 Solution

ID: past-exam-of-the-groups-and-representations-course-of-the-university-of-oxford/2024/2/c/solution

Write

[r_{1}, r_{2}, \dots]

r_{i}

, and depict an empty box by

□

. Appending the box of the fundamental representation to another single box gives a row or a column:

□ \otimes □ = (□ □) \oplus □ □, [1] \otimes [1] = [2] \oplus [1, 1] .

4 \cdot 5/2 = 10

and

4 \cdot 3/2 = 6

, so

4 \otimes 4 = 10_{s} \oplus 6_{a} .

For the conjugate representation, the invariant volume tensor identifies

\overset{ˉ}{4} ≅ Λ^{3} 4

, represented by a column of three boxes. Adding one box gives

[1] \otimes [1, 1, 1] = [2, 1, 1] \oplus [1, 1, 1, 1], □ \otimes □ □ □ = □ □ □ □ \oplus □ □ □ □ .

A column of four boxes is the determinant representation, which is trivial on

S U (4)

(1, 0, 1)

and dimension fifteen. For example the hook-content formula gives

dim [2, 1, 1] = \frac{4 \cdot 5 \cdot 3 \cdot 2}{4 \cdot 1 \cdot 2 \cdot 1} = 15.

Equivalently, a mixed tensor

M^{i}_{j}

M \mapsto U M U^{†}

and splits into its trace and traceless matrix parts. Therefore

4 \otimes \overset{ˉ}{4} = 1 \oplus 15,

where

15

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