Symplectic Geometry – Derive Standard Symplectic Form on 2-Sphere

Question

I have already know that at a point $p\in S^2$ with $u,v$ the tangent vectors at $p$. Then the standard symplectic form is $\omega_p(u,v) :=\langle p,\,u\times v\rangle$ where $u\times v$ is the outer product. My question is: How can one deduce from this that the symplectic form is $dz\wedge d\theta$ in the cylindrical polar coordinates $(z,\theta)$ on $S^2$? Here, $z$ is the height and $\theta$ is the angle. Thanks.

Answer 1

Another way to describe the symplectic form $\omega$ on $S^2 \subset \mathbb{R}^3$ is as the volume form that the standard volume form $\text{vol} = dx \wedge dy \wedge dz$ on $\mathbb{R}^3$ induces on $S^2$ via the inclusion map (using, as usual, the outward-pointing normal). At $p = (x, y, z)$, this normal is $(x \partial_x + y \partial_y + z \partial_z)_p \in T_p \mathbb{R}^3$, which we identify with $p$ in the obvious (in fact, canonical) way. So, $$\omega_p = i_{(x \partial_x + y \partial_y + z \partial_z)_p} \text{vol}_p = i_{(x \partial_x + y \partial_y + z \partial_z)_p} (dx \wedge dy \wedge dz)_p = (x \, dy \wedge dz + y \, dz \wedge dx + z \, dx \wedge dy)_p.$$

Converting $\omega_p$ this to "cylindrical polar coordinates" (or any coordinates) then amounts to pulling this form back by the parameterization $\Phi: U \to S^2$ that defines those coordinates.