Question about area-preserving map

Question

I don't have a strong background of differential geometry. I want to show that the radial projection $P:\mathbb{S}^2 \to \mathbb{S}^1 \times [-1,1]$, $$P(x_1, x_2, x_3) = \left(\frac{x_1}{\sqrt{x_1^2 + x_2^2}},\frac{x_2}{\sqrt{x_1^2 + x_2^2}}, x_3\right)$$ is area-preserving.

So, written in spherical coordinates on $\mathbb{S^2}$ and cylindrical coordinates on the cylinder, this map becomes $P(\theta, \phi)=(\theta, \sin{\phi})$.

Now I don't understand how to proceed. I saw that the volume form on the sphere is given by $dA=\cos{\phi}\, d\theta \wedge d\phi$ and volume form on the cylinder is given by $dC = d\theta \wedge dz$. Now I don't understand how to proceed.

Answer 1

The volume form on $\mathbb{S}^2$ is actually $dA=\sin(\phi) d\phi \wedge d\theta$...

You just need to show that the pullback of the volume form on $\mathbb{S}^1 \times [-1,1]$ along the map $P$ matches the volume form on $\mathbb{S}^2$. For $\mathbb{S}^2$, we have that;

$$x_1=\sin(\phi)\cos(\theta),\space\space x_2=\sin(\phi)\sin(\theta),\space\space x_3=\cos(\phi)$$

Which will then yield our coordinates on $\mathbb{S}^1 \times [-1,1]$ in terms of the original spherical coordinates;

$$\tilde{x_1}=\cos(\theta),\space\space \tilde{x_2}=\sin(\theta),\space\space \tilde{x}_3=\cos(\phi)$$

Therefore, in cylindrical coordinates;

$$r=1, \space\space \theta=\theta,\space\space z= \cos(\phi)$$

Now, the volume element on the cylinder is;

$$dC=d\theta \wedge dz$$

Therefore, the pullback of $dC$ will then be;

$$P^*(dC)=d\theta \wedge d(\cos(\phi))=-\sin(\phi)\space d\theta \wedge d\phi$$

$$=\sin(\phi) \space d\phi \wedge d\theta$$

So the map $P$ does, indeed, preserve the area.