Tensor index question

Question

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???

Answer 1

To flesh out a bit more what @Erik Jörgenfelt answered (I think correctly):

I am not familiar with "raised"/(contravariant?) covariant derivatives but formally it is clear how to construct it: $$\xi_{\mu}^{~~;\nu}=g^{\nu\alpha}\xi_{\mu;\alpha}.$$ I would understand it as the normal covariant derivative with the differentiation index raised. Formally this is not a problem: the covariant derivative of a vector transforms like a tensor of rank two: so raising and lowering indices with the metric are well defined operations. So in that sense it is nothing new. If one accepts this it is straight forward to get from your first to your second equation. So lets first use the Killing equation: $$ \xi_{\mu;\nu\lambda}-\xi_{\mu;\lambda\nu}=-\xi_{\nu;\mu\lambda}-\xi_{\mu;\lambda\nu}=R_{\mu\sigma\lambda\nu}\xi^\sigma. $$ Now before we contract something we need to raise something (lets raise $\mu$):

$$ -{\xi_\nu^{~~;\mu}}_{;\lambda}-\xi^\mu_{~~;\lambda\nu}=R^\mu_{~~\sigma\lambda\nu}\xi^\sigma. $$

Now we sum over $\mu=\lambda$ and use $R^\lambda_{~~\sigma\lambda\nu}=R_{\sigma\nu}=R_{\nu\sigma}$ to get to:

$$ -{\xi_\nu^{~~;\lambda}}_{;\lambda}-\xi^\lambda_{~~;\lambda\nu}=R_{\nu\sigma}\xi^\sigma. $$ So in the last two steps we contracted the indices $\mu$ and $\lambda$. To get to what we want out lets raise $\nu$:

\begin{align} -{\xi^{\nu;\lambda}}_{;\lambda}-{\xi^\lambda_{~~;\lambda}}^{;\nu}&=R^\nu_{~~\sigma}\xi^\sigma\\ {\xi^{\nu;\lambda}}_{;\lambda}+R^\nu_{~~\sigma}\xi^\sigma&=-{\xi^\lambda_{~~;\lambda}}^{;\nu}. \end{align}

This is equivalent to your second equation, it looks odd to me to use $\lambda$ and $\mu$ as summation indices after one contracted them. It is formally ok, since you can chose the labels for summation indices in different terms as you please, but I would stick with one of the two not both. We are left with a covariant derivative of a covariant divergence (RHS), the term @Erik Jörgenfelt explained (LHS 1. term) and the Ricci tensor contracted with $\xi$ (LHS 2. term).

So form a tensor calculus point of view this should be the way to get from your first to second equation.

Answer 2

In short: yes, you can manipulate the indices representing the covariant derivative just like you can any other indices. For example $$ \xi^{\nu;\lambda}{}_{;\lambda} = g^{\lambda\mu}\xi^\nu{}_{;\mu\lambda} = \nabla^2\xi^\nu, $$ is just the covariant derivative taken twice in such a manner that the direction of the second covariant derivative is given by the gradient, i.e. the first covariant derivatve. Notice the similarity with the Laplacian of standard vector analysis.