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A 6-dimensional ( fundamental) representation of $SU(6)$ becomes (3,2) representation in $SU(3)\times SU(2)$. We can decompose $6\times 6$ of $SU(6)$ into 21-dimensional symmetric and 15-dimensional anti-symmetric representations. What can be the symmetric and anti-symmetric parts of that representation for $SU(3)\times SU(2)$? Or is it not possible to decompose in this way?

To find answer I proceed as follows: $(3,2)\times(3,2)=(1+8,1,3)=(1,1)+(1,3)+(8,1)+(8,3)$. But I failed to identify all the symmetric and antisymmetric representations out of them. Alone $(8,3)$ is 24 dimensional and bigger than the 21-dymensional symmetric representation.

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up vote 1 down vote accepted

First we have $3\times 3 = 6+3$ and $2\times 2 = 3+1$. Then combining this we have

$(3,2)\times (3,2) = [ (3,1)+(6,3) ] + [ (3,3)+(6,1) ]$

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Thank you very much. I understand my mistake. – Pritibhajan Jan 29 '13 at 9:53

$6 \otimes 6 \to 18 \oplus 6 \oplus 9 \oplus 3$

$Sym^2(6) \to 18 \oplus 3$

$Ext^2(6) \to 9 \oplus 6$

here $18=(6 \otimes 3)$,$6=(6 \otimes 1)$,$9=(3 \otimes 3)$,$3=(3 \otimes 1)$ of $SU(3) \times SU(2)$.

I was looking at something similar

SU(6) -> SU(3) branching rule

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