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A related, but slightly different question appears here: Upper bound on the number of charts needed to cover a topological manifold

So, we know that for a topological $n$-manifold there is always a covering by $n+1$ charts, but clearly we can sometimes do better. For instance, with $n$-spheres we can always use stereographic projection with two charts. Some manifolds, such as projective spaces (as far as I know), seem to only admit coverings by $n+1$ charts. My question:

Quite obviously a manifold that admits a covering by one chart is homeomorphic to an open subset of $\mathbb{R}^n$. Can we say anything analagous for the case when a manifold can be covered by, at best, $2,3,\ldots,n,n+1$ charts?

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    $\begingroup$ Correction: A manifold that admits a covering by one chart is homeomorphic to an open subset of $\mathbb R^n$, not necessarily to $\mathbb R^n$ itself. $\endgroup$ – Jack Lee Jan 14 '17 at 19:45
  • $\begingroup$ Of course - thank you! $\endgroup$ – Andrew Whelan Jan 14 '17 at 21:14
  • $\begingroup$ Funnily enough, I was browsing through one of your books when I thought of the question - obviously omitting that blunder I made! $\endgroup$ – Andrew Whelan Jan 14 '17 at 21:15
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Note: This answer only addresses the question for charts of the form $\phi : U \to \mathbb{R}^n$ (in particular, $U$ is contractible).


The Lusternik-Schnirelmann category of a topological space $X$ is the smallest integer $k$ such that $X$ admits an open cover $\{U_0, \dots, U_k\}$ with the property that each inclusion $U_i \hookrightarrow X$ is nullhomotopic; denote the Lusternik-Schnirelmann category of $X$ by $LS(X)$.

One of the interesting properties of the Lusternik-Schnirelmann category of a topological space $X$ is its relation to the cup length of its cohomology ring. The $R$ cup-length of $X$ is the largest $m$ such that there are $\alpha_i \in H^{n_i}(X; R)$ with $n_i \geq 1$, such that $\alpha_1\cup\dots\cup\alpha_m \neq 0$. The $R$ cup-length of $X$ is denoted by $\operatorname{cup}_R(X)$ and satisfies $\operatorname{cup}_R(X) \leq LS(X)$.

If $M$ is a topological manifold, and $\phi : U \to \mathbb{R}^n$ is a chart, then the inclusion $U \hookrightarrow M$ is nullhomotopic as $U$ is contractible. Therefore, if $M$ can be covered by $k$ charts, $LS(M) \leq k-1$, and hence $\operatorname{cup}_R(M) \leq k - 1$ for any $R$.

Example $1$: Let $M = (S^1)^n$, and $R = \mathbb{R}$; note that $M$ is smooth. There are classes $\alpha_1, \dots, \alpha_n \in H^1(M; \mathbb{R})$ corresponding to $dx^1, \dots, dx^n$ under the isomorphism $H^1(M; \mathbb{R}) \cong H^1_{\text{dR}}(M)$, and their cup product is non-zero and hence $\operatorname{cup}_{\mathbb{R}}(M) = n$. Therefore $LS(X) \geq n$ and hence $M$ cannot be covered with $n$ or fewer charts.

Example $2$: Suppose $M$ is a $2m$-dimensional compact symplectic manifold and $R = \mathbb{R}$, then $\operatorname{cup}_{\mathbb{R}}(M) \geq m$ (take $\alpha_1 = \dots = \alpha_m = \omega$, the symplectic form). Therefore, $LS(M) \geq m$ and so $M$ cannot be covered by $m$ or fewer charts.

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  • $\begingroup$ Excellent answer - thank you. $\endgroup$ – Andrew Whelan Jan 14 '17 at 18:12

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