Each convex polytope $P$ has a combinatorial type, its so-called face lattice. This lattice is just the poset of all faces of $P$ ordered by inclusion. Given one realization of such a combinatorial type, one can easily get many others. Just apply an arbitrary projective map to the given realization and the image will be combinatorially equivalent.

Now it would be very nice if one could realize every combinatorial type with vertices on the sphere. Unfortunately, this is not possible. For example, consider the octahedron with pyramids stacked on each facet -- this cannot have all vertices on the sphere (and still be convex).

So my questions is:

Is there a convex body in $\mathbb{R}^d$ such that every possible combinatorial type of a $d$-dimensional convex polytope can be realized with vertices on its surface?

  • $\begingroup$ I don't understand. Am octahedron with a pyramid glued to each facet is not a convex polytope, is it? In what sense is it a possible combinatorial type of a convex polytope? $\endgroup$ Jun 8, 2012 at 12:16
  • $\begingroup$ @GerryMyerson: Consider a facet $F$ of the octahedron $O$. Let $x$ be a point outside $O$ but very close to the centroid of $F$. Taking the convex hull of $\{x\} \cup P$ yields a polytope that is an octahedron with a pyramid stacked on top of $F$. Repeat this construction for every facet of $O$. This yields a convex polytope of the kind I meant. $\endgroup$
    – Gregor
    Jun 8, 2012 at 12:29
  • $\begingroup$ Thanks - I was imagining regular tetrahedral pyramids, but you keep things convex by making them much flatter. So now what I don't see is why the vertices can't all be on a sphere. $\endgroup$ Jun 8, 2012 at 13:13
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    $\begingroup$ I'll check out those references when I can, but for now, what's wrong with this: Put a point at the North Pole, and one at the South Pole, and four points on the equator, and join each Pole to each equatorial point, and draw the Equator. That's your octahedron, and it divides the globe into 8 regions. Put a point in each region, and join it to the three vertices of its region. It seems to me that gives you the extended octahedron, with all vertices on the sphere. $\endgroup$ Jun 9, 2012 at 0:58
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    $\begingroup$ @GerryMyerson That wouldn't be convex anymore. One can imagine the new points placed with that construction would create an inscribed cube, whose edges are outside the original octahedron. $\endgroup$
    – Nick Alger
    Sep 24, 2012 at 18:06

1 Answer 1


I asked this question on MathOverflow and Sergei Ivanov gave an answer in the affirmative. See here for his beautiful argument.


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