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Why is Casimir operator to be an invariant of coresponding Lie Algebra?

  1. Jan 4, 2012 #1
    Please teach me this:

    Why is Casimir operator T[itex]^{a}[/itex]T[itex]^{a}[/itex] be an invariant of the coresponding Lie algebra? I know that Casimir operator commutes with all the group generators T[itex]^{a}[/itex].

    Thank you very much for your kind helping.
  2. jcsd
  3. Jan 4, 2012 #2


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    It's not clear what you're asking.

    You said that you know a Casimir operator commutes with all the group generators T[itex]^{a}[/itex], so I guess you're asking why that means the Casimir is invariant under the group action?

    If that's your question, the group action on an arbitrary quantity X is of the form
    X ~\to~ e^{z_a T^a} X e^{-z_a T^a}
    where the [itex]z_a[/itex] are ordinary (scalar) parameters, and I'm using implicit summation over the index [itex]a[/itex].

    Expanding the exponentials as a Taylor series then shows that if X commutes with every [itex]T^a[/itex], then X also commutes with the exponentials. Thus you can pass X through one of the exponentials, and the exponentials cancel. Thus X is unchanged (i.e., invariant).

    Or were you asking how to prove that [itex]T^a T^a[/itex] commutes the other generators?
    If that's your question, please give a specific example of the Lie algebra.
  4. Jan 4, 2012 #3


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    What exactly is your educational qualification? I ask this because you seem to ask question you know the answer to! If you KNOW that
    [tex][T^{a},T^{c}T^{c}] = 0,[/tex]
    then how can’t you know that
    [tex]\delta^{a}\left(T^{c}T^{c}\right) = [T^{a}, T^{c}T^{c}] = 0[/tex]

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