Proposition: Noether identity with switch defect

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 ${\xi }^{\rho }$.