# geodesics in hyperbolic space

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?

• Have you tried to check what it means for the velocity vector to have constant length in the hyperbolic metric? – Ted Shifrin Feb 27 '19 at 20:46
• 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 – Tiff Feb 27 '19 at 21:15
• So, does this check with $\tanh(at)$? – Ted Shifrin Feb 27 '19 at 21:18
• Oh, you're right, $\tanh(at)$ does make it constant! Thanks! @TedShifrin – Tiff Feb 27 '19 at 21:39

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 $$\frac{z+i}{iz+1}$$ $$\frac{iz+1}{z+i}$$