Csaba Vincze (2013)

University of Debrecen

In what follows we reformulate Kirchberger's theorem for spherical separation by the help of the stereographic projection.

**Definition **The sets D and E in the coordinate space of dimension n is strictly separated by a sphere G if one of the following statements holds:

- all the points of A are inside and all thepoints of B are outside of G,

- all the points of B are inside and all the points of A are outside of G.

The method to transfer the separation by hyperplanes into a spherical separation is based on the stereographic projection. It is a widely used process in cartography as a way to make a flat map of the earth. Because the earth is spherical, any map must distort shapes or sizes to some degree. The rule for stereographic projection has a nice geometric description. Think of the earth as a sphere sitting on the plane of the paper. (The south pole touches the paper.) Now imagine a light bulb at the north pole p which shines through the sphere. Each point on the sphere has a "shadow" to determine itsown place on the map. The south pole works as the center point of the map. Latitudes appear as circles around the center (longitudes appear as lines passing through the center of the map). Objects near the south pole are not stretched very much but the equator is twice as big on the map as on the sphere. The north pole gets sent off to infinity. Because the sphere and the plane appearin many areas of mathematics and its applications, so does the stereographic projection. It plays an important role in diverse fields including complex analysis, cartography, geology, and photography. In some applications (see complex analysis) the image of the north pole is taken at the infinity. In terms of a pure topological language the sphere is homeomorphic to the one point compactification of the plane. The stereographic projection has the followingimportant properties:

(i) it is conformal (it preserves the angle at which curves cross each other)

(ii) it is a circle preserving map (lines in the plane are considered circles with infinite radius).

Circles on a sphere come from intersections with (hyper)planes. If the plane contains the pole p of the projection then the image of the circle will be a line which can be also considered as a circle with infinite radius. It is typical in the applications. Sometimes the projection is taken with respect tothe plane H passing through the equator of the sphere. Since the projections on different (but parallel) planes can be transfer into each other by a central similarity through the point p they share the properties (i) and (ii). In what follows we use the projection onto the plane of the equator together with an analytical description to enjoy all the advantages of a dimension-free approach.

**Analytic description**. Let S be the unit sphere centered at the origin in the coordinate space of dimension n+1. The points on S will be denoted by pairs of the form (v,s), where v is in the coordinate plane H spanned by the first n canonical basis vector and s is a real number between - 1 and 1. Let p=(** 0**,1) be the point of
the projection ("north pole"). The condition for the intersection of the line passing through (v,s) and p with H is the vanishing of the last coordinate of a point on the parametric line

$\left(\mathrm{0,1}\right)+t\left(\right(v,s)-(\mathrm{0,1}\left)\right)=(v\mathrm{\text{'}},0)$ |

as t runs through the real numbers. Therefore

$1+ts-t=0\Rightarrow t=\frac{1}{1-s}$ |

and, consequently,

$v\mathrm{\text{'}}=\frac{1}{1-s}v.$ |

Its inverse works formally as

$v\mathrm{\text{'}}\mapsto \left(\right(1-s)v\mathrm{\text{'}},s),$ |

where s can be expressed by taking the norm of the vectors v and v':

$\parallel v\mathrm{\text{'}}\parallel =\frac{1}{(1-s)}\parallel v\parallel =\frac{1}{(1-s)}\sqrt{1-{s}^{2}}$ |

because the element (v, s) has unit length. Therefore

$\parallel v\mathrm{\text{'}}{\parallel}^{2}=\frac{1+s}{1-s}\Rightarrow s=\frac{\parallel v\mathrm{\text{'}}{\parallel}^{2}-1}{\parallel v\mathrm{\text{'}}{\parallel}^{2}+1}$ |

and, consequently, the inverse transformation can be given as

$v\mathrm{\text{'}}\mapsto \frac{2}{1+\parallel v\mathrm{\text{'}}{\parallel}^{2}}(v\mathrm{\text{'}},\frac{\parallel v\mathrm{\text{'}}{\parallel}^{2}-1}{2}).$ |

Especially,

$(x,y)\mapsto \frac{2}{1+{x}^{2}+{y}^{2}}(x,y,\frac{{x}^{2}+{y}^{2}-1}{2}).$ |

**Excercise 8.2.1 **
*Prove that *

$d\left(\right(\mathrm{0,1}),(v,s\left)\right)\cdot d\left(\right(\mathrm{0,1}),(v\mathrm{\text{'}},0\left)\right)=2.$ |

Hint. Recall that

$\parallel v{\parallel}^{2}+{s}^{2}=1\text{}and\text{}v\mathrm{\text{'}}=\frac{1}{1-s}v.$ |

The result says that the stereographic projection is actually the restriction of an inversion with center p. Therefore the properties (i) and (ii) of the stereographic projection can be concluded from the corresponding properties of the inversion. The advantage of proving the properties for inversions is that we can use the standard calculus and linear algebra without adaptation to (hyper)surfaces.

**Excercise 8.2.2 **
*Prove that the Jacobian matrix of the inversion is proportional of the unit matrix at each point of the space except the center. *

**Remark **In the sense of the previous excercise the inversion is conformal because the angles are defined in terms of tangent objects. In order to see the conformal property of the stereographic projection in the classical case we can use the following more elementary argumentation: consider the tangent (hyper)plane H' to the sphere atp. H' is parallel to the (hyper)plane H of the equator and, consequently, for any line l in H we have a parallel line l' in H' by intersecting H' with the plane P spanned by l and the pole p of the projection. The common part of P and S is a circle C'. It is clear that l' is tangential to C' and, consequently, the angle between two (intersecting )lines in the
plane H is the same as the angle between their inverse circles (which is just the angle between their tangent lines passing through the common point p).

**Excercise 8.2.3 **
*Prove that the stereographic projection is a circle/sphere preserving map. *

Hint. Taking the difference of the canonical equations it can be easily seen that the intersection of two spheres is also a sphere although the dimension decreases in general. Therefore itis enough to prove that the inversions are sphere preserving maps.

**Theorem 8.2.4 **
*Let D and E be compact sets in the coordinate space of dimension n. They can be strictly separated by a sphere if and only if for each subset T of at most n+3 elements in D U E the sets *

$T\cap D\text{}and\text{}T\cap E$ |

can be strictly separated by a sphere.

**Proof **Consider the coordinate space of dimension n as a hyperplane

${E}^{n}\subset {E}^{n+1}$ |

in the space of dimension n+1 and let D' and E' be the inverse images of the given sets under the stereographic projection of the unit sphere in the embedding space. The original version of Kirchberger's theorem 8.1.2 says that D' and E' can be strictly separated by a hyperplane L (of dimension n) in the embedding space. Since it is a strict separation we can suppose, without loss of the generality, that the separating hyperplane does not contain the pole of the projection. Therefore the image of the intersection of L and S under the projection is a (non-degenerate) sphere G in the coordinate hyperplane of dimension n. G separates D and E as was to be proved. ▮

The figure shows that the number n+3 can not be reduced. Although we have five circles to separate the points belonging to different letters in all the quadruples

${A}_{1}=\{{p}_{1},{p}_{2},{p}_{3},{q}_{1}\},{A}_{2}=\{{p}_{1},{p}_{2},{p}_{3},{q}_{2}\},{A}_{3}=\{{p}_{1},{p}_{2},{q}_{1},{q}_{2}\},$ |

${A}_{4}=\{{p}_{1},{p}_{3},{q}_{1},{q}_{2}\},{A}_{5}=\{{p}_{2},{p}_{3},{q}_{1},{q}_{2}\},$ |

the sets D={p(1), p(2), p(3)} and E={q(1), q(2)} can not be separated by circles.