full

# Poisson derivations of a semiclassical limit of a family of quantum second
Weyl algebras

S Launois111The first-named author’s research was partly supported by EPSRC
grant EP/R009279/1. and I Oppong222The second-named author’s research was
partly supported by EPSRC grant EP/W522454/1.

###### Abstract

In [15], we studied deformations $A_{\alpha,\beta}$ of the second Weyl algebra
and computed their derivations. In the present paper, we identify the
semiclassical limits $\mathcal{A}_{\alpha,\beta}$ of these deformations and
compute their Poisson derivations. Our results show that the first Hochschild
cohomology group HH${}^{1}(A_{\alpha,\beta})$ is isomorphic to the first
Poisson cohomology group HP${}^{1}(\mathcal{A}_{\alpha,\beta}).$

##  1 Introduction

Throughout this study, $\mathbb{K}$ will denote a field with characteristic
zero.

Given a non-commutative algebra $A$ and its semiclassical limit $\mathcal{A},$
an intriguing question has always been “Do the properties of $A$ always
reflect the (Poisson) properties of $\mathcal{A}$?” For example, given the
centre, automorphisms, endomorphisms, derivations, and prime ideals of $A$,
can one successfully predict/conjecture the Poisson centre, Poisson
automorphisms, Poisson endomorphisms, Poisson derivations and Poisson prime
ideals of $\mathcal{A}$? Suppose for instance that the prime spectrum of $A$
reflects the Poisson prime spectrum of $\mathcal{A}.$ A follow-on question
will be whether they are homeomorphic/isomorphic? The answers to these
questions, in some specific cases and for some specific algebras, are
affirmative. For example, Goodearl [7] has conjectured that the prime and
primitive spectra of the quantized coordinate rings are respectively
homeomorphic to the Poisson prime and Poisson primitive spectra of their
corresponding semiclassical limits when the base field is algebraically closed
and of characteristic zero. This conjecture has been verified for the
following quantized coordinate rings: $\mathcal{O}_{q}(\mathbb{K}^{n})$ (see
[10, Theorem 4.1]), $\mathcal{O}_{q}(SL_{2}(\mathbb{K}))$ (see [7, Example
9.7]), $\mathcal{O}_{q}(SL_{3}(\mathbb{K}))$ (see [6, Theorem 5.21 & Corollary
5.22]) and $\mathcal{O}_{q}(GL_{2})$ (see [6, Corollary 5.23]). Goodearl
further established that the prime and primitive spectra of the enveloping
algebra $U(\mathfrak{g})$ are respectively homeomorphic to the prime and
primitive spectra of its semiclassical limit (see [7, Theorem 8.11, Example
2.6]). From [4], we also have that the Poisson endomorphisms of the Poisson
quantum generalized Weyl algebra are precisely the Poisson analogue of the
endomorphisms of the quantum generalised Weyl algebra. Belov-Kanel and
Kontsevich [1] have also conjectured that the group of automorphisms of an
$n^{\text{th}}$-Weyl algebra $A_{n}(\mathbb{K})$ is isomorphic to the group of
Poisson automorphisms of the corresponding Poisson Weyl algebra in
characteristic zero.

In [15], we studied a family of simple quotients

$A_{\alpha,\beta}:=U_{q}^{+}(G_{2})/\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle\
\ \ \ (\alpha,\beta)\in\mathbb{K}^{2}\setminus\\{(0,0)\\})$

of the positive part of the quantized enveloping algebra $U_{q}^{+}(G_{2})$,
and concluded that the algebra $A_{\alpha,\beta}$ is a $q$-deformation of a
quadratic extension of the second Weyl algebra $A_{2}(\mathbb{K}).$ Since
$A_{\alpha,\beta}$ deforms approximately to $A_{2}(\mathbb{K}),$ it is
considered as a quantum second Weyl algebra. Our goal here is to study a
semiclassical limit $\mathcal{A}_{\alpha,\beta}$ of $A_{\alpha,\beta},$ and
compare its Lie algebra of Poisson derivations to the Lie algebra of
derivations of $A_{\alpha,\beta}$ studied in [15]. Another property that is
also worth investigating is the Belov-Kanel and Kontsevich conjecture [ibid].
Thus, it is natural to ask if the automorphism group of $A_{\alpha,\beta}$ is
isomorphic to the Poisson automorphism group of $\mathcal{A}_{\alpha,\beta}$.
We will return to this problem in the near future after we have successfully
studied the automorphism group of $A_{\alpha,\beta}.$ In the present case and
as already mentioned, we only focus on studying the Lie algebra of Poisson
derivations of $\mathcal{A}_{\alpha,\beta}$, and comparing them to their non-
commutative counterparts in [15].

In the noncommutative world, the knowledge of the derivations of twisted group
algebras, studied by Osborn and Passman [19], has helped in studying the
derivations of other non-commutative algebras such as the quantum second Weyl
algebra (see [15]), quantum matrices (see [13]), generalized Weyl algebras
(see [11]) and some specific examples of quantum enveloping algebras (see
[14], [20], and [21]). In view of this, we also study the Poisson derivations
of the Poisson analogue of the twisted group algebras—called Poisson group
algebras—and apply the results to study the Poisson derivations of a
semiclassical limit $\mathcal{A}_{\alpha,\beta}$ of $A_{\alpha,\beta}$. The
rest of the paper is organised as follows.

In Section 2, we recall some basics on Poisson algebras and semiclassical
limit. We then proceed to study the Poisson derivations of the Poisson group
algebras. Similarly to their non-commutative counterparts in [19], every
Poisson derivation of a Poisson group algebra is the sum of an inner Poisson
derivation and a central/scalar Poisson derivation.

In Section 3, we study a semiclassical limit $\mathcal{A}$ of the quantum
algebra $U_{q}^{+}(G_{2})$ and establish that $\mathcal{A}$ is a Poisson
polynomial $\mathbb{K}$-algebra generated by six indeterminates
$X_{1},\ldots,X_{6}.$ Since $\mathcal{A}=\mathbb{K}[X_{1},\ldots,X_{6}]$
satisfies the conditions in [12, Hypothesis 1.7], we can apply the Poisson
deleting derivations algorithm [12] to study its Poisson centre $\mathcal{A}$:
a polynomial ring $\mathbb{K}[\Omega_{1},\Omega_{2}]$ in two variables. In
Section 4, we study some Poisson $\mathcal{H}$-prime ideals of $\mathcal{A}$
using Goodearl’s $\mathcal{H}$-stratification theory [8], and proceed to study
some family
$(\langle\Omega_{1}-\alpha,\Omega_{2}-\alpha\rangle)_{(\alpha,\beta)\in\mathbb{K}^{2}\setminus\\{(0,0)\\})}$
of maximal and primitive Poisson prime ideals of $\mathcal{A}$. Consequently,
we study their corresponding Poisson simple quotients

$\mathcal{A}_{\alpha,\beta}:=\mathbb{K}[X_{1},\ldots,X_{6}]/\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle,$

and conclude that the Poisson algebra $\mathcal{A}_{\alpha,\beta}$ is a
semiclassical limit of the quantum second Weyl algebra $A_{\alpha,\beta}.$
Having a complete description of the semiclassical limit
$\mathcal{A}_{\alpha,\beta}$ of $A_{\alpha,\beta},$ we proceed to study its
Poisson derivations in the final section of this paper, by following
procedures similar to its non-commutative counterpart $A_{\alpha,\beta}$ (see
[15, §5]). That is, we successively embed $\mathcal{A}_{\alpha,\beta}$ into a
suitable Poisson torus $\mathcal{R}_{3}$ via localization as follows:

$\mathcal{A}_{\alpha,\beta}=\mathcal{R}_{7}\subset\mathcal{R}_{6}=\mathcal{R}_{7}\Sigma_{6}^{-1}\subset\mathcal{R}_{5}=\mathcal{R}_{6}\Sigma_{5}^{-1}\subset\mathcal{R}_{4}=\mathcal{R}_{5}\Sigma_{4}^{-1}\subset\mathcal{R}_{3}.$

These embeddings and localization allow us to extend every Poisson derivation
of $\mathcal{A}_{\alpha,\beta}$ successively and uniquely to a Poisson
derivation of each of the Poisson algebras $\mathcal{R}_{i}$ through to the
Poisson torus $\mathcal{R}_{3}.$ Since a Poisson torus is an example of a
Poisson group algebra, we have that every Poisson derivation of
$\mathcal{R}_{3}$ is the sum of an inner Poisson derivation and a
central/scalar Poisson derivation. Given the Poisson derivations of
$\mathcal{R}_{3},$ we backwardly and successively pull the Poisson derivations
of $\mathcal{R}_{3}$ to $\mathcal{A}_{\alpha,\beta}.$ This gives us a complete
description of the Poisson derivations of $\mathcal{A}_{\alpha,\beta}.$
Similarly to their non-commutative counterparts in [15], every Poisson
derivation of $\mathcal{A}_{\alpha,\beta}$ is an inner Poisson derivations
provided $\alpha\beta\neq 0,$ and the sum of inner Poisson and scalar Poisson
derivations whenever $\alpha$ or $\beta$ is zero. More precisely, the first
Poisson cohomology group HP${}^{1}(\mathcal{A}_{\alpha,\beta})$ is a one-
dimensional vector space in the case where $\alpha$ or $\beta$ is zero.

##  2 Poisson derivations of Poisson group algebras

This section begins with a reminder about Poisson algebras and semiclassical
limit. We will then proceed to introduce Poisson group algebras, and
consequently, study their derivations.

###  2.1 Poisson algebras

A Poisson algebra $\mathcal{A}$ is a commutative algebra over $\mathbb{K}$
endowed with a skew-symmetric $\mathbb{K}$-bilinear map
$\\{-,-\\}:\mathcal{A}\times\mathcal{A}\longrightarrow\mathcal{A}$ which
satisfies the Leibniz rule (i.e., $\\{x,yz\\}=\\{x,y\\}z+y\\{x,z\\};$
$x,y,z\in\mathcal{A}$) and Jacobi identity (i.e.,
$\\{x,\\{y,z\\}\\}+\\{y,\\{z,x\\}\\}+\\{z,\\{x,y\\}\\}=0;$
$x,y,z\in\mathcal{A}$).

The map $\\{-,-\\}$ is called the Poisson bracket. From [16, Prop. 1.7], every
Poisson bracket of $\mathcal{A}$ extends uniquely to the localizations of
$\mathcal{A}.$ A Poisson ideal of $\mathcal{A}$ is any ideal $I$ such that
$\\{\mathcal{A},I\\}\subseteq I.$ Given a Poisson ideal $I$ of $\mathcal{A},$
it is well known that the quotient algebra $\mathcal{A}/I$ is a Poisson
algebra with an induced Poisson bracket defined as
$\\{\bar{x},\bar{y}\\}=\overline{\\{x,y\\}},$ where $\bar{x}:=x+I$ and
$\bar{y}:=y+I,$ note that $x,y\in\mathcal{A}.$ Finally, the subalgebra
$Z_{P}(\mathcal{A}):=\\{a\in\mathcal{A}\mid\\{a,x\\}=0,\ \forall
x\in\mathcal{A}\\}$ is called the  Poisson centre  of $\mathcal{A}.$

######  Remark 2.1.

If $\mathcal{A}$ is a Poisson algebra and $\\{x_{1},\ldots,x_{n}\\}$ is a
generating set for $\mathcal{A}$ (as an algebra), then

  * (1)

it is always enough to define a Poisson bracket $\\{-,-\\}$ on $\mathcal{A}$
by defining it on only the generating set.

  * (2)

for all $f,g\in\mathcal{A},$ we have that
$\\{f,g\\}=\displaystyle\sum_{i,j=1}^{n}\\{x_{i},x_{j}\\}\frac{\partial
f}{\partial x_{i}}\frac{\partial g}{\partial x_{j}}$ (see [7, Example
2.2(a)]).

###  2.2 Semiclassical limit

Given a non-commutative algebra, one can move from the ‘Non-commutative World’
to the ‘Poisson World’ through a process called semiclassical limit, and
reverse this process through quantization. This transformation (semiclassical
limit) and its reverse transformation (quantization) have been widely studied
(for example, see [5, §§1.1.3], [2, Chapter III.5], and [9, §2]). In line with
the presentation in [5, §§1.1.3], we present the following overview of
semiclassical limit. Let $R$ be a commutative principal ideal domain
containing the field $\mathbb{K}$ and $hR$ be a maximal ideal of $R$ for a fix
$h\in R.$ Let $A$ be an algebra which is not necessarily commutative torsion-
free $R$-algebra such that the quotient $\mathcal{A}:=A/hA$ is a commutative
algebra. For $u,v\in A;$ we have that $\bar{u}:=u+hA$ and $\bar{v}:=v+hA$ are
their respective canonical images in $\mathcal{A}$. Since
$\bar{u}\bar{v}=\bar{v}\bar{u},$ we have that $[u,v]:=uv-vu\in hA.$ There
exists a unique element $\gamma(u,v)$ of $A$ such that $[u,v]=h\gamma(u,v).$
It follows that

$\\{\bar{u},\bar{v}\\}:=\gamma(u,v)+hA=\frac{[u,v]}{h}+hA$

defines a Poisson bracket on $\mathcal{A}$ (see [5, §§1.1.3] for further
details). We say that $A$ is a quantization of $\mathcal{A},$ and
$\mathcal{A}$ is a semiclassical limit of $A.$ Fix $\lambda\in\mathbb{K}.$ The
algebra $\mathcal{A}_{\lambda}:=A/(h-\lambda)A$ is a deformation of the
Poisson algebra $\mathcal{A}=\mathcal{A}_{0}$ if the central element
$h-\lambda$ is not invertible in $A.$ We refer the interested reader to [9,
§2] for some known examples of semiclassical limit of some families of quantum
algebras.

###  2.3 Introduction to Poisson group algebras

In [19, §1&2], Osborn and Passman studied the derivations of twisted group
algebras. In line with their results, we also study the Poisson derivations of
Poisson group algebras. The results in this section will be crucial in the
final section of this paper where we study the Poisson derivations of a
semiclassical limit of the quantum second Weyl algebra $A_{\alpha,\beta}.$

Let $G$ represent a finitely generated abelian group and $\lambda:G\times
G\longrightarrow\mathbb{K}$ be a map such that $\lambda(y,x)=-\lambda(x,y)$
and $\lambda(x,yz)=\lambda(x,y)+\lambda(x,z).$ We define a Poisson group
algebra $\mathbb{K}_{P}^{\lambda}[G]$ as a commutative $\mathbb{K}$-algebra
which has a copy $\overline{G}:=\\{\bar{g}\mid g\in G\\}$ of $G$ as a basis
and satisfies the Poisson bracket via
$\\{\bar{x},\bar{y}\\}=\lambda(x,y)\bar{x}\bar{y}=\lambda(x,y)\overline{xy}$
for all $x,y\in G$ (note that $\bar{x}\bar{y}=\overline{xy}$). Observe that
$\lambda(x,y)=0$ if and only if $\\{\bar{x},\bar{y}\\}=0$.

As an example, take the group algebra $\mathbb{K}[\mathbb{Z}^{2}]$ generated
by $x^{\pm 1},y^{\pm 1}$ over $\mathbb{K},$ with basis
$\overline{\mathbb{Z}^{2}}:=\\{x^{i}y^{j}\mid(i,j)\in\mathbb{Z}^{2}\\}.$ A
Poisson structure can be defined on $\mathbb{K}[\mathbb{Z}^{2}]$ via
$\\{x^{i}y^{j},x^{k}y^{l}\\}=\lambda((i,j),(k,l))x^{i+k}y^{j+l}$, where
$\lambda((i,j),(k,l)):=il-jk$, to obtain a rank $2$ Poisson torus
$\mathbb{K}_{P}^{\lambda}[\mathbb{Z}^{2}].$ In general,
$\mathbb{K}_{P}^{\lambda}[\mathbb{Z}^{n}]$ is a Poisson torus of rank $n$ over
the field $\mathbb{K}$ for some
$\lambda:\mathbb{Z}^{n}\times\mathbb{Z}^{n}\rightarrow\mathbb{K},$ where
$\mathbb{Z}^{n}$ is the usual additive group.

Let $\gamma\in\mathbb{K}_{P}^{\lambda}[G].$ One can write $\gamma$ as
$\gamma=\sum_{g}c_{g}\bar{g}$ with $g\in G$ and $c_{g}\in\mathbb{K}.$ Note
that $c_{g}=0$ for almost all $g$. The set supp$(\gamma):=\\{g\in G\mid
c_{g}\neq 0$ in $\gamma\\}$ is called the support of $\gamma.$ Furthermore,
the set $C:=\\{g\in G\mid\\{\bar{g},\bar{x}\\}=0$ for all $x\in G\\}$ and
$\Delta(\bar{x}):=\\{g\in G\mid\\{\bar{g},\bar{x}\\}=0\\}$ are both subgroups
of $G.$ The following remark establishes a relationship between these two
subgroups.

######  Remark 2.2.

Let $\\{g_{1},\ldots,g_{n}\\}$ be a generating set for the group $G.$ Then,
$C=\bigcap_{i=1}^{n}\Delta(g_{i}).$

######  Lemma 2.3.

The Poisson centre $Z_{P}(\mathbb{K}_{P}^{\lambda}[G])$ of
$\mathbb{K}_{P}^{\lambda}[G]$ is $\mathbb{K}_{P}^{\lambda}[C]$.

###### Proof.

Clearly, $\mathbb{K}_{P}^{\lambda}[C]\subseteq
Z_{P}(\mathbb{K}_{P}^{\lambda}[G]).$ For the reverse inclusion, take
$\gamma=\sum_{g}c_{g}\bar{g}\in Z_{P}(\mathbb{K}_{P}^{\lambda}[G]).$ It
follows that
$0=\\{\gamma,\bar{x}\\}=\sum_{g}c_{g}\\{\bar{g},\bar{x}\\}=\sum_{g}c_{g}\lambda(g,x)\bar{g}\bar{x}$
for any $x\in G.$ Consequently, $\lambda(g,x)=0$ for all $g\in$
supp$(\gamma).$ This implies that supp$(\gamma)\subseteq C,$ hence
$\gamma\in\mathbb{K}_{P}^{\lambda}[C].$ ∎

######  Remark 2.4.

Let $e$ be the identity element of $G$. One can easily observe that
$Z_{P}(\mathbb{K}_{P}^{\lambda}[G])=\mathbb{K}$ if and only if $C=\\{e\\}.$

###  2.4 Central and inner Poisson derivations

####  2.4.1 Central Poisson derivations.

Let $\theta:(G,\cdot)\longrightarrow(\mathbb{K}_{P}^{\lambda}[C],+)$ be a
group homomorphism. That is, $\theta(xy)=\theta(x)+\theta(y)$ for all $x,y\in
G.$ Define a $\mathbb{K}$-linear operator $\mathcal{D}:=\mathcal{D}_{\theta}$
by

$\mathcal{D}(\bar{x})=\theta(x)\bar{x}$

for all $x\in G.$

######  Lemma 2.5.

$\mathcal{D}$ is a Poisson derivation of $\mathbb{K}_{P}^{\lambda}[G]$.

###### Proof.

We need to show that
$\mathcal{D}(\bar{x}\bar{y})=\mathcal{D}(\bar{x})\bar{y}+\bar{x}\mathcal{D}(\bar{y})\
\text{ and}\
\mathcal{D}(\\{\bar{x},\bar{y}\\})=\\{\mathcal{D}(\bar{x}),\bar{y}\\}+\\{\bar{x},\mathcal{D}(\bar{y})\\}$
for all $x,y\in G.$ Now,
$\mathcal{D}(\bar{x}\bar{y})=\theta(xy)\overline{xy}=\theta(xy)\bar{x}\bar{y}=\theta(x)\bar{x}\bar{y}+\theta(y)\bar{x}\bar{y}=\mathcal{D}(\bar{x})\bar{y}+\bar{x}\mathcal{D}(\bar{y}).$

Secondly,
$\mathcal{D}(\\{\bar{x},\bar{y}\\})=\lambda{(x,y)}\mathcal{D}(\bar{x}\bar{y})=\theta(xy)\lambda{(x,y)}\bar{x}\bar{y}=\theta(xy)\\{\bar{x},\bar{y}\\}=[\theta(x)+\theta(y)]\\{\bar{x},\bar{y}\\}=\theta(x)\\{\bar{x},\bar{y}\\}+\theta(y)\\{\bar{x},\bar{y}\\}=\\{\theta(x)\bar{x},\bar{y}\\}+\\{\bar{x},\theta(y)\bar{y}\\}=\\{\mathcal{D}(\bar{x}),\bar{y}\\}+\\{\bar{x},\mathcal{D}(\bar{y})\\}$
(note that $\\{\theta(x),\bar{y}\\}=\\{\bar{x},\theta(y)\\}=0$, since
$\theta(x)$ and $\theta(y)$ are Poisson central elements). ∎

Similarly to [19], we will call $\mathcal{D}$ in Lemma 2.5 as a central
Poisson derivation when $Z_{P}(\mathbb{K}_{P}^{\lambda}[G])$ strictly contains
$\mathbb{K},$ and a scalar Poisson derivation when
$Z_{P}(\mathbb{K}_{P}^{\lambda}[G])=\mathbb{K}.$

Observe that
$\mathcal{D}(\bar{x})=\theta(x)\bar{x}\in\mathbb{K}_{P}^{\lambda}[Cx]$ for all
$x\in G.$

####  2.4.2 Inner Poisson derivations.

Let $\gamma=\sum_{g}c_{g}\bar{g}\in\mathbb{K}_{P}^{\lambda}[G],$ where
$c_{g}\in\mathbb{K},$ and ham${}_{\gamma}:=\\{\gamma,-\\}.$ It is well known
that
ham${}_{\gamma}:\mathbb{K}_{P}^{\lambda}[G]\longrightarrow\mathbb{K}_{P}^{\lambda}[G]$
is a derivation called the hamiltonian derivation associated to $\gamma$.
Moreover,
ham${}_{\gamma}(\bar{x})=\\{\gamma,\bar{x}\\}=\sum_{g}\lambda{(g,x)}c_{g}\bar{g}\bar{x}\in\mathbb{K}_{P}^{\lambda}[Gx]$
for all $x\in G.$ As usual, we can assume that $C\ \cap$
supp$(\gamma)=\emptyset.$ Therefore,
ham${}_{\gamma}(\bar{x})\in\mathbb{K}_{P}^{\lambda}[(G\setminus C)x]$ for all
$x\in G.$

We call the hamiltonian derivation hamγ an  inner Poisson derivation.

We can now state our main result in this section in the theorem below.

######  Theorem 2.6.

Every Poisson derivation of the Poisson group algebra
$\mathbb{K}_{P}^{\lambda}[G]$ is uniquely the sum of an inner Poisson
derivation and a central Poisson derivation.

###### Proof.

Let $\mathcal{D}$ be a Poisson derivation of $\mathbb{K}_{P}^{\lambda}[G].$
Then, for $x\in G,$ we have that
$\mathcal{D}(\bar{x})\in\mathbb{K}_{P}^{\lambda}[G].$ Hence,
$\mathcal{D}(\bar{x})=\sum_{h\in G}b_{h}(x)\bar{h}=\sum_{h\in
G}b_{h}(x)\bar{h}\bar{x}^{-1}\bar{x}.$ Now, the map $G\rightarrow G$ with
$h\mapsto hx^{-1}$ is bijective, and so

$\mathcal{D}(\bar{x})=\sum_{g}a_{g}(x)\bar{g}\bar{x},$

where $g:=hx^{-1}$ and $a_{g}(x):=b_{gx}(x).$ Note that
$a_{g}:G\longrightarrow\mathbb{K}$ and $a_{g}(x)=0$ for almost all $x\in G.$

Since $\mathcal{D}$ is a Poisson derivation, we have that
$\mathcal{D}(\bar{x}\bar{y})=\mathcal{D}(\bar{x})\bar{y}+\bar{x}\mathcal{D}(\bar{y})$
for all $x,y\in G.$ As a result,

$\sum_{g}a_{g}(xy)\bar{g}\bar{x}\bar{y}=\sum_{g}a_{g}(x)\bar{g}\bar{x}\bar{y}+\sum_{g}a_{g}(y)\bar{g}\bar{x}\bar{y}=\sum_{g}[a_{g}(x)+a_{g}(y)]\bar{g}\bar{x}\bar{y}.$

Identifying the coefficients in the above equality reveals that

$a_{g}(xy)=a_{g}(x)+a_{g}(y).$

Secondly,
$\mathcal{D}(\\{\bar{x},\bar{y}\\})=\\{\mathcal{D}(\bar{x}),\bar{y}\\}+\\{\bar{x},\mathcal{D}(\bar{y})\\}.$
Now,

$\displaystyle\mathcal{D}(\\{\bar{x},\bar{y}\\})=\lambda{(x,y)}\mathcal{D}(\bar{x}\bar{y})=\sum_{g}\lambda{(x,y)}a_{g}(xy)\bar{g}\bar{x}\bar{y}.$
(1)

On the other hand,

$\displaystyle\\{\mathcal{D}(\bar{x}),\bar{y}\\}+\\{\bar{x},\mathcal{D}(\bar{y})\\}$
$\displaystyle=\sum_{g}a_{g}(x)\\{\bar{g}\bar{x},\bar{y}\\}+\sum_{g}a_{g}(y)\\{\bar{x},\bar{g}\bar{y}\\}$
$\displaystyle=\sum_{g}a_{g}(x)\\{\overline{gx},\bar{y}\\}+\sum_{g}a_{g}(y)\\{\bar{x},\overline{gy}\\}$
$\displaystyle=\sum_{g}[a_{g}(x)\lambda(gx,y)+a_{g}(y)\lambda(x,gy)]\bar{g}\bar{x}\bar{y}$
$\displaystyle=\sum_{g}a_{g}(x)[\lambda(g,y)+\lambda(x,y)]+a_{g}(y)[\lambda(x,g)+\lambda(x,y)]\bar{g}\bar{x}\bar{y}$
$\displaystyle=\sum_{g}[\lambda(x,y)a_{g}(xy)+a_{g}(x)\lambda(g,y)-a_{g}(y)\lambda(g,x)]\bar{g}\bar{x}\bar{y}.$
(2)

Since
$\mathcal{D}(\\{\bar{x},\bar{y}\\}=\\{\mathcal{D}(\bar{x}),\bar{y}\\}+\\{\bar{x},\mathcal{D}(\bar{y})\\},$
comparing (1) to (2) reveals that

$\displaystyle\lambda{(x,y)}a_{g}(xy)$
$\displaystyle=\lambda(x,y)a_{g}(xy)+a_{g}(x)\lambda(g,y)-a_{g}(y)\lambda(g,x).$

This implies that

$a_{g}(x)\lambda(g,y)=a_{g}(y)\lambda(g,x).$ (3)

Suppose that $g\in C$. It follows that $\lambda(g,y)=\lambda(g,x)=0$ for all
$x,y\in G.$ Since $a_{g}(xy)=a_{g}(x)+a_{g}(y),$ the map
$\theta:(G,\cdot)\longrightarrow(\mathbb{K}_{P}^{\lambda}[C],+)$ given by
$\theta(x)=\sum_{g\in C}a_{g}(x)\bar{g}$ is a group homomorphism. Hence,
$\theta$ defines a central Poisson derivation $\mathcal{D}_{\theta}$ of
$\mathbb{K}_{P}^{\lambda}[G],$ where

$\displaystyle\mathcal{D}_{\theta}(\bar{x})=\sum_{g\in
C}a_{g}(x)\bar{g}\bar{x}.$ (4)

Now, let $g\not\in C.$ There exists $y\in G$ such that $\lambda(g,y)\neq 0.$
Fix $y$ and define

$c_{g}:=\frac{a_{g}(y)}{\lambda(g,y)}.$

Take any arbitrary element $x\in G$. It follows that

$c_{g}\lambda(g,x)=\frac{a_{g}(y)\lambda(g,x)}{\lambda(g,y)}.$

From (3), we have that

$c_{g}\lambda(g,x)=\frac{a_{g}(y)\lambda(g,x)}{\lambda(g,y)}=\frac{a_{g}(x)\lambda(g,y)}{\lambda(g,y)}=a_{g}(x),$

for all $x\in G.$

Define $\gamma\in\mathbb{K}_{P}^{\lambda}[G]$ as $\gamma:=\sum_{g\not\in
C}c_{g}\bar{g}.$ Then,

$\displaystyle\text{ham}_{\gamma}(\bar{x})=\\{\gamma,\bar{x}\\}=\sum_{g\not\in
C}c_{g}\lambda(g,x)\bar{g}\bar{x}=\sum_{g\not\in C}a_{g}(x)\bar{g}\bar{x}.$
(5)

From (4) and (5), one can conclude that every Poisson derivation $\mathcal{D}$
of $\mathbb{K}_{P}^{\lambda}[G]$ can be written as
$\mathcal{D}=\mathcal{D}_{\theta}+\text{ham}_{\gamma}.$ This decomposition of
$\mathcal{D}$ into an inner Poisson derivation ($\text{ham}_{\gamma}$) and a
central Poisson derivation ($\mathcal{D}_{\theta}$) is actually unique,
because $\mathbb{K}_{P}^{\lambda}[Gx]$ can be decomposed as
$\mathbb{K}_{P}^{\lambda}[Gx]=\mathbb{K}_{P}^{\lambda}[Cx]\oplus\mathbb{K}_{P}^{\lambda}[(G\setminus
C)x].$ Now, every central Poisson derivation maps $\bar{x}$ to an element of
the subspace $\mathbb{K}_{P}^{\lambda}[Cx],$ and every inner Poisson
derivation maps $\bar{x}$ to an element of the subspace
$\mathbb{K}_{P}^{\lambda}[(G\setminus C)x].$ ∎

######  Corollary 2.7.

Suppose that $C=\\{e\\}$ (equivalently,
$Z_{P}(\mathbb{K}_{P}^{\lambda}[G])=\mathbb{K}$). Then, every Poisson
derivation of $\mathbb{K}_{P}^{\lambda}[G]$ is uniquely the sum of an inner
and a scalar Poisson derivation.

##  3 Poisson prime spectrum and Poisson deleting derivations algorithm of a
semiclassical limit of $U_{q}^{+}(G_{2})$

This section aims to study a semiclassical limit of the positive part of the
quantized enveloping algebra of type $G_{2},$ $U_{q}^{+}(G_{2})$. Given the
semiclassical limit of $U_{q}^{+}(G_{2})$, we will study its Poisson prime
spectrum using Goodearl’s $\mathcal{H}$-stratification theory [8], and its
Poisson deleting derivations algorithm–introduced by Launois and Lecoutre
[12]. Given the data of the Poisson deleting derivations algorithm, we will
study the Poisson centre of the semiclassical limit.

###  3.1 Semiclassical limit of the algebra $U_{q}^{+}(G_{2})$

Recall from [15, §2] that $U_{q}^{+}(G_{2})$ satisfies the following
relations:

$\displaystyle E_{2}E_{1}$ $\displaystyle=q^{-3}E_{1}E_{2}$ $\displaystyle
E_{3}E_{1}$ $\displaystyle=q^{-1}E_{1}E_{3}-(q+q^{-1}+q^{-3})E_{2}$
$\displaystyle E_{3}E_{2}$ $\displaystyle=q^{-3}E_{2}E_{3}$ $\displaystyle
E_{4}E_{1}$ $\displaystyle=E_{1}E_{4}+(1-q^{2})E_{3}^{2}$ $\displaystyle
E_{4}E_{2}$
$\displaystyle=q^{-3}E_{2}E_{4}-\frac{q^{4}-2q^{2}+1}{q^{4}+q^{2}+1}E_{3}^{3}$
$\displaystyle E_{4}E_{3}$ $\displaystyle=q^{-3}E_{3}E_{4}$ $\displaystyle
E_{5}E_{1}$ $\displaystyle=qE_{1}E_{5}-(1+q^{2})E_{3}$ $\displaystyle
E_{5}E_{2}$ $\displaystyle=E_{2}E_{5}+(1-q^{2})E_{3}^{2}$ $\displaystyle
E_{5}E_{3}$ $\displaystyle=q^{-1}E_{3}E_{5}-(q+q^{-1}+q^{-3})E_{4}$
$\displaystyle E_{5}E_{4}$ $\displaystyle=q^{-3}E_{4}E_{5}$ $\displaystyle
E_{6}E_{1}$ $\displaystyle=q^{3}E_{1}E_{6}-q^{3}E_{5}$ $\displaystyle
E_{6}E_{2}$ $\displaystyle=q^{3}E_{2}E_{6}+(q^{4}+q^{2}-1)E_{4}+$
$\displaystyle E_{6}E_{3}$ $\displaystyle=E_{3}E_{6}+(1-q^{2})E_{5}^{2}$
$\displaystyle\qquad\ (q^{2}-q^{4})E_{3}E_{5}$ $\displaystyle E_{6}E_{4}$
$\displaystyle=q^{-3}E_{4}E_{6}-\frac{q^{4}-2q^{2}+1}{q^{4}+q^{2}+1}E_{5}^{3}$
$\displaystyle E_{6}E_{5}$ $\displaystyle=q^{-3}E_{5}E_{6}.$

Set $U_{i}:=(q-1)E_{i}$ for $i=1,3,4,5,$ and $U_{i}:=f(q)(q-1)E_{i}$ for
$i=2,6;$ where $f(q)=q^{4}+q^{2}+1.$ Then, $U_{q}^{+}(G_{2})$ is now generated
by $U_{1},\ldots,U_{6}$ subject to the relations:

$\displaystyle U_{2}U_{1}$ $\displaystyle=q^{-3}U_{1}U_{2}$
$\displaystyle\qquad U_{3}U_{2}$ $\displaystyle=q^{-3}U_{2}U_{3}$
$\displaystyle U_{3}U_{1}$ $\displaystyle=q^{-1}U_{1}U_{3}-q^{-3}(q-1)U_{2}$
$\displaystyle\qquad U_{4}U_{1}$ $\displaystyle=U_{1}U_{4}+(1-q^{2})U_{3}^{2}$
$\displaystyle U_{4}U_{2}$
$\displaystyle=q^{-3}U_{2}U_{4}-(q+1)^{2}(q-1)U_{3}^{3}$ $\displaystyle\qquad
U_{4}U_{3}$ $\displaystyle=q^{-3}U_{3}U_{4}$ $\displaystyle U_{5}U_{1}$
$\displaystyle=qU_{1}U_{5}-(1+q^{2})(q-1)U_{3}$ $\displaystyle\qquad
U_{5}U_{2}$ $\displaystyle=U_{2}U_{5}+f(q)(1-q^{2})U_{3}^{2}$ $\displaystyle
U_{5}U_{3}$ $\displaystyle=q^{-1}U_{3}U_{5}-f(q)(q^{-2}-q^{-3})U_{4}$
$\displaystyle\qquad U_{5}U_{4}$ $\displaystyle=q^{-3}U_{4}U_{5}$
$\displaystyle U_{6}U_{1}$
$\displaystyle=q^{3}U_{1}U_{6}-f(q)(q^{4}-q^{3})U_{5}$ $\displaystyle\qquad
U_{6}U_{2}$ $\displaystyle=q^{3}U_{2}U_{6}+f(q)^{2}(q^{2}-q^{4})U_{3}U_{5}+$
$\displaystyle U_{6}U_{3}$ $\displaystyle=U_{3}U_{6}+f(q)(1-q^{2})U_{5}^{2}$
$\displaystyle\qquad f(q)^{2}(q^{4}+q^{2}-1)(q-1)U_{4}$ $\displaystyle
U_{6}U_{4}$ $\displaystyle=q^{-3}U_{4}U_{6}-(q+1)^{2}(q-1)U_{5}^{3}$
$\displaystyle\qquad U_{6}U_{5}$ $\displaystyle=q^{-3}U_{5}U_{6}.$

We now find a ‘new’ presentation for $U_{q}^{+}(G_{2})$ that allows us to
introduce a quantisation of $U_{q}^{+}(G_{2}).$ Let
$\widehat{U_{q}^{+}(G_{2})}$ be a $\mathbb{K}[z^{\pm 1}]$-algebra generated by
$\widehat{U}_{1},\ldots,\widehat{U}_{6}$ subject to the relations:

$\displaystyle\widehat{U}_{2}\widehat{U}_{1}$
$\displaystyle=z^{-3}\widehat{U}_{1}\widehat{U}_{2}$
$\displaystyle\qquad\widehat{U}_{3}\widehat{U}_{2}$
$\displaystyle=z^{-3}\widehat{U}_{2}\widehat{U}_{3}$
$\displaystyle\widehat{U}_{3}\widehat{U}_{1}$
$\displaystyle=z^{-1}\widehat{U}_{1}\widehat{U}_{3}-z^{-3}(z-1)\widehat{U}_{2}$
$\displaystyle\qquad\widehat{U}_{4}\widehat{U}_{1}$
$\displaystyle=\widehat{U}_{1}\widehat{U}_{4}+(1-z^{2})\widehat{U}_{3}^{2}$
$\displaystyle\widehat{U}_{4}\widehat{U}_{2}$
$\displaystyle=z^{-3}\widehat{U}_{2}\widehat{U}_{4}-(z+1)^{2}(z-1)\widehat{U}_{3}^{3}$
$\displaystyle\qquad\widehat{U}_{4}\widehat{U}_{3}$
$\displaystyle=z^{-3}\widehat{U}_{3}\widehat{U}_{4}$
$\displaystyle\widehat{U}_{5}\widehat{U}_{1}$
$\displaystyle=z\widehat{U}_{1}\widehat{U}_{5}-(1+z^{2})(z-1)\widehat{U}_{3}$
$\displaystyle\qquad\widehat{U}_{5}\widehat{U}_{2}$
$\displaystyle=\widehat{U}_{2}\widehat{U}_{5}+f(z)(1-z^{2})\widehat{U}_{3}^{2}$
$\displaystyle\widehat{U}_{5}\widehat{U}_{3}$
$\displaystyle=z^{-1}\widehat{U}_{3}\widehat{U}_{5}-f(z)(z^{-2}-z^{-3})\widehat{U}_{4}$
$\displaystyle\qquad\widehat{U}_{5}\widehat{U}_{4}$
$\displaystyle=z^{-3}\widehat{U}_{4}\widehat{U}_{5}$
$\displaystyle\widehat{U}_{6}\widehat{U}_{1}$
$\displaystyle=z^{3}\widehat{U}_{1}\widehat{U}_{6}-f(z)(z^{4}-z^{3})\widehat{U}_{5}$
$\displaystyle\qquad\widehat{U}_{6}\widehat{U}_{2}$
$\displaystyle=z^{3}\widehat{U}_{2}\widehat{U}_{6}+f(z)^{2}(z^{2}-z^{4})\widehat{U}_{3}\widehat{U}_{5}+$
$\displaystyle\widehat{U}_{6}\widehat{U}_{3}$
$\displaystyle=\widehat{U}_{3}\widehat{U}_{6}+f(z)(1-z^{2})\widehat{U}_{5}^{2}$
$\displaystyle\qquad f(z)^{2}(z^{4}+z^{2}-1)(z-1)\widehat{U}_{4}$
$\displaystyle\widehat{U}_{6}\widehat{U}_{4}$
$\displaystyle=z^{-3}\widehat{U}_{4}\widehat{U}_{6}-(z+1)^{2}(z-1)\widehat{U}_{5}^{3}$
$\displaystyle\qquad\widehat{U}_{6}\widehat{U}_{5}$
$\displaystyle=z^{-3}\widehat{U}_{5}\widehat{U}_{6},$

where $f(z)=z^{4}+z^{2}+1.$ Fix $\lambda\in\mathbb{K}^{*}.$ Observe that the
element $z-\lambda$ is central and not invertible in
$\widehat{U_{q}^{+}(G_{2})},$ hence we set
$\mathcal{A}_{\lambda}:=\widehat{U_{q}^{+}(G_{2})}/(z-\lambda)\widehat{U_{q}^{+}(G_{2})}.$
Now, $\mathcal{A}_{q}$ is the non-commutative algebra $U_{q}^{+}(G_{2})$ and
$\mathcal{A}_{1}=\mathbb{K}[X_{1},\ldots,X_{6}]$ with
$X_{i}:=\widehat{U}_{i}+(z-1)\widehat{U_{q}^{+}(G_{2})}$ is a Poisson algebra
with the Poisson bracket defined as follows:

$\displaystyle\\{X_{2},X_{1}\\}$ $\displaystyle=-3X_{1}X_{2}$
$\displaystyle\\{X_{3},X_{1}\\}$ $\displaystyle=-X_{1}X_{3}-X_{2}$
$\displaystyle\\{X_{3},X_{2}\\}$ $\displaystyle=-3X_{2}X_{3}$
$\displaystyle\\{X_{4},X_{1}\\}$ $\displaystyle=-2X_{3}^{2}$
$\displaystyle\\{X_{4},X_{2}\\}$ $\displaystyle=-3X_{2}X_{4}-{4}X_{3}^{3}$
$\displaystyle\\{X_{4},X_{3}\\}$ $\displaystyle=-3X_{3}X_{4}$
$\displaystyle\\{X_{5},X_{1}\\}$ $\displaystyle=X_{1}X_{5}-2X_{3}$
$\displaystyle\\{X_{5},X_{2}\\}$ $\displaystyle=-6X_{3}^{2}$
$\displaystyle\\{X_{5},X_{3}\\}$ $\displaystyle=-X_{3}X_{5}-3X_{4}$
$\displaystyle\\{X_{5},X_{4}\\}$ $\displaystyle=-3X_{4}X_{5}$
$\displaystyle\\{X_{6},X_{1}\\}$ $\displaystyle=3X_{1}X_{6}-3X_{5}$
$\displaystyle\\{X_{6},X_{2}\\}$
$\displaystyle=3X_{2}X_{6}+9X_{4}-18X_{3}X_{5}$
$\displaystyle\\{X_{6},X_{3}\\}$ $\displaystyle=-6X_{5}^{2}$
$\displaystyle\\{X_{6},X_{4}\\}$ $\displaystyle=-3X_{4}X_{6}-4X_{5}^{3}$
$\displaystyle\\{X_{6},X_{5}\\}$ $\displaystyle=-3X_{5}X_{6}.$

Therefore, $\mathcal{A}_{1}$ is a semiclassical limit of the non-commutative
algebra $\widehat{U_{q}^{+}(G_{2})},$ and $\mathcal{A}_{q}$ is a deformation
of the Poisson algebra $\mathcal{A}_{1}.$ For simplicity, we set

$\mathcal{A}:=\mathcal{A}_{1}=\mathbb{K}[X_{1},\ldots,X_{6}]$

in the rest of this paper. One can write the Poisson algebra $\mathcal{A}$ as
an iterated Poisson-Ore extension over $\mathbb{K}$ (see [17, Theorem 1.1] for
the definition of iterated Poisson-Ore extension) as follows:

$\mathcal{A}=\mathbb{K}[X_{1}][X_{2};\sigma_{2}]_{P}[X_{3};\sigma_{3},\delta_{3}]_{P}[X_{4};\sigma_{4},\delta_{4}]_{P}[X_{5};\sigma_{5},\delta_{5}]_{P}[X_{6};\sigma_{6},\delta_{6}]_{P};$
(6)

where, $\sigma_{i}$ and $\delta_{i}$ are respectively the Poisson derivations
and Poisson $\sigma_{i}$-derivations of

$\mathbb{K}[X_{1}][X_{2};\sigma_{2}]_{P}[X_{3};\sigma_{3},\delta_{3}]_{P}\ldots[X_{i-1};\sigma_{i-1},\delta_{i-1}]_{P}$

($2\leq i\leq 6$ and $\delta_{2}=0$) defined as follows:

$\displaystyle\sigma_{2}(X_{1})$ $\displaystyle=-3X_{1}$
$\displaystyle\sigma_{3}(X_{1})$ $\displaystyle=-X_{1}$
$\displaystyle\sigma_{3}(X_{2})$ $\displaystyle=-3X_{2}$
$\displaystyle\sigma_{4}(X_{1})$ $\displaystyle=0$
$\displaystyle\sigma_{4}(X_{2})$ $\displaystyle=-3X_{2}$
$\displaystyle\sigma_{4}(X_{3})$ $\displaystyle=-3X_{3}$
$\displaystyle\sigma_{5}(X_{1})$ $\displaystyle=X_{1}$
$\displaystyle\sigma_{5}(X_{2})$ $\displaystyle=0$
$\displaystyle\sigma_{5}(X_{3})$ $\displaystyle=-X_{3}$
$\displaystyle\sigma_{5}(X_{4})$ $\displaystyle=-3X_{4}$
$\displaystyle\sigma_{6}(X_{1})$ $\displaystyle=3X_{1}$
$\displaystyle\sigma_{6}(X_{2})$ $\displaystyle=3X_{2}$
$\displaystyle\sigma_{6}(X_{3})$ $\displaystyle=0$
$\displaystyle\sigma_{6}(X_{4})$ $\displaystyle=-3X_{4}$
$\displaystyle\sigma_{6}(X_{5})$ $\displaystyle=-3X_{5},$

and

$\displaystyle\delta_{3}(X_{1})$ $\displaystyle=-X_{2}$
$\displaystyle\delta_{3}(X_{2})$ $\displaystyle=0$
$\displaystyle\delta_{4}(X_{1})$ $\displaystyle=-2X_{3}^{2}$
$\displaystyle\delta_{4}(X_{2})$ $\displaystyle=-{4}X_{3}^{3}$
$\displaystyle\delta_{4}(X_{3})$ $\displaystyle=0$
$\displaystyle\delta_{5}(X_{1})$ $\displaystyle=-2X_{3}$
$\displaystyle\delta_{5}(X_{2})$ $\displaystyle=-6X_{3}^{2}$
$\displaystyle\delta_{5}(X_{3})$ $\displaystyle=-3X_{4}$
$\displaystyle\delta_{5}(X_{4})$ $\displaystyle=0$
$\displaystyle\delta_{6}(X_{1})$ $\displaystyle=-3X_{5}$
$\displaystyle\delta_{6}(X_{2})$ $\displaystyle=9X_{4}-18X_{3}X_{5}$
$\displaystyle\delta_{6}(X_{3})$ $\displaystyle=-6X_{5}^{2}$
$\displaystyle\delta_{6}(X_{4})$ $\displaystyle=-{4}X_{5}^{3}$
$\displaystyle\delta_{6}(X_{5})$ $\displaystyle=0.$

######  Remark 3.1.

[2, Theorem I.1.13] (Hilbert’s Basis Theorem) Since $\mathbb{K}$ is a
noetherian domain, the Poisson algebra
$\mathcal{A}=\mathbb{K}[X_{1}][X_{2};\sigma_{2}]_{P}[X_{3};\sigma_{3},\delta_{3}]_{P}[X_{4};\sigma_{4},\delta_{4}]_{P}[X_{5};\sigma_{5},\delta_{5}]_{P}[X_{6};\sigma_{6},\delta_{6}]_{P}$
is a noetherian domain.

###  3.2 Poisson prime spectrum of the semiclassical limit of the algebra
$U_{q}^{+}(G_{2})$

We study the Poisson prime spectrum of the Poisson algebra
$\mathcal{A}=\mathbb{K}[X_{1},\ldots,X_{6}]$ in this subsection. Let $P$ be a
proper Poisson ideal of a Poisson algebra $\mathcal{A}$ and $I_{1}$, $I_{2}$
be Poisson ideals of $\mathcal{A}$ such that $P\supseteq I_{1}I_{2}.$ The
ideal $P$ is called a Poisson prime ideal provided $P\supseteq I_{1}$ or
$P\supseteq I_{2}.$ Since the Poisson algebra $\mathcal{A}$ is a noetherian
domain, every Poisson ideal which is also a prime ideal is a Poisson prime
ideal and vice versa (see [8, Lemma 1.1]). The set of all the Poisson prime
ideals of $\mathcal{A}$ is called the Poisson prime spectrum of $\mathcal{A},$
denoted by P.Spec$(\mathcal{A}).$ The largest Poisson prime ideal contained in
a given maximal ideal of $\mathcal{A}$ is called a Poisson primitive ideal.
The set of all these Poisson primitive ideals is called the Poisson primitive
spectrum of $\mathcal{A}$, denoted by P.Prim$(\mathcal{A}).$

Let $\mathcal{H}$ be an algebraic torus acting rationally on $\mathcal{A}$ by
Poisson automorphism (see [8, §2] for the definition of a rational torus
action). A Poisson prime ideal $P$ is $\mathcal{H}$-invariant if $h\cdot P=P$
for all $h\in\mathcal{H}.$ Moreover, $J:=\bigcap_{h\in\mathcal{H}}h\cdot P$ is
the largest Poisson $\mathcal{H}$-invariant prime ideal contained in $P.$

The set

$\text{P.Spec}_{J}(\mathcal{A}):=\\{P\in\text{P.Spec}(\mathcal{A})\mid(P:\mathcal{H})=J\\}$

is called the $J$-stratum of P.Spec$(\mathcal{A}).$ The $\mathcal{H}$-strata
$\text{P.Spec}_{J}(\mathcal{A})$ partitioned $\text{P.Spec}(\mathcal{A})$ into
a disjoint union of strata. Hence,

$\text{P.Spec}(\mathcal{A})=\underset{J\in\mathcal{H}\text{-P.Spec}(\mathcal{A})}{\bigsqcup}\text{P.Spec}_{J}(A),$

where $\mathcal{H}\text{-P.Spec}(\mathcal{A})$ is the collection of all the
Poisson $\mathcal{H}$-invariant prime ideals of $\mathcal{A}.$ This partition
is called an $\mathcal{H}$-stratification of $\text{P.Spec}(\mathcal{A}).$ In
a similar manner, an $\mathcal{H}$-stratification of P.Prim$(\mathcal{A})$ is
obtained as follows:

$\text{P.Prim}(\mathcal{A})=\underset{J\in\mathcal{H}\text{-P.Prim}(\mathcal{A})}{\bigsqcup}\text{P.Prim}_{J}(\mathcal{A}),$

where
$\text{P.Prim}_{J}(\mathcal{A})=\text{P.Spec}_{J}(\mathcal{A})\cap\text{P.Prim}(\mathcal{A})$.

The iterated-Poisson algebra $\mathcal{A}$ has only finitely many Poisson
$\mathcal{H}$-primes by [9, Theorem 1.7], and so we deduce from [8, Theorem
4.3] the following result.

######  Proposition 3.2.

Let $P\in$ P.Spec${}_{J}(\mathcal{A}),$ P is a Poisson primitive ideal of
$\mathcal{A}$ if and only if $P$ is maximal in P.Spec${}_{J}(\mathcal{A}).$

###  3.3 Poisson deleting derivations algorithm of the semiclassical limit of
$U_{q}^{+}(G_{2})$

In [12], the authors studied a Poisson version, called the Poisson deleting
derivations algorithm (PDDA for short), of the well-known Cauchon’s deleting
derivations algorithm (see [3]). In this subsection, we study PDDA of
$\mathcal{A}=\mathbb{K}[X_{1},\ldots,X_{6}].$ From (6), we have that
$\mathcal{A}=\mathbb{K}[X_{1}][X_{2};\sigma_{2},\delta_{2}]_{p}\ldots[X_{6};\sigma_{6},\delta_{6}]_{P}.$
One can easily verify that $\mathcal{A}$ satisfies the conditions in the
hypothesis below.

######  Hypothesis 3.3.

  * (H1)

For all $1\leq j<i\leq 6,$ there exists $\mu_{ij}\in\mathbb{K}$ with
$\mu_{ji}:=-\mu_{ij},$ such that $\sigma_{i}(X_{j})=\mu_{ij}X_{j}.$

  * (H2)

The derivations $\delta_{i}$ are all locally nilpotent and
$\delta_{i}\alpha-\alpha\delta_{i}=\eta_{i}\delta_{i}$ for some non-zero
scalar $\eta_{i},$ for all $2\leq i\leq 6.$

As a result, the PDDA can be used to study the Poisson prime spectrum of
$\mathcal{A}$ (see [12, Hypothesis 1.7]). Let $1\leq i\leq 6$ and $2\leq j\leq
7.$ Using the relation

$X_{i,j}:=\begin{cases}X_{i,j+1}&\text{if $i\geq j$}\\\
\displaystyle\sum_{k=0}^{+\infty}\frac{1}{\eta_{j}^{k}k!}\delta_{j}^{k}(X_{i,j+1})X_{j,j+1}^{-k}&\text{if
$i<j$},\end{cases}$

(note that since $\delta_{j}$ is locally nilpotent, the summation is finite),
one can construct a family $(X_{1,j},\ldots,X_{6,j})$ of elements of the
division ring of fractions Fract$(\mathcal{A})$ of $\mathcal{A}$ as follows:

$\displaystyle X_{1,6}$ $\displaystyle=X_{1}-\frac{1}{2}X_{5}X_{6}^{-1}$
$\displaystyle X_{2,6}$
$\displaystyle=X_{2}+\frac{3}{2}X_{4}X_{6}^{-1}-3X_{3}X_{5}X_{6}^{-1}+X_{5}^{3}X_{6}^{-2}$
$\displaystyle X_{3,6}$ $\displaystyle=X_{3}-X_{5}^{2}X_{6}^{-1}$
$\displaystyle X_{4,6}$ $\displaystyle=X_{4}-\frac{2}{3}X_{5}^{3}X_{6}^{-1}$
$\displaystyle X_{1,5}$
$\displaystyle=X_{1,6}-X_{3,6}X_{5,6}^{-1}+\frac{3}{4}X_{4,6}X_{5,6}^{-2}$
$\displaystyle X_{2,5}$
$\displaystyle=X_{2,6}-3X_{3,6}^{2}X_{5,6}^{-1}+\frac{9}{2}X_{3,6}X_{4,6}X_{5,6}^{-2}-\frac{9}{4}X_{4,6}^{2}X_{5,6}^{-3}$
$\displaystyle X_{3,5}$ $\displaystyle=X_{3,6}-\frac{3}{2}X_{4,6}X_{5,6}^{-1}$
$\displaystyle X_{1,4}$
$\displaystyle=X_{1,5}-\frac{1}{3}X_{3,5}^{2}X_{4,5}^{-1}$ $\displaystyle
X_{2,4}$ $\displaystyle=X_{2,5}-\frac{2}{3}X_{3,5}^{3}X_{4,5}^{-1}$
$\displaystyle X_{1,3}$ $\displaystyle=X_{1,4}-\frac{1}{2}X_{2,4}X_{3,4}^{-1}$
$\displaystyle T_{1}$ $\displaystyle:=X_{1,2}=X_{1,3}$ $\displaystyle T_{2}$
$\displaystyle:=X_{2,2}=X_{2,3}=X_{2,4}$ $\displaystyle T_{3}$
$\displaystyle:=X_{3,2}=X_{3,3}=X_{3,4}=X_{3,5}$ $\displaystyle T_{4}$
$\displaystyle:=X_{4,2}=X_{4,3}=X_{4,4}=X_{4,5}=X_{4,6}$ $\displaystyle T_{5}$
$\displaystyle:=X_{5,2}=X_{5,3}=X_{5,4}=X_{5,5}=X_{5,6}=X_{5}$ $\displaystyle
T_{6}$ $\displaystyle:=X_{6,2}=X_{6,3}=X_{6,4}=X_{6,5}=X_{6,6}=X_{6}.$

For each $2\leq j\leq 7,$ the algebra $\mathcal{A}^{(j)}$ represents the
subalgebra of Fract$(\mathcal{A})$ generated by all the $X_{i,j}.$ That is,
$\mathcal{A}^{(j)}=\mathbb{K}[X_{1,j},\ldots,X_{6,j}].$ Note that
$\mathcal{A}^{(7)}=\mathcal{A}.$ It follows from [12, Prop. 1.11] that

$\displaystyle\mathcal{A}^{(j)}$
$\displaystyle\cong\mathbb{K}[X_{1}][X_{2};\sigma_{2},\delta_{2}]_{P}\cdots[X_{j-1};\sigma_{j-1},\delta_{j-1}]_{P}[X_{j};\tau_{j}]_{P}\cdots[X_{7};\tau_{7}]_{P},$

by an isomorphism that maps $X_{i,j}$ to $X_{i},$ and
$\tau_{j},\ldots,\tau_{7}$ denote the Poisson derivations defined by
$\tau_{l}(X_{i})=\mu_{li}X_{i}$ for all $1\leq i<l\leq 7.$ With a slight abuse
of notation, one can identify $\tau_{j},\ldots,\tau_{7}$ with
$\sigma_{j},\ldots,\sigma_{7}$ respectively.

######  Notation 3.4.

$\overline{\mathcal{A}}:=\mathcal{A}^{(2)}=\mathbb{K}[T_{1},\ldots,T_{6}].$

One can easily check that $\overline{\mathcal{A}}$ is a Poisson affine space
associated to the skew-symmetric matrix

$\displaystyle M:=$ $\displaystyle\begin{bmatrix}0&3&1&0&-1&-3\\\
-3&0&3&3&0&-3\\\ -1&-3&0&3&1&0\\\ 0&-3&-3&0&3&3\\\ 1&0&-1&-3&0&3\\\
3&3&0&-3&-3&0\end{bmatrix}.$

That is, $\overline{\mathcal{A}}$ satisfies the relation
$\\{T_{i},T_{j}\\}=\mu_{ij}T_{j}T_{i}$ for all $1\leq i,j\leq 6,$ where
$\mu_{ij}$ are the entries of $M.$

###  3.4 Canonical embedding

The set $\Sigma:=\\{X_{j,j+1}^{n}\mid n\in\mathbb{N}\\}=\\{X_{j,j}^{n}\mid
n\in\mathbb{N}\\}$ is a multiplicative system of regular elements of
$\mathcal{A}^{(j)}$ and $\mathcal{A}^{(j+1)}.$ Moreover,
$\mathcal{A}^{(j)}\Sigma_{j}^{-1}=\mathcal{A}^{(j+1)}\Sigma_{j}^{-1}$ [12,
Prop. 1.11]. One can use the PDDA to relate P.Spec$(\mathcal{A})$ to
P.Spec$(\overline{\mathcal{A}})$ by constructing an embedding $\psi_{j}:$
P.Spec$(\mathcal{A}^{(j+1)})\hookrightarrow$ P.Spec$(\mathcal{A}^{(j)})$
defined by

$\psi_{j}(P):=\begin{cases}P\Sigma_{j}^{-1}\cap\mathcal{A}^{(j)}&\text{if
$X_{j,j+1}=T_{j}\not\in P,$}\\\ g_{j}^{-1}(P/\langle
X_{j,j+1}\rangle)&\text{if $X_{j,j+1}\in P,$}\end{cases}$

for each $2\leq j\leq 6$ (see [12, Lemma 2.3]). The map $g_{j}$ is a
surjective homomorphism

$g_{j}:\mathcal{A}^{(j)}\rightarrow\mathcal{A}^{(j+1)}/\langle
X_{j,j+1}\rangle$

defined by

$g_{j}(X_{j,j}):=X_{j,j+1}+\langle X_{j,j+1}\rangle$

(further details can be found in [12, §2]). From [12, §2.1], there exists an
increasing homeomorphism from the topological space

$\\{P\in\text{P.Spec}(\mathcal{A}^{(j+1)})\mid X_{j,j+1}\not\in P\\}$

onto the topological space

$\\{Q\in\text{P.Spec}(\mathcal{A}^{(j)})\mid X_{j,j}\not\in P\\}$

whose inverse is also an increasing homeomorphism (the topology being the
Zariski topology). The map $\psi_{j}$ is injective but not necessarily
bijective. However, $\psi_{j}$ induces a bijection between
$\\{P\in\text{P.Spec}(\mathcal{A}^{(j+1)})\mid P\cap\Sigma_{j}=\emptyset\\}$
and $\\{Q\in\text{P.Spec}(\mathcal{A}^{(j)})\mid Q\cap\Sigma_{j}=\emptyset\\}$
[12, Lemma 2.1]. The so-called canonical embedding

$\psi:=\psi_{2}\circ\cdots\circ\psi_{6}=:\text{P.Spec}(\mathcal{A})\hookrightarrow\text{P.Spec}(\overline{\mathcal{A}})$

is obtained by composing all the $\psi_{j}.$ This canonical embedding $\psi$
helps to construct a partition of P.Spec$(\mathcal{A})$ into a disjoint union
of strata known as the canonical partition via the Cauchon diagrams.

The torus invariant Poisson prime spectrum has generally been described in
[12, §2.2] via Cauchon’s diagrams. Based on that, we can describe the set
$\mathcal{H}\text{-P.Spec}(\overline{\mathcal{A}})$ as follows. For any subset
$C$ of $\\{1,\dots,6\\}$, let $K_{C}$ denote the Poisson
$\mathcal{H}$-invariant prime ideal of $\overline{\mathcal{A}}$ generated by
the $T_{i}$ with $i\in C$. We deduce from [12, §2.2] that

$K_{C}=\langle T_{i}\mid i\in C\rangle,$

and

$\mathcal{H}\text{-P.Spec}(\overline{\mathcal{A}})=\\{K_{C}~{}|~{}C\subseteq\\{1,\dots,6\\}\\},$

so that

$\psi(\mathcal{H}\text{-P.Spec}(\mathcal{A}))\subseteq\\{K_{C}~{}|~{}C\subseteq\\{1,\dots,6\\}\\}.$

###  3.5 Poisson centre of $\mathcal{A}$

The monomials $\Omega_{1}:=T_{1}T_{3}T_{5}$ and $\Omega_{2}:=T_{2}T_{4}T_{6}$
are Poisson central elements of the Poisson affine space
$\overline{\mathcal{A}},$ since $\\{\Omega_{i},T_{j}\\}=0$ for all $i=1,2,$
and $1\leq j\leq 6.$ We now want to successively pull $\Omega_{1}$ and
$\Omega_{2}$ from $\overline{\mathcal{A}}$ into ${\mathcal{A}}$ using the data
of the PDDA of ${\mathcal{A}}$. Through a direct computation, one can confirm
that

$\displaystyle\Omega_{1}$ $\displaystyle:=T_{1}T_{3}T_{5}$
$\displaystyle=X_{1,3}X_{3,3}X_{5,3}$
$\displaystyle=X_{1,4}X_{3,4}X_{5,4}-\frac{1}{2}X_{2,4}X_{5,4}$
$\displaystyle=X_{1,5}X_{3,5}X_{5,5}-\frac{1}{2}X_{2,5}X_{5,5}$
$\displaystyle=X_{1,6}X_{3,6}X_{5,6}-\frac{3}{2}X_{1,6}X_{4,6}-\frac{1}{2}X_{2,6}X_{5,6}+\frac{1}{2}X_{3,6}^{2}$
$\displaystyle=X_{1}X_{3}X_{5}-\frac{3}{2}X_{1}X_{4}-\frac{1}{2}X_{2}X_{5}+\frac{1}{2}X_{3}^{2},$

and

$\displaystyle\Omega_{2}$ $\displaystyle:=T_{2}T_{4}T_{6}$
$\displaystyle=X_{2,4}X_{4,4}X_{6,4}$
$\displaystyle=X_{2,5}X_{4,5}X_{6,5}-\frac{2}{3}X_{3,5}^{3}X_{6,5}$
$\displaystyle=X_{2,6}X_{4,6}X_{6,6}-\frac{2}{3}X_{3,6}^{3}X_{6,6}$
$\displaystyle=X_{2}X_{4}X_{6}-\frac{2}{3}X_{3}^{3}X_{6}-\frac{2}{3}X_{2}X_{5}^{3}+2X_{3}^{2}X_{5}^{2}-3X_{3}X_{4}X_{5}+\frac{3}{2}X_{4}^{2}.$

Since
$\Omega_{i}\in\mathcal{A}^{(7)}\subseteq\mathcal{A}^{(j)}\subseteq\overline{\mathcal{A}},$
we have that $\Omega_{1}$ and $\Omega_{2}$ are also Poisson central elements
of $\mathcal{A}^{(j)}$ for each $2\leq j\leq 7.$ We proceed to show that the
Poisson centre of $\mathcal{A}$ is a polynomial ring in two variables:
$\Omega_{1}$ and $\Omega_{2}.$ That is,
$Z_{P}(\mathcal{A})=\mathbb{K}[\Omega_{1},\Omega_{2}].$ The following
discussions will lead us to the proof.

The set $S_{j}:=\\{\lambda T_{j}^{i_{j}}T_{j+1}^{i_{j+1}}\ldots
T_{6}^{i_{6}}\mid i_{j},\ldots,i_{6}\in\mathbb{N}\\}$ is a multiplicative
system of non-zero divisors of $\mathcal{A}^{(j)}$ for each $2\leq j\leq 6.$
One can therefore localize $\mathcal{A}^{(j)}$ at $S_{j}$ as follows:

$\mathfrak{R}_{j}:=\mathcal{A}^{(j)}S_{j}^{-1}.$

Note that the set $\Sigma_{j}:=\\{T_{j}^{n}\mid n\in\mathbb{N}\\}$ is also a
multiplicative set in both $\mathcal{A}^{(j)}$ and $\mathcal{A}^{(j+1)}$ for
each $2\leq j\leq 6.$ It follows from [12, Prop. 1.11] that

$\mathcal{A}^{(j)}\Sigma_{j}^{-1}=\mathcal{A}^{(j+1)}\Sigma_{j}^{-1}.$

One can easily verify that

$\displaystyle\mathfrak{R}_{j}=\mathfrak{R}_{j+1}\Sigma_{j}^{-1},\ \ \text{for
all}\ \ 2\leq j\leq 6.$

Moreover, the localization

$\mathfrak{R}_{1}:=\mathfrak{R}_{2}[T_{1}^{-1}]$

also holds in $\mathfrak{R}_{2},$ since $T_{1}$ generates a multiplicative
system in $\mathfrak{R}_{2}$. In fact, $\mathfrak{R}_{1}$ is the Poisson torus
associated to the Poisson affine space $\overline{\mathcal{A}}.$ As a result,
$\mathfrak{R}_{1}=\mathbb{\mathbb{K}}[T_{1}^{\pm 1},\ldots,T_{6}^{\pm 1}],$
where $\\{T_{i},T_{j}\\}={\mu_{ij}}T_{j}T_{i}$ for all $1\leq i,j\leq 6$. The
PDDA helps to construct the following embeddings:

$\displaystyle\mathcal{A}:=\mathfrak{R}_{7}$
$\displaystyle\subset\mathfrak{R}_{6}=\mathfrak{R}_{7}\Sigma_{6}^{-1}\subset\mathfrak{R}_{5}=\mathfrak{R}_{6}\Sigma_{5}^{-1}\subset\mathfrak{R}_{4}=\mathfrak{R}_{5}\Sigma_{4}^{-1}$
$\displaystyle\subset\mathfrak{R}_{3}=\mathfrak{R}_{4}\Sigma_{3}^{-1}\subset\mathfrak{R}_{2}=\mathfrak{R}_{3}\Sigma_{2}^{-1}\subset\mathfrak{R}_{1}.$
(7)

Note that the family $(X_{1,j}^{k_{1}}\ldots X_{6,j}^{k_{6}}),$ where
$k_{i}\in\mathbb{N}$ if $i<j$ and $k_{i}\in\mathbb{Z}$ otherwise is a PBW-
basis of $\mathfrak{R}_{j}$ for all $2\leq j\leq 7.$ In addition, the family
$(T_{1}^{k_{1}}\ldots T_{6}^{k_{6}})_{k_{1},\ldots,k_{6}\in\mathbb{Z}}$ is a
basis of $\mathfrak{R}_{1}.$

######  Lemma 3.5.

  * 1.

$Z_{P}(\mathfrak{R}_{1})=\mathbb{K}[\Omega_{1}^{\pm 1},\Omega_{2}^{\pm 1}].$

  * 2.

$Z_{P}(\overline{\mathcal{A}})=\mathbb{K}[\Omega_{1},\Omega_{2}].$

  * 3.

$Z_{P}(\mathfrak{R}_{3})=\mathbb{K}[\Omega_{1},\Omega_{2}].$

  * 4.

$Z_{P}(\mathcal{A})=\mathbb{K}[\Omega_{1},\Omega_{2}].$

###### Proof.

1\. Obviously, $\mathbb{K}[\Omega_{1}^{\pm 1},\Omega_{2}^{\pm 1}]\subseteq
Z_{P}(\mathfrak{R}_{1}).$ For the reverse inclusion, let $y\in
Z_{P}(\mathfrak{R}_{1}).$ Then, $y$ can be written in terms of the basis of
$\mathfrak{R}_{1}$ as
$y=\sum_{(i,\ldots,n)\in\mathbb{Z}^{6}}a_{(i,\ldots,n)}T_{1}^{i}T_{2}^{j}T_{3}^{k}T_{4}^{l}T_{5}^{m}T_{6}^{n}.$
One can verify that $\\{y,T_{1}\\}=(-3j-k+m+3n)yT_{1}.$ Since $y\in
Z_{P}(\mathfrak{R}_{1}),$ it follows that $-3j-k+m+3n=0.$ Following the same
pattern for $T_{2},T_{3},T_{4},T_{5}$ and $T_{6}$, one can confirm that
$3i-3k-3l+3n=0,\ $ $i+3j-3l-m=0,\ $ $3j+3k-3m-3n=0,\ $ $-i+k+3l-3n=0,$ and
$-3i-3j+3l+3m=0.$ Solving this system of six equations will reveal that
$i=k=m$ and $j=l=n.$ One can therefore write

$y=\displaystyle\sum_{(i,j)\in\mathbb{Z}^{2}}a_{(i,j)}T_{1}^{i}T_{2}^{j}T_{3}^{i}T_{4}^{j}T_{5}^{i}T_{6}^{j}=\sum_{(i,j)\in\mathbb{Z}^{2}}q^{\bullet}a_{(i,j)}T_{1}^{i}T_{3}^{i}T_{5}^{i}T_{2}^{j}T_{4}^{j}T_{6}^{j}=\sum_{(i,j)\in\mathbb{Z}^{2}}q^{\bullet}a_{(i,j)}\Omega_{1}^{i}\Omega_{2}^{j}\in\mathbb{K}[\Omega_{1}^{\pm
1},\Omega_{2}^{\pm 1}].$

2\. Similar argument as in (1) will prove the result.

3\. Observe that $\mathbb{K}[\Omega_{1},\Omega_{2}]\subseteq
Z_{P}(\overline{\mathcal{A}})\subseteq
Z_{P}(\mathfrak{R}_{3})=\mathbb{K}[\Omega_{1},\Omega_{2}].$ Hence,
$Z_{P}(\mathfrak{R}_{3})=\mathbb{K}[\Omega_{1},\Omega_{2}].$

4\. Since $\mathfrak{R}_{i}$ is a localization of $\mathfrak{R}_{i+1},$ it
follows that $Z_{P}(\mathfrak{R}_{i+1})\subseteq Z_{P}(\mathfrak{R}_{i}).$
From (3.5), we have that $Z_{P}(\mathcal{A})\subseteq
Z_{P}(\mathfrak{R}_{3}).$ Observe that
$\mathbb{K}[\Omega_{1},\Omega_{2}]\subseteq Z_{P}(\mathcal{A})\subseteq
Z_{P}(\mathfrak{R}_{3})=\mathbb{K}[\Omega_{1},\Omega_{2}].$ Hence,
$Z_{P}(\mathcal{A})=\mathbb{K}[\Omega_{1},\Omega_{2}].$ ∎

##  4 Maximal and primitive Poisson ideals, and simple quotients of the
semiclassical limit of $U_{q}^{+}(G_{2})$

This section studies the Poisson $\mathcal{H}$-invariant prime ideals of
$\mathcal{A}=\mathbb{K}[X_{1},\ldots,X_{6}]$ of height one. Obviously,
$\langle 0\rangle$ is the only height zero Poisson $\mathcal{H}$-invariant
prime ideal of $\mathcal{A}.$ Now, the Poisson $\mathcal{H}$-invariant prime
ideals of at most height one will enable us to study a family of primitive and
maximal Poisson ideals of $\mathcal{A},$ and consequently study a family of
simple quotients of $\mathcal{A}$.

###  4.1 Height one Poisson $\mathcal{H}$-invariant prime ideals of
$\mathcal{A}$

In this subsection, we study the Poisson $\mathcal{H}$-invariant prime ideals
of $\mathcal{A}=\mathbb{K}[X_{1},\ldots,X_{6}]$ with height one. We will begin
by showing that $\langle\Omega_{1}\rangle$ and $\langle\Omega_{2}\rangle$ are
Poisson prime ideals. Note that $\langle\Theta\rangle_{R}$ will denote an
ideal generated by the element $\Theta$ in any the ring $R.$ Where no doubt
arises, we will simply write $\langle\Theta\rangle.$

Recall from Subsection 3.4 that there exists a bijection between
$\\{P\in\text{P.Spec}(\mathcal{A}^{(j+1)})\mid P\cap\Sigma_{j}=\emptyset\\}$
and $\\{Q\in\text{P.Spec}(\mathcal{A}^{(j)})\mid
Q\cap\Sigma_{j}=\emptyset\\}$. Observe that $\langle T_{1}\rangle$ and
$\langle T_{2}\rangle$ are both elements of P.Spec$(\overline{\mathcal{A}}).$
The following result shows that $\langle T_{1}\rangle\in$ Im$(\psi)$, and that
$\langle\Omega_{1}\rangle$ is the Poisson prime ideal of $\mathcal{A}$ such
that $\psi(\langle\Omega_{1}\rangle)=\langle T_{1}\rangle.$

######  Lemma 4.1.

$\langle\Omega_{1}\rangle$ is the Poisson prime ideal of $\mathcal{A}$ such
that $\psi(\langle\Omega_{1}\rangle)=\langle T_{1}\rangle$.

###### Proof.

We will prove this result in several steps by showing that:

  * 1.

$\langle T_{1}\rangle\in$ P.Spec$(\mathcal{A}^{(3)}).$

  * 2.

$\langle X_{1,4}T_{3}-\frac{1}{2}T_{2}\rangle\in$ P.Spec$(\mathcal{A}^{(4)})$
[note that $\langle T_{1}\rangle[T_{3}^{-1}]\cap\mathcal{A}^{(4)}=\langle
X_{1,4}T_{3}-\frac{1}{2}T_{2}\rangle$].

  * 3.

$\langle X_{1,5}T_{3}-\frac{1}{2}X_{2,5}\rangle\in$
P.Spec$(\mathcal{A}^{(5)})$ [note that $\langle
X_{1,4}T_{3}-\frac{1}{2}T_{2}\rangle[T_{4}^{-1}]\cap\mathcal{A}^{(5)}=$
$\langle X_{1,5}T_{3}-\frac{1}{2}X_{2,5}\rangle$].

  * 4.

$\langle
X_{1,5}T_{3}-\frac{1}{2}X_{2,5}\rangle[T_{5}^{-1}]\cap\mathcal{A}^{(6)}=\langle\Omega_{1}\rangle_{\mathcal{A}^{(6)}},$
hence $\langle\Omega_{1}\rangle_{\mathcal{A}^{(6)}}\in$
P.Spec$(\mathcal{A}^{(6)}).$

  * 5.

$\langle\Omega_{1}\rangle_{\mathcal{A}^{(6)}}[T_{6}^{-1}]\cap\mathcal{A}=\langle\Omega_{1}\rangle_{\mathcal{A}}$,
hence $\langle\Omega_{1}\rangle_{\mathcal{A}}\in$ P.Spec$(\mathcal{A}).$

We proceed to prove the above claims.

1\. One can easily verify that ${\mathcal{A}^{(3)}}/\langle T_{1}\rangle$ is
isomorphic to a Poisson affine space of rank 5. Hence, $\langle T_{1}\rangle$
is a Poisson prime ideal in $\mathcal{A}^{(3)}.$

2\. Set $I:=\langle X_{1,4}T_{3}-\frac{1}{2}T_{2}\rangle.$ One can verify that
$\\{X_{i,4},I\\}\subseteq I$ for all $i=1,\ldots,6.$ Therefore, $I$ is a
Poisson ideal in $\mathcal{A}^{(4)}$. In addition, ${\mathcal{A}^{(4)}}/I$ is
isomorphic to a polynomial ring in five variables which is a domain, hence $I$
is a prime ideal. Since $I$ is both Poisson and prime ideal, it is a Poisson
prime ideal in $\mathcal{A}^{(4)}.$

3\. Similar to (2).

4\. Observe that
$\Omega_{1}^{\prime}:=X_{1,5}T_{3}-\frac{1}{2}X_{2,5}=\Omega_{1}T_{5}^{-1}$ in
$\mathcal{A}^{(6)}[T_{5}^{-1}].$ Since $\langle\Omega_{1}^{\prime}\rangle\in$
P.Spec$(\mathcal{A}^{(5)}),$ we want to show that
$\langle\Omega_{1}^{\prime}\rangle[T_{5}^{-1}]\cap\mathcal{A}^{(6)}=\langle\Omega_{1}\rangle_{\mathcal{A}^{(6)}}.$
Observe that
$\langle\Omega_{1}\rangle_{\mathcal{A}^{(6)}}\subseteq\langle\Omega_{1}^{\prime}\rangle[T_{5}^{-1}]\cap\mathcal{A}^{(6)}.$
We establish the reverse inclusion. Let
$y\in\langle\Omega_{1}^{\prime}\rangle[T_{5}^{-1}]\cap\mathcal{A}^{(6)}.$
Then, $y\in\langle\Omega_{1}^{\prime}\rangle[T_{5}^{-1}].$ There exists
$i\in\mathbb{N}$ such that $yT_{5}^{i}\in\langle\Omega_{1}^{\prime}\rangle.$
Hence, $yT_{5}^{i}=\Omega_{1}^{\prime}v,$ for some $v\in\mathcal{A}^{(5)}.$
Furthermore, since
$\mathcal{A}^{(5)}[T_{5}^{-1}]=\mathcal{A}^{(6)}[T_{5}^{-1}],$ there exists
$j\in\mathbb{N}$ such that $vT_{5}^{j}=v^{\prime}$ for some
$v^{\prime}\in\mathcal{A}^{(6)}.$ It follows from
$yT_{5}^{i}=\Omega_{1}^{\prime}v$ that
$yT_{5}^{i+j}=\Omega_{1}^{\prime}vT_{5}^{j}=\Omega_{1}^{\prime}v^{\prime}.$
Hence,
$yT_{5}^{\delta}=\Omega_{1}^{\prime}T_{5}v^{\prime}=\Omega_{1}v^{\prime},$
where $\delta=i+j+1$ (note that $\Omega_{1}^{\prime}T_{5}=\Omega_{1}$ in
$\mathcal{A}^{(6)}$). Let $S=\\{s\in\mathbb{N}\mid\ \exists
v^{\prime}\in\mathcal{A}^{(6)}:yT_{5}^{s}=\Omega_{1}v^{\prime}\\}.$ Since
$\delta\in S,$ we have that $S\neq\emptyset.$ Let $s=s_{0}$ be the minimum
element of $S$ such that $yT_{5}^{s_{0}}=\Omega_{1}v^{\prime}$ for some
$v^{\prime}\in\mathcal{A}^{(6)}.$ We want to show that $s_{0}=0.$ Suppose that
$s_{0}>0.$ Since $T_{5}$ is irreducible, $yT_{5}^{s_{0}}=\Omega_{1}v^{\prime}$
implies that $T_{5}$ is a factor of $\Omega_{1}$ or $v^{\prime}.$ Clearly,
$T_{5}$ is not a factor of $\Omega_{1},$ hence, it must be a factor of
$v^{\prime}.$ Now $\mathcal{A}^{(6)}$ can be viewed as a free
$\mathbb{K}[X_{1,6},X_{2,6},X_{3,6},T_{4},T_{6}]-$module with basis
$\left(T_{5}^{\xi}\right)_{\xi\in\mathbb{N}}.$ One can therefore write
$v^{\prime}=\sum_{\xi=1}^{n}\beta_{\xi}T_{5}^{\xi},$ where
$\beta_{\xi}\in\mathbb{K}[X_{1,6},X_{2,6},X_{3,6},T_{4},T_{6}].$ Returning to
$yT_{5}^{s_{0}}=\Omega_{1}v^{\prime},$ we have that
$yT_{5}^{s_{0}}=\Omega_{1}\sum_{\xi=1}^{n}\beta_{\xi}T_{5}^{\xi}.$ This
implies that $yT_{5}^{s_{0}-1}=\Omega_{1}v^{\prime\prime},$ where
$v^{\prime\prime}=\sum_{\xi=1}^{n}\beta_{\xi}T_{5}^{\xi-1}\in\mathcal{A}^{(6)}.$
Consequently, $s_{0}-1\in S,$ a contradiction! Therefore, $s_{0}=0$ and
$y=\Omega_{1}v^{\prime}\in\langle\Omega_{1}\rangle_{\mathcal{A}^{(6)}}.$
Hence,
$\langle\Omega_{1}^{\prime}\rangle[T_{5}^{-1}]\cap\mathcal{A}^{(6)}\subseteq\langle\Omega_{1}\rangle_{\mathcal{A}^{(6)}}$
as expected.

5\. The proof is similar to (4). ∎

Following similar procedures, one can also prove that $\langle
T_{2}\rangle\in$ Im$(\psi),$ and that $\langle\Omega_{2}\rangle$ is the
Poisson prime ideal of $\mathcal{A}$ such that
$\psi(\langle\Omega_{2}\rangle)=\langle T_{2}\rangle$ (the interested reader
can check out the details of the proof in [18]). We state only the result in
the following lemma.

######  Lemma 4.2.

$\langle\Omega_{2}\rangle$ is the Poisson prime ideal of $\mathcal{A}$ such
that $\psi(\langle\Omega_{2}\rangle)=\langle T_{2}\rangle$.

Observe that $\langle T_{1},T_{2}\rangle$ and $\langle T_{2},T_{3}\rangle$ are
Poisson prime ideals of $\mathcal{\overline{A}}.$ In the next lemma, we will
show that $\langle T_{1},T_{2}\rangle,\langle
T_{2},T_{3}\rangle\in\psi(\text{P.Spec}(\mathcal{A}))$.

######  Lemma 4.3.

$\langle T_{1},T_{2}\rangle,\langle
T_{2},T_{3}\rangle\in\psi(\text{P.Spec}(\mathcal{A})).$

###### Proof.

Let $J^{(j)}\in$ P.Spec$(\mathcal{A}^{(j)})$ for all $2\leq j\leq 6.$ We
already know that $J^{(j+1)}=J^{(j)}[T_{j}^{-1}]\cap\mathcal{A}^{(j+1)}\in$
P.Spec$(\mathcal{A}^{(1+j)})$ provided $T_{j}\not\in J^{(j)}.$ Note that
$J^{(7)}=J$ and $\mathcal{A}^{(7)}=\mathcal{A}.$

We begin by showing that $\langle
T_{1},T_{2}\rangle\in\psi(\text{P.Spec}(\mathcal{A})).$
Set $J_{1,2}^{(3)}:=\langle T_{1},T_{2}\rangle\in$
P.Spec$(\mathcal{A}^{(3)})$. Observe that $T_{3}\not\in J_{1,2}^{(3)}.$
Therefore, $J_{1,2}^{(3)}[T_{3}^{-1}]\cap\mathcal{A}^{(4)}=J_{1,2}^{(4)}.$
Suppose that $T_{4}\in J_{1,2}^{(4)}.$ Then, since
$J_{1,2}^{(3)}[T_{3}^{-1}]=J_{1,2}^{(4)}[T_{3}^{-1}],$ we have that $T_{4}\in
J_{1,2}^{(3)}[T_{3}^{-1}]\cap\mathcal{A}^{(4)}=J_{1,2}^{(4)}[T_{3}^{-1}]\cap\mathcal{A}^{(3)}=J_{1,2}^{(3)},$
a contradiction! Therefore, $T_{4}\not\in J_{1,2}^{(4)}.$ Hence,
$J_{1,2}^{(4)}[T_{4}^{-1}]\cap\mathcal{A}^{(5)}=J_{1,2}^{(5)}.$ Suppose that
$T_{5}\in J_{1,2}^{(5)}.$ Then, $T_{5}\in
J_{1,2}^{(4)}[T_{4}^{-1}]\cap\mathcal{A}^{(5)}=J_{1,2}^{(5)}[T_{4}^{-1}]\cap\mathcal{A}^{(4)}=J_{1,2}^{(4)},$
a contradiction! Therefore, $T_{5}\not\in J_{1,2}^{(5)}.$ Hence,
$J_{1,2}^{(5)}[T_{5}^{-1}]\cap\mathcal{A}^{(6)}=J_{1,2}^{(6)}.$ Similarly, one
can show that $T_{6}\not\in J_{1,2}^{(6)}.$ Hence,
$J_{1,2}^{(6)}[T_{6}^{-1}]\cap\mathcal{A}=J_{1,2}.$ Therefore, there exists
$J_{1,2}\in\text{P.Spec}(\mathcal{A})$ such that $\psi(J_{1,2})=\langle
T_{1},T_{2}\rangle.$ Similar arguments will also show that there exists
$J_{2,3}\in\text{P.Spec}(\mathcal{A})$ such that $\psi(J_{2,3})=\langle
T_{2},T_{3}\rangle.$ ∎

Recall that $\Omega_{1}=T_{1}T_{3}T_{5}$ and $\Omega_{2}=T_{2}T_{4}T_{6}$ in
$\overline{\mathcal{A}}.$ Observe that $\Omega_{1},\Omega_{2}$ are both
elements of $\langle T_{1},T_{2}\rangle$ and $\langle T_{2},T_{3}\rangle.$
From Lemma 4.3, we know that $J_{1,2}$ and $J_{2,3}$ are the elements of
P.Spec$(\mathcal{A})$ such that $\psi(J_{1,2})=\langle T_{1},T_{2}\rangle$ and
$\psi(J_{2,3})=\langle T_{2},T_{3}\rangle.$ In the next lemma, we show that
$J_{1,2}$ and $J_{2,3}$ contain $\Omega_{1}$ and $\Omega_{2}.$

######  Lemma 4.4.

$\Omega_{1}$ and $\Omega_{2}$ are elements of $J_{1,2}$ and $J_{2,3}.$

###### Proof.

Recall that $\Omega_{1}$ and $\Omega_{2}$ are central elements of
$\mathcal{A}^{(j)}$ for all $2\leq j\leq 7.$ Given the set-up in the proof of
Lemma 4.3, we know that $\Omega_{1},\Omega_{2}\in$ $J_{1,2}^{(3)}=\langle
T_{1},T_{2}\rangle.$ By an induction on $j$ (where $j=3,4,5,6$), one can
confirm that $\Omega_{1},\Omega_{2}\in
J_{1,2}^{(j)}[T_{j}^{-1}]\cap\mathcal{A}^{(j+1)}=J_{1,2}^{(j+1)}$. Therefore;
$\Omega_{1},\Omega_{2}\in J_{1,2}.$ Similarly, one can easily show that
$\Omega_{1},\Omega_{2}\in J_{2,3}^{(4)}=\langle T_{2},T_{3}\rangle.$ ∎

We now want to find the height one Poisson $\mathcal{H}$-invariant prime
ideals of $\mathcal{A},$ and show that the height $2$ Poisson
$\mathcal{H}$-invariant prime ideals of $\mathcal{A}$ contain those of height
one.

Let the torus $\mathcal{H}:=(\mathbb{K}^{*})^{2}$ acts by Poisson
automorphisms on $\mathcal{A}$ via:

$\displaystyle h\cdot X_{1}$ $\displaystyle=\alpha_{1}X_{1}$ $\displaystyle
h\cdot X_{2}$ $\displaystyle=\alpha_{1}^{3}\alpha_{6}X_{2}$ $\displaystyle
h\cdot X_{3}$ $\displaystyle=\alpha_{1}^{2}\alpha_{6}X_{3}$ $\displaystyle
h\cdot X_{4}$ $\displaystyle=\alpha_{1}^{3}\alpha_{6}^{2}X_{4}$ $\displaystyle
h\cdot X_{5}$ $\displaystyle=\alpha_{1}\alpha_{6}X_{5}$ $\displaystyle h\cdot
X_{6}$ $\displaystyle=\alpha_{6}X_{6},$

for all $h:=(\alpha_{1},\alpha_{6})\in\mathcal{H}.$ This $\mathcal{H}$-action
is rational. Furthermore, $\Omega_{1}$ and $\Omega_{2}$ are Poisson
$\mathcal{H}$-eigenvectors. Hence, $\langle\Omega_{1}\rangle$ and
$\langle\Omega_{2}\rangle$ are Poisson $\mathcal{H}$-invariant prime ideals.

Set $\theta_{1}:=$ id${}_{\text{P.Spec}(\overline{\mathcal{A}})}.$ For all
$2\leq j\leq 6,$ define $\theta_{j}:=\theta_{j-1}\circ\psi_{j}.$ Then,
$\theta_{j}:$ P.Spec$(\mathcal{A}^{(j+1)})\hookrightarrow$
P.Spec$(\overline{\mathcal{A}}).$ The map $\theta_{j}$ is injective [12,
§2.3]. Let $\langle X_{j,j+1}\rangle_{P}$ denote the smallest Poisson ideal in
$\mathcal{A}^{(1+j)}$ containing $X_{j,j+1}.$ It follows from [12, Lemma 2.2]
that there is a surjective Poisson algebra homomorphism
$g_{j}:\mathcal{A}^{(j)}\longrightarrow\mathcal{A}^{(j+1)}/\langle
X_{j,j+1}\rangle_{P}$ given by
$g_{j}(X_{i,j})=\overline{X_{i,j+1}}:=X_{i,j+1}+\langle X_{j,j+1}\rangle_{P}$
for all $1\leq i\leq 6.$ Denote the kernel of $g_{j}$ by ker$(g_{j})$ and the
image of $\psi$ by Im$(\psi)$. We have the following lemma.

######  Lemma 4.5.

[12, Proposition 2.8] Let $P\in$ P.Spec$(\overline{\mathcal{A}}).$ The
following are equivalent:

  * •

$P\in$ Im$(\psi),$

  * •

for all $2\leq j\leq 6,$ we have that $P\in$ Im$(\theta_{j-1})$ and either
$X_{j,j}=X_{j,j+1}\not\in\theta_{j-1}^{-1}(P)$ or
ker$(g_{j})\subseteq\theta_{j-1}^{-1}(P).$

Note that the map $\psi$ induces a canonical embedding from
$\mathcal{H}$-P.Spec$(\mathcal{A})$ to
$\mathcal{H}$-P.Spec$(\overline{\mathcal{A}}).$ Observe that $\\{\langle
T_{i}\rangle\mid i=1,\ldots,6\\}$ is the set of only height one Poisson
$\mathcal{H}$-invariant prime ideals we have in $\overline{\mathcal{A}}.$
Since $\psi$ preserves the height of a prime ideal, if $\psi^{-1}(\langle
T_{i}\rangle)\in$ P.Spec$(\mathcal{A}),$ then it is a height one Poisson
$\mathcal{H}$-invariant prime ideal in $\mathcal{A}$ for all $1\leq i\leq 6.$
For example, we already know that $\psi^{-1}(\langle
T_{1}\rangle)=\langle\Omega_{1}\rangle$ and $\psi^{-1}(\langle
T_{2}\rangle)=\langle\Omega_{2}\rangle$. Therefore, $\langle\Omega_{1}\rangle$
and $\langle\Omega_{2}\rangle$ are height one Poisson $\mathcal{H}$-invariant
prime ideals in $\mathcal{A}.$ We will show in the next lemma that
$\langle\Omega_{1}\rangle$ and $\langle\Omega_{2}\rangle$ are the only height
one Poisson $\mathcal{H}$-invariant prime ideals in $\mathcal{A}.$

######  Lemma 4.6.

For each $j\in\\{3,4,5,6\\},$ $\langle
T_{j}\rangle\not\in\psi(\text{P.Spec}(\mathcal{A})).$

###### Proof.

Suppose that there exists $j\in\\{3,4,5,6\\}$ such that $\langle
T_{j}\rangle\in\psi(\text{P.Spec}(\mathcal{A})).$ Then, there exists $P\in$
P.Spec$(\mathcal{A})$ such that $\psi(P)=\langle
T_{j}\rangle_{\mathcal{A}^{(j)}},$ where
$\psi:=\psi_{j}\circ\cdots\circ\psi_{6}.$ Set $P^{(j)}:=\langle
T_{j}\rangle_{\mathcal{A}^{(j)}}.$ Since $T_{j}\in P^{(j)},$ it follows from
Lemma 4.5 that ker$(g_{j})\subseteq P^{(j)}.$ The rest follows in cases.

  * •

When $j=3,$ then ker$(g_{3})\subseteq P^{(3)}=\langle
T_{3}\rangle_{\mathcal{A}^{(3)}}.$ One can easily deduce from the Poisson
bracket of $\mathcal{A}$ that $\\{X_{3,4},X_{1,4}\\}=-X_{1,4}X_{3,4}-X_{2,4}.$
This implies that $X_{2,4}=-\\{X_{3,4},X_{1,4}\\}-X_{1,4}X_{3,4}\in\langle
X_{3,4}\rangle_{P}=\langle T_{3}\rangle_{P}.$ It follows that
$g_{3}(X_{2,3})=X_{2,4}+\langle T_{3}\rangle_{P}=\overline{0}.$ Hence,
$X_{2,3}=T_{2}\in$ ker$(g_{3})\subseteq\langle
T_{3}\rangle_{\mathcal{A}^{(3)}},$ a contradiction! Consequently, $\langle
T_{3}\rangle\not\in\psi(\text{P.Spec}(\mathcal{A})).$

  * •

When $j=4,$ then ker$(g_{4})\subseteq P^{(4)}=\langle
T_{4}\rangle_{\mathcal{A}^{(4)}}.$ From the Poisson bracket of $\mathcal{A},$
we have that $X_{3,5}^{2}=-\frac{1}{2}\\{X_{4,5},X_{1,5}\\}\in\langle
X_{4,5}\rangle_{P}=\langle T_{4}\rangle_{P}.$ Since $g_{4}$ is a homomorphism,
it follows that $g_{4}(X_{3,4}^{2})=(g_{4}(X_{3,4}))^{2}=X_{3,5}^{2}+\langle
T_{4}\rangle_{P}=\overline{0}.$ Therefore, $X_{3,4}^{2}=T_{3}^{2}\in$
ker$(g_{4})\subseteq\langle T_{4}\rangle_{\mathcal{A}^{(4)}},$ a
contradiction! Hence, $\langle
T_{4}\rangle\not\in\psi(\text{P.Spec}(\mathcal{A})).$

  * •

When $j=5$, then ker$(g_{5})\subseteq P^{(5)}=\langle
T_{5}\rangle_{\mathcal{A}^{(5)}}.$ One can deduce from the Poisson bracket of
$\mathcal{A}$ that
$X_{4,6}=-\frac{1}{3}X_{3,6}X_{5,6}-\frac{1}{3}\\{X_{5,6},X_{3,6}\\}\in\langle
X_{5,6}\rangle_{P}=\langle T_{5}\rangle_{P}.$ It follows that
$g_{5}(X_{4,5})=X_{4,6}+\langle T_{5}\rangle_{P}=\overline{0}.$ Hence,
$X_{4,6}=T_{4}\in$ ker$(g_{5})\subseteq\langle
T_{5}\rangle_{\mathcal{A}^{(5)}},$ a contradiction! Therefore, $\langle
T_{5}\rangle\not\in\psi(\text{P.Spec}(\mathcal{A})).$

  * •

Finally, when $j=6,$ then ker$(g_{6})\subseteq P^{(6)}=\langle
T_{6}\rangle_{\mathcal{A}^{(6)}}.$ Similarly, one can verify that
$X_{5}=X_{1}X_{6}-\frac{1}{3}\\{X_{6},X_{1}\\}\in\langle
X_{6}\rangle_{P}=\langle T_{6}\rangle_{P}.$ Now, $g_{6}(X_{5,6})=X_{5}+\langle
T_{6}\rangle_{P}=\overline{0}.$ Therefore, $X_{5,6}=T_{5}\in$
ker$(g_{6})\subseteq\langle T_{6}\rangle_{\mathcal{A}^{(6)}},$ a
contradiction! Hence, $\langle
T_{6}\rangle\not\in\psi(\text{P.Spec}(\mathcal{A})).$

In conclusion, $\langle T_{j}\rangle\not\in\psi(\text{P.Spec}(\mathcal{A}))$
for all $j\in\\{3,4,5,6\\}$ as expected. ∎

The corollary below is deduced from the proof of Lemma 4.6.

######  Corollary 4.7.

Let $P\in\psi($P.Spec$(\mathcal{A})).$ If $T_{j}\in P,$ then
$T_{j-1},\ldots,T_{2}\in P$ for all $3\leq j\leq 6.$

Recall from Lemma 4.3 that there exist $J_{1,2}$ and $J_{2,3}$ of
$\text{P.Spec}(\mathcal{A})$ such that $\psi(J_{1,2})=\langle
T_{1},T_{2}\rangle$ and $\psi(J_{2,3})=\langle T_{2},T_{3}\rangle.$ As a
result of Corollary 4.7, the Poisson ideals $\langle T_{1},T_{2}\rangle$ and
$\langle T_{2},T_{3}\rangle$ are the only height two Poisson
$\mathcal{H}$-invariant prime ideals of $\psi(\text{P.Spec}(\mathcal{A})).$
Since $\psi$ preserves Poisson $\mathcal{H}$-invariant prime ideals and the
height of a Poisson prime ideal, this implies that $J_{1,2}$ and $J_{2,3}$ are
the only height two Poisson $\mathcal{H}$-invariant prime ideals of
$\mathcal{A}.$ It follows from Lemma 4.4 that the height two Poisson
$\mathcal{H}$-invariant prime ideals of $\mathcal{A}$ contain $\Omega_{1}$ and
$\Omega_{2}.$

######  Remark 4.8.

Since the height two Poisson $\mathcal{H}$-invariant prime ideals of
$\mathcal{A}$ contain $\Omega_{1}$ and $\Omega_{2},$ every non-zero Poisson
$\mathcal{H}$-invariant prime ideal of $\mathcal{A}$ will contain either
$\Omega_{1}$ or $\Omega_{2}.$ Note that those Poisson $\mathcal{H}$-invariant
primes of at least height 2 will contain both $\Omega_{1}$ and $\Omega_{2}.$

###  4.2 Some maximal and primitive Poisson ideals of $\mathcal{A}$

We begin this subsection by finding the $\mathcal{H}$-strata corresponding to
$\langle 0\rangle,$ $\langle\Omega_{1}\rangle$ and $\langle\Omega_{2}\rangle$.
Note that all Poisson ideals in $\mathcal{A}$ shall be written as
$\langle\Theta\rangle,$ where $\Theta\subset\mathcal{A}.$ However, if we want
to refer to a Poisson ideal in any other Poisson algebra, say $R$, then that
Poisson ideal shall be written as $\langle\Theta\rangle_{R},$ where in this
case, $\Theta\subset R.$

######  Proposition 4.9.

Assume that $\mathbb{K}$ is algebraically closed. Let $\mathcal{P}$ be the set
of those monic irreducible polynomials
$P(\Omega_{1},\Omega_{2})\in\mathbb{K}[\Omega_{1},\Omega_{2}]$ with
$P(\Omega_{1},\Omega_{2})\neq\Omega_{1}$ and
$P(\Omega_{1},\Omega_{2})\neq\Omega_{2}$. Then P.Spec${}_{\langle
0\rangle}(\mathcal{A})=\\{\langle 0\rangle\\}\cup\\{\langle
P(\Omega_{1},\Omega_{2})\rangle\mid
P(\Omega_{1},\Omega_{2})\in\mathcal{P}\\}\cup\\{\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle\mid\alpha,\beta\in\mathbb{K}^{*}\\}.$

###### Proof.

We claim that P.Spec${}_{\langle
0\rangle}(\mathcal{A})\subseteq\\{Q\in\text{P.Spec}(\mathcal{A})\mid\Omega_{1},\Omega_{2}\not\in
Q\\}.$ To establish this claim, let us assume that this is not the case.
Suppose that there exists $Q\in$ P.Spec${}_{\langle 0\rangle}(\mathcal{A})$
such that $\Omega_{1}$ or $\Omega_{2}$ belongs to $Q;$ then the product
$\Omega_{1}\Omega_{2}$ which is Poisson $\mathcal{H}$-eigenvector belongs to
$Q.$ Consequently, $\Omega_{1}\Omega_{2}\in\bigcap_{h\in\mathcal{H}}h\cdot
Q=\langle 0\rangle,$ a contradiction! Hence, P.Spec${}_{\langle
0\rangle}(\mathcal{A})\subseteq\\{Q\in\text{P.Spec}(\mathcal{A})\mid\Omega_{1},\Omega_{2}\not\in
Q\\}.$ Conversely, suppose that $Q\in$ P.Spec$(\mathcal{A})$ such that
$\Omega_{1},\Omega_{2}\not\in Q,$ then $\bigcap_{h\in\mathcal{H}}h\cdot Q$ is
a Poisson $\mathcal{H}$-invariant prime ideal of $\mathcal{A},$ which contains
neither $\Omega_{1}$ nor $\Omega_{2}.$ The only possibility for
$\bigcap_{h\in\mathcal{H}}h\cdot Q$ is the zero ideal since every non-zero
Poisson $\mathcal{H}$-invariant prime ideal of $\mathcal{A}$ contains
$\Omega_{1}$ or $\Omega_{2}$ (Remark 4.8). Thus,
$\bigcap_{h\in\mathcal{H}}h\cdot Q=\langle 0\rangle.$ Hence, $Q\in$
P.Spec${}_{\langle 0\rangle}(\mathcal{A}).$ Therefore,
$\\{Q\in\text{P.Spec}(\mathcal{A})\mid\Omega_{1},\Omega_{2}\not\in
Q\\}\subseteq$ P.Spec${}_{\langle 0\rangle}(\mathcal{A})$, and so we have
proved that P.Spec${}_{\langle
0\rangle}(\mathcal{A})=\\{Q\in\text{P.Spec}(\mathcal{A})\mid\Omega_{1},\Omega_{2}\not\in
Q\\}.$

Since $\Omega_{1},\Omega_{2}\in Z_{P}(\mathcal{A}),$ we have that the set
$\\{\Omega_{1}^{i}\Omega_{2}^{j}\mid i,j\in\mathbb{N}\\}$ is a multiplicative
set in $\mathcal{A}.$ We can now localize $\mathcal{A}$ as
$R:=\mathcal{A}[\Omega_{1}^{-1},\Omega_{2}^{-1}].$ Given $Q\in$
P.Spec${}_{\langle 0\rangle}(\mathcal{A}),$ the map $\phi:Q\mapsto
Q[\Omega_{1}^{-1},\Omega_{2}^{-1}]$ is an increasing bijection from
P.Spec${}_{\langle 0\rangle}(\mathcal{A})$ onto P.Spec($R$).

Let us verify that $R$ is Poisson $\mathcal{H}$-simple before we describe
P.Spec($R$). Now, $\phi$ still induces a bijection between the set of those
Poisson $\mathcal{H}$-invariant prime ideals of P.Spec${}_{\langle
0\rangle}(\mathcal{A})$ and the set of Poisson $\mathcal{H}$-invariant prime
ideals of P.Spec$(R)$. It is already known that the set of Poisson
$\mathcal{H}$-invariant prime ideals of $\mathcal{A}$ that contains neither
$\Omega_{1}$ nor $\Omega_{2}$ consists only of the zero ideal $\\{\langle
0\rangle\\}$ (Remark 4.8). This implies that $\langle 0\rangle_{R}$ is the
only Poisson $\mathcal{H}$-invariant prime ideal of $R.$ Every Poisson
$\mathcal{H}$-invariant proper ideal of $R$ is contained in a Poisson
$\mathcal{H}$-invariant prime ideal of $R$. Therefore, $\langle 0\rangle_{R}$
is the only unique Poisson $\mathcal{H}$-invariant proper ideal of $R.$ This
confirms that $R$ is Poisson $\mathcal{H}$-simple. It follows from [8, Theorem
4.2] that the extension and contraction maps provide mutually inverse
bijections between P.Spec($R$) and Spec($Z_{P}(R)$). From Lemma 3.5,
$Z_{P}(\mathcal{A})=\mathbb{K}[\Omega_{1},\Omega_{2}],$ and so
$Z_{P}(R)=\mathbb{K}[\Omega_{1}^{\pm 1},\Omega_{2}^{\pm 1}].$ Since
$\mathbb{K}$ is algebraically closed, we have that Spec$(Z_{P}(R))=\\{\langle
0\rangle_{Z_{P}(R)}\\}\cup\\{\langle
P(\Omega_{1},\Omega_{2})\rangle_{Z_{P}(R)}\mid
P(\Omega_{1},\Omega_{2})\in\mathcal{P}\\}\cup\\{\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle_{Z_{P}(R)}\mid\alpha,\beta\in\mathbb{K}^{*}\\}.$
One can now recover P.Spec($R$) from Spec($Z_{P}(R)$) as follows:
P.Spec($R$)$=\\{\langle 0\rangle_{R}\\}\cup\\{\langle
P(\Omega_{1},\Omega_{2})\rangle_{R}\mid
P(\Omega_{1},\Omega_{2})\in\mathcal{P}\\}\cup\\{\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle_{R}\mid\alpha,\beta\in\mathbb{K}^{*}\\}.$
It follows that P.Spec${}_{\langle 0\rangle}(\mathcal{A})=\\{\langle
0\rangle_{R}\cap\mathcal{A}\\}\cup\\{\langle
P(\Omega_{1},\Omega_{2})\rangle_{R}\cap\mathcal{A}\mid
P(\Omega_{1},\Omega_{2})\in\mathcal{P}\\}\cup\\{\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle_{R}\cap\mathcal{A}\mid\alpha,\beta\in\mathbb{K}^{*}\\}.$

Undoubtedly, $\langle 0\rangle_{R}\cap\mathcal{A}=\langle 0\rangle.$ We now
have to show that $\langle
P(\Omega_{1},\Omega_{2})\rangle_{R}\cap\mathcal{A}=\langle
P(\Omega_{1},\Omega_{2})\rangle,$ $\forall
P(\Omega_{1},\Omega_{2})\in\mathcal{P},$ and
$\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle_{R}\cap\mathcal{A}=\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle,$
$\forall\alpha,\beta\in\mathbb{K}^{*},$ in order to complete the proof.

Fix $P(\Omega_{1},\Omega_{2})\in\mathcal{P}.$ Clearly, $\langle
P(\Omega_{1},\Omega_{2})\rangle\subseteq\langle
P(\Omega_{1},\Omega_{2})\rangle_{R}\cap\mathcal{A}.$ To show the reverse
inclusion, let $y\in\langle
P(\Omega_{1},\Omega_{2})\rangle_{R}\cap\mathcal{A}$. Since $y\in\langle
P(\Omega_{1},\Omega_{2})\rangle_{R},$ this implies that
$y=dP(\Omega_{1},\Omega_{2})$ for some $d\in R.$ Also, $d\in R$ implies that
there exists $i,j\in\mathbb{N}$ such that $d=a\Omega_{1}^{-i}\Omega_{2}^{-j},$
where $a\in\mathcal{A}.$ Therefore,
$y=a\Omega_{1}^{-i}\Omega_{2}^{-j}P(\Omega_{1},\Omega_{2}),$ which implies
that $y\Omega_{1}^{i}\Omega_{2}^{j}=aP(\Omega_{1},\Omega_{2}).$ Choose
$(i,j)\in\mathbb{N}^{2}$ minimal (in the lexicographic order on
$\mathbb{N}^{2}$) such that the equality holds. Without loss of generality,
let us suppose that $i>0,$ then
$aP(\Omega_{1},\Omega_{2})\in\langle\Omega_{1}\rangle.$ Since
$\langle\Omega_{1}\rangle$ is a prime ideal, this implies that
$a\in\langle\Omega_{1}\rangle$ or
$P(\Omega_{1},\Omega_{2})\in\langle\Omega_{1}\rangle.$ Since
$P(\Omega_{1},\Omega_{2})\in\mathcal{P},$ this implies that
$P(\Omega_{1},\Omega_{2})\not\in\langle\Omega_{1}\rangle.$ Hence,
$a\in\langle\Omega_{1}\rangle,$ which implies that $a=t\Omega_{1}$ for some
$t\in\mathcal{A}.$ Returning to
$y\Omega_{1}^{i}\Omega_{2}^{j}=aP(\Omega_{1},\Omega_{2}),$ we have that
$y\Omega_{1}^{i}\Omega_{2}^{j}=t\Omega_{1}P(\Omega_{1},\Omega_{2}).$ Finally,
$y\Omega_{1}^{i-1}\Omega_{2}^{j}=tP(\Omega_{1},\Omega_{2}).$ This clearly
contradicts the minimality of $(i,j),$ hence $(i,j)=(0,0).$ As a result,
$y=aP(\Omega_{1},\Omega_{2})=\langle P(\Omega_{1},\Omega_{2})\rangle.$
Consequently, $\langle
P(\Omega_{1},\Omega_{2})\rangle_{R}\cap\mathcal{A}=\langle
P(\Omega_{1},\Omega_{2})\rangle$ for all
$P(\Omega_{1},\Omega_{2})\in\mathcal{P}$ as desired.

Similarly, we show that
$\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle_{R}\cap\mathcal{A}=\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle;$
$\forall\alpha,\beta\in\mathbb{K}^{*}.$ Fix $\alpha,\beta\in\mathbb{K}^{*}.$
Obviously,
$\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle\subseteq\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle_{R}\cap\mathcal{A}.$
We establish the reverse inclusion. Let
$y\in\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle_{R}\cap\mathcal{A}.$
Since $y\in\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle_{R},$ we have that
$y=m_{0}(\Omega_{1}-\alpha)+n_{0}(\Omega_{2}-\beta),$ where $m_{0},n_{0}\in
R.$ Also, $m_{0},n_{0}\in R$ implies that there exists $i,j\in\mathbb{N}$ such
that $m_{0}=m\Omega_{1}^{-i}\Omega_{2}^{-j}$ and
$n_{0}=n\Omega_{1}^{-i}\Omega_{2}^{-j}$ for some $m,n\in\mathcal{A}.$
Therefore,
$y=m\Omega_{1}^{-i}\Omega_{2}^{-j}(\Omega_{1}-\alpha)+n\Omega_{1}^{-i}\Omega_{2}^{-j}(\Omega_{2}-\beta)$,
which implies that
$y\Omega_{1}^{i}\Omega_{2}^{j}=m(\Omega_{1}-\alpha)+n(\Omega_{2}-\beta).$
Choose $(i,j)\in\mathbb{N}^{2}$ minimal (in the lexicographic order on
$\mathbb{N}^{2}$) such that the equality holds. Without loss of generality,
suppose that $i>0$ and let
$f:\mathcal{A}\longrightarrow{\mathcal{A}}/{\langle\Omega_{2}-\beta\rangle}$
be a canonical surjection. We have that
$f(y)f(\Omega_{1})^{i}f(\Omega_{2})^{j}=f(m)f(\Omega_{1}-\alpha).$ It follows
that $f(m)f(\Omega_{1}-\alpha)\in\langle f(\Omega_{1})\rangle.$ Note that
$f(\Omega_{1}-\alpha)\not\in\langle f(\Omega_{1})\rangle.$ This implies that
$f(m)\in\langle f(\Omega_{1})\rangle.$ Therefore,
$\exists\lambda\in\mathcal{A}$ such that $f(m)=f(\lambda)f(\Omega_{1}).$
Consequently,
$f(y)f(\Omega_{1})^{i}f(\Omega_{2})^{j}=f(\lambda)f(\Omega_{1})f(\Omega_{1}-\alpha).$
Since $f(\Omega_{1})\neq 0,$ this implies that
$f(y)f(\Omega_{1})^{i-1}f(\Omega_{2})^{j}=f(\lambda)f(\Omega_{1}-\alpha).$
Therefore,
$y\Omega_{1}^{i-1}\Omega_{2}^{j}=\lambda(\Omega_{1}-\alpha)+\lambda^{\prime}(\Omega_{2}-\beta),$
for some $\lambda^{\prime}\in\mathcal{A}$. This contradicts the minimality of
$(i,j).$ Hence, $(i,j)=(0,0)$ and so
$y=m(\Omega_{1}-\alpha)+n(\Omega_{2}-\beta)\in\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle.$
In conclusion,
$\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle_{R}\cap\mathcal{A}=\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle,$
$\forall\alpha,\beta\in\mathbb{K}^{*}.$ ∎

One can also proove the following results in a similar manner. The details of
the proof can be deduced from [18, §2.4].

######  Proposition 4.10.

Assume that $\mathbb{K}$ is algebraically closed.

  * 1.

P.Spec${}_{\langle\Omega_{1}\rangle}(\mathcal{A})=\\{\langle\Omega_{1}\rangle\\}\cup\\{\langle\Omega_{1},\Omega_{2}-\beta\rangle\mid\beta\in\mathbb{K}^{*}\\}.$

  * 2.

P.Spec${}_{\langle\Omega_{2}\rangle}(\mathcal{A})=\\{\langle\Omega_{2}\rangle\\}\cup\\{\langle\Omega_{1}-\alpha,\Omega_{2}\rangle\mid\alpha\in\mathbb{K}^{*}\\}.$

######  Corollary 4.11.

The Poisson ideal $\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle$ is
Poisson primitive in $\mathcal{A}$ for each
$(\alpha,\beta)\in\mathbb{K}^{2}\setminus\\{(0,0)\\}.$

###### Proof.

Since the Poisson ideal $\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle$ is
maximal in its respective strata for each
$(\alpha,\beta)\in\mathbb{K}^{2}\setminus\\{(0,0)\\},$ it is also Poisson
primitive (see Prop. 3.2). ∎

While we assumed that $\mathbb{K}$ is algebraically closed in the above
propositions and corollary, Corollary 4.11 is also true when we drop this
assumption (with the same proof).

######  Proposition 4.12.

Let $(\alpha,\beta)\in\mathbb{K}^{2}\setminus\\{(0,0)\\}.$ The Poisson prime
ideal $\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle$ is maximal in
$\mathcal{A}.$

###### Proof.

Suppose that there exists a maximal Poisson ideal $I$ of $\mathcal{A}$ such
that $\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle\varsubsetneq
I\varsubsetneq\mathcal{A}.$ Let $J$ be the Poisson $\mathcal{H}$-invariant
prime ideal in $\mathcal{A}$ such that $I\in$ P.Spec${}_{J}(\mathcal{A}).$ By
Propositions 4.9 and 4.10, $J$ cannot be $\langle 0\rangle,$
$\langle\Omega_{1}\rangle$ or $\langle\Omega_{2}\rangle,$ since either of
these will lead to a contradiction. Every non-zero Poisson
$\mathcal{H}$-invariant prime ideal contains only $\Omega_{1}$ or only
$\Omega_{2}$ or both (Remark 4.8). Since
$J\neq\langle\Omega_{1}\rangle,\langle\Omega_{2}\rangle,$ this implies that
$J$ contains both $\Omega_{1}$ and $\Omega_{2}.$ Moreover, since $J\subseteq
I,$ this implies that $\Omega_{1},\Omega_{2}\in I.$ Given
$\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle\subset I$, we have that
$\Omega_{1}-\alpha,\Omega_{2}-\beta\in I.$ It follows that $\alpha,\beta\in
I,$ hence $I=\mathcal{A},$ a contradiction! This confirms that
$\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle$ is a maximal Poisson ideal
in $\mathcal{A}.$ ∎

###  4.3 Simple quotients of the semiclassical limit of $U_{q}^{+}(G_{2})$

Given that $\Omega_{1}-\alpha$ and $\Omega_{2}-\beta$ generate a maximal
Poisson prime ideal of $\mathcal{A},$ the factor algebra

$\mathcal{A}_{\alpha,\beta}:=\frac{\mathcal{A}}{\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle}$

is a Poisson-simple noetherian domain for all
$(\alpha,\beta)\in\mathbb{K}^{2}\setminus\\{(0,0)\\}.$ Denote the canonical
image of $X_{i}$ by
$x_{i}:=X_{i}+\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle$ for each
$1\leq i\leq 6.$ The algebra $\mathcal{A}_{\alpha,\beta}$ is commutative, and
satisfies the following two relations:

$\displaystyle x_{1}x_{3}x_{5}$
$\displaystyle-\frac{3}{2}x_{1}x_{4}-\frac{1}{2}x_{2}x_{5}+\frac{1}{2}x_{3}^{2}=\alpha,$
(8) $\displaystyle x_{2}x_{4}x_{6}$
$\displaystyle-\frac{2}{3}x_{3}^{3}x_{6}-\frac{2}{3}x_{2}x_{5}^{3}+2x_{3}^{2}x_{5}^{2}-3x_{3}x_{4}x_{5}+\frac{3}{2}x_{4}^{2}=\beta.$
(9)

We also have the following extra relations in $\mathcal{A}_{\alpha,\beta},$
which can be verified through direct computations.

######  Lemma 4.13.

$\displaystyle(1)\ x_{3}^{2}=$ $\displaystyle
2\alpha+3x_{1}x_{4}+x_{2}x_{5}-2x_{1}x_{3}x_{5}.$ $\displaystyle(2)\
x_{4}^{2}=$ $\displaystyle\frac{2}{3}\beta+\frac{8}{9}\alpha
x_{3}x_{6}+\frac{4}{3}x_{1}x_{3}x_{4}x_{6}+\frac{4}{9}x_{2}x_{3}x_{5}x_{6}-\frac{16}{9}\alpha
x_{1}x_{5}x_{6}-\frac{8}{3}x_{1}^{2}x_{4}x_{5}x_{6}$
$\displaystyle+\frac{16}{9}x_{1}^{2}x_{3}x_{5}^{2}x_{6}-\frac{8}{9}x_{2}x_{5}^{3}-\frac{8}{3}\alpha
x_{5}^{2}-4x_{1}x_{4}x_{5}^{2}+\frac{8}{3}x_{1}x_{3}x_{5}^{3}+2x_{3}x_{4}x_{5}-\frac{2}{3}x_{2}x_{4}x_{6}$
$\displaystyle-\frac{8}{9}x_{1}x_{2}x_{5}^{2}x_{6}.$ $\displaystyle(3)\
x_{3}^{2}x_{4}$ $\displaystyle=2\alpha x_{4}+x_{2}x_{4}x_{5}+2\beta
x_{1}+\frac{8}{3}\alpha
x_{1}x_{3}x_{6}+4x_{1}^{2}x_{3}x_{4}x_{6}+\frac{4}{3}x_{1}x_{2}x_{3}x_{5}x_{6}$
$\displaystyle-8x_{1}^{3}x_{4}x_{5}x_{6}-\frac{8}{3}x_{1}^{2}x_{2}x_{5}^{2}x_{6}+\frac{16}{3}x_{1}^{3}x_{3}x_{5}^{2}x_{6}-\frac{8}{3}x_{1}x_{2}x_{5}^{3}-8\alpha
x_{1}x_{5}^{2}-12x_{1}^{2}x_{4}x_{5}^{2}$
$\displaystyle+8x_{1}^{2}x_{3}x_{5}^{3}+4x_{1}x_{3}x_{4}x_{5}-2x_{1}x_{2}x_{4}x_{6}-\frac{16}{3}\alpha
x_{1}^{2}x_{5}x_{6}.$ $\displaystyle(4)\ x_{3}x_{4}^{2}=\frac{2}{3}$
$\displaystyle\beta x_{3}+\frac{16}{9}\alpha^{2}x_{6}+\frac{16}{3}\alpha
x_{1}x_{4}x_{6}+\frac{16}{9}\alpha x_{2}x_{5}x_{6}+\frac{16}{9}\alpha
x_{1}x_{3}x_{5}x_{6}+\frac{4}{9}x_{2}^{2}x_{5}^{2}x_{6}$ $\displaystyle+$
$\displaystyle\frac{8}{9}x_{1}x_{2}x_{3}x_{5}^{2}x_{6}-\frac{64}{9}\alpha
x_{1}^{3}x_{5}x_{6}^{2}-\frac{160}{9}\alpha
x_{1}^{2}x_{5}^{2}x_{6}-\frac{80}{3}x_{1}^{3}x_{4}x_{5}^{2}x_{6}-\frac{64}{9}x_{1}^{2}x_{2}x_{5}^{3}x_{6}$
$\displaystyle-$
$\displaystyle\frac{8}{9}x_{2}x_{3}x_{5}^{3}-\frac{8}{3}\alpha
x_{3}x_{5}^{2}+4x_{1}x_{3}x_{4}x_{5}^{2}+\frac{160}{9}x_{1}^{3}x_{3}x_{5}^{3}x_{6}-16x_{1}^{2}x_{4}x_{5}^{3}-\frac{8}{3}x_{1}x_{2}x_{5}^{4}$
$\displaystyle-$
$\displaystyle\frac{4}{3}x_{1}x_{2}x_{4}x_{5}x_{6}+\frac{8}{3}\beta
x_{1}^{2}x_{6}+\frac{32}{9}\alpha
x_{1}^{2}x_{3}x_{6}^{2}+\frac{16}{3}x_{1}^{3}x_{3}x_{4}x_{6}^{2}+\frac{16}{9}x_{1}^{2}x_{2}x_{3}x_{5}x_{6}^{2}$
$\displaystyle-$
$\displaystyle\frac{32}{3}x_{1}^{4}x_{4}x_{5}x_{6}^{2}-\frac{8}{3}x_{1}^{2}x_{2}x_{4}x_{6}^{2}+4\alpha
x_{4}x_{5}+2x_{2}x_{4}x_{5}^{2}+4\beta
x_{1}x_{5}+\frac{64}{9}x_{1}^{4}x_{3}x_{5}^{2}x_{6}^{2}$ $\displaystyle-$
$\displaystyle\frac{2}{3}x_{2}x_{3}x_{4}x_{6}-\frac{32}{3}\alpha
x_{1}x_{5}^{3}+\frac{32}{3}x_{1}^{2}x_{3}x_{5}^{4}+\frac{32}{3}x_{1}^{2}x_{3}x_{4}x_{5}x_{6}-\frac{32}{9}x_{1}^{3}x_{2}x_{5}^{2}x_{6}^{2}.$

Now, the commutative algebra $\mathcal{A}_{\alpha,\beta}$ is a Poisson
$\mathbb{K}$-algebra with the Poisson bracket defined as follows:

$\displaystyle\\{x_{2},x_{1}\\}$ $\displaystyle=-3x_{1}x_{2}$
$\displaystyle\\{x_{3},x_{1}\\}$ $\displaystyle=-x_{1}x_{3}-x_{2}$
$\displaystyle\\{x_{3},x_{2}\\}$ $\displaystyle=-3x_{2}x_{3}$
$\displaystyle\\{x_{4},x_{1}\\}$ $\displaystyle=-2x_{3}^{2}$
$\displaystyle\\{x_{4},x_{2}\\}$ $\displaystyle=-3x_{2}x_{4}-{4}x_{3}^{3}$
$\displaystyle\\{x_{4},x_{3}\\}$ $\displaystyle=-3x_{3}x_{4}$
$\displaystyle\\{x_{5},x_{1}\\}$ $\displaystyle=x_{1}x_{5}-2x_{3}$
$\displaystyle\\{x_{5},x_{2}\\}$ $\displaystyle=-6x_{3}^{2}$
$\displaystyle\\{x_{5},x_{3}\\}$ $\displaystyle=-x_{3}x_{5}-3x_{4}$
$\displaystyle\\{x_{5},x_{4}\\}$ $\displaystyle=-3x_{4}x_{5}$
$\displaystyle\\{x_{6},x_{1}\\}$ $\displaystyle=3x_{1}x_{6}-3x_{5}$
$\displaystyle\\{x_{6},x_{2}\\}$
$\displaystyle=3x_{2}x_{6}+9x_{4}-18x_{3}x_{5}$
$\displaystyle\\{x_{6},x_{3}\\}$ $\displaystyle=-6x_{5}^{2}$
$\displaystyle\\{x_{6},x_{4}\\}$ $\displaystyle=-3x_{4}x_{6}-4x_{5}^{3}$
$\displaystyle\\{x_{6},x_{5}\\}$ $\displaystyle=-3x_{5}x_{6}.$

######  Remark 4.14.

The Poisson algebra $\mathcal{A}_{\alpha,\beta}$ is the semiclassical limit of
the quantum second Weyl algebra $A_{\alpha,\beta}$ studied in [15].

In the remainder of this section, we study a Poisson torus arising from a
localization of $\mathcal{A}_{\alpha,\beta},$ and a linear basis of
$\mathcal{A}_{\alpha,\beta},$ both of which will be useful in computing the
Poisson derivations of $\mathcal{A}_{\alpha,\beta}$ in the next section.

####  4.3.1 A Poisson torus

Let $\alpha,\beta\neq 0.$ Recall from Subsection 3.5 that
$\Omega_{1}=T_{1}T_{3}T_{5}$ and $\Omega_{2}=T_{2}T_{4}T_{6}$ in
$\overline{\mathcal{A}}.$ From [12, Corollaries 3.3 & 3.4], there exists a
multiplicative set $S_{\alpha,\beta}$ such that

$\mathcal{A}_{\alpha,\beta}S_{\alpha,\beta}^{-1}\cong\mathscr{P}_{\alpha,\beta}:=\frac{\mathfrak{R}_{1}}{\langle
T_{1}T_{3}T_{5}-\alpha,T_{2}T_{4}T_{6}-\beta\rangle},$

where $\mathfrak{R}_{1}=\mathbb{K}[T_{1}^{\pm 1},\ldots,T_{6}^{\pm 1}]$ is a
Poisson torus associated to the Poisson affine space $\overline{\mathcal{A}}.$
Let $t_{i}:=T_{i}+\langle T_{1}T_{3}T_{5}-\alpha,T_{2}T_{4}T_{6}-\beta\rangle$
denote the canonical image of $T_{i}$ in $\mathscr{P}_{\alpha,\beta}$ for each
$1\leq i\leq 6.$ The algebra $\mathscr{P}_{\alpha,\beta}$ is a Poisson torus
generated by $t_{1}^{\pm 1},\ldots,t_{6}^{\pm 1}$ subject to the relations:

$t_{1}=\alpha t_{3}^{-1}t_{5}^{-1}\ \ \text{and}\ \ t_{2}=\beta
t_{4}^{-1}t_{5}^{-1}.$

One can verify that $\mathscr{P}_{\alpha,\beta}\cong\mathbb{K}[t_{3}^{\pm
1},t_{4}^{\pm 1},t_{5}^{\pm 1},t_{6}^{\pm 1}],$ and that the isomorphism holds
whether $\alpha$ or $\beta$ is zero.

####  4.3.2 Linear basis for $\mathcal{A}_{\alpha,\beta}$

Set $\mathcal{A}_{\beta}:=\mathcal{A}/\langle\Omega_{2}-\beta\rangle,\
\beta\in\mathbb{K}.$ Denote the canonical image of $X_{i}$ in
$\mathcal{A}_{\beta}$ by
$\widehat{x_{i}}:=X_{i}+\langle\Omega_{2}-\beta\rangle$ for each $1\leq i\leq
6.$ It can be verified that
$\mathcal{A}_{\alpha,\beta}\cong\mathcal{A}_{\beta}/\langle\widehat{\Omega}_{1}-\alpha\rangle.$
Note that $\mathcal{A}_{\beta}$ satisfies the relation:

$\displaystyle\widehat{x_{4}}^{2}=\frac{2}{3}\beta-\frac{2}{3}\widehat{x_{2}}\widehat{x_{4}}\widehat{x_{6}}+\frac{4}{9}\widehat{x_{3}}^{3}\widehat{x_{6}}+\frac{4}{9}\widehat{x_{2}}\widehat{x_{5}}^{3}-\frac{4}{3}\widehat{x_{3}}^{2}\widehat{x_{5}}^{2}+2\widehat{x_{3}}\widehat{x_{4}}\widehat{x_{5}}.$
(10)

######  Proposition 4.15.

The set
$\mathfrak{F}=\\{\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k}\widehat{x_{4}}^{\xi}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m}\mid(\xi,i,j,k,l,m)\in\\{0,1\\}\times\mathbb{N}^{5}\\}$
is a $\mathbb{K}$-basis of $\mathcal{A}_{\beta}.$

###### Proof.

Since $(\Pi_{s=1}^{6}X_{s}^{i_{s}})_{i_{s}\in\mathbb{N}}$ is a basis of
$\mathcal{A}$ over $\mathbb{K},$ we have that
$(\Pi_{s=1}^{6}\widehat{x_{s}}^{i_{s}})_{i_{s}\in\mathbb{N}}$ is a spanning
set of $\mathcal{A}_{\beta}$ over $\mathbb{K}.$ We want to show that
$\mathfrak{F}$ is a spanning set of $\mathcal{A}_{\beta}.$ It is sufficient to
do that by showing that
$\widehat{x_{1}}^{i_{1}}\widehat{x_{2}}^{i_{2}}\widehat{x_{3}}^{i_{3}}\widehat{x_{4}}^{i_{4}}\widehat{x_{5}}^{i_{5}}\widehat{x_{6}}^{i_{6}}$
can be written as a finite linear combination of the elements of
$\mathfrak{F}$ over $\mathbb{K}$ for all $i_{1},\ldots,i_{6}\in\mathbb{N}.$ We
do this by an induction on $i_{4}.$ The result is clear when $i_{4}=0.$ For
$i_{4}\geq 0,$ suppose that

$\widehat{x_{1}}^{i_{1}}\widehat{x_{2}}^{i_{2}}\widehat{x_{3}}^{i_{3}}\widehat{x_{4}}^{i_{4}}\widehat{x_{5}}^{i_{5}}\widehat{x_{6}}^{i_{6}}=\sum_{(\xi,\underline{v})\in
I}a_{(\xi,\underline{v})}\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k}\widehat{x_{4}}^{\xi}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m},$

where $\underline{v}:=(i,j,k,l,m)\in\mathbb{N}^{5},$ $I$ is a finite subset of
$\\{0,1\\}\times\mathbb{N}^{5},$ and the $a_{(\xi,\underline{v})}$ are
scalars. It follows that

$\widehat{x_{1}}^{i_{1}}\widehat{x_{2}}^{i_{2}}\widehat{x_{3}}^{i_{3}}\widehat{x_{4}}^{i_{4}+1}\widehat{x_{5}}^{i_{5}}\widehat{x_{6}}^{i_{6}}=\sum_{(\xi,\underline{v})\in
I}a_{(\xi,\underline{v})}\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k}\widehat{x_{4}}^{\xi+1}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m}.$

We have to show that
$\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k}\widehat{x_{4}}^{\xi+1}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m}\in$
Span$(\mathfrak{F})$ for all $(\xi,\underline{v})\in I.$ The result is obvious
when $\xi=0.$ For $\xi=1,$ then using (10), one can verify that
$\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k}\widehat{x_{4}}^{2}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m}\in$
Span$(\mathfrak{F}).$ Consequently,
$\widehat{x_{1}}^{i_{1}}\widehat{x_{2}}^{i_{2}}\widehat{x_{3}}^{i_{3}}\widehat{x_{4}}^{i_{4}+1}\widehat{x_{5}}^{i_{5}}\widehat{x_{6}}^{i_{6}}\in$
Span$(\mathfrak{F}).$ Therefore, $\mathfrak{F}$ spans $\mathcal{A}_{\beta}.$
Before we continue the proof, the following ordering $<_{4}$ needs to be
noted.

  * $\clubsuit$

Let $(i^{\prime},j^{\prime},k^{\prime},l^{\prime},m^{\prime},n^{\prime}),$
$(i,j,k,l,m,n)\in$ $\mathbb{N}^{6}.$ We say that
$(i,j,k,l,m,n)<_{4}(i^{\prime},j^{\prime},k^{\prime},l^{\prime},m^{\prime},n^{\prime})$
if $[l<l^{\prime}]$ or $[l=l^{\prime}$ and $i<i^{\prime}]$ or $[l=l^{\prime},\
i=i^{\prime}$ and $j<j^{\prime}]$ or $[l=l^{\prime},\ i=i^{\prime},\
j=j^{\prime}$ and $k<k^{\prime}]$ or $[l=l^{\prime},\ i=i^{\prime},\
j=j^{\prime},\ k=k^{\prime}$ and $m<m^{\prime}]$ or $[l=l^{\prime},\
i=i^{\prime},\ j=j^{\prime},\ k=k^{\prime},\ m=m^{\prime}$ and $n\leq
n^{\prime}].$

Note that the square brackets $[\ ]$ are there to differentiate the options.

We proceed to show that $\mathfrak{F}$ is a linearly independent set. Suppose
that

$\sum_{(\xi,\underline{v})\in
I}a_{(\xi,\underline{v})}\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k}\widehat{x_{4}}^{\xi}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m}=0.$

It follows that

$\sum_{(\xi,\underline{v})\in
I}a_{(\xi,\underline{v})}X_{1}^{i}X_{2}^{j}X_{3}^{k}X_{4}^{\xi}X_{5}^{l}X_{6}^{m}=\nu(\Omega_{2}-\beta),$

where $\nu\in\mathcal{A}.$ Write $\nu=\sum_{(i,\ldots,n)\in
J}b_{(i,\ldots,n)}X_{1}^{i}X_{2}^{j}X_{3}^{k}X_{4}^{l}X_{5}^{m}X_{6}^{n},$
where $J$ is a finite subset of $\mathbb{N}^{6},$ and $b_{(i,\ldots,n)}$ are
scalars. From Subsection 3.5, we have that

$\Omega_{2}=X_{2}X_{4}X_{6}-\frac{2}{3}X_{3}^{3}X_{6}-\frac{2}{3}X_{2}X_{5}^{3}+2X_{3}^{2}X_{5}^{2}-3X_{3}X_{4}X_{5}+\frac{3}{2}X_{4}^{2}.$

It follows that

$\displaystyle\sum_{(\xi,\underline{v})\in
I}a_{(\xi,\underline{v})}X_{1}^{i}X_{2}^{j}X_{3}^{k}X_{4}^{\xi}X_{5}^{l}X_{6}^{m}$
$\displaystyle=\sum_{(i,\ldots,n)\in
J}b_{(i,\ldots,n)}X_{1}^{i}X_{2}^{j}X_{3}^{k}X_{4}^{l}X_{5}^{m}X_{6}^{n}(\Omega_{2}-\beta)$
$\displaystyle=\sum_{(i,\ldots,n)\in
J}\frac{3}{2}b_{(i,\ldots,n)}X_{1}^{i}X_{2}^{j}X_{3}^{k}X_{4}^{l+2}X_{5}^{m}X_{6}^{n}+\text{LT}_{<_{4}},$

where $\text{LT}_{<_{4}}$ contains lower order terms with respect to $<_{4}$
(see item $\clubsuit$). Moreover, $\text{LT}_{<_{4}}$ vanishes when
$b_{(i,\ldots,n)}=0$ for all $(i,\ldots,n)\in J.$ One can easily confirm this
when the previous line of equality (right hand side) is fully expanded.

Suppose that there exists $(i,j,k,l,m,n)\in J$ such that
$b_{(i,j,k,l,m,n)}\neq 0.$ Let
$(i^{\prime},j^{\prime},k^{\prime},l^{\prime},m^{\prime},n^{\prime})$ be the
greatest element of $J$ with respect to $<_{4}$ such that
$b_{(i^{\prime},j^{\prime},k^{\prime},l^{\prime},m^{\prime},n^{\prime})}\neq
0.$ Identifying the coefficients of
$X_{1}^{i^{\prime}}X_{2}^{j^{\prime}}X_{3}^{k^{\prime}}X_{4}^{l^{\prime}+2}X_{5}^{m^{\prime}}X_{6}^{n^{\prime}},$
we have
$\frac{3}{2}b_{(i^{\prime},j^{\prime},k^{\prime},l^{\prime},m^{\prime},n^{\prime})}=0$
(note that the family
$(X_{1}^{i}X_{2}^{j}X_{3}^{k}X_{4}^{l}X_{5}^{m}X_{6}^{n})_{i,\ldots,n\in\mathbb{N}}$
is a basis for $\mathcal{A}$ and $\text{LT}_{<_{4}}$ contains lower order
terms). Therefore,
$b_{(i^{\prime},j^{\prime},k^{\prime},l^{\prime},m^{\prime},n^{\prime})}=0,$ a
contradiction! As a result, $b_{(i,j,k,l,m,n)}=0$ for all $(i,j,k,l,m,n)\in
J,$ and

$\sum_{(\xi,\underline{v})\in
I}a_{(\xi,\underline{v})}X_{1}^{i}X_{2}^{j}X_{3}^{k}X_{4}^{\xi}X_{5}^{l}X_{6}^{m}=0.$

Consequently, $a_{(\xi,i,j,k,l)}=0$ for all $(\xi,i,j,k,l)\in I.$ ∎

We are now ready to find a basis for $\mathcal{A}_{\alpha,\beta}.$

######  Proposition 4.16.

The set
$\mathfrak{P}=\\{x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}x_{5}^{k}x_{6}^{l}\mid(\epsilon_{1},\epsilon_{2},i,j,k,l)\in\\{0,1\\}^{2}\times\mathbb{N}^{4}\\}$
is a $\mathbb{K}$-basis of $\mathcal{A}_{\alpha,\beta}.$

###### Proof.

Since the set
$\mathfrak{F}=\\{\widehat{x_{1}}^{i_{1}}\widehat{x_{2}}^{i_{2}}\widehat{x_{3}}^{i_{3}}\widehat{x_{4}}^{\xi}\widehat{x_{5}}^{i_{5}}\widehat{x_{6}}^{i_{6}}\mid(i_{1},i_{2},i_{3},\xi,i_{5},i_{6})\in\\{0,1\\}\times\mathbb{N}^{5}\\}$
is a $\mathbb{K}$-basis of $\mathcal{A}_{\beta}$ (Prop. 4.15) and
$\mathcal{A}_{\alpha,\beta}$ is identified with
${\mathcal{A}_{\beta}}/{\langle\widehat{\Omega}_{1}-\alpha\rangle},$ it
follows that
$(x_{1}^{i_{1}}x_{2}^{i_{2}}x_{3}^{i_{3}}x_{4}^{\xi}x_{5}^{i_{5}}x_{6}^{i_{6}})_{(i_{1},i_{2},i_{3},\xi,i_{5},i_{6})\in\\{0,1\\}\times\mathbb{N}^{5}}$
is a spanning set of $\mathcal{A}_{\alpha,\beta}$ over $\mathbb{K}.$ We want
to show that $\mathfrak{P}$ spans $\mathcal{A}_{\alpha,\beta}$ by showing that
$e_{1}^{i_{1}}e_{2}^{i_{2}}e_{3}^{i_{3}}e_{4}^{\xi}e_{5}^{i_{5}}e_{6}^{i_{6}}$
can be written as a finite linear combination of the elements of
$\mathfrak{P}$ over $\mathbb{K}$ for all
$(i_{1},i_{2},i_{3},\xi,i_{5},i_{6})\in\\{0,1\\}\times\mathbb{N}^{5}$. By
Proposition 4.15, it is sufficient to do this by an induction on $i_{3}.$ The
result is obvious when $i_{3}=0$ or $1.$ For $i_{3}\geq 1,$ suppose that

$\displaystyle
x_{1}^{i_{1}}x_{2}^{i_{2}}x_{3}^{i_{3}}x_{4}^{\xi}x_{5}^{i_{5}}x_{6}^{i_{6}}$
$\displaystyle=\sum_{(\epsilon_{1},\epsilon_{2},\underline{v})\in
I}a_{(\epsilon_{1},\epsilon_{2},\underline{v})}x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}x_{5}^{k}x_{6}^{l},$

where $\underline{v}:=(i,j,k,l)\in\mathbb{N}^{4},$ and the
$a_{(\epsilon_{1},\epsilon_{2},\underline{v})}$ are all scalars. Moreover, $I$
is a finite subset of $\\{0,1\\}^{2}\times\mathbb{N}^{4}$. It follows from the
inductive hypothesis that

$\displaystyle
x_{1}^{i_{1}}x_{2}^{i_{2}}x_{3}^{i_{3}+1}x_{4}^{\xi}x_{5}^{i_{5}}x_{6}^{i_{6}}=\sum_{(\epsilon_{1},\epsilon_{2},\underline{v})\in
I}a_{(\epsilon_{1},\epsilon_{2},\underline{v})}x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}+1}x_{4}^{\epsilon_{2}}x_{5}^{k}x_{6}^{l}.$

We need to show that
$x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}+1}x_{4}^{\epsilon_{2}}x_{5}^{k}x_{6}^{l}\in$
Span$(\mathfrak{P})$ for all $(\epsilon_{1},\epsilon_{2},\underline{v})\in I.$
The result is obvious when $(\epsilon_{1},\epsilon_{2})=(0,0),(0,1).$ Using
Lemma 4.13(1),(3), one can also show that
$x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}+1}x_{4}^{\epsilon_{2}}x_{5}^{k}x_{6}^{l}\in$
Span$(\mathfrak{P})$ for all $(\epsilon_{1},\epsilon_{2})=(1,0),(1,1);$ and
$i,j,k,l\in\mathbb{N}.$ Therefore,
$x_{1}^{i_{1}}x_{2}^{i_{2}}x_{3}^{i_{3}+1}x_{4}^{\xi}x_{5}^{i_{5}}x_{6}^{i_{6}}\in$
Span$(\mathfrak{P}).$ As a result, $\mathfrak{P}$ spans
$\mathcal{A}_{\alpha,\beta}.$

We proceed to show that $\mathfrak{F}$ is a linearly independent set. Suppose
that

$\displaystyle\sum_{(\epsilon_{1},\epsilon_{2},\underline{v})\in
I}a_{(\epsilon_{1},\epsilon_{2},\underline{v})}x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}x_{5}^{k}x_{6}^{l}=0.$

Then,

$\displaystyle\sum_{(\epsilon_{1},\epsilon_{2},\underline{v})\in
I}a_{(\epsilon_{1},\epsilon_{2},\underline{v})}\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{\epsilon_{1}}\widehat{x_{4}}^{\epsilon_{2}}\widehat{e_{5}}^{k}\widehat{e_{6}}^{l}$
$\displaystyle=(\widehat{\Omega}_{1}-\alpha)\nu$

in $\mathcal{A}_{\beta},$ where $\nu\in\mathcal{A}_{\beta}.$ Set
$\underline{w}:=(i,j,k,l,m)\in\mathbb{N}^{5}$, and let $J_{1},J_{2}$ be finite
subsets of $\mathbb{N}^{5}.$ One can write $\nu$ in terms of the basis
$\mathfrak{F}$ of $\mathcal{A}_{\beta}$ as:

$\nu=\displaystyle\sum_{\underline{w}\in
J_{1}}b_{\underline{w}}\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k}\widehat{x_{4}}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m}+\displaystyle\sum_{\underline{w}\in
J_{2}}c_{\underline{w}}\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m},$

where $b_{\underline{w}}$ and $c_{\underline{w}}$ are scalars. Note that
$\widehat{\Omega}_{1}=\widehat{x_{1}}\widehat{x_{3}}\widehat{x_{5}}-\frac{3}{2}\widehat{x_{1}}\widehat{x_{4}}-\frac{1}{2}\widehat{x_{2}}\widehat{x_{5}}+\frac{1}{2}\widehat{x_{3}}^{2}.$
Given this expression, and the relation (10), one can express the above
equality as follows:

$\displaystyle\sum_{(\epsilon_{1},\epsilon_{2},\underline{v})\in
I}a_{(\epsilon_{1},\epsilon_{2},\underline{v})}\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{\epsilon_{1}}\widehat{x_{4}}^{\epsilon_{2}}\widehat{e_{5}}^{k}\widehat{e_{6}}^{l}=$
$\displaystyle\displaystyle\sum_{\underline{w}\in
J_{1}}b_{\underline{w}}\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k}\widehat{x_{4}}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m}(\widehat{\Omega}_{1}-\alpha)$
$\displaystyle+\displaystyle\sum_{\underline{w}\in
J_{2}}c_{\underline{w}}\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m}(\widehat{\Omega}_{1}-\alpha)$
$\displaystyle=$ $\displaystyle\sum_{\underline{w}\in
J_{1}}\frac{1}{2}b_{\underline{w}}\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k+2}\widehat{x_{4}}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m}$
$\displaystyle-\sum_{\underline{w}\in
J_{1}}\frac{2}{3}b_{\underline{w}}\widehat{x_{1}}^{i+1}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k+3}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m+1}$
$\displaystyle-\sum_{\underline{w}\in
J_{2}}\frac{3}{2}c_{\underline{w}}\widehat{x_{1}}^{i+1}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k}\widehat{x_{4}}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m}$
$\displaystyle+\sum_{\underline{w}\in
J_{2}}\frac{1}{2}c_{\underline{w}}\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k+2}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m}+\Upsilon,$

where

$\displaystyle\Upsilon=$ $\displaystyle\sum_{\underline{w}\in
J_{1}}r_{1}b_{\underline{w}}\widehat{x_{1}}^{i+1}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m}+\sum_{\underline{w}\in
J_{1}}r_{2}b_{\underline{w}}\widehat{x_{1}}^{i+1}\widehat{x_{2}}^{1+j}\widehat{x_{3}}^{k}\widehat{x_{4}}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m+1}$
$\displaystyle+\sum_{\underline{w}\in
J_{1}}r_{3}b_{\underline{w}}\widehat{x_{1}}^{i+1}\widehat{x_{2}}^{j+1}\widehat{x_{3}}^{k}\widehat{x_{5}}^{l+3}\widehat{x_{6}}^{m}+\sum_{\underline{w}\in
J_{1}}r_{4}b_{\underline{w}}\widehat{x_{1}}^{i+1}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k+2}\widehat{x_{5}}^{l+2}\widehat{x_{6}}^{m}$
$\displaystyle+\sum_{\underline{w}\in
J_{1}}r_{5}b_{\underline{w}}\widehat{x_{1}}^{i+1}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k+1}\widehat{x_{4}}\widehat{x_{5}}^{l+1}\widehat{x_{6}}^{m}+\sum_{\underline{w}\in
J_{1}}r_{6}b_{\underline{w}}\widehat{x_{1}}^{i}\widehat{x_{2}}^{j+1}\widehat{x_{3}}^{k}\widehat{x_{4}}\widehat{x_{5}}^{l+1}\widehat{x_{6}}^{m}$
$\displaystyle+\sum_{\underline{w}\in
J_{1}}r_{7}b_{\underline{w}}\beta\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k}\widehat{x_{4}}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m}+\sum_{\underline{w}\in
J_{1}}r_{8}b_{\underline{w}}\widehat{x_{1}}^{i+1}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k+1}\widehat{x_{4}}\widehat{x_{5}}^{l+1}\widehat{x_{6}}^{m}$
$\displaystyle+\sum_{\underline{w}\in
J_{2}}r_{9}c_{\underline{w}}\widehat{x_{1}}^{i+1}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k+1}\widehat{x_{5}}^{l+1}\widehat{x_{6}}^{m}+\sum_{\underline{w}\in
J_{2}}r_{10}c_{\underline{w}}\widehat{x_{1}}^{i}\widehat{x_{2}}^{j+1}\widehat{x_{3}}^{k}\widehat{x_{5}}^{l+1}\widehat{x_{6}}^{m}$
$\displaystyle+\sum_{\underline{w}\in
J_{2}}r_{11}c_{\underline{w}}\alpha\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{k}\widehat{x_{5}}^{l}\widehat{x_{6}}^{m}.$

Note that $r_{1},\ldots,r_{11}$ are some non-zero rational numbers. Before we
continue the proof, the following ordering $<_{3}$ needs to be noted.

  * $\spadesuit$

Let $(\vartheta_{1},\vartheta_{2},\vartheta_{3},\vartheta_{5},\vartheta_{6})$,
$(\varsigma_{1},\varsigma_{2},\varsigma_{3},\varsigma_{5},\varsigma_{6})\in\mathbb{N}^{5}.$
We say that
$(\varsigma_{1},\varsigma_{2},\varsigma_{3},\varsigma_{5},\varsigma_{6})<_{3}(\vartheta_{1},\vartheta_{2},\vartheta_{3},\vartheta_{5},\vartheta_{6})$
if $[\vartheta_{3}>\varsigma_{3}]$ or $[\vartheta_{3}=\varsigma_{3}$ and
$\vartheta_{1}>\varsigma_{1}]$ or $[\vartheta_{3}=\varsigma_{3},\
\vartheta_{1}=\varsigma_{1}$ and $\vartheta_{2}>\varsigma_{2}]$ or
$[\vartheta_{3}=\varsigma_{3},\ \vartheta_{1}=\varsigma_{1},\
\vartheta_{2}=\varsigma_{2}$ and $\vartheta_{5}>\varsigma_{5}]$ or
$[\vartheta_{3}=\varsigma_{3},\ \vartheta_{1}=\varsigma_{1},\
\vartheta_{2}=\varsigma_{2},\ \vartheta_{5}=\varsigma_{5}$ and
$\vartheta_{6}\geq\varsigma_{6}].$

Observe that $\Upsilon$ contains lower order terms with respect to $<_{3}$ in
each monomial type (note that there are two different types of monomials in
the basis of $\mathcal{A}_{\beta};$ one with $\widehat{x_{4}}$ and the other
without $\widehat{x_{4}}$ ). Now, suppose that there exists $(i,j,k,l,m)\in
J_{1}$ and $(i,j,k,l,m)\in J_{2}$ such that $b_{(i,j,k,l,m)}\neq 0$ and
$c_{(i,j,k,l,m)}\neq 0.$ Let $(v_{1},v_{2},v_{3},v_{5},v_{6})$ and
$(w_{1},w_{2},w_{3},w_{5},w_{6})$ be the greatest elements of $J_{1}$ and
$J_{2}$ respectively with respect to $<_{3}$ such that
$b_{(v_{1},v_{2},v_{3},v_{5},v_{6})}$ and
$c_{(w_{1},w_{2},w_{3},w_{5},w_{6})}$ are non-zero. Since $\mathfrak{F}$ is a
linear basis for $\mathcal{A}_{\beta}$ and $\Upsilon$ contains lower order
terms with respect to $<_{3}$, we have the following: if $w_{3}-v_{3}<2,$ then
identifying the coefficients of
$\widehat{x_{1}}^{v_{1}}\widehat{x_{2}}^{v_{2}}\widehat{x_{3}}^{v_{3}+2}\widehat{x_{4}}\widehat{x_{5}}^{v_{5}}\widehat{x_{6}}^{v_{6}}$
implies that $\frac{1}{2}b_{(v_{1},v_{2},v_{3},v_{5},v_{6})}=0,$ a
contradiction! Finally, if $w_{3}-v_{3}\geq 2,$ then identifying the
coefficients of
$\widehat{x_{1}}^{w_{1}}\widehat{x_{2}}^{w_{2}}\widehat{x_{3}}^{w_{3}+2}\widehat{x_{5}}^{w_{5}}\widehat{x_{6}}^{w_{6}}$
implies that $\frac{1}{2}c_{(w_{1},w_{2},w_{3},w_{5},w_{6})}=0$, another
contradiction! Therefore, either all $b_{(i,j,k,m,n)}$ or all
$c_{(i,j,k,m,n)}$ are zero. Without loss of generality, suppose that there
exists $(i,j,k,m,n)\in J_{2}$ such that $c_{(i,j,k,m,n)}$ is not zero. Then,
$b_{(i,j,k,m,n)}$ are all zero. Let $(w_{1},w_{2},w_{3},w_{5},w_{6})$ be the
greatest element of $J_{2}$ such that $c_{(w_{1},w_{2},w_{3},w_{5},w_{6})}\neq
0.$ Identifying the coefficients of
$\widehat{x_{1}}^{w_{1}}\widehat{x_{2}}^{w_{2}}\widehat{x_{3}}^{w_{3}+2}\widehat{x_{5}}^{w_{5}}\widehat{x_{6}}^{w_{6}}$
in the above equality implies that
$\frac{1}{2}c_{(w_{1},w_{2},w_{3},w_{5},w_{6})}=0,$ a contradiction! We can
therefore conclude that $b_{(i,j,k,m,n)}$ and $c_{(i,j,k,m,n)}$ are all zero.
Consequently,

$\displaystyle\sum_{(\epsilon_{1},\epsilon_{2},\underline{v})\in
I}a_{(\epsilon_{1},\epsilon_{2},\underline{v})}\widehat{x_{1}}^{i}\widehat{x_{2}}^{j}\widehat{x_{3}}^{\epsilon_{1}}\widehat{x_{4}}^{\epsilon_{2}}\widehat{x_{5}}^{k}\widehat{x_{6}}^{l}$
$\displaystyle=0.$

Since $\mathfrak{F}$ is a basis for $\mathcal{A}_{\beta}$, this implies that
$a_{(\epsilon_{1},\epsilon_{2},\underline{v})}=0$ for all
$(\epsilon_{1},\epsilon_{2},\underline{v})\in I.$ Therefore, $\mathfrak{P}$ is
a linearly independent set. ∎

######  Corollary 4.17.

Let $\underline{v}=(i,j,k,l)\in\mathbb{N}^{2}\times\mathbb{Z}^{2},\ I$
represent a finite subset of
$\\{0,1\\}^{2}\times\mathbb{N}^{2}\times\mathbb{Z}^{2},$ and
$(a_{(\epsilon_{1},\epsilon_{2},\underline{v})})_{(\epsilon_{1},\epsilon_{2},\underline{v})\in
I}$ be a family of scalars. If

$\sum_{(\epsilon_{1},\epsilon_{2},\underline{v})\in
I}a_{(\epsilon_{1},\epsilon_{2},\underline{v})}x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k}t_{6}^{l}=0,$

then $a_{(\epsilon_{1},\epsilon_{2},\underline{v})}=0$ for all
$(\epsilon_{1},\epsilon_{2},\underline{v})\in I.$

###### Proof.

The result is obvious when $k,l\geq 0$ due to Proposition 4.16. When $k$
(resp. $l$) is negative, then one can multiply the above equality enough times
by $t_{5}$ (resp. $t_{6}$) to kill all the negative powers and then apply
Proposition 4.16 to complete the proof. ∎

##  5  Poisson derivations of the semiclassical limit of the quantum second
Weyl algebra $A_{\alpha,\beta}$

This section focuses on studying the Poisson derivations of the Poisson
algebra $\mathcal{A}_{\alpha,\beta}$.

###  5.1 Preliminaries and strategy

Let $2\leq j\leq 7$ and $(\alpha,\beta)\in\mathbb{K}^{2}\setminus\\{(0,0)\\}.$
Set

$\mathcal{A}_{\alpha,\beta}^{(j)}:=\frac{\mathcal{A}^{(j)}}{\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle},$

where $\mathcal{A}^{(j)}$ is defined in Subsection 3.3, and $\Omega_{1}$ and
$\Omega_{2}$ are the generators of the centre of $\mathcal{A}^{(j)}$ (see
Subsection 3.3). Recall that
$\mathcal{A}^{(7)}=\mathcal{A}=\mathbb{K}[X_{1},\ldots,X_{6}].$ It follows
that $\mathcal{A}_{\alpha,\beta}^{(7)}=\mathcal{A}_{\alpha,\beta}.$ For each
$2\leq j\leq 7,$ denote the canonical images of the generators $X_{i,j}$ of
$\mathcal{A}^{(j)}$ in $\mathcal{A}_{\alpha,\beta}^{(j)}$ by $x_{i,j}$ for all
$1\leq i\leq 6.$ Furthermore, from [12], one can deduce the following data of
$\mathcal{A}_{\alpha,\beta}$ from the PDDA of $\mathcal{A}$ (see Subsection
3.3) as follows:

$\displaystyle x_{1,6}$ $\displaystyle=x_{1}-\frac{1}{2}x_{5}x_{6}^{-1}$
$\displaystyle x_{2,6}$
$\displaystyle=x_{2}+\frac{3}{2}x_{4}x_{6}^{-1}-3x_{3}x_{5}x_{6}^{-1}+x_{5}^{3}x_{6}^{-2}$
$\displaystyle x_{3,6}$ $\displaystyle=x_{3}-x_{5}^{2}x_{6}^{-1}$
$\displaystyle x_{4,6}$ $\displaystyle=x_{4}-\frac{2}{3}x_{5}^{3}x_{6}^{-1}$
$\displaystyle x_{1,5}$
$\displaystyle=x_{1,6}-x_{3,6}x_{5,6}^{-1}+\frac{3}{4}x_{4,6}x_{5,6}^{-2}$
$\displaystyle x_{2,5}$
$\displaystyle=x_{2,6}-3x_{3,6}^{2}x_{5,6}^{-1}+\frac{9}{2}x_{3,6}x_{4,6}x_{5,6}^{-2}-\frac{9}{4}x_{4,6}^{2}x_{5,6}^{-3}$
$\displaystyle x_{3,5}$ $\displaystyle=x_{3,6}-\frac{3}{2}x_{4,6}x_{5,6}^{-1}$
$\displaystyle x_{1,4}$
$\displaystyle=x_{1,5}-\frac{1}{3}x_{3,5}^{2}x_{4,5}^{-1}$ $\displaystyle
x_{2,4}$ $\displaystyle=x_{2,5}-\frac{2}{3}x_{3,5}^{3}x_{4,5}^{-1}$
$\displaystyle x_{1,3}$ $\displaystyle=x_{1,4}-\frac{1}{2}x_{2,4}x_{3,4}^{-1}$
$\displaystyle t_{1}$ $\displaystyle:=x_{1,2}=x_{1,3}$ $\displaystyle t_{2}$
$\displaystyle:=x_{2,2}=x_{2,3}=x_{2,4}$ $\displaystyle t_{3}$
$\displaystyle:=x_{3,2}=x_{3,3}=x_{3,4}=x_{3,5}$ $\displaystyle t_{4}$
$\displaystyle:=x_{4,2}=x_{4,3}=x_{4,4}=x_{4,5}=x_{4,6}$ $\displaystyle t_{5}$
$\displaystyle:=x_{5,2}=x_{5,3}=x_{5,4}=x_{5,5}=x_{5,6}=x_{5}$ $\displaystyle
t_{6}$ $\displaystyle:=x_{6,2}=x_{6,3}=x_{6,4}=x_{6,5}=x_{6,6}=x_{6}.$

For simplicity, we will refer to the above data as the PDDA of
$\mathcal{A}_{\alpha,\beta}.$

Note that the $t_{i}$ are the canonical images of $T_{i}$ in
$\mathcal{A}^{(2)}_{\alpha,\beta}$ for all $1\leq i\leq 6.$ For each $2\leq
j<7,$ set $S_{j}:=\\{\lambda t_{j}^{i_{j}}t_{j+1}^{i_{j+1}}\ldots
t_{6}^{i_{6}}\mid i_{j},\ldots,i_{6}\in\mathbb{N},\
\lambda\in\mathbb{K}^{*}\\}.$ One can observe that $S_{j}$ is a multiplicative
system of non-zero divisors (or regular elements) of
$\mathcal{A}_{\alpha,\beta}^{(j)}.$ As a result, one can localize
$\mathcal{A}_{\alpha,\beta}^{(j)}$ at $S_{j}.$ Let us denote this localization
by $\mathcal{R}_{j}.$ That is,

$\mathcal{R}_{j}:=\mathcal{A}_{\alpha,\beta}^{(j)}S_{j}^{-1}.$

Again, the set $\Sigma_{j}:=\\{t_{j}^{k}\mid k\in\mathbb{N}\\}$ is a
multiplicative set in both $\mathcal{A}_{\alpha,\beta}^{(j)}$ and
$\mathcal{A}_{\alpha,\beta}^{(j+1)}$ for each $2\leq j\leq 6.$ Therefore,

$\mathcal{A}_{\alpha,\beta}^{(j)}\Sigma_{j}^{-1}=\mathcal{A}_{\alpha,\beta}^{(j+1)}\Sigma_{j}^{-1}.$

It follows that

$\displaystyle\mathcal{R}_{j}=\mathcal{R}_{j+1}\Sigma_{j}^{-1},$ (11)

for all $2\leq j\leq 6,$ with the convention that
$\mathcal{R}_{7}:=\mathcal{A}_{\alpha,\beta}.$ Similarly to (3.5), we
construct the following embeddings:

$\displaystyle\mathcal{R}_{7}=\mathcal{A}_{\alpha,\beta}\subset\mathcal{R}_{6}=\mathcal{R}_{7}\Sigma_{6}^{-1}\subset\mathcal{R}_{5}=\mathcal{R}_{6}\Sigma_{5}^{-1}\subset\mathcal{R}_{4}=\mathcal{R}_{5}\Sigma_{4}^{-1}\subset\mathcal{R}_{3}.$
(12)

Observe that
$\mathcal{R}_{3}=\mathcal{A}_{\alpha,\beta}^{(3)}S_{3}^{-1}=\mathcal{R}_{4}\Sigma_{3}^{-1}$
is the Poisson torus $\mathscr{P}_{\alpha,\beta}=\mathbb{K}[t_{3}^{\pm
1},t_{4}^{\pm 1},t_{5}^{\pm 1},t_{6}^{\pm 1}]$ in Subsection 4.3.

Strategy to compute the Poisson derivations of
$\mathcal{A}_{\alpha,\beta}=\mathcal{R}_{7}$. From Corollary 2.7, we know that
the Poisson derivations of the Poisson torus $\mathcal{R}_{3}.$ Therefore, we
will extend the Poisson derivations of $\mathcal{A}_{\alpha,\beta}$ to Poisson
derivations of $\mathcal{R}_{3}.$ We will then ‘pull back’ the Poisson
derivations of $\mathcal{R}_{3}$ sequentially to the Poisson derivations of
$\mathcal{A}_{\alpha,\beta},$ and this will give us a complete description of
the Poisson derivations of $\mathcal{A}_{\alpha,\beta}.$ This process will be
carried out in steps, and the linear basis for $\mathcal{R}_{i}$ will play
crucial role. As a result, we will compute these bases in the subsequent
subsection.

The embeddings in (12) present an opportunity to compute the centre of each of
the algebras $\mathcal{R}_{i},$ which will be helpful in studying the Poisson
derivations of $\mathcal{A}_{\alpha,\beta}.$

######  Lemma 5.1.

As usual, let $Z_{P}(\mathcal{R}_{i})$ denote the Poisson centre of
$\mathcal{R}_{i}.$ Then $Z_{P}(\mathcal{R}_{i})=\mathbb{K}$ for each $3\leq
i\leq 7.$

###### Proof.

It is easy to verify that $Z_{P}(\mathcal{R}_{3})=\mathbb{K}.$ Note that
$\mathcal{R}_{7}=\mathcal{A}_{\alpha,\beta}.$ Since $\mathcal{R}_{i}$ is a
localization of $\mathcal{R}_{i+1},$ we have that $\mathbb{K}\subseteq
Z_{P}(\mathcal{R}_{7})\subseteq Z_{P}(\mathcal{R}_{6})\subseteq\cdots\subseteq
Z_{P}(\mathcal{R}_{3})=\mathbb{K}.$ Therefore,
$Z_{P}(\mathcal{R}_{7})=Z_{P}(\mathcal{R}_{6})=\cdots=Z_{P}(\mathcal{R}_{3})=\mathbb{K}.$
∎

###  5.2 Linear bases for $\mathcal{R}_{3},\mathcal{R}_{4}$ and
$\mathcal{R}_{5}.$

We aim to find a basis for $\mathcal{R}_{j}$ for each $j=3,4,5.$ Since
$\mathcal{R}_{3}$ is a Poisson torus generated by $t_{3}^{\pm
1},\ldots,t_{6}^{\pm 1}$ over $\mathbb{K},$ the set
$\\{t_{3}^{i}t_{4}^{j}t_{5}^{k}t_{6}^{l}\mid i,j,k,l\in\mathbb{Z}\\}$ is a
basis of $\mathcal{R}_{3}.$

For simplicity, we set

$\displaystyle f_{1}:$ $\displaystyle=x_{1,4}$ $\displaystyle F_{1}:$
$\displaystyle=X_{1,4}$ $\displaystyle z_{1}:$ $\displaystyle=x_{1,5}$
$\displaystyle Z_{1}:$ $\displaystyle=X_{1,5}$ $\displaystyle z_{2}:$
$\displaystyle=x_{2,5}$ $\displaystyle Z_{2}:$ $\displaystyle=X_{2,5}.$

####  5.2.1 Basis for $\mathcal{R}_{4}$

Observe that

$\mathcal{A}_{\alpha,\beta}^{(4)}=\frac{\mathcal{A}^{(4)}}{\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle},$

where $\Omega_{1}=F_{1}T_{3}T_{5}-\frac{1}{2}T_{2}T_{5}$ and
$\Omega_{2}=T_{2}T_{4}T_{6}$ in $\mathcal{A}^{(4)}$ (Subsection 3.3). Set

$\mathcal{A}_{\beta}^{(4)}S_{4}^{-1}:=\frac{\mathcal{A}^{(4)}S_{4}^{-1}}{\langle\Omega_{2}-\beta\rangle},$

where $\beta\in\mathbb{K}$. We will denote the canonical images of $X_{i,4}$
(resp. $T_{i}$) in $\mathcal{A}_{\beta}^{(4)}$ by $\widehat{x_{i,4}}$ (resp.
$\widehat{t_{i}}$) for all $1\leq i\leq 6.$ Observe that
$\widehat{t_{2}}=\beta\widehat{t_{6}}^{-1}\widehat{t_{4}}^{-1}$ in
$\mathcal{A}_{\beta}^{(4)}S_{4}^{-1}.$ As usual, one can identify
$\mathcal{R}_{4}$ with
$\mathcal{A}_{\beta}^{(4)}S_{4}^{-1}/\langle\Omega_{1}-\alpha\rangle.$

######  Proposition 5.2.

The set
$\mathfrak{P}_{4}=\\{f_{1}^{i_{1}}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}},t_{3}^{i_{3}}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}}\mid(i_{1},i_{3},i_{4},i_{5},i_{6})\in\mathbb{N}^{2}\times\mathbb{Z}^{3}\\}$
is a $\mathbb{K}$-basis of $\mathcal{R}_{4}.$

###### Proof.

One can easily verify that
$\left(\widehat{f_{1}}^{k_{1}}\widehat{t_{3}}^{k_{3}}\widehat{t_{4}}^{k_{4}}\widehat{t_{5}}^{k_{5}}\widehat{t_{6}}^{k_{6}}\right)_{(k_{1},k_{3},\ldots,k_{6})\in\mathbb{N}^{2}\times\mathbb{Z}^{3}}$
is a basis of $\mathcal{A}_{\beta}^{(4)}S_{4}^{-1}$. Since
$\mathcal{A}_{\beta}^{(4)}S_{4}^{-1}={\mathcal{A}^{(4)}S_{4}^{-1}}/{\langle\Omega_{2}-\beta\rangle},$
the family
$(f_{1}^{k_{1}}t_{3}^{k_{3}}t_{4}^{k_{4}}t_{5}^{k_{5}}t_{6}^{k_{6}})_{(k_{1},k_{3},\ldots,k_{6})\in\mathbb{N}^{2}\times\mathbb{Z}^{3}}$
spans $\mathcal{R}_{4}.$ We show that $\mathfrak{P}_{4}$ is a spanning set of
$\mathcal{R}_{4}$ by showing that
$f_{1}^{k_{1}}t_{3}^{k_{3}}t_{4}^{k_{4}}t_{5}^{k_{5}}t_{6}^{k_{6}}$ can be
written as a finite linear combination of the elements of $\mathfrak{P}_{4}$
for all $(k_{1},k_{3},\ldots,k_{6})\in\mathbb{N}^{2}\times\mathbb{Z}^{3}.$ It
is sufficient to do this by an induction on $k_{1}.$ The result is clear when
$k_{1}=0.$ Assume that the statement is true for $k_{1}\geq 0.$ That is,

$f_{1}^{k_{1}}t_{3}^{k_{3}}t_{4}^{k_{4}}t_{5}^{k_{5}}t_{6}^{k_{6}}=\sum_{\underline{i}\in
I_{1}}{a}_{\underline{i}}f_{1}^{i_{1}}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}}+\sum_{\underline{j}\in
I_{2}}b_{\underline{j}}t_{3}^{i_{3}}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}},$

where $\underline{i}=(i_{1},i_{4},i_{5},i_{6})\in
I_{1}\subset\mathbb{N}\times\mathbb{Z}^{3}$ and
$\underline{j}=(i_{3},i_{4},i_{5},i_{6})\in
I_{2}\subset\mathbb{N}\times\mathbb{Z}^{3}.$ Note that the $a_{\underline{i}}$
and $b_{\underline{j}}$ are all scalars.

$\displaystyle
f_{1}^{k_{1}+1}t_{3}^{k_{3}}t_{4}^{k_{4}}t_{5}^{k_{5}}t_{6}^{k_{6}}$
$\displaystyle=f_{1}(f_{1}^{k_{1}}t_{3}^{k_{3}}t_{4}^{k_{4}}t_{5}^{k_{5}}t_{6}^{k_{6}})=\sum_{\underline{i}\in
I_{1}}{a}_{\underline{i}}f_{1}^{i_{1}+1}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}}+\sum_{\underline{j}\in
I_{2}}b_{\underline{j}}f_{1}t_{3}^{i_{3}}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}}.$

Clearly, the monomial
$f_{1}^{i_{1}+1}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}}\in$
Span($\mathfrak{P}_{4}$). We have to also show that
$f_{1}t_{3}^{i_{3}}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}}\in$
Span($\mathfrak{P}_{4}$) for all $i_{3}\in\mathbb{N}$ and
$i_{4},i_{5},i_{6}\in\mathbb{Z}.$ This can easily be achieved by induction on
$i_{3},$ and the use of the relation $f_{1}t_{3}=\alpha
t_{5}^{-1}+\frac{1}{2}\beta t_{4}^{-1}t_{6}^{-1}.$ Therefore, by the principle
of mathematical induction, $\mathfrak{P}_{4}$ is a spanning set of
$\mathcal{R}_{4}$ over $\mathbb{K}.$

We prove that $\mathfrak{P}_{4}$ is a linearly independent set. Suppose that

$\sum_{\underline{i}\in
I_{1}}a_{\underline{i}}f_{1}^{i_{1}}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}}+\sum_{\underline{j}\in
I_{2}}b_{\underline{j}}t_{3}^{i_{3}}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}}=0.$

It follows that there exists $\nu\in\mathcal{A}_{\beta}^{(4)}S_{4}^{-1}$ such
that

$\displaystyle\sum_{\underline{i}\in
I_{1}}a_{\underline{i}}\widehat{f_{1}}^{i_{1}}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}}+\displaystyle\sum_{\underline{j}\in
I_{2}}b_{\underline{j}}\widehat{t_{3}}^{i_{3}}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}}=\left(\widehat{\Omega}_{1}-\alpha\right)\nu.$

Write $\nu=\displaystyle\sum_{\underline{l}\in
J}c_{\underline{l}}\widehat{f_{1}}^{i_{1}}\widehat{t_{3}}^{i_{3}}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}},$
with $\underline{l}={(i_{1},i_{3},i_{4},i_{5},i_{6})}\in
J\subset\mathbb{N}^{2}\times\mathbb{Z}^{3}$ and
$c_{\underline{l}}\in\mathbb{K}.$ One can easily deduce that
$\widehat{\Omega}_{1}=\widehat{f_{1}}\widehat{t_{3}}\widehat{t_{5}}-\frac{1}{2}\widehat{t_{2}}\widehat{t_{5}}=\widehat{f_{1}}\widehat{t_{3}}\widehat{t_{5}}-\frac{1}{2}\beta\widehat{t_{6}}^{-1}\widehat{t_{4}}^{-1}\widehat{t_{5}}$
(note that $\widehat{t_{2}}=\beta\widehat{t_{6}}^{-1}\widehat{t_{4}}^{-1}$).
It follows that

$\displaystyle\displaystyle\sum_{\underline{i}\in
I_{1}}a_{\underline{i}}\widehat{f_{1}}^{i_{1}}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}}+\displaystyle\sum_{\underline{j}\in
I_{2}}b_{\underline{j}}\widehat{t_{3}}^{i_{3}}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}}=$
$\displaystyle\displaystyle\sum_{\underline{l}\in
J}c_{\underline{l}}\widehat{f_{1}}^{i_{1}+1}\widehat{t_{3}}^{i_{3}+1}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}+1}\widehat{t_{6}}^{i_{6}}$
$\displaystyle-\sum_{\underline{l}\in J}\frac{1}{2}\beta
c_{\underline{l}}\widehat{f_{1}}^{i_{1}}\widehat{t_{3}}^{i_{3}}\widehat{t_{4}}^{i_{4}-1}\widehat{t_{5}}^{i_{5}+1}\widehat{t_{6}}^{i_{6}-1}$
$\displaystyle-\displaystyle\sum_{\underline{l}\in J}\alpha
c_{\underline{l}}\widehat{f_{1}}^{i_{1}}\widehat{t_{3}}^{i_{3}}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}}.$

Suppose that there exists $(i_{1},i_{3},i_{4},i_{5},i_{6})\in J$ such that
$c_{(i_{1},i_{3},i_{4},i_{5},i_{6})}\neq 0.$ Let
$(w_{1},w_{3},w_{4},w_{5},w_{6})\in J$ be the greatest element (in the
lexicographic order on $\mathbb{N}^{2}\times\mathbb{Z}^{3}$) of $J$ such that
$c_{(w_{1},w_{3},w_{4},w_{5},w_{6})}\neq 0.$ Since
$\left(\widehat{f_{1}}^{k_{1}}\widehat{t_{3}}^{k_{3}}\widehat{t_{4}}^{k_{4}}\widehat{t_{5}}^{k_{5}}\widehat{t_{6}}^{k_{6}}\right)_{(k_{1},k_{3},\ldots,k_{6})\in\mathbb{N}^{2}\times\mathbb{Z}^{3}}$
is a basis of $\mathcal{A}^{(4)}S_{4}^{-1},$ this implies that the
coefficients of
$\widehat{f_{1}}^{w_{1}+1}\widehat{t_{3}}^{w_{3}+1}\widehat{t_{4}}^{w_{4}}\widehat{t_{5}}^{w_{5}+1}\widehat{t_{6}}^{w_{6}}$
in the above equality can be identified as:
$c_{(w_{1},w_{3},w_{4},w_{5},w_{6})}=0,$ a contradiction! Therefore,
$c_{(i_{1},i_{3},i_{4},i_{5},i_{6})}=0$ for all
$(i_{1},i_{3},i_{4},i_{5},i_{6})\in J.$ This further implies that

$\displaystyle\sum_{\underline{i}\in
I_{1}}a_{\underline{i}}\widehat{f_{1}}^{i_{1}}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}}+\displaystyle\sum_{\underline{j}\in
I_{2}}b_{\underline{j}}\widehat{t_{3}}^{i_{3}}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}}=0.$

Consequently, $a_{\underline{i}}$ and $b_{\underline{j}}$ are all zero. In
conclusion, $\mathfrak{P}_{4}$ is a linearly independent set. ∎

####  5.2.2 Basis for $\mathcal{R}_{5}.$

We will identify $\mathcal{R}_{5}$ with
${\mathcal{A}_{\alpha}^{(5)}S_{5}^{-1}}/{\langle\widehat{\Omega}_{2}-\beta\rangle},$
where
$\mathcal{A}_{\alpha}^{(5)}S_{5}^{-1}=\mathcal{A}^{(5)}S_{5}^{-1}/\langle\Omega_{1}-\alpha\rangle.$
Note that the canonical images of $X_{i,5}$ (resp. $T_{i}$) in
$\mathcal{A}_{\alpha}^{(5)}$ will be denoted by $\widehat{x_{i,5}}$ (resp.
$\widehat{t_{i}}$) for all $1\leq i\leq 6$. We now find a basis for
$\mathcal{A}_{\alpha}^{(5)}S_{5}^{-1}.$ Recall from Subsection 3.3 that
$\Omega_{1}=Z_{1}T_{3}T_{5}-\frac{1}{2}Z_{2}T_{5}$ and
$\Omega_{2}=Z_{2}T_{4}T_{6}-\frac{2}{3}T_{3}^{3}T_{6}$ in $\mathcal{A}^{(5)}$
(remember that $Z_{1}:=X_{1,5}$ and $Z_{2}:=X_{2,5}$ ). Since
$z_{2}t_{4}t_{6}-\frac{2}{3}t_{3}^{3}t_{6}=\beta$ and
$\widehat{z_{1}}\widehat{t_{3}}\widehat{t_{5}}-\frac{1}{2}\widehat{z_{2}}\widehat{t_{5}}=\alpha$
in $\mathcal{R}_{5}$ and $\mathcal{A}_{\alpha}^{(5)}S_{5}^{-1}$ respectively,
we have the relation
$\widehat{z_{2}}=2\widehat{z_{1}}\widehat{t_{3}}-2\alpha\widehat{t_{5}}^{-1}$
in $A_{\alpha}^{(5)}S_{5}^{-1}$ and, in $R_{5},$ we have the following two
relations:

$\displaystyle z_{2}$ $\displaystyle=2(z_{1}t_{3}-\alpha t_{5}^{-1}).$ (13)
$\displaystyle t_{3}^{3}$ $\displaystyle=\dfrac{3}{2}(z_{2}t_{4}-\beta
t_{6}^{-1})=3z_{1}t_{3}t_{4}-\dfrac{3}{2}\beta t_{6}^{-1}-3\alpha
t_{4}t_{5}^{-1}.$ (14)

######  Proposition 5.3.

The set
$\mathfrak{P}_{5}=\left\\{z_{1}^{i_{1}}t_{3}^{\xi}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}}\mid(\xi,i_{1},i_{4},i_{5},i_{6})\in\\{0,1,2\\}\times\mathbb{N}^{2}\times\mathbb{Z}^{2}\right\\}$
is a $\mathbb{K}$-basis of $\mathcal{R}_{5}$.

###### Proof.

One can easily show that the family
$\left(\widehat{z_{1}}^{k_{1}}\widehat{t_{3}}^{k_{3}}\widehat{t_{4}}^{k_{4}}\widehat{t_{5}}^{k_{5}}\widehat{t_{6}}^{k_{6}}\right)_{(k_{1},k_{3},k_{4},k_{5},k_{6})\in\mathbb{N}^{2}\times\mathbb{Z}^{2}}$
is a basis of
$\mathcal{A}_{\alpha}^{(5)}S_{5}^{-1}/\langle\widehat{\Omega}_{2}-\beta\rangle$.
Since $\mathcal{R}_{5}$ is identified with
$\mathcal{A}_{\alpha}^{(5)}S_{5}^{-1}/{\langle\widehat{\Omega}_{2}-\beta\rangle},$
we show that
$z_{1}^{k_{1}}t_{3}^{k_{3}}t_{4}^{k_{4}}t_{5}^{k_{5}}t_{6}^{k_{6}}$ can be
written as a finite linear combination of the elements of $\mathfrak{P}_{5}$
for all
$(k_{1},k_{3},k_{4},k_{5},k_{6})\in\mathbb{N}^{3}\times\mathbb{Z}^{2}.$ It is
sufficient to do this by an induction on $k_{3}.$ The result is obvious when
$k_{3}=0,1,2.$ For $k_{3}\geq 2,$ suppose that

$z_{1}^{k_{1}}t_{3}^{k_{3}}t_{4}^{k_{4}}t_{5}^{k_{5}}t_{6}^{k_{6}}=\sum_{(\xi,\underline{i})\in
I}a_{(\xi,\underline{i})}z_{1}^{i_{1}}t_{3}^{\xi}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}},$

where $I$ is a finite subset of
$\\{0,1,2\\}\times\mathbb{N}^{2}\times\mathbb{Z}^{2},$ and the
$a_{(\xi,\underline{i})}$ are all scalars. It follows that

$z_{1}^{k_{1}}t_{3}^{k_{3}+1}t_{4}^{k_{4}}t_{5}^{k_{5}}t_{6}^{k_{6}}=\left(z_{1}^{k_{1}}t_{3}^{k_{3}}t_{4}^{k_{4}}t_{5}^{k_{5}}t_{6}^{k_{6}}\right)t_{3}=\sum_{(\xi,\underline{i})\in
I}a_{(\xi,\underline{i})}z_{1}^{i_{1}}t_{3}^{\xi+1}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}}.$

Now, $z_{1}^{i_{1}}t_{3}^{\xi+1}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}}\in$
Span($\mathfrak{P}_{5}$) when $\xi=0,1.$ For $\xi=2,$ one can easily verify
that $z_{1}^{i_{1}}t_{3}^{3}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}}\in$
Span($\mathfrak{P}_{5}$) by using the relation in (14). Therefore, by the
principle of mathematical induction, $\mathfrak{P}_{5}$ spans
$\mathcal{R}_{5}.$

We now prove that $\mathfrak{P}_{5}$ is a linearly independent set. Suppose
that

$\sum_{(\xi,\underline{i})\in
I}a_{(\xi,\underline{i})}z_{1}^{i_{1}}t_{3}^{\xi}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}}=0.$

Since $\mathcal{R}_{5}$ is identified with
${\mathcal{A}_{\alpha}^{(5)}S_{5}^{-1}}/{\langle\widehat{\Omega}_{2}-\beta\rangle},$
we have that

$\sum_{(\xi,\underline{i})\in
I}a_{(\xi.\underline{i})}\widehat{z_{1}}^{i_{1}}\widehat{t_{3}}^{\xi}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}}=\langle\widehat{\Omega}_{2}-\beta\rangle\nu,$

where $\nu\in\mathcal{A}_{\alpha}^{(5)}S_{5}^{-1}.$ Write
$\nu=\sum_{\underline{j}\in
J}b_{\underline{j}}\widehat{z_{1}}^{i_{1}}\widehat{t_{3}}^{i_{3}}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}},$
with $\underline{j}=(i_{1},i_{3},i_{4},i_{5},i_{6})\in
J\subset\mathbb{N}^{3}\times\mathbb{Z}^{2}$ and
$b_{\underline{j}}\in\mathbb{K}.$ Given that
$\Omega_{2}=Z_{2}T_{4}T_{6}-\frac{2}{3}T_{3}^{3}T_{6}$ in $\mathcal{A}^{(5)}$
and the relation (13), one can deduce that

$\widehat{\Omega}_{2}=\widehat{z_{2}}\widehat{t_{4}}\widehat{t_{6}}-\frac{2}{3}\widehat{t_{3}}^{3}\widehat{t_{6}}=2\widehat{z_{1}}\widehat{t_{3}}\widehat{t_{4}}\widehat{t_{6}}-2\alpha\widehat{t_{4}}\widehat{t_{5}}^{-1}\widehat{t_{6}}-\frac{2}{3}\widehat{t_{3}}^{3}\widehat{t_{6}}.$

Therefore,

$\displaystyle\sum_{(\xi,\underline{i})\in
I}a_{(\xi.\underline{i})}\widehat{z_{1}}^{i_{1}}\widehat{t_{3}}^{\xi}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}}=$
$\displaystyle\sum_{\underline{j}\in
J}2b_{\underline{j}}\widehat{z_{1}}^{i_{1}+1}\widehat{t_{3}}^{i_{3}+1}\widehat{t_{4}}^{i_{4}+1}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}+1}$
$\displaystyle-\sum_{\underline{j}\in
J}\frac{2}{3}b_{\underline{j}}\widehat{z_{1}}^{i_{1}}\widehat{t_{3}}^{i_{3}+3}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}+1}$
$\displaystyle-\sum_{\underline{j}\in J}2\alpha
b_{\underline{j}}\widehat{z_{1}}^{i_{1}}\widehat{t_{3}}^{i_{3}}\widehat{t_{4}}^{i_{4}+1}\widehat{t_{5}}^{i_{5}-1}\widehat{t_{6}}^{i_{6}+1}$
$\displaystyle-\sum_{\underline{j}\in J}\beta
b_{\underline{j}}\widehat{z_{1}}^{i_{1}}\widehat{t_{3}}^{i_{3}}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}}.$

Suppose that there exists $(i_{1},i_{3},i_{4},i_{5},i_{6})\in J$ such that
$b_{(i_{1},i_{3},i_{4},i_{5},i_{6})}\neq 0.$ Let
$(w_{1},w_{3},w_{4},w_{5},w_{6})\in J$ be the greatest element (in the
lexicographic order on $\mathbb{N}^{3}\times\mathbb{Z}^{2}$) of $J$ such that
$b_{(w_{1},w_{3},w_{4},w_{5},w_{6})}\neq 0.$ Given that
$\left(\widehat{z_{1}}^{k_{1}}\widehat{t_{3}}^{k_{3}}\widehat{t_{4}}^{k_{4}}\widehat{t_{5}}^{k_{5}}\widehat{t_{6}}^{k_{6}}\right)_{(k_{1},k_{3},\ldots,k_{6})\in\mathbb{N}^{3}\times\mathbb{Z}^{2}}$
is a basis of $\mathcal{A}_{\alpha}^{(5)}S_{5}^{-1},$ one can identify the
coefficients of
$\widehat{z_{1}}^{w_{1}+1}\widehat{t_{3}}^{w_{3}+1}\widehat{t_{4}}^{w_{4}+1}\widehat{t_{5}}^{w_{5}}\widehat{t_{6}}^{w_{6}+1}$
in the above equality as: $2b_{(w_{1},w_{3},w_{4},w_{5},w_{6})}=0.$ Hence,
$b_{(w_{1},w_{3},w_{4},w_{5},w_{6})}=0,$ a contradiction! Therefore,
$b_{(i_{1},i_{3},i_{4},i_{5},i_{6})}=0$ for all
$(i_{1},i_{3},i_{4},i_{5},i_{6})\in J.$ Consequently,

$\sum_{(\xi,\underline{i})\in
I}a_{(\xi.\underline{i})}\widehat{z_{1}}^{i_{1}}\widehat{t_{3}}^{\xi}\widehat{t_{4}}^{i_{4}}\widehat{t_{5}}^{i_{5}}\widehat{t_{6}}^{i_{6}}=0.$

It follows that $a_{(\xi,\underline{i})}=0$ for all $(\xi,\underline{i})\in
I.$ As a result, $\mathfrak{P}_{5}$ is a linearly independent set. ∎

######  Corollary 5.4.

Let $I$ be a finite subset of
$\\{0,1,2\\}\times\mathbb{N}\times\mathbb{Z}^{3}$ and
$(a_{(\xi,\underline{i})})_{\underline{i}\in I}$ be a family of scalars. If

$\sum_{(\xi,\underline{i})\in
I}a_{(\xi.\underline{i})}z_{1}^{i_{1}}t_{3}^{\xi}t_{4}^{i_{4}}t_{5}^{i_{5}}t_{6}^{i_{6}}=0,$

then $a_{(\xi,\underline{i})}=0$ for all $(\xi,\underline{i})\in I.$

###### Proof.

When $i_{4}\geq 0,$ then the result is obvious as a result of Proposition 5.3.
For $i_{4}<0,$ multiply both sides of the equality enough times by $t_{4}$ to
kill all the negative powers of $t_{4},$ and then apply Proposition 5.3 to
complete the proof. ∎

######  Remark 5.5.

We were not successful in finding a basis for $\mathcal{R}_{6}.$ However, this
has no effect on our main results in this section. Since
$\mathcal{R}_{7}=\mathcal{A}_{\alpha,\beta},$ we already have a basis for
$\mathcal{R}_{7}$ (Prop. 4.16).

######  Remark 5.6.

Recall the notations:

$\displaystyle f_{1}:$ $\displaystyle=x_{1,4}$ $\displaystyle F_{1}:$
$\displaystyle=X_{1,4}$ $\displaystyle z_{1}:$ $\displaystyle=x_{1,5}$
$\displaystyle Z_{1}:$ $\displaystyle=X_{1,5}$ $\displaystyle z_{2}:$
$\displaystyle=x_{2,5}$ $\displaystyle Z_{2}:$ $\displaystyle=X_{2,5}.$

Then,

$\displaystyle f_{1}$ $\displaystyle=t_{1}+\frac{1}{2}t_{2}t_{3}^{-1}$
$\displaystyle x_{3,6}$ $\displaystyle=t_{3}+\frac{3}{2}t_{4}t_{5}^{-1}$
$\displaystyle z_{1}$ $\displaystyle=f_{1}+\frac{1}{3}t_{3}^{2}t_{4}^{-1}$
$\displaystyle x_{1}$ $\displaystyle=x_{1,6}+\frac{1}{2}t_{5}t_{6}^{-1}$
$\displaystyle z_{2}$ $\displaystyle=t_{2}+\frac{2}{3}t_{3}^{3}t_{4}^{-1}$
$\displaystyle x_{3}$ $\displaystyle=x_{3,6}+t_{5}^{2}t_{6}^{-1}$
$\displaystyle x_{1,6}$
$\displaystyle=z_{1}+x_{3,6}t_{5}^{-1}-\frac{3}{4}t_{4}t_{5}^{-2}$
$\displaystyle x_{4}$ $\displaystyle=t_{4}+\frac{2}{3}t_{5}^{3}t_{6}^{-1}.$

These selected data of the PDDA of $\mathcal{A}_{\alpha,\beta}$ will be useful
in Subsections 5.3.1.

###  5.3 Poisson derivations of $\mathcal{A}_{\alpha,\beta}$

We are now ready to study the Poisson derivations of
$\mathcal{A}_{\alpha,\beta}.$ We will begin with the case where both $\alpha$
and $\beta$ are non-zero, and subsequently look at the case where either
$\alpha$ or $\beta$ is zero.

####  5.3.1 Poisson derivations of $\mathcal{A}_{\alpha,\beta}\
(\alpha,\beta\neq 0)$

Throughout this subsection, we assume that $\alpha$ and $\beta$ are non-zero.
Let Der${}_{P}(\mathcal{A})$ be the collection of all the Poisson
$\mathbb{K}$-derivations of $\mathcal{A}_{\alpha,\beta}$, and $\mathcal{D}\in$
Der${}_{P}(\mathcal{A}).$ Now, $\mathcal{D}$ extends uniquely to a Poisson
derivation of each of the algebras in (12) via localization. Hence,
$\mathcal{D}$ is a Poisson derivation of the Poisson torus
$\mathcal{R}_{3}=\mathbb{K}[t_{3}^{\pm 1},t_{4}^{\pm 1},t_{5}^{\pm
1},t_{6}^{\pm 1}].$ It follows from Corollary 2.7 that $\mathcal{D}$ can be
written as

$\mathcal{D}=\text{ham}_{x}+\rho,$

where $\rho$ is a scalar Poisson derivation of $\mathcal{R}_{3}$ defined as
$\rho(t_{i})=\lambda_{i}t_{i},\ i=3,4,5,6;$ with $\lambda_{i}\in
Z_{P}(\mathcal{R}_{3})=\mathbb{K}$, and
ham${}_{x}=\\{x,-\\}:\mathcal{R}_{3}\rightarrow\mathcal{R}_{3}$ with
$x\in\mathcal{R}_{3}$ (see Corollary 2.7).

We aim to describe $\mathcal{D}$ as a Poisson derivation of
$\mathcal{A}_{\alpha,\beta}.$ We do this in several steps. We first describe
$\mathcal{D}$ as a Poisson derivation of $\mathcal{R}_{4}.$

######  Lemma 5.7.

  * 1.

$x\in\mathcal{R}_{4}.$

  * 2.

$\lambda_{5}=\lambda_{4}+\lambda_{6},$
$\rho(f_{1})=-(\lambda_{3}+\lambda_{5})f_{1}$ and
$\rho(t_{2})=-\lambda_{5}t_{2}.$

  * 3.

Set $\lambda_{1}:=-(\lambda_{3}+\lambda_{5})$ and $\lambda_{2}:=-\lambda_{5}.$
Then,