# Srednicki 70.4: Group Theory

1. Dec 20, 2013

### RobyVonRintein

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. Dec 21, 2013

### RobyVonRintein

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 ??

3. Dec 21, 2013

### RobyVonRintein

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 ??