# Invariance of curvature under a conformal mapping

Let $\Omega_{1}, \Omega_{2} \subseteq \mathbb{C}$ be bounded domains. Let $\rho$ be a metric on $\Omega_2$ and $h: \Omega_1 \rightarrow \Omega_2$ a conformal mapping. Let $$h^*\rho(z) = \rho(h(z))\cdot|h'(z)|$$ be the pullback metric on $\Omega_1$. According to the definition I have been given, the curvature on $(\Omega_2,\rho)$ is

$$K_{\Omega_2,\rho}(z) = -\frac{\Delta \log(\rho(z))}{\rho(z)^2}$$

It is then claimed that $K_{\Omega_1,h^*\rho}(z) = K_{\Omega_2,\rho}(h(z))$, i.e.

$$-\frac{\Delta \log(\rho(h(z))\cdot|h'(z)|) }{(\rho(h(z))\cdot|h'(z)|)^2} = -\frac{\Delta \log(\rho(h(z)))}{\rho(h(z))^2}$$

I cannot derive this result however, and quite frankly I find it dubious. Is it true?

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1. We can get rid of $|h'|$ under the logarithm, since $\log |h'|$ is harmonic.

2. The composition of a function ($\log \rho$ in this case) with a conformal map $h$ multiplies the Laplacian by $|h'|^2$. This is shown in one line here.

3. By definition of pullback, the denominator is multiplied by $|h'|^2$ too. So $|h'|^2$ cancels, and we are done.

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The calculation is fairly painful. I think that Theorem 2.5.4 of this should do the trick.

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but it hurts so good... – user53153 Jan 28 '13 at 4:53

By using the pullback metric $h^*[\rho ds]$ on $\Omega_1$, you turn the map $h:(\Omega_1, h^*[\rho ds])\to (\Omega_2, \rho ds)$ into a metric space isometry, and the notion of curvature is a geometric invariant. So necessarily the two are equivalent (and there is NO computation to do!:-)

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Welcome to Math.SE! There seem to be two assertions in play here: (i) A $2$-dimensional Riemannian metric possesses a geometric invariant called curvature; (ii) A certain analytic expression in a conformally-flat metric computes the curvature. As I read it, the OP was effectively asked to establish (ii) by showing explicitly that the analytic expression is invariant under a conformal transformation of the domain. Your answer is certainly the elegant conceptual interpretation, however. :) – Andrew D. Hwang Oct 27 '15 at 14:15