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Can the full set of II. Bianchi identities be derived from the symmetries of the action?

Physics Asked by Void on June 20, 2021

In pseudo-Riemannian geometry we can derive the II. Bianchi identities by considering, e.g. the expression of the Riemann tensor in Riemann normal coordinates. They read
$$R_{munukappalambda;gamma} + R_{munugammakappa;lambda} + R_{munulambdagamma;kappa} = 0,.$$
In essence, they are a part of the integrability relations that specify that the curvature corresponds to a Levi-Civita connection (i.e. that it is generated by "a potential", the Christoffel symbols, which themselves are generated by another "potential", the metric $g_{mu nu}$).

Now consider the Einstein-Hilbert action (modulo factors, cosmological constant zero for simplicity)
$$S_{rm EH} = int R sqrt{-g} , mathrm{d^4} x ,. $$
We can show that it is invariant wrt infinitesimal coordinate transforms $x^mu + xi^mu(x^mu)$ and, even more, that this is true before applying field equations. This is an infinite-dimensional Lie symmetry of the action, which implies by the second Noether’s theorem a set of "strong conservation laws" that in this case are the twice contracted II. Bianchi identities
$$(R^{munu} – frac{1}{2}R g^{mu
nu})_{;mu} = 0 ,.$$

That is, a part of the II. Bianchi identities can be derived from the symmetries of the E.-H. action. However, is there a way to derive the full set of II. Bianchi identities in this manner? That is, is there a variational procedure that would allow me to derive a larger set of Bianchi-type identities based on the symmetries of the Einstein-Hilbert action (or similar)?

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