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Fundamental and contravariant representations

  1. Feb 11, 2015 #1
    The invariant of SL(2,C) is proven to be invariant under the action of the group by the following

    [itex]\epsilon'_{\alpha\beta} = N_{\alpha}^{\rho}N_{\beta}^{\sigma}\epsilon_{\rho\sigma}=\epsilon_{\alpha\beta}detN=\epsilon_{\alpha\beta}[/itex]

    The existence of an invariant of this form (with two indices down) tells us that the covariant (here fundamental of SL(2,C)) and contra variant reps aren't independent. This seems to me to follow directly from the fact that detN=1. Is this true? And can anyone bring forward an example where covariant and contra variant reps are truly independent?
  2. jcsd
  3. Feb 12, 2015 #2


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    Yes, covariant and contravariant spinor representations are EQUIVALENT, i.e. they belong to the same representation space of SL(2,C).
    In SL(2,C), lowering or raising the indices by the spinor metric [itex]\epsilon_{ \alpha \beta }[/itex] and [itex]\epsilon^{ \alpha \beta }[/itex] does not produce new representation. However, complex conjugation does lead to new inequivalent representation [itex]\psi^{ \dot{ \alpha } } = \epsilon^{ \dot{ \alpha } \dot{ \beta } } \psi_{ \dot{ \beta } }[/itex].
  4. Feb 12, 2015 #3
    Why do you reply to posts if you don't read the questions properly first?

    1. I know they are EQUIVALENT. I am asking whether this is a direct consequence of the fact that detN=1, i.e. if I could generalise this, for instance, by saying that "every algebra with detN=1 possesses a 2nd rank invariant tensor", or perhaps not, I am missing something.

    2. Again, here I ask whether anybody can bring forward an example where covariant and contravariant representations are truly independent of each other, not whether the conjugate representation is.
  5. Feb 13, 2015 #4


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    Instead of asking MEANINGLESS questions and being RUDE to the person who is trying to help you, you should spend your time learning the ABC of group theory.
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