# Invariants of Unimodular Quadratic Polynomial Poisson Algebras of Dimension
3

CHENGYUAN MA  Department of Mathematics, University of Washington<EMAIL_ADDRESS>
###### Abstract.

Let $P=\Bbbk[x_{1},x_{2},x_{3}]$ be a unimodular quadratic Poisson algebra and
let $G$ be a finite subgroup of the graded Poisson automorphism group of $P$.
In this paper, we prove a variant of the Shephard-Todd-Chevalley Theorem for
the Poisson algebra $P$ and variants of both the Shephard-Todd-Chevalley
Theorem and the Watanabe Theorem for its Poisson enveloping algebra $U(P)$
under the induced action of $G$.

###### Key words and phrases:

Poisson alegbra $\cdot$ Invariant subalgebra $\cdot$ Reflection $\cdot$
Rigidity $\cdot$ Homological Determinant

###### 2020 Mathematics Subject Classification:

17B63, 16R30

##  0\. Introduction

Throughout $\Bbbk$ is an algebraically closed field of characteristic 0. Let
$A=\Bbbk[x_{1},\cdots,x_{n}]$ and let $G$ be a finite subgroup of the graded
automorphism group of $A$. The invariant subalgebra of $A$ under the action of
$G$ is

$A^{G}:=\left\\{a\in A:\phi(a)=a\textnormal{ for all }\phi\in G\right\\}.$

It is natural to ask: what properties, in particular, what homological
properties does the invariant subalgebra $A^{G}$ satisfy? Two of the earliest
answers are encapsulated in the Shephard-Todd-Chevalley Theorem and the
Watanabe Theorem, articulated as follows:

######  Theorem 0.1.

(Shephard-Todd-Chevalley Theorem, [18], [5]) Let $A=\Bbbk[x_{1},\cdots,x_{n}]$
and $G$ be a finite subgroup of the graded automorphism group of $A$. Then the
invariant subalgebra $A^{G}$ is regular (or equivalently, $A^{G}\cong A$ as
$\Bbbk$-algebras) if and only if $G$ is generated by (pseudo-)reflections.

######  Theorem 0.2.

(Watanabe Theorem, [20]) Let $A=\Bbbk[x_{1},\cdots,x_{n}]$ and $G$ be a finite
subgroup of the graded automorphism group of $A$ containing no
(pseudo-)reflections.Then the invariant subalgebra $A^{G}$ is Gorenstein if
and only if $\det(\phi\big{|}_{A_{1}})$ = 1 for all $\phi\in G$.

In the following decades, the Shephard-Todd-Chevalley Theorem and the Watanabe
Theorem have evolved into pivotal motivations for invariant theory,
particularly in non-commutative settings: if $A$ is an Artin-Schelter regular
algebra and $G$ is a finite subgroup of the graded automorphism group of $A$,
under what conditions on $G$ is the invariant subalgebra $A^{G}$ Artin-
Schelter regular or Artin-Schelter Gorenstein? Artin-Schelter regularity,
initially introduced in [2], emerges as a non-commutative adaption of the
polynomial rings because of its fulfillment of a spectrum of properties
inherent to polynomial rings. In a more formal manner:

######  Definition 0.3.

A finitely generated $\Bbbk$-algebra $A$ is called Artin-Schelter regular if

  1. (1)

$A$ is connected $\mathbb{N}$-graded: $A$ admits a $\Bbbk$-vector space
decomposition $A=\displaystyle{\bigoplus_{n\in\mathbb{N}}A_{n}}$ such that
$A_{0}=\Bbbk$ and $A_{i}A_{j}\subseteq A_{i+j}$ for all $i,j\in\mathbb{N}$.

  2. (2)

$A$ has finite Gelfand-Kirillov dimension: $\dim_{\Bbbk}A_{n}$ has polynomial
growth.

  3. (3)

$A$ has finite global dimension $d$.

  4. (4)

$A$ is Gorenstein: $\text{Ext}_{A}^{i}(\Bbbk,A)\cong\begin{cases}0&i\neq d\\\
\Bbbk(l)&i=d\end{cases}$, for some $l\in\mathbb{Z}$.

Returning to our discussion of the non-commutative Shephard-Todd-Chevalley
question and the non-commutative Watanabe question. There are some established
answers, including, but not limited to, universal enveloping algebra of
semisimple Lie algebras and Weyl algebras in [1], non-PI Sklyanin algebras of
global dimension $\geq$ 3 in [10], skew polynomial rings and quantum matrix
algebras in [11], down-up algebras in [12]. Recently, Gaddis, Veerapen, and
Wang have proposed an investigation into these questions within the realm of
Poisson algebras. Broadly speaking, Poisson algebras are a family of
commutative algebras endowed with a non-commutative bracket. In a more
rigorous language:

######  Definition 0.4.

A Poisson algebra is a commutative $\Bbbk$-algebra $P$ together with a
bracket:

$\\{-,-\\}:P^{\otimes 2}\to P$

such that

  1. (1)

$(P,\\{-,-\\})$ is a Lie algebra over $\Bbbk$, namely $\\{-,-\\}$ satisfies
bilinearity, alternativity, anti-commutativity, and the Jacobi identity.

  2. (2)

$\\{-,-\\}$ satisfies Leibniz rule: $\\{a,bc\\}=\\{a,b\\}c+b\\{a,c\\}$ for all
$a,b,c\in P$.

Poisson algebras originally emerged in classical mechanics and subsequently
assumed a significant role in mathematical physics. In recent decades, Poisson
algebras have also garnered attention in pure mathematics. This heightened
interest is partially attributed to their approximity with Artin-Schelter
regular algebras. One instance of such appromixity is Poisson enveloping
algebras.

######  Definition 0.5.

Let $P$ be a Poisson algebra. A (The) Poisson enveloping algebra of $P$ is a
triple $(U,\alpha,\beta)$:

  * •

$U$ is a $\Bbbk$-algebra,

  * •

$\alpha:(P,\cdot)\to U$ is an algebra homomorphism,

  * •

$\beta:(P,\\{-,-\\})\to U_{L}$ is a Lie algebra homomorphism,

subjecting to the following conditions:

  1. (1)

$\alpha(\\{a,b\\})=\beta(a)\alpha(b)-\alpha(b)\beta(a)$ for all $a,b\in P$.

  2. (2)

$\beta(ab)=\alpha(a)\beta(b)+\alpha(b)\beta(a)$ for all $a,b\in P$.

  3. (3)

If $(U^{\prime},\alpha^{\prime},\beta^{\prime})$ is another triple satisfying
(1) and (2), then there exists a unique algebra homomorphism $h:U\to
U^{\prime}$ making the following diagram commutative:

${U}$${P}$${U^{\prime}}$$\scriptstyle{h}$$\scriptstyle{\alpha^{\prime},\beta^{\prime}}$$\scriptstyle{\alpha,\beta}$

If $P=\Bbbk[x_{1},\cdots,x_{n}]$ is a quadratic Poisson algebra, then its
Poisson enveloping algebra $U(P)$ satisifies a range of preferred properties,
including being Artin-Schelter regular. This is one connection between Poisson
algebras and Artin-Schelter regular algebras. Another connection is found in
the context of semiclassical limits and deformation quantizations. However,
this topic will not be discussed in this paper but will be explored in a
forthcoming paper.

Poisson Algebras  Artin-Schelter Regular Algebrasdeformation
quantizationsPoisson Enveloping Algebrassemiclassical limits

Given that Poisson algebras exhibit a pronounced association with Artin-
Schelter regular algebras through semiclassical limit, deformation
quantizations and Poisson enveloping algebras, there is a strong interrelation
between investigations pertaining to the Shephard-Todd-Chevalley question and
the Watanabe question in the context of Poisson algebras and investigations of
these questions in the context of Artin-Schelter regular algebras. In their
study [7], Gaddis, Veerapen, and Wang provided a partial answer to the
Shephard-Todd-Chevalley question by investigating multiple Poisson structures
arising from the semiclassical limits of specific families of Artin-Schelter
regular algebras. Additionally, they offered valuable insights on the Watanabe
question for Poisson enveloping algebras under induced actions. Building upon
their research, we shall further investigate these questions with regard to a
broader range of Poisson structures, with a primary emphasis on quadratic
Poisson structures on the polynomial ring of three variables
$\Bbbk[x_{1},x_{2},x_{3}]$.

In section 2, we provide an algorithm for classifying graded Poisson
automorphisms of a quadratic Poisson structure on $\Bbbk[x_{1},\cdots,x_{n}]$.
We then apply this algorithm to unimodular quadratic Poisson structures on
$\Bbbk[x_{1},x_{2},x_{3}]$ in Proposition 2.2 and Proposition 2.3.

In section 3, we prove one of the main theorem of this paper, a graded
rigidity theorem answering the Shephard-Todd-Chevalley question for unimodular
quadratic Poisson structures on $\Bbbk[x_{1},x_{2},x_{3}]$.

######  Theorem 0.6.

(Theorem 3.1) Let $P=\Bbbk[x_{1},x_{2},x_{3}]$ be a unimodular quadratic
Poisson algebra and let $G\subseteq\text{PAut}_{\text{gr}}(P)$ be a finite
subgroup. Then the invariant subalgebra $P^{G}$ is isomorphic to $P$ as
Poisson algebras if and only if $P$ is trivial.

For Poisson algebras lacking Poisson reflections, Theorem 3.1. is an immediate
consequence of the classical Shephard-Todd-Chevalley Theorem. For the
remaining unimodular quadratic Poisson structures on
$\Bbbk[x_{1},x_{2},x_{3}]$, we provide a case-by-case proof, building upon the
classification in Proposition 2.2 and Proposition 2.3. This result offers
additional instances of graded rigid Poisson algebras beyond those examined in
[7]. As it transpires, most invariant subalgebras of unimodular quadratic
Poisson structures on $\Bbbk[x_{1},x_{2},x_{3}]$ under Poisson reflection
groups fail to preserve unimodularity. There exists two invariant subalgebras
that retain unimodularity; nonetheless, they are not isomorphic to the
original Poisson algebra. Up to this point, we do not have any example of
Poisson algebra that is not graded rigid.

In section 4, we prove the remaining two main theorems of this paper, an
answer to the Shephard-Todd-Chevalley question and a formula for computing the
homological determinant of an induced automorphism for Poisson enveloping
algebras of quadratic Poisson structures on $\Bbbk[x_{1},\cdots,x_{3}]$:

######  Theorem 0.7.

(Theorem 4.4) Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a quadratic Poisson algebra
and $U(P)$ be its Poisson enveloping algebra. Let $G$ be a finite nontrivial
subgroup of the graded Poisson automorphism group of $P$ and let
$\widetilde{G}$ be the corresponding finite nontrivial subgroup of the graded
automorphism group of $U(P)$. The invariant subalgebra $U(P)^{\widetilde{G}}$
is Artin-Schelter regular if and only if $G$ is trivial.

######  Theorem 0.8.

(Theorem 4.7) Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a quadratic Poisson algebra
and let $U(P)$ be its Poisson enveloping algebra. Let $\phi$ be a finite-order
graded Poisson automorphism of $P$ and let $\widetilde{\phi}$ be the induced
graded automorphism of $U(P)$. Then

$\text{hdet}\widetilde{\phi}=(\det\phi\big{|}_{P_{1}})^{2}.$

The foundation for both Theorem 4.4 and Theorem 4.7 lies in the observations
made in Lemma 4.2: if a graded Poisson automorphism $\phi$ of a quadratic
Poisson structure on $\Bbbk[x_{1},\cdots,x_{n}]$ has eigenvalues
$\lambda_{1},\cdots,\lambda_{m}$, with multiplicity $c_{1},\cdots,c_{m}$, then
the induced automorphism $\widetilde{\phi}$ of the Poisson enveloping algebra
has $\lambda_{1},\cdots,\lambda_{m}$ with multiplicity $2c_{1},\cdots,2c_{m}$
(it is noteworthy that this observation provides a generalization of [7,
Theorem 5.6]). For Theorem 4.4, we compare the eigenvalues of
$\widetilde{\phi}$ with the eigenvalues of quasi-reflections of $U(P)$
described in [10, Theorem 3.1] and conclude that $\widetilde{\phi}$ cannot be
a quasi-reflection. For Theorem 4.7, we derive a formula for computing the
trace series of $\widetilde{\phi}$ and relate the trace series to the
homological determinant of $\widetilde{\phi}$ as in [9, Lemma 2.6].
Ultimately, we demonstrate the significance of Theorem 4.7 in Corollary 4.8,
answering the Watanabe question for Poisson enveloping algebras of quadratic
Poisson structures on $\Bbbk[x_{1},\cdots,x_{n}]$.

Before delving further into the results in this paper, I would like to take a
moment to acknowledge and express my sincerest gratitude to my advisor James
Zhang for his invaluable advice and guidance throughout this research project;
in addition, I would like to thank Professor Xingting Wang for his reviews and
remarks on this paper.

##  1\. Preliminaries

Let $P=\Bbbk[x_{1},x_{2},x_{3}]$ be a Poisson algebra under the standard
grading. $P$ is called quadratic if $\\{P_{1},P_{1}\\}\subseteq P_{2}$. Dufour
and Haraki have classified all quadratic Poisson structures on $P$ into 13 + 1
classes [6, Theorem 2]. In this paper, we are primarily interested in the + 1
class: a class consisting of Poisson structures of the following form:

$\\{x_{1},x_{2}\\}=\frac{\partial\Omega}{\partial
x_{3}},\quad\\{x_{2},x_{3}\\}=\frac{\partial\Omega}{\partial
x_{1}},\quad\\{x_{3},x_{1}\\}=\frac{\partial\Omega}{\partial x_{2}},$

for some homogeneous polynomial $\Omega$ of $x_{1},x_{2},x_{3}$ of degree 3.
Such Poisson structures are called Jacobian. In some literature, such Poisson
structures are alternatively referred to as unimodular, owing to [13,
Proposition 2.6]’s demonstration of their equivalence when
$P=\Bbbk[x_{1},x_{2},x_{3}]$. The homogeneous polynomial $\Omega$ were
classified into 9 subclasses up to some scalar in [3]:

Case |  $\boldsymbol{\Omega}$
---|---
1 |  $x_{1}^{3}$
2 |  $x_{1}^{2}x_{2}$
3 |  $2x_{1}x_{2}x_{3}$
4 |  $x_{1}^{2}x_{2}+x_{1}x_{2}^{2}$
5 |  $x_{1}^{3}+x_{2}^{2}x_{3}$
6 |  $x_{1}^{3}+x_{1}^{2}x_{3}+x_{2}^{2}x_{3}$
7 |  $\frac{1}{3}(x_{1}^{3}+x_{2}^{3}+x_{3}^{3})+\lambda x_{1}x_{2}x_{3},\lambda^{3}\neq-1$
8 |  $x_{1}^{3}+x_{1}^{2}x_{2}+x_{1}x_{2}x_{3}$
9 |  $x_{1}^{2}x_{3}+x_{1}x_{2}^{2}$

In this section, we will review the necessary concepts and tools requisite for
the studies of these Poisson algebras.

Let $P,Q$ be Poisson algebras. A map $\phi:P\to Q$ is called a Poisson
homomorphism if $\phi$ is an algebra homomorphism and a Lie algebra
homomorphism.

Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a Poisson algebra under the standard
grading. A graded Poisson automorphism of $P$ is a bijective Poisson
homomorphism $\phi:P\to P$ such that $\phi(P_{i})=P_{i}$ for all $i\geq 0$.
The graded Poisson automorphism group of $P$ will be denoted as
$\text{PAut}_{\text{gr}}(P)$. A Poisson reflection of $P$ is a finite-order
graded Poisson automorphism $\phi:P\to P$ such that $\phi\big{|}_{P_{1}}$ has
the following eigenvalues: $\underbrace{1,\cdots,1}_{n-1},\xi$, for some
primitive $m$th root of unity $\xi$. The set consisting of all Poisson
reflections of $P$ will be denoted as $\text{PR}(P)$.

Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a Poisson algebra. Its Poisson enveloping
algebra $U(P)$ is necessarily unique and can be described by an explicit set
of generators and relations as follows:

######  Theorem 1.1.

[17] Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a Poisson algebra. Then $U(P)$ is
the free $\Bbbk$-algebra generated by $x_{1},\cdots,x_{n},y_{1},\cdots,y_{n}$,
subject to the following relations:

  1. (1)

$[x_{i},x_{j}]=0$,

  2. (2)

$[y_{i},y_{j}]=\displaystyle{\sum_{k=1}^{n}\frac{\partial\\{x_{i},x_{j}\\}}{\partial
x_{k}}y_{k}}$,

  3. (3)

$[y_{i},x_{j}]=\\{x_{i},x_{j}\\}$,

for all $1\leq i,j\leq n$, and $\alpha,\beta$ are defined as the follows:

$\displaystyle\alpha:P\to$ $\displaystyle U(P),\hskip 21.68121pt\beta:P\to
U(P)$ $\displaystyle f\mapsto$ $\displaystyle f,\hskip
54.92479ptf\mapsto\sum_{k=1}^{n}\frac{\partial f}{\partial x_{k}}y_{k}.$

The salience of Poisson enveloping algebras within the research of invariant
theory of Poisson algebras lies in its role as one of the two intermediary
links connecting Poisson algebras to Artin-Schelter regular algebras. Let
$P=\Bbbk[x_{1},\cdots,x_{n}]$ be a quadratic Poisson algebra. Then $U(P)$ is
Artin-Schelter regular [14, Corollary 1.5] and satisfies a range of preferred
qualities:

  1. (1)

$U(P)$ is Noetherian [16, Proposition 9].

  2. (2)

$U(P)$ admits a Poincaré-Birkhoff-Witt basis $\\{x_{1}^{i_{1}}\cdots
x_{n}^{i_{n}}y_{1}^{j_{1}}\cdots y_{n}^{j_{n}}:i_{r},j_{s}\geq 0\\}$ [17,
Theorem 3.7],

  3. (3)

$U(P)$ has global dimension $2n$ [4, Proposition 2.1].

  4. (4)

The Hilbert series $h_{U(P)}(t)=\displaystyle{\frac{1}{(1-t)^{2n}}}$ [7, Lemma
5.4].

Because $U(P)$ is an Artin-Schelter regular algebra, we can extend the
commutative concepts of “reflections” and “determinant” to it.

Let $A$ be a connected $\mathbb{N}$-graded, locally finite $\Bbbk$-algebra and
let $\phi$ be a graded automorphism of $A$. The trace series of $\phi$ is
$\text{Tr}_{A}(\phi,t)=\displaystyle{\sum_{i=0}^{\infty}\text{tr}(\phi\big{|}_{A_{i}})t^{i}}$.
In particular, if $\phi=\text{id}_{A}$, we recover the Hilbert series:
$\text{Tr}_{A}(\text{id}_{A},t)=h_{A}(t)$.

A practical application of the trace series is the computation of the Hilbert
series of invariant subalgebras:

######  Theorem 1.2.

(Molien’s Theorem) Let $A$ be a connected $\mathbb{N}$-graded, locally finite
$\Bbbk$-algebra and let $G\subseteq\text{Aut}_{\text{gr}}(A)$ be a finite
subgroup. Then

$h_{A^{G}}(t)=\frac{1}{|G|}\sum_{\phi\in G}\text{Tr}_{A}(\phi,t).$

Let $A$ be an Artin-Schelter regular algebra with Hilbert series
$h_{A}(t)=\displaystyle{\frac{1}{(1-t)^{n}f(t)}}$ for some $f(t)$ satisfying
$f(1)\neq 0$. A quasi-reflection of $A$ is a finite-order graded automorphism
$\phi:A\to A$ such that
$\text{Tr}_{A}(\phi,t)=\displaystyle{\frac{1}{(1-t)^{n-1}g(t)}}$ for some
$g(t)$ satisfying $g(1)\neq 0$. In particular, if $A$ is a commutative
polynomial ring, we recover the notion of a classical reflection:
$\phi\big{|}_{A_{1}}$ has eigenvalues $\underbrace{1,\cdots,1}_{n-1},\xi$ for
some primitive $m$th root of unity $\xi$. If $A$ is a noncommutative quantum
polynomial ring generated in degree 1, such as the Poisson enveloping algebra
of a quadratic Poisson algebra, [10, Theorem 3.1] proves that a quasi-
reflection $\phi$ of $A$ necessarily takes one of the following form:

  * •

$\phi\big{|}_{A_{1}}$ has eigenvalues $\underbrace{1,\cdots,1}_{n-1},\xi$ for
some primitive $m$th root of unity $\xi$,

  * •

$\phi$ has order 4 and $\phi\big{|}_{A_{1}}$ has eigenvalues
$\underbrace{1,\cdots,1}_{n-2},i,-i$.

Let $A$ be an Artin-Schelter Gorenstein algebra of injective dimension $n$ and
let $\phi$ be a graded automorphism of $A$. [9, Lemma 2.2] proved that the
$\phi$-linear map $H_{{A}_{\geq 1}}^{n}(A)\to\hskip
0.72229pt^{\sigma}H_{{A}_{\geq 1}}^{n}(A)$ necessarily equals to the product
of a nonzero scalar $c$ and the Matlis dual map $(\phi^{-1})^{\prime}$. [9]
defines the homological determinant of $\phi$ to be
$\text{hdet}\phi=\displaystyle{\frac{1}{c}}$. In particular, if $A$ is a
commutative polynomial ring, [9, page 3222] showed that $\text{hdet}\phi$
coincides with $\det\phi\big{|}_{A_{1}}$.

The introduction of the homological determinant, as one might anticipate, aims
to furish a noncommutative counterpart to the Watanabe Theorem:

######  Theorem 1.3.

(Watanabe Theorem, [9, Theorem 3.3]) Let $A$ be a Noetherian, Artin-Schelter
Gorenstein $\Bbbk$-algebra and let $H\subseteq\text{GrAut}(A)$ be a finite
subgroup. If $\text{hdet}h=1$ for all $h\in H$, then $A^{H}$ is Artin-Schelter
Gorenstein.

##  2\. Graded Poisson Automorphisms and Poisson Reflections

In this section, we classify the graded Poisson automorphisms and Poisson
reflections for unimodular quadratic Poisson structures on
$\Bbbk[x_{1},x_{2},x_{3}]$, laying the foundation for variants of the
Shephard-Todd-Chevalley Theorem and a variant of the Watanabe Theorem for such
Poisson algebras.

Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a Poisson algebra under the standard
grading and let $\phi\in\text{PAut}_{\text{gr}}(P)$. The graded Poisson
automorphism $\phi$ is uniquely determined by its action on
$P_{1}=\displaystyle{\bigoplus_{i=1}^{n}\Bbbk x_{i}}$, allowing it to be
represented as an invertible $n\times n$ matrix:
$\phi=\begin{bmatrix}a_{ij}\end{bmatrix}_{1\leq i,j\leq n}$ for some
$a_{ij}\in\Bbbk$. For $i\leq j$,

$\phi(\\{x_{i},x_{j}\\})=\\{\phi(x_{i}),\phi(x_{j})\\}=\bigg{\\{}\sum_{k=1}^{n}a_{ik}x_{k},\sum_{j=1}^{n}a_{jl}x_{l}\bigg{\\}}=\sum_{1\leq
k<l\leq n}(a_{ik}a_{jl}-a_{il}a_{jk})\\{x_{k},x_{l}\\}.$

In particular, when $n=3$, the calculation presented above can be summarized
as the following lemma:

######  Lemma 2.1.

Let $P=\Bbbk[x_{1},x_{2},x_{3}]$ be one of the unimodular quadratic Poisson
algebra and let $\phi\in\text{PAut}_{\text{gr}}(P)$. Then $\phi$ can be
uniquely represented as an invertible $3\times 3$ matrix over $\Bbbk$:
$\begin{bmatrix}a_{11}&a_{12}&a_{13}\\\ a_{21}&a_{22}&a_{23}\\\
a_{31}&a_{32}&a_{33}\end{bmatrix}$ satisfying the following equations:

  1. (1)

$\phi(\\{x_{1},x_{2}\\})=(a_{11}a_{22}-a_{12}a_{21})\\{x_{1},x_{2}\\}+(a_{12}a_{23}-a_{13}a_{22})\\{x_{2},x_{3}\\}+(a_{13}a_{21}-a_{11}a_{23})\\{x_{3},x_{1}\\}$,

  2. (2)

$\phi(\\{x_{2},x_{3}\\})=(a_{21}a_{32}-a_{22}a_{31})\\{x_{1},x_{2}\\}+(a_{22}a_{33}-a_{23}a_{32})\\{x_{2},x_{3}\\}+(a_{23}a_{31}-a_{21}a_{33})\\{x_{3},x_{1}\\}$,

  3. (3)

$\phi(\\{x_{3},x_{1}\\})=(a_{12}a_{31}-a_{11}a_{32})\\{x_{1},x_{2}\\}+(a_{13}a_{32}-a_{12}a_{33})\\{x_{2},x_{3}\\}+(a_{11}a_{33}-a_{13}a_{31})\\{x_{3},x_{1}\\}$.

In the following two propositions, we provide a complete classification of
graded Poisson automorphisms and Poisson reflections of unimodular quadratic
Poisson structures on $\Bbbk[x_{1},x_{2},x_{3}]$.

######  Proposition 2.2.

Let $P=\Bbbk[x_{1},x_{2},x_{3}]$ be one of the unimodular quadratic Poisson
algebra. Then $\text{PAut}_{\text{gr}}(P)$ contains graded Poisson
automorphisms that are uniquely represented by the following matrices:

Case |  $\boldsymbol{\textbf{PAut}_{\textbf{gr}}(\textbf{P})}$
---|---
1 |  $\begin{bmatrix}\pm\sqrt{bf-ce}&0&0\\\ a&b&c\\\ d&e&f\end{bmatrix}$
2 |  $\begin{bmatrix}a&0&0\\\ 0&b&0\\\ c&d&a\end{bmatrix}$
3 |  $\begin{bmatrix}a&0&0\\\ 0&b&0\\\ 0&0&c\end{bmatrix}$, $\begin{bmatrix}0&a&0\\\ 0&0&b\\\ c&0&0\end{bmatrix}$, $\begin{bmatrix}0&0&a\\\ b&0&0\\\ 0&c&0\end{bmatrix}$
4 |  $\begin{bmatrix}0&a&0\\\ -a&-a&0\\\ b&c&a\end{bmatrix},\begin{bmatrix}0&-a&0\\\ -a&0&0\\\ b&c&a\end{bmatrix},\begin{bmatrix}-a&0&0\\\ a&a&0\\\ b&c&a\end{bmatrix}$,
|  $\begin{bmatrix}-a&-a&0\\\ a&0&0\\\
b&c&a\end{bmatrix},\begin{bmatrix}a&a&0\\\ 0&-a&0\\\
b&c&a\end{bmatrix},\begin{bmatrix}a&0&0\\\ 0&a&0\\\ b&c&a\end{bmatrix}$
5 |  $\begin{bmatrix}a&0&0\\\ 0&a&0\\\ 0&0&a\end{bmatrix}$
6 |  $\begin{bmatrix}a&0&0\\\ 0&a&0\\\ 0&0&a\end{bmatrix}$, $\begin{bmatrix}-\frac{1}{2}a&\pm\frac{\sqrt{3}}{2}a&0\\\ \mp\frac{\sqrt{3}}{2}a&-\frac{1}{2}a&0\\\ \frac{9}{8}a&\mp\frac{3\sqrt{3}}{8}a&a\end{bmatrix}$
7 |  $\left(\langle\begin{bmatrix}a&0&0\\\ 0&a&0\\\ 0&0&a\end{bmatrix}:a\neq 0\rangle\times\langle\begin{bmatrix}1&0&0\\\ 0&b&0\\\ 0&0&b^{2}\end{bmatrix}:b=\xi_{3},\xi_{3}^{2}\rangle\right)\rtimes\langle\begin{bmatrix}0&1&0\\\ 0&0&1\\\ 1&0&0\end{bmatrix}\rangle$,
8 |  $\begin{bmatrix}a&0&0\\\ 0&\frac{a^{2}}{b}&0\\\ b-a&0&b\end{bmatrix}$
9 |  $\begin{bmatrix}a&0&0\\\ b&a&0\\\ -\frac{b^{2}}{a}&-b&a\end{bmatrix}$

######  Proposition 2.3.

Let $P=\Bbbk[x_{1},x_{2},x_{3}]$ be one of the unimodular quadratic Poisson
algebra. Then $\text{PR}(P)$ contains Poisson reflections that are uniquely
represented by the following matrices:

Case |  $\textbf{PR}\boldsymbol{(P)}$
---|---
1 |  $\begin{bmatrix}-1&0&0\\\ a&1&0\\\ d&0&1\end{bmatrix}$
2 |  $\begin{bmatrix}1&0&0\\\ 0&\xi&0\\\ 0&d&1\end{bmatrix}$
3 |  $\begin{bmatrix}1&0&0\\\ 0&1&0\\\ 0&0&\xi\end{bmatrix}$, $\begin{bmatrix}1&0&0\\\ 0&\xi&0\\\ 0&0&1\end{bmatrix}$, $\begin{bmatrix}\xi&0&0\\\ 0&1&0\\\ 0&0&1\end{bmatrix}$
4 |  $\begin{bmatrix}0&-1&0\\\ -1&0&0\\\ b&b&1\end{bmatrix}$, $\begin{bmatrix}-1&0&0\\\ 1&1&0\\\ b&0&1\end{bmatrix}$, $\begin{bmatrix}1&1&0\\\ 0&-1&0\\\ 0&c&1\end{bmatrix}$
5 |  $\emptyset$
6 |  $\emptyset$
7 |  $\emptyset$
8 |  $\begin{bmatrix}-1&0&0\\\ 0&1&0\\\ 2&0&1\end{bmatrix}$
9 |  $\emptyset$

###### Proof.

Unimodular 1. $\\{x_{1},x_{2}\\}=0$, $\\{x_{2},x_{3}\\}=3x_{1}^{2}$,
$\\{x_{3},x_{1}\\}=0$.

Let $\phi\in\text{PAut}_{\text{gr}}(P)$. By Lemma 2.1, we have the following
system of equations, with redundant equations omitted:

  1. (1)

$a_{11}^{2}=a_{22}a_{33}-a_{23}a_{32}.$

  2. (2)

$a_{12}^{2}=0$.

  3. (3)

$a_{13}^{2}=0$.

These relations simplify to $a_{11}=\pm\sqrt{a_{22}a_{33}-a_{23}a_{32}}\neq
0$, $a_{12}=a_{13}=0$. In conclusion,

$\text{PAut}_{\text{gr}}(P)=\left\\{\begin{bmatrix}\pm\sqrt{bf-ce}&0&0\\\
a&b&c\\\ d&e&f\end{bmatrix}:bf\neq ce\right\\}.$

Unimodular 2. $\\{x_{1},x_{2}\\}=0$, $\\{x_{2},x_{3}\\}=2x_{1}x_{2}$,
$\\{x_{3},x_{1}\\}=x_{1}^{2}$.

Let $\phi\in\text{PAut}_{\text{gr}}(P)$. By Lemma 2.1, we have the following
system of equations, with redundant equations omitted:

  1. (1)

$a_{13}a_{21}=a_{11}a_{23}.$

  2. (2)

${2a_{11}a_{21}=a_{23}a_{31}-a_{21}a_{33}}$.

  3. (3)

${a_{11}^{2}=a_{11}a_{33}-a_{13}a_{31}}.$

  4. (4)

${a_{12}^{2}=0}.$

  5. (5)

${a_{13}^{2}=0}.$

By (4) and (5), $a_{12}=a_{13}=0$ and $a_{11}\neq 0$. By substituting the
variables in (1) and (3), $a_{23}=0$ and $a_{33}=a_{11}$. Finally, it can be
deduced from (2) that $a_{21}=0$. In conclusion,

$\text{PAut}_{\text{gr}}(P)=\left\\{\begin{bmatrix}a&0&0\\\ 0&b&0\\\
c&d&a\end{bmatrix}:a,b\neq 0\right\\}.$

Unimodular 3. $\\{x_{1},x_{2}\\}=x_{1}x_{2}$, $\\{x_{2},x_{3}\\}=x_{2}x_{3}$,
$\\{x_{3},x_{1}\\}=x_{1}x_{3}$.

Let $\phi\in\text{Aut}_{\text{gr}}(P)$. By Lemma 2.1, we have the following
system of equations, with redundant equations omitted:

  1. (1)

$a_{11}a_{21}=0$.

  2. (2)

$a_{12}a_{22}=0$.

  3. (3)

$a_{13}a_{23}=0$.

  4. (4)

$a_{11}a_{22}+a_{12}a_{21}=a_{11}a_{22}-a_{12}a_{21}$.

  5. (5)

$a_{11}a_{23}+a_{13}a_{21}=a_{13}a_{21}-a_{11}a_{23}$.

  6. (6)

$a_{12}a_{23}+a_{13}a_{22}=a_{12}a_{23}-a_{13}a_{22}$.

  7. (7)

$a_{11}a_{31}=0$.

  8. (8)

$a_{12}a_{32}=0$.

  9. (9)

$a_{13}a_{33}=0$.

  10. (10)

$a_{11}a_{32}+a_{12}a_{31}=a_{12}a_{31}-a_{11}a_{32}$.

  11. (11)

$a_{11}a_{33}+a_{13}a_{31}=a_{11}a_{33}-a_{13}a_{31}$.

  12. (12)

$a_{12}a_{33}+a_{13}a_{32}=a_{13}a_{32}-a_{12}a_{33}$.

We conduct a case-by-case examination:

  * •

Suppose that $a_{11}\neq 0$. First, (1), (5), (7), (10) lead to
$a_{21}=a_{23}=a_{31}=a_{32}=0$ and $a_{22},a_{33}\neq 0$. Expanding upon
this, (2) and (9) imply that $a_{12}=a_{13}=0$.

  * •

Suppose that $a_{11}=0$ and $a_{12}\neq 0$. First, (2), (4), (8), (12) lead to
$a_{21}=a_{22}=a_{32}=a_{33}=0$ and $a_{23},a_{31}\neq 0$. Expanding upon
this, (3) implies that $a_{13}=0$.

  * •

Suppose that $a_{11}=0$ and $a_{12}=0$. First, the invertibility of $\phi$
implies that $a_{13}\neq 0$. The equations (3), (6), (9), (11) result in
$a_{22}=a_{23}=a_{31}=a_{33}=0$ and $a_{21},a_{32}\neq 0$.

In conclusion,

$\text{PAut}_{\text{gr}}(P)=\left\\{\begin{bmatrix}a&0&0\\\ 0&b&0\\\
0&0&c\end{bmatrix},\begin{bmatrix}0&a&0\\\ 0&0&b\\\
c&0&0\end{bmatrix},\begin{bmatrix}0&0&a\\\ b&0&0\\\
0&c&0\end{bmatrix}:a,b,c\neq 0\right\\}.$

Unimodular 4. $\\{x_{1},x_{2}\\}=0$,
$\\{x_{2},x_{3}\\}=2x_{1}x_{2}+x_{2}^{2}$,
$\\{x_{3},x_{1}\\}=x_{1}^{2}+2x_{1}x_{2}$.

Let $\phi\in\text{Aut}_{\text{gr}}(P)$. By Lemma 2.1, we have the following
system of equations, with redundant equations omitted:

  1. (1)

$2a_{11}a_{21}+a_{21}^{2}=a_{23}a_{31}-a_{21}a_{33}$.

  2. (2)

$2a_{12}a_{22}+a_{22}^{2}=a_{22}a_{33}-a_{23}a_{32}$.

  3. (3)

$2a_{13}a_{23}+a_{23}^{2}=0$.

  4. (4)

$a_{11}a_{22}+a_{12}a_{21}+a_{21}a_{22}=a_{22}a_{33}-a_{23}a_{32}+a_{23}a_{31}-a_{21}a_{33}$.

  5. (5)

$a_{11}^{2}+2a_{11}a_{21}=a_{11}a_{33}-a_{13}a_{31}$.

  6. (6)

$a_{12}^{2}+2a_{12}a_{22}=a_{13}a_{32}-a_{12}a_{33}$.

  7. (7)

$a_{13}^{2}+2a_{13}a_{23}=0$.

It can be deduced from (3) and (7) that $a_{13}=a_{23}=0$ and $a_{33}\neq 0$.
Next, we will proceed with a case-by-case discussion.

  * •

Suppose that $a_{21}\neq 0$ and $a_{11}=0$. From invertibility, $a_{12}\neq
0$. From (1), $a_{21}=-a_{33}$. If $a_{22}\neq 0$, the combination of (2) and
(6) leads to $a_{12}=-a_{22}=a_{33}$ and the remaining equations are
nullified. If $a_{22}=0$, (4) states that $a_{12}=-a_{33}$ and the remaining
equations are nullified. Consequently, we have two possible forms for $\phi$:
$\begin{bmatrix}0&a&0\\\ -a&-a&0\\\ b&c&a\end{bmatrix}$,
$\begin{bmatrix}0&-a&0\\\ -a&0&0\\\ b&c&a\end{bmatrix}$, for some $a\neq 0$.

  * •

Suppose that $a_{21}\neq 0$ and $a_{11}\neq 0$. First, a combination of (1)
and (5) leads to $a_{11}=-a_{21}=-a_{33}$. If $a_{22}\neq 0$, a combination of
(2) and (4) implies $a_{12}=0$ and $a_{22}=a_{33}$ and the remaining equations
are nullified. If $a_{22}=0$, (4) states that $a_{12}=-a_{33}$ and the
remaining equations are nullified. Consequently, we have two possible forms
for $\phi$: $\begin{bmatrix}-a&0&0\\\ a&a&0\\\ b&c&a\end{bmatrix}$,
$\begin{bmatrix}-a&-a&0\\\ a&0&0\\\ b&c&a\end{bmatrix}$, for some $a\neq 0$.

  * •

Suppose that $a_{21}=0$. From invertibility, $a_{22}\neq 0$. If $a_{12}\neq
0$, a combination of (2) and (6) results in $a_{12}=-a_{22}=a_{33}$, (4)
results in $a_{11}=a_{33}$, and the remaining equations are nullified. If
$a_{12}=0$, (2) and (4) imply $a_{11}=a_{22}=a_{33}$ and the remaining
equations are nullified. Consequently, we have two possible forms for $\phi$:
$\begin{bmatrix}a&a&0\\\ 0&-a&0\\\ b&c&a\end{bmatrix}$,
$\begin{bmatrix}a&0&0\\\ 0&a&0\\\ b&c&a\end{bmatrix}$, for some $a\neq 0$.

In conclusion,

$\text{PAut}_{\text{gr}}(P)=\left\\{\begin{bmatrix}0&a&0\\\ -a&-a&0\\\
b&c&a\end{bmatrix},\begin{bmatrix}0&-a&0\\\ -a&0&0\\\
b&c&a\end{bmatrix},\begin{bmatrix}-a&0&0\\\ a&a&0\\\
b&c&a\end{bmatrix},\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}\right\\}\\\
$ $\hskip
84.64711pt\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}\left\\}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\begin{bmatrix}-a&-a&0\\\
a&0&0\\\ b&c&a\end{bmatrix},\begin{bmatrix}a&a&0\\\ 0&-a&0\\\
b&c&a\end{bmatrix},\begin{bmatrix}a&0&0\\\ 0&a&0\\\ b&c&a\end{bmatrix}:a\neq
0\right\\}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color<EMAIL_ADDRESS>
Unimodular 5. $\\{x_{1},x_{2}\\}=x_{2}^{2}$, $\\{x_{2},x_{3}\\}=3x_{1}^{2}$,
$\\{x_{3},x_{1}\\}=2x_{2}x_{3}$.

Let $\phi\in\text{Aut}_{\text{gr}}(P)$. By Lemma 2.1, we have the following
system of equations, with redundant equations omitted:

  1. (1)

$a_{21}^{2}=3a_{12}a_{23}-3a_{13}a_{22}$.

  2. (2)

$a_{23}^{2}=0$.

  3. (3)

$a_{11}^{2}=a_{22}a_{33}-a_{23}a_{32}$.

  4. (4)

$a_{13}^{2}=0$.

  5. (5)

$a_{11}a_{12}=0$.

  6. (6)

$2a_{22}a_{32}=a_{12}a_{31}-a_{11}a_{32}$.

  7. (7)

$a_{21}a_{32}+a_{22}a_{31}=0$.

  8. (8)

$a_{22}a_{33}+a_{23}a_{32}=a_{11}a_{33}-a_{13}a_{31}$.

It is immediate from (2) and (4) that $a_{13}=a_{23}=0$ and $a_{33}\neq 0$.
Expanding upon these, (1) and (8) suggest that $a_{21}=0$ and
$a_{11}=a_{22}\neq 0$. Continuing further, (3) and (5) imply that
$a_{11}=a_{33}$ and $a_{12}=0$. Finally, (6) and (7) necessitate that
$a_{31}=a_{32}=0$. In conclusion,

$\text{PAut}_{\text{gr}}(P)=\left\\{\begin{bmatrix}a&0&0\\\ 0&a&0\\\
0&0&a\end{bmatrix}:a\neq 0\right\\}.$

Unimodular 6. $\\{x_{1},x_{2}\\}=x_{1}^{2}+x_{2}^{2}$,
$\\{x_{2},x_{3}\\}=3x_{1}^{2}+2x_{1}x_{3}$, $\\{x_{3},x_{1}\\}=2x_{2}x_{3}$.

Let $\phi\in\text{Aut}_{\text{gr}}(P)$. By Lemma 2.1, we have the following
system of equations, with redundant equations omitted:

  1. (1)

$a_{13}^{2}+a_{23}^{2}=0.$

  2. (2)

$3a_{11}^{2}+2a_{11}a_{31}=a_{21}a_{32}-a_{22}a_{31}+3a_{22}a_{33}-3a_{23}a_{32}$.

  3. (3)

$3a_{12}^{2}+2a_{12}a_{32}=a_{21}a_{32}-a_{22}a_{31}$.

  4. (4)

$3a_{13}^{2}+2a_{13}a_{33}=0.$

  5. (5)

$3a_{11}a_{12}+a_{11}a_{32}+a_{12}a_{31}=0$.

  6. (6)

$3a_{11}a_{13}+a_{11}a_{33}+a_{13}a_{31}=a_{22}a_{33}-a_{23}a_{32}$.

  7. (7)

$3a_{12}a_{13}+a_{12}a_{33}+a_{13}a_{32}=a_{23}a_{31}-a_{21}a_{33}$.

  8. (8)

$2a_{21}a_{31}=a_{12}a_{31}-a_{11}a_{32}+3a_{13}a_{32}-3a_{12}a_{33}$.

  9. (9)

$a_{23}a_{33}=0.$

  10. (10)

$a_{21}a_{32}+a_{22}a_{31}=0.$

Suppose that $a_{33}=0$. Equation (1) and (4) necessitate that
$a_{13}=a_{23}=0$, a contradiction to the invertibility. Therefore, it follows
that $a_{33}\neq 0$, and subsequently, according to equations (1) and (9),
$a_{13}=a_{23}=0$. From (6) and (7), $a_{11}=a_{22}$ and $a_{12}=-a_{21}$.

Let us assume that $a_{12}=0$ (implying $a_{21}=0$ implicitly). From (5),
(10), and the invertibility, $a_{31}=a_{32}=0$. Finally, from (2),
$a_{11}=a_{22}=a_{33}$ and the remaining equations are nullified. This results
in one possible form of $\phi$: $\begin{bmatrix}a&0&0\\\ 0&a&0\\\
0&0&a\end{bmatrix}$ for some $a\neq 0$. Now, let us consider the alternative
scenario $a_{12}\neq 0$. From a combination of (3) and (10), we can derive
that $a_{32}=-\frac{3}{4}a_{12}$. Substituting our results into (5), we obtain
that $a_{31}=-\frac{9}{4}a_{11}$. Examining (8), we observe that
$a_{11}=-\frac{1}{2}a_{33}$. Lastly, from (10), $a_{12}=\pm\sqrt{3}a_{11}$ and
the remaining equations are nullified. This results in one possible form of
$\phi$: $\begin{bmatrix}-\frac{1}{2}a&\pm\frac{\sqrt{3}}{2}a&0\\\
\mp\frac{\sqrt{3}}{2}a&-\frac{1}{2}a&0\\\
\frac{9}{8}a&\mp\frac{3\sqrt{3}}{8}a&a\end{bmatrix}$ for some $a\neq 0$. In
conclusion,

$\text{PAut}_{\text{gr}}(P)=\left\\{\begin{bmatrix}a&0&0\\\ 0&a&0\\\
0&0&a\end{bmatrix},\begin{bmatrix}-\frac{1}{2}a&\pm\frac{\sqrt{3}}{2}a&0\\\
\mp\frac{\sqrt{3}}{2}a&-\frac{1}{2}a&0\\\
\frac{9}{8}a&\mp\frac{3\sqrt{3}}{8}a&a\end{bmatrix}:a\neq 0\right\\}.$

Unimodular 7. $\\{x_{1},x_{2}\\}=x_{3}^{2}+\lambda x_{1}x_{2}$,
$\\{x_{2},x_{3}\\}=x_{1}^{2}+\lambda x_{2}x_{3}$,
$\\{x_{3},x_{1}\\}=x_{2}^{2}+\lambda x_{1}x_{3}$.

This instance has been addressed in [15, Theorem 1]. The graded Poisson
automorphism group

$\text{PAut}_{\text{gr}}(P)=\left(\langle\begin{bmatrix}a&0&0\\\ 0&a&0\\\
0&0&a\end{bmatrix}:a\neq 0\rangle\times\langle\begin{bmatrix}1&0&0\\\ 0&b&0\\\
0&0&b^{2}\end{bmatrix}:b=\xi_{3},\xi_{3}^{2}\rangle\right)\rtimes\langle\begin{bmatrix}0&1&0\\\
0&0&1\\\ 1&0&0\end{bmatrix}\rangle.$

Unimodular 8. $\\{x_{1},x_{2}\\}=x_{1}x_{2}$,
$\\{x_{2},x_{3}\\}=3x_{1}^{2}+2x_{1}x_{2}+x_{2}x_{3}$,
$\\{x_{3},x_{1}\\}=x_{1}^{2}+x_{1}x_{3}$.

Let $\phi\in\text{Aut}_{\text{gr}}(P)$. By Lemma 2.1, we have the following
system of equations, with redundant equations colored omitted:

  1. (1)

$a_{11}a_{21}=3a_{12}a_{23}-3a_{13}a_{22}+a_{13}a_{21}-a_{11}a_{23}$.

  2. (2)

$a_{12}a_{22}=0$.

  3. (3)

$a_{13}a_{23}=0$.

  4. (4)

$a_{11}a_{23}+a_{13}a_{21}=a_{13}a_{21}-a_{11}a_{23}$.

  5. (5)

$3a_{11}^{2}+a_{21}a_{31}+2a_{11}a_{21}=3a_{22}a_{33}-3a_{23}a_{32}+a_{23}a_{31}-a_{21}a_{33}$.

  6. (6)

$3a_{12}^{2}+a_{22}a_{32}+2a_{12}a_{22}=0$.

  7. (7)

$3a_{13}^{2}+a_{23}a_{33}+2a_{13}a_{23}=0$.

  8. (8)

$6a_{11}a_{12}+a_{21}a_{32}+a_{22}a_{31}+2a_{11}a_{22}+2a_{12}a_{21}=a_{21}a_{32}-a_{22}a_{31}+2a_{22}a_{33}-2a_{23}a_{32}$.

  9. (9)

$a_{11}a_{32}+a_{12}a_{31}+2a_{11}a_{12}=a_{12}a_{31}-a_{11}a_{32}+2a_{13}a_{32}-2a_{12}a_{33}.$

Suppose that $a_{12}\neq 0$. By combining (2) and (6), we can deduce that
$a_{12}=0$, a contradiction. Consequently, it follows that $a_{12}=0$. From
(3), (4), and the invertibility, $a_{23}=0$. Immediately, (7) translates to
$a_{13}=0$ and $a_{11},a_{33}\neq 0$. Subsequently, it can be inferred from
(1) and (9) that $a_{21}=a_{32}=0$ and $a_{22}\neq 0$. Based on (5) and (8),
it follows that $a_{22}=\frac{a_{11}^{2}}{a_{33}}$ and $a_{31}=a_{33}-a_{11}$,
respectively. In conclusion,

$\text{PAut}_{\text{gr}}(P)=\left\\{\begin{bmatrix}a&0&0\\\
0&\frac{a^{2}}{b}&0\\\ b-a&0&b\end{bmatrix}:a\neq 0\right\\}.$

Unimodular 9. $\\{x_{1},x_{2}\\}=x_{1}^{2}$,
$\\{x_{2},x_{3}\\}=2x_{1}x_{3}+x_{2}^{2}$, $\\{x_{3},x_{1}\\}=2x_{1}x_{2}$.

Let $\phi\in\text{Aut}_{\text{gr}}(P)$. By Lemma 2.1, we have the following
system of equations, with redundant equations colored omitted:

  1. (1)

$a_{11}^{2}=a_{11}a_{22}-a_{12}a_{21}$.

  2. (2)

$a_{12}^{2}=a_{12}a_{23}-a_{13}a_{22}$.

  3. (3)

$a_{13}^{2}=0$.

  4. (4)

$a_{21}^{2}+2a_{11}a_{31}=a_{21}a_{32}-a_{22}a_{31}$.

  5. (5)

$a_{22}^{2}+2a_{12}a_{32}=a_{22}a_{33}-a_{23}a_{32}$.

  6. (6)

$a_{23}^{2}+2a_{13}a_{33}=0$.

  7. (7)

$2a_{11}a_{21}=a_{12}a_{31}-a_{11}a_{32}.$

The initial step is straightforward. Equations (3), (6), (2) imply
$a_{12}=a_{13}=a_{23}=0$ and $a_{11},a_{33}\neq 0$. Simplify the remaining
equations. From (1) and (5), we conclude that $a_{11}=a_{22}=a_{33}$. From
(7), we deduce that $a_{32}=-2a_{21}$. Finally, by examining (4), we ascertain
that $a_{31}=-\frac{a_{21}^{2}}{a_{11}}$. In conclusion,

$\text{PAut}_{\text{gr}}(P)=\left\\{\begin{bmatrix}a&0&0\\\ b&a&0\\\
-\frac{b^{2}}{a}&-2b&a\end{bmatrix}:a\neq 0\right\\}.$

∎

###### Proof.

Unimodular 5, Unimodular 9.

It is clear that there are no Poisson reflections for these Poisson
structures, as any graded Poisson automorphism has three repeated eigenvalues.

Unimodular 1.

A graded Poisson automorphism $\phi$ has the form
$\phi\big{|}_{P_{1}}=\begin{bmatrix}\pm\sqrt{bf-ce}&0&0\\\ a&b&c\\\
d&e&f\end{bmatrix}$. Its eigenvalues are $\lambda_{1}=\pm\sqrt{bf-ce}$,
$\lambda_{2},\lambda_{3}=\displaystyle{\frac{b+f\pm\sqrt{(b-f)^{2}+4ce}}{2}}$.
Notice that $\lambda_{2}\lambda_{3}=\lambda_{1}^{2}$. If $\phi$ is a Poisson
reflection, $\\{\lambda_{1},\lambda_{2},\lambda_{3}\\}=\\{1,1,\xi\\}$ for some
primitive root of unity $\xi$. If $\lambda_{1}=1$, then
$\lambda_{2}\lambda_{3}=1$, contradicting to
$\\{\lambda_{2},\lambda_{3}\\}=\\{1,\xi\\}$. If $\lambda_{1}=\xi$, then
$\lambda_{2}\lambda_{3}=\xi^{2}$, contradicting to
$\\{\lambda_{2},\lambda_{3}\\}=\\{1,1\\}$ unless $\xi=-1$. In that case,
$\lambda_{1}=\pm\sqrt{bf-ce}=-1$, $\lambda_{2}+\lambda_{3}=b+f=2$, and
$\phi\big{|}_{P_{1}}$ takes the form $\begin{bmatrix}-1&0&0\\\ a&b&c\\\
d&e&2-b\end{bmatrix}$ subject to the constraint $b(2-b)-ce=1$. Upon
computation, it is found that the (2,3)-entry of the matrix, when raised to
the $n$th power, is $nc$. If $\phi$ has finite order, then $c=0$. Given that
$b(2-b)=1$, it follows that $b=1$ and $\phi\big{|}_{P_{1}}$ takes a simpler
form $\begin{bmatrix}-1&0&0\\\ a&1&0\\\ d&e&1\end{bmatrix}$. Again, upon
computation, it is found that the (3,2)-entry of the matrix, when raised to
the $n$th power, is $ne$. If $\phi$ has finite order, then $e=0$. Upon
substituting this value, the resulting matrix has finite order 2. In
conclusion,

$\text{PR}(P)=\left\\{\begin{bmatrix}-1&0&0\\\ a&1&0\\\
d&0&1\end{bmatrix}\right\\}.$

Unimodular 2.

A graded Poisson automorphism $\phi$ has the form
$\phi\big{|}_{P_{1}}=\begin{bmatrix}a&0&0\\\ 0&b&0\\\ c&d&a\end{bmatrix}$. Its
eigenvalues are $a,a,b$. If $\phi$ is a Poisson reflection, $a=1$, $b=\xi$ for
some primitive root of unity $\xi$, and $\phi\big{|}_{P_{1}}$ takes the form
$\begin{bmatrix}1&0&0\\\ 0&\xi&0\\\ c&d&1\end{bmatrix}$. When raised to the
$n$th power, the (3,1)-entry of the matrix $(\phi\big{|}_{P_{1}})^{n}$ is
equal to $nc$, implying that $c=0$. In the meantime, the (2,2)-entry of the
matrix is equal to $\xi^{n}$ and the (3,2)-entry of the matrix is equal to
$\displaystyle{d\frac{(\xi^{n}-1)}{\xi-1}}$. If $n$ is a multiple of the order
of $\xi$, the matrix $(\phi\big{|}_{P_{1}})^{n}$ is equal to the identity
matrix. In conclusion,

$\text{PR}(P)=\left\\{\begin{bmatrix}1&0&0\\\ 0&\xi&0\\\
0&d&1\end{bmatrix}\right\\}.$

Unimodular 3.

A graded Poisson automorphism $\phi$ has three possible forms:
$\phi\big{|}_{P_{1}}=\begin{bmatrix}a&0&0\\\ 0&b&0\\\ 0&0&c\end{bmatrix}$,
$\begin{bmatrix}0&a&0\\\ 0&0&b\\\ c&0&0\end{bmatrix}$,
$\begin{bmatrix}0&0&a\\\ b&0&0\\\ 0&c&0\end{bmatrix}$. It is apparent that the
first matrix encompasses the following Poisson reflections:
$\begin{bmatrix}\xi&0&0\\\ 0&1&0\\\ 0&0&1\end{bmatrix}$,
$\begin{bmatrix}1&0&0\\\ 0&\xi&0\\\ 0&0&1\end{bmatrix}$,
$\begin{bmatrix}1&0&0\\\ 0&1&0\\\ 0&0&\xi\end{bmatrix}$ for some primitive
root of unity $\xi$. For the second and third matrices, their eigenvalues are
$\lambda_{1}=\sqrt[3]{abc},\lambda_{2}=\sqrt[3]{abc}\xi_{3},\lambda_{3}=\sqrt[3]{abc}{\xi_{3}^{2}}$,
where $\xi_{3}$ is a primitive 3rd root of unity. If $\phi$ is a Poisson
reflection, then $\\{\lambda_{1},\lambda_{2},\lambda_{3}\\}=\\{1,1,\xi\\}$ for
some primitive root of unity $\xi$. If $\lambda_{1}=1$, then
$\lambda_{2}\lambda_{3}=\xi$, contradicting to
$\lambda_{2}\lambda_{3}=\lambda_{1}^{2}$. If $\lambda_{1}=\xi$, then
$\displaystyle{\frac{\lambda_{3}}{\lambda_{2}}}=\frac{1}{1}=1$, contradicting
to
$\displaystyle{\frac{\lambda_{3}}{\lambda_{2}}=\frac{\sqrt[3]{abc}\xi_{3}^{2}}{\sqrt[3]{abc}\xi_{3}}}=\xi_{3}$.
Consequently, the second and third matrices cannot be Poisson reflections. In
conclusion,

$\text{PR}(P)=\left\\{\begin{bmatrix}\xi&0&0\\\ 0&1&0\\\
0&0&1\end{bmatrix},\begin{bmatrix}1&0&0\\\ 0&\xi&0\\\
0&0&1\end{bmatrix},\begin{bmatrix}1&0&0\\\ 0&1&0\\\
0&0&\xi\end{bmatrix}\right\\}.$

Unimodular 4.

A graded Poisson automorphism $\phi$ has six possible forms:
$\phi\big{|}_{P_{1}}=\begin{bmatrix}0&a&0\\\ -a&-a&0\\\ b&c&a\end{bmatrix}$,
$\begin{bmatrix}0&-a&0\\\ -a&0&0\\\
b&c&a\end{bmatrix},\begin{bmatrix}-a&0&0\\\ a&a&0\\\
b&c&a\end{bmatrix},\begin{bmatrix}-a&-a&0\\\ a&0&0\\\
b&c&a\end{bmatrix},\begin{bmatrix}a&a&0\\\ 0&-a&0\\\
b&c&a\end{bmatrix},\begin{bmatrix}a&0&0\\\ 0&a&0\\\ b&c&a\end{bmatrix}$. The
eigenvalues of the first and fourth matrices are
$\lambda_{1}=a,\lambda_{2},\lambda_{3}=\displaystyle{\frac{(-1\pm
i\sqrt{3})a}{2}}$. If $\phi$ is a Poisson reflection, then
$\\{\lambda_{1},\lambda_{2},\lambda_{3}\\}=\\{1,1,\xi\\}$ for some primitive
root of unity $\xi$. In any case, $\lambda_{2}+\lambda_{3}\neq-\lambda_{1}$, a
contradiction. The eigenvalues of the sixth matrix are $a,a,a$; consequently,
it is impossible for such a matrix to be a Poisson reflection. The eigenvalues
of the second, third, and fifth matrices are
$\lambda_{1}=-a,\lambda_{2},\lambda_{3}=a$. If $\phi$ is a Poisson reflection,
then $a=1$ and there exists three candidates for $\phi$:
$\begin{bmatrix}0&-1&0\\\ -1&0&0\\\ b&c&1\end{bmatrix}$,
$\begin{bmatrix}-1&0&0\\\ 1&1&0\\\ b&c&1\end{bmatrix}$,
$\begin{bmatrix}1&1&0\\\ 0&-1&0\\\ b&c&1\end{bmatrix}$. By a straightforward
calculation,

$\begin{bmatrix}0&-1&0\\\ -1&0&0\\\
b&c&1\end{bmatrix}^{n}=\begin{cases}\begin{bmatrix}1&0&0\\\ 0&1&0\\\
\frac{n(b-c)}{2}&\frac{n(c-b)}{2}&1\end{bmatrix}&n\text{ is even,}\\\
\vspace{-0.75cm}\\\ \begin{bmatrix}0&-1&0\\\ -1&0&0\\\
\frac{n+1}{2}b-\frac{n-1}{2}c&\frac{n+1}{2}c-\frac{n-1}{2}b&1\end{bmatrix}&n\text{
is odd.}\end{cases}$

In this case, a finite order Poisson reflection $\phi$ satisfies $b=c$.

$\begin{bmatrix}-1&0&0\\\ 1&1&0\\\
b&c&1\end{bmatrix}^{n}=\begin{cases}\begin{bmatrix}1&0&0\\\ 0&1&0\\\
\frac{n}{2}c&nc&1\end{bmatrix}&n\text{ is even,}\\\ \vspace{-0.75cm}\\\
\begin{bmatrix}0&-1&0\\\ -1&0&0\\\ b+\frac{n-1}{2}c&nc&1\end{bmatrix}&n\text{
is odd.}\end{cases}$

In this case, a finite order Poisson reflection $\phi$ satisfies $c=0$.

$\begin{bmatrix}1&1&0\\\ 0&-1&0\\\
b&c&1\end{bmatrix}^{n}=\begin{cases}\begin{bmatrix}1&0&0\\\ 0&1&0\\\
nb&\frac{n}{2}b&1\end{bmatrix}&n\text{ is even,}\\\ \vspace{-0.75cm}\\\
\begin{bmatrix}1&1&0\\\ 0&-1&0\\\ nb&\frac{n-1}{2}b+c&1\end{bmatrix}&n\text{
is odd.}\end{cases}$

In this case, a finite order Poisson reflection $\phi$ satisfies $b=0$.

In conclusion,

$\text{PR}(P)=\left\\{\begin{bmatrix}0&-1&0\\\ -1&0&0\\\
b&b&1\end{bmatrix},\begin{bmatrix}-1&0&0\\\ 1&1&0\\\
b&0&1\end{bmatrix},\begin{bmatrix}1&1&0\\\ 0&-1&0\\\
0&c&1\end{bmatrix}\right\\}.$

Unimodular 6.

A graded Poisson automorphism $\phi$ has two possible forms:
$\phi\big{|}_{P_{1}}=\begin{bmatrix}a&0&0\\\ 0&a&0\\\ 0&0&a\end{bmatrix}$,
$\begin{bmatrix}-\frac{1}{2}a&\pm\frac{\sqrt{3}}{2}a&0\\\
\mp\frac{\sqrt{3}}{2}a&-\frac{1}{2}a&0\\\
\frac{9}{8}a&\mp\frac{3\sqrt{3}}{8}a&a\end{bmatrix}$. The eigenvalues of the
former matrix and the latter matrix are
$\lambda_{1},\lambda_{2},\lambda_{3}=a,a,a$ and
$\lambda_{1}=a,\lambda_{2},\lambda_{3}=\displaystyle{\frac{(-1\pm
i\sqrt{3})a}{2}}$, respectively. Neither can be Poisson reflections, as
discussed in prior instances.

Unimodular 7.

[7, Lemma 4.3] provided a comprehensive analysis of the linear Poisson normal
elements pertaining to this particular instance. Given that $\lambda^{3}\neq
1$, it follows that there are no linear Poisson normal elements, thereby
precluding the existence of Poisson reflections [7, Lemma 2.2].

Unimodular 8.

A graded Poisson automorphism $\phi$ has the form
$\phi\big{|}_{P_{1}}=\begin{bmatrix}a&0&0\\\ 0&\frac{a^{2}}{b}&0\\\
b-a&0&b\end{bmatrix}$. Its eigenvalues are
$\lambda_{1}=a,\lambda_{2}=\displaystyle{\frac{a^{2}}{b}},\lambda_{3}=b$. If
$\phi$ is a Poisson reflection, then
$\\{\lambda_{1},\lambda_{2},\lambda_{3}\\}=\\{1,1,\xi\\}$ for some primitive
root of unity $\xi$. If $\lambda_{1}=1$, then
$\lambda_{2}\lambda_{3}=\xi\neq\lambda_{1}^{2}$, a contradiction. If
$\lambda_{1}=\xi$, then the constraint
$\lambda_{2}\lambda_{3}=\lambda_{1}^{2}$ implies that $\xi=-1$ and
$\phi\big{|}_{P_{1}}$ takes the form $\begin{bmatrix}-1&0&0\\\ 0&1&0\\\
2&0&1\end{bmatrix}$. Such $\phi$ has finite order 2. In conclusion,

$\text{PR}(P)=\left\\{\begin{bmatrix}-1&0&0\\\ 0&1&0\\\
2&0&1\end{bmatrix}\right\\}.$

∎

##  3\. A Variant of The Shephard-Todd-Chevalley Theorem

In this section, we prove a variant of the Shephard-Todd-Chevalley Theorem for
unimodular quadratic Poisson structures on $\Bbbk[x_{1},x_{2},x_{3}]$, stated
as the follows:

######  Theorem 3.1.

Let $P=\Bbbk[x_{1},x_{2},x_{3}]$ be a unimodular quadratic Poisson algebra and
let $G\subseteq\text{PAut}_{\text{gr}}(P)$ be a finite subgroup. Then the
invariant subalgebra $P^{G}$ is isomorphic to $P$ as Poisson algebras if and
only if $P$ is trivial.

Prior to embarking on the proof of Theorem 3.1, we remark that establishing a
Poisson algebra $P$ and another Poisson algebra $Q$ are non-isomorphic Poisson
algebras, by demonstrating the absence of suitable mappings, can be a
challenging task. In this paper, we are primarily relying on the following two
lemmas to distinguish Poisson algebras from their invariant subalgebras.

The first lemma asserts that unimodularity can serve as a distinguishing
characteristic for Poisson algebras.

######  Lemma 3.2.

Let $P=\Bbbk[x_{1},\cdots,x_{n}]$, $Q=\Bbbk[y_{1},\cdots,y_{n}]$ be Poisson
algebras. If $P$ is unimodular and $Q$ is non-unimodular, then $P$ is not
isomorphic to $Q$ as Poisson algebras.

###### Proof.

Let $\underline{m}_{P},\underline{m}_{Q}$ be the modular derivations of $P,Q$,
respectively. Suppose that $P$ is isomorphic to $Q$ as Poisson algebras
through an isomorphism $\phi$. Given that the modular derivation
$\underline{m}_{Q}$ is independent of the generators [19, page 6], we may
compute $\underline{m}_{Q}$ with respect to the generators
$\\{\phi(x_{1}),\cdots,\phi(x_{n})\\}$, instead of the conventional generators
$\\{y_{1},\cdots,y_{n}\\}$. For all $f\in Q$,

$\underline{m}_{Q}(f)=\sum_{i=1}^{n}\frac{\partial\\{\phi(x_{i}),\phi(g)\\}}{\partial\phi(x_{i})}=\sum_{i=1}^{n}\frac{\partial\phi\\{x_{i},g\\}}{\partial\phi(x_{i})}=\phi\left(\sum_{i=1}^{n}\frac{\partial\\{x_{i},g\\}}{\partial
x_{i}}\right)=\phi(m_{P}(g))=0,$

for some $g\in P$. By definition, the Poisson algebra $Q$ is unimodular, a
contradiction. ∎

The second lemma, while less general in scope compared to the former one, have
nonetheless been proven to be highly useful in our paper.

######  Lemma 3.3.

Let $P=\Bbbk[x_{1},x_{2},x_{3}]$ be a quadratic Poisson algebra. If
$Q=\Bbbk[y_{1},y_{2},y_{3}]$ is a Poisson algebra satisfying the following
conditions:

  1. (1)

there exists a pair $(i,j)\in\\{1,2,3\\}^{\oplus 2}$ such that
$\\{y_{i},y_{j}\\}=f(y_{1},y_{2})-ry_{3}$ for some homogeneous $f$ with
respect to the grading $\deg(y_{1})=\deg(y_{2})=1$, and some
$r\in\Bbbk^{\times}$,

  2. (2)

the bracket $\\{y_{k},y_{l}\\}$ is a scalar multiple of a single monomial in
$y_{1}$, $y_{2}$, $y_{3}$, for all $(k,l)\in\\{1,2,3\\}^{\oplus
2}\backslash\\{(i,j)\\}$,

then $P$ is not isomorphic to $Q$ as Poisson algebras.

###### Proof.

Suppose, for the sake of contradiction, that $P$ and $Q$ are isomorphic as
Poisson algebras. Observe that such a Poisson isomorphism passes to a
$\Bbbk$-algebra isomorphism:

$P/(\\{P,P\\})\xrightarrow{\sim}Q/(\\{Q,Q\\})$

For the $\Bbbk$-algebra $P$, the ideal $(\\{P,P\\})$ is generated by
$\\{x_{i},x_{j}\\}$, where $1\leq i,j\leq 3$, because:

$\\{g,h\\}=\sum_{1\leq i,j\leq 3}\frac{\partial g}{\partial
x_{i}}\frac{\partial h}{\partial x_{j}}\\{x_{i},x_{j}\\},$

for all $g,h\in P$. The quotient algebra $P/(\\{P,P\\})$ is a connected
$\mathbb{N}$-graded algebra that is finitely generated in degree 1, as the
Poisson algebra $P$ is quadratic and consequently the ideal $(\\{P,P\\})$
admits a set of homogeneous generators of degree 2.

For the $\Bbbk$-algebra $Q$, by the same reasoning, the ideal $\\{Q,Q\\}$ is
generated by $\\{y_{i},y_{j}\\}$, where $1\leq i,j\leq 3$. Based on the
assumptions, we can write

$\\{y_{a},y_{b}\\}_{Q}=f(y_{1},y_{2})-ry_{1},\quad\\{y_{b},y_{c}\\}_{Q}=sy_{1}^{\alpha_{1}}y_{2}^{\alpha_{2}}y_{3}^{\alpha_{3}},\quad\\{y_{c},y_{a}\\}_{Q}=ty_{1}^{\beta_{1}}y_{2}^{\beta_{2}}y_{3}^{\beta_{3}},$

for some $\\{a,b,c\\}$ that is a relabeling of $\\{1,2,3\\}$, $s,t\in\Bbbk$,
and $\alpha_{1},\alpha_{2},\alpha_{3},\beta_{1},\beta_{2},\beta_{3}\geq 0$.
The quotient algebra

$\displaystyle Q/(\\{Q,Q\\})=$
$\displaystyle\Bbbk[y_{1},y_{2},y_{3}]/(f-ry_{1},sy_{1}^{\alpha_{1}}y_{2}^{\alpha_{2}}y_{3}^{\alpha_{3}},ty_{1}^{\beta_{1}}y_{2}^{\beta_{2}}y_{3}^{\beta_{3}})$
$\displaystyle\cong$
$\displaystyle\Bbbk[y_{1},y_{2}]/(\frac{s}{r^{\alpha_{3}}}y_{1}^{\alpha_{1}}y_{2}^{\alpha_{2}}f^{\alpha_{3}},\frac{t}{r^{\beta_{3}}}y_{1}^{\beta_{1}}y_{2}^{\beta_{2}}f^{\beta_{3}})$

is a connected $\mathbb{N}$-graded algebra that is finitely generated in
degree 1, as $f$ is a homogeneous polynomial with respect to the grading
$\deg(y_{1})=\deg(y_{2})=1$.

Based on the preceding argument, [8, Lemma 4.1] applies. However,

$\dim_{\Bbbk}\big{(}P/(\\{P,P\\})\big{)}=3\neq
2\geq\dim_{\Bbbk}\big{(}Q/(\\{Q,Q\\})\big{)},$

a contradiction. ∎

We are now ready to prove Theorem 3.1.

###### Proof.

It suffices to prove $\Rightarrow$. The classical Shephard-Todd-Chevalley
Theorem states that if $P^{G}\cong P$ as algebras, then $G$ is generated by
reflections. Since $G\subseteq\text{PAut}_{\text{gr}}(P)$, $G$ is generated by
Poisson reflections. By Proposition 2.3, Unimodular 5, Unimodular 6,
Unimodular 7, Unimodular 9 have no Poisson reflections, hence the statement is
trivially true. For Unimodular 1, Unimodular 2, Unimodular 3, Unimodular 4,
Unimodular 8, we proceed with a detailed analysis of each instance.

Unimodular 1. $\\{x_{1},x_{2}\\}=0$, $\\{x_{2},x_{3}\\}=3x_{1}^{2}$,
$\\{x_{3},x_{1}\\}=0$.

A Poisson reflection $\phi$ has the form
$\phi\big{|}_{P_{1}}=\begin{bmatrix}-1&0&0\\\ a&1&0\\\ d&0&1\end{bmatrix}$. If
$G$ contains two distinct Poisson reflections:
$\phi_{1}\big{|}_{P_{1}}=\begin{bmatrix}-1&0&0\\\ a_{1}&1&0\\\
d_{1}&0&1\end{bmatrix}$, $\phi_{2}\big{|}_{P_{1}}=\begin{bmatrix}-1&0&0\\\
a_{2}&1&0\\\ d_{2}&0&1\end{bmatrix}$. However, the product
$(\phi_{1}\phi_{2})\big{|}_{P_{1}}=\begin{bmatrix}1&0&0\\\ a_{2}-a_{1}&1&0\\\
d_{2}-d_{1}&0&1\end{bmatrix}$, a matrix of infinite order unless $a_{2}=a_{1}$
and $d_{2}=d_{1}$. Consequently, we may assume that $G$ is cyclic generated:
$G=\begin{bmatrix}-1&0&0\\\ a&1&0\\\ d&0&1\end{bmatrix}\cong\mathbb{Z}_{2}$.
To compute the invariant subalgebra, we start by observing that the
polynomials $y_{1}=x_{1}^{2}$, $y_{2}=\displaystyle{\frac{a}{2}}x_{1}+x_{2}$,
$y_{3}=\displaystyle{\frac{d}{2}}x_{1}+x_{3}$ are algebraically independent
and remain invariant under the action of $G$. Embed the $\Bbbk$-algebra
generated by $y_{1},y_{2},y_{3}$ into the invariant subalgebra $P^{G}$.
Utilizing Molien’s Theorem to compute $h_{P^{G}}(t)$, we compare the Hilbert
series of these two $\Bbbk$-algebras, and conclude that
$P^{G}=\Bbbk[y_{1},y_{2},y_{3}]$. The Poisson structure on the invariant
subalgebra $P^{G}$ is:

$\\{y_{1},y_{2}\\}=0,\quad\\{y_{2},y_{3}\\}=3y_{1},\quad\\{y_{3},y_{1}\\}=0.$

By invoking Lemma 3.2, it becomes evident that the Poisson algebras $P$ and
$Q$ are non-isomorphic Poisson algebras.

Unimodular 2. $\\{x_{1},x_{2}\\}=0$, $\\{x_{2},x_{3}\\}=2x_{1}x_{2}$,
$\\{x_{3},x_{1}\\}=x_{1}^{2}$.

A Poisson reflection $\phi$ has the form
$\phi\big{|}_{P_{1}}=\begin{bmatrix}1&0&0\\\ 0&\xi&0\\\ 0&d&1\end{bmatrix}$.
If $G$ contains two non-commuting Poisson reflections:
$\phi_{1}\big{|}_{P_{1}}=\begin{bmatrix}1&0&0\\\ 0&\xi_{n_{1}}&0\\\
0&d_{1}&1\end{bmatrix}$ and $\phi_{2}\big{|}_{P_{1}}=\begin{bmatrix}1&0&0\\\
0&\xi_{n_{2}}&0\\\ 0&d_{2}&1\end{bmatrix}$, where $\xi_{n_{1}}$ (resp.
$\xi_{n_{2}}$) is a primitve $n_{1}$th (resp. $n_{2}$th) root of unity. Since
$\phi_{1}\phi_{2}\neq\phi_{2}\phi_{1}$, the product
$(\phi_{1}\phi_{2}\phi_{1}^{n_{1}-1}\phi_{2}^{n_{2}-1})\big{|}_{P_{1}}=\begin{bmatrix}1&0&0\\\
0&1&0\\\ 0&d&1\end{bmatrix}$ for some $d\neq 0$; however, such a matrix has
infinite order, a contradiction. Therefore, $G$ is a finite abelian group.
Keeping $\phi_{1}$ and $\phi_{2}$ as above, the commutativity
$\phi_{1}\phi_{2}=\phi_{2}\phi_{1}$ is equivalent to
$d_{2}=\displaystyle{\frac{\xi_{n_{2}}-1}{\xi_{n_{1}}-1}d_{1}}$.

Decompose $G\cong\mathbb{Z}_{n_{1}}\times\cdots\times\mathbb{Z}_{n_{m}}$ for
some $n_{i}\in\mathbb{N}$ satisfying $n_{i}|n_{i+1}$. According to the
commutativity condition, we may take the cyclic generator of
$\mathbb{Z}_{n_{i}}$ to be $\phi_{i}\big{|}_{P_{1}}=\begin{bmatrix}1&0&0\\\
0&\xi_{n_{i}}&0\\\ 0&\frac{\xi_{n_{i}}-1}{\xi_{n_{1}}-1}d_{1}&1\end{bmatrix}$
for some primitive $n_{i}$th root of unity $\xi_{n_{i}}$, for all $1\leq i\leq
m$. Define $S=\\{\alpha\in\mathbb{N}^{m}:1\leq\alpha_{i}\leq n_{i}\\}$. For
$\alpha=(\alpha_{i})_{1\leq i\leq m}\in S$, we define
$\xi^{\alpha}=\xi_{n_{1}}^{\alpha_{1}}\cdots\xi_{n_{m}}^{\alpha_{m}}$.
According to the Molien’s Theorem,

$h_{P^{G}}(t)=\frac{1}{n_{1}\cdots n_{m}}\sum_{\alpha\in
S}\frac{1}{(1-t)^{2}(1-\xi^{\alpha}t)}=\frac{1}{(1-t)^{2}}\left(\frac{1}{n_{1}\cdots
n_{m}}\sum_{\alpha\in S}\frac{1}{(1-\xi^{\alpha}t)}\right).$

Before computing $\displaystyle{\sum_{\alpha\in
S}\frac{1}{(1-\xi^{\alpha}t)}}$, we claim that $\\{\xi^{\alpha}:\alpha\in
S\\}$ is precisely $n_{1}\cdots n_{m-1}$ copies of
$\\{\xi_{n_{m}}^{\alpha_{m}}:1\leq\alpha_{m}\leq n_{m}\\}$. To prove the
claim, we proceed by induction. When $m=2$, the set $\\{\xi^{\alpha}:\alpha\in
S\\}$ contains the following elements:

$\displaystyle 1$ $\displaystyle\xi_{n_{2}}$ $\displaystyle\xi_{n_{2}}^{2}$
$\displaystyle\cdots$ $\displaystyle\xi_{n_{2}}^{n_{2}-1}$ $\displaystyle 1$
$\displaystyle\xi_{n_{1}}\xi_{n_{2}}$
$\displaystyle\xi_{n_{1}}\xi_{n_{2}}^{2}$ $\displaystyle\cdots$
$\displaystyle\xi_{n_{1}}\xi_{n_{2}}^{n_{2}-1}$ $\displaystyle\vdots$
$\displaystyle\vdots$ $\displaystyle\vdots$ $\displaystyle\ddots$
$\displaystyle\vdots$ $\displaystyle 1$
$\displaystyle\xi_{n_{1}}^{n_{1}-1}\xi_{n_{2}}$
$\displaystyle\xi_{n_{1}}^{n_{1}-1}\xi_{n_{2}}^{2}$ $\displaystyle\cdots$
$\displaystyle\xi_{n_{1}}^{n_{1}-1}\xi_{n_{2}}^{n_{2}-1}$

Each line can be realized as the image of a left multiplication map
$l_{\xi_{n_{1}}^{\alpha_{1}}}:\langle\xi_{n_{2}}\rangle\to\langle\xi_{n_{2}}\rangle$,
for all $0\leq\alpha_{1}\leq n_{1}-1$. Since $n_{1}|n_{2}$, and consequently,
$\xi_{n_{1}}^{\alpha_{1}}$ is an element of $\langle\xi_{n_{2}}\rangle$, each
line is a permutation of the first line. This proves the base case:
$\\{\xi_{n_{1}}^{\alpha_{1}}\xi_{n_{2}}^{\alpha_{2}}|1\leq\alpha_{1}\leq
n_{1},1\leq\alpha_{2}\leq n_{2}\\}$ is precisely $n_{1}$ copies of
$\\{\xi_{n_{2}}^{\alpha_{2}}|1\leq\alpha_{2}\leq n_{2}\\}$. Inductively, when
$m$ is arbitrary, the set $\\{\xi^{\alpha}|\alpha\in S\\}$ contains the
following elements:

$\displaystyle 1$ $\displaystyle E_{m-1}$ $\displaystyle E_{m-1}\xi_{n_{m}}$
$\displaystyle\cdots$ $\displaystyle E_{m-1}\xi_{n_{m}}^{n_{m}-1},$

where
$E_{m-1}=\\{\xi_{n_{1}}^{\alpha_{1}}\cdots\xi_{n_{m-1}}^{\alpha_{n_{m-1}}}:1\leq\alpha_{i}\leq
n_{i}\\}$. By the induction hypothesis, the set
$\\{\xi_{n_{1}}^{\alpha_{1}}\cdots\xi_{n_{m-1}}^{\alpha_{n_{m-1}}}:1\leq\alpha_{i}\leq
n_{i}\\}$ is $n_{1}n_{2}\cdots n_{m-2}$ copies of
$\\{\xi_{n_{m-1}}^{\alpha_{m-1}}:1\leq\alpha_{m-1}\leq n_{m-1}\\}$.
Consequently, we can view the preceding lines of elements as $n_{1}n_{2}\cdots
n_{m-2}$ layers of the subsequent lines of elements:

$\displaystyle 1$ $\displaystyle\xi_{n_{m}}$ $\displaystyle\xi_{n_{m}}^{2}$
$\displaystyle\cdots$ $\displaystyle\xi_{n_{m}}^{n_{m}-1}$ $\displaystyle 1$
$\displaystyle\xi_{n_{m-1}}\xi_{n_{m}}$
$\displaystyle\xi_{n_{m-1}}\xi_{n_{m}}^{2}$ $\displaystyle\cdots$
$\displaystyle\xi_{n_{m-1}}\xi_{n_{m}}^{n_{m}-1}$ $\displaystyle\vdots$
$\displaystyle\vdots$ $\displaystyle\vdots$ $\displaystyle\ddots$
$\displaystyle\vdots$ $\displaystyle 1$
$\displaystyle\xi_{n_{m-1}}^{n_{m-1}-1}\xi_{n_{m}}$
$\displaystyle\xi_{n_{m-1}}^{n_{m-1}-1}\xi_{n_{m}}^{2}$ $\displaystyle\cdots$
$\displaystyle\xi_{n_{m-1}}^{n_{m-1}-1}\xi_{n_{m}}^{n_{m}-1}$

Again, from the induction hypothesis, the above lines of elements are
$n_{1}n_{2}\cdots n_{m-1}$ copies
$\\{\xi_{n_{m}}^{\alpha_{m}}|1\leq\alpha_{m}\leq n_{m}\\}$ as claimed. We can
compute the Hilbert series of the invariant subalgebra $P^{G}$ by employing
the claim:

$\displaystyle h_{P^{G}}(t)=$
$\displaystyle\frac{1}{(1-t)^{2}}\bigg{(}\frac{n_{1}\cdots
n_{m-1}}{n_{1}\cdots n_{m}}\sum_{1\leq\alpha_{m}\leq
n_{m}}{\frac{1}{(1-\xi_{n_{m}}^{\alpha_{m}}t)}}\bigg{)}$ $\displaystyle=$
$\displaystyle\frac{1}{(1-t)^{2}}\bigg{(}\frac{1}{n_{m}}\sum_{1\leq\alpha_{m}\leq
n_{m}}{\frac{1}{(1-\xi_{n_{m}}^{\alpha_{m}}t)}}\bigg{)}$ $\displaystyle=$
$\displaystyle\frac{1}{(1-t)^{2}(1-t^{n_{m}})}.$

Set $l=n_{m}$. Furthermore, set $y_{1}=x_{1}$,
$y_{2}=d_{1}x_{2}+(1-\xi_{n_{1}})x_{3}$, $y_{3}=x_{2}^{l}$. The elements
$y_{1},y_{2},y_{3}$ are three algebraically independent polynomials that are
invariant under the action of $G$. Consequently, we may embed
$\Bbbk[y_{1},y_{2},y_{3}]$ into the invariant subalgebra $P^{G}$. It is
evident that the Hilbert series of $\Bbbk[y_{1},y_{2},y_{3}]$ is
$\displaystyle{\frac{1}{(1-t)^{2}(1-t^{l})}}$, and therefore, the embedding
$\Bbbk[y_{1},y_{2},y_{3}]\hookrightarrow P^{G}$ is surjective because the
cokernel has Hilbert series 0. Accordingly, we conclude that the invariant
subalgebra $P^{G}=\Bbbk[y_{1},y_{2},y_{3}]$ and has the following Poisson
structure:

$\\{y_{1},y_{2}\\}=(\xi_{n_{1}}-1)y_{1}^{2},\quad\\{y_{2},y_{3}\\}=2l(\xi_{n_{1}}-1)y_{1}y_{3},\quad\\{y_{3},y_{1}\\}=0.$

Suppose that $P^{G}\cong P$ as Poisson algebras. Specifically, by applying
(the contrapositive of) Lemma 3.2, the invariant subalgebra is unimodular.
This implies that we can find a superpotential
$\Omega\in\Bbbk[y_{1},y_{2},y_{3}]$ satisfying:

$\frac{\partial\Omega}{\partial
y_{3}}=(\xi_{n_{1}}-1)y_{1}^{2},\quad\frac{\partial\Omega}{\partial
y_{1}}=2l(\xi_{n_{1}}-1)y_{1}y_{3},\quad\frac{\partial\Omega}{\partial
y_{2}}=0.$

It is straightforward to verify no such $\Omega$ exists unless $l=1$, or
equivalently, unless $G$ is trivial.

Unimodular 3. $\\{x_{1},x_{2}\\}=2x_{1}x_{2}$,
$\\{x_{2},x_{3}\\}=2x_{2}x_{3},\\{x_{3},x_{1}\\}=2x_{1}x_{3}$.

This case is addressed in [7, Theorem 4.5]. The conclusion is as follows:
$P^{G}\cong P$ as Poisson algebras if and only if $G$ is trivial.

Unimodular 4. $\\{x_{1},x_{2}\\}=0$,
$\\{x_{2},x_{3}\\}=2x_{1}x_{2}+x_{2}^{2}$,
$\\{x_{3},x_{1}\\}=x_{1}^{2}+2x_{1}x_{2}$.

A Poisson reflection $\phi$ has three possible forms:
$\phi\big{|}_{P_{1}}=\begin{bmatrix}0&-1&0\\\ -1&0&0\\\
b&b&1\end{bmatrix},\begin{bmatrix}-1&0&0\\\ 1&1&0\\\ b&0&1\end{bmatrix}$,
$\begin{bmatrix}1&1&0\\\ 0&-1&0\\\ 0&c&1\end{bmatrix}$. First, notice that the
finite automorphism group $G$ cannot contain two Poisson reflections of the
same form.

  * •

Suppose that $G$ contains two distinct Poisson reflections of the first form:
$\phi_{1}\big{|}_{P_{1}}=\begin{bmatrix}0&-1&0\\\ -1&0&0\\\
b_{1}&b_{1}&1\end{bmatrix}$, $\phi_{2}\big{|}_{P_{1}}=\begin{bmatrix}0&-1&0\\\
-1&0&0\\\ b_{2}&b_{2}&1\end{bmatrix}$. The product
$(\phi_{1}\phi_{2})\big{|}_{P_{1}}=\begin{bmatrix}1&0&0\\\ 0&1&0\\\
b_{2}-b_{1}&b_{2}-b_{1}&1\end{bmatrix}$ is a matrix of infinite order,
contradicting to the finiteness of $G$.

  * •

Suppose that $G$ contains two distinct Poisson reflections of the second form:
$\phi_{1}\big{|}_{P_{1}}=\begin{bmatrix}-1&0&0\\\ 1&1&0\\\
b_{1}&0&1\end{bmatrix}$, $\phi_{2}\big{|}_{P_{1}}=\begin{bmatrix}-1&0&0\\\
1&1&0\\\ b_{2}&0&1\end{bmatrix}$. The product
$(\phi_{1}\phi_{2})\big{|}_{P_{1}}=\begin{bmatrix}1&0&0\\\ 0&1&0\\\
b_{2}-b_{1}&0&1\end{bmatrix}$ is a matrix of infinite order, contradicting to
the finiteness of $G$.

  * •

Suppose that $G$ contains two distinct Poisson reflections of the third form:
$\phi_{1}\big{|}_{P_{1}}=\begin{bmatrix}1&1&0\\\ 0&-1&0\\\
0&c_{1}&1\end{bmatrix}$, $\phi_{2}\big{|}_{P_{1}}=\begin{bmatrix}1&1&0\\\
0&-1&0\\\ 0&c_{2}&1\end{bmatrix}$. The product
$(\phi_{1}\phi_{2})\big{|}_{P_{1}}=\begin{bmatrix}1&0&0\\\ 0&1&0\\\
0&b_{2}-b_{1}&1\end{bmatrix}$ is a matrix of infinite order, contradicting to
the finiteness of $G$.

Consequently, the finite automorphism group $G$ falls into one of the
following three categories:

  1. (1)

The group $G$ is generated by three Poisson reflections of distinct types.

  2. (2)

The group $G$ is generated by two Poisson reflections of distinct types.

  3. (3)

The group $G$ is generated by a single Poisson reflection.

In Case (1), $G=\langle\phi_{1}\big{|}_{P_{1}}=\begin{bmatrix}0&-1&0\\\
-1&0&0\\\ a&a&1\end{bmatrix},\phi_{2}\big{|}_{P_{1}}=\begin{bmatrix}-1&0&0\\\
1&1&0\\\ b&0&1\end{bmatrix},\phi_{3}\big{|}_{P_{1}}=\begin{bmatrix}1&1&0\\\
0&-1&0\\\ 0&c&1\end{bmatrix}\rangle$. By calculation, the product
$(\phi_{1}\phi_{2}\phi_{3})\big{|}_{P_{1}}^{n}$ is not equal to the identity
matrix $I_{3}$ when $n$ is odd, and it equals the following matrix:

$(\phi_{1}\phi_{2}\phi_{3})\big{|}_{P_{1}}^{n}=\begin{bmatrix}1&0&0\\\
0&1&0\\\ \frac{n}{2}(b+c-a)&n(b+c-a)&1\end{bmatrix}$

when $n$ is even. As a result, the finiteness of $G$ necessitates the
condition: $a=b+c$. Given this equality, through further calculations, we
observe that:

$\displaystyle(\phi_{1}\phi_{2}\phi_{1})\big{|}_{P_{1}}=$
$\displaystyle\begin{bmatrix}0&-1&0\\\ -1&0&0\\\
b+c&b+c&1\end{bmatrix}\begin{bmatrix}-1&0&0\\\ 1&1&0\\\
b&0&1\end{bmatrix}\begin{bmatrix}0&-1&0\\\ -1&0&0\\\ b+c&b+c&1\end{bmatrix}=$
$\displaystyle\begin{bmatrix}1&1&0\\\ 0&-1&0\\\ 0&c&1\end{bmatrix}=$
$\displaystyle\phi_{3}\big{|}_{P_{1}},$
$\displaystyle(\phi_{2}\phi_{3}\phi_{2})\big{|}_{P_{1}}=$
$\displaystyle\begin{bmatrix}-1&0&0\\\ 1&1&0\\\
b&0&1\end{bmatrix}\begin{bmatrix}1&1&0\\\ 0&-1&0\\\
0&c&1\end{bmatrix}\begin{bmatrix}-1&0&0\\\ 1&1&0\\\ b&0&1\end{bmatrix}=$
$\displaystyle\begin{bmatrix}0&-1&0\\\ -1&0&0\\\ b+c&b+c&1\end{bmatrix}=$
$\displaystyle\phi_{1}\big{|}_{P_{1}},$
$\displaystyle(\phi_{3}\phi_{1}\phi_{3})\big{|}_{P_{1}}=$
$\displaystyle\begin{bmatrix}1&1&0\\\ 0&-1&0\\\
0&c&1\end{bmatrix}\begin{bmatrix}0&-1&0\\\ -1&0&0\\\
b+c&b+c&1\end{bmatrix}\begin{bmatrix}1&1&0\\\ 0&-1&0\\\ 0&c&1\end{bmatrix}=$
$\displaystyle\begin{bmatrix}-1&0&0\\\ 1&1&0\\\ b&0&1\end{bmatrix}=$
$\displaystyle\phi_{2}\big{|}_{P_{1}}.$

In conclusion, we have established that
$G=\langle\phi_{1},\phi_{2},\phi_{3}\rangle=\langle\phi_{1},\phi_{2}\rangle=\langle\phi_{2},\phi_{3}\rangle=\langle\phi_{3},\phi_{1}\rangle$,
and it is sufficient to consider the case when
$G=\langle\phi_{1},\phi_{2}\rangle$ for both Case (1) and Case (2).

In Case (2), $G=\langle\phi_{1}\big{|}_{P_{1}}=\begin{bmatrix}0&-1&0\\\
-1&0&0\\\ a&a&1\end{bmatrix},\phi_{2}\big{|}_{P_{1}}=\begin{bmatrix}-1&0&0\\\
1&1&0\\\ b&0&1\end{bmatrix}\rangle$. By calculation,

$\phi_{1}^{2}\big{|}_{P_{1}}=\phi_{2}^{2}\big{|}_{P_{1}}=(\phi_{1}\phi_{2})^{3}\big{|}_{P_{1}}=I_{3},\quad\phi_{1}\phi_{2}\neq\phi_{2}\phi_{1}.$

The first equality implies that $G$ is isomorphic to a quotient of the
symmetric group $S_{3}$. Further, the second inequality implies that $G$ is
isomorphic to $S_{3}$, as $S_{3}$ is the smallest finite non-abelian group.
Apply the Molien’s Theorem:

$h_{P^{G}}(t)=\frac{1}{6}\left(\frac{1}{(1-t)^{3}}+\frac{3}{(1-t)^{2}(1+t)}+\frac{2}{(1-t)(1+t+t^{2})}\right)\\\
=\frac{1}{(1-t)(1-t^{2})(1-t^{3})}.$

After calculating
$\displaystyle{\int}f=\frac{1}{|G|}\displaystyle{\sum_{\phi\in G}\phi(f)}$ for
all monomials $f\in\Bbbk[x_{1},x_{2},x_{3}]$ with degree less or equal to 6,
we discover that the elements
$\displaystyle{y_{1}=\frac{1}{3}(a+b)x_{1}+\frac{1}{3}(2a-b)x_{2}+x_{3}}$,
$y_{2}=x_{1}^{2}+x_{2}^{2}$ $+x_{1}x_{2}$,
$\displaystyle{y_{3}=2x_{1}^{3}+3x_{1}^{2}x_{2}-3x_{1}x_{2}^{2}-2x_{2}^{3}}$
are three algebraically independent polynomials that are invariant under the
action of $G$. Embed $\Bbbk[y_{1},y_{2},y_{3}]$ into the invariant subalgebra
$P^{G}$. Since the domain and codomain share the same Hilbert series, the
cokernel is trivial. In other words, the invariant subalgebra
$P^{G}=\Bbbk[y_{1},y_{2},y_{3}]$ and has the following Poisson structure:

$\\{y_{1},y_{2}\\}=y_{3},\quad\\{y_{2},y_{3}\\}=0,\quad\\{y_{3},y_{1}\\}=-6y_{2}^{2}.$

By applying Lemma 3.3, $P^{G}$ is not isomorphic to $P$ as Poisson algebras.

In Case (1), we will concurrently discuss three subcases:
$G_{1}=\langle\phi_{1}\big{|}_{P_{1}}=\begin{bmatrix}0&-1&0\\\ -1&0&0\\\
a&a&1\end{bmatrix}\rangle$,
$G_{2}=\langle\phi_{2}\big{|}_{P_{1}}=\begin{bmatrix}-1&0&0\\\ 1&1&0\\\
b&0&1\end{bmatrix}\rangle$, and
$G_{3}=\langle\phi_{3}\big{|}_{P_{1}}=\begin{bmatrix}1&1&0\\\ 0&-1&0\\\
0&c&1\end{bmatrix}\rangle$. Notice that the finite automorphism group $G_{i}$
is isomorphic to $\mathbb{Z}_{2}$ for all $1\leq i\leq 3$, and the Hilbert
series of the invariant subalgebra is

$h_{P^{G}}(t)=\frac{1}{2}\left(\frac{1}{(1-t)^{3}}+\frac{1}{(1-t)^{2}(1+t)}\right)=\frac{1}{(1-t)^{2}(1-t^{2})}$

by applying Molien’s Theorem. As in our previous analysis, we can find three
algebraically independent polynomials that are invariant under the action of
$G_{i}$:

|  $\boldsymbol{y_{1}}$ |  $\boldsymbol{y_{2}}$ |  $\boldsymbol{y_{3}}$
---|---|---|---
$G_{1}$ |  $x_{1}x_{2}$ |  $-x_{1}+x_{2}$ |  $ax_{1}+x_{3}$
$G_{2}$ |  $x_{1}^{2}$ |  $\frac{b}{2}x_{1}+x_{3}$ |  $\frac{1}{2}x_{1}+x_{2}$
$G_{3}$ |  $x_{2}^{2}$ |  $-cx_{1}+x_{3}$ |  $2x_{1}+x_{2}$

By comparing the Hilbert series of the domain and the codomain of the
embedding $\Bbbk[y_{1},y_{2},y_{3}]\hookrightarrow P^{G}$, we ascertain that
the invariant subalgebra $P^{G}=\Bbbk[y_{1},y_{2},y_{3}]$ and has the
following Poisson structures:

|  $\boldsymbol{\\{y_{1},y_{2}\\}}$ |  $\boldsymbol{\\{y_{2},y_{3}\\}}$ |  $\boldsymbol{\\{y_{3},y_{1}\\}}$
---|---|---|---
$G_{1}$ |  $0$ |  $6y_{1}+y_{2}^{2}$ |  $y_{1}y_{2}$
$G_{2}$ |  $-4y_{1}y_{3}$ |  $\frac{3}{4}y_{1}-y_{3}^{2}$ |  $0$
$G_{3}$ |  $2y_{1}y_{3}$ |  $-\frac{3}{2}y_{1}+\frac{1}{2}y_{3}^{2}$ |  $0$

Suppose that $P^{G}\cong P$ as Poisson algebras. Specifically, by applying
(the contrapositive of) Lemma 3.2, the invariant subalgebra is unimodular.
This implies that we can find a superpotential
$\Omega\in\Bbbk[y_{1},y_{2},y_{3}]$ satisfying:

$\frac{\partial\Omega}{\partial
y_{3}}=\\{y_{1},y_{2}\\},\quad\frac{\partial\Omega}{\partial
y_{1}}=\\{y_{2},y_{3}\\},\quad\frac{\partial\Omega}{\partial
y_{2}}=\\{y_{3},y_{1}\\}.$

It is straightforward to verify no such $\Omega$ exists for $G_{1}$, $G_{2}$,
and $G_{3}$, and therefore, $P^{G}$ is not unimodular, a contradiction.

In summary, if $G$ is not generated by Poisson reflections, then $P^{G}$ is
not isomorphic to $P$ as $\Bbbk$-algebras; if $G$ is generated by Poisson
reflections, then $P^{G}$ is not isomorphic to $P$ as Poisson algebras.
Consequently, the condition $P^{G}\cong P$ as Poisson algebras implies that
$G$ is necessarily trivial.

Unimodular 8. $\\{x_{1},x_{2}\\}=x_{1}x_{2}$,
$\\{x_{2},x_{3}\\}=3x_{1}^{2}+2x_{1}x_{2}+x_{2}x_{3}$,
$\\{x_{3},x_{1}\\}=x_{1}^{2}+x_{1}x_{3}$.

A Poisson reflection $\phi$ has the form
$\phi\big{|}_{P_{1}}=\begin{bmatrix}-1&0&0\\\ 0&1&0\\\ 2&0&1\end{bmatrix}$.
Suppose that $G=\langle\phi\rangle\cong\mathbb{Z}_{2}$. By the Molien’s
Theorem,

$h_{P^{G}}(t)=\frac{1}{2}\left(\frac{1}{(1-t)^{3}}+\frac{1}{(1-t)^{2}(1+t)}\right)=\frac{1}{(1-t)^{2}(1-t^{2})}.$

It’s not difficult to find three elements $y_{1}=x_{2}$, $y_{2}=x_{1}+x_{3}$,
$y_{3}=x_{1}^{2}$ that are algebraically independent and are invariant under
the action of $G$. Consider the natural inclusion
$\Bbbk[y_{1},y_{2},y_{3}]\hookrightarrow P^{G}$. Extend the natural inclusion
into a short exact sequence and apply Hilbert series. Given that the Hilbert
series of a short exact sequence sums up to 0, we deduce that the invariant
subalgebra $P^{G}=\Bbbk[y_{1},y_{2},y_{3}]$. The Poisson structure on the
invariant subalgebra is:

$\\{y_{1},y_{2}\\}=y_{1}y_{2}+3y_{3},\quad\\{y_{2},y_{3}\\}=2y_{2}y_{3},\quad\\{y_{3},y_{1}\\}=2y_{1}y_{3}.$

To prove that $P^{G}$ is not isomorphic to $P$ as Poisson algebras, one can
either invoke Lemma 3.2 or Lemma 3.3. Both approaches are equally
straightforward. ∎

##  4\. A Variant of The Watanabe Theorem

In this section, we prove a variant of the Shephard-Todd-Chevalley Theorem and
a variant of the Watanabe Theorem for Poisson enveloping algebras of quadratic
Poisson structures on $\Bbbk[x_{1},\cdots,x_{n}]$, under the actions induced
by $\text{PAut}_{\text{gr}}(P)\to\text{Aut}_{\text{gr}}(U(P))$.

Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a quadratic Poisson algebra and let
$U(P)$ be its Poisson enveloping algebra. Let
$\phi\in\text{PAut}_{\text{gr}}(P)$. [7, Lemma 5.1] constructed an induced
graded algebra automorphism $\widetilde{\phi}$ of $U(P)$ as follows:

$\displaystyle\widetilde{\phi}(x_{i})=\phi(x_{i}),\quad\widetilde{\phi}(y_{i})=\sum_{j=1}^{n}\frac{\partial
g(x_{i})}{\partial x_{j}}y_{j},$

for all $1\leq i\leq n$. This is the natural action to consider for the
following reasons:

######  Lemma 4.1.

Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a quadratic Poisson algebra. Suppose that
$\phi$ is a graded Poisson automorphism of $P$. Then there exists a unique
graded automorphism $\widetilde{\phi}$ on the Poisson enveloping algebra
$U(P)$ such that

$\alpha\circ\phi=\widetilde{\phi}\circ\alpha,\quad\beta\circ\phi=\widetilde{\phi}\circ\beta.$

###### Proof.

Define $\widetilde{\phi}:U(P)\to U(P)$ as follows:
$\widetilde{\phi}(x_{i})=\phi(x_{i})$ and
$\widetilde{\phi}(y_{i})=\displaystyle{\sum_{j=1}^{n}\frac{\partial\phi(x_{i})}{\partial
x_{j}}y_{j}}$, for all $1\leq i\leq n$. For commutativity,
$\alpha\circ\phi=\widetilde{\phi}\circ\alpha$;

$\beta\circ\phi(x_{i})=\sum_{j=1}^{n}\frac{\partial\phi(x_{i})}{\partial
x_{j}}y_{j},\quad\widetilde{\phi}\circ\beta(x_{i})=\widetilde{\phi}(y_{i})=\sum_{j=1}^{n}\frac{\partial\phi(x_{i})}{\partial
x_{j}}y_{j},$

for all $1\leq i\leq n$, and therefore
$\beta\circ\phi=\widetilde{\phi}\circ\beta$. Suppose that
$\widetilde{\widetilde{\phi}}:U(P)\to U(P)$ is another graded automorphism
such that $\alpha\circ\phi=\widetilde{\widetilde{\phi}}\circ\alpha$ and
$\beta\circ\phi=\widetilde{\widetilde{\phi}}\circ\beta$. From the
commutativity $\alpha\circ\phi=\widetilde{\widetilde{\phi}}\circ\alpha$, we
have
$\widetilde{\widetilde{\phi}}(x_{i})=\phi(x_{i})=\widetilde{\phi}(x_{i})$, for
all $1\leq i\leq n$. From the commutativity
$\beta\circ\phi=\widetilde{\widetilde{\phi}}\circ\beta$, we have

$\widetilde{\widetilde{\phi}}(y_{i})=\widetilde{\widetilde{\phi}}\circ\beta(x_{i})=\beta\circ\phi(x_{i})=\widetilde{\phi}\circ\beta(x_{i})=\widetilde{\phi}(y_{i}),$

for all $1\leq i\leq n$. Since $\widetilde{\widetilde{\phi}}$ and
$\widetilde{\phi}$ agree on the generators of $U(P)$, the graded automorphism
$\widetilde{\widetilde{\phi}}$ coincides with the graded automorphism
$\widetilde{\phi}$. ∎

Retain the above notations. Suppose that $G$ is a subgroup of the graded
Poisson automorphism of $P$. By Lemma 4.1, we can construct a subgroup
$\widetilde{G}=\\{\widetilde{\phi}:\phi\in G\\}$ of the graded automorphism
group of $U(P)$. It is natural to ask: is $\widetilde{G}$ isomorphic to $G$ as
groups? Once again, the answer is affirmative.

######  Lemma 4.2.

Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a quadratic Poisson algebra. Suppose that
$G$ is a subgroup of the graded Poisson automorphism group of $P$ and
$\widetilde{G}=\\{\widetilde{\phi}:\phi\in G\\}$ is the corresponding subgroup
of the graded automorphism group of $U(P)$. Then $\widetilde{G}$ is isomorphic
to $G$ as groups.

###### Proof.

Define $G\to\widetilde{G}:\phi\mapsto\widetilde{\phi}$. First, we claim that
this mapping is a group homomorphism. Let $\phi_{1},\phi_{2}\in G$. It is
clear that $\widetilde{\phi_{1}}\widetilde{\phi_{2}}$ and
$\widetilde{\phi_{1}\phi_{2}}$ agree on the generators $x_{1},\cdots,x_{n}$.
In the meantime, on the generators $y_{1},\cdots,y_{n}$,

$\displaystyle\widetilde{\phi_{1}}\widetilde{\phi_{2}}(y_{i})=$
$\displaystyle\widetilde{\phi_{1}}(\widetilde{\phi_{2}}(y_{i}))\hskip
155.06743pt$ $\displaystyle=$
$\displaystyle\widetilde{\phi_{1}}\left(\sum_{j=1}^{n}\frac{\partial\phi_{2}(x_{i})}{\partial
x_{j}}y_{j}\right)$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{n}\phi_{1}\left(\frac{\partial\phi_{2}(x_{i})}{\partial
x_{j}}\right)\left(\sum_{k=1}^{n}\frac{\partial\phi_{1}(x_{j})}{\partial
x_{k}}y_{k}\right)$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{n}\sum_{k=1}^{n}\phi_{1}\left(\frac{\partial\phi_{2}(x_{i})}{\partial
x_{j}}\right)\left(\frac{\partial\phi_{1}(x_{j})}{\partial x_{k}}y_{k}\right)$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{n}\sum_{k=1}^{n}\frac{\partial\phi_{1}(\phi_{2}(x_{i}))}{\partial\phi_{1}(x_{j})}\frac{\partial\phi_{1}(x_{j})}{\partial
x_{k}}y_{k}$ $\displaystyle=$
$\displaystyle\sum_{k=1}^{n}\frac{\partial\phi_{1}(\phi_{2}(x_{i}))}{\partial
x_{k}}y_{k}$ $\displaystyle=$
$\displaystyle\widetilde{\phi_{1}\phi_{2}}(y_{i}),$

for all $1\leq i\leq n$, in which the fifth equality follows from the
commutativity of the following diagram:

${\Bbbk[x_{1},\cdots,x_{n}]}$${\Bbbk[x_{1},\cdots,x_{n}]}$${\Bbbk[x_{1},\cdots,x_{n}]}$${\Bbbk[x_{1},\cdots,x_{n}]}$$\scriptstyle{\frac{\partial}{\partial
x_{j}}}$$\scriptstyle{\phi_{1}}$$\scriptstyle{\phi_{1}}$$\scriptstyle{\frac{\partial}{\partial\phi_{1}(x_{j})}}$

For injectivity, suppose that $\widetilde{\phi_{1}}=\widetilde{\phi_{2}}$. On
the generators $x_{1},\cdots,x_{n}$,
$\phi_{1}(x_{i})=\widetilde{\phi_{1}}(x_{i})=\widetilde{\phi_{2}}(x_{i})=\phi_{2}(x_{i})$,
for all $1\leq i\leq n$. Consequently, $\phi_{1}=\phi_{2}$. Finally,
surjectivity follows easily from the construction. ∎

Lemma 4.2 and Lemma 4.3 allow us to formulate the following questions, as
generalizations of the Shephard-Todd-Chevalley Theorem and the Watanabe
Theorem:

  1. (1)

Under what conditions on $G$ is $U(P)^{\widetilde{G}}$ Artin-Schelter regular?

  2. (2)

Under what conditions on $G$ is $U(P)^{\widetilde{G}}$ Artin-Schelter
Gorenstein?

The answer to Question (1) entails establishing the absence of quasi-
reflections in $\widetilde{G}$ through the enumeration of the eigenvalues of
its elements, given that the set of eigenvalues of a quasi-reflection of
$U(P)$ exhibits a highly constrained form as elucidated in [10, Theorem 3.1].

######  Lemma 4.3.

Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a quadratic Poisson algebra. Let $\phi$
be a graded Poisson automorphism of $P$ and let $\widetilde{\phi}$ be the
corresponding graded automorphism of $U(P)$. Suppose that
$\phi\big{|}_{P_{1}}$ has eigenvalues $\lambda_{1},\cdots,\lambda_{m}$, with
multiplicity $c_{1},\cdots,c_{m}$, respectively. Then
$\widetilde{\phi}\big{|}_{U(P)_{1}}$ has eigenvalues
$\lambda_{1},\cdots,\lambda_{m}$, with multiplicity $2c_{1},\cdots,2c_{m}$,
respectively.

###### Proof.

The Poisson enveloping algebra $U(P)$ is a quadratic $\Bbbk$-algebra generated
by $x_{1},\cdots,x_{n}$, $y_{1},\cdots,y_{n}$. Fix $1\leq i\leq m$. Let
$\\{v_{i,1},\cdots,v_{i,c_{i}}\\}$ be a basis for the eigenspace of
$\lambda_{i}$ in $P_{1}$. By calculation,

$\widetilde{\phi}(v_{i,j})=\phi(v_{i,j})=\lambda_{i}v_{i,j},$
$\displaystyle\widetilde{\phi}(\sum_{k=1}^{n}\frac{\partial\phi(v_{i,j})}{\partial
x_{k}}y_{k})=\sum_{k=1}^{n}\phi\left(\frac{\partial(\lambda_{i}v_{i,j})}{\partial
x_{k}}\right)\widetilde{\phi}(y_{k})=$
$\displaystyle\lambda_{i}\sum_{k=1}^{n}\phi(\frac{\partial v_{i,j}}{\partial
x_{k}})\widetilde{\phi}(y_{k})$ $\displaystyle=$
$\displaystyle\lambda_{i}\sum_{k=1}^{n}\sum_{l=1}^{n}\frac{\partial\phi(v_{i,j})}{\partial\phi(x_{k})}\frac{\partial\phi(x_{k})}{\partial
x_{l}}y_{l}=\lambda_{i}\sum_{l=1}^{n}\frac{\partial\phi(v_{i,j})}{\partial
x_{l}}y_{l}.$

for all $1\leq j\leq c_{i}$. Given that $\widetilde{\phi}$ is a graded
automorphism, the vectors
$\widetilde{\phi}(v_{i,1}),\cdots,\widetilde{\phi}(v_{i,c_{i}})$,
$\displaystyle{\widetilde{\phi}(\sum_{k=1}^{n}\frac{\partial\phi(v_{i,j})}{\partial
x_{k}}y_{k}})$,
$\cdots,\displaystyle{\widetilde{\phi}(\sum_{k=1}^{n}\frac{\partial\phi(v_{c_{i}})}{\partial
x_{k}}y_{k}})$ are linearly independent in $U(P)_{1}$. Since
$\displaystyle{\sum_{i=1}^{m}c_{i}}=n$,
$\displaystyle{\sum_{i=1}^{m}2c_{i}}=2n$. In light of the fact that
$\displaystyle{\sum_{i=1}^{m}c_{i}}=n$, it follows that
$\displaystyle{\sum_{i=1}^{m}2c_{i}}=2n$. This assertation implies that the
set

$\left\\{\widetilde{\phi}(v_{i,1}),\cdots,\widetilde{\phi}(v_{i,c_{i}}),\displaystyle{\widetilde{\phi}(\sum_{k=1}^{n}\frac{\partial\phi(v_{i,j})}{\partial
x_{k}}y_{k}}),\cdots,\displaystyle{\widetilde{\phi}(\sum_{k=1}^{n}\frac{\partial\phi(v_{c_{i}})}{\partial
x_{k}}y_{k}})\right\\}_{1\leq i\leq m}$

forms an eigenbasis for $U(P)_{1}$. In particular, the multiplicity of
$\lambda_{i}$ in $U(P)_{1}$ equals to twice of multiplicity of $\lambda_{i}$
in $P_{1}$. ∎

Now, we are prepared to establish a variant of the Shephard-Todd-Chevalley
Theorem for the Poisson enveloping algebras of quadratic Poisson structures on
$\Bbbk[x_{1},\cdots,x_{n}]$ under the induced action
$\text{PAut}_{\text{gr}}(P)\to\text{Aut}_{\text{gr}}(U(P))$.

######  Theorem 4.4.

Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a Poisson algebra and $U(P)$ be its
Poisson enveloping algebra. Let $G$ be a finite nontrivial subgroup of the
graded Poisson automorphism group of $P$ and let $\widetilde{G}$ be the
corresponding finite nontrivial subgroup of the graded automorphism group of
$U(P)$. The invariant subalgebra $U(P)^{\widetilde{G}}$ is Artin-Schelter
regular if and only if $G$ is trivial.

###### Proof.

It suffices to assume the graded Poisson automorphism group $G$ is nontrivial.
The Poisson enveloping algebra $U(P)$, according to [7, Lemma 5.4], is a
quantum polynomial ring. Therefore its quasi-reflections, as discussed in [10,
Theorem 3.1], is either a classical reflection or a mystic reflection:

  * •

The eigenvalues of $\widetilde{\phi}$ are
$\underbrace{1,\cdots,1}_{2n-1},\xi$, for some primitive root of unity $\xi$.

  * •

The order of $\widetilde{\phi}$ is 4 and the eigenvalues of $\widetilde{\phi}$
are $\underbrace{1,\cdots,1}_{2n-2}$, $i$, $-i$.

Comparing to the eigenvalues enumerated in Lemma 5.4.1, the induced graded
automorphism group $\widetilde{G}$ contains no quasi-reflections, which, as
indicated in [10, Lemma 6.1], results in the invariant subalgebra
$U(P)^{\widetilde{G}}$ having infinite global dimension. As a consequence, the
invariant subalgebra $U(P)^{\widetilde{G}}$ is not Artin-Schelter regular. ∎

Artin-Schelter regularity, as stated in Theorem 4.4, cannot be attained for
any nontrivial $G$. Artin-Schelter Gorensteinness, on the other hand, can be
achievered in certain instances. In practice, it is exceedingly challenging to
verify Artin-Schelter Gorensteinness through homological approaches due to the
difficulty of describing the generators and relations of
$U(P)^{\widetilde{G}}$ systemmatically. Instead, we shift our attention to [9,
Theorem 3.3], the “non-commutative Watanabe Theorem”:

######  Theorem 4.5.

[9, Theorem 3.3] Let $A$ be a Noetherian Artin-Schelter Gorenstein
$\Bbbk$-algebra and let $H$ be a finite subgroup of the graded automorphism
group of $A$. If $\text{hdet}h=1$ for all $h\in H$, then $A^{H}$ is Artin-
Schelter Gorenstein.

To apply [9, Theorem 3.3], we require a foundational understanding of the
homological determinant of each induced graded automorphism $\widetilde{\phi}$
of $U(P)$, which will be the primary goal of the following lemma.

######  Lemma 4.6.

Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a quadratic Poisson algebra and let
$\phi$ be a finite-order graded Poisson automorphism of $P$. Suppose that
$\phi\big{|}_{P_{1}}$ has eigenvalues $\lambda_{1},\cdots,\lambda_{m}$, with
multiplicity $c_{1},\cdots,c_{m}$, respectively. Then

$\text{Tr}_{U(P)}(\widetilde{\phi},t)=\frac{1}{(1-\lambda_{1}t)^{2c_{1}}\cdots(1-\lambda_{m}t)^{2c_{m}}}.$

###### Proof.

Let $1+\displaystyle{\sum_{i\geq 1}a_{i}t^{i}}$ be the Taylor expansion of
$\text{Tr}_{P}(\phi,t)$, where $a_{i}\in\Bbbk$ for all $i\geq 1$. It is
sufficient to prove that $(1+\displaystyle{\sum_{i\geq
1}a_{i}t^{i}})^{2}=\text{Tr}_{U(P)}(\widetilde{\phi},t)$. Fix $d\geq 1$. By
[17, Theorem 3.7], the degree $d$ component of the Poisson enveloping algebra
$U(P)_{d}$ admits a $\Bbbk$-linear basis $\\{x_{1}^{p_{1}}\cdots
x_{n}^{p_{n}}y_{1}^{q_{1}}\cdots
y_{n}^{q_{n}}:\displaystyle{\sum_{j=1}^{n}(p_{j}+q_{j})}=d\\}$. Let
$r_{1},\cdots,r_{n}\geq 0$, and $b_{r_{1},\cdots,r_{n}}$ be the coefficient of
the term $x_{1}^{r_{1}}\cdots x_{n}^{r_{n}}$ in $\phi(x_{1}^{r_{1}}\cdots
x_{n}^{r_{n}})$. Consider the coefficient of the term $x_{1}^{p_{1}}\cdots
x_{n}^{p_{n}}y_{1}^{q_{1}}\cdots y_{n}^{q_{n}}$ in
$\widetilde{\phi}(x_{1}^{p_{1}}\cdots x_{n}^{p_{n}}y_{1}^{q_{1}}\cdots
y_{n}^{q_{n}})$. There are three observations:

  1. (1)

Given that $[x_{i},x_{j}]=0$ in $U(P)$, the coefficient of the term
$x_{1}^{p_{1}}\cdots x_{n}^{p_{n}}$ in $\widetilde{\phi}(x_{1}^{p_{1}}\cdots
x_{n}^{p_{n}})$ is $b_{p_{1},\cdots,p_{n}}$.

  2. (2)

Given that
$[y_{i},y_{j}]=\displaystyle{\sum_{k=1}^{n}\frac{\partial\\{x_{i},x_{j}\\}}{\partial
x_{k}}y_{k}}$ and $[y_{i},x_{j}]=\\{x_{i},x_{j}\\}$ in $U(P)$, the coefficient
of the term $y_{1}^{q_{1}}\cdots y_{n}^{q_{n}}$ in
$\widetilde{\phi}(y_{1}^{q_{1}}\cdots y_{n}^{q_{n}})$ is
$b_{q_{1},\cdots,q_{n}}$.

  3. (3)

Given that
$\widetilde{\phi}(x_{j})\subseteq\displaystyle{\bigoplus_{k=1}^{n}\Bbbk
x_{k}}$, the coefficient of $x_{1}^{p_{1}}\cdots
x_{n}^{p_{n}}y_{1}^{q_{1}}\cdots y_{n}^{q_{n}}$ in
$\widetilde{\phi}(x_{1}^{p_{1}}\cdots x_{n}^{p_{n}}y_{1}^{q_{1}}\cdots
y_{n}^{q_{n}})$ is $b_{p_{1},\cdots,p_{n}}b_{q_{1},\cdots,q_{n}}$.

Let $1+\displaystyle{\sum_{i\geq 1}c_{i}t^{i}}$ be the Taylor expansion of
$\text{Tr}_{U(P)}(\widetilde{\phi},t)$, where $c_{i}\in\Bbbk$ for all $i\geq
1$. In accordance with the definition of the trace series and the above
argument, the coefficient relating to the dimension of the degree $d$
component $c_{d}$ equals to the summation of all
$b_{p_{1},\cdots,p_{n}}b_{q_{1},\cdots,q_{n}}$ ranging over
$p_{1}+\cdots+p_{n}+q_{1}+\cdots+q_{n}=d$. Equivalently,
$c_{d}=\displaystyle{\sum_{i+j=d}a_{i}a_{j}}$, and therefore,
$(1+\displaystyle{\sum_{i\geq 1}a_{i}t^{i}})^{2}=1+\displaystyle{\sum_{i\geq
1}c_{i}t^{i}}$.

Finally, since $P$ is a commutative polynomial ring,
$\displaystyle{\text{Tr}_{P}(g,t)=\frac{1}{(1-\lambda_{1}t)^{c_{1}}\cdots(1-\lambda_{m}t)^{c_{m}}}}$,
and consequently,

$\text{Tr}_{U(P)}(\widetilde{\phi},t)=\frac{1}{(1-\lambda_{1}t)^{2c_{1}}\cdots(1-\lambda_{m}t)^{2c_{m}}},$

as desired. ∎

With the assistance of Lemma 4.6, we are prepared to state a simple formula
for the computation of the homological determinant of $\widetilde{\phi}$.

######  Theorem 4.7.

Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a quadratic Poisson algebra and let
$U(P)$ be its Poisson enveloping algebra. Let $\phi$ be a finite-order graded
Poisson automorphism of $P$ and let $\widetilde{\phi}$ be the induced graded
automorphism of $U(P)$. Then

$\text{hdet}\widetilde{\phi}=(\det\phi\big{|}_{P_{1}})^{2}.$

###### Proof.

By [9, Proposition 4.2], the graded automorphism $\widetilde{\phi}$ is
rational over $\Bbbk$. Apply [9, Lemma 2.6], the trace series

$\text{Tr}_{U(P)}(\widetilde{\phi},t)=(\text{hdet}\widetilde{\phi})^{-1}t^{-2n}+\textit{lower
terms},$

when written as a Laurent series in $t^{-1}$. By Lemma 4.6,

$\displaystyle\text{Tr}_{U(P)}(\widetilde{\phi},t)=$
$\displaystyle\frac{1}{(1-\lambda_{1}t)^{2c_{1}}\cdots(1-\lambda_{m}t)^{2c_{m}}}$
$\displaystyle\vskip 3.0pt plus 1.0pt minus 1.0pt=$
$\displaystyle\frac{1}{\left(\det(\phi\big{|}_{P_{1}})\right)^{2}t^{2n}+\textit{lower
terms}}$ $\displaystyle\vskip 3.0pt plus 1.0pt minus 1.0pt=$
$\displaystyle\left(\text{det}(\phi\big{|}_{P_{1}})^{2}\right)^{-1}t^{-2n}+\textit{lower
terms}.$

Comparing the leading coefficient,
$\text{hdet}\widetilde{\phi}=(\det\phi\big{|}_{P_{1}})^{2}$. ∎

Combining [9, Theorem 3.3] and Theorem 4.7, we are able to provide an answer
to Question (2) for quadratic Poisson structures on
$\Bbbk[x_{1},\cdots,x_{n}]$:

######  Corollary 4.8.

Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a quadratic Poisson algebra and let
$U(P)$ be its Poisson enveloping algebra. Let $G$ be a finite subgroup of the
graded Poisson automorphism group of $P$ and let
$\widetilde{G}=\\{\widetilde{\phi}:\phi\in G\\}$ be the corresponding finite
subgroup of the graded automorphism group of $U(P)$. If $G$ is generated by
graded Poisson automorphisms $\phi_{1},\cdots,\phi_{m}$ such that
$\text{det}(\phi_{i}\big{|}_{P_{1}})=\pm 1$, then $U(P)^{\widetilde{G}}$ is
Artin-Schelter Gorenstein.

###### Proof.

This is a restatement of [9, Theorem 3.3] when we substitute the value of the
homological determinant as per the formula provided in Theorem 4.7. ∎

We conclude this section with the following example.

######  Example 4.9.

Let $P=\Bbbk[x_{1},x_{2}]$ be the Poisson algebra
$\\{f,g\\}=\displaystyle{\left(\frac{\partial f}{\partial x_{1}}\frac{\partial
g}{\partial x_{2}}-\frac{\partial g}{\partial x_{1}}\frac{\partial f}{\partial
x_{2}}\right)}qx_{1}x_{2}$, for some $q\neq 0$, for all $f,g\in P$. Let
$\xi_{n}$ be a primitive $n$th root of unity. Consider the graded Poisson
automorphism group $G=\langle\phi=[x_{1}\mapsto\xi_{n}x_{1},x_{2}\mapsto
x_{2}]\rangle$ of $P$, and its induced graded automorphism group
$\widetilde{G}=\langle\widetilde{\phi}=[x_{1}\mapsto\xi_{n}x_{1},x_{2}\mapsto
x_{2},y_{1}\mapsto\xi_{n}y_{1},y_{2}\mapsto y_{2}]\rangle$ of $U(P)$. The
invariant subalgebra $U(P)^{\widetilde{G}}$ is isomorphic to the graded
$\Bbbk$-algebra generated by $a_{1},\cdots,a_{n+1}$, $b$, $c$ subjecting to
the following relations:

  * •

$[a_{i},a_{j}]$,

  * •

$a_{i}a_{j}=a_{k}a_{i+j-k}$,

  * •

$a_{i+1}b-\displaystyle{\sum_{j=0}^{i}\binom{i}{j}q^{j}ba_{i+1-j}}$,

  * •

$a_{i+1}c-\displaystyle{\sum_{j=0}^{i}\binom{i}{j}q^{j}(c+npb)a_{i+1-j}}$.

According to Theorem 4.4 and Corollary 4.8, the invariant subalgebra
$U(P)^{\widetilde{G}}$ is not Artin-Schelter regular, and is not Artin-
Schelter Gorenstein except when $n=2$.

##  5\. Future Work

In this section, we remark on some intriguing findings encountered during the
proof of the main results presented in this paper and put forth several
avenues for future research.

######  Question 5.1.

Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a quadratic Poisson algebra and let
$G\subseteq\text{PAut}_{\text{gr}}(P)$ be a finite subgroup. In the case of
all quadratic Poisson structures that have been examined, Theorem 3.1, [7,
Theorem 4.4, 4.11, 4.17, 4.19], $P^{G}\cong P$ as Poisson algebras if and only
if $G$ is trivial. Does this statement hold universally for all quadratic
Poisson structures? Stated differently, is this the Shephard-Todd-Chevalley
Theorem for quadratic Poisson algebras? A good starting place is the 13 non-
unimodular quadratic Poisson structures classified in [[6], Theorem 2].

######  Question 5.2.

Let $P=\Bbbk[x_{1},\cdots,x_{n}]$ be a Poisson algebra with its Poisson
structure derived from the semiclassical limit of an Artin-Schelter regular
algebra $A$. It appears that the invariant subalgebras of $P$ and the
invariant subslagebras of $A$ bear a striking resemblance. One example is [11,
Theorem 4.5] and [7, Theorem 3.8]: these two papers state an identical
Shephard-Todd-Chevalley theorem for the skew polynomial rings and the Poisson
algebras arising from them. Another example is [11, Proposition 5.8] and [7,
4.6-4.11]: these two papers capture some significant similarities between the
quantum matrix algebras and the Poisson algebras arising from them. Naturally,
we inquire if there exists a comprehensive theory linking the invariants of
the Poisson algebras and their corresponding Artin-Schelter regular algebras.

## References

  * AP [95] J. Alev and P. Polo.  A rigidity theorem for finite group actions on enveloping algebras of semisimple lie algebras.  Advances in Mathematics, 111:2:208–226, 1995.
  * AS [87] Michael Artin and William F. Schelter.  Graded algebras of global dimension 3.  Advances in Mathematics, 66:171–216, 1987.
  * BW [79] J. W. Bruce and C. T. C. Wall.  On the classification of cubic surfaces.  Journal of the London Mathematical Society, s2-19(2):245–256, 1979.
  * BZ [18] Kenneth A. Brown and James J. Zhang.  Unimodular graded Poisson Hopf algebras.  Bulletin of the London Mathematical Society, 50:5:887–898, 2018.
  * Che [55] Claude Chevalley.  Invariants of finite groups generated by reflections.  American Journal of Mathematics, 77:4:778–782, 1955.
  * DH [91] J. Dufour and A. Haraki.  Rotationnels et structures de Poisson quadratiques.  Comptes rendus de l’Académie des Sciences, 312 I:137–140, 1991.
  * GVW [23] Jason Gaddis, Padmini Veerapen, and Xingting Wang.  Reflection groups and rigidity of quadratic Poisson algebras.  Algebras and Representation Theory, 26:329–358, 2023.
  * GW [20] Jason Gaddis and Xingting Wang.  The Zariski cancellation problem for Poisson algebras.  Journal of the London Mathematical Society, 101:1250–1279, 2020.
  * JZ [00] Peter Jørgensen and James Zhang.  Gourmet’s guide to Gorensteinness.  Advances in Mathematics, 151:313–345, 2000.
  * KKZ [09] E. Kirkman, J. Kuzmanovich, and J. J. Zhang.  Rigidity of graded regular algebras.  Transactions of the American Mathematical Society, 360:12:6331–6369, 2009.
  * KKZ [10] E. Kirkman, J. Kuzmanovich, and J. Zhang.  Shephard–Todd–Chevalley theorem for skew polynomial rings.  Algebras and Representation Theory, 13:127–158, 2010.
  * KKZ [15] E. Kirkman, J. Kuzmanovich, and J. Zhang.  Invariant theory of finite group actions on down-up algebras.  Transformation Groups, 20:113–165, 2015.
  * LWW [15] J. Luo, Q.-S. Wu, and S.-Q. Wang.  Twisted Poincaré duality between Poisson homology and Poisson cohomology.  Journal of Algebra, 442:484–505, 2015.
  * LWZ [20] Jiafeng Lü, Xingting Wang, and Guangbin Zhuang.  A note on the duality between Poisson homology and cohomology.  Communications in Algebra, 48:10:4170–4175, 2020.
  * MLTU [09] L. Makar-Limanov, U. Turusbekova, and U. Umirbaev.  Automorphisms of elliptical Poisson algebras.  Contemporary Mathematics, 483:169–177, 2009.
  * Oh [99] Sei-Qwon Oh.  Poisson enveloping algebras.  Communications in Algebra, 27:5:2181–2186, 1999.
  * OPS [06] Sei-Qwon Oh, Chun-Gil Park, and Yong-Yeon Shin.  A Poincaré-Birkhoff-Witt theorem for Poisson enveloping algebras.  Communications in Algebra, 30:10, 2006.
  * ST [54] G. C. Shephard and J. A. Todd.  Finite unitary reflection groups.  Canadian Journal of Mathematics, 6:274–304, 1954.
  * TWZ [22] Xin Tang, Xingting Wang, and James J. Zhang.  Twists of graded poisson algebras and related properties.  arXiv:2206.05639, 2022.
  * Wat [74] Keiichi Watanabe.  Certain invariant subrings are Gorenstein II.  Osaka Journal of Mathematics, 11:379–388, 1974.