Check that the image of zero-radius cycle is a zero-radius cycle¶
Exercise I.6.24.ii [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}} $ The image $\zcycle[\tilde]{s}{\sigma }=\cycle{s_2}{\sigma} \overline{\zcycle{s_1}{\sigma}}\cycle{s_2}{\sigma}$ of a $\sigma$-zero-radius cycle $\zcycle[\tilde]{s_1}{\sigma}$ under the cycle similarity is a $\sigma$-zero-radius cycle $\zcycle[\tilde]{s_1}{\sigma}$. The $(s_1s_2)$-centre of $\zcycle[\tilde]{s}{\bs}$ is the linear-fractional transformation of the $(s_2/s_1)$-centre of $\zcycle{s}{\bs}$.
Solution For the conjugated cycle Zn
of a zero-radius cycle Z
we check that that Zn
is a zero-radius as well.
from init_cycle import *
Zn=cycle2D(Z.cycle_similarity(C,e,sign_mat1,sign_mat2,pow(sign_mat1,-1).evalm()))
"Conjugation of a zero-radius cycle is a zero-radius cycle: %s" % \
Zn.hdet(e,sign_mat1).normal().subs({pow(s1,2) : 1},subs_options.algebraic)\
.normal().is_zero()
P
is the h-centre of the cycle Z
.
P=Z.center(diag_matrix([-1,1]))
Then we calculate the library's cycle::mooebius_map()
of P
defined by FSCc matrix of a generic cycle C
. It coincides with the linear-fractional transformation defined by this matrix explicitly.
PM=C.moebius_map(P, e, (sign_mat*sign_mat1).evalm())
For the Möbius transformation we are using the conjugated value PN
of P
, which is the elliptic centre.
PN=P[0] - I*P[1]
"Cycle generates Moebius map: %s" %\
(PM[0] + I*PM[1]-((l+I*s*s1*n)*PN-m)/(k*PN+(-l+I*s*s1*n))).\
subs({sign : -1}).normal().is_zero()
Now, Pc
is the $s_2 s_1$-centre of Zn
and Pc1
is the linear-fractional map defined by the cycle C
. For vector and paravector formalism the sign in calculations need to be adjusted.
Pc=Zn.center(diag_matrix([-1,-s2*s1]))
if (UseVectors):
Pc1=C.moebius_map(Z.center(diag_matrix([-1,-s2/s1])))
else:
Pc1=C.moebius_map(Z.center(diag_matrix([-1,s2/s1])))
Now we check the statement, that is Pc
and Pc1
are the same for the non-zero values of s
and s1
.
"s1*s2-center of the similar cycle is the Moebius transformation of s2/s1-centre: %s" % \
((Pc1[0]-Pc[0]).normal().subs({pow(s1,2) : 1, pow(s2,2) : 1}, subs_options.algebraic).is_zero() \
and (Pc1[1]-Pc[1]).normal().subs({pow(s1,2) : 1, pow(s2,2) : 1}, subs_options.algebraic).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.