Proposition: No compactly supported residual gauge

For a vector field $\xi$, $P_{\mathrm{gr}}(K\xi )=-\frac{1}{4}\mathcal{P}K(\square \xi )$. Hence $K\xi \in \ker P_{\mathrm{gr}}$ iff $\square \xi$ is a Killing field of $\eta$. If $\xi$ is compactly supported, this forces $\xi =0$. Consequently

and the frozen algebra acts on the gauge-fixed free graviton only as a background transformation of $(h_{\mathrm{cl}},k_{\mathrm{cl}})$, never as a residual symmetry of $\hat{h}$.

Proof. $\mathcal{P}K\xi =K\xi -\eta \partial \cdot \xi$ and $\square$ commutes with $K$, $\partial \cdot$, and $\eta$; so $\square \mathcal{P}K\xi =\mathcal{P}K\square \xi$. $\mathcal{P}$ is invertible, so the kernel condition is $K\square \xi =0$. A compactly supported Killing field of Minkowski space vanishes, so $\square \xi =0$ with $\xi$ compactly supported, and Cauchy uniqueness gives $\xi =0$.