Theorem: Certified identification for Gate S-bv

Let $W_0$, $h_0=-W_0$ and ${E_0}^{\mathrm{A}}$ be as in XVI-K, eq:S / lem:families (quoting XI-E, eq:W, eq:h and XVI-J, eq:sineA, eq:step), massless case. Set $S:=h_0+i{E}_0^{\mathrm{A}}$ and $Q(z)=\eta (z,z)$. Then, in ${\mathcal{D}}^{\prime }(\Omega )$,

where $(Q-i0{)}^{-1}$ is the boundary-value distribution $T_1$ of Lemma lem:bv, and $\delta (Q)$ denotes the pullback surface distribution ${\Phi }^{\ast }(\delta (s)\otimes 1)$ in any local chart with first coordinate $Q$. In particular Input (S-bv) holds with

and with the same boundary side $Q-i0$ as $T_1$. Since $S=-d$ (XVI-J, eq:DF), this is XVI-J, eq:rational; the theorem is a written check of it, not a new extension.

Proof. (1) Position-space formulae from the Fourier definitions. Wightman factor. From XVI-K, eq:W-fam / XI-E, eq:W, with $z=x-y$ and ${\omega }_0=|\mathbf{p}|$,

Angular integration yields the radial formula

Insert the positive-frequency damping $z^0\mapsto z^0-i\tau$ ($\tau \downarrow 0$), i.e.\ $\alpha =\tau +i{z}^0$ with $\mathrm{Re}\alpha >0$, and use ${{\int }_0}^{\infty }{e}^{-\alpha p}\sin (pr)\mathrm{d}p=r/({\alpha }^2+r^2)$. (The damping parameter is called $\tau$ throughout; it is unrelated to the preparation width $\epsilon$. The factor $e^{-\tau |\mathbf{p}|}$ only realises the oscillatory integral as a limit in ${\mathcal{D}}^{\prime }({\mathbb{R}}^4)$: for a test function $\phi$, $⟨{W_0}^{(\tau )},\phi ⟩=\int \mathrm{d}{z}^0\int \frac{e^{-(\tau +i{z}^0)|\mathbf{p}|}}{(2\pi {)}^{3}2|\mathbf{p}|}\tilde{\phi }(z^0,-\mathbf{p}){\mathrm{d}}^3\mathbf{p}$, and the integrand is dominated by $|\tilde{\phi }(z^0,-\mathbf{p})|/((2\pi {)}^{3}2|\mathbf{p}|)$, which is integrable; it changes no definition and introduces no $i\epsilon$ prescription.) One obtains the smooth function

Off the cone, ${g_{\tau }}^{-1}\to Q^{-1}$ in $C^{\infty }$; near each nonzero null point, where $z^0$ has a fixed sign, Lemma lem:wpA-varreg gives ${g_{\tau }}^{-1}\to (Q-i0\mathrm{sgn}(z^0){)}^{-1}$. Passing to $\tau \downarrow 0$ therefore gives the distributional identity on $\Omega$

(The Euclidean slice $z^0=-i{\tau }_{\mathrm{E}}$, ${\tau }_{\mathrm{E}}>0$, recovers $W_0=1/(4{\pi }^2({{\tau }_{\mathrm{E}}}^2+r^2))$, matching Remark rem:Sbv-motivation and the check in kernel_relations.py.)

Causal factor. From $W_0-{W_0}^{\mathsf{T}}=-i{\delta }_0$ (XI-E, after eq:W; XVI-J) and (eq:wpA-W), ${\delta }_0=\frac{1}{2\pi }\mathrm{sgn}(z^0)\delta (Q)$ on $\Omega$. XVI-J, eq:step, gives ${E_0}^{\mathrm{A}}=-\theta (-z^0){\delta }_0$, hence

(Equivalently, ${E_0}^{\mathrm{R}}=\theta (z^0){\delta }_0=\frac{1}{2\pi }\theta (z^0)\delta (Q)$ is the standard massless retarded kernel for $\square E=\delta$.)

{(2) Combination: cancellation of $\mathrm{sgn}(z^0)$.} Using $h_0=-W_0$,

On $\{z^0>0\}$ one has $\mathrm{sgn}=+1$, $\theta (-z^0)=0$; on $\{z^0<0\}$, $\mathrm{sgn}=-1$, $\theta (-z^0)=1$. The coefficient of $\delta (Q)$ becomes

on both sheets. (The identity $\theta (z^0)+\theta (-z^0)=1$ holds on $\Omega \cap \{Q=0\}$, because $Q=0$ and $z^0=0$ force $\mathbf{z}=0$, hence $z=0\notin \Omega$.) Therefore

where the last step is the pullback of $(s-i0{)}^{-1}=\mathrm{PV}(1/s)+i\pi \delta (s)$. The $\mathrm{sgn}(z^0)$-dependence of $W_0$ is exactly cancelled by $i{E}_0^{\mathrm{A}}$, which yields the single boundary value $Q-i0$ (not $Q+i0$).

(3) Normalization off the cone. On the open sets $\{Q>0\}$ and $\{Q<0\}$, $\delta (Q)=0$ and ${E_0}^{\mathrm{A}}=0$, so $S=c/Q$ with $c=1/(4{\pi }^2)$; e.g.\ $S=-c$ at $(0,1,0,0)$ and $S=c$ at $(1,0,0,0)$.

(4) Equality across the nonzero null cone. Let $z_{\ast }\in \Omega$ with $Q(z_{\ast })=0$. Then $\mathrm{d}Q(z_{\ast })=2\eta (z_{\ast },\cdot )\ne 0$, so $Q$ is a local coordinate: there is a chart $\Phi$ with ${\Phi }_1=Q$ (proof of Lemma lem:bv). In that chart $(Q-i0{)}^{-1}={\Phi }^{\ast }((s-i0{)}^{-1}\otimes 1)$, and step (2) shows that $-W_0+i{E}_0^{\mathrm{A}}$ equals $c$ times this same pullback on both time sheets. The surface distribution $\delta (Q)$ is well defined on the cone minus the origin by the same submersion, and the sheet-wise boundary values $(Q-i0\mathrm{sgn}{z}^0{)}^{-1}$ used in (eq:wpA-W) are those of Lemma lem:wpA-varreg. Thus (eq:wpA-id) holds near every nonzero null point, and therefore on all of $\Omega$.

(5) Cross-checks. (a) $\square W_0=0$ and $\square {E_0}^{\mathrm{A}}=\delta$ (XI-B, lem:Green), so $\square S=i\delta$ on ${\mathbb{R}}^4$ and $\square S=0$ on $\Omega$, matching XVI-K, lem:grading(3); off the cone $\square (c/Q)=0$ as a rational identity in four dimensions. (b) Both sides are homogeneous of degree $-2$ (XVI-K, lem:homog). (c) $W_0-{W_0}^{\mathsf{T}}=-i{\delta }_0$ and ${E_0}^{\mathrm{R}}-{E_0}^{\mathrm{A}}={\delta }_0$ imply $S^{\mathsf{T}}=S$, consistently with $D_F=-\mathcal{P}S$ (XVI-J, lem:signs). (d) $K{D}_F=i\mathrm{id}$ with $K=-\square \mathcal{P}$ and $D_F=-\mathcal{P}S$ is equivalent to $\square S=i\delta$.