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Is there such an identity about SO(3)?

  1. May 28, 2006 #1
    [tex]T^i K_{ij} = K T^j K^{-1}[/tex]
    repeated indices imply summation.
    [tex]T^i[/tex] are the generators (Lie algebra elements) of SO(3).
    i.e.[tex]T^i_{jk} = - \epsilon_{ijk}[/tex]
    [tex]T^i \in so(3)[/tex]
    [tex]K \in SO(3)[/tex]

    How to show it's true?
    Is there a universal formula for all Lie group?
    Last edited: May 28, 2006
  2. jcsd
  3. May 28, 2006 #2

    matt grime

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    Those elements do not generate SO(3). SO(3) is uncountable, it cannot have a finte set of generators.
  4. May 28, 2006 #3

    George Jones

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    For physicists, generators of a Lie group are elements of the corresponding Lie algebra, or maybe such elements multiplied by i. I don't know what the K's are.
  5. May 28, 2006 #4
    Yes. I have just clarified!
    I also checked a special case, and it's true.
    I don't know how to prove this identity besides brute-force calculation which gives no insight at all.
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