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Let $M$ be the Poincare ball model of the Hyperbolic space, and let $\zeta \in T_0M$. In my lecture notes it is claimed that $$c(t)=\tanh(\Vert \zeta \Vert t )\zeta/\Vert \zeta \Vert$$ is the geodesic that satisfies the initial conditions $c(0)=0$ and $c'(0)=\zeta$.

I know that lines through the origin are geodesics, and this is clearly a line through the origin that also satisfies the initial conditions. But my question is, where does this particular parametrization come from? How can I verify if it is correct?

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  • $\begingroup$ Have you tried to check what it means for the velocity vector to have constant length in the hyperbolic metric? $\endgroup$ – Ted Shifrin Feb 27 '19 at 20:46
  • $\begingroup$ Hm. I'm probably doing it wrong. If we just start by $c(t)=r(t)\zeta$, the velocity vector is $r'(t) \zeta$. Plugged in to the metric this gives me $(2r'(t)/(1-r(t)^2))^2$, which should be constant...? @TedShifrin $\endgroup$ – Tiff Feb 27 '19 at 21:15
  • $\begingroup$ So, does this check with $\tanh(at)$? $\endgroup$ – Ted Shifrin Feb 27 '19 at 21:18
  • $\begingroup$ Oh, you're right, $\tanh(at)$ does make it constant! Thanks! @TedShifrin $\endgroup$ – Tiff Feb 27 '19 at 21:39
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Begin with the two dimensional upper half plane. One type of geodesic is, with constant $A,$ $$ x = A, \; \; y = e^t $$

The other type is a semicircle, now with constant $B > 0,$ $$ x = A + B \tanh t \; , \; \; y = B \operatorname{sech} t $$

The simplest way to map these, back and forth, to the unit disc is to regard the $y$ direction as imaginary, then use the Moebius transformations $$ \frac{z+i}{iz+1} $$ $$ \frac{iz+1}{z+i} $$

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