Theorem: Free-shell tangency is exactly Killing symmetry

For smooth $\xi$, the following are equivalent:

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

2. $K\xi =0$;

3. $${\xi }^{\mu }(x)=a^{\mu }+B^{\mu }{}_{\nu }{x}^{\nu }$$

   for constants with $B_{\mu \nu }=-B_{\nu \mu }$.

A field satisfying these conditions and vanishing on a nonempty open set is identically zero.

Proof. Assume (i). At a fixed point, (eq:commutator) and the jet lemma imply

for every allowed jet, where

Take $a=0,H=0$ and arbitrary $b$, obtaining ${\square }_{\eta }\xi =0$.

Take $a=0,b=0$ and arbitrary trace-free symmetric $H$. The annihilator of that codimension-one space is the span of the trace functional, so

at the point. Finally take $a=1,b=0$ and a symmetric $H$ of trace $m^2$. Then $c{m}^2=0$, hence $c=0$. The point was arbitrary, so $K\xi =0$.

Differentiating the Killing equation gives

Thus $\xi$ is affine with antisymmetric linear part.

Conversely, an affine Killing field has ${\square }_{\eta }\xi =0$ and vanishing symmetric first derivative. The commutator identity gives (i). An affine field vanishing on an open set is zero.