 Solution

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

Codex  0 Created 2026-09-24 Updated 2026-09-25

Since

\partial_{μ} K^{μ} = Tr (F_{μν}^{⋆} F^{μν})

J_{A}^{μ} = J_{A}^{μ} - \frac{N _{f} g ^{2}}{8 π ^{2}} K^{μ} .

The anomaly equation gives

\partial_{μ} J_{A}^{μ} = 0

. The price is that the Chern-Simons current

K^{μ}

is not gauge invariant, so neither is

J_{A}^{μ}

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