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Given the following Young tableaux

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for $SU(3)$, how can I deduce that it corresponds to the adjoint representation? I was thinking that the dimension of this representation is 8, as in the case of the adjoint representation. Is this sufficient?

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  • $\begingroup$ For SU(3), you know a height -2 tower is an antiquark ($\bar 3$) and a plain box a quark (3), so this is $q\bar q$, the adjoint. Can you now write down the YT for the adjoint of SU(5)? You need not compute the dimensionality. $\endgroup$ – Cosmas Zachos Oct 27 '20 at 19:09
  • $\begingroup$ I think is the same YT, am I wrong? $\endgroup$ – dfgoe55 Oct 27 '20 at 19:53
  • $\begingroup$ The first column is 4-high, the second 1. $\endgroup$ – Cosmas Zachos Oct 27 '20 at 19:57
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One can compute the the weights by filling in the numbers 1,2,3 according to the rule for semi-standard tableaux (not decreasing along the rows, strictly increasing down the columns). Each of the eight possible tableaux gives the eigenvalues of $\lambda_3$ (the number of 1's minus the number of 2's) and $\lambda_8$ (number of 1's plus number of 2's minus twice the number of 3's all divided by $\sqrt 3$). If you plot them you will recognise the weight diagram of the octet (adjoint) rep.

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  • $\begingroup$ But following another path... if I start from the action of the group on a three-indices tensor: $\psi^{ijk} \rightarrow U^i_l U^j_m U^k_n \psi^{lmn}$ the I can pass to the action of the algebra on the same tensor: $\psi^{ijk} \rightarrow u^i_l \psi^{ljk} + u^j_l \psi^{ilk} + u^k_l \psi^{ijl}$. Now is it possible to prove the the action of the algebra on $\psi$ is the adjoint action? Maybe writing explicitely the form of $\psi$...some hints? $\endgroup$ – dfgoe55 Oct 27 '20 at 19:40
  • $\begingroup$ @dfgoe55 You need to properly symmetrize your tensor first, else what you have lives in $(3,0)\oplus (1,1)\oplus (0,0)$. It so happens that the only irrep of $su(3)$ with dim=8 is the adjoint, but such luck does not necessarily hold in general. The solution of this answer is unambiguous and would work in general. $\endgroup$ – ZeroTheHero Oct 27 '20 at 20:03
  • $\begingroup$ The structrure of the tensor is the following $\psi^{ijk} = \phi^{ijk} - \phi^{kji} + \phi^{jik} - \phi^{kij}$...is it wrong? $\endgroup$ – dfgoe55 Oct 27 '20 at 20:08

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