An equation for Killing vector fields

differential-geometryriemannian-geometry

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)
\begin{equation}
\nabla_{\mu}X_{\nu} + \nabla_{\nu}X_{\mu} = 0
\end{equation}

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
\begin{equation}
\nabla^{\lambda}\nabla^{\mu}X_{\mu} = \nabla^{\lambda}(\nabla^{\mu}X_{\mu}) = \nabla^{\lambda}\text{div}X
\end{equation}

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.

Best Answer

Yes, $\nabla^{\lambda}\nabla^{\mu}X_{\mu}$ means $\nabla^{\lambda}(\nabla^{\mu}X_{\mu})$. It equals $0$ since $\nabla^{\mu}X_{\mu} = 0$.

(I think that is your only question? Everything in your derivation looks correct.)

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