# Nijenhuis operators on low dimensional pre-Lie algebras and solutions of
CYBE

Xiang Gao  School of Mathematical Sciences
Zhejiang Normal University
Jinhua 321004
China<EMAIL_ADDRESS>,  Chuangchuang Kang  School of Mathematical
Sciences
Zhejiang Normal University
Jinhua 321004
China<EMAIL_ADDRESS>,  Jiafeng Lü  School of Mathematical Sciences
Zhejiang Normal University
Jinhua 321004
China<EMAIL_ADDRESS>and  Shizhuo Yu  School of Mathematical Sciences
and LPMC
Nankai University
Tianjin 300317
China<EMAIL_ADDRESS>
###### Abstract.

We first classify all Nijenhuis operators on 2-dimensional complex pre-Lie
algebras and 3-dimensional complex associative algebras. Moreover, applying
the obtained classification, we provide solutions of classical Yang-Baxter
equation of the semi-direct product sub-adjacent Lie algebras.

###### Key words and phrases:

Nijenhuis operators, pre-Lie algebras, sub-adjacent Lie algebras, classical
Yang-Baxter equation

###### 2010 Mathematics Subject Classification:

17B60, 16T25, 17B81

##  1\. Introduction

Nijenhuis torsion was first introduced by Nijenhuis in 1951 ([20]). Let $M$ be
a smooth $n$-dimensional manifold, $N$ a smooth operator field on it, and $v$,
$u$ smooth vector fields. Then the Nijenhuis torsion
$(\mathcal{T}_{N})_{~{}jk}^{i}$ of $N$ is a tensor of type (1, 2) given by the
formula

$\mathcal{T}_{N}(v,u)=[Nv,Nu]+N^{2}[v,u]-N[Nv,u]-N[v,Nu],$

or in local coordinates $\mathbf{x}=(x^{1},...,x^{n})$,

$(\mathcal{T}_{N})_{~{}jk}^{i}=N_{~{}j}^{l}\frac{\partial
N_{~{}k}^{i}}{\partial x^{l}}-N_{~{}k}^{l}\frac{\partial
N_{~{}j}^{i}}{\partial x^{l}}-N_{~{}l}^{i}\frac{\partial
N_{~{}k}^{l}}{\partial x^{j}}+N_{~{}l}^{i}\frac{\partial
N_{~{}j}^{l}}{\partial x^{k}}.$

An operator $N$ is called Nijenhuis operator if $\mathcal{T}_{N}=0$. The
vanishing of the Nijenhuis torsion is a necessary (but not sufficient)
condition for integrability of an operator field ([4]). Moreover, the
Nijenhuis operator plays a central role in various fields such as complex
differential geometry and complex Lie algebra theory ([27]), deformation of
associative (resp. Lie) algebra ([13, 21]) and pairs of Dirac structures ([10,
8]). And numerous interesting generalizations have been made regarding the
Nijenhuis operator, including its application to pre-Lie algebras ([25]),
$n$-Lie algebras ([18]), Hom-Lie algebras ([9]) and Courant algebroids with
Nijenhuis structures ([16]).

Pre-Lie algebras, also known as left-symmetric algebras, were first introduced
by Cayley in the context of rooted tree algebras ([7]), which have been
applied in study homogeneous convex cones ([24]), affine manifolds and affine
structures on Lie group ([19, 17]). The author proved in [26] that an
operator, which linearly depends on the local coordinates, is a Nijenhuis
operator if and only if the coefficients are structure constants of a pre-Lie
algebra.

In [25], the authors defined the concept of a Nijenhuis operator on pre-Lie
algebras, which generates a trivial deformation of pre-Lie algebras. The local
behavior of Nijenhuis operators at singular points plays an important role in
the study of complex integrable systems ([5]). Up to now, only a few Nijenhuis
operators on (simple) pre-Lie algebras are explicitly known (see [6, 2, 15,
14] for references) and a complete classification of simple pre-Lie algebras
has not yet been achieved.

Let $A$ be a pre-Lie algebra (product of $x$ and $y$ is denoted by $xy$) over
a field $\mathbb{C}$, a linear map $R:A\rightarrow A$ is called a Rota-Baxter
operator of weight $\lambda\in\mathbb{C}$ on $A$ if

(1.1) $R(x)R(y)=R(R(x)y+xR(y)+\lambda xy),~{}\forall~{}x,y\in A.$

If $N$ is a Nijenhuis operator on a pre-Lie algebra that satisfies $N^{2}=0$,
then $N$ is a Rota-Baxter operator of weight zero on $A$ (see Proposition
5.1). Let $\mathfrak{g}$ be a Lie algebra, a linear map
$R:\mathfrak{g}\rightarrow\mathfrak{g}$ is called an operator form of the
classical Yang-Baxter equation (CYBE) if,

(1.2) $[R(x),R(y)]=R([R(x),y]+[x,R(y)]),~{}\forall~{}x,y\in\mathfrak{g},$

which is also called a Rota-Baxter operator on ${\mathfrak{g}}$. If $R$ is a
Rota-Baxter operator of weight zero on a pre-Lie algebra $A$, then $R$ is a
Rota-Baxter operator of weight zero on the sub-adjacent Lie algebra
${\mathfrak{g}}(A)$ (see Proposition 5.2). Therefore, we can obtain a Rota-
Baxter operator of weight zero on its sub-adjacent Lie algebra from a
Nijenhuis operator on a pre-Lie algebra.

The CYBE originated from the study of inverse scattering theory, which is
recognized as the “semi-classical limit” of the quantum Yang-Baxter equation
([11, 12, 3]). Recall that $r=\sum\limits_{i}a_{i}\otimes
b_{i}\in{\mathfrak{g}}\otimes{\mathfrak{g}}$ is called a solution of CYBE in
$\mathfrak{g}$ if

(1.3) $[r_{12},r_{13}]+[r_{12},r_{23}]+[r_{13},r_{23}]=0\;\;{\rm
in}\;U({\mathfrak{g}}),$

where $U({\mathfrak{g}})$ is the universal enveloping algebra of
${\mathfrak{g}}$ and

$r_{12}=\sum\limits_{i}a_{i}\otimes b_{i}\otimes 1,\quad
r_{13}=\sum\limits_{i}a_{i}\otimes 1\otimes b_{i},\quad
r_{23}=\sum\limits_{i}1\otimes a_{i}\otimes b_{i}.$

Semonov-Tian-Shansky ([23]) proved that if a nondegenerate symmetric invariant
bilinear form exists on a Lie algebra $\mathfrak{g}$ and if $r$ is a skew-
symmetric solution of the CYBE, then (1.2) is equals to (1.3). C. Bai ([1])
proved that a linear map $R:\mathfrak{g}\rightarrow\mathfrak{g}$ is a Rota-
Baxter operator if and only if $r=R-R^{21}$ is a skew-symmetric solution of
CYBE in $\mathfrak{g}\ltimes_{{\rm ad}^{\ast}}\mathfrak{g}^{\ast}$. In [22],
the authors determined all Rota-Baxter operators (of weight zero) on
$\mathfrak{sl}(2,\mathbb{C})$ under the Cartan-Weyl basis and found solutions
of the classical Yang-Baxter equation in the six-dimensional Lie algebra
$\mathfrak{sl}(2,\mathbb{C})\ltimes_{\operatorname{{\mathrm{ad}}}^{*}}\mathfrak{sl}(2,\mathbb{C})^{*}$.
In this paper we obtain solutions of CYBE in a semidirect product sub-adjacent
Lie algebra from a Nijenhuis operator on a pre-Lie algebra.

The goal of this paper is to classify Nijenhuis operators on 2-dimensional
complex pre-Lie algebras and 3-dimensional complex associative algebras.
Applying the classification, we also provide solutions of the classical Yang-
Baxter equation in the semi-direct product sub-adjacent Lie algebras.

This paper is organized as follows. In Section 2, we recall some basic
concepts and known results about pre-Lie algebras and Nijenhuis operators ([6,
2]). In Section 3, we present all Nijenhuis operators on 2-dimensional complex
pre-Lie algebras (Proposition 3.1 and 3.2). Section 4 provides all Nijenhuis
operators on 3-dimensional complex associative algebras (Proposition 4.1 and
4.2 ). In Section 5, we give Rota-Baxter operators on Lie algebras
${\mathfrak{g}}(A)$ that are derived from Nijenhuis operators on pre-Lie
algebras $A$. Furthermore, for these Rota-Baxter operators on
${\mathfrak{g}}(A)$, we give the induced solutions of CYBE in the Lie algebras
${\mathfrak{g}}(A)\ltimes_{{\rm ad}^{\ast}}{\mathfrak{g}}(A)^{\ast}$.

Throughout this paper, all pre-Lie algebras are of finite dimension over
$\mathbb{C}$ and all associative algebras are of finite (Krull) dimension over
$\mathbb{C}$.

##  2\. Preliminaries

In this section, we will provide some preliminaries and classification results
of 2-dimensional and 3-dimensional pre-Lie algebras in [6] and [2].

######  Definition 2.1.

Let $A$ be a vector space over $\mathbb{C}$ with a bilinear product
$\cdot:A\times A\rightarrow A$. Then $(A,\cdot)$ is called a pre-Lie algebra
if for any $x,y,z\in A$,

(2.4) $(x\cdot y)\cdot z-x\cdot(y\cdot z)=(y\cdot x)\cdot z-y\cdot(x\cdot z).$

For a pre-Lie algebra $(A,\cdot)$, the commutator

(2.5) $[x,y]=x\cdot y-y\cdot x,$

defines a Lie algebra ${\mathfrak{g}}={\mathfrak{g}}(A)$, which is called the
sub-adjacent Lie algebra of $(A,\cdot)$.

######  Proposition 2.2.

([6]) Let $\left(A,\cdot\right)$ be a 2-dimensional commutative pre-Lie
algebra and $\left\\{e_{1},e_{2}\right\\}$ be a basis of $A$. Then $(A,\cdot)$
must be isomorphic to one of the following algebras:

$(A_{1})$ $e_{1}\cdot e_{1}=e_{1},~{}e_{2}\cdot e_{2}=e_{2}$;

$(A_{2})$ $e_{1}\cdot e_{1}=e_{1},~{}e_{1}.e_{2}=e_{2}\cdot e_{1}=e_{2}$;

$(A_{3})$ $e_{1}\cdot e_{1}=e_{1}$;

$(A_{4})$ $e_{i}\cdot e_{j}=0,~{}i,j=1,2$;

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

######  Proposition 2.3.

([6]) Let $\left(B,\cdot\right)$ be a 2-dimensional non commutative pre-Lie
algebra and $\left\\{e_{1},e_{2}\right\\}$ be a basis of $B$. Then $(B,\cdot)$
must be isomorphic to one of the following algebras:

$(B_{1})$ $e_{2}\cdot e_{1}=-e_{1},~{}e_{2}\cdot e_{2}=e_{1}-e_{2}$;

$(B_{2})$ $e_{2}\cdot e_{1}=-e_{1},~{}e_{2}\cdot e_{2}=-e_{2}$;

$(B_{3})$ $e_{2}\cdot e_{1}=-e_{1},~{}e_{2}\cdot e_{2}=ke_{2},~{}k\neq-1$;

$(B_{4})$ $e_{1}\cdot e_{2}=e_{1},~{}e_{2}\cdot e_{2}=e_{2}$;

$(B_{5})$ $e_{1}\cdot e_{2}=le_{1},~{}e_{2}\cdot
e_{1}=(l-1)e_{1},~{}e_{2}\cdot e_{2}=e_{1}+le_{2},~{}l\neq 0$;

$(B_{6})$ $e_{1}\cdot e_{1}=2e_{1},~{}e_{1}\cdot e_{2}=e_{2},~{}e_{2}\cdot
e_{2}=e_{1}$.

######  Proposition 2.4.

([2]) Let $\left(C,\cdot\right)$ be a 3-dimensional commutative associative
algebra and $\left\\{e_{1},e_{2},e_{3}\right\\}$ be a basis of $C$. Then
$(C,\cdot)$ must be isomorphic to one of the following algebras:

$(C_{1})$ $e_{3}\cdot e_{3}=e_{1}$;

$(C_{2})$ $e_{i}\cdot e_{j}=0,~{}i,j=1,2,3$;

$(C_{3})$ $e_{2}\cdot e_{2}=e_{1},~{}e_{3}\cdot e_{3}=e_{1}$;

$(C_{4})$ $e_{2}\cdot e_{3}=e_{3}\cdot e_{2}=e_{1},~{}e_{3}\cdot e_{3}=e_{2}$;

$(C_{5})$ $e_{1}\cdot e_{1}=e_{1},~{}e_{2}\cdot e_{2}=e_{2},~{}e_{3}\cdot
e_{3}=e_{3}$;

$(C_{6})$ $e_{2}\cdot e_{2}=e_{2},~{}e_{3}\cdot e_{3}=e_{3}$;

$(C_{7})$ $e_{1}\cdot e_{3}=e_{3}\cdot e_{1}=e_{1},~{}e_{2}\cdot
e_{2}=e_{2},~{}e_{3}\cdot e_{3}=e_{3}$;

$(C_{8})$ $e_{3}\cdot e_{3}=e_{3}$;

$(C_{9})$ $e_{1}\cdot e_{3}=e_{3}\cdot e_{1}=e_{1},~{}e_{3}\cdot e_{3}=e_{3}$;

$(C_{10})$ $e_{1}\cdot e_{3}=e_{3}\cdot e_{1}=e_{1},~{}e_{2}\cdot
e_{3}=e_{3}\cdot e_{2}=e_{2},~{}e_{3}\cdot e_{3}=e_{3}$;

$(C_{11})$ $e_{1}\cdot e_{1}=e_{2},~{}e_{3}\cdot e_{3}=e_{3}$;

$(C_{12})$ $e_{1}\cdot e_{1}=e_{2},~{}e_{1}\cdot e_{3}=e_{3}\cdot
e_{1}=e_{1},~{}e_{2}\cdot e_{3}=e_{3}\cdot e_{2}=e_{2},~{}e_{3}\cdot
e_{3}=e_{3}$.

######  Proposition 2.5.

([2]) Let $\left(D,\cdot\right)$ be a 3-dimensional non-commutative
associative algebra and $\left\\{e_{1},e_{2},e_{3}\right\\}$ be a basis of
$D$. Then $(D,\cdot)$ must be isomorphic to one of the following algebras:

$(D_{1})$ $e_{1}\cdot e_{2}=\frac{1}{2}e_{3},~{}e_{2}\cdot
e_{1}=-\frac{1}{2}e_{3}$;

$(D_{2})$ $e_{2}\cdot e_{1}=-e_{3}$;

$(D_{3})$ $e_{1}\cdot e_{1}=e_{3},~{}e_{1}\cdot e_{2}=e_{3},~{}e_{2}\cdot
e_{2}=\lambda e_{3},~{}\lambda\neq 0$;

$(D_{4})$ $e_{3}\cdot e_{2}=e_{2},~{}e_{3}\cdot e_{3}=e_{3}$;

$(D_{5})$ $e_{2}\cdot e_{3}=e_{2},~{}e_{3}\cdot e_{3}=e_{3}$;

$(D_{6})$ $e_{1}\cdot e_{1}=e_{1},~{}e_{3}\cdot e_{2}=e_{2},~{}e_{3}\cdot
e_{3}=e_{3}$;

$(D_{7})$ $e_{1}\cdot e_{1}=e_{1},~{}e_{2}\cdot e_{3}=e_{2},~{}e_{3}\cdot
e_{3}=e_{3}$;

$(D_{8})$ $e_{1}\cdot e_{3}=e_{1},~{}e_{3}\cdot e_{1}=e_{1},~{}e_{3}\cdot
e_{2}=e_{2},~{}e_{3}\cdot e_{3}=e_{3}$;

$(D_{9})$ $e_{1}\cdot e_{1}=e_{1},~{}e_{1}\cdot e_{2}=e_{2}\cdot
e_{1}=e_{2},~{}e_{1}\cdot e_{3}=e_{3}\cdot e_{1}=e_{3},~{}e_{3}\cdot
e_{2}=e_{2},~{}e_{3}\cdot e_{3}=e_{3}$;

$(D_{10})$ $e_{3}\cdot e_{1}=e_{1},~{}e_{3}\cdot e_{2}=e_{2},~{}e_{3}\cdot
e_{3}=e_{3}$;

$(D_{11})$ $e_{1}\cdot e_{3}=e_{1},~{}e_{2}\cdot e_{3}=e_{2},~{}e_{3}\cdot
e_{3}=e_{3}$;

$(D_{12})$ $e_{3}\cdot e_{1}=e_{1},~{}e_{2}\cdot e_{3}=e_{2},~{}e_{3}\cdot
e_{3}=e_{3}$.

######  Definition 2.6.

Let $(A,\cdot)$ be a pre-Lie algebra. A linear map $N:A\rightarrow A$ is
called a Nijenhuis operator if

(2.6) $N(x)\cdot N(y)=N(N(x)\cdot y+x\cdot N(y)-N(x\cdot
y)),\quad\forall~{}x,y\in A.$

Let $(A,\cdot)$ be a pre-Lie algebra and $\left\\{e_{1},\cdots,e_{n}\right\\}$
be a basis of $A$. A linear map $N:A\rightarrow A$ is a Nijenhuis operator if

(2.7) $N(e_{i})\cdot N(e_{j})=N(N(e_{i})\cdot e_{j}+e_{i}\cdot
N(e_{j})-N(e_{i}\cdot e_{j})),\quad\forall~{}1\leq i,j\leq n.$

######  Proposition 2.7.

Let $(A,\cdot)$ be a pre-Lie algebra and
$\left\\{e_{1},e_{2},\cdots,e_{n}\right\\}$ be a basis of $A$. For all
positive integers $i,j,t=1,2,\cdots,n$ and structural constants
$C_{ij}^{t}\in\mathbb{C}$, set

(2.8) $e_{i}\cdot e_{j}=\sum_{t=1}^{n}C_{ij}^{t}e_{t}.$

If $N:A\rightarrow A$ is a linear map given by

(2.9)
$N\left(e_{i}\right)=\sum_{j=1}^{n}n_{ij}e_{j},\quad~{}n_{ij}\in\mathbb{C},$

then $N$ is a Nijenhuis operator on $A$ if and only if $n_{ij}$ satisfies the
following equations:

(2.10)
$\sum_{k,l,m,t=1}^{n}\left(C_{kl}^{m}n_{ik}n_{jl}+C_{ij}^{t}n_{tl}n_{lm}-C_{kj}^{l}n_{ik}n_{lm}-C_{il}^{k}n_{jl}n_{km}\right)=0.$

###### Proof.

For all $e_{i},e_{j}\in\left\\{e_{1},\cdots,e_{n}\right\\}$, let

$N\left(e_{i}\right)=\sum_{k=1}^{n}n_{ik}e_{k},~{}N\left(e_{j}\right)=\sum_{l=1}^{n}n_{jl}e_{l},~{}e_{i}\cdot
e_{j}=\sum_{t=1}^{n}C_{ij}^{t}e_{t}.$

By calculating the left-hand side of (2.7), we have:

$N(e_{i})\cdot N(e_{j})=\sum_{k,l=1}^{n}n_{ik}n_{jl}e_{k}\cdot
e_{l}=\sum_{k,l,m=1}^{n}C_{kl}^{m}n_{ik}n_{jl}e_{m}.$

By calculating the right-hand side of (2.7), we have:

$\displaystyle N(N(e_{i})\cdot e_{j}+e_{i}\cdot N(e_{j})-N(e_{i}\cdot e_{j}))$
$\displaystyle=$ $\displaystyle N(\sum_{k=1}^{n}n_{ik}e_{k}\cdot
e_{j}+e_{i}\cdot\sum_{l=1}^{n}n_{jl}e_{l}-N(\sum_{t=1}^{n}C_{ij}^{t}e_{t}))$
$\displaystyle=$ $\displaystyle
N(\sum_{k,l=1}^{n}C_{kj}^{l}n_{ik}e_{l}+\sum_{l,k=1}^{n}C_{il}^{k}n_{jl}e_{k}-\sum_{t,l=1}^{n}C_{ij}^{t}n_{tl}e_{l})$
$\displaystyle=$
$\displaystyle\sum_{k,l,m,t=1}^{n}(C_{kj}^{l}n_{ik}n_{lm}e_{m}+C_{il}^{k}n_{jl}n_{km}e_{m}-C_{ij}^{t}n_{tl}n_{lm}e_{m}).$

If $N$ is a Nijenhuis operator, then

$\displaystyle N(e_{i})\cdot N(e_{j})-N(N(e_{i})\cdot e_{j}+e_{i}\cdot
N(e_{j})-N(e_{i}\cdot e_{j}))$ $\displaystyle=$
$\displaystyle\sum_{k,l,m=1}^{n}C_{kl}^{m}n_{ik}n_{jl}e_{m}-\sum_{k,l,m,t=1}^{n}(C_{kj}^{l}n_{ik}n_{lm}e_{m}+C_{il}^{k}n_{jl}n_{km}e_{m}-C_{ij}^{t}n_{tl}n_{lm}e_{m})$
$\displaystyle=$
$\displaystyle\sum_{k,l,m,t=1}^{n}(C_{kl}^{m}n_{ik}n_{jl}e_{m}+C_{ij}^{t}n_{tl}n_{lm}e_{m}-C_{kj}^{l}n_{ik}n_{lm}e_{m}-C_{il}^{k}n_{jl}n_{km}e_{m})$
$\displaystyle=$ $\displaystyle 0.$

Therefore, (2.10) holds. ∎

##  3\. Nijenhuis operators on 2-dimensional pre-Lie algebras

Using the classification results of 2-dimensional commutative and non-
commutative complex pre-Lie algebras given in Propositions 2.2 and 2.3, this
section aims to determine all Nijenhuis operators on $(A_{i},\cdot)$ and
$(B_{i},\cdot)$, respectively.

###  3.1. Nijenhuis operators on 2-dimensional commutative pre-Lie algebras

The purpose of this subsection is to determine all Nijenhuis operators on
2-dimensional commutative pre-Lie algebras given by Proposition 2.2. We use
$N_{A_{i}}^{j}$ to represent the $j$-th Nijenhuis operator on the pre-Lie
algebras $(A_{i},\cdot)$.

######  Proposition 3.1.

Let $N:A_{i}\rightarrow A_{i}$ be a linear map on pre-Lie algebras
$(A_{i},\cdot)$ defined by (2.9). Then the Nijenhuis operators on
$2$-dimensional commutative pre-Lie algebras are given as follows:

$(1)$ The Nijenhuis operators on the pre-Lie algebra $(A_{1},\cdot)$ are:

  * $N_{A_{1}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{A_{1}}^{1}(e_{2})=n_{21}e_{1}+(n_{11}+n_{21})e_{2},~{}n_{21}\neq 0.$

  * $N_{A_{1}}^{2}(e_{1})=n_{11}e_{1},~{}~{}N_{A_{1}}^{2}(e_{2})=n_{22}e_{2}.$

  * $N_{A_{1}}^{3}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{A_{1}}^{3}(e_{2})=(n_{11}-n_{12})e_{2},~{}n_{12}\neq 0.$

$(2)$ The Nijenhuis operators on the pre-Lie algebra $(A_{2},\cdot)$ are:

  * $N_{A_{2}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{A_{2}}^{1}(e_{2})=n_{22}e_{2}.$

  * $N_{A_{2}}^{2}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{A_{2}}^{2}(e_{2})=n_{11}e_{2},~{}n_{12}\neq 0.$

$(3)$ The Nijenhuis operators on the pre-Lie algebra $(A_{3},\cdot)$ are:

  * $N_{A_{3}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{A_{3}}^{1}(e_{2})=n_{22}e_{2}.$

  * $N_{A_{3}}^{2}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{A_{3}}^{2}(e_{2})=n_{11}e_{2},~{}n_{12}\neq 0.$

$(4)$ The Nijenhuis operators on the pre-Lie algebra $(A_{4},\cdot)$ are:

  * $N_{A_{4}}^{1}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{A_{4}}^{1}(e_{2})=n_{21}e_{1}+n_{22}e_{2}.$

$(5)$ The Nijenhuis operators on the pre-Lie algebra $(A_{5},\cdot)$ are:

  * $N_{A_{5}}^{1}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{A_{5}}^{1}(e_{2})=n_{11}e_{2}.$

###### Proof.

Let $\left\\{e_{1},e_{2}\right\\}$ be a basis of 2-dimensional commutative
pre-Lie algebras $(A_{i},\cdot)$. Set

$N(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}N(e_{2})=n_{21}e_{1}+n_{22}e_{2}.$

In order to show that $N$ is a Nijenhuis operator, we only need to check

(3.11) $\displaystyle N(e_{1})\cdot N(e_{1})$ $\displaystyle=$ $\displaystyle
N(N(e_{1})\cdot e_{1}+e_{1}\cdot N(e_{1})-N(e_{1}\cdot e_{1})),$ (3.12)
$\displaystyle N(e_{1})\cdot N(e_{2})$ $\displaystyle=$ $\displaystyle
N(N(e_{1})\cdot e_{2}+e_{1}\cdot N(e_{2})-N(e_{1}\cdot e_{2})),$ (3.13)
$\displaystyle N(e_{2})\cdot N(e_{1})$ $\displaystyle=$ $\displaystyle
N(N(e_{2})\cdot e_{1}+e_{2}\cdot N(e_{1})-N(e_{2}\cdot e_{1})),$ (3.14)
$\displaystyle N(e_{2})\cdot N(e_{2})$ $\displaystyle=$ $\displaystyle
N(N(e_{2})\cdot e_{2}+e_{2}\cdot N(e_{2})-N(e_{2}\cdot e_{2})).$

Calculating the left-hand side of (3.11) for the pre-Lie algebra
$(A_{1},\cdot)$, we have:

(3.15) $\displaystyle N(e_{1})\cdot N(e_{1})$ $\displaystyle=$
$\displaystyle(n_{11}e_{1}+n_{12}e_{2})(n_{11}e_{1}+n_{12}e_{2})$
$\displaystyle=$ $\displaystyle{n_{11}}^{2}e_{1}+{n_{12}}^{2}e_{2}.$

Calculating the right-hand side of (3.11), we have:

(3.16) $\displaystyle N(N(e_{1})\cdot e_{1}+e_{1}\cdot N(e_{1})-N(e_{1}\cdot
e_{1}))$ $\displaystyle=$ $\displaystyle N((n_{11}e_{1}+n_{12}e_{2})\cdot
e_{1}+e_{1}\cdot(n_{11}e_{1}+n_{12}e_{2})-N(e_{1}))$ $\displaystyle=$
$\displaystyle N(2n_{11}e_{1}-n_{11}e_{1}-n_{12}e_{2})$ $\displaystyle=$
$\displaystyle N(n_{11}e_{1}-n_{12}e_{2})$ $\displaystyle=$
$\displaystyle{n_{11}}^{2}e_{1}+n_{11}n_{12}e_{2}-n_{12}n_{21}e_{1}-n_{12}n_{22}e_{2}$
$\displaystyle=$
$\displaystyle(n_{11}^{2}-n_{12}n_{21})e_{1}+(n_{11}n_{12}-n_{12}n_{22})e_{2}.$

Comparing the coefficients in (3.15) and (3.16), we have:

(3.17)
$n_{{11}}^{2}=n_{11}^{2}-n_{12}n_{21},~{}{n_{12}}^{2}=n_{11}n_{12}-n_{12}n_{22}.$

Similarly, by calculating (3.12)-(3.14), we have:

(3.18) $\displaystyle
n_{11}n_{21}=n_{11}n_{21}+n_{12}n_{21},~{}n_{12}n_{22}=n_{12}n_{21}+n_{12}n_{22}.$
(3.19)
$\displaystyle{n_{21}}^{2}=n_{21}n_{22}-n_{11}n_{21},~{}{n_{22}}^{2}={n_{22}}^{2}-n_{12}n_{21}.$

Simplifying (3.17)-(3.19), and we obtain the following three equations:

(3.20) $\left\\{\begin{array}[]{l}n_{12}n_{21}=0,\\\
{n_{12}}^{2}=n_{11}n_{12}-n_{12}n_{22},\\\
{n_{21}}^{2}=n_{21}n_{22}-n_{11}n_{21}.\end{array}\right.$

To solve the quadratic equations, we distinguish the two cases depending on
whether or not $n_{12}=0$.

$\mathbf{Case~{}1}$: If $n_{12}=0$, then (3.20) implies
$n_{21}(n_{21}-n_{22}+n_{11})=0$. There are two subcases: $n_{21}=0$ and
$n_{21}\neq 0$. If $n_{21}\neq 0$, we have $n_{11}+n_{21}=n_{22}$. Then we get
Nijenhuis operators $N_{A_{1}}^{1}$:

  * $N_{A_{1}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{A_{1}}^{1}(e_{2})=n_{21}e_{1}+(n_{11}+n_{21})e_{2},~{}n_{21}\neq 0.$

If $n_{21}=0$, then we get Nijenhuis operators $N_{A_{1}}^{2}$:

  * $N_{A_{1}}^{2}(e_{1})=n_{11}e_{1},~{}~{}N_{A_{1}}^{1}(e_{2})=n_{22}e_{2}.$

$\mathbf{Case~{}2}$: If $n_{12}\neq 0$, we have
$n_{21}=0,~{}n_{22}=n_{11}-n_{12}$. Then we get Nijenhuis operators
$N_{A_{1}}^{3}$:

  * $N_{A_{1}}^{3}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{A_{1}}^{3}(e_{2})=(n_{11}-n_{12})e_{2},~{}n_{12}\neq 0.$

Therefore, we obtain all Nijenhuis operators on the pre-Lie algebra
$(A_{1},\cdot)$. Similarly, we can obtain all Nijenhuis operators on other
2-dimensional commutative pre-Lie algebras. ∎

###  3.2. Nijenhuis operators on 2-dimensional non-commutative pre-Lie
algebras

The classification of 2-dimensional non-commutative pre-Lie algebras is given
by Proposition 2.3. We use $N_{B_{i}}^{j}$ to represent the $j$-th Nijenhuis
operator on the pre-Lie algebras $(B_{i},\cdot)$. Then we obtain the following
results through direct computation.

######  Proposition 3.2.

Let $N:B_{i}\rightarrow B_{i}$ be a linear map on pre-Lie algebras
$(B_{i},\cdot)$ defined by (2.9). Then the Nijenhuis operators on
2-dimensional non-commutative pre-Lie algebras are given as follows:

$(1)$ The Nijenhuis operators on the pre-Lie algebra $(B_{1},\cdot)$ are:

  * $N_{B_{1}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{B_{1}}^{1}(e_{2})=n_{21}e_{1}+n_{11}e_{2}.$

$(2)$ The Nijenhuis operators on the pre-Lie algebra $(B_{2},\cdot)$ are:

  * $N_{B_{2}}^{1}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{B_{2}}^{1}(e_{2})=n_{21}e_{1}+n_{22}e_{2}.$

$(3)$ The Nijenhuis operators on the pre-Lie algebra $(B_{3},\cdot)$ are:

  * $N_{B_{3}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{B_{3}}^{1}(e_{2})=n_{21}e_{1}+n_{11}e_{2},~{}n_{21}\neq 0.$

  * $N_{B_{3}}^{2}(e_{1})=n_{11}e_{1},~{}~{}N_{B_{3}}^{2}(e_{2})=n_{22}e_{2}.$

$(4)$ The Nijenhuis operators on the pre-Lie algebra $(B_{4},\cdot)$ are:

  * $N_{B_{4}}^{1}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{B_{4}}^{1}(e_{2})=n_{21}e_{1}+n_{22}e_{2}.$

$(5)$ The Nijenhuis operators on the pre-Lie algebra $(B_{5},\cdot)$ are:

  * $N_{B_{5}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{B_{5}}^{1}(e_{2})=n_{21}e_{1}+n_{11}e_{2}.$

  * $N_{B_{5}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{B_{5}}^{1}(e_{2})=\frac{n_{11}-n_{22}}{l-1}e_{1}+n_{22}e_{2},~{}n_{11}\neq n_{22},~{}l\neq 1.$

$(6)$ The Nijenhuis operators on the pre-Lie algebra $(B_{6},\cdot)$ are:

  * $N_{B_{6}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{B_{6}}^{1}(e_{2})=n_{21}e_{1}+(n_{11}\pm n_{21}\sqrt{-1})e_{2}.$

  * $N_{B_{6}}^{2}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{B_{6}}^{2}(e_{2})=-n_{12}e_{1}+n_{11}e_{2},~{}n_{12}\neq 0.$

###### Proof.

The proof is similar to that in Proposition 3.1. ∎

##  4\. Nijenhuis operators on 3-dimensional associative algebras

As associative algebras belong to the class of special pre-Lie algebras, in
this section we will use the classification results of commutative and non-
commutative associative algebras (Proposition 2.4 and 2.5) to determine all
Nijenhuis operators on $(C_{i},\cdot)$ and $(D_{i},\cdot)$, respectively.

###  4.1. Nijenhuis operators on 3-dimensional commutative associative
algebras

The purpose of this subsection is to determine all Nijenhuis operators on
3-dimensional commutative associative algebras. Proposition 2.4 gives the
classification of 3-dimensional complex commutative associative algebras. We
use $N_{C_{i}}^{j}$ to represent the $j$-th Nijenhuis operator on the pre-Lie
algebras $(C_{i},\cdot)$. Then we have the following results.

######  Proposition 4.1.

Let $N:C_{i}\rightarrow C_{i}$ be a linear map on commutative associative
algebras $(C_{i},\cdot)$ defined by (2.9). Then the Nijenhuis operators on
3-dimensional commutative associative algebras are given as follows:

$(1)$ The Nijenhuis operators on the associative algebra $(C_{1},\cdot)$ are:

  * $N_{C_{1}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{1}}^{1}(e_{2})=n_{21}e_{1}+n_{22}e_{2},~{}~{}N_{C_{1}}^{1}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{11}e_{3}.$

  * $N_{C_{1}}^{2}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{C_{1}}^{2}(e_{2})=-\frac{(n_{11}-n_{33})^{2}}{n_{12}}e_{1}+(2n_{33}-n_{11})e_{2},~{}~{}N_{C_{1}}^{2}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{33}e_{3},~{}n_{12}\neq 0.$

$(2)$ The Nijenhuis operators on the associative algebra $(C_{2},\cdot)$ are:

  * $N_{C_{2}}^{1}(e_{1})=n_{11}e_{1}+n_{12}e_{2}+n_{13}e_{3},~{}~{}N_{C_{2}}^{1}(e_{2})=n_{21}e_{1}+n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{C_{2}}^{1}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{33}e_{3}.$

$(3)$ The Nijenhuis operators on the associative algebra $(C_{3},\cdot)$ are:

  * $N_{C_{3}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{3}}^{1}(e_{2})=n_{21}e_{1}+n_{11}e_{2},~{}~{}N_{C_{3}}^{1}(e_{3})=n_{31}e_{1}+n_{11}e_{3}.$

  * $N_{C_{3}}^{2}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{3}}^{2}(e_{2})=n_{21}e_{1}+n_{11}e_{2},N_{C_{3}}^{2}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+(n_{11}\pm n_{32}\sqrt{-1})e_{3},\\\ ~{}~{}n_{32}\neq 0.$

  * $N_{C_{3}}^{3}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{3}}^{3}(e_{2})=n_{21}e_{1}+(n_{11}\pm n_{23}\sqrt{-1})e_{2}+n_{23}e_{3},~{}N_{C_{3}}^{3}(e_{3})=n_{31}e_{1}+n_{11}e_{3},~{}n_{23}\neq 0.$

  * $N_{C_{3}}^{4}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{3}}^{4}(e_{2})=n_{21}e_{1}+(n_{11}-n_{23}\sqrt{-1})e_{2}+n_{23}e_{3},~{}~{}N_{C_{3}}^{4}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+(n_{11}+n_{32}\sqrt{-1})e_{3},~{}n_{23},~{}n_{32}\neq 0.$

  * $N_{C_{3}}^{5}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{3}}^{5}(e_{2})=n_{21}e_{1}+(n_{11}+n_{23}\sqrt{-1})e_{2}+n_{23}e_{3},~{}~{}N_{C_{3}}^{5}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+(n_{11}-n_{32}\sqrt{-1})e_{3},~{}n_{23},~{}n_{32}\neq 0.$

$(4)$ The Nijenhuis operators on the associative algebra $(C_{4},\cdot)$ are:

  * $N_{C_{4}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{4}}^{1}(e_{2})=n_{21}e_{1}+n_{11}e_{2},~{}~{}N_{C_{4}}^{1}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{11}e_{3}.$

$(5)$ The Nijenhuis operators on the associative algebra $(C_{5},\cdot)$ are:

  * $N_{C_{5}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{5}}^{1}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{5}}^{1}(e_{3})=n_{33}e_{3}.$

  * $N_{C_{5}}^{2}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{5}}^{2}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{5}}^{2}(e_{3})=n_{31}e_{1}+(n_{11}+n_{31})e_{3},~{}n_{31}\neq 0.$

  * $N_{C_{5}}^{3}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{5}}^{3}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{5}}^{3}(e_{3})=n_{32}e_{2}+(n_{22}+n_{32})e_{3},~{}n_{32}\neq 0.$

  * $N_{C_{5}}^{4}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{5}}^{4}(e_{2})=n_{21}e_{1}+(n_{11}+n_{21})e_{2},~{}~{}N_{C_{5}}^{4}(e_{3})=n_{33}e_{3},~{}n_{21}\neq 0.$

  * $N_{C_{5}}^{5}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{5}}^{5}(e_{2})=(n_{23}+n_{33})e_{2}+n_{23}e_{3},~{}~{}N_{C_{5}}^{5}(e_{3})=n_{33}e_{3},~{}n_{23}\neq 0.$

  * $N_{C_{5}}^{6}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{5}}^{6}(e_{2})=n_{21}e_{1}+(n_{11}+n_{21})e_{2},~{}~{}N_{C_{5}}^{6}(e_{3})=n_{32}e_{1}+n_{32}e_{2}+(n_{11}+n_{21}+n_{32})e_{3},~{}n_{21},~{}n_{32}\neq 0.$

  * $N_{C_{5}}^{7}(e_{1})=(n_{22}+n_{32}-n_{31})e_{1},~{}~{}N_{C_{5}}^{7}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{5}}^{7}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+(n_{22}+n_{32})e_{3},~{}n_{31},~{}n_{32}\neq 0.$

  * $N_{C_{5}}^{8}(e_{1})=n_{11}e_{1}+n_{13}e_{3},~{}~{}N_{C_{5}}^{8}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{5}}^{8}(e_{3})=(n_{11}+n_{13})e_{3},~{}n_{13}\neq 0.$

  * $N_{C_{5}}^{9}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{C_{5}}^{9}(e_{2})=(n_{11}-n_{12})e_{2},~{}~{}N_{C_{5}}^{9}(e_{3})=n_{33}e_{3},~{}n_{12}\neq 0.$

  * $N_{C_{5}}^{10}(e_{1})=n_{11}e_{1}+n_{13}e_{3},~{}~{}N_{C_{5}}^{10}(e_{2})=n_{23}e_{1}+(n_{11}+n_{23})e_{2}+n_{23}e_{3},~{}~{}N_{C_{5}}^{10}(e_{3})=(n_{11}-n_{13})e_{3},~{}n_{13},~{}n_{23}\neq 0.$

  * $N_{C_{5}}^{11}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{C_{5}}^{11}(e_{2})=(n_{11}-n_{12})e_{2},~{}~{}N_{C_{5}}^{11}(e_{3})=n_{31}e_{1}+n_{31}e_{2}+(n_{31}+n_{11})e_{3},~{}n_{12},~{}n_{31}\neq 0.$

  * $N_{C_{5}}^{12}(e_{1})=(n_{23}+n_{33}-n_{21})e_{1},~{}~{}N_{C_{5}}^{12}(e_{2})=n_{21}e_{1}+(n_{23}+n_{33})e_{2}+n_{23}e_{3},~{}~{}N_{C_{5}}^{12}(e_{3})=n_{33}e_{3},~{}n_{21},~{}n_{23}\neq 0.$

  * $N_{C_{5}}^{13}(e_{1})=(n_{22}-n_{23}-n_{31})e_{1},~{}~{}N_{C_{5}}^{13}(e_{2})=n_{23}e_{1}+n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{C_{5}}^{13}(e_{3})=n_{31}e_{1}+(n_{22}-n_{23})e_{3},~{}n_{31},~{}n_{23}\neq 0.$

  * $N_{C_{5}}^{14}(e_{1})=n_{11}e_{1}+n_{12}e_{2}+n_{13}e_{3},~{}~{}N_{C_{5}}^{14}(e_{2})=(n_{11}-n_{12})e_{2},~{}~{}N_{C_{5}}^{14}(e_{3})=(n_{11}-n_{13})e_{3},~{}n_{12},~{}n_{13}\neq 0.$

  * $N_{C_{5}}^{15}(e_{1})=n_{11}e_{1}+n_{12}e_{2}+n_{12}e_{3},~{}~{}N_{C_{5}}^{15}(e_{2})=(n_{11}-n_{12}-n_{32})e_{2},~{}~{}N_{C_{5}}^{15}(e_{3})=n_{32}e_{2}+(n_{11}-n_{12})e_{3},~{}n_{12},~{}n_{13},~{}n_{32}\neq 0.$

  * $N_{C_{5}}^{16}(e_{1})=n_{11}e_{1}+n_{12}e_{2}+n_{12}e_{3},~{}~{}N_{C_{5}}^{16}(e_{2})=(n_{11}-n_{12})e_{2}+n_{23}e_{3},~{}~{}N_{C_{5}}^{16}(e_{3})=(n_{11}-n_{12}-n_{23})e_{3},~{}n_{12},~{}n_{13},~{}n_{23}\neq 0.$

$(6)$ The Nijenhuis operators on the associative algebra $(C_{6},\cdot)$ are:

  * $N_{C_{6}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{6}}^{1}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{6}}^{1}(e_{3})=n_{33}e_{3}.$

  * $N_{C_{6}}^{2}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{6}}^{2}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{6}}^{2}(e_{3})=n_{31}e_{1}+n_{11}e_{3},~{}n_{31}\neq 0.$

  * $N_{C_{6}}^{3}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{6}}^{3}(e_{2})=n_{21}e_{1}+n_{11}e_{2},~{}~{}N_{C_{6}}^{3}(e_{3})=n_{33}e_{3},~{}n_{21}\neq 0.$

  * $N_{C_{6}}^{4}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{6}}^{4}(e_{2})=n_{21}e_{1}+n_{11}e_{2},~{}~{}N_{C_{6}}^{4}(e_{3})=n_{31}e_{1}+n_{11}e_{3},~{}n_{21},~{}n_{31}\neq 0.$

  * $N_{C_{6}}^{5}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{6}}^{5}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{6}}^{5}(e_{3})=n_{32}e_{2}+(n_{22}+n_{32})e_{3},~{}n_{32}\neq 0.$

  * $N_{C_{6}}^{6}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{6}}^{6}(e_{2})=(n_{23}+n_{33})e_{2}+n_{23}e_{3},~{}~{}N_{C_{6}}^{6}(e_{3})=n_{33}e_{3},~{}n_{23}\neq 0.$

  * $N_{C_{6}}^{7}(e_{1})=(n_{22}+n_{32})e_{1},~{}~{}N_{C_{6}}^{7}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{6}}^{7}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+(n_{22}+n_{32})e_{3},~{}n_{31},~{}n_{32}\neq 0.$

  * $N_{C_{6}}^{8}(e_{1})=(n_{23}+n_{33})e_{1},N_{C_{6}}^{8}(e_{2})=n_{21}e_{1}+(n_{23}+n_{33})e_{2}+n_{23}e_{3},N_{C_{6}}^{8}(e_{3})=n_{33}e_{3},n_{21},n_{23}\neq 0.$

$(7)$ The Nijenhuis operators on the associative algebra $(C_{7},\cdot)$ are:

  * $N_{C_{7}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{7}}^{1}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{7}}^{1}(e_{3})=n_{33}e_{3}.$

  * $N_{C_{7}}^{2}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{7}}^{2}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{7}}^{2}(e_{3})=n_{31}e_{1}+n_{11}e_{3},~{}n_{31}\neq 0.$

  * $N_{C_{7}}^{3}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{7}}^{3}(e_{2})=n_{21}e_{1}+n_{11}e_{2},~{}~{}N_{C_{7}}^{3}(e_{3})=n_{31}e_{1}+n_{11}e_{3},~{}n_{21}\neq 0.$

  * $N_{C_{7}}^{4}(e_{1})=(n_{22}+n_{32})e_{1},~{}~{}N_{C_{7}}^{4}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{7}}^{4}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+(n_{22}+n_{32})e_{3},~{}n_{31},~{}n_{32}\neq 0.$

  * $N_{C_{7}}^{5}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{7}}^{5}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{7}}^{5}(e_{3})=n_{32}e_{2}+(n_{22}+n_{32})e_{3},~{}n_{32}\neq 0.$

  * $N_{C_{7}}^{6}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{7}}^{6}(e_{2})=(n_{23}+n_{33})e_{2}+n_{23}e_{3},~{}~{}N_{C_{7}}^{6}(e_{3})=n_{33}e_{3},~{}n_{23}\neq 0.$

  * $N_{C_{7}}^{7}(e_{1})=n_{22}e_{1},~{}~{}N_{C_{7}}^{7}(e_{2})=n_{21}e_{1}+n_{22}e_{2},~{}~{}N_{C_{7}}^{7}(e_{3})=-n_{21}e_{1}+n_{32}e_{2}+(n_{22}+n_{32})e_{3},~{}n_{21},~{}n_{32}\neq 0.$

  * $N_{C_{7}}^{8}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{7}}^{8}(e_{2})=n_{21}e_{1}+(n_{23}+n_{11})e_{2}+n_{23}e_{3},~{}~{}N_{C_{7}}^{8}(e_{3})=-n_{21}e_{1}+n_{11}e_{3},~{}n_{21},~{}n_{23}\neq 0.$

$(8)$ The Nijenhuis operators on the associative algebra $(C_{8},\cdot)$ are:

  * $N_{C_{8}}^{1}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{C_{8}}^{1}(e_{2})=n_{21}e_{1}+n_{22}e_{2},~{}~{}N_{C_{8}}^{1}(e_{3})=n_{33}e_{3}.$

  * $N_{C_{8}}^{2}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{C_{8}}^{2}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{8}}^{2}(e_{3})=n_{32}e_{2}+n_{22}e_{3},~{}n_{32}\neq 0.$

  * $N_{C_{8}}^{3}(e_{1})=(n_{33}-\frac{n_{21}n_{32}}{n_{31}})e_{1}+\frac{n_{32}n_{33}-n_{22}n_{32}}{n_{31}}e_{2},~{}~{}N_{C_{8}}^{3}(e_{2})=n_{21}e_{1}+n_{22}e_{2},~{}~{}N_{C_{8}}^{3}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{33}e_{3},~{}n_{31}\neq 0.$

$(9)$ The Nijenhuis operators on the associative algebra $(C_{9},\cdot)$ are:

  * $N_{C_{9}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{9}}^{1}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{9}}^{1}(e_{3})=n_{33}e_{3}.$

  * $N_{C_{9}}^{2}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{9}}^{2}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{9}}^{2}(e_{3})=n_{32}e_{2}+n_{22}e_{3},~{}n_{32}\neq 0.$

  * $N_{C_{9}}^{3}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{9}}^{3}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{9}}^{3}(e_{3})=n_{31}e_{1}+n_{11}e_{3},~{}n_{31}\neq 0.$

  * $N_{C_{9}}^{4}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{9}}^{4}(e_{2})=n_{11}e_{2},~{}~{}N_{C_{9}}^{4}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{11}e_{3},~{}n_{31},~{}n_{32}\neq 0.$

  * $N_{C_{9}}^{5}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{9}}^{5}(e_{2})=n_{21}e_{1}+n_{22}e_{2},~{}~{}N_{C_{9}}^{5}(e_{3})=n_{31}e_{1}+n_{11}e_{3},~{}n_{21}\neq 0.$

  * $N_{C_{9}}^{6}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{C_{9}}^{6}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{9}}^{6}(e_{3})=n_{32}e_{2}+n_{22}e_{3},~{}n_{12}\neq 0.$

$(10)$ The Nijenhuis operators on the associative algebra $(C_{10},\cdot)$
are:

  * $N_{C_{10}}^{1}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{C_{10}}^{1}(e_{2})=n_{21}e_{1}+n_{22}e_{2},~{}~{}N_{C_{10}}^{1}(e_{3})=n_{33}e_{3}.$

  * $N_{C_{10}}^{2}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{C_{10}}^{2}(e_{2})=n_{22}e_{2},~{}~{}N_{C_{10}}^{2}(e_{3})=n_{32}e_{2}+n_{22}e_{3},~{}n_{32}\neq 0.$

  * $N_{C_{10}}^{3}(e_{1})=(n_{33}-\frac{n_{21}n_{32}}{n_{31}})e_{1}+\frac{n_{32}n_{33}-n_{22}n_{32}}{n_{31}}e_{2},~{}~{}N_{C_{10}}^{3}(e_{2})=n_{21}e_{1}+n_{22}e_{2},~{}~{}N_{C_{10}}^{3}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{33}e_{3},~{}n_{31}\neq 0.$

$(11)$ The Nijenhuis operators on the associative algebra $(C_{11},\cdot)$
are:

  * $N_{C_{11}}^{1}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{C_{11}}^{1}(e_{2})=n_{11}e_{2},~{}~{}N_{C_{11}}^{1}(e_{3})=n_{33}e_{3}.$

  * $N_{C_{11}}^{2}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{C_{11}}^{2}(e_{2})=n_{11}e_{2},~{}~{}N_{C_{11}}^{2}(e_{3})=n_{32}e_{2}+n_{11}e_{3},~{}n_{32}\neq 0.$

  * $N_{C_{11}}^{3}(e_{1})=n_{11}e_{1}-n_{31}e_{2},~{}~{}N_{C_{11}}^{3}(e_{2})=n_{11}e_{2},~{}~{}N_{C_{11}}^{3}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{11}e_{3},~{}n_{31}\neq 0.$

$(12)$ The Nijenhuis operators on the associative algebra $(C_{12},\cdot)$
are:

  * $N_{C_{12}}^{1}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{C_{12}}^{1}(e_{2})=n_{11}e_{2},~{}~{}N_{C_{12}}^{1}(e_{3})=n_{33}e_{3}.$

  * $N_{C_{12}}^{2}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{C_{12}}^{2}(e_{2})=n_{11}e_{2},~{}~{}N_{C_{12}}^{2}(e_{3})=n_{32}e_{2}+n_{11}e_{3},~{}n_{32}\neq 0.$

  * $N_{C_{12}}^{3}(e_{1})=n_{11}e_{1}+n_{31}e_{2},~{}~{}N_{C_{12}}^{3}(e_{2})=n_{11}e_{2},~{}~{}N_{C_{12}}^{3}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{11}e_{3},~{}n_{31}\neq 0.$

###### Proof.

Let $\left\\{e_{1},e_{2},e_{3}\right\\}$ be a basis of 3-dimensional
associative algebras $(C_{i},\cdot)$, and let $N$ be a linear map on $C_{i}$.
For all $n_{ij}\in\mathbb{F},~{}1\leqslant i,j\leqslant 3$, set

$N(e_{1})=n_{11}e_{1}+n_{12}e_{2}+n_{13}e_{3},~{}N(e_{2})=n_{21}e_{1}+n_{22}e_{2}+n_{23}e_{3},~{}N(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{33}e_{3}.$

The non-zero structural constant for associative algebra $(C_{1},\cdot)$ is
given by the following:

$C_{33}^{1}=1.$

By Proposition 2.7, $N$ is a Nijenhuis operator on $C_{1}$ if and only if
$n_{ij}$ satisfies the following equations:

(4.21) $\left\\{\begin{array}[]{l}n_{13}=0,\\\ n_{23}=0,\\\
({n_{11}}-n_{33})^{2}=-n_{12}n_{21},\\\ n_{12}(2n_{33}-n_{11}-n_{22})=0.\\\
\end{array}\right.$

To solve (4.21), we distinguish the two cases depending on whether $n_{12}$ is
equal to zero or not.

$\mathbf{Case~{}1}$: If $n_{12}=0$, then we have $n_{11}=n_{33}$. As a result,
we obtain Nijenhuis operators $N_{C_{1}}^{1}$:

  * $N_{C_{1}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{C_{1}}^{1}(e_{2})=n_{21}e_{1}+n_{22}e_{2},~{}~{}N_{C_{1}}^{1}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{11}e_{3}.$

$\mathbf{Case~{}2}$: If $n_{12}\neq 0$, then we have
$n_{21}=-\frac{(n_{11}-n_{33})^{2}}{n_{12}},~{}n_{22}=2n_{33}-n_{11}$. Thus we
get Nijenhuis operators $N_{C_{1}}^{2}$:

  * $N_{C_{1}}^{2}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{C_{1}}^{2}(e_{2})=-\frac{(n_{11}-n_{33})^{2}}{n_{12}}e_{1}+(2n_{33}-n_{11})e_{2},~{}~{}N_{C_{1}}^{2}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{33}e_{3},~{}n_{12}\neq 0.$

Similarly, we can obtain all Nijenhuis operators on other 3-dimensional
commutative associative algebras. ∎

###  4.2. Nijenhuis operators on 3-dimensional non-commutative associative
algebras

We use $N_{D_{i}}^{j}$ to represent the $j$-th Nijenhuis operator on the pre-
Lie algebras $(D_{i},\cdot)$. Then we have the following results.

######  Proposition 4.2.

Let $N:D_{i}\rightarrow D_{i}$ be a linear map on non-commutative associative
algebras $(D_{i},\cdot)$ defined by (2.9). Then the Nijenhuis operators on
3-dimensional non-commutative associative algebras are given as follows:

$(1)$ The Nijenhuis operators on the associative algebra $(D_{1},\cdot)$ are:

  * $N_{D_{1}}^{1}(e_{1})=n_{11}e_{1}+n_{13}e_{3},~{}~{}N_{D_{1}}^{1}(e_{2})=n_{21}e_{1}+n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{1}}^{1}(e_{3})=n_{11}e_{3}.$

  * $N_{D_{1}}^{2}(e_{1})=n_{11}e_{1}+n_{12}e_{2}+n_{13}e_{3},~{}~{}N_{D_{1}}^{2}(e_{2})=n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{1}}^{2}(e_{3})=n_{11}e_{3},~{}n_{12}\neq 0.$

  * $N_{D_{1}}^{3}(e_{1})=n_{11}e_{1}+n_{12}e_{2}+n_{13}e_{3},~{}~{}N_{D_{1}}^{3}(e_{2})=n_{21}e_{1}+(n_{33}-\frac{n_{12}n_{21}}{n_{33}-n_{11}})e_{2}+n_{23}e_{3},~{}~{}N_{D_{1}}^{3}(e_{3})=n_{33}e_{3},~{}n_{33}\neq n_{11}.$

$(2)$ The Nijenhuis operators on the associative algebra $(D_{2},\cdot)$ are:

  * $N_{D_{2}}^{1}(e_{1})=n_{11}e_{1}+n_{13}e_{3},~{}~{}N_{D_{2}}^{1}(e_{2})=n_{21}e_{1}+n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{2}}^{1}(e_{3})=n_{22}e_{3},~{}n_{21}\neq 0.$

  * $N_{D_{2}}^{2}(e_{1})=n_{11}e_{1}+n_{12}e_{2}+n_{13}e_{3},~{}~{}N_{D_{2}}^{2}(e_{2})=n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{2}}^{2}(e_{3})=n_{11}e_{3},~{}n_{12}\neq 0.$

  * $N_{D_{2}}^{3}(e_{1})=n_{11}e_{1}+n_{13}e_{3},~{}~{}N_{D_{2}}^{3}(e_{2})=n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{2}}^{3}(e_{3})=n_{22}e_{3}.$

  * $N_{D_{2}}^{4}(e_{1})=n_{33}e_{1}+n_{13}e_{3},~{}~{}N_{D_{2}}^{4}(e_{2})=n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{2}}^{4}(e_{3})=n_{33}e_{3},~{}n_{33}\neq n_{22}.$

$(3)$ The Nijenhuis operators on the associative algebra $(D_{3},\cdot)$ are:

  * $N_{D_{3}}^{1}(e_{1})=(n_{33}+\frac{(-1+\sqrt{1-4\lambda})n_{12}}{2})e_{1}+n_{12}e_{2}+n_{13}e_{3},~{}~{}N_{D_{3}}^{1}(e_{2})=\frac{(-1+\sqrt{1-4\lambda})(n_{22}-n_{33})}{2}e_{1}+n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{3}}^{1}(e_{3})=n_{33}e_{3},~{}n_{12},~{}\lambda\neq 0.$

  * $N_{D_{3}}^{2}(e_{1})=(n_{33}+\frac{(-1-\sqrt{1-4\lambda})n_{12}}{2})e_{1}+n_{12}e_{2}+n_{13}e_{3},~{}~{}N_{D_{3}}^{2}(e_{2})=\frac{(-1-\sqrt{1-4\lambda})(n_{22}-n_{33})}{2}e_{1}+n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{3}}^{2}(e_{3})=n_{33}e_{3},~{}n_{12},~{}\lambda\neq 0.$

  * $N_{D_{3}}^{3}(e_{1})=n_{33}e_{1}+n_{13}e_{3},~{}~{}N_{D_{3}}^{3}(e_{2})=\frac{(-1\pm\sqrt{1-4\lambda})(n_{22}-n_{33})}{2}e_{1}+n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{3}}^{3}(e_{3})=n_{33}e_{3},~{}\lambda\neq 0.$

$(4)$ The Nijenhuis operators on the associative algebra $(D_{4},\cdot)$ are:

  * $N_{D_{4}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{4}}^{1}(e_{2})=n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{4}}^{1}(e_{3})=n_{32}e_{2}+n_{33}e_{3}.$

  * $N_{D_{4}}^{2}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{4}}^{2}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{4}}^{2}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{11}e_{3},~{}n_{31}\neq 0.$

  * $N_{D_{4}}^{3}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{4}}^{3}(e_{2})=n_{21}e_{1}+n_{22}e_{2},~{}~{}N_{D_{4}}^{3}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{11}e_{3},~{}n_{21}\neq 0.$

  * $N_{D_{4}}^{4}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{D_{4}}^{4}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{4}}^{4}(e_{3})=n_{32}e_{2}+n_{22}e_{3},~{}n_{12}\neq 0.$

$(5)$ The Nijenhuis operators on the associative algebra $(D_{5},\cdot)$ are:

  * $N_{D_{5}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{5}}^{1}(e_{2})=n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{5}}^{1}(e_{3})=n_{32}e_{2}+n_{33}e_{3}.$

  * $N_{D_{5}}^{2}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{5}}^{2}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{5}}^{2}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{11}e_{3},~{}n_{31}\neq 0.$

  * $N_{D_{5}}^{3}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{5}}^{3}(e_{2})=n_{21}e_{1}+n_{22}e_{2},~{}~{}N_{D_{5}}^{3}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{11}e_{3},~{}n_{21}\neq 0.$

  * $N_{D_{5}}^{4}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{D_{5}}^{4}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{5}}^{4}(e_{3})=n_{32}e_{2}+n_{22}e_{3},~{}n_{12}\neq 0.$

$(6)$ The Nijenhuis operators on the associative algebra $(D_{6},\cdot)$ are:

  * $N_{D_{6}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{6}}^{1}(e_{2})=n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{6}}^{1}(e_{3})=n_{32}e_{2}+n_{33}e_{3}.$

  * $N_{D_{6}}^{2}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{D_{6}}^{2}(e_{2})=n_{11}e_{2},~{}~{}N_{D_{6}}^{2}(e_{3})=n_{32}e_{2}+n_{11}e_{3},~{}n_{12}\neq 0.$

  * $N_{D_{6}}^{3}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{6}}^{3}(e_{2})=n_{21}e_{1}+n_{22}e_{2}+n_{21}e_{3},~{}~{}N_{D_{6}}^{3}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+(n_{11}+n_{31})e_{3},~{}n_{21}\neq 0.$

  * $N_{D_{6}}^{4}(e_{1})=n_{11}e_{1}+n_{13}e_{3},~{}~{}N_{D_{6}}^{4}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{6}}^{4}(e_{3})=(n_{11}-n_{13})e_{3},~{}n_{13}\neq 0.$

  * $N_{D_{6}}^{5}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{6}}^{5}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{6}}^{5}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+(n_{11}+n_{31})e_{3},~{}n_{31}\neq 0.$

  * $N_{D_{6}}^{6}(e_{1})=n_{11}e_{1}+n_{12}e_{2}+n_{13}e_{2},~{}~{}N_{D_{6}}^{6}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{6}}^{6}(e_{3})=(\frac{n_{11}n_{12}-n_{12}n_{13}-n_{12}n_{22}}{n_{13}})e_{2}+(n_{11}-n_{13})e_{3},~{}n_{12},~{}n_{13}\neq 0.$

$(7)$ The Nijenhuis operators on the associative algebra $(D_{7},\cdot)$ are:

  * $N_{D_{7}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{7}}^{1}(e_{2})=n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{7}}^{1}(e_{3})=n_{32}e_{2}+n_{33}e_{3}.$

  * $N_{D_{7}}^{2}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{D_{7}}^{2}(e_{2})=n_{11}e_{2},~{}~{}N_{D_{7}}^{2}(e_{3})=n_{32}e_{2}+n_{11}e_{3},~{}n_{12}\neq 0.$

  * $N_{D_{7}}^{3}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{7}}^{3}(e_{2})=n_{21}e_{1}+n_{22}e_{2}+n_{21}e_{3},~{}~{}N_{D_{7}}^{3}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+(n_{11}+n_{31})e_{3},~{}n_{21}\neq 0.$

  * $N_{D_{7}}^{4}(e_{1})=n_{11}e_{1}+n_{13}e_{3},~{}~{}N_{D_{7}}^{4}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{7}}^{4}(e_{3})=(n_{11}-n_{13})e_{3},~{}n_{13}\neq 0.$

  * $N_{D_{7}}^{5}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{7}}^{5}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{7}}^{5}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+(n_{11}+n_{31})e_{3},~{}n_{31}\neq 0.$

  * $N_{D_{7}}^{6}(e_{1})=n_{11}e_{1}+n_{12}e_{2}+n_{13}e_{2},~{}~{}N_{D_{7}}^{6}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{7}}^{6}(e_{3})=(\frac{n_{11}n_{12}-n_{12}n_{13}-n_{12}n_{22}}{n_{13}})e_{2}+(n_{11}-n_{13})e_{3},~{}n_{12},~{}n_{13}\neq 0.$

$(8)$ The Nijenhuis operators on the associative algebra $(D_{8},\cdot)$ are:

  * $N_{D_{8}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{8}}^{1}(e_{2})=n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{8}}^{1}(e_{3})=n_{32}e_{2}+n_{33}e_{3}.$

  * $N_{D_{8}}^{2}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{8}}^{2}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{8}}^{2}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{11}e_{3},~{}n_{31}\neq 0.$

  * $N_{D_{8}}^{3}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{8}}^{3}(e_{2})=n_{21}e_{1}+n_{22}e_{2},~{}~{}N_{D_{8}}^{3}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{11}e_{3},~{}n_{21}\neq 0.$

  * $N_{D_{8}}^{4}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{D_{8}}^{4}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{8}}^{4}(e_{3})=n_{32}e_{2}+n_{22}e_{3},~{}n_{12}\neq 0.$

$(9)$ The Nijenhuis operators on the associative algebra $(D_{9},\cdot)$ are:

  * $N_{D_{9}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{9}}^{1}(e_{2})=n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{9}}^{1}(e_{3})=n_{32}e_{2}+n_{33}e_{3}.$

  * $N_{D_{9}}^{2}(e_{1})=(n_{31}+n_{33})e_{1},~{}~{}N_{D_{9}}^{2}(e_{2})=n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{9}}^{2}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{33}e_{3},~{}n_{31}\neq 0.$

  * $N_{D_{9}}^{3}(e_{1})=(n_{31}+n_{33})e_{1},~{}~{}N_{D_{9}}^{3}(e_{2})=n_{21}e_{1}+n_{22}e_{2}-n_{21}e_{3},~{}~{}N_{D_{9}}^{3}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{33}e_{3},~{}n_{21}\neq 0.$

  * $N_{D_{9}}^{4}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{D_{9}}^{4}(e_{2})=n_{11}e_{2},~{}~{}N_{D_{9}}^{4}(e_{3})=n_{32}e_{2}+n_{11}e_{3},~{}n_{12}\neq 0.$

  * $N_{D_{9}}^{5}(e_{1})=n_{11}e_{1}+n_{12}e_{2}+n_{13}e_{2},~{}~{}N_{D_{9}}^{5}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{9}}^{5}(e_{3})=(\frac{n_{11}n_{12}+n_{12}n_{13}-n_{12}n_{22}}{n_{13}})e_{2}+(n_{11}+n_{13})e_{3},~{}n_{13}\neq 0.$

$(10)$ The Nijenhuis operators on the associative algebra $(D_{10},\cdot)$
are:

  * $N_{D_{10}}^{1}(e_{1})=n_{11}e_{1}+n_{12}e_{2}+n_{13}e_{3},~{}~{}N_{D_{10}}^{1}(e_{2})=n_{21}e_{1}+n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{10}}^{1}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{33}e_{3}.$

$(11)$ The Nijenhuis operators on the associative algebra $(D_{11},\cdot)$
are:

  * $N_{D_{11}}^{1}(e_{1})=n_{11}e_{1}+n_{12}e_{2}+n_{13}e_{3},~{}~{}N_{D_{11}}^{1}(e_{2})=n_{21}e_{1}+n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{11}}^{1}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{33}e_{3}.$

$(12)$ The Nijenhuis operators on the associative algebra $(D_{12},\cdot)$
are:

  * $N_{D_{12}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{12}}^{1}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{12}}^{1}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{33}e_{3}.$

  * $N_{D_{12}}^{2}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{12}}^{2}(e_{2})=n_{21}e_{1}+n_{22}e_{2},~{}~{}N_{D_{12}}^{2}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{11}e_{3},~{}n_{21}\neq 0.$

  * $N_{D_{12}}^{3}(e_{1})=n_{11}e_{1}+n_{12}e_{2},~{}~{}N_{D_{12}}^{3}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{12}}^{3}(e_{3})=n_{31}e_{1}+n_{32}e_{2}+n_{22}e_{3},~{}n_{12}\neq 0.$

  * $N_{D_{12}}^{4}(e_{1})=n_{11}e_{1},~{}~{}N_{D_{12}}^{4}(e_{2})=n_{21}e_{1}+n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{12}}^{4}(e_{3})=\frac{n_{21}n_{33}-n_{11}n_{21}}{n_{23}}e_{1}+n_{32}e_{2}+n_{33}e_{3},~{}n_{23}\neq 0.$

  * $N_{D_{12}}^{5}(e_{1})=n_{11}e_{1}+n_{12}e_{2}+n_{13}e_{3},~{}~{}N_{D_{12}}^{5}(e_{2})=n_{22}e_{2},~{}~{}N_{D_{12}}^{5}(e_{3})=n_{31}e_{1}+\frac{n_{12}n_{33}-n_{12}n_{22}}{n_{13}}~{}e_{2}+n_{33}e_{3},~{}n_{13}\neq 0.$

###### Proof.

The proof is similar to that in Proposition 4.1. ∎

##  5\. Rota-Baxter operators on sub-adjacent Lie algebras ${\mathfrak{g}}(A)$
and solutions of CYBE

In the previous sections, we classified Nijenhuis operators $N$ on
2-dimensional pre-Lie algebras and some 3-dimensional pre-Lie algebras
$(A,\cdot)$. Using these results, we first obtain Rota-Baxter operators $R$ of
weight zero on the sub-adjacent Lie algebras ${\mathfrak{g}}(A)$ in this
section. Then, the solutions of CYBE in the Lie algebras
${\mathfrak{g}}(A)\ltimes_{{\rm ad}^{\ast}}{\mathfrak{g}}(A)^{\ast}$ are
induced by these Rota-Baxter operators.

###  5.1. Rota-Baxter operators on sub-adjacent Lie algebras
${\mathfrak{g}}(A)$

######  Proposition 5.1.

Let $\left(A,\cdot\right)$ be a pre-Lie algebra and
$N:\mathfrak{g}\rightarrow\mathfrak{g}$ a linear map. If $N^{2}=0$ , then $N$
is a Nijenhuis operator if and only if $N$ is a Rota-Baxter operator of weight
zero on $(A,\cdot)$.

###### Proof.

Let $N:\mathfrak{g}\rightarrow\mathfrak{g}$ be a linear map and $N^{2}=0$. If
$N$ is a Nijenhuis operator, then

$\displaystyle N(x)\cdot N(y)$ $\displaystyle=$ $\displaystyle N(N(x)\cdot
y+x\cdot N(y)-N(x\cdot y))$ $\displaystyle=$ $\displaystyle N(N(x)\cdot
y+x\cdot N(y))-N^{2}(x\cdot y)$ $\displaystyle=$ $\displaystyle N(N(x)\cdot
y+x\cdot N(y)),\quad\forall~{}x,y\in A.$

Therefore $N$ is a Rota-Baxter operator of weight zero on $(A,\cdot)$. Proof
of the converse part is similar. ∎

######  Proposition 5.2.

If R is a Rota-Baxter operator of weight zero on a pre-Lie algebra
$(A,\cdot)$, then $R$ is a Rota-Baxter operator of weight zero on its sub-
adjacent Lie algebra ${\mathfrak{g}}(A)$.

###### Proof.

Suppose $R$ is a Rota-Baxter operator of weight zero on a pre-Lie algebra
$(A,\cdot)$. Then for all $x,y\in A$, we have

$\displaystyle R(x)\cdot R(y)=R(R(x)\cdot y+x\cdot R(y)),$ $\displaystyle
R(y)\cdot R(x)=R(R(y)\cdot x+y\cdot R(x)).$

Therefore,

$\displaystyle[R(x),R(y)]$ $\displaystyle=$ $\displaystyle R(x)\cdot
R(y)-R(y)\cdot R(x)$ $\displaystyle=$ $\displaystyle R(R(x)\cdot y+x\cdot
R(y))-R(R(y)\cdot x+y\cdot R(x))$ $\displaystyle=$ $\displaystyle R(R(x)\cdot
y-y\cdot R(x)+x\cdot R(y)-R(y)\cdot x)$ $\displaystyle=$ $\displaystyle
R([R(x),y]+[x,R(y)]),\quad\forall~{}x,y\in A.$

Hence $R$ is a Rota-Baxter operator of weight zero on its sub-adjacent Lie
algebra ${\mathfrak{g}}(A).$ ∎

######  Proposition 5.3.

Let ${\mathfrak{g}}(B)$ be a non-trivial sub-adjacent Lie algebra of
$2$-dimensional pre-Lie algebras $(B,\cdot)$. Then ${\mathfrak{g}}(B)$ has a
basis $\left\\{e_{1},e_{2}\right\\}$ with respect to which the non-zero
product is given by one of the following:

$\displaystyle{\mathfrak{g}}(B_{1})~{}[e_{1},e_{2}]=e_{1};\quad\quad{\mathfrak{g}}(B_{2})~{}[e_{1},e_{2}]=e_{1};$
$\displaystyle{\mathfrak{g}}(B_{3})~{}[e_{1},e_{2}]=e_{1};\quad\quad{\mathfrak{g}}(B_{4})~{}[e_{1},e_{2}]=e_{1};$
$\displaystyle{\mathfrak{g}}(B_{5})~{}[e_{1},e_{2}]=e_{1};\quad\quad{\mathfrak{g}}(B_{6})~{}[e_{1},e_{2}]=e_{2}.$

###### Proof.

The pre-Lie algebra $(B_{1},\cdot)$ is

$e_{2}\cdot e_{1}=-e_{1},~{}e_{2}\cdot e_{2}=e_{1}-e_{2}.$

By (2.5), the Lie brackets in $B_{1}$ are

$\displaystyle[e_{1},e_{1}]=e_{1}\cdot e_{1}-e_{1}\cdot
e_{1}=0,\quad\quad[e_{1},e_{2}]=e_{1}\cdot e_{2}-e_{2}\cdot e_{1}=e_{1},$
$\displaystyle[e_{2},e_{1}]=e_{2}\cdot e_{1}-e_{1}\cdot
e_{2}=-e_{1},\quad\quad[e_{2},e_{2}]=e_{2}\cdot e_{2}-e_{2}\cdot e_{2}=0.$

Therefore the non-zero product of Lie algebra ${\mathfrak{g}}(B_{1})$ is

${\mathfrak{g}}(B_{1})~{}[e_{1},e_{2}]=e_{1}.$

The other sub-adjacent Lie algebras are obtained for the same step. ∎

By Proposition 5.3 we know ${\mathfrak{g}}(B_{1})$ is equal to
${\mathfrak{g}}(B_{2})$, ${\mathfrak{g}}(B_{3})$, ${\mathfrak{g}}(B_{4})$,
${\mathfrak{g}}(B_{5})$. We denote them by ${\mathfrak{g}_{1}}$.

######  Proposition 5.4.

Let ${\mathfrak{g}}(D)$ be a non-trivial sub-adjacent Lie algebra of
$3$-dimensional associative algebras $(D,\cdot)$. Then ${\mathfrak{g}}(D)$ has
a basis $\left\\{e_{1},e_{2},e_{3}\right\\}$ with respect to which the non-
zero product is given by one of the following:

$\displaystyle{\mathfrak{g}}(D_{1})~{}[e_{1},e_{2}]=e_{3};\quad\quad{\mathfrak{g}}(D_{2})[e_{1},e_{2}]=e_{3};\quad\quad{\mathfrak{g}}(D_{3})~{}[e_{1},e_{2}]=e_{3};$
$\displaystyle{\mathfrak{g}}(D_{4})[e_{2},e_{3}]=-e_{2};\quad\quad{\mathfrak{g}}(D_{5})~{}[e_{2},e_{3}]=e_{2};\quad\quad{\mathfrak{g}}(D_{6})[e_{2},e_{3}]=-e_{2};$
$\displaystyle{\mathfrak{g}}(D_{7})~{}[e_{2},e_{3}]=e_{2};\quad\quad{\mathfrak{g}}(D_{8})[e_{2},e_{3}]=-e_{2};\quad\quad{\mathfrak{g}}(D_{9})~{}[e_{2},e_{3}]=-e_{2};$
$\displaystyle{\mathfrak{g}}(D_{10})[e_{1},e_{3}]=-e_{1},~{}[e_{2},e_{3}]=-e_{2};\quad\quad{\mathfrak{g}}(D_{11})~{}[e_{1},e_{3}]=e_{1},~{}[e_{2},e_{3}]=e_{2};$
$\displaystyle{\mathfrak{g}}(D_{12})[e_{1},e_{3}]=-e_{1},~{}[e_{2},e_{3}]=e_{2}.$

###### Proof.

The proof is similar to that of Proposition 5.3 ∎

By Proposition 5.4, we have ${\mathfrak{g}}(D_{1})$ is equal to
${\mathfrak{g}}(D_{2})$, ${\mathfrak{g}}(D_{3})$. We denote them by
${\mathfrak{g}_{2}}$. ${\mathfrak{g}}(D_{4})$ is equal to
${\mathfrak{g}}(D_{6})$, ${\mathfrak{g}}(D_{8})$, ${\mathfrak{g}}(D_{9})$. We
denote them by ${\mathfrak{g}_{3}}$. ${\mathfrak{g}}(D_{5})$ is equal to
${\mathfrak{g}}(D_{7})$. We denote them by ${\mathfrak{g}_{4}}$.

######  Remark 5.5.

The sub-adjacent Lie algebras of the pre-Lie algebras $(A,\cdot)$ and
$(C,\cdot)$ are trivial. Therefore, all linear maps on their sub-adjacent Lie
algebras are Rota-Baxter operators of weight zero.

Let ${\mathfrak{g}}(A)$ be a 2-dimensiona sub-adjacent Lie algebra of
$(A,\cdot)$ and $\left\\{e_{1},e_{2}\right\\}$ be a basis of
${\mathfrak{g}}(A)$. Let $R:{\mathfrak{g}}(A)\rightarrow{\mathfrak{g}}(A)$ be
a linear map. For all $1\leq i,j\leq 2$, set

$R(e_{1})=r_{11}e_{1}+r_{12}e_{2},~{}~{}R(e_{2})=r_{21}e_{1}+r_{22}e_{2},~{}\quad~{}r_{ij}\in\mathbb{C}.$

Then the linear map $R:{\mathfrak{g}}(A)\rightarrow{\mathfrak{g}}(A)$ is
determined by

(5.22) $\left(\begin{array}[]{l}R(e_{1})\\\
R(e_{2})\end{array}\right)=\left(\begin{array}[]{lll}r_{11}&r_{12}\\\
r_{21}&r_{22}\end{array}\right)\left(\begin{array}[]{l}e_{1}\\\
e_{2}\end{array}\right).$

In Proposition 3.1 and Proposition 3.2, the classification of Nijenhuis
operators $N$ on 2-dimensional pre-Lie algebras $(A,\cdot)$ was given. By
Proposition 5.1 and 5.2, if $N^{2}=0$, then we can get Rota-Baxter operators
of weight zero on sub-adjacent Lie algebras ${\mathfrak{g}}(A)$. We use
$R_{{\mathfrak{g}}(A)}^{j}$ to represent the $j$-th Rota-Baxter operator of
weight zero on sub-adjacent Lie algebras ${\mathfrak{g}}(A)$. Then we have the
following results.

######  Proposition 5.6.

Let $R:{\mathfrak{g}_{1}}\rightarrow{\mathfrak{g}_{1}}$ be a linear map on the
sub-adjacent Lie algebra ${\mathfrak{g}_{1}}$ defined by (5.22). Then Rota-
Baxter operators of weight zero on the sub-adjacent Lie algebra
${\mathfrak{g}_{1}}$ derived by Nijenhuis operators $N_{B}$ are given as
follows:

Nijenhuis operators on $B$ | Rota-Baxter operators of weight zero on ${\mathfrak{g}_{1}}$
---|---
$N_{B_{1}}^{1},~{}N_{B_{5}}^{1}$ | $R_{\mathfrak{g}_{1}}^{1}=\left(\begin{array}[]{cc}0&0\\\ n_{21}&0\\\ \end{array}\right)$
$N_{B_{3}}^{2}$ | $R_{\mathfrak{g}_{1}}^{2}=0$
$N_{B_{2}}^{1},~{}N_{B_{4}}^{1}$ | $R_{\mathfrak{g}_{1}}^{3}=\\{N|N^{2}=0\\}$
$N_{B_{3}}^{1}$ | $R_{\mathfrak{g}_{1}}^{4}=\left(\begin{array}[]{cc}0&0\\\ n_{21}&0\\\ \end{array}\right)n_{21}\neq 0$

###### Proof.

The Nijenhuis operators $N_{B_{1}}^{1}$ on the pre-Lie algebra $(B_{1},\cdot)$
are

$N_{B_{1}}^{1}(e_{1})=n_{11}e_{1},~{}~{}N_{B_{1}}^{1}(e_{2})=n_{21}e_{1}+n_{11}e_{2}.$

Then we have

$({N_{B_{1}}^{1}})^{2}(e_{1})={n_{11}}^{2}e_{1},~{}~{}({N_{B_{1}}^{1}})^{2}(e_{2})=2n_{11}n_{21}e_{1}+{n_{11}}^{2}e_{2}.$

If $({N_{B_{1}}^{1}})^{2}=0$, then $n_{11}=0$. Therefore, we get Nijenhuis
operators $N$:

$N(e_{1})=0,~{}~{}N(e_{2})=n_{21}e_{1}.$

By Proposition 5.1 and 5.2, the operators $N=R_{\mathfrak{g}_{1}}^{1}$ are
Rota-Baxter operators of weight zero on the sub-adjacent Lie algebra
${\mathfrak{g}_{1}}$. Then the Rota-Baxter operators
$R_{\mathfrak{g}_{1}}^{1}$ derived by Nijenhuis operators $N_{B_{1}}^{1}$ are:

$R_{\mathfrak{g}_{1}}^{1}=\left(\begin{array}[]{cc}0&0\\\ n_{21}&0\\\
\end{array}\right).$

The other Rota-Baxter operators are obtained for the same step. ∎

######  Remark 5.7.

The Rota-Baxter operator of weight zero on the sub-adjacent Lie algebra
${{\mathfrak{g}}(B_{6})}$ derived by Nijenhuis operators $N_{B}$ is 0.

######  Proposition 5.8.

Let $R:{\mathfrak{g}_{2}}\rightarrow{\mathfrak{g}_{2}}$ be a linear map on the
sub-adjacent Lie algebra ${\mathfrak{g}_{2}}$. Then the matrices of Rota-
Baxter operators of weight zero on the sub-adjacent Lie algebra
${\mathfrak{g}_{2}}$ derived by Nijenhuis operators $N$ are given as follows:

Nijenhuis operators on $D$ | Rota-Baxter operators of weight zero on ${\mathfrak{g}_{2}}$
---|---
$N_{D_{1}}^{1}$ | $R_{\mathfrak{g}_{2}}^{1}=\left(\begin{array}[]{ccc}0&0&0\\\ n_{21}&0&n_{23}\\\ 0&0&0\\\ \end{array}\right)$
| $R_{\mathfrak{g}_{2}}^{2}=\left(\begin{array}[]{ccc}0&0&n_{13}\\\
0&0&n_{23}\\\ 0&0&0\\\ \end{array}\right)n_{13}\neq 0$
$N_{D_{1}}^{2},~{}N_{D_{2}}^{2}$ | $R_{\mathfrak{g}_{2}}^{3}=\left(\begin{array}[]{ccc}0&n_{12}&n_{13}\\\ 0&0&0\\\ 0&0&0\\\ \end{array}\right)n_{12}\neq 0$
$N_{D_{1}}^{3}$ | $R_{\mathfrak{g}_{2}}^{4}=\left(\begin{array}[]{ccc}n_{11}&n_{12}&-\frac{n_{12}n_{23}}{n_{11}}\\\ -\frac{{n_{11}}^{2}}{n_{12}}&-n_{11}&n_{23}\\\ 0&0&0\\\ \end{array}\right)n_{11},n_{12}\neq 0$
$N_{D_{2}}^{1}$ | $R_{\mathfrak{g}_{2}}^{5}=\left(\begin{array}[]{ccc}0&0&0\\\ n_{21}&0&n_{23}\\\ 0&0&0\\\ \end{array}\right)n_{21}\neq 0$
$N_{D_{2}}^{3},~{}N_{D_{3}}^{3}$ | $R_{\mathfrak{g}_{2}}^{6}=\left(\begin{array}[]{ccc}0&0&n_{13}\\\ 0&0&n_{23}\\\ 0&0&0\\\ \end{array}\right)$
$N_{D_{3}}^{1}$ | $R_{\mathfrak{g}_{2}}^{7}=\left(\begin{array}[]{ccc}k_{1}n_{12}&n_{12}&n_{13}\\\ -{k_{1}}^{2}n_{12}&-k_{1}n_{12}&-k_{1}n_{13}\\\ 0&0&0\\\ \end{array}\right)\begin{array}[]{c}k_{1}=\frac{(-1+\sqrt{1-4\lambda})}{2},\\\ n_{12},\lambda\neq 0\end{array}$
$N_{D_{3}}^{2}$ | $R_{\mathfrak{g}_{2}}^{8}=\left(\begin{array}[]{ccc}k_{2}n_{12}&n_{12}&n_{13}\\\ -{k_{2}}^{2}n_{12}&-k_{2}n_{12}&-k_{2}n_{13}\\\ 0&0&0\\\ \end{array}\right)\begin{array}[]{c}k_{2}=\frac{(-1-\sqrt{1-4\lambda})}{2},\\\ n_{12},\lambda\neq 0\end{array}$

###### Proof.

The Nijenhuis operators $N_{D_{1}}^{1}$ on the pre-Lie algebra $(D_{1},\cdot)$
are

$N_{D_{1}}^{1}(e_{1})=n_{11}e_{1}+n_{13}e_{3},~{}~{}N_{D_{1}}^{1}(e_{2})=n_{21}e_{1}+n_{22}e_{2}+n_{23}e_{3},~{}~{}N_{D_{1}}^{1}(e_{3})=n_{11}e_{3}.$

Then we have

$\displaystyle({N_{D_{1}}^{1}})^{2}(e_{1})$ $\displaystyle=$
$\displaystyle{n_{11}}^{2}e_{1}+n_{11}n_{13}e_{2},$
$\displaystyle({N_{D_{1}}^{1}})^{2}(e_{2})$ $\displaystyle=$ $\displaystyle
n_{21}(n_{11}+n_{22})e_{1}+{n_{22}}^{2}e_{2}+(n_{13}n_{21}+n_{22}n_{23}+n_{11}n_{23})e_{3},$
$\displaystyle({N_{D_{1}}^{1}})^{2}(e_{3})$ $\displaystyle=$
$\displaystyle{n_{11}}^{2}e_{3}.$

If $({N_{D_{1}}^{1}})^{2}=0$, then $n_{11}=0$, $n_{22}=0$, $n_{13}n_{21}=0$.

$\mathbf{Case~{}1}$: If $n_{13}=0$, then we get Nijenhuis operators $N$:

$N(e_{1})=0,~{}~{}N(e_{2})=n_{21}e_{1}+n_{23}e_{3},~{}~{}N(e_{3})=0.$

By Proposition 5.1 and 5.2, the operators $N=R_{\mathfrak{g}_{2}}^{1}$ are
Rota-Baxter operators of weight zero on the sub-adjacent Lie algebra
${\mathfrak{g}_{2}}$. The Rota-Baxter operators $R_{\mathfrak{g}_{2}}^{1}$
defined by (5.22) are:

$R_{\mathfrak{g}_{2}}^{1}=\left(\begin{array}[]{ccc}0&0&0\\\
n_{21}&0&n_{23}\\\ 0&0&0\\\ \end{array}\right).$

$\mathbf{Case~{}2}$: If $n_{13}\neq 0$, then $n_{21}=0$. Thus we get Nijenhuis
operators $N$:

$N(e_{1})=n_{13}e_{3},~{}~{}N(e_{2})=n_{23}e_{3},~{}~{}N(e_{3})=0,~{}n_{13}\neq
0.$

Then the Rota-Baxter operators $R_{\mathfrak{g}_{2}}^{1}$ defined by (5.22)
are:

$R_{\mathfrak{g}_{2}}^{2}=\left(\begin{array}[]{ccc}0&0&n_{13}\\\
0&0&n_{23}\\\ 0&0&0\end{array}\right),~{}n_{13}\neq 0.$

The other Rota-Baxter operators are obtained for the same step. ∎

######  Proposition 5.9.

Let $R:{\mathfrak{g}_{3}}\rightarrow{\mathfrak{g}_{3}}$ be a linear map on the
sub-adjacent Lie algebra ${\mathfrak{g}_{3}}$ defined by (5.22). Then Rota-
Baxter operators of weight zero on the sub-adjacent Lie algebra
${\mathfrak{g}_{3}}$ derived by Nijenhuis operators $N$ are given as follows:

Nijenhuis operators on $D$ | Rota-Baxter operators of weight zero on ${\mathfrak{g}_{3}}$
---|---
$N_{D_{4}}^{1},~{}N_{D_{6}}^{1},~{}N_{D_{8}}^{1},~{}N_{D_{9}}^{1}$ | $R_{\mathfrak{g}_{3}}^{1}=\left(\begin{array}[]{ccc}0&0&0\\\ 0&0&0\\\ 0&n_{32}&0\\\ \end{array}\right)$
| $R_{\mathfrak{g}_{3}}^{2}=\left(\begin{array}[]{ccc}0&0&0\\\
0&n_{22}&n_{23}\\\ 0&-\frac{{n_{22}}^{2}}{n_{23}}&-n_{22}\\\
\end{array}\right)n_{23}\neq 0$
$N_{D_{4}}^{2},~{}N_{D_{8}}^{2}$ | $R_{\mathfrak{g}_{3}}^{3}=\left(\begin{array}[]{ccc}0&0&0\\\ 0&0&0\\\ n_{31}&n_{32}&0\end{array}\right)n_{31}\neq 0$
$N_{D_{4}}^{3},~{}N_{D_{8}}^{3}$ | $R_{\mathfrak{g}_{3}}^{4}=\left(\begin{array}[]{ccc}0&0&0\\\ n_{21}&0&0\\\ n_{31}&0&0\end{array}\right)n_{21}\neq 0$
$N_{D_{4}}^{4},~{}N_{D_{6}}^{2},~{}N_{D_{8}}^{4},~{}N_{D_{9}}^{4}$ | $R_{\mathfrak{g}_{3}}^{5}=\left(\begin{array}[]{ccc}0&n_{12}&0\\\ 0&0&0\\\ 0&n_{32}&0\end{array}\right)n_{12}\neq 0$
$N_{D_{6}}^{3}$ | $R_{\mathfrak{g}_{3}}^{6}=\left(\begin{array}[]{ccc}0&0&0\\\ n_{21}&-n_{31}&n_{21}\\\ n_{31}&-\frac{{n_{31}}^{2}}{n_{21}}&n_{31}\end{array}\right)n_{21}\neq 0$
$N_{D_{9}}^{3}$ | $R_{\mathfrak{g}_{3}}^{7}=\left(\begin{array}[]{ccc}0&0&0\\\ n_{21}&-n_{33}&-n_{21}\\\ -n_{33}&\frac{{n_{33}}^{2}}{n_{21}}&n_{33}\end{array}\right)n_{21}\neq 0$

###### Proof.

The proof is similar to that of Proposition 5.8. ∎

######  Proposition 5.10.

Let $R:{\mathfrak{g}_{4}}\rightarrow{\mathfrak{g}_{4}}$ be a linear map on the
sub-adjacent Lie algebra ${\mathfrak{g}_{4}}$ defined by (5.22). Then Rota-
Baxter operators of weight zero on the sub-adjacent Lie algebra
${\mathfrak{g}_{4}}$ derived by Nijenhuis operators $N$ are given as follows:

Nijenhuis operators on $D$ | Rota-Baxter operators of weight zero on ${\mathfrak{g}_{4}}$
---|---
$N_{D_{5}}^{1},~{}N_{D_{7}}^{1}$ | $R_{\mathfrak{g}_{4}}^{1}=\left(\begin{array}[]{ccc}0&0&0\\\ 0&0&0\\\ 0&n_{32}&0\\\ \end{array}\right)$
| $R_{\mathfrak{g}_{4}}^{2}=\left(\begin{array}[]{ccc}0&0&0\\\
0&n_{22}&n_{23}\\\ 0&-\frac{{n_{22}}^{2}}{n_{23}}&-n_{22}\\\
\end{array}\right)n_{23}\neq 0$
$N_{D_{5}}^{2}$ | $R_{\mathfrak{g}_{4}}^{3}=\left(\begin{array}[]{ccc}0&0&0\\\ 0&0&0\\\ n_{31}&n_{32}&0\end{array}\right)n_{31}\neq 0$
$N_{D_{5}}^{3}$ | $R_{\mathfrak{g}_{4}}^{4}=\left(\begin{array}[]{ccc}0&0&0\\\ n_{21}&0&0\\\ n_{31}&0&0\end{array}\right)n_{21}\neq 0$
$N_{D_{5}}^{4},~{}N_{D_{7}}^{2}$ | $R_{\mathfrak{g}_{4}}^{5}=\left(\begin{array}[]{ccc}0&n_{12}&0\\\ 0&0&0\\\ 0&n_{32}&0\end{array}\right)n_{12}\neq 0$
$N_{D_{7}}^{3}$ | $R_{\mathfrak{g}_{4}}^{4}=\left(\begin{array}[]{ccc}0&0&0\\\ n_{21}&-n_{31}&n_{21}\\\ n_{31}&-\frac{{n_{31}}^{2}}{n_{21}}&n_{31}\end{array}\right)n_{21}\neq 0$

###### Proof.

The proof is similar to that of Proposition 5.8. ∎

######  Remark 5.11.

All linear maps on $(D_{10},\cdot)$ and $(D_{11},\cdot)$, whose squares are 0,
are Rota-Baxter operators of weight zero on the sub-adjacent Lie algebras
${{\mathfrak{g}}(D_{10})}$ and ${{\mathfrak{g}}(D_{11})}$.

######  Proposition 5.12.

Let $R:{{\mathfrak{g}}(D_{12})}\rightarrow{{\mathfrak{g}}(D_{12})}$ be a
linear map on the sub-adjacent Lie algebra ${{\mathfrak{g}}(D_{12})}$ defined
by (5.22). Then Rota-Baxter operators of weight zero on the sub-adjacent Lie
algebra ${{\mathfrak{g}}(D_{12})}$ derived by Nijenhuis operators $N$ are
given as follows:

Nijenhuis operators on $D$ | Rota-Baxter operators of weight zero on ${\mathfrak{g}}(D_{12})$
---|---
$N_{D_{12}}^{1}$ | $R_{{\mathfrak{g}}_{(D_{12})}}^{1}=\left(\begin{array}[]{ccc}0&0&0\\\ 0&0&0\\\ n_{31}&n_{32}&0\end{array}\right)$
$N_{D_{12}}^{2}$ | $R_{{\mathfrak{g}}_{(D_{12})}}^{2}=\left(\begin{array}[]{ccc}0&0&0\\\ n_{21}&0&0\\\ n_{31}&0&0\end{array}\right)n_{21}\neq 0$
$N_{D_{12}}^{3}$ | $R_{{\mathfrak{g}}_{(D_{12})}}^{3}=\left(\begin{array}[]{ccc}0&n_{12}&0\\\ 0&0&0\\\ 0&n_{32}&0\end{array}\right)n_{12}\neq 0$
$N_{D_{12}}^{4}$ | $R_{{\mathfrak{g}}_{(D_{12})}}^{4}=\left(\begin{array}[]{ccc}0&0&0\\\ n_{21}&-n_{33}&n_{23}\\\ \frac{n_{21}n_{33}}{n_{23}}&-\frac{{n_{33}}^{2}}{n_{23}}&n_{33}\end{array}\right)n_{23}\neq 0$
$N_{D_{12}}^{5}$ | $R_{{\mathfrak{g}}_{(D_{12})}}^{5}=\left(\begin{array}[]{ccc}-n_{33}&n_{12}&n_{13}\\\ 0&0&0\\\ -\frac{{n_{33}}^{2}}{n_{13}}&\frac{n_{12}n_{33}}{n_{13}}&n_{33}\end{array}\right)n_{13}\neq 0$

###### Proof.

The proof is similar to that of proposition 5.8. ∎

######  Remark 5.13.

There are the same Nijenhuis operators on different pre-Lie algebras. For
example, $(1)~{}N_{D_{1}}^{2}$ is equal to $N_{D_{2}}^{2}$.
$(2)~{}N_{D_{4}}^{1}$ is equal to $N_{D_{5}}^{1}$, $N_{D_{6}}^{1}$,
$N_{D_{7}}^{1}$, $N_{D_{8}}^{1}$, $N_{D_{9}}^{1}$. $(3)~{}N_{D_{4}}^{2}$ is
equal to $N_{D_{5}}^{2}$, $N_{D_{8}}^{2}$. $(4)~{}N_{D_{4}}^{3}$ is equal to
$N_{D_{5}}^{3}$, $N_{D_{8}}^{3}$. $(5)~{}N_{D_{4}}^{4}$ is equal to
$N_{D_{5}}^{4}$, $N_{D_{8}}^{4}$. $(6)~{}N_{D_{6}}^{2}$ is equal to
$N_{D_{7}}^{2}$, $N_{D_{9}}^{4}$. $(7)~{}N_{D_{6}}^{3}$ is equal to
$N_{D_{7}}^{3}$. The sub-adjacent Lie algebras ${\mathfrak{g}}(D_{4})$ and
${\mathfrak{g}}(D_{12})$ are different, but the Rota-Baxter operators derived
by $N_{D_{4}}^{4}$ and $N_{D_{12}}^{3}$ are the same. Therefore, the Rota-
Baxter operators of weight zero on the different sub-adjacent Lie algebras
derived by Nijenhuis operators may be identical.

###  5.2. Solutions of CYBE in Lie algebras ${\mathfrak{g}}(A)\ltimes_{{\rm
ad}^{\ast}}{\mathfrak{g}}(A)^{\ast}$

Let $\rho:\mathfrak{g}\rightarrow\mathfrak{gl}(V)$ be a representation of a
Lie algebra $\mathfrak{g}$. On the vector space $\mathfrak{g}\oplus V$, we can
define a Lie algebra structure (denoted by $\mathfrak{g}\ltimes_{\rho}V$ )
given by

(5.23)
$[x_{1}+v_{1},x_{2}+v_{2}]=[x_{1},x_{2}]+\rho(x_{1})v_{2}-\rho(x_{2})v_{1},\quad\forall~{}x_{1},x_{2}\in\mathfrak{g},v_{1},v_{2}\in
V.$

Define a linear map
$\rho^{\ast}:{\mathfrak{g}}\rightarrow\mathfrak{gl}({\mathfrak{g}}^{\ast})$ by

(5.24)
$\langle\rho_{x}^{*}(\xi),y\rangle=-\langle\xi,\rho_{x}(y)\rangle,\;\;\forall~{}x,y\in\mathfrak{g},\xi\in{\mathfrak{g}}^{*}.$

Then $\rho^{\ast}$ is a representation of a Lie algebra $\mathfrak{g}$. It is
the dual representation of $\rho$.

######  Proposition 5.14.

([1]) If $R:V\rightarrow\mathfrak{g}$ is a linear map, then $R$ can be viewed
as an element in $\mathfrak{g}\otimes
V^{\ast}\subset(\mathfrak{g}\ltimes_{\rho^{\ast}}V^{\ast})\otimes(\mathfrak{g}\ltimes_{\rho^{\ast}}V^{\ast})$.

###### Proof.

Let $\\{e_{1},\cdots,e_{n}\\}$ be a basis of $\mathfrak{g}$. Let
$\\{v_{1},\cdots,v_{m}\\}$ be a basis of $V$ and
$\\{v^{\ast}_{1},\cdots,v^{\ast}_{m}\\}$ be its dual basis, that is
$v^{\ast}_{i}(v_{j})=\delta_{ij}$. Set
$R(v_{i})=\sum\limits_{j=1}^{n}a_{ij}e_{j},$ $i=1,2,\cdots,n$. Since as vector
spaces, ${\rm Hom}(V,\mathfrak{g})\cong\mathfrak{g}\otimes V^{\ast}$, we have

(5.25) $R=\sum_{i=1}^{m}R(v_{i})\otimes
v_{i}^{\ast}=\sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij}e_{j}\otimes
v_{i}^{\ast}\in\mathfrak{g}\otimes
V^{\ast}\subset(\mathfrak{g}\ltimes_{\rho^{\ast}}V^{\ast})\otimes(\mathfrak{g}\ltimes_{\rho^{\ast}}V^{\ast}).$

Hence $R$ can be viewed as an element in $\mathfrak{g}\otimes
V^{\ast}\subset(\mathfrak{g}\ltimes_{\rho^{\ast}}V^{\ast})\otimes(\mathfrak{g}\ltimes_{\rho^{\ast}}V^{\ast})$.
∎

For any tensor element $r=\sum_{i}a_{i}\otimes b_{i}\in V\otimes V$, denote
$r^{21}=\sum_{i}b_{i}\otimes a_{i}$.

######  Lemma 5.15.

([1]) Let $\mathfrak{g}$ be a Lie algebra. A linear map
$R:\mathfrak{g}\rightarrow\mathfrak{g}$ is a Rota-Baxter operator if and only
if $r=R-R^{21}$ is a skew-symmetric solution of CYBE in
$\mathfrak{g}\ltimes_{{\rm ad}^{\ast}}\mathfrak{g}^{\ast}$.

######  Proposition 5.16.

Let $\\{e_{1},e_{2}\\}$ be a basis of ${\mathfrak{g}_{1}}$ and
$\\{e^{\ast}_{1},e^{\ast}_{2}\\}$ be its dual basis. Then
${\mathfrak{g}_{1}}\ltimes_{{\rm ad}^{\ast}}{\mathfrak{g}_{1}}^{\ast}$ has a
basis $\\{e_{1},e_{2},e_{1}^{\ast},e_{2}^{\ast}\\}$ with respect to which the
non-zero products are given by the following:

$[e_{1},e_{2}]=e_{1},\quad\quad[e_{1},e^{\ast}_{1}]=-e^{\ast}_{2},\quad\quad[e_{2},e^{\ast}_{1}]=e^{\ast}_{1}.$

The solutions of CYBE in the Lie algebra ${\mathfrak{g}_{1}}\ltimes_{{\rm
ad}^{\ast}}{\mathfrak{g}_{1}}^{\ast}$ are given by

$r_{1}=n_{21}e_{1}\otimes e_{2}^{\ast}-e_{2}^{\ast}\otimes n_{21}e_{1}.$

###### Proof.

Let $\rho=\operatorname{{\mathrm{ad}}}^{\ast}$. By (5.24), the dual
representation
$\operatorname{{\mathrm{ad}}}^{\ast}:{\mathfrak{g}_{1}}\rightarrow\mathfrak{gl}({\mathfrak{g}_{1}}^{\ast})$
is defined by

(5.26)
$\langle\operatorname{{\mathrm{ad}}}_{x}^{*}(\xi),y\rangle=-\langle\xi,\operatorname{{\mathrm{ad}}}_{x}(y)\rangle,\;\;\forall~{}x,y\in\mathfrak{g}_{1},~{}\xi\in{\mathfrak{g}_{1}}^{*}.$

By (5.23) we have

$[e_{1},e_{1}]=0,\quad\quad[e_{1},e_{2}]=-[e_{2},e_{1}]=e_{1},\quad\quad[e_{2},e_{2}]=0.$
$[e^{\ast}_{1},e^{\ast}_{1}]=0,\quad\quad[e^{\ast}_{1},e^{\ast}_{2}]=-[e^{\ast}_{2},e^{\ast}_{1}]=0,\quad\quad[e^{\ast}_{2},e^{\ast}_{2}]=0.$
$[e_{1},e^{\ast}_{1}]=-[e^{\ast}_{1},e_{1}]=\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}e^{\ast}_{1},\quad\quad[e_{1},e^{\ast}_{2}]=-[e^{\ast}_{2},e_{1}]=\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}e^{\ast}_{2},$
$[e_{2},e^{\ast}_{1}]=-[e^{\ast}_{1},e_{2}]=\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}e^{\ast}_{1},\quad\quad[e_{2},e^{\ast}_{2}]=-[e^{\ast}_{2},e_{2}]=\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}e^{\ast}_{2}.$

For $\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{1})$, set
$\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{1})=k_{11}e^{\ast}_{1}+k_{12}e^{\ast}_{2},~{}k_{11},~{}k_{12}\in\mathbb{C}.$
Then

$\langle\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{1}),{e_{1}}\rangle=-\langle
e^{\ast}_{1},\operatorname{{\mathrm{ad}}}_{e_{1}}(e_{1})\rangle=-\langle
e^{\ast}_{1},0\rangle=0,\;\;$
$(k_{11}e^{\ast}_{1}+k_{12}e^{\ast}_{2})({e_{1}})=k_{11}.$

This implies that $k_{11}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{1}),{e_{2}}\rangle=-\langle
e^{\ast}_{1},\operatorname{{\mathrm{ad}}}_{e_{1}}(e_{2})\rangle=-\langle
e^{\ast}_{1},{e_{1}}\rangle=-1,\;\;$
$(k_{11}e^{\ast}_{1}+k_{12}e^{\ast}_{2})({e_{2}})=k_{12}.$

We obtain $k_{12}=-1$. Therefore,
$\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{1})=-e^{\ast}_{2}$.

For $\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{2})$, set
$\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{2})=k_{21}e^{\ast}_{1}+k_{22}e^{\ast}_{2},~{}k_{21},~{}k_{22}\in\mathbb{C}.$
We have

$\langle\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{2}),{e_{1}}\rangle=-\langle
e^{\ast}_{2},\operatorname{{\mathrm{ad}}}_{e_{1}}(e_{1})\rangle=-\langle
e^{\ast}_{2},0\rangle=0,\;\;$
$(k_{21}e^{\ast}_{1}+k_{22}e^{\ast}_{2})({e_{1}})=k_{21}.$

This implies that $k_{21}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{2}),{e_{2}}\rangle=-\langle
e^{\ast}_{2},\operatorname{{\mathrm{ad}}}_{e_{1}}(e_{2})\rangle=-\langle
e^{\ast}_{2},{e_{1}}\rangle=0,\;\;$
$(k_{21}e^{\ast}_{1}+k_{22}e^{\ast}_{2})({e_{2}})=k_{22}.$

We obtain $k_{22}=0$. Hence
$\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{2})=0$.

For $\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{1})$, set
$\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{1})=k_{31}e^{\ast}_{1}+k_{32}e^{\ast}_{2},~{}k_{31},~{}k_{32}\in\mathbb{C}.$
We have

$\langle\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{1}),{e_{1}}\rangle=-\langle
e^{\ast}_{1},\operatorname{{\mathrm{ad}}}_{e_{2}}(e_{1})\rangle=-\langle
e^{\ast}_{1},-{e_{1}}\rangle=1,\;\;$
$(k_{31}e^{\ast}_{1}+k_{32}e^{\ast}_{2})({e_{1}})=k_{31}.$

Then, we obtain $k_{31}=1$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{1}),{e_{2}}\rangle=-\langle
e^{\ast}_{1},\operatorname{{\mathrm{ad}}}_{e_{2}}(e_{2})\rangle=-\langle
e^{\ast}_{2},0\rangle=0,\;\;$
$(k_{31}e^{\ast}_{1}+k_{32}e^{\ast}_{2})({e_{2}})=k_{32}.$

Thus $k_{32}=0$. Therefore, we have
$\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{1})=e^{\ast}_{1}$.

For $\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{2})$, set
$\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{2})=k_{41}e^{\ast}_{1}+k_{42}e^{\ast}_{2},~{}k_{41},~{}k_{42}\in\mathbb{C}.$
Then

$\langle\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{2}),{e_{1}}\rangle=-\langle
e^{\ast}_{2},\operatorname{{\mathrm{ad}}}_{e_{2}}(e_{1})\rangle=-\langle
e^{\ast}_{2},-{e_{1}}\rangle=0,\;\;$
$(k_{41}e^{\ast}_{1}+k_{42}e^{\ast}_{2})({e_{1}})=k_{41}.$

This implies that $k_{41}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{2}),{e_{2}}\rangle=-\langle
e^{\ast}_{2},\operatorname{{\mathrm{ad}}}_{e_{2}}(e_{2})\rangle=-\langle
e^{\ast}_{2},0\rangle=0,\;\;$
$(k_{41}e^{\ast}_{1}+k_{42}e^{\ast}_{2})({e_{2}})=k_{42}.$

We have $k_{42}=0$. Hence
$\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{2})=0$. The Rota-Baxter
operators $R_{\mathfrak{g}_{1}}^{1}$ of weight zero on the sub-adjacent Lie
algebra ${\mathfrak{g}_{1}}$ are

$R_{\mathfrak{g}_{1}}^{1}=\left(\begin{array}[]{cc}0&0\\\ n_{21}&0\\\
\end{array}\right).$

By Proposition 5.14, we have $R_{\mathfrak{g}_{1}}^{1}=n_{21}e_{1}\otimes
e_{2}^{\ast}$ and $(R_{\mathfrak{g}_{1}}^{1})^{21}=e_{2}^{\ast}\otimes
n_{21}e_{1}$. By Lemma 5.15, the solutions of CYBE in the Lie algebra
${\mathfrak{g}_{1}}\ltimes_{{\rm ad}^{\ast}}{\mathfrak{g}_{1}}^{\ast}$ are

$r_{1}=R_{\mathfrak{g}_{1}}^{1}-(R_{\mathfrak{g}_{1}}^{1})^{21}=n_{21}e_{1}\otimes
e_{2}^{\ast}-e_{2}^{\ast}\otimes n_{21}e_{1}.$

∎

######  Proposition 5.17.

Let $\\{e_{1},e_{2},e_{3}\\}$ be a basis of ${\mathfrak{g}_{2}}$ and
$\\{e^{\ast}_{1},e^{\ast}_{2},e^{\ast}_{3}\\}$ be its dual basis, then
${\mathfrak{g}_{2}}\ltimes_{{\rm ad}^{\ast}}{\mathfrak{g}_{2}}^{\ast}$ has a
basis $\\{e_{1},e_{2},e_{3},e_{1}^{\ast},e_{2}^{\ast},e^{\ast}_{3}\\}$ with
respect to which the non-zero products are given by the following:

$[e_{1},e_{2}]=e_{3},\quad\quad[e_{1},e^{\ast}_{3}]=-e^{\ast}_{2},\quad\quad[e_{2},e^{\ast}_{3}]=e^{\ast}_{1}$

The solutions of CYBE in the Lie algebra ${\mathfrak{g}_{2}}\ltimes_{{\rm
ad}^{\ast}}{\mathfrak{g}_{2}}^{\ast}$ are given by the following:

  * $r_{1}=(n_{21}e_{1}+n_{23}e_{2})\otimes e_{2}^{\ast}-e_{2}^{\ast}\otimes(n_{21}e_{1}+n_{23}e_{2}).$

  * $r_{2}=(n_{11}e_{1}+n_{12}e_{2}-\frac{n_{12}n_{23}}{n_{11}}e_{3})\otimes e_{1}^{\ast}+(-\frac{{n_{11}}^{2}}{n_{12}}e_{1}-n_{11}e_{2}+n_{23}e_{3})\otimes e_{2}^{\ast}\\\ -e_{1}^{\ast}\otimes(n_{11}e_{1}+n_{12}e_{2}-\frac{n_{12}n_{23}}{n_{11}}e_{3})-e_{2}^{\ast}\otimes(-\frac{{n_{11}}^{2}}{n_{12}}e_{1}-n_{11}e_{2}+n_{23}e_{3})~{}n_{11},n_{12}\neq 0.$

  * $r_{3}=(n_{12}e_{2}+n_{13}e_{3})\otimes e_{1}^{\ast}-e_{1}^{\ast}\otimes(n_{12}e_{2}+n_{13}e_{3}),~{}n_{12}\neq 0.$

  * $r_{4}=n_{13}e_{3}\otimes e_{1}^{\ast}+n_{23}e_{3}\otimes e_{2}^{\ast}-e_{1}^{\ast}\otimes n_{13}e_{3}-e_{2}^{\ast}\otimes n_{23}e_{3}.$

  * $r_{5}=(\frac{(-1+\sqrt{1-4\lambda})}{2}n_{12}e_{1}+n_{12}e_{2}+n_{13}e_{3})\otimes e_{1}^{\ast}-\frac{(-1+\sqrt{1-4\lambda})}{2}(\frac{(-1+\sqrt{1-4\lambda})}{2}n_{12}e_{1}+n_{12}e_{2}+n_{13}e_{3})\otimes e_{2}^{\ast}\\\ -e_{1}^{\ast}\otimes(\frac{(-1+\sqrt{1-4\lambda})}{2}n_{12}e_{1}+n_{12}e_{2}+n_{13}e_{3})+e_{2}^{\ast}\otimes\frac{(-1+\sqrt{1-4\lambda})}{2}(\frac{(-1+\sqrt{1-4\lambda})}{2}n_{12}e_{1}+n_{12}e_{2}+n_{13}e_{3}),~{}n_{12},\lambda\neq 0.$

  * $r_{6}=(\frac{(-1-\sqrt{1-4\lambda})}{2}n_{12}e_{1}+n_{12}e_{2}+n_{13}e_{3})\otimes e_{1}^{\ast}-\frac{(-1-\sqrt{1-4\lambda})}{2}(\frac{(-1-\sqrt{1-4\lambda})}{2}n_{12}e_{1}+n_{12}e_{2}+n_{13}e_{3})\otimes e_{2}^{\ast}\\\ -e_{1}^{\ast}\otimes(\frac{(-1-\sqrt{1-4\lambda})}{2}n_{12}e_{1}+n_{12}e_{2}+n_{13}e_{3})+e_{2}^{\ast}\otimes\frac{(-1-\sqrt{1-4\lambda})}{2}(\frac{(-1-\sqrt{1-4\lambda})}{2}n_{12}e_{1}+n_{12}e_{2}+n_{13}e_{3}),~{}n_{12},\lambda\neq 0.$

###### Proof.

Let $\rho=\operatorname{{\mathrm{ad}}}^{\ast}$. By (5.24), the dual
representation
$\operatorname{{\mathrm{ad}}}^{\ast}:{\mathfrak{g}_{2}}\rightarrow\mathfrak{gl}({\mathfrak{g}_{2}}^{\ast})$
is defined by

(5.27)
$\langle\operatorname{{\mathrm{ad}}}_{x}^{*}(\xi),y\rangle=-\langle\xi,\operatorname{{\mathrm{ad}}}_{x}(y)\rangle,\;\;\forall~{}x,y\in\mathfrak{g}_{2},~{}\xi\in{\mathfrak{g}_{2}}^{*}.$

By (5.23) we have

$[e_{1},e_{1}]=0,\quad[e_{1},e_{2}]=-[e_{2},e_{1}]=e_{3},\quad[e_{1},e_{3}]=-[e_{3},e_{1}]=0,$
$[e_{2},e_{2}]=0,\quad[e_{2},e_{3}]=-[e_{3},e_{2}]=0,\quad[e_{3},e_{3}]=0,$
$[e^{\ast}_{1},e^{\ast}_{1}]=0,\quad[e^{\ast}_{1},e^{\ast}_{2}]=-[e^{\ast}_{2},e^{\ast}_{1}]=0,\quad[e^{\ast}_{1},e^{\ast}_{3}]=-[e^{\ast}_{3},e^{\ast}_{1}]=0,$
$[e^{\ast}_{2},e^{\ast}_{2}]=0,\quad[e^{\ast}_{2},e^{\ast}_{3}]=-[e^{\ast}_{3},e^{\ast}_{2}]=0,\quad[e^{\ast}_{3},e^{\ast}_{3}]=0,$
$[e_{1},e^{\ast}_{1}]=-[e^{\ast}_{1},e_{1}]=\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}e^{\ast}_{1},\quad[e_{1},e^{\ast}_{2}]=-[e^{\ast}_{2},e_{1}]=\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}e^{\ast}_{2},\quad[e_{1},e^{\ast}_{3}]=-[e^{\ast}_{3},e_{1}]=\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}e^{\ast}_{3},$
$[e_{2},e^{\ast}_{1}]=-[e^{\ast}_{1},e_{2}]=\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}e^{\ast}_{1},\quad[e_{2},e^{\ast}_{2}]=-[e^{\ast}_{2},e_{2}]=\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}e^{\ast}_{2},\quad[e_{2},e^{\ast}_{3}]=-[e^{\ast}_{3},e_{2}]=\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}e^{\ast}_{3},$
$[e_{3},e^{\ast}_{1}]=-[e^{\ast}_{1},e_{3}]=\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}e^{\ast}_{1},\quad[e_{3},e^{\ast}_{2}]=-[e^{\ast}_{2},e_{3}]=\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}e^{\ast}_{2}.\quad[e_{3},e^{\ast}_{3}]=-[e^{\ast}_{3},e_{3}]=\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}e^{\ast}_{3}.$

For $\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{1})$, set
$\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{1})=k_{11}e^{\ast}_{1}+k_{12}e^{\ast}_{2}+k_{13}e^{\ast}_{3},~{}k_{11},~{}k_{12},~{}k_{13}\in\mathbb{C}.$
Then

$\langle\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{1}),{e_{1}}\rangle=-\langle
e^{\ast}_{1},\operatorname{{\mathrm{ad}}}_{e_{1}}(e_{1})\rangle=-\langle
e^{\ast}_{1},0\rangle=0,\;\;$
$(k_{11}e^{\ast}_{1}+k_{12}e^{\ast}_{2}+k_{13}e^{\ast}_{3})({e_{1}})=k_{11}.$

This implies that $k_{11}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{1}),{e_{2}}\rangle=-\langle
e^{\ast}_{1},\operatorname{{\mathrm{ad}}}_{e_{1}}(e_{2})\rangle=-\langle
e^{\ast}_{1},{e_{3}}\rangle=0,\;\;$
$(k_{11}e^{\ast}_{1}+k_{12}e^{\ast}_{2}+k_{13}e^{\ast}_{3})({e_{2}})=k_{12}.$

We obtain $k_{12}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{1}),{e_{3}}\rangle=-\langle
e^{\ast}_{1},\operatorname{{\mathrm{ad}}}_{e_{1}}(e_{3})\rangle=-\langle
e^{\ast}_{1},0\rangle=0,\;\;$
$(k_{11}e^{\ast}_{1}+k_{12}e^{\ast}_{2}+k_{13}e^{\ast}_{3})({e_{3}})=k_{13}.$

We obtain $k_{13}=0$. Therefore,
$\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{1})=0$.

For $\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{2})$, set
$\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{2})=k_{21}e^{\ast}_{1}+k_{22}e^{\ast}_{2}+k_{23}e^{\ast}_{3},~{}k_{21},~{}k_{22},~{}k_{23}\in\mathbb{C}.$
Then

$\langle\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{2}),{e_{1}}\rangle=-\langle
e^{\ast}_{2},\operatorname{{\mathrm{ad}}}_{e_{1}}(e_{1})\rangle=-\langle
e^{\ast}_{2},0\rangle=0,\;\;$
$(k_{21}e^{\ast}_{1}+k_{22}e^{\ast}_{2}+k_{23}e^{\ast}_{3})({e_{1}})=k_{21}.$

This implies that $k_{21}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{2}),{e_{2}}\rangle=-\langle
e^{\ast}_{2},\operatorname{{\mathrm{ad}}}_{e_{1}}(e_{2})\rangle=-\langle
e^{\ast}_{2},{e_{3}}\rangle=0,\;\;$
$(k_{21}e^{\ast}_{1}+k_{22}e^{\ast}_{2}+k_{23}e^{\ast}_{3})({e_{2}})=k_{22}.$

We obtain $k_{22}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{2}),{e_{3}}\rangle=-\langle
e^{\ast}_{2},\operatorname{{\mathrm{ad}}}_{e_{1}}(e_{3})\rangle=-\langle
e^{\ast}_{2},0\rangle=0,\;\;$
$(k_{21}e^{\ast}_{1}+k_{22}e^{\ast}_{2}+k_{23}e^{\ast}_{3})({e_{3}})=k_{23}.$

We obtain $k_{23}=0$. Therefore,
$\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{2})=0$.

For $\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{3})$, set
$\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{3})=k_{31}e^{\ast}_{1}+k_{32}e^{\ast}_{2}+k_{33}e^{\ast}_{3},~{}k_{31},~{}k_{32},~{}k_{33}\in\mathbb{C}.$
Then

$\langle\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{3}),{e_{1}}\rangle=-\langle
e^{\ast}_{3},\operatorname{{\mathrm{ad}}}_{e_{1}}(e_{1})\rangle=-\langle
e^{\ast}_{3},0\rangle=0,\;\;$
$(k_{31}e^{\ast}_{1}+k_{32}e^{\ast}_{2}+k_{33}e^{\ast}_{3})({e_{1}})=k_{31}.$

This implies that $k_{31}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{3}),{e_{2}}\rangle=-\langle
e^{\ast}_{3},\operatorname{{\mathrm{ad}}}_{e_{1}}(e_{2})\rangle=-\langle
e^{\ast}_{3},{e_{3}}\rangle=-1,\;\;$
$(k_{31}e^{\ast}_{1}+k_{32}e^{\ast}_{2}+k_{33}e^{\ast}_{3})({e_{2}})=k_{32}.$

We obtain $k_{32}=-1$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{3}),{e_{3}}\rangle=-\langle
e^{\ast}_{3},\operatorname{{\mathrm{ad}}}_{e_{1}}(e_{3})\rangle=-\langle
e^{\ast}_{3},0\rangle=0,\;\;$
$(k_{31}e^{\ast}_{1}+k_{32}e^{\ast}_{2}+k_{33}e^{\ast}_{3})({e_{3}})=k_{33}.$

We obtain $k_{33}=0$. Therefore,
$\operatorname{{\mathrm{ad}}}_{e_{1}}^{*}(e^{\ast}_{3})=-e^{\ast}_{2}$.

For $\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{1})$, set
$\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{1})=k_{41}e^{\ast}_{1}+k_{42}e^{\ast}_{2}+k_{43}e^{\ast}_{3},~{}k_{41},~{}k_{42},~{}k_{43}\in\mathbb{C}.$
Then

$\langle\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{1}),{e_{1}}\rangle=-\langle
e^{\ast}_{1},\operatorname{{\mathrm{ad}}}_{e_{2}}(e_{1})\rangle=-\langle
e^{\ast}_{1},-e_{3}\rangle=0,\;\;$
$(k_{41}e^{\ast}_{1}+k_{42}e^{\ast}_{2}+k_{43}e^{\ast}_{3})({e_{1}})=k_{41}.$

This implies that $k_{41}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{1}),{e_{2}}\rangle=-\langle
e^{\ast}_{1},\operatorname{{\mathrm{ad}}}_{e_{2}}(e_{2})\rangle=-\langle
e^{\ast}_{1},0\rangle=0,\;\;$
$(k_{41}e^{\ast}_{1}+k_{42}e^{\ast}_{2}+k_{43}e^{\ast}_{3})({e_{2}})=k_{42}.$

We obtain $k_{42}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{1}),{e_{3}}\rangle=-\langle
e^{\ast}_{1},\operatorname{{\mathrm{ad}}}_{e_{2}}(e_{3})\rangle=-\langle
e^{\ast}_{1},0\rangle=0,\;\;$
$(k_{41}e^{\ast}_{1}+k_{42}e^{\ast}_{2}+k_{43}e^{\ast}_{3})({e_{3}})=k_{43}.$

We obtain $k_{43}=0$. Therefore,
$\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{1})=0$.

For $\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{2})$, set
$\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{2})=k_{51}e^{\ast}_{1}+k_{52}e^{\ast}_{2}+k_{53}e^{\ast}_{3},~{}k_{51},~{}k_{52},~{}k_{53}\in\mathbb{C}.$
Then

$\langle\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{2}),{e_{1}}\rangle=-\langle
e^{\ast}_{2},\operatorname{{\mathrm{ad}}}_{e_{2}}(e_{1})\rangle=-\langle
e^{\ast}_{2},-e_{3}\rangle=0,\;\;$
$(k_{51}e^{\ast}_{1}+k_{52}e^{\ast}_{2}+k_{53}e^{\ast}_{3})({e_{1}})=k_{51}.$

This implies that $k_{51}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{2}),{e_{2}}\rangle=-\langle
e^{\ast}_{2},\operatorname{{\mathrm{ad}}}_{e_{2}}(e_{2})\rangle=-\langle
e^{\ast}_{2},0\rangle=0,\;\;$
$(k_{51}e^{\ast}_{1}+k_{52}e^{\ast}_{2}+k_{53}e^{\ast}_{3})({e_{2}})=k_{52}.$

We obtain $k_{52}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{2}),{e_{3}}\rangle=-\langle
e^{\ast}_{2},\operatorname{{\mathrm{ad}}}_{e_{2}}(e_{3})\rangle=-\langle
e^{\ast}_{2},0\rangle=0,\;\;$
$(k_{51}e^{\ast}_{1}+k_{52}e^{\ast}_{2}+k_{53}e^{\ast}_{3})({e_{3}})=k_{53}.$

We obtain $k_{53}=0$. Therefore,
$\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{2})=0$.

For $\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{3})$, set
$\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{3})=k_{61}e^{\ast}_{1}+k_{62}e^{\ast}_{2}+k_{63}e^{\ast}_{3},~{}k_{61},~{}k_{62},~{}k_{63}\in\mathbb{C}.$
Then

$\langle\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{3}),{e_{1}}\rangle=-\langle
e^{\ast}_{3},\operatorname{{\mathrm{ad}}}_{e_{2}}(e_{1})\rangle=-\langle
e^{\ast}_{3},-e_{3}\rangle=1,\;\;$
$(k_{61}e^{\ast}_{1}+k_{62}e^{\ast}_{2}+k_{63}e^{\ast}_{3})({e_{1}})=k_{61}.$

This implies that $k_{61}=1$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{3}),{e_{2}}\rangle=-\langle
e^{\ast}_{3},\operatorname{{\mathrm{ad}}}_{e_{2}}(e_{2})\rangle=-\langle
e^{\ast}_{3},0\rangle=0,\;\;$
$(k_{61}e^{\ast}_{1}+k_{62}e^{\ast}_{2}+k_{63}e^{\ast}_{3})({e_{2}})=k_{62}.$

We obtain $k_{62}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{3}),{e_{3}}\rangle=-\langle
e^{\ast}_{3},\operatorname{{\mathrm{ad}}}_{e_{2}}(e_{3})\rangle=-\langle
e^{\ast}_{3},0\rangle=0,\;\;$
$(k_{61}e^{\ast}_{1}+k_{62}e^{\ast}_{2}+k_{63}e^{\ast}_{3})({e_{3}})=k_{63}.$

We obtain $k_{63}=0$. Therefore,
$\operatorname{{\mathrm{ad}}}_{e_{2}}^{*}(e^{\ast}_{3})=e^{\ast}_{1}$.

For $\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{1})$, set
$\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{1})=k_{71}e^{\ast}_{1}+k_{72}e^{\ast}_{2}+k_{73}e^{\ast}_{3},~{}k_{71},~{}k_{72},~{}k_{73}\in\mathbb{C}.$
Then

$\langle\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{1}),{e_{1}}\rangle=-\langle
e^{\ast}_{1},\operatorname{{\mathrm{ad}}}_{e_{3}}(e_{1})\rangle=-\langle
e^{\ast}_{1},0\rangle=0,\;\;$
$(k_{71}e^{\ast}_{1}+k_{72}e^{\ast}_{2}+k_{73}e^{\ast}_{3})({e_{1}})=k_{71}.$

This implies that $k_{71}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{1}),{e_{2}}\rangle=-\langle
e^{\ast}_{1},\operatorname{{\mathrm{ad}}}_{e_{3}}(e_{2})\rangle=-\langle
e^{\ast}_{1},0\rangle=0,\;\;$
$(k_{71}e^{\ast}_{1}+k_{72}e^{\ast}_{2}+k_{73}e^{\ast}_{3})({e_{2}})=k_{72}.$

We obtain $k_{72}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{1}),{e_{3}}\rangle=-\langle
e^{\ast}_{1},\operatorname{{\mathrm{ad}}}_{e_{3}}(e_{3})\rangle=-\langle
e^{\ast}_{1},0\rangle=0,\;\;$
$(k_{71}e^{\ast}_{1}+k_{72}e^{\ast}_{2}+k_{73}e^{\ast}_{3})({e_{3}})=k_{73}.$

We obtain $k_{73}=0$. Therefore,
$\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{1})=0$.

For $\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{2})$, set
$\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{2})=k_{81}e^{\ast}_{1}+k_{82}e^{\ast}_{2}+k_{83}e^{\ast}_{3},~{}k_{81},~{}k_{82},~{}k_{83}\in\mathbb{C}.$
Then

$\langle\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{2}),{e_{1}}\rangle=-\langle
e^{\ast}_{2},\operatorname{{\mathrm{ad}}}_{e_{3}}(e_{1})\rangle=-\langle
e^{\ast}_{2},0\rangle=0,\;\;$
$(k_{81}e^{\ast}_{1}+k_{82}e^{\ast}_{2}+k_{83}e^{\ast}_{3})({e_{1}})=k_{81}.$

This implies that $k_{81}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{2}),{e_{2}}\rangle=-\langle
e^{\ast}_{2},\operatorname{{\mathrm{ad}}}_{e_{3}}(e_{2})\rangle=-\langle
e^{\ast}_{2},0\rangle=0,\;\;$
$(k_{81}e^{\ast}_{1}+k_{82}e^{\ast}_{2}+k_{83}e^{\ast}_{3})({e_{2}})=k_{82}.$

We obtain $k_{82}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{2}),{e_{3}}\rangle=-\langle
e^{\ast}_{2},\operatorname{{\mathrm{ad}}}_{e_{3}}(e_{3})\rangle=-\langle
e^{\ast}_{2},0\rangle=0,\;\;$
$(k_{81}e^{\ast}_{1}+k_{82}e^{\ast}_{2}+k_{83}e^{\ast}_{3})({e_{3}})=k_{83}.$

We obtain $k_{83}=0$. Therefore,
$\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{2})=0$.

For $\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{3})$, set
$\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{3})=k_{91}e^{\ast}_{1}+k_{92}e^{\ast}_{2}+k_{93}e^{\ast}_{3},~{}k_{91},~{}k_{92},~{}k_{93}\in\mathbb{C}.$
Then

$\langle\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{3}),{e_{1}}\rangle=-\langle
e^{\ast}_{3},\operatorname{{\mathrm{ad}}}_{e_{3}}(e_{1})\rangle=-\langle
e^{\ast}_{3},0\rangle=0,\;\;$
$(k_{91}e^{\ast}_{1}+k_{92}e^{\ast}_{2}+k_{93}e^{\ast}_{3})({e_{1}})=k_{91}.$

This implies that $k_{91}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{3}),{e_{2}}\rangle=-\langle
e^{\ast}_{3},\operatorname{{\mathrm{ad}}}_{e_{3}}(e_{2})\rangle=-\langle
e^{\ast}_{3},0\rangle=0,\;\;$
$(k_{91}e^{\ast}_{1}+k_{92}e^{\ast}_{2}+k_{93}e^{\ast}_{3})({e_{2}})=k_{92}.$

We obtain $k_{92}=0$.

$\langle\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{3}),{e_{3}}\rangle=-\langle
e^{\ast}_{3},\operatorname{{\mathrm{ad}}}_{e_{3}}(e_{3})\rangle=-\langle
e^{\ast}_{3},0\rangle=0,\;\;$
$(k_{91}e^{\ast}_{1}+k_{92}e^{\ast}_{2}+k_{93}e^{\ast}_{3})=k_{93}.$

We obtain $k_{93}=0$. Therefore,
$\operatorname{{\mathrm{ad}}}_{e_{3}}^{*}(e^{\ast}_{3})=0$.

The Rota-Baxter operators $R_{\mathfrak{g}_{2}}^{1}$ of weight zero on the
sub-adjacent Lie algebra ${\mathfrak{g}_{2}}$ are

$R_{\mathfrak{g}_{2}}^{1}=\left(\begin{array}[]{ccc}0&0&0\\\
n_{21}&0&n_{23}\\\ 0&0&0\\\ \end{array}\right).$

By Proposition 5.14, we have
$R_{\mathfrak{g}_{2}}^{1}=(n_{21}e_{1}+n_{23}e_{3})\otimes e_{2}^{\ast}$ and
$(R_{\mathfrak{g}_{1}}^{1})^{21}=e_{2}^{\ast}\otimes(n_{21}e_{1}+n_{23}e_{3})$.
By Lemma 5.15, the solutions of CYBE in the Lie algebra
${\mathfrak{g}_{2}}\ltimes_{{\rm ad}^{\ast}}{\mathfrak{g}_{2}}^{\ast}$ are

$r_{1}=R_{\mathfrak{g}_{2}}^{1}-(R_{\mathfrak{g}_{2}}^{1})^{21}=(n_{21}e_{1}+n_{23}e_{3})\otimes
e_{2}^{\ast}-e_{2}^{\ast}\otimes(n_{21}e_{1}+n_{23}e_{3}).$

The other solutions of CYBE in the Lie algebra
${\mathfrak{g}_{2}}\ltimes_{{\rm ad}^{\ast}}{\mathfrak{g}_{2}}^{\ast}$ are
obtained for the same step.

∎

######  Proposition 5.18.

Let $\\{e_{1},e_{2},e_{3}\\}$ be a basis of ${\mathfrak{g}_{3}}$ and
$\\{e^{\ast}_{1},e^{\ast}_{2},e^{\ast}_{3}\\}$ be its dual basis, then
${\mathfrak{g}_{3}}\ltimes_{{\rm ad}^{\ast}}{\mathfrak{g}_{3}}^{\ast}$ has a
basis $\\{e_{1},e_{2},e_{3},e_{1}^{\ast},e_{2}^{\ast},e^{\ast}_{3}\\}$ with
respect to which the non-zero products are given by the following:

$[e_{2},e_{3}]=-e_{2},\quad\quad[e_{2},e^{\ast}_{2}]=e^{\ast}_{3},\quad\quad[e_{3},e^{\ast}_{2}]=-e^{\ast}_{2}.$

The solutions of CYBE in the Lie algebra ${\mathfrak{g}_{3}}\ltimes_{{\rm
ad}^{\ast}}{\mathfrak{g}_{3}}^{\ast}$ are given by the following:

  * $r_{1}=n_{32}e_{2}\otimes e_{3}^{\ast}-e_{3}^{\ast}\otimes n_{32}e_{2}.$

  * $r_{2}=(n_{22}e_{2}+{n_{23}}e_{3})\otimes e_{2}^{\ast}-(\frac{{n_{22}}^{2}}{n_{23}}e_{2}+n_{22}e_{3})\otimes e_{3}^{\ast}\\\ -e_{2}^{\ast}\otimes(n_{22}e_{2}+{n_{23}}e_{3})+e_{3}^{\ast}\otimes(\frac{{n_{22}}^{2}}{n_{23}}e_{2}+n_{22}e_{3}),~{}n_{23}\neq 0.$

  * $r_{3}=(n_{31}e_{1}+n_{32}e_{2})\otimes e_{3}^{\ast}-e_{3}^{\ast}\otimes(n_{31}e_{1}+n_{32}e_{2}),~{}n_{31}\neq 0.$

  * $r_{4}=n_{21}e_{1}\otimes e_{2}^{\ast}+n_{31}e_{1}\otimes e_{3}^{\ast}-e_{2}^{\ast}\otimes n_{21}e_{1}-e_{3}^{\ast}\otimes n_{31}e_{1},~{}n_{21}\neq 0.$

  * $r_{5}=n_{12}e_{2}\otimes e_{1}^{\ast}+n_{32}e_{2}\otimes e_{3}^{\ast}-e_{1}^{\ast}\otimes n_{12}e_{2}-e_{3}^{\ast}\otimes n_{32}e_{2},~{}n_{12}\neq 0.$

  * $r_{6}=(n_{21}e_{1}-n_{31}e_{2}+{n_{21}}e_{3})\otimes e_{2}^{\ast}+(n_{31}e_{1}-\frac{{n_{31}}^{2}}{n_{21}}e_{2}+n_{31}e_{3})\otimes e_{3}^{\ast}\\\ -e_{2}^{\ast}\otimes(n_{21}e_{1}-n_{31}e_{2}+{n_{21}}e_{3})-e_{3}^{\ast}\otimes(n_{31}e_{1}-\frac{{n_{31}}^{2}}{n_{21}}e_{2}+n_{31}e_{3}),~{}n_{21}\neq 0.$

  * $r_{7}=(n_{21}e_{1}-n_{33}e_{2}-{n_{21}}e_{3})\otimes e_{2}^{\ast}+(-n_{33}e_{1}+\frac{{n_{33}}^{2}}{n_{21}}e_{2}+n_{33}e_{3})\otimes e_{3}^{\ast}\\\ -e_{2}^{\ast}\otimes(n_{21}e_{1}-n_{33}e_{2}-{n_{21}}e_{3})-e_{3}^{\ast}\otimes(-n_{33}e_{1}+\frac{{n_{33}}^{2}}{n_{21}}e_{2}+n_{33}e_{3}),~{}n_{21}\neq 0.$

###### Proof.

By using a similar proof as that in Proposition 5.17. ∎

######  Proposition 5.19.

Let $\\{e_{1},e_{2},e_{3}\\}$ be a basis of ${\mathfrak{g}_{4}}$ and
$\\{e^{\ast}_{1},e^{\ast}_{2},e^{\ast}_{3}\\}$ be its dual basis, then
${\mathfrak{g}_{4}}\ltimes_{{\rm ad}^{\ast}}{\mathfrak{g}_{4}}^{\ast}$ has a
basis $\\{e_{1},e_{2},e_{3},e_{1}^{\ast},e_{2}^{\ast},e^{\ast}_{3}\\}$ with
respect to which the non-zero products are given by the following:

$[e_{2},e_{3}]=e_{2},\quad\quad[e_{2},e^{\ast}_{2}]=-e^{\ast}_{3},\quad\quad[e_{3},e^{\ast}_{2}]=e^{\ast}_{2}.$

The solutions of CYBE in the Lie algebra ${\mathfrak{g}_{4}}\ltimes_{{\rm
ad}^{\ast}}{\mathfrak{g}_{4}}^{\ast}$ are given by the following:

  * $r_{1}=n_{32}e_{2}\otimes e_{3}^{\ast}-e_{3}^{\ast}\otimes n_{32}e_{2}.$

  * $r_{2}=(n_{22}e_{2}+{n_{23}}e_{3})\otimes e_{2}^{\ast}-(\frac{{n_{22}}^{2}}{n_{23}}e_{2}+n_{22}e_{3})\otimes e_{3}^{\ast}\\\ -e_{2}^{\ast}\otimes(n_{22}e_{2}+{n_{23}}e_{3})+e_{3}^{\ast}\otimes(\frac{{n_{22}}^{2}}{n_{23}}e_{2}+n_{22}e_{3}),~{}n_{23}\neq 0.$

  * $r_{3}=(n_{31}e_{1}+n_{32}e_{2})\otimes e_{3}^{\ast}-e_{3}^{\ast}\otimes(n_{31}e_{1}+n_{32}e_{2}),~{}n_{31}\neq 0.$

  * $r_{4}=n_{21}e_{1}\otimes e_{2}^{\ast}+n_{31}e_{1}\otimes e_{3}^{\ast}-e_{2}^{\ast}\otimes n_{21}e_{1}-e_{3}^{\ast}\otimes n_{31}e_{1},~{}n_{21}\neq 0.$

  * $r_{5}=n_{12}e_{2}\otimes e_{1}^{\ast}+n_{32}e_{2}\otimes e_{3}^{\ast}-e_{1}^{\ast}\otimes n_{12}e_{2}-e_{3}^{\ast}\otimes n_{32}e_{2},~{}n_{12}\neq 0.$

  * $r_{6}=(n_{21}e_{1}-n_{31}e_{2}+{n_{21}}e_{3})\otimes e_{2}^{\ast}+(n_{31}e_{1}-\frac{{n_{31}}^{2}}{n_{21}}e_{2}+n_{31}e_{3})\otimes e_{3}^{\ast}\\\ -e_{2}^{\ast}\otimes(n_{21}e_{1}-n_{31}e_{2}+{n_{21}}e_{3})-e_{3}^{\ast}\otimes(n_{31}e_{1}-\frac{{n_{31}}^{2}}{n_{21}}e_{2}+n_{31}e_{3}),~{}n_{21}\neq 0.$

###### Proof.

By a similar proof to that in Proposition 5.17. ∎

######  Proposition 5.20.

Let $\\{e_{1},e_{2},e_{3}\\}$ be a basis of ${\mathfrak{g}}(D_{12})$ and
$\\{e^{\ast}_{1},e^{\ast}_{2},e^{\ast}_{3}\\}$ be its dual basis, then
${\mathfrak{g}}(D_{12})$ $\ltimes_{{\rm
ad}^{\ast}}{\mathfrak{g}}(D_{12})^{\ast}$ has a basis
$\\{e_{1},e_{2},e_{3},e_{1}^{\ast},e_{2}^{\ast},e^{\ast}_{3}\\}$ with respect
to which the non-zero products are given by the following:

$[e_{1},e_{3}]=-e_{1},\quad[e_{1},e^{\ast}_{1}]=e^{\ast}_{3},\quad[e_{2},e_{3}]=e_{2},\quad[e_{2},e^{\ast}_{2}]=-e^{\ast}_{3},\quad[e_{3},e^{\ast}_{1}]=-e^{\ast}_{1},\quad[e_{3},e^{\ast}_{2}]=e^{\ast}_{2}.$

The solutions of CYBE in the Lie algebra ${\mathfrak{g}}(D_{12})\ltimes_{{\rm
ad}^{\ast}}{\mathfrak{g}}(D_{12})^{\ast}$ are given by the following:

  * $r_{1}=(n_{31}e_{1}+n_{32}e_{2})\otimes e_{3}^{\ast}-e_{3}^{\ast}\otimes(n_{31}e_{1}+n_{32}e_{2}).$

  * $r_{2}=n_{21}e_{1}\otimes e_{2}^{\ast}+n_{31}e_{1}\otimes e_{3}^{\ast}-e_{2}^{\ast}\otimes n_{21}e_{1}-e_{3}^{\ast}\otimes n_{31}e_{1},~{}n_{21}\neq 0.$

  * $r_{3}=n_{12}e_{2}\otimes e_{1}^{\ast}+n_{32}e_{2}\otimes e_{3}^{\ast}-e_{1}^{\ast}\otimes n_{12}e_{2}-e_{3}^{\ast}\otimes n_{32}e_{2},~{}n_{12}\neq 0.$

  * $r_{4}=(n_{21}e_{1}-n_{33}e_{2}+{n_{23}}e_{3})\otimes e_{2}^{\ast}+(\frac{n_{21}n_{33}}{n_{23}}e_{1}-\frac{{n_{33}}^{2}}{n_{23}}e_{2}+n_{33}e_{3})\otimes e_{3}^{\ast}\\\ -e_{2}^{\ast}\otimes(n_{21}e_{1}-n_{33}e_{2}+{n_{23}}e_{3})-e_{3}^{\ast}\otimes(\frac{n_{21}n_{33}}{n_{23}}e_{1}-\frac{{n_{33}}^{2}}{n_{23}}e_{2}+n_{33}e_{3}),~{}n_{23}\neq 0.$

  * $r_{5}=(-n_{33}e_{1}+n_{12}e_{2}+{n_{13}}e_{3})\otimes e_{1}^{\ast}+(-\frac{{n_{33}}^{2}}{n_{13}}e_{1}+\frac{n_{12}n_{33}}{n_{13}}e_{2}+n_{33}e_{3})\otimes e_{3}^{\ast}\\\ -e_{1}^{\ast}\otimes(-n_{33}e_{1}+n_{12}e_{2}+{n_{13}}e_{3})-e_{3}^{\ast}\otimes(-\frac{{n_{33}}^{2}}{n_{13}}e_{1}+\frac{n_{12}n_{33}}{n_{13}}e_{2}+n_{33}e_{3}),~{}n_{13}\neq 0.$

###### Proof.

The proof is similar to that of Proposition 5.17. ∎

Acknowledgements: The authors would like to thank the referees for helpful
comments. The fourth author acknowledges support from the NSF China
(12101328).

## References

  * [1] C. Bai, A unified algebraic approach to the classical Yang-Baxter equation, _J. Phy. A: Math. Theor_., 40 (2007), 11073-11082.
  * [2] C. Bai, Bijective 1-cocycles and classification of 3-dimensional left-symmetric algebras, _Comm. Algebra_ , 37 (2009), 1016-1057.
  * [3] A. A. Belavin, Dynamical symmetry of integrable quantum systems, _Nucl. Phys. B_ , 180 (1981), 189-200.
  * [4] A. V. Bolsinov, A. Y. Konyaev and V. S. Matveev, Nijenhuis Geometry, _Adv. Math_., 394 (2022), 108001.
  * [5] A. V. Bolsinov, V. S. Matveev, E. Miranda and S. Tabachnikov, Open problems, questions and challenges in finite-dimensional integrable systems, _Philos. Trans. R. Soc. A_ , 376 (2018), 20170430.
  * [6] D. Burde, Simple left-symmetric algebras with solvable Lie algebra, _Manuscr Math._ , 95 (1998), 397-411.
  * [7] A. Cayley, On the theory of analytic forms called trees, _Cambridge. Univ. Press_. 3 (1890), 242-246.
  * [8] J. Clemente-Gallardo and J. M. Nunes da Costa, Dirac-Nijenhuis Structures, _J. Phy. A_ , 37 (2004), 7267-7296.
  * [9] A. Das and S. Sen, Nijenhuis operators on Hom-Lie algebras, _Comm. Algebra_ , 50 (2022), 1038-1054.
  * [10] I. Dorfman, Dirac structures and integrability of nonlinear evolution equations, _Nonlinear Sci. Theory Appl._ , John Wiley & Sons, Ltd., Chichester, 1993. xii+176 pp.
  * [11] L. D. Faddeev and L. Takhtajan, The quantum inverse scattering method of the inverse problem and the Heisenberg XYZ model, _Russ. Math. Surv_., 34 (1979), 11-68.
  * [12] L. D. Faddeev and L. Takhtajan, Hamiltonian methods in the theory of solitons, _Classics Math._ , Springer, Berlin, 2007. x+592 pp.
  * [13] M. Gerstenhaber, On the deformation of rings and algebras, _Ann. of Math._ , 79 (1964), 59-104.
  * [14] H. Kim, Complete left-invariant affine structures on nilpotent Lie groups, _J. Diff. Geom._ , 24 (1986), 373-394.
  * [15] K. Kong, C. Bai and D. Meng, On real left-symmetric algebras, _Comm. Algebra_ , 40 (2012), 1641-1668.
  * [16] Y. Kosmann-Schwarzbach, Nijenhuis structures on Courant algebroids, _Bull. Braz. Math. Soc.(N.S.)_ , 42 (4) (2011), 625-649.
  * [17] J. L. Koszul, Domaines born$\acute{e}$s homog$\grave{e}$nes et orbites de groupes de transformations affines, _Ann. of Math._ , 78 (1963), 267-288.
  * [18] J. Liu, Y. Sheng, Y. Zhou and C. Bai, Nijenhuis operators on $n$-Lie algebras, _Commun. Theor. Phys._ , 65 (6) (2016), 659-670.
  * [19] A. Medina, Flat left-invariant connections adapted to the automorphism structure of a Lie group, _J. Diff. Geom._ , 16 (1981), 445-474.
  * [20] A. Nijenhuis, $X_{n-1}$-forming sets of eigenvectors, _Indag. Math._ , 54 (1951), 200-212.
  * [21] A. Nijenhuis and R. W. Jr. Richardson, Deformations of Lie algebra structures, _J. Math. Mech._ , 171 (1967), 89-105.
  * [22] J. Pei, C. Bai and L. Guo, Rota-Baxter on ${\rm sl}(2,\mathbb{C})$ and solution of the classical Yang-Baxter equation, _J. Math. Phys._ , 55 (2014), 021701.
  * [23] M. A. Semonov-Tian-Shansky, What is a classical R-matrix, _Funktsional. Anal. i Prilozhen._ , 17 (1983), 259-272.
  * [24] E. B. Vinberg, Convex homogeneous cones, _Trudy Moskov. Mat. Obsc._ , 12 (1963), 340-403.
  * [25] Q. Wang, Y. Sheng, C. Bai and J. Liu, Nijenhuis operators on pre-Lie algebras, _Commun. Contemp. Math._ , 7 (2019), 1850050.
  * [26] A. Winterhalder, Linear Nijenhuis-tensors and the construction of integrable systems, 1997, _arXiv.org:_ 9709008\.
  * [27] R. O. Wells, Differential analysis on complex manifolds, Graduate Texts, Vol. 65, Springer-Verlag, Berlin-Heidelberg-New York, 1980.