I'm taking a course on General Relativity and I'm trying to prove that for a Killing vector field $\xi^\mu$ the following equation holds:
$$\nabla^a \nabla_a \xi^\mu = -R^\mu_a \xi^a$$
Where $R_{ab}$ is the Ricci tensor. To prove this I thought of applying the operator $\nabla^a$ to the equation that $\xi$ satisfies due to being a Killing vector field. Then I get:
$$\nabla^a \nabla_a \xi_b = -\nabla^a \nabla_b \xi_a$$
And then I wanted to prove somehow that the RHS is very closely related to the expression that I want to obtain. A little problem is that in the LHS appears $\xi_a$ instead of $\xi^a$ so I would have to raise the index using the metric tensor.
However, I haven't been able to prove this result because I end up with a lot of terms involving different covariant derivatives. I've also tried using Ricci's identity and even the definition of the Riemann tensor but to no avail thus far.
Using the Killing identity and the definition of the Riemann tensor I got to:
$$ \nabla^a \nabla_v \xi_a = -R_{\gamma v} \xi^\gamma + \nabla_v \nabla_a \xi^a$$
But I don't know how to justify that the second term in the RHS is 0.
How should I proceed to prove this result?
Best Answer
$$ \nabla^a \nabla_a \xi^\mu = g^{ab}g^{\mu\nu} \nabla_a \nabla_b \xi_\nu=\\ =-g^{ab}g^{\mu\nu} \nabla_a \nabla_\nu \xi_b=-g^{\mu\nu} \nabla_a \nabla_\nu \xi^a=\\ =-g^{\mu\nu} \nabla_a \nabla_\nu \xi^a +g^{\mu\nu} \nabla_\nu \nabla_a \xi^a =\\ =-g^{\mu\nu} [\nabla_a, \nabla_\nu] \xi^a=\\ =-R^{a}{}_{\sigma a\nu}\xi^\sigma=-R_{\sigma\nu}\xi^\sigma $$ which is what you want to prove. At one moment, in order to complete a "commutator", I added a term involving a derivative of $\nabla_a\xi^a=g^{ab}\nabla_a\xi_b =(1/2) \nabla_a\xi^a=g^{ab}\nabla_{(a}\xi_{b)} = 0$. It's zero because the metric is symmetric but the symmetrization (symmetric part) of the covariant derivative of the Killing vector field vanishes.