8
$\begingroup$

We all have seen the polar coordinate system back in high school. Recall that $S^2$ may be parametrized by $(\theta,\phi)$ with the correspondence $$\begin{align} x(\theta,\phi) &= \sin(\theta)\cos(\phi) \\ y(\theta,\phi) &= \sin(\theta)\sin(\phi) \\ z(\theta,\phi) &= \cos(\theta) \end{align}$$

where $\theta\in [0,\pi]$ and $\phi\in [0,2\pi)$. There's nothing complicated about this. However, as I began to study some differential geometry and manifold, I find this to be confusing.

Viewing $S^2$ as a $2$-dimensional manifold, we want to think of this parametrization as a coordinate chart $(U,\psi)$, where $\psi$ maps $(x,y,z)$ to the corresponding $(\theta,\phi)$ in $\Bbb R^2$. Then $\psi^{-1}$ would be our parametrization of the sphere.

However, the definition requires that $\psi$ is a diffeomorphism from the open set $U$ onto its image. Clearly $[0,\pi]\times [0,2\pi)$ is not open (this is related to the fact that there's no global chart for $S^2$). Yes, we may restrict to $U\subsetneq S^2$ but that is not my question.

In the language of manifold theory, what does it mean to talk about the global chart or the global parametrization of a sphere?

Even though I know that it's an abuse of languge, strictly speaking, but it appears a lot in the literature. For example, I am reading a book on Riemannian manifold and there's a part that says

We therefore introduce on $\Bbb R^d$ the standard polar coordinates $$ (r,\varphi^1,\dots,\varphi^{d-1}) $$ where $\varphi=(\varphi^1,\dots,\varphi^{d-1})$ parametrizes the unit sphere $S^{d-1}$. ... Express the (Riemannian) metric in polar coordinates ... in these coordinates at $0\in T_pM$, $0$ corresponds to $p\in M$, $$ g_{rr}(0)=1,\ g_{r\varphi}(0)=0. $$

Why are we allowed to use regular rules of taking derivative and such in this "global coordinate"? How do we rigorously justify the use of this global coordinate system?

$\endgroup$
  • $\begingroup$ You seem to be bothered by the phrase "parameteizes the unit sphere." But this is no more problematic than saying $x=\cos\theta$, $y=\sin\theta$ for $\theta\in[0,2\pi)$ "parametrizes the unit circle." Of course in manifold theory the circle in fact requires two coordinate charts and cannot be parametrized globally, because we must cut out a point for our charts to be defined on open sets. In practical applications (e.g. integration), this doesn't pose a problem; issues at sets of measure zero don't matter. $\endgroup$ – symplectomorphic Nov 15 '17 at 15:21
  • $\begingroup$ @symplectomorphic I decided to make my question more precise. If you wouldn't mind could you take a look at this? mathoverflow.net/questions/286182/… $\endgroup$ – BigbearZzz Nov 15 '17 at 22:28
3
$\begingroup$

You should think of this parametrization as a ramified cover of the $2$-sphere $\mathbf{S}^2$ by $\mathbf{R}^2$. Namely, instead of restricting it to get a bijection from $[0,\pi] \times [0,2\pi)$ onto $\mathbf{S}^2$ (which requires a non-canonical choice of endpoints), consider it as a quotient map $$\mathbf{R}^2 \rightarrow \mathbf{S}^2$$ which a calculation of the Jacobian matrix shows is ramified precisely over the points $(0,0,\pm 1) \in \mathbf{S}^2$, when $\theta$ is an integer multiple of $\pi$. Away from these points, the map is a covering projection, and so you obtain charts by taking local sections. You can think of polar coordinates on $\mathbf{R}^n$ in the same way.

Given a metric $g$ on the codomain of a differentiable map between manifolds, you can pull it back to the domain to obtain a metric except on the fibres over the ramified points; this is the formalization of expressing $g$ in polar coordinates. Without knowing precisely what conclusions you wish to draw about $g$ itself it is difficult to say more about how to use this formalism.

$\endgroup$
  • $\begingroup$ I believe one of the main aim of representing $g$ in polar coordinate in the book is to calculate the volume form and do some integration. So basically you're saying that everything works just fine except for the inverse images of the poles? Would that affect integration? (I'm guessing no since the points have 0 measure, but I'm not sure if there's any blow-up phenomenon that I should be careful of or not.) $\endgroup$ – BigbearZzz Nov 15 '17 at 13:58
  • $\begingroup$ Integration is only one of the problems. I decided to make my question a bit more precise. I you wouldn't mind, could you please take a look? mathoverflow.net/questions/286182/… $\endgroup$ – BigbearZzz Nov 15 '17 at 22:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.