I am looking at the solution in the book "Problem book in Relativity and Gravitation" for problem 10.6. I don't think I need to go into the details of the problem (I will do so if need be) because I am only confused about one step (the tensor stuff is still fairly new to me). The book can be found for free online here: http://apps.nrbook.com/relativity/index.html
So I start with the following equation
\begin{equation} \xi_{\mu;\nu\lambda}-\xi_{\mu;\lambda\nu} = R_{\mu\sigma\lambda\nu}\xi^{\sigma} \end{equation} where $R_{\mu\sigma\lambda\nu}$ is the Riemann curvature tensor, and the $\xi$ are killing vectors. The solutions manual says the following:
"Now we use the Killing equation $\xi_{\mu;\nu}=-\xi_{\nu;\mu}$ and contract $\mu$ and $\lambda$." \begin{equation} \xi^{\nu;\lambda}_{\;\;\;;\lambda} + R^{\nu}_{\;\sigma}\xi^{\sigma}=-(\xi^{\mu}_{\;;\mu})^{;v} \end{equation}
I am having trouble getting this. So okay, I use the Killing equation, and I believe that gives me the following: \begin{equation} -\xi_{\nu;\mu\lambda}-\xi_{\mu;\lambda\nu} = R_{\mu\sigma\lambda\nu}\xi^{\sigma} \end{equation} So now I need to contract $\mu$ and $\lambda$. So to do that, I need to raise the appropriate indices. But what is tripping me up is dealing with the covariant derivative. Is the covariant derivative invariant through index contractions? Am I allowed to do something like this? \begin{equation} \xi^{\nu}_{\;;\mu\lambda}=g^{\nu\delta}\xi_{\delta;\mu\lambda} \end{equation}
What does it mean to even raise or lower the covariant derivative? Since $\xi_{\nu;\mu}=\nabla_{\mu}\xi_\nu$. How does this differ for $\xi_{\nu}^{\;;\nu}$? Can I raise this in the same way as any other index???