Every ridge in the convex polytope is an intersection of exactly two facets

Question

This statement is used without proof in various sources (incl. Ziegler's "Lectures on polytopes" and Gruber's "Convex and discrete geometry"), yet I don't have any idea how to prove it, or have intuition about it in $n>3$ dimensions. Someone suggested me to use a concept of dual polytope to prove it, I have not explored it thoroughly and maybe someone can help with understanding how duality is convenient here.

Answer 1

You do not need to know what the dual poltype is in detail, but it is helpful to known about the face-lattice. All you need is that the dual polytope has the same face-lattice as the original polytope, but turned up-side-down, in particular, every $\delta$-dimensional face of the original polytope becomes a $(d-\delta-1)$-dimensional face of the dual.

The facets are the faces of dimension $d-1$, and the ridges are the faces of dimension $d-2$. In the dual polytope, every facet becomes a vertex (face of dimension $0$), and every ridge becomes an edge (face of dimension $1$).

Obviously, every edge of the dual contains exactly two vertices. Since the face-lattice of the dual and the face-lattice of the original are so similar, this statement translates to "every ridge $R$ of the original polytope is contained in exactly two facets, say $F_1$ and $F_2$".

You known that the intersection of faces of a polytope is again a face of that polytope. So what could the intersection $F_1\cap F_2$ be? This intersection is a face that contains the ridge $R$ since both facets contain this ridge. Since the ridge is of dimension $d-2$, the only faces that can contain $R$ are $R$, $F_1$, $F_2$ and maybe another facet (and the polytope itself). But we have discussed previously that $F_1$ and $F_2$ are the only facets that contain $R$. Hence $F_1\cap F_2=R$.

Answer 2

Your quest on proving that statement is just like proving that there is just a single parallel line to a given line. The latter seems obviously true within euclidean spaces, so one might be tempted to prove it, but it is neither true in spherical (less than 1) nor in hyperbolical ones (more than 1). In fact, in history right that trial to prove it by means of a contradictional attempt directly lead to the discovery of those other space types.

And in fact quite similarily, there are polytopes, which have 2 facets per ridge (the usual real space polytopes), but there are realizations of according abstract polytopal structures with 3 or more facets per ridge. This then is the realm of so called complex polytopes.

Cf. to https://en.wikipedia.org/wiki/Complex_polytope