The inversion in a sphere is defined the same way as we define the inversion in a circle.
Formally, let Σ be the sphere with the center O and radius r. The inversionin Σof a pointP is the pointP′ ∈[OP)such that
OP·OP′ =r2.
In this case, the sphereΣ will be called the sphere of inversion and its center is called thecenter of inversion.
We also add∞ to the space and assume that the center of inversion is mapped to∞and the other way around. The spaceR3with the point
∞will be calledinversive space.
The inversion of the space has many properties of the inversion of the plane. Most important for us are the analogs of theorems 10.6, 10.7, 10.25 which can be summarized as follows:
16.3. Theorem. The inversion in the sphere has the following proper- ties:
(a) Inversion maps a sphere or a plane into a sphere or a plane. (b) Inversion maps a circle or a line into a circle or a line.
(c) Inversion preserves the cross-ratio; that is, if A′, B′, C′, and D′ are the inverses of the pointsA,B,C andD correspondingly, then
AB·CD BC·DA =
A′B′·C′D′
B′C′·D′A′. (d) Inversion maps arcs into arcs.
(e) Inversion preserves the absolute value of the angle measure between tangent half-lines to the arcs.
We do not present the proofs here, but they nearly repeat the corre- sponding proofs in plane geometry. To prove(a), you will need in addition the following lemma; its proof is left to the reader.
16.4. Lemma. Let Σ be a subset of the Euclidean space that contains at least two points. Fix a pointO in the space.
ThenΣis a sphere if and only if for any plane Πpassing thruO, the intersectionΠ∩Σis either empty set, one point set or a circle.
The following observation helps to reduce part(b) to part(a). 16.5. Observation. Any circle in the space is an intersection of two spheres.
Let us define acircular cone as a set formed by line segments from a fixed point, called thetip of the cone, to all the points on a fixed circle, called thebaseof the cone; we always assume that the base does not lie in the same plane as the tip. We say that the cone isright if the center of the base circle is the foot point of the tip on the base plane; otherwise we call itoblique.
16.6. Exercise. Let K be an oblique circular cone. Show that there is a plane Π that is not parallel to the base plane of K such that the intersectionΠ∩K is a circle.
Stereographic projection
N O P P′ S Σ ΥThe plane thruP, O, and S. Consider the unit sphereΣcentered at
the origin (0,0,0). This sphere can be described by the equationx2+y2+z2=
= 1.
LetΠdenotes thexy-plane; it is de- fined by the equationz= 0. Clearly,Π runs thru the center ofΣ.
LetN = (0,0,1)andS = (0,0,−1) denote the “north” and “south” poles of Σ; these are the points on the sphere that have extremal distances toΠ. Let Ω denotes the “equator” of Σ; it is the intersectionΣ∩Π.
For any pointP 6=S onΣ, consider the line(SP)in the space. This line intersectsΠin exactly one point, denoted byP′. SetS′=∞.
The mapξs:P 7→P′ is called thestereographic projection fromΣ to Π with respect to the south pole. The inverse of this mapξ−1
called thestereographic projection fromΠ toΣ with respect to the south pole.
The same way, one can define thestereographic projectionsξnandξn−1 with respect to the north pole N.
Note thatP =P′ if and only ifP ∈Ω.
Note that ifΣandΠare as above, then the composition of the stere- ographic projectionsξs: Σ→Π andξ−s1: Π→Σare the restrictions of the inversion in the sphereΥwith the centerS and radius √2 to Σand Πcorrespondingly.
From above and Theorem 16.3, it follows that the stereographic pro- jection preserves the angles between arcs; more preciselythe absolute value of the angle measurebetween arcs on the sphere.
This makes it particularly useful in cartography. A map of a big region of earth cannot be done in 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 Σ, Π, Υ, Ω, O, S, andN
are as above.
16.7. Exercise. Show that ξn ◦ξ−s1, the composition of stereographic projections from ΠtoΣ fromS, and from ΣtoΠ fromN is the inverse of the planeΠin Ω.
16.8. Exercise. Show that a stereographic projectionΣ→Πsends the great circles to plane circlines that intersectΩat opposite points.
The following exercise is analogous to Lemma 13.10.
16.9. Exercise. Fix a pointP ∈Πand letQbe another point inΠ. Let
P′ and Q′ denote their stereographic projections to Σ. Set x=P Qand
y=P′Q′ s. Show that lim x→0 y x= 2 1 +OP2.
Central projection
The central projection is analogous to the projective model of hyperbolic plane which is discussed in Chapter 17.
LetΣbe the unit sphere centered at the origin which will be denoted byO. LetΠ+denotes the plane defined by the equationz= 1. This plane
is parallel to thexy-plane and it passes thru the north poleN = (0,0,1) ofΣ.
N O P P′ Σ+ Π+
Recall that the northern hemisphere ofΣ, is the subset of points(x, y, z)∈Σsuch that
z >0. The northern hemisphere will be de- noted byΣ+.
Given a pointP ∈Σ+, consider the half-
line [OP). Let P′ denotes the intersection of [OP) andΠ+. Note that if P = (x, y, z),
then P′ = (x z,
y
z,1). It follows that P ↔ P′ is a bijection between Σ
+
andΠ+.
The described bijection Σ+ ↔Π+ is called the central projection of
the hemisphereΣ+.
Note that the central projection sends the intersections of the great circles with Σ+ to the lines in Π+. The latter follows since the great
circles are intersections of Σ with planes passing thru the origin as well as the lines inΠ+ are the intersection ofΠ+ with these planes.
The following exercise is analogous to Exercise 17.5 in hyperbolic ge- ometry.
16.10. Exercise. Let △sABC be a nondegenerate spherical triangle. Assume that the plane Π+ is parallel to the plane passing thru A, B,
andC. LetA′,B′, and C′ denote the central projections ofA,B andC. (a) Show that the midpoints of [A′B′], [B′C′], and [C′A′] are central projections of the midpoints of[AB]s,[BC]s, and[CA]scorrespond- ingly.
(b) Use part (a) to show that the medians of a spherical triangle inter- sect at one point.