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Technibaryon transformation

  1. Sep 7, 2015 #1
    Consider a model with an exact ##SU(N_{TC})## techni-color symmetry and a ##SU(N_{TF})_L\otimes SU(N_{TF})_R## global techni-flavour symmetry which is spontaneously broken to the diagonal sub-group ##SU(N_{TF})## by condensates producing techni-pions (TC\pi) and techni-baryons(TCb).
    What i'm trying to understand is how the various TCb transform under ##SU(N_{TF})## and ##SU(2)_{spin}##.
    Because the wave-function is totally antisymmetric in techni-color we expect that TCb must be fully symmetric in spin and techni-flavour. Following the line of thought of Georgi's chapter 15 I have embedded ##SU(N_{TF})## and ##SU(2)_{spin}## in a ##SU(2N_{TF})##. The TCb transform under this ##SU(2N_{TF})## like a ##N_{TC}## completely symmetric combination, or in Young tableaux notation like a tableaux with ##N_{TC}## horizontal box. Now Georgi says that to understand what representation of ##SU(N_{TF})## and ##SU(2)_{spin}## are contained in the fully symmetric representation of ##SU(2N_{TF})## we must take the tensor product of all tableaux with ##N_{TC}## boxes of ##SU(N_{TF})## and ##SU(2)_{spin}## and see if they contain the fully symmetric representation. The results (see for example Chivukula, R.S. & Walker, T.P., 1989. Technicolor cosmology, Boston, MA: Boston Univ.) should be the following:
    -For ##N_{TC}=3## Schermata 2015-09-07 alle 17.55.43.png

    -For ##N_{TC}=4## Schermata 2015-09-07 alle 17.55.28.png
    The problem is that, except for the ##N=3## case, I'm not able to obtain the fully symmetric representation by taking the tensor product of this tableaux.
     
  2. jcsd
  3. Sep 7, 2015 #2

    fzero

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    I don't think we necessarily need to have the totally symmetric representation of ##SU(2N_{TF})##, just a representation that reduces to the trivial representation of ##S_{N_{TC}}##. Any diagram that is trivial after deleting all rows with ##N_{TC}## or more boxes from the diagram will work. The totally symmetric representation for ##SU(2N_{TF}## will work, but so will others.

    Looking at the products of the representations in your reference, we have
    young1.JPG
    where in the middle line we actually have a few copies of the trivial ##S_4## representation appearing. For the missing representations like
    young2.JPG
    we can see that we don't have enough boxes in the rows to make a trivial ##S_4## representation.
     
  4. Sep 8, 2015 #3
    Yes I agree with this.
    Why I can delete all the rows with ##N_{TC}## or more boxes?
    Am I wrong or the trivial representation of ##S_4## is a Young tableaux with 4 horizontal boxes?
     
  5. Sep 8, 2015 #4

    fzero

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    Yes it is, but a Young tableaux with 4 horizontal columns of any equal size is also the trivial representation. So I mean delete in the same way that you would if you had a column with ##N## boxes for ##SU(N)##.
     
  6. Sep 8, 2015 #5
    Could you suggest some reference where this result is derived?
     
  7. Sep 8, 2015 #6

    fzero

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    Georgi discusses the Young tableaux for ##S_n## in sects 1.21-24. The trouble is that we already know that irreps of ##S_4## never have more than 4 boxes so not all of the rules there apply (for instance we can't naively compute the dimension using 8 as the number of boxes). But we can assign ##j##-cycles to the the boxes, so we get 4 2-cycles on 4 objects, which should be easy to show is reducible to 4 1-cycles.
     
  8. Sep 8, 2015 #7
    Are you sure? For example the four 2-cycles (12)(23)(14)(23) do not seem to be equivalent to four 1-cycles.
     
  9. Sep 8, 2015 #8

    fzero

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    The rules of the Young Tableau are that the columns are the ##j##-cycles and that we can't repeat an index in the same row. If we had a valid diagram, we would never repeat an index in any box, but with more than ##n## boxes we have to relax something.

    With the above rules we only get elements like (13)(21)(34)(42) that are always reducible.
     
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