Invariants of Unimodular Quadratic Polynomial Poisson Algebras of Dimension 3

TL;DR

This paper extends classical theorems like Shephard-Todd-Chevalley and Watanabe to the setting of unimodular quadratic Poisson algebras of dimension 3 and their Poisson enveloping algebras, analyzing invariants under automorphisms.

Contribution

It proves a variant of the Shephard-Todd-Chevalley theorem for 3-dimensional unimodular quadratic Poisson algebras and extends related results to their Poisson enveloping algebras.

Findings

Established a Shephard-Todd-Chevalley type theorem for the Poisson algebra.

Extended the theorem to the Poisson enveloping algebra under induced group actions.

Provided structural insights into invariants of unimodular quadratic Poisson algebras.

Abstract

Let $P = \Bbbk[x1,x2,x3]$ 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 $P$ and variants the Shephard-Todd-Chevalley theorem and the Watanabe theorem for its Poisson enveloping algebra $U(P)$ under the induced group $\widetilde{G}$.

Equations

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

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

$$U$$

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

$$\{x_1,x_2\}=\frac{\partial \Omega }{\partial x_3},\{x_2,x_3\}=\frac{\partial \Omega }{\partial x_1},\{x_3,x_1\}=\frac{\partial \Omega }{\partial x_2},$$

$$\alpha :P\to$$

$$f\mapsto$$

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

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

$${\text{PAut}}_{\text{gr}}(\text{P})$$

$$\text{PR}(\mathbf{P})$$

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

$${\text{PAut}}_{\text{gr}}(P)=\left\{\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ c& d& a\end{array}\right]:a,b\ne 0\right\}.$$

$${\text{PAut}}_{\text{gr}}(P)=\left\{\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right],\left[\begin{array}{ccc}0& a& 0\\ 0& 0& b\\ c& 0& 0\end{array}\right],\left[\begin{array}{ccc}0& 0& a\\ b& 0& 0\\ 0& c& 0\end{array}\right]:a,b,c\ne 0\right\}.$$

$$\text{PAut}_{\text{gr}}(P)=\left\{\begin{bmatrix}0&a&0\\ -a&-a&0\\ b&c&a\end{bmatrix},\begin{bmatrix}0&-a&0\\ -a&0&0\\ b&c&a\end{bmatrix},\begin{bmatrix}-a&0&0\\ a&a&0\\ b&c&a\end{bmatrix},\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}\right\}\\$$

$$\hskip 84.64711pt\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}\left\}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\begin{bmatrix}-a&-a&0\\ a&0&0\\ b&c&a\end{bmatrix},\begin{bmatrix}a&a&0\\ 0&-a&0\\ b&c&a\end{bmatrix},\begin{bmatrix}a&0&0\\ 0&a&0\\ b&c&a\end{bmatrix}:a\neq 0\right\}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}.$$

$${\text{PAut}}_{\text{gr}}(P)=\left\{\left[\begin{array}{ccc}a& 0& 0\\ 0& a& 0\\ 0& 0& a\end{array}\right]:a\ne 0\right\}.$$

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

$${\text{PAut}}_{\text{gr}}(P)=\left(⟨\left[\begin{array}{ccc}a& 0& 0\\ 0& a& 0\\ 0& 0& a\end{array}\right]:a\ne 0⟩\times ⟨\left[\begin{array}{ccc}1& 0& 0\\ 0& b& 0\\ 0& 0& b^2\end{array}\right]:b={\xi }_3,{\xi }_3^2⟩\right)⋊⟨\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]⟩.$$

$${\text{PAut}}_{\text{gr}}(P)=\left\{\left[\begin{array}{ccc}a& 0& 0\\ 0& \frac{a^2}{b}& 0\\ b-a& 0& b\end{array}\right]:a\ne 0\right\}.$$

$${\text{PAut}}_{\text{gr}}(P)=\left\{\left[\begin{array}{ccc}a& 0& 0\\ b& a& 0\\ -\frac{b^2}{a}& -2b& a\end{array}\right]:a\ne 0\right\}.$$

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

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

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

$${\left[\begin{array}{ccc}0& -1& 0\\ -1& 0& 0\\ b& c& 1\end{array}\right]}^n=\{\begin{array}{ll}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ \frac{n(b-c)}{2}& \frac{n(c-b)}{2}& 1\end{array}\right]& n\text{ is even,}\\ \text{vspace}-0.75cm& \\ \left[\begin{array}{ccc}0& -1& 0\\ -1& 0& 0\\ \frac{n+1}{2}b-\frac{n-1}{2}c& \frac{n+1}{2}c-\frac{n-1}{2}b& 1\end{array}\right]& n\text{ is odd.}\end{array}$$

$${\left[\begin{array}{ccc}-1& 0& 0\\ 1& 1& 0\\ b& c& 1\end{array}\right]}^n=\{\begin{array}{ll}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ \frac{n}{2}c& nc& 1\end{array}\right]& n\text{ is even,}\\ \text{vspace}-0.75cm& \\ \left[\begin{array}{ccc}0& -1& 0\\ -1& 0& 0\\ b+\frac{n-1}{2}c& nc& 1\end{array}\right]& n\text{ is odd.}\end{array}$$

$${\left[\begin{array}{ccc}1& 1& 0\\ 0& -1& 0\\ b& c& 1\end{array}\right]}^n=\{\begin{array}{ll}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ nb& \frac{n}{2}b& 1\end{array}\right]& n\text{ is even,}\\ \text{vspace}-0.75cm& \\ \left[\begin{array}{ccc}1& 1& 0\\ 0& -1& 0\\ nb& \frac{n-1}{2}b+c& 1\end{array}\right]& n\text{ is odd.}\end{array}$$

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

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

$${\underline{m}}_Q(f)=\sum _{i=1}^n\frac{\partial \{\varphi (x_i),\varphi (g)\}}{\partial \varphi (x_i)}=\sum _{i=1}^n\frac{\partial \varphi \{x_i,g\}}{\partial \varphi (x_i)}=\varphi \left(\sum _{i=1}^n\frac{\partial \{x_i,g\}}{\partial x_i}\right)=\varphi (m_P(g))=0,$$

$$P/(\{P,P\})\stackrel{\sim }{\to }Q/(\{Q,Q\})$$

Full text

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

[TABLE]

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:

[TABLE]

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:

[TABLE]

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.

[TABLE]

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

[TABLE]

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:

[TABLE]

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]:

[TABLE]

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:

[TABLE]

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

[TABLE]

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 $i\leq j$,

[TABLE]

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:

[TABLE]

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:

[TABLE]

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,

[TABLE]

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,

[TABLE]

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,

[TABLE]

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,

[TABLE]

[TABLE]

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,

[TABLE]

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,

[TABLE]

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

[TABLE]

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,

[TABLE]

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,

[TABLE]

∎

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,

[TABLE]

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,

[TABLE]

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,

[TABLE]

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,

[TABLE]

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

[TABLE]

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

[TABLE]

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

In conclusion,

[TABLE]

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,

[TABLE]

∎

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 $f\in Q$,

[TABLE]

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:

[TABLE]

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

[TABLE]

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

[TABLE]

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

[TABLE]

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,

[TABLE]

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:

[TABLE]

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,

[TABLE]

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:

[TABLE]

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:

[TABLE]

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:

[TABLE]

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:

[TABLE]

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:

[TABLE]

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:

[TABLE]

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:

[TABLE]

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:

[TABLE]

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,

[TABLE]

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:

[TABLE]

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:

[TABLE]

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

[TABLE]

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

[TABLE]

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:

[TABLE]

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:

[TABLE]

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,

[TABLE]

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:

[TABLE]

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:

[TABLE]

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

[TABLE]

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, $\alpha\circ\phi=\widetilde{\phi}\circ\alpha$;

[TABLE]

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

[TABLE]

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 $y_{1},\cdots,y_{n}$,

[TABLE]

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

[TABLE]

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,

[TABLE]

[TABLE]

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

[TABLE]

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

[TABLE]

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,

[TABLE]

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

[TABLE]

Proof.

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

[TABLE]

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

[TABLE]

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.