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\noindent{\Large
Transposed Poisson structures on quasi-filiform Lie algebras of \\
maximum length}


 \bigskip



\begin{center}

 {\bf
Jobir Adashev \footnote{Institute of Mathematics Academy of Sciences of Uzbekistan, Tashkent, Uzbekistan; \ adashevjq@mail.ru},
Azamat Saydaliyev \footnote{Institute of Mathematics Academy of Sciences of Uzbekistan, Tashkent, Uzbekistan; \ azamatsaydaliyev6@gmail.com},
Feruza Toshtemirova \footnote{Chirchik State Pedagogical University, Chirchik, Uzbekistan; \ feruzaisrailova45@gmail.com}


}

\end{center}


\

\medskip




\noindent{\bf Abstract}:
{\it In this work, we explore the $\frac{1}{2}$-derivations of quasi-filiform Lie algebras of maximum length, focusing on dimensions 7 and higher. Using these derivations, we then construct non-trivial transposed Poisson algebras. Additionally, we develop a commutative associative multiplication method to build a transposed Poisson algebra associated with each Lie algebra under consideration.}

\medskip

\bigskip

\noindent {\bf Keywords}:
{\it   Lie algebra, transposed Poisson algebra, $\frac12$-derivation.}

\bigskip
\noindent {\bf MSC2020}:  17A30, 17B40, 17B61, 17B63.


 \bigskip

\section*{Introduction}

Bai et.al.~\cite{Bai} introduced a dual notion of the Poisson algebra, which is called a \textit{transposed Poisson algebra}, by changing the roles of the two multiplications in the Leibniz rule.
Poisson algebra is introduced to commutative associative algebras by their derivation. Similarly, on a Lie algebra, the concept of a transposed Poisson algebra is defined by the $\frac{1}{2}$-derivation. This way of defining not only shares some properties of a Poisson algebra, such as the closedness under tensor products and the Koszul self-duality as an operation, but also admits a rich class of identities \cite{kms,Bai,fer23,bfk22,lb23,bl23,conj}.

The description of all Poisson algebras with a fixed Lie or associative part~\cite{YYZ07,jawo,kk21} is one of the natural tasks in the theory.
In this paper, we present transposed Poisson algebras that demonstrate quasi-filiform Lie algebras of maximum length.
Abdurasulov et.al.~\cite{Kobiljon}, obtained the description of all transposed Poisson algebra structures on solvable Lie algebras with filiform nilradical.

 It is worth noting that any unital transposed Poisson algebra is a particular case of a ``contact bracket'' algebra	and a quasi-Poisson algebra \cite{bfk22}. Every transposed Poisson algebra is a commutative  Gelfand-Dorfman algebra \cite{kms}, and is also an algebra of Jordan brackets \cite{fer23}. Ferreira et. al. \cite{FKL} established a relation between transposed Poisson algebras and $\frac{1}{2}$-derivations of Lie algebras. These ideas played an active role in describing all transposed Poisson structures on  Witt and Virasoro algebras in \cite{FKL}.  Twisted Heisenberg-Virasoro,   Schr\"odinger-Virasoro and extended Schr\"odinger-Virasoro algebras were studied in \cite{yh21}. The fact that the original deformative Schrödinger-Virasoro algebras have nontrivial $\frac{1}{2}$-derivations, indicating that they possess nontrivial transposed Poisson structures is shown in \cite{SZ}. In \cite{klv221} Transposed Poisson structures on deformative Schrödinger-Witt algebras, non-finitely graded Witt algebras, and non-finitely graded Heisenberg-Witt algebras are classified. Moreover, in \cite{klv222} authors computed $\frac{1}{2}$-derivations on the deformed generalized Heisenberg-Virasoro algebras and on not-finitely graded Heisenberg-Virasoro algebras, and classified all transposed Poisson structures on such algebras. $\delta$-derivations of simple Jordan algebras with values in irreducible bimodules studied in \cite{zz} Meanwhile, Schr\"odinger algebra in $(n+1)$-dimensional space-time was discussed in \cite{ytk}. Then, in \cite{kk23} Witt type Lie algebras were discussed and \cite{kkg23} studied generalized Witt algebras. The papers \cite{kk22,kkg23} studied Block Lie algebras, while \cite{KK7} examined Lie algebras of upper triangular matrices and  a transposed Poisson structures on Lie algebra of upper triangular matrices. Additionally, \cite{kkinc} was devoted to the study of Lie incidence algebras. In \cite{MF} considered a new family of modified double
Poisson brackets and mixed double Poisson algebras. It was proved in \cite{klv22} that any complex finite-dimensional solvable Lie algebra admits a non-trivial transposed Poisson structure. In \cite{bfk23} authors gave the algebraic and geometric classification of three-dimensional transposed Poisson algebras. The list of actual open questions on transposed Poisson algebras can be seen in \cite{bfk22}. Recently, all transposed Poisson algebra structures on oscillator Lie algebras, i.e., on one-dimensional solvable extensions of  $(2n+1)$-dimensional Heisenberg algebra; on solvable Lie algebras with naturally graded filiform nilpotent radical; on $(n+1)$-dimensional solvable extensions of the $(2n + 1)$-dimensional Heisenberg algebra; and on $n$-dimensional solvable extensions of $n$-dimensional algebra with the trivial multiplication were described in  \cite{KKh}. This paper also gave an example of a finite-dimensional Lie algebra with non-trivial $\frac{1}{2}$-derivations without non-trivial transposed Poisson algebra structures. In \cite{bfk23}, the algebraic and geometric classification of all complex $3$-dimensional transposed Poisson algebras was obtained, whereas \cite{ch24} focused on the algebraic classification of all complex $3$-dimensional transposed Poisson $3$-Lie algebras.

The purpose of this article is to find all transposed Poisson algebras that demonstrate quasi-filiform Lie algebras of maximum length. Building on our previous work \cite{Kobiljon1}, where we addressed the case of quasi-filiform Lie algebras in dimensions 5 and 6, this paper extends the results to dimensions 7 and higher, thereby providing a general solution to the problem. In order to achieve our goal, we have organized this paper as follows: in Section 2, we describe $\frac{1}{2}$-derivations of quasi-filiform Lie algebras of maximum length and Section 3, describes all non-trivial transposed Poisson algebras with quasi-filiform Lie algebras of maximum length. Next, by using descriptions of $\frac{1}{2}$-derivations of Lie algebras, we establish commutative associative multiplication to construct a transposed Poisson algebra associated with given Lie algebra.


\bigskip

\section{PRELIMINARIES}

This section discusses the concepts and known results on algebras over the field $\mathbb{C}$ unless the otherwise is stated.

A Poisson algebra consists of a vector space \( L \) equipped with two bilinear operations, denoted \( \cdot \) and \( [\cdot, \cdot] \), which map \( L \times L \) to \( L \). In this structure, \( (L, \cdot) \) is a commutative associative algebra, while \( (L, [\cdot, \cdot]) \) forms a Lie algebra. The operations must satisfy a compatibility condition known as the Leibniz rule, expressed as

\[
[x, y \cdot z] = [x, y] \cdot z + y \cdot [x, z],
\]

where the adjoint operators of the Lie algebra act as derivations on the commutative associative algebra.

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.
Below we give the definition
of transposed Poisson algebra, which was given in the paper by Bai, Bai, Guo, and Wu \cite{Bai}.


\begin{definition}\label{TPA}
Let $\mathfrak{L}$ be a vector space equipped with two bilinear operations
$$
\cdot,\; [-,-] :\mathfrak{L}\otimes \mathfrak{L}\to \mathfrak{L}.$$
The triple $(\mathfrak{L},\cdot,[-,-])$ is called a \textbf{transposed Poisson algebra} if $(\mathfrak{L},\cdot)$ is a commutative associative algebra and $(\mathfrak{L},[-,-])$ is a Lie algebra which satisfies the following compatibility condition
\begin{equation}
2z\cdot [x,y]=[z\cdot x,y]+[x,z\cdot y].\label{eq:dualp}
\end{equation}
\end{definition}

Eq.~\eqref{eq:dualp} is called {\bf transposed
Leibniz rule} because the roles of 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.


%
%We show that the
%compatibility relations of the Poisson algebra and those of the Transposed Poisson algebra are independent in the following sense.
%
%\begin{proposition} \cite{Bai} Let $(\mathfrak{L},\cdot)$ be a commutative associative algebra and $(\mathfrak{L},[-,-])$ be a Lie algebra. Then $(\mathfrak{L},\cdot,[-,-])$ is both a
%Poisson algebra and a Transposed Poisson algebra if and only
%if
%\begin{equation}
%x\cdot [y,z]=[x\cdot y,z]=0.
%\label{eq:inter0}
%\end{equation}
%\end{proposition}

\begin{definition}\label{halfderiv}
  Let $(\mathfrak{L}, [-,-])$ be an algebra with a multiplication $[-,-],$ and $\varphi$ be
a bilinear map. Then $\varphi$ is a $\frac12$-\textbf{derivation} if it satisfies:
\begin{equation}\label{halfderiv1}
\varphi([x, y]) = \frac12([\varphi(x), y] + [x, \varphi(y)]).
\end{equation}
\end{definition}

Note that $\frac{1}{2}$-derivations are a particular case of $\delta$-derivations introduced by Filippov in \cite{fil1}
(see, \cite{k07,k10} and references therein).  It is easy to see from Definition \ref{halfderiv} that $[\mathfrak{L},\mathfrak{L}]$ and $Ann(\mathfrak{L})$ are invariant under any $\frac 12$-derivation of $\mathfrak{L}$. Definitions \ref{TPA} and \ref{halfderiv} immediately imply the following lemma.

\begin{lemma}\label{lemma1}
  Let $(\mathfrak{L}, [-,-])$ be a Lie algebra and $\cdot$ a new binary (bilinear) operation on $\mathfrak{L}$. Then
$(\mathfrak{L}, \cdot, [-,-])$ is a transposed Poisson algebra if and only if $\cdot$ is commutative and associative, and for
every $z \in \mathfrak{L}$ the multiplication by $z$ in $(\mathfrak{L}, \cdot)$ is a $\frac12$-derivation of $(\mathfrak{L}, [-,-])$.
\end{lemma}


%For an arbitrary Lie algebra $L$ we define the \textbf{derived} and \textbf{central series} as follows:
%$$\mathfrak{L}^{[1]}=\mathfrak{L}, \ \mathfrak{L}^{[s+1]}=[\mathfrak{L}^{[s]},\mathfrak{L}^{[s]}], \ s\geq 1,$$
%$$\mathfrak{L}^{1}=\mathfrak{L}, \ \mathfrak{L}^{k+1}=[\mathfrak{L}^{k},\mathfrak{L}], \ k\geq 1.$$
%
%\begin{definition} An $n$-dimensional Lie algebra $\mathfrak{L}$ is
%called \textbf{solvable (nilpotent)} if there exists $s\in {\mathbb{N}}$ ($k\in \mathbb{N}$) such that $\mathfrak{L}^{[s]}=\{0\}$ ($\mathfrak{L}^{k}=\{0\}$).
%Such minimal number is called \textbf{index of solvability} (\textbf{nilpotency}).
%\end{definition}
%
%The maximal nilpotent ideal of a Lie algebra is said to be the nilradical of the algebra.


Lower central series for a given Lie algebra $\mathfrak{L}$ is  defined as follows:
\[\mathfrak{L}^1=\mathfrak{L}, \ \mathfrak{L}^{k+1}=[\mathfrak{L}^k,\mathfrak{L}],  \ k \geq 1.\]




\begin{definition}
An $n$-dimensional Lie algebra is called \textbf{quasi-filiform} if $\mathfrak{L}^{n-2}\neq \{0\}$ and $\mathfrak{L}^{n-1}=\{0\}$.
\end{definition}



A Lie algebra $\mathfrak{L}$ is $\mathbb{Z}$-graded, if $\mathfrak{L} =\bigoplus_{i\in \mathbb{Z}}\mathbf{V}_i$, where $[V_i, V_j]\subseteq V_{i+j}$ for any $i, j \in \mathbb{Z}$ with a finite number of non-null spaces $V_i$. We say that a nilpotent Lie algebra $\mathfrak{L}$ admits \textbf{the connected gradation} $\mathfrak{L} = V_{k_1}\oplus\dots\oplus V_{k_t}$, if $V_{k_i}\neq\{0\}$ for any $i \  (1\leq i\leq t)$.

\begin{definition}
The number $l(\oplus \mathfrak{L})=l(V_{k_1}\oplus\dots\oplus V_{k_t})=k_t-k_1 + 1$ is called \textbf{the
length of gradation}. A gradation is said to be \textbf{of maximum length}, if $l(\oplus \mathfrak{L}) = \dim(\mathfrak{L})$.

\end{definition}

We denote $l(\mathfrak{L}) = \max\{l(\oplus L)\ \text{such that}\  L = V_{k_1}\oplus\dots\oplus V_{k_t}\  \text{is a connected gradation}\}$ of \textbf{the length of an algebra} $\mathfrak{L}$.

\begin{definition}
 A Lie algebra $\mathfrak{L}$ is called of \textbf{maximum length} if $l(\mathfrak{L}) = \dim(\mathfrak{L})$.
\end{definition}

Since we covered the dimensions $5$ and $6$ in our previous work \cite{Kobiljon1}, we will consider $n \geq 7$ in this work.
In the following theorem we present the classification of quasi-filiform Lie algebras of maximum length given in \cite{Gomez}.



\begin{theorem} \label{thm26}
 Let $L$ be an $n$-dimensional ($n \geq 7$) quasi-filiform Lie algebra
of maximum length. Then the algebra $L$ is isomorphic to one of the following pairwise non-isomorphic algebras:

\[g^1_{(n,1)}:\begin{cases}
[e_1,e_i]=e_{i+1},& 2\leq i\leq n-2,\\[1mm]
[e_i,e_{n-i}]=(-1)^ie_{n},& 2\ \leq i\leq  \frac{n-1}{2}, \ n \  is \  odd;\\[1mm]
\end{cases}\]

\[g^2_{(n,1)}:\begin{cases}
[e_1,e_i]=e_{i+1},& 2\leq i\leq n-2,\\[1mm]
[e_i,e_n]=e_{i+2},& 2\leq i\leq n-3; \\[1mm]
\end{cases} \quad g^3_{(n,1)}:\begin{cases}
[e_1,e_i]=e_{i+1},& 2\leq i\leq {n-2},\\[1mm]
[e_i,e_{n}]=e_{i+2},& 2\leq i\leq {n-3}\\[1mm]
[e_2,e_i]=e_{i+3}, & 3\leq i \leq {n-4};
\end{cases}\]
\[g^1_{7}:\begin{cases}
[e_1,e_i]=e_{i+1},& 2\leq i\leq 5,\\[1mm]
[e_2,e_i]=e_{i+2},& 3\leq i\leq 4,\\[1mm]
[e_i,e_{7-i}]=(-1)^ie_{7},& 2\leq i\leq 3;\\[1mm]
\end{cases} \quad g^2_{9}:\begin{cases}
[e_1,e_i]=e_{i+1},& 2\leq i\leq 7,\\[1mm]
[e_2,e_i]=e_{i+2},& 3\leq i\leq 4 ,\\[1mm]
[e_2,e_5]=3e_7,\\[1mm]
[e_2,e_6]=5e_8,\\[1mm]
[e_3,e_i]=-2e_{i+3},&4\leq i\leq 5,\\[1mm]
[e_i,e_{9-i}]=(-1)^ie_{9},&2\leq i \leq 4;
\end{cases}\]
\[g^3_{11}:\begin{cases}
[e_1,e_i]=e_{i+1},& 2\leq i\leq 9,\\[1mm]
[e_2,e_i]=e_{i+2},& 3\leq i\leq 4 ,\\[1mm]
[e_2,e_i]=-e_{i+2},& 6\leq i\leq 7\\[1mm]
[e_3,e_7]=-e_{10},\\[1mm]
[e_3,e_i]=e_{i+3},&4\leq i\leq 5,\\[1mm]
[e_4,e_i]=e_{i+4},&5\leq i\leq 6,\\[1mm]
[e_i,e_{11-i}]=(-1)^ie_{11},&2\leq i \leq 5,
\end{cases}\]
where $\{e_1,e_2,\dots,e_{n}\}$ is a basis of the algebra.
\end{theorem}





\section{$\frac12$-derivation of  quasi-filiform Lie algebras of maximum length}


In this section, we will derive and calculate $\frac12$-derivation of  quasi-filiform Lie algebras of maximum length with dimension $n\geq 7$.
We start with the following theorem.

\begin{theorem}\label{halfderiv1} Any $\frac 1 2$-derivations of the algebra $g^1_{(n,1)}$
has the form


$$\begin{cases}
\varphi(e_1)=\sum\limits_{i=1}^{n}\alpha_ie_i, \\ \varphi(e_2)=\alpha_1e_2+\beta_{n-2}e_{n-2}+\beta_{n-1}e_{n-1}+\beta_ne_n, \\  \varphi(e_3)=\alpha_1e_3+\frac{1}{2}\beta_{n-2}e_{n-1}-\frac{1}{2}\alpha_{n-2}e_{n}, \\
\varphi(e_i)=\alpha_1e_i+\frac{(-1)^i}{2}\alpha_{n-i+1}e_n, \  4\leq i\leq n-1,\\
\varphi(e_n)=\alpha_1e_n.\end{cases}$$
\end{theorem}

\begin{proof}
It is easy to see that $g^1_{(n,1)}$ has $2$ generators. We use these generators to calculate $\frac12$-derivation.
$$\varphi(e_1)=\sum\limits_{i=1}^{n}\alpha_ie_i,\quad \varphi(e_2)=\sum\limits_{i=1}^{n}\beta_ie_i.$$

Now consider the condition of $\frac 1 2$-derivation for the elements $e_1$ and $e_2:$
\begin{center}
$\varphi(e_3)=\varphi([e_1,e_2])=\frac{1}{2}([\varphi(e_1),e_2]+[e_1,\varphi(e_2)])$
\end{center}
\begin{center}
$=\frac{1}{2}([\sum\limits_{i=1}^{n}\alpha_ie_i,e_2]+[e_1,\sum\limits_{i=1}^{n}\beta_ie_i])
=\frac{1}{2}((\alpha_1+\beta_2)e_3+\sum\limits_{i=4}^{n-1}\beta_{i-1}e_i-\alpha_{n-2}e_n).$
\end{center}

We prove the following equality for $3\leq i\leq n-1$ by induction:
\begin{center}
$\varphi(e_i)=\frac{1}{2^{i-2}}\Big((2^{i-2}-1)\alpha_{1}+\beta_2\Big)e_{i}+\frac{1}{2^{i-2}}\sum\limits_{t=i+1}^{n-1}\beta_{t-i+2}e_t+\frac{(-1)^{i}}{2}\alpha_{n-i+1}e_n.$
\end{center}
If $i=3$, the relationship holds according to the above equality. Now, we prove that it is true for $i$ and $i+1$.
By considering the condition of $\frac 1 2$-derivation for the elements $e_1, e_i$ we have
\begin{center}
$\varphi(e_{i+1})=\varphi([e_1,e_i])=\frac{1}{2}([\varphi(e_1),e_i]+[e_1,\varphi(e_i)])$
\end{center}
\begin{center}
$=\frac{1}{2}\Big([\sum\limits_{k=1}^{n}\alpha_ke_k,e_i]+[e_1,\frac{1}{2^{i-2}}\Big((2^{i-2}-1)\alpha_{1}+\beta_2\Big)e_{i}+\frac{1}{2^{i-2}}\sum\limits_{t=i+1}^{n-1}\beta_{t-i+2}e_t+\frac{(-1)^{i}}{2}\alpha_{n-i+1}e_n]\Big)$
\end{center}
\begin{center}
$=\frac{1}{2}\Big(\alpha_{1}e_{i+1}+(-1)^{i+1}\alpha_{n-i}e_n+\frac{1}{2^{i-2}}\Big((2^{i-2}-1)\alpha_{1}+\beta_2\Big)e_{i+1}+\frac{1}{2^{i-2}}\sum\limits_{t=i+1}^{n-2}\beta_{t-i+2}e_{t+1}\Big)$
\end{center}
\begin{center}
$=\frac{1}{2^{i-1}}\Big((2^{i-1}-1)\alpha_{1}+\beta_2\Big)e_{i+1}+\frac{1}{2^{i-1}}\sum\limits_{t=i+2}^{n-1}\beta_{t-i+1}e_t+\frac{(-1)^{i+1}}{2}\alpha_{n-i}e_n.$
\end{center}


Now, consider the condition of $\frac 1 2$-derivation for the elements $e_2, e_{n-2}:$
\begin{center}
$\varphi(e_{n})=\varphi([e_2,e_{n-2}])=\frac{1}{2}([\varphi(e_2),e_{n-2}]+[e_2,\varphi(e_{n-2})])$
\end{center}
\begin{center}
$=\frac{1}{2}\Big([\sum\limits_{i=1}^{n}\beta_ie_i,e_{n-2}]+[e_2,\frac{1}{2^{n-4}}\Big((2^{n-4}-1)\alpha_{1}+\beta_2\Big)e_{n-2}+\frac{1}{2^{n-4}}\beta_{3}e_{n-1}+\frac{(-1)^{n-2}}{2}\alpha_{3}e_n]\Big)
$
\end{center}
\begin{center}
$=\frac{1}{2}\Big(\beta_{1}e_{n-1}+\beta_{2}e_n+\frac{1}{2^{n-4}}\Big((2^{n-4}-1)\alpha_{1}+\beta_2\Big)e_{n}\Big)=\frac{1}{2}\beta_{1}e_{n-1}+\frac{1}{2^{n-3}}\Big((2^{n-4}-1)\alpha_{1}+(2^{n-4}+1)\beta_2\Big)e_{n}.$
\end{center}
Thus, we obtain that
\begin{equation}\label{fien1}
\varphi(e_{n})=\frac{1}{2}\beta_{1}e_{n-1}+\frac{1}{2^{n-3}}\Big((2^{n-4}-1)\alpha_{1}+(2^{n-4}+1)\beta_2\Big)e_{n}.
\end{equation}
Now, we consider the condition of $\frac 1 2$-derivation for the elements $e_2$ and $e_i$, $3\leq i\leq n-3:$
\begin{center}
$0=\varphi([e_2,e_i])=\frac{1}{2}([\varphi(e_2),e_i]+[e_2,\varphi(e_i)])$
\end{center}
\begin{center}
$=\frac{1}{2}\Big([\sum\limits_{i=1}^{n}\beta_ie_i,e_i]+[e_2,\frac{1}{2^{i-2}}\Big((2^{i-2}-1)\alpha_{1}+\beta_2\Big)e_{i}+\frac{1}{2^{i-2}}\sum\limits_{t=i+1}^{n-1}\beta_{t-i+2}e_t+\frac{(-1)^{i}}{2}\alpha_{n-i+1}e_n]\Big)$
\end{center}
\begin{center}
$=\frac{1}{2}\Big([\beta_{1}e_{i+1}-(-1)^i\beta_{n-i}e_n+\frac{1}{2^{i-2}}\beta_{n-i}e_{n}\Big).$
\end{center}
By the comparison of coefficients at the basis elements we obtain that
$$\beta_1=\beta_i=0, \ 3\leq i\leq n-3.$$

By considering the condition of $\frac 1 2$-derivation for the elements $e_3$ and $e_{n-3}$ we have
\begin{center}
$\varphi([e_3,e_{n-3}])=\frac{1}{2}([\varphi(e_3),e_{n-3}]+[e_3,\varphi(e_{n-3})])$
\end{center}
\begin{center}
$=\frac{1}{2}\Big([\frac{1}{2}((\alpha_1+\beta_2)e_3+\beta_{n-2}e_{n-1}-\alpha_{n-2}e_n),e_{n-3}]+
[e_3,\frac{1}{2^{n-5}}\Big((2^{n-5}-1)\alpha_{1}+\beta_2\Big)e_{n-3}+\frac{1}{2}\alpha_{4}e_n]\Big)$
\end{center}
\begin{center}
$=\frac{1}{2}\Big(-\frac{1}{2}((\alpha_1+\beta_{2})e_{n}-\frac{1}{2^{n-5}}\Big((2^{n-5}-1)\alpha_{1}+\beta_2\Big)e_{n}\Big).$
\end{center}
On the other hand,
\begin{equation}\label{fien2}
\varphi([e_3,e_{n-3}])=-\varphi(e_n)=-\Big(\frac{1}{2^2}((\alpha_1+\beta_{2})+\frac{1}{2^{n-4}}\Big((2^{n-5}-1)\alpha_{1}+\beta_2\Big)\Big)e_{n}.
\end{equation}
By comparing the coefficients at the basis elements in expressions (\ref{fien1}) and (\ref{fien2}) we obtain that
$\alpha_{1}=\beta_2.$

This completes the proof of the theorem. \end{proof}

Now we study the $\frac 1 2$-derivation of the algebra $g^2_{(n,1)}.$

\begin{theorem}\label{halfderiv2} Any $\frac 1 2$-derivations of the algebra $g^2_{(n,1)}$ has the form

$$\begin{cases}
\varphi(e_1)=\sum\limits_{i=1}^{n-1}\alpha_ie_i, \\
\varphi(e_2)=\alpha_1e_2+\beta_{n-2}e_{n-2}+\beta_{n-1}e_{n-1}, \\
\varphi(e_3)=\alpha_1e_3+\frac{1}{2}\beta_{n-2}e_{n-1}, \\
\varphi(e_i)=\alpha_1e_i, \  4\leq i\leq n-1,\\
\varphi(e_n)=-\sum\limits_{i=3}^{n-2}\alpha_{i-1}e_i+\gamma_{n-1}e_{n-1}+\alpha_1e_n.\end{cases}$$
\end{theorem}

\begin{proof} From the multiplication table of the algebra  $g^2_{(n,1)}$ we conclude that $e_1, e_2$ and $e_{n}$ are the generator basis elements of
the algebra. We use these generators to calculate $\frac12$-derivation:
\begin{center}
$\varphi(e_1)=\sum\limits_{i=1}^{n}\alpha_ie_i,\ \varphi(e_2)=\sum\limits_{i=1}^{n}\beta_ie_i,\ \varphi(x)=\sum\limits_{i=1}^{n}\gamma_ie_i.$
\end{center}
Now consider the condition of $\frac 1 2$-derivation for the elements $e_1$ and $e_2:$
\begin{center}
$\varphi(e_3)=\varphi([e_1,e_2])=\frac{1}{2}([\varphi(e_1),e_2]+[e_1,\varphi(e_2)])$
\end{center}
\begin{center}
$=\frac{1}{2}([\sum\limits_{i=1}^{n}\alpha_ie_i,e_2]+[e_1,\sum\limits_{i=1}^{n}\beta_ie_i])
=\frac{1}{2}\Big((\alpha_1+\beta_2)e_3+(\beta_3-\alpha_n)e_4+\sum\limits_{i=5}^{n-1}\beta_{i-1}e_i\Big).$
\end{center}


We prove the following equality for $3\leq i\leq n-2$ by induction:
\begin{center}
$\varphi(e_i)=\frac{1}{2^{i-2}}\Big((2^{i-2}-1)\alpha_{1}+\beta_2\Big)e_{i}+\frac{1}{2^{i-2}}\Big(\beta_{3}-(2^{i-2}-1)\alpha_n\Big)e_{i+1}+\frac{1}{2^{i-2}}\sum\limits_{t=i+2}^{n-1}\beta_{t-i+2}e_t$
\end{center}
and
\begin{center}
$\varphi(e_{n-1})=\frac{1}{2^{n-3}}\Big((2^{n-3}-1)\alpha_{1}+\beta_2\Big)e_{n-1}.$
\end{center}

If $i=3$, the relationship holds according to the above equality. Now, we prove that it is true for $i$ and $i+1$.
By considering the condition of $\frac 1 2$-derivation for the elements $e_1, e_i$ , we have the following:
\begin{center}
$\varphi(e_{i+1})=\varphi([e_1,e_i])=\frac{1}{2}([\varphi(e_1),e_i]+[e_1,\varphi(e_i)])$
\end{center}
\begin{center}
$=\frac{1}{2}\Big([\sum\limits_{k=1}^{n}\alpha_ke_k,e_i]+[e_1,\frac{1}{2^{i-2}}\Big((2^{i-2}-1)\alpha_{1}+\beta_2\Big)e_{i}+\frac{1}{2^{i-2}}\Big(\beta_{3}-(2^{i-2}-1)\alpha_n\Big)e_{i+1}+\frac{1}{2^{i-2}}\sum\limits_{t=i+2}^{n-1}\beta_{t-i+2}e_t\Big)$
\end{center}
\begin{center}
$=\frac{1}{2}\Big(\alpha_{1}e_{i+1}-\alpha_{n}e_{i+2}+\frac{1}{2^{i-2}}\Big((2^{i-2}-1)\alpha_{1}+\beta_2\Big)e_{i+1}+\frac{1}{2^{i-2}}\Big(\beta_{3}-(2^{i-2}-1)\alpha_n\Big)e_{i+2}$ \
$+\frac{1}{2^{i-2}}\sum\limits_{t=i+2}^{n-2}\beta_{t-i+2}e_{t+1}\Big)$
\end{center}
\begin{center}
$=\frac{1}{2^{i-1}}\Big((2^{i-1}-1)\alpha_{1}+\beta_2\Big)e_{i+1}+\frac{1}{2^{i-1}}\Big(\beta_3-(2^{i-1}-1)\alpha_{n}\Big)e_{i+2}+\frac{1}{2^{i-1}}\sum\limits_{t=i+3}^{n-1}\beta_{t-i+1}e_t.$
\end{center}

From $$0=\varphi([e_1,e_n])=\frac{1}{2}([\varphi(e_1),e_n]+[e_1,\varphi(e_n)])=\frac{1}{2}\Big([\sum\limits_{k=1}^{n}\alpha_ke_k,e_n]+[e_1,\sum\limits_{k=1}^{n}\gamma_ke_k]\Big)$$
$$=\frac{1}{2}\Big(\sum\limits_{k=2}^{n-3}\alpha_ke_{k+2}+\sum\limits_{k=2}^{n-2}\gamma_ke_{k+1}\Big)=\frac{1}{2}\Big(\sum\limits_{k=4}^{n-1}\alpha_{k-2}e_{k}+\sum\limits_{k=3}^{n-1}\gamma_{k-1}e_{k}\Big),$$ we obtain,
\begin{center}
$\gamma_2=0,\ \gamma_i=-\alpha_{i-1},\ 3\leq i\leq n-2.$
\end{center}

Now consider the condition of $\frac 1 2$-derivation for the elements $e_2, e_{n}:$
\begin{center}
$\varphi([e_2,e_{n}])=\frac{1}{2}([\varphi(e_2),e_{n}]+[e_2,\varphi(e_{n})])=\frac{1}{2}\Big([\sum\limits_{i=1}^{n}\beta_{i}e_i,e_{n}]+
[e_2,\sum\limits_{i=1}^{n}\gamma_{i}e_i]\Big)$
\end{center}
\begin{center}
$=\frac{1}{2}\Big(\sum\limits_{i=2}^{n-3}\beta_{i}e_{i+2}-\gamma_{1}e_3+\gamma_{n}e_4\Big)=\frac{1}{2}\Big(-\gamma_{1}e_3+(\beta_{2}+\gamma_{n})e_4+\sum\limits_{i=5}^{n-1}\beta_{i-2}e_{i}\Big).$
\end{center}
Thus,
\begin{center}
$\varphi([e_2,e_{n}])=\varphi(e_{4}).$
\end{center}
By Comparing the coefficients at the basis elements, we obtain that
$$\gamma_{1}=0, \ 2\gamma_n=3\alpha_1-\beta_2; \ \beta_3=-3\alpha_n, \ \beta_i=0, \ 4\leq i\leq n-3.$$


By using the property of $\frac 1 2$-derivation for the products $[e_2,e_3]=0$ and $[e_3,e_n]=e_5$, we have

\begin{center}$\beta_1=0;  \ \beta_n=0, \ \beta_2=\alpha_1; \ \beta_3=\alpha_n=0.$\end{center}



This completes the proof of the theorem. \end{proof}


Now, we will study the $\frac 1 2$-derivation of the algebra $g^3_{(n,1)}.$

\begin{theorem}\label{halfderiv3} Any $\frac 1 2$-derivations of the algebra $g^3_{(n,1)}$ has the form
\begin{itemize}
   \item for $n=7:$
  $$\begin{cases}
\varphi(e_1)=\alpha_1e_1+\alpha_2e_2+\alpha_3e_3+\alpha_4e_4+\alpha_5e_5+\alpha_6e_6, \\
\varphi(e_2)=\alpha_1e_2+4\alpha_2e_4+\beta_5e_5+\beta_6e_6,\\
\varphi(e_3)=\alpha_1e_3+2\alpha_2e_5+\frac{1}{2}(\beta_5-\alpha_3)e_6, \\
\varphi(e_4)=\alpha_1e_4+\frac{3}{2}\alpha_2e_6, \\
\varphi(e_5)=\alpha_1e_5, \ \varphi(e_6)=\alpha_1e_6, \\
\varphi(e_7)=-\alpha_2e_3-\alpha_3e_4-\alpha_4e_5+\gamma_6e_6+\alpha_1e_7, \\
  \end{cases}$$
  \item for $n\geq8:$
$$\begin{cases}
\varphi(e_1)=\alpha_1e_1+\sum\limits_{i=3}^{n-1}\alpha_ie_i, \\
\varphi(e_2)=\alpha_1e_2+\sum\limits_{i=5}^{n-3}\alpha_{i-2}e_i+\beta_{n-2}e_{n-2}+\beta_{n-1}e_{n-1}, \\
\varphi(e_3)=\alpha_1e_3+\frac{1}{2}(\beta_{n-2}-\alpha_{n-4})e_{n-1}, \\
\varphi(e_i)=\alpha_1e_i, \  4\leq i\leq n-1,\\
\varphi(e_n)=-\sum\limits_{i=4}^{n-2}\alpha_{i-1}e_i+\gamma_{n-1}e_{n-1}+\alpha_1e_n.\end{cases}$$\end{itemize}
\end{theorem}

\begin{proof}
The algebra $g^3_{(n,1)}$ has generators $e_1, e_2$ and $e_n$. Then, we have the following
\begin{center}
$\varphi(e_1)=\sum\limits_{i=1}^{n}\alpha_ie_i,\
\varphi(e_2)=\sum\limits_{i=1}^{n}\beta_ie_i,\ \varphi(e_n)=\sum\limits_{i=1}^{n}\gamma_ie_i.$
\end{center}
Now, consider the condition of $\frac 1 2$-derivation for the elements $e_2$ and $e_n$ we obtain
\begin{center}
$\varphi(e_3)=\varphi([e_1,e_2])=\frac{1}{2}([\varphi(e_1),e_2]+[e_1,\varphi(e_2)])=\frac{1}{2}([\sum\limits_{i=1}^{n}\alpha_ie_i,e_2]+[e_1,\sum\limits_{i=1}^{n}\beta_ie_i])$
\end{center}
\begin{center}
$=\frac{1}{2}\Big(\alpha_1e_3-\sum\limits_{i=3}^{n-4}\alpha_{i}e_{i+3}-\alpha_ne_4+\sum\limits_{i=2}^{n-2}\beta_{i}e_{i+1}\Big)$
\end{center}
\begin{center}
$=\frac{1}{2}\Big((\alpha_1+\beta_2)e_3+(\beta_3-\alpha_n)e_4+\beta_4e_5+\sum\limits_{i=6}^{n-1}(\beta_{i-1}-\alpha_{i-3})e_{i}\Big).$
\end{center}

We can prove the following equalities by induction:
\begin{center}
$\varphi(e_i)=\frac{1}{2^{i-2}}\Big((2^{i-2}-1)\alpha_{1}+\beta_2\Big)e_{i}+\frac{1}{2^{i-2}}\Big(\beta_{3}-(2^{i-2}-1)\alpha_n\Big)e_{i+1}$\end{center}
\begin{center}
$+\frac{1}{2^{i-2}}\Big((2^{i-2}-2)\alpha_2+\beta_{4}\Big)e_{i+2}+\frac{1}{2^{i-2}}\sum\limits_{t=i+3}^{n-1}(\beta_{t-i+2}-\alpha_{t-i})e_t$
\end{center}
for $3\leq i\leq n-4$ and
\begin{center}
$\varphi(e_{n-3})=\frac{1}{2^{n-5}}\Big((2^{n-5}-1)\alpha_{1}+\beta_2\Big)e_{n-3}+\frac{1}{2^{n-5}}\Big(\beta_{3}-(2^{n-5}-1)\alpha_n\Big)e_{n-2}$\end{center}
\begin{center}
$+\frac{1}{2^{n-5}}\Big((2^{n-5}-2)\alpha_2+\beta_{4}\Big)e_{n-1},$
\end{center}
\begin{center}
$\varphi(e_{n-2})=\frac{1}{2^{n-4}}\Big((2^{n-4}-1)\alpha_{1}+\beta_2\Big)e_{n-2}+\frac{1}{2^{n-4}}\Big(\beta_{3}-(2^{n-4}-1)\alpha_n\Big)e_{n-1},$
\end{center}
\begin{center}
$\varphi(e_{n-1})=\frac{1}{2^{n-3}}\Big((2^{n-3}-1)\alpha_{1}+\beta_2\Big)e_{n-1}.$
\end{center}

From $$0=\varphi([e_1,e_n])=\frac{1}{2}([\varphi(e_1),e_n]+[e_1,\varphi(e_n)])=\frac{1}{2}\Big([\sum\limits_{k=1}^{n}\alpha_ke_k,e_n]+[e_1,\sum\limits_{k=1}^{n}\gamma_ke_k]\Big)$$
$$=\frac{1}{2}\Big(\sum\limits_{k=2}^{n-3}\alpha_ke_{k+2}+\sum\limits_{k=2}^{n-2}\gamma_ke_{k+1}\Big)=\frac{1}{2}\Big(\sum\limits_{k=4}^{n-1}\alpha_{k-2}e_{k}+\sum\limits_{k=3}^{n-1}\gamma_{k-1}e_{k}\Big),$$ we obtain that
\begin{center}
$\gamma_2=0,\ \gamma_i=-\alpha_{i-1},\ 3\leq i\leq n-2.$
\end{center}

Now, we consider the condition of $\frac 1 2$-derivation for the elements $e_2, e_{n}:$
\begin{center}
$\varphi([e_2,e_{n}])=\frac{1}{2}([\varphi(e_2),e_{n}]+[e_2,\varphi(e_{n})])=\frac{1}{2}\Big([\sum\limits_{i=1}^{n}\beta_{i}e_i,e_{n}]+
[e_2,\sum\limits_{i=1}^{n}\gamma_{i}e_i]\Big)$
\end{center}
\begin{center}
$=\frac{1}{2}\Big(\sum\limits_{i=2}^{n-3}\beta_{i}e_{i+2}-\gamma_{1}e_3+\sum\limits_{i=3}^{n-4}\gamma_{i}e_{i+3}+\gamma_{n}e_4\Big)$
\end{center}
\begin{center}
$=\frac{1}{2}\Big(-\gamma_{1}e_3+(\beta_{2}+\gamma_{n})e_4+\beta_{3}e_5+\sum\limits_{i=6}^{n-1}(\beta_{i-2}-\gamma_{i-3})e_{i}\Big).$
\end{center}
Thus, we have
\begin{center}
$\varphi([e_2,e_{n}])=\varphi(e_{4})=\frac{1}{4}(3\alpha_{1}+\beta_2)e_{4}+\frac{1}{4}(\beta_{3}-3\alpha_n)e_{5}+\frac{1}{4}(2\alpha_2+\beta_{4})e_{6}+\frac{1}{4}\sum\limits_{t=7}^{n-1}(\beta_{t-2}-\alpha_{t-4})e_t.$
\end{center}
Comparing the coefficients at the basis elements, we obtain
$$\gamma_{1}=0, \ 2\gamma_n=3\alpha_1-\beta_2, \  \beta_3=-3\alpha_n, \ \beta_4=4\alpha_2, \  \beta_i=\alpha_{i-2}, \ 5\leq i\leq n-3.$$

By using the property of $\frac{1}{2}$-derivation for the products $[e_2,e_3]=e_6$ and $[e_3,e_n]=e_5$, we have

\begin{center}$\beta_1=0, \ \beta_n=0, \ \beta_2=\alpha_1;$\end{center}
\begin{center}$\beta_3=\alpha_n=0, \ \mbox{and for} \ n\geq 8 \ \mbox{we have} \ \beta_4=\alpha_2=0.$\end{center}

This completes the proof of the theorem. \end{proof}


The following theorems, we present an analogous descriptions of $\frac 1 2$-derivation for quasi-filiform Lie algebras of maximum length $g^1_{7}, g^2_{9}$ and $g^3_{11}$.

\begin{theorem}\label{halfderiv4} Any $\frac 1 2$-derivations of the algebra $g^1_{7}$ has the form
  $$\begin{cases}
\varphi(e_1)=\alpha_1e_1+\alpha_3e_3+\alpha_4e_4+\alpha_5e_5+\alpha_6e_6+\alpha_7e_7, \\
\varphi(e_2)=\alpha_1e_2-\frac{1}{3}\alpha_3e_4+\beta_5e_5+\beta_6e_6+\beta_7e_7,\\
\varphi(e_3)=\alpha_1e_3-\frac{2}{3}\alpha_3e_5+\frac{1}{2}(\beta_5-\alpha_4)e_6-\frac{1}{2}\alpha_5e_7, \\
\varphi(e_4)=\alpha_1e_4-\frac{1}{3}\alpha_3e_6+\frac{1}{2}\alpha_4e_7, \\
\varphi(e_5)=\alpha_1e_5-\frac{1}{2}\alpha_3e_7, \\
\varphi(e_6)=\alpha_1e_6, \\
\varphi(e_7)=\alpha_1e_7. \\
  \end{cases}$$
\end{theorem}


\begin{theorem}\label{2halfderiv12} Any $\frac 1 2$-derivations of the algebra $g^2_{9}$ has the form
  $$\begin{cases}
\varphi(e_1)=\alpha_1e_1+\alpha_5e_5+\alpha_6e_6+\alpha_7e_7+\alpha_8e_8+\alpha_9e_9, \\
\varphi(e_2)=\alpha_1e_2+\frac{1}{3}\alpha_5e_6+\beta_7e_7+\beta_8e_8+\beta_9e_9,\\
\varphi(e_3)=\alpha_1e_3-\frac{4}{3}\alpha_5e_7+\frac{1}{2}(\beta_7-5\alpha_6)e_8-\frac{1}{2}\alpha_7e_9, \\
\varphi(e_4)=\alpha_1e_4+\frac{1}{3}\alpha_5e_8+\frac{1}{2}\alpha_6e_9, \\
\varphi(e_5)=\alpha_1e_5-\frac{1}{2}\alpha_5e_9, \\
\varphi(e_i)=\alpha_1e_i, \ \ 6\leq i\leq 9. \\
  \end{cases}$$
\end{theorem}

\begin{theorem}\label{2halfderiv3} Any $\frac 1 2$-derivations of the algebra $g^3_{11}$ has the form
  $$\begin{cases}
\varphi(e_1)=\alpha_1e_1+\alpha_6e_6+\alpha_7e_7+\alpha_8e_8+\alpha_9e_9+\alpha_{10}e_{10}+\alpha_{11}e_{11}, \\
\varphi(e_2)=\alpha_1e_2-\alpha_6e_7-\alpha_7e_8+\beta_9e_9+\beta_{10}e_{10}+\beta_{11}e_{11},\\
\varphi(e_3)=\alpha_1e_3+\frac{1}{2}\beta_9e_{10}-\frac{1}{2}\alpha_9e_{11}, \\
\varphi(e_4)=\alpha_1e_4+\frac{1}{2}\alpha_7e_{10}+\frac{1}{2}\alpha_8e_{11}, \\
\varphi(e_5)=\alpha_1e_5-\frac{1}{2}\alpha_6e_{10}-\frac{1}{2}\alpha_7e_{11}, \\
\varphi(e_6)=\alpha_1e_6+\frac{1}{2}\alpha_6e_{11}, \\
\varphi(e_i)=\alpha_1e_i, \ \ 7\leq i\leq 11. \\
  \end{cases}$$
\end{theorem}

\section{Transposed Poisson structure for  quasi-filiform Lie algebras of maximum length}

In this section, we describe all transposed Poisson algebra structures on quasi-filiform Lie algebras of maximum length.




\begin{theorem} Let $(g^1_{(n,1)}, \cdot, [-,-])$ be a transposed Poisson algebra structure defined on the Lie algebra
$g^1_{(n,1)}$. Then, the multiplication of $(g^1_{(n,1)}, \cdot)$ has the following form:

${\bf TP}_1(g^1_{(n,1)}):\ \begin{cases}
e_1\cdot e_1=\sum\limits_{j=4}^{n}\alpha_{j}e_j, \ e_1\cdot e_2=\beta_{1}e_{n-2}+\beta_{2}e_{n-1}+\beta_{3}e_n, \\[1mm]
e_1\cdot e_3=\frac{1}{2}\beta_{1}e_{n-1}-\frac{1}{2}\alpha_{n-2}e_{n}, \ e_1\cdot e_j=\frac{(-1)^j}{2}\alpha_{n-j+1}e_n, \  4\leq j\leq n-3, \\[1mm]
e_2\cdot e_2=\beta_{4}e_{n-2}+\beta_{5}e_{n-1}+\beta_{6}e_n, \ e_2\cdot e_3=\frac{1}{2}\beta_{4}e_{n-1}-\frac{1}{2}\beta_{1}e_{n};
\end{cases}
$

${\bf TP}_2(g^1_{(n,1)}):\ \begin{cases}
e_1\cdot e_1=e_3+ \sum\limits_{j=4}^{n}\alpha_{j}e_j, \
e_1\cdot e_2=\beta_{1}e_{n-2}+\beta_{2}e_{n-1}+\beta_{3}e_n, \\[1mm]
e_1\cdot e_3=\frac{1}{2}\beta_{1}e_{n-1}-\frac{1}{2}\alpha_{n-2}e_{n}, \
e_1\cdot e_j=\frac{(-1)^j}{2}\alpha_{n-j+1}e_n, \  4\leq j\leq n-3, \\[1mm]
e_1\cdot e_{n-2}=-\frac{1}{2}e_n, \
e_2\cdot e_2=\beta_{4}e_{n-1}+\beta_{5}e_n, \
e_2\cdot e_3=-\frac{1}{2}\beta_{1}e_{n};
\end{cases}
$

${\bf TP}_3(g^1_{(n,1)}):\ \begin{cases}
e_1\cdot e_1=e_2+ \sum\limits_{j=3}^{n}\alpha_{j}e_j, \
e_1\cdot e_2=2\beta_{1}e_{n-1}+\beta_{2}e_n, \\[1mm]
e_1\cdot e_3=-\frac{1}{2}\alpha_{n-2}e_{n}, \
e_1\cdot e_j=\frac{(-1)^j}{2}\alpha_{n-j+1}e_n, \  4\leq j\leq n-2, \\
e_1\cdot e_{n-1}=\frac{1}{2}e_n, \
e_2\cdot e_2=\beta_{1}e_n;
\end{cases}
$\\
where the transposed Poisson algebra has its products with respect to the bracket $[-,-]$, and the remaining products are equal to zero.

\end{theorem}
\begin{proof} Let $(g^1_{(n,1)}, \cdot, [-,-])$ be a transposed Poisson algebra structure defined on the Lie algebra $g^1_{(n,1)}$. Then,
for any element $x \in g^1_{(n,1)} $, the operator of multiplication $\varphi_x(y) = x \cdot y$ is a $\frac12$-derivation. Hence, by Theorem \ref{halfderiv1}, we derive the following relations for $1 \leq i \leq n$:

$$\begin{cases}
\varphi_{e_i}(e_1)=\sum\limits_{j=1}^{n}\alpha_{i,j}e_j, \\
\varphi_{e_i}(e_2)=\alpha_{i,1}e_2+\beta_{i,n-2}e_{n-2}+\beta_{i,n-1}e_{n-1}+\beta_{i,n}e_n, \\
\varphi_{e_i}(e_3)=\alpha_{i,1}e_3+\frac{1}{2}\beta_{i,n-2}e_{n-1}-\frac{1}{2}\alpha_{i,n-2}e_{n}, \\
\varphi_{e_i}(e_j)=\alpha_{i,1}e_j+\frac{(-1)^j}{2}\alpha_{i,n-j+1}e_n, \  4\leq j\leq n-1, \\
\varphi_{e_i}(e_n)=\alpha_{i,1}e_n. \\
\end{cases}$$

By considering equality  $\varphi_{e_i}(e_j)=e_i\cdot e_j=e_j\cdot e_i=\varphi_{e_j}(e_i)$, we have following restrictions:

\begin{longtable}{lllllllll}
    $ \{e_1, e_2\},$ & $\Rightarrow$ & $\alpha_{2,1}=0, \ \alpha_{2,2}=\alpha_{1,1}, \ \alpha_{2,t}=0,\ 3\leq t\leq n-3,$ \\
& & $\alpha_{2,n-2}=\beta_{1,n-2}, \ \alpha_{2,n-1}=\beta_{1,n-1}, \ \alpha_{2,n}=\beta_{1,n},$ \\

$ \{e_1, e_3\},$ & $\Rightarrow $ & $\alpha_{3,1}=\alpha_{3,2}=0, \ \alpha_{3,3}=\alpha_{1,1},  \ \alpha_{3,t}=0,\ 4\leq t\leq n-2,$ \\

& & $\alpha_{3,n-1}=\frac{1}{2}\beta_{1,n-2}, \ \alpha_{3,n}=-\frac12\beta_{1,n},$ \\

$ \{e_1, e_j\},$ & $\Rightarrow$ & $\alpha_{j,t}=0,\ 1\leq t\leq j-1, \ \alpha_{j,j}=\alpha_{1,1},$ \\
$4\leq j\leq n-1,$ & & $\alpha_{j,n}=\frac{(-1)^j}{2}\alpha_{1,n-j+1}, \  \alpha_{j,t}=0,\ j+1\leq t\leq n-1,$ \\

$\{e_1, e_n\},$ & $\Rightarrow$ & $\alpha_{n,t}=0,\ 1\leq t\leq n-1, \ \alpha_{n,n}=\alpha_{1,1}, $ \\

$\{e_2, e_3\},$ & $\Rightarrow$ & $\beta_{3,n-2}=0, \ \beta_{3,n-1}=\frac{1}{2}\beta_{2,n-2}, \ \beta_{3,n}=-\frac{1}{2}\beta_{1,n-2},$  \\
$ \{e_2, e_j\},$ & $\Rightarrow$ & $\beta_{j,n-2}=\beta_{j,n-1}=\beta_{j,n}=0, $ \\
$ 4\leq j\leq n-1,$ & & \\
$ \{e_2, e_n\},$ & $\Rightarrow$ & $\beta_{n,n-2}=\beta_{n,n-1}=\beta_{n,n}=0,$\\
 $ \{e_3, e_j\},$ & $\Rightarrow$ & $\alpha_{3,t}=0, \ 2\leq t\leq n-3, $\\
$ 4\leq j\leq n-1.$& &\\
\end{longtable}


%For $i=1, \ j=2$ we have
%$$\alpha_{1,1}e_2+\beta_{1,n-2}e_{n-2}+\beta_{1,n-1}e_{n-1}+\beta_{1,n}e_n=
%\sum\limits_{j=1}^{n}\alpha_{2,j}e_j.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{2,1}=0, \ \alpha_{2,2}=\alpha_{1,1}, \  \alpha_{2,n-2}=\beta_{1,n-2}, \ \alpha_{2,n-1}=\beta_{1,n-1}, \ \alpha_{2,n}=\beta_{1,n}, \ \alpha_{2,t}=0,\ 3\leq t\leq n-3.$$

%For $i=1, \ j=3$ we have
%$$\alpha_{1,1}e_3+\frac{1}{2}\beta_{1,n-2}e_{n-1}-\frac{1}{2}\alpha_{1,n-2}e_{n}=
%\sum\limits_{j=1}^{n}\alpha_{3,j}e_j.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{3,1}=\alpha_{3,2}=0, \ \alpha_{3,3}=\alpha_{1,1}, \  \alpha_{3,n-1}=\frac{1}{2}\beta_{1,n-2}, \ \alpha_{3,n}=-\frac12\beta_{1,n}  , \ \alpha_{3,t}=0,\ 4\leq t\leq n-2.$$

%
%For $i=1, \ 4\leq j\leq n-1$ we have
%$$\alpha_{1,1}e_j+\frac{(-1)^j}{2}\alpha_{1,n-j+1}e_n=
%\sum\limits_{t=1}^{n}\alpha_{n,t}e_t.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{j,t}=0,\ 1\leq t\leq j-1, \ \alpha_{j,j}=\alpha_{1,1}, \  \alpha_{j,n}=\frac{(-1)^j}{2}\alpha_{1,n-j+1}, \  \alpha_{j,t}=0,\ j+1\leq t\leq n-1.$$
%
%For $i=1, \ j=n$ we have
%$$\alpha_{1,1}e_n=
%\sum\limits_{t=1}^{n}\alpha_{n,t}e_t.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{n,t}=0,\ 1\leq t\leq n-1, \ \alpha_{n,n}=\alpha_{1,1}.$$


%
%For $i=2, \ j=3$ we have
%$$\frac{1}{2}\beta_{2,n-2}e_{n-1}-\frac{1}{2}\beta_{1,n-2}e_{n}=
%\beta_{3,n-2}e_{n-2}+\beta_{3,n-1}e_{n-1}+\beta_{3,n}e_n.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{3,n-2}=0, \ \beta_{3,n-1}=\frac{1}{2}\beta_{2,n-2}, \ \beta_{3,n}=-\frac{1}{2}\beta_{1,n-2}.$$


%For $i=2, \ 4\leq j\leq n-1$ we have
%$$0=\beta_{j,n-2}e_{n-2}+\beta_{j,n-1}e_{n-1}+\beta_{j,n}e_n.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{j,n-2}=\beta_{j,n-1}=\beta_{j,n}=0, \ 4\leq j\leq n-1.$$


%For $i=2, \ j=n$ we have
%$$0=\beta_{n,n-2}e_{n-2}+\beta_{n,n-1}e_{n-1}+\beta_{n,n}e_n.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{n,n-2}=\beta_{n,n-1}=\beta_{n,n}=0.$$

%
%For $i=3, \ 4\leq j\leq n-1$ we have
%$$\frac{(-1)^j}{2}\alpha_{3,n-j+1}e_n=0.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{3,t}=0, \ 2\leq t\leq n-3.$$

Thus, we have the following table of commutative multiplications of the transposed Poisson algebra structure $(g^1_{(n,1)},, \cdot, [-,-])$:

\begin{equation}\label{gn1}
\begin{cases}
e_1\cdot e_1=\sum\limits_{j=2}^{n}\alpha_{j}e_j, \\
e_1\cdot e_2=\beta_{1}e_{n-2}+\beta_{2}e_{n-1}+\beta_{3}e_n, \\
e_1\cdot e_3=\frac{1}{2}\beta_{1}e_{n-1}-\frac{1}{2}\alpha_{n-2}e_{n}, \\
e_1\cdot e_j=\frac{(-1)^j}{2}\alpha_{n-j+1}e_n, \  4\leq j\leq n-1, \\
e_2\cdot e_2=\beta_{4}e_{n-2}+\beta_{5}e_{n-1}+\beta_{6}e_n, \\
e_2\cdot e_3=\frac{1}{2}\beta_{4}e_{n-1}-\frac{1}{2}\beta_{1}e_{n}.\\
\end{cases}
\end{equation}

Considering the identity $x\cdot (y\cdot z)=(x\cdot y)\cdot z$ in \eqref{gn1}, we obtain the following restrictions on structure constants:
$$\begin{array}{lll}
\{e_1,e_2,e_3\}, & \Rightarrow & \alpha_{2}\beta_{4}=0,  \\[1mm]
\{e_1,e_2,e_2\}, & \Rightarrow & \alpha_{3}\beta_{4}-\alpha_{2}\beta_{5}=0,  \\[1mm]
\{e_1,e_1,e_3\}, & \Rightarrow & \alpha_{2}\beta_{1}=0,  \\[1mm]
\{e_1,e_1,e_2\}, & \Rightarrow & 2\alpha_{2}\beta_{5}+\alpha_{3}\beta_{4}=0, \ \alpha_{2}(2\beta_{6}-\beta_{2})=0,  \\[1mm]
\{e_1,e_1,e_i\}, & \Rightarrow & \alpha_{2}\beta_{4}=0, \ \alpha_{3}\beta_{4}=0, \ \alpha_{2}\beta_{5}=0,\ \alpha_{2}\beta_{1}=0, \  \alpha_{2}(2\beta_{6}-\beta_{2})=0, \\[1mm]
4\leq i\leq n-1.& &
\end{array}$$

%
%Endi \eqref{gn1} algebra uchun assatsiativlik ayniyatini
%$x\cdot (y\cdot z)=(x\cdot y)\cdot z$ tekshiramiz.

%Let $x=e_1, y=e_2,\ z=e_3.$
%$$0=(e_1\cdot e_2)\cdot e_3=e_1\cdot (e_2\cdot e_3)=e_1\cdot (\frac{1}{2}\beta_{4}e_{n-1}-\frac{1}{2}\beta_{1}e_{n})=\frac{1}{4}\alpha_{2}\beta_{4}e_{n}.$$
%Comparing coefficients at the basis elements we obtain that $\alpha_{2}\beta_{4}=0.$

%
%Let $x=e_1, y=e_2,\ z=e_2.$
%$$0=(e_1\cdot e_2)\cdot e_2=e_1\cdot (e_2\cdot e_2)=e_1\cdot (\beta_{4}e_{n-2}+\beta_{5}e_{n-1}+\beta_{6}e_n)$$
%$$=(-\frac{1}{4}\alpha_{3}\beta_{4}+\frac{1}{4}\alpha_{2}\beta_{5})e_{n}.$$
%Comparing coefficients at the basis elements we obtain that $\alpha_{3}\beta_{4}-\alpha_{2}\beta_{5}=0.$

%Let $x=e_1, y=e_1,\ z=e_3.$
%$$\alpha_{2}(\frac{1}{2}\beta_{4}e_{n-1}-\frac{1}{2}\beta_{1}e_{n})=(e_1\cdot e_1)\cdot e_3$$
%$$=e_1\cdot (e_1\cdot e_3)=e_1\cdot (\frac{1}{2}\beta_{1}e_{n-1}-\frac{1}{2}\alpha_{n-2}e_{n})
%=\frac{1}{4}\alpha_{2}\beta_{1}e_{n}.$$
%Comparing coefficients at the basis elements we obtain that $\alpha_{2}\beta_{1}=0.$

%Let $x=e_1, y=e_1,\ z=e_2.$
%$$\alpha_{2}(\beta_{5}e_{n-1}+\beta_{6}e_n)+
%\alpha_{3}(\frac{1}{2}\beta_{4}e_{n-1}-\frac{1}{2}\beta_{1}e_{n})=(e_1\cdot e_1)\cdot e_2$$
%$$=e_1\cdot (e_1\cdot e_2)=e_1\cdot (\beta_{1}e_{n-2}+\beta_{2}e_{n-1}+\beta_{3}e_n)
%=\frac{1}{2}(-\alpha_{3}\beta_{1}+\alpha_{2}\beta_{2})e_{n}.$$
%Comparing coefficients at the basis elements we obtain that
%$$2\alpha_{2}\beta_{5}+\alpha_{3}\beta_{4}=0, \ \alpha_{2}(2\beta_{6}-\beta_{2})=0.$$

%Let $x=e_1, y=e_1,\ z=e_i, \ 4\leq i\leq n-1.$
%$$0=(e_1\cdot e_1)\cdot e_i=e_1\cdot (e_1\cdot e_i)=0.$$
%
%Biz quyidagi tengliklarni olamiz.
%$$\alpha_{2}\beta_{4}=0, \ \ \alpha_{3}\beta_{4}=0, \ \ \alpha_{2}\beta_{5}=0,\ \ \alpha_{2}\beta_{1}=0, \ \  \alpha_{2}(2\beta_{6}-\beta_{2})=0.$$

Now we consider the general change of basis:
%
%$TP(\alpha_2, \dots \alpha_n,  \beta_1, \dots, \beta_6)$ algebrasida biz quyidagicha umumiy almashtirishni qaraylik.

$$e_1'=\sum\limits_{j=1}^{n}A_je_j, \ e_2'=\sum\limits_{j=1}^{n}B_je_j, \ e_{i+1}'=[e_1',e_{i}'], \ 2\leq i\leq n-2, \ e_n'=[e_2',e_{n-2}'],$$
where $B_1=0$ and $A_1B_2\neq0$. Then, from the multiplication $e_1\cdot e_1=\sum\limits_{j=2}^{n}\alpha_{j}e_j,$ we discover that the structure constant $\alpha_2$ changes as follows:
$$\alpha_2'=\frac{A_1^2}{B_2}\alpha_2.$$

Thus, we have the following cases.
\begin{enumerate}
    \item If $\alpha_{2}=0.$  Again, by using a change of basis and from the product  $e_1\cdot e_1=\sum\limits_{j=3}^{n}\alpha_{j}e_j,$ we obtain the following relation:
$$\alpha_3'=\frac{A_1}{B_2}\alpha_3.$$

Now, we consider the following subcases.
\begin{enumerate}

 \item If $\alpha_{3}=0.$ Then, we have the algebra ${\bf TP}_1(g^1_{(n,1)}).$


 \item If $\alpha_{3}\neq 0,$ then, we have $\beta_{4}=0,$ and via transformation
$$\phi(e_1)=e_1, \ \phi(e_i)=\alpha_{3}e_i, \ 2\leq i\leq n-1,  \ \phi(e_{n})=\alpha_{3}^2e_n,$$
we get the algebra ${\bf TP}_2(g^1_{(n,1)})$.
\end{enumerate}

\item If $\alpha_{2}\neq0,$ then we derive $\beta_{4}=\beta_{5}=\beta_{1}=0, \ \beta_{2}=2\beta_{6},$ and via the transformation
$$\phi(e_1)=e_1, \ \phi(e_i)=\alpha_{2}e_i, \ 2\leq i\leq n-1,  \ \phi(e_{n})=\alpha_{2}^2e_n,$$
we obtain the algebra  ${\bf TP}_3(g^1_{(n,1)})$.

\end{enumerate}


\end{proof}







\begin{theorem} Let $(g^2_{(n,1)}, \cdot, [-,-])$ be a transposed Poisson algebra structure defined on Lie algebra
$g^2_{(n,1)}$. Then, the multiplication of $(g^2_{(n,1)}, \cdot)$ has the following form:

${\bf TP}_1(g^2_{(n,1)}):\ \begin{cases}
e_1\cdot e_1=e_2+\sum\limits_{i=3}^{n-1}\alpha_{i}e_i, \ e_1\cdot e_2=2\alpha_{1}e_{n-2}+\alpha_{n}e_{n-1}, \ e_1\cdot e_3=\alpha_{1}e_{n-1}, \\[1mm]
e_1\cdot e_n=-e_3-\sum\limits_{i=4}^{n-2}\alpha_{i-1}e_i+\alpha_{n+1}e_{n-1},\ e_2\cdot e_n=-\alpha_1\alpha_{3}e_{n-1},\\[1mm] e_n\cdot e_n=\sum\limits_{i=4}^{n-2}\alpha_{i-1}e_i +\alpha_{n+2}e_{n-1};
\end{cases}$

${\bf TP}_2(g^2_{(n,1)}):\ \begin{cases}
e_1\cdot e_1=e_3+\sum\limits_{i=4}^{n-1}\alpha_{i}e_i, \ e_1\cdot e_2=\alpha_{n}e_{n-1}, \ e_1\cdot e_n=-e_4-\sum\limits_{i=5}^{n-2}\alpha_{i-1}e_i+\alpha_{n+1}e_{n-1},\\[1mm]
e_2\cdot e_2=\alpha_{n+3}e_{n-1}, \ e_2\cdot e_n=\alpha_{n+4}e_{n-1},\ e_n\cdot e_n=e_4+\sum\limits_{i=5}^{n-2}\alpha_{i-1}e_i +\alpha_{n+5}e_{n-1};
\end{cases}$

${\bf TP}_3(g^2_{(n,1)}):\ \begin{cases}
e_1\cdot e_1=\sum\limits_{i=4}^{n-1}\alpha_{i}e_i, \ e_1\cdot e_2=2\alpha_{1}e_{n-2}+\alpha_{n}e_{n-1}, \ e_1\cdot e_3=\alpha_{1}e_{n-1}, \\[1mm]
e_1\cdot e_n=-\sum\limits_{i=5}^{n-2}\alpha_{i-1}e_i+\alpha_{n+1}e_{n-1},\ e_2\cdot e_2=2\alpha_{n+2}e_{n-2}+\alpha_{n+3}e_{n-1}, \\[1mm]
e_2\cdot e_3=\alpha_{n+2}e_{n-1}, \ e_2\cdot e_n=\alpha_{n+4}e_{n-1},\ e_n\cdot e_n=\sum\limits_{i=5}^{n-2}\alpha_{i-1}e_i +\alpha_{n+5}e_{n-1};
\end{cases}$\\
where the transposed Poisson algebra has its products with respect to the bracket $[-,-]$, and the remaining products are equal to zero.
\end{theorem}

\begin{proof}
Let $(g^2_{(n,1)}, \cdot, [-,-])$ be a transposed Poisson algebra structure defined on Lie algebra $g^2_{(n,1)}$. Then,
for any element $x \in g^2_{(n,1)} $, we have operator of multiplication $\varphi_x(y) = x \cdot y$ is a $\frac12$-derivation. Hence, for $1 \leq i \leq n$, we derive

$$\begin{cases}
\varphi_{e_i}(e_1)=\sum\limits_{j=1}^{n-1}\alpha_{i,j}e_j, \\
\varphi_{e_i}(e_2)=\alpha_{i,1}e_2+\beta_{i,n-2}e_{n-2}+\beta_{i,n-1}e_{n-1}, \\
\varphi_{e_i}(e_3)=\alpha_{i,1}e_3+\frac{1}{2}\beta_{i,n-2}e_{n-1}, \\
\varphi_{e_i}(e_j)=\alpha_{i,1}e_j, \  4\leq j\leq n-1,\\
\varphi_{e_i}(e_n)=-\sum\limits_{j=3}^{n-2}\alpha_{i,j-1}e_j+\gamma_{i,n-1}e_{n-1}+\alpha_{i,1}e_n.
 \end{cases}$$

It is known that $\varphi_{e_i}(e_j)=e_i\cdot e_j=e_j\cdot e_i=\varphi_{e_j}(e_i).$ Then for some pair elements, we derive the following restrictions:

\begin{longtable}{llllll}
    $\{e_1, e_2\}, $ & $\Rightarrow$ &$\alpha_{2,1}=0, \  \alpha_{2,2}=\alpha_{1,1}, \  \alpha_{2,j}=0, 3\leq j\leq n-3, $ \\

    & &$\alpha_{2,n-2}=\beta_{1,n-2}, \ \alpha_{2,n-1}=\beta_{1,n-1}, $\\

    $\{e_1, e_3\},$ & $\Rightarrow$ & $\alpha_{3,1}=\alpha_{3,2}=0, \  \alpha_{3,3}=\alpha_{1,1}, \  \alpha_{3,j}=0, 4\leq j\leq n-2,$ \\
      & & $\alpha_{3,n-1}=\frac{1}{2}\beta_{1,n-2},$  \\
    $\{e_1, e_j\}, $ & $\Rightarrow$ &$\alpha_{j,j}=\alpha_{1,1},\  \alpha_{j,t}=0, \  1\leq t\neq j\leq n-1,$  \\
   $4\leq j\leq n-1, $ & & \\
    $\{e_1, e_n\}, $ & $\Rightarrow$ & $\alpha_{n,1}=\alpha_{n,2}=0, \  \alpha_{n,t}=-\alpha_{1,t-1}, \ 3\leq t\leq n-2, $ \\
     & & $\alpha_{n,n-1}=\gamma_{1,n-1},\  \alpha_{1,1}=0,$ \\

 $\{e_2, e_3\}, $ & $\Rightarrow$ & $\beta_{3,n-2}=0, \ \beta_{3,n-1}=\frac{1}{2}\beta_{2,n-2},$  \\
 $\{e_2, e_j\},$ & $\Rightarrow $ & $\beta_{j,n-2}=\beta_{j,n-1}=0,$ \\
 $4\leq j\leq n-1,$ & &\\
 $\{e_2, e_n\},$ & $\Rightarrow $ & $\beta_{n,n-2}=0, \  \beta_{n,n-1}=\gamma_{2,n-1},$\\
 $ \{e_i, e_n\},$ & $\Rightarrow $ & $\gamma_{i,n-1}=0, \quad 3\leq i\leq n-1.$
\end{longtable}


%For $i=1, \ j=2$ we have
%$$\alpha_{1,1}e_2+\beta_{1,n-2}e_{n-2}+\beta_{1,n-1}e_{n-1}=\sum\limits_{j=1}^{n-1}\alpha_{2,j}e_j.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{2,1}=0, \  \alpha_{2,2}=\alpha_{1,1}, \  \alpha_{2,j}=0, 3\leq j\leq n-3, \ \alpha_{2,n-2}=\beta_{1,n-2}, \ \alpha_{2,n-1}=\beta_{1,n-1}.$$
%
%For $i=1, \ j=3$ we have
%$$\alpha_{1,1}e_3+\frac{1}{2}\beta_{1,n-2}e_{n-1}=\sum\limits_{j=1}^{n-1}\alpha_{3,j}e_j.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{3,1}=\alpha_{3,2}=0, \  \alpha_{3,3}=\alpha_{1,1}, \  \alpha_{3,j}=0, 4\leq j\leq n-2, \ \alpha_{3,n-1}=\frac{1}{2}\beta_{1,n-2}.$$
%
%
%For $i=1, \ 4\leq  j\leq n-1$ we have
%$$\alpha_{1,1}e_j=\sum\limits_{t=1}^{n-1}\alpha_{j,t}e_t.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{j,j}=\alpha_{1,1},\  \alpha_{j,t}=0, \  1\leq t\neq j\leq n-1.$$
%
%For $i=1, \ j=n$ we have
%$$-\sum\limits_{j=3}^{n-2}\alpha_{1,j-1}e_j+\gamma_{1,n-1}e_{n-1}+\alpha_{1,1}e_n =\sum\limits_{t=1}^{n-1}\alpha_{n,t}e_t.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{n,1}=\alpha_{n,2}=0, \  \alpha_{n,t}=-\alpha_{1,t-1}, \ 3\leq t\leq n-2, \  \alpha_{n,n-1}=\gamma_{1,n-1},\  \alpha_{1,1}=0.$$
%
%
%For $i=2, \ j=3$ we have
%$$\frac{1}{2}\beta_{2,n-2}e_{n-1}=\beta_{3,n-2}e_{n-2}+\beta_{3,n-1}e_{n-1}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{3,n-2}=0, \ \beta_{3,n-1}=\frac{1}{2}\beta_{2,n-2}.$$
%
%For $i=2, \ 4\leq  j\leq n-1$ we have
%$$0=\beta_{j,n-2}e_{n-2}+\beta_{j,n-1}e_{n-1}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{j,n-2}=\beta_{j,n-1}=0.$$
%
%For $i=2, \ j=n$ we have
%$$\gamma_{2,n-1}e_{n-1}=\beta_{n,n-2}e_{n-2}+\beta_{n,n-1}e_{n-1}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{n,n-2}=0, \  \beta_{n,n-1}=\gamma_{2,n-1}.$$
%
%
%For $3\leq i\leq n-1, \ j=n$ we have $\gamma_{i,n-1}=0.$
%

Thus, we have the following table of commutative multiplications of the transposed Poisson algebra structure defined on  $g^2_{(n,1)}$:

\begin{equation}\label{2klas}
\begin{cases}
e_1\cdot e_1=\sum\limits_{i=2}^{n-1}\alpha_{i}e_i, \ e_1\cdot e_2=2\alpha_{1}e_{n-2}+\alpha_{n}e_{n-1}, \ e_1\cdot e_3=\alpha_{1}e_{n-1}, \\
e_1\cdot e_n=-\sum\limits_{i=3}^{n-2}\alpha_{i-1}e_i+\alpha_{n+1}e_{n-1},\ e_2\cdot e_2=2\alpha_{n+2}e_{n-2}+\alpha_{n+3}e_{n-1}, \\
e_2\cdot e_3=\alpha_{n+2}e_{n-1}, \ e_2\cdot e_n=\alpha_{n+4}e_{n-1},\ e_n\cdot e_n=\sum\limits_{i=4}^{n-2}\alpha_{i-1}e_i +\alpha_{n+5}e_{n-1}.
\end{cases}\end{equation}

Considering the associative identity $x\cdot (y\cdot z)=(x\cdot y)\cdot z$ in \eqref{2klas}, we obtain the following restrictions on structure constants:
$$\begin{array}{lll}
\{e_1,e_1,e_2\}, & \Rightarrow & \alpha_{2}\alpha_{n+2}=0, \ \alpha_{2}\alpha_{n+3}+ \alpha_{3}\alpha_{n+2}=0,  \\[1mm]
\{e_1,e_1,e_n\}, & \Rightarrow & \alpha_{1}\alpha_{3}+\alpha_{2}\alpha_{n+4}=0.
\end{array}$$


Similarly, by using the multiplication of Lie algebra $g^2_{(n,1)}$ we consider the general basis change:
$$e_1'=\sum\limits_{t=1}^{n}A_te_t, \ e_2'=\sum\limits_{t=1}^{n}B_te_t, \ e_n'=\sum\limits_{t=1}^{n}C_te_t.$$
Then the product $e_1'\cdot e_1'=\sum\limits_{i=2}^{n-1}\alpha_{i}'e_i'$ gives
$$\alpha_2'=\frac{A_1^2}{B_2}\alpha_2.$$
We have the following cases.

\begin{enumerate}
    \item Let $\alpha_2\neq0.$ Then by choosing $B_2=A_1^2\alpha_2$, we put $\alpha_2'=1.$ In this case we obtain the algebra ${\bf TP}_1(g^2_{(n,1)}).$

    \item Let $\alpha_2=0$. Then from above restrictions we have $\alpha_3\alpha_{n+2}=\alpha_1\alpha_3=0.$ If we apply a general change of basis, then we have
    $$\alpha_3'=\frac{A_1}{B_2}\alpha_3.$$

    \begin{enumerate}
        \item If $\alpha_3\neq 0,$ then $\alpha_3'=1$ and we have the algebra ${\bf TP}_2(g^2_{(n,1)}).$
        \item If $\alpha_3=0,$ then we have the algebra ${\bf TP}_3(g^2_{(n,1)}).$
    \end{enumerate}

\end{enumerate}

\end{proof}




\begin{theorem}
Let $(g^3_{(7,1)}, \cdot, [-,-])$ be a transposed Poisson algebra structure defined on Lie algebra
$g^3_{(7,1)}$. Then, the multiplication of $(g^3_{(7,1)}, \cdot)$ has the following form:

${\bf TP}_1(g^3_{(7,1)}):\ \begin{cases}
e_1\cdot e_1=\alpha_{4}e_4+ \alpha_{5}e_5, \ e_1\cdot e_2=2\alpha_{6}e_5 +\alpha_{7}e_6,\ e_1\cdot e_3=\alpha_{6}e_6, \\[1mm]
e_1\cdot e_7=-\alpha_{4}e_5+ \alpha_{8}e_6, \ e_2\cdot e_2=2\alpha_{1}e_5+\alpha_{9}e_6,\ e_2\cdot e_3=\alpha_{1}e_6, \\[1mm]
e_2\cdot e_7=\alpha_{10}e_6,\ e_7\cdot e_7=\alpha_{11}e_6; \end{cases}$

${\bf TP}_2(g^3_{(7,1)}):\ \begin{cases}
e_1\cdot e_1=2e_3+\alpha_{4}e_4+\alpha_{5}e_5, \ e_1\cdot e_2=2\alpha_{6}e_5 +\alpha_{7}e_6,\ e_1\cdot e_3=(\alpha_{6}-1)e_6, \\[1mm]
e_1\cdot e_7=-2e_4-\alpha_{4}e_5+\alpha_{8}e_6, \ e_2\cdot e_2=\alpha_{9}e_6,\ e_2\cdot e_7=\alpha_{10}e_6,\\[1mm]
e_7\cdot e_7=2e_5+\alpha_{11}e_6;
\end{cases}$\\
where the transposed Poisson algebra has its products with respect to the bracket $[-,-]$, and remaining products are equal to zero.

\end{theorem}
\begin{proof} Let $(g^3_{(7,1)}, \cdot, [-,-])$ be a transposed Poisson algebra structure defined on Lie algebra $g^3_{(7,1)}$. Then
for any element $x \in g^3_{(7,1)} $, the operator of multiplication $\varphi_x(y) = x \cdot y$ is a $\frac12$-derivation. Hence, for $1 \leq i \leq 7$ by Theorem \ref{halfderiv3} we derive the following:

$$\begin{cases}
\varphi_{e_i}(e_1)=\alpha_{i,1}e_1+\alpha_{i,2}e_2+ \alpha_{i,3}e_3+\alpha_{i,4}e_4+\alpha_{i,5}e_5+\alpha_{i,6}e_6, \\
\varphi_{e_i}(e_2)=\alpha_{i,1}e_2+4\alpha_{i,2}e_4+ \beta_{i,5}e_5+\beta_{i,6}e_6,\\
\varphi_{e_i}(e_3)=\alpha_{i,1}e_3+2\alpha_{i,2}e_5+ \frac{1}{2}(\beta_{i,5}-\alpha_{i,3})e_6, \\
\varphi_{e_i}(e_4)=\alpha_{i,1}e_4+\frac{3}{2}\alpha_{i,2}e_6, \\
\varphi_{e_i}(e_5)=\alpha_{i,1}e_5, \\
\varphi_{e_i}(e_6)=\alpha_{i,1}e_6, \\
\varphi_{e_i}(e_7)=-\alpha_{i,2}e_3-\alpha_{i,3}e_4-\alpha_{i,4}e_5+\gamma_{i,6}e_6+\alpha_{i,1}e_7, \\
  \end{cases}$$

Considering the property $\varphi_{e_i}(e_j)=e_i\cdot e_j=e_j\cdot e_i=\varphi_{e_j}(e_i)$, we obtain the following restrictions:

 $$\begin{array}{lll}
    \{e_1, e_2\}, & \Rightarrow & \alpha_{2,1}=0, \  \alpha_{2,2}=\alpha_{1,1}, \  \alpha_{2,3}=0, \ \alpha_{2,4}=4\alpha_{1,2}, \ \alpha_{2,5}=\beta_{1,5}, \ \alpha_{2,6}=\beta_{1,6},\\[1mm]
    \{e_1, e_3\}, & \Rightarrow & \alpha_{3,1}=\alpha_{3,2}=0, \  \alpha_{3,3}=\alpha_{1,1}, \  \alpha_{3,4}=-2\alpha_{1,6}, \ \alpha_{3,5}=2\alpha_{1,2}, \ \alpha_{3,6}=\frac{1}{2}(\beta_{1,5}-\alpha_{1,3}),  \\[1mm]
    \{e_1, e_4\}, & \Rightarrow & \alpha_{4,1}=\alpha_{4,2}=\alpha_{4,3}=0, \  \alpha_{4,4}=\alpha_{1,1}, \  \alpha_{4,5}=0, \ \alpha_{4,6}=\frac{3}{2}\alpha_{1,2},  \\[1mm]
    \{e_1, e_5\}, & \Rightarrow & \alpha_{5,1}=\alpha_{5,2}=\alpha_{5,3}=\alpha_{5,4}=0, \ \alpha_{5,5}=\alpha_{1,1}, \ \alpha_{5,6}=0,  \\[1mm]
    \{e_1, e_6\}, & \Rightarrow & \alpha_{6,1}=\alpha_{6,2}=\alpha_{6,3}=\alpha_{6,4}=\alpha_{6,5}=0, \ \alpha_{6,6}=\alpha_{1,1},  \\[1mm]
   \{e_1, e_7\}, & \Rightarrow & \alpha_{1,1}=0, \ \alpha_{7,1}=\alpha_{7,2}=0, \ \alpha_{7,3}=-\alpha_{1,2}, \ \alpha_{7,4}=-\alpha_{1,3}, \ \alpha_{7,5}=-\alpha_{1,4}, \ \alpha_{7,6}=\gamma_{1,6},  \\[1mm]
\{e_2, e_3\}, & \Rightarrow & \beta_{3,5}=0, \ \beta_{3,6}=\frac{1}{2}\beta_{2,5},  \\[1mm]
 \{e_2, e_4\}, & \Rightarrow & \beta_{4,5}=\beta_{4,6}=0,  \\[1mm]
 \{e_2, e_5\}, & \Rightarrow & \beta_{5,5}=\beta_{5,6}=0,\\[1mm]
 \{e_2, e_6\}, & \Rightarrow & \beta_{6,5}=\beta_{6,6}=0,\\[1mm]
 \{e_2, e_7\}, & \Rightarrow & \beta_{7,5}=-4\alpha_{1,2}, \  \beta_{7,6}=\gamma_{2,6},\\[1mm]
\{e_3, e_7\}, & \Rightarrow & \alpha_{1,6}=0, \ \gamma_{3,6}=-\frac{3}{2}\alpha_{1,2},\\[1mm]
\{e_4, e_7\}, & \Rightarrow & \gamma_{4,6}=0,\\[1mm]
\{e_5, e_7\}, & \Rightarrow & \gamma_{5,6}=0,\\[1mm]
\{e_6, e_7\}, & \Rightarrow & \gamma_{6,6}=0.\\[1mm]
    \end{array}$$
%
%For $i=1, \ j=2$ we have
%$$\alpha_{1,1}e_2+4\alpha_{1,2}e_4+ \beta_{1,5}e_5+\beta_{1,6}e_6=\alpha_{2,1}e_1+\alpha_{2,2}e_2+ \alpha_{2,3}e_3+\alpha_{2,4}e_4+\alpha_{2,5}e_5+\alpha_{2,6}e_6.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{2,1}=0, \  \alpha_{2,2}=\alpha_{1,1}, \  \alpha_{2,3}=0, \ \alpha_{2,4}=4\alpha_{1,2}, \ \alpha_{2,5}=\beta_{1,5}, \ \alpha_{2,6}=\beta_{1,6}.$$
%
%For $i=1, \ j=3$ we have
%$$\alpha_{1,1}e_3+2\alpha_{1,2}e_5+ \frac{1}{2}(\beta_{1,5}-\alpha_{1,3})e_6=\alpha_{3,1}e_1+\alpha_{3,2}e_2+ \alpha_{3,3}e_3+\alpha_{3,4}e_4+\alpha_{3,5}e_5+\alpha_{3,6}e_6.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{3,1}=\alpha_{3,2}=0, \  \alpha_{3,3}=\alpha_{1,1}, \  \alpha_{3,4}=-2\alpha_{1,6}, \ \alpha_{3,5}=2\alpha_{1,2}, \ \alpha_{3,6}=\frac{1}{2}(\beta_{1,5}-\alpha_{1,3}).$$
%
%For $i=1, \ j=4$ we have
%$$\alpha_{1,1}e_4+\frac{3}{2}\alpha_{1,2}e_6= \alpha_{4,1}e_1+\alpha_{4,2}e_2+\alpha_{4,3}e_3+ \alpha_{4,4}e_4+\alpha_{4,5}e_5+\alpha_{4,6}e_6.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{4,1}=\alpha_{4,2}=\alpha_{4,3}=0, \  \alpha_{4,4}=\alpha_{1,1}, \  \alpha_{4,5}=0, \ \alpha_{4,6}=\frac{3}{2}\alpha_{1,2}.$$
%
%For $i=1, \ j=5$ we have
%$$\alpha_{1,1}e_5=
%\alpha_{5,1}e_1+\alpha_{5,2}e_2+\alpha_{5,3}e_3+ \alpha_{5,4}e_4+\alpha_{5,5}e_5+\alpha_{5,6}e_6.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{5,1}=\alpha_{5,2}=\alpha_{5,3}=\alpha_{5,4}=0, \ \alpha_{5,5}=\alpha_{1,1}, \ \alpha_{5,6}=0.$$
%
%For $i=1, \ j=6$ we have
%$$\alpha_{1,1}e_6=\alpha_{6,1}e_1+\alpha_{6,2}e_2+\alpha_{6,3} e_3+\alpha_{6,4}e_4+\alpha_{6,5}e_5+\alpha_{6,6}e_6.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{6,1}=\alpha_{6,2}=\alpha_{6,3}=\alpha_{6,4}=\alpha_{6,5}=0, \ \alpha_{6,6}=\alpha_{1,1}.$$
%
%For $i=1, \ j=7$ we have
%$$-\alpha_{1,2}e_3-\alpha_{1,3}e_4-\alpha_{1,4}e_5 +\gamma_{1,6}e_6+\alpha_{1,1}e_7
%=\alpha_{7,1}e_1+\alpha_{7,2}e_2+\alpha_{7,3} e_3+\alpha_{7,4}e_4+\alpha_{7,5}e_5+\alpha_{7,6}e_6.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{1,1}=0, \ \alpha_{7,1}=\alpha_{7,2}=0, \ \alpha_{7,3}=-\alpha_{1,2}, \ \alpha_{7,4}=-\alpha_{1,3}, \ \alpha_{7,5}=-\alpha_{1,4}, \ \alpha_{7,6}=\gamma_{1,6}.$$
%
%For $i=2, \ j=3$ we have
%$$\frac{1}{2}\beta_{2,5}e_6=\beta_{3,5}e_5+\beta_{3,6}e_6.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{3,5}=0, \ \beta_{3,6}=\frac{1}{2}\beta_{2,5}.$$
%
%For $i=2, \ j=4$ we have
%$$0=\beta_{4,5}e_5+ \beta_{4,6}e_6.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{4,5}=\beta_{4,6}=0.$$
%
%For $i=2, \ j=5$ we have
%$$0=\beta_{5,5}e_5+\beta_{5,6}e_6.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{5,5}=\beta_{5,6}=0.$$
%
%For $i=2, \ j=6$ we have
%$$0=\beta_{6,5}e_5+\beta_{6,6}e_6.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{6,5}=0, \  \beta_{6,6}=0.$$
%
%For $i=2, \ j=7$ we have
%$$-4\alpha_{1,2}e_5+\gamma_{2,6}e_6 = \beta_{7,5}e_5+\beta_{7,6}e_6.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{7,5}=-4\alpha_{1,2}, \  \beta_{7,6}=\gamma_{2,6}.$$
%
%For $i=3, \ j=7$ we have
%$$2\alpha_{1,6}e_5+\gamma_{3,6}e_6=-\frac{3}{2}\alpha_{1,2}e_6.$$
%Comparing coefficients at the basis elements we obtain that $\alpha_{1,6}=0, \ \gamma_{3,6}=-\frac{3}{2}\alpha_{1,2}.$
%
%For $i=4, \ j=7$ we have $\gamma_{4,6}=0.$
%
%For $i=5, \ j=7$ we have $\gamma_{5,6}=0.$
%For $i=6, \ j=7$ we have $\gamma_{6,6}=0.$
%
%$$\begin{cases}
%\varphi_{e_1}(e_1)=\alpha_{1,2}e_2+ \alpha_{1,3}e_3+ \alpha_{1,4}e_4+\alpha_{1,5}e_5, \\
%\varphi_{e_1}(e_2)=4\alpha_{1,2}e_4+ \beta_{1,5}e_5 +\beta_{1,6}e_6,\\
%\varphi_{e_1}(e_3)=2\alpha_{1,2}e_5+ \frac{1}{2}(\beta_{1,5}-\alpha_{1,3})e_6, \\
%\varphi_{e_1}(e_4)=\frac{3}{2}\alpha_{1,2}e_6, \\
%\varphi_{e_1}(e_7)=-\alpha_{1,2}e_3-\alpha_{1,3}e_4-\alpha_{1,4}e_5+\gamma_{1,6}e_6, \\
%\varphi_{e_2}(e_2)=\beta_{2,5}e_5+\beta_{2,6}e_6,\\
%\varphi_{e_2}(e_3)=\frac{1}{2}\beta_{2,5}e_6, \\
%\varphi_{e_2}(e_7)=-4\alpha_{1,2}e_5+\gamma_{2,6}e_6,\\
%\varphi_{e_3}(e_7)=-\frac32\alpha_{1,2}e_6, \\
%\varphi_{e_7}(e_7)=\alpha_{1,2}e_4+\alpha_{1,3}e_5 +\gamma_{7,6}e_6. \\
%\end{cases}$$
%

Thus, we have the following table of commutative multiplications of the transposed Poisson algebra structure $(g^3_{(7,1)},, \cdot, [-,-])$:

$$\begin{cases}
e_1\cdot e_1=2\alpha_{2}e_2+2\alpha_{3}e_3+ \alpha_{4}e_4+\alpha_{5}e_5, \ e_1\cdot e_2=8\alpha_{2}e_4+2\alpha_{6}e_5 +\alpha_{7}e_6,\\[1mm]
e_1\cdot e_3=4\alpha_{2}e_5+(\alpha_{6}-\alpha_{3})e_6, \ e_1\cdot e_4=3\alpha_{2}e_6, \ e_1\cdot e_7=-2\alpha_{2}e_3-2\alpha_{3}e_4-\alpha_{4}e_5+\alpha_{8}e_6, \\[1mm]
e_2\cdot e_2=2\alpha_{1}e_5+\alpha_{9}e_6,\ e_2\cdot e_3=\alpha_{1}e_6, \ e_2\cdot e_7=-8\alpha_{2}e_5+\alpha_{10}e_6,\\[1mm]
e_3\cdot e_7=-3\alpha_{2}e_6, \ e_7\cdot e_7=2\alpha_{2}e_4+2\alpha_{3}e_5 +\alpha_{11}e_6.
\end{cases}$$

Considering the associative identity  for the triples $\{e_1,e_1,e_7\}$ and $\{e_1,e_1,e_2\}$, we obtain the following restrictions, respectively:
$$\alpha_2=0, \ \alpha_{1}\alpha_{3}=0.$$

Using  multiplication of Lie algebra $g^3_{(7,1)}$ and multiplication $e_1\cdot e_1=2\alpha_{3}e_3+ \alpha_{4}e_4+\alpha_{5}e_5$, and considering a general change of basis, we can show that $\alpha_3$ is an invariant.
%
%Agar bu algebrada quyidagicha umumiy almashtirish bajarsak
%$$e_1'=\sum\limits_{t=1}^{7}A_te_t, \quad e_2'=\sum\limits_{t=1}^{7}B_te_t, \quad e_7'=\sum\limits_{t=1}^{7}C_te_t,$$
%quyidagi tenglikni olamiz.
%$$\alpha_3'=\frac{A_1}{B_2}\alpha_3.$$
%

If $\alpha_{3}=0.$ Then, we have the algebra ${\bf TP}_1(g^3_{(7,1)}).$

If $\alpha_{3}\neq 0.$ Then, we obtain $\alpha_{1}=0$. By choosing change of basis, we can assume that $\alpha_3=1$ and obtain the algebra ${\bf TP}_2(g^3_{(7,1)}).$
\end{proof}


\begin{theorem}\label{thm3klas}
Let $(g^3_{(n,1)}, \cdot, [-,-])$ be a transposed Poisson algebra structure defined on Lie algebra $g^3_{(n,1)}$ and $n\geq 8$. Then, the multiplication of $(g^3_{(n,1)}, \cdot)$ has the following form:

${\bf TP}(g^3_{(8,1)}):\ \begin{cases}
e_1\cdot e_1=\sum\limits_{t=4}^{7} \alpha_{t}e_t, \ e_1\cdot e_2=\beta_{1}e_{6}+\beta_{2}e_{7}, \ e_1\cdot e_3=\frac{1}{2}(\beta_{1}-\alpha_{4})e_{7}, \\[1mm]
e_1\cdot e_8=-\alpha_{4}e_5-\alpha_{5}e_6 +\gamma_{1}e_{7},\ e_2\cdot e_2=\beta_{3}e_{6}+\beta_{4}e_{7}, \\[1mm]
e_2\cdot e_3=\frac{1}{2}\beta_{3}e_{7}, \ e_2\cdot e_8=\gamma_{2}e_{7},\ e_8\cdot e_8=\alpha_{4}e_6+\gamma_{3}e_{7};\end{cases}$

${\bf TP}_1(g^3_{(9,1)}):\ \begin{cases}
e_1\cdot e_1=\sum\limits_{t=4}^{8} \alpha_{t}e_t, \ e_1\cdot e_2=\beta_{1}e_{7}+\beta_{2}e_{8}, \ e_1\cdot e_3=\frac{1}{2}(\beta_{1}-\alpha_{5})e_{8}, \\[1mm]
e_1\cdot e_9=-\sum\limits_{t=5}^{7}\alpha_{t-1}e_t +\gamma_{1}e_{8},\ e_2\cdot e_2=\beta_{3}e_{7}+\beta_{4}e_{8}, \ e_2\cdot e_3=\frac{1}{2}\beta_{3}e_{8}, \\[1mm]
e_2\cdot e_9=-\alpha_{4}e_7+\gamma_{2}e_{8},\ e_3\cdot e_9=\frac12\alpha_{4}e_{8}, \ e_9\cdot e_9=\sum\limits_{t=6}^{7}\alpha_{t-2}e_t+\gamma_{3}e_{8};\end{cases}$

${\bf TP}_2(g^3_{(9,1)}):\ \begin{cases}
e_1\cdot e_1=e_3+\sum\limits_{t=5}^{8} \alpha_{t}e_t, \ e_1\cdot e_2=e_5+\beta_{1}e_{7}+ \beta_{2}e_{8}, \\[1mm]
e_1\cdot e_3=\frac{1}{2}(\beta_{1}-\alpha_{5})e_{8}, \ e_1\cdot e_9=-e_4-\sum\limits_{t=6}^{7}\alpha_{t-1}e_t +\gamma_{1}e_{8},\\[1mm]
e_2\cdot e_2=e_{7}+\beta_{4}e_{8}, \ e_2\cdot e_9=-e_6+\gamma_{2}e_{8},\ e_9\cdot e_9=e_5+\alpha_{5}e_7 +\gamma_{3}e_{8};
\end{cases}$


${\bf TP}_1(g^3_{(n,1)}):\ \begin{cases}
e_1\cdot e_1=\sum\limits_{t=4}^{n-1} \alpha_{t}e_t, \ e_1\cdot e_2=\sum\limits_{t=6}^{n-3}\alpha_{t-2}e_t +\beta_{1}e_{n-2}+\beta_{2}e_{n-1}, \\[1mm]
e_1\cdot e_3=\frac{1}{2}(\beta_{1}-\alpha_{n-4})e_{n-1}, \ e_1\cdot e_n=-\sum\limits_{t=5}^{n-2}\alpha_{t-1}e_t +\gamma_{1}e_{n-1},\\[1mm]
e_2\cdot e_2=\sum\limits_{t=8}^{n-3}\alpha_{t-4}e_t +\beta_{3}e_{n-2}+\beta_{4}e_{n-1}, \ e_2\cdot e_3=\frac{1}{2}(\beta_{3}-\alpha_{n-6})e_{n-1}, \\[1mm]
e_2\cdot e_n=-\sum\limits_{t=7}^{n-2}\alpha_{t-3}e_t +\gamma_{2}e_{n-1},\ e_3\cdot e_n=\frac12\alpha_{n-5}e_{n-1}, \
e_n\cdot e_n=\sum\limits_{t=6}^{n-2}\alpha_{t-2}e_t+\gamma_{3}e_{n-1}; \end{cases}$



${\bf TP}_2(g^3_{(n,1)}):\ \begin{cases}
e_1\cdot e_1=e_3+\sum\limits_{t=4}^{n-6} \alpha_{t}e_t+\sum\limits_{t=n-4}^{n-1} \alpha_{t}e_t, \ e_1\cdot e_2=e_5+\sum\limits_{t=5}^{n-4}\alpha_{t-2}e_t +\beta_{1}e_{n-2}+\beta_{2}e_{n-1}, \\[1mm]
e_1\cdot e_3=\frac{1}{2}(\beta_{1}-\alpha_{n-4})e_{n-1}, \ e_1\cdot e_n=-e_4-\sum\limits_{t=5}^{n-5}\alpha_{t-1}e_t-\sum\limits_{t=n-3}^{n-2}\alpha_{t-1}e_t +\gamma_{1}e_{n-1},\\[1mm]
e_2\cdot e_2=e_7+\sum\limits_{t=8}^{n-2}\alpha_{t-4}e_t+\beta_{4}e_{n-1}, \ e_2\cdot e_n=-e_6-\sum\limits_{t=7}^{n-3}\alpha_{t-3}e_t +\gamma_{2}e_{n-1},\\[1mm]
e_n\cdot e_n=e_5+\sum\limits_{t=6}^{n-4}\alpha_{t-2}e_t+\alpha_{n-4}e_{n-2}+\gamma_{3}e_{n-1};\end{cases}$\\
where the transposed Poisson algebra has its products with respect to the bracket $[-,-]$, and remaining products are equal to zero.
\end{theorem}

\begin{proof}
Let $(g^3_{(n,1)}, \cdot, [-,-])$ be a transposed Poisson algebra structure defined on Lie algebra $g^3_{(n,1)}$. Then
for any element $x \in g^3_{(n,1)} $ the operator of multiplication $\varphi_x(y) = x \cdot y$ is a $\frac12$ -derivation. Hence, for $1 \leq i \leq n$ we derive

$$\begin{cases}
\varphi_{e_i}(e_1)=\alpha_{i,1}e_1+\sum\limits_{t=3}^{n-1}\alpha_{i,t}e_t, \\
\varphi_{e_i}(e_2)=\alpha_{i,1}e_2+\sum\limits_{t=5}^{n-3}\alpha_{i,t-2}e_t+\beta_{i,n-2}e_{n-2}+\beta_{i,n-1}e_{n-1}, \\
\varphi_{e_i}(e_3)=\alpha_{i,1}e_3+\frac{1}{2}(\beta_{i,n-2}-\alpha_{i,n-4})e_{n-1}, \\
\varphi_{e_i}(e_j)=\alpha_{i,1}e_j, \  4\leq j\leq n-1,\\
\varphi_{e_i}(e_n)=-\sum\limits_{t=4}^{n-2}\alpha_{i,t-1}e_t+\gamma_{i,n-1}e_{n-1}+\alpha_{i,1}e_n.
 \end{cases}$$

It is known that $\varphi_{e_i}(e_j)=e_i\cdot e_j=e_j\cdot e_i=\varphi_{e_j}(e_i).$ Then for some pairs of elements, we derive the following restrictions:

\begin{longtable}{lllll}
    $\{e_1, e_2\},$ & $\Rightarrow$ &$\alpha_{1,1}=\alpha_{2,1}=\alpha_{2,3}=\alpha_{2,4}=0, \ \alpha_{2,t}=\alpha_{1,t-2},\  5\leq t\leq n-3, $\\
    & &$\alpha_{2,n-2}=\beta_{1,n-2}, \ \alpha_{2,n-1}=\beta_{1,n-1},$ \\
    $\{e_1, e_3\},$ & $\Rightarrow$ & $\alpha_{3,1}=0, \ \alpha_{3,t}=0, \ 3\leq t\leq n-2,\   \alpha_{3,n-1}=\frac{1}{2}(\beta_{1,n-2}-\alpha_{1,n-4}), $\\
    $ \{e_1, e_j\}, $ & $\Rightarrow $& $\alpha_{j,1}=0, \ \alpha_{j,t}=0, \  3\leq t\leq n-1, $ \\
   $4\leq j\leq n-1,$ & & \\
   $ \{e_1, e_n\},$ & $\Rightarrow$ & $\alpha_{n,1}=\alpha_{n,3}=0, \  \alpha_{n,t}=-\alpha_{1,t-1}, \ 4\leq t\leq n-2, \  \alpha_{n,n-1}=\gamma_{1,n-1}, $ \\
 $\{e_2, e_3\},$ & $\Rightarrow$ & $\beta_{3,n-2}=0, \ \beta_{3,n-1}=\frac{1}{2}(\beta_{2,n-2}-\alpha_{1,n-6}), $ \\
 $\{e_2, e_j\},$ & $\Rightarrow$ & $\beta_{j,n-2}=\beta_{j,n-1}=0,$ \\
 $4\leq j\leq n-1,$& &\\
 $\{e_2, e_n\},$ & $\Rightarrow$ &$ \beta_{n,n-2}=-\alpha_{1,n-5}, \  \beta_{n,n-1}=\gamma_{2,n-1},$\\
 $ \{e_3, e_n\},$ & $\Rightarrow$ & $\gamma_{3,n-1}=\frac{1}{2}\alpha_{1,n-5}, $ \\
$   \{e_i, e_n\},$ & $\Rightarrow $& $\gamma_{i,n-1}=0, $\\
 $4\leq i\leq n-1.$
\end{longtable}

%
%For $i=1, \ j=2$ we have
%$$\alpha_{1,1}e_2+\sum\limits_{t=5}^{n-3}\alpha_{1,t-2}e_t+\beta_{1,n-2}e_{n-2}+\beta_{1,n-1}e_{n-1}= \alpha_{2,1}e_1+\sum\limits_{t=3}^{n-1}\alpha_{2,t}e_t.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{1,1}=\alpha_{2,1}=\alpha_{2,3}=\alpha_{2,4}=0, \ \alpha_{2,t}=\alpha_{1,t-2}, 5\leq t\leq n-3, \ \alpha_{2,n-2}=\beta_{1,n-2}, \ \alpha_{2,n-1}=\beta_{1,n-1}.$$
%
%For $i=1, \ j=3$ we have
%$$\frac{1}{2}(\beta_{1,n-2}-\alpha_{1,n-4})e_{n-1}= \alpha_{3,1}e_1+\sum\limits_{t=3}^{n-1}\alpha_{3,t}e_t.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{3,1}=0, \ \alpha_{3,t}=0, \ 3\leq t\leq n-2,  \alpha_{3,n-1}=\frac{1}{2}(\beta_{1,n-2}-\alpha_{1,n-4}).$$
%
%
%For $i=1, \ 4\leq  j\leq n-1$ we have
%$$0=\alpha_{j,1}e_1+\sum\limits_{t=3}^{n-1} \alpha_{j,t}e_t.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{j,1}=0, \ \alpha_{j,t}=0, \  3\leq t\leq n-1.$$
%
%For $i=1, \ j=n$ we have
%$$-\sum\limits_{t=4}^{n-2}\alpha_{1,t-1}e_t+
%\gamma_{1,n-1}e_{n-1}= \alpha_{n,1}e_1+ \sum\limits_{t=3}^{n-1} \alpha_{n,t}e_t.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{n,1}=\alpha_{n,3}=0, \  \alpha_{n,t}=-\alpha_{1,t-1}, \ 4\leq t\leq n-2, \  \alpha_{n,n-1}=\gamma_{1,n-1}.$$
%
%
%For $i=2, \ j=3$ we have
%$$\frac{1}{2}(\beta_{2,n-2}-\alpha_{1,n-6})e_{n-1}= \beta_{3,n-2}e_{n-2}+\beta_{3,n-1}e_{n-1}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{3,n-2}=0, \ \beta_{3,n-1}=\frac{1}{2}(\beta_{2,n-2}-\alpha_{1,n-6}).$$
%
%For $i=2, \ 4\leq  j\leq n-1$ we have
%$$0=\beta_{j,n-2}e_{n-2}+\beta_{j,n-1}e_{n-1}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{j,n-2}=\beta_{j,n-1}=0.$$
%
%For $i=2, \ j=n$ we have
%$$-\sum\limits_{t=6}^{n-2}\alpha_{1,t-3}e_t+\gamma_{2,n-1}e_{n-1}=-\sum\limits_{t=6}^{n-3}\alpha_{1,t-3}e_t +\beta_{n,n-2}e_{n-2}+\beta_{n,n-1}e_{n-1}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{n,n-2}=-\alpha_{1,n-5}, \  \beta_{n,n-1}=\gamma_{2,n-1}.$$
%
%For $i=3, \ j=n$ we have
%$$\gamma_{3,n-1}e_{n-1}=\frac{1}{2}\alpha_{1,n-5}e_{n-1}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\gamma_{3,n-1}=\frac{1}{2}\alpha_{1,n-5}.$$
%
%For $4\leq i\leq n-1, \ j=n$ we have $\gamma_{i,n-1}=0.$
%
%
%$$\begin{cases} \varphi_{e_1}(e_1)=\sum\limits_{t=3}^{n-1} \alpha_{1,t}e_t, \\ \varphi_{e_1}(e_2)=\sum\limits_{t=5}^{n-3}\alpha_{1,t-2}e_t+\beta_{1,n-2}e_{n-2}+\beta_{1,n-1}e_{n-1}, \\ \varphi_{e_1}(e_3)=\frac{1}{2}(\beta_{1,n-2}-\alpha_{1,n-4})e_{n-1}, \\ \varphi_{e_1}(e_n)=-\sum\limits_{t=4}^{n-2}\alpha_{1,t-1}e_t+\gamma_{1,n-1}e_{n-1},\\ \varphi_{e_2}(e_2)=\sum\limits_{t=7}^{n-3}\alpha_{1,t-4}e_t+\beta_{2,n-2}e_{n-2}+\beta_{2,n-1}e_{n-1}, \\ \varphi_{e_2}(e_3)=\frac{1}{2}(\beta_{2,n-2}-\alpha_{1,n-6})e_{n-1}, \\ \varphi_{e_2}(e_n)=-\sum\limits_{t=6}^{n-2}\alpha_{1,t-3}e_t+\gamma_{2,n-1}e_{n-1},\\ \varphi_{e_3}(e_n)=\frac12\alpha_{1,n-5}e_{n-1}.\\ \varphi_{e_n}(e_n)=\sum\limits_{t=5}^{n-2}\alpha_{1,t-2}e_t+\gamma_{n,n-1}e_{n-1}.\\  \end{cases}$$

So, we have the following table of commutative multiplications of the transposed Poisson algebra structure $(g^3_{(n,1)},, \cdot, [-,-])$

For $n=8:$

$$\begin{cases}
e_1\cdot e_1=\sum\limits_{t=3}^{7} \alpha_{t}e_t, \ e_1\cdot e_2=\alpha_{3}e_5+ \beta_{1}e_{6}+\beta_{2}e_{7}, \ e_1\cdot e_3=\frac{1}{2}(\beta_{1}-\alpha_{4})e_{7}, \\[1mm]
e_1\cdot e_8=-\sum\limits_{t=4}^{6}\alpha_{t-1}e_t +\gamma_{1}e_{7},\ e_2\cdot e_2=\beta_{3}e_{6}+\beta_{4}e_{7}, \ e_2\cdot e_3=\frac{1}{2}\beta_{3}e_{7}, \\[1mm]
e_2\cdot e_8=-\alpha_{3}e_6 +\gamma_{2}e_{7},\ e_3\cdot e_8=\frac12\alpha_{3}e_{7},\ e_8\cdot e_8=\sum\limits_{t=5}^{6}\alpha_{t-2}e_t+\gamma_{3}e_{7};
\end{cases} $$

For $n\geq9:$
$$\begin{cases}
e_1\cdot e_1=\sum\limits_{t=3}^{n-1} \alpha_{t}e_t, \ e_1\cdot e_2=\sum\limits_{t=5}^{n-3}\alpha_{t-2}e_t +\beta_{1}e_{n-2}+\beta_{2}e_{n-1}, \ e_1\cdot e_3=\frac{1}{2}(\beta_{1}-\alpha_{n-4})e_{n-1}, \\[1mm]
e_1\cdot e_n=-\sum\limits_{t=4}^{n-2}\alpha_{t-1}e_t +\gamma_{1}e_{n-1},\ e_2\cdot e_2=\sum\limits_{t=7}^{n-3}\alpha_{t-4}e_t +\beta_{3}e_{n-2}+\beta_{4}e_{n-1}, \\[1mm]
e_2\cdot e_3=\frac{1}{2}(\beta_{3}-\alpha_{n-6})e_{n-1}, \ e_2\cdot e_n=-\sum\limits_{t=6}^{n-2}\alpha_{t-3}e_t +\gamma_{2}e_{n-1},\\
e_3\cdot e_n=\frac12\alpha_{n-5}e_{n-1}, \ e_n\cdot e_n=\sum\limits_{t=5}^{n-2}\alpha_{t-2}e_t+\gamma_{3}e_{n-1}.\\
\end{cases}
$$

Considering the associative identity  for the triples $\{e_1,e_1,e_2\}$ and $\{e_1,e_1,e_n\}$, we obtain the following restrictions, respectively:

\begin{equation}\label{gn3}\alpha_{3}(\beta_{3}-\alpha_{n-6})=0, \ \alpha_{3}\alpha_{n-5}=0.\end{equation}

We have the following cases.

\begin{enumerate}
    \item Let $n=8.$ Then we get $\alpha_3=0$ and we have the algebra ${\bf TP}(g^3_{(8,1)})$.

    \item Let $n\geq 9$. Then by using the multiplication of the Lie algebra $g^3_{(n,1)}$ we consider the general basis change:
$$e_1'=\sum\limits_{t=1}^{n}A_te_t, \ e_2'=\sum\limits_{t=1}^{n}B_te_t, \ e_n'=\sum\limits_{t=1}^{n}C_te_t.$$
The product $e_1'\cdot e_1'=\sum\limits_{i=3}^{n-1}\alpha_{i}'e_i'$ gives
$$\alpha_3'=\frac{A_1}{B_2}\alpha_3.$$
\begin{enumerate}
        \item If $\alpha_{3}=0.$ Then for $n=9$ we derive the algebra ${\bf TP}_1(g^3_{(9,1)})$ and for the case when $n\geq 10$ we obtain ${\bf TP}_1(g^3_{(n,1)})$.
        \item If $\alpha_{3}\neq 0.$ Then  by choosing $B_2=A_1\alpha_3$ we can assume $\alpha_{3}'=1$ and from the restrictions \eqref{gn3} we get $\alpha_{n-5}=0, \beta_3'=\alpha_{n-6}$. In this case for the $n=9$ we obtain ${\bf TP}_2(g^3_{(9,1)})$ and for $n\geq 10$ we have the algebra ${\bf TP}_2(g^3_{(n,1)})$.
    \end{enumerate}
\end{enumerate}

\end{proof}



\begin{theorem}\label{thm4klas} Let $(g^1_{7}, \cdot, [-,-])$ be a transposed Poisson algebra structure defined on Lie algebra
$g^1_{7}$. Then, the multiplication of $(g^1_{7}, \cdot)$ has the following form:

${\bf TP}_1(g^1_{7}): \ \begin{cases}
e_1\cdot e_1=\alpha_{4}e_4 +\alpha_{5}e_5 +\alpha_{6}e_6+\alpha_{7}e_7, \ e_1\cdot e_2=\alpha_{1}e_5+\alpha_{2}e_6+\alpha_{8}e_7,\\[1mm]
e_1\cdot e_3=\frac{1}{2}(\alpha_{1}-\alpha_{4})e_6-\frac{1}{2}\alpha_{5}e_7, \ e_1\cdot e_4=\frac{1}{2}\alpha_{4}e_7, \\[1mm]
e_2\cdot e_2=\alpha_{9}e_5+\alpha_{10}e_6+ \alpha_{11}e_7,\ e_2\cdot e_3=\frac{1}{2}\alpha_{9}e_6-\frac{1}{2}\alpha_{1}e_7;
\end{cases}$

${\bf TP}_2(g^1_{7}): \ \begin{cases}
e_1\cdot e_1=6e_3+\alpha_{4}e_4 +\alpha_{5}e_5+\alpha_{6}e_6+\alpha_{7}e_7, \ e_1\cdot e_2=-2e_4+ \alpha_{1}e_5+ \alpha_{2}e_6+\alpha_{8}e_7,\\[1mm]
e_1\cdot e_3=-4e_5+\frac{1}{2}(\alpha_{1}-\alpha_{4})e_6-\frac{1}{2}\alpha_{5}e_7, \ e_1\cdot e_4=-2e_6+ \frac{1}{2}\alpha_{4}e_7, \ e_1\cdot e_5=-3e_7, \\[1mm]
e_2\cdot e_2=-\frac23e_5+\alpha_{10}e_6+ \alpha_{11}e_7,\ e_2\cdot e_3=\frac{2}{3}e_6-\frac{1}{2}\alpha_{1}e_7, \  e_2\cdot e_4=-e_7, \  e_3\cdot e_3=2e_7;
\end{cases}$\\
where the transposed Poisson algebra has its products with respect to the bracket $[-,-]$, and remaining products are equal to zero.
\end{theorem}

\begin{proof}
Let $(g^1_{7}, \cdot, [-,-])$ be a transposed Poisson algebra structure defined on the Lie algebra $g^3_{(n,1)}$. Then
for any element of $x \in g^1_{7} $, the operator of multiplication $\varphi_x(y) = x \cdot y$ is a $\frac12$-derivation. Hence, for $1 \leq i\leq 7$ by using Theorem \ref{halfderiv4} we derive

 $$\begin{cases}
\varphi_{e_i}(e_1)=\alpha_{i,1}e_1+\alpha_{i,3}e_3+ \alpha_{i,4}e_4+\alpha_{i,5}e_5+\alpha_{i,6}e_6+\alpha_{i,7}e_7, \\
\varphi_{e_i}(e_2)=\alpha_{i,1}e_2-\frac{1}{3}\alpha_{i,3}e_4+ \beta_{i,5}e_5+\beta_{i,6}e_6+\beta_{i,7}e_7,\\
\varphi_{e_i}(e_3)=\alpha_{i,1}e_3-\frac{2}{3}\alpha_{i,3}e_5+ \frac{1}{2}(\beta_{i,5}-\alpha_{i,4})e_6-\frac{1}{2}\alpha_{i,5}e_7, \\
\varphi_{e_i}(e_4)=\alpha_{i,1}e_4-\frac{1}{3}\alpha_{i,3}e_6+ \frac{1}{2}\alpha_{i,4}e_7, \ \varphi_{e_i}(e_5)=\alpha_{i,1}e_5-\frac{1}{2}\alpha_{i,3}e_7, \\ \varphi_{e_i}(e_6)=\alpha_{i,1}e_6, \ \varphi_{e_i}(e_7)=\alpha_{i,1}e_7.
  \end{cases}$$

By considering equality  $\varphi_{e_i}(e_j)=e_i\cdot e_j=e_j\cdot e_i=\varphi_{e_j}(e_i)$ for the some elements ${e_i}$ and $e_j$, we have the following restrictions:

 $$\begin{array}{lll}
    \{e_1, e_2\}, & \Rightarrow & \alpha_{1,1}=\alpha_{2,1}=\alpha_{2,3}=0, \ \alpha_{2,4}=-\frac{1}{3}\alpha_{1,3}, \ \alpha_{2,5}=\beta_{1,5}, \ \alpha_{2,6}=\beta_{1,6}, \ \alpha_{2,7}=\beta_{1,7},\\[1mm]
    \{e_1, e_3\}, & \Rightarrow & \alpha_{3,1}=\alpha_{3,3}=\alpha_{3,4}=0, \  \alpha_{3,5}=-\frac{2}{3}\alpha_{1,3}, \ \alpha_{3,6}=\frac{1}{2}(\beta_{1,5}-\alpha_{1,4}), \ \alpha_{3,7}=-\frac{1}{2}\alpha_{1,5},  \\[1mm]
    \{e_1, e_4\}, & \Rightarrow & \alpha_{4,1}=\alpha_{4,3}=\alpha_{4,4}=\alpha_{4,5}=0, \  \alpha_{4,6}=-\frac{1}{3}\alpha_{1,3}, \ \alpha_{4,7}=\frac{1}{2}\alpha_{1,4},  \\[1mm]
    \{e_1, e_5\}, & \Rightarrow & \alpha_{5,1}=\alpha_{5,3}=\alpha_{5,4}=\alpha_{5,5}=\alpha_{5,6}=0, \ \alpha_{5,7}=-\frac{1}{2}\alpha_{1,3},  \\[1mm]
    \{e_1, e_6\}, & \Rightarrow & \alpha_{6,1}=\alpha_{6,3}=\alpha_{6,4}=\alpha_{6,5}=\alpha_{6,6}=\alpha_{6,7}=0,  \\[1mm]
   \{e_1, e_7\}, & \Rightarrow & \alpha_{7,1}=\alpha_{7,3}=\alpha_{7,4}=\alpha_{7,5}=\alpha_{7,6}=\alpha_{7,7}=0,  \\[1mm]
\{e_2, e_3\}, & \Rightarrow & \beta_{3,5}=0, \ \beta_{3,6}=\frac{1}{6}(3\beta_{2,5}+\alpha_{1,3}), \ \beta_{3,7}=-\frac{1}{2}\beta_{1,5},  \\[1mm]
 \{e_2, e_4\}, & \Rightarrow & \beta_{4,5}=\beta_{4,6}=0, \ \beta_{4,7}=-\frac{1}{6}\alpha_{1,3},  \\[1mm]
 \{e_2, e_5\}, & \Rightarrow & \beta_{5,5}=\beta_{5,6}=\beta_{5,7}=0,\\[1mm]
 \{e_2, e_6\}, & \Rightarrow & \beta_{6,5}=\beta_{6,6}=\beta_{6,7}=0,\\[1mm]
 \{e_2, e_7\}, & \Rightarrow & \beta_{7,5}=\beta_{7,6}=\beta_{7,7}=0.    \end{array}$$
%
%For $i=1, \ j=2$ we have
%$$\alpha_{1,1}e_2-\frac{1}{3}\alpha_{1,3}e_4+ \beta_{1,5}e_5+\beta_{1,6}e_6+\beta_{1,7}e_7=\alpha_{2,1}e_1+\alpha_{2,3}e_3+ \alpha_{2,4}e_4+ \alpha_{2,5}e_5+ \alpha_{2,6}e_6+\alpha_{2,7}e_7.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{1,1}=0, \ \alpha_{2,1}=\alpha_{2,3}=0, \ \alpha_{2,4}=-\frac{1}{3}\alpha_{1,3}, \ \alpha_{2,5}=\beta_{1,5}, \ \alpha_{2,6}=\beta_{1,6}, \ \alpha_{2,7}=\beta_{1,7}.$$
%
%For $i=1, \ j=3$ we have
%$$-\frac{2}{3}\alpha_{1,3}e_5+ \frac{1}{2}(\beta_{1,5}-\alpha_{1,4})e_6-\frac{1}{2}\alpha_{1,5}e_7 =\alpha_{3,1}e_1+\alpha_{3,3}e_3+ \alpha_{3,4}e_4+ \alpha_{3,5}e_5+ \alpha_{3,6}e_6+\alpha_{3,7}e_7.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{3,1}=\alpha_{3,3}=\alpha_{3,4}=0, \  \alpha_{3,5}=-\frac{2}{3}\alpha_{1,3}, \ \alpha_{3,6}=\frac{1}{2}(\beta_{1,5}-\alpha_{1,4}), \ \alpha_{3,7}=-\frac{1}{2}\alpha_{1,5}.$$
%%
%For $i=1, \ j=4$ we have
%$$-\frac{1}{3}\alpha_{1,3}e_6+ \frac{1}{2}\alpha_{1,4}e_7 =\alpha_{4,1}e_1+\alpha_{4,3}e_3+ \alpha_{4,4}e_4+ \alpha_{4,5}e_5+ \alpha_{4,6}e_6+\alpha_{4,7}e_7.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{4,1}=\alpha_{4,3}=\alpha_{4,4}=\alpha_{4,5}=0, \  \alpha_{4,6}=-\frac{1}{3}\alpha_{1,3}, \ \alpha_{4,7}=\frac{1}{2}\alpha_{1,4}.$$
%
%For $i=1, \ j=5$ we have
%$$-\frac{1}{2}\alpha_{1,3}e_7=
%\alpha_{5,1}e_1+\alpha_{5,3}e_3+ \alpha_{5,4}e_4+ \alpha_{5,5}e_5+ \alpha_{5,6}e_6+\alpha_{5,7}e_7.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{5,1}=\alpha_{5,3}=\alpha_{5,4}=\alpha_{5,5}=
%\alpha_{5,6}=0, \ \alpha_{5,7}=-\frac{1}{2}\alpha_{1,3}.$$
%
%For $i=1, \ j=6$ we have
%$$0=\alpha_{6,1}e_1+\alpha_{6,3}e_3+ \alpha_{6,4}e_4+ \alpha_{6,5}e_5+ \alpha_{6,6}e_6+\alpha_{6,7}e_7.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{6,1}=\alpha_{6,3}=\alpha_{6,4}=\alpha_{6,5}=\alpha_{6,6}=\alpha_{6,7}=0.$$
%
%For $i=1, \ j=7$ we have
%$$0=\alpha_{7,1}e_1+\alpha_{7,3}e_3+ \alpha_{7,4}e_4+ \alpha_{7,5}e_5+ \alpha_{7,6}e_6+\alpha_{7,7}e_7.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{7,1}=\alpha_{7,3}=\alpha_{7,4}=\alpha_{7,5}=\alpha_{7,6}=\alpha_{7,7}=0.$$
%
%For $i=2, \ j=3$ we have
%$$\frac{1}{2}(\beta_{2,5}+\frac13\alpha_{1,3})e_6-\frac{1}{2}\beta_{1,5}e_7=\beta_{3,5}e_5+ \beta_{3,6}e_6 +\beta_{3,7}e_7.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{3,5}=0, \ \beta_{3,6}=\frac{1}{6}(3\beta_{2,5}+\alpha_{1,3}), \ \beta_{3,7}=-\frac{1}{2}\beta_{1,5}.$$
%
%For $i=2, \ j=4$ we have
%$$-\frac{1}{6}\alpha_{1,3}e_7=\beta_{4,5}e_5+ \beta_{4,6}e_6+\beta_{4,7}e_7.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{4,5}=\beta_{4,6}=0, \ \beta_{4,7}=-\frac{1}{6}\alpha_{1,3}.$$
%
%
%For $i=2, \ j=5$ we have
%$$0=\beta_{5,5}e_5+ \beta_{5,6}e_6+\beta_{5,7}e_7.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{5,5}=\beta_{5,6}=\beta_{5,7}=0.$$
%
%For $i=2, \ j=6$ we have
%$$0=\beta_{6,5}e_5+ \beta_{6,6}e_6+\beta_{6,7}e_7.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{6,5}=\beta_{6,6}=\beta_{6,7}=0.$$
%
%For $i=2, \ j=7$ we have
%$$0=\beta_{7,5}e_5+ \beta_{7,6}e_6+\beta_{7,7}e_7.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{7,5}=\beta_{7,6}=\beta_{7,7}=0.$$
%
%$$\begin{cases}
%\varphi_{e_1}(e_1)=\alpha_{1,3}e_3+ \alpha_{1,4}e_4 +\alpha_{1,5}e_5+\alpha_{1,6}e_6+\alpha_{1,7}e_7, \\
%\varphi_{e_1}(e_2)=-\frac{1}{3}\alpha_{1,3}e_4+ \beta_{1,5}e_5+\beta_{1,6}e_6+\beta_{1,7}e_7,\\
%\varphi_{e_1}(e_3)=-\frac{2}{3}\alpha_{1,3}e_5+ \frac{1}{2}(\beta_{1,5}-\alpha_{1,4})e_6-\frac{1}{2}\alpha_{1,5}e_7, \\
%\varphi_{e_1}(e_4)=-\frac{1}{3}\alpha_{1,3}e_6+ \frac{1}{2}\alpha_{1,4}e_7, \\
%\varphi_{e_1}(e_5)=-\frac{1}{2}\alpha_{1,3}e_7, \\
%\varphi_{e_2}(e_2)= \beta_{2,5}e_5+ \beta_{2,6}e_6+ \beta_{2,7}e_7,\\
%\varphi_{e_2}(e_3)=\frac{1}{6}(3\beta_{2,5}+ \alpha_{1,3})e_6-\frac{1}{2}\beta_{1,5}e_7, \\
%\varphi_{e_2}(e_4)=-\frac{1}{6}\alpha_{1,3}e_7, \\
%\varphi_{e_3}(e_3)=\frac{1}{3}\alpha_{1,3}e_7, \\
%\end{cases}$$

Thus, we have the following table of commutative multiplications of the transposed Poisson algebra structure $(g^1_{(7)},, \cdot, [-,-])$


$$\begin{cases}
e_1\cdot e_1=\alpha_{3}e_3+\alpha_{4}e_4 +\alpha_{5}e_5+\alpha_{6}e_6+\alpha_{7}e_7, \ e_1\cdot e_2=-\frac{1}{3}\alpha_{3}e_4+ \alpha_{1}e_5+\alpha_{2}e_6+\alpha_{8}e_7,\\
e_1\cdot e_3=-\frac{2}{3}\alpha_{3}e_5+ \frac{1}{2}(\alpha_{1}-\alpha_{4})e_6-\frac{1}{2}\alpha_{5}e_7, \ e_1\cdot e_4=-\frac{1}{3}\alpha_{3}e_6+ \frac{1}{2}\alpha_{4}e_7, \ e_1\cdot e_5=-\frac{1}{2}\alpha_{3}e_7, \\
e_2\cdot e_2=\alpha_{9}e_5+\alpha_{10}e_6+ \alpha_{11}e_7,\ e_2\cdot e_3=\frac{1}{6}(3\alpha_{9}+\alpha_{3})e_6-\frac{1}{2}\alpha_{1}e_7, \\
e_2\cdot e_4=-\frac{1}{6}\alpha_{3}e_7, \ e_3\cdot e_3=\frac{1}{3}\alpha_{3}e_7, \\
\end{cases}$$

The equality $e_1\cdot (e_1\cdot e_2)=(e_1\cdot e_1)\cdot e_2$ gives $\alpha _3 \left(\alpha _3+9 \alpha _9\right)=0.$ Using a general change of basis in a similar way as above, we can show that $\alpha_3$ is an invariant.

\begin{enumerate}
    \item If $\alpha_3=0,$ then we have the algebra ${\bf TP}_1(g^1_{7})$;
    \item If $\alpha_3\neq0,$ then we get the algebra ${\bf TP}_2(g^1_{7})$.
\end{enumerate}
\end{proof}


\begin{theorem}\label{thm4klas} Let $(g^2_{9}, \cdot, [-,-])$ be a transposed Poisson algebra structure defined on the Lie algebra
$g^2_{9}$. Then the multiplication of $(g^2_{9}, \cdot)$ has the following form:
$${\bf TP}(g^2_{9}):\ \begin{cases}
e_1\cdot e_1=\alpha_{1}e_5+ \alpha_{2}e_6+ \alpha_{3}e_7+\alpha_{4}e_8+\alpha_{5}e_9, \ e_1\cdot e_2=\frac{1}{3}\alpha_{1}e_6+\alpha_{6}e_7+ \alpha_{7}e_8+\alpha_{8}e_9,\\[1mm]
e_1\cdot e_3=-\frac{4}{3}\alpha_{1}e_7+ \frac{1}{2}(\alpha_{6}- 5\alpha_{2})e_8-\frac{1}{2}\alpha_{3}e_9, \ e_1\cdot e_4=\frac{1}{3}\alpha_{1}e_8+ \frac{1}{2}\alpha_{2}e_9, \\[1mm]
e_1\cdot e_5=-\frac{1}{2}\alpha_{1}e_9, \ e_2\cdot e_2=\alpha_{9}e_7+\alpha_{10}e_8+ \alpha_{11}e_9,\\[1mm]
e_2\cdot e_3=\frac{1}{6}(3\alpha_{9}- 5\alpha_{1})e_8-\frac{1}{2}\alpha_{6}e_9, \ e_2\cdot e_4=\frac{1}{6}\alpha_{1}e_9, \ e_3\cdot e_3=\frac{2}{3}\alpha_{1}e_9,\end{cases}$$
where the transposed Poisson algebra has its products with respect to the bracket $[-,-]$, and remaining products are equal to zero.
\end{theorem}

\begin{proof}
Let $(g^2_{9}, \cdot, [-,-])$ be a transposed Poisson algebra structure defined on Lie algebra $g^2_{9}$. Then
for any element $x \in g^2_{9} $, the operator of multiplication $\varphi_x(y) = x \cdot y$ is a $\frac12$-derivation. Hence, for $1 \leq i\leq 9$ we derive

$$\begin{cases}
\varphi_{e_i}(e_1)=\alpha_{i,1}e_1+\alpha_{i,5}e_5+ \alpha_{i,6}e_6+\alpha_{i,7}e_7+\alpha_{i,8}e_8+\alpha_{i,9}e_9, \\
\varphi_{e_i}(e_2)=\alpha_{i,1}e_2+\frac{1}{3}\alpha_{i,5}e_6+ \beta_{i,7}e_7+ \beta_{i,8}e_8+\beta_{i,9}e_9,\\
\varphi_{e_i}(e_3)=\alpha_{i,1}e_3-\frac{4}{3}\alpha_{i,5}e_7+ \frac{1}{2}(\beta_{i,7}- 5\alpha_{i,6})e_8-\frac{1}{2}\alpha_{i,7}e_9, \\
\varphi_{e_i}(e_4)=\alpha_{i,1}e_4+\frac{1}{3}\alpha_{i,5}e_8+ \frac{1}{2}\alpha_{i,6}e_9, \ \varphi_{e_i}(e_5)=\alpha_{i,1}e_5-\frac{1}{2}\alpha_{i,5}e_9, \ \varphi_{e_i}(e_j)=\alpha_{i,1}e_j, \ \ 6\leq j\leq 9. \\
\end{cases}$$


It is known that $\varphi_{e_i}(e_j)=e_i\cdot e_j=e_j\cdot e_i=\varphi_{e_j}(e_i).$ Then for some pairs of elements, we derive the following restrictions:

$$\begin{array}{lll}
    \{e_1, e_2\}, & \Rightarrow & \alpha_{1,1}=\alpha_{2,1}=\alpha_{2,5}=0, \ \alpha_{2,6}=\frac{1}{3}\alpha_{1,5}, \ \alpha_{2,7}=\beta_{1,7}, \ \alpha_{2,8}=\beta_{1,8}, \ \alpha_{2,9}=\beta_{1,9},\\[1mm]
    \{e_1, e_3\}, & \Rightarrow & \alpha_{3,1}=\alpha_{3,5}=\alpha_{3,6}=0, \  \alpha_{3,7}=-\frac{4}{3}\alpha_{1,5}, \ \alpha_{3,8}=\frac{1}{2}(\beta_{1,7}- 5\alpha_{1,6}), \ \alpha_{3,9}=-\frac{1}{2}\alpha_{1,7},  \\[1mm]
    \{e_1, e_4\}, & \Rightarrow & \alpha_{4,1}=\alpha_{4,5}=\alpha_{4,6}=\alpha_{4,7}=0, \  \alpha_{4,8}=\frac{1}{3}\alpha_{1,5}, \ \alpha_{4,9}=\frac{1}{2}\alpha_{1,6},  \\[1mm]
    \{e_1, e_5\}, & \Rightarrow & \alpha_{5,1}=\alpha_{5,5}=\alpha_{5,6}=\alpha_{5,7}=\alpha_{5,8}=0, \ \alpha_{5,9}=-\frac{1}{2}\alpha_{1,5},  \\[1mm]
    \{e_1, e_j\}, & \Rightarrow & \alpha_{j,1}=\alpha_{j,5}=\alpha_{j,6}=\alpha_{j,7}=\alpha_{j,8}=\alpha_{j,9}, \ 6\leq j\leq 9, \\[1mm]
 \{e_2, e_3\}, & \Rightarrow & \beta_{3,7}=0, \ \beta_{3,8}=\frac{1}{6}(3\beta_{2,7}- 5\alpha_{1,5}), \ \beta_{3,9}=-\frac{1}{2}\beta_{1,7},  \\[1mm]
 \{e_2, e_4\}, & \Rightarrow & \beta_{4,7}=\beta_{4,8}=0, \ \beta_{4,9}=\frac{1}{6}\alpha_{1,3},  \\[1mm]
 \{e_2, e_j\}, & \Rightarrow & \beta_{j,7}=\beta_{j,8}=\beta_{j,9}=0, \ 5\leq j\leq 9.\\[1mm]
   \end{array}$$
%
%For $i=1, \ j=2$ we have
%$$\alpha_{1,1}e_2+\frac{1}{3}\alpha_{1,5}e_6+ \beta_{1,7}e_7+\beta_{1,8}e_8+\beta_{1,9}e_9 =\alpha_{2,1}e_1+\alpha_{2,5}e_5+ \alpha_{2,6}e_6+\alpha_{2,7}e_7+\alpha_{2,8}e_8 +\alpha_{2,9}e_9.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{1,1}=0, \ \alpha_{2,1}=\alpha_{2,5}=0, \ \alpha_{2,6}=\frac{1}{3}\alpha_{1,5}, \ \alpha_{2,7}=\beta_{1,7}, \ \alpha_{2,8}=\beta_{1,8}, \ \alpha_{2,9}=\beta_{1,9}.$$
%
%For $i=1, \ j=3$ we have
%$$-\frac{4}{3}\alpha_{1,5}e_7+ \frac{1}{2}(\beta_{1,7}- 5\alpha_{1,6})e_8-\frac{1}{2}\alpha_{1,7}e_9 =\alpha_{3,1}e_1+\alpha_{3,5}e_5+ \alpha_{3,6}e_6+\alpha_{3,7}e_7+\alpha_{3,8}e_8+\alpha_{3,9}e_9.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{3,1}=\alpha_{3,5}=\alpha_{3,6}=0, \  \alpha_{3,7}=-\frac{4}{3}\alpha_{1,5}, \ \alpha_{3,8}=\frac{1}{2}(\beta_{1,7}- 5\alpha_{1,6}), \ \alpha_{3,9}=-\frac{1}{2}\alpha_{1,7}.$$
%
%For $i=1, \ j=4$ we have
%$$\frac{1}{3}\alpha_{1,5}e_8+\frac{1}{2}\alpha_{1,6}e_9 =\alpha_{4,1}e_1+\alpha_{4,5}e_5+ \alpha_{4,6}e_6+ \alpha_{4,7}e_7+\alpha_{4,8}e_8+\alpha_{4,9}e_9.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{4,1}=\alpha_{4,5}=\alpha_{4,6}=\alpha_{4,7}=0, \  \alpha_{4,8}=\frac{1}{3}\alpha_{1,5}, \ \alpha_{4,9}=\frac{1}{2}\alpha_{1,6}.$$
%
%For $i=1, \ j=5$ we have
%$$-\frac{1}{2}\alpha_{1,5}e_9=
%\alpha_{5,1}e_1+\alpha_{5,5}e_5+ \alpha_{5,6}e_6+ \alpha_{5,7}e_7+\alpha_{5,8}e_8+\alpha_{5,9}e_9.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{5,1}=\alpha_{5,5}=\alpha_{5,6}=\alpha_{5,7}=
%\alpha_{5,8}=0, \ \alpha_{5,9}=-\frac{1}{2}\alpha_{1,5}.$$
%
%For $i=1, \ 6\leq j\leq 9$ we have
%$$0=\alpha_{j,1}e_1+\alpha_{j,5}e_5+ \alpha_{j,6}e_6+ \alpha_{j,7}e_7+\alpha_{j,8}e_8+\alpha_{j,9}e_9.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{j,1}=\alpha_{j,5}=\alpha_{j,6}=
%\alpha_{j,7}=\alpha_{j,8}=\alpha_{j,9}=0.$$
%%
%For $i=2, \ j=3$ we have
%$$\frac{1}{6}(3\beta_{2,7}- 5\alpha_{1,5})e_8-\frac{1}{2}\beta_{1,7}e_9 =\beta_{3,7}e_7+ \beta_{3,8}e_8+ \beta_{3,9}e_9.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{3,7}=0, \ \beta_{3,8}=\frac{1}{6}(3\beta_{2,7}- 5\alpha_{1,5}), \ \beta_{3,9}=-\frac{1}{2}\beta_{1,7}.$$
%
%For $i=2, \ j=4$ we have
%$$\frac{1}{6}\alpha_{1,3}e_9=\beta_{4,7}e_7+ \beta_{4,8}e_8+\beta_{4,9}e_9.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{4,7}=\beta_{4,8}=0, \ \beta_{4,9}=\frac{1}{6}\alpha_{1,3}.$$
%
%For $i=2, \ 5\leq j \leq 9$ we have
%$$0=\beta_{j,7}e_7+ \beta_{j,8}e_8+\beta_{j,9}e_9..$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{j,7}=\beta_{j,8}=\beta_{j,9}=0.$$
%
%$$\begin{cases}
%\varphi_{e_1}(e_1)=\alpha_{1,5}e_5+ \alpha_{1,6}e_6+ \alpha_{1,7}e_7+\alpha_{1,8}e_8+\alpha_{1,9}e_9, \\
%\varphi_{e_1}(e_2)=\frac{1}{3}\alpha_{1,5}e_6+ \beta_{1,7}e_7+ \beta_{1,8}e_8+\beta_{1,9}e_9,\\
%\varphi_{e_1}(e_3)=-\frac{4}{3}\alpha_{1,5}e_7+ \frac{1}{2}(\beta_{1,7}- 5\alpha_{1,6})e_8-\frac{1}{2}\alpha_{1,7}e_9, \\
%\varphi_{e_1}(e_4)=\frac{1}{3}\alpha_{1,5}e_8+ \frac{1}{2}\alpha_{1,6}e_9, \\
%\varphi_{e_1}(e_5)=-\frac{1}{2}\alpha_{1,5}e_9, \\
%\varphi_{e_2}(e_2)=\beta_{2,7}e_7+ \beta_{2,8}e_8+ \beta_{2,9}e_9,\\
%\varphi_{e_2}(e_3)=\frac{1}{6}(3\beta_{2,7}- 5\alpha_{1,5})e_8-\frac{1}{2}\beta_{1,7}e_9, \\
%\varphi_{e_2}(e_4)=\frac{1}{6}\alpha_{1,5}e_9, \\
%\varphi_{e_3}(e_3)=\frac{2}{3}\alpha_{1,5}e_9, \\
%\end{cases}$$
%
%
%
%

So, we have the following table of multiplications of the transposed Poisson algebra structure $(g^2_{(9)},, \cdot, [-,-])$. Note that the identity of  associativity holds for any triple.

$$\begin{cases}
e_1\cdot e_1=\alpha_{1}e_5+ \alpha_{2}e_6+ \alpha_{3}e_7+\alpha_{4}e_8+\alpha_{5}e_9, \ e_1\cdot e_2=\frac{1}{3}\alpha_{1}e_6+\alpha_{6}e_7+ \alpha_{7}e_8+\alpha_{8}e_9,\\
e_1\cdot e_3=-\frac{4}{3}\alpha_{1}e_7+ \frac{1}{2}(\alpha_{6}- 5\alpha_{2})e_8-\frac{1}{2}\alpha_{3}e_9, \ e_1\cdot e_4=\frac{1}{3}\alpha_{1}e_8+ \frac{1}{2}\alpha_{2}e_9, \\
e_1\cdot e_5=-\frac{1}{2}\alpha_{1}e_9, \ e_2\cdot e_2=\alpha_{9}e_7+\alpha_{10}e_8+ \alpha_{11}e_9,\\
e_2\cdot e_3=\frac{1}{6}(3\alpha_{9}- 5\alpha_{1})e_8-\frac{1}{2}\alpha_{6}e_9, \ e_2\cdot e_4=\frac{1}{6}\alpha_{1}e_9, \ e_3\cdot e_3=\frac{2}{3}\alpha_{1}e_9.
\end{cases}$$
\end{proof}
\begin{theorem}\label{thm5klas} Let $(g^3_{11}, \cdot, [-,-])$ be a transposed Poisson algebra structure defined on the Lie algebra
$g^3_{11}$. Then the multiplication of $(g^3_{11}, \cdot)$ has the following form:
$${\bf TP}(g^3_{11}):\ \begin{cases}
e_1\cdot e_1=\alpha_{6}e_6+ \alpha_{7}e_7+ \alpha_{8}e_8+ \alpha_{9}e_9+ \alpha_{10}e_{10}+ \alpha_{11}e_{11}, \\
e_1\cdot e_2=-\alpha_{6}e_7-\alpha_{7}e_8+\alpha_{1}e_9+ \alpha_{2}e_{10}+\alpha_{3}e_{11},\ e_1\cdot e_3=\frac{1}{2}\alpha_{1}e_{10}-\frac{1}{2}\alpha_{9}e_{11}, \\ e_1\cdot e_4=\frac{1}{2}\alpha_{7}e_{10}+ \frac{1}{2}\alpha_{8}e_{11}, \ e_1\cdot e_5=-\frac{1}{2}\alpha_{6}e_{10}-\frac{1}{2}\alpha_{7}e_{11}, \ e_1\cdot e_6=\frac{1}{2}\alpha_{6}e_{11}, \\
e_2\cdot e_2=\alpha_{6}e_8+\alpha_{4}e_9+ \alpha_{5}e_{10}+\alpha_{12}e_{11},\ e_2\cdot e_3=\frac{1}{2}\alpha_{4}e_{10}-\frac{1}{2}\alpha_{1}e_{11}, \\
e_2\cdot e_4=-\frac{1}{2}\alpha_{6}e_{10}- \frac{1}{2}\alpha_{7}e_{11}, \ e_2\cdot e_5=\frac{1}{2}\alpha_{6}e_{11},
\end{cases}$$
where it is taken into account that the transposed Poisson algebra has its products with respect to the bracket $[-,-]$, and the remaining products are equal to zero.
\end{theorem}

\begin{proof}
Let $(g^3_{11}, \cdot, [-,-])$ be a transposed Poisson algebra structure defined on Lie algebra $g^3_{11}$. Then
for any element of $x \in g^3_{11} $ the operator of multiplication $\varphi_x(y) = x \cdot y$ is a $\frac12$-derivation. Further, using Theorem \ref{2halfderiv3} for $1 \leq i\leq 11$ we derive
$$\begin{cases}
\varphi_{e_i}(e_1)=\alpha_{i,1}e_1+\alpha_{i,6}e_6+ \alpha_{i,7}e_7+ \alpha_{i,8}e_8+ \alpha_{i,9}e_9+ \alpha_{i,10}e_{10}+\alpha_{i,11}e_{11}, \\

\varphi_{e_i}(e_2)=\alpha_{i,1}e_2-\alpha_{i,6}e_7-\alpha_{i,7}e_8+ \beta_{i,9}e_9+ \beta_{i,10}e_{10}+ \beta_{i,11}e_{11},\\

\varphi_{e_i}(e_3)=\alpha_{i,1}e_3+\frac{1}{2}\beta_{i,9}e_{10}-\frac{1}{2}\alpha_{i,9}e_{11}, \\

\varphi_{e_i}(e_4)=\alpha_{i,1}e_4+\frac{1}{2}\alpha_{i,7}e_{10}+ \frac{1}{2}\alpha_{i,8}e_{11}, \\

\varphi_{e_i}(e_5)=\alpha_{i,1}e_5-\frac{1}{2}\alpha_{i,6}e_{10}-\frac{1}{2}\alpha_{i,7}e_{11}, \\

\varphi_{e_i}(e_6)=\alpha_{i,1}e_6+\frac{1}{2}\alpha_{i,6}e_{11}, \\

\varphi_{e_i}(e_j)=\alpha_{i,1}e_j, \ \ 7\leq j\leq 11. \\
\end{cases}$$

Considering the property $\varphi_{e_i}(e_j)=e_i\cdot e_j=e_j\cdot e_i=\varphi_{e_j}(e_i)$, we obtain the following restrictions:

 $$\begin{array}{lll}
    \{e_1, e_2\}, & \Rightarrow & \alpha_{1,1}=\alpha_{2,1}=\alpha_{2,6}=0, \\[1mm]
     & &\alpha_{2,7}=-\alpha_{1,6}, \ \alpha_{2,8}=-\alpha_{1,7}, \ \alpha_{2,9}=\beta_{1,9}, \ \alpha_{2,10}=\beta_{1,10}, \ \alpha_{2,11}=\beta_{1,11},\\[1mm]
    \{e_1, e_3\}, & \Rightarrow & \alpha_{3,1}=\alpha_{3,6}=\alpha_{3,7}=\alpha_{3,8}=\alpha_{3,9}=0, \ \alpha_{3,10}=\frac{1}{2}\beta_{1,9}, \ \alpha_{3,11}=-\frac{1}{2}\alpha_{1,9},  \\[1mm]
    \{e_1, e_4\}, & \Rightarrow & \alpha_{4,1}=\alpha_{4,6}=\alpha_{4,7}=\alpha_{4,8}=\alpha_{4,9}=0, \ \alpha_{4,10}=\frac{1}{2}\alpha_{1,7}, \ \alpha_{4,11}=\frac{1}{2}\alpha_{1,8},  \\[1mm]
    \{e_1, e_5\}, & \Rightarrow & \alpha_{5,1}=\alpha_{5,6}=\alpha_{5,7}=\alpha_{5,8}=\alpha_{5,9}=0, \ \alpha_{5,10}=-\frac{1}{2}\alpha_{1,6}, \ \alpha_{5,11}=-\frac{1}{2}\alpha_{1,7},  \\[1mm]
   \{e_1, e_6\}, & \Rightarrow & \alpha_{6,1}=\alpha_{6,6}=\alpha_{6,7}= \alpha_{6,8}=\alpha_{6,9}=\alpha_{6,10}=0, \ \alpha_{6,11}=\frac{1}{2}\alpha_{1,6},  \\[1mm]
    \{e_1, e_j\}, & \Rightarrow & \alpha_{j,1}=\alpha_{j,6}=\alpha_{j,7}= \alpha_{j,8}=\alpha_{j,9}=\alpha_{j,10}=\alpha_{j,11}=0, \ 7\leq j\leq 9,  \\[1mm]

 \{e_2, e_3\}, & \Rightarrow & \beta_{3,9}=0, \ \beta_{3,10}=\frac{1}{2}\beta_{2,9}, \ \beta_{3,11}=-\frac{1}{2}\beta_{1,9},  \\[1mm]
 \{e_2, e_4\}, & \Rightarrow & \beta_{4,9}=0, \ \beta_{4,10}=-\frac{1}{2}\alpha_{1,6}, \ \beta_{4,11}=-\frac{1}{2}\alpha_{1,7},  \\[1mm]
 \{e_2, e_5\}, & \Rightarrow & \beta_{5,9}=\beta_{5,10}=0, \ \beta_{5,11}=\frac{1}{2}\alpha_{1,6},\\[1mm]
 \{e_2, e_j\}, & \Rightarrow & \beta_{j,9}=\beta_{j,10}=\beta_{j,11}=0, \ 6\leq j\leq 11.
    \end{array}$$
%
%For $i=1, \ j=2$ we have
%$$\alpha_{1,1}e_2-\alpha_{1,6}e_7-\alpha_{1,7}e_8+ \beta_{1,9}e_9+ \beta_{1,10}e_{10}+ \beta_{1,11}e_{11}$$
%$$ =\alpha_{2,1}e_1+\alpha_{2,6}e_6+ \alpha_{2,7}e_7+ \alpha_{2,8}e_8+ \alpha_{2,9}e_9+ \alpha_{2,10}e_{10}+\alpha_{2,11}e_{11}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{1,1}=0, \ \alpha_{2,1}=\alpha_{2,6}=0, \ \alpha_{2,7}=-\alpha_{1,6}, \ \alpha_{2,8}=-\alpha_{1,7}, \ \alpha_{2,9}=\beta_{1,9}, \ \alpha_{2,10}=\beta_{1,10}, \ \alpha_{2,11}=\beta_{1,11}.$$
%
%For $i=1, \ j=3$ we have
%$$\frac{1}{2}\beta_{1,9}e_{10}-\frac{1}{2}\alpha_{1,9}e_{11}=\alpha_{3,1}e_1+\alpha_{3,6}e_6+ \alpha_{3,7}e_7+ \alpha_{3,8}e_8+ \alpha_{3,9}e_9+ \alpha_{3,10}e_{10}+\alpha_{3,11}e_{11}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{3,1}=\alpha_{3,6}=\alpha_{3,7}= \alpha_{3,8}=\alpha_{3,9}=0, \ \alpha_{3,10}=\frac{1}{2}\beta_{1,9}, \ \alpha_{3,11}=-\frac{1}{2}\alpha_{1,9}.$$
%
%
%For $i=1, \ j=4$ we have
%$$\frac{1}{2}\alpha_{1,7}e_{10}+ \frac{1}{2}\alpha_{1,8}e_{11}=\alpha_{4,1}e_1+\alpha_{4,6}e_6+ \alpha_{4,7}e_7+ \alpha_{4,8}e_8+\alpha_{4,9}e_9+ \alpha_{4,10}e_{10}+\alpha_{4,11}e_{11}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{4,1}=\alpha_{4,6}=\alpha_{4,7}= \alpha_{4,8}=\alpha_{4,9}=0, \ \alpha_{4,10}=\frac{1}{2}\alpha_{1,7}, \ \alpha_{4,11}=\frac{1}{2}\alpha_{1,8}.$$
%%
%For $i=1, \ j=5$ we have
%$$-\frac{1}{2}\alpha_{1,6}e_{10}-\frac{1}{2}\alpha_{1,7}e_{11}=\alpha_{5,1}e_1+\alpha_{5,6}e_6+ \alpha_{5,7}e_7+ \alpha_{5,8}e_8+\alpha_{5,9}e_9+ \alpha_{5,10}e_{10}+\alpha_{5,11}e_{11}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{5,1}=\alpha_{5,6}=\alpha_{5,7}= \alpha_{5,8}=\alpha_{5,9}=0, \ \alpha_{5,10}=-\frac{1}{2}\alpha_{1,6}, \ \alpha_{5,11}=-\frac{1}{2}\alpha_{1,7}.$$
%
%
%For $i=1, \ j=6$ we have
%$$\frac{1}{2}\alpha_{1,6}e_{11}=\alpha_{6,1}e_1+ \alpha_{6,6}e_6+ \alpha_{6,7}e_7+ \alpha_{6,8}e_8+ \alpha_{6,9}e_9+\alpha_{6,10}e_{10}+ \alpha_{6,11}e_{11}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{6,1}=\alpha_{6,6}=\alpha_{6,7}= \alpha_{6,8}=\alpha_{6,9}=\alpha_{6,10}=0, \ \alpha_{6,11}=\frac{1}{2}\alpha_{1,6}.$$
%
%
%For $i=1, \ 7\leq j\leq 9$ we have
%$$0=\alpha_{j,1}e_1+ \alpha_{j,6}e_6+ \alpha_{j,7}e_7+ \alpha_{j,8}e_8+ \alpha_{j,9}e_9+\alpha_{j,10}e_{10}+ \alpha_{j,11}e_{11}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\alpha_{j,1}=\alpha_{j,6}=\alpha_{j,7}= \alpha_{j,8}=\alpha_{j,9}=\alpha_{j,10}=\alpha_{j,11}=0.$$
%
%For $i=2, \ j=3$ we have
%$$\frac{1}{2}\beta_{2,9}e_{10}-\frac{1}{2}\beta_{1,9}e_{11} =\beta_{3,9}e_9+ \beta_{3,10}e_{10}+ \beta_{3,11}e_{11}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{3,9}=0, \ \beta_{3,10}=\frac{1}{2}\beta_{2,9}, \ \beta_{3,11}=-\frac{1}{2}\beta_{1,9}.$$
%
%
%For $i=2, \ j=4$ we have
%$$-\frac{1}{2}\alpha_{1,6}e_{10}-\frac{1}{2}\alpha_{1,7}e_{11}=\beta_{4,9}e_9+ \beta_{4,10}e_{10}+ \beta_{4,11}e_{11}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{4,9}=0, \ \beta_{4,10}=-\frac{1}{2}\alpha_{1,6}, \ \beta_{4,11}=-\frac{1}{2}\alpha_{1,7}.$$
%
%For $i=2, \ j=5$ we have
%$$\frac{1}{2}\alpha_{1,6}e_{11}=\beta_{5,9}e_9+ \beta_{5,10}e_{10}+ \beta_{5,11}e_{11}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{5,9}=\beta_{5,10}=0, \ \beta_{5,11}=\frac{1}{2}\alpha_{1,6}.$$
%
%
%For $i=2, \ 6\leq j\leq 11$ we have
%$$0=\beta_{j,9}e_9+ \beta_{j,10}e_{10}+ \beta_{j,11}e_{11}.$$
%Comparing coefficients at the basis elements we obtain that
%$$\beta_{j,9}=\beta_{j,10}=\beta_{j,11}=0.$$
%
%
Thus, from the above restrictions we obtain the transposed Poisson algebra structure ${\bf TP}(g^3_{11})$ defined on Lie algebra  $g^3_{11}.$
%$$\begin{cases}
%\varphi_{e_1}(e_1)=\alpha_{1,6}e_6+ \alpha_{1,7}e_7+ \alpha_{1,8}e_8+ \alpha_{1,9}e_9+ \alpha_{1,10}e_{10}+ \alpha_{1,11}e_{11}, \\
%\varphi_{e_1}(e_2)=-\alpha_{1,6}e_7-\alpha_{1,7}e_8+ \beta_{1,9}e_9+ \beta_{1,10}e_{10}+ \beta_{1,11}e_{11},\\
%\varphi_{e_1}(e_3)=\frac{1}{2}\beta_{1,9}e_{10}-\frac{1}{2}\alpha_{1,9}e_{11}, \\
%\varphi_{e_1}(e_4)=\frac{1}{2}\alpha_{1,7}e_{10}+ \frac{1}{2}\alpha_{1,8}e_{11}, \\
%\varphi_{e_1}(e_5)=-\frac{1}{2}\alpha_{1,6}e_{10}-\frac{1}{2}\alpha_{1,7}e_{11}, \\
%\varphi_{e_1}(e_6)=\frac{1}{2}\alpha_{1,6}e_{11}, \\
%\varphi_{e_2}(e_2)=\alpha_{1,6}e_8+\beta_{2,9}e_9+ \beta_{2,10}e_{10}+\beta_{2,11}e_{11},\\
%\varphi_{e_2}(e_3)=\frac{1}{2}\beta_{2,9}e_{10}-\frac{1}{2}\beta_{1,9}e_{11}, \\
%\varphi_{e_2}(e_4)=-\frac{1}{2}\alpha_{1,6}e_{10}- \frac{1}{2}\alpha_{1,7}e_{11}, \\
%\varphi_{e_2}(e_5)=\frac{1}{2}\alpha_{1,6}e_{11}, \\
%\end{cases}$$
%
%
%We have the algebra:
%
%$$\begin{cases}
%e_1\cdot e_1=\alpha_{6}e_6+ \alpha_{7}e_7+ \alpha_{8}e_8+ \alpha_{9}e_9+ \alpha_{10}e_{10}+ \alpha_{11}e_{11}, \\
%e_1\cdot e_2=-\alpha_{6}e_7-\alpha_{7}e_8+\alpha_{1}e_9+ \alpha_{2}e_{10}+\alpha_{3}e_{11},\\
%e_1\cdot e_3=\frac{1}{2}\alpha_{1}e_{10}-\frac{1}{2}\alpha_{9}e_{11}, \\
%e_1\cdot e_4=\frac{1}{2}\alpha_{7}e_{10}+ \frac{1}{2}\alpha_{8}e_{11}, \\
%e_1\cdot e_5=-\frac{1}{2}\alpha_{6}e_{10}-\frac{1}{2}\alpha_{7}e_{11}, \\
%e_1\cdot e_6=\frac{1}{2}\alpha_{6}e_{11}, \\
%e_2\cdot e_2=\alpha_{6}e_8+\alpha_{4}e_9+ \alpha_{5}e_{10}+\alpha_{12}e_{11},\\
%e_2\cdot e_3=\frac{1}{2}\alpha_{4}e_{10}-\frac{1}{2}\alpha_{1}e_{11}, \\
%e_2\cdot e_4=-\frac{1}{2}\alpha_{6}e_{10}- \frac{1}{2}\alpha_{7}e_{11}, \\
%e_2\cdot e_5=\frac{1}{2}\alpha_{6}e_{11}. \\
%\end{cases}$$
%
\end{proof}

\bigskip
\begin{large}
\textbf{Declarations}
\end{large}

\bigskip
\textbf{Conflict of interest} There is no conflict of interest.

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