1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Then the above theorem holds with $diag(-1,1)$ for any two-dimensinal "pseudo"-Riemannian or "Lorentzian" manifold? $\endgroup$ – Keith May 18 '18 at 3:22
  • $\begingroup$ Because the context of the theorem above is ion.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/… (the theorem on the bottom) $\endgroup$ – Keith May 18 '18 at 3:23
  • $\begingroup$ 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. $\endgroup$ – Anthony Carapetis May 18 '18 at 6:44
  • $\begingroup$ 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? $\endgroup$ – Keith May 18 '18 at 7:30

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.