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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?

mangrove84
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2 Answers2

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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.

Didier
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  • thanks for this, but some things are unclear to me: a) How did you get the second equality? Because $g(\partial_\mu, \partial_\nu)$ is a simple function and then directional derivative and Lie derivative coincide? b) If you can show that $\partial_\lambda g_{\mu\nu}$ is a tensorial quantity, then also the Christoffel symbols should be a tensorial quantity, but they're not. c) With your proof, I can show that any partial derivative of any tensor is a tensor; where does it go wrong? – mangrove84 Feb 15 '23 at 13:04
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    @mangrove84 a) You're right, it relies on the fact that Lie derivative acts on functions as vector fields.b) and c) I removed all reference to tensoriality to avoid confusion. You're right saying that this is not a tensor on the whole manifold, but there is an interpretation of it as a tensor on the domain of the coordinates system (as well as for Christoffel symbols). But this is not the point there. – Didier Feb 15 '23 at 13:09
  • @mangrove84 Hopefully, this still answers your original question – Didier Feb 15 '23 at 13:15
  • I'm still trying to understand $\mathcal{L}{\partial{\xi}}\mathcal{L}{\partial{\lambda}} = \mathcal{L}{\partial{\lambda}}\mathcal{L}{\partial{\xi}}$ since in fact $\xi$ is my (random) Lagrangian displacement and not a coordinate derivative, i.e. I don't expect $[\partial_\lambda, \xi]$ to vanish. – mangrove84 Feb 15 '23 at 13:39
  • @mangrove84 Could you elaborate on what you call a random Langrangian displacement? I'm not a physicist (and lots of people there aren't) so it might be valuable to give details on its construction – Didier Feb 15 '23 at 13:49
  • The Lagrangian displacement, in my case, traces the location of a fluid element when it is moved away from its equilibrium. By "random" I wanted to say that $\xi$ could be any vector, it is not necessarily parallel to any of the coordinates. – mangrove84 Feb 15 '23 at 14:04
  • @mangrove84 So, basically, $\xi$ can be any vector field on $M$ right? – Didier Feb 15 '23 at 14:08
  • Yes, $\xi$ could be any vector field. PS: I'm trying to address you via the "@" sign, but this always gets removed from my comment. – mangrove84 Feb 15 '23 at 14:09
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    @mangrove84 Great, then I'll try to think about it and come back to you later. Regarding the "@": this is because you are replying on my answer where only the two of us posted comments, and there is no ambiguity. If a third person were to intervene, then there will be no auto-deletion of this "@" sign. However, I can reply to you with this sign because I could be commenting my own answer and not replying to your comment (I think) – Didier Feb 15 '23 at 14:12
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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): $$ \mathcal{L}_\xi \partial_\lambda g_{\mu\nu} = \partial_\lambda \mathcal{L}_\xi g_{\mu\nu}. $$

mangrove84
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