# Factorizable Lie bialgebras, quadratic Rota-Baxter Lie algebras and Rota-
Baxter Lie bialgebras

Honglei Lang  College of Science, China Agricultural University, Beijing,
100083, China<EMAIL_ADDRESS>and  Yunhe Sheng  Department of Mathematics,
Jilin University, Changchun 130012, Jilin, China<EMAIL_ADDRESS>
###### Abstract.

In this paper, first we introduce the notion of quadratic Rota-Baxter Lie
algebras of arbitrary weight, and show that there is a one-to-one
correspondence between factorizable Lie bialgebras and quadratic Rota-Baxter
Lie algebras of nonzero weight. Then we introduce the notions of matched
pairs, bialgebras and Manin triples of Rota-Baxter Lie algebras of arbitrary
weight, and show that Rota-Baxter Lie bialgebras, Manin triples of Rota-Baxter
Lie algebras and certain matched pairs of Rota-Baxter Lie algebras are
equivalent. The coadjoint representations and quadratic Rota-Baxter Lie
algebras play important roles in the whole study. Finally we generalize some
results to the Lie group context. In particular, we show that there is a one-
to-one correspondence between factorizable Poisson Lie groups and quadratic
Rota-Baxter Lie groups.

###### Key words and phrases:

factorizable Lie bialgebra, quadratic Rota-Baxter Lie algebra, Rota-Baxter Lie
bialgebra, factorizable Poisson Lie group
2020 _Mathematics Subject Classification._ 17B38, 17B62, 22E60, 53D17

###### Contents

  1. 1 Introduction
  2. 2 Factorizable Lie bialgebras and quadratic Rota-Baxter Lie algebras
    1. 2.1 Preliminaries on factorizable Lie bialgebras
    2. 2.2 Factorizable Lie bialgebras and quadratic Rota-Baxter Lie algebras
  3. 3 Matched pairs of Rota-Baxter Lie algebras
    1. 3.1 Rota-Baxter Lie algebras and matched pairs of Lie algebras
    2. 3.2 Matched pairs of Rota-Baxter Lie algebras
  4. 4 Rota-Baxter Lie bialgebras
    1. 4.1 Rota-Baxter Lie bialgebras and matched pairs of Rota-Baxter Lie algebras
    2. 4.2 Rota-Baxter Lie bialgebras and Manin triples of Rota-Baxter Lie algebras
  5. 5 Rota-Baxter Lie groups and factorizable Poisson Lie groups
    1. 5.1 Factorizable Poisson Lie groups and quadratic Rota-Baxter Lie groups
    2. 5.2 Rota-Baxter Lie groups and matched pairs of Lie groups

##  1\. Introduction

Poisson Lie groups, as semi-classical counterparts of Hopf algebras, or
quantum groups, are Lie groups with compatible Poisson structures in the sense
that the group multiplication is a Poisson map. The infinitesimals of Poisson
Lie groups are Lie bialgebras. A quasitriangular Hopf algebra [10] is a Hopf
algebra $\mathcal{A}$ with an invertible element
$R\in\mathcal{A}\otimes\mathcal{A}$, called quantum $R$-matrices, satisfying
some conditions. Particularly, $R$ satisfies the quantum Yang-Baxter equation.
The quasiclassical limits of quantum $R$-matrices are classical $r$-matrices
([33]), which give rise to the notion of quasitriangular Lie bialgebras.
Factorizable Lie bialgebras and factorizable Hopf algebras, introduced in
[32], are important classes of quasitriangular Lie bialgebras and
quasitriangular Hopf algebras. For instance, the double of an arbitrary Lie
bialgebra is a factorizable Lie bialgebra. As the name suggests, factorizable
Lie bialgebras (Hopf algebras) are used to establish the relation between
classical $r$-matrices (quantum $R$-matrices) and certain factorization
problems in Lie algebras (Hopf algebras). A key feature of the factorizable
case is that the symmetric part of a classical $r$-matrix
$r\in\otimes^{2}\mathfrak{g}$ (quantum $R$-matrix) identifies the underlying
vector space of the Lie algebra $\mathfrak{g}$ (the Hopf algebra) with its
dual space $\mathfrak{g}^{*}$. Then a Lie bialgebra
$(\mathfrak{g},\mathfrak{g}^{*})$ gives two compatible Lie algebra structures
on $\mathfrak{g}$, which leads to a double Lie algebra. Such double Lie
algebras generate Hamiltonian systems and can be solved by the factorization;
see [33, 34].

Rota-Baxter associative algebras were introduced in the probability study of
G. Baxter. They have important applications in various areas, including the
Connes-Kreimer’s algebraic approach to renormalization of quantum field theory
[9], noncommutative symmetric functions [11, 40], splitting of operads [3,
30], quantum analogue of Poisson geometry [37] and double Poisson algebras [1,
19, 35]. A linear map $B:\mathfrak{g}\,\rightarrow\,\mathfrak{g}$ is called a
Rota-Baxter operator of weight $\lambda$ on a Lie algebra $\mathfrak{g}$ if

$[B(x),B(y)]_{\mathfrak{g}}=B([B(x),y]_{\mathfrak{g}}+[x,B(y)]_{\mathfrak{g}}+\lambda[x,y]_{\mathfrak{g}}),\quad\forall
x,y\in\mathfrak{g}.$

There is a close relationship between Rota-Baxter operators on Lie algebras
and solutions of the (modified) classical Yang-Baxter equation. More
precisely, a Rota-Baxter operator of weight 0 on a Lie algebra is naturally
the operator form of a classical $r$-matrix [33] under certain conditions;
Rota-Baxter operators of weight 1 on a Lie algebra one-to-one correspond to
solutions of the modified classical Yang-Baxter equation. To better understand
such connections, the notion of an $\mathcal{O}$-operator (also called a
relative Rota-Baxter operator [30] or a generalized Rota-Baxter operator [38])
on a Lie algebra $\mathfrak{g}$ with respect to a representation $(V,\rho)$
was introduced by Kupershmidt in [26], which can be traced back to Bordemann’s
earlier work [8]. An $\mathcal{O}$-operator on a Lie algebra $\mathfrak{g}$
with respect to a representation $(V,\rho)$ is a linear operator
$T:V\to\mathfrak{g}$ satisfying

$[Tu,Tv]_{\mathfrak{g}}=T(\rho(Tu)v-\rho(Tv)u),\quad\forall u,v\in V.$

A skew-symmetric classical $r$-matrix $r\in\wedge^{2}\mathfrak{g}^{*}$
naturally gives rise to an $\mathcal{O}$-operator
$r^{\sharp}:\mathfrak{g}^{*}\to\mathfrak{g}$ on $\mathfrak{g}$ with respect to
the coadjoint representation. On the other hand, any $\mathcal{O}$-operator
$T:V\to\mathfrak{g}$ gives rise to a skew-symmetric solution of the classical
Yang-Baxter equation in the semidirect product Lie algebra
$\mathfrak{g}\ltimes_{\rho}V$, see [2]. Rota-Baxter operators and
$\mathcal{O}$-operators also play important roles in the study of integrable
systems [4, 33]. For further details on Rota-Baxter operators, see [20].

Even though the operator forms of skew-symmetric solutions of the classical
Yang-Baxter equation are very clear, namely $\mathcal{O}$-operators with
respect to the coadjoint representation (equivalently Rota-Baxter operators of
weight zero [7, 33] under certain conditions), the operator forms of non-skew-
symmetric solutions of the classical Yang-Baxter equation are still
mysterious. Recently, Goncharov established a correspondence between non-skew-
symmetric solutions of the classical Yang-Baxter equation and Rota-Baxter
operators of nonzero weight for certain Lie algebras [15, 16]. On the other
hand, Kosmann-Schwarzbach clarified the relation between factorizable Lie
bialgebras and double Lie algebras [24], while the latter are closely related
to Rota-Baxter Lie algebras of weight 1. Thus it is natural to expect a direct
relation between factorizable Lie bialgebras and Rota-Baxter Lie algebras.
This serves as the first purpose of the paper. We introduce the notion of a
quadratic Rota-Baxter Lie algebra of weight $\lambda$, and establish a one-to-
one correspondence between factorizable Lie bialgebras and quadratic Rota-
Baxter Lie algebras of nonzero weight.

It is well known that quadratic Lie algebras play important roles in the
theories of Manin triples of Lie algebras and Lie bialgebras. Since we have
defined quadratic Rota-Baxter Lie algebras of arbitrary weight, it is natural
to develop the bialgebra theory, the Manin triple theory as well as the
matched pair theory for Rota-Baxter Lie algebras. To define matched pairs, we
introduce the notion of representations of Rota-Baxter Lie algebras of weight
$\lambda$. A Rota-Baxter Lie algebra of weight $\lambda$ admits the adjoint
representation, as well as the coadjoint representation, while the latter is
quite tricky. For a quadratic Rota-Baxter Lie algebra of weight $\lambda$, its
adjoint representation and coadjoint representation are isomorphic. The data
in the coadjoint representation also serves as a guidance on how to define a
Rota-Baxter Lie bialgebra of weight $\lambda$. We introduce the notions of
matched pairs, bialgebras and Manin triples of Rota-Baxter Lie algebras of
weight $\lambda$ and show that they are equivalent.

Recently, Rota-Baxter operators are defined in the categories of Lie groups
[22] and cocommutative Hopf algebras [17], and have found some applications
[5, 6]. Taking the differentiation, a Rota-Baxter operator on a Lie group
gives a Rota-Baxter operator on the corresponding Lie algebra. A Rota-Baxter
operator on a cocommutative Hopf algebra gives rise to a Rota-Baxter operator
on the Lie algebra of primitive elements and a Rota-Baxter operator on the
group of group-like elements. We further generalize some results in the
algebraic setting to the Lie group context. In particular, we show that
factorizable Poisson Lie groups are in one-to-one correspondence with
quadratic Rota-Baxter Lie groups, and a Rota-Baxter Lie group gives a matched
pair of Lie groups. It is still mysterious to combine a Rota-Baxter operator
and a Poisson structure on a Lie group to define a Rota-Baxter Poisson Lie
group, and we will study it in the future. Another topic that we would like to
explore in the future is the operator forms of solutions of the classical
dynamical Yang-Baxter equation, which were first considered in [14] and
further studied in [13].

The paper is organized as follows. In Section 2, we introduce the notion of a
quadratic Rota-Baxter Lie algebra of weight $\lambda$, and show that
factorizable Lie bialgebras one-to-one correspond to quadratic Rota-Baxter Lie
algebras of nonzero weight (Theorem 2.5 and Theorem 2.10). In Section 3, first
we show that a Rota-Baxter Lie algebra gives a matched pair of Lie algebras.
Then we introduce the notion of a representation of a Rota-Baxter Lie algebra
of weight $\lambda$. A Rota-Baxter Lie algebra has the adjoint representation
and the coadjoint representation, and for a quadratic Rota-Baxter Lie algebra
they are isomorphic (Theorem 3.10). Finally we introduce the notion of a
matched pair of Rota-Baxter Lie algebras and show that it gives rise to a
descendent matched pair of Lie algebras (Theorem 3.15). In Section 4, first we
introduce the notion of a Rota-Baxter Lie bialgebra of weight $\lambda$, and
show that they are equivalent to certain matched pairs of Rota-Baxter Lie
algebras (Theorem 4.3). The double of a Rota-Baxter Lie bialgebra is still a
Rota-Baxter Lie bialgebra (Proposition 4.6). We show that factorizable Lie
bialgebras naturally give Rota-Baxter Lie bialgebras (Theorem 4.7). Then we
introduce the notion of Manin triples of Rota-Baxter Lie algebras and show
that they are equivalent to Rota-Baxter Lie bialgebras. In Section 5, we show
that factorizable Poisson Lie groups one-to-one correspond to quadratic Rota-
Baxter Lie groups (Theorem 5.4 and Theorem 5.7), and a Rota-Baxter Lie group
gives rise to a matched pair of Lie groups (Theorem 5.11).

Acknowledgements.  This research is supported by NSFC (11922110, 11901568). We
give our warmest thanks to Maxim Goncharov for helpful comments.

##  2\. Factorizable Lie bialgebras and quadratic Rota-Baxter Lie algebras

In this section, we establish a one-to-one correspondence between factorizable
Lie bialgebras and quadratic Rota-Baxter Lie algebras of nonzero weight.

###  2.1. Preliminaries on factorizable Lie bialgebras

We first briefly recall the definitions of quasitriangular Lie bialgebras and
factorizable Lie bialgebras following [10, 12, 24, 27, 28, 32, 39].

######  Definition 2.1.

A Lie bialgebra is a pair of Lie algebras
$(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}})$ and
$(\mathfrak{g}^{*},[\cdot,\cdot]_{\mathfrak{g}^{*}})$ such that

$\Delta[x,y]_{\mathfrak{g}}=[\Delta(x),y]_{\mathfrak{g}}+[x,\Delta(y)]_{\mathfrak{g}},\quad\forall
x,y\in\mathfrak{g},$

where $\Delta:\mathfrak{g}\to\wedge^{2}\mathfrak{g}$ is defined by
$\langle\Delta(x),\xi\wedge\eta\rangle:=\langle
x,[\xi,\eta]_{\mathfrak{g}^{*}}\rangle$, and called the cobracket.

A Lie bialgebra is denoted by $(\mathfrak{g},\mathfrak{g}^{*})$ or
$(\mathfrak{g},\Delta)$. For such a Lie bialgebra, there is a Lie algebra
structure $[\cdot,\cdot]_{\bowtie}$ on the double
$\mathfrak{g}\oplus\mathfrak{g}^{*}$ such that $\mathfrak{g}$ and
$\mathfrak{g}^{*}$ are Lie subalgebras and

$[x,\xi]_{\bowtie}=-\mathfrak{a}\mathfrak{d}_{\xi}^{*}x+\mathrm{ad}_{x}^{*}\xi,\quad\forall
x\in\mathfrak{g},\xi\in\mathfrak{g}^{*},$

where $\mathfrak{a}\mathfrak{d}_{\xi}^{*}x\in\mathfrak{g}$ and
$\mathrm{ad}_{x}^{*}\xi\in\mathfrak{g}^{*}$ are given by

$\langle\mathfrak{a}\mathfrak{d}_{\xi}^{*}x,\eta\rangle=-\langle
x,[\xi,\eta]_{\mathfrak{g}^{*}}\rangle,\quad\langle\mathrm{ad}_{x}^{*}\xi,y\rangle=-\langle\xi,[x,y]_{\mathfrak{g}}\rangle.$

Denote this Lie algebra by $\mathfrak{g}\bowtie\mathfrak{g}^{*}$. We refer the
readers to [27, 28] for the equivalent description of Lie bialgebras by using
Manin triples of Lie algebras. One important observation is that if
$(\mathfrak{g},\mathfrak{g}^{*})$ is a Lie bialgebra, then
$(\mathfrak{g}^{*},\mathfrak{g})$ is also a Lie bialgebra.

Let $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}})$ be a Lie algebra. An element
$r=\sum_{i}x_{i}\otimes y_{i}\in\mathfrak{g}\otimes\mathfrak{g}$ is called a
solution of the classical Yang-Baxter equation if it satisfies

$[r_{12},r_{13}]+[r_{13},r_{23}]+[r_{12},r_{23}]=0,$

where $r_{12}=\sum_{i}x_{i}\otimes y_{i}\otimes 1$, $r_{23}=\sum_{i}1\otimes
x_{i}\otimes y_{i}$ and $r_{13}=\sum_{i}x_{i}\otimes 1\otimes y_{i}$, and the
bracket $[r_{12},r_{13}]$ is defined by

$[r_{12},r_{13}]=[\sum_{i}x_{i}\otimes y_{i}\otimes 1,\sum_{j}x_{j}\otimes
1\otimes y_{j}]=\sum_{i,j}[x_{i},x_{j}]_{\mathfrak{g}}\otimes y_{i}\otimes
y_{j},$

and similarly for $[r_{13},r_{23}]$ and $[r_{12},r_{23}]$.

Any $r\in\mathfrak{g}\otimes\mathfrak{g}$ induces a linear operator
$r_{+}:\mathfrak{g}^{*}\to\mathfrak{g}$ defined by $r_{+}(\xi)=r(\xi,\cdot)$
for all $\xi\in\mathfrak{g}^{*}.$ Define $r_{-}:=-r_{+}^{*}$. Let us introduce
a bracket $[\cdot,\cdot]_{r}$ on $\mathfrak{g}^{*}$:

$[\xi,\eta]_{r}=\mathrm{ad}_{r_{+}\xi}^{*}\eta-\mathrm{ad}^{*}_{r_{-}\eta}\xi,\quad\forall\xi,\eta\in\mathfrak{g}^{*}.$

If $r$ satisfies the $\mathrm{ad}$-invariant condition

(1) $[r+\sigma(r),x]_{\mathfrak{g}}=0,\quad\forall x\in\mathfrak{g},$

where $\sigma$ is the flip operator which interchanges the components in
$\mathfrak{g}\otimes\mathfrak{g}$, and it is a solution of the classical Yang-
Baxter equation, then $(\mathfrak{g}^{*},[\cdot,\cdot]_{r})$ is a Lie algebra,
which is denoted by $\mathfrak{g}_{r}^{*}$. Moreover,
$(\mathfrak{g},\mathfrak{g}^{*}_{r})$ constitutes a Lie bialgebra, which is
called a quasitriangular Lie bialgebra. If $r$ is skew-symmetric, it is called
a triangular Lie bialgebra.

######  Theorem 2.2.

([24, 32]) Let $r\in\mathfrak{g}\otimes\mathfrak{g}$ satisfy the
$\mathrm{ad}$-invariant condition (1). Then $r$ satisfies the classical Yang-
Baxter equation if and only if $(\mathfrak{g}^{*},[\cdot,\cdot]_{r})$ is a Lie
algebra and the linear maps
$r_{+},r_{-}:(\mathfrak{g}^{*},[\cdot,\cdot]_{r})\to(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}})$
are both Lie algebra homomorphisms.

Denote by $I$ the operator

(2) $\displaystyle I=r_{+}-r_{-}:\mathfrak{g}^{*}\to\mathfrak{g}.$

Note that $I^{*}=I$, and the $\mathrm{ad}$-invariant condition (1) is
equivalent to

(3) $I\circ\mathrm{ad}_{x}^{*}=\mathrm{ad}_{x}\circ I,\quad\forall
x\in\mathfrak{g}.$

Actually $\frac{1}{2}I$ is the symmetric part of $r$. If $r$ is skew-
symmetric, then $I=0$. Factorizable Lie bialgebras are however concerned with
the opposite case that $I$ is nondegenerate; see [32].

######  Definition 2.3.

A quasitriangular Lie bialgebra $(\mathfrak{g},\mathfrak{g}^{*}_{r})$ defined
by $r\in\mathfrak{g}\otimes\mathfrak{g}$ is called factorizable if the linear
map $I:\mathfrak{g}^{*}\to\mathfrak{g}$ defined in (2) is a linear isomorphism
of vector spaces.

###  2.2. Factorizable Lie bialgebras and quadratic Rota-Baxter Lie algebras

Let $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}})$ be a Lie algebra and
$B:\mathfrak{g}\,\rightarrow\,\mathfrak{g}$ a Rota-Baxter operator of weight
$\lambda$ on $\mathfrak{g}$. Then there is a new Lie bracket
$[\cdot,\cdot]_{B}$ on $\mathfrak{g}$ defined by

$[x,y]_{B}=[B(x),y]_{\mathfrak{g}}+[x,B(y)]_{\mathfrak{g}}+\lambda[x,y]_{\mathfrak{g}}.$

The Lie algebra $(\mathfrak{g},[\cdot,\cdot]_{B})$ is called the descendent
Lie algebra, and denoted by $\mathfrak{g}_{B}$. It is obvious that $B$ is a
Lie algebra homomorphism from $\mathfrak{g}_{B}$ to $\mathfrak{g}$:

$B[x,y]_{B}=[B(x),B(y)]_{\mathfrak{g}}.$

Recall that a nondegenerate symmetric bilinear form
$S\in\otimes^{2}\mathfrak{g}^{*}$ on a Lie algebra $\mathfrak{g}$ is said to
be invariant if

(4) $\displaystyle S([x,y]_{\mathfrak{g}},z)+S(y,[x,z]_{\mathfrak{g}})$
$\displaystyle=$ $\displaystyle 0,\quad\forall x,y\in\mathfrak{g}.$

A quadratic Lie algebra $(\mathfrak{g},S)$ is a Lie algebra $\mathfrak{g}$
equipped with a nondegenerate symmetric invariant bilinear form
$S\in\otimes^{2}\mathfrak{g}^{*}$.

######  Definition 2.4.

Let $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},B)$ be a Rota-Baxter Lie
algebra of weight $\lambda$, and $S\in\otimes^{2}\mathfrak{g}^{*}$ a
nondegenerate symmetric bilinear form. The triple $(\mathfrak{g},B,S)$ is
called a quadratic Rota-Baxter Lie algebra of weight $\lambda$ if
$(\mathfrak{g},S)$ is a quadratic Lie algebra and the following compatibility
condition holds:

(5) $\displaystyle S(x,{B}y)+S({B}x,y)+\lambda S(x,y)$ $\displaystyle=$
$\displaystyle 0,\quad\forall x,y\in\mathfrak{g}.$

A factorizable Lie bialgebra naturally gives rise to a quadratic Rota-Baxter
Lie algebra of nonzero weight.

######  Theorem 2.5.

Let $(\mathfrak{g},\mathfrak{g}^{*}_{r})$ be a factorizable Lie bialgebra with
$I=r_{+}-r_{-}$. Then $(\mathfrak{g},B,S_{I})$ is a quadratic Rota-Baxter Lie
algebra of weight $\lambda$, where the linear map
$B:\mathfrak{g}\to\mathfrak{g}$ and $S_{I}\in\otimes^{2}\mathfrak{g}^{*}$ are
defined by

(6) $\displaystyle B$ $\displaystyle=$ $\displaystyle\lambda r_{-}\circ
I^{-1},$ (7) $\displaystyle S_{I}(x,y)$ $\displaystyle=$ $\displaystyle\langle
I^{-1}x,y\rangle,\quad\forall x,y\in\mathfrak{g}.$

###### Proof.

Since $r_{+},r_{-}$ are both Lie algebra homomorphisms, for
$x,y\in\mathfrak{g}$, we have

$\displaystyle I([I^{-1}x,I^{-1}y]_{r})$ $\displaystyle=$
$\displaystyle(r_{+}-r_{-})[I^{-1}x,I^{-1}y]_{r}$ $\displaystyle=$
$\displaystyle[(I+r_{-})I^{-1}x,(I+r_{-})I^{-1}y]_{\mathfrak{g}}-[r_{-}I^{-1}x,r_{-}I^{-1}y]_{\mathfrak{g}}$
$\displaystyle=$
$\displaystyle[r_{-}I^{-1}x,y]_{\mathfrak{g}}+[x,r_{-}I^{-1}y]_{\mathfrak{g}}+[x,y]_{\mathfrak{g}}.$

Therefore, we have

$\displaystyle
B([Bx,y]_{\mathfrak{g}}+[x,By]_{\mathfrak{g}}+\lambda[x,y]_{\mathfrak{g}})$
$\displaystyle=$
$\displaystyle\lambda^{2}r_{-}I^{-1}\big{(}[r_{-}I^{-1}x,y]_{\mathfrak{g}}+[x,r_{-}I^{-1}y]_{\mathfrak{g}}+[x,y]_{\mathfrak{g}}\big{)}$
$\displaystyle=$ $\displaystyle\lambda^{2}r_{-}[I^{-1}x,I^{-1}y]_{r}$
$\displaystyle=$
$\displaystyle\lambda^{2}[r_{-}I^{-1}x,r_{-}I^{-1}y]_{\mathfrak{g}}$
$\displaystyle=$ $\displaystyle[Bx,By]_{\mathfrak{g}},$

which implies that $B$ is a Rota-Baxter operator of weight $\lambda$ on
$\mathfrak{g}$.

Then, we show that $(\mathfrak{g},B,S_{I})$ is a quadratic Rota-Baxter Lie
algebra. Since $I^{*}=I$, it is obvious that
$S_{I}\in\otimes^{2}\mathfrak{g}^{*}$ is symmetric. It suffices to check (4)
and (5). In fact, by the invariant property (3), we have
$I^{-1}\circ\mathrm{ad}_{x}=\mathrm{ad}_{x}^{*}\circ I^{-1}$. So

$\displaystyle S_{I}([x,y]_{\mathfrak{g}},z)+S_{I}(y,[x,z]_{\mathfrak{g}})$
$\displaystyle=$ $\displaystyle\langle
I^{-1}[x,y]_{\mathfrak{g}},z\rangle+\langle
I^{-1}y,[x,z]_{\mathfrak{g}}\rangle$ $\displaystyle=$ $\displaystyle\langle
I^{-1}\circ\mathrm{ad}_{x}(y)-\mathrm{ad}_{x}^{*}\circ I^{-1}(y),z\rangle$
$\displaystyle=$ $\displaystyle 0.$

Moreover, by using $r_{-}^{*}=-r_{+}$ and $I=r_{+}-r_{-}$, we have

$\displaystyle S_{I}(x,By)+S_{I}(Bx,y)+\lambda S_{I}(x,y)$ $\displaystyle=$
$\displaystyle\lambda(\langle I^{-1}x,r_{-}\circ I^{-1}(y)\rangle+\langle
I^{-1}\circ r_{-}\circ I^{-1}(x),y\rangle+\langle I^{-1}x,y\rangle)$
$\displaystyle=$ $\displaystyle\lambda\langle(-I^{-1}\circ r_{+}\circ
I^{-1}+I^{-1}\circ r_{-}\circ I^{-1}+I^{-1})(x),y\rangle$ $\displaystyle=$
$\displaystyle 0.$

Therefore, $(\mathfrak{g},B,S_{I})$ is a quadratic Rota-Baxter Lie algebra of
weight $\lambda$. ∎

######  Remark 2.6.

Let $(\mathfrak{g},B,S)$ be a quadratic Rota-Baxter Lie algebra of weight
$\lambda$. Define a linear operator $B^{*}:\mathfrak{g}\to\mathfrak{g}$ by
$S(x,By)=S(B^{*}x,y)$ for all $x,y\in\mathfrak{g}$. Then (5) is equivalent to
$B+B^{*}=-\lambda\mathsf{id}$. Then by Theorem 2.5, a factorizable Lie
bialgebra $(\mathfrak{g},\mathfrak{g}_{r}^{*})$ gives rise to a Rota-Baxter
operator $B$ on $\mathfrak{g}$ such that $B+B^{*}=-\lambda\mathsf{id}$.
Similar results for simple Lie algebras and quadratic Lie algebras were given
earlier by Goncharov; see [15, Theorem 4] and [16, Theorems 1, 3].

It is well-known that if $B:\mathfrak{g}\to\mathfrak{g}$ is a Rota-Baxter
operator of weight $\lambda$ on a Lie algebra, then

(9) $\widetilde{B}:=-\lambda\mathsf{id}-B$

is also a Rota-Baxter operator of weight $\lambda$. So as a consequence of
Theorem 2.5, we have the following result.

######  Corollary 2.7.

Let $(\mathfrak{g},\mathfrak{g}^{*}_{r})$ be a factorizable Lie bialgebra with
$I=r_{+}-r_{-}$. Then $(\mathfrak{g},\widetilde{B},S_{I})$ is also a quadratic
Rota-Baxter algebra of weight $\lambda$, where
$\widetilde{B}=-\lambda\mathsf{id}-B=-\lambda r_{+}\circ I^{-1}$ and
$S_{I}\in\otimes^{2}\mathfrak{g}^{*}$ is defined in (7).

###### Proof.

It is obvious that $\widetilde{B}$ is a Rota-Baxter operator of weight
$\lambda$. Since $B$ satisfies the invariant condition (5), we have

$\displaystyle S_{I}(x,\widetilde{B}y)+S_{I}(\widetilde{B}x,y)+\lambda
S_{I}(x,y)$ $\displaystyle=$ $\displaystyle-S_{I}(x,{B}y)-\lambda
S_{I}(x,y)-S_{I}({B}x,y)-\lambda S_{I}(x,y)+\lambda S_{I}(x,y)$
$\displaystyle=$ $\displaystyle-S_{I}(x,{B}y)-S_{I}({B}x,y)-\lambda
S_{I}(x,y)$ $\displaystyle=$ $\displaystyle 0,$

which implies that $(\mathfrak{g},\widetilde{B},S_{I})$ is also a quadratic
Rota-Baxter algebra of weight $\lambda$. ∎

######  Corollary 2.8.

Let $(\mathfrak{g},\mathfrak{g}^{*}_{r})$ be a factorizable Lie bialgebra with
$I=r_{+}-r_{-}$, and $B$ the induced Rota-Baxter operator of weight $\lambda$
on $\mathfrak{g}$. Then
$\big{(}\mathfrak{g}_{B},(\mathfrak{g}^{*},[\cdot,\cdot]_{I})\big{)}$ is a Lie
bialgebra, where

$[\xi,\eta]_{I}:=\lambda
I^{-1}([\frac{1}{\lambda}I\xi,\frac{1}{\lambda}I\eta]_{\mathfrak{g}}),\quad\forall\xi,\eta\in\mathfrak{g}^{*},~{}\lambda\neq
0.$

Moreover, $\frac{1}{\lambda}I:\mathfrak{g}^{*}\to\mathfrak{g}$ gives a Lie
bialgebra isomorphism from $\big{(}\mathfrak{g}^{*}_{r},\mathfrak{g}\big{)}$
to $\big{(}\mathfrak{g}_{B},(\mathfrak{g}^{*},[\cdot,\cdot]_{I})\big{)}$.

###### Proof.

First we check that
$\frac{1}{\lambda}I:\mathfrak{g}_{r}^{*}\to\mathfrak{g}_{B}$ is a Lie algebra
isomorphism. In fact, for any $\xi,\eta\in\mathfrak{g}^{*}$, taking
$x=I\xi\in\mathfrak{g}$ and $y=I\eta\in\mathfrak{g}$, the equation (2.2) tells
us that

$\displaystyle\frac{1}{\lambda}I[\xi,\eta]_{r}=\frac{1}{\lambda^{2}}\big{(}[BI\xi,I\eta]_{\mathfrak{g}}+[I\xi,BI\eta]_{\mathfrak{g}}+\lambda[I\xi,I\eta]_{\mathfrak{g}}\big{)}=[\frac{1}{\lambda}I\xi,\frac{1}{\lambda}I\eta]_{B}.$

So $\frac{1}{\lambda}I$ is a Lie algebra isomorphism.

It is obvious that the map
$(\frac{1}{\lambda}I)^{*}=\frac{1}{\lambda}I:(\mathfrak{g}^{*},[\cdot,\cdot]_{I})\to\mathfrak{g}$
is also Lie algebra isomorphism. Since $(\mathfrak{g}_{r}^{*},\mathfrak{g})$
is a Lie bialgebra, it follows that the pair
$(\mathfrak{g}_{B},(\mathfrak{g}^{*},[\cdot,\cdot]_{I}))$ is also a Lie
bialgebra and $\frac{1}{\lambda}I:\mathfrak{g}_{r}^{*}\to\mathfrak{g}_{B}$ is
a Lie bialgebra isomorphism. ∎

######  Example 2.9.

Let $(\mathfrak{g},\mathfrak{g}^{*})$ be an arbitrary Lie bialgebra. Its
Drinfeld double $\mathfrak{d}=\mathfrak{g}\bowtie\mathfrak{g}^{*}$ with
$r=\sum_{i}\xi_{i}\otimes x_{i}\in\mathfrak{d}\otimes\mathfrak{d}$ is a
quasitriangular Lie bialgebra ([32]), where $\\{x_{i}\\}$ is a basis of
$\mathfrak{g}$ and $\\{\xi_{i}\\}$ is its dual basis for $\mathfrak{g}^{*}$.
Then $r_{+},r_{-}:\mathfrak{d}^{*}\to\mathfrak{d}$ are given by

$r_{+}(x,\xi)=(x,0),\quad r_{-}(x,\xi)=(0,-\xi),\quad\forall
x\in\mathfrak{g},\xi\in\mathfrak{g}^{*}.$

Observe that $I(x,\xi)=(x,\xi)$. So $(\mathfrak{d},\mathfrak{d}^{*}_{r})$ is a
factorizable Lie bialgebra. By Theorem 2.5, there associates a quadratic Rota-
Baxter Lie algebra $(\mathfrak{d},B,S_{I})$, where

$\displaystyle B(x,\xi)$ $\displaystyle=$ $\displaystyle\lambda r_{-}\circ
I^{-1}(x,\xi)=-\lambda(0,\xi),$ $\displaystyle S_{I}(x+\xi,y+\eta)$
$\displaystyle=$ $\displaystyle\langle
I^{-1}(x+\xi),y+\eta\rangle=\xi(y)+\eta(x).$

In this case, direct calculation shows that
$\mathfrak{d}^{*}=\mathfrak{g}\oplus\overline{\mathfrak{g}^{*}}$, namely, the
direct sum of the Lie algebra $\mathfrak{g}$ and the Lie algebra
$\overline{\mathfrak{g}^{*}}$, where $\overline{\mathfrak{g}^{*}}$ denotes the
Lie algebra with the Lie bracket $-[\cdot,\cdot]_{\mathfrak{g}^{*}}$.

At the end of this section, we show that a quadratic Rota-Baxter Lie algebra
of nonzero weight also gives rise to a factorizable Lie bialgebra.

######  Theorem 2.10.

Let $(\mathfrak{g},B,S)$ be a quadratic Rota-Baxter Lie algebra of weight
$\lambda$ ($\lambda\neq 0)$, and
$\mathcal{I}_{S}:\mathfrak{g}^{*}\to\mathfrak{g}$ the induced linear
isomorphism given by $\langle\mathcal{I}_{S}^{-1}x,y\rangle:=S(x,y)$. Then
$r\in\mathfrak{g}\otimes\mathfrak{g}$ determined by

$r_{+}:=\frac{1}{\lambda}(B+\lambda\mathsf{id})\circ\mathcal{I}_{S}:\mathfrak{g}^{*}\to\mathfrak{g},\quad
r_{+}(\xi)=r(\xi,\cdot),\quad\forall\xi\in\mathfrak{g}^{*}$

satisfies the classical Yang-Baxter equation, and gives rise to a
quasitriangular Lie bialgebra $(\mathfrak{g},\mathfrak{g}_{r}^{*})$, which is
factorizable.

###### Proof.

Since $S$ is symmetric, it follows that $\mathcal{I}_{S}^{*}=\mathcal{I}_{S}$.
By the fact that $S(x,By)+S(Bx,y)+\lambda S(x,y)=0$, we have

$\langle\mathcal{I}_{S}^{-1}x,By\rangle+\langle\mathcal{I}_{S}^{-1}\circ
B(x),y\rangle+\lambda\langle\mathcal{I}_{S}^{-1}x,y\rangle=0,$

which implies that $B^{*}\circ\mathcal{I}_{S}^{-1}+\mathcal{I}_{S}^{-1}\circ
B+\lambda\mathcal{I}_{S}^{-1}=0$, and then

$\mathcal{I}_{S}\circ B^{*}+B\circ\mathcal{I}_{S}+\lambda\mathcal{I}_{S}=0.$

Hence

$r_{-}:=-r_{+}^{*}=-\frac{1}{\lambda}(\mathcal{I}_{S}\circ
B^{*}+\lambda\mathcal{I}_{S})=\frac{1}{\lambda}B\circ\mathcal{I}_{S},$

and $\mathcal{I}_{S}=r_{+}-r_{-}$. Define a bracket operation
$[\cdot,\cdot]_{r}$ on $\mathfrak{g}^{*}$ by

$[\xi,\eta]_{r}=\mathrm{ad}_{r_{+}\xi}^{*}\eta-\mathrm{ad}_{r_{-}\eta}^{*}\xi,\quad\forall\xi,\eta\in\mathfrak{g}^{*}.$

We first check that

(10)
$\displaystyle\frac{1}{\lambda}\mathcal{I}_{S}[\xi,\eta]_{r}=[\frac{1}{\lambda}\mathcal{I}_{S}\xi,\frac{1}{\lambda}\mathcal{I}_{S}\eta]_{B},$

which would indicate that $[\cdot,\cdot]_{r}$ is a Lie bracket and
$\frac{1}{\lambda}\mathcal{I}_{S}$ is a Lie algebra isomorphism from
$(\mathfrak{g}^{*},[\cdot,\cdot]_{r})$ to $(\mathfrak{g},[\cdot,\cdot]_{B})$.
Actually, following from
$S([x,y]_{\mathfrak{g}},z)+S(y,[x,z]_{\mathfrak{g}})=0$, we have

$\langle\mathcal{I}_{S}^{-1}\circ\mathrm{ad}_{x}(y),z\rangle-\langle\mathrm{ad}_{x}^{*}\circ\mathcal{I}_{S}^{-1}(y),z\rangle=0,$

which yields the invariance of $\mathcal{I}_{S}$:

$\mathcal{I}_{S}\circ\mathrm{ad}_{x}^{*}=\mathrm{ad}_{x}\circ\mathcal{I}_{S}.$

Then we have

$\displaystyle\mathcal{I}_{S}[\xi,\eta]_{r}$ $\displaystyle=$
$\displaystyle\mathcal{I}_{S}(\mathrm{ad}_{r_{+}\xi}^{*}\eta-\mathrm{ad}_{r_{-}\eta}^{*}\xi)$
$\displaystyle=$
$\displaystyle[r_{+}\xi,\mathcal{I}_{S}\eta]_{\mathfrak{g}}-[r_{-}\eta,\mathcal{I}_{S}\xi]_{\mathfrak{g}}$
$\displaystyle=$ $\displaystyle[r_{+}\xi,r_{+}\eta-
r_{-}\eta]_{\mathfrak{g}}-[r_{-}\eta,r_{+}\xi-r_{-}\xi]_{\mathfrak{g}}$
$\displaystyle=$
$\displaystyle[r_{+}\xi,r_{+}\eta]_{\mathfrak{g}}-[r_{-}\xi,r_{-}\eta]_{\mathfrak{g}}.$

On the other hand, we have

$\displaystyle[\mathcal{I}_{S}\xi,\mathcal{I}_{S}\eta]_{B}$ $\displaystyle=$
$\displaystyle[B\mathcal{I}_{S}\xi,\mathcal{I}_{S}\eta]_{\mathfrak{g}}+[\mathcal{I}_{S}\xi,B\mathcal{I}_{S}\eta]_{\mathfrak{g}}+\lambda[\mathcal{I}_{S}\xi,\mathcal{I}_{S}\eta]_{\mathfrak{g}}$
$\displaystyle=$ $\displaystyle\lambda[r_{-}\xi,r_{+}\eta-
r_{-}\eta]_{\mathfrak{g}}+\lambda[r_{+}\xi-
r_{-}\xi,r_{-}\eta]_{\mathfrak{g}}+\lambda[r_{+}\xi-r_{-}\xi,r_{+}\eta-
r_{-}\eta]_{\mathfrak{g}}$ $\displaystyle=$
$\displaystyle\lambda\big{(}[r_{+}\xi,r_{+}\eta]_{\mathfrak{g}}-[r_{-}\xi,r_{-}\eta]_{\mathfrak{g}}\big{)},$

which implies that (10) holds. Finally since
$B+\lambda\mathsf{id},B:(\mathfrak{g},[\cdot,\cdot]_{B})\to(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}})$
are both Lie algebra homomorphisms, we deduce that

$r_{+}:=\frac{1}{\lambda}(B+\lambda\mathsf{id})\circ\mathcal{I}_{S}~{}\mbox{and}~{}r_{-}=\frac{1}{\lambda}B\circ\mathcal{I}_{S}:(\mathfrak{g}^{*},[\cdot,\cdot]_{r})\to(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}})$

are both Lie algebra homomorphisms. Therefore, by Theorem 2.2,
$(\mathfrak{g},\mathfrak{g}_{r}^{*})$ is a quasitriangular Lie bialgebra.
Since $\mathcal{I}_{S}=r_{+}-r_{-}$ is an isomorphism, the Lie bialgebra
$(\mathfrak{g},\mathfrak{g}_{r}^{*})$ is factorizable. ∎

######  Remark 2.11.

The relation between factorizable Lie bialgebras and double Lie algebras
([33]) is explained clearly in [24]. Double Lie algebras relate closely with
Rota-Baxter Lie algebras. Let $\mathfrak{g}$ be a Lie algebra and
$R:\mathfrak{g}\to\mathfrak{g}$ be a linear map. Define
$[x,y]_{R}=[Rx,y]_{\mathfrak{g}}+[x,Ry]_{\mathfrak{g}}$. If
$[\cdot,\cdot]_{R}$ is also a Lie bracket, the pair $(\mathfrak{g},R)$ is
called a double Lie algebra. In particular, if $R$ satisfies the following
modified Yang-Baxter equation:

$[Rx,Ry]_{\mathfrak{g}}-R([Rx,y]_{\mathfrak{g}}+[x,Ry]_{\mathfrak{g}})+[x,y]_{\mathfrak{g}}=0,$

$[\cdot,\cdot]_{R}$ satisfies the Jacobi identity and $(\mathfrak{g},R)$ is a
double Lie algebra. Indeed, $B$ is a Rota-Baxter operator of weight $1$ on
$\mathfrak{g}$ if and only if $R=\mathsf{id}+2B:\mathfrak{g}\to\mathfrak{g}$
is a solution of the modified Yang-Baxter equation. Here we relate directly
factorizable Lie bialgebras with Rota-Baxter Lie algebras of nonzero weight.

##  3\. Matched pairs of Rota-Baxter Lie algebras

In this section, first we show that a Rota-Baxter Lie algebra gives rise to a
matched pair of Lie algebras. Then we introduce the notion of matched pairs of
Rota-Baxter Lie algebras of weight $\lambda$ and show that a matched pair of
Rota-Baxter Lie algebras induces a descendent matched pair of Lie algebras.

###  3.1. Rota-Baxter Lie algebras and matched pairs of Lie algebras

Matched pairs of Lie algebras are also known as twilled Lie algebras [25] or
double Lie algebras [28].

A matched pair of Lie algebras ([29, 36]) consists of a pair of Lie algebras
$(\mathfrak{g},\mathfrak{h})$, a representation
$\rho:\mathfrak{g}\to\mathfrak{gl}(\mathfrak{h})$ of $\mathfrak{g}$ on
$\mathfrak{h}$ and a representation
$\mu:\mathfrak{h}\to\mathfrak{gl}(\mathfrak{g})$ of $\mathfrak{h}$ on
$\mathfrak{g}$ such that

(11) $\displaystyle\rho(x)[\xi,\eta]_{\mathfrak{h}}$ $\displaystyle=$
$\displaystyle[\rho(x)\xi,\eta]_{\mathfrak{h}}+[\xi,\rho(x)\eta]_{\mathfrak{h}}+\rho\big{(}(\mu(\eta)x\big{)}\xi-\rho\big{(}\mu(\xi)x\big{)}\eta,$
(12) $\displaystyle\mu(\xi)[x,y]_{\mathfrak{g}}$ $\displaystyle=$
$\displaystyle[\mu(\xi)x,y]_{\mathfrak{g}}+[x,\mu(\xi)y]_{\mathfrak{g}}+\mu\big{(}\rho(y)\xi\big{)}x-\mu\big{(}\rho(x)\xi\big{)}y,$

for all $x,y\in\mathfrak{g}$ and $\xi,\eta\in\mathfrak{h}$. We will denote a
matched pair of Lie algebras by $(\mathfrak{g},\mathfrak{h};\rho,\mu)$, or
simply by $(\mathfrak{g},\mathfrak{h})$.

The following alternative description of matched pairs of Lie algebras is
well-known.

######  Proposition 3.1.

Let $(\mathfrak{g},\mathfrak{h};\rho,\mu)$ be a matched pair of Lie algebras.
Then there is a Lie algebra structure on the direct sum space
$\mathfrak{g}\oplus\mathfrak{h}$ with the Lie bracket
$[\cdot,\cdot]_{\bowtie}$ given by

$[x+\xi,y+\eta]_{\bowtie}=\big{(}[x,y]_{\mathfrak{g}}+\mu(\xi)y-\mu(\eta)x\big{)}+\big{(}[\xi,\eta]_{\mathfrak{h}}+\rho(x)\eta-\rho(y)\xi).$

Denote this Lie algebra by $\mathfrak{g}\bowtie\mathfrak{h}$.

Conversely, if $(\mathfrak{g}\oplus\mathfrak{h},[\cdot,\cdot])$ is a Lie
algebra such that $\mathfrak{g}$ and $\mathfrak{h}$ are Lie subalgebras, then
$(\mathfrak{g},\mathfrak{h};\rho,\mu)$ is a matched pair of Lie algebras,
where the representations $\rho:\mathfrak{g}\to\mathfrak{gl}(\mathfrak{h})$
and $\mu:\mathfrak{h}\to\mathfrak{gl}(\mathfrak{g})$ are determined by

$[x,\xi]=\rho(x)\xi-\mu(\xi)x,\quad\forall
x\in\mathfrak{g},\xi\in\mathfrak{h}.$

Let $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}})$ be a Lie algebra. We denote
by
$(\mathfrak{g}\oplus\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}\oplus\mathfrak{g}})$
the direct sum Lie algebra with the bracket given by

$[(x_{1},x_{2}),(y_{1},y_{2})]_{\mathfrak{g}\oplus\mathfrak{g}}=([x_{1},y_{1}]_{\mathfrak{g}},[x_{2},y_{2}]_{\mathfrak{g}}).$

######  Proposition 3.2.

Let $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},B)$ be a Rota-Baxter Lie
algebra of weight $\lambda$ ($\lambda\neq 0$) with the descendent Lie algebra
$\mathfrak{g}_{B}$. Then there is a Lie algebra structure $[\cdot,\cdot]_{D}$
on the vector space $\mathfrak{g}\oplus\mathfrak{g}$ given by

$\displaystyle[(0,x),(0,y)]_{D}$ $\displaystyle=$
$\displaystyle(0,[x,y]_{\mathfrak{g}});$
$\displaystyle{[(\xi,0),(\eta,0)]_{D}}$ $\displaystyle=$
$\displaystyle([\xi,\eta]_{B},0);$ $\displaystyle{[(0,x),(\xi,0)]_{D}}$
$\displaystyle=$
$\displaystyle-[(\xi,0),(0,x)]_{D}=([x,\xi]_{\mathfrak{g}},[x,B\xi]_{\mathfrak{g}}-B[x,\xi]_{\mathfrak{g}}).$

Moreover, there is a Lie algebra isomorphism from
$(\mathfrak{g}\oplus\mathfrak{g},[\cdot,\cdot]_{D})$ to the direct sum Lie
algebra $\mathfrak{g}\oplus\mathfrak{g}$ defined by

(13)
$\displaystyle\phi:\mathfrak{g}\oplus\mathfrak{g}\to\mathfrak{g}\oplus\mathfrak{g},\quad(\xi,x)\mapsto(B\xi+\lambda\xi+x,B\xi+x).$

###### Proof.

First, it is direct to show that $\phi$ is an isomorphism between vector
spaces whose inverse is

$\phi^{-1}:\mathfrak{g}\oplus\mathfrak{g}\to\mathfrak{g}\oplus\mathfrak{g},\qquad(x,y)\mapsto\frac{1}{\lambda}(x-y,\lambda
y-B(x-y)).$

In the sequel, we show that the bracket $[\cdot,\cdot]_{D}$ is exactly the
pull-back of the direct sum Lie algebra structure on
$\mathfrak{g}\oplus\mathfrak{g}$ by $\phi$, i.e.

$[e_{1},e_{2}]_{D}=\phi^{-1}[\phi(e_{1}),\phi(e_{2})]_{\mathfrak{g}\oplus\mathfrak{g}},\quad\forall
e_{1},e_{2}\in\mathfrak{g}\oplus\mathfrak{g}.$

Consequently, $(\mathfrak{g}\oplus\mathfrak{g},[\cdot,\cdot]_{D})$ is a Lie
algebra and $\phi$ is a Lie algebra isomorphism.

By the fact that $B[\xi,\eta]_{B}=[B\xi,B\eta]_{\mathfrak{g}}$, we have

$\displaystyle\phi^{-1}[\phi(\xi,0),\phi(\eta,0)]_{\mathfrak{g}\oplus\mathfrak{g}}$
$\displaystyle=$
$\displaystyle\phi^{-1}[(B\xi+\lambda\xi,B\xi),(B\eta+\lambda\eta,B\eta)]_{\mathfrak{g}\oplus\mathfrak{g}}$
$\displaystyle=$
$\displaystyle\phi^{-1}([B\xi+\lambda\xi,B\eta+\lambda\eta]_{\mathfrak{g}},[B\xi,B\eta]_{\mathfrak{g}})$
$\displaystyle=$
$\displaystyle\phi^{-1}(B[\xi,\eta]_{B}+\lambda[\xi,\eta]_{B},B[\xi,\eta]_{B})$
$\displaystyle=$ $\displaystyle[(\xi,0),(\eta,0)]_{D},$

and

$\displaystyle\phi^{-1}[\phi(0,x),\phi(\xi,0)]_{\mathfrak{g}\oplus\mathfrak{g}}$
$\displaystyle=$
$\displaystyle\phi^{-1}[(x,x),(B\xi+\lambda\xi,B\xi)]_{\mathfrak{g}\oplus\mathfrak{g}}$
$\displaystyle=$
$\displaystyle\phi^{-1}([x,B\xi+\lambda\xi]_{\mathfrak{g}},[x,B\xi]_{\mathfrak{g}})$
$\displaystyle=$
$\displaystyle\phi^{-1}(B[x,\xi]_{\mathfrak{g}}+\lambda[x,\xi]_{\mathfrak{g}}+[x,B\xi]_{\mathfrak{g}}-B[x,\xi]_{\mathfrak{g}},B[x,\xi]_{\mathfrak{g}}+[x,B\xi]_{\mathfrak{g}}-B[x,\xi]_{\mathfrak{g}})$
$\displaystyle=$
$\displaystyle([x,\xi]_{\mathfrak{g}},[x,B\xi]_{\mathfrak{g}}-B[x,\xi]_{\mathfrak{g}})$
$\displaystyle=$ $\displaystyle[(0,x),(\xi,0)]_{D}.$

It is obvious that
$[(0,x),(0,y)]_{D}=\phi^{-1}[\phi(0,x),\phi(0,y)]_{\mathfrak{g}\oplus\mathfrak{g}}$.
Therefore the bracket $[\cdot,\cdot]_{D}$ is exactly the pull-back of the
direct sum Lie bracket $[\cdot,\cdot]_{\mathfrak{g}\oplus\mathfrak{g}}$ by
$\phi$. ∎

Parallel results of Proposition 3.2 for associative algebras can be found in
[18].

######  Theorem 3.3.

Let $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},B)$ be a Rota-Baxter Lie
algebra of weight $\lambda$ ($\lambda\neq 0$) with the descendent Lie algebra
$\mathfrak{g}_{B}$. Then $(\mathfrak{g}_{B},\mathfrak{g};\rho,\mu)$ is a
matched pair of Lie algebras, where
$\rho:\mathfrak{g}_{B}\to\mathfrak{gl}(\mathfrak{g})$ and
$\mu:\mathfrak{g}\to\mathfrak{gl}(\mathfrak{g}_{B})$ are given by

(14)
$\rho(\xi)(x)=B[x,\xi]_{\mathfrak{g}}-[x,B\xi]_{\mathfrak{g}},\quad\mu(x)(\xi)=[x,\xi]_{\mathfrak{g}},\quad\forall
x\in\mathfrak{g},\xi\in\mathfrak{g}_{B}.$

Moreover, the corresponding Lie algebra $\mathfrak{g}_{B}\bowtie\mathfrak{g}$
is exactly the Lie algebra
$(\mathfrak{g}\oplus\mathfrak{g},[\cdot,\cdot]_{D})$ given in Proposition 3.2.

###### Proof.

By Proposition 3.2, it is obvious that both $\mathfrak{g}_{B}$ and
$\mathfrak{g}$ are Lie subalgebras of the Lie algebra
$(\mathfrak{g}\oplus\mathfrak{g},[\cdot,\cdot]_{D})$, and

${[(\xi,0),(0,x)]_{D}}=(\rho(\xi)(x),-\mu(x)(\xi)).$

Thus, $(\mathfrak{g}_{B},\mathfrak{g};\rho,\mu)$ is a matched pair of Lie
algebras. The other conclusion is obvious. ∎

Since $\phi$ is a Lie algebra isomorphism and $\mathfrak{g}_{B}$ is a Lie
subalgebra, it follows that
$\operatorname{Im}(\phi|_{\mathfrak{g}_{B}})\subset\mathfrak{g}\oplus\mathfrak{g}$
is a Lie subalgebra, which is isomorphic to $\mathfrak{g}_{B}$. It gives an
alternative approach to prove the factorization theorem of Rota-Baxter Lie
algebras. We refer to [33, 22] for more details for the factorization theorem
of Rota-Baxter Lie algebras of weight $1$.

######  Corollary 3.4.

Let $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},B)$ be a Rota-Baxter Lie
algebra of weight $\lambda$ $(\lambda\neq 0)$. Then for any
$x\in\mathfrak{g}$, there exists a unique decomposition $x=x_{+}-x_{-}$ with
$(x_{+},x_{-})\in\operatorname{Im}(\phi|_{\mathfrak{g}_{B}})\subset\mathfrak{g}\oplus\mathfrak{g}$.

###### Proof.

We have $x=\frac{1}{\lambda}(Bx+\lambda x)-\frac{1}{\lambda}Bx=x_{+}-x_{-}$,
where $(x_{+},x_{-})=\phi(\frac{1}{\lambda}x,0)$. The decomposition is unique
since $\phi:\mathfrak{g}_{B}\to\operatorname{Im}(\phi|_{\mathfrak{g}_{B}})$ is
a Lie algebra isomorphism. ∎

Let $(\mathfrak{g},\mathfrak{g}^{*}_{r})$ be a factorizable Lie bialgebra with
$I=r_{+}-r_{-}$. Then $(\mathfrak{g},B=\lambda r_{-}\circ I^{-1})$ is a Rota-
Baxter Lie algebra of weight $\lambda$. By Theorem 3.3, we have the Lie
algebra $\mathfrak{g}_{B}\bowtie\mathfrak{g}$, which is the double of the
matched pair $(\mathfrak{g}_{B},\mathfrak{g};\rho,\mu)$. On the other hand,
there is a Lie algebra $\mathfrak{g}^{*}_{r}\bowtie\mathfrak{g}$, which is the
double of the Lie bialgebra $(\mathfrak{g}^{*}_{r},\mathfrak{g})$.

######  Proposition 3.5.

With the above notations, we have a commutative diagram of Lie algebra
isomorphisms:

(19)

where $\phi$ is defined by (13) and $\psi$ is given by

$\psi(\xi,x)=(r_{+}\xi+x,r_{-}\xi+x),\quad\forall
x\in\mathfrak{g},\xi\in\mathfrak{g}^{*}.$

###### Proof.

We first show that
$\tilde{I}:=(\frac{1}{\lambda}I)\oplus\mathsf{id}:\mathfrak{g}_{r}^{*}\bowtie\mathfrak{g}\to\mathfrak{g}_{B}\bowtie\mathfrak{g}$
is a Lie algebra isomorphism. We only need to check the relation

(20)
$\displaystyle\tilde{I}([x,\xi]_{\bowtie})=[\tilde{I}(x),\tilde{I}(\xi)]_{D},\quad\forall
x\in\mathfrak{g},\xi\in\mathfrak{g}^{*}.$

In fact, by the $\mathrm{ad}$-invariant condition (3), we have

$\tilde{I}([x,\xi]_{\bowtie})=(\frac{1}{\lambda}I(\mathrm{ad}_{x}^{*}\xi),-\mathfrak{a}\mathfrak{d}_{\xi}^{*}x)=(\frac{1}{\lambda}[x,I\xi]_{\mathfrak{g}},-\mathfrak{a}\mathfrak{d}_{\xi}^{*}x),$

and

${[\tilde{I}(x),\tilde{I}(\xi)]_{D}}=[(0,x),(\frac{1}{\lambda}I\xi,0)]_{D}=(\frac{1}{\lambda}[x,I\xi]_{\mathfrak{g}},\frac{1}{\lambda}[x,BI\xi]_{\mathfrak{g}}-\frac{1}{\lambda}B[x,I\xi]_{\mathfrak{g}}).$

Taking pairing with any $\eta\in\mathfrak{g}^{*}$ and since
$r_{-}=-r_{+}^{*}$, one finds

$\displaystyle\langle-\mathfrak{a}\mathfrak{d}_{\xi}^{*}x,\eta\rangle$
$\displaystyle=$ $\displaystyle-\langle
x,\mathrm{ad}_{r_{+}\eta}^{*}\xi-\mathrm{ad}_{r_{-}\xi}^{*}\eta\rangle$
$\displaystyle=$
$\displaystyle\langle[x,r_{-}\xi]_{\mathfrak{g}},\eta\rangle+\langle[r_{+}\eta,x]_{\mathfrak{g}},\xi\rangle$
$\displaystyle=$
$\displaystyle\langle\frac{1}{\lambda}[x,BI\xi]_{\mathfrak{g}}-r_{-}(\mathrm{ad}_{x}^{*}\xi),\eta\rangle$
$\displaystyle=$
$\displaystyle\langle\frac{1}{\lambda}[x,BI\xi]_{\mathfrak{g}}-\frac{1}{\lambda}BI(\mathrm{ad}_{x}^{*}\xi),\eta\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda}\langle[x,BI\xi]_{\mathfrak{g}}-B[x,I\xi]_{\mathfrak{g}},\eta\rangle,$

which implies that (20) holds.

Then, for $(\xi,x)\in\mathfrak{g}_{r}^{*}\bowtie\mathfrak{g}$, we have

$\phi\circ(\frac{1}{\lambda}I\oplus\mathsf{id})(\xi,x)=\phi(\frac{1}{\lambda}I\xi,x)=(\frac{1}{\lambda}BI\xi+I\xi+x,\frac{1}{\lambda}BI\xi+x)=(r_{+}\xi+x,r_{-}\xi+x)=\psi(\xi,x).$

Since both $\phi$ and $\frac{1}{\lambda}I\oplus\mathsf{id}$ are Lie algebra
isomorphisms, it follows that $\psi$ is also a Lie algebra isomorphism. ∎

###  3.2. Matched pairs of Rota-Baxter Lie algebras

In this subsection, we introduce the notion of matched pairs of Rota-Baxter
Lie algebras of weight $\lambda$, and show that a matched pair of Rota-Baxter
Lie algebras of weight $\lambda$ gives rise to a descendent matched pair of
Rota-Baxter Lie algebras.

Since in the definition of a matched pair of Lie algebras, we need the concept
of representations of Lie algebras. For the purpose of defining matched pairs
of Rota-Baxter Lie algebras, we introduce the notion of representations of
Rota-Baxter Lie algebras of weight $\lambda$. Note that the concept of
representations of Rota-Baxter Lie algebras of weight 0 was already given in
[21] in the study of cohomologies of Rota-Baxter Lie algebras, and the notion
of representations of Rota-Baxter associative algebras was introduced in [23],
and further studied in [31].

######  Definition 3.6.

A representation of a Rota-Baxter Lie algebra $(\mathfrak{g},B)$ of weight
$\lambda$ on a vector space $W$ with respect to a linear transformation
$T\in\mathfrak{gl}(W)$ is a representation $\rho$ of the Lie algebra
$\mathfrak{g}$ on $W$, satisfying

$\rho(Bx)(Tu)=T\Big{(}\rho(Bx)u+\rho(x)(Tu)+\lambda\rho(x)u\Big{)},\quad\forall
x\in\mathfrak{g},u\in W.$

We will denote a representation by $(W,T,\rho).$

Let $(W,T,\rho)$ be a representation of a Rota-Baxter Lie algebra
$(\mathfrak{g},B)$ of weight $\lambda$. Since $(W,\rho)$ is a representation
of the Lie algebra $\mathfrak{g}$, we have the semidirect product Lie algebra
$\mathfrak{g}\ltimes W$. Then define the map

$B\oplus T:\mathfrak{g}\ltimes W\to\mathfrak{g}\ltimes W,\qquad x+u\mapsto
Bx+Tu.$

The following conclusion is obvious.

######  Proposition 3.7.

With above notations, $(\mathfrak{g}\ltimes W,B\oplus T)$ is a Rota-Baxter Lie
algebra of weight $\lambda$, called the semidirect product of
$(\mathfrak{g},B)$ and the representation $(W,T,\rho)$.

######  Example 3.8.

It is straightforward to see that $(\mathfrak{g},B,\mathrm{ad})$ is a
representation of a Rota-Baxter Lie algebra $(\mathfrak{g},B)$ of weight
$\lambda$, which is called the adjoint representation of $(\mathfrak{g},B)$.

######  Definition 3.9.

Let $(W,T,\rho)$ and $(W^{\prime},T^{\prime},\rho^{\prime})$ be two
representations of a Rota-Baxter Lie algebra $(\mathfrak{g},B)$ of weight
$\lambda$. A homomorphism from $(W,T,\rho)$ to
$(W^{\prime},T^{\prime},\rho^{\prime})$ is a linear map $\phi:W\to W^{\prime}$
such that

$\displaystyle\phi\circ\rho(x)$ $\displaystyle=$
$\displaystyle\rho^{\prime}(x)\circ\phi,\quad\forall x\in\mathfrak{g},$
$\displaystyle\phi\circ T$ $\displaystyle=$ $\displaystyle
T^{\prime}\circ\phi.$

Rota-Baxter Lie algebras of weight $\lambda$ not only admit adjoint
representations, but also coadjoint representations. The following result
plays an important role in our later study of Rota-Baxter Lie bialgebras of
weight $\lambda$ (see Theorem 4.3).

######  Theorem 3.10.

Let $(\mathfrak{g},B)$ be a Rota-Baxter Lie algebra of weight $\lambda$. Then
$(\mathfrak{g}^{*},-\lambda\mathsf{id}-B^{*},\mathrm{ad}^{*})$ is a
representation, which is called the coadjoint representation of
$(\mathfrak{g},B)$.

Moreover, if $(\mathfrak{g},B,S)$ is a quadratic Rota-Baxter Lie algebra of
weight $\lambda$, then the linear map
$S^{\sharp}:\mathfrak{g}\to\mathfrak{g}^{*}$ defined by $\langle
S^{\sharp}(x),y\rangle=S(x,y)$ for all $x,y\in\mathfrak{g}$, is an isomorphism
from the adjoint representation $(\mathfrak{g},B,\mathrm{ad})$ to the
coadjoint representation
$(\mathfrak{g}^{*},-\lambda\mathsf{id}-B^{*},\mathrm{ad}^{*})$.

###### Proof.

For all $\xi\in\mathfrak{g}^{*}$ and $x,y\in\mathfrak{g}$, since $B$ is a
Rota-Baxter operator of weight $\lambda$ on $\mathfrak{g}$, we have

$\displaystyle\langle\mathrm{ad}^{*}_{Bx}(-\lambda\mathsf{id}-B^{*})\xi-(-\lambda\mathsf{id}-B^{*})\Big{(}\mathrm{ad}^{*}_{Bx}\xi+\mathrm{ad}^{*}_{x}(-\lambda\mathsf{id}-B^{*})\xi+\lambda\mathrm{ad}^{*}_{x}\xi\Big{)},y\rangle$
$\displaystyle=$
$\displaystyle\langle\xi,\lambda[Bx,y]_{\mathfrak{g}}+B[Bx,y]_{\mathfrak{g}}-\lambda[Bx,y]_{\mathfrak{g}}-[Bx,By]_{\mathfrak{g}}+\lambda
B[x,y]_{\mathfrak{g}}+B[x,By]_{\mathfrak{g}}\rangle$ $\displaystyle=$
$\displaystyle 0,$

which implies that
$(\mathfrak{g}^{*},-\lambda\mathsf{id}-B^{*},\mathrm{ad}^{*})$ is a
representation.

Let $(\mathfrak{g},B,S)$ be a quadratic Rota-Baxter Lie algebra of weight
$\lambda$. By (4), we have

$S^{\sharp}\circ\mathrm{ad}_{x}=\mathrm{ad}_{x}^{*}\circ S^{\sharp},$

By (5), we have

$S^{\sharp}\circ B=(-\lambda\mathsf{id}-B^{*})\circ S^{\sharp}.$

Therefore, $S^{\sharp}:\mathfrak{g}\to\mathfrak{g}^{*}$ is a homomorphism from
the adjoint representation $(\mathfrak{g},B,\mathrm{ad})$ to the coadjoint
representation $(\mathfrak{g}^{*},-\lambda\mathsf{id}-B^{*},\mathrm{ad}^{*})$.
∎

######  Remark 3.11.

We emphasize that in general $(\mathfrak{g}^{*},B^{*},\mathrm{ad}^{*})$ is not
a representation of $(\mathfrak{g},B)$.

Now we introduce the notion of matched pairs of Rota-Baxter Lie algebras of
weight $\lambda$.

######  Definition 3.12.

A matched pair of Rota-Baxter Lie algebras of weight $\lambda$ consists of a
pair of Rota-Baxter Lie algebras $((\mathfrak{g},B),(\mathfrak{h},C))$ of
weight $\lambda$, a representation
$\rho:\mathfrak{g}\to\mathfrak{gl}(\mathfrak{h})$ of the Rota-Baxter Lie
algebra $(\mathfrak{g},B)$ on $(\mathfrak{h},C)$ and a representation
$\mu:\mathfrak{h}\to\mathfrak{gl}(\mathfrak{g})$ of the Rota-Baxter Lie
algebra $(\mathfrak{h},C)$ on $(\mathfrak{g},B)$ such that
$(\mathfrak{g},\mathfrak{h};\rho,\mu)$ is a matched pair of Lie algebras.

We will denote a matched pair of Rota-Baxter Lie algebras of weight $\lambda$
by $((\mathfrak{g},B),(\mathfrak{h},C);\rho,\mu)$, or simply by
$((\mathfrak{g},B),(\mathfrak{h},C))$.

It is straightforward to obtain the following alternative characterization of
matched pairs of Rota-Baxter Lie algebras.

######  Proposition 3.13.

Let $(\mathfrak{g},B)$ and $(\mathfrak{h},C)$ be Rota-Baxter Lie algebras of
weight $\lambda$, $\rho:\mathfrak{g}\to\mathfrak{gl}(\mathfrak{h})$ a
representation of the Lie algebra $\mathfrak{g}$ on $\mathfrak{h}$ and
$\mu:\mathfrak{h}\to\mathfrak{gl}(\mathfrak{g})$ a representation of the Lie
algebra $\mathfrak{h}$ on $\mathfrak{g}$. Then
$((\mathfrak{g},B),(\mathfrak{h},C);\rho,\mu)$ is a matched pair of Rota-
Baxter Lie algebras if and only if $(\mathfrak{g},\mathfrak{h};\rho,\mu)$ is a
matched pair of Lie algebras and the following equalities hold:

(21) $\displaystyle C(\rho(Bx)\xi+\rho(x)(C\xi)+\lambda\rho(x)\xi)$
$\displaystyle=$ $\displaystyle\rho(Bx)(C\xi);$ (22) $\displaystyle
B(\mu(C\xi)x+\mu(\xi)(Bx)+\lambda\mu(\xi)x)$ $\displaystyle=$
$\displaystyle\mu(C\xi)(Bx),$

for all $x\in\mathfrak{g}$ and $\xi\in\mathfrak{h}$.

On the double of a matched pair of Lie algebras, there is a Lie algebra
structure. There is also a Rota-Baxter Lie algebra structure on the double of
a matched pair of Rota-Baxter Lie algebras as expected.

######  Proposition 3.14.

Let $(\mathfrak{g},B)$ and $(\mathfrak{h},C)$ be Rota-Baxter Lie algebras of
weight $\lambda$, $\rho:\mathfrak{g}\to\mathfrak{gl}(\mathfrak{h})$ and
$\mu:\mathfrak{h}\to\mathfrak{gl}(\mathfrak{g})$ are linear maps. Then
$((\mathfrak{g},B),(\mathfrak{h},C);\rho,\mu)$ is a matched pair of Rota-
Baxter Lie algebras of weight $\lambda$ if and only if
$(\mathfrak{g},\mathfrak{h};\rho,\mu)$ is a matched pair of Lie algebras and

$B\oplus
C:\mathfrak{g}\oplus\mathfrak{h}\to\mathfrak{g}\oplus\mathfrak{h},\qquad
x+\xi\mapsto Bx+C\xi,$

is a Rota-Baxter operator of weight $\lambda$ on the Lie algebra
$\mathfrak{g}\bowtie\mathfrak{h}$.

###### Proof.

Let $((\mathfrak{g},B),(\mathfrak{h},C);\rho,\mu)$ be a matched pair of Rota-
Baxter Lie algebras of weight $\lambda$. By definition,
$(\mathfrak{g},\mathfrak{h};\rho,\mu)$ is a matched pair of Lie algebras. By
(21) and (22), we have

$[Bx,C\xi]_{\bowtie}=(B\oplus
C)\big{(}[Bx,\xi]_{\bowtie}+[x,C\xi]_{\bowtie}+\lambda[x,\xi]_{\bowtie}\big{)}.$

Therefore $B\oplus C$ is a Rota-Baxter operator of weight $\lambda$ on the Lie
algebra $\mathfrak{g}\bowtie\mathfrak{h}$.

The converse part can be proved similarly. ∎

Let $((\mathfrak{g},B),(\mathfrak{h},C);\rho,\mu)$ be a matched pair of Rota-
Baxter Lie algebras of weight $\lambda$. Then there are three descendent Lie
algebras $\mathfrak{g}_{B},\mathfrak{h}_{C}$ and
$(\mathfrak{g}\bowtie\mathfrak{h})_{B\oplus C}$ coming from the three Rota-
Baxter operators $B:\mathfrak{g}\to\mathfrak{g},C:\mathfrak{h}\to\mathfrak{h}$
and $B\oplus
C:\mathfrak{g}\oplus\mathfrak{h}\to\mathfrak{g}\oplus\mathfrak{h}$ of weight
$\lambda$, respectively. In the following theorem, their relation is clarified
and a descendent matched pair of Lie algebras is obtained.

######  Theorem 3.15.

Let $((\mathfrak{g},B),(\mathfrak{h},C);\rho,\mu)$ be a matched pair of Rota-
Baxter Lie algebras of weight $\lambda$. Then
$(\mathfrak{g}_{B},\mathfrak{h}_{C};\rho_{(B,C)},\mu_{(B,C)})$ is a matched
pair of Lie algebras, where $\rho_{(B,C)}$ and $\mu_{(B,C)}$ are given by

(23) $\displaystyle\rho_{(B,C)}(x)\xi$ $\displaystyle=$
$\displaystyle\rho(Bx)\xi+\rho(x)(C\xi)+\lambda\rho(x)\xi,$ (24)
$\displaystyle\mu_{(B,C)}(\xi)x$ $\displaystyle=$
$\displaystyle\mu(C\xi)x+\mu(\xi)(Bx)+\lambda\mu(\xi)x.$

Moreover, we have

$\mathfrak{g}_{B}\bowtie\mathfrak{h}_{C}=(\mathfrak{g}\bowtie\mathfrak{h})_{B\oplus
C}$

as Lie algebras.

The matched pair
$(\mathfrak{g}_{B},\mathfrak{h}_{C};\rho_{(B,C)},\mu_{(B,C)})$ is called the
descendent matched pair.

###### Proof.

By Proposition 3.14, there is a descendent Lie algebra structure on
$\mathfrak{g}\oplus\mathfrak{h}$, denoted by
$(\mathfrak{g}\bowtie\mathfrak{h})_{B\oplus C}$, which contains
$\mathfrak{g}_{B}$ and $\mathfrak{h}_{C}$ as Lie subalgebras. Let us examine
this Lie bracket on the crossing term. For $x\in\mathfrak{g}$ and
$\xi\in\mathfrak{h}$, we have

$\displaystyle[x,\xi]_{(\mathfrak{g}\bowtie\mathfrak{h})_{B\oplus C}}$
$\displaystyle=$
$\displaystyle[Bx,\xi]_{\bowtie}+[x,C\xi]_{\bowtie}+\lambda[x,\xi]_{\bowtie}$
$\displaystyle=$
$\displaystyle\rho(Bx)\xi+\rho(x)(C\xi)+\lambda\rho(x)\xi-\big{(}\mu(C\xi)x+\mu(\xi)(Bx)+\lambda\mu(\xi)x\big{)}$
$\displaystyle=$ $\displaystyle\rho_{(B,C)}(x)\xi-\mu_{(B,C)}(\xi)x.$

Then by Proposition 3.1,
$(\mathfrak{g}_{B},\mathfrak{h}_{C};\rho_{(B,C)},\mu_{(B,C)})$ forms a matched
pair of Lie algebras. Moreover, the induced Lie algebra on its double
$\mathfrak{g}_{B}\bowtie\mathfrak{h}_{C}$ coincides with
$(\mathfrak{g}\bowtie\mathfrak{h})_{B\oplus C}$. ∎

######  Corollary 3.16.

Let $((\mathfrak{g},B),(\mathfrak{h},C);\rho,\mu)$ be a matched pair of Rota-
Baxter Lie algebras of weight $\lambda$. Then
$((\mathfrak{g}_{B},B),(\mathfrak{h}_{C},C);\rho_{(B,C)},\mu_{(B,C)})$ is a
matched pair of Rota-Baxter Lie algebras of weight $\lambda$.

###### Proof.

Using the well-known fact that a Rota-Baxter operator of weight $\lambda$ on a
Lie algebra is also a Rota-Baxter operator on the descendent Lie algebra, it
follows that $B\oplus C$ is a Rota-Baxter operator of weight $\lambda$ on
$(\mathfrak{g}\bowtie\mathfrak{h})_{B\oplus C}$. By Theorem 3.15,
$\mathfrak{g}_{B}\bowtie\mathfrak{h}_{C}=(\mathfrak{g}\bowtie\mathfrak{h})_{B\oplus
C}$. Therefore, $B\oplus C$ is a Rota-Baxter operator of weight $\lambda$ on
$\mathfrak{g}_{B}\bowtie\mathfrak{h}_{C}$. By Proposition 3.14,
$((\mathfrak{g}_{B},B),(\mathfrak{h}_{C},C);\rho_{(B,C)},\mu_{(B,C)})$ is a
matched pair of Rota-Baxter Lie algebras of weight $\lambda$. ∎

######  Example 3.17.

Let $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},B)$ and
$(\mathfrak{h},[\cdot,\cdot]_{\mathfrak{h}},C)$ be two Rota-Baxter Lie
algebras of weight $\lambda$. Then
$((\mathfrak{g},B),(\mathfrak{h},C);\rho=0,\mu=0)$ becomes a matched pair of
Rota-Baxter Lie algebras.

######  Proposition 3.18.

Let $(\mathfrak{g},B)$ be a Rota-Baxter Lie algebra of weight $\lambda$. Then
$((\mathfrak{g}_{B},B),(\mathfrak{g},B);\rho,\mu)$ is a matched pair of Rota-
Baxter Lie algebras of weight $\lambda$, where $\rho$ and $\mu$ are given by
(14).

###### Proof.

By Proposition 3.2, $(\mathfrak{g}_{B},\mathfrak{g};\rho,\mu)$ is a matched
pair of Lie algebras. As $B$ is a Rota-Baxter operator of weight $\lambda$ on
both $\mathfrak{g}$ and $\mathfrak{g}_{B}$, it suffices to check the
Conditions (21) and (22). In fact, Condition (21) follows directly from the
fact that $B$ is a Rota-Baxter operator of weight $\lambda$ on the Lie algebra
$\mathfrak{g}$. In order to show Condition (22), we first note that

$\displaystyle B(\rho(B\xi)x+\rho(\xi)(Bx)+\lambda\rho(\xi)x)$
$\displaystyle=$ $\displaystyle
B(B[x,B\xi]_{\mathfrak{g}}-[x,B^{2}\xi]_{\mathfrak{g}}+B[Bx,\xi]_{\mathfrak{g}}-[Bx,B\xi]_{\mathfrak{g}}+\lambda
B[x,\xi]_{\mathfrak{g}}-\lambda[x,B\xi]_{\mathfrak{g}})$ $\displaystyle=$
$\displaystyle
B(-[x,B^{2}\xi]_{\mathfrak{g}}-\lambda[x,B\xi]_{\mathfrak{g}}).$

On the other hand, as $B$ is a Rota-Baxter operator on $\mathfrak{g}$, we have

$\displaystyle\rho(B\xi)(Bx)$ $\displaystyle=$ $\displaystyle
B[Bx,B\xi]_{\mathfrak{g}}-[Bx,B^{2}\xi]_{\mathfrak{g}}$ $\displaystyle=$
$\displaystyle
B([Bx,B\xi]_{\mathfrak{g}}-[Bx,B\xi]_{\mathfrak{g}}-[x,B^{2}\xi]_{\mathfrak{g}}-\lambda[x,B\xi]_{\mathfrak{g}})$
$\displaystyle=$ $\displaystyle
B(-[x,B^{2}\xi]_{\mathfrak{g}}-\lambda[x,B\xi]_{\mathfrak{g}}).$

So Condition (22) is obtained. Therefore,
$((\mathfrak{g}_{B},B),(\mathfrak{g},B),\rho,\mu)$ is a matched pair of Rota-
Baxter Lie algebras of weight $\lambda$. ∎

##  4\. Rota-Baxter Lie bialgebras

In this section, we introduce the notion of Rota-Baxter Lie bialgebras of
weight $\lambda$, and show that Rota-Baxter Lie bialgebras, certain matched
pairs of Rota-Baxter Lie algebras and Manin triples of Rota-Baxter Lie
algebras are equivalent.

###  4.1. Rota-Baxter Lie bialgebras and matched pairs of Rota-Baxter Lie
algebras

We introduce the following definition of Rota-Baxter operators on a Lie
bialgebra and show that Rota-Baxter Lie bialgebras and certain matched pairs
of Rota-Baxter Lie algebras are equivalent. In particular, we show that the
Drinfeld double of a Rota-Baxter Lie bialgebra is still a Rota-Baxter Lie
bialgebra, and factorizable Lie bialgebras also give rise to Rota-Baxter Lie
bialgebras.

######  Definition 4.1.

A Rota-Baxter operator of weight $\lambda$ on a Lie bialgebra
$(\mathfrak{g},\mathfrak{g}^{*})$ is a linear map
$B:\mathfrak{g}\to\mathfrak{g}$ such that

  * (i)

$B$ is a Rota-Baxter operator of weight $\lambda$ on $\mathfrak{g}$;

  * (ii)

$\widetilde{B}^{*}:=-\lambda\mathsf{id}-B^{*}$ is a Rota-Baxter operator of
weight $\lambda$ on $\mathfrak{g}^{*}$.

A Lie bialgebra with a Rota-Baxter operator of weight $\lambda$ is called a
Rota-Baxter Lie bialgebra of weight $\lambda$.

We denote a Rota-Baxter Lie bialgebra of weight $\lambda$ by
$(\mathfrak{g},\mathfrak{g}^{*},B)$.

As a well-known fact of Rota-Baxter operators,
$\widetilde{B}^{*}=-\lambda\mathsf{id}-B^{*}:\mathfrak{g}^{*}\to\mathfrak{g}^{*}$
is a Rota-Baxter operator of weight $\lambda$ on $\mathfrak{g}^{*}$ if and
only if $B^{*}:\mathfrak{g}^{*}\to\mathfrak{g}^{*}$ is a Rota-Baxter operator
of weight $\lambda$. The descendent Lie brackets of $\widetilde{B}^{*}$ and
$B^{*}$ on $\mathfrak{g}^{*}$ are related by

$\displaystyle[\xi,\eta]_{\widetilde{B}^{*}}=-[B^{*}\xi,\eta]_{\mathfrak{g}^{*}}-[\xi,B^{*}\eta]_{\mathfrak{g}^{*}}-\lambda[\xi,\eta]_{\mathfrak{g}^{*}}=-[\xi,\eta]_{B^{*}}.$

The reason why we adopt $\widetilde{B}^{*}$ instead of $B^{*}$ will become
clear from the following Theorem 4.3, Remark 4.4 and Proposition 4.6.

The following result is straightforward.

######  Proposition 4.2.

If $B$ is a Rota-Baxter operator of weight $\lambda$ on the Lie bialgebra
$(\mathfrak{g},\mathfrak{g}^{*})$. Then $\widetilde{B}^{*}$ is a Rota-Baxter
operator of weight $\lambda$ on the Lie bialgebra
$(\mathfrak{g}^{*},\mathfrak{g})$.

A Lie bialgebra $(\mathfrak{g},\mathfrak{g}^{*})$ is naturally a matched pair
of Lie algebras. In this case, the representation of $\mathfrak{g}$ on
$\mathfrak{g}^{*}$ is given by the coadjoint representation $\mathrm{ad}^{*}$
of the Lie algebra $\mathfrak{g}$ on $\mathfrak{g}^{*}$ and the representation
of $\mathfrak{g}^{*}$ on $\mathfrak{g}$ is given by the coadjoint
representation $\mathfrak{a}\mathfrak{d}^{*}$ of the Lie algebra
$\mathfrak{g}^{*}$ on $\mathfrak{g}$.

######  Theorem 4.3.

With above notations, $(\mathfrak{g},\mathfrak{g}^{*},B)$ is a Rota-Baxter Lie
bialgebra of weight $\lambda$ if and only if
$((\mathfrak{g},B),(\mathfrak{g}^{*},\widetilde{B}^{*});\mathrm{ad}^{*},\mathfrak{a}\mathfrak{d}^{*})$
is a matched pair of Rota-Baxter Lie algebras of the same weight.

###### Proof.

By Theorem 3.10, $(\mathfrak{g}^{*},\widetilde{B}^{*},\mathrm{ad}^{*})$ is a
representation of the Rota-Baxter Lie algebra $(\mathfrak{g},B)$, and
$(\mathfrak{g},B,\mathfrak{a}\mathfrak{d}^{*})$ is a representation of the
Rota-Baxter Lie algebra $(\mathfrak{g}^{*},\widetilde{B}^{*})$. Moreover, it
is well-known that $(\mathfrak{g},\mathfrak{g}^{*})$ is a Lie bialgebra if and
only if
$(\mathfrak{g},\mathfrak{g}^{*};\mathrm{ad}^{*},\mathfrak{a}\mathfrak{d}^{*})$
is a matched pair of Lie algebras. Therefore,
$(\mathfrak{g},\mathfrak{g}^{*},B)$ is a Rota-Baxter Lie bialgebra of weight
$\lambda$ if and only if
$((\mathfrak{g},B),(\mathfrak{g}^{*},\widetilde{B}^{*});\mathrm{ad}^{*},\mathfrak{a}\mathfrak{d}^{*})$
is a matched pair of Rota-Baxter Lie algebras of the same weight. ∎

######  Remark 4.4.

Let $(\mathfrak{g},\mathfrak{g}^{*})$ be a Lie bialgebra,
$B:\mathfrak{g}\to\mathfrak{g}$ and
$B^{*}:\mathfrak{g}^{*}\to\mathfrak{g}^{*}$ Rota-Baxter operators of weight
$\lambda$. Then in general
$((\mathfrak{g},B),(\mathfrak{g}^{*},B^{*});\mathrm{ad}^{*},\mathfrak{a}\mathfrak{d}^{*})$
is not a matched pair of Rota-Baxter Lie algebras of weight $\lambda$. In
fact, let $C=B^{*}$ in (21) and take pairing with $y\in\mathfrak{g}$. The left
hand side of (21) equals

$\langle
B^{*}(\mathrm{ad}_{Bx}^{*}\xi+\mathrm{ad}_{x}^{*}B^{*}\xi+\lambda\mathrm{ad}_{x}^{*}\xi),y\rangle=-\langle\xi,[Bx,By]_{\mathfrak{g}}+B[x,By]_{\mathfrak{g}}+\lambda[x,By]_{\mathfrak{g}}\rangle,$

and the right hand side of (21) amounts to

$\langle\mathrm{ad}_{Bx}^{*}B^{*}\xi,y\rangle=-\langle\xi,B[Bx,y]_{\mathfrak{g}}\rangle.$

So in general, the relation (21) does not hold, and
$((\mathfrak{g},B),(\mathfrak{g}^{*},B^{*});\mathrm{ad}^{*},\mathfrak{a}\mathfrak{d}^{*})$
is not a matched pair of Rota-Baxter Lie algebras of weight $\lambda$.

######  Remark 4.5.

Let $(\mathfrak{g},\mathfrak{g}^{*},B)$ be a Rota-Baxter Lie bialgebra of
weight $\lambda$. By Theorem 4.3 and Theorem 3.15, it induces a descendent
matched pair
$(\mathfrak{g}_{B},\mathfrak{g}^{*}_{\widetilde{B}^{*}};\rho_{(B,\widetilde{B}^{*})},\mu_{(B,\widetilde{B}^{*})})$
of Lie algebras. By (23) and (24), the representation
$\rho_{(B,\widetilde{B}^{*})}:\mathfrak{g}_{B}\to\mathfrak{gl}(\mathfrak{g}^{*})$
and
$\mu_{(B,\widetilde{B}^{*})}:\mathfrak{g}^{*}_{\widetilde{B}^{*}}\to\mathfrak{gl}(\mathfrak{g})$
are given by

$\rho_{(B,\widetilde{B}^{*})}(x)(\xi)=\mathrm{ad}_{Bx}^{*}\xi-\mathrm{ad}_{x}^{*}B^{*}\xi,\quad\mu_{(B,\widetilde{B}^{*})}(\xi)(x)=\mathfrak{a}\mathfrak{d}_{\xi}^{*}Bx-\mathfrak{a}\mathfrak{d}_{B^{*}\xi}^{*}x.$

It is straightforward to check that $\rho_{(B,\widetilde{B}^{*})}$ and
$\mu_{(B,\widetilde{B}^{*})}$ are not the coadjoint representations of
$\mathfrak{g}_{B}$ and $\mathfrak{g}^{*}_{\widetilde{B}^{*}}$. Therefore, the
matched pair
$(\mathfrak{g}_{B},\mathfrak{g}^{*}_{\widetilde{B}^{*}};\rho_{(B,\widetilde{B}^{*})},\mu_{(B,\widetilde{B}^{*})})$
does not come from a Lie bialgebra.

The following result tells us that the Drinfeld double of a Rota-Baxter Lie
bialgebra is still a Rota-Baxter Lie bialgebra of the same weight.

######  Proposition 4.6.

Let $(\mathfrak{g},\mathfrak{g}^{*},B)$ be a Rota-Baxter Lie bialgebra of
weight $\lambda$. Then $(\mathfrak{d},\mathfrak{d}^{*},\mathcal{B})$ is a
Rota-Baxter Lie bialgebra of weight $\lambda$, where
$(\mathfrak{d},\mathfrak{d}^{*})$ is the Lie bialgebra given by Example 2.9,
and
$\mathcal{B}:\mathfrak{g}\oplus\mathfrak{g}^{*}\to\mathfrak{g}\oplus\mathfrak{g}^{*}$
is the linear map defined by

(25) $\mathcal{B}(x+\xi)=Bx-\lambda\xi-B^{*}\xi,\quad
x\in\mathfrak{g},~{}\xi\in\mathfrak{g}^{*}.$

###### Proof.

First by Theorem 4.3,
$((\mathfrak{g},B),(\mathfrak{g}^{*},\widetilde{B}^{*});\mathrm{ad}^{*},\mathfrak{a}\mathfrak{d}^{*})$
is a matched pair of Rota-Baxter Lie algebras. By Proposition 3.14, the linear
map $\mathcal{B}$ is a Rota-Baxter operator of weight $\lambda$ on the
Drinfeld double $\mathfrak{d}:=\mathfrak{g}\bowtie\mathfrak{g}^{*}$.

Since $\widetilde{B}^{*}=-\lambda\mathsf{id}-B^{*}$ is a Rota-Baxter operator
of weight $\lambda$ on $\mathfrak{g}^{*}$, it is also a Rota-Baxter operator
on the Lie algebra
$\overline{\mathfrak{g}^{*}}=(\mathfrak{g}^{*},-[\cdot,\cdot]_{\mathfrak{g}^{*}})$.
Since $\mathfrak{d}^{*}$ is a direct sum Lie algebra, it is obvious that
$-\lambda\mathsf{id}-\mathcal{B}^{*}=\mathcal{B}$ is a Rota-Baxter operator of
weight $\lambda$ on the dual Lie algebra
$\mathfrak{d}^{*}=\mathfrak{g}\oplus\overline{\mathfrak{g}^{*}}$.
Consequently, $(\mathfrak{d},\mathfrak{d}^{*},\mathcal{B})$ is a Rota-Baxter
Lie bialgebra of weight $\lambda$. ∎

Factorizable Lie bialgebras provide a class of examples of Rota-Baxter Lie
bialgebras.

######  Theorem 4.7.

Let $(\mathfrak{g},\mathfrak{g}^{*}_{r})$ be a factorizable Lie bialgebra with
$I=r_{+}-r_{-}$. Then $(\mathfrak{g},\mathfrak{g}_{r}^{*},B)$ is a Rota-Baxter
Lie bialgebra of weight $\lambda$, where $B:=\lambda r_{-}\circ I^{-1}$ is
given by (6).

###### Proof.

It is obvious that $\widetilde{B}^{*}=-\lambda\mathsf{id}-B^{*}=\lambda
I^{-1}\circ r_{-}$. Moreover, by the facts that
$\frac{1}{\lambda}I:\mathfrak{g}_{r}^{*}\to(\mathfrak{g},[\cdot,\cdot]_{B})$
is an isomorphism of Lie algebras (see Corollary 2.8) and
$r_{-}:\mathfrak{g}^{*}_{r}\to\mathfrak{g}$ is a Lie algebra homomorphism, we
have

$\displaystyle\lambda I^{-1}r_{-}([\lambda
I^{-1}r_{-}\xi,\eta]_{r}+[\xi,\lambda
I^{-1}r_{-}\eta]_{r}+\lambda[\xi,\eta]_{r})$ $\displaystyle=$
$\displaystyle\lambda I^{-1}([\lambda
r_{-}I^{-1}r_{-}\xi,r_{-}\eta]_{\mathfrak{g}}+[r_{-}\xi,\lambda
r_{-}I^{-1}r_{-}\eta]_{\mathfrak{g}}+\lambda[r_{-}\xi,r_{-}\eta]_{\mathfrak{g}})$
$\displaystyle=$ $\displaystyle\lambda I^{-1}([r_{-}\xi,r_{-}\eta]_{B})$
$\displaystyle=$ $\displaystyle[\lambda I^{-1}r_{-}\xi,\lambda
I^{-1}r_{-}\eta]_{r},$

which implies that $\widetilde{B}^{*}$ is a Rota-Baxter operator of weight
$\lambda$ on $\mathfrak{g}^{*}_{r}$. Therefore,
$(\mathfrak{g},\mathfrak{g}_{r}^{*},B)$ is a Rota-Baxter Lie bialgebra of
weight $\lambda$. ∎

######  Corollary 4.8.

Let $(\mathfrak{g},\mathfrak{g}^{*}_{r})$ be a factorizable Lie bialgebra with
$I=r_{+}-r_{-}$. Then have the following commutative diagram of Lie algebra
homomorphisms:

$\textstyle{\cdots\mathfrak{g}^{*}_{\widetilde{B^{*}}^{k}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\frac{1}{\lambda}I}$$\scriptstyle{\cong}$$\scriptstyle{-\lambda\mathsf{id}-B^{*}}$$\scriptstyle{r_{-}}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathfrak{g}^{*}_{\widetilde{B^{*}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\frac{1}{\lambda}I}$$\scriptstyle{\cong}$$\scriptstyle{-\lambda\mathsf{id}-B^{*}}$$\scriptstyle{r_{-}}$$\textstyle{\mathfrak{g}_{r}^{*}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\frac{1}{\lambda}I}$$\scriptstyle{\cong}$$\scriptstyle{-\lambda\mathsf{id}-B^{*}}$$\scriptstyle{r_{-}}$$\textstyle{\mathfrak{g}_{I}^{*}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\frac{1}{\lambda}I}$$\scriptstyle{\cong}$$\textstyle{\cdots\mathfrak{g}_{B^{k+1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{B}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathfrak{g}_{B^{2}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{B}$$\textstyle{\mathfrak{g}_{B}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{B}$$\textstyle{\mathfrak{g},}$

where $B=\lambda r_{-}\circ I^{-1}$ and $\widetilde{B^{*}}=\lambda I^{-1}\circ
r_{-}$ and $\mathfrak{g}_{B^{k}}$ is the descendent Lie algebra of the Rota-
Baxter operator $B$ on $\mathfrak{g}_{B^{k-1}}$.

###### Proof.

We prove that
$\frac{1}{\lambda}I:\mathfrak{g}^{*}_{\widetilde{B^{*}}^{k}}\to\mathfrak{g}_{B^{k+1}}$
is a Lie algebra isomorphism by induction on $k$ $(k\geq 0)$. First observe
that $\frac{1}{\lambda}I:\mathfrak{g}^{*}_{r}\to\mathfrak{g}_{B}$ is a Lie
algebra isomorphism, which is the case of $k=0$. Assuming the claim holds for
$k-1$, then we have

$\displaystyle\frac{1}{\lambda}I[\xi,\eta]_{\widetilde{B^{*}}^{k}}$
$\displaystyle=$ $\displaystyle\frac{1}{\lambda}I\big{(}[\lambda
I^{-1}r_{-}\xi,\eta]_{\widetilde{B^{*}}^{k-1}}+[\xi,\lambda
I^{-1}r_{-}\eta]_{\widetilde{B^{*}}^{k-1}}+\lambda[\xi,\eta]_{\widetilde{B^{*}}^{k-1}}\big{)}$
$\displaystyle=$
$\displaystyle[r_{-}\xi,\frac{1}{\lambda}I\eta]_{B^{k}}+[\frac{1}{\lambda}I\xi,r_{-}\eta]_{B^{k}}+\lambda[\frac{1}{\lambda}I\xi,\frac{1}{\lambda}I\eta]_{B^{k}}$
$\displaystyle=$
$\displaystyle[\frac{1}{\lambda}I\xi,\frac{1}{\lambda}I\eta]_{B^{k+1}},$

which implies that it holds for $k$. All the other facts are obvious. ∎

######  Example 4.9.

Let $(\mathfrak{g},\mathfrak{g}^{*})$ be a Lie bialgebra. Then
$(\mathfrak{g},\mathfrak{g}^{*},B)$ is a Rota-Baxter Lie bialgebra of weight
$\lambda$, where the linear map $B:\mathfrak{g}\to\mathfrak{g}$ is defined by
$B(x)=-\lambda x$. Note that $-\lambda\mathsf{id}-B^{*}=0$.

######  Example 4.10.

Consider the Lie bialgebra $(\mathfrak{d},\mathfrak{d}^{*})$ given in Example
2.9, where $\mathfrak{d}=\mathfrak{g}\bowtie\mathfrak{g}^{*}$. Then the linear
map

$B:\mathfrak{g}\bowtie\mathfrak{g}^{*}\mapsto\mathfrak{g}\bowtie\mathfrak{g}^{*},\qquad
x+\xi\mapsto-\lambda\xi,$

is a Rota-Baxter operator of weight $\lambda$ on the Lie bialgebra
$(\mathfrak{d},\mathfrak{d}^{*})$. In fact, it is obvious that $B$ is a Rota-
Baxter operator of weight $\lambda$ on the Lie algebra $\mathfrak{d}$. On the
other hand, we have

$(-\lambda\mathsf{id}-B^{*})(x+\xi)=-\lambda\xi,$

which implies that $-\lambda\mathsf{id}-B^{*}$ is also a Rota-Baxter operator
of weight $\lambda$ on the Lie algebra $\mathfrak{d}^{*}$. Therefore,
$(\mathfrak{d},\mathfrak{d}^{*},B)$ is a Rota-Baxter Lie bialgebra of weight
$\lambda$.

###  4.2. Rota-Baxter Lie bialgebras and Manin triples of Rota-Baxter Lie
algebras

A well-known result regrading Lie bialgebras is that there is a one-one
correspondence between Lie bialgebras and Manin triples. A Manin triple of Lie
algebras is a triple $((\mathfrak{d},S),\mathfrak{g},\mathfrak{h})$, where
$(\mathfrak{d},S)$ is a quadratic Lie algebra, $\mathfrak{g}$ and
$\mathfrak{h}$ are Lie algebras such that

  * (i)

$\mathfrak{g}$ and $\mathfrak{h}$ are Lie subalgebras;

  * (ii)

$\mathfrak{d}=\mathfrak{g}\oplus\mathfrak{h}$ as vector spaces;

  * (iii)

both $\mathfrak{g}$ and $\mathfrak{h}$ are isotropic with respect to the
nondegenerate invariant symmetric bilinear form $S$.

Given a Lie bialgebra $(\mathfrak{g},\mathfrak{g}^{*})$, the triple
$((\mathfrak{g}\bowtie\mathfrak{g}^{*},S),\mathfrak{g},\mathfrak{g}^{*})$ is a
Manin triple, where $S$ is given by

(26) $S(x+\xi,y+\eta)=\xi(y)+\eta(x).$

Conversely, given a Manin triple
$((\mathfrak{d},S),\mathfrak{g},\mathfrak{h})$, identifying $\mathfrak{h}$
with $\mathfrak{g}^{*}$ by using the nondegenerate invariant symmetric
bilinear form $S$, we obtain a Lie bialgebra
$(\mathfrak{g},\mathfrak{g}^{*})$.

Now we introduce the notion of Manin triples of Rota-Baxter Lie algebras of
weight $\lambda$ using quadratic Rota-Baxter Lie algebras of weight $\lambda$
given in Definition 2.4.

######  Definition 4.11.

A Manin triple of Rota-Baxter Lie algebras of weight $\lambda$ consists of a
triple $((\mathcal{G},\mathcal{B},S),(\mathfrak{g},B),(\mathfrak{h},C))$,
where $(\mathcal{G},\mathcal{B},S)$ is a quadratic Rota-Baxter Lie algebra of
weight $\lambda$, $(\mathfrak{g},B)$ and $(\mathfrak{h},C)$ are Rota-Baxter
Lie algebras of weight $\lambda$ such that

  * (i)

$(\mathfrak{g},B)$ and $(\mathfrak{h},C)$ are Rota-Baxter Lie subalgebras,
i.e. $\mathfrak{g}$ and $\mathfrak{h}$ are Lie subalgebras of $\mathcal{G}$
and $\mathcal{B}|_{\mathfrak{g}}=B,~{}\mathcal{B}|_{\mathfrak{h}}=C$;

  * (ii)

$\mathcal{G}=\mathfrak{g}\oplus\mathfrak{h}$ as vector spaces;

  * (iii)

both $\mathfrak{g}$ and $\mathfrak{h}$ are isotropic with respect to the
nondegenerate symmetric invariant bilinear form $S$.

Similar to the classical case, we have the following result.

######  Theorem 4.12.

There is a one-one correspondence between Manin triples of Rota-Baxter Lie
algebras of weight $\lambda$ and Rota-Baxter Lie bialgebras of the same
weight.

###### Proof.

Let $(\mathfrak{g},\mathfrak{g}^{*},B)$ be a Rota-Baxter Lie bialgebra. Then
$(\mathfrak{g},B)$ and $(\mathfrak{g}^{*},\widetilde{B}^{*})$ are Rota-Baxter
Lie algebras of weight $\lambda$. By Proposition 4.6,
$(\mathfrak{g}\bowtie\mathfrak{g}^{*},\mathcal{B})$ is a Rota-Baxter Lie
algebra of weight $\lambda$, where $\mathcal{B}$ is defined by (25). Moreover,
it is straightforward to deduce that
$(\mathfrak{g}\bowtie\mathfrak{g}^{*},\mathcal{B},S)$ is a quadratic Rota-
Baxter Lie algebra of weight $\lambda$, where $S$ is given by (26).
Consequently,
$((\mathfrak{g}\bowtie\mathfrak{g}^{*},\mathcal{B},S),(\mathfrak{g},B),(\mathfrak{g}^{*},-\lambda\mathsf{id}-B^{*}))$
is a Manin triple of Rota-Baxter Lie algebras of weight $\lambda$.

Conversely, let
$((\mathcal{G},\mathcal{B},S),(\mathfrak{g},B),(\mathfrak{h},C))$ be a Manin
triple of Rota-Baxter Lie algebras of weight $\lambda$. Then similar to the
classical argument, first we can identify $\mathfrak{h}$ with
$\mathfrak{g}^{*}$ by the nondegenerate bilinear form $S$, and we obtain a Lie
bialgebra $(\mathfrak{g},\mathfrak{g}^{*})$. Then using the invariant
condition (5), we identify $C$ with $-\lambda\mathsf{id}-B^{*}$. Consequently,
both $(\mathfrak{g},B)$ and $(\mathfrak{g}^{*},-\lambda\mathsf{id}-B^{*})$ are
Rota-Baxter Lie algebras of weight $\lambda$, i.e.
$(\mathfrak{g},\mathfrak{g}^{*},B)$ is a Rota-Baxter Lie bialgebra. ∎

For a Manin triple $((\mathfrak{d},S),\mathfrak{g},\mathfrak{h})$ of Lie
algebras, define $\mathcal{B}:\mathfrak{d}\to\mathfrak{d}$ by
$\mathcal{B}(x+\xi)=-\lambda\xi$, where
$x\in\mathfrak{g},~{}\xi\in\mathfrak{h}$. Then $\mathcal{B}$ is a Rota-Baxter
operator of weight $\lambda$ on $\mathfrak{d}$. Furthermore, it is direct to
show that
$\big{(}(\mathfrak{d},\mathcal{B},S),(\mathfrak{g},0),(\mathfrak{h},-\lambda\mathsf{id})\big{)}$
is a Manin triple of Rota-Baxter Lie algebras of weight $\lambda$. Two
explicit examples coming from this construction are given as follows.

######  Example 4.13.

Let $sl(n,\mathbb{C})$ be the Lie algebra of $n\times n$ traceless complex
matrices. Consider its Iwasawa decomposition

$sl(n,\mathbb{C})=su(n)\oplus sb(n,\mathbb{C}),$

where $su(n):=\\{X\in sl(n,\mathbb{C});X+\bar{X}^{T}=0\\}$ and
$sb(n,\mathbb{C})$ is the Lie algebra of all $n\times n$ traceless upper
triangular complex matrices with real diagonal entries. With the scalar
product $S(X,Y)=\mathrm{Im}(tr(XY))$ on the Lie algebra $sl(n,\mathbb{C})$,
i.e. the imaginary part of the trace of $XY$, the triple
$((sl(n,\mathbb{C}),S),su(n),sb(n,\mathbb{C}))$ forms a Manin triple ([28]).

The linear map

$\mathcal{B}:sl(n,\mathbb{C})\to sl(n,\mathbb{C}),\qquad
X=(x_{ij})\mapsto-\lambda A=-\lambda(a_{ij}),$

where
$a_{ii}=\frac{x_{ii}+\overline{x_{ii}}}{2},a_{ij}=x_{ij}+\overline{x_{ji}},i<j$
and $a_{ij}=0,i>j$, is a Rota-Baxter operator of weight $\lambda$ on
$sl(n,\mathbb{C})$. In fact, we have

$\mathcal{B}(X)=-\lambda A,$

if $X\in sl(n,\mathbb{C})$ has the decomposition $X=C+A$ for $C\in su(n)$ and
$A\in sb(n,\mathbb{C})$. In fact, $C=(c_{ij})$ is defined by
$c_{ii}=\frac{x_{ii}-\overline{x_{ii}}}{2},c_{ij}=-\overline{x_{ji}},i<j$ and
$c_{ij}=x_{ij},i>j$. It is obvious $\mathcal{B}|_{su(n)}=0$ and
$\mathcal{B}|_{sb(n,\mathbb{C})}=-\lambda\mathsf{id}$ are Rota-Baxter
operators of weight $\lambda$. Therefore,
$((sl(n,\mathbb{C}),\mathcal{B},S),(su(n),0),(sb(n,\mathbb{C}),-\lambda\mathsf{id}))$
is a Manin triple of Rota-Baxter Lie algebras of weight $\lambda$. This
example can be generalized to the Iwasawa decomposition of an arbitrary semi-
simple Lie algebra.

######  Example 4.14.

Let $\mathfrak{g}=sl(n,\mathbb{C})$ and consider the direct sum Lie algebra
$\mathfrak{g}\oplus\mathfrak{g}=sl(n,\mathbb{C})\oplus sl(n,\mathbb{C})$.
Define

$\displaystyle\mathfrak{g}_{diag}$ $\displaystyle=$
$\displaystyle\\{(X,X);X\in sl(n,\mathbb{C})\\};$
$\displaystyle\mathfrak{g}^{*}_{st}$ $\displaystyle=$
$\displaystyle\\{(Y+X_{+},-Y+X_{-});Y\in\mathfrak{h},X_{+}\in n_{+},X_{-}\in
n_{-}\\},$

where $\mathfrak{h},n_{+},n_{-}$ are the Lie algebras of diagonal, strictly
upper and strictly lower triangular matrices in $sl(n,\mathbb{C})$. With
respect to the scalar product

$\langle(X_{1},Y_{1}),(X_{2},Y_{2})\rangle_{\mathfrak{g}\oplus\mathfrak{g}}=\mathrm{Im}(tr(X_{1}X_{2}))-\mathrm{Im}(tr(Y_{1}Y_{2})),$

the triple
$(\mathfrak{g}\oplus\mathfrak{g},\mathfrak{g}_{diag},\mathfrak{g}^{*}_{st})$
consists a Manin triple. Moreover, the map

$\displaystyle\mathcal{B}:sl(n,\mathbb{C})\oplus sl(n,\mathbb{C})$
$\displaystyle\to$ $\displaystyle sl(n,\mathbb{C})\oplus sl(n,\mathbb{C}),$
$\displaystyle(X,Y)$ $\displaystyle\mapsto$
$\displaystyle-\lambda\big{(}\frac{1}{2}(X-Y)_{0}+(X-Y)_{+},-\frac{1}{2}(X-Y)_{0}-(X-Y)_{-}\big{)},$

is a Rota-Baxter operator of weight $\lambda$ on this Manin triple, where
$(X-Y)_{0},(X-Y)_{+}$ and $(X-Y)_{-}$ are the diagonal part, strictly upper
triangular part and strictly lower triangular part of the matrix $X-Y$,
respectively. In fact, taking into consideration of the decomposition
$\mathfrak{g}\oplus\mathfrak{g}=\mathfrak{g}_{diag}\oplus\mathfrak{g}_{st}^{*}$,
the linear map $\mathcal{B}$ is actually the projection to the
$\mathfrak{g}_{st}^{*}$\- component multiplied by $-\lambda$.

##  5\. Rota-Baxter Lie groups and factorizable Poisson Lie groups

In this section, we show that factorizable Poisson Lie groups are in one-to-
one correspondence with quadratic Rota-Baxter Lie groups. Moreover, a Rota-
Baxter Lie group gives rise to a matched pair of Lie groups.

###  5.1. Factorizable Poisson Lie groups and quadratic Rota-Baxter Lie groups

The notion of Rota-Baxter Lie groups was introduced in [22], whose
differentiations are Rota-Baxter Lie algebras of weight 1.

######  Definition 5.1.

([22]) A Rota-Baxter operator on a Lie group $G$ is a smooth map
$\mathfrak{B}:G\longrightarrow G$ such that

(27)
$\displaystyle\mathfrak{B}(g_{1})\mathfrak{B}(g_{2})=\mathfrak{B}(g_{1}\mathrm{Ad}_{\mathfrak{B}(g_{1})}g_{2}),\quad\forall
g_{1},g_{2}\in G.$

The pair $(G,\mathfrak{B})$ is called a Rota-Baxter Lie group.

Let $(G,\mathfrak{B})$ be a Rota-Baxter Lie group. Then there is a new group
structure $\star$ on $G$ given by

$g\star h=g\mathrm{Ad}_{\mathfrak{B}(g)}h.$

The new Lie group $(G,\star)$ is called the descendent Lie group, and denoted
by $G_{\mathfrak{B}}$. Moreover, $\mathfrak{B}$ is a Lie group homomorphism:

(28) $\mathfrak{B}(g\star h)=\mathfrak{B}(g)\mathfrak{B}(h).$

######  Theorem 5.2.

([22]) Let $(G,\mathfrak{B})$ be a Rota-Baxter Lie group, and $\mathfrak{g}$
the Lie algebra of the Lie group $G$. Then
$(\mathfrak{g},B:=\mathfrak{B}_{*e})$ is a Rota-Baxter Lie algebra of weight
$1$, where $e$ is the identity of $G.$

Recall from Theorem 2.10 that given a quadratic Rota-Baxter Lie algebra
$(\mathfrak{g},B,S)$, one has a linear isomorphism
$\mathcal{I}_{S}:\mathfrak{g}^{*}\to\mathfrak{g}$ defined by
$\langle\mathcal{I}_{S}^{-1}(x),y\rangle=S(x,y)$. Using $\mathcal{I}_{S}$, we
pull-back the Lie algebra structure $[\cdot,\cdot]_{B}$ on $\mathfrak{g}$ to
define a Lie algebra structure on $\mathfrak{g}^{*}$. Denote it by
$\mathfrak{g}^{*}_{S}$. Then
$\mathcal{I}_{S}:\mathfrak{g}^{*}_{S}\to\mathfrak{g}_{B}$ becomes a Lie
algebra isomorphism. Let $G^{*}$ be the simply connected Lie group whose Lie
algebra is $\mathfrak{g}^{*}_{S}$. Now we introduce the notion of quadratic
Rota-Baxter Lie groups, which will be used to characterize factorizable
Poisson Lie groups.

######  Definition 5.3.

A quadratic Rota-Baxter Lie group is a triple $(G,\mathfrak{B},S)$, where
$(G,\mathfrak{B})$ is a Rota-Baxter Lie group,
$S\in\otimes^{2}\mathfrak{g}^{*}$ such that
$(\mathfrak{g},B:=\mathfrak{B}_{*e},S)$ is a quadratic Rota-Baxter Lie algebra
and the linear isomorphism
$\mathcal{I}_{S}:\mathfrak{g}^{*}_{S}\to\mathfrak{g}$ can be lifted to a
global diffeomorphism $J_{S}:G^{*}\to G$.

A Poisson Lie group $G$ is called quasitriangular ([39]) if its corresponding
Lie bialgebra $(\mathfrak{g},\mathfrak{g}^{*})$ is quasitriangular. Let
$R_{+},R_{-}:G^{*}\to G$ be the lifted Lie group homomorphisms of the Lie
algebra homomorphisms $r_{+},r_{-}:\mathfrak{g}^{*}\to\mathfrak{g}$, where
$G^{*}$ is the simply connected dual Poisson Lie group of $G$. For simplicity,
we always suppose $G$ is connected and simply connected.

Since $r_{+}^{*}=-r_{-}$ is a Lie algebra anti-homomorphism, the map $R_{+}$
is an anti-Poisson map. In summary, both $R_{+}$ and $R_{-}$ are Lie group
homomorphisms and anti-Poisson maps.

Define

(29) $\displaystyle J:G^{*}\to G,\quad u\mapsto R_{+}(u)R_{-}(u)^{-1},\quad
u\in G^{*}.$

A quasitriangular Poisson Lie group $G$ is called factorizable if $J$ is a
global diffeomorphism. Denote a factorizable Poisson Lie group by $(G,J)$.

######  Theorem 5.4.

Let $(G,J)$ be a factorizable Poisson Lie group,
$(\mathfrak{g},\mathfrak{g}^{*}_{r})$ the corresponding factorizable Lie
bialgebra with $I=r_{+}-r_{-}$. Then $(G,\mathfrak{B}:=R_{-}\circ
J^{-1},S_{I})$ is a quadratic Rota-Baxter Lie group, where
$S_{I}\in\otimes^{2}\mathfrak{g}^{*}$ is defined by $S_{I}(x,y):=\langle
I^{-1}x,y\rangle$ for $x,y\in\mathfrak{g}$.

Moreover, $J:G^{*}\to G_{\mathfrak{B}}$ is a Lie group isomorphism.

###### Proof.

Since $R_{+},R_{-}:G^{*}\to G$ are Lie group homomorphisms, we have

$\displaystyle J(J^{-1}gJ^{-1}h)$ $\displaystyle=$ $\displaystyle
R_{+}(J^{-1}gJ^{-1}h)R_{-}(J^{-1}gJ^{-1}h)^{-1}$ $\displaystyle=$
$\displaystyle
R_{+}(J^{-1}g)R_{+}(J^{-1}h)R_{-}(J^{-1}h)^{-1}R_{-}(J^{-1}g)^{-1}$
$\displaystyle=$ $\displaystyle
J(J^{-1}g)R_{-}(J^{-1}g)J(J^{-1}h)R_{-}(J^{-1}h)R_{-}(J^{-1}h)^{-1}R_{-}(J^{-1}g)^{-1}$
$\displaystyle=$ $\displaystyle gR_{-}(J^{-1}g)hR_{-}(J^{-1}g)^{-1}.$

Then we have

$\displaystyle\mathfrak{B}(g\mathfrak{B}(g)h\mathfrak{B}(g)^{-1})$
$\displaystyle=$ $\displaystyle
R_{-}J^{-1}(gR_{-}(J^{-1}g)hR_{-}(J^{-1}g)^{-1})$ $\displaystyle=$
$\displaystyle R_{-}(J^{-1}gJ^{-1}h)$ $\displaystyle=$ $\displaystyle
R_{-}(J^{-1}g)R_{-}(J^{-1}h)$ $\displaystyle=$
$\displaystyle\mathfrak{B}(g)\mathfrak{B}(h).$

Therefore, $\mathfrak{B}$ is Rota-Baxter operator on $G$. The fact that
$(G,\mathfrak{B},S_{I})$ is a quadratic Rota-Baxter Lie group (Definition 5.3)
follows from Theorem 2.5.

Moreover, for any $u,v\in G^{*}$, setting $g=Ju$ and $h=Jv$, by (5.1) we get

$J(uv)=Ju\mathfrak{B}(Ju)Jv\mathfrak{B}(Ju)^{-1}=(Ju)\star(Jv).$

Since $J$ is a global diffeomorphism, it is a Lie group isomorphism from the
Lie group $G^{*}$ to the descendent Lie group $G_{\mathfrak{B}}$. ∎

By [22, Proposition 2.4], if $\mathfrak{B}:G\to G$ is a Rota-Baxter operator
on the Lie group $G$, then $\widetilde{\mathfrak{B}}:G\to G$ defined by
$\widetilde{\mathfrak{B}}(g)=g^{-1}\mathfrak{B}(g^{-1})$ is also a Rota-Baxter
operator on $G$.

######  Corollary 5.5.

Let $(G,J)$ be a factorizable Poisson Lie group. Then the map
$\widetilde{\mathfrak{B}}:G\to G$ given by

$\widetilde{\mathfrak{B}}(g)=R_{+}(J^{-1}(g^{-1}))$

is a Rota-Baxter operator on $G$.

###### Proof.

We have

$\displaystyle\widetilde{\mathfrak{B}}(g)$ $\displaystyle=$ $\displaystyle
g^{-1}R_{-}(J^{-1}(g^{-1}))$ $\displaystyle=$ $\displaystyle
J(J^{-1}(g^{-1}))R_{-}(J^{-1}(g^{-1}))$ $\displaystyle=$ $\displaystyle
R_{+}(J^{-1}(g^{-1}))R_{-}(J^{-1}(g^{-1}))^{-1}R_{-}(J^{-1}(g^{-1}))$
$\displaystyle=$ $\displaystyle R_{+}(J^{-1}(g^{-1})).$

By Theorem 5.4, $\mathfrak{B}$ is a Rota-Baxter operator. Therefore
$\widetilde{\mathfrak{B}}:G\to G$ is also a Rota-Baxter operator. ∎

######  Corollary 5.6.

Let $(G,J)$ be a factorizable Poisson Lie group, and $\mathfrak{B}$ the
induced Rota-Baxter operator on $G$. Then the descendent Lie group
$G_{\mathfrak{B}}$ is a Poisson Lie group with the Poisson structure
$\pi_{G_{\mathfrak{B}}}=J_{*}(\pi_{G^{*}})$. Its infinitesimal Lie bialgebra
is $\big{(}\mathfrak{g}_{B},(\mathfrak{g}^{*},[\cdot,\cdot]_{I})\big{)}$ given
in Corollary 2.8.

At the end of this section, we show that a quadratic Rota-Baxter Lie group
gives rise to a factorizable Poisson Lie group.

######  Theorem 5.7.

Let $(G,\mathfrak{B},S)$ be a quadratic Rota-Baxter Lie group and
$J_{S}:G^{*}\to G$ the induced diffeomorphism. Then $G$ is a factorizable
Poisson Lie group with

$R_{-}:=\mathfrak{B}\circ J:G^{*}\to G,\qquad R_{+}:=\mathfrak{B}_{+}\circ
J:G^{*}\to G,$

where $\mathfrak{B}_{+}(g)=g\mathfrak{B}(g)$.

###### Proof.

By definition, the infinitesimal $(\mathfrak{g},B:=\mathfrak{B}_{*e},S)$ is a
quadratic Rota-Baxter Lie algebra of weight 1. By Theorem 2.10, we obtain that
$(\mathfrak{g},\mathfrak{g}^{*}_{r})$ is a factorizable Lie bialgebra. Note
that

$\displaystyle(R_{+})_{*e}$ $\displaystyle=$
$\displaystyle(B+\mathsf{id})\circ\mathcal{I}_{S}=r_{+},$
$\displaystyle(R_{-})_{*e}$ $\displaystyle=$ $\displaystyle
B\circ\mathcal{I}_{S}=r_{-},$ $\displaystyle(J_{S})_{*e}$ $\displaystyle=$
$\displaystyle\mathcal{I}_{S}.$

So integrating this factorizable Lie bialgebra ([28]), we get a factorizable
Poisson Lie group structure on $G$. ∎

As the Lie group level of Example 2.9, we have

######  Example 5.8.

With notations in Example 2.9, let $G$ and $G^{*}$ be the simply connected Lie
groups of $\mathfrak{g}$ and $\mathfrak{g}^{*}$. Suppose that $G$ is complete.
Then the Lie group integrating the Lie algebra $\mathfrak{d}$ is $D=G\bowtie
G^{*}$. Denote by $D^{*}$ the simply connected dual Poisson Lie group. In
fact, $D^{*}=G\times\overline{G^{*}}$, the direct Lie group of $G$ and
$\overline{G^{*}}$, where $\overline{G^{*}}$ is the manifold $G^{*}$ with the
group structure being $s\cdot_{\overline{G^{*}}}t=ts$ for $s,t\in G^{*}$. Note
that $r_{+},r_{-}$ lift naturally to $R_{+},R_{-}:D^{*}\to D$, which is given
by

$R_{+}(g,u)=g,\qquad R_{-}(g,u)=u^{-1},\qquad\forall g\in G,u\in G^{*}.$

So $D$ is a factorizable Poisson Lie group. The induced Rota-Baxter operator
$\mathfrak{B}$ on the Lie group $D$ is given by

$\mathfrak{B}(g,u)=u^{-1}.$

###  5.2. Rota-Baxter Lie groups and matched pairs of Lie groups

In this subsection, we obtain a matched pair of Lie groups from a Rota-Baxter
Lie group. Matched pairs of Lie groups, also called double Lie groups, were
explored in [28, 29]. For matched pair of groups, we refer the readers to
[36].

A pair of Lie groups $(P,Q)$ is called a matched pair of Lie groups ([29]) if
there is a left action of $P$ on $Q$ and a right action of $Q$ on $P$:

$P\times Q\to Q,\qquad(p,q)\mapsto p\triangleright q;\qquad P\times Q\to
P\qquad(p,q)\mapsto p\triangleleft q,$

such that

(31) $\displaystyle p\triangleright(q_{1}q_{2})$ $\displaystyle=$
$\displaystyle(p\triangleright q_{1})\big{(}(p\triangleleft
q_{1})\triangleright q_{2}\big{)};$ (32)
$\displaystyle(p_{1}p_{2})\triangleleft q$ $\displaystyle=$
$\displaystyle\big{(}p_{1}\triangleleft(p_{2}\triangleright
q)\big{)}(p_{2}\triangleleft q).$

Denote a Lie group matched pair by $(P,Q;\triangleright,\triangleleft)$.

The following equivalent characterization of matched pairs of Lie groups is
well-known.

######  Proposition 5.9.

Let $(P,Q;\triangleright,\triangleleft)$ be a matched pair of Lie groups. Then
there is a Lie group structure $\cdot_{\bowtie}$ on $Q\times P$ defined by

$(q_{1},p_{1})\cdot_{\bowtie}(q_{2},p_{2})=\big{(}q_{1}(p_{1}\triangleright
q_{2}),(p_{1}\triangleleft q_{2})p_{2}\big{)}.$

Conversely, if $(Q\times P,\cdot)$ is a Lie group such that $Q$ and $P$ are
Lie subgroups, then $(P,Q;\triangleright,\triangleleft)$ is a matched pair of
Lie groups, where the left action $\triangleright:P\times Q\to Q$ of $P$ on
$Q$ and the right action $\triangleleft:P\times Q\to P$ of $Q$ on $P$ are
determined by

$(1,p)\cdot(q,1)=(p\triangleright q,p\triangleleft q).$

Denote the Lie group $(Q\times P,\cdot_{\bowtie})$ by $Q\bowtie P$.

Let $G$ be a Lie group. Denote by $(G\times G,\cdot_{G\times G})$ the direct
product Lie group with the group structure given by

$(g_{1},g_{2})\cdot_{G\times G}(h_{1},h_{2})=(g_{1}h_{1},g_{2}h_{2}).$

######  Proposition 5.10.

Let $(G,\mathfrak{B})$ be a Rota-Baxter Lie group. Then there is a Lie group
structure $\cdot_{D}$ on $G\times G$ determined by

$\displaystyle(1,g)\cdot_{D}(1,h)$ $\displaystyle=$ $\displaystyle(1,gh);$
$\displaystyle(s,1)\cdot_{D}(t,1)$ $\displaystyle=$ $\displaystyle(s\star
t,1)=(s\mathfrak{B}(s)t\mathfrak{B}(s)^{-1},1);$
$\displaystyle(s,1)\cdot_{D}(1,g)$ $\displaystyle=$ $\displaystyle(s,g);$
$\displaystyle(1,g)\cdot_{D}(s,1)$ $\displaystyle=$
$\displaystyle(gsg^{-1},\mathfrak{B}(gsg^{-1})^{-1}g\mathfrak{B}(s)),$

where $s,t$ are elements in the first component, and $g,h$ are elements in the
second component. Moreover, there is a Lie group isomorphism $\Phi$ from
$(G\times G,\cdot_{D})$ to the direct product Lie group $(G\times
G,\cdot_{G\times G})$ defined by

(33) $\displaystyle\Phi(s,g)=(s\mathfrak{B}(s)g,\mathfrak{B}(s)g).$

###### Proof.

It is simple to see that $\Phi$ is a diffeomorphism whose inverse is given by

$\Phi^{-1}(g,h)=(gh^{-1},\mathfrak{B}(gh^{-1})^{-1}h).$

Then we claim that the product $\cdot_{D}$ on $G\times G$ is the pull-back of
the direct product Lie group structure on $G\times G$ by $\Phi$, i.e.,

$e_{1}\cdot_{D}e_{2}=\Phi^{-1}\big{(}\Phi(e_{1})\cdot_{G\times
G}\Phi(e_{2})\big{)},\quad\forall e_{1},e_{2}\in G\times G.$

First, it is obvious that
$(1,g)\cdot_{D}(1,h)=\Phi^{-1}\big{(}\Phi(1,g)\cdot_{G\times
G}\Phi(1,h)\big{)}$. By (28), we have

$\displaystyle\Phi^{-1}\big{(}\Phi(s,1)\cdot_{G\times G}\Phi(t,1)\big{)}$
$\displaystyle=$
$\displaystyle\Phi^{-1}\big{(}(s\mathfrak{B}(s),\mathfrak{B}(s))\cdot_{G\times
G}(t\mathfrak{B}(t),\mathfrak{B}(t))\big{)}$ $\displaystyle=$
$\displaystyle\Phi^{-1}(s\mathfrak{B}(s)t\mathfrak{B}(s)^{-1}\mathfrak{B}(s)\mathfrak{B}(t),\mathfrak{B}(s)\mathfrak{B}(t))$
$\displaystyle=$ $\displaystyle\Phi^{-1}(s\star t\mathfrak{B}(s\star
t),\mathfrak{B}(s\star t))$ $\displaystyle=$
$\displaystyle(s,1)\cdot_{D}(t,1).$

Moreover, by definition, we have

$\displaystyle\Phi^{-1}\big{(}\Phi(s,1)\cdot_{G\times G}\Phi(1,g)\big{)}$
$\displaystyle=$
$\displaystyle\Phi^{-1}\big{(}(s\mathfrak{B}(s),\mathfrak{B}(s))\cdot_{G\times
G}(g,g)\big{)}$ $\displaystyle=$
$\displaystyle\Phi^{-1}(s\mathfrak{B}(s)g,\mathfrak{B}(s)g)=(s,g)$
$\displaystyle=$ $\displaystyle(s,1)\cdot_{D}(1,g),$

and

$\displaystyle\Phi^{-1}\big{(}\Phi(1,g)\cdot_{G\times
G}\Phi(s,1)\big{)}=\Phi^{-1}(gs\mathfrak{B}(s),g\mathfrak{B}(s))=(gsg^{-1},\mathfrak{B}(gsg^{-1})^{-1}g\mathfrak{B}(s))=(1,g)\cdot_{D}(s,1).$

Therefore, the Lie group structure $\cdot_{D}$ on $G\times G$ is the pull-back
of the direct product Lie group structure on $G\times G$ by $\Phi$, and $\Phi$
is a Lie group isomorphism. ∎

By using the associativity, the group structure $\cdot_{D}$ is given by

$\displaystyle(s,g)\cdot_{D}(t,h)$ $\displaystyle=$
$\displaystyle(s,1)\cdot_{D}(1,g)\cdot_{D}(t,1)\cdot_{D}(1,h)$
$\displaystyle=$
$\displaystyle(s,1)\cdot_{D}(gtg^{-1},\mathfrak{B}(gtg^{-1})^{-1}g\mathfrak{B}(t))\cdot_{D}(1,h)$
$\displaystyle=$
$\displaystyle(s,1)\cdot_{D}(gtg^{-1},1)\cdot_{D}(1,\mathfrak{B}(gtg^{-1})^{-1}g\mathfrak{B}(t))\cdot_{D}(1,h)$
$\displaystyle=$
$\displaystyle(sgtg^{-1},\mathfrak{B}(gtg^{-1})^{-1}g\mathfrak{B}(t)h).$

######  Theorem 5.11.

Let $(G,\mathfrak{B})$ be a Rota-Baxter Lie group. Then we have a matched pair
of Lie groups $(G,G_{\mathfrak{B}};\triangleright,\triangleleft)$, where the
actions are given by

$g\triangleright s=gsg^{-1},\quad g\triangleleft
s=\mathfrak{B}(gsg^{-1})^{-1}g\mathfrak{B}(s),\quad\forall g\in G,s\in
G_{\mathfrak{B}}.$

Moreover, the corresponding Lie group $G_{\mathfrak{B}}\bowtie G$ is exactly
the Lie group $(G\times G,\cdot_{D})$ given in Proposition 5.10.

###### Proof.

By (5.2), we get

$(s,g)\cdot_{D}(t,h)=(s(g\triangleright t),(g\triangleleft t)h).$

By Proposition 5.9, we deduce that
$(G,G_{\mathfrak{B}};\triangleright,\triangleleft)$ is a matched pair of Lie
groups. ∎

Let $(G,J)$ be a factorizable Poisson Lie group. Then
$(G,\mathfrak{B}=R_{-}\circ J^{-1})$ is a Rota-Baxter Lie group. By Theorem
5.11, we have a Lie group $G_{\mathfrak{B}}\bowtie G$. On the other hand,
there is a Lie group $G^{*}\bowtie G$ on the double of a Poisson Lie group.

######  Proposition 5.12.

With the notations above, we have a commutative diagram of Lie group
isomorphisms:

(39)

where $\Phi$ is defined by (33) and $\Psi$ is given by

$\Psi(s,g)=(R_{+}(s)g,R_{-}(s)g),\qquad\forall s\in G^{*},g\in G.$

Moreover, the differential of this diagram is the diagram (19) in the case of
$\lambda=1$.

###### Proof.

It is known that $\Phi$ is a Lie group homomorphism that integrates $\phi$. By
Theorem 5.4, $J:G^{*}\to G_{\mathfrak{B}}$ is the unique Lie group isomorphism
integrating $I:\mathfrak{g}^{*}_{r}\to\mathfrak{g}_{B}$, as $G^{*}$ is simply
connected. So it is direct to see $J\times\mathsf{id}:G^{*}\bowtie G\to
G_{\mathfrak{B}}\bowtie G$ is a Lie group isomorphism whose infinitesimal is
$I\oplus\mathsf{id}:\mathfrak{g}^{*}_{r}\bowtie\mathfrak{g}\to\mathfrak{g}_{B}\bowtie\mathfrak{g}$
in (19). Moreover, by the calculation

$\Phi\circ(J\times\mathsf{id})(s,g)=\Phi(Js,g)=(JsR_{-}J^{-1}(Js)g,R_{-}J^{-1}(Js)g)=(R_{+}(s)g,R_{-}(s)g)=\Psi(s,g),$

we see that $\Phi\circ(J\times\mathsf{id})=\Psi$, which is also a Lie group
isomorphism such that $\Psi_{*e}=\psi$. Thus we conclude that diagram (39) is
the integration of the diagram (19) in the case of $\lambda=1$. ∎

Since $\Phi$ is a Lie group isomorphism, it follows that
$\operatorname{Im}(\Phi|_{G_{\mathfrak{B}}})\subset G\times G$ is a Lie
subgroup, which is isomorphic to $G_{\mathfrak{B}}$. It gives an alternative
approach to the factorization of Rota-Baxter Lie groups given in [22].

######  Corollary 5.13.

Let $(G,\mathfrak{B})$ be a Rota-Baxter Lie group. Then for any $g\in G$,
there exists a unique decomposition $g=g_{+}g_{-}^{-1}$ with
$(g_{+},g_{-})\in\operatorname{Im}(\Phi|_{G_{\mathfrak{B}}})\subset G\times
G$.

###### Proof.

We have $g=g\mathfrak{B}(g)\mathfrak{B}(g)^{-1}$, where
$(g\mathfrak{B}(g),\mathfrak{B}(g))=\Phi(g,1)$ for $g\in G_{\mathfrak{B}}$ and
this decomposition is unique since $\Phi$ is a diffeomorphism. ∎

At the end of this subsection, we establish the relation between the Lie group
$G_{\mathfrak{B}}\bowtie G$ and the Lie algebra
$\mathfrak{g}_{B}\bowtie\mathfrak{g}$, where $\mathfrak{g}$ is the Lie algebra
of $G$ and $B=\mathfrak{B}_{*e}$ is the differentiated Rota-Baxter operator of
weight 1 on $\mathfrak{g}.$

######  Proposition 5.14.

Let $(G,\mathfrak{B})$ be a Rota-Baxter Lie group whose infinitesimal Rota-
Baxter Lie algebra is $(\mathfrak{g},B=\mathfrak{B}_{*e})$. Then the Lie
algebra of the Lie group $G_{\mathfrak{B}}\bowtie G$ given in Theorem 5.11 is
the Lie algebra $\mathfrak{g}_{B}\bowtie\mathfrak{g}$ given in Theorem 3.3 and
$\Phi_{*e}=\phi$.

###### Proof.

Denote by $\left\llbracket\cdot,\cdot\right\rrbracket$ the induced Lie bracket
on $\mathfrak{g}\oplus\mathfrak{g}$ of the Lie group $G_{\mathfrak{B}}\bowtie
G$. It is simple to see that

$\left\llbracket(0,x),(0,y)\right\rrbracket=(0,[x,y]_{\mathfrak{g}}),\qquad\left\llbracket(\xi,0),(\eta,0)\right\rrbracket=([\xi,\eta]_{B},0).$

Then, by (5.2), we have

$\displaystyle\left\llbracket(0,x),(\xi,0)\right\rrbracket$ $\displaystyle=$
$\displaystyle\frac{d}{d\epsilon}\frac{d}{d\epsilon^{\prime}}|_{\epsilon,\epsilon^{\prime}=0}(1,\exp^{\epsilon
x})\cdot_{D}(\exp^{\epsilon^{\prime}\xi},1)\cdot_{D}(1,\exp^{-\epsilon x})$
$\displaystyle=$
$\displaystyle\frac{d}{d\epsilon}\frac{d}{d\epsilon^{\prime}}|_{\epsilon,\epsilon^{\prime}=0}(1,\exp^{\epsilon
x})\cdot_{D}(\exp^{\epsilon^{\prime}\xi},\exp^{-\epsilon x})$ $\displaystyle=$
$\displaystyle\frac{d}{d\epsilon}\frac{d}{d\epsilon^{\prime}}|_{\epsilon,\epsilon^{\prime}=0}(\exp^{\epsilon
x}\exp^{\epsilon^{\prime}\xi}\exp^{-\epsilon x},\mathfrak{B}(\exp^{\epsilon
x}\exp^{\epsilon^{\prime}\xi}\exp^{-\epsilon x})^{-1}\exp^{\epsilon
x}\mathfrak{B}(\exp^{\epsilon^{\prime}\xi})\exp^{-\epsilon x})$
$\displaystyle=$
$\displaystyle([x,\xi]_{\mathfrak{g}},\frac{d}{d\epsilon}\frac{d}{d\epsilon^{\prime}}|_{\epsilon,\epsilon^{\prime}=0}\mathfrak{B}(\exp^{\epsilon
x}\exp^{\epsilon^{\prime}\xi}\exp^{-\epsilon
x})^{-1}+\frac{d}{d\epsilon}\frac{d}{d\epsilon^{\prime}}|_{\epsilon,\epsilon^{\prime}=0}\exp^{\epsilon
x}\mathfrak{B}(\exp^{\epsilon^{\prime}\xi})\exp^{-\epsilon x})$
$\displaystyle=$
$\displaystyle([x,\xi]_{\mathfrak{g}},-B[x,\xi]_{\mathfrak{g}}+[x,B\xi]_{\mathfrak{g}})$
$\displaystyle=$
$\displaystyle[(0,x),(\xi,0)]_{\mathfrak{g}_{B}\bowtie\mathfrak{g}}.$

So we proved
$\left\llbracket\cdot,\cdot\right\rrbracket=[\cdot,\cdot]_{\mathfrak{g}_{B}\bowtie\mathfrak{g}}$.
Namely, the Lie algebra of $G_{\mathfrak{B}}\bowtie G$ is
$\mathfrak{g}_{B}\bowtie\mathfrak{g}$. It is obvious that $\Phi_{*e}=\phi$. ∎

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