The fermion in a chiral multiplet has <R-charge> $R[\Phi]-1$, while the gluino has charge one. Cancellation of the $Sp(N_c)^2U(1)_R$ anomaly gives
$$
I(\mathrm{adj})+2N_fI(\square)(R[\Phi]-1)=0.
$$
Using $I(\mathrm{adj})=2(N_c+1)$ and $I(\square)=1$ yields
$$
\boxed{R[\Phi]=1-\frac{N_c+1}{N_f}
=\frac{N_f-N_c-1}{N_f}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 307 / 4 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-307/4/i/solution