Irreducible SU(3) tensors

[PineApple2]

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 $\epsilon_{ijk}$?

 

[DrSuage]

Hi - yes you are along the right lines. Just remember that anything contracted over all the indices of epsilon will be a singlet because of the determinant condition on SU(n) groups... and that all tensors can be decomposed into a symmetric and an antisymmetric part...(without giving away the answer oops).

However there is another way of doing these types of computation that is much more elegant - have you looked into Young Tableaux?