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

A colleague of mine gave me the hint, that is as crucial as it is simple, on how to calculate $$ \mathcal{L}_\xi \partial_\lambda g_{\mu\nu}. $$ Express the partial derivative using the covariant derivative (with the handy fact that the metric is compatible) plus some terms including the Christoffel symbols, i.e. $$ \partial_\lambda g_{\mu\nu} = \nabla_\lambda g_{\mu\nu} + \Gamma^\sigma {}_{\mu\lambda} g_{\sigma\nu} + \Gamma^\sigma {}_{\nu\lambda} g_{\mu\sigma}. $$ After using the Leibniz-rule on the Lie derivative, one also needs an expression for the Lie derivative of Christoffel symbols but this can be found also here on SE-Mathematics, and we have $$ \mathcal{L}_\xi \Gamma^\mu {}_{\nu\lambda} = \xi^\sigma R^\mu {}_{\lambda\sigma\nu} + \nabla_\nu \nabla_\lambda \xi^\mu. $$ Using these relations and the properties of the Riemann tensor, it takes only a few lines of algebraic reformulations to arrive at Eq. (1): $$ \mathcal{L}_\xi \partial_\lambda g_{\mu\nu} = \partial_\lambda \mathcal{L}_\xi g_{\mu\nu}. $$

Answer 2

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.