Stereographic Projection of \( S^1 \) into \( \mathbb{R} \)

Let \( (x, y) \) denote coordinates in \( \mathbb{R}^2 \) and suppose that \( x^2 + y^2 = 1 \). We can parametrize the unit circle centered at the origin \( S^1 \) by the equations
\begin{equation}
x = \sin t,\quad y = \cos t.
\end{equation}
This parametrization goes around the circle clockwise starting at the north pole \( N = (0, 1) \), where \( t = 0 \). Geometrically, stereographic projection of \( S^1 \setminus\{N\} \) onto \( \mathbb{R} \) is the map defined by extending a line through the points \( N \) and \( (x, y) \) to the \( x \)-axis. The point \( (\xi, 0) \) where this line and the \( x \)-axis meet is the stereographic projection of \( (x, y) \) onto \( \mathbb{R} \) (after dropping the \( y \)-coordinate). This map is given by
\begin{equation}\label{eq:Projection}
\xi = \frac{x}{1 – y} = \frac{\sin t}{1 – \cos t}.
\end{equation}


Figure 1. Stereographic projection of the circle \( S^1 \) into the real number line \( \mathbb{R} \).

Stereographic projection of the circle \( S^1 \) onto the real number line \( \mathbb{R} \).

To obtain (\ref{eq:Projection}), we proceed as follows. Fix a value of \(t \in (0, 2\pi) \) and let \( \gamma: \mathbb{R} \rightarrow \mathbb{R}^2 \) be the parametrization of the line through \( N \) and the point \( (\sin t, \cos t) \) given by
\begin{equation}\label{eq:ParametricLine}
\gamma(u) = (1-u)\,(0,1) + u\,(\sin t, \cos t).
\end{equation}
If we write \( \gamma(u) = ( \xi, \eta ) \) for some value of \( u \), we see that (\ref{eq:ParametricLine}) is equivalent to the pair of equations
\begin{equation}\label{eq:XiEta}
\xi = u\,\sin t,\quad \eta = 1 – u + u\,\cos t.
\end{equation}
The point \( ( \xi, \eta ) \) lies on the \( x \)-axis if \( \eta = 0 \). Solving for \( u \), we obtain
\begin{equation}
u = \frac{1}{1 – \cos t}.
\end{equation}
Substituting this into the expression for \( \xi \) in (\ref{eq:XiEta}) gives (\ref{eq:Projection}).

We now turn this calculation around to obtain inverse stereographic projection, which provides us with a parametrization of \( S^1 \) by \( \xi \). To that end, we define
\begin{equation}\label{eq:INverseParametricLine}
\rho:\mathbb{R}\rightarrow S^1\setminus\{N\},\quad \rho(u) = (1-u)\,(0,1) + u\,(\xi, 0).
\end{equation}
Writing out the components (as before) we obtain
\begin{equation}\label{eq:XY}
x = u\,\xi, \quad y = 1 – u.
\end{equation}
In order for \( (x, y) \) to be a point on \( S^1 \), we must have
\begin{equation}
x^2 + y^2 = u^2\xi^2 + (1 – u)^2 = 1.
\end{equation}
This simplifies to
\begin{equation}\label{eq:U}
u\,[u\,(1 + \xi^2) – 2] = 0.
\end{equation}
It then follows that
\begin{equation}\label{eq:USolution}
u = 0\quad {\rm or}\quad u = \frac{2}{1 + \xi^2}.
\end{equation}
Now, a line from \( N \) to the point \( (\xi, 0) \) on the \( x \)-axis intersects \( S^1 \) in one other point, which is the point we seek. This point corresponds to the nonzero value of \( u \) in (\ref{eq:USolution}). Substituting this value for \( u \) into (\ref{eq:XY}), we find
\begin{equation}\label{eq:ISP}
x = u\,\xi = \frac{2\xi}{1 + \xi^2},\quad y = 1 – u = \frac{-1 + \xi^2}{1 + \xi^2}.
\end{equation}

Remark. The map defined by (\ref{eq:ISP}) is a rational map, that is, the coordinates \( x \) and \( y \) of a point on \( S^1 \) are rational functions of the coordinate \( \xi \) on \( \mathbb{R} \). Since the rational numbers \( \mathbb{Q} \) are dense in \( \mathbb{R} \), an immediate consequence of this is that the points in \( S^1 \) with rational coordinates are dense in \( S^1 \). This follows from the continuity of the map
\begin{equation}
\varphi : \mathbb{R} \rightarrow S^1\setminus\{N\} ,\quad \varphi(\xi) = \left(\frac{2\xi}{1 + \xi^2}, \frac{-1 + \xi^2}{1 + \xi^2} \right).
\end{equation}