$SU(3)$ structures and branching rules I am reading this paper http://arxiv.org/abs/hep-th/0211102 and I would like to understand better about the branching rule $SO(6) \equiv SU(4) \rightarrow SU(3)$ used for eq. C.11 in the Appendix. I understand the branching rules of the 15 since it's the adjoint of $su(4)$ basically, but I don't really get which "20" decomposes as illustrated. So far (by some Young Tableaux computations) I found only inequivalent 20's with the following branchings:
$$
20 \rightarrow 8+3+\bar 3+ 6 \\
20' \rightarrow 10+6+3+1 \\
20'' \rightarrow 6+6+8 
$$ 
I am wondering whether I am missing some other inequivalent 20 or whether I am doing something wrong with the computation. So, how to understand the branching C.11 for that 20?
 A: The sentence above C.11 explicitly says that they talk about 3-forms under $SO(6)$, i.e. antisymmetric tensors $T_{[abc]}$ where $a,b,c=1,2,3,4,5,6$. Those have
$$ \frac{6\times 5\times 4}{3\times 2 \times 1} = 20 $$
components. By the Dirac matrix calculus, all differential forms may be obtained from the tensor product of two spinors and the Dirac spinor is simply $4\oplus \bar 4$ under $SU(4)$. Because 3 (number of indices) is odd, we need the product of two spinors of the same chirality, i.e. $4\otimes 4\oplus \bar 4\otimes \bar 4$. The tensor product have a symmetric and antisymmetric part. The antisymmetric part $4\wedge 4$ is obviously the vector $6$, so it's the symmetric parts
$$(4\otimes_+ 4 )\oplus (\bar 4 \otimes_+ \bar 4) $$
that is equivalent to the 3-form of $SO(6)$, or $10\oplus \bar 10$. The decomposition of this rep under $SU(3)$ is simply obtained by dividing the range of indices $1,2,3,4$ of these symmetric spintensors to $3+1$. $10$ gets decomposed to $1+3+6$ (components where $0,1,2$ indices among the four take the value 4, respectively) and similarly with bars, so the total 20 is decomposed as
$$ 1+1+3+\bar 3+6+\bar 6$$
under $SU(3)$ where $+=\oplus$.
