How to prove This Equation between Riemann Tensor and Killing Vector?
$$ [\nabla_\mu, \nabla_\rho]\xi_\sigma = R_{\sigma\nu\mu\rho}\xi^\nu $$
I know
$$ R(\vec{X},\vec{Y},\vec{Z})=[\nabla_{\vec{X}}, \nabla_{\vec{Y}}]\vec{Z} - \nabla_{[\vec{X}, \vec{Y}]}\vec{Z} $$
but why the second part equals zero?
The fifth equation here is $$\nabla_\mu\nabla_\sigma \mathcal{K}_\rho-\nabla_\mu\nabla_\rho \mathcal{K}_\sigma+[\nabla_\mu,\nabla_\rho]\mathcal{K}_\sigma-[\nabla_\rho,\nabla_\sigma]\mathcal{K}_\mu+[\nabla_\sigma,\nabla_\mu]\mathcal{K}_\rho=0$$
Here is an outline of how I would reach to the next step shown there. From Killing's equation, we can write $$ \nabla_\mu\nabla_\sigma \mathcal{K}_\rho = - \nabla_\mu\nabla_\rho \mathcal{K}_\sigma $$ This reduces the first two terms to $$ \nabla_\mu\nabla_\sigma \mathcal{K}_\rho - \nabla_\mu\nabla_\rho \mathcal{K}_\sigma = 2 \nabla_\mu\nabla_\sigma\mathcal{K}_\rho$$
For the rest, you can use the equation defining the Reimann tensor, i.e., $$[\nabla_\mu ,\nabla_\sigma ]\mathcal{K}^\rho=R^\rho_{\mu\sigma\nu}\mathcal{K}^\nu$$
To lower the index on $R^\rho_{\sigma\mu\nu}$ you can use the metric, i.e., $$ R_{\sigma\delta\mu\nu} = g_{\sigma \rho}R^\rho_{\delta\mu\nu}$$
This gives
$$[\nabla_\mu,\nabla_\sigma]\mathcal{K}_\nu=g_{\nu \rho}R^\rho_{\mu\sigma\delta}\mathcal{K}^\delta = R_{\mu\sigma \delta\nu}\mathcal{K}^\delta$$
the rest goes as shown in the link.
