Let $C$ be a $d$-dimensional (abstract) polytopal complex. Most of what I say below could be asked in this general setting, but for a start, let's further restrict to simple polytopal spheres, that is, $C$ is homeomorphic to the $d$-sphere, and at each vertex meet exactly $d+1$ facets.

My goal is to formalize and then prove a certain kind of statement. For a warmup, here are two examples of what I am after:

  • Assume that each vertex of $C$ is incident to $d+1$ facets, each one combinatorially equivalent to the $d$-cube. From that I can already deduce that $C$ is combinatorially equivalent to the (boundary of the) $(d+1)$-cube.
  • Assume that $d=2$ and each vertex is incident to a 4-gonal cell and two 6-gonal cells. This already suffices to conclude that $C$ is combinatorially equivalent to the (boundary of the) permutahedron.

In general, I want to prove something like this:

If each vertex of a simple polytopal sphere $C$ looks the same locally (that is, is incident to the same type of facets), then $C$ is already uniquely determined.

So how to prove this?

Intuitively, this is easy. Take, for example, the second example from above (and see the image below): if you try to draw its edge graph, you will start with a single vertex (the white dot) surrounded by a 4-gon and two 6-gons. You then see that there are several vertices for which only a single face is missing (the black dots), and you add these. You go on until you realize that now all the vertices satisfy the constraint. You never made a choice, and so the result must be unique.

My question is the following:

Question: How can this proof be formalized?

There seem to be some subtleties, some are hinting at certain generalizations:

  • You need to use that $C$ is a polytopal sphere, otherwise the complex might not be unique. For example, there are infinitely many possibilities to realize a polytopal torus in which every vertex is incident to exactly three 6-gons. Maybe "sphere" can be replaced by simply connected, though.
  • We do not necessarily need simple, but it is the "simplest" of several conditions that ensure that we never have to make a choice when placing the next facet. Here is a case where above reasoning fails because we dropped simplicity: if we want every vertex of a 2-dimensional complex to be surrounded by a 3-gon and three 4-gons, there are two solutions: this one and this one.
  • How do I know that there is always a vertex at which the next facet is determined?
  • How can one be sure that the order in which I add new facets (which are forced on me) does not make a difference. Or does it?
  • 1
    $\begingroup$ Re: the last question - "How can one be sure that the order in which I add new facets (which are forced on me) does not make a difference. Or does it?" - since a facet can never stop being forced once it is forced, I think a routine application of the 'diamond lemma' proves confluence, i.e., that it does not matter which order you carry out the forced facet additions. $\endgroup$ Mar 27, 2020 at 14:09
  • $\begingroup$ @SamHopkins This was already a very helpful hint. I wasn't aware of this lemma. Just a note: one might want to apply above reasoning to tilings of $\Bbb R^n$ too. In that case, the process of adding further facets isn't finite. If I understood correctly, this finiteness is an important condition in the diamond lemma. Is there away around this? $\endgroup$
    – M. Winter
    Mar 27, 2020 at 14:50
  • $\begingroup$ In general there is no way around some kind of finiteness condition to apply the diamond lemma. $\endgroup$ Mar 27, 2020 at 15:21

1 Answer 1


Sorry, but there are counterexamples to your local to global quest:

Consider the uniform small rhombicuboctahedron versus the elongated square gyrobicupola (Miller's solid, one of the Johnson solids). - Both show up the same vertex figures all over, still they are globally different (i.e. non-unique) outcomes (polyhedra).

--- rk

  • $\begingroup$ This counterexample is already contained in my post in the third to last bullet point. This is why I need the rule for adding one more tile that the placement must be unique, as it is in the simple case. $\endgroup$
    – M. Winter
    Dec 18, 2020 at 12:53
  • $\begingroup$ Just want to let you know that similar finds to Miller's solid lately were found within 4D, cf. polytope.net/hedrondude/polychora.htm#pseudouniform $\endgroup$ Jul 18, 2021 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.