# Irreducible SU(3) tensors

• A
Suppose that in the tensor component ##T^a_b ## the upper index is the ## \bf{3}## component and the lower index is the ##\bf{\bar{3}} ## component. To be concrete, consider the decomposition
$$u^iv_j= \left( u^iv_j-\frac{1}{3}\delta^i_j u^kv_k \right) +\frac{1}{3}\delta^i_j u^kv_k$$
which corresponds to
$$\bf{3}\otimes\bf{\bar{3}}=\bf{8}\oplus\bf{1}$$
I want to see that indeed the transformation of ##u^iv_j ## is reducible, but the transformations of ## \left( u^iv_j-\frac{1}{3}\delta^i_j u^kv_k \right)## and ##\left(\frac{1}{3}\delta^i_j u^kv_k \right) ## are irreducible. My thought was to contract each of the supposedly irreducible tensors with the only possible invariant tensors, namely ##\delta^i_j ##, ##\epsilon^{ijk} ##, ##\epsilon_{ijk} ##, and to see that in each case I get zero, which means there are no invariant subspaces. So, taking ## \left( u^iv_j-\frac{1}{3}\delta^i_j u^kv_k \right)##
$$\delta^j_i \left( u^iv_j-\frac{1}{3}\delta^i_j u^kv_k \right) = u^iv_i - \frac{1}{3}\delta^i_i u^kv_k =0$$
is that a correct way to check irreducibility? and if so, how do I check this for ##\epsilon^{ijk} ## and ##\epsilon_{ijk} ##?