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Commutator SO(3) - what am I doing wrong?

  1. Jul 13, 2012 #1

    BWV

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    reading that the commutator of rotations on two orthogonal axes is i * the rotation matrix for the third axis

    but if I commute this

    \begin{pmatrix}\mathrm{cos}\left( \theta\right) & -\mathrm{sin}\left( \theta\right) & 0\cr \mathrm{sin}\left( \theta\right) & \mathrm{cos}\left( \theta\right) & 0\cr 0 & 0 & 1\end{pmatrix}

    with this

    \begin{pmatrix}1 & 0 & 0\cr 0 & \mathrm{cos}\left( \theta\right) & -\mathrm{sin}\left( \theta\right) \cr 0 & \mathrm{sin}\left( \theta\right) & \mathrm{cos}\left( \theta\right) \end{pmatrix}

    I get (using innerproduct function in Maxima)
    \begin{pmatrix}0 & \mathrm{cos}\left( \theta\right) \,\mathrm{sin}\left( \theta\right) -\mathrm{sin}\left( \theta\right) & -{\mathrm{sin}\left( \theta\right) }^{2}\cr \mathrm{cos}\left( \theta\right) \,\mathrm{sin}\left( \theta\right) -\mathrm{sin}\left( \theta\right) & 0 & \mathrm{cos}\left( \theta\right) \,\mathrm{sin}\left( \theta\right) -\mathrm{sin}\left( \theta\right) \cr {\mathrm{sin}\left( \theta\right) }^{2} & \mathrm{cos}\left( \theta\right) \,\mathrm{sin}\left( \theta\right) -\mathrm{sin}\left( \theta\right) & 0\end{pmatrix}

    which has a determinant of zero and therefore not part of SO(3)

    obviously I am not getting something, but don't see it. Any help is much appreciated
     
  2. jcsd
  3. Jul 13, 2012 #2
    you are confusing the Lie group and the Lie algebra. The commutation
    relation applies to the Lie algebra (ie the generators of the Lie group).
     
  4. Jul 15, 2012 #3

    BWV

    User Avatar

    ah, thanks.
     
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