7
$\begingroup$

Let $D$ be the unit disk in the complex plane, and assume that $g$ is a Riemannian metric on $D$ which is complete and conformal to the standard Euclidean metric. Can it be the case that the Gaussian curvature of $g$ approaches zero as we approach $\partial D$?

$\endgroup$

1 Answer 1

6
$\begingroup$

Yes. Take the metric with length element $\rho(z)|dz|$ where $\rho(z)=(1-|z|)^{-2}$. It is complete since $\int^1\rho(t)dt=\infty$, and the curvature $$-\rho^{-2}\Delta\log\rho=\rho^{-4}({\rho'}^2-\rho\rho'')=-2(1-r)^2\to 0,$$ where $r=|z|$ and the primes indicate differentiation with respect to $r$.

$\endgroup$
4
  • $\begingroup$ Great example! Is there a limit on the rate at which the curvature can approach zero? $\endgroup$
    – user160856
    Jul 12, 2020 at 23:13
  • $\begingroup$ @user160856: since there is no flat complete metric, there must be some such limit on the rate, but I do not see how to determine it at this moment, even for the metrics depending only on $r$. $\endgroup$ Jul 13, 2020 at 11:11
  • $\begingroup$ Why does the fact that there is no flat complete metric on the unit disk imply that there must be some limit on the rate? $\endgroup$
    – user160856
    Jul 13, 2020 at 15:32
  • $\begingroup$ @user160856: If I knew I would answer your question in the comment. $\endgroup$ Jul 13, 2020 at 17:38

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.