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My intuition tells me that the shortest distance between two points on the surface corresponds to a line segment joining the two points on the map of said surface, because, the path on the surface is same as the shortest path in the map. However, this turns out to be wrong.

Take for instance, the Beltrami-Poincare half-plane model of $\mathbb{H}^2$, the shortest path between two points seems to be an arc of a semi circle centered at somewhere on the horizon. Picture:

enter image description here

Why is the shortest distance not a straight line in the map here?

Probably I am missing something quite basic, but I just can't seem to figure it out.

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2 Answers 2

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Here's one way to think of the issue: Suppose $$ ds^{2} = \lambda(x, y)^{2} (dx^{2} + dy^{2}) $$ for some smooth, positive function $\lambda$. Given two points $p$ and $q$, a path from $p$ to $q$ may "shorten itself" by veering off the Euclidean segment in order to travel in a region where $\lambda$ is smaller.

In the hyperbolic plane, $\lambda(x, y) = 1/y$, which grows rapidly as $y \to 0^{+}$ and which drops off as $y$ grows. As you know, the shortest paths are arcs of circles meeting the boundary $y = 0$ at right angles. Conceptually, these arcs of circles "climb just enough" into a region of larger $y$ to minimize their length. (That's not a proof, of course, just a conceptual interpretation. You're invited to sharpen the interpretation, e.g., explaining why circles meet the boundary at right angles, or why if $y_{0}$ is large the geodesic connecting two points $p = (x_{0}, y_{0})$ and $q = (x_{0} + 1, y_{0})$ is nearly a Euclidean segment.)

More vividly, we could construct a function $\lambda$ that is close to zero in some disk and very larger elsewhere. In the resulting conformal metric, a path might deviate considerably from a Euclidean segment in order to pass through the disk where $\lambda$ is small.

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  • $\begingroup$ I am not able to understand this at all. In the real world, I can draw a line using a pencil between two points, but if I had a hyperbolic plane to draw it, would what I draw conciously as a line turn into a semi circle? $\endgroup$
    – Babu
    Commented Nov 29, 2021 at 12:08
  • $\begingroup$ And secondly, it seems like using a ruler the shortest distance should still be a straight line in the plane.. or can't we use the ruler anymore? $\endgroup$
    – Babu
    Commented Nov 29, 2021 at 12:08
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    $\begingroup$ A metric more or less defines what is a ruler by specifying the hypotenuse of an infinitesimal right triangle. My guess is you may be seeing Cartesian coordinates in the hyperbolic plane and thinking geometrically of "distance between $(x_1,y_1)$ and $(x_2,y_2)$" using the Pythagorean theorem,$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$That formula works only for the Euclidean metric. For a metric conformal to the Euclidean metric, a Cartesian displacement $(dx, dy)$ has length $\lambda(x, y) \sqrt{dx^2 + dy^2}$. [...] $\endgroup$ Commented Nov 29, 2021 at 22:26
  • $\begingroup$ Said another way, a hyperbolic ruler of fixed length shrinks in Cartesian coordinates as the ruler approaches the line $y = 0$. Dually, a small Cartesian displacement $(dx, dy)$ has hyperbolic length depending on its location, and this length grows without bound as $y \to 0^+$. $\endgroup$ Commented Nov 29, 2021 at 22:28
  • $\begingroup$ Exactly, let me put my doubt another way, the metric (normed) is something like $ ds' = \frac{ds}{y}$ I had thought ds' is the distance on the surface of pseudo sphere but is it really that it is distance on the plane as well? $\endgroup$
    – Babu
    Commented Nov 30, 2021 at 3:21
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What you might be missing out is that the natural state of Hyperbolic plane is as an abstract space, independent of any embedding/immersions in Euclidean space. The half plane model attempts to bring it in to the Euclidean plane, but there are distortions involved.

As an analogy, we can try mapping the Earth's surface, to a map, a 2D plane. Inevitably, there will be distortions involved. You might be able to preserve some of your preferred properties depending on the map you choose, but it might be that the the straight lines on the globe is no longer the straight lines on a map.

Here, in the Beltrami Poincare Half-plane model, the model preserves the angles, but the straight lines are no longer straight in the usual, Euclidean sense, due to the distortions involved.

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  • $\begingroup$ I mean striaght lines between two points would be still straight lines, only in the map to the surface does it look different. But, at the same itme straight lines ( shortest distance in map) between two points may not be shortest path on spere.. argh... this is so confusing $\endgroup$
    – Babu
    Commented Nov 29, 2021 at 7:59
  • $\begingroup$ Yeah exactly. The (image of the) shortest path on the sphere is not the same as the shortest path on the 2-dimensional map of it. In a similar way, hyperbolic plane exists in itself, like a sphere does, independent of the way you choose to picture it on a plane. $\endgroup$ Commented Nov 29, 2021 at 8:05
  • $\begingroup$ The reason why it might look confusing is because, you are able to look at/visualise sphere independently as it lies in $\mathbb{R}^3$ without its lines/geodesics being distorted. However, this can't be done to the hyperbolic plane, and we might find it difficult to visualise initially. $\endgroup$ Commented Nov 29, 2021 at 8:08
  • $\begingroup$ I think I got what you mean, but I am still confused. I think some examples will help $\endgroup$
    – Babu
    Commented Nov 29, 2021 at 8:11
  • $\begingroup$ I am not sure what examples can be given. It might be easier if you can let me know what the confusion is. First of all, are you familiar with the fact that a space can exist in itself, without considering an ambient space? Or is that a confusion? $\endgroup$ Commented Nov 29, 2021 at 9:20

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