1
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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: $\Gamma^{\theta}_{r\theta} \neq 0$

$\endgroup$
1
  • $\begingroup$ Though I wish you had elaborated a bit further, you pointed out exactly where my mistake is. Thanks! $\endgroup$
    – Hector
    Feb 13 at 8:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.