Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is it true that any complete metric of nonnegative (Gauss) curvature on $\mathbb R^2$ is conformally equivalent to the standard Euclidean metric on $\mathbb R^2$?

Remarks: locally any Riemannian metric on a surface is conformally equivalent to the standard $\mathbb R^2$, due to existence of isothermal coordinates. Thus a Riemannian metric defines a conformal structure on the surface, and the uniformization theorem says that a simply-connected open surface is conformally equivalent to the complex plane or the unit disk. Thus the question is whether a complete nonnegatively curved plane is conformally equivalent to the complex plane.

share|improve this question
    
@Igor:Does'nt the volume grow at most like the euclidean plane by Bishop-Gromov or am I making a mistake?If the volume growth is atmost quadratic it is easy to show parabolicity. –  Mohan Ramachandran Mar 16 '11 at 18:56
    
Mohan, you are right that by Bishop-Gromov the volume growth is at most quadratic, and also at least linear (both things are true for complete open manifolds of nonnegative Ricci curvature). Does it imply parabolicity? I do not know much of these matters. –  Igor Belegradek Mar 16 '11 at 19:12
1  
Yes it implies parabolicity.In case of complete manifolds with non-negative ricci curvature and atmost quadratic volume growth parabolicity seems to have been prove by Cheng and Yau.In case of surfaces this is much older I believe you can find an argument in a paper of Milnor American Math Monthly vol 84 no 1 Jan 1977 pages 43-46 –  Mohan Ramachandran Mar 16 '11 at 19:18
1  
Also see the paper of A Grigoryan Bulletin of AMS vol 36 pages 135 to 249 .Look at pages 172-173 . –  Mohan Ramachandran Mar 16 '11 at 19:36
    
Thanks, Mohan! I think I see how Cheng-Yau's result does the job, but I cannot figure out why Milnor's paper is relevant; where does he talk about volume, and does he really treat the non-rotationally symmetric case? Anyway, if you care to copy this comment into an answer, I would be happy to accept it. –  Igor Belegradek Mar 16 '11 at 19:44

2 Answers 2

up vote 7 down vote accepted

Cheng-Yau proved that: A complete Riemannian manifold with non-negative Ricci curvature and at most quadratic growth for volumes of balls as the radius goes to infinity is parabolic.

EDIT (by Igor Belegradek). Various criteria for parabolicity are found in the survey of Grigoryan. In particular, on page 177 it is mentioned that a a complete Riemannian manifold with at most quadratic volume growth is parabolic. For complete open nonnegatively curved surfaces the volume growth is at most quadratic by Bishop-Gromov. On the other hand, there are parabolic complete manifolds with arbitrary fast volume growth (see page 180 of the same survey). Finally, the very first proof of parabolicity of complete nonnegatively curved plane seems to be due to Blanc-Fiala (1941); the reference is in Huber's paper mentioned in my comment to Anton's answer.

share|improve this answer

Assume your surface is conformally equivalent to a disc $D$ and $e^\phi$ be the conformal factor. From completeness, $\phi(x)\to\infty$ as $x\to \partial D$. Gauss curvature can be expressed as $K=-\frac12{\cdot} e^{-\phi}{\cdot}\Delta\phi$. Thus, $\Delta\phi\le 0$. The later contradicts maximum principle.

P.S. As Igor noticed, the argument has a gap: we only have that upper limit of $\phi(x)$ is $\infty$ as $x$ converge to any point on the boundary. He also give a ref with a complete proof. The argument would work if for any superharmonic function $\phi$ on $D$ there is a curve $\gamma$ from $0$ to the boundary such that $$\int\limits_\gamma e^\phi<\infty.$$ The later is proved by Fedja Nazarov here.

share|improve this answer
    
Neat! I kept thinking of the unit disk as carrying a hyperbolic metric, which made it harder to see superharmonicity of $\phi$. –  Igor Belegradek Oct 26 '10 at 3:07
    
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. –  Igor Belegradek Mar 15 '11 at 20:14
    
@Igor, Right, I will correct it. –  Anton Petrunin Mar 15 '11 at 23:43
5  
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. –  Igor Belegradek Mar 16 '11 at 15:45
    
Anton, thanks for updating this. I must confess I have not yet worked through all the details of Fedja Nazarov's proof. –  Igor Belegradek Sep 7 '11 at 0:28

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.