The full source stress closes the cubic bootstrap

5. The bootstrap on the projective quotient

5.1 Recover the action and its variation domain

Let $A[T]$ be the generator reconstructed in Theorem 3.1. The source action uses endpoint and short-path operations:

$$S_{\mathrm{src}}[T,B,\chi]=\lambda^{-2}\int_U \left[(B-\eta)^{\mu\nu}D_{\mu\nu}(A[T]) +B^{\mu\nu}Q_{\mu\nu}(A[T])\right]\,d^4x+S_m[B,\chi]. \tag{30}$$

This is the chosen variational law in raw relation (4). The free quadratic part is Deser's spin-two starting law. The cubic part will be checked against total-current self-coupling. Neither is a condition for source admission.

Projection by (2) gives

$$S_2=\int(h^{\mu\nu}D_{\mu\nu}(c)+\eta^{\mu\nu}Q_{\mu\nu}(c)),\qquad S_3=\lambda\int h^{\mu\nu}Q_{\mu\nu}(c). \tag{31}$$

The inverse is $B=\eta+\lambda h$ and $T=\operatorname{Hol}(\lambda c)$, by the transport ODE. Every density and generator variation in the declared domain lifts through (11). Thus the action's variational operations are recovered along with its values.

5.2 Construct the quotient inverse explicitly

Let $c_I=c^\alpha{}_{\mu\beta}$ range over all 64 components in four dimensions. Define

$$q_z(c)=z^{\mu\nu}Q_{\mu\nu}(c),\qquad (H_z)_{IJ}=\frac{\partial^2q_z}{\partial c_I\partial c_J}, \qquad J(z)=D^*z. \tag{32}$$

Compact variations fix the adjoint: $\int z^{\mu\nu}D_{\mu\nu}(u)=\int J(z)_Iu_I$ after the matched boundary integration. Explicitly,

$$J(h)^\ell{}_{pq}=-\partial_\ell h^{qp}+\delta^\ell_p\partial_n h^{qn}. \tag{33}$$

These definitions have no inverse differential operator. Let $P\omega$ have components $\delta^\alpha_\beta\omega_\mu$. In the fixed component basis $P^TP=4I$ and $\Pi=PP^T/4$ projects onto $\ker H$ with $H=H_\eta$. Theorem 4.1 makes $H+\Pi$ invertible. Set

$$G=(H+\Pi)^{-1}-\Pi,\qquad HG=GH=I-\Pi,\qquad G^T=G. \tag{34}$$

This is a coordinate inverse on the quotient. A different valid inverse changes the relevant outputs only by a projective vector, which $D$ annihilates. No physical Euclidean metric is selected by this use of component space.

For symmetric $z$, $H_zP=0$ by projective invariance of $q_z$ and $P^TJ(z)=0$ by $D(P\omega)=0$. Define

$$L(h)=-GJ(h),\qquad K(h,c)=GH_hc,\qquad \tau_{\mu\nu}(h,c)=Q_{\mu\nu}(c)-D_{\mu\nu}K(h,c). \tag{35}$$

The free equation $Hc+J(h)=0$ gives $c=L(h)+P\omega$. The symmetric Ricci readout is independent of $\omega$. The linearized Levi–Civita connection of

$$\varphi_{\mu\nu}=-h_{\mu\nu}+\tfrac12\eta_{\mu\nu}\operatorname{tr}_\eta h \tag{36}$$

solves this equation and differs from $L(h)$ only projectively. Thus $r(h)=DL(h)$ is exactly Deser's metric spin-two operator:

$$2r_{\mu\nu}(h)=\Box h_{\mu\nu} -\partial_\mu\partial^\alpha h_{\alpha\nu} -\partial_\nu\partial^\alpha h_{\alpha\mu} -\tfrac12\eta_{\mu\nu}\Box\operatorname{tr}_\eta h . \tag{37}$$

Its trace-adjusted Einstein form has identically zero flat divergence. Substitution yields the linear gauge transformations $\delta h^{\mu\nu}=\eta^{\mu\nu}\partial_\alpha\xi^\alpha-\partial^\mu\xi^\nu-\partial^\nu\xi^\mu$ and $\delta c^\alpha{}_{\mu\nu}=\partial_\mu\partial_\nu\xi^\alpha$, together with projective freedom. The gauge identities follow from the chosen quadratic law.

5.3 Vary the auxiliary background horizontally

Introduce auxiliary inverse metric density $\psi$ near $\eta$. Here $h$ is a weight-one tensor density and $c$ a general $(1,2)$ tensor. With $C(\psi)=\mathrm{LC}(g(\psi))$, covariantize

$$S_2(\psi)=\int\left[ h^{\mu\nu}(\nabla_\alpha c^\alpha{}_{\nu\mu}-\nabla_\nu c^\alpha{}_{\alpha\mu}) +\psi^{\mu\nu}Q_{\mu\nu}(c)\right]. \tag{38}$$

The auxiliary background is torsion free as part of the computation. The dynamical $c$ is unrestricted. Its direct connection variation is

$$\begin{aligned} \delta_C S_{2,\mathrm{kin}}=\int h^{\mu\nu}\big[ &\delta C^\alpha{}_{\alpha\rho}c^\rho{}_{\nu\mu} -\delta C^\rho{}_{\alpha\nu}c^\alpha{}_{\rho\mu}\\ &-\delta C^\rho{}_{\alpha\mu}c^\alpha{}_{\nu\rho} +\delta C^\rho{}_{\nu\mu}c^\alpha{}_{\alpha\rho}\big]. \end{aligned} \tag{39}$$

It also equals $\int(H_hc)^T\delta C$. Check the equality directly, or expand $BR(A)$ around $(\psi,C(\psi))$ and take the quadratic coefficient.

Theorem 5.1 (Off-shell source stress). Under the auxiliary tensor/density variation,

$$\left.\frac{\delta S_2(\psi)}{\delta\psi^{\mu\nu}}\right|_{\psi=\eta} =Q_{\mu\nu}(c)-D_{\mu\nu}(GH_hc)=\tau_{\mu\nu}(h,c). \tag{40}$$

The identity holds off shell for all smooth $h,c$ in the domain, including unrestricted torsion components.

Proof. The variation $\delta C$ is the linearized Levi–Civita connection of $\delta\psi$. It solves $H\delta C+J(\delta\psi)=0$, so $\delta C=-GJ(\delta\psi)+P\zeta$ for a one-form $\zeta$. Since $H_hP=0$,

$$\int(H_hc)^T\delta C=-\int(GH_hc)^T J(\delta\psi) =-\int\delta\psi^{\mu\nu}D_{\mu\nu}(GH_hc). \tag{41}$$

The explicit $\psi Q$ term supplies $Q$. The integration uses the same compact-variation boundary contract as (32). This proves every component of (40), with derivatives of both fields retained.

The source operation is

$$B(\psi)=\psi+\lambda h,\qquad T(\psi)=\operatorname{Hol}[C(\psi)+\lambda c],\qquad c_\psi=\frac{A[T]-C(\psi)}{\lambda}. \tag{42}$$

Use the relative tensor generator in (38). Holding $c_\psi$ fixed is different from freezing the physical transport. If $B$ follows (42) while $T$ is frozen, the chain-rule defect is

$$\tau_{\mathrm{fixed}\ T}=\tau+\lambda^{-1}D(GE_c^{(2)}),\qquad E_c^{(2)}=J(h)+Hc . \tag{43}$$

Indeed $\delta c_\psi=-\delta C/\lambda$, and the adjoint calculation gives (43). For $h=0$ and only $c^0{}_{11}=x^0$ nonzero, the correct $\tau_{11}$ is zero and the defect is $1/\lambda$. Dropping the kinetic covariant variation altogether is a different failure: it keeps $Q$ and loses $-DK$. Equation (42) implements the intended operation.

5.4 The exact residual closes the current

Theorem 5.2 (Closed bootstrap and off-shell recovery). The first-order Euler residuals of $S_2+S_3$ are

$$E_h=D(c)+\lambda Q(c),\qquad E_c=J(h)+Hc+\lambda H_hc . \tag{44}$$

The exact identity is

$$r(h)+\lambda\tau(h,c)=E_h-D(GE_c). \tag{45}$$

On source solutions $r(h)=-\lambda\tau$. The interaction supplies no further auxiliary stress, so the self-coupling closes in one step.

Proof. Apply $G$ to the second residual: $GE_c=-L(h)+(I-\Pi)c+\lambda K(h,c)$. Since $D\Pi c=0$, subtracting its $D$ image from $E_h$ gives (45). On a connection solution, $c=L(h)-\lambda K(h,c)+P\omega$ exactly, and its derivative contributes the derivative stress in the $h$ equation. Holding the relative tensor $c$ fixed under (42), $\lambda hQ$ has no explicit auxiliary-background dependence or explicit derivatives. Its auxiliary stress is zero, so the current already equals the total self-current in this prescription.

Equation (45) gives $\|r+\lambda\tau\|\le\|E_h\|+\|DGE_c\|$ in matching norms. It exposes the derivative cost of transferring an error in the connection equation. Small undifferentiated connection residuals alone do not certify a small Einstein residual.

The completed action follows after the current has been verified:

$$\lambda^{-2}B^{\mu\nu}R_{\mu\nu}(A) =\lambda^{-1}\eta^{\mu\nu}D_{\mu\nu}(c)+h^{\mu\nu}D_{\mu\nu}(c) +\eta^{\mu\nu}Q_{\mu\nu}(c)+\lambda h^{\mu\nu}Q_{\mu\nu}(c). \tag{46}$$

The first term is the explicit constant-density divergence. No convergence of an infinite series is assumed. The current construction and nonlinear closure recover Deser's equations throughout the declared domain.