Shape made by Beltrami with constant $K<0$

Question

Beltrami made (out of thin paper or stiff or starched cloth not mentioned) a model of a surface of constant negative Gauss curvature $ K=-1/c^2$. The original might have resembled a large saddle shaped Pringles chip, and frills might have developed by sagging with time, if one were to make a guess.

Also it is now easy to guess that the mold he used had dimensions comparable to the length and breadth of the table on which this is/was placed. The narrow neck he obtained is/was by hand rolling to a tight paper scroll, imo.

Is there a higher definition photograph with more details available?

More importantly, does anyone know its modern day parametrization?

The original photograph is given on page 133 as Fig 56, Roberto Bonola Non-Euclidean Geometry, 1955 Edition, Dover, ISBN 0-486-60027-0. (Thanks in advance to anyone for scanning and posting it here).

The following is inserted from Daina Tamania's blog ( http://hyperbolic-crochet.blogspot.co.uk/2010/07/story-about-origins-of-model-of.html ):

It may be recalled all such shapes can be scrolled up tight somewhat like a table napkin rolled by hand into a smaller ring.. maintaing parallelsim among previous parallels due to isometry or contant $K$ property, into newer smaller scroll.

Answer 1

Here is a parametrization of the Pseudosphere:

$$ x (u,v) = a \frac {cos(v)}{cosh(u)} $$

$$ y (u,v) = a \frac {sin(v)}{cosh(u)} $$

$$ z(u,v) = a(u - tanh(u) ) $$

with $ v \in [0,2\pi) $ , $ u \in (-\infty, \infty) $

And, a useful link to the original model of Beltrami : http://www.mathcurve.com/surfaces/pseudosphere/pseudosphere.shtml

(bottom of page , on the right)

Answer 2

Lets start at the beginning

There is no 3 dimensional Euclidean surface that represents the complete hyperbolic plane (Hilberts theorem )

There are Surfaces of revolution that have (outside some boundaries ) a constant negative curvature $K$ .

All surfacesof revolution revolving around the $x_3 $ axis can be formulated as:

$ x_1 = \phi(v) \cos u $ , $ x_2 = \phi(v) \sin u $ and $ x_3 = \psi(v) $

For a revolution surfaces that have (outside some boundaries ) a constant negative curvature of $K = -1 $ also needs:

• $(\phi')^2+(\psi')^2 = 1 $ and
• $\phi'' = \phi $

The meridian curve (generatrix ) $ x_1 = \phi(v) , x_2 = 0 $ and $ x_3 = \psi(v) $ is a geodesic on the surface.

combining this with the hyperbolic plane the meridian curve is an hyperbolic line

There are 3 types of these surfaces

• the "pseudo spherical surfaces of the elliptic type":

also called the "conic type" $\phi(v)=C\sinh v$ and $\psi(v)=\int_0^v \sqrt{1-C^2\cosh^2v} dv$

The meridians are hyperbolic lines trough the vertex and the parallels are arcs of hyperbolic circles with the vertex as center.

• the "pseudo spherical surface of the parabolic type":

This is the one that is also known as the pseudosphere or the tractoid

$\phi(v)=e^v$ and $\psi(v)=\int_0^v \sqrt{1-e^{2v}} dv$ or

$\phi(v)= \frac{1}{\cosh v} $ and $\psi(v) = v -\tanh v $

(there are also other formula's)

The meridians are hyperbolic lines asymtopic to each other horoparallel) and the parallels are arcs of horocycles.

• the "pseudo spherical surface of the hyperbolic type":

$\phi(v)=C\cosh v$ and $\psi(v)=\int_0^v \sqrt{1-C^2\sinh^2v} dv$

PS this is not an one sheeted hyperboliod (an hyperboloid doesn't have an constant curvature.)

The meridians are hyperbolic lines athat share a common perpendicular and the parallels are the common perpendicular and arcs of hypercycles equidistant to this common perpendicular) .

I think this is the surface than Riemann describes in his famous Habilationsvortrag, (section II $ 5):

The theory of surfaces of constant curvature will serve for a geometric illustration. It is easy to see that surface whose curvature is positive may always be rolled on a sphere whose radius is unity divided by the square root of the curvature; but to review the entire manifoldness of these surfaces, let one of them have the form of a sphere and the rest the form of surfaces of revolution touching it at the equator. [...] The surface with curvature zero will be a cylinder standing on the equator; the surfaces with negative curvature will touch the cylinder externally and be formed like the inner portion (towards the axis) of the surface of a ring.

pictures of all three types of "pseudosperical surfaces" can be seen at:

virtualmathmuseum.org/Surface/gallery_o.html#PseudosphericalSurfaces

and

demonstrations.wolfram.com/SurfacesOfRevolutionWithConstantGaussianCurvature/

• So now back to the question which did Beltrami use?

• First of all does it matter?

Surfaces that have the same constant curvature are isometric, and you can take this very literal:

a triangle were the sides have lengths a,b, and c on one surface is congruent to an triangle on one another surface where the sides have lengths. (theorem of Minding 1938))

the different surfaces are just different "cut-outs" of the hyperbolic plane

But besides that I am not sure about it.

Reading Stillwell's translation of Beltrami's "Saggio" in "Sources of hyperbolic geometry" it first reads like he is using the hyperbolic type (this is the only surface where there are easy identifiable orthogonal hyperbolic lines) but later he changes to the tractoid.

Maybe he changed to the tractoid because this is the only one with a easy formulation.

But again does it really matter?

• About the pictures:

The rolled out version to me looks like a small part of the the hyperbolic type if only it was rolled out over an curved surface (a cylinder orthogonal to the lenght of the surface bump up) so that the frills disappear. and become a side of a ring or indeed a big pringle.

Maybe best compare it to a small part of the surface generated in the wolfram demonstration "negative curvature , first type" with the slider at the right and then the yellow side up and taking only a small part of it (around $30^o $ of the big circle)

Hopes this helps