Theorem: Smooth compact representative of a conserved distribution

Let $C$ be a compact symmetric contravariant tensor distribution with $K^{\ast }C=0$, and suppose

is smooth. Choose a temporal cutoff $\chi$, zero sufficiently far in the past and one sufficiently far in the future. Then

is smooth, compact, and conserved, and

Proof. Commutation of trace reversal and constant-coefficient operators gives

Thus $L_{\mathrm{E}}q=0$ by (eq:LEidentities). The tensor $L_{\mathrm{E}}(\chi q)$ is supported in the transition slab of $\chi$. Since $q$ is spacelike compact, this support is compact. Conservation follows from the Bianchi identity in (eq:LEidentities).

We prove equality first for $h\in {\mathcal{S}}_{\mathrm{dD}}$. Put $d=D(\chi q)$, which is smooth and compact. Taking trace reversal in (eq:LEidentities) gives

The last term pairs to zero with $h$, because $D(h)=0$ and $d$ is compact.

Write $\tilde{C}={\left(C\right)}^{♭♭}$, so $\bar{q}={\mathsf{E}}_2\tilde{C}$, and set $F=\square (\chi \bar{q})$, a smooth compact tensor. Uniqueness of the retarded and advanced Cauchy problems gives

Therefore ${\mathsf{E}}_{2}F=-\bar{q}$ and

By (eq:compactprimitive), $F+\tilde{C}=\square A_2$ with $A_2$ compact. Since $\square h=0$ on ${\mathcal{S}}_{\mathrm{dD}}$,

Consequently $F_{J_C}(h)=-⟨F,h⟩=⟨C,h⟩$.

For general $h\in \mathcal{S}$, the zero-Cauchy compensator gives $h^{\mathrm{dD}}=h-K{w}_0(h)\in {\mathcal{S}}_{\mathrm{dD}}$. Both $C$ and $J_C$ annihilate smooth pure gauges by compact integration by parts. The equality for $h^{\mathrm{dD}}$ therefore implies equality for $h$.