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Does an action of the form $$S=\int f(\text{tensors})g^{ij}\nabla_kA^k_{ij}\ d^4x$$ where $A$ is antisymmetric in the lower indices produce, upon variation, any nontrivial equation?

Given that the covariant derivative annihilates the metric, one can bring it to the right and therefore take the "trace" of the antisymmetric tensor, making it zero. However, naïve variation and simplification of a specific Lagrangian that I was exploring led to what looked like a legitimate field equation. In other words: Is it thus necessarily true that there is some (tensor) identity which would reduce the "field equations" to something trivial?

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OP's action $S$ is manifestly zero. Hence the variation $\delta S$ is also zero.

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  • $\begingroup$ In retrospect, I'm not sure why I would think otherwise... $\endgroup$ – user195162 Sep 30 '18 at 17:07

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