Help for proving a mapping is a diffeomorphism

  • Thread starter whattttt
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  • #1
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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 continous 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
 

Answers and Replies

  • #2
Fredrik
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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.
 
  • #3
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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>.
 
  • #4
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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
 
  • #5
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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?
 
  • #6
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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
 
  • #7
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The function is its own inverse on the plane minus the origin.
 
  • #8
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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?
 
  • #9
quasar987
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mapping
 
  • #10
how about going polar in this sence (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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