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# 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.