Review of deep learning approaches in solving rock fragmentation problems

Mikhail V. Ronkin; Elena N. Akimova; Vladimir E. Misilov; Mikhail V. Ronkin; Elena N. Akimova; Vladimir E. Misilov

doi:10.3934/math.20231219

AIMS Mathematics

2023, Volume 8, Issue 10: 23900-23940. doi: 10.3934/math.20231219

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Review Special Issues

Review of deep learning approaches in solving rock fragmentation problems

1.
Department of Information Technologies and Control Systems, Institute of Radioelectronics and Information Technology, Ural Federal University, Mira Street 19, 620002 Ekaterinburg, Russia
2.
Krasovskii Institute of Mathematics and Mechanics, Ural Branch of RAS, S. Kovalevskaya Street 16, 620108 Ekaterinburg, Russia

Received: 21 June 2023 Revised: 20 July 2023 Accepted: 27 July 2023 Published: 07 August 2023
MSC : 68T07, 68W10

One of the most significant challenges of the mining industry is resource yield estimation from visual data. An example would be identification of the rock chunk distribution parameters in an open pit. Solution of this task allows one to estimate blasting quality and other parameters of open-pit mining. This task is of the utmost importance, as it is critical to achieving optimal operational efficiency, reducing costs and maximizing profits in the mining industry. The mentioned task is known as rock fragmentation estimation and is typically tackled using computer vision techniques like instance segmentation or semantic segmentation. These problems are often solved using deep learning convolutional neural networks. One of the key requirements for an industrial application is often the need for real-time operation. Fast computation and accurate results are required for practical tasks. Thus, the efficient utilization of computing power to process high-resolution images and large datasets is essential. Our survey is focused on the recent advancements in rock fragmentation, blast quality estimation, particle size distribution estimation and other related tasks. We consider most of the recent results in this field applied to open-pit, conveyor belts and other types of work conditions. Most of the reviewed papers cover the period of 2018-2023. However, the most significant of the older publications are also considered. A review of publications reveals their specificity, promising trends and best practices in this field. To place the rock fragmentation problems in a broader context and propose future research topics, we also discuss state-of-the-art achievements in real-time computer vision and parallel implementations of neural networks.

Keywords:

Citation: Mikhail V. Ronkin, Elena N. Akimova, Vladimir E. Misilov. Review of deep learning approaches in solving rock fragmentation problems[J]. AIMS Mathematics, 2023, 8(10): 23900-23940. doi: 10.3934/math.20231219

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Abstract

1. Introduction

A Poisson algebra is a triple, $(L, \cdot, \left[{-, - } \right])$ , where $(L, \cdot)$ is a commutative associative algebra and $(L, \left[{-, - } \right])$ is a Lie algebra that satisfies the following Leibniz rule:

$\left[ {x, y\cdot z} \right] = \left[ {x, y} \right]\cdot z + y\cdot \left[ {x, z} \right], \forall x, y, z \in L.$

Poisson algebras appear naturally in the study of Hamiltonian mechanics and play a significant role in mathematics and physics, such as in applications of Poisson manifolds, integral systems, algebraic geometry, quantum groups, and quantum field theory (see ^[7,11,24,25]). Poisson algebras can be viewed as the algebraic counterpart of Poisson manifolds. With the development of Poisson algebras, many other algebraic structures have been found, such as Jacobi algebras ^[1,9], Poisson bialgebras ^[20,23], Gerstenhaber algebras, Lie-Rinehart algebras ^[16,17,26], $F$ -manifold algebras ^[12], Novikov-Poisson algebras ^[28], quasi-Poisson algebras ^[8] and Poisson $n$ -Lie algebras ^[10].

As a dual notion of a Poisson algebra, the concept of a transposed Poisson algebra was recently introduced by Bai et al. ^[2]. A transposed Poisson algebra $(L, \cdot, \left[{-, - } \right])$ is defined by exchanging the roles of the two binary operations in the Leibniz rule defining the Poisson algebra:

$2z \cdot \left[ {x, y} \right] = \left[ {z \cdot x, y} \right] + \left[ {x, z \cdot y} \right], \forall x, y, z \in L,$

where $(L, \cdot)$ is a commutative associative algebra and $(L, \left[{-, - } \right])$ is a Lie algebra.

It is shown that a transposed Poisson algebra possesses many important identities and properties and can be naturally obtained by taking the commutator in the Novikov-Poisson algebra ^[2]. There are many results on transposed Poisson algebras, such as those on transposed Hom-Poisson algebras ^[18], transposed BiHom-Poisson algebras ^[21], a bialgebra theory for transposed Poisson algebras ^[19], the relation between $\frac{1}{2}$ -derivations of Lie algebras and transposed Poisson algebras ^[14], the relation between $\frac{1}{2}$ -biderivations and transposed Poisson algebras ^[29], and the transposed Poisson structures with fixed Lie algebras (see ^[6] for more details).

The notion of an $n$ -Lie algebra (see Definition 2.1), as introduced by Filippov ^[15], has found use in many fields in mathematics and physics ^[4,5,22,27]. The explicit construction of $n$ -Lie algebras has become one of the important problems in this theory. In ^[3], Bai et al. gave a construction of $(n+1)$ -Lie algebras through the use of $n$ -Lie algebras and some linear functions. In ^[13], Dzhumadil′daev introduced the notion of a Poisson $n$ -Lie algebra which can be used to construct an ( $n+1$ )-Lie algebra under an additional strong condition. In ^[2], Bai et al. showed that this strong condition for $n = 2$ holds automatically for a transposed Poisson algebra, and they gave a construction of 3-Lie algebras from transposed Poisson algebras with derivations. They also found that this constructed $3$ -Lie algebra and the commutative associative algebra satisfy the analog of the compatibility condition for transposed Poisson algebras, which is called a transposed Poisson $3$ -Lie algebra. This motivated them to introduce the concept of a transposed Poisson $n$ -Lie algebra (see Definition 2.2) and propose the following conjecture:

Conjecture 1.1. ^[2] Let $n\geq 2$ be an integer and $(L, \cdot, \mu_n)$ a transposed Poisson $n$ -Lie algebra. Let $D$ be a derivation of $(L, \cdot)$ and $(L, \mu_n)$ . Define an $(n+1)$ -ary operation:

$\mu_{n+1}(x_1, \cdots, x_{n+1}): = \sum\limits_{i = 1}^{n+1} (-1)^{i-1}D(x_i) \mu_n(x_1, \cdots, \hat{x}_i, \cdots, x_{n+1}), \quad \forall x_1, \cdots, x_{n+1}\in L,$

where $\hat{x}_i$ means that the $i$ -th entry is omitted. Then, $(L, \cdot, \mu_{n+1})$ is a transposed Poisson $(n+1)$ -Lie algebra.

In this paper, based on the identities for transposed Poisson $n$ -Lie algebras given in Section 2, we prove that Conjecture 1.1 holds under a certain strong condition described in Section 3 (see Definition 2.3 and Theorem 3.2).

Throughout the paper, all vector spaces are taken over a field of characteristic zero. To simplify notations, the commutative associative multiplication $(\cdot)$ will be omitted unless the emphasis is needed.

2. Identities in transposed Poisson $n$ -Lie algebras

In this section, we first recall some definitions, and then we exhibit a class of identities for transposed Poisson $n$ -Lie algebras.

Definition 2.1. ^[15] Let $n\geq 2$ be an integer. An $n$ -Lie algebra is a vector space $L$ , together with a skew-symmetric linear map $\left[{-, \cdots, - } \right]:{ \otimes ^n}L \to L$ , such that, for any $x_i, y_j \in L, 1 \le i \le n-1, 1 \le j \le n,$ the following identity holds:

$\begin{equation} {\left[ {{{\left[ {{y_1}, \cdots , {y_n}} \right]}}, {x_1}, \cdots , {x_{n - 1}}} \right]} = \sum\limits_{i = 1}^n {{{( - 1)}^{i - 1}}{{\left[ {{{\left[ {{y_i}, {x_1}, \cdots , {x_{n - 1}}} \right]}}, {y_1}, \cdots , {{\hat y}_i}, \cdots , {y_n}} \right]}}}. \end{equation}$

(2.1)

Definition 2.2. ^[2] Let $n\geq 2$ be an integer and $L$ a vector space. The triple $(L, \cdot, \left[{-, \cdots, - } \right])$ is called a transposed Poisson $n$ -Lie algebra if $(L, \cdot)$ is a commutative associative algebra and $(L, \left[{-, \cdots, - } \right])$ is an $n$ -Lie algebra such that, for any $h, {x_i} \in L, 1 \le i \le n$ , the following identity holds:

$\begin{equation} nh\left[ {{x_1}, \cdots , {x_n}} \right] = \sum\limits_{i = 1}^n {\left[ {{x_1}, \cdots , h{x_i}, \cdots , {x_n}} \right]}. \end{equation}$

(2.2)

Some identities for transposed Poisson algebras in ^[2] can be extended to the following theorem for transposed Poisson $n$ -Lie algebras.

Theorem 2.1. Let $(L, \cdot, \left[{-, \cdots, - } \right])$ be a transposed Poisson $n$ -Lie algebra. Then, the following identities hold:

(1) For any $x_i\in L, 1\leq i\leq n+1$ , we have

$\begin{align} \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}{x_i}\left[ {{x_1}, \cdots , {{\hat x}_i}, \cdots , {x_{n + 1}}} \right]} = 0; \end{align}$

(2.3)

(2) For any $h, x_i, y_j\in L, 1\leq i\leq n-1, 1\leq j\leq n$ , we have

$\begin{align} \sum\limits_{i = 1}^n {{{( - 1)}^{i - 1}}\left[ {h\left[ {{y_i}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {y_n}} \right] = } \left[ {h\left[ {{y_1}, \cdots , {y_n}} \right], {x_1}, \cdots , {x_{n - 1}}} \right]; \end{align}$

(2.4)

(3) For any $x_i, y_j\in L, 1\leq i\leq n-1, 1\leq j\leq n+1$ , we have

$\begin{align} \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}\left[{y_i}, {{x_1}, \cdots , {x_{n-1}}} \right]\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} = 0; \end{align}$

(2.5)

(4) For any $x_1, x_2, y_i\in L, 1\leq i\leq n$ , we have

$\begin{align} \sum\limits_{i = 1}^n {\sum\limits_{j = 1, j \ne i}^n {\left[ {{y_1}, \cdots , {y_i}{x_1}, \cdots , {y_j}{x_2}, \cdots , {y_n}} \right]} } = n(n - 1){x_1}{x_2}\left[ {{y_1}, {y_2}, \cdots , {y_n}} \right]. \end{align}$

(2.6)

Proof. (1) By Eq (2.2), for any $1\leq i\leq n+1$ , we have

$n{x_i}\left[ {{x_1}, \cdots , {x_{i - 1}}, {x_{i + 1}}, \cdots , {x_{n+1}}} \right] = \sum\limits_{j \ne i}^{} {\left[ {{x_1}, \cdots , {x_{i - 1}}, {x_{i + 1}}, \cdots , {x_i}{x_j}, \cdots , {x_{n+1}}} \right]}.$

Thus, we obtain

$\sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}n{x_i}\left[ {{x_1}, \cdots , {{\hat x}_i}, \cdots , {x_{n + 1}}} \right]} = \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {{{( - 1)}^{i - 1}}\left[ {{x_1}, \cdots , {{\hat x}_i}, \cdots , {x_i}{x_j}, \cdots , {x_{n + 1}}} \right]} }.$

Note that, for any $i > j$ , we have

$\begin{eqnarray*} &&{( - 1)^{i - 1}}\left[ {{x_1}, \cdots , {x_{j - 1}}, {x_i}{x_j}, {x_{j + 1}}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right] \\ &&+ {( - 1)^{j - 1}}\left[ {{x_1}, \cdots , {{\hat x}_j}, \cdots , {x_{i - 1}}, {x_j}{x_i}, {x_{i + 1}}, \cdots , {x_n}} \right]\\ & = & {( - 1)^{i - 1 + (i - j - 1)}}\left[ {{x_1}, \cdots , {x_{j - 1}}, {x_{j + 1}}, \cdots , {x_{i - 1}}, {x_i}{x_j}, {x_{i + 1}}, \cdots , {x_n}} \right] \\ && + {( - 1)^{j - 1}}\left[ {{x_1}, \cdots , {{\hat x}_j}, \cdots , {x_{i - 1}}, {x_j}{x_i}, {x_{i + 1}}, \cdots , {x_n}} \right]\\ & = & \big( {{{( - 1)}^{ - j - 2}} + {{( - 1)}^{j - 1}}} \big)\left[ {{x_1}, \cdots , {x_{j - 1}}, {x_{j + 1}}, \cdots , {x_{i - 1}}, {x_i}{x_j}, {x_{i + 1}}, \cdots , {x_n}} \right]\\ & = & 0, \end{eqnarray*}$

which gives $\sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {{{(- 1)}^{i - 1}}\left[{{x_1}, \cdots, {{\hat x}_i}, \cdots, {x_i}{x_j}, \cdots, {x_{n + 1}}} \right]} } = 0.$

Hence, we get

$\sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}n{x_i}\left[ {{x_1}, \cdots , {{\hat x}_i}, \cdots , {x_{n + 1}}} \right]} = 0.$

(2) By Eq (2.2), we have

and, for any $1\leq j\leq n$ ,

$\begin{eqnarray*} &&{( - 1)^{j - 1}}(\left[ {h\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , {\hat{y}_j}, \cdots , {y_{n - 1}}} \right] \\ &&+\sum\limits_{i = 1, i\neq j}^{n } {\left[ {\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , h{y_i}, \cdots , \hat{y}_j, \cdots , {y_{n - 1}}} \right]}) \\ & = &{( - 1)^{j - 1}}nh\left[ {\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , {\hat{y}_j}, \cdots , {y_{n - 1}}} \right]. \end{eqnarray*}$

By taking the sum of the above $n+1$ identities and applying Eq (2.1), we get

$\begin{eqnarray*} &&- \left[ {h\left[ {{y_1}, \cdots , {y_n}} \right], {x_1}, \cdots , {x_{n - 1}}} \right]- \sum\limits_{i = 1}^{n - 1} {\left[ {\left[ {{y_1}, \cdots , {y_n}} \right], {x_1}, \cdots , h{x_i}, \cdots , {x_{n - 1}}} \right]}\\ &&+\sum\limits_{j = 1}^{n} {( - 1)^{j - 1}}(\left[ {h\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , {\hat{y}_j}, \cdots , {y_{n - 1}}} \right] \\ &&+\sum\limits_{i = 1, i\neq j}^{n } {\left[ {\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , h{y_i}, \cdots , \hat{y}_j, \cdots , {y_{n - 1}}} \right]}) \\ & = & -nh\left[ {\left[ {{y_1}, \cdots , {y_n}} \right], {x_1}, \cdots , {x_{n - 1}}} \right]+\\ &&nh\sum\limits_{j = 1}^{n} {( - 1)^{j - 1}}\left[ {\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , {\hat{y}_j}, \cdots , {y_{n - 1}}} \right]\\ & = & 0. \end{eqnarray*}$

We denote

$\begin{eqnarray*} {{A_j}} & : = &\sum\limits_{i = 1, i \ne j}^n {{{( - 1)}^{i - 1}}\left[ {\left[ {{y_i}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , h{y_j}, \cdots , {{\hat y}_i}, \cdots , {y_n}} \right]}, 1\leq j\leq {n}, \\ {B_i} &: = & \left[ {\left[ {{y_1}, \cdots , {y_n}} \right], {x_1}, \cdots , h{x_i}, \cdots , {x_{n - 1}}} \right], 1\leq i\leq {n-1}. \end{eqnarray*}$

Then, the above equation can be rewritten as

$\begin{equation} \begin{array}{l} \sum\limits_{i = 1}^n {{{( - 1)}^{i - 1}}\left[ {h\left[ {{y_i}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {y_n}} \right]} - \left[ {h\left[ {{y_1}, \cdots , {y_n}} \right], {x_1}, \cdots , {x_{n - 1}}} \right]\\ + \sum\limits_{j = 1}^n {{A_j}} - \sum\limits_{i = 1}^{n - 1} {{B_i}} = 0. \end{array} \end{equation}$

(2.7)

By applying Eq (2.1) to ${{A_j}}, 1\leq j\leq n$ , we have

$\begin{eqnarray*} {{A_j}} & = & \sum\limits_{i = 1, i \ne j}^n {{{( - 1)}^{i - 1}}\left[ {\left[ {{y_i}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , h{y_j}, \cdots , {{\hat y}_i}, \cdots , {y_n}} \right]}\\ & = & \left[ {\left[ {{y_1}, \cdots , h{y_j}, \cdots , {y_n}} \right], {x_1}, \cdots , {x_{n - 1}}} \right] \\ &&+ {( - 1)^j}\left[ {\left[ {h{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , {{\hat y}_j}, \cdots , {y_n}} \right]. \end{eqnarray*}$

Thus, we get

$\begin{eqnarray*} \sum\limits_{j = 1}^n { {{A_j}} } & = &\sum\limits_{j = 1}^n {\left[ {\left[ {{y_1}, \cdots , h{y_j}, \cdots , {y_n}} \right], {x_1}, \cdots , {x_{n - 1}}} \right]} \\ && + \sum\limits_{j = 1}^n {( - 1)^j}{\left[ {\left[ {h{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , {{\hat y}_j}, \cdots , {y_n}} \right]}\\ & = & n\left[ {h\left[ {{y_1}, \cdots , {y_n}} \right], {x_1}, \cdots , {x_{n - 1}}} \right]\\ && + \sum\limits_{j = 1}^n {( - 1)^j}{\left[ {\left[ {h{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , {{\hat y}_j}, \cdots , {y_n}} \right]}. \end{eqnarray*}$

By applying Eq (2.1) to ${{B_i}}, 1\leq i\leq n-1$ , we have

$\begin{eqnarray*} B_i & = &\left[ {\left[ {{y_1}, \cdots , {y_n}} \right], {x_1}, \cdots , h{x_i}, \cdots , {x_{n - 1}}} \right] \\ & = & \sum\limits_{j = 1}^n {{{( - 1)}^{j - 1}}\left[ {\left[ {{y_j}, {x_1}, \cdots , h{x_i}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots {{\hat y}_j}, \cdots , {y_n}} \right]}. \end{eqnarray*}$

Thus, we get

$\begin{eqnarray*} \sum\limits_{i = 1}^{n - 1} {{B_i}} & = & \sum\limits_{i = 1}^{n - 1} {\sum\limits_{j = 1}^n {{{( - 1)}^{j - 1}}\left[ {\left[ {{y_j}, {x_1}, \cdots , h{x_i}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots {{\hat y}_j}, \cdots , {y_n}} \right]} } \\ & = &\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^{n - 1} {{{( - 1)}^{j - 1}}\left[ {\left[ {{y_j}, {x_1}, \cdots , h{x_i}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots {{\hat y}_j}, \cdots , {y_n}} \right]} }. \end{eqnarray*}$

Note that, by Eq (2.2), we have

$\begin{eqnarray*} &&\sum\limits_{i = 1}^{n - 1} {{{( - 1)}^{j-1}}\left[ {\left[ {{y_j}, {x_1}, \cdots , h{x_i}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots {{\hat y}_j}, \cdots , {y_n}} \right]} \\ & = &{( - 1)^{j-1}}n\left[ {h\left[ {{y_j}, {x_1}, \cdots , {x_i}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots {{\hat y}_j}, \cdots , {y_n}} \right] \\ &&+{( - 1)^{j}} \left[ {\left[ {h{y_j}, {x_1}, \cdots , {x_i}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots {{\hat y}_j}, \cdots , {y_n}} \right]. \end{eqnarray*}$

Thus, we obtain

$\begin{eqnarray*} \sum\limits_{i = 1}^{n - 1} {{B_i}} & = &\sum\limits_{j = 1}^n {{{( - 1)}^{j - 1}}n\left[ {h\left[ {{y_j}, {x_1}, \cdots , {x_i}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots {{\hat y}_j}, \cdots , {y_n}} \right]} \\ && + \sum\limits_{j = 1}^n {{{( - 1)}^j}\left[ {\left[ {h{y_j}, {x_1}, \cdots , {x_i}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots {{\hat y}_j}, \cdots , {y_n}} \right]}. \end{eqnarray*}$

By substituting these equations into Eq (2.7), we have

$\begin{eqnarray*} &&\sum\limits_{i = 1}^n {{{( - 1)}^{i - 1}}\left[ {h\left[ {{y_i}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {y_n}} \right]} - \left[ {h\left[ {{y_1}, \cdots , {y_n}} \right], {x_1}, \cdots , {x_{n - 1}}} \right]\\ &&+ n\left[ {h\left[ {{y_1}, \cdots , {y_n}} \right], {x_1}, \cdots , {x_{n - 1}}} \right] + \sum\limits_{j = 1}^n {( - 1)^j}{\left[ {\left[ {h{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , {{\hat y}_j}, \cdots , {y_n}} \right]}\\ &&- \sum\limits_{j = 1}^n {{{( - 1)}^{j - 1}}n\left[ {h\left[ {{y_j}, {x_1}, \cdots , {x_i}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots {{\hat y}_j}, \cdots , {y_n}} \right]} \\ &&-\sum\limits_{j = 1}^n {{{( - 1)}^j}\left[ {\left[ {h{y_j}, {x_1}, \cdots , {x_i}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots {{\hat y}_j}, \cdots , {y_n}} \right]} \\ & = &0, \end{eqnarray*}$

which implies that

$(n - 1)(\sum\limits_{i = 1}^n {{{( - 1)}^i}\left[ {h\left[ {{y_i}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {y_n}} \right]} + \left[ {h\left[ {{y_1}, \cdots , {y_n}} \right], {x_1}, \cdots , {x_{n - 1}}} \right]) = 0.$

Therefore, the proof of Eq (2.4) is completed.

(3) By Eq (2.2), for any $1\leq j\leq n+1$ , we have

$\begin{eqnarray*} &&{( - 1)^{j - 1}}n\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right]\left[ {{y_1}, \cdots , {{\hat y}_j}, \cdots , {y_{n + 1}}} \right]\\ & = & \sum\limits_{i = 1, i \ne j}^{n + 1} {{{( - 1)}^{j - 1}}\left[ {{y_1}, \cdots , {y_i}\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], \cdots , {{\hat y}_j}, \cdots , {y_{n + 1}}} \right]}. \end{eqnarray*}$

By taking the sum of the above $n+1$ identities, we obtain

$\begin{eqnarray*} &&\sum\limits_{j = 1}^{n + 1} {{{( - 1)}^{j - 1}}n\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right]\left[ {{y_1}, \cdots , {{\hat y}_j}, \cdots , {y_{n + 1}}} \right]} \\ & = & \sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {{{( - 1)}^{j - 1}}\left[ {{y_1}, \cdots , {y_i}\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], \cdots , {{\hat y}_j}, \cdots , {y_{n + 1}}} \right]} }. \end{eqnarray*}$

Thus, we only need to prove the following equation:

$\sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {{{( - 1)}^{j - 1}}\left[ {{y_1}, \cdots , {y_i}\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], \cdots , {{\hat y}_j}, \cdots , {y_{n + 1}}} \right]} } = 0.$

Note that

$\begin{eqnarray*} &&\sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {{{( - 1)}^{j - 1}}\left[ {{y_1}, \cdots , {y_i}\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], \cdots , {{\hat y}_j}, \cdots , {y_{n + 1}}} \right]} } \\ & = &\sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {{{( - 1)}^{j - 1}}\left[ {{y_1}, \cdots , {y_i}\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], \cdots , {{\hat y}_j}, \cdots , {y_{n + 1}}} \right]} } \\ & = & \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1}^{i - 1} {{{( - 1)}^{i + j - 1}}\left[ {{y_i}\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , {{\hat y}_j}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } \\ && + \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = i + 1}^{n + 1} {{{( - 1)}^{i + j}}\left[ {{y_i}\left[ {{y_j}, {x_1}, \cdots , {x_{n - 1}}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_j}, \cdots , {y_{n + 1}}} \right]} } \\ &\mathop = \limits^{(2.4)}& \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^i}\left[ {{y_i}\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right], {x_1}, \cdots , {x_{n - 1}}} \right]} \\ &\mathop = \limits^{(2.3)}& 0. \end{eqnarray*}$

Hence, the conclusion holds.

(4) By applying Eq (2.2), we have

$\begin{eqnarray*} &&{n^2}{x_1}{x_2}\left[ {{y_1}, {y_2}, \cdots , {y_n}} \right] = n{x_1}\sum\limits_{j = 1}^n {\left[ {{y_1}, \cdots , {y_j}{x_2}, \cdots , {y_n}} \right]} \\ & = & \sum\limits_{i = 1}^n {\sum\limits_{j = 1, j \ne i}^n {\left[ {{y_1}, \cdots , {y_i}{x_1}, \cdots , {y_j}{x_2}, \cdots , {y_n}} \right]} } + \sum\limits_{j = 1}^n {\left[ {{y_1}, \cdots , {y_j}{x_1}{x_2}, \cdots , {y_n}} \right]}\\ & = &\sum\limits_{i = 1}^n {\sum\limits_{j = 1, j \ne i}^n {\left[ {{y_1}, \cdots , {y_i}{x_1}, \cdots , {y_j}{x_2}, \cdots , {y_n}} \right]} } + n{x_1}{x_2}\left[ {{y_1}, \cdots , {y_n}} \right], \end{eqnarray*}$

which gives

$n(n - 1){x_1}{x_2}\left[ {{y_1}, {y_2}, \cdots , {y_n}} \right] = \sum\limits_{i = 1}^n {\sum\limits_{j = 1, j \ne i}^n {\left[ {{y_1}, \cdots , {y_i}{x_1}, \cdots , {y_j}{x_2}, \cdots , {y_n}} \right]} } .$

Hence, the proof is completed.

To prove Conjecture 1.1, we need the following extra condition.

Definition 2.3. A transposed Poisson $n$ -Lie algebra $(L, \cdot, \left[{-, \cdots, - } \right])$ is called strong if the following identity holds:

$\begin{equation} \begin{array}{l} {y_1}\left[ {h{y_2}, {x_1}, \cdots , {x_{n - 1}}} \right] - {y_2}\left[ {h{y_1}, {x_1}, \cdots , {x_{n - 1}}} \right] + \sum\limits_{i = 1}^{n - 1} {{{( - 1)}^{i - 1}}h{x_i}\left[ {{y_1}, {y_2}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {x_{n - 1}}} \right]} \\ = 0 \end{array} \end{equation}$

(2.8)

for any ${y_1}, {y_2}, {x_i}\in L, 1\leq i\leq n - 1.$

Remark 2.1. When $n = 2$ , the identity is

${y_1}\left[ {h{y_2}, {x_1}} \right] + {y_2}\left[ {{x_1}, h{y_1}} \right] + h{x_1}\left[ {{y_1}, {y_2}} \right] = 0,$

which is exactly Theorem 2.5 (11) in ^[2]. Thus, in the case of a transposed Poisson algebra, the strong condition always holds. So far, we cannot prove that the strong condition fails to hold for $n\geq 3.$

Proposition 2.1. Let $(L, \cdot, \left[{-, \cdots, - } \right])$ be a strong transposed Poisson $n$ -Lie algebra. Then,

$\begin{equation} {y_1}\left[ {h{y_2}, {x_1}, \cdots , {x_{n - 1}}} \right] - h{y_1}\left[ {{y_2}, {x_1}, \cdots , {x_{n - 1}}} \right] = {y_2}\left[ {h{y_1}, {x_1}, \cdots , {x_{n - 1}}} \right] - h{y_2}\left[ {{y_1}, {x_1}, \cdots , {x_{n - 1}}} \right] \end{equation}$

(2.9)

for any ${y_1}, {y_2}, {x_i} \in L, 1\leq i\leq n - 1$ .

Proof. By Eq (2.3), we have

$- h{y_1}\left[ {{y_2}, {x_1}, \cdots , {x_{n - 1}}} \right] + h{y_2}\left[ {{y_1}, {x_1}, \cdots , {x_{n - 1}}} \right] = \sum\limits_{i = 1}^{n - 1} {{{( - 1)}^{i - 1}}h{x_i}\left[ {{y_1}, {y_2}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {x_{n - 1}}} \right]}.$

Then, the statement follows from Eq (2.8).

3. Proof of the conjecture for strong transposed Poisson $n$ -Lie algebras

In this section, we will prove Conjecture 1.1 for strong transposed Poisson $n$ -Lie algebras. First, we recall the notion of derivations of transposed Poisson $n$ -Lie algebras.

Definition 3.1. Let $(L, \cdot, \left[{-, \cdots, - } \right])$ be a transposed Poisson $n$ -Lie algebra. The linear operation $D: L\rightarrow L$ is called a derivation of $(L, \cdot, \left[{-, \cdots, - } \right])$ if the following holds for any $u, v, x_{i}\in L, 1\leq i\leq n$ :

(1) $D$ is a derivation of $\left({L, \cdot} \right)$ , i.e., $D\left({uv} \right) = D\left(u \right)v + uD\left(v \right)$ ;

(2) $D$ is a derivation of $(L, \left[{-, \cdots, - } \right])$ , i.e.,

$D\left( \left[ {{x_1}, \cdots , x_n} \right] \right) = \sum\limits_{i = 1}^n {\left[ {{x_1}, \cdots , {x_{i - 1}}, D\left( {{x_i}} \right), {x_{i + 1}}, \cdots , {x_n}} \right]}.$

Lemma 3.1. Let $(L, \cdot, \left[{-, \cdots, - } \right])$ be a transposed Poisson $n$ -Lie algebra and $D$ a derivation of $(L, \cdot, \left[{-, \cdots, - } \right])$ . For any $y_i\in L, 1\leq i\leq n+1$ , we have the following:

(1)

$\begin{align} &\sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}D({y_i})} D\left( {\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} \right)\\ = &\sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {{{( - 1)}^{i - 1}}D({y_i})\left[ {{y_1}, \cdots , D({y_j}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} }; \end{align}$

(3.1)

(2)

$\begin{align} &\sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}D({y_i})} D\left( {\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} \right)\\ = &\sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = j + 1, k \ne i}^{n + 1} {{{( - 1)}^i}{y_i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } }, \end{align}$

(3.2)

where, for any $i > j$ , $\sum\limits_{i}^{j}$ denotes the empty sum, which is equal to zero.

Proof. (1) The statement follows immediately from Definition 3.1.

(2) By applying Eq (3.1), we need to prove the following equation:

$\begin{eqnarray*} & &\sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {{{( - 1)}^{i - 1}}nD({y_i})\left[ {{y_1}, \cdots , D({y_j}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } \\ & = & \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = j + 1, k \ne i}^{n + 1} {{{( - 1)}^i}n{y_i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } }. \end{eqnarray*}$

For any $1\leq i\leq n+1$ , denote $A_i : = n\sum\limits_{j = 1, j \ne i}^{n + 1} {{{(- 1)}^{i - 1}}D({y_i})\left[{{y_1}, \cdots, D({y_j}), \cdots, {{\hat y}_i}, \cdots, {y_{n + 1}}} \right]}$ . Then, we have

$\sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {{{( - 1)}^{i - 1}}nD({y_i})\left[ {{y_1}, \cdots , D({y_j}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } = \sum\limits_{i = 1}^{n + 1} { {{A_i}} }.$

Note that

$\begin{eqnarray*} &&{A_i} = {( - 1)^{i - 1}}(nD({y_i})\left[ {D({y_1}), {y_2}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right] + nD({y_i})\left[ {{y_1}, D({y_2}), {y_3}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]\\ &&+ \cdots + nD({y_i})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_n}, D({y_{n + 1}})} \right])\\ & = & {( - 1)^{i - 1}}(\left[ {D({y_i})D({y_1}), {y_2}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right] \\ &&+ \sum\limits_{k = 2, k \ne i}^{n + 1} {\left[ {D({y_1}), {y_2}, \cdots , {y_k}D({y_i}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} \\ &&+ \left[ {{y_1}, D({y_i})D({y_2}), {y_3}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]\\ && + \sum\limits_{k = 1, k \ne 2, i}^{n + 1} {\left[ {{y_1}, D({y_2}), {y_3}, \cdots , {y_k}D({y_i}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} \\ &&+ \cdots + \left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_n}, D({y_i})D({y_{n + 1}})} \right] \\ &&+ \sum\limits_{k = 1, k \ne i}^n {\left[ {{y_1}, \cdots , {y_k}D({y_i}), \cdots , {{\hat y}_i}, \cdots , {y_n}, D({y_{n + 1}})} \right]} )\\ & = & {( - 1)^{i - 1}}\sum\limits_{j = 1, j \ne i}^{n + 1} {\left[ {{y_1}, \cdots , D({y_i})D({y_j}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} \\ &&+ {( - 1)^{i - 1}}\sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = 1, k \ne j, i}^{n + 1} {\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_k}D({y_i}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} }. \end{eqnarray*}$

Thus, we have

$\begin{eqnarray*} \sum\limits_{i = 1}^{n + 1} { {{A_i}} } & = &\sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {{{( - 1)}^{j - 1}}\left[ {{y_1}, \cdots , D({y_j})D({y_i}), \cdots , {{\hat y}_j}, \cdots , {y_{n + 1}}} \right]} } \\ &&+ \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = 1, k \ne i, j}^{n + 1} {{( - 1)^{i-1}}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_k}D({y_i}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } } \\ & = & {T_1} + {T_2}, \end{eqnarray*}$

where

$\begin{eqnarray*} {T_1} &: = &\sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {{{( - 1)}^{j - 1}}\left[ {{y_1}, \cdots , D({y_j})D({y_i}), \cdots , {{\hat y}_j}, \cdots , {y_{n + 1}}} \right]} }, \\ {T_2}& : = & \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = 1, k \ne i, j}^{n + 1} {{( - 1)^{i-1}}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_k}D({y_i}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } }. \\ \end{eqnarray*}$

Note that

$\begin{eqnarray*} {T_1} = \sum\limits_{j, i = 1}^{n + 1} {{B_{ji}}} , \end{eqnarray*}$

where ${B_{ji}} = {(- 1)^{j - 1}}\left[{{y_1}, \cdots, D({y_j})D({y_i}), \cdots, {{\hat y}_j}, \cdots, {y_{n + 1}}} \right]$ for any $1\leq j\neq i\leq n+1$ , and ${B_{ii}} = 0$ for any $1\leq i\leq n+1$ .

For any $1\leq i, j\leq n+1,$ without loss of generality, assume that $i < j$ ; then, we have

$\begin{eqnarray*} &&{B_{ji}} + {B_{ij}}\\ & = & {( - 1)^{j - 1}}\left[ {{y_1}, \cdots , D({y_j})D({y_i}), \cdots , {{\hat y}_j}, \cdots , {y_{n + 1}}} \right]\\ &&+ {( - 1)^{i - 1}}\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , D({y_i})D({y_j}), \cdots , {y_{n + 1}}} \right]\\ & = & {( - 1)^{j - 1}}\left[ {{y_1}, \cdots , D({y_j})D({y_i}), \cdots , {{\hat y}_j}, \cdots , {y_{n + 1}}} \right]\\ &&+ {( - 1)^{i - 1 + j - i + 1}}\left[ {{y_1}, \cdots , D({y_j})D({y_i}), \cdots , {{\hat y}_j}, \cdots , {y_{n + 1}}} \right]\\ & = & 0, \end{eqnarray*}$

which implies that ${T_1} = \sum\limits_{j, i = 1}^{n + 1} {{B_{ji}}} = 0.$

Thus, we get that $\sum\limits_{i = 1}^{n + 1} { {{A_i}} } = {T_2}$ .

We rewrite

$\begin{eqnarray*} &&\sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n+1} {\sum\limits_{k = j + 1, k \ne i}^{n + 1} {{{( - 1)}^i}n{y_i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } }\\ & = & \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n+1} {\sum\limits_{k = j + 1, k \ne i}^{n + 1} {\sum\limits_{t = 1, t \ne j, k, i}^{n + 1} {{{( - 1)}^i}}} } }\\ &&\cdot\left[ {{y_1}, \cdots , D({y_j}), \cdots , D({y_k}), \cdots , {y_t}{y_i}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right] \\ &&+ \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n+1} {\sum\limits_{k = j + 1, k \ne i}^{n + 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , {y_i}D({y_j}), \cdots , D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } } \\ &&+ \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n+1} {\sum\limits_{k = j + 1, k \ne i}^{n + 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_i}D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } } \\ & = & {M_1} + {M_2} + {M_3}, \end{eqnarray*}$

where

$\begin{eqnarray*} {M_1}& : = & \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = j + 1, k \ne i}^{n + 1} {\sum\limits_{t = 1, t \ne j, k, i}^{n + 1} {{{( - 1)}^i}}} } }\\ &&\cdot\left[ {{y_1}, \cdots , D({y_j}), \cdots , D({y_k}), \cdots , {y_t}{y_i}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right], \\ {M_2}& : = & \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = j + 1, k \ne i}^{n + 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , {y_i}D({y_j}), \cdots , D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } }, \\ {M_3} &: = & \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = j + 1, k \ne i}^{n + 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_i}D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } }. \end{eqnarray*}$

Note that

$\begin{eqnarray*} {M_1}& = & \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n+1} {\sum\limits_{k = j + 1, k \ne i}^{n + 1} {\sum\limits_{t = 1, t \ne j, k, i}^{n + 1} {{{( - 1)}^i}}} } }\\ &&\cdot\left[ {{y_1}, \cdots , D({y_j}), \cdots , D({y_k}), \cdots , {y_t}{y_i}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right] \\ & = &\sum\limits_{i, j, k, t = 1}^{n + 1} {{B_{ijkt}}}, \end{eqnarray*}$

where

$\begin{gather*} {B_{ijkt}} = \begin{cases} 0, { \quad } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text{if any two indices are equal or } k < j; \\ {( - 1)^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , D({y_k}), \cdots , {y_t}{y_i}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right], { \quad } \text{otherwise}. \end{cases} \end{gather*}$

For any $1\leq j, k\leq n+1,$ without loss of generality, assume that $t < i;$ then, we have

$\begin{eqnarray*} &&{B_{ijkt}} + {B_{tjki}}\\ & = & {( - 1)^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , D({y_k}), \cdots , {y_t}{y_i}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]\\ &&+ {( - 1)^t}\left[ {{y_1}, \cdots , D({y_j}), \cdots , D({y_k}), \cdots , {{\hat y}_t}, \cdots , {y_t}{y_i}, \cdots , {y_{n + 1}}} \right]\\ & = & {( - 1)^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , D({y_k}), \cdots , {y_t}{y_i}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]\\ &&+ {( - 1)^{t + i - t - 1}}\left[ {{y_1}, \cdots , D({y_j}), \cdots , D({y_k}), \cdots , {y_t}{y_i}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]\\ & = & 0, \end{eqnarray*}$

which implies that ${M_1} = 0.$

Therefore, we only need to prove the following equation:

$\begin{eqnarray*} &&{M_2} + {M_3} \\ & = & \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = 1, k \ne i, j}^{n + 1} {{{( - 1)}^{i - 1}}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_k}D({y_i}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } }. \end{eqnarray*}$

First, we have

$\begin{eqnarray*} && \sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = j + 1, k \ne i}^{n + 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , {y_i}D({y_j}), \cdots , D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } \\ &&+ \sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = j + 1, k \ne i}^{n + 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_i}D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } \\ & = & \sum\limits_{k = 1, k \ne i}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{k - 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , {y_i}D({y_j}), \cdots , D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } \\ &&+ \sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = j + 1, k \ne i}^{n + 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_i}D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } \\ & = & \sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = 1, k \ne i}^{j - 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , {y_i}D({y_k}), \cdots , D({y_j}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } \\ &&+ \sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = j + 1, k \ne i}^{n + 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_i}D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} }\\ & = &\sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = 1, k \ne i, j}^{n + 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_i}D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } . \end{eqnarray*}$

Thus,

$\begin{eqnarray*} &&M_2+M_3 \\ & = & \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = 1, k \ne i, j}^{n + 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_i}D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } } \\ & = & \sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {\sum\limits_{k = 1, k \ne i, j}^{n + 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_i}D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } }\\ & = & \sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {\sum\limits_{k = 1, k \ne j}^{i - 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_i}D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } } \\ &&+ \sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {\sum\limits_{k = i + 1, k \ne j}^{n + 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {{\hat y}_i}, \cdots , {y_i}D({y_k}), \cdots , {y_{n + 1}}} \right]} } } . \end{eqnarray*}$

Note that, for any $1\leq j\leq n+1,$ we have

$\begin{eqnarray*} &&\sum\limits_{i = 1, i \ne j}^{n + 1} {\sum\limits_{k = 1, k \ne j}^{i - 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_i}D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } \\ & = & \sum\limits_{i = 1, i \ne j}^{n + 1} {\sum\limits_{k = 1, k \ne j}^{i - 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_{k - 1}}, {y_i}D({y_k}), {y_{k + 1}}, \cdots , {{\hat y}_i} \cdots , {y_{n + 1}}} \right]} } \\ & = & \sum\limits_{i = 1, i \ne j}^{n + 1} {\sum\limits_{k = 1, k \ne j}^{i - 1} {{{( - 1)}^{k - 1}}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {{\hat y}_k}, \cdots , {y_{i - 1}}, {y_i}D({y_k}), {y_{i + 1}}, \cdots , {y_{n + 1}}} \right]} } \\ & = & \sum\limits_{i = 1, i \ne j}^{n + 1} {\sum\limits_{k = 1, k \ne j}^{i - 1} {{{( - 1)}^{k - 1}}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {{\hat y}_k}, \cdots , {y_i}D({y_k}), \cdots , {y_{n + 1}}} \right]} }. \end{eqnarray*}$

Similarly, we have

$\begin{eqnarray*} &&\sum\limits_{i = 1, i \ne j}^{n + 1} {\sum\limits_{k = i + 1, k \ne j}^{n + 1} {{{( - 1)}^i}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {{\hat y}_i}, \cdots , {y_i}D({y_k}), \cdots , {y_{n + 1}}} \right]} } \\ & = & \sum\limits_{i = 1, i \ne j}^{n + 1} {\sum\limits_{k = i + 1, k \ne j}^{n + 1} {{{( - 1)}^{k-1}}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_i}D({y_k}), \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} }. \end{eqnarray*}$

Thus,

$\begin{eqnarray*} &&{M_2} + {M_3}\\ & = & \sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {\sum\limits_{k = 1, k \ne j}^{i - 1} {{{( - 1)}^{k - 1}}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {{\hat y}_k}, \cdots , {y_i}D({y_k}), \cdots , {y_{n + 1}}} \right]} } } \\ &&+ \sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {\sum\limits_{k = i + 1, k \ne j}^{n + 1} {{{( - 1)}^{k-1}}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_i}D({y_k}), \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} } } \\ & = & \sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {\sum\limits_{k = 1, k \ne i, j}^{n + 1} {{{( - 1)}^{k - 1}}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_i}D({y_k}), \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} } }\\ & = & \sum\limits_{k = 1}^{n + 1} {\sum\limits_{j = 1, j \ne k}^{n + 1} {\sum\limits_{i = 1, i \ne j, k}^{n + 1} {{{( - 1)}^{k - 1}}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_i}D({y_k}), \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} } }\\ & = &\sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {\sum\limits_{k = 1, k \ne j, i}^{n + 1} {{{( - 1)}^{i - 1}}\left[ {{y_1}, \cdots , D({y_j}), \cdots , {y_k}D({y_i}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } } . \end{eqnarray*}$

The proof is completed.

Theorem 3.1. Let $(L, \cdot, \left[{-, \cdots, - } \right])$ be a strong transposed Poisson $n$ -Lie algebra and $D$ a derivation of $(L, \cdot, \left[{-, \cdots, - } \right])$ . Define an $(n + 1)$ -ary operation:

$\begin{equation} {\mu _{n + 1}}({x_1}, \cdots , {x_{n + 1}}): = \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}}D({x_i})[ {x_1}, \cdots , {{\hat x}_i}, \cdots , {x_{n + 1}}] \end{equation}$

(3.3)

for any ${x_i}\in L, 1\leq i\leq n+1.$ Then, $(L, \mu_{n+1})$ is an $(n+1)$ -Lie algebra.

Proof. For convenience, we denote

${\mu _{n + 1}}\left( {{x_1}, \cdots , {x_{n + 1}}} \right): = \left[ {{x_1}, \cdots , {x_{n + 1}}} \right].$

On one hand, we have

$\begin{eqnarray*} && {\left[ {\left[ {{y_1}, \cdots , {y_{n + 1}}} \right], {x_1}, \cdots , {x_n}} \right]}\\ &\mathop = \limits^{(3.3)}& \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}\left[ {D({y_i})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right], {x_1}, \cdots , {x_n}} \right]} \\ &\mathop = \limits^{(3.3)}& \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}D(D({y_i})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right])\left[ {{x_1}, \cdots , {x_n}} \right]} \\ &&{ + \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1}^n {{{( - 1)}^{i + j - 1}}D({x_j})} } \left[ {D({y_i})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right], {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]}\\ & = &\sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}{D^2}({y_i})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]\left[ {{x_1}, \cdots , {x_n}} \right]} \\ &&+ \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}D({y_i})D(\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right])\left[ {{x_1}, \cdots , {x_n}} \right]}\\ &&+ \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1}^n {{{( - 1)}^{i + j - 1}}D({x_j})} } \left[ {D({y_i})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right], {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]\\ &\mathop = \limits^{(3.1)}& \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}{D^2}({y_i})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]\left[ {{x_1}, \cdots , {x_n}} \right]}\\ &&+ \sum\limits_{i = 1}^{n + 1} {\sum\limits_{k = 1, k \ne i}^{n + 1} {{{( - 1)}^{i - 1}}D({y_i})\left[ {{y_1}, \cdots , D({y_k}), \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]\left[ {{x_1}, \cdots , {x_n}} \right]} } \\ &&+ \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1}^n {{{( - 1)}^{i + j - 1}}D({x_j})} } \left[ {D({y_i})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right], {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]\\ & = & \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}{D^2}({y_i})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]\left[ {{x_1}, \cdots , {x_n}} \right]} \\ &&+ \sum\limits_{k = 1}^{n + 1} {\sum\limits_{i = 1}^{k - 1} {{{( - 1)}^{k + i - 1}}D({y_i})\left[ {D({y_k}), {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]\left[ {{x_1}, \cdots , {x_n}} \right]} }\\ &&+ \sum\limits_{k = 1}^{n + 1} {\sum\limits_{i = k + 1}^{n + 1} {{{( - 1)}^{i + k}}D({y_i})\left[ {D({y_k}), {y_1}, \cdots , {{\hat y}_k}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]\left[ {{x_1}, \cdots , {x_n}} \right]} } \\ &&+ \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1}^n {{{( - 1)}^{i + j - 1}}D({x_j})} } \left[ {D({y_i})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right], {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]. \end{eqnarray*}$

On the other hand, for any $1\leq k\leq n$ , we have

$\begin{eqnarray*} &&{( - 1)^{k - 1}}\left[ {\left[ {{y_k}, {x_1}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]\\ &\mathop = \limits^{(3.3)}& {( - 1)^{k - 1}}\left[ {D({y_k})\left[ {{x_1}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]\\ &&+ \sum\limits_{j = 1}^n {{{( - 1)}^{j + k - 1}}\left[ {D({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} \\ &\mathop = \limits^{(3.3)}& {( - 1)^{k - 1}}D(D({y_k})\left[ {{x_1}, \cdots , {x_n}} \right])\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]\\ &&+ \sum\limits_{i = 1}^{k - 1} {{{( - 1)}^{i + k - 1}}D({y_i})\left[ {D({y_k})\left[ {{x_1}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} \\ && + \sum\limits_{i = k + 1}^{n + 1} {{{( - 1)}^{i + k}}D({y_i})\left[ {D({y_k})\left[ {{x_1}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_k}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} \\ &&+ \sum\limits_{j = 1}^n {{{( - 1)}^{j + k - 1}}D(D({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right])\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} \\ && +\sum\limits_{j = 1}^n {\sum\limits_{i = k + 1}^{n + 1} {{{(( - 1)}^{ i + j}}D({y_i}) } }\\ &&\cdot\left[ {D({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_k}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right] )\\ &&+ \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^{k - 1} {{{(( - 1)}^{i + j - 1}}D({y_i})} }\\ &&\cdot\left[ {D({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right])\\ & = & {( - 1)^{k - 1}}{D^2}({y_k})\left[ {{x_1}, \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]\\ &&+ \sum\limits_{i = 1}^{k - 1} {{{( - 1)}^{i + k - 1}}D({y_i})\left[ {D({y_k})\left[ {{x_1}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} \\ &&+ \sum\limits_{i = k + 1}^{n + 1} {{{( - 1)}^{i + k}}D({y_i})\left[ {D({y_k})\left[ {{x_1}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_k}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} \\ &&+ \sum\limits_{j = 1}^n {{{( - 1)}^{j + k - 1}}{D^2}({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} \\ &&+ \sum\limits_{j = 1}^n {\sum\limits_{i = k + 1}^{n + 1} {{{(( - 1)}^{ i + j}}D({y_i}) } }\\ &&\cdot\left[ {D({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_k}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right] )\\ &&+\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^{k - 1} {{{(( - 1)}^{ i + j - 1}}D({y_i}) } }\\ &&\cdot\left[ {D({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right])\\ &&+ {( - 1)^{k - 1}}D({y_k})D(\left[ {{x_1}, \cdots , {x_n}} \right])\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]\\ &&+ \sum\limits_{j = 1}^n {{{( - 1)}^{j + k - 1}}D({x_j})D(\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right])\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} \\ &\mathop = \limits^{(3.2)}& {( - 1)^{k - 1}}{D^2}({y_k})\left[ {{x_1}, \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]\\ &&+ \sum\limits_{i = 1}^{k - 1} {{{( - 1)}^{i + k - 1}}D({y_i})\left[ {D({y_k})\left[ {{x_1}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} \\ &&+ \sum\limits_{i = k + 1}^{n + 1} {{{( - 1)}^{i + k}}D({y_i})\left[ {D({y_k})\left[ {{x_1}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_k}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} \\ &&+ \sum\limits_{j = 1}^n {{{( - 1)}^{j + k - 1}}{D^2}({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} \\ && +\sum\limits_{j = 1}^n {\sum\limits_{i = k + 1}^{n + 1} {{{(( - 1)}^{ i + j}}D({y_i}) } }\\ &&\cdot\left[ {D({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_k}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right] )\\ &&+ \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^{k - 1} {{{(( - 1)}^{ i + j - 1}}D({y_i}) } }\\ &&\cdot\left[ {D({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right])\\ &&+ \sum\limits_{j = 1}^n {\sum\limits_{t = j + 1}^n {{{( - 1)}^k}{y_k}\left[ {{x_1}, \cdots , D({x_j}), \cdots , D({x_t}), \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} } \\ &&+ \sum\limits_{i = 1}^n {\sum\limits_{j = 1, j \ne i}^n {{{( - 1)}^{k + i}}{x_i}\left[ {D({y_k}), {x_1}, \cdots , D({x_j}), \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} } \\ &&+ \sum\limits_{i = 1}^n {\sum\limits_{j = 1, j \ne i}^n {\sum\limits_{t = j + 1, t \ne i}^n {{{( - 1)}^{k + i}}{x_i}\left[ {{y_k}, {x_1}, \cdots , D({x_j}), D({x_t}), \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]} } } \\ &&\cdot \left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]. \end{eqnarray*}$

We denote

$\sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}\left[ {\left[ {{y_i}, {x_1}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} = \sum\limits_{i = 1}^7 {{A_i}},$

where

$\begin{eqnarray*} {A_1} &: = & \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}{D^2}({y_i})\left[ {{x_1}, \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]}, \\ {A_2} &: = & \sum\limits_{k = 1}^{n + 1} {\sum\limits_{j = 1}^n {{{( - 1)}^{k + j - 1}}} {D^2}({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]}, \\ {A_3} &: = & \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1}^n {\sum\limits_{k = j + 1}^n {{{( - 1)}^i}{y_i}\left[ {{x_1}, \cdots , D({x_j}), \cdots , D({x_k}), \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} } }, \\ {A_4} &: = &\sum\limits_{k = 1}^{n + 1} {\sum\limits_{i = 1}^n {\sum\limits_{j = 1, j \ne i}^n {\sum\limits_{t = j + 1, t \ne i}^n {({{( - 1)}^{k + i}}{x_i}\left[ {{y_k}, {x_1}, \cdots , D({x_j}), D({x_t}), \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]} } } } \\ && \cdot \left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]), \\ {A_5} &: = & \sum\limits_{k = 1}^{n + 1} {\sum\limits_{i = 1}^{k - 1} {{{( - 1)}^{k + i - 1}}D({y_i})\left[ {D({y_k})\left[ {{x_1}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} } \\ &&+ \sum\limits_{k = 1}^{n + 1} {\sum\limits_{i = k + 1}^{n + 1} {{{( - 1)}^{i + k}}D({y_i})\left[ {D({y_k})\left[ {{x_1}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_k}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} }, \\ {A_6}&: = & \sum\limits_{k = 1}^{n + 1} {\sum\limits_{i = 1}^n {\sum\limits_{j = 1, j \ne i}^n {{{(( - 1)}^{k + i}}{x_i}\left[ {D({y_k}), {x_1}, \cdots , D({x_j}), \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]} } } \\ && \cdot\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right] ), \\ {A_7}&: = & \sum\limits_{k = 1}^{n + 1} {\sum\limits_{j = 1}^n {\sum\limits_{i = k + 1}^{n + 1} {{{(( - 1)}^{k + i + j}}D({y_i})} } }\\ &&\cdot\left[ {D({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_k}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right] )\\ &&+ \sum\limits_{k = 1}^{n + 1} {\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^{k - 1} {{{(( - 1)}^{k + i + j - 1}}D({y_i})} } }\\ &&\cdot\left[ {D({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]) . \end{eqnarray*}$

By Eq (2.5), for fixed $j$ , we have

$\sum\limits_{k = 1}^{n + 1} {{{( - 1)}^{k + j - 1}}{D^2}({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} = 0.$

So, we obtain that ${{A_2}} = 0.$

By Eq (2.3), for fixed $j$ and $k$ , we have

$\sum\limits_{i = 1}^{n + 1} {{{( - 1)}^i}{y_i}\left[ {{x_1}, \cdots , D({x_j}), \cdots , D({x_k}), \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} = 0.$

So, we obtain that ${{A_3}} = 0.$

By Eq (2.5), for fixed $j$ and $t$ , we have

$\sum\limits_{k = 1}^{n + 1} {{{( - 1)}^{k + i}}{x_i}\left[ {{y_k}, {x_1}, \cdots , D({x_j}), D({x_t}), \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} = 0.$

So, we obtain that ${{A_4}} = 0.$

By Eq (2.9), for fixed $i$ and $k$ , we have

$\begin{eqnarray*} &&{( - 1)^{k + i - 1}}D({y_i})\left[ {D({y_k})\left[ {{x_1}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]\\ &&+ {( - 1)^{i + k}}D({y_k})\left[ {D({y_i})\left[ {{x_1}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]\\ & = & {( - 1)^{k + i - 1}}D({y_i})\left[ {D({y_k}), {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]\left[ {{x_1}, \cdots , {x_n}} \right]\\ &&+ {( - 1)^{i + k}}D({y_k})\left[ {D({y_i}), {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]\left[ {{x_1}, \cdots , {x_n}} \right]. \end{eqnarray*}$

Thus, we obtain

$\begin{eqnarray*} {{A_5}} & = &\sum\limits_{k = 1}^{n + 1} {\sum\limits_{i = 1}^{k - 1} {{{( - 1)}^{k + i - 1}}D({y_i})\left[ {D({y_k}), {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]\left[ {{x_1}, \cdots , {x_n}} \right]} } \\ && + \sum\limits_{k = 1}^{n + 1} {\sum\limits_{i = k + 1}^{n + 1} {{{( - 1)}^{i + k}}D({y_i})\left[ {D({y_k}), {y_1}, \cdots , {{\hat y}_k}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]\left[ {{x_1}, \cdots , {x_n}} \right]} }. \end{eqnarray*}$

By Eq (2.3), for fixed $j$ and $k$ , we have

$\begin{eqnarray*} &&\sum\limits_{i = 1}^n {{{( - 1)}^{k + i}}{x_i}\left[ {D({y_k}), {x_1}, \cdots , D({x_j}), \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]} \\ & = & {( - 1)^{k - 1}}D({y_k})\left[ {{x_1}, \cdots , D({x_j}), \cdots , {x_n}} \right] + {( - 1)^{k + j - 1}}D({x_j})\left[ {D({y_k}), {x_1}, \cdots , {x_n}} \right]\\ & = & {( - 1)^{k + j}}D({y_k})\left[ {D({x_j}), {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right] + {( - 1)^{k + j - 1}}D({x_j})\left[ {D({y_k}), {x_1}, \cdots , {x_n}} \right]. \end{eqnarray*}$

Thus, we get

$\begin{eqnarray*} {{A_6}} & = &\sum\limits_{k = 1}^{n + 1} {\sum\limits_{j = 1}^n {{{( - 1)}^{k + j}}D({y_k})\left[ {D({x_j}), {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} } \\ &&+ \sum\limits_{k = 1}^{n + 1} {\sum\limits_{j = 1}^n {{{( - 1)}^{k + j - 1}}D({x_j})\left[ {D({y_k}), {x_1}, \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} } . \end{eqnarray*}$

By Eq (2.4), for fixed $j$ and $i$ , we have

$\begin{eqnarray*} && \sum\limits_{k = i + 1}^{n + 1} {{{( - 1)}^{k + i + j - 1}}D({y_i})\left[ {D({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_i}, \cdots , {{\hat y}_k}, \cdots , {y_{n + 1}}} \right]} \\ && + \sum\limits_{k = 1}^{i - 1} {{{( - 1)}^{k + i + j}}D({y_i})\left[ {D({x_j})\left[ {{y_k}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right], {y_1}, \cdots , {{\hat y}_k}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]} \\ & = & {( - 1)^{j + i - 1}}D({y_i})\left[ {D({x_j})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right], {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]. \end{eqnarray*}$

So, we obtain

${{{A_7}}} = \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{j + i - 1}}D({y_i})\left[ {D({x_j})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right], {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]} } .$

By Eq (2.9), we have

$\begin{eqnarray*} &&{( - 1)^{i + j}}D({y_i})\left[ {D({x_j}), {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]\\ &&+ {( - 1)^{i + j - 1}}D({x_j})\left[ {D({y_i}), {x_1}, \cdots , {x_n}} \right]\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right]\\ &&+ {( - 1)^{j + i - 1}}D({y_i})\left[ {D({x_j})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right], {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]\\ & = & {( - 1)^{j + i - 1}}D({x_j})\left[ {D({y_i})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right], {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]. \end{eqnarray*}$

So, we get

${{{A_6}}} + {{{A_7}}} = \sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1}^n {{{( - 1)}^{j + i - 1}}D({x_j})\left[ {D({y_i})\left[ {{y_1}, \cdots , {{\hat y}_i}, \cdots , {y_{n + 1}}} \right], {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right].} }$

Thus, we have

$\sum\limits_{i = 1}^7 {{A_i}} = {{A_1}} +{{A_5}} + {{A_6}} + {{A_7}} = \left[ {\left[ {{y_1}, \cdots , {y_{n + 1}}} \right], {x_1}, \cdots , {x_n}} \right].$

Therefore, $(L, \mu_{n+1})$ is an $(n+1)$ -Lie algebra.

Now, we can prove Conjecture 1.1 for strong transposed Poisson $n$ -Lie algebras.

Theorem 3.2. With the notations in Theorem 3.1, $(L, \cdot, \mu_{n+1})$ is a strong transposed Poisson $(n + 1)$ -Lie algebra.

Proof. For convenience, we denote ${\mu _{n + 1}}\left({{x_1}, \cdots, {x_{n + 1}}} \right): = \left[{{x_1}, \cdots, {x_{n + 1}}} \right].$ According to Theorem 3.1, we only need to prove Eqs (2.2) and (2.8).

Proof of Eq (2.2). By Eq (3.3), we have

$\begin{eqnarray*} &&\sum\limits_{i = 1}^{n + 1} {\left[ {{x_1}, \cdots , h{x_i}, \cdots , {x_{n + 1}}} \right]} \\ & = & D(h{x_1})\left[ {{x_2}, \cdots , {x_{n + 1}}} \right] + \sum\limits_{j = 2}^{n + 1} {{{( - 1)}^{j - 1}}D({x_j})\left[ {h{x_1}, {x_2}, \cdots , {{\hat x}_j}, \cdots , {x_{n + 1}}} \right]} \\ &&- D(h{x_2})\left[ {{x_1}, {x_3}, \cdots , {x_{n + 1}}} \right] + \sum\limits_{j = 1, j \ne 2}^{n + 1} {{{( - 1)}^{j - 1}}D({x_j})\left[ {{x_1}, h{x_2}, {x_3}, \cdots , {{\hat x}_j}, \cdots , {x_{n + 1}}} \right]} \\ &&+ \cdots + {( - 1)^n}D(h{x_n})\left[ {{x_1}, \cdots , {x_n}} \right] + \sum\limits_{j = 1}^n {{{( - 1)}^{j - 1}}D({x_j})\left[ {{x_1}, \cdots , {{\hat x}_j}, \cdots , {x_n}, h{x_{n + 1}}} \right]}\\ & = & \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}D(h{x_i})\left[ {{x_1}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]} \\ &&+\sum\limits_{i = 1}^{n + 1} {\sum\limits_{j = 1, j \ne i}^{n + 1} {{{( - 1)}^{j - 1}}D({x_j})\left[ {{x_1}, , \cdots , h{x_i}, \cdots , {{\hat x}_j}, \cdots , {x_{n + 1}}} \right]} } \\ & = & \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}D(h{x_i})} \left[ {{x_1}, \cdots , {{\hat x}_i}, \cdots , {x_{n + 1}}} \right]\\ &&+ \sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {{{( - 1)}^{j - 1}}D({x_j})\left[ {{x_1}, \cdots , h{x_i}, \cdots , {{\hat x}_j}, \cdots , {x_{n + 1}}} \right]} } \\ & = & \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}hD({x_i})} \left[ {{x_1}, \cdots , {{\hat x}_i}, \cdots , {x_{n + 1}}} \right] \\ &&+ \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}{x_i}D(h)} \left[ {{x_1}, \cdots , {{\hat x}_i}, \cdots , {x_{n + 1}}} \right]\\ && + \sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {{{( - 1)}^{j - 1}}D({x_j})\left[ {{x_1}, \cdots , h{x_i}, \cdots , {{\hat x}_j}, \cdots , {x_{n + 1}}} \right]} } \\ & \mathop = \limits^{(2.3)}& \sum\limits_{i = 1}^{n + 1} {{{( - 1)}^{i - 1}}hD({x_i})} \left[ {{x_1}, \cdots , {{\hat x}_i}, \cdots , {x_{n + 1}}} \right] \\ &&+\sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {{{( - 1)}^{j - 1}}D({x_j})\left[ {{x_1}, \cdots , h{x_i}, \cdots , {{\hat x}_j}, \cdots , {x_{n + 1}}} \right]} } \\ &\mathop = \limits^{(3.3)}& h\left[ {{x_1}, \cdots , {x_{n + 1}}} \right] + \sum\limits_{j = 1}^{n + 1} {\sum\limits_{i = 1, i \ne j}^{n + 1} {{{( - 1)}^{j - 1}}D({x_j})\left[ {{x_1}, \cdots , h{x_i}, \cdots , {{\hat x}_j}, \cdots , {x_{n + 1}}} \right]} } \\ &\mathop = \limits^{(2.2)}& h\left[ {{x_1}, \cdots , {x_{n + 1}}} \right] + nh\sum\limits_{j = 1}^{n + 1} {{{( - 1)}^{j - 1}}D({x_j})\left[ {{x_1}, \cdots , {{\hat x}_j}, \cdots , {x_{n + 1}}} \right]} \\ &\mathop = \limits^{(3.3)}& h\left[ {{x_1}, \cdots , {x_{n + 1}}} \right] + nh\left[ {{x_1}, \cdots , {x_{n + 1}}} \right]\\ & = & (n + 1)h\left[ {{x_1}, \cdots , {x_{n + 1}}} \right]. \end{eqnarray*}$

Proof of Eq (2.8). By Eq (3.3), we have

$\begin{eqnarray*} &&{y_1}\left[ {h{y_2}, {x_1}, \cdots , {x_n}} \right] - {y_2}\left[ {h{y_1}, {x_1}, \cdots , {x_n}} \right] + \sum\limits_{i = 1}^n {{{( - 1)}^{i - 1}}h{x_i}\left[ {{y_1}, {y_2}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]} \\ & = &{y_1}{y_2}D(h)\left[ {{x_1}, \cdots , {x_n}} \right] + {y_1}hD({y_2})\left[ {{x_1}, \cdots , {x_n}} \right] - {y_1}D({x_1})\left[ {h{y_2}, {x_2}, \cdots , {x_n}} \right]\\ &&+ {y_1}D({x_2})\left[ {h{y_2}, {x_1}, {x_3}, \cdots , {x_n}} \right] + \cdots + {( - 1)^n}{y_1}D({x_n})\left[ {h{y_2}, {x_1}, \cdots , {x_{n - 1}}} \right]\\ &&- {y_2}{y_1}D(h)\left[ {{x_1}, \cdots , {x_n}} \right] - {y_2}hD({y_1})\left[ {{x_1}, \cdots , {x_n}} \right] + {y_2}D({x_1})\left[ {h{y_1}, {x_2}, \cdots , {x_n}} \right]\\ &&- {y_2}D({x_2})\left[ {h{y_1}, {x_1}, {x_3}, \cdots , {x_n}} \right] + \cdots + {( - 1)^{n - 1}}{y_2}D({x_n})\left[ {h{y_1}, {x_1}, \cdots , {x_{n - 1}}} \right]\\ && + h{x_1}D({y_1})\left[ {{y_2}, {x_2}, \cdots , {x_n}} \right] - h{x_1}D({y_2})\left[ {{y_1}, {x_2}, \cdots , {x_n}} \right]\\ &&+ h{x_1}D({x_2})\left[ {{y_1}, {y_2}, {x_3}, \cdots , {x_n}} \right]+ \cdots + {( - 1)^{n + 1}}h{x_1}D({x_n})\left[ {{y_1}, {y_2}, {x_2}, \cdots , {x_{n - 1}}} \right]\\ &&- h{x_2}D({y_1})\left[ {{y_2}, {x_1}, {x_3}, \cdots , {x_n}} \right] + h{x_2}D({y_2})\left[ {{y_1}, {x_1}, {x_3}, \cdots , {x_n}} \right] \\ &&- h{x_2}D({x_1})\left[ {{y_1}, {y_2}, {x_3}, \cdots , {x_n}} \right]+ \cdots + {( - 1)^{n + 2}}h{x_2}D({x_n})\left[ {{y_1}, {y_2}, {x_1}, {x_3}, \cdots , {x_{n - 1}}} \right]\\ &&+ \cdots + {( - 1)^{n - 1}}h{x_n}D({y_1})\left[ {{y_2}, {x_1}, \cdots , {x_{n - 1}}} \right] + {( - 1)^n}h{x_n}D({y_2})\left[ {{y_1}, {x_1}, \cdots , {x_{n - 1}}} \right]\\ &&+ {( - 1)^{n + 1}}h{x_n}D({x_1})\left[ {{y_1}, {y_2}, {x_2}, \cdots , {x_{n - 1}}} \right] \\ &&+ \cdots + {( - 1)^{2n - 1}}h{x_n}D({x_{n - 1}})\left[ {{y_1}, {y_2}, {x_1}, \cdots , {x_{n - 2}}} \right]\\ & = & - {y_2}hD({y_1})\left[ {{x_1}, \cdots , {x_n}} \right] + h{x_1}D({y_1})\left[ {{y_2}, {x_2}, \cdots , {x_n}} \right]\\ &&+ \sum\limits_{i = 2}^n {{{( - 1)}^{i - 1}}h{x_i}D({y_1})\left[ {{y_2}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]} \\ &&+ {y_1}hD({y_2})\left[ {{x_1}, \cdots , {x_n}} \right] - h{x_1}D({y_2})\left[ {{y_1}, {x_2}, \cdots , {x_n}} \right] \\ &&+ \sum\limits_{i = 2}^n {{{( - 1)}^i}h{x_i}D({y_2})\left[ {{y_1}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]}\\ &&- {y_1}D({x_1})\left[ {h{y_2}, {x_2}, \cdots , {x_n}} \right] + {y_2}D({x_1})\left[ {h{y_1}, {x_2}, \cdots , {x_n}} \right]\\ &&+ \sum\limits_{i = 2}^n {{{( - 1)}^{i - 1}}h{x_i}D({x_1})\left[ {{y_1}, {y_2}, {x_2}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]} \\ &&+ {y_1}D({x_2})\left[ {h{y_2}, {x_1}, {x_3}, \cdots , {x_n}} \right] - {y_2}D({x_2})\left[ {h{y_1}, {x_1}, {x_3}, \cdots , {x_n}} \right]\\ &&+ h{x_1}D({x_2})\left[ {{y_1}, {y_2}, {x_3}, \cdots , {x_n}} \right] + \sum\limits_{i = 3}^n {{{( - 1)}^i}h{x_i}D({x_2})\left[ {{y_1}, {y_2}, {x_1}, {x_3}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]} \\ &&\cdots+ {( - 1)^n}{y_1}D({x_n})\left[ {h{y_2}, {x_1}, \cdots , {x_{n - 1}}} \right] + {( - 1)^{n - 1}}{y_2}D({x_n})\left[ {h{y_1}, {x_1}, \cdots , {x_{n - 1}}} \right]\\ &&+ \sum\limits_{j = 1}^{n - 1} {{{( - 1)}^{n + j - 1}}h{x_j}D({x_n})\left[ {{y_1}, {y_2}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {x_{n - 1}}} \right]} \\ & = & {A_1} + {A_2} + \sum\limits_{i = 1}^n {{B_i}}, \end{eqnarray*}$

where

$\begin{eqnarray*} {A_1} &: = & - {y_2}hD({y_1})\left[ {{x_1}, \cdots , {x_n}} \right] + \sum\limits_{i = 1}^n {{{( - 1)}^{i -1}}h{x_i}D({y_1})\left[ {{y_2}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]}, \\ {A_2} &: = & {y_1}hD({y_2})\left[ {{x_1}, \cdots , {x_n}} \right] + \sum\limits_{i = 1}^n {{{( -1)}^i}h{x_i}D({y_2})\left[ {{y_1}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]}, \\ \end{eqnarray*}$

and, for any $1\leq i\leq n,$

$\begin{eqnarray*} {B_i} &: = &{( - 1)^i}{y_1}D({x_i})\left[ {h{y_2}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right] + {( - 1)^{i - 1}}{y_2}D({x_i})\left[ {h{y_1}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]\\ &&+ \sum\limits_{j = 1}^{i - 1} {{{( - 1)}^{i + j - 1}}h{x_j}D({x_i})\left[ {{y_1}, {y_2}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]} \\ &&+ \sum\limits_{j = i + 1}^n {{{( - 1)}^{i + j}}h{x_j}D({x_i})\left[ {{y_1}, {y_2}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]} . \end{eqnarray*}$

By Eq (2.3), we have

$\begin{eqnarray*} {A_1}& = & hD({y_1})\left( { - {y_2}\left[ {{x_1}, \cdots , {x_n}} \right] + \sum\limits_{i = 1}^n {{{( - 1)}^{i - 1}}{x_i}\left[ {{y_2}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]} } \right) = 0. \end{eqnarray*}$

Similarly, we have that ${A_2} = 0.$

By Eq (2.8), for any $1\leq i\leq n$ , we have

$\begin{eqnarray*} {B_i} & = &{( - 1)^i}D({x_i})({y_1}\left[ {h{y_2}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right] - {y_2}\left[ {h{y_1}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]\\ &&+ \sum\limits_{j = 1}^{i - 1} {{{( - 1)}^{j - 1}}h{x_j}\left[ {{y_1}, {y_2}, {x_1}, \cdots , {{\hat x}_j}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]} \\ &&+ \sum\limits_{j = i + 1}^n {{{( - 1)}^j}h{x_j}\left[ {{y_1}, {y_2}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {{\hat x}_j}, \cdots , {x_n}} \right]} )\\ & = & 0. \end{eqnarray*}$

Thus, we get

${y_1}\left[ {h{y_2}, {x_1}, \cdots , {x_n}} \right] - {y_2}\left[ {h{y_1}, {x_1}, \cdots , {x_n}} \right] + \sum\limits_{i = 1}^n {{{( - 1)}^{i - 1}}h{x_i}\left[ {{y_1}, {y_2}, {x_1}, \cdots , {{\hat x}_i}, \cdots , {x_n}} \right]} = 0.$

The proof is completed.

Example 3.1. The commutative associative algebra $L = k[x_1, x_2, x_3]$ , together with the bracket

$\begin{eqnarray*} [x, y]: = x \cdot {D_1}(y) - y \cdot {D_1}(x), \forall x, y \in L. \end{eqnarray*}$

gives a transposed Poisson algebra $(L, \cdot, [-, -])$ , where ${D_1} = \partial_{x_1}$ ([, Proposition 2.2]). Note that the transposed Poisson algebra $(L, \cdot, [-, -])$ is strong according to Remark 2.5. Now, let ${D_2} = \partial_{x_2}$ ; one can check that $D_2$ is a derivation of $(L, \cdot, [-, -])$ . Then, there exists a strong transposed Poisson 3-Lie algebra defined by

$\begin{eqnarray*} [x, y, z]: = {D_2}(x)(y{D_1}(z) - z{D_1}(y)) + {D_2}(y)(z{D_1}(x) - x{D_1}(z))+{D_2}(z)(x{D_1}(y) - y{D_1}(x)), \ \forall x, y, z \in L. \end{eqnarray*}$

We note that $[{x_1}, {x_2}, {x_3}] = {x_3}$ , which is non-zero. The strong condition can be checked as follows:

For any $h, y_1, y_2, z_1, z_2\in L,$ by a direct calculation, we have

$\begin{eqnarray*} {y_1}[h{y_2}, {z_1}, {z_2}] & = & {y_1}{z_1}h{D_1}({z_2}){D_2}({y_2}) - {y_1}{z_2}h{D_1}({z_1}){D_2}({y_2})\\ &&+ {y_1}{y_2}{z_1}{D_1}({z_2}){D_2}({h}) - {y_1}{y_2}{z_2}{D_1}({z_1}){D_2}({h})\\ &&- {y_1}{y_2}h{D_1}({z_2}){D_2}({z_1}) + {y_1}{z_2}h{D_1}({y_2}){D_2}({z_1}) + {y_1}{y_2}{z_2}{D_1}(h){D_2}({z_1})\\ &&+ {y_1}{y_2}h{D_1}({z_1}){D_2}({z_2}) - {y_1}{z_1}h{D_1}({y_2}){D_2}({z_2}) - {y_1}{y_2}{z_1}{D_1}(h){D_2}({z_2}), \end{eqnarray*}$

$\begin{eqnarray*} - {y_2}[h{y_1}, {z_1}, {z_2}] & = & - {y_2}{z_1}h{D_1}({z_2}){D_2}({y_1}) + {y_2}{z_2}h{D_1}({z_1}){D_2}({y_1})\\ &&- {y_1}{y_2}{z_1}{D_1}({z_2}){D_2}(h) + {y_1}{y_2}{z_2}{D_1}({z_1}){D_2}(h)\\ &&+ {y_1}{y_2}h{D_1}({z_2}){D_2}({z_1}) - {y_2}{z_2}h{D_1}({y_1}){D_2}({z_1}) - {y_1}{y_2}{z_2}{D_1}(h){D_2}({z_1})\\ &&- {y_1}{y_2}h{D_1}({z_1}){D_2}({z_2}) + {y_2}{z_1}h{D_1}({y_1}){D_2}({z_2}) + {y_1}{y_2}{z_1}{D_1}(h){D_2}({z_2}), \end{eqnarray*}$

$\begin{eqnarray*} h{z_1}[{y_1}, {y_2}, {z_2}] & = & h{y_2}{z_1}{D_1}({z_2}){D_2}({y_1}) - h{z_1}{z_2}{D_1}({y_2}){D_2}({y_1})- h{y_1}{z_1}{D_1}({z_2}){D_2}({y_2})\\ &&+ h{z_1}{z_2}{D_1}({y_1}){D_2}({y_2})+ h{y_1}{z_1}{D_1}({y_2}){D_2}({z_2}) - h{y_2}{z_1}{D_1}({y_1}){D_2}({z_2}), \end{eqnarray*}$

$\begin{eqnarray*} - h{z_2}[{y_1}, {y_2}, {z_1}]& = & - h{y_2}{z_2}{D_1}({z_1}){D_2}({y_1}) + h{z_1}{z_2}{D_1}({y_2}){D_2}({y_1})+ h{y_1}{z_2}{D_1}({z_1}){D_2}({y_2}) \\ &&- h{z_1}{z_2}{D_1}({y_1}){D_2}({y_2})- h{y_1}{z_2}{D_1}({y_2}){D_2}({z_1}) + h{y_2}{z_2}{D_1}({y_1}){D_2}({z_1}). \end{eqnarray*}$

Thus, we get

${y_1}[h{y_2}, {z_1}, {z_2}] - {y_2}[h{y_1}, {z_1}, {z_2}] + h{z_1}[{y_1}, {y_2}, {z_2}] - h{z_2}[{y_1}, {y_2}, {z_1}] = 0.$

4. Conclusions

We have studied transposed Poisson $n$ -Lie algebras. We first established an important class of identities for transposed Poisson $n$ -Lie algebras, which were subsequently used throughout the paper. We believe that the identities developed here will be useful in investigations of the structure of transposed Poisson $n$ -Lie algebras in the future. Then, we introduced the notion of a strong transposed Poisson $n$ -Lie algebra and derived an $(n+1)$ -Lie algebra from a strong transposed Poisson $n$ -Lie algebra with a derivation. Finally, we proved the conjecture of Bai et al. ^[2] for strong transposed Poisson $n$ -Lie algebras.

Use of AI tools declaration

The authors declare that they have not used artificial intelligence tools in the creation of this article.

Acknowledgments

Ming Ding was supported by the Guangdong Basic and Applied Basic Research Foundation (2023A1515011739) and the Basic Research Joint Funding Project of University and Guangzhou City under grant number 202201020103.

Conflict of interest

The authors declare that there is no conflict of interest.

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