Set of points in S^3, way to show spaced equal or not?

1. Mar 20, 2015

Spinnor

In an earlier post here I wanted to chop up a three-sphere into cubes, Ben suspected it was not possible and I have no reason to think otherwise. From earlier help by Fezro, here, I may be able to move this forward. Assuming the posts by Fezro are correct I think I can come up with a set of points in S^3 that a computer program could check the angles and distances between nearby points. Let a set of points in S^3 be given by the following

P(a,b,c)

= $exp(i[σ_1(2πa/n)+σ_2(2πb/n)+σ_3(2πc/n)]) \begin{pmatrix} 1 \\ 0\\ \end{pmatrix} = \begin{pmatrix} z_1(a,b,c) \\ z_2(a,b,c)\\ \end{pmatrix}$

where a, b, and c are integers and vary independently from 0 to n (n is some integer), $σ_1,σ_2,σ_3$ are the Pauli matrices, and $z_1(a,b,c)$ and $z_2(a,b,c)$ give us the coordinates x, z, y, and w of the point P(a,b,c) in R^4.

Should it be straightforward to use the above with a computer to determine angles and distances between nearest neighbors? Can the exponential above be simplified or can a computer program easily handle the exponential? Is there a clever way to show that in fact nearest neighbor points above are not equally spaced?

Thanks for any help!

Last edited: Mar 20, 2015
2. Mar 21, 2015

Spinnor

If I had defined my set of points using the spin down spinor,

\begin{pmatrix}
0 \\
1\\
\end{pmatrix}

would there be a fundamental difference between the two sets of points? Could one set of points be rotated into the other set?

Thanks for any help!