Theorem: Local free-shell tangency is Killing symmetry

For smooth $\xi$ on ${\mathcal{D}}_0$, the following are equivalent:

1. $P(X_{\xi }\varphi )=0$ on ${\mathcal{D}}_0$ for every $\varphi \in {\mathcal{S}}_{\varphi }$;

2. $K\xi =0$ on ${\mathcal{D}}_0$;

3. in developing coordinates,

   $${\xi }^{\mu }(x)=a^{\mu }+B^{\mu }{}_{\nu }{x}^{\nu },B_{\mu \nu }=-B_{\nu \mu },$$

   for constants $a,B$.

A nonzero smooth vector compactly supported in ${\mathcal{D}}_0$ therefore fails the tangency condition.

Proof. At a fixed point, Lemma lem:jets and (eq:commutator) imply

for every allowed jet. Take $A=0,H=0$, and arbitrary $b$, obtaining $\square \xi =0$. Next take $A=0,b=0$, and arbitrary trace-free symmetric $H$. The symmetric derivative must be a multiple of ${\eta }^{\mu \nu }$. Finally choose $A=1,b=0$, and $H$ of trace $m^2$. Since $m>0$, that multiple is zero.

Differentiating the Killing equation and permuting indices gives

Connectedness gives the affine form. Conversely such an affine Killing field makes the commutator zero. An affine field vanishing on a nonempty open set is zero.