ] c = 0. RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 41
Finally [[[ x , x ] c , [ x , x ] c ] c , x ] c = 0 in B by Lemma 3.23 and x = 0 holds in B by Lemma 3.3 since q q = q − = 1. (cid:3) The diagram (5.36 f) . A presentation of e B q when N = 4 , x = 0; x = 0; x = 0; x = 0; x = 0; x = 0; x = 0; x = 0; [[ x (14) , x ] c , x ] c = q ( q − q )[[ x (14) , x ] c , x ] c . (5.48)If N = 6, a set of defining relations of the distinguished pre-Nichols algebra is the following: x = 0; x = 0; x = 0; x = 0; x = 0; x = 0; x = 0; x = 0; x = 0; [[ x (14) , x ] c , x ] c = q ( q − q )[[ x (14) , x ] c , x ] c . (5.49)When N = 4 the relations defining e B q are x = 0; x = 0; x = 0; x = 0; x = 0; [ x , x (13) ] c = 0; x = 0; x = 0; x = 0; [[ x (14) , x ] c , x ] c = q ( q − q )[[ x (14) , x ] c , x ] c . (5.50) Proposition 5.43. If q has Dynkin diagram (5.36 f ), then e B q is eminent.Proof. Let B be a pre-Nichols algebra of q with GKdim B < ∞ . • The subdiagram spanned by the vertices { , , } is of type D ( q |{ } ), hence the definingrelations with support contained in 1 , , B by Proposition 5.29. • The vertices { , , } determine a subdiagram of type (5.31), hence all the definingrelations with support contained in x , x , x hold in B by Proposition 5.35.Finally [[ x (14) , x ] c , x ] c = q ( q − q )[[ x (14) , x ] c , x ] c in B by Lemma 3.24 and x = 0holds in B by Lemma 3.3 since q q = q − = 1. (cid:3) Type G (3) . Here N = ord q > − ◦ q − q ◦ q − q ◦ , − ◦ q − ◦ q − q ◦ , − q − ◦ q − ◦ q − q ◦ , − ◦ q ◦ q − ☞☞☞☞☞ q − − ◦ q ✹✹✹✹✹ (5.51)see [AA, § Theorem 5.44.
Let q of type G (3) . Then the distinguished pre-Nichols algebra is eminent. The diagram (5.51 a) . A presentation of e B q when N = 4 , x = 0; x = 0; x = 0; x = 0; x = 0 . (5.52)For N = 6 the defining relations of the distinguished pre-Nichols algebra are x = 0; x = 0; [ x , x (321) ] c = 0; x = 0; x = 0; x = 0; [[ x , x ] c , x ] c = 0 . (5.53) If N = 4, then a presentation of e B q is x = 0; x = 0; [[[ x (13) , x ] c , x ] c , x ] c = 0; x = 0; x = 0; x = 0; [ x , x ] c = 0 . (5.54) Proposition 5.45. If q has Dynkin diagram (5.51 a), then e B q is eminent.Proof. Let B be a pre-Nichols algebra of q such that GKdim B < ∞ . • The vertices { , } determine a subdiagram of Cartan type G , hence those relations in(5.52), (5.53) and (5.54) involving only x and x hold in B by Theorem 4.6. • The subdiagram spanned by { , } is of type super A, so the defining relations supportedin x , x hold in B by Theorem 5.4 .Consider the relations involving x , x and x . If N = 6 we have [ x , x ] = 0 by Lemma3.20, and if N = 4 then [[[ x (13) , x ] c , x ] c , x ] c = 0 by Lemma 3.25. Finally, x = 0 eitherby Lemma 3.3 if N = 6 or by Lemma 3.4 if N = 6. (cid:3) The diagram (5.51 b) . A set of defining relations of e B q is x = 0; x = 0; x = 0; x = 0; [[ x , [ x , x (13) ] c ] c , x ] c = 0 . (5.55) Proposition 5.46. If q has Dynkin diagram (5.51 b), then e B q is eminent.Proof. Given a finite GKdim pre-Nichols algebra B , Lemma 3.1 gives x = x = 0. Also, • If N = 6, then x = 0 by Lemma 3.5 and x = 0 by Lemma 3.3. • If N = 6, then x = 0 by Lemma 3.10 and x = 0 by Lemma 3.4.Finally, the relations of B q of degree lower than that of x u = [[ x , [ x , x (13) ] c ] c , x ] c aresatisfied in B , so x u ∈ P ( B ). Then x u = 0 by Lemma 3.26. (cid:3) The diagram (5.51 c) . A set of defining relations of e B q when N = 6 is x = 0; x = 0; x = 0; x = 0;[ x , [ x , x ] c ] c = q q q [ x , x ] c − ( q − − q − ) q q x x . (5.56)If N = 6 the distinguished pre-Nichols algebra is presented by x = 0; [ x , x ] c = 0; x = 0; x = 0; x = 0;[ x , [ x , x ] c ] c = q q q [ x , x ] c − ( q − − q − ) q q x x . (5.57) Proposition 5.47. If q has Dynkin diagram (5.51 c), then e B q is eminent.Proof. Let B be a finite GKdim pre-Nichols algebra of q . Then x = 0 by Lemma 3.1,and x = 0 by Lemma 3.3 if N = 4 or by Lemma 3.3 if N = 4. Now • If N = 6, then x = x = 0 by Lemma 3.5. • If N = 6, then x = 0 by Lemma 3.10, x = 0 by Lemma 3.9 and [ x , x ] c = 0by Lemma 3.15. RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 43
Finally, the defining relations of B q of degree lower than that of x u = [ x , [ x , x ] c ] c = q q q [ x , x ] c − ( q − − q − ) q q x x are satisfied in B , so x u ∈ P ( B ). Then x u = 0 by Lemma 3.27. (cid:3) The diagram (5.51 d) . A presentation of e B q is x = 0; x = 0; x = 0; x = 0; x (13) + q − q − q − q [ x , x ] c − q (1 − q ) x x = 0 . (5.58) Proposition 5.48. If q has Dynkin diagram (5.51,d), then e B q is eminent.Proof. This follows from Lemmas 3.1, 3.5 and 3.22. (cid:3) Eminent pre-Nichols algebras of standard type
Here we consider pre-Nichols algebras of standard (non-super) type [AA, § B θ or G , we study each case in a separate subsection.The main results can be summarized as follows: Theorem 6.1.
Let q be a braiding matrix of standard type. Then the distinguished pre-Nichols algebra e B q is eminent. Type B θ . Here q is a braiding of standard stype B θ . Up to relabeling I θ [AA, § A θ − ( − ζ ; J ) − ζ ζ ◦ , (6.1)where ζ ∈ G ′ . A set of defining relations of the distinguished pre-Nichols algebra is x ij = 0 , i < j −
1; [ x ( i − i +1) , x i ] c = 0 , i ∈ J ; x ii ( i ± = 0 , i ∈ I θ − − J ; [ x θθ ( θ − θ − , x θ ( θ − ] c = 0; x i = 0 , i ∈ J ; [ x θθ ( θ − , x θ ( θ − ] c = 0 , θ − ∈ J ; x θ = 0 . (6.2)In the following lemmas B denotes a finite GKdim pre-Nichols algebra of q . Lemma 6.2. If i, j ∈ I θ are such that i < j − , then x ij = 0 in B .Proof. Assume first i +1 < j < θ . If q ii q jj = 1 we have x ij = 0 by Lemma 3.3. If q ii q jj = 1,since e q ij +1 e q jj +1 = e q jj +1 = 1, we get x ij = 0 from Lemma 3.4.If i ∈ I θ − , since q θθ = −
1, Lemma 3.3 gives x iθ = 0. (cid:3) Lemma 6.3. If i ∈ I θ − − J , then x iii ± = 0 in B .Proof. This follows from Lemma 3.5, as c ij = − q ii = 6 = 2 , (cid:3) Lemma 6.4. If i ∈ J , then [ x ( i − ,i +1) , x i ] c = 0 in B . Proof.
Notice that Lemmas 3.12, 6.2 and 6.3 assure x u := [ x ( i − ,i +1) , x i ] c ∈ P ( B ). Supposethat x u = 0. By Lemma 3.13we have i ± / ∈ J . Moreover, if i < θ − i + 2 ∈ I θ , q i − i − q i +1 i +1 = 1, e q i +1 i +2 = 1, e q i − i +2 = e q ii +2 = 1 and e q i − i = e q − ii +1 = ±
1, whichcontradicts Lemma 3.14. Thus i = θ −
1. Since θ − / ∈ J we have q θ − θ − = − ζ .Thus q uu = − e q u ( θ − = 1, e q u ( θ − = 1, e q θu = q − = 1. Summarizing, P ( B ) contains k x θ − ⊕ k x θ − ⊕ k x θ ⊕ k x u which has Dynkin diagram − ζ ◦ θ − − ζ − ◦ θ − − ζ ζ ◦ θ ζ ζ ◦ u . This diagram is not in [H2, Table 3] since • there are three vertices with labels = −
1, and • these vertices have pairwise different labels.So x u = 0 in B . (cid:3) Lemma 6.5. (a) Si θ − ∈ J , entonces [ x θθ ( θ − , x θ ( θ − ] c = 0 en B . (b) [ x θθ ( θ − θ − , x θ ( θ − ] c = 0 en B .Proof. From 3.1, 6.2, 6.3 and [An1, Lemma 5.9], it follows that[ x θθ ( θ − θ − , x θ ( θ − ] c ∈ P ( B ) , [ x θθ ( θ − , x θ ( θ − ] c ∈ P ( B ) si θ − ∈ J . Thus the desired relations follow from Lemmas 3.15 and 3.16. (cid:3)
Theorem 6.6.
Let q be a braiding of standard type B θ . Then the distinguished pre-Nicholsalgebra is eminent.Proof. Follows from Lemmas 3.1, 6.2, 6.3, 6.4 and 6.5. (cid:3)
Type G . There are three different possibilities for the Dynkin diagram of a braiding q with root system of standard type G , namely q ◦ q q − ◦ , q ◦ q − ◦ , q ◦ q − ◦ , (6.3)where q is a root of unity of order 8 [AA, § The generalized Dynkin diagram (6.3 a) . Lemma 6.7.
Assume q is of type q ◦ q q − ◦ with q ∈ G ′ . (a) The Nichols algebra is minimally presented by generators x , x and relations x , x , [ x , x ] c , x , x . (6.4)(b) The distinguished pre-Nichols algebra is eminent.Proof. (a) By [An2, Theorem 3.1], B q is presented by the relations (6.4) and[ x , x ] c + q q x , [ x , x ] c , [ x , x ] c , RE-NICHOLS ALGEBRAS OF SUPER AND STANDARD TYPE 45
We verify using
GAP that this last bunch of relations can be deduced from x , x and[ x , x ] c . For the minimality, it is enough to show that [ x , x ] c does not belongto the ideal of T ( V ) generated by x and x , which is again checked with GAP .(b) Notice first that the previous computation provides a minimal presentation: e B q = T ( V ) / h x , x , [ x , x ] c i . If B is a finite GKdim pre-Nichols of q , then x = 0 and x = 0 by Lemmas 3.1 and 3.5,respectively. Thus [ x , x ] c ∈ P ( B ). It must be [ x , x ] c = 0 by Lemmas 2.7 and2.6, because the Dynkin diagram of k [ x , x ] c + k x ⊂ P ( B ) is ◦ q q ◦ . (cid:3) The generalized Dynkin diagram (6.3 b) . By [AA, § e B q = T ( V ) / h x , x , [ x , x ] c + q (1 − q ) − x i . Lemma 6.8. If q is of type q ◦ q − ◦ with q ∈ G ′ , then e B q is eminent.Proof. If B is a finite GKdim pre-Nichols, then x = x = 0 in B by Lemma 3.1. Then x u := [ x , x ] c + q (1 − q ) − x is primitive in B , and it must vanish because theDynkin diagram of k x u + k x ⊂ P ( B ) is ◦ q q ◦ . (cid:3) The generalized Dynkin diagram (6.3 c) . By [AA, § e B q = T ( V ) / h x , x , [ x , x ] c i . Lemma 6.9. If q is of type q ◦ q − ◦ with q ∈ G ′ , then e B q is eminent.Proof. If B is a finite GKdim pre-Nichols, then x = x = 0 in B by Lemmas 3.5 and3.1. Then [ x , x ] c is primitive in B , and it must vanish because the Dynkin diagramof k [ x , x ] c + k x ⊂ P ( B ) is ◦ q q ◦ . (cid:3) References [A1] N. Andruskiewitsch.
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FaMAF-CIEM (CONICET), Universidad Nacional de C´ordoba, Medina Allende s/n, CiudadUniversitaria, 5000 C´ordoba, Rep´ublica Argentina
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