0

I've been self-teaching myself differential geometry and still somewhat new. As I understand it, a local coordinate chart on an $n$-dimensional, real, smooth manifold, $M$, is a pair $(M_{(q)},q)$ where $M_{(q)}\subset M$ is some region (the coordinate domain) and

$$ q=(q^1,\dots,q^n):(M_{(q)}\subset M)\to (\mathbb{R}^{n}_{(q)} \subset \mathbb{R}^n) $$

is a local diffeomorphism from the coordinate domain to its image $\mathbb{R}^{n}_{(q)}:=q(M_{(q)})\subset \mathbb{R}^n$.

When trying to conceptualize various things about manifolds and geometry, I often use the 2-sphere $S^2$ as my go-to playground for working through ideas and, when using coordinates, I always use the "usual" coordinates $q=(\theta,\phi)$. However, the 2-sphere is usually also one of the first examples given in any introductory source on manifolds and they always provide the stereographic projection coordinates as an example of a local coordinate chart (that's fine, I'm not confused about that). But I don't think I've ever seen the usual $(\theta,\phi)$ given as an example of local coordinates and it seems like that would be the obvious first example. Is this just because authors assume it's already familiar to everyone and not worth mentioning?

Perhaps I have not been exposed to sources that do give $(\theta,\phi)$ as local coordinates on $S^2$, but then there was this shocker: I saw in Abraham and Marsden's "Foundations of Mechanics'' when they use $(\theta,\phi)$ in the usual way but briefly mention that they are "not true coordinate functions" (I don't remember the exact quote). If that is the case, then I have misunderstood the hell out of everything from the beginning and need to backtrack a lot.

So, what's the deal? Are $(\theta,\phi)$ just too-obvious-to-mention local coordinates on $S^2$ or have I misunderstood everything?

J Peterson
  • 363
  • I would guess that the problem is what happens to $(\theta, \phi)$ coordinates at the poles. – Misha Lavrov Oct 16 '24 at 06:13
  • 2
    They are/can be a true local coordinate chart; however, they are much more easily defined via the inverse which is a local parametrization, so people just don’t want to explicitly write out the domain of the chart (the sphere minus a certain arc of a great circle joining the north and south pole say). Also, people often blur the distinction between the chart (open subset of manifold to open set in $\Bbb{R}^n$) vs its inverse, so perhaps that’s what Abraham and Marsden are alluding to? No way to tell unless you give a precise reference. – peek-a-boo Oct 16 '24 at 06:16
  • 1
    You may want to see my answer here as well where I go into all these details. – peek-a-boo Oct 16 '24 at 06:18
  • @peek-a-boo Ok, I had guessed (and hoped) that the case is along the lines of what you described but I had to make sure that I wasn't misunderstanding something so fundamental. The remark in Abraham and Marsden has been haunting me ever since I encountered it some time ago. I'll see if I can find the exact quote (I think it was in their treatment of the Delaunay variables). You're linked answer is staggeringly thorough, as usual. – J Peterson Oct 16 '24 at 06:58
  • @MishaLavrov The poles are not an issue as such singularities are assumed in the definition of a local coordinate chart. That is, for $S^2$ the domain $S^2_{(\theta,\phi)}\subset S^2$ of $(\theta,\phi)$ is not required to be all of $S^2$. – J Peterson Oct 16 '24 at 07:08
  • 1
    I think I see what you mean (page 636). Previously they define an angular variable to be a mapping with values in $S^1$. And part of this new Kepler map $K$ they’re discussing has the first two component functions taking values in $S^1$ rather than $\Bbb{R}$, so in this sense they’re not coordinates/chart maps. So it’s obvious these aren’t the usual definition (they’re a different type of object) but it’s also clear that this isn’t a huge issue, so they’re brief with it. – peek-a-boo Oct 16 '24 at 07:11
  • @peek-a-boo Oh yes I found it now, thanks. I don't understand their goal in distinguishing an "angular variable" as something different than a function (I don't see the difference since they then regard $S^1\subset\mathbb{R}$). But it doesn't seem like it is significant (for me anyway) so I'll not fret over it any longer. – J Peterson Oct 16 '24 at 19:01
  • Let us continue this discussion in chat. – peek-a-boo Oct 16 '24 at 19:23
  • @peek-a-boo , Apologies, I somehow just now noticed your invitation to continue the discussion in a chat and the link no longer works. – J Peterson May 04 '25 at 14:00
  • 1
    @JPeterson But you can see the stuff I wrote there right? That should be good enough – peek-a-boo May 04 '25 at 17:08

0 Answers0