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Oct 7, 2013 at 0:44 comment added user6818 @WillieWong I have given what I think is the right answer in this new question, mathoverflow.net/questions/144159/… [...it would be great to see you there...]
Oct 7, 2013 at 0:43 comment added user6818 @WillieWong [....now that I think of it the basic issue is that since I want to transform the asymptotics of eigenfunctions of the Laplacian, it probably is not true that if I change coordinate systems then the new coordinate substituted into the old function will still keep them eigenfunctions of the Laplacian...]
Oct 7, 2013 at 0:42 comment added user6818 @WillieWong There are physics reasons to believe as to why a polynomial behaviour is the right answer. From other discussions that I had, I learnt that the way to convert the answer to my $z$ coordinates is to realize that the dependency on $y$ should be geometrically thought of as a dependency on the geodesic distance from the vertex of the hyperboloid. So one transforms that distance into the $z$ coordinates and then asks the question about the functions.
Sep 24, 2013 at 9:40 comment added Willie Wong I actually don't see why $e^{1/z}$ "makes no sense". Why do you think there should be harmonics that scale like power law near the conformal boundary? After all, in hyperbolic models the area of the spheres near infinity grows exponentially. (Also, check your signs; isn't the large $y$ behaviour in the paper you cited $e^{-\rho y}$ where $\rho > 0$?)
Sep 24, 2013 at 9:27 history closed Will Jagy
Andrés E. Caicedo
Todd Trimble
Ryan Budney
Willie Wong
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Sep 22, 2013 at 0:25 comment added user6818 Can you tell as to what is the metric in the x, y and the z coordinates of the transformations given in that answer? Then I can be sure that it at least gets the start and the end metric as the two metrics which I have listed. [..can you give an example of the isometries of the Poincare disk that you mention?..is there a way to just write them down?...and hope that one of them does something in the exponent - I would have thought that there is some function $f$ such the $e^y = f(z)$...)
Sep 21, 2013 at 22:16 review Close votes
Sep 24, 2013 at 9:27
Sep 21, 2013 at 22:07 comment added Will Jagy There are many, many isometries on the Poincare disk, or ball in more dimensions. You are free to apply those before the map to upper half space to get your desired conditions. what is your actual background?
Sep 21, 2013 at 22:03 comment added user6818 @WillJagy As far as I can see that answer is not helping - in the Mobius transformation described there in the large $y$ limit at fixed $y_n$ the asymptotic relation is $y \sim \frac{1}{z}$ - this paper says that the large-y asymptotics are of the form $e^y$ and hence in $z$ coordinates they will look like $e^{\frac{1}{z}}$ - and this makes no sense in small $z$ - I would like to isolate all those harmonics which near $z=0$ scale as some say $z^a$ for a given number $a$ - but that can't be done in this form!
Sep 21, 2013 at 21:59 comment added Will Jagy you got a good answer at math.stackexchange.com/questions/499042/…
Sep 21, 2013 at 21:55 history asked user6818 CC BY-SA 3.0