XX-C State, Subtraction, and Spatial-Tail Audit for the Original-Rod Lapse Local Hadamard Subtraction, Native Newton Bounds, and Explicit Same-Operator Obstructions

On the identical power-law Einstein–scalar background and original-rod prescription displayed in XX-B, we construct a local bundle-valued subtraction parametrix with the actual action-Hessian normalization. Its difference from a normalized Feynman inverse is smooth near the diagonal. We prove uniform dipole estimates for the original Newton inverse acting on compact-source retarded or advanced responses, and give a precise sufficient decreasing-source condition for analogous one-sided covariance estimates.

The supplied free Hadamard class does not select a physical state or impose an infrared condition. We exhibit, on the same coupled operator, a smooth finite-rank modification whose value on a nonzero gauge-invariant tensor observable is negative. This excludes a positive physical state with that exact covariance, without excluding positive representatives elsewhere in the class. A separate polynomial-solution modification shows that the class does not even ensure that the decaying rod inverse is defined.

We also compute an intrinsic obstruction to a frequently overstated tail requirement: the native two-rod commutator has a nonzero inverse-distance spatial separation term. Thus an inverse-square bound cannot be demanded of both ordered two-rod covariances without changing the canonical data. This obstruction does not remove the one-rod term in the quadratic lapse.

The smooth subsidiary contraction defect is retained and its exact correction problem is written explicitly. No positive full physical state, unconditional one-sided native covariance-tail theorem, or nonlinear lapse contraction is released. Accordingly, the requested Q2 package remains incomplete; its boundary is proved and not concealed by a renamed closure.

The predecessor XX-B supplies a complete gauge-fixed free operator and a nonempty class of algebraic Hadamard contractions XXB. It does not supply one uniquely selected physical covariance, a physical quotient state, or global spatial estimates for an arbitrary member of that class.

The present manuscript keeps:

1. the same invariant action and on-shell background;

2. the same matter-scalar lapse, clock calibration, positive classical branch, and factorial expansion;

3. the same slice-harmonic original rods and decaying spatial boundary condition;

4. the same gauge condition, action-Hessian normalization, causal commutator, and compact-test field domains.

The equations needed for the proofs are reproduced below. No unavailable text of XX-A is represented as having been inspected.

A fixed representative is denoted $W_{\mathcal{B}}$. The normalized Hadamard class is denoted ${\mathfrak{W}}_{\mathcal{B}}$. Membership in the class does not assert positivity, subsidiary invariance, decay, or existence of the Newton compositions. A change within the class is a new normalization choice, not a verification of the original representative.

An indefinite gauge-fixed fiber pairing does not by itself obstruct positivity on the physical gauge quotient. Conversely, positivity of one scalar or tensor subsector does not construct a state on the complete quotient. These distinctions matter in linearized Einstein–Klein–Gordon theory Hack.

We explicitly decline to announce a full physical state for the unspecified $W_{\mathcal{B}}$. Instead, we prove:

1. an exact obstruction to positivity for an explicitly constructed representative in the same native class;

2. exact local subtraction;

3. unconditional compact-source rod-response tails;

4. a conditional one-sided covariance-tail theorem with its hypothesis stated as a genuine additional infrared requirement;

5. an unconditional two-rod commutator obstruction.

This is a complete mathematical audit of the stated data. It is not a complete native Q2 state-and-tail package.

We use:

1. Cauchy well-posedness and causal Green operators for normally hyperbolic bundle operators BGP.

2. Distinguished bundle-valued Feynman parametrices and their uniqueness modulo smooth kernels IS.

3. The distinction between a Feynman construction and the additional conditions required for a state FS.

The hypotheses needed for A1 and A2 are checked in Sections sec:freeze and sec:parametrix. No positivity theorem for positive-definite field bundles is applied to the indefinite gauge-fixed bundle.

The Cheung records are version-qualified documentary comparisons CheungXIXD,CheungXVIIIA,CheungXVIIIB, CheungXIXA,CheungXIXC,CheungEm02E. Their unprinted calculations are not substituted for the native proofs.

Use signature $(-,+,+,+)$ and

Only compactly supported variations of the local action density are required.

Fix $p>1$, $t_{\ast }>0$, and

The spacetime is

No endpoint boundary condition is imposed on this open slab.

Writing $H=\dot{a}/a=p/t$ and $v_{\mu }={\partial }_{\mu }\bar{\varphi }$, the background equations are

They follow directly from $A^2=2p{M}^2$.

Each $t=\mathrm{constant}$ surface is Cauchy. Indeed, along a causal curve

so a causal curve with a time endpoint inside $I$ has a finite spatial endpoint and is extendible. The background is therefore globally hyperbolic.

Let

Use the nondegenerate integrated pairings

The invariant Euler equations are

The action Hessian is defined without an unspecified normalization by

All variations in (eq:H) are taken at $(\bar{g},\bar{\varphi })$.

For completeness, the local variation rules are

The gauge generator and its adjoint are

On the on-shell background,

Retain

Explicitly,

Define four-dimensional trace reversal

and

The second-derivative metric part of $\mathcal{H}$ is

The second term is cancelled by $r{\mathcal{F}}^{†}\mathcal{F}$. Mixed entries have at most one derivative. Consequently

It is formally self-adjoint for $(z,w{)}_{\mathcal{A}}=(z,\mathcal{A}w{)}_0$. This verifies the normally hyperbolic hypotheses, but the pairing is not positive definite.

The compact-spacetime residual condition $Q\xi =0$ has only the zero compactly supported solution by Cauchy uniqueness. This does not eliminate nonzero homogeneous solutions with general Cauchy data.

Let ${G_{\mathcal{L}}}^{\mathrm{ret},\mathrm{adv}}$ and ${G_Q}^{\mathrm{ret},\mathrm{adv}}$ be the causal inverses supplied by the normally hyperbolic Green theorem BGP, and set

The actual inverse equations use

The exact intertwining identities are

For example, the retarded version of the first Green identity follows because its two sides solve the same inhomogeneous equation and have the same causal support; uniqueness then applies.

The seed contractions obey

Here ${\mathrm{WF}}^{\prime }$ reverses the second covector, and ${\mathcal{N}}_{+}$ is the positive-frequency null bicharacteristic relation in the XX-B convention.

In a real field trivialization, (eq:Wdata) is equivalently

No sign condition on ${\mu }_P$ is imposed.

The auxiliary and ghost contractions are reconstructed by

Thus a stated bosonic normalization change also changes the auxiliary blocks; they are not silently kept fixed.

Retain

The raw lapse is

not the differently rescaled lapse calibrated to $T_s$.

The first inverse clock displacement is

The first induced spatial metric is

Writing

define the local linear rod-source matrix

Then $B_1=\mathcal{C}z$.

The spatial inverse and first rod are

Equivalently, after integration by parts,

The underlying rods still solve

No lapse multiplier occurs in this equation.

Define

The denominator lies in the numerator because $K^{†}\mathcal{H}=0$. Further quotienting by a presymplectic radical, if required, is not assumed to have been completed here.

The causal form induced by $E_{\mathcal{B}}$ descends to (eq:quotient). Indeed, on a compact test $a$,

which pairs to zero against a test in $\ker K^{†}$.

A symmetric contraction need not descend merely because its causal part does. Its descent and positivity are additional conditions.

If a positive physical state has exact two-point restriction $W_{\mathcal{B}}$ on the invariant observables, then

Its value must also be independent of the representative of the class in (eq:quotient).

Proof. The field associated with $f$ is a physical observable. Positivity applied to its adjoint times itself gives (eq:physicalpositive). Quotient relations require representative independence.

Indefiniteness of $(\cdot ,\cdot {)}_0$ is not a proof that (eq:physicalpositive) is impossible.

There are a real smooth solution $U=(h,0)$ and a real compact test $f=(j,0)$ such that

The solution has uniformly compact spatial support on compact time intervals.

Proof. First choose a nonzero compact smooth symmetric tensor $q_{ij}$ on ${\mathbb{R}}^3$ satisfying

An explicit local construction is as follows. Put

choose a real constant symmetric matrix $S$ and a compact real smooth $\chi$, and set

The Fourier symbol of $D$ is $|\mathbf{k}{|}^2{\delta }_{ij}-k_i{k}_j$. It makes the displayed $q$ divergence-free and trace-free. Taking $S=\mathrm{diag}(1,-1,0)$ and a suitable nonzero $\chi$ makes $q\ne 0$.

Fix $t_0\in I$, and solve

The scalar normally hyperbolic Cauchy theorem applies componentwise. Divergence and trace satisfy the same equation with zero data and hence remain zero.

Set $h_{ij}=a^2{\gamma }_{ij}$, $h_{00}=h_{0i}=0$, and $\phi =0$. Substitution in (eq:curvature)–(eq:scalar) gives $\mathcal{H}U=0$ precisely when (eq:tensorequation) holds. The scalar variation vanishes because the background spatial Hessian of $\bar{\varphi }$ is proportional to ${\delta }_{ij}$. The gauge condition vanishes by the same trace and divergence identities.

Equivalently, this substitution reduces the quadratic action to

whose Euler equation is (eq:tensorequation). The background Euler identities remove the terms without derivatives. This provides a direct check of the normalization.

Choose a nonnegative time bump $\rho$ supported sufficiently close to $t_0$, and set

Its spacetime covariant divergence vanishes: the spatial components use ${\partial }_i{q}_{ij}=0$, and the temporal component uses ${\delta }^{ij}{q}_{ij}=0$. Therefore $K^{†}f=0$. Also

is strictly positive for a sufficiently small support of $\rho$. Finite propagation proves the support claim.

This witness is constructed on the displayed native operator. The tensor-sector discussion of Hack Hack is a comparison, not a replacement of this calculation.

For a real smooth solution $U$, write

This denotes a symmetric real smooth bisolution.

Let $W_{\mathcal{B}}\in {\mathfrak{W}}_{\mathcal{B}}$, and choose $U,f$ as in Lemma lem:TT. Put

and define

Then:

1. ${W_{\mathcal{B}}}^{\mathrm{bad}}$ has the same field equations, canonical antisymmetric part, adjoint condition, and Hadamard singularities;

2. ${W_{\mathcal{B}}}^{\mathrm{bad}}(f,f)=-1$;

3. no positive physical state can have this exact restriction on the invariant observable $f$.

The class of $f$ in (eq:quotient) is nonzero.

Proof. The added kernel is smooth, symmetric, real, and a bisolution of $\mathcal{B}$. Thus it changes none of the singular or antisymmetric data. The definition of $\lambda$ gives $q+\lambda c^2=-1$.

If $f=\mathcal{H}a$ for compact $a$, formal self-adjointness and $\mathcal{H}U=0$ would give $(U,f{)}_0=0$, a contradiction. Hence $f$ is a nonzero invariant observable. Proposition prop:necessary excludes the exact positive restriction.

This is not a proof that the entire class lacks positive physical representatives. It is a proof that the class data do not authenticate positivity of a representative. The same distinction appears in Cheung's covariance audit CheungXIXD.

Because $\mathcal{F}U=0$, the modification (eq:bad) does not change the subsidiary defect considered in Section sec:subsidiary. Its compact-support-on-time-compacts property also means that it does not alter an already proved one-sided asymptotic tail at spatial infinity.

Every normally hyperbolic operator has a unique connection-plus-potential representation

Here $\mathcal{D}$ is a smooth bundle connection and $\mathcal{V}$ is a smooth endomorphism. In local coordinates, they are obtained by matching the first- and zeroth-order coefficients of (eq:hyperbolic); thus no new operator is introduced.

The hypotheses for the bundle parametrix construction are:

1. a smooth finite-rank bundle;

2. a smooth Lorentzian metric and time orientation;

3. a scalar normally hyperbolic principal symbol;

4. smooth lower-order coefficients;

5. for the global distinguished construction, global hyperbolicity.

All hold by Section sec:freeze. Positivity of the fiber pairing is not required for this local subtraction construction BGP,IS.

Choose a geodesically convex, causally convex neighborhood $O⋐\mathcal{M}$. Let $\sigma (x,y)$ be one half of the signed squared geodesic distance, and let ${\Delta }_{\mathrm{vV}}$ be the van Vleck determinant.

Define

where $\mathcal{P}$ is $\mathcal{D}$-parallel transport from $y$ to $x$. Equivalently,

The logarithmic coefficients are determined by

These are regular transport equations along the unique geodesic in $O$. Their smooth solutions are fixed by regularity at coincidence. In particular,

Let

be a smooth asymptotic summation with cutoffs in $\sigma$, chosen so that the residual of the transport construction vanishes to all orders on the local characteristic set. Fix one such summation. It is not claimed unique modulo smooth terms.

For the positive-frequency boundary value, use

The length $\ell >0$ is a declared subtraction scale. Set

A cutoff equal to one near the diagonal may be added. Only its near-diagonal germ is used.

The native action-Hessian subtraction is

The multiplication in the source slot is essential.

The sign in (eq:HF) matches the XX-B convention

For the flat scalar action operator $\square =-{{\partial }_t}^2+\Delta$, the corresponding massless positive-frequency singularity is

This fixes the factor and boundary-value convention.

The construction (eq:Utransport)–(eq:Hnormalized) is a local subtraction parametrix for the actual operator $\mathcal{B}$. For any normalized native Feynman inverse $F_{\mathcal{B}}$ in the same distinguished class,

near the diagonal in $O\times O$. Likewise,

A sufficiently long finite truncation gives a remainder of any preassigned finite differentiability order.

Proof. Applying ${\mathcal{L}}_x$ to the inverse-distance and logarithmic terms gives the transport equations (eq:Utransport)–(eq:Vrecursion). They cancel the successive non-smooth coefficients. The all-order asymptotic summation makes the residual smooth. Multiplication by ${{\mathcal{A}}_y}^{-1}$ gives the correct delta normalization for $\mathcal{B}=\mathcal{A}\mathcal{L}$.

This is a distinguished Feynman parametrix. Uniqueness of distinguished parametrices modulo smooth kernels IS gives the first difference. The corresponding positive-frequency splitting has the same normalized singular coefficients, giving (eq:smoothW). Finite truncation leaves only sufficiently high powers of $\sigma$ times logarithms and their boundary values; their differentiability increases with the number of retained coefficients.

Cancellation is proved using the transport equations, inverse normalization, and distinguished-parametrix uniqueness. Equality of wavefront-set upper bounds alone would not prove it. The theorem is local. It fixes neither a state nor spatial tails. The smooth remainder is not required to decrease.

The same construction applies to $Q$. Differential application of $\mathcal{F}$ in the full contraction matrix produces the corresponding auxiliary subtraction germs. This statement concerns free contractions, not an interacting lapse renormalization.

Changing $\ell$ changes the smooth local subtraction germ. It does not suppress a nondecaying smooth global bisolution. Local Wick-product normalization and global state selection are distinct obligations HWlocal.

For a compact interval $J⋐I$, define the uniform Schwartz seminorms

They use coordinate components in a fixed smooth background trivialization.

Suppose $b^i(t,\mathbf{x})$ is smooth and either:

1. has uniformly compact spatial support for $t\in J$, with all required derivatives bounded; or

2. has finite seminorms (eq:Schwartz) for every $N,m,\alpha$.

Then

is a smooth decaying solution of $\Delta u=-{\partial }_i{b}^i$, and

outside a fixed ball. It is the unique smooth solution with $u\to 0$ at spatial infinity.

Proof. Integration by parts gives

For compact support, the kernel and its spatial derivatives are bounded by constants times $|\mathbf{x}{|}^{-2-|\alpha |}$ once $|\mathbf{x}|$ is twice the support radius. Time differentiation acts only on $b$.

For Schwartz sources, split the integral into $|\mathbf{y}|\le |\mathbf{x}|/2$ and its complement. The first part has the same kernel bound, using a uniform $L^1$ bound on the source. In the complementary part, use a smooth partition around $\mathbf{y}=\mathbf{x}$; transfer derivatives of the singular kernel onto the smooth rapidly decreasing source when necessary. The source and its derivatives decrease to arbitrary order in that region, while the dipole kernel is locally integrable. This bounds that part to an arbitrarily high inverse power after increasing the Schwartz seminorms used in the estimate.

The same argument proves smoothness and permits time differentiation. A difference of two decaying solutions is an entire harmonic function tending to zero. The maximum principle gives uniqueness.

The inverse-square rate is a consequence of the divergence structure. It is not a general bound for a Newton potential with nonzero monopole.

Let $f\in {C_c}^{\infty }(\mathcal{M};\mathcal{V})$, and put

For every $J⋐I$, $z^{\pm }$ has uniformly compact spatial support on $J$. Its original first rod

satisfies (eq:dipolebound), uniformly on $J$.

Thus the distributional kernels ${\mathcal{R}}_x{G}_{\mathcal{B}}^{\mathrm{ret},\mathrm{adv}}(x,y)$, interpreted by testing the input slot with compact $f$, have a proved inverse-square output tail.

Proof. Causal support gives

If the spatial projection of $\mathrm{supp}f$ is contained in a ball of radius $R_f$, then on $J$ the support radius is at most

The supremum is finite on a compact interval inside $I$. The solution is smooth and its required derivatives are bounded on that compact spatial-time region.

The matrix $\mathcal{C}$ is local with smooth time coefficients. Therefore ${\text{calCz}}^{\pm }$ has the same support property. Apply Theorem thm:dipole.

The rod correction destroys strict spacetime causal support: it is instantaneous on a spatial slice. The theorem proves a tail, not a causal replacement of the rods.

For classical coefficient fields with common compact spatial support on $J$, the second rod equation is

Here $B_1,B_2$ are the induced-metric coefficients. Both $B_2$ and $B_1\partial Y_1$ are compact in space. Theorem thm:dipole therefore gives the same inverse-square estimate for $Y_2$.

The complete quadratic lapse retains

The absence of $Y_2$ follows from ${\partial }_i{N}_0=0$, not from dropping its construction.

The tail theorem controls the classical rod term on this domain. It does not define its quantum Wick contraction.

For a compact field test $f$, let

The section $w_f$ is smooth and solves $\mathcal{B}{w}_f=0$. No spatial decay follows from (eq:Wdata)–(eq:Wwavefront) alone.

Proof. The field equation is immediate. The wavefront relation has no point with zero covector in the tested slot. The standard pushforward criterion after compact testing therefore makes $w_f$ smooth. Spatial decay is not a microlocal conclusion. The counterexample below proves the final assertion on the native operator.

Suppose a selected representative satisfies

for each compact test $f$. Then

is well defined and obeys (eq:dipolebound). The estimate is uniform for bounded test families whenever the seminorm bounds in (eq:IR) are uniform on those families.

Proof. Apply Theorem thm:dipole to $b^{ia}=({\text{calCw}}_f{)}^{ia}$.

Condition (eq:IR) is a sufficient infrared condition, not a consequence of being Hadamard. A weaker weighted domain may suffice, but it must be specified with the required derivative and moment bounds before it replaces (eq:IR).

There is no assertion that the XX-B class has been normalized to satisfy this condition. Theorem thm:conditional is therefore not the missing unconditional native covariance theorem.

Let $\alpha$ be a fixed smooth classical inverse coordinate map from the retained perturbation domain. On a compact time interval, assume that outside a ball it has the form

with

and that its finite derivative orders needed below are bounded on the remaining region. The original compact classical perturbations and their decaying rods have precisely this asymptotic form for a sufficiently small invertible map.

If a smooth section $u$ satisfies (eq:dipolebound), then the fixed pullback $u\circ \alpha$, with the corresponding bounded bundle transformation, has the same rate. A rapidly decreasing smooth source remains rapidly decreasing under such a fixed pullback.

Proof. For sufficiently large $|\mathbf{x}|$,

The chain rule expresses each fixed derivative of the pullback as a finite sum of derivatives of $u$ times bounded derivatives of $\alpha$. The displayed asymptotic derivative bounds give the same spatial rates. The argument with arbitrarily large weights proves the Schwartz statement.

A two-point distribution can be pulled back by a fixed diffeomorphism in both slots. This is an ordinary distributional pullback and preserves the transformed wavefront relation.

A field-dependent quantum coordinate substitution is not that operation. Its expansion includes products of fields and coordinate corrections. No such product admission is supplied by Proposition prop:pullback.

The class ${\mathfrak{W}}_{\mathcal{B}}$ contains pairs of representatives for which the one-sided decaying rod composition cannot be simultaneously defined. Consequently the XX-B algebraic and microlocal conditions do not imply the Q2 spatial-tail requirement.

Proof. Fix $t_0\in I$. Because all coefficients of $\mathcal{B}$ depend only on time in a spatially homogeneous frame, the space of spatial polynomials of degree at most one is preserved by $\mathcal{L}$. The equation reduces on that space to a finite system of smooth ordinary differential equations.

Choose a real solution $U_{\mathrm{aff}}$ with Cauchy data

and any fixed smooth polynomial time-derivative data, for example zero coordinate derivatives. This gives a smooth global solution on $I$; no spatial decay is asserted.

At $t_0$,

The first rod equation in component $a=1$ would therefore require

There is no smooth solution tending to zero at spatial infinity. Its spherical mean satisfies

as follows by averaging the Poisson equation over spheres.

Choose a compact test $f$ with $(U_{\mathrm{aff}},f{)}_0\ne 0$. Both

satisfy the same Hadamard, field-equation, adjoint, and canonical conditions.

If both had linear decaying rod compositions after testing with $f$, their difference would supply a decaying solution of (eq:constantPoisson) times the nonzero constant $(U_{\mathrm{aff}},f{)}_0$. This is impossible. Thus at least one of the pair is outside that decaying composition domain.

This theorem does not declare the affine solution a permissible classical rod configuration. It is a permissible smooth ambiguity of the compact-test algebraic covariance class. That distinction is exactly the missing domain condition.

The quadratic lapse term in (eq:lapse) contains one rod multiplied by a local derivative. A covariance of two first rods is a different composition. It involves two Newton inverses.

It would be incorrect to require the inverse-square dipole estimate of Theorem thm:GreenTail for both ordered two-rod covariances. Their commutator already has a slower native tail.

This can be proved without selecting any state.

Use the orthonormal coframe $\mathrm{d}t,a(t)\mathrm{d}{x}^i$, and write

In these field components the rod-source matrix is time independent:

Spatial translation invariance and uniqueness make the causal kernel translation invariant:

Its restriction to fixed $t,t^{\prime }$ exists as a spatial distribution: null wavefront covectors have nonzero spatial parts and do not meet the conormal bundle of that time restriction.

For fixed $t,t^{\prime }$, the distribution $e$ is supported in the ball

Let

This is a well-defined finite-dimensional matrix, smooth in the two times. The Cauchy jump normalization gives

The minus sign comes from the principal differential part $-{{\partial }_t}^2$ of ${\square }_{\bar{g}}$.

In three dimensions define

The Fourier multiplier of $K_{ij}$ is $k_i{k}_j/|\mathbf{k}{|}^4$. It is locally integrable at $\mathbf{k}=0$. No infrared subtraction is needed for this multiplier.

The native double-rod causal distribution is

The convolution is defined because its second factor is compactly supported in space. This prescription is the two-rod composition of the canonical kernel.

For a compact distribution $d$, Taylor expansion of $K_{ij}(\mathbf{z}-\mathbf{y})$ outside its support gives

The statement holds for distributions of finite order: derivatives of the Taylor remainder on the compact support retain the inverse-square bound.

For $\mathbf{k}\ne 0$,

Proof. The metric functional contracted with $k_i$ has the spatial symmetric coefficient

Its spatial trace is $k_a/2$, and

Four-dimensional trace reversal in the metric inverse subtracts one half of the product of these traces. Therefore the metric contribution is

The scalar contribution is