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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.

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2 Answers 2

up vote 4 down vote accepted

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

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.

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You mean all cells are truncated cubes? I cannot quite imagine how they could fit together –  მამუკა ჯიბლაძე May 7 at 21:06
Could it be this one? –  მამუკა ჯიბლაძე May 7 at 21:09
I've actually found description of the original (simplicial) guy on Wikipedia, it is called Disphenoidal 288-cell there –  მამუკა ჯიბლაძე May 7 at 21:17
Yes, all cells are truncated cubes, and you found the correct picture. It's easiest to understand the truncated octahedral tiling first. –  Dylan Thurston May 7 at 21:38

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