Pointed Hopf Algebras as Crossed Product*①

REN Bei-shang,WEI Yu-qiu,ZHANG Xiao-ting,CHEN Yi-jie

(1.Software Engineering Institute of Guangzhou,Guangzhou 510990,China;2.Nanning Normal University,Nanning 530001,China;3.Guangxi University Foreign Languages,Nanning 530222,China)

Abstract：This paper mainly constructed pointed Hopf algebra H(1)#σK[G/N]from grouplike coalgebra G(H)of Hopf algebra H,group algebra and crossed product.And it further characterized structure of indecomposable components for a grouplike element of pointed Hopf algebra and discussed indecomposable properties of pointed Hopf algebras.

Key words：pointed Hopf algebra;crossed product;wedge product;indecomposable component;filtrations of coalgebras

1 Introduction

Throughout this paperKis a field andHis aK-Hopf algebra andCis aK-coalgebra.There is a convenient adaptation of the Heyneman-Sweedler[1]singma notation for coalgebras and comodules asΔ(c)=∑c(1)?c(2)andρ(c)=∑c(-1)?c(0),?c∈C.Crossed products can be viewed as generalizations of smash products and provide interesting examples of algebras in their own right.Pointed Hopf algebras have a crossed product decomposition into an indecomposable Hopf subalgebra and a group algebra.Here we go back to cross products and establish the decomposition result in the next.

LetCbe a coalgebra andAan algebra overK.A measuring ofCandAis a linear mapC?A→A(Herec?a→c·a) such that

c·1=ε(c)1andc·(ab)=(c(1)·a)(c(2)·b)for allc∈Canda,b∈A.

When there is such a mapCmeasuresAto itself.

LetHbe a bialgebra overK.SupposeH?A→Ais a measuring andσ:H×H→Ais bilinear.Consider the vector spaceA#σH=A?HoverKwith product defined by

(a?h)(b?l)=a(h(1)·b)σ(h(2),l(1))?h(3)l(2)

(1.1)

for alla,b∈Aandh,l∈H.

Remark1.1 SupposeAis a leftH-module algebra.The module action is a measuring.Ifσ(h,l)=ε(h)ε(l)for allh,l∈H,then (1.1) is the smash product multiplication.If in additionh·a=ε(h)afor alla∈Aandh∈H, then (1.1) is the tensor product of algebras multiplication.By [2] we would like to know whenA#σHwith its product is an associative algebra with unity 1?1=1A?1A.

Theorem1.2[3]LetHbe a bialgebra andAbe an algebra overK.SupposeH?A→Ais a measuring andσ:H×H→Ais bilinear.ThenA#σHis an associative algebra with product given by (1.1) and unity1A?1Hif and only if:

(1)σ(h,1)=ε(h)1=σ(1,h);

(2) 1·a=a;

(3) (h(1)·σ(l(1),m(1)))σ(h(2),l(2)m(2))=σ(h(1),l(1))σ(h(2)l(2),m);

(4) (h(1)·(l(1)·a))σ(h(2),l(2))=σ(h(1),l(1))((h(2)l(2))·a)，

for allh,l,m∈H,a∈A.Whenσis convolution invertible the condition (4) is equivalent to

(5)h·(l·a)=σ(h(1),l(1))((h(2)l(2))·a)σ-1(h(3),l(3)) for allh,l,m∈H,a∈A.

The conditions of parts (1) and (3) of the preceding theorem can be rephrased “σis a 2-cycle” and parts (2) and (5) can be expressed “Ais a twisted leftH-module”.

We apply Theorem 1.2 whenAis a Hopf algebra andHis a group algebra,hence is cocommutative.In our applicationA#σHis a Hopf algebra with the tensor product coalgebra structure.WhenAis a bialgebra overKandHis cocommutative there are natural necessary and sufficient conditions forA#σHto be a bialgebra with the tensor product coalgebra structure.

Corollary1.3[4]LetA,Hbe bialgebras overKwhereHis cocommutative.Suppose the hypothesis and parts (1)-(4) of Theorem 1.2 are satisfied and letA#σHbe the algebra described therein.ThenA#σHis a bialgebra with the tensor product coalgebra structure if and only if

(1)ε(h·a)=ε(h)ε(a)andε(σ(h,l))=ε(h)ε(l);

(2)Δ(h·a)=h(1)·a(1)?h(2)·a(2);

(3)Δ(σ(h,l))=σ(h(1),l(1))?σ(h(2),l(2)),

for allh,l∈Handa∈A.

The wedge product of subspaces ofCcorresponds to multiplication of subspaces in the dual algebraC*.In the fact,we have the following definition .

Definition1.4[5]Subspace thatCis a coalgebra overK.The wedge product of subspacesUandVofCisU∧V=Δ-1(U?C+C?V).

LetU,Vbe subspaces ofCand letπU:C→C/U,πV:C→C/Vbe the projections.

An equivalent description of the wedge product ofUandVisU∧V=Ker((πU?πV)°Δ).

IfUandVare subcoalgebras ofC.Then we haveU,V?V∧Uand henceU,V?U∧V.It’s more than thatU+V?U∧VandU+V?V∧U.Considering whether or not one of these inclusions is proper leads to an interesting circle of ideas related to the indecomposable subcoalgebras ofC.

Definition1.5[5]Suppose thatU,Vare subcoalgebras of a coalgebraC.ThenUandVare directly linked inCifU+Vis a proper subset ofU∧V+V∧U.

Two simple subcoalgebrasS,S′ofCare said to be linked inCifS=S′or for somen≥1 there are simple subcoalgebrasS=S0,…,Sn=S′ofCsuch thatSiandSi+1are directly linked inCfor all 0≤i

Definition1.6[1，6](1) A grouplike elements ofCis ac∈Cwhich satisfies the following conditions

Δ(c)=c?candε(c)=1.

(1.2)

The set of grouplike elements ofCis denoted byG(C).

(2) A skew primitive element ofCis ac∈Cwhich satisfiedΔ(c)=g?c+c?h,whereg,h∈G(C).Then set of skew primitive elements ofCis denoted byPg,h(C).

WhenCis pointed its simple subcoalgebras are theKg′s,whereg∈G(C).

2 Main results

Lemma2.1 Suppose thatCandDare coalgebras ofKandU,Vare subspaces ofC.Iff:C→Dis a map of coalgebras overK.Then

(1)f(U∧V)?f(U)∧f(V).

(2) Suppose thatfis onto and Ker(f)?U∩V.Thenf(U∧V)=f(U)∧f(V).

ProofPart (1) is easy to see.To show part (2) we need show thatf(U)∧f(V)?f(U∧V).Letd∈f(U)∧f(V).Sincefis ontoΔ(d)=(f?f)(v) for somev∈U?C+C?Vandd=f(c) for somec∈C.Sincefis a coalgebra map (f?f)Δ(c)=Δf(c)=Δ(d)=(f?f)(v) which means thatΔ(c)-v∈Ker(f?f).By assumption Ker(f)?U∩V.Thus Ker(f?f)=Ker(f)?C+C? Ker (f)?U⊕C+C?Vfrom which we deduceΔ(c)=C?U+C?V.We have shownc∈U∧Vand thusf(U)∧f(V)?f(U∧V).■

Lemma2.2 Suppose thatf:C→Corf:C→Ccopis a coalgebra isomorphism.LetSandS′be simple subcoalgebras ofC.Thenf(S),f(S′) are simple subcoalgebras ofD(whereD=CorD=Ccop) andS～S′impliesf(S)～f(S′).

ProofThatf(S)，f(S′) are simple subcoalgebras ofDis easy to see.LetU,Vbe subspaces ofC.Thenf(U∧V)?f(U)∧f(V) by part (1) of Lemma 2.1.Sincef-1is a coalgebra isomorphismf(U)∧f(V)=f(f-1(f(U)∧f(V)))?f(f-1(f(U))∧f-1(f(V)))=f(U∧V).Thereforef(U∧V)=f(U)∧f(V).Note thatV∧CcopU=U∧CV.Thus iff:C→Dcop,or equivalentlyf:Ccop→D,is a coalgebra isomorphism,f(U∧V)=f(V)∧f(U).The lemma follows from these general observations. ■

Lemma2.3 Supposeg,h∈G(C).ThenKgandKhare directly linked inCif and only if there is anx?K(g-h) such thatΔ(x)=g?x+x?horΔ(x)=h?x+x?g.

ProofKg∧Kh=Kg⊕Pg,h=Pg,h⊕KhbyKg∧Kh=Pg,h(C)⊕Kg=Pg,h(C)⊕Kh.■

Forg∈G(H)letH(g)be the indecomposable component containingKg.

Theorem2.4 LetHbe a pointed Hopf algebra with antipodeSoverKandG=G(H).Then we have the following properties.

(i)G(Hg)={h∈G|h～g}for allg∈G.

(ii)N=G(H(1))is a normal subproup ofG.

(iii)gH(h)=H(gh)=H(g)hfor allg,h∈G.

(iv)S(H(g))=H(g-1)for allg,h∈G.

(v)H(g)H(h)?H(gh)for allg∈G.

Proof(1) We have noted that the set of simple subcoalgebras in an indecomposable component of a coalgebra is the equivalence class of any of the simples in the component.

SinceKg?H(g)part (i) follows.

(2) We first observe that left or right multiplication by a grouplike element ofHis a coalgebra automorphism ofHand thus preserves the equivalence relation ～ by Lemma 2.2.

Letg,h,s∈G.As a consequence

g～himpliessg～shandgs～hs.

(2.1)

NowS:H→Hcopis a coalgebra anti-automorphism.Thusg～himpliesS(g)～S(h);hence

g～himpliesg-1～h-1.

(2.2)

Part (ii) now follows from part (1),(2.1),and (2.2).

(3) We have noted that the coalgebra automorphisms,in particular left and right multiplication by grouplike elements,and coalgebra anti-automorphisms,of whichSis one,permute the indecomposable components ofH.AsKghis a subset ofgH(h),H(g)h, andH(gh), these three indecomposable components are the same and part (iii) follows.

(4) SinceKg-1is a subset of the indecomposable componentS(H(g)),Part (iv) follows.

(5) First of all note that the restriction of multiplicationπ:H(g)?H(h)→H(g)H(h)is an onto coalgebra map.NowH(g)?H(h)is pointed and

G(H(g)?H(h))={s?t|s∈G(H(g)),t∈G(H(h))}

by part (i) of the sameπ((H(g)?H(h))0)?(H(g)H(h))0.Thus

G(H(g)H(h))={st|s∈G(H(g)),t∈G(H(h))}.

by part (i) and (2.1) all of the grouplike elements ofH(g)H(h)belong to the same equivalence class.ThusH(g)H(h)is link indecomposable which means it is a subspace of an indecomposable component ofC.SinceKgh∈H(g)H(h),H(gh), necessarilyH(g)H(h)?H(gh).■

ProofLetg,h∈G.ThenH(g)=H(h)iffg～h.The latter is the case iffg-1h～1 by(2.1).Then

gN={h∈G|H(h)=H(g)}

(2.3)

(2.4)

for alla∈H(1)andi∈I.Observe thatfis a coalgebra structure.Alsof(1)=1H(1)?1K[G/N].

By part (iv) and (v) of Theorem 2.4,H(1)is a Hopf subalgebra ofH.By part (iii) again,gH(1)g-1=H(1).Obverse thatK[G/N] measuresH(1)to itself by

(2.5)

for alli∈Ianda∈H(1).

(2.6)

ConsiderH(1)#σK[G/N] =H(1)?K[G/N]with the product defined by (1.1).Fora,b∈H(1)the calculation

shows thatfis multiplicative.ThusH(1)#σK[G/N]is a Hopf algebra. ■

Lemma2.6 LetHbe a pointed Hopf algebra overK.ThenN=G(H(1)) is generated by thoseg∈G(H) such that there is anx?K(g-1) such thatΔ(x)=1?x+x?g.

ProofSupposeg∈G(H) satisfies the condition of the lemma,thenKgandK1 are directly linked inH.Thereforeg～1 which meansg∈NwhereN={g∈G(H)|g～1} by part (i) of Theorem 2.4.Thus the subgroupLofG(H) generated by theg∈G(H) which satisfy the condition of the lemma is a subgroup ofN.

Now supposeg,h∈G(H) are directly linked inH.Thenhg-1orgh-1satisfies the condition of the lemma by Lemma 2.3.Thereforehg-1∈Lorgh-1∈L; consequentlyhg-1,gh-1∈LsinceLis a subgroup.Our conclusion: eitherg,h∈Lorg?L,h?L.

Letg∈N.Theng=1 or for somen≥1 there are grouplike elements 1=g0,…,gn=gsuch thatKgiandKgi+1are directly linked inHfor all 0≤i

Theorem2.7 LetHbe a pointed Hopf algebra overKwhich is generated as an algebra byg1,…,gn∈G(H),their inverses,andx1,…,xn∈Hwhich satisfyΔ(xi)=1?xi+xi?giandxi?K(gi-1) for all 1≤i≤n.ThenHis indecomposable.

ProofLetV0be the span ofg1,…,gnand their inverses and letV1be the span ofV0andx1,…,xn.ThenV0,V1are subcoalgebras ofHandΔ(V1)?V0?V1+V1?V0.ThereforeV0andV1are the first two terms in a filtration ofC=V1.ThusC0?V0by Lemma 1.7 which meansC0=V0.SinceCgeneratesHas an algebraH0,is a subspace of the subalgebra generated byC0.ThereforeG(H) is generated byg1,…,gnas a group.Nowg1,…,gn∈G(H(1)) by Lemma 2.6.ThereforeG(H)=G(H(1))=Nwhich means thatHhas one indecomposable component by (2.3).