# Decomposition of group representation using tensor method

I am dealing with the decomposition of the representation $5\otimes5$ of $SU(5)$:

$$5\otimes5=15\oplus10$$

demonstration:

$$u^iv^j=\frac{1}{2}(u^iv^j+u^jv^i)+\frac{1}{2}(u^iv^j-u^jv^i)=$$

$$=\frac{1}{2}(u^iv^j+u^jv^i)+\frac{1}{2}\epsilon^{ijxyk}\epsilon_{xyklm}u^lv^m$$

where the term $\frac{1}{2}(u^iv^j+u^jv^i)$ has 15 independent components and the other has 10 components.

My question is: being the $\epsilon^{ijxyk}$ invariant in $SU(5)$, shouldn't the tensor $\epsilon_{xyklm}u^lv^m$ transform under the $\overline{10}$ representation, having 3 low free index?

(according to my notation an upper index transform under the $D$ representation while a lower index transforms under the $\overline{D}$ representation).

It's a convention which of these reps is called ${\bf 10}$ and which is called $\overline{\bf 10}$. The convention that the people choose is arguably the simpler one among the two: ${\bf 10}$ is the antisymmetric product of two ${\bf 5}$, i.e. ${\bf 5}\wedge{\bf 5}$, which are also without bars.
With that choice, one can prove that $\overline{\bf 10}$ which is defined as the complex conjugate representation is either an analogous representation where the upper indices are replaced with lower ones or vice versa; or it is ${\bf 5}\wedge {\bf 5}\wedge{\bf 5}$.
It is simply a mathematical fact (which you have implicitly proved, using the epsilon symbol) that ${\bf 5}\wedge{\bf 5}$ is the complex conjugate of the representation ${\bf 5}\wedge {\bf 5}\wedge{\bf 5}$ – so if one of them is called without the bar, the other one must be with the bar, and the dominant convention is one written above.