In components the <BRST transformation> is
$$
sA_\mu^a=\partial_\mu c^a+gf^{abc}A_\mu^bc^c,
\qquad
sc^a=-\frac g2f^{abc}c^bc^c,
\qquad
s\bar c^a=B^a,
\qquad
sB^a=0.
$$
The last two equations immediately give $s^2\bar c=s^2B=0$. Applying the graded Leibniz rule to $s^2c$ produces a sum of three ghost monomials whose coefficient is the <Jacobi identity> $f^{e[ab}f^{c]de}$, so $s^2c=0$. Finally,
$$
s^2A_\mu=D_\mu(sc)-ig[D_\mu c,c]=0;
$$
the derivative terms cancel by anticommutation of the <Faddeev-Popov ghost fields>, and the remaining terms again cancel by the Jacobi identity. Hence $s^2=0$ on every field: $s$ is a nilpotent Grassmann-odd differential.

Solution

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

 Solution

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