Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 3 down vote accepted
  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.

share|improve this answer

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

share|improve this answer
    
but it hurts so good... –  user53153 Jan 28 '13 at 4:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.