Power of point through the cycle inner product¶
Exercise I.5.21 [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 $\zcycle{}{\sigma}$ be the zero-radius cycle defined by $z=u-\alli\sigma v$, that is, with the $\sigma$-centre at the point $(u, v)$. Show that, in the elliptic and hyperbolic (but not parabolic) cases, a power of a point, $(u,v)\in\Space[\sigma]{R}{}$ with respect to a cycle $\cycle{}{\sigma }$, is equal to the cycle product $\scalar{\zcycle{}{\sigma }}{\cycle{}{\sigma }}$, where both cycles are $k$-normalised.
Solution. We will verify it in a loop:
from init_cycle import *
case=["E", "P", "H"]
Zn=cycle2D([u,-sign1*v],e,0)
for i in range(0,3,2):
print("Power of point is inner product in %s case: %s" % \
(case[i], (C.val(P)-C.cycle_product(Zn,e).subs({sign1 : sign})).subs({sign : i-1}).is_zero()))
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.