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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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    $\begingroup$ ... And the Riemannian mapping doesn't take vertices to vertices, so you don't want to call it conformal. Is that what you mean? $\endgroup$
    – Alexander Shamov
    Commented Jul 20, 2013 at 8:42
  • $\begingroup$ I think (s)he thinks of the two objects as marked Riemann surfaces (with boundary), as is common in in the subject. $\endgroup$
    – Igor Rivin
    Commented Jul 20, 2013 at 23:34
  • $\begingroup$ Yes, that is correct. I am thinking of the two objects as Riemann surfaces with boundary. $\endgroup$
    – Ritwik
    Commented Jul 21, 2013 at 1:29

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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.

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