# From the Young tableaux to the algebra

Given the following Young tableaux

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?

• 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. – Cosmas Zachos Oct 27 '20 at 19:09
• I think is the same YT, am I wrong? – dfgoe55 Oct 27 '20 at 19:53
• The first column is 4-high, the second 1. – Cosmas Zachos Oct 27 '20 at 19:57

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.
• 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? – dfgoe55 Oct 27 '20 at 19:40
• @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. – ZeroTheHero Oct 27 '20 at 20:03
• The structrure of the tensor is the following $\psi^{ijk} = \phi^{ijk} - \phi^{kji} + \phi^{jik} - \phi^{kij}$...is it wrong? – dfgoe55 Oct 27 '20 at 20:08