If $V$ is a Killing vector field, I need to prove that

$$V^{m}\nabla_{m}R = 0$$

where $R$ is the Ricci scalar $R = g^{mn}R_{mn}$. I´m having some trouble with this, I already showed that $$\nabla_{n}\nabla_{a}V^{r} = R^{r}_{ans}V^{s}$$

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    – quid
    Aug 30 '15 at 20:05

Killing fields generate isometries and the scalar curvature is isometry-invariant, so moving along a Killing field does not change the curvature.

More precisely: since the scalar curvature is just a function, the covariant derivative $V^m \nabla_m R$ is equal to the Lie derivative $$\mathcal L_V R = \frac{d}{dt}\Big|_{t=0}(R\circ \Phi^V_t)$$ where $\Phi^V$ is the flow of $V$. Since the curvature is isometry-invariant, we have $R\circ \Phi^V_t=R$ and thus the derivative is zero.

If you're looking for a more computational answer, I think the twice-contracted second Bianchi identity $V^m \nabla_m R = V^i \nabla^j R_{ij}$ should be a good starting point. From a quick try I couldn't see how to make it pop out, but all you should need are the symmetries of the curvature tensor and the defining equation of Killing fields.


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