Put matter on the same realization

Geometry and stress must share one calibration, comparison and normalization. Joint sections and variations recover their coupling, with trace reversal preserving information that trace deletion would lose.

9. Matter on the same realization and the general field equation

Einstein’s §16 introduces a matter energy array whose covariant form is symmetric and replaces the gravitational energy by gravitational plus matter energy in Einstein (51). He explicitly acknowledges that the relativity postulate alone does not justify this coupling. We inherit that premise and construct its operations on the same calibration, connection, coordinate domain and normalization.

Add a raw calibrated two-array $\Sigma$, retaining its antisymmetric part as a discrepancy until the symmetric sector is selected. Define

$$T=E^{\mathsf T}\Sigma E,\qquad T^\mu{}_{\nu}=(E^{-1}\eta^{-1}\Sigma E)^\mu{}_{\nu},\qquad \tau=\operatorname{tr}(qT)=\operatorname{tr}(\eta^{-1}\Sigma). \tag{9.1}$$

For a fixed $E$, the reverse is $\Sigma=E^{-\mathsf T}TE^{-1}$. Hence every smooth local metric and symmetric-stress pair has a joint realization. Equal pairs are represented precisely by

$$E'=\Lambda E,\qquad \Sigma'=\Lambda^{-\mathsf T}\Sigma\Lambda^{-1}, \qquad \Lambda^{\mathsf T}\eta\Lambda=\eta, \tag{9.2}$$

with the previously determined comparison transformation. This follows from Theorem 3.1 and cancellation of $E$ in the equal stress readouts. Changing only the calibration or only the stress is generally a different physical preparation.

Joint tangent coverage is equally explicit. For desired $h=\delta g$ and $K_T=\delta T$, let $H=\delta E=Eqh/2$ and $A=HE^{-1}$. Then

$$\delta\Sigma=E^{-\mathsf T}K_TE^{-1}-A^{\mathsf T}\Sigma-\Sigma A. \tag{9.3}$$

Differentiating (9.1) proves both prescribed responses. Holding the coordinate stress fixed corresponds to $K_T=0$, not generally to $\delta\Sigma=0$. This is a readout-response identity, not an assumed universal matter action.

Trace reversal, not trace deletion

For any covariant two-array define

$$\mathsf J_g(A)=A-\tfrac12g\operatorname{tr}(qA),\qquad \mathsf J_\eta(S)=S-\tfrac12\eta\operatorname{tr}(\eta^{-1}S). \tag{9.4}$$

In four dimensions, $\operatorname{tr}_q\mathsf J_g(A)=-\operatorname{tr}_q A$ and $\mathsf J_g^2(A)=A$. Substitution gives

$$E^{\mathsf T}\mathsf J_\eta(\Sigma)E=\mathsf J_g(T). \tag{9.5}$$

It is an invertible operation, unlike the trace-free projection $A-g\operatorname{tr}(qA)/4$.

Retain the source-action current $t$ of (8.4); it is not separately fitted to matter. Set

$$\mathcal J^\sigma{}_{\mu}=t^\sigma{}_{\mu}+T^\sigma{}_{\mu}, \tag{9.6}$$

and retain the off-shell coupled residuals

$$U=W+\kappa\mathsf J_g(T),\quad Z=\operatorname{Ric}-\kappa\mathsf J_g(T),\quad \mathcal K=\mathsf J_g(Z)=G-\kappa T, \qquad U=-Z-S. \tag{9.7}$$

Adding the same stress to (8.5) proves, on the unrestricted chart domain,

$$\boxed{\partial_\alpha(q^{\sigma\beta}\Gamma_{\!\mathrm E}^\alpha{}_{\mu\beta}) +\kappa\left(\mathcal J^\sigma{}_{\mu} -\tfrac12\delta^\sigma_\mu\operatorname{tr}\mathcal J\right) =q^{\sigma\beta}U_{\mu\beta}.} \tag{9.8}$$

Einstein’s total-energy replacement selects the zero of the left side in a unit-volume chart. Invertibility of $q$ gives $U=0$, recovering Einstein (52) and Einstein (53). Since $S=0$ there, the equivalent generally covariant equation is

$$\boxed{\operatorname{Ric}_{\mu\nu}=\kappa(T_{\mu\nu}-\tfrac12g_{\mu\nu}\tau) \quad\Longleftrightarrow\quad G_{\mu\nu}=\kappa T_{\mu\nu}.} \tag{9.9}$$

The scalar equation is $\mathcal R=-\kappa\tau$, so no trace component has been lost. In the calibrated representation,

$$\boxed{\mathscr G_{ab}=\kappa\Sigma_{ab},\qquad \mathscr G=\mathscr R-\tfrac12\eta\operatorname{tr}(\eta^{-1}\mathscr R).} \tag{9.10}$$

Equations (7.2) and (9.5) prove both directions. Every target solution pair lifts using the local metric section and the stress inverse; every source solution reads to such a pair. This is conditional solution-set recovery, not independent selection of the action or matter coupling.