# Twists of graded Poisson algebras
and related properties

Xin Tang, Xingting Wang and James J. Zhang  Tang: Department of Mathematics &
Computer Science, Fayetteville State University, Fayetteville, NC 28301, USA
<EMAIL_ADDRESS>Wang: Department of Mathematics, Howard University,
Washington, DC, 20059, USA<EMAIL_ADDRESS>Zhang: Department of
Mathematics, Box 354350, University of Washington, Seattle, Washington 98195,
USA<EMAIL_ADDRESS>
###### Abstract.

We introduce a Poisson version of the graded twist of a graded associative
algebra and prove that every graded Poisson structure on a connected graded
polynomial ring $A:=\Bbbk[x_{1},\ldots,x_{n}]$ is a graded twist of a
unimodular Poisson structure on $A$, namely, if $\pi$ is a graded Poisson
structure on $A$, then $\pi$ has a decomposition

$\pi~{}=~{}\pi_{unim}+\frac{1}{\sum_{i=1}^{n}\deg x_{i}}E\wedge{\mathbf{m}}$

where $E$ is the Euler derivation, $\pi_{unim}$ is the unimodular graded
Poisson structure on $A$ corresponding to $\pi$, and ${\mathbf{m}}$ is the
modular derivation of $(A,\pi)$. This result is a generalization of the same
result in the quadratic setting. The rigidity of graded twisting,
$PH^{1}$-minimality, and $H$-ozoneness are studied. As an application, we
compute the Poisson cohomologies of the quadratic Poisson structures on the
polynomial ring of three variables when the potential is irreducible, but not
necessarily having isolated singularities.

###### Key words and phrases:

modular derivation, Poisson algebra, graded twist

###### 2010 Mathematics Subject Classification:

Primary 17B63, 17B40, 16S36, 16W20

## Introduction

Poisson algebras have recently been studied extensively by many researchers,
see e.g., [Ba2, Ba3, Go1, Go2, GLa, GLe, JO, LS, LuWW1, LuWW2, LvWZ1], with
topics related to (twisted) Poincaré duality and the modular derivation,
Poisson Dixmier-Moeglin equivalences, Poisson enveloping algebras and so on.
Poisson algebras have been used in the representation theory of PI Sklyanin
algebras [WWY1, WWY2]. Isomorphism problem and cancellation problem in the
Poisson setting have been investigated in [GW1, GW2].

Let $\Bbbk$ be a base field. Except for Sections 1 and 2 we further assume
that $\Bbbk$ is of characteristic zero. Quadratic Poisson structures on
$\Bbbk[x_{1},\ldots,x_{n}]$ with $\deg(x_{i})=1$ for all $i=1,\cdots,n$ have
been playing an important role in several other subjects, see papers [LX] by
Liu-Xu, [Bo] by Bondal, and [Py] by Pym. Deformation quantizations of such
Poisson structures are homogeneous coordinate rings of quantum
${\mathbb{P}}^{n-1}$s. In general, such a deformation quantization is skew-
Calabi-Yau; while it is Calabi-Yau if and only if the Poisson structure on
$\Bbbk[x_{1},\ldots,x_{n}]$ is unimodular [Do].

In additional to the quadratic case, we are interested in weighted Poisson
structures on $\Bbbk[x_{1},\ldots,x_{n}]$ where $\deg x_{i}>0$ for all
$i=1,\ldots,n$. Note that deformation quantizations of weighted Poisson
structures are homogeneous coordinate rings of weighted quantum
${\mathbb{P}}^{n-1}$s. If $\pi$ is a graded Poisson structure on
$\Bbbk[x_{1},\ldots,x_{n}]$ where $\sum_{i=1}\deg x_{i}\neq 0$ in the base
field $\Bbbk$, we prove that $\pi$ has a decomposition

(E0.0.1) $\pi~{}=~{}\pi_{unim}+\frac{1}{\sum_{i=1}^{n}\deg
x_{i}}E\wedge{\mathbf{m}}$

where $E$ is the Euler derivation, $\pi_{unim}$ is the unimodular graded
Poisson structure on $\Bbbk[x_{1},\ldots,x_{n}]$ corresponding to $\pi$, and
${\mathbf{m}}$ is the modular derivation of $(\Bbbk[x_{1},\ldots,x_{n}],\pi)$.
If $\deg x_{i}=1$ for all $i$, (E0.0.1) was observed by Bondal [Bo], Liu-Xu
[LX], and in the book [LPV, Theorem 8.26]. Similar to the ideas in [Py], to
classify all graded Poisson structures on polynomial rings where $\deg
x_{i}>0$, it is a good idea to first classify unimodular ones.

To prove (E0.0.1), we will use a Poisson version of the graded twist [Zh]. Let
$A$ be a ${\mathbb{Z}}$-graded Poisson algebra such that both the commutative
multiplication $\cdot$ and the Poisson bracket $\pi:=\\{-,-\\}$ are graded of
degree 0. If $a\in A$ is homogeneous, we use $|a|$ to denote its degree.
Define the Euler derivation $E$ of $A$ by

$E(a)~{}=~{}|a|\,a$

for all homogeneous element $a\in A$. Let $\delta$ be a graded Poisson
derivation of $A$. We define a new Poisson structure, denoted by
$\pi_{new}:=\\{-,-\\}_{new}$, to be

(E0.0.2) $\\{a,b\\}_{new}~{}:=~{}\\{a,b\\}+E(a)\delta(b)-\delta(a)E(b)$

for all homogeneous elements $a,b\in A$, or equivalently

$\pi_{new}~{}:=~{}\pi+E\wedge\delta.$

We will show that $(A,\cdot,\\{-,-\\}_{new})$ (or $(A,\pi_{new})$) is a graded
Poisson algebra in Section 2 and it is denoted by $A^{\delta}$.

Now we state some results. Let $A$ be a polynomial algebra
$\Bbbk[x_{1},\ldots,x_{n}]$ and let $\delta$ be a derivation of $A$. By [LPV,
(4.21)], the divergence of $\delta$ is defined to be

(E0.0.3)
$\operatorname{div}(\delta)~{}:=~{}\sum_{i=1}^{n}\frac{\partial\delta(x_{i})}{\partial
x_{i}},$

which is independent of the choices of generators $\\{x_{1},\ldots,x_{n}\\}$
[Definition 1.1 and Lemma 1.2]. For a more general Poisson algebra, the
definition of $\operatorname{div}(\delta)$ will be given in Definition 1.1
which is dependent on the volume form. Recall a Hamiltonian derivation of a
Poisson algebra $A$ is given by $H_{a}:=\\{a,-\\}$ for any $a\in A$. The
modular derivation of $A$ is defined by

(E0.0.4) ${\mathbf{m}}(a)~{}:=~{}-\operatorname{div}(H_{a})$

for all $a\in A$ [Definition 1.3]. We need the following lemma that concerns
the divergence of the modular derivation.

######  Lemma 0.1.

[Wa, Corollary 3.10] [LPV, Proposition 4.17] Let $A$ be a Poisson algebra with
volume form $\nu$ and ${\mathbf{m}}$ be the modular derivation of $A$
corresponding to $\nu$. Then $\operatorname{div}({\mathbf{m}})=0$.

###### Proof.

Following the notation of [Wa, Theorem 3.5], we denote ${\mathbf{m}}$ by
$\phi$ and $\nu$ by vol. By the proof of [Wa, Corollary 3.10],
${\mathcal{L}}_{{\mathbf{m}}}(\nu)=0$. Then, by Definition 1.1,
$\operatorname{div}({\mathbf{m}})=0$. ∎

According to the ideas of Dolgushev [Do], the modular derivation of a Poisson
algebra is corresponding to the Nakayama automorphism of a noetherian AS
regular algebra. Hence the above lemma is a Poisson version of [RRZ2,
Corollary 5.5] which says that the Nakayama automorphism of a noetherian AS
regular algebra has the homological determinant 1.

When $A$ is a polynomial algebra $\Bbbk[x_{1},\ldots,x_{n}]$ with any Poisson
structure, the definitions of the divergence $\operatorname{div}$ and the
modular derivation $\mathbf{m}$ are independent of choices of the volume form.
Here is one of main results of this paper.

######  Theorem 0.2.

Let $\delta$ be a graded Poisson derivation of a ${\mathbb{Z}}$-graded Poisson
polynomial algebra $A:=\Bbbk[x_{1},\ldots,x_{n}]$. Let ${\mathbf{n}}$ be the
modular derivation of $A^{\delta}$. Then

${\mathbf{n}}={\mathbf{m}}+(\sum_{i=1}^{n}\deg
x_{i})\delta-\operatorname{div}(\delta)E.$

Note that Theorem 0.2 holds even when ${\text{char}}\;\Bbbk>0$, see Remark
3.7.

If we consider the analogy between the modular derivation of a Poisson algebra
and the Nakayama automorphism of a graded skew Calabi-Yau algebra [Do],
Theorem 0.2 is a Poisson version of [RRZ1, Theorem 0.3]. Combining Theorem 0.2
with Lemma 0.1, we obtain

######  Corollary 0.3.

Let $A$ is a weighted graded Poisson algebra $\Bbbk[x_{1},\ldots,x_{n}]$ with
$\deg x_{i}>0$ for all $i$. Let $\delta=-{\frac{1}{\mathfrak{l}}}{\mathbf{m}}$
where ${\mathfrak{l}}=\sum_{i=1}^{n}\deg(x_{i})$. Then $A^{\delta}$ is
unimodular. As a consequence, (E0.0.1) holds.

Let $A$ be a ${\mathbb{Z}}$-graded Poisson algebra. Suppose $\delta$ is a
derivation of $A$ and $a,b,c$ are homogeneous elements of $A$. Let

$\displaystyle p(\\{-,-\\},\delta;a,b,c):=$
$\displaystyle|a|a[\delta(\\{b,c\\})-\\{\delta(b),c\\}-\\{b,\delta(c)\\}]$
(E0.3.1)
$\displaystyle-|b|b[\delta(\\{a,c\\})-\\{\delta(a),c\\}-\\{a,\delta(c)\\}]$
$\displaystyle+|c|c[\delta(\\{a,b\\})-\\{\delta(a),b\\}-\\{a,\delta(b)\\}].$

######  Definition 0.4.

Let $A$ be a ${\mathbb{Z}}$-graded Poisson algebra. A derivation $\delta$ of
$A$ is called semi-Poisson if $p(\\{-,-\\},\delta,a,b,c)=0$ for all
homogeneous elements $a,b,c$ in $A$.

It is clear that

${\text{Poisson derivation}}\Rightarrow{\text{ semi-Poisson
derivation}}\Rightarrow{\text{ derivation}}$

and opposite implications are not true [Example 2.6]. Let $Gspd(A)$ (resp.
$Gpd(A)$) be the set of graded semi-Poisson derivations (resp. graded Poisson
derivations) of degree 0. We prove the following

######  Theorem 0.5.

Let $A$ be a graded Poisson algebra $\Bbbk[x_{1},\ldots,x_{n}]$ with $\deg
x_{i}>0$ for all $i$.

  1. (1)

If $A$ is unimodular, then $Gspd(A)=Gpd(A)$.

  2. (2)

If $B$ is a twist of $A$, then $Gspd(A)=Gspd(B)$.

  3. (3)

$Gspd(A)$ is a finite-dimensional Lie algebra.

Now we introduce the rigidity of graded twisting of $A$, denoted by $rgt(A)$
(see Definition 4.3), to measure the complexity/rigidity of a Poisson
structure on $A$. We relate the rigidity with other properties. We say a
Poisson derivation $\phi$ of $A$ is ozone if $\phi(z)=0$ for all $z$ in the
Poisson center of $A$. It is obvious that every Hamiltonian derivation is
ozone, but the converse is not true in general. Recall that $E$ denotes the
Euler derivation.

Let $M$ be a ${\mathbb{Z}}$-graded $\Bbbk$-vector space. The Hilbert series of
$M$ is defined to be

(E0.5.1) $h_{M}(t)=\sum_{i\in{\mathbb{Z}}}(\dim M_{i})t^{i}.$

Let $PH^{i}(A)$ denote the $i$th Poisson cohomology of $A$ (E1.5.3). We have
the following result.

######  Theorem 0.6.

Let $\Bbbk$ be algebraically closed. Let $A$ be the Poisson algebra
$\Bbbk[x_{1},x_{2},x_{3}]$ with $\deg(x_{i})=1$ for $i=1,2,3$. Let $Z$ be the
Poisson center of $A$. Then the following are equivalent.

  1. (1)

$rgt(A)=0$.

  2. (2)

Any graded twist of $A$ is isomorphic to $A$.

  3. (3)

The Hilbert series of the graded vector space of Poisson derivations of $A$ is
$\frac{1}{(1-t)^{3}}$.

  4. (4)

$h_{PH^{1}(A)}(t)$ is $\frac{1}{1-t^{3}}$.

  5. (5)

$h_{PH^{1}(A)}(t)$ is equal to $h_{Z}(t)$.

  6. (6)

Every Poisson derivation $\phi$ has a decomposition

$\phi=zE+H_{a}$

where $z\in Z$ and $a\in A$. Here $z$ is unique and $a$ is unique up to a
central element.

  7. (7)

Every ozone derivation is Hamiltonian.

  8. (8)

$A$ is unimodular and the potential is irreducible.

  9. (9)

$h_{PH^{3}(A)}(t)-h_{PH^{2}(A)}(t)=t^{-3}$.

Some partial generalizations of the above theorem to the higher dimensional
case are given in Section 7. As an application, we have the following result.

######  Corollary 0.7.

Let $\Bbbk$ be algebraically closed. Let $A$ be the unimodular quadratic
Poisson structure on $\Bbbk[x,y,z]$ with irreducible potential $\Omega$. Then

  1. (1)

$\displaystyle h_{PH^{0}(A)}(t)=\frac{1}{1-t^{3}}$.

  2. (2)

$\displaystyle h_{PH^{1}(A)}(t)=\frac{1}{1-t^{3}}$.

  3. (3)

$\displaystyle h_{PH^{2}(A)}(t)=\frac{1}{t^{3}}(\frac{(1+t)^{3}}{1-t^{3}}-1)$.

  4. (4)

$\displaystyle h_{PH^{3}(A)}(t)=\frac{(1+t)^{3}}{t^{3}(1-t^{3})}$.

When the potential $\Omega$ has isolated singularities, the Poisson
cohomologies have been computed by several authors, see [Pe1, Pe2, Pi1, Pi2,
VdB] and the references therein. The above corollary is probably the first
computation of the Poisson cohomologies when $\Omega$ is irreducible, but does
not have isolated singularities.

The paper is organized as follows. Section 1 recalls some basic definitions
such as divergence and modular derivation. In Section 2 we introduce the
Poisson version of a graded twist. The proofs of Theorem 0.2 and Corollary 0.3
are given in Section 3. The rigidity of graded twisting is introduced in
Section 4 and Theorem 0.5 is proven there. In Sections 5 and 6 we compute the
rigidity of some Poisson structures on polynomial rings. Theorem 0.6 and
Corollary 0.7 are proved in Section 7.

##  1\. Preliminaries

In this section we recall several definitions such as divergence, modular
derivation, and Poisson cohomology. Other basic definitions about Poisson
algebras can be found in the book [LPV]. Everything in this section is well
known.

In Sections 1 and 2, let $\Bbbk$ be a base field of any characteristic. Let
$\Omega^{1}(A)$ be the module of Kähler differentials over $A$ [LPV, Sect.
3.2.1]. For each $k\geq 0$, let $\Omega^{k}(A)$ be
$\wedge^{p}_{A}\Omega^{1}(A)$ [LPV, Sect. 3.2.2]. Let $d=\operatorname{Kdim}A$
where $\operatorname{Kdim}$ denotes the Krull dimension. If $A$ is smooth and
$\Omega^{d}(A)$ is a free $A$-module with a generator $\nu$, then $\nu$ is
called a volume form of $A$. The differential $d:A\to\Omega^{1}(A)$ extends to
a well-defined differential of the complex $\Omega^{\bullet}(A)$ and the
complex $(\Omega^{\bullet},d)$ is called the algebraic de Rham complex of $A$.

For each $k\geq 0$, let ${\mathfrak{X}}^{k}(A)$ be the set of skew-symmetric
$k$-derivations of $A$. It is also true that

(E1.0.1) ${\mathfrak{X}}^{p}(A)\cong\operatorname{Hom}_{A}(\Omega^{p}(A),A)$

for all $p\geq 0$ [LPV, (3.15)].

For every element $P\in{\mathfrak{X}}^{p}(A)$, the internal product with
respect to $P$, denoted by $\iota_{P}$, is an $A$-module map

$\iota_{P}:\Omega^{\bullet}(A)\to\Omega^{\bullet-p}(A)$

which is determined by

(E1.0.2) $\iota_{P}(dF_{1}\wedge dF_{2}\wedge\cdots\wedge
dF_{k})=\begin{cases}0&k<p,\\\
\sum_{\sigma\in{\mathbb{S}}_{p,k-p}}sgn(\sigma)P[F_{\sigma(1)},\ldots,F_{\sigma(p)}]\\\
\qquad\qquad dF_{\sigma(p+1)}\wedge\cdots\wedge
dF_{\sigma(k)}\in\Omega^{k-p}(A)&k\geq p\end{cases}$

for all $dF_{1}\wedge dF_{1}\wedge\cdots\wedge dF_{k}\in\Omega^{k}(A)$. Here
${\mathbb{S}}_{p,q}\subset{\mathbb{S}}_{k}$ is the set of $(p,q)$-shuffles
with $p+q=k$.

For every $P\in{\mathfrak{X}}^{p}(A)$, the Lie derivative with respective to
$P$ is defined to be

(E1.0.3)
${\mathcal{L}}_{P}=[\iota_{P},d]:\Omega^{\bullet}(A)\to\Omega^{\bullet-p+1}(A),$

see [LPV, (3.49)]. Below is the definition of the divergence of a derivation.
In several definitions in this paper we assume that $A$ is a smooth Poisson
algebra with a fixed volume form $\nu$.

######  Definition 1.1.

[LPV, (4.20)] Let $\delta$ be a derivation of $A$, namely,
$\delta\in{\mathfrak{X}}^{1}(A)$. The divergence of $\delta$, denoted by
$\operatorname{div}(\delta)$, is an element in $A$ defined by the equation

(E1.1.1) ${\mathcal{L}}_{\delta}(\nu)=\operatorname{div}(\delta)\nu.$

It is clear that the divergence of $\delta$ is dependent on the volume form
$\nu$, but independent of the Poisson structure of $A$. The definition of the
divergence of a skew-symmetric $k$-derivation, for $k\geq 2$, can be found in
[LPV, Sect. 4.3.3].

Part (1) of the following lemma justifies the definition of the divergence
given in (E0.0.3). Let $G$ be an abelian group (or semigroup). A $G$-graded
algebra with a Poisson structure is called a $G$-graded Poisson structure if
$\deg(\\{a,b\\})=\deg a+\deg b$ for all homogeneous elements $a,b\in A$.

######  Lemma 1.2.

Let $A$ be a Poisson polynomial algebra $\Bbbk[x_{1},\ldots,x_{n}]$.

  1. (1)

(E1.2.1)
$\operatorname{div}(\delta)=\sum_{i=1}^{n}\frac{\partial\delta(x_{i})}{\partial
x_{i}}.$

  2. (2)

If $A$ is a ${\mathbb{Z}}$-graded Poisson polynomial algebra with $x_{i}$
homogeneous for all $i$ and if $\delta$ is graded (of degree 0), then
$\operatorname{div}(\delta)\in A_{0}$.

  3. (3)

If $A$ is a connected ${\mathbb{N}}$-graded Poisson polynomial algebra with
$\deg x_{i}>0$ for all $i$ and if $\delta$ is graded (of degree 0), then
$\operatorname{div}(\delta)\in\Bbbk$.

  4. (4)

Suppose, in addition to (3), $\deg x_{i}=1$ for all $i$. Let $\delta$ be a
derivation of $A$ (of degree 0). Write

$\delta(x_{i})=\sum_{j=1}^{n}c_{ij}x_{j}$

where $c_{ij}\in\Bbbk$ for all $1\leq i,j\leq n$. Then

(E1.2.2) $\operatorname{div}(\delta)=\sum_{i=1}^{n}c_{ii}.$

  5. (5)

Let $A$ be a ${\mathbb{Z}}$-graded Poisson algebra with
$\deg(x_{i})\in{\mathbb{Z}}$ and let $E$ be the Euler derivation of $A$
defined in the introduction. Then $\operatorname{div}(E)=\deg(\nu)$.

###### Proof.

(1) Since $\Bbbk[x_{1},\ldots,x_{n}]$, $\nu:=dx_{1}\wedge\cdots\wedge dx_{n}$
is a volume form. By the definition of the Lie derivative
${\mathcal{L}}_{\delta}$,

$\displaystyle{\mathcal{L}}_{\delta}\nu$
$\displaystyle=d(\sum_{i=1}(-1)^{i-1}\delta(x_{i})dx_{1}\wedge\cdots\wedge\widehat{dx_{i}}\wedge\cdots\wedge
dx_{n})$
$\displaystyle=\sum_{i=1}^{n}(-1)^{i-1}(\sum_{j=1}^{n}\frac{\partial\delta(x_{i})}{\partial
x_{j}}dx_{j})\wedge dx_{1}\wedge\cdots\wedge\widehat{dx_{i}}\wedge\cdots\wedge
dx_{n}$ $\displaystyle=(\sum_{i=1}^{n}\frac{\partial\delta(x_{i})}{\partial
x_{i}})dx_{1}\wedge\cdots\wedge dx_{n}$
$\displaystyle=(\sum_{i=1}^{n}\frac{\partial\delta(x_{i})}{\partial
x_{i}})\nu.$

Then the assertion follows.

(2) Since $\deg\delta=0$, $\deg\delta(x_{i})=\deg x_{i}$. As a consequence,
$\deg(\frac{\partial\delta(x_{i})}{\partial x_{i}})=0$. By part (1),
$\deg(\operatorname{div}\delta)=0$. The assertion follows.

(3) This follows from part (3) and the fact that $A_{0}=\Bbbk$.

(4) This follows from part (1) and the fact that
$\frac{\partial\delta(x_{i})}{\partial x_{i}}=c_{ii}$ for all $i$.

(5) In this case, $\nu=dx_{1}\wedge\cdots\wedge dx_{n}$ and $E(x_{i})=(\deg
x_{i})x_{i}$. The assertion follows from (E1.2.1). ∎

We recall the following definition.

######  Definition 1.3.

Let $A$ be a Poisson algebra with volume form $\nu$.

  1. (1)

[LPV, Definition 4.10] The modular derivation (or modular vector field) of $A$
associated to $\nu$ is defined to be

${\mathbf{m}}(a):=-\operatorname{div}H_{a}$

for all $a\in A$, or equivalently,

${\mathcal{L}}_{H_{a}}(\nu)=-{\mathbf{m}}(a)\nu.$

  2. (2)

[LPV, Definition 4.12] If ${\mathbf{m}}=0$ for some volume form $\nu$, then
$A$ is called unimodular.

If $A=\Bbbk[x_{1},\ldots,x_{n}]$, then ${\mathbf{m}}$ is independent of the
choice of the volume forms $\nu$.

Let us give an easy example.

######  Example 1.4.

Let $A$ be the Poisson polynomial algebra $\Bbbk[x_{1},x_{2}]$ with
$\\{x_{1},x_{2}\\}=x_{1}^{n}$ for some integer $n\geq 0$. It is easy to check
that ${\mathbf{m}}(x_{1})=0$ and that ${\mathbf{m}}(x_{2})=nx_{1}^{n-1}$. If
${\rm{char}}\;\Bbbk=p>0$ and $p\mid n$, then $A$ is unimodular. This is an
interesting fact. Now suppose $n=2$. Since
$\\{x_{1}x_{2},x_{2}\\}=x_{1}^{2}x_{2}$,
${\mathbf{m}}(\\{x_{1}x_{2},x_{2}\\})={\mathbf{m}}(x_{1}^{2}x_{2})=2x_{1}^{3}$.
As a consequence, $\operatorname{div}([\delta_{1},\delta_{2}])$ is in general
nonzero for any two derivations $\delta_{1},\delta_{2}$.

Next we review the Poisson cohomology. Let $(A,\pi)$ be a Poisson algebra. For
each $k$, ${\mathfrak{X}}^{k}(A)$ is the space of skew-symmetric
$k$-derivations of $A$. The Poisson coboundary map
$d_{\pi}:{\mathfrak{X}}^{\bullet}(A)\to{\mathfrak{X}}^{\bullet+1}(A)$ is
defined as follows. For any $Q\in{\mathfrak{X}}^{q}(A)$, where
$q\in{\mathbb{N}}$, we define

(E1.5.1) $\displaystyle d_{\pi}^{q}(Q)[F_{0},\ldots,F_{q}]:=$
$\displaystyle\sum_{i=0}^{q}(-1)^{i}\\{F_{i},Q[F_{0},\ldots,\widehat{F_{i}},\ldots,F_{q}]\\}$
$\displaystyle+\sum_{0\leq i\leq j\leq
q}(-1)^{i+j}Q[\\{F_{i},F_{j}],F_{0},\ldots,\widehat{F_{i}},\ldots,\widehat{F_{j}},\ldots,F_{q}],$

for all $F_{0},\cdots,F_{q}\in A$. In particular,

$\displaystyle d_{\pi}^{0}(Q)[F_{0}]$ $\displaystyle=\\{F_{0},Q\\},$
$\displaystyle d_{\pi}^{1}(Q)[F_{0},F_{1}]$
$\displaystyle=\\{F_{0},Q[F_{1}]\\}-\\{F_{1},Q[F_{0}]\\}-Q[\\{F_{0},F_{1}\\}],$
$\displaystyle d_{\pi}^{2}(Q)[F_{0},F_{1},F_{2}]$
$\displaystyle=\\{F_{0},Q[F_{1},F_{2}]\\}-\\{F_{1},Q[F_{0},F_{2}]\\}+\\{F_{2},Q[F_{0},F_{1}]\\}$
$\displaystyle\quad-Q[\\{F_{0},F_{1}\\},F_{2}]+Q[\\{F_{0},F_{2}\\},F_{1}]-Q[\\{F_{1},F_{2}\\},F_{0}]$

For $P\in{\mathfrak{X}}^{p}(A)$ and $Q\in{\mathfrak{X}}^{q}(A)$, the wedge
product $P\wedge Q\in{\mathfrak{X}}^{p+q}(A)$ is the skew-symmetric
$(p+q)$-derivation of $A$, defined by

$\displaystyle(P\wedge Q)$ $\displaystyle[F_{1},\cdots,F_{p+q}]:=$
$\displaystyle\sum_{\sigma\in{\mathbb{S}}_{p,q}}sgn(\sigma)P[F_{\sigma(1)},\ldots,F_{\sigma(p)}]\,Q[F_{\sigma(p+1)},\ldots,F_{\sigma(p+q)}],$

for all $F_{1},\cdots,F_{p+q}\in A$. In particular, if
$P\in{\mathfrak{X}}^{1}(A)$ and $Q\in{\mathfrak{X}}^{2}(A)$, then we have

(E1.5.2) $(P\wedge
Q)[F_{1},F_{2},F_{3}]=P[F_{1}]Q[F_{2},F_{3}]-P[F_{2}]Q[F_{1},F_{3}]+P[F_{3}]Q[F_{1},F_{2}].$

Therefore $({\mathfrak{X}}^{\ast}(A),\wedge,d)$ is a dga (differential graded
algebra). For each $q\geq 0$, the $q$-th Poisson cohomology of $A$ is defined
to be

(E1.5.3) $PH^{q}(A):=\frac{\ker d_{\pi}^{q}}{\text{im}d^{q-1}_{\pi}}.$

It is clear from the definition that the 1st Poisson cohomology of $A$ is

(E1.5.4) $PH^{1}(A):=\frac{{\text{the set of Poisson derivations}}}{{\text{the
set of Hamiltonian derivations}}}.$

If $A$ is a quadratic Poisson algebra $\Bbbk[x,y,z]$ (with
$\deg(x)=\deg(y)=\deg(z)=1$), then the complex
$({\mathfrak{X}}^{\bullet}(A),d_{\pi})$ is [Pi1, (15)]

$0\to A\to(A[1])^{\oplus 3}\to(A[2])^{\oplus 3}\to A[3]\to 0.$

By the additivity of Hilbert series, we have

(E1.5.5) $\sum_{i=0}^{3}(-1)^{i}h_{PH^{i}(A)}(t)=-t^{-3}.$

It is easy to check that

  1. (a)

the lowest degree of nonzero elements in $PH^{0}(A)$ is 0 and
$PH^{0}(A)_{0}=\Bbbk$.

  2. (b)

the lowest degree of nonzero elements in $PH^{1}(A)$ is $\geq-1$.

  3. (c)

the lowest degree of nonzero elements in $PH^{2}(A)$ is $\geq-2$.

  4. (d)

the lowest degree of nonzero elements in $PH^{3}(A)$ is $-3$ and
$PH^{3}(A)_{-3}=\Bbbk$.

If $A$ is further unimodular, then

  1. (e)

the lowest degree of nonzero elements in $PH^{2}(A)$ is $-2$ and
$PH^{2}(A)_{-2}=\Bbbk^{\oplus 3}$.

A natural operation on ${\mathfrak{X}}^{\bullet}(A)$ is the Schouten bracket

$[\cdot,\cdot]_{S}:{\mathfrak{X}}^{p}(A)\times{\mathfrak{X}}^{q}(A)\to{\mathfrak{X}}^{p+q-1}(A)$

for all $p,q\geq 0$. We refer to [LPV, Section 3.3.2] for the precise
definition. By [LPV, (4.5)],

$d_{\pi}(\cdot)=-[\cdot,\pi]_{S}.$

By [LPV, Proposition 3.7],
$({\mathfrak{X}}^{\bullet}(A),\wedge,[\cdot,\cdot]_{S})$ is a Gerstenhaber
algebra .

Let $A$ or $(A,\pi)$ be a Poisson algebra with Poisson bracket $\pi$. Let
$\xi$ be any nonzero scalar. We define a new Poisson bracket
$\pi_{\xi}:=\xi\pi$ or $\\{-,-\\}_{\xi}:=\xi\\{-,-\\}$. Then it is easy to see
that $A^{\prime}:=(A,\pi_{\xi})$ is indeed a Poisson algebra. In general,
$A^{\prime}$ is not isomorphic to $A$, but they are closely related as
follows.

######  Lemma 1.5.

Retain the notations as above. Let $d_{\pi}^{q}$ (resp.
$d_{\pi^{\prime}}^{q}$) be the differential of ${\mathfrak{X}}^{\bullet}(A)$
(resp. ${\mathfrak{X}}^{\bullet}(A^{\prime})$) as defined in (E1.5.1).

  1. (1)

$d_{\pi^{\prime}}^{q}=\xi d_{\pi}^{q}$ for all $q$.

  2. (2)

$\ker d_{\pi^{\prime}}^{q}=\ker d_{\pi}^{q}$ for all $q$.

  3. (3)

$\text{im}d_{\pi^{\prime}}^{q}=\text{im}d_{\pi}^{q}$ for all $q$.

  4. (4)

$PH^{q}(A)=PH^{q}(A^{\prime})$ for all $q$.

##  2\. Twists of graded Poisson algebras

Let $G$ be an abelian group and $A$ be a $G$-graded Poisson algebra (namely,
both the multiplication $\cdot$ and the Poisson bracket $\\{-,-\\}$ of $A$ are
graded of degree 0). We use $g$ for elements in $G$. If $a$ is a homogeneous
element in $A$, we use $|a|$ to denote its degree in $G$.

The aim of this section is to define a Poisson version of the graded twist of
graded associative algebras [Zh].

######  Definition 2.1.

Let $\delta:=\\{\delta_{g}\mid g\in G\\}$ be a set of graded derivations of
$A$ (of degree 0). We say $\delta$ is a Poisson twisting system if it
satisfies the following conditions

  1. (1)

For all $g,h\in G$,

(E2.1.1) $\delta_{g}\delta_{h}~{}=~{}\delta_{h}\delta_{g}.$

  2. (2)

For homogeneous elements $a,b\in A$,

(E2.1.2) $\delta_{|ab|}~{}=~{}\delta_{|a|}+\delta_{|b|}.$

  3. (3)

For homogeneous elements $a,b,c,\in A$,

(E2.1.3) $\displaystyle
a[\delta_{|a|}(\\{b,c\\})-\\{\delta_{|a|}(b),c\\}-\\{b,\delta_{|a|}(c)\\}]-b$
$\displaystyle[\delta_{|b|}(\\{a,c\\})-\\{\delta_{|b|}(a),c\\}-\\{a,\delta_{|b|}(c)\\}]$
$\displaystyle+c[\delta_{|c|}(\\{a,b\\})-\\{\delta_{|c|}(a),b\\}-\\{a,\delta_{|c|}(b)\\}]$
$\displaystyle=0.$

######  Remark 2.2.

  1. (1)

The definition of a Poisson twisting system is a “translation” of the twisting
system in the setting of graded associative algebras given in [Zh, Definition
2.1].

  2. (2)

Let

(E2.2.1) $\displaystyle p(\\{-,-\\},\delta\;a,b,c)~{}:=~{}$ $\displaystyle
a[\delta_{|a|}(\\{b,c\\})-\\{\delta_{|a|}(b),c\\}-\\{b,\delta_{|a|}(c)\\}]$
$\displaystyle-b[\delta_{|b|}(\\{a,c\\})-\\{\delta_{|b|}(a),c\\}-\\{a,\delta_{|b|}(c)\\}]$
$\displaystyle+c[\delta_{|c|}(\\{a,b\\})-\\{\delta_{|c|}(a),b\\}-\\{a,\delta_{|c|}(b)\\}].$

Then (E2.1.3) is equivalent to $p(\\{-,-\\};\delta;a,b,c)=0$. If each
$\delta_{g}$ is a Poisson derivation, it is automatic that
$p(\\{-,-\\},\delta,;a,b,c)=0$. The converse is not true, see Example 2.6.

  3. (3)

Suppose $G={\mathbb{Z}}$ and let $\phi=\delta_{1}$. By (E2.1.2),
$\delta_{n}=n\phi$. It is clear that

$p(\\{-,-\\},\delta\;a,b,c)=(E\wedge d_{\pi}^{1}(\phi))(a,b,c),$

which implies that (E2.1.3) is equivalent to $E\wedge d_{\pi}^{1}(\phi)=0$. By
[LPV, Sect. 4.3] and the fact that $d_{\pi}^{1}(E)=0$, the equation $E\wedge
d_{\pi}^{1}(\phi)=0$ is equivalent to $d_{\pi}^{2}(E\wedge\phi)=0$. Assume
that $a,b,c$ are homogeneous of degree one. Then

(E2.2.2) $\displaystyle p(\\{-,-\\},\delta;a,b,c)~{}=~{}$
$\displaystyle|a|a[\phi(\\{b,c\\})-\\{\phi(b),c\\}-\\{b,\phi(c)\\}]$
$\displaystyle-|b|b[\phi(\\{a,c\\})-\\{\phi(a),c\\}-\\{a,\phi(c)\\}]$
$\displaystyle+|c|c[\phi(\\{a,b\\})-\\{\phi(a),b\\}-\\{a,\phi(b)\\}]$

which agrees with $p(\\{-,-\\},\phi;a,b,c)$ as defined in (E0.3.1).

  4. (4)

Let $G={\mathbb{Z}}$ and $A$ be a ${\mathbb{Z}}$-graded Poisson algebra. A
convenient Poisson twisting system is constructed as follows. Let $\phi$ be a
graded Poisson derivation of $A$ (namely, $d_{\pi}^{1}(\phi)=0$). For each
$n\in{\mathbb{Z}}$, let $\delta_{n}:=n\phi$ and $\delta:=\\{\delta_{n}\mid
n\in{\mathbb{Z}}\\}$. Then (E2.1.1) and (E2.1.2) are obvious and (E2.1.3)
follows from the fact that $\delta_{n}$ is a Poisson derivation, see part (2)
or (3).

######  Example 2.3.

Let $G={\mathbb{Z}}/(n)$ for some positive integer $n$. Let $A$ be a
$G$-graded Poisson algebra and $\delta$ be a graded Poisson derivation of $A$.
Suppose $p:={\text{char}}\;\Bbbk$ is positive. If $p\mid n$, let
$\delta_{i}=i\delta$ for all $i\in G$. Then $\\{\delta_{i}\mid i\in G\\}$ is a
Poisson twisting system. If $p\nmid n$, then there is no nontrivial Poisson
twisting system for $A$.

Let $A$ be a $G$-graded Poisson algebra and let $\delta:=\\{\delta_{g}\mid
g\in G\\}$ be a system of derivations of $A$. We define

(E2.3.1) $\langle a,b\rangle:=\\{a,b\\}+a\delta_{|a|}(b)-b\delta_{|b|}(a)$

for all homogeneous elements $a,b\in A$.

######  Theorem 2.4.

Let $\delta:=\\{\delta_{g}\mid g\in G\\}$ be a set of graded derivations of
$A$ satisfying (E2.1.1) and (E2.1.2). Then $(A,\langle\cdot,\cdot\rangle)$ is
a Poisson algebra if and only if (E2.1.3) holds.

###### Proof.

If $G={\mathbb{Z}}$, there is a shorter proof using the Schouten bracket. For
a general abelian group $G$, we make the following direct computation.

Claim 1: $\langle\cdot,\cdot\rangle$ is skew-symmetric. This claim follows
immediately from (E2.3.1).

Claim 2: for every homogeneous element $a$, $\langle a,-\rangle$ is a
derivation.

For homogeneous elements $a,b,c$ in $A$, we have

$\displaystyle\langle a,bc\rangle$
$\displaystyle=\\{a,bc\\}+a\delta_{|a|}(bc)-bc\delta_{|bc|}(a)$
$\displaystyle=\\{a,b\\}c+\\{a,c\\}b+a(b\delta_{|a|}(c)+\delta_{|a|}(b)c)-bc\delta_{|bc|}(a)$
$\displaystyle\langle a,b\rangle c$
$\displaystyle=(\\{a,b\\}+a\delta_{|a|}(b)-b\delta_{|b|}(a))c$ $\displaystyle
b\langle a,c\rangle$
$\displaystyle=b(\\{a,c\\}+a\delta_{|a|}(c)-c\delta_{|c|}(a)).$

By the above and (E2.1.2), we obtain that

$\langle a,bc\rangle=\langle a,b\rangle c+b\langle a,c\rangle.$

Claim 3: $\langle\cdot,\cdot\rangle$ satisfies the Jacobi identity if and only
if (E2.1.3) holds. As a consequence, $(A,\langle\cdot,\cdot\rangle)$ is a
Poisson algebra if and only if (E2.1.3) holds.

For homogeneous elements $a,b,c$ in $A$, we have

$\displaystyle\langle a,\langle b,c\rangle\rangle$ $\displaystyle=\\{a,\langle
b,c\rangle\\}+a\delta_{|a|}(\langle b,c\rangle)-\langle
b,c\rangle\delta_{|bc|}(a)$
$\displaystyle=\\{a,(\\{b,c\\}+b\delta_{|b|}(c)-c\delta_{|c|}(b))\\}$
$\displaystyle\qquad+a\delta_{|a|}(\\{b,c\\}+b\delta_{|b|}(c)-c\delta_{|c|}(b))$
$\displaystyle\qquad\quad-(\\{b,c\\}+b\delta_{|b|}(c)-c\delta_{|c|}(b))\delta_{|bc|}(a)$
$\displaystyle=\\{a,\\{b,c\\}\\}$
$\displaystyle\quad+\\{a,b\\}\delta_{|b|}(c)+b\\{a,\delta_{|b|}(c)\\}$
$\displaystyle\quad-\\{a,c\\}\delta_{|c|}(b)-c\\{a,\delta_{|c|}(b)\\}$
$\displaystyle\qquad+a\delta_{|a|}(\\{b,c\\})$
$\displaystyle\qquad+a\delta_{|a|}(b)\delta_{|b|}(c)+ab\delta_{|a|}\delta_{|b|}(c)$
$\displaystyle\qquad-a\delta_{|a|}(c)\delta_{|c|}(b)-ac\delta_{|a|}\delta_{|c|}(b)$
$\displaystyle\qquad\quad-\\{b,c\\}\delta_{|bc|}(a)-b\delta_{|b|}(c)\delta_{|bc|}(a)+c\delta_{|c|}(b)\delta_{|bc|}(a)$

and

$\displaystyle\langle\langle a,b\rangle,c\rangle$ $\displaystyle=\langle
c,\langle b,a\rangle\rangle$ $\displaystyle=\\{c,\\{b,a\\}\\}$
$\displaystyle\quad+\\{c,b\\}\delta_{|b|}(a)+b\\{c,\delta_{|b|}(a)\\}$
$\displaystyle\quad-\\{c,a\\}\delta_{|a|}(b)-a\\{c,\delta_{|a|}(b)\\}$
$\displaystyle\qquad+c\delta_{|c|}(\\{b,a\\})$
$\displaystyle\qquad+c\delta_{|c|}(b)\delta_{|b|}(a)+cb\delta_{|c|}\delta_{|b|}(a)$
$\displaystyle\qquad-c\delta_{|c|}(a)\delta_{|a|}(b)-ca\delta_{|c|}\delta_{|a|}(b)$
$\displaystyle\qquad\quad-\\{b,a\\}\delta_{|ba|}(c)-b\delta_{|b|}(a)\delta_{|ba|}(c)+a\delta_{|a|}(b)\delta_{|ba|}(c)$

and

$\displaystyle\langle b,\langle a,c\rangle\rangle$
$\displaystyle=\\{b,\\{a,c\\}\\}$
$\displaystyle\quad+\\{b,a\\}\delta_{|a|}(c)+a\\{b,\delta_{|a|}(c)\\}$
$\displaystyle\quad-\\{b,c\\}\delta_{|c|}(a)-c\\{b,\delta_{|c|}(a)\\}$
$\displaystyle\qquad+b\delta_{|b|}(\\{a,c\\})$
$\displaystyle\qquad+b\delta_{|b|}(a)\delta_{|a|}(c)+ba\delta_{|b|}\delta_{|a|}(c)$
$\displaystyle\qquad-b\delta_{|b|}(c)\delta_{|c|}(a)-bc\delta_{|b|}\delta_{|c|}(a)$
$\displaystyle\qquad\quad-\\{a,c\\}\delta_{|ac|}(b)-a\delta_{|a|}(c)\delta_{|ac|}(b)+c\delta_{|c|}(a)\delta_{|ac|}(b).$

Using the Jacobi identity

$-\\{a,\\{b,c\\}\\}+\\{\\{a,b\\},c\\}+\\{b,\\{a,c\\}\\}=0,$

(E2.1.1), and (E2.1.2), we can simplify

$-\langle a,\langle b,c\rangle\rangle+\langle\langle
c,b\rangle,a\rangle+\langle b,\langle a,c\rangle\rangle$

to

$p(\\{-,-\\},\delta;a,b,c).$

Therefore $\langle\cdot,\cdot\rangle$ satisfies the Jacobi identity if and
only if $p(\delta;a,b,c)=0$. Claim 3 follows. The consequence is clear. ∎

######  Definition 2.5.

Let $\delta:=\\{\delta_{g}\mid g\in G\\}$ be a Poisson twisting system of a
$G$-graded Poisson algebra $A$. Then the new Poisson algebra
$(A,\langle-,-\rangle)$ given in Theorem 2.4 is called the twist of $A$ by
$\delta$ and denoted by $A^{\delta}$.

######  Example 2.6.

Let $A=\Bbbk[x,y]$ be a ${\mathbb{Z}}$-graded Poisson algebra defined by
$\\{x,y\\}=x^{2}$. Let $\phi$ be the derivation sending $x\to-x$ and $y\to
y-x$. Let $\delta_{n}=n\phi$. It is easy to see that

$\displaystyle d_{\pi}^{1}(\phi)[x,y]:$
$\displaystyle=-\phi[\\{x,y\\}]+\\{x,\phi[y]\\}+\\{\phi[x],y\\}$
$\displaystyle=-\phi[x^{2}]+\\{x,y-x\\}+\\{-x,y\\}=2x^{2}\neq 0$

which implies that $\phi$ is not a Poisson derivation.

We claim that $\delta:=\\{\delta_{n}\\}$ is a Poisson twisting system. Let $f$
be the derivation of $A$ determined by

$f(x)=0,\quad{\text{and}}\quad f(y)=-x.$

It is easy to verify that $f$ is a Poisson derivation. By Remark 2.2(4),
$f^{\prime}:=\\{nf\mid n\in{\mathbb{Z}}\\}$ is a Poisson twisting system, and
by Theorem 2.4, $A^{f^{\prime}}$ is equipped with a Poisson structure such
that

$\langle x,y\rangle=\\{x,y\\}+xf(y)-yf(x)=x^{2}-x^{2}-0=0.$

Therefore $A^{f^{\prime}}$ has trivial Poisson structure. Let $g$ be the
Poisson derivation of $A^{f^{\prime}}$ determined by

$g(x)=-x,\quad{\text{and}}\quad g(y)=y.$

Let $g^{\prime}=\\{ng\mid n\in{\mathbb{Z}}\\}$. By Remark 2.2(4), $g^{\prime}$
is a Poisson twisting system of $A^{f^{\prime}}$ and the Poisson structure of
$(A^{f^{\prime}})^{g^{\prime}}$ is determined by, for all homogeneous elements
$a,b\in A$,

$\displaystyle\\{a,b\\}_{new}:$ $\displaystyle=\langle
a,b\rangle+|a|ag(b)-|b|bg(a)$
$\displaystyle=\\{a,b\\}+|a|af(b)-|b|bf(a)+|a|ag(b)-|b|bg(a)$
$\displaystyle=\\{a,b\\}+ah_{|a|}(b)-bh_{|b|}(a)$

where $h_{n}=nf+ng$ for all $n\in{\mathbb{Z}}$. Since $f+g$ is a derivation of
$A$, by Theorem 2.4, $h^{\prime}:=\\{h_{n}\mid n\in{\mathbb{Z}}\\}$ is a
Poisson twisting system of $A$. It is clear that $\delta=h^{\prime}$. So
$\delta$ is a Poisson twisting system.

Since $\delta$ is a Poisson twisting system, by Remark 2.2(3), $\phi$ is a
graded semi-Poisson derivation. By the first paragraph, $\phi$ is not a
Poisson derivation.

######  Lemma 2.7.

Suppose $G$ is cyclic. Then the set of Poisson twisting systems of $A$ is a
$\Bbbk$-vector space.

###### Proof.

Let $\delta$ and $\varphi$ be two Poisson twisting systems. It is clear that
$c\delta$ is a Poisson twisting system for all $c\in\Bbbk$. It remains to show
$h:=\delta+\varphi$ is a Poisson twisting system.

Since $G$ is cyclic, $h_{n}=nh_{1}$. So (E2.1.1) is clear. Now (E2.1.2) and
(E2.1.3) hold as these are “linear” in terms of $\delta$. ∎

######  Remark 2.8.

If $G$ is ${\mathbb{Z}}^{2}$, then the set of Poisson twisting systems of
$(\Bbbk[x_{1},x_{2},x_{3}],0)$ with $\deg x_{1}=\deg x_{2}=(1,0)$ and $\deg
x_{3}=(0,1)$ is not a $\Bbbk$-vector space. To see this, we consider two
graded Poisson derivations $\delta_{1}$ and $\phi_{1}$ that are not commuting
(for example, $\delta_{1}:x_{1}\to x_{1},x_{2}\to 0,x_{3}\to 0$ and
$\phi_{1}:x_{1}\to x_{2},x_{2}\to 0,x_{3}\to 0$). Let
$\delta_{(n,m)}=n\delta_{1}$ and $\phi_{(n,m)}=m\phi_{1}$. It is easy to see
that both $\delta$ and $\phi$ are twisting systems of the $G$-graded Poisson
algebra $(\Bbbk[x_{1},x_{2},x_{3}],0)$. We define $\delta+\phi$ by
$(\delta+\phi)_{(n,m)}=n\delta_{1}+m\phi_{1}$ for all
$(n,m)\in{\mathbb{Z}}^{2}$. Since $\delta_{1}$ and $\phi_{1}$ are not
commuting, we see that (E2.1.1) fails for $\delta+\phi$.

As noted before a derivation $\delta$ of $A$ is Poisson if and only if
$d_{\pi}^{1}(\delta)=0$. By Definition 0.4, a graded derivation $\delta$ of a
${\mathbb{Z}}$-graded Poisson algebra $A$ is semi-Poisson if $E\wedge
d_{\pi}^{1}(\delta)=0$.

Next we show that the Poisson twisting systems induce an equivalence relation.
Let $A$ be a $G$-graded commutative algebra. Two graded Poisson structures
$\pi$ and $\pi^{\prime}$ on $A$ are called equivalent if $(A,\pi^{\prime})$ is
a graded twist of $(A,\pi)$. In this case we write
$(A,\pi)\sim(A,\pi^{\prime})$.

######  Proposition 2.9.

Suppose $G$ is cyclic. Then $\sim$ is an equivalence relation.

###### Proof.

It is clear that $(A,\pi)\sim(A,\pi)$ by taking the trivial Poisson twisting
system $\delta$. So $\sim$ is reflexive.

To prove the symmetry of $\sim$, we suppose that $(A,\pi^{\prime})$ is a
graded twist of $(A,\pi)$ by $\delta$. We claim that $-\delta$ is a Poisson
twisting system of $(A,\pi^{\prime})$. If this claim is proved, then it is
obvious that $(A,\pi^{\prime})^{-\delta}=(A,\pi)$ as desired. It remains to
show that $-\delta$ satisfies (E2.1.1), (E2.2.1), and (E2.1.3). The first two
are easy. For the last one, we compute

$\displaystyle\delta_{|a|}(\langle b,c\rangle)$
$\displaystyle-\langle\delta_{|a|}(b),c\rangle-\langle
b,\delta_{|a|}(c)\rangle$
$\displaystyle=\delta_{|a|}(\\{b,c\\}+b\delta_{|b|}(c)-c\delta_{|c|}(b))$
$\displaystyle\quad-[\\{\delta_{|a|}(b),c\\}+\delta_{|a|}(b)\delta_{|b|}(c)-c\delta_{|c|}\delta_{|a|}(b)]$
$\displaystyle\quad-[\\{b,\delta_{|a|}(c)\\}+b\delta_{|b|}\delta_{|a|}(c)-\delta_{|a|}(c)\delta_{|c|}(b)]$
$\displaystyle=\delta_{|a|}(\\{b,c\\})-\\{\delta_{|a|}(b),c\\}-\\{b,\delta_{|a|}(c)\\}$
$\displaystyle\quad+\delta_{|a|}(b)\delta_{|b|}(c)+b\delta_{|a|}\delta_{|b|}(c)-\delta_{|a|}(c)\delta_{|c|}(b)-c\delta_{|a|}\delta_{|c|}(b)$
$\displaystyle\quad-\delta_{|a|}(b)\delta_{|b|}(c)+c\delta_{|c|}\delta_{|a|}(b)-b\delta_{|b|}\delta_{|a|}(c)+\delta_{|a|}(c)\delta_{|c|}(b)$
$\displaystyle=\delta_{|a|}(\\{b,c\\})-\\{\delta_{|a|}(b),c\\}-\\{b,\delta_{|a|}(c)\\}$

which implies that

$\displaystyle p(\langle-,-\rangle,-\delta;a,b,c)$
$\displaystyle=p(\\{-,-\\},-\delta;a,b,c)=0.$

Therefore $-\delta$ is a Poisson twisting system of $A^{\delta}$ and
$(A^{\delta})^{-\delta}=A$. So $\sim$ is symmetric.

The proof above does not use the hypothesis that $G$ is cyclic. The following
part of the proof uses that hypothesis.

To prove the transitivity of $\sim$, we use the idea given in Example 2.6.
Suppose $\delta$ is a Poisson twisting system of $A$ and $\phi$ a Poisson
twisting system of $A^{\delta}$. Let
$\sigma:=\\{\sigma_{g}:=\delta_{g}+\phi_{g}\mid g\in G\\}$.

Since $G$ is cyclic, $\sigma_{n}=n\sigma_{1}$ by definition for all $n\in G$
(and there might be some restriction on ${\text{char}}\;\Bbbk$ when $G$ is
finite). Therefore (E2.1.1)-(E2.1.2) are obvious. Define

$\\{a,b\\}_{new}=\\{a,b\\}+a\sigma_{|a|}(b)-b\sigma_{|b|}(a)=\langle
a,b\rangle+a\phi_{|a|}(b)-b\phi_{|b|}(a).$

Then $\\{-,-\\}_{new}$ is the Poisson bracket of $(A^{\delta})^{\phi}$. By
Theorem 2.4, $\sigma$ is a Poisson twisting system of $A$ and
$A^{\sigma}=(A^{\delta})^{\phi}$. Therefore $\sim$ is transitive. ∎

######  Remark 2.10.

Let $G$ be ${\mathbb{Z}}^{2}$ and $A:=\Bbbk[x_{0},x_{1},x_{2},x_{3}]$ with
$\deg x_{i}=(1,0)$ for $i=0,1,2$ and $\deg x_{3}=(0,1)$. We claim that $\sim$
is not an equivalence relation among the Poisson structures on $A$. We use $Y$
for the ${\mathbb{Z}}^{2}$-graded Poisson algebra with trivial Poisson
structure.

Let $\delta_{1}$ be the Poisson derivation of $Y$ sending $x_{0}\to 0$,
$x_{1}\to x_{2}$, $x_{2}\to 0$, and $x_{3}\to 0$. Let $\phi_{1}$ be the
Poisson derivation of $Y$ sending $x_{0}\to 0$, $x_{1}\to x_{1}$, $x_{2}\to 0$
and $x_{3}\to 0$. We define two Poisson twisting systems as follows. Let
$\delta:=\\{\delta_{(n,m)}=n\delta_{1}\\}$ and
$\phi:=\\{\phi_{(n,m)}=m\phi_{1}\\}$. Since $\delta_{1}$ and $\phi_{1}$ are
Poisson derivations, it is easy to verify that $\delta$, $-\delta$ and $\phi$
are Poisson twisting systems. By Theorem 2.4, $X:=Y^{-\delta}$ is a Poisson
algebra and by the first part of the proof of Proposition 2.9, $\delta$ is a
Poisson twisting system of $X$ and $Y=X^{\delta}$. So we have $X\sim Y$. Let
$Z=Y^{\phi}$. Then $Y\sim Z$. We claim that $X\not\sim Z$. Suppose on the
contrary that $X\sim Z$. Then $Z=X^{h}$ for some Poisson twisting system $h$
of $X$. Applying (E2.3.1) to pairs of elements of the form $(x_{i},x_{j})$ for
all $0\leq i,j\leq 3$, we see that $h=\delta+\phi$. But it is clear that
(E2.1.1) fails for $\delta+\phi$ as
$\delta_{1}\phi_{1}\neq\phi_{1}\delta_{1}$. Therefore there is no Poisson
twisting system $h$ such that $Z=X^{h}$ as desired.

This example suggests that there should be a more general definition of
twisting systems that induce an equivalence relation $\sim$.

We conclude this section with some examples.

######  Example 2.11.

Here are three examples of twists of graded Poisson algebras.

  1. (1)

Let $A$ be the Poisson polynomial ring $\Bbbk[x_{1},\ldots,x_{n}]$ with
trivial Poisson bracket. Consider $A$ as a ${\mathbb{Z}}^{n}$-graded algebra
with $\deg(x_{i})=e_{i}$ where
$e_{i}=(0,\ldots,0,1,0,\ldots,0)\in{\mathbb{Z}}^{n}$ with $1$ in the $i$th
position. Let $\\{p_{ij}\mid 1\leq i<j\leq n\\}$ be a subset of $\Bbbk$. For
each $i$, define a ${\mathbb{Z}}^{n}$-graded Poisson derivation $\delta_{i}$
by

$\delta_{i}(x_{j})=\begin{cases}p_{ij}x_{j}&j>i,\\\ 0&j\leq i.\end{cases}$

For each $(a_{1},\ldots,a_{n})\in{\mathbb{Z}}^{n}$, let
$\delta_{(a_{1},\ldots,a_{n})}=\sum_{i=1}^{n}a_{i}\delta_{i}$. Since each
$\delta_{(a_{1},\ldots,a_{n})}$ is a graded Poisson derivation of $A$, it is
easy to see that

$\delta:=\\{\delta_{(a_{1},\ldots,a_{n})}\mid(a_{1},\ldots,a_{n})\in{\mathbb{Z}}^{n}\\}$

is a twisting system of $A$. By (E2.3.1), the Poisson bracket of the new
Poisson algebra $A^{\delta}$ is determined by

$\langle
x_{i},x_{j}\rangle=x_{i}\delta_{i}(x_{j})-x_{j}\delta_{j}(x_{i})=p_{ij}x_{i}x_{j}\quad{\text{for
all $i<j$.}}$

  2. (2)

Let $A$ be the Poisson polynomial ring $\Bbbk[x_{1},\ldots,x_{n}]$ with
trivial Poisson bracket. Consider $A$ as a ${\mathbb{Z}}$-graded algebra with
$\deg(x_{i})=1$ for all $i$. Let $\delta_{1}$ be a Poisson derivation of $A$
determined by

$\delta_{1}(x_{i})=\begin{cases}-x_{i-1}&i>1,\\\ 0&i=1.\end{cases}$

Let $\delta:=\\{\delta_{d}:=d\delta_{1}\mid\forall\;d\in{\mathbb{Z}}\\}.$
Since $\delta_{1}$ is a graded Poisson derivation, $\delta$ is a twisting
system of $A$. By (E2.3.1), the Poisson bracket of the new Poisson algebra
$A^{\delta}$ is determined by

$\langle
x_{i},x_{j}\rangle=x_{i}\delta(x_{j})-x_{j}\delta(x_{i})=-x_{i}x_{j-1}+x_{j}x_{i-1}\quad{\text{for
all $i<j$.}}$

When $n=2$, the Poisson bracket of $A^{\delta}$ is determined by

$\langle x_{2},x_{1}\rangle=x_{1}^{2}.$

  3. (3)

Suppose that ${\text{char}}\;\Bbbk=0$. Let $A$ be the $n$th Weyl Poisson
algebra

$\Bbbk[x_{1},\ldots,x_{n},y_{1},\ldots,y_{n}]$

with Poisson bracket determined by

$\\{x_{i},y_{j}\\}=\delta_{ij},\quad\\{x_{i},x_{j}\\}=0,\quad\\{y_{i},y_{j}\\}=0$

for all $1\leq i,j\leq n$. Here $\delta_{ij}$ is the usual Kronecker delta.
Let $M=(m_{ij})_{n\times n}$ be any $n\times n$-matrix over $\Bbbk$. We
consider $A$ as a ${\mathbb{Z}}$-graded Poisson algebra with $\deg(x_{i})=1$
and $\deg(y_{i})=-1$ for all $1\leq i\leq n$. Let $\delta_{1}$ be the graded
Poisson derivation of $A$ determined by

$\delta_{1}(x_{i})=\sum_{j=1}^{n}(-m_{ij})x_{j},\quad\delta_{1}(y_{i})=\sum_{j=1}^{n}m_{ji}y_{j}$

for all $i$. Since $\delta_{1}$ is a graded Poisson derivation,
$\delta:=\\{n\delta_{1}\mid n\in{\mathbb{Z}}\\}$ is a twisting system of $A$.
By (E2.3.1), the Poisson bracket of the new Poisson algebra $A^{\delta}$ is
determined by $\langle x_{i},x_{j}\rangle=0=\langle y_{i},y_{j}\rangle$ and

(E2.11.1) $\langle
x_{i},y_{j}\rangle=\delta_{ij}+x_{i}\delta_{1}(y_{j})-y_{j}\delta_{-1}(x_{i})=\delta_{ij}+x_{i}(\sum_{s=1}^{n}m_{sj}y_{s})-y_{j}(\sum_{s=1}^{n}m_{is}x_{s})$

for all $i<j$.

We claim that $A^{\delta}$ is Poisson simple. To avoid some tedious analysis,
we only give a sketch proof. The first step is to assume $\Bbbk$ is
algebraically closed by replacing $\Bbbk$ by its algebraic closure. Since
$\Bbbk$ is algebraically closed, we can assume $M$ is upper triangular,
namely, $m_{ij}=0$ for all $i>j$. Let $M^{\prime}$ be the matrix after
deleting the diagonal entries of $M$ and we can similarly define a twisting
system $\delta^{\prime}$ using $M^{\prime}$. By (E2.11.1), one sees that
$A^{\delta}=A^{\delta^{\prime}}$. In other words, we can assume that $M$ is
strictly upper triangular. Now we can assume that $M$ is a block matrix with
diagonal block equal to

$\begin{pmatrix}0&1&0&\cdots&0\\\ 0&0&1&\cdots&0\\\
\cdots&\cdots&\cdots&\cdots&\cdots\\\ 0&0&0&\cdots&1\\\
0&0&0&\cdots&0\end{pmatrix}.$

To illustrate the idea, we assume $n=2$ and $M$ is given as above. (For
general $M$, the proof is more complicated.) The Poisson structure of
$A^{\delta}$ is determined by

$\displaystyle\langle x_{1},x_{2}\rangle$ $\displaystyle=0,$
$\displaystyle\langle y_{1},y_{2}\rangle$ $\displaystyle=0,$
$\displaystyle\langle x_{1},y_{1}\rangle$ $\displaystyle=1-x_{2}y_{1}$
$\displaystyle\langle x_{1},y_{2}\rangle$
$\displaystyle=x_{1}y_{1}-x_{2}y_{2}$ $\displaystyle\langle
x_{2},y_{1}\rangle$ $\displaystyle=0$ $\displaystyle\langle
x_{2},y_{2}\rangle$ $\displaystyle=1+x_{2}y_{1}$

Now we are ready to show that every nonzero Poisson ideal $I$ of $A^{\delta}$
contains $1$. Let $f$ be a nonzero element in $I$. We assume that the
$y_{2}$-degree of $f$ is minimal and that $f\not\in\Bbbk$. Write
$f=\sum_{i\geq 0}f_{i}y_{2}^{i}$ where $f_{i}\in\Bbbk[x_{1},x_{2},y_{1}]$ for
all $i$. If $f_{i}\neq 0$ for some $i>0$, then $H_{x_{2}}(f)=\sum_{i\geq
0}f_{i}iy_{2}^{i-1}(1+x_{2}y_{1})\neq 0$. It is clear that the $y_{2}$-degree
of $H_{x_{2}}(f)$ is smaller than the $y_{2}$-degree of $f$, yielding a
contradiction. Therefore $y_{2}$-degree of $f$ is zero, namely, $0\neq f\in
I\cap\Bbbk[x_{1},x_{2},y_{1}]$. Similarly, by using $H_{y_{1}}$, one can show
that there is an element $0\neq f\in\Bbbk[x_{2},y_{1}]\cap I$. Write
$f=\sum_{i=0}^{s}g_{i}y_{1}^{i}$ where $g_{i}\in\Bbbk[x_{2}]$ for all $i$.
Assume that $f$ has minimal $y_{1}$-degree. If the $y_{1}$-degree of $f$, say
$s$, is positive, then

$H_{x_{1}}(f)+sfx_{2}=\sum_{i=0}^{s}g_{i}iy_{1}^{i-1}+\sum_{i=0}^{s}(s-i)g_{i}y_{1}^{i}x_{2}$

which has $y_{1}$-degree $s-1$, yielding a contradiction. Therefore $0\neq
f\in\Bbbk[x_{2}]\cap I$. Write $f=\sum_{i=a}^{b}c_{i}x_{2}^{i}$ where
$c_{a}c_{b}\neq 0$. Assume that $b-a\geq 0$ is minimal. Next we show that
$b-a=0$. Suppose to the contrary that $a<b$. Define
$\Phi(f)=-H_{y_{2}}(f)-\deg(f)fy_{1}$. Then

$\Phi(f)=\sum_{i=a}^{b}c_{i}ix_{2}^{i-1}+\sum_{i=a}^{b}c_{i}(i-b)x_{2}^{i}y_{1}.$

Then

$\displaystyle H_{x_{1}}(\Phi(f))+\Phi(f)x_{2}$
$\displaystyle=(\sum_{i=a}^{b}c_{i}(i-b)x_{2}^{i})(1-x_{2}y_{1})$
$\displaystyle\quad+\sum_{i=a}^{b}c_{i}ix_{2}^{i}+\sum_{i=a}^{b}c_{i}(i-b)x_{2}^{i}x_{2}y_{1}$
$\displaystyle=\sum_{i=a}^{b}c_{i}(i-b)x_{2}^{i}+\sum_{i=a}^{b}c_{i}ix_{2}^{i}$
$\displaystyle=\sum_{i=a}^{b}c_{i}(2i-b)x_{2}^{i}$
$\displaystyle=2f^{\prime}(x_{2})x_{2}-bf\in\Bbbk[x_{2}]\cap I.$

This implies that $x_{2}f^{\prime}\in I$ and a possible linear combination of
$f$ and $x_{2}f^{\prime}$ gives a nonzero polynomial with smaller $b-a$,
yielding a contradiction. Therefore we can assume $f=x^{a}$. In this case,
$-H_{y_{2}}(f)=ax_{2}^{a-1}(1+x_{2}y_{1})=ax_{2}^{a-1}+afy_{1}$. This implies
that $x^{a-1}\in I$ when $a\neq 0$. An induction argument shows that $1\in I$
as required.

##  3\. Proofs of Theorem 0.2 and Corollary 0.3

For the rest of the paper we assume that ${\rm{char}}\;\Bbbk=0$. First we
recall the definition of divergence of a skew symmetric $k$-derivation for
$k\geq 0$ [LPV, Sect. 4.4.3]. A special case is given in Definition 1.1. For
$P\in{\mathfrak{X}}^{p}(A)$, the internal product $\iota_{P}$ is defined at
the beginning of Section 1.

Let $\nu$ be a volume form of $A$ which is in $\Omega^{d}(A)$ with $d$ being
the top degree of nonzero $\Omega^{d}(A)$. We define the star operator

$\star(=\star_{A}):{\mathfrak{X}}^{\bullet}(A)\to\Omega^{d-\bullet}(A)$

as follows: for each $k\geq 0$ and $Q\in{\mathfrak{X}}^{k}$, we set

$\star_{A}Q:=\iota_{Q}\nu.$

So $\star_{A}$ is a $\Bbbk$-linear map from
${\mathfrak{X}}^{k}(A)\to\Omega^{d-k}(A)$ for each $k$. It follows from
(E1.0.2) that $\star_{A}$ is an $A$-linear map.

######  Lemma 3.1.

Let $B$ be a smooth affine domain of dimension $g$ with volume form $\nu$.
Then $\star_{B}$ is an isomorphism.

###### Proof.

To prove that $\star$ is an isomorphism, it suffices to show that
$\star_{B}\otimes_{B}B_{\mathfrak{m}}$ is an isomorphism for all maximal
ideals ${\mathfrak{m}}$ of $B$. Let $A$ be the local ring $B_{\mathfrak{m}}$.
Then $A$ is a regular local ring of global dimension, Krull dimension and
transcendence degree $n$. Since all the operations commute with the
localization,

$\star_{B}\otimes_{B}B_{\mathfrak{m}}=\star_{B}\otimes_{B}A=\star_{A}.$

Now we assume that $A$ is local with maximal ideal ${\mathfrak{m}}$ generated
by $\\{x_{1},\ldots,x_{n}\\}$. Then $\Omega^{1}(A)$ is a free module of rank
$g$. Write $\Omega^{1}(A)=\oplus_{i=1}^{n}Adx_{i}$. Then for each $k\geq 0$,
$\Omega^{k}(A)$ is a free $A$-module with basis

$\\{d_{i_{1},\cdots,i_{k}}:=dx_{i_{1}}\wedge\cdots\wedge dx_{i_{k}}\mid 1\leq
i_{1}<\cdots<i_{k}\leq n\\}$

and that, via (E1.0.1), ${\mathfrak{X}}^{k}(A)$ is a free $A$-module with
basis as in (E1.0.1)

$\\{\partial_{i_{1},\ldots,i_{k}}:={\frac{\partial\;}{\partial
x_{i_{1}}}}\wedge\cdots\wedge{\frac{\partial\;}{\partial x_{i_{k}}}}\mid 1\leq
i_{1}<\cdots<i_{k}\leq n\\}.$

Recall that $\nu=a\,d_{1,2,\ldots,n}$ for an invertible element $a\in A$.
(Usually we write $\nu_{0}=d_{1,2,\ldots,n}$). By definition,

$\displaystyle\star_{A}\partial_{i_{1},\ldots,i_{k}}$
$\displaystyle=\iota_{\partial_{i_{1},\ldots,i_{k}}}(a\nu_{0})=a\iota_{d_{i_{1},\ldots,i_{k}}}(d_{1,2,\cdots,n})$
$\displaystyle=\sum_{\sigma\in{\mathbb{S}}_{k,n-k}}sgn(\sigma)a\partial_{i_{1},\ldots,i_{k}}[x_{\sigma(1)},\ldots,x_{\sigma(k)}]d_{1,\ldots,\widehat{\sigma(1)},\ldots,\widehat{\sigma(k)},\ldots,n}$
$\displaystyle=\pm
a\,d_{1,\ldots,\widehat{i_{1}},\cdots,\widehat{i_{k}},\ldots,n}$

where $\pm
1=sgn(\\{i_{1},\ldots,i_{k},1,\ldots,\widehat{i_{1}},\ldots,\widehat{i_{k}},\ldots,n\\})$.
Therefore $\star_{A}$ is an isomorphism as desired. ∎

######  Definition 3.2.

We say $A$ is a standard Poisson algebra if $A$ is an affine smooth
${\mathbb{Z}}$-graded Poisson domain with a homogeneous volume form $\nu$
(with $\deg(\nu)$ not necessarily zero) and $\star$ is an isomorphism.

Note that every polynomial Poisson algebra $\Bbbk[x_{1},\ldots,x_{n}]$ is
standard (even when ${\rm{char}}\;\Bbbk>0$). Lemma 3.1 provides another class
of such algebras. Now we assume that $A$ is standard of dimension $n$. By
[LPV, Sect. 4.4.3], the divergence operator with respect to the volume form
$\nu$ is a graded $\Bbbk$-linear map of degree $-1$,

$\operatorname{div}:{\mathfrak{X}}^{\bullet}(A)\to{\mathfrak{X}}^{\bullet-1}(A),$

which makes the following diagram commutes

$\begin{CD}{\mathfrak{X}}^{\bullet}(A)@>{\star}>{}>\Omega^{n-\bullet}(A)\\\
@V{\operatorname{div}}V{}V@V{}V{d}V\\\
{\mathfrak{X}}^{\bullet-1}(A)@>{\star}>{}>\Omega^{n-\bullet+1}(A)\end{CD}$

Since $A$ is standard, the star operator $\star$ is an $A$-linear isomorphism.
Now we have the following lemmas which were proved in [LPV, Sect. 4.4.3].

######  Lemma 3.3.

[LPV, Proposition 4.16] Suppose $(A,\pi)$ is standard with volume form $\nu$.
Let $\delta$ and $\phi$ be two derivations of $A$. Then

$\operatorname{div}(\delta\wedge\phi)=\operatorname{div}(\phi)\delta-\operatorname{div}(\delta)\phi-[\delta,\phi].$

The following lemma gives another proof of Lemma 0.1.

######  Lemma 3.4.

[LPV, Proposition 4.17] Suppose $(A,\pi)$ is standard with volume form $\nu$.
Let ${\mathbf{m}}$ be the modular derivation of $A$. Then

${\mathbf{m}}=-\operatorname{div}(\pi).$

As a consequence, the divergence of ${\mathbf{m}}$ is zero.

######  Question 3.5.

It is not clear how to handle nonaffine smooth domain $A$ as the proof of
Lemma 3.1 uses the fact $A$ is affine.

######  Theorem 3.6.

Suppose $(A,\pi)$ is standard with volume form $\nu$. Let $\delta$ be a graded
semi-Poisson derivation of $A$. Let ${\mathbf{m}}$ (respectively,
${\mathbf{n}}$) be the modular derivation of $A$ (respectively $A^{\delta}$).
Then

${\mathbf{n}}={\mathbf{m}}+(\operatorname{div}E)\delta-(\operatorname{div}\delta)E.$

###### Proof.

Let $\pi^{\prime}$ be the Poisson structure of $A^{\delta}$. By (E2.3.1),

$\pi^{\prime}=\pi+E\wedge\delta.$

By Lemmas 3.3 and 3.4, we have

$\displaystyle{\mathbf{n}}$
$\displaystyle=-\operatorname{div}(\pi^{\prime})=-\operatorname{div}(\pi)-\operatorname{div}(E\wedge\delta)$
$\displaystyle={\mathbf{m}}+\operatorname{div}(E)\delta-\operatorname{div}(\delta)E-[\delta,E].$

The assertion follows as $[\delta,E]=0$ for each graded derivation $\delta$. ∎

###### Proof of Theorem 0.2.

Let $A$ be a ${\mathbb{Z}}$-graded Poisson algebra $\Bbbk[x_{1},\ldots,x_{n}]$
with each $x_{i}$ homogeneous. By Lemma 1.2(5),
$\operatorname{div}(E)=\deg(\nu)=\sum_{i=1}^{n}\deg(x_{i})=:{\mathfrak{l}}$.
Now the assertions follow from Theorem 3.6. ∎

######  Remark 3.7.

There is a different proof of Theorem 0.2 without using Theorem 3.6 (details
are omitted). In fact, the different proof does not use the hypothesis that
${\text{char}}\;\Bbbk=0$.

The following result is an immediate consequence of Theorem 3.6.

######  Theorem 3.8.

Suppose $(A,\pi)$ is standard with volume $d$-form $\nu$. Assume that
$\operatorname{div}(E)\in\Bbbk$ is nonzero. Let ${\mathbf{m}}$ be the modular
derivation of $A$ and let
$\delta=-{\frac{1}{\operatorname{div}(E)}}{\mathbf{m}}$. Then
$(A^{\delta},\pi^{\prime})$ is unimodular and

$\pi=\pi^{\prime}+\frac{1}{\operatorname{div}(E)}\,E\wedge{\mathbf{m}.}$

In particular, we have $\mathcal{L}_{\delta}(\alpha)=0$ where
$\alpha=\star\pi^{\prime}$ is the closed differential $(d-2)$-form associated
with the unimodular Poisson structure $\pi^{\prime}$ on $A$ .

###### Proof.

Let ${\mathbf{n}}$ be the modular derivation of $(A^{\delta},\pi^{\prime})$.
By Theorem 3.6 and the fact
$\delta=-\frac{1}{\operatorname{div}(E)}{\mathbf{m}}$,

${\mathbf{n}}={\mathbf{m}}+\operatorname{div}(E)\delta-(\operatorname{div}\delta)E=-(\operatorname{div}\delta)E=0$

where the last equation follows from $\operatorname{div}{\mathbf{m}}=0$ [Lemma
0.1]. Therefore $(A^{\delta},\pi^{\prime})$ is unimodular. By (E2.3.1), for
all $a,b\in A$,

$\displaystyle\pi^{\prime}(a,b)$ $\displaystyle=\langle
a,b\rangle=\\{a,b\\}+a\delta_{|a|}(b)-b\delta_{|b|}(a)$
$\displaystyle=\pi(a,b)+|a|a\delta(b)-|b|b\delta(a)=\pi(a,b)+E(a)\delta(b)-\delta(a)E(b)$

which implies that

$\pi^{\prime}=\pi+E\wedge\delta$

which is equivalent to the above assertion. Finally by [LPV, Proposition
3.11(2)] we have

$\displaystyle\mathcal{L}_{\delta}(\alpha)=$
$\displaystyle\mathcal{L}_{\delta}(\iota_{\pi^{\prime}}(\nu))=\iota_{\pi^{\prime}}(\mathcal{L}_{\delta}(\nu))+\iota_{[\delta,\pi^{\prime}]_{S}}(\nu)$
$\displaystyle=$
$\displaystyle\iota_{\pi^{\prime}}(\operatorname{div}(\delta)\nu)+\iota_{[\delta,\pi^{\prime}]_{S}}(\nu)=\iota_{[\delta,\pi^{\prime}]_{S}}(\nu)$

One can easily check that $\delta$ is also a Poisson derivation of
$(A^{\delta},\pi^{\prime})$. So
$[\delta,\pi^{\prime}]_{S}=-d_{\pi^{\prime}}(\delta)=0$ and we get
$\mathcal{L}_{\delta}(\alpha)=0$. ∎

###### Proof of Corollary 0.3.

Let $A$ be a ${\mathbb{Z}}$-graded Poisson algebra $\Bbbk[x_{1},\cdots,x_{n}]$
with each $x_{i}$ homogeneous. By lemma 1.2(5),
${\mathfrak{l}}=\sum_{i=1}^{n}\deg(x_{i})$. Now the assertions follow from
Theorem 3.8. ∎

##  4\. Rigidity of graded twisting

Let $A$ be a ${\mathbb{Z}}$-graded Poisson algebra. Recall that the set of
graded semi-Poisson derivations (resp. graded Poisson derivations) of $A$ of
degree 0 is denoted by $Gspd(A)$ (resp. $Gpd(A)$). We first prove Theorem 0.5.

######  Lemma 4.1.

Let $A$ be a ${\mathbb{Z}}$-graded Poisson algebra $\Bbbk[x_{1},\ldots,x_{n}]$
with $\deg(x_{i})>0$ for every $i$. Suppose that $A$ is unimodular and that
${\mathfrak{l}}:=\sum_{i=1}^{n}\deg(x_{i})$ is a nonzero element in $\Bbbk$.
If $\delta$ is a semi-Poisson derivation of $A$, then $\delta$ is a Poisson
derivation of $A$. Namely, $Gspd(A)=Gpd(A)$.

###### Proof.

Since $\deg(x_{i})>0$ for all $i$, by Lemma 1.2(1,3), both
$\operatorname{div}(\delta)$ and ${\mathfrak{l}}=\operatorname{div}(E)$ are in
$\Bbbk$. Let $B$ be the twist $A^{\delta}$ with modular derivation
${\mathbf{n}}$. By Theorem 0.2,

${\mathbf{n}}={\mathbf{m}}+{\mathfrak{l}}\delta-\operatorname{div}(\delta)E={\mathfrak{l}}\delta-\operatorname{div}(\delta)E.$

Since ${\mathbf{n}}$ and $E$ (and $\operatorname{div}(\delta)E$) are Poisson
derivations of $B$, we have that ${\mathfrak{l}}\delta$ (and hence $\delta$)
is a Poisson derivation of $B$. Let $\langle-,-\rangle^{\prime}$
(respectively, $\\{-,-\\}$) be the Poisson structure of $B$ (respectively,
$A$). By (E2.3.1), we have

$\\{-,-\\}=\langle-,-\rangle-E\wedge\delta.$

Then, for all homogeneous elements $a,b\in A$,

$\displaystyle\delta(\\{a,b\\})-$
$\displaystyle\\{\delta(a),b\\}-\\{a,\delta(b)\\}$
$\displaystyle=\delta(\langle a,b\rangle)-\langle\delta(a),b\rangle-\langle
a,\delta(b)\rangle$
$\displaystyle\quad-\delta((E\wedge\delta)[a,b])+E\wedge\delta[\delta(a),b]+E\wedge\delta[a,\delta(b)]$
$\displaystyle=0-\delta(|a|a\delta(b)-|b|b\delta(a))$
$\displaystyle\quad+(|a|\delta(a)\delta(b)-|b|b\delta^{2}(a))+(|a|a\delta^{2}(b)-|b|\delta(b)\delta(a))$
$\displaystyle=-|a|\delta(a)\delta(b)-|a|a\delta^{2}(b)+|b|\delta(b)\delta(a)+|b|b\delta^{2}(a)$
$\displaystyle\quad+(|a|\delta(a)\delta(b)-|b|b\delta^{2}(a))+(|a|a\delta^{2}(b)-|b|\delta(b)\delta(a))$
$\displaystyle=0.$

Therefore $\delta$ is a Poisson derivation of $A$. ∎

######  Lemma 4.2.

Let $B$ be a twist of $A$. Then $Gspd(A)=Gspd(B)$.

###### Proof.

Write $B=A^{\delta}$. So $B=A$ as a commutative algebra. Let $\pi$
(respectively, $\pi^{\prime}$) be the Poisson bracket of $A$ (respectively,
$B$).

For every derivation $\phi$ of $A$, $[E,\phi]_{S}=[E,\phi]=0$. For any two
derivations $\phi_{1},\phi_{2}$, we have

$\displaystyle\;[E\wedge\phi_{1},E\wedge\phi_{2}]_{S}$
$\displaystyle=\pm[E,E\wedge\phi_{2}]_{S}\wedge\phi_{1}\pm
E\wedge[\phi_{1},E\wedge\phi_{2}]_{S}$
$\displaystyle=\pm([E,E]_{S}\wedge\phi_{2}\pm[E,\phi_{2}]_{S}\wedge
E)\wedge\phi_{1}$ $\displaystyle\qquad\pm
E\wedge([\phi_{1},E]_{S}\wedge\phi_{2}\pm[\phi_{1},\phi_{2}]_{S}\wedge E)$
$\displaystyle=0.$

Let $\phi$ be a semi-Poisson derivation of $A$. By definition,

$[E\wedge\phi,\pi]_{S}=0.$

Then

$\displaystyle\;[E\wedge\phi,\pi^{\prime}]_{S}$
$\displaystyle=[E\wedge\phi,\pi+E\wedge\delta]_{S}$
$\displaystyle=[E\wedge\phi,\pi]_{S}+[E\wedge\phi,E\wedge\delta]_{S}$
$\displaystyle=0.$

Therefore $\phi$ is a semi-Poisson derivation of $B$. ∎

###### Proof of Theorem 0.5.

Part (1) is Lemma 4.1 and part (2) is Lemma 4.2.

(3) By Corollary 0.3 and part (2), we may assume that $A$ is unimodular. By
part (1), $Gspd(A)$ is the $\Bbbk$-vector space of graded Poisson derivations
$A$. It is well-known that it is a Lie algebra. Let $\phi$ be any Poisson
derivation of $A$. It is clear that $\phi$ is determined by
$\\{\phi(x_{i})\\}_{i=1}^{n}$. Therefore $Gspd(A)$ is finite dimensional. ∎

One of the main definitions in this paper is the following.

######  Definition 4.3.

Let $A$ be a ${\mathbb{Z}}$-graded Poisson algebra.

  1. (1)

The rigidity of graded twisting (or simply rigidity) of $A$ is defined to be

$rgt(A)=1-\dim_{\Bbbk}Gspd(A).$

  2. (2)

We say $A$ is rigid if $rgt(A)=0$.

  3. (3)

We say $A$ is $(-1)$-rigid if $rgt(A)=-1$.

Note that this notion of rigidity is different from the rigidity defined in
[GVW, Definition 0.1] and other papers.

It follows from Lemma 4.2 that

(E4.3.1) $rgt(A)=rgt(A^{\delta}).$

Other basic facts about $rgt(A)$ are listed in the following lemma.

######  Lemma 4.4.

Let $A$ be a ${\mathbb{Z}}$-graded Poisson algebra with $A_{i}\neq 0$ for some
$i\neq 0$. In parts (2)-(6), we further assume that $A$ is
$\Bbbk[x_{1},\ldots,x_{n}]$ with $\deg x_{i}>0$ for all $i$.

  1. (1)

Suppose that $rgt(A)=0$. Then every graded twist of $A$ is isomorphic to $A$.

  2. (2)

If $A$ is not unimodular, then $rgt(A)\leq-1$.

  3. (3)

If $rgt(A)=0$, then $A$ is unimodular.

  4. (4)

If $rgt(A)\neq 0$, then $\dim_{\Bbbk}Gspd(A)\geq\dim_{\Bbbk}Gpd(A)\geq 2$.

  5. (5)

If $rgt(A)=-1$, then $\dim_{\Bbbk}Gspd(A)=\dim_{\Bbbk}Gpd(A)=2$.

  6. (6)

Let $\\{A(a)\\}_{a\in\Bbbk}$ be a family of Poisson polynomial algebras such
that (i) $A(a)$ is a Poisson twist of $A(a^{\prime})$ for all
$a,a^{\prime}\in\Bbbk$ and that (ii) there is an $a_{0}$ such that $A(a_{0})$
is unimodular. If $\dim_{\Bbbk}Gpd(A(a_{0}))=2$, then $rgt(A(a))=-1$ for all
$a$.

  7. (7)

If $rgt(A)=0$ and $A$ is a connected graded domain, then every Poisson normal
element of $A$ is Poisson central.

  8. (8)

If $\dim_{\Bbbk}Gpd(A)=1$ and $\operatorname{div}(E)\neq 0$, then $rgt(A)=0$.

###### Proof.

(1) Since $A_{i}\neq 0$ for some $i$, the Euler derivation $E$ is not zero.
Since $rgt(A)=0$, $Gspd(A)=\Bbbk E$. Let $B$ be a graded twist of $A$. Then
$B=A^{\delta}$ where $\delta\in Gspd(A)$. Let $\langle-,-\rangle$ be the
Poisson bracket of $B$. Since $\delta=\alpha E$ for some $\alpha\in\Bbbk$, one
sees from (E2.3.1) that $\langle a,b\rangle=\\{a,b\\}$ where $\\{a,b\\}$ is
the original Poisson bracket of $A$. The assertion follows.

(2) Since $A$ is not unimodular, the modular derivation ${\mathbf{m}}$ is not
in $\Bbbk E$, as $\operatorname{div}(E)=\sum_{i=1}^{n}\deg x_{i}\neq 0$ and
$\operatorname{div}({\mathbf{m}})=0$ [Lemma 0.1]. Therefore
$\dim_{\Bbbk}Gspd(A)\geq 2$. The assertion follows.

(3) This is equivalent to (2).

(4) By definition, it is clear that
$\dim_{\Bbbk}Gspd(A)\geq\dim_{\Bbbk}Gpd(A)$. It remains to show
$\dim_{\Bbbk}Gpd(A)\geq 2$. If $A$ is unimodular, then, by Lemma 4.1,

$\dim_{\Bbbk}Gpd(A)=\dim_{\Bbbk}Gspd(A)=1-rgt(A)\geq 2.$

Now we assume that $A$ is not unimodular with nonzero modular derivation
${\mathbf{m}}$. Since $\operatorname{div}(E)\neq 0$ and
$\operatorname{div}({\mathbf{m}})=0$, the $\Bbbk$-dimension of $Gpd(A)$ is at
least 2 as desired.

(5) By definition, $\dim_{\Bbbk}Gspd(A)=1-rgt(A)=2$. The assertion follows
from part (4).

(6) It follows from Lemmas 4.1 and 4.2 that we have

$rgt(A(a))=rgt(A(a_{0}))=1-\dim_{\Bbbk}Gspd(A(a_{0}))=1-\dim_{\Bbbk}Gpd(A(a_{0}))=-1.$

The assertion follows from (4).

(7) We only need to consider a homogeneous Poisson normal element $f$ of
positive degree. Note the log-Hamiltonian derivation $LH_{f}:=f^{-1}\\{f,-\\}$
is a Poisson derivation of degree $0$. Suppose $f$ is not central. Then
$LH_{f}(f)\neq 0$. Note that $LH_{f}$ is clearly not the Euler derivation.
Therefore $rgt(A)\leq-1$, yielding a contradiction. The assertion follows.

(8) We know $Gpd(A)$ is spanned by the Euler derivation $E$, which is not the
modular derivation by Lemma 0.1. This implies that $A$ is unimodular and the
result follows by Lemma 4.2. ∎

Examples of $rgt(A)$ will be given in the next 2 sections.

##  5\. Examples and comments

Note that the graded twists in the associative algebra setting has an
important property, namely, a graded algebra $R$ and its twist have isomorphic
corresponding graded module categories [Zh]. So we are wondering if a similar
result holds in the Poisson setting. The following example shows that this is
not the case.

######  Example 5.1.

Let $A$ be the Poisson algebra $\Bbbk[x_{1},x_{2}]$ with trivial Poisson
structure. Let $\delta$ be the Poisson derivation of $A$ determined by

$\delta(x_{1})=0,\quad{\text{and}}\quad\delta(x_{2})=x_{2}.$

Let $B$ be the graded twist of $A$ by $\delta$, namely, $B=A^{\delta}$. By
definition, $B$ is a Poisson algebra $\Bbbk[x_{1},x_{2}]$ with Poisson bracket
determined by

$\\{x_{1},x_{2}\\}=x_{1}x_{2}.$

Let $U(A)$ denote the Poisson enveloping algebra of $A$ [Ba3]. Since $A$ has
the trivial Poisson structure, $U(A)$ is the commutative polynomial ring
$\Bbbk[x_{1},x_{2},y_{1},y_{2}]$. Let $U(B)$ be the Poisson enveloping algebra
of $B$. We claim that $U(B)$ is not a graded twist of $U(A)$ in the sense of
[Zh].

Suppose on the contrary that $U(B)$ is a graded twist of $U(A)$ in the sense
of [Zh]. Then, by [Zh, Theorem 1.1],

(E5.1.1) ${\text{GrMod-}}U(A)\cong{\text{GrMod-}}U(B).$

Let $D(A)$ (respectively, $D(B)$) be the degree zero part of graded quotient
ring of $U(A)$ (respectively, $U(B)$). Then it follows from (E5.1.1) that
$D(A)\cong D(B)$. Since $U(A)$ is commutative, $D(A)$ is commutative. Thus
$D(B)$ is commutative. Next we prove that $D(B)$ is not commutative, so we
obtain a contradiction. By [Ba3, Theorem 2.2], $U(B)$ is generated by four
elements $x_{1},x_{2},\delta_{1},\delta_{2}$ and subject to 6 relations

$\displaystyle x_{1}x_{2}$ $\displaystyle=x_{2}x_{1},$
$\displaystyle\delta_{1}x_{1}$ $\displaystyle=x_{1}\delta_{1},$
$\displaystyle\delta_{1}x_{2}$ $\displaystyle=x_{2}\delta_{1}+x_{1}x_{2},$
$\displaystyle\delta_{2}x_{1}$ $\displaystyle=x_{1}\delta_{2}-x_{1}x_{2},$
$\displaystyle\delta_{2}x_{2}$ $\displaystyle=x_{2}\delta_{2},$
$\displaystyle\delta_{2}\delta_{1}$
$\displaystyle=\delta_{1}\delta_{2}+x_{2}\delta_{1}+x_{1}\delta_{2}.$

Let $a=x_{1}x_{2}^{-1}$ and $b=\delta_{2}x_{2}^{-1}$ which are elements in
$D(B)$. It follows from the six relations that

$ba=ab-a.$

So $D(B)$ is not commutative, yielding a contradiction. Therefore $U(B)$ is
not a graded twist of $U(A)$.

As a consequence, the category of graded Poisson modules over $A^{\delta}$,
denoted by ${\text{GrPMod-}}A^{\delta}$ is not equivalent to the category of
graded Poisson modules over $A$, denoted by ${\text{GrPMod-}}A$. That is,

${\text{GrPMod-}}A^{\delta}\not\cong{\text{GrPMod-}}A.$

######  Remark 5.2.

When $A$ is a connected graded Poisson algebra with $A_{i}\neq 0$ for some
$i>0$. Then $PH^{1}(A)$ is also graded. Since $(PH^{1}(A))_{0}\cong Gpd(A)$,
we have $rgt(A)\leq 1-\dim_{\Bbbk}(PH^{1}(A))_{0}$. If
$\dim_{\Bbbk}(PH^{1}(A))_{0}=1$, then $rgt(A)=0$ by Lemma 4.4(8). Therefore we
can obtain information about $rgt(A)$ if we know $PH^{1}(A)$.

######  Remark 5.3.

Let $A$ be the quadratic Poisson algebra corresponding to the $4$-dimensional
Sklyanin algebra. The Poisson (co)homologies of $A$ have been computed in
[Pe1]. By the Hilbert series of Poisson homologies of $A$ given in [Pe1,
p.1154] and the Poincaré duality, both $PH^{0}(A)$ and $PH^{1}(A)$ have
Hilbert series $\frac{1}{(1-t^{2})^{2}}$. By Remark 5.2, $rgt(A)=0$. Therefore
$A$ is rigid of graded twisting. By the correspondence between quadratic
Poisson algebras and the homogeneous coordinate rings of quantum spaces, the
corresponding Sklyanin algebra is considered as rigid in some sense. Further,
since $h_{PH^{1}(A)}(t)=h_{PH^{0}(A)}(t)=h_{Z}(t)=\frac{1}{(1-t^{2})^{2}}$,
$A$ is $PH^{1}$-minimal in the sense of Definition 7.3(1).

Note that the Poisson (co)homologies of the quadratic Poisson algebra
corresponding to the $3$-dimensional Sklyanin algebra were computed in [Pe2,
Pi1, VdB], an argument similar to the above shows that $rgt(A)=0$ (namely, $A$
is rigid of graded twisting) and $A$ is $PH^{1}$-minimal. We will give an
elementary and direct computation of this $rgt(A)$ in Example 6.6(Case 3).

######  Remark 5.4.

Let $n\geq 2$ and $a\in\mathbb{C}$. Let
$A(n,a)=\mathbb{C}[x_{1},\ldots,x_{n}]$ be the family of Poisson polynomial
algebras studied in [LS]. Then one can check for each fixed $n\geq 2$ the
family $A(n,a)$ satisfies all the assumptions stated in Lemma 4.4(6) with
$a_{0}=\frac{(n+2)(1-n)}{2(n+1)}$ such that $rgt(A(n,a))=-1$. Computations are
omitted. In particular for any $a,a^{\prime}\in\mathbb{C}$, $A(n,a)$ is a
Poisson twist of $A(n,a^{\prime})$, which is a Poisson version of [LS, Theorem
4.2].

We will compute $rgt$ for some classes of Poisson algebras. Here is a warmup.

######  Example 5.5.

Let $\deg x_{i}=i$ for $i=1,2,3$. Let $\Omega=x^{6}+y^{3}+z^{2}+\lambda xyz$
where $\lambda\in\Bbbk$. For any $\lambda$, $\Omega$ is irreducible. Define a
Poisson structure on $A:=\Bbbk[x,y,z]$ by

$\\{f,g\\}:=\det\begin{pmatrix}\Omega_{x}&f_{x}&g_{x}\\\
\Omega_{y}&f_{y}&g_{y}\\\ \Omega_{z}&f_{z}&g_{z}\end{pmatrix}$

for all $f,g\in A$. It is easy to see that

(E5.5.1) $\displaystyle\\{x,y\\}$ $\displaystyle=\Omega_{z}=2z+\lambda xy,$
(E5.5.2) $\displaystyle\\{x,z\\}$ $\displaystyle=-\Omega_{y}=-(3y^{2}+\lambda
xz),$ (E5.5.3) $\displaystyle\\{y,z\\}$
$\displaystyle=\Omega_{x}=6x^{5}+\lambda yz$

and that $A$ is unimodular. If $\lambda^{6}\neq 6^{3}$, then
$A_{sing}:=A/(\Omega_{x},\Omega_{y},\Omega_{z})$ is finite dimensional. In
this case, $\Omega$ has isolated singularity. Let $\delta$ be a graded Poisson
derivation. Then

$\displaystyle\delta(x)$ $\displaystyle=c_{1}x,$ $\displaystyle\delta(y)$
$\displaystyle=c_{2}y+c_{3}x^{2},$ $\displaystyle\delta(z)$
$\displaystyle=c_{4}z+c_{5}x^{3}+c_{6}xy.$

Subtracting by $c_{1}E$, we may assume that $c_{1}=0$. Applying $\delta$ (with
$c_{1}=0$) to (E5.5.1), we obtain that

$c_{2}(2z+\lambda xy)=2(c_{4}z+c_{5}x^{3}+c_{6}xy)+\lambda
x(c_{2}y+c_{3}x^{2}),$

which implies that $c_{2}=c_{4}$, $c_{6}=0$, and $2c_{5}+\lambda c_{3}=0$.
Applying $\delta$ (with $c_{1}=0$ and $c_{6}=0$) to (E5.5.2), we obtain that

$-c_{4}(3y^{2}+\lambda xz)=-6y(c_{2}y+c_{3}x^{2})-\lambda
x(c_{4}z+c_{5}x^{3}),$

which implies that $c_{2}=c_{3}=c_{4}=c_{5}=0$. Therefore $\delta=0$. This
means that $rgt(A)=0$. By Lemma 4.4(1), $A$ does not have any non-trivial
twists.

In general, when $\Omega$ has isolated singularity, the fact that $rgt(A)=0$
also follows from the Poisson cohomology computation given in [Pi1,
Proposition 4.5] (after matching up the notations). The same idea applies to
the algebra in Example 6.6(Case 3).

##  6\. Some computations of $rgt$

In this section we compute $rgt$ for all quadratic Poisson structures on
$\Bbbk[x,y,z]$. Some of the computations have been done by other researchers
in different language (for example, some are hidden inside in Poisson
cohomology computation), but we provide all details of computations of $rgt$
for completeness. The classification of all quadratic Poisson structures on
$\Bbbk[x,y,z]$ were given in [DH, DML, LX].

First we fix some notations. Let $\Bbbk$ be an algebraically closed field of
characteristic zero (one might assume $\Bbbk={\mathbb{C}}$ if necessary). Let
$V=A_{1}=\Bbbk x+\Bbbk y+\Bbbk z$ and let $\\{-,-\\}$ be a quadratic Poisson
bracket of $A:=\Bbbk[x,y,z]=\Bbbk[V]$. Let $f$ be a graded Poisson derivation
of $(A,\\{-,-\\})$. Let $W=\\{V,V\\}$. It is clear that

(E6.0.1)
$f(W)=f(\\{V,V\\})\subseteq\\{f(V),V\\}+\\{V,f(V)\\}\subseteq\\{V,V\\}=W.$

Write

(E6.0.2)
$f(x)=a_{1}x+a_{2}y+a_{3}z,f(y)=b_{1}x+b_{2}y+b_{3}z,f(z)=c_{1}x+c_{2}y+c_{3}z.$

After replacing $f$ by $f-a_{1}E$, we can further assume that

(E6.0.3) $a_{1}\ {\text{in \eqref{E6.0.2} is zero.}}$

If $\\{-,-\\}$ is unimodular, then it comes from a “potential” $\Omega\in
A_{3}$. One can classify $\Omega$ as follows: (a): $\Omega$ is a product of
three linear terms, (b): $\Omega$ is a product of a linear term and an
irreducible polynomial of degree $2$, and (c): $\Omega$ is irreducible of
degree $3$. This classification is classical and well-known, see for example
[BM, KM, Ri].

The following four examples deal with the first case, namely, $\Omega$ is a
product of three linear terms.

Define $A_{sing}$ to be $A/(\Omega_{x},\Omega_{y},\Omega_{z})$.

######  Example 6.1.

Let $\Omega=x^{3}$. Then

$\displaystyle\\{x,y\\}$ $\displaystyle=\Omega_{z}=0,$
$\displaystyle\\{z,x\\}$ $\displaystyle=\Omega_{y}=0,$
$\displaystyle\\{y,z\\}$ $\displaystyle=\Omega_{x}=3x^{2}.$

It is clear that $\operatorname{Kdim}(A_{sing})=2$. Let $f$ be a Poisson
derivation of $A$. By (E6.0.1)

$2xf(x)=f(x^{2})\in f(W)\subseteq W=\Bbbk x^{2}.$

Then $f(x)=ax$ for some $a\in\Bbbk$. By (E6.0.3), we may assume that $f(x)=0$.
Retain the notations in (E6.0.2). Applying $f$ to $\\{y,z\\}=3x^{2}$ implies
that $b_{2}+c_{3}=0$ with $b_{1},b_{3},c_{1},c_{2}$ free. Therefore
$rgt(A)=-5$. One can check that every Poisson normal element of $A$ is Poisson
central.

######  Example 6.2.

Let $\Omega=x^{2}y$. Then

$\displaystyle\\{x,y\\}$ $\displaystyle=\Omega_{z}=0,$ (E6.2.1)
$\displaystyle\\{z,x\\}$ $\displaystyle=\Omega_{y}=x^{2},$
$\displaystyle\\{y,z\\}$ $\displaystyle=\Omega_{x}=2xy.$

It is clear that $\operatorname{Kdim}(A_{sing})=2$. Let $f$ be a Poisson
derivation of $A$. By (E6.0.1), we have

$\Bbbk x^{2}+\Bbbk xy=W\supseteq f(W)=\Bbbk(2xf(x))+\Bbbk(f(x)y+xf(y)).$

Then $f(x)y$ does not have terms $y^{2}$ and $yz$. So $f(x)=ax$ for some
$a\in\Bbbk$, and by (E6.0.3), we may assume that $f(x)=0$. Using the notations
in (E6.0.2), then (E6.2.1) implies that $b_{1}=b_{3}=c_{3}=0$ with
$b_{2},c_{1},c_{2}$ free. Therefore $rgt(A)=-3$.

######  Example 6.3.

Let $\Omega=xyz$. Then

$\displaystyle\\{x,y\\}$ $\displaystyle=\Omega_{z}=xy,$ (E6.3.1)
$\displaystyle\\{z,x\\}$ $\displaystyle=\Omega_{y}=xz,$
$\displaystyle\\{y,z\\}$ $\displaystyle=\Omega_{x}=yz.$

As before we assume that $a_{1}=0$. Note that $W:=\Bbbk xy+\Bbbk yz+\Bbbk xz$
which does not contain term $x^{2}$ and $y^{2}$. By (E6.0.1), we have

$(a_{1}x+a_{2}y+a_{3}z)y+(b_{1}x+b_{2}y+b_{3}z)x=f(x)y+xf(y)=f(xy)\in W$

which implies that $a_{2}=b_{1}=0$. Similarly, using $f(xz),f(yz)\in W$, we
obtain that $a_{3}=c_{1}=b_{3}=c_{2}=0$. Thus $f(x)=0$, $f(y)=b_{2}y$ and
$f(z)=c_{3}z$. Therefore $rgt(A)=-2$.

One can check that $\operatorname{Kdim}(A_{sing})=1$.

######  Example 6.4.

Let $\Omega=xy(x+y)$. Then

$\displaystyle\\{x,y\\}$ $\displaystyle=\Omega_{z}=0,$ (E6.4.1)
$\displaystyle\\{z,x\\}$ $\displaystyle=\Omega_{y}=x^{2}+2xy,$
$\displaystyle\\{y,z\\}$ $\displaystyle=\Omega_{x}=2xy+y^{2}.$

Again we may assume that $a_{1}=0$. By (E6.0.1), we have

$\displaystyle f(x^{2}+2xy)$
$\displaystyle=2xf(x)+2xf(y)+2yf(x)\in\Bbbk(x^{2}+2xy)+\Bbbk(2xy+y^{2})=:W,$
$\displaystyle f(2xy+y^{2})$
$\displaystyle=2xf(y)+2yf(x)+2yf(y)\in\Bbbk(x^{2}+2xy)+\Bbbk(2xy+y^{2}).$

As a consequence, both $f(x)$ and $f(y)$ do not have the $z$ term, namely,
$a_{3}=b_{3}=0$. Furthermore, by the above and a little bit of linear algebra,
we have

$b_{1}=b_{2}=-a_{2}.$

Now we can write $f(x)=a_{2}y$ and $f(y)=-a_{2}x-a_{2}y$. Applying $f$ to the
second equation of (E6.4.1), we obtain that $c_{3}=0$ and $a_{2}=0$ with
$c_{1}$ and $c_{2}$ free. Therefore $rgt(A)=-2$.

One can check that $\operatorname{Kdim}(A_{sing})=1$.

Next we consider the second case. Some linear algebra details will be omitted
in the next two examples.

######  Example 6.5.

Suppose $\Omega=x\omega(x,y,z)$ where $\omega$ is an irreducible polynomial of
degree $2$. Since $\Bbbk$ is algebraically closed, up to a linear change of
variables, we can assume that $\omega$ is either $yz+ax^{2}+bxy+cxz$ (with
further change of basis, we can assume that $c=0$) or $y^{2}+ax^{2}+bxy+cxz$
(with further change of basis, we can assume that $b=0$). So we need to
consider the following two cases:

Case 1: $\omega=yz+ax^{2}+bxy$. We can further assume $a=1$ since $\omega$ is
irreducible. So $\omega=yz+x^{2}+bxy$. Replacing $z$ by $z-bx$, we have
$\omega=yz+x^{2}$ and $\Omega=xyz+x^{3}$. In this case, the Poisson bracket of
$A$ is determined by

$\displaystyle\\{x,y\\}$ $\displaystyle=\Omega_{z}=xy,$ (E6.5.1)
$\displaystyle\\{z,x\\}$ $\displaystyle=\Omega_{y}=xz,$
$\displaystyle\\{y,z\\}$ $\displaystyle=\Omega_{x}=yz+3x^{2}.$

One can check that $\operatorname{Kdim}(A_{sing})=1$. Recall that $W$ is
$\\{V,V\\}=\Bbbk xy+\Bbbk xz+\Bbbk(yz+3x^{2})$ which does not involve either
$y^{2}$ or $z^{2}$. By the second equation of (E6.5.1), we have

$f(x)z+xf(z)\in W$

which implies that $f(x)$ does not have the $z$ term, or $a_{3}=0$. Similarly,
$b_{3}=0$ by the third equation of (E6.5.1). By using the first equation of
(E6.5.1), we obtain that $a_{2}=0$. By (E6.0.3), we can assume that $f(x)=0$.
Now the first equation of (E6.5.1) implies that $\\{x,f(y)\\}=xf(y)$. So
$f(y)\in\Bbbk y$ or $f(y)=b_{2}y$.

Using the second equation of (E6.5.1), one can show that $c_{1}=c_{2}=0$. By
using the third equation of (E6.5.1) and the fact that $W$ does not contain
the term $y^{2}$, we obtain that $c_{2}=0$. So $f(z)=c_{3}z$. From this we can
derive that $b_{2}+c_{3}=0$. Therefore $rgt(A)=-1$.

Case 2: $\omega=y^{2}+ax^{2}+cxz$. If $c=0$, then it is covered in case 1. If
$c\neq 0$, we can assume $a=0$ and $c=1$. So $\omega=y^{2}+xz$ and
$\Omega=xy^{2}+x^{2}z$. Then we have

$\displaystyle\\{x,y\\}$ $\displaystyle=\Omega_{z}=x^{2},$ (E6.5.2)
$\displaystyle\\{z,x\\}$ $\displaystyle=\Omega_{y}=2xy,$
$\displaystyle\\{y,z\\}$ $\displaystyle=\Omega_{x}=y^{2}+2xz.$

One can check that $\operatorname{Kdim}(A_{sing})=1$. By definition $W=\Bbbk
x^{2}+\Bbbk xy+\Bbbk(y^{2}+2xz)$, and it does not contain terms $z^{2}$ and
$yz$. Using the third equation of (E6.5.2), we have

$f(y^{2}+2xz)=2yf(y)+2xf(z)+2zf(x)\in W.$

Therefore $f(x)z$ does not contain a $z^{2}$ term. So $f(x)=a_{1}x+a_{2}y$ and
with (E6.0.3) we can assume that $f(x)=a_{2}y$. Now we apply $f$ to the first
equation of (E6.5.2), we obtain that $b_{2}=0$ and $b_{3}=-a_{2}$. (Some
calculations are omitted.) Applying $f$ to the second equation of (E6.5.2), we
obtain that $a_{2}=0$, $c_{3}=0$ and $c_{2}=-2b_{1}$. Finally applying $f$ to
the third equation of (E6.5.2), we obtain that $c_{1}=0$ with $b_{1}$ free.
Therefore $rgt(A)=-1$.

The final example deals with the irreducible cubic $\Omega$.

######  Example 6.6.

Suppose $\Omega$ is an irreducible cubic function in $x,y,z$. By
classification (see, for example, [KM, Theorems 1 and 2] and [BM, Theorem
2.12]), there are following two singular ones and one non-singular.

Case 1: $\Omega=x^{3}+y^{2}z$. Then we have

$\displaystyle\\{x,y\\}$ $\displaystyle=\Omega_{z}=y^{2},$ (E6.6.1)
$\displaystyle\\{z,x\\}$ $\displaystyle=\Omega_{y}=2yz,$
$\displaystyle\\{y,z\\}$ $\displaystyle=\Omega_{x}=3x^{2}.$

So $W:=\Bbbk y^{2}+\Bbbk yz+\Bbbk x^{2}$ does not have terms $z^{2}$, $xy$ and
$xz$. Then $f(x^{2})=2xf(x)\in W$ implies that $f(x)\in\Bbbk x$. By (E6.0.3),
we have $f(x)=0$. Applying $f$ to the first equation of (E6.6.1), we obtain
that $f(y)=0$. Applying $f$ to the last two equations of (E6.6.1), we obtain
that $f(z)=0$. So $rgt(A)=0$.

One can check that $\operatorname{Kdim}(A_{sing})=1$. By a Gröbner Basis
argument, one sees that the Hilbert series of $A_{sing}$ is
$\frac{2}{(1-t)}+t^{2}+t-1$.

Case 2: $\Omega=x^{3}+x^{2}z+y^{2}z$. Then we have

$\displaystyle\\{x,y\\}$ $\displaystyle=\Omega_{z}=x^{2}+y^{2},$ (E6.6.2)
$\displaystyle\\{z,x\\}$ $\displaystyle=\Omega_{y}=2yz,$
$\displaystyle\\{y,z\\}$ $\displaystyle=\Omega_{x}=3x^{2}+2xz.$

So $W$ does not have terms $z^{2}$ and $xy$. By the second equation of
(E6.6.2), we have

$f(yz)=yf(z)+zf(y)\in W$

which implies that $f(y)$ has no $z$ term and $f(z)$ has no $x$ term. By the
third relation of (E6.6.2), we obtain that $f(x)$ has no $z$ term. By
(E6.0.3), one can assume that $f(x)=a_{2}y$. Applying $f$ to the first
equation, we obtain that $b_{2}=0$ and $b_{1}=-a_{2}$ (namely, $f(x)=a_{2}y$
and $f(y)=-a_{2}x$). Applying $f$ to the second equation, we obtain that
$c_{2}=a_{2}=0$ (so $f(x)=f(y)=0$). Then applying $f$ to the third equation of
(E6.6.2) yields that $f(z)=0$. Therefore $rgt(A)=0$.

One can check that $\operatorname{Kdim}(A_{sing})=1$. By a Gröbner Basis
argument, one sees that the Hilbert series of $A_{sing}$ is
$\frac{2}{(1-t)}+t^{2}+t-1$.

Case 3: $\Omega=\frac{1}{3}(x^{3}+y^{3}+z^{3})+\lambda xyz$ where
$\lambda^{3}\neq-1$ (which is the Hesse normal form given in [BM, Theorem,
2.12]). One can check that $A_{sing}$ is finite dimensional or
$\operatorname{Kdim}(A_{sing})=0$. Consequently, $\Omega$ has an isolated
singularity at zero. As mentioned at the end of Example 5.5, we have
$rgt(A)=0$ which follows from the Poisson cohomology computation given in
[Pi1, Proposition 4.5] (and [VdB, Theorem 5.1]). Here we will give a direct
computation. By definition,

$\displaystyle\\{x,y\\}$ $\displaystyle=\Omega_{z}=z^{2}+\lambda xy,$ (E6.6.3)
$\displaystyle\\{z,x\\}$ $\displaystyle=\Omega_{y}=y^{2}+\lambda xz,$
$\displaystyle\\{y,z\\}$ $\displaystyle=\Omega_{x}=x^{2}+\lambda yz.$

Note that $W=\Bbbk(z^{2}+\lambda xy)+\Bbbk(y^{2}+\lambda
xz)+\Bbbk(x^{2}+\lambda yz)$. This means that in $W$, $z^{2}$ (respectively,
$y^{2}$ and $x^{2}$) appears together with $\lambda xy$ (respectively, $xz$
and $yz$). By the first equation of (E6.6.3), we have $f(z^{2}+\lambda xy)\in
W$. Using the notation in (E6.0.2), we compute

$\displaystyle f(z^{2}+\lambda xy)$
$\displaystyle=2zf(z)+\lambda(xf(y)+yf(x))$
$\displaystyle=2z(c_{1}x+c_{2}y+c_{3}z)+\lambda[x(b_{1}x+b_{2}y+b_{3}z)+y(a_{1}x+a_{2}y+a_{3}z)]$
$\displaystyle\equiv 2c_{1}xz+2c_{2}yz+2c_{3}(-\lambda
xy)+\lambda[b_{1}(-\lambda yz)+b_{2}xy+b_{3}xz$
$\displaystyle\qquad\qquad+a_{1}xy+a_{2}(-\lambda xz)+a_{3}yz]\mod W$
$\displaystyle\equiv(2c_{1}+\lambda
b_{3}-\lambda^{2}a_{2})xz+(2c_{2}-\lambda^{2}b_{1}+\lambda a_{3})yz$
$\displaystyle\qquad\qquad+(-2\lambda c_{3}+\lambda b_{2}+\lambda a_{1})xy\mod
W.$

So we have

$\displaystyle 2c_{1}+\lambda b_{3}-\lambda^{2}a_{2}$ $\displaystyle=0,$
$\displaystyle 2c_{2}+\lambda a_{3}-\lambda^{2}b_{1}$ $\displaystyle=0,$
$\displaystyle-2\lambda c_{3}+\lambda b_{2}+\lambda a_{1}$ $\displaystyle=0.$

From now on we assume that $\lambda\neq 0$ (if $\lambda=0$, the proof is
slightly simpler and is omitted to save the space). With this assumption, we
can remove $\lambda$ from the third equation of the above system. Similarly,
by working with the last two equations in (E6.6.3), we obtain the following

$\displaystyle 2b_{1}+\lambda c_{2}-\lambda^{2}a_{3}$ $\displaystyle=0,$
$\displaystyle 2b_{3}+\lambda a_{2}-\lambda^{2}c_{1}$ $\displaystyle=0,$
$\displaystyle-2b_{2}+c_{3}+a_{1}$ $\displaystyle=0,$ $\displaystyle
2a_{2}+\lambda c_{1}-\lambda^{2}b_{3}$ $\displaystyle=0,$ $\displaystyle
2a_{3}+\lambda b_{1}-\lambda^{2}c_{2}$ $\displaystyle=0,$
$\displaystyle-2a_{1}+b_{2}+c_{3}$ $\displaystyle=0.$

By (E6.0.3), we may assume that $a_{1}=0$. Then by $3$ equations involving
$a_{1}$, we obtain that $b_{2}=c_{3}=0$. Applying $f$ to three equations in
(E6.6.3) with some linear algebra computations, we obtain that
$f(x)=f(y)=f(z)=0$ (but this is true only if $\lambda^{3}\neq-1$). Therefore
$rgt(A)=0$.

Since $\\{\Omega_{x},\Omega_{y},\Omega_{z}\\}$ is a regular sequence, the
Hilbert series of $A_{sing}$ is $(1+t)^{3}$.

Based on the above examples, we have the following classification.

######  Corollary 6.7.

Let $A$ be a quadratic Poisson polynomial ring $\Bbbk[x,y,z]$.

  1. (1)

Suppose $A$ is unimodular. Then $(A,\Omega,rgt(A))$ is listed as follows up to
isomorphisms

$\begin{array}[]{|c|l|l|l|l|l|l|l|}\hline\cr\Omega&\;0&x^{3}&x^{2}y&xyz&xy(x+y)&xyz+x^{3}&xy^{2}+x^{2}z\\\
\hline\cr rgt(A)&-8&-5&-3&-2&-2&-1&-1\\\ \hline\cr&&{\text{Ex.
6.1}}&{\text{Ex. 6.2}}&{\text{Ex. 6.3}}&{\text{Ex. 6.4}}&{\text{Ex.
6.5(1)}}&{\text{Ex. 6.5(2)}}\\\ \hline\cr\end{array}$

$\begin{array}[]{|c|l|l|l|l|}\hline\cr\Omega&x^{3}+y^{2}z&x^{3}+x^{2}z+y^{2}z&\frac{1}{3}(x^{3}+y^{3}+z^{3})+\lambda
xyz,\lambda^{3}\neq-1\\\ \hline\cr rgt(A)&0&0&0\\\ \hline\cr&{\text{Ex.
6.6(1)}}&{\text{Ex. 6.6(2)}}&{\text{Ex. 6.6(3)}}\\\ \hline\cr\end{array}$

  2. (2)

$Gpd(A)$ is one-dimensional if and only if $A$ is unimodular with $\Omega$ in
the second table.

######  Definition 6.8.

A Poisson derivation $\phi$ of a Poisson algebra $A$ is called ozone if
$\phi(z)=0$ for all $z$ in the Poisson center of $A$.

By Definition 1.3(1), the modular derivation ${\mathbf{m}}$ is always ozone.

######  Lemma 6.9.

Let $A$ be the quadratic Poisson algebra in Example 6.6(1) with
$\Omega=x^{3}+y^{2}z$. Then every ozone derivation of $A$ is Hamiltonian.

###### Proof.

By definition,

(E6.9.1) $\displaystyle\\{x,y\\}$ $\displaystyle=y^{2},$ (E6.9.2)
$\displaystyle\\{z,x\\}$ $\displaystyle=2yz,$ (E6.9.3)
$\displaystyle\\{y,z\\}$ $\displaystyle=3x^{2}.$

We define a new grading on the polynomial ring $A=\Bbbk[x,y,z]$. Let $G$ be
${\mathbb{Z}}$ and define $\deg_{G}(x)=0$, $\deg_{G}(y)=1$ and
$\deg_{G}(z)=-2$. For example, $\deg_{G}(\Omega)=0$. Every element $f\in A$
can be written as $\sum_{i\in{\mathbb{Z}}}f_{(i)}$ where $f_{(i)}$ is
homogeneous of $G$-degree $i$. Then $f=f_{(i)}$ if and only if $f$ is
homogeneous of $G$-degree $i$. By (E6.9.1) and (E6.9.2) the Hamiltonian
derivation $H_{x}$ has $G$-degree 1.

Claim 1: If $f$ is homogeneous of $G$-degree $i$, then $H_{x}(f)=ify$.

Proof: Let $f$ be a linear combination of monomials $x^{a}y^{b}z^{c}$. Since
$\deg_{G}(f)=i$, we have $b-2c=i$. Then

$\displaystyle H_{x}(x^{a}y^{b}z^{c})$
$\displaystyle=bx^{a}y^{b-1}y^{2}z^{c}+cx^{a}y^{b}z^{c-1}(-2yz)$
$\displaystyle=(b-2c)(x^{a}y^{b}z^{c})y=i(x^{a}y^{b}z^{c})y.$

So the claim follows.

Let $\phi$ denote an ozone Poisson derivation of $A$.

Claim 2: Up to a Hamiltonian derivation, $\phi(x)=yw_{(0)}$ where
$\deg_{G}(w_{(0)})=0$.

Proof: Since $\Omega$ is Poisson central, $\phi(\Omega)=0$. Then

(E6.9.4) $0=\phi(\Omega)=3x^{2}\phi(x)+2yz\phi(y)+y^{2}\phi(z).$

This implies that $y\mid\phi(x)$. Let $\phi(x)=yw$ where
$w=\sum_{i\in{\mathbb{Z}}}w_{(i)}$ where $\deg_{G}(w_{(i)})=i$. By Claim 1,
$H_{\sum_{i\neq 0}\frac{-1}{i}w_{(i)}}(x)=H_{x}(\sum_{i\neq
0}\frac{1}{i}w_{(i)})=\sum_{i\neq 0}w_{(i)}y$. After replacing $\phi$ by
$\phi-H_{\sum_{i\neq 0}\frac{-1}{i}w_{(i)}}$, we obtain that
$\phi(x)=yw_{(0)}$ as required.

Claim 3: If $\deg_{G}(w)=0$, then $w$ is a polynomial in $x$ and $\Omega$.

Proof: Since $w$ has $G$-degree 0, $w=\sum_{i,j\geq
0}\alpha_{i,j}x^{i}y^{2j}z^{j}$. The assertion follows after replacing
$y^{2}z$ by $\Omega-x^{3}$.

From now on, we assume that $\phi(x)=yw_{(0)}$.

Claim 4: $\deg_{G}(\\{\phi(x),y\\})=3$ or
$\\{\phi(x),y\\}=\\{\phi(x),y\\}_{(3)}$.

Proof: Write $\phi(x)=y\sum_{i,k\geq 0}\alpha_{i,k}x^{i}\Omega^{k}$. We
compute

$\displaystyle\\{\phi(x),y\\}$ $\displaystyle=\\{y\sum_{k\geq
0}\alpha_{k}x^{n-3k}\Omega^{k},y\\}$ $\displaystyle=y\\{\sum_{i,k\geq
0}\alpha_{i,k}x^{i}\Omega^{k},y\\}$ $\displaystyle=y\sum_{i,k\geq
0}\alpha_{i,k}ix^{i-1}y^{2}\Omega^{k}=y^{3}\sum_{i,k\geq
0}\alpha_{i,k}ix^{i-1}\Omega^{k}$

which has $G$-degree 3.

Claim 5: $y^{3}\mid\phi(x)$.

Proof: By Claim 2, $\phi(x)=y\sum_{i,k\geq 0}\alpha_{i,k}x^{i}y^{2k}z^{k}$. If
$\alpha_{i,0}\neq 0$ for some $i$, we have a nonzero term $yx^{i+2}$ in
$3x^{2}\phi(x)$. But $yx^{i+2}$ cannot appear in $2yz\phi(y)+y^{2}\phi(z)$ for
any $i$, which contradicts (E6.9.4). Therefore $\alpha_{i,0}=0$ for all $i$
and $y^{3}\mid\phi(x)$.

Claim 6: $y\mid\phi(y)$.

Proof: This follows from (E6.9.4) and Claim 5.

It follows from (E6.9.4), Claim 5, and Claim 6 that

(E6.9.5) $\displaystyle\phi(x)$ $\displaystyle=yw_{0}=y^{3}zv_{0},$ (E6.9.6)
$\displaystyle\phi(y)$ $\displaystyle=yf,$ (E6.9.7) $\displaystyle\phi(z)$
$\displaystyle=-2zf-3x^{2}yzv_{0}.$

where $v_{0}$ has $G$-degree 0 and $f\in A$. Next we will apply $\phi$ to the
relations given in (E6.9.1) and (E6.9.2). We compute

$\displaystyle 0$
$\displaystyle=\phi(\\{x,y\\}-y^{2})=\\{\phi(x),y\\}+\\{x,\phi(y)\\}-2y\phi(y)$
$\displaystyle=\\{yw_{(0)},y\\}+\\{x,yf\\}-2y\phi(y)$
$\displaystyle=y\\{w_{(0)},y\\}_{(2)}+y^{2}f+y\\{x,f\\}-2y^{2}f$
$\displaystyle=y\\{w_{(0)},y\\}_{(2)}-y^{2}(\sum_{i\in{\mathbb{Z}}}f_{(i)})+y\\{x,\sum_{i\in{\mathbb{Z}}}f_{(i)}\\}$
$\displaystyle=y\\{w_{(0)},y\\}_{(2)}-y^{2}(\sum_{i\in{\mathbb{Z}}}f_{(i)})+y^{2}(\sum_{i\in{\mathbb{Z}}}if_{(i)})$
$\displaystyle=y\\{w_{(0)},y\\}_{(2)}+y^{2}(\sum_{i\in{\mathbb{Z}}}(i-1)f_{(i)}).$

Therefore

(E6.9.8) $f=f_{(1)}\quad{\text{and}}\quad\\{w_{(0)},y\\}=0.$

As a consequence, $f=yq_{(0)}$ where $\deg_{G}(q_{(0)})=0$. So we have
$\phi(y)=y^{2}q_{(0)}$ and $\phi(z)=-2yzq_{(0)}-3x^{2}yzv_{(0)}$.

Applying $\phi$ to (E6.9.2), we have

$\displaystyle 0$
$\displaystyle=\phi(\\{z,x\\}-2yz)=\\{\phi(z),x\\}+\\{z,\phi(x)\\}-2\phi(y)z-2y\phi(z)$
$\displaystyle=(-2q_{(0)}-3x^{2}v_{(0)})\\{yz,x\\}+zv_{(0)}\\{z,y^{3}\\}-2y^{2}q_{(0)}z-2y(-2yzq_{(0)}-3x^{2}yzv_{(0)})$
$\displaystyle=(-2q_{(0)}-3x^{2}v_{(0)})y^{2}z+zv_{(0)}(-9x^{2}y^{2})-2y^{2}q_{(0)}z+4y^{2}zq_{(0)}+6x^{2}y^{2}zv_{(0)}$
$\displaystyle=-6x^{2}y^{2}zv_{(0)}.$

Therefore $v_{(0)}=0$ and

(E6.9.9) $\displaystyle\phi(x)$ $\displaystyle=0,$ (E6.9.10)
$\displaystyle\phi(y)$ $\displaystyle=y^{2}q_{(0)},$ (E6.9.11)
$\displaystyle\phi(z)$ $\displaystyle=-2yzq_{(0)}.$

Write $q_{(0)}=\sum_{i,k\geq 0}\alpha_{i,k}x^{i}\Omega^{k}$. Let
$q^{\prime}_{(0)}=\sum_{i,k\geq 0}\beta_{i,k}x^{i}\Omega^{k}$ where
$\beta_{i,k}:=\frac{\alpha_{i,k}}{i+1}$. It is easy to check that
$H_{xq^{\prime}_{(0)}}=\phi$. Therefore $\phi$ is Hamiltonian as desired. ∎

######  Lemma 6.10.

Let $A$ be the quadratic Poisson algebra in Example 6.6(2) with potential
$\Omega=x^{3}+x^{2}z+y^{2}z$. Then every ozone derivation of $A$ is
Hamiltonian.

###### Proof.

The Jacobian Poisson structure on $A={\Bbbk}[x,y,z]$ is explicitly given by

$\\{x,y\\}=x^{2}+y^{2},\ \\{y,z\\}=3x^{2}+2xz,\ \\{z,x\\}=2yz.$

We show that every ozone derivation of $A$ is Hamiltonian, which is based on a
tedious computation. Since $A$ is graded, it suffices to check every graded
Poisson derivation $\phi$ of degree $n$ vanishing on the Poisson center $Z$ is
Hamiltonian. So we can write

$\phi(x)=\sum_{i=0}^{n+1}\phi_{i}x^{i}\in{\Bbbk}[x,y,z]\text{ of degree $n+1$
with $\phi_{i}\in{\Bbbk}[y,z]_{n+1-i}$. }$

Claim 1: By subtracting a Hamiltonian derivation $H_{g}=\\{g,-\\}$ from $\phi$
for some suitable $g\in{\Bbbk}[x,y,z]_{n}$, we can assume that
$\phi_{i}\in{\Bbbk}[z]\bigoplus y^{3}z({\Bbbk}[z,y^{2}z])$ for all $0\leq
i\leq n-1$.

Proof: For simplicity, we denote the ${\Bbbk}$-linear map
$T_{m}:{\Bbbk}[y,z]_{m}\to{\Bbbk}[y,z]_{m+1}$ by

$T_{m}(f):=2yz\frac{\partial{f}}{\partial z}-y^{2}\frac{\partial{f}}{\partial
y}\text{ for any $m\geq 0$.}$

It is clear to check that

  * •

${\rm ker}(T_{m})=0$ if $3\nmid m$ and ${\rm
ker}(T_{m})={\Bbbk}(y^{2}z)^{\frac{m}{3}}$ if $3\mid m$;

  * •

${\rm img}(T_{m})\bigoplus y({\rm ker}(T_{m}))=y({\Bbbk}[y,z]_{m})$. In
particular $y(y^{2}z)^{\frac{m}{3}}\not\in{\rm img}(T_{m})$ if $3\mid m$.

For any homogeneous polynomial $g=\sum g_{i}x^{i}\in{\Bbbk}[x,y,z]_{n}$ with
$g_{i}\in{\Bbbk}[y,z]_{n-i}$, we get

$\displaystyle\\{g,x\\}$ $\displaystyle=\frac{\partial g}{\partial
y}\\{y,x\\}+\frac{\partial g}{\partial z}\\{z,x\\}=-\frac{\partial g}{\partial
y}(x^{2}+y^{2})+\frac{\partial g}{\partial z}(2yz)$
$\displaystyle=-\frac{\partial g_{n-1}}{\partial y}x^{n+1}-\frac{\partial
g_{n-2}}{\partial y}x^{n}+(T_{1}(g_{n-1})-\frac{\partial g_{n-3}}{\partial
y})x^{n-1}+\cdots$ $\displaystyle\quad+(T_{n-2}(g_{2})-\frac{\partial
g_{0}}{\partial y})x^{2}+T_{n-1}(g_{1})x+T_{n}(g_{0}).$

Hence by choosing $g_{0},g_{1},\ldots,g_{n-1}$ for the coefficients of
$x^{0},x^{1},\ldots,x^{n-1}$ inductively, we can achieve Claim 1.

Claim 2: Up to modulo a Hamiltonian derivation, we can set

$\displaystyle\phi(x)$ $\displaystyle=xzf(x,z)+y^{3}zg(x,y^{2}z)$
$\displaystyle\phi(y)$
$\displaystyle=(\frac{3}{2}xy+yz)f(x,z)+(\frac{3}{2}y^{4}-xy^{2}z)g(x,y^{2}z)+(x^{2}+y^{2})p$
$\displaystyle\phi(z)$
$\displaystyle=-(3xz+2z^{2})f(x,z)-3y^{3}zg(x,y^{2}z)-2yzp$

where $f(x,z)\in{\Bbbk}[x,z]_{n-1}$, $g(x,y^{2}z)\in{\Bbbk}[x,y^{2}z]_{n-3}$
and $p\in{\Bbbk}[x,y,z]_{n-1}$. Furthermore, we can assume $p$ does not
contain $x^{n-1}$.

Proof: By Claim 1, up to a Hamiltonian derivation, we can write

$\displaystyle\phi(x)=xzf(x,z)+y^{3}zg(x,y^{2}z)+az^{n+1}+bx^{n}y+cx^{n+1}$

for some coefficients $a,b,c\in{\Bbbk}$. Since ${\Bbbk}[\Omega]\subseteq Z$,
we have

$\displaystyle\phi(\Omega)=(3x^{2}+2xz)\phi(x)+2yz\phi(y)+(x^{2}+y^{2})\phi(z)=0.$

We have three terms $2axz^{n+2}$, $3cx^{n+3}$ and $3bx^{n+2}y$ appearing in
$(3x^{2}+2xz)\phi(x)$, which can only be canceled from the remaining terms in
$2yz\phi(y)+(x^{2}+y^{2})\phi(z)$. We get $a=0$ since $xz^{n+2}$ is certainly
not contained there. For the term $3cx^{n+3}$, it has to be canceled by the
terms in $(x^{2}+y^{2})\phi(z)$, which implies that
$\phi(z)=-3cx^{n+1}+\cdots$. But the another term $-3cx^{n+1}y^{2}$ appearing
in $(x^{2}+y^{2})\phi(z)$ has to be canceled inside the same product
$(x^{2}+y^{2})\phi(z)$. Repeating the argument, we get there is some
$v\in{\Bbbk}[x,y]$ such that $3cx^{n+3}+(x^{2}+y^{2})v=0$. This is absurd
unless $c=0$. The same argument shows that $b=0$ as well. So we can write
$\phi(x)$ as in Claim 2 and the expressions of $\phi(y)$ and $\phi(z)$ follow
immediately. Finally, by further subtracting $\\{ax^{n},-\\}$ from $\phi$, we
can replace $p$ with $p-anx^{n-1}$. So by choosing a suitable scalar $a$, we
can assume $p$ does not contain $x^{n-1}$.

Claim 3: we have

$\displaystyle\\{p,x\\}$
$\displaystyle=-(2z+\frac{3}{2}x)f(x,z)-(3xz+2z^{2})\frac{\partial
f(x,z)}{\partial z}+zx\frac{\partial f(x,z)}{\partial x}$
$\displaystyle\quad+(3y^{3}-2xyz)g(x,y^{2}z)+y^{3}z\frac{\partial
g(x,y^{2}z)}{\partial x}-2xy^{3}z^{2}\frac{\partial
g(x,y^{2}z)}{\partial(y^{2}z)}.$

Proof: Since $\phi$ is a Poisson derivation, we have

$\displaystyle\phi(\\{x,y\\})=\\{\phi(x),y\\}+\\{x,\phi(y)\\}.$

A long and tedious calculation yields Claim 3.

Now we write

$f=\sum_{i=1}^{n}a_{i}x^{n-i}z^{i-1}\quad\text{and}\quad
g=\sum_{i=1}^{\lfloor{\frac{n}{3}}\rfloor}b_{i}x^{n-3i}(y^{2}z)^{i-1}$

for some $a_{i},b_{i}\in{\Bbbk}$ where we set $a_{i}=0$ if $i\not\in[1,n]$ and
$b_{i}=0$ if $i\not\in[1,\lfloor\frac{n}{3}\rfloor]$. Define

$c_{i}:=(n-3i)a_{i}-(\frac{3}{2}+3i)a_{i+1}.$

Then Claim 3 can be rewritten as

$\displaystyle\\{p,x\\}$
$\displaystyle=\sum_{i=0}^{n}c_{i}z^{i}x^{n-i}+\sum_{i=1}^{\lfloor\frac{n}{3}\rfloor}(3b_{i}y^{2i+1}z^{i-1})x^{n-3i}$
$\displaystyle\quad-\sum_{i=1}^{\lfloor\frac{n}{3}\rfloor}(2ib_{i}y^{2i-1}z^{i})x^{n+1-3i}+\sum_{i=1}^{\lfloor\frac{n}{3}\rfloor}((n-3i)b_{i}y^{2i+1}z^{i})x^{n-1-3i}.$

Write $p=\sum_{i=0}^{n-1}p_{i}x^{i}$ with $p_{i}\in{\Bbbk}[y,z]_{n-1-i}$ and
$p_{n-1}=0$. As in Claim 1, we get

$\displaystyle\\{p,x\\}=\sum_{i=-1}^{n-1}\left(T_{i}(p_{n-1-i})-\frac{\partial
p_{n-3-i}}{\partial y}\right)x^{n-1-i},$

where we set $T_{-1}=0$ and $p_{-2}=p_{-1}=p_{n}=0$. Hence we have

$\displaystyle T_{3i-1}(p_{n-3i})-\frac{\partial p_{n-3i-2}}{\partial y}$
$\displaystyle=c_{3i}z^{3i}+3b_{i}y^{2i+1}z^{i-1}$ (E6.10.1) $\displaystyle
T_{3i-2}(p_{n+1-3i})-\frac{\partial p_{n-1-3i}}{\partial y}$
$\displaystyle=c_{3i-1}z^{3i-1}-2ib_{i}y^{2i-1}z^{i}$ $\displaystyle
T_{3i}(p_{n-1-3i})-\frac{\partial p_{n-3-3i}}{\partial y}$
$\displaystyle=c_{3i+1}z^{3i+1}+(n-3i)b_{i}y^{2i+1}z^{i}$

for all $1\leq i\leq\lfloor\frac{n}{3}\rfloor$ together with $-\frac{\partial
p_{n-2}}{\partial y}=c_{0}$ and $-\frac{\partial p_{n-3}}{\partial y}=c_{1}z$.
In particular, if $n\equiv 2\pmod{3}$, then $T_{n-1}(p_{0})=c_{n}z^{n}$.

Claim 4: We have $f=0$.

Proof: We show that $c_{0}=\cdots=c_{n}=0$, which implies that
$a_{1}=\cdots=a_{n}=0$. By the above equations, it suffices to show that
$T_{n-i-1}(p_{i})-\frac{\partial p_{i-2}}{\partial y}\in y{\Bbbk}[y,z]$ for
$0\leq i\leq n$. We claim that $p_{i}\in{\Bbbk}[y^{2},z]$ for $0\leq i\leq n$.
It is clear when $i=0$. Suppose it works for $p_{i}$ for all $i\leq m$. Then
inductively, the above equations imply that $T_{n-m-2}(p_{m+1})\in{\rm
span}_{\Bbbk}\\{y^{2i+1}z^{j}\,|\,\text{for all possible}\ i,j\\}$. Note that
${\rm ker}(T_{n-m-2})\in{\Bbbk}[y^{2},z]$. Our claim follows by the definition
of $T_{n-m-2}$. Since ${\rm img}(T_{n-i-1})\in(y)$, we get
$T_{n-i-1}(p_{i})-\frac{\partial p_{i-2}}{\partial y}\in y{\Bbbk}[y,z]$ for
$0\leq i\leq n$.

Claim 5: We have $g=0$.

Proof: By Claim 4, we can take $f=0$. We will only treat the case $n\equiv
0\pmod{3}$ here and other cases will follow in a similar manner. We write
$n=3s$ and group the equations (E6.10.1) into $s+1$ parts named by (Ei) with
$0\leq i\leq s$. In details, (E0) is given by

(E0.1) $\displaystyle T_{n-1}(p_{0})$ $\displaystyle=3b_{s}y^{2s+1}z^{s-1}$
(E0.2) $\displaystyle T_{n-2}(p_{1})$ $\displaystyle=-2sb_{s}y^{2s-1}z^{s}$
(E0.3) $\displaystyle T_{n-3}(p_{2})-\frac{\partial p_{0}}{\partial y}$
$\displaystyle=3b_{s-1}y^{2s-1}z^{s-1}$

For $1\leq i\leq s-2$, (Ei) is given by

(Ei.1) $\displaystyle T_{n-3i-1}(p_{3i})-\frac{\partial p_{3i-2}}{\partial y}$
$\displaystyle=3b_{s-i}y^{2s-2i+1}z^{s-i-1}$ (Ei.2) $\displaystyle
T_{n-3i-2}(p_{3i+1})-\frac{\partial p_{3i-1}}{\partial y}$
$\displaystyle=-2(s-i)b_{s-i}y^{2s-2i-1}z^{s-i}$ (Ei.3) $\displaystyle
T_{n-3i-3}(p_{3i+2})-\frac{\partial p_{3i}}{\partial y}$
$\displaystyle=(3s-3(s-i-1))b_{s-i-1}y^{2s-2i-1}z^{s-i-1}$

Moreover, (E(s-1)) and (Es) are given by

(E(s-1).1) $\displaystyle T_{2}(p_{n-3})-\frac{\partial p_{n-5}}{\partial y}$
$\displaystyle=3b_{1}y^{3}$ (E(s-1).2) $\displaystyle
T_{1}(p_{n-2})-\frac{\partial p_{n-4}}{\partial y}$ $\displaystyle=-2b_{1}yz$
(E(s-1).3) $\displaystyle-\frac{\partial p_{n-3}}{\partial y}$
$\displaystyle=0$ (Es) $\displaystyle-\frac{\partial p_{n-2}}{\partial y}$
$\displaystyle=0.$

As in Claim 4, we know $p_{i}\in{\Bbbk}[y^{2},z]_{n-1-i}$. Assign the
lexicographic order with $y>z$ on all monomials in ${\Bbbk}[y,z]$. We prove
the following statement inductively for all $0\leq i\leq s-2$ with $n=3s$:

$\displaystyle p_{3i}$
$\displaystyle=-\frac{3}{2}b_{s-i}(y^{2})^{s-i}z^{s-i-1}+\text{lower terms in
${\Bbbk}[y^{2},z]_{n-3i-1}$}$ $\displaystyle p_{3i+1}$
$\displaystyle=\alpha_{i}(y^{2})^{s-i-1}z^{s-i}+\text{lower terms in
${\Bbbk}[y^{2},z]_{n-3i-2}$}$ $\displaystyle p_{3i+2}$
$\displaystyle=\beta_{i}(y^{2})^{s-i-1}z^{s-i-1}+\text{lower terms in
${\Bbbk}[y^{2},z]_{n-3i-3}$}$

for some $\alpha_{i},\beta_{i}\in{\Bbbk}$ and $(i+1)b_{s-i-1}=(s-i)b_{s-i}$.
When $i=0$, we use (E0.1) and (E0.2) to get

$p_{0}=-\frac{3}{2}b_{s}y^{2s}z^{s-1},\quad p_{1}=-sb_{s}y^{2s-2}z^{s}$

since $T_{n-1},T_{n-2}$ are injective. So (E0.3) implies that

$T_{n-3}(p_{2})=3b_{s-1}y^{2s-1}z^{s-1}+\frac{\partial p_{0}}{\partial
y}=3(b_{s-1}-sb_{s})y^{2s-1}z^{s-1}.$

Since $3\mid n-3$, we have ${\rm ker}(T_{n-3})={\Bbbk}y^{2s-2}z^{s-1}$ and
$y^{2s-1}z^{s-1}\not\in{\rm img}(T_{n-3})$. We get $b_{s-1}=sb_{s}$ and
$p_{2}=\beta_{0}y^{2s-2}z^{s-1}$ for some $\beta_{0}\in{\Bbbk}$. Suppose the
statement holds for $p_{3i},p_{3i+1},p_{3i+2}$. Then (E(i+1).1) implies that

$\displaystyle T_{n-3i-4}(p_{3i+3})$ $\displaystyle=\,$ $\displaystyle
3b_{s-i-1}y^{2s-2i-1}z^{s-i-2}+\frac{\partial}{\partial
y}\left(\alpha_{i}(y^{2})^{s-i-1}z^{s-i}+\text{lower terms in
${\Bbbk}[y^{2},z]_{n-3i-2}$}\right)$ $\displaystyle=\,$ $\displaystyle
3b_{s-i-1}y^{2s-2i-1}z^{s-i-2}+\text{lower terms in
$y({\Bbbk}[y^{2},z]_{n-3i-4})$}.$

Since $T_{n-3i-4}$ is injective, we get

$\displaystyle
p_{3i+3}=-\frac{3}{2}b_{s-i-1}(y^{2})^{s-i-1}z^{s-i-2}+\text{lower terms in
${\Bbbk}[y^{2},z]_{n-3i-4}$}$

Similarly from (E(i+1).2) we get

$\displaystyle p_{3i+4}=\alpha_{i+1}(y^{2})^{s-i-2}z^{s-i-1}+\text{lower terms
in ${\Bbbk}[y^{2},z]_{n-3i-5}$}$

for some $\alpha_{i+1}\in{\Bbbk}$. Finally, (E(i+1).3) implies that

$\displaystyle T_{n-3i-6}(p_{3i+5})=$
$\displaystyle\,(3s-3(s-i-2))b_{s-i-2}y^{2s-2i-3}z^{s-i-2}$
$\displaystyle\,+\frac{\partial}{\partial
y}(-\frac{3}{2}b_{s-i-1}(y^{2})^{s-i-1}z^{s-i-2}+\text{lower terms in
${\Bbbk}[y^{2},z]_{n-3i-4}$})$ $\displaystyle=$
$\displaystyle\,3((i+2)b_{s-i-2}-(s-i-1)b_{s-i-1})y^{2s-2i-3}z^{s-i-2}$
$\displaystyle\,+\text{lower terms in $y({\Bbbk}[y^{2},z]_{n-3i-6})$}$

Note that ${\rm ker}(T_{n-3i-6})={\Bbbk}y^{2s-2i-4}z^{s-i-2}$ and
$y^{2s-2i-3}z^{s-i-2}\not\in{\rm img}(T_{n-3i-6})$ for $3\mid n-3i-6$. So we
get $(i+2)b_{s-i-2}=(s-i-1)b_{s-i-1}$ and we can write

$p_{3i+5}=\beta_{i+1}(y^{2})^{s-i-2}z^{s-i-2}+\text{lower terms in
${\Bbbk}[y^{2},z]_{n-3i-6}$}$

for some $\beta_{i+1}\in{\Bbbk}$. This completes our induction argument. From
the above result, we have

$p_{n-5}=p_{3(s-2)+1}=\alpha_{s-2}y^{2}z^{2}+\text{lower terms in
${\Bbbk}[y^{2},z]_{4}$}$

From (E(s-1).1): $T_{2}(p_{n-3})-\frac{\partial p_{n-5}}{\partial
y}=3b_{1}y^{3}$, we get $p_{n-3}=-\frac{3}{2}b_{1}y^{2}+\lambda z^{2}$ for
some $\lambda\in{\Bbbk}$. Moreover, (E(s-1).3): $-\frac{\partial
p_{n-3}}{\partial y}=0$ implies that $b_{1}=0$. Again from the above
statement, we have all $b_{i}=0$ and $g=0$.

Finally, we can show that $\phi$ is Hamiltonian. By all the above claims, up
to a proper Hamiltonian derivation, we can take a Poisson derivation $\phi$ of
degree $n$ as

$\phi(x)=0,\quad\phi(y)=(x^{2}+y^{2})p,\quad\phi(y)=-2yzp$

for some $p\in{\Bbbk}[x,y,z]_{n-1}$. From
$\\{\phi(x),y\\}+\\{x,\phi(y)\\}=\phi(x^{2}+y^{2})$, we get $\\{p,x\\}=0$ or
$(x^{2}+y^{2})p_{y}=2yzp_{z}$. We show that $p=p(x,\Omega)$ by induction on
the degree of $f$. It is clear that we can write $p_{y}=2yzq$ and
$p_{z}=(x^{2}+y^{2})q$ for some $q\in{\Bbbk}[x,y,z]$ of degree ${\rm
deg}(p)-3$. Then $p_{zy}=p_{yz}$ implies that $(x^{2}+y^{2})q_{y}=2yzq_{z}$.
So our induction hypothesis implies that $q=q(x,\Omega)$. Take any polynomial
$h(x,\Omega)$ such that $\frac{\partial h(x,\Omega)}{\partial\Omega}=q$. An
easy calculation shows that $h_{y}=p_{y}$ and $h_{z}=p_{z}$. So
$p-h\in{\Bbbk}[x]$. This proves our claim. Now take any $Q(x,\Omega)$ such
that $\frac{\partial Q(x,\Omega)}{\partial x}=p(x,\Omega)$. Then one checks
that $\phi=\\{Q,-\\}$ and $\phi$ is Hamiltonian. ∎

######  Remark 6.11.

If $A$ is a non-unimodular quadratic Poisson polynomial algebra
$\Bbbk[x_{1},\cdots,x_{n}]$, then by Corollary 0.3, $A^{\delta}$ is unimodular
for some graded Poisson derivation $\delta$ of $A$ (in fact
$\delta=\frac{1}{\sum_{i}\deg x_{i}}{\mathbf{m}}$). By (E4.3.1),
$rgt(A)=rgt(A^{\delta})$. If one can calculate $rgt$ for all unimodular
quadratic Poisson structures on $\Bbbk[x_{1},\cdots,x_{n}]$, then the above
formula provides a way of computing $rgt(A)$ when $A$ is not unimodular.

Note that all 13 classes of non-unimodular quadratic Poisson structures on
$\Bbbk[x_{1},x_{2},x_{3}]$ were listed explicitly in [DH] (also see [DML,
LX]). For each class, the modular derivation ${\mathbf{m}}$ is easy to
compute. Therefore $rgt$ can be calculated by the method mentioned in the
above paragraph.

##  7\. Rigidity, $H$-ozoneness, and $PH^{1}$-minimality

In this section we will study some connections between rigidity of graded
twisting, ozone derivations and the first Poisson cohomology.

Let $A$ be a general Poisson algebra with Poisson center $Z$. Let $Pd(A)$ be
the Lie algebra of all Poisson derivations of $A$ and let $Hd(A)$ be the Lie
ideal of $Pd(A)$ of all Hamiltonian derivations. Recall from (E1.5.4) that the
first Poisson cohomology of $A$ is defined to be

(E7.0.1) $PH^{1}(A):=Pd(A)/Hd(A).$

If $A$ is ${\mathbb{Z}}$-graded, then so is $PH^{1}(A)$. Part (1) of the
following definition is Definition 6.8.

######  Definition 7.1.

Let $A$ be a Poisson algebra.

  1. (1)

A Poisson derivation $\phi$ of $A$ is called ozone if $\phi(z)=0$ for all
$z\in Z$.

  2. (2)

Let $Od(A)$ denote the Lie algebra of all ozone Poisson derivations of $A$.

  3. (3)

We say $A$ is $H$-ozone if $Od(A)=Hd(A)$, namely, if every ozone Poisson
derivation is Hamiltonian.

It is clear that $Od(A)$ is a Lie ideal of $Pd(A)$ and

$Hd(A)\subseteq Od(A)\subseteq Pd(A).$

In general, not every ozone Poisson derivation is Hamiltonian.

For the rest of this section, we only consider locally finite connected
${\mathbb{N}}$-graded Poisson algebras $A$ with $A_{i}\neq 0$ for some $i>0$.
And later we will only consider $A=\Bbbk[x,y,z]$ where
$\deg(x)=\deg(y)=\deg(z)=1$. In this case the Euler derivation, denoted by
$E$, sending $a\to(\deg a)a(=:|a|a)$, is a nonzero Poisson derivation.

######  Lemma 7.2.

Let $A$ be a connected graded Poisson algebra with center $Z$. Suppose $Z$ is
a domain. Then $ZE\cap Od(A)=0$ if $Z\neq\Bbbk$. As a consequence, $ZE\cap
Hd(A)=0$ and the canonical map $ZE\to PH^{1}(A)$ is a graded injective
$Z$-module map.

###### Proof.

Let $f$ be any homogeneous element in $Z$. It is easy to check that $fE$ is a
Poisson derivation. So $ZE$ is an abelian Lie subalgebra of $Pd(A)$. Moreover,
one can check that $Pd(A)$ is a $Z$-module.

Next we assume that $Z\neq\Bbbk$. Let $\phi$ be in $ZE\cap Od(A)$ and we can
write is as $\phi=fE$ for some $f\in Z$. Let $z\in Z$ be a nonzero element of
positive degree. Then $\phi(z)=0$ as $\phi\in Od(A)$. Since $\phi=fE$, we
obtain that $0=f(\deg z)z$. This implies that $f=0$ or $\phi=0$. Hence $ZE\cap
Od(A)=0$.

Since $Hd(A)\subseteq Od(A)$, $ZE\cap Hd(A)=0$. So the map

$ZE\to Pd(A)/Hd(A)=:PH^{1}(A)$

is injective.

Finally if $Z=\Bbbk$, then it is trivial since $E\not\in Hd(A)$. ∎

By the above lemma, the minimal possibility of $PH^{1}(A)$ is $ZE$. This
motivates the following definition.

######  Definition 7.3.

Let $A$ be a nontrivial connected graded Poisson algebra with Poisson center
$Z$. Suppose $Z$ is a domain.

  1. (1)

We say A is $PH^{1}$-minimal if $PH^{1}(A)\cong ZE$.

  2. (2)

We say $A$ has an Euler-ozone decomposition if

$Pd(A)=ZE\rtimes Od(A).$

  3. (3)

We say $A$ has an Euler-Hamiltonian decomposition if

$Pd(A)=ZE\rtimes Hd(A).$

By Remark 5.3, if $A$ is the Poisson polynomial algebra corresponding to the
$4$-dimensional (resp. $3$-dimensional) Sklyanin algebra, then it is
$PH^{1}$-minimal. Note that the dimension of $Gpd(A)$ is the constant term of
$h_{Pd(A)}(t)$. So $A$ is rigid of graded twisting if and only if the constant
term of $h_{Pd(A)}(t)$ is $1$ [Remark 5.2]. Therefore we have

(E7.3.1) ${\text{$A$ is
$PH^{1}$-minimal}}\quad\Rightarrow\quad{\text{$rgt(A)=0$}}.$

######  Proposition 7.4.

Let $A$ be a connected graded Poisson algebra. Then the following are
equivalent.

  1. (i)

$A$ is $PH^{1}$-minimal.

  2. (ii)

$h_{PH^{1}(A)}(t)=h_{Z}(t)$ provided $Z$ is a domain.

  3. (iii)

$h_{Pd(A)}(t)=h_{A}(t)$.

###### Proof.

(i) $\Leftrightarrow$ (ii) Follows from the definition.

(ii) $\Leftrightarrow$ (iii) It is clear that the map $A\to Hd(A)$ sending
$a\to H_{a}$ is surjective. The kernel is the center $Z$. So
$h_{Hd(A)}(t)=h_{A}(t)-h_{Z}(t)$. By (E7.0.1),
$h_{PH^{1}(A)}(t)=h_{Pd(A)}(t)-h_{Hd(A)}(t)$. Therefore

$\displaystyle h_{Pd(A)}(t)-h_{A}(t)$
$\displaystyle=h_{PH^{1}(A)}(t)+h_{Hd(A)}(t)-h_{A}(t)$
$\displaystyle=h_{PH^{1}(A)}(t)+h_{A}(t)-h_{Z}(t)-h_{A}(t)$
$\displaystyle=h_{PH^{1}(A)}(t)-h_{Z}(t).$

The assertion follows. ∎

Let $A$ be the Poisson algebra in Example 6.6(Case 3). This Poisson algebra is
corresponding to the $3$-dimensional Sklyanin algebra. The potential $\Omega$
has isolated singularity. By [Pi1, Proposition 4.5] (and [VdB, Theorem 5.1]),
$PH^{1}(A)=ZE$, consequently, $A$ is $PH^{1}$-minimal.

######  Lemma 7.5.

Let $A$ be a connected graded Poisson algebra. Assume that $Z$ is a non-
trivial domain. If $A$ is $PH^{1}$-minimal, then $A$ is $H$-ozone and has an
Euler-Hamiltonian decomposition.

###### Proof.

Since $A$ is $PH^{1}$-minimal, $ZE\cong PH^{1}(A)=Pd(A)/Hd(A)$. This implies
that $Pd(A)=ZE\rtimes Hd(A)$. So $A$ has an Euler-Hamiltonian decomposition.

For a graded Poisson derivation $\phi$ of degree $d$, by the Euler-Hamiltonian
decomposition, we have

$\phi=fE+H_{a}.$

Let $z$ be a nonzero central element of positive degree. Then
$\phi(z)=|z|fz+H_{a}(z)=|z|fz$. If $\phi$ is ozone, $0=\phi(z)=|z|fz$ which
implies that $f=0$ and $\phi=H_{a}$ as required. ∎

######  Lemma 7.6.

Let $A$ be a connected graded Poisson domain. Suppose $A$ is $H$-ozone.

  1. (1)

Every Poisson normal element in $A$ is Poisson central.

  2. (2)

Suppose $A$ is a Poisson polynomial ring. Then $A$ is unimodular.

###### Proof.

(1) Let $x$ be a nonzero Poisson normal element. Then it is the sum of
homogeneous Poisson normal elements. So we can assume that $x$ is homogeneous.
Let $\phi$ be the log-Hamilton derivation $x^{-1}H_{x}$. Since $H_{x}(z)=0$
for all $z$ in the center $Z$, $\phi$ is ozone. By the hypothesis, $\phi$ is
Hamiltonian, namely, $\phi=H_{y}$ for some element $y$. Since $\phi$ has
degree 0, $\deg y=0$ (or $y\in\Bbbk$) and consequently, $\phi=0$. This implies
that $x$ is central.

(2) Let ${\mathbf{m}}$ be the modular derivation of $A$. It follows from the
definition that it is ozone. Since $\deg{\mathbf{m}}=0$, by the hypothesis,
${\mathbf{m}}=H_{y}$ for some element $y$ of degree 0. Hence $y\in\Bbbk$ and
consequently, ${\mathbf{m}}=0$. The assertion follows. ∎

We have proved the following diagram

$\begin{CD}{\text{$A$ is $PH^{1}$-minimal}}@>{\text{Lemma
\ref{xxlem7.5}}}>{}>{\text{$A$ is $H$-ozone}}\\\
@V{\eqref{E7.3.1}}V{}V@V{}V{{\text{Lemma \ref{xxlem7.6}(2)}}}V\\\
{\text{$rgt(A)=0$}}@>{}>{\text{Lemma \ref{xxlem4.4}(3)}}>{\text{$A$ is
unimodular}}\end{CD}$

Next we show that some of the conditions are equivalent under extra
hypotheses.

######  Lemma 7.7.

Suppose $A$ is a connected graded Poisson algebra. Let $Z$ be the center of
$A$ and assume that $Z=\Bbbk[z]$ where $z$ is homogeneous with $\deg z>0$.

  1. (1)

If $Pd(A)_{\leq-1}=0$, $A$ has an Euler-ozone decomposition.

  2. (2)

Suppose $A$ is $H$-ozone. Then $rgt(A)=0$.

###### Proof.

(1) Let $\phi$ be a Poisson derivation of $A$ of degree $i$. By the
hypothesis, $i\geq 0$.

Case 1: $\deg(z)\mid i$. Since $\phi(z)$ is central, $\phi(z)=az^{n}$ for some
$a\in\Bbbk$ and $n\geq 0$. Then $\phi^{\prime}:=\phi-\frac{a}{|z|}z^{n-1}E$
satisfies $\phi^{\prime}(z)=0$. So $\phi^{\prime}$ is ozone. Therefore
$\phi=\frac{a}{|z|}z^{n-1}E+\phi^{\prime}$.

Case 2: $\deg(z)\nmid i$. Since $\phi(z)$ is central, it must be $0$.
Therefore $\phi$ is ozone.

Combining these two cases, every Poisson derivation is the sum of $fE$ for
some $f\in Z$ and an ozone derivation.

(2) By Lemma 4.4(4), it suffices to show that $\dim_{\Bbbk}Gpd(A)=1$, or
equivalently, $Gpd(A)=\Bbbk E$. Let $\phi\in Gpd(A)$ and $\phi(z)=az$ for some
$a\in\Bbbk$. Let $\delta$ be $\phi-\frac{a}{|z|}E$. Then $\delta\in Gpd(A)$ is
ozone. By the hypothesis, $\delta$ is Hamiltonian, say $\delta=H_{f}$ for some
homogeneous element $f\in A$. Since $\deg(\delta)=0$, $\deg(f)=0$. Since $A$
is connected graded, $H_{f}=0$ and consequently, $\delta=0$. Thus
$\phi=\frac{a}{|z|}E$ and $Gpd(A)=\Bbbk E$. ∎

Now we are ready to prove Theorem 0.6.

###### Proof of Theorem 0.6.

(1) $\Rightarrow$ (2): Since $rgt(A)=0$, every Poisson derivation $\delta$ is
of the form $cE$. Then $E\wedge\delta=0$. By (E2.3.1), $\langle
a,b\rangle=\\{a,b\\}$. So $A=A^{\delta}$. The assertion follows.

(2) $\Rightarrow$ (1): By Corollary 0.3, there is a Poisson derivation
$\delta$ such that $A^{\delta}$ is unimodular. Since $A^{\delta}\cong A$ for
all $\delta$, $A$ is unimodular. Suppose to the contrary that $A$ is not
rigid. Then there is a Poisson derivation $\delta$ not in $ZE$. Thus, by
Theorem 0.2, the modular derivation of $A^{\delta}$ is

${\mathbf{n}}=0+(\sum_{i=1}^{3}\deg(x_{i}))\delta-\operatorname{div}(\delta)E$

which cannot be zero as $\operatorname{div}(\delta)\in\Bbbk$ [Lemma 1.2(3)].
Therefore $A^{\delta}$ is not isomorphic to $A$, yielding a contradiction.

(5) $\Leftrightarrow$ (6): Under the hypothesis of (5), $A$ is
$PH^{1}$-minimal. One implication follows by Lemma 7.5 and the other is clear.

(6) $\Rightarrow$ (7): See the proof of Lemma 7.5.

(7) $\Rightarrow$ (1): This is Lemma 7.7(2).

(3) $\Leftrightarrow$ (5): This is Proposition 7.4.

(1) $\Leftrightarrow$ (8): This is Corollary 6.7.

(8) $\Rightarrow$ (6,7): If $\Omega=\frac{1}{3}(x^{3}+y^{3}+z^{3})+\lambda
xyz$ with $\lambda^{3}\neq-1$, then by the comments before Lemma 7.5, $A$ is
$PH^{1}$-minimal. Hence $A$ is $H$-ozone since (5) $\Leftrightarrow$ (6). If
$\Omega$ is $x^{3}+y^{2}z$ or $x^{3}+x^{2}z+y^{2}z$, it follows by Lemmas 6.9
and 6.10 that $A$ is $H$-ozone. In all three cases in Example 6.6, one can
check easily that $Pd(A)_{\leq-1}=0$. By Lemma 7.7(1), $A$ has an Euler-ozone
decomposition. Since $A$ is $H$-ozone, $A$ has an Euler-Hamiltonian
decomposition.

(5) $\Rightarrow$ (1): The assertion follows from Remark 5.2.

(4) $\Rightarrow$ (1): The assertion follows from Remark 5.2

(8) $\Rightarrow$ (4): In all three cases, $Z$ is $\Bbbk[\Omega]$ (this is a
well-known fact and a special case of it is [MTU, Lemma 1], also see Lemmas
7.8 and 7.9 later), which has Hilbert series $\frac{1}{1-t^{3}}$. The
assertion follows from (5) since (5) is equivalent to (8).

(4) $\Leftrightarrow$ (9): It follows from (E1.5.5) that
$h_{PH^{3}(A)}(t)-h_{PH^{2}(A)}(t)=h_{PH^{0}(A)}(t)-h_{PH^{1}(A)}(t)+t^{-3}$.
We know that $PH^{0}(A)=Z$. So the assertion follows from the fact that
$h_{PH^{1}(A)}(t)=h_{PH^{0}(A)}(t)=\frac{1}{1-t^{3}}$ if and only if
$h_{PH^{3}(A)}(t)-h_{PH^{2}(A)}(t)=t^{-3}$. ∎

Before we prove Corollary 0.7, we will need to calculate the Hilbert series of
$PH_{0}(A)$ for $A$ in the first two cases of Example 6.6. By definition and
[Pe2, p.2357], the 0th Poisson homology of the Poisson polynomial algebra
$\Bbbk[x,y,z]$ is

$PH_{0}(A)\cong\frac{A}{\\{A,A\\}}=\frac{A}{(H_{x}(A)+H_{y}(A)+H_{z}(A))}.$

Case 1: $\Omega=x^{3}+y^{2}z$. We use the $G$-grading introduced in the proof
of Lemma 6.9, namely, $\deg_{G}(x)=0,\deg_{G}(y)=1$ and $\deg_{G}(z)=-2$. Let

$\displaystyle a_{i,j,0,l}$ $\displaystyle=x^{i}y^{j}z^{0}\Omega^{l},\quad
i,j,l\geq 0,$ $\displaystyle b_{i,0,k,l}$
$\displaystyle=x^{i}y^{0}z^{k}\Omega^{l},\quad i,l\geq 0,k\geq 1,$
$\displaystyle c_{i,1,k,l}$ $\displaystyle=x^{i}yz^{k}\Omega^{l},\quad i,l\geq
0,k\geq 1,$ $\displaystyle{\mathbb{A}}$ $\displaystyle:=\\{a_{i,j,0,l}\mid
i,j,l\geq 0\\},$ $\displaystyle{\mathbb{B}}$
$\displaystyle:=\\{b_{i,0,k,l}\mid i,l\geq 0,k\geq 1\\},$
$\displaystyle{\mathbb{C}}$ $\displaystyle:=\\{c_{i,1,k,l}\mid i,l\geq 0,k\geq
1\\}.$

If $X$ is a subset of elements in $A$, we use $\Bbbk X$ to denote the
$\Bbbk$-linear span of $X$.

######  Lemma 7.8.

Retain the above notations.

  1. (1)

${\mathbb{A}}\cup{\mathbb{B}}\cup{\mathbb{C}}$ is a $\Bbbk$-linear basis of
$A$.

  2. (2)

$\displaystyle H_{x}(a_{i,j,0,l})$ $\displaystyle=ja_{i,j+1,0,l},$
$\displaystyle H_{x}(b_{i,0,k,l})$ $\displaystyle=(-2k)c_{i,1,k,l},$
$\displaystyle H_{x}(c_{i,1,k,l})$
$\displaystyle=\begin{cases}(1-2k)(b_{i,0,k-1,l+1}-b_{i+3,0,k-1,l})&k>1,\\\
(1-2k)(a_{i,0,0,l+1}-a_{i+3,0,0,l})&k=1.\end{cases}$

  3. (3)

$\displaystyle H_{y}(a_{i,j,0,l})$ $\displaystyle=(-i)a_{i-1,j+2,0,l},$
$\displaystyle H_{y}(b_{i,0,k,l})$
$\displaystyle=\begin{cases}(-i)b_{i-1,0,k-1,l+1}+(i+3k)b_{i+2,0,k-1,l}&k>1,\\\
(-i)a_{i-1,0,0,l+1}+(i+3)a_{i+2,0,0,l}&k=1,\end{cases}$ $\displaystyle
H_{y}(c_{i,1,k,l})$
$\displaystyle=\begin{cases}(-i)c_{i-1,1,k-1,l+1}+(i+3k)c_{i+2,1,k-1,l}&k>1,\\\
(-i)a_{i-1,1,0,l+1}+(i+3)a_{i+2,1,0,l}&k=1.\end{cases}$

  4. (4)

$\displaystyle H_{z}(a_{i,j,0,l})$
$\displaystyle=\begin{cases}2ia_{i-1,j-1,0,l+1}+(-2i-3j)a_{i+2,j-1,0,l}&j>0,\\\
2ic_{i-1,1,1,l}&j=0,\end{cases}$ $\displaystyle H_{z}(b_{i,0,k,l})$
$\displaystyle=2ic_{i-1,1,k+1,l},$ $\displaystyle H_{z}(c_{i,1,k,l})$
$\displaystyle=2ib_{i-1,0,k,l+1}+(-3-2i)b_{i+2,0,k,l}.$

  5. (5)

$A/(H_{x}(A)+H_{y}(A)+H_{z}(A))$ has a $\Bbbk$-linear basis

$\\{a_{0,0,0,0},a_{1,0,0,0}\\}\cup\\{a_{0,1,0,l}\\}_{l\geq
0}\cup\\{a_{1,1,0,l}\\}_{l\geq 0}\cup\\{b_{0,0,k,0}\\}_{k\geq
1}\cup\\{b_{1,0,k,0}\\}_{k\geq 1}.$

  6. (6)

The Hilbert series of $PH_{0}(A)$ is

$\frac{(1+t)^{3}}{1-t^{3}}.$

  7. (7)

The Hilbert series of $PH^{0}(A)=h_{Z}(t)$ is

$\frac{1}{1-t^{3}}.$

###### Proof.

This follows from a tedious and direct computation. ∎

The proof of the above lemma is routine and long, but very elementary, only
using easy linear algebra arguments. To save space the details are omitted
here. Note that Lemma 7.8(7) is a well-known fact.

Case 2: $\Omega=x^{3}+x^{2}z+y^{2}z$. We need to prove a lemma similar to
Lemma 7.8. We use the same notations as in Case 1 except that $\Omega$ is
$x^{3}+x^{2}z+y^{2}z$ instead of $x^{3}+y^{2}z$.

######  Lemma 7.9.

Let $A$ be as in Example 6.6(Case 2) with potential
$\Omega=x^{3}+x^{2}z+y^{2}z$.

  1. (1)

${\mathbb{A}}\cup{\mathbb{B}}\cup{\mathbb{C}}$ is a $\Bbbk$-linear basis of
$A$.

  2. (2)

$\displaystyle H_{x}(a_{i,j,0,l})$
$\displaystyle=ja_{i,j+1,0,l}+ja_{i+2,j-1,0,l},$ $\displaystyle
H_{x}(b_{i,0,k,l})$ $\displaystyle=(-2k)c_{i,1,k,l},$ $\displaystyle
H_{x}(c_{i,1,k,l})$
$\displaystyle=\begin{cases}(1-2k)(b_{i,0,k-1,l+1}-b_{i+3,0,k-1,l})+2kb_{i+2,0,k,l}&k>1,\\\
-(a_{i,0,0,l+1}-a_{i+3,0,0,l})+2b_{i+2,0,1,l}&k=1.\end{cases}$

  3. (3)

$\displaystyle H_{y}(a_{i,j,0,l})$
$\displaystyle=(-i)a_{i-1,j+2,0,l}+(-i)a_{i+1,j,0,l},$ $\displaystyle
H_{y}(b_{i,0,k,l})$
$\displaystyle=\begin{cases}(-i)b_{i-1,0,k-1,l+1}+(i+3k)b_{i+2,0,k-1,l}+2kb_{i+1,0,k,l}&k>1,\\\
(-i)a_{i-1,0,0,l+1}+(i+3)a_{i+2,0,0,l}+2b_{i+1,0,1,l}&k=1,\end{cases}$
$\displaystyle H_{y}(c_{i,1,k,l})$
$\displaystyle=\begin{cases}(-i)c_{i-1,1,k-1,l+1}+(i+3k)c_{i+2,1,k-1,l}+2kc_{i+1,1,k,l}&k>1,\\\
(-i)a_{i-1,1,0,l+1}+(i+3)a_{i+2,1,0,l}+2c_{i+1,1,1,l}&k=1.\end{cases}$

  4. (4)

$\displaystyle H_{z}(a_{i,j,0,l})$
$\displaystyle=\begin{cases}2ic_{i-1,1,1,l}&j=0,\\\
2ia_{i-1,0,0,l+1}+(-2i-3)a_{i+2,0,0,l}+(-2i-2)b_{i+1,0,1,l}&j=1,\\\
2ia_{i-1,1,0,l+1}+(-2i-6)a_{i+2,1,0,l}+(-2i-4)c_{i+1,1,1,l}&j=2,\\\
2ia_{i-1,j-1,0,l+1}+(-2i-3j)a_{i+2,j-1,0,l}&\\\
\qquad\qquad+(-2i-2j)x^{i+1}y^{j-1}z\Omega^{l}&j\geq 3,\\\ \end{cases}$
$\displaystyle H_{z}(b_{i,0,k,l})$ $\displaystyle=2ic_{i-1,1,k+1,l},$
$\displaystyle H_{z}(c_{i,1,k,l})$
$\displaystyle=2ib_{i-1,0,k,l+1}+(-3-2i)b_{i+2,0,k,l}+(-2-2i)b_{i+1,0,k+1,l}.$

  5. (5)

$A/(H_{x}(A)+H_{y}(A)+H_{z}(A))$ has a $\Bbbk$-linear basis

$\\{a_{0,0,0,0},a_{1,0,0,0}\\}\cup\\{a_{3i,1,0,0}\\}_{i\geq
0}\cup\\{a_{1+3i,1,0,0}\\}_{i\geq 0}\cup\\{b_{0,0,k,0}\\}_{k\geq
1}\cup\\{b_{1,0,k,0}\\}_{k\geq 1}.$

  6. (6)

The Hilbert series of $PH_{0}(A)$ is

$\frac{(1+t)^{3}}{1-t^{3}}.$

  7. (7)

The Hilbert series of $PH^{0}(A)=h_{Z}(t)$ is

$\frac{1}{1-t^{3}}.$

###### Proof.

This follows from a tedious and direct computation. ∎

Similar to Lemma 7.8, the proof of Lemma 7.9 is routine and long (even longer
than the proof of Lemma 7.8), but still very elementary. To save space the
details are omitted here. Note that Lemma 7.9(7) is a well-known fact.

Finally we prove Corollary 0.7.

###### Proof of Corollary 0.7.

(1) This is clear since $Z=\Bbbk[\Omega]$ by Lemmas 7.8 and 7.9.

(2) Follows from Theorem 0.6((8)$\Rightarrow$(4)).

(4) By Poincaré duality [LuWW1, Theorem 3.5],
$h_{PH^{3}(A)}(t)=t^{-3}h_{PH_{0}(A)}$. Then the assertion follows from Lemmas
7.8 and 7.9.

(3) Follows from Parts (1,2,4) and (E1.5.5). ∎

### Acknowledgments

Wang was partially supported by Simons collaboration grant #688403 and Air
Force Office of Scientific Research grant FA9550-22-1-0272. Zhang was
partially supported by the US National Science Foundation (No. DMS-1700825 and
DMS-2001015). Part of this research work was done during the first and second
authors’ visit to the Department of Mathematics at University of Washington in
January 2022. They are grateful for the third author’s invitation and wish to
thank University of Washington for its hospitality.

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