Prove that the set:
$$\mathbb{S}=\left \{ x \in \mathbb{R}^n \bigg|\;\; \left \| x \right \|_{2}=1 \right \}$$
defines a manifold.
Solution
There is nothing to discuss if $n=1$. So, we suppose that $n\geq 2$.
Let $N=\left(0,0,...,1\right), \; P=\left(0,0,...,-1\right)\in \mathbb{S}$ . The set $\mathbb{S}$ is equipped with the topology $\mathbb{A}=\left\{\mathbb{S} \cap U\subseteq \mathbb{R}^{n}: U\in\mathbb{T}\right\}$, where $\mathbb{T}$ is the usual topology of $\mathbb{R}^{n}$, which is a Haussdorf topological space with countable base for this topology.
Consider the sets $U_{N}=\left(\mathbb{R}^{n} \setminus \left\{N\right\}\right)\cap \mathbb{S} \,,U_{P}=\left(\mathbb{R}^{n} \setminus \left\{P\right\}\right)\cap \mathbb{S} \in\mathbb{A}$ and the functions
$$f_1:U_{N}\longrightarrow \mathbb{R}^{n-1}\,,f_1(x_1,x_2,...,x_n)=\left(\dfrac{x_1}{1-x_{n}},\dfrac{x_2}{1-x_{n}},...,\dfrac{x_{n-1}}{1-x_{n}}\right) \\
f_2:U_{P}\longrightarrow \mathbb{R}^{n-1}\,,f_1(x_1,x_2,...,x_n)=\left(\dfrac{x_1}{1+x_{n}},\dfrac{x_2}{1+x_{n}},...,\dfrac{x_{n-1}}{1+x_{n}}\right)$$
which are well defined, bijections and continuous . Also the functions $f_1^{-1}, \; f_2^{-1}$ are continuous too.
Furthermore, $U_{N}\cup U_{P}=\mathbb{S}$ and $U_{N}\cap U_{\mathbb{S}}=\mathbb{S} \setminus \left\{N,P\right\}$ and the functions $f_{i}^{-1}\circ f_{j}\,,i\,,j\in\left\{1,2\right\}\,,i\neq j$ are differentiable. The set $\left\{\left(U_{N},f_{1}\right)\,,\left(U_{P},f_{2}\right)\right\}$ is a parametrization of $\mathbb{S}$ and $\mathbb{S}$ becomes a differentiable manifold with dimension $\dim \mathbb{S} = n-1$.
$$\mathbb{S}=\left \{ x \in \mathbb{R}^n \bigg|\;\; \left \| x \right \|_{2}=1 \right \}$$
defines a manifold.
Solution
There is nothing to discuss if $n=1$. So, we suppose that $n\geq 2$.
Let $N=\left(0,0,...,1\right), \; P=\left(0,0,...,-1\right)\in \mathbb{S}$ . The set $\mathbb{S}$ is equipped with the topology $\mathbb{A}=\left\{\mathbb{S} \cap U\subseteq \mathbb{R}^{n}: U\in\mathbb{T}\right\}$, where $\mathbb{T}$ is the usual topology of $\mathbb{R}^{n}$, which is a Haussdorf topological space with countable base for this topology.
Consider the sets $U_{N}=\left(\mathbb{R}^{n} \setminus \left\{N\right\}\right)\cap \mathbb{S} \,,U_{P}=\left(\mathbb{R}^{n} \setminus \left\{P\right\}\right)\cap \mathbb{S} \in\mathbb{A}$ and the functions
$$f_1:U_{N}\longrightarrow \mathbb{R}^{n-1}\,,f_1(x_1,x_2,...,x_n)=\left(\dfrac{x_1}{1-x_{n}},\dfrac{x_2}{1-x_{n}},...,\dfrac{x_{n-1}}{1-x_{n}}\right) \\
f_2:U_{P}\longrightarrow \mathbb{R}^{n-1}\,,f_1(x_1,x_2,...,x_n)=\left(\dfrac{x_1}{1+x_{n}},\dfrac{x_2}{1+x_{n}},...,\dfrac{x_{n-1}}{1+x_{n}}\right)$$
which are well defined, bijections and continuous . Also the functions $f_1^{-1}, \; f_2^{-1}$ are continuous too.
Furthermore, $U_{N}\cup U_{P}=\mathbb{S}$ and $U_{N}\cap U_{\mathbb{S}}=\mathbb{S} \setminus \left\{N,P\right\}$ and the functions $f_{i}^{-1}\circ f_{j}\,,i\,,j\in\left\{1,2\right\}\,,i\neq j$ are differentiable. The set $\left\{\left(U_{N},f_{1}\right)\,,\left(U_{P},f_{2}\right)\right\}$ is a parametrization of $\mathbb{S}$ and $\mathbb{S}$ becomes a differentiable manifold with dimension $\dim \mathbb{S} = n-1$.
Let \(\displaystyle{\left(t_1,...,t_{n-1}\right)\in f_{1}(U_{N})}\) . Then, there exists \(\displaystyle{\left(x_1,...,x_n\right)\in U_{N}}\)
ReplyDeletesuch that \(\displaystyle{f_1(x_1,...,x_n)=\left(\dfrac{x_1}{1-x_{n}},\dfrac{x_{2}}{1-x_{n}},...,\dfrac{x_{n-1}}{1-x_{n}}\right)=\left(t_1,...,t_{n-1}\right)}\) .
Now, \(\displaystyle{\sum_{i=1}^{n-1}t_{i}^2=\dfrac{1}{\left(1-x_{n}\right)^2}\,\sum_{i=1}^{n}x_{i}^2}\), so :
\(\displaystyle{g(t_1,...,t_{n-1})=\sum_{i=1}^{n-1}t_{i}^2=\dfrac{1-x_{n}^2}{(1-x_{n})^2}=\dfrac{1+x_{n}}{1-x_{n}}}\) or :
\(\displaystyle{x_{n}=\dfrac{g(t_1,...,t_{n-1})-1}{1+g(t_1,...,t_{n-1})}}\) .
Now, \(\displaystyle{x_{i}=t_{i}\,(1-x_{n})=\dfrac{2\,t_{i}}{1+g(t_1,...,t_{n-1})}\,\,,1\leq i\leq n-1}\) . So,
\(\displaystyle{f_1^{-1}(t_1,...,t_{n-1})=\left(\dfrac{2\,t_1}{1+g},...,\dfrac{g-1}{1+g}\right)\,\,,(t_1,...,t_{n-1})\in f_1(U_{N})}\) .
SImilarly,
\(\displaystyle{f_{2}^{-1}(s_1,...,s_{n-1})=\left(\dfrac{2\,s_1}{1+k},...,\dfrac{1-k}{1+k}\right)\,,(s_1,...,s_{n-1})\in f_{2}(U_{S})}\) .
\(\displaystyle{k=\sum_{i=1}^{n-1}s_{i}^2}\).
ReplyDeleteOne way to find the functions \(\displaystyle{f_1\,,f_2}\) is the following :
ReplyDeleteFor \(\displaystyle{f_1}\) : (Similarly, for \(\displaystyle{f_2}\).
Let \(\displaystyle{A\left(x_1,...,x_n\right)\in U_{N}}\) . If \(\displaystyle{A\left(x_1,...,x_n\right)=P}\), then :
\(\displaystyle{f_1(P)=\left(0,0,...,0\right)\in\mathbb{R}^{n-1}}\) . Let \(\displaystyle{A\left(x_1,...,x_n\right)\neq P}\).
One parametrization of the line curve passing through the points \(\displaystyle{A\,,N}\) is
\(\displaystyle{c:\mathbb{R}\longrightarrow \mathbb{R}^{n}\,,t\mapsto c(t)=t\,A+(1-t)\,N=\left(t\,x_1,t\,x_2,...,t\,x_n+1-t\right)}\) .
Obviously, \(\displaystyle{c\left(\left[0,1\right]\right)=AN}\) . For each \(\displaystyle{t\in\mathbb{R}}\) holds :
\(\displaystyle{t\,x_n+1-t=0\iff t\,(x_{n}-1)=-1\iff t=\dfrac{1}{1-x_{n}}}\) .
So, the line \(\displaystyle{c(\mathbb{R}}\) meets \(\displaystyle{R^{n-1}}\) in a unique point :
\(\displaystyle{\left(\dfrac{x_1}{1-x_{n}},\dfrac{x_2}{1-x_{n}},...,\dfrac{x_{n-1}}{1-x_{n}},0\right)}\) .
We define :
\(\displaystyle{f_1(x_1,...,x_n)=\left(\dfrac{x_1}{1-x_{n}},\dfrac{x_2}{1-x_{n}},...,\dfrac{x_{n-1}}{1-x_{n}}\right)\,,\forall\,\left(x_1,...,x_n\right)\in U_{N}}\) .