# 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.

(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.

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I guess I wasn't specific enough. I was hoping for a way to figure out what this polytope "looks like": how many faces in each dimension and what they are made of, etc. In the best case, I want a simplical complex for this polytope. –  Joseph Victor Jun 7 '12 at 19:56
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
thanks a bunch :) –  Joseph Victor Jun 7 '12 at 21:11
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