Landmark IX-Free Actual-functional binary factorization of one sharp ordered word

We advance the actual-functional factorization problem for $B_f,W_g,B_{f^{\prime }}$, retaining the IX-A carrier and the IX-C interfaces. We identify the concrete presentation-relation kernels and prove that the fixed mixed kernel $K$ has full distributional support. A compact-primitive argument then proves injectivity of both partial-contraction presentation maps. Consequently the algebraic scalar relation tests impose no restriction on the real parent parameter: ${\mathcal{C}}_{\mathrm{desc}}=\mathbb{R}$.

An explicit leaf-covector argument constructs the three canonical two-edge distributions. Their compact polynomial kernels exist, and both metric-quadratic outputs satisfy the full carrier admission test, including directional descendants. This constructs the second-stage coefficient maps on the degree-two intermediate blocks. Reversal and fixed-support test-family continuity are proved on these blocks.

The release remains unresolved. Descent of the prescribed degree-four second-stage coefficients and agreement on actual-pair and old-product overlaps are not established. Thus neither complete sharp association is released. The triangle calibration remains conditional on the explicitly copied analytical premise reported in IX-C. No universal obstruction, global quantum algebra, or IX-D construction is claimed.

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

Figure fig:scope-IX-Free records that separation. The drawing adds no hypothesis.

and proof boundary{ Release: Unresolved. This manuscript proves the new partial results stated above. It does not supply the remaining degree-four descent and overlap proofs by declaring a common graph sum to be a binary product.

A missing map is not a nonexistence theorem. The results below are relative to the explicitly stated inherited premises. The full IX-A, IX-B, and VII-D proofs are not independently certified here.}

The selected word is

The target is binary factorization through order ${\hslash }^3$ on explicitly named actual-functional subspaces, with constant joint-test kernel ${u_{\mu }}^{\rho }+cr$ for one fixed $c\in \mathbb{R}$.

A release is Constructed only after every required interface is discharged. It is Proved obstructed only if an explicit contradiction excludes every construction satisfying those interfaces. Otherwise it is Unresolved. These alternatives refer to this sector and contract, not to every enlargement or normalization.

Let

Write $x=(t,\mathbf{x})$, $y=(s,\mathbf{y})$, and $x^{\prime }=(t^{\prime },{\mathbf{x}}^{\prime })$. Set

The operator ${\mathsf{E}}_0$ is the zero-value, zero-first-time-derivative Cauchy inverse at $t=0$. Write $w=Lh$. Configurations $\psi =(h,\varphi )$ and configuration directions are smooth and off shell.

The selected functionals are

Repeated bundle indices are summed.

We use the ordinary smooth mapping properties of these fixed Cauchy and Green operators. Smooth compact sources give smooth solutions; restriction to a fixed compact set is continuous in the source; and finite propagation over bounded time intervals gives compact propagation regions. In particular, applying $L$ in one variable of a smooth joint function is continuous in the smooth topology, including derivatives in its parameter variables. These are smooth linear PDE properties, not singular-extension assumptions.

The supplied source is IX-C IXC. Its statements concerning IX-A, IX-B, and VII-D are used below in the following explicit form. They are premises, not newly verified companion proofs.

The carrier consists of actual finite polynomial functionals with compact symmetric distributional kernels satisfying

Here

The space ${\mathcal{M}}_1$ contains compact smooth linear metric kernels and has no nonzero nonspacelike wavefront covectors.

The topology ${\tau }_{\Sigma }$ records strong compact-distribution coefficients and every intrinsic trace coordinate

Classical multiplication, configuration-direction differentiation, and conjugation preserve ${\mathcal{P}}_{\Sigma }$. Classical multiplication is separately continuous. No quantum multiplication is included.

The generating test maps are injective. Their fixed-support test-family continuity and the fixed-support first-stage continuity reported in IX-C are inherited. For nonzero real $f$,

The sharp adjoint identity is

For the prescribed regulator placement,

There are distributions ${u_{\mu }}^{\rho },r\in {\mathcal{D}}^{\prime }(X)$ with

For each fixed compact $K_0\subset X$,

The estimate includes compacts meeting $\mathrm{Rel}\cap {\Delta }_{13}$. Parent conversion is exactly $c_{ab}r$, without an additional normal derivative, nonconstant weight, or intersection term.

The parent distributions satisfy dagger symmetry. Both regulated associations equal the same regulated word and have the same subtracted constant limit.

The following first stages are already constructed:

The three one-edge outputs in (eq:oneedges) are also already constructed and admitted.

Fix $\rho ,\mu$ once and define

The same $c$ is used for every test, presentation, association, and declared observation derivative. The scalar class $\{U_c:c\in \mathbb{R}\}$ is not an ambiguity classification for nonconstant outputs.

A completed construction must satisfy:

1. Every output belongs to ${\mathcal{P}}_{\Sigma }$, with all required metric derivatives and directional descendants.

2. Order zero is classical pointwise multiplication.

3. Every specified old-bulk overlap agrees with ${\star }_H$.

4. The first stages are locked to (eq:Q) and (eq:QR).

5. The constant kernel is $U_c$, with one real parent parameter transported by declared observation compatibility.

6. No subtraction changes an already fixed old bulk product.

7. Unchanged approximants need not converge sharply in ${\tau }_{\Sigma }$.

8. Every finite relation between actual presentations is respected.

9. Every continuity claim has a named domain and proof. Test-family continuity is not carrier-stage hypocontinuity.

The inherited preservation obligations, where a product has actually been defined, are

Here $\mathrm{cl}$ denotes the order-zero projection. These are obligations, not conclusions about an unconstructed product.

Let

and set

Define

The direct sums are by actual homogeneous field degree, four and two. The underlying spaces ${\mathcal{L}}_4,{\mathcal{R}}_4$ coincide, but their ordered roles differ.

All presentation tensor products are algebraic. Put

and define

The factorization diagram is

where $v=p_L$ or $q_L$. There is a reversed diagram on the right. The dashed arrow exists exactly when the labelled prescription annihilates the displayed relation space.

Overlap conditions occur inside ${\mathcal{P}}_{\Sigma }\otimes {\mathcal{P}}_{\Sigma }$, on

No presentation map is assumed to be a topological quotient.

The fixed mixed kernel is

The low-frequency amplitudes are locally integrable in three spatial dimensions. The high-frequency terms are oscillatory distributions.

For $v={\sum }_i{\alpha }_i{f}_i\otimes g_i\in {\mathcal{E}}_L$, write

This identifies the algebraic tensor product with a subspace of ${C_c}^{\infty }(M_{+}^2)$. Injectivity follows by choosing a representation with linearly independent first factors and applying linear functionals on their finite-dimensional span to isolate each second factor.

Because $v$ has support compactly inside $M_{+}^2$, distributional observation-variable integration by parts gives

A cutoff equal to one on the coefficient support is understood when the configuration is not compactly supported.

For $v\in {\mathcal{E}}_L$,

For $v_R\in {\mathcal{E}}_R$, the right equation is

Proof. The scalar slot and the $00$-metric slot are independent. Differentiating the mixed polynomial once in each field direction recovers its coefficient distribution on product tests, with the normalization $2$ in (eq:qkernel). Finite sums of product tests are dense in the joint test space, so these pairings determine the distribution. The converse is immediate. The right argument is identical.

There is no surface term in (eq:qkernel), since its observation coefficient is supported away from $t=0$. This does not authorize omission of the delta term in the different operation (eq:boundary).

Let ${J^{\mu }}_{\alpha \beta }(x,z)$ denote the distribution kernel of $L$:

Let ${e_{00}}^{\alpha \beta }{h}_{\alpha \beta }=h_{00}$. For $v\in {\mathcal{E}}_L$, define

The symmetrization is the average over the three metric slots. Before smooth multiplication, $J(x,z_1)$ and $\delta (z_2-z_3)$ are tensor factors in disjoint variables. Thus (eq:pkernel) is a defined distribution.

Observation support, finite propagation, and the delta relation give a common compact support for its field variables.

With the coefficient convention

one has

For $v_R(y,x^{\prime })$, let $\tilde{v}(x^{\prime },y)=v_R(y,x^{\prime })$. Then

Proof. Insert the kernel of $L$ in (eq:generators), multiply by $W_g$, and integrate the scalar observation derivative by parts. The original insertion sign changes the coefficient to the positive derivative in (eq:pkernel).

Differentiating once in the scalar direction and three times in independent metric directions recovers the symmetric coefficient kernel, multiplied by $3!$. Polarization and density of product tests prove uniqueness. The right formula uses only pointwise commutativity of the original classical factors.

This characterizes $\ker p_L,\ker p_R$ concretely. It does not prove that either kernel is zero.

The next argument evaluates the actual scalar relation tests. It does not merely repeat their affine classification.

The distribution in (eq:K) satisfies

Proof. Write $a=t-s$ and $\mathbf{z}=\mathbf{x}-\mathbf{y}$. The Fourier normalization is

Applying ${\square }_x$ to (eq:K) is legitimate distributionally. The two terms without the explicit factor $t$ are homogeneous wave solutions. Moreover,

Consequently

For completeness, damping the radial integral by $e^{-\delta |p|}$, $\delta >0$, and performing its angular integration gives

On the open set $a\ne 0$, $|\mathbf{z}{|}^2\ne a^2$, its limiting derivative is

which is nonzero. That open set is dense. Its inverse image in $M_{+}^2$ is also dense, since the difference map is open.

Thus the right-hand side of (eq:boxK) has full support on $M_{+}^2$. A differential operator cannot enlarge distributional support, so $K$ has full support.

Let $I=(0,\infty )$, let $N$ be a manifold, and let $T\in {\mathcal{E}}^{\prime }(I\times N)$. If ${\partial }_{t}T=0$, then $T=0$.

Proof. A distribution with zero derivative in a connected interval is constant in that variable. Here is the product-test argument. Choose $\chi \in {C_c}^{\infty }(I)$ with $\int \chi =1$. For every $\eta \in {C_c}^{\infty }(I)$, $\eta -(\int \eta )\chi$ has integral zero and is the derivative of a compactly supported smooth function in $I$. Hence

Choose $\eta$ with integral one and support in an interval disjoint from the time projection of $\mathrm{supp}T$. The left-hand side is zero, so every pairing on the right is zero. Product-test density gives $T=0$.

The maps

are linear isomorphisms onto their actual-functional images.

Proof. If $q_L(v)=0$, Proposition prop:relations gives ${\partial }_t(vK)=0$. The distribution $vK$ is compactly supported in $M_{+}^2$. Lemma lem:primitive gives $vK=0$.

If the smooth function $v$ were nonzero at a point, it would have a smooth reciprocal on a nonempty neighborhood. There $K=v^{-1}(vK)=0$, contradicting Lemma lem:fullsupport. Thus $v=0$, and the algebraic tensor embedding proves injectivity. Surjectivity onto the named image is its definition.

For the right map, use (eq:rightrelation); conjugation and exchange of variables preserve full support.

Define the combined scalar relation space

Pairing its joint tests with ${u_{\mu }}^{\rho },r$ defines $\alpha ,\beta$, respectively.

For the actual presentation maps of this manuscript,

For every fixed real $c$, the calibrated left and right scalar maps exist uniquely as algebraic bilinear maps:

Proof. Theorem thm:qinjective makes the two relation kernels zero. The sharp generating map is injective by the inherited premise. Multilinearity of distributional pairing therefore defines unique bilinear maps on the actual images. Constants belong to ${\mathcal{P}}_{\Sigma }$.

The conclusion is algebraic scalar descent, not permission to use every $c$ in a completed bulk-relative product. Old-product overlap conditions may impose additional restrictions. No such restriction is evaluated here.

Put

We use the standard product, submersion-pullback, and proper pushforward calculus for distributions Hormander,Demanet. The graph-specific covector checks are proved below.

Every wavefront element of $K$, $\overline{K}$, or $S_{\mu \nu }$ has nonzero covectors in both endpoint slots. For the Euclidean norm on coordinate covectors, the two endpoint norms are equal.

Proof. Split each Fourier integral into low and high frequencies with a smooth radial cutoff. The low-frequency part of $K$ is smooth: its amplitudes are locally integrable, and every coordinate derivative inserts an integrable bounded-frequency factor. The same observation applies to ${\Delta }_m^{+}$ and its finite derivatives.

The high-frequency phases of $K$ are

Their endpoint covectors are, respectively,

They are nonzero at both endpoints and have equal norms. The usual integration-by-parts estimate for oscillatory integrals excludes covectors outside these phase-generated cones. Smooth amplitudes and differentiation do not enlarge the resulting wavefront bound.

For the massive scalar kernel the high-frequency phase is

Its conic limiting endpoint directions are opposite nonzero null covectors with equal norms. Differentiation to obtain $S_{\mu \nu }$ does not enlarge this bound. Conjugation reverses covectors and preserves the assertion.

The comparison of endpoint norms also excludes spurious one-slot directions arising from taking closures of sums: if one endpoint contribution tends to zero, so does the other contribution of that edge.

Pull back two endpoint kernels satisfying Lemma lem:endpoints to the three-vertex variables of a two-edge tree. Their distributional product exists.

After multiplication by a compact smooth joint coefficient, pushforward integrating all vertices except any one vertex is smooth.

Proof. Each of the two edges has a leaf belonging to no other edge. A nonzero covector from that edge has a nonzero component at that leaf, whereas the other edge has zero component there. Thus the two pulled-back wavefront covectors cannot sum to zero. The distributional product criterion applies.

For the second assertion, the pushforward wavefront bound requires zero covectors in all integrated slots. Remove an integrated leaf different from the retained vertex. Its covector forces its incident edge to be inactive, meaning that the edge contributes the zero covector in the augmented wavefront bound. Repeat on the remaining edge. Both edges must be inactive, so no nonzero covector remains at the retained vertex.

If the retained vertex is the central vertex, remove the two integrated leaves immediately. If it is a leaf, remove the other leaf first and then the integrated central vertex. The endpoint norm comparison justifies the same reasoning for closed conic limits. Compact multiplication makes the pushforward proper. Its wavefront set is empty, so it is smooth.

This is the leaf-slot argument required by the forest structure. The statement “the graph is a tree” alone would not suffice.

Let $V=f\otimes g\otimes f^{\prime }$. Define compact distributions on $X$:

Each product in (eq:E1223)–(eq:E1323) exists canonically as a distribution on all of $X$. Each has assignment multiplicity $2$, positive insertion sign, and residual field degree two.

Proof. For (eq:E1223), the leaves $x,x^{\prime }$ give the product criterion. For (eq:E1213), the leaves are $y,x^{\prime }$. For (eq:E1323), they are $x,y$. Lemma lem:forest applies in each case.

Both sharp insertions contribute a minus sign, whose product is positive. There are two assignments of the middle metric slots, hence the displayed factor $2$, not $4$. Two edges remove four slots from the initial six.

Let ${\pi }_{ab}$ denote projection onto the indicated vertices. The outputs are the actual quadratic functionals

For (eq:G1223), the compact scalar coefficient kernel is

There is no sign change from two scalar integrations by parts.

For (eq:G1213), first form the compact residual kernel

then transpose $L$ in its second variable and symmetrize the two metric slots. For (eq:G1323), use

and transpose $L$ in its first variable.

These transpositions are defined on compact distributions. Indeed, applying $L$ to a smooth joint test function in one variable is continuous in the joint smooth topology. Its transpose therefore acts on compact distributions. Finite propagation supplies a common compact field support. This construction does not complete any presentation space or invoke a quotient topology.

The products just constructed need no extension across an excluded set: their wavefront products exist on all of $X$. This selects a canonical graph prescription. It does not classify all nonconstant terms permitted by the copied interfaces, nor impose a new prohibition on alternative contact terms.

The following failure language remains in force:

1. Failure of a sufficient wavefront criterion means that proof route fails.

2. An unestablished regulated limit is unresolved.

3. Failure of one extension excludes that extension.

4. Nonexistence requires a separate argument.

5. Failure of one graph prescription is not, by itself, incompatibility with all copied interfaces.

The functionals (eq:G1223)–(eq:G1323) belong to ${\mathcal{P}}_{\Sigma }$, including every metric admission test after configuration-direction differentiation.

Proof. The scalar-only output ${\mathsf{G}}_{12,23}$ has the compact scalar kernel already constructed. Its metric derivatives and their directional descendants vanish.

Consider $G={\mathsf{G}}_{12,13}$. Its polarized bilinear form is

Thus $G(h)=b_G(h,h)$ and ${G^{\prime }}_h(\psi )[k]=2b_G(h,k)$.

Define

These formulas mean the corresponding proper distributional pushforwards. By Lemma lem:forest, both are smooth: they integrate the compact weighted tree except for a single vertex. They are compactly supported in $\mathrm{supp}g$ and $\mathrm{supp}{f}^{\prime }$, respectively.

The metric derivative is therefore the actual linear kernel

Its first summand is compact smooth and belongs to ${\mathcal{M}}_1$; its second belongs to ${\mathcal{Q}}_{\Sigma }$.

For one configuration direction $\zeta$, the same formulas hold with $h$ replaced by ${\zeta }_h$ and $w$ replaced by $L{\zeta }_h$. This is the configuration-independent metric derivative of ${\mathrm{D}}_{\zeta }G$. Two directional derivatives of $G$ are constant functionals, whose metric derivative is zero. Higher derivatives vanish. Scalar directions have no effect. This checks every finite direction list.

For $G={\mathsf{G}}_{13,23}$, the corresponding smooth compact functions are

Again its derivative is $e_{00}{m}_{\psi }+q_{a_{\psi }}$, and the same finite-degree argument checks all descendants.

No inference from a single nonspacelike covector was used. Membership was proved by exhibiting the required decomposition. Conversely, an admission obstruction would have to exclude every ${\mathcal{M}}_1+{\mathcal{Q}}_{\Sigma }$ decomposition, not merely one proposed decomposition.

The inherited one-edge outputs are

where

For each fixed real $c$, the following prescriptions define unique bilinear maps on actual functionals:

Their domains are ${\mathcal{L}}_2\times \mathcal{B}$ and $\mathcal{B}\times {\mathcal{R}}_2$, respectively, and their targets are ${\mathcal{P}}_{\Sigma }$.

Proof. Theorem thm:qinjective gives a unique algebraic presentation tensor for each intermediate argument. The sharp generating map gives a unique test for the other argument. The displayed graph sums are linear in the presentation tensor and in the sharp test, so they descend to bilinear maps. In particular, the coefficient sum, not merely its individual terms, is well defined.

Theorem thm:treeadmission places the positive-order nonconstant outputs in ${\mathcal{P}}_{\Sigma }$. Corollary cor:cdesc handles the scalar coefficient. Classical closure handles order zero.

These are constructed coefficient maps on named blocks. They are not yet a bulk-relative binary product on the sum of all requested coefficient domains.

The degree-four prescriptions remain

Their values now exist and are admitted. The unresolved question is whether they factor through $p_L,p_R$.

The prescriptions (eq:lambdaL) define maps on ${\mathcal{L}}_4\times \mathcal{B}$ exactly when

The right prescriptions define maps on $\mathcal{B}\times {\mathcal{R}}_4$ exactly when

Here the relation kernels are precisely the distributional equations in Proposition prop:prelations.

Proof. Necessity follows by evaluating a proposed map at a zero actual argument. For sufficiency, choose a presentation of the actual intermediate argument. Two choices differ by the relevant relation kernel, which the hypothesis annihilates. Bilinearity and uniqueness then follow.

These tests have not been discharged here. In particular, injectivity of $q_L,q_R$ does not imply injectivity of $p_L,p_R$.

For vector subspaces $U,U^{\prime },V,V^{\prime }$, all contained in fixed ambient spaces, algebraic tensor products satisfy

One proof is to extend bases of the intersections to bases of the sums and compare the resulting tensor coordinates.

Field-degree bigrading shows that a left degree-four block and a right degree-four block have zero intersection: their argument degrees are $(4,2)$ and $(2,4)$. The mixed degree-four/degree-two intersections are also zero.

The potentially nontrivial left/right degree-two intersection is

No vanishing of the two factors is proved here.

For each new block $U\otimes V$, its old-product overlap is

Thus the old-product obligation is a concrete value-preservation test, not a convergence assertion.

Linear coefficient prescriptions on two subspaces glue to their sum if and only if they agree on the intersection. For three subspaces, each next prescription must agree on its intersection with the sum of the preceding domains. Pairwise agreement alone need not suffice.

Proof. For two domains define the proposed value on $d_1+d_2$ as the sum of the prescribed values. Independence of decomposition is equivalent to agreement on the intersection. For three domains, apply the two-domain statement to the already glued sum and the third domain.

The last qualification prevents a hidden multi-domain relation from being omitted. The requisite agreement on (eq:newoverlap), (eq:oldoverlap), and the relevant sum intersections remains unresolved.

Write

Under the selected contraction-degree prescription, a completed candidate would have

The residual degrees are $6,4,2,0$.

The absence of a third contraction against a degree-two sharp factor is the selected graph prescription. It is not a theorem forbidding every possible additional contact term.

Suppose the tests in Proposition prop:remainingdescent hold and the coefficient prescriptions glue consistently with their specified overlaps. With the same fixed real $c$, the two associations then agree through order ${\hslash }^3$. Their common coefficients are

Proof. Order zero is classical associativity. At the three positive orders, substitute (eq:Ldegree2)–(eq:Rdegree2scalar) and the descended versions of (eq:lambdaL)–(eq:lambdaR). This gives, respectively,

Each is an equality of the displayed actual polynomial functionals. Equality of all four coordinates is equality in ${{\mathcal{V}}_{\Sigma }}^{[3]}$.

This proposition does not construct its degree-four maps. Writing the common graph sum cannot replace the missing factorization and gluing proofs.

A nonconstant discrepancy cannot be absorbed into $cr$: the order-one and order-two coordinates are separate from the constant order-three coordinate.

For functionals, star means complex conjugation. For a joint distribution, dagger conjugates and exchanges $x,x^{\prime }$:

The two-edge outputs satisfy

The constructed degree-two coefficient maps satisfy

Proof. The free scalar kernel has

Conjugation of a defined wavefront product is the product of the conjugated factors. Exchanging the endpoint variables therefore gives the two displayed graph identities.

The locked first-stage coefficients obey

The graph identities give reversal at order one. At order two, the inherited parent symmetry and real $c$ give