Help for proving a mapping is a diffeomorphism

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    Diffeomorphism Mapping
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Discussion Overview

The discussion revolves around proving whether a specific mapping from R² to R², defined as (x/(x²+y²), y/(x²+y²)), is a diffeomorphism. Participants explore the continuity and differentiability of the mapping, particularly in relation to the origin, and consider potential restrictions to make it a diffeomorphism.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that the mapping is not defined at the origin, suggesting it cannot be a diffeomorphism on R².
  • Another participant proposes that the mapping is smooth away from the origin and hints at the existence of a smooth inverse.
  • A participant suggests finding the inverse by expressing u and v in terms of x and y, indicating a method to derive the inverse mapping.
  • One participant emphasizes understanding the geometric interpretation of the mapping, relating it to polar coordinates and the transformation of points based on their distance from the origin.
  • Another participant notes that the function behaves as its own inverse when excluding the origin.
  • A later reply discusses the potential of using polar coordinates to analyze the mapping further, suggesting a method to derive equations for r' and u'.

Areas of Agreement / Disagreement

Participants generally agree that the mapping is not a diffeomorphism on R² due to its undefined nature at the origin. However, there is no consensus on the specific values to restrict it to in order to establish a diffeomorphism, nor on the exact method to prove it.

Contextual Notes

Some participants express uncertainty about the continuity and differentiability of the mapping, particularly in relation to the origin, and the steps required to find the inverse are not fully resolved.

whattttt
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Hi, does anyone have any idea how to prove the mapping

R^2->R^2
(x/(x^2+y^2), y/(x^2+y^2)

is a diffeomorphism, and if it is not restrict the values so it is one

I am fairly sure it is not over R^2 as it is not continuous at 0, but I don't know what values to restrict it over. I have tried to find the inverse but to no avail, thanks in advance
 
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Is this the exact same question that you asked in your other thread (and then edited out)? If it's exactly the same, then the hint I gave you there is no good. One thread would still be enough, so I have asked the moderators to merge the two.
 
It's not even defined at the origin, so immediately you know it's not a diffeomorphism on R^2. Away from the origin it's smooth by inspection, so it would be enough to find a smooth inverse. You won't have to look very far to find one <wink>.
 
Ok, I didn't think it was defined at the origin. On order to find the inverse do I just set for example u=(x/x^2+y^2) and v=(y/x^2+y^2) and then try and get x and y in terms of u and v. Thanks
 
It will be more helpful to think geometrically about what this map is doing: it's dividing the coordinates of a point by the square of its distance from the origin. In polar coordinates, it would map (r, theta) to (1/r, theta). What would the inverse of that be?
 
I see how it is mapping all the points into a circle centred around the origin bit I'm not sure how the polar co-ords fit in. Do I just make x=rcos(theta) and y =rsin(theta). If as you said what the mapping is doing is dividing by the square of it's distance does that mean the inverse is multiplying by the square of it's distance. Sorry for all the questions
 
The function is its own inverse on the plane minus the origin.
 
So in order to prove that it is diffeomorphic all I have to do is show that (x/x^2+y^2,y/x^2*y^2) is continuous, differentiable and state that minus the origin it is it's own inverse?
 
mapping
 
  • #10
how about going polar in this sense (x,y)=( r cos(u),r sin(u) )|-> ( r' cos (u') /r'^2 , rsin(u')/ r'^2 ) and solving for r',u' ? This gives you two equations write them both down.
 

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