Lie derivative and partial derivative commute when applied to metric

Question

I am currently trying to find an expression for the Lagrangian variation of the Christoffel symbols $\Delta \Gamma^\lambda {}_{\mu\nu}$. For the Eulerian variation $\delta \Gamma^\lambda {}_{\mu\nu}$ everything is a lot simpler since the Eulerian variation $\delta$ commutes with the partial derivative $\partial$; but the Lagrangian variation $\Delta = \delta + \mathcal{L}_\xi$ (where $\xi$ is the Lagrangian displacement, i.e. some arbitrary vector field) does not obey such simple commutation relations.

My hope is that the expression for the Lagrangian variation of the Christoffel symbol is
$$
\Delta \Gamma^\mu_{\nu\lambda} = \frac{1}{2} g^{\mu\kappa}
\left(
\nabla_\lambda \Delta g_{\kappa\nu}
+ \nabla_\nu \Delta g_{\kappa\lambda}
– \nabla_\kappa \Delta g_{\nu\lambda}
\right),
$$
where $\Delta g_{\mu\nu}$ is the Lagrangian variation of the metric.

I can derive this expression if I could show that
$$
\partial_\lambda \mathcal{L}_\xi g_{\mu\nu} = \mathcal{L}_\xi \partial_\lambda g_{\mu\nu}. \qquad (1)
$$
But $\partial_\lambda g_{\mu\nu}$ is not a tensorial quantity, so I'm not sure how to calculate the Lie derivative of it. If I pretend that these are the components of a tensor and write down the expression for the Lie derivative (right-hand side of (1)), I end up with the additional terms $(\partial_\lambda \partial_\mu \xi^\sigma) g_{\sigma\nu} + (\partial_\lambda \partial_\nu \xi^\sigma) g_{\mu\sigma}$; I believe they appear only because I wrongly use the tensor formula for $\mathcal{L}_\xi$ when applying it to non-tensor objects.

Is (1) correct and if so, how could this be shown?

Answer 1

The expression $\partial_{\lambda}g_{\mu\nu}$ in fact a Lie derivative: indeed, for functions, the Lie derivative in the direction of a vector field and the action of the vector field itself coincide, so that \begin{align} \partial_{\lambda}g_{\mu\nu} &= \partial_{\lambda}(g(\partial_{\mu},\partial_{\nu}))\\ &= \mathcal{L}_{\partial_{\lambda}}(g(\partial_{\mu},\partial_{\nu})) \\ &= (\mathcal{L}_{\partial_{\lambda}}g)(\partial_{\mu},\partial_{\nu}) + g([\partial_{\lambda},\partial_{\mu}],\partial_{\nu}) + g(\partial_{\mu},[\partial_{\lambda},\partial_{\nu}])\\ &= (\mathcal{L}_{\partial_{\lambda}}g)(\partial_{\mu},\partial_{\nu}), \end{align} where the last equality comes from the fact that for a coordinates system, $[\partial_{\lambda},\partial_{\nu}]=0$.

Now, recall that $\mathcal{L}_{\partial_{\xi}}\mathcal{L}_{\partial_{\lambda}} = \mathcal{L}_{\partial_{\lambda}}\mathcal{L}_{\partial_{\xi}}$ because for any two vector fields, $\mathcal{L}_X\mathcal{L}_Y - \mathcal{L}_Y\mathcal{L}_X = \mathcal{L}_{[X,Y]}$. The result follows immediately.