Proving the Mapping f:S^2\times \mathbb{R} \rightarrow S^2

[cristo]

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 $\mathbb{R}^3 \times \mathbb{R} \rightarrow \mathbb{R}^3$ is defined by

$$((x,y,z),t) \longmapsto \left( \frac{x}{z\sinh t+\cosh t},\frac{y}{z\sinh t+\cosh t},\frac{z\cosh t+\sinh t}{z\sinh t+\cosh t}\right)$$

Show that this determines a mapping $f:S^2\times \mathbb{R} \rightarrow S^2$.

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 $z=\sqrt{1-x^2-y^2}$. However, I can't get this to work, and so it's probably not the correct method!

Any help/hints would be greatly appreciated!

 

[Chris Hillman]

Well, the target space of your mapping is ${\mathbold R}^3$. 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 $(x,y,z)(t)$, how could you show that the curve lies on a sphere $x^2 +y^2 + z^2 = 1$, if it does?

 

[cristo]

Well, if the curve lies on the sphere, then it must satisfy the equation $x^2 +y^2 + z^2 = 1$. 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!

 

[cristo]

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

So, $$x^2+y^2+(z\cosh t+\sinh t)^2=(z\sinh t+\cosh t)^2$$

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

Is this right?

 

[Hurkyl]

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

So, $$x^2+y^2+(z\cosh t+\sinh t)^2=(z\sinh t+\cosh t)^2$$

Expanding and simplifying this gives $x^2+y^2+z^2=1$, 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?

 

[cristo]

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 $x^2+y^2+z^2=1$. How would I show this?

 

[Hurkyl]

I think so. Surely a curve lies on S² if and only if it satisfies the equation $x^2+y^2+z^2=1$. 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.

 

[cristo]

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!