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Direct product of two representations

  1. May 30, 2015 #1
    Hi their,

    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.
     
    Last edited: May 30, 2015
  2. jcsd
  3. May 30, 2015 #2

    fzero

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    I'd suggest working the ##SU(5)## product out in components and then reducing indices along the branching rules. This should let you match up to the ##SU(3)\times SU(2)\times U(1)## irreps the way that you want.
     
  4. May 30, 2015 #3
    May you give me further clarification or an example, because I'm not familiar with branching rules to know what do you mean..

    Thanx
     
  5. May 30, 2015 #4

    fzero

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    A branching rule is the description of how the representation of a group decomposes into irreps of a subgroup (usually maximal). In your second equation you have the branching rules for the ##\mathbf{10}## and ##\mathbf{5^*}##. You need the corresponding branching rules for the ##\mathbf{5}## and ##\mathbf{45}## (they should be in the Slansky review). By following indices through the products you should be able to do the matching that you want to do.
     
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