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I have been studying compact topological manifolds lately, in particular the $n$-sphere, $S^n$. The reason $S^2$ cannot be covered by one chart is because it is closed and bounded (and hence, by Heine-Borel, compact). That is to say, there is no OPEN subset of $S^2$ that covers the entire manifold.

$S^1$ is also closed and bounded and contains no single open subset to $\mathbb{R}$ which covers the whole space.

Is it true that any $n$-sphere, $S^n$ needs at least two charts to cover the whole space?

Furthermore, is there any compact manifold which can be covered by a single chart, or does the definition of a chart, which requires an open subset of the manifold, automatically rule out this case?

Thanks.

Johnver
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    Yes, if any manifold $M$ can by covered by a single chart $(M, \phi)$, by definition is homeomorphic to the open subset $\phi(M) \subseteq \Bbb R^n$ for some $n$. The only connected open compact subsets of $R^n$ are the empty set and, when $n = 0$, $\Bbb R^0$ itself, which are hence the only such manifolds. – Travis Willse Oct 24 '15 at 20:01
  • For any n, S^n is an open subset of S^n that covers the entire manifold. – Dan Asimov Sep 09 '23 at 18:16

2 Answers2

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Yes, it's possible to have a manifold with a single chart (note that if $X$ is a space, then $X \subset X$ is open! so it's not a problem that the definition of a chart requires an open subset).

The Euclidean space itself $\mathbb{R}^n$ is such a manifold. More generally, any open subset $U \subset \mathbb{R}^n$ is (almost by definition) a submanifold of $\mathbb{R}^n$ which is covered by a single chart: itself. Conversely, if $M$ is a manifold which is covered by a single chart $(U, \phi : U \to M)$, then it's by definition homeomorphic to $U$ (via $\phi$), an open subset of $\mathbb{R}^n$. So open subsets of $\mathbb{R}^n$ are the only manifold which are covered by only one chart.

(They can be rather wild: by a theorem of Whitney any manifold $M$ embeds in some $\mathbb{R}^N$ for $N$ big enough, then an $\epsilon$-neighborhood of $M \subset \mathbb{R}^N$ is homotopy equivalent to $M$ for small enough $\epsilon$. So essentially you can get all manifolds up to homotopy like this.)

The proof that $S^n$ cannot be covered by a single chart is exactly the same as for $S^2$: it's compact, but a nonempty (thanks @Travis) open subset of $\mathbb{R}^n$ is never compact. More generally a nonempty closed $n$-manifold cannot embed in $\mathbb{R} ^n$.

Najib Idrissi
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A compact topological manifold has not a global chart. Assume that $M$ is a compact topological manifold and $\phi: M \to \Bbb{R}^n$ is a global chart for $M$, since $\phi$ is a homeomorphism then $\phi (M)=\Bbb{R}^n$, i. e. $\Bbb{R}^n$ is compact. This is a contradiction.

Morteza Hasanvand
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