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If I have a simply-connected compact domain $ \Omega $ in $ \mathbb{R}^2 $, endowed with a Riemannian metric $ g $, does there exist a green's function on $ \Omega $ for the laplace operator induced by $ g $?

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What connecttion is there between the two questions? –  Mariano Suárez-Alvarez Jul 21 '13 at 1:56
    
in any case, it is very easy to obtain simply connected $\Omega$s with pairs of cnjugate points. For example, take the complement of a small closed disc in the sphere: that is diffeomorphic to an open disk in the plane. Put on the latter the metric which makes that diffeo an isometry. –  Mariano Suárez-Alvarez Jul 21 '13 at 1:58
    
You're absolutely right, stupid question that I didn't think through. –  user81327 Jul 21 '13 at 2:12
    
What is "compact domain"? (I usually think of domains as being open connected sets). Green function for simply connected domain exists iff the domain is conformally equivalent to the disk. If $g$ is the Euclidean metric, this holds for all proper simply connected domains. But the presence of $g$ (about which you said nothing) makes it impossible to give a concrete answer: I can use diffeomorphism to open disk to represent any simply connected domain as $(\mathbb D,g)$. –  40 votes Jul 21 '13 at 2:58
    
I meant the closure is compact. And $ g $ is not the euclidean metric, it is some arbitrary Riemannian metric. I have seen results that the green's function will exist as long as the manifold is complete, but I am simply not sure. Thank you for your help, Mariano and 40 votes. –  user81327 Jul 21 '13 at 3:04

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