5

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.

tryst with freedom
  • 11,538

2 Answers2

4

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.

Andrew D. Hwang
  • 78,195
  • 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? – tryst with freedom Nov 29 '21 at 12:08
  • 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? – tryst with freedom Nov 29 '21 at 12:08
  • 1
    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 $\lambda(x, y) \sqrt{dx^2 + dy^2}$. [...] – Andrew D. Hwang Nov 29 '21 at 22:26
  • 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 $y \to 0^+$. – Andrew D. Hwang Nov 29 '21 at 22:28
  • 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? – tryst with freedom Nov 30 '21 at 03:21
  • " The metric defines the ruler" that point you wrote, that really helped , now the last thing I'm stuck at is the previous comment – tryst with freedom Nov 30 '21 at 11:01
  • If $ds$ denotes the Euclidean length of a Cartesian displacement, then $ds'$ in your comment is the hyperbolic length. This relation quantifies the sense in which the conformal factor $1/y$ "stretches a ruler": A small Euclidean length near $y = 0$ is a large hyperbolic length, and far from $y = 0$ is a small hyperbolic length. To travel from $p$ to $q$ in the shortest distance, there is an "incentive" for a path to bend upward, making $y$ larger and $1/y$ smaller; and it turns out arcs of circles "optimize the tension." – Andrew D. Hwang Nov 30 '21 at 13:48
  • Wait, you haven't answered my point still, is ds' the distance on the chart or the surface? – tryst with freedom Nov 30 '21 at 14:40
  • The distinction between the chart and the surface is sort of illusory: The chart equipped with the induced metric is isometric to (a piece of) the surface. (If it matters, to return to a few comments back, a pseudosphere is not the whole hyperbolic plane, but merely a thin, finite-area wedge with edges identified to give a surface homeomorphic to a punctured disk.) – Andrew D. Hwang Nov 30 '21 at 16:52
  • where did this punctured disk thing come from? – tryst with freedom Nov 30 '21 at 17:27
  • I have sent an email to you from the linked one in your post – tryst with freedom Nov 30 '21 at 18:06
  • Let us continue this discussion in chat. – tryst with freedom Nov 30 '21 at 18:10
  • IIRC, there are a couple of further comments in chat. – Andrew D. Hwang Dec 04 '21 at 13:30
  • Hi Andrew D.Hwang, thanks for your continued support. I'll have a check on the messages in few hours or tomorrow morning (IST) – tryst with freedom Dec 04 '21 at 13:38
  • Hi, Andrew D.Hwang, could you please check my answer here P.s: It's not really a rigorous answer. – tryst with freedom Dec 08 '21 at 11:51
3

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.

Reader Manifold
  • 1,297
  • 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 – tryst with freedom Nov 29 '21 at 07:59
  • 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. – Reader Manifold Nov 29 '21 at 08:05
  • 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. – Reader Manifold Nov 29 '21 at 08:08
  • I think I got what you mean, but I am still confused. I think some examples will help – tryst with freedom Nov 29 '21 at 08:11
  • 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? – Reader Manifold Nov 29 '21 at 09:20
  • well I mean if it couldn't, we'd have an infinite tower of spaces because in each space on could ask "what contains this?" and then ask what contains the container and so on like russian dolls – tryst with freedom Nov 29 '21 at 09:23
  • Yeah, that's right. If the space can be embedded in $\mathbb{R}^n$, then it can be done so in any $\mathbb{R}^m$ for $m\geq n$. There can be a lot of different ambient spaces possible, so it's useful to think of properties that are intrinsic to the space itself, and not the embedding in $\mathbb{R}^n$. (Sorry for the late reply by the way..) – Reader Manifold Dec 03 '21 at 12:15