A cycle passing the centre of inversion has a straight line as its image¶

Exercise I.6.32 [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}{⋋} $ Check the following:

Let the $\sigma$-implementation of a cycle $\cycle{s}{\sigma}$ pass the $\sigma$-centre of a cycle $\cycle[\tilde]{s}{\sigma}$. Then, the $\sigma$-reflection of $\cycle{s}{\sigma}$ in $\cycle[\tilde]{s}{\sigma}$ is a straight line.
Let two cycles $\cycle{s}{\sigma}$ and $\cycle[\tilde]{s}{\sigma}$ intersect in two points $P$, $P'\in\Space[\sigma]{R}{}$ such that $P-P'$ is not a divisor of zero in the respective number system. Then, there is an inversion which maps the pencil of cycles orthogonal to $\cycle{s}{\sigma}$ and $\cycle[\tilde]{s}{\sigma}$ (see Exercise I.6.10.ii) into a pencil of concentric cycles.

Hint. Make an inversion into a cycle with $\sigma$-centre $P$, then $\cycle{s}{\sigma}$ and $\cycle[\tilde]{s}{\sigma}$ will be transformed into straight lines due to the previous item. These straight lines will intersect in a finite point $P''$ which is the image of $P'$ under the inversion. The pencil orthogonal to $\cycle{s}{\sigma}$ and $\cycle[\tilde]{s}{\sigma}$ will be transformed to a pencil orthogonal to these two straight lines. A CAS calculations shows that all cycles from the pencil have $\sigma$-centre at $P''$.

Solution Cycle C2 defines the inversion, C3 passes the centre of C2.

from init_cycle import *
C2=cycle2D([u,v],e,m)
C3=C.subject_to(C.passing(C2.center()))
"Image of a cycle passing the centre of inversion, is a straight line: %s" %\
(C3.cycle_similarity(C2).get_k()).subs({pow(sign,3) : sign}).is_zero()

            Python wrappers for MoibInv Library
     ---------------------------------------------
Please cite this software as
V.V. Kisil, MoebInv: C++ libraries for manipulations in non-Euclidean geometry, SoftwareX, 11(2020),100385. doi:10.1016/j.softx.2019.100385.
     ---------------------------------------------

Using vector formalism and idx

'Image of a cycle passing the centre of inversion, is a straight line: True'

Straight line passing $(u,-\sigma v)$

P2=[u,-sign*v]
C4=C.subs(k==0)
C6=C4.subject_to(C4.passing(P2))

Another straight line passing $(u,-\sigma v)$

C5=C1.subs(k1==0)
C7=C5.subject_to(C5.passing(P2))

Some more variables are defined for a new cycle.

k0=realsymbol("k0")
l0=realsymbol("l0")
n0=realsymbol("n0")
m0=realsymbol("m0")

A cycle orthogonal to those straight lines

C0=cycle2D(k0,[l0,n0],m0,e)
Cf=C0.subject_to([C0.is_orthogonal(C6),C0.is_orthogonal(C7)])
"Orthogonal cycle have centre at the intersection point: %s" %\
((Cf.center()[0]-P2[0]).is_zero() and (Cf.center()[1]-P2[1]).is_zero())

'Orthogonal cycle have centre at the intersection point: True'

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.

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