## How are the snub polyhedra uniform?

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

### How are the snub polyhedra uniform?

The snub cuboctahedron and snub icosidodecahedron have triangles on the edges of the squares or pentagons which don't seem to be vertex-transitive, as there are three separate edges connecting to a square or pentagon, a triangle of the same type, and a triangle of a different type.
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ubersketch
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### Re: How are the snub polyhedra uniform?

Elements of a uniform polytope don't have to be vertex-transitive under the symmetry of the polytope, only their own.
Mecejide
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### Re: How are the snub polyhedra uniform?

Mecejide wrote:Elements of a uniform polytope don't have to be vertex-transitive under the symmetry of the polytope, only their own.

Don't you mean the other way around?
Marek14
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### Re: How are the snub polyhedra uniform?

When I said their own, i meant on their own, as in not part of the polytope. So no.
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### Re: How are the snub polyhedra uniform?

In three dimensions there are three variables that represent the three different edges of the girco or grid's verf (a scalene triangle in general).
When alternating into a snic or snid, the vertex figure becomes a face of the polyhedron, and since you have to solve three equations for three edges, there is a guaranteed solution to get an equilateral result.

This is not the case in general in 4D where the solution involves solving four equations for six edges, meaning that there are cases in which no solution can be found for uniformity.
Mercurial, the Spectre
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### Re: How are the snub polyhedra uniform?

Snubbing as such happens to be a sequence of 2 different operations!
• One is the to be applied alternation (e.g. vertex alternation applied onto omnitruncates), seen as a mere faceting of the seed polytope. - This one clearly is possible always.
• The other one is a topolgical variation, which tries to head for all unit edges only (thus bringing back to uniformity). - This is the one, which is not always possible in general.

A short intro into snubbing theory can be found on my Incmats webpage here.
A detailed article on snubbing theory is also available, cf. "Snubs, Alternated Facetings, and Stott-Coxeter-Dynkin Diagrams", by Dr. R. Klitzing, in "Symmetry: Culture and Science", vol. 21, no.4, 329-344, 2010 - or downloadable from Research Gate or even from my website too.

--- rk
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### Re: How are the snub polyhedra uniform?

Wait it actually was the other way around.
Mecejide
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### Re: How are the snub polyhedra uniform?

Mecejide wrote:Wait it actually was the other way around.

What are you refering by "it"? - I'm assuming you are refering to the 2 processes involved by snubbing.

Snubbing indeed in the past was considered to be a variation (of some Wythoffian polytope) first and applying a faceting thereafter.
But even then the same point remains: the process of (appropriate) variation is not generally possible as that one has to bow to an according degree of freedom. Whereas the mere faceting is clearly applicable in every case.

And in fact this was the ingenious idea here, simply to switch the applications of those 2 operations. So that the faceting part becomes applicable to ANY Wythoffian seed polytope, and it is up to a subordinate consideration, whether a further variation towards uniformity can be achieved or not.

And infact, as I worked out in that cited paper, the decorations of the Dynkin diagram with node marks "o" (unringed node), "x" (ringed node), and "s" (snub node) comes out to rather describe the mere faceting process, as it is applicable generally, and so has nothing to do with the post-poned process of topological variation. I.e. it not truely describes the WHOLE snubbing process (even if such symbols like s3s4s usually are being considered to describe the uniform variant of the snub cube).

The genuine idea, which was also outlined in that paper, was that alternation needs not be restricted to vertices only. Even other polytopal elements, like edges, faces, cells, whatever, could be alternatedly replaced by the according sectioning facet underneath ("sefa"), thereby generalysing the vertex figure ("verf"). Keeping this in mind, it becomes obvious that indeed ANY decoration of the Dynkin symbol with any "powdering" by "s"-, "o"-, and "x"-nodes perfectly makes sense! (Then describing the mere faceted figure.)

--- rk
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### Re: How are the snub polyhedra uniform?

Klitzing wrote:
Mecejide wrote:Wait it actually was the other way around.

What are you refering by “it”?
--- rk

I was referring to my answer to Marek14’s question.
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### Re: How are the snub polyhedra uniform?

ubersketch wrote:The snub cuboctahedron and snub icosidodecahedron have triangles on the edges of the squares or pentagons which don't seem to be vertex-transitive, as there are three separate edges connecting to a square or pentagon, a triangle of the same type, and a triangle of a different type.

Vertex-transitivity means that any vertex of the object can be sent to any other vertex by a symmetry of the same object. A regular triangle is vertex-transitive by itself; the surrounding faces are irrelevant. The snub cube is vertex-transitive; this makes no reference to its triangular faces.
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mr_e_man
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### Re: How are the snub polyhedra uniform?

mr_e_man wrote:
ubersketch wrote:The snub cuboctahedron and snub icosidodecahedron have triangles on the edges of the squares or pentagons which don't seem to be vertex-transitive, as there are three separate edges connecting to a square or pentagon, a triangle of the same type, and a triangle of a different type.

Vertex-transitivity means that any vertex of the object can be sent to any other vertex by a symmetry of the same object. A regular triangle is vertex-transitive by itself; the surrounding faces are irrelevant. The snub cube is vertex-transitive; this makes no reference to its triangular faces.

He means the triangles that share an edge with the squares or pentagons, which are topologically scalene. Scalene triangles obviously are not vertex transitive. The fact is that the snub cube/dodecahedron has two different triangles: one always equilateral, and the other scalene, owing to them being s{3,4} and s{3,5}. The snub {4,5}, for example, has squares, pentagons, and scalene triangles. Of course it can be made into unit edges, but the topology does not change, unless it is of the form s{p,p}, which is just an alternated t{p,4}.
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### Re: How are the snub polyhedra uniform?

The triangle is "topologically scalene" only when we consider it as a part of the snub cube; its vertices cannot be sent to each other by symmetries of the snub cube. They can be sent to each other by symmetries of the triangle: reflections and rotations that map the triangle onto itself, regardless of what they do to the snub cube.

Uniformity of the snub cube only requires the triangle to have this "self-symmetry".
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### Re: How are the snub polyhedra uniform?

The issue is that the subgroup of symmetries that are distinguished by leaving the triangular face in the same plane are not sufficient to map any of its vertices onto any other. Thus, the triangular face can be considered to be non-uniform in the context of the polyhedron. However, the definition of uniformity does not ask for any such requirements, and instead just requires that the faces be uniform under their own symmetries.

I noticed this issue when I was younger. I found it nonintuitive that uniformity required a polytope's facets to be uniform yet did not also require the subsymmetry generated by keeping the facet in the same hyperplane to map any of the facet's vertices onto any other. Thus, I defined a property of polytopes called "local uniformity," requiring that the subsymmetry restricting any element to its own hyperplane must act uniformly on that element. Most uniform polytopes are locally uniform, and the few that are not have an impression of being uniform only "by coincidence" that all facets happen to be uniform. In terms of subsymmetries, they act more like non-uniform scaliforms.

All Wythoffian polytopes are locally uniform, and as far as I can tell, every regiment member of a Wythoffian polytope is also locally uniform. Non-Wythoffian locally uniform regients are rare; the only ones that come to mind are gidrid, the sabbadipady regiment, the regiments of prismatic honeycombs, and hyperbolic tilings from non-tetragonal domains. There may be some intimate connection between local uniformity and a kaleidoscopic construction, because prismatic honeycombs (and even the sabbadipady regiment, I think) can also be constructed kaleidoscopically from domains in the shape of scalene triangular prisms. I am currently working on a nomenclature for locally uniform regiments.

Climbing method and elemental naming scheme are good.
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### Re: How are the snub polyhedra uniform?

Topologically, snubs occur in all dimensions. The trouble is that one has n(n-1)/2 equations in n variables. This is because the full vertex-figure is a simplex, and you are trying to make all the edges equal by placing the vertices. It can be done for 3d, but not generally for higher dimensions.

s3s3s3s topologically has 10 icosahedra, 20 octahedra, and 60 tetrahedra. But you can't make all of these regular at once.

s3s4o3o reduces to two variables in two unknowns. It equates to two kind of vertex (base and apex of a pyramid), and two kinds of edges. It exists as a uniform figure.

o4s3s4o would have icosahedra and tetrahedra as cells, but this can not be made regular, since it's two unknowns in one equation.
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### Re: How are the snub polyhedra uniform?

I'm pretty sure gidrid wouldn't be locally uniform since its snub squares aren't vertex transitive under the symmetries of gidrid. There is a chance that something similar to the symmetries of the squares of the rhombichoron (the squares are regular but they don't have square symmetry) but I don't think that happens.
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### Re: How are the snub polyhedra uniform?

I should mention that taking a truncated cuboctahedron/truncated icosidodecahedron and transforming the edge lengths while preserving the symmetry always gives a semiuniform polytope, while taking the subset of the truncated cuboctahedron/truncated icosidodecahedron that produces the snub cuboctahedron/snub icosadodecahedron and transforming the same way as before almost always leads to a non-semiuniform polyhedron with non-equilateral triangles (in fact, there is only one case out of the uncountably many snub cuboctahedra/snub icosadodecahedra that doesn't lead to non-equilateral triangles.)
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