# Decomposition of group representation using tensor method

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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).

This post imported from StackExchange Physics at 2015-03-12 12:20 (UTC), posted by SE-user Caos

retagged Mar 12, 2015

I want to attract your attention to the fact that this "decomposition" is kind of illusory. Indeed, if $u$ and $v$ belong to different 5D spaces and the indices $i$ and $j$ are fixed, the "direct" product $u^iv^j$ is simply proportional to each of the fixed elements, say, to $u^3$ and $v^5$ whereas the first decomposed expression "involves" other elements, namely $u^5$ and $v^3$. Thus the second addendum in the decomposition formula serves to subtract extra terms from the first addendum :-)

The true decomposition is only valid for $u^i u^j$.

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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.

This post imported from StackExchange Physics at 2015-03-12 12:20 (UTC), posted by SE-user Luboš Motl
answered Mar 11, 2015 by (10,278 points)

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