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Lemma: eplicit formula for the distance between two points

Lemma I.7.5 [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}{⋋} $ The {distance} between two points $P=u+\alli v$ and $P'=u'+\alli v'$ in the elliptic or hyperbolic spaces is \begin{equation} d_{\sigma,\bs}^2(P, P') = \frac{ \bs ((u-u')^2-\sigma(v- v')^2) +4(1-\sigma\bs) v v'} {(u- u')^2 \bs-(v-v')^2} ((u-u')^2 -\sigma(v- v')^2), \end{equation} and, in parabolic space, it is (see [Yaglom79, (5), p.~38]) \begin{equation} d_{p,\bs}^2(y, y') = (u-u')^2. \end{equation}

Solution. Define a cycle template

In [1]:
from init_cycle import *
C10 = cycle2D(numeric(1), [l, n], m, 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

Specialise by the condition passing two points

In [2]:
C10 = C10.subject_to([C10.passing(P), C10.passing(P1)], [m, n, l])
Latex(f"${C10.string()}$")
Out[2]:
$(1, {\left(\begin{array}{cc}{l}&-\frac{1}{2} \frac{{u'}^{2}}{v-{v'}}-\frac{ {l} u}{v-{v'}}+\frac{1}{2} \frac{ \sigma {v'}^{2}}{v-{v'}}-\frac{1}{2} \frac{ v^{2} \sigma}{v-{v'}}+\frac{1}{2} \frac{u^{2}}{v-{v'}}+\frac{ {u'} {l}}{v-{v'}}\end{array}\right)}^{{symbol113} }, \frac{ v \sigma {v'}^{2}+ {v'} u^{2}+2 v {u'} {l}- v {u'}^{2}- v^{2} \sigma {v'}-2 {l} {v'} u}{v-{v'}})$

Define square of the diameter

In [3]:
D = 4*C10.radius_sq(es)

Find l providing the extremal value of the diameter

In [4]:
S=lsolve([D.diff(l) == 0], [l])
Latex("Extremum is achived for $%s$" % S[l])
Out[4]:
Extremum is achived for $-\frac{1}{2} \frac{ {u'} u^{2} \sigma_1- v^{2} {u'} \sigma \sigma_1- u^{3} \sigma_1- \sigma {v'}^{2} u \sigma_1+ {u'}^{2} u \sigma_1-2 v {u'} {v'}- {u'}^{3} \sigma_1+ v^{2} \sigma u \sigma_1+2 {v'}^{2} u-2 v {v'} u+ {u'} \sigma {v'}^{2} \sigma_1+2 v^{2} {u'}}{ u^{2} \sigma_1-v^{2}+ {u'}^{2} \sigma_1-2 {u'} u \sigma_1+2 v {v'}-{v'}^{2}}$

Find the extremal value

In [5]:
Len_c = D.subs(S).normal()

Write a regularised expresion for it

In [6]:
dist = (sign1*(pow(u-u1,2)-sign*pow(v-v1,2))+4*(1-sign*sign1)*v*v1)*(pow(u-u1,2)\
-sign*pow(v-v1,2))/(pow(u-u1,2)*sign1-pow(v-v1,2))
Latex("Distance between $(u,v)$ and $(u_1,v_1)$ is $%s$: $%s$" %\
(dist, (Len_c-dist).normal().is_zero()))
Out[6]:
Distance between $(u,v)$ and $(u_1,v_1)$ is $-\frac{ {( {( {(v-{v'})}^{2} \sigma-{({u'}-u)}^{2})} \sigma_1+4 v {v'} {(-1+ \sigma \sigma_1)})} {( {(v-{v'})}^{2} \sigma-{({u'}-u)}^{2})}}{{(v-{v'})}^{2}- {({u'}-u)}^{2} \sigma_1}$: $True$

Parabolic case, the extremum is at l=(u_u1)/2

In [7]:
Len_c = D.subs({sign : 0, l : (u+u1)*half}).normal()
Latex("Value at the middle point (parabolic point space): $%s$" % Len_c)
Out[7]:
Value at the middle point (parabolic point space): ${u'}^{2}-2 {u'} u+u^{2}$

The similar check for the case v=v', the the value of l is now fixed to $(u+u_1)/2$. Specialise by the condition passing two points

In [8]:
C10 = cycle2D(numeric(1), [l, n], m, e)
C10 = C10.subject_to([C10.passing(P), C10.passing([u1,v])])

Define square of the diameter

In [9]:
D = 4*C10.radius_sq(es)

Find l providing the extremal value of the diameter

In [10]:
S=lsolve([D.diff(n) == 0], [n])
Latex("Extremum is achived for $%s$" % S[n])
Out[10]:
Extremum is achived for $-\frac{v}{\sigma_1}$

Find the extremal value

In [11]:
Len_c = D.subs(S).normal()

Write a regularised expresion for it

In [12]:
dist = (sign1*(pow(u-u1,2)-sign*pow(v-v1,2))+4*(1-sign*sign1)*v*v1)*(pow(u-u1,2)\
-sign*pow(v-v1,2))/(pow(u-u1,2)*sign1-pow(v-v1,2))
Latex("Distance between $(u,v)$ and $(u1,v)$ is $%s$: %s" %\
(dist.subs({v1: v}), (Len_c-dist.subs({v1 : v})).normal().is_zero()))
Out[12]:
Distance between $(u,v)$ and $(u1,v)$ is $\frac{ {({u'}-u)}^{2} \sigma_1-4 v^{2} {(-1+ \sigma \sigma_1)}}{\sigma_1}$: 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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