The number of $(d-1)$-faces in a $d$-polytope is at least $(d+1)$

Question

I have to try to prove the following:

Let $V = {v_1,...,v_n} \subset \mathbb{R}^d$ be a point configuration affinely spanning $\mathbb{R}$ (i.e., $\operatorname{aff}(V) = \mathbb{R}^d)$. Let $H$ be the collection of hyperplanes spanned by V.

Show that every $d$-polytope has a $(d-1)$-face and show that every $d$-polytope has at least $(d+1)$ faces of dimension $(d-1)$.

I don't really know what I have to do in this exercise, because it seems to be clear but I cannot write a proof.. I know for example a cube ($d=3$) has ($d-1$)-faces, so we get a $2$-face, but how can I show that for all $d$-polytopes?

Answer 1

I'll assume that the definitions we're using are these:$\def\R{\mathbb{R}} \DeclareMathOperator{\aff}{aff}$

• The dimension of a subset of $\R^d$ is the dimension of its affine hull.
• A polytope is the convex hull of finitely many points in $\R^d$, and a $d$-polytope is whose dimension is $d$.
• A supporting hyperplane of a polytope $P$ is a hyperplane $H$ with a nonempty intersection with $P$ such that $P$ is contained in one of the two closed half-spaces determined by $H$.
• A face of $P$ is any intersection of $P$ with a supporting hyperplane, and an $i$-face is a face whose dimension is $i$.

There is a $(d-1)$-face.

Let $P$ be a $d$-polytope, and $F = P \cap H$ be a $j$-face for the maximum possible $j$. (By our definition of face, $ j \leq(d-1)$.) Here $H$ is a supporting hyperplane.

If $j \leq d-2$, then $\aff(F)$ is a codimension-2 subspace contained in $H$. Consider a family of hyperplanes $H'$ obtained from $H$ by rotating it about the axis $\aff(F)$. Every such hyperplane contains $\aff(F)$ (and so contains $F$).

Suppose that we rotate just until $H'$ first contains another vertex $v$ of $P$ besides those in $F$. Then $H' \cap P$ is a face of $P$, containing $F \cup \{v\}$. Since $v \notin \aff(F)$, $F \cup \{v\}$ has strictly larger dimension than $F$ does, contradicting our choice of $F$.

Otherwise, we can rotate $H$ all the way around without encountering another vertex of $P$, which means that $P$ is contained in $\aff(F)$; but this contradicts that $P$ is $d$-dimensional.

Therefore, we must have $j = d-1$, and $F$ is a $(d-1)$-face (or facet.)

At least $d+1$ of them.

We need this fact:

Lemma. Each vertex of $P$ is the intersection of all the facets containing it.

Perhaps you already know this. If not, here's a sketch of a proof.

Let $v$ be a vertex of $P$. Obviously, $v$ is contained in the intersection of all the facets containing $v$. Suppose that this intersection also contains some point $w$, $w \neq v$. Then let $H$ be a supporting hyperplane of $P$ such that $H \cap P = \{v\}$ (such a hyperplane exists by our definition of face; a vertex is a 0-face.) We can rotate $H$ away from $w$ until it meets another point of $P$; we can keep rotating, away from $w$, until the hyperplane meets $P$ in a facet. But this facet cannot contain $w$. $\Box$

The intersection of $j$ hyperplanes has dimension $d-j$ (if it's not empty), so a vertex is the intersection of at least $d$ hyperplanes. (Equivalently, you need at least $d$ equations in $d$ unknowns for there to be a unique solution.) So there must be at least $d$ facets containing each vertex, hence at least $d$ facets of $P$. But $P$ has more than one vertex, so there have to be at least two distinct sets of $d$ facets. Hence $P$ has at least $d+1$ facets.