# A Irreducible SU(3) tensors

1. May 12, 2016

### 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 $\epsilon_{ijk}$?

2. May 17, 2016

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. May 18, 2016

### 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?