# Metric tensor under diffeomorphisms

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?

• 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. Dec 14, 2023 at 13:49
• 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. Dec 14, 2023 at 13:50
• 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. Dec 14, 2023 at 13:58
• @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? Dec 14, 2023 at 14:15
• @PhysicsKoan This is exactly what I said, so the Lie derivative $\mathcal{L}_{\epsilon} g_{\mu \nu}$ should not vanish in general Dec 14, 2023 at 14:20

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