# More index manipulation on Killing vectors

On the solution of problem 10.6 of the book "Problem Book in Relativity and Gravitation by A. Lightman, R. H. Price" they mention using the Killing equation:

$$\xi^{}_{\mu;\nu}=-\xi^{}_{\nu;\mu}$$

and contracting both the $$\mu$$ and $$\lambda$$ indicies to arrive at the equation:

$$\xi^{\nu;\lambda}_{\space\space\space\space\space\space;\lambda}+R^{\nu}_{\space\space\sigma}\xi^{\sigma}=-(\xi^{\mu}_{\space\space;\mu})^{;\nu}$$

Here is my attempt:

Killing equation:

$$\nabla^{}_{\nu}\xi^{}_{\mu}+\nabla{}_{\mu}\xi^{}_{\nu}=\nabla^{}_{\lambda}\nabla^{}_{\nu}\xi^{}_{\mu}+\nabla^{}_{\lambda}\nabla{}_{\mu}\xi^{}_{\nu}=0$$

Definition of the Riemann curvature tensor that they give us:

$$\nabla^{}_{\lambda}\nabla{}_{\nu}\xi^{}_{\mu}-\nabla^{}_{\nu}\nabla{}_{\lambda}\xi^{}_{\mu}= R^{}_{\mu\sigma\lambda\nu}\xi^{\sigma}$$

I subtracted , in the Killing equation, the second term in the above equation and got:

$$\nabla^{}_{\lambda}\nabla^{}_{\nu}\xi^{}_{\mu}-\nabla^{}_{\nu}\nabla{}_{\lambda}\xi^{}_{\mu}+\nabla^{}_{\lambda}\nabla{}_{\mu}\xi^{}_{\nu}=-\nabla^{}_{\nu}\nabla{}_{\lambda}\xi^{}_{\mu}$$

Substituded the definition of the Riemann tensor for the first two terms and used $$g^{\mu\lambda}$$ for the contraction witch resulted in:

$$R^{}_{\sigma\nu}\xi^{\sigma}+\nabla^{\mu}\nabla{}_{\mu}\xi^{}_{\nu}=-\nabla^{}_{\nu}\nabla^{\mu}\xi^{}_{\mu}$$

I am almost at the final result but can´t quite get it.

You should use the definition of the curvature tensor $$[\nabla_\mu, \nabla_\nu]\xi_\kappa = R_{\mu\nu\kappa}^{\ \ \ \ \ \ \lambda}\xi_\lambda.$$ Then by differentiating the Killing condition we obtain $$\nabla_\kappa\nabla_\mu\xi_\nu + \nabla_\kappa\nabla_\nu\xi_\mu = 0.$$ One can then add a term to both sides to get a commutator and insert the above definition of the curvature tensor: $$R_{\kappa\mu\nu}^{\ \ \ \ \ \ \lambda}\xi_\lambda + \nabla_\kappa\nabla_\nu\xi_\mu = -\nabla_\mu\nabla_\kappa\xi_\nu.$$ Finally by contracting using the metric $$g^{\nu\kappa}$$ (and using the fact that the Levi-Civita connection is metric-compatible) you arrive at your formula.
• I made some small mistakes (this has now been corrected but my apologies if it resulted in confusion). The result, however, remains the same. The only thing you miss is the symmetry of the Ricci curvature and raising the $\nu$-index. Oct 6, 2020 at 10:13