Convex vertex-transitive-faced polyhedra

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Convex vertex-transitive-faced polyhedra

Postby mr_e_man » Tue Mar 03, 2020 2:37 am

What are some examples of polyhedra that are not required to have regular faces (as in the Johnson solids), but only vertex-transitive faces?

Any face with an odd number of edges would have to be regular. A face with an even number of edges, say 2n, would be a truncation of a regular n-gon. (This includes rectangles, which can be considered truncated digons.) So there may be different edge lengths, but the angles are the same as in regular polygons.

If the polyhedron itself is vertex-transitive, there's nothing really new; Wythoff's construction gives us deformations of the (uniform) prisms and Archimedean solids.
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Re: Convex vertex-transitive-faced polyhedra

Postby Klitzing » Tue Mar 03, 2020 6:11 am

Two of the Gévay polychora would come rather close here:
Their cells are (in this scaling) unit cubes, N-gonal variants of the cuboctahedron (ie. the hull of a tip-2-tip cycle of N unit squares), and (if its polar edge then has size c) c-sized o3oNc (N=3: tetrahedra, N=5: dodecahedra). By construction all cells, faces, but also the polychora themselves are circumscribable. But of course neither those exceptional cells nor their x-x-c triangles are vertex transitive, rather both use 2 vertex classes instead…

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Re: Convex vertex-transitive-faced polyhedra

Postby Mercurial, the Spectre » Tue Mar 03, 2020 2:36 pm

If you mean isogonal faces, then you need to make sure that the triangles and pentagons have to be regular (since heptagons and above only occur in prisms) as per symmetry group. Therefore the set of such polyhedra is equal to the isogonal polyhedra because any meetup of three rectangles implies a triangular leftover (four rectangles will give a flat tiling) and the rectangles have to be identical for the triangle to be equilateral. The same applies to hexagons, with the limit being two, and that there must be no other configuration. The only solutions are those whose symmetries are polyhedral (the Archimedean solids) and dihedral (prisms and antiprisms).

In 4D the examples Klitzing provided contain cells that are rectified prisms, which have isosceles triangles and hence not a solution.
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Re: Convex vertex-transitive-faced polyhedra

Postby Klitzing » Tue Mar 03, 2020 8:02 pm

Mercurial, the Spectre wrote:In 4D the examples Klitzing provided contain cells that are rectified prisms, which have isosceles triangles and hence not a solution.

As I already told those only come close to the original idea, they weren't meant to follow it!

As to the question "What are some examples...?" it is obvious that any uniform polytope clearly would be hierarchically vertex transitive. Therefore those won't be variants of Wythoffians only, rather snubs and other odd uniforms like gidrid, gap etc. do belong here as well. Sure those aren't the examples what was aimed for.

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Re: Convex vertex-transitive-faced polyhedra

Postby mr_e_man » Wed Mar 04, 2020 6:32 pm

To clarify, I'm thinking of polyhedra such that any (2D) face is equiangular and alternates two edge lengths, or has one edge length (and thus is regular). This includes all (convex) uniforms and Johnson solids.

Mercurial, the Spectre wrote:If you mean isogonal faces, then you need to make sure that the triangles and pentagons have to be regular (since heptagons and above only occur in prisms) as per symmetry group. Therefore the set of such polyhedra is equal to the isogonal polyhedra because any meetup of three rectangles implies a triangular leftover (four rectangles will give a flat tiling) and the rectangles have to be identical for the triangle to be equilateral. The same applies to hexagons, with the limit being two, and that there must be no other configuration. The only solutions are those whose symmetries are polyhedral (the Archimedean solids) and dihedral (prisms and antiprisms).

What do you mean by "triangular leftover"? A rectangular box is certainly allowed, which has no triangles anywhere (except as vertex figures, which are irrelevant).

Mercurial, the Spectre wrote:(since heptagons and above only occur in prisms)

Octagons occur in the truncated cube... :\
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Re: Convex vertex-transitive-faced polyhedra

Postby mr_e_man » Wed Mar 04, 2020 6:39 pm

I think we could generalize to "convex equiangular-faced polyhedra" without too much trouble. I'm reading Johnson's paper, "The Faces of a Regular-Faced Polyhedron", and most of the arguments only depend on angles, not edge lengths. I've verified Lemma 1: Any convex equiangular-faced polyhedron, other than a prism or antiprism, cannot have an n-gon with n >= 12. I'm still working on Lemma 2, which deals with 11-gons.
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Re: Convex vertex-transitive-faced polyhedra

Postby Klitzing » Wed Mar 04, 2020 10:44 pm

But the use of Johnson solids wouldn't be allowed for 3D faces within 4D vertex transitive faced polytopes, just because those aren't vertex transitive themselves.

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Re: Convex vertex-transitive-faced polyhedra

Postby mr_e_man » Thu Mar 05, 2020 1:23 am

The idea was to generalize CRF, "convex regular-faced" polytopes, to "convex (something else)-faced" polytopes. When talking about CRFs, we understand that "face" means "2D face", not "3D face"; we don't require CRF polychora to use only Platonic solids. So also here I mean "2D face", not "3D face" (which I would call a "cell", or in 4D a "facet").
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Re: Convex vertex-transitive-faced polyhedra

Postby ndl » Thu Mar 05, 2020 2:31 am

Are you talking about what Bowers calls "semi-uniform"? It's on the bottom of the following page:
http://www.polytope.net/hedrondude/polyhedra.htm
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Re: Convex vertex-transitive-faced polyhedra

Postby mr_e_man » Thu Mar 05, 2020 4:08 am

These polyhedra are not required to have any symmetries. But if they are vertex-transitive (isogonal), then indeed they're semi-uniform.
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Re: Convex vertex-transitive-faced polyhedra

Postby mr_e_man » Thu Mar 05, 2020 4:42 pm

Here's an example of a polyhedron that's vertex-transitive but not semi-uniform. It's "between" a prism and an antiprism, with trapezoids instead of triangles. The polyhedron is vertex-transitive, but the trapezoid faces are not. That's the opposite of what I want to consider here.

Here's an example of a vertex-transitive-faced polyhedron that's not vertex-transitive. Take an octahedron, and expand it by different amounts along the three axes, placing rectangles between the triangles. The result is combinatorially the same as the small rhombicuboctahedron, and it has the same dihedral angles (135° and 144.74°), but it has less symmetry.

This illustrates something interesting. Conjecture: Any convex equiangular-faced polyhedron can be continuously deformed into a convex regular-faced polyhedron (of the same abstract combinatorial type), while keeping all face angles, dihedral angles, and solid angles constant. And a similar statement for higher-dimensional polytopes. What do you think? This would imply that CRF polytopes are equivalent to CEF polytopes, and that edge lengths are completely irrelevant.

Note, no such thing happens with the "dual" concept, "convex edge-transitive-faced" or "convex equilateral-faced" polyhedra, as shown by the rhombic dodecahedron. (A rhombus is edge-transitive.) It has four 4-gons around a vertex, but no convex regular-faced or equiangular-faced polyhedron can have four 4-gons around a vertex.
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Re: Convex vertex-transitive-faced polyhedra

Postby Mercurial, the Spectre » Thu Mar 05, 2020 5:56 pm

mr_e_man wrote:Here's an example of a polyhedron that's vertex-transitive but not semi-uniform. It's "between" a prism and an antiprism, with trapezoids instead of triangles. The polyhedron is vertex-transitive, but the trapezoid faces are not. That's the opposite of what I want to consider here.

Here's an example of a vertex-transitive-faced polyhedron that's not vertex-transitive. Take an octahedron, and expand it by different amounts along the three axes, placing rectangles between the triangles. The result is combinatorially the same as the small rhombicuboctahedron, and it has the same dihedral angles (135° and 144.74°), but it has less symmetry.

This illustrates something interesting. Conjecture: Any convex equiangular-faced polyhedron can be continuously deformed into a convex regular-faced polyhedron (of the same abstract combinatorial type), while keeping all face angles, dihedral angles, and solid angles constant. And a similar statement for higher-dimensional polytopes. What do you think? This would imply that CRF polytopes are equivalent to CEF polytopes, and that edge lengths are completely irrelevant.

Note, no such thing happens with the "dual" concept, "convex edge-transitive-faced" or "convex equilateral-faced" polyhedra, as shown by the rhombic dodecahedron. (A rhombus is edge-transitive.) It has four 4-gons around a vertex, but no convex regular-faced or equiangular-faced polyhedron can have four 4-gons around a vertex.


The prismatic polyhedra you mean are actually called trapezoprisms (= bialternatosnub 2n-prisms). You take a 4n-prism, then take two adjacent vertices of a 4n-gon face, skip two other vertices, and repeat until you get back to the vertices, halving the vertex count. Now do the process on the other 4n-gon face ensuring that the resulting 2n-gons are antialigned. In fact it has n-antiprismatic symmetry.

Expanding an octahedron gives a rhombicuboctahedron, which has a pyritohedral subsymmetry. Of course it has half the symmetry, and can be formed from a great rhombicuboctahedron by the process of a bialternatosnub (aka. snubbing the octagons into rectangles in a way that the hexagons alternate into triangles). You're correct on this. PS: I made mistakes earlier (for heptagons and higher, I mean odd-sided polygons, and for the three rectangles, it's a cuboid if the three rectangles form no gap, otherwise you get a rhombicuboctahedron with three squares and one equilateral triangle per vertex.)

For this conjecture, it's true since the set equals to a convex uniform polyhedron (Archimedeans + prisms/antiprisms).

The converse is basically equal to asking for a polyhedron that only has one edge length, though this only applies to faces that are equilateral triangles or rhombi (triambi and above are unlikely to close into a specific vertex configuration unless regular). In fact, there do exist the snub disphenoid and any other Johnson solid. If we allow for rhombi we can have a variant of the propello tetrahedron with equal edges.
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Re: Convex vertex-transitive-faced polyhedra

Postby mr_e_man » Thu Mar 05, 2020 7:41 pm

Mercurial, the Spectre wrote:Expanding an octahedron gives a rhombicuboctahedron, which has a pyritohedral subsymmetry. Of course it has half the symmetry, and can be formed from a great rhombicuboctahedron by the process of a bialternatosnub (aka. snubbing the octagons into rectangles in a way that the hexagons alternate into triangles). You're correct on this. PS: I made mistakes earlier (for heptagons and higher, I mean odd-sided polygons, and for the three rectangles, it's a cuboid if the three rectangles form no gap, otherwise you get a rhombicuboctahedron with three squares and one equilateral triangle per vertex.)


I said "expand by different amounts". If one or two of these amounts is 0, we get a square orthobicupola or an elongated square bipyramid. The expanded polyhedron doesn't necessarily have cubic or pyritohedral symmetry.

Mercurial, the Spectre wrote:For this conjecture, it's true since the set equals to a convex uniform polyhedron (Archimedeans + prisms/antiprisms).


No. I've already said that the polyhedron doesn't need to be vertex-transitive, and that this set contains the Johnson solids as a subset.
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Re: Convex vertex-transitive-faced polyhedra

Postby Mercurial, the Spectre » Thu Mar 05, 2020 9:05 pm

mr_e_man wrote:
Mercurial, the Spectre wrote:Expanding an octahedron gives a rhombicuboctahedron, which has a pyritohedral subsymmetry. Of course it has half the symmetry, and can be formed from a great rhombicuboctahedron by the process of a bialternatosnub (aka. snubbing the octagons into rectangles in a way that the hexagons alternate into triangles). You're correct on this. PS: I made mistakes earlier (for heptagons and higher, I mean odd-sided polygons, and for the three rectangles, it's a cuboid if the three rectangles form no gap, otherwise you get a rhombicuboctahedron with three squares and one equilateral triangle per vertex.)


I said "expand by different amounts". If one or two of these amounts is 0, we get a square orthobicupola or an elongated square bipyramid. The expanded polyhedron doesn't necessarily have cubic or pyritohedral symmetry.

Mercurial, the Spectre wrote:For this conjecture, it's true since the set equals to a convex uniform polyhedron (Archimedeans + prisms/antiprisms).


No. I've already said that the polyhedron doesn't need to be vertex-transitive, and that this set contains the Johnson solids as a subset.

Please, don't make vague statements about your constructions; a thorough explanation will help us understand.
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Re: Convex vertex-transitive-faced polyhedra

Postby mr_e_man » Thu Mar 05, 2020 11:38 pm

Maybe some pictures will help.

The first one is an expanded octahedron (or a rhombicuboctahedron). The triangular faces are regular. The quadrilateral faces are rectangles, not squares. All faces are vertex-transitive.

The second one is a triangular cupola. The triangular faces are regular, but have two different sizes. The hexagonal face on the bottom has 120° angles and two different edge lengths, and 3-fold symmetry. All faces are vertex-transitive.

The third one is a hexagonal prism. The hexagonal faces have 120° angles and two different edge lengths, and 2-fold symmetry; they're not vertex-transitive, but they are equiangular.

(I mean "vertex-transitive" according to the face's own symmetry, not the whole polyhedron's symmetry. We've discussed this before: viewtopic.php?f=25&t=2440. But such confusion about symmetry becomes irrelevant if we just talk about angle measures.)
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CEFPolyhedra.png
CEFPolyhedra.png (13.42 KiB) Viewed 2362 times
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Re: Convex vertex-transitive-faced polyhedra

Postby ndl » Fri Mar 06, 2020 1:03 am

Any Johnson or uniform which is a full or partial Stott expansion can be reduced some amount to create one of your CEFs. Also the expanded-kaleido facetings could be expanded a little more to make one of these.
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Re: Convex vertex-transitive-faced polyhedra

Postby mr_e_man » Fri Mar 06, 2020 6:02 am

Judging from this (and similar animations for J91 and J92), if an EKF is expanded any more, it will have trapezoid faces, which are not equiangular.
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Re: Convex vertex-transitive-faced polyhedra

Postby mr_e_man » Mon Mar 23, 2020 9:04 pm

Here's an argument for my conjecture.

While constructing a CEF polyhedron, we only need to solve equations (essentially the spherical law of cosines) relating the face angles and dihedral angles; indeed the edge lengths are irrelevant. So, given any CEF polyhedron with different edge lengths, we can instead set the lengths to 1, to make a CRF polyhedron with the same combinatorial type and the same angles.

As for "continuous deformation", consider the set of all polyhedra with a given combinatorial type and given angles. Suppose the number of edges is E, and faces F. Since the only variables are the edge lengths, each such polyhedron can be considered as a point in an E-dimensional space (RE). But not every point represents a valid polyhedron. The edges around each face must form a closed loop.

CEFPolygon.png
CEFPolygon.png (10.34 KiB) Viewed 1888 times


Let's say the edges around an n-gon have lengths e0, e1, e2, ... , en-1. (I often use the notation ei for vectors, but here they're just numbers.) We set up a coordinate system on this face, with one vertex as the origin, and an incident edge as the x-axis, and the next edge at an angle θ=360°/n from the first. To enforce the loop condition, we require this sum of vectors (as shown in the image with n=8 ) to return to the origin:

e0*(1, 0) + e1*(cos θ, sin θ) + e2*(cos 2θ, sin 2θ) + ... + en-1*(cos (n-1)θ, sin (n-1)θ) = (0, 0),

that is,

e0 + e1cos θ + e2cos 2θ + ... + en-1cos (n-1)θ = 0,
0 + e1sin θ + e2sin 2θ + ... + en-1sin (n-1)θ = 0.

These are linear equations, thus representing hyperplanes in RE. There will be 2 such equations for each face, so 2F in total. Of course we also require ei > 0, restricting to the "positive orthant" or a "corner of a cube" in RE. The intersection of this with all the hyperplanes is (E - 2F)-dimensional, or larger, if some of the equations are redundant. It is connected (it's not split into several pieces), so we may move continuously between any two points in it. And one such point is (1, 1, ... , 1), representing a polyhedron with all edge lengths equal.
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Re: Convex vertex-transitive-faced polyhedra

Postby mr_e_man » Mon Mar 23, 2020 9:08 pm

According to the link given in this post, convex equiangular-faced polyhedra have already been considered by Pryahkin.
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Re: Convex vertex-transitive-faced polyhedra

Postby quickfur » Thu Apr 30, 2020 9:04 pm

Klitzing wrote:But the use of Johnson solids wouldn't be allowed for 3D faces within 4D vertex transitive faced polytopes, just because those aren't vertex transitive themselves.

--- rk

A vertex that's not transitive in the 3D facet itself may actually be transitive in the 4D polytope containing said facets. For example, the vertices of a tridiminished icosahedron are not transitive, but in bidex, which has a surface of 48 transitive tridiminished icosahedra, the vertices are transitive.

IOW, the transitivity of a vertex depends on global properties of the entire polytope, not only the local properties (transitivity over a single facet).
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Re: Convex vertex-transitive-faced polyhedra

Postby Klitzing » Fri May 01, 2020 1:16 pm

Right you are, Quickfur. But you're pulling the citation out of context. The original idea was to promote hierarchically transitive figures only. And thereto I was replying, saying that this is kind too restrictive to be of any further interest.
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Re: Convex vertex-transitive-faced polyhedra

Postby quickfur » Fri May 01, 2020 5:47 pm

Ahh I see. Sorry!
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Re: Convex vertex-transitive-faced polyhedra

Postby Klitzing » Sat May 02, 2020 11:06 am

:D you are welcome!
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