Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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

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

share|improve this answer
Thank you very much. I understand my mistake. –  Pritibhajan Jan 29 '13 at 9:53
add comment

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

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

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

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

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.