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Roughly, a cubical complex is like a simplicial complex except all the pieces glued together are combinatorial cubes of various dimensions. A cubical sphere is a cubical complex that is homeomorphic to a sphere. I have encountered papers that distinguish between cubical spheres and cubical polytopes, but I do not understand the distinction. Is there a distinction already in $\mathbb{R}^3$? If so, could anyone provide an example? A reference to clear definitions would suffice as well. Thanks!

My understanding is that, say, the rhombic triacontahedron is both a cubical polytope and a cubical sphere in $\mathbb{R}^3$:
           Rhombic triacontahedron
          Image from Wikipedia article

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Nice image. XD +1 – Patrick Da Silva Jul 21 '11 at 0:57
I think the distinction is that you don't allow "holes" in a cubical sphere. So, for example (I hope I make myself clear): take nine dice and arrange them as a $3 \times 3$-square. Remove the one in the middle. The surface of this certainly gives you a cubical complex that doesn't deserve the name sphere, as it is homeomorphic to a torus. – t.b. Jul 21 '11 at 1:03
@Theo: So is your torus then classified as a cubical polytope? – Joseph O'Rourke Jul 21 '11 at 1:04
I would say so. The square faces can be glued to a polytope along their edges. (A polytope doesn't include convexity assumptions, as far as I know). – t.b. Jul 21 '11 at 1:06
How do you define "combinatorial cubes"? – Michael Lugo Jul 21 '11 at 1:12

2 Answers 2

(Too long for a comment): Here are some thoughts on this circle of ideas though I am not sure what distinctions have been made with these terms in the literature. The diagram you show is a 3-polytope whose surface is built up of 2-dimensional combinatorial cubes, namely 4--gons. However, it is not clear that this 3-polytope or similar 3-polytopes including their interior points can be always be decomposed as 3-cubes that meet along faces. The note on this page which talks about combinatorial cubes: shows a diagram of as 4-cube but it also can be thought of a 3-cube whose interior has been cut up into other combinatorial 3-cubes. As regards the torus, there are some graphs which will not embed on a sphere but will embed on a torus. Now one can ask if one can embed in 3-space a surface (topologically a torus) with flat faces so that the vertex edge graph of this surface is the given graph. When this is possible it is common to call the resulting surface a toroidal polytope. The adjective toroidal overcomes the usage of polytope to be something convex.

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Thanks, Joe, I think you are absolutely correct in focusing on building the 2-surface from glued 3-cubes, as opposed to just looking at the surface alone. (I think in my example it accidentally works out, because it is a zonotope...) – Joseph O'Rourke Jul 21 '11 at 17:34
Joe: The dual of any 4-valent 3-polytope will have all of its faces 4-gons but I would be surprised if all of these 3-polytopes can be decomposed into 3-cubes. – Joseph Malkevitch Jul 22 '11 at 2:53
Perhaps this thesis might be of interest: – Joseph Malkevitch Jul 31 '11 at 13:19
Thanks, Joe, I will look at it! (Sorry for not revisiting this in so long...) – Joseph O'Rourke Aug 28 '11 at 19:58

I am assuming the cubical polytopes are convex - otherwise it's easy to come up with counterexamples - say cubulated tori.

In this case the distinction can be framed as follows. Say you have a cubical convex polytope - this is homeomorphic to a ball so its boundary always gives you a cubical sphere. So the set of cubical polytopes are contained in the set of cubical spheres (in the sense of combinatorial equivalence of their face lattice). The more interesting question is whether each cubical sphere is combinatorially equivalent to the boundary of a convex cubical polytope.

In three dimensions (or equivalently for $2$-spheres) cubical polytopes and cubical spheres coincide. Given any cubical $2$-sphere, its $1$-skeleton is a 3-connected planar graph so we can use Steinitz's Theorem to construct a polytope with the same combinatorial type.

Things get trickier in higher dimensions. I do not know of an explicit example of a non-polytopal $3$-sphere in the cubical case but the simplicial case has been studied extensively. There are many examples of simplicial $3$-spheres that are not combinatorially equivalent to the boundary of any $4$ simplicial polytope. The typical construction uses a knot on $3$ edges inside the simplicial sphere. If the knot is non-trivial the corresponding sphere will be non-shellable and thus not polytopal since every polytope is shellable. See Theorem 1 and 2 here. In the same paper Lutz gives a small example of a non-polytopal sphere on $13$ vertices. I think one could use the same ideas starting from Furch's knotted cube to get a non-shellable cubical 3-sphere.

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