Timeline for 4-polytope with vertices at the binary octahedral group
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 7, 2012 at 23:00 | vote | accept | Joseph Victor | ||
| Jun 7, 2012 at 23:00 | comment | added | Joseph Victor | I wrote a program to build this, got the same number of cells, and then computed the simplical homology of the complex and saw that it was the same as the three sphere, so I think this works. Thanks again. | |
| Jun 7, 2012 at 21:10 | comment | added | Will Sawin | If that's all correct then this is a simplicial complex with $48$ vertices, $336$ edges, $576$ faces, and $288$ cells. | |
| Jun 7, 2012 at 21:06 | comment | added | Will Sawin | 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 $(1/\sqrt{2},0,1/\sqrt{2},0)$,$(1/2,1/2,1/2,1/2)$. | |
| Jun 7, 2012 at 20:59 | comment | added | Will Sawin | 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. | |
| Jun 7, 2012 at 19:56 | comment | added | Joseph Victor | 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. | |
| Jun 7, 2012 at 19:17 | history | answered | Will Sawin | CC BY-SA 3.0 |