Formula for the intersection of a sphere with regards to stereographic projection

Question

So I have this question for an assignment and am just completely lost.

Let $S^n$ be the unit sphere with centre at $0$ in the space $R^{n+1}$. Let $N=(0,...,0,1)$ in such a space. Define the stereographic projection $p:S^n\setminus\{N\} \rightarrow R^n = R^n \times \{0\} \subset R^{n+1}$.

For each x in the sphere, the point $p(x)$ is the intersection of the line and the point $x,$ with the hyperplane $x_{n+1} =0.$

I need to find an explicit formula for such $p(x)$ and also its inverse, and then prove $p$ is a homeomorphism. I genuinely have no clue where to start.

From wikipedia and some videos, I have found some equations for $R^3$ but the $n+1$ has me completely lost. How do I derive such an equation, both in $R^3$ and particularly for any $R^{n+1}$?

Answer 1

The Power of a Point

The Power of the point $p$ with respect to the unit circle centered at $O$ is $$ (p-x)\cdot(p-N)=|p|^2-1\tag1 $$ The Pythagorean Theorem says $$ |p-N|^2=|p|^2+1\tag2 $$ Therefore, $$ \begin{align} (x-N)\cdot(p-N) &=\left((p-N)-(p-x)\right)\cdot(p-N)\\ &=|p-N|^2-(p-x)\cdot(p-N)\\ &=2\vphantom{N^2}\tag3 \end{align} $$ Since $x-N\parallel p-N$ we get $$ p-N=\frac2{|x-N|^2}(x-N)\tag4 $$ and $$ x-N=\frac2{|p-N|^2}(p-N)\tag5 $$ Equations $(4)$ and $(5)$ give formulas to compute $p$ given $x$ and vice-versa.

Answer 2

The line $\mathcal L_x$ through $N$ and $x \in S^n \setminus \{N\}$ can be parameterized by $$l_x(t) = N + t(x -N) .$$ The coordinate functions of $l_x$ are $$l^i_x(t) = \begin{cases} tx_i & i = 1,\ldots,n \\ 1 + t(x_{n+1} -1) & i = n+1 \end{cases}$$ $\mathcal L_x$ intersects $\mathbb R^n \times \{0\}$ in the point $p(x)$. Thus we have to determine $t_0$ such that $l^{n+1}_x(t_0) = 1 + t_0(x_{n+1}-1) = 0$. We get $$t_0 = \dfrac{1}{1-x_{n+1}}$$ and therefore $$p(x) = l_x(t_0) = \left(\dfrac{x_1}{1-x_{n+1}},\ldots,\dfrac{x_n}{1-x_{n+1}},0\right) .$$

The inverse of $p$ can be determined as follows. Given $u = (u_1,\ldots,u_n,0) \in \mathbb R^n \times \{0\}$, we want to find $x \in S^n$ such that $p(x) = u$, i.e. $$(\dfrac{x_1}{1-x_{n+1}},\ldots,\dfrac{x_n}{1-x_{n+1}}) = (u_1,\ldots,u_n).$$ This implies $$\lVert u \rVert^2 = \sum_{i=1}^n u_i^2 = \dfrac{1}{(1-x_{n+1})^2}\sum_{i=1}^n x_i^2 .$$ We require $x \in S^n$, i.e. $\sum_{i=1}^{n+1} x_i^2 = 1$. Therefore $$\lVert u \rVert^2 = \dfrac{1}{(1-x_{n+1})^2}(1-x_{n+1}^2) = \dfrac{1+x_{n+1}}{1-x_{n+1}}$$ which gives $$x_{n+1} = \dfrac{\lVert u \rVert^2 -1}{\lVert u \rVert^2 +1} $$ and $$x_i = \dfrac{2u_i}{\lVert u \rVert^2 +1}, i =1, \ldots, n. $$ Thus $$p^{-1}(u) = \left(\dfrac{2u_1}{\lVert u \rVert^2 +1}, \ldots, \dfrac{2u_n}{\lVert u \rVert^2 +1},\dfrac{\lVert u \rVert^2 -1}{\lVert u \rVert^2 +1}\right) .$$ This approach was purely formal. Alternatively we can do it geometrically. The line $\mathcal L_x$ through $N$ and $x$ is also a line through $N$ and $u = p(x)$ which can be parameterized by $$g_x(t) = N + t(u -N) .$$ The coordinate functions of $g_x$ are $$g^i_x(t) = \begin{cases} tu_i & i = 1,\ldots,n \\ 1 + t(u_{n+1} -1) = 1 -t & i = n+1 \end{cases}$$ It is clear that $\mathcal L_x$ intersects $S^n$ in the two points $N$ and $x$. Thus we have to determine $t$ such that $$1 = \lVert g_x(t) \rVert^2 = \sum_{i=1}^n t^2u_i^2 + (1-t)^2 =t^2\lVert u \rVert^2 + (1-t)^2. $$ This can be simplified to $$0 = ((\lVert u \rVert^2 +1)t -2 )t .$$ Solutions are $t_0=0$ (which gives $g_x(0) = N$) and $$t_1 = \dfrac{2}{\lVert u \rVert^2 + 1} .$$ We obtain $$x = g_x(t_1) = \left(\dfrac{2u_1}{\lVert u \rVert^2 +1}, \ldots, \dfrac{2u_n}{\lVert u \rVert^2 +1},\dfrac{\lVert u \rVert^2 -1}{\lVert u \rVert^2 +1}\right) .$$