Index permutations on Killing equation

Question

Problem 10.7 of the book "Problem Book in Relativity and Gravitation” by A. Lightman and R. H. Price reads:

If $\xi$ is a Killing vector, prove that $\xi_{\mu;\alpha\beta}=R_{\gamma\beta\alpha\mu}\xi^\gamma$.

The enclosed solution includes:

$0=\xi_{\sigma;\rho\mu}-\xi_{\sigma;\mu\rho}+\xi_{\mu;\sigma\rho}-\xi_{\mu;\rho\sigma}+\xi_{\rho;\mu\sigma}-\xi_{\rho;\sigma\mu}$ (2)

$0=\xi_{\sigma;\rho\mu}-\xi_{\sigma;\mu\rho}-\xi_{\mu;\rho\sigma}(3)$

They use the Killing equation to obtain (3) from (2) but I don't understand what they did exactly. I used the Killing equation on the last term in equation (2) to switch the first two indices and make it a positive term and the terms:

$\xi{}_{\mu;\sigma\rho}+\xi{}_{\rho;\mu\sigma}+\xi{}_{\sigma;\rho\mu}$

must be equal to zero so that we are left with equation (3). Is there an identity (like the Bianchi identity) for the permutation of these 3 indices that makes the sum of those three terms above zero? Or is there something else that I’m missing?

Answer 1

Apply the Killing equation $\xi_{\alpha;\,\beta}=-\xi_{\beta;\,\alpha}$ to the third, fifth and sixth terms in (2):$$\begin{align}0&=\xi_{\sigma;\rho\mu}-\xi_{\sigma;\mu\rho}+\xi_{\mu;\sigma\rho}-\xi_{\mu;\rho\sigma}+\xi_{\rho;\mu\sigma}-\xi_{\rho;\sigma\mu}\\&=\xi_{\sigma;\rho\mu}-\xi_{\sigma;\mu\rho}-\xi_{\sigma;\mu\rho}-\xi_{\mu;\rho\sigma}-\xi_{\mu;\rho\sigma}+\xi_{\sigma;\rho\mu}.\end{align}$$These are (3)'s terms twice each.