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Given any $n$-dimensional convex polyhedron in $\Bbb R^n$.

Question: Is it always possible to find a combinatorially equivalent polyhedron with vertices only on the sphere $S^{n-1}\subset\Bbb R^n$, i.e. an equivalent sphere inscribed polyhedron?

One can also ask it this way: can I build (from a combinatorial point of view) any polyhedron by choosing some points on a sphere and taking the convex hull?

I mean, this is easy to see for polyhedra whose faces are all simplices, but it seems pretty hard in general. You cannot just project all vertices to the sphere because "coplanar" vertices might be no longer "coplanar" afterwards, hence cannot describe an equivalent face.

Extra:

  • If not in general, is it possible for $n=3$?
  • Can this be made easier by allowing the vertices to be on any sphere $S^m$ with $m\geq n-1$.

Update:

I should have been more precise. The sphere inscribed polyhedron should be convex too. Also, preferrably no two distinct faces of the polyhedron should be coplanar, otherwise they can be combined to a single face. In this sense, the Triakis tetrahedron is indeed a counter-example to my question. There are inscribed polyhedra equivalent to it, but they are either not convex or have coplanar faces.

M. Winter
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  • As you say, this is easy to see for polyhedra whose faces are all simplices, but then you mention the triakis tetrahedron in the update and all of its faces are triangles! Apparently, I cannot just project its vertices to a sphere and obtain that all dihedral angles are strictly less than $180^\circ$? – Jeppe Stig Nielsen Mar 16 '24 at 13:00
  • Yes, my statement about simplicial polyhedra was simply wrong. I learned of the Triakis tetrahedron only later. – M. Winter Mar 18 '24 at 10:25
  • I learn that every convex polyhedron can be brought on a form where all edges are tangents to a unit sphere, and additionally the barycenter of all the points where the edges touch the sphere is the center of that sphere. Is also mentioned on this question which mentions en passant the impossibility of what you originally asked. See also Wikipedia: Midsphere § Canonical polyhedron etc. – Jeppe Stig Nielsen Mar 18 '24 at 17:17

1 Answers1

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This is not possible in general, or even in particular when $n=3$. See this paper of Ziegler for a reference.

I would guess that this is possible if I allow the sphere to have arbitrarily high dimension (maybe you can take $S^n$ as the one point compactification of $\mathbb{R}^n$ and use stereographic projection to inscribe your $n$-polytope in $S^n$, but I'm not sure.)

Tachyon
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    Quick quote for the curious: "For example, the triakis tetrahedron, a convex polytope obtained by stacking a tetrahedron onto each facet of a tetrahedron, is not inscribable." –  Jun 05 '17 at 01:11
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    @Rahul It is not inscribable as a strictly convex polytope, but there is a combinatorially equivalent polytope with its vertices on $S^2$. Do you know of an example where no such inscribed combinatorial equivalent polytope exists, convex or not? – M. Winter Jun 14 '17 at 11:09