stereographic projectionStereographicProjection2005-05-24 11:34:372005-05-25 11:58:27Definitionnorth polesouth poleprojectionmap% this is the default PlanetMath preamble. as your knowledge
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The $n$-dimensional Euclidean \PMlinkid{unit sphere}{186} $S^n$ is
defined as a subset
of $\mbb{R}^{n+1}$:
\begin{equation*}
S^n=\biggl\{(x_1,\ldots,x_{n+1})\in\mbb{R}^{n+1}\mid\Sum_{k=1}^{n+1}x_k^2=1\biggr\}.
\end{equation*}
The \emph{stereographic projection} maps all points of $S^n$ to
the $n$-dimensional Euclidean space $\mbb{R}^n$ except one. Let
$N:=(0,\ldots,0,1)\in S^n$ be this point (it is usually called the
\emph{north pole}). Then the stereographic projection is defined by
\begin{equation*}
\sigma\colon S^n\setminus
N\to\mbb{R}^n,\quad(x_1,\ldots,x_{n+1})\mapsto\frac{c-1}{x_{n+1}-1}(x_1,\ldots,x_n).
\end{equation*}
Here, $c$ is an arbitrary real number. If $c=1$, the projection
degenerates; in all other cases, however, $\sigma$ is a smooth
bijective mapping.
The image $P'$ of a point $P$ under $\sigma$ can be geometrically
constructed as follows. Embed $\mbb{R}^n$ into
$\mbb{R}^{n+1}$ as a hyperplane at $x_{n+1}=c$. Unless $c=1$, the
straight line defined by $N$ and $P$ intersects with $\mbb{R}^n$ in
precisely one point, $P'$. The most common values for $c$ are $c=-1$
and $c=0$, see figures~\ref{fig1} and~\ref{fig2}.
\begin{figure}
\label{fig1}
\begin{center}
\includegraphics{StereographicProjection.1.eps}
\end{center}
\sf\caption{Stereographic projection of the one dimensional unit
sphere for $c=-1$}
\end{figure}
\begin{figure}
\label{fig2}
\begin{center}
\includegraphics{StereographicProjection.2.eps}
\end{center}
\sf\caption{Stereographic projection of the one dimensional unit
sphere for $c=0$}
\end{figure}
Let $-\id\colon\mbb{R}^{n+1}\to\mbb{R}^{n+1}$ be the map $x\mapsto
-x$, then $\tilde{\sigma}:=\sigma\circ(-\id)$ (a suitably restricted composition) maps all points of $S^n$
except the \emph{south pole} $S:=(0,\ldots,0,-1)$ smoothly and
bijectively to $\mbb{R}^n$. Together, $\sigma$ and $\tilde{\sigma}$
form an atlas of $S^n$, so $S^n$ is an $n$-dimensional smooth
manifold.