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The following question was asked by me on Mathematics.SE. Unfortunately, no one answered it so I thought I might give it a try one level higher. Below the line you can find the slightly edited question, the original one can be found here.


Every 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 combinatorial type of a d-dimensional convex polytope can be realized with vertices on its surface?

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up vote 15 down vote accepted

Yes, there is such a body. Actually there is one very close to the standard unit ball and containing disjoint representatives of each combinatorial type (but these representatives are very small).

Indeed, every combinatorial type of a $d$-polytope has a realization which looks as follows: there is a "large" $(d-1)$-dimensional facet and the remaining surface is a graph over this facet. To construct such a realization, choose a $(d-1)$-facet, pick a hyperplane parallel to it and very close to it (but not intersectiong the polytope), and apply a projective map which sends this hyperplane to infinity.

We can further "flatten" this realization so that it is very close to its large face. Choose a very small $\varepsilon>0$, apply a homothety such that the diameter of the polytope becomes less than $\varepsilon$, and place the resulting tiny polytope so that it touches the sphere by a point on its "large" face. Then consider the convex hull of the sphere and the polytope. All vertices will be on the boundary of this convex hull if the polytope is sufficiently "flattened". And the convex hull diverges from the ball only in a neighborhood of size $\sim\sqrt\varepsilon$.

Now pick another combinatorial type of a polytope and repeat the procedure with a much smaller $\varepsilon$ and a location on the sphere chosen so that the neighborhood affected by the second polytope does not interfere with the first one. And so on. Since there are only countably many combinatorial types, they all can be packed into the sphere, provided that $\varepsilon$ goes to 0 sufficiently fast.

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What a beautiful argument!!! –  Joseph O'Rourke Sep 13 '12 at 19:33
    
A lovely argument, indeed. Thanks a lot! –  Gregor Samsa Sep 13 '12 at 21:21
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