4-polytope with vertices at the binary octahedral group Does anybody know if there is a convex polytope in $R^4$ with vertices at the binary octahedral group (identifying $H$ with $R^4$).
The binary tetrahedral group lies at the vertices of the so-called 24-cell, and the binary octahedral group is just a direct some of two binary tetrahedral groups, but it is not clear how to interpret that geometrically.
Experimentally, I have found that, for each octahedron in the 24-cell, each vertex in that octahedron is equidistance from exactly one point in binoct not in bintet.  I don't know if this is relevant at all.
 A: Can't you just take the convex hull of the points? Specifically, if you have any set of points on the unit sphere of $\mathbb R^n$, then those points will be the vertices of their own convex hull.
(Proof: A point is in the vertex of the polyhedron if and only if it is not in the convex hull of the other points. Thus no points but the original points are vertices. Since the polyhedron is contained in the unit sphere, each point on the boundary of the sphere must be a vertex, which includes all the original points.)
So of course such a polytope exists.
A: I just came across this old question, which I happened to think about earlier. Let me give a more explicit description. As you noted, the binary tetrahedral group forms the vertices of a 24-cell. The 24-cell is self-dual, made up of 24 octahedra, with three meeting around a vertex.
The 48 points of the binary octahedral group are the union of the 24 vertices of the 24-cell and the 24 vertices of the dual 24-cell (corresponding to faces of the 24-cell). You can see the faces of the convex hull of these 48 points as corresponding to the faces of the spherical Delaunay triangulation of these points. The obvious thing to do is divide each octahedron of the 24-cell into 8 tetrahedra (by coning to the center), but that turns out to be wrong (i.e., it is not the Delaunay triangulation). Instead, you subdivide the octahedra into 8 tetrahedra, and then for each pair of tetrahedra glued to each other from neighboring octahedra, you redivide them into 3 tetrahedra (by adding the edge of the dual 24-cell connecting the two centers). You can get the count of vertices, edges, faces, and cells that Will Sawin gave from this.
I like to think about the dual to this simplicial complex, which is a simple polyhedron. ("Simple" here just means "dual to simplicial". More concretely, it is a polyhedral decomposition where the polyhedra meet in three around an edge and in four around a vertex, locally like a soap bubble.) This dual polyhedron is a decomposition of $S^3$ into 48 truncated cubes.
This whole construction is entirely analogous to the way you can get the truncated-octahedron tiling of $\mathbb{R}^3$ by taking a cubical decomposition of $\mathbb{R}^3$ and its dual.
A: I believe this is the disphenoidal 288-cell. A spherical code that combines two 24-cells has the same minimal Euclidean distance as the vertices of the 288-cell, and those same vertices form the 2O group when written as unit quaternions. See https://errorcorrectionzoo.org/c/disphenoidal288cell for references.
