Geometry 1 - TU Berlin · 1 Projective Geometry 1.1 Introduction to projective spaces Letusstartwithageneraldeﬁnitionforanarbitraryprojectivespace. Yetinthislecture

Geometry 1

LECTURE NOTES

Thilo Rörig

April 10, 2014

Contents

0 Introduction 10.1 Historical Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Course Content . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1 Projective Geometry 51.1 Introduction to projective spaces . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Homogeneous coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 The theorems of Pappus and Desargues . . . . . . . . . . . . . . . . . . . . 91.4 Projective transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.5 Cross-ratio (Doppelverhältnis) . . . . . . . . . . . . . . . . . . . . . . . . . 161.6 Complete quadrilateral and quadrangle . . . . . . . . . . . . . . . . . . . . . 191.7 Moebius pairs of tetrahedra . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.8 The Fundamental Theorem of Real Projective Geometry . . . . . . . . . . . 251.9 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.10 Conic sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.11 Polarity: Pole - Polar Relationship . . . . . . . . . . . . . . . . . . . . . . . 391.12 Quadrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481.13 Orthogonal Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2 Hyperbolic Geometry 502.1 Lorentz Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502.2 Hyperbolic Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.3 Projective model of Hn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 552.4 Lines in the hyperbolic plane . . . . . . . . . . . . . . . . . . . . . . . . . . 582.5 Distances in the hyperbolic plane . . . . . . . . . . . . . . . . . . . . . . . . 612.6 Intersections of H2 with planes . . . . . . . . . . . . . . . . . . . . . . . . . 632.7 Stereographic projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632.8 Sphere inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 642.9 Poincaré disc/ball model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 682.10 Relation between Kleins and Poincaré model . . . . . . . . . . . . . . . . . 712.11 The Poincaré half-space(/-plane) model . . . . . . . . . . . . . . . . . . . . 732.12 Hyperbolic triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 772.13 Hyperbolic motions in H2

+ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

3 Klein’s Erlangen Program 83

4 Möbius Geometry 834.1 Two-dimensional Möbius Geometry . . . . . . . . . . . . . . . . . . . . . . . 844.2 Relation between projective and Möbius transformations . . . . . . . . . . . 854.3 Spherical model of Möbius geometry . . . . . . . . . . . . . . . . . . . . . . 85

4.4 Projective model of Möbius geometry . . . . . . . . . . . . . . . . . . . . . . 85

Index 87

0 Introduction

0.1 Historical Remarks

The word geometry is of Greek origin (γεωμετρία) and means “measurement of the earth”.But the first accounts were found in Egypt and Mesopotamia (today Iraq) around 2000BC. It was known how to calculate simple areas and volumes and people of that timealready had an approximation of π. But no general statements/theorems and proofs werefound. Around 600 BC in Greece, Thales was the first to deduce theorems from a setof axioms. At about 300 BC, Euclid wrote the book Elements in which he states thefive axioms of what is today called Euclidean geometry. His fifth axioms is known as theparallel postulate and can be phrased as follows:

“Given a line and a point not on this line, there is a unique line through thepoint parallel to the line.”

In the following centuries people tried to prove this postulate from the other four butwere not successful. As we know today, there exist other geometries that violate this pos-tulate. These geometries are called Non-Euclidean Geometries: e.g. spherical geometry,hyperbolic geometry (developed independently by Lobachevsky and Bolyai; 19th century),projective geometry (Poncelet; 19th century). Another important development was theintroduction of coordinates in the 17th century by Descartes and Fermat. The geometrystarting from axioms is called synthetic geometry and the geometry based of coordinatesanalytic geometry.

0.2 Course Content

In this course we will consider projective, hyperbolic and Moebius geometry. And if timepermits we might also have a look at Lie, Laguerre, or Pluecker geometry. The aim is toprovide you with lots of geometric images and their analytic counterparts.The continuation of this course is a speciality of the TU Berlin and is called Geometry II:Discrete Differential Geomtry. It is concerned with intelligent discretizations of smooththeories, e.g. from differential geometry and dynamical systems. The recently establishedSFB/TR Discretization in Geometry and Dynamics will enlarge the opportunities to dofurther research in this direction.Example 0.1 (Perspective example). We start with an example painters have to deal withwhen capturing a scene in a painting, cf. “Albrecht Dürer: Mann beim Zeichnen einerLaute, 1525”.

Albrecht Dürer: Mann beim Zeichnen einer Laute, 1525

1

http://www.discretization.de

0 Introduction

We will consider the projection of a spatial scene onto a plane but the central projectionof one plane onto another. So consider two planes E,E′ ⊂ R3 that are not parallel and apoint P ∈ R3 that does not lie on the planes as shown in the following picture.

Fig. 1: Central projection of the plane E to the plane E’ through the point P.The projection π : E → E′ does the following: let A be a point on the plane E. Then

A′ = π(A) = line(P,A) ∩ E′

Problem: We would like to have a bijective projection map, but the points on the line Èare not projected to any point on E′, since the plane spanned by È and P is parallel toE′. Further the points on È′ lack a preimage, since the plane spanned by È′ and P isparallel to E. The lines È resp. È′ are the vanishing lines of E resp. E′.Idea: Introduce two additional lines `∞ and `′∞ to E and E′, respectively. The points onÈ will be mapped to `′∞ and the points on `∞ are mapped to the points on È′ . The lines`∞ and `∞′ are called the lines at infinity.Done in the right way, this yields a bijective projection π : E ∪ `∞ → E′ ∪ `∞′ . Parallellines in E are mapped to concurrent lines in E′. The intersection point Q′ of the images ofparallel lines lies on the vanishing line È′ . The point Q′ is the image of the “intersectionpoint at infinity” Q ∈ `∞ of the original parallel lines.Example. We are now able to construct a perspective drawing of a chessboard given theimage of the first square. The chessboard contains two families of obvious parallel lines,namely the horizontal and vertical directions. But the diagonals of the chessboard are alsoparallel. Hence we obtain four points on the vanishing line on E′, one for each parallelfamily. These allow us to construct the perspective projection.

Fig. 2: Construction of a perspective image of a chessboard.

2

0.2 Course Content

Question. Can you construct the perspective image of the first square given the positionof the chessboard, your position (the “eye” point), and the position of the plane projectedonto?Now let us turn to the analytic treatment of the above example. Let the two planes andthe point be the following:

E =

x1x2x3

∈ R3 |x3 = 0

, E′ =

x1x2x3

∈ R3 |x1 = 0

, P =

−101

.As the planes are two dimensional subspaces we assign two coordinates to each of thepoints on either of the planes, i.e. A =

(a1a2

)and A′ =

(a′1a′

2

). We calculate the coordinates

of A′ from A by considering the line through A and P and its intersection point with E′.Thus we need to solve: a1

a20

+ λ

−1

01

−a1a20

=

0a′1a′2

⇒ a1 + λ(−1− a1) = 0 ⇒ λ = a1

1 + a1.

Hence we obtain

a′1 = a2 + a11 + a1

(−a2) = a21 + a1

a′2 = 0 + a11 + a1

(1− 0) = a11 + a1

and thus

A′ = π(A) = 11 + a1

(a2a1

).

(0.1)

Whenever there is a fractional linear transformation it is worth to introduce homogeneouscoordinates to linearize the equations, i.e. introduce

(b1b2b3

)with b3 6= 0 for ( a1

a2 ) (unique

up to non-zero scaling) with a1 = b1b3

and a2 = b2b3. Similarly,

(b′

1b′

2b′

3

)with b′3 6= 0 for

(a′

1a′

2

)with a′1 = b′

1b′

3and a′2 = b′

2b′

3. If we then look back at Equation 0.1 we obtain the following

relation between the homogeneous coordinates:b′1b′3

= a′1 = a21 + a1

= b2/b31 + b1/b3

= b2b1 + b3

and

b′2b′3

= a′2 = a11 + a1

= b1/b31 + b1/b3

= b1b1 + b3

⇒ b′1 = b2, b′2 = b1, b

′3 = b1 + b3 .

Hence in homogeneous coordinates the projection has become a linear (!) map:b1b2b3

7→b′1b′2b′3

=

0 1 01 0 01 0 1

b1b2b3

.

But what happens to the vanishing line È? The points on the vanishing line have coordi-nates

(−1a2

). Introducing homogeneous coordinates this amounts to

(b1b2b3

)with b3 6= 0 and

3

0 Introduction

b1b3

= −1. Thus b1 + b3 = 0 and in particular b1 6= 0. For the image of such a point underthe projection these conditions imply:0 1 0

1 0 01 0 1

b1b2b3

=

b2b1

b1 + b3

=

b2b10

.

The image point does not correspond to a point on the plane, because the last coordinateis zero contradicting the assumption b′3 6= 0 for the homogeneous coordinates of points onE′. But since b1 6= 0 these coordinates may be interpreted as homogeneous coordinates ofa line, namely, the line at infinity `′∞ added to the plane E′.

Question. What are the images of the line `∞ if we assign homogeneous coordinates(b1b20

)with b1 6= 0 to the points on the line?

4

1 Projective Geometry

1.1 Introduction to projective spaces

Let us start with a general definition for an arbitrary projective space. Yet in this lecturewe will almost entirely deal with real projective spaces.

Definition 1.1.1. Let V be a vector space over an arbitrary field F . The projective spaceP (V ) is the set of 1-dimensional vector subspaces of V . If dim(V ) = n + 1, then thedimension of the projective space is n.

(a) A 1-dimensional projective space is a projective line.

(b) A 2-dimensional projective space is a projective plane.

Canonical analytic description:Two vectors v and v span the same 1-dimensional subspace if there exist λ 6= 0 such thatv = λv. This yields an equivalence relation on V \ {0}:

v ∼ v :⇔ ∃λ 6= 0 : v = λv.

The points of the projective space P (V ) are equivalence classes of ∼:

P (V ) = (V \ {0})/∼.

We write [v] ∈ P (V ) and call v a representative vector of the line {λv |λ ∈ F}.Let us have a look at different models for real projective space RP2 = P (R3).Subspace modelConsider all 1-dimensional subspaces ` = {λv | v ∈ R3, v 6= 0, λ ∈ R} of R3.

Fig. 3: Subspace model

Sphere modelWe may normalize non-zero vectors to become unit vectors. Then there are only twovectors representing the same line. If we now introduce an equivalence relation

v ∼ v :⇔ v = ±v for v, v ∈ Sn

we obtain RPn = Sn/ ∼.

5


Fig. 4: Sphere model

Hemisphere modelConsider the upper hemisphere H2

+ ={( x1

x2x3

)∈ R3 | ‖x‖ = 1, x3 ≥ 0

}. Then all non-

horizontal lines have a unique representative vector in H2+ and we only need to identify

points on the equator H2+ ∩ {x |x3 = 0}.

Fig. 5: Hemisphere model

Examples of lower dimension:

• RP0 = P (R1) = S0/∼ = {∗} is just a single point.

• RP1 = P (R2) = S1/∼ = S1. This is most obvious using the hemisphere model, sincethe hemisphere is just a line and the two points on the equator are identified (bottomof the picture below). But we can also twist an S1 in a way such that opposite pointscome to lie on top of each other (top of picture below).

6

1.2 Homogeneous coordinates

Fig. 6: Real projective line

1.2 Homogeneous coordinates

Consider a basis {b1, . . . , bn+1} of the vector space V and a vector v ∈ V , v 6= 0 withcoordinates (x1, . . . , xn+1), i.e. v = ∑n+1

i=0 xibi. Then (x1, . . . , xn+1) are homogeneouscoordinates of [v] ∈ P (V ). These coordinates are not unique. They depend on the choiceof basis and are then only unique up to a scalar factor:

[v] = [x1, . . . , xn+1] = [λx1, . . . , λxn+1], for λ 6= 0.

Homogeneous coordinates of RPn. Let [v] ∈ RPn with coordinates [v] = [x1, . . . , xn+1].Since the homogeneous coordinates are only unique up to a scalar factor we do the follow-ing:

• if xn+1 6= 0 then

[v] = [x1, . . . , xn+1] =[x1xn+1

, . . . ,xnxn+1

, 1]

= [y1, . . . , yn+1] ' Rn.

The coordinates (y1, . . . , yn+1) are affine coordinates for [v] with xn+1 6= 0.

• if xn+1 = 0 then[v] = [x1, . . . , xn, 0] ' RPn−1.

So we obtain the following decomposition RPn = Rn ∪ RPn−1. This can be iterated andRPn−1 may again be decomposed into Rn−1 and RPn−2, and so on. The part isomorphicto Rn (e.g. xn+1 6= 0) is called the affine part of RPn and the other part isomorphic toRPn−1 (e.g. xn+1 = 0) is called the part at infinity. So in case of RP1 we have a point atinfinity, in case of RP2 there is an entire projective line at infinity. The following pictureshows the decomposition of RP2 into parts R2 (the red x3 = 1 plane) and the line atinfinity RP1 corresponding to the lines in the green x3 = 0 plane. Of course, the x3 = 0plane may itself be decomposed into an affine part, the dark blue line x2 = 1 (and x3 = 0)and the point at infinity represented by the light blue line x2 = 0 (and x3 = 0).

7


Fig. 7: Affine and infinite part of the real projective plane

Remark 1.2.1.

• To introduce an affine coordinate on a projective space we may choose an arbitrarylinear form `(x) = ∑n+1

i=1 aixi on V and decompose the points of the projective spacedepending on whether `(x) = 0 or `(x) 6= 0. In the decomposition above the linearform is `(x) = xn+1.

• A projective space should be thought of as a “homogeneous object” (to be madeprecise). So there are no preferred directions (points of projective space) as thehemisphere model or the introduction of an affine coordinate might suggest.

• The same definitions are used in the complex case and yield complex projectivespaces: CP1 = C ∪ {∞} = C is the Riemann sphere or complex projective line.

Definition 1.2.2. A projective subspace of the projective space P (V ) is a projective spaceP (U) where U is a vector subspace of V . If the dimension of P (U) is k (the correspondingvector subspace has dimension k + 1), the we call P (U) a k-plane. A 0-plane is a point,a 1-plane a line, a 2-plane a plane, and if P (V ) has dimension n a (n − 1)-plane is ahyperplane.

Proposition 1.2.3. Through any two distinct points in a projective space there passes aunique projective line.

Proposition 1.2.4. In a projective plane two distinct projective lines intersect in a uniquepoint.

These two propositions can be proved using linear algebra. Further properties of vectorsubspaces may be transferred to properties of projective subspaces:

• The intersection of two projective subspaces P (U1) and P (U2) is P (U1) ∩ P (U2) =P (U1 ∩ U2).

• The projective span or join of two projective subspaces P (U1) and P (U2) is P (U1 +U2).

• From the dimension formula of linear algebra we obtain a dimension formula forprojective subspaces:

dim(U1 + U2) = dimU1 + dimU2 − dim(U1 ∩ U2) ,dimP (U1 + U2) = dimP (U1) + dimP (U2)− dimP (U1) ∩ P (U2) .

8

1.3 The theorems of Pappus and Desargues


Definition 1.3.1. Let P (V ) be a projective space of dimension n. Then n+ 2 points inP (V ) are said to be in general position if no n + 1 of them are contained in a (n − 1)-dimensional projective subspace. In terms of linear algebra this implies that no n + 1representative vectors are linearly dependend, i.e. every n+ 1 are linearly independent.

Example 1.3.2. If n = 1 we have to consider n+ 2 = 3 points on a projective line. Thesethree points are in general position as long as they are disjoint.

Fig. 8: Points in general position on a projective line

If n = 2 we have to consider n + 2 = 4 points in a projective plane. The four points arein general position if no three are collinear.

Fig. 9: Points in general position in a projective plane

Question. The above pictures only show an affine part of the respective projective spaces.What happens if some of the points lie at infinity?For points in general position there exists a canonical choice for the representative vectorsdescribed in the following lemma.

Lemma 1.3.3. Let A1, . . . , An+2 be n+ 2 points in general position in an n-dimensionalprojective space P (V ). Then there exist representative vectors vi ∈ V with Ai = [vi] fori = 1, . . . , n+ 2 such that:

n+2∑i=1

vi = 0.

Moreover, this choice is unique up to a common scalar multiplicative factor, i.e., if vi =µivi with µi 6= 0 for i = 1, . . . , n+ 2 such that

∑n+2i=1 vi = 0 then µ1 = µ2 = . . . = µn+2.

Proof. Existence: Let wi with Ai = [wi] be representative vectors of the points Ai fori = 1, . . . , n + 2. Since dimV = n + 1 the n + 2 vectors are linearly dependent. So thereexist λi for i = 1, . . . , n+ 2 not all equal to zero such that

n+2∑i=1

λiwi = 0.

9


Since the points Ai are in general position all λi are non-zero. So setting vi = λiwi weobtain Ai = [vi] with

n+2∑i=1

vi = 0.

Uniqueness: Let vi = µivi with µi 6= 0 be a different set of representative vectors with

n+2∑i=1

vi = 0.

From this equation we obtain the following system of n + 1 homogeneous equations inn+ 2 variables:

n+2∑i=1

µivi = 0.

Since the points Ai are in general position the above system has rank n + 1 and hence a1-dimensional set of solutions. But (1, . . . , 1) is a solution of the system and hence thereexists µ 6= 0 with

(µ1, . . . , µn+2) = µ(1, . . . , 1).

So µ1 = . . . = µn+2 = µ.

Theorem 1.3.4 (Pappus ∼290-350 AC). Let A, B, C and A′, B′, C ′ be two collineartriples of distinct points in the real projective plane. Then the points

A′′ = (BC ′) ∩ (B′C) B′′ = (AC ′) ∩ (A′C) C ′′ = (A′B) ∩ (AB′)

are collinear.

Fig. 10

Proof. Without loss of generality let A, B, B′ and C ′ be four points in general position.(Question: What happens if the points are not in general position?). Choose representativevectors [a] = A, [b] = B, [b′] = B′, and [c′] = C ′ such that a+ b+ b′ + c′ = 0. Since a, b,c′ are linearly independent we obtain homogeneous coordinates:

A =

100

, B =

010

, C ′ =

001

, B′ =

−1−1−1

.

10


Then

C =

1y0

with y 6= 0 since C 6= A, and A′ =

11z

with z 6= 1 since A′ 6= B′.

The lines spanned by the six points are the following:

AB′ =

x1x2x3

∣∣∣∣∣∣∣ x2 = x3

, A′B =

x1x2x3

∣∣∣∣∣∣∣ s1

1z

+ t

010

=

ss+ tsz

,

AC ′ =

x1x2x3

∣∣∣∣∣∣∣ x2 = 0

, A′C =

x1x2x3

∣∣∣∣∣∣∣ s1

1z

+ t

1y0

=

s+ ts+ tysz

,

BC ′ =

x1x2x3

∣∣∣∣∣∣∣ x1 = 0

, B′C =

x1x2x3

∣∣∣∣∣∣∣ s−1−1−1

+ t

1y0

=

−s+ t−s+ ty−s

.

For the points of intersection A′′, B′′, and C ′′ this implies:

C ′′ = AB′ ∩A′B =

1zz

, B′′ = AC ′ ∩A′C =

1− y0−yz

, A′′ = BC ′ ∩B′C =

0y − 1−1

.Now we construct a linear dependence to show that the three points are collinear:

z

0y − 1−1

−1− y

0−yz

+ (1− y)

1zz

=

y − 1 + 1− yz(y − 1) + (1− y)z−z + yz + (1− y)z

= 0 .

So A′′, B′′, and C ′′ are collinear.

Remark 1.3.5.

• The Pappus configuration is very symmetric in the following perspective: It consistsof nine lines and nine points and each line contains three points and each point lieson three lines.

• Hilbert showed in 1899 that Pappus Theorem “corresponds” to the commutativityof multiplication. That is, a synthetic geometry that satisfies Pappus Theoremcorresponds to the geometry of a projective space that is the projectivation of avectorspace over a field.

Definition 1.3.6. Two triangles 41 = 4(A1, B1, C1) and 42 = 4(A2, B2, C2) are inperspective w.r.t. a point S if

S = (A1A2) ∩ (B1B2) ∩ (C1C2)

The triangles are in perspective w.r.t. a line ` if

C ′ = (A1B1) ∩ (A2B2) B′ = (A1C1) ∩ (A2C2) A′ = (B1C1) ∩ (B2C2)

lie on `.

11


Fig. 11: Perspective triangles

Theorem 1.3.7 (Desargues). Let 41 = 4(A1, B1, C1) and 42 = 4(A2, B2, C2) be twotriangles. Then 41 and 42 are in perspective w.r.t. a point if and only if they are inperspective w.r.t. a line.

Fig. 12: Desargues configuration

Proof. ⇒: We may assume that A1, A2, S and B1, B2, S and C1, C2, S are in general po-sition on their respective lines. By a Lemma from Lecture 3 we may choose representativevectors

A1 =[a1], A2 =

[a2], S =

[s]

such that a1 + a2 + s = 0

B1 =[b1], B2 =

[b2]

such that b1 + b2 + s = 0

C1 =[c1], C2 =

[c2]

such that c1 + c2 + s = 0

We have: a1 + a2 = b1 + b2 = c1 + c2 = −s. The points of intersection of correspondingsides are:

(A1C1) ∩ (A2C2) =[a1 − c1

]=[c2 − a2

](A1B1) ∩ (A2B2) =

[b1 − a1

]=[a2 − b2

](B1C1) ∩ (B2C2) =

[c1 − b1

]=[b2 − c2

]These three points lie on one line, because

(a1 − c1) + (b1 − a1) + (c1 − b1) = 0

12


Fig. 13: Desargues configuration

⇐: Starting with triangles 41 = 4(A1, B1, C1) and 42 = 4(A2, B2, C2) in perspectivew.r.t. the line ` = (A3B3) = (A3C3) = (B3C3) we observe that the triangles4(A1, A2, B3)and 4(B1, B2, A3) are in perspective w.r.t. C3. Then the forward direction implies that

C1 = (A1B3) ∩ (B1A3) C2 = (B2A3) ∩ (A2B3) S = (A1A2) ∩ (B1B2)

are collinear. Hence the triangles 41 and 42 are in perspective w.r.t. S.

3d picture: If we consider two triangles in RP3 in perspective w.r.t. a point P , theneach triangle spans a plane and these planes intersect in a line. The intersection points ofcorresponding sides of the triangles have to lie on the intersection line of the planes. Andhence the triangles are in perspective w.r.t. this line.

Fig. 14: Desargues configuration in 3D

Remark 1.3.8 (Symmetry of Desargues’ configuration). Consider 5 points A1, A2, . . . , A5 inRP3 in general position. Further let ìj = (AiAj) be the ten lines spanned by these pointsand let Eijk = (AiAjAk) be the ten planes spanned by these points. Every line is containedin three planes and every plane contains three lines. Now intersect this configuration witha plane E different from Eijk and introduce the following labels:

pij = ìj ∩ E gij = Eklm ∩ E with {i, j, k, l,m} = {1, 2, 3, 4, 5} .

13


Fig. 15: Symmetric labelling of Desargues configuration

This labelling shows that there are(52)pairs of perspective triangles.

1.4 Projective transformations

Let V , W be two vectorspaces over the same field and of the same dimension and F : V →W a linear isomorphism. In particular ker(F ) = {0}, so F maps 1-dimensional subspacesto 1-dimensional subspaces.Hence F induces a map from P (V ) to P (W ).

Definition 1.4.1. A projective transformation f from P (V ) to P (W ) is a map definiedby a linear isomorphism F : V →W such that

f ([v]) = [F (v)]

for all [v] ∈ P (V ).

Proposition 1.4.2. Two linear isomorphisms F, F : V → W induce the same projectivetransformation if and only if F = λF for some λ 6= 0.

Proof. “⇐”: If F = λF then

f([v]) := [F (v)] = [λF (v)] = [F (v)] =: f([v])

for all [v] ∈ P (V ).“⇒”: Let f : P (V ) → P (W ) with f ([v]) = [F (v)] =

[F (v)

]. Hence for every [v] ∈ P (V )

there exists λv 6= 0 such that F (v) = λvF (v). To show: λv = λ for some λ 6= 0 and all[v] ∈ P (V ). Let {b1, b2, . . . , bn+1} be a basis of V . Then there exists λi 6= 0 with

F (bi) = λi(F )(bi) ∀i = 1, 2, . . . , n+ 1 .

For[n+1∑i=1

bi

]there exists λ 6= 0 such that

n+1∑i=1

λiF (bi) =n+1∑i=1

F (bi) = F

(n+1∑i=1

bi

)= λF

(n+1∑i=1

bi

)=

n+1∑i=1

λF (bi)

14

1.4 Projective transformations

which is equivalent ton+1∑i=1

(λi − λ)F (bi) = 0.

Since F is a linear isomorphism, the set {F (bi) : i = 1, 2, . . . , n+ 1} is a basis of V and inparticular linearly independent.It follows λi = λ for all i = 1, 2, . . . , n+ 1 and F = λF .

Definition 1.4.3. The real projective linear group PGl(n,R) is definied as the quotientof the group of linear isomorphisms (the general linear group)

Gl(n,R) = {F : Rn → Rn : F linear isomorphism}= {A : Rn → Rn : n× n matrices with det(A) 6= 0} .

by the normal subgroup of non-zero multiples of the identity matrix:

PGl(n,R) = Gl(n,R)/{λ · In : λ 6= 0} .

This is the group of projective transformations from RPn−1 to RPn−1.Example 1.1.

• In terms of affine coordinates the projective transformations from RP1 to RP1 arefractional linear transformations.

Fig. 16: Projective transformations of the projective line

• Central projections are projective transformations.Theorem 1.4.4. Let A1, A2, . . . , An+2 and B1, B2, . . . , Bn+2 be points in general positionin the n-dimensional projective spaces P (V ) und P (W ). There exists a unique projectivetransformation f : P (V )→ P (W ) such that f(Ai) = Bi.

Proof. Existence: Let ai ∈ V such that Ai = [ai] for all i = 1, . . . , n + 1 and An+2 =[n+1∑i=1

ai

]. Analogously there are bi ∈ V such that Bi = [bi] for all i = 1, . . . , n + 1 and

Bn+2 =[n+1∑i=1

bi

]. Define a linear isomorphism F : V → W using the bases {ai} and {bi}

by setting F (ai) = bi for all i = 1, . . . , n+ 1. Then

F (an+2) = F

(n+1∑i=1

ai

)=

n+1∑i=1

F (ai) =n+1∑i=1

bi = bn+2 .

Hence f([v]) = [F (v)] is a projective transformation with f(Ai) = Bi ∀i = 1, . . . , n+ 2Uniqueness: . . .

15


1.5 Cross-ratio (Doppelverhältnis)

We are looking for invariants with respect to projective transformations.

Definition 1.5.1. Let Pi = [vi] =[(xiyi

)], i = 1, . . . , 4, be four distinct points on a

projective line RP1. Then the cross-ratio of these points is

cr(P1, P2, P3, P4) = det(v1v2)det(v2v3)

det(v3v4)det(v4v1) = x1y2 − x2y1

x2y3 − x3y2

x3y4 − x4y3x4y1 − x1y4

=y1y2(x1

y1− x2

y2)

y2y3(x2y2− x3

y3)y3y4(x3

y3x4y4

)y4y1(x4

y4− x1

y1) .

• the cross-ratio depends on the choice of representative vector.

• If yi 6= 0, introduce affine coordinates ui = xiyi. This yields

cr(P1, P2, P3, P4) = u1 − u2u2 − u3

u3 − u4u4 − u1

.

If for example y1 = 0 (hence u1 =∞), then

cr(P1, P2, P3, P4) = ∞− u2u2 − u3

u3 − u4u4 −∞

= −u3 − u4u2 − u3

.

Cancelling ∞ yields the correct result, since

cr(P1, P2, P3, P4) = x1y2(x3y4 − x4y3)(x2y3 − x3y2)(−x1y4) = x1y2y3y4(u3 − u4)

−x1y2y3y4(u1 − u3) = −(u3 − u4)u2 − u3

.

Lemma 1.5.2. The cross-ratio of four distinct points in RP1 does not depend on thechoice of bases in the vector space R2 (i.e. it does not depend on the choice of homogeneouscoordinates).

Proof. Let Pi =[(xiyi

)]with

(xiyi

)= A

(xiyi

)for some A ∈ GL(2,R). Then in our

formular for the cross-ratio we can pull out and cancel the determinant of A. Thus thecross-ratio of these points is given by

cr(P1, P2, P3, P4) =det

(x1 x2y1 y2

)det

(x3 x4y3 y4

)

det(x2 x3y3 y4

)det

(x4 x1y4 y1

)

=det

(A

(x1 x2y1 y2

))det

(A

(x3 x4y3 y4

))

det(A

(x2 x3y3 y4

))det

(A

(x4 x1y4 y1

))

=det

(x1 x2y1 y2

)det

(x3 x4y3 y4

)

det(x2 x3y3 y4

)det

(x4 x1y4 y1

) .

16

1.5 Cross-ratio (Doppelverhältnis)

Corollary 1.5.3. If f : RP1 → RP1 is a projective transformation, then

cr(P1, P2, P3, P4) = cr(f(P1), f(P2), f(P3), f(P4))

for pairwise distinct Pi ∈ RP1.

Proof. As above, f([v]) = [Av] with A ∈ GL(n,R).

Theorem 1.5.4. Projective transformations map projective lines to projective lines andpreserve cross-ratios for arbitrary four distinct points mapped from one projective line toanother.

Proof.

• Linear isomorphisms map 2-dim subspaces to 2-dim subspaces.

• If we restrict the projective transformation to a line we can use the above lemmaand corollary.

Definition 1.5.5. Let `1, `2, `3, `4 be four concurrent lines (`1 ∩ `2 ∩ `3 ∩ `4 = P ) in RP2

that are distinct. Then the cross-ratio of the four lines is

cr(`1, `2, `3, `4) = cr(P1, P2, P3, P4) ,

where ` is a projective line with P /∈ ` and Pi = ì ∩ `.

Fig. 17

The cross-ratio does not depend on the line `.

Proposition 1.5.6. Let P1, P2, P3, P4 be four distinct points on a projective line andf : RP1 → RP1 a projective transformation, that maps P2, P3, P4 to points with affinecoordinates 0, 1, ∞, respectively. Then the affine coordinate of f(P1) is cr(P1, P2, P3, P4).

Proof. Let q1 be the affine coordinate of f(P1). then

q1 = q1 − 00− 1

1−∞∞− q1

= cr(q1, 0, 1,∞) = cr(f(P1), f(P2), f(P3), f(P4)) = cr(P1, P2, P3, P4) .

17


Remark 1.5.7. The cross-ratio of four distinct points is never 0, 1 or ∞.

Proposition 1.5.8. Let P1, P2, P3, P4 and Q1, Q2, Q3, Q4 be two sets of distinct pointson a projective line (or two projective lines). Then there exists a projective transformationf : RP1 → RP1 with f(Pi) = Qi if and only if cr(P1, P2, P3, P4) = cr(Q1, Q2, Q3, Q4).

Proof. “ ⇒” projective transformations preserve cross-ratios.“⇐” Let f : RP1 → RP1 be a projective transformation with f(P2) = Q2, f(P3) = Q3 andf(P4) = Q4. Then f(P1) = Q1. Show Q1 = Q1. We have

cr(P1, P2, P3, P4) = cr(f(P1), f(P2), f(P3), f(P4)) = cr(Q1, Q2, Q3, Q4) .

But that impliescr(Q1, Q2, Q3, Q4) = cr(Q1, Q2, Q3, Q4) . (1.1)

If we introduce coordinates

Q2 =[(

01

)], Q3 =

[(11

)], Q4 =

[(10

)]

andQ1 =

[(q11

)], Q1 =

[(q11

)],

from (1.1) we obtain q1 = q1. This implies Q1 = Q1.

The cross-ratio depends on the order of the points!

cr(P1, P2, P3, P4) = u1 − u2u2 − u3

u3 − u4u4 − u1

.

The value of the cross-ratio does not change, if we swap two distinct pairs of points:

cr(P1, P2, P3, P4) = cr(P2, P1, P4, P3) = cr(P3, P4, P1, P2) = cr(P4, P3, P2, P1)

From the 24 permutations of the two points we only need to calculate the cross-ratio forpermutations of the last three entries:

• cr(P1, P2, P3, P4) = cr(q, 0, 1,∞) = q,

• cr(P1, P2, P4, P3) = cr(q, 0,∞, 1) = qq−1 ,

• cr(P1, P3, P2, P4) = cr(q, 1, 0,∞) = 1− q,

• cr(P1, P3, P4, P2) = cr(q, 1,∞, 0) = q−1q ,

• cr(P1, P4, P2, P3) = cr(q,∞, 0, 1) = 11−q and

• cr(P1, P4, P3, P2) = cr(q,∞, 1, 0) = 1q .

18

1.6 Complete quadrilateral and quadrangle


Definition 1.6.1 (complete quadrilateral). A configuration consisting of four lines in theprojective plane – no three through one point – and the six intersection points, one foreach pair of lines, form a complete quadrilateral.

Fig. 18: Complete quadrilateral

Definition 1.6.2 (complete quadrangle). Let A, B, C and D be four points in generalposition in a projective plane. The complete quadrilateral consists of these four points andsix lines – one for each pair of points. The lines spanned by disjoint sets of points arecalled opposite.

Fig. 19: A complete quadrilateral with opposite lines shown in the same color.

Definition 1.6.3 (quadrangular set). Let ` be a line not through any of the points A, B,C and D. We obtain three pairs of intersection points Pi, Qi for i = 1, 2, 3 with the threepairs of opposite lines. The set {P1, Q1;P2, Q2;P3, Q3} is called a quadrangular set.Note the symmetry of the configuration with respect to the exchange of Pi and Qi andthe permutation of the indices 1, 2, 3.

Fig. 20: Complete quadrangle and quadrangular set

19


Definition 1.6.4 (multi-ratio). Let P1, P2, . . . , P6 be six points on a projective line withrepresentative vectors Pi = [vi] for i = 1, . . . , 6. The multi-ratio of these six points isdefined by

m(P1, P2, P3, P4, P5, P6) = det(v1v2)det(v2v3)

det(v3v4)det(v4v5)

det(v5v6)det(v6v1) .

Remark 1.6.5. For the multi-ratio we have the following properties:

• The multi-ratio is well-defined and independent of the choice of homogeneous coor-dinates.

• The multi-ratio is a projective invariant.

• For the calculation we may use affine coordinates and if one of the coordinates isinfinite, we may cancel them as we did in the calculation of the cross-ratio.

• The multi-ratio can be calculated via cross-ratios:

m(P1, P2, P3, P4, P5, P6) = (−1) cr(P1, P2, P3, P4) cr(P1, P4, P5, P6) .

Theorem 1.6.6. A quadrangular set of points is characterized by the relation

m(P1, Q2, P3, Q1, P2, Q3) = −1 .

Any five points of a quadrangular set determine the sixth point.

Proof. We will use the relation between croos-ratio and multi-ratio to prove the relation.So

m(P1, Q2, P3, Q1, P2, Q3) = (−1) cr(P1, Q2, P3, Q1) cr(P1, Q1, P2, Q3) .

The central projection of the line CD to the line ` from A and D yields the following twoequations:

CDA−→ ` : cr(D,C,Q1, R) = cr(P1, Q3, P2, Q1) = 1

cr(P1, Q1, P2, Q3)CD

B−→ ` : cr(R,D,C,Q1) = cr(P1, Q2, P3, Q1) .

Thus we obtain for the multiratio:

m(P1, Q2, P3, Q1, P2, Q3) = (−1) cr(P1, Q2, P3, Q1) cr(P1, Q1, P2, Q3)

= − cr(D,C,Q1, R) 1cr(D,C,Q1, R) = −1 .

If five points are given we can easily see for example using affine coordinates that therelation m(P1, Q2, P3, Q1, P2, Q3) = −1 determines the affine coordinate of the sixth point,and hence the point itself.

Theorem 1.6.7 (Pappus’ Theorem on quadrangular sets). Let {P1, Q1;P2, Q2;P3, Q3} bea quadrangular set of points on a projective line. Let A, B and C be three distinct pointssuch that ABP1, ACP2 and BCP3 are collinear. Then the lines Q1C, Q2B and Q3Aintersect in one point. In other words, we may construct a complete quadrilateral for aquadrangular set.

20


Proof. Let D = Q1C ∩ Q2B and Q3 = AD ∩ `. Then both {P1, Q1;P2, Q2;P3, Q3} and{P1, Q1;P2, Q2;P3, Q3} are quadrangular sets. But since five of the points coincide wehave Q3 = Q3 and hence the points A, D, and Q3 lie on one line.

Fig. 21: Pappus’ theorem on quadrangular sets

Theorem 1.6.8 (complete quadrilateral). Let `1, `2, `3 and `4 be four lines in the pro-jective plane such that no three go through one point. Further let A = `1 ∩ `2, B = `2 ∩ `3,C = `3 ∩ `4, D = `4 ∩ `1 and P = `1 ∩ `3 and Q = `2 ∩ `4 be the six points of the completequadrilateral. Define ` = PQ and X = AC ∩ ` and Y = BD ∩ `. Then

cr(P,X,Q, Y ) = −1 .

Fig. 22: Complete quadrilateral

Remark 1.6.9. We say that the pair {P,Q} seperates the pair {X,Y } harmonically or Yis the harmonic conjugate of X with respect to {P,Q} if cr(P,X,Q, Y ) = −1.

Computational proof. We can apply a projective transformation of the plane to map thefour points A, B, C and D onto the unit square with vertices0

01

,1

01

,1

11

,0

11

.Then the line ` is the line at infinity and we can easily calculate the cross-ratio and see itis equal to −1.

21


Fig. 23: Normalized complete quadrilateral

Proof via multi-ratio. Consider the multi-ratio with

P1 = Q1 = P , P3 = Q3 = Q , Q2 = X , P2 = Y ,

where p is the affine coordinate of P , q the one of Q etc. Then we have

−1 = m(P,X,Q, P, Y,Q) = p− xx− q

q − pp− y

y − qq − p

= p− xx− q

q − yy − p

= cr(P,X,Q, Y ) .

Fig. 24: Complete quadrilateral and quadrangular set

To give a third proof, we introduce the notion of projective involutions.

Definition 1.6.10. Consider a projective map f : RP1 → RP1. f is called a projectiveinvolution if f 6= id but f2 = id.

f can be defined by two pairs of points A, B and C, D such that:

f(A) = B , f(B) = A , f(C) = D , f(D) = C .

Any projective involution f has either 0 or 2 fixed points. (homework)

22

1.7 Moebius pairs of tetrahedra

If f is a projective involution with two fixed points P , Q, then

cr(P,X,Q, f(X)) = cr(P, f(X), Q,X) = 1cr(P,X,Q, f(X)) .

Hence cr(P,X,Q, f(X)) = ±1 for all X 6= P,Q.Since a cross-ratio of four distinct points cannot become zero, we obtain a characteristicequation for a projective involution with two fixed points:

cr(P,X,Q, f(X)) = −1 for all X 6= P,Q .

Proof of the theorem on complete quadrilateral via projective involutions. Consider a pro-jective transformation f : RP1 → RP1 with

Af7−→ B B

f7−→ A Cf7−→ D D

f7−→ C .

This map is a projective involution. So f∣∣ìs also an involution with fixed points P and

Q. Further f(X) = Y . This implies cr(P,X,Q, Y ) = −1.

1.7 Moebius pairs of tetrahedra

Definition 1.7.1. A pair of tetrahedra is called a Moebius pair if the vertices of onetetrahedron lie in the planes spanned by the faces of the second and the vertices of thesecond lie in the face-planes of the first.

Theorem 1.7.2. If four vertices of the first tetrahedron lie in the four face-planes of thesecond and three of the vertices of the second tetrahedron lie in the three face-planes ofthe first then the fourth vertex lies in the fourth plane.

(; Geometry II, consistency of nets. Bobenko, Suris: Discrete Differential Geometry)

Proof. Let us introduce the following labelling motivated by a (combinatorial) cube:

Fig. 25: Labelling of a Moebius pair of tetrahedra

The vertices of two tetrahedra T1 and T2 are the following

T1 : P, P12, P13, P23 T2 : P1, P2, P3, P123 .

23


The face-planes are labelled in the following way:

T1 : Π1 3 P, P12, P13 T2 : Π12 3 P1, P2, P123

Π2 3 P, P12, P23 Π13 3 P1, P3, P123

Π3 3 P, P13, P23 Π23 3 P2, P3, P123

Π123 3 P12, P23, P13 Π 3 P1, P2, P3 .

Now seven vertices lie on seven planes:

Pi ∈ Πi , P ∈ Π , Pij ∈ Πij .

To show: P123 ∈ Π123. The plane Π contains the vertices P , P1, P2 and P3. The in-tersections of the planes Πi and Πij with Π form a complete quadrangle. This yields aquadrangular set on the line Π ∩Π123.

Fig. 26: Quadrangular set

Now we have a look at the intersection of the planes Πi and Πij with Π123. The linesΠi ∩Π123 yield a triangle on the vertices P12, Π13, and Π23.

Fig. 27

By the Theorem of Pappus on quadrangular sets the lines Πij ∩ Π123 through the pointsΠij intersect in one point on Π123. This point is P123 and it lies on Π123.

24

1.8 The Fundamental Theorem of Real Projective Geometry

Fig. 28: Moebius pair of tetrahedra

Definition 1.7.3. A Koenigs cube is a combinatorial cube with six flat faces (i.e. eachsets P , Pi, Pj , Pij and Pi, Pij , Pik, P123 lie in a plane) such that the vertices P , P12, P13,P23 and P1, P2, P3, P123 each lie in a plane.

Theorem 1.7.4. If the red vertices (P , P12, P13, P23) lie in a plane and all the faces areflat, then the blue vertices (P1, P2, P3, P123) lie in a plane.

Proof. Homework.

1.8 The Fundamental Theorem of Real Projective Geometry

Theorem 1.8.1 (Fundamental Theorem of real projective geometry). Let f : RPn → RPn,n ≥ 2, be a bijective map that maps lines to lines. Then f is a projective transformation.

Remark 1.8.2. The theorem does not hold for arbitrary fields. For example f : CPn → CPn(n ≥ 2) with

f

z1

...zn+1

=

z1...

zn+1

is a bijective map, mapping complex projective lines to complex projective lines but it isnot a projective transformation of CPn.

Generaliziation for arbitrary fields: A bijective map f : P (V ) → P (V ), where V is avectorspace over some field that maps lines to lines is induced by an “almost linear” mapF : V → V , i.e.

i) F (v + w) = F (v) + F (w) for all v, w ∈ V ,

ii) F (λv) = α(λ)F (v) for all λ ∈ F, v ∈ V , where α : F→ F is a field automorphism.

The only field automorphism of R is the identity and of C the complex conjugation.

25


Lemma 1.8.3. Let f : RP2 → RP2 be a bijective map that maps lines to lines. Then ifA,B,C,D ∈ ` ⊆ RP2 with cr(A,B,C,D) = −1 then cr(f(A), f(B), f(C), f(D)) = −1.

Proof. Use theorem on complete quadrilateral.

Fig. 29

This implies cr(f(A), f(B), f(C), f(D)) = −1.

Lemma 1.8.4. Let f : RP1 → RP1 be a bijective map such that f(0) = 0, f(1) = 1,f(∞) = ∞ and cr(A,B,C,D) = −1 implies cr(f(A), f(B), f(C), f(D)) = −1. Thenf = id.

Corollary 1.8.5. Let f : RP1 → RP1 bijective, such that cr(A,B,C,D) = −1 impliescr(f(A), f(B), f(C), f(D)) = −1. Then f is a projective transformation.

Proof. Let g : RP1 → RP1 be a projective transformation with g(f(0)) = 0, g(f(1)) =1, g(f(∞)) = ∞. Then by Lemma 1.8.4, g ◦ f = id. Thus f = g−1 is a projectivetransformation.

Proof of Lemma 1.8.4. We look at different points on RP1 that have a cross-ratio of −1:

i) cr(x, x+y2 , y,∞) = −1 implies cr(f(x), f(x+y

2 ), f(y),∞) = −1 and by that we have−f(x)−f( x+y

2 )f( x+y

2 )−f(y) = −1. This gives us f(x+y2 )−f(x) = −f(x+y

2 )+f(y) and thus f(x+y2 ) =

12(f(x) + f(y)).

ii) cr(0, x, 2x,∞) = −1 implies cr(0, f(x), f(2x),∞) = −1, i.e. f(x)f(x)−f(2x) = −1 and

therefore f(2x) = 2f(x).

iii) (i) and (ii) give us f(x+ y) = f(x) + f(y).

iv) We have 0 = f(0) = f(x−x) = f(x+(−x)) = f(x)+f(−x) and hence f(−x) = −f(x).

v) Now, by (iii) and (iv) we obtain f(kx) = kf(x) for all k ∈ Z.

vi) (v) implies f(qx) = qf(x) for all q ∈ Q. In particular, for x = 1 we have f(q) = q forall q ∈ Q. To show f(r) = r for all r ∈ R, we will prove the monotonicity of f on R.

vii) By cr(−x, 1, x, x2) = −1 we obtain cr(f(−x), 1, f(x), f(x2)) = −1.

viii) For f(x2) = f(x)2 for x > 0. This yields f(x) > 0 using (vii)

ix) If y−x > 0, then f(y−x) > 0 which implies f(y)− f(x) > 0 and hence f(y) > f(x).

Since f is a monotonically increasing function which is the identity on Q, f is the identity.

26

1.9 Duality

Proof of the fundamental theorem (for n = 2). Let f : RP2 → RP2 be a bijective map,mapping lines to lines. Let P1 = [1, 0, 0]T, P2 = [0, 1, 0]T, P3 = [0, 0, 1]T, P4 = [1, 1, 1]Tand g : RP2 → RP2 be a projective transformation such that g(f(Pi)) = Pi.

Fig. 30

Let X ∈ RP2 not on the line P1P2. Then the lines PiPj are mapped onto themselvessince g ◦ f fixes the Pi and maps lines to lines. In particular, g(f(P2P3)) = P2P3 andg(f(P1P3)) = P1P3. Let Q1 = P1P4 ∩P2P3 and Q2 = P2P4 ∩P1P3. By Lemma 1.8.3 g ◦ fpreserves cross-ratio of −1 and by Corollary 1.8.5 the map g ◦ f restricted to the linesP1P3 and P2P3, resp., is a projective transformation. Since g ◦ f has three fixed points oneach of the lines P1P3 and P2P3 we see that g ◦ f

∣∣P1P3

and g ◦ f |P2P3 is the identity.Let X1 = P1X ∩ P2P3 and X2 = P2X ∩ P1P3 then X = P1X1 ∩ P2X2 and g(f(Xi)) =Xi, i = 1, 2 and X = P1X1 ∩ P2X2. So g(f(X)) = X since g(f(P1X1)) = P1X1 andg(f(P2X2)) = P2X2. Hence f = g−1 is a projective transformation (if X ∈ P1P2 choosedifferent affine coordinates or note that the point P3P4 ∩ P1P2 is fixed by g ◦ f and g ◦ frestricted to the line P1P2 is again the identity).

Theorem 1.8.6. Let U ⊆ RPn be a subset that contains an open ball B ⊂ Rn ⊂ RPn (forthe decomposition RPn = Rn ∪ RPn−1 for some affine coordinate) and f : U → RPn aninjective map mapping lines to lines, that is, if ` is a line with `∩U 6= ∅ then there existsa line `′ such that f(U ∩ `) ⊂ `′. Then f is the restriction of a projective transformation.

1.9 Duality

Let ` be a line in RP2. Then the line can be described by one homogeneous equation:x1x2x3

∈ `⇔ a1x1 + a2x2 + a3x3 = 0 ,

where the ai are unique up to a scalar multiple λ 6= 0. We can take in a way thatwill be explained in detail, the point [a1, a2, a3]T as homogeneous coordinates for theline `. The lines in RP2 yield another projective plane with homogeneous coordinates[a1, a2, a3]T. This is what we call the dual projective plane (RP2)∗. If we fix one point[x1, x2, x3]T ∈ RP2, the set of lines through this point corresponds to a line in (RP2)∗.

27


Fig. 31: Duality in the real projective plane

Note: This is the intuition, let’s see the formal definition.

Definition 1.9.1. Let V be a vector space of finite dimension over some field F. The dualvector space is defined as

V ∗ := {ϕ : V → F : ϕ is linear} .

Note: If dim(V ) =∞, we have to add continuity as a property for the linear function ϕ.Let us now consider bases. If {b1, . . . , bn} is a basis for V , then {b∗1, . . . , b∗n} is a basis forV ∗ (this is called dual basis) where

b∗i (bj) = δij .

It is easy to check that this is a basis for V ∗. This yields dim(V ) = dim(V ∗).Remark 1.9.2. To identify V with (V ∗)∗ we don’t need a basis of V . A vector v ∈ V canbe interpreted as a linear functional of V ∗ as

V → (V ∗)∗ , v 7→ v(·)

with v(ϕ) = ϕ(v) for any ϕ ∈ V ∗. But this construction cannot be made in infinitedimensional vector spaces and in general V = (V ∗)∗ does not hold.

Definition 1.9.3. If f : V →W is a linear map, then the dual map f∗ is defined as

f∗ : W ∗ → V ∗ , ψ 7→ f∗(ψ)

with f∗(ψ)(v) = ψ(f(v)) for v ∈ V .

Definition 1.9.4. Let U ⊂ V a vector subspace, then the annihilator of U is defined as

U◦ := {ϕ ∈ V ∗ : ϕ(u) = 0 for all u ∈ U}

and is a subspace of V ∗.

For the dimensions of U and U◦ we have that

dim(U) + dim(U◦) = dim(V ) .

Remark 1.9.5. The following properties hold:

28

1.9 Duality

• (U1 ∩ U2)◦ = U◦1 + U◦2 ,

• (U1 + U2)◦ = U◦1 ∩ U◦2 .

In an euclidean vector space (a vector space of finite dimension over R endowed withsome inner product) one can identify V and V ∗ canonically and up to these identificationU◦ = U⊥.

Definition 1.9.6. Let P (V ) a projective space. Then we define the dual projective spaceas P (V ∗).

Example 1.9.7.

1. A point [v] ∈ P (V ) corresponds to a hyperplane P (span(v)◦) ⊂ P (V ∗).

2. Points [ψ] ∈ P (span(v)◦) correspond to hyperplanes through [v] ∈ P (V ).

3. More generally, the dual of a k-plane in P (V ) corresponds to an (n − k − 1)-plane ifdimP (V ) = n.

4. Points in the dual (n − k − 1)-plane correspond to hyperplanes containing the givenk-plane.

Dictionary for Duality of RP3

RP3 (RP3)∗point planeline lineplane pointline in a plane line through a pointintersection of line and plane plane spanned by a line and a point

Remark 1.9.8. To every incidence configuration we may associate a dual configuration.Example 1.9.9. Desargues configuration.

Fig. 32: Desargues configuration and its dual

Cross-ratio of lines through a point (again).

29


Fig. 33: Cross-ratio of four lines through a point

Proposition 1.2. The map which maps a point g∗ ∈ p∗ to the intersection point of g ∩ ìs a projective map.

Proof. Let {b1, b2, b3} be a basis of R3, such that P = [b1] and ` = P (span{b2, b3}). Thisyields a dual basis {b∗1, b∗2, b∗3} of R3 with b∗i (bj) = δij . The line in (RP2)∗ correspondingto the point P is P ∗ = P ({b1}◦) = P (span{b∗2, b∗3}).Let g∗ = [sb∗2 + tb∗3] ∈ p∗ be a point on the line P ∗. A point x ∈ ` can be written asQ = [xb2 + yb3]. To have Q ∈ g, we need

0 = (sb∗2 + tb∗3)(xb2 + yb3) = sx+ ty .

Solving this for x and y yields the following point of intersection Q = g ∩ ` = [−tb2 + sb3],so the map P ∗ → ` is given by

g ∩ ` =[(

0 −11 0

)(st

)]

for g = [sb∗2 + tb∗3]. Hence, it is a projective transformation and preserves cross-ratio.

1.10 Conic sections

Conic sections are solution sets of the following quadratic equations{(uv

)∈ R2|au2 + 2buv + cv2 + du+ ev + f = 0

}.

Theorem 1.3. Any quadratic equation in Euclidean coordinates can be transformed by anEuclidean motion ( xy ) = A ( uv ) + b, A ∈ O(2,R), b ∈ R2 to one of the following forms:

1. (xa )2 + (yb )2 = 1, ellipse

Fig. 34

30

1.10 Conic sections

2. (xa )2 − (yb )2 = 1, hyperbola

Fig. 35

3. y = ax2, parabola

4. (xa )2 − (yb )2 = 0, pair of crossing lines

Fig. 36

5. x2 = a2, two parallel lines

Fig. 37

6. x2 = 0, a double line

7. (xa )2 + (yb )2 = 0, a point

8. (xa )2 + (yb )2 = −1, empty set

Concept of the proof. Diagonalize the matrix(a bb c

)describing the quadratic part of the

above equation and then calculate translation vector b.

31


These are called conic sections, since they arise as intersections of cone of revolution (ora cylinder) with an affine plane. The cone of revolution is defined by

xyz

∈ R3|x2 + y2 − z2 = 0

.

Fig. 38

Some facts about conic sections:

1. Ellipse(a > b),

• principle axes a and b.• F1,2 = (±

√a2 − b2, 0), F1,2 are focal points.

• r1 + r2 = constant.

2. Hyperbola

• F12 = (±√a2 + b2, 0), F1,2 are focal points.

• |r1 − r2| = constant.

3. Parabola

• F = (0, 11+a), F is focal point.

• r1r2

= 1

Proposition 1.4. A plane cutting through the upper half of the cone of revolution and allits generating lines yields a ellipse.

32

1.10 Conic sections

Proof. Consider two spheres S1, S2 touching the cone and the plane, let p be a point ofintersection points of the plane and the cone. We have |F1P | = |Y1P | and |F2P | = |Y2P |,since the tangents here have the equal length, hence

|F1P |+ |F2P | = |Y1P |+ |Y2P | = |Y1Y2| = const .

Fig. 39

Theorem 1.5 (Optical properties). Rays emitted from one focal point of the ellipse arriveat the other focal point of the ellipse.

Fig. 40

33


Proof. Consider an ellipse with foci F1 and F2, and a point on the ellipse P . Then constructthe point F ′1 on the line F2P such that |F ′1P | = |F1P |. Now let ` be the angle bisector ofthe angle ∠F1PF

′1.

Fig. 41

We will show that ` is the tangent line. Let X ∈ `. Then |F1X| = |F ′1X|. But by thetriangle inequality we obtain

|F2X|+ |F ′1X| ≥ |F2F′1| = |F1P |+ |F2P |

with equality if and only if X = P . So ` is the tangent and the angles are equal.

Rays emitted from one focal point of the hyperbola are reflected such that the reflectedlines go through the other focal point.

Fig. 42

Rays emitted from the focal point of the parabola become parallel after reflection.

34

1.10 Conic sections

Fig. 43

In the following, we shall consider conic sections from the projective point of view.

Definition 1.10.1. Let V be a vector space over R (or C). A map b : V × V → R is asymmetric bilinear form, if

i) b(v, w) = b(w, v) ∀v, w ∈ V

ii) b(α1v1 + α2v2, w) = α1b(v1, w) + α2b(v2, w) for all α1, α2 ∈ R, v1, v2, w ∈ V .

b is non-degenerate, ifb(v, w) = 0 ∀w ∈ V ⇒ v = 0 .

The corresponding quadratic form is defined by the

q(v) = b(v, v) ∀v ∈ V .

The symmetric bilinear form is determined by the corresponding quadratic form via thepolarisation identity

b(v, w) = 12(q(v + w)− q(v)− q(w)) .

If {v1, . . . , vn} is a basis of V, then we can associate a matrix to a bilinear form by

B = (bij)i,j=1,...,n with bij := b(vi, vj) .

This yields

b(v, w) = b

n∑i=1

xivi ,n∑j=1

yjvj

=n∑i=1

n∑j=1

b(vi, vj)xiyj =n∑i=1

n∑j=1

bijxiyj = vTBw ,

where v = (x1, . . . , xn)T and w = (y1, . . . , yn)T.For the quadratic form this implies

q(v) = vTBv =n∑

i,j=1bijxixj .

Remark 1.10.2. Quadratic forms correspond to homogeneous polynomial of degree two.

35


Theorem 1.10.3 (Sylvester). For a given quadratic form q on a real vector space thereis a basis such that

q(v) =p∑i=1

x2i −

p+q∑i=p+1

x2i .

If a quadratic form qc is given on a complex vector space, there is a basis such that

qc(v) =p∑i=1

z2i .

The triple (p, q, n − p − q) is called the signature of the quadratic form q. The signatureis invariant with respect to change of basis. q is non-degenerate, if and only if p+ q = n.

Definition 1.10.4. If q is a non-zero quadratic form on R3. Then

C = {[x] ∈ RP 2 | q(x) = 0}

is a conic.

This definition does not depend on the choice of representative vectors.Every quadratic form on R3 corresponds to a symmetric 3× 3-matrix. So it is defined bysix real values. Every quadratic form corresponds to a point in R6. As q and λq definethe same conic C, a conic corresponds to a point in RP5. According to the Sylvester’sTheorem there exists a basis, such that

q(v) = λ1x21 + λ2x

22 + λ3x

23

for some λi ∈ {−1, 0, 1}.Non-degenerate conics:

i) q(v) = x21 + x2

2 + x23, i.e. signature (+++). In this case the conic defined by q is the

empty set:

{[x] ∈ RP 2 | q(x) = 0} = {[x] ∈ RP 2 |x21 + x2

2 = −1} = ∅ .

ii) q(v) = x21 + x2

2 − x23, i.e. signatur (++-). The corresponding conic is an ellipse, a

hyperbola or a parabola.

Degenerate conics:

iii) q(v) = x21 +x2

2, i.e. signature (++0). The correspondig conic is a single point in RP2.

iv) q(v) = x21 − x2

2, i.e. signature (+-0). The corresponding conic consists of two lines inRP2 (x1 = x2, x1 = −x2).

v) q(v) = x21, i.e. signature (+00). Here we have a single line in RP2.

Depending on the affine coordinate we obtain different affine images of the conic withsignature (++-):

36

1.10 Conic sections

Fig. 44

Theorem 1.10.5. Let P1, P2, P3, P4, P5 be five points in RP2, then there exists a conicthrough P1, . . . , P5. Moreover:

i) If no four points lie on a line, the conic is unique.

ii) If no three points lie on a line, the conic is non-degenerate.

Lemma 1.10.6. If three collinear points are on a conic, then the conic contains the wholeline spanned by these points.

Proof. Let P1 = [v1], P2 = [v2], P3 = [v1 + v2] be three collinear points and q a quadraticform defining some conic containig these points. Thus 0 = b(v1, v1) = q(v1) and 0 =q(v2) = b(v2, v2) as P1 and P2 lie on the conic. Also, 0 = q(v1 + v2) = b(v1 + v2, v1 + v2) =b(v1, v1) + 2b(v1, v2) + b(v2, v2), since P2 lies on it as well. This yields b(v1, v2) = 0. So forQ = [sv1 + tv2] we obtain

q(sv1 + tv2) = s2b(v1, v1) + 2stb(v1, v2) + t2b(v2, v2) = 0

and Q lies on the conic.

Remark 1.10.7. If a conic in RP 2 contains a line, then it is degenerate.

Proof of Theorem 1.10.5. Existence:Let Pi = [vi] with vi ∈ R3 for all i = 1, . . . , 5. Let q : R3 → R3 be a quadratic form. Thenan incidence Pi ∈ C := {[x] ∈ RP 2|q (x) = 0} yields an homogeneous equation for thecorresponding matrix entries. q (vi) = 0 is equivalent to vTi Bvi = 0. The system q (vi) = 0for all i = 1, . . . , 5 has at least a one dimensional space of solutions. So there exists aquadratic form, such that the conic contains all Pi.First statement: If four of the Pi lie on a line, then the conic contains a line and isdegenerate. In this case there exists a one parameter family of conics containing the Pi.If three points P1, P2, P3 lie on a line `, but P4, P5 /∈ `. Then by Lemma 1.10.6, ` ⊂ C andC is degenerate. So the conic consists of the lines l and P4P5. In particular C is unique.Second statement and second part of (i): Now let no three of the points be colinear andq1 and q2 are two quadratic forms with q1 (vi) = q2 (v2) = 0. But then the quadratic formq = q1 + λq2 also satisfies q (vi) = 0 for all i = 1, . . . , 5. The determinant det (q1 + λq2)is a polynomial in λ of degree 3. So it has a zero, i.e. there exists some λ0 such that

37


det (q1 + λ0q2) = 0 and q1 + λ0q2 is degenerate. Thus three points must be colinear. Thus q1 = q2 is unique and non-degenerate.

Fig. 45

Definition 1.10.8. A pencil of conics is a line in RP5, which is the space of conics in RP2.Let [q1] and [q2] be two conics, then the conic [q] is in the pencil, if there exist homogeneouscoordinates

(λ1 λ2

)T, such that [q] = [λ1q1 + λ2q2]. Let P1, P2, P3, P4 ∈ RP2 be 4 points

in general position. Then the conics through these four points build a pencil.

Fig. 46

The pencil contains three degenerate conics: q1 = `12`34, q2 = `13`24, q3 = `14`23, whereìj : R3 → R is the linear function vanishing on the lines defining Pi and Pj . Now for anarbitrary fifth point P5 6= Pi, i = 1, . . . , 4, there exists a unique conic through P1, . . . , P5by the last theorem. The conics of the pencil are given by homogeneous coordinates (λ1λ2)with [q] = [λ1q1 + λ2q2]. The fifth point P5 = [v5] lies on the conic if

q (v5) = 0 ⇒ λ1q1 (v5) + λ2q2 (v5) = 0 ⇒ λ1 = q2 (v5) ∧ λ2 = −q1 (v5) .

The conic containing P5 in the pencil is [q] = [q2 (v5) q1 − q1 (v5) q2].Summary: We now have found the following identifications:Quadratic forms ↔ symmetric bilinear forms ↔ symmetric matrices ↔ points inR6

38

1.11 Polarity: Pole - Polar Relationship

Conis ↔ quadratic forms up to a skalar multiple ↔ points in P(R6) = RP5

Pencils of conics ↔ lines in RP5.

Theorem 1.10.9 (Pascal’s Theorem). Let A, B, C, D, E and F be six points on a conic.Consider the hexagon defined by these vertices. Then the intersection points of oppositesides are colinear. I.e. there exists a line containing G = AB ∩ DE, I = BC ∩ EF ,H = CD ∩AF .

Fig. 47

Proof. Consider the two pencils of the conics through A, B, C, D and A, D, E, F . Thenboth pencils contain the conic as there holds

q = λ1ÀB`CD + λ2ÀD`BC = µ1ÀF `DE + µ2ÀDÈF

which is equivalent to

λ1ÀB`CD − µ1ÀF `DE = ÀD (µ2ÈF − λ2`BC) .

Claim: The line containing the intersection points s given by µ2ÈF − λ2`BC .

1. G = [vG] is on the line, since ÀD (vG) 6= 0, but ÀB (vG) = 0 and `DE (vG) = 0. Hence

(µ2ÈF − λ2`BC) (vG) = 0 .

2. Analogously, H is on the line.

3. I = [vi] is on the line, since ÈF (vI) = 0 and `BC (vI) = 0.

Remark 1.10.10. Pappus theorem corresponds to Pascal’s theorem on a degenerated conic.


Let b : V × V → F be a symmetric non-degenerate bilinear form.

Definition 1.11.1. Let U ≤ V be a vector subspace. Then

U⊥ := {v ∈ V | b(u, v) = 0 , ∀u ∈ U} .

39


If U = {u1, ..., uk} then

U⊥ := (spanU)⊥ = {v ∈ V | b(ui, v) = 0 , for all i = 1, . . . , k} .

We call U⊥ the orthogonal complement of U .

Remark 1.11.2. dimU⊥ = dimU0 = n− k, if dimU = k and dimV = n.Example 1.11.3.

(i) Define b : R2 × R2 → R by

b

((x1y1

),

(x2y2

))= x1x2 + y1y2 .

This yields usual orthogonality and scalar product.

(ii) Define b : R2 × R2 → R by

b

((x1y1

),

(x2y2

))= x1x2 − y1y2 .

In this case, orthogonality can be visualized as follows:

Fig. 48

We have (10

)⊥={(

xy

)∈ R2

∣∣∣∣∣ 0 = b

((10

),

(xy

))= x

}(

11

)⊥={(

xy

)∣∣∣∣∣ x− y = 0⇔ x = y

}(

12

)⊥={(

xy

)∣∣∣∣∣ x− 2y = 0}.

For example, take x = 2, y = 1.

40


• If b is indefinite, (i.e. signature contains +’s and −’s), then there exist vectors vwith b(v, v) = 0. These are called isotropic vectors.

• In example 2, if we restrict to the subspace {λ ( 11 ) | λ ∈ R}, the non-degenerate

bilinear form restricted to this subspace is degenerate.

Polarity in RP2.Using the orthogonal complement, we define a map that maps points in RP2 to a line via[p] 7→ [p⊥] and lines to points, i.e. ` 7→ [`⊥].Definition 1.11.4. The line [p⊥] corresponding to a point [p] is the polar (or polar line)of [p] and [p] is the pole of its polar [p⊥].

Geometry of pole-polar relationship: Up to projective transformations, there is onlyone non-degenerate, non-empty conic in RP2. It corresponds to a non-degenerate sym-metric indefinite bilinear form with signature (+ + -), (- - +).Definition 1.11.5. A line in RP2 is a tangent to a conic if it has exactly one point incommon with the conic or lies entirely in the conic.

Fig. 49

Proposition 1.11.6. The polar line of a point [p] on a conic is a tangent at this point.

Fig. 50

41


Proof. Let C = {[x] ∈ RP 2 | b(x, x) = 0}. The polar line to [p] is

P ({p}⊥) = {[x] ∈ RP 2 | b(p, x) = 0} .

• p lies on the polar, since b(p, p) = 0⇐ [p] ∈ C.

• now let [q] ∈ RP 2 with b(p, q) = 0.

The polar line is given by: [λp+ q] (except [p]).

b(λp+ q, λp+ q) = λ2b(p, p) + 2λb(p, q) + b(q, q) = 0 + 0 + b(q, q) .

If b(q, q) = 0, then b(λp + q, λp + q) = 0 for all λ ∈ R and thus the polar is contained inC. If b(q, q) 6= 0, then b(λp+ q, λp+ q) 6= 0 and the polar is tangent at [p].

Example 1.11.7.

b

x1x2x3

),

y1y2y3

= x1y1 + x2y2 − x3y3 .

In affine coordinatesu1 = x1

x3, u2 = x2

x3

we get the conic

C ={(

u1u2

) ∣∣∣ u21 + u2

2 = 1}.

A parametrization is given by

γ(α) =(u1u2

)=(

cosαsinα

).

The tangent at γ(α0) is

{γ(α0) + tγ′(α0)} ={γ(α0) + t

(− sinα0cosα0

)},

for setting

p =

cosαsinα

1

yields

U⊥ =

xyz

∣∣∣∣∣∣∣ bcosα

sinα1

,xyz

= 0

and the condition is equivalent to x cosα + y sinα − z = 0. Hence the tangent line to [p]is given by

=

λcosα

sinα1

+ µ

− sinαcosα

0

∣∣∣∣∣∣∣λ, µ ∈ R

.

42


Fig. 51

Proposition 1.11.8. Let P1, P2 ∈ RP2 and ` non-degenerate conic, then the polar linesof P1 and P2 intersect in the pole of the line P1P2.

Proof. Let P1 = [p1] , P2 = [p2] and P = [p]. Then we have b (p1, p) = 0 and b (p2, p) = 0,since P ∈ P

(p⊥1

)∩ P

(p⊥2

). Thus:

b(λp1 + µp2, p) = 0,∀λ, µ ∈ R.

So the line P1P2 is the polar line of P .

Construction of the polar line for points “outside” and “inside” the conicsection.

a) If P is outside the conic, then the polar line can be constructed using the two tangentstouching the conic as shown in the picture.

Fig. 52

43


b) If P is inside the conic, then we consider two arbitrary lines `1 and `2 through P . Thepolar line ` of P is the line through P1 and P2 that are the poles of `1 and `2.

Fig. 53

Theorem 1.11.9 (Polar Triangle). Let A, B, C, D be four points in general position ona non-degenerate conic in RP2. Then the intersection points of pairs of opposite sides ofthe complete quadrangle formed by A, B, C and D form a polar triangle, i.e. X is thepole of Y Z, Y is the pole of XZ and Z is the pole of XY .

Fig. 54

If we choose the basis B = {x, y, z} with X = [x], Y = [y], Z = [z], then the above isequivalent to the matrix of the corresponding bilinear form being diagonal. Additionally,the statement holds for arbitrary non-degenerate conics in the pencil through A, B, C andD.

Proof. Normalize:

44


Fig. 55

Now,

A =

−111

B =

111

C =

1−11

D =

−1−11

.

Then

X =

100

Y =

001

Z =

010

.

If

B =

b11 b12 b13b12 b22 b23b13 b23 b33

then we obtain by evaluating the bilinear form on the points the following equations:

1. For A on the conic, we get b11 + b22 + b33 − 2b12 − 2b13 + 2b23 = 0.

2. For B on the conic, we get b11 + b22 + b33 + 2b12 + 2b13 + 2b23 = 0.

3. For C on the conic, we get b11 + b22 + b33 − 2b12 + 2b13 − 2b23 = 0.

4. For D on the conic, we get b11 + b22 + b33 + 2b12 − 2b13 − 2b23 = 0.

By subtracting the second, third and the fourth equation, resp., from the first, we obtain

−4b12 − 4b13 = 0 −4b13 − 4b23 = 0 −4b12 − 4b23 = 0

Thus, b12 = b13 = b23 = 0. We now have

B =

b11 0 00 b22 00 0 b33

.

45


By summing up all four equations, we get b11 + b22 + b33 = 0, so B w.r.t basis {x, y, z} isdiagonal.In the above calculation we only used that the points A, B, C and D are on the conic. SoX, Y , Z build a polar triangle for an arbitrary non-degenerate conic in the pencil. Thepencil can be generated by

B1 =

−1 0 00 1 00 0 0

and B2 =

0 0 00 −1 00 0 1

.

Corollary 1.11.10. Consider a non-degenerate conic C, a point P not on C and a line `through P intersection C in two points X and Y . Let Q be the intersection of ` with thepolar line of P , then

CR (P,X,Q, Y ) = 1 .

Proof. Use the theorem on the polar triangle and the theorem on the complete quadrilat-eral to obtain

cr(P,X ′, Q′, Y ′) = −1.

Fig. 56

Then a central projection with center Z yields the desired result.

Dual conics and Brianchon’s Theorem Let C be a non-degenerate non-empty conicin RP2.

46


Fig. 57

Then the set of tangents to C yields a conic in(RP2)∗. (homework)

Theorem 1.11.11 (Brianchon’s Theorem). Let A, B, C, D, E, F be a hexagon cirum-scribed around a conic (i.e. AB,BC, . . . are tangents),then the diagonals AD, BE, CFintersect in one point.

Proof. Dualize it and then use Pascal’s theorem.

Fig. 58

Note that the cyclic order of the points on the conic is preserved. In the above picturethis is not the case! The order of the points/tangents was intentionally changed to obtaina nice picture for both of the theorems.Use your favourite interactive geometry software to study what happens when you changethe order of the points in the hexagons.

47


1.12 Quadrics

Definition 1.12.1. Let q be a quadratic form on Rn or Cn. Then the correspondingquadric is Q = {[v] ∈ RPn : q(v) = 0}A conic is a quadric in RP2.

How many indefinite non-degenerate quadratic forms exists in Rn+1 up to change a basis?Classification by the signature: (+++ . . .+), (+++ . . .+−), . . . , (+−. . .−), (−−− . . .−)– the assumption of indefiniteness leaves n indefinite quadratic forms (+ + + . . .+−), . . ., (+− . . .−).The number of non-empty non-degenerate quadrics in RPn is dn2 eExample 1.12.2. n = 2: In RP2 there exist only one indefinite non-degenerate quadric/conicup to projective transformations. It has signature (+ +−) or (−−+), resp.n = 3: In RP3 there exist d3

2e = 2 different quadrics up to projective transformations:

• (+ + +−), (−−−+), Q1 = {[v] ∈ RP 3 : v21 + v2

2 + v23 − v2

4 = 0},

• (+ +−−), (−−++), Q2 = {[v] ∈ RP 3 : v21 + v2

2 − v23 − v2

4 = 0}.

Fig. 591. is a unit sphere (or ellipse, paraboloid or 2-sheeted hyperboloid) depending on thechoice of affine coordinate. To obtain a sphere as affine image of the quadric we canchoose the affine coordinates u1 = v1

v4, u2 = v2

v4, and u3 = v3

v4.

2. is one-sheeted hyperboloid or hyperbolic paraboloid.

Degenerate Quadratic forms/quadrics Let b be a degenerate bilinear form. Thenthe space {u ∈ U : b(u, v) = 0∀v ∈ V } = ker(b) is a subspace of V . Consider a subspaceU1 ⊂ V such that V = ker(b) ⊕ U1, then b :U1 defines a non-degenerate quadric Q1in P (U1). The quadric Q defined by b is the union of lines through points in the non-degenerate quadric Q1 defined by b|U1 and points in P (ker(b)) if Q1 6= ∅ (see Exercise8.1).Example 1.12.3 (degenerate conic). Consider the following bilinear form in RP 2:

b(

x1x2x3

,x1x2x3

) = x12 − x2

2 .

48

1.13 Orthogonal Transformations

Then ker(b) = span{e3} and R3 = ker(b)⊕ span{e1, e2}︸︷︷︸U1

. Projectively, P (ker(b)) is a point

and the quadric defined by b|U1 in P (U1) consists of two points. So the degenerate/singularconic definined by b in RP2 consists of two crossing lines.

Fig. 60

Proposition 1.12.4. A non-degenerate non-empty quadric Q ⊂ RPn determines thecorresponding bilinear form up to a non-zero scalar multiple.

Proof. Let b and b be two linear forms defining Q. Consider an orthonormal basis w.r.t.b, i.e. {e1, . . . , ep, f1, . . . , fq}, such that:

• b(ei, ei) = 1 for all i = 1, . . . , p;

• b(fj , fj) = −1 for all j = 1, . . . , q; and

• b(ei, ek) = b(fj , fl) = b(ei, fj) = 0 for all i, k = 1, . . . , p with i 6= k and j, l = 1, . . . , qwith j 6= l.

Consider the vectors ei ± fj . Then

b(ei ± fj , ei ± fj) = b(ei, ei)︸︷︷︸=1

± 2b(ei, fj)︸︷︷︸=0

+ b(fj , fj)︸︷︷︸−1

= 0

⇒0 = b(ei ± fj , ei ± fj) = b(ei, ei)± 2b(ei, fj) + b(fj , fj)⇔± 2b(ei, fj) = −b(ei, ei)− b(fj , fj)⇒b(ei, fj) = 0⇒b(ei, ei) = −b(fj , fj) .

So setting b(e1, e1) = λ 6= 0 implies that b(ei, ei) = λ for all i = 1, . . . , p and b(fj , fj) = −λfor all j = 1, . . . , q.

1.13 Orthogonal Transformations

Definition 1.13.1. Let b be a non-degenerate symmetric bilinear form on Rn+1. ThenF : Rn+1 → Rn+1 is orthogonal (wrt. b) if b(F (v), F (w)) = b(v, w)∀v, w ∈ Rn+1.

49

2 Hyperbolic Geometry

The group of orthogonal transformations for a bilinear form of signature (p, q) with p +q = n + 1 is denoted by O(p, q). If q = 0 we obtain the "usual" group of orthogonaltransformations O(n+ 1) = O(n+ 1, 0).

If Q ⊂ RPn is a non-degenerate quadric defined by b and F : Rn+1 → Rn+1 is anorthogonal transformation, then the map f : RPn → RPn with [x] 7→ f([x]) = [F (x)]maps the quadric onto itself: f(Q) = Q.Proposition 1.13.2. If the signature of a non-degenerate quadric Q is (p, q) with p 6= qand p 6= 0 6= q, then any projective transformation with f(Q) = Q is induced by anorthogonal transformation F w.r.t. b that defines Q.

Proof. Let b be the bilinear form defining Q and F the linear transformation definingthe projective transformation f . Then b(v, w) := b(F (v), F (w)) is another bilinear formdefining Q.According to the previous proposition there exists λ 6= 0 such that

b = λb.

If λ > 0 then 1√λF is an orthogonal transformation, since

b( 1√λF (v), 1√

λF (w)) = 1

λb(F (v), F (w)) = 1

λb(v, w) = b(v, w) .

So we need to show that λ > 0.If {e1, . . . , en+1} is an orthogonal basis of Rn+1 w.r.t. b then {F (e1), . . . , F (en+1} is alsoorthogonal with

• If b(ei, ei) = 1 then b(ei, ei) = λ

• b(ei, ei) = −1⇒ b(ei, ei) = −λ.

But the signature is invariant w.r.t. to change of basis (i.e. w.r.t. F). So since p 6= q wehave that the λ is positive.

If p = q (neutral signature), for example p = q = 2 for a quadric in RP3, then thereexists a projective transformation preserving the quadric not induced by an orthogonaltransformation. Let b(x, x) = x1

2 + x22 − x3

2 − x42. Then the map:

f : RP3 → RP3,

x1x2x3x4

7→x3x4x1x2

yields b(F (v), F (w)) = −b(x, x) = x3

2 + x42 − x1

2 − x22. So f preserves the quadric but

it is not induced by an orthogonal transformation F .


2.1 Lorentz Spaces

The vector space Rp+q with the bilinear form

〈x, y〉p,q =p∑i=1

xiyi −p+q∑i=p+1

xiyi

50

2.2 Hyperbolic Spaces

is the Lorentz space Rp,q. For our purposes the space Rn,1 with scalar product (non-degenerate symmetric bilinear form)

〈x, y〉n,1 =n∑i=1

xiyi − xn+1yn+1

is the most important case.Reminder The orthogonal transformations are denoted by

O(p, q) = {f ∈ GL(p+ q,R) | 〈f(v), f(w)〉p,q = 〈v, w〉 ∀v, w ∈ Rp+q} .

Example 2.1.1. R2,1 ; R3 with 〈x, y〉2,1 = x1y1 + x2y2 − x3y3.

Fig. 61

• 〈x, x〉2,1 = x21 + x2

2 − x23 < 0 (= −1 in picture) – time-like vectors

• 〈x, x〉2,1 = 0 – light-like vectors

• 〈x, x〉2,1 > 0 (= 1 in picture) – space-like vectors


Definition 2.2.1. The n-dimensional hyperbolic space is

Hn := {x ∈ Rn,1 | 〈x, x〉n,1 =n∑i=1

x2i − x2

n+1 = −1, xn+1 > 0} .

51


The length of a curve γ : [a, b]→ Hn is defined by

length(γ) =ˆ b

a

√〈γ′(t), γ′(t)〉n,1 dt .

Why is 〈γ′(t), γ′(t)〉n,1 > 0?The tangent space at a point p ∈ Hn is the orthogonal complement p⊥ of p, γ : [a, b]→ Hn

with γ(t0) = p.

γ ∈ Hn =⇒ 〈γ(t), γ(t)〉n,1 = −1=⇒ 2

⟨γ(t), γ′(t)

⟩n,1 = 0 .

The scalar product of p is 〈p, p〉n,1 = −1. Construct a basis {p, b1, . . . , bn} with 〈p, bi〉n,1 =0. Then the matrix of 〈·, ·〉n,1 has the form(

−1 00 ( )

)=⇒ signature(〈·, ·〉n,1 |p⊥) is (n, 0).

; On the tangent spaces of the hyperbolic space embedded in Rn,1 we obtain a positive-definite scalar product by restricting 〈·, ·〉n,1.−→ Riemannian manifolds.We can measure angles between curves:

Fig. 62

cosα =〈γ′1(0), γ′2(0)〉n,1√

〈γ′1(0), γ′1(0)〉n,1 〈γ′2(0), γ′2(0)〉n,1for γ1, γ2 in hyperbolic space.The Hyperbolic Line The hyperbolic line H1 ⊂ R1,1 is given by {x ∈ R1,1 | x2

1 − x22 =

−1, x2 > 0}. We want to measure the length of the curve γ in H1 connecting two pointsp, q.

52


Fig. 63

A parametrization of H1 ⊂ R1,1 is given by γ : R −→ H1, t 7−→(sinh tcosh t

). Then

〈γ(t), γ(t)〉1,1 = sinh2 t− cosh2 t = −1 .

Let p = γ(s1) and q = γ(s2) where s1 < s2. Then, since γ′(t) =(cosh t

sinh t),

d(p, q) = length(γ|[s1,s2]) =ˆ s2

s1

√〈γ′(t), γ′(t)〉1,1 dt =

ˆ s2

s1

√cosh2 t− sinh2 t︸︷︷︸

=1

dt

= s2 − s1.

; arc-length parametrization of H1.For p and q we have

〈p, q〉1,1 = 〈γ(s1), γ(s2)〉1,1= sinh s1 sinh s2 − cosh s1 cosh s2

= − cosh(s2 − s1)

=⇒ cosh(d(p, q)) = −〈p, q〉1,1d(p, q) = arcosh(−〈p, q〉1,1) .

Hyperbolic lines in Hn

Definition 2.2.2. A hyperbolic line in Hn is the non-empty intersection of Hn with a2-dim. subspace U of Rn,1.

53


Fig. 64

Proposition 2.2.3. The restriction of 〈·, ·〉n,1 to U has signature (+−).

Proof. Since U ∩Hn 6= ∅ there exists u ∈ U with 〈u, u〉 = −1. Let {u, v} be a basis of U .Define

w = v + 〈v, u〉u = v − 〈v, u〉〈u, u〉

u.

Then

〈u,w〉 = 〈u, v〉+ 〈v, u〉〈u, u〉︸︷︷︸−1

= 0

=⇒w ∈ u⊥

=⇒〈w,w〉 > 0

since the signature of 〈·, ·〉 |u⊥ is (+ + . . .+).

So we can identify Hn ∩ U ⊂ Rn,1 naturally with H1 ⊂ R1,1.

Theorem 2.2.4. The shortest piecewise continously differentiable curve in Hn connectingtwo points p and q is the hyperbolic line segment between them. It’s length is

d(p, q) = arcosh(−〈p, q〉).

Proof. We have to show that for an arbitrary cuve γ : [a, b] → Hn, with γ(a) = p andγ(b) = q we have:

length(γ) =ˆ b

a

√〈γ′(t), γ′(t)〉dt ≥ d(p, q),

with equality if and only if γ is the hyperbolic line segment. Let

f(x) = arcosh(−〈p, x〉).

Then the gradient of f is given by

∇xf = − 1√〈p, x〉2 − 1

p

54

2.3 Projective model of Hn

We are only interested in the part of the gradient that is in the tangent space of Hn. Sowe project the gradient onto the tangent space at x and obtain

π∇(x) = ∇xf −〈gradxf, x〉〈x, x〉

x

= ∇xf + 〈∇xf, x〉x

= − 1√〈p, x〉2 − 1

(p− 〈p, x〉x).

Observe that the projected gradient is always in the plane spanned by p and x and itpoints "away" from p. A simple calculation shows that 〈π∇(x), π∇(x)〉 = 1. Then usingthe fundamental theorem of calculus and the chain rule we obtain

d(p, q) = f(q)− f(p)

=ˆ b

a〈∇γ(t)f, γ

′(t)〉dt

=ˆ b

a〈π∇(γ(t)), γ′(t)〉dt the part in x⊥ does not matter!

≤ˆ b

a‖π∇(γ(t))‖︸︷︷︸

=1

‖γ′(t)‖dt Cauchy-Schwarz inequality

= length(γ).

Equality in the Cauchy-Schwarz inequality holds if and only if the tangent vector γ′ andthe projected gradient are parallel. With the above observation on the projected gradientwe obtain equality if and only if the curve is exactly the hyperbolic line segment.

Remark 2.2.5. Hyperbolic lines can be considered analogs of straight lines in Euclideanspace. They obviously violate Euclid’s parallel axiom, since for a given line and a pointnot on the line there exist many hyperbolic lines through the point not intersecting theline.


We derive a projective model of hyperbolic space in the following way:

Fig. 65

55


We define a map:

σ : Hn → Hnpr = {[x] ∈ RPn : 〈x, x〉n,1 < 0}

p 7→ [p] .

This map is a bijection and we can define a metric on Hnpr from the metric in Hn that

turns σ into an isometry:

cosh dpr([x], [y]) = |〈x, y〉|√〈x, x〉〈y, y〉

.

If we normalize the vectors x ∈ Rn,1 with 〈· , ·〉x, x = 0 such that xn+1 = 1, then we obtain⟨(u1

),

(u1

)⟩= 0⇔ (u, u)︸︷︷︸

Eucl. scalar product

= 1 .

So the affine image of the quadric Q = {[x] ∈ Rn,1 : 〈x, x〉n,1 = 0} is the unit sphereSn−1 = {u ∈ Rn : (u, u) = 1}. The points in Hn resp. Hn

pr are then naturally identifiedwith the inside of the unit sphere:

D = {u ∈ Rn : (u, u) < 1}.

If 〈x, x〉 < 0 then xn+1 6= 0. So we can normalize the representatives and define a mapfrom the projective model of hyperbolic space to the open ball D

Hnpr 3 [

(x

xn+1

)] = [

(u1

)] 7−→ u ∈ Rn with (u, u) < 1.

This is a bijective map and can be turned into an isometry by defining the following metricon the open ball:

cosh dK(u, v) = 1− (u, v)√(1− (u, u))(1− (v, v))

= cosh dpr([(u1

)], [(v1

)]).

Summing up until now we have three isometric models for hyperbolic space

(Hn, d)︸︷︷︸hyperboloid model

←→ (Hnpr, dpr)︸︷︷︸

projective model

←→ (D, dK)︸︷︷︸Klein (disc) model

Definition 2.3.1. A projective transformation f : RPn → RPn (n > 1) with f(Q) = Q,where Q = {[x] ∈ RPn : 〈x, x〉n,1 = 0} is a hyperbolic motion.

Theorem 2.3.2. Hyperbolic motions are isometries of Hnpr, i.e. they preserve the distances

dpr([x], [y]) = dpr(f([x]), f([y])), for all [x], [y] ∈ Hnpr.

They build the group PO(n, 1) = O(n, 1)/{± Id}.

Proof. The signature of the scalar product is not neutral for n > 1 so we can use thetheorem relating the preservation of the quadric with orthogonal transformations. Sothe hyperbolic motions are induced by orthogonal transformations which form the groupO(n, 1). Further these transformations map hyperbolic space to hyperbolic space, sincethey preserve the entire bilinear form and in particular its sign. As the metric is definedvia the scalar product (bilinear form) the hyperbolic motions are isometries.

56


Theorem 2.3.3. For two points A,B ∈ Hnpr let X, Y be the points of intersection of the

line AB with the quadric Q = {[x] : 〈x, x〉n,1 = 0}, such that {A, Y } and {B,X} separateeach other. Then

dpr(A,B) = 12 log cr(B,X,A, Y ) .

Fig. 66

Proof. Remember that cr(q, 1, 0,∞) = q. If we choose an affine coordinate such thatX = 0, A = 1 and Y =∞, then we see that cr(B,X,A, Y ) > 1, i.e. log cr(B,X,A, Y ) > 0.Parametrize AB by [a+ tb] where A = [a] and B = [b]. Solving

〈a+ tb, a+ tb〉n,1 = 0 (2.1)

yields the two solutions t1, t2 such that X = [a+ t1b] and Y = [a+ t2b]. We have t1t2 > 0and: If t1, t2 > 0 then t2 > t1, i.e. t2

t1> 1 – if t1, t2 < 0 then t2 < t1, i.e. t2

t1> 1. Looking

back at (2.1) gives us

0 = 〈a+ tb, a+ tb〉 = 〈b, b〉t2 + 2t〈a, b〉+ 〈a, a〉 = 〈b, b〉(t− t1)(t− t2) .

Comparing the coefficients yields

2〈a, b〉 = −〈b, b〉(t1 + t2) 〈a, a〉 = 〈b, b〉t1t2 ,

i.e

t1 + t2 = −2〈a, b〉〈b, b〉

t1t2 = 〈a, a〉〈b, b〉

.

Thuscosh dpr(A,B) = |〈a, b〉|√

〈a, a〉〈b, b〉= |t1 + t2|

2√t1 + t2

.

The cross ratio is

cr(B,X,A, Y ) = cr(∞, t1, 0, t2) = ∞− t1t1 − 0

0− t2t2 −∞

= t2t1.

57


This implies

cosh 12 log cr(B,X,A, Y ) = 1

2

(e

12 log t2

t1 + e−12 log t2

t1

)= 1

2

(√t2t1

+√t1t2

)

= 12|t2|+ |t1|√

t1t2= 1

2|t1 + t2|√

t1t2,

since t1t2 > 0.

2.4 Lines in the hyperbolic plane

Hyperbolic lines are secants of the unit ballObviously, to a line and a point in H2 there exist many lines through the point that donot intersect the line: This contradicts the Parallel Axiom.Using polarity, we can assign unit normal vectors to hyperbolic lines in H2. With ` =P (U), where U is a two-dimensional subspace with signature (+−) (where by the signatureof a subspace V we denote the signature of 〈· , ·〉 restricted to V ) we have n ∈ P (U⊥),where U⊥ has signature (+) – so we can choose n with 〈n, n〉 = 1. For every line thereexist two unit normal vectors defined via polarity with respect to 〈· , ·〉n,1. There is abijection between half planes and vectors n ∈ R2,1 with 〈n, n〉 = 1.

Proposition 2.4.1. Let `1, `2 ∈ H2 be two distinct lines in the hyperbolic plane with unitnormals n1, n2 (i.e. 〈n1, n1〉 = 1 and 〈n1, p〉 = 0 for all p ∈ `1). Then the following areequivalent:

(i) The two lines intersect.

(ii) span(n1, n2) has signature (++).

(iii) |〈n1, n2〉2,1| < 1.

Proof. (i)⇒(ii): Let `1 ∩ `2 =: x. Then span{x} has signature (−), since 〈x, x〉 = −1.Thus span{x}⊥ has signature (++) and n1, n2 ∈ span{x}⊥. This implies span{n1, n2} =span{x}⊥ has signature (++).(ii)⇒(i) can be shown similarly.(ii)⇔(iii): The matrix for 〈· , ·〉

∣∣span{n1,n2} is given by

A =(〈n1, n1〉〈n1, n2〉〈n1, n2〉〈n2, n2〉

)=(

1 〈n1, n2〉〈n1, n2〉 1

).

A is positive definite if and only if span{n1, n2} is (++), which is (ii). But also, positivedefiniteness if equivalent to 1 > 0 and 1− 〈n1, n2〉2 > 0, which is (iii).

Definition 2.4.2. Let `1, `2 be two intersecting lines defined by unit normals n1, n2. Theexterior angle α of these two lines is given by

cos α = 〈n1, n2〉 .

The interior angle α = π − α is given by

cosα = −〈n1, n2〉 .

58

2.4 Lines in the hyperbolic plane

Fig. 67

In particular, two lines are perpendicular, if 〈n1, n2〉 = 0.

Remark 2.4.3. The Euclidean angle of intersection differs from the hyperbolic angle definedabove.

Proposition 2.4.4.

(i) Given x ∈ H2 and a line ` ⊂ H2, there exists a unique line through x perpendicularto `.

(ii) If `1, `2 ∈ H2 are two non-intersecting lines (|〈n1, n2〉| > 1), then there exists aunique line perpendicular to `1 and `2.

Fig. 68

Proof. Homework.

Let `1, `2 ∈ H2 be two hyperbolic lines with unit normal vectors n1, n2: If |〈n1, n2〉| < 1,then `1 and `2 intersect.

59


Fig. 69

If |〈n1, n2〉| = 1, we call `1 and `2 parallel.

Fig. 70

If |〈n1, n2〉| > 1, we call `1 and `2 ultra-parallel.

60

2.5 Distances in the hyperbolic plane

Fig. 71

2.5 Distances in the hyperbolic plane

Proposition 2.5.1. The point on a line ` ⊂ H2 closest to a point x ∈ H2 is the intersec-tion point of ` with its unique perpendicular line through x.

Proof.

Fig. 72

W.l.o.g. assume x /∈ `. We have

〈np, x〉 = 〈np, xp〉 = 0 〈n, xp〉 = 0 〈np, n〉 = 0 .

We normalize np and xp such that 〈np, np〉 = 1 and 〈xp, xp〉 = −1. Now γ(s) = cosh(s)xp+sinh(s)np is an arclength parametrization of `. The distance from x to γ(s) is

cosh d(x, γ(s)) = −〈x, γ(s)〉 = −〈x, cosh(s)xp〉 − 〈x, sinh(s)np〉 = − cosh(s)〈x, xp〉 .

This is minimal for s = 0, γ(0) = xp. Thus the closest point is xp.

Proposition 2.5.2.

61


(i) For the distance of a point x ∈ H2 to a line ` ⊂ H2 defined by a unit normal n, wehave

sinh d(x, `) = |〈x, n〉| ,

where d(x, `) = d(x, xp) and xp is closest to x on `.

(ii) The distance between two non-intersecting lines `1, `2 ⊂ H2 with unit normals n1,n2 is

cosh d(`1, `2) = |〈n1, n2〉| ,

where d(`1, `2) is the minimal distance of two points on `1, `2, resp., or d(`1, `2) =d(x1, x2), where xi = ì ∩ `p and `p ⊥ `1, `2.

Proof. (i). Consider the arclength parametrization

γp(s) = cosh(s)xp + sinh(s)n

of `p.

Fig. 73

Nowx = γp(s0) with s0 = ±d(x, xp) .

Thus|〈n, x〉| = |〈n, cosh(s0)xp + sinh(s0)n〉| = |sinh(s0)| ,

as n is normalized to 〈n, n〉 = 1. This yields

sinh d(x, `) = sinh d(x, xp) = |〈n, x〉| .

(ii). Parametrize `1 as follows:

γ1(s) = cosh(s)x1 + sinh(s)np .

The distance of γ1(s) to `2 is

sinh d(γ(s), `2) = |〈γ1(s), n2〉| = cosh(s)|〈x1, n2〉|

62

2.6 Intersections of H2 with planes

since 〈n2, np〉 = 0. This is minimal for s = 0, i.e. x1 is closest to `2 on `1 Similarly, x2 isthe point closest to `1 on `2. Now parametrize `p.

x1 = cosh(s0)x2 + sinh(s0)n2 x2 = cosh(s0)x1 + sinh(s0)n1 .

Here, s0 = ±d(x1, x2) = ±s0. Thus with 〈n1, n1〉 = 1

0 = 〈n, x1〉 = cosh(s0)〈n1, x2〉+ sinh(s0)〈n1, n2〉 = cosh(s0) sinh(s0) + sinh(s0)〈n1, n2〉= sinh(s0)(cosh(s0)± 〈n1, n2〉)

and hence cosh d(`1, `2) = |〈n1, n2〉|.

2.6 Intersections of H2 with planes

Definition 2.6.1. An affine plane is determined by one equation

〈n, x〉2,1 = b

where n ∈ R2,1, n 6= 0, and b ∈ R.

Let n be time-like, 〈n, n〉 < 0, and n3 > 0. If 〈n, n〉 = −1, then n ∈ H2 and −〈n, x〉 =cosh d(x, n) shall be equivalent to 〈n, x〉 = b, i.e. the intersection of the plane given by〈n, x〉 = b is a hyperbolic circle around n with radius r such that cosh r = −b (b < 0).

Fig. 74

2.7 Stereographic projection

Definition 2.7.1 (stereographic projection). Denote the unit sphere in Rn+1 by Sn ={x ∈ Rn+1|(x, x) = 1

}and its north pole by N = en+1. The stereographic projection is a

map σ : Sn → Rn ∪ {∞} from N to the plane E ={x ∈ Rn+1|xn+1 = 0

} ∼= Rn throughthe equator:

σ(X) ={∞ , X = NlNX ∩ E , X 6= N

63


Analytically σ is given by

σ(X) = σ

x1

...xn+1

= 1

1− xn+1

x1...

xn+1

since

N + λ(X −N) = en+1 + λ

x1...

xn+1

=

y1...yn0

⇒ λ = − 1xn+1 − 1 = 1

1− xn+1, X 6= N.

Fig. 75

2.8 Sphere inversion

Definition 2.8.1 (sphere inversion). The sphere inversion iM,r : Rn ∪ {∞} → Rn ∪ {∞}in the sphere SM,r = {x ∈ Rn| ‖x−M‖ = r} is given by the following conditions:

1. X ′ lies on the ray −−→MX,

2. ‖M −X‖ ‖M −X ′‖ = r2,

3. M ←→∞.

64


Fig. 76

Properties of sphere inversions

(i) iM,r is an involution,

(ii) SM,r is fixed,

(iii) affine subspaces containing M are mapped onto themselves.

Lemma 2.8.2.

Fig. 77

Proof. Since

‖M −A‖∥∥M −A′∥∥ = r2 = ‖M −B‖

∥∥M −B′∥∥ ⇐⇒ ‖M −A‖‖M −B‖

= ‖M −B′‖

‖M −A′‖

we get that 4AMB and 4B′MA′ are similar. In particular:

]MA′B′ = ]ABM.]MAB = ]A′B′M.

65


Corollary 2.8.3. Circles through M are mapped to lines. Inparticular, the line is parallelto the tangent to the circle at M.

Fig. 78

Remark 2.8.4. All lines are thought to contain the point ∞ and are considered as circlesof infinte radius.

Corollary 2.8.5. The stereographic projection is the restriction of the sphere inversioniM,r with M = en+1 and r =

√2.

Proof.

Fig. 79

1. The equator of Sn is fixed.

2. Sn is mapped to a plane (Corollary) through the equator (1).

3. Both inversion and stereographic projection preserve the rays.

4. The intersection points of a ray with Sn and Rn are unique, hence iM,r

∣∣∣Sn

= σ.

5. N = M −→∞ by iM,r.

66


Theorem 2.8.6. An inversion in a sphere:

(i) maps spheres/hyperplanes to spheres/hyperplanes,

(ii) is a conformal map, i.e. it preserves angles.

Proof. (i) Consider the following sketch:

Fig. 80

Hence the angles α and β satisfy

(π − α) + β + π

2 = π ⇒ α− β = π

2 .

(ii)

Fig. 81

Remark 2.8.7. In general, the center of a sphere is not mapped to the center of its imageby an inversion.

67


Corollary 2.8.8. (i) The stereographic projection maps spheres through the north poleto hyperplanes and spheres not through the north pole to spheres.

(ii) All spheres/hyperplanes are images of spheres on Sn under σ (because i0,1 is aninvolution).

(iii) σ is conformal.

Formula for the sphere inversion Consider the second figure showing the general ideaof a sphere inversion. From the defining conditions we get that:

• X ′ = M + λ(X −M), λ < 0

•∥∥X ′ −M∥∥︸︷︷︸

X′−M=λ(X−M)

‖X −M‖ = r2

⇒ λ ‖X −M‖ ‖X −M‖ = r2

⇒ λ = r2

‖X−M‖2

Hence the sphere inversion is given by

iM,r = M + r2

‖X −M‖2(X −M).

2.9 Poincaré disc/ball model

Fig. 82

Consider the central projection with center s of Hn onto the plane xn+1 = 0. This mapcan be described by

σ : Hn → D = {u ∈ Rn : (u, u) < 1}

x 7→ 1xn+1 + 1

x1...xn

.

68

2.9 Poincaré disc/ball model

The inverse is given by

σ−1(u) = 11− (u, u)

(2u

1 + (u, u)

).

For s+ t

((u0

)− s

)=(

tu−1 + t

)is a point on ~su and

−�1 =⟨(

tu−1 + t

),

(tu−1 + t

)⟩n,1

= t2(u, u)− (t− 1)2 = t2(u, u)− t2 + 2t− �1

yields t = 0 or t(u, u) + 2 − t = 0 which is equivalent to t = 21−(u,u) . This implies

X =( 2

1−(u,u)u

−1 + 21−(u,u)

)= 1

1−(u,u)

(2u

1 + (u, u)

).

The map σ is the restriction of the following map to Hn:

x 7→ s− 2〈x− s, x− s〉n,1

(x− s) , for 〈x− s, x− s〉n,1 6= 0 .

Fig. 83

A point x with 〈x− s, x− s〉 < 0 is mapped to a point on the ray ~sx.The plane xn+1 = 0 is the orthogonal complement of s:⟨

s− 2〈x− s, x− s〉

(x− s), s⟩

= 〈s, s〉 − 2〈x− s, s〉〈x− s, x− s〉

= −1− 2(〈x, s〉 − 〈s, s〉)〈x, x〉 − 2〈x, s〉+ 〈s, s〉

= −1− 2(〈x, s〉+ 1)−2− 2〈x, s〉 = 0 .

This map is an involution on the set {x ∈ Rn,1 : 〈x − s, x − s〉 < 0 , xn+1 > −1} withσ(Hn) = D and σ(D) = Hn, where D ∼= D × {0}.

69


Fig. 84

Theorem 2.9.1. Consider two curves γ1, γ2 ⊆ D with γ1(0) = γ2(0) = p ∈ D anda := γ′1(0), b := γ′2(0). Let γ1 := σ ◦ γ1 and γ2 := σ ◦ γ2 be the corresponding curves inHn with γ1(0) = γ2(0) = P = σ(p) and A := γ′1(0), B := γ′2(0). Then:

〈A,B〉n,1 = 4(1− (p, p))2 (a, b) .

This impliescoshα = 〈A,B〉√

〈A,A〉〈B,B〉= (a, b)√

(a, a)(b, b),

where α is the angle of intersection of γ1 and γ2.

Proof. Calculate derivatives:

A = γ′1(0) = ddt

(s− 2〈γ1(t)− s, γ1(t)〉(γ1(0)− s)

)∣∣∣∣ t = 0

= −2 a

〈p− s, p− s〉+ 2 2〈p− s, a〉〈p− s, p− s〉2

(p− s) .

Similarly,B = −2 b

〈p− s, p− s〉+ 2 2〈p− s, b〉〈p− s, p− s〉2

(p− s) .

Now,

〈A,B〉 = 4〈a, b〉〈p− s, p− s〉2

−��

��

��

2 · 8〈p− s, a〉〈p− s, b〉〈p− s, p− s〉3

+(((

(((((((

(((((((

16〈p− s, a〉〈p− s, b〉〈p− s, p− s〉〈p− s, p− s〉4

= 4(〈p, p〉 − 2〈p, s〉+ 〈s, s〉)2 (a, b) = 4

((p, p)− 1)2 (a, b) .

Definition 2.9.2. The open unit ball D = {u ∈ Rn : (u, u) < 1} endowed with theRiemannian metric

gu(a, b) = 4(1− (u, u))2 (a, b) ,

70

2.10 Relation between Kleins and Poincaré model

where a, b are tangent vectors to D at u, is the Poincaré ball model of the hyperbolicspace. It is conformal, i.e. the hyperbolic angles equal the Euclidean angles.The length of a curve γ : [0, 1]→ D is given by

length(γ) =ˆ 1

0

√gγ(t)(γ′(t), γ′(t))dt .

2.10 Relation between Kleins and Poincaré model

Fig. 85

Claim: A is the vertical projection along en+1 of c(x) onto the hemisphere:

Ai = c(x)i for all i = 1, . . . , n .

For this, observe that A = s+ t(x− s) for some t and (A,A) = 1. Thuss+ t

x1...

xn+1 + 1

, s+ t

x1...

xn+1 + 1

= 1

yields

1 = 1 + 2

s, t x1

...xn+1 + 1

+ t2

∥∥∥∥∥∥∥ x1

...xn+1 + 1

∥∥∥∥∥∥∥

2

and hence0 = −2t(xn+1 + 1) + t2

(n∑i=1

x2i + x2

n+1 + 2xn+1 + 1).

Now by 〈x, x〉 = −1 we obtain

t2n∑i=1

x2i = t2(x2

n+1 − 1)

which gives us t = 0 ort = 2(xn+1 + 1)

2xn+1(xn+1 + 1) = 1xn+1

.

71


So A =(

0−1

)+ 1

xn+1

(x

xn+1 + 1

)∈ Rn,1. Thus

(A)i=1,...,n = 1xn+1

x .

The coordinates of c(x) are given by

c(x) =( 1xn+1

x

1

).

This proves the claim.

Proposition 2.10.1. We can identify the Poincaré and the Klein model by the followingconstruction:

Fig. 86

Consider the Klein and Poincaré disc and embed it in R3 at x3 = 0 inside the unit sphere(cf. second picture). To obtain a point in the Poincaré model from a point in the Kleinmodel, we do the following:

1. Project vertically, along e3, onto the upper hemisphere.

2. Use the stereographic projection from −e3 (south pole) to project the upper hemisphereback onto the disc.

72

2.11 The Poincaré half-space(/-plane) model

Corollary 2.10.2.

Fig. 87

The geodesics in the Poincaré model are circular arcs that intersect the unit sphere/circleorthogonally.

Proof. In the Klein model geodesics are secants. They are mapped to semicircles thatintersect the boundary orthogonally in the hemisphere. Since the steographic projectionis conformal and maps circles (or lines) the geodesics in the Poincaré disc are of the givenform.

Example 2.10.3. For two lines construct the unique perpendicular.

Fig. 88Remark 2.10.4. On the way we constructed another conformal model of the hyperbolicspace on the hemisphere.


Consider the following map:

73


Fig. 89

σ : x 7→ σ(x) = 2e2 + 4(x− 2e2, x− 2e2)(x− 2e2) .

We have σ(S1 + e2) = (R× {0}) ∪∞. Further, σ(D + e2) = H2−, where H2

− := {(x, y)T ∈R2 : y < 0}. By a similar calculation as the one in the proof of conformality of the Poincarémodel we can show the following:

Fig. 90

(A,B) = 16(p− e, p− e)2 (a, b)

by considering the map

σ : D → H21 , p 7→ 2e2 + 4

(p− e2, p− e2)(p− e2) = σ(p+ e2) .

For the hyperbolic metric on the Poincaré model we had

gp(a, b) = 4(1− (p, p))2 (a, b) = 4(p− e2, p− e2)2

16(1− (p, p))2 (A,B) = 1q2

2(A,B) ,

74


where q = σ(p).

Definition 2.11.1. The upper half space Hn+ := {x ∈ Rn : xn > 0} endowed with the

Riemannian metricgq(A,B) = 1

q2n

(A,B)

where q ∈ Hn+ and A,B are tangent vectors to Hn

+ at q is the Poincaré half-space modelof the hyperbolic space.

Properties: This model is conformal, i.e. the angles between curves correspond to theEuclidean angles. The geodesics are semi-circles or lines that intersect the boundaryorthogonally.

Fig. 91

Lengths/distances in H2+.

Two points P1 and P2 in the upper half-plane model either have the same x coordinateor not. If they share the x coordinate the geodesic through two points is a straight line.Otherwise it is a semi-circle. Let us calculate the hyperbolic distance between two pointsdepending on the two cases.Case 1.

Fig. 92

75


γ : [t1, t2]→ H2+ , γ(t) =

(0t

).

d(p1, p2) =ˆ t2

t1

√√√√gγ(t)

((01

),

(01

))dt =

ˆ t2

t1

1t

√√√√((01

),

(01

))dt = log t2

t1.

So the length/distance only depends on the ratio t2t1, not on the values.

Case 2.

Fig. 93

P1 = r

(cosα1sinα1

)P2 = r

(cosα2sinα2

).

Here,

γ : [α1, α2]→ H2+ , γ(α) = r

(cosαsinα

).

By this we get

d(p1, p2) =ˆ α1

α2

1sin tdt = log tan t

2

∣∣∣∣α2

α1

= logtan α1

2tan α2

2,

i.e. the distance depends on the angle, not on the radius.Remark 2.11.2. The following maps are isometries of H2

+:

• Horizontal translations,(xy

)7→(x+ cy

),

• scalings,(xy

)7→ λ

(xy

)for λ > 0.

76

2.12 Hyperbolic triangles

Proposition 2.11.3. The distance in the Poincaré halfplane model H2+ ⊂ C between two

points z1, z2 is given by

cosh d(z1, z2)− 1 = |z1 − z2|2

2 Im z1 Im z2.

Proof. "‘Pfff"’.


Let A, B, C be three points in the hyperbolic plane H2 not all on one line. Then thehyperbolic triangle is given by

4ABC = {λA+ µB + νC : λ, µ, ν > 0} ∩H2 .

Fig. 94

The triangle defined by A, B, C in the Klein model is4ABC = {λA+ µB + νC : λ, µ, ν ≥ 0, λ+ µ+ ν = 1} .

We label a triangle the following way:

Fig. 95

77


Observation: α+β+γ < π for any hyperbolic triangle. Consider a triangle in the Poincarédisc: Use a hyperbolic motion to move A to the center of the disc.

Fig. 96

The angles between the hyperbolic lines at B and C are smaller than the angles of theEuclidean triangle, hence α+ β + γ < π.Proposition 2.12.1. For α, β, γ > 0 with α+β+ γ < π there exists a hyperbolic trianglewith angles α, β, γ. Let δ > 0 such that α+ β + γ + δ = π. Then there exists a Euclideantriangle with angles α, β := β + δ

2 and γ := δ2 .

Proof.

Fig. 97

Construct a circle S through B and C such that ∠OBC = δ and O on the other side of athan A. The angles between S and b, c are β and γ, resp. Now, construct a circle centeredin A which intersects S orthogonally. Scaling this cunstruction such that the second circlehas radius 1 yields a triangle with the prescribed angles in the Poincaré model.

Theorem 2.12.2. The area of a hyperbolic triangle with angles α, β, γ is π− (α+β+γ).

Proof. Consider a triangle with one point at infinity, i.e. on the boundary of the disc inthe Poincaré model.

78


Fig. 98

The area of the triangle is

T (α, β, 0) =ˆ cosβ

− cosα

ˆ ∞√

1−x2

1y2 dydx =

ˆ cosβ

− cosα− 1y

∣∣∣∣∞√1−x2

dx =ˆ cosβ

cos(π−α)

1√1− x2

dx = π−(α+β) .

Fig. 99

Now,

T (α, β, γ) = T (α, γ+γ1, 0)−T (γ1, π−β, 0) = π−α−γ−γ1−(π−γ1−π+β) = π−(α+β+γ) .

Theorem 2.12.3 (Hyperbolic Pythagoras). Let 4ABC be a hyperbolic triangle with an-gles α, β and γ = π

2 . Then the lengths of the edges a, b and c satisfy

cosh c = cosh a · cosh b .

79


Proof.

Fig. 100

Calculate lengths:

cosh d(z1, z2)− 1 = |z1 − z2|2

2 Im z1 Im z2,

i.e.

cosh a = 1 + |B − C|2

2 ImB ImC= 1 + (k − 1)2

2k = k2 + 12k

cosh c = 1 + |eiϕ − ki|22k sinϕ = 1 +

(eiϕ − ki

) (e−iϕ + ki

)2k sinϕ = 1 + k2 + 1− 2k sinϕ

2k sinϕ = k2 + 12k sinϕ

cosh b = 12 + 12 · 1 · sinϕ = 1

sinϕ .

What is the relation of hyperbolic and triangles?Looking at "‘small"’ hyperbolic triangles (i.e. triangles with small edge lengths and area)hyperbolic triangles recovers Euclidean triangles. If the area is very small, then the sumof its interior angles is almost π. The hyperbolic Pythagoras’ Theorem becomes theEuclidean Pythagoras’ Theorem for small edge lengths: The Taylor series for cosh iscosh x = ∑ f (k)(0)

k! xk = 1 + 12x

2 + · · · . Thus cosh c = cosh a · cosh b becomes

1 + c2

2 + · · · =(

1 + a2

2 + · · ·)(

1 + b2

2 + · · ·),

hence1 + c2

2 = 1 + a2

2 + b2

2 + a2b2

4 + · · ·︸︷︷︸≈0

which gets c2 = a2 + b2 for small a, b and c.

80

2.13 Hyperbolic motions in H2+

2.13 Hyperbolic motions in H2+

Consider the following maps:

g : C→ C , z 7→ az + b

cz + d,

where ad− bc > 0 and a, b, c, d ∈ R. Here, g maps ∞ to ac and −d

c to ∞. Also,

g : C→ C , z 7→ −az + b

−cz + d.

• The maps g and g are bijective (the inverse is just the transformation corresponding

to(a bc d

)−1

).

• These maps preserve the upper half-plane: We have

Im g(z) = 12i

(az + b

cz + d− az + b

cz + d

)=

12i|cz + d|2

(ac|z|2 + adz − ac|z|2 − adz + bcz + bd− bcz − bd

)= ad− bc

2i|cz + d|2(z − z) = ad− bc

|cz + d|2Im z > 0 .

(similarly for g).

• These maps are isometries:

|g(z1)− g(z2)|22 Im g(z1) Im g(z2) = |cz1 + d|2|cz2 + d|2

(ad− bc)21

Im z1 Im z2

(ad− bc)2|z2 − z1|2

2|cz1 + d|2|cz2 + d|2

= |z1 − z2|2

2 Im z1 Im z2.

(again, similarly for g).

• Hyperbolic lines are mapped to hyperbolic lines.

Theorem 2.13.1. An arbitrary hyperbolic isometry of H2+ is of the form

z 7→ az + b

cz + dor z 7→ −az + b

−cz + d,

where, again, ad− bc > 0 and a, b, c, d ∈ R.

Proof. Let ϕ : H2+ → H2

+ be an isometry.

81


Fig. 101

We can construct a map g which is one of the following, such that g◦ϕ maps the imaginaryaxis onto itself and g ◦ ϕ(i) = i and g ◦ ϕ((0, i)) = (0, i):

• z 7→ z−uv with ϕ(i) = u+ iv,

• z 7→ vz−u ,

• z 7→ α z−xz−y where ϕ(0) = x, ϕ(∞) = y.

g ◦ϕ is an isometry, since g and ϕ are, also g ◦ϕ is the identity on the imaginary axis, as itis isometic and maps the imaginary axis preserving (0, i) and (i,∞). Now consider x+ iywith g ◦ ϕ(x+ iy) = u+ iv. We have d(ki, x+ iy) = d(g ◦ ϕ(ki), u+ iv) = d(ki, u+ iv) forall k ∈ R+. Thus

|ki− (x+ iy)|22ky = |ki− (u+ iv)|2

2kvand hence

(x2 + (k − y)2)v = (u2 + (v − k)2)y

for all k. This implies y = v and u = ±x. So g ◦ϕ is either the identity or it the reflectionwith respect to the imaginary axis, z 7→ −z. As ϕ = g−1 ◦g ◦ϕ, ϕ is of the given type.

82

3 Klein’s Erlangen Program

Geometry should be considered the study of invariants under a group of transformations.Different geometries with their spaces, transformation groups and invariants:Geometry space group invariantsEuclidean Rn x 7→ Ax+ b, A ∈ O(n), b ∈ Rn distances, angles, . . .Similarity Rn x 7→ λAx + b, λ 6= 0, A ∈ O(n),

b ∈ Rnangles, ratios of distances,. . .

Affine Rn x 7→ Ax+b, A ∈ GL(n,R), b ∈ Rn parallelity, ratios of dis-tances on a line

Projectivy RPn PGL(n+ 1) incidences of points andlines, k-planes, cross ratios

Hyperbolic Hnpr PO(n, 1): orthogonal transforma-

tions w.r.t. 〈· , ·〉n,1hyperbolic distances andangles

4 Möbius Geometry

Consider Rn ∪ {∞} and the following transformations:

Reflection in the hyperplane {x ∈ Rn : (x− p, n) = 0}.(missing picture)

x 7→ x′ = x− 2(p− x, n)(n, n) n .

Inversion in a sphere {x ∈ Rn : (x−m,x−m) = r}(missing picture)

x 7→ x′ = m+ r2

‖x−m‖2(x−m) .

Definition 4.0.2. A Möbius transformation is a composition of reflections in hyper-planes and inversions in hyperspheres. The group of Möbius tranformations is denoted byMoeb(n).

Example 4.0.3. Write a given transformation as a Möbius transformation.

1. Translation. This is the composition of two reflections.(missing picture)

2. Rotation. This is again the composition of two reflections.(missing picture)

3. Scaling. This is the composition of two sphere inversions in concentric spheres.(missing picture)

Remark 4.0.4. Inversions become reflections for r →∞. Hyperplanes can be considered ashyperspheres of infinite radius. Euclidean transformations and similarity transformationsare Möbius transformations.

Theorem 4.0.5. Möbius transformations are conformal and map hyperspheres and hy-perplanes to hyperspheres/-planes.

83

4 Möbius Geometry

Questions: What is Moeb(n)? Can it be represented as a subgroup of some PGL(N,R)?

Theorem 4.0.6. Let f : Rn∪{∞} → Rn∪{∞} be a bijective map that maps hyperspheres/-planes to hyperpheres/-planes. Then f is a Möbius transformation.

Proof. Case 1. Hyperplanes are mapped to hyperplanes only, in particular f(∞) = ∞.Then f maps lines to lines, as lines are the intersections of hyperplanes. Hence, f isan affine transformation. Further, f maps hyperspheres to hyperspheres and is thus asimilarity transformation. By that, f ∈ Moeb(n).Case 2. There exists p ∈ Rn such that f(∞) = p. Then apply a sphere inversion g in asphere with center p. Now, g ◦ f(∞) = ∞ and thus g ◦ f is a Möbius transformation byCase 1. Then also f = g ◦ g ◦ f ∈ Moeb(n).

4.1 Two-dimensional Möbius Geometry

The space is R2 ∪ {∞} ∼= S2 ∼= C ∪ {∞} ∼= CP1. This is the complex projective line. Thegroup of projective transformations is PGL(2,C). With respect to an affine coordinatethese projective transformations are of the form

z 7→ az + b

cz + d,

where ad− bc 6= 0, a, b, c, d ∈ C, −dc 7→ ∞ and ∞ 7→ a

c .Similarity transformations are of the form z 7→ az + b or z 7→ az + b, a, b ∈ C. Sphereinversion and reflection are of the form

z 7→ m+ r2

|m− z|2(m− z) = m+ r2

m− z= −mz + |m|2 + r2

m− z

andz 7→ −nz + p+ p

n

n,

resp.Möbius transformations are projective transformations of CP1 composed with z 7→ z.

Proposition 4.1.1. The Möbius transformations of CP1 = R2 ∪ {∞} are precisely themaps of the form

z 7→ az + b

cz + dor z 7→ az + b

cz + d

with a, b, c, d ∈ C and ad− bc 6= 0.

Proof. We have seen already that Möbius transformations are of the given form, since theabove transformations form a group. If we rewrite z 7→ az+b

cz+d it is obviously a Möbiustransformation:

z 7→ az + b

cz + d= a

c+ bc− adc(cz + d)

which is

zsimilarity7−−−−−−→ cz + d

inversion,reflection7−−−−−−→ 1

cz + d

similarity7−−−−−−→ bc− adc(cz + d) + a

c.

(for c = 0, the map is merely a similarity).

84

4.2 Relation between projective and Möbius transformations

4.2 Relation between projective and Möbius transformations

By identifying R2 ∪ {∞} with CP1 we can use some facts on projective transformationsto obtain properties of Möbius transformations.The cross ratio is preserved by projective transformations, hence for orientation preservingMöbius transformations f we have

cr(z1, z2, z3, z4) = cr(f(z1), f(z2), f(z3), f(z4)) .

For orientation reversing transformations g we obtain

cr(z1, z2, z3, z4) = cr(g(z1), g(z2), g(z3), g(z4)) .

For arbitrary three pairs zi 7→ wi, i = 1, 2, 3, there exists a unique projective transforma-tion f ∈ PGL(2,C) such that f(zi) = wi, i = 1, 2, 3. Smilarly for orientation reversingMöbius transformations. An arbitrary circle can be mapped to the real axis in C by aMöbius transformation, since three points define a unique circle (or line). If four pointslie on a circle, then they can be mapped to the real axis by a Möbius transformation.But Möbius transformations preserve cross ratios (up to complex conjugation) and thecross ratio of four real values is real. So the cross ratio of four points lying on a circle isreal, too. Conversely, for four points z1, z2, z3 and z4 with real cross ratio, there existsa unique Möbius transformation f with f(z2) = 0, f(z3) = 1 and f(z4) = ∞. So byf(z1) = cr(z1, z2, z3, z4) ∈ R, the four points f(z1), f(z2), f(z3) and f(z4) lie on a line andf−1 maps these points to a circle through z1, z2, z3 and z4.

4.3 Spherical model of Möbius geometry

Unify hyperspheres and hyperplanes by a stereographic projection

σ : Rn ∪ {∞} → Sn .

(missing picture)The stereographic projection is the restriction of a sphere inversion σ : Rn+1 → Rn+1.Hyperspheres in Rn are intersections of Rn × {0} and a unique orthogonal hypersphereS1. Rn × {0} is mapped to the unit sphere and S1 is mapped to a sphere orthogonal toSn. Hyperplanes in Rn are mapped to hyperspheres through the north pole on Sn.(missing picture)We will prove later that the Möbius transformations on Sn are generated by inversions inspheres orthogonal to Sn.Hyperspheres on Sn are also intersections of Sn with hyperplanes:(missing picture)

4.4 Projective model of Möbius geometry

The unit sphere Sn is the affine image of the quadric P (Rn+1,1) ⊇ Q = {[x] : 〈x, x〉n+1,1 =0}.(missing picture)We can identify hyperspheres in Sn with points outside the quadric. The hyperspherecorresponding to p with 〈p, p〉 > 0 is Sn ∩ p⊥.

85

4 Möbius Geometry

Great hyperspheres correspond to points p with pn+2 = 0.What about the Euclidean model?Introduce new coordinates:

e0 = 12(en+2 − en+1) e∞ = 1

2(en+2 + en+1) .

(missing picture)Then {e0, e1, . . . , en, e∞} is a basis, such that 〈e0, e0〉 = 〈e∞, e∞〉 = 0, 〈e0, ei〉 = 〈e∞, ei〉 =0, 〈ei, ej〉 = δij for all i, j ∈ {1, . . . , n}. Also,

〈e0, e∞〉 = 14(〈en+2, en+2〉 − 〈en+1, en+1〉) = 1

4(−1− 1) = −12 .

The intersection of the plane

{e0 + x+ x∞e∞ : x ∈ Rn, x∞ ∈ R}

with the light cone Ln+1,1 is a paraboloid, since

〈e0 + x+ x∞e∞, e0 + x+ x∞e∞〉 = 0

is equivalent to(x, x)n + 2x∞〈e0, e∞〉 = (x, x)n − x∞ = 0 .

So we have a bijective map Rn → Qn0 := Ln+1,1 ∩ {e0 + x+ x∞e∞ : x ∈ Rn, x∞ ∈ R}.

Proposition 4.4.1. The identification of Qn0 and Sn = Ln+1,1 ∩ {xn+2 = 1} along thegenerators of the light cone is the stereographic projection.

Proof. x ∈ Rn corresponds to [e0 + x + ‖x‖2e∞] and y =(

y′

yn+1

)∈ Sn corresponds to

[y + en+2]. Now,

[e0 + x+ ‖x‖2e∞] != [y + en+2] = [y′ + yn+1en+1 + en+2]= [y′ + yn+1(e∞ − e0) + e∞ + e0]= [y′ + (1− yn+1)e0 + (1 + yn+1)e∞]

=[e0 + y′

1− yn+1+ 1 + yn+1

1− yn+1e∞

]=[e0 + y′

1− yn+1+ 1− y2

n+1(1− yn+1)2 e∞

]

=[e0 + y′

1− yn+1+ ‖y′‖2

1− yn+1e∞

],

as ‖y‖2 = 1 implies ‖y′‖2 = 1 − y2n+1. But stereographics projection is given by y 7→

0 y′

1−yn+1.

The north pole N = (0, . . . , 0, 1) is mapped to

[N + en+2] = [en+1 + en+2] = [e∞] .

86

Indexanalytic geometry . . . . . . . . . . . . . . . . . . . . . . . . 1

Brianchon’s Theorem . . . . . . . . . . . . . . . . . . . 47

completequadrangle . . . . . . . . . . . . . . . . . . . . . . . . . 19quadrilateral . . . . . . . . . . . . . . . . . . . . . . . .19

conic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30, 36cross-ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Desargues’ Theorem . . . . . . . . . . . . . . . . . . . . 12dual projective space. . . . . . . . . . . . . . . . . . . .29

Fundamental Theorem of real projective ge-ometry . . . . . . . . . . . . . . . . . . . . . . . . . 25

general position . . . . . . . . . . . . . . . . . . . . . . . . . . 9

homogeneous coordinates . . . . . . . . . . . . . . . . 7

Koenigs cube . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Laguerre geometry . . . . . . . . . . . . . . . . . . . . . . . 1Lie geometry. . . . . . . . . . . . . . . . . . . . . . . . . . . . .1

Moebius pair . . . . . . . . . . . . . . . . . . . . . . . . . . . 23multi-ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

orthogonal complement . . . . . . . . . . . . . . . . . 40

Pappus’ Theorem. . . . . . . . . . . . . . . .10, 20, 39Pascal’s Theorem . . . . . . . . . . . . . . . . . . . . . . . 39pencil of conics . . . . . . . . . . . . . . . . . . . . . . . . . 38Pluecker geometry . . . . . . . . . . . . . . . . . . . . . . . 1polar line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41polar triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . 44pole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41projective

line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5subspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8transformation. . . . . . . . . . . . . . . . . . . . . .14

quadrangular set . . . . . . . . . . . . . . . . . . . . . . . . 19

signature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36Sylvester’s Theorem . . . . . . . . . . . . . . . . . . . . 36synthetic geometry . . . . . . . . . . . . . . . . . . . . . . . 1

tangent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41triangles in perspective . . . . . . . . . . . . . . . . . 11

87

Documents

Geometry 1 - TU Berlin · 1 Projective Geometry 1.1 Introduction to projective spaces Letusstartwithageneraldeﬁnitionforanarbitraryprojectivespace. Yetinthislecture