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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.

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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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  • $\begingroup$ 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. $\endgroup$
    – sz3
    Sep 21, 2020 at 4:05
  • $\begingroup$ @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 "$\nabla_2 X^3$". $\endgroup$ Sep 21, 2020 at 4:58

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