# 4-polytope with vertices at the binary octahedral group

Hey everybody,

Does anybody know if there is a convex polytope in $R^4$ with vertices at the binary octahedral group (identitfying $H$ 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.

Thanks so much!

-Joseph

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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.
Given a set of points, you can check if they form a vertex/edge/face/cell by checking if there is an inequality that they satisfy that the rest of the vertices do not. For instance it's clear that the edges out of $(1,0,0,0)$ go to the $14$ other vertices with a positive $x$ component. Then rotate that around the other vertices to get a complete edge graph. But computing this sort of thing is probably much better done by an appropriate computer program than by people. –  Will Sawin Jun 7 '12 at 20:59
There are faces of the form $(1,0,0,0)$, $(1/\sqrt{2},1/\sqrt{2},0,0)$ and $(1/\sqrt{2},0,1/\sqrt{2},0)$ and faces of the form $(1,0,0,0)$, $(1/\sqrt{2},1/\sqrt{2},0,0)$, $(1/2,1/2,1/2,1/2)$ Those should be all the faces coming out of $(1,0,0,0)$, thus rotating them should get all faces. Cells should only be of the form $(1,0,0,0)$, $(1/\sqrt{2},1/\sqrt{2},0,0)$ and $(1/\sqrt{2},0,1/\sqrt{2},0)$,$(1/2,1/2,1/2,1/2)$. –  Will Sawin Jun 7 '12 at 21:06
If that's all correct then this is a simplicial complex with $48$ vertices, $336$ edges, $576$ faces, and $288$ cells. –  Will Sawin Jun 7 '12 at 21:10