I'm studying the tensor methods in $SU(N)$ and decided to work through Georgi book, namely chapter 10. As far as I can tell, the point is to decompose a tensor to irreducible components; symmetric and asymmetric (traceless) part. For that we can use the fact that $\delta^j_i$, $\varepsilon_{ijk}$ and $\varepsilon^{ijk}$ are invariant tensors under $SU(3)$.
The first example is decomposing a tensor product of $(1,0) \otimes (1,0) := \mathbf{3} \otimes \mathbf{3}$. Using the notation, that lower indices in states $|_i\rangle$ indicate the states in $\mathbf{3}$, we construct general tensor as $u^iv^j$, which we can decompose to symmetric and antisymmetric part:
$$ u^iv^j = \frac{1}{2}\left(u^iv^j + u^jv^i\right) - \frac{1}{2}\varepsilon^{ijk}\varepsilon_{jlm}u^lv^m. $$
Now, if I understand correctly, the first part $\left(u^iv^j + u^jv^i\right)$ is irreducible because it's symmetric and the second part is irreducible because it's antisymmetric. But they're not traceless, so that's where I'm unsure. Are we looking to construct a sum of just symmetric and antisymmetric tensors? Because in the next example $\mathbf{3} \otimes \mathbf{\bar{3}}$, where we decompose $u^iv_j$ as
$$ u^iv_j = \left( u^iv_j - \frac{1}{3}\delta^i_ju^kv_k \right) - \frac{1}{3}\delta^i_ju^kv_k $$ that is $u^iv_j = $ 'traceless' + 'invariant tensor'.
To recap; what exactly are we looking for when decomposing a tensor in $SU(3)$ or $SU(N)$ for that matter? Why don't we subtract the trace in the first example, but we do exactly that in the second.
I have some more follow up questions, but I believe the common agreement is not to pose multiple questions in one thread.
