# Verification that $S^{n}$ is a differentiable manifold.

Setting $S^{n} := \{x\in\mathbb{R}^{n+1}: \|x\| = 1\}$, and labelling the north and south poles as $N:= (0,\ldots,0,1)$, $S:=(0,\ldots,0,-1)$, I can set the coordinate charts up as follows:

Let $U_N = S^n - N$ and $U_S = S^n - S$.

Taking the usual stereographical projections $\phi_{N}:U_{N}\to\mathbb{R}^{n}$ and $\phi_S : U_S \to\mathbb{R}^n$, we turn $S^n$ into a topological manifold of dimension $n$.\ \ I'm having trouble verifying that this structure satisfies the definition of a differentiable manifold, as I do not know how to check that $f:=\phi_{N}\circ \phi_{S}^{-1}:\mathbb{R}^{n} - 0$ is $C^{\infty}$. It is clear how to show (once I draw the picture) that $f$ is a bijection.

Do I need to derive a formula for the $\phi_{N}$ and $\phi_{S}^{-1}$ map in order to show that it is $C^{\infty}$? Any references I found this in usually display it as an example and mention casually that it is easy to show that $f$ is $C^\infty$, but I'm not even sure how to begin.

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Yes, just write down a formula for the composition and then verify that it is infinitely differentiable. Once you have a formula for the $\phi$ you should be able to write down a formula for the composition; this involves knowing what stereographic projection is. – Jason Polak Oct 28 '12 at 0:41

Let $y = \phi_N (x)$, then this is equivalent to $N+t(x-N) = (y,-1)$, with slight abuse of notation. Since $1+t(x_{n+1}-1) = -1$, it follows that $y = \frac{2}{1-x_{n+1}} (x_1,...,x_n)$. Hence $\phi_N$ is smooth on $\mathbb{R}^{n+1} \setminus \{N\} \supset U_N$ (ie, smooth as a mapping $\mathbb{R}^{n+1} \setminus \{N\} \to \mathbb{R}^n$).

Now suppose $x = \phi_N^{-1}(y)$, again equivalent to $N+t(x-N) = (y,-1)$. We wish to find the $x \in \partial B(0,1)$ that satisfies this equation. Rearranging this equation gives $t x = (y, t-2)$ (again a slight abuse of notation). Summing the squares of the components gives $t^2 \sum_{i=1}^{n+1} x_i^2 = \sum_{i=1}^{n} y_i^2+(t-2)^2 = \sum_{i=1}^{n} y_i^2 + t^2 -4t +4$. If $x \in \partial B(0,1)$, this gives $t = 1 + \frac{1}{4}\sum_{i=1}^{n} y_i^2$, from which it follows that $\phi_N^{-1}(y) = \frac{1}{1 + \frac{1}{4}\sum_{i=1}^{n} y_i^2} (y, -1+\frac{1}{4}\sum_{i=1}^{n} y_i^2)$.

It is easy to see that $\phi_N^{-1}$ is smooth on $\mathbb{R}^n$ (ie, smooth as a mapping $\mathbb{R}^n \to \mathbb{R}^{n+1}$).

Let $\sigma(x) = (x_1,...,x_n,-x_{n+1})$ (just reverses the last coordinate). Note that $\sigma$ is a diffeomorphism between $\mathbb{R}^{n+1}$ and $\mathbb{R}^{n+1}$, $\sigma(U_N) = U_S$ and $\sigma = \sigma^{-1}$. Then it is easy to see that $\phi_S = \phi_N \circ \sigma$. Furthermore, $\phi_S^{-1} = \sigma \circ \phi_N^{-1}$. The smoothness properties of $\phi_S$ and $\phi_S^{-1}$ are the same as those of $\phi_N$, $\phi_N^{-1}$ with appropriate changes (ie, replace $N$ by $S$).

It follows that $\phi_N \circ \phi_S^{-1}$ is smooth (as a mapping $\mathbb{R}^n\setminus \{0\} \to \mathbb{R}^n$, noting that $\phi_S^{-1}(0) = N$).

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thank you very much! – roo Oct 28 '12 at 6:14
You are very welcome! – copper.hat Oct 28 '12 at 6:20