# Tag Info

This is not a complete answer, but fills in some of the missing pieces Trimok asked about in the comments to Stan's answer: Note that I did not verify that Stan's proof actually works. Re 1)  \begin{align} \nabla_Z \nabla_Y X &= Z^\lambda(Y^\mu X^\nu_{;\mu})_{;\lambda} \partial_\nu \\&= Z^\lambda(Y^\mu_{;\lambda} X^\nu_{;\mu} + Y^\mu ...
Edit: Note: I have posted another proof of this in another question, here. Those who prefer coordinates may find it slightly more palatable. I gather from your comments that you can do this if you have $\mathcal{L}_X\text{Ric} = 0$. Thus I will outline a somewhat more general result, assuming a certain identity connecting Killing vectors and Riemann ...