Cyclic $d$-polytope -- does every facet border every other facet?

A cyclic $$d$$-polytope is "neighborly," where every set of vertices sized $$d/2$$ together make up a face. What I'm interested in: in a cyclic $$d$$-polytope with $$N$$ facet ($$d-1$$ dimensional faces), does every facet "border" every other facet (i.e., share a $$d-2$$ dimensional face)?

Why I think it might: we can use the upper bound theorem in dual form (https://www.cs.mcgill.ca/~fukuda/soft/polyfaq/node12.html) to find the total number of $$d-2$$ dimensional faces for a $$d$$-polytope with $$N$$ facets. If you use the formula in the provided link, we find that there are $$N(N-1)/2$$ such $$d-2$$ dimensional faces. Since every such face borders $$2$$ facets, this total number of $$d-2$$ dimensional faces suggests that the property holds.

2 Answers

The cyclic polytope C(4,6) has nine tetrahedron facets. Each of these facets can only border four other facets, so not every facet borders all other facets.

This unfortunately fails in even the simplest case: when $$d = 2$$.

Consider the cyclic polytope $$C_{2}(T) \subset \mathbb{R}^{2}$$ where $$T = \{ t_{1}, t_{2}, t_{3}, t_{4} \}_{<} \subset \mathbb{Z}$$. It has facets $$F_{i, i + 1} = \mathrm{conv}\{(t_{i}, t_{i}^{2}), (t_{i + 1}, t_{i + 1}^{2}) \}$$, and it's not hard to see that facets $$F_{i, j}, F_{k, \ell}$$ border each other if and only if $$\{i, j \} \cap \{ k, \ell \}$$ is a singleton set.

Consider the facets $$F_{1, 4}, F_{2, 3}$$. They do not meet at a vertex of $$C_{2}(T)$$ since their indexing sets are disjoint.