# Volume form on $\mathbb{S}^2$

Let $\omega = x_1 \,dx_2 \wedge dx_3 + x_2 \,dx_3 \wedge dx_1 + x_3\, dx_1 \wedge dx_2$, with $(x_1,x_2,x_3) \in \mathbb{S}^2$, be a volume form on $\mathbb{S}^2$, and let $f : \mathbb{S}^1 \times ]-\pi/2,\pi/2[ \rightarrow \mathbb{S}^2 \setminus \{ (0,0,1),(0,0,-1) \}$ be the function defined by $f((u,v),\theta) = (u \cos \theta, v \cos \theta, \sin \theta)$.

• $f$ is a diffeomorphism ;
• $\omega = \frac{dx_1 \wedge dx_2}{x_3}$ on $\mathbb{S}^2 \setminus \{ x_3 = 0 \}$.

Also, $f^*\omega = \sin \theta \cos^2 \theta \,du \wedge dv - v \cos \theta \,du \wedge d\theta + u \cos \theta \,dv \wedge d\theta$. (not entirely sure about that)

I want to show that $\omega$ is exact on $\mathbb{S}^2 \setminus \{ (0,0,1),(0,0,-1) \}$. To do that, I'm supposed to prove that $f^*\omega$ is exact and use the fact that $f$ is a diffeomorphism. Unfortunately, I can't find a "primitive" $\beta$ such that $d\beta = f^*\omega$.

1. Does the given expression for $f^*\omega$ is correct?
2. How can I prove that $f^*\omega$ is exact, without using Poincaré's lemma?

You have to be careful as $(u,v)$ are not local coordinate of $\mathbb S^1$. – John Ma Dec 13 '13 at 3:03
Indeed, thank you! Setting $u = \cos \phi$ and $v = \sin \phi$, we have $f^*\omega = \cos \theta d\theta \wedge d\phi$, so that $f^*\omega$ is exact, and so is $\omega$. This conclusion holds on $\mathbb{S}^2 \setminus \{ (0,0,1),(0,0,-1) \}$. But what about $\mathbb{S}^2 \setminus \{ (0,0,1) \}$ and $\mathbb{S}^2$ ? In these cases, $f$ is no longer a diffeomorphism since its Jacobian matrix's determinant vanishes at $(0,0,1)$ and $(0,0,-1)$. Thanks! – Gatz' Dec 13 '13 at 7:00
Actually $w$ is not exact on $\mathbb S^2$. Did you learn Stokes theorem? – John Ma Dec 13 '13 at 7:14
I know it is not exact on $\mathbb{S}^2$ by Poincaré's lemma. But since I'm supposed to avoid it, I have no idea how to proceed. But I do know Stokes theorem. – Gatz' Dec 13 '13 at 7:17
The hint is to calculate $\int_{\mathbb S^2} w$. – John Ma Dec 13 '13 at 7:19