 Solution

ID: past-exam-of-the-groups-and-representations-course-of-the-university-of-oxford/2025/2/d/solution

c (8) = 1

D_{4}

λ_{adj} = (0, 1, 0, 0)

, the highest root. The supplied Gram matrix has second row

(1, 2, 1, 1)

, whose sum is five. Hence

C_{2} (adj) = \frac{1}{2} (2 + 2 \cdot 5) = 6.

c (adj) = 6

. The

k

k

vector representations; additivity of the trace gives

c (r_{W}) = k

Substitution into the given one-loop renormalization-group beta function yields

β (g) = - \frac{g ^{3}}{16 π ^{2}} (22 - \frac{2 k}{3}) = - \frac{33 - k}{24 π ^{2}} g^{3} .

For positive

g

, asymptotic freedom at this order requires a negative beta function, equivalently

k < 33; for a nonnegative integer number of fermions, k = 0, 1, \dots, 32.

Indeed, if

b_{0} = 22 - 2 k /3 > 0

, integrating

μ d g / d μ = - b_{0} g^{3} / (16 π^{2})

gives

\frac{1}{g ^{2} ( μ )} = \frac{1}{g ^{2} ( μ _{0} )} + \frac{b _{0}}{8 π ^{2}} lo g \frac{μ}{μ _{0}},

so the gauge coupling decreases towards zero as the energy scale increases. At

k = 33

the one-loop coefficient vanishes and the given equation cannot decide asymptotic freedom;

k > 33

has the opposite one-loop running.

With the alternative convention

C (λ) = λ^{T} G (λ + 21)

described in c,

c (8) = 2

c (adj) = 12

, and the same source formula reads

β (g) = - (44 - 4 k /3) g^{3} / (16 π^{2})

. Rescaling the gauge coupling together with the Lie algebra generators relates the conventions; the bound

k < 33

is identical.

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