 Solution

Keep the color matrix intact: ${\mathcal{A}}_{\mu }=A_{\mu }^a{T}^a$. Under parity, invariance of its contraction with the vector current requiresUnder charge conjugation, the fermionic reordering in part (b) transposes the color matrix as well as the spinor matrix. For a fixed color matrix $M$, $\bar{\psi }M{\gamma }^{\mu }\psi$ becomes $-\bar{\psi }{M}^T{\gamma }^{\mu }\psi$. Hence invariance of the interaction requiresThis is not a rule that changes the sign of every individual color component: transposition also acts on the generators of the fundamental representation.

Write the gauge field strength as ${\mathcal{F}}_{\mu \nu }={\partial }_{\mu }{\mathcal{A}}_{\nu }-{\partial }_{\nu }{\mathcal{A}}_{\mu }+ig[{\mathcal{A}}_{\mu },{\mathcal{A}}_{\nu }]$. Since $[A^T,B^T]=-[A,B{]}^T$, the charge conjugation rule gives ${\mathcal{F}}_{\mu \nu }\mapsto -{\mathcal{F}}_{\mu \nu }^T$. Including the coordinate reflection in CP symmetry,For generators normalized by $\mathrm{Tr}(T^a{T}^b)=\kappa {\delta }^{ab}$, the color contraction equals ${\kappa }^{-1}\mathrm{Tr}({\mathcal{F}}_{\mu \nu }{\mathcal{F}}_{\rho \sigma })$. The two charge-conjugation minus signs cancel, and transposition reverses the order inside the trace without changing it. The Levi-Civita symbol acquires $\det \Lambda =-1$ under the remaining reflection. ThusThe Yang-Mills theta term is CP-odd; a generic fixed nonzero $\theta$ produces CP violation. Setting $\theta =0$ removes this source. Although the density is a total derivative, it can affect the quantum theory through nontrivial topological sectors; the local transformation test is not rendered irrelevant by that fact.