# Question about differential forms spherical

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!

• 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 $\Delta A=\sin\phi_0 \, \Delta\phi \, \Delta\theta.$ Nov 23, 2021 at 18:17

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 $$(x,y)\rightarrow (0,0)$$.
• 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. Nov 23, 2021 at 19:07