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\title[On Rota---Baxter operators on $gl_2(\mathbb C)$]{The description of Rota-Baxter operators of nonzero weight on complex general linear Lie algebra of order 2.}
\author{{M. Goncharov}}%
\address{Maxim Goncharov
\newline\hphantom{iii} Sobolev Institute of Mathematics,
\newline\hphantom{iii} Novosibirsk State University, Department of Mechanics and
Mathematics, Russia}%
\email{goncharov.gme@gmail.com}%


\thanks{\sc Goncharov M.E.,
The description of Rota-Baxter operators of nonzero weight on general linear complex Lie algebra of order 2.}
\thanks{\copyright \ 2022 Goncharov M.E.}
\thanks{\rm The work was supported RAS Fundamental Research Program, project  FWNF-2022-0002}

\vspace{1cm}
\maketitle {\small
\begin{quote}
\noindent{\sc Abstract. } In the paper, a classification of Rota-Baxter operators of weight 1 on  general linear complex Lie algebra of order 2 is given. The description was made up to the action of $Aut(gl_2(\mathbb C))$.\medskip

\noindent{\bf Keywords:} Lie algebra, Rota---Baxter operator, general linear Lie algebra
 \end{quote}
}




\section{Introduction}
The Rota---Baxter operators first appeared in the work of F. Tricomi \cite{TRI}. These operators became  popular after the work of G. Baxter \cite{Baxter}, where these operators were independently defined as a formalism for the study of integral operators. For a long period of time, Rota Baxter operators were studied in combinatorics and probability theory mainly (see \cite{Atkinson}, \cite{Cartier}, \cite{Rota}). A new impulse the theory of Rota---Baxter operators received in the 80s of the last century, when a deep connection between Rota---Baxter operators of weight 0 and skew-symmetric solutions of the classical Yang---Baxter equation was found \cite{BD,STS}. In the recent years, there were found connections between Rota-Baxter operators and various objects of mathematics such as pre- and post- Lie algebras, double Lie algebras,   shuffle algebras, associative Yang-Baxter equation (AYBE) etc. 



One of the important and interesting areas of research here is the description of the Rota---Baxter operators on the most important classes of algebras. To date, Rota---Baxter operators of zero and nonzero weights on the Lie algebra $sl_2(\mathbb{C})$ were studied in \cite{Kolesnikov,KonovDissert,sl2,sl2-0}, operators on the matrix algebra $M_2(\mathbb C )$ were studied in \cite{BGP,Mat2}. The classification of Rota---Baxter operators of nonzero weight defined on the algebra $sl_3(\mathbb{C})$ was given by V.\,V.\,Sokolov \cite{Sokolov}. In \cite{M3}, the classification of nonsplitting Rota---Baxter operators of nonzero weight on the matrix algebra $M_3(F)$, where $F$ is an algebraically closed field of characteristic 0, was given. Note that all the mentioned classes of algebras are simple.

In this article, we classify Rota---Baxter operators on $gl_2(\mathbb C)$, the general linear Lie algebra of order 2 over the field of complex numbers $\mathbb C$. The classification is given up to the action of the group of automorphisms $Aut(gl_2(\mathbb C))$. It is worth noting that the algebra $gl_2(\mathbb C)$ has a non-zero center, so it is not even a semisimple Lie algebra.


\section{Preliminary results.}


Let $A$ be an algebra over a field $F$, $\lambda\in F$.

\begin{definition}
A linear map $R: A\mapsto A$ is called a Rota-Baxter operator of weight $\lambda$, if for all  $a,b\in A$:
\begin{equation}\label{rbdef}
R(a)R(b)=R(R(a)b+aR(b)+\lambda ab).
\end{equation}\end{definition} 

If $R$ is a Rota-Baxter operator of weight $\lambda$ on an algebra $A$ and $\varphi$ is an automorphism or an anti-automorphism of $A$, then a map $R'=\varphi^{-1}\circ R\circ \varphi $ is also a Rota-Baxter operator of the same weight $\lambda$ \cite{BGP}. This remark allows us to define an equivalence relation on the set of Rota---Baxter operators of weight $\lambda$  and to make the classification up to the action of the group generated by automorphisms and anti-automorphisms of the algebra $A$. In this case, we will say that operators $R$ and $R'$ are {\it conjugate}.

If  $0\neq\alpha\in F$, then a map $\alpha R$ is a Rota-Baxter operator of weight $\alpha\lambda$. Therefore, up to the multiplication by a nonzero scalar, we have only two different cases: $\lambda=0$ и $\lambda =1$.

{\bf Example.} If $A$ is an arbitrary algebra and $0\neq \lambda\in F$, then maps $R_0$ and $R_1$, defined as $R_0(x)=0$ or $R_1(x)=-\lambda x$ for all $x\in A$, are Rota-Baxter operators of weight $\lambda$. Operators of these types are called trivial.
 
Let $A=gl_2(\mathbb C)=(M_2(\mathbb C),[\cdot,\cdot])$ be the general linear Lie algebra over the field of complex numbers $\mathbb C$ with the Lie product
$$
[x,y]=xy-yx.
$$

In this work, we describe Rota-Baxter operators of weight 1 on $gl_2(\mathbb C)$. Note, that if a map  $\varphi$ is an anti-automorphism of a Lie algebra, then a map $-\varphi$ is an automorphism of the same algebra. Thus, in the case of Lie algebras, the classification should be made up to the action of the group of automorphisms.

We will use the following notations: $\mathrm E\in gl_2(\mathbb C)$ is the identity map, $e_{ij}$  is the usual matrix unit, $h=e_{11}-e_{22}$. As a basis of $gl_2(\mathbb C)$ we will take a set $\{\mathrm{E},h,e_{12},e_{21}\}$. We will denote by $sl_2(\mathbb C)$ an ideal in  $gl_2(\mathbb C)$, that consist of matrices with trace zero.

Note that for any $\theta\neq 0$, a map $\varphi_{\theta}$ defined as
\begin{equation}\label{aut}
\varphi_{\theta}(\mathrm{E})=\theta E,\ \varphi_{\theta}(z)=z\ \text{for all}\ z\in sl_2(\mathbb C),
\end{equation}
is an automorphism of  $gl_2(\mathbb C)$. 


Let $R$ be a Rota-Baxter operator on $gl_2(\mathbb C)$. Note, that if  $R(\mathrm E)_J$ is the Jordan form of the matrix $R(\mathrm E)$ and $T$ is an invertible matrix such that 
$$R(\mathrm E)_J=T^{-1}R(\mathrm E) T,
$$
then a map $\varphi_T:~gl_2(\mathbb C)\mapsto gl_2(\mathbb C)$, defined as
$$
\varphi_T(a)=T^{-1}aT,$$
for all  $a\in gl_2(\mathbb C)$, is an automorphism of $gl_2(\mathbb C)$. Moreover,
$$
\varphi_T\circ R\circ \varphi_{T^{-1}}(\mathrm E)=T^{-1}R(T\mathrm E T^{-1})T=T^{-1}R(\mathrm E)T=R(\mathrm E)_J.
$$

Thus, up to the action of $Aut(gl_2(\mathbb C))$, we can assume that $R(\mathrm E)$ is a Jordan matrix. We have the following possibilities for $R( \mathrm E)_J$:\\
1. $R(\mathrm E)=\lambda E+e_{12}$, $\lambda\in \mathbb C$, a Jordan $2\times 2$ block. \\
2. $R(\mathrm E)=\lambda_1 e_{11}+\lambda_2 e_{22}$, $\lambda_1\neq \lambda_2\in \mathbb C$, a diagonal matrix with different eigenvalues.\\
3. $R(\mathrm E)=\lambda \mathrm E$, $\lambda\in \mathbb C$, a scalar matrix.

 We will consider each of these cases separately.



\section{The main part.}

The aim of this section is to give a classification of Rota-Baxter operators of weight 1 on $gl_2(\mathbb C)$. We will break the classification into cases, depending on the Jordan form of the matrix $R(\mathrm{E})$.

\begin{lemma}
Let  $R:gl_2(\mathbb C)\mapsto gl_2(\mathbb C)$ be a Rota-Baxter operator of weight 1 such that  $R(\mathrm{E})=\lambda \mathrm{E}+e_{12}$. Then $R$ is one of the following:
\begin{gather} \label{r1}
\  R(\mathrm E)=\lambda \mathrm E+e_{12},\ R(h)=R(e_{12})=R(e_{21})=0,\ \lambda\in \mathbb C,
\end{gather}
\begin{gather} \label{r2}
 R(\mathrm E)=\lambda \mathrm E+e_{12},\ R(h)=-h,\  R(e_{12})=-e_{12},\  R(e_{21})=-e_{21}\ \lambda\in \mathbb C.
\end{gather}
\end{lemma} 

\begin{proof} Consider
\begin{equation}\label{t1}
[R(h),e_{12}]=[R(h),R(\mathrm{E})]=R([R(h),\mathrm{E}]+[h,R(\mathrm{E})]+[h,\mathrm{E}])=2R(e_{12}).
\end{equation}

Therefore, $R(e_{12})\in [gl_2(\mathbb C),e_{12}]$, that is $R(e_{12})=\alpha_1 h+\alpha_2 e_{12}$ for some $\alpha_1,\alpha_2\in \mathbb C$.

From
\begin{equation}\label{t2}
[R(e_{12}),e_{12}]=[R(e_{12}),R(\mathrm{E})]=R([R(e_{12}),\mathrm{E}]+[e_{12},R(\mathrm{E})]+[e_{12},\mathrm{E}])=0
\end{equation}
we get that $\alpha_1=0$ and $R(e_{12})=\alpha e_{12}$ for some $\alpha\in \mathbb C$.

Similarly,
\begin{equation}\label{t4}
[e_{12},R(e_{21})]=[R(\mathrm{E}),R(e_{21})]=R([R(\mathrm{E}),e_{21}])=R(h).    
\end{equation}

This means that
\begin{equation}\label{t3}
R(h)=\beta_1 h+\beta_2 e_{12}
\end{equation}
for some $\beta_1,\beta_2\in \mathbb C$.

Suppose that $\alpha=0$. Then  $R(e_{12})=0$. In this case, from \eqref{t1} and \eqref{t3} it follows that $R(h)=\beta_2 e_{12}$. From \eqref{t4} it follows that $R(e_{21})$ lies in a subspace spanned by  $e_{12},h, \mathrm{E}$. Consider
\begin{equation}\label{t5}
[R(e_{21}),R(h)]=[R(e_{21}),\beta_2 e_{12}]=\gamma e_{12},\ \gamma\in \mathbb C.
\end{equation}
On the other hand, since $R([R(e_{21}),h])=\delta R(e_{12})=0$ for some $\delta\in \mathbb C$ and $R([e_{21},R(h)]=\beta_2 R(h)=\beta_2^2 e_{12}$, we have that
\begin{equation}\label{t6}
R([R(e_{21}),h]+[e_{21},R(h)]+[e_{21},h])=2R(e_{21})+\beta_2^2 e_{12}.
\end{equation}
From the last two equalities, we get that
$$
R(e_{21})=\theta e_{12},\ \theta \in \mathbb C.
$$

Consider
$$
0=[R(e_{12}),R(e_{21})]=R([R(e_{12}),e_{21}]+[e_{12},R(e_{21})]+[e_{12},e_{21}])=R(h).
$$
Hence, $R(h)=0$. From \eqref{t5} and \eqref{t6} we finally get that $R(e_{21})=0$ and we obtain the operator \eqref{r1}.

Suppose that $\alpha\neq 0$. From \eqref{t1} and \eqref{t3} it follows that $\beta_1=\alpha$, that is $R(h)=\alpha h+\beta_2 e_{12}$.

From
\begin{multline*}
2\alpha^2 e_{12}=[R(h),R(e_{12})]\\
=R([R(h),e_{12}]+[h,R(e_{12})]+[h,e_{12}])=(4\alpha+2)R(e_{12})=(4\alpha+2)\alpha e_{12}
\end{multline*}
we get that $\alpha=-1$. Using \eqref{t4}, we obtain:
\begin{multline*}
-R(h)=-[e_{12},R(e_{21})]\\
=[R(e_{12}),R(e_{21})]=R([R(e_{12}),e_{21}]+[e_{12},R(e_{21})]+[e_{12},e_{21}])=R(R(h)).
\end{multline*}
That is, $R(R(h))=-R(h)$. Since  $R(h)=- h+\beta_2 e_{12}$ and $R(e_{12})=-e_{12}$, we obtain $\beta_2=0$ and $R(h)=-h$. From  \eqref{t4} it follows that $R(e_{21})=-e_{21}+\alpha_1 e_{12}+\alpha_2 \mathrm{E}$ for some $\alpha_1,\alpha_2\in \mathbb C$.

Finally, from
\begin{multline*}
    -[R(e_{21}),h]=[R(e_{21}),R(h)] \\ 
    = R([R(e_{21}),h]+[e_{21},R(h)]+[e_{21},h])=R([R(e_{21}),h]),
\end{multline*}
it follows that $R(e_{21})=-e_{21}$, and we get that $R$ is equal to the operator  \eqref{r2}. \end{proof}




\begin{lemma}

  Let  $R:gl_2(\mathbb C)\mapsto gl_2(\mathbb C)$ be a Rota-Baxter operator of weight 1 such that  $R(\mathrm{E})=\lambda_1 e_{11}+\lambda_2 e_{22}$, where $\lambda_1\neq \lambda_2$.
Then $R$ is conjugate to one of the following operators:
\begin{gather} \label{r3}
\  R(\mathrm{E})=\lambda \mathrm{E}+h,\ R(h)=0,\ R(e_{12})=R(e_{21})=0,\ \lambda\in\mathbb C ;
\end{gather}
\begin{gather} \label{r4}
 \  R(\mathrm{E})=\lambda \mathrm{E}+h,\ R(h)=-h,\ R(e_{12})=-e_{12},\ R(e_{21})=-e_{21},\ \lambda\in\mathbb C;
\end{gather}
\begin{gather} \label{r5}
 \  R(\mathrm{E}){=}\lambda \mathrm{E}{+}h,\ R(h){=}\alpha_1\mathrm{E}{+}\alpha_2h,\ R(e_{12}){=}-e_{12},\ R(e_{21})=0,\ \lambda,\alpha_1,\alpha_2\in\mathbb C.
\end{gather}
\end{lemma}

\begin{proof}
Let $R(\mathrm{E})=\lambda_1 e_{11}+\lambda_2 e_{22},\ \lambda_1\neq \lambda_2$.  Then $R(\mathrm{E})=\alpha_1 E+\alpha_2 h$, where $\alpha_1,\alpha_2\in \mathbb C$. Since $\lambda_1\neq \lambda_2$, then $\alpha_2\neq 0$. Define $\theta=\frac{1}{\alpha_2}$. We have
$$
\varphi_{\theta}^{-1}\circ R\circ \varphi_{\theta} (\mathrm{E})=\theta \varphi_{\theta}^{-1}(\alpha_1 \mathrm{E}+\alpha_2 h)=\alpha_1\mathrm{E}+\frac{\alpha_2}{\alpha_2}h=\alpha_1\mathrm{E}+h.
$$

Therefore, up to the action of $Aut(gl_2(\mathbb C))$, we can assume that $R(\mathrm{E})=\lambda E+h$.

 Let $R(e_{12})=\begin{pmatrix}
\beta_{11} & \beta_{12}\\
\beta_{21} & \beta_{22} 
\end{pmatrix}$. Then we have
$$
[R(e_{12}),R(\mathrm{E})]=\left [\begin{pmatrix}
\beta_{11} & \beta_{12}\\
\beta_{21} & \beta_{22} 
\end{pmatrix}, \lambda \mathrm{E}+h \right]=-2\beta_{12}e_{12}+2\beta_{21}e_{21}.
$$
On the other hand, 
$$
R([R(e_{12}),\mathrm{E}]+[e_{12},R(\mathrm{E})]+[e_{12},\mathrm{E}])=R([e_{12},\lambda\mathrm{E}+h])=-2R(e_{12}).
$$
 From the last two equalities, we can conclude that $\beta_{11}=\beta_{22}=0$. Moreover, since the characteristic of the ground field in not equal to 2, then $\beta_{21}=0$. Thus,
 $$
 R(e_{12})=\beta_{12}e_{12}.
 $$

Similarly,  $R(e_{21})=\beta_{21}e_{21}$ for some $\beta_{21}\in \mathbb C$.


Consider $R(h)$. Let us note that
$$
[R(h),\mathrm{E}]+[h,R(\mathrm{E})]+[h,\mathrm{E}]=0.
$$
Hence, $[R(h),R(\mathrm{E})]=0$ and $R(h)=\begin{pmatrix} \delta_1 & 0 \\
0 & \delta_2\end{pmatrix}$, $\delta_i\in \mathbb C$.

Let $e_{ij}\in \{e_{12},e_{21}\}$. Then
\begin{multline*}
  (\delta_i-\delta_j)\beta_{ij}e_{ij}   = [R(h),R(e_{ij})]\\
  = R([R(h),e_{ij}]+[h,R(e_{ij})]+[h,e_{ij}])=(\delta_i-\delta_j+(-1)^j(2\beta_{ij}+2))\beta_{ij}e_{ij},
\end{multline*}
where $\beta_{ij}$ are defined from  conditions $R(e_{ij})=\beta_{ij}e_{ij}$, $i \neq j$. Thus, we have two options: $\beta_{ij}=0$ or $\beta_{ij}=-1$.

If $\beta_{12}=\beta_{21}=0$, then from
$$
0=[R(e_{12}),R(e_{21})]=R([R(e_{12}),e_{21}]+[e_{12},R(e_{21})]+[e_{12},e_{21}])=R(h)
$$
it follows that $R(h)=0$ and we obtain the operator \eqref{r3}. Similarly, if $\beta_{12}=\beta_{21}=-1$, then we get that $R(h)=-h$ and $R$ is equal to the operator \eqref{r4}.

Finally, if $\beta_{12}\neq \beta_{21}$, then up to the action of an automorphism  $\varphi: z\mapsto -z^{\tau}$ (where $\tau$ is the transpose) for all $z\in gl_2(\mathbb C)$, we can assume that $\beta_{12}=0$, $\beta_{21}=-1$. Thus, in this case we obtain the operator \eqref{r5}. 
\end{proof}

\begin{lemma}
 Let  $R:gl_2(\mathbb C)\mapsto gl_2(\mathbb C)$ be a Rota-Baxter operator of weight 1 such that  $R(\mathrm{E})=\lambda\mathrm{E}$. 
Then $R$ is conjugate to one of the following operators:

\begin{gather} \label{r6}
\  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=0,\ R(e_{12})=-e_{12}+th;\ R(e_{21})=0,\ t\in\{0,1\}
\end{gather}
\begin{gather} \label{r7}
 \  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=0,\ R(e_{12})=-e_{12}+th+E;\ R(e_{21})=0,\ t\in\{0,1\}
\end{gather}
\begin{gather} \label{r8}
 \  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=E,\ R(e_{12})=-e_{12}+h+\alpha \mathrm{E};\ R(e_{21})=0,\ \alpha\in\mathbb C
 \end{gather}
\begin{gather} \label{r81}
 \  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=E,\ R(e_{12})=-e_{12}+\mathrm{E};\ R(e_{21})=0
\end{gather}
\begin{gather} \label{r9}
 \  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=th,\ R(e_{12})=-e_{12},\ R(e_{21})=0,\ t\in \mathbb C,\ t\neq 0
\end{gather}
\begin{gather} \label{r10}
 \  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=th+\mathrm{E},\ R(e_{12})=-e_{12},\ R(e_{21})=0,\ t\in \mathbb C,\ t\neq 0
\end{gather}
\begin{gather} \label{r11}
 \  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=-h+\alpha \mathrm{E},\ R(e_{12})=-e_{12},\ R(e_{21})=\mathrm{E},\ \alpha\in \mathbb C
\end{gather}
\begin{gather}\label{r12}
    \ R(\mathrm{E})=\lambda\mathrm{E}, R(h)=th,\ R(e_{12})=te_{12},\ R(e_{21})=te_{21},\ t\in\{0,-1\}.
\end{gather}
\end{lemma}
\begin{proof}
In this case, the center of the algebra $gl_2(\mathbb C)$ is $R$-invariant. Hence, $R$ induces a Rota-Baxter operator $\bar{R}$ on the quotient algebra $g_1=gl_2(\mathbb{C})/Z(gl_2(\mathbb{C}))\cong sl_2(\mathbb{C})$. Given an element $a\in gl_2(\mathbb C)$,   $\bar{a}$ denotes the coset generated by $a$.


The classification of Rota-Baxter operators of nonzero weight 1 on $sl_2(\mathbb{C})$ was already made in  \cite{sl2}. Unfortunately, we can't use this classification since it was made without taking into account the action of the group $Aut(sl_2(\mathbb C))$. The description of Rota-Baxter operators up to the action of $Aut(sl_2(\mathbb C))$ was made in a preprint of V. Gubarev and R. Kozlov \cite{GK}. Besides the trivial ones, there are two non-conjugate to each other operators on $sl_2(\mathbb C)$:

I. $R(h)=R(e_{21})=0,\ R(e_{12})=-e_{12}+th,\ t\in\{0,1\}$,

II. $R(h)=th,\ R(e_{12})=-e_{12},\ R(e_{21})=0$, $t\in\mathbb C,\ t\neq 0$.

Let $\bar{\varphi}$ be an automorphism of $g_1$. For a coset $\bar{a}\in g_1$ we can uniquely choose a generator $a\in gl_2({\mathbb{C}})$ such that $tr(a)=0$. Define a map $\varphi:gl_2(\mathbb C)\mapsto gl_2(\mathbb C)$ as $\varphi(\mathrm{E})=\mathrm{E}$ and $\varphi(p)=q$ if $tr(p)=tr(q)=0$ and $\bar{\varphi}(\bar{p})=\bar{q}$. It is easy to see that  the map $\varphi$ is an automorphism of the algebra $gl_2(\mathbb C)$. Moreover, for all $x\in gl_2(\mathbb C)$ we have $\bar{\varphi}(\bar{x})=\overline{\varphi(x)}$. Note that if $R$ is an arbitrary Rota-Baxter operator (of any weight) on $gl_2(\mathbb{C})$ satisfying $R(\mathrm{E})=\theta \mathrm{E}$,  $\bar{R}:g_1\mapsto g_1$ is the corresponding Rota-Baxter operator on $g_1$ and $\bar{\varphi}$ is an arbitrary automorphism of  $g_1$, then the composition $R_1=\varphi^{-1}\circ R\circ \varphi $ is a Rota-Baxter operator of the same weight satisfying $R_1(\mathrm{E})=\theta \mathrm{E}$ and $\overline{R_1}=\overline{\varphi}^{-1}\circ \bar{R}\circ \overline{\varphi}$. This allows us to consider that $\bar{R}$ is a trivial Rota-Baxter operator or one of the operators I or II defined above. We will consider these situations one by one.


{\bf Case 1.} Let $\bar{R}(\bar{h})=\bar{R}(\overline{e_{21}})=0,\ \bar{R}(\overline{e_{12}})=-\overline{e_{12}}+t\bar{h},\ t\in\{0,1\}$.

Going back to the operator $R$, we can say that
$$
R(\mathrm{E}){=}\lambda \mathrm{E},\ R(h){=}\alpha_1 \mathrm{E},\ R(e_{21}){=}\alpha_2 \mathrm{E},\ R(e_{12}){=}{-}e_{12}{+}th{+}\alpha_3 \mathrm{E},\ t\in\{0,1\},\ \alpha_i\in \mathbb C.
$$
From
$$0=[R(e_{21}),R(h)]=R([R(e_{21}),h]+[e_{21},R(h)]+[e_{21},h])=2R(e_{21})$$
it follows that $\alpha_2=0$. 

If $\alpha_1=\alpha_3=0$, then we obtain \eqref{r6}. Suppose that $\alpha_1=0, \alpha_3\neq0$. Let $R_1=\varphi_{\alpha_3^{-1}}^{-1}\circ R\circ \varphi_{\alpha_3^{-1}}$. Then direct computations show that
$$
R_1(\mathrm{E})=\lambda \mathrm{E},\ R_1(e_{12})=-e_{12}+th+\mathrm{E},\ R_1(e_{21})=0,\ R_1(h)=0.
$$
We obtain the operator \eqref{r7}. 

Suppose that $\alpha_1\neq 0$. 
Using the conjugation with the automorphism $\varphi_{\alpha_1^{-1}}$, we obtain an operator
\begin{gather}
 \  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=E,\ R(e_{12})=-e_{12}+th+\frac{\alpha_3}{\alpha_1} \mathrm{E},\ R(e_{21})=0.
 \end{gather}
 
 If $t=1$, then we get $\eqref{r8}$. Let $t=0$. Consider an automorphism  $\psi_{\alpha_3^{-1}}$. If $R_1=\varphi_{\alpha_3^{-1}}^{-1}\circ R\circ \varphi_{\alpha_3^{-1}}$, then direct computations show that
 $R_1(\mathrm{E})=\lambda \mathrm{E}$, $R_1(h)=\mathrm{E}$, $R_1(e_{21})=0$. Moreover, 
 $$R_1(e_{12})=\varphi_{\alpha_3^{-1}}^{-1}(R(\alpha_3^{-1}e_{12}))=\alpha_3^{-1}\varphi_{\alpha_3^{-1}}^{-1}(-e_{12}+\alpha_3\mathrm{E})=-e_{12}+\mathrm{E}.$$

Hence, we obtain  \eqref{r81}.


{\bf Case 2.} In this case, we have that
$$
R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=th+\alpha_1 \mathrm{E},\ R(e_{21})=\alpha_2 \mathrm{E},\ R(e_{12})=-e_{12}+\alpha_3 \mathrm{E},\ t,\alpha_i\in \mathbb C,\ t\neq 0.
$$
From 
$$0=[R(e_{21}),R(h)]=R([R(e_{21}),h]+[e_{21},R(h)]+[e_{21},h])=2(t+1)R(e_{21})$$
it follows that a condition $t\neq -1$ implies $\alpha_2=0$. At the same time, if $t=-1$, then $\alpha_2$ is arbitrary.

Note that an equality
$$
[R(e_{12}),R(e_{21})]=R([R(e_{12}),e_{21}]+[e_{12},R(e_{21})]+[e_{12},e_{21}])
$$
holds and gives no restrictions for parameters $\alpha_i$. Consider
\begin{gather*}
     [R(e_{12}),R(h)]  = [-e_{12}+\alpha_3\mathrm{E},th+\alpha_1\mathrm{E}]=2te_{12},\\
     R([R(e_{12}),h]{+}[e_{12},R(h)]{+}[e_{12},h])=R(-[e_{12},h]{+}t[e_{12},h]{+}[e_{12},h]){=}2te_{12}{+}2t\alpha_3\mathrm E.
\end{gather*}
Since $t\neq 0$, then $\alpha_3=0$.

Let $\alpha_2=0$. If, at the same time, $\alpha_1=0$, then we obtain \eqref{r9}. If  $\alpha_1\neq 0$, then the conjugation with $\varphi_{\alpha_1^{-1}}$ gives us the operator \eqref{r10}.

If  $t=-1$  and $\alpha_2\neq 0$, then the conjugation with $\varphi_{\alpha_2^{-1}}$ gives us \eqref{r11}.

{\bf Case 3.} It remains to consider the cases of trivial operator $\overline{R}$, that is $\overline{R}(\overline{v})=t\overline{v}$, $\overline{v}\in gl_2(\mathbb{C})/Z(gl_2(\mathbb C))$, $t\in\{0,-1\}$. Note that in these cases for all $a,b\in sl_2(\mathbb C)$:
$$
t^2[a,b]=[R(a),R(b)]=R([R(a),b]+[a,R(b)]+[a,b])=(-1)^t R([a,b]).
$$
Therefore, $R(sl_2(\mathbb C))\subset sl_2(\mathbb C)$ and $R$ is equal to \eqref{r12}.\end{proof}

{\bf Remark 1.} In Lemmas 1-3, in order to obtain conditions for operators \eqref{r1}-\eqref{r2} and \eqref{r3}-\eqref{r12}, we've checked that these operators satisfy the condition \eqref{rbdef} for the case when $a$ and $b$ are vectors from the basis of $gl_2(\mathbb C)$. Thus, these operators are Rota-Baxter operators of weight 1.

\section{The conjugation check.}

In this section, we will prove that operators from lemmas 1-3 are not conjugate. 

\begin{lemma}
 Let $R_i$, ($i=1,2,3$) be Rota-Baxter operators of weight 1 on $gl_2(\mathbb C)$, such that $R_1(\mathrm{E})=\lambda_1\mathrm{E}+e_{12}$, $R_2(\mathrm{E})=\lambda_2\mathrm{E}+h$, $R_3(\mathrm{E})=\lambda_3\mathrm{E}$. Then $R_i$ are not conjugate.
\end{lemma}
\begin{proof}
 First we note that $R_3$ can't be conjugate to $R_1$ or $R_2$. Indeed, if $\varphi$ is an automorphism of the algebra $gl_2(\mathbb C)$, then $\varphi(\mathrm{E})=\theta \mathrm{E}$ for some $\theta\in \mathbb C\backslash \{0\}$. Hence,
 $$
 (\varphi^{-1}\circ R_3\circ\varphi)(\mathrm{E})=(\varphi^{-1}\circ R)(\theta\mathrm{E})=\theta\lambda_3\varphi^{-1}(\mathrm{E})=\lambda_3\mathrm{E}.
 $$

Consider $R_1$ and $R_2$. Let $\varphi\in Aut(gl_2(\mathbb C))$. Then
$$
(\varphi^{-1}\circ R_1\circ\varphi)(\mathrm{E})=(\varphi^{-1}\circ R)(\theta\mathrm{E})=\theta\varphi^{-1}(\lambda_1\mathrm{E}+e_{12})=\lambda_1 \mathrm{E}+ \varphi^{-1}(e_{12}).
$$

Since $e_{12}$ is an ad-nilpotent element of $gl_2(\mathbb C)$, then $\varphi(e_{12})$ can't be semisimple.  Therefore,  $R_1$ and  $R_2$ are not conjugate.\end{proof}
{\bf Remark 2.} All operators from lemmas 1-3 depend on the parameter $\lambda$. From the proof of Lemma 4, it follows that two Rota-Baxter operators from lemmas 1-3 corresponding to different parameters $\lambda_1\neq \lambda_2$, can't be sconjugate. 

\begin{lemma}
Operators \eqref{r1}  and \eqref{r2} are not conjugate.
\end{lemma}
\begin{proof}
Indeed, the image of the operator \eqref{r1} is a one-dimensional subspace spanned by $R(\mathrm{E})$ while in the image of \eqref{r2} there are at least two linear independent vectors $e_{12}$ and $e_{21}$.
\end{proof}

\begin{lemma}
Operators $\eqref{r3}$, $\eqref{r4}$ and $\eqref{r5}$ are not conjugate. Moreover, operators of the type \eqref{r5} corresponding to different pairs of parameters $(\alpha_1,\alpha_2)$ are not conjugate as well.
\end{lemma}
\begin{proof}
The image of an operator of type  $\eqref{r3}$ is one dimensional while the image of an operator from  $\eqref{r4}$ or $\eqref{r5}$ is at least two. Thus, \eqref{r3} can't be conjugate to an operator from $\eqref{r4}$ or $\eqref{r5}$.

Let $R_1$ and $R_2$ be Rota-Baxter operators of types \eqref{r4} and $\eqref{r5}$ respectively. Then $[Im(R_1),Im(R_1)]=sl_2(\mathbb C)$ while $Im(R_2)$ is a solvable algebra. Hence, operators $R_1$ and $R_2$ are not conjugate.

It remains to prove that different pairs of parameters ($\alpha_1$,$\alpha_2$) in \eqref{r5} gives us non-conjugate Rota-Baxter operators.  Let $R$ be a Rota-Baxter operator of type  $\eqref{r5}$, $\varphi$ be an automorphism of $gl_2(\mathbb C))$. Suppose that an operator $R_1=\varphi^{-1}\circ R\circ\varphi$ is again a Rota-Baxter operator of type  $\eqref{r5}$. Let $\varphi(\mathrm{E})=\theta\mathrm{E}$, then

$$
(\varphi^{-1}\circ R\circ\varphi)(\mathrm{E})=(\varphi^{-1}\circ R)(\theta\mathrm{E})=\theta\varphi^{-1}(\lambda_1\mathrm{E}+h)=\lambda \mathrm{E}+ \theta\varphi^{-1}(h).
$$

Therefore,  $\varphi(h)=\theta h$. Then, 
$$
(\varphi^{-1}\circ R\circ\varphi)(h)=(\varphi^{-1}\circ R)(\theta h)=\theta\varphi^{-1}(\alpha_1\mathrm{E}+\alpha_2h)=\alpha_1 \mathrm{E}+ \alpha_2 h.
$$

Hence, the conjugation by an automorphism can't transform $R$ into a Rota-Baxter operator of the same type \eqref{r5} with another pair of parameters $(\alpha_1,\alpha_2)$. \end{proof}


\begin{lemma}
 Operators $\eqref{r6}-\eqref{r12}$ from Lemma 3, including operators of the same type corresponding to different parameters,  are not conjugate. 
\end{lemma}
\begin{proof}
Note that an operator $R_1$ of a type  $\eqref{r6}-\eqref{r81}$ and an operator $R_2$ of a type \eqref{r9}--\eqref{r11} can't be conjugate since they can't be conjugate by an automorphism of type  $\varphi_{\alpha}$ and the induced operators $\overline{R_i}:gl_2(\mathbb C)/Z(gl_2(\mathbb C))\mapsto gl_2(\mathbb C)/Z(gl_2(\mathbb C))$ are not conjugate as operators defined on $g/Z(gl_2(\mathbb C))=sl_2(\mathbb C)$. Similar arguments show that different parameters $t$ give us non-conjugate operators. 


An operator of type \eqref{r12} can't be conjugate to an operator of another type from Lemma 3 since only in \eqref{r12} we have that $Im(R)\subset Z(gl_2(\mathbb C))$ (if $t=0$) and $sl_2(\mathbb C)\subset Im(R)$ (if $t=-1)$. Note that similar arguments show that the case $t=0$ is not conjugate to the case when $t=-1$ in \eqref{r12}.

Consider operators $\eqref{r6}-\eqref{r81}$.  If  $R$ is a Rota-Baxter operator of type $\eqref{r6}$, then for all $x\in sl_2(\mathbb C)$,  $R(x)\in sl_2(\mathbb C)$. Since $sl_2(\mathbb C)$ is $Aut(gl_2(\mathbb C))$-invariant, then $sl_2(\mathbb C)$ is $\varphi^{-1}\circ R\circ\varphi$-invariant. Hence, $R$ can't be conjugate to an operator from  \eqref{r7}, \eqref{r8} and \eqref{r81}. Similar arguments show that an operator from \eqref{r9} can't be conjugate to an operator from \eqref{r10} or \eqref{r11}.

Let $R$ be a Rota-Baxter operator of type \eqref{r7}. Then $R(h)=0$. Let $\varphi$ be an automorphism of $gl_2(\mathbb C)$ such that $\varphi^{-1}\circ R\circ \varphi (h)=\mathrm{E}$. Then $R(\varphi(h))=\theta \mathrm{E}$, $\theta\neq 0$ that is not possible since $\mathrm{tr} (\varphi(h))=0$. Hence, $R$ can't be conjugate to an operator from $\eqref{r8}$ or \eqref{r81}.

Operator \eqref{r8} can't be conjugate to \eqref{r81} since the induced operators on the quotient algebra $gl_2(\mathbb C)/Z(gl_2(\mathbb C)$ are not conjugate (they correspond to different values of the parameter $t$, see the proof of Lemma 3). 

Let $R$ be a Rota-Baxter operator from \eqref{r8} and $\varphi$ be an automorphism of  $gl_2(\mathbb C)$ such that $R_1=\varphi^{-1}\circ R\circ \varphi$ is again a Rota-Baxter operator of type \eqref{r8}. Suppose that $\varphi(\mathrm{E})=\theta \mathrm{E}$. Since $R_1(h)=\mathrm{E}$, then
$R(\varphi(h))=\theta\mathrm{E}$. Note that $\mathrm{tr}(\varphi(h))=0$. Hence, $\varphi(h)=\theta h$. Condition $R_1(e_{21})=0$ implies that $R(\varphi(e_{12}))=0$. Therefore, $\varphi(e_{21})=\delta e_{21}$ which in turn implies that $\varphi(e_{12})=\gamma e_{12}$. From  $[\varphi(h),\varphi(e_{12})]=\varphi([h,e_{12}])$ we get that $\theta=1$.


Suppose that $R_1(e_{12})=-e_{12}+h+\alpha_1 \mathrm{E}$. Then 
$$R(\varphi(e_{12}))=-\varphi(e_{12})+\theta h+\theta \alpha_1 \mathrm{E}=-\gamma e_{12}+h+\alpha_1 \mathrm{E}.
$$
Thus,
$$
\gamma(-e_{12}+h+\alpha\mathrm{E})=-\gamma e_{12}+h+\alpha_1 \mathrm{E}.
$$
This means that $\gamma=1$ and $\alpha=\alpha_1$. In other words, Rota-Baxter operators of type \eqref{r8}, corresponding to different values of the parameter $\alpha$, are not conjugate. 

It remains to consider operators \eqref{r9}--\eqref{r11}. We've already noted that an operator from \eqref{r9} can't be conjugate to an operator from \eqref{r10} or \eqref{r11}.

If $R$ is a Rota-Baxter operator of type \eqref{r10}, then $e_{21}\in ker(R)$. At the same time, if $R_1$ is an operator of type \eqref{r11}, then $ker(R_1)\cap sl_2(\mathbb C)=\{0\}$. Note that for every $\varphi\in Aut(gl_2(\mathbb C))$ we have that $\varphi(e_{21})\in sl_2(\mathbb C)$. Therefore,  $\varphi^{-1}\circ R_1\circ \varphi(e_{21})\neq 0$ for every $\varphi\in Aut(gl_2(\mathbb C))$. Hence, operators from \eqref{r10} and \eqref{r11} lie in different orbits under the action of $Aut(gl_2(\mathbb C))$.

What is left is to prove that different values of the parameter $\alpha$ in \eqref{r11} give non-conjugate Rota-Baxter operators. Let $R$ be a Rota-Baxter operator of type \eqref{r11} and $\varphi$ be an automorphism of $gl_2(\mathbb C)$ such that $R_1=\varphi^{-1}\circ R\circ \varphi$ is again a Rota-Baxter operator from \eqref{r11}. Suppose that $\varphi(\mathrm{E})=\theta \mathrm{E}$. From  $R(e_{21})=R_1(e_{21})=\mathrm{E}$ we get that $\varphi(e_{21})=\theta e_{21}$. From $R_1(e_{12})=-e_{12}$ it follows that $\varphi(e_{12})=\delta_1 e_{12}+\delta_2 h$ for some $\delta_1,\delta_2\in \mathbb C$. Since $R_1(h)=-h+\alpha_1 \mathrm{ E}$ for some $\alpha_1\in \mathbb C$,  we have
\begin{equation}\label{ut3}
\varphi(h)=-R(\varphi(h))+\alpha_1\theta \mathrm{ E}.
\end{equation}

On the other hand, 
\begin{equation}\label{ut4}
\varphi(h)=[\varphi(e_{12}),\varphi(e_{21})]\in \mathbb{C}  h+\mathbb{C} e_{21}.
\end{equation}

Note that $e_{21}\notin Im(R)$. Hence, from \eqref{ut3} and \eqref{ut4} it follows that $\varphi(h)=\gamma h$ for some $\gamma\in \mathbb C$. Since $\varphi$ is an automorphism of $gl_2(\mathbb C)$, then $\gamma=1$. Then \eqref{ut3} implies that $\theta=1$. Therefore,
$$R_1(h)=\varphi^{-1}\circ R\circ \varphi (h)=-h+\alpha \mathrm{E}.$$
That completes the proof.\end{proof}

Lemmas 1-6 imply the following

\begin{theorem}
Let $R$ be a Rota-Baxter operator of weight 1 defined on $gl_2(\mathbb C)$. Then, up to the action of  $Aut(gl_2(\mathbb C))$, $R$ is one of the following:

\begin{gather*}
1.\  R(\mathrm E)=\lambda \mathrm E+e_{12},\ R(h)=R(e_{12})=R(e_{21})=0,\ \lambda\in\mathbb C;
\end{gather*}
\begin{gather*} 
2. R(\mathrm E)=\lambda \mathrm E+e_{12},\ R(h)=-h,\  R(e_{12})=-e_{12},\  R(e_{21})=-e_{21},\ \lambda\in\mathbb C;
\end{gather*}
\begin{gather*}
3.\  R(\mathrm{E})=\lambda \mathrm{E}+h,\ R(h)=0,\ R(e_{12})=R(e_{21})=0,\ \lambda\in\mathbb C ;
\end{gather*}
\begin{gather*}
4. \  R(\mathrm{E})=\lambda \mathrm{E}+h,\ R(h)=-h,\ R(e_{12})=-e_{12},\ R(e_{21})=-e_{21},\ \lambda\in\mathbb C;
\end{gather*}
\begin{gather*}
 5.\  R(\mathrm{E})=\lambda \mathrm{E}+h,\ R(h)=\alpha_1\mathrm{E}+\alpha_2h,\ R(e_{12})=-e_{12},\ R(e_{21})=0,\ \lambda,\alpha_1,\alpha_2\in\mathbb C;
\end{gather*}
\begin{gather*}
6.\  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=0,\ R(e_{12})=-e_{12}+th;\ R(e_{21})=0,\ \lambda\in\mathbb C,\ t\in\{0,1\};
\end{gather*}
\begin{gather*}
7. \  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=0,\ R(e_{12})=-e_{12}+th+E;\ R(e_{21})=0,\ \lambda\in\mathbb C,\ t\in\{0,1\};
\end{gather*}
\begin{gather*}
 8.\  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=E,\ R(e_{12})=-e_{12}+h+\alpha \mathrm{E};\ R(e_{21})=0,\ \lambda,\alpha\in\mathbb C;
 \end{gather*}
\begin{gather*}
9. \  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=E,\ R(e_{12})=-e_{12}+\mathrm{E};\ R(e_{21})=0,\ \lambda\in\mathbb C;
\end{gather*}
\begin{gather*}
10. \  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=th,\ R(e_{12})=-e_{12},\ R(e_{21})=0,\ \lambda\in\mathbb C,\ t\in \mathbb C,\ t\neq 0;
\end{gather*}
\begin{gather*}
 11.\  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=th+\mathrm{E},\ R(e_{12})=-e_{12},\ R(e_{21})=0,\ \lambda\in\mathbb C,\ t\in \mathbb C,\ t\neq 0;
\end{gather*}
\begin{gather*}
 12.\  R(\mathrm{E})=\lambda \mathrm{E},\ R(h)=-h+\alpha \mathrm{E},\ R(e_{12})=-e_{12},\ R(e_{21})=\mathrm{E},\ \lambda,\alpha\in \mathbb C;
\end{gather*}
\begin{gather*}
   13. \ R(\mathrm{E})=\lambda\mathrm{E}, R(h)=th,\ R(e_{12})=te_{12},\ R(e_{21})=te_{21},\ \lambda\in\mathbb C,\ t\in\{0,-1\}.
\end{gather*}

Operators 1-13 (including operators of the same type satisfying different values of parameters) lie in different orbits under the action of the group $Aut(gl_2(\mathbb C))$.   
\end{theorem}



\bigskip
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