Does anybody know if there is a convex polytope in $R^4$ with vertices at the [binary octahedral group](https://en.wikipedia.org/wiki/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.



4-polytope with vertices at the binary octahedral group
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.


