# Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint

If we let $\mathbb{H}^2$ be the hyperbolic plane and we let $\gamma_1,\gamma_2$ be geodesics which do not intersect. I have a question which asks me to show that either $\gamma_1$ and $\gamma_2$ have a unique common perpendicular or that they have a common endpoint in $S_\infty$ where we define $S_\infty$ to be the boundary of $\mathbb{H}$ and $\mathbb{H}$ to be the interior of the unit disc $D$ in $\mathbb{R}$

I am not really sure how to approach this?

Thanks for any help

HINT: Are you using the half-plane model? If so, without loss of generality, take $\gamma_1$ to be a vertical ray and $\gamma_2$ to be a disjoint semicircle centered on the real axis. Now, you want another such semicircle perpendicular to them both. Can you use basic algebra/geometry to find it?

• No I am using the interior of the unit disc in $\mathbb{R}^2$ sorry I should have mentioned that I will edit my question May 19, 2014 at 17:51
• A bit more awkward, but same idea. Reduce to the case where $\gamma_1$ goes through the origin. May 19, 2014 at 18:03
• Ok so I say that $\gamma_1$ is the horizontal line through the origin and take $\gamma_2$ such that they don't share endpoints on $S_\infty$. Now I need to find geodesic intersecting them both perpendicularly. So any vertical line will intersect $\gamma_1$ perpendicularly can I then show that it intersects $\gamma_2$ perpendicularly at some point? May 19, 2014 at 18:43
• No, you need to use another semicircle. This common perpendicular needs to be a geodesic as well. So you need to solve for a semicircle centered on $S_\infty$ that will be perpendicular to both. It is easy in the half-plane model to solve by algebra. Here you might want to resort to an intermediate value theorem argument, rather than doing it explicitly. I don't know the flavor of your course. May 19, 2014 at 18:48
• My geodesics are circles that intersect the boundary perpendicularly so I was being stupid taking straight line not through the center as it does not intersect perpendicularly sorry. So do I take $\gamma_3$ which intersects $\gamma_2$ perpendicularly. Then vary the other endpoint and by IVT there will be a point which intersects $\gamma_1$ perpendicularly. May 19, 2014 at 18:55

I think you are using the Poincare disk model of the hyperbolic plane. (see http://en.wikipedia.org/wiki/Poincare_disk_model )

This model is conformal (the angles between "hyperbolic" geodesics are the angle between the "Euclidean" circles.)

Then you can just solve it as if you are doing normal Euclidean geometry and have to prove that:

• Given a circle $d$ (The boundary circle of the hyperbolic plane )
• Given 2 circles $\gamma_1$ and $\gamma_2$ perpendiclular to Circle $d$ that do not intersect (The 2 non-intersecting "hyperbolic" geodesics)

Prove that circles $\gamma_1$ and $\gamma_2$ either:

• Meet at the boundary circle (Have a common endpoint at $S_\infty$ ) or
• There is a circle $c$ that is perpendicular to $d$ , $\gamma_1$ and $\gamma_2$ (Have a common perpendicular geodesic)

(and in all this also any line trough centre of circle $d$ also counts as a circle.)

OK, but how do I go about doing this?

You can construct circle $c$ if $\gamma_1$ and $\gamma_2$ don't intersect in Euclidean geometry as follows:

• Point $D$ is the centre of circle $d$
• Points $A_1$ and $B_1$ are the points where $\gamma_1$ intersects with circle $d$
• Points $A_2$ and $B_2$ are the points where $\gamma_2$ intersects with circle $d$
• Draw line $l_1$ trough $A_1$ and $B_1$
• Draw line $l_2$ trough $A_2$ and $B_2$
• point $C$ is where line $l_1$ and line $l_2$ intersect
• Point E is the midpoint of segment $CD$
• Draw circle $e$ with centre $E$ going trough $C$ and $D$
• Point F is one of the points where circle $e$ intersects circle $d$
• Circle $c$ is the circle with centre $C$ going trough $F$

The part of circle $c$ that is inside circle $d$ is the hyperbolic geodesic you are looking for.

That is how you construct your geodesic under ideal situations.

Now there can be complications:

• What if line $l_1$ and line $l_2$ do not intersect (Hint then the common perpendicular of $\gamma_1$ and $\gamma_2$ is a straight euclidean line trough $D$)
• What $\gamma_1$ and $\gamma_2$ have a common endpoint on circle $d$? (circle $c$ becomes a single point)

and if you want to be pedantic also:

• What if $\gamma_1$, $\gamma_2$ or both are Euclidean lines (trough poind $D$ ) instead of perpendicular circles?

Also you need to prove:

• that circle $e$ and circle $d$ intersect. (to warm up)
• that cirle $c$ is perpendicular to circle $d$ (do this one first)
• that cirle $c$ is perpendicular to circle $\gamma_1$
• that cirle $c$ is perpendicular to circle $\gamma_2$
• that there is only one cirle $c$

GOOD LUCK

• OK, but how do I go about doing this? May 19, 2014 at 22:38