Problem 1-7 (d) in John M. Lee's Introduction To Smooth Manifolds asks to

... verify that the atlas consisting of two charts $(\mathbb{S}^n \setminus \{N\}, \sigma)$ and $(\mathbb{S}^n \setminus \{S\}, \tilde{\sigma})$ defines a smooth structure on $\mathbb{S}^n$.

, where $N = (0,\ldots,0,1) \in \mathbb{R}^{n+1}$ and $S = (0,\ldots,0,-1) \in \mathbb{R}^{n+1}$, $\sigma$ is the stereographic projection from $N$ and $\tilde{\sigma}$ is the stereographic projection from $S$. Lee defines a smooth structure to be a maximal smooth atlas on a topological manifold. I have managed to prove that $\sigma$ and $\tilde{\sigma}$ are bijective, that the transition map $\tilde{\sigma} \circ \sigma^{-1}$ is a diffeomorphism, and hence the charts $(\mathbb{S}^n \setminus \{N\}, \sigma)$ and $(\mathbb{S}^n \setminus \{S\}, \tilde{\sigma})$ are smoothly compatible. Thus, the atlas consisting of these two charts is smooth.

I have struggled to prove that these two charts form a maximal smooth atlas on $\mathbb{S}^n$. After some frustration, I checked this site and found that this smooth atlas is not actually maximal, according to the following post:

How to show this atlas is maximal on the sphere $S^n$?

So is the term smooth structure in the question a typo? Should it have instead asked just to prove that the two charts form a smooth atlas on $\mathbb{S}^n$?



1 Answer 1


As you might recall, every smooth atlas $\mathcal{A}$ for a topological manifold $M$ is contained in a $\textbf{unique maximal smooth atlas}$ ($\textit{Proposition 1.17}$ ). So by showing that $\mathcal{A} = \{(\Bbb{S}^n \smallsetminus N,\sigma), (\Bbb{S}^n\smallsetminus S,\tilde{\sigma})\}$ is a smooth atlas for $\Bbb{S}^n$, you automatically have a unique smooth structure determined by $\mathcal{A}$. Put it another way, $\mathcal{A}$ defines a smooth structure on $\Bbb{S}^n$. So it's not a typo.

  • $\begingroup$ Thanks for the clarification. $\endgroup$ Aug 9, 2018 at 15:36
  • $\begingroup$ @FredericChopin You’re welcome. $\endgroup$ Aug 9, 2018 at 15:39

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.