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Let $D$ be the Poincaré disk endowed with the metric $g=4\dfrac{dx^2+dy^2}{(1-x^2-y^2)^2}$. I want to find a sample equation for the geodesics between two points $p$ and $q$ in the disk. We know that the geodesics are portions of circles that intersect the boundary of the disk at right angles and the lines through the origin. The problem is that if we transform geodesics from the Hyperbolic space the formulas becomes complicated !!
Is there any sample way to find them ?

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  • $\begingroup$ The statement "The problem is that if we transform geodesics from the Hyperbolic space the formulas becomes complicated!!" makes it sound as if you know parametrizations of hyperbolic geodesics but are hoping for them to be something other than (namely, simpler than) what they are. Can you please clarify what you're asking? $\endgroup$ Apr 7 at 15:51
  • $\begingroup$ A geodesic in the hyperbolic plane has the form $\gamma_{p,T}(t)=cos(t)p+\sin(t)T$ and if we use the stereographic projection from $H \to D$, defined by : $\phi(x, y, z)=\left(\frac{x}{z}, \frac{y}{z}, 0\right)$, we will get $\phi(\gamma(t))=\left(\frac{\cosh (t) x+\sinh (t) u}{\cosh (t) z+\sinh (t) w}, \frac{\cosh (t) y+\sinh (t) v}{\cosh (t) z+\sinh (t) w}\right)$, for $p=(x,y,z)$ and $T=(u,v,w) \in T_p \mathbb H$. $\endgroup$
    – M-S
    Apr 7 at 15:58
  • $\begingroup$ I don't know the formula for two points but I think it would be similar. $\endgroup$
    – M-S
    Apr 7 at 16:00
  • $\begingroup$ Here's one possible approach: Rotate the two points until the two points are equidistant from the $y$-axis. Do another linear fractional transformation to the upper half plane, where I think the calculations might be easier. Find a parameterization or equation of the half circle that passes through the two points. Now use the inverse transformations to bring it all back to the original position of the points. $\endgroup$
    – Deane
    Apr 9 at 20:02
  • $\begingroup$ Do you want an equation for the circle or do you want en expression for the position of the geodesic at time $t\in{\Bbb R}$? $\endgroup$
    – H. H. Rugh
    Apr 13 at 7:05

1 Answer 1

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The easiest is probably to write it down in complex coordinates. A unit-speed geodesic passing through the origin $0\in{\Bbb D}$ at time zero in the direction of $n\in {\Bbb C}$ with $|n|=1$ is given by $$ w_{n}(t) = n \tanh \frac{t}{2}.$$ It is a straight line-segment keeping the angle $\theta$ with the real axis. Now, given $p,q\in{\Bbb D}$, the Möbius transformation: $$ w=R(z) = \frac{z-p}{1-\bar{p}z} \Leftrightarrow z=R^{-1}(w) = \frac{w+p}{1+\bar{p}w}$$ is an isometry that maps $p$ to 0. Let $q'=R(q)=\frac{q-p}{1-\bar{p}q}$ and set $n=q'/|q'|$. To get the unit speed geodesic trough $p$ and $q$ (passing $p$ at time zero) simply take the preimage of $w_n(t)$: $$ z(t) = R^{-1}(w_n(t)).$$ You may here insert the various expressions to get a rather messy looking explicit formula (it becomes a fractional linear map in $\tanh(t/2)$ but with rather complicated coefficients). Depending on what you want to do with the geodesic you might not want to use this explicit formula.

Möbius transformations preserves the collection of disks union lines as well as angles (being conformal) so as you mention the trajectory of $z(t)$ must be part of a circle perpendicular to the unit circle.

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  • $\begingroup$ I really appreciate your help Sir ! I already have a similar formula (I have considered the geodesic $\gamma(t)=\tanh(\frac{t}{2}d)$ with $d$ the hyperbolic distance between $0$ and a the point $q'$ which I suppose it is in the real axis) and get a similar result ! The equation $z(t)=\frac{\tanh(t/2)n+p}{1+\tanh(t/2)\bar pn}$ still messy so I don't know if can we do better ! The goal is to have simple equation for geodesics to find barycentric coordinates in the Poincaré and Klein disk ! $\endgroup$
    – M-S
    Apr 13 at 14:42
  • $\begingroup$ For barycentric coordinates you think of the hyperbolic distances to 3 fixed (non-aligned) points? If so, I am not sure why you need the geodesics? $\endgroup$
    – H. H. Rugh
    Apr 13 at 15:48
  • $\begingroup$ @Rugh, sorry. I don't think I meant to post my comment to your answer. I meant to post it as a comment to the problem itself. $\endgroup$
    – Deane
    Apr 13 at 16:36
  • $\begingroup$ @H.H.Rugh To be honest Sir, even me I have this remark but my teacher insisted me on finding a simple equation for geodesics between $p$ and $q$. $\endgroup$
    – M-S
    Apr 13 at 16:44
  • $\begingroup$ @M-S The equation for the geodesic is simpler than describing the trajectory: The condition is that ${\rm Im}\frac{1}{n} w_n(t) = 0$ or equivalently ${\rm Im} R(z)/R(q) = 0$ which you may develop in $z$ (It becomes an equation for a circle/line as it should). Perhaps this helps? $\endgroup$
    – H. H. Rugh
    Apr 14 at 7:25

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