Em02-E Relational Cosmological Lapse: Classical Identification and a Renormalization Audit Full-tensor quadratic kernels, nonlocal support, and the boundary of the quantum regulator-removal question

We identify one relational lapse in an explicitly specified Einstein–scalar theory on a power-law inflationary background. The clock is the scalar field. Spatial labels are harmonic coordinates on its level sets, with the asymptotic Cartesian boundary condition. We use the original inflaton-slice harmonic rods, rather than the modified rods employed in some simplified cosmological formulas.

The complete classical expansion through second order is derived in tensor components. It retains arbitrary metric perturbations, second-order field data, the clock displacement, and the spatial coordinate correction. At this order the lapse depends on the first spatial rod correction but not on the second. The resulting quadratic functional has an explicit local part and an instantaneous spatial-Green part.

We prove that compact external smearing does not generally make the functional derivatives spatially compact. We also calculate the off-diagonal wavefront set of the instantaneous Green kernel and show, by an exact distributional example, that an internal diagonal correction can have a non-diagonal observable-level image. Quantitative tail and regulator-comparison criteria separate spatial integration from ultraviolet removal.

The quantum input gate is not passed: a complete gauge-fixed covariance, a state or parametrix with the required spatial control, an admissible counterterm classification, and the retained-order quantum gauge identity are not constructed. Accordingly, the release is an incomplete quantum audit, not a regulator-removal theorem or a universal obstruction. The classical formulas and auxiliary distributional results are proved independently of the declared models in Em02-C and Em02-D.

Release boundary. This paper completes a classical identification and a mathematical audit of the proposed quantum calculation. It does not complete the quantum regulator-removal problem. In particular, the title does not assert regulator removal.

The intended quantum question is:

For the relational lapse specified here, at a stated quantum order, does the regulated expression admit a distributional limit after corrections satisfying its renormalization and quantum gauge requirements?

A meaningful answer requires an actual regulated quantum family and an independently determined correction class. Neither can be replaced by a model coefficient from another paper.

The relational construction follows the geometric prescription of Brunetti, Fredenhagen, Hack, Pinamonti, and Rejzner BFHPR (Sections 3–4). Its general form is

Here $f$ is a compact test density on the background label space. It is fixed under active gauge transformations. It is not the scalar potential, an interaction switch, or a state.

We choose the scalar clock and the original inflaton-slice harmonic spatial coordinates described in BFHPR (equations (28)–(31)). We denote the latter by ${Y_{\Gamma }}^i$, and set

The modified spatial rods of equation (35) of that reference are not used. The two choices generally give different second-order observables. This distinction is part of the identification, not an approximation.

All calculations below use cosmic time and the action normalization in Section sec:theory. We do not import the reference's on-shell scalar formulas without converting their conventions.

The parameter $\lambda$ counts field perturbations:

It is not $\hslash$, a smoothing length, or an interaction coupling.

The concrete classical target is the complete coefficient through ${\lambda }^2$. The first prospective quantum calculation is the order-$\hslash$ contraction associated with quantizing its quadratic part in a chosen free representation. Section sec:quantum derives this contraction count.

An interacting order-$\hslash$ observable can require additional action vertices and observable coefficients, depending on the coupling grading. No complete interacting inventory is claimed from the free quadratic count.

The present quantum outcome is the third. The positive results are the classical identification, quadratic kernel bridge, and distributional support and tail calculations proved below.

Use signature $(-+++)$, units $c=1$, and the action

The curvature convention is

The field equations are

All variations in the classical observable calculation are off shell. The background alone is required to solve (eq:Einstein)–(eq:KG).

Fix

and define

Work on

A dot denotes cosmic-time differentiation. Put

There is no limit toward $t=0$ in this paper.

The pair $(g_0,{\varphi }_0)$ solves (eq:Einstein)–(eq:KG). It is expanding and accelerating, and its scalar clock is nondegenerate on every compact subinterval of $I$.

Proof. For a spatially flat homogeneous background, the equations obtained from (eq:Einstein)–(eq:KG) are

Substitution gives

which proves the first equation. The second follows from $\dot{H}=-p/t^2$. Writing $c=\sqrt{2p}{M}_{\mathrm{Pl}}$, the third equation has terms

whose sum is zero.

Finally, $\dot{a}>0$, $\ddot{a}/a=p(p-1)/t^2>0$, and $v>0$ has a positive lower bound on each compact time interval.

For the calculations through second order, take smooth, spatially compact perturbations with a common compact support:

The tensors $h,k$ are arbitrary symmetric covariant tensors. No scalar, vector, or tensor metric component is discarded.

The formulas are formal Taylor coefficients. For an actual nonperturbative coordinate map, one must additionally restrict to configurations for which the scalar level sets are spacelike, the harmonic-coordinate problem has the stated boundary solution, and the resulting coordinate map is invertible on the region used by $f$. We do not assert such global properties for every smooth configuration.

For the formal coefficients considered here, all elliptic sources are smooth and spatially compact, and the required decaying solutions are constructed explicitly.

Use $s\in I$ as a background clock label. The perturbed slice carrying that label is

Write it as

The unique formal solution of the clock equation is

All quantities on the right are evaluated at $(s,\mathbf{x})$.

Proof. Expand ${\tilde{\varphi }}_{\lambda }({\tau }_{\lambda },\mathbf{x})$ through second order:

Setting the coefficients to zero proves the result because $v\ne 0$.

Let

The induced positive spatial metric on the clock slice is

Write

The coefficients are

Proof. The spatial tangent vectors are

Taylor-expand the spatial-spatial metric component at $t=s$. The mixed components first contribute at degree two, as $\lambda h_{0i}\lambda {\partial }_j{u}_1$. The background time-time component is $-1$, so its degree-two contribution is $-{\lambda }^2{\partial }_i{u}_1{\partial }_j{u}_1$. With the factorial convention of (eq:field-jet), these are exactly (eq:q1)–(eq:q2).

Set

where spatial indices on $Q_1,Q_2$ are raised using ${\delta }_{ij}$. Define

These matrices are unrelated to the component maps denoted by $\mathsf{A}$ in Em02-A–D.

For a scalar $F$, the slice Laplacian is

One has

where

In particular,

Proof. The determinant and inverse expansions are

Their product has coefficients $A_1,A_2$. The inverse determinant factor is $1-\lambda b_1+{\lambda }^2(2{b_1}^2-b_2)/2$. Multiplication proves (eq:P1)–(eq:P2). Applying $P_1$ to $x^i$ gives (eq:B).

Fix

Then $\Delta G_{\Delta }F=F$, and $G_{\Delta }F\to 0$ at spatial infinity. For compact smooth $F$, also $G_{\Delta }\Delta F=F$. The sign in (eq:Coulomb) is fixed by $\Delta (-1/(4\pi |\mathbf{x}|))={\delta }^{(3)}$.

Seek harmonic rods on the clock slice in the form

with ${Y_{\lambda }}^i-x^i\to 0$ at spatial infinity coefficientwise.

The unique decaying coefficients are

They are smooth. On compact time intervals, $Y_1=O(|\mathbf{x}{|}^{-2})$ and $Y_2=O(|\mathbf{x}{|}^{-1})$ at spatial infinity.

Proof. The harmonic equation at the two successive orders is

All coefficients of $P_1,P_2$ have spatially compact support. Hence both sources are smooth and compactly supported, and (eq:Coulomb) constructs the solutions.

Any difference is a smooth harmonic function tending to zero at infinity. The maximum principle on successively larger balls gives zero, proving uniqueness.

The second estimate is the usual direct $1/|\mathbf{x}|$ bound for the displayed compact-source integral. For the first, (eq:B) is a spatial divergence, so its integral vanishes. Subtract $|\mathbf{x}{|}^{-1}$ from the kernel in (eq:Coulomb) outside a ball containing the source. The remaining kernel is $O(|\mathbf{x}{|}^{-2})$. The same estimates are uniform on compact time intervals.

The background slice inverse is $a(s{)}^2{G}_{\Delta }$. It is an elliptic spatial inverse, not a retarded or advanced spacetime inverse.

Let $Z_{\lambda }(s,\cdot )$ be the formal spatial inverse of $Y_{\lambda }(s,\cdot )$. Then

The relational map is

Consequently its first and second displacement coefficients are

These identities follow by composing the Taylor series. They implement (eq:observable) with $X_{{\Gamma }_0}(s,\mathbf{x})=({\varphi }_0(s),\mathbf{x})$.

Define

The timelike-clock branch has $W_0=v^2>0$.

In this section, raising a spacetime index on $h,k$ uses $g_0$, not the Euclidean spatial metric. Set

Write $W_{\lambda }=W_0+\lambda W_1+{\lambda }^2{W}_2/2$.

The coefficients are

For $N_{\varphi ,\lambda }=N_0+\lambda N_1+{\lambda }^2{N}_2/2$,

Proof. The inverse metric expansion is

Multiply it by the two scalar-gradient expansions and retain the minus sign in (eq:lapse). This gives (eq:W1)–(eq:W2). The first two derivatives of $W^{-1/2}$ give (eq:N).

This tensor formula includes the mixed metric components and all scalar spatial gradients. No scalar-sector display is used as a substitute for the full coefficient.

The coefficients are

Every quantity on the right is given explicitly by (eq:u), (eq:q1), (eq:B), (eq:Y), and (eq:W1)–(eq:N).

The second spatial rod coefficient $Y_2$ does not occur in $n_2$.

Proof. The composition rule gives

Since $N_0$ depends only on time, substitution of (eq:alpha) leaves

Use $z_1=-Y_1$ and ${\partial }_i(N_1+{\dot{N}}_0{u}_1)={\partial }_i{n}_1$. The spatial part of ${\alpha }_2$ multiplies ${\partial }_i{N}_0=0$, which proves the last assertion.

In cosmic coordinates, $h^{00}=h_{00}$ and ${g_0}^{00}=-1$. Thus

To compare with conformal time, let $\mathrm{d}t=a\mathrm{d}\tau$. Then ${\varphi }_0^{\prime }=av$, $\mathcal{H}=aH$, and the conventional scalar lapse perturbation satisfies $A=-h_{00}/2$. Consequently

This matches the off-shell first-order relational lapse construction in BFHPR (equation (36)). No on-shell Bardeen-potential identity is needed here.

Take formal pullbacks generated by smooth vector fields compactly supported in $I\times {\mathbb{R}}^3$. They preserve the prescribed asymptotic spatial labels. The background label map $X_{{\Gamma }_0}$ remains fixed.

Through the retained order,

and therefore

Proof. The scalar clock transforms by pullback. A diffeomorphism maps its level hypersurfaces and their induced metrics into the corresponding level hypersurfaces and metrics of the transformed configuration.

The pullback of each harmonic spatial coordinate is harmonic for the pulled-back slice metric. Because $\chi$ is the identity at spatial infinity, it has the same asymptotic Cartesian boundary data. The coefficientwise uniqueness proved in Proposition prop:rods identifies it with the coordinate constructed for ${\chi }^{\ast }\Gamma$.

Invert the resulting scalar-coordinate identity to obtain the transformation law for $\alpha$. The lapse is a scalar built from the metric and scalar gradient, so $N_{\varphi ,{\chi }^{\ast }\Gamma }=N_{\varphi ,\Gamma }\circ \chi$. The two compositions cancel. Integration against the fixed label-space density $f$ preserves the identity.

This proposition is an off-shell classical statement. It is neither a quantum BRST identity nor a statement that a particular gauge-fixed state or regulator is invariant. It also does not assert invariance under diffeomorphisms that change the boundary data at spatial infinity.

Fix a real compact test density

For an affine perturbation $\eta =(h,\phi )$, set $k=\psi =0$ in the coefficient formulas and define

Then

Define the linear differential operations

Set

This is a local quadratic expression. Every coefficient in it is fixed by the preceding formulas.

The quadratic observable is exactly

For compact perturbations this is a continuous quadratic functional. It has a symmetric distributional Hessian ${\mathcal{H}}_f$ satisfying

Its bilocal part is obtained by polarizing the second line of (eq:kernel-bridge). There are no additional classical quadratic terms for this coordinate prescription.

Proof. The last term of (eq:n2), divided by two, is $-{Y_1}^i{\partial }_i{n}_1$. Since ${Y_1}^i=-G_{\Delta }{B}^i\eta$, it is precisely the second line of (eq:kernel-bridge). All other terms are (eq:Llocal).

On any fixed compact input-support class, the kernel $|\mathbf{x}-\mathbf{y}{|}^{-1}$ is locally integrable. The differential operations $B,P$ have smooth coefficients on compact time intervals. Direct integral estimates give continuity in finite smooth seminorms.

Polarization gives the symmetric bilinear form

The local part is represented by derivatives of the diagonal delta distribution. The nonlocal part is represented by $G_{\Delta }(\mathbf{x}-\mathbf{y})\delta (t-u)$ with the displayed differential operations transferred to its two field slots. These are defined distributions before any quantum contraction. Their sum is ${\mathcal{H}}_f$.

The bridge is a classical Hessian calculation, not a graph renormalization theorem. Its metric field space has ten symmetric components plus the scalar component.

For a compact vector field $\xi$, put

Classical invariance implies

Proof. Differentiate

first at ${\Gamma }_0$, and then in direction $\eta$. The first derivative gives the first identity. The direction derivative gives the second, since the derivative of ${\mathcal{L}}_{\xi }\Gamma$ is ${\mathcal{L}}_{\xi }\eta$.

Thus linear gauge transformations of the quadratic coefficient couple to the nonlinear transformation of the linear coefficient. Dropping the latter is not a valid quadratic gauge audit.

As a direct first-order check,

makes (eq:linear-explicit) zero. For a pure spatial gauge perturbation, $Q_{1,ij}={\partial }_i{\xi }_j+{\partial }_j{\xi }_i$, so

These checks fix the displacement signs.

There are compact test densities $f$ for which ${\mathcal{H}}_f$ has noncompact spatial support in a field slot. This occurs even though the external smearing is compact.

Proof. Use two independent perturbation directions.

In the first direction take

Then $P\eta =0$, $Q_1=2D1$, and $B^i\eta ={\partial }_{i}D$.

In the second direction take

Then $P\theta =u$, $B^i\theta =0$. The local mixed term is zero: with both scalar perturbations zero, the lapse's local quadratic metric term has no mixed spatial-trace/time-time contribution.

Consequently the polarized quadratic form is

Integrating the compact $D$ source by parts gives its kernel coefficient

Choose $f_0(t,\mathbf{x})={\eta }_0(t)F(\mathbf{x})$, with $F\ge 0$, $\int F>0$, and let $u(t,\mathbf{x})=\beta (t)x^1$ on a neighborhood of its support, using a compact spatial cutoff outside that neighborhood. Then for a unit vector $\mathbf{n}$,

At times where ${\eta }_0\beta \ne 0$, this is nonzero in cones with $n_1\ne 0$ for arbitrarily large $R$. A compact $D$ supported near any such point detects the coefficient. Hence the kernel support is not spatially compact.

This does not prevent testing the Hessian on compact field directions. It does prevent inferring the compact functional support required by some local-functional constructions.

The background spatial inverse, considered as a spacetime kernel with coordinate-volume pairing, is

On the region $\mathbf{x}\ne \mathbf{y}$,

In relative variables $(\sigma ,\mathbf{r})=(t-u,\mathbf{x}-\mathbf{y})$, the principal kernel

is homogeneous of degree $-2$ and has scaling degree two at the origin.

Proof. For $\mathbf{x}\ne \mathbf{y}$, the spatial factor in (eq:instantaneous) is smooth and nowhere zero. Multiplication by it does not change the wavefront set of $\delta (t-u)$. In local coordinates $(t-u,u,\mathbf{x},\mathbf{y})$, this is the tensor product of a one-dimensional delta with a smooth nonvanishing factor. Its nonzero covectors are exactly the conormal covectors to $t=u$, proving (eq:WF).

Under positive dilation of all four relative coordinates,

The distribution is nonzero, so its scaling degree is exactly two. The smooth nonzero background coefficient preserves the leading scaling degree.

A spacetime instantaneous Green kernel is not singular only on the full spacetime diagonal. Equation (eq:WF) exhibits equal-time singularities at spatially separated points. Accordingly, a shorthand comparison with a full diagonal delta cannot be used as a literal wavefront-set identity.

The explicit instantaneous kernel in BFHPR (Section 5.2) must be analyzed together with the other graph factors. The present proposition concerns that kernel type alone; it does not assert that a particular complete graph has an off-diagonal divergence.

For a distribution $t$ in ${\mathbb{R}}^n$, use

The scaling-degree extension theorems apply after an actual distribution is defined away from the specified coincidence set BF (Theorems 5.2–5.3).

Suppose two extensions $t,\tilde{t}$ agree away from the origin and have scaling degrees at most $s<\infty$. Then

when $s\ge n$. If $s<n$, they agree.

Proof. Their difference is supported at the origin. A finite-order distribution supported at one point is a finite linear combination of delta derivatives. A nonzero highest derivative of order $m$ has scaling degree $n+m$. The bound $n+m\le s$ gives the result. For $s<n$, no nonzero point-supported term satisfies it.

For a diagonal with tangential variables, the corresponding coefficients can depend on those variables. Finite normal order alone does not make the complete ambiguity space finite-dimensional.

No scaling degree of a complete quantum lapse graph has been calculated here. The degree two of the single kernel (eq:instantaneous) is not the degree of its product with a differentiated Feynman distribution.

The admissible correction space cannot be obtained by simply naming the final attachment diagonal. It must be transported through the actual observable construction.

The following calculation is a distributional support test. It does not declare any particular counterterm physically admissible.

Let $c\in {C_c}^{\infty }({\mathbb{R}}^3)$ be nonzero and define

Applying a spatial Green kernel in the first slot gives the well-defined distribution

For every $\mathbf{z}$ where $c(\mathbf{z})\ne 0$, it is nonzero on open sets with $\mathbf{x}\ne \mathbf{z}$. Its support is therefore not contained in the output diagonal.

Proof. The right side is locally integrable because $|\mathbf{x}-\mathbf{z}{|}^{-1}$ is locally integrable in three relative spatial dimensions. For a compact smooth test $\Phi (\mathbf{x},\mathbf{z})$, its action is

This is also the distributional evaluation of

At separated points the kernel is smooth and nonzero, proving the support assertion.

Time can be retained as a common parameter in (eq:image). No product of two time deltas is required.

For a source derivative acting on the internal incidence, the precise formula is

interpreted distributionally. This retains derivatives of both the coefficient and the Green kernel.

For the quantum calculation, one would first classify graph-level corrections on the actual singular subgraphs. If the derived observable construction defines a map $\Pi$ on that class, its image is

Equation (eq:physical-image) is an interface to be instantiated. It is not an evaluated physical correction space in this paper.

Proposition prop:image rules out the inference

It does not prove that every extended support is allowed. Covariance, scaling, field content, gauge compatibility, and normalization still constrain ${\mathfrak{C}}_{\mathrm{graph}}$.

Before a subtraction is used to decide the quantum limit, the chosen theory and order must determine:

1. the graph distributions and their coincidence sets;

2. the tensor structures and background/clock dependence permitted in their extensions;

3. the derivative and scaling bounds;

4. causal and local covariance conditions in the applicable renormalization framework;

5. quantum BRST/BV conditions and the relevant equation ideal;

6. allowed finite normalizations;

7. existence and tail control for their observable-level images.

This list is an admissibility contract, not a completed classification of ${\mathfrak{C}}_{\mathrm{adm}}$.

Neither successful cancellation nor failure of attachment-diagonal support alone settles admissibility.

Let $K\subset {\mathbb{R}}^3$ be compact, and suppose

uniformly in $\delta$. For sufficiently large $R$,

For a first spatial derivative of $G_{\Delta }$,