Define the open set {$O_\alpha$} that covers the $n$-dimensional sphere and the charts $\psi_\alpha:O_\alpha \rightarrow U_\alpha \subset R^n$.

Question

That is precisely the question. I thought I would just take n-dimensional spherical coordinates, but somehow that doesn't seem to work. The tasks that build on this make it impossible. Maybe I'm just too stupid. Ultimately, the point is to show that the $n$-dimensional sphere is a manifold. Do you have another possible set $O_\alpha$ with charts $\psi_\alpha$ for me to use?

Answer 1

There are many ways to do it.

1. There is an atlas given by the two stereographic projections $\chi^n_N$ and $\chi^n_S$ from north pole $N = (0,\dots,0,1)$ and south pole $S = (0,\dots,0,-1)$. Explicitly $$\chi^n_N : S^n \setminus \{ N \} \to \mathbb R^n, \chi^n_N((x_1,\dots,x_{n+1})) = \frac{1}{1-x_{n+1}}(x_1,\dots,x_n) , $$ $$\chi^n_S : S^n \setminus \{ S \} \to \mathbb R^n, \chi^n_S((x_1,\dots,x_{n+1})) = \frac{1}{1+x_{n+1}}(x_1,\dots,x_n) . $$

2. There are atlases which are simpler, but have more charts. As an example take the $2(n+1)$ charts living on $U_i^\epsilon = \{ (x_1,\dots,x_{n+1}) \in S^n \mid \epsilon x_i > 0 \}$, with $i = 1,\dots,n+1$ and $\epsilon =\pm 1$, and projecting $U_i^\epsilon$ onto the standard open $n$-ball $B^n = \{ (x_1,\dots,x_n) \in \mathbb R^n \mid \sum_{j=1}^n x_j^2 < 1 \}$ by omitting the $i$-th coordinate. Explicitly $$\psi_i : U^\epsilon_i \to B^n, \psi_i((x_1,\dots,x_{n+1})) = (x_1,\dots,x_{i-1},x_{i+1},\dots, x_{n+1}) .$$

Note that a single chart cannot form an atlas: There cannot exist a homeomorphism from $S^n$ to an open $V \subset \mathbb R^n$ because $S^n$ is compact, but no open subset of $\mathbb R^n$ is compact.