I'm currently trying to do something that sounds doable, but on which I've now been stuck for a while.
The idea is that, if you consider an $n$-hypercube, the polygon found by doing a $2d$ cross-section of it can have between 3 and $2n$ edges, with a mean value at $4$ for random cuts.
In particular, it seems that you should be able to find a regular $2n$-gon for a family of very specific cross-sections, in the same way that a cut of the $3d$ cube gives a regular hexagon if you do a central cut perpendicular to a long diagonal.
However, in general dimension $n$, it is not obvious to me that there is a simple construction to find these sections: there is no obvious $n-2$ dimensional generalization of the long diagonal of the cube.
My question is: is there a general way to find the $2d$ planes that contain these special sections for $n>3$?
Best Answer
Let $U$ be the $2 \times n$ matrix whose rows are ${\big(} \cos \tfrac{\pi k}{n}, \sin \tfrac{\pi k}{n} {\big)}$ for $1 \leq k \leq n$. I claim that $\sqrt{\tfrac{2}{n}} U$ is an orthogonal embedding $\mathbb{R}^2 \to \mathbb{R}^n$.
Once I check this, my answer is to slice the $n$-cube by the image of the map $U$. Take our cube to be $\{ (z_1, z_2, \ldots, z_n) : -1 \leq z_1, z_2, \ldots, z_n \leq 1 \}$, we see that $\sqrt{\tfrac{2}{n}} U \left[ \begin{smallmatrix} x\\y \end{smallmatrix} \right]$ is in the unit cube if and only if $\left|x \cos \tfrac{\pi k}{n} + y \sin \tfrac{\pi k}{n} \right| \leq \sqrt{\tfrac{n}{2}}$ for $1 \leq k \leq n$. This is a regular $2n$-gon, as requested.
Okay, so let's just check that this map is an orthogonal embedding. We need to compute that $U^T U = \sqrt{\tfrac{n}{2}} \text{Id}_2$. In other words, we need to check that $$ \sum_{k=1}^n \cos^2 \frac{k \pi}{n} = \sum_{k=1}^n \sin^2 \frac{k \pi}{n} = \frac{n}{2} \qquad \text{and} \qquad \sum_{k=1}^n \cos \frac{k \pi}{n}\sin \tfrac{k \pi}{n} = 0.$$ Using the identities $\cos^2 \theta = \tfrac{1+\cos(2 \theta)}{2}$, $\sin^2(\theta) = \tfrac{1-\sin(2 \theta)}{2}$ and $\sin \theta \cos \theta = \tfrac{\sin(2 \theta)}{2}$, we have $$ \sum_{k=1}^n \cos^2 \frac{k \pi}{n} = \frac{n}{2} + \frac{1}{2} \sum_{k=1}^n \cos \frac{2k \pi}{n} = \frac{n}{2}.$$ $$ \sum_{k=1}^n \sin^2 \frac{k \pi}{n} = \frac{n}{2} - \frac{1}{2} \sum_{k=1}^n \cos \frac{2k \pi}{n} = \frac{n}{2}.$$ $$\sum_{k=1}^n \cos \frac{k \pi}{n}\sin \frac{k \pi}{n} = \frac{1}{2} \sum_{k=1}^n \sin \frac{2k \pi}{n} = 0$$ as desired.
The way to think about something like this is that, if you have a polygon $P$ in $\mathbb{R}^d$ that you want to achieve as a section of a polytope $Q$ in $\mathbb{R}^n$, then you should figure out which $n$ affine linear functions on $\mathbb{R}^d$ you want to have as the pullback of the coordinate functions from $\mathbb{R}^n$. Since a $2n$-gon of width $2w$ lies in the $n$-strips $\left|x \cos \tfrac{\pi k}{n} + y \sin \tfrac{\pi k}{n} \right| \leq w$, it makes sense to take the $n$-coordinate functions to be $(x,y) \mapsto x \cos \tfrac{\pi k}{n} + y \sin \tfrac{\pi k}{n}$.