Nowhere $0$ form on the sphere?

Question

Consider the differential form on $\mathbb R^3$ given by $ x dy \wedge dz + y dz \wedge dx + z dx \wedge dy$. I converted this to spherical coordinates using a laborious calculation, and when I'm done, by some miracle (which would be cool if someone could explain exactly how that works), I get something really compact and elegant: $\rho^3 \sin \phi d\phi \wedge d\theta$.

If we call this form $\Omega$, my task is to compute $i^*\Omega,$ where $i$ represents the inclusion from $\mathbb S^2 \rightarrow \mathbb R^3$, which seems to be pretty easy also - on the unit sphere we just have $\rho = 1$.

Now why is this nowhere $0$ on the sphere? It looks like it's $0$ along the half plane where $\phi = 0$!

Answer 1

First of all, there's just a point on the unit sphere where $\phi=0$ — the north pole (but there's also the south pole, where $\phi=\pi$, to worry about). But remember that spherical coordinates actually fail to give a coordinate system at these points (and we can debate what happens when $\theta = 0$ or $2\pi$).

In the original cartesian coordinates, you see that at the poles the $2$-form is given by $z\,dx\wedge dy = \pm dx\wedge dy$, and, since the tangent plane to the sphere is the $xy$-plane, this $2$-form is definitely non-zero on the sphere at those points.

Answer 2

If $$T(\rho,\theta,\phi) = (\rho\cos\theta\cos\phi, \rho\sin\theta\cos\phi, \rho\sin \phi), $$ we wish to compute $\Omega = T^\ast\omega$, where $\omega$ is our initial $2-$form. This is the formal way to do it - the pull-back by the change of coordinates. We do this as follows:

\begin{align} T^\ast\omega &= T^\ast(x\, dy \wedge dz + y\, dz \wedge dx + z\, dx \wedge dy) \\ &= (x \circ T) T^\ast(dy\wedge dz) + (y\circ T) T^\ast(dz \wedge dx) +(z\circ T) T^\ast(dx \wedge dy) \\ &= \rho \cos \theta \cos \phi \,T^\ast(dy \wedge dz) + \rho\sin \theta \cos \phi\,T^\ast(dz \wedge dx) + \rho\sin \phi\,T^\ast(dx \wedge dy) \\ &= \rho \cos \theta \cos \phi \,T^\ast(dy) \wedge T^\ast(dz) + \rho\sin \theta \cos \phi\,T^\ast(dz) \wedge T^\ast(dx) + \rho\sin \phi\,T^\ast(dx) \wedge T^\ast(dy) \\ &=\rho \cos \theta \cos \phi \,d(T^\ast y) \wedge d(T^\ast z) + \rho\sin \theta \cos \phi\,d(T^\ast z) \wedge d(T^\ast x) + \rho\sin \phi\,d(T^\ast x) \wedge d(T^\ast y) \\ &= \rho \cos \theta \cos \phi \,d(y \circ T) \wedge d(z \circ T) + \rho\sin \theta \cos \phi\,d(z \circ T) \wedge d(x \circ T) + \rho\sin \phi\,d(x \circ T) \wedge d(y \circ T) \end{align}

Try to understand which property of the pull-back I used in each step. Now we compute as usual:

\begin{align} d(x \circ T) &= \cos\theta \cos \phi \,d\rho - \rho \sin \theta \cos \phi\,d\theta - \rho \cos \theta \sin \phi\,d\phi \\ d(y \circ T) &= \sin \theta \cos \phi \,d\rho + \rho \cos \theta \cos \phi\,d\theta - \rho \sin \theta \sin\phi\,d\phi \\ d(z \circ T) &= \sin \phi\,d\rho + \rho \cos \phi\,d\phi \end{align}

and plug in the expression to get $T^\ast\omega = \rho^3\sin\phi\,d\phi\wedge d\theta$. I skipped these last steps because now we're only simplifying stuff using properties of $\wedge$.

Then:

\begin{align} i^\ast\Omega &= i^\ast(\rho^3\sin\phi\,d\phi\wedge d\theta) \\ &= (\rho \circ i)^3\sin (\phi\circ i) i^\ast(d\phi \wedge d\theta) \\ &= 1^3 \sin \phi \,i^\ast(d\phi) \wedge i^\ast(d\theta) \\ &= \sin\phi\, d(\phi \circ i)\wedge d(\theta \circ i) \\ &= \sin \phi\, d\phi \wedge d\theta\end{align}

We only have problems at $\phi = 0$ and $\phi = \pi$. But in these points, our change of coordinates is invalid (say, $DT(1,\theta,0)$ and $DT(1,\theta,\pi)$ don't have full rank).

Answer 3

Let me remark that there is no need to convert to spherical coordinates. The coordinates $x,y,z$ are well defined smooth functions on $S^2$, so their exterior derivatives are well defined one-forms on $S^2$. Naturality of the exterior derivative and the operations on differential forms shows that the pullback of your form to $S^2$ is still given by $xdy\wedge dz+ydz\wedge dx+zdx\wedge dy$. To show that this is nowhere vanishing, observe that on $S^2$, one has $x^2+y^2+z^2=1$, so applying the exterior derivative (and dividing by 2) we get $0=xdx+ydy+zdz$. Now on the open subset $\{x\neq 0\}$ you can use this equation to rewrite your form as $\frac1x(x^2+y^2+z^2)dy\wedge dz$. At the same time, on this open subset you can use $y$ and $z$ as local coordinates, which shows that $dy\wedge dz$ and hence your form is nowhere vanishing on $\{x\neq 0\}$. Similar arguments apply to the sets $\{y\neq 0\}$ and $\{z\neq 0\}$, and together the three sets cover $S^2$.