 Solution

ID: past-exam-of-the-groups-and-representations-course-of-the-university-of-oxford/2022/3/e/solution

Sym^{2} (A \oplus B) = Sym^{2} A \oplus (A \otimes B) \oplus Sym^{2} B .

Now

Sym^{2} 2 = 3

and

Sym^{2} 3 = 6

, rather than the singlet and antitriplet arising from exterior squares. Thus

15 ↓= (3, 1)_{6} \oplus (2, 3)_{1} \oplus (1, 6)_{- 4} .

3 + 6 + 6 = 15

By contrast, solution gives the following left-handed Weyl fermion content of

\overline{5} \oplus 10

L : (2, 1)_{- 3}, d^{c} : (1, \overline{3})_{2}, Q : (2, 3)_{1}, u^{c} : (1, \overline{3})_{- 4}, e^{c} : (1, 1)_{6} .

Y = q_{c} /6

. The

15

contains

Q

but lacks

L

d^{c}

u^{c}

and

e^{c}

; instead it contains an SU(2) triplet and an SU(3) sextet. Equality of total dimensions therefore does not give the same representation content, and

15

cannot supply the family.

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