2

The regular $n$-crosspolytope is the $n$-dimensional generalization of the regular octahedron. I know that all of them can be realized as $01$-polytopes, that is, as a polytope with coordinates in $\{0,1\}^d$ for some $d\ge n$.

Question: Which of these can be realized as a $01$-polytope with edge-length $\sqrt 2$?

I know that this is possible for $n\in\{1,2,3,4\}$:

  • $n=1$ (a line segment) is realized with coordinates $\{01, 10\}$.
  • $n=2$ is the square which can be realized with coordinates $$\{\; 0101\quad 0110\quad 1001\quad 1010\;\}.$$
  • $n=3$ is the regular octahedron and can be realized with coordinates $$\{\;1100\quad 0110\quad 0011\quad 1001\quad 0101\quad 1010\;\}.$$
  • $n=4$ is the $4$-crosspolytope, or 16-cell, which can be realized with coordinates $$\{\;0000\quad 1100\quad 0110\quad 0011\quad 1001\quad 0101\quad 1010\quad 1111\;\}.$$
M. Winter
  • 30,997
  • Are you interested in an isometric embedding of the entire solid polytope, or in a length-preserving embedding of its surface, or in a $\sqrt 2$-unit-distance graph realizing its 1-skeleton? – Magma Aug 07 '19 at 12:41
  • @Magma I want the embedded crosspolytope to still be regular. So I suppose this is an isometric embedding. – M. Winter Aug 07 '19 at 12:43

1 Answers1

3

A realization of the $n$-dimensional cross polytope as a $01$-polytope with side length $\sqrt{2}$ is only possible for $n \leq 4$:

Suppose there's an embedding of the $n$-dimensional cross polytope with vertices in $\{0,1\}^d$. Let $V$ and $V'$ be two opposite vertices of the embedded cross polytope.

The distance between $V$ and $V'$ is equal to $\sqrt{2}\sqrt{2} = 2$, so the coordinate vectors of $V$ and $V'$ must differ in exactly four places (wlog. the first four places).

Let $W$ be any other vertex of the embedded cross polytope. Its distances to $V$ and to $V'$ are both $\sqrt{2}$, so its coordinate vector must differ from the coordinate vectors of $V$ and $V'$ in two places each. But since $V$ and $V'$ differ in the first four places already, this means that $W$'s coordinates must differ from the coordinates of $V$ only in two of the first four places, then must differ from the coordinates of $V'$ in the other two of the first four places.

Hence, all the vertices of the polytope (and thus the entire polytope) must lie in the $4$-dimensional subspace through $V$ spanning the first four coordinates.

Since the entire cross polytope fits into a $4$-dimensional space isometrically, it must have no more than $4$ dimensions itself, so $n \leq 4$.

You've found correct embeddings for $n = 2$ to $n = 4$ yourself, but the $n=1$ embedding should have vertices at $0000$ and $1111$, since the two vertices of the $1$-dimensional cross polytope have distance $2$ since they're combinatorially not adjacent.

Magma
  • 6,450