# Theorem on any 2-dimensional riemannian manifold conformally flat

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?

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.
• Then the above theorem holds with $diag(-1,1)$ for any two-dimensinal "pseudo"-Riemannian or "Lorentzian" manifold? – Keith May 18 '18 at 3:22
• 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? – Keith May 18 '18 at 7:30