# Co-Poisson structures on polynomial Hopf algebras

Qi Lou  and  Quanshui Wu  School of Mathematical Sciences, Fudan University,
Shanghai 200433, China<EMAIL_ADDRESS>School of Mathematical Sciences,
Fudan University, Shanghai 200433, China<EMAIL_ADDRESS>
###### Abstract.

The Hopf dual $H^{\circ}$ of any Poisson Hopf algebra $H$ is proved to be a
co-Poisson Hopf algebra provided $H$ is noetherian. Without noetherian
assumption, unlike it is claimed in literature, the statement does not hold.
It is proved that there is no nontrivial Poisson Hopf structure on the
universal enveloping algebra of a non-abelian Lie algebra. So the polynomial
Hopf algebra, viewed as the universal enveloping algebra of a finite-
dimensional abelian Lie algebra, is considered. The Poisson Hopf structures on
polynomial Hopf algebras are exactly linear Poisson structures. The co-Poisson
structures on polynomial Hopf algebras are characterized. Some correspondences
between co-Poisson and Poisson structures are also established.

###### Key words and phrases:

Poisson algebra, co-Poisson coalgebra, Poisson Hopf algebra, co-Poisson Hopf
algebra

###### 2010 Mathematics Subject Classification:

Primary 17B63, 16W10, 16S30

##  1\. Introduction

Poisson structure naturally appears in classical/quantum mechanics, in
mathematical physics, and in deformation theory. It is an important algebra
structure in Poisson geometry, algebraic geometry and non-commutative
geometry. There are lots of research in the related subjects.

Co-Poisson structure is a dual concept of Poisson structure in categorial
point of view. It arises also in mathematics and mathematical physics
naturally as explained in the next two paragraphs.

Let $G$ be a Lie group and $\mathcal{O}(G)$ be its algebra of functions. A Lie
group $G$ is said to be a Poisson Lie group if $\mathcal{O}(G)$ is a Poisson
Hopf algebra. It is well-know that the category of connected and simply-
connected Lie groups is equivalent to the category of finite-dimensional Lie
algebras. In this case, $\mathcal{O}(G)$ is identified with the Hopf dual
$U(\mathfrak{g})^{\circ}$ of the universal enveloping algebra
$U(\mathfrak{g})$, where $\mathfrak{g}$ is the corresponding Lie algebra of
$G$. The Poisson counterpart of this fact holds also, namely, the category of
connected and simply-connected Poisson Lie groups is equivalent to the
category of finite-dimensional Lie bialgebras ([KS, Theorem 3.3.1] or [Dr,
Theorem 1]). However, the Lie bialgebra structures on any Lie algebra
$\mathfrak{g}$ is in one-to-one correspondence with the co-Poisson Hopf
structures on $U(\mathfrak{g})$ ([CP, Proposition 6.2.3]).

On the other hand, to quantize a Lie group or Lie algebra one should equip it
with an extra structure, namely, a Poisson Lie group structure or Lie
bialgebra structure, respectively. Therefore co-Poisson structure naturally
appears in the theory of quantum groups and in mathematical physics.

If $G$ is a connected and simply-connected Poisson Lie group and
$\mathfrak{g}$ is the corresponding Lie bialgebra, then the Poisson Hopf
structure on $U(\mathfrak{g})^{\circ}\cong\mathcal{O}(G)$ is the dual of the
co-Poisson Hopf structures on $U(\mathfrak{g})$. In [OP], the authors proved
that the dual Hopf algebra $U(\mathfrak{g})^{\circ}$ of $U(\mathfrak{g})$ is a
Poisson Hopf algebra for any finite-dimensional Lie bialgebra $\mathfrak{g}$.
In fact, in general, as stated in [KS, Proposition 3.1.5] earlier, the dual
Hopf algebra of any co-Poisson Hopf algebra is a Poisson Hopf algebra. A
complete proof is given in a recent paper by Oh [Oh, Theorem 2.2]. The dual
proposition that the dual Hopf algebra of any Poisson Hopf algebra is a co-
Poisson Hopf algebra is also stated in [KS, Proposition 3.1.5]. Unfortunately,
this claim is not true in general as showed in our Example 3.6. Under an
additional assumption that the algebra is noetherian, we prove the statement
is true in Proposition 3.5.

We prove that there is no nontrivial Poisson Hopf structure on the universal
enveloping algebra of a non-abelian Lie algebra in Proposition 2.5. So, we
turn to consider in latter sections the abelian case, i.e., the polynomial
Hopf algebra $A=k[x_{1},x_{2},\cdots,x_{d}]$, viewed as the universal
enveloping algebra of an abelian Lie algebra of dimension $d$. The Poisson
Hopf structures on $A=k[x_{1},x_{2},\cdots,x_{d}]$ are exactly linear Poisson
structures on $A$ (see Proposition 5.3). By establishing a reciprocity law
between two linear maps of $A\to A\otimes A$ for
$A=k[x_{1},x_{2},\cdots,x_{d}]$ (see Proposition 4.3), all co-Poisson
coalgebra and co-Poisson Hopf algebra structures on
$A=k[x_{1},x_{2},\cdots,x_{d}]$ are described in Theorem 4.7 and Proposition
5.4 respectively. In particular, the co-Poisson coalgebra structures on
$A=k[x,y]$ are given by the linear maps $I:A\to k(x\otimes y-y\otimes x)$ (see
Proposition 4.8). By using the algebra of divided power series, the co-Poisson
coalgebra structures on $A=k[x_{1},x_{2},\cdots,x_{d}]$ are showed to be in
one-to-one correspondence with the Poisson algebra structures on
$\tilde{A}=k[[x_{1},x_{2},\cdots,x_{d}]]$, the algebra of formal power series
in Theorem 5.8.

The paper is organized as follows. The definitions of (co-)Poisson
(co)algebras are recalled in Section 2. Some preliminary results and examples
are also given in Section 2. In Section 3, we establish some dual properties
between co-Poisson structures and Poisson structures. In Section 4, we
characterize co-Poisson coalgebra structures on polynomial Hopf algebra. In
Section 5, we characterize co-Poisson Hopf structures on polynomial Hopf
algebras.

Convention. Let $k$ be a base field. All vector spaces, algebras, coalgebras,
and Hopf algebras are over $k$. All linear maps mean $k$-linear. Unadorned
$\otimes$ means $\otimes_{k}$.

Let $V$ be a vector space. Let $t_{n}:V^{\otimes n}\to V^{\otimes
n}\,(n\in\mathbb{N}^{+})$ be the linear map given by
$v_{1}\otimes\cdots\otimes v_{n}\mapsto v_{n}\otimes v_{1}\otimes\cdots\otimes
v_{n-1}.$ For convenience, let $\circlearrowleft=1+t_{3}+t_{3}^{2}$.

Suppose $(C,\Delta,\varepsilon)$ is a coalgebra where $\Delta$ is the
comultiplication and $\varepsilon$ is the counit. We frequently use the sigma
notation

$\Delta(c)=\sum c_{1}\otimes c_{2}\,\,\textrm{and}\,\,(\Delta\otimes
1)\Delta(c)=\sum c_{1}\otimes c_{2}\otimes c_{3},$

where $\sum$ is often omitted in the computations.

Let $\Delta^{(2)}=(\Delta\otimes
1)\circ\Delta=(1\otimes\Delta)\circ\Delta:C\to C\otimes C\otimes C$, and
$\Delta^{\prime}=\Delta-t_{2}\circ\Delta$ be the cocommutator.

Suppose $(A,\mu,\eta)$ is an algebra where $\mu$ is the multiplication and
$\eta$ is the unit. For any $a,b\in A$, $[a,b]=ab-ba$ is the commutator.

##  2\. Poisson structures and co-Poisson structures

###  2.1. Poisson algebras and Poisson Hopf algebras

######  Definition 2.1.

[Li, Wei] An algebra $A$ equipped with a linear map $\\{-,-\\}:A\otimes A\to
A$ is called a Poisson algebra if

  1. (1)

$A$ with $\\{-,-\\}:A\otimes A\to A$ is a Lie algebra.

  2. (2)

$\\{-,c\\}:A\to A$ is a derivation with respect to the multiplication of $A$
for all $c\in A$, that is, $\\{ab,c\\}=a\\{b,c\\}+\\{a,c\\}b$ for all $a,b\in
A$.

It should be noted that we don’t assume that $A$ is commutative in general. As
showed in [FL, Theorem 1.2], if $A$ is prime and not commutative, then any
nontrivial Poisson structure $\\{-,-\\}$ on $A$ is the commutator bracket
$[-,-]$ up to some scalar $\alpha$ in the center of the Martindale right
quotient ring of $A$, where $\alpha$ is essentially determined by a bimodule
morphism from some nonzero ideal to $A$.

######  Proposition 2.2.

[FL, Theorem 1.2] Let $A$ be a Poisson algebra with $\\{A,A\\}\neq 0$ and
$[A,A]\neq 0$. If $A$ is prime, then

  1. (1)

for any $a,b\in A$, there is an isomorphism of $A$-bimodules

$f_{(a,b)}:A[a,b]A\to A\\{a,b\\}A,[a,b]\mapsto\\{a,b\\};$

  2. (2)

in the Martindale right quotient ring of $A$, $\\{-,-\\}=\alpha[-,-]$, where
$\alpha$ is represented by $f_{(a,b)}$ for any $a,b\in A$ such that $[a,b]\neq
0.$

###### Proof.

(1) By using Leibniz rule and calculating $\\{ac,bd\\}$ in both variables (see
[Vo, Theorem 1] or [FL, Lemma 1.1]), it follows that, for all $a,b,c,d\in A$,

$[a,b]\\{c,d\\}=\\{a,b\\}[c,d].$ (2.1)

Then, for all $a,b,c,d,e\in A$, $[a,b]\\{ec,d\\}=\\{a,b\\}[ec,d]$, which
implies that

$[a,b]e\\{c,d\\}=\\{a,b\\}e[c,d].$ (2.2)

Since $[A,A]\neq 0$, $\\{A,A\\}\neq 0$ and $A$ is prime, then $[a,b]=0$ if and
only if $\\{a,b\\}=0$ for any $a,b\in A$.

Suppose $[a,b]\neq 0$. If $\sum_{i=1}^{n}x_{i}[a,b]y_{i}=0$, then for any
$c\in A$,

$0=\sum_{i=1}^{n}(\\{a,b\\}c)x_{i}[a,b]y_{i}=\sum_{i=1}^{n}[a,b]c(x_{i}\\{a,b\\}y_{i}).$

Then $\sum_{i=1}^{n}x_{i}\\{a,b\\}y_{i}=0$ as $A$ is prime and $[a,b]\neq 0$.
This shows that the map

$f_{(a,b)}:A[a,b]A\to
A\\{a,b\\}A,~{}~{}\sum_{i=1}^{n}x_{i}[a,b]y_{i}\mapsto\sum_{i=1}^{n}x_{i}\\{a,b\\}y_{i}$

is well-defined. So, $f_{(a,b)}$ is an $A$-bimodule morphism, and is in fact
an isomorphism.

(2) If $[a,b]\neq 0$, then the bimodule morphism $f_{(a,b)}:A[a,b]A\to
A,[a,b]\mapsto\\{a,b\\}$ represents an element in the center of the Martindale
right quotient ring of $A$ ([MR, 10.3.5] or [He, Chapter 1, §3]). It follows
from (2.2) that $f_{(a,b)}$ and $f_{(c,d)}$ represent the same element, say
$\alpha$, in the Martindale right quotient ring of $A$ if both $[a,b]$ and
$[c,d]$ are nonzero. Then $\\{-,-\\}=\alpha[-,-]$ in the Matindale right
quotient ring of $A$. ∎

By Proposition 2.2, any Poisson algebra structure $\\{-,-\\}$ on the Weyl
algebra $A_{n}(k)$ or the matrix algebra $M_{n}(k)$ is $\alpha[-,-]$ for some
$\alpha\in k$.

Commutative Poisson algebras appear naturally in geometry and algebra. For
more examples of commutative Poisson algebras, see [LPV], [KS] and [LWW].

######  Definition 2.3.

([CP, Definition 6.2.1]) Let $(H,\mu,\eta,\Delta,\varepsilon,S)$ be a Hopf
algebra with a linear map $\\{-,-\\}:H\otimes H\to H$. Then $H$ is called a
Poisson Hopf algebra if

  1. (1)

$(H,\\{-,-\\})$ is a Poisson algebra.

  2. (2)

The structures are compatible in the sense that, for all $a,b\in H$,

$\displaystyle\Delta\left(\\{a,b\\}\right)=\sum\\{a_{1},b_{1}\\}\otimes
a_{2}b_{2}+\sum a_{1}b_{1}\otimes\\{a_{2},b_{2}\\}.$ (2.3)

Let $A$ and $B$ be two Poisson algebras. An algebra morphism $f:A\to B$ is
called a Poisson algebra morphism if
$f\left(\\{a,b\\}_{A}\right)=\\{f(a),f(b)\\}_{B}$ for all $a,b\in A$.

Equation (2.3) means that $\Delta:H\to H\otimes H$ is a Poisson algebra
morphism when $H$ is commutative by the following lemma.

######  Lemma 2.4.

Let $A$ and $B$ be two commutative Poisson algebras. There is naturally a
Poisson structure $\\{-,-\\}_{A\otimes B}$ on the tensor product algebra
$A\otimes B$ given by

$\\{a\otimes b,a^{\prime}\otimes b^{\prime}\\}_{A\otimes
B}=\\{a,a^{\prime}\\}_{A}\otimes
bb^{\prime}+aa^{\prime}\otimes\\{b,b^{\prime}\\}_{B}.$

It is easy to see that there is no nontrivial Poisson Hopf algebra structure
on any group algebra $k[G]$. There is also no nontrivial Poisson Hopf
structure on the universal enveloping algebra $U(\mathfrak{g})$ if
$\mathfrak{g}$ is a non-abelian Lie algebra.

######  Proposition 2.5.

Let $\mathfrak{g}$ be a non-abelian Lie algebra over a field of characteristic
$\neq 2$. Then there is no nontrivial Poisson Hopf structure on
$U(\mathfrak{g})$.

###### Proof.

Suppose $\\{-,-\\}:U(\mathfrak{g})\otimes U(\mathfrak{g})\to U(\mathfrak{g})$
is a non-trivial Poisson Hopf structure on $U(\mathfrak{g})$. Since
$\mathfrak{g}$ is non-abelian and $U(\mathfrak{g})$ is prime, then by
Proposition 2.2, for any $a,b\in\mathfrak{g}$, $[a,b]=0$ if and only if
$\\{a,b\\}=0$. Recall the compatible condition (2.3), i.e., for all $x,y\in
U(\mathfrak{g})$,

$\Delta\\{x,y\\}=x_{1}y_{1}\otimes\\{x_{2},y_{2}\\}+\\{x_{1},y_{1}\\}\otimes
x_{2}y_{2}.$

This implies that if both $x$ and $y$ are primitive then so is $\\{x,y\\}$. If
we take $x=ab$ and $y=cd$ for $a,b,c,d\in\mathfrak{g}$, the compatible
condition (2.3) will imply that

$[b,c]\otimes\\{a,d\\}+[b,d]\otimes\\{a,c\\}$

is skew-symmetric for all $a,b,c,d\in\mathfrak{g}$. So,
$[b,c]\otimes\\{a,b\\}$ is skew-symmetric by taking $d=b$. Suppose $[b,c]\neq
0$ as $[\mathfrak{g},\mathfrak{g}]\neq 0$. Since both $[b,c]$ and $\\{a,b\\}$
are primitive, then $\\{a,b\\}=0$ for all $a\in\mathfrak{g}$ as
$\operatorname{char}k\neq 2$. In particular, $\\{c,b\\}=0$, and so, $[c,b]=0$,
which is a contradiction. ∎

As pointed out in [KS, Remark 3.1.4], if $H$ is a commutative Poisson Hopf
algebra, then the counit $\varepsilon:H\to k$ is a Poisson algebra morphism,
and the antipode $S:H\to H$ is a Poisson algebra anti-morphism. Here is a
proof of the facts (see also [Oh, Lemma 4.2]).

######  Lemma 2.6.

Let $H$ be a Poisson Hopf algebra. Then the counit $\varepsilon:H\to k$ is a
Poisson algebra morphism. The antipode $S:H\to H$ is a Poisson algebra anti-
morphism provided $H$ is commutative.

###### Proof.

We need to show $\varepsilon\left(\\{a,b\\}\right)=0$ and
$S\left(\\{a,b\\}\right)=\\{S(b),S(a)\\}$ for all $a,b\in H$. Since
$\varepsilon(h)=\varepsilon(h_{1})\varepsilon(h_{2})$ for all $h\in H$, by
(2.3),

$\varepsilon\left(\\{a,b\\}\right)=\varepsilon\left(\\{a_{1},b_{1}\\}\right)\varepsilon(a_{2}b_{2})+\varepsilon(a_{1}b_{1})\varepsilon\left(\\{a_{2},b_{2}\\}\right)=2\varepsilon\left(\\{a,b\\}\right).$

Thus $\varepsilon\left(\\{a,b\\}\right)=0$. For the second assertion,

$\displaystyle\\{S(b),S(a)\\}=$
$\displaystyle\\{S(b_{1}),S(a)\\}b_{2}S(b_{3})$ $\displaystyle=$
$\displaystyle-S(b_{1})\\{b_{2},S(a)\\}S(b_{3})$ $\displaystyle=$
$\displaystyle-S(b_{1})\\{b_{2},S(a_{1})\\}a_{2}S(a_{3})S(b_{3})$
$\displaystyle=$ $\displaystyle
S(b_{1})S(a_{1})\\{b_{2},a_{2}\\}S(a_{3})S(b_{3})$ $\displaystyle=$
$\displaystyle-S(b_{1})S(a_{1})b_{2}a_{2}S\left(\\{b_{3},a_{3}\\}\right)$
$\displaystyle=$ $\displaystyle S\left(\\{a,b\\}\right),$

where the second last $``="$ follows from the fact
$\\{a_{1},b_{1}\\}S(a_{2}b_{2})+a_{1}b_{1}S\left(\\{a_{2},b_{2}\\}\right)=0$
for all $a,b\in H$, and the last $``="$ follows from the commutativity of $H$.
∎

###  2.2. Co-Poisson coalgebras and co-Poisson Hopf algebras

The concepts of co-Poisson coalgebra and co-Poisson Hopf algebra are
introduced in [CP, Chapter 6] to study quantizations of Lie bialgebras. They
are dual to Poisson algebra and Poisson Hopf algebra.

######  Definition 2.7.

([CP, Definition 6.2.2]) A coalgebra $(C,\Delta,\varepsilon)$ equipped with a
linear map $q:C\to C\otimes C$ is called a co-Poisson coalgebra if

  1. (1)

$C$ with $q:C\to C\otimes C$ is a Lie coalgebra, that is,

$\displaystyle(1+t_{2})\circ q$ $\displaystyle=0,$ (skew-symmetric)
$\displaystyle(1+t_{3}+t_{3}^{2})\circ(q\otimes 1)\circ q$ $\displaystyle=0.$
(co-Jacobi identity)

  2. (2)

$(\Delta\otimes 1)q=(1\otimes q)\Delta-t_{3}^{2}(q\otimes 1)\Delta$. (co-
Leibniz rule)

In co-Poisson coalgebra $C$, we use the sigma notations

$\Delta(c)=\sum c_{1}\otimes c_{2},\,q(c)=\sum c_{(1)}\otimes
c_{(2)}\,\,\textrm{and}\,\,(q\otimes 1)q(c)=\sum c_{(1)}\otimes c_{(2)}\otimes
c_{(3)}$

where $\sum$ is often omitted in the computations. Then,

$(1\otimes q)q(c)=(1\otimes q)(-c_{(2)}\otimes c_{(1)})=-c_{(3)}\otimes
c_{(1)}\otimes c_{(2)}=c_{(3)}\otimes c_{(2)}\otimes c_{(1)}.$

######  Remark 2.8.

By using the sigma notation, the co-Leibniz rule reads as

$\sum c_{(1)1}\otimes c_{(1)2}\otimes c_{(2)}=\sum c_{1}\otimes
c_{2(1)}\otimes c_{2(2)}-\sum c_{1(2)}\otimes c_{2}\otimes c_{1(1)}$

for all $c\in C$. It is equivalent to

$(1\otimes\Delta)q=(q\otimes 1)\Delta-t_{3}(1\otimes q)\Delta.$ (2.4)

If the coalgebra $C$ is cocommutative, then the co-Leibniz rule is also
equivalent to

$(\Delta\otimes 1)q=(1-t_{3})(1\otimes q)\Delta.$ (2.5)

The cocommutator $\Delta^{\prime}$ gives a co-Poisson structure on any
coalgebra $(C,\Delta,\varepsilon)$.

It follows from the co-Leibniz rule that there is no nontrivial co-Poisson
coalgebra structure on any group coalgebra $k[G]$ .

######  Definition 2.9.

Let $(C,\Delta,\varepsilon)$ be a coalgebra. A linear map $d:C\to C$ is called
a coderivation if, for all $c\in C$,

$\Delta(d(c))=d(c_{1})\otimes c_{2}+c_{1}\otimes d(c_{2}).$

######  Example 2.10.

Let $C$ be a cocommutative coalgebra, $d_{1}$ and $d_{2}$ be two coderivations
of $C$ such that $d_{1}d_{2}=d_{2}d_{1}$. Let $q:C\to C\otimes C$ be the map
$c\mapsto q(c)$, where

$q(c)=d_{1}(c_{1})\otimes d_{2}(c_{2})-d_{2}(c_{1})\otimes d_{1}(c_{2}).$

Then $(C,q)$ is a co-Poisson coalgebra.

Dual to the fact $\\{a,-\\}:A\to A$ is a derivation in any Poisson algebra $A$
is the following.

######  Lemma 2.11.

Let $(C,\Delta,\varepsilon,q)$ be a co-Poisson coalgebra. Then for any $f\in
C^{*}$, $(f\otimes 1)q:C\to C$ and $(1\otimes f)q:C\to C$ are coderivations of
$C$.

Dual to the fact $\\{1_{A},a\\}=\\{a,1_{A}\\}=0$ in any Poisson algebra $A$ is
the following.

######  Lemma 2.12.

Let $(C,\Delta,\varepsilon,q)$ be a co-Poisson coalgebra. Then

$(\varepsilon\otimes 1)q=(1\otimes\varepsilon)q=0,\,\textrm{that
is},\,\varepsilon(c_{(1)})c_{(2)}=c_{(1)}\varepsilon(c_{(2)})=0\,\,\textrm{for
all}\,\,c\in C.$

######  Definition 2.13.

[CP, Definition 6.2.2] A Hopf algebra $(H,\mu,\eta,\Delta,\varepsilon,S)$
equipped with a linear map $q:H\to H\otimes H$ is called a co-Poisson Hopf
algebra if

  1. (1)

$H$ with $q:H\to H\otimes H$ is a co-Poisson coalgebra.

  2. (2)

$q$ is a $\Delta$-derivation, that is, for all $a,b\in H$,

$q(ab)=q(a)\Delta(b)+\Delta(a)q(b).$ (2.6)

Co-Poisson Hopf structures on $U(\mathfrak{g})$ are known.

######  Proposition 2.14.

[CP, Proposition 6.2.3] Let $\mathfrak{g}$ be a Lie algebra over a field $k$
of characteristic zero. Then the co-Poisson Hopf structures on the universal
enveloping algebra $U(\mathfrak{g})$ are determined uniquely by the Lie
bialgebra structures on $\mathfrak{g}$.

Let $C$ and $D$ be two co-Poisson coalgebras. A coalgebra morphism $g:C\to D$
is called a co-Poisson coalgebra morphism if $(g\otimes
g)(q_{C}(c))=q_{D}\left(g(c)\right)$ for all $c\in C$.

Equation (2.6) means that $\mu:H\otimes H\to H$ is a co-Poisson coalgebra
morphism provided $H$ is cocommutative by Lemma 2.15.

If $C$ and $D$ are two coalgebras, then $C\otimes D$ is a coalgebra with

$\Delta_{C\otimes D}=(1\otimes\tau\otimes
1)(\Delta_{C}\otimes\Delta_{D}),\,\textrm{ i.e., }\,\Delta_{C\otimes
D}(a\otimes b)=a_{1}\otimes b_{1}\otimes a_{2}\otimes b_{2}.$

######  Lemma 2.15.

Let $(C,q_{C})$ and $(D,q_{D})$ be two cocommutative co-Poisson coalgebras.
Then $(C\otimes D,q_{C\otimes D})$ is a co-Poisson coalgebra, with
$q_{C\otimes D}$ being the composition

$C\otimes D\xrightarrow{q_{C}\otimes\Delta_{D}+\Delta_{C}\otimes
q_{D}}C\otimes C\otimes D\otimes D\xrightarrow{1\otimes t_{2}\otimes
1}C\otimes D\otimes C\otimes D,$

that is,

$q_{C\otimes D}(a\otimes b)=a_{(1)}\otimes b_{1}\otimes a_{(2)}\otimes
b_{2}+a_{1}\otimes b_{(1)}\otimes a_{2}\otimes b_{(2)}.$

The dual form of Lemma 2.6 is the following.

######  Lemma 2.16.

Let $(H,\mu,\eta,\Delta,\varepsilon,S,q)$ be a co-Poisson Hopf algebra.

  1. (1)

$\eta$ is a co-Poisson coalgebra morphism.

  2. (2)

If $H$ is cocommutative, then $S$ is a co-Poisson coalgebra anti-morphism.

###### Proof.

(1) This is trivial as $q(1_{H})=0$ by (2.6).

(2) We need to show $q\left(S(h)\right)=S(h_{(2)})\otimes S(h_{(1)})$ for all
$h\in H$. It follows from the co-Leibniz rule that for any $h\in H$,

$h_{1(1)1}\otimes h_{1(1)2}\otimes h_{2}\otimes h_{1(2)}=h_{1}\otimes
h_{2(1)}\otimes h_{3}\otimes h_{2(2)}-h_{1(2)}\otimes h_{2}\otimes
h_{3}\otimes h_{1(1)}.$

Since $S(h_{1(1)1})h_{1(1)2}S(h_{2})\otimes S(h_{1(2)})=0$ by Lemma 2.12, then

$S(h_{1})h_{2(1)}S(h_{3})\otimes
S(h_{2(2)})=S(h_{1(2)})\varepsilon(h_{2})\otimes S(h_{1(1)})=S(h_{(2)})\otimes
S(h_{(1)}),$

where the last $``="$ is derived from the co-Leibniz rule (2.4) and Lemma
2.12.

On the other hand, by the co-Leibniz rule,

$\displaystyle h_{1}\otimes h_{2}\otimes h_{3(1)1}\otimes h_{3(1)2}\otimes
h_{3(2)}\otimes h_{4}$ $\displaystyle=$ $\displaystyle h_{1}\otimes
h_{2}\otimes h_{3}\otimes h_{4(1)}\otimes h_{4(2)}\otimes h_{5}-h_{1}\otimes
h_{2}\otimes h_{3(2)}\otimes h_{4}\otimes h_{3(1)}\otimes h_{5}.$

By Lemma 2.12, $S(h_{1})h_{3(2)}S(h_{4})\otimes
S(h_{2})h_{3(1)1}S(h_{3(1)2})=0.$ Thus

$\displaystyle S(h_{1})h_{3(1)}S(h_{5})\otimes S(h_{2})h_{3(2)}S(h_{4})$
$\displaystyle=$ $\displaystyle S(h_{1})h_{4(2)}S(h_{5})\otimes
S(h_{2})h_{3}S(h_{4(1)})$ $\displaystyle=$ $\displaystyle
S(h_{1})h_{2(2)}S(h_{3})\otimes S(h_{2(1)})$ $\displaystyle=$ $\displaystyle
S(h_{(1)})\otimes S(h_{(2)}).$

It remains to show that $S(h_{1})h_{3(2)}S(h_{5})\otimes
S(h_{2})h_{3(1)}S(h_{4})=q\left(S(h)\right)$. Since
$0=q(h_{1}S(h_{2}))=q(h_{1})\Delta\left(S(h_{2})\right)+\Delta(h_{1})q\left(S(h_{2})\right)$
for any $h\in H$, then

$h_{1(1)}S(h_{3})\otimes h_{1(2)}S(h_{2})+h_{1}S(h_{3})_{(1)}\otimes
h_{2}S(h_{3})_{(2)}=0.$

Hence

$h_{1}\otimes h_{3}S(h_{5})_{(1)}\otimes h_{2}\otimes
h_{4}S(h_{5})_{(2)}+h_{1}\otimes h_{3(1)}S(h_{5})\otimes h_{2}\otimes
h_{3(2)}S(h_{4})=0.$

Thus

$\displaystyle S(h_{1})h_{3(2)}S(h_{5})\otimes S(h_{2})h_{3(1)}S(h_{4})$
$\displaystyle=$ $\displaystyle S(h_{1})h_{3}S(h_{5})_{(1)}\otimes
S(h_{2})h_{4}S(h_{5})_{(2)}$ $\displaystyle=$ $\displaystyle
S(h_{1})h_{2}S(h_{5})_{(1)}\otimes S(h_{3})h_{4}S(h_{5})_{(2)}$
$\displaystyle=$ $\displaystyle S(h)_{(1)}\otimes S(h)_{(2)}$ $\displaystyle=$
$\displaystyle q\left(S(h)\right),$

where the cocommutativity is used for the second $``="$. ∎

The dual form of (2.1) for co-Poisson coalgebras is the following proposition.

######  Proposition 2.17.

Let $(C,\Delta,\varepsilon)$ be a co-Poisson coalgebra with co-Poisson
structure $q$. Then

$(q\otimes\Delta^{\prime})\circ\Delta=(\Delta^{\prime}\otimes q)\circ\Delta.$

###### Proof.

It follows by using the co-Leibniz rule to calculate

$(\Delta\otimes\Delta)q(a)=(\Delta\otimes 1\otimes
1)(1\otimes\Delta)q(a)=(1\otimes 1\otimes\Delta)(\Delta\otimes 1)q(a).$

∎

The dual form of (2.2) for co-Poisson coalgebras is the following.

######  Corollary 2.18.

Let $(C,\Delta,\varepsilon)$ be a co-Poisson coalgebra with co-Poisson
structure $q$. Then

$(\Delta^{\prime}\otimes(1\otimes
q)\Delta)\circ\Delta=(q\otimes(1\otimes\Delta^{\prime})\Delta)\circ\Delta,$

i.e, for any $a\in C$,

$\displaystyle\,\,\,\,\,\,\,\,a_{1}\otimes a_{2}\otimes a_{3}\otimes
q(a_{4})-a_{2}\otimes a_{1}\otimes a_{3}\otimes q(a_{4})$
$\displaystyle=q(a_{1})\otimes a_{2}\otimes a_{3}\otimes a_{4}-q(a_{1})\otimes
a_{2}\otimes a_{4}\otimes a_{3}.$

Poisson and co-Poisson structures on 4-dimensional Sweedler Hopf algebra are
clear.

######  Example 2.19.

Let $H_{4}=k\langle x,g\mid x^{2}=0,g^{2}=1,xg=-gx\rangle$ be the
4-dimensional Sweedler Hopf algebra, where the coalgebra structure is given by
$\Delta(g)=g\otimes g$, $\Delta(x)=x\otimes 1+g\otimes x$, $\varepsilon(g)=1$,
$\varepsilon(x)=0$; and the antipode is given by $S(g)=g=g^{-1}$, $S(x)=-gx$.
Assume $\operatorname{char}k\neq 2$.

  1. (1)

Any Poisson structure on $H_{4}$ is given by $\\{g,x\\}=\lambda x+\mu gx$ for
some $\lambda,\mu\in k$.

  2. (2)

There is no nontrivial Poisson Hopf structure on $H_{4}$. In fact, if the
Poisson structure given by $\\{g,x\\}=\lambda x+\mu gx$ is a Poisson Hopf
structure, then by applying (2.3) with $a=g,b=x$ and $a=x,b=gx$ we get
$\lambda=\mu=0$.

  3. (3)

Any co-Poisson structure $q$ on $H_{4}$ is given by

$\displaystyle q(1)$ $\displaystyle=q(g)=0,$ $\displaystyle q(x)$
$\displaystyle=\alpha(1\otimes x-x\otimes 1+x\otimes g-g\otimes x),$
$\displaystyle q(gx)$ $\displaystyle=\beta(1\otimes gx-gx\otimes 1+gx\otimes
g-g\otimes gx)$

for some $\alpha,\beta\in k$. In fact, for any $h\in H_{4}$, we may assume

$\displaystyle q(h)$ $\displaystyle=\alpha_{1}(1\otimes x-x\otimes
1)+\alpha_{2}(1\otimes g-g\otimes 1)+\alpha_{3}(1\otimes gx-gx\otimes 1)$
$\displaystyle+\alpha_{4}(x\otimes g-g\otimes x)+\alpha_{5}(x\otimes gx-
gx\otimes x)+\alpha_{6}(g\otimes gx-gx\otimes g)$

as $q(h)$ is skew-symmetric. It follows from $(\varepsilon\otimes 1)q=0$
(Lemma 2.12) that

$\alpha_{1}=\alpha_{4},\alpha_{2}=0,\alpha_{3}=-\alpha_{6}.$

Then, the co-Leibniz rule implies that $q(1)=q(g)=0$,

$q(x)=\alpha(1\otimes x-x\otimes 1+x\otimes g-g\otimes x),$
$q(gx)=\beta(1\otimes gx-gx\otimes 1+gx\otimes g-g\otimes gx)$

for some $\alpha,\beta\in k$.

  4. (4)

There is no nontrivial co-Poisson Hopf structure on $H_{4}$. In fact, suppose
a co-Poisson structure $q$ as given in (3) is a co-Poisson Hopf structure on
$H_{4}$, then $\alpha=\beta=0$ by the equations

$q(gx)=q(g)\Delta(x)+\Delta(g)q(x)\,\textrm{ and
}\,0=q(x^{2})=q(x)\Delta(x)+\Delta(x)q(x).$

##  3\. Dual properties between Poisson and co-Poisson Hopf algebras

As the vector space dual of any coalgebra is an algebra, the dual of any co-
Poisson coalgebra is a Poisson algebra.

######  Proposition 3.1.

  1. (1)

Suppose $(C,\Delta,\varepsilon)$ is a coalgebra, $q:C\to C\otimes C$ is a
linear map. Then $(C,q)$ is a co-Poisson coalgebra if and only if
$(C^{*},q^{*})$ is a Poisson algebra.

  2. (2)

Suppose $\sigma:C\to D$ is a linear map between co-Poisson coalgebras. Then
$\sigma:C\to D$ is a co-Poisson coalgebra morphism if and only if
$\sigma^{*}:D^{*}\to C^{*}$ is a Poisson algebra morphism.

###### Proof.

(1) Note that $q^{*}$ is the map $q^{*}:C^{*}\otimes C^{*}\subseteq(C\otimes
C)^{*}\to C^{*},f\otimes g\mapsto\\{f,g\\}$, where $\\{f,g\\}\in C^{*}$ is the
map $C\to k,\,c\mapsto(f\otimes g)q(c)=f(c_{(1)})g(c_{(2)})$.

Note that $(\\{f,g\\}+\\{g,f\\})(c)=(f\otimes g)(1+t_{2})q(c)$. It is easy to
see that $\\{-,-\\}_{C^{*}}:C^{*}\times C^{*}\to C^{*}$ is skew-symmetric if
and only if $q$ is skew-symmetric.

Since $\\{\\{f,g\\},h\\}(c)=(f\otimes g\otimes h)(q\otimes 1)q(c)$, then

$\circlearrowleft\\{\\{f,g\\},h\\}(c)=(f\otimes g\otimes
h)(1+t_{3}+t_{3}^{2})(q\otimes 1)q(c).$

It follows that $\\{-,-\\}_{C^{*}}$ satisfies the Jacobi identity if and only
if $q$ satisfies the co-Jacobi identity.

Similarly, since $\\{f*g,h\\}(c)=(f\otimes g\otimes h)\left((\Delta\otimes
1)q(c)\right)$ and

$((f*\\{g,h\\})-(\\{h,f\\}*g))(c)=(f\otimes g\otimes h)((1\otimes q)\Delta-
t_{3}^{2}(q\otimes 1)\Delta)(c),$

then $\\{f*g,h\\}=f*\\{g,h\\}+\\{f,h\\}*g$ if and only if $(\Delta\otimes
1)q=(1\otimes q)\Delta-t_{3}^{2}(q\otimes 1)\Delta$.

(2) It is well known that $\sigma:C\to D$ is a coalgebra morphism if and only
if $\sigma^{*}:D^{*}\to C^{*}$ is an algebra morphism. Since

$(\sigma^{*}\\{f,g\\}_{D^{*}}-\\{\sigma^{*}f,\sigma^{*}g\\}_{C^{*}})(c)=(f\otimes
g)(\Delta_{D}\sigma-(\sigma\otimes\sigma)\Delta_{C})(c),$

then
$\sigma^{*}\\{-,-\\}_{D^{*}}=\\{-,-\\}_{C^{*}}(\sigma^{*}\otimes\sigma^{*})$
if and only if $\Delta_{D}\sigma=(\sigma\otimes\sigma)\Delta_{C}$. The proof
is finished. ∎

If $A$ is an algebra, then $A^{\circ}=\\{f\in A^{*}\mid\ker
f\,\textrm{contains a cofinite ideal of}\,A\\}$ is a coalgebra, which is
called the finite dual of $A$. Suppose $(A,p=\\{-,-\\})$ is a Poisson algebra.
Example 3.2 shows that $(A^{\circ},q=p^{*})$ may not be a co-Poisson coalgebra
because $p^{*}:A^{*}\to(A\otimes A)^{*}$ may not be restricted to a map
$p^{*}_{A^{\circ}}:A^{\circ}\to A^{\circ}\otimes A^{\circ}$.

######  Example 3.2.

Let $A=k[x_{1},x_{2},\cdots,x_{n},\cdots]$ be a polynomial algebra with
infinitely many variables $\\{x_{i}\mid i\geq 1\\}$. Let $p(x_{i}\otimes
x_{j})=\\{x_{i},x_{j}\\}=1$ for all $i<j$. Then $p$ gives a Poisson algebra
structure on $A$. Suppose $\varepsilon:A\to k,x_{i}\mapsto 0$, is the
augmentation map. Then $\varepsilon\in A^{\circ}$, but
$p^{*}(\varepsilon)=\varepsilon p\notin(A\otimes A)^{\circ}$.

In fact, if $\varepsilon p\in(A\otimes A)^{\circ}$, then there is a cofinite
ideal $I$ of $A$ such that $I\otimes A+A\otimes I\subseteq\ker(\varepsilon
p)$. Then, $\\{I,A\\}\subseteq\ker\varepsilon$. Since $A/I$ is finite-
dimensional, there exists a nonzero linear polynomial in $I$, which will imply
that $1\in\\{I,A\\}$. It contradicts to $\\{I,A\\}\subseteq\ker\varepsilon$.

But if $A$ is left or right noetherian, then we have the following positive
conclusion.

######  Proposition 3.3.

Let $(A,p=\\{-,-\\})$ be a Poisson algebra, and $f:A\to B$ be a Poisson
algebra morphism.

  1. (1)

If $A$ is a (left or right) noetherian algebra, then $(A^{\circ},q=p^{*})$ is
a co-Poisson coalgebra.

  2. (2)

If both $A$ and $B$ are (left or right) noetherian, then $f^{*}:B^{\circ}\to
A^{\circ}$ is a co-Poisson coalgebra morphism.

###### Proof.

It suffices to show that $p^{*}:A^{*}\to(A\otimes A)^{*}$ restricts to a map

$p^{*}|_{A^{\circ}}:A^{\circ}\to A^{\circ}\otimes A^{\circ}\cong(A\otimes
A)^{\circ}.$

Suppose $f\in A^{\circ}$ and $I\subseteq\ker f$ is a cofinite ideal of $A$.
Since $A$ is left or right noetherian, then $I/I^{2}$ is a finitely generated
left or right $A/I$-module. Hence $\dim_{k}I/I^{2}<\infty$ as
$\dim_{k}A/I<\infty$. So, $I^{2}$ is also a cofinite ideal of $A$. Let
$J=I^{2}\otimes A+A\otimes I^{2}$. Then $J$ is a cofinite ideal of $A\otimes
A$, and

$p^{*}(f)(J)=fp(J)\subseteq f(I)=\\{0\\}.$

It follows that $p^{*}(f)\in(A\otimes A)^{\circ}$. ∎

Note that $A^{\circ}=0$ for the Weyl algebra $A=A_{n}(k)$. So, even in
noetherian case $(A,p)$ may not be a Poisson algebra when
$(A^{\circ},q=p^{*})$ is a co-Poisson coalgebra.

In [OP], the authors prove that the Hopf dual $H^{\circ}$ of a co-Poisson Hopf
algebra $H$ is a Poisson Hopf algebra when $H$ is an almost normalizing
extension over $k$. In fact, this is true in general as stated in [KS,
Proposition 3.1.5]. We give a proof here. Oh gives a proof also in a recent
paper ([Oh, Theorem 2.2]).

######  Proposition 3.4.

Let $(H,\mu,\eta,\Delta,\varepsilon,S,q)$ be a co-Poisson Hopf algebra. Then
the Hopf dual $H^{\circ}$ is a Poisson Hopf algebra.

###### Proof.

It is well-know that $H^{\circ}$ is a Hopf algebra ([Sw, Section 6.2], [Mo,
Theorem 9.1.3]). By Proposition 3.1, to show $H^{\circ}$ is a Poisson algebra
it suffices to show that $q^{*}:(H\otimes H)^{*}\to H^{*}$ restricts to a map
$q^{\circ}:H^{\circ}\otimes H^{\circ}\cong(H\otimes H)^{\circ}\to H^{\circ},$
that is, $q^{*}\left((H\otimes H)^{\circ}\right)\subseteq H^{\circ}$.

Suppose $F\in(H\otimes H)^{\circ}$ and $J\subseteq\ker F$ is a cofinite ideal
in $H\otimes H$. Since $\Delta$ is an algebra morphism, $\Delta^{-1}(J)$ is an
ideal of $H$. It follows from $q(ab)=q(a)\Delta(b)+\Delta(a)q(b)$ that
$I=q^{-1}(J)\cap\Delta^{-1}(J)$ is an ideal of $H$. Since the linear map

$H/I\to\frac{H\otimes H}{J}\oplus\frac{H\otimes
H}{J},\quad\overline{h}\mapsto\left(\overline{q(h)},\overline{\Delta(h)}\right)$

is injective, $I$ is a cofinite ideal of $H$. Note that
$q^{*}(F)(I)=F(q(I))\subseteq F(J)=0$, i.e., $I\subseteq\ker q^{*}(F)$. It
follows that $q^{*}(F)\in H^{\circ}$.

To finish the proof, we need to show that $\mu^{*}:H^{\circ}\to
H^{\circ}\otimes H^{\circ}$ satisfies the compatible condition (2.3), that is,
for all $f,g\in H^{\circ}$,

$\mu^{*}(\\{f,g\\}_{H^{\circ}})=\\{f_{1},g_{1}\\}_{H^{\circ}}\otimes(f_{2}*g_{2})+(f_{1}*g_{1})\otimes\\{f_{2},g_{2}\\}_{H^{\circ}}.$

This is true, because, for any $x,y\in H$, on one hand, by (2.6),

$\displaystyle\mu^{*}(\\{f,g\\}_{H^{\circ}})(x\otimes y)$ $\displaystyle=$
$\displaystyle\\{f,g\\}_{H^{\circ}}(xy)=(f\otimes
g)q(xy)=f((xy)_{(1)})g((xy)_{(2)})$ $\displaystyle=$ $\displaystyle
f((x_{(1)}y_{1})g((x_{(2)}y_{2})+f((x_{1}y_{(1)})g((x_{2}y_{(2)}),$

on the other hand,

$\displaystyle(\\{f_{1},g_{1}\\}_{H^{\circ}}\otimes(f_{2}*g_{2})+(f_{1}*g_{1})\otimes\\{f_{2},g_{2}\\}_{H^{\circ}})(x\otimes
y)$ $\displaystyle=$
$\displaystyle(\\{f_{1},g_{1}\\}_{H^{\circ}}(x)\,(f_{2}*g_{2})(y)+(f_{1}*g_{1})(x)\,\\{f_{2},g_{2}\\}_{H^{\circ}}(y)$
$\displaystyle=$ $\displaystyle
f_{1}(x_{(1)})g_{1}(x_{(2)})f_{2}(y_{1})g_{2}(y_{2})+f_{1}(x_{1})g_{1}(x_{2})f_{2}(y_{(1)})g_{2}(y_{(2)})$
$\displaystyle=$ $\displaystyle
f((x_{(1)}y_{1})g((x_{(2)}y_{2})+f((x_{1}y_{(1)})g((x_{2}y_{(2)}).$

∎

The following is also stated in [KS, Proposition 3.1.5] without noetherian
hypothesis. Without this hypothesis, it is not true as showed in Example 3.6.

######  Proposition 3.5.

Let $(H,\mu,\eta,\Delta,\varepsilon,S,p)$ be a left or right noetherian
Poisson Hopf algebra. Then the Hopf dual $H^{\circ}$ is a co-Poisson Hopf
algebra.

###### Proof.

By Proposition 3.3, $H^{\circ}$ is a co-Poisson coalgebra. We only need to
check the compatible condition (2.6), that is, for all $f,g\in H^{\circ}$,

$p^{*}(f*g)=p^{*}(f)\mu^{*}(g)+\mu^{*}(f)p^{*}(g).$

This is true, because for any $x,y\in H$,

$\displaystyle p^{*}(f*g)(x\otimes y)=(f*g)(\\{x,y\\})$ $\displaystyle=$
$\displaystyle
f(\\{x_{1},y_{1}\\})\,g(x_{2}y_{2})+f(x_{1}y_{1})\,g(\\{x_{2},y_{2}\\}),$

and

$\displaystyle(p^{*}(f)\mu^{*}(g)+\mu^{*}(f)p^{*}(g))(x\otimes y)$
$\displaystyle=$ $\displaystyle(f_{(1)}*g_{1}\otimes f_{(2)}*g_{2})(x\otimes
y)+(f_{1}*g_{(1)}\otimes f_{2}*g_{(2)})(x\otimes y)$ $\displaystyle=$
$\displaystyle
f_{(1)}(x_{1})g_{1}(x_{2})f_{(2)}(y_{1})g_{2}(y_{2})+f_{1}(x_{1})g_{(1)}(x_{2})f_{2}(y_{1})g_{(2)}(y_{2})$
$\displaystyle=$ $\displaystyle
f(\\{x_{1},y_{1}\\})\,g(x_{2}y_{2})+f(x_{1}y_{1})\,g(\\{x_{2},y_{2}\\}).$

∎

######  Example 3.6.

Let $A=k[x_{1},\cdots,x_{d},\cdots]$ be a polynomial algebra with infinitely
many variables $\\{x_{i}\mid i\geq 1\\}$, which is a Hopf algebra viewed as
the enveloping algebra of the abelian Lie algebra $\mathfrak{g}=kx_{1}\oplus
kx_{2}\oplus\cdots\oplus kx_{d}\oplus\cdots.$ Let

$\displaystyle\\{x_{1},x_{i}\\}$ $\displaystyle=0\textrm{ for all
}i\in\mathbb{N},$ $\displaystyle\\{x_{i},x_{j}\\}$
$\displaystyle=\begin{cases}x_{1},\quad&j=i+1,\\\
0,&\textrm{otherwise}\end{cases}\,\,\textrm{for all }1<i<j\in\mathbb{N}.$

Then $\\{\\{x_{i},x_{j}\\},x_{k}\\}=0$ for all $i,j,k$. As in [LPV,
Proposition 1.8], $A$ is endowed with a Poisson algebra structure. It is easy
to check by induction on the degree of homogeneous elements that (2.3) holds.
So, $A$ is a Poisson Hopf algebra.

We assert that $A^{\circ}$ is not a co-Poisson Hopf algebra by showing that
$\\{-,-\\}^{*}(A^{\circ})\nsubseteq A^{\circ}\otimes A^{\circ}$. Let $f:A\to
k$ be the linear map given by

$\begin{cases}f(1_{A})=f(x_{1})=1_{k},\\\ f(a)=0\,\textrm{ for all other monic
monomials }a\in A.\end{cases}$

Then $I=\langle\\{x_{1}^{2},x_{2},x_{3},\cdots\\}\rangle\subseteq\ker f$. Note
that $I$ is a cofinite ideal of $A$, and so, $f\in A^{\circ}$. Suppose
$\\{-.-\\}^{*}(f)\in A^{\circ}\otimes A^{\circ}$ and

$\\{-.-\\}^{*}(f)=\sum_{i=1}^{n}g_{i}\otimes h_{i},$

with $I_{i}\in\ker g_{i}$ and $J_{i}\in\ker h_{i}$ are all cofinite ideals of
$A$. Then

$J=\bigcap_{i=1}^{n}(I_{i}\cap J_{i}).$

is a cofinite ideal of $A$, and $\\{-,-\\}^{*}(f)(J\otimes
A)=\sum_{i=1}^{n}(g_{i}\otimes h_{i})(J\otimes A)=0$. It follows that
$\\{J,A\\}\subseteq\ker f$.

Since $J$ is cofinite in $A$, $\\{x_{2}+J,x_{3}+J,\cdots\\}$ is linearly
dependent in $A/J$. Then there exists $i\geq 2$ and
$\lambda_{2},\cdots,\lambda_{i-1},\lambda_{i}\in K$ such that $\lambda_{i}\neq
0$ and

$\lambda_{2}x_{2}+\lambda_{3}x_{3}+\cdots+\lambda_{i}x_{i}\in J.$

Now

$\\{\lambda_{2}x_{2}+\lambda_{3}x_{3}+\cdots+\lambda_{i}x_{i},x_{i+1}\\}=\lambda_{i}x_{1}\notin\ker
f,$

which contradicts to $\\{J,A\\}\subseteq\ker f$.

##  4\. Co-Poisson coalgebra structures on $k[x_{1},x_{2},\cdots,x_{d}]$

Suppose $\mathfrak{g}=kx_{1}\oplus kx_{2}\oplus\cdots\oplus kx_{d}$ is a
$d$-dimensional Lie algebra. The co-Poisson Hopf structures on
$U(\mathfrak{g})$ are in one-to-one correspondence with the Lie bialgebra
structures on $\mathfrak{g}$. If $\mathfrak{g}$ is non-abelian, then there is
no nontrivial Poisson Hopf structure on $U(\mathfrak{g})$ (see Proposition
2.5). So, we turn to consider the case when $\mathfrak{g}$ is abelian from now
on. Then $A=U(\mathfrak{g})=k[x_{1},\cdots,x_{d}]$. We first characterize co-
Poisson coalgebra structures on $k[x_{1},x_{2},\cdots,x_{d}]$ in this section.

Suppose $(C,\Delta,\varepsilon)$ is a coalgebra. Let $P(C)$ be the subspace of
$C$ consisting of all primitive elements of $C$. Assume $P(C)=\oplus_{i\in
I}ke_{i}$ as a vector space, with $(I,<)$ being well ordered by using Well
Ordering Principle. Let

$\mathcal{I}=\bigoplus_{i,j\in I,i<j}k(e_{i}\otimes e_{j}-e_{j}\otimes
e_{i}).$

######  Lemma 4.1.

Retain the notations above. Suppose $X\in C\otimes C$. Then $X\in\mathcal{I}$
if and only if

$(1+t_{2})(X)=0\,\textrm{ and }\,(\Delta\otimes 1)(X)=(1-t_{3})(1\otimes X).$

###### Proof.

“$\Rightarrow$” Trivial.

“$\Leftarrow$” Assume $0\neq X=\sum_{i=1}^{n}a_{i}\otimes b_{i}$ with $n$
minimal. Then $\\{a_{1},\cdots,a_{n}\\}$ and $\\{b_{1},\cdots,b_{n}\\}$ are
both linearly independent. Since $(1+t_{2})(X)=0$, i.e.,
$\sum_{i=1}^{n}a_{i}\otimes b_{i}=-\sum_{i=1}^{n}b_{i}\otimes a_{i}$, then

$(1-t_{3})(1\otimes X)=\sum_{i=1}^{n}(1\otimes a_{i}\otimes b_{i}-b_{i}\otimes
1\otimes a_{i})=\sum_{i=1}^{n}(1\otimes a_{i}+a_{i}\otimes 1)\otimes b_{i}.$

Since $(\Delta\otimes 1)(X)=(1-t_{3})(1\otimes X)$, then
$\sum_{i=1}^{n}\Delta(a_{i})\otimes b_{i}=\sum_{i=1}^{n}(1\otimes
a_{i}+a_{i}\otimes 1)\otimes b_{i}$. Thus $a_{i}\in P(C)$ by the independence
of the $b_{i}$’s.

By using $\sum_{i=1}^{n}a_{i}\otimes b_{i}=-\sum_{i=1}^{n}b_{i}\otimes a_{i}$
and a similar discussion, we have $b_{i}\in P(C)$ as well. The assertion
follows. ∎

The following facts are obvious.

######  Lemma 4.2.

Let $B$ be a bialgebra. Then

  1. (1)

If $X\in B\otimes B$ is skew-symmetric, then so is $X\Delta(x)$ for any $x\in
P(B)$.

  2. (2)

For any $a\in B$ and $X\in B\otimes B$,

$(\Delta\otimes 1)(X\Delta(a))=(\Delta\otimes 1)(X)\cdot\Delta^{(2)}(a).$

In the following, $A=U(\mathfrak{g})=k[x_{1},x_{2},\cdots,x_{d}]$. Let
$\mathcal{H}(A)$ be the set of all monic monomials of $A$. For any
$a\in\mathcal{H}(A)$, $\Delta(a)=\sum a_{1}\otimes a_{2}$ is always assumed to
be the expression by the standard $k$-basis of $k[x_{1},x_{2},\cdots,x_{d}]$.
For any $a\in\mathcal{H}(A)$, $|a|$ is the degree of $a$.

First, we establish a reciprocity law for two linear maps from $A$ to
$A\otimes A$, which is a key step to characterize the co-Poisson structures.

######  Proposition 4.3.

Let $q:A\to A\otimes A$ and $I:A\to A\otimes A$ be two linear maps. Then
$I(a)=(-1)^{|a_{2}|}q(a_{1})\Delta(a_{2})$ for all $a\in A$ if and only if
$q(a)=I(a_{1})\Delta(a_{2})$ for all $a\in A$.

###### Proof.

First note that in our case (the algebra is generated by primitive elements),
for any $1\neq a\in\mathcal{H}(A)$,

$(-1)^{|a_{2}|}\Delta(a_{1}a_{2})=0.$ (4.1)

Since $a\otimes 1$ is one of the terms of $\Delta(a)$ for any
$a\in\mathcal{H}(A)$, then

$\sum_{a_{2}\neq 1}(-1)^{|a_{2}|+1}\Delta(a_{1}a_{2})=\Delta(a).$ (4.2)

“$\Rightarrow$” Obviously, $I(1)=q(1)$, and so $q(a)=I(a_{1})\Delta(a_{2})$
for $a=1$. We prove $q(a)=I(a_{1})\Delta(a_{2})$ holds for any
$a\in\mathcal{H}(A)$ by induction on the degree of $a$. Suppose
$q(a)=I(a_{1})\Delta(a_{2})$ holds for all $a\in\mathcal{H}(A)$ of degree no
more than $n$. To finish the proof, it suffices to show that
$q(ax)=I(a_{1}x)\Delta(a_{2})+I(a_{1})\Delta(a_{2}x)$ for any $x\in P(A)$.
Since, by assumption,

$\displaystyle
I(ax)=(-1)^{|a_{2}|}q(a_{1}x)\Delta(a_{2})+(-1)^{|a_{2}|+1}q(a_{1})\Delta(a_{2}x)$
$\displaystyle=$ $\displaystyle q(ax)+\sum_{a_{3}\neq
1}(-1)^{|a_{3}|}(I(a_{1}x)\Delta(a_{2})+I(a_{1})\Delta(a_{2}x))\Delta(a_{3})$
$\displaystyle+(-1)^{|a_{3}|+1}I(a_{1})\Delta(a_{2})\Delta(a_{3}x),$

then

$I(ax)+I(a_{1})\Delta(a_{2}x)=q(ax)+\sum_{a_{3}\neq
1}(-1)^{|a_{3}|}I(a_{1}x)\Delta(a_{2}a_{3}).$

Thus $q(ax)=I(a_{1}x)\Delta(a_{2})+I(a_{1})\Delta(a_{2}x)$ by (4.2).

“$\Leftarrow$” Obviously, $I(a)=(-1)^{|a_{2}|}q(a_{1})\Delta(a_{2})$ holds for
$a=1$ as $q(1)=I(1)$ by $q(a)=I(a_{1})\Delta(a_{2})$. It suffices to show that
if $I(a)=(-1)^{|a_{2}|}q(a_{1})\Delta(a_{2})$ then for any $x\in P(A)$,

$I(ax)=(-1)^{|a_{2}|}q(a_{1}x)\Delta(a_{2})+(-1)^{|a_{2}|+1}q(a_{1})\Delta(a_{2}x).$

This is equivalent to that

$I(ax)=(-1)^{|a_{3}|}(I(a_{1}x)\Delta(a_{2})+I(a_{1})\Delta(a_{2}x))\Delta(a_{3})+(-1)^{|a_{3}|+1}I(a_{1})\Delta(a_{2})\Delta(a_{3}x),$

i.e.,

$I(ax)=(-1)^{|a_{3}|}I(a_{1}x)\Delta(a_{2}a_{3}),$

which is always true by (4.1). ∎

To prove Proposition 4.5, we need the following lemma.

######  Lemma 4.4.

Let $A=U(\mathfrak{g})=k[x_{1},\cdots,x_{d}]$. Then for any linear map $q:A\to
A\otimes A$ and $a\in\mathcal{H}(A)$,

$(-1)^{|a_{2}|}(1\otimes
q)\Delta(a_{1})\cdot\Delta^{(2)}(a_{2})=(-1)^{|a_{2}|}1\otimes
q(a_{1})\Delta(a_{2}).$

###### Proof.

We claim first for any $a\in\mathcal{H}(A)$,

$\sum(-1)^{|a_{1}|+|a_{2}|}a_{1}a_{3}\otimes a_{2}\otimes
a_{4}=\sum(-1)^{|a_{1}|}1\otimes a_{1}\otimes a_{2}.$ (4.3)

It is obviously true for $a=1$. Now assume (4.3) holds for $a$. We show (4.3)
holds for $ax$ for any $x\in P(A)$.

Since $\Delta^{(3)}(x)=1\otimes 1\otimes 1\otimes x+1\otimes 1\otimes x\otimes
1+1\otimes x\otimes 1\otimes 1+x\otimes 1\otimes 1\otimes 1,$ then

$\displaystyle(-1)^{|(ax)_{1}|+|(ax)_{2}|}(ax)_{1}(ax)_{3}\otimes(ax)_{2}\otimes(ax)_{4}$
$\displaystyle=$ $\displaystyle(-1)^{|a_{1}|+|a_{2}|}a_{1}a_{3}\otimes
a_{2}\otimes a_{4}x+(-1)^{|a_{1}|+|a_{2}|}a_{1}a_{3}x\otimes a_{2}\otimes
a_{4}$ $\displaystyle-(-1)^{|a_{1}|+|a_{2}|}a_{1}a_{3}\otimes a_{2}x\otimes
a_{4}-(-1)^{|a_{1}|+|a_{2}|}a_{1}xa_{3}\otimes a_{2}\otimes a_{4}$
$\displaystyle=$ $\displaystyle(-1)^{|a_{1}|+|a_{2}|}a_{1}a_{3}\otimes
a_{2}\otimes a_{4}x-(-1)^{|a_{1}|+|a_{2}|}a_{1}a_{3}\otimes a_{2}x\otimes
a_{4}$ $\displaystyle=$ $\displaystyle(-1)^{|a_{1}|}1\otimes a_{1}\otimes
a_{2}x-(-1)^{|a_{1}|}1\otimes a_{1}x\otimes a_{2}$ $\displaystyle=$
$\displaystyle(-1)^{|(ax)_{1}|}1\otimes(ax)_{1}\otimes(ax)_{2}.$

Hence (4.3) holds for all $a\in\mathcal{H}(A)$. By applying $1\otimes
q\otimes\Delta$ to (4.3), then

$(-1)^{|a_{1}|+|a_{2}|}a_{1}a_{3}\otimes
q(a_{2})\otimes\Delta(a_{4})=(-1)^{|a_{1}|}1\otimes
q(a_{1})\otimes\Delta(a_{2}).$

Thus

$(-1)^{|a_{1}|}(1\otimes
q)\Delta(a_{1})\cdot\Delta^{(2)}(a_{2})=(-1)^{|a_{1}|}1\otimes
q(a_{1})\Delta(a_{2}),$

and the proof is finished. ∎

######  Proposition 4.5.

Let $A=U(\mathfrak{g})=k[x_{1},\cdots,x_{d}]$ and
$\mathcal{I}=\oplus_{i<j}k(x_{i}\otimes x_{j}-x_{j}\otimes x_{i}).$ Then, for
any linear map $q:A\to A\otimes A$, the following are equivalent.

  1. (1)

$q$ is skew-symmetric and satisfies the co-Leibniz rule.

  2. (2)

$I(a)=(-1)^{|a_{2}|}q(a_{1})\Delta(a_{2})\in\mathcal{I}$ for all $a\in A$.

###### Proof.

“(1) $\Rightarrow$ (2)” By Lemma 4.1, it suffice to show
$(1+t_{2})\left((-1)^{|a_{2}|}q(a_{1})\Delta(a_{2})\right)=0$ and
$(\Delta\otimes
1)((-1)^{|a_{2}|}q(a_{1})\Delta(a_{2}))=(1-t_{3})\left(1\otimes(-1)^{|a_{2}|}q(a_{1})\Delta(a_{2})\right).$

Note that $A$ is generated by primitive elements. It follows from Lemma 4.2
that $(1+t_{2})((-1)^{|a_{2}|}q(a_{1})\Delta(a_{2}))=0$ for all
$a\in\mathcal{H}(A)$.

By Lemmas 4.2, 4.4 and the co-Leibniz rule (2.5),

$\displaystyle(\Delta\otimes 1)((-1)^{|a_{2}|}q(a_{1})\Delta(a_{2}))$
$\displaystyle=$ $\displaystyle(-1)^{|a_{2}|}(\Delta\otimes
1)q(a_{1})\cdot\Delta^{(2)}(a_{2})$ $\displaystyle=$
$\displaystyle(1-t_{3})(-1)^{|a_{2}|}(1\otimes
q)\Delta(a_{1})\cdot\Delta^{(2)}(a_{2})$ $\displaystyle=$
$\displaystyle(1-t_{3})\left((-1)^{|a_{2}|}(1\otimes
q)\Delta(a_{1})\cdot\Delta^{(2)}(a_{2})\right)$ $\displaystyle=$
$\displaystyle(1-t_{3})\left(1\otimes(-1)^{|a_{2}|}q(a_{1})\Delta(a_{2})\right).$

“(2) $\Rightarrow$ (1)” Suppose
$(-1)^{|a_{2}|}q(a_{1})\Delta(a_{2})\in\mathcal{I}$ for all
$a\in\mathcal{H}(A)$. Then $q(1)\in\mathcal{I}$. We check the skew symmetric
property of $q$ and the co-Leibniz rule

$(\Delta\otimes 1)q(a)=(1-t_{3})(1\otimes q)\Delta(a)$

by induction on the degree of $a$. They are true for $a=1$ by Lemma 4.1.
Assume they are true for $a$ of degree no more than $n$. Now for any $a$ with
$\deg a=n+1$. Since $I(a)=(-1)^{|a_{2}|}q(a_{1})\Delta(a_{2})\in\mathcal{I}$,

$q(a)=\sum_{a_{2}\neq 1}(-1)^{|a_{2}|+1}q(a_{1})\Delta(a_{2})+I(a).$

Since $q(a_{1})$ is skew symmetric for $a_{2}\neq 1$ by induction hypothesis
and $I(a)\in\mathcal{I}$, then $q(a)$ is skew symmetric by Lemma 4.2. By Lemma
4.4,

$\displaystyle(\Delta\otimes 1)(q(a))$ $\displaystyle=$
$\displaystyle(\Delta\otimes 1)\left(\sum_{a_{2}\neq
1}(-1)^{|a_{2}|+1}q(a_{1})\Delta(a_{2})+I(a)\right)$ $\displaystyle=$
$\displaystyle(-1)^{|a_{2}|+1}\sum_{a_{2}\neq 1}(\Delta\otimes
1)q(a_{1})\cdot\Delta^{(2)}(a_{2})+(\Delta\otimes 1)(I(a))$ $\displaystyle=$
$\displaystyle(1-t_{3})\left(\sum_{a_{2}\neq 1}(-1)^{|a_{2}|+1}(1\otimes
q)\Delta(a_{1})\right)\cdot\Delta^{(2)}(a_{2})+(1-t_{3})(1\otimes I(a))$
$\displaystyle=$ $\displaystyle(1-t_{3})\left(\sum_{a_{2}\neq
1}(-1)^{|a_{2}|+1}(1\otimes q)\Delta(a_{1})\cdot\Delta^{(2)}(a_{2})\right)$
$\displaystyle+(1-t_{3})\left(\sum(-1)^{|a_{2}|}(1\otimes
q(a_{1})\Delta(a_{2}))\right)$ $\displaystyle=$
$\displaystyle(1-t_{3})\left((1\otimes q)\Delta(a)\right).$

The proof is finished. ∎

######  Proposition 4.6.

For any $a\in\mathcal{H}(A)$, let $I(a)=\sum_{1\leq i,j\leq
d}\lambda_{a}^{ij}x_{i}\otimes x_{j}\in\mathcal{I}$ with
$(\lambda_{a}^{ij})_{d\times d}\in M_{d}(k)$ skew-symmetric. Then the linear
map $q:A\to A\otimes A,a\mapsto I(a_{1})\Delta(a_{2})$ defines a co-Poisson
structure on the coalgebra $A$ if and only if for all $1\leq i<j<k\leq d$ and
$a\in A$,

$\sum_{s=1}^{d}\left(\lambda_{a_{1}}^{sk}\lambda_{x_{s}a_{2}}^{ij}+\lambda_{a_{1}}^{si}\lambda_{x_{s}a_{2}}^{jk}+\lambda_{a_{1}}^{sj}\lambda_{x_{s}a_{2}}^{ki}\right)=0.$
(4.4)

###### Proof.

By Proposition 4.5, we need only to care for the co-Jacobi identity. For any
$a\in\mathcal{H}(A)$,

$\displaystyle(q\otimes 1)q(a)=(q\otimes 1)(I(a_{1})\Delta(a_{2}))$
$\displaystyle=$ $\displaystyle(q\otimes
1)(\sum_{s,t}\lambda_{a_{1}}^{st}(x_{s}\otimes
x_{t})\Delta(a_{2}))=\sum_{s,t}\lambda_{a_{1}}^{st}q(x_{s}a_{2})\otimes
x_{t}a_{3}$ $\displaystyle=$
$\displaystyle\sum_{s,t}\lambda_{a_{1}}^{st}(I(x_{s}a_{2})\Delta(a_{3})+I(a_{2})\Delta(x_{s}a_{3}))\otimes
x_{t}a_{4}$ $\displaystyle=$
$\displaystyle\sum_{s,t,i,j}\lambda_{a_{1}}^{st}\lambda_{x_{s}a_{2}}^{ij}(x_{i}\otimes
x_{j}\otimes x_{t})(a_{3}\otimes a_{4}\otimes a_{5})$
$\displaystyle+\sum_{s,t,i,j}\lambda_{a_{1}}^{st}\lambda_{a_{2}}^{ij}(x_{i}x_{s}\otimes
x_{j}\otimes x_{t}+x_{i}\otimes x_{j}x_{s}\otimes x_{t})(a_{3}\otimes
a_{4}\otimes a_{5})$ $\displaystyle=$ $\displaystyle U+V,$

where $U$ is the first term, i.e.,

$U=\sum_{s,t,i,j}\lambda_{a_{1}}^{st}\lambda_{x_{s}a_{2}}^{ij}(x_{i}\otimes
x_{j}\otimes x_{t})(a_{3}\otimes a_{4}\otimes a_{5}),$

and $V$ is the second term, i.e.,

$V=\sum_{s,t,i,j}\lambda_{a_{1}}^{st}\lambda_{a_{2}}^{ij}(x_{i}x_{s}\otimes
x_{j}\otimes x_{t}+x_{i}\otimes x_{j}x_{s}\otimes x_{t})(a_{3}\otimes
a_{4}\otimes a_{5}).$

Then, by the skew-symmetric property of $\\{\lambda_{a}^{ij}\\}_{d\times d}$
and the cocommutativity of $A$,

$\displaystyle\circlearrowleft V=$
$\displaystyle\sum_{s,t,i,j}(\lambda_{a_{1}}^{st}\lambda_{a_{2}}^{ij}+\lambda_{a_{1}}^{ij}\lambda_{a_{2}}^{ts})(x_{i}x_{s}\otimes
x_{j}\otimes x_{t})(a_{3}\otimes a_{4}\otimes a_{5})$
$\displaystyle+\sum_{s,t,i,j}(\lambda_{a_{1}}^{st}\lambda_{a_{2}}^{ij}+\lambda_{a_{1}}^{ji}\lambda_{a_{2}}^{st})(x_{i}\otimes
x_{j}x_{s}\otimes x_{t})(a_{3}\otimes a_{4}\otimes a_{5})$
$\displaystyle+\sum_{s,t,i,j}(\lambda_{a_{1}}^{st}\lambda_{a_{2}}^{ij}+\lambda_{a_{1}}^{ji}\lambda_{a_{2}}^{st})(x_{t}\otimes
x_{i}\otimes x_{j}x_{s})(a_{3}\otimes a_{4}\otimes a_{5})$ $\displaystyle=$
$\displaystyle 0.$

So, the co-Jacobi identity holds if and only if $\circlearrowleft U=0$. Note
that

$\displaystyle\circlearrowleft U=$
$\displaystyle\sum_{s,t,i,j}\lambda_{a_{1}}^{st}\lambda_{x_{s}a_{2}}^{ij}(x_{i}\otimes
x_{j}\otimes x_{t}+x_{t}\otimes x_{i}\otimes x_{j}+x_{j}\otimes x_{t}\otimes
x_{i})(a_{3}\otimes a_{4}\otimes a_{5})$ $\displaystyle=$
$\displaystyle\sum_{i,j,k}\sum_{s}\left(\lambda_{a_{1}}^{sk}\lambda_{x_{s}a_{2}}^{ij}+\lambda_{a_{1}}^{si}\lambda_{x_{s}a_{2}}^{jk}+\lambda_{a_{1}}^{sj}\lambda_{x_{s}a_{2}}^{ki}\right)(x_{i}\otimes
x_{j}\otimes x_{k})(a_{3}\otimes a_{4}\otimes a_{5}).$

If $\circlearrowleft U=0$, then, by considering the coefficients of elements
in degree 3 in $\circlearrowleft U$ (i.e., when $a_{3}\otimes a_{4}\otimes
a_{5}=1\otimes 1\otimes 1$),

$c_{ijk}=\sum_{s}\left(\lambda_{a_{1}}^{sk}\lambda_{x_{s}a_{2}}^{ij}+\lambda_{a_{1}}^{si}\lambda_{x_{s}a_{2}}^{jk}+\lambda_{a_{1}}^{sj}\lambda_{x_{s}a_{2}}^{ki}\right)=0$

for all $1\leq i,j,k\leq d$ and $a\in\mathcal{H}(A)$.

Conversely, if $c_{ijk}=0$ for all $1\leq i,j,k\leq d$ and
$a\in\mathcal{H}(A)$, then $\circlearrowleft U=0$. ∎

In summary, we have the following result.

######  Theorem 4.7.

Let $A=U(\mathfrak{g})=k[x_{1},\cdots,x_{d}]$. Then a linear map $q:A\to
A\otimes A$ gives a co-Poisson coalgebra structure on $A$ if and only if there
is a linear map $I:A\to A\otimes A$ such that for all $a\in\mathcal{H}(A)$,

  1. (1)

$q(a)=I(a_{1})\Delta(a_{2})$.

  2. (2)

$I(a)=\sum_{1\leq i,j\leq d}\lambda_{a}^{ij}x_{i}\otimes x_{j}\in\mathcal{I}$
with $(\lambda_{a}^{ij})_{d\times d}\in M_{d}(k)$ skew-symmetric.

  3. (3)

For all $1\leq i<j<k\leq d$,
$\sum_{s=1}^{d}\left(\lambda_{a_{1}}^{sk}\lambda_{x_{s}a_{2}}^{ij}+\lambda_{a_{1}}^{si}\lambda_{x_{s}a_{2}}^{jk}+\lambda_{a_{1}}^{sj}\lambda_{x_{s}a_{2}}^{ki}\right)=0.$

The following proposition shows that the co-Jacobi identity holds trivially in
two variables case. So, the co-Poisson coalgebra structures on $k[x,y]$ are
given by the linear maps $I:k[x,y]\to k(x\otimes y-y\otimes x)$.

######  Proposition 4.8.

Let $A=k[x,y]$. Then there is an one-to-one correspondence between the co-
Poisson coalgebra structures on $A$ and the linear maps
$A\to\mathcal{I}=k(x\otimes y-y\otimes x)$, given by

$(q:A\to A\otimes
A)\mapsto(I:A\to\mathcal{I},a\mapsto(-1)^{|a_{2}|}q(a_{1})\Delta(a_{2}),$

with the inverse map $(I:A\to\mathcal{I})\mapsto(q:A\to A\otimes A,a\mapsto
I(a_{1})\Delta(a_{2})).$

###### Proof.

We only need to show the co-Jacobi identity always holds in this case. Assume
$I(a)=l_{a}(x\otimes y-y\otimes x)$ with $l_{a}\in k$. For any $a\in A$,

$q(a)=I(a_{1})\Delta(a_{2})=l_{a_{1}}(x\otimes y-y\otimes x)(a_{2}\otimes
a_{3})=l_{a_{1}}(xa_{2}\otimes ya_{3}-ya_{2}\otimes xa_{3}).$

Then $\circlearrowleft(q\otimes 1)q(a)=\circlearrowleft
l_{a_{1}}\left(q(xa_{2})\otimes ya_{3}-q(ya_{2})\otimes xa_{3}\right).$ Now

$\displaystyle\circlearrowleft l_{a_{1}}q(xa_{2})\otimes ya_{3}$
$\displaystyle=$ $\displaystyle\circlearrowleft
l_{a_{1}}\left(I(xa_{2})\Delta(a_{3})+I(a_{2})\Delta(xa_{3})\right)\otimes
ya_{4}$ $\displaystyle=$ $\displaystyle\circlearrowleft
l_{a_{1}}l_{xa_{2}}(x\otimes y-y\otimes x)(a_{3}\otimes a_{4})\otimes ya_{5}$
$\displaystyle+\circlearrowleft l_{a_{1}}l_{a_{2}}(x\otimes y-y\otimes
x)(xa_{3}\otimes a_{4}+a_{3}\otimes xa_{4})\otimes ya_{5}$ $\displaystyle=$
$\displaystyle\circlearrowleft l_{a_{1}}l_{xa_{2}}(xa_{3}\otimes ya_{4}\otimes
ya_{5}-ya_{3}\otimes xa_{4}\otimes ya_{5})$ $\displaystyle+\circlearrowleft
l_{a_{1}}l_{a_{2}}(x^{2}a_{3}\otimes ya_{4}\otimes ya_{5}+xa_{3}\otimes
xya_{4}\otimes ya_{5}$ $\displaystyle-xya_{3}\otimes xa_{4}\otimes
ya_{5}-ya_{3}\otimes x^{2}a_{4}\otimes ya_{5})$ $\displaystyle=$
$\displaystyle\circlearrowleft l_{a_{1}}l_{a_{2}}(xa_{3}\otimes xya_{4}\otimes
ya_{5}-xya_{3}\otimes xa_{4}\otimes ya_{5}).$

Similarly,

$\circlearrowleft l_{a_{1}}q(ya_{2})\otimes xa_{3}=\circlearrowleft
l_{a_{1}}l_{a_{2}}(ya_{3}\otimes xya_{4}\otimes xa_{5}-xya_{3}\otimes
ya_{4}\otimes xa_{5}).$

Thus $\circlearrowleft(q\otimes 1)q(a)=0$ for all $a\in A=k[x,y]$. ∎

##  5\. (Co-)Poisson Hopf structures on $k[x_{1},x_{2},\cdots,x_{d}]$

###  5.1. Poisson Hopf structures

As proved in [LPV, Proposition 1.8], any Poisson structure on the polynomial
algebra $A=k[x_{1},x_{2},\cdots,x_{d}]$ is given by $\\{x_{i},x_{j}\\}=f_{ij}$
where $\\{f_{ij}\\}_{d\times d}$ is a skew-symmetric matrix over $A$ such that
for all $1\leq i<j<k\leq d$,

$\sum_{l=1}^{d}\left(f_{lk}\frac{\partial f_{ij}}{\partial
x_{l}}+f_{li}\frac{\partial f_{jk}}{\partial x_{l}}+f_{lj}\frac{\partial
f_{ki}}{\partial x_{l}}\right)=0.$ (5.1)

If $A=U(\mathfrak{g})=k[x_{1},x_{2},\cdots,x_{d}]$ is viewed as a Hopf
algebra, then the Poisson algebra structures on $A$ can be described in a form
dual to Theorem 4.7. The following is a reciprocity law for linear maps
$A\otimes A\to A$.

######  Lemma 5.1.

Let $p:A\otimes A\to A$ and $J:A\otimes A\to A$ be two linear maps. Then
$p(a\otimes b)=J(a_{1}\otimes b_{1})a_{2}b_{2}$ for all $a,b\in A$ if and only
if $J(a\otimes b)=(-1)^{|a_{2}|+|b_{2}|}p(a_{1}\otimes b_{1})a_{2}b_{2}$ for
all $a,b\in A$.

###### Proof.

Similar to that of Proposition 4.3. ∎

######  Proposition 5.2.

Let $A=U(\mathfrak{g})=k[x_{1},\cdots,x_{d}]$. Then a linear map $p:A\otimes
A\to A$ gives a Poisson structure on $A$ if and only if there is a linear map
$J:A\otimes A\to A$ such that for all $a,b\in A$,

  1. (1)

$p(a\otimes b)=J(a_{1}\otimes b_{1})a_{2}b_{2}.$

  2. (2)

$J$ is skew-symmetric, and $J(a\otimes b)=0$ except both $a$ and $b$ are of
degree $1$.

  3. (3)

The Jacobi identity holds for $J$.

Actually, Proposition 5.2 is exactly [LPV, Proposition 1.8]. If $p:A\otimes
A\to A$ is a Poisson algebra structure on $A$, and $J(a\otimes
b)=(-1)^{|a_{2}|+|b_{2}|}p(a_{1}\otimes b_{1})a_{2}b_{2}$, then $J$ satisfies
conditions (2) and (3) in Proposition 5.2. In this case, condition (1) is the
same as

$\\{f,g\\}=\sum_{1\leq i,j\leq d}\frac{\partial f}{\partial
x_{i}}\frac{\partial g}{\partial x_{j}}\\{x_{i},x_{j}\\}.$

The Poisson Hopf structures on $A$ are classified in the following
proposition. They are exactly linear Poisson structures on $A$.

######  Proposition 5.3.

Any Poisson Hopf structure on $A=k[x_{1},x_{2},\cdots,x_{d}]$ is given by

$\\{x_{i},x_{j}\\}=\sum_{l=1}^{d}\lambda^{ij}_{l}x_{l}\qquad(1\leq i,j\leq
d),$

where $\lambda^{ij}_{l}=-\lambda^{ji}_{l}$, subject to the relations, for all
$1\leq i<j<k\leq d$ and all $1\leq s\leq d$,

$\sum_{l=1}^{d}\left(\lambda^{ij}_{l}\lambda^{lk}_{s}+\lambda^{jk}_{l}\lambda^{li}_{s}+\lambda^{ki}_{l}\lambda^{lj}_{s}\right)=0.$

###### Proof.

Suppose $\\{-,-\\}$ is a Poisson Hopf structure on $A$. Then, by (2.3),

$\Delta\left(\\{x_{i},x_{j}\\}\right)=1\otimes\\{x_{i},x_{j}\\}+\\{x_{i},x_{j}\\}\otimes
1,$

that is, $\\{x_{i},x_{j}\\}$ is a primitive element of $A$. Hence
$\\{x_{i},x_{j}\\}=\sum_{l=1}^{d}\lambda^{ij}_{l}x_{l}$ for some
$\lambda^{ij}_{l}\in k$ such that $\lambda^{ij}_{l}=-\lambda^{ji}_{l}$ for all
$l$. Then (5.1) is equivalent to, for all $1\leq i<j<k\leq d$ and all $1\leq
s\leq d$,

$\sum_{l=1}^{d}\left(\lambda^{ij}_{l}\lambda^{lk}_{s}+\lambda^{jk}_{l}\lambda^{li}_{s}+\lambda^{ki}_{l}\lambda^{lj}_{s}\right)=0.$

Conversely, any such a Poisson algebra structure is in fact a Poisson Hopf
structure on $A$. We need to check that
$\Delta\left(\\{a,b\\}\right)=a_{1}b_{1}\otimes\\{a_{2},b_{2}\\}+\\{a_{1},b_{1}\\}\otimes
a_{2}b_{2}$ for all $a,b\in\mathcal{H}(A)$, which can be done by induction. ∎

###  5.2. Co-Poisson Hopf structures

Next we discuss co-Poisson Hopf structures on $A$. We fix some notations here.
For any $a=x_{1}^{n_{1}}x_{2}^{n_{2}}\cdots x_{d}^{n_{d}}\in\mathcal{H}(A)$,
we denote by $a!=n_{1}!\cdots n_{d}!$ and $a^{(i)}=n_{i}$ for $1\leq i\leq d$.
For any $b=x_{1}^{m_{1}}x_{2}^{m_{2}}\cdots x_{d}^{m_{d}}\in\mathcal{H}(A)$,
if $b\mid a$, we denote by

${a\choose b}={n_{1}\choose m_{1}}{n_{2}\choose m_{2}}\cdots{n_{d}\choose
m_{d}};$

if $b\nmid a$, we set ${a\choose b}=0$.

######  Proposition 5.4.

Let $A=U(\mathfrak{g})=k[x_{1},x_{2},\cdots,x_{d}]$. Then a linear map $q:A\to
A\otimes A$ gives a co-Poisson Hopf algebra structure on $A$ if and only if
there is a linear map $I:A\to A\otimes A$ such that for all
$a\in\mathcal{H}(A)$,

  1. (1)

$q(a)=I(a_{1})\Delta(a_{2})$.

  2. (2)

For all $1\leq s\leq d$, $I(a)=0$ if $a\neq x_{s}$ and $I(x_{s})=\sum_{1\leq
i,j\leq d}\lambda_{s}^{ij}x_{i}\otimes x_{j}\in\mathcal{I}$ with
$(\lambda_{s}^{ij})_{d\times d}\in M_{d}(k)$ skew-symmetric.

  3. (3)

For all $0\leq i<j<k\leq d$, $1\leq s\leq d$,
$\sum_{l=1}^{d}\left(\lambda_{s}^{lk}\lambda_{l}^{ij}+\lambda_{s}^{li}\lambda_{l}^{jk}+\lambda_{s}^{lj}\lambda_{l}^{ki}\right)=0$.

###### Proof.

“$\Rightarrow$” Let $I:A\to A\otimes A$ be the map
$I(a)=(-1)^{|a_{2}|}q(a_{1})\Delta(a_{2})$ for all $a\in A$. Then, by
Proposition 4.3, $q(a)=I(a_{1})\Delta(a_{2})$, i.e., (1) holds.

Since $q(ab)=q(a)\Delta(b)+\Delta(a)q(b)$ for all $a,b\in A$, $I(1)=q(1)=0$.
Thus $q(x_{i})=I(x_{i})$ for all $1\leq i\leq d$. Then, by induction and
(2.6), for any $a\in\mathcal{H}(A)$,

$q(a)=\sum_{l=1}^{d}{a\choose
x_{l}}I(x_{l})\Delta\left(\frac{a}{x_{l}}\right).$

If $a\in\mathcal{H}(A)$ with $\deg a=2$, then $I(a)=0$ by
$q(a)=I(a_{1})\Delta(a_{2})$. By induction, we see $I(a)=0$ for any
$a\in\mathcal{H}(A)$ of degree $\geq 3$. By Proposition 4.5, (2) holds.

In this case, the equation (4.4) is equivalent to
$\sum_{l=1}^{d}\left(\lambda_{s}^{lk}\lambda_{l}^{ij}+\lambda_{s}^{li}\lambda_{l}^{jk}+\lambda_{s}^{lj}\lambda_{l}^{ki}\right)=0,$
i.e., (3) holds.

“$\Leftarrow$” If $I(a)=0$ for all $x_{i}\neq a\in\mathcal{H}(A)$, then
$q(a)=I(a_{1})\Delta(a_{2})$ becomes

$q(a)=\sum_{l=1}^{d}{a\choose
x_{l}}I(x_{l})\Delta\left(\frac{a}{x_{l}}\right).$

Now it is easy to check $q(ab)=q(a)\Delta(b)+\Delta(a)q(b)$ for all
$a,b\in\mathcal{H}(A)$. ∎

######  Example 5.5.

Let $A=k[x,y]$. Then there is an one-to-one correspondence between the co-
Poisson Hopf structures on $A$ and the set $\\{(I(x),I(y))\mid I(x),I(y)\in
k(x\otimes y-y\otimes x)\\}$, given by

$\displaystyle q$ $\displaystyle\mapsto(q(x),q(y)),\,\textrm{and}$
$\displaystyle(I(x),I(y))$ $\displaystyle\mapsto q:x^{n}y^{m}\mapsto
nI(x)\Delta(x^{n-1}y^{m})+mI(y)\Delta(x^{n}y^{m-1}).$

######  Example 5.6.

Let $A=k[x_{1},x_{2},x_{3}]$. Then any Poisson Hopf structure on $A$ is given
by

$\\{x_{1},x_{2}\\}=\lambda_{12}^{1}x_{1}+\lambda_{12}^{2}x_{2}+\lambda_{12}^{3}x_{3},$
$\\{x_{2},x_{3}\\}=\lambda_{23}^{1}x_{1}+\lambda_{23}^{2}x_{2}+\lambda_{23}^{3}x_{3},$
$\\{x_{3},x_{1}\\}=\lambda_{31}^{1}x_{1}+\lambda_{31}^{2}x_{2}+\lambda_{31}^{3}x_{3},$

subject to the relations,

$\lambda_{12}^{1}\lambda_{31}^{l}+\lambda_{23}^{2}\lambda_{12}^{l}+\lambda_{31}^{3}\lambda_{23}^{l}=\lambda_{12}^{2}\lambda_{23}^{l}+\lambda_{23}^{3}\lambda_{31}^{l}+\lambda_{31}^{1}\lambda_{12}^{l},\,\,\textrm{for
all}\,\,1\leq l\leq 3.$

Any co-Poisson Hopf structure on $A$ is given by

$q(x_{1})=\lambda_{1}^{12}(x_{1}\otimes x_{2}-x_{2}\otimes
x_{1})+\lambda_{1}^{23}(x_{2}\otimes x_{3}-x_{3}\otimes
x_{2})+\lambda_{1}^{31}(x_{3}\otimes x_{1}-x_{1}\otimes x_{3}),$
$q(x_{2})=\lambda_{2}^{12}(x_{1}\otimes x_{2}-x_{2}\otimes
x_{1})+\lambda_{2}^{23}(x_{2}\otimes x_{3}-x_{3}\otimes
x_{2})+\lambda_{2}^{31}(x_{3}\otimes x_{1}-x_{1}\otimes x_{3}),$
$q(x_{3})=\lambda_{3}^{12}(x_{1}\otimes x_{2}-x_{2}\otimes
x_{1})+\lambda_{3}^{23}(x_{2}\otimes x_{3}-x_{3}\otimes
x_{2})+\lambda_{3}^{31}(x_{3}\otimes x_{1}-x_{1}\otimes x_{3}),$

subject to the relations,

$\lambda_{k}^{12}\lambda_{1}^{31}+\lambda_{k}^{23}\lambda_{2}^{12}+\lambda_{k}^{31}\lambda_{3}^{23}=\lambda_{k}^{31}\lambda_{1}^{12}+\lambda_{k}^{12}\lambda_{2}^{23}+\lambda_{k}^{23}\lambda_{3}^{31},\,\,\textrm{for
any}\,\,1\leq k\leq 3.$

###  5.3. Dual (co-)Poisson structures

It is well-known that the dual algebra $A^{*}$ of the coalgebra
$A=k[x_{1},x_{2},\cdots,x_{d}]$ is the algebra of formal divided power series
(it is called Hurwitz series in [Ke, Proposition 2.4] in one variable), which
is isomorphic to the algebra of formal power series
$k[[x_{1},x_{2},\cdots,x_{d}]]$. By Cartier-Gabriel-Kostant-Milnor-Moore
Theorem ([Sw, Theorem 8.1.5] and [MM, $\S$ 6]), the finite dual $A^{\circ}$ of
the polynomial Hopf algebra $A=k[x_{1},x_{2},\cdots,x_{d}]$ is isomorphic to
$A\ltimes k^{n}$ as Hopf algebras, where $k^{n}$ carries the additive group
structure (the group-like elements in the Hopf dual). Similar to [LPV,
Proposition 1.8], the following lemma holds.

######  Lemma 5.7.

Let $\tilde{A}=k[[x_{1},\cdots,x_{d}]]$ be the algebra of formal power series.
Any Poisson structure on $\tilde{A}$ is given by $\\{x_{i},x_{j}\\}=f_{ij}$,
where $(f_{ij})_{d\times d}$ is a skew-symmetric matrix over $\tilde{A}$ such
that for all $1\leq i<j<k\leq d$,

$\\{\\{x_{i},x_{j}\\},x_{k}\\}+\\{\\{x_{j},x_{k}\\},x_{i}\\}+\\{\\{x_{k},x_{i}\\},x_{j}\\}=0.$

In this case, for all $f,g\in\tilde{A}$,
$\\{f,g\\}=\sum_{i,j=1}^{d}\frac{\partial f}{\partial x_{i}}\frac{\partial
g}{\partial x_{j}}f_{ij}.$

######  Theorem 5.8.

Suppose $\operatorname{char}k=0$. Let
$\tilde{A}=k[[x_{1},x_{2},\cdots,x_{d}]]$ be the algebra of formal power
series and $A=k[x_{1},x_{2},\cdots,x_{d}]$. Then there is an one-to-one
correspondence between the Poisson algebra structures on $\tilde{A}$ and the
co-Poisson coalgebra structures on $A$.

###### Proof.

Suppose
$\\{x_{i},x_{j}\\}=f_{ij}=\sum_{a\in\mathcal{H}(A)}\lambda^{ij}_{a}a\,(1\leq
i,j\leq d)$ gives a Poisson algebra structure on $\tilde{A}$. Let
$\alpha^{ij}_{a}=a!\lambda^{ij}_{a}$ and

$I(a)=\sum_{1\leq i,j\leq d}\alpha^{ij}_{a}x_{i}\otimes x_{j},\quad
q(a)=I(a_{1})\Delta(a_{2})\,\textrm{for all}\,a\in\mathcal{H}(A).$

Then $q:A\to A\otimes A$ is a co-Poisson coalgebra structure on $A$. In fact,
for all $1\leq i<j<k\leq d$,

$\\{\\{x_{i},x_{j}\\},x_{k}\\}=\sum_{a,b\in\mathcal{H}(A)}\sum_{l=1}^{d}\lambda^{ij}_{a}\lambda^{lk}_{b}\frac{\partial
a}{\partial
x_{l}}b=\sum_{c\in\mathcal{H}(A)}\sum^{ab=c}_{a,b\in\mathcal{H}(A)}\sum_{l=1}^{d}(a^{(l)}+1)\lambda^{ij}_{ax_{l}}\lambda^{lk}_{b}c.$

Then $\circlearrowleft\\{\\{x_{i},x_{j}\\},x_{k}\\}=0$ if and only if for all
$c\in\mathcal{H}(A)$ and all $1\leq i<j<k\leq d$,

$\sum^{ab=c}_{a,b\in\mathcal{H}(A)}\sum_{l=1}^{d}(a^{(l)}+1)\left(\lambda^{ij}_{ax_{l}}\lambda^{lk}_{b}+\lambda^{jk}_{ax_{l}}\lambda^{li}_{b}+\lambda^{ki}_{ax_{l}}\lambda^{lj}_{b}\right)=0.$

Note that $\alpha^{ij}_{a}=a!\lambda^{ij}_{a}$. It is equivalent to

$\sum^{ab=c}_{a,b\in\mathcal{H}(A)}\sum_{l=1}^{d}\frac{1}{a!b!}\left(\alpha^{ij}_{ax_{l}}\alpha^{lk}_{b}+\alpha^{jk}_{ax_{l}}\alpha^{li}_{b}+\alpha^{ki}_{ax_{l}}\alpha^{lj}_{b}\right)=0.$

By multiplying it with $c!$, it is easy to see that it is equivalent to

$\sum_{l=1}^{d}\left(\alpha^{ij}_{c_{1}x_{l}}\alpha^{lk}_{c_{2}}+\alpha^{jk}_{c_{1}x_{l}}\alpha^{li}_{c_{2}}+\alpha^{ki}_{c_{1}x_{l}}\alpha^{lj}_{c_{2}}\right)=0$

for all $c\in\mathcal{H}(A)$ and $1\leq i<j<k\leq d$.

On the other hand, suppose $q:A\to A\otimes A$ is a co-Poisson coalgebra
structure on $A$. Let
$I(a)=(-1)^{|a_{2}|}q(a_{1})\Delta(a_{2}):=\alpha^{ij}_{a}x_{i}\otimes x_{j}$,
and $\lambda^{ij}_{a}=\frac{1}{a!}\alpha^{ij}_{a}$. Then

$\\{x_{i},x_{j}\\}=f_{ij}:=\sum_{a\in\mathcal{H}(A)}\lambda^{ij}_{a}a.$

gives a Poisson algebra structure on $\tilde{A}$. ∎

######  Remark 5.9.

A co-Poisson coalgebra structure $q$ on $A$ given by $I:A\to A\otimes A$ is
called rational if there is an integer $n$ such that $I(a)=0$ for all
$a\in\mathcal{H}(A)$ with $\deg a\geq n$. Then there is an one-to-one
correspondence between Poisson algebra structures on $A$ and rational co-
Poisson coalgebra structures on $A$.

Combining Propositions 5.3 and 5.4, we have the following.

######  Theorem 5.10.

There is an one-to-one correspondence between Poisson Hopf structures on $A$
and co-Poisson Hopf structures on $A$. More precisely, assume

$\\{x_{i},x_{j}\\}=\lambda^{ij}_{1}x_{1}+\cdots+\lambda^{ij}_{d}x_{d}$

is a Poisson Hopf structure on $A$. Let

$I(x_{s})=\sum_{1\leq i,j\leq d}\lambda^{ij}_{s}x_{i}\otimes x_{j}$

for all $1\leq s\leq d$ and $I(a)=0$ for all other $a\in\mathcal{H}(A)$. Then
$q(a)=I(a_{1})\Delta(a_{2})$ defines a co-Poisson Hopf structure on $A$.

## Acknowledgments

The authors thank Ruipeng Zhu for useful discussions. This research is
supported by NSFC key project 11331006 and NSFC project 11171067.

## References

  * [CP] V. Chari, A. Pressley, A Guide to Quantum Groups, Cambridge University Press, Providence, 1994.
  * [Dr] V. G. Drinfeld, Quantum groups, Proc. Internat. Congr. Math. (Berkeley, 1986), Amer. Math. Soc., Providence, RI, 1987, 798–820.
  * [FL] D. R. Farkas, G. Letzter, Ring theory from symplectic geometry, J. Pure Appl. Algebra 125 (1998), 155–190.
  * [He] I. N. Herstein, Rings with Involution, Chicago Lecture in Math., Univ. of Chicago Press, Chicago, 1976.
  * [Ke] W. F. Keigher, On the ring of Hurwitz series, Comm. Algebra 25 (1997), 1845–1859.
  * [KS] L. I. Korogodski, Y. S. Soibelman, Algebras of Functions on Quantum Groups, Part I, Mathematical surveys and monographs, V. 56, Amer. Math. Soc., Providence, 1998.
  * [Li] A. Lichnerowicz, Les varieties de Poisson et leurs algebras de Lie associees (French), J. Differential Geometry 12 (1977), 253–300.
  * [LPV] C. Laurent-Gengoux, A. Pichereau and P. Vanhaecke, Poisson Structures, Grundlehren der Mathematischen Wissenschaften 347, Springer, Heidelberg, 2013.
  * [LWW] J. Luo, S.-Q. Wang, Q.-S. Wu, Twisted Poincaré duality between Poisson homology and cohomology, J. Algebra 442 (2015), 484–505.
  * [MM] J. W. Milnor, J. C. Moore, On the structure of Hopf algebras, Ann. of Math., 81 (1965), 211–264.
  * [Mo] S. Montgomery, Hopf Algebras and Their Actions on Rings, CBMS Reg. Conf. Ser. Math. 82, Amer. Math. Soc., Providence, RI, 1993.
  * [MR] J. C. McConnell and J. C . Robson, Noncommutative Noetherian Rings, Wiley, Chichester, 1987.
  * [Oh] S.-Q. Oh, A Poisson Hopf algebra related to a twisted quantum group, Comm. Algebra 45 (2017), 76–104.
  * [OP] S.-Q. Oh, H.-M. Park, Duality of co-Poisson Hopf algebras, Bull. Korean Math. Soc. 48 (2011), 17–21.
  * [Sw] M. E. Sweedler, Hopf Algebras, Benjamin, New York, 1969.
  * [Vo] T. Voronov, On the Poisson envelope of a Lie algebra, “Noncommutative” moment space, Funct. Anal. Appl. 29 (1995), 196–199.
  * [Wei] A. Weinstein, Lecture on Symplectic Manifolds, CBMS Conference series in Math. 29, 1977.