SCIENCE CHINA Mathematics . . ARTICLES January 2015 Vol.58 No.1: 1–xx doi:xx.xxxx/sxxxxx-xxx-xxxx-x Dual Lie Bialgebra Structures of Poisson Types 5 1 0 Guang’ai Song1 & Yucai Su2,∗ 2 n a J 1College of Mathematics and Information Science, Shandong Institute of Businessand Technology , Yantai 264005, China; 2Department of Mathematics, Tongji University,Shanghai 200092,China 1 Email: [email protected], [email protected] 3 ReceivedOctober11,2014;accepted January22,2015;publishedonline2015 ] A Q Abstract Let A = F[x,y] be the polynomial algebra on two variables x,y over an algebraically closed field . F of characteristic zero. Under the Poisson bracket, A is equipped with a natural Lie algebra structure. It is h t proven thatthe maximalgoodsubspace of A∗ induced fromthemultiplicationofthe associativecommutative a algebraAcoincideswiththemaximalgoodsubspaceofA∗ inducedfromthePoissonbracketofthePoissonLie m algebraA. Basedonthis,structuresofdualLiebialgebrasofthePoissontypeareinvestigated. Asby-products, [ fiveclassesofnewinfinitedimensionalLiealgebrasareobtained. 1 v Keywords Poissonalgebra,Virasoro-likealgebra,Liebialgebra,dualLiebialgebra,goodsubspace 9 4 MSC(2010) 17B62,17B05,17B06 0 0 0 Citation: SongGA,SuYC. ScienceChina: Mathematicstitle. SciChinaMath,2014, 57, doi: 10.1007/s11425- . 000-0000-0 2 0 5 1 Introduction 1 : v i Lie bialgebras,having close relationswith Yang-Baxterequations [6], are important ingredients in quan- X tum groups, which have drawn more and more attentions in literature (e.g., [2,3,5,8–17,26,28,29]). r Michaelis [10] investigated structures of Witt type Lie bialgebras. Ng and Taft [16] gave a classification a of this type Lie bialgebras, and obtained that all structures of Lie bialgebras on the one sided Witt algebra, the Witt algebra and the Virasoro algebra are coboundary triangular (cf. [15]). For the cases of generalized Witt type Lie algebras and generalized Virasoro-like Lie algebras, the authors of [17,28] proved that all structures of Lie bialgebras on them are coboundary triangular. Similar results hold for some other kinds of Lie algebras (cf., e.g., [28,29]). FromtheexamplesofinfinitedimensionalLiebialgebrasconstructedin[10],manyinfinitedimensional Lie bialgebra structures we know are coboundary triangular. It may sound that coboundary triangular Lie bialgebras are relatively simple. However, they are not trivial in the sense that many natural prob- lems associated with them remain open (see, also Remark 4.7). For example, even for the (two-sided) Witt algebra and the Virasoro algebra, a complete classification of coboundary triangular Lie bialgebra structures on them is still an open problem. Nevertheless, not much on representations of infinite di- mensional Lie bialgebras is known. From the viewpoint of Lie bialgebras,considering dual Lie bialgebra structuresmayhelpusunderstandmoreoninfinitedimensionalLiebialgebrastructures. Forinstance,by considering structures of dual Lie bialgebras of Witt and Virasoro types, the authors of [18] surprisingly ∗Correspondingauthor (cid:13)c Science ChinaPressandSpringer-Verlag BerlinHeidelberg2014 math.scichina.com link.springer.com 2 SongGA,SuYC Sci ChinaMath January 2015 Vol.58 No.1 obtained some new series of infinite dimensional Lie algebras. In the present paper, we study structures of dual Lie bialgebras of Poisson type. One may have noticed that the dual of a finite dimensional Lie bialgebra is naturally a Lie bialgebra,and so one would not predict anything new in this case. However, for the cases of infinite dimensional Lie bialgebras, the situations become quite different, which can be seen in the following contents. Let us recall the definition of Poisson algebras here: a Poisson algebra is a triple (P,[·,·],·) such that (P,[·,·]) is a Lie algebra, (P,·) is an associative algebra, and the following Leibniz rule holds: [a,bc]=[a,b]c+b[a,c] for a,b,c∈P. (1.1) In particular, for any commutative associative algebra (A,·), and any commutative derivations ∂ ,∂ of 1 2 A, we obtain a Poisson algebra (A,[·,·],·) with Lie bracket [·,·] defined as follows. [a,b]=∂ (a)∂ (b)−∂ (a)∂ (b) for a,b∈A. (1.2) 1 2 2 1 If we take A = F[x±1,y±1] (where F is an algebraically closed field of characteristic zero) and ∂ = 1 x ∂ , ∂ = y ∂ , then we obtain the Virasoro-like algebra (A,[·,·]) with basis {xiyj|i,j ∈ Z} and Lie ∂x 2 ∂y bracket [xiyj,xkyℓ]=(iℓ−jk)xi+kyj+ℓ for i,j,k,ℓ∈Z. (1.3) TheVirasoro-likealgebra(1.3)canbegeneralizedasfollows: ForanynondegenerateadditivesubgroupΓ ofF2 (i.e.,ΓcontainsanF-basisofF2),wehavethe groupalgebraA=F[Γ]withbasis{L |α∈Γ}and α multiplication defined by µ(L ,L ) = L for α,β ∈ Γ. Then we have the (generalized) Virasoro-like α β α+β algebra (A,ϕ) with Lie bracket ϕ defined by ϕ(L ,L )=(α β −β α )L for α=(α ,α ), β =(β ,β )∈Γ. (1.4) α β 1 2 1 2 α+β 1 2 1 2 Furthermore, if we take A=F[x,y] and ∂ = ∂ , ∂ = ∂ , then we obtain the classical Poisson algebra 1 ∂x 2 ∂y (F[x,y],[·,·],·), whose Lie bracket is given by [f,g]=J(f,g) for f,g ∈F[x,y], (1.5) ∂f ∂f where J(f,g):= ∂x ∂y is the Jacobian determinant of f and g. ∂g ∂g ∂x ∂y The reason we(cid:12)have(cid:12)a special interest in the classical Poisson algebra also lies in the fact that this (cid:12) (cid:12) algebra is closely(cid:12)relate(cid:12)d to the distinguished Jacobian conjecture (e.g., [30,31]), which can be stated as “any non-zero endomorphism of (F[x,y],[·,·],·) is an isomorphism”. One observes that a Jacobi pair (f,g) (i.e., f,g ∈F[x,y] satisfying J(f,g)∈F\{0})correspondsto a solution r=f⊗fg−fg⊗f of the classical Yang-Baxter Equation (cf. (2.1)), thus gives rise to a Lie bialgebra structure on F[x,y]. The paper is organized as follows. Some definitions and preliminary results are briefly recalled in Section 2. Then in Section 3, structures of dual coalgebrasof F[x,y] are addressed. Finally in Section 4, structures of dual Lie bialgebras of Poissontype are investigated. The main results of the present paper are summarized in Theorems 3.2, 4.4, 4.6, 4.8 and 4.9. 2 Definitions and preliminary results Throughout the paper, all vector spaces are assumed to be over an algebraically closed field F of char- acteristic zero. As usual, we use Z to denote the set of nonnegative integers. We briefly recall some + notions on Lie bialgebras, for details, we refer readers to, e.g., [6,17]. Definition 2.1. 1. A Lie bialgebra is a triple (L,[·,·],δ) such that (L,[·,·]) is a Lie algebra, (L,δ) is a Lie coalgebra, and δ :L→L⊗L is a derivation, namely, δ[x,y]=x·δ(y)−y·δ(x) for x,y ∈L, where x·(y⊗z)=[x,y]⊗z+y⊗[x,z] for x,y,z ∈L. 2. A Lie bialgebra(L,[·,·],δ) is coboundary if δ is coboundaryin the sense that there exists r∈L⊗L written as r = r[1]⊗r[2], such that δ(x)=x·r for x∈L. P SongGA,SuYC SciChina Math January 2015 Vol.58 No.1 3 3. AcoboundaryLiebialgebra(L,[·,·],δ)istriangularifr satisfiesthefollowingclassical Yang-Baxter Equation (CYBE), C(r)=[r ,r ]+[r ,r ]+[r ,r ]=0, (2.1) 12 13 12 23 13 23 where r = r[1] ⊗ r[2] ⊗ 1, r = r[1] ⊗ 1 ⊗ r[2], r = r[1] ⊗ 1 ⊗ r[2] are elements in 12 13 23 U(L)⊗U(L)⊗U(L),and U(L) is the universal enveloping algebra of L. P P P Two Lie bialgebras(g,[·,·],δ) and (g′,[·,·]′,δ′) are saidto be dually paired if their bialgebrastructures are related via h[f,h]′,ξi=hf ⊗h,δξi, hδ′f,ξ⊗ηi=hf,[ξ,η]i for f,h∈g′, ξ,η ∈g, (2.2) where h·,·i is a nondegenerate bilinear form on g′×g, which is naturally extended to a nondegenerate bilinear form on (g′⊗g′)×(g⊗g). In particular, if g′ = g as a vector space, then g is called a self-dual Lie bialgebra. The following result whose proof is straightforwardcan be found in [9]. Proposition 2.2. Let (g,[·,·],δ) be a finite dimensional Lie bialgebra, then so is the linear dual space g∗ :=HomF(g,F) by dualisation, namely (g∗,[·,·]′,δ′) is the Lie bialgebra defined by (2.2) with g′ =g∗. In particular, g and g∗ are dually paired. Thus a finite dimensional Lie biallgebra (g,[·,·],δ) is always self-dual as there exists a vector space isomorphism g → g∗ which pulls back the bialgebra structure on g∗ to g to obtain another bialgebra structure on g to make it to be self-dual. However, in sharp contrast to the finite dimensional case, infinite dimensional Lie bialgebras are not self-dual in general. For convenience, we denote by ϕ the Lie bracket of Lie algebra (g,[·,·]), which can be regarded as a linear map ϕ:g⊗g→g. Let ϕ∗ :g∗ →(g⊗g)∗ be the dual of ϕ. Definition 2.3. [11] Let (g,ϕ) be a Lie algebra over F. A subspace V of g∗ is called a good subspace if ϕ∗(V) ⊂ V ⊗V. Denote ℜ = {V |V is a good subspace of g∗}. Then g◦ = V, is also a good V∈ℜ subspace of g∗, which is obviously the maximal good subspace of g∗. P It is clear that if g is a finite dimensional Lie algebra,then g◦ =g∗. Proposition 2.4. [11]ForanygoodsubspaceV ofg∗, the pair(V,ϕ∗) isa Liecoalgebra. Inparticular, (g◦,ϕ∗) is a Lie coalgebra. For any Lie algebrag, the dual space g∗ has a naturalright g-module structure defined for f ∈g∗ and x∈g by (f ·x)(y)=f([x,y]) for y ∈g. We denote f ·g=span{f ·x|x∈g}, the space of translates of f by elements of g. We summarize some results of [2,3,5,8]as follows. Proposition 2.5. Let g be a Lie algebra. Then 1. g◦ ={f ∈g∗|f ·g is finite dimensional}. 2. g◦ =(ϕ∗)−1(g∗⊗g∗), the preimage of g∗⊗g∗ in g∗. The notion of good subspaces of an associative algebra can be defined analogously. In the next two sections, we shall investigate g◦ for some associative or Lie algebras g. 3 The structure of F[x,y]◦ Let (A,µ,η) be an associative F-algebra with unit, where µ and η are respectively the multiplication µ:A⊗A→A and the unit η :F →A, satisfying µ◦(id⊗µ)=µ◦(µ⊗id)): A⊗A⊗A→A, (η⊗id)(k⊗a)=(id⊗η)(a⊗k): F ⊗A∼=A⊗F ∼=A, 4 SongGA,SuYC Sci ChinaMath January 2015 Vol.58 No.1 for k ∈ F, a ∈ A. Then a coassociative coalgebra is a triple (C,∆,η), which is obtained by conversing arrowsinthedefinitionofanassociativealgebra. Namely,∆:C →C⊗C andη :F →C arerespectively comultiplication and counit of C, satisfying (∆⊗id)◦∆=(id⊗∆)◦∆: C →C⊗C⊗C, (η⊗id)◦∆=(id⊗η)◦∆: C →C⊗C →F ⊗C ∼=C⊗F ∼=C. ForanyvectorspaceA,thereexistsanaturalinjectionρ:A∗⊗A∗ →(A⊗A)∗ definedbyρ(f,g)(a,b)= hf,aihg,bi for f,g ∈ A∗ and a,b,∈ A. In case A is finite dimensional, ρ is an isomorphism. If (A,µ) is associative,themultiplicationµinducesthemapµ∗ :A∗ →(A⊗A)∗. IfAisfinitedimensional,thenthe isomorphism ρ insures that (A∗,µ∗,η∗) is a coalgebra, where for simplicity, µ∗ denotes the composition of the maps: A∗ →µ∗ (A⊗A)∗ (ρ→)−1 A∗⊗A∗. Now let (A,µ) be a commutative associative algebra. Then A◦ = (µ∗)−1(A∗ ⊗ A∗) (cf. [26] and Proposition 2.5). For ∂ ∈Der(A) and f ∈A◦, using ∂µ=µ(id⊗∂+∂⊗id), µ∗∂∗(f)=(id⊗∂∗+∂∗⊗id)µ∗(f)∈A∗⊗A∗, weobtain∂∗(A◦)⊂A◦. Thus,weobservethattherearetwonaturalapproachestoproduceLiecoalgebras from some subspaces of A∗. One is induced from the associativestructure of A as follows: First we have the cocommutative coassociativecoalgebra(A◦,µ◦) with µ◦ :=µ∗|A◦. Then we obtion the Lie coalgebra A◦ := (A◦,∆) with cobracket, induced from cocommutative coassociative coalgebra structure, defined µ by ∆(f)=(∂◦⊗∂◦−∂◦⊗∂◦)µ◦(f) for f ∈A◦, (3.1) 1 2 2 1 where∂ ,∂ ∈DerAaretwofixedderivationssatisfying∂ ∂ =∂ ∂ .Hereandbelow,forany∂ ∈Der(A), 1 2 1 2 2 1 we denote ∂◦ =∂∗|A◦. Anotherapproachisasfollows: LetA =(A,[·,·])betheLiealgebradefinedin(1.2)(whereϕ=[·,·]). ϕ The Lie coalgebra induced from A is A◦ = (A◦,ϕ◦), where the subspace A◦ of A∗ is determined by ϕ ϕ ϕ ϕ Proposition 2.5 with cobracketdefined by ϕ◦(f)=(µ(∂ ⊗∂ −∂ ⊗∂ ))∗(f)=(∂ ⊗∂ −∂ ⊗∂ )∗µ∗(f) for f ∈A◦. (3.2) 1 2 2 1 1 2 2 1 ϕ Proposition 3.1. Let (A,µ) be a commutative associative algebra with unit, and ∂ ,∂ ∈ Der(A) are 1 2 commutative. Then the Lie coalgebra A◦ is a Lie subcoalgebra of A◦. µ ϕ Proof. For f ∈ A◦, we have ϕ◦(f) = (∂ ⊗ ∂ − ∂ ⊗ ∂ )∗µ∗(f) = (∂ ⊗ ∂ − ∂ ⊗ ∂ )∗µ◦(f) = µ 1 2 2 1 1 2 2 1 (∂◦⊗∂◦−∂◦∂◦)µ◦(f)=∆(f),wherethelastequalityfollowsfrom(3.1). Thus,A◦ isaLiesubcoalgebra 1 2 2 1 µ of A◦. ✷ ϕ Theorem 3.2. Let (A,µ) be a commutative associative algebra, and (A,ϕ) the Poisson Lie algebra definedin(1.2). Ifthereexistsh∈AsuchthattheidealI of(A,ϕ)generatedbyhhasfinitecodimension, then A◦ =A◦. In particularly, if A=F[x,y], then A◦ =A◦. µ ϕ µ ϕ Proof. Denote·and⋆theactionsof(A,µ)and(A,ϕ)onA∗ respectively,i.e.,(f·a)(b)=f(µ(a,b)),and (f⋆a)(b)=f(ϕ(a,b)) for a,b∈A,f ∈A∗. Fromthe relationϕ(a,bc)=ϕ(a,c)b+ϕ(a,b)c fora,b,c∈A, we have (f ⋆a)·b−(f ·b)⋆a=f ·ϕ(a,b), ∀a,b∈A. If f ∈ A◦, i.e. f ⋆A is finite dimensional, then f ·ϕ(b,A) is finite dimensional. Thus if the ideal I ϕ has finite codimension, and f ·I is finite dimensional, it follows that f ·A is finite dimensional. From Proposition 2.5, we have f ∈A◦. ✷ µ Remark 3.3. The difference between A◦ and A◦ is that A◦, as a Lie coalgebra, is induced from µ ϕ µ coassociative coalgebra (A◦,µ◦), and A◦ is determined by (µ∗)−1(A∗ ⊗A∗) as a vector subspace of A∗ µ (a good subspace of the dual of (A,µ)), but A◦ is the dual of the Lie algebra (A,ϕ) (determined by ϕ Proposition 2.5). SongGA,SuYC SciChina Math January 2015 Vol.58 No.1 5 4 Dual Lie bialgebras of Poisson type Poissonalgebras(cf.(1.1))haveimportantalgebrastructures,whichhavecloserelationswiththeVirasoro algebra and vertex operator (super)algebras (e.g., [1]). They can be also regardedas special cases of Lie algebrasofBlocktype. Therefore,someattentionshavebeenpaidonthemandsomerelatedLiealgebras (e.g., [4,7,19–25,27,28]). In this section, we consider the dual structures of Poissontype Lie bialgebras. The following result can be found in [26]. Proposition 4.1. Let A,B be commutative associative algebras, regarding A∗⊗B∗ ⊂ (A⊗B)∗, then A◦⊗B◦ =(A⊗B)◦. Recall from [17] that the dual space of F[x] can be identified with the space F[[ε]]. From [14,18], and Proposition 4.1, we have Proposition 4.2. 1. Let f = ∞ f εi ∈F[[ε]] with f ∈F. Then i=0 i i f∈F[x]◦ ⇐⇒ fn=h1Pfn−1+h2fn−2+···+hrfn−r for some r ∈N, hi∈F and all n>r g(ε) ⇐⇒ f ∈ g(ε),h(ε)∈F[ε], h(0)6=0 . h(ε) n (cid:12) o (cid:12) 2. Denote A=F[x,y], the polynom(cid:12)ial algebra on tow variables x,y. Then A◦ =F[x,y]◦ ∼=F[x]◦⊗F[y]◦. Let (g,ϕ,δ) be a Lie bialgebra. The map ϕ∗ induces a map ϕ◦ := ϕ∗|g◦ : g◦ → g◦ ⊗g◦, making (g◦,ϕ◦) to be a Lie coalgebra. By [15, Proposition 3], the map δ∗ : g∗⊗g∗ ֒→ (g⊗g)∗ →δ∗ g∗ induces a map δ◦ := δg∗◦⊗g◦ : g◦⊗g◦ → g◦, making (g◦,δ◦) to be a Lie algebra. Thus we obtain a Lie bialgebra (g◦,δ◦,ϕ◦), the dual Lie bialgebra of (g,ϕ,δ). NowtakeA=F[x,y]. Letεi,ηi bedualsofxi,yi ∈Arespectively,namely,hεiηj,xkyli=εiηj(xkyl)= δ δ for i,j,k,l ∈ Z . Any element u ∈ A∗ can be written as u = u εiηj (possibly an infinite i,k j,l + i,j i,j sum). Let g = g xkyl ∈A (a finite sum). Then k,l k,l P Phu,gi=u(g)= u g hεiηj,xkyli= u g δ δ (a finite sum). (4.1) i,j k,l i,j k,l i,k j,l i,j,k,l i,j,k,l P P Let∂ = ∂ , ∂ = ∂ . Thenwehavethe PoissonLie algebra(A,ϕ) definedby (1.5). FromTheorem3.2, 1 ∂x 2 ∂y it is easy to check that A◦ =A◦. ϕ µ Convention 4.3. (1) If an undefined notation appears in an expression,we treat it zero; for instance εiηj =0 if i<0 or j <0. (2) When there is no confusion, we use [·,·] to denote the bracket in g or g◦, i.e., [·,·] = ϕ or δ◦. We also use ∆ to denote the cobracket in g or g◦, i.e., ∆=δ or ϕ◦. Let m,n ∈ Z , and take a = xmyn, b = xy ∈ A. Then [a,b] = (m−n)a. Thus by [10], the triple + (A,[·,·],∆ ) with r = a⊗b−b⊗a is a coboundary triangular Lie bialgebra whose cobracket is defined r by ∆ (f)=f ·r =[f,a]⊗b+a⊗[f,b]−[f,b]⊗a−b⊗[f,a] for f ∈A. (4.2) r Theorem 4.4. Let(A,[·,·],∆ )bethe coboundarytriangularLiebialgebradefinedabove. ThedualLie r bialgebra of A is (A◦,[·,·],∆), where A◦ is described by Proposition 4.2(2) with cobracket∆ defined by ∆(εmηn)= (kt−ls)εkηl⊗εsηt, (4.3) k+s=m+1, l+t=n+1 P and bracket [·,·] uniquely determined by the skew-symmetry and the following (m(t+1)−n(s+1))εs+1−mηt+1−n if (i,j)=(1,1), (s,t)6=(1,1), [εiηj,εsηt]= (s−t)εsηt if (i,j)=(m,n)6=(s,t)6=(1,1), (4.4)  0 otherwise.  6 SongGA,SuYC Sci ChinaMath January 2015 Vol.58 No.1 Proof. Assume µ◦(εmηn)= c εkηl⊗εsηt for some c ∈F. Then k,l,.s,t∈Z+ k,l,s,t k,l,s,t P c =µ◦(εmηn)(xiyj ⊗xpyq)=hεmηn,µ(xiyj ⊗xpyq)i=hεmηn,xi+pyj+qi=δ δ . i,j,p,q m,i+p n,j+q Thus, µ◦(εmηn)= εkηl⊗εsηt. Assume ∂◦(εiηj)= c εsηt. Then k+s=m,l+t=n 1 s,t s,t ckP,l =∂1◦(εiηj)(xkyl)=εiηj(∂1(xkyl))=kδi,k−1δPj,l =(i+1)δi+1,kδj,l, i.e., ∂◦(εiηj)=(i+1)εi+1ηj. Similarly, ∂◦(εiηj)=(j+1)εiηj+1. From (3.1), we obtain 1 2 ∆(εmηn) = (∂◦⊗∂◦−∂◦⊗∂◦)µ◦(εmηj)= (kt−ls)εkηl⊗εsηt. 1 2 2 1 k+s=m+1,l+t=n+1 P Therefore, (4.3) holds. Next, we verify (4.4). We have h[εiηj,εsηt],xkyli=hεiηj ⊗εsηt,(kn−lm)xk+m−1yl+n−1⊗xy+(k−l)xmyn⊗xkyl −(kn−lm)xy⊗xk+m−1yl+n−1−(k−l)xkyl⊗xmyni =h(n(i+1)−m(j+1))δ δ εi+1−mηj+1−n+(s−t)δ δ εsηt s,1 t,1 i,m j,n −(n(s+1)−m(t+1))δ δ εs+1−mηt+1−n−(i−j)δ δ εiηj,xkyli. (4.5) i,1 j,1 s,m t,n If (i,j)=(1,1) and (s,t)6=(1,1), then (4.5) gives (note that (m,n)6=(1,1)) (m(t+1)−n(s+1))εs+1−mηt+1−n if s+1−m>0, t+1−m>0, [εη,εsηt]= ( 0 otherwise. Thus we obtain the first case of (4.4) (cf. Convention 4.3(1)). If (i,j) = (m,n) 6= (s,t) 6= (1,1), then (4.5) gives [εmηn,εsηt]=(s−t)εsηt, which is the second case of (4.4). It remains to verify the last case of (4.4). By (4.5), we have [εiηj,εsηt] = 0 if i,s 6= 1,m or j,t 6= 1,n. We discuss the situations in two subcases. Subcase 1. Assume i=1 (thus j 6=1). Then (4.5) becomes [εηj,εsηt] = 2n−m(j+1) δ δ ε2−mηj+1−n+(s−t)δ δ εsηt−(1−j)δ δ εηj. (4.6) s,1 t,1 1,m j,n s,m t,n Note that ε2−m =(cid:0)0 if m > 2, (cid:1)in this case, (4.6) becomes [εηj,εsηt] = −(1−j)δ δ εηj, and we s,m t,n have (4.4). Now assume m = 0. Then (4.6) gives [εηj,εsηt] = 2nδ δ ε2ηj+1−n −(1−j)δ δ εηj, s,1 t,1 s,0 t,n and we see that the last case of (4.4) holds in this case. Next assume m = 1. Then (4.6) becomes [εηj,εsηt]=(2n−(j+1))δ δ εηj+1−n+(s−t)δ εsηt−(1−j)δ δ εηj. Hence the lastcaseof(4.4) s,1 t,1 j,n s,1 t,n holds. Finally assume m=2. By (4.6), we have [εηj,εsηt] = (2n−2(j+1))δ δ ηj+1−n−(1−j)δ δ εηj, (4.7) s,1 t,1 s,2 t,n and the last case of (4.4) holds again. Subcase 2. Assume i=m6=16=s. We have[εmηj,εsηt]=(s−t)δ εsηt−(m−j)δ δ εmηj by (4.5), j,n s,m t,n i.e., [εmηj,εsηt]=(s−t)εsηt if j = n, or (j−m)εmηj if (s,t)=(m,n), or 0 otherwise. This completes the proof of the theorem. ✷ Proposition 4.5. Let f(x,y) = m n a xiyj ∈ F[x,y] with a 6= 0, and k,l ∈ Z , c ∈ F. i=0 j=0 i,j m,n + Denote Suppf ={(i,j)∈Z2 |a 6=0}. Then + ij P P [xkyl,f(x,y)]=cf(x,y)6=0 ⇐⇒ k =l=1, j−i=c, ∀(i,j)∈Suppf, (4.8) [xkyl,f(x,y)]=0 ⇐⇒ kj−li=0, ∀(i,j)∈Suppf. (4.9) In particular, if denote r = A⊗B−B⊗A, where either A = xy, B = n a xiyc+i for some c ∈ Z i=0 i + and a ∈F, or A=xkyl, B =f(x,y)= m n a xiyj such that kj−li=0 for all (i,j)∈Suppf, i i=0 j=0 i,j P then r is a solution of classical Yang-Baxter equation. P P SongGA,SuYC SciChina Math January 2015 Vol.58 No.1 7 Proof. We have [xkyl,f(x,y)]= a (kj −li)xk+i−1yl+j−1. By comparing the coefficients of the i,j i,j highest term (i.e., xmyn) and xiyj for all i,j, one immediately obtains (4.8) and (4.9). ✷ P First consider A = xy, B = n a xiyi+m with a 6= 0 and m ∈ Z \{0}. Then [A,B] = mB 6= 0. i=0 i n + The triple (A,[·,·],∆ ) with r =A⊗B−B⊗A is a coboundarytriangularLie bialgebraofPoissontype r P with bracket defined by (1.5) and cobracket defined by ∆ (g)=[g,A]⊗B+A⊗[g,B]−B⊗[g,A]−[g,B]⊗A, ∀g ∈A. r The following is one of the main results of the present paper. Theorem 4.6. Let (A,[·,·],∆ ) be the coboundary triangular Lie bialgebra defined as above. The dual r Lie bialgebraof(A,[·,·],∆ )is (A◦,[·,·],∆), whereA◦ is describedbyProposition4.2(2),with cobracket r ∆ defined by (4.3) and bracket uniquely determined by the skew-symmetry and the following. (1) In case m6=0, we have (cf. Convention 4.3(1)), for (s,t)6=(1,1) and p6=q, n [εpηp,εsηt]=δ a (si+sm+m−ti)εs−i+1ηt−i−m+1, (4.10) p,1 i i=0 P n n [εpηq,εsηt]=(p−q) a δ δ εpηq−(s−t) a δ δ εsηt. (4.11) i s,i t,i+m i p,i q,i+m i=0 i=0 P P (2) In case m=0, we have, for (s,t)6=(1,1), n [εη,εsηt]= a (s−t)iεs−i+1ηt−i+1, (4.12) i i=2 [εpηp,εsηt]=P(t−s)a εsηt if p∈{0,2,3,··· ,n}, (4.13) p [εpηq,εsηt]=0 if p6=q, s6=t. (4.14) Proof. Denote C=[xkyl,xy]=(k−l)xkyl and D=[xkyl,B]= n a (ki+km−li)xi+k−1yi+l+m−1. We i=0 i have P h[εpηq,εsηt],xkyli=hεpηq⊗εsηt,C⊗B+A⊗D−B⊗C−D⊗Ai=Pp,q−Ps,t, (4.15) s,t p,q where (regarding k,l as fixed) n n Psp,,tq = (k−l)δp,kδq,l aiδs,iδt,i+m+δp,1δq,1 ai(ki+km−li)δs,i+k−1δt,i+m+l−1 i=0 i=0 P P n n = (p−q)δp,kδq,l aiδs,iδt,i+m+δp,1δq,1 ai(si+sm+m−ti)δs−i+1,kδt−i−m+1,l i=0 i=0 P P = hHp,q,xkyli, and where, s,t n n Hp,q =(p−q) a δ δ εpηq+δ δ a (si+sm+m−ti)εs−i+1ηt−i−m+1. (4.16) s,t i s,i t,i+m p,1 q,1 i i=0 i=0 Thus P P [εpηq,εsηt]=Hp,q−Hs,t. (4.17) s,t p,q Assume (s,t)6= (1,1). First suppose m 6= 0. Then (4.15)–(4.17) give [εpηp,εsηt]= 0 for p= q 6= 1, and [εη,εsηt]= n a (si+sm+m−ti)εs−i+1ηt−i−m+1 for p= q =1. We have (4.10). If p 6=q, we have i=0 i [εpηq,εsηt]=(p−q) n a δ δ εpηq−(s−t) n a δ δ εsηt, by (4.15)–(4.17), and we have P i=0 i s,i t,i+m i=0 i p,i q,i+m (4.11). Now suppose m=0. Then (4.16) becomes P P n n Hp,q =(p−q) a δ δ εpηq +δ δ a (s−t)iεs−i+1ηt−i+1. (4.18) s,t i s,i t,i p,1 q,1 i i=0 i=0 P P We have [εη,εsηt] = n a (s−t)iεs−i+1ηt−i+1 for (p,q) = (1,1), i.e., we have (4.12). If p = q ∈ i=2 i {0,2,3,...,n}, it is easy to see from (4.18) and (4.17) that we have (4.13). Finally assume p 6= q. One P can easily obtain (4.14). ✷ 8 SongGA,SuYC Sci ChinaMath January 2015 Vol.58 No.1 Remark 4.7. Theorems4.4and4.6andthe followingtheoremprovideussomeexamplesofthe nontriv- ialityofcoboundarytriangularLiebialgebras,eveninthe caseofverytrivialsolutionr=A⊗B−B⊗A of CYBE with [A,B]=0. Theorem 4.8. Let(k,l)∈Z2 befixedandA=xkyl,B =f(x,y)= a xiyj ∈Awitha ∈F, + (i,j)∈S i,j i,j where S =Suppf is some subset of Z2 such that kj−li=0 for (i,j)∈S. Denote r =A⊗B−B⊗A. + P Then (A,[·,·],∆ ) is a coboundary triangular Lie bialgebra of Poisson type. The dual Lie bialgebra of r (A,[·,·],∆ )is(A◦,[·,·],∆)withcobracket∆definedasinTheorem4.4andbracketuniquelydetermined r by the skew-symmetry and the following. (1) If (s,t)6=(k,l), then (l−k)a εη+ a (j−i)εs−i+1ηt−j+1−(l−k)a εs−k+1ηt−l+1, s,t i,j k,l (i,j)∈S  P [εkηl,εsηt]=(i(,Pjl)−∈Ska)ia,js(,tjε(ηs++1)−i(ta+i,j1()j)−εsi−)iε+s1−ηi+t−1jη+t−1−j+(1l,(s+1)−k(t+1))ak,lεs−k+1ηt−l+1, (4.19) (i,j)∈S P according to thefoll(oi,wPj)i∈nSgafio,ju(rj(csas+es1)−i(t+1))εs−i+1ηt−j+1, (i) (s,t),(k,l)∈S, (ii) (s,t)∈/S,(k,l)∈S, (iii) (s,t)∈S,(k,l)∈/S or (iv) (s,t),(k,l)∈/ S. (2) If (p,q)6=(k,l), (s,t)6=(k,l), then (l−k)a εp−k+1ηq−l+1−(l−k)a εs−k+1ηt−l+1 if (p,q),(s,t)∈S, s,t p,q [εpηq,εsηt]= (l(p+1)−k(q+1))a εp−k+1ηq−l+1 if (p,q)∈/S,(s,t)∈S, (4.20)  s,t 0 if (p,q)∈/ S,(s,t)∈/ S. Proof. Since [A,B]=0, the triple (A,[·,·],∆ ) is obviously a coboundary triangular Lie bialgebra. We r only need to determine the bracket relations. Note that h[εpηq,εsηt],xm′yn′i equals εpηq⊗εsηt,[xm′yn′,A]⊗B+A⊗[xm′yn′,B]−B⊗[xm′yn′,A]−[xm′yn′,B]⊗A =D(m′l−n′k)δp,m′+k−1δq,n′+l−1 ai,jδs,iδt,j +δp,kδq,l ai,j(m′j−n′i)δs,i+m′−E1δt,j+n′−1 (i,j)∈S (i,j)∈S −(m′l−n′k)δs,m′+k−1δt,n′+l−1 P ai,jδp,iδq,j−δs,kδt,l P ai,j(m′j−n′i)δp,i+m′−1δq,j+n′−1 (i,j)∈S (i,j)∈S =hHp,q−Hs,t,xm′yn′i, P P (4.21) s,t p,q where (noting that a =0 if (s,t)∈/ S) s,t Hp,q= l(p+1)−k(q+1) a εp−k+1ηq−l+1+δ δ a j(s+1)−i(t+1) εs−i+1ηt−j+1. s,t s,t p,k q,l i,j (i,j)∈S (cid:0) (cid:1) P (cid:0) (cid:1) If (p,q)=(k,l) and (s,t)6=(k,l), then (4.21) gives (noting that a =0 if (k,l)∈/ S) k,l [εkηl,εsηt] = (l−k)a εη+ a j(s+1)−i(t+1) εs−i+1ηt−j+1 s,t i,j (i,j)∈S P (cid:0) (cid:1) − l(s+1)−k(t+1) a εs−k+1ηt−l+1. k,l (cid:0) (cid:1) In particular, if (s,t),(k,l)∈S, then (using the fact that kj−li=0 for (i,j)∈S) [εkηl,εsηt]=(l−k)a εη+ a (j−i)εs−i+1ηt−j+1−(l−k)a εs−k+1ηt−l+1, s,t i,j k,l (i,j)∈S P SongGA,SuYC SciChina Math January 2015 Vol.58 No.1 9 which gives the first case of (4.19). If (s,t)∈/ S,(k,l)∈S, then [εkηl,εsηt]= a j(s+1)−i(t+1) εs−i+1ηt−j+1− l(s+1)−k(t+1) a εs−k+1ηt−l+1, i,j k,l (i,j)∈S P (cid:0) (cid:1) (cid:0) (cid:1) which gives the second case of (4.19). If (s,t) ∈ S,(k,l) ∈/ S, then [εkηl,εsηt] = (l − k)a εη + s,t a (j − i)εs−i+1ηt−j+1, which gives the third case of (4.19). If (s,t) ∈/ S,(k,l) ∈/ S, then (i,j)∈S i,j [εkηl,εsηt]= a (j(s+1)−i(t+1))εs−i+1ηt−j+1, which completes the proof of (4.19). P (i,j)∈S i,j Now assume (p,q)6=(k,l),(s,t)6=(k,l). Then (4.21) gives P [εpηq,εsηt]= l(p+1)−k(q+1) a εp−k+1ηq−l+1− l(s+1)−k(t+1) a εs−k+1ηt−l+1. s,t p,q If (p,q),(s,t)∈S, th(cid:0)en [εpηq,εsηt]=(l−(cid:1)k)a εp−k+1ηq−l+1−(cid:0)(l−k)a εs−k+1ηt−(cid:1)l+1, giving the first case s,t p,q of (4.20). If (p,q)∈/S,(s,t)∈S, then [εpηq,εsηt]= l(p+1)−k(q+1) a εp−k+1ηq−l+1, giving the second s,t case of (4.20). In case (p,q) ∈/ S,(s,t) ∈/ S, we have [εpηq,εsηt] = 0, which completes the proof of the (cid:0) (cid:1) theorem. ✷ In the final part of the paper, we will present an example of a dual Lie bialgebra which has different feature from the previous dual Lie bialgebras (Theorems 4.4, 4.6 and 4.8). Denote A = x(1 + y)(2 + y), B = x2(1 + y)3(2 + y) ∈ A. Then (1.5) shows [A,B] = B. Take r =A⊗B−B⊗A. We obtain a coboundary triangular Lie bialgebra (A,[·,·],∆) with bracket defined by (1.5), and cobracket defined by ∆(f)=f ·r =[f,A]⊗B+A⊗[f,B]−[f,B]⊗A−B⊗[f,A]. Theorem 4.9. Let(A,[·,·],∆)betheLiebialgebradefinedabove. ThedualLiebialgebraof(A,[·,·],∆) is (A◦,[·,·],∆), where A◦ and ∆ are defined in Theorem 4.4, and the bracket is uniquely determined by the skew-symmetry and the following. First, denote c =2, c =3, c =1, and c =0 for j 6=0,1,2, 0 1 2 j k =2, k =7, k =9, k =5, k =1, and k =0 for 4<t∈Z . 0 1 2 3 4 t + Then [εηj,εsηt] = c (4s−2t+2)εs−1ηt−3+(15s−10t+5)εs−1ηt−2 j (4.22) +1(cid:16)8(s−t)εs−1ηt−1+(7s−14t−7)εs−1ηt+4(t+1)εs−1ηt+1 if s6=1,2, (cid:17) [εηj,εηt] = c (6−2t)ηt−3+(20−10t)ηt−2+18(1−t)ηt−1−14tηt+4(t+1)ηt+1 j (4.23) −c(cid:16) (6−2j)ηj−3+(20−10j)ηj−2+18(1−j)ηj−1−14jηj+4(j+1)η(cid:17)j+1 , t [εηj,ε2ηt] = k (cid:16)(3−j)εηj−1+3(1−j)εηj −2(j+1)εηj+1 (cid:17) t +c(cid:16) (10−2t)εηt−3 (cid:17) (4.24) j (cid:16)+(35−10t)εηt−2+18(2−t)εηt−1+(7−14t)εηt+4(t+1)εηt+1 , [ε2ηj,εsηt]=−k (2s−t+1)εsηt−1+3(s−t)εsηt−2(t+1)εsηt+1 if s6=1,(cid:17)2, (4.25) j [ε2ηj,ε2ηt] = k (cid:16)(5−j)ε2ηj−1+3(2−j)ε2ηj −2(j+1)ε2ηj+1 (cid:17) t (4.26) −k(cid:16) (5−t)ε2ηt−1+3(2−t)ε2ηt−2(t+1)ε2ηt+1(cid:17) , j [εiηj,εsηt]=0 if (cid:16)i6=1,2,s6=1,2. (cid:17) (4.27) Proof. Denote C =[xmyn,A],D =[xmyn,B]. By (1.5), we have C =[xmyn,x(y+1)(y+2)]=(2m−n)xmyn+1+3(m−n)xmyn−2nxmyn−1, D =[xmyn,x2(y+1)3(y+2)] 10 SongGA,SuYC SciChinaMath January 2015 Vol.58 No.1 =(4m−2n)xm+1yn+3+(15m−10n)xm+1yn+2+18(m−n)xm+1yn+1 +(7m−14n)xm+1yn−4nxm+1yn−1. For i,j,s,t∈Z , note that h[εiηj,εsηt],xmyni is equal to + hεiηj ⊗εsηt,[xmyn,A]⊗B+A⊗[xmyn,B]−[xmyn,B]⊗A−B⊗[xmyn,A]i =hεiηj ⊗εsηt,C⊗B+A⊗D−D⊗A−B⊗Ci=Pi,j −Ps,t, (4.28) s,t i,j where Pi,j =hεiηj,Cihεsηt,Bi+hεiηj,Aihεsηt,Di, which is equal to s,t (2m−n)δi,mδj,n+1+3(m−n)δi,mδj,n−2nδi,mδj,n−1 (cid:16) (cid:17) × 2δ δ +7δ δ +9δ δ +5δ δ +δ δ s,2 t,0 s,2 t,1 s,2 t,2 s,2 t,3 s,2 t,4 (cid:16) (cid:17) + 2δ δ +3δ δ +δ δ i,1 j,0 i,1 j,1 i,1 j,2 (cid:16) (cid:17) × (4m−2n)δ δ +(15m−10n)δ δ +18(m−n)δ δ s,m+1 t,n+3 s,m+1 t,n+2 s,m+1 t,n+1 (cid:16) +(7m−14n)δs,m+1δt,n−4nδs,m+1δt,n−1 (cid:17) = 2δ δ +7δ δ +9δ δ +5δ δ +δ δ s,2 t,0 s,2 t,1 s,2 t,2 s,2 t,3 s,2 t,4 D(cid:16) × (2i−j+1)εiηj−1+3(i−j)εiηj −2(j+1)εiη(cid:17)j+1) (cid:16) + 2δ δ +3δ δ +δ δ i,1 j,0 i,1 j,1 i,1 j,2 ×(cid:16) (4s−2t+2)εs−1ηt−3+(1(cid:17)5s−10t+5)εs−1ηt−2+18(s−t)εs−1ηt−1 (cid:16)+(7s−14t−7)εs−1ηt+4(t+1)εs−1ηt+1 ,xmyn . (cid:17) E We obtain [εiηj,εsηt]=Hi,j −Hs,t, (4.29) s,t i,j where Hi,j is equal to s,t 2δ δ +7δ δ +9δ δ +5δ δ +δ δ (2i−j+1)εiηj−1+3(i−j)εiηj−2(j+1)εiηj+1 s,2 t,0 s,2 t,1 s,2 t,2 s,2 t,3 s,2 t,4 +(cid:16) 2δ δ +3δ δ +δ δ (cid:17)(cid:16) (cid:17) i,1 j,0 i,1 j,1 i,1 j,2 (cid:16)× (4s−2t+2)εs−1ηt−3+(15s(cid:17)−10t+5)εs−1ηt−2+18(s−t)εs−1ηt−1 (cid:16)+(7s−14t−7)εs−1ηt+4(t+1)εs−1ηt+1 . (cid:17) If i6=1,2, s6=1,2, then (4.29) gives that [εiηj,εsηt] = 0, which is (4.27). Assume i = 1, s 6= 1,2. Then [εηj,εsηt]=H1,j, which gives (4.22). If i=1,s=1, then [εηj,εηt] is equal to s,t 2δ +3δ +δ (6−2t)ηt−3+(20−10t)ηt−2+18(1−t)ηt−1−14tηt+4(t+1)ηt+1 j,0 j,1 j,2 −(cid:16) 2δ +3δ +δ (cid:17)(cid:16)(6−2j)ηj−3+(20−10j)ηj−2+18(1−j)ηj−1−14jηj+4(j+1)ηj+1 , (cid:17) t,0 t,1 t,2 (cid:16) (cid:17)(cid:16) (cid:17) which implies (4.23). Assume i=1,s=2. Then (4.29) implies that [εηj,ε2ηt] is equal to 2δ +7δ +9δ +5δ +δ (3−j)εηj−1+3(1−j)εηj −2(j+1)εηj+1 t,0 t,1 t,2 t,3 t,4 +(cid:16) 2δ +3δ +δ (10−2t)εηt−(cid:17)3(cid:16)+(35−10t)εηt−2+18(2−t)εηt−1 (cid:17) j,0 j,1 j,2 (cid:16) (cid:17)(cid:16)+(7−14t)εηt+4(t+1)εηt+1 , (cid:17) which gives (4.24). If i=2,s6=1,2, then (4.29) shows that [ε2ηj,εsηt] is equal to −(2δ +7δ +9δ +5δ +δ ) (2s−t+1)εsηt−1+3(s−t)εsηt−2(t+1)εsηt+1 , j,0 j,1 j,2 j,3 j,4 (cid:16) (cid:17)