Identity surrounding Killing vector field on a spacetime $\nabla_a \nabla_b w_c = -{R_{bca}}^d w_d$

Question

Let $w^a$ be a Killing vector field on a spacetime $(M, g_{ab})$, i.e., $w^a$ satisfies $\nabla_{(a}w_{b)} = 0$. I hypothesize that$$\nabla_a \nabla_b w_c = -{R_{bca}}^d w_d,$$but I am not sure how I would be able to see this. Could anybody help?

Answer 1

Since $\nabla_{(a}\omega_{b)}=0$, $\nabla_a\omega_b$ is skew symmetric and thus equals the two-form $d\omega$. But then of course $dd\omega=0$, which reads as $0=\nabla_a\nabla_b\omega_c+\nabla_c\nabla_a\omega_b+\nabla_b\nabla_c\omega_a$. Using skew symmetry in $a$ and $b$ in the middle summand, one obtains $\nabla_a\nabla_b\omega_c=\nabla_c\nabla_b\omega_a-\nabla_b\nabla_c\omega_a$, and the result follows from the definition of curvature.

Answer 2

You might want to look at Weinberg's Book on Gravitation and Cosmology. The equation you wrote above is (13.1.9) of the book.