Landmark X-C The missing second-order retarded maps for the ramp relational scalar, with the metric kept classical

At the fixed smooth regulator of Landmark X-B-v3, we construct the second-vertex scalar map and the iterated first-vertex scalar map on the frozen off-shell linear insertion tests. The scalar alone is quantum; the metric coefficients remain classical and off shell. The maps have smooth compact retarded output, and their sum equals the second scalar response copied from Landmark X-A. They are the corresponding coefficients of a test-level retarded Møller map, characterized by a Yang–Feldman recursion.

The completed relational coefficient through ${\kappa }^2$ is background-gauge invariant in the on-shell free algebra on the unchanged ramp plateau algebra. The proof of record is a three-term variation formula: a compact free-equation term, a dressed coordinate defect, and a dressed switch defect. The switch derivatives are displayed before imposing the plateau conditions. The second-order variation retains propagation of the first-order edge gap.

A complete pure-gauge two-jet verifies the operator signs and closes the remainder left in X-B-v3, with the bare second dressing retained. A conformal massless diagnostic gives a classical number and checks the sourced signs. The maps have no positive-$\hslash$ remainder in the specified linear rooted prescription. No general retarded-product hierarchy, sharp limit, quantum metric, or two-point response is constructed.

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

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

{Release: Constructed, through ${\kappa }^2$ at fixed $\epsilon$}{ On the X-B-v3 sheet, the maps

exist on the frozen off-shell linear tests, have smooth compact retarded output, and carry no positive power of $\hslash$. Their sum on $F_f$ is the copied response $\int f{\rho }_{2,g}$, where

The completed coefficient

is invariant in ${\mathfrak{A}}_0$ on ${\mathfrak{g}}_U$, with the free field held fixed. Together with the copied ${\hat{A}}_1$, this gives background invariance through ${\kappa }^2$. The pure-gauge remainder is cancelled with ${\mathcal{O}}^{(2)}$ retained.

Not claimed: the sharp limit, quantum metric, graviton tree, triangle, ${\mathsf{T}}_2$, ${\mathsf{T}}_3$, nonlinear-insertion $R_2$, or two-point responses.}

The target is reassigned to the missing scalar dynamics in X-B-v3's truncation. It is not the sharp-limit decision of an earlier pilot table. Taking $\epsilon \to 0$ before supplying the displayed ${\rho }_{2,g}$ would leave that truncation incomplete.

Three endings are fixed:

1. Constructed. Both maps exist on the frozen representatives, their sum is the copied response, and the completed coefficient is invariant.

2. Proved obstructed. A nonzero defect survives on the frozen algebra after both maps are included, or the required response normalization fails.

3. Unresolved. A vertex, sign, map, or insertion representative remains unnamed.

The ending reached is (1). A missing map is not a no-go. The previously displayed remainder is not repaired by a contact, by deleting a dressing term, or by shrinking the gauge algebra.

Both $\kappa$ and $\hslash$ are formal. Throughout the paper, operator and field identities involving the background expansion are understood modulo ${\kappa }^3$. Support assertions for coefficients of formal pullbacks are coefficientwise assertions, not claims about the support of an unexpanded finite pullback.

The supplied X-A.tex XA and X-B-v3.tex XB supply the classical ramp prescription, its covariance, admission, and the first scalar map. Those covariance and cone proofs are not repeated. The conventions needed to state the new construction are recorded here so that its equations and domains are explicit.

Let

The quantum field is defined on $M$. The slab $M_T=[0,T]\times {\mathbb{R}}^3$ and its initial surface belong to the classical coordinate prescription.

The classical backgrounds are arbitrary smooth real symmetric tensors $h,k$, and

Indices are moved with $\eta$. Background topology means convergence of all derivatives on compact subsets of $M$.

Fix $0<\epsilon <T$ and the same ramp

Thus ${\sigma }_{\epsilon }$ is smooth and flat at zero. Fix the same $g\in {C_c}^{\infty }((0,T)\times {\mathbb{R}}^3;[0,1])$, with $P_g=\mathrm{int}\{g=1\}$, and

The formal group ${\mathcal{G}}_U$ consists of pullbacks generated by $\kappa \xi +{\kappa }^2\zeta$, $\xi ,\zeta \in {\mathfrak{g}}_U$.

For the corresponding flow $F$, the complete metric law is

The second line is essential; $k=0$ does not describe a second-order pure-gauge jet when $h=K\xi$.

The coordinate coefficients copied from X-A are

The density coefficients are displayed in Section sec:sheet. Retarded propagation in (eq:coordinates) reproduces the zero-data Cauchy prescription on the slab.

The copied coordinate laws are

In particular, ${\mathsf{X}}_1[K\xi ]=\xi$ on ${\mathfrak{g}}_U$. X-A's coordinate-only intertwiner also applies to compact generators supported in $\{\epsilon <t<T\}$, without requiring an interaction plateau.

Fix a real $f\in {C_c}^{\infty }((0,T)\times {\mathbb{R}}^3)$, put $K=\mathrm{supp}f$, and

For $B^{\mu }={{\mathsf{X}}_1}^{\nu }{\partial }_{\nu }{{\mathsf{X}}_1}^{\mu }-{{\mathsf{X}}_2}^{\mu }$, the bare off-shell tests are

Write $F_c(u)=\int cu{\mathrm{d}}^{4}x$. Then ${\mathcal{O}}^{(j)}=F_{c_j}$, and ${\mathcal{O}}_{\kappa }=F_f+\kappa F_{c_1}+{\kappa }^2{F}_{c_2}$. The tests are smooth, supported in $K$, and continuously dependent on the backgrounds at fixed $\epsilon$. Their admission is copied from X-B-v3.

Use X-B-v3's free polynomial Wick algebra ${\mathcal{A}}_{\mathrm{off}}$ and its equation ideal ${\mathfrak{J}}_0$, generated by $F_{P_{m}v}$ for $v\in \mathcal{D}(M)={C_c}^{\infty }(M;\mathbb{C})$. Set

The relevant relations are

where ${{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}},{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}$ are the Green operators of $P_m$. The linear output topology is the quotient test topology. The insertion domain is instead

identified with actual off-shell tests.

This distinction cannot be suppressed. The copied first map satisfies

and the second expression need not vanish. Consequently no iteration in this paper is performed through an on-shell input class.

The copied positive Euler coefficients are ${\mathsf{L}}_{1,g},{\mathsf{L}}_{2,g}$, and for a smooth free input $u$,

The switched Euler operator ${\mathsf{E}}_g[q]=P_m+\kappa {\mathsf{L}}_{1,g}+{\kappa }^2{\mathsf{L}}_{2,g}$ is covariant under ${\mathcal{G}}_U$. The unswitched curved density operator satisfies naturality under every compactly supported diffeomorphism. These are copied premises from X-A.

X-B-v3 constructs ${\hat{A}}_1=\hat{\varphi }(c_1-{\mathsf{L}}_{1,g}{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}f)$ and proves its invariance. Its truncation is

The mixed term has the fixed minus insertion sign:

Its copied finite transformation on ${\mathfrak{g}}_U$ is

This is not an invariance statement.

The classical second coefficient has the additional $\int f{\rho }_{2,g}$. On the complete pure-gauge jet $h=K\xi ,k=\frac{1}{2}{\mathcal{L}}_{\xi }K\xi$, the incomplete coefficient in (eq:old-truncation) is $-\frac{1}{2}\int f{X}_{\xi }^2\hat{\varphi }{\mathrm{d}}^{4}x$. The missing classical response is $\frac{1}{2}{X_{\xi }}^{2}u$. Constructing that response is the present target.

The convolution coefficient and X-Enter are not inputs. X-Enter's Option A concerns readout of the convolution coefficient alone. Its Option B reference is not supplied by the ramp observable, and the ramp edge integral is not the convolution quantity $c_{\epsilon }$. Reading ${\hat{A}}_2$ would preserve an upstream identity, not prove it.

VII-A, VI, VII-C, IX-E-2, and IX-E-3 remain comparison. No cone, token, sharp insertion, or earlier kernel is identified with the ramp coordinate kernel. VII-C's hypotheses (E1)–(E4) are not assumed.

Branch B, with a quantum metric, is a different construction. There is no gravitational action, gauge-fixed graviton propagator, or metric contraction in the present sheet. A graviton tree or mixed triangle would require its own definition and analysis. Neither existence nor obstruction of those graphs follows from the scalar-only maps below.

No time-ordered hierarchy is used to fill the missing coefficient. The restricted maps are defined directly and then checked against their dynamics, support, and equation-ideal variation. This limited construction is sufficient for the stated target; it is not a substitute for a product programme on nonlinear insertions.

Write $H={\mathrm{tr}}_{\eta }h$. The copied density coefficients are

Set

These are the ${\kappa }^2$ coefficients of $\sqrt{-q}{q}^{-1}$ and $\sqrt{-q}$, respectively.

The frozen switched density is $(1-g){\mathfrak{L}}_{\eta }+g{\mathfrak{L}}_q$. Its two interaction coefficients are

The second vertex is therefore fixed by the supplied density before either new map is defined.

The divergence-form coefficients, including switch derivatives, are

They are formally self-adjoint, with smooth compact coefficients. Neither is obtained by multiplying an unswitched Euler equation after variation by $g$.

Compact integration by parts gives the sign ledger

The positive Euler coefficients are not the variational Hessians. This resolves the inconsistent Hessian wording in X-A's handoff in accordance with its explicit Euler-sign audit.

{}1.12

Put

Every composition below is coefficientwise in the formal ring.

Define

For actual $c\in \mathcal{D}(M)$, the two missing maps are

The notation $\frac{1}{2}{R}_2$ names the second Taylor coefficient for the first vertex alone, not a general retarded product.

The advanced operator in these formulas is the transpose of retarded propagation of the scalar. Thus the terminology “retarded map” is consistent with backward causal support of its test.

The inner test ${\mathsf{L}}_{1,g}{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}c$ is an actual smooth compact test. Iteration may be written as applying the first-map rule to $F_{-{\mathsf{L}}_{1,g}{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}c}$, but it must not be written as applying that rule to an unspecified representative of its on-shell class. Equation (eq:first-map) explains why: changing an insertion by $F_{P_{m}v}$ can change its response.

The two maps (eq:second-map)–(eq:iteration) have smooth compact output tests supported in

They are continuous in compact-test and smooth background topologies. For every smooth $u$, and hence for every smooth free $u$,

Their sum on $F_f$ is therefore $\int f{\rho }_{2,g}$, understood as the free field with the displayed compact test. No diagonal extension or subtraction is used.

Proof. The Green operators act continuously on compact smooth tests, with their respective causal supports; transposition exchanges the retarded and advanced operators BGP (Chapter 3). Consequently ${{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}c$ is smooth, and applying ${\mathsf{L}}_{2,g}$ produces a compact test supported in $\mathrm{supp}g\cap J^{-}(\mathrm{supp}c)$.

The intermediate support $K_1=\mathrm{supp}({\mathsf{L}}_{1,g}{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}c)$ is contained in the same set. The iterated output is supported in

Continuous differentiation and multiplication give continuity. For input tests in a fixed compact support, every output lies in a fixed compact subset of $\mathrm{supp}g$.

Formal self-adjointness and transposition give

All source factors are compact because of $g$. Every Green operation is applied to a smooth source or test; there is no singular product requiring extension.

For arbitrary smooth input $u$, the scalar response polynomial is

It satisfies

For a free input it reproduces the copied retarded coefficients.

With the free Hessian inverse ${{\Delta }_0}^{\mathrm{R}}=-{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}$, the first response is

No further minus sign is inserted. The second vertex enters once with the same negative response sign. Two successive first responses give a positive iteration.

For the first vertex alone, replacing $\kappa$ by $\lambda$ in $(1+\lambda {{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}{\mathsf{L}}_{1,g}{)}^{-1}u$ shows that the second derivative at zero is twice ${{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}{\mathsf{L}}_{1,g}{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}{\mathsf{L}}_{1,g}u$. This explains the coefficient $\frac{1}{2}{R}_2$. The normalization is dynamical, not a contact chosen after a gauge calculation.

The second-vertex map is linear in its quadratic vertex. The diagonal iteration is quadratic, not additive, in $V_{1,g}$.

For two specified quadratic vertices $V,W$, with positive Euler coefficients $L_V,L_W$, its polarization is the bookkeeping formula

It is bilinear in these two coefficients and reduces on the diagonal to twice (eq:iteration). Replacing $V$ by $V+W$ therefore retains the mixed terms. Dropping them under a claim of diagonal additivity would be incorrect.

Formula (eq:polarization) does not construct a hierarchy on arbitrary entries. The causal statements made here are the displayed chain support, linearity or polarization, and vanishing when the vertex supports cannot form a causal chain into the insertion support. In particular, both new maps vanish if $\mathrm{supp}g\cap J^{-}(\mathrm{supp}c)=\varnothing$. No full causal-factorization hierarchy is claimed.

Define the linear interacting field and formal Green operators by

These are finite formal expansions at the declared order. No assertion about numerical inversion of the perturbed operator is needed.

The following identities hold:

1. ${\mathcal{R}}_q=1-{\mathcal{R}}_q\mathsf{L}[q]{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}$. It is the unique formal series with leading term $1$ satisfying this recursion. Equivalently, ${\hat{\varphi }}_q(c)=\hat{\varphi }(c)-{\hat{\varphi }}_q(\mathsf{L}[q]{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}c)$.

2. ${\hat{\varphi }}_q({\mathsf{E}}_g[q]v)=0$ for every $v\in \mathcal{D}(M)$.

3. If $\mathrm{supp}c\cap J^{+}(\mathrm{supp}g)=\varnothing$, then ${\hat{\varphi }}_q(c)=\hat{\varphi }(c)$.

4. The algebraic commutator is

   $$[{\hat{\varphi }}_q(c),{\hat{\varphi }}_q(d)]=i\hslash ⟨c,({{\mathsf{E}}_{\mathsf{q}}}^{\mathrm{A}}-{{\mathsf{E}}_{\mathsf{q}}}^{\mathrm{R}})d⟩1.$$

Proof. The inverse identity $(1+A{)}^{-1}=1-(1+A{)}^{-1}A$, with $A=\mathsf{L}[q]{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}$, proves (i). Comparing coefficients determines $-{\mathsf{L}}_{1,g}{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}$ at first order and ${\mathsf{L}}_{1,g}{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}{\mathsf{L}}_{1,g}{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}-{\mathsf{L}}_{2,g}{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}$ at second order uniquely.

For compact $v$, ${{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}{P}_{m}v=v$, so

Its class vanishes, proving (ii). The support condition in (iii) gives $\mathsf{L}[q]{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}c=0$.

For (iv), transposition gives ${{\mathcal{R}}_q}^{\mathsf{T}}=(1+{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}\mathsf{L}[q]{)}^{-1}=1-{{\mathsf{E}}_{\mathsf{q}}}^{\mathrm{R}}\mathsf{L}[q]$, while ${\mathcal{R}}_q=1-\mathsf{L}[q]{{\mathsf{E}}_{\mathsf{q}}}^{\mathrm{A}}$. Using the free commutator,

The operator in the middle equals

This proves the commutator identity.

The theorem characterizes the declared linear-field construction. It does not produce a state or an interacting two-point response. The commutator is an algebraic consistency check only.

The maps (eq:second-map) and (eq:iteration) contain no positive power of $\hslash$ and no Hadamard contraction kernel. They are tree-exact in the specified linear rooted prescription.

Proof. Algebraically this follows from (eq:moller): its coefficients are ordinary test operators, with neither $\hslash$ nor a Hadamard kernel. The quantum statement concerns these coefficients, not the $\hslash$ in the field commutator.

The local contraction count confirms the scope. Use the fixed polynomial Wick presentation of X-B-v3. In that presentation each cross-contraction differentiates both entries once and supplies one factor of $\hslash$ FR (Sections 4–5). A quadratic vertex with a linear insertion admits only one cross-contraction, as in X-B-v3's first-map verification.

For two quadratic vertices and one linear insertion there are five scalar legs. A connected rooted graph joining the three entries needs two edges, consumes four legs, and leaves one scalar leg. Since the insertion has degree one, the graph is a chain ending at that insertion. Three contractions would require six legs and are impossible. The two possible vertex orderings give the bookkeeping polarization (eq:polarization).

The two contraction factors are removed by the second-order retarded prefactor. A vacuum component between the two vertices disconnected from the insertion is not part of the rooted relative response. Self-contractions are absent in the fixed Wick presentation. Thus the rooted coefficients have no positive-$\hslash$ remainder.

The leg count does not construct time-ordered products or license unanalysed diagonal extensions. The maps are defined by (eq:moller) and characterized by Theorem thm:yf; the count explains why that restricted construction has the stated quantum order. No cancellation theorem for an unconstructed ${\mathsf{T}}_2,{\mathsf{T}}_3$ programme is used.

Scalar-independent changes of quadratic Wick ordering have zero first field derivative and do not change these linear responses. Nor does tree-exactness make ${\mathcal{R}}_q$ a map on free on-shell inputs:

whose class can be nonzero. The warning about actual intermediate tests remains in force.

For $c[q]=f+\kappa c_1+{\kappa }^2{c}_2$, set

The last line is the coefficient recursion obtained from $(1+\mathsf{L}[q]{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}})e=c$. Expanding it gives

Consequently

has precisely the four second-order terms in the release box.

Every $e_j$ is a smooth compact test, with

The background dependence is continuous at fixed $\epsilon$. By transposition and the copied bare expansion, $⟨e_2,u⟩$ is X-A's complete classical second coefficient for every smooth free input $u$. This observation is a normalization check, not yet the Ward proof.

For arbitrary compact $\xi ,\zeta$ in the open slab, put

The first acts on scalars; the second is density pullback, including the Jacobian.

For compact formal tests $c$ and smooth formal scalars $\psi$,

Proof. ${D_{\xi }}^{\mathsf{T}}=-X_{\xi }$. Thus $(F^{♯}{)}^{\mathsf{T}}$ is the inverse scalar pullback $1-\kappa X_{\xi }+{\kappa }^2(-X_{\zeta }+\frac{1}{2}{X_{\xi }}^2)$. Multiplying by $F^{\ast }$ gives the identity through second order.

For $q^{\prime }=F^{\ast }q$, define

Write ${\Theta }_F=\kappa {\Theta }^{(1)}+{\kappa }^2{\Theta }^{(2)}$ and ${\Sigma }_F=\kappa {\Sigma }^{(1)}+{\kappa }^2{\Sigma }^{(2)}$.

The first defect is a compact test. The second is a differential operator with coefficients supported on the compact generator support.

Coefficientwise, $\mathrm{supp}{\Theta }^{(j)}\subset K$, and

For compact generators supported in $\{\epsilon <t<T\}$, ${\Theta }_F=0$.

Proof. Differential operators in a formal pullback do not enlarge the support of each test coefficient; hence the support statement. Expansion of the definition gives (eq:theta).

On the coordinate plateau, the copied off-shell coordinate identity gives $⟨c[q^{\prime }],F^{\ast }\psi ⟩=⟨c[q],\psi ⟩$ for every scalar jet. Formal duality makes the right side $⟨F^{♯}c[q],F^{\ast }\psi ⟩$. Since $F^{\ast }$ is formally invertible, the tests agree. This uses the copied intertwiner, not a new proof of ramp covariance.

Let ${\mathsf{Q}}_j$ denote the unswitched positive Euler coefficients, and set ${\mathsf{N}}_j={\mathsf{Q}}_j-{\mathsf{L}}_{j,g}$. Then

These operators need not have compact coefficients. Their transformation differences below do.

The switch defect has coefficients supported in $C=\mathrm{supp}\xi \cup \mathrm{supp}\zeta$, and

If $g=1$ on a neighbourhood of $C$, then ${\Sigma }_F=0$.

Proof. The copied naturality of the unswitched curved operator ${\mathsf{E}}_1[q]$, and ${\mathsf{E}}_g[q]={\mathsf{E}}_1[q]-\kappa {\mathsf{N}}_1-{\kappa }^2{\mathsf{N}}_2$, give

Expansion gives (eq:sigma). Outside $C$, the transformation is locally the identity, so the operator coefficients vanish there.

For an explicit display, put $X=X_{\xi }$, $D=D_{\xi }$, and $\Delta A_2=A_2[h^{\prime },k^{\prime }]-A_2[h,k]$. The second line of (eq:sigma), acting on $u$, is