Hopf-Galois algebras and their Poisson structures
aa r X i v : . [ m a t h . QA ] M a y Hopf-Galois algebras and their Poisson structures ∗ Huihui Zheng, Liangyun Zhang
College of Science, Nanjing Agricultural University, Nanjing 210095, China
Abstract : As is known to all, Hopf-Galois objects have a significant research value for analyzing tensor cate-gories of comodules and classification questions of pointed Hopf algebras, and are natural generalizations of Hopfalgebras with a Galois-theoretic flavour. In this paper, we mainly prove a criterion for an Ore extension of aHopf-Galois algebra to be a Hopf-Galois algebra, and introduce the conception of Poisson Hopf-Galois algebras,and establish the relationship between Poisson Hopf-Galois algebras and Poisson Hopf algebras. Moreover, westudy Poisson Hopf-Galois structures on Poisson polynomial algebras, and mainly give a necessary and sufficientcondition for the Poisson enveloping algebra of a Poisson Hopf-Galois algebra to be a Hopf-Galois algebra.
KeyWords : Hopf algebra, Hopf-Galois algebra, Poisson polynomial algebra, Poisson Hopf-Galois algebra, Pois-son enveloping algebra. : 16T05, 17B63 § A (right) Hopf-Galois object [5] R over a Hopf algebra H is a (right) comodule algebra R over H with a trivial coinvariants R coH = k (a field) and such that the Galois map can : R ⊗ R → R ⊗ H, can ( r ⊗ s ) = rs (0) ⊗ s (1) , for any r, s ∈ R is bijective, where ρ ( s ) = s (0) ⊗ s (1) ∈ R ⊗ H represents the comodule structure of R .Hopf-Galois objects are important for analyzing tensor categories of comodules [17, 18] andclassification questions of pointed Hopf algebras [1, 2], and are natural generalizations of Hopfalgebras with a Galois-theoretic flavour.Quantum torsors were introduced by Grunspan [4] with more axioms and simplified later bySchauenburg [16]. Since there exists a relationship between this quantum torsor and some Hopf-Galois object, the authors call such a structure a Hopf-Galois algebra in [3]. The following resultrelates what we called Hopf-Galois algebras and classical Hopf-Galois objects. Theorem
Let R be an algebra. The following assertion given in [3] are equivalent:(1) There exists an algebra map µ : R → R ⊗ R op ⊗ R making R into a Hopf-Galois algebra.(2) R is a right Hopf-Galois object over some Hopf algebra.The result ensure that there exists a definition of Hopf-Galois objects that does not use anyHopf algebra, in terms of an algebra map µ : R → R ⊗ R op ⊗ R , subject to certain axioms.In [3], the authors have proved that eery Hopf algebra is Hopf-Galois algebra, and a Hopf-Galoisalgebra R arises from a Hopf algebra if and only if there exists an algebra map α : R → k . So, ∗ This work is supported by Natural Science Foundation (11571173).1opf-Galois algebras, like Hopf-Galois objects, are a generalization of Hopf algebras. In addition,Hopf-Galois structures on a generalized ambiskew ring were discussed, and some examples of Hopf-Galois algebras rather than Hopf algebras were provided in [3].In [7], the authors have found that any pointed Poisson Hopf algebra has a Hopf-Galois struc-ture. Naturally, we can consider the Poisson version of a Hopf-Galois algebra. As we expected,in section 3, we establish the correspondence relation between Poisson Hopf algebras and PoissonHopf-Galois algebras.Ore extensions of Hopf algebras (HA) and Poisson Hopf algebras (PHA) have been consideredin [14, 7], so, we have a question as follows, that is, how we study Ore extensions of Hopf-Galoisalgebras (HGA) and Poisson Hopf-Galois algebras (PHGA). In section 2, we give the necessary andsufficient conditions for the Ore extension of a Hopf-Galois algebra to be a Hopf-Galois algebra.Similarly, in section 4, the Poisson Ore extension of a Poisson Hopf-Galois algebra is also studied.HAHA HGAHGA ✲✲❄ ❄
OE PHA PHGAPHA PHGA ❄ ✲ ❄✲
OEOE ? OE ?Here OE denotes the Ore extension of an algebra.In [10], the author has proved that the Poisson enveloping algebra of a Poisson Hopf algebrahas a Hopf algebra structure. Naturally, we can consider whether Poisson enveloping algebra of aPoisson Hopf-Galois algebra has a Hopf-Galois algebra structure. In section 5, we find a necessaryand sufficient condition for the Poisson enveloping algebra U ( R ) of a Poisson Hopf-Galois algebra R to be a Hopf-Galois algebra. PHAHAUE ❄ PHGAHGA ❄✲✲
UE ?Here UE denotes the Poisson enveloping algebra of a Poisson algebra.
Notation
Throughout this paper, let k be a fixed field. Unless otherwise specified, linearity,modules and ⊗ are all meant over k . And we freely use the Hopf algebras terminology introducedin [15, 19]. For a coalgebra C , we write its comultiplication ∆( c ) with c ⊗ c , for any c ∈ C ; fora Hopf-Galois algebra R , we denote its structure by µ ( x ) = x (1) ⊗ x (2) ⊗ x (3) , for any x ∈ R , inwhich we omit the summation symbols for convenience. § In this section, we recall the conception of Hopf-Galois algebras, and mainly prove a criterionfor an Ore extension of a Hopf-Galois algebra to be a Hopf-Galois algebra.
Definition 2.1
A Hopf-Galois algebra [3] is a non-zero algebra R together with an algebra2ap µ : R → R ⊗ R op ⊗ R, where R ⊗ R op ⊗ R is a tensor product algebra, and such that( µ ⊗ id ⊗ id ) ◦ µ = ( id ⊗ id ⊗ µ ) ◦ µ, ( m ⊗ id ) ◦ µ = η ⊗ id, ( id ⊗ m ) ◦ µ = id ⊗ η, where m : R ⊗ R → R and η : k → R denote the respective multiplication and unit of R . We shallwrite, for any r ∈ R, µ ( r ) = r (1) ⊗ r (2) ⊗ r (3) . Hence, for any r ∈ R , r ⊗ r (1) ⊗ r (2) r (3) , ⊗ r = r (1) r (2) ⊗ r (3) ; (2 . r (1) ⊗ r (2) ⊗ r (3) ⊗ r (4) ⊗ r (5) := µ ( r (1) ) ⊗ r (2) ⊗ r (3) = r (1) ⊗ r (2) ⊗ µ ( r (3) ) . (2 . R , we call it a Hopf-Galois commutative algebra if it is commutative.In what follows, we call the above algebra map µ a Hopf-Galois map. Remark 2.2 (1) Let R f −→ B → R isa Hopf-Galois algebra with the Hopf-Galois map µ R . Then, B is also a Hopf-Galois algebra withthe Hopf-Galois map: µ B : B −→ B ⊗ B op ⊗ B, b ≡ f ( r ) f ( r (1) ) ⊗ f ( r (2) ) ⊗ f ( r (3) ) , where µ R ( b ) = r (1) ⊗ r (2) ⊗ r (3) ∈ R ⊗ R op ⊗ R. In particular, for any (two-sided) ideal I of the algebra R , we know that the quotient algebra R/I is also a Hopf-Galois algebra if R is a Hopf-Galois algebra.(2) Let R be a Hopf-Galois commutative algebra with a Hopf-Galois map µ . Define µ ′ ( r ) = r (3) ⊗ r (2) ⊗ r (1) for r ∈ R . Then, µ ′ is also a Hopf-Galois map on R . Proof. (1) It is straightforward by Definition 2.1.(2) Denote µ ′ ( r ) by r (1) ′ ⊗ r (2) ′ ⊗ r (3) ′ for r ∈ R . Then, we easily know that r (1) ′ ⊗ r (2) ′ r (3) ′ = r (3) ⊗ r (2) r (1) = r ⊗ ,r (1) ′ r (2) ′ ⊗ r (3) ′ = r (3) r (2) ⊗ r (1) = 1 ⊗ r r ∈ R . Moreover µ ′ ( r (1) ′ ) ⊗ r (2) ′ ⊗ r (3) ′ = µ ′ ( r (3) ) ⊗ r (2) ⊗ r (1) = r (3)(1) ′ ⊗ r (3)(2) ′ ⊗ r (3)(3) ′ ⊗ r (2) ⊗ r (1) = r (3)(3) ⊗ r (3)(2) ⊗ r (3)(1) ⊗ r (2) ⊗ r (1) = r (5) ⊗ r (4) ⊗ r (3) ⊗ r (2) ⊗ r (1) ,r (1) ′ ⊗ r (2) ′ ⊗ µ ′ ( r (3) ′ ) = r (3) ⊗ r (2) ⊗ µ ′ ( r (1) )= r (3) ⊗ r (2) ⊗ r (1)(1) ′ ⊗ r (1)(2) ′ ⊗ r (1)(3) ′ = r (3) ⊗ r (2) ⊗ r (1)(3) ⊗ r (1)(2) ⊗ r (1)(1) = r (5) ⊗ r (4) ⊗ r (3) ⊗ r (2) ⊗ r (1) , so µ ′ ( r (1) ′ ) ⊗ r (2) ′ ⊗ r (3) ′ = r (1) ′ ⊗ r (2) ′ ⊗ µ ′ ( r (3) ′ ).In addition, for any x, y ∈ R , we have µ ′ ( xy ) = ( xy ) (1) ′ ⊗ ( xy ) (2) ′ ⊗ ( xy ) (3) ′ = ( xy ) (3) ⊗ ( xy ) (2) ⊗ ( xy ) (1) = x (3) y (3) ⊗ x (2) y (2) ⊗ x (1) y (1) = µ ′ ( x ) µ ′ ( y ) , so µ ′ is an algebra map.We recall the concept of group-like elements in Hopf-Galois algebras. Definition 2.3
A group-like element [3] in a Hopf-Galois algebra R is an invertible element g ∈ R such that µ ( g ) = g ⊗ g − ⊗ g. In the following, we denote by G ( R ) the set of group-like elements in R . If g ∈ G ( R ), wewrite g · r := grg − .We recall the following result and its proof in [3]. Proposition 2.4
Let H be a Hopf-Galois algebra. Then, there is an algebra map α : H → k if and only if H is a Hopf algebra. Proof.
If there is an algebra map α : H → k , we can prove that H is a Hopf algebra, whosecounit is α and the other structure maps defined by∆( x ) = α ( x (2) ) x (1) ⊗ x (3) , S ( x ) = α ( x (1) x (3) ) x (2) . (2 . H is a Hopf algebra, define, for any x ∈ R , µ ( x ) = x ⊗ S ( x ) ⊗ x , (2 . H is a Hopf-Galois algebra. (cid:3) Example 2.5
Assume H = k { x, g | x = 0 , g = 1 , xg = − gx } is the 4-dimensional SweedlerHopf algebra in [9]. Then, it is easy to check that H is a Hopf-Galois algebra with a Hopf-Galoismap as folows µ ( x ) = x ⊗ ⊗ − g ⊗ gx ⊗ g ⊗ g ⊗ x ; µ ( g ) = g ⊗ g ⊗ g. k = 2.In what follows, we discuss Hopf-Galois algebra structures on given Ore extension.Let A be an algebra and τ an algebra endomorphism of A . Let δ be a τ -derivation of A . Thatis, δ is a k -linear endomorphism A such that δ ( ab ) = τ ( a ) δ ( b ) + δ ( a ) b for all a, b ∈ A . The Oreextension A [ z ; τ, δ ] of the algebra A is an algebra generated by the variable z and the algebra A with the relation za = τ ( a ) z + δ ( a )for all a ∈ A .If { a i | i ∈ I } is a k-basis of A , then, by [8], { a i z j | i ∈ I, j ∈ N } is a k -basis of the Oreextension A [ z ; τ, δ ].Furthermore, assume that A has a Hopf-Galois algebra structure. Then, we can define a Hopf-Galois algebra structure on the Ore extension A [ z ; τ, δ ]. Definition 2.6
Let A be a Hopf-Galois algebra and H = A [ z ; τ, δ ] an Ore extension of A . If H = A [ z ; τ, δ ] is a Hopf-Galois algebra with A as a Hopf-Galois subalgebra, then we say H = A [ z ; τ, δ ] is a Hopf-Galois Ore extension of A . Proposition 2.7
Let A be a Hopf-Galois algebra and H = A [ z ; τ, δ ] an Ore extension of A . If H = A [ z ; τ, δ ] is a Hopf-Galois Ore extension of A , and determined by µ ( z ) = z ⊗ ⊗ g ⊗ g − ⊗ z − g ⊗ g − z ⊗ , for some invertible element g ∈ A . Then g is a group like element of Hopf-Galois algebra A . Proof.
Assume that there is an extension µ : A → A ⊗ A op ⊗ A such that µ ( z ) = z ⊗ ⊗ g ⊗ g − ⊗ z − g ⊗ g − z ⊗ . It is obvious that H is a free left A -module under the left multiplication with the basis { z i | i ≥ } . By comparing( µ ⊗ id ⊗ id ) µ ( z ) = z ⊗ ⊗ ⊗ ⊗ g ⊗ g − ⊗ z ⊗ ⊗ − g ⊗ g − z ⊗ ⊗ ⊗ µ ( g ) ⊗ g − ⊗ z − µ ( g ) ⊗ g − z ⊗ id ⊗ id ⊗ µ ) µ ( z ) = z ⊗ ⊗ ⊗ ⊗ g ⊗ g − ⊗ z ⊗ ⊗ g ⊗ g − ⊗ g ⊗ g − ⊗ z − g ⊗ g − ⊗ g ⊗ g − z ⊗ − g ⊗ g − z ⊗ ⊗ ⊗ µ ( g ) = g ⊗ g − ⊗ g . (cid:3) The next theorem gives a sufficient and necessary condition for an Ore extension of a Hopf-Galois algebra to have a Hopf-Galois algebra structure.
Theorem 2.8
Let A be a Hopf-Galois algebra, H = A [ z ; τ, δ ] and g a group like element ofHopf-Galois algebra A . If τ is an algebra automorphism, then H = A [ z ; τ, δ ] is a Hopf-Galois Oreextension of A , and determined by µ ( z ) = z ⊗ ⊗ g ⊗ g − ⊗ z − g ⊗ g − z ⊗ , if and only if the following conditions are satisfied: for all a ∈ A ,51) µ ( τ ( a )) = τ ( a (1) ) ⊗ a (2) ⊗ a (3) = τ ( a (1) ) ⊗ τ ( a (2) ) ⊗ τ ( a (3) ),(2) g · a (1) ⊗ g · a (2) ⊗ a (3) = τ ( a (1) ) ⊗ τ ( a (2) ) ⊗ a (3) ,(3) δ ( a (1) ) ⊗ a (2) ⊗ a (3) + ga (1) ⊗ a (2) g − ⊗ δ ( a (3) ) + ga (1) ⊗ g − δ ( τ − ( g · a (2) )) ⊗ a (3) = µ ( δ ( a )) . Proof.
Assume that µ : A → A ⊗ A op ⊗ A can be extended to H by µ ( z ) = z ⊗ ⊗ g ⊗ g − ⊗ z − g ⊗ g − z ⊗ . Since za = τ ( a ) z + δ ( a ), we have µ ( za ) = µ ( z ) µ ( a )= ( z ⊗ ⊗ g ⊗ g − ⊗ z − g ⊗ g − z ⊗ a (1) ⊗ a (2) ⊗ a (3) )= za (1) ⊗ a (2) ⊗ a (3) + ga (1) ⊗ a (2) g − ⊗ za (3) − ga (1) ⊗ a (2) g − z ⊗ a (3) = τ ( a (1) ) z ⊗ a (2) ⊗ a (3) + δ ( a (1) ) ⊗ a (2) ⊗ a (3) + ga (1) ⊗ a (2) g − ⊗ τ ( a (3) ) z + ga (1) ⊗ a (2) g − ⊗ δ ( a (3) ) − ga (1) ⊗ a (2) g − z ⊗ a (3) = τ ( a (1) ) z ⊗ a (2) ⊗ a (3) + δ ( a (1) ) ⊗ a (2) ⊗ a (3) + ga (1) ⊗ a (2) g − ⊗ τ ( a (3) ) z + ga (1) ⊗ a (2) g − ⊗ δ ( a (3) ) − ga (1) ⊗ g − ( g · a (2) ) z ⊗ a (3) = ( τ ( a (1) ) ⊗ a (2) ⊗ a (3) )( z ⊗ ⊗
1) + δ ( a (1) ) ⊗ a (2) ⊗ a (3) + ( g · a (1) ⊗ g · a (2) ⊗ τ ( a (3) ))( g ⊗ g − ⊗ z ) + ga (1) ⊗ a (2) g − ⊗ δ ( a (3) ) − ga (1) ⊗ g − zτ − ( g · a (2) ) ⊗ a (3) + ga (1) ⊗ g − δ ( τ − ( g · a (2) )) ⊗ a (3) = ( τ ( a (1) ) ⊗ a (2) ⊗ a (3) )( z ⊗ ⊗
1) + δ ( a (1) ) ⊗ a (2) ⊗ a (3) + ( g · a (1) ⊗ g · a (2) ⊗ τ ( a (3) ))( g ⊗ g − ⊗ z ) + ga (1) ⊗ a (2) g − ⊗ δ ( a (3) ) − ( g · a (1) ⊗ τ − ( g · a (2) ) ⊗ a (3) )( g ⊗ g − z ⊗
1) + ga (1) ⊗ g − δ ( τ − ( g · a (2) )) ⊗ a (3) ,µ ( za ) = µ ( τ ( a ) z + δ ( a )) = µ ( τ ( a )) µ ( z ) + µ ( δ ( a ))= µ ( τ ( a ))( z ⊗ ⊗
1) + µ ( τ ( a ))( g ⊗ g − ⊗ z ) − µ ( τ ( a ))( g ⊗ g − z ⊗
1) + µ ( δ ( a )) . If H = A [ z ; τ, δ ] has a Hopf-Galois algebra structure, so we get τ ( a ) (1) ⊗ τ ( a ) (2) ⊗ τ ( a ) (3) = τ ( a (1) ) ⊗ a (2) ⊗ a (3) = g · a (1) ⊗ g · a (2) ⊗ τ ( a (3) ) ,τ ( a ) (1) ⊗ τ ( a ) (2) ⊗ τ ( a ) (3) = g · a (1) ⊗ τ − ( g · a (2) ) ⊗ a (3) ,µ ( δ ( a )) = δ ( a (1) ) ⊗ a (2) ⊗ a (3) + ga (1) ⊗ a (2) g − ⊗ δ ( a (3) ) + ga (1) ⊗ g − δ ( τ − ( g · a (2) )) ⊗ a (3) . The last equation coincides with the condition (3). Combining the first above equation withthe second above equation, we can show that g · a (1) ⊗ g · a (2) ⊗ a (3) = τ ( a (1) ) ⊗ τ ( a (2) ) ⊗ a (3) .Furthermore, by the above equations, we get µ ( τ ( a )) = τ ( a (1) ) ⊗ a (2) ⊗ a (3) = g · a (1) ⊗ g · a (2) ⊗ τ ( a (3) ) = τ ( a (1) ) ⊗ τ ( a (2) ) ⊗ τ ( a (3) ) . Conversely, if the conditions (1) ∼ (3) hold, then, H has a Hopf-Galois structure by the previouscomputation. §
6n this section, we introduce the conception of Poisson Hopf-Galois algebra, and establish therelationship between Poisson Hopf-Galois algebras and Poisson Hopf algebras.
Definition 3.1
A Poisson algebra in [6] is a commutative algebra A equipped with a bilinearmap {· , ·} : A ⊗ A → A such that A is a Lie algebra under {· , ·} and { a, bc } = { a, b } c + b { a, c } (3 . a, b, c ∈ A . Remark 3.2 If A and B are Poisson algebras, then, the tensor product algebra A ⊗ B has aPoisson structure defined by { a ⊗ b, a ′ ⊗ b ′ } A ⊗ B = aa ′ ⊗ { b, b ′ } + { a, a ′ } ⊗ bb ′ (3 . a, a ′ ∈ A , and b, b ′ ∈ B . Definition 3.3
A Poisson algebra A is said to be a Poisson Hopf algebra in [6] if A is also aHopf algebra ( A, , m, ∆ , ε, S ) such that∆( { a, b } ) = { ∆( a ) , ∆( b ) } A ⊗ A (3 . a, b ∈ A .For Poisson algebras A and B , an algebra homomorphism φ : A → B is said to be a Poissonhomomorphism, anti-homomorphism) if φ satisfies the rule φ ( { a, b } ) = { φ ( a ) , φ ( b ) } ( φ ( { a, b } ) = { φ ( b ) , φ ( a ) } )for all a, b ∈ A . Lemma 3.4
Let A be a Poisson Hopf algebra. Then, by [7], ε ( { a, b } ) = 0 and the antipode S is a Poisson anti-automorphism.In what follows, we introduce the conception of a Poisson Hopf-Galois algebra by the sameaxioms like Grunspan’s Poisson quantum torsors, but without endomorphism θ in [4]. Definition 3.5
A Poisson Hopf-Galois algebra R is a Poisson algebra and a Hopf-Galois algebra,such that the Hopf-Galois map µ on R is a Poisson homomorphism, that is, µ is an algebra map,and for any a, b ∈ R, µ ( { a, b } ) = { µ ( a ) , µ ( b ) } . (3 . R ⊗ R op ⊗ R is given by { x ⊗ y ⊗ z, x ′ ⊗ y ′ ⊗ z ′ } = { x, x ′ } ⊗ yy ′ ⊗ zz ′ − xx ′ ⊗ { y, y ′ } ⊗ zz ′ + xx ′ ⊗ yy ′ ⊗ { z, z ′ } , (3 . x, y, z, x ′ , y ′ , z ′ ∈ R . So, the equality (3.4) can be written into { a, b } (1) ⊗ { a, b } (2) ⊗ { a, b } (3) = { a (1) , b (1) } ⊗ a (2) b (2) ⊗ a (3) b (3) − a (1) b (1) ⊗ { a (2) , b (2) } ⊗ a (3) b (3) + a (1) b (1) ⊗ a (2) b (2) ⊗ { a (3) , b (3) } . emark 3.6 (1) Assume that R f −→ B → R is a Poisson Hopf-Galois algebra with the Hopf-Galois map µ R , then, B is also a PoissonHopf-Galois algebra with the Hopf-Galois map: µ B : B −→ B ⊗ B op ⊗ B, b ≡ f ( r ) f ( r (1) ) ⊗ f ( r (2) ) ⊗ f ( r (3) ) , where µ R ( b ) = r (1) ⊗ r (2) ⊗ r (3) ∈ R ⊗ R op ⊗ R. In particular, for any ideal I of the Poisson algebra R (that is, RI ⊆ I and { R, I } ⊆ I ), we knowthat the quotient algebra R/I is also a Poisson Hopf-Galois algebra if R is a Poisson Hopf-Galoisalgebra.(2) Let R be a Poisson Hopf-Galois algebra with a Hopf-Galois map µ . Define µ ′ ( r ) = r (3) ⊗ r (2) ⊗ r (1) for r ∈ R . Then, ( R, µ ′ ) is also a Poisson Hopf-Galois algebra with the Hopf-Galois map µ ′ . Proof. (1) On the basis of Remark 2.2, we know that (
B, µ B ) is a Hopf-Galois algebra withthe Hopf-Galois map µ B .So, by Definition 3.5, we only need to prove that µ B ( { b, b ′ } ) = { µ B ( b ) , µ B ( b ′ ) } for any b, b ′ ∈ B in order to prove that ( B, µ B ) is a Poisson Hopf-Galois algebra.In fact, since there are elements r, r ′ ∈ R such that b = f ( r ) and b ′ = f ( r ′ ), we get µ B ( { b, b ′ } ) = µ B ( { f ( r ) , f ( r ′ ) } ) = µ B f ( { r, r ′ } )= f ( { r, r ′ } (1) ) ⊗ f ( { r, r ′ } (2) ) ⊗ f ( { r, r ′ } (3) ) (3 . = f ( { r (1) , r ′ (1) } ) ⊗ f ( r (2) r ′ (2) ) ⊗ f ( r (3) r ′ (3) ) − f ( r (1) r ′ (1) ) ⊗ f ( { r (2) , r ′ (2) } ) ⊗ f ( r (3) r ′ (3) )+ f ( r (1) r ′ (1) ) ⊗ f ( r (2) r ′ (2) ) ⊗ f ( { r (3) , r ′ (3) } ) , { µ B ( b ) , µ B ( b ′ ) } = { µ B ( f ( r )) , µ B ( f ( r ′ )) } = { f ( r (1) ) ⊗ f ( r (2) ) ⊗ f ( r (3) ) , f ( r ′ (1) ) ⊗ f ( r ′ (2) ) ⊗ f ( r ′ (3) ) } (3 . = { f ( r (1) ) , f ( r ′ (1) ) } ⊗ f ( r (2) ) f ( r ′ (2) ) ⊗ f ( r (3) ) f ( r ′ (3) ) − f ( r (1) ) f ( r ′ (1) ) ⊗ { f ( r (2) ) , f ( r ′ (2) ) } ⊗ f ( r (3) ) f ( r ′ (3) )+ f ( r (1) ) f ( r ′ (1) ) ⊗ f ( r (2) ) f ( r ′ (2) ) ⊗ { f ( r (3) ) , f ( r ′ (3) ) } = f ( { r (1) , r ′ (1) } ) ⊗ f ( r (2) r ′ (2) ) ⊗ f ( r (3) r ′ (3) ) − f ( r (1) r ′ (1) ) ⊗ f ( { r (2) , r ′ (2) } ) ⊗ f ( r (3) r ′ (3) )+ f ( r (1) r ′ (1) ) ⊗ f ( r (2) r ′ (2) ) ⊗ f ( { r (3) , r ′ (3) } ) , so µ B ( { b, b ′ } ) = { µ B ( b ) , µ B ( b ′ ) } .(2) It is straightforward.In what follows, we establish the relationship between Poisson Hopf-Galois algebras and PoissonHopf algebras. Proposition 3.7
Let H be a Poisson algebra. Then the following conclusions hold.(1) Suppose that H is a Hopf-Galois algebra with the Hopf-Galois map µ . If there is an algebramap α : H → k such that { H, H } ⊆
Ker α , then, H has a structure of Poisson Hopf algebra.(2) If H is a Poisson Hopf algebra, then, H is also a Poisson Hopf-Galois algebra.8 roof. (1) By Proposition 2.3, we know that H is a Hopf algebra whose counit is α and theother structure maps still defined by Eq.(2.3).So, in order to prove that H has a structure of Poisson Hopf algebra, we need to prove that∆( { x, y } ) = { ∆( x ) , ∆( y ) } for any x, y ∈ H .In fact, since H is a Poisson Hopf-Galois algebra, we have µ ( { x, y } ) = { µ ( x ) , µ ( y ) } , that is,Eq.(3.4) holds.Hence, we get∆( { x, y } ) = α ( { x, y } (2) ) { x, y } (1) ⊗ { x, y } (3) = α ( x (2) y (2) ) { x (1) , y (1) } ⊗ x (3) y (3) − α ( { x (2) , y (2) } ) x (1) y (1) ⊗ x (3) y (3) + α ( x (2) y (2) ) x (1) y (1) ⊗ { x (3) , y (3) } = α ( x (2) y (2) ) { x (1) , y (1) } ⊗ x (3) y (3) + α ( x (2) y (2) ) x (1) y (1) ⊗ { x (3) , y (3) } , { ∆( x ) , ∆( y ) } = x y ⊗ { x , y } + { x , y } ⊗ x y = α ( x (2) ) x (1) α ( y (2) ) y (1) ⊗ { x (3) , y (3) } + { α ( x (2) ) x (1) , α ( y (2) ) y (1) } ⊗ x (3) y (3) = α ( x (2) y (2) ) x (1) y (1) ⊗ { x (3) , y (3) } + α ( x (2) y (2) ) { x (1) , y (1) } ⊗ x (3) y (3) . So, ∆( { x, y } ) = { ∆( x ) , ∆( y ) } , and H is a Poisson Hopf algebra.(2) Assume that H is a Poisson Hopf algebra. Then, there exist an algebra map α = ε : H → k such that { H, H } ⊆
Ker α by Lemma 3.4.Define the map µ by Eq.(2.4), we get H is a Hopf-Galois algebra by Proposition 2.3. In whatfollows, we only need to check µ ( { x, y } ) = { µ ( x ) , µ ( y ) } in order to prove that H is a PoissonHopf-Galois algebra.As a matter of fact, since ∆( { x, y } ) = { ∆( x ) , ∆( y ) } , we have { x, y } ⊗ { x, y } ⊗ { x, y } = x y ⊗ x y ⊗ { x , y } + { x , y } ⊗ x y ⊗ x y + x y ⊗ { x , y } ⊗ x y . Hence we get µ ( { x, y } ) = { x, y } ⊗ S ( { x, y } ) ⊗ { x, y } = x y ⊗ S ( x y ) ⊗ { x , y } + { x , y } ⊗ S ( x y ) ⊗ x y + x y ⊗ S ( { x , y } ) ⊗ x y = x y ⊗ S ( x y ) ⊗ { x , y } + { x , y } ⊗ S ( x y ) ⊗ x y − x y ⊗ { S ( x ) , S ( y ) } ⊗ x y , { µ ( x ) , µ ( y ) } = { x (1) ⊗ x (2) ⊗ x (3) , y (1) ⊗ y (2) ⊗ y (3) } = { x ⊗ S ( x ) ⊗ x , y ⊗ S ( y ) ⊗ y } = { x , y } ⊗ S ( x ) S ( y ) ⊗ x y − x y ⊗ { S ( x ) , S ( y ) } ⊗ x y + x y ⊗ S ( x ) S ( y ) ⊗ { x , y } = { x , y } ⊗ S ( x y ) ⊗ x y − x y ⊗ { S ( x ) , S ( y ) } ⊗ x y + x y ⊗ S ( x y ) ⊗ { x , y } . So, µ ( { x, y } ) = { µ ( x ) , µ ( y ) } , and H is a Poisson Hopf-Galois algebra. (cid:3) xample 3.8 Let B = k [ g, g − , x ] be the polynomial algebra. Define a bracket product on B as follows { x, g } = λgx, where 0 = λ ∈ k . Then, there is a Poisson Hopf-Galois algebra on B with the Hopf-Galois map µ ( g ) = g ⊗ g − ⊗ g,µ ( x ) = 1 ⊗ ⊗ x − ⊗ xg ⊗ g − + x ⊗ g ⊗ g − . Proof.
Assume that ( B, {· , ·} ) is a Poisson algebra. Then, { g, g − } = 0 by the equality0 = { g, g − g } = { g, g − } g + g − { g, g } . Since { x, g − g } = { x, g − } g + g − { x, g } , we get { x, g − } = − λg − x. In what follows, we check that B is a Hopf-Galois algebra. Because µ ( g ) = g ⊗ g − ⊗ g , we get µ ( g − ) = g − ⊗ g ⊗ g − . By the above discussion, we easily get that g ⊗ g (1) ⊗ g (2) g (3) , ⊗ g = g (1) g (2) ⊗ g (3) ,g − ⊗ g − ) (1) ⊗ ( g − ) (2) ( g − ) (3) , ⊗ g − = ( g − ) (1) ( g − ) (2) ⊗ ( g − ) (3) ,x (1) ⊗ x (2) x (3) = 1 ⊗ x − ⊗ xgg − + x ⊗ x ⊗ ,x (1) x (2) ⊗ x (3) = 1 ⊗ x − xg ⊗ g − + xg ⊗ g − = 1 ⊗ x,µ ( g (1) ) ⊗ g (2) ⊗ g (3) = g (1) ⊗ g (2) ⊗ µ ( g (3) ) ,µ ( x (1) ) ⊗ x (2) ⊗ x (3) = 1 ⊗ ⊗ ⊗ ⊗ x − ⊗ ⊗ ⊗ xg ⊗ g − + 1 ⊗ ⊗ x ⊗ g ⊗ g − = x (1) ⊗ x (2) ⊗ µ ( x (3) ) , so, B is a Hopf-Galois algebra with the Hopf-Galois map µ .In the following, we prove that µ is a Poisson map in order to explain B is a Poisson Hopf-Galoisalgebra.As a matter of fact, we have no difficulty to prove { µ ( g ) , µ ( g − ) } = { g ⊗ g − ⊗ g, g − ⊗ g ⊗ g − } = 0= µ ( { g, g − } ) , { µ ( x ) , µ ( g ) } = { ⊗ ⊗ x − ⊗ xg ⊗ g − + x ⊗ g ⊗ g − , g ⊗ g − ⊗ g } = { ⊗ ⊗ x, g ⊗ g − ⊗ g } − { ⊗ xg ⊗ g − , g ⊗ g − ⊗ g } + { x ⊗ g ⊗ g − , g ⊗ g − ⊗ g } = g ⊗ g − ⊗ { x, g } + g ⊗ { xg, g − } ⊗ g ⊗ x ⊗ { g − , g } + { x, g } ⊗ ⊗ − xg ⊗ { g, g − } ⊗ xg ⊗ ⊗ { g − , g } g ⊗ g − ⊗ { x, g } + g ⊗ { xg, g − } ⊗ { x, g } ⊗ ⊗ g ⊗ g − ⊗ { x, g } − g ⊗ { g − , x } g ⊗ { x, g } ⊗ ⊗ g ⊗ g − ⊗ λgx − g ⊗ λx ⊗ λgx ⊗ ⊗ λ ( g ⊗ g − ⊗ gx − g ⊗ x ⊗ gx ⊗ ⊗ λ ( g ⊗ g − ⊗ g )(1 ⊗ ⊗ x − ⊗ xg ⊗ g − + x ⊗ g ⊗ g − )= λµ ( g ) µ ( x ) = µ ( { x, g } ) . In a similar way, we can prove µ ( { g − , x } ) = { µ ( g − ) , µ ( x ) } . § The Poisson polynomial algebra introduced in [12] is a Poisson version of the Ore extension.In this section, we study the Poisson Hopf-Galois structure on a Poisson polynomial algebra.
Definition 4.1
Let B be a Poisson algebra. A Poisson derivation on B is a k -linear map α from B to B which is a derivation with respect to both multiplication and the Poisson bracket,that is, for any a, b ∈ B , α ( ab ) = α ( a ) b + aα ( b ) , (4 . α ( { a, b } ) = { α ( a ) , b } + { a, α ( b ) } . (4 . Remark 4.2
Let B be a Poisson algebra and α a Poisson derivation on B . Suppose that δ isa derivation with respect to multiplication on B , such that for any a, b ∈ B , δ ( { a, b } ) = { δ ( a ) , b } + { a, δ ( b ) } + α ( a ) δ ( b ) − δ ( a ) α ( b ) . (4 . B can be extended uniquely toa Poisson structure on the polynomial algebra R = B [ x ] via { x, b } = α ( b ) x + δ ( b ) (4 . b ∈ B .In the following, we write R = B [ x ; α, δ ] p for this Poisson algebra B and call it a Poisson Oreextension of B . Definition 4.3
Let B be a Poisson Hopf-Galois algebra with the Hopf-Galois map µ , and B [ x ; α, δ ] p a Poisson Ore extension of B . If B [ x ; α, δ ] p is a Poisson Hopf-Galois algebra with B asa Poisson Hopf-Galois subalgebra, then we call B [ x ; α, δ ] P a Poisson Hopf-Galois Ore extension of B . Naturally, we have the following question: when is B [ x ; α, δ ] p a Poisson Hopf-Galois Ore ex-tension of B ?In what follows, we always assume that B [ x ; α, δ ] p is a Poisson Hopf-Galois algebra with B asa Poisson Hopf-Galois subalgebra, whose Hopf-Galois map is given by µ ( x ) = x ⊗ ⊗ − g ⊗ g − x ⊗ g ⊗ g − ⊗ x, (4 . g is a group-like element in B . 11 heorem 4.4 Let B be a Poisson Hopf-Galois algebra with a group-like element g . Then B [ x ; α, δ ] p is a Poisson Hopf-Galois Ore extension of B , if and only if(1) the equalities (4.6) ∼ (4.8) hold for any a, b ∈ B , α ( b ) = g − { g, b } , (4 . µ ( α ( b )) = α ( b (1) ) ⊗ b (2) ⊗ b (3) , (4 . b (1) ⊗ b (2) ⊗ α ( b (3) ) = b (1) ⊗ g { g − , b (2) } ⊗ b (3) , (4 . { g − , b } α ( a ) = { g − , a } α ( b ) , (4 . δ satisfies µ ( δ ( b )) = δ ( b (1) ) ⊗ b (2) ⊗ b (3) + gb (1) ⊗ { g − x, b (2) } ⊗ b (3) + gb (1) ⊗ g − b (2) ⊗ δ ( b (3) ) (4 . b ∈ B . Proof.
For any b ∈ B , we have µ ( { x, b } ) = µ ( α ( b ) x + δ ( b )) = µ ( α ( b )) µ ( x ) + µ ( δ ( b )) (4 . = µ ( α ( b ))( x ⊗ ⊗ − µ ( α ( b ))( g ⊗ g − x ⊗
1) + µ ( α ( b ))( g ⊗ g − ⊗ x ) + µ ( δ ( b )) , { µ ( x ) , µ ( b ) } = { x ⊗ ⊗ − g ⊗ g − x ⊗ g ⊗ g − ⊗ x, b (1) ⊗ b (2) ⊗ b (3) } = { x ⊗ ⊗ , b (1) ⊗ b (2) ⊗ b (3) } − { g ⊗ g − x ⊗ , b (1) ⊗ b (2) ⊗ b (3) } + { g ⊗ g − ⊗ x, b (1) ⊗ b (2) ⊗ b (3) } (3 . = { x, b (1) } ⊗ b (2) ⊗ b (3) − { g, b (1) } ⊗ g − xb (2) ⊗ b (3) + gb (1) ⊗ { g − x, b (2) } ⊗ b (3) + { g, b (1) } ⊗ g − b (2) ⊗ xb (3) − gb (1) ⊗ { g − , b (2) } ⊗ xb (3) + gb (1) ⊗ g − b (2) ⊗ { x, b (3) } (4 . = ( α ( b (1) ) x + δ ( b (1) )) ⊗ b (2) ⊗ b (3) − { g, b (1) } ⊗ g − xb (2) ⊗ b (3) + gb (1) ⊗ { g − x, b (2) } ⊗ b (3) + { g, b (1) } ⊗ g − b (2) ⊗ xb (3) − gb (1) ⊗ { g − , b (2) } ⊗ xb (3) + gb (1) ⊗ g − b (2) ⊗ ( α ( b (3) ) x + δ ( b (3) ))= α ( b (1) ) x ⊗ b (2) ⊗ b (3) + δ ( b (1) ) ⊗ b (2) ⊗ b (3) − { g, b (1) } ⊗ g − xb (2) ⊗ b (3) + gb (1) ⊗ { g − x, b (2) } ⊗ b (3) + { g, b (1) } ⊗ g − b (2) ⊗ xb (3) − gb (1) ⊗ { g − , b (2) } ⊗ xb (3) + gb (1) ⊗ g − b (2) ⊗ α ( b (3) ) x + gb (1) ⊗ g − b (2) ⊗ δ ( b (3) )= ( α ( b (1) ) ⊗ b (2) ⊗ b (3) )( x ⊗ ⊗
1) + δ ( b (1) ) ⊗ b (2) ⊗ b (3) − ( g − { g, b (1) } ⊗ b (2) ⊗ b (3) )( g ⊗ g − x ⊗
1) + gb (1) ⊗ { g − x, b (2) } ⊗ b (3) + ( g − { g, b (1) } ⊗ b (2) ⊗ b (3) )( g ⊗ g − ⊗ x ) − ( b (1) ⊗ g { g − , b (2) } ⊗ b (3) )( g ⊗ g − ⊗ x )+ ( b (1) ⊗ b (2) ⊗ α ( b (3) ))( g ⊗ g − ⊗ x ) + gb (1) ⊗ g − b (2) ⊗ δ ( b (3) ) . Assume that B [ x ; α, δ ] p is a Poisson Hopf-Galois algebra. Then, µ ( { x, b } ) = { µ ( x ) , µ ( b ) } ).12hus, by the above computation, we get µ ( α ( b )) = α ( b (1) ) ⊗ b (2) ⊗ b (3) ,µ ( α ( b )) = g − { g, b (1) } ⊗ b (2) ⊗ b (3) ,µ ( α ( b )) = b (1) ⊗ b (2) ⊗ α ( b (3) ) − b (1) ⊗ g { g − , b (2) } ⊗ b (3) + g − { g, b (1) } ⊗ b (2) ⊗ b (3) ,µ ( δ ( b )) = δ ( b (1) ) ⊗ b (2) ⊗ b (3) + gb (1) ⊗ { g − x, b (2) } ⊗ b (3) + gb (1) ⊗ g − b (2) ⊗ δ ( b (3) ) . By the above second and third equations, we can get b (1) ⊗ b (2) ⊗ α ( b (3) ) = b (1) ⊗ g { g − , b (2) } ⊗ b (3) , so, by Eq. (2.1) and the above second equation, we have α ( b ) = g − { g, b } . Hence, we obtain that α ( { a, b } ) = g − { g, { a, b }} = − g − { a, { b, g }} − g − { b, { g, a }} , { α ( a ) , b } + { a, α ( b ) } = { g − { g, a } , b } + { a, g − { g, b }} = −{ b, g − { g, a }} + { a, g − { g, b }} = −{ b, g − }{ g, a } − g − { b, { g, a }} + { a, g − }{ g, b } − g − { a, { b, g }} . Again by Eq. (4.2), we get { b, g − }{ g, a } = { a, g − }{ g, b } , that is, { g − , b } α ( a ) = { g − , a } α ( b ).Conversely, if conditions (1) and (2) hold, it is easy to see µ ( { x, b } ) = { µ ( x ) , µ ( b ) } ) by a directcomputation. Hence B [ x ; α, δ ] p has a Poisson Hopf-Galois algebra structure with B as a PoissonHopf-Galois subalgebra. (cid:3) § In section, we mainly give a necessary and sufficient condition for the Poisson enveloping algebraof a Poisson Hopf-Galois algebra to be a Hopf-Galois algebra.
Definition 5.1
Let A be a Poisson algebra, and ( U ( A ) , α, β ) a triple, where U ( A ) is an algebra, α : A → U ( A ) is an algebra homomorphism, and β : A → U ( A ) L is a Lie homomorphism suchthat α ( { a, b } ) = [ β ( a ) , α ( b )] , β ( ab ) = α ( a ) β ( b ) + α ( b ) β ( a ) , (5 . a, b ∈ A , is called the Poisson enveloping algebra in [11] for A if ( U ( A ) , α, β ) satisfies thefollowing: for any given algebra B , if γ is an algebra homomorphism from A into B , and δ is a Liehomomorphism from ( A, {· , ·} ) into B L such that γ ( { a, b } ) = [ δ ( a ) , γ ( b )] , δ ( ab ) = γ ( a ) δ ( b ) + γ ( b ) δ ( a ) , (5 . a, b ∈ A , then, there exists a unique algebra homomorphism h from U ( A ) into B such that hα = γ and hβ = δ , that is, we have the following commutative diagrams: A AU ( A ) B ❅❅❅■ (cid:0)(cid:0)(cid:0)✠❄❅❅❅❅❘(cid:0)(cid:0)(cid:0)✒ α βγ δh Here B L denotes the induced Lie algebra by B . Remark 5.2 (1) Note that there exists a unique Poisson enveloping algebra ( U ( A ) , α, β ) upto isomorphic, that U ( A ) is generated by α ( A ) and β ( A ) (see [11], proof of Theorem 5) and that β (1) = 0 by Eq. (5.1), for every Poisson algebra A .(2) Let ( U ( A ) , α, β ) be the Poisson enveloping algebra for a Poisson algebra A . In what follows,we call the two maps α, β Poisson enveloping maps.
Lemma 5.3
Suppose that γ and δ are two linear maps from a Poisson algebra A into an algebra B such that γ ( { a, b } ) = [ δ ( a ) , γ ( b )] , δ ( ab ) = γ ( a ) δ ( b ) + γ ( b ) δ ( a ) , for all a, b ∈ A , then, by [10], γ ( { a, b } ) = [ γ ( a ) , δ ( b )] , δ ( ab ) = δ ( a ) γ ( b ) + δ ( b ) γ ( a ) . (5 . B is given by [ x, y ] = xy − yx for x, y ∈ B . Remark 5.4 (1) By Lemma 5.3, for every Poisson enveloping algebra ( U ( A ) , α, β ) of a Poissonalgebra A , Eq. (5.1) can also be written by α ( { a, b } ) = [ α ( a ) , β ( b )] , β ( ab ) = β ( a ) α ( b ) + β ( b ) α ( a ) . (5 . ′ )(2) Let A = ( A, · , {· , ·} ) be a Poisson algebra. Define a bilinear map {· , ·} on A by { a, b } = { b, a } for all a, b ∈ A . Then, by [10], A = ( A, · , {· , ·} ) is a Poisson algebra. Thus, according toProposition 9 in [10], ( U ( A ) op , α, β ) is also the Poisson enveloping algebra for A , where ( U ( A ) op , ◦ )denotes the opposite algebra of U ( A ).(3) Let ( U ( A ) , α, β ) be the Poisson enveloping algebra for a Poisson algebra A . Then, by (2),we know β : A → U ( A ) opL is a Lie anti-homomorphism, that is, β ( { a, b } ) = β ( b ) ◦ β ( a ) − β ( a ) ◦ β ( b ),and α : A → U ( A ) op is an algebra homomorphism.Again by (5.1) and (5 . ′ ), we have β ( ab ) = α ( a ) ◦ β ( b ) + α ( b ) ◦ β ( a ) = β ( a ) ◦ α ( b ) + β ( b ) ◦ α ( a ) . (5 . α ( { a, b } ) = α ( b ) ◦ β ( a ) − β ( a ) ◦ α ( b ) = β ( b ) ◦ α ( a ) − α ( a ) ◦ β ( b ) . (5 . Lemma 5.5
Let ( U ( A ) , α, β ) be the Poisson enveloping algebra for a Poisson algebra A . Then141) α ⊗ α ⊗ α : A ⊗ A ⊗ A −→ U ( A ) ⊗ U ( A ) op ⊗ U ( A ) is an algebra homomorphism.Here A ⊗ A ⊗ A and U ( A ) ⊗ U ( A ) op ⊗ U ( A ) denote the usual tensor product algebras.(2) ξ : A ⊗ A ⊗ A −→ ( U ( A ) ⊗ U ( A ) op ⊗ U ( A )) L is a Lie homomorphism.Here ξ = ( id ⊗ τ + id ⊗ id ⊗ id + τ ⊗ id )( α ⊗ β ⊗ α ), that is, ξ = α ⊗ α ⊗ β + α ⊗ β ⊗ α + β ⊗ α ⊗ α ,and the Lie structure of A ⊗ A ⊗ A is given by Eq. (3.5). Proof. (1) Since ( U ( A ) , α, β ) is the Poisson enveloping algebra for a Poisson algebra A , themap α from A to U ( A ) is an algebra map. Hence α ⊗ α ⊗ α : A ⊗ A ⊗ A −→ U ( A ) ⊗ U ( A ) op ⊗ U ( A )is an algebra homomorphism.(2) In order to prove ξ to be a Lie homomorphism, we need to prove the following equality ξ ( { a ⊗ b ⊗ c, a ′ ⊗ b ′ ⊗ c ′ } ) = [ ξ ( a ⊗ b ⊗ c ) , ξ ( a ′ ⊗ b ′ ⊗ c ′ )] (5 . a, b, c, a ′ , b ′ , c ′ ∈ A .As a matter of fact, by Remark 5.4, we get ξ ( { a ⊗ b ⊗ c, a ′ ⊗ b ′ ⊗ c ′ } )= ξ ( { a, a ′ } ⊗ bb ′ ⊗ cc ′ − aa ′ ⊗ { b, b ′ } ⊗ cc ′ + aa ′ ⊗ bb ′ ⊗ { c, c ′ } )= ( α ⊗ α ⊗ β )( { a, a ′ } ⊗ bb ′ ⊗ cc ′ − aa ′ ⊗ { b, b ′ } ⊗ cc ′ + aa ′ ⊗ bb ′ ⊗ { c, c ′ } )+( α ⊗ β ⊗ α )( { a, a ′ } ⊗ bb ′ ⊗ cc ′ − aa ′ ⊗ { b, b ′ } ⊗ cc ′ + aa ′ ⊗ bb ′ ⊗ { c, c ′ } )+( β ⊗ α ⊗ α )( { a, a ′ } ⊗ bb ′ ⊗ cc ′ − aa ′ ⊗ { b, b ′ } ⊗ cc ′ + aa ′ ⊗ bb ′ ⊗ { c, c ′ } )= α ( { a, a ′ } ) ⊗ α ( bb ′ ) ⊗ β ( cc ′ ) − α ( aa ′ ) ⊗ α ( { b, b ′ } ) ⊗ β ( cc ′ )+ α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ β ( { c, c ′ } ) + α ( { a, a ′ } ) ⊗ β ( bb ′ ) ⊗ α ( cc ′ ) − α ( aa ′ ) ⊗ β ( { b, b ′ } ) ⊗ α ( cc ′ ) + α ( aa ′ ) ⊗ β ( bb ′ ) ⊗ α ( { c, c ′ } )+ β ( { a, a ′ } ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − β ( aa ′ ) ⊗ α ( { b, b ′ } ) ⊗ α ( cc ′ )+ β ( aa ′ ) ⊗ α ( bb ′ ) ⊗ α ( { c, c ′ } )= α ( { a, a ′ } ) ⊗ α ( bb ′ ) ⊗ β ( cc ′ ) − α ( aa ′ ) ⊗ α ( { b, b ′ } ) ⊗ β ( cc ′ )+ α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ ( β ( c ′ ) ◦ β ( c ) − β ( c ) ◦ β ( c ′ )) + α ( { a, a ′ } ) ⊗ β ( bb ′ ) ⊗ α ( cc ′ ) − α ( aa ′ ) ⊗ ( β ( b ′ ) ◦ β ( b ) − β ( b ) ◦ β ( b ′ )) ⊗ α ( cc ′ ) + α ( aa ′ ) ⊗ β ( bb ′ ) ⊗ α ( { c, c ′ } )+( β ( a ′ ) ◦ β ( a ) − β ( a ) ◦ β ( a ′ )) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − β ( aa ′ ) ⊗ α ( { b, b ′ } ) ⊗ α ( cc ′ )+ β ( aa ′ ) ⊗ α ( bb ′ ) ⊗ α ( { c, c ′ } )= α ( { a, a ′ } ) ⊗ α ( bb ′ ) ⊗ β ( cc ′ ) | {z } − α ( aa ′ ) ⊗ α ( { b, b ′ } ) ⊗ β ( cc ′ ) | {z } + α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ β ( c ′ ) ◦ β ( c ) − α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ β ( c ) ◦ β ( c ′ )+ α ( { a, a ′ } ) ⊗ β ( bb ′ ) ⊗ α ( cc ′ ) | {z } − α ( aa ′ ) ⊗ β ( b ′ ) ◦ β ( b ) ⊗ α ( cc ′ )+ α ( aa ′ ) ⊗ β ( b ) ◦ β ( b ′ ) ⊗ α ( cc ′ ) + α ( aa ′ ) ⊗ β ( bb ′ ) ⊗ α ( { c, c ′ } ) | {z } + β ( a ′ ) ◦ β ( a ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − β ( a ) ◦ β ( a ′ ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − β ( aa ′ ) ⊗ α ( { b, b ′ } ) ⊗ α ( cc ′ ) | {z } + β ( aa ′ ) ⊗ α ( bb ′ ) ⊗ α ( { c, c ′ } ) | {z } (5 . . ′ ) = α ( { a, a ′ } ) ⊗ α ( bb ′ ) ⊗ β ( c ) α ( c ′ ) + α ( { a, a ′ } ) ⊗ α ( bb ′ ) ⊗ β ( c ′ ) α ( c ) | {z } − α ( aa ′ ) ⊗ α ( { b, b ′ } ) ⊗ α ( c ) β ( c ′ ) − α ( aa ′ ) ⊗ α ( { b, b ′ } ) ⊗ α ( c ′ ) β ( c ) | {z } α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ β ( c ′ ) ◦ β ( c ) − α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ β ( c ) ◦ β ( c ′ )+ α ( { a, a ′ } ) ⊗ β ( b ) α ( b ′ ) ⊗ α ( cc ′ ) + α ( { a, a ′ } ) ⊗ β ( b ′ ) α ( b ) ⊗ α ( cc ′ ) | {z } − α ( aa ′ ) ⊗ β ( b ′ ) ◦ β ( b ) ⊗ α ( cc ′ ) + α ( aa ′ ) ⊗ β ( b ) ◦ β ( b ′ ) ⊗ α ( cc ′ )+ α ( aa ′ ) ⊗ β ( b ) α ( b ′ ) ⊗ α ( { c, c ′ } ) + α ( aa ′ ) ⊗ β ( b ′ ) α ( b ) ⊗ α ( { c, c ′ } ) | {z } + β ( a ′ ) ◦ β ( a ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − β ( a ) ◦ β ( a ′ ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − α ( a ) β ( a ′ ) ⊗ α ( { b, b ′ } ) ⊗ α ( cc ′ ) − α ( a ′ ) β ( a ) ⊗ α ( { b, b ′ } ) ⊗ α ( cc ′ ) | {z } + β ( a ) α ( a ′ ) ⊗ α ( bb ′ ) ⊗ α ( { c, c ′ } ) + β ( a ′ ) α ( a ) ⊗ α ( bb ′ ) ⊗ α ( { c, c ′ } ) | {z } (5 . . ′ ) = ( α ( a ) β ( a ′ ) − β ( a ′ ) α ( a )) | {z } ⊗ α ( bb ′ ) ⊗ β ( c ) α ( c ′ )+ ( β ( a ) α ( a ′ ) − α ( a ′ ) β ( a )) | {z } ⊗ α ( bb ′ ) ⊗ β ( c ′ ) α ( c ) − α ( aa ′ ) ⊗ ( β ( b ) α ( b ′ ) − α ( b ′ ) β ( b )) | {z } ⊗ α ( c ) β ( c ′ ) − α ( aa ′ ) ⊗ ( α ( b ) β ( b ′ ) − β ( b ′ ) α ( b )) | {z } ⊗ α ( c ′ ) β ( c )+ α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ β ( c ′ ) ◦ β ( c ) − α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ β ( c ) ◦ β ( c ′ )+ ( α ( a ) β ( a ′ ) − β ( a ′ ) α ( a )) | {z } ⊗ β ( b ) α ( b ′ ) ⊗ α ( cc ′ )+ ( β ( a ) α ( a ′ ) − α ( a ′ ) β ( a )) | {z } ⊗ β ( b ′ ) α ( b ) ⊗ α ( cc ′ ) − α ( aa ′ ) ⊗ β ( b ′ ) ◦ β ( b ) ⊗ α ( cc ′ ) + α ( aa ′ ) ⊗ β ( b ) ◦ β ( b ′ ) ⊗ α ( cc ′ )+ α ( aa ′ ) ⊗ β ( b ) α ( b ′ ) ⊗ ( α ( c ) β ( c ′ ) − β ( c ′ ) α ( c )) | {z } + α ( aa ′ ) ⊗ β ( b ′ ) α ( b ) ⊗ ( β ( c ) α ( c ′ ) − α ( c ′ ) β ( c )) | {z } + β ( a ′ ) ◦ β ( a ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − β ( a ) ◦ β ( a ′ ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − α ( a ) β ( a ′ ) ⊗ ( β ( b ) α ( b ′ ) − α ( b ′ ) β ( b )) | {z } ⊗ α ( cc ′ ) − α ( a ′ ) β ( a ) ⊗ ( α ( b ) β ( b ′ ) − β ( b ′ ) α ( b )) | {z } ⊗ α ( cc ′ )+ β ( a ) α ( a ′ ) ⊗ α ( bb ′ ) ⊗ ( α ( c ) β ( c ′ ) − β ( c ′ ) α ( c )) | {z } + β ( a ′ ) α ( a ) ⊗ α ( bb ′ ) ⊗ ( β ( c ) α ( c ′ ) − α ( c ′ ) | {z } β ( c ))= α ( a ) β ( a ′ ) ⊗ α ( bb ′ ) ⊗ β ( c ) α ( c ′ ) − α ( a ′ ) β ( a ) ⊗ α ( bb ′ ) ⊗ β ( c ′ ) α ( c )+ α ( aa ′ ) ⊗ β ( b ) ◦ α ( b ′ ) ⊗ α ( c ) β ( c ′ ) − α ( aa ′ ) ⊗ β ( b ′ ) ◦ α ( b ) ⊗ α ( c ′ ) β ( c )+ α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ β ( c ) β ( c ′ ) − α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ β ( c ′ ) β ( c ) − β ( a ′ ) α ( a ) ⊗ α ( b ′ ) ◦ β ( b ) ⊗ α ( cc ′ ) + β ( a ) α ( a ′ ) ⊗ α ( b ) ◦ β ( b ′ ) ⊗ α ( cc ′ ) − α ( aa ′ ) ⊗ β ( b ′ ) ◦ β ( b ) ⊗ α ( cc ′ ) + α ( aa ′ ) ⊗ β ( b ) ◦ β ( b ′ ) ⊗ α ( cc ′ ) − α ( aa ′ ) ⊗ α ( b ′ ) ◦ β ( b ) ⊗ β ( c ′ ) α ( c ) + α ( aa ′ ) ⊗ α ( b ) ◦ β ( b ′ ) ⊗ β ( c ) α ( c ′ )+ β ( a ) β ( a ′ ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − β ( a ′ ) β ( a ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ )+ α ( a ) β ( a ′ ) ⊗ β ( b ) ◦ α ( b ′ ) ⊗ α ( cc ′ ) − α ( a ′ ) β ( a ) ⊗ β ( b ′ ) ◦ α ( b ) ⊗ α ( cc ′ )+ β ( a ) α ( a ′ ) ⊗ α ( bb ′ ) ⊗ α ( c ) β ( c ′ ) − β ( a ′ ) α ( a ) ⊗ α ( bb ′ ) ⊗ α ( c ′ ) β ( c ) , [ ξ ( a ⊗ b ⊗ c ) , ξ ( a ′ ⊗ b ′ ⊗ c ′ )]16 ξ ( a ⊗ b ⊗ c ) ξ ( a ′ ⊗ b ′ ⊗ c ′ ) − ξ ( a ′ ⊗ b ′ ⊗ c ′ ) ξ ( a ⊗ b ⊗ c )= ( α ( a ) ⊗ α ( b ) ⊗ β ( c ) + α ( a ) ⊗ β ( b ) ⊗ α ( c ) + β ( a ) ⊗ α ( b ) ⊗ α ( c ))( α ( a ′ ) ⊗ α ( b ′ ) ⊗ β ( c ′ ) + α ( a ′ ) ⊗ β ( b ′ ) ⊗ α ( c ′ ) + β ( a ′ ) ⊗ α ( b ′ ) ⊗ α ( c ′ )) − ( α ( a ′ ) ⊗ α ( b ′ ) ⊗ β ( c ′ ) + α ( a ′ ) ⊗ β ( b ′ ) ⊗ α ( c ′ ) + β ( a ′ ) ⊗ α ( b ′ ) ⊗ α ( c ′ ))( α ( a ) ⊗ α ( b ) ⊗ β ( c ) + α ( a ) ⊗ β ( b ) ⊗ α ( c ) + β ( a ) ⊗ α ( b ) ⊗ α ( c ))= α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ β ( c ) β ( c ′ ) + α ( aa ′ ) ⊗ α ( b ) ◦ β ( b ′ ) ⊗ β ( c ) α ( c ′ )+ α ( a ) β ( a ′ ) ⊗ α ( bb ′ ) ⊗ β ( c ) α ( c ′ ) + α ( aa ′ ) ⊗ β ( b ) ◦ α ( b ′ ) ⊗ α ( c ) β ( c ′ )+ α ( aa ′ ) ⊗ β ( b ) ◦ β ( b ′ ) ⊗ α ( cc ′ ) + α ( a ) β ( a ′ ) ⊗ β ( b ) ◦ α ( b ′ ) ⊗ α ( cc ′ )+ β ( a ) α ( a ′ ) ⊗ α ( bb ′ ) ⊗ α ( c ) β ( c ′ ) + β ( a ) α ( a ′ ) ⊗ α ( b ) ◦ β ( b ′ ) ⊗ α ( cc ′ )+ β ( a ) β ( a ′ ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − α ( a ′ a ) ⊗ α ( b ′ b ) ⊗ β ( c ′ ) β ( c ) − α ( a ′ a ) ⊗ α ( b ′ ) ◦ β ( b ) ⊗ β ( c ′ ) α ( c ) − α ( a ′ ) β ( a ) ⊗ α ( b ′ b ) ⊗ β ( c ′ ) α ( c ) − α ( a ′ a ) ⊗ β ( b ′ ) ◦ α ( b ) ⊗ α ( c ′ ) β ( c ) − α ( a ′ a ) ⊗ β ( b ′ ) ◦ β ( b ) ⊗ α ( c ′ c ) − α ( a ′ ) β ( a ) ⊗ β ( b ′ ) ◦ α ( b ) ⊗ α ( c ′ c ) − β ( a ′ ) α ( a ) ⊗ α ( b ′ b ) ⊗ α ( c ′ ) β ( c ) − β ( a ′ ) α ( a ) ⊗ α ( b ′ ) ◦ β ( b ) ⊗ α ( c ′ c ) − β ( a ′ ) β ( a ) ⊗ α ( b ′ b ) ⊗ α ( c ′ c ) . By the above proof, we know that ξ ( { a ⊗ b ⊗ c, a ′ ⊗ b ′ ⊗ c ′ } ) = [ ξ ( a ⊗ b ⊗ c ) , ξ ( a ′ ⊗ b ′ ⊗ c ′ )]. So, ξ is a Lie homomorphism. (cid:3) Lemma 5.6
Let A and B be Poisson algebras and C an algebra. If φ : A → B is a Poissonhomomorphism, α : B → C an algebra homomorphism and β : B → C L a Lie homomorphism suchthat α ( { b, b ′ } ) = [ β ( b ) , α ( b ′ )] , β ( bb ′ ) = α ( b ) β ( b ′ ) + α ( b ′ ) β ( b )for all b, b ′ ∈ B , then, according to [10], αφ : A → C is an algebra homomorphism and βφ : A → C L a Lie homomorphism such that αφ ( { a, a ′ } ) = [ βφ ( a ) , αφ ( a ′ )] , βφ ( aa ′ ) = αφ ( a ) βφ ( a ′ ) + αφ ( a ′ ) βφ ( a )for all a, a ′ ∈ A . Lemma 5.7
Let ( U ( A ) , α, β ) be the Poisson enveloping algebra for a Poisson algebra A . Then( U ( A ) ⊗ U ( A ) op ⊗ U ( A ) , α ⊗ α ⊗ α, ξ ) is the Poisson enveloping algebra for A ⊗ A ⊗ A .Here ξ = ( id ⊗ τ + id ⊗ id ⊗ id + τ ⊗ id )( α ⊗ β ⊗ α ), that is, ξ = α ⊗ α ⊗ β + α ⊗ β ⊗ α + β ⊗ α ⊗ α . Proof.
In order to prove that ( U ( A ) ⊗ U ( A ) op ⊗ U ( A ) , α ⊗ α ⊗ α, ξ ) is the Poisson envelopingalgebra for A ⊗ A ⊗ A , the Proof is divided into the following steps by Definition 5.1. • Step 1, we prove that α ⊗ α ⊗ α is an algebra homomorphism and ξ is a Lie homomorphism.It is obvious by Lemma 5.5. • Step 2, we prove the equalities (5 .
1) hold for the Poisson enveloping map α ⊗ α ⊗ α .For any a, b, c, a ′ , b ′ , c ′ ∈ A , we get( α ⊗ α ⊗ α )( { a ⊗ b ⊗ c, a ′ ⊗ b ′ ⊗ c ′ } )= ( α ⊗ α ⊗ α )( { a, a ′ } ⊗ bb ′ ⊗ cc ′ − aa ′ ⊗ { b, b ′ } ⊗ cc ′ + aa ′ ⊗ bb ′ ⊗ { c, c ′ } )17 α ( { a, a ′ } ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − α ( aa ′ ) ⊗ α ( { b, b ′ } ) ⊗ α ( cc ′ ) + α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ α ( { c, c ′ } ) (5 . = ( β ( a ) α ( a ′ ) − α ( a ′ ) β ( a ) | {z } ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − α ( aa ′ ) ⊗ ( α ( b ′ ) ◦ β ( b ) − β ( b ) ◦ α ( b ′ ) | {z } ) ⊗ α ( cc ′ )+ α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ ( β ( c ) α ( c ′ ) − α ( c ′ ) β ( c ) | {z } )= β ( a ) α ( a ′ ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − α ( a ′ ) β ( a ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − α ( aa ′ ) ⊗ α ( b ′ ) ◦ β ( b ) ⊗ α ( cc ′ )+ α ( aa ′ ) ⊗ β ( b ) ◦ α ( b ′ ) ⊗ α ( cc ′ ) + α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ β ( c ) α ( c ′ ) − α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ α ( c ′ ) β ( c ) , [ ξ ( a ⊗ b ⊗ c ) , ( α ⊗ α ⊗ α )( a ′ ⊗ b ′ ⊗ c ′ )]= ξ ( a ⊗ b ⊗ c )( α ( a ′ ) ⊗ α ( b ′ ) ⊗ α ( c ′ )) − ( α ( a ′ ) ⊗ α ( b ′ ) ⊗ α ( c ′ )) ξ ( a ⊗ b ⊗ c )= ( α ( a ) ⊗ α ( b ) ⊗ β ( c ) + α ( a ) ⊗ β ( b ) ⊗ α ( c ) + β ( a ) ⊗ α ( b ) ⊗ α ( c ) | {z } )( α ( a ′ ) ⊗ α ( b ′ ) ⊗ α ( c ′ )) − ( α ( a ′ ) ⊗ α ( b ′ ) ⊗ α ( c ′ ))( α ( a ) ⊗ α ( b ) ⊗ β ( c ) + α ( a ) ⊗ β ( b ) ⊗ α ( c ) + β ( a ) ⊗ α ( b ) ⊗ α ( c ) | {z } )= α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ β ( c ) α ( c ′ ) + α ( aa ′ ) ⊗ β ( b ) ◦ α ( b ′ ) ⊗ α ( cc ′ )+ β ( a ) α ( a ′ ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) − α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ α ( c ′ ) β ( c ) − α ( aa ′ ) ⊗ α ( b ′ ) ◦ β ( b ) ⊗ α ( cc ′ ) − α ( a ′ ) β ( a ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) , so, ( α ⊗ α ⊗ α )( { a ⊗ b ⊗ c, a ′ ⊗ b ′ ⊗ c ′ } ) = [ ξ ( a ⊗ b ⊗ c ) , ( α ⊗ α ⊗ α )( a ′ ⊗ b ′ ⊗ c ′ )]. ξ ( aa ′ ⊗ bb ′ ⊗ cc ′ )= α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ β ( cc ′ ) + α ( aa ′ ) ⊗ β ( bb ′ ) ⊗ α ( cc ′ ) + β ( aa ′ ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) (5 . . ′ ) = α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ ( α ( c ) β ( c ′ ) + α ( c ′ ) β ( c ) | {z } ) + α ( aa ′ ) ⊗ ( α ( b ′ ) ◦ β ( b ) + α ( b ) ◦ β ( b ′ ) | {z } ) ⊗ α ( cc ′ )+( α ( a ) β ( a ′ ) + α ( a ′ ) β ( a ) | {z } ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) , ( α ⊗ α ⊗ α )( a ⊗ b ⊗ c ) ξ ( a ′ ⊗ b ′ ⊗ c ′ ) + ( α ⊗ α ⊗ α )( a ′ ⊗ b ′ ⊗ c ′ ) ξ ( a ⊗ b ⊗ c )= ( α ( a ) ⊗ α ( b ) ⊗ α ( c ))( α ( a ′ ) ⊗ α ( b ′ ) ⊗ β ( c ′ ) + α ( a ′ ) ⊗ β ( b ′ ) ⊗ α ( c ′ ) + β ( a ′ ) ⊗ α ( b ′ ) ⊗ α ( c ′ ) | {z } )+( α ( a ′ ) ⊗ α ( b ′ ) ⊗ α ( c ′ ))( α ( a ) ⊗ α ( b ) ⊗ β ( c ) + α ( a ) ⊗ β ( b ) ⊗ α ( c ) + β ( a ) ⊗ α ( b ) ⊗ α ( c ) | {z } )= α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ α ( c ) β ( c ′ ) + α ( aa ′ ) ⊗ α ( b ) ◦ β ( b ′ ) ⊗ α ( cc ′ ) + α ( a ) β ( a ′ ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ )+ α ( aa ′ ) ⊗ α ( bb ′ ) ⊗ α ( c ′ ) β ( c ) + α ( aa ′ ) ⊗ α ( b ′ ) ◦ β ( b ) ⊗ α ( cc ′ ) + α ( a ′ ) β ( a ) ⊗ α ( bb ′ ) ⊗ α ( cc ′ ) , so, ξ ( aa ′ ⊗ bb ′ ⊗ cc ′ ) = ( α ⊗ α ⊗ α )( a ⊗ b ⊗ c ) ξ ( a ′ ⊗ b ′ ⊗ c ′ ) + ( α ⊗ α ⊗ α )( a ′ ⊗ b ′ ⊗ c ′ ) ξ ( a ⊗ b ⊗ c ). • Step 3, we prove the universal property of ( U ( A ) ⊗ U ( A ) op ⊗ U ( A ) , α ⊗ α ⊗ α, ξ ).Given an algebra B , let m B be the multiplication map on B . If γ is an algebra homomorphismfrom A ⊗ A ⊗ A into B and δ is a Lie homomorphism from A ⊗ A ⊗ A into B L such that γ ( { a ⊗ b ⊗ c, a ′ ⊗ b ′ ⊗ c ′ } ) = [ δ ( a ⊗ b ⊗ c ) , γ ( a ′ ⊗ b ′ ⊗ c ′ )] , (5 . δ (( a ⊗ b ⊗ c )( a ′ ⊗ b ′ ⊗ c ′ )) = γ ( a ⊗ b ⊗ c ) δ ( a ′ ⊗ b ′ ⊗ c ′ ) + γ ( a ′ ⊗ b ′ ⊗ c ′ ) δ ( a ⊗ b ⊗ c ) (5 . a, b, c, a ′ , b ′ , c ′ ∈ A .Let i , i and i be the Poisson homomorphisms defined by i : A → A ⊗ A ⊗ A, i ( a ) = a ⊗ ⊗ ,i : A → A ⊗ A ⊗ A, i ( a ) = 1 ⊗ a ⊗ ,i : A → A ⊗ A ⊗ A, i ( a ) = 1 ⊗ ⊗ a a ∈ A .Then, by Lemma 5.6 and the universal property of a Poisson algebra A , there exist algebrahomomorphisms f, θ : U ( A ) → B , and g : U ( A ) op → B such that f α = γi , f β = δi , gα = γi , gβ = δi , θα = γi , θβ = δi , that is, we have the following commutative diagrams: AU ( A ) B ✲ ✲✻ ✻ A ⊗ A ⊗ Ai α, β f γ, δ A A ⊗ A ⊗ AU ( A ) op B ✲✻ ✻✲ α, β γ, δi g AU ( A ) ✻ ✻✲ ✲ A ⊗ A ⊗ ABα, β γ, δi θ Moreover, we have δi ( a ) γi ( b ) = γi ( b ) δi ( a ) , (5 . δ ( a ⊗ b ⊗ γ (1 ⊗ ⊗ c ) = γ (1 ⊗ ⊗ c ) δ ( a ⊗ b ⊗
1) (5 . a, b ∈ A . This is because[ δi ( a ) , γi ( b )] (5 . = γ ( { a ⊗ ⊗ , ⊗ b ⊗ } ) (3 . = 0 , [ δ ( a ⊗ b ⊗ , γ (1 ⊗ ⊗ c )] (5 . = γ ( { a ⊗ b ⊗ , ⊗ ⊗ c } ) (3 . = 0 . In a similar way, by Eq. (5 . δi m ( a ) γi n ( b ) = γi n ( b ) δi m ( a ) (5 . ′ )for any 1 m = n δ is a Lie homomorphism from A ⊗ A ⊗ A to B , we can directly prove that δi m ( a ) δi n ( b ) = δi n ( b ) δi m ( a ) (5 . m = n a, b, c ∈ A , we have m B ( id ⊗ m B )( f ⊗ g ⊗ θ )( α ⊗ α ⊗ α )( a ⊗ b ⊗ c )= f ( α ( a )) g ( α ( b )) θ ( α ( c )) = γi ( a ) γi ( a ) γi ( a )= γ ( i ( a ) i ( b ) i ( c )) = γ ( a ⊗ b ⊗ c ) ,m B ( id ⊗ m B )( f ⊗ g ⊗ θ ) ξ ( a ⊗ b ⊗ c )= f ( α ( a )) g ( α ( b )) θ ( β ( c )) + f ( α ( a )) g ( β ( b )) θ ( α ( c )) + f ( β ( a )) g ( α ( b )) θ ( α ( c ))= γi ( a ) γi ( b ) δi ( c ) + γi ( a ) δi ( b ) γi ( c ) + δi ( a ) γi ( b ) γi ( c ) (5 . = γi ( a ) γi ( b ) δi ( c ) + γi ( a ) δi ( b ) γi ( c ) + γi ( b ) δi ( a ) γi ( c )= γi ( a ) γi ( b ) δi ( c ) + ( γi ( a ) δi ( b ) + γi ( b ) δi ( a )) | {z } γi ( c ) (5 . = γi ( a ) γi ( b ) δi ( c ) + δ ( i ( a ) i ( b )) γi ( c )= γ ( a ⊗ b ⊗ δ (1 ⊗ ⊗ c ) + δ ( a ⊗ b ⊗ γ (1 ⊗ ⊗ c ) (5 . = γ ( a ⊗ b ⊗ δ (1 ⊗ ⊗ c ) + γ (1 ⊗ ⊗ c ) δ ( a ⊗ b ⊗ (5 . = δ ( a ⊗ b ⊗ c ) , a, b, c ∈ A , that is, we have m B ( id ⊗ m B )( f ⊗ g ⊗ θ )( α ⊗ α ⊗ α ) = γ, (5 . m B ( id ⊗ m B )( f ⊗ g ⊗ θ ) ξ = δ. (5 . m B ( id ⊗ m B )( f ⊗ g ⊗ θ ) is an algebra homomorphism from U ( A ) ⊗ U ( A ) op ⊗ U ( A ) to B .Since γ is an algebra homomorphism, we are easy to see that γ ( a ⊗ b ⊗ c ) γ ( a ′ ⊗ b ′ ⊗ c ′ ) = γ ( a ′ ⊗ b ′ ⊗ c ′ ) γ ( a ⊗ b ⊗ c ) (5 . a, b, c, a ′ , b ′ , c ′ ∈ A .For any a, b, c, a ′ , b ′ , c ′ ∈ A , we have, m B ( id ⊗ m B )( f ⊗ g ⊗ θ )(( α ( a ) ⊗ β ( b ) ⊗ α ( c ))( α ( a ′ ) ⊗ α ( b ′ ) ⊗ β ( c ′ )))= m B ( id ⊗ m B )( f ⊗ g ⊗ θ )( α ( a ) α ( a ′ ) ⊗ β ( b ) ◦ α ( b ′ ) ⊗ α ( c ) β ( c ′ ))= m B ( id ⊗ m B )( f ( α ( a ) α ( a ′ )) ⊗ g ( β ( b ) ◦ α ( b ′ )) ⊗ θ ( α ( c ) β ( c ′ )))= m B ( id ⊗ m B )( f ( α ( a )) f ( α ( a ′ )) ⊗ g ( β ( b )) g ( α ( b ′ )) ⊗ θ ( α ( c )) θ ( β ( c ′ )))= f ( α ( a )) f ( α ( a ′ )) g ( β ( b )) g ( α ( b ′ )) θ ( α ( c )) θ ( β ( c ′ ))= γi ( a ) γi ( a ′ ) δi ( b ) | {z } γi ( b ′ ) γi ( c ) δi ( c ′ ) (5 . ′ ) = γi ( a ) δi ( b ) γi ( a ′ ) γi ( b ′ ) γi ( c ) | {z } δi ( c ′ ) (5 . = γi ( a ) δi ( b ) γi ( a ′ ) γi ( c ) | {z } γi ( b ′ ) δi ( c ′ ) = γi ( a ) δi ( b ) γi ( c ) γi ( a ′ ) γi ( b ′ ) δi ( c ′ )= m B ( id ⊗ m B )( f ⊗ g ⊗ θ )(( α ( a ) ⊗ β ( b ) ⊗ α ( c ))) m B ( id ⊗ m B )( f ⊗ g ⊗ θ )(( α ( a ′ ) ⊗ α ( b ′ ) ⊗ β ( c ′ ))) . Similarly, we can prove the other cases are satisfied, so, m B ( id ⊗ m B )( f ⊗ g ⊗ θ ) is an algebrahomomorphism such that m B ( id ⊗ m B )( f ⊗ g ⊗ θ )( α ⊗ α ⊗ α ) = γ, m B ( id ⊗ m B )( f ⊗ g ⊗ θ ) ξ = δ by the equalities (5.10) and (5.11).Finally, we prove that the algebra homomorphism from U ( A ⊗ A ⊗ A ) to B is unique.Suppose that there exists a map h : U ( A ⊗ A ⊗ A ) → B is an algebra homomorphism such that h ( α ⊗ α ⊗ α ) = γ, hξ = δ. Then, for all a ∈ A , we have m B ( id ⊗ m B )( f ⊗ g ⊗ θ )( α ( a ) ⊗ ⊗
1) = m B ( id ⊗ m B )( f ⊗ g ⊗ θ )( α ⊗ α ⊗ α )( a ⊗ ⊗ (5 . = γ ( a ⊗ ⊗
1) = h ( α ⊗ α ⊗ α )( a ⊗ ⊗ h ( α ( a ) ⊗ ⊗ . In a similar way, we can prove that m B ( id ⊗ m B )( f ⊗ g ⊗ θ )(1 ⊗ α ( a ) ⊗
1) = h (1 ⊗ α ( a ) ⊗ ,m B ( id ⊗ m B )( f ⊗ g ⊗ θ )(1 ⊗ ⊗ α ( a )) = h (1 ⊗ ⊗ α ( a )) . Moreover, we know that m B ( id ⊗ m B )( f ⊗ g ⊗ θ )( β ( a ) ⊗ ⊗
1) = m B ( id ⊗ m B )( f ⊗ g ⊗ θ ) ξ ( a ⊗ ⊗ (5 . = δ ( a ⊗ ⊗
1) = hξ ( a ⊗ ⊗ h ( β ( a ) ⊗ ⊗ .
20n a similar way, we can prove that m B ( id ⊗ m B )( f ⊗ g ⊗ θ )(1 ⊗ β ( a ) ⊗
1) = h (1 ⊗ β ( a ) ⊗ ,m B ( id ⊗ m B )( f ⊗ g ⊗ θ )(1 ⊗ ⊗ β ( a )) = h (1 ⊗ ⊗ β ( a )) . Hence, by the above proof and the fact that U ( A ) is generated by α ( A ) and β ( A ), we obtainthat m B ( id ⊗ m B )( f ⊗ g ⊗ θ ) = h , which completes the proof. (cid:3) The following lemma can be found in [10].
Lemma 5.8
Let ( U ( A ) , α A , β A ) and ( U ( B ) , α B , β B ) be Poisson enveloping algebras for Poissonalgebras A and B , respectively. If φ : A → B is a Poisson homomorphism, then, there exists aunique algebra homomorphism U ( φ ) : U ( A ) → U ( B ) such that U ( φ ) α A = α B φ, U ( φ ) β A = β B φ, (5 . U ( A ) A U ( B ) B ✲✲✻ ✻ α A , β A α B , β B U ( φ ) φ According to the above lemmas, we obtain a main result in this section.
Theorem 5.9
Let (
A, µ ) be a Poisson Hopf-Galois algebra, and ( U ( A ) , α, β ) its Poisson en-veloping algebra. Then ( U ( A ) , U ( µ )) satisfies the following conditions: U ( µ ) α = ( α ⊗ α ⊗ α ) µ, U ( µ ) β = ξµ. (5 . ξ = ( id ⊗ τ + id ⊗ id ⊗ id + τ ⊗ id )( α ⊗ β ⊗ α ), that is, ξ = α ⊗ α ⊗ β + α ⊗ β ⊗ α + β ⊗ α ⊗ α .Thus, ( U ( A ) , U ( µ )) is a Hopf-Galois algebra if and only if β + m U ( A ) ( id ⊗ m U ( A ) )( α ⊗ β ⊗ α ) µ = 0 , (5 . β ( a ) = − α ( a (1) ) β ( a (2) ) α ( a (3) ) for any a ∈ A . Proof.
Since µ is a Poisson homomorphism and ( U ( A ) ⊗ U ( A ) op ⊗ U ( A ) , α ⊗ α ⊗ α, ξ ) is thePoisson enveloping algebra of A ⊗ A ⊗ A , there exists an algebra homomorphism U ( µ ) : U ( A ) → U ( A ) ⊗ U ( A ) op ⊗ U ( A ) by Lemma 5.8, such that the equalities (5.16) hold.“ ⇐ =” According to the equalities (5.16), we have α ( a ) (1) ⊗ α ( a ) (2) ⊗ α ( a ) (3) = α ( a (1) ) ⊗ α ( a (2) ) ⊗ α ( a (3) ) , (5 . β ( a ) (1) ⊗ β ( a ) (2) ⊗ β ( a ) (3) = ξ ( a (1) ) ⊗ ξ ( a (2) ) ⊗ ξ ( a (3) ) (5 . a ∈ A , where U ( µ )( α ( a )) is denoted by α ( a ) (1) ⊗ α ( a ) (2) ⊗ α ( a ) (3) , and U ( µ )( β ( a )) denotedby β ( a ) (1) ⊗ β ( a ) (2) ⊗ β ( a ) (3) .By Eq. (5.18) and µ being a Hopf-Galois map for A , we are easy to check that U ( µ )( α ( a ) (1) ) ⊗ α ( a ) (2) ⊗ α ( a ) (3) = α ( a ) (1) ⊗ α ( a ) (2) ⊗ U ( µ )( α ( a ) (3) ) . (5 . a ∈ A , we have U ( µ )( β ( a ) (1) ) ⊗ β ( a ) (2) ⊗ β ( a ) (3)(5 . = U ( µ )( α ( a (1) )) ⊗ α ( a (2) ) ⊗ β ( a (3) ) + U ( µ )( α ( a (1) )) ⊗ β ( a (2) ) ⊗ α ( a (3) )+ U ( µ )( β ( a (1) )) ⊗ α ( a (2) ) ⊗ α ( a (3) ) (5 . , (5 . = α ( a (1)(1) ) ⊗ α ( a (1)(2) ) ⊗ α ( a (1)(3) ) ⊗ α ( a (2) ) ⊗ β ( a (3) )+ α ( a (1)(1) ) ⊗ α ( a (1)(2) ) ⊗ α ( a (1)(3) ) ⊗ β ( a (2) ) ⊗ α ( a (3) )+ α ( a (1)(1) ) ⊗ α ( a (1)(2) ) ⊗ β ( a (1)(3) ) | {z } ⊗ α ( a (2) ) ⊗ α ( a (3) )+ α ( a (1)(1) ) ⊗ β ( a (1)(2) ) ⊗ α ( a (1)(3) ) | {z } ⊗ α ( a (2) ) ⊗ α ( a (3) )+ β ( a (1)(1) ) ⊗ α ( a (1)(2) ) ⊗ α ( a (1)(3) ) | {z } ⊗ α ( a (2) ) ⊗ α ( a (3) ) ,β ( a ) (1) ⊗ β ( a ) (2) ⊗ U ( µ )( β ( a ) (3) ) (5 . = α ( a (1) ) ⊗ α ( a (2) ) ⊗ U ( µ )( β ( a (3) )) + α ( a (1) ) ⊗ β ( a (2) ) ⊗ U ( µ )( α ( a (3) ))+ β ( a (1) ) ⊗ α ( a (2) ) ⊗ U ( µ )( α ( a (3) ))= α ( a (1) ) ⊗ α ( a (2) ) ⊗ α ( a (3)(1) ) ⊗ α ( a (3)(2) ) ⊗ β ( a (3)(3) ) | {z } + α ( a (1) ) ⊗ α ( a (2) ) ⊗ α ( a (3)(1) ) ⊗ β ( a (3)(2) ) ⊗ α ( a (3)(3) ) | {z } + α ( a (1) ) ⊗ α ( a (2) ) ⊗ β ( a (3)(1) ) ⊗ α ( a (3)(2) ) ⊗ α ( a (3)(3) ) | {z } + α ( a (1) ) ⊗ β ( a (2) ) ⊗ α ( a (3)(1) ) ⊗ α ( a (3)(2) ) ⊗ α ( a (3)(3) )+ β ( a (1) ) ⊗ α ( a (2) ) ⊗ α ( a (3)(1) ) ⊗ α ( a (3)(2) ) ⊗ α ( a (3)(3) ) (2 . = α ( a (1)(1) ) ⊗ α ( a (1)(2) ) ⊗ α ( a (1)(3) ) ⊗ α ( a (2) ) ⊗ β ( a (3) )+ α ( a (1)(1) ) ⊗ α ( a (1)(2) ) ⊗ α ( a (1)(3) ) ⊗ β ( a (2) ) ⊗ α ( a (3) )+ α ( a (1)(1) ) ⊗ α ( a (1)(2) ) ⊗ β ( a (1)(3) ) ⊗ α ( a (2) ) ⊗ α ( a (3) )+ α ( a (1)(1) ) ⊗ β ( a (1)(2) ) ⊗ α ( a (1)(3) ) ⊗ α ( a (2) ) ⊗ α ( a (3) )+ β ( a (1)(1) ) ⊗ α ( a (1)(2) ) ⊗ α ( a (1)(3) ) ⊗ α ( a (2) ) ⊗ α ( a (3) ) , so, we have U ( µ )( β ( a ) (1) ) ⊗ β ( a ) (2) ⊗ β ( a ) (3) = β ( a ) (1) ⊗ β ( a ) (2) ⊗ U ( µ )( β ( a ) (3) ) . (5 . .
1) of an algebra map U ( µ ) holds.Assume that β ( a ) = − α ( a (1) ) β ( a (2) ) α ( a (3) ) for any a ∈ A . Then, we have β ( a ) (1) ⊗ β ( a ) (2) β ( a ) (3)(5 . = α ( a (1) ) ⊗ α ( a (2) ) β ( a (3) ) + α ( a (1) ) ⊗ β ( a (2) ) α ( a (3) ) + β ( a (1) ) ⊗ α ( a (2) ) α ( a (3) ) (2 . = α ( a (1) ) ⊗ α ( a (2) ) β ( a (3) ) + α ( a (1) ) ⊗ β ( a (2) ) α ( a (3) ) + β ( a ) ⊗ − α ( a (1) ) ⊗ α ( a (2) ) α ( a (3)(1) ) β ( a (3)(2) ) α ( a (3)(3) ) | {z } + α ( a (1) ) ⊗ β ( a (2) ) α ( a (3) ) + β ( a ) ⊗ − α ( a (1)(1) ) ⊗ α ( a (1)(2) ) α ( a (1)(3) ) β ( a (2) ) α ( a (3) ) + α ( a (1) ) ⊗ β ( a (2) ) α ( a (3) ) + β ( a ) ⊗ − α ( a (1) ) ⊗ β ( a (2) ) α ( a (3) ) + α ( a (1) ) ⊗ β ( a (2) ) α ( a (3) ) + β ( a ) ⊗ β ( a ) ⊗ . In a similar way, we can prove that β ( a ) (1) β ( a ) (2) ⊗ β ( a ) (3) = 1 ⊗ β ( a )for any a ∈ A . Hence ( U ( A ) , U ( µ )) is a Hopf-Galois algebra.“ ⇐ = ” If ( U ( A ) , U ( µ )) is a Hopf-Galois algebra, then β ( a ) (1) ⊗ β ( a ) (2) β ( a ) (3) = β ( a ) ⊗ a ∈ A .According to Eq. (5.19), we know that β ( a ) (1) ⊗ β ( a ) (2) ⊗ β ( a ) (3) = α ( a (1) ) ⊗ α ( a (2) ) ⊗ β ( a (3) )+ α ( a (1) ) ⊗ β ( a (2) ) ⊗ α ( a (3) ) + β ( a (1) ) ⊗ α ( a (2) ) ⊗ α ( a (3) ) . By applying m U ( A ) ( id ⊗ m U ( A ) ) to the above equation, we get β ( a ) = − α ( a (1) ) β ( a (2) ) α ( a (3) ) , which completes this proof. (cid:3) eferenceseferences