Any two-dimensional (Euclidean) space is locally conformal to flat space; that is, given a two-dimensional manifold $M$ equipped with a (Euclidean) metric $g_{ab}$, it is always possible to locally find a function $\sigma$ such that

$g_{ab} = e^{2\sigma} \delta_{ab}$,

where $\delta_{ab}$ is the metric of flat Euclidean space.

Of course, this is not true globally; for example, if $(M,g_{ab})$ is the round two-sphere, then the function $\sigma$ above must blow up somewhere. Intuitively, this must be the case because the plane and the two-sphere have different topologies. My question is basically whether this topological obstruction is the only problem: that is, if $M$ is a topological two-sphere, is it always possible to find a globally well-behaved function $\sigma$ such that

$g_{ab} = e^{2\sigma} g^\mathrm{(round)}_{ab}$,

where $g^\mathrm{(round)}_{ab}$ is the usual round metric on the two-sphere?

(More generally, is it always possible to find a globally well-behaved $\sigma$ such that $g_{ab} = e^{2\sigma} g^\mathrm{(sym)}_{ab}$, where $g^\mathrm{(sym)}_{ab}$ is the maximally symmetric metric on $M$?)

  • 3
    $\begingroup$ As stated this is not quite right - for example, if you choose a coordinate patch in which the round metric is diagonal, any metric $g$ which is not already diagonal cannot be made diagonal by a conformal change. What is true is that you can find a $\sigma$ such that $e^{2 \sigma} g$ is isometric to the round metric - this the uniformization theorem. $\endgroup$ – Anthony Carapetis May 18 '17 at 13:55
  • $\begingroup$ Oh, sure - I only care about the geometry, not the metric, so I'm only interested in conformal equivalence up to isometries. But yes, the uniformization theorem was exactly what I wanted! If you write up your comment as an answer I'll accept it. $\endgroup$ – Sebastian May 19 '17 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.