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 $\partial D$?
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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 $r$.
answered Jul 9 at 12:12
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$\begingroup$ Great example! Is there a limit on the rate at which the curvature can approach zero? $\endgroup$ – user160856 Jul 12 at 23:13
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$\begingroup$ @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 $r$. $\endgroup$ – Alexandre Eremenko Jul 13 at 11:11
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$\begingroup$ 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 at 15:32
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$\begingroup$ @user160856: If I knew I would answer your question in the comment. $\endgroup$ – Alexandre Eremenko Jul 13 at 17:38
