 Solution

ID: past-exam-of-the-groups-and-representations-course-of-the-university-of-oxford/2020/3/c/solution

Under the stated SU(2) embedding, the first two components of the defining SU(3) representation form a doublet, while the third is fixed. Hence

3 ↓_{S U (2)} = 2 \oplus 1

. Its dual representation restricts in the same way because the SU(2) doublet is self-dual.

8 = 3 \otimes \overline{3} - 1

2 \otimes 2 = 3 \oplus 1

gives

8 ↓_{S U (2)} = (2 \oplus 1) \otimes (2 \oplus 1) - 1 = 3 \oplus 2 \oplus 2 \oplus 1 .

Here subtraction denotes removal of the trivial representation generated by the identity matrix; it is not a negative-dimensional summand. The first SU(3) factor is unchanged, so the requested branching of a representation rules are

(3, 3) (\overline{3}, 1) (1, 3) (1, 8) ↓_{H} = (3, 2) \oplus (3, 1), ↓_{H} = (\overline{3}, 1), ↓_{H} = (1, 2) \oplus (1, 1), ↓_{H} = (1, 3) \oplus 2 (1, 2) \oplus (1, 1) .

The dimensions on the two sides are respectively

9 = 6 + 3

3 = 3

3 = 2 + 1

and

8 = 3 + 2 + 2 + 1

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