Help with a function on S^2

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  • #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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  • #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?
 
  • #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!
 
  • #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?
 
  • #5
Hurkyl
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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?

Not quite -- that's the converse of what you want to show. Is this proof reversible?
 
  • #6
cristo
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Not quite -- that's the converse of what you want to show. Is this proof reversible?

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?
 
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  • #7
Hurkyl
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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?
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.
 
  • #8
cristo
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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.


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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