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Given a rank-2 $\mathrm{SU}(N)$ tensor $X^{ab}$, it transforms as $X'^{ab} = U^a{}_c U^b{}_d X^{cd}$, where $U \in \mathrm{SU}(N)$. We can decompose it into a symmetric and an anti-symmetric part $$ X^{ab} = X_+^{ab} + X_-^{ab} \,, \qquad X_\pm^{ab} = \frac{1}{2}(X^{ab} \pm X^{ba}) $$ and since the permutation operation commutes with the $\mathrm{SU}(N)$ transformation, the two parts do not mix under these transformations, i.e., the tensor $X^{ab}$ is reducible.

My question is now: How can we show that the tensors $X_\pm^{ab}$ cannot be reduced any further?

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closed as off-topic by AccidentalFourierTransform, ZeroTheHero, Kyle Kanos, ahemmetter, user191954 Dec 2 '18 at 3:44

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    $\begingroup$ I'm voting to close this question as off-topic because it belongs on Math.SE. $\endgroup$ – AccidentalFourierTransform Nov 29 '18 at 17:03
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The symmetric part is actually reducible into $\frac{Tr(X)\delta^{ab}}{N}$ and $(\frac{X^{ab}+X^{ba}}{2} - \frac{Tr(X)\delta^{ab}}{N})$

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    $\begingroup$ No, this is incorrect. $\delta^{ab}$ is not an $\mathrm{SU}(N)$-invariant symbol (it is an $\mathrm{SO}(N)$-invariant symbol; the $\mathrm{SU}(N)$-invariant symbols are $\delta^a{}_b$ and $\epsilon$). The symmetric representation of $\mathrm{SU}(N)$ is irreducible. The traceless representation of $\mathrm{SU}(N)$ is the adjoint representation, not the symmetric one. $\endgroup$ – AccidentalFourierTransform Nov 29 '18 at 17:00

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