# Polytope where each vertex belongs to all but two facets

Let $$P$$ be a (convex, bounded) polytope with the following property: for every vertex $$v$$, there are exactly two facets which do not contain $$v$$. Does it follow that $$P$$ is (combinatorially) a Cartesian product of two simplices?

Some remarks

• by "facet" I mean "face of codimension $$1$$"
• a Cartesian product of two simplices has the desired property
• this is trivially true for $$\dim P = 2$$
• this is true for $$\dim P = 3$$: by playing with the Euler formula and the list of polyhedra with small $$f$$-vectors, one checks that only the triangular prism satisfies the condition
• it is a classical fact that a $$n$$-dimensional simple polytope with $$n+2$$ facets is a Cartesian product of two simplices. So the answer is positive if $$P$$ is assumed to be simple.
• The triangular prism (=cartesian product of a triangle and an interval) has 5 facets. Any given vertex is contained in 3 facets and therefore in all facets except 2. Commented Sep 1, 2021 at 18:14
• Just curious, is it the case that the product of $n$ simplices (of any dimensions) has each vertex not belong to exactly $n$ facets? Commented Sep 1, 2021 at 18:25
• Yes this is true. Also note that the converse to your statement is false for $n \geq 3$: for $n=3$ the pentagon (in dimension $2$) and the "pentagonal wedge" depicted in en.wikipedia.org/wiki/Hexahedron#/media/File:Hexahedron4.svg (in dimension 3) are counterexamples. Commented Sep 1, 2021 at 18:38
• fyi: A quick check with sage reveals that this is also true for all polytopes of dimension 4 with less than 10 vertices (or less than 10 facets). "playing with the Euler formula" might then also give the result for all polytopes in dimension 4. Commented Sep 1, 2021 at 19:49
• Is it equivalent to say: If P is a convex polytope in which every vertex is contained in all but two facets, then P must be simple. Commented Sep 1, 2021 at 20:07

There are other polytopes with this property that can be obtained via the free join construction.

Given two polytopes $$P_1\subset\Bbb R^{d_1}$$ and $$P_2\subset\Bbb R^{d_2}$$, the free join $$P_1\bowtie P_2$$ is obtained by embedding $$P_1$$ and $$P_2$$ into skew affine subspaces of $$\Bbb R^{d_1+d_2+1}$$ and taking the convex hull.

The claim is that if $$P_1$$ and $$P_2$$ have your property, then so does $$P_1\bowtie P_2$$. To see this, note that the facets of $$P_1\bowtie P_2$$ are exactly of the form $$P_1\bowtie f_2$$ and $$f_1\bowtie P_2$$, where $$f_i$$ is a facet of $$P_i$$. Now, if $$v$$ is a vertex in, say, $$P_1\subset P_1\bowtie P_2$$ that is in all facets of $$P_1$$ except for $$f,f'\subset P_1$$, then $$v$$ is in all facets of $$P_1\bowtie P_2$$ except for $$f\bowtie P_2$$ and $$f'\bowtie P_2$$. Equivalently for vertices in $$P_2$$.

Example. Take the free join of two squares, which is a 5-dimensional polytope with 8 vertices, 8 facets and vertex degree 6. This polytope is not simple and can thus not be the cartesian product of two simplices.

This construction yields counterexamples in dimension $$d\ge 5$$. Concerning $$d=4$$, there should not be any counterexamples in this dimension.

Suppose that $$P\subset\Bbb R^4$$ is a counterexample. According to Moritz Firsching's comment we can assume that it has $$n\ge 10$$ facets. Each vertex is then contained in exactly $$n-2\ge 8$$ of them. More generally, each set of $$k$$ vertices is contained in at least $$n-2k$$ facets. But let $$f\subset P$$ be a 2-face of $$P$$ and $$v_1,v_2,v_3\in f$$ three affinely independent vertices, then this set of vertices, and thus also $$f$$, is contained in $$n-6\ge 4$$ facets. This cannot be, since each 2-face of $$P$$ is contained in exactly two facets.

I believe this generalizes to show that any polytope with your property must have $$\le 2d$$ facets.

• Given that this construction is also "canonical" like the one with the Cartesian product, one might ask a modified question whether these are the only possible cases. Starting in the reverse direction: Maybe there are some highly/strongly regular (etc.) graphs which can be casted as polytopes in a suitable number of dimensions without being any kind of "product"? Commented Sep 3, 2021 at 6:30
• In fact, certain eigenspaces of association schemes could be the place to look for. Commented Sep 3, 2021 at 6:51
• The free join of two squares also has $8$ faces, as opposed to $7$ for both $S_1\times S_4$ and $S_2\times S_3$. Commented Sep 3, 2021 at 6:56
• Great! That constructions gives counter-examples for dimension 5 and up! In dimension 3, there aren't any, which leaves the question open for dimension 4, I guess... Commented Sep 3, 2021 at 11:32
• @MoritzFirsching - my failed attempt: in dimension $4$ (as in all others) cutting a symplex with a generic hyperplane creates creates pairs of polytopes all with the required property (except in case of a half with only one old vertex), but they all turn out to be likely products of $2$ simplexes, in that they have exactly the same number of edges of each dimension as those. Commented Sep 3, 2021 at 12:13