I found a theorem stating that "Any two-dimensional Riemannian manifold is conformally flat". Here the flat metric seems like the metric $g_{ab}=diag(-1,1)$. However, the normal Euclidean space $\mathbb{R}^2$ with the metric $g_{ab}=diag(1,1)$ cannot be conformally flat. So what is the exact meaning of this theorem? Could anyone please explain?
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In the context of Riemannian (not pseudo-Riemannian or Lorentzian) geometry, flat means locally isometric to the Euclidean metric $\mathrm{diag}(1,1).$ The indefinite metric $\mathrm{diag}(-1,1)$ is not Riemannian, so of course no Riemannian metric will be locally isometric to it.
answered May 18, 2018 at 2:35
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$\begingroup$ Then the above theorem holds with $diag(-1,1)$ for any two-dimensinal "pseudo"-Riemannian or "Lorentzian" manifold? $\endgroup$– KeithCommented May 18, 2018 at 3:22
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$\begingroup$ Because the context of the theorem above is ion.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/… (the theorem on the bottom) $\endgroup$– KeithCommented May 18, 2018 at 3:23
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$\begingroup$ Yes, this result is also true for Lorentzian surfaces. The source you link is confusing, since it defines conformal flatness only in the context of Lorentzian geometry, but then states the theorem for a Riemannian metric. $\endgroup$ Commented May 18, 2018 at 6:44
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$\begingroup$ OK, let us summarize. The Riemannian surfaces are conformally flat where the flat metric is $diag(1,1)$. The Lorentzian surfaces are conformally flat where the flat metric is $diag(-1,1)$. Is this right? $\endgroup$– KeithCommented May 18, 2018 at 7:30