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Invariant of degree p

  1. Nov 25, 2011 #1
    ...

    I think I've misunderstood their definition of an invariant. The pth power of the trace function seems to be homogeneous of degree p rather than linear:

    [tex]I(\lambda A) = J((\lambda A)\otimes (\lambda A)) = (\text{Tr}\, \lambda A)^2[/tex]

    [tex]=\lambda\lambda A^i_kA^r_m\delta^k_i\delta^m_r=\lambda\lambda A^k_kA^m_m=\lambda^2 (\text{Tr}\, A)^2= \lambda^2 IA \neq \lambda IA,[/tex]

    where [itex]\lambda[/itex] is a scalar. (Summing over like indices.)
     
  2. jcsd
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