# Platonic Hopf. Given the vertices of a tetrahedron circumscribed by unit sphere, find the stereographic projection of inverse Hopf fibers

I am trying to find the equations in $\mathbb{R}^3$ for the fibers of the four vertices of the tetrahedron circumscribed by the unit sphere.

I want to find $s\circ h^{-1}$, where $s$ is the stereographic projection $\mathbb{S}^3\setminus(1,0,0,0)\to\mathbb{R}^3$ and $h$ is the Hopf fibration $\mathbb{S}^3\to\mathbb{S}^2$.

I have mostly been referencing David Lyons "An Elementary Inroduction to the Hopf Fibration" http://mas.lvc.edu/~lyons/pubs/hopf_paper_preprint.pdf

I have also done a lot of online searching, but I can't find a resource that clearly explains going from points in $\mathbb{S}^2$ to circles in $\mathbb{R}^3$.

My points on $\mathbb{S}^2$ are $(0,0,1)$, $(\frac{\sqrt{8}}{3},0,-\frac{1}{3})$, $(-\frac{\sqrt{2}}{3},\sqrt{\frac{2}{3}},-\frac{1}{3})$, and $(-\frac{\sqrt{2}}{3},-\sqrt{\frac{2}{3}},-\frac{1}{3})$.

On p. 12 of Lyons article, Investigation (L), he states that "$s\circ h^{-1}((1,0,0))$ is the $x$-axis, $s\circ h^{-1}((-1,0,0))$ is the unit circle in the $yz$-plane, and for any other point...$s\circ h^{-1}(P)$ is a circle in $\mathbb{R}^3$ that intersects the $yz$-plane in exactly two points." Maybe if I understood how he gets the second point in this statement, I could figure out how to find the fibers for my points.

I'm not quite sure where to go from here. Do I need to find the equations of the fibers in $\mathbb{S}^3$ first, and then use stereographic projection to find the circles in $\mathbb{R}^3$, or is there a way to find the circles in $\mathbb{R}^3$ directly?

• for each vertex $p$ in the 2-sphere, the sets $h^{-1}p$ are fibers on $S^3$ which can be projected to $\Bbb R^3$, right? – janmarqz Jan 2 '16 at 15:51
• Yes, each fiber is a circle in $\mathbb{S}^3$. – paygem Jan 2 '16 at 17:49