Uniform polyhedra with "hidden faces"

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

Uniform polyhedra with "hidden faces"

Postby mr_e_man » Fri Oct 08, 2021 6:28 am

Supposedly, there are 4 regular and 53 other uniform non-convex polyhedra, excluding prisms and antiprisms. But this count relies on certain definitions. One requirement in particular seems unreasonable:

Hidden faces. Some polyhedra have faces that are hidden, in the sense that no points of their interior can be seen from the outside. These are usually not counted as uniform polyhedra.

This doesn't make sense, because a self-intersecting polytope is not a solid....

So which polyhedra would be counted as uniform, when this requirement is dropped?

I define a (non-degenerate, Euclidean) polyhedron as a realization F of an abstract polyhedron, where each abstract k-face X is realized as a k-dimensional subspace F(X) in Euclidean space, and the partial order is realized as the subspace relation (so X≤Y in the abstract poset if and only if F(X)⊆F(Y) in Euclidean space). Note, this means that an edge is realized as an entire line, not a segment; but that's not important. I think this is not too different from McMullen & Schulte's definition.

Of course we should focus on finite polyhedra; no apeirohedra.
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Re: Uniform polyhedra with "hidden faces"

Postby ubersketch » Fri Oct 08, 2021 6:45 pm

I don't think uniform polyhedroids with hidden faces, but otherwise fulfill every other condition for uniformity, exist. They do exist in 5D though and we usually consider those cases are uniform despite the hidden faces.
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Re: Uniform polyhedra with "hidden faces"

Postby mr_e_man » Mon Oct 11, 2021 3:30 am

What's a polyhedroid?


Let's try to prove that there can't be hidden faces in any uniform polyhedron P.

All vertices of P are on a sphere (by symmetry). All vertices adjacent to one given vertex are on another sphere (because the edges have the same length), and thus on the intersection of these two spheres which is a circle. So the verf (call it Q, a spheric polygon) has its vertices on a circle.

Points on a circle are convexly independent; none of them is inside the convex polygon formed by the other points. So any vertex of Q is visible from outside, not surrounded and hidden by edges of Q. It follows that any edge of Q has some part (near a vertex) visible from outside.

Returning to 3D, this means that any face of P has some part (near an edge and a vertex) visible from outside.


But why would they refer to hidden faces at all, if they never occur anyway (in 3D)?

I doubt it's based on the existence of higher-dimensional uniform polytopes with hidden parts, because higher dimensions are hard to visualize, and having hidden parts is a visual property.

I think it's based on degenerate polyhedra like these five. Are those even valid abstract polyhedra? Or are they compounds? In any case, I would say they're degenerate because of the coincident edges, not because of the hidden faces.


In fact some of the 53 accepted polyhedra are degenerate according to my definition: seside, sirsid, gidrid, and gisdid have coplanar faces. Any others?

Degeneracy can happen in several ways: coincident vertices, collinear edges, coplanar faces, vertices on the wrong edges (impossible for uniforms because a line intersects a sphere in at most 2 points), vertices in the wrong faces (planes), or edges in the wrong faces (planes).
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Re: Uniform polyhedra with "hidden faces"

Postby galoomba » Tue Nov 09, 2021 10:13 pm

mr_e_man wrote:I think it's based on degenerate polyhedra like these five. Are those even valid abstract polyhedra? Or are they compounds? In any case, I would say they're degenerate because of the coincident edges, not because of the hidden faces.

Those polyhedra are considered "exotic" because they have 4 faces at some edges, the definition of a polytope requires there to be exactly 2 facets at each ridge.
mr_e_man wrote:In fact some of the 53 accepted polyhedra are degenerate according to my definition: seside, sirsid, gidrid, and gisdid have coplanar faces. Any others?

Degeneracy can happen in several ways: coincident vertices, collinear edges, coplanar faces, vertices on the wrong edges (impossible for uniforms because a line intersects a sphere in at most 2 points), vertices in the wrong faces (planes), or edges in the wrong faces (planes).

Coplanar faces are not a problem, though there is an issue in higher dimensions when those facets share elements, polytopes with those are called "coincidic" and not included in lists. To me that sounds kinda arbitrary, Bowers actually uses an even more arbitrary version of the rule (coplanar facets may share vertices, but not edges).
I think with "edges in the wrong faces (planes)" you're describing what Bowers calls "wild" or "intercepting" cases, they happen in 4d already, he considers them valid uniforms.
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Re: Uniform polyhedra with "hidden faces"

Postby mr_e_man » Thu Nov 11, 2021 7:56 pm

galoomba wrote:
mr_e_man wrote:I think it's based on degenerate polyhedra like these five. Are those even valid abstract polyhedra? Or are they compounds? In any case, I would say they're degenerate because of the coincident edges, not because of the hidden faces.

Those polyhedra are considered "exotic" because they have 4 faces at some edges, the definition of a polytope requires there to be exactly 2 facets at each ridge.

Well, you're considering those as single edges. They can instead be considered as coincident pairs of edges, with 2 faces meeting at one edge, and another 2 faces meeting at the other edge. Of course the 4 faces appear to touch both edges; that's an example of what I meant by "edges in the wrong faces".

galoomba wrote:Coplanar faces are not a problem

What do you mean?

I can see that polyhedra with coplanar faces are commonly accepted as non-degenerate. I do not accept them.

galoomba wrote:I think with "edges in the wrong faces (planes)" you're describing what Bowers calls "wild" or "intercepting" cases, they happen in 4d already, he considers them valid uniforms.

I'm describing an edge contained in the plane of a face, which is not supposed to be incident with that face. This violates my requirement that the incidence relation be exactly the same as the subspace-containment relation.

Indeed, any wild polytope violates this requirement. For, if it satisfies this requirement, then there's a pair of elements of dimensions d+1 and d-1, both incident with at least three elements of dimension d, which violates the dyadic property.

I have this requirement because, otherwise, it's difficult or impossible to tell which elements are supposed to be incident with each other, without "cheating" and looking at the corresponding abstract polytope. Such difficulty is what characterizes degenerate polytopes.
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