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Group of rotations of S^n

  1. Jan 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Group of rotations of S^1 = SO(2)=S^1 conincidently
    Group of rotations of S^2 = SO(3)
    Group of rotations of S^3 = SO(4)


    3. The attempt at a solution
    SO(3) is the group of all rotations in R^3 so it can rotate all elements of S^2 which is part of R^3. Although I can't show it rigorously.

    Would it be enough to say since SO(3) consists of all rotations in R^3, it can offcouse rotate all elements having unit distance from the origin.

    Similar argument made to S^1 and S^3.
    Last edited: Jan 26, 2008
  2. jcsd
  3. Jan 27, 2008 #2

    George Jones

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    No, the last two are not correct.

    [tex]S^2 \cong SO \left( 3 \right) / SO \left( 2 \right)[/tex]

    [tex]S^3 \cong SO \left( 4 \right) / SO \left( 3 \right) \cong SU \left( 2 \right)[/tex]

    In general,

    [tex]S^n \cong SO \left( n+1 \right) / SO \left( n \right)[/tex]

    [tex]S^{2n + 1} \cong SU \left( n+1 \right) / SU \left( n \right).[/tex]

    [edit] Sorry, I think I misunderstood what you wrote. Note that "=" should not be substituted for the word "is."

    I think what you meant is something like the following.

    Consider [itex]S^n[/itex] as a subset of [itex]\mathbb{R}^{n+1}[/itex]. Show that each element of [itex]SO \left( n+1 \right)[/itex] maps [itex]S^n[/itex] to [itex]S^n[/itex]. A general mapping from [itex]\mathbb{R}^{n+1}[/itex] to [itex]\mathbb{R}^{n+1}[/itex] might map an element of [itex]S^n[/itex] to an element of [itex]\mathbb{R}^{n+1}[/itex] that is not in [itex]S^n[/itex], so there is something "special" going on here.[/edit]
    Last edited: Jan 27, 2008
  4. Jan 27, 2008 #3

    George Jones

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    Continuing, let [itex]x[/itex] be an element of [itex]S^n[/itex], so [itex]x[/itex] is an element of [itex]\mathbb{R}^{n+1}[/itex] such that

    [tex]x^T x = 1.[/tex]

    Set [itex]y = Ax[/itex], with [itex]A[/itex] in [itex]SO \left( n+1 \right)[/itex], and calculate the length of [itex]y[/itex].
  5. Jan 30, 2008 #4

    With this one, [tex]S^{2n + 1} \cong SU \left( n+1 \right) / SU \left( n \right).[/tex]

    does it mean each coset representation is a matrix in SU(2)?
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