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Atlas on S^2

  1. Feb 3, 2015 #1
    Now, this is kind of embarrassing, but I've been trying to do this for too long now and failed: I want to construct an atlas for ##S^2##, but I want to use spherical coordinates rather than stereographic projection.
    Of course the first chart image is simply ##\theta \in (0, \pi), \varphi \in (0,2 \pi)##, which is in a sense the sphere without a line along ##\varphi = 0## going from ##\theta =0## to ##\theta = \pi##.
    All I want now is a second chart that is in a sense "complementary" to the first one, that is I want to cut the sphere along the ##\theta= \pi /2##, ##\varphi \in ( \pi/2 , 3 \pi /2)## line and use a spherical coordinate system for the rest with a simple transition function that just shifts the angles. Is this even possible? If so, how?
     
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  3. Feb 4, 2015 #2

    Orodruin

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    Unless you change the axes that define your spherical coordinates you will not be able to do this as the poles will always be left out. However, you can do it by using spherical coordinates based on a different axis, but this will not be related to the first by a simple shift of the angles.
     
  4. Feb 4, 2015 #3
    Thanks. I kind of expected that it wouldn't work, but did not want to believe it, because it makes things way more complicated...
    Then I have to solve
    $$ -\cos \varphi' \sin \theta' = \cos \varphi \sin \varphi $$
    $$ \cos \theta' = \sin \varphi \sin \theta $$
    $$ \sin \varphi' \sin \theta' = \cos \theta$$
    for ##\theta, \theta' \in (0, \pi); \varphi, \varphi' \in (0, 2 \pi)##, which is a PITA.

    But I think I know the way out now: I just use 3 charts
    $$ \theta_0=\theta; \theta_1= \theta + \pi/2 ; \theta_2 = \theta - \pi /2 $$
    $$ \varphi_0 = \varphi ; \varphi_1 = \varphi - \pi /2 ; \varphi_2 = \varphi - \pi/2 $$.
     
  5. Feb 4, 2015 #4

    Orodruin

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    It is not clear to me what you intend by these charts. The big problem in using different parameter ranges for one definition of spherical coordinates is that the poles will be singular in that definition and therefore you will not be able to cover them using any chart without rotating the axis wrt which you are defining the spherical coordinates. This is why stereographic coordinates are very helpful as you can easily cover all of the sphere except one single point using one chart and then you just take the stereographic projection based on the antipodal point for the next chart and you have an atlas with two charts where each chart is only missing a single point.
     
  6. Feb 4, 2015 #5
    Yes, I know, but the problem is that my sphere is not really embedded in ##\mathbb R ^3##, I actually started with the chart image and a metric on that chart image in something that is analogous to spherical coordinates (Schwarzschild coordinates on a yet to construct Schwarzschild spacetime) and then glued that image into a sphere and now I'm just looking for a simple way to turn this thing into a manifold by defining charts with simple transition functions.

    EDIT: The transformation ##\theta \to \theta \pm \pi/2## is an axis rotation.
     
  7. Feb 4, 2015 #6

    Orodruin

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    Well, you do not need to embed your sphere into R3 in order to parameterise it using a stereographic projection. The stereographic projection is still a parameterisation and the only thing is that it is relatively easy to visualise it using the embedding.
     
  8. Feb 4, 2015 #7

    Ben Niehoff

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  9. Feb 5, 2015 #8
    I guess I'll just use stereographic projection then, because I don't want to overcomplicate things. Still I have a sense that it is possible to do what I said employing the spherical symmetry, but oh well...
     
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