Proposition: Arbitrary pure-trace polarization

For arbitrary smooth scalar functions ${\psi }_1,{\psi }_2$, with $p={\psi }_1{\psi }_2$, the imported correction satisfies

The same identity holds under complex bilinear extension. The coefficient of ${\sigma }^{\prime \prime }$ is zero on this sector, including its polarization.

Proof. Since $\sigma$ depends only on $t$,

Equation (eq:trace-input) becomes

Apply (eq:polarization). The diagonal terms cancel, leaving

Each summand of ${\mathcal{T}}^{\mu \nu }$ is a first derivative with constant coefficient, so the product rule identifies this with the right side of (eq:trace-identity).

The formula contains no conjugation and is bilinear over $\mathbb{C}$ after complexification.

The reported $C_2$ coefficient vanishes for every pure-trace input. The sum of two pure-trace inputs is pure trace. Consequently its polarization also vanishes.