It's easy to prove that the convex hull of any vertex-transitive polyhedron is vertex-transitive. Specifically, any symmetry of the original polyhedron that moves any vertex to another will also move the same vertices on the convex hull, but since it will preserve the general shape of the original polyhedron (and therefore the set of vertices) it must preserve the convex hull.

However, my question is the following: Are the convex hulls of edge-transitive polyhedra edge-transitive, and are the convex hulls of face-transitive polyhedra face-transitive?

I have gathered evidence backing up this, mostly from uniform polyhedra, but I can't use the same technique as before to support my claim. Does anyone know why is it true or does anyone have a counterexample?

  • $\begingroup$ Very nice question, but you should modify your proof in the vertex-transitive case. The reason it holds is that the vertex set of the convex hull is a subset of the original vertex set. Thus fails for edges and faces, which makes the problem interesting. +1. $\endgroup$ – Moishe Kohan Mar 22 '17 at 0:33
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    $\begingroup$ Isn't it true that a non-convex polyhedron is neither vertex-transitive, edge-transitive or face-transitive? (For example, if it is non-convex, some of vertices will be in the convex hull, some not, and similarly for edges and faces.) In particular, assuming some kind of transitivity, the polyhedron is convex and thus equal to its own convex hull. $\endgroup$ – verret Mar 22 '17 at 0:57
  • $\begingroup$ I don't get the problem. There can be non-convex polyhedra that are vertex, edge and face-transitive, Kepler-Poinsot polyhedra are a well known example. In fact, my original intent in answering this question was to find all of those polyhedra. So, what's wrong with what I asked? $\endgroup$ – Anonymous Pi Mar 22 '17 at 1:05
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    $\begingroup$ Could you clarify which definition of polyhedron you are using? (In the non-convex case, there are many competing ones.) $\endgroup$ – verret Mar 22 '17 at 1:48
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    $\begingroup$ @verret For me, a polyhedron is a set of faces and edges, such that exactly two faces meet at each edge. With some extra conditions, particularly not being a compound and having no two elements as coincident. I know this isn't the most orthodox definition out there but it's the one that most suits my needs. $\endgroup$ – Anonymous Pi Mar 22 '17 at 1:53

I'm going to assume that we're dealing with polyhedra whose faces are planar polygons (possibly self-intersecting, like a pentagram, but contained in a plane), with no two co-planar.

Here's a very sketchy proof sketch.

Now, in Grünbaum and Shephard's article Duality of Polyhedra, they are attempting to grind the axe "Duals are not as well-defined as everyone seems to think they are", but what they actually say is that for polyhedra with planar, possibly self-intersecting faces, a dual can always be constructed by polarity with matching symmetry/transitivity/regularity properties.

In particular, a face-transitive polyhedron has all its face centroids lying on a sphere, the center of which is a fixed point of its symmetry group, which is the natural choice for the center of the sphere of reciprocation.

So, a face-transitive polyhedron $P$ has a dual $P^*$ which is vertex-transitive. As you say, $\operatorname{conv}(P^*)$ is a vertex-transitive convex polytope, and so its dual, $\operatorname{conv}(P^*)^*$ is a face-transitive convex polytope.

Now, the question is, is the dual of the convex hull of the dual the same as the convex hull of the original polytope? I.E. does $\operatorname{conv}(P^*)^* = \operatorname{conv}(P)$? And my claim is "Sure, what else would it be?"

That leaves the case of edge-transitive polyhedra. But every edge-transitive polyhedron is also either vertex-transitive or face-transitive.

You can find this claim on the Wikipedia page Isotoxal figure ("An isotoxal polyhedron or tiling must be either isogonal (vertex-transitive) or isohedral (face-transitive) or both."), and also in Peter Cromwell's 1999 book Polyhedra, in both cases appearing to apply to star polyhedra as well as convex ones. I'm sure that I once dug up a proof of this, but I forget where.

Perhaps you can make some progress closing up the gaps I left you. ;-)

  • $\begingroup$ That's actually a pretty nice answer. I'll think about it :) $\endgroup$ – Anonymous Pi May 11 '17 at 15:23

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