Dimension of a polytope cut with a hyperplane

Question

Let $P \subseteq \mathbb{R}^d$ be a convex polytope and let $H \subseteq \mathbb{R}^d$ be a hyperplane with two sides $H^+$ and $H^-$. Let $V$ be the vertex set of $P$. Suppose $V = A \cup B$ where $A$ and $B$ are non-empty and disjoint, and where $A \subseteq H^+$ and $B \subseteq H^-$. I want to show that $\dim(P \cap H) = \dim(P) - 1$.

Basically I'm cutting a polytope in half and I want to show the dimension decreases by one. This should be a simple question but after many many hours thinking about it I have made no progress. Any help?

Answer 1

The problem you are asking is somwhat connected to the quite old one, which relates a (convex) polytope's vertex definition to its according hyperplane (aka facet) description.

Suppose that one to be solved already. Then your polytope $P$ will be described by its set $S$ of hyperplanes (spans of facets). Now consider the intersections $h_i = s_i \cap H$ for each $s_i\in S$ (possibly empty). Obviously the set of the non-empty $h_i$ then represents a subdimensional hyperplane description of $H\cap P$.

Again you might want to convert that one in turn into its vertex description. It then will be happen that every vertex from that new vertex set $V'\subset H$ clearly must be a point on one of those former edges of $P$ between a point of $A$ and a point of $B$, i.e. the intersection of that edge with $H$.

--- rk

Answer 2

A convex polytope $P$ in $\mathbb R ^d$ is by definition the convex hull of a finite number of points. It is usual to show that this is equivalent to the intersection of a finite number of half-spaces $S_{i, i=1\dots k}$, where this intersection is non-empty and bounded.

We assume WLOG that $P$ has dimension $n$, and we can work in $\mathbb R ^n$: either $n=d$, or the polytope is "flat" in one or more dimensions, and we place ourselves in an affine subspace $E$ that contains $P$ and has same dimension. Then the new $S'_i$ are the intersection of the previous $S_i$ and $E$.

When we cut $P$ with an hyperplane $H$ of $\mathbb R ^d$, there are $2$ cases: $H \cap E = E$, or $H \cap E \subsetneq E$.
If $H \cap E = E$, as $P\subset E, P \subset H$, so the vertex set of $P$ cannot be decomposed into $A$ and $B$ non-empty with $A \subseteq H^+$ and $B \subseteq H^-$. So this case does not happen.

We are left with $H \cap E \subsetneq E$. In this case, as $E$ is an $n$-dimensional affine subspace, $H' = H \cap E$ is an $n-1$ dimensional subspace, i.e. an hyperplane of $E$.

For each point $b \in P' = P \cap H'$, we know that depending upon where $b$ is situated in $P$, its neighborood in $P$ will be homeomorphic to $\mathbb R ^n$ (if $b$ is in the interior of $P$, "interior" being relative to $\mathbb R^n$), or to a half-space of $\mathbb R ^n$ (if $b$ is on the frontier of $P$).
If $b$ is in the interior of $P$, we can find an open neighborood $N$ of $b$ such that $N \subset P$. Then $N$ is homeomorphic to $\mathbb R^n$, it has (topological) dimension $n$. In $P'$, $N' = N \cap H'$ has then dimension $n-1$, as $H'$ is an hyperplane of $E$. This shows that dimension is reduced by $1$.