Past exam of the groups and representations course of the University of Oxford / 2025 / 2 / d / Solution

In the normalization $c(8)=1$ of c, the Adjoint representation of $D_4$ has highest weight ${\lambda }_{\mathrm{adj}}=(0,1,0,0)$, the highest root. The supplied Gram matrix has second row $(1,2,1,1)$, whose sum is five. HenceThe Adjoint representation has the same dimension as the Lie algebra, so its Dynkin index is also $c(\mathrm{adj})=6$. The $k$ Weyl fermions form a direct sum of $k$ vector representations; additivity of the trace gives $c(r_W)=k$.

Substitution into the given one-loop renormalization-group beta function yieldsFor positive $g$, asymptotic freedom at this order requires a negative beta function, equivalentlyIndeed, if $b_0=22-2k/3>0$, integrating $\mu dg/d\mu =-b_0{g}^3/(16{\pi }^2)$ givesso 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 $C(\lambda )={\lambda }^{T}G(\lambda +21)$ described in c, $c(8)=2$, $c(\mathrm{adj})=12$, and the same source formula reads $\beta (g)=-(44-4k/3)g^3/(16{\pi }^2)$. Rescaling the gauge coupling together with the Lie algebra generators relates the conventions; the bound $k<33$ is identical.