How to prove This Equation between Riemann Tensor and Killing Vector?

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?

• Does this help physics.stackexchange.com/q/512281
– S.G
Commented Sep 19, 2023 at 3:11
• I use the same method as physics.stackexchange.com/a/512311/356911 and my question is just how the 5th equation there goes to the 6th equation which uses $[\nabla_\mu, \nabla_\rho]\xi_\sigma=R_{\sigma\nu\mu\rho}\xi^\nu$ Commented Sep 19, 2023 at 6:19

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.

• @Ghoster ah thanks for that.
– S.G
Commented Sep 19, 2023 at 5:21
• So $[\nabla_\mu, \nabla_\sigma]\xi^\rho = R^\rho_{\mu\sigma\nu}\xi^\nu$ is the definition of the Riemann tensor? I just want to know how this is proved. I know $R(\vec{X},\vec{Y},\vec{Z})=[\nabla_{\vec{X}}, \nabla_{\vec{Y}}]\vec{Z} - \nabla_{[\vec{X}, \vec{Y}]}\vec{Z}$ Commented Sep 19, 2023 at 6:15
• @Firestar-Reimu Yours is similar to mine. The last term that has [X,Y] has been set to zero as the vector fields in the chosen coordinate induced basis commute.
– S.G
Commented Sep 19, 2023 at 6:37