Timeline for Is every nonnegatively curved plane conformal to the complex plane?
Current License: CC BY-SA 3.0
12 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Sep 7, 2011 at 0:28 | comment | added | Igor Belegradek | Anton, thanks for updating this. I must confess I have not yet worked through all the details of Fedja Nazarov's proof. | |
| Sep 6, 2011 at 22:24 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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| Mar 16, 2011 at 18:34 | history | edited | Anton Petrunin | CC BY-SA 2.5 |
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| Mar 16, 2011 at 15:45 | comment | added | Igor Belegradek | The reference is [Huber. "On subharmonic functions and differential geometry in the large", Comment. Math. Helv. 32 1957 13–72], where it is proved that given a complete Riemannian metric on a noncompact surface $S$, if the negative part of the total curvature is finite, then $S$ is parabolic. In particular, if $S$ is simply-connected, it is conformal to the plane. I wish you pretty argument worked, but things seem to be more subtle; apparently, completeness is not easy to read off the conformal factor. | |
| Mar 15, 2011 at 23:43 | comment | added | Anton Petrunin | @Igor, Right, I will correct it. | |
| Mar 15, 2011 at 20:14 | comment | added | Igor Belegradek | Anton, in your argument I no longer understand why $\phi(x)$ tends to infinity when $x$ approaches the boundary. Completeness implies that $\phi$ isn't bounded above, but it might oscillates from $+\infty$ to $-\infty$. On a bright note, the desired assertion is true: apparently Huber proved that if the negative part of the total curvature of a complete surface is finite, then the surface is conformal to a plane. | |
| Oct 26, 2010 at 3:45 | history | edited | Autumn Kent | CC BY-SA 2.5 |
fixed tex
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| Oct 26, 2010 at 3:08 | vote | accept | Igor Belegradek | ||
| Mar 16, 2011 at 19:54 | |||||
| Oct 26, 2010 at 3:07 | comment | added | Igor Belegradek | Neat! I kept thinking of the unit disk as carrying a hyperbolic metric, which made it harder to see superharmonicity of $\phi$. | |
| Oct 26, 2010 at 2:46 | history | edited | Anton Petrunin | CC BY-SA 2.5 |
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| Oct 26, 2010 at 2:17 | history | answered | Anton Petrunin | CC BY-SA 2.5 |