Symplectic structure on $\mathbb{S}^{2}$

Question

This question has been asked several times but I cannot find a satisfactory answer. Consider $\mathbb{S}^{2} \subseteq \mathbb{R}^{3}$ and define, for every $p \in \mathbb{S}^{2}$ and every $u,v \in T_{p}\mathbb{S}^{2}$, the $\mathbb{R}$-bilinear form $\omega_{p}(u,v) := \langle p, u \times v \rangle = \det(p,u,v)$. It is clear that the assignment $p \longmapsto \omega_{p}$ defines a symplectic form on $\mathbb{S}^{2}$. What I want to show is that, in cylindrical polar coordinates $(\theta,z)$, $w$ can be written as $\omega = d\theta \wedge dz$.

I guess that the idea is to write first $\omega$ in the canonical coordinates $(x,y,z)$ in $\mathbb{R}^{3}$ and then do a change of coordinates. But, how can I write $w$ in standard coordinates first?

Here it is answered, but I don't understand how $\omega$ is written in standard coordinates.

Answer 1

We write everything in terms of $(x, y, z$-coordinates. Let $p = (x, y, z)$ and let $\Omega_p (v, w) = \det (p \ v \ w)$. In this standard coordinates,

\begin{align} \frac{\partial}{\partial x} &= (1,0,0), \\ \frac{\partial}{\partial y} &= (0,1,0),\\ \frac{\partial}{\partial z} &= (0,0,1). \end{align}

So

\begin{align} \Omega_p \left(\frac{\partial}{\partial x} , \frac{\partial}{\partial y}\right) &= \det \begin{bmatrix} x & 1 & 0 \\ y &0 & 1\\ z & 0 & 0\end{bmatrix}=z \end{align}

and similarly

$$ \Omega_p \left(\frac{\partial}{\partial y} , \frac{\partial}{\partial z}\right) = x, \ \ \ \Omega_p \left(\frac{\partial}{\partial z} , \frac{\partial}{\partial x}\right) = y$$

Thus

$$ \Omega = z dx\wedge dy + x dy \wedge dz + y dz\wedge dx.$$

Answer 2

Another proof: $\omega$ is the area form of $\Bbb S^2$, so if $$X(\theta,z)= (\sqrt{1-z^2}\cos\theta,\sqrt{1-z^2}\sin\theta,z),$$then $$\omega = \|X_\theta\times X_z\|\,{\rm d}\theta\wedge {\rm d}z.$$A direct computation gives that the coefficient above is $1$.