Prove that any face of a convex polyhedron is a convex polyhedron in the affine space spanned by it

Question

As the question states above, I can understand the concept and why this is true visually but I have no idea how that would translate into a proof, its geometrically obvious but how would we mathematically prove such a statement?

Answer 1

Sorry for the wordy reply, but here goes. I assume you mean polytope instead of polyhedron (otherwise, if you are in $\mathbb{R}^3$, it should be polyhedron and polygon. But this is just semantics.

So let us assume you have a polytope in $\mathbb{R}^d$. This can be defined from a collection of $n$ hyperplanes as the intersection of the 'positive half-spaces' of each of the $n$ hyperplanes.

To be a bit more formal, consider the generic equation of a hyperplane $H_{a,b}$ in $\mathbb{R}^d$, $\langle a,x\rangle +b=0$, where $a\in\mathbb{R}^d$ and $b$ is just a scalar. The 'positive half-space' of this hyperplane can be defined as $H_{a,b}^+:=\{x\in\mathbb{R}^d: \langle a,x\rangle +b\geq 0\}$. Note that the restriction to the positive part is conventional, since if you would want the inequality to read $\leq 0$, you can just swap $a$ with $-a$ and $b$ with $-b$.

So with this notation, for every polygon $P$ there is some number $n=n(P)$ (the number of faces essentially) and $n$ hyperplanes given by the pairs $(a_1,b_1),\dots,(a_n,b_n)$ such that $P=\cap_{j=1}^n H_{a_j,b_j}^+$. To see that this is equivalent to the perhaps more common definition of a polytope, note that from this we can also get $P$ as the convex hull of the points obtained from intersections between the hyperplanes $H_{a_j,b_j}$.

With this definition, a particular face of $P$ can be defined as $H_{a_k,b_k}\cap P$ or equivalently as $H_{a_k,b_k}\cap\big(\cap_{j\neq k}H_{a_j,b_j}^+\big)$. Let us call this face $F_k$ and let $J:=\{j\neq k: H_{a_k,b_k}\cap H_{a_j,b_j}\neq \varnothing\}$.

Let us have another look at the equation of the hyperplane $H_{a_k,b_k}$. Written out, this would be $a_{k,1} x_1+a_{k,2}x_2+\dots+a_{k,d}x_d+b_k=0$. Since we want this to be a proper hyperplane (and not the whole of $\mathbb{R}^d$), at least one of the coefficients $a_{k,\ell}$ needs to be non-zero. For simplicity and wlog, let us assume that $a_{k,d}\neq 0$. This means that any point in $H_{a_k,b_k}$ will have the form $(x_1,x_2,\dots,x_{d-1},-\frac{b_k+\sum_{\ell=1}^{d-1} a_{k,\ell}x_\ell}{a_{k,d}})$ with $x_1,\dots,x_{d-1}$ arbitrary reals.

Now fix some arbitrary $j\in J$ and let us look at an arbitrary point $x=(x_1,\dots,x_{d-1},x_d)\in H_{a_k,b_k}\cap H_{a_j,b_j}^+$. In view of the last paragraph, this is equivalent to \begin{equation} \begin{split} 0 & \leq \langle a_j,(x_1,\dots,x_{d-1},-\frac{b_k+\sum_{\ell=1}^{d-1} a_{k,\ell}x_\ell}{a_{k,d}})\rangle +b_j \\ & = \sum_{\ell=1}^{d-1}\big(a_{j,\ell}-a_{k,\ell}\frac{a_{j,d}}{a_{k,d}}\big)x_\ell+\big(b_j-b_k\frac{a_{j,d}}{a_{k,d}}\big)\\ & =: \langle a_{j'},(x_1,\dots,x_{d-1})\rangle + b_{j'}. \end{split} \end{equation}

So $F_k= H_{a_k,b_k}\cap \big(\cap_{j\in J} H_{a_j,b_j}^+\big) = \cap_{j\in J} \big(H_{a_k,b_k}\cap H_{a_j,b_k}^+\big) =\cap_{j\in J} H_{a_{j'},b_{j'}}^+$, where the $a_{j'}$ are appropriate vectors in $\mathbb{R}^{d-1}$ and the $b_{j'}$ are appropriate scalars as discussed above. This means that $F_k$ is a $d-1$ dimensional convex polytope by definition. It lies in the hyperplane (or affine subspace if you prefer) $H_{a_k,b_k}$.