Two diagnostics

Let $\xi \in {\mathfrak{g}}_U$, $\zeta =0$, and take

Write $X=X_{\xi }$, $D=D_{\xi }$, $L={\mathsf{L}}_{1,g}[K\xi ]$, and $M_2={\mathsf{L}}_{2,g}[K\xi ,\frac{1}{2}{\mathcal{L}}_{\xi }K\xi ]$. Expanding the copied plateau Euler covariance on the flat configuration gives

Using $P_m{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}f=f$ and ${{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}{P}_{m}v=v$ for compact $v$,

Therefore

The function inside $P_m$ is compact because of $X$. The new maps consequently sum in ${\mathfrak{A}}_0$ to

The copied bare second dressing is $\frac{1}{2}\int f{X}_{\xi }^2\hat{\varphi }$, and the mixed term is $-\int f{X}_{\xi }^2\hat{\varphi }$. Thus

The new maps cancel the incomplete remainder with ${\mathcal{O}}^{(2)}$ retained. Taking $k=0$ instead would not test a pure-gauge two-jet.

to the reversed-sign shorthand. Reversing only the second-vertex sign gives $L{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}L{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}f+M_2{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}f$, whose class is

The second term is not generally an equation zero. Thus a universal assertion that the reversed new sum is simply $\frac{3}{2}\int f{X}_{\xi }^2\hat{\varphi }$ is incorrect. What is universal is that reversal adds $2\hat{\varphi }(M_2{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}f)$ to the correctly completed coefficient. The operator computation above, rather than the shorthand, tests the sign.

For an explicit sign-failure diagnostic, choose admissible massless data with a sufficiently large plateau cylinder, $\xi =a(t){\partial }_t$ near the relevant causal compact, a spatial cutoff outside it, and nonzero compact $a$ supported before the observation times. On the free solution $u=t$,

After the support of $a$, reversal therefore produces $2\int (a^{\prime }{)}^2\mathrm{d}s>0$, although $X^{2}t=0$ there. A nonnegative observation bump detects this nonzero classical evaluation. This also shows why the extra term cannot be discarded.

Let $m=0$, $h=\psi \eta$, and $k=0$. Then

Indeed, for the formal conformal metric the density $\sqrt{-q}{q}^{-1}=(1+\kappa \psi )\eta$ is exactly linear. Thus ${\mathsf{L}}_{2,g}=0$ in this massless example: only the iteration supplies the second scalar response.