$\mathcal{D}(x_{\kappa,4})=\mathrm{ham}_{x}(x_{\kappa,4})+\lambda_{\kappa}x_{\kappa,4}$
for all $\kappa\in\\{1,\ldots,6\\}.$

###### Proof.

1\. Observe that $\mathcal{Q}:=\mathbb{K}[t_{4}^{\pm 1},t_{5}^{\pm
1},t_{6}^{\pm 1}]$ is a subalgebra of both $\mathcal{R}_{3}$ and
$\mathcal{R}_{4}.$ Furthermore, $\mathcal{R}_{3}=\mathcal{R}_{4}[t_{3}^{-1}].$
One can easily verify that $z:=t_{4}t_{5}^{-1}t_{6}$ is a Poisson central
element of $\mathcal{Q}.$ Since $\mathcal{R}_{3}$ is a Poisson torus, it can
be presented as a free $\mathcal{Q}$-module with basis
$(t_{3}^{j})_{j\in\mathbb{Z}}.$ One can therefore write $x\in\mathcal{R}_{3}$
as $x=\sum_{j\in\mathbb{Z}}b_{j}t_{3}^{j},\ \text{where}\
b_{j}\in\mathcal{Q}.$ Decompose $x$ as $x=x_{-}+x_{+},$ where
$x_{+}:=\sum_{j\geq 0}b_{j}t_{3}^{j}$ and $x_{-}:=\sum_{j<0}b_{j}t_{3}^{j}.$
Clearly, $x_{+}\in\mathcal{R}_{4}.$ We now want to show that
$x_{-}\in\mathcal{R}_{4}.$ Write $x_{-}=\sum_{j=-1}^{-m}b_{j}t_{3}^{j}$ for
some $m\in\mathbb{N}_{>0}.$

Now, $\mathcal{D}(z)=$
ham${}_{x}(z)+\rho(z)=\mathrm{ham}_{x_{-}}(z)+\text{ham}_{x_{+}}(z)+(\lambda_{4}-\lambda_{5}+\lambda_{6})z\in\mathcal{R}_{4}.$
We have that
$\text{ham}_{x_{+}}(z)+(\lambda_{4}-\lambda_{5}+\lambda_{6})z\in\mathcal{R}_{4},$
hence $\text{ham}_{x_{-}}(z)\in\mathcal{R}_{4}.$ Note that
$\\{t_{3},z\\}=2zt_{3},$ and $\\{\gamma,z\\}=0$ for all $\gamma\in\mathcal{Q}$
since $z$ is Poisson central in $\mathcal{Q}.$ One can therefore express
$\text{ham}_{x_{-}}(z)$ as follows:

$\text{ham}_{x_{-}}(z)=\\{x_{-},z\\}=\sum_{j=-1}^{-m}b_{j}\\{t_{3}^{j},z\\}=\sum_{j=-1}^{-m}2jb_{j}zt_{3}^{j}\in\mathcal{R}_{4}.$

Let $n\in\mathbb{N}_{>0},$ and set

$\ell^{(n)}:=\underbrace{\\{\\{\ldots\\{}_{n-\text{times}}x_{-},\underbrace{z\\},z\\},\ldots,z\\}}_{n-\text{times}}\in\mathcal{R}_{4}.$

We claim that

$\ell^{(n)}=\sum_{j=-1}^{-m}(2j)^{n}z^{n}b_{j}t_{3}^{j},$

for all $n\in\mathbb{N}_{>0}.$ Observe that

$\ell^{(1)}=\text{ham}_{x_{-}}(z)=\sum_{j=-1}^{-m}2jb_{j}zt_{3}^{j},$

hence the result is true for $n=1.$ Suppose that the result is true for $n\geq
1$. Then,

$\ell^{(n+1)}=\\{\ell^{(n)},z\\}=\sum_{j=-1}^{-m}(2j)^{n}z^{n}b_{j}\\{t_{3}^{j},z\\}=\sum_{j=-1}^{-m}(2j)^{n+1}z^{n+1}b_{j}t_{3}^{j}$

as expected. By the principle of mathematical induction, the claim is proved.

Given that $\ell^{(n)}=\sum_{j=-1}^{-m}(2j)^{n}z^{n}b_{j}t_{3}^{j},$ it
follows that

$\mu_{n}=\sum_{j=-1}^{-m}(2j)^{n}b_{j}t_{3}^{j},\ \ \text{where}\ \
\mu_{n}:=\ell^{(n)}z^{-n}\in\mathcal{R}_{4}.$

The above equality can be written as a matrix equation:

$\begin{bmatrix}-2&-4&-6&\cdots&-2m\\\
(-2)^{2}&(-4)^{2}&(-6)^{2}&\cdots&(-2m)^{2}\\\
(-2)^{3}&(-4)^{3}&(-6)^{3}&\cdots&(-2m)^{3}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
(-2)^{m}&(-4)^{m}&(-6)^{m}&\cdots&(-2m)^{m}\\\
\end{bmatrix}\begin{bmatrix}b_{-1}t_{3}^{-1}\\\ b_{-2}t_{3}^{-2}\\\
b_{-3}t_{3}^{-3}\\\ \vdots\\\ b_{-m}t_{3}^{-m}\\\
\end{bmatrix}=\begin{bmatrix}\mu_{1}\\\ \mu_{2}\\\ \mu_{3}\\\ \vdots\\\
\mu_{m}\\\ \end{bmatrix}.$

One can observe that the coefficient matrix

$\begin{bmatrix}-2&-4&-6&\cdots&-2m\\\
(-2)^{2}&(-4)^{2}&(-6)^{2}&\cdots&(-2m)^{2}\\\
(-2)^{3}&(-4)^{3}&(-6)^{3}&\cdots&(-2m)^{3}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
(-2)^{m}&(-4)^{m}&(-6)^{m}&\cdots&(-2m)^{m}\\\ \end{bmatrix}$

is similar to a Vandermonde matrix (since the terms in each column form a
geometric sequence) which is well known to be invertible. This therefore
implies that each $b_{j}t_{3}^{j}$ is a linear combination of the
$\mu_{n}\in\mathcal{R}_{4}.$ As a result, $b_{j}t_{3}^{j}\in\mathcal{R}_{4}$
for all $j\in\\{-1,\ldots,-m\\}.$ Consequently,
$x_{-}=\sum_{j=-1}^{-m}b_{j}t_{3}^{j}\in\mathcal{R}_{4}$ as desired.

2\. Recall that $\rho(t_{\kappa})=\lambda_{\kappa}t_{\kappa}$ for all
$\kappa\in\\{3,4,5,6\\}$ and $\lambda_{\kappa}\in\mathbb{K}.$ From Remark 5.6,
we have that $f_{1}=t_{1}+\frac{1}{2}t_{2}t_{3}^{-1}.$ Again, recall from
Subsection 4.3.1 that $t_{1}=\alpha t_{3}^{-1}t_{5}^{-1}$ and $t_{2}=\beta
t_{4}^{-1}t_{6}^{-1}$ in $\mathcal{R}_{3}=\mathscr{P}_{\alpha,\beta}.$ As a
result, $f_{1}=\alpha t_{3}^{-1}t_{5}^{-1}+\frac{1}{2}\beta
t_{3}^{-1}t_{4}^{-1}t_{6}^{-1}.$ Therefore,

$\displaystyle\rho(f_{1})=$ $\displaystyle-(\lambda_{3}+\lambda_{5})\alpha
t_{3}^{-1}t_{5}^{-1}-\frac{1}{2}(\lambda_{3}+\lambda_{4}+\lambda_{6})\beta
t_{3}^{-1}t_{4}^{-1}t_{6}^{-1}.$ (15)

Also, $\rho(f_{1})\in\mathcal{R}_{4}$ implies that $\rho(f_{1})$ can be
written in terms of the basis $\mathfrak{P}_{4}$ of $\mathcal{R}_{4}$ (Prop.
5.2) as

$\displaystyle\rho(f_{1})=$
$\displaystyle\displaystyle\sum_{r>0}a_{r}f_{1}^{r}+\displaystyle\sum_{s\geq
0}b_{s}t_{3}^{s},$ (16)

where $a_{r}$ and $b_{s}$ belong to $\mathcal{Q}=\mathbb{K}[t_{4}^{\pm
1},t_{5}^{\pm 1},t_{6}^{\pm 1}].$ Note that

$\displaystyle f_{1}^{r}$ $\displaystyle=\left(\alpha
t_{3}^{-1}t_{5}^{-1}+\frac{1}{2}\beta
t_{3}^{-1}t_{4}^{-1}t_{6}^{-1}\right)^{r}=\sum_{i=0}^{r}{r\choose
i}(\alpha)^{i}(\beta/2)^{r-1}t_{3}^{-r}t_{4}^{i-r}t_{5}^{-i}t_{6}^{i-r}$
$\displaystyle=c_{r}t_{3}^{-r},$ (17)

where

$\displaystyle c_{r}=\displaystyle\sum_{i=0}^{r}{r\choose
i}(\alpha)^{i}(\beta/2)^{r-i}t_{4}^{i-r}t_{5}^{-i}t_{6}^{i-r}\in\mathcal{Q}\setminus\\{0\\}.$
(18)

Substitute (17) into (16) to obtain

$\displaystyle\rho(f_{1})=$
$\displaystyle\displaystyle\sum_{r>0}a_{r}c_{r}t_{3}^{-r}+\displaystyle\sum_{s\geq
0}b_{s}t_{3}^{s}.$ (19)

One can rewrite (15) as

$\rho(f_{1})=dt_{3}^{-1},$ (20)

where $d=-(\lambda_{5}+\lambda_{3})\alpha
t_{5}^{-1}-\frac{1}{2}(\lambda_{6}+\lambda_{4}+\lambda_{3})\beta
t_{4}^{-1}t_{6}^{-1}\in\mathcal{Q}.$ Comparing (19) to (20) shows that
$b_{s}=0$ for all $s\geq 0,$ and $a_{r}c_{r}=0$ for all $r\neq 1$. Therefore,
$\rho(f_{1})=a_{1}c_{1}t_{3}^{-1}.$ Moreover, from (18),
$c_{1}=\frac{1}{2}\beta t_{4}^{-1}t_{6}^{-1}+\alpha t_{5}^{-1}.$ Hence,

$\displaystyle\rho(f_{1})=a_{1}c_{1}t_{3}^{-1}$
$\displaystyle=a_{1}\left(\frac{1}{2}\beta t_{4}^{-1}t_{6}^{-1}+\alpha
t_{5}^{-1}\right)t_{3}^{-1}=a_{1}\alpha
t_{3}^{-1}t_{5}^{-1}+\frac{1}{2}a_{1}\beta t_{3}^{-1}t_{4}^{-1}t_{6}^{-1}.$
(21)

Comparing (21) to (15) reveals that
$a_{1}=-(\lambda_{5}+\lambda_{3})=-(\lambda_{6}+\lambda_{4}+\lambda_{3}).$
Consequently, $\lambda_{5}=\lambda_{6}+\lambda_{4}.$ Hence,
$\rho(f_{1})=-(\lambda_{5}+\lambda_{3})\alpha
t_{3}^{-1}t_{5}^{-1}-\frac{1}{2}(\lambda_{5}+\lambda_{3})\beta
t_{3}^{-1}t_{4}^{-1}t_{6}^{-1}=-(\lambda_{5}+\lambda_{3})f_{1}.$ Finally,
since $t_{2}=\beta t_{4}^{-1}t_{6}^{-1}$ in $\mathcal{R}_{4},$ it follows that

$\rho(t_{2})=-(\lambda_{6}+\lambda_{4})\beta
t_{4}^{-1}t_{6}^{-1}=-(\lambda_{6}+\lambda_{4})t_{2}=-\lambda_{5}t_{2}.$

3\. Set $\lambda_{1}:=-(\lambda_{3}+\lambda_{5})$ and
$\lambda_{2}:=-\lambda_{5}.$ Remember that $f_{1}=x_{1,4}$ and $t_{i}=x_{i,4}$
($2\leq i\leq 6$). It follows from points (1) and (2) that
$\mathcal{D}(x_{\kappa,4})=\text{ham}_{x}(x_{\kappa,4})+\rho(x_{\kappa,4})=\text{ham}_{x}(x_{\kappa,4})+\lambda_{\kappa}x_{\kappa,4}$
for all $\kappa\in\\{1,\ldots,6\\}.$

In conclusion, $\mathcal{D}=\text{ham}_{x}+\rho,$ with $x\in\mathcal{R}_{4}.$
∎

We now proceed to describe $\mathcal{D}$ as a Poisson derivation of
$\mathcal{R}_{5}.$

######  Lemma 5.8.

  * 1.

$x\in\mathcal{R}_{5}.$

  * 2.

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

  * 3.

Set $\lambda_{1}:=-(\lambda_{3}+\lambda_{5})$ and $\lambda_{2}:=-\lambda_{5},$
then
$\mathcal{D}(x_{\kappa,5})=\mathrm{ham}_{x}(x_{\kappa,5})+\lambda_{\kappa}x_{\kappa,5}$
for all $\kappa\in\\{1,\ldots,6\\}.$

###### Proof.

In this proof, we denote
$\underline{\upsilon}:=(i,j,k,l)\in\mathbb{N}\times\mathbb{Z}^{3}.$

1\. We already know that $x\in\mathcal{R}_{4}=\mathcal{R}_{5}[t_{4}^{-1}].$
Given the basis $\mathfrak{P}_{5}$ of $\mathcal{R}_{5}$ (Prop. 5.3), $x$ can
be written as $x=\displaystyle\sum_{(\xi,\underline{\upsilon})\in
I}a_{(\xi,\underline{\upsilon})}z_{1}^{i}t_{3}^{\xi}t_{4}^{j}t_{5}^{k}t_{6}^{l},$
where $I$ is a finite subset of
$\\{0,1,2\\}\times\mathbb{N}\times\mathbb{Z}^{3}$ and the
$a_{(\xi,\underline{\upsilon})}$ are scalars. Write $x=x_{-}+x_{+},$ where

$x_{+}=\sum_{\begin{subarray}{c}(\xi,\underline{\upsilon})\in I\\\ j\geq
0\end{subarray}}a_{(\xi,{\underline{\upsilon})}}z_{1}^{i}t_{3}^{\xi}t_{4}^{j}t_{5}^{k}t_{6}^{l}\
\ \text{and}\ \
x_{-}=\displaystyle\sum_{\begin{subarray}{c}(\xi,\underline{\upsilon})\in I\\\
j<0\end{subarray}}a_{(\xi,\underline{\upsilon})}z_{1}^{i}t_{3}^{\xi}t_{4}^{j}t_{5}^{k}t_{6}^{l}.$

Suppose that $x_{-}\neq 0.$ Then, there exists a minimum $j_{0}<0$ such that
$a_{(\xi,i,j_{0},k,l)}\neq 0$ for some $(\xi,i,j_{0},k,l)\in I$ and
$a_{(\xi,i,j,k,l)}=0$ for all $(\xi,i,j_{0},k,l)\in I$ with $j<j_{0}.$ Given
this assumption, write

$\displaystyle
x_{-}=\displaystyle\sum_{\begin{subarray}{c}(\xi,\underline{\upsilon})\in I\\\
j_{0}\leq
j\leq-1\end{subarray}}a_{(\xi,\underline{\upsilon})}z_{1}^{i}t_{3}^{\xi}t_{4}^{j}t_{5}^{k}t_{6}^{l}.$

Contrary to our assumption, we aim to show that $x_{-}=0.$ Let $s=3,6.$ Then,
$\mathcal{D}(t_{s})=\text{ham}_{x_{+}}(t_{s})+\text{ham}_{x_{-}}(t_{s})+\rho(t_{s})\in\mathcal{R}_{5}$
for each $s=3,6.$ This implies that
$\text{ham}_{x_{-}}(t_{s})\in\mathcal{R}_{5},$ since
$\text{ham}_{x_{+}}(t_{s})+\rho(t_{s})=\text{ham}_{x_{+}}(t_{s})+\lambda_{s}t_{s}\in\mathcal{R}_{5}.$
Set $\underline{w}:=(i,j,k,l)\in\mathbb{N}^{2}\times\mathbb{Z}^{2}.$ One can
therefore write $\text{ham}_{x_{-}}(t_{s})\in\mathcal{R}_{5}$ in terms of the
basis $\mathfrak{P}_{5}$ of $\mathcal{R}_{5}$ as:

$\displaystyle\text{ham}_{x_{-}}(t_{s})$
$\displaystyle=\sum_{(\xi,\underline{w})\in
J}b_{(\xi,{\underline{w}})}z_{1}^{i}t_{3}^{\xi}t_{4}^{j}t_{5}^{k}t_{6}^{l},$
(22)

where $J$ is a finite subset of
$\\{0,1,2\\}\times\mathbb{N}^{2}\times\mathbb{Z}^{2}$ and the
$b_{(\xi,\underline{w})}$ are scalars.

When $s=6,$ then using Remark 2.1(2), one can also express
$\text{ham}_{x_{-}}(t_{6})$ as:

$\displaystyle\text{ham}_{x_{-}}(t_{6})$
$\displaystyle=\sum_{\begin{subarray}{c}(\xi,\underline{\upsilon})\in I\\\
j_{0}\leq
j\leq-1\end{subarray}}3(k+j-i)a_{(\xi,\underline{\upsilon})}z_{1}^{i}t_{3}^{\xi}t_{4}^{j}t_{5}^{k}t_{6}^{l+1}.$

Comparing this expression for $\text{ham}_{x_{-}}(t_{6})$ to (22) (when
$s=6$), we have that

$\sum_{\begin{subarray}{c}(\xi,\underline{\upsilon})\in I\\\ j_{0}\leq
j\leq-1\end{subarray}}3(k+j-i)a_{(\xi,\underline{\upsilon})}z_{1}^{i}t_{3}^{\xi}t_{4}^{j}t_{5}^{k}t_{6}^{l+1}=\sum_{(\xi,\underline{w})\in
J}b_{(\xi,{\underline{w}})}z_{1}^{i}t_{3}^{\xi}t_{4}^{j}t_{5}^{k}t_{6}^{l}.$

As $\mathfrak{P}_{5}$ is a basis for $\mathcal{R}_{5}$ (Prop. 5.3), we deduce
from Corollary 5.4 that
$\left(z_{1}^{i}t_{3}^{\xi}t_{4}^{j}t_{5}^{k}t_{6}^{l}\right)_{(i\in\mathbb{N};j,k,l\in\mathbb{Z};\xi\in\\{0,1,2\\})}$
is a basis for $\mathcal{R}_{5}[t_{4}^{-1}].$ Now, at $j=j_{0},$ denote
$\underline{\upsilon}=(i,j,k,l)$ by $\underline{\upsilon}_{0}:=(i,j_{0},k,l).$
Since $\underline{v}_{0}\in\mathbb{N}\times\mathbb{Z}^{3}$ (with $j_{0}<0$)
and $\underline{w}=(i,j,k,l)\in\mathbb{N}^{2}\times\mathbb{Z}^{2}$ (with
$j\geq 0$), it follows from the above equality that, at
$\underline{\upsilon}_{0},$ we must have

$\displaystyle k=i-j_{0},$ (23)

for some $(\xi,\underline{v}_{0})\in I.$

Similarly, when $s=3,$ then using Remark 2.1(2), one can also express
$\text{ham}_{x_{-}}(t_{3})$ as:

$\displaystyle\text{ham}_{x_{-}}(t_{3})=$
$\displaystyle-\sum\left[\frac{3}{2}\beta(3i-k-3j_{0})a_{2,i,j_{0},k,l+1}+2(i+1)\alpha
a_{(0,i+1,j_{0},k+1,l)}\right]z_{1}^{i}t_{4}^{j_{0}}t_{5}^{k}t_{6}^{l}$
$\displaystyle+\sum\left[(3i-k-3j_{0})a_{(0,i,j_{0},k,l)}-2(i+1)\alpha
a_{(1,i+1,j_{0},k+1,l)}\right]z_{1}^{i}t_{3}t_{4}^{j_{0}}t_{5}^{k}t_{6}^{l}$
$\displaystyle+\sum\left[(3i-k-3j_{0})a_{(1,i,j_{0},k,l)}-2(i+1)\alpha
a_{(2,i+1,j_{0},k+1,l)}\right]z_{1}^{i}t_{3}^{2}t_{4}^{j_{0}}t_{5}^{k}t_{6}^{l}+\mathcal{K},$

where
$\mathcal{K}\in\text{Span}\left(\mathfrak{P}_{5}\setminus\\{z_{1}^{i}t_{3}^{\xi}t_{4}^{j_{0}}t_{5}^{k}t_{6}^{l}\mid(\xi,i,j_{0},k,l)\in\\{0,1,2\\}\times\mathbb{N}\times\mathbb{Z}^{3}\\}\right)$
(note that one will need the following two expressions
$z_{2}=2(z_{1}t_{3}-\alpha t_{5}^{-1})\ \ \text{and}\ \
t_{3}^{3}=3z_{1}t_{3}t_{4}-3\alpha
t_{4}t_{5}^{-1}-\dfrac{3\beta}{2}t_{6}^{-1}$ from (13) and (14) to express
some of the monomials in terms of the basis $\mathfrak{P}_{5}$ of
$\mathcal{R}_{5}$). Comparing this expression for $\text{ham}_{x_{-}}(t_{3})$
to (22) (when $s=3$) reveals that

$\displaystyle\sum_{(\xi,\underline{w})\in J}$ $\displaystyle
b_{(\xi,{\underline{w}})}z_{1}^{i}t_{3}^{\xi}t_{4}^{j}t_{5}^{k}t_{6}^{l}=$
$\displaystyle-\sum\left[\frac{3}{2}\beta(3i-k-3j_{0})a_{(2,i,j_{0},k,l+1}+2(i+1)\alpha
a_{(0,i+1,j_{0},k+1,l)}\right]z_{1}^{i}t_{4}^{j_{0}}t_{5}^{k}t_{6}^{l}$
$\displaystyle+\sum\left[(3i-k-3j_{0})a_{(0,i,j_{0},k,l)}-2(i+1)\alpha
a_{(1,i+1,j_{0},k+1,l)}\right]z_{1}^{i}t_{3}t_{4}^{j_{0}}t_{5}^{k}t_{6}^{l}$
$\displaystyle+\sum\left[(3i-k-3j_{0})a_{(1,i,j_{0},k,l)}-2(i+1)\alpha
a_{(2,i+1,j_{0},k+1,l)}\right]z_{1}^{i}t_{3}^{2}t_{4}^{j_{0}}t_{5}^{k}t_{6}^{l}+\mathcal{K}.$

We have already established that
$\left(z_{1}^{i}t_{3}^{\xi}t_{4}^{j}t_{5}^{k}t_{6}^{l}\right)_{(i\in\mathbb{N};j,k,l\in\mathbb{Z};\xi\in\\{0,1,2\\})}$
is a basis for $\mathcal{R}_{5}[t_{4}^{-1}].$ Since
$\underline{v}_{0}=(i,j_{0},k,l)\in\mathbb{N}\times\mathbb{Z}^{3}$ (with
$j_{0}<0$) and $\underline{w}=(i,j,k,l)\in\mathbb{N}^{2}\times\mathbb{Z}^{2}$
(with $j\geq 0$), it follows from the above equality that, at
$\underline{\upsilon}_{0},$ we must have

$\displaystyle\frac{3}{2}\beta(3i-k-3j_{0})a_{(2,i,j_{0},k,l+1)}+2(i+1)\alpha
a_{(0,i+1,j_{0},k+1,l)}$ $\displaystyle=0,$ (24)
$\displaystyle(3i-k-3j_{0})a_{(0,i,j_{0},k,l)}-2(i+1)\alpha
a_{(1,i+1,j_{0},k+1,l)}$ $\displaystyle=0,$ (25)
$\displaystyle(3i-k-3j_{0})a_{(1,i,j_{0},k,l)}-2(i+1)\alpha
a_{(2,i+1,j_{0},k+1,l)}$ $\displaystyle=0.$ (26)

Suppose that there exists $(\xi,i,j_{0},k,l)\in I$ such that $3i-k-3j_{0}=0.$
Then,

$\displaystyle k=3(i-j_{0}).$ (27)

Comparing (27) to (23) clearly shows that $i-j_{0}=0$ which implies that
$i=j_{0}<0,$ a contradiction (note that $i\geq 0$). Therefore,
$3i-k-3j_{0}\neq 0$ for all $(\xi,i,j,k)\in I.$

Now, observe that if there exists $\xi\in\\{0,1,2\\}$ such that
$a_{(\xi,i,j_{0},k,l)}=0$ for all
$(i,j_{0},k,l)\in\mathbb{N}\times\mathbb{Z}^{3}$, then one can easily deduce
from equations (24), (25) and (26) that $a_{(\xi,i,j_{0},k,l)}=0$ for all
$(\xi,i,j_{0},k,l)\in I.$ This contradicts our initial assumption. Therefore,
for each $\xi\in\\{0,1,2\\},$ there exists some
$(i,j_{0},k,l)\in\mathbb{N}\times\mathbb{Z}^{3}$ such that
$a_{(\xi,i,j_{0},k,l)}\neq 0.$ Without loss of generality, let $(u,j_{0},v,w)$
be the greatest element in the lexicographic order on
$\mathbb{N}\times\mathbb{Z}^{3}$ such that $a_{(0,u,j_{0},v,w)}\neq 0$ and
$a_{(0,i,j_{0},k,l)}=0$ for all $i>u.$

From (25), at $(i,j_{0},k,l)=(u,j_{0},v,w)$, we have:

$\displaystyle(3u-v-3j_{0})a_{(0,u,j_{0},v,w)}-2(u+1)\alpha
a_{(1,u+1,j_{0},v+1,w)}$ $\displaystyle=0.$

From (26), at $(i,j_{0},k,l)=(u+1,j_{0},v+1,w)$, we have:

$\displaystyle(3u-v-3j_{0})a_{(1,u+1,j_{0},v+1,w)}-2(u+1)\alpha
a_{(2,u+2,j_{0},v+2,w)}$ $\displaystyle=0.$

Finally, from (24), at $(i,j_{0},k,l)=(u+2,j_{0},v+2,w-1),$ we have:

$\displaystyle\frac{3}{2}\beta(3u-v-3j_{0})a_{(2,u+2,j_{0},v+2,w)}+2(u+1)\alpha
a_{(0,u+3,j_{0},v+3,w-1)}$ $\displaystyle=0.$

Since $3i-k-3j_{0}\neq 0$ for all $(i,j_{0},k,l)\in I$ and $u+3>u,$ it follows
from the above three displayed equations (beginning from the last one) that

$a_{(0,u+3,j_{0},v+3,w-1)}=0\Rightarrow a_{(2,u+2,j_{0},v+2,w)}=0\Rightarrow
a_{(1,u+1,j_{0},v+1,w)}=0\Rightarrow a_{(0,u,j_{0},v,w)}=0,$

a contradiction! Hence, $a_{(0,i,j_{0},k,l)}=0$ for all
$(i,j_{0},k,l)\in\mathbb{N}\times\mathbb{Z}^{3}.$ From (24), (25) and (26),
one can then conclude that $a_{(\xi,i,j_{0},k,l)}=0$ for all
$(\xi,i,j_{0},k,l)\in I.$ This is a contradiction to our assumption, hence
$x_{-}=0.$ Consequently, $x=x_{+}\in\mathcal{R}_{5}$ as desired.

2\. It follows from Remark 5.6 that
$z_{2}=t_{2}+\frac{2}{3}t_{3}^{3}t_{4}^{-1}.$ Since
$\rho(t_{\kappa})=\lambda_{\kappa}t_{\kappa},\ \kappa\in\\{2,3,4,5,6\\},$ with
$\lambda_{2}=-\lambda_{5}$ (see Lemma 5.7), it follows that

$\displaystyle\rho(z_{2})=$
$\displaystyle-\lambda_{5}t_{2}+\frac{2}{3}(3\lambda_{3}-\lambda_{4})t_{3}^{3}t_{4}^{-1}=-\lambda_{5}z_{2}+\frac{2}{3}(3\lambda_{3}-\lambda_{4}+\lambda_{5})t_{3}^{3}t_{4}^{-1}.$

Furthermore,

$\displaystyle\mathcal{D}(z_{2})$
$\displaystyle=\text{ham}_{x}(z_{2})+\rho(z_{2})=\text{ham}_{x}(z_{2})-\lambda_{5}z_{2}+\frac{2}{3}(3\lambda_{3}-\lambda_{4}+\lambda_{5})t_{3}^{3}t_{4}^{-1}\in\mathcal{R}_{5}.$

We have that
$(3\lambda_{3}-\lambda_{4}+\lambda_{5})t_{3}^{3}t_{4}^{-1}\in\mathcal{R}_{5},$
since $\text{ham}_{x}(z_{2})-\lambda_{5}z_{2}\in\mathcal{R}_{5}.$ This implies
that
$(3\lambda_{3}-\lambda_{4}+\lambda_{5})t_{3}^{3}\in\mathcal{R}_{5}{t}_{4}.$
Set $w:=3\lambda_{3}-\lambda_{4}+\lambda_{5}.$ Suppose that $w\neq 0.$ From
(14), we have:

$t_{3}^{3}=3z_{1}t_{3}t_{4}-\dfrac{3}{2}\beta t_{6}^{-1}-3\alpha
t_{4}t_{5}^{-1}.$

It follows that

$wt_{3}^{3}=3wz_{1}t_{3}t_{4}-3w\alpha t_{4}t_{5}^{-1}-\frac{3}{2}w\beta
t_{6}^{-1}\in\mathcal{R}_{5}t_{4}.$

Since $t_{3}^{3},\ t_{4}t_{5}^{-1}$ and $z_{1}t_{3}t_{4}$ are all elements of
$\mathcal{R}_{5}t_{4},$ this implies that $t_{6}^{-1}\in\mathcal{R}_{5}t_{4}.$
Hence, $1\in\mathcal{R}_{5}t_{4}t_{6},$ a contradiction (see the basis
$\mathfrak{P}_{5}$ of $\mathcal{R}_{5}$ (Prop. 5.3)). Therefore,
$w=3\lambda_{3}-\lambda_{4}+\lambda_{5}=0,$ and so
$\lambda_{4}=3\lambda_{3}+\lambda_{5}.$ This further implies that
$\rho(z_{2})=-\lambda_{5}z_{2}$ as desired.

Again, from Lemma 5.7, we have that
$\rho(f_{1})=-(\lambda_{3}+\lambda_{5})f_{1}.$ Recall from Remark 5.6 that
$z_{1}=f_{1}+\frac{1}{3}t_{3}^{2}t_{4}^{-1}.$ It follows that

$\displaystyle\rho(z_{1})=$
$\displaystyle-(\lambda_{3}+\lambda_{5})f_{1}+\frac{1}{3}(2\lambda_{3}-\lambda_{4})t_{3}^{2}t_{4}^{-1}=-(\lambda_{3}+\lambda_{5})z_{1}+\frac{1}{3}(3\lambda_{3}-\lambda_{4}+\lambda_{5})t_{3}^{2}t_{4}^{-1}$
$\displaystyle=$
$\displaystyle-(\lambda_{3}+\lambda_{5})z_{1}+\frac{1}{3}(3\lambda_{3}-(3\lambda_{3}+\lambda_{5})+\lambda_{5})t_{3}^{2}t_{4}^{-1}=-(\lambda_{3}+\lambda_{5})z_{1}.$

Finally, we know that $\rho(t_{6})=\lambda_{6}t_{6}.$ From the relation (14),
we have:

$t_{3}^{3}=3z_{1}t_{3}t_{4}-3\alpha
t_{4}t_{5}^{-1}-\dfrac{3\beta}{2}t_{6}^{-1}.$

This implies that

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

Apply $\rho$ to this relation to obtain

$-\lambda_{6}t_{6}^{-1}=3\lambda_{3}\left(\frac{2}{3\beta}\left(3z_{1}t_{3}t_{4}-3\alpha
t_{4}t_{5}^{-1}-t_{3}^{3}\right)\right).$

Clearly, $\lambda_{6}=-3\lambda_{3}$ as desired.

3\. Set $\lambda_{1}:=-(\lambda_{3}+\lambda_{5})$ and
$\lambda_{2}:=-\lambda_{5}.$ Remember that $z_{1}=x_{1,5}$ and
$z_{2}=x_{2,5}.$ It follows from points (1) and (2) that
$\mathcal{D}(x_{\kappa,5})=\mathrm{ham}_{x}(x_{\kappa,5})+\rho(x_{\kappa,5})=\mathrm{ham}_{x}(x_{\kappa,5})+\lambda_{\kappa}x_{\kappa,5}$
for all $\kappa\in\\{1,\ldots,6\\}.$

In conclusion, $\mathcal{D}=\text{ham}_{x}+\rho$ with $x\in\mathcal{R}_{5}.$ ∎

We are now ready to describe $\mathcal{D}$ as a Poisson derivation of
$\mathcal{A}_{\alpha,\beta}.$

######  Lemma 5.9.

  * 1.

$x\in\mathcal{A}_{\alpha,\beta}.$

  * 2.

$\rho(x_{\kappa})=0$ for all $\kappa\in\\{1,\ldots,6\\}.$

  * 3.

$\mathcal{D}=\mathrm{ham}_{x}.$

###### Proof.

In this proof, we denote
$\underline{\upsilon}:=(i,j,k,l)\in\mathbb{N}^{2}\times\mathbb{Z}^{2}.$ Also,
recall from the PDDA of $\mathcal{A}_{\alpha,\beta}$ at the beginning of this
section that $t_{5}=x_{5}$ and $t_{6}=x_{6}.$

1\. Given the basis $\mathfrak{P}$ of $\mathcal{A}_{\alpha,\beta}$ (Prop.
4.16), one can write
$x\in\mathcal{R}_{5}=\mathcal{A}_{\alpha,\beta}[t_{5}^{-1},t_{6}^{-1}]$ as

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

where $I$ is a finite subset of
$\\{0,1\\}^{2}\times\mathbb{N}^{2}\times\mathbb{Z}^{2}$ and
$a_{(\epsilon_{1},\epsilon_{2},\underline{\upsilon})}$ are scalars. Write
$x=x_{-}+x_{+},$ where

$x_{+}=\sum_{\begin{subarray}{c}(\epsilon_{1},\epsilon_{2},\underline{\upsilon})\in
I\\\ k,\ l\geq
0\end{subarray}}a_{(\epsilon_{1},\epsilon_{2},\underline{\upsilon})}x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k}t_{6}^{l},$

and

$x_{-}=\displaystyle\sum_{\begin{subarray}{c}(\epsilon_{1},\epsilon_{2},\underline{\upsilon})\in
I\\\ k<0\ \text{or}\
l<0\end{subarray}}a_{(\epsilon_{1},\epsilon_{2},\underline{\upsilon})}x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k}t_{6}^{l}.$

Suppose that $x_{-}\neq 0.$ Then, there exists a minimum negative integer
$k_{0}$ or $l_{0}$ such that $a_{(\epsilon_{1},\epsilon_{2},i,j,k_{0},l)}\neq
0$ or $a_{(\epsilon_{1},\epsilon_{2},i,j,k,l_{0})}\neq 0$ for some
$(\epsilon_{1},\epsilon_{2},i,j,k_{0},l),(\epsilon_{1},\epsilon_{2},i,j,k,l_{0})\in
I,$ and $a_{(\epsilon_{1},\epsilon_{2},i,j,k,l)}=0$ whenever $k<k_{0}$ or
$l<l_{0}.$ Write

$x_{-}=\displaystyle\sum_{\begin{subarray}{c}(\epsilon_{1},\epsilon_{2},\underline{\upsilon})\in
I\\\ k_{0}\leq k\leq-1\ \text{or}\ l_{0}\leq
l\leq-1\end{subarray}}a_{(\epsilon_{1},\epsilon_{2},\underline{\upsilon})}x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k}t_{6}^{l}.$

Now
$\mathcal{D}(x_{3})=\text{ham}_{x_{+}}(x_{3})+\text{ham}_{x_{-}}(x_{3})+\rho(x_{3})\in\mathcal{A}_{\alpha,\beta}.$
From Remark 5.6, we have that $x_{3}=x_{3,6}+t_{5}^{2}t_{6}^{-1}$ and
$x_{3,6}=t_{3}+\frac{3}{2}t_{4}t_{5}^{-1}.$ Putting these two together gives

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

Again, from Remark 5.6, we also have that
$t_{4}=x_{4}-\frac{2}{3}t_{5}^{3}t_{6}^{-1}.$ Note that
$\rho(t_{\kappa})=\lambda_{\kappa}t_{\kappa},\ \kappa=3,4,5,6.$

Now,

$\displaystyle\rho(x_{3})$
$\displaystyle=\lambda_{3}t_{3}+\frac{3}{2}(\lambda_{4}-\lambda_{5})t_{4}t_{5}^{-1}+(2\lambda_{5}-\lambda_{6})t_{5}^{2}t_{6}^{-1}$
$\displaystyle=\lambda_{3}\left(x_{3,6}-\frac{3}{2}t_{4}t_{5}^{-1}\right)+\frac{3}{2}(\lambda_{4}-\lambda_{5})t_{4}t_{5}^{-1}+(2\lambda_{5}-\lambda_{6})t_{5}^{2}t_{6}^{-1}$
$\displaystyle=\lambda_{3}x_{3,6}-\frac{3}{2}(\lambda_{3}-\lambda_{4}+\lambda_{5})t_{4}t_{5}^{-1}+(2\lambda_{5}-\lambda_{6})t_{5}^{2}t_{6}^{-1}$
$\displaystyle=\lambda_{3}(x_{3}-t_{5}^{2}t_{6}^{-1})-\frac{3}{2}(\lambda_{3}-\lambda_{4}+\lambda_{5})\left(x_{4}-\frac{2}{3}t_{5}^{3}t_{6}^{-1}\right)t_{5}^{-1}+(2\lambda_{5}-\lambda_{6})t_{5}^{2}t_{6}^{-1}$
$\displaystyle=\lambda_{3}x_{3}+\alpha_{1}x_{4}t_{5}^{-1}+\alpha_{2}t_{5}^{2}t_{6}^{-1},$
(28)

where $\alpha_{1}=\frac{3}{2}(\lambda_{4}-\lambda_{3}-\lambda_{5})$ and
$\alpha_{2}=(3\lambda_{5}-\lambda_{4}-\lambda_{6}).$ Therefore,
$\mathcal{D}(x_{3})=\text{ham}_{x_{+}}(x_{3})+\text{ham}_{x_{-}}(x_{3})+\lambda_{3}x_{3}+\alpha_{1}x_{4}t_{5}^{-1}+\alpha_{2}t_{5}^{2}t_{6}^{-1}\in\mathcal{A}_{\alpha,\beta}.$
It follows that
$\mathcal{D}(x_{3})t_{5}t_{6}=\text{ham}_{x_{+}}(x_{3})t_{5}t_{6}+\text{ham}_{x_{-}}(x_{3})t_{5}t_{6}+\lambda_{3}x_{3}t_{5}t_{6}+\alpha_{1}x_{4}t_{6}+\alpha_{2}t_{5}^{3}\in\mathcal{A}_{\alpha,\beta}.$
Hence, $\text{ham}_{x_{-}}(x_{3})t_{5}t_{6}\in\mathcal{A}_{\alpha,\beta},$
since
$\text{ham}_{x_{+}}(x_{3})t_{5}t_{6}+\lambda_{3}x_{3}t_{5}t_{6}+\alpha_{1}x_{4}t_{6}+\alpha_{2}t_{5}^{3}\in\mathcal{A}_{\alpha,\beta}.$

One can also verify that

$\displaystyle\text{ham}_{x_{-}}(x_{3})t_{5}t_{6}=\sum_{(\epsilon_{1},\epsilon_{2},\underline{v})\in
I}$ $\displaystyle
a_{(\epsilon_{1},\epsilon_{2},\underline{v})}\left((i+3j-3\epsilon_{2}-k)x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}+1}x_{4}^{\epsilon_{2}}t_{5}^{k+1}t_{6}^{l+1}\right.$
$\displaystyle-3kx_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}+1}t_{5}^{k}t_{6}^{l+1}+ix_{1}^{i-1}x_{2}^{j+1}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k+1}t_{6}^{l+1}$
$\displaystyle\left.-6lx_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k+3}t_{6}^{l}\right).$
(29)

Assume that there exists $l<0$ such that
$a_{(\epsilon_{1},\epsilon_{2},i,j,k,l)}\neq 0.$ It follows from our initial
assumption that $a_{(\epsilon_{1},\epsilon_{2},i,j,k,l_{0})}\neq 0.$ Now, at
$l=l_{0},$ denote $\underline{\upsilon}=(i,j,k,l)$ by
$\underline{\upsilon}_{0}:=(i,j,k,l_{0}).$ From (29), we have that

$\displaystyle\text{ham}_{x_{-}}(x_{3})t_{5}t_{6}=$
$\displaystyle-\sum_{(\epsilon_{1},\epsilon_{2},\underline{\upsilon}_{0})\in
I}6l_{0}a_{(\epsilon_{1},\epsilon_{2},\underline{\upsilon}_{0})}x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k+3}t_{6}^{l_{0}}+\mathcal{J}_{1},$

where
$\mathcal{J}_{1}\in\text{Span}\left(\mathfrak{P}\setminus\\{x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k}t_{6}^{l_{0}}\mid\epsilon_{1},\epsilon_{2}\in\\{0,1\\},\
k\in\mathbb{Z}\ \text{and}\ i,j\in\mathbb{N}\\}\right).$

Set $\underline{w}:=(i,j,k,l)\in\mathbb{N}^{4}.$ One can also write
$\text{ham}_{x_{-}}(x_{3})t_{5}t_{6}\in\mathcal{A}_{\alpha,\beta}$ in terms of
the basis $\mathfrak{P}$ of $\mathcal{A}_{\alpha,\beta}$ (Prop. 4.16) as:

$\displaystyle\text{ham}_{x_{-}}(x_{3})t_{5}t_{6}=$
$\displaystyle\sum_{(\epsilon_{1},\epsilon_{2},\underline{w})\in
J}b_{(\epsilon_{1},\epsilon_{2},\underline{w})}x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k}t_{6}^{l},$
(30)

where $J$ is a finite subset of $\\{0,1\\}^{2}\times\mathbb{N}^{4}$ and
$b_{(\epsilon_{1},\epsilon_{2},\underline{w})}\in\mathbb{K}.$ It follows that

$\sum_{(\epsilon_{1},\epsilon_{2},\underline{w})\in
J}b_{(\epsilon_{1},\epsilon_{2},\underline{w})}x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k}t_{6}^{l}=-\sum_{(\epsilon_{1},\epsilon_{2},\underline{\upsilon}_{0})\in
I}6l_{0}a_{(\epsilon_{1},\epsilon_{2},\underline{\upsilon}_{0})}x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k+3}t_{6}^{l_{0}}+\mathcal{J}_{1}.$

As $\mathfrak{P}$ is a basis for $\mathcal{A}_{\alpha,\beta},$ we deduce from
Corollary 4.17 that
$\left(x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k}t_{6}^{l}\right)_{((\epsilon_{1},\epsilon_{2},\underline{v})\in\\{0,1\\}^{2}\times\mathbb{N}^{2}\times\mathbb{Z}^{2})}$
is a basis for $\mathcal{A}_{\alpha,\beta}[t_{5}^{-1},t_{6}^{-1}].$ Since
$\underline{v}_{0}=(i,j,k,l_{0})\in\mathbb{N}^{2}\times\mathbb{Z}^{2}$ (with
$l_{0}<0$) and $\underline{w}=(i,j,k,l)\in\mathbb{N}^{4}$ (with $l\geq 0$) in
the above equality, we must have

$6l_{0}a_{(\epsilon_{1},\epsilon_{2},\underline{\upsilon}_{0})}=0.$

Note that $l_{0}\neq 0,$ it follows that
$a_{(\epsilon_{1},\epsilon_{2},\underline{\upsilon}_{0})}=a_{(\epsilon_{1},\epsilon_{2},i,j,k,l_{0})}$
are all zero, a contradiction! Therefore, $l\geq 0$ (i.e. there is no negative
exponent of $t_{6}$).

Given that $l\geq 0,$ it follows from our initial assumption that there exists
$k=k_{0}<0$ such that $a_{(\epsilon_{1},\epsilon_{2},i,j,k_{0},l)}\neq 0.$ The
rest of the proof will show that this assumption cannot also hold.

Set
$\underline{\upsilon}_{0}:=(i,j,k_{0},l)\in\mathbb{N}^{2}\times\mathbb{Z}\times\mathbb{N}.$
From (29), we have that

$\displaystyle\text{ham}_{x_{-}}(x_{3})t_{5}t_{6}=$
$\displaystyle-\displaystyle\sum_{(\epsilon_{1},\epsilon_{2},\underline{\upsilon}_{0})\in
I}3ka_{(\epsilon_{1},\epsilon_{2},\underline{\upsilon}_{0})}x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}+1}t_{5}^{k_{0}}t_{6}^{l+1}+V,$

where
$V\in\mathcal{J}_{2}:=\text{Span}\left(\mathfrak{P}\setminus\\{x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k_{0}}t_{6}^{l}\mid\epsilon_{1},\epsilon_{2}\in\\{0,1\\}\
\text{and}\ i,j,l\in\mathbb{N}\\}\right).$ It follows that

$\displaystyle\text{ham}_{x_{-}}(x_{3})t_{5}t_{6}=$
$\displaystyle-\sum_{(\epsilon_{1},\epsilon_{2},\underline{v})\in I}$
$\displaystyle
3k_{0}a_{(0,0,\underline{v})}x_{1}^{i}x_{2}^{j}x_{4}t_{5}^{k_{0}}t_{6}^{l+1}-\sum_{(\epsilon_{1},\epsilon_{2},\underline{v})\in
I}3k_{0}a_{(1,0,\underline{v})}x_{1}^{i}x_{2}^{j}x_{3}x_{4}t_{5}^{k_{0}}t_{6}^{l+1}$
$\displaystyle-\sum_{(\epsilon_{1},\epsilon_{2},\underline{v})\in I}$
$\displaystyle
3k_{0}a_{(0,1,\underline{v})}x_{1}^{i}x_{2}^{j}x_{4}^{2}t_{5}^{k_{0}}t_{6}^{l+1}-\sum_{(\epsilon_{1},\epsilon_{2},\underline{v})\in
I}3k_{0}a_{(1,1,\underline{v})}x_{1}^{i}x_{2}^{j}x_{3}x_{4}^{2}t_{5}^{k_{0}}t_{6}^{l+1}+V.$
(31)

Write the relations in Lemma 4.13(2),(4) as follows:

$\displaystyle x_{4}^{2}=$
$\displaystyle\frac{2}{3}\beta-\frac{2}{3}x_{2}x_{4}x_{6}+\frac{8}{9}\alpha
x_{3}x_{6}+\frac{4}{3}x_{1}x_{3}x_{4}x_{6}+L_{1},$ (32) $\displaystyle
x_{3}x_{4}^{2}=$ $\displaystyle\frac{2}{3}\beta
x_{3}-\frac{2}{3}x_{2}x_{3}x_{4}x_{6}+\frac{16}{9}\alpha^{2}x_{6}+\frac{16}{3}\alpha
x_{1}x_{4}x_{6}+\frac{8}{3}\beta x_{1}^{2}x_{6}$
$\displaystyle-\frac{8}{3}x_{1}^{2}x_{2}x_{4}x_{6}^{2}+\frac{32}{9}\alpha
x_{1}^{2}x_{3}x_{6}^{2}+\frac{16}{3}x_{1}^{3}x_{3}x_{4}x_{6}^{2}+L_{2},$ (33)

where $L_{1}$ and $L_{2}$ are some elements of the ideal
$\mathcal{A}_{\alpha,\beta}t_{5}\subseteq\mathcal{J}_{2}.$ Substitute (32) and
(33) into (31) and simplify to obtain:

$\displaystyle\text{ham}_{x_{-}}(e_{3})t_{5}t_{6}=\sum$
$\displaystyle[\lambda_{1,1}\beta
a_{(0,1,i,j,k_{0},l-1)}+\lambda_{1,2}\alpha^{2}a_{(1,1,i,j,k_{0},l-2)}$
$\displaystyle+\lambda_{1,3}\beta
a_{(1,1,i-2,j,k_{0},l-2)}]x_{1}^{i}x_{2}^{j}t_{5}^{k_{0}}t_{6}^{l}$
$\displaystyle+$ $\displaystyle\sum[\lambda_{2,1}\alpha
a_{(0,1,i,j,k_{0},l-2)}+\lambda_{2,2}\beta a_{(1,1,i,j,k_{0},l-1)}$
$\displaystyle+\lambda_{2,3}\alpha
a_{(1,1,i-2,j,k_{0},l-3)}]x_{1}^{i}x_{2}^{j}x_{3}t_{5}^{k_{0}}t_{6}^{l}$
$\displaystyle+$
$\displaystyle\sum[\lambda_{3,1}a_{(0,1,i,j-1,k_{0},l-2)}+\lambda_{3,2}\alpha
a_{(1,1,i-1,j,k_{0},l-2)}$
$\displaystyle+\lambda_{3,3}a_{(1,1,i-2,j-1,k_{0},l-3)}+\lambda_{3,4}a_{(0,0,i,j,k_{0},l-1)}]x_{1}^{i}x_{2}^{j}x_{4}t_{5}^{k_{0}}t_{6}^{l}$
$\displaystyle+$
$\displaystyle\sum[\lambda_{4,1}a_{(0,1,i-1,j,k_{0},l-2)}+\lambda_{4,2}a_{(1,1,i,j-1,k_{0},l-2)}$
$\displaystyle+\lambda_{4,3}a_{(1,1,i-3,j,k_{0},l-3)}+\lambda_{4,4}a_{(1,0,i,j,k_{0},l-1)}]x_{1}^{i}x_{2}^{j}x_{3}x_{4}t_{5}^{k_{0}}t_{6}^{l}+V^{\prime},$
(34)

where $V^{\prime}\in\mathcal{J}_{2}$. Also,
$\lambda_{s,t}:=\lambda_{s,t}(j,k_{0},l)$ are some families of scalars which
are non-zero for all $s,t\in\\{1,2,3,4\\}$ and $j,l\in\mathbb{N}$, except
$\lambda_{1,4}$ and $\lambda_{2,4}$ which are assumed to be zero since they do
not exist in the above expression. Note that although each $\lambda_{s,t}$
depends on $j,k_{0},l,$ we have not made this dependency explicit in the above
expression since the minimum requirement we need to complete the proof is for
all the $\lambda_{s,t}$ existing in the above expression to be non-zero, which
we already have.

Observe that (34) and (30) are equal, hence

$\displaystyle\sum_{(\epsilon_{1},\epsilon_{2},\underline{w})\in
J}b_{(\epsilon_{1},\epsilon_{2},\underline{w})}x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k}t_{6}^{l}=$
$\displaystyle\sum[\lambda_{1,1}\beta
a_{(0,1,i,j,k_{0},l-1)}+\lambda_{1,2}\alpha^{2}a_{(1,1,i,j,k_{0},l-2)}$
$\displaystyle+\lambda_{1,3}\beta
a_{(1,1,i-2,j,k_{0},l-2)}]x_{1}^{i}x_{2}^{j}t_{5}^{k_{0}}t_{6}^{l}$
$\displaystyle+$ $\displaystyle\sum[\lambda_{2,1}\alpha
a_{(0,1,i,j,k_{0},l-2)}+\lambda_{2,2}\beta a_{(1,1,i,j,k_{0},l-1)}$
$\displaystyle+\lambda_{2,3}\alpha
a_{(1,1,i-2,j,k_{0},l-3)}]x_{1}^{i}x_{2}^{j}x_{3}t_{5}^{k_{0}}t_{6}^{l}$
$\displaystyle+$
$\displaystyle\sum[\lambda_{3,1}a_{(0,1,i,j-1,k_{0},l-2)}+\lambda_{3,2}\alpha
a_{(1,1,i-1,j,k_{0},l-2)}$
$\displaystyle+\lambda_{3,3}a_{(1,1,i-2,j-1,k_{0},l-3)}+\lambda_{3,4}a_{(0,0,i,j,k_{0},l-1)}]x_{1}^{i}x_{2}^{j}x_{4}t_{5}^{k_{0}}t_{6}^{l}$
$\displaystyle+\sum$
$\displaystyle[\lambda_{4,1}a_{(0,1,i-1,j,k_{0},l-2)}+\lambda_{4,2}a_{(1,1,i,j-1,k_{0},l-2)}$
$\displaystyle+\lambda_{4,3}$ $\displaystyle
a_{(1,1,i-3,j,k_{0},l-3)}+\lambda_{4,4}a_{(1,0,i,j,k_{0},l-1)}]x_{1}^{i}x_{2}^{j}x_{3}x_{4}t_{5}^{k_{0}}t_{6}^{l}+V^{\prime}.$

We have previously established that
$\left(x_{1}^{i}x_{2}^{j}x_{3}^{\epsilon_{1}}x_{4}^{\epsilon_{2}}t_{5}^{k}t_{6}^{l}\right)_{((\epsilon_{1},\epsilon_{2},\underline{v})\in\\{0,1\\}^{2}\times\mathbb{N}^{2}\times\mathbb{Z}^{2})}$
is a basis of $\mathcal{A}_{\alpha,\beta}[t_{5}^{-1},t_{6}^{-1}]$ (note that
$l\geq 0$ in this part of the proof).

Since
$\underline{v}_{0}=(i,j,k_{0},l)\in\mathbb{N}^{2}\times\mathbb{Z}\times\mathbb{N}$
(with $k_{0}<0$) and $\underline{w}=(i,j,k,l)\in\mathbb{N}^{4}$ (with $k\geq
0$) in the above equality, it follows that

$\displaystyle\lambda_{1,1}$ $\displaystyle\beta
a_{(0,1,i,j,k_{0},l-1)}+\lambda_{1,2}\alpha^{2}a_{(1,1,i,j,k_{0},l-2)}+\lambda_{1,3}\beta
a_{(1,1,i-2,j,k_{0},l-2)}=0,$ (35) $\displaystyle\lambda_{2,1}$
$\displaystyle\alpha a_{(0,1,i,j,k_{0},l-2)}+\lambda_{2,2}\beta
a_{(1,1,i,j,k_{0},l-1)}+\lambda_{2,3}\alpha a_{(1,1,i-2,j,k_{0},l-3)}=0,$ (36)
$\displaystyle\lambda_{3,1}$ $\displaystyle
a_{(0,1,i,j-1,k_{0},l-2)}+\lambda_{3,2}\alpha
a_{(1,1,i-1,j,k_{0},l-2)}+\lambda_{3,3}a_{(1,1,i-2,j-1,k_{0},l-3)}$
$\displaystyle+\lambda_{3,4}a_{(0,0,i,j,k_{0},l-1)}=0,$ (37)
$\displaystyle\lambda_{4,1}$ $\displaystyle
a_{(0,1,i-1,j,k_{0},l-2)}+\lambda_{4,2}a_{(1,1,i,j-1,k_{0},l-2)}+\lambda_{4,3}a_{(1,1,i-3,j,k_{0},l-3)}$
$\displaystyle+\lambda_{4,4}a_{(1,0,i,j,k_{0},l-1)}=0.$ (38)

From (35) and (36), one can easily deduce that

$\displaystyle
a_{(0,1,i,j,k_{0},l)}=-\frac{\alpha^{2}\lambda_{1,2}}{\beta\lambda_{1,1}}a_{(1,1,i,j,k_{0},l-1)}-\frac{\lambda_{1,3}}{\lambda_{1,1}}a_{(1,1,i-2,j,k_{0},l-1)},$
(39) $\displaystyle
a_{(1,1,i,j,k_{0},l)}=-\frac{\alpha\lambda_{2,1}}{\beta\lambda_{2,2}}a_{(0,1,i,j,k_{0},l-1)}-\frac{\alpha\lambda_{2,3}}{\beta\lambda_{2,2}}a_{(1,1,i-2,j,k_{0},l-2)}.$
(40)

Note that $a_{(\epsilon_{1},\epsilon_{2},i,j,k_{0},l)}:=0$ whenever $i<0$ or
$j<0$ or $l<0$ for all $\epsilon_{1},\epsilon_{2}\in\\{0,1\\}.$

Claim. The coefficients $a_{(0,1,i,j,k_{0},l)}$ and $a_{(1,1,i,j,k_{0},l)}$
are all zero for all $l\geq 0$. We now justify the claim by an induction on
$l.$ From (39) and (40), the result is obviously true when $l=0.$ For $l\geq
0,$ assume that $a_{(0,1,i,j,k_{0},l)}=a_{(1,1,i,j,k_{0},l)}=0.$ Then, it
follows from (39) and (40) that

$a_{(0,1,i,j,k_{0},l+1)}=-\frac{\alpha^{2}\lambda_{1,2}}{\beta\lambda_{1,1}}a_{(1,1,i,j,k_{0},l)}-\frac{\lambda_{1,3}}{\lambda_{1,1}}a_{(1,1,i-2,j,k_{0},l)},$
$a_{(1,1,i,j,k_{0},l+1)}=-\frac{\alpha\lambda_{2,1}}{\beta\lambda_{2,2}}a_{(0,1,i,j,k_{0},l)}-\frac{\alpha\lambda_{2,3}}{\beta\lambda_{2,2}}a_{(1,1,i-2,j,k_{0},l-1)}.$

From the inductive hypothesis,
$a_{(1,1,i,j,k_{0},l)}=a_{(1,1,i-2,j,k_{0},l)}=a_{(0,1,i,j,k_{0},l)}=a_{(1,1,i-2,j,k_{0},l-1)}$
$=0.$ Hence, $a_{(1,1,i,j,k_{0},l+1)}=a_{(0,1,i,j,k_{0},l+1)}=0.$ By the
principle of mathematical induction,
$a_{(0,1,i,j,k_{0},l)}=a_{(1,1,i,j,k_{0},l)}=0$ for all $l\geq 0$ as desired.
Given that the families $a_{(0,1,i,j,k_{0},l)}$ and $a_{(1,1,i,j,k_{0},l)}$
are all zero, it follows from (37) and (38) that $a_{(0,0,i,j,k_{0},l)}$ and
$a_{(1,0,i,j,k_{0},l)}$ are also zero for all
$(i,j,k_{0},l)\in\mathbb{N}^{2}\times\mathbb{Z}\times\mathbb{N}.$ This
contradicts our assumption. Hence, $x_{-}=0.$ Consequently,
$x=x_{+}\in\mathcal{A}_{\alpha,\beta}$ as desired.

2\. From Remark 5.6, we have that
$x_{4}=x_{4,6}+\frac{2}{3}t_{5}^{3}t_{6}^{-1}=t_{4}+\frac{2}{3}t_{5}^{3}t_{6}^{-1}$.
Again, from Lemma 5.9, we have that $\lambda_{4}=3\lambda_{3}+\lambda_{5}$ and
$\lambda_{6}=-3\lambda_{3}.$ Therefore,

$\displaystyle\rho(x_{4})$
$\displaystyle=\lambda_{4}t_{4}+\frac{2}{3}(3\lambda_{5}-\lambda_{6})t_{5}^{3}t_{6}^{-1}$
$\displaystyle=(3\lambda_{3}+\lambda_{5})x_{4,6}+2(\lambda_{3}+\lambda_{5})t_{5}^{3}t_{6}^{-1}$
$\displaystyle=(3\lambda_{3}+\lambda_{5})\left(x_{4}-\frac{2}{3}t_{5}^{3}t_{6}^{-1}\right)+2(\lambda_{3}+\lambda_{5})t_{5}^{3}t_{6}^{-1}$
$\displaystyle=(3\lambda_{3}+\lambda_{5})x_{4}+\frac{4}{3}\lambda_{5}t_{5}^{3}t_{6}^{-1}.$

Hence,

$\displaystyle\mathcal{D}(x_{4})$
$\displaystyle=\text{ham}_{x}(x_{4})+\rho(x_{4})=\text{ham}_{x}(x_{4})+(3\lambda_{3}+\lambda_{5})x_{4}+\frac{4}{3}\lambda_{5}t_{5}^{3}t_{6}^{-1}\in\mathcal{A}_{\alpha,\beta}.$

It follows that $\lambda_{5}t_{5}^{3}t_{6}^{-1}\in\mathcal{A}_{\alpha,\beta},$
since
$\text{ham}_{x}(x_{4})+(3\lambda_{3}+\lambda_{5})x_{4}\in\mathcal{A}_{\alpha,\beta}.$
Consequently, $\lambda_{5}t_{5}^{3}\in\mathcal{A}_{\alpha,\beta}t_{6}.$ Using
the basis of $\mathcal{A}_{\alpha,\beta}$ (Prop. 4.16), we easily have that
$\lambda_{5}=0$. Therefore, $\rho(x_{4})=3\lambda_{3}x_{4}$ and
$\rho(t_{5})=0.$ We already know from Lemma 5.9 that
$\rho(t_{6})=-3\lambda_{3}t_{6}.$ From (28), we have
$\rho(x_{3})=\lambda_{3}x_{3}+\frac{3}{2}(\lambda_{4}-\lambda_{3}-\lambda_{5})x_{4}t_{5}^{-1}+(3\lambda_{5}-\lambda_{4}-\lambda_{6})t_{5}^{2}t_{6}^{-1}.$
Given that $\lambda_{4}=3\lambda_{3},\ \lambda_{5}=0$ and
$\lambda_{6}=-3\lambda_{3},$ we have that
$\rho(x_{3})=\lambda_{3}x_{3}+3\lambda_{3}x_{4}t_{5}^{-1}.$ Now,
$\mathcal{D}(x_{3})=\text{ham}_{x}(x_{3})+\rho(x_{3})=\text{ham}_{x}(x_{3})+\lambda_{3}x_{3}+3\lambda_{3}x_{4}t_{5}^{-1}\in\mathcal{A}_{\alpha,\beta}.$
Observe that
$\text{ham}_{x}(x_{3}),\lambda_{3}x_{3}\in\mathcal{A}_{\alpha,\beta}.$ Hence,
$\lambda_{3}x_{4}t_{5}^{-1}\in\mathcal{A}_{\alpha,\beta}$ which implies that
$\lambda_{3}x_{4}\in\mathcal{A}_{\alpha,\beta}t_{5}.$ Similarly,
$\lambda_{3}=0.$ We now have that
$\rho(x_{3})=\rho(x_{4})=\rho(x_{5})=\rho(x_{6})=0.$ We finish the proof by
showing that $\rho(x_{1})=\rho(x_{2})=0.$ Recall from (9) that

$x_{2}x_{4}x_{6}-\frac{2}{3}x_{3}^{3}x_{6}-\frac{2}{3}x_{2}x_{5}^{3}+2x_{3}^{2}x_{5}^{2}-3x_{3}x_{4}x_{5}+\frac{3}{2}x_{4}^{2}=\beta.$

Apply $\rho$ to this relation to obtain
$\rho(x_{2})x_{4}x_{6}-\frac{2}{3}\rho(x_{2})x_{5}^{3}=0.$ This implies that
$\rho(x_{2})\left(x_{4}x_{6}-\frac{2}{3}x_{5}^{3}\right)=0.$ Since
$x_{4}x_{6}-\frac{2}{3}x_{5}^{3}\neq 0,$ it follows that $\rho(x_{2})=0.$
Similarly, from (8), we have that

$x_{1}x_{3}x_{5}-\frac{3}{2}x_{1}x_{4}-\frac{1}{2}x_{2}x_{5}+\frac{1}{2}x_{3}^{2}=\alpha.$

Apply $\rho$ to this relation to obtain
$\rho(x_{1})\left(x_{3}x_{5}-\frac{3}{2}x_{4}\right)=0.$ Since
$x_{3}x_{5}-\frac{3}{2}x_{4}\neq 0,$ we must have: $\rho(x_{1})=0.$ In
conclusion, $\rho(x_{\kappa})=0$ for all $\kappa\in\\{1,\ldots,6\\}.$

3\. As a result of (1) and (2), we have that
$\mathcal{D}(x_{\kappa})=\text{ham}_{x}(x_{\kappa}).$ Consequently,
$\mathcal{D}=\text{ham}_{x}$ as desired. ∎

####  5.3.2 Poisson derivations of $\mathcal{A}_{\alpha,0}$ and
$\mathcal{A}_{0,\beta}$

Following procedures similarly to the previous case (i.e.
$\mathcal{A}_{\alpha,\beta}$ with $\alpha\beta\neq 0$), one can also compute
the Poisson derivations of $\mathcal{A}_{\alpha,0}$ and
$\mathcal{A}_{0,\beta}.$ The computations have been done, however, for the
avoidance of redundancy, we are not going to include them here. We only
summarize the results. Before we do that, we compute explicitly the scalar
Poisson derivations of $\mathcal{A}_{\alpha,0}$ and $\mathcal{A}_{0,\beta}.$

######  Lemma 5.10.

Let $(\alpha,\beta)\in\mathbb{K}^{2}\setminus\\{(0,0)\\}.$ Suppose that
$\vartheta$ and $\tilde{\vartheta}$ are linear maps on
$\mathcal{A}_{\alpha,0}$ and $\mathcal{A}_{0,\beta}$ respectively, and are
defined by:

$\vartheta(x_{1})=-x_{1},\ \ \vartheta(x_{2})=-x_{2},\ \ \vartheta(x_{3})=0,\
\ \vartheta(x_{4})=x_{4},\ \ \vartheta(x_{5})=x_{5},\ \ \
\vartheta(x_{6})=2x_{6},$

and

$\tilde{\vartheta}(x_{1})=-2x_{1},\ \ \tilde{\vartheta}(x_{2})=-3x_{2},\ \
\tilde{\vartheta}(x_{3})=-x_{3},\ \ \tilde{\vartheta}(x_{4})=0,\ \
\tilde{\vartheta}(x_{5})=x_{5},\ \ \ \tilde{\vartheta}(x_{6})=3x_{6}.$

Then, $\vartheta$ and $\tilde{\vartheta}$ extended to $\mathcal{A}_{\alpha,0}$
and $\mathcal{A}_{0,\beta}$ respectively using the Leibniz rule are
$\mathbb{K}$-Poisson derivations of $\mathcal{A}_{\alpha,0}$ and
$\mathcal{A}_{0,\beta}$ respectively.

###### Proof.

We need to show that $\vartheta$ satisfies the following two relations (see
(8) and (9)):

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

and the Poisson bracket of $\mathcal{A}_{\alpha,\beta}$ (see Subsection 4.3)
when $\alpha\neq 0$ and $\beta=0,$ and do the same thing for
$\tilde{\vartheta}$ when $\alpha=0$ and $\beta\neq 0.$ We will only do this
for the relation
$x_{1}x_{3}x_{5}-\frac{3}{2}x_{1}x_{4}-\frac{1}{2}x_{2}x_{5}+\frac{1}{2}x_{3}^{2}=\alpha$
and the Poisson bracket $\\{x_{6},x_{2}\\}=3x_{2}x_{6}+9x_{4}-18x_{3}x_{5}$ in
$\mathcal{A}_{\alpha,0}$, and leave the remaining ones for the reader to
verify. We have:

$\displaystyle\vartheta(x_{1})x_{3}x_{5}+x_{1}\vartheta(x_{3})x_{5}+x_{1}x_{3}\vartheta(x_{5})$
$\displaystyle-\frac{3}{2}[\vartheta(x_{1})x_{4}+x_{1}\vartheta(x_{4})]$
$\displaystyle-\frac{1}{2}[\vartheta(x_{2})x_{5}$
$\displaystyle+x_{2}\vartheta(x_{5})]+\vartheta(x_{3})x_{3}=0,$

and

$\displaystyle\vartheta(\\{x_{6},x_{2}\\})$
$\displaystyle=\vartheta(3x_{2}x_{6}+9x_{4}-18x_{3}x_{5})$
$\displaystyle=3[\vartheta(x_{2})x_{6}+x_{2}\vartheta(x_{6})]+9\vartheta(x_{4})-18[\vartheta(x_{3})x_{5}+x_{3}\vartheta(x_{5})]$
$\displaystyle=3(-x_{2}x_{6}+2x_{2}x_{6})+9x_{4}-18x_{3}x_{5}$
$\displaystyle=3x_{2}x_{6}+9x_{4}-18x_{3}x_{5}$
$\displaystyle=\\{x_{6},x_{2}\\}$
$\displaystyle=2\\{x_{6},x_{2}\\}-\\{x_{6},x_{2}\\}$
$\displaystyle=\\{2x_{6},x_{2}\\}+\\{x_{6},-x_{2}\\}$
$\displaystyle=\\{\vartheta(x_{6}),x_{2}\\}+\\{x_{6},\vartheta(x_{2})\\}.$

∎

We summarize our main results in this section in the theorem below. Recall
that HP${}^{1}(\Delta)$ denote the first Poisson cohomology group of the
Poisson algebra $\Delta$.

######  Theorem 5.11.

Given
$\mathcal{A}_{\alpha,\beta}=\mathbb{K}[X_{1},\ldots,X_{6}]/\langle\Omega_{1}-\alpha,\Omega_{2}-\beta\rangle,$
with $(\alpha,\beta)\in\mathbb{K}^{2}\setminus\\{(0,0)\\},$ we have the
following results:

  * 1.

if $\alpha,\beta\neq 0;$ then every Poisson derivation $\mathcal{D}$ of
$\mathcal{A}_{\alpha,\beta}$ can uniquely be written as
$\mathcal{D}=\mathrm{ham}_{x},$ where $x\in\mathcal{A}_{\alpha,\beta}.$

  * 2.

if $\alpha\neq 0$ and $\beta=0,$ then every Poisson derivation $\mathcal{D}$
of $\mathcal{A}_{\alpha,0}$ can uniquely be written as
$\mathcal{D}=\mathrm{ham}_{x}+\lambda\vartheta,$ where $\lambda\in\mathbb{K}$
and $x\in\mathcal{A}_{\alpha,0}.$

  * 3.

if $\alpha=0$ and $\beta\neq 0,$ then every Poisson derivation $\mathcal{D}$
of $\mathcal{A}_{0,\beta}$ can uniquely be written as
$\mathcal{D}=\mathrm{ham}_{x}+\lambda\tilde{\vartheta},$ where
$\lambda\in\mathbb{K}$ and $x\in\mathcal{A}_{0,\beta}.$

  * 4.

$HP^{1}(\mathcal{A}_{\alpha,0})=\mathbb{K}[\vartheta]$ and
$HP^{1}(\mathcal{A}_{0,\beta})=\mathbb{K}[\tilde{\vartheta}],$ where
$[\vartheta]$ and $[\tilde{\vartheta}]$ respectively denote the classes of
$\vartheta$ and $\tilde{\vartheta}$ modulo the space of inner Poisson
derivations.

  * 5.

if $\alpha,\beta\neq 0;$ then $HP^{1}(\mathcal{A}_{\alpha,\beta})=\\{[0]\\},$
where $[0]$ denotes the class of $0$ modulo the space of inner Poisson
derivations.

Let $(\alpha,\beta)\in\mathbb{K}^{2}\setminus\\{(0,0)\\}$. One can easily
conclude that the first Poisson cohomology group
HP${}^{1}(\mathcal{A}_{\alpha,\beta})$ is isomorphic to the first Hochschild
cohomology group HH${}^{1}(A_{\alpha,\beta})$ studied in [15, Theorem 5.12].

It is natural to ask whether HP${}^{i}(\mathcal{A}_{\alpha,\beta})$ is
isomorphic to HH${}^{i}(A_{\alpha,\beta})$ for all $i$.

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S Launois
School of Mathematics, Statistics and Actuarial Science,
University of Kent
Canterbury, Kent, CT2 7FS,
UK
email<EMAIL_ADDRESS>

I Oppong
School of Mathematics, Statistics and Actuarial Science,
University of Kent
Canterbury, Kent, CT2 7FS,
UK
email<EMAIL_ADDRESS>