Classical manifold

Differential Geometry
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Papapetros Vaggelis
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Classical manifold

#1

Post by Papapetros Vaggelis »

Prove that the set \(\displaystyle{S^n=\left\{x=\left(x_1,...,x_{n+1}\right)\in\mathbb{R}^{n+1}: ||x||_{2}=1\right\}\,,n\in\mathbb{N}}\)

is a differentiable manifold of dimension $\dim S^n=n$ .
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Grigorios Kostakos
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Re: Classical manifold

#2

Post by Grigorios Kostakos »

We consider that the sphere \[\displaystyle{S^n=\left\{x=\left(x_1,...,x_{n+1}\right)\in\mathbb{R}^{n+1}: \|x\|_{2}=1\right\}\,,\quad n\in\mathbb{N}}\,,\] is equipped with the induced topology of \(\mathbb{R}^{n+1}\). Let $N=(0,0,\ldots,0,1)\in\mathbb{R}^{n+1}$ and $S=(0,0,\ldots,0,-1)\in\mathbb{R}^{n+1}$ be the north and the south pole of the sphere, respectively. The mappings (stereographic projections) \begin{align*}
p_N:U_N=S^n\setminus\{N\}&\longrightarrow\mathbb{R}^{n}\,;\\
x=(x_1,x_2,\ldots,x_{n+1})&\longmapsto p_N(x)=\Big(\tfrac{2x_1}{1-x_{n+1}},\tfrac{2x_2}{1-x_{n+1}},\ldots,\tfrac{2x_n}{1-x_{n+1}}\Big)\\
p_S:U_S=S^n\setminus\{S\}&\longrightarrow\mathbb{R}^{n}\,;\\
x=(x_1,x_2,\ldots,x_{n+1})&\longmapsto p_S(x)=\Big(\tfrac{2x_1}{1+x_{n+1}},\tfrac{2x_2}{1+x_{n+1}},\ldots,\tfrac{2x_n}{1+x_{n+1}}\Big)
\end{align*}
are homomorfisms from \(U_N\) and \(U_S\), respectively, onto $\mathbb{R}^{n}$.
Stereographic projection $p_N$
Stereographic projection $p_N$
[/centre]
The mappings
\begin{align*}
p_N\circ p_S^{-1},p_S\circ p_N^{-1}:\mathbb{R}^{n}\setminus\{0\}\longrightarrow\mathbb{R}^{n}\setminus\{0\}\,,\\
p_N\circ p_S^{-1}(v)=p_S\circ p_N^{-1}(v)=\frac{4v}{|v|^2}
\end{align*} are differentiable and so the set $\big\{(U_N,p_N),(U_S,p_S)\big\}$ is an atlas of $S^N$, by which $S^N$ becomes a differentiable manifold of dimension \(n\).
Grigorios Kostakos
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