Landmark XI-F Admission of the cubic Einstein response, and the plateau sum of the second-order variations

Landmark XI-E named, and excluded from its release, the classical cubic Einstein response $u_3=\mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}}(2{\mathcal{H}}_2[\gamma ])$ and its scalar-completed insertion ${\mathcal{C}}_3={\mathcal{A}}_1[u_3;\varphi ]$. We decide two gates for that object at fixed regulator width, in the normalization ${G_{\mathrm{gr}}}^{\mathrm{R}}=-\mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}}$.

Admission. The second metric coefficient carries the frozen zero Cauchy data on $\Sigma$, and the Einstein–Hilbert density carries neither the switch $g$ nor the ramp ${\sigma }_{\epsilon }$. The slab kernel of ${\mathcal{C}}_3$ therefore has the form $\theta (y^0)\beta (y)j_3(y;\cdot ,\cdot )$: a sharp onset at the initial surface in a metric slot. For every frozen switch and every mass we exhibit admissible observation data for which this kernel has a timelike conormal covector $\pm \mathrm{d}t$ in its wavefront set, so it leaves the polynomial microcausal class of XI-E. The admission map of XI-E does not apply to ${\mathcal{C}}_3$. The failure is not the finite-rank mismatch and not a loop; it is the sharp-Cauchy singularity of VII-B, now in the gravitational slot, which the ramp regulates only in the coordinate operator. Whether the free-shell restriction of ${\mathcal{C}}_3$ has some other microcausal representative is not decided.

Plateau sum. From XI-B's gauge fermion, the order-$\kappa$ variation of the de Donder term under the single-field law is $2D^{\ast }Ds$, with divergence $-\square Ds$; its retarded response through the frozen inverse is the pure gauge $-2\mathcal{R}{{\mathsf{E}}_0}^{\mathrm{R}}Ds=K{\zeta }_{\mathrm{gf}}$, with ${\zeta }_{\mathrm{gf}}=-{{\mathsf{E}}_0}^{\mathrm{R}}Ds$ the de Donder second-order parameter. Hence $u_3[\gamma +K\xi ]-u_3[\gamma ]=s_{\xi }+K{\zeta }_{\mathrm{gf}}[s_{\xi }]+\mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}}{{\mathcal{L}}^{\mathrm{dens}}}_{\xi }(A\gamma )$, $s_{\xi }={\mathcal{L}}_{\xi }\gamma +\frac{1}{2}{\mathcal{L}}_{\xi }K\xi$: the complete second-order metric law of X-A, with $\zeta$ selected by gauge fixing. The cubic-completed second coefficient then shifts, for a free scalar and $A\gamma =0$ on $U$, by ${\mathcal{A}}_1[K{\zeta }_{\mathrm{gf}};{\varphi }_0]$: a pure switching term carrying $1-g$ or $\partial g$, with no ramp term, vanishing whenever $J^{-}(\mathrm{supp}f)\cap J^{+}(\mathrm{supp}\xi )$ lies in the plateau. On the XI-E witness the ledger reads $-I$ (held fixed) $+I$ (covariant cubic) $+0$ (gauge-fixing response) $+0+0$ (quadratic terms) $=0$. The production sheet's literal sum ${\delta }_{\mathrm{split}}+{\delta }_0{\mathcal{C}}_3+{\delta }_{\mathrm{gf}}=0+I+0$ double counts: XI-E's by-hand co-transformation and the covariant part of the cubic response are one mechanism. The corrected plateau sum vanishes; ${\delta }_{\mathrm{gf}}$ is not $-I$, it is $0$.

Neither result constructs an interacting differential, a quantum retarded product of the cubic source, a loop value, a sharp limit, or an Einstein-jet repair.

Figure fig:split-XI-F records one view of the same separation. Figure fig:chain-XI-F records one view of the same separation. Figure fig:pair-XI-F records one view of the same separation. Figure fig:gate-XI-F records one view of the same separation. The drawings add no hypothesis.

Figure fig:scope-XI-F records that separation. The drawing adds no hypothesis.

: two gates, three endings, worst ending governs{

Gate 1 (F1, admission): Proved obstructed.

The frozen representative of ${\mathcal{C}}_3$ has a timelike conormal covector at $\Sigma$ in a metric slot and is not in ${\mathcal{F}}_{\mathrm{mc},\mathrm{pol}}$. XI-E's admission map does not apply. The on-shell class question is open.

Gate 2 (F2–F3, plateau sum): Constructed.

The de Donder companion is computed from XI-B's fermion with the coefficient $-1$; the finite transformation of $u_3$ is the X-A second-order law with $\zeta ={\zeta }_{\mathrm{gf}}$; the plateau sum is zero when $A\gamma =0$ on $U$, with the off-plateau remainder a switching term supported in $(\mathrm{supp}\mathrm{d}g\cup \mathrm{supp}((1-g)f))\cap J^{+}(\mathrm{supp}\xi )$.

Overall ending: Proved obstructed (Gate 1). No quantum version of the plateau sum exists in ${\mathcal{E}}_H$, because ${\mathcal{C}}_3$ is not admitted. No ${\hat{s}}_{\mathrm{int}}$, loop, $\epsilon \to 0$ limit, or repair of ${\mathcal{J}}_g({\varphi }_m)$.}

XI-E closed three routes and left one object outside all of them: the cubic Einstein response $u_3$ and its insertion ${\mathcal{C}}_3$ XIE. XI-E computed the classical variation of ${\mathcal{C}}_3$ on a witness and found a second nonzero number $I$, distinct from the $I$ that its split cancellation had already removed. It did not admit ${\mathcal{C}}_3$ to its Hadamard algebra, and it did not compute the gauge-fixing companion of the cubic variation. Those are the two gates here.

1. Admit ${\mathcal{C}}_3$ into the polynomial on-shell Hadamard algebra ${\mathcal{E}}_H$ of XI-E, or exhibit a wavefront direction that leaves its polynomial microcausal class.

2. Compute the order-$\kappa$ variation of the de Donder term under the single-field transformation law, from XI-B's gauge fermion, and its retarded response through ${G_{\mathrm{gr}}}^{\mathrm{R}}=-\mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}}$.

3. Assemble the second-order variations on the XI-E witness and decide whether their sum vanishes on the plateau, with the off-plateau support stated.

4. Say what the result is not.

The three endings of the series apply to each gate: Constructed, Proved obstructed, Unresolved. The worst ending governs the release.

The sheet anticipated an admission proof by composition of retarded kernels, a gauge-fixing companion equal to $-I$, and a three-term sum ${\delta }_{\mathrm{split}}+{\delta }_0{\mathcal{C}}_3+{\delta }_{\mathrm{gf}}$. Each of these must be corrected before the gates can be decided.

1. The source $J_3=2{\mathcal{H}}_2[\gamma ]$ is local, unswitched and unramped, and the second coefficient has zero data on $\Sigma$. Its retarded response is ${{\mathsf{E}}_0}^{\mathrm{R}}{J}_3$ with the zero-data inverse, which truncates the source at $t=0$. The resulting kernel has a sharp onset at $\Sigma$ in a metric slot. That onset, not the finite-rank mismatch, is the obstruction (Section sec:F1).

2. The gauge-fixing companion is the retarded response to the order-$\kappa$ de Donder variation $2D^{\ast }D{s}_{\xi }$. It equals $K{\zeta }_{\mathrm{gf}}[s_{\xi }]$ with ${\zeta }_{\mathrm{gf}}=-{{\mathsf{E}}_0}^{\mathrm{R}}D{s}_{\xi }$: pure gauge, zero data, supported in $J^{+}(\mathrm{supp}\xi )$. Its ${\mathcal{A}}_1$-pairing on the plateau is zero, not $-I$ (Section sec:F2).

3. XI-E's cancellation theorem restored $\gamma \mapsto \gamma +\kappa {\mathcal{L}}_{\xi }\gamma$ by hand in a parametrization with an external second coefficient $k$. Once $k=u_3[\gamma ]$ is generated by the response, the covariant part of $u_3[\gamma +K\xi ]-u_3[\gamma ]$ is that shift. Adding ${\delta }_{\mathrm{split}}=0$, which already contains the by-hand $+I$, to ${\delta }_0{\mathcal{C}}_3=I$ counts the same term twice (Section sec:F3).

4. The quadratic flow term $\frac{1}{2}{\mathcal{L}}_{\xi }K\xi$ of the single-field law is supplied by the response through the flatness identity $J_3[K\xi ]=-A(\frac{1}{2}{\mathcal{L}}_{\xi }K\xi )$. It is not added separately.

5. The second-order parameter selected by gauge fixing is not in ${\mathfrak{g}}_U$. Its support is $J^{+}(\mathrm{supp}\xi )\cap M_T$. The resulting remainder is a switching term of X-A's and VI's type, with no ramp layer, and it vanishes under an explicit causal plateau condition.

6. The identity is stated with the Einstein-shell term $\mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}}{{\mathcal{L}}^{\mathrm{dens}}}_{\xi }(A\gamma )$ retained. On the gauge-fixed free shell $A\gamma =-2D^{\ast }b$; the term vanishes when $D^{\ast }b=0$ on $U$, which holds for the witness and for every linearized Einstein solution. XI-A's distinction between the gauge-fixed and the Einstein shell is not erased.

X-E's normalization ${\Delta }_{\mathrm{R}}=-4\mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}}$ with dressing $-2$, and X-F's three-tree cancellation in that normalization, are not premises. Their mechanism, that a held-fixed split can be a truncation, is the same mechanism that appears in Section sec:F3; no factor, flat-gauge residual, or sum-zero lemma is copied. VII-C's word, reflected residue, and hypotheses (E1)–(E4) stay out. The empty fixed-input Einstein jet ${\mathcal{J}}_g({\varphi }_m)=\varnothing$ remains XI-A's result XIA.

Indices are moved with $\eta$; parentheses include the factor $\frac{1}{2}$. Classical fields are smooth on a neighbourhood of the closed slab; no spatial decay is imposed. The ramp ${\sigma }_{\epsilon }(t)=\chi (t/\epsilon )$, $0<\epsilon <T$, with the explicit $\chi$ of X-A, satisfies ${\sigma }_{\epsilon }=0$ on $\{t\le 0\}$, ${\sigma }_{\epsilon }=1$ on $\{t\ge \epsilon \}$, ${\sigma }_{\epsilon }>0$ on $\{t>0\}$, and $\chi (1-s)=1-\chi (s)$. The switch $g\in {C_c}^{\infty }({M_T}^{\circ };[0,1])$ has $P_g=\mathrm{int}\{g=1\}\ne \varnothing$. The plateau $U$ is open, relatively compact, with $\overline{U}\subset P_g\cap \{\epsilon <t<T\}$, and ${\mathfrak{g}}_U={C_c}^{\infty }(U;TM)$. The observation test is a real $f\in {C_c}^{\infty }({M_T}^{\circ })$ with

a compact set that is closed under $J^{-}$ inside $M_T$.

For a symmetric tensor $r$,

With $A=2G_{\mathrm{lin}}$,

Here ${{\mathsf{E}}_0}^{\mathrm{R}}$ is the zero-initial-data inverse of $\square$ on the slab, ${{\mathsf{E}}_0}^{\mathrm{A}}$ the zero-final-data inverse, and ${{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}},{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}$ the Klein–Gordon analogues. For $a\in {C_c}^{\infty }({M_T}^{\circ })$ and $s$ smooth on $M_T$, $\int a{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}s=\int ({{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}a)s$ (X-A, transposition lemma); the proof only needs ${{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}a$ to vanish near $t=T$, so it applies verbatim to $a$ smooth on $M_T$ with $\mathrm{supp}a\subset {\mathcal{D}}_f$. For $s$ smooth with zero Cauchy data and $\mathrm{supp}s\subset \{t>0\}$, ${{\mathsf{E}}_0}^{\mathrm{R}}\square s=s$ and ${{\mathsf{E}}_0}^{\mathrm{R}}{\partial }_{\mu }s={\partial }_{\mu }{{\mathsf{E}}_0}^{\mathrm{R}}s$.

For $s$ smooth, compactly supported in $U$,

Proof. ${G_{\mathrm{gr}}}^{\mathrm{R}}As=(-\mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}})(-\square \mathcal{P}s)+(-\mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}})(-2D^{\ast }Ds)=\mathcal{PP}s+2\mathcal{P}{D}^{\ast }{{\mathsf{E}}_0}^{\mathrm{R}}Ds=s-2\mathcal{R}{{\mathsf{E}}_0}^{\mathrm{R}}Ds$, using (eq:algebra), ${{\mathsf{E}}_0}^{\mathrm{R}}\square s=s$, and ${{\mathsf{E}}_0}^{\mathrm{R}}{D}^{\ast }=D^{\ast }{{\mathsf{E}}_0}^{\mathrm{R}}$ on $Ds$, which has zero data.

${{\mathcal{A}}_2}^{\mathrm{sc}}[H,k;\varphi ]$ is X-A's complete on-shell second coefficient, written out in XI-D: with ${\rho }_1=-{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}{\mathsf{V}}_{1,g}[H]\varphi$, ${\rho }_2={{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}{\mathsf{V}}_{1,g}[H]{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}{\mathsf{V}}_{1,g}[H]\varphi -{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}{\mathsf{V}}_{2,g}[H,k]\varphi$,

Two copied facts about it:

for every free ${\varphi }_0$, every smooth $(h,k)$, and

The free input ${\varphi }_0$ is not transformed in (eq:XAv); it is held fixed, exactly as XI-E held the scalar fixed in its cubic variation.

The independent multiplet is $(\gamma ,\varphi ,c,\bar{c},b)$ with the free rule $s_0\gamma =\mathcal{R}c$, $s_0\varphi =s_{0}c=s_{0}b=0$, $s_0\bar{c}=ib$. The gauge fermion of XI-B is

and elimination of $b=D\gamma$ gives $S_{\mathrm{gf}}=\int D_{\nu }\gamma D^{\nu }\gamma$ with Euler derivative $E_{\mathrm{gf}}[\gamma ]=2D^{\ast }D\gamma$ XIB,XIA. The free shell ${\mathcal{C}}_{\mathrm{sol}}$ is $-\square \mathcal{P}\gamma =0$, $b=D\gamma$, $P_m\varphi =0$, $\square c=\square \bar{c}=0$.

XI-E's Route 1 constructed the polynomial on-shell Hadamard algebra ${\mathcal{E}}_H=({{\mathcal{F}}_{\mathrm{mc},\mathrm{pol}}}^{\mathrm{sol}}[[\hslash ,\kappa ]],{\star }_H)$ with block data $H_{\gamma \gamma }=-\mathcal{P}{h}_0$, $H_{\gamma b}=D_y{H}_{\gamma \gamma }$, $H_{\varphi \varphi }=-h_m$, $H_{c\bar{c}}=-i{h}_0$, graded antisymmetric part equal to XI-B's causal matrix, and no state selected. Its domain is:

${\mathcal{F}}_{\mathrm{mc},\mathrm{pol}}$ consists of finite graded polynomials $F(\Phi )={\sum }_n⟨u_n,{\Phi }^{\otimes n}⟩$ on ambient configurations on $M={\mathbb{R}}^4$, with compact, graded symmetric distributional kernels $u_n$ on $M^n$ satisfying

where ${\tilde{V}}_{+},{\tilde{V}}_{-}$ are the closed future and past causal covector cones with zero included, and the total zero covector is never a wavefront point. Field indices are finite and do not change the criterion. Admission is kernel inclusion in ${\mathcal{F}}_{\mathrm{mc},\mathrm{pol}}$ followed by restriction to ${\mathcal{C}}_{\mathrm{sol}}$: $Q_H(F)=[F{|}_{{\mathcal{C}}_{\mathrm{sol}}}]\in {\mathcal{E}}_H$. Slab functionals are viewed on $M$ through their slab restriction; for the inherited symbols the ramp is flat at $t=0$, so no boundary conormal appears.

The inherited symbols ${\mathcal{B}}_f=-\int f{\mathsf{X}}_1[\gamma ]\cdot \partial \varphi$ and $\mathcal{T}(f,g)[\varphi ]=\int f({\mathsf{X}}_1[r_g[\varphi ]])\cdot \partial \varphi$, $r_g=-\frac{1}{2}\mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}}(g{T}_0[\varphi ])$, are admitted in ${\mathcal{E}}_H$; the physical one-graviton response of ${\mathcal{B}}_f$ is $-\mathcal{T}$ at coupling degree ${\kappa }^2$ XIC. Their admission is not reopened here.

XI-A's unswitched action $S_{\mathrm{EH}}=-\frac{2}{{\kappa }^2}\int \sqrt{-q}R[q]$ with $q=\eta +\kappa \gamma$ gives $\sqrt{-q}{G}^{\mu \nu }(q)=\kappa {\mathcal{H}}_1+{\kappa }^2{\mathcal{H}}_2+{\kappa }^3{\mathcal{H}}_3+O({\kappa }^4)$, ${\mathcal{H}}_1=G_{\mathrm{lin}}(\gamma )=\frac{1}{2}A\gamma$, with the finite formulas of XI-A's appendix. The cubic vertex has metric Euler derivative $J_3[\gamma ]=2{\mathcal{H}}_2[\gamma ]$, a quadratic local differential polynomial. XI-E defined

of field degree $(2,1)$, and proved, for ${\delta }_0\gamma =K\xi$, $\xi \in {\mathfrak{g}}_U$,

where $({{\mathcal{L}}^{\mathrm{dens}}}_{\xi }v{)}^{\mu \nu }={\xi }^{\rho }{\partial }_{\rho }{v}^{\mu \nu }-v^{\rho \nu }{\partial }_{\rho }{\xi }^{\mu }-v^{\mu \rho }{\partial }_{\rho }{\xi }^{\nu }+({\partial }_{\rho }{\xi }^{\rho })v^{\mu \nu }$.

The witness (XI-C, XI-D, XI-E): $m=0$,

in a causal cylinder on which $g=1$, with a compact spatial cutoff on $\xi$ whose derivatives lie outside the observation dependence region, the plateau containing $\mathrm{supp}\xi$, and $f\ge 0$, $f\not\equiv 0$ observed after $\mathrm{supp}a$. Then $\square \varphi =0$, $-\square \mathcal{P}\gamma =0$, $b=D\gamma =(-2,0,0,0)$ is constant, $A\gamma =0$, and

XI-E also proved the split cancellation $-{\mathcal{A}}_1[{\mathcal{L}}_{\xi }\gamma ]+{\mathcal{A}}_1[{\mathcal{L}}_{\xi }\gamma ]=0$ in the $(h,\gamma ,k)$ parametrization with an external $k$, and recorded that the two $I$'s are different $I$'s. The two affine numbers, $S_{\epsilon }(t)$ for the bare insertion and $\epsilon /2$ for the complete word, are not touched here.

The coefficient of $\mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}}$ is $-1$ throughout. Neither $-2\mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}}$ nor $-4\mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}}$ is used, and no graviton two-point function is selected.

Fix $f$, $g$, and a smooth $\varphi$ on the slab. There is a smooth symmetric contravariant tensor $\alpha =\alpha [f,g,\varphi ]$ on $M_T$ with $\mathrm{supp}\alpha \subset {\mathcal{D}}_f$ such that

Explicitly, with $b={{\mathsf{E}}_0}^{\mathrm{A}}(f\mathrm{d}\varphi )$, $b^{\prime }={{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}f$, and $(C^{\ast }b{)}^{\rho \sigma }=-{\partial }^{(\rho }{b}^{\sigma )}+\frac{1}{2}{\eta }^{\rho \sigma }{\partial }_{\mu }{b}^{\mu }$,

Proof. The coordinate term $-\int f{\mathsf{X}}_1[r]\cdot \partial \varphi$ is $-\text{Jc}(f\mathrm{d}\varphi )[r]$ in X-A's notation, whose kernel is ${\sigma }_{\epsilon }{C}^{\ast }({{\mathsf{E}}_0}^{\mathrm{A}}(f\mathrm{d}\varphi ))$ by X-A's kernel lemma; this gives the first term. For the scalar term, transposition gives $\int f{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}{\mathsf{V}}_{1,g}[r]\varphi =\int b^{\prime }{\mathsf{V}}_{1,g}[r]\varphi$, and integrating by parts in the divergence (legitimate because $g$ has compact interior support),

Inserting ${a_1}^{\mu \nu }[r]=\frac{1}{2}{\eta }^{\mu \nu }\mathrm{tr}r-r^{\mu \nu }$, $b_1[r]=\frac{1}{2}\mathrm{tr}r$, and symmetrizing gives the second term of (eq:alphaexplicit). Supports: $b,b^{\prime }$ are supported in ${\mathcal{D}}_f$, and ${\mathcal{D}}_f$ is past-closed in $M_T$.

Let $\beta :={{\mathsf{E}}_0}^{\mathrm{A}}(\mathcal{P}\alpha )$, a smooth symmetric tensor on $M_T$ with $\mathrm{supp}\beta \subset {\mathcal{D}}_f$. Then

Viewed as a functional of ambient configurations through slab restriction, the degree-$(2,1)$ kernel $u$ of ${\mathcal{C}}_3$ is supported in $\{y_1=y_2=y\in {\mathcal{D}}_f\cap \{y^0\ge 0\}\}\times {\mathcal{D}}_f$ and has the form

where $j_3$ is the constant-coefficient bilinear local kernel of $J_3$ (a finite sum of products of derivatives of $\delta (y-y_1)$ and $\delta (y-y_2)$) and $\hat{\beta }$ is the $\varphi$-slot kernel of $\beta$. The factor $\theta (y^0)$ is forced: the frozen Cauchy prescription gives the second coefficient zero data on $\Sigma$, so ${{\mathsf{E}}_0}^{\mathrm{R}}$ truncates every source at $t=0$, and the Einstein–Hilbert density carries no $g$ and no ${\sigma }_{\epsilon }$ that could make $J_3[\gamma ]$ vanish near $\Sigma$.

Proof. ${\mathcal{C}}_3=\int \alpha \cdot u_3=\int \alpha \cdot \mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}}{J}_3$. Trace reversal is self-adjoint under the $\eta$-contraction, $⟨\alpha ,\mathcal{P}r⟩=⟨\alpha ,r⟩-\frac{1}{2}\mathrm{tr}\alpha \mathrm{tr}r=⟨\mathcal{P}\alpha ,r⟩$, and transposition (applicable because $\mathrm{supp}\mathcal{P}\alpha \subset {\mathcal{D}}_f$ vanishes near $t=T$ and ${{\mathsf{E}}_0}^{\mathrm{R}}{J}_3$ has zero data) gives (eq:C3beta). The $y$-integral runs over the slab only, and $J_3[\gamma ](y)$ depends on the jet of $\gamma$ at $y$; this is (eq:ukernel). Compactness: $y\in {\mathcal{D}}_f$, and the $\varphi$-slot of $\alpha$ is supported in ${\mathcal{D}}_f$ by Lemma lem:alpha.

The onset $\theta (y^0)$ is the sharp step $\theta (t)$ that X-A replaced by ${\sigma }_{\epsilon }$ in the coordinate operator (X-A, kernel lemma), and that VII-B found to carry a timelike conormal covector. Here it reappears in a metric slot, where the ramp does not act. Whether it is harmless depends on whether the coefficient $\beta$ vanishes on $\Sigma$. It does not.

Let $u$ be a compactly supported tensor-valued distribution on $M^3$ and $v,\psi$ smooth on $M$. Put $\ell :=⟨u,\cdot \otimes v\otimes \psi ⟩$, a compactly supported distribution on $M$. Then

In particular, if $u$ satisfies (eq:mc) with $n=3$, then $\mathrm{WF}(\ell )$ contains no nonzero covector in ${\tilde{V}}_{+}\cup {\tilde{V}}_{-}$.

Proof. Smooth multiplication by $v\otimes \psi$ does not enlarge the wavefront set, and integration over $(y_2,y_3)$ of a compactly supported distribution is the pushforward estimate of Hormander (Theorem 8.2.12). If $(y_1,p;y_2,0;y_3,0)\in \mathrm{WF}(u)$ with $0\ne p\in {\tilde{V}}_{+}$, then the triple lies in ${\tilde{V}}_{+}^3$, contradicting (eq:mc); similarly for ${\tilde{V}}_{-}$.

Let ${\gamma }_{0\mu }=0$ and ${\gamma }_{ij}={\gamma }_{ij}(t)$ on an open set $W\subset M_T$. Then on $W$,

with spatial traces and contractions taken with ${\delta }_{ij}$, and ${{\mathcal{H}}_1}^{00}[\gamma ]=0$.

Proof. On $W$ the metric is $q=\mathrm{d}{t}^2-h_{ij}(t)\mathrm{d}{x}^i\mathrm{d}{x}^j$ with $h=\delta -\kappa \gamma (t)$. The nonzero Christoffel symbols of XI-A's convention are ${{\Gamma }^0}_{ij}=\frac{1}{2}{\dot{h}}_{ij}$ and ${{\Gamma }^i}_{0j}=K^i{}_j:=\frac{1}{2}{h}^{ik}{\dot{h}}_{kj}$. Direct substitution into $R_{\mu \nu }={\partial }_{\rho }{{\Gamma }^{\rho }}_{\mu \nu }-{\partial }_{\nu }{{\Gamma }^{\rho }}_{\mu \rho }+{{\Gamma }^{\rho }}_{\rho \lambda }{{\Gamma }^{\lambda }}_{\mu \nu }-{{\Gamma }^{\rho }}_{\nu \lambda }{{\Gamma }^{\lambda }}_{\mu \rho }$ gives $R_{0i}=0$, $R_{00}=-\dot{K}-K^i{}_j{K}^j{}_i$ with $K=K^i{}_i$, and $h^{ij}{R}_{ij}=\dot{K}+K^2$, so that $R=R_{00}-h^{ij}{R}_{ij}$ and

Since $q^{00}=1$ and $q^{0i}=0$, $G^{00}=G_{00}$ and $G^{0i}=0$ for every $\kappa$; hence ${J_3}^{0i}=0$. Expanding $h^{ik}={\delta }^{ik}+O(\kappa )$ and $\dot{h}=-\kappa \dot{\gamma }$ gives $K^i{}_j=-\frac{1}{2}\kappa {\dot{\gamma }}_{ij}+O({\kappa }^2)$, $K=-\frac{1}{2}\kappa \mathrm{tr}\dot{\gamma }+O({\kappa }^2)$, and $\sqrt{-q}=1+O(\kappa )$, so $\sqrt{-q}{G}^{00}=\frac{1}{8}{\kappa }^2[(\mathrm{tr}\dot{\gamma }{)}^2-{\dot{\gamma }}_{ij}{\dot{\gamma }}_{ij}]+O({\kappa }^3)$. Thus ${{\mathcal{H}}_1}^{00}=0$ and ${J_3}^{00}=2{{\mathcal{H}}_2}^{00}=q(\dot{\gamma },\dot{\gamma })$. The signature does not affect $G_{\mu \nu }$; the Friedmann check $G_{00}=3H^2$ for $h=a(t{)}^2\delta$ confirms the sign.

Fix $m\ge 0$, the frozen $g$, and the frozen ${\sigma }_{\epsilon }$. Let $t_g:=\min \{t:(t,\mathbf{x})\in \mathrm{supp}g\}>0$. Choose $f\in {C_c}^{\infty }({M_T}^{\circ })$, $f\ge 0$, $f\not\equiv 0$, with $\mathrm{supp}f\subset (0,t_g)\times {\mathbb{R}}^3$ and $\cos (mt)>0$ on $\mathrm{supp}f$. Let ${\varphi }_m$ be XI-A's input, ${\varphi }_m=\sin (mt)/m$ for $m>0$ and ${\varphi }_m=t$ for $m=0$. Then the cubic kernel $u$ of ${\mathcal{C}}_3(f,g)$ in (eq:ukernel) satisfies

More precisely, with $\ell :=⟨u,\cdot \otimes v_2\otimes {\varphi }_m⟩$ for the explicit $v_2$ below, there are points $y$ in an arbitrarily thin collar around $\Sigma$ with $(y,\pm \mathrm{d}t)\in \mathrm{WF}(\ell )$. Consequently the ambient representative of ${\mathcal{C}}_3(f,g)$ is not in ${\mathcal{F}}_{\mathrm{mc},\mathrm{pol}}$, and XI-E's admission map $Q_H$ does not apply to it.

Proof. {Step 1: the kernel for this $f$.} Since ${\mathcal{D}}_f\subset \{t<t_g\}$ and $\mathrm{supp}g\subset \{t\ge t_g\}$, we have ${\mathcal{D}}_f\cap \mathrm{supp}g=\varnothing$, so the $g$-term of (eq:alphaexplicit) vanishes and $\alpha =-{\sigma }_{\epsilon }{C}^{\ast }b$ with $b={{\mathsf{E}}_0}^{\mathrm{A}}(f\mathrm{d}{\varphi }_m)$. Here ${\partial }_{\mu }{\varphi }_m=\cos (mt){{\delta }^0}_{\mu }$, so $b^0={{\mathsf{E}}_0}^{\mathrm{A}}(f\cos (mt))$, $b^i=0$, and

Hence $\mathrm{tr}\alpha =-{\sigma }_{\epsilon }{\partial }_t{b}^0$ and

Therefore $\beta ={{\mathsf{E}}_0}^{\mathrm{A}}\mathcal{P}\alpha$ has ${\beta }^{ij}=0$. Put

Because ${{\mathsf{E}}_0}^{\mathrm{A}}$ acts on sources of compact spatial support per slice and ${\int }_{{\mathbb{R}}^3}\Delta (\cdot ){\mathrm{d}}^{3}x=0$, spatial integration commutes with ${{\mathsf{E}}_0}^{\mathrm{A}}$ and reduces it to its zero-frequency form: $\int {{\mathsf{E}}_0}^{\mathrm{A}}F{\mathrm{d}}^{3}x={{\int }_s}^T(t^{\prime }-s)\bar{F}(t^{\prime })\mathrm{d}{t}^{\prime }$ with $\bar{F}=\int F{\mathrm{d}}^{3}x$. Thus

Then ${\partial }_t\bar{b}(t)=-{{\int }_t}^T\bar{F}\le 0$, strictly negative for $t<t_{-}:=\inf \{t:(t,\mathbf{x})\in \mathrm{supp}f\}$, while ${\sigma }_{\epsilon }(t^{\prime })>0$ for $t^{\prime }>0$. Hence