Cycle product of two zero-radius cycles¶
Exercise I.5.17.iv [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}{⋋} $ Let $\zcycle{s}{\bs}$ and $\zcycle[\tilde]{s}{\bs}$ be two \CAS{ex-zero-radius-product}% $k$-normalised $\bs$-zero-radius cycles with e-centres $(u_0,v_0)$ and $(u_1,v_1)$. Then \begin{equation} \label{eq:zero-cycle-product} \scalar{\zcycle{s}{\bs}}{\zcycle[\tilde]{s}{\bs}}= (u_0-u_1)^2-\bs(v_0-v_1)^2. \end{equation}
Solution. A direct check
from init_cycle import *
Latex("Cycle product of $%s$ and $%s$ is:\n $%s$" %\
(Z.string(), Z1.string(), Z.cycle_product(Z1,es,sign_mat)))
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.