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It's a group theory question .. it's known that

## 10 \otimes 5^* = 45 \oplus 5, ##

Make the direct product by components:

##[ (1,1)^{ab}_{1} \oplus (3,2)^{ib}_{1/6} \oplus (3^*,1)^{ij}_{-2/3} ] \otimes [ (1,2)_{ c~-1/2} \oplus (3^*,1)_{ k~1/3} ] = (1,2)^{ab}_{ c~1/2} \oplus (3^*,1)^{ab}_{ k~4/3} \oplus (3,1)^{ib}_{ c ~ -1/3} \oplus (3,3)^{ib}_{ c~-1/3} \oplus (1,2)^{ib}_{ k~1/2} \oplus (8,2)^{ib}_{ k ~ 1/2} \oplus (3^*,2)^{ij}_{ c~-7/6} \oplus (3,1)^{ij}_{ k~-1/3} \oplus (6^*,1)^{ij}_{ k ~ -1/3} ##,

Where a,b,c =1,2, I,j,k= 1,..,3 and ## [5 \otimes 5]_{antisymmetric } = 10 ##. Now there are in the 10 x 5* product two doublets (1,2) and two triplets (3,1), in which have different indices and so different interactions...

According to the number of the degrees of freedom, one doublet and one triplet should go to 5 representation ( in 10 x 5* product ), while the last scalars goes to 45 representation..

The Question is how to know which doublet or each triplet should belongs to 5 or 45 to avoid redandunce in the degrees of freedom?

Or we say for example a doublet ## (1,2,1/2) \equiv (1,2)^{ab}_{ c~1/2} \oplus (1,2)^{ib}_{ k~1/2} ## is found in both 5 and 45 decompositions ? won't be here redundancy..

Bests,

S.

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# Direct product of two representations

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