None ex.I.7.23
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Check basic properties of the infinitesimal radius cycle

Exercise I.7.23 [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}} $ Let $\bs$ and $\rs$ be two metric signs and let a point $(u_0,v_0)\in\Space[p]{R}{}$ with $v_0>0$. Consider a cycle $\cycle{s}{\bs}$ defined by \begin{equation} \label{eq:inf-cycle} \cycle{s}{\bs}=(1,\, u_0,\, n,\, u_0^2 -\rs n^2+\epsilon^2), \end{equation} where \begin{equation} \label{eq:inf-cycle-n-value} n=\left\{ \begin{array}{ll} \displaystyle \frac{v_0-\sqrt{v_0^2-(\rs-\bs)\epsilon^2}}{\rs-\bs},&\text{ if } \rs\neq \bs,\\ \displaystyle \frac{\epsilon^2}{2v_0},&\text{ if } \rs=\bs. \end{array}\right. \end{equation} Then,

  1. The point $(u_0, v_0)$ is $\rs$-focus of the cycle.
  2. The square of the $\bs$-radius is exactly $-\epsilon^2$, i.e. defines an infinitesimal-radius cycle.
  3. The {focal length} of the cycle is an infinitesimal number of order $\epsilon^2$.

Hint. Combining two quadratic equations (one defines the squared $\bs$-radius, another---$v$-coordinate of the focus), we found that $n$ satisfies the equation: \begin{equation} (\rs-\bs)n^2-2v_0n+\epsilon^2=0. \end{equation} Moreover, only the root from of the quadratic case $\rs-\bs\neq 0$ gives an infinitesimal focal length. Then, we can find the $m$ component of the cycle. The answer is also supported by CAS calculations.

Solution. We define usual substitutions for infinitesimal cycles simplification.

In [1]:
from init_cycle import *
sign_cube = {pow(sign, 3) : sign}
sign1_cube = {pow(sign1, 3) : sign1}
vp=possymbol("vp")

            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

This is a procedure which checks the statements from the exercise

In [2]:
def infinitesimal_calculations(nval):
    C10 = cycle2D(1, [u, n],  pow(u,2)-pow(n,2)*sign1+pow(epsilon,2), e).subs(nval)
    display(Latex("Checking infinitesimal cycle: $%s$" % C10.string()))
    display(Latex("  Square of radius of the infinitesimal cycle is: $%s$" %\
    C10.radius_sq(es).subs(sign_cube, subs_options.algebraic)\
    .subs(sign1_cube, subs_options.algebraic).normal()))
    display(Latex("  Focus of infinitesimal cycle is: $%s$" % C10.focus(er,True).subs(nval).normal()))
    return Latex("  Focal length is: $%s$" % C10.focal_length().series(relational(epsilon, 0, eq), 3).normal())

The check for the first type of infinitesimal cycles

In [3]:
infinitesimal_calculations({n : (vp-(pow(pow(vp,2)-pow(epsilon,2)*(sign2-sign1),half)))/(sign2-sign1)})
Checking infinitesimal cycle: $(1, {\left(\begin{array}{cc}u&\frac{\sqrt{ {(\sigma_1-\sigma_2)} \epsilon^{2}+vp^{2}}-vp}{\sigma_1-\sigma_2}\end{array}\right)}^{{symbol114} }, u^{2}+\epsilon^{2}-\frac{ \sigma_1 {(\sqrt{ {(\sigma_1-\sigma_2)} \epsilon^{2}+vp^{2}}-vp)}^{2}}{{(\sigma_1-\sigma_2)}^{2}})$
Square of radius of the infinitesimal cycle is: $- \epsilon^{2}$
Focus of infinitesimal cycle is: $\left(\begin{array}{c}u\\vp\end{array}\right)$
Out[3]:
Focal length is: ${(\frac{1}{4} \frac{1}{vp})} \epsilon^{2}+\mathcal{O}(\epsilon^{3})$

The check for the parabolic focus case.

In [4]:
infinitesimal_calculations({n : pow(epsilon,2)/2/vp, sign2 : sign1})
Checking infinitesimal cycle: $(1, {\left(\begin{array}{cc}u&\frac{1}{2} \frac{\epsilon^{2}}{vp}\end{array}\right)}^{{symbol234} }, u^{2}+\epsilon^{2}-\frac{1}{4} \frac{ \sigma_1 \epsilon^{4}}{vp^{2}})$
Square of radius of the infinitesimal cycle is: $- \epsilon^{2}$
Focus of infinitesimal cycle is: $\left(\begin{array}{c}u\\vp\end{array}\right)$
Out[4]:
Focal length is: ${(\frac{1}{4} \frac{1}{vp})} \epsilon^{2}$

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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