Since $\partial_\mu K^\mu=\operatorname{Tr}(F_{\mu\nu}{}^\star F^{\mu\nu})$, define
$$
\widetilde J_A^\mu=J_A^\mu-\frac{N_fg^2}{8\pi^2}K^\mu.
$$
The anomaly equation gives $\partial_\mu\widetilde J_A^\mu=0$. The price is that the <Chern-Simons current> $K^\mu$ is not gauge invariant, so neither is $\widetilde J_A^\mu$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 305 / 2 / vi / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-305/2/vi/solution