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I am having issues with the Lie derivative of the metric tensor with respect to a basis vector. I know we can prove that the Lie derivative of the metric tensor with respect to the vector $\boldsymbol{\xi}$ satisfies \begin{align} \mathcal{L}_{\boldsymbol{\xi}} g_{\alpha\beta} &= \xi^\mu \partial_\mu g_{\alpha\beta} + g_{\mu\beta} \partial_\alpha \xi^\mu + g_{\alpha\mu} \partial_\beta \xi^\mu \\ &= \xi^\mu \nabla_\mu g_{\alpha\beta} + g_{\mu\beta} \nabla_\alpha \xi^\mu + g_{\alpha\mu} \nabla_\beta \xi^\mu \\ &= g_{\mu\beta} \nabla_\alpha \xi^\mu + g_{\alpha\mu} \nabla_\beta \xi^\mu \end{align} which can be verified simply by subtracting and using the condition of torsion free metric and metric compatibility.

The problem comes when I substitute the vector $\boldsymbol{\xi}$ with the basis vector $\boldsymbol{e_\gamma}$. Clearly, $\boldsymbol{e_\gamma} = \delta^\lambda_\gamma \boldsymbol{e_\lambda}$, so $(\boldsymbol{e_\gamma})^\mu = \delta^\mu_\gamma$. Substituting this back in the equations of the Lie derivative, I get from the first line \begin{equation} \mathcal{L}_{\boldsymbol{e_\gamma}} g_{\alpha\beta} = \partial_\gamma g_{\alpha\beta} \end{equation} and from the last line, however, \begin{equation} \mathcal{L}_{\boldsymbol{e_\gamma}} g_{\alpha\beta} = 0 \end{equation} since the (covariant) derivatives of $\delta^\mu_\gamma$ are zero. But $\partial_\gamma g_{\alpha\beta}=0$ cannot be true in general. Where did my argument go wrong?

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2 Answers 2

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The covariant derivative of $\delta_{\gamma}^{\mu}$ is not zero in general. For example, in the $(R,\theta)$ plane polar coordinates, the covariant derivative of $1e_r + 0e_{\theta}$ in the direction of $e_{\theta}$ is surely not zero. In this case: $\Gamma^{\theta}_{r\theta} \neq 0$

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  • $\begingroup$ Though I wish you had elaborated a bit further, you pointed out exactly where my mistake is. Thanks! $\endgroup$
    – Hector
    Feb 13, 2022 at 8:52
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It's a bit late to say this, but beware: On writing ${\bf e}_\gamma= \delta^\lambda_\gamma {\bf e}_\lambda$ you need to note that that in $\delta^\lambda_\gamma $ the superscript is an index describing a component and the subscript a label labelling the basis vector. Quite different things! For coordinate basis ${\bf e}_\mu = \partial_\mu$ it is not safe to write $\nabla_{{\partial}_\mu}= \nabla_\mu$ for this reason. When you act again $\nabla_X \nabla_{{\partial}_\mu}$ you do not need a connection term for the label $\mu$, but when you act on $\nabla_X \nabla_{\mu}$ the $\mu$ is an index and needs a Christoffel symbol,.

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