We define the Moebius transformation with the K subgroup¶
Exercise I.3.13.i [1] $ \newcommand{\scalar}[3][]{\left\langle #2,#3 \right\rangle_{#1}} \newcommand{\Space}[3][]{\mathbb{#2}^{#3}_{#1}} \newcommand{\Space}[3][]{\mathbb{#2}^{#3}_{#1}} \newcommand{\cycle}[3][]{{#1 C^{#2}_{#3}}} \newcommand{\realline}[3][]{#1 R^{#2}_{#3}} \newcommand{\bs}{\breve{\sigma}} \newcommand{\zcycle}[3][]{#1 Z^{#2}_{#3}} \newcommand{\SL}[1][2]{\mathrm{SL}_{#1}(\Space{R}{})} \newcommand{\rs}{\sigma_r} \newcommand{\lvec}[1]{\overrightarrow{#1}} \newcommand{\rmi}{\mathrm{i}} \newcommand{\alli}{\iota} \newcommand{\rme}{\mathrm{e}} \newcommand{\rmd}{\mathrm{d}} \newcommand{\rmh}{\mathrm{j}} \newcommand{\rmp}{\varepsilon} \newcommand{\modulus}[2][]{\left| #2 \right|_{#1}} \newcommand{\sperp}{⋋} $ The derived action of the subgroup (K) is given by: \begin{equation} \label{eq:K-action-derived} K^d_\sigma (u,v)=(1+u^2+\sigma v^2)\partial_u+2uv\partial_v, \qquad \sigma=\alli^2. \end{equation} Hint. Use the explicit formula for Möbius transformation of the components. An alternative with CAS is provided as well.
Solution. Define the K-transformation
from init_cycle import *
[U,V]=clifford_moebius_map(sl2_clifford([cos(t),sin(t),-sin(t),cos(t)],e),[u,v],e)
Derived action is:
Latex("Derived action of the K subgroup: $(%s, %s)$" % \
(U.diff(t).subs({t : 0}), V.diff(t).subs({t : 0})))
This notebook is a part of the MoebInv notebooks project [2] .
References¶
Vladimir V. Kisil. Geometry of Möbius Transformations: Elliptic, Parabolic and Hyperbolic Actions of $SL_2(\mathbb{R})$. Imperial College Press, London, 2012. Includes a live DVD.
Vladimir V. Kisil, MoebInv notebooks, 2019.