We verify the expression of the parabolic Cayley transform on the cycles¶
Exercise I.10.9 [1] Let $\newcommand{\Space}[3][]{\mathbb{#2}^{#3}_{#1}} \newcommand{\cycle}[3][]{{#1 C^{#2}_{#3}}} \newcommand{\realline}[3][]{#1 R^{#2}_{#3}} \newcommand{\bs}{\breve{\sigma}}$ $\cycle{s}{a}$ be a cycle in $\Space[\sigma]{R}{}$. Check that:
(e,h) In the elliptic or hyperbolic cases, the Cayley transform of the cycle $\cycle{}{\sigma}$ is $\realline{}{\sigma}\cycle[\hat]{}{\sigma }\cycle{}{\sigma}\cycle[\hat]{}{\sigma }\realline{}{\sigma}$, i.e. the composition of the similarity (I.6.10) by the cycle $\cycle[\hat]{s}{\sigma }=(\sigma ,0,1,1)$ and the similarity by the real line (see the first and last drawings in Fig.I.10.2.
(p) In the parabolic case, the Cayley transform maps a cycle $(k,l,n,m)$ to the cycle $(k-2\bs n, l, n ,m-2 n)$.
Hint We can follow a similar path to the proof of Theorem I.4.13. Alternatively, for the first part, we notice that the matrix $Y_\sigma $ of the Cayley transform and the FSCc matrix of the cycle $\cycle[\hat]{s}{\sigma }$ are different by a constant factor. The reflection in the real line compensates the effect of complex conjugation in the similarity (I.6.10).
Solution. Take the Cayley transform of a point...
from init_cycle import *
P0=(u+I*v-I)/(-I*(u+I*v)+1)
... and its Möbius transformation
Pc=clifford_moebius_map(one, -e1, sign1*e1, one, P, e)
Compare both outputs:
print("Clifford calculations coinsides with complex numbers: %s" %\
((P0+P0.conjugate()-2*Pc[0]).subs({sign : -1,sign1 : -1}).normal().is_zero() and
(P0-P0.conjugate()-2*I*Pc[1]).subs({sign : -1,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.