Lemma: Punctured retarded kernel

The distribution (eq:punctured-response) is defined on $M\setminus \{0\}$, has scaling degree four at the omitted point, and equals

Here ${\delta }^{\prime }$ denotes differentiation with respect to the scalar argument $s$.

Proof. The nonzero wavefront covectors of $W$ have one time orientation. There is no opposing pair within $W\cdot W$, so this square is a defined distributional product. The same holds for $W(-z{)}^2$.

On the nonzero future null cone, $t>0$ and $\mathrm{d}s\ne 0$. The one-dimensional boundary-value identity

therefore pulls back to $s$. Using (eq:W-boundary) gives

Substitution proves (eq:cone-kernel).

Away from the origin, the commutator is supported on the nonzero null cones, where the time cut is locally smooth. At $t=0$ and $\mathbf{x}\ne 0$ it is locally zero. Hence the time cut creates no additional punctured product problem.

Under positive dilation, $s$ has degree two and ${\delta }^{\prime }(s)$ has degree minus four. The punctured kernel is a nonzero homogeneous distribution of degree minus four, giving scaling degree four.