A Riemann surface is a manifold of one complex dimension. If we have an orientable, connected Riemannian manifold of two real dimensions, what further conditions (if any) are needed for it be equipped with a complex structure and therefore seen as a Riemann surface?

I've read something on Wikipedia about isothermal coordinates, which seem to be a way of locally constructing a complex structure from the metric. But when and how can these local structures be stitched together to form a global complex structure?

  • 1
    $\begingroup$ Always. Hint: Conformal maps of open subsets in $R^2$ are holomorphic. $\endgroup$ – Moishe Kohan Nov 8 '17 at 3:29
  • $\begingroup$ when it dimension is two it is called surface $\endgroup$ – Guy Fsone Nov 8 '17 at 3:29
  • $\begingroup$ A conformal map can never be Holomorphic. but rather it can be anti-holomorphic $\endgroup$ – Guy Fsone Nov 8 '17 at 3:30
  • $\begingroup$ @GuyFsone: It depends on your definition of a conformal map: Mine requires conformal maps to preserve oriented angles between tangent vectors. $\endgroup$ – Moishe Kohan Nov 9 '17 at 3:11

OK, I think I have this figured out how.

Isothermal coordinates can be constructed in a neighborhood of each point on the manifold, as long as the manifold isn't too “weird” (according to one reference, it suffices for the metric to be Hölder continuous). This shows that the manifold is conformally flat, since it has conformal coordinate maps to/from flat Euclidean space.

Where two of these neighborhoods intersect, the transition maps between their coordinates are then also conformal (since they're the composition of two conformal maps—from one chart to the manifold, and back to the other chart).

Thus we can promote the $\mathbb{R}^2$ coordinate domains to $\mathbb{C}$ with its usual complex structure, and now all the transition maps are holomorphic, so the manifold is a complex manifold. We can also pull back the complex structure from $\mathbb{C}$ to explicitly construct the complex structure on the manifold. This will be consistent across coordinate charts, since a complex structure is conformally invariant.

  • $\begingroup$ The last thing for you to figure out is how orientability of your surface is used. $\endgroup$ – Moishe Kohan Nov 9 '17 at 1:02
  • $\begingroup$ Also, Chern's assumptions are not optimal, the theorem works even for measurable Riemannian metrics (under a certain $L_\infty$ condition). He even gives a reference to the more general result (Morrey). $\endgroup$ – Moishe Kohan Nov 9 '17 at 3:14

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.