None ex.I.5.23.ii
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Power of intersecting cycles and cosine of intersection angle

Exercise I.5.23.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}} $ Check for $\sigma=\pm 1$, that the $(\sigma,\sigma)$-power of two intersecting $\sigma$-cycles is the $\sigma$-cosine of the angle between the tangents to cycles at an intersecting point.

Solution. First we define the tangent of a cycle at a given point using implict derivatives. The routine defines an equation on a cycle to have given tangent at $(0,1)$.

In [1]:
from init_cycle import *
def has_tan(C,u,v,U,V):
    return C.val([u,v]).diff(u)*U+C.val([u,v]).diff(v)*V==0

            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

Some variables to use later.

In [2]:
U=realsymbol("U")
V=realsymbol("V")
U1=realsymbol("U1", "U_1")
V1=realsymbol("V1", "V_1")

Take two cycles both passing $(u,v)$ with given tangents $(U,V)$ and $(U_1,V_1)$.

In [3]:
C0=C.subject_to([C.passing([u,v]),has_tan(C,u,v,U,V)]).normalize_det()
C10=C1.subject_to([C1.passing([u,v]),has_tan(C1,u,v,U1,V1)]).normalize_det()

Just out of curiosity take a look on the explicit formula of C0.

In [4]:
Latex(f"${C0.string()}$")
Out[4]:
$( k \sqrt{\frac{U^{2}}{ k^{2} \sigma^{2} V^{2} v^{2}- k^{2} \sigma U^{2} v^{2}+ n^{2} V^{2}-2 k U^{2} n v+2 k \sigma n V^{2} v- \sigma U^{2} n^{2}}}, {\left(\begin{array}{cc}-\frac{ \sqrt{\frac{U^{2}}{ k^{2} \sigma^{2} V^{2} v^{2}- k^{2} \sigma U^{2} v^{2}+ n^{2} V^{2}-2 k U^{2} n v+2 k \sigma n V^{2} v- \sigma U^{2} n^{2}}} n V}{U}+ k \sqrt{\frac{U^{2}}{ k^{2} \sigma^{2} V^{2} v^{2}- k^{2} \sigma U^{2} v^{2}+ n^{2} V^{2}-2 k U^{2} n v+2 k \sigma n V^{2} v- \sigma U^{2} n^{2}}} u-\frac{ k \sqrt{\frac{U^{2}}{ k^{2} \sigma^{2} V^{2} v^{2}- k^{2} \sigma U^{2} v^{2}+ n^{2} V^{2}-2 k U^{2} n v+2 k \sigma n V^{2} v- \sigma U^{2} n^{2}}} \sigma V v}{U}& \sqrt{\frac{U^{2}}{ k^{2} \sigma^{2} V^{2} v^{2}- k^{2} \sigma U^{2} v^{2}+ n^{2} V^{2}-2 k U^{2} n v+2 k \sigma n V^{2} v- \sigma U^{2} n^{2}}} n\end{array}\right)}^{{symbol50} }, \frac{ \sqrt{\frac{U^{2}}{ k^{2} \sigma^{2} V^{2} v^{2}- k^{2} \sigma U^{2} v^{2}+ n^{2} V^{2}-2 k U^{2} n v+2 k \sigma n V^{2} v- \sigma U^{2} n^{2}}} {( k \sigma U v^{2}-2 k \sigma u V v+ k U u^{2}+2 U n v-2 u n V)}}{U})$

Tangent calculations:

In [5]:
T=V/U
T1=V1/U1

We check the formula $\cos^{-2}t +\mathrm{sign}(\sigma) \tan^{-2} t =1$ in elliptic and hyperbolic cases.

In [6]:
"Cycle product det-normalized cycles gives cosine of the angle in the elliptic case: %s" %\
(pow(C0.cycle_product(C10).normal()/2,-2)+sign*pow((T1-T)/(1-sign*T*T1),2)-1).normal().subs({sign : -1}).normal().is_zero()
Out[6]:
'Cycle product det-normalized cycles gives cosine of the angle in the elliptic case: True'
In [7]:
"Cycle product det-normalized cycles gives cosine of the angle in the hyperbolic case: %s" %\
(pow(C0.cycle_product(C10).normal()/2,-2)+sign*pow((T1-T)/(1-sign*T*T1),2)-1).normal().subs({sign : 1}).normal().is_zero()
Out[7]:
'Cycle product det-normalized cycles gives cosine of the angle in the hyperbolic case: True'

In parabolic cases we just output the value.

In [8]:
Latex("Par: $%s$" % pow(C0.cycle_product(C10).normal()/2,-2).subs({sign : 0}).normal())
Out[8]:
Par: $\frac{ U_1^{2} U^{2}}{{( U_1 {k'} U \sqrt{-\frac{U_1^{2}}{2 U_1^{2} {k'} v {n'}- V_1^{2} {n'}^{2}}} n v \sqrt{\frac{U^{2}}{ n^{2} V^{2}-2 k U^{2} n v}}- V_1 \sqrt{-\frac{U_1^{2}}{2 U_1^{2} {k'} v {n'}- V_1^{2} {n'}^{2}}} n V \sqrt{\frac{U^{2}}{ n^{2} V^{2}-2 k U^{2} n v}} {n'}+ U_1 k U \sqrt{-\frac{U_1^{2}}{2 U_1^{2} {k'} v {n'}- V_1^{2} {n'}^{2}}} v \sqrt{\frac{U^{2}}{ n^{2} V^{2}-2 k U^{2} n v}} {n'})}^{2}}$
In [9]:
Latex(f"Par: ${(pow(C0.cycle_product(C10).normal()/2,-2)+sign*pow((T1-T)/(1-sign*T*T1),2)-1).normal().subs({sign : 0}).normal()}$")
Out[9]:
Par: $\frac{2 U_1^{4} {k'} U^{2} n^{2} V^{2} v {n'}+ U_1^{4} k^{2} U^{4} v^{2} {n'}^{2}-2 U_1^{3} k V_1 U^{3} n V v {n'}^{2}-2 U_1^{3} {k'} V_1 U^{3} n^{2} V v {n'}+ U_1^{4} {k'}^{2} U^{4} n^{2} v^{2}-2 U_1^{4} k {k'} U^{4} n v^{2} {n'}+2 U_1^{2} k V_1^{2} U^{4} n v {n'}^{2}}{ {( V_1^{2} {n'}-2 U_1^{2} {k'} v)} {(2 k U^{2} v- n V^{2})} {( U_1 {k'} U \sqrt{-\frac{U_1^{2}}{2 U_1^{2} {k'} v {n'}- V_1^{2} {n'}^{2}}} n v \sqrt{\frac{U^{2}}{ n^{2} V^{2}-2 k U^{2} n v}}- V_1 \sqrt{-\frac{U_1^{2}}{2 U_1^{2} {k'} v {n'}- V_1^{2} {n'}^{2}}} n V \sqrt{\frac{U^{2}}{ n^{2} V^{2}-2 k U^{2} n v}} {n'}+ U_1 k U \sqrt{-\frac{U_1^{2}}{2 U_1^{2} {k'} v {n'}- V_1^{2} {n'}^{2}}} v \sqrt{\frac{U^{2}}{ n^{2} V^{2}-2 k U^{2} n v}} {n'})}^{2} n {n'}}$

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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