The permutations of (1,1,0,0), (-1,1,0,0), (-1,-1,0,0) are vertices of a polytope.

Question

Here's my attempt at the proof. If possible please assess its validity. Thanks.

Answer 1

Why this sort of argumentation? Consider this:

Cartesian coordinates are orthogonal. Thus the points (1,1,1,1) and all its sign changes define a tesseract of edge size 2. It has as its Dynkin symbol o3o3o4x.

Next consider therefrom the hull of its edge centers. This would be the set of vertices (1,1,1,0) and all permutations and sign changes. It is the rectified tesseract with Dynkin symbol o3o3x4o.

Now let's turn to the face centers instead (i.e. of its 2D boundaries). Then this polychoron would be the hull of (1,1,0,0) and all permutations and sign changes, i.e. your problem. By construction this is now the birectified tesseract with Dynkin symbol o3x3o4o.

Alternatively o3x3o4o also could be read as rectified 16-cell (by progression of the special node from the other end of the symbol). Or, as it would turn out, it also is equivalent to the 24-cell, showing up additional symmetry, Thus yielding as alternate Dynkin symbol x3o4o3o as well.

--- rk

Answer 2

The set of all those points are on the sphere of radius $\sqrt2$ centered at the origin. To show that they are vertices of their convex hull it is enough to find a hyperplane through each of them that has that convex hull on a side, and the tangent plane to that sphere does the job.