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It's probably a stupid question but I can't understand where I am wrong.

I have a manifold with metric $g(X,Y)$ and I know that under infinitesimal diffeomorphism ($x'^\mu = x^\mu + \varepsilon^\mu$) the components of the metric have to transform as $$ g_{\mu\nu}=\frac{dx'^\sigma}{dx^\mu}\frac{dx'^\rho}{dx^\nu}g'_{\sigma\rho} $$ so that the metric tensor $g = g_{\mu\nu}dx^\mu dx^\nu$ is left invariant (the metric tensor $g$ is the same in all local maps, and the components $g_{\mu\nu}$ relative to different maps can be mapped one into another with the above diffeormophispm). Therefore, plugging my equation for the infinitesimal diffeomorphism $x'^\mu = x^\mu + \varepsilon^\mu$, I get $$ g_{\mu\nu} = (\delta_\mu^\sigma + \partial_\mu \varepsilon^\sigma)(\delta_\nu^\rho + \partial_\nu \varepsilon^\rho)(g_{\sigma\rho}+\partial_\tau g_{\sigma\rho}\varepsilon^\tau) = g_{\mu\nu}+L_\varepsilon (g_{\mu\nu}) \; \rightarrow\; L_\epsilon(g_{\mu\nu}) = 0 $$ But I know that under diffeomorphisms the metric tensor components should transform as $$ g'_{\mu\nu} = g_{\mu\nu}+L_\varepsilon(g_{\mu\nu}) $$ So I should have $L_\varepsilon(g_{\mu\nu})\neq 0 $ in general, but why? And how can I obtain the above formula?

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  • $\begingroup$ The metric is not invariant but covariant, so the (components) do not satisfy $g_{\mu \nu} = g_{\mu \nu}^{,}$. You have the correct transformation law above. $\endgroup$
    – Eletie
    Dec 14, 2023 at 13:49
  • $\begingroup$ I think what your question is more about isometries than diffeomorfisms. And what you might be looking for is the Killing vector field condition which reads $\mathcal{L}_X(g)=0$ being $X = \epsilon^{\mu} \partial_{\mu}$ the Killing vector field in your example. If $X$ is not a Killing vector field (no isometry) then the isomorfism is not and isometry. Not sure I helped. $\endgroup$ Dec 14, 2023 at 13:50
  • $\begingroup$ The metric $g(X,Y)$ is a scalar and thus invariant. The only mapping between manifolds that preserves the metric tensor components is the identity map. $\endgroup$ Dec 14, 2023 at 13:58
  • $\begingroup$ @Eletie The metric tensor $g=g_{\mu\nu}dx^\mu dx^\nu$ is invariant under change of coordinates, the components of such tensor $g_{\mu\nu}$ are the ones which are covariant, and indeed I wrote the trasformation relations for covariant components. Is there something that I am missing? $\endgroup$ Dec 14, 2023 at 14:15
  • $\begingroup$ @PhysicsKoan This is exactly what I said, so the Lie derivative $\mathcal{L}_{\epsilon} g_{\mu \nu}$ should not vanish in general $\endgroup$
    – Eletie
    Dec 14, 2023 at 14:20

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So we have the metric $g$ in some coordinates $x^\mu$ is given by \begin{equation} g = g_{\mu\nu}(x)\mathrm{d}x^\mu \otimes \mathrm{d}x^\nu \end{equation} As you are correcly saying the tensor $g$ is independent of the coordinates you choose for spacetime, thus if you pick different coordinates $x^{\prime\mu}$ you have \begin{equation} g = g^\prime_{\mu\nu}(x^\prime)\mathrm{d}x^{\prime\mu} \otimes \mathrm{d}x^{\prime\nu} \end{equation} The differential transforms as \begin{equation} \mathrm{d}x^{\prime\mu} = \frac{\partial x^{\prime\mu}}{\partial x^\nu}\mathrm{d}x^\nu \end{equation} thus you get \begin{equation} g_{\mu\nu}(x) = \frac{\partial x^{\prime\lambda}}{\partial x^\mu}\frac{\partial x^{\prime\sigma}}{\partial x^\nu}g^\prime_{\lambda\sigma}(x^\prime) \end{equation} as you correctly write. Now let us consider infinitesimal coordinate transformations $x^{\prime\mu} = x^\mu - \epsilon^\mu(x)$ we have \begin{equation} \frac{\partial x^{\prime\lambda}}{\partial x^\mu} = {\delta^\lambda} _\mu -\frac{\partial \epsilon^\lambda}{\partial x^\mu} \end{equation} Thus \begin{equation} g_{\mu\nu}(x) = g^\prime_{\mu\nu}(x^\prime) - \frac{\partial \epsilon^\lambda}{\partial x^\mu} g^\prime_{\lambda\nu}(x^\prime) - \frac{\partial \epsilon^\sigma}{\partial x^\nu} g^\prime_{\mu\sigma}(x^\prime) + \mathcal{O}(\epsilon^2) \end{equation} note they evaluated at different points. The Lie derivative is basically the variation \begin{equation} \delta g_{\mu\nu} = g^\prime_{\mu\nu}(x) - g_{\mu\nu}(x) \end{equation} note they are evaluated at the same point. We have \begin{equation} g^\prime_{\mu\nu}(x^\prime) = g^\prime_{\mu\nu}(x) - \epsilon^\lambda \frac{\partial g^\prime_{\mu\nu}(x)}{\partial x^\lambda} + \mathcal{O}(\epsilon^2) \end{equation} thus \begin{equation} g_{\mu\nu}(x) = g^\prime_{\mu\nu}(x) - \epsilon^\lambda \frac{\partial g^\prime_{\mu\nu}(x)}{\partial x^\lambda} - \frac{\partial \epsilon^\lambda}{\partial x^\mu} g^\prime_{\lambda\nu}(x) - \frac{\partial \epsilon^\sigma}{\partial x^\nu} g^\prime_{\mu\sigma}(x) + \mathcal{O}(\epsilon^2) \end{equation} And you get \begin{equation} \delta g_{\mu\nu}(x) = \mathcal{L}_\epsilon g_{\mu\nu} = \epsilon^\lambda \frac{\partial g^\prime_{\mu\nu}(x)}{\partial x^\lambda} + \frac{\partial \epsilon^\lambda}{\partial x^\mu} g^\prime_{\lambda\nu}(x) + \frac{\partial \epsilon^\sigma}{\partial x^\nu} g^\prime_{\mu\sigma}(x) \end{equation}

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    $\begingroup$ Thank you! This is exactly what I was looking for! $\endgroup$ Dec 14, 2023 at 18:04
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    $\begingroup$ I used to get confused over this stuff too in the past $\endgroup$
    – lucabtz
    Dec 14, 2023 at 18:10

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