II-B: Finite contact-span certificates on a declared conical Enter spacetime A Grade S operational adaptation

We give a finite operational adaptation of the contact-span certificate of Manuscript II on the prescribed conical spacetime of I-B. The classical background, the flat-space response prescription, and the contact class remain separate inputs. We reproduce the finite algebraic proof: the three-label target is $m_S=(0,16/3,16/3{)}^T$, the complete redundant catalogue of $15120$ parity-even two-derivative quartic contacts has evaluated rank $2$, two original columns have minor $64$, and the constant coefficients $2/3,2/3$ fit the target.

A complete destructive two-qubit clock instrument supplies a classical code. An explicit dictionary assigns its four outcomes to the three external labels, with the fourth outcome completely specified rather than discarded. We prove preservation of the finite certificate under this dictionary. We also construct a bounded finite response-table probe whose centered binary instruments realize the evaluated response space on arbitrary reference-entangled inputs. This is a lookup-table realization, not a measurement construction for a gravitational field.

A target-adapted SWAP chain transports the surviving payload with exact reference preservation and a finite resource ledger. Complete scheduled dependencies lie inside the prescribed metric cone. The resulting model satisfies the explicitly defined Grade S contract. The finite fit remains $\mathrm{SEP}$-finite: no global contact identity, derived geometry, sourced Einstein dynamics, continuum field limit, or inherited quantum coefficient is obtained.

release{ The construction consists of:

Here $\mathsf{Grade}\mathsf{S}$ is a contract defined in this paper, not an external accreditation. The algebraic status is $\mathrm{SEP}$-finite.}

boundary Enter consumes a finite clock encoding, not the geometric time axis. A clock record becomes a contact label only after the dictionary is declared. A table-probe residual is not a stress tensor. The flat-space contact catalogue is not silently replaced by a curved-background or weighted catalogue.

Manuscript II solves an exact finite contact-membership problem for one two-species massless spin-two tree CheungII. Its conclusion concerns response functions evaluated on three specified external labels. I-B supplies a different object: a finite experiment with destructive clock readout on a prescribed conical Lorentzian product CheungIB.

The present paper constructs an explicit interface between these objects. It does not replace the action of II by an action on the cone. In particular, it does not assert that the globally defined plane waves or propagator of the flat-space calculation are scattering data on a conical spacetime.

The interface has three parts:

1. a classical dictionary from a clock code to an external label;

2. a pullback of the already defined response functions along that dictionary;

3. a finite operational probe realizing the resulting numerical table as a bounded response signature.

The third part makes the operational model nonempty without pretending to construct a gravitational apparatus.

Cheung's Enter calculus treats quantum names as access rights and destructive readout as an operation that consumes the measured names while returning a classical result CheungEnter,CheungI. The introductory philosophy distinguishes destructive readout, quantum recovery, implementation, and geometric identification CheungPhil.

We retain that discipline. Agreement of record probabilities is weaker than agreement of instruments; a channel alone does not specify a thermal machine; and naming a register “clock” does not identify it with a continuum coordinate. We adopt finite Born and projective-update semantics. We do not rederive the separate measurement-characterization theorem from an abbreviated hypothesis list.

The instrument terminology is consistent with the operational framework of Davies and Lewis and with finite completely positive maps DaviesLewis,Choi. Every channel identity used below is proved directly.

A $\mathsf{Grade}\mathsf{S}$ adaptation in this paper consists of:

1. a prescribed Lorentzian background and a regular local rod chart;

2. a finite named experiment, specified accessible algebras, a complete instruction tree, and an explicit resource ledger;

3. a complete destructive clock instrument and a declared record, rod, and causal dictionary;

4. exact evaluation of the frozen finite response certificate, together with a stated operational realization of its numerical signature;

5. a reference-preserving transport channel and a proof that all admitted scheduled dependencies are causally embedded.

The contract does not require the geometry, the response prescription, or the contact class to be reconstructed from records.

The stronger phrase $\mathsf{Grade}\mathsf{E}$ will denote the unconstructed emergence target in which suitable operational data determine the relevant geometric and dynamical structures, with the comparison and uniqueness obligations explicitly supplied. Merely satisfying Definition def:grade does not satisfy that target.

No companion theorem is needed as an unstated premise. The finite algebraic and operational assertions used in the release are defined and proved below.

Fix

Let

with spatial metric

Its metric completion has an apex $o$. Set

and prescribe

The singular completion locus is

It is a worldsheet. Fixing $z$ gives one worldline.

We work on the smooth regular part $M$. We do not assign an ordinary pointwise curvature tensor at the apex. This is the prescribed-background convention of I-B CheungIB.

There is a regular spatial neighborhood $U$ with a developing chart

such that $V$ is convex and

Choose ${\partial }_T$ future directed. Then:

1. a future-causal tangent $v=a{\partial }_T+w\ne 0$ satisfies

   $$a>0,|w{|}_h\le ca;$$

2. events $(T,x)$, $(T+\Delta T,x^{\prime })$ in this chart are joined by a future-causal straight segment whenever

   $$\Delta T>0,{\Vert x^{\prime }-x\Vert }_2\le c\Delta T;$$

3. along a static observer, $\mathrm{d}\tau =\mathrm{d}T$.

Proof. On a sufficiently small lifted angular sector put

Its spatial metric is Euclidean. Restrict to the inverse image of a sufficiently small convex ball avoiding the apex and the sector boundary.

The causal inequality follows from

The straight segment has constant developed velocity $(x^{\prime }-x)/\Delta T$, proving the second assertion. For a static curve, $g=-c^2\mathrm{d}{T}^2$, proving the third.

The natural orthonormal frame on this patch is

Thus the auxiliary Minkowski convention used below is compatible with the local frame:

This compatibility identifies tensor components at declared ports. It does not construct global field solutions on the cone.

The circumference of $r=R$ is $2\pi \alpha R$. Hence the global deficit is $\mu$. Locally, every sufficiently small regular patch is flat. Neither fact is inferred from a quantum record.

No source map, metric-update rule, or matter action on $(M,g)$ is included in the geometric specification. In particular, the two species of the following algebraic problem do not become two dynamically coupled metrics on $M$.

All response functions in this section are defined on an auxiliary flat-space domain $\mathcal{X}$. Fourier waves are $e^{ik\cdot x}$, all momenta are incoming, and index operations use $\eta$.

A label specifies:

1. species $J_i\in \{1,2\}$, $i=1,\dots ,4$;

2. nonzero real null momenta satisfying

   $$\sum _i{k}_i=0,s=(k_1+k_2{)}^2,t=(k_1+k_3{)}^2,u=(k_1+k_4{)}^2,stu\ne 0;$$

3. one replaced leg $j$ and a real vector ${\xi }_j$;

4. real symmetric physical tensors ${ϵ}_i$, $i\ne j$, and timelike references $t_i$, satisfying

   $${k_i}^{\mu }{ϵ}_{i\mu \nu }=0,{t_i}^{\mu }{ϵ}_{i\mu \nu }=0,{\eta }^{\mu \nu }{ϵ}_{i\mu \nu }=0.$$

The replaced tensor is

No condition $k\cdot \xi =0$ is imposed. Simultaneous permutations of all external data are allowed.

These are formal vertex-evaluation labels. Their later assignment to local rods does not assert that they are compact wave packets, conical asymptotic states, or apparatus preparations.

For one species define

This is the massless spin-two quadratic normalization FierzPauli. Its Hessian is

Use the fixed propagator coefficient

For $p^2\ne 0$ and $p^{\mu }{J}_{\mu \nu }=0$,

Also $K_{0}G(k,\xi )=0$.

Proof. Write $J_{\mathrm{tr}}={\eta }^{\mu \nu }{J}_{\mu \nu }$. For $h=DJ$,

Substitution into the Fourier transform of (eq:K0) cancels the trace and longitudinal terms. The remainder is $J$. Direct substitution of (eq:G) proves the second identity.

No inverse identity on arbitrary nonconserved tensors is used.

Fix the curvature sign and coefficients by the first-derivative density

whose integrated expansion, modulo divergences, is

For homogeneous $E_n$, set

Thus $v_n(u_1,\dots ,u_n)$ is the coefficient of $z_1\cdots z_n$ in $E_n[{\sum }_i{z}_i{u}_i]$. There is no additional factorial.

Let $d_{IJK}$ be real and completely symmetric. Define

and

The reference quartic is

With an interaction parameter $\lambda$, the locked action is

Every ${\theta }_{\ell }$ is constant on $\mathcal{X}$.

Einstein self-coupling and local deformation theory provide context Deser,BarnichHenneaux,BoulangerEtAl; no classification theorem from those works replaces the calculation below.

For a gauge parameter $\zeta$, let

The variation $\delta h=G_{\zeta }+\kappa R_{\zeta }h$ is the scaled Lie derivative of the metric. Invariance of the integrated Einstein density modulo a divergence gives, after polarization,

The pairings mean integrated coefficients or stripped conserved-momentum coefficients. The identities follow by differentiating ${\delta }_0{E}_3+{\delta }_1{S}_2=0$ and ${\delta }_0{E}_4+{\delta }_1{E}_3=0$, respectively.

Define the cubic current by

For free $v,w$, (eq:ward3) with arbitrary $\zeta$ implies conservation of $J(v,w)$. Lemma lem:inverse therefore applies at the physical end of each exchange.

The stripped tree prescription is

The common momentum delta, ${\lambda }^2$, and overall Feynman factor $i$ are removed.

Take leg one replaced. Put

and define

The momentum of ${\mathcal{R}}_{1i}$ is $k_1+k_i$; that insertion is generally off shell.

Write

and $\overline{D}=(D_s+D_t+D_u)/3$.

The single-longitudinal reference response is

Proof. Equations (eq:ward3) and Lemma lem:inverse give, channel by channel,

The tree minus sign therefore makes the exchange contributions $D_s{W}_2+D_t{W}_3+D_u{W}_4$. The quartic contributes $-\overline{D}(W_2+W_3+W_4)$ by (eq:ward4).

Define

and let $d_{\ast }$ have $d_{112}=1$ and all other independent components zero. Then $q(d_{\ast })=1$. Put $m=F(d_{\ast })$.

For species $1122$ or $2211$,

so

Other two-and-two placements permute the channel coefficients. For $1111,2222,1112,1222$, up to permutation, the three channel coefficients agree, so $F(d)=0$. Consequently the complete response has the global factorization

This factorization is not a global contact identity.

Let $\mathcal{V}$ be the real vector space of local quartic density classes with exactly two derivatives, constant coefficients, no extra fields, no inverse operators, and no background tensors other than $\eta$ and the orientation tensor. The densities are translation invariant, Lorentz scalar, and parity even. The quotient is by total derivatives and genuine tensor identities, not by equations of motion or field redefinitions.

For a species word $\mathbf{I}=(I_1,I_2,I_3,I_4)\in \{1,2{\}}^4$ and a perfect matching $P$ of ten slots, define

There are

raw labels.

A response catalogue is complete when:

1. every allowed response change is a listed linear combination with coefficients constant on the evaluation domain;

2. every listed direction is allowed, or its permitted coefficient subspace is explicitly specified;

3. admitted representative changes are either covered by those combinations or proved response-null under the same prescription.

The raw contacts span $\mathcal{V}$, every raw contact is allowed, and their evaluation descends to $\mathcal{V}$. Thus the catalogue satisfies all three clauses of Definition def:complete, with unrestricted coefficient space ${\mathbb{R}}^{15120}$.

Proof. If both derivatives act on one factor, integrate one derivative by parts. This gives a sum with derivatives on distinct factors, modulo a divergence. Relabel those factors into positions one and two. The remaining ten tensor slots are completely paired by metrics and hence occur in (eq:raw). Every even collection of orientation tensors reduces to metric pairings by repeated use of the determinant identity. Species words exhaust all assignments.

Conversely each raw expression has the declared locality, field degree, derivative degree, covariance, and parity.

After differentiation, a total derivative has the overall factor $i{\sum }_j{k}_{j\mu }=0$. Tensor identities vanish before and after evaluation. Hence action-equivalent representatives have identical responses.

The response map $\mathcal{R}:\mathcal{V}\to {\mathbb{R}}^{\mathcal{X}}$ need not be injective. Its image is

Action equivalence and equality in the response quotient ${\mathbb{R}}^{\mathcal{X}}/\mathcal{C}$ are different notions.

For covariant momenta $p,q$, define

The response of a raw contact is

One $e_i$ is $G$. Substitution of four distinctly labelled waves into (eq:raw) and extraction of their product gives this formula. The two derivatives contribute $i^2=-1$.

Use upper-index momenta

and covariant tensors

Leg one is replaced, the physical tensors are $A,B,B$, and their references are $t_0=(1,0,0,0)$.

Set $S=({\beta }_0,{\beta }_{+},{\beta }_{-})$.

Each momentum is nonzero and null; they sum to zero, and

The tensors are symmetric, trace free, temporally zero, and transverse to their corresponding momenta. Thus all three labels belong to $\mathcal{X}$.

At ${\beta }_0$,

Proof. The reflection

fixes $k_1,k_2,A,t_0$, interchanges $k_3,k_4$, preserves $B$, and negates the gauge vector. Follow the reflection by interchange of all data of legs three and four. Their species and physical tensors agree.

A parity-even scalar response is unchanged by these operations but is linear in the replaced tensor, which has been negated. It therefore equals its negative. This applies to each raw contact and to the complete reference tree.

On the linear subspace

the cubic coefficient is, modulo a divergence,

Consequently, for incoming $p+q+r=0$,

No on-shell hypothesis on $p,q,r$ is required.

Proof. Set $q_i=1+\kappa H_i$, $q_0=1$, and ${\varphi }_i=\frac{1}{2}\log q_i$, ${\varphi }_0=0$. For a diagonal metric the nonzero Christoffel patterns are

Substitution into (eq:GG) gives

The coefficient of $\kappa$ in a prefactor is

For spatial $a$ this is zero. For $a=0$ it is the remaining $H_k$. Since ${\eta }^{00}=-1$, the asserted cubic follows.

Polarize it with one $A$ wave and two separately labelled $B$ waves. For each choice of undifferentiated wave the summed tensor coefficient is $-2$; for example,

Each time derivative contributes $-i{k}^0$, giving the first expression in (eq:ABB). The second uses energy conservation.

At ${\beta }_{+}$ and ${\beta }_{-}$, temporal entries of the physical tensors vanish, so

Equation (eq:ABB) gives

At $d_{\ast }$, the species channel factors are $(D_s,D_t,D_u)=(0,1,1)$. Therefore

This also checks the tree signs: the quartic longitudinal response is $2$, the pre-minus-sign exchange contractions are $(6,-2,-2)$, and

At ${\beta }_{+}$, the covariant momenta are the rows of

and

Let $c_{\mathbf{I},P}$ be (eq:evaluator) evaluated using these integer data and $J=(1,1,2,2)$. With

the entire evaluated matrix is

The first row is Lemma lem:blind, and the last row is species exchange. Equation (eq:encoding), together with (eq:evaluator), specifies all $45360$ entries.

Every contact entry in (eq:encoding) has absolute value at most $768$.

Proof. Only species words with two occurrences of each species contribute. For each such word there are four compatible external assignments. The two differentiated momenta each have two nonzero components. The tensors have respectively $3,2,2,2$ nonzero entries, so at most

component products occur per assignment. Each product has absolute value at most $2$. Matching constraints can only remove terms. Thus $4\cdot 96\cdot 2=768$ is a valid bound.

Choose

Let ${\ell }_a=((1,1,2,2),P_0)$ and ${\ell }_b=((2,2,1,1),P_0)$. Their actions are

where an unindexed $h^I$ is the Lorentz trace.

Proof. At ${\beta }_{+}$, only the longitudinal tensor can occupy the trace in ${\mathcal{O}}_a$. The other species-one factor receives $A$; the species-two factors receive $B,B$ in two orders. Since

each order contributes

The total is $8$. The trace of a physical species-two tensor makes ${\mathcal{O}}_b$ vanish there. Species exchange gives the other row, and blindness gives the first entries.

These contractions use a traceful longitudinal insertion. Imposing $k\cdot \xi =0$ would change the locked domain.

The complete evaluated contact image is

and

The original minor on rows two and three and columns ${\ell }_a,{\ell }_b$ is $64$.

If ${\alpha }_0$ has only

nonzero, then

Moreover,

and

Proof. All columns have zero first entry, so the rank is at most two. The two original columns of Lemma lem:pivots have last-two-row minor $\det \mathrm{diag}(8,8)=64$, giving rank at least two over both fields and spanning the asserted plane.

Their combination with coefficients $2/3,2/3$ is $m_S$. Appending the target therefore does not increase rank. An annihilator $z=(z_0,z_{+},z_{-})$ must satisfy $8z_{+}=8z_{-}=0$; conversely every $(z_0,0,0)$ annihilates the matrix. The target has the same blind first entry.

Subtracting ${\alpha }_0$ identifies the full affine fit family. Rank-nullity gives $15120-2=15118$.

The raw-coordinate dimension is not a dimension of inequivalent actions. No separator exists on $S$ against the declared class. The result is precisely $\mathrm{SEP}$-finite.

For a matrix $N\in K^{s\times L}$, a vector $u\in K^s$, and a field $K$, exactly one of the following holds:

or

The second case is equivalent to $\mathrm{rank}[N\mid u]=\mathrm{rank}N+1$.

Proof. A fit makes every left annihilator of $N$ annihilate $u$. If $u\notin \mathrm{im}N$, append $u$ to a basis of $\mathrm{im}N$, extend to a basis of $K^s$, and define a linear functional equal to one on $u$ and zero on the other basis vectors. Appending a single column increases rank exactly when that column lies outside the original image.

For a complete catalogue, a finite separator excludes a global constant-coefficient identity. A finite fit has no converse global implication. For example, on $\mathbb{R}$, for any prescribed finite $S^{\prime }=\{t_1,\dots ,t_s\}$,

agree on $S^{\prime }$ but not globally.

One sufficient finite-to-global route is injective evaluation on

If evaluation on a family is injective on $U$, the difference of a fitted identity has zero evaluation and hence is the zero function. Such injectivity is not proved for $S$. A direct global proof would be another route.

The following distinctions hold.

1. Under the same response prescription, replacing $m$ by $m+{\sum }_{\ell }{\rho }_{\ell }{n}_{\ell }$, with constant allowed coefficients, preserves global membership and finite augmented rank.

2. A compatible enlargement of the contact class preserves the existing finite fit. Its complete rank and left kernel must be assessed anew.

3. Multiplication of a density by a nonconstant weight need not descend to the unweighted integration-by-parts quotient.

Proof. For (i), the change lies in $\mathcal{C}$; finitely it is an elementary augmented-column operation. Fitting vectors change by $\rho$, and every old separator annihilates the change.

For (ii), the same allowed combination still fits the target. Any annihilator of the enlarged image also annihilates that combination, but the image may be larger.

For (iii), for compactly supported $B^{\mu }$,

For $B^{\mu }={\phi }^3{\partial }^{\mu }\phi$,

which is not generally a divergence. One may take $\phi$ to be the trace of a compactly supported tensor field.

Classical weighting of sampled record probabilities, considered later, is a different operation from weighting an action density.

A finite experiment specifies:

1. $(\overline{M},M,g,\mathcal{W},U,\chi ,c)$;

2. finite live quantum names, their finite qubit tensor factors, and port locations in $M$;

3. accessible matrix algebras, with identity action on inaccessible factors;

4. a finite instruction tree including allocations, unitaries, destructive readouts, classical controls, causal record wires, retained outputs, auxiliary systems, schedules, and storage reservations.

No controller reads an undeclared state description or an inaccessible spectator. This is the four-structure discipline used in I-B and En03-A CheungIB,CheungEn03.

An untouched reference $E$ is permitted. It has no control wire into the experiment, and its storage is accounted for separately.

Let

Choose an orthonormal clock basis $\{|{\tau }_k⟩:k=0,1,2,3\}$ and put $P_k=|{\tau }_k⟩⟨{\tau }_k|$. Apply the unitary $B_C|{\tau }_k⟩=|k⟩$, followed by computational-basis Enter on both clock names.

Let $R$ be a classical register, with observable algebra

Define

The instrument is

It is completely positive and trace preserving. On an input state $\rho$, its subnormalized blocks are