Universal Enveloping Algebras of Poisson Superalgebras

arXiv:2102.06895v1 [math.RA] 13 Feb 2021 oino i leris nhsppr ieatdfie h universa the defined Rinehart paper, ( his algebra In algebroids. a an Lie as of viewed notion be can Lie- and of [15], generalization Rinehart by super treatment a comprehensive is former The superalgebras. Poisson superalgeb Poisson of case the the PBW to partial results and such several [6], extend extensions We Poisson-Ore and var 11] a algeb in [5, studied enveloping algebras been universal have algebras its enveloping study universal then, to Since is algebras Poisson supermanifolds. of and theory superalgebras Lie as such us structures increased the super witnessed qua also and has geometry, one symplectic theories, or supersymmetry Poisson as such areas in use found rnltst eurn h ¨he ieetasΩ Kähler differentials the requiring to translates ermv hspoetvt eurmn ysoigta if that showing by requirement e.g. projectivity see algebras, this Poisson remove other We for hold to known was theorem PBW h aefrPisnsuperalgebras. Poisson for particula case In the holds. still (super)algebras Lie-Rinehart for theorem u praht rv h B hoe ssterlto between relation the uses theorem PBW the prove to approach Our Hamilto of study the in naturally arises algebra Poisson a of notion The 2010 hsppri raie sfollows. as organized is paper This e od n phrases. and words Key NVRA NEOIGAGBA FPISNSUPERALGEBRAS POISSON OF ALGEBRAS ENVELOPING UNIVERSAL ahmtc ujc Classification. Subject Mathematics neoigagbafreape uha udai polynomi superalgebras. quadratic as such examples for algebra itera enveloping (resp. superalgebra Hopf envelopin a is universal extension) the Poisson-Ore superalge show we Lie-Rinehart superalgebras; for Poisson theorem for PBW new a particular, In Abstract. ,L A, n rvdta if that proved and ) nti ae,w en n td h nvra neoigal enveloping universal the study and define we paper, this In oso ueagba neoigagba i-ieats Lie-Rinehart algebra, enveloping superalgebra, Poisson L a rjciea an as projective was 76,1B0 73,16T05. 17B35, 17B60, 17B63, 0. HMSLAMKIN THOMAS Introduction A ob rjcieoe h oso algebra Poisson the over projective be to 1 e r xeso) n edtrieteuniversal the determine we and extension); Ore ted ler faPisnHp ueagba(resp. superalgebra Hopf Poisson a of algebra g lPisnsprlersadPisnsymmetric Poisson and superalgebras Poisson al A mdl.I em fPisnagba,this algebras, Poisson of terms In -module. A rsi rvd edn oaPWtheorem PBW a to leading proved, is bras fPisnsprlers ln ihother with along superalgebras, Poisson of e n ehdo tdigterepresentation the studying of method One a,adoti ulPWtheorem. PBW full a obtain and ras, and ,ti od ftebs igi ed sis as field, a is ring base the if holds this r, tmgop.Det h eeomn of development the to Due groups. ntum eyo otxs uha o oso Hopf Poisson for as such contexts, of iety ieatagba hc eefis given first were which algebras Rinehart gbacaaou ftemr geometric more the of analogue lgebraic 3,tegnrlpolmwssilopen. still was problem general the [3], neoigagbao Lie-Rinehart a of algebra enveloping l a nie nrdcdi 1]b Oh. by [10] in introduced idea an ra, L rm aebe band[,5 13]. 5, [3, obtained been have orems r re(ue)oue,tePBW the (super)modules, free are prler,PWTheorem. PBW uperalgebra, er fPisnsuperalgebras. Poisson of gebra i-ieatsprlersand superalgebras Lie-Rinehart inmcais n a since has and mechanics, nian A hl the While . In section 1, we recall the basic definitions in supermathematics. In section 2, we define and construct the universal enveloping algebra, as well as proving standard theorems such as the correspondence between Poisson modules over a Poisson superalgebra, and modules over its universal enveloping algebra. Section 3 is dedicated to proving a new PBW theorem for Lie-Rinehart superalgebras:

Theorem 1 (Theorem 3.5). The canonical A-superalgebra morphism SA(L) → gr(V (A, L)) is an isomorphism.

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 study the universal enveloping algebra of Poisson Hopf superalgebras. The following result is a super generalization of [11, Theorem 10].

Theorem 2 (Theorem 5.9). If (A, η, ∇, ε, ∆,S) is a Poisson Hopf superalgebra, then U(A) is a Hopf superalgebra.

Section 6 shows that for a quadratic Poisson polynomial algebra, its universal enveloping algebra is a quadratic algebra and further, that its quadratic dual is the universal enveloping algebra of an exterior algebra with appropriate dual bracket. Finally, in section 7, we deﬁne Poisson-Ore extensions for Poisson superalgebras 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 funding this project. The author would also like to thank Jason Gaddis for suggesting the project idea, as well as for 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 work over a commutative ring of characteristic 0. All (super)algebras, (super) tensor products, etc. are assumed to be taken over the base field (or ring) unless specified otherwise. We recall the basic definitions for supermathematics, referring the reader to [7] for further details.

Deﬁnition 1.1. A superring is a Z2-graded ring R = R0 ⊕ R1 such that RiRj ⊆ Ri+J where the subscripts are taken mod 2. A superring is supercommutative if

rs = (−1)|r||s|sr for all homogeneous r, s ∈ R.

We shall assume all superrings are supercommutative. Further, any deﬁnition which applies to superrings applies to any ring S if we give S the trivial grading S = S ⊕ 0. 2 Deﬁnition 1.2. A (left) supermodule over a superring R = R0 ⊕ R1 is a Z2-graded (left) module M =

M0 ⊕ M1 such that RiMj ⊆ Mi+j . Elements of M0 are called even, elements of M1 are called odd, and an element 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 = M0 ⊕M1 is a free supermodule if M has a basis consisting of homogeneous elements. 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 = R0 ⊕R1 is a supermodule A = A0 ⊕A1 together with an associative multiplication that admits a unit and satisﬁes AiAj ⊆ Ai+j . A superalgebra is supercommutative if it is supercommutative as a superring; that is,

ab = (−1)|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, is the superalgebra with grading b (A ⊗ B)0 = (A0 ⊗ B0) ⊕ (A1 ⊗ B1) b (A ⊗ B)1 = (A0 ⊗ B1) ⊕ (A1 ⊗ B0) b with multiplication deﬁned on homogeneous elements by

(a ⊗ b)(c ⊗ d) = (−1)|b||c|(ac ⊗ bd).

Remark 1.6. In this paper, we shall make a distinction between supermodules (resp. superalgebras) and modules (resp. algebras). The reason for this distinction is most readily seen in the diﬀerent multiplication for the super tensor product than the normal tensor product. Therefore, we treat the superalgebra A and the 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 then define and construct the universal enveloping algebra, prove standard theorems such as the correspondence of module categories, and provide an example computation for the universal enveloping algebra. 3 Definition 2.1. A Poisson superalgebra is a supercommutative superalgebra R with a bracket {·, ·} such that (R, {·, ·}) is a Lie superalgebra, and {·, ·} satisfies the Leibniz rule:

{a,bc} = (−1)|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 supercommutator bracket. Also, any Poisson algebra can be considered a Poisson superalgebra with trivial odd part.

Example 2.3. Let Pn = Λ(x1,...,xn,y1,...,yn) be an exterior algebra with grading |xi| = |yj | = 1 for all

1 ≤ i, j ≤ n. Then Pn becomes a Poisson superalgebra via the bracket

{xi, xj } =0= {yi,yj }, {xi,yj} = δij where δij is the Kronecker delta.

Deﬁnition 2.4. Let R be a Poisson superalgebra. A vector space M is a (left) Poisson R-module if there is an algebra homomorphism α : R → End M and a linear map β : R → End M such that

β({x, y})= β(x)β(y) − (−1)|x||y|β(y)β(x)

α({x, y})= β(x)α(y) − (−1)|x||y|α(y)β(x)

β(xy)= α(x)β(y) + (−1)|x||y|α(y)β(x).

In order to simplify the deﬁnition of Poisson modules, as well as the deﬁnition of the universal enveloping algebra, 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 respect to R) if

(1) U is an algebra, (2) α : R → U is an algebra homomorphism, and (3) β : R → U is a linear map such that

β({x, y})= β(x)β(y) − (−1)|x||y|β(y)β(x)

α({x, y})= β(x)α(y) − (−1)|x||y|α(y)β(x)

β(xy)= α(x)β(y) + (−1)|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 Z2-graded. As such, 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]. 4 Definition 2.6. Let R be a Poisson superalgebra. The universal enveloping algebra (also called Poisson enveloping algebra or just enveloping algebra) of R is a triple (U(R),α,β) that is universal with respect to property 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:

α R // U(R) ❇❇❇ β ✤ ❇❇❇❇ ❇❇❇❇ ✤ ❇❇❇❇γ ❇❇❇❇ ✤ ϕ δ ❇❇❇❇ ✤ ❇❇❇❇ ❇❇ ✤ ❇❇❇! ! B

Note that we will sometimes use the notation Re 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 = {mr | r ∈ R} and H = {mr | r ∈ R} be two vector space copies of R with the obvious linear isomorphisms. Let T be the free algebra generated by the set M ∪ H, and let J be the ideal of T generated by elements of the form

• mxmy − mxy |x||y| • hxmy − (−1) myhx − m{x,y}

• m1 − 1 |x||y| • hxhy − (−1) hyhx − h{x,y} |x||y| • mxhy + (−1) myhx − hxy

for x, y ∈ R. We claim (T/J,α,β), where α(x) = mx and β(x) = hx, is the universal enveloping algebra of R. That the above triple satisﬁes property P is immediate from the deﬁnition, so suppose (B,γ,δ) is

another triple satisfying property P. Deﬁne an algebra homomorphism ϕ : T → B by ϕ(mx) = γ(x) and

ϕ(hx)= δ(x) for x ∈ R. Then

ϕ(m1 − 1) = γ(1) − 1=0

ϕ(mxmy − mxy)= γ(x)γ(y) − γ(xy)=0

|x||y| |x||y| ϕ(hxmy − (−1) myhx − m{x,y})= δ(x)γ(y) − (−1) γ(y)δ(x) − γ({x, y})=0

|x||y| |x||y| ϕ(hxhy − (−1) hyhx − h{x,y})= δ(x)δ(y) − (−1) δ(y)δ(x) − δ({x, y})=0

|x||y| |x||y| ϕ(mxhy + (−1) myhx − hxy)= γ(x)δ(y) + (−1) γ(y)δ(x) − δ(xy)=0

so 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. 5 Remark 2.8. The free algebra T in the above proof is naturally Z2-graded with |mx| = |hx| = |x|, and since the ideal J is generated by homogeneous elements, U(R) inherits this grading. Thus U(R) has a canonical

Z2-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 homomorphism 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.

Corollary 2.10. If (U(R),α,β) is the universal enveloping algebra of a Poisson superalgebra R, then α is injective.

Proof. For a ∈ R, let γa denote left multiplication by a, and let δa denote the adjoint map δa(b) = {a,b}. Deﬁne γ,δ : R → End R by

γ(a)= γa, δ(a)= δa.

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), then

0= ϕα(x)= γx

so γx(1) = x =0.

Example 2.11. Consider the Poisson superalgebra Pn defined in Example 1.3. We show U(Pn) =∼ C2n, where Cm is the free supercommutative superalgebra generated by odd variables Xi, Yj , 1 ≤ i, j ≤ m, subject to the relations XiYj + Yj Xi = δij . Indeed, first note that by Theorem 2.7, U(Pn) is generated by mxi ,myj ,hxi ,hyj and satisfies

[mxi ,mxj ] = [mxi ,myj ] = [myi ,myj ]=0, [hxi ,hxj ] = [hxi ,hyj ] = [hyi ,hyj ]=0

[hxi ,mxj ] = [hyi ,myj ]=0, [hxi ,myj ]= δij = [hyi ,mxj ]

where [·, ·] is the supercommutator. Therefore, there is a surjective algebra homomorphism ϕ : C2n → U(Pn) deﬁned by

Xi 7→ mxi , Xi+n 7→ myi , Yi 7→ hyi , Yi+n 7→ hxi for 1 ≤ i ≤ n. The following proposition then proves ϕ is an isomorphism.

Proposition 2.12. The algebra Cm is simple for all m. 6 Proof. Observe every element of Cm can be written as a linear combination of elements of the form

α1 αm β1 βm X1 ··· Xm Y1 ··· Ym for αi,βi ∈ {0, 1}. Order the set of such monomials via the degree lexicographic order such that X1 < ··· < Xm < Y1 < ··· < Ym, so that we may select a leading term for any nonzero element of Cm.

α1 αm β1 βm Consider now a monomial f = X1 ··· Xm Y1 ··· Ym and let r ∈{1, 2,...,m}. Observe

γ1 α1 αr αm β1 βm yrf + fyr = (−1) X1 ··· YrXr · Xm Y1 ··· Ym

γ2 α1 αr αm β1 βm + (−1) X1 ··· Xr Yr ··· Xm Y1 ··· Ym for some γ1,γ2 using the fact that Cm is supercommutative. Now, if αr = 0 then df (−1)γ1 − (−1)γ2 =0= dXr where df is the formal derivative of f, whereas if α = 1 then dXr r

γ1 γ2 α1 αm β1 βm (−1) + (−1) = X1 ··· (YrXr + XrYr) ··· Xm Y1 ··· Ym

α1 αr−1 αr+1 αm β1 βm = X1 ··· Xr−1 Xr+1 ··· Xm Y1 ··· Ym df = . dXr Therefore, regardless of the value of α , we have that df ∈ C fC for all 1 ≤ r ≤ m. A similar argument r dXr m m df works for as well as higher order derivatives. Hence by repeatedly differentiating, we find 1 ∈ CmfCm, dYr so CmfCm = Cm. For a nonzero polynomial p in Cm, we can differentiate such that all non-leading terms are annihilated, once again obtaining 1 ∈ CmpCm. It follows that any ideal of Cm is either the zero ideal, or

Cm.

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-Rinehart algebras. We wish to follow the same approach, so in this section we prove a PBW theorem for Lie-Rinehart superalgebras. 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-supermodule morphism ρ : L → Der(A), such that for x, y ∈ L,a ∈ A,

[x,ay] = (−1)|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. Recall that for two Lie superalgebras g and h together with a Lie superalgebra morphism ϕ : h → Der(g), we may 7 form their semidirect product g ⋊ϕ h as follows: the underlying super vector space is g ⊕ h, with the natural grading, with bracket

[a + x, b + y]= [a,b]+ ϕ(x)(b) − (−1)|a||y|ϕ(y)(a) + [x, y] for a,b ∈ g, and x, y ∈ h. In particular, for a Lie-Rinehart superalgebra (A, L), we may form a semidirect product 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 Z2 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 map

such that

g([x, y]) = g(x)g(y) − (−1)|x||y|g(y)g(x)

f(x(a)) = g(x)f(a) − (−1)|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 the sense of Lie theory, as well as an A-module, with compatibility conditions between the actions.

Proposition 3.4. The triple (V (A, L), ιA, ιL) is universal with respect to property R, where ιA, ιL are the natural 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 with property R. Since the underlying vector space of A ⋊ L is the direct sum of A and L, we have a unique linear map 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) − (−1)|a||y|y(a)) + g([x, y]) 8 = g(x)f(b) − (−1)|b||x|g(x)f(b) − (−1)|a||b|g(y)f(a)+ f(a)g(y)

+ g(x)g(y) − (−1)|x||y|g(y)g(x) while

h(a + x)h(b + y)−(−1)|a||b|h(b + y)h(a + x)

= (f(a)+ g(x))(f(b)+ g(y)) − (−1)|a||b|(f(b)+ g(y))(f(a)+ g(x))

= g(x)f(b) − (−1)|b||x|g(x)f(b) − (−1)|a||b|g(y)f(a)+ f(a)g(y)

+ g(x)g(y) − (−1)|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 / U(A ⋊ L) ❍❍ ✤ ❍❍ ❍❍ ✤ ❍❍ ❍❍ ✤ ϕ h ❍❍ ❍❍ ✤ ❍❍ ❍❍$ ✤ 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 ψιA = f,ψιL = g, and it is easy to see this map is unique.

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 ≥ 0, let Vp denote the left A-subsupermodule of V (A, L) generated by products of at most p elements of L, and let V−1 =0. Then {Vp} deﬁnes a ﬁltration of V (A, L). Denote by gr(V (A, L)) |a||z| the associated graded A-module and note az − (−1) za ∈ Vp−1 for a ∈ A and z ∈ Vp. Hence gr(V (A, L)) can be regarded as an A-superalgebra. Further, one can easily show gr(V (A, L)) is supercommutative. We will prove the following PBW theorem, where SA(L) denotes the supersymmetric A-algebra on L.

Theorem 3.5. The canonical A-superalgebra morphism SA(L) → gr(V (A, L)) is an isomorphism. 9 To 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 Wp 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 = (Wp + J/J) (see e.g. [2, Example I.6.5]). Note that this filtration coincides with the PBW filtration defined above, so ∞ gr(V (A, L)) = W /(W + W ∩ J) M p p−1 p p=0 with multiplication defined by (x + Tn−1)(y + Tm−1)= xy + Tn+m−1, where Tn = Wn + Wn−1 ∩ J. One can ′ + ⋊ ′ ′ ′ similarly define W = SR (A L) with ideal I = (a ⊗ z − az) to form the quotient V = W /I. Then W has an analogous filtration to W, with associated graded algebra ∞ gr V ′ = W ′ /(W ′ + W ′ ∩ I) M p p−1 p p=0 ′ ′ ′ ′ ′ ′ having multiplication (x + Tn−1)(y + Tm−1)= xy + Tn+m−1, where Tn = Wn + Wn−1 ∩ I.

Lemma 3.6. The A-superalgebras gr(V (A, L)) and gr V ′ are isomorphic.

Proof. Let {zi} be a totally ordered homogeneous R-basis of A ⋊ L. By the PBW theorem for Lie superalgebras, the map

+ ⋊ + ⋊ ϕ : SR (A L) → U (A L)

r1 rn r1 rn z1 ··· zn 7→ z1 ··· zn

′ is an R-linear isomorphism. Consider the (map induced by) the restriction ϕn : Wn → Wn/(Wn−1 +Wn ∩J). ′ ′ ′ We claim ker(ϕn)= Wn−1 + Wn ∩ I. Indeed, consider y = w1 ⊗ w2 ⊗···⊗ (a ⊗ wi − awi) ⊗···⊗ wm ∈ Wn ∩ I ⋊ ′ where a ∈ A and each wj ∈ A L. Since ϕ(Wn−1)= Wn−1, we may assume exactly n of the wj have a nonzero

L component. Decompose y into a linear combination of elements of the form z1 ⊗···⊗(a⊗zi −azi)⊗···⊗zm,

and further decompose azi into a linear combination of basis elements, say azi = cikzik. Rearranging the Pk z’s so they are in order and writing a = z0 for notational convenience, we see

ϕ(z ⊗···⊗ (a ⊗ z − az ) ⊗···⊗ z ) = (−1)αz ⊗···⊗ z − c z ⊗···⊗ z 1 i i m π(0) π(m) X ik σk(1) σk (m) k

for some permutations π, σk and some α ∈{0, 1}. On the other hand, as an element of Wn, we have

z1 ⊗···⊗ (a ⊗ zi − azi) ⊗···⊗ zm

− (−1)αz ⊗···⊗ z + c z ⊗···⊗ z ∈ W . π(0) π(m) X ik σk (1) σk (m) n−1 k Hence

ϕ(z1 ⊗···⊗ (a ⊗ zi − azi) ⊗···⊗ zm) ∈ Wn−1 + Wn ∩ J,

from which it follows ϕ(y) ∈ Wn−1 + Wn ∩ J as well, proving

′ ′ ϕ(Wn−1 + Wn ∩ I) ⊆ Wn−1 + Wn ∩ J. 10 ′ Also, since ϕ(Wn) = Wn, the above argument also shows the reverse inclusion. Finally, since ϕ is an isomorphism,

−1 −1 ′ ′ ′ ′ ϕ (Wn−1 + Wn ∩ J)= ϕ ϕ(Wn−1 + Wn ∩ I)= Wn−1 + Wn ∩ I

′ ∼ ′ from which we obtain the desired ker(ϕn) = Wn−1 + Wn ∩ 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 an isomorphism of R-algebras and A-supermodules.

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 = SA(L) by showing SA(L) satisﬁes the universal property of V ′.

Lemma 3.7. The triple (SA(L),iA,iL), where iA : A → SA(L) and iL : L → SA(L) are the inclusion maps for 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-algebra homomorphism ϕ : SA(L) → B such that ϕiA = f and ϕil = 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 − (−1)|x||y|yx)=0 for x, y ∈ L. Indeed,

ϕ(xy − (−1)|x||y|yx)= ϕ(x)ϕ(y) − (−1)|x||y|ϕ(y)ϕ(x)

= g(x)g(y) − (−1)|x||y|g(y)g(x)

= g([x, y])

=0 since L has trivial bracket.

Before proving Theorem 3.5, we justify our claim that it is independent of [15, Theorem 3.1]. To see this, 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 suited for proving the PBW theorem for Poisson superalgebras.

′ Proof of Theorem 3.5. By Lemma 3.7, the map ϕ : SA(L) → V sending a 7→ a and x 7→ 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 SA(L) is graded, V is also graded, with grading equivalent to the previously deﬁned ﬁltration. Therefore, as A-superalgebras,

′ SA(L) =∼ gr V =∼ gr(V (A, L)). 11 Finally, since there exists isomorphisms

′ ′ SA(L) → gr V gr V → gr(V (A, L))

′ ′ a 7→ a + T−1 a + T−1 7→ a + T−1

′ ′ x 7→ x + T0 x + T0 7→ x + T0 we see the PBW map from SA(L) to gr(V (A, L)) is indeed an isomorphism.

4. 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 way that the enveloping algebra in the Poisson sense is the same as the enveloping algebra in the Lie-Rinehart sense, thus leading to a PBW theorem for Poisson algebras. In this section, via the same approach, we prove the PBW theorem for Poisson superalgebras. We first need to recall the construction of the even Kähler superdifferentials over a superalgebra [8].

Deﬁnition 4.1. Let A be a superalgebra, and let M be an A-supermodule. An even A-superderivation is an even linear map D : A → M such that D(ab)= aD(b) + (−1)|a||b|bD(a).

ev Proposition 4.2 ([8, Lemma 3.1]). There is an A-supermodule ΩA and an even superderivation dev : A → ev ΩA such that for any A-supermodule M, composition with dev gives an isomorphism (of abelian groups)

∼ ev Derk(A, M)0 = HomA(ΩA ,M)

ev where Derk(A, M)0 denotes the group of even superderivations from A to M, and HomA(ΩA ,M) denotes ev the group of A-supermodule morphisms from ΩA to M.

ev The supermodule ΩA can be constructed as follows. Let S be the free A-supermodule generated by the ev set {df | f ∈ A, f homogeneous}, with grading |df| = |f|. Then ΩA is the quotient of S by the relations d(rf + sg) = rdf + sdg and d(fg) = fdg + (−1)|f||g|gdf, where r, s ∈ k,f,g ∈ A, and the corresponding superderivation is dev(f) = df.

Remark 4.3. If M is just an A-module, not necessarily graded, then one still has a 1-1 correspondence ev between A-module homomorphisms ΩA → M and linear maps D : A → M satisfying D(ab) = aD(b)+ (−1)|a||b|bD(a).

ev Example 4.4. Let A be a Poisson superalgebra over k. Then the pair (A, ΩA ) becomes a Lie-Rinehart ev superalgebra over A where ΩA is given the bracket

[adf,bdg] = (−1)|b||f|abd{f,g} + a{f,b}dg − (−1)|af||bg|b{g,a}df and the anchor map is ρ(df)= {f, −}. 12 ev Now consider the Lie-Rinehart enveloping algebra V (A, ΩA ) together with the two maps

⋊ ev ev α : A → A ΩA → V (A, ΩA )

dev ev ⋊ ev ev β : A −−→ ΩA −→ A ΩA −→ V (A, ΩA ).

ev We show the triple (V (A, ΩA ),α,β) is universal with respect to property P, thereby inducing a unique ev isomorphism V (A, ΩA ) → U(A).

ev Lemma 4.5. The triple (V (A, ΩA ),α,β) satisﬁes property P. That is, α is an algebra map and β is a linear map satisfying the relations

β({a,b})= β(a)β(b) − (−1)|a||b|β(b)β(a)

α({a,b})= β(a)α(b) − (−1)|a||b|α(b)β(a)

β(ab)= α(a)β(b) + (−1)|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} = [da, db] = [β(a),β(b)] = β(a)β(b) − (−1)|a||b|β(b)β(a)

α({a,b})= {a,b} = ρ(da)(b) = [da,b] = [β(a), α(b)] = β(a)α(b) − (−1)|a||b|α(b)β(a)

β(ab) = d(ab)= adb + (−1)|a||b|bda = α(a)β(b) + (−1)|a||b|α(b)β(a).

Proposition 4.6. Let (B,γ,δ) be a triple satisfying property P. Then there is a unique algebra map ev λ : V (A, ΩA ) → B such that the following diagram commutes:

α β / ev o A V (A, ΩA ) A ■■ ✉ ■■ ✉✉ ■■ ✉✉ ■■ ✉✉ ■■ λ ✉✉ m ■■ ✉✉ ■■ ✉✉ h ■■ ✉✉ ■■ ✉✉ ■$ ✉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 |a||b| ev δ(ab) = γ(a)δ(b) + (−1) γ(b)δ(a), there is a unique A-module homomorphism θ : ΩA → B such that

δ = θdev by Remark 4.3. Further, the triple (B,γ,θ) satisﬁes property R, since

γ((bdf)(a)) = γ(b{f,a})

= γ(b)δ(f)γ(a) − (−1)|a||f|γ(b)γ(a)δ(f)

= γ(b)θ(df)γ(a) − (−1)|a||bf|γ(ab)δ(df)

= θ(bdf)γ(a) − (−1)|a||bf|γ(a)θ(bdf)

13 ev with the other two equalities being similarly easy to verify. Therefore, by the universal property of V (A, ΩA ), ev we have a unique algebra map λ : V (A, ΩA ) → B such that λα = γ, and λβ = θdev = δ.

Proposition 4.7. Let A be a Poisson superalgebra, and let U(A) be its enveloping algebra. Then there is ev a unique isomorphism Λ : U(A) → V (A, ΩA ) such that the following diagram commutes:

α β A / V (A, Ωev) o A ❍ A ❍❍ ✈✈ ❍❍ ✈✈ ❍❍ ✈✈ ❍❍ Λ ✈✈ ❍❍ ✈✈ m ❍❍ ✈✈ h ❍❍ ✈✈ ❍❍ ✈✈ ❍$ ✈z ✈ U(A)

Proof. This follows immediately from Proposition 4.6.

Corollary 4.8. Let A be a Poisson superalgebra, U(A) its enveloping algebra, and consider the ﬁltration

{Fn} on U(A) induced by the isomorphism Λ. Then there is an A-superalgebra isomorphism

ev ∼ SA(ΩA ) = 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[x1,...,xn | y1,...,ym] on even generators xi, ev and odd generators yj . It’s clear ΩA is generated as an A-supermodule by dx1,..., dxn, dy1,..., dym. We ev claim this is in fact a (homogeneous) A-basis of ΩA . Indeed, for i ≤ n, deﬁne an even derivation Di : A → A by Di(xj ) = δij and Di(yj )=0. Similarly, for n

Di(xj )=0, and Di(yj ) = δij yi. One easily veriﬁes each Di preserves the grading, and so they are indeed ev even derivations. Hence by Proposition 4.2, there are induced A-supermodule morphisms δi :ΩA → A such that δidev = Di, from which it follows the dxi, dyj are linearly independent. Therefore, by Corollary 4.8, for any Poisson bracket on A, U(A) has a k-basis consisting of elements of the form

r1 rn rn+1 rn+m s1 sn sn+1 sn+m mx1 ··· mxn my1 ··· mym hx1 ··· hxn hy1 ··· hym where each ri,si is a nonnegative integer for 1 ≤ i ≤ n, and rj ,sj ∈{0, 1} for n +1 ≤ j ≤ n + m. We remark that 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 {xi}, then for any Poisson bracket on S(V ), U(S(V )) has a k-basis consisting of elements of the form

r1 rn s1 sm mx1 ··· mxn hy1 ··· hym for all nonnegative n,m, where the ri,si are nonnegative (resp. 0 or 1) for xi,yi even (resp. xi,yi odd), and xi < xi+1, yi

In [11], Oh showed that the enveloping algebra of a Poisson Hopf algebra is itself a Hopf algebra. In this section, we show an analogous result holds for Poisson Hopf superalgebras. Throughout this section, the sym- bol ⊗ denotes the super tensor product, and we will use sumless Sweedler notation for the comultiplication of ab (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 Hopf superalgebra (A, η, ∇, ε, ∆,S) such that

∆({a,b})= {∆(a), ∆(b)}A ⊗ A b for all a,b ∈ A, where the bracket on A ⊗ A is b ′ ′ |a′||b| ′ ′ ′ ′ {a ⊗ a ,b ⊗ b }A ⊗ A = (−1) ({a,b} ⊗ a b + ab ⊗{a ,b }). b b b b b Lemma 5.2. If (A, η, ∇, ε, ∆,S) is a Poisson Hopf superalgebra, then the counit ε is a super Poisson homomorphism, 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 is

true if and only if ε({a,b}) = 0 for all a,b ∈ A. Since ε is the counit, we have ε(x1)ε(x2)= ε(x), and thus

|b1||a2| ε({a,b}) = (−1) (ε({a1,b1})ε(a2b2)+ ε(a1b1)ε({a2,b2}))

|b1||a2| = (−1) (ε({a1ε(a2),b1ε(b2)})+ ε({ε(a1)a2,ε(b1)b2}))

= ε({a1ε(a2),b1ε(b2)})+ ε({ε(a1)a2,ε(b1)b2})

= ε({a,b})+ ε({a,b})

where the third equality uses the fact that if b1 is odd, then ε(b1)=0, and likewise for a2. Now consider the antipode S. That S is a superalgebra morphism follows from the fact that it is a superalgebra anti-morphism, and the fact that Poisson superalgebras are supercommutative. That S is bijective also follows from supercommutativity of A. Now, from the equalities

|a2||b| 0= {ε(a),b} = {S(a1)a2,b} = S(a1){a2,b} + (−1) {S(a1),b}a2

|a||b1| 0= {a,ε(b)} = {a,S(b1)b2} = {a,S(b1)}b2 + (−1) S(b1){a,b2}

we obtain

|a2||b| S(a1){a2,b} = −(−1) {S(a1),b}a2 15 and

|a||b1| {a,S(b1)}b2 = −(−1) S(b1){a,b2}.

Also, by the ﬁrst paragraph,

|b1||a2| 0= ε({a,b})= S({a,b}1){a,b}2 = (−1) (S({a1,b1})a2b2 + S(a1b1){a2,b2}) from which we get

S({a1,b1})a2b2 = −S(a1b1){a2,b2}

|a1||b1| |a2||b2| = (−1) (−1) S(b1){S(a1),b2}a2

= −{S(a1),S(b1)}a2b2.

Finally, we have

S({a,b})= S({a1ε(a2),b1ε(b2)})

= S({a1,b1})ε(a2)ε(b2)

= S({a1,b1})a2S(a3)b2S(b3)

|a3||b2| = (−1) S({a1,b1})a2b2S(a3)S(b3)

|a3||b2| = −(−1) {S(a1),S(b1)}a2b2S(a3)S(b3)

= −{S(a1),S(b1)}a2S(a3)b2S(b3)

= −{S(a1ε(a2)),S(b2ε(b2))}

= −{S(a),S(b)} completing the proof.

The following lemma satisﬁes, in particular, triples (B,γ,δ) satisfying property P (with respect to some Poisson 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 maps satisfying

γ({a,b})= δ(a)γ(b) − (−1)|a||b|γ(b)δ(a), δ(ab)= γ(a)δ(b) + (−1)|a||b|γ(b)δ(a) for all a,b ∈ A. Then

γ({a,b})= γ(a)δ(b) − (−1)|a||b|δ(b)γ(a), δ(ab)= δ(a)γ(b) + (−1)|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)

16 we have

2δ(ab)= δ(a)γ(b)+ γ(a)δ(b) + (−1)|a||b|(δ(b)γ(a)+ γ(b)δ(a))

= δ(ab)+ δ(a)γ(b) + (−1)|a||b|δ(b)γ(a)

2γ({a,b})= δ(a)γ(b)+ γ(a)δ(b) − (−1)|a||b|(δ(b)γ(a)+ γ(b)δ(a))

= γ({a,b})+ γ(a)δ(b) − (−1)|a||b|δ(b)γ(a) from which the conclusions follow.

Lemma 5.4. Let (U(A),α,β) be the universal enveloping algebra of a Poisson superalgebra A. Then

(i) α ⊗ α : A ⊗ A → U(A) ⊗ U(A) is a superalgebra morphism, and (ii) α ⊗b β + βb⊗ α is a Lie superalgebrab morphism. b b Proof. (i) That α ⊗ α is a superalgebra morphism follows from the fact that α is. (ii) By Lemma 5.3,b for a,a′,b,b′ ∈ A,

(α ⊗ β + β ⊗ α)({a ⊗ a′,b ⊗ b′}) − [(α ⊗ β + β ⊗ α)(a ⊗ a′)(α ⊗ β + β ⊗ α)(b ⊗ b′) b b b b b b b b b b |a ⊗ a′||b ⊗ b′| ′ ′ − (−1) b b (α ⊗ β + β ⊗ α)(b ⊗ b )(α ⊗ β + β ⊗ α)(a ⊗ a )]

′ b b b b b b = (−1)|a ||b|(α({a,b}) ⊗ β(a′b′)+ α(ab) ⊗ β({a′,b′})+ β({a,b}) ⊗ α(a′b′)+ β(ab) ⊗ α({a′,b′})) b b b b ′ ′ ′ ′ |a ⊗ a′||b ⊗ bk ′ − (α(a) ⊗ β(a )+ β(a) ⊗ α(a ))(α(b) ⊗ β(b )+ β(b) ⊗ α(b )) + (−1) b b (α(b) ⊗ β(b ) b b b b b + β(b) ⊗ α(b′))(α(a) ⊗ β(a′)+ β(a) ⊗ α(a′)) b b b |a′||b| ′ ′ |a′||b′| ′ ′ |a||b| ′ ′ = (−1) hα(ab) ⊗ (β(a )β(b ) − (−1) β(b )β(a )) + (β(a)α(b) − (−1) α(b)β(a)) ⊗ (α(a )β(b )

′ b′ ′ b′ + (−1)|a ||b |α(b′)β(a′)) + (α(a)β(b) + (−1)|a||b|α(b)β(a)) ⊗ (β(a′)α(b′) − (−1)|a ||b |α(b′)β(a′)) b + (β(a)β(b) − (−1)|a||b|β(b)β(a)) ⊗ α(a′b′) − α(ab) ⊗ β(a′)β(b′)+ α(a)β(b) ⊗ β(a′)α(b′) b b b ′ ′ ′ ′ |a ⊗ a′||b ⊗ b′| ′ ′ + β(a)α(b) ⊗ α(a )β(b )+ β(a)β(b) ⊗ α(a b ) + (−1) b b hα(b)α(ba) ⊗ β(b )β(a ) b b b ′ ′ ′ ′ ′ ′ + α(b)β(a) ⊗ β(b )α(a )+ β(b)α(a) ⊗ α(b )β(a )+ β(b)β(a) ⊗ α(b a )ii

′ b b ′ ′ b ′ ′ ′ = −(−1)|a ||b|+|a||b|α(b)β(a) ⊗ (α(a′)β(b′) − (−1)|a ||b |β(b′)α(a′)) + (−1)|a ||b|+|a ||b |(β(a)α(b)

b ′ ′ ′ − (−1)|a||b|α(b)β(a)) ⊗ α(b′)β(a′) − (−1)|a ||b|+|a ||b |(α(a)β(b) − (−1)|a||b|β(b)α(a)) ⊗ α(b′)β(a′)

′ b ′ ′ b + (−1)|a ||b|+|a||b|α(b)β(a) ⊗ (β(a′)α(b′) − (−1)|a ||b |α(b′)β(a′))

′ b ′ ′ ′ = −(−1)|a ||b|+|a||b|α(b)β(a) ⊗ α({a′,b′}) + (−1)|a ||b|+|a ||b |α({a,b}) ⊗ α(b′)β(a′)

′ ′ ′ b ′ b − (−1)|a ||b|+|a ||b |α({a,b}) ⊗ α(b′)β(a′) + (−1)|a ||b|+|a||b|α(b)β(a) ⊗ α({a′,b′}) b b =0.

Clearly α ⊗ β + β ⊗ α preserves the grading, so it is indeed a Lie superalgebra morphism. b b 17 Lemma 5.5. Let A, B be Poisson superalgebras, and let C be an algebra. If ϕ : A → B is a super Poisson homomorphism and (C,α,β) satisﬁes property P, then so does (C, αϕ, βϕ).

Proof. Straightforward.

Lemma 5.6. Let (U(A),α,β) be the Poisson enveloping algebra of a Poisson superalgebra A. Then (U(A) ⊗ U(A), α ⊗ α, α ⊗ β + β ⊗ α) is the Poisson enveloping algebra of A ⊗ A. b b b b b Proof. It is simple to verify

(α ⊗ α)({a ⊗ a′,b ⊗ b′}) = (α ⊗ β + β ⊗ α)(a ⊗ a′)(α ⊗ α)(b ⊗ b′) b b b b b b b b |a ⊗ a′||b ⊗ b′| ′ ′ − (−1) b b (α ⊗ α)(b ⊗ b )(α ⊗ β + β ⊗ α)(a ⊗ a ) b b b b b (α ⊗ β + β ⊗ α)((a ⊗ a′)(b ⊗ b′)) = (α ⊗ α)(a ⊗ a′)(α ⊗ β + β ⊗ α)(b ⊗ b′) b b b b b b b b b |a ⊗ a′||b ⊗ b′| ′ ′ + (−1) b b (α ⊗ α)(b ⊗ b )(α ⊗ β + β ⊗ α)(a ⊗ a ). b b b b b ′ ′ for all a,a ,b,b ∈ A. Let i1,i2 be the super Poisson homomorphisms from A to A ⊗ A deﬁned by b i1(a)= a ⊗ 1 b i2(a)=1 ⊗ a b for a ∈ A. Let B be an algebra with multiplication map µB . If γ,δ : A ⊗ A → B are maps such that (B,γ,δ) satisﬁes property P, then, by Lemma 5.5, there exist algebra maps f,gb : U(A) → B such that the diagrams

α,β α,β A / U(A) A / U(A)

i1 f i2 g

 A ⊗ A / B A ⊗ A / B γ,δ γ,δ b b commute. Moreover, we have

′ |a||a′| ′ ′ |a||a′| ′ δi1(a)γi2(a ) − (−1) γi2(a )δi1(a)= δ(a ⊗ 1)γ(1 ⊗ a ) − (−1) γ(1 ⊗ a )δ(a ⊗ 1) b b b b = γ({a ⊗ 1, 1 ⊗ a′}) b b = γ({a, 1} ⊗ a′ + a ⊗{1,a′})=0. b b Therefore,

′ ′ ′ ′ µB(f ⊗ g)(α ⊗ α)(a ⊗ a )= fα(a)gα(a )= γ(i1(a)i2(a )) = γ(a ⊗ a ) b b b b ′ ′ ′ µB(f ⊗ g)(α ⊗ β + β ⊗ α)(a ⊗ a )= fα(a)gβ(a )+ gβ(a)gα(a ) b b b b ′ ′ = γi1(a)δi2(a )+ δi1(a)γi2(a )

′ |a||a′| ′ = γi1(a)δi2(a ) + (−1) γi2(a )δi1(a) 18 ′ = δ(i1(a)i2(a ))

= δ(a ⊗ a′). b

Thus µB(f ⊗ g): U(A) ⊗ U(A) → B is an algebra map such that b b

µB(f ⊗ g)(α ⊗ α)= γ, µB(f ⊗ g)(α ⊗ β + β ⊗ α)= δ. b b b b b Now, if h : U(A) ⊗ U(A) → B is another algebra map such that b h(α ⊗ α)= γ, h(α ⊗ β + β ⊗ α)= δ b b b then

µB(f ⊗ g)(α(a) ⊗ 1) = γ(a ⊗ 1) = h(α(a) ⊗ 1) b b b b µB(f ⊗ g)(1 ⊗ α(a)) = γ(1 ⊗ a)= h(a ⊗ α(a)) b b b b µB(f ⊗ g)(β(a) ⊗ 1) = δ(a ⊗ 1) = h(β(a) ⊗ 1) b b b b µB(f ⊗ g)(1 ⊗ β(a)) = δ(1 ⊗ a)= h(1 ⊗ β(a)) b b b b

for all a ∈ A. Finally, since U(A) is generated by α(A) and β(A), we have µB(f ⊗ g) = h. Lemma 5.4 completes the proof. b

The following result shows that sending a Poisson superalgebra to its enveloping algebra deﬁnes a functor from the category of Poisson superalgebras to the category of associative (super)algebras.

Lemma 5.7. Let (UA, αA,βA) and (UB, αB,βB) be the universal enveloping algebras for Poisson superalgebras A, B respectively. If ϕ : A → B is a super Poisson homomorphism, then there is a unique algebra map U(ϕ): U(A) → U(B) such that the diagram

αA,βA A / U(A)

ϕ U(ϕ)

 B / U(B) αB ,δB

commutes.

Proof. This follows immediately from Lemma 5.5.

Let (A, ·, {·, ·}) be a Poisson superalgebra. Deﬁne another bracket {·, ·}′ on A by {a,b}′ = −{a,b}. Then A′ = (Aop, ◦, {·, ·}′) is also a Poisson superalgebra, where for a (super)algebra B, we denote Bop = (B, ◦) the opposite (super)algebra of B. 19 Proposition 5.8. Let (U(A),α,β) be the universal enveloping algebra for a Poisson superalgebra A and treat 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 triple satisfying property P with respect to A′. Then

γ({a,b})= −γ({a,b}′)= −δ(a)γ(b) + (−1)|a||b|γ(b)δ(a)

= (−δ)(a)γ(b) − (−1)|a||b|γ(b)(−δ)(a)

−δ({a,b})= δ({a,b}′)= δ(a)δ(b) − (−1)|a||b|δ(b)δ(a)

= (−δ)(a)(−δ)(b) − (−1)|a||b|(−δ)(b)(−δ)(a)

−δ(ab)= −γ(a)δ(b) − (−1)|a||b|γ(b)δ(a)= γ(a)(−δ)(b) + (−1)|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 respect to A, then (B, γ, −δ) satisfies property P with respect to A′. That (U(A), α, −β) is a universal enveloping algebra of A′ follows immediately. Consider now the algebra map

ϕ : U(A) → U(A)op

mx 7→ mx

hx 7→ −hx.

Note |x||y| |x||y| ϕ(hxmy − (−1) myhx − m{x,y})= −hx ◦ my + (−1) my ◦ hx − m{x,y} =0 with the other relations of U(A) similarly mapping to 0, proving ϕ is indeed well-deﬁned. Similarly, one can deﬁne an algebra map

ψ : U(A)op → U(A)

mx → mx

hx →−hx

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.

Theorem 5.9. If (A, η, ∇, ε, ∆,S) is a Poisson Hopf superalgebra, then

(U(A), ηU(A), ∇U(A),U(ε),U(∆),U(S))

is a Hopf superalgebra such that

U(∆)α = (α ⊗ α)∆,U(∆)β = (α ⊗ β + β ⊗ α)∆, b 20 b b U(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) ⊗ U(A), α ⊗ α, α ⊗ β + β ⊗ α) is the Poisson enveloping algebra of A ⊗ A by Lemma 5.6, there is a uniqueb algebra mapb U(∆)b : U(Ab) → U(A) ⊗ U(A) such that b b

U(∆)α = (α ⊗ α)∆,U(∆)β = (α ⊗ β + β ⊗ α)∆ b b b 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, idk, 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 A1 by Lemma 5.2, there is an algebra homomorphism U(S): U(A) → U(A)op such that U(S)α = αS,U(S)β = βS by Lemma 5.7 and Proposition 5.8. It is routinely veriﬁed that (U(A), ηU(A), ∇U(A),U(ε),U(∆),U(S)) is indeed a Hopf superalgera.

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 via the Leibniz rule. Also, S(L) becomes a Hopf superalgebra via

∆(a)= a ⊗ 1+ a ⊗ a, ε(a)=0, S(a)= −a b b 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 superalgebra by 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 a natural Hopf superalgebra structure. Denoting B = PS(L), we show Be =∼ U(V ⋊ L) as Hopf superalgebras. e Indeed, consider the map ϕ : V ⋊ L → B sending v + x to mv + hx for v ∈ V, x ∈ L. One easily veriﬁes ϕ is a Lie superalgebra morphism, so there is an induced algebra map Φ : U(V ⋊ L) → Be. It is simple to show Φ is a Hopf superalgebra map, and to show it is bijective, recall that from Example 4.9 if B has homogeneous e basis {xi}, then B has a basis consisting of elements of the form

r1 rn s1 sm mx1 ··· mxn hy1 ··· hym for all nonnegative n,m, where the ri,si are nonnegative (resp. 0 or 1) for xi,yi even (resp. xi odd), and xi < xi+1, yi

r1 rn s1 sm x1 ··· xn y1 ··· ym

e with the same restrictions on ri,si. Thus Φ send a basis of U(V ⋊L) onto a basis of B , proving it is bijective, and hence a Hopf superalgebra isomorphism. 21 6. Enveloping algebras of quadratic Poisson algebras and quadratic duality

Let P = k[x1, x2,...xn] be a polynomial algebra. We say that P is a quadratic Poisson polynomial algebra if for each xi, xj , the bracket satisﬁes

{x , x } = Ci,j x x i j X k,l k l k,l

i,j ! for some scalars Ck,l [16]. It is well known that P and the exterior algebra P = Λ(θ1,θ2 ...,θn) are quadratic duals of each other. For quadratic Poisson polynomial algebras, we deﬁne a dual bracket on P ! by

{θ ,θ } = Ci,j θ θ . k l X k,l j i i,j

! ! If we grade P so that every θi 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 ,θ } = Ci,j θ θ . k l X k,l i j i,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 over the set X.

Proposition 6.1. Let P = k[x1,...,xn] be a quadratic Poisson polynomial algebra. The triple (R,α,β) is the Poisson enveloping algebra of P , where R is the quadratic algebra T (mxi ,hxi )/I, I is the ideal

I = m m − m m , h m − m h − Ci,j m m , xi xj xj xi xi xj xj xi X k,l xk xl k,l

h h − h h − Ci,j (m h + m h ) , xi xj xj xi X k,l xk xl xl xk k,l

α : P → R is an algebra homomorphism sending each generator xi to mxi , and β : P → R is a linear map satisfying β(xy)= α(x)β(y)+ α(y)β(x) that sends 1 and 0, and xi to hxi .

Proof. It suﬃces to show this triple satisﬁes the universal property of the Poisson enveloping algebra. That is, 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 is a unique algebra map Φ : R → B so that Φα = γ, Φβ = δ. To prove (4.1), we proceed by induction on the degree (note it suﬃces to show for monomials). Indeed, the equality is trivial if one (or both) of x, y is a scalar, and by the relations on R,

α({x , x })= Ci,j m m = [h ,m ] = [β(x ), α(x )]. i j X k,l xk xl xi xj i j k,l 22 Suppose now that equation (1) holds for all monomials of degree at most N, and let y,z be two such elements. Then

α({yxi,z})= α(y{xi,z} + {y,z}xi)= α(y)[β(xi), α(z)] + α(xi)[β(y), α(z)]

= α(y)[β(xi), α(z)] + α(xi)[β(y), α(z)]

+ [α(y), α(z)]β(xi) + [α(xi), α(z)]β(y)

= [α(y)β(xi)+ α(xi)β(y), α(z)]

= [β(yxi), α(z)] since α(P ) is a commuative subalgebra. Therefore,

α({yxi, zxj})= α({yxi,z}xj + z{yxi, xj })

= [β(yxi), α(z)]α(xj )+ α(z)[β(yxi), α(xj )]

= [β(yxi), α(z)α(xj )].

Similarly, we show β is a Lie algebra map via induction on the degree. The proofs for degree 0 and 1 are simple, so suppose β({x, y}) = [β(x),β(y)] for all (basis) elements y,z of degree at most N. Then

β({yxi,z})= β(y{xi,z} + {y,z}xi)

= α(y)[β(xi),β(z)] + [α(xi),β(z)]β(y) + [α(y),β(z)]β(xi)+ α(xi)[β(y),β(z)]

= [β(yxi),β(z)].

Note that in the second to last equality we use [11, Lemma 4] to write α({xi,z}) = [α(xi),β(z)] and likewise for α({y,z}). Hence

β({yxi, zxj })= β(z{yxi, xj } + {yxi,z}xj)

= α(z)[β(yxi),β(xj )] + [β(yxi), α(xj )]β(z) + [β(yxi), α(z)]β(xj )

+ α(xj )[β(yxi),β(z)]

= [β(yxi),β(zxj )].

It remains to show the uniqueness and existence of Φ. Indeed, it is clear that such a function is unique, since Φ(mxi ) = γ(xi), Φ(hxj ) = δ(xj ), and the mxi ,hxj generate R as an algebra, while existence follows from a routine veriﬁcation showing Φ(I) = 0.

Proposition 6.2. The triple (Q, α, β) is the Poisson enveloping algebra of P !, where Q is the quadratic algebra T (mθk ,hθl )/J, J is the ideal

J = m m + m m , h m + m h + Ci,j m m , θk θl θl θk θk θl θl θk X k,l θi θj i,j 23 h h + h h + Ci,j (m h − m h ) , θk θl θl θk X k,l θi θj θj θi i,j

|x||y| α is an algebra map sending θi to mθi , and β is a linear map satisfying β(xy)= α(x)β(y)+(−1) α(y)β(x)

that sends 1 to 0, and θi to hθi .

Proof. The proof is similar to Proposition 6.1. Note that Lemma 5.3 is used rather than [11, Lemma 4].

! ∗ ∗ Proposition 6.3. U(P ) is generated by mxi and hxi subject to the relations

∗ ∗ ∗ ∗ 0= hxk hxl + hxl hxk ,

0= h∗ m∗ + m∗ h∗ + Ci,j h∗ h∗ , xk xl xl xk X k,l xi xj i,j

0= m∗ m∗ + m∗ m∗ + Ci,j (h∗ m∗ − h∗ m∗ ). xk xl xl xk X k,l xi xj xj xi i,j 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 listed act as 0 on the generators of I. Thus we need only show these are the only such relations. Indeed, denote the ⊥ ⊥ ideal described in the proposition statement by K. Since K ⊆ I , it suﬃces to show dim(K2) = dim(I2 ), ∗ ∗ ∗ ∗ ⊥ ! or equivalently, dim(V ⊗ V /K2) = dim(V ⊗ V /I2 ). Indeed, [17] shows that the Hilbert series of U(P ) is (1 + t)2n, so dim(V ∗ ⊗ V ∗/I⊥) = 2n = 2n2 − n. Now, consider V ∗ ⊗ V ∗/K, which is generated by 2 2 ∗ ∗ 2 length 2 products of the mxk and hxl , 4n elements in total. From the ﬁrst relation of K, we may remove ∗ ∗ all elements of the form hxk hxl , k ≥ l, and still have our set be spanning. Similarly, the second and third ∗ ∗ ∗ ∗ equations allow us to remove the elements mxk hxl , for all k, l, and mxk mxl , k ≥ l, respectively. Hence we are left with n2 +2=2 n =2n2 − 2 elements in our spanning set. Therefore, 2 ∗ ∗ 2 ∗ ∗ ⊥ ∗ ∗ dim(V ⊗ V /K2) ≤ 2n − n = dim(V ⊗ V /I2 ) ≤ dim(V ⊗ V /K2),

which is the desired result.

Example 6.4. If P = k[x1,...,xn] is given the commutator bracket, then U(P ) is a polynomial algebra on 2n generators by Proposition 6.1; note this is consistent with Example 4.9. Then the induced bracket on P ! is also trivial, so U(P !) is an exterior algebra on 2n generators by Proposition 6.2. Hence the duality described 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 studied in 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 Re is an iterated Ore extension of A. In this section, we extend Poisson-Ore extensions to Poisson superalgebras, and show the enveloping algebra is also an iterated Ore extension. 24 Deﬁnition 7.1. Let R be a Poisson superalgebra and let α : R → R be a linear map. Then α is a Poisson superderivation if

(i) α(ab)= α(a)b + (−1)|a||α|aα(b) (ii) α({a,b})= {α(a),b} + (−1)|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 polynomial superring 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}} + (−1)|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

i j |a||b| i+j |a||b| i+j−1 {ax , bx } = ({a,b}R − (−1) jbα(a)+ iaα(b))x + (iaδ(b) − (−1) jbδ(a))x for all monomials axi, bxj in R[x]. Note this bracket satisﬁes (5.1) and {f,g} = −(−1)|f||g|{g,f} for homogeneous f, g. Also, {f, −} is a superderivation of degree |f|, since α and δ are even superderivations. It remains to check the Jacobi identity: for monomials axi, bxj , cxk ∈ R[x],

(−1)|a||c|{axi, {bxj, cxk}} + (−1)|a||b|{bxj, {cxk, axi}} + (−1)|b||c|{cxk, {axi, bxj }} =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

(−1)|a||c|{axi+1, {bxj , cxk}} + (−1)|a||b|{bxj , {cxk, axi+1}} + (−1)|b||c|{cxk, {axi+1, bxj }}

|a||c| i |a||b| i |b||c| i = (−1) {ax , {b,c}} + (−1) {b, {c,ax }} + (−1) {c, {ax ,b}}x 25 + (−1)|a||c|axi({x, {b,c}} + {b, {c, x}} + (−1)|b||c|{c, {x, b}})

=0 by 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.

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; that is, δ is a homomorphism of abelian groups satisfying

δ(r1r2)= σ(r1)δ(r2)+ δ(r1)r2

for r1, r2 ∈ 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 Poisson e superalgebra and let A = R[x; α, δ]p be a Poisson-Ore extension of R. We wish to show A is an iterated Ore e e e extension of R . In particular, for appropriate σ1, σ2, η1, η2, we wish to show A =∼ R [mx; σ1, η1][hx; σ2, η2]. e To determine what σ1, η1 should be, observe that in A ,

mxmr = mrmx and mxhr = hrmx − m{r,x} = hrmx + mα(r)mx + mδ(r)

for r ∈ R, so

σ1(mr)= mr, σ1(hr)= hr + mα(r)

η1(mr)=0, η1(hr)= mδ(r).

Similarly, one has

σ2(mr)= mr, σ2(hr)= hr + mα(r), σ2(mx)= mx,

η2(mr)= mα(r)mx + mδ(r), η2(hr) = (hα(r) + mα2(r))mx + mδα(r) + hδ(r), η2(mx)=0.

To extend to general elements, we declare σ1, σ2 to be algebra homomorphisms, and for η1 (resp. η2) to be

a σ1-derivation (resp. σ2-derivation).

e Lemma 7.4. The maps σ1 and σ2 are automorphisms of R , and η1 (resp. η2) is a σ1-derivation (resp. e σ2-derivation) of R . 26 Proof. To show σ1 is well-deﬁned, consider the following diagram

m R // Re ❅ ✤ ❅❅❅❅ h ❅ ❅ ✤ ❅❅❅❅ f ❅ ❅ ✤ ❅❅❅❅ σ1 g ❅❅❅❅ ✤ ❅❅❅❅ ❅❅❅ ✤ Re where f(r)= mr and g(r)= mα(r) + hr for r ∈ R. Hence the diagram is commutative and one easily shows e e (R ,f,g) satisfies property P, so σ1 is well-defined by the universal property of R . Note the fact that σ2 is e e n e well-defined then follows immediately since R [mx; σ1, η1] =∼ R m as left R -modules, and σ2|Re = σ1. Ln≥0 x −1 −1 That σ1 is an automorphism follows by construction of the inverse via σ1 (mr)= mr, σ1 (hr)= hr −mα(r). −1 e −1 Similarly, σ2 has inverse identical to that of σ1 on R (as σ2|Re = σ1), while σ2 (mx)= mx. e For η1, it is simple to show η1 sends each relation in R to 0. Hence η1 can be extended to a well-defined e e σ1-derivation on R . Similarly, one can show, albeit more tediously, that η2 sends each relation of R to 0, e e and so η2 is well-defined on R . That η2 is then well-defined on all of R [mx; σ1, η1] follows from the same argument as σ2.

e e Next, we will show the algebra homomorphism ϕ : A → R [mx; σ1, η1][hx; σ2, η2] deﬁned by

ϕ(mr)= mr, ϕ(hr)= hr, ϕ(mx)= mx, ϕ(hx)= hx, (1) is an isomorphism. We ﬁrst need to show ϕ is well-deﬁned. To do this, recall Ae is the quotient F/I, where

F is the free algebra generated by {ma,hA | a ∈ A}, and I is the ideal consisting of the usual relations. We ∞ i then extend ϕ to {ma,ha | a ∈ A} by declaring that for a = cix ∈ A = R[x], where coﬁnitely many ci Pi=0 are 0, ∞ ∞ ϕ(m )= m xi and ϕ(h )= (im mi−1h + mi h ). a X ci a X ci x x x ci i=0 i=0 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 Ae, ϕ is well-deﬁned.

|a||b| Proof. We show ϕ(m{a,b} +(−1) mbha −hamb) = 0, with the other generators mapping to 0 via a similar ∞ i ∞ j argument. Indeed, let a = cix ,b = dj x and note |a| = |ci|, |b| = |dj | for all i, j. Recalling, from Pi=0 Pj=0 the proof of Theorem 6.2, that the bracket on A is

i j |ci||dj | i+j |ci||cj | i+j−1 {cix , dj x } = ({ci, dj }− (−1) jdj α(ci)+ iciα(dj ))x + (iciδ(dj ) − (−1) jdj δ(ci))x , we have

|a||b| ϕ(m{a,b}+(−1) mbha − hamb) 27 = ϕ m ∞ |a||b| i+j |a||b| i+j−1 P {ci,dj }−(−1) jdj α(ci)+iciα(dj ) x +(iciδ(dj )− −1) jdj δ(ci) x i,j=0 |a||b| + (−1) ϕ(m ∞ j h ∞ i ) − ϕ(h ∞ i m ∞ j ) Pj=0 dj x Pi=0 cix Pi=0 cix Pj=0 dj x ∞ i+j i+j−1 = m |a||b| m + m |a||b| m X h {ci,dj }−(−1) jdj α(ci)+iciα(dj ) x iciδ(dj )−(−1) jdj δ(ci) x i i,j=0 ∞ ∞ + (−1)|a||b| m mj (im mi−1h + mi h ) X dj x X ci x x x ci j=0 i=0 ∞ ∞ − (im mi−1h + mi h ) m mj X ci x x x ci X dj x i=0 j=0 ∞ = m mi+j − (−1)|a||b|jm mi+j + im mi+j + im mi+j−1 X h {ci,dj } x dj α(ci) x ciα(dj ) x ciδ(dj ) x i,j=0

|a||b| i+j−1 |a||b| i+j−1 |a||b| i+j − (−1) jmdj δ(ci)mx + (−1) imdj ci mx hx + (−1) mdj mx hci

i−1 j i j − imci mx hxmdj mx − mxhci mdj mxi ∞ = m mi+j − (−1)|a||b|jm mi+j + im mi+j + im mi+j−1 X h {ci,dj } x dj α(ci) x ciα(dj ) x ciδ(dj ) x i,j=0

|a||b| i+j−1 |a||b| i+j−1 |a||b| i+j − (−1) jmdj δ(ci)mx + (−1) imdj ci mx hx + (−1) mdj mx hci

i−1 j i |a||b| j − imc m md hx + m mx + m m − m m + (−1) md hc m i x j α(dj ) δ(dj ) x x {ci,dj } j i xi ∞ = m mi+j − (−1)|a||b|jm mi+j + im mi+j + im mi+j−1 X h {ci,dj } x dj α(ci) x ciα(dj ) x ciδ(dj ) x i,j=0

|a||b| i+j−1 |a||b| i+j−1 |a||b| i+j − (−1) jmdj δ(ci)mx + (−1) imdj ci mx hx + (−1) mdj mx hci

− im mi−1 m h + m m + m mj − mi m mj ci x dj x α(dj ) x δ(dj ) x x {ci,dj } x |a||b| i j j j−1 − (−1) m md m hc − jm m − jm m x j x i α(ci) x δ(ci) x i =0.

Note that in the second to last equality, we used the fact that

j j j j−1 hci mx = mxhci − jmα(ci)mx − jmδ(ci)mx which one can prove via induction.

e e Theorem 7.6. The algebra homomorphism ϕ : A → R [mx; σ1, η1][hx; σ2, η2] deﬁned by (1) is an isomorphism.

Proof. By Lemma 7.5, ϕ is well-defined, so it remains to show it is bijective. To do this, we construct an e e e inverse map ψ. Indeed, consider first the triple (A ,f,g), where f : R → A maps r to mr, and g : R → A e maps r to hr. It is easy to verify (A ,f,g) satisfies property P with respect to R, so there is an induced 28 algebra homomorphism θ : Re → Ae: m R // Re ❅ ✤ ❅❅❅❅ h ❅❅❅❅ ✤ ❅❅❅❅ f ❅❅❅❅ ✤ θ g ❅❅❅❅ ✤ ❅❅❅❅ ❅❅❅ ✤ Ae i j i j e We thus set ψ(ymxhx)= θ(y)mxhx for y ∈ R . It follows immediately from the definitions of ϕ, ψ that they are inverse functions, and hence are an inverse pair of algebra isomorphisms.

Corollary 7.7. Let R be a Poisson superalgebra and let A be an iterated Poisson-Ore extension of R. Then Ae inherits the following properties from Re:

(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) comes from [4].

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 ﬁnitely generated as a superalgebra, then U(R) is left/right Noetherian.

Proof. Suppose R is generated by the even variables x1,...,xn and the odd variables y1,...,ym. Then U(R) is generated by α(R),β(R), and since

α(xy)= α(x)α(y)

β(xy)= α(x)β(y) + (−1)|x||y|α(y)β(x) for x, y ∈ R, it follows U(R) is generated by α(xi), α(yj ),β(xi),β(yj ), 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[X1, ··· ,Xn | Y1, ··· , Ym]. Since R is ﬁnitely generated over k, it follows from the discussion after Proposition 3.3.7 in [18] that R[X1, ··· ,Xn | Y1, ··· , Ym], and hence gr U, is left/right Noetherian. By [9, Theorem 1.6.9], U(R) is left/right Noetherian.

Example 7.9. Let R be a polynomial superalgebra with arbitrary Poisson bracket. Then U(R) is Noetherian by the above proposition, so if A is any iterated Poisson-Ore extension of R, then U(A) is also Noetherian by Corollary 7.7. 29 References

[1] Damien Calaque, Giovanni Felder, Andrea Ferrario, and Carlo A. Rossi. Bimodules and branes in deformation quantization. Compos. Math., 147(1):105–160, 2011. [2] Christian Kassel. Quantum groups, volume 155 of Graduate Texts in Mathematics. Springer-Verlag, New York, 1995. [3] Thierry Lambre, Cyrille Ospel, and Pol Vanhaecke. Poisson enveloping algebras and the Poincaré-Birkhoff-Witt theorem. J. Algebra, 485:166–198, 2017. [4] Liyu Liu, Shengqiang Wang, and Quanshui Wu. Twisted Calabi-Yau property of Ore extensions. J. Noncommut. Geom., 8(2):587–609, 2014. [5] Jiafeng Lü, Xingting Wang, and Guangbin Zhuang. Universal enveloping algebras of Poisson Hopf algebras. J. Algebra, 426:92–136, 2015. [6] Jiafeng Lü, Xingting Wang, and Guangbin Zhuang. Universal enveloping algebras of Poisson Ore extensions. Proc. Amer. Math. Soc., 143(11):4633–4645, 2015. [7] Yuri I. Manin. Gauge field theory and complex geometry, volume 289 of Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences]. Springer-Verlag, Berlin, second edition, 1997. Translated from the 1984 Russian original by N. Koblitz and J. R. King, With an appendix by Sergei Merkulov. [8] A. Masuoka and A. N. Zubkov. On the notion of Krull super-dimension. J. Pure Appl. Algebra, 224(5):106245, 2020. [9] J. C. McConnell and J. C. Robson. Noncommutative Noetherian rings, volume 30 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, revised edition, 2001. With the cooperation of L. W. Small. [10] Sei-Qwon Oh. Poisson enveloping algebras. Comm. Algebra, 27(5):2181–2186, 1999. [11] Sei-Qwon Oh. Hopf structure for Poisson enveloping algebras. Beiträge Algebra Geom., 44(2):567–574, 2003. [12] Sei-Qwon Oh. Poisson polynomial rings. Comm. Algebra, 34(4):1265–1277, 2006. [13] Sei-Qwon Oh, Chun-Gil Park, and Yong-Yeon Shin. A Poincaré-Birkhoff-Witt theorem for Poisson enveloping algebras. Comm. Algebra, 30(10):4867–4887, 2002. [14] A. Polishchuk. Algebraic geometry of Poisson brackets. volume 84, pages 1413–1444. 1997. Algebraic geometry, 7. [15] George S. Rinehart. Differential forms on general commutative algebras. Trans. Amer. Math. Soc., 108:195–222, 1963. [16] E. K. Sklyanin. Some algebraic structures connected with the Yang-Baxter equation. Funktsional. Anal. i Prilozhen., 16(4):27–34, 96, 1982. [17] Matthew Towers. Poisson and Hochschild cohomology and the semiclassical limit. J. Noncommut. Geom., 9(3):665–696, 2015. [18] Dennis Bouke Westra. Superrings and Supergroups. PhD dissertation, Universität Wien, 2009.

Miami University, Department of Mathematics, 301 S. Patterson Ave., Oxford, Ohio 45056 Email address: [email protected]