# Geodesic Uniqueness in the Hyperbolic Plane

I am studying Hyperbolic Geometry. At this part, I have proved that semicircles and straight lines orthogonals to the real axis are geodesics in the hyperbolic plane. But how I proof that this geodesics are uniques? That it does not exist others geometric places between two different points that minimizes the hyperbolic length?

I have seen a topic related to this, but I do not know nothing about Riemannian Geometry. If someone could indicate to me some references to this proof(or the idea behind) I will be very grateful!

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There is a general fact in Riemannian geometry which follows from "standard" existence and uniqueness theorems in the study of ODEs. Mainly, given any point $p$ in a Riemannian manifold $M$ and given any $v\in T_pM$, there is a unique geodesic defined for some open neighborhood of time around 0 starting at $p$ in the direction of $v$.

Assuming this, it's not too hard to show that you've found all the geodesics. You just need to verify that for any point $p$ in the hyperbolic plane and tangent vector, either the vector points up/down (so a vertical line geodesic has $v$ as tangent vector), or some semicircle through $p$ hitting the $x$-axis perpendicularly does.

Finally, to show there is a unique geodesic between any two points, one argues as follows:

Given $p$ and $q$, if $q$ and $p$ have the same $x$ coordinate, a vertical line passes through them. Further, no semicircle hitting the $x$-axis perpendicularly can go through both $p$ and $q$ since the semicircle is the graph of a function, so passes the vertical line test.

Next, given $p$ and $q$ with different $x$ coordinates, it's clear that a vertical line can't connect them, so we need only find all semicircles passing through both perpendicular to the $x$-axis. Algebraically, proceed by noting the equation of such a circle is $(x-a)^2 + y^2 = r^2$ for some $a$ and $r$. Plugging in $(x,y) = (p_1,p_2)$ and $(x,y) = (q_1,q_2)$ respectively and setting the two equal to each other (since they both equal $r^2$), one gets $$(p_1-a)^2 + p_2^2 = (q_1-a)^2 + q_2^2.$$

By simplifying this, one sees the $a^2$ cancels out and so one gets an equation linear in $a$, so $a$ is uniquely determined. (The assumption that $p_1\neq q_1$ means that the equation has the linear piece). Knowing $a$ easily implies that $r$ is uniquely determined (keeping in mind that $r \geq 0$). This shows that there is a unique semicircle perpendicular to the $x$-axis going through $p$ and $q$.

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I think I followed your argument in the first two paragraphs. But where I use the fact that the metric of this space is the hyperbolic one? I can proceed like that in the Euclidian plane and obtain the same results, but semicircles are not geodesic in this space. –  Pierre May 21 '11 at 11:25
ignore the previous post, now I understood all, thank you. I've already proved that this curves are geodesics, so if I can construct, for a given point p and tangent vector v, a vertical line or semicircle orthogonal to the real axis, the theorem that you told me guarantee that, necessarily, those are the unique geodesics. And I already have calculate the radius and center of the semicircle and, in fact, they are unique. Thank you a lot! p.s. You can give me some references to this theorem that you told me? Thank you in advance! –  Pierre May 21 '11 at 11:40
I don't have any references currently available, but if I remember to, I can look it up at school on Monday. –  Jason DeVito May 21 '11 at 23:13
You can argue "geometrically" as follows: For two points $p$, $q$ on the imaginary axis it is pretty obvious that the segment $[p,q]$ is the unique shortest curve connecting $p$ and $q$. Now if $p$, $q\in H$ are arbitrary there is a Moebius transformation $T: H\to H$ sending $p$ and $q$ to two points $p'$, $q'$ on the imaginary axis. Since $T$ is a hyperbolic isometry it follows that $p$ and $q$ have a unique shortest connection, and since $T$ (now considered as a map $\bar {\mathbb C}\to \bar {\mathbb C}$) maps circles to circles and preserves angles between curves it follows that this shortest connection lies on a circle intersecting the real axis orthogonally.