Proposition: Non-covariance of the gauge-fixed operator at order κ

Let $E_{\mathrm{tot}}[\gamma ]=A\gamma +E_{\mathrm{gf}}[\gamma ]+\kappa J_3[\gamma ]=-\square \mathcal{P}\gamma +\kappa J_3[\gamma ]$ be the gravitational part of the gauge-fixed Euler derivative through order $\kappa$. For $\xi \in {\mathfrak{g}}_U$ and ${\gamma }^{\prime }$ as above,

Proof. $-\square \mathcal{P}(K\xi +\kappa s_{\xi })=(A+2D^{\ast }D)(K\xi +\kappa s_{\xi })=2D^{\ast }\square \xi +\kappa (A{s}_{\xi }+2D^{\ast }D{s}_{\xi })$ by (eq:algebra). Next, $J_3$ is quadratic, so $J_3[\gamma +K\xi ]=J_3[\gamma ]+{\delta }_0{J}_3[\gamma ]+J_3[K\xi ]$, with ${\delta }_0{J}_3={{\mathcal{L}}^{\mathrm{dens}}}_{\xi }(A\gamma )-A{\mathcal{L}}_{\xi }\gamma$ by (eq:XIEvar). For $J_3[K\xi ]$: the pullback $F_{\kappa }^{\ast }\eta =\eta +\kappa K\xi +{\kappa }^2(K\zeta +\frac{1}{2}{\mathcal{L}}_{\xi }K\xi )+O({\kappa }^3)$ is flat for every $\kappa$, so $\sqrt{-q}G(F_{\kappa }^{\ast }\eta )\equiv 0$; its ${\kappa }^2$ coefficient is ${\mathcal{H}}_1[K\zeta +\frac{1}{2}{\mathcal{L}}_{\xi }K\xi ]+{\mathcal{H}}_2[K\xi ]=0$, and $AK\zeta =0$ gives

Hence $\kappa J_3[{\gamma }^{\prime }]=\kappa J_3[\gamma ]+\kappa ({{\mathcal{L}}^{\mathrm{dens}}}_{\xi }(A\gamma )-A{s}_{\xi })+O({\kappa }^2)$, and the $A{s}_{\xi }$ terms cancel.