# An equation for Killing vector fields

Let $$(M,g)$$ be Riemannian manifold with Levi-Civita Connection $$\nabla$$. We know that a vector field $$X$$ is a Killing vector field if and only if it satisfies the Killing equation (written in abstract index notation) $$$$\nabla_{\mu}X_{\nu} + \nabla_{\nu}X_{\mu} = 0$$$$ Now I'd like to show that $$X$$ also satisfies the equation $$\begin{equation*} \Delta_{g}X^{\mu} + {R^{\mu}}_{\nu}X^{\nu} = 0 \tag{\heartsuit} \end{equation*}$$ where $$\Delta_{g} = \nabla^{\mu}\nabla_{\mu}$$ is the Laplace-Beltrami operator and $$R_{\mu \nu}$$ is the Ricci tensor. The derivation should be straightforward. Indeed, if we apply $$g^{\lambda \nu} \nabla^{\mu}$$ to both sides of the Killing equation, we can commute the order of covariant differentiation and get \begin{align*} g^{\lambda \nu}\nabla^{\mu}\nabla_{\mu}X_{\nu} + g^{\lambda \nu}\nabla^{\mu}\nabla_{\nu}X_{\mu} &= \Delta_{g}X^{\lambda} + \nabla^{\mu}\nabla^{\lambda}X_{\mu}\\ & = \Delta_{g}X^{\lambda} + \nabla^{\lambda}\nabla^{\mu}X_{\mu} + {R^{\mu \lambda}}_{\mu \nu}X^{\nu}\\ & = \Delta_{g}X^{\lambda}+ \nabla^{\lambda}\text{div}X + {R^{\lambda}}_{\nu}X^{\nu}\\ & = \Delta_{g}X^{\lambda}+ {R^{\lambda}}_{\nu}X^{\nu}\\ & = 0 \end{align*} where the second to last equality follows from the fact that a Killing vector field is divergence free. However, I'm not sure about the third equality, that $$$$\nabla^{\lambda}\nabla^{\mu}X_{\mu} = \nabla^{\lambda}(\nabla^{\mu}X_{\mu}) = \nabla^{\lambda}\text{div}X$$$$ The main confusion comes from whether we can evaluate the term $$\nabla^{\mu}X_{\mu}$$ first, and then apply the outer convariant differentiation. On the other hand, I'm pretty sure $$(\heartsuit)$$ holds, since it will serve as a key step to prove the fact that $$\Delta_{g}$$ commutes with Killing vector fields on Riemannian manifolds.

Yes, $$\nabla^{\lambda}\nabla^{\mu}X_{\mu}$$ means $$\nabla^{\lambda}(\nabla^{\mu}X_{\mu})$$. It equals $$0$$ since $$\nabla^{\mu}X_{\mu} = 0$$.
• Yes, this is my only question. Thank you! When doing this type of computation, I'm under the expression that all covariant derivatives are taken before indices are applied. So I'm a bit unsure why in this case we can first evaluate $\nabla^{\mu}X_{\mu}$ and then take the outer covariant derivative.
• @ShaoyangZhou Ok I think I see the confusion. Let's take an example like $\nabla_\mu \nabla_\nu X^\rho$. It means $\nabla_\mu (\nabla_\nu X^\rho)$: first evaluate $\nabla_\nu X^\rho$. And I means the whole tensor, not just one component. Think of the indices as abstract. If you want a concrete component w.r.t. some coordinates, that's done at the end. For example, to compute the "$\mu=1, \nu=2, \rho=3$" component of $\nabla_\mu \nabla_\nu X^\rho$, you need to compute all $n^2$ components of $\nabla_\nu X^\rho$, not just "$\nabla_2 X^3$". Sep 21, 2020 at 4:58