Generalization of Steffen's polyhedron?

[quickfur]

Today I stumbled across Steffen's polyhedron, which is a non-convex, closed polyhedron that has flexible dihedral angles in spite of being a closed shape. (It is proven that no convex polyhedron can be flexible.)

It makes me wonder, is it possible for a non-convex 4D polytope to have flexible vertices, in spite of Marek's proof? Or is this only possible in 3D? An initial idea I have is to assemble a polychoron out of Steffen's polyhedra. Since the dihedral angles of these cells would then be flexible, the vertex figure involving such a polyhedron also ought to be flexible. Not sure if such a shape can be closed up into a polychoron while maintaining flexibility, though!

[Klitzing]

Indeed, it is possible to build flexible polychora. - Cf. http://www.geometrie.tuwien.ac.at/stachel/cross.pdf.
(That paper even is mentioned on https://en.wikipedia.org/wiki/Flexible_polyhedron#Generalizations.)
--- rk

[quickfur]

Hmm. So how does this relate to Marek's proof of 4D vertex rigidity? Does the proof only apply when convexity holds? Even though on the surface convexity does not seem to be a requirement for the proof. Or is the "loophole" because the vertex figure itself is a flexible polyhedron?

[student91]

I do not yet understand the point he is trying to make; He takes a cross-polytope, divides the vertices into two groups, then changes the squares in these planes into cyclic antiparrallelograms, and then slightly later states that "All 2-faces of [the cross-polytope] are triangles", but by changing the squares some of your 2-faces become (regular) squares, which by the motion described will fold themselves over the edge in the antiparallelograms. So either he makes the triangles by splitting up the squares into two triangles, or he does not yet understand how the polytope he described is constructed. Maybe we want an incidence-matrix of some sort sometimes here...

So to answer your question, it seems most vertices have a vertex-figure like in bricards octahedron, which is flexible, and thus the whole polygon is flexible (Or to say otherwise, the length of the shortchords used in marek's proof is not constant)

[Klitzing]

At least I just managed to dig out the contact infos for both, if that helps:

• Prof. Hellmuth Stachel : cf. http://www.geometrie.tuwien.ac.at/stachel/
• Prof. Anke Walz : cf. https://www.kutztown.edu/academics/coll ... e-walz.htm

Both do mention flexibility research as one of their main topics. Anke even did her PHD on that topic.

--- rk

[quickfur]

[...]
So to answer your question, it seems most vertices have a vertex-figure like in bricards octahedron, which is flexible, and thus the whole polygon is flexible (Or to say otherwise, the length of the shortchords used in marek's proof is not constant)

Yes, I think that's the key point here. The vertex figure is a flexible polyhedron, so its shortchords are non-constant, therefore the rest of Marek's proof does not apply to it. Of course, since it has been proven that flexible polyhedra cannot be convex, it follows that any such flexible polychora also cannot be convex. So we cannot expect any CRFs to come from this.

OTOH, an interesting question to explore is, is it possible to retain the unit-edge constraint and construct a unit-edged polychoron that contains one of the Johnson crown jewels as cells? The idea being, to somehow use the flexibility of a flexible polyhedron to bridge the unusual chord lengths of a crown jewel such that everything closes up with unit edges. Though that would also require the flexible polyhedron to have unit edges; I'm unsure if such exists. But it might be possible to substitute faces with non-unit edges with unit-edged polygon complexes, I'm not 100% sure.