First order and scalar-only comparison

To state a background transformation, introduce a prescribed classical tensor $h$, a second coefficient $k$, and an independent smooth tensor configuration $\gamma$. Put $H=h+\gamma$.

For this subsection the copied scalar-only completion is a classical polynomial functional:

where

For free $\varphi$ and $\xi \in {\mathfrak{g}}_U$,

More explicitly, the off-shell difference is the scalar equation test $⟨P_m(X_{\xi }{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}f),\varphi ⟩$. This is the first-order background identity proved in X-B-v3 XB; it is also verified by ${\mathsf{L}}_{1,1}[K\xi ]=D_{\xi }{P}_m-P_m{X}_{\xi }$, with $D_{\xi }v={\partial }_{\mu }({\xi }^{\mu }v)$.

Replacing the quantum metric by a fixed classical $r$ before taking the scalar variation gives

These are exactly the scalar-only maps. Setting $r=0$ gives their zero specialization. It does not recover a nonzero background response without a retained background.

Conversely, $\mathcal{T}$ is independent of an external $\gamma$. Setting $\gamma =0$ after taking the graviton response does not remove $\mathcal{T}$. Restriction of a configuration and taking its response do not commute in that fashion.