Consider a compact Riemannian manifold $M$ with a metric $g$ and a geometric functional on it, i.e. a functional of only $g$:
$$J[g]$$
This can be the volume, the Einstein-Hilbert action, the Green's function of the Laplace operator, the zeta function of $(M,g)$ etc.
Define the variation $\Pi_{ij}$ of this functional by
$$\delta J(g)[\delta g]=\intop_M \Pi_{ij}(x)\delta g^{ij}(x)\mathrm{d}V(x).$$
My question is if there exists a theorem that says
$$\nabla^i \Pi _{ij}=0.$$
The basic idea for why this could be true is that since $J$ is a functional of only $g$, its variation is also a functional of only $g$. Therefore the $x$-dependence in $\Pi_{ij}(x)$ can come in only through $g_{ij}(x)$. However, covariant derivatives of the metric are zero.
This becomes an actual proof for functionals of explicitly local quantities like the Einstein-Hilbert action $J=\int R\mathrm{d}V$. I would like to understand if this is extended to non-local functionals like $J=G(x,y)$, where $G$ is the Green's function of $\Delta$.
I haven't been able to come up with a functional whose variation is known or easily computable for which this is not true. In physics these "conservation laws" are related to diffeomorphism invariance, but I wonder if this simple version of it can be made rigorous.
One way to look at this is to say that the variation of the metric can always be written as $\delta g_{ij}=\nabla_i \xi_j+\nabla_j \xi_i$ for some vector field $\xi^i$. Then
$$\delta J[g]=-2 \intop_M \nabla^i \Pi _{ij} \xi^j \mathrm{d}V.$$
If this variation were to be zero, we would have our conservation law. However, I see no reason for it to be zero. Somehow it looks suspiciously believable because, say, the Einstein tensor is divergenceless, and even the variation of the Green's function $G(x,y)$ seems to have this property away from $x$ and $y$ (based on some variational formulas I have found in the mathematical literature).