Uniqueness of the Hilbert stress-energy tensor

(The difference is orthogonal to all variations.) Put $D:=T-T^{\mathrm{\prime }}$, a smooth symmetric covariant $2$-tensor field. For any smooth metric family $s\mapsto g_s$ at $g$ with compactly supported variation, [L1] gives the same derivative of $S$ computed with $T$ and with $T^{\mathrm{\prime }}$, so ${\int }_M{D}_{ab}{q}^{ab}{\mathrm{v}\mathrm{o}\mathrm{l}}_g=0$, where $q^{ab}=\frac{d}{ds}{g}_s^{ab}{\mathrm{\mid }}_{s=0}$.

(Realizing perturbations.) Fix a smooth symmetric compactly supported field $h$. By [L2], choose ${\lambda }_0>0$ with $g+\mu h$ Lorentzian for $\mathrm{\mid }\mu \mathrm{\mid }<{\lambda }_0$; shrink ${\lambda }_0$ also below the domain-admission radius ${\epsilon }_h$ of the defining hypotheses. For $0<\lambda <{\lambda }_0$ the map $s\mapsto g+s\lambda h$ (with $\mathrm{\mid }s\mathrm{\mid }<1$) is a smooth metric family at $g$ with compactly supported variation. Differentiating $g_s^{ac}{g}_{scb}={\delta }^a{}_b$ in $s$ by [L4], $q^{ac}{g}_{cb}+g^{ac}\lambda h_{cb}=0$, so $q^{ab}=-\lambda g^{ac}{g}^{bd}{h}_{cd}$. Substituting into step 1.1 and dividing by $-\lambda \mathrm{\ne }0$, ${\int }_M{D}_{ab}{h}^{ab}{\mathrm{v}\mathrm{o}\mathrm{l}}_g=0$ for every smooth symmetric compactly supported $h$, where $h^{ab}=g^{ac}{g}^{bd}{h}_{cd}$.

(Coefficient vanishing on charts.) As $h$ ranges over all smooth symmetric compactly supported $(0,2)$-fields, $h^{ab}$ ranges over all smooth symmetric compactly supported $(2,0)$-fields by [L4]; equivalently, step 2.1 states ${\int }_M{D}_{ab}{q}^{ab}{\mathrm{v}\mathrm{o}\mathrm{l}}_g=0$ for every smooth symmetric compactly supported $q$ written in covariant form. Fix a chart $(U,x)$ and restrict to $q$ supported in $U$; by [L5] the integral is the chart integral ${\int }_U{D}_{ab}{q}^{ab}w{d}^{4}x$. Since $D$ is smooth, hence continuous, [L3] gives $D=0$ on $U$.

(Conclusion.) Every point of $M$ lies in a chart, so step 3.1 gives $D=0$ on all of $M$, that is, $T=T^{\mathrm{\prime }}$.