Landmark XVI-K The full-Hessian vertex within the selected root-attached C 2 two-edge sector

We extend the constant-slot construction of XVI-J by retaining the full second variation of the declared unintegrated cubic density, including derivatives on both contracted tensor slots. The source remains restricted to the polarization of $2C_2\partial \partial {\sigma }_{\epsilon }$; the plateau source and the $M_2$ summand are not included.

The density's jet structure proves that the numerical edge products contain at most two derivatives in total. Their punctured distributions are homogeneous of degrees $-4,-5,-6$, and admit extensions with subtraction orders $0,1,2$. A single subtraction function and grade-dependent zero moments define a linear extension map on their finite-dimensional span. Constant-coefficient relations are preserved, while the possible defects of differentiation are retained as explicit contact distributions.

Under the imported compactness and $Z_2$ wavefront hypotheses for the root, pullback, multiplication, and proper pushforward construct a jointly continuous bilinear selected-sector coefficient. Its constant-slot truncation agrees with XVI-J, including the chosen extension at coincidence. Changes of prescription propagate derivatives of diagonal contact terms through the root.

The analytic existence proof does not depend on the supplied numerical coefficient tables. Those tables are documented separately as exact-arithmetic computational verification. The result is not a complete interaction coefficient, a retarded interacting field, or a BRST/BV observable.

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

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

.{ Here “full” means the full second variation of the fixed unintegrated cubic density, not the complete interaction. All derivative placements on its two contracted slots are retained. The overall sector still contains only the $C_2$ source and two graviton edges.

The preparation width $0<\epsilon <T$, preparation data $f,g$, and ramp ${\sigma }_{\epsilon }$ are fixed. The independent interaction switch is ${\chi }_{\mathrm{I}}\in {C_c}^{\infty }(M;\mathbb{R})$.

The five levels are the Hessian coefficients, punctured scalar kernels, their selected extensions, smooth source-decorated coefficients, and the root-attached functional. Statements at one level do not imply statements at another.

No sharp-width limit, state, all-orders time ordering, ultraviolet completion, or interacting gauge identity is constructed. }

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

All indices are moved with $\eta$, and parentheses on indices include the factor $1/2$. The symmetric-tensor fiber is

All tensor and distributional pairings are bilinear, with no complex conjugation.

Trace reversal and the free tensor operator are

For multi-indices $p,q,\gamma \in {\mathbb{N}}^4$,

The zero multi-index is denoted by $0$, and $e_{\alpha }$ is the $\alpha$-th unit multi-index.

The notation is deliberately separated:

A positive dilation parameter is written $a>0$. It is not a witness frequency. The scalar configuration is $\phi$, and a compact distributional test is usually $\psi$.

Fix $0<\epsilon <T$, preparation data $f,g$, and a smooth time-only ramp satisfying

Fix independently ${\chi }_{\mathrm{I}}\in {C_c}^{\infty }(M;\mathbb{R})$.

The supplied XII-C source interface is

Only the middle summand is retained here. Its finite algebraic definition is given in (eq:C2)–(eq:source). The identification with that summand is inherited; no audit of the complete cubic Euler source is claimed.

The root $B_{f,g,\mu \nu }(y,x)$ satisfies:

1. Its designated composition exists as a tensor-valued distribution on $M_y\times M_x$.

2. Its components have compact support: $B_{f,g,\mu \nu }\in {\mathcal{E}}^{\prime }(M_y\times M_x)$.

3. Componentwise, in the fixed Cartesian trivialization,

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

These are the root-composition hypotheses imported from XII-D, version 1, lem:root-kernel, as retained by XVI-J. They are assumed, not derived in this manuscript. The root's singular support is not removed.

The supplied root is described by the inherited composition

where $\vartheta$ is the compact ambient time collar. This formula records provenance only. Its internal variable $z$ and the distribution $A_{f,g}$ are not reopened in the proof. The collar $\vartheta$ is not the Heaviside function $\theta$.

Use the ordinary second tensor derivative of the supplied quadratic insertion:

Use the symmetric quadratic source polarization

and count the symmetry factor $1/2!$ for the two identical edges once. These conventions match the retained XVI-J sector. Coupling and loop weights are not included in the coefficient defined below.

Define

The integral is distributional. The ambient advanced inverse of $\square$ is specified by the spatial Fourier multiplier

with the continuous value of the quotient at zero momentum. Equivalently, for compact sources,

For sources compactly supported in the slab interior, this restricts to the supplied upper limit $T$. The graph edges use the ambient kernel, not an arbitrary slab operator.

Set

The scalar $d$ is the massless positive-frequency time-ordered kernel in this normalization. It is the scalar kernel denoted by ${{\Delta }_F}^{\mathrm{FR}}$ in XVI-J.

For

the two summands are defined by

Both pairings converge absolutely. Moreover,

Proof. The partial Fourier transform of a compact smooth test is rapidly decreasing in $\mathbf{p}$, uniformly for $t$ in a compact interval. The weight $1/|\mathbf{p}|$ is locally integrable in three dimensions, and

This proves absolute convergence. The time restriction in (eq:E-fam) is a restriction of this integral; it does not require multiplying distributions at coincidence.

The temporal Fourier families of $d$ are

For $t<0$, this follows from $e^{-\mathrm{i}\omega t}+2\mathrm{i}\sin (\omega t)=e^{\mathrm{i}\omega t}$. The jump of its temporal derivative is $-\mathrm{i}$, so $\square d=-\mathrm{i}\delta$ and $\square \mathcal{S}=\mathrm{i}\delta$. The two branches, together with spatial parity, give symmetry under $z\mapsto -z$. Finally,

Neither $\mathcal{S}$ nor $D_F$ is an object of the later extension problem. They are already global distributions.

For a symmetric tensor $\gamma$, put

For $n=1,2,3$, define

These are the supplied XII-C unintegrated density atoms. They specify the density itself, not merely its Euler equivalence class.

The interaction used in this sector is

No integration by parts changes this density representative. The usual definition of distributional derivatives used later does not change that convention.

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

A diagonal basis matrix has its indicated entry equal to one. An off-diagonal matrix has both symmetric entries equal to one. Thus

Second jets are indexed by multi-indices, not by a redundant ordered pair of derivative indices. In particular, a mixed second derivative is one independent jet coordinate.

For the declared unintegrated density, there are unique coefficients ${L^{pq}}_{ij}(w;h)$ such that

The identity is an identity in independent field jets, before any integration by parts. Its properties are:

1. Each ${L^{pq}}_{ij}$ is linear, with constant rational coefficients, in residual jets of order exactly $2-|p|-|q|$.

2. Every nonzero monomial of the full Hessian has total derivative order two over $h,A,C$.

3. The slot symmetry is ${L^{pq}}_{ij}={L^{qp}}_{ji}$.

4. The constant-slot coefficient is

   $${L^{00}}_{ij}(w;h)=D^2{\ell }_3(h)[B_i,B_j](w)=L_{ij}(w;h),$$

   the constant-slot form retained by XVI-J.

5. The original vertex decomposition contains no derivative on ${\chi }_{\mathrm{I}}$.

These structural claims do not depend on a numerical coefficient list.

Proof. The atoms $u_i,b_i$ have field degree $i$ and derivative order zero. The atom ${\Gamma }_n$ has field degree $n$ and total derivative order one. Both $\partial {\Gamma }_n$ and the products in $R_n$ have field degree $n$ and derivative order two. Hence $r_n$ has field degree $n$, and each term of $r_3+b_1{r}_2+b_2{r}_1$ is cubic with total derivative order two.

The mixed variation of a cubic monomial replaces two field-jet factors by the corresponding $A$- and $C$-jets and leaves one residual $h$-jet. It preserves the total derivative count. The residual jet therefore has order $2-|p|-|q|$, proving the first two statements and the range in (eq:Lpq).

Uniqueness follows from independence of the symmetric field components and their multi-index jets. Symmetry of the mixed variation gives the third statement. Constant increments eliminate all terms with $|p|+|q|>0$, proving the fourth. The fifth follows because the switch multiplies the density and is not differentiated in this representation.

For a fully finite specification, let $\mathcal{L}((g_{i,r}{)}_{|r|\le 2})$ be the jet polynomial obtained from (eq:Gamma)–(eq:ell3). Then

This specifies every coefficient without an external table. No extra factorial is associated with a mixed derivative jet.

For functions depending only on $s=w^1$,

For $h=\psi \eta$, $A=u\eta$, and $C=v\eta$,

On $\psi =s^2$, this becomes

Proof. Substitution in the finite atoms gives

with $b_1=2\psi$ and $b_2={\psi }^2$. The ${\psi }^2{\psi }^{\prime \prime }$ terms cancel in $r_3+b_1{r}_2+b_2{r}_1$, leaving $\frac{3}{2}\psi ({\psi }^{\prime }{)}^2$. This proves the density formula. Taking the mixed variation of $-3(\psi +ϵu+\delta v)({\psi }^{\prime }+ϵu^{\prime }+\delta v^{\prime }{)}^2$ proves (eq:tracehessian).

This is a hand-derived check of differentiated slots and their signs. It does not verify all general component entries.

The general coefficient list and numerical regressions are documented in Appendix app:script as computational verification. They are not premises of the analytic existence theorem.

For a symmetric matrix $H$, evaluate $u_0,u_1,u_2,b_1,b_2$ on $H$, and introduce the formal polynomial

Define

Coefficient extraction in (eq:C2) is a finite polynomial operation; it is the retained XII-C coefficient.

For a symmetric covariant jet $\Sigma$, put

This polarization has no derivatives on $X,Y$. The actual jet is always $\Sigma (y)=\partial \partial {\sigma }_{\epsilon }(y)$.

The Gram matrix and its inverse are

For $B^i=(G^{-1}{)}^{ij}{B}_j$, one has

After removal of its scalar factor $\mathcal{S}$, the edge fiber tensor is

Proof. A diagonal basis element has squared pairing one. An off-diagonal $B_{(ab)}$ contributes two entries, giving $2{\eta }^{aa}{\eta }^{bb}$. Distinct basis matrices have disjoint component support. This proves the Gram matrix. The inverse-pairing identity $b^{-1}={\sum }_i{B}^i\otimes B_i$ then gives (eq:edgefiber).

The Hessian slots and source polarization are covariant bilinear forms on field directions. Each edge contracts their covector slots using the inverse fiber pairing. Thus the source receives $Z_i=-\mathcal{P}{B}^i$, while the vertex is expanded on $B_i$. The field basis is not self-dual.

Put $M_{\times }={\mathbb{R}}^4\setminus \{0\}$. The standard off-diagonal Feynman wavefront relation gives, at each singular $z\in M_{\times }$, only one time orientation for the nonzero covectors of $\mathcal{S}$. This is the relation stated after equation (77) in FR (Section 7); see also BDH (Proposition 26). No wavefront statement at $z=0$ is needed here.

For all multi-indices $p,q$,

exists, and ${q^{\circ }}_{pq}={q^{\circ }}_{qp}$.

For $r_2(y,w)=y-w$ and any distribution $u$ on the relative space,

Consequently, on $y\ne w$,

Proof. Derivatives do not enlarge wavefront sets. At a nonzero null point, two nonzero covectors in the relevant wavefront fibers have the same time orientation, and hence cannot sum to zero. At nonsingular points there is no obstruction. Hörmander's product criterion therefore defines (eq:qpq); see Hormander (Theorem 8.2.10) and BDH (Theorem 13). That product is commutative.

The relative map is a linear submersion. In coordinates $(y,z)$, $z=y-w$, its pullback is a distribution constant in the extra $y$-variable. Differentiation at fixed $y$ therefore gives ${\partial }_w=-{\partial }_z$, proving the distributional identities. Pullback by a submersion is compatible with the products when their wavefront criterion holds, yielding the final assertion.

For completeness, the use of the relative-space wavefront relation can be justified from the two-point relation: the two-point kernel is the pullback by $(x,y)\mapsto x-y$. Its wavefront covectors have the shape $(k,-k)$. Restriction to $(x,y)=(z,0)$ is permitted because the restriction conormal has shape $(0,k^{\prime })$, which cannot meet that set for nonzero $k$. The quoted off-diagonal orientation therefore transfers to $\mathcal{S}(z)$.

For $|p|+|q|\le 2$, set

The interaction switch appears exactly once in $a_{pq}$.

Slot symmetry and symmetry of $\mathfrak{s}$ imply ${\mathcal{K}}_{pq}={\mathcal{K}}_{qp}$. Let $\mathcal{N}$ be the set of unordered placements $\mathsf{k}=\{p,q\}$ with $|p|+|q|\le 2$, and put

There are $25$ unordered labels. This is an inventory of spanning labels, not a claim of linear independence of the scalar distributions.

The selected off-diagonal two-edge distribution is

It contains the full Hessian of the declared density, but only the $C_2$ source. Its overall numerical factor is $2/2!=1$. It is invariant under constant changes of fiber basis.

Proof. Contract two copies of (eq:edgefiber) with the source polarization and the full jet Hessian. The vertex-end derivatives fall on the scalar edge factors at $w$. Lemma lem:sign produces the sign in (eq:apq); the remaining fiber contraction is (eq:Kpq).

Assumption ass:normalization supplies the insertion factor two and the single identical-edge factor $1/2!$. The change of the vertex does not alter either factor. Grouping the two equal ordered terms when $p\ne q$ gives (eq:ak).

For a constant basis change ${\tilde{B}}_i=B_j{A}_{ji}$, the dual vectors transform as ${\tilde{B}}^i=(A^{-1}{)}_{ij}{B}^j$, while each Hessian matrix transforms as ${\tilde{L}}^{pq}=A^{\mathsf{T}}{L}^{pq}A$. The two inverse matrices in the source slots cancel the two matrices in the Hessian. Equivalently, the contraction is defined intrinsically by the fiber tensor (eq:edgefiber).

This proposition does not evaluate a smooth Hessian directly on singular fields without justification: the finite jet formula specifies the distributional expression, and Lemma lem:sign establishes every required punctured product.

The numerical kernels are translation invariant. Their coefficients generally depend on $y,w$ separately. The source jet depends on $y$; the residual field and interaction switch depend on $w$.

For a distribution $t$, write

The scaling degree at zero is

On the punctured domain, tests are restricted accordingly. This is the scaling-degree convention of BF (Section 5.1).

For positive dilations,

and therefore

Proof. For the scaled test,

In (eq:W-fam), substitute $t=au$, $\mathbf{p}=\mathbf{k}/a$. The time and momentum measures, the factor $1/|\mathbf{p}|$, and the test contribute $a$, $a^{-3}$, $a$, and $a^{-1}$, respectively. Their product is $a^{-2}$. The same substitution in (eq:E-fam) preserves $t<0$ and turns its sine quotient into $a\sin (-u|\mathbf{k}|)/|\mathbf{k}|$. It gives the same power. The derivative assertion follows from

For every $\mathsf{k}=\{p,q\}\in \mathcal{N}$,

For a nonzero kernel the scaling degree is exactly $4+N_{\mathsf{k}}$. The declared subtraction order is always

For a zero kernel, this declared upper bound is retained and its selected extension is zero; its scaling degree is not assigned a finite value.

Proof. Dilations are diffeomorphisms of the punctured domain and commute with the defined distributional product. Apply Lemma lem:homog to its two factors. A nonzero homogeneous distribution of degree $-4-N_{\mathsf{k}}$ has scaling degree $4+N_{\mathsf{k}}$. The zero distribution satisfies the upper bound.

The cardinalities count allowable placements. The supplied computational record further reports that every ordered placement has a nonzero Hessian coefficient; that stronger occurrence statement is not needed for the analytic bound.

No numerical kernel has more than two vertex derivatives. The preparation jet $\Sigma (y)$ stays in the coefficient and contributes no additional edge derivative.

For $k=4,5,6$, define

Then

On the punctured domain,

All three graded subspaces are nonzero.

Proof. Each element of ${\mathcal{H}}_k$ is homogeneous of degree $-k$. If $u_4+u_5+u_6=0$, dilation gives $a^{-4}{u}_4+a^{-5}{u}_5+a^{-6}{u}_6=0$ for every $a>0$. Three distinct dilations give an invertible Vandermonde system, proving directness.

The products satisfy the distributional product rule where their wavefront criterion holds, giving (eq:Leibniz1)–(eq:Leibniz2). By Lemma lem:families, $\square \mathcal{S}=\mathrm{i}\delta$; hence $\square \mathcal{S}=0$ on the punctured domain, which proves (eq:trace).

The massless time-ordered square is nonzero and has scaling degree four; this is also stated in FR (Section 7, following Theorem 7.2). Thus ${\mathcal{H}}_4=\mathrm{span}\{{\mathcal{S}}^2\}\ne 0$. If ${\mathcal{H}}_5=0$, all first derivatives of ${\mathcal{S}}^2$ vanish by (eq:Leibniz1). On the connected punctured space it is then constant, incompatible with nonzero homogeneity of degree $-4$. If ${\mathcal{H}}_6=0$, equation (eq:Leibniz2) similarly makes every $\mathcal{S}{\partial }_{\alpha }\mathcal{S}$ constant, and its degree $-5$ makes it zero. This contradicts ${\mathcal{H}}_5\ne 0$.

The inventory is not a basis. For example, (eq:trace) is a constant-coefficient relation within ${\mathcal{H}}_6$. No independence claim for the other spanning kernels is required below. The scaling bounds are numerical-kernel bounds, not a sharp scaling-degree statement for the complete decorated sum or the root-attached functional.

The prescription consists of:

1. the intrinsic dilation grading (eq:grading);

2. one fixed $\varpi \in {C_c}^{\infty }({\mathbb{R}}^4;\mathbb{R})$, $\varpi (0)=1$, identical to the subtraction function chosen in XVI-J;

3. the projections built from that function, one for each order $\rho =0,1,2$;

4. zero moments on the corresponding finite Taylor complement.

The extension acts only in $z=y-w$. Smooth coefficients are restored afterwards.

For $|\alpha |,|\beta |\le \rho$, set

Order multi-indices by increasing total degree. The matrix is triangular with diagonal entries one, because its entry vanishes unless $\alpha \le \beta$. Define

Then

Thus ${\mathsf{W}}_{\rho }$ projects onto

In particular,

The sans-serif projection ${\mathsf{W}}_0$ is distinct from the positive-frequency distribution $W_0$.

Choose any auxiliary cutoff $\zeta \in {C_c}^{\infty }({\mathbb{R}}^4)$ equal to one near zero. For $u\in {\mathcal{H}}_k$, $\rho =k-4$, define

Each test in the limit vanishes near zero, so its pairing uses only the punctured distribution.

The limit (eq:operational) exists, is independent of the auxiliary cutoff, and is continuous in the induced test topology. It is linear in $u$. It equals the action on ${\mathcal{D}}_{\rho }$ of any extension of $u$ with scaling degree at most $k$.

Proof. Fix a compact annulus separated from zero. The restriction of $u$ has a finite-order seminorm bound there. For $\psi \in {\mathcal{D}}_{\rho }$, Taylor's theorem and the chain rule give, on this fixed annulus,

for every fixed finite $m$, with bounds in finitely many seminorms of $\psi$. After scaling an annular pairing to this annulus, homogeneity supplies the factor $a^{4-k}$. The pairing is therefore

A dyadic decomposition near zero has summable bounds ${\sum }_{j\ge 0}O(2^{-j})$. This proves convergence and continuity. Differences of two scaled cutoff choices have annular support at scale $a$, so their action is also $O(a)$, proving independence.

The point-extension theorem provides an extension with scaling degree at most $k$ BF (Theorems 5.2–5.3). Any two such extensions differ by derivatives of $\delta$ of order at most $k-4=\rho$; these annihilate ${\mathcal{D}}_{\rho }$. Equivalently, for $|\beta |=\rho +1$, the distribution $z^{\beta }u$ has scaling degree at most $3<4$ and has a unique extension. Its action reconstructs the action on jet-vanishing tests, as in the proof of BF (Theorem 5.3). The cutoff construction realizes that same action.

For $u\in {\mathcal{H}}_k$, $\rho =k-4$, define

For $u=u_4+u_5+u_6\in \mathcal{V}$, set ${\mathfrak{R}}_{\mathrm{K}}(u)={{\sum }_{k=4}}^6{\mathfrak{R}}_{\mathrm{K}}(u_k)$.

The cutoff limit is a distributional formula; it is not an ordinary integral of the singular kernel over its light cone. The auxiliary cutoff is not an additional normalization datum. The fixed $\varpi$ and the moments are.

The map ${\mathfrak{R}}_{\mathrm{K}}:\mathcal{V}\to {\mathcal{D}}^{\prime }({\mathbb{R}}^4)$ is linear, well defined, and satisfies

For $u\in {\mathcal{H}}_k$,

These conditions determine the extension uniquely.

Every constant-coefficient relation among the punctured kernels is preserved:

In particular, the extended version of (eq:trace) is zero.

Finally,

the selected square extension of XVI-J, with the same $\varpi$, order-zero projection, and zero moment.

Proof. Lemma lem:vanishing and the continuous projection give a continuous distributional action, linear in $u$. Tests supported away from zero have zero Taylor jet and recover the punctured distribution. The direct grading makes the definition on $\mathcal{V}$ unambiguous.

The construction is the zero-moment point extension with projection (eq:Wrho); the scaling bound follows from BF (Theorem 5.3 and equations (38)–(39)). The moments hold because ${\mathsf{W}}_{\rho }{w^{(\rho )}}_{\alpha }=0$. For a nonzero $u\in {\mathcal{H}}_k$, its restriction already has scaling degree $k$, so the extension has degree exactly $k$. For $u=0$, the selected extension is zero.

Differences of extensions with the stated bound are point-supported distributions

Pairing with the dual Taylor tests makes all their coefficients zero if the moments agree, proving uniqueness. Linearity proves (eq:coherence), independently of any basis chosen to compute the finite span.

For ${\mathcal{S}}^2$, the data are exactly (eq:W0) and $⟨{\mathfrak{R}}_{\mathrm{K}}({\mathcal{S}}^2),\varpi ⟩=0$. Uniqueness at scaling degree four therefore proves (eq:Jextension) on all of ${\mathbb{R}}^4$.

A basis adapted to ${\mathcal{H}}_4,{\mathcal{H}}_5,{\mathcal{H}}_6$ may be used for bookkeeping. It is not a further choice in the prescription.

This coherence concerns constant linear relations on $\mathcal{V}$. It does not assert compatibility with arbitrary smooth-multiplier identities or with alternative integration-by-parts presentations of the decorated sector. The density, derivative placement, and order of operations are fixed.

Define

It is supported at zero and has the form

Similarly,

is supported at zero, and