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Lie groups question

  1. Jan 22, 2008 #1
    Let [itex]\rho : \mathbb{H} \to \mathbb{H}; q \mapsto u^{-1}q u[/itex]
    where u is any unit quaternion. Then [itex]\rho[/itex] is a continuous automorphism of H.

    I'm asked to show that [itex]\rho[/itex] preserves the inner product and cross product on the subspace [itex]\mathbf{i}\mathbb{R} + \mathbf{j}\mathbb{R} + \mathbf{k}\mathbb{R}[/itex] consisting of purely imaginary quaternions.

    The only thing I can think of is that [itex]\rho[/itex] acts on [itex]\mathbf{i}\mathbb{R} + \mathbf{j}\mathbb{R} + \mathbf{k}\mathbb{R}[/itex] by rotating that subspace (for which I know a proof), and rotations preserve angles and orientation.

    Is there a more direct method which avoids using the fact that [itex]\rho[/itex] rotates [itex]\mathbf{i}\mathbb{R} + \mathbf{j}\mathbb{R} + \mathbf{k}\mathbb{R}[/itex] ?
     
  2. jcsd
  3. Jan 22, 2008 #2

    HallsofIvy

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    The obvious method would be to apply the inner product and cross product to [itex]u^{-1}qu[/itex] and show that you get the same thing.
     
  4. Jan 23, 2008 #3
    Hi HallsofIvy,

    I tried that but there are 4 x 3 x 4 =48 terms when we come to calculate e.g.

    [itex]u^{-1}q u = \bar{u} q u = (u_0 - u_1i-u_2j-u_2k)(p_1 i + p_2j+p_3k)(u_0 + u_1i+u_2j+u_2k)[/itex]. Is it really necessary to expand this whole thing out and then take the dot product with another one?
     
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