Non-Euclidean CRFs

[mr_e_man]

Are the Johnson solids the same in spherical or hyperbolic space? I would expect that the curvature, changing the angles, adds or removes some possibilities. Of course, the more symmetric solids (like the cube) exist in all three spaces, though the shape depends on size.

In hyperbolic space, a horosphere is both convex and intrinsically flat. So Euclidean plane tilings with regular tiles can be made convex. I suppose these should be excluded from CRFs, simply because they're infinite. Not only does each tiling have infinitely many tiles, but there are (uncountably!*) infinitely many different types of tilings.

Hyperbolic n-space can be embedded in (n,1) psEuc space (as the hyperboloid model); then a horosphere becomes a circular paraboloid. And the square tiling becomes a paraboloidal polyhedron: take z = w - 1 = (x² + y²)/2 with x and y integers, to generate the vertices, then take the convex hull. So these Euclidean tilings can be made convex in flat space, though the angles and symmetries are non-Euclidean. In particular, a translation becomes a parabolic rotation.

We could also consider tilings with vertices on the other unit hyperboloid (the single-sheet one), but those would not be convex.

I'm also interested in more general psEuc spaces like (2,2). This is a 4-dimensional space with a null cone made of rays from the origin through a Clifford torus: x² + y² - z² - w² = 0. I wonder what types of symmetries a polytope could have here. The orthogonal group O(2,2) contains the general linear group GL(2) which is represented by all invertible 2x2 matrices.

*A Euclidean tiling can be made by stacking rows on each other, where each row is a strip of triangles or squares with the same edge lengths. A strip of squares represents a 0, and a strip of triangles represents a 1; the whole tiling represents a number in binary, with infinitely many digits. There are uncountably many such numbers, thus uncountably many such tilings. See https://en.wikipedia.org/wiki/Cantor%27 ... l_argument .

[Marek14]

In spherical space, Johnson solids are the same, but hyperbolic space has infinitely many.

One way to see it is to consider stacking cubes in a tower. In Euclidean space, these are not Johnson solids because double of the dihedral angle of cube is 180 degrees. But in hyperbolic space, any cube will have smaller dihedral angle and so a tower of cube of any height will still be convex.

Another example are pyramids and bipyramids. In Euclidean space, you can only have pentagonal pyramid as a Johnson solid, but in hyperbolic space, you are not limited -- any pyramid can be made into a Johnson solid.

Finally, you can make "lenses": take a portion of a horosphere or pseudosphere, tile it in any way you want, and then just glue two such portions with matching vertices together.

[mr_e_man]

The tower of cubes can also be thought of as a tiling (or truncation) of a cylinder.

...So the number of finite hyperbolic CRFs is at least countably infinite. I believe it must be countable: For any number n, we can make a finite list of all possible polyhedra with at most n faces, each face having at most n edges; then augment the list with that for n+1, etc. This list will eventually reach any such polyhedron, so they are countable.

And I've already shown, by "Euclidean" tilings of a horosphere, that the number of infinite hyperbolic CRFs is uncountably infinite.

The 3D Euclidean regular-celled honeycombs, which have only 3 types of cells (cube, tet, oct), are also uncountable. The argument is similar to the 2D case. Instead of rows of triangles and squares, we use the planes between layers of the octet honeycomb; a plane where octs connect to tets is a 0, and a plane where octs connect to octs is a 1.

Finally, you can make "lenses": take a portion of a horosphere or pseudosphere, tile it in any way you want, and then just glue two such portions with matching vertices together.

Could you show a picture of this, or further explanation? I think it would be rare to be able to match the vertices.

[Marek14]

:nod: The tower of cubes can also be thought of as a tiling (or truncation) of a cylinder.

...So the number of finite hyperbolic CRFs is at least countably infinite. I believe it must be countable: For any number n, we can make a finite list of all possible polyhedra with at most n faces, each face having at most n edges; then augment the list with that for n+1, etc. This list will eventually reach any such polyhedron, so they are countable.

And I've already shown, by "Euclidean" tilings of a horosphere, that the number of infinite hyperbolic CRFs is uncountably infinite.

The 3D Euclidean regular-celled honeycombs, which have only 3 types of cells (cube, tet, oct), are also uncountable. The argument is similar to the 2D case. Instead of rows of triangles and squares, we use the planes between layers of the octet honeycomb; a plane where octs connect to tets is a 0, and a plane where octs connect to octs is a 1.

Finally, you can make "lenses": take a portion of a horosphere or pseudosphere, tile it in any way you want, and then just glue two such portions with matching vertices together.

Could you show a picture of this, or further explanation? I think it would be rare to be able to match the vertices.

Well, bipyramids and cupolas are example (hexagonal cupola would have its triangles/squares/hexagons inscribed into horosphere, higher cupolas into pseudosphere).

And I thought of this: If you take a piece of convex triangular pseudohedron, doesn't it have enough degrees of freedom to press the edges along a plane and then reflect it and join with a second copy?

[wendy]

To get an idea of the infiniteness of hyperbolic CRF, consider something like {5,3,5}. If you join two cells together, the resulting polytope has 11 pentagons, and is convex.

As long as you don't have three dodecahedra at an edge, you can keep adding dodecahedra to your heart's content, and they would be convex-regular-face polytopes. Every imaginable way of stacking these in this way, gives rise to a convex polyhedron.

[mr_e_man]

And I thought of this: If you take a piece of convex triangular pseudohedron, doesn't it have enough degrees of freedom to press the edges along a plane and then reflect it and join with a second copy?

Maybe, but it's not clear to me.

To get an idea of the infiniteness of hyperbolic CRF...

And yet it's the smallest possible infinity!

Anyway, thanks for the examples: the dodecs and pyramids and such.

I'm also interested in more general psEuc spaces like (2,2). This is a 4-dimensional space with a null cone made of rays from the origin through a Clifford torus: x² + y² - z² - w² = 0. I wonder what types of symmetries a polytope could have here. The orthogonal group O(2,2) contains the general linear group GL(2) which is represented by all invertible 2x2 matrices.

Is anyone familiar with these "indefinite orthogonal groups"? A discrete subgroup could take one point and generate the vertices of a polytope; or take one CRF and possibly generate a higher-dimensional CRF, if the generated edges and faces align correctly.

[mr_e_man]

A simple example is the group generated by two orthogonal hyperbolic rotations, just like a duoprism is generated by two orthogonal circular rotations. This would produce vertices at (x,y,z,w) = (A sinh mα, B sinh nβ, A cosh mα, B cosh nβ), where m and n vary through the integers. Using the psEuc metric (which gives the dot product some minus signs), the distance between the adjacent vertices (m,n) and (m+1,n) is 2A sinh α/2. If we want all edge lengths to be equal, then this requires 2A sinh α/2 = 2B sinh β/2.

For simplicity, let's take A=B=sinh α=sinh β=1. It follows that cosh α=sqrt2, which I'll call r to make the coordinates more legible. The hyperbolic functions decompose by sinh(a+b) = sinh a cosh b + cosh a sinh b, and cosh(a+b) = cosh a cosh b + sinh a sinh b, so all the vertices (m,n) are

(0,0): (0, 0, 1, 1). (0,±1): (0, ±1, 1, r). (0,±2): (0, ±2r, 1, 3). (0,±3): (0, ±7, 1, 5r). (0,±4): (0, ±12r, 1, 17). (0,±5): (0, ±41, 1, 29r). ...
(1,0): (1, 0, r, 1). (1,±1): (1, ±1, r, r). (1,±2): (1, ±2r, r, 3). (1,±3): (1, ±7, r, 5r). (1,±4): (1, ±12r, r, 17). (1,±5): (1, ±41, r, 29r). ...
(2,0): (2r, 0, 3, 1). (2,±1): (2r, ±1, 3, r). (2,±2): (2r, ±2r, 3, 3). (2,±3): (2r, ±7, 3, 5r). (2,±4): (2r, ±12r, 3, 17). (2,±5): (2r, ±41, 3, 29r). ...
...

In general, if vertex (m,n) has coordinates (x,y,z,w), then vertex (m+1,n) has coordinates (xr+z, y, x+zr, w), and vertex (m,n+1) has coordinates (x, yr+w, z, y+wr).

Of course, there are edges and square faces between these adjacent vertices. The other faces are "horizontal" apeirogons with varying m and fixed n, and "vertical" apeirogons with fixed m and varying n. The cells are apeirogonal prisms.

[mr_e_man]

Let's go back to 2D.

The metric/quadratic form x² + y² produces Euclidean geometry, which we're all familiar with.

The metric x² - y² produces Lorentzian geometry, with hyperbolic rotations as symmetries. A regular "polygon" is an apeirogon, with vertices on a hyperbola.

The metric x² is degenerate, with vertical shear transformations and parabolic rotations as symmetries. An apeirogon with shear symmetry is not strictly convex; it has vertices along a vertical line. We could have a convex regular apeirogon with vertices on a parabola.

The metric 0 is completely degenerate, with general (or special, area-preserving) linear transformations as symmetries. A regular polygon is a linear transformation of any of the previous types, including Euclidean. For example, a rectangle is regular, because the metric doesn't tell us that the edges have different lengths. But if we require that all edges have not only the same length (0 in this case), but length exactly 1, then nothing is regular here.

So the only pseudo-Euclidean plane with finite, positive-length, regular polygons is Euclidean.

Now we can consider 3D polyhedra. With the above restrictions on the faces, we must have a Euclidean plane somewhere; thus the metric is either x² + y² + z², or x² + y² - z², or x² + y².

The Euclidean metric x² + y² + z² produces the Platonic, Archimedean, and Johnson solids.

The degenerate metric x² + y² produces a few new CRFs, such as the hexagonal cupola. This would ordinarily be planar, but now we can separate the hexagon and the dodecagon into two parallel planes without changing the edge lengths, by translating in the null direction. (I think the planes don't even need to be parallel.) We can similarly construct the hexagonal pyramid, bipyramid, orthobicupola, gyrobicupola. Anything else?

The Lorentzian metric x² + y² - z² produces the infinite families of n-gonal (bi)pyramids and (bi)cupolas with n=7,8,9.... Anything else? (Again I emphasize that this is in flat space. The previous examples, the tower of cubes and the dodecahedron tree, are in curved space.)

I suppose somebody should write a program for visualizing and manipulating things in these spaces. This goes along with quickfur's desire for a "4D Lego" polytope building program.

[mr_e_man]

With the degenerate metric, the view of the polyhedron from above would look like a region of a Euclidean plane tiling with regular polygons. The view from below would look like the same region, perhaps tiled differently. The region must be strictly convex. It should be easy to enumerate the possibilities.

...We can get 5 more, total 10, by augmenting with hexagonal pyramids: augmented hexagonal cupola, (bi)augmented hexagonal (ortho,gyro)bicupola.

Another one is based on a region with a triangle in the middle, 3 squares surrounding it, and 3 pairs of triangles filling the gaps between the squares.

Another one is based on a region with 2 squares in the middle, and 2 strips of 3 triangles bracketing the squares; see these pictures. Now the total is 12.

These last 2 resemble the spheno- and luna- Johnson solids. Any suggestions for names?

How about "trisphenodihedron" and "disphenotetrahedron"? (Each spheno- complex wraps over the rim, with one lune above and one lune below. "Dihedron" and "tetrahedron" refer to the remaining triangular faces.)

Looking back from the next post, I guess they should just be called "3-expanded" and "2-expanded 6-bipyramid".

[mr_e_man]

With the Lorentzian metric, there are three types of directions: upward (pointing to the upper sheet of the hyperboloid), downward (pointing to the lower sheet), and sideways (pointing to the single-sheet hyperboloid). The polyhedron's edges must all point sideways, because the faces are Euclidean. The faces' normal vectors point upward or downward.

Any vertex on the upper or lower half of the polyhedron has an angle excess (a negative angle defect). Any vertex on the rim connecting the two halves has a positive angle defect. The sum of the angle defects is always 4pi.

...Again we can get more CRFs by augmenting. The n-gonal cupolas can be augmented with 2n-pyramids (on the bottom) but not with n-pyramids (on the top).

If my calculations are correct, an n-cupola can be stacked on top of a 2n-cupola, and the result is convex! This, in turn, can be stacked on top of a 4n-cupola, on top of an 8n-cupola, etc. Let's call this a "stratocupola", or an "m-stratic n-cupola"; the top face is an n-gon, and the bottom face is a 2^mn-gon. Two of these can be combined like a bicupola, and rotated in not just 2 ways (ortho- and gyro-), but 2^m different ways (counting chirality, and assuming the upper and lower halves are the same).

The two "special" polyhedra from the degenerate space generalize to an infinite family of Lorentzian CRFs. The upper half is an n-gon surrounded by n squares, with m triangles between each pair of squares. It's an expansion of the mn-bipyramid. (It can also be augmented with one or two pyramids or stratocupolas.)

[mr_e_man]

There are only a few possible vertex types on the rim of a Lorentzian CRF. The sum of the faces' angles must be less than 2pi, but that doesn't mean we can use the same vertex types as in Euclidean geometry. There, if 3 faces met at a vertex, we would have a triangle inequality (c < a + b) relating the angles in the faces, which are the edges of a triangle on the surface of a sphere centred on the vertex. But here the triangle inequality is reversed (c > a + b), if A and B are upper faces and C is a lower face (or vice versa). The angles are the edges of a triangle on the surface of a single-sheet hyperboloid centred on the vertex.

Here are the possible rim vertex configurations:

3.3.n, n > 6; seen in pyramids
3.4.n, n > 12; seen in cupolas
3.5.n, n > 30; not seen yet
3.3.3.3; seen in bipyramids
3.4.3.3; seen in augmented cupolas
3.4.4.3; seen in orthobicupolas
3.4.3.4; seen in gyrobicupolas
3.5.3.3; not seen yet
3.5.4.3; not seen yet
3.5.3.4; not seen yet
3.5.5.3; not seen yet
3.5.3.5; not seen yet

(These must have 2 upper faces and 2 lower faces, not 3 upper faces and 1 lower face, as the lower angle would have to exceed the sum of the three upper angles, which is impossible with regular polygon angles.)

It appears that we're missing some generalization or modification of the pentagonal rotunda.

[Klitzing]

It appears that we're missing some generalization or modification of the pentagonal rotunda.

Isn't that

3.5.n, n > 30; not seen yet

exactly that very generalization?
--- rk

[mr_e_man]

But what would be its other vertices, not on the rim? They need an angle excess, so 3.5.3.5 doesn't work.

Going around the rim, the obvious choice for the upper faces is 5, 3, 5, 3, 5, 3, etc. But instead we could have 5, 3, 3, 5, 3, 3, etc, repeating n times (so the bottom face is a 3n-gon, with n > 10). Then the upper vertices could be 3.3.5.3.5 and 3.5.3.n .

...But that doesn't work either. The vertices with 3.3.3n are rigid, forming part of a 3n-pyramid, so the adjacent triangles reach to the central axis, leaving no room for pentagons or an n-gon. Indeed, anything with a 3.3.n vertex must be an n-pyramid.

Likewise, the vertices 5.3.n and 3.4.n are incompatible; the pentagon and the square give the triangle two different heights above the n-gon. Thus the faces around the rim cannot contain 5, 3, 3, nor 5, 3, 4; it must be 5, 3, 5, 3, etc, if any such thing exists. We still need to find how to close up the top.

[mr_e_man]

(I edited a post to add "stratocupolas". Should I have made a new post instead? Editing helps to organize, but it may be overlooked by people who only read the newest posts and don't expect the old ones to change.)

[mr_e_man]

It turns out that 3.5.n cannot be used after all!

When attempting to build a rotunda, we get a vertex with one triangle between two pentagons, and an opening opposite the triangle. The opening between the pentagons has an angle α, with

cos α = (sqrt5 + 1)/4 + (sqrt5 - 1) cos(2π/n) - 2 cos²(2π/n)

or equivalently

sin(α/2) = cos(2π/n) - (sqrt5 - 1)/4.

(This is valid for both Euclidean and Lorentzian spaces.)

The case n = 5 corresponds to the sharp end of an ursahedron, which has 3 pentagons around a triangle, and no space between the pentagons; α = 0.

For the pentagonal rotunda, n = 10 and cos(2π/10) = (sqrt5+1)/4, and this formula gives α = π/3 (or 60^o) which means that another triangle fits exactly between the pentagons.

For the flat case n = 30, this gives α = 7π/15 (or 84^o).

If we send n to infinity, then cos(2π/n) approaches 1, and this gives α = 87.41^o.

In any case with n > 30, whatever polygons we try to fit between the two pentagons, the sum of their angles must be between 84^o and α, which is smaller than 87.41^o. This is clearly impossible with regular polygon angles.

[mr_e_man]

In hyperbolic space, a horosphere is both convex and intrinsically flat. So Euclidean plane tilings with regular tiles can be made convex. I suppose these should be excluded from CRFs, simply because they're infinite. Not only does each tiling have infinitely many tiles, but there are (uncountably!*) infinitely many different types of tilings.

Furthermore, even a single type of tiling is not necessarily rigid. Consider the regular {6,4} hyperbolic tiling, embedded in 2+1D Lorentzian space, or perhaps in 3D hyperbolic space. The vertex figure is a square, which can be deformed into a rhombus, with one degree of freedom. The whole tiling also has one degree of freedom.

More specifically, if we call a vertex V and the four surrounding vertices W,X,Y,Z in a cycle, then the vector dot products are:

(W - V)² = (X - V)² = (Y - V)² = (Z - V)² = 1
(W - V).(X - V) = (X - V).(Y - V) = (Y - V).(Z - V) = (Z - V).(W - V) = cos 120° = -1/2
(W - V).(Y - V) = p , (X - V).(Z - V) = q

By default, the "diagonals" or "chords" are equal, p = q = -2. In general, they can be anything below -1, related by (1 + p)*(1 + q) = 1. The diagonals at one vertex get transferred to the surrounding vertices, so choosing p uniquely determines the shape of the whole tiling.

Similarly, any {n,4} hyperbolic tiling with even n has one degree of freedom. (The diagonals are related by (1 + p)*(1 + q) = 4*r², where r = cos((n - 2)π/n) is the dot product of adjacent edges.) Any {n,4} tiling with odd n has no degrees of freedom. I guess an {n,5} tiling would have infinite degrees of freedom; given a vertex with 2 faces in place, we have one degree of freedom in placing the 3 other faces; and we'll see infinitely many such vertices while constructing the tiling.

Now I wonder what the deforming {6,4} tiling looks like. It should be similar to a regular {6,4} tiling with coloured edges, alternating two edge types.

[mr_e_man]

Are the Johnson solids the same in spherical or hyperbolic space? I would expect that the curvature, changing the angles, adds or removes some possibilities. Of course, the more symmetric solids (like the cube) exist in all three spaces, though the shape depends on size.

In spherical space, Johnson solids are the same, but hyperbolic space has infinitely many.

Actually, I think bilbiro and thawro exist only in Euclidean space, not in hyperbolic or spheric space.

And there is at least one CRF polyhedron that exists only in spheric space, and only with a specific edge length s. It is the "cubicosahedron" (number 4 here, also here), which has 3 square faces and 16 trigonal faces. The angle in a trigon is 2arcsin(x/2) = 62.1200°, and the angle in a square is 2arcsin(x/√2) = 93.7129°, where x = sec(s/2) = 1.031870889193 .

There are also CRF polyhedra existing only in hyperbolic space with a specific edge length. Take some convex configuration of polygons in a 2D hyperbolic tiling with edge length t (for example 4.4.3.4.3 which has cosh(t) = 9/7), and extend lines from the vertices into 3D, perpendicular to the plane, for a distance s/2 in both directions, to make the polygons into prisms with edge length s. We need the prisms to have all edge lengths equal, to have square sides; the square should have inradius t/2 and edge length s. This results in the equation

cosh(s) = (1 + cosh(t)) / (3 - cosh(t)).

(Notice that this requires cosh(t) < 3, or equivalently sech(t/2) > 1/√2.) This construction doesn't prove that the polyhedron doesn't exist with other edge lengths; indeed, applying it to the {3,7} vertex gives the elongated heptagonal dipyramid, which exists in general, whenever the heptagonal pyramid exists. But at least the 4.4.3.4.3 example (the polyhedron being made of two trigonal prisms and three cubes around an edge) does have a unique edge length, according to my calculator.

Of course, there are many CRF polyhedra existing in several spaces, with a continuum of possible edge lengths s. Let x = sec(s/2) or sech(s/2) or 1, depending on the curvature of space. It seems that if a polyhedron exists for two different values x₁ < x₂, then it exists for all x ≤ x₂ (with x > 0). In other words, either it exists only for a single value x = x₁, or it exists for an entire interval (0, x₃) or (0, x₃]. (I think x=x₃ would give a dihedral angle 0° or 180° somewhere. And x=0 is for an ideal polyhedron (vertices at infinity); those aren't rigid, so I exclude them.) I haven't found any counter-examples, but I don't know how to prove it either.

[Klitzing]

Any spherical polytope can be put into euclidean space (and potentially could be used as a building block for euclidean honeycombs)
and likewise any euclidean honeycomb can be reinterpreted as a holotope (adding its halfspace below) and thus being used within hyperbolic geometry.
Thus in total any spherical polytope, even such things like J91 and J92, well can be put into hyperbolic geometry as well.
(Whether it would become possible to continue an according hyperbolic tiling, is a completely different story, for sure.)

--- rk

[mr_e_man]

Yes, but the embedding doesn't preserve straight lines. The CRF cubicosahedron is not a Euclidean polyhedron, regardless of embedding 3D spheric space in 4D. You can take the convex hull of the vertices in 4D, to make the trigon angle 60° instead of 62°, but the result is a skew polyhedron, or rather a polychoron because it has 4D volume.

Similarly, a bilbiro or thawro embedded in 4D hyperbolic space would be either curved or skew.

[mr_e_man]

And x=0 is for an ideal polyhedron (vertices at infinity); those aren't rigid, so I exclude them.

An ideal polyhedron is flexible, only because a face can slide along an edge while the edge remains fixed. The edge has 2 degrees of freedom: the dihedral angle, and the sliding distance.

But there is a way to define ideal CRF polyhedra such that they are rigid.

Each face, being regular, has a line or plane of symmetry that bisects an edge. So the edge has a midpoint, even though it has infinite length. We should require the two midpoints, given to the edge by the two incident faces, to coincide. This eliminates the sliding.

Given a vertex (at infinity), expand a horosphere outward until it hits the midpoint of an incident edge. Since the faces are regular, the horosphere will touch all of the incident edges' midpoints at the same time. Define the vertex figure as the intersection of this horosphere and the polyhedron. The result is essentially a Euclidean polygon, whose angles are the polyhedron's dihedral angles. The intersection of the horosphere and an incident n-gon is an arc of a horocycle, with length 2cos(π/n).

Since Cauchy's "arm lemma" applies as well to Euclidean polygons as to spheric polygons, the rigidity theorem applies to the polyhedron.

Indeed, by similar reasoning, at least some types of ultra-ideal CRF polyhedra are also rigid. If the vertices are beyond infinity, but the edge midpoints are still finite, then the vertex figures are finite hyperbolic polygons. The arm lemma applies here as well.

[mr_e_man]

According to this paper, Lorentzian convex polyhedra are rigid; the dihedral (hyperbolic) angles are uniquely determined by the configuration and shapes of faces. And ultra-ideal hyperbolic convex polyhedra are also rigid. Convex polyhedra in Lorentzian space-forms - Jean-Marc Schlenker