Two diagnostics

Take $\xi \in {\mathfrak{g}}_U$ and the complete pure-gauge two-jet

The copied coordinate laws give

On the free field,

Therefore ${\hat{A}}_1[K\xi ]=0$.

The retained mixed term is

The minus sign follows from the normalization and insertion sign. The rejected plus sign would not be compatible with them.

On this complete jet, the bare second dressing is

Thus the incomplete second coefficient in (eq:truncation) is

which need not vanish. For example, in the massless theory choose a compact time-component parameter in $U$ for which ${X_{\xi }}^{2}t$ is nonzero, and an observation test supported where it has fixed sign.

Classically, the missing response is ${\rho }_{2,g}=\frac{1}{2}{X_{\xi }}^{2}u$, and it cancels this remainder. No corresponding second-order quantum map is constructed here. The diagnostic demonstrates the order boundary, not a defect in the first-order theorem and not a contact-normalization problem.

Let $m=0$ and $h=\psi \eta$, with smooth compact $\psi$. Then

using $\square \hat{\varphi }=0$. The retarded correction is

This is an element of ${\mathfrak{A}}_0$, with a smooth compact test. The coordinate time component is ${{\mathsf{X}}_1}^0=-{{\mathsf{E}}_0}^{\mathrm{R}}{\partial }_t({\sigma }_{\epsilon }\psi )$.

For a time-dependent illustration, select admissible data with $\psi =\psi (t)$ and $g=g_0(t)$ near the relevant causal compact, and cut them off spatially outside it. Define

On the smooth free solution $u=t$,

Classical evaluation of the specified linear representatives gives

and

These are classical numbers, not elements of ${\mathfrak{A}}_0$, quantum expectation values, or states on the Wick algebra. They check the mismatch of the two switches and the mixed-term sign. The time-dependent cutoffs do not replace the general construction or its support proof.