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The Hopf fibration $h:S^{3}\rightarrow S^{2}$ is given by $h(a,b,c,d)=(a^{2}+b^{2}-c^{2}-d^{2},2(ad+bc),2(bd-ac))$.

A stereographic projection is a map $s:S^{3}\backslash (1,0,0,0)\rightarrow \mathbb{R}^{3}$ given by $s(w,x,y,z)=(\frac{x}{1-w},\frac{y}{1-w},\frac{z}{1-w})$.

I am supposed to find out what is $s\circ h^{-1}((1,0,0))$ and $s\circ h^{-1}((-1,0,0))$. These are fine for me and I got $s\circ h^{-1}((1,0,0))$ a line (x-axis) and $s\circ h^{-1}((-1,0,0))$ is a circle on the yz-plane where $h^{-1}((-1,0,0))$ is $\{ke^{it}\}_{0\le t\le 2\pi}$ and $k$ here is the quaternion $0+0i+0j+1k$.

Here is the problem, I would like to show that in general for any $P=(p_{1},p_{2},p_{3})\in S^{2}$ and $P$ not equal to $(1,0,0)$ and $(-1,0,0)$, it is a circle that intersects the yz-plane in exactly two points.

The fiber over $P$ can be written as $\{r_{p}e^{it}\}_{0\le t\le 2\pi}$ where $r_{p}=\frac{1}{\surd(2(1+p_{1}))}((1+p_{1})i+p_{2}j+p_{3}k)$. I found $s\circ h^{-1}(P)$ to be $x=\frac{-(1+p_{1})sint}{\surd(2(1+p_{1}))}$, $y=\frac{(1+p_{1})cost}{\surd(2(1+p_{1}))}$, $z=\frac{p_{2}cost+p_{3}sint}{\surd(2(1+p_{1}))}$. To show that this is a circle, I try to prove that it is on a plane since the intersection of a plane and a unit sphere is a circle.

With some calculation, I manage to find this plane (tedious calculation) which is given by $(p^{2}_{2}+p^{2}_{3})x-p_{2}(1+p_{1})y-p_{3}(1+p_{1})z=0$

Is there an easier way to do this? that is to show that it is a circle. From what I know we don't have a formula for circle in $\mathbb{R}^{3}$ am I right?

Also, what are the strategies to show the following geometry problems? How should I approach these problems?

  1. Show that this circle is linked with the circle from $s\circ h^{-1}((-1,0,0))$

  2. Show that the $x$-axis passes through the interior of $s\circ h^{-1}((-1,0,0))$

  3. Show that the images under $s$ of any two fiber circles are linked

Thank you for reading my problem :)

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