Classification of Four-Dimensional Complex Poisson Algebras

Abstract

This paper provides a complete classification of four-dimensional complex Poisson algebras, extending previous work on lower-dimensional cases. We develop a systematic approach that combines the classification of four-dimensional complex commutative associative algebras with the classification of four-dimensional complex Lie algebras, allowing us to determine all compatible Poisson structures up to isomorphism. Our main result presents 71 distinct types: 60 isolated non-isomorphic Poisson algebras with non-trivial commutative product and 11 one-parameter families of non-isomorphic Poisson algebras. In addition to these, we include the known four-dimensional complex Lie algebras, viewed as Poisson algebras with trivial commutative product. This purely algebraic classification provides a foundation for future studies, including the geometric classification of Poisson algebras, their degenerations, and the analysis of orbit closures in the corresponding algebraic variety.

1. Introduction

Poisson algebras are vector spaces equipped with two binary operations, a commutative product and a Lie bracket, which satisfy a certain compatibility condition known as the Leibniz rule. Introduced in Hamiltonian mechanics as the dual of the category of classical mechanical systems, Poisson algebras have a fundamental role in the study of quantum groups, differential geometry, noncommutative geometry, integrable systems, quantum field theory, or vertex operator algebras (see [1,2,3,4,5,6,7]). Additionally, this class of algebras plays an important role in many areas of mathematics including symplectic geometry, representation theory, quantum field theory, and algebraic geometry. Poisson algebras can be thought of as the algebraic counterpart of Poisson manifolds, which are smooth manifolds M whose commutative algebra $C^{\infty }(M,\mathbb{R})$ of real smooth functions is endowed with a Lie bracket $[-,-]$, satisfying the Leibniz rule; i.e., $C^{\infty }(M,\mathbb{R})$ is a Poisson algebra. Here, they are used to study geometric structures that preserve certain properties under deformation.

Classifying algebras is important in several areas of physics and mathematics. Understanding the classification of algebras of small dimensions is often a first step to obtain the corresponding to larger algebras. Concretely, by classifying Poisson algebras we can identify their underlying symmetries and study their geometric properties, which can lead to new insights into the behavior of physical systems.

In [8] the authors developed a method to obtain the algebraic classification of Poisson algebras defined on a commutative associative algebra, and they applied it to obtain the classification of three-dimensional complex Poisson algebras. In addition, they also studied the algebraic classification of the Poisson algebras defined on a commutative associative null-filiform or filiform algebra. This method focuses on presenting a procedure to classify all the Poisson algebras associated with a given commutative associative algebra. We briefly explain the method. Pick an arbitrary commutative associative algebra $\mathcal{P}$. Further, compute the set $Z^2(\mathcal{P},\mathcal{P})$ of all skew symmetric bilinear maps on $\mathcal{P}$, satisfying some adequate conditions (Definition 3). It is proven that for any Poisson algebra $(\mathcal{P},·,[-,-\left]\right)$ there exists $\theta \in Z^2(\mathcal{P},\mathcal{P})$, such that the algebra is isomorphic to a Poisson algebra $(\mathcal{P},·,{[-,-]}_{\theta })$ associated with the given commutative associative algebra $\mathcal{P}$. Find the orbits of the automorphism group $\mathrm{Aut}\left(\mathcal{P}\right)$ on $Z^2(\mathcal{P},\mathcal{P})$ by a proper action (3). It turns out that by choosing a representative $\theta$ from each orbit, all Poisson algebras are obtained $(\mathcal{P},·,{[-,-]}_{\theta })$, up to isomorphism. So, all the Poisson algebras associated with a given commutative associative algebra are obtained.

In the present paper, the authors produce a complete classification of four-dimensional complex Poisson algebras, by applying the above results, taking into account the classification, up to isomorphism, of the complex commutative associative algebras of dimension 4 presented in [9], as well as, using the Lie algebra classification, up to isomorphism, given in [10].

The novelty of this work lies in providing the first comprehensive algebraic classification of all four-dimensional complex Poisson algebras. While the classifications of the underlying commutative associative and Lie algebra structures were already established, a unified framework encompassing all compatible Poisson brackets had not been developed before. Our results identify and distinguish new Poisson algebra classes, including parameter-dependent families. In particular, we emphasize that this study presents the first complete algebraic classification in dimension 4, extending the known results for the three-dimensional case.

The paper is organized as follows: Section 2 recalls the classification method developed in [8]. Section 3 presents our main classification result, including the complete list of four-dimensional complex Poisson algebras. Section 4 contains the detailed proof of the classification theorem.

2. The Algebraic Classification Method

In this section, we recall the method to obtain the algebraic classification of the Poisson algebras over an arbitrary field $\mathbb{F}$ of characteristic zero present in [8], to make this work self-contained. Let us recall some basic definitions needed in the sequel.

Definition 1.

A Poisson algebra is a vector space $\mathcal{P}$ endowed with two bilinear operations:

(1)

An commutative associative multiplication denoted by $-·-:\mathcal{P}\times \mathcal{P}\to \mathcal{P}$;

(2)

A Lie algebra multiplication denoted by $[-,-]:\mathcal{P}\times \mathcal{P}\to \mathcal{P}$.

• These two operations are compatible in the sense that they satisfy the following Leibniz identity

$$[x·y,z]=[x,z]·y+x·[y,z],$$

(1)

for any $x,y,z\in \mathcal{P}$. The dimension of a Poisson algebra is its dimension as vector space.

Condition (1) ensures that the Lie bracket behaves like a derivation with respect to the commutative associative multiplication.

Definition 2.

Consider two Poisson algebras $({\mathcal{P}}_1,{·}_1,{[·,·]}_1)$ and $({\mathcal{P}}_2,{·}_2,{[·,·]}_2)$. A Poisson algebras homomorphism (or just homomorphism, when the context is clear), is a linear map $\varphi :{\mathcal{P}}_1\to {\mathcal{P}}_2$ preserving the two products, that is,

$$\varphi \left(x{·}_{1}y\right)=\varphi \left(x\right){·}_2\varphi \left(y\right),\varphi \left({[x,y]}_1\right)={[\varphi \left(x\right),\varphi \left(y\right)]}_2,$$

for all $x,y\in {\mathcal{P}}_1$.

Now, picking any arbitrary commutative associative algebra, we may consider all the Poisson structures defined over this algebra. This notion is captured in the following definition.

Definition 3.

Let $(\mathcal{P},·)$ be a commutative associative algebra. Define $Z^2(\mathcal{P},\mathcal{P})$ to be the set of all skew-symmetric bilinear maps $\theta :\mathcal{P}\times \mathcal{P}\to \mathcal{P}$ such that

$$\begin{array}{ccc}& & \theta \left(\theta \right(x,y),z)+\theta \left(\theta \right(y,z),x)+\theta \left(\theta \right(z,x),y)=0,\hfill \\ & & \theta (x·y,z)-\theta (x,z)·y-x·\theta (y,z)=0,\hfill \end{array}$$

for all $x,y,z$ in $\mathcal{P}$. Then $Z^2(\mathcal{P},\mathcal{P})\ne \varnothing$ since $\theta =0\in Z^2(\mathcal{P},\mathcal{P})$.

Observe that, for $\theta \in Z^2(\mathcal{P},\mathcal{P})$, we may define on $\mathcal{P}$ a bracket ${[-,-]}_{\theta }:\mathcal{P}\times \mathcal{P}\to \mathcal{P}$ by

$${[x,y]}_{\theta }:=\theta (x,y),$$

(2)

for any $x,y$ in $\mathcal{P}$.

Lemma 1.

Let $(\mathcal{P},·)$ be a commutative associative algebra and $\theta \in Z^2(\mathcal{P},\mathcal{P})$. Then $(\mathcal{P},·,{[-,-]}_{\theta })$ is a Poisson algebra endowed with the product defined in (2).

Proof. 

Let $(\mathcal{P},·)$ be a commutative associative algebra and let $\theta \in Z^2(\mathcal{P},\mathcal{P})$. Then $(\mathcal{P},{[·,·]}_{\theta })$ is an anticommutative algebra. Moreover, since $\theta \in Z^2(\mathcal{P},\mathcal{P})$, we have

$$\begin{array}{cc}\hfill & {[{[x,y]}_{\theta },z]}_{\theta }+{[{[y,z]}_{\theta },x]}_{\theta }+{[{[z,x]}_{\theta },y]}_{\theta }=\theta (\theta (x,y),z)+\theta (\theta (y,z),x)+\theta (\theta (z,x),y)=0,\hfill \\ \hfill & {[x·y,z]}_{\theta }-{[x,z]}_{\theta }·y-x·{[y,z]}_{\theta }=\theta (x·y,z)-\theta (x,z)·y-x·\theta (y,z)=0,\hfill \end{array}$$

for $x,y,z$ in $\mathcal{P}$, as desired. □

On the other hand, we may prove that if $(\mathcal{P},·,[-,-\left]\right)$ is a Poisson algebra, then there exists $\theta \in Z^2(\mathcal{P},\mathcal{P})$ such that $(\mathcal{P},·,{[-,-]}_{\theta })\cong (\mathcal{P},·,[-,-])$. Indeed, let us consider the skew-symmetric bilinear map $\theta :\mathcal{P}\times \mathcal{P}\to \mathcal{P}$ defined by $\theta (x,y):=[x,y]$ for $x,y\in \mathcal{P}$. Then $\theta \in Z^2(\mathcal{P},\mathcal{P})$ and $(\mathcal{P},·,{[-,-]}_{\theta })=(\mathcal{P},·,[-,-])$.

Now, let $(\mathcal{P},·)$ be a commutative associative algebra and $\mathrm{Aut}\left(\mathcal{P}\right)$ be the automorphism group of $\mathcal{P}$. Then we can define an action of $\mathrm{Aut}\left(\mathcal{P}\right)$ on $Z^2(\mathcal{P},\mathcal{P})$ by

$$(\theta \ast \varphi )(x,y):={\varphi }^{-1}\left(\theta (\varphi \left(x\right),\varphi \left(y\right))\right),$$

(3)

for any $\varphi \in \mathrm{Aut}\left(\mathcal{P}\right)$ and $\theta \in Z^2(\mathcal{P},\mathcal{P})$, with $x,y$ in $\mathcal{P}$.

Lemma 2.

Let $(\mathcal{P},·)$ be a commutative associative algebra and $\theta ,\vartheta \in Z^2(\mathcal{P},\mathcal{P})$. Then $(\mathcal{P},·,{[-,-]}_{\theta })$ and $(\mathcal{P},·,{[-,-]}_{\vartheta })$ are isomorphic if and only if there exists $\varphi \in \mathrm{Aut}\left(\mathcal{P}\right)$ such that $\theta \ast \varphi =\vartheta$.

Proof. 

If there exists $\varphi \in \mathrm{Aut}\left(\mathcal{P}\right)$ such that $\theta \ast \varphi =\vartheta$, then $\varphi :(\mathcal{P},·,{[-,-]}_{\vartheta })\to$$(\mathcal{P},·,{[-,-]}_{\theta })$ is an isomorphism since $\varphi \left(\vartheta \right(x,y)=\theta (\varphi \left(x\right),\varphi \left(y\right))$, with $x,y$ in $\mathcal{P}$. On the other hand, if $\varphi :(\mathcal{P},·,{[-,-]}_{\vartheta })\to$$(\mathcal{P},·,{[-,-]}_{\theta })$ is an isomorphism of Poisson algebras, then $\varphi \in \mathrm{Aut}\left(\mathcal{P}\right)$ and $\varphi \left({[x,y]}_{\vartheta }\right)={[\varphi \left(x\right),\varphi \left(y\right)]}_{\theta }$, with $x,y$ in $\mathcal{P}$. Hence $\vartheta (x,y)={\varphi }^{-1}\left(\theta (\varphi \left(x\right),\varphi \left(y\right))\right)=(\theta \ast \varphi )(x,y)$ and therefore $\theta \ast \varphi =\vartheta$. □

Hence, we have a procedure to classify all the Poisson algebras associated with a given commutative associative algebra $(\mathcal{P},·)$. It is performed in three steps:

(1)

Compute $Z^2(\mathcal{P},\mathcal{P})$.

(2)

Find the orbits of $\mathrm{Aut}\left(\mathcal{P}\right)$ on $Z^2(\mathcal{P},\mathcal{P})$.

(3)

Choose a representative $\theta$ from each orbit and then construct the Poisson algebra $(\mathcal{P},·,{[-,-]}_{\theta })$.

Remark 1.

Similarly, we can construct an analogous method for classifying the four-dimensional Poisson algebras from the classification of Lie algebras of dimension four.

Let us denote the following notation. Let $\{e_1,e_2,\dots ,e_n\}$ be a fixed basis of a commutative associative algebra $(\mathcal{P},·)$. We define ${\mathrm{\Lambda }}^2(\mathcal{P},\mathbb{F}):={\mathrm{span}}_{\mathbb{F}}\{{\Delta }_{i,j}:1\le i<j\le n\}$, where each ${\Delta }_{i,j}:\mathcal{P}\times \mathcal{P}\to \mathbb{F}$ is the skew-symmetric bilinear form defined by

$${\Delta }_{i,j}(e_l,e_m):=\left\{\begin{array}{cc}\hfill ll1,& if(i,j)=(l,m),\\ \hfill -1,& if(i,j)=(m,l),\\ \hfill 0,& otherwise.\end{array}\right.$$

Now, if $\theta \in Z^2(\mathcal{P},\mathcal{P})$, then $\theta$ can be uniquely written as $\theta (x,y)={\sum }_{i=1}^n{B}_i(x,y)e_i$ where $B_1,B_2,\dots ,B_n$ is a sequence of skew-symmetric bilinear forms on $\mathcal{P}$. Also, we may write $\theta =(B_1,B_2,\dots ,B_n)$ . Let ${\varphi }^{-1}\in \mathrm{Aut}\left(\mathcal{P}\right)$ be given by the matrix $\left(b_{ij}\right)$. If $(\theta \ast \varphi )(x,y)={\sum }_{i=1}^n{B}_i^{\prime }(x,y)e_i$, then $B_i^{\prime }={\sum }_{j=1}^n{b}_{ij}{\varphi }^t{B}_j\varphi$.

Remark 2.

Note that if $({\mathcal{P}}_1,{·}_1,{[·,·]}_1)$ and $({\mathcal{P}}_2,{·}_2,{[·,·]}_2)$ are two isomorphic Poisson algebras, then the commutative associative algebras $({\mathcal{P}}_1,{·}_1)$ and $({\mathcal{P}}_2,{·}_2)$ are isomorphic. So, given two non-isomorphic commutative associative algebras $\mathcal{P}$ and ${\mathcal{P}}^{\prime }$, we have that any Poisson structure on $\mathcal{P}$ is not isomorphic to any Poisson structure on ${\mathcal{P}}^{\prime }$.

Remark 3.

Let $X=\left(\begin{array}{cc}\alpha & \beta \end{array}\right)\in {\mathcal{M}}_{1\times 2}\left(\mathbb{F}\right)$ and $X\ne 0$. Then there exists an invertible matrix $A\in {\mathcal{M}}_{2\times 2}\left(\mathbb{F}\right)$ such that $XA=\left(\begin{array}{cc}1& 0\end{array}\right)$. In fact, first assume that $\alpha \ne 0$. Then $\left(\begin{array}{cc}\alpha & \beta \end{array}\right)\left(\begin{array}{cc}{\alpha }^{-1}& -\beta \\ 0& \alpha \end{array}\right)=\left(\begin{array}{cc}1& 0\end{array}\right)$. Now, if $\alpha =0$ then $\left(\begin{array}{cc}0& \beta \end{array}\right)\left(\begin{array}{cc}0& 1\\ {\beta }^{-1}& 0\end{array}\right)=\left(\begin{array}{cc}1& 0\end{array}\right)$.

3. Poisson Algebras of Dimension 4

From now, we present the classification, up to isomorphism, of the Poisson algebras for dimension 4 over the field of complex numbers $\mathbb{C}$. For simplicity, every time we write the multiplication table of a Poisson algebra, the products of basic elements whose values are zero or can be recovered by the commutativity, in the case of $-·-$, or by the anticommutativity, in the case of $[-,-]$, are omitted. First we recall the classification, up to isomorphism, of the complex commutative associative algebras of dimension 4 presented in [9]:

Theorem 1

([9]). Let $\mathcal{A}$ be a complex commutative associative algebra of dimension 4. Then $\mathcal{A}$ is isomorphic to one of the following algebras:

• ${\mathcal{A}}_{01}:$ trivial algebra.
• ${\mathcal{A}}_{02}:e_1^2=e_2.$
• ${\mathcal{A}}_{03}:e_1^2=e_3,e_2^2=e_3.$
• ${\mathcal{A}}_{04}:e_1^2=e_2,e_1·e_2=e_3.$
• ${\mathcal{A}}_{05}:e_1^2=-e_3,e_1·e_2=e_4,e_2^2=e_3.$
• ${\mathcal{A}}_{06}:e_1·e_2=e_4,e_2^2=e_3.$
• ${\mathcal{A}}_{07}:e_1^2=e_4,e_2^2=e_4,e_3^2=e_4.$
• ${\mathcal{A}}_{08}:e_1^2=e_2,e_1·e_2=e_4,e_3^2=e_4.$
• ${\mathcal{A}}_{09}:e_1^2=e_2,e_1·e_2=e_3,e_1·e_3=e_4,e_2^2=e_4.$
• ${\mathcal{A}}_{10}:e_1^2=e_1,e_2^2=e_2,e_3^2=e_3,e_4^2=e_4.$
• ${\mathcal{A}}_{11}:e_1^2=e_1,e_2^2=e_2,e_3^2=e_3,e_3·e_4=e_4.$
• ${\mathcal{A}}_{12}:e_1^2=e_1,e_1·e_2=e_2,e_3^2=e_3,e_3·e_4=e_4.$
• ${\mathcal{A}}_{13}:e_1^2=e_1,e_2^2=e_2,e_2·e_3=e_3,e_2·e_4=e_4.$
• ${\mathcal{A}}_{14}:e_1^2=e_1,e_2^2=e_2,e_2·e_3=e_3,e_2·e_4=e_4,e_3^2=e_4.$
• ${\mathcal{A}}_{15}:e_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_1·e_4=e_4.$
• ${\mathcal{A}}_{16}:e_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_1·e_4=e_4,e_2^2=e_3.$
• ${\mathcal{A}}_{17}:e_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_1·e_4=e_4,e_2^2=e_3,e_2·e_3=e_4.$
• ${\mathcal{A}}_{18}:e_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_1·e_4=e_4,e_2^2=e_4,e_3^2=e_4.$
• ${\mathcal{A}}_{19}:e_1^2=e_1,e_2^2=e_2,e_3^2=e_3.$
• ${\mathcal{A}}_{20}:e_1^2=e_1,e_2^2=e_2,e_2·e_3=e_3.$
• ${\mathcal{A}}_{21}:e_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3.$
• ${\mathcal{A}}_{22}:e_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_2^2=e_3.$
• ${\mathcal{A}}_{23}:e_1^2=e_1,e_2^2=e_2.$
• ${\mathcal{A}}_{24}:e_1^2=e_1,e_2^2=e_2,e_3^2=e_4.$
• ${\mathcal{A}}_{25}:e_1^2=e_1,e_1·e_2=e_2.$
• ${\mathcal{A}}_{26}:e_1^2=e_1,e_1·e_2=e_2,e_3^2=e_4.$
• ${\mathcal{A}}_{27}:e_1^2=e_1.$
• ${\mathcal{A}}_{28}:e_1^2=e_1,e_2·e_3=e_4.$
• ${\mathcal{A}}_{29}:e_1^2=e_1,e_2^2=e_3.$
• ${\mathcal{A}}_{30}:e_1^2=e_1,e_2^2=e_3,e_2·e_3=e_4.$

Now, we present the main result of the paper.

Theorem 2.

Let $\mathcal{P}$ be a complex Poisson algebra of dimension 4. Then $\mathcal{P}$ is isomorphic to one of the Lie algebras given in [10] (Lemma 3) or to one of the following algebras:

• ${\mathcal{P}}_{4,1}:e_1^2=e_2.$
• ${\mathcal{P}}_{4,2}:\left\{\begin{array}{c}{e}_1^2=e_2,\hfill \\ \left[e_1,e_3\right]=e_2.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,3}:\left\{\begin{array}{c}{e}_1^2=e_2,\hfill \\ \left[e_3,e_4\right]=e_2.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,4}^{\alpha }:\left\{\begin{array}{c}{e}_1^2=e_2,\hfill \\ \left[e_1,e_3\right]=e_3,\left[e_1,e_4\right]=\alpha e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,5}:\left\{\begin{array}{c}{e}_1^2=e_2,\hfill \\ \left[e_1,e_4\right]=e_2,\left[e_1,e_3\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,6}:\left\{\begin{array}{c}{e}_1^2=e_2,\hfill \\ \left[e_3,e_4\right]=e_2,\left[e_1,e_3\right]=e_3,\left[e_1,e_4\right]=-e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,7}:\left\{\begin{array}{c}{e}_1^2=e_2,\hfill \\ \left[e_1,e_3\right]=e_3,\left[e_1,e_4\right]=e_3+e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,8}:\left\{\begin{array}{c}{e}_1^2=e_2,\hfill \\ \left[e_1,e_4\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,9}:\left\{\begin{array}{c}{e}_1^2=e_2,\hfill \\ \left[e_1,e_3\right]=e_2,\left[e_1,e_4\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,10}:\left\{\begin{array}{c}{e}_1^2=e_2,\hfill \\ \left[e_3,e_4\right]=e_2,\left[e_1,e_4\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,11}:\left\{\begin{array}{c}{e}_1^2=e_2,\hfill \\ \left[e_3,e_4\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,12}:\left\{\begin{array}{c}{e}_1^2=e_2,\hfill \\ \left[e_1,e_4\right]=e_2,\left[e_3,e_4\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,13}^{\alpha }:\left\{\begin{array}{c}{e}_1^2=e_3,e_2^2=e_3,\hfill \\ \left[e_1,e_2\right]=\alpha e_3,\left[e_1,e_4\right]=e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,14}:\left\{\begin{array}{c}{e}_1^2=e_3,e_2^2=e_3,\hfill \\ \left[e_1,e_4\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,15}:\left\{\begin{array}{c}{e}_1^2=e_3,e_2^2=e_3,\hfill \\ \left[e_1,e_4\right]=e_3,\left[e_1,e_2\right]=e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,16}^{\alpha }:\left\{\begin{array}{c}{e}_1^2=e_3,e_2^2=e_3,\hfill \\ \left[e_1,e_2\right]=\alpha e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,17}:\left\{\begin{array}{c}{e}_1^2=e_3,e_2^2=e_3,\hfill \\ \left[e_1,e_2\right]=e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,18}:\left\{\begin{array}{c}{e}_1^2=e_3,e_2^2=e_3,\hfill \\ \left[e_1,e_4\right]=e_3,\left[e_2,e_4\right]=i{e}_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,19}:\left\{\begin{array}{c}{e}_1^2=e_3,e_2^2=e_3,\hfill \\ \left[e_1,e_4\right]=e_3,\left[e_2,e_4\right]=i{e}_3,\left[e_1,e_2\right]=e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,20}^{\alpha }:\left\{\begin{array}{c}{e}_1^2=e_3,e_2^2=e_3,\hfill \\ \left[e_1,e_2\right]=\alpha e_3,\left[e_1,e_4\right]=e_4,\left[e_2,e_4\right]=i{e}_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,21}:e_1^2=e_2,e_1·e_2=e_3.$
• ${\mathcal{P}}_{4,22}:\left\{\begin{array}{c}{e}_1^2=e_2,e_1·e_2=e_3,\hfill \\ \left[e_1,e_4\right]=e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,23}:\left\{\begin{array}{c}{e}_1^2=e_2,e_1·e_2=e_3,\hfill \\ \left[e_1,e_4\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,24}:e_1^2=e_3,e_2^2=e_4.$
• ${\mathcal{P}}_{4,25}^{\alpha }:\left\{\begin{array}{c}{e}_1^2=e_3,e_2^2=e_4,\hfill \\ \left[e_1,e_2\right]=e_3+\alpha e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,26}:\left\{\begin{array}{c}{e}_1·e_2=e_4,e_2^2=e_3,\hfill \\ \left[e_1,e_2\right]=e_1,\left[e_1,e_3\right]=2e_4,\left[e_2,e_4\right]=-e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,27}:\left\{\begin{array}{c}{e}_1·e_2=e_4,e_2^2=e_3,\hfill \\ \left[e_1,e_2\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,28}^{\alpha }:\left\{\begin{array}{c}{e}_1·e_2=e_4,e_2^2=e_3,\hfill \\ \left[e_1,e_2\right]=\alpha e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,29}:\left\{\begin{array}{c}{e}_1·e_2=e_4,e_3^2=e_4,\hfill \\ \left[e_1,e_3\right]=-e_1,\left[e_2,e_3\right]=e_2,\left[e_1,e_2\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,30}:\left\{\begin{array}{c}{e}_1·e_2=e_4,e_3^2=e_4,\hfill \\ \left[e_1,e_3\right]=e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,31}^{\alpha }:\left\{\begin{array}{c}{e}_1·e_2=e_4,e_3^2=e_4,\hfill \\ \left[e_1,e_2\right]=\alpha e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,32}:e_1^2=e_2,e_1·e_2=e_4,e_3^2=e_4.$
• ${\mathcal{P}}_{4,33}:\left\{\begin{array}{c}{e}_1^2=e_2,e_1·e_2=e_4,e_3^2=e_4,\hfill \\ \left[e_1,e_3\right]=e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,34}:e_1^2=e_2,e_1·e_2=e_3,e_1·e_3=e_4,e_2^2=e_4.$
• ${\mathcal{P}}_{4,35}:e_1^2=e_1,e_2^2=e_2,e_3^2=e_3,e_4^2=e_4.$
• ${\mathcal{P}}_{4,36}:e_1^2=e_1,e_2^2=e_2,e_3^2=e_3,e_3·e_4=e_4.$
• ${\mathcal{P}}_{4,37}:e_1^2=e_1,e_1·e_2=e_2,e_3^2=e_3,e_3·e_4=e_4.$
• ${\mathcal{P}}_{4,38}:e_1^2=e_1,e_2^2=e_2,e_2·e_3=e_3,e_2·e_4=e_4.$
• ${\mathcal{P}}_{4,39}:\left\{\begin{array}{c}{e}_1^2=e_1,e_2^2=e_2,e_2·e_3=e_3,e_2·e_4=e_4,\hfill \\ \left[e_3,e_4\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,40}:e_1^2=e_1,e_2^2=e_2,e_2·e_3=e_3,e_2·e_4=e_4,e_3^2=e_4.$
• ${\mathcal{P}}_{4,41}^{\alpha }:\left\{\begin{array}{c}{e}_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_1·e_4=e_4,\hfill \\ \left[e_2,e_4\right]=\alpha e_2,\left[e_3,e_4\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,42}:\left\{\begin{array}{c}{e}_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_1·e_4=e_4,\hfill \\ \left[e_2,e_4\right]=e_2,\left[e_3,e_4\right]=e_2+e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,43}:e_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_1·e_4=e_4.$
• ${\mathcal{P}}_{4,44}:\left\{\begin{array}{c}{e}_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_1·e_4=e_4,\hfill \\ \left[e_3,e_4\right]=e_2.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,45}:\left\{\begin{array}{c}{e}_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_1·e_4=e_4,\hfill \\ \left[e_3,e_4\right]=e_2,\left[e_2,e_3\right]=e_3,\left[e_2,e_4\right]=-e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,46}:e_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_1·e_4=e_4,e_2^2=e_3.$
• ${\mathcal{P}}_{4,47}:\left\{\begin{array}{c}{e}_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_1·e_4=e_4,e_2^2=e_3,\hfill \\ \left[e_2,e_4\right]=e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,48}:\left\{\begin{array}{c}{e}_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_1·e_4=e_4,e_2^2=e_3,\hfill \\ \left[e_2,e_4\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,49}:e_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_1·e_4=e_4,e_2^2=e_3,e_2·e_3=e_4.$
• ${\mathcal{P}}_{4,50}^{\alpha }:\left\{\begin{array}{c}{e}_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_1·e_4=e_4,e_2^2=e_4,e_3^2=e_4,\hfill \\ \left[e_2,e_3\right]=\alpha e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,51}:e_1^2=e_1,e_2^2=e_2,e_3^2=e_3.$
• ${\mathcal{P}}_{4,52}:e_1^2=e_1,e_2^2=e_2,e_2·e_3=e_3.$
• ${\mathcal{P}}_{4,53}:e_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3.$
• ${\mathcal{P}}_{4,54}:\left\{\begin{array}{c}{e}_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,\hfill \\ \left[e_2,e_3\right]=e_2.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,55}:e_1^2=e_1,e_1·e_2=e_2,e_1·e_3=e_3,e_2^2=e_3.$
• ${\mathcal{P}}_{4,56}:e_1^2=e_1,e_2^2=e_2.$
• ${\mathcal{P}}_{4,57}:\left\{\begin{array}{c}{e}_1^2=e_1,e_2^2=e_2,\hfill \\ \left[e_3,e_4\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,58}:e_1^2=e_1,e_2^2=e_2,e_3^2=e_4.$
• ${\mathcal{P}}_{4,59}:e_1^2=e_1,e_1·e_2=e_2.$
• ${\mathcal{P}}_{4,60}:\left\{\begin{array}{c}{e}_1^2=e_1,e_1·e_2=e_2,\hfill \\ \left[e_3,e_4\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,61}:e_1^2=e_1,e_1·e_2=e_2,e_3^2=e_4.$
• ${\mathcal{P}}_{4,62}^{\alpha }:\left\{\begin{array}{c}{e}_1^2=e_1,\hfill \\ \left[e_2,e_4\right]=\alpha e_2,\left[e_3,e_4\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,63}:\left\{\begin{array}{c}{e}_1^2=e_1,\hfill \\ \left[e_2,e_4\right]=e_2,\left[e_3,e_4\right]=e_2+e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,64}:e_1^2=e_1.$
• ${\mathcal{P}}_{4,65}:\left\{\begin{array}{c}{e}_1^2=e_1,\hfill \\ \left[e_3,e_4\right]=e_2.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,66}:\left\{\begin{array}{c}{e}_1^2=e_1,\hfill \\ \left[e_3,e_4\right]=e_2,\left[e_2,e_3\right]=e_3,\left[e_2,e_4\right]=-e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,67}^{\alpha }:\left\{\begin{array}{c}{e}_1^2=e_1,e_2·e_3=e_4,\hfill \\ \left[e_2,e_3\right]=\alpha e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,68}:\left\{\begin{array}{c}{e}_1^2=e_1,e_2^2=e_3,\hfill \\ \left[e_2,e_4\right]=e_4.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,69}:\left\{\begin{array}{c}l{e}_1^2=e_1,e_2^2=e_3,\hfill \\ \left[e_2,e_4\right]=e_3.\hfill \end{array}\right.$
• ${\mathcal{P}}_{4,70}:e_1^2=e_1,e_2^2=e_3.$
• ${\mathcal{P}}_{4,71}:e_1^2=e_1,e_2^2=e_3,e_2·e_3=e_4.$

• Between these algebras there are precisely the following isomorphisms:

• ${\mathcal{P}}_{4,4}^{\alpha }\cong {\mathcal{P}}_{4,4}^{\beta }$ if and only if $(\alpha -\beta )(\alpha \beta -1)=0$.
• ${\mathcal{P}}_{4,13}^{\alpha }\cong {\mathcal{P}}_{4,13}^{\beta }$ if and only if ${\alpha }^2={\beta }^2$.
• ${\mathcal{P}}_{4,16}^{\alpha }\cong {\mathcal{P}}_{4,16}^{\beta }$ if and only if ${\alpha }^2={\beta }^2$.
• ${\mathcal{P}}_{4,20}^{\alpha }\cong {\mathcal{P}}_{4,20}^{\beta }$ if and only if ${\alpha }^2={\beta }^2$.
• ${\mathcal{P}}_{4,25}^{\alpha }\cong {\mathcal{P}}_{4,25}^{\beta }$ if and only if $\alpha =\beta$.
• ${\mathcal{P}}_{4,28}^{\alpha }\cong {\mathcal{P}}_{4,28}^{\beta }$ if and only if $\alpha =\beta$.
• ${\mathcal{P}}_{4,31}^{\alpha }\cong {\mathcal{P}}_{4,31}^{\beta }$ if and only if ${\alpha }^2={\beta }^2$.
• ${\mathcal{P}}_{4,41}^{\alpha }\cong {\mathcal{P}}_{4,41}^{\beta }$ if and only if $(\alpha -\beta )(\alpha \beta -1)=0$.
• ${\mathcal{P}}_{4,50}^{\alpha }\cong {\mathcal{P}}_{4,50}^{\beta }$ if and only if ${\alpha }^2={\beta }^2$.
• ${\mathcal{P}}_{4,62}^{\alpha }\cong {\mathcal{P}}_{4,62}^{\beta }$ if and only if $(\alpha -\beta )(\alpha \beta -1)=0$.
• ${\mathcal{P}}_{4,67}^{\alpha }\cong {\mathcal{P}}_{4,67}^{\beta }$ if and only if ${\alpha }^2={\beta }^2$.

4. The Proof of Theorem 2

$\left(\mathcal{P},·\right)={\mathcal{A}}_{01}$. Then $\mathcal{P}$ is a Lie algebra. So $\mathcal{P}$ is isomorphic to one of the Lie algebras given in [10] (Lemma 3).

$\left(\mathcal{P},·\right)={\mathcal{A}}_{02}$. The automorphism group of ${\mathcal{A}}_{02}$, $\mathrm{Aut}\left({\mathcal{A}}_{02}\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\varphi =\left(\begin{array}{cccc}{a}_{11}& 0& 0& 0\\ a_{21}& a_{11}^2& a_{23}& a_{24}\\ a_{31}& 0& a_{33}& a_{34}\\ a_{41}& 0& a_{43}& a_{44}\end{array}\right).$$

Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then

$$\begin{array}{ccc}\hfill B_1& =& 0,\hfill \\ \hfill B_2& =& {\alpha }_1{\Delta }_{1,3}+{\alpha }_2{\Delta }_{1,4}+{\alpha }_3{\Delta }_{3,4},\hfill \\ \hfill B_3& =& {\alpha }_4{\Delta }_{1,3}+{\alpha }_5{\Delta }_{1,4}+{\alpha }_6{\Delta }_{3,4},\hfill \\ \hfill B_4& =& {\alpha }_7{\Delta }_{1,3}+{\alpha }_8{\Delta }_{1,4}+{\alpha }_9{\Delta }_{3,4},\hfill \end{array}$$

such that

$$\begin{array}{ccc}\hfill {\alpha }_4{\alpha }_9-{\alpha }_6{\alpha }_7& =& 0,\hfill \\ \hfill {\alpha }_5{\alpha }_9-{\alpha }_6{\alpha }_8& =& 0,\hfill \\ \hfill {\alpha }_1{\alpha }_6+{\alpha }_2{\alpha }_9-{\alpha }_3{\alpha }_4-{\alpha }_3{\alpha }_8& =& 0,\hfill \end{array}$$

for some ${\alpha }_1,\dots ,{\alpha }_9\in \mathbb{C}$. Let $\varphi =\left(a_{ij}\right)\in$$\mathrm{Aut}\left({\mathcal{A}}_{02}\right)$. Write

$$\theta \ast \varphi =\left(0,{\beta }_1{\Delta }_{1,3}+{\beta }_2{\Delta }_{1,4}+{\beta }_3{\Delta }_{3,4},{\beta }_4{\Delta }_{1,3}+{\beta }_5{\Delta }_{1,4}+{\beta }_6{\Delta }_{3,4},{\beta }_7{\Delta }_{1,3}+{\beta }_8{\Delta }_{1,4}+{\beta }_9{\Delta }_{3,4}\right).$$

Then

$$\begin{array}{ccc}\hfill {\beta }_6& =& {\alpha }_6{a}_{44}-{\alpha }_9{a}_{34},\hfill \\ \hfill {\beta }_9& =& {\alpha }_9{a}_{33}-{\alpha }_6{a}_{43}.\hfill \end{array}$$

By Remark 3, we may assume $\left({\alpha }_6,{\alpha }_9\right)\in \left\{\left(0,0\right),\left(1,0\right)\right\}$. Assume first that $\left({\alpha }_6,{\alpha }_9\right)=\left(0,0\right)$. Then, we have ${\alpha }_3\left({\alpha }_4+{\alpha }_8\right)=0$. Moreover, we have

$$\begin{array}{ccc}\hfill {\beta }_4& =& {\displaystyle \frac{a_{11}}{a_{33}{a}_{44}-a_{34}{a}_{43}}}\left({\alpha }_4{a}_{33}{a}_{44}+{\alpha }_5{a}_{43}{a}_{44}-{\alpha }_7{a}_{33}{a}_{34}-{\alpha }_8{a}_{34}{a}_{43}\right),\hfill \\ \hfill {\beta }_5& =& {\displaystyle \frac{a_{11}}{a_{33}{a}_{44}-a_{34}{a}_{43}}}\left({\alpha }_5{a}_{44}^2-{\alpha }_7{a}_{34}^2+{\alpha }_4{a}_{34}{a}_{44}-{\alpha }_8{a}_{34}{a}_{44}\right),\hfill \\ \hfill {\beta }_7& =& -{\displaystyle \frac{a_{11}}{a_{33}{a}_{44}-a_{34}{a}_{43}}}\left({\alpha }_5{a}_{43}^2-{\alpha }_7{a}_{33}^2+{\alpha }_4{a}_{33}{a}_{43}-{\alpha }_8{a}_{33}{a}_{43}\right),\hfill \\ \hfill {\beta }_8& =& -{\displaystyle \frac{a_{11}}{a_{33}{a}_{44}-a_{34}{a}_{43}}}\left({\alpha }_4{a}_{34}{a}_{43}+{\alpha }_5{a}_{43}{a}_{44}-{\alpha }_7{a}_{33}{a}_{34}-{\alpha }_8{a}_{33}{a}_{44}\right).\hfill \end{array}$$

Whence $\left(\begin{array}{cc}{\beta }_4& {\beta }_5\\ {\beta }_7& {\beta }_8\end{array}\right)=a_{11}{\left(\begin{array}{cc}{a}_{33}& a_{34}\\ a_{43}& a_{44}\end{array}\right)}^{-1}\left(\begin{array}{cc}{\alpha }_4& {\alpha }_5\\ {\alpha }_7& {\alpha }_8\end{array}\right)\left(\begin{array}{cc}{a}_{33}& a_{34}\\ a_{43}& a_{44}\end{array}\right)$. Thus we may assume:

$$\left(\begin{array}{cc}{\alpha }_4& {\alpha }_5\\ {\alpha }_7& {\alpha }_8\end{array}\right)\in \left\{\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}1& 0\\ 0& \alpha \end{array}\right),\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right),\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\right\}.$$

So, the following cases arise:

• $\left(\begin{array}{cc}{\alpha }_4& {\alpha }_5\\ {\alpha }_7& {\alpha }_8\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$.

  ❖

  ${\alpha }_3=0$. If $\left({\alpha }_1,{\alpha }_2\right)=\left(0,0\right)$, then $\theta =0$, and we get the algebra ${\mathcal{P}}_{4,1}$. Otherwise, let $\varphi$ be the first of the following matrices if ${\alpha }_2=0$ or the second if ${\alpha }_2\ne 0$:

  $$\left(\begin{array}{cccc}{\alpha }_1& 0& 0& 0\\ 0& {\alpha }_1^2& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 1\\ 0& 0& {\displaystyle \frac{1}{{\alpha }_2}}& -{\displaystyle \frac{{\alpha }_1}{{\alpha }_2}}\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,{\Delta }_{1,3},0,0\right)$. So we obtain the representative $\left(0,{\Delta }_{1,3},0,0\right)$. Hence we get the Poisson algebra ${\mathcal{P}}_{4,2}$.

  ❖

  ${\alpha }_3\ne 0$. We define $\varphi$ to be the following automorphism:

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ -{\displaystyle \frac{{\alpha }_2}{{\alpha }_3}}& 0& 1& 0\\ {\displaystyle \frac{{\alpha }_1}{{\alpha }_3}}& 0& 0& {\displaystyle \frac{1}{{\alpha }_3}}\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,{\Delta }_{3,4},0,0\right)$. So we get the Poisson algebra ${\mathcal{P}}_{4,3}$.

• $\left(\begin{array}{cc}{\alpha }_4& {\alpha }_5\\ {\alpha }_7& {\alpha }_8\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& \alpha \end{array}\right)$.

  ❖

  ${\alpha }_3=0$. If $\alpha \ne 0$, we choose $\varphi$ as follows:

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\alpha }_1& {\displaystyle \frac{{\alpha }_2}{\alpha }}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,0,{\Delta }_{1,3},\alpha {\Delta }_{1,4}\right)$. Hence we get the Poisson algebras ${\mathcal{P}}_{4,4}^{\alpha \ne 0}$. Furthermore, the Poisson algebras ${\mathcal{P}}_{4,4}^{\alpha \ne 0}$ and ${\mathcal{P}}_{4,4}^{\beta \ne 0}$ are isomorphic if and only if $\left(\alpha -\beta \right)\left(\alpha \beta -1\right)=0$. If $\alpha =0$, we define $\varphi$ to be the first of the following matrices if ${\alpha }_2=0$ or the second if ${\alpha }_2\ne 0$:

  $$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\alpha }_1& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\alpha }_1& 0\\ 0& 0& 1& 0\\ 0& 0& 0& {\displaystyle \frac{1}{{\alpha }_2}}\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,0,{\Delta }_{1,3},0\right)$ if ${\alpha }_2=0$ or $\theta \ast \varphi =\left(0,{\Delta }_{1,4},{\Delta }_{1,3},0\right)$ if ${\alpha }_2\ne 0$. So we get the Poisson algebra ${\mathcal{P}}_{4,4}^{\alpha =0}$ if ${\alpha }_2=0$ or the Poisson algebra ${\mathcal{P}}_{4,5}$ if ${\alpha }_2\ne 0$.

  ❖

  ${\alpha }_3\ne 0$. Since ${\alpha }_3\left({\alpha }_4+{\alpha }_8\right)=0$, we have $\alpha =-1$. Choose $\varphi$ as follows:

  $$\varphi ={\displaystyle \frac{1}{{\alpha }_3}}\left(\begin{array}{cccc}{\alpha }_3& 0& 0& 0\\ 0& {\alpha }_3& 0& 0\\ -{\alpha }_2& 0& {\alpha }_3& 0\\ {\alpha }_1& 0& 0& 1\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,{\Delta }_{3,4},{\Delta }_{1,3},-{\Delta }_{1,4}\right)$. So we get the Poisson algebra ${\mathcal{P}}_{4,6}$.

• $\left(\begin{array}{cc}{\alpha }_4& {\alpha }_5\\ {\alpha }_7& {\alpha }_8\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$.
  Since ${\alpha }_3\left({\alpha }_4+{\alpha }_8\right)=0$, we have ${\alpha }_3=0$. Let $\varphi$ be the following matrix:

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\alpha }_1& {\alpha }_2-{\alpha }_1\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right).$$
  Then $\theta \ast \varphi =\left(0,0,{\Delta }_{1,3}+{\Delta }_{1,4},{\Delta }_{1,4}\right)$. Therefore, we get the Poisson algebra ${\mathcal{P}}_{4,7}$.
• $\left(\begin{array}{cc}{\alpha }_4& {\alpha }_5\\ {\alpha }_7& {\alpha }_8\end{array}\right)=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$.

  ❖

  ${\alpha }_3=0$. Let $\varphi$ be the first of the following matrices if ${\alpha }_1=0$ or the second if ${\alpha }_1\ne 0$:

  $$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\alpha }_2& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}{\alpha }_1& 0& 0& 0\\ 0& {\alpha }_1^2& {\alpha }_2& 0\\ 0& 0& 1& 0\\ 0& 0& 0& {\displaystyle \frac{1}{{\alpha }_1}}\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,0,{\Delta }_{1,4},0\right)$ if ${\alpha }_1=0$ or $\theta \ast \varphi =\left(0,{\Delta }_{1,3},{\Delta }_{1,4},0\right)$ if ${\alpha }_1\ne 0$. Therefore we obtain the Poisson algebras ${\mathcal{P}}_{4,8}$ and ${\mathcal{P}}_{4,9}$.

  ❖

  ${\alpha }_3\ne 0$. Choose $\varphi$ as follows:

  $$\varphi =\left(\begin{array}{cccc}{\alpha }_3& 0& 0& 0\\ 0& {\alpha }_3^2& {\alpha }_2{\alpha }_3& 0\\ 0& 0& {\alpha }_3& 0\\ {\alpha }_1& 0& 0& 1\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,{\Delta }_{3,4},{\Delta }_{1,4},0\right)$ and we get the algebra ${\mathcal{P}}_{4,10}$.

Let us assume now that $\left({\alpha }_6,{\alpha }_9\right)=\left(1,0\right)$. Then ${\alpha }_7={\alpha }_8=0$ and ${\alpha }_1={\alpha }_3{\alpha }_4$. Set $\lambda ={\alpha }_2-{\alpha }_3{\alpha }_5$. Let $\varphi$ be the first of the following matrices if $\lambda =0$ or the second if $\lambda \ne 0$:

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\alpha }_3& 0\\ -{\alpha }_5& 0& 1& 0\\ {\alpha }_4& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}\lambda & 0& 0& 0\\ 0& {\lambda }^2& {\alpha }_3& 0\\ -\lambda {\alpha }_5& 0& 1& 0\\ \lambda {\alpha }_4& 0& 0& 1\end{array}\right).$$

Then $\theta \ast \varphi =\left(0,0,{\Delta }_{3,4},0\right)$ if $\lambda =0$ or $\theta \ast \varphi =\left(0,{\Delta }_{1,4},{\Delta }_{3,4},0\right)$ if $\lambda \ne 0$. Thus, we obtain the Poisson algebras ${\mathcal{P}}_{4,11}$ and ${\mathcal{P}}_{4,12}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{03}$. Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then

$$\begin{array}{ccc}\hfill B_1& =& B_2=0,\hfill \\ \hfill B_3& =& {\alpha }_1{\Delta }_{1,2}+{\alpha }_2{\Delta }_{1,4}+{\alpha }_3{\Delta }_{2,4},\hfill \\ \hfill B_4& =& {\alpha }_4{\Delta }_{1,2}+{\alpha }_5{\Delta }_{1,4}+{\alpha }_6{\Delta }_{2,4},\hfill \end{array}$$

such that

$${\alpha }_2{\alpha }_6-{\alpha }_3{\alpha }_5=0,$$

for some ${\alpha }_1,\dots ,{\alpha }_6\in \mathbb{C}$. Moreover, the automorphism group of ${\mathcal{A}}_{03}$, $\mathrm{Aut}\left({\mathcal{A}}_{03}\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\left(\begin{array}{cccc}{a}_{11}& \epsilon a_{21}& 0& 0\\ a_{21}& ϵa_{11}& 0& 0\\ a_{31}& a_{32}& a_{11}^2+a_{21}^2& a_{34}\\ a_{41}& a_{42}& 0& a_{44}\end{array}\right):\left(\epsilon ,ϵ\right)\in \left\{\left(1,-1\right),\left(-1,1\right)\right\}.$$

Let $\varphi =\left(a_{ij}\right)\in$$\mathrm{Aut}\left({\mathcal{A}}_{03}\right)$. Then $\theta \ast \varphi =(0,0,{\beta }_1{\Delta }_{1,2}+{\beta }_2{\Delta }_{1,4}+{\beta }_3{\Delta }_{2,4},{\beta }_4{\Delta }_{1,2}+{\beta }_5{\Delta }_{1,4}+{\beta }_6{\Delta }_{2,4})$ where

$$\begin{array}{ccc}\hfill {\beta }_1& =& {\displaystyle \frac{1}{a_{44}\left(a_{11}^2+a_{21}^2\right)}}\left(\begin{array}{c}{\alpha }_2{a}_{11}{a}_{42}{a}_{44}+{\alpha }_3{a}_{21}{a}_{42}{a}_{44}-{\alpha }_5{a}_{11}{a}_{42}{a}_{34}-{\alpha }_6{a}_{21}{a}_{42}{a}_{34}+ϵ{\alpha }_1{a}_{11}^2{a}_{44}-\epsilon {\alpha }_1{a}_{21}^2{a}_{44}\\ -ϵ{\alpha }_4{a}_{11}^2{a}_{34}+\epsilon {\alpha }_4{a}_{21}^2{a}_{34}-ϵ{\alpha }_3{a}_{11}{a}_{41}{a}_{44}-\epsilon {\alpha }_2{a}_{21}{a}_{41}{a}_{44}+ϵ{\alpha }_6{a}_{11}{a}_{41}{a}_{34}+\epsilon {\alpha }_5{a}_{21}{a}_{41}{a}_{34}\end{array}\right),\hfill \\ \hfill {\beta }_2& =& {\displaystyle \frac{1}{a_{11}^2+a_{21}^2}}\left({\alpha }_2{a}_{11}{a}_{44}+{\alpha }_3{a}_{21}{a}_{44}-{\alpha }_5{a}_{11}{a}_{34}-{\alpha }_6{a}_{21}{a}_{34}\right),\hfill \\ \hfill {\beta }_3& =& {\displaystyle \frac{1}{a_{11}^2+a_{21}^2}}\left(ϵ{\alpha }_3{a}_{11}{a}_{44}+\epsilon {\alpha }_2{a}_{21}{a}_{44}-ϵ{\alpha }_6{a}_{11}{a}_{34}-\epsilon {\alpha }_5{a}_{21}{a}_{34}\right),\hfill \\ \hfill {\beta }_4& =& {\displaystyle \frac{1}{a_{44}}}\left(ϵ{\alpha }_4{a}_{11}^2-\epsilon {\alpha }_4{a}_{21}^2+{\alpha }_5{a}_{11}{a}_{42}+{\alpha }_6{a}_{21}{a}_{42}-ϵ{\alpha }_6{a}_{11}{a}_{41}-\epsilon {\alpha }_5{a}_{21}{a}_{41}\right),\hfill \\ \hfill {\beta }_5& =& {\alpha }_5{a}_{11}+{\alpha }_6{a}_{21},\hfill \\ \hfill {\beta }_6& =& ϵ{\alpha }_6{a}_{11}+\epsilon {\alpha }_5{a}_{21}.\hfill \end{array}$$

Assume first that ${\alpha }_5^2+{\alpha }_6^2\ne 0$. Let $\varphi$ be the following automorphism:

$$\varphi ={\displaystyle \frac{1}{{\alpha }_5^2+{\alpha }_6^2}}\left(\begin{array}{cccc}{\alpha }_5& {\alpha }_6& 0& 0\\ {\alpha }_6& -{\alpha }_5& 0& 0\\ 0& 0& 1& {\alpha }_2{\alpha }_5+{\alpha }_3{\alpha }_6\\ 0& {\alpha }_4& 0& {\alpha }_5^2+{\alpha }_6^2\end{array}\right).$$

Then $\theta \ast \varphi =\left(0,0,\alpha {\Delta }_{1,2},{\Delta }_{1,4}\right)$ for some $\alpha \in \mathbb{C}$. So we get the representatives ${\theta }^{\alpha }=\left(0,0,\alpha {\Delta }_{1,2},{\Delta }_{1,4}\right)$. Further, the representatives ${\theta }^{\alpha }$ and ${\theta }^{\beta }$ are in the same orbit if and only if ${\alpha }^2={\beta }^2$. So we get the Poisson algebras ${\mathcal{P}}_{4,13}^{\alpha }$. Assume now that ${\alpha }_5^2+{\alpha }_6^2=0$ (i.e., ${\alpha }_6=\pm i{\alpha }_5$ where $i=\sqrt{-1}$).

• ${\alpha }_5=0$. Then so is ${\alpha }_6=0$. Assume first that ${\alpha }_2^2+{\alpha }_3^2\ne 0$. If ${\alpha }_4=0$, we define $\varphi$ to be the following automorphism:

  $$\varphi ={\displaystyle \frac{1}{{\alpha }_2^2+{\alpha }_3^2}}\left(\begin{array}{cccc}{\alpha }_2& {\alpha }_3& 0& 0\\ {\alpha }_3& -{\alpha }_2& 0& 0\\ 0& 0& 1& 0\\ 0& {\alpha }_1& 0& 1\end{array}\right).$$
  Then $\theta \ast \varphi =\left(0,0,{\Delta }_{1,4},0\right)$. So we get the Poisson algebra ${\mathcal{P}}_{4,14}$. If ${\alpha }_4\ne 0$, we define $\varphi$ to be the following automorphism:

  $$\varphi ={\displaystyle \frac{1}{{\alpha }_4^2\left({\alpha }_2^2+{\alpha }_3^2\right)}}\left(\begin{array}{cccc}{\alpha }_2{\alpha }_4& -{\alpha }_3{\alpha }_4& 0& 0\\ {\alpha }_3{\alpha }_4& {\alpha }_2{\alpha }_4& 0& 0\\ 0& 0& 1& 0\\ 0& -{\alpha }_1{\alpha }_4& 0& {\alpha }_4\end{array}\right).$$
  Then $\theta \ast \varphi =\left(0,0,{\Delta }_{1,4},{\Delta }_{1,2}\right)$. So we get the Poisson algebra ${\mathcal{P}}_{4,15}$. Assume now that ${\alpha }_3=\pm i{\alpha }_2$ where $i=\sqrt{-1}$.

  ❖

  ${\alpha }_2=0$. Then ${\alpha }_3=0$. If ${\alpha }_4=0$, then $\theta =\left(0,0,\alpha {\Delta }_{1,2},0\right)$ for some $\alpha \in \mathbb{C}$. So we have the representatives ${\vartheta }^{\alpha }=\left(0,0,\alpha {\Delta }_{1,2},0\right)$. Moreover, for any $\varphi =\left(a_{ij}\right)\in$ $\mathrm{Aut}\left({\mathcal{A}}_{03}\right)$, we have ${\vartheta }^{\alpha }\ast \varphi =\left(0,0,\beta {\Delta }_{1,2},0\right)$ with ${\alpha }^2={\beta }^2$. Thus the representatives ${\vartheta }^{\alpha }, {\vartheta }^{\beta }$ are in the same orbit if and only if ${\alpha }^2={\beta }^2$. Hence we get the algebras ${\mathcal{P}}_{4,16}^{\alpha }$. If ${\alpha }_4\ne 0$, we choose $\varphi$ as follows:

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& {\alpha }_1\\ 0& 0& 0& {\alpha }_4\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,0,0,{\Delta }_{1,2}\right)$. So we get the Poisson algebra ${\mathcal{P}}_{4,17}$.

  ❖

  ${\alpha }_2\ne 0$. Then ${\alpha }_3\ne 0$. If ${\alpha }_4=0$, we choose $\varphi$ to be the first of the following matrices when ${\alpha }_3=i{\alpha }_2$ or the second when ${\alpha }_3=-i{\alpha }_2$:

  $$\left(\begin{array}{cccc}{\alpha }_2& 0& 0& 0\\ 0& {\alpha }_2& 0& 0\\ 0& 0& {\alpha }_2^2& 0\\ -i{\alpha }_1& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}{\alpha }_2& 0& 0& 0\\ 0& -{\alpha }_2& 0& 0\\ 0& 0& {\alpha }_2^2& 0\\ i{\alpha }_1& 0& 0& 1\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,0,{\Delta }_{1,4}+i{\Delta }_{2,4},0\right)$. So we get the Poisson algebra ${\mathcal{P}}_{4,18}$.

  If ${\alpha }_4\ne 0$, we choose $\varphi$ to be the first of the following matrices if ${\alpha }_3=i{\alpha }_2$ or the second if ${\alpha }_3=-i{\alpha }_2$:

  $${\displaystyle \frac{1}{{\alpha }_2^2{\alpha }_4^2}}\left(\begin{array}{cccc}{\alpha }_2{\alpha }_4& 0& 0& 0\\ 0& {\alpha }_2{\alpha }_4& 0& 0\\ 0& 0& 1& {\alpha }_1\\ 0& 0& 0& {\alpha }_4\end{array}\right),{\displaystyle \frac{1}{{\alpha }_2^2{\alpha }_4^2}}\left(\begin{array}{cccc}-{\alpha }_2{\alpha }_4& 0& 0& 0\\ 0& {\alpha }_2{\alpha }_4& 0& 0\\ 0& 0& 1& -{\alpha }_1\\ 0& 0& 0& -{\alpha }_4\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,0,{\Delta }_{1,4}+i{\Delta }_{2,4},{\Delta }_{1,2}\right)$. So we get the Poisson algebra ${\mathcal{P}}_{4,19}$.

• ${\alpha }_5\ne 0$. Then ${\alpha }_6\ne 0$. Let $\varphi$ be the first of the following matrices when ${\alpha }_6=i{\alpha }_5$ or the second when ${\alpha }_6=-i{\alpha }_5$:

  $${\displaystyle \frac{1}{{\alpha }_5^2}}\left(\begin{array}{cccc}{\alpha }_5& 0& 0& 0\\ 0& {\alpha }_5& 0& 0\\ 0& 0& 1& {\alpha }_2{\alpha }_5\\ 0& -{\alpha }_4& 0& {\alpha }_5^2\end{array}\right),{\displaystyle \frac{1}{{\alpha }_5^2}}\left(\begin{array}{cccc}{\alpha }_5& 0& 0& 0\\ 0& -{\alpha }_5& 0& 0\\ 0& 0& 1& {\alpha }_2{\alpha }_5\\ 0& {\alpha }_4& 0& {\alpha }_5^2\end{array}\right).$$
  Then $\theta \ast \varphi =\left(0,0,\alpha {\Delta }_{1,2},{\Delta }_{1,4}+i{\Delta }_{2,4}\right)$ for some $\alpha \in \mathbb{C}$. So we have the representatives ${\eta }^{\alpha }=\left(0,0,\alpha {\Delta }_{1,2},{\Delta }_{1,4}+i{\Delta }_{2,4}\right)$. Moreover, the representatives ${\eta }^{\alpha }, {\eta }^{\beta }$ are in the same orbit if and only if ${\alpha }^2={\beta }^2$. So we get the Poisson algebras ${\mathcal{P}}_{4,20}^{\alpha }$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{04}$. Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then $\theta =\left(0,0,{\alpha }_1{\Delta }_{1,4},{\alpha }_2{\Delta }_{1,4}\right)$ for some ${\alpha }_1,{\alpha }_2\in \mathbb{C}$. The automorphism group of ${\mathcal{A}}_{04}$, $\mathrm{Aut}\left({\mathcal{A}}_{04}\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\left(\begin{array}{cccc}{a}_{11}& 0& 0& 0\\ a_{21}& a_{11}^2& 0& 0\\ a_{31}& 2a_{11}{a}_{21}& a_{11}^3& a_{34}\\ a_{41}& 0& 0& a_{44}\end{array}\right).$$

Let $\varphi =\left(a_{ij}\right)\in$$\mathrm{Aut}\left({\mathcal{A}}_{04}\right)$. Then $\theta \ast \varphi =\left(0,0,{\beta }_1{\Delta }_{1,4},{\beta }_2{\Delta }_{1,4}\right)$ where

$$\begin{array}{ccc}\hfill {\beta }_1& =& {\displaystyle \frac{1}{a_{11}^2}}\left({\alpha }_1{a}_{44}-{\alpha }_2{a}_{34}\right),\hfill \\ \hfill {\beta }_2& =& {\alpha }_2{a}_{11}.\hfill \end{array}$$

If $\theta =0$, we get the Poisson algebra ${\mathcal{P}}_{4,21}$. Assume now that $\theta \ne 0$. If ${\alpha }_2\ne 0$, we choose $\varphi$ as follows:

$$\varphi ={\displaystyle \frac{1}{{\alpha }_2^3}}\left(\begin{array}{cccc}{\alpha }_2^2& 0& 0& 0\\ 0& {\alpha }_2& 0& 0\\ 0& 0& 1& {\alpha }_1{\alpha }_2^2\\ 0& 0& 0& {\alpha }_2^3\end{array}\right).$$

Then $\theta \ast \varphi =\left(0,0,0,{\Delta }_{1,4}\right)$. Hence we get the Poisson algebra ${\mathcal{P}}_{4,22}$. If ${\alpha }_2=0$, we choose $\varphi$ as follows:

$$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& {\displaystyle \frac{1}{{\alpha }_1}}\end{array}\right).$$

Then $\theta \ast \varphi =\left(0,0,{\Delta }_{1,4},0\right)$. So we get the Poisson algebra ${\mathcal{P}}_{4,23}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{05}$. It is easy to see that the algebra ${\mathcal{A}}_{05}$ is isomorphic to the following algebra

$${\mathcal{A}}_{05}^{\prime }:e_1^2=e_3,e_2^2=e_4.$$

So we may assume $\left(\mathcal{P},·\right)={\mathcal{A}}_{05}^{\prime }$. Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then $\theta =\left(0,0,{\alpha }_1{\Delta }_{1,2},{\alpha }_2{\Delta }_{1,2}\right)$ for some ${\alpha }_1,{\alpha }_2\in \mathbb{C}$. The automorphism group of ${\mathcal{A}}_{05}^{\prime }$, $\mathrm{Aut}\left({\mathcal{A}}_{05}^{\prime }\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\left(\begin{array}{cccc}ϵa_{11}& \left(1-ϵ\right)a_{12}& 0& 0\\ \left(1-ϵ\right)a_{21}& ϵa_{22}& 0& 0\\ a_{31}& a_{32}& ϵa_{11}^2& \left(1-ϵ\right)a_{12}^2\\ a_{41}& a_{42}& \left(1-ϵ\right)a_{21}^2& ϵa_{22}^2\end{array}\right):ϵ\in \left\{0,1\right\}.$$

Let $\varphi =\left(a_{ij}\right)\in$$\mathrm{Aut}\left({\mathcal{A}}_{05}^{\prime }\right)$. Then $\theta \ast \varphi =\left(0,0,{\beta }_1{\Delta }_{1,2},{\beta }_2{\Delta }_{1,2}\right)$ where ${\beta }_1={\displaystyle \frac{a_{22}}{a_{11}}}{\alpha }_1, {\beta }_2={\alpha }_2{\displaystyle \frac{a_{11}}{a_{22}}}$ if $ϵ=1$ or ${\beta }_1=-{\alpha }_2{\displaystyle \frac{a_{12}}{a_{21}}}, {\beta }_2=-{\displaystyle \frac{{\alpha }_1}{a_{12}}}{a}_{21}$ if $ϵ=0$. If $\theta =0$, we get the Poisson algebra ${\mathcal{P}}_{4,24}$. If $\theta \ne 0$, then we may assume without any loss of generality that ${\alpha }_1\ne 0$. Let $\varphi$ be the following automorphism:

$$\varphi =\left(\begin{array}{cccc}{\alpha }_1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& {\alpha }_1^2& 0\\ 0& 0& 0& 1\end{array}\right).$$

Then $\theta \ast \varphi =\left(0,0,{\Delta }_{1,2},\alpha {\Delta }_{1,2}\right)$ for some $\alpha \in \mathbb{C}$. Hence we get the representatives ${\theta }^{\alpha }=\left(0,0,{\Delta }_{1,2},\alpha {\Delta }_{1,2}\right)$. Moreover, ${\theta }^{\alpha }$ and ${\theta }^{\beta }$ are in the same orbit if and only if $\alpha =\beta$. Hence we get the Poisson algebras ${\mathcal{P}}_{4,25}^{\alpha }$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{06}$. Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then

$$\theta =({\alpha }_1{\Delta }_{1,2},0,{\alpha }_2{\Delta }_{1,2},{\alpha }_3{\Delta }_{1,2}+2{\alpha }_1{\Delta }_{1,3}-{\alpha }_1{\Delta }_{2,4})$$

for some ${\alpha }_1,{\alpha }_2,{\alpha }_3\in \mathbb{C}$. The automorphism group of ${\mathcal{A}}_{06}$, $\mathrm{Aut}\left({\mathcal{A}}_{06}\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\left(\begin{array}{cccc}{a}_{11}& a_{12}& 0& 0\\ 0& a_{22}& 0& 0\\ a_{31}& a_{32}& a_{22}^2& 0\\ a_{41}& a_{42}& 2a_{12}{a}_{22}& a_{11}{a}_{22}\end{array}\right).$$

Let $\varphi =\left(a_{ij}\right)\in$ Aut$\left({\mathcal{A}}_{06}\right)$. Then $\theta \ast \varphi =({\beta }_1{\Delta }_{1,2},0,{\beta }_2{\Delta }_{1,2},{\beta }_3{\Delta }_{1,2}+2{\beta }_1{\Delta }_{1,3}-{\beta }_1{\Delta }_{2,4})$ where

$$\begin{array}{ccc}\hfill {\beta }_1& =& {\alpha }_1{a}_{22},\hfill \\ \hfill {\beta }_2& =& {\displaystyle \frac{1}{a_{22}}}\left({\alpha }_2{a}_{11}-{\alpha }_1{a}_{31}\right),\hfill \\ \hfill {\beta }_3& =& {\displaystyle \frac{1}{a_{22}}}\left(2{\alpha }_1{a}_{32}-2{\alpha }_2{a}_{12}+{\alpha }_3{a}_{22}\right).\hfill \end{array}$$

Let us consider the following cases:

• ${\alpha }_1\ne 0$. Let $\varphi$ be the following automorphism:

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& {\displaystyle \frac{1}{{\alpha }_1}}& 0& 0\\ {\displaystyle \frac{{\alpha }_2}{{\alpha }_1}}& -{\displaystyle \frac{1}{2{\alpha }_1^2}}{\alpha }_3& {\displaystyle \frac{1}{{\alpha }_1^2}}& 0\\ 0& 0& 0& {\displaystyle \frac{1}{{\alpha }_1}}\end{array}\right).$$
  Then $\theta \ast \varphi =({\Delta }_{1,2},0,0,2{\Delta }_{1,3}-{\Delta }_{2,4})$. So we obtain the Poisson algebra ${\mathcal{P}}_{4,26}$.
• ${\alpha }_1=0,{\alpha }_2\ne 0$. Let $\varphi$ be the following automorphism:

  $$\varphi =\left(\begin{array}{cccc}1& {\displaystyle \frac{1}{2}}{\alpha }_3& 0& 0\\ 0& {\alpha }_2& 0& 0\\ 0& 0& {\alpha }_2^2& 0\\ 0& 0& {\alpha }_2{\alpha }_3& {\alpha }_2\end{array}\right).$$
  Then $\theta \ast \varphi =(0,0,{\Delta }_{1,2},0)$. So we obtain the Poisson algebra ${\mathcal{P}}_{4,27}$.
• ${\alpha }_1={\alpha }_2=0$. Then $\theta \ast \varphi =\theta$ for any $\varphi \in$ Aut$\left({\mathcal{A}}_{06}\right)$. Thus we have the representatives ${\theta }^{\alpha }=\left(0,0,0,\alpha {\Delta }_{1,2}\right)$, and ${\theta }^{\alpha },$${\theta }^{\beta }$ are in the same orbit if and only if $\alpha =\beta$. Therefore, we obtain the Poisson algebras ${\mathcal{P}}_{4,28}^{\alpha }$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{07}$. Since ${\mathcal{A}}_{07}$ is isomorphic to ${\mathcal{A}}_{07}^{\prime }:e_1·e_2=e_4,e_3^2=e_4$, we may assume $\left(\mathcal{P},·\right)={\mathcal{A}}_{07}^{\prime }$. Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then

$$\theta =(-{\alpha }_1{\Delta }_{1,3},{\alpha }_1{\Delta }_{2,3},{\alpha }_1{\Delta }_{1,2},{\alpha }_2{\Delta }_{1,2}+{\alpha }_3{\Delta }_{1,3}+{\alpha }_4{\Delta }_{2,3})$$

for some ${\alpha }_1,\dots ,{\alpha }_4\in \mathbb{C}$. The automorphism group of ${\mathcal{A}}_{07}^{\prime }$, $\mathrm{Aut}\left({\mathcal{A}}_{07}^{\prime }\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\varphi =\left(\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& 0\\ a_{21}& a_{22}& a_{23}& 0\\ a_{31}& a_{32}& a_{33}& 0\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right)$$

such that

$$\left(\begin{array}{ccc}{a}_{23}& a_{13}& a_{33}\\ a_{22}& a_{12}& a_{32}\\ a_{21}& a_{11}& a_{31}\end{array}\right)\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right)=\left(\begin{array}{ccc}0& 0& a_{44}\\ a_{44}& 0& 0\\ 0& a_{44}& 0\end{array}\right).$$

Let us consider the following cases:

• ${\alpha }_1\ne 0$. Let us define $\varphi$ as follows:

  $$\varphi ={\displaystyle \frac{1}{{\alpha }_1^2}}\left(\begin{array}{cccc}{\alpha }_1& 0& 0& 0\\ 0& {\alpha }_1& 0& 0\\ 0& 0& {\alpha }_1& 0\\ -{\alpha }_3& {\alpha }_4& {\alpha }_2& 1\end{array}\right).$$
  Then $\theta \ast \varphi =\left(-{\Delta }_{1,3},{\Delta }_{2,3},{\Delta }_{1,2},0\right)$. So we get the Poisson algebra ${\mathcal{P}}_{4,29}$.
• ${\alpha }_1=0,{\alpha }_3\ne 0$. Set $\lambda ={\alpha }_2^2-2{\alpha }_3{\alpha }_4$. Let $\varphi$ be the first of the following matrices if $\lambda =0$ or the second if $\lambda \ne 0$:

  $$\left(\begin{array}{cccc}{\displaystyle \frac{1}{{\alpha }_3}}& -{\alpha }_4& {\displaystyle \frac{{\alpha }_2}{{\alpha }_3}}& 0\\ 0& {\alpha }_3& 0& 0\\ 0& -{\alpha }_2& 1& 0\\ 0& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}-{\displaystyle \frac{1}{{\alpha }_3^2}}\left({\alpha }_2^2-{\alpha }_3{\alpha }_4+\sqrt{\lambda }{\alpha }_2\right)& -{\displaystyle \frac{1}{2\lambda }}\left({\alpha }_3{\alpha }_4-{\alpha }_2^2+\sqrt{\lambda }{\alpha }_2\right)& {\displaystyle \frac{1}{\sqrt{\lambda }}}{\alpha }_4& 0\\ 1& -{\displaystyle \frac{1}{2\lambda }}{\alpha }_3^2& -{\displaystyle \frac{1}{\sqrt{\lambda }}}{\alpha }_3& 0\\ -{\displaystyle \frac{1}{{\alpha }_3}}\left({\alpha }_2+\sqrt{\lambda }\right)& -{\displaystyle \frac{1}{2{\lambda }^{\frac{3}{2}}}}{\alpha }_3\left(\lambda -\sqrt{\lambda }{\alpha }_2\right)& {\displaystyle \frac{1}{\sqrt{\lambda }}}{\alpha }_2& 0\\ 0& 0& 0& 1\end{array}\right).$$
  Then $\theta \ast \varphi =\left(0,0,0,{\Delta }_{1,3}\right)$ if $\lambda =0$ or $\theta \ast \varphi =\left(0,0,0,\alpha {\Delta }_{1,2}\right)$ with $\alpha \ne 0$ if $\lambda \ne 0$. Thus we obtain the Poisson algebras ${\mathcal{P}}_{4,30}$ and ${\mathcal{P}}_{4,31}^{\alpha \ne 0}$. Moreover, the algebras ${\mathcal{P}}_{4,31}^{\alpha \ne 0}$ and ${\mathcal{P}}_{4,31}^{\alpha \ne 0}$ are isomorphic if and only if ${\alpha }^2={\beta }^2$.
• ${\alpha }_1=0,{\alpha }_3=0$.

  ❖

  ${\alpha }_2\ne 0$. If we choose $\varphi$ as

  $$\varphi =\left(\begin{array}{cccc}1& -{\displaystyle \frac{1}{2{\alpha }_2^2}}{\alpha }_4^2& {\displaystyle \frac{1}{{\alpha }_2}}{\alpha }_4& 0\\ 0& 1& 0& 0\\ 0& -{\displaystyle \frac{1}{{\alpha }_2}}{\alpha }_4& 1& 0\\ 0& 0& 0& 1\end{array}\right),$$

  then $\theta \ast \varphi =\left(0,0,0,{\alpha }_2{\Delta }_{1,2}\right)$ and so we again get the Poisson algebras ${\mathcal{P}}_{4,31}^{\alpha \ne 0}$.

  ❖

  ${\alpha }_2=0$. If ${\alpha }_4=0$, then $\theta =0$, and we get the algebra ${\mathcal{P}}_{4,31}^{\alpha =0}$. If ${\alpha }_4\ne 0$, we choose $\varphi$ to be the following automorphism:

  $$\varphi =\left(\begin{array}{cccc}0& -{\alpha }_4& 0& 0\\ -{\displaystyle \frac{1}{{\alpha }_4}}& 0& 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& 1\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,0,0,{\Delta }_{1,3}\right)$ and we have again the algebra ${\mathcal{P}}_{4,30}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{08}$. Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then $\theta =\left(0,0,0,\alpha {\Delta }_{1,3}\right)$ for some $\alpha \in \mathbb{C}$. If $\alpha =0$, we then get the algebra ${\mathcal{P}}_{4,32}$. If $\alpha \ne 0$, we define $\varphi$ to be the following diagonal matrix:

$$\varphi =\left(\begin{array}{cccc}{\alpha }^2& 0& 0& 0\\ 0& {\alpha }^4& 0& 0\\ 0& 0& {\alpha }^3& 0\\ 0& 0& 0& {\alpha }^6\end{array}\right).$$

Then $\varphi \in \mathrm{Aut}\left({\mathcal{A}}_{08}\right)$ and $\theta \ast \varphi =\left(0,0,0,{\Delta }_{1,3}\right)$. Hence we get the Poisson algebra ${\mathcal{P}}_{4,33}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{09}$. Then $Z^2\left(\mathcal{P},\mathcal{P}\right)=\left\{0\right\}$. So we get the algebra ${\mathcal{P}}_{4,34}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{10}$. Then $Z^2\left(\mathcal{P},\mathcal{P}\right)=\left\{0\right\}$. So we get the algebra ${\mathcal{P}}_{4,35}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{11}$. Then $Z^2\left(\mathcal{P},\mathcal{P}\right)=\left\{0\right\}$. So we get the algebra ${\mathcal{P}}_{4,36}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{12}$. Then $Z^2\left(\mathcal{P},\mathcal{P}\right)=\left\{0\right\}$. So we get the algebra ${\mathcal{P}}_{4,37}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{13}$. Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then $\theta =\left(0,0,{\alpha }_1{\Delta }_{3,4},{\alpha }_2{\Delta }_{3,4}\right)$ for some ${\alpha }_1,{\alpha }_2\in \mathbb{C}$. The automorphism group of ${\mathcal{A}}_{13}$, $\mathrm{Aut}\left({\mathcal{A}}_{13}\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& a_{33}& a_{34}\\ 0& 0& a_{43}& a_{44}\end{array}\right).$$

Let $\varphi =\left(a_{ij}\right)\in$$\mathrm{Aut}\left({\mathcal{A}}_{13}\right)$. Then $\theta \ast \varphi =\left(0,0,{\beta }_1{\Delta }_{3,4},{\beta }_2{\Delta }_{3,4}\right)$ where

$$\begin{array}{ccc}\hfill {\beta }_1& =& {\alpha }_1{a}_{44}-{\alpha }_2{a}_{34},\hfill \\ \hfill {\beta }_2& =& {\alpha }_2{a}_{33}-{\alpha }_1{a}_{43}.\hfill \end{array}$$

Then, by Remark 3, we may assume $\left({\alpha }_1,{\alpha }_2\right)\in \left\{\left(0,0\right),\left(1,0\right)\right\}$. So we get the algebras ${\mathcal{P}}_{4,38}$ and ${\mathcal{P}}_{4,39}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{14}$. Then $Z^2\left(\mathcal{P},\mathcal{P}\right)=\left\{0\right\}$. So we get the algebra ${\mathcal{P}}_{4,40}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{15}$. The automorphism group of ${\mathcal{A}}_{15}$, $\mathrm{Aut}\left({\mathcal{A}}_{15}\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& a_{23}& a_{24}\\ 0& a_{32}& a_{33}& a_{34}\\ 0& a_{42}& a_{43}& a_{44}\end{array}\right).$$

Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then

$$\begin{array}{ccc}\hfill B_1& =& 0,\hfill \\ \hfill B_2& =& {\alpha }_1{\Delta }_{2,3}+{\alpha }_2{\Delta }_{2,4}+{\alpha }_3{\Delta }_{3,4},\hfill \\ \hfill B_3& =& {\alpha }_4{\Delta }_{2,3}+{\alpha }_5{\Delta }_{2,4}+{\alpha }_6{\Delta }_{3,4},\hfill \\ \hfill B_4& =& {\alpha }_7{\Delta }_{2,3}+{\alpha }_8{\Delta }_{2,4}+{\alpha }_9{\Delta }_{3,4},\hfill \end{array}$$

such that

$$\begin{array}{ccc}\hfill {\alpha }_1{\alpha }_8-{\alpha }_2{\alpha }_7+{\alpha }_4{\alpha }_9-{\alpha }_6{\alpha }_7& =& 0,\hfill \\ \hfill {\alpha }_1{\alpha }_6+{\alpha }_2{\alpha }_9-{\alpha }_3{\alpha }_4-{\alpha }_3{\alpha }_8& =& 0,\hfill \\ \hfill {\alpha }_1{\alpha }_5-{\alpha }_2{\alpha }_4-{\alpha }_5{\alpha }_9+{\alpha }_6{\alpha }_8& =& 0,\hfill \end{array}$$

for some ${\alpha }_1,\dots ,{\alpha }_9\in \mathbb{C}$. Let $\varphi =\left(a_{ij}\right)\in$$\mathrm{Aut}\left({\mathcal{A}}_{15}\right)$. Write

$$\theta \ast \varphi =\left(0,{\beta }_1{\Delta }_{2,3}+{\beta }_2{\Delta }_{2,4}+{\beta }_3{\Delta }_{3,4},{\beta }_4{\Delta }_{2,3}+{\beta }_5{\Delta }_{2,4}+{\beta }_6{\Delta }_{3,4},{\beta }_7{\Delta }_{2,3}+{\beta }_8{\Delta }_{2,4}+{\beta }_9{\Delta }_{3,4}\right).$$

Now if we define $\varphi$ to be the matrix

$$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& a_{23}& a_{24}\\ 0& 0& a_{33}& a_{34}\\ 0& 0& a_{43}& a_{44}\end{array}\right),$$

then $\left(\begin{array}{cc}{\beta }_4& {\beta }_5\\ {\beta }_7& {\beta }_8\end{array}\right)=a_{22}{\left(\begin{array}{cc}{a}_{33}& a_{34}\\ a_{43}& a_{44}\end{array}\right)}^{-1}\left(\begin{array}{cc}{\alpha }_4& {\alpha }_5\\ {\alpha }_7& {\alpha }_8\end{array}\right)\left(\begin{array}{cc}{a}_{33}& a_{34}\\ a_{43}& a_{44}\end{array}\right)$. So we may assume that

$$\left(\begin{array}{cc}{\alpha }_4& {\alpha }_5\\ {\alpha }_7& {\alpha }_8\end{array}\right)\in \left\{\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}1& 0\\ 0& \alpha \end{array}\right),\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right),\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\right\}.$$

Then we have the following cases:

• $\left(\begin{array}{cc}{\alpha }_4& {\alpha }_5\\ {\alpha }_7& {\alpha }_8\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$. Again, if we define $\varphi$ to be the matrix

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& a_{23}& a_{24}\\ 0& 0& a_{33}& a_{34}\\ 0& 0& a_{43}& a_{44}\end{array}\right),$$

  then

  $$\begin{array}{ccc}\hfill {\beta }_6& =& {\alpha }_6{a}_{44}-{\alpha }_9{a}_{34},\hfill \\ \hfill {\beta }_9& =& {\alpha }_9{a}_{33}-{\alpha }_6{a}_{43}.\hfill \end{array}$$

  So, by Remark 3, we may assume $\left({\alpha }_6,{\alpha }_9\right)\in \left\{\left(1,0\right),\left(0,0\right)\right\}$.

  ❖

  $\left({\alpha }_6,{\alpha }_9\right)=\left(1,0\right)$. Then ${\alpha }_1=0$ since otherwise $\theta \notin Z^2\left(\mathcal{P},\mathcal{P}\right)$. If ${\alpha }_2\ne 1$, we define $\varphi$ to be the following matrix:

  $$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\displaystyle \frac{{\alpha }_3}{1-{\alpha }_2}}& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,{\alpha }_2{\Delta }_{2,4},{\Delta }_{3,4},0\right)$. So we have the representatives ${\theta }^{\alpha \ne 1}=\left(0,\alpha {\Delta }_{2,4},{\Delta }_{3,4},0\right)$. If ${\alpha }_2=1$ and ${\alpha }_3=0$, we obtain the representative ${\theta }^{\alpha =1}=\left(0,{\Delta }_{2,4},{\Delta }_{3,4},0\right)$. Morover, the representatives ${\theta }^{\alpha }, {\theta }^{\beta }$ are in the same orbit if and only if $\left(\alpha -\beta \right)\left(\alpha \beta -1\right)=0$. So we get the Poisson algebras ${\mathcal{P}}_{4,41}^{\alpha }$. If ${\alpha }_2=1$ and ${\alpha }_3\ne 0$, we define $\varphi$ to be the following matrix:

  $$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& {\alpha }_3& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right).$$

  Then $\theta \ast \varphi =$$\left(0,{\Delta }_{2,4}+{\Delta }_{3,4},{\Delta }_{3,4},0\right)$. So we get the Poisson algebra ${\mathcal{P}}_{4,42}$.

  ❖

  $\left({\alpha }_6,{\alpha }_9\right)=\left(0,0\right)$. If ${\alpha }_1={\alpha }_2={\alpha }_3=0$, we get the algebra ${\mathcal{P}}_{4,43}$. Assume now that $\left({\alpha }_1,{\alpha }_2,{\alpha }_3\right)\ne \left(0,0,0\right)$. If $\left({\alpha }_1,{\alpha }_2\right)\ne \left(0,0\right)$, we define $\varphi$ to be the first of the following matrices if ${\alpha }_1\ne 0$ or the second if ${\alpha }_1=0$:

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& {\displaystyle \frac{1}{{\alpha }_1}}{\alpha }_3& {\alpha }_1& 0\\ 0& -{\displaystyle \frac{1}{{\alpha }_1}}{\alpha }_2& 0& {\displaystyle \frac{1}{{\alpha }_1}}\\ 0& 1& 0& 0\end{array}\right),\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& {\alpha }_3& 1& 0\\ 0& -{\alpha }_2& 0& 0\\ 0& 0& 0& {\displaystyle \frac{1}{{\alpha }_2}}\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,0,{\Delta }_{3,4},0\right)$ and we get the Poisson algebra ${\mathcal{P}}_{4,41}^{\alpha =0}$. If $\left({\alpha }_1,{\alpha }_2\right)=\left(0,0\right)$, we define $\varphi$ to be the following matrix:

  $$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& {\alpha }_3& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,{\Delta }_{3,4},0,0\right)$. So we get the Poisson algebra ${\mathcal{P}}_{4,44}$.

• $\left(\begin{array}{cc}{\alpha }_4& {\alpha }_5\\ {\alpha }_7& {\alpha }_8\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& \alpha \end{array}\right)$. Then

  $$\begin{array}{ccc}\hfill {\alpha }_1\alpha +{\alpha }_9& =& 0,\hfill \\ \hfill {\alpha }_1{\alpha }_6+{\alpha }_2{\alpha }_9-{\alpha }_3-{\alpha }_3\alpha & =& 0,\hfill \\ \hfill {\alpha }_2-{\alpha }_6\alpha & =& 0.\hfill \end{array}$$

  Assume first that $\alpha \ne 0$. Set $\lambda =-{\displaystyle \frac{1}{\alpha }}\left({\alpha }_2{\alpha }_9-\alpha {\alpha }_3+\alpha {\alpha }_1{\alpha }_6\right)$. Then $\lambda \left(\alpha +1\right)=0$. If $\lambda =0$, we choose $\varphi$ as follows:

  $$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& {\alpha }_6& -{\displaystyle \frac{1}{\alpha }}{\alpha }_9& -1\\ 0& 0& 1& 0\\ 0& 1& 0& 0\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,\alpha {\Delta }_{2,4},{\Delta }_{3,4},0\right)$ . So we obtain the Poisson algebras ${\mathcal{P}}_{4,41}^{\alpha }$. If $\lambda \ne 0$, then $\alpha =-1$. Further, if we choose $\varphi$ as

  $$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\alpha }_9& {\displaystyle \frac{1}{\lambda }}{\alpha }_6\\ 0& 0& 1& 0\\ 0& 0& 0& {\displaystyle \frac{1}{\lambda }}\end{array}\right),$$

  then $\theta \ast \varphi =\left(0,{\Delta }_{3,4},{\Delta }_{2,3},-{\Delta }_{2,4}\right)$. Hence we get the Poisson algebras ${\mathcal{P}}_{4,45}$. Assume now that $\alpha =0$. Then ${\alpha }_2={\alpha }_9=0$ and ${\alpha }_3={\alpha }_1{\alpha }_6$. Moreover, if we choose $\varphi$ as

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& {\alpha }_6& {\alpha }_1& -1\\ 0& 0& 1& 0\\ 0& 1& 0& 0\end{array}\right),$$

  then $\theta \ast \varphi =\left(0,0,{\Delta }_{3,4},0\right)$. Hence we get the Poisson algebra ${\mathcal{P}}_{4,41}^{\alpha =0}$.
• $\left(\begin{array}{cc}{\alpha }_4& {\alpha }_5\\ {\alpha }_7& {\alpha }_8\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$. Then ${\alpha }_1+{\alpha }_9=0,{\alpha }_3+{\alpha }_9^2=0$ and ${\alpha }_2-{\alpha }_6+2{\alpha }_9=0$. Choose $\varphi$ as follows:

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& -i{\alpha }_9& i{\alpha }_6-\left(1+i\right){\alpha }_9& -1\\ 0& i& 1& 0\\ 0& 0& i& 0\end{array}\right):i=\sqrt{-1}.$$
  Then $\theta \ast \varphi =\left(0,{\Delta }_{2,4}+{\Delta }_{3,4},{\Delta }_{3,4},0\right)$. So we get the algebra ${\mathcal{P}}_{4,42}$.
• $\left(\begin{array}{cc}{\alpha }_4& {\alpha }_5\\ {\alpha }_7& {\alpha }_8\end{array}\right)=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. Then ${\alpha }_1={\alpha }_9$ and ${\alpha }_9\left({\alpha }_2+{\alpha }_6\right)=0$. Let us consider the following cases:

  ❖

  ${\alpha }_9={\alpha }_2+{\alpha }_6=0$. Let $\varphi$ be the first of the following matrices if ${\alpha }_3+{\alpha }_6^2\ne 0$ or the second if ${\alpha }_3+{\alpha }_6^2=0$:

  $$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& -{\alpha }_6-\sqrt{{\alpha }_3+{\alpha }_6^2}& \sqrt{{\alpha }_3+{\alpha }_6^2}-{\alpha }_6& 0\\ 0& 1& 1& 0\\ 0& 0& 0& {\displaystyle \frac{1}{\sqrt{{\alpha }_3+{\alpha }_6^2}}}\end{array}\right),\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& -{\alpha }_6& 1& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 1\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,-{\Delta }_{2,4},{\Delta }_{3,4},0\right)$ if ${\alpha }_3+{\alpha }_6^2\ne 0$ or $\theta \ast \varphi =\left(0,{\Delta }_{3,4},0,0\right)$ if ${\alpha }_3+{\alpha }_6^2=0$. So we get the algebras ${\mathcal{P}}_{4,41}^{\alpha =-1}$ and ${\mathcal{P}}_{4,44}$.

  ❖

  ${\alpha }_9\ne 0,{\alpha }_2+{\alpha }_6=0$. Let $\varphi$ be the following automorphism:

  $$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& {\displaystyle \frac{{\alpha }_6}{{\alpha }_9}}& -{\displaystyle \frac{1}{{\alpha }_9}}& {\displaystyle \frac{1}{2}}{\displaystyle \frac{{\alpha }_6^2}{{\alpha }_9}}+{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\alpha }_3}{{\alpha }_9}}\\ 0& -{\displaystyle \frac{1}{{\alpha }_9}}& 0& 0\\ 0& 0& 0& 1\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,{\Delta }_{3,4},{\Delta }_{2,3},-{\Delta }_{2,4}\right)$. So we get the algebra ${\mathcal{P}}_{4,45}$.

  ❖

  ${\alpha }_9=0, {\alpha }_2+{\alpha }_6\ne 0$. Consider the following automorphism:

  $$\varphi =\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& {\alpha }_2+{\alpha }_6& -{\alpha }_6& 0\\ 0& 0& 1& 0\\ 0& 0& 0& {\displaystyle \frac{1}{{\alpha }_2+{\alpha }_6}}\end{array}\right).$$

  Then $\theta \ast \varphi =\left(0,{\Delta }_{2,4}+{\beta }_3{\Delta }_{3,4},{\Delta }_{2,4},0\right)$ with ${\beta }_3={\displaystyle \frac{{\alpha }_3-{\alpha }_2{\alpha }_6}{{\left({\alpha }_2+{\alpha }_6\right)}^2}}$.

  ♦

  ${\beta }_3=0$. Then $\theta \ast \varphi {\varphi }^{\prime }=\left(0,0,{\Delta }_{3,4},0\right)$ where

  $${\varphi }^{\prime }=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 1& 1& 0\\ 0& 0& 0& 1\end{array}\right).$$

  So we get the algebra ${\mathcal{P}}_{4,41}^{\alpha =0}$.

  ♦

  ${\beta }_3=-{\displaystyle \frac{1}{4}}$. Then $\theta \ast \varphi {\varphi }^{\prime }=\left(0,{\Delta }_{2,4}+{\Delta }_{3,4},{\Delta }_{3,4},0\right)$ where

  $${\varphi }^{\prime }=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& -{\displaystyle \frac{1}{2}}& 0& 0\\ 0& -1& 1& 0\\ 0& 0& 0& 2\end{array}\right).$$

  So we get the algebra ${\mathcal{P}}_{4,42}$.

  ♦

  ${\beta }_3\ne 0, -{\displaystyle \frac{1}{4}}$. Then $\theta \ast \varphi {\varphi }^{\prime }=\left(0,\beta {\Delta }_{2,4},{\Delta }_{3,4},0\right)$ for some $\beta \in \mathbb{C}$ where

  $${\varphi }^{\prime }=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 1& 0\\ 0& {\displaystyle \frac{-1-\sqrt{1+4{\beta }_3}}{2{\beta }_3}}& {\displaystyle \frac{-1+\sqrt{1+4{\beta }_3}}{2{\beta }_3}}& 0\\ 0& 0& 0& {\displaystyle \frac{-1+\sqrt{1+4{\beta }_3}}{2{\beta }_3}}\end{array}\right).$$

  So we get the algebra ${\mathcal{P}}_{4,41}^{\alpha =\beta }$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{16}$. Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then $\theta =\left(0,0,{\alpha }_1{\Delta }_{2,4},{\alpha }_2{\Delta }_{2,4}\right)$ for some ${\alpha }_1,{\alpha }_2\in \mathbb{C}$. The automorphism group of ${\mathcal{A}}_{16}$, $\mathrm{Aut}\left({\mathcal{A}}_{16}\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& 0& 0\\ 0& a_{32}& a_{22}^2& a_{34}\\ 0& a_{42}& 0& a_{44}\end{array}\right).$$

Let $\varphi =\left(a_{ij}\right)\in$$\mathrm{Aut}\left({\mathcal{A}}_{16}\right)$. Then $\theta \ast \varphi =\left(0,0,{\beta }_1{\Delta }_{2,4},{\beta }_2{\Delta }_{2,4}\right)$ where

$$\begin{array}{ccc}\hfill {\beta }_1& =& {\displaystyle \frac{1}{a_{22}}}\left({\alpha }_1{a}_{44}-{\alpha }_2{a}_{34}\right),\hfill \\ \hfill {\beta }_2& =& {\alpha }_2{a}_{22}.\hfill \end{array}$$

If $\theta =0$, we get the algebra ${\mathcal{P}}_{4,46}$. Otherwise, let $\varphi$ be the first of the following matrices if ${\alpha }_2\ne 0$ or the second if ${\alpha }_2=0$:

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& {\displaystyle \frac{1}{{\alpha }_2}}& 0& 0\\ 0& 0& {\displaystyle \frac{1}{{\alpha }_2^2}}& {\displaystyle \frac{{\alpha }_1}{{\alpha }_2}}\\ 0& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& {\displaystyle \frac{1}{{\alpha }_1}}\end{array}\right).$$

Then $\theta \ast \varphi =\left(0,0,0,{\Delta }_{2,4}\right)$ if ${\alpha }_2\ne 0$ while $\theta \ast \varphi =\left(0,0,{\Delta }_{2,4},0\right)$ if ${\alpha }_2=0$. So we get the algebras ${\mathcal{P}}_{4,47}$ and ${\mathcal{P}}_{4,48}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{17}$. Then $Z^2\left(\mathcal{P},\mathcal{P}\right)=\left\{0\right\}$. So we get the algebra ${\mathcal{P}}_{4,49}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{18}$. Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then $\theta =\left(0,0,0,\alpha {\Delta }_{2,3}\right)$ for some $\alpha \in \mathbb{C}$. The automorphism group of ${\mathcal{A}}_{18}$, $\mathrm{Aut}\left({\mathcal{A}}_{18}\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& -ϵa_{32}& 0\\ 0& a_{32}& ϵa_{22}& 0\\ 0& a_{42}& a_{43}& a_{22}^2+a_{32}^2\end{array}\right):{ϵ}^2=1.$$

Let $\varphi =\left(a_{ij}\right)\in$$\mathrm{Aut}\left({\mathcal{A}}_{18}\right)$. Then $\theta \ast \varphi =\left(0,0,0,\beta {\Delta }_{2,3}\right)$ where $\beta =ϵ\alpha$. So we get the representatives ${\theta }^{\alpha }=\left(0,0,0,\alpha {\Delta }_{2,3}\right)$. Moreover, the representatives ${\theta }^{\alpha }$ and ${\theta }^{\beta }$ are in the same orbit if and only if ${\alpha }^2={\beta }^2$. Hence we get the algebras ${\mathcal{P}}_{4,50}^{\alpha }$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{19}$. Then $Z^2\left(\mathcal{P},\mathcal{P}\right)=\left\{0\right\}$. So we get the algebra ${\mathcal{P}}_{4,51}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{20}$. Then $Z^2\left(\mathcal{P},\mathcal{P}\right)=\left\{0\right\}$. So we get the algebra ${\mathcal{P}}_{4,52}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{21}$. Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then $\theta =\left(0,{\alpha }_1{\Delta }_{2,3},{\alpha }_2{\Delta }_{2,3},0\right)$ for some ${\alpha }_1,{\alpha }_2\in \mathbb{C}$. The automorphism group of ${\mathcal{A}}_{21}$, $\mathrm{Aut}\left({\mathcal{A}}_{21}\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& a_{23}& 0\\ 0& a_{32}& a_{33}& 0\\ 0& 0& 0& a_{44}\end{array}\right).$$

Let $\varphi =\left(a_{ij}\right)\in$$\mathrm{Aut}\left({\mathcal{A}}_{21}\right)$. Then $\theta \ast \varphi =\left(0,{\beta }_1{\Delta }_{2,3},{\beta }_2{\Delta }_{2,3},0\right)$ where

$$\begin{array}{ccc}\hfill {\beta }_1& =& {\alpha }_1{a}_{33}-{\alpha }_2{a}_{23},\hfill \\ \hfill {\beta }_2& =& {\alpha }_2{a}_{22}-{\alpha }_1{a}_{32}.\hfill \end{array}$$

Then, by Remark 3, we may assume $\left({\alpha }_1,{\alpha }_2\right)\in \left\{\left(0,0\right),\left(1,0\right)\right\}$. So we get the algebras ${\mathcal{P}}_{4,53}$ and ${\mathcal{P}}_{4,54}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{22}$. Then $Z^2\left(\mathcal{P},\mathcal{P}\right)=\left\{0\right\}$. So we get the algebra ${\mathcal{P}}_{4,55}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{23}$. Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then $\theta =\left(0,0,{\alpha }_1{\Delta }_{3,4},{\alpha }_2{\Delta }_{3,4}\right)$ for some ${\alpha }_1,{\alpha }_2\in \mathbb{C}$. The automorphism group of ${\mathcal{A}}_{23}$, $\mathrm{Aut}\left({\mathcal{A}}_{23}\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\left(\begin{array}{cccc}ϵ& 1-ϵ& 0& 0\\ 1-ϵ& ϵ& 0& 0\\ 0& 0& a_{33}& a_{34}\\ 0& 0& a_{43}& a_{44}\end{array}\right):ϵ\in \left\{0,1\right\}.$$

Let $\varphi =\left(a_{ij}\right)\in$$\mathrm{Aut}\left({\mathcal{A}}_{23}\right)$. Then $\theta \ast \varphi =\left(0,0,{\beta }_1{\Delta }_{3,4},{\beta }_2{\Delta }_{3,4}\right)$ where

$$\begin{array}{ccc}\hfill {\beta }_1& =& {\alpha }_1{a}_{44}-{\alpha }_2{a}_{34},\hfill \\ \hfill {\beta }_2& =& {\alpha }_2{a}_{33}-{\alpha }_1{a}_{43}.\hfill \end{array}$$

Then, by Remark 3, we may assume $\left({\alpha }_1,{\alpha }_2\right)\in \left\{\left(0,0\right),\left(1,0\right)\right\}$. So we get the algebras ${\mathcal{P}}_{4,56}$ and ${\mathcal{P}}_{4,57}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{24}$. Then $Z^2\left(\mathcal{P},\mathcal{P}\right)=\left\{0\right\}$. So we get the algebra ${\mathcal{P}}_{4,58}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{25}$. Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then $\theta =\left(0,0,{\alpha }_1{\Delta }_{3,4},{\alpha }_2{\Delta }_{3,4}\right)$ for some ${\alpha }_1,{\alpha }_2\in \mathbb{C}$. The automorphism group of ${\mathcal{A}}_{25}$, $\mathrm{Aut}\left({\mathcal{A}}_{25}\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& 0& 0\\ 0& 0& a_{33}& a_{34}\\ 0& 0& a_{43}& a_{44}\end{array}\right).$$

Let $\varphi =\left(a_{ij}\right)\in$ $\mathrm{Aut}\left({\mathcal{A}}_{25}\right)$. Then $\theta \ast \varphi =\left(0,0,{\beta }_1{\Delta }_{3,4},{\beta }_2{\Delta }_{3,4}\right)$ where

$$\begin{array}{ccc}\hfill {\beta }_1& =& {\alpha }_1{a}_{44}-{\alpha }_2{a}_{34},\hfill \\ \hfill {\beta }_2& =& {\alpha }_2{a}_{33}-{\alpha }_1{a}_{43}.\hfill \end{array}$$

Then, by Remark 3, we may assume $\left({\alpha }_1,{\alpha }_2\right)\in \left\{\left(0,0\right),\left(1,0\right)\right\}$. So we get the algebras ${\mathcal{P}}_{4,59}$ and ${\mathcal{P}}_{4,60}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{26}$. Then $Z^2\left(\mathcal{P},\mathcal{P}\right)=\left\{0\right\}$. So we get the algebra ${\mathcal{P}}_{4,61}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{27}$. The automorphism group of ${\mathcal{A}}_{27}$, $\mathrm{Aut}\left({\mathcal{A}}_{27}\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& a_{23}& a_{24}\\ 0& a_{32}& a_{33}& a_{34}\\ 0& a_{42}& a_{43}& a_{44}\end{array}\right).$$

Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then

$$\begin{array}{ccc}\hfill B_1& =& 0,\hfill \\ \hfill B_2& =& {\alpha }_1{\Delta }_{2,3}+{\alpha }_2{\Delta }_{2,4}+{\alpha }_3{\Delta }_{3,4},\hfill \\ \hfill B_3& =& {\alpha }_4{\Delta }_{2,3}+{\alpha }_5{\Delta }_{2,4}+{\alpha }_6{\Delta }_{3,4},\hfill \\ \hfill B_4& =& {\alpha }_7{\Delta }_{2,3}+{\alpha }_8{\Delta }_{2,4}+{\alpha }_9{\Delta }_{3,4},\hfill \end{array}$$

such that

$$\begin{array}{ccc}\hfill {\alpha }_1{\alpha }_8-{\alpha }_2{\alpha }_7+{\alpha }_4{\alpha }_9-{\alpha }_6{\alpha }_7& =& 0,\hfill \\ \hfill {\alpha }_1{\alpha }_6+{\alpha }_2{\alpha }_9-{\alpha }_3{\alpha }_4-{\alpha }_3{\alpha }_8& =& 0,\hfill \\ \hfill {\alpha }_1{\alpha }_5-{\alpha }_2{\alpha }_4-{\alpha }_5{\alpha }_9+{\alpha }_6{\alpha }_8& =& 0,\hfill \end{array}$$

for some ${\alpha }_1,\dots ,{\alpha }_9\in \mathbb{C}$. Since Aut$\left({\mathcal{A}}_{15}\right)=$ Aut$\left({\mathcal{A}}_{27}\right)$ and $Z^2\left({\mathcal{A}}_{15},{\mathcal{A}}_{15}\right)=Z^2\left({\mathcal{A}}_{27},{\mathcal{A}}_{27}\right)$, we obtain the algebras ${\mathcal{P}}_{4,62}^{\alpha },$${\mathcal{P}}_{4,63},$${\mathcal{P}}_{4,64}, {\mathcal{P}}_{4,65}, {\mathcal{P}}_{4,66}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{28}$. Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then $\theta =\left(0,0,0,\alpha {\Delta }_{2,3}\right)$ for some $\alpha \in \mathbb{C}$. The automorphism group of ${\mathcal{A}}_{28}$, $\mathrm{Aut}\left({\mathcal{A}}_{28}\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& ϵa_{22}& \left(1-ϵ\right)a_{23}& 0\\ 0& \left(1-ϵ\right)a_{32}& ϵa_{33}& 0\\ 0& a_{42}& a_{43}& ϵa_{22}{a}_{33}+\left(1-ϵ\right)a_{23}{a}_{32}\end{array}\right):ϵ\in \left\{0,1\right\}.$$

Let $\varphi =\left(a_{ij}\right)\in$ $\mathrm{Aut}\left({\mathcal{A}}_{28}\right)$. Then $\theta \ast \varphi =\left(0,0,0,\beta {\Delta }_{2,3}\right)$ where

$$\beta ={\displaystyle \frac{\alpha }{a_{23}{a}_{32}+ϵa_{22}{a}_{33}-ϵa_{23}{a}_{32}}}\left(ϵa_{22}{a}_{33}-a_{23}{a}_{32}+ϵa_{23}{a}_{32}\right)$$

Whence ${\beta }^2={\alpha }^2$. Hence we get the algebras ${\mathcal{P}}_{4,67}^{\alpha }$. Moreover, the algebras ${\mathcal{P}}_{4,67}^{\alpha }$ and ${\mathcal{P}}_{4,67}^{\beta }$ are isomorphic if and only if ${\alpha }^2={\beta }^2$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{29}$. Let $\theta =\left(B_1,B_2,B_3,B_4\right)$ be an arbitrary element of $Z^2\left(\mathcal{P},\mathcal{P}\right)$. Then $\theta =\left(0,0,{\alpha }_1{\Delta }_{2,4},{\alpha }_2{\Delta }_{2,4}\right)$ for some ${\alpha }_1,{\alpha }_2\in \mathbb{C}$. The automorphism group of ${\mathcal{A}}_{29}$, $\mathrm{Aut}\left({\mathcal{A}}_{29}\right)$, consists of the automorphisms $\varphi$ given by a matrix of the following form:

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& 0& 0\\ 0& a_{32}& a_{22}^2& a_{34}\\ 0& a_{42}& 0& a_{44}\end{array}\right).$$

Let $\varphi =\left(a_{ij}\right)\in$ $\mathrm{Aut}\left({\mathcal{A}}_{29}\right)$. Then $\theta \ast \varphi =\left(0,0,{\beta }_1{\Delta }_{2,4},{\beta }_2{\Delta }_{2,4}\right)$ where

$$\begin{array}{ccc}\hfill {\beta }_1& =& {\displaystyle \frac{1}{a_{22}}}\left({\alpha }_1{a}_{44}-{\alpha }_2{a}_{34}\right),\hfill \\ \hfill {\beta }_2& =& {\alpha }_2{a}_{22}.\hfill \end{array}$$

If ${\alpha }_2\ne 0$, we choose $\varphi$ as follows:

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& {\displaystyle \frac{1}{{\alpha }_2}}& 0& 0\\ 0& 0& {\displaystyle \frac{1}{{\alpha }_2^2}}& {\displaystyle \frac{{\alpha }_1}{{\alpha }_2}}\\ 0& 0& 0& 1\end{array}\right).$$

Then $\theta \ast \varphi =\left(0,0,0,{\Delta }_{2,4}\right)$. Hence we get the algebra ${\mathcal{P}}_{4,68}$. If ${\alpha }_2=0$ and ${\alpha }_1\ne 0$, we choose $\varphi$ as follows:

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& {\alpha }_1& 0& 0\\ 0& 0& {\alpha }_1^2& 0\\ 0& 0& 0& 1\end{array}\right).$$

Then $\theta \ast \varphi =\left(0,0,{\Delta }_{2,4},0\right)$. So we get the algebra ${\mathcal{P}}_{4,69}$. If ${\alpha }_1={\alpha }_2=0$, then $\theta =0$, and we get the algebra ${\mathcal{P}}_{4,70}$.

$\left(\mathcal{P},·\right)={\mathcal{A}}_{30}$. Then $Z^2\left(\mathcal{P},\mathcal{P}\right)=\left\{0\right\}$. So we get the algebra ${\mathcal{P}}_{4,71}$.