# Metric and Curvature on a Riemann Surface

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.

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.
• @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. – Moishe Cohen Mar 24 '14 at 23:09