Lemma: Off-origin form and oriented conormal bound

On $r\ne 0$,

and

Away from the nonzero light cone, $d$ is smooth. On a spacelike region,

Proof. The angular Fourier integral in (eq:W) gives the massless rational two-point distribution with the positive frequency temporal boundary condition. Selecting the two temporal branches in (eq:timeordered) gives (eq:rational) off the origin.

At a nonzero null point, $\mathrm{d}q\ne 0$, so $q$ is a local coordinate transverse to the cone. For the stated Fourier convention, $(s-\mathrm{i}0{)}^{-1}$ has the single negative-frequency conormal ray. Pulling it back by $q$ gives (eq:WFd). At $q\ne 0$, the reciprocal is smooth.

This is also the massless specialization of the Feynman wavefront description in BDHintro (Proposition 26). That paper uses the opposite Fourier sign in its physical wavefront convention; reversing the covector sign gives (eq:WFd). The one-ray product test is independent of that sign choice.