Yanhua Wang
(School of Mathematics, Shanghai University of Finance and Economics)
LetHbe a Hopf algebra (bialgebra),a left-left Yetter-Drinfeld module over Hopf algebra (bialgebra)His ak-linear spaceVwhich is a leftH-module,a leftH-comodule and satisfies a certain compatibility condition.Yetter-Drinfeld modules were introduced by Yetter in [1] under the name of "crossed bimodule".Radford proved that pointed Hopf algebras can be decomposed into two tensor factors,one factor of the two factors is no longer a Hopf algebra,but a rather a Yetter-Drinfeld Hopf algebra over the other factor[2].Subsequently,Schauenburg proved that the category of Yetter-Drinfeld module overHwas equivalent to the category of left module over Drinfeld double,and also to the category of Hopf module overH[3],and Sommerhauser studied Yetter-Drinfeld Hopf algebra over groups of prime order[4].
Some conclusions of Hopf algebras can be applied to Yetter-Drinfeld Hopf algebras.For example: Doi considered the Hopf module theory of Yetter-Drinfeld Hopf algebras in [5],Scharfschwerdt proved the Nichols Zoeller theorem for Yetter-Drinfeld Hopf algebras in [6],and Andruskiewitsch and Schneider gave the trace formula for Yetter-Drinfeld Hopf algebras in [7].
In this paper,we generalized the antipode properties of Hopf algebras to Yetter-Drinfeld Hopf algebras.We proved the antipode of a Yetter-Drinfeld Hopf algebra is an anti-algebra and anti-coalgebra map,see Proposition 1 and Proposition 2.We study the tensor algebra of Yetter-Drinfeld module,and show that the tensor algebra of Yetter-Drinfeld module is a Yetter-Drinfeld Hopf algebra under a tensor multiplication and a "twisted" comultiplication,see Theorem 4.
In the following,kwill be a field.All algebras and coalgebras are overk.All unadorned ? are taken overk.
(ab) →v=a→ (b→v), 1 →v=v.
The arrow → denotes left module action.The category of leftA-module is denoted byAM.
Let (C,△,) be a coalgebra.A leftC-comodule is ak-vector spaceVtogether with ak-linear mapρ:VC?V:v∑v-1?v0such that
∑v-2?v-1?v0=∑v-1?(v0)-1?(v0)0, ∑(v-1)v0=v.
The category of leftC-module is denoted byCM.
Let (H,m,u,△,,S) be a Hopf algebra with antipodeS.A left Yetter-Drinfeld module overHis ak-vector spaceVwhich is both a leftH-module and leftH-comodule and satisfies the compatibility condition
∑(h→v)-1?(h→v)0=∑h1v-1Sh3?h2→v0,
(1)
(a1)Ais a leftH-module algebra,i.e.,
h→(ab)=∑(h1→a)(h2→b),h→1A=(h)1A.
(a2)Ais a leftH-comodule algebra,i.e.,
ρ(ab)=∑(ab)-1?(ab)0=∑a-1b-1?a0b0,
ρ(1A)=1H?1A.
(a3)Ais a leftH-module coalgebra,i.e.,
△(h→a)=∑(h1→a1)?(h2→a2),(h→a)=H(h)A(a).
(a4)Ais a leftH-comodule coalgebra,i.e.,
∑a-1?(a0)1?(a0)2=∑a1-1a2-1?a10?a20,
∑a-1A(a0)=A(a)1H.
△°m(a?b)=(m?m)(id?τ?id)(△?△)(a?b)=
∑a1(a2-1→b1)?a20b2,
Ais called a Yetter-Drinfeld Hopf algebra or Hopf algebras in Yetter-Drinfeld category if it has an antipodeSthat is a convolution inverse to id,i.e.,
One easily see thatSis bothH-linear andH-colinear.In general,Yetter-Drinfeld Hopf algebras are not ordinary Hopf algebras because the bialgebra axiom asserts that they obey (a5).However,it may happen that Yetter-Drinfeld Hopf algebras are ordinary Hopf algebras when the pre-braiding is trivial,for details see [4].
Next,we give a basic property of Yetter-Drinfeld Hopf algebra.It is well know that the antipode of a Hopf algebra is an anti-algebra and anti-coalgebra map,see [8-10].This is also true for Yetter-Drinfeld Hopf algebra.The following lemma give the character.
S1)S(ab)=∑(a-1→S(b))S(a0),andS(1)=1.
ProofIfSis an anti-algebra automorphism,then we haveSm=m(s?s)τands(1)=1.We take the idea of Sweedler in [10,P.74].
We have
(Sm*m)(a?b)=
m(Sm?m)△(a?b)=
m(Sm?m)(∑a1?a2-1→b1?a20?b2)=
S(∑a1(a2-1→b1))?a20b2=
∑S((ab)1)(ab)2=
u(a?b).
On the other hand
(m*m(S?S)τ)(a?b)=
m(m?m(S?S)τ)△(a?b)=
m(m?m(S?S)τ)(∑a1?a2-1→b1?a20?b2)=
m∑[a1(a2-1→b1)?m(S?S)τ(a20?b2)]=
m∑[a1(a2-2→b1)?m(S?S)(a2-1→b2?a20)]=
m∑[a1(a2-2→b1)?(a2-1→S(b2))S(a20)]=
∑a1(a2-2→b1)(a2-1→S(b2))S(a20)]=
∑a1[a2-1→(b1S(b2))]S(a20)=
∑a1(a2-1→(b))S(a20)=
∑a1(a2-1)(b)S(a20)=
∑u(b)a1S(a2)=u(b)u(a)=u(a?b) .
ThusSm*m=m*(m(s?s)τ)=u,henceSm=m(s?s)τ.By(1)=1 and △(1)=1?1,we have (S*id)(1)=S(1)1=u(1)=1.SoS(1)=1.
The proof ofSis an anti-coalgebra automorphism is similar to the proof ofSis an anti-algebra automorphism.
ProofBy assumption,it suffice to prove that if (id*S)(a)=u(a) and (S*id)(b)=u(b),then (id*S)(ab)=u(ab),a,b∈X.
By △(ab)=∑a1(a2-1→b1)?a20b2, we have
(id*S)(ab)=∑a1(a2-1→b1)S(a20b2)=
∑a1S(a2)
u(a)u(b)=
u(ab) .
This complete the proof.
Assume thatVis a vector space,thenTV=V?V?V?… ?V?n=T0V?T1V?T2V?…TnVbecomes an algebra with the connected multiplication
(v1?v2?…?vs)(w1?w2?…?wt)=v1?v2?…?vs?w1?w2?…?wt.
(2)
We have the following lemma about the tensor algebra of a Yetter-Drinfeld module.
Lemma3LetVbe a left Yetter-Drinfeld module,thenTVis also a left Yetter-Drinfeld module.
ProofDefine the module action and comodule action as follows
h→(v1?v2?…?vn)=∑(h1→v1)?(h2→v2)?…?(hn→vn)
(3)
and
ρ(v1?v2?…?vn)=∑(v1-1v2-1…vn-1)?v10?v20?…?vn0,
(4)
?h∈H,v1?v2?…?vn∈TV.
It is easy to proveTVis a leftH-module.For anyh,g∈H,v1?v2?…?vn∈TV.We have 1→(v1?v2?…?vn)=v1?v2?…?vnand
h→(g→(v1?v2?…?vn))=
∑(h1→g1→v1)?(h2→g2→v2)?…?(hn→gn→vn)=
∑(h1g1→v1)?(h2g2→v2)?…?(hngn→vn)=
(hg)→(v1?v2?…?vn).
TVis a leftH-comodule.Since
(?id)ρ(v1?v2?…?vn)=
v1?v2?…?vn
and
(△?id)ρ(v1?v2?…?vn)=
(△?id)(∑v1-1v2-1…vn-1?v10?v20?…?vn0)=
∑(v1-2v2-2…vn-2)?(v1-1v2-1…vn-1)?(v10?v20?…?vn0)=
(id?ρ)ρ(v1?v2?…?vn)
∑(h→(v1?v2?…?vn))-1?(h→(v1?v2?…?vn))0=
∑(h1→v1?h2→v2?…?hn→vn)-1?(h1→v1?h2→v2?…?hn→vn)0=
∑(h1→v1)-1?(h2→v2)-1?…?(hn→vn)-1?(h1→v1)0?(h2→v2)0?…?(hn→vn)0=
∑(h1v1-1S(h3)h4v2-1S(h6)…S(h3n+1)vn-1S(h3n+3)?h2→v10?h5→v20?…?h3n+2→vn0=
∑(h1v1-1v2-1…vn-1S(hn+2))?h2→v10?h3→v20?…?hn+1→vn0=
∑(h1v1-1v2-1…vn-1S(h3))?h2→(v10?v20?…?vn0)=
∑(h1(v1?v2?…?vn)-1S(h3))?h2→(v1?v2?…?vn)0.
This complete the proof.
Theorem4IfVis a Yetter-Drinfeld module over Hopf algebraH,then the tensor algebraTVofVis a Yetter-Drinfeld Hopf algebra overH.
ProofFor anyx=v1?v2?…?vs∈TsV,y=w1?w2?… ?wt∈TtV,define the multiplication ofxyas the tensor multiplication
xy=v1?v2?…?vs?w1?w2?…?wt.
which show that △ is coassociative.
By the comultiplication ofV,forv?w∈T2V,we have
△(v?w)=(m?m)(id?τ?id)(△(v)?△(w))=
(m?m)(id?τ?id)((1?v+v?1)?(1?w+w?1)=
Forv?w?t∈T3V,the comultiplication ofT3Vis
In general,we denote
?v2?…?vn) =
Using the above notation,the comultiplication ofTVis
DefineS(v)=-v,v∈V,and
For anyv∈V,we havem(S?id)△(v)=s(1)v+vS(1)=0=(v) andm(id?S)△(v)=s(v)+vS(1)=0=(v).Therefore,the property of antipode is satisfied for generators ofTV.By Lemma 2,Sis the antipode ofTV.This completes the proof.
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