# Can one use Brownian motion to prove that two manifolds are not conformally equivalent?

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).

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... 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