# $SU(3)$ 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?

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 $+=\oplus$.
• Ok, basically one uses the Fierz decomposition, right? A couple of things are not so clear: 1) why the anti-symm part is the vector 6 ? You mean $6+\bar 6$? 2) (stupid thing, just to clarify) shouldn't the 10 get decomposed as 6+3+1 when where 0,1,2 indices among the four take the value 4, respectively ? I mean, 6 is the symmetric for $SU(3)$, the 3 is the fundamental and then the symmetric piece with indices (4,4) is the trivial wrt $SU(3)$. – BLS Jun 23 '16 at 15:55
• The vector 6 is $4\wedge 4$ of SU(4) which is equivalent to $\bar 4\wedge \bar 4$ by the Hodge dualization. I haven't made any error of writing 6 instead of $6\oplus \bar 6$. Also: Under the SU(3) subgroup, this vector $6$ obviously decomposes as $3\oplus \bar 3$, not as the irreducible symmetric spintensor. This is obvious if you embed the $SU(3)$ into $SO(6)$ by simply pairing the 6 real coordinates of $SO(6)$ into 3 complex coordinates (pairs). So the 6 become the 3 complex and their complex conjugates, something that one does all the time if he studies Calabi-Yau three-folds. – Luboš Motl Jun 23 '16 at 18:10
• There are trivial ways to see that basically all your guesses are just plain wrong. For example, the symmetric tensor 6 of $SU(3)$ is a complex rep - inequivalent to its complex conjugate $\bar 6$ - so it just can't be $6$ of $SO(6)$ which is a real rep. BTW concerning terminology, yes, by "usual Dirac matrix calculus", I meant the Fierz identity. – Luboš Motl Jun 23 '16 at 18:16
• Indeed the $4 \wedge 4$ has the same Young Tableaux of its conjugate (so it's equivalent to $\bar 4 \wedge \bar 4$ and it's a real irrep of $SU(4)$). I was simply referring to the wrong Tableaux yesterday and I couldn't figure out what I was doing wrong. Btw, if you can tell me some very good references where to learn branching rules in general, it would be very helpful! – BLS Jun 24 '16 at 3:24
• And the initial confusion stemmed from the fact that I thought that that 20 should be an irrep for $SU(4)$ while actually it's not, as Fierz tells us. Thank you! – BLS Jun 24 '16 at 3:28