I came across this statement in one of the introductions to differential geometry that some of the manifolds cannot be expressed with a single global coordinate system and one of the examples is the surface of a sphere.

A global coordinate system is one where we have one-to-one mapping from all points on manifold (S) to $\mathbb R^n$. I can map each point on a sphere to $\mathbb R^3$, hence I should have a global coordinate system?

Am I missing something very obvious?

  • $\begingroup$ The dimension of the manifold must be the same as the dimension $n$ of $\Bbb R^n$ containing the image of the coordinate chart(s). $\endgroup$ Dec 16, 2019 at 4:32
  • $\begingroup$ But then I can just say the dimension of surface of the sphere is 3? $\endgroup$ Dec 16, 2019 at 4:39
  • $\begingroup$ No, the chart must be a local homeomorphism, which is why the above condition holds. $\endgroup$ Dec 16, 2019 at 4:44

1 Answer 1


A chart for a manifold $M$ ($\dim M = n$) is a pair $(U,\varphi)$ where $U\subseteq M$ is open and $\varphi\colon U \to \varphi[U]\subseteq \Bbb R^n$ is a homeomorphism onto the open set $\varphi[U] \subseteq \Bbb R^n$. If $M$ is compact, the existence of a global chart $(M,\varphi)$ means that the image $\varphi[M] \subseteq \Bbb R^n$ is open, non-empty, and compact (by continuity of $\varphi$). This is impossible.

  • $\begingroup$ Thanks! I think I understand your argument and on reading further I figured it out too but it makes me wonder if there is a particular intuition or reason why we define the range space to be open set and not closed? $\endgroup$ Dec 16, 2019 at 4:42
  • 2
    $\begingroup$ The same reason why when dealing with Calculus in general, one considers open sets: when computing the limit of a function as $x$ tends to some point $p$ in the open set, one can approach $p$ from all possible directions (as the open set contains some ball centered at $p$). And derivatives are, in the end of the day, limits. Our goal is to do Calculus on the manifold, by transferring everything from $\Bbb R^n$ via charts. $\endgroup$
    – Ivo Terek
    Dec 16, 2019 at 4:44

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.