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

Varying the Yang-Mills theory field strength and using $\delta A_{\mu }=D_{\mu }\alpha$ givesThus $F_{\mu \nu }$ transforms covariantly in the adjoint representation.

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With matrix-valued fields, charge conjugation acts asUnder parity, $A_0$ is even and $A_i$ is odd:Consequently $F_{0i}$ is odd and $F_{ij}$ is even. Under time-reversal symmetry, which is antiunitary, $A_0$ is even and $A_i$ is odd after $t\mapsto -t$; hence $F_{0i}$ is even and $F_{ij}$ is odd. Equivalently, the non-Abelian electric and magnetic fields transform as

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Since $\delta F_{\mu \nu }=-ig[F_{\mu \nu },\alpha ]$, both $F_{\mu \nu }{F}^{\mu \nu }$ and $F_{\mu \nu }{}^{\star }{F}^{\mu \nu }$ vary by a commutator. Cyclicity of the matrix trace therefore makes both terms gauge invariant. In electric and magnetic variables,The Yang-Mills kinetic term is even under $C$, $P$, and $T$. The Yang-Mills theta term is even under $C$ because both fields change sign and transpose, but odd under $P$ and under $T$ because exactly one of $\mathbf{E},\mathbf{B}$ changes sign.

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Expanding $F_{\mu \nu }$, using antisymmetry of ${ϵ}^{\mu \nu \rho \sigma }$, and cyclically rearranging the matrix trace giveswhere the Chern-Simons current isThe theta contribution to the action is therefore a boundary term. For variations that vanish at the spacetime boundary, its variation vanishes, so it contributes nothing to the classical Euler-Lagrange field equations. Its integral can still distinguish topologically nontrivial gauge configurations in the quantum theory.

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For massless fermions the classical axial rotation has Noether currentPromote $\beta$ to a spacetime-dependent parameter. The classical variation, after integration by parts, is $-\int d^{4}x\beta {\partial }_{\mu }{J}_A^{\mu }$. The chiral anomaly is equivalently encoded by the stated shift $\theta \mapsto \theta +2N_f\beta$, which changes the theta action byThe anomalous Ward identity is thereforeso $\alpha =N_f{g}^2/(8{\pi }^2)$ in the trace convention of the question.

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Since ${\partial }_{\mu }{K}^{\mu }=\mathrm{Tr}(F_{\mu \nu }{}^{\star }{F}^{\mu \nu })$, defineThe anomaly equation gives ${\partial }_{\mu }{\tilde{J}}_A^{\mu }=0$. The price is that the Chern-Simons current $K^{\mu }$ is not gauge invariant, so neither is ${\tilde{J}}_A^{\mu }$.

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