TITLE: Can one use Brownian motion to prove that two manifolds are not conformally equivalent?
QUESTION [10 upvotes]: 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).

REPLY [11 votes]: 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.