Confusion about differential forms and spherical coordinates

Question

Let $x,y,z$ denote the standard coordinates of $\Bbb R^3$, and let $r,\phi,\theta$ denote the spherical coordinates on $\Bbb R^3-\{0\}$. It is well-known that $dx \wedge dy\wedge dz=r^2 \sin \phi ~dr\wedge d\phi \wedge d\theta$. But, since $\phi=0$ on the positive $z$-axis and $\phi=\pi$ on the negative $z$-axis, this means that $dx\wedge dy\wedge dz$ vanishes on the $z$-axis. However, $dx\wedge dy\wedge dz$ is a top form on $\Bbb R^3$ and it must be a basis of $\bigwedge^* (T_p\Bbb R^3)$ for each point $p\in \Bbb R^3$. What went wrong?

Answer 1

Spherical coordinates don't cover all of $\mathbb{R}^3\setminus\{0\}$. Typcally, one restricts to the ranges $$ r\in(0,\infty),\ \ \ \ \ \theta\in(0,2\pi),\ \ \ \ \ \phi\in(0,\pi) $$ which maps to a domain (which notably does not include the $z$-axis) $$ U=\left\{(x,y,z)\in\mathbb{R}^3:y\neq0\text{ or }x<0\right\} $$ It isn't possible to extend the coordinates further without running into discontinuities. The volume form $\omega=r^2\sin(\phi)dr\wedge d\phi\wedge d\theta$ is only defined on $U$ as written, where it is equal to $dx\wedge dy\wedge dz$. While it's possible to smoothly extend $\omega$ to all of $\mathbb{R}^3$, one cannot assume that the coordiante functions $r,\theta,\phi$ (or their derivatives) can be extended: if you want to describe the behavior outside of the domain $U$, it would be best to change to a new set of coordinates which cover the points of interest.

The behavior around the $z$-axis gives an example of the pitfalls of assuming $r,\theta,\phi$ extend smoothly beyond their domains: even though $\sin(\phi)$ approaches zero, $d\theta$ has a singularity, and it's not obvious in polar coordinates that these two divergences "cancel out" to produce a nonzero limit.