The Laplace-Beltrami operator is invertible on the space of 1-forms on $S^2$ (since $S^2$ has zero first betti number). Therefore it has an inverse, the Green's function. Now let the radius $r$ of $S^2$ go to infinity, or more precisely, we can identify $S^2\setminus\{\textrm{north pole}\}$ with $\mathbb{R}^2$ and send the radius $r$ round metric on the former to the flat metric on the latter. Do the sequence of Green's operators $G_r$ (restricted to $S^2\setminus\{\textrm{north pole}\}$) converge to the usual Green's operator $G$ on $\mathbb{R}^2$ (which would be $G(x,y) = -\frac{1}{2\pi}\log(|x-y|)$ times the identity operator on $1$-forms)?
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$\begingroup$ Is this true on 0-forms (using geodesic polar coordinates with respect to north pole)? That answer may follow from Cohl's computations in arxiv.org/pdf/1108.3679.pdf, since we get $G=-\frac{1}{2\pi}\log|\cot\frac{\theta}{2}|$ where $\theta$ is geodesic distance between points on the sphere. Now, there is a formula for $\nabla^2$ on 1-forms using this normal coordinate system, in terms of the Laplace operator applied to the divergence and curl of the 1-form (viewed as a vector field via the metric). I wonder if this therefore follows from knowledge about the Green's function on 0-forms. $\endgroup$– Chris GerigCommented Dec 28, 2017 at 21:54
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$\begingroup$ I was wondering would you like to explain why the Laplace-Beltrame operator is invertible on the space of 1-form on $\mathbb S^2$ due to a reason of, $\mathbb S^2$ has a first betti number but not just $\Delta$ is, as itself symbol, not generate expect 0, to give such a result, but this seems to be the definition of $\Delta$ to be a elliptic operator... $\endgroup$– Hu xiyuCommented Dec 29, 2017 at 0:09
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$\begingroup$ So if I do not misunderstanding, the meaning of Laplace-Beltrame operator is invertible should be laplace-beltrami $\Delta$'s Dirichlet boundary problem (there is no boundary on course) could be solved and the solution is unique (up to a constant) ? $\endgroup$– Hu xiyuCommented Dec 29, 2017 at 0:11
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$\begingroup$ @Chris: Yes, good idea to look at 0-forms (somehow I mistakenly thought this to be true and jumped to 1-forms). In that case, the Laplacian has kernel (constant functions), so the Green's function is ambiguous up to a constant. The natural Green's function to consider is one whose average value is zero (since then it's the inverse of the Laplacian on the orthogonal complement to constants). This is equivalent to choice the author uses (in d = 2 at least), and the result is actually false (Green's functions do not converge as r -> infinity)! This suggests the answer to my question may be false. $\endgroup$– Tim NguyenCommented Dec 29, 2017 at 1:10
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