Consider the 2-sphere $S^2$inside $\mathbb R^3$, and let \begin{align} S:(0,\infty)\times(0,\pi)\times(0,2\pi)&\to\mathbb R^3\setminus\{(0,0,0)\} \\ (r,\phi,\theta)&\mapsto(r\sin\phi\cos\theta,r\sin\phi\sin\theta,r\cos\phi). \end{align} be the map defining spherical coordinates. The form $\omega=r^{-3}(x dy\wedge dz+ydz\wedge dx+zdx\wedge dy)$ on $\mathbb R^3$ restricted to the $S^2$ is in spherical coordinates equal to $F=\sin\phi\,d\phi\wedge d\theta$, over $U$. Of course, the form $\omega$ is defined on the entire 2-sphere. The two 1-forms $A_N=(1-\cos\phi)d\theta$ and $A_S=-(1+\cos\phi)d\theta$ defined on $U$ satisfy $dA_N=F=dA_S$. My question is the following. In the books that I am reading now where I came across this (Nakahara, Geometry, Topology and Physics, Naber, Topology, Geometry and Gauge Fields, Foundations & Interactions), they write $F$ on the whole of $S^2$ as $\sin\phi\,d\phi\wedge d\theta$, and $A_N$ and $A_S$ they say to be defined on $U_N=S^2-{(0,0,-1)}$ and $U_S=S^2\setminus\{(0,0,1)\}$, respectively. But, spherical coordinates are ill-defined on the north and south pole. Can we still make sense of the forms there in these coordinates with some adjustment/argument, or is this simply abuse of the fact that it's only ill-defined at one point that we then don't worry about anymore?

What's more, the forms $A_N$ and $A_S$ are of course pullbacks of the form $Im(\overline z_1\,dz_1+\overline z_2\,dz_2)$ on $S^3=\{(z_1,z_2)\in\mathbb C^2\mid|z_1|^2+|z_2|^2=1\}$, by the canonical sections $\sigma_N:U_N\to S^3$ and $\sigma_S:U_S\to S^3$ (restricted to $U_N\cap U_S$) defined by the local trivialisations of the Hopf bundle $S^1\to S^3\to S^2$. Again, the authors pullback the form to $U_N$ ($U_S$, respectively) instead of $U_N\cap U_S$). Same question applies, really. How to make sense of this, or is it just sloppy notation?

  • $\begingroup$ Note that even though $A_N$ is a priori not defined at the north pole, it does in fact vanish there. For example, near the north pole we have $A_N = (1-z)\frac{-y\,dx+x\,dy}{x^2+y^2} = \frac1{1+z}(-y\,dx+x\,dy)$, which is clearly well-defined, smooth, and vanishes at $(0,0,1)$. $\endgroup$ – Ted Shifrin May 5 '18 at 17:12
  • $\begingroup$ @TedShifrin Should $A_N = (1-\cos \phi )d\theta $ on $U_N$ be understood as $A_N = (1-\cos \phi )d\theta$ on $U_N \smallsetminus (0,0,1)$ and $A_N = 0 = \lim_{\phi \to 0} A_N$ at $(0,0,1)$ ? $\endgroup$ – Sou May 5 '18 at 17:36
  • $\begingroup$ @TedShifrin The same cannot be said for the two-form F, right? That is just $dx\wedge dy$ at for example the north pole. Still, they do write that $dA_N=F$ with $F=\sin\phi d\phi\wedge d\theta$ on $U_N$ and similarly for $A_S$..I find this strange. $\endgroup$ – B. Pasternak May 6 '18 at 13:12
  • $\begingroup$ Why is it strange? Note that $d\big(\frac12(-y\,dx+x\,dy)\big) = dx\wedge dy$. If a $1$-form vanishes at a point, its exterior derivative certainly needn't. $\endgroup$ – Ted Shifrin May 6 '18 at 15:17
  • $\begingroup$ @TedShifrin Of course not, I didn't mean that the vanishing/non-vanishing is strange. I meant that $F$ does not vanish at the north pole (south pole), but that the expression $\sin\phi d\phi\wedge d\theta$, if one were to plug in $\phi=0,\pi$, does vanish. So, it seems strange to me to write $F$ as such on $U_N$ or $U_S$, let alone on $S^2$ itself. $\endgroup$ – B. Pasternak May 6 '18 at 15:21

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