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

  1. Mar 20, 2015 #1
    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)

    = [itex]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}[/itex]

    where a, b, and c are integers and vary independently from 0 to n (n is some integer), [itex]σ_1,σ_2,σ_3[/itex] are the Pauli matrices, and [itex]z_1(a,b,c)[/itex] and [itex]z_2(a,b,c)[/itex] 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. jcsd
  3. Mar 21, 2015 #2
    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!
     
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