taking derivatives componentwise and also it’s an invariant tensor of SU(N)
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It's an invariant tensor of SL(3). The invariant symbol of SL(N) is epsilon with N indices. SU(N) inherits as a subgroup.
I'm pretty sure the epsilon tensor is only an invariant of SU(N) not SL(N) .. The δ_{i \bar{j}} is an SL(N) invariant
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Because there are two invaraint tensors of SU(N), and i know the delta is inherited and they can't have the groups can't have the same invariant tensors i thought
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The delta comes from the unitary part, not the special part. The epsilon comes from the special part. See my previous comment.
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No. Special linear is matrices with unit determinant. How do we write the determinant of matrix M_{ij} in index notation? det(M) = epsilon^{i_1 i_2 ... i_n} M_{1 i_1} M_{2 i_2} ... M_{n i_n}. You need the epsilon to do this -- it is the invariant symbol.
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good point i thought for some reason the proof involved unitarity
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