# Uniquely describing a polytopal complex by prescribing the local structure around its vertices

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?
• 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. – Sam Hopkins Mar 27 '20 at 14:09
• @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? – M. Winter Mar 27 '20 at 14:50
• In general there is no way around some kind of finiteness condition to apply the diamond lemma. – Sam Hopkins Mar 27 '20 at 15:21