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

Let me start by a very simple example; consider the following question:

"Let D1 be a square and D2 a rectangle (boundary included). View them as subsets of the complex plane. Does there exist a conformal map (extending to the boundary) taking D1 to D2?"

Of course, the answer is no, but I want to point out an unusual "proof" of this assertion. Suppose the answer was yes. I think we can assume that the center gets mapped to the center. Start a brownian motion from the center of D1. The probability that this brownian motion hits any of the four sides is equal. However the probability that a brownian motion hits any of the four sides starting from the center of of D2 is not equal. And this is a contradiction, because brownian motion is conformally invariant (which is a non trivial fact, but its true).

I believe this "proof" can be made rigorous. My question is the following:

Can this same idea be used to show for instance two complex manifolds are not biholomorphic to each other? Of course there maybe a simpler proof using more direct methods, but I am still curious to know if the idea of using brownian motion can be used to answer such a question (ie are two manifolds conformally equivalent).

share|improve this question
1  
... And the Riemannian mapping doesn't take vertices to vertices, so you don't want to call it conformal. Is that what you mean? –  Alexander Shamov Jul 20 '13 at 8:42
    
I think (s)he thinks of the two objects as marked Riemann surfaces (with boundary), as is common in in the subject. –  Igor Rivin Jul 20 '13 at 23:34
    
Yes, that is correct. I am thinking of the two objects as Riemann surfaces with boundary. –  Ritwik Jul 21 '13 at 1:29
add comment

1 Answer 1

An easy example is the proof that the open disk is not conformally equivalent to the plane, since the tail sigma-field of the Brownian motion on the disk is nontrivial (it contains information about the boundary point) whereas the tail sigma-field of the BM in the plane is trivial. I guess this observation should have generalizations in terms of boundary theory of Markov processes.

share|improve this answer
add comment

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.