I read in at least one paper and in the wiki below


<http://en.wikipedia.org/wiki/Quark_model>


that the 56 symmetric irrep of SU(6) breaks down into 10^{3/2} + 8^{1/2}
irreps of SU(3)xSU(2). Here the first is 40 dimensional (10 of SU(3) x 4 of SU(2))
and the second is 16 dimensional (8 of SU(3) x 2 of SU(2)).


The problem is that this is simply not true. I checked the branching rules and this
just doesn't show up. Can someone please double check if the above decomposition is
correct or not.


Thanks.



SU(6) -> SU(3) branching rule
You appear to have made a mistake in your calculation of the branching rules. The answer given in the wiki is correct, but it seems that you are using the 'wrong' subgroup of $\mathrm{SU}(6)$. Perhaps you are using the subgroup isomorphic to $\mathrm{SU}(2)\times\mathrm{SU}(3)$ under which the fundamental $\mathrm{SU}(6)$-representation $\mathbb{C}^6$ breaks up as $\mathbb{C}\oplus\mathbb{C}^2\oplus\mathbb{C}^3$. 


In the wiki page you cite, the author is using the subgroup under which the fundamental $\mathrm{SU}(6)$-representation is still irreducible, but is a tensor product of $V$, the $2$-dimensional representation of $\mathrm{SU}(2)$, and $W$, the $3$-dimensional representation of $\mathrm{SU}(3)$. Then the statement (which is correct) is that, as $\bigl(\mathrm{SU}(2)\times\mathrm{SU}(3)\bigr)$-representations, one has
$$
\mathsf{S}^3(V\otimes W)\simeq \bigl(\mathsf{S}^3(V)\otimes \mathsf{S}^3(W)\bigr) \oplus \bigl(V\otimes (W\otimes W^\*)\_0\bigr).
$$
These are the two irreducible subspaces that you see in the answer.


This is a special case of the general formula for $\mathsf{S}^3(V\otimes W)$ as a sum of tensor products of representations of $\mathrm{SL}(V)$ and $\mathrm{SL}(W)$. [One term is missing because $\Lambda^3(V)=0$, and I have used other identifications that hold because the ranks of the two factor groups are so small.]


