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Srednicki 70.4: Group Theory

  1. Dec 20, 2013 #1
    1. The problem statement, all variables and given/known data

    (a) For SU(N), we have:
    N ⊗ N = A_A + S_S
    where A corresponds to a field with two antisymetric fundamental SU(N) in- dices φij = −φji, and S corresponds to a field with two symmetric fundamental SU(N) indices φij = φji. By considering an SU(2) subgroup of SU(N), compute T(A) and T(S).

    2. Relevant equations

    Under the SU(2) subgroup, N transforms as:

    [2 ⊕ (N − 2)1S] ⊗ [2 ⊕ (N − 2)1s] = A + S.

    3. The attempt at a solution

    Distributing, we get, among other terms, 2 ⊗ (N − 2)1S ⊕ 2 ⊗ (N − 2)1S. It's then just a matter of separating the symmetric and anti-symmetric terms.

    What I don't understand is why one of these 2 ⊗ (N − 2)1S terms is symmetric and the otherone is anti-symmetric. They seem identical to me..... I think part of the problem is I don't understand Srednicki's notation 1s (sometimes 1's) -- does this just mean the singlet is symmetrical?

    I understand Young Tableaux, but not sure how to use them here, the only diagram I can work out is adjoint, and therefore neither symmetrical nor anti-symmetrical.

    This is the only difficulty, if we can figure out this, I know how to solve the problem from there.....

    Thanks!
     
    Last edited: Dec 20, 2013
  2. jcsd
  3. Dec 21, 2013 #2
    To phrase the question a different way, consider

    4 x 4 = 10_S + 6_A.

    Now if we consider the SU(3) subgroup, we have:

    (3 + 1) * (3 + 1) = 3 * 3 + 3 * 1 + 1 * 3 + 1 * 1

    Now clearly 1 * 1 = 1 and 3 * 3 = 6 + 3_A. So we have:

    10 + 6_A = 6 + 3_A + 1 +3 * 1 + 1 * 3.

    which gives:

    3 + 3_A = 3 * 1 + 1 * 3

    But how can this be? Whenever I do young Tableaux, I get that 3 * 1 = 1 and 1 * 3 = 3 + 2_S, which sum to 4 + 2_S ??
     
  4. Dec 21, 2013 #3
    To phrase the question a different way, consider

    4 x 4 = 10_S + 6_A.

    Now if we consider the SU(3) subgroup, we have:

    (3 + 1) * (3 + 1) = 3 * 3 + 3 * 1 + 1 * 3 + 1 * 1

    Now clearly 1 * 1 = 1 and 3 * 3 = 6 + 3_A. So we have:

    10 + 6_A = 6 + 3_A + 1 +3 * 1 + 1 * 3.

    which gives:

    3 + 3_A = 3 * 1 + 1 * 3

    But how can this be? Whenever I do young Tableaux, I get that 3 * 1 = 1 and 1 * 3 = 3 + 2_S, which sum to 4 + 2_S ??
     
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