# Invariant of degree p

1. Nov 25, 2011

### Rasalhague

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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:

$$I(\lambda A) = J((\lambda A)\otimes (\lambda A)) = (\text{Tr}\, \lambda A)^2$$

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

where $\lambda$ is a scalar. (Summing over like indices.)

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