What is this 4D shape with Radial equilateral symmetry property?

Question

The 2D hexagon can be embedded in $\mathbb{R}^3$ with vertices at coordinates $(1, -1, 0),(-1,1,0),(0,1,-1),(0,1,-1),(1,0,-1),(-1,0,1)$.

Consider the general set of vertices of this form, $\mathcal{Q}=\{\mathbf{q}=\langle q_1,q_2,...,q_d\rangle : q_i = 1, q_j = -1, q_{\not = i \not = j} = 0, \: \forall \: i,j \in [1,d] \; | \; i \neq j \}$

The set of vertices $\mathcal{Q}$ when d=4 defines a 3D cuboctahedron embedded in $\mathbb{R}^4$.

The hexagon and cuboctahedron and indeed all the (d-1)-Dimensional shapes embedded in $\mathbb{R}^{d-1}$ described by $\mathcal{Q}$ have radial equilateral symmetry.

I'm confused about the 4D shape embedded in $\mathbb{R}^5$. I thought it would be the 24-cell judging by what it says in the wiki article about radial equilateral symmetry (https://en.wikipedia.org/wiki/Cuboctahedron)? However, $|Q| = 2{d \choose 2}$ which means when $d=5$, Q only has 20 vertices rather than 24...

To clarify my question, I would like to know what the name of the 4D shape described by $\mathcal{Q}$ when $d=5$ is.

Any help is much appreciated, thanks :)

Answer 1

Well, the four vertices (1,-1,0,0,0), (1,0,-1,0,0), (1,0,0,-1,0), (1,0,0,0,-1) are all the same distance apart and hence form a tetrahedron. Your figure has 10 of these; besides, it has some other cells which are harder to identify. Also, the permutations of coordinates give it the symmetry of a 4-simplex, AKA 5-cell. By searching among all the friends and relatives of the latter, we finally discover the Runcinated 5-cell.

Indeed, your construction is mentioned someplace midway through the huge list of its properties.

So it goes.

Answer 2

From your very definition of the vertex set, i.e. $$\mathcal{Q}_d=\{\mathbf{q}=\langle q_1,q_2,...,q_d\rangle : q_i = +1, q_j = -1, q_{\not = i \not = j} = 0, \: \forall \: i,j \in [1,d] \; | \; i \neq j \}$$ you directly can deduce that all these polytopes have to be exactly 3 vertex layers:

1. the vertex layer with $q_1=+1$, then having just a single $q_j=-1$ somewhere in the remainder coordinates, therefore describing simply a simplex facet,

2. the vertex layer with $q_1=0$, i.e. where both $q_i=-1$ and $q_j=+1$ still occur in the remainder coordinates, and which thus describes a midsection which is exactly the same polytope again, just within one dimension less,

3. the vertex layer with $q_1=-1$, then having just a single $q_j=+1$ somewhere in the remainder coordinates, therefore describing again a simplex facet, which however now is oriented within dual positioning wrt. to the former (opposite) one.

You also already mentioned the low dimensional examples of this polytope $\mathcal{Q}_d$. Infact we generally have the following sequence of maximally expanded simplices (which are aka the root polytopes of the lattice $A_d$):

• $\mathcal{Q}_2 =$ x3x (hexagon)
• $\mathcal{Q}_3 =$ x3o3x (cuboctahedron) = x3o || x3x || o3x
• $\mathcal{Q}_4 =$ x3o3o3x (runcinated 5-cell, aka small prismated pentachoron) = x3o3o || x3o3x || o3o3x
• $\mathcal{Q}_5 =$ x3o3o3o3x (stericated hexateron, aka small cellated dodecateron) = x3o3o3o || x3o3o3x || o3o3o3x
• etc. ...

It furthermore is notworthly that all these polytopes have the property that their edge length equals their circumradius.

--- rk

Cf. https://bendwavy.org/klitzing/explain/analog.htm#exp-simplex