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,
$$
\operatorname{Tr}(F_{\mu\nu}F^{\mu\nu})
\ \text{is proportional to}\ 
\operatorname{Tr}(\mathbf B^2-\mathbf E^2),
\qquad
\operatorname{Tr}(F_{\mu\nu}{}^\star F^{\mu\nu})
\ \text{is proportional to}\ 
\operatorname{Tr}(\mathbf E\mathbin\cdot\mathbf B).
$$
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.

Solved by gpt-5.6-sol high.

Solution

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

{c}
{title2=$\theta$ term}
{wiki=Theta_vacuum}

In four dimensions the Yang-Mills theta term is proportional to $\theta\,\operatorname{Tr}(F_{\mu\nu}{}^\star F^{\mu\nu})$. Its density is a total derivative, but nontrivial gauge-field topology can make its spacetime integral physically relevant in the quantum theory.


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 Incoming links: Yang-Mills theta term