# Deriving $\nabla_\mu \nabla_\sigma \mathcal{K}^\rho=R^\rho_{\sigma\mu\nu}\mathcal{K}^\nu$

I want to derive this equation from Carroll's book.

$$\nabla_\mu \nabla_\sigma \mathcal{K}^\rho=R^\rho_{\sigma\mu\nu}\mathcal{K}^\nu$$

We know that $$\mathcal{K}^\nu$$ is a killing vector and satisfies $$\nabla_{(\mu} \mathcal{K}_{\nu)}=0$$ and that the geometry is torsionfree and metric compatible. I understand that I have to use the definition of Riemann tensor and the Bianchi identity but I don't see how.

## 1 Answer

The definition of the covariant derivative is $$[\nabla_\mu,\nabla_\sigma]V^\nu=R^\nu_{\hphantom{\nu}\rho\mu\nu}V^\rho$$ Now, we start with the Killing equation $$\nabla_\sigma \mathcal{K}_\rho+\nabla_\rho \mathcal{K}_\sigma=0$$ and take a covariant derivative of this equation to obtain $$\nabla_\mu\nabla_\sigma \mathcal{K}_\rho+\nabla_\mu\nabla_\rho \mathcal{K}_\sigma=0$$ We can relabel the indices of the equation and the equation remains the same. So we can add and subtract the following permutation of indices done in a convenient way so that the equation remains the same. $$(\nabla_\mu\nabla_\sigma \mathcal{K}_\rho+\overbrace{\nabla_\mu\nabla_\rho \mathcal{K}_\sigma}^1)-(\overbrace{\nabla_\rho\nabla_\mu \mathcal{K}_\sigma}^1+\overbrace{\nabla_\rho\nabla_\sigma \mathcal{K}_\mu}^2)+(\overbrace{\nabla_\sigma\nabla_\rho \mathcal{K}_\mu}^2+\overbrace{\nabla_\sigma\nabla_\mu \mathcal{K}_\rho}^3)-(\overbrace{\nabla_\mu\nabla_\sigma \mathcal{K}_\rho}^3+\nabla_\mu\nabla_\rho \mathcal{K}_\sigma)=0$$ The first two terms inside the brackets are the same as the previous equation, whereas the other six terms inside the brackets are suitable permutations of the first two terms. The two terms inside each bracket add to zero according to the Killing equation. Now we can group some terms (shown by the numbers) as commutators as follows: $$\nabla_\mu\nabla_\sigma \mathcal{K}_\rho-\nabla_\mu\nabla_\rho \mathcal{K}_\sigma+[\nabla_\mu,\nabla_\rho]\mathcal{K}_\sigma-[\nabla_\rho,\nabla_\sigma]\mathcal{K}_\mu+[\nabla_\sigma,\nabla_\mu]\mathcal{K}_\rho=0$$ Using the Killing equation in the first two terms, we obtain $$2\nabla_\mu\nabla_\sigma\mathcal{K}_\rho=[-R_{\sigma\nu\mu\rho}+R_{\mu\nu\rho\sigma}-R_{\rho\nu\sigma\mu}]\mathcal{K}^\nu$$ We can rearrange this equation using the symmetries of the Riemann tensor as $$\nabla_\mu\nabla_\sigma\mathcal{K}_\rho=\frac{1}{2}[-R_{\nu\sigma\rho\mu}+R_{\nu\mu\sigma\rho}-R_{\nu\rho\mu\sigma}]\mathcal{K}^\nu$$ Since we know that $$R_{\nu\sigma\rho\mu}+R_{\nu\mu\sigma\rho}+R_{\nu\rho\mu\sigma}=0$$, we can write $$\nabla_\mu\nabla_\sigma\mathcal{K}_\rho=R_{\nu\mu\sigma\rho}\mathcal{K}^\nu=R_{\rho\sigma\mu\nu}\mathcal{K}^\nu$$ Now we can raise the index $$\rho$$ to obtain the final result $$\nabla_\mu\nabla_\sigma\mathcal{K}^\rho=R^\rho_{\hphantom{\rho}\sigma\mu\nu}\mathcal{K}^\nu$$

• Could you explain how you added all those terms with the equation remaining the same? I don't understand how all the extra terms are 0. – redhood Nov 6 '19 at 14:22
• Alright, I went through it. Thanks a ton! – redhood Nov 6 '19 at 15:28
• @fielder Please note that some error was incorporated in the fourth equation of the answer. I had corrected them and reformatted the equation so that the grouping of the terms of the equation is clear enough to understand. – Richard Nov 7 '19 at 7:01
• Yes, once I saw you need to add the killing equation with different indices till you get the Riemann tensors you want it was quite straight forward, thanks! – redhood Nov 7 '19 at 10:52