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Help with a function on S^2

  1. Jan 6, 2007 #1

    cristo

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    I wonder if anyone can help with this question. It's a part of a Differential Geometry exam question which I can't get!

    A map [itex]\mathbb{R}^3 \times \mathbb{R} \rightarrow \mathbb{R}^3 [/itex] is defined by

    [tex] ((x,y,z),t) \longmapsto \left( \frac{x}{z\sinh t+\cosh t},\frac{y}{z\sinh t+\cosh t},\frac{z\cosh t+\sinh t}{z\sinh t+\cosh t}\right) [/tex]

    Show that this determines a mapping [itex] f:S^2\times \mathbb{R} \rightarrow S^2 [/itex].

    I tried substituiting polar coordinates for S2 in place of (x,y,z) in the above function. Then I figured that if the function's image is a subset of S2, then the coordinates must satisfy [itex]z=\sqrt{1-x^2-y^2}[/itex]. However, I can't get this to work, and so it's probably not the correct method!

    Any help/hints would be greatly appreciated!
     
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  3. Jan 6, 2007 #2

    Chris Hillman

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    Well, the target space of your mapping is [itex]{\mathbold R}^3[/itex]. Can you show that the range of your mapping is an ordinary round sphere in that space?

    If you don't see it yet, try this: I give you the coordinates of a curve [itex](x,y,z)(t)[/itex], how could you show that the curve lies on a sphere [itex]x^2 +y^2 + z^2 = 1[/itex], if it does?
     
  4. Jan 6, 2007 #3

    cristo

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    Well, if the curve lies on the sphere, then it must satisfy the equation [itex]x^2 +y^2 + z^2 = 1[/itex]. So would I just square the components on the RHS of the expression for the mapping above, and show that they sum to 1? However, the terms contain x,y,z- i.e. it is not expressed explicitly in t so I'm not sure this would work.

    Sorry, I'm sure there's something really basic here that I'm just not seeing!
     
  5. Jan 6, 2007 #4

    cristo

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    ACtually, I think I've got it. We want [tex]\left(\frac{x}{z\sinh t+\cosh t}\right)^2 + \left(\frac{y}{z\sinh t+\cosh t}\right)^2+\left(\frac{z\cosh t+\sinh t}{z\sinh t+\cosh t}\right)^2=1 [/tex]

    So, [tex]x^2+y^2+(z\cosh t+\sinh t)^2=(z\sinh t+\cosh t)^2[/tex]

    Expanding and simplifying this gives [itex]x^2+y^2+z^2=1[/itex], and so the range of the mapping is on the sphere.

    Is this right?
     
  6. Jan 6, 2007 #5

    Hurkyl

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    Not quite -- that's the converse of what you want to show. Is this proof reversible?
     
  7. Jan 6, 2007 #6

    cristo

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    I think so. Surely a curve lies on S2 if and only if it satisfies the equation [itex]x^2+y^2+z^2=1[/itex]. How would I show this?
     
    Last edited: Jan 6, 2007
  8. Jan 6, 2007 #7

    Hurkyl

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    I think you have misunderstood.

    What you have shown is that if f(P, t) lies on the sphere, then P lies on the sphere.

    But your goal was to prove that if P lies on the sphere, then f(P, t) lies on the sphere.
     
  9. Jan 6, 2007 #8

    cristo

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    Ahh, ok I understand what you mean now. I think the proof probably is reversible, but have no clue as to how to show it! Can you give a hint? Sorry to be a pain!
     
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