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To project a generic tensor to an irreducible module of SO(N) one has to (anti)symmetrize the indices and then subtract traces, e.g. for symmetric traceless 2-tensors

$$ \frac{1}{2} (\delta_{I_1 J_1} \delta_{I_2 J_2} + \delta_{I_1 J_2} \delta_{I_2 J_1}) - \frac{1}{N} \delta_{I_1 I_2} \delta_{J_1 J_2} . $$

It is explained in Cvitanovic's book [1] how to find such a projector for examples, but no general result is given. Is a general formula including the prefactors of the trace parts known in the literature? If not, how would you derive it? I'm interested in the general case of mixed symmetry.

[1] http://birdtracks.eu/version9.0/GroupTheory.pdf

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    $\begingroup$ See Symmetry, Representations, and Invariants by Goodman and Wallach. The decomposition of tensor you are looking for are in Appendix F, which is freely available on Goodman's website: math.rutgers.edu/~goodman/repbook.html $\endgroup$
    – Vít Tuček
    Commented Sep 4, 2014 at 16:23
  • $\begingroup$ Thanks, but I did not see an explicit formula for the projectors or a way to extract them from this. Am I missing something? $\endgroup$
    – Grobi Grobsen
    Commented Sep 5, 2014 at 13:28
  • $\begingroup$ It is a bit unclear what exactly are you asking for. For $\mathrm{GL}(n, \mathbb{C})$ the theory of Schur functors give projectors to all irreducible submodules of $\otimes^k \mathbb{C}^n$. For the orthogonal or symplectic group it works in the same way on the space of harmonic tensors which are the subspace of of the full tensor power on which all traces are zero. So you can first project to harmonic tensors and then use a Schur functor. $\endgroup$
    – Vít Tuček
    Commented Sep 6, 2014 at 17:53

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