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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.

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  • $\begingroup$ The question needs to include the definition of this $R_{\gamma,\theta}$, as defined in the paper. $\endgroup$
    – Ben McKay
    Commented Aug 16, 2021 at 11:43
  • $\begingroup$ Thank you. $R_{\left(\gamma,\theta\right)}$ is an action of $SE\left(2\right)$ with the translation $\gamma$ and the rotation $\theta$. $\endgroup$
    – Chivul
    Commented Aug 17, 2021 at 2:28
  • $\begingroup$ Since the automated sizing of $\mathrm{SE}\left(2\right)$ \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. $\endgroup$
    – LSpice
    Commented Aug 17, 2021 at 2:52
  • $\begingroup$ Thank you for your help, @LSpice <3 $\endgroup$
    – Chivul
    Commented Aug 17, 2021 at 3:20
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    $\begingroup$ @Vichakh: The map $R$ is acting on an infinite dimensional space of functions, not on the plane, so $R$ is not $\mathcal{A}$. You might define $R$ for the benefit of mathoverflow users. $\endgroup$
    – Ben McKay
    Commented Aug 17, 2021 at 9:25

1 Answer 1

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The representation $R$ can be seen as a map: $$R:\operatorname{SE} (2)\to\operatorname{Diff}(\mathbb R^2) $$

and the derivations above correctly allows to compute the corresponding infinitesimal action

$$\mathfrak{se} (2)\to \mathfrak X(\mathbb R^2) $$

which is an injective map having as image the three-dimensional vector space of vector fields on the plane spanned by $J_x$, $J_y$ and $J_\phi$. Here the Lie bracket is just the commutator of vector fields $$ [X, Y] =X\circ Y-Y\circ X$$ and $\partial_x$ (resp. $y$) is just the vector field of constant horizontal (resp. vertical) translations that acts, as derivations on functions, exactly as the partial derivative along $x$ (resp. $y$).

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  • $\begingroup$ No, the representation is not a map to diffeomorphisms. Every representation is a morphism to linear transformations of some vector space. In this problem, the vector space is a space of square integrable functions, as you seen in the paper which the OP references in his question. $\endgroup$
    – Ben McKay
    Commented Aug 17, 2021 at 9:19
  • $\begingroup$ Can you explain more about the map $R:\operatorname{SE}(2)\rightarrow \operatorname{Diff}(\mathbb{R}^2)$? I have not yet caught up with the idea. $\endgroup$
    – Chivul
    Commented Aug 17, 2021 at 11:49
  • $\begingroup$ @Vichakh: the idea is that every rigid motion of the plane is a diffeomorphism of the plane, so the group $SE(2)$ of orientation preserving rigid motions of the plane is a group of diffeomorphisms of the plane. So Chik67 is suggesting that the map $R$ is the inclusion map. But he is wrong, that is the map $\mathcal{A}$. The map $R$ is the action now indicated in the question. $\endgroup$
    – Ben McKay
    Commented Aug 17, 2021 at 13:47

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