When I read the paper Universal approximations of invariant maps by neural networks of Dmitry Yarotsky, it happens on page 36 that he used some concepts about the representation of Lie algebra of the Lie group $\operatorname{SE}(2)$.
Describe rigid motions of $\mathbb{R}^2$ by identifying it with $\mathbb{C}$. An element in $\operatorname{SE}(2)$ can be written as $\left(\gamma,\theta\right)=\left(x+iy,e^{i\phi}\right)$ with some $x,y\in\mathbb{R}$ and $\phi\in\left[0,2\pi\right)$. The action of $\operatorname{SE}(2)$ on $\mathbb{R}^2\cong\mathbb{C}$:
\begin{equation*}
\mathcal{A}_{\left(x+iy,e^{i\theta}\right)}z=x+iy+e^{i\theta}z,\quad z\in \mathbb{C}.
\end{equation*}
Consider the generators of the representation:
\begin{equation*}
J_x=i\lim_{\delta x\to 0}\dfrac{R_{\left(\delta x,1\right)}-1}{\delta x}, \quad J_y=i\lim_{i\delta y\to 0}\dfrac{R_{\left(\delta y,1\right)}-1}{\delta x}, \quad J_\phi=i\lim_{\delta \phi\to 0}\dfrac{R_{\left(0,e^{i\delta \phi}\right)}-1}{\delta \phi}
\end{equation*}
where $R_{\left(\gamma,\theta\right)}$ is the action of $\operatorname{SE}(2)$ given by
$$R_{\left(\gamma,\theta\right)}\Phi=\Phi\circ\mathcal{A}^{-1}.$$
The generators can be explicitly written as
\begin{equation*}
J_x=-i\partial_x, \quad J_y=-i\partial_y,\quad J_\phi=-i\partial_\phi=-i\left(x\partial_y-y\partial_x\right)
\end{equation*}
and obey the commutation relations
\begin{equation}\label{eq41}
\left[J_x,J_y\right]=0,\quad \left[J_x,J_\phi\right]=-iJ_y, \quad \left[J_y,J_\phi\right]=iJ_x.
\end{equation}
I am a newbie in Lie representation. Can you explain to me the definition of $J_x, \partial_x$, the Lie brackets, and how these equations hold here? Or can you give me some books/papers defining these concepts? I looked upon the internet about the representation of Lie algebra but met nothing like these.
\left(2\right)
doesn't seem necessary here, and creates weird spacing between the name of the group and the parenthesis, I edited it to $\operatorname{SE}(2)$\operatorname{SE}(2)
, which looks better to me. I hope that was all right. $\endgroup$