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.


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.

  • $\begingroup$ What operation is $i_p \operatorname{vol}_p$? The only thing that I know to do with forms is to pull them back, but this is clearly not the case here. What is it, then? You seem to feed that normal vector to each factor present in the volume form, but I don't see why. $\endgroup$ – Alex M. Apr 18 '16 at 19:58
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    $\begingroup$ There was a typo, which explains the inclusion. The map $i_X$ is interior multiplication, that sends the $k$-form $\omega$ to the $(k - 1)$-form, which is defined by $(\iota_X \omega)(Y_1, \ldots, Y_{k - 1}) := \omega(X, Y_1, \ldots, Y_{k - 1})$. In short, the operation just feeds $X$ into the first slot. $\endgroup$ – Travis Willse Apr 18 '16 at 20:07

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