# Irreducible SU(3) tensors

• A
• PineApple2
In summary, the conversation discusses how to check the reducibility and irreducibility of tensors in a decomposition. The method proposed involves contracting the tensor with invariant tensors, such as the Kronecker delta and epsilon tensors. The conversation also mentions the use of Young Tableaux as a more elegant method for these types of computations.

#### 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} ##?

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?