aa r X i v : . [ m a t h . R A ] F e b UNIVERSAL ENVELOPING ALGEBRAS OF POISSON SUPERALGEBRAS
THOMAS LAMKIN
Abstract.
In this paper, we define and study the universal enveloping algebra of Poisson superalgebras.In particular, a new PBW theorem for Lie-Rinehart superalgebras is proved, leading to a PBW theoremfor Poisson superalgebras; we show the universal enveloping algebra of a Poisson Hopf superalgebra (resp.Poisson-Ore extension) is a Hopf superalgebra (resp. iterated Ore extension); and we determine the universalenveloping algebra for examples such as quadratic polynomial Poisson superalgebras and Poisson symmetricsuperalgebras. Introduction
The notion of a Poisson algebra arises naturally in the study of Hamiltonian mechanics, and has sincefound use in areas such as Poisson or symplectic geometry, and quantum groups. Due to the development ofsupersymmetry theories, one has also witnessed the increased use of Poisson superalgebras, along with othersuper structures such as Lie superalgebras and supermanifolds. One method of studying the representationtheory of Poisson algebras is to study its universal enveloping algebra, an idea introduced in [10] by Oh.Since then, universal enveloping algebras have been studied in a variety of contexts, such as for Poisson Hopfalgebras [5, 11] and Poisson-Ore extensions [6], and partial PBW theorems have been obtained [3, 5, 13].We extend several such results to the case of Poisson superalgebras, and obtain a full PBW theorem.Our approach to prove the PBW theorem uses the relation between Lie-Rinehart superalgebras andPoisson superalgebras. The former is a super generalization of Lie-Rinehart algebras which were first givencomprehensive treatment by Rinehart [15], and can be viewed as an algebraic analogue of the more geometricnotion of Lie algebroids. In his paper, Rinehart defined the universal enveloping algebra of a Lie-Rinehartalgebra (
A, L ) and proved that if L was projective as an A -module. In terms of Poisson algebras, thistranslates to requiring the K¨ahler differentials Ω A to be projective over the Poisson algebra A . While thePBW theorem was known to hold for other Poisson algebras, see e.g. [3], the general problem was still open.We remove this projectivity requirement by showing that if A and L are free (super)modules, the PBWtheorem for Lie-Rinehart (super)algebras still holds. In particular, this holds if the base ring is a field, as isthe case for Poisson superalgebras.This paper is organized as follows. Mathematics Subject Classification.
Key words and phrases.
Poisson superalgebra, enveloping algebra, Lie-Rinehart superalgebra, PBW Theorem. n section 1, we recall the basic definitions in supermathematics. In section 2, we define and constructthe universal enveloping algebra, as well as proving standard theorems such as the correspondence betweenPoisson modules over a Poisson superalgebra, and modules over its universal enveloping algebra. Section 3is dedicated to proving a new PBW theorem for Lie-Rinehart superalgebras: Theorem 1 (Theorem 3.5) . The canonical A -superalgebra morphism S A ( L ) → gr( V ( A, L )) is an isomor-phism.In particular, this holds if the base ring of the Lie-Rinehart pair (
A, L ) is a field. In section 4, we follow [5]and use the results of section 3 to prove the PBW theorem for Poisson superalgebras. In section 5, we studythe universal enveloping algebra of Poisson Hopf superalgebras. The following result is a super generalizationof [11, Theorem 10].
Theorem 2 (Theorem 5.9) . If (
A, η, ∇ , ε, ∆ , S ) is a Poisson Hopf superalgebra, then U ( A ) is a Hopf super-algebra.Section 6 shows that for a quadratic Poisson polynomial algebra, its universal enveloping algebra is aquadratic algebra and further, that its quadratic dual is the universal enveloping algebra of an exterioralgebra with appropriate dual bracket. Finally, in section 7, we define Poisson-Ore extensions for Poissonsuperalgebras and show their universal enveloping algebras are iterated Ore extensions, generalizing [6,Theorem 0.1]. Acknowledgements.
The author would like to thank Miami University and the USS program for fundingthis project. The author would also like to thank Jason Gaddis for suggesting the project idea, as well asfor many helpful discussions and advice throughout.1.
Background on Supermathematics
Throughout this paper, we work over a field k of characteristic 0, except in section 3 where we workover a commutative ring of characteristic 0. All (super)algebras, (super) tensor products, etc. are assumedto be taken over the base field (or ring) unless specified otherwise. We recall the basic definitions forsupermathematics, referring the reader to [7] for further details. Definition 1.1. A superring is a Z -graded ring R = R ⊕ R such that R i R j ⊆ R i + J where the subscriptsare taken mod 2. A superring is supercommutative if rs = ( − | r || s | sr for all homogeneous r, s ∈ R .We shall assume all superrings are supercommutative. Further, any definition which applies to superringsapplies to any ring S if we give S the trivial grading S = S ⊕ efinition 1.2. A (left) supermodule over a superring R = R ⊕ R is a Z -graded (left) module M = M ⊕ M such that R i M j ⊆ M i + j . Elements of M are called even, elements of M are called odd, and anelement which is either even or odd is called homogeneous. We denote parity of a homogeneous element x by | x | .We will say a supermodule M = M ⊕ M is a free supermodule if M has a basis consisting of homogeneouselements. For example, a super vector space is always a free supermodule over its base field. Definition 1.3. A morphism from an R -supermodule M to an R -supermodule N is an R -linear map ϕ : M → N . If ϕ preserves the grading, we say ϕ is an even linear map, whereas if ϕ reserves the grading,we say ϕ is an odd linear map. Definition 1.4. A superalgebra over a superring R = R ⊕ R is a supermodule A = A ⊕ A together with anassociative multiplication that admits a unit and satisfies A i A j ⊆ A i + j . A superalgebra is supercommutativeif it is supercommutative as a superring; that is, ab = ( − | a || b | for all a, b ∈ A .As superalgebras are superrings, we shall assume all superalgebras are supercommutative. Definition 1.5. If A and B are two superalgebras, then the super tensor product of A and B , denoted A b ⊗ B , is the superalgebra with grading( A b ⊗ B ) = ( A ⊗ B ) ⊕ ( A ⊗ B )( A b ⊗ B ) = ( A ⊗ B ) ⊕ ( A ⊗ B )with multiplication defined on homogeneous elements by( a ⊗ b )( c ⊗ d ) = ( − | b || c | ( ac ⊗ bd ) . Remark 1.6.
In this paper, we shall make a distinction between supermodules (resp. superalgebras) andmodules (resp. algebras). The reason for this distinction is most readily seen in the different multiplicationfor the super tensor product than the normal tensor product. Therefore, we treat the superalgebra A andthe underlying algebra A as (slightly) different entities.2. The Universal Enveloping algebra of a Poisson Superalgebra
In this section, we first recall the definition of a Poisson superalgebra and provide some examples. We thendefine and construct the universal enveloping algebra, prove standard theorems such as the correspondenceof module categories, and provide an example computation for the universal enveloping algebra. efinition 2.1. A Poisson superalgebra is a supercommutative superalgebra R with a bracket {· , ·} suchthat ( R, {· , ·} ) is a Lie superalgebra, and {· , ·} satisfies the Leibniz rule: { a, bc } = ( − | a || b | b { a, c } + { a, b } c for all a, b, c ∈ R. That is, { x, ·} is a degree | x | superderivation for all x ∈ R. Example 2.2.
Any associative superalgebra becomes a Poisson superalgebra via the supercommutatorbracket. Also, any Poisson algebra can be considered a Poisson superalgebra with trivial odd part.
Example 2.3.
Let P n = Λ( x , . . . , x n , y , . . . , y n ) be an exterior algebra with grading | x i | = | y j | = 1 for all1 ≤ i, j ≤ n. Then P n becomes a Poisson superalgebra via the bracket { x i , x j } = 0 = { y i , y j } , { x i , y j } = δ ij where δ ij is the Kronecker delta. Definition 2.4.
Let R be a Poisson superalgebra. A vector space M is a (left) Poisson R -module if thereis an algebra homomorphism α : R → End M and a linear map β : R → End M such that β ( { x, y } ) = β ( x ) β ( y ) − ( − | x || y | β ( y ) β ( x ) α ( { x, y } ) = β ( x ) α ( y ) − ( − | x || y | α ( y ) β ( x ) β ( xy ) = α ( x ) β ( y ) + ( − | x || y | α ( y ) β ( x ) . In order to simplify the definition of Poisson modules, as well as the definition of the universal envelopingalgebra, we introduce the following notation.
Definition 2.5.
Let R be a Poisson superalgebra. We say a triple ( U, α, β ) satisfies property P (with respectto R ) if(1) U is an algebra,(2) α : R → U is an algebra homomorphism, and(3) β : R → U is a linear mapsuch that β ( { x, y } ) = β ( x ) β ( y ) − ( − | x || y | β ( y ) β ( x ) α ( { x, y } ) = β ( x ) α ( y ) − ( − | x || y | α ( y ) β ( x ) β ( xy ) = α ( x ) β ( y ) + ( − | x || y | α ( y ) β ( x ) . In particular, a vector space M is a Poisson R -module if (End M, α, β ) satisfies property P for some α, β .Note that since we do not assume the algebra U in a triple satisfying property P to be Z -graded. Assuch, we cannot simplify the definition by saying β is a Lie superalgebra morphism.We are now ready to define and construct the universal enveloping algebra following [10]. efinition 2.6. Let R be a Poisson superalgebra. The universal enveloping algebra (also called Poissonenveloping algebra or just enveloping algebra) of R is a triple ( U ( R ) , α, β ) that is universal with respect toproperty P . That is, ( U ( R ) , α, β ) satisfies property P , and if ( B, γ, δ ) is another triple satisfying property P , then there is a unique algebra homomorphism ϕ : U ( R ) → B such that the following diagram commutes: R α / / β / / δ ! ! ❇❇❇❇❇❇❇❇❇❇❇❇❇❇❇❇ γ ! ! ❇❇❇❇❇❇❇❇❇❇❇❇❇❇❇❇ U ( R ) ϕ (cid:15) (cid:15) ✤✤✤✤✤ B Note that we will sometimes use the notation R e to denote the universal enveloping algebra. Theorem 2.7.
Every Poisson superalgebra R has a unique universal enveloping algebra. Proof.
Uniqueness follows from the standard argument for universal constructions. To construct U ( R ), let M = { m r | r ∈ R } and H = { m r | r ∈ R } be two vector space copies of R with the obvious linearisomorphisms. Let T be the free algebra generated by the set M ∪ H, and let J be the ideal of T generatedby elements of the form • m x m y − m xy • h x m y − ( − | x || y | m y h x − m { x,y } • m − • h x h y − ( − | x || y | h y h x − h { x,y } • m x h y + ( − | x || y | m y h x − h xy for x, y ∈ R. We claim (
T /J, α, β ) , where α ( x ) = m x and β ( x ) = h x , is the universal enveloping algebraof R. That the above triple satisfies property P is immediate from the definition, so suppose ( B, γ, δ ) isanother triple satisfying property P . Define an algebra homomorphism ϕ : T → B by ϕ ( m x ) = γ ( x ) and ϕ ( h x ) = δ ( x ) for x ∈ R. Then ϕ ( m −
1) = γ (1) − ϕ ( m x m y − m xy ) = γ ( x ) γ ( y ) − γ ( xy ) = 0 ϕ ( h x m y − ( − | x || y | m y h x − m { x,y } ) = δ ( x ) γ ( y ) − ( − | x || y | γ ( y ) δ ( x ) − γ ( { x, y } ) = 0 ϕ ( h x h y − ( − | x || y | h y h x − h { x,y } ) = δ ( x ) δ ( y ) − ( − | x || y | δ ( y ) δ ( x ) − δ ( { x, y } ) = 0 ϕ ( m x h y + ( − | x || y | m y h x − h xy ) = γ ( x ) δ ( y ) + ( − | x || y | γ ( y ) δ ( x ) − δ ( xy ) = 0so the map ϕ = ϕπ, where π : T → T /J is the projection map, is a well-defined algebra homomorphism.That ϕm = γ and ϕh = δ is obvious. Finally, if ψ is another algebra homomorphism such that ψm = γ and ψh = δ, then ϕ = ψ on the generators of U ( R ) , so ϕ = ψ on all of U ( R ) . That is, ϕ is unique. (cid:3) emark 2.8. The free algebra T in the above proof is naturally Z -graded with | m x | = | h x | = | x | , and sincethe ideal J is generated by homogeneous elements, U ( R ) inherits this grading. Thus U ( R ) has a canonical Z -grading. Corollary 2.9.
There is a 1-1 correspondence between Poisson R -modules and U ( R )-modules. Proof.
Suppose (
M, γ, δ ) is a Poisson R -module. Then the universal property induces an algebra homomor-phism ϕ : U ( R ) → End M which turns M into a U ( R )-module.Conversely, suppose M is a U ( R )-module via an algebra homomorphism ϕ : U ( R ) → End M. Then since( U ( R ) , α, β ) satisfies property P , so does the triple (End M, ϕα, ϕβ ) , thus making M a Poisson R -module. (cid:3) Corollary 2.10.
If ( U ( R ) , α, β ) is the universal enveloping algebra of a Poisson superalgebra R, then α isinjective. Proof.
For a ∈ R , let γ a denote left multiplication by a , and let δ a denote the adjoint map δ a ( b ) = { a, b } .Define γ, δ : R → End R by γ ( a ) = γ a , δ ( a ) = δ a . One easily verifies (End
R, γ, δ ) satisfies property P , so there is an induced algebra homomorphism ϕ : U ( R ) → End R. Therefore, if x ∈ ker( α ) , then0 = ϕα ( x ) = γ x so γ x (1) = x = 0 . (cid:3) Example 2.11.
Consider the Poisson superalgebra P n defined in Example 1.3. We show U ( P n ) ∼ = C n , where C m is the free supercommutative superalgebra generated by odd variables X i , Y j , ≤ i, j ≤ m, subject to the relations X i Y j + Y j X i = δ ij . Indeed, first note that by Theorem 2.7, U ( P n ) is generated by m x i , m y j , h x i , h y j and satisfies[ m x i , m x j ] = [ m x i , m y j ] = [ m y i , m y j ] = 0 , [ h x i , h x j ] = [ h x i , h y j ] = [ h y i , h y j ] = 0[ h x i , m x j ] = [ h y i , m y j ] = 0 , [ h x i , m y j ] = δ ij = [ h y i , m x j ]where [ · , · ] is the supercommutator. Therefore, there is a surjective algebra homomorphism ϕ : C n → U ( P n )defined by X i m x i , X i + n m y i , Y i h y i , Y i + n h x i for 1 ≤ i ≤ n. The following proposition then proves ϕ is an isomorphism. Proposition 2.12.
The algebra C m is simple for all m. roof. Observe every element of C m can be written as a linear combination of elements of the form X α · · · X α m m Y β · · · Y β m m for α i , β i ∈ { , } . Order the set of such monomials via the degree lexicographicorder such that X < · · · < X m < Y < · · · < Y m , so that we may select a leading term for any nonzeroelement of C m . Consider now a monomial f = X α · · · X α m m Y β · · · Y β m m and let r ∈ { , , . . . , m } . Observe y r f + f y r = ( − γ X α · · · Y r X α r r · X α m m Y β · · · Y β m m + ( − γ X α · · · X α r r Y r · · · X α m m Y β · · · Y β m m for some γ , γ using the fact that C m is supercommutative. Now, if α r = 0 then( − γ − ( − γ = 0 = dfdX r where dfdX r is the formal derivative of f, whereas if α r = 1 then( − γ + ( − γ = X α · · · ( Y r X r + X r Y r ) · · · X α m m Y β · · · Y β m m = X α · · · X α r − r − X α r +1 r +1 · · · X α m m Y β · · · Y β m m = dfdX r . Therefore, regardless of the value of α r , we have that dfdX r ∈ C m f C m for all 1 ≤ r ≤ m. A similar argumentworks for dfdY r as well as higher order derivatives. Hence by repeatedly differentiating, we find 1 ∈ C m f C m , so C m f C m = C m . For a nonzero polynomial p in C m , we can differentiate such that all non-leading termsare annihilated, once again obtaining 1 ∈ C m pC m . It follows that any ideal of C m is either the zero ideal, or C m . (cid:3) The PBW Theorem for Lie-Rinehart Superalgebras
In [5], the authors prove a PBW theorem for Poisson algebras via the PBW theorem for Lie-Rinehartalgebras. We wish to follow the same approach, so in this section we prove a PBW theorem for Lie-Rinehartsuperalgebras. In this section alone, we work over a commutative ring R rather than the field k. Definition 3.1. A Lie-Rinehart superalgebra is a pair (
A, L ), where A is a supercommutative superalgebra, L is a Lie superalgebra as well as an A -supermodule, together with a Lie superalgebra and A -supermodulemorphism ρ : L → Der( A ), such that for x, y ∈ L, a ∈ A ,[ x, ay ] = ( − | a || x | a [ x, y ] + ρ ( x )( a ) y. In the future, we will denote ρ ( x )( a ) by x ( a ) for simplicity, and we call ρ the anchor map.By analogy with [15], we describe the universal enveloping algebra of a Lie-Rinehart superalgebra. Recallthat for two Lie superalgebras g and h together with a Lie superalgebra morphism ϕ : h → Der( g ) , we may orm their semidirect product g ⋊ ϕ h as follows: the underlying super vector space is g ⊕ h, with the naturalgrading, with bracket [ a + x, b + y ] = (cid:0) [ a, b ] + ϕ ( x )( b ) − ( − | a || y | ϕ ( y )( a ) (cid:1) + [ x, y ]for a, b ∈ g, and x, y ∈ h. In particular, for a Lie-Rinehart superalgebra (
A, L ) , we may form a semidirectproduct A ⋊ L via ρ. Since the result is a Lie superalgebra, we can consider its universal enveloping algebra U = U ( A ⋊ L ) , with inclusion map i : A ⋊ L → U. Let U + denote the subalgebra generated by i ( A ⋊ L ) . Lastly, let P denote the ideal of U + generated by the elements i ( az ) − i ( a ) i ( z ) , for all a ∈ A, z ∈ A ⋊ L. Then the universal enveloping algebra of A ⋊ L is the quotient V ( A, L ) = U + /P. We remark that since the ideal P is generated by homogeneous elements, with respect to the natural Z grading on U, we may regard V ( A, L ) as an R -superalgebra (or an A -supermodule).As with Lie-Rinehart algebras, V ( A, L ) has a universal property for which we need the following defini-tions.
Definition 3.2.
For a Lie-Rinehart algebra (
A, L ) , we say a triple ( U, f, g ) satisfies property R if(1) U is an algebra(2) f : A → U is an algebra homomorphism(3) g : L → U is an R-linear mapsuch that g ([ x, y ]) = g ( x ) g ( y ) − ( − | x || y | g ( y ) g ( x ) f ( x ( a )) = g ( x ) f ( a ) − ( − | a || x | f ( a ) g ( x ) g ( ax ) = f ( a ) g ( x )for all a ∈ A, x ∈ L. Definition 3.3. A Lie-Rinehart ( A, L ) -module is an R -module M together with maps f : A → End( M )and g : L → End( M ) so that (End( M ) , f, g ) satisfies property R . In other words, M is an L -module, in thesense of Lie theory, as well as an A -module, with compatibility conditions between the actions. Proposition 3.4.
The triple ( V ( A, L ) , ι A , ι L ) is universal with respect to property R , where ι A , ι L are thenatural inclusions of A, L in V ( A, L ) , respectively. Proof.
That the above triple has property R is easy to verify. Let ( B, f, g ) denote another triple withproperty R . Since the underlying vector space of A ⋊ L is the direct sum of A and L, we have a unique linearmap h : A ⋊ L → B that restricts to f, g on A, L, respectively. Also, for homogeneous a + x, b + y ∈ A ⋊ L,h ([ a + x, b + y ]) = f ( x ( b ) − ( − | a || y | y ( a )) + g ([ x, y ]) g ( x ) f ( b ) − ( − | b || x | g ( x ) f ( b ) − ( − | a || b | g ( y ) f ( a ) + f ( a ) g ( y )+ g ( x ) g ( y ) − ( − | x || y | g ( y ) g ( x )while h ( a + x ) h ( b + y ) − ( − | a || b | h ( b + y ) h ( a + x )= ( f ( a ) + g ( x ))( f ( b ) + g ( y )) − ( − | a || b | ( f ( b ) + g ( y ))( f ( a ) + g ( x ))= g ( x ) f ( b ) − ( − | b || x | g ( x ) f ( b ) − ( − | a || b | g ( y ) f ( a ) + f ( a ) g ( y )+ g ( x ) g ( y ) − ( − | x || y | g ( y ) g ( x )where we used the supercommutativity of A in the last equality. Therefore, by the universal property of U ( A ⋊ L ) , there is a unique algebra map ϕ making the following diagram commute: A ⋊ L / / h $ $ ❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍ U ( A ⋊ L ) ϕ (cid:15) (cid:15) ✤✤✤✤✤ B In addition, the ideal P is in the kernel of ϕ, since for a ∈ A and z = b + x ∈ A ⋊ L, we have ϕ ( i ( a ) i ( z ) − i ( az )) = ϕ ( i ( a )) ϕ ( i ( z )) − ϕ ( i ( az ))= h ( a ) h ( z ) − h ( az )= f ( a )( f ( b ) + g ( x )) − h ( az )= f ( ab ) + g ( ax ) − h ( az )= h ( a ( b + x )) − h ( ax )= 0 . Therefore, there is a well-defined algebra map ψ : V ( A, L ) → B satisfying ψι A = f, ψι L = g, and it is easyto see this map is unique. (cid:3) For the remainder of this section, we assume that for a Lie-Rinehart pair (
A, L ) , both A and L are free R -supermodules. For p ≥ , let V p denote the left A -subsupermodule of V ( A, L ) generated by products ofat most p elements of L, and let V − = 0 . Then { V p } defines a filtration of V ( A, L ) . Denote by gr( V ( A, L ))the associated graded A -module and note az − ( − | a || z | za ∈ V p − for a ∈ A and z ∈ V p . Hence gr( V ( A, L ))can be regarded as an A -superalgebra. Further, one can easily show gr( V ( A, L )) is supercommutative. Wewill prove the following PBW theorem, where S A ( L ) denotes the supersymmetric A -algebra on L . Theorem 3.5.
The canonical A -superalgebra morphism S A ( L ) → gr( V ( A, L )) is an isomorphism. o this end, denote W = U + ( A ⋊ L ) and let J be the ideal J = ( a ⊗ z − az ), so V = V ( A, L ) =
W/J.
Filter W by defining W p to be the A -subsupermodule generated by products of at most p elements of L. Then the quotient V = W/J is naturally filtered by (
W/J ) p = ( W p + J/J ) (see e.g. [2, Example I.6.5]).Note that this filtration coincides with the PBW filtration defined above, sogr( V ( A, L )) = ∞ M p =0 W p / ( W p − + W p ∩ J )with multiplication defined by ( x + T n − )( y + T m − ) = xy + T n + m − , where T n = W n + W n − ∩ J. One cansimilarly define W ′ = S + R ( A ⋊ L ) with ideal I = ( a ⊗ z − az ) to form the quotient V ′ = W ′ /I. Then W ′ hasan analogous filtration to W, with associated graded algebragr V ′ = ∞ M p =0 W ′ p / ( W ′ p − + W ′ p ∩ I )having multiplication ( x + T ′ n − )( y + T ′ m − ) = xy + T ′ n + m − , where T ′ n = W ′ n + W ′ n − ∩ I . Lemma 3.6.
The A -superalgebras gr( V ( A, L )) and gr V ′ are isomorphic. Proof.
Let { z i } be a totally ordered homogeneous R -basis of A ⋊ L. By the PBW theorem for Lie superal-gebras, the map ϕ : S + R ( A ⋊ L ) → U + ( A ⋊ L ) z r · · · z r n n z r · · · z r n n is an R -linear isomorphism. Consider the (map induced by) the restriction ϕ n : W ′ n → W n / ( W n − + W n ∩ J ) . We claim ker( ϕ n ) = W ′ n − + W ′ n ∩ I. Indeed, consider y = w ⊗ w ⊗ · · · ⊗ ( a ⊗ w i − aw i ) ⊗ · · · ⊗ w m ∈ W ′ n ∩ I where a ∈ A and each w j ∈ A ⋊ L. Since ϕ ( W ′ n − ) = W n − , we may assume exactly n of the w j have a nonzero L component. Decompose y into a linear combination of elements of the form z ⊗· · ·⊗ ( a ⊗ z i − az i ) ⊗· · ·⊗ z m , and further decompose az i into a linear combination of basis elements, say az i = P k c ik z ik . Rearranging the z ’s so they are in order and writing a = z for notational convenience, we see ϕ ( z ⊗ · · · ⊗ ( a ⊗ z i − az i ) ⊗ · · · ⊗ z m ) = ( − α z π (0) ⊗ · · · ⊗ z π ( m ) − X k c ik z σ k (1) ⊗ · · · ⊗ z σ k ( m ) for some permutations π, σ k and some α ∈ { , } . On the other hand, as an element of W n , we have z ⊗ · · · ⊗ ( a ⊗ z i − az i ) ⊗ · · · ⊗ z m − ( − α z π (0) ⊗ · · · ⊗ z π ( m ) + X k c ik z σ k (1) ⊗ · · · ⊗ z σ k ( m ) ∈ W n − . Hence ϕ ( z ⊗ · · · ⊗ ( a ⊗ z i − az i ) ⊗ · · · ⊗ z m ) ∈ W n − + W n ∩ J, from which it follows ϕ ( y ) ∈ W n − + W n ∩ J as well, proving ϕ ( W ′ n − + W ′ n ∩ I ) ⊆ W n − + W n ∩ J. lso, since ϕ ( W ′ n ) = W n , the above argument also shows the reverse inclusion. Finally, since ϕ is anisomorphism, ϕ − ( W n − + W n ∩ J ) = ϕ − ϕ ( W ′ n − + W ′ n ∩ I ) = W ′ n − + W ′ n ∩ I from which we obtain the desired ker( ϕ n ) = W n − + W ′ n ∩ I. Thus gr( V ( A, L )) ∼ = gr V ′ as R -modules.One easily shows this isomorphism is also a ring homomorphism and preserves the grading, and so it is anisomorphism of R -algebras and A -supermodules. (cid:3) Consider the quotient V ′ = W ′ /I and observe V ′ is the universal enveloping algebra of the pair ( A, L ) , where L has the trivial bracket and the anchor map is zero. We show V ′ = S A ( L ) by showing S A ( L ) satisfiesthe universal property of V ′ . Lemma 3.7.
The triple ( S A ( L ) , i A , i L ) , where i A : A → S A ( L ) and i L : L → S A ( L ) are the inclusion mapsfor A and L , respectively, is universal with respect to property R . Proof.
It is simple to verify the above triple satisfies property R , so it remains to show the triple is universal.Indeed, let ( B, f, g ) be another triple satisfying property R . We need to show there is a unique R -algebrahomomorphism ϕ : S A ( L ) → B such that ϕi A = f and ϕi l = g. Uniqueness of ϕ is clear as ϕ ( a ) = f ( a )and ϕ ( x ) = g ( x ) for a ∈ A, x ∈ L. To show ϕ is well-defined, it suffices to show ϕ ( xy − ( − | x || y | yx ) = 0 for x, y ∈ L. Indeed, ϕ ( xy − ( − | x || y | yx ) = ϕ ( x ) ϕ ( y ) − ( − | x || y | ϕ ( y ) ϕ ( x )= g ( x ) g ( y ) − ( − | x || y | g ( y ) g ( x )= g ([ x, y ])= 0since L has trivial bracket. (cid:3) Before proving Theorem 3.5, we justify our claim that it is independent of [15, Theorem 3.1]. To seethis, note that our theorem does not apply to, say, the Lie-Rinehart pair (
R, L ) , where L is any projective,non-free Lie algebra over R. On the other hand, as we will see in the next section, our result is more suitedfor proving the PBW theorem for Poisson superalgebras.
Proof of Theorem 3.5.
By Lemma 3.7, the map ϕ : S A ( L ) → V ′ sending a a and x x, for a ∈ A and x ∈ L, is an R -algebra isomorphism. It is easy to see ϕ is also an A -supermodule isomorphism.In addition, since S A ( L ) is graded, V ′ is also graded, with grading equivalent to the previously definedfiltration. Therefore, as A -superalgebras, S A ( L ) ∼ = gr V ′ ∼ = gr( V ( A, L )) . inally, since there exists isomorphisms S A ( L ) → gr V ′ gr V ′ → gr( V ( A, L )) a a + T ′− a + T ′− a + T − x x + T ′ x + T ′ x + T we see the PBW map from S A ( L ) to gr( V ( A, L )) is indeed an isomorphism. (cid:3) The PBW theorem for Poisson superalgebras
In [5], the authors showed that a Poisson algebra can be viewed as a Lie-Rinehart algebra in such a waythat the enveloping algebra in the Poisson sense is the same as the enveloping algebra in the Lie-Rinehartsense, thus leading to a PBW theorem for Poisson algebras. In this section, via the same approach, we provethe PBW theorem for Poisson superalgebras. We first need to recall the construction of the even K¨ahlersuperdifferentials over a superalgebra [8].
Definition 4.1.
Let A be a superalgebra, and let M be an A -supermodule. An even A -superderivation isan even linear map D : A → M such that D ( ab ) = aD ( b ) + ( − | a || b | bD ( a ) . Proposition 4.2 ([8, Lemma 3.1]) . There is an A -supermodule Ω ev A and an even superderivation d ev : A → Ω ev A such that for any A -supermodule M , composition with d ev gives an isomorphism (of abelian groups)Der k ( A, M ) ∼ = Hom A (Ω ev A , M )where Der k ( A, M ) denotes the group of even superderivations from A to M, and Hom A (Ω ev A , M ) denotesthe group of A -supermodule morphisms from Ω ev A to M. The supermodule Ω ev A can be constructed as follows. Let S be the free A -supermodule generated by theset { d f | f ∈ A, f homogeneous } , with grading | d f | = | f | . Then Ω ev A is the quotient of S by the relationsd( rf + sg ) = r d f + s d g and d( f g ) = f d g + ( − | f || g | g d f, where r, s ∈ k, f, g ∈ A, and the correspondingsuperderivation is d ev ( f ) = d f. Remark 4.3. If M is just an A -module, not necessarily graded, then one still has a 1-1 correspondencebetween A -module homomorphisms Ω ev A → M and linear maps D : A → M satisfying D ( ab ) = aD ( b ) +( − | a || b | bD ( a ) . Example 4.4.
Let A be a Poisson superalgebra over k. Then the pair ( A, Ω ev A ) becomes a Lie-Rinehartsuperalgebra over A where Ω ev A is given the bracket[ a d f, b d g ] = ( − | b || f | ab d { f, g } + a { f, b } d g − ( − | af || bg | b { g, a } d f and the anchor map is ρ (d f ) = { f, −} . ow consider the Lie-Rinehart enveloping algebra V ( A, Ω ev A ) together with the two maps α : A → A ⋊ Ω ev A → V ( A, Ω ev A ) β : A d ev −−→ Ω ev A −→ A ⋊ Ω ev A −→ V ( A, Ω ev A ) . We show the triple ( V ( A, Ω ev A ) , α, β ) is universal with respect to property P , thereby inducing a uniqueisomorphism V ( A, Ω ev A ) → U ( A ). Lemma 4.5.
The triple ( V ( A, Ω ev A ) , α, β ) satisfies property P . That is, α is an algebra map and β is a linearmap satisfying the relations β ( { a, b } ) = β ( a ) β ( b ) − ( − | a || b | β ( b ) β ( a ) α ( { a, b } ) = β ( a ) α ( b ) − ( − | a || b | α ( b ) β ( a ) β ( ab ) = α ( a ) β ( b ) + ( − | a || b | α ( b ) β ( a )for a, b ∈ A . Proof.
That α is an algebra map is trivial, whereas for the above three relations, we have β ( { a, b } ) = d { a, b } = [d a, d b ] = [ β ( a ) , β ( b )] = β ( a ) β ( b ) − ( − | a || b | β ( b ) β ( a ) α ( { a, b } ) = { a, b } = ρ (d a )( b ) = [d a, b ] = [ β ( a ) , α ( b )] = β ( a ) α ( b ) − ( − | a || b | α ( b ) β ( a ) β ( ab ) = d( ab ) = a d b + ( − | a || b | b d a = α ( a ) β ( b ) + ( − | a || b | α ( b ) β ( a ) . (cid:3) Proposition 4.6.
Let (
B, γ, δ ) be a triple satisfying property P . Then there is a unique algebra map λ : V ( A, Ω ev A ) → B such that the following diagram commutes: A m $ $ ■■■■■■■■■■■■■■■■■■■ α / / V ( A, Ω ev A ) λ (cid:15) (cid:15) A β o o h z z ✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉ B Proof.
Observe first that B is an A -module via the action a · m = γ ( a ) m, for any a ∈ A, m ∈ B. Since δ ( ab ) = γ ( a ) δ ( b ) + ( − | a || b | γ ( b ) δ ( a ) , there is a unique A -module homomorphism θ : Ω ev A → B such that δ = θd ev by Remark 4.3. Further, the triple ( B, γ, θ ) satisfies property R , since γ (( b d f )( a )) = γ ( b { f, a } )= γ ( b ) δ ( f ) γ ( a ) − ( − | a || f | γ ( b ) γ ( a ) δ ( f )= γ ( b ) θ (d f ) γ ( a ) − ( − | a || bf | γ ( ab ) δ (d f )= θ ( b d f ) γ ( a ) − ( − | a || bf | γ ( a ) θ ( b d f ) ith the other two equalities being similarly easy to verify. Therefore, by the universal property of V ( A, Ω ev A ) , we have a unique algebra map λ : V ( A, Ω ev A ) → B such that λα = γ, and λβ = θd ev = δ. (cid:3) Proposition 4.7.
Let A be a Poisson superalgebra, and let U ( A ) be its enveloping algebra. Then there isa unique isomorphism Λ : U ( A ) → V ( A, Ω ev A ) such that the following diagram commutes: A m $ $ ❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍ α / / V ( A, Ω ev A ) Λ (cid:15) (cid:15) A β o o h z z ✈✈✈✈✈✈✈✈✈✈✈✈✈✈✈✈✈✈ U ( A ) Proof.
This follows immediately from Proposition 4.6. (cid:3)
Corollary 4.8.
Let A be a Poisson superalgebra, U ( A ) its enveloping algebra, and consider the filtration { F n } on U ( A ) induced by the isomorphism Λ . Then there is an A -superalgebra isomorphism S A (Ω ev A ) ∼ = gr U ( A ) . By Example 1.2, Corollary 4.8 in particular applies to any Poisson algebra, hence the promised strength-ening of [5, Corollary 5.8].
Example 4.9.
Consider the polynomial superalgebra A = k [ x , . . . , x n | y , . . . , y m ] on even generators x i , and odd generators y j . It’s clear Ω ev A is generated as an A -supermodule by d x , . . . , d x n , d y , . . . , d y m . Weclaim this is in fact a (homogeneous) A -basis of Ω ev A . Indeed, for i ≤ n, define an even derivation D i : A → A by D i ( x j ) = δ ij and D i ( y j ) = 0 . Similarly, for n < i ≤ n + m, define an even derivation D i : A → A by D i ( x j ) = 0 , and D i ( y j ) = δ ij y i . One easily verifies each D i preserves the grading, and so they are indeedeven derivations. Hence by Proposition 4.2, there are induced A -supermodule morphisms δ i : Ω ev A → A suchthat δ i d ev = D i , from which it follows the d x i , d y j are linearly independent. Therefore, by Corollary 4.8, forany Poisson bracket on A, U ( A ) has a k -basis consisting of elements of the form m r x · · · m r n x n m r n +1 y · · · m r n + m y m h s x · · · h s n x n h s n +1 y · · · h s n + m y m where each r i , s i is a nonnegative integer for 1 ≤ i ≤ n, and r j , s j ∈ { , } for n + 1 ≤ j ≤ n + m . We remarkthat a similar argument works for any supersymmetric algebra S ( V ) over a super vector space V ; that is, if S ( V ) has a (totally ordered) homogeneous basis { x i } , then for any Poisson bracket on S ( V ), U ( S ( V )) hasa k -basis consisting of elements of the form m r x · · · m r n x n h s y · · · h s m y m for all nonnegative n, m , where the r i , s i are nonnegative (resp. 0 or 1) for x i , y i even (resp. x i , y i odd), and x i < x i +1 , y i < y i +1 . . Enveloping algebra of Poisson Hopf superalgebras
In [11], Oh showed that the enveloping algebra of a Poisson Hopf algebra is itself a Hopf algebra. In thissection, we show an analogous result holds for Poisson Hopf superalgebras. Throughout this section, the sym-bol b ⊗ denotes the super tensor product, and we will use sumless Sweedler notation for the comultiplicationof a (super)coalgebra.For Poisson superalgebras A, B, a superalgebra morphism ϕ : A → B is a super Poisson homomorphism (respectively, super Poisson anti-homomorphism ) if ϕ ( { a, b } ) = { ϕ ( a ) , ϕ ( b ) } (respectively, ϕ ( { a, b } ) = −{ ϕ ( a ) , ϕ ( b ) } )for all a, b ∈ A. Definition 5.1.
A Poisson superalgebra A is said to be a Poisson Hopf superalgebra if A is also a Hopfsuperalgebra ( A, η, ∇ , ε, ∆ , S ) such that∆( { a, b } ) = { ∆( a ) , ∆( b ) } A b ⊗ A for all a, b ∈ A, where the bracket on A b ⊗ A is { a b ⊗ a ′ , b b ⊗ b ′ } A b ⊗ A = ( − | a ′ || b | ( { a, b } b ⊗ a ′ b ′ + ab b ⊗ { a ′ , b ′ } ) . Lemma 5.2.
If (
A, η, ∇ , ε, ∆ , S ) is a Poisson Hopf superalgebra, then the counit ε is a super Poissonhomomorphism, and the antipode S is a super Poisson anti-automorphism. Proof.
We first show ε is a super Poisson homomorphism. Indeed, since the bracket on k is trivial, this istrue if and only if ε ( { a, b } ) = 0 for all a, b ∈ A. Since ε is the counit, we have ε ( x ) ε ( x ) = ε ( x ) , and thus ε ( { a, b } ) = ( − | b || a | ( ε ( { a , b } ) ε ( a b ) + ε ( a b ) ε ( { a , b } ))= ( − | b || a | ( ε ( { a ε ( a ) , b ε ( b ) } ) + ε ( { ε ( a ) a , ε ( b ) b } ))= ε ( { a ε ( a ) , b ε ( b ) } ) + ε ( { ε ( a ) a , ε ( b ) b } )= ε ( { a, b } ) + ε ( { a, b } )where the third equality uses the fact that if b is odd, then ε ( b ) = 0 , and likewise for a . Now consider the antipode S. That S is a superalgebra morphism follows from the fact that it is asuperalgebra anti-morphism, and the fact that Poisson superalgebras are supercommutative. That S isbijective also follows from supercommutativity of A. Now, from the equalities0 = { ε ( a ) , b } = { S ( a ) a , b } = S ( a ) { a , b } + ( − | a || b | { S ( a ) , b } a { a, ε ( b ) } = { a, S ( b ) b } = { a, S ( b ) } b + ( − | a || b | S ( b ) { a, b } we obtain S ( a ) { a , b } = − ( − | a || b | { S ( a ) , b } a nd { a, S ( b ) } b = − ( − | a || b | S ( b ) { a, b } . Also, by the first paragraph,0 = ε ( { a, b } ) = S ( { a, b } ) { a, b } = ( − | b || a | ( S ( { a , b } ) a b + S ( a b ) { a , b } )from which we get S ( { a , b } ) a b = − S ( a b ) { a , b } = ( − | a || b | ( − | a || b | S ( b ) { S ( a ) , b } a = −{ S ( a ) , S ( b ) } a b . Finally, we have S ( { a, b } ) = S ( { a ε ( a ) , b ε ( b ) } )= S ( { a , b } ) ε ( a ) ε ( b )= S ( { a , b } ) a S ( a ) b S ( b )= ( − | a || b | S ( { a , b } ) a b S ( a ) S ( b )= − ( − | a || b | { S ( a ) , S ( b ) } a b S ( a ) S ( b )= −{ S ( a ) , S ( b ) } a S ( a ) b S ( b )= −{ S ( a ε ( a )) , S ( b ε ( b )) } = −{ S ( a ) , S ( b ) } completing the proof. (cid:3) The following lemma satisfies, in particular, triples (
B, γ, δ ) satisfying property P (with respect to somePoisson superalgebra A ) and is needed in the proof of Lemma 5.4. Lemma 5.3.
Let A be a Poisson superalgebra, let B be a k -algebra, and let γ, δ : A → B be linear mapssatisfying γ ( { a, b } ) = δ ( a ) γ ( b ) − ( − | a || b | γ ( b ) δ ( a ) , δ ( ab ) = γ ( a ) δ ( b ) + ( − | a || b | γ ( b ) δ ( a )for all a, b ∈ A. Then γ ( { a, b } ) = γ ( a ) δ ( b ) − ( − | a || b | δ ( b ) γ ( a ) , δ ( ab ) = δ ( a ) γ ( b ) + ( − | a || b | δ ( b ) γ ( a ) . Proof.
Since γ ( { a, b } ) + δ ( ab ) = δ ( a ) γ ( b ) + γ ( a ) δ ( b ) γ ( { b, a } ) + δ ( ba ) = δ ( b ) γ ( a ) + γ ( b ) δ ( a ) e have 2 δ ( ab ) = δ ( a ) γ ( b ) + γ ( a ) δ ( b ) + ( − | a || b | ( δ ( b ) γ ( a ) + γ ( b ) δ ( a ))= δ ( ab ) + δ ( a ) γ ( b ) + ( − | a || b | δ ( b ) γ ( a )2 γ ( { a, b } ) = δ ( a ) γ ( b ) + γ ( a ) δ ( b ) − ( − | a || b | ( δ ( b ) γ ( a ) + γ ( b ) δ ( a ))= γ ( { a, b } ) + γ ( a ) δ ( b ) − ( − | a || b | δ ( b ) γ ( a )from which the conclusions follow. (cid:3) Lemma 5.4.
Let ( U ( A ) , α, β ) be the universal enveloping algebra of a Poisson superalgebra A. Then(i) α b ⊗ α : A b ⊗ A → U ( A ) b ⊗ U ( A ) is a superalgebra morphism, and(ii) α b ⊗ β + β b ⊗ α is a Lie superalgebra morphism. Proof. (i) That α b ⊗ α is a superalgebra morphism follows from the fact that α is.(ii) By Lemma 5.3, for a, a ′ , b, b ′ ∈ A, ( α b ⊗ β + β b ⊗ α )( { a b ⊗ a ′ , b b ⊗ b ′ } ) − [( α b ⊗ β + β b ⊗ α )( a b ⊗ a ′ )( α b ⊗ β + β b ⊗ α )( b b ⊗ b ′ ) − ( − | a b ⊗ a ′ || b b ⊗ b ′ | ( α b ⊗ β + β b ⊗ α )( b b ⊗ b ′ )( α b ⊗ β + β b ⊗ α )( a b ⊗ a ′ )]= ( − | a ′ || b | ( α ( { a, b } ) b ⊗ β ( a ′ b ′ ) + α ( ab ) b ⊗ β ( { a ′ , b ′ } ) + β ( { a, b } ) b ⊗ α ( a ′ b ′ ) + β ( ab ) b ⊗ α ( { a ′ , b ′ } )) − ( α ( a ) b ⊗ β ( a ′ ) + β ( a ) b ⊗ α ( a ′ ))( α ( b ) b ⊗ β ( b ′ ) + β ( b ) b ⊗ α ( b ′ )) + ( − | a b ⊗ a ′ || b b ⊗ b k ( α ( b ) b ⊗ β ( b ′ )+ β ( b ) b ⊗ α ( b ′ ))( α ( a ) b ⊗ β ( a ′ ) + β ( a ) b ⊗ α ( a ′ ))= ( − | a ′ || b | h α ( ab ) b ⊗ ( β ( a ′ ) β ( b ′ ) − ( − | a ′ || b ′ | β ( b ′ ) β ( a ′ )) + ( β ( a ) α ( b ) − ( − | a || b | α ( b ) β ( a )) b ⊗ ( α ( a ′ ) β ( b ′ )+ ( − | a ′ || b ′ | α ( b ′ ) β ( a ′ )) + ( α ( a ) β ( b ) + ( − | a || b | α ( b ) β ( a )) b ⊗ ( β ( a ′ ) α ( b ′ ) − ( − | a ′ || b ′ | α ( b ′ ) β ( a ′ ))+ ( β ( a ) β ( b ) − ( − | a || b | β ( b ) β ( a )) b ⊗ α ( a ′ b ′ ) − α ( ab ) b ⊗ β ( a ′ ) β ( b ′ ) + α ( a ) β ( b ) b ⊗ β ( a ′ ) α ( b ′ )+ β ( a ) α ( b ) b ⊗ α ( a ′ ) β ( b ′ ) + β ( a ) β ( b ) b ⊗ α ( a ′ b ′ ) + ( − | a b ⊗ a ′ || b b ⊗ b ′ | h α ( b ) α ( ba ) b ⊗ β ( b ′ ) β ( a ′ )+ α ( b ) β ( a ) b ⊗ β ( b ′ ) α ( a ′ ) + β ( b ) α ( a ) b ⊗ α ( b ′ ) β ( a ′ ) + β ( b ) β ( a ) b ⊗ α ( b ′ a ′ ) ii = − ( − | a ′ || b | + | a || b | α ( b ) β ( a ) b ⊗ ( α ( a ′ ) β ( b ′ ) − ( − | a ′ || b ′ | β ( b ′ ) α ( a ′ )) + ( − | a ′ || b | + | a ′ || b ′ | ( β ( a ) α ( b ) − ( − | a || b | α ( b ) β ( a )) b ⊗ α ( b ′ ) β ( a ′ ) − ( − | a ′ || b | + | a ′ || b ′ | ( α ( a ) β ( b ) − ( − | a || b | β ( b ) α ( a )) b ⊗ α ( b ′ ) β ( a ′ )+ ( − | a ′ || b | + | a || b | α ( b ) β ( a ) b ⊗ ( β ( a ′ ) α ( b ′ ) − ( − | a ′ || b ′ | α ( b ′ ) β ( a ′ ))= − ( − | a ′ || b | + | a || b | α ( b ) β ( a ) b ⊗ α ( { a ′ , b ′ } ) + ( − | a ′ || b | + | a ′ || b ′ | α ( { a, b } ) b ⊗ α ( b ′ ) β ( a ′ ) − ( − | a ′ || b | + | a ′ || b ′ | α ( { a, b } ) b ⊗ α ( b ′ ) β ( a ′ ) + ( − | a ′ || b | + | a || b | α ( b ) β ( a ) b ⊗ α ( { a ′ , b ′ } )= 0 . Clearly α b ⊗ β + β b ⊗ α preserves the grading, so it is indeed a Lie superalgebra morphism. (cid:3) emma 5.5. Let
A, B be Poisson superalgebras, and let C be an algebra. If ϕ : A → B is a super Poissonhomomorphism and ( C, α, β ) satisfies property P , then so does ( C, αϕ, βϕ ) . Proof.
Straightforward. (cid:3)
Lemma 5.6.
Let ( U ( A ) , α, β ) be the Poisson enveloping algebra of a Poisson superalgebra A. Then( U ( A ) b ⊗ U ( A ) , α b ⊗ α, α b ⊗ β + β b ⊗ α ) is the Poisson enveloping algebra of A b ⊗ A. Proof.
It is simple to verify( α b ⊗ α )( { a b ⊗ a ′ , b b ⊗ b ′ } ) = ( α b ⊗ β + β b ⊗ α )( a b ⊗ a ′ )( α b ⊗ α )( b b ⊗ b ′ ) − ( − | a b ⊗ a ′ || b b ⊗ b ′ | ( α b ⊗ α )( b b ⊗ b ′ )( α b ⊗ β + β b ⊗ α )( a b ⊗ a ′ )( α b ⊗ β + β b ⊗ α )(( a b ⊗ a ′ )( b b ⊗ b ′ )) = ( α b ⊗ α )( a b ⊗ a ′ )( α b ⊗ β + β b ⊗ α )( b b ⊗ b ′ )+ ( − | a b ⊗ a ′ || b b ⊗ b ′ | ( α b ⊗ α )( b b ⊗ b ′ )( α b ⊗ β + β b ⊗ α )( a b ⊗ a ′ ) . for all a, a ′ , b, b ′ ∈ A . Let i , i be the super Poisson homomorphisms from A to A b ⊗ A defined by i ( a ) = a b ⊗ i ( a ) = 1 b ⊗ a for a ∈ A. Let B be an algebra with multiplication map µ B . If γ, δ : A b ⊗ A → B are maps such that ( B, γ, δ )satisfies property P , then, by Lemma 5.5, there exist algebra maps f, g : U ( A ) → B such that the diagrams A α,β / / i (cid:15) (cid:15) U ( A ) f (cid:15) (cid:15) A b ⊗ A γ,δ / / B A α,β / / i (cid:15) (cid:15) U ( A ) g (cid:15) (cid:15) A b ⊗ A γ,δ / / B commute. Moreover, we have δi ( a ) γi ( a ′ ) − ( − | a || a ′ | γi ( a ′ ) δi ( a ) = δ ( a b ⊗ γ (1 b ⊗ a ′ ) − ( − | a || a ′ | γ (1 b ⊗ a ′ ) δ ( a b ⊗ γ ( { a b ⊗ , b ⊗ a ′ } )= γ ( { a, } b ⊗ a ′ + a b ⊗ { , a ′ } ) = 0 . Therefore, µ B ( f b ⊗ g )( α b ⊗ α )( a b ⊗ a ′ ) = f α ( a ) gα ( a ′ ) = γ ( i ( a ) i ( a ′ )) = γ ( a b ⊗ a ′ ) µ B ( f b ⊗ g )( α b ⊗ β + β b ⊗ α )( a b ⊗ a ′ ) = f α ( a ) gβ ( a ′ ) + gβ ( a ) gα ( a ′ )= γi ( a ) δi ( a ′ ) + δi ( a ) γi ( a ′ )= γi ( a ) δi ( a ′ ) + ( − | a || a ′ | γi ( a ′ ) δi ( a ) δ ( i ( a ) i ( a ′ ))= δ ( a b ⊗ a ′ ) . Thus µ B ( f b ⊗ g ) : U ( A ) b ⊗ U ( A ) → B is an algebra map such that µ B ( f b ⊗ g )( α b ⊗ α ) = γ, µ B ( f b ⊗ g )( α b ⊗ β + β b ⊗ α ) = δ. Now, if h : U ( A ) b ⊗ U ( A ) → B is another algebra map such that h ( α b ⊗ α ) = γ, h ( α b ⊗ β + β b ⊗ α ) = δ then µ B ( f b ⊗ g )( α ( a ) b ⊗
1) = γ ( a b ⊗
1) = h ( α ( a ) b ⊗ µ B ( f b ⊗ g )(1 b ⊗ α ( a )) = γ (1 b ⊗ a ) = h ( a b ⊗ α ( a )) µ B ( f b ⊗ g )( β ( a ) b ⊗
1) = δ ( a b ⊗
1) = h ( β ( a ) b ⊗ µ B ( f b ⊗ g )(1 b ⊗ β ( a )) = δ (1 b ⊗ a ) = h (1 b ⊗ β ( a ))for all a ∈ A. Finally, since U ( A ) is generated by α ( A ) and β ( A ) , we have µ B ( f b ⊗ g ) = h. Lemma 5.4completes the proof. (cid:3)
The following result shows that sending a Poisson superalgebra to its enveloping algebra defines a functorfrom the category of Poisson superalgebras to the category of associative (super)algebras.
Lemma 5.7.
Let ( U A , α A , β A ) and ( U B , α B , β B ) be the universal enveloping algebras for Poisson superal-gebras A, B respectively. If ϕ : A → B is a super Poisson homomorphism, then there is a unique algebramap U ( ϕ ) : U ( A ) → U ( B ) such that the diagram A α A ,β A / / ϕ (cid:15) (cid:15) U ( A ) U ( ϕ ) (cid:15) (cid:15) B α B ,δ B / / U ( B )commutes. Proof.
This follows immediately from Lemma 5.5. (cid:3)
Let ( A, · , {· , ·} ) be a Poisson superalgebra. Define another bracket {· , ·} ′ on A by { a, b } ′ = −{ a, b } . Then A ′ = ( A op , ◦ , {· , ·} ′ ) is also a Poisson superalgebra, where for a (super)algebra B , we denote B op = ( B, ◦ )the opposite (super)algebra of B . roposition 5.8. Let ( U ( A ) , α, β ) be the universal enveloping algebra for a Poisson superalgebra A andtreat U ( A ) as a superalgebra. Then ( U ( A ) op , α, β ) is the universal enveloping algebra of A ′ . Proof.
We first show ( U ( A ) , α, − β ) is a universal enveloping algebra of A ′ . Indeed, let ( B, γ, δ ) be a triplesatisfying property P with respect to A ′ . Then γ ( { a, b } ) = − γ ( { a, b } ′ ) = − δ ( a ) γ ( b ) + ( − | a || b | γ ( b ) δ ( a )= ( − δ )( a ) γ ( b ) − ( − | a || b | γ ( b )( − δ )( a ) − δ ( { a, b } ) = δ ( { a, b } ′ ) = δ ( a ) δ ( b ) − ( − | a || b | δ ( b ) δ ( a )= ( − δ )( a )( − δ )( b ) − ( − | a || b | ( − δ )( b )( − δ )( a ) − δ ( ab ) = − γ ( a ) δ ( b ) − ( − | a || b | γ ( b ) δ ( a ) = γ ( a )( − δ )( b ) + ( − | a || b | γ ( b )( − δ )( a )so ( B, γ, − δ ) satisfies property P with respect to A. Similarly, if (
B, γ, δ ) satisfies property P with respectto A, then ( B, γ, − δ ) satisfies property P with respect to A ′ . That ( U ( A ) , α, − β ) is a universal envelopingalgebra of A ′ follows immediately.Consider now the algebra map ϕ : U ( A ) → U ( A ) op m x m x h x
7→ − h x . Note ϕ ( h x m y − ( − | x || y | m y h x − m { x,y } ) = − h x ◦ m y + ( − | x || y | m y ◦ h x − m { x,y } = 0with the other relations of U ( A ) similarly mapping to 0, proving ϕ is indeed well-defined. Similarly, one candefine an algebra map ψ : U ( A ) op → U ( A ) m x → m x h x → − h x which is likewise well-defined. Then the pair ϕ, ψ are inverses to each other, proving U ( A ) ∼ = U ( A ) op . Therefore, we have that ( U ( A ) op , α, β ) is also a universal enveloping algebra for A ′ , completing the proof. (cid:3) Theorem 5.9.
If (
A, η, ∇ , ε, ∆ , S ) is a Poisson Hopf superalgebra, then( U ( A ) , η U ( A ) , ∇ U ( A ) , U ( ε ) , U (∆) , U ( S ))is a Hopf superalgebra such that U (∆) α = ( α b ⊗ α )∆ , U (∆) β = ( α b ⊗ β + β b ⊗ α )∆ , ( ε ) α = ε, U ( ε ) β = 0 ,U ( S ) α = αS, U ( S ) β = βS. Proof.
Since ∆ is a super Poisson homomorphism and ( U ( A ) b ⊗ U ( A ) , α b ⊗ α, α b ⊗ β + β b ⊗ α ) is the Poissonenveloping algebra of A b ⊗ A by Lemma 5.6, there is a unique algebra map U (∆) : U ( A ) → U ( A ) b ⊗ U ( A )such that U (∆) α = ( α b ⊗ α )∆ , U (∆) β = ( α b ⊗ β + β b ⊗ α )∆by Lemma 5.7. Similarly, there is an algebra homomorphism U ( ε ) : U ( A ) → k such that U ( ε ) α = ε, U ( ε ) β =0 since ( k, id k ,
0) is the enveloping algebra of k, and ε is a super Poisson homomorphism by Lemma 5.2.Since the antipode S is a super Poisson homomorphism from A to A by Lemma 5.2, there is an algebrahomomorphism U ( S ) : U ( A ) → U ( A ) op such that U ( S ) α = αS, U ( S ) β = βS by Lemma 5.7 and Proposition5.8. It is routinely verified that ( U ( A ) , η U ( A ) , ∇ U ( A ) , U ( ε ) , U (∆) , U ( S )) is indeed a Hopf superalgera. (cid:3) Example 5.10.
Let ( L, [ · , · ]) be a Lie superalgebra over k and let S ( L ) be the supersymmetric algebra over L . Then S ( L ) becomes a Poisson superalgebra via the bracket { a, b } = [ a, b ] for a, b ∈ L, and extending viathe Leibniz rule. Also, S ( L ) becomes a Hopf superalgebra via∆( a ) = a b ⊗ a b ⊗ a, ε ( a ) = 0 , S ( a ) = − a for a ∈ L. One easily verifies via induction on degree that ∆( { x, y } ) = { ∆( x ) , ∆( y ) } for x, y ∈ S ( L ) , so S ( L )is a Poisson Hopf superalgebra, the Poisson symmetric superalgebra. We denote this Poisson superalgebraby PS( L ) . Let V be a super vector space copy of L with trivial Lie bracket. Then we can form the semidirect product V ⋊ L via the adjoint action and consider its universal enveloping algebra U ( V ⋊ L ) . Recall U ( V ⋊ L ) has anatural Hopf superalgebra structure. Denoting B = PS( L ), we show B e ∼ = U ( V ⋊ L ) as Hopf superalgebras.Indeed, consider the map ϕ : V ⋊ L → B e sending v + x to m v + h x for v ∈ V, x ∈ L. One easily verifies ϕ isa Lie superalgebra morphism, so there is an induced algebra map Φ : U ( V ⋊ L ) → B e . It is simple to show Φis a Hopf superalgebra map, and to show it is bijective, recall that from Example 4.9 if B has homogeneousbasis { x i } , then B e has a basis consisting of elements of the form m r x · · · m r n x n h s y · · · h s m y m for all nonnegative n, m , where the r i , s i are nonnegative (resp. 0 or 1) for x i , y i even (resp. x i odd), and x i < x i +1 , y i < y i +1 . On the other hand, by the PBW theorem for Lie superalgebras, U ( V ⋊ L ) has basis x r · · · x r n n y s · · · y s m m with the same restrictions on r i , s i . Thus Φ send a basis of U ( V ⋊ L ) onto a basis of B e , proving it is bijective,and hence a Hopf superalgebra isomorphism. . Enveloping algebras of quadratic Poisson algebras and quadratic duality
Let P = k [ x , x , . . . x n ] be a polynomial algebra. We say that P is a quadratic Poisson polynomial algebra if for each x i , x j , the bracket satisfies { x i , x j } = X k,l C i,jk,l x k x l for some scalars C i,jk,l [16]. It is well known that P and the exterior algebra P ! = Λ( θ , θ . . . , θ n ) are quadraticduals of each other. For quadratic Poisson polynomial algebras, we define a dual bracket on P ! by { θ k , θ l } = X i,j C i,jk,l θ j θ i . If we grade P ! so that every θ i is odd, then the above bracket gives P ! the structure of a Poisson superalgebra;we point out the authors in [1] define a similar bracket: { θ k , θ l } = X i,j C i,jk,l θ i θ j . We will show that the resulting enveloping algebras of P and P ! are both quadratic algebras, and further,that they are quadratic duals of each other. Throughout this section, T ( X ) denotes the free algebra overthe set X . Proposition 6.1.
Let P = k [ x , . . . , x n ] be a quadratic Poisson polynomial algebra. The triple ( R, α, β ) isthe Poisson enveloping algebra of P , where R is the quadratic algebra T ( m x i , h x i ) /I , I is the ideal I = (cid:16) m x i m x j − m x j m x i , h x i m x j − m x j h x i − X k,l C i,jk,l m x k m x l ,h x i h x j − h x j h x i − X k,l C i,jk,l ( m x k h x l + m x l h x k ) (cid:17) ,α : P → R is an algebra homomorphism sending each generator x i to m x i , and β : P → R is a linear mapsatisfying β ( xy ) = α ( x ) β ( y ) + α ( y ) β ( x ) that sends 1 and 0, and x i to h x i . Proof.
It suffices to show this triple satisfies the universal property of the Poisson enveloping algebra. Thatis, we need to show that α ( { x, y } ) = [ β ( x ) , α ( y )] (4.1)for x, y ∈ P, that β is a Lie algebra map, and that for every triple ( B, γ, δ ) satisfying property P , there isa unique algebra map Φ : R → B so that Φ α = γ, Φ β = δ. To prove (4.1), we proceed by induction on thedegree (note it suffices to show for monomials). Indeed, the equality is trivial if one (or both) of x, y is ascalar, and by the relations on R , α ( { x i , x j } ) = X k,l C i,jk,l m x k m x l = [ h x i , m x j ] = [ β ( x i ) , α ( x j )] . uppose now that equation (1) holds for all monomials of degree at most N , and let y, z be two such elements.Then α ( { yx i , z } ) = α ( y { x i , z } + { y, z } x i ) = α ( y )[ β ( x i ) , α ( z )] + α ( x i )[ β ( y ) , α ( z )]= α ( y )[ β ( x i ) , α ( z )] + α ( x i )[ β ( y ) , α ( z )]+ [ α ( y ) , α ( z )] β ( x i ) + [ α ( x i ) , α ( z )] β ( y )= [ α ( y ) β ( x i ) + α ( x i ) β ( y ) , α ( z )]= [ β ( yx i ) , α ( z )]since α ( P ) is a commuative subalgebra. Therefore, α ( { yx i , zx j } ) = α ( { yx i , z } x j + z { yx i , x j } )= [ β ( yx i ) , α ( z )] α ( x j ) + α ( z )[ β ( yx i ) , α ( x j )]= [ β ( yx i ) , α ( z ) α ( x j )] . Similarly, we show β is a Lie algebra map via induction on the degree. The proofs for degree 0 and 1 aresimple, so suppose β ( { x, y } ) = [ β ( x ) , β ( y )] for all (basis) elements y, z of degree at most N . Then β ( { yx i , z } ) = β ( y { x i , z } + { y, z } x i )= α ( y )[ β ( x i ) , β ( z )] + [ α ( x i ) , β ( z )] β ( y ) + [ α ( y ) , β ( z )] β ( x i ) + α ( x i )[ β ( y ) , β ( z )]= [ β ( yx i ) , β ( z )] . Note that in the second to last equality we use [11, Lemma 4] to write α ( { x i , z } ) = [ α ( x i ) , β ( z )] and likewisefor α ( { y, z } ) . Hence β ( { yx i , zx j } ) = β ( z { yx i , x j } + { yx i , z } x j )= α ( z )[ β ( yx i ) , β ( x j )] + [ β ( yx i ) , α ( x j )] β ( z ) + [ β ( yx i ) , α ( z )] β ( x j )+ α ( x j )[ β ( yx i ) , β ( z )]= [ β ( yx i ) , β ( zx j )] . It remains to show the uniqueness and existence of Φ . Indeed, it is clear that such a function is unique,since Φ( m x i ) = γ ( x i ) , Φ( h x j ) = δ ( x j ) , and the m x i , h x j generate R as an algebra, while existence followsfrom a routine verification showing Φ( I ) = 0. (cid:3) Proposition 6.2.
The triple (
Q, α, β ) is the Poisson enveloping algebra of P ! , where Q is the quadraticalgebra T ( m θ k , h θ l ) /J , J is the ideal J = (cid:16) m θ k m θ l + m θ l m θ k , h θ k m θ l + m θ l h θ k + X i,j C i,jk,l m θ i m θ j , θ k h θ l + h θ l h θ k + X i,j C i,jk,l ( m θ i h θ j − m θ j h θ i ) (cid:17) ,α is an algebra map sending θ i to m θ i , and β is a linear map satisfying β ( xy ) = α ( x ) β ( y )+( − | x || y | α ( y ) β ( x )that sends 1 to 0 , and θ i to h θ i . Proof.
The proof is similar to Proposition 6.1. Note that Lemma 5.3 is used rather than [11, Lemma 4]. (cid:3)
Proposition 6.3. U ( P ) ! is generated by m ∗ x i and h ∗ x i subject to the relations0 = h ∗ x k h ∗ x l + h ∗ x l h ∗ x k , h ∗ x k m ∗ x l + m ∗ x l h ∗ x k + X i,j C i,jk,l h ∗ x i h ∗ x j , m ∗ x k m ∗ x l + m ∗ x l m ∗ x k + X i,j C i,jk,l ( h ∗ x i m ∗ x j − h ∗ x j m ∗ x i ) . Therefore, U ( P ) ! ∼ = U ( P ! ) . Proof.
Let U ( P ) ∼ = R as in Proposition 6.1, and denote R = T ( V ) /I . It is simple to check the relations listedact as 0 on the generators of I . Thus we need only show these are the only such relations. Indeed, denote theideal described in the proposition statement by K. Since K ⊆ I ⊥ , it suffices to show dim( K ) = dim( I ⊥ ) , or equivalently, dim( V ∗ ⊗ V ∗ /K ) = dim( V ∗ ⊗ V ∗ /I ⊥ ) . Indeed, [17] shows that the Hilbert series of U ( P ) ! is (1 + t ) n , so dim( V ∗ ⊗ V ∗ /I ⊥ ) = (cid:0) n (cid:1) = 2 n − n . Now, consider V ∗ ⊗ V ∗ /K, which is generated bylength 2 products of the m ∗ x k and h ∗ x l , n elements in total. From the first relation of K, we may removeall elements of the form h ∗ x k h ∗ x l , k ≥ l, and still have our set be spanning. Similarly, the second and thirdequations allow us to remove the elements m ∗ x k h ∗ x l , for all k, l, and m ∗ x k m ∗ x l , k ≥ l, respectively. Hence weare left with n + 2 = 2 (cid:0) n (cid:1) = 2 n − V ∗ ⊗ V ∗ /K ) ≤ n − n = dim( V ∗ ⊗ V ∗ /I ⊥ ) ≤ dim( V ∗ ⊗ V ∗ /K ) , which is the desired result. (cid:3) Example 6.4. If P = k [ x , . . . , x n ] is given the commutator bracket, then U ( P ) is a polynomial algebraon 2 n generators by Proposition 6.1; note this is consistent with Example 4.9. Then the induced bracketon P ! is also trivial, so U ( P ! ) is an exterior algebra on 2 n generators by Proposition 6.2. Hence the dualitydescribed in the above proposition is immediate.7. Poisson-Ore extensions of Poisson superalgebras and their enveloping algebras
In [12], Oh generalized the Ore extension construction to Poisson algebras, though they were also studiedin a limited way earlier by Polishchuk [14]. Later in [6], the authors showed that for a Poisson-Ore extension R of a Poisson algebra A, the enveloping algebra R e is an iterated Ore extension of A. In this section, weextend Poisson-Ore extensions to Poisson superalgebras, and show the enveloping algebra is also an iteratedOre extension. efinition 7.1. Let R be a Poisson superalgebra and let α : R → R be a linear map. Then α is a Poissonsuperderivation if(i) α ( ab ) = α ( a ) b + ( − | a || α | aα ( b )(ii) α ( { a, b } ) = { α ( a ) , b } + ( − | a || α | { a, α ( b ) } for a, b ∈ R. Theorem 7.2.
Let α, δ be linear maps on a Poisson superalgebra R with bracket {· , ·} R . Then the polynomialsuperring R [ x ] , where we declare x to be even, becomes a Poisson superalgebra with bracket satisfying { a, b } = { a, b } R , { x, a } = α ( a ) x + δ ( a ) (5.1)for a, b ∈ R if and only if α is an even Poisson superderivation and δ is an even superderivation such that δ ( { a, b } R ) = α ( a ) δ ( b ) − δ ( a ) α ( b ) + { a, δ ( b ) } R + { δ ( a ) , b } R (5.2)for a, b ∈ R. In this case, we denote the Poisson superalgebra R [ x ] by R [ x ; α, δ ] p . Proof. If R [ x ] is a Poisson superalgebra, then |{ x, a }| = | a | = | α ( a ) | = | δ ( a ) | , so α, δ are even linear maps.Also, { x, ab } = α ( ab ) x + δ ( ab ) a { x, b } + { x, a } b = ( aα ( b ) + α ( a ) b ) x + aδ ( b ) + δ ( a ) b for a, b ∈ R, so α, δ are both even superderivations on R. Moreover, by the Jacobi identity,0 = { x, { a, b }} + { a, { b, x }} + ( − | a || b | { b, { x, a }} = ( α ( { a, b } R ) − { a, α ( b ) } − { α ( a ) , b } ) x + δ ( { a, b } R ) − { a, δ ( b ) } R − { δ ( a ) , b } R + δ ( a ) α ( b ) − α ( a ) δ ( b ) . Hence α is an even Poisson superderivation and δ is an even derivation satisfying (5.2).Conversely, we suppose α is a Poisson superderivation and δ satisfies (5.2). Define a bracket on R [ x ] by { ax i , bx j } = ( { a, b } R − ( − | a || b | jbα ( a ) + iaα ( b )) x i + j + ( iaδ ( b ) − ( − | a || b | jbδ ( a )) x i + j − for all monomials ax i , bx j in R [ x ] . Note this bracket satisfies (5.1) and { f, g } = − ( − | f || g | { g, f } for ho-mogeneous f, g. Also, { f, −} is a superderivation of degree | f | , since α and δ are even superderivations. Itremains to check the Jacobi identity: for monomials ax i , bx j , cx k ∈ R [ x ] , ( − | a || c | { ax i , { bx j , cx k }} + ( − | a || b | { bx j , { cx k , ax i }} + ( − | b || c | { cx k , { ax i , bx j }} = 0 . (5.3)We proceed by induction on i, j, k. The case i = j = k = 0 is trivial, so suppose (5.3) holds for i, j = k = 0 . Then ( − | a || c | { ax i +1 , { bx j , cx k }} + ( − | a || b | { bx j , { cx k , ax i +1 }} + ( − | b || c | { cx k , { ax i +1 , bx j }} = (cid:16) ( − | a || c | { ax i , { b, c }} + ( − | a || b | { b, { c, ax i }} + ( − | b || c | { c, { ax i , b }} (cid:17) x ( − | a || c | ax i ( { x, { b, c }} + { b, { c, x }} + ( − | b || c | { c, { x, b }} )= 0by the Leibniz rule and the induction hypothesis. One can similarly perform induction on j for the case k = 0 , then perform induction on k to complete the proof. (cid:3) We briefly recall the definition of an (associative) Ore extension.
Definition 7.3.
Let R be a ring, σ : R → R a ring endomorphism, and δ : R → R a σ -derivation of R ; thatis, δ is a homomorphism of abelian groups satisfying δ ( r r ) = σ ( r ) δ ( r ) + δ ( r ) r for r , r ∈ R . The Ore extension R [ x ; σ, δ ] is generated over R by the indeterminate x with relations xr = σ ( r ) x + δ ( r )for all r ∈ R .We now proceed to studying the universal enveloping algebra of Poisson-Ore extensions. Let R be a Poissonsuperalgebra and let A = R [ x ; α, δ ] p be a Poisson-Ore extension of R. We wish to show A e is an iterated Oreextension of R e . In particular, for appropriate σ , σ , η , η , we wish to show A e ∼ = R e [ m x ; σ , η ][ h x ; σ , η ] . To determine what σ , η should be, observe that in A e ,m x m r = m r m x and m x h r = h r m x − m { r,x } = h r m x + m α ( r ) m x + m δ ( r ) for r ∈ R, so σ ( m r ) = m r , σ ( h r ) = h r + m α ( r ) η ( m r ) = 0 , η ( h r ) = m δ ( r ) . Similarly, one has σ ( m r ) = m r , σ ( h r ) = h r + m α ( r ) , σ ( m x ) = m x ,η ( m r ) = m α ( r ) m x + m δ ( r ) , η ( h r ) = ( h α ( r ) + m α ( r ) ) m x + m δα ( r ) + h δ ( r ) , η ( m x ) = 0 . To extend to general elements, we declare σ , σ to be algebra homomorphisms, and for η (resp. η ) to bea σ -derivation (resp. σ -derivation). Lemma 7.4.
The maps σ and σ are automorphisms of R e , and η (resp. η ) is a σ -derivation (resp. σ -derivation) of R e . roof. To show σ is well-defined, consider the following diagram R m / / h / / f ❅❅❅❅❅❅❅❅❅❅❅❅❅❅ g ❅❅❅❅❅❅❅❅❅❅❅❅❅❅ R eσ (cid:15) (cid:15) ✤✤✤✤✤ R e where f ( r ) = m r and g ( r ) = m α ( r ) + h r for r ∈ R. Hence the diagram is commutative and one easily shows( R e , f, g ) satisfies property P , so σ is well-defined by the universal property of R e . Note the fact that σ iswell-defined then follows immediately since R e [ m x ; σ , η ] ∼ = L n ≥ R e m nx as left R e -modules, and σ | R e = σ . That σ is an automorphism follows by construction of the inverse via σ − ( m r ) = m r , σ − ( h r ) = h r − m α ( r ) . Similarly, σ − has inverse identical to that of σ on R e (as σ | R e = σ ), while σ − ( m x ) = m x . For η , it is simple to show η sends each relation in R e to 0. Hence η can be extended to a well-defined σ -derivation on R e . Similarly, one can show, albeit more tediously, that η sends each relation of R e to 0,and so η is well-defined on R e . That η is then well-defined on all of R e [ m x ; σ , η ] follows from the sameargument as σ . (cid:3) Next, we will show the algebra homomorphism ϕ : A e → R e [ m x ; σ , η ][ h x ; σ , η ] defined by ϕ ( m r ) = m r , ϕ ( h r ) = h r , ϕ ( m x ) = m x , ϕ ( h x ) = h x , (1)is an isomorphism. We first need to show ϕ is well-defined. To do this, recall A e is the quotient F/I, where F is the free algebra generated by { m a , h A | a ∈ A } , and I is the ideal consisting of the usual relations. Wethen extend ϕ to { m a , h a | a ∈ A } by declaring that for a = P ∞ i =0 c i x i ∈ A = R [ x ] , where cofinitely many c i are 0, ϕ ( m a ) = ∞ X i =0 m c i x i and ϕ ( h a ) = ∞ X i =0 ( im c i m i − x h x + m ix h c i ) . Finally, we extend ϕ to all of F by via the homomorphism condition. Lemma 7.5.
Extending ϕ to all of F as described above, we have ϕ ( I ) = 0 . Therefore, as a map from A e ,ϕ is well-defined. Proof.
We show ϕ ( m { a,b } + ( − | a || b | m b h a − h a m b ) = 0, with the other generators mapping to 0 via a similarargument. Indeed, let a = P ∞ i =0 c i x i , b = P ∞ j =0 d j x j and note | a | = | c i | , | b | = | d j | for all i, j. Recalling, fromthe proof of Theorem 6.2, that the bracket on A is { c i x i , d j x j } = ( { c i , d j } − ( − | c i || d j | jd j α ( c i ) + ic i α ( d j )) x i + j + ( ic i δ ( d j ) − ( − | c i || c j | jd j δ ( c i )) x i + j − , we have ϕ ( m { a,b } +( − | a || b | m b h a − h a m b ) ϕ (cid:16) m P ∞ i,j =0 (cid:2)(cid:0) { c i ,d j }− ( − | a || b | jd j α ( c i )+ ic i α ( d j ) (cid:1) x i + j +( ic i δ ( d j ) − (cid:0) − | a || b | jd j δ ( c i ) (cid:1) x i + j − (cid:3)(cid:17) + ( − | a || b | ϕ ( m P ∞ j =0 d j x j h P ∞ i =0 c i x i ) − ϕ ( h P ∞ i =0 c i x i m P ∞ j =0 d j x j )= ∞ X i,j =0 h m { c i ,d j }− ( − | a || b | jd j α ( c i )+ ic i α ( d j ) m i + jx + m ic i δ ( d j ) − ( − | a || b | jd j δ ( c i ) m i + j − x i + ( − | a || b | (cid:16) ∞ X j =0 m d j m jx (cid:17)(cid:16) ∞ X i =0 ( im c i m i − x h x + m ix h c i ) (cid:17) − (cid:16) ∞ X i =0 ( im c i m i − x h x + m ix h c i ) (cid:17)(cid:16) ∞ X j =0 m d j m jx (cid:17) = ∞ X i,j =0 h m { c i ,d j } m i + jx − ( − | a || b | jm d j α ( c i ) m i + jx + im c i α ( d j ) m i + jx + im c i δ ( d j ) m i + j − x − ( − | a || b | jm d j δ ( c i ) m i + j − x + ( − | a || b | im d j c i m i + j − x h x + ( − | a || b | m d j m i + jx h c i − im c i m i − x h x m d j m jx − m ix h c i m d j m jx i = ∞ X i,j =0 h m { c i ,d j } m i + jx − ( − | a || b | jm d j α ( c i ) m i + jx + im c i α ( d j ) m i + jx + im c i δ ( d j ) m i + j − x − ( − | a || b | jm d j δ ( c i ) m i + j − x + ( − | a || b | im d j c i m i + j − x h x + ( − | a || b | m d j m i + jx h c i − im c i m i − x (cid:0) m d j h x + m α ( d j ) m x + m δ ( d j ) (cid:1) m jx − m ix (cid:0) m { c i ,d j } + ( − | a || b | m d j h c i (cid:1) m jx i = ∞ X i,j =0 h m { c i ,d j } m i + jx − ( − | a || b | jm d j α ( c i ) m i + jx + im c i α ( d j ) m i + jx + im c i δ ( d j ) m i + j − x − ( − | a || b | jm d j δ ( c i ) m i + j − x + ( − | a || b | im d j c i m i + j − x h x + ( − | a || b | m d j m i + jx h c i − im c i m i − x (cid:0) m d j h x + m α ( d j ) m x + m δ ( d j ) (cid:1) m jx − m ix m { c i ,d j } m jx − ( − | a || b | m ix m d j (cid:0) m jx h c i − jm α ( c i ) m jx − jm δ ( c i ) m j − x (cid:1)i = 0 . Note that in the second to last equality, we used the fact that h c i m jx = m jx h c i − jm α ( c i ) m jx − jm δ ( c i ) m j − x which one can prove via induction. (cid:3) Theorem 7.6.
The algebra homomorphism ϕ : A e → R e [ m x ; σ , η ][ h x ; σ , η ] defined by (1) is an isomor-phism. Proof.
By Lemma 7.5, ϕ is well-defined, so it remains to show it is bijective. To do this, we construct aninverse map ψ. Indeed, consider first the triple ( A e , f, g ) , where f : R → A e maps r to m r , and g : R → A e maps r to h r . It is easy to verify ( A e , f, g ) satisfies property P with respect to R, so there is an induced lgebra homomorphism θ : R e → A e : R m / / h / / f ❅❅❅❅❅❅❅❅❅❅❅❅❅❅ g ❅❅❅❅❅❅❅❅❅❅❅❅❅❅ R eθ (cid:15) (cid:15) ✤✤✤✤✤ A e We thus set ψ ( ym ix h jx ) = θ ( y ) m ix h jx for y ∈ R e . It follows immediately from the definitions of ϕ, ψ that theyare inverse functions, and hence are an inverse pair of algebra isomorphisms. (cid:3)
Corollary 7.7.
Let R be a Poisson superalgebra and let A be an iterated Poisson-Ore extension of R. Then A e inherits the following properties from R e :(1) being a domain;(2) being Noetherian;(3) having finite global dimension;(4) having finite Krull dimension;(5) being twisted Calabi-Yau. Proof.
Properties (1)-(4) are well known properties for Ore extensions, see [9] for example, while (5) comesfrom [4]. (cid:3)
To give an example where the above corollary can be applied, consider the following proposition.
Proposition 7.8.
Let ( U ( R ) , α, β ) be the enveloping algebra of a Poisson superalgebra R. If R is finitelygenerated as a superalgebra, then U ( R ) is left/right Noetherian. Proof.
Suppose R is generated by the even variables x , . . . , x n and the odd variables y , . . . , y m . Then U ( R )is generated by α ( R ) , β ( R ) , and since α ( xy ) = α ( x ) α ( y ) β ( xy ) = α ( x ) β ( y ) + ( − | x || y | α ( y ) β ( x )for x, y ∈ R, it follows U ( R ) is generated by α ( x i ) , α ( y j ) , β ( x i ) , β ( y j ), 1 ≤ i ≤ n , 1 ≤ j ≤ m as a superalgebra.Recall that with the PBW filtration on U ( R ) , the superalgebra gr U is supercommutative. Therefore,gr U is a homomorphic image of the polynomial superalgebra R [ X , · · · , X n | Y , · · · , Y m ] . Since R is finitelygenerated over k, it follows from the discussion after Proposition 3.3.7 in [18] that R [ X , · · · , X n | Y , · · · , Y m ] , and hence gr U, is left/right Noetherian. By [9, Theorem 1.6.9], U ( R ) is left/right Noetherian. (cid:3) Example 7.9.
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Miami University, Department of Mathematics, 301 S. Patterson Ave., Oxford, Ohio 45056
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