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I would like to know the $4\times 4$ matrices that correspond to the antisymmetric representation of $SU(4)$. I know that there should be a relation with a $6$ representation of $SO(6)$. Is there a way to obtain a matrix representation of these antisymmetric rep of $SU(4)$. Where do I start?

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  • $\begingroup$ By the antisymmetric rep do you mean $SU(4)$ acting on $\Lambda^2\Bbb C^4$? Are you asking how to turn matrices in $SU(4)$ into matrices in $GL(6,\Bbb C)$, or what are you asking? $\endgroup$ – anon Dec 23 '17 at 20:26
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I'm pretty convinced you misunderstood the 4×4 bit. You wanted to say ${\mathbf 4} \otimes_A {\mathbf 4} ={\mathbf 6} $, so the antisymmetrized Kronecker product of two fundamental reps of su(4), the one whose Young tableau is two boxes stacked on top of each other. (For su(3) this would have been the $\bar{\mathbf 3}$, the antiquark.)

This 6 is also the fundamental irrep of so(6), whilst the 4 is the spinor of so(6), all the while being the fundamental of su(4).

So you want the 15 generators of su(4) ~ so(6) represented as 6×6 matrices. The simplest basis is 6×6 real antisymmetric matrices, that is the real antisymmetric basis of so(6): The diagonal has to vanish from antisymmetry, and the 15 real entries in the lower left triangle are the opposite real numbers of the upper right triangle. Multiplying these 15 matrices by i to get to the physics notation, you have 15 Hermitian generators of the su(4) algebra. Try an example.

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