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- Feb 29, 2012
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Hello! 
I'm trying to do a problem from John Lee's book Introduction to Smooth Manifolds where it is asked to work with stereographic coordinates.
It defines $N = (0,\ldots,0,1)$ to be the north pole in $\mathbb{S}^n \subset \mathbb{R}^{n+1}$ and $S = -N$ the south pole. The stereographic projection $\sigma: \mathbb{S}^n \backslash \{ N \} \to \mathbb{R}$ is defined as
$$\sigma(x^1, \ldots, x^{n+1}) = \frac{(x^1, \ldots, x^n)}{1 - x^{n+1}},$$
with $\tilde{\sigma}(x) = - \sigma(-x) = \frac{(x^1, \ldots, x^n)}{1 + x^{n+1}}$ working as the projection from the south pole, $\tilde{\sigma}: \mathbb{S}^n \backslash \{ S \} \to \mathbb{R}^n$.
I'm having troubles when asked to show that $\sigma$ is bijective and that its inverse is given by
$$\sigma^{-1} (x^1, \ldots, x^n) = \frac{(2x^1, \ldots, 2x^n, |x|^2 -1)}{|x|^2 +1}.$$
Well, one way to show is that $\sigma^{-1} \circ \sigma (x) = x$ for all $x \in \mathbb{S}^n \backslash \{ N \}$ and $\sigma \circ \sigma^{-1} (y) = y$ for all $y \in \mathbb{R}^n$, but I feel I keep missing something or getting things wrong in my calculations.
For example, when doing $\sigma^{-1} \circ \sigma (x)$ we have
$$\sigma^{-1} \circ \sigma (x) = \sigma^{-1} (\sigma (x))$$
$$= \sigma^{-1} \left( \frac{(x^1, \ldots, x^n)}{1-x^{n+1}} \right)$$
$$= \frac{ \left( \frac{2x^1}{1-x^{n+1}}, \ldots, \frac{2x^n}{1-x^{n+1}}, \frac{|x'|^2}{(1-x^{n+1})^2} -1 \right)}{\frac{|x'|^2}{(1-x^{n+1})^2} +1},$$
where I'm denoting $x' = |(x^1, \ldots, x^n)|$. Then, it follows that $\frac{1}{\frac{|x'|^2}{(1-x^{n+1})^2} +1} = \frac{(1-x^{n+1})^2}{|x'|^2 + (1-x^{n+1})^2}$, allowing us to cancel some of the denominators. The above expression turns into
$$\frac{ \left( \frac{2x^1}{1-x^{n+1}}, \ldots, \frac{2x^n}{1-x^{n+1}}, \frac{|x'|^2}{(1-x^{n+1})^2} -1 \right)}{\frac{|x'|^2}{(1-x^{n+1})^2} +1} = \frac{ \left( 2x^1 (1-x^{n+1}), \ldots, 2x^n (1-x^{n+1}), |x'|^2 - (1-x^{n+1})^2 \right)}{|x'|^2 + (1-x^{n+1})^2}.$$
This is where I get stuck. I don't see a way out of this to make it become a simple $(x^1, \ldots, x^{n+1})$, and that could be for two reasons: pure ignorance or I'm taking the wrong path.
Making use of the post, I also did the calculations required for the transition map $\tilde{\sigma} \circ \sigma^{-1}$, I would also appreciate someone performing a second check.
$$\tilde{\sigma} \circ \sigma^{-1}(x) = \tilde{\sigma} \left( \frac{(2x^1, \ldots, 2x^n, |x'|^2 -1)}{|x'|^2 +1} \right)$$
$$= \frac{\frac{(2x^1, \ldots, 2x^n)}{|x'|^2 +1}}{(|x'|^2 -1) +1}$$
$$= \frac{(2x^1, \ldots, 2x^n)}{|x'|^2(|x'|^2 +1)}.$$
Thanks for all your troubles reading this overly long post.
All help is appreciated!
P.S.: If anybody knows a more effective way of aligning these questions here in MHB, please post (or PM me, in case moderators find it off-topic). In usual LaTeX I use \begin{flalign} but I haven't tried it here.
I'm trying to do a problem from John Lee's book Introduction to Smooth Manifolds where it is asked to work with stereographic coordinates.It defines $N = (0,\ldots,0,1)$ to be the north pole in $\mathbb{S}^n \subset \mathbb{R}^{n+1}$ and $S = -N$ the south pole. The stereographic projection $\sigma: \mathbb{S}^n \backslash \{ N \} \to \mathbb{R}$ is defined as
$$\sigma(x^1, \ldots, x^{n+1}) = \frac{(x^1, \ldots, x^n)}{1 - x^{n+1}},$$
with $\tilde{\sigma}(x) = - \sigma(-x) = \frac{(x^1, \ldots, x^n)}{1 + x^{n+1}}$ working as the projection from the south pole, $\tilde{\sigma}: \mathbb{S}^n \backslash \{ S \} \to \mathbb{R}^n$.
I'm having troubles when asked to show that $\sigma$ is bijective and that its inverse is given by
$$\sigma^{-1} (x^1, \ldots, x^n) = \frac{(2x^1, \ldots, 2x^n, |x|^2 -1)}{|x|^2 +1}.$$
Well, one way to show is that $\sigma^{-1} \circ \sigma (x) = x$ for all $x \in \mathbb{S}^n \backslash \{ N \}$ and $\sigma \circ \sigma^{-1} (y) = y$ for all $y \in \mathbb{R}^n$, but I feel I keep missing something or getting things wrong in my calculations.
For example, when doing $\sigma^{-1} \circ \sigma (x)$ we have
$$\sigma^{-1} \circ \sigma (x) = \sigma^{-1} (\sigma (x))$$
$$= \sigma^{-1} \left( \frac{(x^1, \ldots, x^n)}{1-x^{n+1}} \right)$$
$$= \frac{ \left( \frac{2x^1}{1-x^{n+1}}, \ldots, \frac{2x^n}{1-x^{n+1}}, \frac{|x'|^2}{(1-x^{n+1})^2} -1 \right)}{\frac{|x'|^2}{(1-x^{n+1})^2} +1},$$
where I'm denoting $x' = |(x^1, \ldots, x^n)|$. Then, it follows that $\frac{1}{\frac{|x'|^2}{(1-x^{n+1})^2} +1} = \frac{(1-x^{n+1})^2}{|x'|^2 + (1-x^{n+1})^2}$, allowing us to cancel some of the denominators. The above expression turns into
$$\frac{ \left( \frac{2x^1}{1-x^{n+1}}, \ldots, \frac{2x^n}{1-x^{n+1}}, \frac{|x'|^2}{(1-x^{n+1})^2} -1 \right)}{\frac{|x'|^2}{(1-x^{n+1})^2} +1} = \frac{ \left( 2x^1 (1-x^{n+1}), \ldots, 2x^n (1-x^{n+1}), |x'|^2 - (1-x^{n+1})^2 \right)}{|x'|^2 + (1-x^{n+1})^2}.$$
This is where I get stuck. I don't see a way out of this to make it become a simple $(x^1, \ldots, x^{n+1})$, and that could be for two reasons: pure ignorance or I'm taking the wrong path.
Making use of the post, I also did the calculations required for the transition map $\tilde{\sigma} \circ \sigma^{-1}$, I would also appreciate someone performing a second check.
$$\tilde{\sigma} \circ \sigma^{-1}(x) = \tilde{\sigma} \left( \frac{(2x^1, \ldots, 2x^n, |x'|^2 -1)}{|x'|^2 +1} \right)$$
$$= \frac{\frac{(2x^1, \ldots, 2x^n)}{|x'|^2 +1}}{(|x'|^2 -1) +1}$$
$$= \frac{(2x^1, \ldots, 2x^n)}{|x'|^2(|x'|^2 +1)}.$$
Thanks for all your troubles reading this overly long post.
All help is appreciated!P.S.: If anybody knows a more effective way of aligning these questions here in MHB, please post (or PM me, in case moderators find it off-topic). In usual LaTeX I use \begin{flalign} but I haven't tried it here.