Consider the $2$-form on $\mathbb{R}^3$ given by $$\omega = x\,dy\wedge dz+y\,dz\wedge dx+z\,dx\wedge dy.$$ We can restrict this to the sphere. If I use spherical coordinates with $\phi$ the polar angle from the $z$-axis and $\theta$ the azimuthal angle from the $x$-axis, the resulting form I get from computing pullbacks is $\sin\phi\,d\phi\wedge d\theta$.

This can only be computed where spherical coordinates work, which turns out to be everywhere on $S^2$ except for the north and south pole.

I know for a fact that the form is nonzero everywhere on $S^2$. But it seems like $\sin\phi\,d\phi\wedge d\theta$ gets very "small" near the north pole: the scaling by $\sin\phi$ approaches zero. So it seems for the form to be continuous, we would need it to be zero at the north pole, but I know this is not the case.

How am I to reconcile these facts, geometrically? It seems like $d\phi$ and $d\theta$ don't "shrink" near the poles either--I am looking for some geometric intuition. Thanks!

  • $\begingroup$ On the unit sphere, consider a small "square" with corners in $(\phi_0,\theta_0),$ $(\phi_0+\Delta\phi, \theta_0),$ $(\phi_0+\Delta\phi, \theta_0+\Delta\theta)$ and $(\phi_0, \theta_0+\Delta\theta).$ The area of it is approximately $\Delta A=\sin\phi_0 \, \Delta\phi \, \Delta\theta.$ $\endgroup$
    – md2perpe
    Nov 23, 2021 at 18:17

1 Answer 1


Indeed, $d\theta$ does not shrink near the poles. In fact, it blows up in such a say that $\sin\phi d\phi\wedge d\theta$ is finite and non-zero at the poles.

One way to see this is to convert $d\theta$ into a linear combination of $dx,dy$, and $dz$. Since $\theta = \arctan(y/x)$, it is easy to compute that $$d\theta =\frac{1}{x^2 + y^2}\left( -y dx + x dy\right).$$

Considering $\{dx, dy, dz\}$ to be an orthornomal basis for the cotangent space at each point of $\mathbb{R}^3$, it follows that $|d\theta|\rightarrow \infty$ as $(x,y)\rightarrow (0,0)$.

  • $\begingroup$ Ah, that makes sense. Since the tangent vector $\partial/\partial\theta$ is so small near the north pole, $d\theta$ can take a minuscule tangent vector and still evaluate it to be big, so the $\sin\phi$ essentially "offsets" it. $\endgroup$
    – Vasting
    Nov 23, 2021 at 19:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .