Landmark X-B Fixed-regulator admission and one retarded correction for the ramp relational scalar, with the metric kept classical

At fixed $\epsilon >0$, with $\varphi$ the free quantum Klein–Gordon field and $h,k$ classical backgrounds, the bare ramp dressing is admitted through ${\kappa }^2$. We distinguish joint microcausal admission of its polynomial kernels from the scalar-only linear test domain actually used for the quantum construction.

The single map $R_1(V_{1,g};\cdot )$ is constructed on specified off-shell insertion tests. Its output is a free field with a smooth compact test, it has the stated retarded support, and it is tree-exact: all positive-$\hslash$ corrections vanish. Its action on the first dressing coefficient equals the mixed term of the classical ramp expansion.

The completed coefficient ${\hat{A}}_1$ inherits the ramp background-gauge identity on the algebra frozen in Landmark X-A. The bare ${\kappa }^2$ dressing is admitted and its classical transformation is retained, but no full quantum identity at that order is claimed: ${\rho }_{2,g}$ is displayed and is not supplied by a constructed map. Edge and switch defects remain outside the claimed algebra. The dressing survives at zero switch. No sharp limit, graviton tree contribution, loop triangle, or two-point response is constructed.

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

Figure fig:scope-X-B records that separation. The drawing adds no hypothesis.

: Constructed, with a fixed order boundary{ At fixed $\epsilon >0$, with $\varphi$ the free quantum field and $h,k$ classical, the bare dressing is admitted through ${\kappa }^2$. The single map $R_1(V_{1,g};\cdot )$ is constructed on the frozen off-shell tests, is tree-exact, and has the stated retarded support. The completed coefficient ${\hat{A}}_1$ inherits the ramp identity on ${\mathfrak{g}}_U$.

The ${\kappa }^2$ dressing is admitted and its classical transformation is retained, but no full quantum identity at that order is claimed: ${\rho }_{2,g}$ is displayed and not supplied by a constructed map. The edge and switch defects stay outside ${\mathfrak{g}}_U$. No sharp limit, graviton tree, or loop triangle is constructed.}

Three endings are available:

1. Constructed. The named admission condition, retarded map, support, and stated-order gauge identity are proved.

2. Proved obstructed. An inadmissible joint wavefront set or a nonzero defect blocks a claimed construction on the frozen algebra.

3. Unresolved. Field content, insertion domain, normalization, or an essential map is unspecified.

The ending reached is (1), precisely within the release box. A missing higher-order map is not a no-go. An incomplete second-order expression is not an invariant. A defect cannot be removed by shrinking the frozen algebra after the calculation, or by selecting a contact afterward.

Here $\kappa$ counts background perturbation order and $\hslash$ counts quantum contractions. Both are formal. Equality modulo ${\kappa }^3$ means coefficient equality, not a numerical remainder estimate. Admission through ${\kappa }^2$ and covariance through ${\kappa }^1$ are deliberately different statements.

A, frozen here. The scalar is the free quantum field of $P_m=\square +m^2$ on Minkowski space, $m\ge 0$. The metric coefficients are smooth, classical, and off shell. There is no quantum metric propagator and no gravitational action.

The slab and initial surface belong to the classical ramp prescription. Its coordinate sources vanish for $t\le 0$, so the zero-data coordinate inverse agrees with the retarded inverse on those sources. No quantum Cauchy data are prescribed. The quantum field is defined on all of Minkowski space.

B, named and not opened. Quantizing $h$ would require a specified quadratic graviton action, gauge fixing, and a different contraction domain. None is present in the classical source manuscript. An additional graviton response would not follow from the scalar-only normalization below. Its Ward formulation and possible loop graphs would require separate analysis.

The present theorem makes no assertion that every such successor graph exists or fails. In particular, a schematic triangle is not a proved obstruction merely because it contains a partial diagonal. Branch B is a different construction, not a correction to a missing term inside Branch A.

The classical source is the supplied manuscript {Landmark X-A: Classical covariance of one fixed-regulator switched relational scalar through order ${\kappa }^2$}, X-A.tex XA. The following are copied premises from that ramp manuscript. Its covariance proof is not repeated.

Let

Indices are moved with $\eta$. Fix $0<\epsilon <T$ and

Thus ${\sigma }_{\epsilon }$ is smooth and flat at the initial surface, and ${\sigma }_{\epsilon }^{\prime }>0$ on $0<t<\epsilon$.

Fix $g\in {C_c}^{\infty }((0,T)\times {\mathbb{R}}^3;[0,1])$, $P_g=\mathrm{int}\{g=1\}$, and the same nonempty relatively compact open $U$ as in X-A:

Fix a real label-space test $f\in {C_c}^{\infty }((0,T)\times {\mathbb{R}}^3)$. Put

The set ${\mathcal{D}}_f$ is compact.

The formal metric is $q=\eta +\kappa h+{\kappa }^{2}k$, with arbitrary smooth real symmetric $h,k$. Write $h={\mathrm{tr}}_{\eta }h$ inside the following coefficient formulas:

Set

The ramp auxiliary density is

It does not contain the interaction switch $g$.

The coordinate coefficients are

The retarded operator is understood on the smooth past-compact coordinate sources, or locally by finite propagation. These formulas reproduce the zero-data Cauchy prescription.

With

the linear coordinate kernel is

Equivalently, the derivative of the argument $x-z$ gives the same formula with a minus sign. The coefficient maps are smooth and continuous in smooth background topologies at fixed $\epsilon$. Smeared first-derivative kernels are smooth, with no initial hypersurface delta.

For a scalar jet $\phi ={\phi }_0+\kappa {\phi }_1+{\kappa }^2{\phi }_2$, put

The copied expansion is

Its smeared coefficients depend on backgrounds only in neighborhoods of ${\mathcal{D}}_f$, and continuously at fixed $\epsilon$. Setting ${\phi }_1={\phi }_2=0$ defines the bare dressing used here. At $g=0$ this dressing survives: the coordinate prescription is unchanged.

Write

For the formal flow of $\kappa \xi +{\kappa }^2\zeta$, $\xi ,\zeta \in {\mathfrak{g}}_U$,

In particular ${\mathsf{X}}_1[K\xi ]=\xi$ on ${\mathfrak{g}}_U$.

For the excluded edge parameter $\xi =\alpha (t)\beta (\mathbf{x}){\partial }_1$, where $\alpha \ge 0$, $\alpha \not\equiv 0$, $\mathrm{supp}\alpha \subset (0,\epsilon )$, and $\int \beta {\mathrm{d}}^{3}x=1$, X-A proves

The coordinate identity fails in general outside the frozen algebra. This witness is not a defect on ${\mathfrak{g}}_U$.

The first switched scalar vertex is

Its positive Euler coefficient, denoted ${\mathsf{V}}_{1,g}$ in X-A, is here

The second coefficient is

For a free smooth solution $u$, the retarded scalar coefficients are

X-A proves simultaneous-pullback covariance off shell, and background covariance of the on-shell relational observable at fixed free input, through second classical order on ${\mathcal{G}}_U$. This is classical input, not a quantum product theorem.

-inputs. The obstructed convolution coefficient and its X-Enter readout do not enter any proof. Neither supplies a cancellation for the ramp observable. VII-A, VI, VII-B, VII-C, IX-E-2, and IX-E-3 are comparison only. The cone condition associated with VII-B can be named as a test; its sharp exclusion is not a premise about (eq:Lambda). No earlier boundary topology or kernel identification is transferred.

{}1.12

Let $\mathcal{D}(M)={C_c}^{\infty }(M;\mathbb{C})$ with its standard LF topology. On a fixed compact $K$, use the Fréchet space ${\mathcal{D}}_K(M)$ with all smooth seminorms. Backgrounds belong to the ambient smooth space

with convergence of all derivatives on compact sets. References to dependence on a closed causal set mean dependence on restrictions to its neighborhoods; no unmentioned topology on a nonsmooth closed set is imposed.

For $c\in \mathcal{D}(M)$, set

This off-shell insertion domain is identified with $\mathcal{D}(M)$.

Integration by parts in the bare expansion fixes the tests:

Thus

After free-field evaluation, the same notation denotes $\hat{\varphi }(c_0+\kappa c_1+{\kappa }^2{c}_2)$. The distinction between representative and class will always be made when applying $R_1$.

Use the polynomial microcausal Wick presentation of the free scalar algebra. Choose a Hadamard bisolution ${\Delta }_{+}$ once, with

For the polynomial entries used here, the Wick product is

The sum is finite for each polynomial pair. This is the standard free presentation FR (Sections 4–5); no closure assertion for an arbitrary class of nonpolynomial smooth functionals is needed.

Let ${\mathcal{A}}_{\mathrm{off}}$ denote this off-shell polynomial algebra. Since ${\Delta }_{+}$ is a bisolution,

The pointwise ideal generated by all $F_{P_{m}u}$, $u\in \mathcal{D}(M)$, is therefore also a star ideal, denoted ${\mathfrak{J}}_0$. Put

Elements of ${\mathfrak{J}}_0$ vanish on smooth free solutions. In particular,

The linear output topology is the quotient test topology on $\mathcal{D}(M)/P_m\mathcal{D}(M)$.

The map constructed below satisfies

which need not vanish. Therefore it is not a map on free on-shell classes in its input. It is a map on the frozen representatives (eq:tests), with output in ${\mathfrak{A}}_0$.

This warning can be witnessed explicitly. For $m=0$, take $h=\psi \eta$, set $r=g\psi$, and choose a compact $r$ with ${\partial }_{t}r\not\equiv 0$. For $u={\partial }_{t}r$, evaluation of the output test against the free solution $t$ gives

Thus even a free-equation-zero insertion can have a nonzero retarded correction. There is no ambiguity here: the geometric prescription supplies the insertion test before the map is applied.

Include zero covectors in the closed future and past causal cones ${\overline{V}}_{+},{\overline{V}}_{-}$, and define

The polynomial microcausal test requires compact derivative kernels and

This is the standard cone condition FR (Section 4). We first apply it to the multiplet $(h,k,\phi )$. That joint mathematical test does not quantize the backgrounds.

A distribution on $M^n$ is of split type if every wavefront covector satisfies

The sum uses the canonical identification of cotangent fibers in Minkowski coordinates. It is not proposed as a coordinate-free definition on arbitrary curved spacetimes.

A split-type distribution satisfies (eq:cone).

Proof. If every $p_i$ is future causal, each time component is nonnegative. Their sum can vanish only if all time components vanish, which forces all causal $p_i$ to vanish. The zero total covector is excluded from a wavefront set. The past-cone case is identical.

The following operations preserve split type:

1. Translation kernels $u(x-z)$ are split, as are diagonal delta kernels.

2. Smooth multiplication, differentiation, and adjoining unused variables preserve split type.

3. If split distributions depend on variable sets $I,J$ with one shared slot, their product exists and is split.

4. Proper integration of internal variables preserves split type.

Proof. A translation kernel has endpoint covectors $(p,-p)$. A diagonal delta has covectors whose sum is zero. Smooth multiplication and differentiation do not enlarge the wavefront set, while unused variables carry zero covectors.

For (iii), lift both factors to the union of their variable sets. A violation of the distributional product criterion would require opposite wavefront covectors whose sum vanishes in every slot. All unshared slots would then be zero. Split type of either factor forces its shared covector to be zero as well, contradicting a nonzero wavefront covector. The wavefront bound for the product consists of sums of factor covectors, allowing a zero contribution from one factor; these sums remain split.

For (iv), the pushforward wavefront bound retains only covectors with zero internal components. The surviving external components therefore still sum to zero. Properness is required on the support over each compact external set. These are the usual product and pushforward rules Hormander (Chapter 8).

The bare coefficients $F_{c_0},F_{c_1},F_{c_2}$, regarded as polynomial functionals of $(h,k,\phi )$, have compact derivative kernels of split type. Consequently they satisfy (eq:cone).

Proof. The zeroth coefficient has a smooth derivative. The first is bilinear in $h,\phi$; its mixed kernel is a derivative of

By (eq:Lambda), this is a smooth multiplier times a derivative of a translation kernel, hence split.

The second coefficient is linear in $\phi$, quadratic in $h$, and linear in $k$. Its highest kernels are generated by:

with fixed tensors, further derivatives, and permutations of background slots suppressed. The $k$-linear term is the corresponding two-slot kernel.

For the first expression, the two coordinate factors share only $x$. For the local quadratic expression, the two delta factors share only $w$, and the resulting distribution meets the Green kernel only at $w$. In the nested expression, attach the delta factor to the inner coordinate kernel at $w$, then attach the outer Green kernel at that same single slot. Each step is covered by Lemma lem:split-calculus. No square of a propagator on identified endpoints has been introduced.

Integrate the internal variable $w$. Retarded support and the factors ${\sigma }_{\epsilon }$, together with the compact observation support, confine every relevant vertex to ${\mathcal{D}}_f$. Thus the pushforwards are proper; a smooth compact cutoff equal to one there can be inserted if desired. Lower derivatives are obtained by smearing remaining slots against smooth configurations, with the same compact localization.

All kernels are therefore split and compact. The flat ramp introduces no distributional boundary multiplier at $t=0$. Lemma lem:split-cone proves the claimed admission.

The preceding proof uses smooth multiplication by ${\sigma }_{\epsilon }$. A sharp step has $\mathrm{WF}(\theta (t))=\{(t=0;\lambda \mathrm{d}t):\lambda \ne 0\}$; it is not a smooth multiplier in the split calculus.

The two sharp prescriptions must also be distinguished. The sharp action-level adjoint corresponding to the divergence-form coordinate source is schematically $\theta C^{\ast }b$. The multiplied-equation prescription has adjoint $C^{\ast }(\theta b)=\theta C^{\ast }b+\delta (t)\mathsf{B}(b{|}_{t=0})$. A nonzero boundary coefficient can therefore introduce a timelike conormal; the latter prescription explicitly has a delta term. The fixed ramp has neither sharp multiplier.

This explains the removal of the sharp initial-conormal mechanism at fixed $\epsilon$. It does not identify the ramp object with VII-B's insertion, prove a sharp-limit theorem, or construct a retarded product with a quantum metric.

For fixed classical backgrounds, every $c_j$ in (eq:tests) is smooth and supported in $K=\mathrm{supp}f$. The maps $(h,k)\mapsto c_j\in {\mathcal{D}}_K(M)$ are continuous polynomial maps at fixed $\epsilon$. Each $F_{c_j}$ belongs to ${\mathcal{L}}_{\mathrm{off}}$, and its class $\hat{\varphi }(c_j)$ is well defined.

Proof. The coordinate coefficients are smooth and continuous by the handoff. Formula (eq:tests) uses smooth products and derivatives multiplied by $f$, so its support is contained in $K$, with continuous dependence in all compact-test seminorms. As a functional of the only quantum configuration, $F_{c_j}$ has smooth first derivative and all higher derivatives zero.

The scalar support is contained in $K$, whereas background dependence may extend throughout ${\mathcal{D}}_f$. Neither inclusion is a claim of equality for every configuration.

The retarded construction uses only:

1. A quadratic insertion $V=-\frac{1}{2}⟨\phi ,\mathsf{L}\phi ⟩$, where $\mathsf{L}$ is formally self-adjoint with smooth compact coefficients.

2. A specified linear representative $F_c$, with $c\in \mathcal{D}(M)$.

Joint admission is additional information, not a substitute for this named quantum domain.

Compactly supported integration by parts gives

The free action has $S_0^{\prime \prime }=-P_m$, hence

The positive Euler coefficient is not the variational Hessian. Using $+{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}$ as the inverse of $S_0^{\prime \prime }$ would reverse the response and violate the normalization sheet.

For a quadratic insertion as in (A1), the classical retarded response of $F_c$ is

This also follows by differentiating the retarded solution of $(S_0+\lambda V{)}^{\prime }=0$.

For the specified representative $F_c$, define

The notation $-⟨c,{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}{\mathsf{L}}_{1,g}[h]\hat{\varphi }⟩$ means this transposed expression. It does not mean applying an inverse to an unsmeared quantum field without first defining its test action.

The map (eq:Rone) is bilinear in the quadratic insertion and the linear representative. Its output test is smooth and obeys

It is continuous in the smooth background and compact-test topologies. It satisfies

It is additive in the vertex and vanishes if $\mathrm{supp}g\cap J^{-}(\mathrm{supp}c)=\varnothing$. No new diagonal extension or subtraction is used.

Proof. The advanced Green operator maps compact smooth tests to smooth functions supported in their causal past. The coefficients of ${\mathsf{L}}_{1,g}[h]$ are smooth and supported in $\mathrm{supp}g$. Therefore $-{\mathsf{L}}_{1,g}[h]{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}c$ is a smooth compact test with (eq:retarded-support).

Formal symmetry of ${\mathsf{L}}_{1,g}[h]$ and $({{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}{)}^{\mathsf{T}}={{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}$ give

Compact vertex coefficients justify the integrations by parts. Green-operator continuity on compact causal regions BGP (Chapter 3), followed by continuous smooth multiplication and differentiation, proves continuity. For fixed input support, the output has a fixed compact support. Additivity and retarded vanishing are immediate.

Only one Green distribution is transposed against a smooth test. There is no singular product requiring extension on a diagonal.

For all admitted bare coefficients,

This is the unweighted map. The physical first-vertex correction has an additional factor $\kappa$. These two notions are not interchangeable.

The support theorem places the correction's scalar test in $\mathrm{supp}g\cap J^{-}(K)$. It does not remove the original dressing's background dependence on ${\mathcal{D}}_f$. At this order, the causal statement consists of additivity and retarded vanishing; no complete causal-factorization hierarchy is claimed.

To verify the quantum interpretation without constructing a second product programme, introduce the auxiliary kernel

It is symmetric with these conventions. Both ${\Delta }_{\mathrm{F}}c$ and ${\Delta }_{+}c$ are smooth. For the pair (A1)–(A2), consider only the auxiliary contraction expression

It is well defined because pairing $V^{\prime }$ with a smooth function requires only compact vertex coefficients.

The Wick expression terminates exactly:

Consequently

For $V^{\prime }=-\mathsf{L}\phi$, this is exactly $-F_{\mathsf{L}{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}c}$. The auxiliary expression supplies the usual one-contraction retarded interpretation DF; it does not constitute a claimed family of time-ordered maps.

On the named scalar-only domain, $R_1(V_{1,g};F_c)$ is independent of $\hslash$ and of the Hadamard contraction kernel:

In particular, the first quantum correction vanishes identically.

Proof. A cross-contraction differentiates each entry once. Since ${F_c}^{(2)}=0$, no term with two cross-contractions exists. The zero-contraction product cancels in (eq:contraction-check). Its single contraction contributes one $\hslash$, removed by $i/\hslash$, and ${\Delta }_{+}$ cancels. There is no remaining positive power of $\hslash$.

The vertex is represented in a fixed free Wick ordering. Changing that ordering for a quadratic vertex changes it only by a scalar-independent term, whose first derivative is zero. Thus it cannot alter this map.

This is a theorem about the declared field content. It does not hide a missing graviton contraction or loop. It also makes no uniqueness claim among arbitrary additional contact normalizations: the map is the explicit normalized one-contraction prescription above, with no contact selected to repair an identity.

Applying the map to the first frozen test gives

Indeed, $⟨c_1,{\rho }_{1,g}⟩=⟨{\partial }_{\mu }(f{{\mathsf{X}}_1}^{\mu }),{\rho }_{1,g}⟩$. Equation (eq:mixed) is the copied classical mixed term, with the scalar replaced by the free field through its test action. The minus insertion sign is fixed; the plan's plus sign is rejected.

Let

and write ${\mathsf{Q}}_h={\mathsf{L}}_{1,g=1}[h]$.

For every smooth vector field,

Proof. The density coefficients are

Expanding ${\mathsf{Q}}_{K\xi }u$, the derivatives of $\partial \cdot \xi$ cancel, leaving

Expanding the right-hand side of (eq:Qidentity) gives the same expression.

Define the off-shell representative

and its free-field class

For a compact parameter $\xi$ in the slab,

where

The first term lies in ${\mathfrak{J}}_0$. The second is the coordinate edge term. The last displays the factors $1-g$ and $\partial g$.

Proof. The changed test is

Write ${\mathsf{L}}_{1,g}[K\xi ]={\mathsf{Q}}_{K\xi }+{\Sigma }_{g,\xi }$. Using $P_m{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}f=f$, Lemma lem:Q gives

Subtract this from

The resulting identity is (eq:variation). The function $X_{\xi }{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}f$ is smooth and compactly supported, so its $P_m$-test belongs to the equation ideal.

The switch coefficients are supported in $\mathrm{supp}\xi \cap (\mathrm{supp}(1-g)\cup \mathrm{supp}\mathrm{d}g)$. This closed-support statement includes plateau boundaries; it does not incorrectly identify support with the open set $\{g\ne 1\}$.

For every classical background $h$ and every $\xi \in {\mathfrak{g}}_U$,

Thus $\hat{\varphi }(f)+\kappa {\hat{A}}_1[h]$ is invariant modulo ${\kappa }^2$, with the free field held fixed.

Proof. The first term of (eq:variation) vanishes in ${\mathfrak{A}}_0$. The copied preflight identity kills the second. On $\mathrm{supp}\xi \subset U$, $g=1$ and $\mathrm{d}g=0$; hence ${\Sigma }_{g,\xi }=0$. These are exactly the declared support conditions, not restrictions introduced after the defect calculation.

Off shell, the plateau difference is precisely $F_{P_m(X_{\xi }{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}f)}$, not zero in general. General pullback of the scalar is not an automorphism of the free algebra: a non-isometric diffeomorphism need not preserve $P_m\varphi =0$. The theorem is instead background covariance at fixed free field, corresponding to the first completed coefficient of X-A's on-shell identity.

The physical first-vertex correction is $\kappa R_1$, so

The classical coefficient is instead

with ${\rho }_{2,g}$ displayed in (eq:rhotwo).

The last term is not supplied by the present map. A full quantum completion would require the second-vertex insertion and the iterated first-vertex insertion. Neither is constructed here. Their classical sum is displayed, not replaced by a contact and not assigned the name of a constructed quantum map.

The bare ${\mathcal{O}}^{(2)}$ is already admitted. Its classical transformation is retained through the copied metric and coordinate laws. That fact does not complete the second-order dynamics.

Put

For $\xi \in {\mathfrak{g}}_U$, its exact finite change in ${\mathfrak{A}}_0$ is

Equivalently, it is

The first two terms are linear in the shift parameter and the last is quadratic.

Proof. Both $c_1[h]$ and ${\mathsf{L}}_{1,g}[h]$ are linear in $h$, while $c_1[K\xi ]=D_{\xi }f$. Expanding $-{\mathsf{L}}_{1,g}[h+K\xi ]{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}(c_1[h]+D_{\xi }f)$ produces the first term in (eq:mixed-transformation) and two terms containing $-{\mathsf{Q}}_{K\xi }{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}$. For a compact test $u$,

The last summand is zero in ${\mathfrak{A}}_0$. Apply this identity with $u=c_1[h]$ and $u=D_{\xi }f$. The integral form follows from ${D_{\xi }}^{\mathsf{T}}=-X_{\xi }$.

This is the transformation of the classical mixed term evaluated on the free field. It is generally nonzero. X-A cancels it in the complete classical coefficient together with the transformations of the remaining terms. Equation (eq:truncation) alone is not given that invariant status.

For arbitrary compact slab parameters, passing (eq:variation) to the free algebra gives

Both mechanisms are visible. They must not be conflated.

If an excluded edge parameter is also supported in the already chosen plateau $P_g$, the switch operator vanishes and the coordinate edge term is isolated. If that plateau condition is absent, the switch term remains. The sheet does not assume that every chosen $g$ has an edge plateau.

For compact $\xi$ supported in $0<t<\epsilon$, the coordinate gap can also be written

The nonzero integral (eq:edge-gap) proves a coordinate failure. It does not, by itself, prove a nonzero total smeared on-shell response for every fixed $f,g$. The observation test can miss the witness, a free-field test can be an equation zero, or scalar switch terms can also be present. No identity is claimed for these excluded parameters, and ${\mathfrak{g}}_U$ is neither enlarged nor shrunk.

Bare dressings alone are not generally background invariant. For example, in a fixed quasi-free Hadamard representation their pairing involves $\hslash ⟨c_1[h],{\Delta }_{+}{c}_1[h^{\prime }]⟩$, which can change when $c_1[h]$ is shifted by $D_{\xi }f$. This does not contradict the invariance of the completed ${\hat{A}}_1$. No interacting two-point response is constructed or inferred.

At $g=0$,