On a convex polytope

Question

Let $e_i$-s denote the standard unit vectors of $\mathbb{R}^n.$ Denote $$\mathcal{C}_k = \left\{ \sum_{i \in S} \pm e_i \colon S \subseteq \{ 1,2,..., n\} \mbox{ and } |S| \leq k \right\}$$ the set of the $k$-termed sums. Let us form the convex hull of the points in $\mathcal{C}_k:$ $$ \mathcal{K}_k := \mbox{conv } \mathcal{C}_k.$$
What is known about the polytope $\mathcal{K}_k?$ For instance, it is regular? What is the graph of $\mathcal{K}_k$? Every vertex has the same degree?

Answer 1

First of all, any point $\sum_{i \in S} \pm e_i$ with $|S|<k$ is in the convex hull of two points with the same entries for $i \in S$, and +1 or -1 respectively in some $k - |S|$ other entries. So you might as well restrict to sets with $|S| = k$.

Then, if $k = n$, you just have the $n$-cube.

If $k = 1$, you have the $n$-crosspolytope.

For other values of $k$, you get various truncations of the $n$-cube, or equivalently of the $n$-crosspolytope. Specifically, your $\mathcal{K}_k$ would be denoted $\mathrm{t}_{k-1}\{3,\dotsc,3,4\}$ in Coxeter's notation (where there are $n-2$ $3$'s), which means that the vertices of the $n$-crosspolytope are truncated to the $(k-1)$-face midpoints. Equivalently, it is $\mathrm{t}_{n-k}\{4,3,\dotsc,3\}$, truncating the $n$-cube to its $(n-k)$-face midpoints.

For $n = 3$ and $k = 2$, you get the cuboctahedron, which is the truncation of either the cube or the octahedron (the 3-crosspolytope) to their edge midpoints.

Some more examples: With $n = 4$: $k = 2$ is the rectified 16-cell, which is actually a 24-cell. $k = 3$ gives the rectified 4-cube.

With $n = 5$: $k = 2$ is the Rectified 5-crosspolytope, $k = 3$ is the Birectified 5-cube, and $k = 4$ is the Rectified 5-cube.

In general, these are all uniform polytopes, which means that they are vertex transitive. In particular, all vertices are in the same number of edges. Except for $k = 1$ and $k = n$, it is never a regular polytope, except in the special case $n = 4$ and $k = 2$.