Lemma: Covector transport

For $w\in {\mathcal{E}}^{\prime }$, every point of $\mathrm{WF}({\mathsf{W}}_w)$ has the form $(z_1,{\xi }_1;z_2,{\xi }_2;z_3,{\xi }_3)$ with $(y_1,{\xi }_1;y_2,{\xi }_2;y_3,{\xi }_3)\in \mathrm{WF}(w)$ for some $(y_1,y_2,y_3)$, and each ${\xi }_i$ null or zero.

Proof. By (eq:Ecaus), ${\mathsf{E}}_{\mu }$ is the pullback of ${\mathcal{E}}_{\mu }$ under the submersion $(z,y)\mapsto z-y$, so by Hormander (Theorem 8.2.4) $\mathrm{WF}({\mathsf{E}}_{\mu })\subset \{(z,\xi ;y,-\xi ):\xi \in {\mathrm{WF}}_{z-y}({\mathcal{E}}_{\mu })\}$, and by Hormander (Theorem 8.3.1) and $(\square +{\mu }^2){\mathcal{E}}_{\mu }=0$ every such $\xi$ is null and nonzero. The kernel of $w\mapsto {\mathsf{W}}_w$ on $M^3\times M^3$ is the tensor product ${\mathsf{E}}_0\otimes {\mathsf{E}}_0\otimes {\mathsf{E}}_m$, whose wavefront set Hormander (Theorem 8.2.9) consists of triples in which each factor is either in $\mathrm{WF}({\mathsf{E}}_{\mu })$ or carries zero covectors in both of its variables. In particular the $z$-covector is zero exactly when the $y$-covector is zero in each factor, so the kernel has empty ${\mathrm{WF}}_{M^3}$ and ${\mathrm{WF}}_{M^3}^{\prime }$, and Hormander (Theorem 8.2.13) gives $\mathrm{WF}({\mathsf{W}}_w)\subset {\mathrm{WF}}^{\prime }({\mathsf{E}}_0\otimes {\mathsf{E}}_0\otimes {\mathsf{E}}_m)\circ \mathrm{WF}(w)$, which is the stated set: the covector triple is transported unchanged and each entry is null or zero.