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 Quote by paweld I cannot work out the following functional derivative: $$\frac{\delta}{\delta g_{\mu\nu}} \int d^4 x f^a_{\phantom{a}b} \nabla_a h^b$$ Where f is a tensor density $$f= \sqrt{\det g} \tilde{f}$$ ($$\tilde{f}$$ is an ordinary tensor) and should be consider as independent of g. In my opinion this is not 0 because the connection (namly Christoffel symbols) depend on metric. One can easily express this derivative in terms of partial derivatives of metric in some coordinates. But this expression is rather messy and doesn't look like a tensor. Is it possible to express it using only geometrical objects (tensors, tensor densities, ...)?
You better take $$h$$ to be $$\tilde{h}$$ because I guess it is a tensor not weighted tensor (in agreement with your notation). Of course the result will still be a tensor density because the weight factor would by no means disappear in the end. In fact one can show that the best stance for the variation of this integral wrt metric tensor will be

$$\int d^4 x (\frac{1}{2}g^{\mu\nu}f^a_b\nabla_a \tilde{h}^b+\frac{\delta \tilde{f}^{\alpha a}}{\delta g_{\mu \nu}} \nabla_a h_{\alpha}+f^{\mu a} \nabla_a \tilde {h}^{\nu}+\frac{\delta [\nabla_a\tilde{h}^b]}{\delta g_{\mu \nu}} f^a_b),$$

where h is a tensor density. Here we have no idea what $$\tilde{h}$$ and $$\tilde {f}$$ are except their being tensor density. If they were given, we could decide on whether their variation wrt metric tensor would vanish or not! If they don't depend on the metric tensor, the second term apparantly vanishes and the last and third terms give together

$$\int d^4 x f^{\mu a}\partial_a \tilde{h}^{\nu}$$

So the result will be

$$\int d^4 x (f^{\mu a}\partial_a \tilde{h} + \frac{1}{2} f^a_b\nabla_a \tilde {h}^b).$$

This is indeed a non-vanishing expression and the most reduced form of the initial integral.

AB