- #1

- 49

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[tex]

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

[/tex]

which corresponds to

[tex]

\bf{3}\otimes\bf{\bar{3}}=\bf{8}\oplus\bf{1}

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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)##

[tex]

\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

[/tex]

is that a correct way to check irreducibility? and if so, how do I check this for ##\epsilon^{ijk} ## and ##\epsilon_{ijk} ##?