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Let $p < q < r < s$ be real numbers. Let $l$ be the geodesic with endpoints at $p$ and $q$ and let $m$ be the geodesic with endpoints at $r$ and $s$. (a) Prove that there is a unique geodesic segment from $l$ to $m$ that is perpendicular to both. (We sometimes say $l$ and $m$ are ultraparallel since they are not only parallel but they do not share endpoints also.)

(b) Now suppose $p < q < r$ are real numbers and $l$ is a geodesic whose endpoints are $p$ and $q$ and $m$ is a geodesic whose endpoints are $q$ and $r$. Then $l$ and $m$ are parallel, but prove that there is not geodesic segment from $m$ to $l$ that is perpendicular to both. That is, unlike the ultraparallel case, there are no points on $l$ and $m$ that realize the shortest distance from $l$ to $m$.

Stahl
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Timothy Bade
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  • Are you talking about the upper half plane model of the hyperbolic plane? I guess so... – Will Jagy Apr 21 '13 at 23:18
  • I'm visualizing this as two half-circles in the plane, resting on the x-axis (yes, I'm using the upper half-plane model). I can see that the unique perpendicular is the x-axis between the two semicircles, but i'm a little lost as to how to prove this. – Timothy Bade Apr 21 '13 at 23:29
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    You are confused. The common perpendicular is a third geodesic, meaning a third semicircle in the model. There are many possible proofs. Indeed, Bolyai gave an intrinsic construction with compass and straightedge. – Will Jagy Apr 21 '13 at 23:52
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    As your other question is about Mobius transformations, you can transform this so that $p_1+s_1 = q_1+r_1,$ at which point the common perpendicular is a vertical ray, then map back. – Will Jagy Apr 21 '13 at 23:56

1 Answers1

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Here are some hints for how I'd proceed.

First, stare at this picture for a while. These are Apollonian circles: two families of circles, so that each circle in one family is orthogonal to every circle in the other family. The corresponding Wikipedia page is here.

For (a), do you see how your two circles generate a blue family? How will you select a circle in the red family?

For (b), check up on the limiting (parabolic) case. What can you say about the complementary pencil of circles and its behavior with respect to the real axis?

Thinking in terms of inversive geometry has been my preferred method of dealing with the geodesics of $\mathbb{H}^2$ ever since I read Thurston's 3 Dimensional Topology and Geometry. There are many other ways to approach this problem. For example, as Will Jagy suggests in the comments, you might use a Möbius transformation to map your two circles to two concentric circles, take the vertical line through their common center, and transform it back. This same method should suggest why you cannot find a common perpendicular to two fellow-traveling geodesics (for example, take their common vertex to $\infty$).

azimut
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Neal
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