The retarded pole is an existing line, not a new product

For $\square ={{\partial }_t}^2-{\Delta }_{\mathbf{x}}$, the fixed retarded kernel

has spatial Fourier transform $\Theta (t)\sin (\rho t)/\rho$, $\rho =|\xi |$. With transform $e^{-i(t\omega +\mathbf{x}\cdot \xi )}$, positive damping gives

Thus, on $\rho >0$,

Near sheet $s\in \{1,-1\}$,

The pole boundary value is fixed by the original retarded kernel. This computation agrees with the supplied XIV-I internal normalization CheungXIVI (:pole). It computes one existing distribution, not its product with unprovided quantum cross-lines.

If an internal reconstruction convention is

then

Indeed $q{\delta }^{\prime }(q)=-\delta (q)$. The two summands are separately nonzero on a nontrivial chart. Therefore polynomial degree of a presentation is not automatically an intrinsic symbol filtration on the reconstructed object CheungXIVJ (:null).

The identity does not invalidate internal estimates. It does require any later coefficient-level construction to respect reconstruction descent.