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What are the vertex coordinates of a regular Hexadecachoron?

I suspect it's just all 8 combinations of $(0,0,0, \pm\frac{1}{\sqrt{2}})$ but I'm not sure how to be sure for sure.

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    $\begingroup$ Um. The Wikipedia Hexadecachoron article you linked to says "The eight vertices of the 16-cell are $(\pm 1, 0, 0, 0)$, $(0, \pm 1, 0, 0)$, $(0, 0, \pm 1, 0)$, $(0, 0, 0, \pm 1)$. All vertices are connected by edges except opposite pairs." It is centered at origin, so you can simply use any real constant $r$ instead of $1$. $\endgroup$ – Nominal Animal Jan 10 at 1:27
  • $\begingroup$ @NominalAnimal yep. I failed at reading. $\endgroup$ – guest Jan 10 at 10:07
  • $\begingroup$ No worries, it happens to all of us. Me fail English often. $\endgroup$ – Nominal Animal Jan 10 at 18:06
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Well, they are exactly the 8 ones obtained as permutations (instead of your said "16 combinations") of that given vertex pair.

Just as the vertices of the 3D cross-polytope, the octahedron, are the 6 ones obtained as permutations of $(0, 0, \pm\frac1{\sqrt2})$, or the vertices of the 2D cross-polytope, the square, are the 4 ones obtained as permutations of $(0, \pm\frac1{\sqrt2})$.

In fact, the general cross-polytope is just defined to be the convex hull of all points on all Cartesian axes (both directions each), which are $\frac1{\sqrt2}$ off.

Just as the general hypersphere is defined by the set of points, which bow to $||x||_2=\sqrt{\sum_i x_i^2}=const$, so the surface of the general cross-polytope would follow to $||x||_1=\sum_i |x_i|=const$. Within your chosen scaling the latter constant happens to be dimension dependent, $const(D)=\frac D{\sqrt2}$. Thereby it is chosen such that the edge length of the thus constructed cross-polytope in any dimension would become unity. Btw. the surface of the general measure-polytope similarily would follow $||x||_{\infty}=\max_i |x_i|=const$.

--- rk

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