SU(3) Transformation behavior of hadronic wave function We were given a “baryon” wave function
$$ \sum_{i,j,k=1}^3 \epsilon_{ijk} q_i q_j q_k$$
which is supposedly invariant under SU(3) gauge transformations. The structure of those transformations was given as
$$\psi \to \psi' = U(x) \psi \,\quad U(x) = \exp(\mathrm i g \chi_a(x) T_a) \,.$$
Then we had to show that this wave function is invariant under infinitesimal transformations. In later problems it was asked to show that a “meson” wave function $\sum_{i=1}^3 \bar q_i q_i$ is invariant whereas the wave function $\bar q_1 q_1$ is not.
The best answer that I was able to come up with is that in the “baryon” case the $\epsilon$-symbol is a top-dimensional form. The space of top-dimensional forms only has one dimension, therefore it cannot transform into anything else, it is a singlet.
For the “mesons” in the first case I have (summation implicit)
$$\bar q_i q_i \to \bar q_j U^\dagger(x)_{ji} U(x)_{ik} q_k
= \bar q_j \delta_{jk} q_k = \bar q_j q_j$$
due to the unitarity of the transformation. In the second argument this breaks because the summation between the unitary transformation is not completely, therefore not yielding an inverse:
$$\bar q_1 q_1 \to \bar q_j U^\dagger(x)_{j1} U(x)_{1k} q_k \,.$$
My tutor and the tutor of the other group were not able to really give some convincing argument, so I am asking here. Are those arguments the correct ones for the invariance and non-invariance under SU(3) color transformations?
 A: In the baryon case you can use a direct proof similar to the one you used for the meson. If you make a $SU(3)$ transformation you get (from now on a summation over repeated indices is assumed)
$$
\varepsilon_{ijk}q_i q_j q_k \rightarrow \varepsilon_{ijk} U_{i\ell} U_{jm} U_{kp} q_\ell q_m q_p
$$
The expression 
$$
\varepsilon_{ijk} U_{i\ell} U_{jm} U_{kp}
$$
is completely antisymmetric in the indices $\ell,m,p$, so it must be proportional to $\varepsilon_{\ell m p}$ and we obtain 
$$
\varepsilon_{ijk} U_{i\ell} U_{jm} U_{kp} =
\lambda \varepsilon_{\ell m p}
$$ 
By contracting both sides with $\varepsilon_{\ell m p}$ we get on the left side the determinant of $U$, on the right side the number of nonzero elements of $\varepsilon$, 
$$
3!\det U = \varepsilon_{\ell m p} \varepsilon_{ijk} U_{i\ell} U_{jm} U_{kp} =
\varepsilon_{\ell m p}\lambda \varepsilon_{\ell m p} = 3! \lambda
$$
and as in $SU(3)$ the determinant is one, $\lambda=1$, so $
\varepsilon_{ijk} U_{i\ell} U_{jm} U_{kp} =\varepsilon_{\ell m p}$. Inserting this expression in the first one we wrote we get the proof of the invariance.
