16.4: Stereographic projection

Consider the unit sphere $\Sigma$ centered at the origin $(0,0,0)$. This sphere can be described by the equation $x^2+y^2+z^2=1$.

Suppose that $\Pi$ denotes the $xy$-plane; it is defined by the equation $z = 0$. Clearly, $\Pi$ runs thru the center of $\Sigma$.

Let $N = (0, 0, 1)$ and $S=(0, 0, -1)$ denote the "north" and "south" poles of $\Sigma$; these are the points on the sphere that have extremal distances to $\Pi$. Suppose that $\Omega$ denotes the “equator” of $\Sigma$; it is the intersection $\Sigma \cap \Pi$.

For any point $P\ne S$ on $\Sigma$, consider the line $(SP)$ in the space. This line intersects $\Pi$ in exactly one point, denoted by $P'$. Set $S'=\infty$.

The map $\xi_s\: P\mapsto P'$ is called the stereographic projection from $\Sigma$ to $\Pi$ with respect to the south pole. The inverse of this map $\xi^{-1}_s\: P' \mapsto P$ is called the stereographic projection from $\Pi$ to $\Sigma$ with respect to the south pole.

The same way, one can define the stereographic projections $\xi_n$ and $\xi^{-1}_n$ with respect to the north pole $N$.

Note that $P=P'$ if and only if $P\in\Omega$.

Note that if $\Sigma$ and $\Pi$ are as above, then the composition of the stereographic projections $\xi_s: \Sigma\to\Pi$ and $\xi^{-1}_s: \Pi\to\Sigma$ are the restrictions to $\Sigma$ and $\Pi$ respectively of the inversion in the sphere $\Upsilon$ with the center $S$ and radius $\sqrt{2}$.

From above and Theorem 16.3.1, it follows that the stereographic projection preserves the angles between arcs; more precisely the absolute value of the angle measure between arcs on the sphere.

This makes it particularly useful in cartography. A map of a big region of earth cannot be done on a constant scale, but using a stereographic projection, one can keep the angles between roads the same as on earth.

In the following exercises, we assume that $\Sigma$, $\Pi$, $\Upsilon$, $\Omega$, $O$, $S$, and $N$ are as above.

Exercise $\PageIndex{1}$

Show that $\xi_n \circ \xi^{-1}_s$, the composition of stereographic projections from $\Pi$ to $\Sigma$ from $S$, and from $\Sigma$ to $\Pi$ from $N$ is the inverse of the plane $\Pi$ in $\Omega$.

Hint

Note that points on $\Omega$ do not move. Moreover, the points inside $\Omega$ are mapped outside of $\Omega$ and the other way around.

Further, note that this map sends circles to circles; moreover, the perpendicular circles are mapped to perpendicular circles. In particular, the circles perpendicular to $\Omega$ are mapped to themselves.

Consider arbitrary point $P \not\in \Omega$. Suppose that $P'$ denotes the inverse of $P$ in $\Omega$. Choose two distinct circles that pass thru $P$ and $P'$. According to Corollary 10.5.2, $\Gamma_1 \perp \Omega$ and $\Gamma_2 \perp \Omega$.

Therefore, the inverse in $\Omega$ sends $\Gamma_1$ to itself and the same holds for $\Gamma_2$.

The image of $P$ has to lie on $\Gamma_1$ and $\Gamma_2$. Since the image of $P$ is distinct from $P$, we get that it has to be $P'$.

Exercise $\PageIndex{2}$

Show that a stereographic projection $\Sigma\to\Pi$ sends the great circles to plane circlines that intersect $\Omega$ at opposite points.

Hint

Apply Theorem 16.3.1(b).

The following exercise is analogous to Lemma 13.5.1.

Exercise $\PageIndex{3}$

Fix a point $P\in \Pi$ and let $Q$ be another point in $\Pi$. Let $P'$ and $Q'$ denote their stereographic projections to $\Sigma$. Set $x=PQ$ and $y=P'Q'_s$. Show that

$\lim_{x\to 0} \dfrac{y}{x}=\dfrac{2}{1+OP^2}.$

Hint

Set $z = P'Q'$. Note that $\dfrac{y}{x} \to 1$ as $x \to 0$.

It remains to shwo that

$\lim_{x \to 0} \dfrac{z}{x} = \dfrac{2}{OP^2}$

Recall that the stereographic projection is the inversion in the sphere $Upsilon$ with the center at the south pole $S$ restricted to the plane $\Pi$. Show that there is a plane $\Lambda$ passing thru $S, P, Q, P'$, and $Q'$. In the plane $\Lambda$, the map $Q \mapsto Q'$ is an inversion in the circle $\Upsilon \cap \Lambda$.

This reduces the problem to Euclidean plane geometry. The remaining calculations in $\Lambda$ are similar to those in the proof of Lemma 13.5.1.