Finite contact-span certificates for a two-species spin-two tree

We solve an exact finite contact-membership problem for a fixed two-species Pauli–Fierz tree. On an explicit three-label family, the single-longitudinal reference response is $m_S=(0,16/3,16/3{)}^T$. A complete redundant catalogue of $15120$ local, parity-even, two-derivative quartic contacts has evaluated rank $2$, equal to the rank of its target augmentation. Two original raw columns are $(0,8,0{)}^T$ and $(0,0,8{)}^T$; their minor is $64$, and the constant coefficients $2/3,2/3$ fit the target simultaneously on all three labels. The complete left kernel is $\mathrm{span}\{(1,0,0)\}$, so no separator exists on this family against the declared class. We derive the required Einstein cubic contraction, prove the analytic controls, and provide an integer encoding of every contact-matrix entry together with an optional exact checker. The status is $\mathrm{SEP}$-finite: the finite-family problem is closed, while global membership or separation remains undecided here. We distinguish action representatives, response classes, affine fitting freedom, class enlargement, and weighted response prescriptions. Companion results are cited only for comparison; none supplies a contact column or a premise of the certificate.

{Principal certificate: $\mathrm{SEP}$-finite}{ For $S=({\beta }_0,{\beta }_{+},{\beta }_{-})$ and the complete declared contact catalogue,

This is one simultaneous finite fit, not a global contact identity.}

{ No global $\mathrm{SEP}$ decision, interacting-gravity construction, physical no-go theorem, quantum coefficient, field instrument, or measurement-uniqueness theorem is claimed. The unweighted certificate is not transferred to a weighted source class, a changed gauge prescription, or an unidentified counterterm space. A record interpretation requires the separately stated coherent-matching hypothesis.}

A nonzero response of one specified tree need not survive the addition of every allowed local contact. Survival modulo contacts is a membership question in a space of response functions. Evaluating that question on a finite family gives a finite linear-algebra problem, but its interpretation depends on admissibility, normalization, and complete coverage of the allowed response changes.

We solve that finite problem for

The first label is blind by symmetry. The other two have reference response $16/3$, but the complete contact image on this family contains both responses.

Theorem thm:certificate gives the central result. Its proof derives four principal quantities:

They are respectively a reference-tree response, a raw contact contraction, an original matrix minor, and fitting coefficients.

The free starting point is the massless spin-two quadratic action in the Pauli–Fierz tradition FierzPauli. Einstein self-interaction and gauge consistency provide historical context Deser. The consistency problem for collections of massless spin-two fields has been studied using local deformation and BRST methods BarnichHenneaux,BoulangerEtAl.

Those works address hypotheses and conclusions different from the finite response-space calculation below. In particular, an action-level consistency or classification theorem is not used as a substitute for evaluating this contact matrix. No equivalence between the present global response-membership question and a full deformation problem is assumed.

Cheung's Enter-calculus result concerns a measurement strategy in a declared quantum-language class CheungEnter. It supplies neither the Einstein contraction nor the contact coverage used here. Related companion manuscripts are discussed only after the certificate has been proved.

The development separates:

1. Locked response functions. The action, external domain, and contact class determine $m,n_1,\dots ,n_{15120}$.

2. Evaluation contract. Exact inputs, quotient conventions, analytic controls, and complete coverage determine the meaning of an evaluated matrix.

3. Finite certificate. A complete encoding, two original columns, an original minor, and a fitting vector settle membership on the stated family.

None of these layers constructs an interacting quantum field theory or an apparatus.

Let

Global membership means

The coefficients are constant across external labels.

Our certificate replaces $\mathcal{X}$ by the particular family $S$. It uses one coefficient vector for all three labels, not a separately chosen vector at each row. Nevertheless, it does not establish (eq:global-quantifier).

The signature is

All momenta are incoming. Displayed vector components have upper indices. All index operations use $\eta$. Fourier waves are $e^{ik\cdot x}$, so

For symmetric tensors, $H:K=H_{\mu \nu }{K}^{\mu \nu }$.

The longitudinal tensor is

It is generated by the Fourier gauge parameter $-i\xi e^{ik\cdot x}$. No restriction $k\cdot \xi =0$ is imposed.

A label $\beta \in \mathcal{X}$ specifies:

1. four species $J_1,\dots ,J_4\in \{1,2\}$;

2. four nonzero real null momenta such that

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

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

4. for every other leg, a real symmetric tensor ${ϵ}_i$ and a timelike reference vector $t_i$ 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 $G(k_j,{\xi }_j)$. Polarization normalization is not imposed. Simultaneous permutations of all external data are admitted.

The physical tensors are TT relative to their declared references. The longitudinal insertion need not be trace free.

For $k\ne 0$, the map $\xi \mapsto G(k,\xi )$ is injective: choose $a$ with $k_a\ne 0$. If $G=0$, its $aa$ component gives ${\xi }_a=0$, and its $ab$ components then give all ${\xi }_b=0$. In particular,

There is no quotient $\xi \sim \xi +\lambda k$ in this domain.

For one species,

For two species,

The Hessian is

The fixed propagator coefficient is

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

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

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

Substitution in the Fourier transform of (eq:K0) cancels the $p_{\mu }{p}_{\nu }{J}_{\mathrm{tr}}$ terms and the remaining trace terms, leaving $J_{\mu \nu }$. Direct substitution of (eq:gauge-tensor) proves the second identity.

No inverse identity on arbitrary nonconserved tensors is asserted.

Fix the curvature sign and normalization by

modulo total derivatives. Equivalently, expand the first-derivative density

For a homogeneous functional $E_n$, put

Thus $v_n$ is exactly the coefficient of $z_1\cdots z_n$ in $E_n[{\sum }_i{z}_i{u}_i]$. No further factorial multiplies that coefficient.

Plane-wave formulas mean the coefficient after removing the common momentum-conservation delta. Integration by parts is interpreted algebraically at conserved vertex momentum. No convergent integral of four unlocalized plane waves is required.

Let $d_{IJK}$ be real and completely symmetric, with

Define

and

The action through order $g^2$ is

Every ${\theta }_{\ell }$ is constant across the external domain.

The ordered species sums and normalized polarizations ensure that a differentiated external vertex has coefficient $d_{J_1{J}_2{J}_3}{v}_3$ or $T_{J_1{J}_2{J}_3{J}_4}{v}_4$. Repeated species do not introduce an additional factorial: all assignments to the distinct bookkeeping variables are already included.

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

Equivalently, the declared tensor-contraction class uses complete metric contractions and even numbers of Levi-Civita tensors. The metric-pairing description is the standard orthogonal invariant-tensor framework Brauer; no independence of those pairings is assumed.

Let ${\mathfrak{P}}_{10}$ be the perfect matchings of $\{1,\dots ,10\}$. For $I=(I_1,I_2,I_3,I_4)\in \{1,2{\}}^4$ and $P\in {\mathfrak{P}}_{10}$, define

There are

raw labels.

The classes of the raw functionals (eq:raw) span $\mathcal{V}$. Every raw functional is an allowed contact.

Proof. If both derivatives act on one factor, integration by parts expresses the density modulo a divergence as a sum with derivatives on two distinct factors. Relabel those factors into positions one and two. There are then ten tensor slots: two derivative indices and eight field indices. Every complete metric contraction pairs those slots and is one of the perfect matchings.

A pair of Levi-Civita tensors reduces, by the determinant identity, to a signed sum of metric contractions. Iterating removes every even number of orientation tensors. This covers the declared parity-even contraction class. Species choices give every word in $\{1,2{\}}^4$.

Conversely, every raw expression has the declared field degree, derivative degree, locality, covariance, parity, and constant coefficients.

The number $15120$ counts redundant raw coordinates. It is not the dimension of inequivalent actions or global responses.

For lower-index momenta $p,q$, define

The differentiated response of a raw contact is

One $e_i$ is the prescribed $G$. The momenta in ${\mathrm{Contr}}_P$ have lower indices.

To derive this formula, substitute

in (eq:raw) and extract $z_1{z}_2{z}_3{z}_4$. All assignments to the factor positions appear. The two derivatives contribute $i^2=-1$.

The response map

is well defined. Its image is precisely

Proof. The differentiated coefficient of a total derivative has an overall factor $i{\sum }_j{k}_{j\mu }$, which vanishes on $\mathcal{X}$. A genuine tensor identity vanishes before and after differentiation. Thus equivalent density representatives have identical responses.

Raw coverage gives every allowed response as a constant combination of the $n_{I,P}$, and every such combination comes from allowed raw contacts.

Action equivalence and response equivalence remain different: $\mathcal{R}$ need not be injective.

On the internal Euclidean space with basis $e_1,e_2$, set

The left multiplication matrices are

Define

Then

For

the nonzero possible components are

Hence

When $q=0$, the symmetric matrices $L_1,L_2$ commute. In a common orthonormal eigenbasis $f_1,f_2$, the product $f_1\circ f_2$ is simultaneously proportional to $f_1$ and to $f_2$, and is therefore zero. Inner-product invariance then gives $f_i\circ f_i={\lambda }_i{f}_i$. Consequently

for an orthogonal matrix $O$. The cubic and reference quartic in that basis are

This elementary decomposition is used as a control, not as an imported multi-graviton classification.

For a gauge parameter $\zeta$, write

and

Under the coordinate variation scaled by $\kappa$,

The Einstein density varies by a total derivative. The first two coefficient identities are

Polarizing them gives

Here pairings include spacetime integration or, for waves, the corresponding stripped conserved-momentum coefficient. The factors agree with the definition of $v_n$: taking two or three distinct field derivatives introduces no extra factorial.

Define the cubic current by

If $K_{0}v=K_{0}w=0$, equation (eq:ward3) with arbitrary $\zeta$ implies conservation of $J(v,w)$. Thus Lemma lem:inverse applies at the physical end of each exchange.

The stripped four-point prescription is

The common momentum delta, $g^2$, and overall Feynman factor $i$ have been removed.

For free $u,v,w$,

Together with the exchange minus sign in (eq:tree), this cancels the corresponding term of (eq:ward4).

Take leg one replaced. For $\zeta =-i{\xi }_1{e}^{i{k}_1\cdot x}$, define

Its momentum is $k_1+k_i$. Put

The composite insertion is generally off shell.

Write

and $\overline{D}=(D_s+D_t+D_u)/3$. If $X_i$ denotes the unweighted exchange contraction before its tree minus sign, the preceding identity gives

The response $F(d)$ of the reference action is

Proof. The exchanges contribute $-D_s{X}_2-D_t{X}_3-D_u{X}_4=D_s{W}_2+D_t{W}_3+D_u{W}_4$. The reference quartic coefficient is $\overline{D}$. Its response is $-\overline{D}(W_2+W_3+W_4)$ by (eq:ward4).

Each longitudinal exchange has become a polynomial contraction. The inverse pole cancels separately in each channel.

For species $1122$,

and therefore

The same formula holds for $2211$. Other two-and-two placements permute the three coefficients.

For the remaining species patterns, all three $D$'s agree:

Their reference response is zero.

Define

On $\mathcal{X}$,

Proof. The displayed species cases exhaust all four-leg words. They establish $F(d)=q(d)F(d_{\ast })$ when leg one is replaced. Simultaneous relabeling moves any replaced leg to position one. The additional quartic contacts contribute linearly by definition.

This proof uses the locked vertices and direct species algebra. No companion obstruction theorem is involved.

For a response problem on a domain $\mathcal{B}$, a list $n_1,\dots ,n_L$ is complete when:

1. every allowed response change is ${\sum }_{\ell }{c}_{\ell }{n}_{\ell }$, with coefficients constant on $\mathcal{B}$;

2. every listed direction is allowed, or the permitted coefficient subspace $W\subseteq {\mathbb{R}}^L$ is explicitly named;

3. all admitted representative variations are covered, either by those combinations or by a proof that they are response-null under the same evaluation prescription.

If a proper $W$ is declared, the allowed image is $N_{S}W$, not automatically the span of all columns.

For the present class, $W={\mathbb{R}}^{15120}$. Lemma lem:coverage proves constant-coefficient spanning and that every generator is allowed. Proposition prop:quotient proves that quotient-representative changes are response-null. Thus all three clauses hold.

The catalogue count alone proves none of these clauses. It records the size of the spanning construction after coverage has been established.

There are two quotients:

The response map connects them but need not be injective. A response test of a density representative tests its action class because evaluation descends to $\mathcal{V}$. A test of contact-surviving information instead tests $[m]$, not one arbitrary reference representative.

In particular, no equation-of-motion quotient, source weighting, or characteristic-product ambiguity is silently added to $\mathcal{V}$.

Every row specifies species, momenta, reference vectors, polarizations, replaced leg, and gauge vector. The central evaluation is rational; contact coefficients may still be arbitrary real constants.

The analytic controls are:

$C_1$

One-species Einstein longitudinal cancellation.

$C_2$

Associative two-species cancellation with the canonical reference quartic.

$C_3$

Rank-one species factorization with its canonical longitudinal law.

$C_4$

Simultaneous permutation of all external labels.

$C_5$

Channelwise equality of direct and Ward-reduced exchanges.

$C_6$

Zero target and zero response of every raw contact at the designated blind seed.

These are proved mathematical identities, not reports of executed regression tests. An analytic zero on a control domain is stronger than a finite list of sampled zeros. Conversely, an analytic identity does not certify an unrun software implementation.

For a finite ordered family $S=({\beta }_1,\dots ,{\beta }_s)$, put

If the catalogue is complete and

then $m\notin \mathcal{C}$. Moreover,

for every allowed constant contact vector.

Proof. A global identity $m={\sum }_{\ell }{\alpha }_{\ell }{n}_{\ell }$ would imply $m_S=N_S\alpha$, contradicting $z{m}_S\ne 0$. Multiply the evaluated factorization by $z$ for the second claim.

Without complete coverage, annihilation of the displayed columns establishes blindness only to the displayed correction subspace.

A finite fit has a different direction of implication: $m_S=N_S\alpha$ proves finite membership. Completeness is needed for the claimed full image, rank, and left kernel; existence of this particular fit requires only that its used columns be allowed.

The result here is $\mathrm{SEP}$-finite. The global question is not answered by that status.

Use upper-index momenta

and covariant tensors

The physical tensors on legs $2,3,4$ are $A,B,B$. Their references are all $t_0=(1,0,0,0)$. Leg one is replaced.

The row order is

All three labels belong to $\mathcal{X}$, with

Proof. The momenta are nonzero, null, and sum to zero. Their pair sums give the stated invariants. Both tensors are real, symmetric, temporally zero, and Lorentz trace free. The spatial support of $k_2$ is in direction three, where $A$ has a zero row. The spatial supports of $k_3,k_4$ are in direction one, where $B$ has a zero row. Thus the momentum and reference transversality conditions hold. Finally ${t_0}^2=-1$, and both gauge vectors are admitted.

The normalized blind seed uses $A/\sqrt{2},B/\sqrt{2},B/\sqrt{2}$. It has the same symmetry blindness, but the finite matrix below uses the rational unnormalized tensors.

On the linear subspace

the cubic coefficient is, modulo a divergence,

It may be polarized using any three fields in this subspace, including an off-shell insertion.

Proof. Set

The nonzero diagonal-metric Christoffel patterns are

Three-distinct-index entries vanish. Substituting these patterns and collecting each derivative direction gives

Since ${\partial }_a{\varphi }_i=\kappa {\partial }_a{H}_i/(2q_i)$, the density (eq:GammaGamma) becomes

The coefficient of $\kappa$ in the prefactor of the term indexed by $a,i,j$ is

For spatial $a$, the three indices exhaust $1,2,3$, so this coefficient vanishes. For $a=0$, it equals the remaining $H_k$. The factor ${\eta }^{00}=-1$ proves (eq:diagonal-cubic).

As a normalization check, the time-derivative quadratic part is

on the trace-free subspace, in agreement with (eq:PF). Restriction to a linear subspace commutes with polarization. No field equation for the inserted fields was used.

For incoming momenta $p+q+r=0$,

Proof. Polarize (eq:diagonal-cubic). For each choice of undifferentiated labeled wave, the sum of its spatial positions and the two assignments of differentiated waves has coefficient $-2$. For instance, with $A$ undifferentiated,

A time derivative contributes $i{k}_0=-i{k}^0$. The two derivative factors and the density sign yield the first formula. The second follows from $p^0+q^0+r^0=0$.

At both ${\beta }_{+}$ and ${\beta }_{-}$,

and

Proof. Because the physical tensors have zero temporal entries, (eq:R) gives

The temporal momenta in $W_2$ are $2,-1,-1$. Those in $W_3,W_4$ are $0,1,-1$, up to permutation. Lemma lem:ABB therefore gives

At $d_{\ast }$, both species channels have $(D_s,D_t,D_u)=(0,1,1)$. Equation (eq:m1122) gives

The tree signs can also be displayed directly: