Killing vector index manipulation I was doing some problems of the book "Problem Book in Relativity and Gravitation by A. Lightman, R. H. Price" and on problem 10.14 I dont understand why they say:
$\xi^{}_{\gamma;\beta}\xi^{\gamma}\xi^{\beta}=\xi^{}_{(\gamma;\beta)}\xi^{\gamma}\xi^{\beta}=0$
Also on their final step they use the killing equation in order to switch the indicies but there is a minus sign involved with that and it doesn´t appear in the expression:
a$^{}_{a}=\frac{\xi^{}_{\beta;\alpha}\xi^{\beta}}{\xi^{}_{\gamma}\xi^{\gamma}}$
What am I missing?
 A: If $\xi^a$ is a Killing vector then Killing's equation says that $\xi_{a;b}$ is antisymmetric. It follows that, when completely contracted with a symmetric tensor, it will give zero. But $\xi^a \xi^b$ is a symmetric tensor. Thus I do both steps in one go. But if you want to separate the two steps, then the argument is as follows. Step 1: any second rank tensor can be written as a sum of symmetric and antisymmetric parts:
$$
\xi_{a;b} = \xi_{(a;b)} + \xi_{[a;b]}
$$
so
$$
\xi_{\mu;\nu} \xi^\mu \xi^\nu = 
\xi_{(\mu;\nu)}\xi^\mu \xi^\nu + \xi_{[\mu;\nu]}\xi^\mu \xi^\nu
= \xi_{(\mu;\nu)}\xi^\mu \xi^\nu
$$
where the second step is an example of the general fact that
when you contract something (here $\xi_{a;b}$) with a symmetric tensor (here $\xi^a \xi^b$), then the antisymmetric part of the first thing will not contribute (its contribution to the total will be zero). Now step 2: if $\xi^a$ is a Killing vector then its covariant derivative is antisymmetric so $\xi_{(a;b)} = 0$.
I think the above should settle your question about the $a_a$ equation too.
