Elliptic centre of the image of a zero radius cycle with the centre (u,v)¶
Exercise I.5.18.v [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}{})} $ The e-centre of the transformation $g\zcycle{s}{\bs}g^{-1}$, $g\in\SL$ of the $\bs$-zero-radius cycle $\zcycle{s}{\bs}$ (I.5.8) coincides with the Möbius action $g\cdot z$ in $\Space{R}{\bs}$, where $z$ is the e-centre of $\zcycle{s}{\bs}$.
Hint. The result can be obtained along the lines from Subsection I.4.4.1.
Solution. Here is computer-added proof. We define the P3
as the centre of the transformed zero-radius cycle.
from init_cycle import *
P3=Z.sl2_similarity(a,b,c,d,e).center(es)
On the other hand we calculate Möbius transformation of the point $(u,v)$
P2=clifford_moebius_map(sl2_clifford(a,b,c,d,e),Z.center(es),e)
Now we check are they the same in the elliptic
"Elliptic centre of zero radius cycle is transformed by the Moebius map: %s" %\
((P3[0]-P2[0]).subs({sign1 : -1}).normal().is_zero() \
and (P3[1]-P2[1]).subs({sign1 : -1}).normal().is_zero())
and hyperbolic cases:
"Hyperbolic centre of zero radius cycle is transformed by the Moebius map: %s" %\
((P3[0]-P2[0]).subs({sign1 : 1}).normal().is_zero() \
and (P3[1]-P2[1]).subs({sign1 : 1}).normal().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.