Landmark X-E A specified gravitational action for the ramp relational scalar, and the first graviton tree

We freeze one gravitational action, one gauge condition, and one quadratic graviton operator for the ramp relational scalar of Landmarks X-A to X-D, in signature $(+,-,-,-)$, at fixed regulator width.

The action is the Einstein–Hilbert density, $S_{\mathrm{gr}}=-(2{\kappa }^2{)}^{-1}\int \sqrt{-q}R$, with the sign fixed by positivity of the transverse-traceless kinetic term and confirmed by Newtonian attraction. Its first variation is derived from the Palatini identity. The metric Euler equation is displayed with every switch factor, and the Noether identity of the total action is proved with its explicit switch defect. On the plateau algebra the total Lagrangian is exactly invariant, so the metric Euler operator is covariant and the classical identity of X-A to X-C holds through ${\kappa }^2$ on the coupled constraint surface. That surface is generically empty: the Bianchi identity forces $({\partial }_{\mu }g)T^{\mu }{}_{\rho }[{\varphi }_0]=0$.

The gauge-fixed operator is $P_{\mathrm{gr}}=-\frac{1}{4}\mathcal{P}\square$ with retarded inverse $-4\mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}}$, $\mathcal{P}$ the trace reversal. Its residual gauge algebra meets the compactly supported plateau algebra only in zero, so the frozen algebra acts on classical backgrounds alone.

One tree is constructed: the metric variation of the switched scalar density, propagated by ${{\Delta }_{\mathrm{R}}}^{\mathrm{gr}}$ and inserted into the affine slot of the completed observable. It is a totally Wick-ordered cubic in the scalar, supported in $J^{-}(\mathrm{supp}f)$, microcausally admitted, and normalized by the classical retarded response, whose relational dressing is $-2{{\mathsf{E}}_0}^{\mathrm{R}}{{\mathsf{E}}_0}^{\mathrm{R}}(({\partial }_{\nu }g)T^{\nu \mu }[{\varphi }_0])$. The one-tree observable is invariant under the frozen algebra through ${\kappa }^1$. At ${\kappa }^2$, the background variation with the graviton held fixed equals $-{\kappa }^2{\hat{A}}_1[{\mathcal{L}}_{\xi }\hat{h}]$ modulo the equation ideals, and this class is nonzero.

The gravitational self-interaction tree, the mixed one-loop term, the second-order quantum metric, and VII-C's word $B_f\star W_g\star B_{f^{\prime }}$ are named and not constructed. The regulator is not removed.

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

Figure fig:scope-X-E records that separation. The drawing adds no hypothesis.

: Constructed, at fixed regulator and fixed input, with a stated order boundary{ Frozen: $S_{\mathrm{gr}}=-(2{\kappa }^2{)}^{-1}\int \sqrt{-q}R$; de Donder gauge fixing $\frac{1}{4}\int C_{\mu }{C}^{\mu }$ for the quantum graviton; $P_{\mathrm{gr}}=-\frac{1}{4}\mathcal{P}\square$; ${{\Delta }_{\mathrm{R}}}^{\mathrm{gr}}=-4\mathcal{P}{{\mathsf{E}}_0}^{\mathrm{R}}$.

Proved. E1: the metric Euler equation (eq:metric-euler)–(eq:metric-euler-expanded) and the Noether identity (eq:noether) with its switch defect. E2: exact ${\mathcal{G}}_U$-invariance of the total Lagrangian, covariance of the metric Euler operator, the classical identity through ${\kappa }^2$ on the coupled constraint surface, and the Bianchi obstruction (eq:bianchi-constraint). E3: the operator, its retarded inverse, the projector, and the residual algebra. E4: one tree ${\mathsf{T}}_{\mathrm{tree}}$, Definition def:tree; kernel, support, admission, normalization. The one-tree observable is invariant on ${\mathfrak{g}}_U$ through ${\kappa }^1$. Its ${\kappa }^2$ background variation with $\hat{h}$ fixed is $-{\kappa }^2{\hat{A}}_1[{\mathcal{L}}_{\xi }\hat{h}]$ modulo the ideals, Theorem thm:ward; this class is nonzero, Proposition prop:witness. E5: no loop, no $W_g$, no subtraction.

Not opened: the self-interaction tree, the mixed loop, a second-order quantum metric, ghosts, gauge-parameter dependence, $\epsilon \to 0$, X-Enter.}

Three endings are available:

1. Constructed. The action, gauge fixing, and one tree are named, and the completed observable transforms correctly on the frozen algebra through the frozen order.

2. Proved obstructed. The gauge identity acquires a nonzero defect on that algebra at the frozen order, or the tree kernel leaves the declared domain.

3. Unresolved. The action, the gauge condition, or the propagator is unnamed.

The ending reached is (1) for the gates as frozen in Section sec:frozen: classical identity through ${\kappa }^2$, quantum-metric background identity through ${\kappa }^1$, one tree at ${\kappa }^2$. The ${\kappa }^2$ quantum-metric identity is not frozen; its defect is displayed. A sheet that froze that identity on this field content would reach ending (2) with Proposition prop:witness as the witness. The defect is not removed by shrinking the algebra.

A missing object is recorded as missing, not as a no-go. Nothing in this manuscript depends on a state for the graviton.

The coordinate prescription, density coefficients, and complete two-jet gauge law are copied from X-A.tex XA; the bare tests, multiplet Wick algebra, and fixed-regulator admission from X-B-v3.tex XB; the completed tests, recursion, and fixed-regulator identity from X-C-v3.tex XC; the separation sheet and explicit equation-ideal representatives from X-D-v3.tex XD. Their proofs are not repeated.

Indices are moved with $\eta$ unless $q$ is written. ${{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}},{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}$ are the Klein–Gordon Green operators, ${{\mathsf{E}}_0}^{\mathrm{R}},{{\mathsf{E}}_0}^{\mathrm{A}}$ the massless ones; $⟨v,{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}f⟩=⟨{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}v,f⟩$. Write $n_{\mu }={\partial }_{\mu }t$, $n^{\mu }={{\delta }^{\mu }}_0$.

Fix $f\in {C_c}^{\infty }((0,T)\times {\mathbb{R}}^3;\mathbb{R})$, $g\in {C_c}^{\infty }((0,T)\times {\mathbb{R}}^3;[0,1])$, a relatively compact plateau $U$ with $\overline{U}\subset P_g=\mathrm{int}\{g=1\}$, ${\mathfrak{g}}_U={C_c}^{\infty }(U;TM)$, and $\delta >0$ with

Fix $0<\epsilon <\delta$ once and for all. The ramp ${\sigma }_{\epsilon }(t)=\chi (t/\epsilon )$ is X-D's, flat at both ends. ${\mathcal{G}}_U$ is the group generated formally by $\kappa \xi +{\kappa }^2\zeta$, $\xi ,\zeta \in {\mathfrak{g}}_U$; every $F\in {\mathcal{G}}_U$ equals the identity outside a compact subset of $U$.

With $H={\mathrm{tr}}_{\eta }h$,

so that $\sqrt{-q}{q}^{\mu \nu }={\eta }^{\mu \nu }+\kappa a^{\mu \nu }+{\kappa }^2{A}^{\mu \nu }+O({\kappa }^3)$ and $\sqrt{-q}=1+\kappa b+{\kappa }^{2}B+O({\kappa }^3)$.

The ramp coordinates ${\mathsf{X}}_1,{\mathsf{X}}_2$ have zero Cauchy data at $\Sigma =\{t=0\}$ and solve

For sources vanishing on $t\le 0$ this agrees with retarded propagation, ${\mathsf{X}}_1=-{\partial }_{\mu }{{\mathsf{E}}_0}^{\mathrm{R}}({\sigma }_{\epsilon }{a}^{\mu \rho })$. The bare tests $c_1={\partial }_{\mu }(f{{\mathsf{X}}_1}^{\mu })$, $c_2$, the Euler coefficients

and the completed tests

are X-D's, at the fixed $\epsilon$. The completed scalar-only observable is $\hat{\Phi }(f)=\hat{\varphi }(f)+\kappa \hat{\varphi }(d_1)+{\kappa }^2\hat{\varphi }(d_2)$. We write

$d_1[h]$ is linear in $h$. There is a symmetric bilinear $Q[h,h^{\prime }]$ with

Proof. $A$ and $B$ are linear in $k$ with coefficients $a_1[k]$, $b_1[k]$; so ${\mathsf{L}}_{2,g}[h,k]={\mathsf{L}}_{2,g}[h,0]+{\mathsf{L}}_{1,g}[k]$. By (eq:coordinate-equations) and the zero data, ${\mathsf{X}}_2[h,k]={\mathsf{X}}_2[h,0]+{\mathsf{X}}_1[k]$. In $c_2=-{\partial }_{\mu }(f{B}^{\mu })+\frac{1}{2}{\partial }_{\mu }{\partial }_{\nu }(f{{\mathsf{X}}_1}^{\mu }{{\mathsf{X}}_1}^{\nu })$, $B^{\mu }={{\mathsf{X}}_1}^{\nu }{\partial }_{\nu }{{\mathsf{X}}_1}^{\mu }-{{\mathsf{X}}_2}^{\mu }$, the only $k$-dependence is $+{\partial }_{\mu }(f{{\mathsf{X}}_1}^{\mu }[k])=c_1[k]$. Collecting the $k$-linear terms of (eq:completed) gives $c_1[k]-{\mathsf{L}}_{1,g}[k]{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}f=d_1[k]$. The remainder is quadratic in $h$; polarize.

{}1.12

The ramp kernel of X-A is not VII-C's $B_f$; no identification is made. No cone, token, or Enter record is an output. The new propagator ${{\Delta }_{\mathrm{R}}}^{\mathrm{gr}}$ is used at fixed $\epsilon$ only; its sharp limit is not taken. X-D's convergence theorem concerns smooth scalar tests at classical backgrounds; it is not admission of a mixed graviton–scalar kernel, and is not used as such.

VII-C VIIC is audited in Section sec:E5. Its assumptions are copied, not inferred.

A gravitational state is not an input. Every statement of this paper at order ${\hslash }^0$ uses only the retarded inverse ${{\Delta }_{\mathrm{R}}}^{\mathrm{gr}}$; the Hadamard bidistribution named in Section sec:algebra enters only the definition of Wick ordering of graviton-quadratic bare terms and no computed quantity.

For a Lorentzian metric $q_{\mu \nu }$ of signature $(+,-,-,-)$,

Under $q\mapsto -q$ the Christoffel symbols, $R^{\rho }{}_{\sigma \mu \nu }$, $R_{\mu \nu }$, and $G_{\mu \nu }$ are unchanged; $R$ and $\sqrt{-q}R$ change sign. Signs therefore cannot be copied across signatures and are fixed below by a positivity check.

with ${\mathfrak{L}}_{\eta }={\mathfrak{L}}_q{|}_{q=\eta }$. Both actions are Lagrangian densities; their Euler–Lagrange derivatives are defined through compactly supported variations, and no decay of $h,k$ is assumed. $\kappa$ is formal. The dictionary ${\kappa }^2=8\pi G_{\mathrm{N}}$ is used in Diagnostic diag:sign only.

For a smooth compactly supported variation $\delta q_{\mu \nu }$,

Consequently

Proof. $\delta {{\Gamma }^{\rho }}_{\mu \nu }$ is a tensor. Both sides of $\delta R^{\rho }{}_{\sigma \mu \nu }={\nabla }_{\mu }\delta {{\Gamma }^{\rho }}_{\nu \sigma }-{\nabla }_{\nu }\delta {{\Gamma }^{\rho }}_{\mu \sigma }$ are tensors, and the identity holds in normal coordinates at any point, where $\Gamma =0$ and $\nabla =\partial$. Contracting, $\delta R_{\sigma \nu }={\nabla }_{\rho }\delta {{\Gamma }^{\rho }}_{\nu \sigma }-{\nabla }_{\nu }\delta {{\Gamma }^{\rho }}_{\rho \sigma }$, so $q^{\sigma \nu }\delta R_{\sigma \nu }={\nabla }_{\rho }{W}^{\rho }$ and $\sqrt{-q}{\nabla }_{\rho }{W}^{\rho }={\partial }_{\rho }(\sqrt{-q}{W}^{\rho })$. With $\delta \sqrt{-q}=\frac{1}{2}\sqrt{-q}{q}^{\mu \nu }\delta q_{\mu \nu }$ and $\delta q^{\mu \nu }=-q^{\mu \alpha }{q}^{\nu \beta }\delta q_{\alpha \beta }$,

The chain rule $\delta q=\kappa \delta h={\kappa }^2\delta k$ gives the remaining formulas.

Set $\Gamma =\kappa {\Gamma }_1[h]+{\kappa }^2({\Gamma }_1[k]+{\Gamma }_2[h,h])+O({\kappa }^3)$ with

Then

and, raising with $\eta$,

The linearized Bianchi identity ${\partial }_{\mu }{G_{\mathrm{lin}}}^{\mu \nu }[h]\equiv 0$ holds for every smooth $h$, by direct computation from (eq:G2). No later proof uses the explicit form of $G_2$ or of any higher coefficient; they are displayed so that (eq:metric-euler-expanded) is a definite equation.

With

one has $\partial {\mathfrak{L}}_q/\partial q_{\mu \nu }=-\frac{1}{2}\sqrt{-q}{T}^{\mu \nu }[q,\varphi ]$ (no $\partial q$ enters ${\mathfrak{L}}_q$), hence

The flat stress tensor obeys

X-A's first-order switched vertex is the metric variation of the switched density:

so $\delta V_{1,g}/\delta h_{\mu \nu }=-\frac{1}{2}g{T}^{\mu \nu }[\varphi ]$ and $\delta V_{1,g}/\delta \varphi =-{\mathsf{L}}_{1,g}[h]\varphi$.

Proof. $-\frac{1}{2}{h}_{\mu \nu }{T}^{\mu \nu }=-\frac{1}{2}{h}^{\mu \nu }{\partial }_{\mu }\varphi {\partial }_{\nu }\varphi +\frac{1}{4}H(\partial \varphi \cdot \partial \varphi -m^2{\varphi }^2)=\frac{1}{2}{a}^{\mu \nu }{\partial }_{\mu }\varphi {\partial }_{\nu }\varphi -\frac{1}{2}{m}^{2}b{\varphi }^2$ by (eq:density). The $\varphi$-derivative is X-C's Hessian ledger $V^{\prime \prime }=-\mathsf{L}$, extended here to the mixed block.

The metric equation is ${\mathsf{E}}^{\mu \nu }=0$; the $k$-equation is the same equation. Expanded,

where $[\sqrt{-q}{G}^{\mu \nu }{]}_{(3)}$ is the cubic coefficient of the closed form, defined by Lemma lem:palatini and the conventions above.

The switch $g$ multiplies the matter source and nothing else; $\partial g$ does not appear in $\mathsf{E}$. It appears in the divergence of $\mathsf{E}$, which is the content of E1's identity. The scalar Euler operator ${\mathsf{E}}_{\varphi }:=\delta S/\delta \varphi =\delta {S^{\mathrm{sc}}}_g/\delta \varphi$ is X-A's switched operator, with ${\mathsf{E}}_{\varphi }=-P_m\varphi +O(\kappa )$.

Let $E^{\mu \nu }=\delta S/\delta q_{\mu \nu }={\kappa }^{-1}{\mathsf{E}}^{\mu \nu }$, a symmetric tensor density, and $\nabla$ the Levi-Civita connection of $q$. For all smooth $(q,\varphi )$, pointwise on $M$,

On the plateau $P_g$ the right side is ${\mathsf{E}}_{\varphi }{\partial }_{\rho }\varphi$. The gravitational action contributes nothing to the right side.

Proof. Let $\xi \in {C_c}^{\infty }(M;TM)$ and vary ${\delta }_{\xi }q={\mathcal{L}}_{\xi }q$, ${\delta }_{\xi }\varphi ={\xi }^{\mu }{\partial }_{\mu }\varphi$, with $g$ held fixed.

Direct computation. $\sqrt{-q}R$ is a natural density, so ${\delta }_{\xi }(\sqrt{-q}R)={\partial }_{\mu }({\xi }^{\mu }\sqrt{-q}R)$ and ${\delta }_{\xi }{S}_{\mathrm{gr}}=0$. Likewise ${\delta }_{\xi }{\mathfrak{L}}_q={\partial }_{\mu }({\xi }^{\mu }{\mathfrak{L}}_q)$, so ${\delta }_{\xi }\int g{\mathfrak{L}}_q=-\int ({\partial }_{\mu }g){\xi }^{\mu }{\mathfrak{L}}_q$. For the flat term, write naturality of ${\mathfrak{L}}_q$ at $q=\eta$:

The bracket is ${\delta }_{\xi }{\mathfrak{L}}_{\eta }[\varphi ]$, and $\partial {\mathfrak{L}}_q/\partial q_{\mu \nu }{|}_{\eta }=-\frac{1}{2}{T}^{\mu \nu }[\varphi ]$, $({\mathcal{L}}_{\xi }\eta {)}_{\mu \nu }={\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$. Hence ${\delta }_{\xi }{\mathfrak{L}}_{\eta }={\partial }_{\mu }({\xi }^{\mu }{\mathfrak{L}}_{\eta })+T^{\mu \nu }[\varphi ]{\partial }_{\mu }{\xi }_{\nu }$, and

Euler–Lagrange form. ${\delta }_{\xi }S=\int E^{\mu \nu }{\mathcal{L}}_{\xi }{q}_{\mu \nu }+\int {\mathsf{E}}_{\varphi }{\xi }^{\rho }{\partial }_{\rho }\varphi$, and for a symmetric tensor density $\int E^{\mu \nu }{\mathcal{L}}_{\xi }{q}_{\mu \nu }=-2\int {\xi }^{\rho }{q}_{\rho \nu }{\nabla }_{\mu }{E}^{\mu \nu }$, by ${\mathcal{L}}_{\xi }{q}_{\mu \nu }={\xi }^{\lambda }{\partial }_{\lambda }{q}_{\mu \nu }+2q_{\lambda (\nu }{\partial }_{\mu )}{\xi }^{\lambda }$, one integration by parts, and $2q_{\rho \nu }{{\Gamma }^{\nu }}_{\mu \sigma }{E}^{\mu \sigma }=E^{\mu \sigma }(2{\partial }_{\mu }{q}_{\rho \sigma }-{\partial }_{\rho }{q}_{\mu \sigma })$.

Comparison. Integrate the last term of the direct computation by parts, use (eq:T-divergence), and equate coefficients of the arbitrary ${\xi }^{\rho }$.

The tensor $({\partial }_{\mu }g)T^{\mu \nu }[\varphi ]$ is the same tensor that controls the varied-action defect of X-A's Route II off the plateau. Proposition prop:noether is its Bianchi form: it measures the failure of the switched matter source to be conserved, and it is independent of any gauge parameter and of $\epsilon$.

Let $F\in {\mathcal{G}}_U$, and let $\mathcal{L}[q,\varphi ]=(1-g){\mathfrak{L}}_{\eta }[\varphi ]+g{\mathfrak{L}}_q[q,\varphi ]-(2{\kappa }^2{)}^{-1}\sqrt{-q}R[q]$. For all smooth $(q,\varphi )$, the density $\mathcal{L}[F^{\ast }q,F^{\ast }\varphi ]-\mathcal{L}[q,\varphi ]$ is supported in a compact subset of $U$ and integrates to zero.

Proof. $F$ is the identity outside a compact $L\subset U$, and $F(L)=L$. Outside $L$ all pulled-back fields and their derivatives coincide with the originals, which gives the support statement.

$\sqrt{-q}R$ and ${\mathfrak{L}}_q$ are natural: $\omega [F^{\ast }q,F^{\ast }\varphi ]=F^{\ast }(\omega [q,\varphi ])$ as densities. Hence $\int (F^{\ast }\omega -\omega )=0$ for $\omega =\sqrt{-q}R$, since $F$ is a diffeomorphism of $M$ onto itself, and $\int g{F}^{\ast }\omega =\int (g\circ F^{-1})\omega =\int g\omega$ for $\omega ={\mathfrak{L}}_q$, since $g\circ F^{-1}=g$: on $L$ both sides equal $1$, and off $L$$F^{-1}=\mathrm{id}$. Finally $\mathrm{supp}(1-g)\subset M\setminus P_g\subset M\setminus L$, where $F^{\ast }\varphi =\varphi$ with all derivatives; so the $(1-g){\mathfrak{L}}_{\eta }$ term is unchanged pointwise.

For $F\in {\mathcal{G}}_U$ and all smooth $(q,\varphi )$,

as a contravariant symmetric tensor density and a scalar density.

Proof. Apply Lemma lem:invariance to $(q+s\delta q,\varphi +s\delta \varphi )$ with compactly supported $(\delta q,\delta \varphi )$ and differentiate at $s=0$:

Change variables by $F$ on the left; the pairing of a tensor density with a tensor is invariant. Since $\delta q,\delta \varphi$ are arbitrary, the densities agree.

This is X-A's plateau covariance lemma for the scalar Euler operator, rerun with the gravitational term and extended to the metric Euler operator. The gravitational term contributes no new defect: Lemma lem:invariance is exact, not an inheritance.

$\mathcal{C}$ is the set of formal triples $(h,k,\varphi )$, $\varphi ={\varphi }_0+\kappa {\varphi }_1+{\kappa }^2{\varphi }_2+\dots$, with ${\mathsf{E}}^{\mu \nu }[q,\varphi ]=0$ and ${\mathsf{E}}_{\varphi }[q,\varphi ]=0$ through ${\kappa }^2$ in (eq:metric-euler-expanded), with the ramp coordinates (eq:coordinate-equations) unchanged.

1. $\mathcal{C}$ is ${\mathcal{G}}_U$-invariant: if $(h,k,\varphi )\in \mathcal{C}$ and $(h^{\prime },k^{\prime })$ is the complete two-jet transform $h^{\prime }=h+K\xi$, $k^{\prime }=k+{\mathcal{L}}_{\xi }h+K\zeta +\frac{1}{2}{\mathcal{L}}_{\xi }K\xi$, $K\xi ={\mathcal{L}}_{\xi }\eta$, with ${\varphi }^{\prime }=F^{\ast }\varphi$, then $(h^{\prime },k^{\prime },{\varphi }^{\prime })\in \mathcal{C}$.

2. X-A's relational scalar and X-C's completed classical observable are ${\mathcal{G}}_U$-invariant through ${\kappa }^2$ for all smooth $(h,k)$, in particular on $\mathcal{C}$. The gravitational action enters neither object.

3. The varied-action identity of X-A's Route II holds for the total action with zero defect on the plateau. Off the plateau the defect is the right side of (eq:noether), recorded for all $(h,k)$, on or off $\mathcal{C}$.

Proof. (i) is Proposition prop:covariance: $F^{\ast }$ is invertible order by order and maps zero to zero. (ii) is X-A (iv) and X-C's fixed-regulator identity, both proved off shell in the metric; X-D's Corollary on the fixed algebra gives the explicit equation-ideal representatives. (iii) is Lemma lem:invariance together with Proposition prop:noether.

The relational scalar has no on-shell/off-shell distinction on ${\mathfrak{g}}_U$. The new content of E2 is that the constraint surface itself is invariant, and the following fact about it.

If $(h,k,\varphi )\in \mathcal{C}$, then

where ${\varphi }_0$ is the free solution $P_m{\varphi }_0=0$ at order ${\kappa }^0$. For $m=0$, ${\varphi }_0=t$, and any nonconstant $g$, $({\partial }_{\mu }g)T^{\mu }{}_{\rho }[t]=({\partial }_{t}g)n_{\rho }-\frac{1}{2}{\partial }_{\rho }g$ does not vanish on $\mathrm{supp}\mathrm{d}g$; hence $\mathcal{C}=\varnothing$ for such data.

Proof. Insert $E^{\mu \nu }=0$, ${\mathsf{E}}_{\varphi }=0$ into (eq:noether): the right side vanishes identically in $\kappa$. At order ${\kappa }^0$, ${\mathfrak{L}}_{\eta }-{\mathfrak{L}}_q=O(\kappa )$, $\varphi ={\varphi }_0$, and ${\mathsf{E}}_{\varphi }=0$ at ${\kappa }^0$ is $P_m{\varphi }_0=0$, which kills the last term. What remains is (eq:bianchi-constraint). For ${\varphi }_0=t$, $T^{\mu \nu }[t]=n^{\mu }{n}^{\nu }-\frac{1}{2}{\eta }^{\mu \nu }$.