None ex.I.6.24.iv
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There is no a suitable point space map in parabolic case for cycle conjugation

Exercise I.6.24.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}{⋋} $ There is a cycle $\cycle{s}{\sigma}$ such that neither map of the parabolic point space $\Space[p]{R}{}$ represents similarity with $\cycle{s}{\sigma}$.

Hint. Consider $\cycle[\tilde]{s}{\sigma}=(1,0,\frac{1}{2},-1)$ and a cycle $\cycle{s}{\sigma}$ passing point $(u,v)$. Then the similarity of $\cycle{s}{\sigma}$ with $\cycle[\tilde]{s}{\sigma}$ passes the point $T(u,v)=(\frac{1+v}{u},\frac{v+v^2}{u^2})$ if and only if either

  • $\cycle{s}{\sigma}$ is a straight line, or
  • $(u,v)$ belongs to $\cycle[\tilde]{s}{\sigma}$ and is fixed by the above map $T$.

That is, the map $T$ of the point space $\Space[p]{R}{}$ serves flat cycles and $\cycle[\tilde]{s}{\sigma}$ but no others. Thus, there is no map of the point space which is compatible with the cycle similarity for an arbitrary cycle.

Solution. We take an inversion in this parabolic cycle

In [1]:
from init_cycle import *
Cp=cycle2D(1,[0,half],-1,e)

            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

The image of the parabolic iversion

In [2]:
Pn=[(1+v)/u,v/u*(1+v)/u]

A cycle passing Pn

In [3]:
Cn=C.subject_to(C.passing(P))
In [4]:
print("Passing original: %s" % bool(Cn.passing(P).subs({sign : 0})))
Latex("Pass image if and only if: $%s$" %\
Cn.cycle_similarity(Cp,es).val(Pn).subs({sign : 0}).normal())

Passing original: True
Out[4]:
Pass image if and only if: $\frac{1}{4} \frac{4 v^{2} k+2 {l} u \sigma_1 v-2 \sigma_1 v k-4 u^{2} v k+2 \sigma_1 n v^{2}+2 {l} u \sigma_1+6 \sigma_1 n v- \sigma_1 k-4 u^{2} \sigma_1 n-2 {l} u^{3} \sigma_1+4 \sigma_1 n-2 u^{2} \sigma_1 n v+ u^{4} \sigma_1 k+4 v k- \sigma_1 v^{2} k}{u^{2}}$

We calculate images of two lines from the grid passing (u,v)

In [5]:
Cu=cycle2D(0,[half,0],u,e).cycle_similarity(Cp)
Cv=cycle2D(0,[0,half],v,e).cycle_similarity(Cp)
Latex("Parabolic inversion maps $(u,v)$ to $(%s, %s)$: %s" %
(Pn[0],Pn[1],\
bool(Cu.subs(sign==0).passing(Pn)) and bool(Cv.subs(sign==0).passing(Pn))))
Out[5]:
Parabolic inversion maps $(u,v)$ to $(\frac{1+v}{u}, \frac{ {(1+v)} v}{u^{2}})$: True

This notebook is a part of the MoebInv notebooks project [2] .

References

  1. 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.

  2. Vladimir V. Kisil, MoebInv notebooks, 2019.

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