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The pseudosphere is a surface of constant negative Gaussian curvature, and it has an isomorphism with a part of the hyperbolic plane. I have a point $(x,y)$ in the appropriate part of the hyperbolic upper half plane, it maps to a point $(X, Y, Z)$ on the surface of the pseudosphere by $$ \begin{align} X &= \operatorname{sech}(\operatorname{arcosh}(y))\cos x,\\ Y &= \operatorname{sech}(\operatorname{arcosh}(y))\sin x,\\ Z &= \operatorname{arcosh}(y) - \tanh(\operatorname{arcosh}(y)). \end{align} $$ This mapping preserves both angles and areas.

There are other surfaces of constant negative curvature, such as the "surface of revolution of constant Gaussian curvature of hyperboloid type". This link (and others) gives parametric equations for the surface itself, but it does not give an explicit mapping between the hyperbolic plane and the surface.

It's claimed in this answer that such "locally isometric" mappings always exist between surfaces of constant negative curvature and some part of the hyperbolic plane. So my question is:

  • What are the equations for an explicit mapping from part of the hyperbolic plane to the "hyperboloid type" surface in 3D space? What about other surfaces of constant negative curvature?

This is for the purpose of building a physical object, so I am hoping for an explicit set of equations like the ones above that I can just put into a computer, with a minimum of extra terminology or 'deep' concepts.

I am most interested in the "hyperbolid type" surface, but there are also surfaces of constant negative curvature of conic type, as well as Dini's surface and many others, shown for example in this paper by Robert McLachlan. If explicit mappings of this kind can be given for any of those as well, that would also be very useful.

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  • $\begingroup$ This is false: "The pseudosphere is a surface of constant negative Gaussian curvature, and it has an isomorphism with a part of the hyperbolic plane." What is true is that there is a locally isometric, regular covering map from a part of the hyperbolic plane to the pseudo-sphere, such that that deck transformation group of that covering map is infinite cyclic. The relevant portion of the hyperbolic plane is a certain horoball. $\endgroup$
    – Lee Mosher
    Commented Sep 3, 2018 at 19:42
  • $\begingroup$ @LeeMosher I do not understand those terms, nor why I cannot simply refer to "a locally isometric, regular covering map ... such that that deck transformation group of that covering map is infinite cyclic" as an isomorphism. The transformation between the "certain horoball" you mention and the pseudosphere is invertible, and is obviously an isomorphism in the category of mappings that preserve some appropriate notions of angle and area. $\endgroup$
    – N. Virgo
    Commented Sep 4, 2018 at 14:14
  • $\begingroup$ (Well I suppose technically the isomorphism is between part of the hyperbolic plane and only part of the pseudosphere. I think possibly that part consists of all of it except for a single vertically oriented cut, from the bottom to +infinity. But that isn't important for my application.) $\endgroup$
    – N. Virgo
    Commented Sep 4, 2018 at 14:19

1 Answer 1

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I have been interested in the problem of mapping some hyperbolic tesselations to surfaces of constant negative curvature, and needed to solve this problem for this purpose. See here for my results, or live in HyperRogue. Source code is available too, though rather complicated.

For the hyperboloid-like surface of revolution, I have found the relevant formulas here, see formula (15.22). The straight line on the surface between the point $x(u,0)$ and $x(u,v)$ is simply $x(u,t)$ for $t \in [0, v]$, and it is easy to show that the distance is simply $v$. Also the straight line from $x(0,0)$ to $x(u,0)$, where $u \in (-\pi, \pi)$, is given similarly by $x(t,0)$, and it has length $bu$. There is also a right angle at $x(u,0)$ between these two straight lines. Thus, the point $x(u,v)$ is mapped to a point in $\mathbb{H}^2$ obtained by going distance $bu$ in a fixed direction from a fixed point, turning 90 degrees, and going distance $v$. Of course you still need to compute the elliptic integral (I have done this myself numerically, but some libraries exist too). My implementation is here, see the function "run_hyperlike", point p->h (given in the Minkowski hyperboloid model coordinates) is mapped to the point p->flat (given in the $\mathbb{R}^3$ coordinates). I think a similar method should work for the conic type too.

For Dini surface or Kuen surface, my solution was to compute the geodesics by solving the differential equation numerically; there is some solution for the Dini surface here but I have not tried to understand and implement it. Select a point $x_0 \in \mathbb{H}^2$ and map it to point $f(x_0)$ on the Dini surface $M$. Also find an isometric mapping $f$ between the tangent plane $T_{x_0}(\mathbb{H}^2)$ and the tangent plane $T_{f(x_0)}(M)$. To find $f(x)$ for $x \in \mathbb{H}^2$, represent $x$ as $\exp_{x_0}(v)$ for $v \in T_{x_0}(\mathbb{H}^2)$ (basically, find the direction and distance from $x_0$ to $x$), and $f(x)$ will be $\exp_{f(x_0)}(f(v))$ (numerically find the geodesic of the given length and starting in the given direction). (exp is the exponential map.) (The default precision setting for solving this equation in HyperRogue is too low -- it is clear from the end picture that the areas are not mapped isometrically everywhere -- but the result looks fine after increasing the precision.)

Kuen surface is harder, because it turns out that $f(x)$ computed used the method above is not continuous and does not cover the whole Kuen surface. One reason for this is that there is no connection between the parts of the surface on the opposite sides of the singular curve. Another reason is that, intuitively, $f(x)$ will depend on the path from $x_0$ to $x$ you take -- another path will go on another side of a singular point and thus end up in a different place. This is the reason why on this picture there is a blue part and a green part (no connection between them), and two tortoises (they map to the same point of the hyperbolic plane). My approach here was to map various parts separately, so that each point of the Kuen surface is covered.

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  • $\begingroup$ Your first and second links are the same, is it intentional? $\endgroup$
    – N. Virgo
    Commented Aug 29, 2018 at 17:59
  • $\begingroup$ The point about distances sounds really useful, but how does one calculate the constants $k$ and $l$? $\endgroup$
    – N. Virgo
    Commented Aug 29, 2018 at 18:04
  • $\begingroup$ Fixed, I think the editor got confused by the links somehow. For the constant $k$: $x(u,0)$ equals $(b\cos u, b\sin u, 0)$ so we are walking on a circle of radius $b$, so $k$ equals $b$. The constant $l$ is actually 1: $(\frac{d\phi}{dt}(v))^2 + (\frac{d\psi}{dt}(v))^2 = b^2 \sinh^2(v/a) / a^2 + (a^2 - b^2 \sinh^2(v/a))/a^2 = 1.$ $\endgroup$
    – Zeno Rogue
    Commented Aug 29, 2018 at 20:15
  • $\begingroup$ I have decided to remove the constants $k$ and $l$ from the answer, and replace them with their actual values. (They were multiplicative factors used to compute distances based on parameter differences.) $\endgroup$
    – Zeno Rogue
    Commented Aug 29, 2018 at 20:23

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