Proposition: No compact residual transformation

If $\xi$ is a compactly supported smooth vector field and ${\mathsf{P}}_{\mathrm{gr}}K\xi =0$, then $\xi =0$. In particular,

Proof. Since constant derivatives commute,

Invertibility of ${\mathsf{P}}_{\mathrm{tr}}$ implies $K(\square \xi )=0$. A Minkowski Killing field has affine coefficients: differentiating its Killing equation and permuting indices gives ${\partial }_{\alpha }{\partial }_{\beta }{v}_{\gamma }=0$. A compactly supported Killing field is therefore zero. Thus $\square \xi =0$. Its Cauchy data vanish before its compact support, so Cauchy uniqueness gives $\xi =0$.