The algebraic and geometric classification of
transposed Poisson algebras111 The first part of this work is supported by the
Spanish Government through the Ministry of Universities grant ‘Margarita
Salas’, funded by the European Union - NextGenerationEU; by FCT
UIDB/00212/2020 and UIDP/00212/2020. The second part of this work is supported
by RSF 22-11-00081.

Patrícia Damas Beitesa, Amir Fernández Ouaridib,c & Ivan Kaygorodova,d,e

a Departamento de Matemática and Centro de Matemática e Aplicações,
Universidade da Beira Interior, Covilhã, Portugal

b Centro de Matemática, Universidade de Coimbra, Coimbra, Portugal

c University of Cadiz, Puerto Real, Spain

d Saint Petersburg University, Russia

e Moscow Center for Fundamental and Applied Mathematics, Moscow, Russia

E-mail addresses:

Patrícia Damas Beites<EMAIL_ADDRESS>
Amir Fernández Ouaridi<EMAIL_ADDRESS>
Ivan Kaygorodov<EMAIL_ADDRESS>
Abstract. The algebraic and geometric classification of all complex
$3$-dimensional transposed Poisson algebras is obtained. Also, we discuss
strong special $3$-dimensional transposed Poisson algebras.

Keywords: Lie algebra, transposed Poisson algebra, $\delta$-derivation,
algebraic classification, geometric classification.

MSC2020: 17A30, 17B40, 17B63.

## Introduction

Poisson algebras arose from the study of Poisson geometry in the 1970s and
have appeared in an extremely wide range of areas in mathematics and physics,
such as Poisson manifolds, algebraic geometry, operads, quantization theory,
quantum groups, and classical and quantum mechanics. The study of Poisson
algebras also led to other algebraic structures, such as $F$-manifold
algebras, Novikov-Poisson algebras, Double Poisson algebras, Poisson $n$-Lie
algebras, etc [35, 6, 13]. The study of all possible Poisson structures with a
certain Lie or associative part is an important problem in the theory of
Poisson algebras [38]. Recently, a dual notion of the Poisson algebra
(transposed Poisson algebra), by exchanging the roles of the two binary
operations in the Leibniz rule defining the Poisson algebra, has been
introduced in the paper [4] of Bai, Bai, Guo, and Wu. They have shown that the
transposed Poisson algebra defined in this way not only shares common
properties with the Poisson algebra, including the closure undertaking tensor
products and the Koszul self-duality as an operad but also admits a rich class
of identities. More significantly, a transposed Poisson algebra naturally
arises from a Novikov-Poisson algebra by taking the commutator Lie algebra of
the Novikov algebra. Unital transposed Poisson algebras are studied in [5].
The ${\rm Hom}$\- and ${\rm BiHom}$-versions of transposed Poisson algebras
are considered in [29, 30]. Some new examples of transposed Poisson algebras
are constructed by applying the Kantor product of multiplications on the same
vector space [14]. More recently, in a paper by Ferreira, Kaygorodov and
Lopatkin, a relation between $\frac{1}{2}$-derivations of Lie algebras and
transposed Poisson algebras has been established [17].

The algebraic classification (up to isomorphism) of algebras of dimension $n$
from a certain variety defined by a certain family of polynomial identities is
a classic problem in the theory of non-associative algebras. There are many
results related to the algebraic classification of small-dimensional algebras
in many varieties of associative and non-associative algebras. So, algebraic
classifications of $2$-dimensional algebras [31], $3$-dimensional evolution
algebras [7], $3$-dimensional anticommutative algebras [27, 26],
$4$-dimensional division algebras [12, 11], $5$-dimensional commutative
nilpotent algebras [24] and $8$-dimensional dual Mock Lie algebras [8] have
been given. Section 1 is devoted to the complete algebraic classification of
non-isomorphic complex $3$-dimensional transposed Poisson algebras. To obtain
this classification, we will use the algebraic classification of suitable Lie
algebras and associative commutative algebras; and the method of describing
all transposed Poisson algebra structures on a given Lie algebras (the present
method has been developed in [17]).

The study of special and non-special algebras starts from the theory of Jordan
algebras. It is known, that the class of special Jordan algebras (i.e.
embedded into associative algebras relative to the multiplication $x\circ
y=xy+yx$) is a quasivariety, but it is not a variety of algebras [34]. It is
known that each Novikov-Poisson algebra under commutator product on non-
associative multiplication gives a transposed Poisson algebra [4]. Let us say
that a transposed Poisson algebra is special if it can be embedded into a
Novikov-Poisson algebra relative to the commutator bracket. Similarly, let us
say that a transposed Poisson algebra is $D$-special (from “differentially”)
if it embeds into a commutative algebra with a derivation relative to the
bracket $[x,y]=\mathfrak{D}(x)y-x\mathfrak{D}(y).$ Obviously, every
$D$-special transposed Poisson algebra is a special one. On the other hand,
the class of all special Gelfand-Dorfman algebras (i.e. embedded into Poisson
algebras with derivation relative to the multiplication $x\circ y=xd(y)$) is
closed with respect to homomorphisms and thus forms a variety [28]. Also known
as each two-generated Jordan algebra is special [33]; each one-generated
Jordan dialgebra is special [37]; each $2$-dimensional Gelfand-Dorfman algebra
is special [28]. Section 2 is devoted to the description of all complex strong
$D$-special (i.e. isomorphic to a commutative algebra with a derivation and
the bracket given by $[x,y]=\mathfrak{D}(x)y-x\mathfrak{D}(y)$) $2$\- and
$3$-dimensional transposed Poisson algebras.

Geometric properties of a variety of algebras defined by a family of
polynomial identities have been an object of study since the 1970s (see, [36,
10, 16, 19, 9, 32, 23]). Gabriel described the irreducible components of the
variety of $4$-dimensional unital associative algebras [19]. Cibils considered
rigid associative algebras with $2$-step nilpotent radical [9]. Grunewald and
O’Halloran computed the degenerations for the variety of $5$-dimensional
nilpotent Lie algebras [23]. All irreducible components of $2$-step nilpotent
commutative and anticommutative algebras have been described in [32, 25].
Chouhy proved that in the case of finite-dimensional associative algebras, the
$N$-Koszul property is preserved under the degeneration relation [10]. The
study of degenerations of algebras is very rich and closely related to
deformation theory, in the sense of Gerstenhaber [20]. The geometric
classification is given for many varieties of non-associative algebras (see,
for example, [16, 8, 15, 2, 3, 22, 16, 26] and references therein).
Degenerations have also been used to study a level of complexity of an algebra
[36, 21]. Section 3 is devoted to the complete geometric classification of
complex $3$-dimensional transposed Poisson algebras.

##  1\. The algebraic classification of $3$-dimensional transposed Poisson
algebras

###  1.1. Preliminaries

Although all algebras and vector spaces are considered over the complex field.
The definition of transposed Poisson algebra was given in a paper by Bai, Bai,
Guo, and Wu [4]. The concept of $\frac{1}{2}$-derivations as a particular case
of $\delta$-derivations was presented in an paper of Filippov [18] (see also
[40] and references therein).

######  Definition 1.

Let ${\mathfrak{L}}$ be a vector space equipped with two nonzero bilinear
operations $-\cdot-$ and $[\cdot,\cdot].$ The triple
$({\mathfrak{L}},\cdot,[\cdot,\cdot])$ is called a transposed Poisson algebra
if $({\mathfrak{L}},\cdot)$ is a commutative associative algebra and
$({\mathfrak{L}},[\cdot,\cdot])$ is a Lie algebra that satisfies the following
compatibility condition

$2z\cdot[x,y]=[z\cdot x,y]+[x,z\cdot y].$

The last relation is called the transposed Leibniz rule because the roles
played by the two binary operations in the Leibniz rule in a Poisson algebra
are switched. Further, the resulting operation is rescaled by introducing a
factor 2 on the left-hand side.

######  Definition 2.

Let $({\mathfrak{L}},[\cdot,\cdot])$ be an algebra with multiplication
$[\cdot,\cdot],$ $\varphi$ be a linear map and $\phi$ be a bilinear map. Then
$\varphi$ is a $\frac{1}{2}$-derivation if it satisfies

$\varphi[x,y]=\frac{1}{2}\big{(}[\varphi(x),y]+[x,\varphi(y)]\big{)};$

$\phi$ is a $\frac{1}{2}$-biderivation if it satisfies

$\phi([x,y],z)$ | $=$ | $\frac{1}{2}\big{(}[\phi(x,z),y]+[x,\phi(y,z)]\big{)},$
---|---|---
$\phi(x,[y,z])$ | $=$ | $\frac{1}{2}\big{(}[\phi(x,y),z]+[y,\phi(x,z)]\big{)}.$

Summarizing Definitions 1 and 2, we have the following key Remark.

######  Remark 3.

Let $({\mathfrak{L}},\cdot,[\cdot,\cdot])$ be a transposed Poisson algebra and
$z$ an arbitrary element from ${\mathfrak{L}}.$ Then the right multiplication
$R_{z}$ in the associative commutative algebra $({\mathfrak{L}},\cdot)$ gives
a $\frac{1}{2}$-derivation of the Lie algebra $({\mathfrak{L}},[\cdot,\cdot])$
and $-\cdot-$ gives a $\frac{1}{2}$-biderivation of
$({\mathfrak{L}},[\cdot,\cdot])$ satisfying the identities $x\cdot y=y\cdot x$
and $x\cdot(y\cdot z)=(x\cdot y)\cdot z$ for any $x,y,z.$ Reciprocally, for
any $\frac{1}{2}$-biderivation
$D:{\mathfrak{L}}\times{\mathfrak{L}}\to{\mathfrak{L}}$ satisfying
$D(x,y)=D(y,x)$ and $D(x,D(y,z))=D(D(x,y),z)$ one gets a Poisson transposed
algebra in an obvious way.

The main example of $\frac{1}{2}$-derivations is the multiplication by an
element from the ground field. Let us call such $\frac{1}{2}$-derivations as
trivial $\frac{1}{2}$-derivations. As a consequence of the following Remark,
we are not interested in trivial $\frac{1}{2}$-derivations.

######  Remark 4.

Let ${\mathfrak{L}}$ be a Lie algebra without non-trivial
$\frac{1}{2}$-derivations. Then every transposed Poisson algebra structure
defined on ${\mathfrak{L}}$ is trivial.

###  1.2. Isomorphism problem for transposed Poisson algebras on a certain Lie
algebra

######  Definition 5.

Let $({\mathfrak{L}}_{1},\cdot_{1},[\cdot,\cdot]_{1})$ and
$({\mathfrak{L}}_{2},\cdot_{2},[\cdot,\cdot]_{2})$ be two transposed Poisson
algebras. Then $({\mathfrak{L}}_{1},\cdot_{1},[\cdot,\cdot]_{1})$ and
$({\mathfrak{L}}_{2},\cdot_{2},[\cdot,\cdot]_{2})$ are isomorphic if and only
if there exists a liner map $\varphi$ such that

$\varphi([x,y]_{1})=[\varphi(x),\varphi(y)]_{2},\
\varphi(x\cdot_{1}y)=\varphi(x)\cdot_{2}\varphi(y).$

Our main strategy for classifying all non-isomorphic transposed Poisson
algebra structures on a certain Lie algebra $({\mathfrak{L}},[\cdot,\cdot])$
is as follows.

  1. (1)

Find all automorphisms ${\rm Aut}({\mathfrak{L}},[\cdot,\cdot])$.

  2. (2)

Consider the multiplication table of $({\mathfrak{L}},\cdot)$ under the action
of elements from ${\rm Aut}({\mathfrak{L}},[\cdot,\cdot])$ and separate all
non-isomorphic cases.

###  1.3. $\frac{1}{2}$-derivations of $3$-dimensional Lie algebras and
transposed Poisson algebras

To describe all $3$-dimensional transposed Poisson algebras we are using the
standard way: obtain the classification of all $\frac{1}{2}$-derivations of
$3$-dimensional Lie algebras and construct all possible transposed Poisson
structures on these algebras by using Remark 3.

####  1.3.1. Classification of $3$-dimensional Lie algebras

In the subsequent result, the classification of $3$-dimensional complex Lie
algebras is recalled (the presented classification was used in [26]).

######  Theorem 6.

Let $({\mathfrak{L}},[\cdot,\cdot])$ be a $3$-dimensional complex nonzero Lie
algebra, then one and only one of the following possibilities holds up to
isomorphism:

$\mathfrak{h}$ | $:$ | $[e_{1},e_{2}]=e_{3},$ |  |
---|---|---|---|---
$\mathfrak{g}_{1}$ | $:$ | $[e_{1},e_{3}]=e_{1},$ | $[e_{2},e_{3}]=e_{2},$ |
$\mathfrak{g}_{2}^{\alpha}$ | $:$ | $[e_{1},e_{3}]=e_{1}+e_{2},$ | $[e_{2},e_{3}]=\alpha e_{2},$ |
$\mathfrak{sl}_{2}$ | $:$ | $[e_{1},e_{2}]=e_{3},$ | $[e_{1},e_{3}]=-e_{2},$ | $[e_{2},e_{3}]=e_{1}.$

Between these algebras, the only non-trivial isomorphisms are
$\mathfrak{g}_{2}^{\alpha}\cong\mathfrak{g}_{2}^{\beta}$ if and only if
$\alpha=\beta^{-1}$.

Ferreira, Kaygorodov and Lopatkin proved that there are no non-trivial
transposed Poisson algebra structures defined on a complex semisimple finite-
dimensional Lie algebra, result which applies to $\mathfrak{sl}_{2}$. Denote
by ${\rm T}_{01}$ the trivial transposed Poisson algebra defined on
$\mathfrak{sl}_{2}$. The transposed Poisson algebra structures defined on
$\mathfrak{h}$ were studied, using $\frac{1}{2}$-biderivations, by Yuan and
Hua in [39, Theorem 4.5], obtaining the algebras:

  1. (1)

${\rm T}_{02}:\left\\{\begin{tabular}[]{l}$e_{2}\cdot e_{2}=e_{3},$\\\
$[e_{1},e_{2}]=e_{3}.$\end{tabular}\right.$

  2. (2)

${\rm T}_{03}^{\alpha}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{2}=\alpha
e_{3},$\\\ $[e_{1},e_{2}]=e_{3}.$\end{tabular}\right.$

  3. (3)

${\rm T}_{04}^{\alpha}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{2}=\alpha
e_{3},e_{2}\cdot e_{2}=e_{1},$\\\ $[e_{1},e_{2}]=e_{3}.$\end{tabular}\right.$

  4. (4)

${\rm T}_{05}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{3},e_{1}\cdot
e_{2}=e_{1},e_{2}\cdot e_{2}=e_{2},e_{2}\cdot e_{3}=e_{3},$\\\
$[e_{1},e_{2}]=e_{3}.$\end{tabular}\right.$

  5. (5)

${\rm T}_{06}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{2}=e_{1},e_{2}\cdot
e_{2}=e_{2},e_{2}\cdot e_{3}=e_{3},$\\\
$[e_{1},e_{2}]=e_{3}.$\end{tabular}\right.$

Hence, only the classification of the transposed Poisson algebras defined on
$\mathfrak{g}_{1}$ and $\mathfrak{g}_{2}^{\alpha}$ are missing.

####  1.3.2. Description of transposed Poisson algebra structure defined on
$\mathfrak{g}_{1}$

######  Remark 7.

Let $\varphi$ be a $\frac{1}{2}$-derivation of $\mathfrak{g}_{1}$. Then

$\varphi(e_{1})=\beta_{33}e_{1},\ \varphi(e_{2})=\beta_{33}e_{2},\
\varphi(e_{3})=\beta_{31}e_{1}+\beta_{32}e_{2}+\beta_{33}e_{3}.$

######  Remark 8.

Let $\phi$ be an automorphism of $\mathfrak{g}_{1}$. Then

$\phi(e_{1})=\lambda_{11}e_{1}+\lambda_{21}e_{2},\
\phi(e_{2})=\lambda_{12}e_{1}+\lambda_{22}e_{2},\ \
\phi(e_{3})=\lambda_{13}e_{1}+\lambda_{23}e_{2}+e_{3},$

where $\lambda_{11}\lambda_{22}\neq\lambda_{21}\lambda_{12}.$

######  Proposition 9.

Let $(\mathfrak{L},\cdot,[\cdot,\cdot])$ be a transposed Poisson algebra
structure defined on $\mathfrak{g}_{1}$. Then
$(\mathfrak{L},\cdot,[\cdot,\cdot])$ is not a Poisson algebra and it is
isomorphic to only one of the following algebras:

  1. (1)

${\rm T}_{07}^{\alpha}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{3}=\alpha
e_{1}$, $e_{2}\cdot e_{3}=\alpha e_{2}$, $e_{3}\cdot e_{3}=\alpha e_{3},$\\\
$[e_{1},e_{3}]=e_{1},$ $[e_{2},e_{3}]=e_{2}.$\end{tabular}\right.$

  2. (2)

${\rm T}_{08}:\left\\{\begin{tabular}[]{l}$e_{3}\cdot e_{3}=e_{1},$\\\
$[e_{1},e_{3}]=e_{1},$ $[e_{2},e_{3}]=e_{2}.$\end{tabular}\right.$

where the parameter $\alpha\in\mathbb{C}$.

###### Proof.

We aim to describe the multiplication $-\cdot-$. By Remark 3, for every
$e_{k}$ there is an associated $\frac{1}{2}$-derivation $\varphi_{k}$ of
$(\mathfrak{L},[\cdot,\cdot])$ such that $\varphi_{j}(e_{i})=e_{i}\cdot
e_{j}=\varphi_{i}(e_{j})$. From Remark 7,

$\varphi_{i}(e_{1})=\beta_{33}^{i}e_{1}$,
$\varphi_{i}(e_{2})=\beta_{33}^{i}e_{2}$,
$\varphi_{i}(e_{3})=\beta_{31}^{i}e_{1}+\beta_{32}^{i}e_{2}+\beta_{33}^{i}e_{3}$

and

$\beta_{33}^{1}e_{2}$ | $=$ | $\varphi_{1}(e_{2})$ | $=$ | $\varphi_{2}(e_{1})$ | $=$ |  $\beta_{33}^{2}e_{1}$,
---|---|---|---|---|---|---
$\beta_{31}^{1}e_{1}+\beta_{32}^{1}e_{2}+\beta_{33}^{1}e_{3}$ | $=$ | $\varphi_{1}(e_{3})$ | $=$ | $\varphi_{3}(e_{1})$ | $=$ | $\beta_{33}^{3}e_{1},$
$\beta_{33}^{3}e_{2}$ | $=$ | $\varphi_{2}(e_{3})$ | $=$ | $\varphi_{3}(e_{2})$ | $=$ | $\beta_{31}^{2}e_{1}+\beta_{32}^{2}e_{2}+\beta_{33}^{2}e_{3}.$

Hence, the commutative multiplication $-\cdot-$ is defined by

$e_{1}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{1},$
---|---|---
$e_{2}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{2},$
$e_{3}\cdot e_{3}$ | $=$ | $\beta_{31}^{3}e_{1}+\beta_{32}^{3}e_{2}+\beta_{33}^{3}e_{3}.$

Through straightforward calculations, it is possible to conclude that
$-\cdot-$ is associative too. Let us now separate all non-isomorphic cases.
Under the action of an automorphism of the Lie algebra
$(\mathfrak{L},[\cdot,\cdot]),$ given in Remark 8, we rewrite the
multiplication table of $(\mathfrak{L},\cdot)$ by the following way:

$e_{1}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{1},$
---|---|---
$e_{2}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{2},$
$e_{3}\cdot e_{3}$ | $=$ | $\frac{\beta^{3}_{31}\lambda_{22}-\beta^{3}_{32}\lambda_{12}+\beta^{3}_{33}\lambda_{13}\lambda_{22}-\beta^{3}_{33}\lambda_{12}\lambda_{23}}{\lambda_{11}\lambda_{22}-\lambda_{21}\lambda_{12}}e_{1}-\frac{\beta^{3}_{31}\lambda_{21}-\beta^{3}_{32}\lambda_{11}+\beta^{3}_{33}\lambda_{13}\lambda_{21}-\beta^{3}_{33}\lambda_{11}\lambda_{23}}{\lambda_{11}\lambda_{22}-\lambda_{21}\lambda_{12}}e_{2}+\beta_{33}^{3}e_{3}.$

Suppose $(\beta_{31}^{3},\beta_{32}^{3},\beta_{33}^{3})\neq 0$, otherwise we
have the zero algebra. Let us consider two cases.

  1. (1)

If $\beta_{33}^{3}\neq 0,$ then by choosing

$\lambda_{13}=-\frac{\beta_{31}^{3}}{\beta_{33}^{3}},$
$\lambda_{23}=-\frac{\beta_{32}^{3}}{\beta_{33}^{3}},$
$\lambda_{11}=\lambda_{22}=1$ and $\lambda_{12}=\lambda_{21}=0,$

we have

$e_{1}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{1},$ | $e_{2}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{2},$ | $e_{3}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{3}.$
---|---|---|---|---|---|---|---|---
  2. (2)

If $\beta_{33}^{3}=0$ and $(\beta_{31}^{3},\beta_{32}^{3})\neq 0,$ then by
choosing some suitable $\lambda_{ij}$ we have

$e_{3}\cdot e_{3}$ | $=$ | $e_{1}.$
---|---|---

It is easy to see that the obtained algebras are not isomorphic. From these
cases, we obtain the algebras ${\rm T}_{07}^{\alpha}$ and ${\rm T}_{08}$,
respectively. ∎

####  1.3.3. Description of transposed Poisson algebra structure defined on
$\mathfrak{g}_{2}^{\alpha}$

######  Remark 10.

Let $\varphi$ be a $\frac{1}{2}$-derivation of $\mathfrak{g}_{2}^{\alpha}$.

  1. (1)

If $\alpha\neq 0,\frac{1}{2},2$, then

$\varphi(e_{1})=\beta_{33}e_{1},\ \varphi(e_{2})=\beta_{33}e_{2},\
\varphi(e_{3})=\beta_{31}e_{1}+\beta_{32}e_{2}+\beta_{33}e_{3}.$

  2. (2)

If $\alpha=\frac{1}{2}$, then

$\varphi(e_{1})=(\beta_{33}-2\beta_{21})e_{1}-4\beta_{21}e_{2},\
\varphi(e_{2})=\beta_{21}e_{1}+(2\beta_{21}+\beta_{33})e_{2},$

$\varphi(e_{3})=\beta_{31}e_{1}+\beta_{32}e_{2}+\beta_{33}e_{3}.$

  3. (3)

If $\alpha=2$, then

$\varphi(e_{1})=\beta_{33}e_{1}+\beta_{12}e_{2},\
\varphi(e_{2})=\beta_{33}e_{2},\
\varphi(e_{3})=\beta_{31}e_{1}+\beta_{32}e_{2}+\beta_{33}e_{3}.$

  4. (4)

If $\alpha=0$, then

$\varphi(e_{1})=\beta_{33}e_{1}-(\beta_{22}-\beta_{33})e_{2},\
\varphi(e_{2})=\beta_{22}e_{2},\
\varphi(e_{3})=\beta_{31}e_{1}+\beta_{32}e_{2}+\beta_{33}e_{3}.$

###### Proof.

Let $\varphi(e_{i})=\beta_{i1}e_{1}+\beta_{i2}e_{2}+\beta_{i3}e_{3}$. From
here and

$0$ | $=$ | $[\varphi(e_{1}),e_{2}]+[e_{1},\varphi(e_{2})]=\beta_{23}e_{1}-(\alpha\beta_{13}-\beta_{23})e_{2},$
---|---|---
$2\varphi(e_{1}+e_{2})$ | $=$ | $[\varphi(e_{1}),e_{3}]+[e_{1},\varphi(e_{3})]=(\beta_{11}+\beta_{33})e_{1}+(\beta_{11}+\alpha\beta_{12}+\beta_{33})e_{2},$
$2\alpha\varphi(e_{2})$ | $=$ | $[\varphi(e_{2}),e_{3}]+[e_{2},\varphi(e_{3})]=\beta_{21}e_{1}+(\beta_{21}+\alpha\beta_{22}+\alpha\beta_{33})e_{2},$

we obtain the following system of equations

$\left\\{\begin{array}[]{lll}\beta_{23}=0\\\ \beta_{13}=0\\\
\beta_{11}+\beta_{33}=2(\beta_{11}+\beta_{21})\\\
\beta_{11}+\alpha\beta_{12}+\beta_{33}=2(\beta_{12}+\beta_{22})\\\
2\alpha\beta_{21}=\beta_{21}\\\
2\alpha\beta_{22}=\beta_{21}+\alpha\beta_{22}+\alpha\beta_{33}\\\
\end{array}\right..$

That gives the proof of the statement. ∎

######  Remark 11.

Let $\phi$ be an automorphism of $\mathfrak{g}_{2}^{\alpha}$.

  1. (1)

If $\alpha\neq-1,$ then

$\phi(e_{1})=\lambda_{11}e_{1}+\lambda_{21}e_{2},\
\phi(e_{2})=(\lambda_{11}+\lambda_{21}(\alpha-1))e_{2},\ \
\phi(e_{3})=\lambda_{13}e_{1}+\lambda_{23}e_{2}+e_{3},$

where $\lambda_{11}(\lambda_{11}+\lambda_{21}(\alpha-1))\neq 0.$

  2. (2)

If $\alpha=-1,$ then

$\phi(e_{1})=\lambda_{11}e_{1}+\lambda_{21}e_{2},\
\phi(e_{2})=(\lambda_{11}-2\lambda_{21})e_{2},\ \
\phi(e_{3})=\lambda_{13}e_{1}+\lambda_{23}e_{2}+e_{3},$

where $\lambda_{11}(\lambda_{11}-2\lambda_{21})\neq 0;$ or

$\phi(e_{1})=\lambda_{11}e_{1}+\lambda_{21}e_{2},\
\phi(e_{2})=-2\lambda_{11}e_{1}-\lambda_{11}e_{2},\ \
\phi(e_{3})=\lambda_{13}e_{1}+\lambda_{23}e_{2}-e_{3},$

where $\lambda_{11}^{2}\neq 2\lambda_{11}\lambda_{21}.$

We aim to describe the multiplication $-\cdot-$. By Remark 3, for every
$e_{k}$ there is an associated $\frac{1}{2}$-derivation $\varphi_{k}$ of
$(\mathfrak{L},[\cdot,\cdot])$ such that $\varphi_{j}(e_{i})=e_{i}\cdot
e_{j}=\varphi_{i}(e_{j})$. Now we have to consider $3$ cases, by Remark 10 and
using that $\mathfrak{g}_{2}^{\frac{1}{2}}\cong\mathfrak{g}_{2}^{2}$.

######  Proposition 12.

Let $(\mathfrak{L},\cdot,[\cdot,\cdot])$ be a transposed Poisson algebra
structure defined on $\mathfrak{g}_{2}^{\alpha\neq 0,\frac{1}{2},2}$. Then
$(\mathfrak{L},\cdot,[\cdot,\cdot])$ is isomorphic to only one of the
following algebras:

  1. (1)

${\rm T}_{09}^{\alpha\neq
0,\frac{1}{2},2,\beta}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{3}=\beta
e_{1},\ e_{2}\cdot e_{3}=\beta e_{2},\ e_{3}\cdot e_{3}=\beta e_{3},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2},$ $[e_{2},e_{3}]=\alpha
e_{2}.$\end{tabular}\right.$

  2. (2)

${\rm T}_{10}^{\alpha\neq
0,\frac{1}{2},2}:\left\\{\begin{tabular}[]{l}$e_{3}\cdot e_{3}=e_{2},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2},$ $[e_{2},e_{3}]=\alpha
e_{2}.$\end{tabular}\right.$

  3. (3)

${\rm T}_{11}^{\alpha\neq
0,\frac{1}{2},2}:\left\\{\begin{tabular}[]{l}$e_{3}\cdot e_{3}=e_{1},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2},$ $[e_{2},e_{3}]=\alpha
e_{2}.$\end{tabular}\right.$

where the parameter $\beta\in\mathbb{C}$. Between these algebras there are
precisely the following non-trivial isomorphisms:

  * •

${\rm T}_{09}^{\alpha_{1},\beta_{1}}\cong{\rm T}_{09}^{\alpha_{2},\beta_{2}}$
if and only if $(\alpha_{2},\beta_{2})=(\alpha_{1},\beta_{1})$ or
$(\alpha_{2},\beta_{2})=(\frac{1}{\alpha_{1}},\frac{\beta_{1}}{\alpha_{1}})$.

  * •

${\rm T}_{11}^{\beta_{1}}\cong{\rm T}_{11}^{\beta_{2}}$ if and only if
$\beta_{2}=\beta_{1}$ or $\beta_{2}=\frac{1}{\beta_{1}}$.

###### Proof.

From Remark 10,

$\varphi_{i}(e_{1})=\beta_{33}^{i}e_{1}$,
$\varphi_{i}(e_{2})=\beta_{33}^{i}e_{2}$,
$\varphi_{i}(e_{3})=\beta_{31}^{i}e_{1}+\beta_{32}^{i}e_{2}+\beta_{33}^{i}e_{3}$

and

$\beta_{33}^{1}e_{2}$ | $=$ | $\varphi_{1}(e_{2})$ | $=$ | $\varphi_{2}(e_{1})$ | $=$ |  $\beta_{33}^{2}e_{1}$,
---|---|---|---|---|---|---
$\beta_{31}^{1}e_{1}+\beta_{32}^{1}e_{2}+\beta_{33}^{1}e_{3}$ | $=$ | $\varphi_{1}(e_{3})$ | $=$ | $\varphi_{3}(e_{1})$ | $=$ | $\beta_{33}^{3}e_{1},$
$\beta_{33}^{3}e_{2}$ | $=$ | $\varphi_{2}(e_{3})$ | $=$ | $\varphi_{3}(e_{2})$ | $=$ | $\beta_{31}^{2}e_{1}+\beta_{32}^{2}e_{2}+\beta_{33}^{2}e_{3}.$

Hence, the commutative multiplication $-\cdot-$ is defined by

$\displaystyle e_{1}\cdot e_{3}$ $\displaystyle=$
$\displaystyle\beta_{33}^{3}e_{1},$ $\displaystyle e_{2}\cdot e_{3}$
$\displaystyle=$ $\displaystyle\beta_{33}^{3}e_{2},$ $\displaystyle e_{3}\cdot
e_{3}$ $\displaystyle=$
$\displaystyle\beta_{31}^{3}e_{1}+\beta_{32}^{3}e_{2}+\beta_{33}^{3}e_{3}.$

Through straightforward calculations, it is possible to conclude that
$-\cdot-$ is associative too. We are interested in
$(\beta_{31}^{3},\beta_{32}^{3},\beta_{33}^{3})\neq 0.$ Let us now separate
all non-isomorphic cases.

  1. First,

we will consider only the case $\alpha\neq-1,0,1,\frac{1}{2},2.$ Under the
action of an automorphism of the Lie algebra $(\mathfrak{L},[\cdot,\cdot]),$
given in Remark 11, we rewrite the multiplication table of
$(\mathfrak{L},\cdot)$ by the following way:

$e_{1}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{1},$
---|---|---
$e_{2}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{2},$
$e_{3}\cdot e_{3}$ | $=$ | $\frac{\beta^{3}_{31}+\beta^{3}_{33}\lambda_{13}}{\lambda_{11}}e_{1}+$
|  | $\frac{\beta^{3}_{31}(\lambda_{11}-\lambda_{22})-\beta^{3}_{32}\lambda_{11}+\alpha\beta^{3}_{32}\lambda_{11}+\beta^{3}_{33}(\lambda_{11}\lambda_{13}-\lambda_{13}\lambda_{22}-\lambda_{11}\lambda_{23})+\alpha\beta^{3}_{32}\lambda_{11}\lambda_{23}}{(\alpha-1)\lambda_{11}\lambda_{22}}e_{2}+\beta_{33}^{3}e_{3}.$

Let us consider the following cases.

    1. (a)

If $\beta_{33}^{3}\neq 0,$ then by choosing

$\lambda_{13}=-\frac{\beta_{31}^{3}}{\beta_{33}^{3}},$
$\lambda_{23}=-\frac{\beta_{32}^{3}}{\beta_{33}^{3}},$
$\lambda_{11}=\lambda_{22}=1,$

we have

$e_{1}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{1},$ | $e_{2}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{2},$ | $e_{3}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{3}.$
---|---|---|---|---|---|---|---|---
    2. (b)

If $\beta_{33}^{3}=0$ and $\beta_{31}^{3}=(1-\alpha)\beta_{32}^{3},$ then by
choosing $\lambda_{11}=\beta_{32}^{3},$ we have

$e_{3}\cdot e_{3}$ | $=$ | $(1-\alpha)e_{1}+e_{2}.$
---|---|---
    3. (c)

$\beta_{33}^{3}=0$ and $\beta_{31}^{3}\neq(1-\alpha)\beta_{32}^{3},$ then by
choosing some suitable $\lambda_{ij},$ we have two following opportunities

$e_{3}\cdot e_{3}=e_{2};$ or $e_{3}\cdot e_{3}=e_{1}.$

Denote the corresponding families of transposed Poisson algebras by ${\rm
T}_{09}^{\alpha,\beta}$, ${\rm T}_{10*}^{\alpha}$, ${\rm T}_{10}^{\alpha}$ and
${\rm T}_{11}^{\alpha}$, respectively.

  2. Second,

we will consider only the case $\alpha=1.$ Under the action of an automorphism
of the Lie algebra $(\mathfrak{L},[\cdot,\cdot]),$ given in Remark 11, we
rewrite the multiplication table of $(\mathfrak{L},\cdot)$ by the following
way:

$e_{1}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{1},$
---|---|---
$e_{2}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{2},$
$e_{3}\cdot e_{3}$ | $=$ | $\frac{\beta^{3}_{31}+\beta^{3}_{33}\lambda_{13}}{\lambda_{11}}e_{1}+\frac{\beta^{3}_{32}\lambda_{11}-\beta^{3}_{31}\lambda_{21}-\beta^{3}_{33}\lambda_{13}\lambda_{21}+\beta^{3}_{33}\lambda_{11}\lambda_{23}}{\lambda_{11}^{2}}e_{2}+\beta_{33}^{3}e_{3}.$

Let us consider the following cases.

    1. (a)

$\beta_{33}^{3}\neq 0,$ then by choosing

$\lambda_{13}=-\frac{\beta_{31}^{3}}{\beta_{33}^{3}},$
$\lambda_{23}=-\frac{\beta_{32}^{3}}{\beta_{33}^{3}},$
$\lambda_{11}=1,\lambda_{21}=0,$

we have

$e_{1}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{1},$ | $e_{2}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{2},$ | $e_{3}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{3}.$
---|---|---|---|---|---|---|---|---
    2. (b)

$\beta_{33}^{3}=0,$ then by choosing some suitable $\lambda_{ij},$ we have two
following opportunities

$e_{3}\cdot e_{3}=e_{2};$ or $e_{3}\cdot e_{3}=e_{1}.$

From here, we obtain the transposed Poisson algebras ${\rm T}_{09}^{1,\beta}$,
${\rm T}_{10}^{1}$ and ${\rm T}_{11}^{1}$, respectively.

  3. Third,

we will consider only the case $\alpha=-1.$ Under the action of an
automorphism of the first type of the Lie algebra
$(\mathfrak{L},[\cdot,\cdot]),$ given in Remark 11, we rewrite the
multiplication table of $(\mathfrak{L},\cdot)$ by the following way:

$e_{1}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{1},$
---|---|---
$e_{2}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{2},$
$e_{3}\cdot e_{3}$ | $=$ | $\frac{\beta^{3}_{31}+\beta^{3}_{33}\lambda_{13}}{\lambda_{11}}e_{1}-$
|  | $\frac{\beta^{3}_{31}\lambda_{11}-2\beta^{3}_{32}\lambda_{11}+\beta^{3}_{33}\lambda_{11}\lambda_{13}-\beta^{3}_{31}\lambda_{22}-\beta^{3}_{33}\lambda_{13}\lambda_{22}-2\beta^{3}_{33}\lambda_{11}\lambda_{23}}{2\lambda_{11}\lambda_{22}}e_{2}+\beta_{33}^{3}e_{3}.$

Let us consider the following cases.

    1. (a)

If $\beta_{33}^{3}\neq 0,$ then by choosing

$\lambda_{13}=-\frac{\beta_{31}^{3}}{\beta_{33}^{3}},$
$\lambda_{23}=-\frac{\beta_{32}^{3}}{\beta_{33}^{3}},$
$\lambda_{11}=1,\lambda_{21}=0,$

we have

$e_{1}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{1},$ | $e_{2}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{2},$ | $e_{3}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{3}.$
---|---|---|---|---|---|---|---|---
    2. (b)

If $\beta_{33}^{3}=0$ and $\beta_{31}^{3}=2\beta_{32}^{3},$ then by choosing
$\lambda_{11}=\beta_{32}^{3},$ we have

$e_{3}\cdot e_{3}$ | $=$ | $2e_{1}+e_{2}.$
---|---|---
    3. (c)

If $\beta_{33}^{3}=0$ and $\beta_{31}^{3}\neq 2\beta_{32}^{3},$ then by
choosing some suitable $\lambda_{ij},$ we have two following opportunities

$e_{3}\cdot e_{3}=e_{2};$ or $e_{3}\cdot e_{3}=e_{1}.$

From here, we obtain the transposed Poisson algebras ${\rm
T}_{09}^{-1,\beta}$, ${\rm T}_{10}^{-1}$ and ${\rm T}_{11}^{-1}$,
respectively.

It is easy to see that under the action of an automorphism of the second type,
the following algebras

$e_{3}\cdot e_{3}=2e_{1}+e_{2}$ and $e_{3}\cdot e_{3}=e_{2}$

are isomorphic.

Lastly, since $\mathfrak{g}_{2}^{\alpha}\cong\mathfrak{g}_{2}^{\beta}$ if and
only if $\alpha=\beta^{-1}$, we may find isomorphisms between the transposed
Poisson algebras that follow from this study. These isomorphisms are the
following:

  * •

${\rm T}_{09}^{\alpha_{1},\beta_{1}}\cong{\rm T}_{09}^{\alpha_{2},\beta_{2}}$
if and only if $(\alpha_{2},\beta_{2})=(\alpha_{1},\beta_{1})$ or
$(\alpha_{2},\beta_{2})=(\frac{1}{\alpha_{1}},\frac{\beta_{1}}{\alpha_{1}})$.
For the non-trivial isomorphisms, choose the change of basis:

$E_{1}=\frac{1}{\alpha_{1}-1}e_{1},\
E_{2}=e_{1}+\frac{\alpha_{1}}{\alpha_{1}-1}e_{2},\ E_{3}=\alpha_{1}e_{3}.$

  * •

${\rm T}_{10*}^{\beta_{1}}\cong{\rm T}_{10}^{\beta_{2}}$ if and only if
$\beta_{2}=\frac{1}{\beta_{1}}$, using the change of basis:

$E_{1}=\frac{1}{\beta_{1}}e_{1},\
E_{2}=\frac{1-\beta_{1}}{\beta_{1}^{2}}e_{1}+\frac{1}{\beta_{1}^{2}}e_{2},\
E_{3}=\frac{1}{\beta_{1}}e_{3}.$

  * •

${\rm T}_{11}^{\beta_{1}}\cong{\rm T}_{11}^{\beta_{2}}$ if and only if
$\beta_{2}=\beta_{1}$ or $\beta_{2}=\frac{1}{\beta_{1}}$. For the non-trivial
isomorphisms, choose the change of basis:

$E_{1}=\beta_{1}^{2}e_{1},\
E_{2}=(\beta_{1}-1)\beta_{1}^{2}e_{1}+\beta_{1}^{3}e_{2},\
E_{3}=\beta_{1}e_{3}.$

∎

######  Proposition 13.

Let $(\mathfrak{L},\cdot,[\cdot,\cdot])$ be a transposed Poisson algebra
structure defined on $\mathfrak{g}_{2}^{2}$. Then
$(\mathfrak{L},\cdot,[\cdot,\cdot])$ is isomorphic to only one of the
following algebras:

  1. (1)

${\rm T}_{09}^{2,\beta}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{3}=\beta
e_{1},\ e_{2}\cdot e_{3}=\beta e_{2},\ e_{3}\cdot e_{3}=\beta e_{3},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2},$ $[e_{2},e_{3}]=2e_{2}.$\end{tabular}\right.$

  2. (2)

${\rm T}_{10}^{2}:\left\\{\begin{tabular}[]{l}$e_{3}\cdot e_{3}=e_{2},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2},$ $[e_{2},e_{3}]=2e_{2}.$\end{tabular}\right.$

  3. (3)

${\rm T}_{11}^{2}:\left\\{\begin{tabular}[]{l}$e_{3}\cdot e_{3}=e_{1},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2},$ $[e_{2},e_{3}]=2e_{2}.$\end{tabular}\right.$

  4. (4)

${\rm T}_{12}^{\beta}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{2},\
e_{1}\cdot e_{3}=\beta e_{1},\ e_{2}\cdot e_{3}=\beta e_{2},\ e_{3}\cdot
e_{3}=\beta e_{3},$\\\ $[e_{1},e_{3}]=e_{1}+e_{2},$
$[e_{2},e_{3}]=2e_{2}.$\end{tabular}\right.$

  5. (5)

${\rm T}_{13}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{2},\ e_{3}\cdot
e_{3}=e_{2},$\\\ $[e_{1},e_{3}]=e_{1}+e_{2},$
$[e_{2},e_{3}]=2e_{2}.$\end{tabular}\right.$

  6. (6)

${\rm T}_{14}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{3}=e_{2},\ e_{3}\cdot
e_{3}=e_{1},$\\\ $[e_{1},e_{3}]=e_{1}+e_{2},$
$[e_{2},e_{3}]=2e_{2}.$\end{tabular}\right.$

  7. (7)

${\rm T}_{15}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{3}=e_{2},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2},$ $[e_{2},e_{3}]=2e_{2}.$\end{tabular}\right.$

where the parameter $\beta\in\mathbb{C}$.

###### Proof.

From Remark 10,

$\varphi_{i}(e_{1})=\beta_{33}^{i}e_{1}+\beta_{12}^{i}e_{2}$,
$\varphi_{i}(e_{2})=\beta_{33}^{i}e_{2}$,
$\varphi_{i}(e_{3})=\beta_{31}^{i}e_{1}+\beta_{32}^{i}e_{2}+\beta_{33}^{i}e_{3}$

and

$\beta_{33}^{1}e_{2}$ | $=$ | $\varphi_{1}(e_{2})$ | $=$ | $\varphi_{2}(e_{1})$ | $=$ |  $\beta_{33}^{2}e_{1}+\beta_{12}^{2}e_{2}$,
---|---|---|---|---|---|---
$\beta_{31}^{1}e_{1}+\beta_{32}^{1}e_{2}+\beta_{33}^{1}e_{3}$ | $=$ | $\varphi_{1}(e_{3})$ | = | $\varphi_{3}(e_{1})$ | $=$ | $\beta_{33}^{3}e_{1}+\beta_{12}^{3}e_{2},$
$\beta_{31}^{2}e_{1}+\beta_{32}^{2}e_{2}+\beta_{33}^{2}e_{3}$ | $=$ | $\varphi_{2}(e_{3})$ | = | $\varphi_{3}(e_{2})$ | $=$ | $\beta_{33}^{3}e_{2}.$

Hence, the commutative multiplication $-\cdot-$ is defined by

$\displaystyle e_{1}\cdot e_{1}$ $\displaystyle=$
$\displaystyle\beta_{12}^{1}e_{2},$ $\displaystyle e_{1}\cdot e_{3}$
$\displaystyle=$ $\displaystyle\beta_{33}^{3}e_{1}+\beta_{32}^{1}e_{2},$
$\displaystyle e_{2}\cdot e_{3}$ $\displaystyle=$
$\displaystyle\beta_{33}^{3}e_{2},$ $\displaystyle e_{3}\cdot e_{3}$
$\displaystyle=$
$\displaystyle\beta_{31}^{3}e_{1}+\beta_{32}^{3}e_{2}+\beta_{33}^{3}e_{3}.$

This multiplication is associative if and only if
$\beta_{31}^{3}\beta_{12}^{1}=\beta_{32}^{1}\beta_{33}^{3}$. Under the action
of an automorphism of the Lie algebra $(\mathfrak{L},[\cdot,\cdot]),$ given in
Remark 11, we rewrite the multiplication table of $(\mathfrak{L},\cdot)$ by
the following way:

$e_{1}\cdot e_{1}$ | $=$ | $\frac{\beta_{12}^{1}\lambda_{11}^{2}}{\lambda_{22}}e_{2},$
---|---|---
$e_{1}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{1}+\frac{\lambda_{11}(\beta^{1}_{32}+\beta^{1}_{12}\lambda_{13})}{\lambda_{22}}e_{2},$
$e_{2}\cdot e_{3}$ | $=$ | $\beta^{3}_{33}e_{2},$
$e_{3}\cdot e_{3}$ | $=$ | $\frac{\beta^{3}_{31}+\beta^{3}_{33}\lambda_{13}}{\lambda_{11}}e_{1}+\frac{\lambda_{11}(\beta^{3}_{31}+\beta^{3}_{32}+2\beta^{1}_{32}\lambda_{13}+\beta^{3}_{33}\lambda_{13}+\beta^{1}_{12}\lambda^{2}_{13}+\beta^{3}_{33}\lambda_{23})-(\beta^{3}_{31}+\beta^{3}_{33}\lambda_{13})\lambda_{22}}{\lambda_{11}\lambda_{22}}e_{2}+\beta_{33}^{3}e_{3}.$

Let us consider the following cases.

  1. (1)

$\beta^{3}_{33}\neq 0,$
$\beta_{32}^{1}=\frac{\beta^{3}_{31}\beta^{1}_{12}}{\beta^{3}_{33}}$ and
$\beta^{1}_{12}\neq 0,$ then by choosing

$\lambda_{11}=1,$ $\lambda_{22}=\beta^{1}_{12},$
$\lambda_{13}=-\frac{\beta^{3}_{31}}{\beta^{3}_{32}},$
$\lambda_{23}=\frac{\beta^{1}_{12}(\beta^{3}_{31})^{2}-\beta^{3}_{32}(\beta^{3}_{33})^{2}}{(\beta_{33}^{3})^{3}},$

we have

$e_{1}\cdot e_{1}$ | $=$ | $e_{2},$ | $e_{1}\cdot e_{3}$ | $=$ | $\beta^{3}_{33}e_{1},$ | $e_{2}\cdot e_{3}$ | $=$ | $\beta^{3}_{33}e_{2},$ | $e_{3}\cdot e_{3}$ | $=$ | $\beta^{3}_{33}e_{3}.$
---|---|---|---|---|---|---|---|---|---|---|---
  2. (2)

$\beta^{3}_{33}\neq 0,$
$\beta_{32}^{1}=\frac{\beta^{3}_{31}\beta^{1}_{12}}{\beta^{3}_{33}}$ and
$\beta^{1}_{12}=0,$ then by choosing

$\lambda_{11}=1,$ $\lambda_{22}=1,$
$\lambda_{13}=-\frac{\beta^{3}_{31}}{\beta^{3}_{33}},$
$\lambda_{23}=-\frac{\beta^{3}_{32}}{\beta^{3}_{32}},$

we have

$e_{1}\cdot e_{3}$ | $=$ | $\beta^{3}_{33}e_{1},$ | $e_{2}\cdot e_{3}$ | $=$ | $\beta^{3}_{33}e_{2},$ | $e_{3}\cdot e_{3}$ | $=$ | $\beta^{3}_{33}e_{3}.$
---|---|---|---|---|---|---|---|---
  3. (3)

$\beta^{3}_{33}=0,$ $\beta_{31}^{3}=0,$ $\beta^{1}_{12}\neq 0$ and
$\beta^{1}_{12}\beta^{3}_{32}\neq(\beta^{1}_{32})^{2},$ then by choosing

$\lambda_{11}=\frac{\sqrt{\beta^{1}_{12}\beta^{3}_{32}-(\beta^{1}_{32})^{2}}}{\beta^{1}_{12}},$
$\lambda_{22}=-\frac{(\beta^{1}_{32})^{2}}{\beta^{1}_{12}}+\beta^{3}_{32},$
$\lambda_{13}=-\frac{\beta^{1}_{32}}{2\beta^{1}_{12}},$ $\lambda_{23}=0,$

we have

$e_{1}\cdot e_{1}$ | $=$ | $e_{2},$ | $e_{3}\cdot e_{3}$ | $=$ | $e_{2}.$
---|---|---|---|---|---
  4. (4)

$\beta^{3}_{33}=0,$ $\beta_{31}^{3}=0,$ $\beta^{1}_{12}\neq 0$ and
$\beta^{1}_{12}\beta^{3}_{32}=(\beta^{1}_{32})^{2},$ then by choosing

$\lambda_{11}=1,$ $\lambda_{22}=\beta^{1}_{12},$
$\lambda_{13}=-\frac{\beta^{1}_{32}}{2\beta^{1}_{12}},$ $\lambda_{23}=0,$

we have

$e_{1}\cdot e_{1}$ | $=$ | $e_{2}.$
---|---|---
  5. (5)

$\beta^{3}_{33}=0,$ $\beta_{12}^{1}=0,$ $\beta^{1}_{32}\neq 0$ and
$\beta^{3}_{31}\neq 0,$ then by choosing

$\lambda_{11}=\beta^{3}_{31},$ $\lambda_{22}=\beta^{1}_{32}\beta^{3}_{31},$
$\lambda_{13}=\frac{\beta^{1}_{32}\beta^{3}_{31}-\beta^{3}_{31}-\beta^{3}_{32}}{2\beta^{1}_{32}},$
$\lambda_{23}=0,$

we have

$e_{1}\cdot e_{3}$ | $=$ | $e_{2},$ | $e_{3}\cdot e_{3}$ | $=$ | $e_{1}.$ |
---|---|---|---|---|---|---
  6. (6)

$\beta^{3}_{33}=0,$ $\beta_{12}^{1}=0,$ $\beta^{1}_{32}\neq 0$ and
$\beta^{3}_{31}=0,$ then by choosing

$\lambda_{11}=1,$ $\lambda_{22}=\beta^{1}_{32},$
$\lambda_{13}=\frac{\beta^{3}_{32}}{2\beta^{1}_{32}},$ $\lambda_{23}=0,$

we have

$e_{1}\cdot e_{3}$ | $=$ | $e_{2}.$
---|---|---
  7. (7)

$\beta^{3}_{33}=0,$ $\beta_{12}^{1}=0,$ $\beta^{1}_{32}=0$ and
$\beta^{3}_{31}=-\beta^{3}_{32},$ then by choosing then by choosing

$\lambda_{11}=\beta^{3}_{32},$ $\lambda_{22}=1,$
$\lambda_{13}=\frac{\beta^{3}_{32}}{2\beta^{1}_{32}},$ $\lambda_{23}=0,$

we have

$e_{3}\cdot e_{3}=-e_{1}+e_{2}.$

  8. (8)

$\beta^{3}_{33}=0,$ $\beta_{12}^{1}=0,$ $\beta^{1}_{32}=0$ and
$\beta^{3}_{31}\neq-\beta^{3}_{32},$ then by choosing some suitable
$\lambda_{ij},$ we have two following opportunities

$e_{3}\cdot e_{3}=e_{1};$ or $e_{3}\cdot e_{3}=e_{2}.$

These cases produce the algebras ${\rm T}_{12}^{\beta\neq 0}$, ${\rm
T}_{09}^{2,\beta}$, ${\rm T}_{13}$, ${\rm T}_{12}^{0}$, ${\rm T}_{14}$, ${\rm
T}_{15}$, ${\rm T}_{10*}^{2}$, ${\rm T}_{10}^{2}$, respectively. Again, recall
that ${\rm T}_{10*}^{2}\cong{\rm T}_{10}^{\frac{1}{2}}$.

∎

######  Proposition 14.

Let $(\mathfrak{L},\cdot,[\cdot,\cdot])$ be a transposed Poisson algebra
structure defined on $\mathfrak{g}_{2}^{0}$. Then
$(\mathfrak{L},\cdot,[\cdot,\cdot])$ is isomorphic to only one of the
following algebras:

  1. (1)

${\rm T}_{09}^{0,\beta}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{3}=\beta
e_{1},\ e_{2}\cdot e_{3}=\beta e_{2},\ e_{3}\cdot e_{3}=\beta e_{3},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2}$.\end{tabular}\right.$

  2. (2)

${\rm T}_{10}^{0}:\left\\{\begin{tabular}[]{l}$e_{3}\cdot e_{3}=e_{2},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2}$.\end{tabular}\right.$

  3. (3)

${\rm T}_{11}^{0}:\left\\{\begin{tabular}[]{l}$e_{3}\cdot e_{3}=e_{1},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2}$.\end{tabular}\right.$

  4. (4)

${\rm T}_{16}:\left\\{\begin{tabular}[]{l}$e_{3}\cdot e_{3}=e_{1}+e_{2},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2}$.\end{tabular}\right.$

  5. (5)

${\rm T}_{17}^{\beta}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{2},$ \
$e_{1}\cdot e_{2}=-e_{2},$ \ $e_{1}\cdot e_{3}=\beta e_{1},$ \ $e_{2}\cdot
e_{2}=e_{2},$ \ $e_{2}\cdot e_{3}=\beta e_{2},$ \ $e_{3}\cdot e_{3}=\beta
e_{3},$\\\ $[e_{1},e_{3}]=e_{1}+e_{2}$.\end{tabular}\right.$

  6. (6)

${\rm T}_{18}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{2},$ \
$e_{1}\cdot e_{2}=-e_{2},$ \ $e_{2}\cdot e_{2}=e_{2},$ \ $e_{3}\cdot
e_{3}=e_{1}+e_{2},$\\\ $[e_{1},e_{3}]=e_{1}+e_{2}$.\end{tabular}\right.$

  7. (7)

${\rm T}_{19}^{\gamma}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{3}=\gamma
e_{1}+\gamma e_{2},$ \ $e_{3}\cdot e_{3}=\gamma e_{3},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2}$.\end{tabular}\right.$

where the parameters $\beta\in\mathbb{C}$ and $\gamma\in\mathbb{C}^{*}$.

###### Proof.

From Remark 10,

$\varphi_{i}(e_{1})=\beta_{33}^{i}e_{1}+(\beta_{33}^{i}-\beta_{22}^{i})e_{2}$,
$\varphi_{i}(e_{2})=\beta_{22}^{i}e_{2}$,
$\varphi_{i}(e_{3})=\beta_{31}^{i}e_{1}+\beta_{32}^{i}e_{2}+\beta_{33}^{i}e_{3}$

and

$\beta_{22}^{1}e_{2}$ | $=$ | $\varphi_{1}(e_{2})$ | $=$ | $\varphi_{2}(e_{1})$ | $=$ |  $\beta_{33}^{2}e_{1}+(\beta_{33}^{2}-\beta_{22}^{2})e_{2}$,
---|---|---|---|---|---|---
$\beta_{31}^{1}e_{1}+\beta_{32}^{1}e_{2}+\beta_{33}^{1}e_{3}$ | $=$ | $\varphi_{1}(e_{3})$ | $=$ | $\varphi_{3}(e_{1})$ | $=$ | $\beta_{33}^{3}e_{1}+(\beta_{33}^{3}-\beta_{22}^{3})e_{2},$
$\beta_{31}^{2}e_{1}+\beta_{32}^{2}e_{2}+\beta_{33}^{2}e_{3}$ | $=$ | $\varphi_{2}(e_{3})$ | $=$ | $\varphi_{3}(e_{2})$ | $=$ | $\beta_{22}^{3}e_{2}.$

Hence, the commutative multiplication $-\cdot-$ is defined by

$\displaystyle e_{1}\cdot e_{1}$ $\displaystyle=$
$\displaystyle\beta_{22}^{2}e_{2},$ $\displaystyle e_{1}\cdot e_{2}$
$\displaystyle=$ $\displaystyle-\beta_{22}^{2}e_{2},$ $\displaystyle
e_{1}\cdot e_{3}$ $\displaystyle=$
$\displaystyle\beta_{33}^{3}e_{1}-(\beta_{22}^{3}-\beta_{33}^{3})e_{2},$
$\displaystyle e_{2}\cdot e_{2}$ $\displaystyle=$
$\displaystyle\beta_{22}^{2}e_{2},$ $\displaystyle e_{2}\cdot e_{3}$
$\displaystyle=$ $\displaystyle\beta_{22}^{3}e_{2},$ $\displaystyle e_{3}\cdot
e_{3}$ $\displaystyle=$
$\displaystyle\beta_{31}^{3}e_{1}+\beta_{32}^{3}e_{2}+\beta_{33}^{3}e_{3}.$

This multiplication is associative if and only if
$(\beta_{22}^{3})^{2}-\beta_{33}^{3}\beta_{22}^{3}+(\beta_{31}^{3}-\beta_{32}^{3})\beta_{22}^{2}=0$.

Under the action of an automorphism of the Lie algebra
$(\mathfrak{L},[\cdot,\cdot]),$ given in Remark 11, we rewrite the
multiplication table of $(\mathfrak{L},\cdot)$ by the following way:

$e_{1}\cdot e_{1}$ | $=$ | $\beta_{22}^{2}\lambda_{22}e_{2},$
---|---|---
$e_{1}\cdot e_{2}$ | $=$ | $-\beta_{22}^{2}\lambda_{22}e_{2},$
$e_{1}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{1}+(\beta^{3}_{33}-\beta^{3}_{22}+\beta^{2}_{22}\lambda_{13}-\beta^{2}_{22}\lambda_{23})e_{2},$
$e_{2}\cdot e_{2}$ | $=$ | $\beta^{2}_{22}\lambda_{22}e_{2},$
$e_{2}\cdot e_{3}$ | $=$ | $(\beta^{3}_{22}-\beta^{2}_{22}\lambda_{13}+\beta^{2}_{22}\lambda_{23})e_{2},$
$e_{3}\cdot e_{3}$ | $=$ | $\frac{\beta^{3}_{31}+\beta^{3}_{33}\lambda_{13}}{\lambda_{11}}e_{1}+$
|  | $\frac{(\beta^{3}_{31}+\beta^{3}_{33}\lambda_{13})\lambda_{22}+\lambda_{11}(\beta^{3}_{32}-\beta^{3}_{31}+\beta^{2}_{22}\lambda_{13}^{2}+(2\beta^{3}_{22}-\beta^{3}_{33})\lambda_{23}+\beta^{2}_{22}\lambda_{23}^{2}+\lambda_{13}(\beta^{3}_{33}-2\beta^{3}_{22}-2\beta^{2}_{22}\lambda_{23}))}{\lambda_{11}\lambda_{22}}e_{2}+\beta_{33}^{3}e_{3}.$

Let us consider the following cases.

  1. (1)

$\beta_{22}^{2}\neq 0,$
$\beta_{31}^{3}=\frac{\beta_{33}^{3}\beta_{22}^{3}+\beta_{32}^{3}\beta_{22}^{2}-(\beta_{22}^{3})^{2}}{\beta_{22}^{2}}$
and $\beta^{3}_{33}\neq 0,$ then by choosing

$\lambda_{11}=1,$ $\lambda_{22}=\frac{1}{\beta^{2}_{22}},$
$\lambda_{13}=\frac{(\beta_{32}^{3})^{2}-\beta^{2}_{22}\beta^{3}_{32}-\beta^{3}_{22}\beta^{3}_{33}}{\beta^{2}_{22}\beta_{33}^{3}},$
$\lambda_{23}=\frac{(\beta_{32}^{3})^{2}-\beta^{2}_{22}\beta^{3}_{32}-2\beta^{3}_{22}\beta^{3}_{33}+(\beta^{3}_{33})^{2}}{\beta^{2}_{22}\beta_{33}^{3}},$

we have

$e_{1}\cdot e_{1}$ | $=$ | $e_{2},$ | $e_{1}\cdot e_{2}$ | $=$ | $-e_{2},$ | $e_{1}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{1},$
---|---|---|---|---|---|---|---|---
$e_{2}\cdot e_{2}$ | $=$ | $e_{2},$ | $e_{2}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{2},$ | $e_{3}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{3}.$
  2. (2)

$\beta_{22}^{2}\neq 0,$
$\beta_{31}^{3}=\frac{\beta_{33}^{3}\beta_{22}^{3}+\beta_{32}^{3}\beta_{22}^{2}-(\beta_{22}^{3})^{2}}{\beta_{22}^{2}},$
$\beta^{3}_{33}=0$ and $\beta^{2}_{22}\beta^{3}_{32}\neq(\beta^{3}_{22})^{2},$
then by choosing

$\lambda_{11}=-\frac{(\beta^{3}_{22})^{2}}{\beta^{2}_{22}}+\beta^{3}_{32},$
$\lambda_{22}=\frac{1}{\beta^{2}_{22}},$ $\lambda_{13}=0,$
$\lambda_{23}=\frac{\beta_{22}^{3}}{\beta^{2}_{22}},$

we have

$e_{1}\cdot e_{1}$ | $=$ | $e_{2},$ | $e_{1}\cdot e_{2}$ | $=$ | $-e_{2},$ | $e_{2}\cdot e_{2}$ | $=$ | $e_{2},$ | $e_{3}\cdot e_{3}$ | $=$ | $e_{1}+e_{2}.$
---|---|---|---|---|---|---|---|---|---|---|---
  3. (3)

$\beta_{22}^{2}\neq 0,$
$\beta_{31}^{3}=\frac{\beta_{33}^{3}\beta_{22}^{3}+\beta_{32}^{3}\beta_{22}^{2}-(\beta_{22}^{3})^{2}}{\beta_{22}^{2}},$
$\beta^{3}_{33}=0$ and $\beta^{2}_{22}\beta^{3}_{32}=(\beta^{3}_{22})^{2},$
then by choosing

$\lambda_{11}=1,$ $\lambda_{22}=\frac{1}{\beta^{2}_{22}},$ $\lambda_{13}=0,$
$\lambda_{23}=\frac{\beta_{22}^{3}}{\beta^{2}_{22}},$

we have

$e_{1}\cdot e_{1}$ | $=$ | $e_{2},$ | $e_{1}\cdot e_{2}$ | $=$ | $-e_{2},$ | $e_{2}\cdot e_{2}$ | $=$ | $e_{2}.$
---|---|---|---|---|---|---|---|---
  4. (4)

$\beta_{22}^{2}=0,$ $\beta^{3}_{22}=0$ and $\beta^{3}_{33}\neq 0,$ then by
choosing

$\lambda_{11}=1,$ $\lambda_{22}=1,$
$\lambda_{13}=-\frac{\beta^{3}_{31}}{\beta^{3}_{32}},$
$\lambda_{23}=\frac{\beta^{3}_{32}-2\beta_{31}^{3}}{\beta^{3}_{33}},$

we have

$e_{1}\cdot e_{3}$ | $=$ | $\beta^{3}_{33}e_{1}+\beta^{3}_{33}e_{2},$ | $e_{3}\cdot e_{3}$ | $=$ | $\beta^{3}_{33}e_{3}.$
---|---|---|---|---|---
  5. (5)

$\beta_{22}^{2}=0,$ $\beta^{3}_{22}=0,$ $\beta^{3}_{33}0$ and
$\beta^{3}_{31}=\beta^{3}_{32},$ then by choosing

$\lambda_{11}=\beta^{3}_{32},$ $\lambda_{22}=1,$ $\lambda_{13}=0,$
$\lambda_{23}=0,$

we have

$e_{3}\cdot e_{3}$ | $=$ | $e_{1}+e_{2}.$
---|---|---
  6. (6)

$\beta_{22}^{2}=0,$ $\beta^{3}_{22}=0,$ $\beta^{3}_{33}0$ and
$\beta^{3}_{31}\neq\beta^{3}_{32},$ then by choosing some suitable
$\lambda_{ij},$ we have two following opportunities

$e_{3}\cdot e_{3}=e_{1};$ or $e_{3}\cdot e_{3}=e_{2}.$

  7. (7)

$\beta_{22}^{2}=0$ and $\beta^{3}_{22}=\beta^{3}_{33}\neq 0,$ then by choosing

$\lambda_{11}=1,$ $\lambda_{22}=1,$
$\lambda_{13}=-\frac{\beta^{3}_{31}}{\beta^{3}_{32}},$
$\lambda_{23}=-\frac{\beta^{3}_{32}}{\beta^{3}_{33}},$

we have

$e_{1}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{1},$ | $e_{2}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{2},$ | $e_{3}\cdot e_{3}$ | $=$ | $\beta_{33}^{3}e_{3}.$
---|---|---|---|---|---|---|---|---

After constructing the corresponding transposed Poisson algebras from these
cases, we obtain the algebras ${\rm T}_{17}^{\beta\neq 0}$, ${\rm T}_{18}$,
${\rm T}_{17}^{0}$, ${\rm T}_{19}^{\gamma\neq 0}$, ${\rm T}_{16}$, ${\rm
T}_{11}^{0}$, ${\rm T}_{10}^{0}$, ${\rm T}_{09}^{0,\beta}$, respectively.

∎

###  1.4. Classification theorem

The results of the previous subsection together with the classification of the
commutative associative algebras of dimension three (we recall the
classification that was used in [22]), give us the following classification
theorem.

######  Theorem A.

Let $(\mathfrak{L},\cdot,[\cdot,\cdot])$ be a nonzero complex $3$-dimensional
transposed Poisson algebra, then $(\mathfrak{L},\cdot,[\cdot,\cdot])$ is
isomorphic to one and only one transposed Poisson algebra listed below:

  * •

${\rm T}_{01}:$ $[e_{1},e_{2}]=e_{3},$ $[e_{1},e_{3}]=-e_{2},$
$[e_{2},e_{3}]=e_{1}.$

  * •

${\rm T}_{02}:\left\\{\begin{tabular}[]{l}$e_{2}\cdot e_{2}=e_{3},$\\\
$[e_{1},e_{2}]=e_{3}.$\end{tabular}\right.$

  * •

${\rm T}_{03}^{\beta}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{2}=\beta
e_{3},$\\\ $[e_{1},e_{2}]=e_{3}.$\end{tabular}\right.$

  * •

${\rm T}_{04}^{\beta}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{2}=\beta
e_{3},e_{2}\cdot e_{2}=e_{1},$\\\ $[e_{1},e_{2}]=e_{3}.$\end{tabular}\right.$

  * •

${\rm T}_{05}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{3},e_{1}\cdot
e_{2}=e_{1},e_{2}\cdot e_{2}=e_{2},e_{2}\cdot e_{3}=e_{3},$\\\
$[e_{1},e_{2}]=e_{3}.$\end{tabular}\right.$

  * •

${\rm T}_{06}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{2}=e_{1},e_{2}\cdot
e_{2}=e_{2},e_{2}\cdot e_{3}=e_{3},$\\\
$[e_{1},e_{2}]=e_{3}.$\end{tabular}\right.$

  * •

${\rm T}_{07}^{\beta}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{3}=\beta
e_{1}$, $e_{2}\cdot e_{3}=\beta e_{2}$, $e_{3}\cdot e_{3}=\beta e_{3},$\\\
$[e_{1},e_{3}]=e_{1},$ $[e_{2},e_{3}]=e_{2}.$\end{tabular}\right.$

  * •

${\rm T}_{08}:\left\\{\begin{tabular}[]{l}$e_{3}\cdot e_{3}=e_{1},$\\\
$[e_{1},e_{3}]=e_{1},$ $[e_{2},e_{3}]=e_{2}.$\end{tabular}\right.$

  * •

${\rm T}_{09}^{\alpha,\beta}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot
e_{3}=\beta e_{1},\ e_{2}\cdot e_{3}=\beta e_{2},\ e_{3}\cdot e_{3}=\beta
e_{3},$\\\ $[e_{1},e_{3}]=e_{1}+e_{2},$ $[e_{2},e_{3}]=\alpha
e_{2}.$\end{tabular}\right.$

  * •

${\rm T}_{10}^{\alpha}:\left\\{\begin{tabular}[]{l}$e_{3}\cdot
e_{3}=e_{2},$\\\ $[e_{1},e_{3}]=e_{1}+e_{2},$ $[e_{2},e_{3}]=\alpha
e_{2}.$\end{tabular}\right.$

  * •

${\rm T}_{11}^{\alpha}:\left\\{\begin{tabular}[]{l}$e_{3}\cdot
e_{3}=e_{1},$\\\ $[e_{1},e_{3}]=e_{1}+e_{2},$ $[e_{2},e_{3}]=\alpha
e_{2}.$\end{tabular}\right.$

  * •

${\rm T}_{12}^{\beta}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{2},\
e_{1}\cdot e_{3}=\beta e_{1},\ e_{2}\cdot e_{3}=\beta e_{2},\ e_{3}\cdot
e_{3}=\beta e_{3},$\\\ $[e_{1},e_{3}]=e_{1}+e_{2},$
$[e_{2},e_{3}]=2e_{2}.$\end{tabular}\right.$

  * •

${\rm T}_{13}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{2},\ e_{3}\cdot
e_{3}=e_{2},$\\\ $[e_{1},e_{3}]=e_{1}+e_{2},$
$[e_{2},e_{3}]=2e_{2}.$\end{tabular}\right.$

  * •

${\rm T}_{14}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{3}=e_{2},\ e_{3}\cdot
e_{3}=e_{1},$\\\ $[e_{1},e_{3}]=e_{1}+e_{2},$
$[e_{2},e_{3}]=2e_{2}.$\end{tabular}\right.$

  * •

${\rm T}_{15}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{3}=e_{2},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2},$ $[e_{2},e_{3}]=2e_{2}.$\end{tabular}\right.$

  * •

${\rm T}_{16}:\left\\{\begin{tabular}[]{ll}$e_{3}\cdot e_{3}=e_{1}+e_{2},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2}.$\end{tabular}\right.$

  * •

${\rm T}_{17}^{\beta}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{2},$ \
$e_{1}\cdot e_{2}=-e_{2},$ \ $e_{1}\cdot e_{3}=\beta e_{1},$ \ $e_{2}\cdot
e_{2}=e_{2},$ \ $e_{2}\cdot e_{3}=\beta e_{2},$ \ $e_{3}\cdot e_{3}=\beta
e_{3},$\\\ $[e_{1},e_{3}]=e_{1}+e_{2}$.\end{tabular}\right.$

  * •

${\rm T}_{18}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{2},$ \
$e_{1}\cdot e_{2}=-e_{2},$ \ $e_{2}\cdot e_{2}=e_{2},$ \ $e_{3}\cdot
e_{3}=e_{1}+e_{2},$\\\ $[e_{1},e_{3}]=e_{1}+e_{2}$.\end{tabular}\right.$

  * •

${\rm T}_{19}^{\gamma}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{3}=\gamma
e_{1}+\gamma e_{2},$ \ $e_{3}\cdot e_{3}=\gamma e_{3},$\\\
$[e_{1},e_{3}]=e_{1}+e_{2}$.\end{tabular}\right.$

  * •

${\rm T}_{20}:$ $e_{1}\cdot e_{1}=e_{1},e_{2}\cdot e_{2}=e_{2},e_{3}\cdot
e_{3}=e_{3}.$

  * •

${\rm T}_{21}:$ $e_{1}\cdot e_{1}=e_{1},e_{2}\cdot e_{2}=e_{2},e_{1}\cdot
e_{3}=e_{3}.$

  * •

${\rm T}_{22}:$ $e_{1}\cdot e_{1}=e_{1},e_{2}\cdot e_{2}=e_{2}.$

  * •

${\rm T}_{23}:$ $e_{1}\cdot e_{1}=e_{1},e_{1}\cdot e_{2}=e_{2},e_{1}\cdot
e_{3}=e_{3},e_{2}\cdot e_{2}=e_{3}.$

  * •

${\rm T}_{24}:$ $e_{1}\cdot e_{1}=e_{1},e_{1}\cdot e_{2}=e_{2},e_{1}\cdot
e_{3}=e_{3}.$

  * •

${\rm T}_{25}:$ $e_{1}\cdot e_{1}=e_{1},e_{1}\cdot e_{2}=e_{2}.$

  * •

${\rm T}_{26}:$ $e_{1}\cdot e_{1}=e_{1},e_{2}\cdot e_{2}=e_{3}.$

  * •

${\rm T}_{27}:$ $e_{1}\cdot e_{1}=e_{1}.$

  * •

${\rm T}_{28}:$ $e_{1}\cdot e_{1}=e_{2},e_{1}\cdot e_{2}=e_{3}.$

  * •

${\rm T}_{29}:$ $e_{1}\cdot e_{2}=e_{3}.$

  * •

${\rm T}_{30}:$ $e_{1}\cdot e_{1}=e_{2}.$

where the parameters $\beta\in\mathbb{C}$ and $\gamma\in\mathbb{C}^{*}$.
Between these algebras there are precisely the following non-trivial
isomorphisms:

  * •

${\rm T}_{03}^{\beta_{1}}\cong{\rm T}_{03}^{\beta_{2}}$ if and only if
$\beta_{1}=\pm\beta_{2}$.

  * •

${\rm T}_{09}^{\alpha_{1},\beta_{1}}\cong{\rm T}_{09}^{\alpha_{2},\beta_{2}}$
if and only if $(\alpha_{2},\beta_{2})=(\alpha_{1},\beta_{1})$ or
$(\alpha_{2},\beta_{2})=(\frac{1}{\alpha_{1}},\frac{\beta_{1}}{\alpha_{1}})$.

  * •

${\rm T}_{11}^{\beta_{1}}\cong{\rm T}_{11}^{\beta_{2}}$ if and only if
$\beta_{2}=\beta_{1}$ or $\beta_{2}=\frac{1}{\beta_{1}}$.

##  2\. Strong special transposed Poisson algebras

It is known that each Novikov-Poisson algebra under commutator product on non-
associative multiplication gives a transposed Poisson algebra [4]. Let us say
that a transposed Poisson algebra is special (strong special) if it can be
embedded into (it is isomorphic to) a Novikov-Poisson algebra relative to the
commutator bracket. Similarly, let us say that a transposed Poisson algebra is
$D$-special (strong special) (from “differentially”) if it embeds into (it is
isomorphic to) a commutative algebra with a derivation relative to the bracket
$[x,y]=\mathfrak{D}(x)\cdot y-x\cdot\mathfrak{D}(y).$ Obviously, every
$D$-special transposed Poisson algebra is a special one. Our main strategy for
classifying all non-isomorphic strong $D$-special transposed Poisson algebras
with a given associative commutative algebras $(\mathfrak{A},\cdot)$ is as
follows.

  1. (1)

Find all derivations $\mathfrak{Der}({\mathfrak{A}},\cdot)$;

  2. (2)

Describe the multiplication of the family of transposed Poisson algebras given
by $({\mathfrak{A}},\cdot,[x,y]=\mathfrak{D}(x)\cdot
y-x\cdot\mathfrak{D}(y))$;

  3. (3)

Consider the multiplication table of $({\mathfrak{A}},[\cdot,\cdot])$ under
the action of elements from ${\rm Aut}({\mathfrak{A}},\cdot)$ and separate all
non-isomorphic cases.

###  2.1. Strong $D$-special $2$-dimensional transposed Poisson algebras

Let us remember, that the classification of complex $2$-dimensional transposed
Poisson algebras is given in [4].

####  2.1.1. The algebraic classification of $2$-dimensional associative
commutative algebras

  * •

${\mathcal{A}}_{01}:$ $e_{1}\cdot e_{1}=e_{1},e_{2}\cdot e_{2}=e_{2}.$

  * •

${\mathcal{A}}_{02}:$ $e_{1}\cdot e_{1}=e_{1},e_{1}\cdot e_{2}=e_{2}.$

  * •

${\mathcal{A}}_{03}:$ $e_{1}\cdot e_{1}=e_{1}.$

  * •

${\mathcal{A}}_{04}:$ $e_{1}\cdot e_{1}=e_{2}.$

By straightforward calculations we have the following result:

######  Lemma 15.

Given
$\mathcal{A}\in\left\\{{\mathcal{A}}_{01},{\mathcal{A}}_{03},{\mathcal{A}}_{04}\right\\}$,
then $[x,y]=\mathfrak{D}(x)\cdot y-x\cdot\mathfrak{D}(y)=0$ for any derivation
$\mathfrak{D}\in\mathfrak{Der}(\mathcal{A})$.

####  2.1.2. Strong $D$-special transposed Poisson algebras on
${\mathcal{A}}_{02}$

Let $\mathfrak{D}$ be a derivation of ${\mathcal{A}}_{02},$ then

$\mathfrak{D}(e_{2})=-\alpha e_{2}.$

Which gives

${\mathcal{D}}_{01}^{\alpha}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot
e_{1}=e_{1},$ \ $e_{1}\cdot e_{2}=e_{2},$\\\ $[e_{1},e_{2}]=\alpha
e_{2}$.\end{tabular}\right.$

All algebras from the family ${\mathcal{D}}_{01}^{\alpha}$ are non-isomorphic
and they exhaust all strong $D$-special $2$-dimensional transposed Poisson
algebras.

####  2.1.3. Non-strong-$D$-special $2$-dimensional transposed Poisson
algebras

Thanks to subsubsection 2.1.2 and [4], we are concluding that there are only
two non-strong-$D$-special transposed Poisson algebras:

${\mathcal{N}}_{01}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{2},$\\\
$[e_{1},e_{2}]=e_{2}.$\par\end{tabular}\right.$ and
${\mathcal{N}}_{02}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot
e_{2}=e_{1},e_{2}\cdot e_{2}=e_{2},$\\\
$[e_{1},e_{2}]=e_{2}.$\par\end{tabular}\right.$

On the one hand, an straightforward verification shows that
${\mathcal{N}}_{01}$ is strong special using the Novikov-Poisson algebra:

${\bf N}_{01}:\left\\{\begin{tabular}[]{l}$e_{2}\cdot e_{2}=e_{1},$\\\
$e_{2}\circ e_{1}=-e_{1}.$\par\end{tabular}\right.$

On the other hand, one can verify that the Novikov-Poisson algebras defined on
$({\mathcal{N}}_{02},\cdot)$ are:

${\bf N}_{02}^{\alpha,\beta,\gamma}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot
e_{2}=e_{1},e_{2}\cdot e_{2}=e_{2},$\\\ $e_{1}\circ e_{2}=\alpha
e_{1},e_{2}\circ e_{1}=\beta e_{1},e_{2}\circ e_{2}=\gamma e_{1}+\alpha
e_{2}.$\par\end{tabular}\right.$

Then, observe that $e_{1}\circ e_{2}-e_{2}\circ e_{1}=(\alpha-\beta)e_{1}$ and
$[e_{1},e_{2}]=e_{2}$. Since the automorphisms of $({\mathcal{N}}_{02},\cdot)$
verify that $\phi(e_{1})=\lambda_{11}e_{1},\phi(e_{2})=e_{2}$ for
$\lambda_{11}\neq 0$, we conclude that ${\mathcal{N}}_{02}$ is not strong
special.

######  Corollary 16.

All complex $2$-dimensional transposed Poisson algebras are strong special.

###  2.2. Strong $D$-special $3$-dimensional transposed Poisson algebras

####  2.2.1. The algebraic classification of $3$-dimensional associative
commutative algebras

The classification of the $3$-dimensional associative commutative algebras was
extracted from [22].

  * •

${\mathfrak{A}}_{01}:$ $e_{1}\cdot e_{1}=e_{1},e_{2}\cdot
e_{2}=e_{2},e_{3}\cdot e_{3}=e_{3}.$

  * •

${\mathfrak{A}}_{02}:$ $e_{1}\cdot e_{1}=e_{1},e_{2}\cdot
e_{2}=e_{2},e_{1}\cdot e_{3}=e_{3}.$

  * •

${\mathfrak{A}}_{03}:$ $e_{1}\cdot e_{1}=e_{1},e_{2}\cdot e_{2}=e_{2}.$

  * •

${\mathfrak{A}}_{04}:$ $e_{1}\cdot e_{1}=e_{1},e_{1}\cdot
e_{2}=e_{2},e_{1}\cdot e_{3}=e_{3},e_{2}\cdot e_{2}=e_{3}.$

  * •

${\mathfrak{A}}_{05}:$ $e_{1}\cdot e_{1}=e_{1},e_{1}\cdot
e_{2}=e_{2},e_{1}\cdot e_{3}=e_{3}.$

  * •

${\mathfrak{A}}_{06}:$ $e_{1}\cdot e_{1}=e_{1},e_{1}\cdot e_{2}=e_{2}.$

  * •

${\mathfrak{A}}_{07}:$ $e_{1}\cdot e_{1}=e_{1},e_{2}\cdot e_{2}=e_{3}.$

  * •

${\mathfrak{A}}_{08}:$ $e_{1}\cdot e_{1}=e_{1}.$

  * •

${\mathfrak{A}}_{09}:$ $e_{1}\cdot e_{1}=e_{2},e_{1}\cdot e_{2}=e_{3}.$

  * •

${\mathfrak{A}}_{10}:$ $e_{1}\cdot e_{2}=e_{3}.$

  * •

${\mathfrak{A}}_{11}:$ $e_{1}\cdot e_{1}=e_{2}.$

By straightforward calculations we have the following result:

######  Lemma 17.

Given
$\mathfrak{A}\in\left\\{{\mathfrak{A}}_{01},{\mathfrak{A}}_{03},{\mathfrak{A}}_{07},{\mathfrak{A}}_{08},{\mathfrak{A}}_{11}\right\\}$,
then $[x,y]=\mathfrak{D}(x)\cdot y-x\cdot\mathfrak{D}(y)=0$ for any derivation
$\mathfrak{D}\in\mathfrak{Der}(\mathfrak{A})$.

####  2.2.2. Strong $D$-special transposed Poisson algebras on
${\mathfrak{A}}_{02}$

Let $\mathfrak{D}$ be a derivation of ${\mathfrak{A}}_{02},$ then

$\mathfrak{D}(e_{3})=-\alpha e_{3}.$

Which gives

${\rm A}_{01}^{\alpha}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{1},$ \
$e_{2}\cdot e_{2}=e_{2},$ \ $e_{1}\cdot e_{3}=e_{3},$\\\ $[e_{1},e_{3}]=\alpha
e_{3}$.\end{tabular}\right.$

All algebras from the family ${\rm A}_{01}^{\alpha}$ are non-isomorphic and
under the following change of basis:

$E_{1}=e_{3}-e_{2},$ $E_{2}=e_{2},$ $E_{3}=-\alpha^{-1}(e_{1}+e_{2}),$

we have that ${\rm A}_{01}^{\alpha}\cong{\rm T}_{17}^{-\alpha^{-1}}$ for
$\alpha\neq 0$.

####  2.2.3. Strong $D$-special transposed Poisson algebras on
${\mathfrak{A}}_{04}$

Let $\mathfrak{D}$ be a derivation of ${\mathfrak{A}}_{04},$ then

$\mathfrak{D}(e_{2})=-\alpha e_{2}-\beta e_{3},$ $\mathfrak{D}(e_{3})=-2\alpha
e_{3}.$

Which gives

${\rm A}_{02}^{\alpha,\beta}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot
e_{1}=e_{1},e_{1}\cdot e_{2}=e_{2},e_{1}\cdot e_{3}=e_{3},e_{2}\cdot
e_{2}=e_{3},$\\\ $[e_{1},e_{2}]=\alpha e_{2}+\beta e_{3},[e_{1},e_{3}]=2\alpha
e_{3}$.\end{tabular}\right.$

Let us consider two separate cases:

  1. $\bullet$

$\alpha=0,$ then under the following changing

$E_{1}=-\beta e_{2},$ $E_{2}=e_{1},$ $E_{3}=-\beta^{2}e_{3},$

we have that ${\rm A}_{02}^{0,\beta}\cong{\rm T}_{05}$ for $\beta\neq 0$.

  2. $\bullet$

$\alpha\neq 0,$ then under the following changing

$E_{1}=e_{2}+(1-\beta\alpha^{-1})e_{3},$ $E_{2}=e_{3},$
$E_{3}=-\alpha^{-1}e_{1},$

we have that ${\rm A}_{02}^{\alpha,\beta}\cong{\rm T}_{12}^{-\alpha^{-1}}.$

####  2.2.4. Strong $D$-special transposed Poisson algebras on
${\mathfrak{A}}_{05}$

Let $\mathfrak{D}$ be a derivation of ${\mathfrak{A}}_{05},$ then

$\mathfrak{D}(e_{2})=-\alpha e_{2}-\beta e_{3},$ $\mathfrak{D}(e_{3})=-\gamma
e_{2}-\delta e_{3}.$

Which gives

${\rm
A}_{03}^{\alpha,\beta,\gamma,\delta}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot
e_{1}=e_{1},e_{1}\cdot e_{2}=e_{2},e_{1}\cdot e_{3}=e_{3},$\\\
$[e_{1},e_{2}]=\alpha e_{2}+\beta e_{3},[e_{1},e_{3}]=\gamma e_{2}+\delta
e_{3}$.\end{tabular}\right.$

It is easy to see that if $({\rm T}_{j},\cdot)\cong{\mathfrak{A}}_{05}$ then
$j=06,07,09.$ Obtaining the following mutually exclusive cases for suitable
parameters $\alpha^{\prime}$ and $\beta^{\prime}$:

${\rm T}_{06}\cong{\rm
A}_{03}^{\alpha,\beta,-\frac{\alpha^{2}}{\beta},-\alpha},\ {\rm
T}_{07}^{\alpha}\cong{\rm A}_{03}^{-\alpha^{-1},0,0,-\alpha^{-1}},{\rm
T}_{09}^{\alpha^{\prime},\beta^{\prime}}\cong{\rm
A}_{03}^{\alpha,\beta,\gamma,\delta}.$

####  2.2.5. Strong $D$-special transposed Poisson algebras on
${\mathfrak{A}}_{06}$

Let $\mathfrak{D}$ be a derivation of ${\mathfrak{A}}_{06},$ then

$\mathfrak{D}(e_{2})=-\alpha e_{2},\mathfrak{D}(e_{3})=\beta e_{3}.$

Which gives

${\rm A}_{04}^{\alpha}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot
e_{1}=e_{1},e_{1}\cdot e_{2}=e_{2},$\\\ $[e_{1},e_{2}]=\alpha
e_{2}$.\end{tabular}\right.$

All algebras from the family ${\rm A}_{04}^{\alpha}$ are non-isomorphic and
under the following changing

$E_{1}=e_{2}-e_{3},$ $E_{2}=e_{3},$ $E_{3}=-\alpha^{-1}e_{1},$

we have that ${\rm A}_{04}^{\alpha}\cong{\rm T}_{19}^{-\alpha^{-1}}$ for
$\alpha\neq 0$.

####  2.2.6. Strong $D$-special transposed Poisson algebras on
${\mathfrak{A}}_{09}$

Let $\mathfrak{D}$ be a derivation of ${\mathfrak{A}}_{09},$ then

$\mathfrak{D}(e_{1})=-\alpha e_{1}+\beta e_{2}+\gamma e_{3}$,
$\mathfrak{D}(e_{2})=-2\alpha e_{2}+2\beta e_{3}$,
$\mathfrak{D}(e_{3})=-3\alpha e_{3}$,

Which gives

${\rm A}_{05}^{\alpha}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot
e_{1}=e_{2},e_{1}\cdot e_{2}=e_{3},$\\\ $[e_{1},e_{2}]=\alpha
e_{3}$.\end{tabular}\right.$

All algebras from the family ${\rm A}_{05}^{\alpha}$ are non-isomorphic and
under the following changing

$E_{1}=e_{2},$ $E_{2}=e_{1},$ $E_{3}=-\alpha e_{3},$

we have that ${\rm A}_{05}^{\alpha}\cong{\rm T}_{04}^{-\alpha^{-1}}$ for
$\alpha\neq 0$.

####  2.2.7. Strong $D$-special transposed Poisson algebras on
${\mathfrak{A}}_{10}$

Let $\mathfrak{D}$ be a derivation of ${\mathfrak{A}}_{10},$ then

$\mathfrak{D}(e_{3})=\alpha e_{1}+\delta e_{3},$ $\mathfrak{D}(e_{2})=\beta
e_{2}+\gamma e_{3},$ $\mathfrak{D}(e_{3})=(\alpha+\beta)e_{3}.$

Denoting $\epsilon=\alpha-\beta$, we have

${\rm A}_{06}^{\epsilon}:\left\\{\begin{tabular}[]{l}$e_{1}\cdot
e_{2}=e_{3},$\\\ $[e_{1},e_{2}]=\epsilon e_{3}$.\end{tabular}\right.$

All algebras from the family ${\rm A}_{06}^{\epsilon}$ are isomorphic only on
the case ${\rm A}_{06}^{\epsilon}\cong{\rm A}_{06}^{-\epsilon}$ and under the
following change of basis

$E_{1}=e_{2},$ $E_{2}=e_{1},$ $E_{3}=-\epsilon e_{3},$

we have that ${\rm A}_{06}^{\epsilon}\cong{\rm T}_{03}^{-\epsilon^{-1}}$ for
$\epsilon\neq 0$.

###  2.3. Classification theorem

The results of the previous subsection together with the classification of the
commutative associative algebras of dimension three (we recall the
classification that was used in [22]), give us the following classification
theorem.

######  Theorem B.

Let $(\mathfrak{L},\cdot,[\cdot,\cdot])$ be a nontrivial complex
$3$-dimensional transposed Poisson algebra (i.e. $\cdot\neq 0$ and
$[\cdot,\cdot]\neq 0$). If $(\mathfrak{L},\cdot,[\cdot,\cdot])$ is a strong
$D$-special transposed Poisson algebra then it is isomorphic to one and only
one listed below:

${\rm D}_{01}^{\alpha}$ | $:$ | ${\rm A}_{01}^{\alpha}$ | $:$ | $\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{1},$ \ $e_{2}\cdot e_{2}=e_{2},$ \ $e_{1}\cdot e_{3}=e_{3},$\\\ $[e_{1},e_{3}]=\alpha e_{3}$.\end{tabular}\right.$
---|---|---|---|---
${\rm D}_{02}$ | $:$ | ${\rm A}_{02}^{0,1}$ | $:$ | $\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{1},e_{1}\cdot e_{2}=e_{2},e_{1}\cdot e_{3}=e_{3},e_{2}\cdot e_{2}=e_{3},$\\\ $[e_{1},e_{2}]=e_{3}$.\end{tabular}\right.$
${\rm D}_{03}^{\alpha}$ | $:$ | ${\rm A}_{02}^{\alpha,0}$ | $:$ | $\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{1},e_{1}\cdot e_{2}=e_{2},e_{1}\cdot e_{3}=e_{3},e_{2}\cdot e_{2}=e_{3},$\\\ $[e_{1},e_{2}]=e_{2},[e_{1},e_{3}]=2e_{3}$.\end{tabular}\right.$
${\rm D}_{04}$ | $:$ | ${\rm A}_{03}^{0,1,0,0}$ | $:$ | $\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{1},e_{1}\cdot e_{2}=e_{2},e_{1}\cdot e_{3}=e_{3},$\\\ $[e_{1},e_{2}]=e_{3}$.\end{tabular}\right.$
${\rm D}_{05}^{\alpha}$ | $:$ | ${\rm A}_{03}^{\alpha,0,0,\alpha}$ | $:$ | $\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{1},e_{1}\cdot e_{2}=e_{2},e_{1}\cdot e_{3}=e_{3},$\\\ $[e_{1},e_{2}]=\alpha e_{2},[e_{1},e_{3}]=\alpha e_{3}$.\end{tabular}\right.$
${\rm D}_{06}^{\alpha,\beta}$ | $:$ | ${\rm A}_{03}^{\alpha,0,\beta,\beta}$ | $:$ | $\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{1},e_{1}\cdot e_{2}=e_{2},e_{1}\cdot e_{3}=e_{3},$\\\ $[e_{1},e_{2}]=\alpha e_{2},[e_{1},e_{3}]=\beta e_{2}+\beta e_{3}$.\end{tabular}\right.$
${\rm D}_{06}^{\alpha}$ | $:$ | ${\rm A}_{04}^{\alpha}$ | $:$ | $\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{1},e_{1}\cdot e_{2}=e_{2},$\\\ $[e_{1},e_{2}]=\alpha e_{2}$.\end{tabular}\right.$
${\rm D}_{07}^{\alpha}$ | $:$ | ${\rm A}_{05}^{\alpha}$ | $:$ | $\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{1}=e_{2},e_{1}\cdot e_{2}=e_{3},$\\\ $[e_{1},e_{2}]=\alpha e_{3}$.\end{tabular}\right.$
${\rm D}_{08}^{\alpha}$ | $:$ | ${\rm A}_{06}^{\alpha}$ | $:$ | $\left\\{\begin{tabular}[]{l}$e_{1}\cdot e_{2}=e_{3},$\\\ $[e_{1},e_{2}]=\alpha e_{3}$.\end{tabular}\right.$

Where

${\rm D}_{06}^{\alpha,\beta}\cong{\rm D}_{06}^{\beta,\alpha}$ and ${\rm
D}_{08}^{\alpha}\cong{\rm D}_{08}^{-\alpha}.$

If $(\mathfrak{L},\cdot,[\cdot,\cdot])$ is a non-strong-$D$-special transposed
Poisson algebra then it is isomorphic to one and only one listed below (see,
also Theorem A):

${\rm T}_{02},\ {\rm T}^{\alpha\neq 0}_{03},\ {\rm T}_{08},\ {\rm
T}_{10}^{\alpha},\ {\rm T}_{11}^{\alpha},\ {\rm T}_{13},\ {\rm T}_{14},\ {\rm
T}_{15},\ {\rm T}_{16},\ {\rm T}_{18}.$

##  3\. The geometric classification of transposed Poisson algebras

Given a vector space ${\mathbb{V}}$ of dimension $n$, the set of bilinear maps

$\textrm{Bil}({\mathbb{V}}\times{\mathbb{V}},{\mathbb{V}})\cong\textrm{Hom}({\mathbb{V}}^{\otimes
2},{\mathbb{V}})\cong({\mathbb{V}}^{*})^{\otimes 2}\otimes{\mathbb{V}}$

is a vector space of dimension $n^{3}$. The set of pairs of bilinear maps (or
bilinear pairs)

$\textrm{Bil}({\mathbb{V}}\times{\mathbb{V}},{\mathbb{V}})\oplus\textrm{Bil}({\mathbb{V}}\times{\mathbb{V}},{\mathbb{V}})\cong({\mathbb{V}}^{*})^{\otimes
2}\otimes{\mathbb{V}}\oplus({\mathbb{V}}^{*})^{\otimes 2}\otimes{\mathbb{V}}$

which is a vector space of dimension $2n^{3}$. This vector space has the
structure of the affine space $\mathbb{C}^{2n^{3}}$ in the following sense:
fixed a basis $e_{1},\ldots,e_{n}$ of ${\mathbb{V}}$, then any pair with
multiplication $(\mu,\mu^{\prime})$, is determined by some parameters
$c_{ij}^{k},c_{ij}^{\prime k}\in\mathbb{C}$, called structural constants, such
that

$\mu(e_{i},e_{j})=\sum_{p=1}^{n}c_{ij}^{k}e_{k}\textrm{ and
}\mu^{\prime}(e_{i},e_{j})=\sum_{p=1}^{n}c_{ij}^{\prime k}e_{k}$

which corresponds to a point in the affine space $\mathbb{C}^{2n^{3}}$. Then a
set of bilinear pairs $\mathcal{S}$ corresponds to an algebraic variety, i.e.,
a Zariski closed set if there are some polynomial equations in variables
$c_{ij}^{k},c_{ij}^{\prime k}$ with zero locus equal to the set of structural
constants of the bilinear pairs in $\mathcal{S}$. Since given the identities
defining transposed Poisson algebras we can obtain a set of polynomial
equations in variables $c_{ij}^{k},c_{ij}^{\prime k}$, the class of
$n$-dimensional transposed Poisson algebras is a variety, denote it by
$\mathcal{T}_{n}$. Now, consider the following action of
$\textrm{GL}({\mathbb{V}})$ on $\mathcal{T}_{n}$:

$(g*(\mu,\mu^{\prime}))(x,y):=(g\mu(g^{-1}x,g^{-1}y),g\mu^{\prime}(g^{-1}x,g^{-1}y))$

for $g\in\textrm{GL}({\mathbb{V}})$, $(\mu,\mu^{\prime})\in\mathcal{T}_{n}$
and for any $x,y\in{\mathbb{V}}$. Observe that the
$\textrm{GL}({\mathbb{V}})$-orbit of $(\mu,\mu^{\prime})$, denoted
$O((\mu,\mu^{\prime}))$, contains all the structural constants of the bilinear
pairs isomorphic to the transposed Poisson algebra with structural constants
$(\mu,\mu^{\prime})$.

The purpose of this section is to describe the irreducible components of the
variety ${\mathcal{T}_{3}}$. Recall that any affine variety can be represented
as a finite union of its irreducible components in a unique way. Additionally,
describing the irreducible components of ${\mathcal{T}_{3}}$ gives us the
rigid bilinear pairs of the variety, which are those bilinear pairs with an
open $\textrm{GL}(\mathbb{V})$-orbit. This is because a bilinear pair is rigid
in a variety if and only if the closure of its orbit is an irreducible
component of the variety. For this, we have to introduce the following notion.
Denote by $\overline{O((\mu,\mu^{\prime}))}$ the closure of the orbit of
$(\mu,\mu^{\prime})\in{\mathcal{T}_{n}}$.

######  Definition 18.

Let ${\rm T}$ and ${\rm T}^{\prime}$ be two $n$-dimensional transposed Poisson
algebras and $(\mu,\mu^{\prime}),(\lambda,\lambda^{\prime})\in\mathcal{T}_{n}$
be their representatives in the affine space, respectively. We say ${\rm T}$
degenerates to ${\rm T}^{\prime}$, and write ${\rm T}\to{\rm T}^{\prime}$, if
$(\lambda,\lambda^{\prime})\in\overline{O((\mu,\mu^{\prime}))}$. If ${\rm
T}\not\cong{\rm T}^{\prime}$, then we call it a proper degeneration.

Conversely, if
$(\lambda,\lambda^{\prime})\not\in\overline{O((\mu,\mu^{\prime}))}$ then we
call it a non-degeneration and we write ${{\rm T}}\not\to{{\rm T}}^{\prime}$.

Furthermore, for a parametric family of transposed Poisson algebras we have
the following notion.

######  Definition 19.

Let ${{\rm T}}(*)=\\{{{\rm T}}(\alpha):{\alpha\in I}\\}$ be a family of
$n$-dimensional transposed Poisson algebras and let ${{\rm T}}^{\prime}$ be
another transposed Poisson algebra. Suppose that ${{\rm T}}(\alpha)$ is
represented by the structure
$(\mu(\alpha),\mu^{\prime}(\alpha))\in{\mathcal{T}}_{n}$ for $\alpha\in I$ and
${{\rm T}}^{\prime}$ is represented by the structure
$(\lambda,\lambda^{\prime})\in{\mathcal{T}}_{n}$. We say the family ${{\rm
T}}(*)$ degenerates to ${{\rm T}}^{\prime}$, and write ${{\rm T}}(*)\to{{\rm
T}}^{\prime}$, if
$(\lambda,\lambda^{\prime})\in\overline{\\{O((\mu(\alpha),\mu^{\prime}(\alpha)))\\}_{\alpha\in
I}}$.

Conversely, if
$(\lambda,\lambda^{\prime})\not\in\overline{\\{O((\mu(\alpha),\mu^{\prime}(\alpha)))\\}_{\alpha\in
I}}$ then we call it a non-degeneration, and we write ${{\rm
T}}(*)\not\to{{\rm T}}^{\prime}$.

Observe that ${\rm T}^{\prime}$ corresponds to an irreducible component of
$\mathcal{T}_{3}$ (more precisely, $\overline{{\rm T}^{\prime}}$ is an
irreducible component) if and only if ${{\rm T}}\not\to{{\rm T}}^{\prime}$ for
any 3-dimensional transposed Poisson algebra ${\rm T}$ and ${{\rm
T}(*)}\not\to{{\rm T}}^{\prime}$ for any parametric family of 3-dimensional
transposed Poisson algebras ${\rm T}(*)$. To prove this, we will use the next
ideas.

Firstly, since
$\mathrm{dim}\,O((\mu,\mu^{\prime}))=n^{2}-\mathrm{dim}\,\mathfrak{Der}({\rm
T})$, then if ${\rm T}\to{\rm T}^{\prime}$ and ${\rm T}\not\cong{\rm
T}^{\prime}$, we have that $\mathrm{dim}\,\mathfrak{Der}({\rm
T})<\mathrm{dim}\,\mathfrak{Der}({\rm T}^{\prime})$, where
$\mathfrak{Der}({\rm T})$ denotes the Lie algebra of derivations of ${\rm T}$.

Secondly, let ${\rm T}$ and ${\rm T}^{\prime}$ be two transposed Poisson
algebras represented by the structures $(\mu,\mu^{\prime})$ and
$(\lambda,\lambda^{\prime})$ from $\mathcal{T}_{n}$, respectively. If there
exist a parametrized change of basis
$g:\mathbb{C}^{*}\to\textrm{GL}({\mathbb{V}})$ such that:

$\lim\limits_{t\to 0}g(t)*(\mu,\mu^{\prime})=(\lambda,\lambda^{\prime}),$

then ${\rm T}\to{\rm T}^{\prime}$. To prove primary degenerations, we will
provide the map $g$.

Thirdly, to prove non-degenerations we may use a remark that follows from this
lemma (see [1]).

######  Lemma 20.

Consider two transposed Poisson algebras ${\rm T}$ and ${\rm T}^{\prime}$.
Suppose ${\rm T}\to{\rm T}^{\prime}$. Let C be a Zariski closed in
${\mathcal{T}}_{n}$ that is stable by the action of the invertible upper
(lower) triangular matrices. Then if there is a representation
$(\mu,\mu^{\prime})$ of ${\rm T}$ in C, then there is a representation
$(\lambda,\lambda^{\prime})$ of ${\rm T}^{\prime}$ in C.

######  Remark 21.

Moreover, let ${{\rm T}}$ and ${{\rm T}}^{\prime}$ be two transposed Poisson
algebras represented by the structures $(\mu,\mu^{\prime})$ and
$(\lambda,\lambda^{\prime})$ from ${{\mathcal{T}}}_{n}$. Suppose ${\rm
T}\to{\rm T}^{\prime}$. Then if $\mu,\mu^{\prime},\lambda,\lambda^{\prime}$
represent algebras ${\rm T}_{0},{\rm T}_{1},{\rm T}^{\prime}_{0},{\rm
T}^{\prime}_{1}$ in the affine space $\mathbb{C}^{n^{3}}$ of algebras with a
single multiplication, respectively, we have ${\rm T}_{0}\to{\rm
T}^{\prime}_{0}$ and ${\rm T}_{1}\to{\rm T}^{\prime}_{1}$. So for example,
$(0,\mu)$ can not degenerate in $(\lambda,0)$ unless $\lambda=0$.

Fourthly, to prove ${\rm T}(*)\to{\rm T}^{\prime}$, suppose that ${\rm
T}(\alpha)$ is represented by the structure
$(\mu(\alpha),\mu^{\prime}(\alpha))\in\mathcal{T}_{n}$ for $\alpha\in I$ and
${\rm T}^{\prime}$ is represented by the structure
$(\lambda,\lambda^{\prime})\in\mathcal{T}_{n}$. If there exists a pair of maps
$(f,g)$, where $f:\mathbb{C}^{*}\to I$ and
$g:\mathbb{C}^{*}\to\textrm{GL}({\mathbb{V}})$ are such that:

$\lim\limits_{t\to
0}g(t)*(\mu\big{(}f(t)\big{)},\mu^{\prime}\big{(}f(t)\big{)})=(\lambda,\lambda^{\prime}),$

then ${\rm T}(*)\to{\rm T}^{\prime}$.

Lastly, to prove ${{\rm T}}(*)\not\to{{\rm T}}^{\prime}$, we may use an
analogous of Remark 21 for parametric families that follows from Lemma 22.

######  Lemma 22.

Consider the family of transposed Poisson algebras ${\rm T}(*)$ and the
transposed Poisson algebra ${\rm T}^{\prime}$. Suppose ${\rm T}(*)\to{\rm
T}^{\prime}$. Let C be a Zariski closed in $\mathcal{T}_{n}$ that is stable by
the action of the invertible upper (lower) triangular matrices. Then if there
is a representation $(\mu(\alpha),\mu^{\prime}(\alpha))$ of ${\rm T}(\alpha)$
in C for every $\alpha\in I$, then there is a representation
$(\lambda,\lambda^{\prime})$ of ${\rm T}^{\prime}$ in C.

The following result by [22] will be used.

######  Theorem 23.

The variety of $3$-dimensional commutative associative algebras has a single
irreducible component corresponding to the algebra $({\rm T}_{20},\cdot)$.
Moreover, $\textrm{dim}(\overline{({\rm T}_{20},\cdot)})=9$.

By this theorem, to find the irreducible components of the variety of
$3$-dimensional transposed Poisson algebras we only have to study the
degenerations and non-degenerations between the algebras ${\rm T}_{i}$, with
$i=01,\ldots,20$. The geometric classification of the variety
$\mathcal{T}_{3}$ is given in Theorem Theorem C.

######  Theorem C.

The variety of $3$-dimensional transposed Poisson algebras has five
irreducible components corresponding to the rigid algebras ${\rm T}_{01}$ and
${\rm T}_{20}$ and the parametric families ${\rm T}_{09}^{\alpha,\beta}$,
${\rm T}_{12}^{\beta}$ and ${\rm T}_{17}^{\beta}$.

###### Proof.

Our strategy to proof the result consists in first showing that every
transposed Poisson can be obtained through a degeneration of one algebra from
one of the five irreducible components proposed. These degenerations follow
from the table below. Then from the orbit dimensions, we just miss the non-
degenerations between ${\rm T}\in\left\\{{\rm T}_{09}^{\alpha,\beta},{\rm
T}_{12}^{\beta},{\rm T}_{17}^{\beta},{\rm T}_{20}\right\\}$ and ${\rm
T}_{01}$. Since $\mathfrak{sl}_{2}$ is an irreducible component of the variety
of $3$-dimensional Lie algebras and ${\rm T}_{01}$ is the only algebra such
that $({\rm T}_{01},[-,-])\cong\mathfrak{sl}_{2}$, then ${\rm T}_{01}$ is an
irreducible component.

∎

Degeneration | Parametrized basis
---|---
${\rm T}_{05}$ | $\to$ | ${\rm T}_{02}$ | $g_{1}(t)=t^{-4}e_{1},$ | $g_{2}(t)=-t^{-3}e_{1}+t^{-2}e_{2},$ | $g_{3}(t)=t^{-6}e_{3}.$
${\rm T}_{04}^{\alpha}$ | $\to$ | ${\rm T}_{03}^{\alpha}$ | $g_{1}(t)=e_{1},$ | $g_{2}(t)=t^{-1}e_{2},$ | $g_{3}(t)=t^{-1}e_{3}.$
${\rm T}_{05}$ | $\to$ | ${\rm T}_{04}^{\beta\neq 0}$ | $g_{1}(t)=t^{-2}e_{1},$ | $g_{2}(t)=-\beta t^{-2}e_{1}+\sqrt{\beta}t^{-1}e_{2}+\sqrt{\beta^{5}}t^{-3}e_{3},$ | $g_{3}(t)=\sqrt{\beta}t^{-3}e_{3}.$
${\rm T}_{05}$ | $\to$ | ${\rm T}_{04}^{0}$ | $g_{1}(t)=t^{-4}e_{1},$ | $g_{2}(t)=-t^{-2}e_{1}+t^{-1}e_{2},$ | $g_{3}(t)=t^{-5}e_{3}.$
${\rm T}_{17}^{\frac{1}{t}}$ | $\to$ | ${\rm T}_{05}$ | $g_{1}(t)=t^{-1}e_{1},$ | $g_{2}(t)=\frac{1}{2}e_{2}+t^{-2}e_{3},$ | $g_{3}(t)=t^{-1}e_{2}.$
${\rm T}_{05}$ | $\to$ | ${\rm T}_{06}$ | $g_{1}(t)=t^{-1}e_{1},$ | $g_{2}(t)=e_{2},$ | $g_{3}(t)=t^{-1}e_{3}.$
${\rm T}_{09}^{1,\beta}$ | $\to$ | ${\rm T}_{07}^{\beta}$ | $g_{1}(t)=e_{1},$ | $g_{2}(t)=te_{2},$ | $g_{3}(t)=e_{3}.$
${\rm T}_{10}^{1}$ | $\to$ | ${\rm T}_{08}$ | $g_{1}(t)=2t^{-2}e_{1},$ | $g_{2}(t)=e_{1}+te_{2},$ | $g_{3}(t)=e_{3}.$
${\rm T}_{11}^{\alpha}$ | $\to$ | ${\rm T}_{10}^{\alpha}$ | $g_{1}(t)=te_{1}+e_{2},$ | $g_{2}(t)=(\alpha+t-1)e_{2},$ | $g_{3}(t)=e_{3}.$
${\rm T}_{09}^{t}$ | $\to$ | ${\rm T}_{11}^{\alpha}$ | $g_{1}(t)=e_{1},$ | $g_{2}(t)=e_{2},$ | $g_{3}(t)=-t^{-1}e_{1}+e_{3}.$
${\rm T}_{12}^{t}$ | $\to$ | ${\rm T}_{13}$ | $g_{1}(t)=e_{1},$ | $g_{2}(t)=e_{2},$ | $g_{3}(t)=-t^{-1}e_{2}+e_{3}.$
${\rm T}_{12}^{t}$ | $\to$ | ${\rm T}_{14}$ | $g_{1}(t)=t^{-1}e_{1},$ | $g_{2}(t)=t^{-1}e_{2},$ | $g_{3}(t)=-t^{-1}+t^{-2}e_{2}+e_{3}.$
${\rm T}_{14}$ | $\to$ | ${\rm T}_{15}$ | $g_{1}(t)=te_{1},$ | $g_{2}(t)=te_{2},$ | $g_{3}(t)=e_{3}.$
${\rm T}_{11}^{\alpha}$ | $\to$ | ${\rm T}_{16}$ | $g_{1}(t)=(t-\alpha+1)e_{1}+e_{2},$ | $g_{2}(t)=te_{2},$ | $g_{3}(t)=e_{3}.$
${\rm T}_{18}$ | $\to$ | ${\rm T}_{17}^{0}$ | $g_{1}(t)=te_{1}+(t-1)e_{2},$ | $g_{2}(t)=e_{2},$ | $g_{3}(t)=e_{3}.$
${\rm T}_{17}^{t}$ | $\to$ | ${\rm T}_{18}$ | $g_{1}(t)=e_{1},$ | $g_{2}(t)=e_{2},$ | $g_{3}(t)=-t^{-1}-t^{-1}e_{2}+e_{3}.$
${\rm T}_{17}^{\gamma}$ | $\to$ | ${\rm T}_{19}^{\gamma}$ | $g_{1}(t)=t^{-1}e_{1},$ | $g_{2}(t)=t^{-1}e_{2},$ | $g_{3}(t)=\gamma t^{-1}e_{2}+e_{3}.$

## References

  * [1] Abdelwahab H., Fernández Ouaridi A., Martín González C., Degenerations of Poisson algebras, Journal of Algebra and Its Applications, 2023, DOI:10.1142/S0219498825500872.
  * [2] Alvarez M.A., Kaygorodov I., The algebraic and geometric classification of nilpotent weakly associative and symmetric Leibniz algebras, Journal of Algebra, 588 (2021), 278–314.
  * [3] Alvarez M., Fehlberg Júnior R., Kaygorodov I., The algebraic and geometric classification of Zinbiel algebras, Journal of Pure and Applied Algebra, 226 (2022), 11, 107106.
  * [4] Bai C., Bai R., Guo L., Wu Y., Transposed Poisson algebras, Novikov-Poisson algebras, and 3-Lie algebras, Journal of Algebra, 632 (2023), 535–566.
  * [5] Beites P. D., Ferreira B. L. M., Kaygorodov I., Transposed Poisson structures, arXiv:2207.00281
  * [6] Bell J., Launois S., Sánchez O., Moosa R., Poisson algebras via model theory and differential-algebraic geometry, Journal of the European Mathematical Society (JEMS), 19 (2017), 7, 2019–2049.
  * [7] Cabrera Casado Yo., Siles Molina M., Velasco M., Classification of three-dimensional evolution algebras, Linear Algebra and Its Applications, 524 (2017), 68–108.
  * [8] Camacho L., Kaygorodov I., Lopatkin V., Salim M., The variety of dual Mock-Lie algebras, Communications in Mathematics, 28 (2020), 2, 161–178.
  * [9] Cibils C., $2$-nilpotent and rigid finite-dimensional algebras, Journal of the London Mathematical Society (2), 36 (1987), 2, 211–218.
  * [10] Chouhy S., On geometric degenerations and Gerstenhaber formal deformations, Bulletin of the London Mathematical Society, 51 (2019), 5, 787–797.
  * [11] Darpö E., Rochdi A., Classification of the four-dimensional power-commutative real division algebras, Proceedings of the Royal Society of Edinburgh, Section A, 141 (2011), 6, 1207–1223.
  * [12] Dieterich E., Öhman J., On the classification of $4$-dimensional quadratic division algebras over square-ordered fields, Journal of the London Mathematical Society (2), 65 (2002), 2, 285–302.
  * [13] Dotsenko V., Algebraic structures of $F$-manifolds via pre-Lie algebras, Annali di Matematica Pura ed Applicata, 198 (2019), 2, 517–527.
  * [14] Fehlberg Júnior R., Kaygorodov I., On the Kantor product, II, Carpathian Mathematical Publications, 14 (2022), 2, 543–563.
  * [15] Fehlberg Júnior R., Kaygorodov I., Kuster C., The algebraic and geometric classification of antiassociative algebras, Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales. Serie A. Matemáticas, 116 (2022), 2, 78.
  * [16] Fernández Ouaridi A., Kaygorodov I., Khrypchenko M., Volkov Yu., Degenerations of nilpotent algebras, Journal of Pure and Applied Algebra, 226 (2022), 3, 106850.
  * [17] Ferreira B. L. M., Kaygorodov I., Lopatkin V., $\frac{1}{2}$-derivations of Lie algebras and transposed Poisson algebras, Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales. Serie A. Matemáticas, 115 (2021), 3, 142.
  * [18] Filippov V., $\delta$-Derivations of Lie algebras, Siberian Mathematical Journal, 39 (1998), 6, 1218–1230.
  * [19] Gabriel P., Finite representation type is open, Proceedings of the International Conference on Representations of Algebras (Carleton Univ., Ottawa, Ont., 1974), pp. 132–155.
  * [20] Gerstenhaber M., On the deformation of rings and algebras, Annals of Mathematics (2), 79 (1964), 59–103.
  * [21] Gorbatsevich V., Anticommutative finite-dimensional algebras of the first three levels of complexity, St. Petersburg Mathematical Journal, 5 (1994), 3, 505–521.
  * [22] Gorshkov I., Kaygorodov I., Popov Yu., Degenerations of Jordan algebras and ”Marginal” algebras, Algebra Colloquium, 28 (2021), 2, 281–294.
  * [23] Grunewald F., O’Halloran J., Varieties of nilpotent Lie algebras of dimension less than six, Journal of Algebra, 112 (1988), 2, 315–325.
  * [24] Jumaniyozov D., Kaygorodov I., Khudoyberdiyev A., The algebraic classification of nilpotent commutative algebras, Electronic Research Archive, 29 (2021), 6, 3909–3993.
  * [25] Ignatyev M., Kaygorodov I., Popov Yu., The geometric classification of $2$-step nilpotent algebras and applications, Revista Matemática Complutense, 35 (2022), 3, 907–922.
  * [26] Ismailov N., Kaygorodov I., Volkov Yu., Degenerations of Leibniz and anticommutative algebras, Canadian Mathematical Bulletin, 62 (2019), 3, 539–549.
  * [27] Kobayashi Yu., Shirayanagi K., Takahasi S.-Ei., Tsukada M., Classification of three-dimensional zeropotent algebras over an algebraically closed field, Communications in Algebra, 45 (2017), 12, 5037–5052.
  * [28] Kolesnikov P., Sartayev B., On the special identities of Gelfand-Dorfman algebras, Experimental Mathematics, 2022, DOI: 10.1080/10586458.2022.2041134.
  * [29] Laraiedh I., Silvestrov S., Transposed ${\rm Hom}$-Poisson and ${\rm Hom}$-pre-Lie Poisson algebras and bialgebras, arXiv:2106.03277
  * [30] Ma T., Li B., Transposed ${\rm BiHom}$-Poisson algebras, Communications in Algebra, 51 (2023), 2, 528–551.
  * [31] Petersson H., The classification of two-dimensional nonassociative algebras, Results in Mathematics, 37 (2000), 1–2, 120–154.
  * [32] Shafarevich I., Deformations of commutative algebras of class $2,$ Leningrad Mathematical Journal, 2 (1991), 6, 1335–1351.
  * [33] Shirshov A., Selected works of A. I. Shirshov, Contemporary Mathematicians. Birkhäuser Verlag, Basel, 2009. viii+242 pp.
  * [34] Sverchkov S., A quasivariety of special Jordan algebras, Algebra and Logic, 22 (1983), 5, 563–573.
  * [35] Van den Bergh M., Double Poisson algebras, Transactions of the American Mathematical Society, 360 (2008), 11, 5711–5769.
  * [36] Volkov Yu., $n$-ary algebras of the first level, Mediterranean Journal of Mathematics, 19 (2022), 1, 2.
  * [37] Voronin V., Special and exceptional Jordan dialgebras, Journal of Algebra and its Applications, 11 (2012), 2, 1250029, 23 pp.
  * [38] Yao Y., Ye Y., Zhang P., Quiver Poisson algebras, Journal of Algebra, 312 (2007), 2, 570–589.
  * [39] Yuan L., Hua Q., $\frac{1}{2}$-(bi)derivations and transposed Poisson algebra structures on Lie algebras, Linear and Multilinear Algebra, 70 (2022), 22, 7672–7701.
  * [40] Zusmanovich P., On $\delta$-derivations of Lie algebras and superalgebras, Journal of Algebra, 324 (2010), 12, 3470–3486.