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The setting for hyperbolic space in this question will be the upper half plane.

Now I know that to measure the distance between two points $p$ and $q$ in the upper half plane, we take

$ \inf \int_\gamma \frac{1}{\Im(z)} |dz| $,

where $\gamma$ is any $C^1$ curve joining these $p$ and $q$. So if $\gamma$ is a curve that realises this infimum, it is a geodesic meaning the triangle inequality for any triple of points $p,q,r$ on $\gamma$, where $q$ is between $p$ and $r$ the triangle inequality is really just an equality.

I was just wondering if there is any difference (in terms of technicalities) between the terms "geodesic" and "hyperbolic line" in the upper half plane, or in any setting of hyperbolic geometry in general.

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up vote 4 down vote accepted

The word "geodesic" could be used to refer to a hyperbolic line, a hyperbolic ray, or a hyperbolic line segment in the hyperbolic plane. Also, the word "geodesic" is used throughout geometry for shortest-length paths, while the notion of "hyperbolic line" is confined to hyperbolic geometry.

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Minor nitpick: The word "geodesic" is used for locally shortest paths, but there is no assumption that they minimize distance between their endpoints. Think of lines of longitude on a sphere. Such things are called geodesics, but if you go too far along one (more than halfway around the sphere), they are no longer minimizing. – Jason DeVito Jun 3 '11 at 17:16
@Jason: This is in fact a matter of culture, background and context: You're certainly right in a Riemannian setting. But I know many metric geometers who define geodesics to be isometries from an interval to a space (and don't bother to add locally or don't even allow linear re-parameterizations). Moreover, I'm sure that Jim is perfectly aware of this :) – t.b. Jun 3 '11 at 17:36
@Theo: Thanks for the clarification - my background is Riemannian geometry so that explains where I'm coming from. Also, I didn't mean to imply Jim was unaware of this, but rather intended the comment to be for the benefit of the OP. I can delete it (and this comment as well) if you (that is, either of you) think it's best. – Jason DeVito Jun 4 '11 at 3:13
@Jason: I certainly didn't take any offense. I think your comment and this discussion would be helpful for someone looking at this question. – Jim Belk Jun 4 '11 at 17:40

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