The <Yang-Mills action> is BRST invariant because its field strength transforms covariantly, and the gauge-fixing contribution is BRST exact. Nilpotence therefore gives
$$
s\mathcal L=s\mathcal L_{\rm YM}+s^2\!\left(\bar c^aL[A^a]-\frac\xi2\bar c^aB^a\right)=0.
$$
The gauge-fixing functional itself is generally not closed: $sL[A]=L[Dc]$. Applying the graded Leibniz rule gives
$$
\boxed{\mathcal L=\frac14F_{\mu\nu}^aF^{\mu\nu,a}
+B^aL[A^a]-\frac\xi2B^aB^a-\bar c^aL[(D c)^a]}.
$$
The <Nakanishi-Lautrup field> $B^a$ is auxiliary. Its algebraic equation $B^a=L[A^a]/\xi$ turns the middle terms into $L[A]^2/(2\xi)$, while the final term is the <Faddeev-Popov ghost field> action.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 304 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-304/4/c/solution