We are given a smooth conformal metric $\rho=\rho(z)\left|dz\right|$ on a Riemann surface $X$. I have a few questions relating to this:

(a) The local formula $R(\rho)=\Delta \mathrm{log}\rho dx\,dy$ gives rise to a natural 2-form (the Ricci form) on $X$. This makes sense, since by the Hodge star operator, we know that the Laplacian gives a map $\Delta: (\textrm{functions on X}) \rightarrow (\textrm{2-forms on X})$. To show that this is the case, we want coordinate independence of this local formula. How can we show this?

(b) We have the curvature of the metric on $X$ defined as $K(\rho)=-\rho^{2}\Delta\mathrm{log}\rho$. I have computed this for various metrics: $\rho=\left|dz\right|$, $\rho=\left|dz|\right/\mathrm{Im}(z)$, and $\rho=2\left|dz|\right/(1+(\left|z|\right)^{2})$. I have found these to be $0,-1,+1$ respectively. Have I computed these correctly?

(c) Finally, I want to show that if $K(\rho)=0$, then in local coordinates our metric is precisely $\rho=\left|dz\right|$. Does this statement follow directly from the formula for curvature or is something else required here? There doesn't seem to be much content to this question.

Any help would be appreciated! Thanks.


As a general suggestion, you would benefit from reading a book on Riemannian geometry which deals with Riemannian manifolds of arbitrary dimension to see what's behind the seemingly arbitrary formulae that you have in the surface case. For instance do Carmo's "Riemannian Geometry" or Petersen's "Riemannian Geometry".

Then you will see a coordinate-free definition of Riemann's curvature tensor, Ricci curvature and scalar curvature; the definition you are using then will become an exercise in computing scalar curvature in conformal coordinates. The statement about metrics with zero curvature becomes a special case of Riemann's theorem that zero curvature metrics are given by $\delta_{ij}$ in suitable coordinates. (This is, in fact, a special case of Cartan's theorem which can be loosely stated as "curvature determines the metric" - more precise statement requires further definitions.) In your case, the coordinate change will be given by a conformal mapping (since your metric was conformal to begin with). For question 2, yes your computations are correct.

  • $\begingroup$ Thank you for the post. I will be sure to consult these sources for further edification. Might I see how (c) works explicitly though? After reading about these various theorems, it is still not elucidated very much. $\endgroup$ – user 3462 Mar 24 '14 at 19:57
  • $\begingroup$ @user3462: This amounts to solving explicitly some differential equation, something I am not particularly good at. Geometrically, what you do is consider the (local) inverse to the exponential map $T_p(X)\to X$. This inverse will be a locally defined isometry from $X$ to the tangent plane $T_pX$ with Euclidean metric. That will be your map (you would have to learn about Jacobi fields to find out why). As I said, read do Carmo or Petersen. $\endgroup$ – Moishe Kohan Mar 24 '14 at 23:09

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