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Finding inverse for a homeomorphism on the sphere (compactification)

  1. Oct 23, 2012 #1
    hi there

    I'd like to show that the sphere
    [tex]\mathbb{S}^n := \{ x \in \mathbb{R}^{n+1} : |x|=1 \}[/tex] is the one-point-compactification of [tex]\mathbb{R}^n[/tex] (*)

    After a lot of trying I got this function:

    [tex]f: \mathbb{S}^n \setminus \{(0,...,0,1)\} \rightarrow \mathbb{R}^n [/tex]
    [tex](x_1,...,x_{n+1}) \mapsto (\frac{x_1}{1-x_{n+1}},...,\frac{x_n}{1-x_{n+1}}) [/tex]

    This is a continuous function, its image is the whole [itex]\mathbb{R}^n[/itex]. If I find its inverse [itex]f^{-1} [/itex] now and show that this one is continuous as well with [itex]image(f^{-1})=\mathbb{S}^n \setminus \{(0,...,0,1)\}[/itex] I have shown (*).


    But I don't find the inverse. [itex] y_i=\frac{x_i}{1-x_{n+1}} [/itex] so [itex] x_i=(1-x_{n+1})*y_1 [/itex] but there is no [itex]x_{n+1}[/itex] here y is a n-dimensional vector...?

    How can I find the inverse of f?

    Regards
     
  2. jcsd
  3. Oct 23, 2012 #2

    mathwonk

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    first try solving for xn+1 in terms of x1,..xn.
     
  4. Oct 23, 2012 #3
    I don't get it:

    If I consider the norms of those vectors I have with [itex] x=(x_1,...,x_n) [/itex]

    [tex]\vec{x}*\frac{1}{1-x_{n+1}} =\vec{y} [/tex] so

    [tex]x_{n+1}=\frac{\|y\|-\|x\|}{\|y\|} [/tex] but then I still have this y. What the trick here to get those [itex]x_{n+1}[/itex]?
     
  5. Oct 23, 2012 #4

    mathwonk

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    can you use the equation for the sphere to solve for xn+1 in terms on x1,..,xn?

    start with case n=1.
     
  6. Oct 24, 2012 #5

    Bacle2

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  7. Oct 30, 2012 #6

    lavinia

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    Steroegraphic projection is conformal so its inverse is also conformal. Is a congormal map continuous?

    Is a conformal mapping an open mapping? That is, does it map open sets onto open sets?
    If so, does that mean that its inverse is continuous?

    This is to show you that you do not need to actually write down the inverse to check whether it is continuous. If you want write the inverse down think of solving for the intersection point on the sphere of straight line from a point in the plane to the north pole.
     
  8. Nov 2, 2012 #7

    Bacle2

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    Still, you cannot treat a map into the Riemann sphere as you would a map from the

    complex plane to itself. Even continuity is tricky, since the sphere is a manifold.

    Or you can treat it as a map from C to C\/{oo}. And then the inverse function

    theorem would tell you when the map is a homeomorphims/diffeomorphism.
     
  9. Nov 3, 2012 #8

    lavinia

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    The sphere minus a point maps confomally onto the plane. Add the north pol and you compactify it.
     
  10. Nov 3, 2012 #9

    Bacle2

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    Yes, I know, this is what I said in post #7 . But conformality is usually defined for

    maps from C to C , where there is a clear meaning of preservation of angles. If you

    want to talk about preservation of angles for curves lying on the sphere, this is

    a whole different story: do you use the tangent space of the sphere? How about

    showing a map into the sphere into C is analytic (since conformal is equivalent to analytic

    with non-zero derivative)? I'm not saying it is wrong; just that it needs an argument. What happens

    with the image vectors near the north-pole? I don't see it; I may be wrong, but it does not seem automatic.

    To talk about analytic in the sphere, you need to bring up charts.
     
    Last edited: Nov 3, 2012
  11. Nov 3, 2012 #10

    lavinia

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    The sphere as a subset of Euclidean space inherits the Riemannian metric directly. You can do charts if you like but you need only to look at vectors in Euclidean space.

    But preserving infinitesimal angles in the plane is no different than on the sphere. Its just that the metrics are different.
     
  12. Nov 3, 2012 #11

    Bacle2

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    Well, I think you would have to show/argue that the stereographic projection T as a map

    from C to the sphere preserves the metric/inner-product in the sense that

    <a,b>_C= <T(a),T(b)>_S^1 . I never saw this argument in your post.
     
  13. Nov 3, 2012 #12

    Bacle2

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    Moreover: if the stereo T preserved the metric: wouldn't this imply that C and S^1

    are isometric?
     
  14. Nov 3, 2012 #13

    lavinia

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    the metric isn't preserved but the map is conformal, angles are preserved infinitesimally. The same thing is true for analytic maps of the complex plane.
    I am beginning to think that I misunderstood your point. i am sorry if I did.
     
  15. Nov 4, 2012 #14

    Bacle2

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    No problem; you may be right and we may be talking about different things. I think it may take too long to untangle ; I think it's run its course.
     
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