Landmark XVI-J A selected extension and a root-attached coefficient for the C 2/constant-slot two-edge sector Constructed under the explicitly stated root hypotheses

We construct a selected-sector coefficient containing only the polarized source $2C_2\partial \partial {\sigma }_{\epsilon }$, the constant-slot part of the declared cubic Hessian, and two undifferentiated graviton edges. The plateau source, the $M_2$ source term, and vertex terms differentiating contracted fields are excluded.

The tensor contraction is derived from the inverse fiber pairing and is independent of the chosen symmetric-tensor basis. The Hessian and source-decorated coefficient on a specified pure-trace background are proved by explicit polynomial identities, without reliance on an unavailable script. An appendix supplies a second, exact-arithmetic verification through curvature atoms and two implementations of the source contraction.

The numerical square has scaling degree four in four relative dimensions. A subtraction function and a zero-moment condition select its extension; an explicit distributional action is given. Under the imported compactness and $Z_2$ wavefront bound for the root, pullback, multiplication, and proper pushforward construct a continuous bilinear functional of the residual metric and scalar test. Its prescription dependence is a diagonal contact term propagated through the root. This is not the complete two-edge interaction coefficient, a retarded interacting field, or a BRST/BV observable.

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

Figure fig:scope-XVI-J records that separation. The drawing adds no hypothesis.

and the four levels.{ The preparation width $0<\epsilon <T$, preparation data $f,g$, and smooth time-only ramp ${\sigma }_{\epsilon }$ are fixed. The interaction switch ${\chi }_{\mathrm{I}}\in {C_c}^{\infty }({\mathbb{R}}^4;\mathbb{R})$ is independent of these data.

The four levels are:

A nonzero coefficient at one level does not establish nonvanishing at a later level.

The construction is conditional on the explicitly stated root hypotheses. No sharp-width limit, global adiabatic limit, ultraviolet completion, all-orders time ordering, or interacting gauge identity is claimed. }

Work on ambient Minkowski space $M={\mathbb{R}}^4$, with

The symmetric-tensor fiber is

Its fixed nondegenerate bilinear pairing is

All raised indices use $\eta$; all pairings are bilinear, without complex conjugation. Parentheses on indices denote normalized symmetrization.

Write

The map $\mathcal{P}$ is self-adjoint for $b$.

The scalar test is denoted by $\phi$. The relative-coordinate map will be denoted by $r$, never by $\phi$.

The following are the only manuscript-specific inputs. Versions identify the supplied excerpts, not an independently verified full corpus.

1. The preparation is fixed, with

   $${\sigma }_{\epsilon }(t)=0(t\le 0),{\sigma }_{\epsilon }(t)=1(t\ge \epsilon ),{\sigma }_{\epsilon }\in C^{\infty }(\mathbb{R}).$$

   The independent interaction switch is ${\chi }_{\mathrm{I}}\in {C_c}^{\infty }(M;\mathbb{R})$.

2. The supplied XII-C, version 1, source formula is

   $${{\mathcal{J}}_{\mathrm{D}}}^{\mu \nu }={\sigma }_{\epsilon }{J_3}^{\mu \nu }+2{C_2}^{\mu \nu \alpha \beta }{\partial }_{\alpha }{\partial }_{\beta }{\sigma }_{\epsilon }-2{M_2}^{\mu \nu \rho }{\partial }_{\rho }{\sigma }_{\epsilon }.$$

   Only its $C_2$ summand is retained. The finite definitions of $C_2$ and of the unintegrated cubic density are given below. They are mathematical definitions in this manuscript, as well as matching the supplied XII-C representative.

3. The matching off-shell insertion has the symmetric quadratic normalization

   $$F_t(h,\phi )=⟨t_{\mathrm{D}},h\otimes h\otimes \phi ⟩,{D_h}^2{F}_t[A,C]=2⟨t_{\mathrm{D}},A\otimes C\otimes \phi ⟩.$$

   The source polarization uses

   $$\mathfrak{s}(X,Y)=\frac{1}{2}(Q(X+Y)-Q(X)-Q(Y)).$$

   These are the supplied XVI-H representation and the revised XVI-I normalization. They are used only to identify the selected-sector graph factor.

4. The root is a covariant tensor-valued distribution

   $$B_{f,g,\mu \nu }\in {\mathcal{E}}^{\prime }(M_y\times M_x).$$

   Each component is compactly supported and satisfies

   $$\mathrm{WF}(B_{f,g,\mu \nu })\subset Z_2:=\{(y,x;\xi ,-\xi ):\xi \ne 0\}.$$

   This is the root-kernel interface attributed in the supplied XVI-J to XII-D, version 1, lem:root-kernel. Its internal advanced-kernel composition is not reopened. Compactness and the wavefront inclusion are assumed here, not proved.

The ambient propagator is defined and checked in Section sec:propagator; its signs are not left as an unverified external input.

No theorem about admission, the frozen XI-G obstruction, or the entire quantum carrier is needed for the construction below.

The notation ${{\mathcal{C}}_{2e}}^{\mathfrak{R}}$ will always mean the selected-sector coefficient. No coupling or $\hslash$ weight is included in that notation.

Define the massless positive-frequency distribution by

The integral is distributional. Set $h_0=-W_0$.

The ambient advanced fundamental solution has spatial Fourier kernel

with the continuous value at zero momentum. It satisfies $\square {E_0}^{\mathrm{A}}=\delta$. For a source compactly supported in the interior of the slab, its restriction is

Thus the supplied slab sine formula is recovered on that source class. The graph itself uses the ambient, translation-invariant kernel.

Put

The spatial Fourier branches of $d$ are

The distribution fixed by (eq:DF) is symmetric under $z\mapsto -z$ and satisfies

Proof. For $t<0$,

For $t>0$, the advanced summand vanishes. The temporal derivative of the two branches has jump $-\mathrm{i}$. Hence the wave operator gives $-\mathrm{i}\delta$. Using ${\mathcal{P}}^2=\mathrm{id}$ proves the tensor inverse identity. The two branches also prove symmetry.

The time-ordered expression is understood through this distributional definition, not through an unspecified value of a step function at coincidence. Neither $d$ nor $D_F$ is extended in the later square-extension problem.

Let

On $z\ne 0$, direct Fourier evaluation gives

At a nonzero null point, $Q_{\mathrm{geom}}$ is a submersion. The boundary value in (eq:rational) has one, rather than both, conormal orientations. Consequently its wavefront covectors at that point cannot cancel pairwise. This is the usual off-diagonal Feynman relation; the product criterion and that relation are described in BDH (Theorem 13 and Proposition 26).

The distribution

exists on ${\mathbb{R}}^4\setminus \{0\}$, including nonzero null points. It is homogeneous of degree $-4$, has scaling degree four at the origin, and is nonzero.

Proof. The single conormal orientation just described satisfies Hörmander's product criterion. At non-null points both factors are smooth. The product is therefore defined on the entire punctured domain.

The massless $d$ is homogeneous of degree $-2$, so its defined punctured square is homogeneous of degree $-4$. It is not zero: at a spacelike point,

A nonzero homogeneous distribution of degree $-4$ has scaling degree four.

Only $q^{\circ }$ is translation invariant. The tensor and switch coefficients restored below are not.

For a symmetric matrix $H$, define

The tensor $C_2$ can be specified compactly and completely by the coefficient identity

Here $[{\lambda }^2]$ means polynomial coefficient extraction.

For a symmetric covariant source jet $J=(J_{\alpha \beta })$, set

Thus $Q(H;J)=\mathfrak{s}(H,H;J)$.

A second, matrix form useful for hand calculations is the following. Write

Then

Equation (eq:Qmatrix) follows by expanding

It agrees term by term with (eq:C2).

The actual source jet is always

The following finite formulas fix the density without integration by parts:

The $u_i,b_i$ in these equations are evaluated on $h$.

Every monomial is cubic in field jets and has total derivative order two. This is the supplied XII-C unintegrated representative XIIC; changing it by a total divergence would change the constant-slot sector unless a separate equivalence proof were supplied.

For constant tensor fields $A,C\in V$, define

This is precisely the part of the Hessian with no derivatives on its two increments.

The map $h\mapsto L_h$ is a linear differential operator depending only on second derivatives of $h$. In particular, it is continuous on smooth configurations.

Proof. A cubic monomial of type $hh{\partial }^{2}h$ contributes only when its two undifferentiated factors are varied. Its remaining coefficient is linear in ${\partial }^{2}h$. A monomial of type $h\partial h\partial h$ has only one undifferentiated factor. Varying twice in constant directions therefore gives zero. This proves the assertion directly from the total derivative count.

Let $B_0,\dots ,B_9$ be the constant symmetric matrices in the order

with the indicated entry equal to one and, for an off-diagonal pair, its symmetric partner also equal to one.

The Gram matrix is

Let $B^i=(G^{-1}{)}^{ij}{B}_j$. Then

as a contravariant tensor on the fiber.

Proof. A diagonal basis matrix has pairing with itself equal to one. An off-diagonal $B_{(ab)}$ contributes two entries, giving $2{\eta }^{aa}{\eta }^{bb}$. Distinct basis matrices have disjoint component support. This gives (eq:Gram). The remaining statements are the defining inverse-matrix identities.

The vertex Hessian is a covariant bilinear form on $V$. A field derivative at either end of an edge is a covector in $V^{\ast }$. The propagator takes such a covector to a field vector after the identification by $b^{-1}$. Its fiber tensor, with the scalar factor $S$ removed, is

Set

The reduced selected-sector distribution on $y\ne w$ is

Equation (eq:Kdef) is independent of the basis of $V$. It is the contraction of the source polarization and vertex Hessian by the two fiber edge tensors (eq:edgefiber).

For the ordinary, non-divided second derivatives used here, its numerical graph factor is $2/2!=1$.

Proof. Applying the source bilinear map to the first vectors of two copies of $\mathsf{E}$, and the vertex bilinear form to their second vectors, gives exactly (eq:Kdef).

For an arbitrary invertible basis change ${\tilde{B}}_i=B_j{A}_{ji}$,

The two inverse matrices from the dual vectors cancel the two matrices from the Hessian. The double sum is unchanged. Equivalently, the fiber tensor $\mathsf{E}$ is intrinsically defined.

The second tensor derivative of the quadratic insertion is $2\mathfrak{s}$, whereas the vertex derivative is the ordinary Hessian $L_h$. The second term of the cross-contraction exponential has coefficient $1/2!$. Thus the product of these two numerical factors is one. No additional division for symmetric spacetime indices is made: the Gram inverse already accounts for the independent components. The identical-edge convention is the one displayed in FR (Eq. (82)).

In particular,

Neither is obtained by treating the field basis as self-dual. Trace reversal acts on these fiber vectors, not on $q^{\circ }$.

Consider the test background

By Lemma lem:Llinear, its constant-slot Hessian is independent of $w$.

For $N_A=\eta A$, $N_C=\eta C$,

Equivalently, if $H={\sum }_i{z}_i{B}_i$, its quadratic polynomial is

Proof. Since only the residual second derivative contributes, evaluate at $w^1=0$, where the residual field and its first derivatives vanish. For

one has there

The Ricci formula (eq:Ricci) gives, with $U=(\eta +\lambda H{)}^{-1}$,

Write $N=\eta H$, $t=\mathrm{tr}N$, $v=\mathrm{tr}(N^2)$, and $m=(N^2{)}^1{}_1$. The expression in parentheses is

Multiplication by

leaves second-order coefficient $m+v/4-t^2/8$. The signed cubic density is therefore the quadratic expression $-4m-v+t^2/2$ at this residual jet. Differentiating it twice proves both formulas.

Expanding (eq:F) gives the full quadratic polynomial

Consequently the complete Hessian is

This proves every entry, including every zero, without a script or a numerical approximation.

For $A=\eta =B_0-B_4-B_7-B_9$,

Also

Proof. The first identity follows from (eq:Ltable). Each zero follows either from (eq:Lclosed) or the absence of the corresponding mixed monomial in (eq:Fexpanded). Thus the value six is checked independently of the other entries and is not used to certify them.

The following auxiliary operation is finite-dimensional polynomial algebra, not a probability expectation:

Let

Then

Indeed, the coefficient of $a_i{a}_j$ in the quadratic source is its polarization on $Z_i,Z_j$.

The diagonal components of $H(a)$ are

Its off-diagonal components are

Thus the diagonal quadratic contraction table, in order $00,11,22,33$, is

The off-diagonal squares are

and all mixed contractions involving two different off-diagonal components, or one diagonal and one off-diagonal component, vanish.

These tables follow by multiplying the displayed rational change-of-basis matrix with (eq:Ltable). They provide a contraction calculation independent of the double polarization sum.

For $h=h_{\ast }$ and a general symmetric jet $J$, the tensor is

The remaining entries follow by symmetry.

Proof. Apply (eq:covdiag)–(eq:covoff) to (eq:Qmatrix). A compact form of the resulting identity is

where $E$ is the contravariant matrix with $E^{11}=1$ and all other entries zero. Both sides are linear in $J$. Substitution of the ten symmetric component jets, using the complete quadratic contraction tables, proves this polynomial identity. Expanding its diagonal and off-diagonal entries gives (eq:Ktable).

For additional explicit arithmetic checks, the next lemma evaluates three component jets directly.

The contraction gives

All other jet components are zero in each check.

Proof. The contraction tables give

Writing $t=\mathrm{tr}(\eta H)$, they also give

For $J_{11}=1$, equation (eq:Qmatrix) yields

Its contraction is

so ${\mathcal{K}}^{00}=14$.

For $J_{01}=J_{10}=1$,

The three groups contract to $9,-2,-14$, respectively. Thus the result is $2(9-2-14)=-14$.

For $J_{00}=1$,

Its contraction is

so ${\mathcal{K}}^{22}=10$.

On $h_{\ast }$, the time-only preparation jet gives

with every other component zero. This tensor is not identically zero.

Proof. Only $J_{00}$ survives, so the substitution in (eq:Ktable) gives the displayed formula. If ${\sigma }_{\epsilon }^{\prime \prime }$ were identically zero, the smooth ramp would be affine. Its constant value on $t\le 0$ would then force it to be constant everywhere, contradicting its value one for $t\ge \epsilon$.

Choose an auxiliary smooth radial cutoff $\zeta \in {C_c}^{\infty }({\mathbb{R}}^4)$ equal to one near zero. For $\psi \in {C_c}^{\infty }({\mathbb{R}}^4)$ with $\psi (0)=0$, define

For each positive $\delta$, the test vanishes near the origin, so this pairing is already defined on the punctured domain. This cutoff limit defines the action on a larger test subspace; it is not a change of the preparation width.

The limit (eq:T0) exists, is independent of the radial cutoff used, and is a continuous linear functional on compact tests satisfying $\psi (0)=0$.

Proof. Use Euclidean polar coordinates $z=\rho \omega$, with $\omega \in S^3$. Homogeneity and (eq:rational) give the polar form

The angular distribution is well defined: at its zeros the argument is a submersion on $S^3$.

The compact sphere has a finite-order distributional estimate for $A$. Since $\psi (0)=0$, the function $\psi (\rho \omega )$ and each fixed finite set of its angular derivatives are $O(\rho )$, uniformly on the sphere. The integral is therefore absolutely convergent in its radial distributional estimate near zero. It vanishes for sufficiently large $\rho$.

The omitted radial interval contributes $O(\delta )$, with constants controlled by finitely many test seminorms. Consequently all such cutoffs give the same limit, and the same estimates prove continuity.

Fix, as part of the prescription, a real $\varpi \in {C_c}^{\infty }({\mathbb{R}}^4)$ with $\varpi (0)=1$. Define

The prescription $\mathfrak{R}={\mathfrak{R}}_{\varpi }$ is

for every compact smooth test $\psi$.

The formula defines an extension of $q^{\circ }$ with scaling degree four, satisfying

It is the unique scaling-degree-four extension with this moment. Every other extension of scaling degree at most four has the form

Proof. Continuity follows from Lemma lem:T0 and continuity of $W_{\varpi }$. If a test is supported away from zero, its value at zero vanishes and (eq:extension) returns the original punctured distribution. Also $W_{\varpi }\varpi =0$, giving the stated moment.

The radial estimate gives a logarithmic scaling correction: under shrinking of the test, the subtracted action is bounded by a constant times ${\lambda }^{-4}(1+|\log \lambda |)$ in fixed test seminorms. Thus its scaling degree is at most four. Its nonzero homogeneous restriction already has degree four, so the degree is exactly four.

The difference of any two such extensions is supported at zero. A point-supported distribution is a sum of derivatives of $\delta$, and $\mathrm{sd}({\partial }^{\alpha }\delta )=4+|\alpha |$. Only a multiple of $\delta$ is permitted. The moment fixes that multiple because $\varpi (0)=1$. This also agrees with the point-extension theorem BF (Theorem 5.3).

For two allowed functions ${\varpi }_1,{\varpi }_2$,

Proof. Their difference vanishes at zero. Subtracting the two instances of (eq:extension) proves the identity on every test.

A fixed subtraction function is an actual normalization datum. It may encode a coordinate length or shape. No Lorentz covariance, scaling covariance, or compatibility with a separately selected derivative extension follows merely from the zero-moment condition. Derivatives of $q^{\circ }$ are not extended here.

The extension concerns one point in four relative dimensions. After smooth tangential coefficients are restored, this prescription is a specific extension of the selected sector; it is not a classification of all extensions with arbitrary tangential coefficient functions.

Use throughout the variable order

Define

Under the root hypotheses (eq:rootinput):

1. The pullbacks $p^{\ast }{B}_{f,g,\mu \nu }$ and $r^{\ast }{\bar{q}}^{\mathfrak{R}}$ exist.

2. Their products exist on all $X$, including $y=w$.

3. For smooth $h$, the distribution

   $$U^{\mathfrak{R}}(h)=\sum _{\mu ,\nu }{p}^{\ast }{B}_{f,g,\mu \nu }{r}^{\ast }{\bar{q}}^{\mathfrak{R}}{\chi }_{\mathrm{I}}(w){\mathcal{K}}^{\mu \nu }(y,w;h)$$

   is compactly supported, and ${\pi }_x$ is proper on its support. Thus $({\pi }_x{)}_{\ast }{U}^{\mathfrak{R}}(h)\in {\mathcal{E}}^{\prime }(M)$ is defined.

Proof. Both maps $p,r$ are submersions, so their pullbacks exist for every distribution on their respective target spaces.

The lifted root covectors and relative-kernel covectors have the respective forms

For the root, $\xi \ne 0$ follows from its $Z_2$ bound. For the relative kernel, $\kappa \ne 0$, with no restriction on its orientation needed. Their sum is

Vanishing of the $x$- and $w$-components forces $\xi =\kappa =0$, impossible for opposing nonzero wavefront covectors. Hörmander's product criterion therefore applies, including at coincidence. This argument uses the root wavefront premise, not its compactness.

The tensor coefficient is smooth. Multiplication by it and finite component contraction preserve existence. The support is contained in

which is compact. Compactness in $(y,x)$ comes from the root; compactness in $w$ comes from the interaction switch. The projection is therefore proper on the support, proving the pushforward claim.

The lifted shapes are enough; no complete diagonal wavefront calculation for ${\bar{q}}^{\mathfrak{R}}$ is required. The root's singular support is retained. Its internal composition is not independently multiplied again.

Assume Input input:all, choose the ambient propagator (eq:DF), and fix ${\mathfrak{R}}_{\varpi }$. For the $C_2$/constant-slot sector only,

is defined for every real smooth residual metric $h$ and real smooth scalar $\phi$.

It is jointly continuous and bilinear for their $C^{\infty }$ topologies. Its values are generally complex. There are compact sets $K_h,K_{\phi }$, a finite integer $m$, and a constant $C$ such that

The integer $m$ is not evaluated.

The coefficient agrees with the original punctured sector when localized away from $y=w$. Nonvanishing of (eq:C) is not asserted.

Proof. Existence and the pairing with noncompact smooth functions follow from Theorem thm:attachment and compact support. Constants may equivalently be replaced by cutoffs equal to one on that support.

Fix the compactly supported component distributions