NOTES ON NON-EUCLIDEAN GEOMETRIESmoussong.web.elte.hu/bsm/neg_notes_19.pdf · ently) in...

NOTES ON NON-EUCLIDEAN GEOMETRIES

Gabor Moussong

Budapest Semesters in Mathematics

2019

Contents

0 Introduction 2

Part One: Preparatory Chapters 5

1 Affine Geometry 5

2 Spherical Geometry 15

3 Inversive Geometry 22

Part Two: Projective Geometry 38

4 Projective Space and Incidence 38

5 Coordinates in Projective Geometry 43

6 Cross-ratio and Projective Geometry of the Line 48

7 Conics 56

Part Three: Hyperbolic Geometry 64

8 The Projective Model 64

9 The Poincare Models 70

10 The Hyperboloid Model 75

11 The Hyperbolic Plane 82

1

2 GABOR MOUSSONG

0 Introduction

0.1 What is geometry?

In very general and vague terms, geometry is the study of space and shapes withmathematical rigor. It is a very broad subject within mathematics, and in theframework of a one-semester course we can only touch upon some selected topics,and present these from a specific point of view.

One of the goals in this course is to show how geometry finds its place among themany abstract structures of modern mathematics. We shall treat geometry throughthe powerful methods of other chapters of abstract mathematics: linear algebra,calculus, and groups. This point of view opens up several channels through whichclassical geometry is linked with today’s research in advanced mathematics, notablyin differential geometry, in topology, and in group theory.

The history of mathematics has produced many different geometric systems, theoldest of which is the familiar Euclidean geometry. Typically, such a geometricsystem – Euclidean or non-Euclidean – has the following types of characteristicfeatures:– basic objects, like points, lines, planes, circles, etc.,– transformations which move around these objects, and– measurements, like distance, area, angle, which remain invariant under these

transformations.Our treatment of geometry will investigate these features in certain classical geome-tries. Special emphasis will be given to the role of transformations, and to the wayhow the structure of these transformations can distinguish between varios types ofgeometry.

0.2 What makes Euclidean geometry Euclidean?

More than two thousand years ago Euclid presented the whole body of geometricknowledge of his era in a systematic way. His axiomatic treatment of geometry seta standard for treating mathematics ever since. The general structure of such anaxiomatic theory starts with some undefined basic notions, called primitive terms(in geometry, such are ponts, lines, and the like), and with some statements, calledaxioms, about these objects, which are accepted as truth without proof. The theoryis then developed through definitions, theorems, and rigorous proofs, all based eitheron the axioms or on earlier theorems.

The theory built on Euclid’s axioms eventually results in an essentially uniquemathematical structure, the so-called Euclidean space. This is the mathematicalcounterpart of our physical space given by everyday experience. A geometric systemis Euclidean if it is isomorphic to the one defined by Euclid’s axioms.

The most famous example of non-Euclidean geometry was discovered in connectionwith one particular axiom of Euclidean geometry, the so-called parallel postulate(stated here in an equivalent form which is different from Euclid’s original):

• Given any line L and any point P not on L, there exists only one line throughP , within the plane containing L and P , which does not intersect L.

NOTES ON NON-EUCLIDEAN GEOMETRIES 3

It was naturally expected that Euclid’s axioms express unquestionable truth aboutthe physical space around us. Since this particular axiom seemed a lot less obviousthan the others, many geometers through the ages tried to deduce the parallelpostulate from the rest of the axioms, thereby making it unnecessary to use. Allthese attempts failed, and finally (in the last third of nineteenth century) it turnedout that such a proof is impossible. Its negation, taken as a substitute of the parallelpostulate, plus the rest of Euclid’s axioms, form a new system of axioms, and definea meaningful geometric system, now called hyperbolic geometry.

Clearly, the parallel postulate makes it possible to define the concept of parallellines in Euclidean geometry. If it fails, then the whole issue of parallelism is com-pletely different. Therefore, it is useful to be aware what parts, which theorems oftraditional elementary Euclidean geometry really depend on the parallel postulate.Here are some notable examples:

– The angle sum theorem. Recall that the standard proof that the angles of atriangle add up to 180 degrees starts with drawing a line parallel to one side.In fact, the angle sum theorem turns out equivalent to the parallel postulate.Therefore, if angles make sense in a non-Euclidean geometric system, then thesum of angles within a triangle is expected different from 180 degrees.

– The concept of vectors, more precisely, of ‘free’ vectors. In Euclidean geometry,vectors are freely translated anywhere; this cannot be done without parallel lines.

– Translations in Euclidean geometry usually are defined using parallel lines, orsimply by vectors. It is possible to define translations under non-Euclideancircumstances but this must be done carefully avoiding any reference to vectorsor parallelism. The resulting non-Euclidean translations behave quite differentfrom what we are used to in elementary geometry.

– The concept of similarity. In Euclidean geometry there exist similar figuresof different size. The standard procedure to construct two similar but non-congruent triangles uses the intercept theorem (on two lines intercepted by apair of parallel lines), therefore, existence of non-congruent similar figures alsodepends on parallelism.

– Use of Cartesian coordinates. Finding coordinates of a point involves drawingparallels to the axes, therefore coordinates cannot be used the same way if thegeometry is non-Euclidean.

– Some further theorems and methods of elementary Euclidean geometry dependon the above, therefore are not expected to work (or are expected to work differ-ently) in non-Euclidean geometry. Such are, for example, the theorem of Thaleson angles inscribed in a semicircle, the theorem of Pythagoras, the trigonometriclaws, and some area formulas for triangles.

With all these concepts missing, or different from usual, non-Euclidean geometriesare quite different from the familiar Euclidean geometry. Even so, much of our workwill be devoted to showing that large part of traditional geometry can be salvagedin the non-Euclidean setting.

0.3 Topics covered in the course and in these notes

Our main goal is to learn the methods, formalism, and mathematical machinery thatis necessary to understand various geometric systems as mathematical structures.

4 GABOR MOUSSONG

We shall not use the axiomatic method at all; actually, no axiom other than theparallel postulate will ever be mentioned. As a starting point, we rely on theintuitive concept of Euclidean space and plane, and we take the usual theorems ofelementary Euclidean geometry for granted. Other types of geometries will laterbe based on concrete definitions and constructions carried out on the basis of, ordirectly within, Euclidean geometry.

The first part of these notes basically stays in the realm of Euclidean geometrywhile introduces some background material for later use. Parts of this are affinegeometry and inversive geometry, which themselves may be regarded as examplesof non-Euclidean geometry. Spherical geometry is briefly introduced as anothereasy non-Euclidean example. The second part covers projective geometry which,besides being an interesting, entertaining, classical piece of mathematics on its ownright, serves as the main technical background for hyperbolic geometry. The thirdpart is devoted to hyperbolic geometry which is introduced through three types ofmodels, each using a different type of mathematical apparatus.

All these types of geometries could be worked out in arbitrary dimensions usingessentially the same methods. For simplicity, we restrict ourselves to the two-dimensional case, that is, we only work with the affine plane, the inversive plane,the two-dimensional sphere, the projective plane, and the hyperbolic plane.


Part One: Preparatory Chapters

1 Affine Geometry

Affine geometry collects those features of Euclidean space which are immediate ap-plications of the linear algebra of vectors in space. Only the vector space operations(that is, addition of vectors, and multiplication of vectors by scalars) are considered.Most notably, distance and angle play no role whatsoever in affine geometry.

1.1 Vectors in Euclidean geometry

We start with a brief review of the definition and the structure of vectors in Eu-clidean space.

Let E denote Euclidean space as the set of points. Ordered pairs (A,B) of points(that is, elements of E×E) are naturally identified with directed straight segments.Two such are equivalent if a suitable translation takes one to the other.

• A vector in E is an equivalence class of directed segments. Notation: write

v =−→AB if the vector v is represented by (A,B). The set of vectors is denoted V .

• Operations on vectors: u + v and λu (where u,v ∈ V , λ ∈ R) are defined in astraightforward way (choosing representatives).

• Fact: these operations turn V into a real vector space of dimension 3.

Remarks: (1) If E were a plane or a line only, we would get dimV = 2 or dimV = 1,respectively.

(2) At the moment we are not interested in other operations on vectors (products,notably). We want to study the part of geometry which relies on the vector spacestructure of V only.

• Summary: We have a set E, a real vector space V , and a “structure map”

E × E → V , denoted as (A,B) 7→ −→AB, subject to the following two conditions:

(1) For any fixed O ∈ E the map E → V , A 7→ −→OA, is bijective;

(2)−→AB +

−→BC =

−→AC for all A,B,C ∈ E.

Remark: These properties serve as the basis for the concept of an abstract affinespace. The set E is called affine line, affine plane, or affine space according asdimV = 1, 2, or 3. Note that distance and angle play no role here.

In what follows, for simplicity, everything is formulated, defined, stated, etc., for theplane. As an exercise, the reader can modify the relevant data and constructionsto get analogous concepts and theorems for three-dimensional or one-dimensionalgeometry.

6 GABOR MOUSSONG

1.2 Coordinate systems

Distance and angle should play no role. Therefore, when a coordinate system isintroduced in the plane, the two axes need not be perpendicular, and the units maybe chosen independently for each axis.

Formally speaking, a coordinate system is a map x which assigns coordinates to eachpoint: x : E → R2, P 7→ x(P ) = (x1, x2). (We call R2 the standard coordinateplane.)

A coordinate system in E is equivalent to a choice of an origin in E plus a basisin V .

In practice, when a coordinate system x is chosen, a point P usually is identifiedwith its image x(P ), i.e., we write P = (x1, x2).

Given two coordinate systems x, y, either in the same plane or in two distinctplanes, there is a natural map associated with them: two points correspond if theyhave the same coordinates. That is, the point P (with coordinates x(P )) of thefirst plane is mapped to the point Q of the second plane for which y(Q) = x(P )holds. (In formal terms, this is the map y−1 ◦ x.)Exercise: if the two planes coincide (and x is the the coordinate system taken first),verify that in terms of x this map is given by a formula x 7→ Mx + v where M isa 2 × 2 invertible matrix, and v is a fixed vector (independent of x). Such mapsserve as prototypical examples of the type of transformations we discuss next.

1.3 Affine transformations

• Definition (affinity). Let f be a map between two (not necessarily distinct)planes. The map f is called an affine transformation (or affinity) if f is expressedin terms of coordinates (that is, as a map f : R2 → R2) as f(x) = Mx + v,where M is any invertible matrix, and v is any vector.

(Strictly speaking, the formula Mx + v only defines R2 → R2 affinities. If Pand Q are planes equipped with coordinate systems x and y, respectively, thenf : P → Q is an affinity if the triple composition y ◦ f ◦ x−1 : R2 → R2 is.)

Note that an affine transformation f(x) = Mx + v has a “linear part”, given bythe matrix M , and a “translation part”, given by v. The calculation f(b)−f(a) =(Mb+ v)− (Ma+ v) = M(b− a) shows that the way f acts on vectors betweenpairs of points is determined only by the linear part, i.e., by the matrix M .

• Definition (affine group). It is routine to verify that the inverse of an affinetransformation is an affine transformation, and that the composition of two affinetransformations is an affine transformation. It follows that affine transformationsof a plane E to itself form a group under compositon. This group is called theaffine group of E, and is usually denoted as Aff(E). If E is the coordinateplane R2, we simply write Aff(2) for Aff(R2). Aff(1) and Aff(3) are definedanalogously.

Remark: The fact whether a map is an affine transformation or not does not de-pend on the choice of coordinates. This follows from the above observations aboutcompositions.


• Examples of affine transformations:

– All Euclidean congruences are affine transformations. Indeed, if the map f be-tween two planes is a Euclidean congruence (that is, a bijective map whichpreserves all relevant geometric properties such as collinearity, distance, angle,area, etc.), then a coordinate system is taken to another coordinate system byf , and the map f simply coincides with the natural map associated with thispair of coordinate systems.

– All homotheties are affine transformations. A homothety (or “central similarity”)h : E → E with center at point O ∈ E takes any point A ∈ E to the point

A′ = h(A) for which−−→OA′ = λ

−→OA with a fixed nonzero scalar λ ∈ R. If the

origin of a coordinate system is at O, then the formula for h is h(x) = λx. Thelinear part of h is given by the scalar matrix λI (where I denotes the identitymatrix).

– Corollary: all similarity transformations are affine. Indeed, any similarity trans-formation can be represented as a composition of a Euclidean congruence witha homothety.

– Parallel projections:

Given two planes E and E′ in three-dimensional space, and a line L not parallelto either, one can define a map E → E′ as follows. For any point A ∈ E there

exists a unique point A′ ∈ E′ with−−→AA′ ‖ L. The map E → E′, A 7→ A′ (A ∈ E),

is called parallel projection from E to E′ in the direction of L. It is not hard toverify that any parallel projection is an affinity. If E ∦ E′, then this affinity ingeneral is not a Euclidean congruence between the planes (not even a similarity).

1.4 Affine invariants

Roughly speaking, affine geometry studies properties which are invariant (i.e., re-main unchanged) under affine transformations. For example, distance, angle, areaare not affine invariants, but collinearity, parallelism, simple ratio (see below) are.

• Theorem. Any affine transformation takes lines to lines, parallel lines to parallellines.

Proof: Any line has a parametric representation tu + a (where u 6= 0). Underan affinity Mx+ v its image is t(Mu) + (Ma+ v) which is a line again. If twolines are parallel, then the same direction vector u can be used for both; thenthe image lines also have the same direction vector Mu.

• Definition (simple ratio). Let A, B, C be three collinear points with A 6=B 6= C. Write

−→AC = λ

−→CB, then this uniquely determined real number λ is

called the simple ratio of points A, B, C. Notation: (ABC) = λ.

8 GABOR MOUSSONG

Remark: If C is between A and B, then (ABC) is indeed the ratio in which Cdivides the segment AB. For instance, (ABC) = 1 if C is the midpoint. If C isoutside the interval from A to B, then (ABC) is negative.

Immediate observations:(1) If A and B are fixed, then (ABC) can take all real number values except −1.(2) (ABC1) = (ABC2) =⇒ C1 = C2.

Both follow from the formula (ABC) = x/(1− x) where A = 0, B = 1, and C = x.

• Lemma. Let a =−→OA and b =

−→OB be linearly independent, and consider

c = αa+ βb =−→OC. If C is collinear with A and B, then α+ β = 1. If, further,

C 6= A, then (ABC) = β/α.

Proof:−→AC = (α − 1)a + βb,

−→CB = −αa + (1 − β)b. If these two vectors are

linearly dependent, then (α − 1)(1− β)− β(−α) = 0, which implies α + β = 1.

Writing 1 − α for β we have−→AC = −βa + βb and

−→CB = −αa + αb, which

imply−→AC = (β/α)

−→CB.

• Theorem. Any affine transformation preserves simple ratio.

Proof: Given A, B, C, choose O off the line, and apply the lemma.

• Corollary. Let f be an affine transformation of a plane E to itself.(1) If f fixes two distinct points A and B, then it fixes all points of line AB.(2) If f fixes three non-collinear points, then f is the identity map of E.

Proof: (1) follows immediately from the preceding theorem. (2) follows from (1)since by (1) all three pairwise connecting lines are pointwise fixed, and throughany given point of E there can be drawn a line which intersects at least two ofthese lines.

1.5 The fundamental theorem

• Definition (affine frame). An affine frame in a plane E is a choice of threenon-collinear points A0, A1, A2 ∈ E.

There is a straightforward bijective correspondence between affine frames and co-

ordinate systems in E: let A0 be the origin,−−−→A0A1 and

−−−→A0A2 be the basis vectors:

The following theorem is sometimes called the fundamental theorem of affine ge-ometry (in dimension two).

• Theorem. If E and E′ are (not necessarily distinct) planes, A0, A1, A2 ∈ Eand A′

0, A′1, A

′2 ∈ E′ are affine frames in each, then there exists a unique affine

transformation f : E → E′ such that f(Ai) = A′i (i = 0, 1, 2).


Proof: Existence: the natural map associated with the two coordinate systems isjust a required affinity. Uniqueness: suppose both f and g are as required, theng−1 ◦ f : E → E is an affinity fixing three non-collinear points. By Corollary (2)above, g−1 ◦ f = idE , that is, f = g.

Remark: In the case E = E′ the statement of the theorem is often phrased asfollows: the affine group Aff(E) acts simply transitively on the set of affine framesin E.

Two figures are called affinely equivalent if a suitable affine transformation mapsone onto the other. In this case the two figures are not distinguishable by affinemeans. For example, any two triangles are affinely equivalent. Affinely equivalentfigures are considered “equal” in affine geometry, just like congruent figures areconsidered equal in Euclidean geometry.

1.6 Example: axial affinities

Throughout this section, E denotes a plane. We consider a special type of affinetransformations in E. When an affinity E → E is given, we follow the generalpractice that the symbol P ′ denotes the image of any point P ∈ E.

• Definition (axis, axial affinity). Let f : E → E be an affine transformation.A line L ⊂ E is called an axis of f , if it is pointwise fixed, i.e., A′ = A for allA ∈ L. An affinity is called axial if it has an axis.

By preservation of simple ratio, if an affinity of E fixes two distinct points, thenit is axial. By the fundamental theorem, a non-identity axial affinity can have nofixed points outside its axis. (In particular, the axis is unique.)

• Lemma. Let f : E → E be an axial affinity with axis L, f 6= idE . Then alllines PP ′, where P /∈ L, are parallel to each other.

Proof: Suppose P,Q /∈ L, P 6= Q. If line PQ meets L at a point, say A, thenby preservation of simple ratio, (APQ) = (AP ′Q′). Now PP ′ ‖ QQ′ follows byelementary geometry (say, similar triangles). If PQ ‖ L, then use a third pointR not on PQ or L to deduce PP ′ ‖ RR′ ‖ QQ′.

• Definition (shear, strain). Given a non-identity axial affinity, by the lemmaabove the parallelism class of lines PP ′ does not depend on the actual choice ofthe point P , which means that it is characteristic of the transformation itself.This parallelism class is called the direction of the axial affinity. If this directionis parallel to the axis, then the axial affinity is called a shear, otherwise it iscalled a strain.

10 GABOR MOUSSONG

If an axial affinity with axis L is a strain, then for P /∈ L and B = L∩PP ′ there

is a uniquely defined λ ∈ R such that−−→BP ′ = λ

−→BP . This λ does not depend on

the choice of P (again, by elementary geometry), and is called the ratio of thestrain. (Note that shears do not have a ratio.)

If a coordinate system is chosen so that the first coordinate axis coincides with the

axis of the affinity, then a shear is given by a matrix

(1 a0 1

)(where a ∈ R), and

a strain with ratio λ is given by

(1 00 λ

)(where λ ∈ R, λ 6= 0). As a familiar

example, any Euclidean reflection is a strain with ratio −1. In this example thedirection of the strain is perpendicular to the axis.

1.7 Structure of the affine group

Recall that Aff(2) denotes the affine group in dimension 2, that is, the group of allaffine transformations R2 → R2 under the operation of composition of maps.

We can recognize two important subgroups of Aff(2). One is the subgroup oftranslations. An affinity f(x) = Mx+ v is a translation if and only if M = I, theidentity matrix. Translations form a subgroup isomorphic to the additive groupR2 of vectors. Under this isomorphism the translation f(x) = x + v correspondsto the vector v ∈ R2. It is common practice to identify translations with vectors,therefore we may consider R2 ≤ Aff(2) as the subgroup of translations.

The other subgroup is formed by all linear maps within Aff(2), that is, affinitiesgiven by a matrix only (with zero translation part). These are precisely thoseelements of Aff(2) which keep the origin fixed. This subgroup simply is the mul-tiplicative group of invertible 2 × 2 matrices, called the general linear group indimension 2, denoted as GL(2,R).

It is clear that these two subgroups are disjoint, that is, R2 ∩ GL(2,R) consistsonly of the identity map. Further, since by definition any affinity is a product ofa linear map and a translation, these two subgroups together generate the wholeaffine group.

• Claim: R2 is a normal subgroup in Aff(2).

Indeed, consider the map from Aff(2) to GL(2,R) which assigns its linear part toany affine transformation. Direct calculation shows that this map is a homomor-phism of groups. The translation subgroup is the kernel of this homomorphism,therefore it is normal. (Exercise: check that selecting the translation part of anaffinity is not a homomorphism of groups, and that GL(2,R) is not a normal sub-group in Aff(2).)

Remark: These observations show that Aff(2) is a so-called semidirect productof the two subgroups R2 and GL(2,R). In general, we say that a group G is asemidirect product of its subgroups N and H (notation: G = N ⋊ H) if (i) N isa normal subgroup, (ii) N ∩ H = 1, and (iii) NH = G. (Note that a semidirectproduct specializes to the direct product if H is also normal.) Affine geometryyields a nontrivial example of a semidirect product: Aff(2) = R2 ⋊GL(2,R).

• An affine transformation f is called orientation preserving if, when written asf(x) =Mx+ v, the matrix M has positive determinant. (Orientation reversingotherwise.)


Remark: Whether an affinity of a plane to itself is orientation preserving or notdoes not depend on the choice of the coordinate system. Indeed, if matrix T is thechange of basis, then the linear part of the affinity with respect to the new basis isTMT−1, and det(TMT−1) = detM .

Orientation preserving affinities form an index two subgroup in Aff(2). All transla-tions, all homotheties, and all shears, are examples of orientation preserving affini-ties of the plane. Among Euclidean congruences translations and rotations areorientation preserving, reflections are orientation reversing.

1.8 Euclidean congruences as affinities

Some affine transformations f(x) = Mx + v are actually Euclidean congruences.It is natural to ask how this fact is seen from the algebraic data of f , i.e., from Mand v.

• Throughout this section we assume that the coordinate system x is Cartesian,that is, the associated basis in V is orthonormal. (Recall that a basis is calledorthonormal if it consists of pairwise ortogonal, unit length, vectors.)

Clearly the map f is a Euclidean congruence if and only if the image of the coordi-nate system under f is also a Cartesian system. Since the image basis vectors areprecisely the column vectors of the matrixM , this condition simply means that thecolumn vectors of M form an orthonormal basis.

The following lemma gives several equivalent characterizations of this property ofmatrices. For our purposes only the n = 2 and n = 3 cases are necessary, but it isworthwhile to state te lemma for general n. Recall that for n-dimensional vectorsx = (x1, . . . , xn), y = (y1, . . . , yn) ∈ Rn dot product and norm are defined as

x · y = x1y1 + . . .+ xnyn , and ‖x‖ =√x · x =

√x21 + . . .+ x2n .

• Lemma. LetM be an n×n real matrix. The following conditions are equivalent:

(i) Column vectors of M form an orthonormal basis;

(ii) M⊤M = I;

(iii) M preserves dot products: (Mx) · (My) = x · y for all x, y ∈ Rn;

(iv) M preserves norms: ‖Mx‖ = ‖x‖ for all x ∈ Rn.

The proof is mostly a routine exercise in linear algebra.

• Definition (orthogonal matrix). A real square matrix is called orthogonal ifeither one (therefore all) conditions of the lemma hold for M . The set of n× northogonal matrices is denoted O(n). Clearly O(n) is a group under matrixmultiplication.If no coordinate system is specified in a Euclidean vector space V , conditions(iii) or (iv) of the lemma can still be used to define the coordinate-free conceptof orthogonal linear maps from V to V , and the orthogonal group O(V ). Alinear map is orthogonal if and only if, with respect to an orthonormal basis, itis represented by an orthogonal matrix.

Using this terminology we can say that an affine transformation f(x) =Mx+ v isa Euclidean congruence if and only if the matrix M is orthogonal.

12 GABOR MOUSSONG

Remark: It follows that the group E(n) of Euclidean congruences of Rn is asemidirect product of the translation subgroup and the orthogonal group: E(n) =Rn ⋊O(n).

If M is an orthogonal matrix, (detM)2 = (detM⊤)(detM) = det(M⊤M) =det I = 1, that is, detM is either +1 or −1. Those with positive determinantare called special orthogonal, and form a subgroup SO(n) of index 2 in O(n).

Orthogonal matrices in dimension 2: any 2× 2 orthogonal matrix is

either

(cosα − sinαsinα cosα

)or

(cosβ sinβsinβ − cosβ

)where α, β ∈ R .

By inspection, the first matrix (with determinant 1) is rotation about the originthrough angle α, the second (with determinant −1) is reflection in the line throughthe origin which makes angle β/2 with the first axis.

1.9 Conics

A general quadratic equation in two real variables x and y is

Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0 ,

where the first three of the six coefficients are not all zero. Curves defined by suchequations are called quadratic curves in the plane.

• Fact: In a Cartesian system of coordinates a quadratic curve typically is a conic(i.e., an ellipse, a parabola, or a hyperbola). It may also “degenerate” to a line,to a pair of lines, to a point, or to the empty set.

If the system of coordinates is moved so that it fits the conic the best, then theequation simplifies as either

x2

a2+y2

b2= 1 (ellipse), y2 = 2px (parabola), or

x2

a2− y2

b2= 1 (hyperbola) .

Suppose an affine transformation f is applied to a quadratic curve. Coordinates ofa point x satisfy the equation of the transformed curve if and only if the coordinatesof f−1(x) satisfy the original equation. Therefore, the equation of the transformedcurve is obtained by substituting the coordinates of f−1(x) into the equation. Sincef−1 is an affine transformation, (inhomogeneous) linear expressions of the variablesare substituted into the original quadratic equation. So, the resulting equationis again quadratic. Corollary: affine transformations always transform quadraticcurves into quadratic curves.


A quadratic curve is a conic if it has at least two points, and contains no line.Therefore, the image of a conic under an affine transformation is always a conic.One may refine this even further by observing that among conics precisely theellipses are bounded, and precisely the hyperbolas are disconnected. Both theseproperties are preserved by affinities, therefore we have the following corollary:

• Affinities transform ellipses to ellipses, parabolas to parabolas, and hyperbolasto hyperbolas.

For instance, an ellipse is always mapped to an ellipse by any shear mapping of theplane, or by any parallel projection between two planes.

1.10 Complex affine line

Let E be a Euclidean plane. We assume that E is equipped with orientation, thatis, it is determined which rotations are considered positive. Note that this structureis precisely what is needed to call E an affine line over the complex field C. Indeed,any choice of two distinct points in E, called 0 and 1, identifies E with the fieldof complex numbers the usual way. (For instance, i ∈ C is obtained by rotating

1 about 0 through +90 degrees.) Vectors−→AB, with A,B ∈ E, become complex

numbers. Thus, by fixing 0 and 1 (an affine frame), we introduce a one-dimensionalcoordinate system in E where any coordinate now is a complex number.

Complex affinities of E = C are the maps f : C → C of the form f(z) = mz + vwhere m, v ∈ C, m 6= 0.

Some special cases:

When m = 1, the map f(z) = z + v is a translation.

When v = 0, the map f(z) = mz is a “complex homothety”. If the coefficient m isreal, then this transformation is a homothety of E in the Euclidean sense. If m ∈ Cand |m| = 1, then f is a rotation. In the general case, f is a combination of thetwo, called a “spiral similarity”.

It follows that all complex affinities of the complex affine line E are actually ori-entation preserving similarity transformations of the Euclidean plane E. It is nothard to see that the converse is also true.

Remark: For a plane E we now have two different ways to interpret E in theframework of affine geometry: it may be a (two-dimensional) affine plane overthe reals, or a complex affine line (which is one-dimensional over C). These twointerpretations differ in several aspects. For example, there are considerably moreaffinities in the first case than in the second.

1.11 Exercises

1. An affine transformation f : R2 → R2, (x′, y′) = f(x, y), is given by the formulas

x′ = 3x+ y + 2

y′ = x+ y + 1 .Draw the image of the unit square of the coordinate system. What is the image of the rectan-gular grid formed by the lines x = integer, y = integer?

2. Given two affine transformations f(x) = Mx+ v and g(x) = Nx+w, where

M =

(

1 −2

−4 3

)

, v = (3,−1) and N =

(

−3 0

1 1

)

, w = (2, 0),

find the linear part and translation part of the composition g ◦ f .

14 GABOR MOUSSONG

3. Given the affine transformation f(x) = Mx+ v, where

M =

(

−1 1−3 2

)

, v = (3,−1),

find the linear part and translation part of f−1.

4. Determine which of the following types of objects is taken to the same type of objects by any

affine transformation:

(a) a half-line, (g) a convex set,

(b) a half-plane, (h) a circle,(c) an interval, (i) a rectangle,(d) a right angle, (j) a parallelogram,

(e) an acute angle, (k) a trapezoid,(f) a polygon, (l) a rhombus.

5. If an affinity is applied to a triangle, show that the centroid (a. k. a. median point) is mapped

to the centroid of the image triangle.

(Recall that a median line connects a vertex of the triangle to the midpoint of the oppositeside, and the centroid is the common point of the three median lines, dividing all of them in

ratio 2 : 1.)

6. Is the statement of Exercise 5 true for other notable points of the triangle (orthocenter, incenter,circumcenter)?

7. Suppose that the composition of three homotheties is the identity. Show that the three centers

are collinear.

8. Let A, B, C be three distinct collinear points. Show (ABC)(BCA)(CAB) = 1.

9. Let P , Q, R, S be four distinct collinear points. Prove (PQS)(QRS)(RPS) = −1.

10. Let a =−→OA, b =

−→OB, and c =

−→OC be linearly independent vectors in three-dimensional

Euclidean space, and consider a vector x = αa + βb + γc =−−→OX. Prove that the point X is

incident to the plane ABC if and only if α+ β + γ = 1.

11. Suppose that, under an affinity of the plane, vertex A of triangle ABC is kept fixed, vertexB is mapped to C, and C is mapped to B. Show that this affinity is axial, and find its axis,direction, and ratio.

12. Given a parallelogram, find a shear mapping which takes it to a rhombus. (Define the shearthrough an affine frame and its image.)

13. Given a triangle ABC and a line L, show that there exists an axial affinity with axis L whichtakes ABC to an equilateral triangle.

14. An affine transformation of the coordinate plane is given by

x′ = −x+ 4y − 2

y′ = −2x+ 5y − 2 .

Prove that this is an axial affinity, and find its axis, direction, and ratio.

15. Find an affinity f of a plane to itself such that A, f(A), and f(

f(A))

are non-collinear for anypoint A.

16. An affine transformation of the coordinate plane R2 is given by the affine frame A0, A1, A2

and its image A′0, A′

1, A′

2. Find the image of point P = (−3,−5) if:

A0 = (1,−1), A1 = (2, 0), A2 = (−1, 3),

A′0 = (−4, 4), A′

1 = (−5, 5), A′2 = (4, 2).

17. Let f : R2 → R2 be rotation of the plane through +90 degrees about the center (2, 1). Write

f in the form f(x) = Mx+ v, and find the matrix M and the vector v.

18. Prove that a square matrix is orthogonal if and only if its row vectors form an orthonormalbasis.


19. Fill in the missing entries of the following 3× 3 matrices to get orthogonal matrices. (Several

solutions may exist.) In the second case try to use rational numbers only.

√3

2

−1

2

7

− 6

73

7

20. Let u ∈ Rn be any nonzero vector. Define the map σ : Rn → Rn by the formula

σ(x) = x− 2u · xu · u u (x ∈ Rn).

Show that σ is an orthogonal linear map. If n = 2 or 3, what well-known Euclidean congruenceis σ? (Note that σ(x) = x whenever x ⊥ u).

21. How many n× n orthogonal matrices are there such that all entries are integers? Check that

these matrices form a subgroup of O(n).

22. Show that the group O(n) contains Sn, the symmetric group on n letters, as a subgroup.(Recall that Sn is the group of all permutations of the set {1, 2, . . . , n}.)

23. (a) Given two points A and B of Euclidean plane, and their images A′ and B′ under some

affine transformation, construct (with ruler and compass) the image of an arbitrarily givenfurther point of the line AB.

(b) An affinity of Euclidean plane is given through an affine frame and its image. Find a way

to construct the image of an arbitrarily given further point of the plane.

24. Prove that any orientation preserving similarity transformation of the plane to itself is eithera translation or a spiral similarity.

(Hint: Work in the complex affine line, and look for a fixed point.)

25. Let A, B, A′, B′ be four points in Euclidean plane given in terms of Cartesian coordinates as

A = (1, 0), B = (2, 0), A′ = (0, 1), B′ = (0, 3).

Find the center of the spiral similarity of the plane which takes A to A′ and B to B′.(Hint: Use complex numbers, write the map as mz + v, use the given points to determine mand v, and solve for the fixed point.)

2 Spherical Geometry

We study the intrinsic geometry of the surface of the unit sphere in Euclidean space.We use the center of the sphere as origin for the vectors in space. Thus, Euclideanspace is identified with the three-dimensional vector space V . When coordinatesare needed, we assume V = R3 via a choice of a Cartesian system of coordinates.An essential element of structure now is the dot product (and norm) in V .

2.1 Basic Concepts

• Definition (unit sphere). The unit sphere in V is the set S = {x ∈ V :‖x‖ = 1}.Circles in S are the subsets of the form S ∩ P where P is any plane such thatthe origin is less than distance 1 from P . A circle S ∩ P is called a great circleif 0 ∈ P ( ⇐⇒ radius is maximal (= 1)).

• Definition (spherical segment). If x,y ∈ S are distinct and are not antipodal(i.e., x 6= ±y, or equivalently, the vectors x and y are linearly independent), thenthere exists a unique great circle C through x and y. The shorter of the two

16 GABOR MOUSSONG

arcs into which x and y divide the great circle C is called the spherical segmentconnecting the points x and y.

Remark. Great circles play the role of straight lines in spherical geometry. Notethat if two points are antipodal, then a great circle passing through them is notunique. But if two (distinct) points are not antipodal, then they uniquely determinethe great circle and the spherical segment connecting them.

• Definition (tangent vector, tangent plane). A vector v ∈ V is calledtangent to S at the point x ∈ S if v ⊥ x ( ⇐⇒ x · v = 0). The tangent planeat x ∈ S is TxS = x⊥ ≤ V , a 2-dimensional linear subspace.

Remark. When visualizing tangent vectors, TxS is customarily identified with itstranslate x+ TxS which really is tangent to the sphere S. The advantage of usingour definition for the tangent plane is that it is now a linear space with respect tothe natural operations on vectors.

• Example. Given a spherical segment with endpoints x and y, there is a uniqueunit vector v ∈ TxS which is determined by the requirement that y = αx+ βvwith β > 0. This vector v is called the unit tangent vector at x to the segmentfrom x to y.

• Definition (spherical distance). We define the spherical distance (“anglemetric”) of two points by the formula ds(x,y) = cos−1(x ·y) (x,y ∈ S). (Notethat ds(x,y) equals the arc length of the spherical segment from x to y.)

Remark. The distance function ds : S×S → R turns the set S into a metric space.Proof of the triangle inequality for ds is postponed to Section 2.2.

• Definition (angle). We define two kinds of angles in spherical geometry.First, the angle between two great circles is defined as the angle between tangentlines drawn at a point of intersection (which happens to equal the angle betweenthe planes containing the circles).


Second, the angle formed by two spherical segments with a common endpoint:for any x,y, z ∈ S, x 6= y 6= z we define ∡xyz as the angle between tangentvectors drawn at y to the spherical segments from y to x and from y to z.

Great circles in S can be conveniently parameterized as follows. A great circle Cis uniquely determined by a point x ∈ C and a unit vector v ∈ TxS tangent to Cat x. (Namely, C = S ∩ P where P is the linear 2-plane spanned by x and v.)

Working in the plane P , and using coordinates with respect to the orthonormalbasis x,v in P , one may parameterize the circle C as

r(t) = (cos t)x+ (sin t)v .

This is arc length parameterization, i.e., for any t1, t2 (with |t1 − t2| ≤ π) we haveds(r(t1), r(t2)) = |t1 − t2|. This follows from the calculation cos ds(r(t1), r(t2)) =r(t1) · r(t2) = cos t1 cos t2 + sin t1 sin t2 = cos(t1 − t2).

2.2 The spherical Law of Cosines

We define the concept of spherical triangles, and prove the most basic formula ofspherical trigonometry.

• Definition (spherical triangle). If x,y, z ∈ S are linearly independent ( ⇐⇒do not all lie on a great circle), then they determine a spherical triangle ∆,namely:∆ = S∩ (convex hull of the three rays from the origin through x, y and z), orequivalently,∆ = S∩ (intersection of the three half-spaces bounded by the linear 2-planesspanned by two of x, y, z and containing the third as interior point).Vertices of ∆ are the points x, y, z; sides of ∆ are the pairwise connectingspherical segments between the vertices. The angles of ∆ are formed by thepairs of its sides.

Let α, β, γ denote the angles at vertices x, y, z, respectively, of a spherical triangle,and let a, b, c denote the side lengths opposite x, y, z, respectively.

• Theorem (spherical Law of Cosines)

cos a = cos b cos c+ sin b sin c cosα

Proof:

18 GABOR MOUSSONG

Choose unit tangent vectors u,v ∈ TxS at the endpoint x to the sphericalsegments from x to y and z, respectively.Parameterize the side xy and substitute the parameter value c, and get theformula y = (cos c)x+ (sin c)u.Similarly, parameterize xz, substitute b, and get z = (cos b)x+ (sin b)v.Dot product of the two yields y · z = (cos b)(cos c) (x · x) + (sin b)(sin c) (u · v).Here y · z = cos a, x · x = 1, u · v = cosα, and the Law of Cosines follows.

Remark. If the size of the triangle converges to zero, then this formula converges tothe ordinary (Euclidean) Law of Cosines. To see this, replace cos a, cos b, cos c, sin band sin c with their Taylor series about 0, and ignore all terms of degree greaterthan two.

• Corollary. For any spherical triangle with side lengths a, b, c the strict triangleinequality a < b+ c holds.

Proof: In the Law of Cosines formula one has sin b > 0, sin c > 0 and cosα > −1;therefore, one gets the inequality

cos a > cos b cos c− sin b sin c .

The right hand side is just cos(b + c). Using that the cosx function is strictlymonotone decreasing on the interval [0, π], the triangle inequality a < b + cfollows.

Remark. This implies that the angle metric introduced in Section 2.1 is indeed ametric.

2.3 Area of spherical triangles

Recall that the total surface area of the unit sphere equals 4π. There is a remarkableformula (the so-called Girard formula, see below) which expresses the surface areaof spherical triangles in terms of their angles. Our goal is to prove the Girardformula.

• Definition (spherical bigon). Connect a pair of antipodal points in S withtwo great semicircles. The union of the two semicircles divides S into two regionseach of which is called a spherical bigon.

Any spherical bigon is uniquely determined up to congruence by its angle ϕ(equal at both vertices). A hemisphere is a special case of a spherical bigon(with ϕ = π), and the whole S may also be regarded a bigon with ϕ = 2π.


• Lemma. The surface area of a spherical bigon with angle ϕ equals 2ϕ.

Proof: The surface area in question only depends on ϕ, therefore we may denoteit as A(ϕ). This is a monotone and additive, therefore linear, function of thevariable ϕ; that is, A(ϕ) = cϕ with some constant c. Total surface area of S is4π, therefore A(2π) = 4π and c = 2. Thus, A(ϕ) = 2ϕ for all 0 ≤ ϕ ≤ 2π.

• Theorem (Girard formula). The surface area of a spherical triangle withangles α, β, γ equals α+ β + γ − π.

Proof: Let A denote the surface area of the spherical triangle in question.

Draw the three great circles containing the three sides of the triangle. Eachpair of them determine two congruent (oppositely placed) spherical bigons oneof which containing the triangle. The angle of such a bigon equals an angle of thetriangle. The six bigons thus obtained cover the whole sphere. The triangle itselfand its antipodal (centrally symmetric) image is triply covered, and the rest ofS is singly covered. This gives the equation 2A(α) + 2A(β) + 2A(γ) = 4π + 4A.Taking into account that A(ϕ) = 2ϕ, the equation A = α+ β + γ − π follows.

• Corollary. Any spherical triangle has angle sum strictly greater than π.

Remark. The concepts we introduced and the theorems we proved in Sections 2.1–2.3 about the geometry of the sphere all have interesting counterparts in hyperbolicgeometry, see 10.5, 11.6. Our brief study of spherical geometry actually serves as awarmup for later parts of the course.

2.4 Congruences in spherical geometry

Congruence transformations of S are restrictions to S of Euclidean congruencesof V which keep the origin fixed. These are precisely the orthogonal linear trans-formations of V . Thus, the group of spherical congruences is O(V ). If a Cartesiancoordinate system is given, then this group is O(3). Orientation preserving congru-ences form the subgroup SO(V ) (or SO(3)).

For an example of a sperical congruence, consider a rotation of the ambient Eu-clidean space about an axis through origin, restricted to S. Such a map is calleda spherical rotation. The special case when the rotation angle equals π is called ahalf turn.

• Theorem. All orientation preserving congruences of S are spherical rotations.

Proof: Given M ∈ SO(3), we first show that 1 is an eigenvalue of M . All realeigenvalues of M have absolute value 1, since M preserves norms of vectors.The characteristic polynomial has odd degree, therefore M has at least one real

20 GABOR MOUSSONG

eigenvalue. If all three are real, at least one of them must equal 1, since theirproduct is detM = 1. If there are non-real eigenvalues, then they are a pair ofcomplex conjugate numbers whose product is positive. Then the real eigenvaluemust also be positive since the product of the three is 1.

Let u be an eigenvector with eigenvalue 1, then the line L spanned by u ispointwise fixed by M . By preservation of dot product, M keeps the plane u⊥

invariant. Restriction of M to this plane must be an orientation preserving two-dimensional Euclidean congruence, that is, a rotation about the origin withinthis plane. Therefore, M itself is a three-dimensional rotation about axis L.

Remarks: (1) Since all elements of SO(3) are rotations, this group is often calledthe rotation group in dimension 3.(2) It is well-known that orientation preserving congruences of Euclidean planeare either rotations or translations. The theorem says that the second type ismissing from spherical geometry. Actually, translations do exist, but they happento coincide with rotations. This is explained the following way. Clearly one maytranslate great circles within themselves through any given spherical distance. It isintuitively clear that, if the sphere is attached to such a moving great circle, then thewhole sphere moves along together with the great circle. This is how a translationcan be defined in spherical geometry. It is easy to see that this congruence isactually a rotation about the axis orthogonal to the plane of the great circle.

Our goal now is to show that the group of orientation preserving congruences inspherical geometry is a simple group. Recall that a group is called simple if it hasno normal subgroups other than the trivial subgroup and the whole group. Thisfeature of spherical geometry is in sharp contrast with affine or Euclidean planegeometry where translations form a proper normal subgroup (cf. 1.7, 1.8).

Recall that a subgroup H of a group G is normal if and only if it is closed underconjugation, that is, ghg−1 ∈ H whenever h ∈ H and g ∈ G. (Two elements,h1, h2 ∈ G are called conjugate if there exists an element g ∈ G such that h2 =gh1g

−1.)

• Theorem. SO(3) is a simple group, that is, it has no nontrivial normal sub-groups.

Proof: We start with a series of claims about SO(3).

Claim 1: All half turns are conjugate in SO(3).Indeed, it suffices to find a rotation that takes a given line through 0 to an-other given line through 0. Such rotations clearly exist, and any such rotationconjugates the half turn about the first line into the half turn about the second.

Claim 2: SO(3) is generated by half turns.Indeed, it is easy to see that the product of two half turns about axes intersectingat angle ϕ is rotation through 2ϕ about the common perpendicular line of theaxes. Clearly any given rotation is obtained this way.

Claim 3: If an element of SO(3) reverses some line through the origin, that is,maps all points of this line to their negatives, then it is a half turn.Indeed, if the rotation angle is different from π, then clearly no line is reversed.

Suppose now that 1 6= G E SO(3). We prove that G contains at least onehalf-turn. Then, by Claim 1, all half turns are in G, and so G = SO(3) byClaim 2.


Choose a nontrivial element f ∈ G; if f itself is not a half turn, then, by replacingf with a suitable power, we may assume that f is rotation through angle α whereπ/2 ≤ α < π. Find a line L through 0 such that f(L) ⊥ L. (Such a line exists bythe intermediate value theorem, since the angle between L and f(L) (for variousL) varies continuously between 0 and α.) Let g denote the half turn about theline f(L). Consider the isometry h = g−1 ◦ f−1 ◦ g ◦ f . Then g−1 ◦ f−1 ◦ g ∈ Gbecause f ∈ G and G is a normal subgroup, and so h ∈ G. The map h reversesthe line L, therefore h is a half turn by Claim 3.

2.5 Exercises

1. Given two distinct points x,y ∈ S, find the locus of all points of S which are at equal sphericaldistance from x and y.

2. Given a spherical line (= great circle) C in S and a point x ∈ S which is not a pole to C(i.e., the vector x is not orthogonal to the plane containing C), show that there exists a uniquespherical line through x which intersects C at a right angle. Prove that the shortest spherical

segment connecting x with C is an arc of this perpendicular great circle.

3. Given two distinct great circles in S, show that the locus of all points at equal spherical distancefrom them is the union of the two angle bisector great circles.

4. Given two distinct great circles in S, show that there exists a unique third great circle whichintersects both at a right angle.

5. Define inscribed and circumscribed circles for spherical triangles. Show that their center is acommon point of the three bisectors of interior angles, and the three perpendicular bisectorsof sides, respectively.

6. Define the medians of a spherical triangle as the spherical segments connecting a vertex withthe midpoint of the opposite side. Apply central projection with center at origin to the flattriangle spanned by the vertices of a spherical triangle, and show that medians of the flat

triangle project to the medians of the spherical triangle. Deduce that the three medians of anyspherical triangle are concurrent.

7. Suppose that a spherical triangle has two right angles. Show that two sides have length π/2.

8. Prove: if a spherical triangle has unique altitudes, then the three altitudes are concurrent.(Note that altitudes are not always uniquely defined: if two of the sides have spherical length

π/2, then one can drop infinitely many perpendiculars to the third side from the opposite vertex(cf. 7). But if the triangle has at most one side of length π/2, then altitudes are unique.)

9. Prove the spherical Law of Sines:

sinα

sin a=

sinβ

sin b=

sin γ

sin c,

where a, b, c are the sides of a spherical triangle, and α, β, γ are the corresponding angles.

(Hint: Express cosα from the Law of Cosines and get a formula for sin2 α/ sin2 a which isinvariant under permutations of a, b and c.)

10. Consider a right spherical triangle with legs a and b and hypotenuse c < π/2. Suppose thatthe altitude h divides c into parts p and q (with p next to a). Prove:

sin2 h = (tan p)(tan q) and tan2 a = (tan p)(tan c).

(Keeping in mind that sinx ≈ tanx ≈ x for small x, these are spherical analogues of well-known

Euclidean formulas.)

11. Prove that the perimeter of any spherical triangle is less than 2π.

(Hint: Extend the triangle to a spherical bigon.)

12. The side length of an equilateral spherical triangle is π/3. Find its area. (Use calculator.)

22 GABOR MOUSSONG

13. Suppose that an equilateral spherical triangle has surface area equal to π/2. Find its side

length.

14. Given a spherical triangle ∆, its polar triangle ∆∗ is the spherical triangle whose verticesare the inward pointing unit normal vectors to the planes containing the sides of ∆. Show(∆∗)∗ = ∆.

15. For a spherical triangle ∆ express sides and angles of the polar triangle ∆∗ in terms of those

of ∆. Deduce that the surface area of ∆ plus the perimeter of ∆∗ equals 2π.

3 Inversive Geometry

We consider circles, certain families of circles, and certain transformations relatedto circles. Most of this takes place in a Euclidean plane which, throughout thischapter, is denoted E.

3.1 Power of a point

• Definition (power of a point). Let C ⊂ E be a circle with center O andradius r. If A is any point of the plane, its power with respect to C is thenumber

pC(A) = d2 − r2 ,

where d is the distance of A from O.

Clearly pC(A) is = 0, < 0, or > 0, according as A ∈ C, A is in the interior, or inthe exterior of C, respectively.

Recall the Intersecting Secants Theorem of elementary geometry: if two straightlines intersect each other at A, and one of them intersects the circle C at P and Q,the other at R and S, then AP · AQ = AR · AS. (Here P = Q or R = S is alsoallowed.) Apply this to the case when the second secant is through O, and get

AP ·AQ = (d+ r) |d− r| = |pC(A)| .

In particular, if A is exterior, then pC(A) equals the square of the length of thetangent drawn to C from A.

• Definition (angle of circles). The angle between two intersecting circles isdefined as the angle of the tangents at either point of intersection. In particular,two circles C1 and C2 are called ortogonal (denoted C1 ⊥ C2), when their angleis π/2. (Or, equivalently, when the two radii drawn at a point of intersection areperpendicular.)


• Lemma. Let C1 and C2 be circles with centers O1 and O2, and radii r1 andr2, respectively. Let d denote the distance O1O2. The following conditions areequivalent:

(i) C1 ⊥ C2; (iii) pC1(O2) = r22;

(ii) pC2(O1) = r21; (iv) r21 + r22 = d2.

Proof: Implications (i) =⇒ (ii), (i) =⇒ (iii), (ii) =⇒ (iv), (iii) =⇒ (iv) are allimmediate from the definition of the power of a point.(iv) =⇒ (i): From (iv) it follows that |r1 − r2| < d < r1 + r2 which means thatthe two circles intersect. Now the converse of the Pythagorean theorem can beapplied to the triangle spanned by a point of intersection and the two centers.

3.2 Radical axis

Given two circles, we consider the locus of points in the plane whose powers withrespect to the circles are equal. In particular, if such a point A is outside bothcircles, then equal tangents can be drawn from A to the two circles.

If C1 and C2 are concentric circles with distinct radii r1 and r2, then no point Aexists with pC1

(A) = pC2(A) since d2 − r21 6= d2 − r22. But as soon as the centers

are distinct, the locus is nonempty, as the following theorem says.

• Theorem. If two non-concentric circles are given, the locus of points havingequal power with respect to the circles is a straight line perpendicular to the lineof the centers.

Proof: Let C1, C2 denote the circles, O1, O2 the centers, r1, r2 the radii, d thedistance O1O2.

Let us first restrict to points of the line O1O2 only, and use a real coordinatealong this line. Let o1, o2 ∈ R be the coordinates of O1, O2. A point of this line,with coordinate x, belongs to the locus if and only if (x−o1)2−r21 = (x−o2)2−r22.This equation reduces to a linear equation with a unique solution, which meansthat there is a single point B of this line which belongs to the locus.Let A be an arbitrary point of the plane. Drop a perpendicular from A to lineO1O2, and let F be the foot of this perpendicular. By the Pythagorean theorem,pCi

(A) = AO2i −r2i = AF 2+FO2

i −r2i , therefore the condition pC1(A) = pC2

(A)is equivalent to pC1

(F ) = pC2(F ). This means that A belongs to the locus if

and only if F does. Therefore, the locus is the straight line erected at point Borthogonally to line O1O2.

• Definition (radical axis). For two non-concentric circles the line given by thepreceding theorem is called the radical axis of the circles.

If the two circles intersect, then their radical axis is the line through their pointsof intersection. If the two circles are tangent, then the radical axis is their tangent

24 GABOR MOUSSONG

line drawn at the common point. Both these follow from the observation that anypoint common to the two circles has zero power with respect to both circles.

• Theorem. If the centers of three circles are non-collinear, then the radical axes,taken in pairs, pass through a common point.

Proof: By non-collinearity of the centers, no two of the radical axes can beparallel. Take the point of intersection of two of them. This point has equalpower with respect to all three circles, therefore it must belong to the thirdradical axis as well.

• Definition (radical center). The common point of the radical axes for threecircles with non-collinear centers is called the radical center of the circles.

3.3 Inversion in a circle

• Definition (inversion). Let C ⊂ E be a circle of radius r, centered at O ∈ E.Two points P, P ′ ∈ E are said to be inverse to each other with respect to C ifP and P ′ belong to the same ray emanating from O, and OP · OP ′ = r2. Thisrelation uniquely determines a map

σC : E − {O} → E − {O}, σC(P ) = P ′ ,

called inversion in C. (Note that point O has no inverse.)

The following are immediate consequences of the definition:

– σC ◦ σC = idE−{O}.

– C = Fix (σC), that is, σC(P ) = P if and only if P ∈ C.

– If L ⊂ E is a straight line and O ∈ L, then σC(L− {O}

)= L− {O}.

– Two inversions in concentric circles only differ by a homothety. More precisely:if circles C1, C2 with radii r1, r2, are both centered at O, then

−−−−−−→OσC2

(P ) =

(r2r1

)2 −−−−−−→OσC1

(P ) .

Two figures (i.e., subsets of E) are called inverse figures with respect to C, if themap σC transforms one onto the other. Now we focus on finding inverses of linesand circles. We reserve the “ ′ ” notation for inverse figures with respect to C, thatis, for any point or subset H of E − {O}, the symbol H ′ stands for σC(H).

• Theorem. Let L ⊂ E be a line. If O ∈ L, then (L − {O})′ = L − {O}, and ifO /∈ L, then the set L′ ∪ {O} is a circle through point O whose tangent line atO is parallel to L.


Proof: This is obvious in the case O ∈ L. Assume now O /∈ L. Drop a perpen-dicular line from O to L, and let F denote its foot. Consider the point F ′ andthe circle D with diameter OF ′. We claim that the set L′ is this circle D withpoint O removed.

Let P ∈ L, P 6= F arbitrary. Then OP · OP ′ = r2 = OF · OF ′, that is,OP/OF = OF ′/OP ′. Thus, triangles OPF and OF ′P ′ are similar, as they havea common angle at O. The former has a right angle at F , therefore the latter hasa right angle at P ′. By Thales’ theorem this means that P ′ ∈ D. Conversely,any point Q ∈ D, Q 6= O is the inverse of some point of L, namely, the point ofintersection of line OQ with L.

• Lemma. If D ⊂ E is a circle which intersects C orthogonally, then D′ = D.Conversely, if some circle D 6= C is its own inverse, then D ⊥ C.

Proof: Consider the power of point O with respect to circleD. ConditionD′ = Dis equivalent to pD(O) = r2, which, using the characterization of orthogonalitygiven in 3.1, is equivalent to C ⊥ D.

• Theorem. Let D ⊂ E be any circle. If O ∈ D, then (D − {O})′ is a straightline parallel to the line tangent to D at O. If O /∈ D, then D′ is a circle.

Proof: The case O ∈ D is a consequence of the previous theorem. Assume nowthat O /∈ D.

If O is in the exterior of D, then pD(O) > 0. Let C1 be the circle of radius√pD(O), concentric with C. Then by 3.1 C1 ⊥ D, so by the lemma, the inver-

sion σC1maps D to itself. But then the concentric inversion σC maps D to a

homothetic image of D, which is a circle.If O is in the interior of D, then pD(O) < 0, and we use the circle C2 about

O of radius r2 =√−pD(O). Now we claim that inversion in C2 takes D to its

centrally symmetric image with respect to point O as center of symmetry. LetP ∈ D arbitrary, and let Q denote the other point of intersection of line OP

26 GABOR MOUSSONG

with D. Then OP · OQ = −pD(O) = r22, and O is between P and Q. Point P ′

is on the same ray from O as P , and also OP · OP ′ = r22. Therefore, P ′ and Qare symmetric with respect to O. This proves our claim. Finally, D′ must be acircle as it is a homothetic image of the circle σC2

(D).

Our next goal now is to prove that the angle at which circles or lines intersect isinvariant under inversion.

• Lemma. If two circles, or a circle and a line, are tangent to each other at apoint different from O, then their inverses are also tangent.

Proof: Inversion is a bijective map of the set E − {O} to itself. Therefore, thelemma follows from the fact that two circles, or a circle and a line, are tangentif and only if they have a single point in common.

Remark: If the point of tangency is O, then the inverse figures clearly are twoparallel lines.

• Theorem. Suppose that two circles, or a circle and a line, or two lines, intersectat angle α. Then their inverses also intersect at angle α.

Proof: We assume that neither of the circles or lines pass through O. The readermay easily modify the arguments to cover the cases when O is a point on one orboth of the figures.

Consider first the case of two lines L1 and L2. Their inverses, L′1 and L′

2, arecircles through O (minus O itself), with tangents at O being parallel to L1 andL2. Due to parallelism, the angle between these tangents equals α. Now O is oneof the points of intersection of circles L′

1 and L′2, and at this point, by definition,

the angle of L′1 and L′

2 equals the angle α of the tangents.Let D1 and D2 be two circles (or a circle and a line), which intersect at angle α.Draw tangent lines L1, L2 at one point of intersection. By the preceding lemmathe angle of D′

1 and D′2 equals the angle of L′

1 and L′2, which, by the case of two

lines above, equals α.

In particular, orthogonal circles are always inverted to orthogonal circles. Thisobservation is the basis of the following characterization of inverse pairs of points.

• Theorem. Let P,Q ∈ E be two distinct points. The following conditions areequivalent:

(i) P and Q are inverse to each other with respect to C;(ii) Any circle or straight line through P and Q intersects C orthogonally.

Proof: (i) ⇒ (ii): If D is a circle, P,Q ∈ D, then pD(O) = OP ·OQ = r2, so by3.1, C ⊥ D. If L is a line with P,Q ∈ L, then O ∈ L, therefore C ⊥ L.


(ii) ⇒ (i): Since line PQ is orthogonal to C, the center O must be collinearwith P and Q. Choose two distinct circles, D1 and D2, through P and Q. SinceD1, D2 ⊥ C, we know that D′

1 = D1 and D′2 = D2. It follows that the set

{P,Q} = D1 ∩ D2 is also its own inverse. The points P and Q cannot bothbelong to C since then C itself would contradict (ii). Therefore, P and Q aremutually inverse points.

If C is a straight line in E (instead of a circle), then condition (ii) of the theoremis easily seen equivalent to the two points being symmetric with respect to line C.This suggests that inversions in circles and reflections in lines are closely relatedtransformations. Let us use a unified notation for them: from now on the symbolσC denotes inversion if C is a circle, and reflection in line C if C is a line.

• Corollary. If two points P and Q are mutually inverse with respect to somecircle D, then their images P ′ and Q′ under inversion in C are mutually inversewith respect to circle D′ (or, are symmetric with respect to D′ if it is a straightline). Similarly, the images of any pair of points symmetric with respect to aline L are mutually inverse (or symmetric) with respect to the inverse of L. Inother words, if σ denotes inversion in C, and D is either a circle or a line, thenone has

σ ◦ σD ◦ σ = σσ(D) .

The same holds when C is a straight line and σ denotes reflection in C.

Proof: This follows immediately from the preceding theorems.

3.4 Inversive plane and Mobius transformations

We want to study compositions of various inversions of the plane E. The problemwith this is that inversions are not defined everywhere. This is remedied by attach-ing a new point called ∞ to the plane, thereby making all inversions operate on thesame set.

• Definition (inversive plane). We extend the plane E with a (single) pointat infinity denoted by the symbol ∞. The point at infinity is assumed differentfrom any ordinary point we use in geometry (e. g., it is certainly not an elementof E). With the point ∞ attached to E, the set E+ = E ∪ {∞} is called aninversive plane, or the inversive extension of E.The point ∞ is assumed to belong to all straight lines in E. More precisely, ifL ⊂ E is any line, then the set L+ = L∪{∞} is considered a line (inversive line)in E+. Thus, all straight lines in E+ share the common point ∞.If f : E → E is any Euclidean congruence (or more generally, any similaritytransformation), we extend f to a map E+ → E+ by the rule f(∞) = ∞. For

28 GABOR MOUSSONG

example, the point ∞ becomes a fixed point of any reflection σL in a line L (inaccordance with ∞ ∈ L+).Finally, if C ⊂ E is any circle, then we extend the inversion σC to a bijectivemap E+ → E+ by declaring σC(O) = ∞ and σC(∞) = O, where O is the centerof C.

It is easy to see that all theorems of the preceding section remain valid after theextension. Actually, some of the statements become simpler because there is no needto separate cases, or because some exceptions are eliminated. For example, if weconsider two parallel straight lines to be tangent to each other at ∞, then inversionpreserves the tangency relation among circles and lines (with no exception).

• Definition (Mobius transformation). A Mobius transformation of the planeE is any map µ : E+ → E+ which can be written as a composition

µ = σk ◦ . . . ◦ σ2 ◦ σ1

where each σi is either a reflection or an inversion of the plane E (extendedto E+). It is clear that all Mobius transformations of E are bijective maps,and form a group under composition. This group, denoted M(E), is called theMobius group, or inversive group, of E.In other words, the Mobius group is the subgroup in the group of all bijectivemaps E+ → E+, generated by all the σC where C is either a circle or a linein E.

It is clear from the results of the preceding section that all Mobius transformationstake circles or lines to circles or lines, preserve tangency, and preserve angle betweencircles or lines.

• Examples

(1) All Euclidean congruences of E, extended to E+, are Mobius transforma-tions. Indeed, in terms of coordinates, any such map is an orthogonal lineartransformation followed by a translation. In dimension 2 we know that ortho-gonal maps are either rotations or reflections. Any rotation or a translation iseasily represented as a composition of two reflections. Therefore, any Euclideancongruence is a product of reflections.

(2) More generally, all similarity transformations of E, extended to E+, areMobius transformations. Indeed, a similarity can always be written as the com-position of a Euclidean congruence and a homothety, and any homothety can berepresented as the composition of two concentric inversions.

Note that all Mobius transformations in these examples keep the point at infinityfixed. The converse is also true: if µ ∈ M(E) and µ(∞) = ∞, then µ|E is aEuclidean similarity. We omit the proof since it is a little harder.

• Theorem. Suppose that a Mobius transformation µ ∈ M(E) keeps a circle orline C ⊂ E pointwise fixed, then either µ = σC , or µ = idE+ .

Proof: Given any point A ∈ E, A /∈ C, choose two circles D1 and D2 so thatA ∈ Di and Di ⊥ C (i = 1, 2).Since a circle orthogonally intersecting C at two prescribed points is unique, bothD1 and D2 must be mapped to themselves by µ. Circles D1 and D2 intersect


each other at two points one of which is A, the other is B = σC(A). Thenµ must map the set {A,B} = D1 ∩ D2 to itself, that is, either µ(A) = A, orµ(A) = σC(A). By continuity, as A varies, the two possibilities cannot mix.Therefore, either µ(A) = A for all A ∈ E, or µ(A) = σC(A) for all A ∈ E.

3.5 Inversive geometry and complex numbers

In this section we assume E = C, that is, Euclidean plane is identified with theplane of complex numbers. Inversive plane now is C+ = C ∪ {∞}. Some well-known transformations of Euclidean plane are easily described in terms of complexnumbers:

– with b ∈ C fixed, the map z 7→ z + b is a translation,

– with a ∈ R, a 6= 0 fixed, the map z 7→ az is a homothety with center at theorigin,

– with a ∈ C, |a| = 1, the map z 7→ az is rotation about the origin,

– the complex conjugation map z 7→ z is reflection in the real axis,

– if L ⊂ C is any straight line, and a Euclidean congruence f : C → C takes thereal axis to L, then reflection in L is given by the formula z 7→ f

(f−1(z)

),

which clearly is a composition of the above.

In addition to these, it is straightforward to check that the formula

z 7→ 1

z(z ∈ C)

defines inversion in the unit circle about origin. If C ⊂ C is an arbitrary circle ofradius r, centered at p ∈ C, then it follows that σC(z)− p = r2

(1/z − p

), that is,

σC(z) =r2

z − p+ p .

With the agreement 1/0 = ∞ and σC(p) = ∞ this formula extends to the bijectivemap σC : C+ → C+ as in 3.4.

• Definition (linear/semilinear fractional map). A linear fractional trans-formation over C is a map f of the form

f(z) =az + b

cz + d,

where a, b, c, d ∈ C and ad − bc 6= 0. This formula naturally extends over theinversive plane C+ as follows. If c = 0, then f is an affine map so we define

30 GABOR MOUSSONG

f(∞) = ∞. If c 6= 0, then on one hand, as f is not defined at z = −d/c, weput f(−d/c) = ∞, and on the other hand, as a/c is the limit of f as z → ∞, wedefine f(∞) = a/c.A map g of the form

g(z) =az + b

cz + d,

where a, b, c, d ∈ C and ad−bc 6= 0, is called a semilinear fractional map, consid-ered again as a transformation of C+ with the same conventions regarding ∞.

It is easily checked that both linear and semilinear fractional maps are C+ → C+

bijective maps. They together form a group under composition of maps, withinwhich the linear fractional transformations form a subgroup of index 2.

• Theorem. A map f : C+ → C+ is a Mobius transformation of the inversiveplane C+ if and only if it is linear fractional or semilinear fractional over C.

Proof: Mobius transformations by definition are products of reflections and in-versions. Both are semilinear fractional, as the examples in the beginning of thissection show.Conversely, suppose first that f(z) = (az+ b)/(cz+ d) is linear fractional. Thenf can be written as a composition of maps that appear among the examples, allof which are Mobius transformations. If f is semilinear fractional, then it is alinear fractional map composed with complex conjugation, therefore, is a Mobiustransformation.

Remarks: Those Mobius transformations which are linear fractional maps are calledorientation preserving. A Mobius transformation is orientation preserving if andonly if it is a product of an even number of inversions or reflections. Odd products,examples of which are inversions and reflections themselves, are orientation revers-ing. Warning: in the literature on inversive geometry often only the orientationpreserving maps are called Mobius transformations.

3.6 Poincare extension

We define Mobius groups for lines and circles, and discuss how they are embeddedin the Mobius group of the plane.

• Definition (one-dimensional Mobius transformation). Let E+ be an in-versive plane. If L+ ⊂ E+ is any line, one can reflect L (within itself) in a pointof L, and perform inversion of L with respect to a pair of points in a straight-forward way. Compositions of these maps (understood as L+ → L+) are calledMobius transformations of L, and form the Mobius group M(L) of L.Let C ⊂ E be a circle, and A,B ∈ C be two distinct points. By inversion of Cwith respect to the pair {A,B} we mean the following map of C to itself. If Aand B are antipodal, then we simply take the reflection in line AB, restrictedto C. If A and B are not antipodal, then the tangent lines to C drawn at A andB meet at some point O ∈ E. Points A and B are kept fixed, and the image ofany point P ∈ C, P 6= A,B, is the unique point Q ∈ C for which O, P , and Qare collinear.Note that this map is the restriction to C of the inversion of the plane E withrespect to the circle centered at O and through A and B. Again, by definition,


composing such maps one gets the Mobius transformations of C, and the Mobiusgroup M(C) of C.

• Definition (natural extension). To each generator of the group M(L) orM(C) we can associate a generator of M(E) in a natural way.

Consider first the case of a line L+ ⊂ E+. Generators of M(L) have naturalextensions in M(E), namely, a reflection in a point in L extends as reflection inthe line orthogonal to L at the point, and an inversion with respect to a pair ofpoints in L extends as inversion with respect to the circle for which the pair isantipodal.

Now let C be a circle in E. Again, generators of the group M(C) have naturalextensions in M(E) by their very definition; a reflection if A, B are antipodalin C, and an inversion otherwise.

• Theorem. In both cases of the preceding definition, the maps given on gener-ators extend to homomorphisms M(L) → M(E) and M(C) → M(E) betweenthe corresponding Mobius groups.

Proof: Let σ 7→ σ denote the natural extension of generators, given in thepreceding definition. The only way to extend this assignment to a group ho-momorphism is to let the product σk . . . σ1 ∈ M(E) correspond to an arbitraryproduct σk . . . σ1 of generators in M(L) (or in M(C)).

One must check that this map is correctly defined, that is, σk+1 . . . σm = σk . . . σ1whenever σk+1 . . . σm = σk . . . σ1. Equivalently, if a product σm . . . σ1 of gener-ators is the identity of L+ (or of C), then the product σm . . . σ1 of the naturalextensions must be the identity of E+.

This follows from the last theorem of Section 3.4. For if µ = σm . . . σ1, then theMobius transformation µ keeps L+ (or C) pointwise fixed, so by the theoremit is either σL (or σC , respectively), or it is the identity of E+. But it cannotbe σL or σC as neither of the σi can interchange the two sides of L (or of C,respectively). Therefore, µ = idE+ indeed.

• Definition (Poincare extension). The homomorphisms PEL : M(L) → M(E)

and PEC : M(C) → M(E), defined above, are called the Poincare extension of

Mobius transformations from L (or from C, respectively) to E.

Remark: Let f : R+ → R+ be a linear fractional map f(x) = (ax + b)/(cx + d)with real coefficients a, b, c, d ∈ R. Then f ∈ M(R). If f is orientation preserving,then its Poincare extension is the same f considered as a function of a complexvariable. If f is orientation reversing, then composing f with complex conjugationmakes it orientation preserving, therefore the Poincare extension of f is semilinearfractional:

32 GABOR MOUSSONG

PC

R (f)(z) =

az + b

cz + dif f is orientation preserving,

az + b

cz + dif f is orientation reversing.

3.7 Coaxal circles

• Definition (coaxal circles). A system of circles in E is called a coaxal systemif each two of the circles have the same radical axis.

There exist obvious examples of coaxal systems: any family of circles passingthrough two distinct common points, or any family of circles having the sametangent line at a common point. A less obvious example is given below.

• Example. Let A and B be two distinct, fixed, points in E. If two further points,P and Q, of the line AB, are inverse with respect to the pair {A,B}, then thecircle on PQ as diameter is called an Apollonian circle with respect to A and B.The power of the midpoint M of AB with respect to any of these circles equalsMP ·MQ =MA2. Therefore, the family of Apollonian circles is coaxal, and theradical axis is the perpendicular bisector of AB.

Our goal is to find maximal coaxal systems of circles in the plane. It is convenientto include the radical axis, and possibly some “null-circles”, i. e., points regardedas circles of zero radius, in the system. Such maximal systems are called “pencils”.

• Definition (pencil). A pencil of circles in E is any one of the following fourtypes of families:– all circles centered at a point A ∈ E, including the null-circle {A};– all circles tangent to a line L at a point A ∈ L, including the line L, and the

null-circle {A};– all circles through a pair of points A,B ∈ E, including the line AB;


– all Apollonian circles with respect to a pair of points A,B ∈ E, including theperpendicular bisector of AB, and the null-circles {A} and {B}.

It is not hard to verify that any two given circles are included in a unique pencilof circles, therefore, these are indeed the maximal coaxal systems. Since linesare also present, it will be convenient if certain families of lines are also calledpencils: all lines through a common point form an intersecting pencil of lines,and all lines parallel to a given line form a parallel pencil of lines.Two pencils are called orthogonal if all members of one are orthogonal to allmembers of the other. For any pencil of circles or lines there exists a uniqueorthogonal pencil. The reader is encouraged to verify this for each case.Pencils of circles or lines naturally fall into three categories: elliptic pencils (con-centric or Apollonian), parabolic pencils (tangent or parallel), and hyperbolicpencils (intersecting). The reason is explained by the next theorem.

• Theorem. All Mobius transformations take pencils of circles or lines to pencilsof circles or lines, preserving their elliptic, parabolic, or hyperbolic character.

Proof: This is clear in the parabolic or hyperbolic case. Elliptic case followsfrom orthogonality between elliptic and hyperbolic pencils.

3.8 Stereographic projection

In Section 3.4 Mobius transformations were defined as maps of the inversive planeE+ to itself. There is a way to interpret Mobius transformations as maps actingon a sphere S. This is achieved using one particular map p : S → E+ to copytransformations of E+ to those of S.

• Definition (stereographic projection). Let S be any sphere in Euclideanspace, O ∈ S be an arbitrarily chosen point, and E be the plane tangent to Sat the point opposite O. For any point A ∈ S, A 6= O, we define p(A) as thepoint of intersection of the line OA with the plane E. (One may say that p(A)is obtained by centrally projecting the point A onto E with center at O.) Nowp : S−{O} → E clearly is a bijective map; we extend it to the whole S by letting

34 GABOR MOUSSONG

p(O) = ∞. Thus, p : S → E+ is bijective. This map is called the stereographicprojection of S to E+.

Remark: Once the point O ∈ S is selected, the plane E may be chosen as anyplane parallel to, and different from, the tangent plane at O. For concretenesswe have made the more or less standard (but arbitrary) choice of selecting theopposite tangent plane. In some applications (see, for instance, Section 9.1) otherchoices may prove useful. It is clear that stereographic projections that correspondto different choices of the plane E only differ by a homothety centered at O.

It is a remarkable feature of stereographic projection that it exhibits some importantproperties very similar to those of Mobius transformations of the plane: takes circlesto circles or lines, and preserves angle. We now prove these properties.

• Theorem. Let D ⊂ S be a circle. If O ∈ D, then p(D) is an inversive line inE+, if not, then it is a circle.

Proof: The proof is easy when O ∈ D: projection of D takes place within theplane F containing D. Therefore, in this case p(D) is the inversive extension ofthe straight line E ∩ F .The general case (when D does not necessarily pass through O) is harder.We shall first realize that stereographic projection is the restriction of a three-dimensional inversion to S, then the proof will follow from the circle-preservingnature of three-dimensional inversions.Given a sphere Σ in three-dimensional Euclidean space, inversion in Σ is de-fined by the same formula as in the plane. Observe that restricting to anyplane containing the center C of Σ we get the familiar two-dimensional inver-sion. Three-dimensional inversion therefore may be regarded as two-dimensionalinversions done simultaneously with respect to all great circles of Σ.Inverse of any plane or sphere P with respect to Σ is again a plane or sphere.To see this, observe first that there is an axis L of rotational symmetry of Pthrough C. (To get L, drop a perpendicular from C to P if P is a plane, orsimply connect C with the center of P if P is a sphere.)

Within any plane containing L the picture is a two-dimensional inversion. Thethree-dimensional picture is obtained by rotating about L, therefore, the resultis either a plane or a sphere (according as the inverse under the two-dimensionaltransformation is a line or a circle).It follows now that the image of any circle under a three-dimensional inversionis always a circle or a line. To see this, one may represent the circle as theintersection of two spheres. Inverses of those are spheres or planes, which indeedintersect along a circle or a line.


Let us return now to the stereographic projection p : S → E+. Consider thesphere Σ centered at O with radius equal to the diameter of S. Observe thatthree-dimensional inversion with respect to Σ maps the sphere S to the planeE+. In other words, p is just restriction of the three-dimensional inversion to S.Therefore, p takes circles to circles or lines as claimed.

The angle of two intersecting circles lying on a sphere is defined just like in thecase of the plane: it is the angle of the two tangent lines drawn at a common point.Due to symmetry it makes no difference which of the two points of intersection isconsidered. If two circles lying on the sphere have a single point in common, thentheir tangent lines at this point coincide. In this case the two circles are tangent toeach other, and their angle may be cosidered 0.

• Theorem. Stereographic projection preserves angles: if two circles D1, D2 ⊂ Smeet at angle α, then the angle of p(D1) and p(D2) in the plane E+ is also α.

Proof: The argument given in Section 3.3 for the case of inversion of two circlesin the plane also works here with minor modifications.

First, one observes that tangency is preserved since it is characterized by havinga single point in common. Second, the theorem is easily proved in the specialcase when D1 and D2 meet at O. Indeed, in this case p(D1) and p(D2) are linesparallel to the tangent lines drawn to D1 and D2, respectively, at the point O.When D1 and D2 do not both pass through O, and meet at some point A ∈ S,consider the (uniquely determined) two circles C1, C2 ⊂ S that both pass throughboth points O and A, and Ci has the same tangent line at A as Di (i = 1, 2).Then the angle of C1 and C2 equals α, and, by the special case above, is preservedby p. Finally, the angle of p(C1) and p(C2) (= α) at point p(A) equals the angleof p(D1) and p(D2), by preservation of tangency.

In order to define Mobius transformations of the sphere S, we simply use the bi-jective map p to copy those of E+ to S. Due to the theorems just proved, thestandard properties of these transformations automatically follow.

• Definition (Mobius transformation of a sphere). By a Mobius transfor-mation of the sphere S we understand any map of the form

p−1 ◦ µ ◦ p : S → S

where µ : E+ → E+ is a Mobius transformation of the plane E. All Mobiustransformations of S form a group, denoted M(S), the Mobius group of S,

36 GABOR MOUSSONG

which clearly is isomorphic to the Mobius group M(E). (The isomorphismM(E) → M(S) is just the assignment µ 7→ p−1 ◦ µ ◦ p.)

All Mobius transformations of the sphere take circles to circles, and preserve anglesof intersection. Actually, a converse theorem, which is harder to prove, is alsotrue: any bijective map S → S that takes circles to circles is necessarily a Mobiustransformation.

When using complex numbers, the inversive plane C+ is often visualized using itsinverse image under a stereographic projection S → C+. The sphere S in this caseis called the Riemann sphere. Linear fractional and semilinear fractional maps thenact as Mobius transformations of the Riemann sphere.

3.9 Exercises

1. Suppose a quadrilateral ABCD has right angles at two opposite vertices A and C. Let Pdenote the point of intersection of the diagonals. Prove AP · PC = BP · PD.

2. Let ABC be an isosceles triangle with AC = BC = a. For a point P of base AB we havep = AP , q = PB, and b = PC. Prove pq = a2 − b2.

(Even though this can be routinely solved using Pythagoras’ theorem, find a quicker solutionwhich refers to the power of a point with respect to a suitably chosen circle.)

3. Given two concentric circles of different radii, show that no circle can intersect both of them

orthogonally.

4. Given three circles with non-collinear centers, find a fourth circle which intersects all of them

orthogonally. Does such a fourth circle always exist? If it exists, is it unique?

5. Given an arbitrary triangle, draw circles on its sides as diameters. Which notable point of thetriangle is the radical center of these three circles?

6. Suppose that x2 + y2 + ax+ by+ c = 0 is the equation of a circle C in R2. Show that for anypoint P = (x0, y0) the power of P with respect to C is obtained by substituting the coordinates

of P into the left hand side of the equation, that is, pC(P ) = x20+ y2

0+ ax0 + by0 + c.

7. For i = 1, 2 put Fi(x, y) = x2 + y2 + aix + biy + ci, and let Fi(x, y) = 0 be the equation ofcircle Ci in R2. Use the result of Exercise 6 to show that, if C1 and C2 are non-concentric,then the equation of the radical axis of C1 and C2 is

F1(x, y)− F2(x, y) = 0 .

(Note that this gives an alternative proof that the radical axis actually is a straight line.)

8. Let C be a circle and L be a line such that C ∩L = ∅. Show that there exists a point Q in theinterior of C such that, for all P ∈ L, the distance PQ equals the length of the tangent drawnfrom P to C.

9. Find coordinates for the inverse of the point (x, y) ∈ R2 with respect to the circle x2+y2 = r2.

10. Let P and P ′ ( 6= P ) be inverse points with respect to circle C of radius r. Let A be the point

of intersection of C with line PP ′ which is farther from P . Prove

1

AP+

1

AP ′ =1

r.

11. Three circles, neither two of which are tangent to each other, pass through a common point P .Consider the region of the plane bounded by the three arcs of the circles connecting their pointsof intersection different from P . What is the sum of interior angles of this region?

(Hint: Apply inversion centered at P .)

12. Let A, B, C, D be four points in the plane, not all on a circle, and no three collinear. Let αdenote the angle between the circles ABC and ABD, and let β denote the angle between the

circles ACD and BCD. Show that α = β.

(Hint: Apply inversion centered at one of the points.)


13. Let C and D denote two distinct circles or lines in the inversive plane. Prove that σCσD =

σDσC holds in the Mobius group if and only if C and D intersect orthogonally.(Hint: For the “if” part apply inversion to take the circles to lines. For the “only if” part

observe that two transformations can commute only if the set of fixed points of one is invariantunder the other.)

14. Prove that for any pair of circles in E there exists an inversion that takes them to congruent

circles.

15. The center of the inverse of a circle, in general, is different from the inverse of the center. Moreprecisely: let O and P be the centers of circles C and D respectively, prove that P ′ = σD′ (O)

(where prime denotes inversion with respect to C). Note that this justifies the first sentencesince the center of D′ is σD′ (∞).(Hint: Either calculate distances, or (shorter) apply σD′ = σCσDσC .)

16. Let A,B, P,Q be four distinct points on a line L. If P and Q are inverse to each other withrespect to the pair {A,B}, then show that A and B are inverse with respect to {P,Q}.

17. How are circles C and D related if the Mobius transformation σDσC keeps the point at infinity

fixed? What if C or D is a line?

18. What can be the image of a half-plane, or of a circular disk, under a Mobius transformationof the plane?

19. Let ε ∈ {−1, 1}. For all nonzero a ∈ R draw the circle Ca on the interval [a, ε/a] of the firstcoordinate axis as diameter. Show that the family of all circles Ca, together with the second

axis, form a pencil of circles.

20. Let C be an Apollonian circle with respect to points A and B. Prove that for a variable pointP of the plane, the ratio PA/PB is constant along C.

21. Prove that for any pair of disjoint circles or lines in E+ there exists an inversion that takesthem to concentric circles.(Hint: Include the two circles in an Apollonian pencil of circles.)

22. Show that points A,B ∈ E are inverse of each other with respect to circle C if and only if Cis an Apollonian circle with respect to A and B.

23. Let C denote the circle of radius 1 about i ∈ C. Write the inversion σC as a semilinear

fractional function.

24. Work out the coefficients of the product of two given linear fractional functions. Compare theresult with multiplication of 2× 2 matrices.

25. Consider the linear fractional function f(z) = (z − i)/(z + i). What is the image of the upperhalf-plane Im(z) > 0 under f? Find the inverse of f in linear fractional form.

26. Show that the semilinear fractional map f(z) =3z − 5

z − 3defines an inversion of the complex

plane. Find the center and radius of this inversion.

27. An inversion of the complex plane, centered at a real number, takes the point A = i to the

point A′ = 2 + 3i. Write this inversion in a semilinear fractional form.

28. Suppose that the semilinear fractional map f(z) =az + b

cz + ddefines an inversion of the complex

plane. Prove |a| = |d|.

38 GABOR MOUSSONG

Part Two: Projective Geometry

4 Projective Space and Incidence

One can project one plane of three-dimensional space to another from a pointchosen as center of projection. This “perspective mapping” or “perspectivity” as atransformation does not fit into the framework of Euclidean or affine geometry asnone of the usual invariants (distance, angle, parallelism, simple ratio) is preserved.If the two planes are not parallel (which is the typical case), the map is not evendefined everywhere. The reader is encouraged to draw sketches to locate the pointsfor which central projection is undefined. The classical construction of projectivespace resolves this deficiency by extending ordinary space with new elements: so-called “ideal” points, lines, etc., at infinity.

4.1 Projective plane: the classical approach

First we attach an ideal point (i.e., point at infinity) to each straight line. At themoment we assume each line L gets its own ideal point ∞L, and shortly we shalldiscuss the relation between various ideal points.

For the extended line L∪ {∞L} we use the notation L. Any point of extended lineL is either ordinary (that is, an element of L), or the ideal point ∞L.

Let us turn to extending planes. Given an arbitrary plane P , all lines in P areassumed to have their ideal points. Now we introduce the following convention:

• For any two lines L,M ⊂ P we consider ∞L = ∞M if and only if L ‖M .

The extended plane P is now defined as

P = P ∪ {∞L : L ⊂ P is a line} .

Formally speaking, an ideal point is attached to P for each parallelism class of linesin P . Points in P come in two varieties: either ordinary (that is, an element of P ),or ideal (of the form ∞L where L is a line in P ).

Lines in P are defined as follows. First, we have ordinary lines: by definition, theyare the extended lines L where L ⊂ P is a line in the usual sense of the word. Inaddition, the set ∞P = P − P of all ideal points of P is declared to be the idealline of P . So, P has infinitely many ideal points, and a single ideal line.

This finishes up the definition of our first “model” of projective plane, the extendedEuclidean plane. The structure of this model is its “incidence structure” whichtells us what subsets of projective plane are considered lines. The content of thisincidence structure is nothing more than it is given which points are incident towhich lines. Later we shall see different, but isomorphic, models for the sameconcept of projective plane.

The following theorem states the two most important incidence properties of pointsand lines in extended plane.


• Theorem. In any extended plane(1) for any two distinct points there exists a unique line passing through both

points;(2) any two distinct lines have a unique point in common.

Proof: One has various cases to consider depending on whether the given pointsor lines are ordinary or ideal. In each case the claims easily follow either fromstandard Euclidean geometry facts, or directly from the definitions.

4.2 Incidence structure of projective 3-space

We may continue in a similar fashion to define 3-dimensional extended space. If Edenotes Euclidean 3-space, we define the points, lines, and planes of E as follows.

As a set, E = E ∪ {∞L : L ⊂ E is a line}. Note that no further convention onideal points is needed since any two parallel lines of E are contained in some planeP ⊂ E which forces the parallel lines to have the same ideal point. As before,elements of E are called ordinary points, and the rest (all of the form ∞L whereL ⊂ E is a line) are the ideal points of E.

There are two kinds of lines in E. By definition, they are either extended linesL where L ⊂ E is a line of Euclidean space, or ∞P where P ⊂ E is a plane inEuclidean sense. These lines are called ordinary and ideal lines of E, respectively.The ideal lines therefore are certain proper subsets of the set of all ideal points.Note that for two planes P,Q ⊂ E one has ∞P = ∞Q if and only if P ‖ Q.

Finally, planes in E are either ordinary planes: extended planes P where P ⊂ E isa Euclidean plane, or ideal: there is a single ideal plane ∞E = E − E, the set ofall ideal points.

Remark: The words “point”, “line”, and “plane”, when referring to extended planeor space, will mean whatever these last definitions allow. That is, they may beeither ordinary or ideal. We shall see later that from the point of view of projectivegeometry there is no essential difference between the nature of ordinary and idealelements.

Again, extended space with its incidence structure (that is, the system of points,lines and planes) is considered a model for three-dimensional projective space. Someproperties of this incidence structure are collected in the following theorem.

• Theorem. In extended space(1) for any two distinct points there exists a unique line incident to both points;(2) for any three non-collinear points there exists a unique plane incident to all

three points;(3) for any line and non-incident point there exists a unique plane containing

both;(4) for any pair of distinct, intersecting lines there exists a unique plane containing

both;(5) the common part of any two distinct planes is a line;(6) if a line is not contained in a plane, then they have a unique point in common.

Proof: For each statement a case-by-case analysis can be done depending onwhich of the given objects are ordinary or ideal. In each case the claim followseither from standard Euclidean geometry facts, or from our definitions.

40 GABOR MOUSSONG

Remarks: (1) Once these incidence properties are established, when dealing withfurther questions concerning incidence, we no longer have to handle ordinary andideal cases separately.

(2) The classical extension definition of projective space can be carried furtherwith increasing dimension. By induction on dimension, one can continue with thedefinition the projective extension of 4-, 5-, etc., n-dimensional Euclidean spaces,together with their incidence structure. In each step the only new feature to bedefined is the (n− 1)-dimensional subspace which consists of all ideal points.

4.3 Central projection

Since ordinary and ideal objects display the same properties as far as incidence isconcerned, from now on we suppress the “bar” notation. Now simply E denotesextended 3-space, and A, B, L, M , P , Q, etc., may denote either ordinary or idealpoints, lines or planes.

• Definition (central projection). Let P, P ′ ⊂ E be planes, and C ∈ E bea point not on P or P ′. We define a map P → P ′, sending points A ∈ P toA′ ∈ P ′ as follows. Given A ∈ P , we use statements (1) and (6) of the theoremin 4.2 to conclude that there exists a unique point A′ ∈ P ′ for which C, A, andA′ are collinear. This map is called the central projection (or perspectivity) ofP to P ′ with center at C.

In certain special cases one may recognize some well-known affine maps (restrictingto ordinary points only). If both P and P ′ are ordinary, and C is ideal, then wehave a parallel projection from P to P ′. If P ‖ P ′, then central projection is asimilarity if C is ordinary, and a translation if C is ideal.

The following theorem shows that central projections establish isomorphisms be-tween the incidence structures of the two planes.

• Theorem. Central projections between planes are bijective, and take lines tolines.

Proof: Each central projection has an inverse, namely, a central projection withthe same center, but with source and target interchanged. This justifies bijec-tivity.Given the central projection P → P ′ with center at C, and a line L ⊂ P , we canapply (3) and (5) of the theorem in 4.2 to conclude that L′ = Q ∩ P ′ is a linewhere Q is the plane containing C and L.

In particular, since it is possible to centrally project an ordinary plane onto theideal plane, the incidence structure of ideal lines in the ideal plane is isomorphic tothe incidence structure of the ordinary plane.


Central projections are capable of moving ordinary lines out to infinity, and viceversa. This is an often used feature of projective geometry. The following lemmais the basis of such tricks.

• Lemma. Given any ordinary plane P , and any ordinary line L ⊂ P , there existsa suitable central projection of P onto another ordinary plane P ′ such that theimage of L is the ideal line of P ′.

Proof:

Choose an arbitrary ordinary point C /∈ P as center, and let the auxiliary planeQ contain C and L. If the image plane P ′ is chosen as an arbitrary ordinaryplane parallel to and different from Q, then all projection lines, through C andany point of L, stay parallel to plane P ′. Therefore, central projection withcenter at C maps line L to ∞P ′ .

4.4 Projective spaces: algebraic approach

Modern treatment of projective geometry has its foundations in linear algebra. Thedefinition of projective spaces is essentially the same in all dimensions, therefore weshall work in an n-dimensional setting. From the point of view of classical Euclideanand non-Euclidean geometry, the cases n = 1, 2, 3 are of primary interest.

• Definition (projectivization of a vector space). Let W be a finite dimen-sional real vector space. Introduce the following binary relation ∼ on nonzerovectors in W :

u ∼ v if v = λu with some λ ∈ R .

It is easily checked that ∼ is indeed an equivalence relation on the set W −{0}.Equivalence classes are the straight lines through origin (with the origin itselfremoved). The set of equivalence classes is called the projective space associatedwith W (or the projectivization of W ), and is denoted as P (W ). The dimensionof P (W ) is understood as dimW −1. That is, an n-dimensional projective spaceis obtained as projectivization of an (n+ 1)-dimensional vector space.

For w ∈ W , w 6= 0, the class containing w is denoted [w]. We say that thepoint [w] is represented by w.

If W = Rn+1, then P (W ) = P (Rn+1) is called the standard projective n-space.For simplicity, it is denoted Pn.

• Definition (projective subspace). If U ≤W is an arbitrary linear subspace,then P (U) clearly is a subset of P (W ). Subsets of P (W ) thus obtained are calledprojective subspaces of P (W ).The empty set is a projective subspace in any projective space, and it is theonly subspace of dimension −1. The 0-dimensional subspaces are precisely the

42 GABOR MOUSSONG

one-element subsets (which naturally are identified with the points of P (W )).One-dimensional and two-dimensional projective subspaces are called projectivelines and planes, respectively. The (n − 1)-dimensional projective subspaces ofa projective n-space are usually called projective hyperplanes.

It is clear that the intersection of an arbitrary family of projective subspaces is itselfa projective subspace. For any subset H ⊆ P (W ) there exists a smallest projectivesubspace containing H (namely, the intersection of all projective subspaces contain-ing H). This is called the subspace generated by H, and is denoted 〈H〉. A pointA = [w] ∈ P (W ) belongs to 〈H〉 if and only if vector w is a linear combinationof vectors representing elements of H. For instance, points A and B are distinct ifand only if their representative vectors are linearly independent. In this case 〈A,B〉denotes the unique projective line through A and B.

The classical extension approach and the linear algebraic approach result in iso-morphic incidence structures. Now we verify this in the two-dimensional case. Thesame method works in all dimensions.

Choose an ordinary plane P , and an ordinary point O /∈ P in E. Use O as theorigin for vectors in three-dimensional space, and letW denote the space of vectors.Define the map f : P → P (W ) by f(A) = [w] where w is any direction vector forthe line connecting O with A.

• Theorem. The map f is bijective, and makes the lines of P correspond preciselyto the lines of P (W ).

Proof: For [w] ∈ P (W ) define the point g([w]) ∈ P as the point of intersectionof the line with direction vector w through O with the plane P . (Note that thispoint is ideal if w is parallel to P .) Clearly g : P (W ) → P is the inverse of f ,showing that f is bijective.If L is any line in P , let Q denote the plane containing O and L, and let U bethe set of vectors in W parallel to Q. Then f(L) = P (U), and all projectivelines in P (W ) are obtained this way.

4.5 Exercises

1. Verify all incidence properties stated in the theorems of 4.1 and 4.2.

2. Given three distinct planes in extended space, show that their common part is either a pointor a line.

3. Given two coplanar ordinary lines, show that a suitable central projection can take them to apair of parallel lines.


4. Show that any convex quadrilateral in an Euclidean plane can be photographed in such a way

that in the picture we see a parallelogram. (We think of the camera as a tool for making centralprojections. The center of the projection is the lens of the camera, and the image plane is the

plane containing the film/sensor.)

5. Show that any convex quadrilateral different from a parallelogram can be photographed insuch a way that in the picture we get a square.

6. Let L1 and L2 be two disjoint projective lines in 3-dimensional projective space P 3. Provethat for any given point A ∈ P 3 − (L1 ∪ L2) there exists a unique projective line through A

which intersects both L1 and L2.(Hint: Do not use Euclidean geometry in separate cases; instead, rely on known incidenceproperties of points, lines and planes in extended space.)

7. Show that the complete graph on five vertices embeds into P 2. That is, connect each pair offive points in P 2 with continuous arcs (e.g., straight segments) that do not cross one another.

(Note: it is well-known that this is impossible in Euclidean plane.)

8. Prove dim〈S ∪ T 〉+dim(S ∩ T ) = dimS +dimT , where S, T ⊆ P (W ) are arbitrary projective

subspaces.

9. Three pairwise disjoint projective lines are given in the four-dimensional projective space P 4

so that they are not contained in any projective hyperplane. Prove that there exists a uniquefourth projective line in P 4 which intersects all three given projective lines.

5 Coordinates in projective geometry

5.1 Homogeneous coordinates and equations

We want to introduce coordinates in projective space. Ordinary affine coordinatescannot be extended over ideal points. Therefore, we use the algebraic model.

• Definition (homogeneous coordinates). Let W be an (n + 1)-dimensionalvector space, and P (W ) its projectivization. Choose a basis a1, a2, . . . , an+1

in W . If A = [w] ∈ P (W ) is an arbitrary point, then write w =∑n+1

i=1 xiai.The numbers x1, x2, . . . , xn+1 ∈ R are called homogeneous coordinates of Awith respect to the chosen basis. The choice of the basis identifies W with thecoordinate space Rn+1, so we use the notations w = (x1, x2, . . . , xn+1) andA = [x1 : x2 : . . . : xn+1].

Clearly no point of P (W ) exists with all homogeneous coordinates equal to 0. Apoint does not determine uniquely its homogeneous coordinates, only their ratios.For instance, it makes no sense to talk about the concrete value of the ith homoge-neous coordinate of a point (unless it is zero).

Since a point of P (W ) has various different sets of homogeneous coordinates, arandom equation written in homogeneous coordinates may not make sense in P (W ).For instance, consider the two-variable equation 3x1 − x22 +1 = 0. It is satisfied by(1, 2) but not by (2, 4) which represent the same point. Clearly the condition forcorrectness of the equation is that (λx1, λx2, . . . , λxn+1) must satisfy the equationwhenever (x1, x2, . . . , xn+1) does. Homogeneous polynomial equations are easilyseen to have this property. This is why in projective geometry one uses homogeneousequations.

The simplest type of homogeneous equations are homogeneous linear equations.General form of a homogeneous linear equation in n+ 1 variables is

a1x1 + a2x2 + . . .+ an+1xn+1 = 0 ,

44 GABOR MOUSSONG

where not all of the coefficients ai are zero. These are precisely the equations ofprojective hyperplanes (that is, lines in a projective plane (n = 2), or points in aprojective line (n = 1)). We shall consider homogeneous quadratic equations later.

5.2 Affine vs. homogeneous coordinates

When dealing with extended Euclidean plane or space as a model for projectivegeometry, we may have two coordinate systems present at the same time. On onehand, we can use the usual affine (or Cartesian) coordinates for ordinary points.On the other hand, if we identify the extended plane or space with P (W ) as in 4.4,we can introduce homogeneous coordinates for all points (both ordinary and ideal).We now discuss the relation between the two types of coordinates. For simplicity,we restrict to the case n = 2. The general case follows the same pattern.

We shall use the following setup (and keep this choice of coordinates fixed). Welet W = R3 equipped with standard coordinates x1, x2, x3, and identify R2 withthe plane x3 = 1 in R3. (That is, this plane is parallel to, and different from thex1x2 coordinate plane.) In this copy of R2 we use the notation x and y for thecoordinates x = x1 and y = x2. We use the method of 4.4 to explicitly identify theextended plane R2 with the projective plane P (R3), and from now on, we consider

them equal, that is, R2 = P (R3) = P 2.

A point A = [x1 : x2 : x3] ∈ P 2 is ideal if and only if x3 = 0. (In fact, x3 = 0 is theequation of the ideal line in this setting; note that this is indeed a homogeneouslinear equation.) Suppose now that A is ordinary, that is, x3 6= 0. Then A also hasordinary x, y coordinates. The relation between the two sets of coordinates clearlyis [x1 : x2 : x3] = [x : y : 1], which leads to the following conversion formulas:

x =x1x3

and y =x2x3

.

If an equation of a figure in R2 is given in terms of ordinary coordinates x andy, we can use these formulas to set up a homogeneus equation for the same figurein R2. We now carry out this “homogenization” procedure for the case of ordinarylines in R2.

Let ax+ by + c = 0 be equation of an ordinary line L ⊂ R2. (Here a and b cannotboth equal 0 as (a, b) is a normal vector to line L.) Using the conversion formulasand multiplying through by x3, we end up with the homogeneous linear equationax1 + bx2 + cx3 = 0. This equation is equivalent to the original one as long as werestrict to ordinary points, but multiplication by the variable x3 may result in new


solutions when this variable takes the value 0. New solutions are found by solvingthe system of equations

ax1 + bx2 + cx3 = 0

x3 = 0 .

All solutions are proportional to x1 = −b, x2 = a, x3 = 0. Therefore, the idealpoint [−b : a : 0] is the single new solution in P 2 of the homogenized equation of theline L. Note that the vector (−b, a, 0) representing this point is a direction vectorfor line L in R3, that is, [−b : a : 0] = ∞L. Therefore, the homogeneous equationdefines L in P 2, and homogenization of linear equations may be regarded as theprecise algebraic counterpart of the cassical extension definition of projective plane.

5.3 Projective transformations: definitions

Projective transformations are best defined in the algebraic setting, since linearalgebra supplies the well-developed context of linear maps and matrices.

• Definition (projective transformation). Let ϕ : W → W ′ be an ar-bitrary invertible linear map between vector spaces W and W ′. The map[ϕ] : P (W ) → P (W ′) defined by the formula [ϕ][w] = [ϕ(w)] is called theprojective transformation induced by ϕ.

Note that under a projective transformation all projective subspaces are mappedto projective subspaces of the same dimension. Therefore, all projective transfor-mations are isomorphisms between the incidence structures.

Clearly [idW ] = idP (W ), [ϕ]−1 = [ϕ−1], and if ϕ : W → W ′ and ϕ′ : W ′ → W ′′ are

invertible, then [ϕ′ ◦ ϕ] = [ϕ′] ◦ [ϕ].In case of projective transformations of a projective space to itself, fixed points areprecisely the points represented by eigenvectors of the inducing linear map.

• Theorem. Two invertible linear maps ϕ,ψ : W → W ′ induce the same projec-tive transformation if and only if ψ = λϕ for some λ ∈ R, λ 6= 0.

Proof: The “if” part is obvious. For the converse suppose [ϕ] = [ψ], and considerthe linear map ψ−1 ◦ ϕ ∈ GL(W ). Now [ψ−1 ◦ ϕ] = [ψ]−1 ◦ [ϕ] = idP (W ). This

means that all nonzero vectors in W are eigenvectors for the map ψ−1 ◦ ϕ,therefore, this map must be scalar times the identity.

By the theorem, the projective transformation [ϕ] may be regarded as the equiva-lence class of ϕ under an equivalence relation similar to the one used in the definitionof projectivization of vector spaces.

• Definition (projective group). All projective transformations P (W ) →P (W ) form a group under the operation of composition. This group is called theprojective group of the vector space W (or of P (W )), and is denoted PGL(W ).By the preceding theorem this group is the quotient of the general linear groupGL(W ) by the normal subgroup formed by all homotheties at the origin.When coordinates are chosen, that is, W = Rn+1, the notation PGL(n+ 1,R)is also in use for the projective group. This is the quotient of the group ofinvertible (n+ 1)× (n+ 1) matrices by the normal subgroup of scalar matrices.For M ∈ GL(n+ 1,R) the equivalence class of the matrix M is denoted [M ].

46 GABOR MOUSSONG

5.4 Projective transformations: examples

There is a natural way to consider affine transformations of Euclidean space asprojective transformations, restricted to ordinary points, of extended space. Wediscuss this in the case of the projective plane. General (n-dimensional) case isdone analogously.

• Definition (projective extension of affinity). Let P and P ′ be Euclideanplanes, and f : P → P ′ be an affine transformation. We define the projectiveextension f : P → P ′ of f between the extended planes as follows. If A ∈ P isan ordinary point, then f(A) = f(A). If A ∈ ∞P is an ideal point, and A = ∞L

for some line L ⊂ P , then we put f(A) = ∞f(L). (Note that correctness of thisdefinition follows from affinities preserving parallelism.)Now we check that f indeed is a projective transformation. This is done usingthe setup in 5.2. We may assume P = P ′ = R2, and R2 = P (R3). As f now isan affine transformation R2 → R2, it can be written as f(x) = Mx + v whereM is a 2× 2 invertible matrix. Define the 3× 3 matrix M as

M =

M v

0 0 1

.

Then M is invertible, and is easily seen to induce f . Therefore, f is projective.

Projective extensions of affinities clearly take ordinary points to ordinary, and idealpoints to ideal. The converse also holds: if a projective transformation g : P → P ′

takes ordinary points to ordinary, then its restriction to P necessarily is an affinemap P → P ′, therefore, g = g|P . To see this, (again assuming P = P ′ = R2) notethat the linear subspace spanned by the first two coordinates in R3 must be keptinvariant by the linear map inducing g, therefore the first two entries in the thirdrow of its matrix must equal zero. Multiplying by a suitable scalar if necessary, thismatrix is then of the form M above.

A different type of transformation is central projection: it can take ordinary pointsto ideal and vice versa. Now we are in the position to show that these maps arealso projective transformations.

• Theorem. Any central projection between two planes of a 3-dimensional pro-jective space is a projective transformation.

Proof: Given a central projection, we want to find an invertible linear mapwhich induces it. Now dimW = 4 as we work in the projective 3-space P (W ).Let the two planes be given as P = P (U) and P ′ = P (U ′) where U and U ′

are 3-dimensional linear subspaces in W , and let C = [c] be the center of theprojection. Then vector c ∈W is linearly independent both from U and from U ′.


Let ϕ denote parallel projection in W from U to U ′ in direction of c. That is,we decompose any x ∈ U as x = x′ + λc where x′ ∈ U ′ and λ ∈ R. (Such adecomposition exists and is unique as W is the direct sum of subspaces U ′ andRc.) Using this decomposition, ϕ is defined as ϕ(x) = x′. Then ϕ : U → U ′ isa linear map, and, since c is independent also from U , it is invertible. We claimthat the projective map [ϕ] induced by ϕ is the central projection P → P ′ withcenter at C. Indeed, for an arbitrary nonzero vector x ∈ U its image ϕ(x) = x′

is contained in the 2-dimensional linear subspace spanned by c and x, that is,[c], [x], and [ϕ(x)] are collinear.

5.5 Exercises

1. Determine α and β if both triples (1, 1−β, αβ) and (−β, 2β,−9) are homogeneous coordinates

of the same point.

2. Are the following triples of points collinear?

(a) [2 : −1 : 3], [4 : 3 : −2], [8 : 11 : −12]

(b) [2 : 0 : 3], [4 : 0 : 3], [1 : −1 : 0]

3. For what value of λ are the following three points collinear: [2 : 1 : 0], [3 : 4 : 2], [5 : −6 : λ] ?

4. Find homogeneous coordinates for the ideal point of the line through points [5 : −3 : 2] and

[0 : 4 : −3] in P 2.

5. Verify that the equation of the line joining the two distinct points [a1 : a2 : a3] and [b1 : b2 : b3]in P 2 is

det

x1 x2 x3

a1 a2 a3b1 b2 b3

= 0 .

6. Write the following equations in homogeneous (polynomial) form, and determine which ideal

points satisfy them:

(a) 2x− 4y + 5 = 0, (b) x2

9+ y2

5= 1, (c) x2

9− y2

16= 1,

(d) x2 = −4y, (e) xy = 1, (f) y = 2x3,

(g) x2 = y3, (h) x2 = y2.

7. Prove that if n is even, then every projective transformation of Pn to itself has a fixed point.

If n is odd, show that this is false.

8. Prove that if n is even, then every projective transformation of Pn to itself has an invariantprojective hyperplane. If n is odd, show that this is false.

9. Prove or disprove: If a projective transformation of the projective plane fixes the vertices ofsome triangle, then it is the identity.

10. Call four points in a projective plane a projective basis, if no three of them are collinear.Given a projective basis A0, A1, A2, A3 in P (W ) (where dimW = 3), show that one can find

representative vectors a0,a1,a2,a3 ∈ W for these points (that is, Ai = [ai], i = 0, 1, 2, 3) suchthat a0 = a1+a2+a3. Prove that any two such choices of vectors are proportional. Generalizeto higher dimensions.

11. Given two projective planes P and Q with projective bases A0, A1, A2, A3 and B0, B1, B2, B3

respectively, use Exercise 10 to prove that there exists a unique projective transformationf : P → Q such that f(Ai) = Bi (i = 0, 1, 2, 3). Generalize to higher dimensions.

12. Let f be a (non-identity) congruence of Euclidean plane, and let f : P 2 → P 2 be the projective

extension of f . Show that the set of fixed points of f in P 2 is either a single point, two points,a single line, or the union of a line and a point. Give an example for each case.

48 GABOR MOUSSONG

6 Cross-ratio and Projective Geometry of the Line

6.1 Cross-ratio

When studying transformations, one often uses quantities which are unchanged ifa transformation is applied. Such quantities – usually called invariants – belongto the most important features of the type of geometry considered. For example,distance, angle, area, volume, etc., are familiar invariants of Euclidean geometryas they are indeed invariant under Euclidean congruences, while simple ratio isinvariant under affine transformations. None of these, however, are invariant underall projective transformations. Projective invariants are harder to find.

• Definition (cross-ratio). Let A, B, C, D be four distinct points of a projectiveline L = P (W ), dimW = 2. We define the cross-ratio of these points, a realnumber denoted (ABCD), as follows.First we choose representative vectors for the points: A = [a], B = [b], C = [c],D = [d]. Now as A 6= B, vectors a and b form a basis in W , therefore we canwrite c = λ1a + µ1b and d = λ2a + µ2b with some real coefficients λ1, µ1, λ2,µ2. Neither of these numbers can equal zero since all four points are distinct, sowe put

(ABCD) =µ1

λ1

/µ2

λ2.

Note that (ABCD) only depends on the points themselves, and not on theparticular choice of representative vectors. This is easiest to see for the pointsone by one: if one of the vectors is replaced with a scalar multiple of itself, thenthe effect of this on the coefficients eventually cancels out in the formula.

Some properties of cross-ratio can be immediately seen as consequences of thedefinition:

– It cannot equal 0 or 1, but all other real numbers may occur as its value.

– If (ABCD1) = (ABCD2), then D1 = D2.

– The two pairs {A,B} and {C,D} separate each other if and only if (ABCD) < 0.(Note that two points always divide a line into two different straight segments.Two pairs separate if members of one pair fall in distinct segments into whichthe line is divided by the other pair.)

– In general, the value of the cross-ratio depends on the order of the four pointsin the formula. The identities (ABDC) = (BACD) = 1/(ABCD) easily followfrom the definition, and by manipulating with linear combinations one can prove(ACBD) = 1− (ABCD). Then, as these three transpositions generate all possi-ble permutations of the four letters, one can work out effects of all permutationson the value of the cross-ratio.

The most important property of cross-ratio is given in the next theorem and itscorollary.

• Theorem. Cross-ratio is invariant under projective transformations. That is, ifA, B, C, D are four distinct points of a projective line L, and f : L → L′ is aprojective transformation, then

(f(A) f(B) f(C) f(D)

)= (ABCD).

Proof: Let L = P (W ), L′ = P (W ′), and f = [ϕ], where ϕ : W →W ′ is a linearisomorphism between the 2-dimensional vector spaces W and W ′.


Choose representatives a, b, c, d ∈W for points A, B, C, D, respectively, thenvectors ϕ(a), ϕ(b), ϕ(c), ϕ(d) ∈ W ′ will represent f(A), f(B), f(C), f(D).Write c = λ1a + µ1b and d = λ2a + µ2b as above. As ϕ is a linear map,vectors ϕ(c) and ϕ(d) are expressed with the same coefficients λ1, µ1, λ2, µ2 ascombinations of ϕ(a) and ϕ(b). Therefore, the two cross-ratios are equal.

• Corollary (Pappus–Steiner theorem). If a central projection takes fourdistinct collinear points A, B, C, D to points A′, B′, C ′, D′, respectively, then(A′B′C ′D′) = (ABCD).

If the projective line containing the four points is an extended Euclidean line, andthe points are all ordinary, then for their cross-ratio there is a formula based ondistances only.

• Theorem. Let A, B, C, D be four distinct, collinear, ordinary, points of aEuclidean line. Then

(ABCD) =(ABC)

(ABD)=AC

CB

/AD

DB,

where, in the last formula, distances are understood with sign.

Proof: If a point O not on the line containing the points is used as origin, and

position vectors a =−→OA, b =

−→OB, c =

−→OC, d =

−→OD are used as representative

vectors, then the claim immediately follows from a lemma in 1.4 which expressessimple ratio as the ratio of coefficients.

Remarks: (1) If we work in the coordinate line R, and choose four distinct pointsa, b, c, d ∈ R, then the formula in the theorem translates to real numbers as

(abcd) =c− a

b− c

/d− a

b− d.

(2) Interestingly, the incidence structure of projective plane completely determinescross ratio. If a bijective map between two projective planes takes lines to lines,then, by the so-called fundamental theorem of projective geometry, it necessarily isa projective transformation. In particular, it must preserve cross-ratio.

6.2 Harmonic separation

• Definition (harmonic relation). Two pairs of points, {A,B} and {C,D}, aresaid to be in harmonic relation if (ABCD) = −1.

50 GABOR MOUSSONG

Since their cross-ratio is negative, harmonic pairs separate each other in theline containing them. Writing the four points in a different order the value ofthe cross-ratio remains −1 as long as separation is maintained between the firstpair and the last pair of the four points. (That is, (BACD) = (ABDC) =(BADC) = (CDAB) = (DCAB) = (CDBA) = (DCBA) = (ABCD) = −1.)Therefore, harmonic separation is a symmetric relation on the set of unorderedpairs of distinct points of a line.If one of the two pairs, say {A,B}, is fixed, for any third point C of the linethere exists a unique fourth point D with (ABCD) = −1. This point is calledthe harmonic conjugate of C with respect to the pair {A,B}.

• Examples

– LetM be the midpoint between distinct ordinary points A and B of the ordinaryline L. Then harmonic conjugate of M with respect to the pair {A,B} is theideal point ∞L. To verify this, choose a point O off the line L as origin, and use

position vectors a =−→OA, b =

−→OB to represent A and B. Then a+ b represents

M , and a− b represents ∞L, therefore (ABM∞L) = (1/1)/(−1/1) = −1.

– Let ABC be a triangle in Euclidean plane. Let the interior and exterior anglebisector at vertex C intersect line AB at points P and Q, respectively. Then thepairs {A,B} and {P,Q} separate harmonically.

This specializes to the previous example if the triangle is isosceles with AC =BC. The general case follows using the Pappus–Steiner theorem: project thefour points with center at C to a line orthogonal to CP .

– Let A, B, C, D be distinct collinear ordinary points in Euclidean plane, andassume that C and D are inverse with respect to the circle with diameter AB.Then (ABCD) = −1.

This is easily verified by direct calculation. For example, the harmonic conjugateof x ∈ R with respect to the pair of points {−1, 1} of the real line is 1/x.


– If a, b ∈ R are distinct positive numbers, then the harmonic conjugate of 0 withrespect to pair {a, b} is the harmonic mean between a and b, that is, the number2ab/(a+ b) (= reciprocal of the arithmetic mean of reciprocals).

Now we define a projective transformation of projective plane which will play animportant role later in hyperbolic geometry.

• Definition (harmonic homology). Let P be a projective plane, L ⊂ P bea line, and A ∈ P be a point with A /∈ L. The pair (L,A) gives rise to a maph = h(L,A) : P → P , called harmonic homology with center at A and axis L,defined the following way.

The point A and all points of L are fixed by h. For any other point X ∈ Pits image h(X) is uniquely determined by the requirement (ABXh(X)) = −1where B ∈ L is the point collinear with A and X.It is clear that h is a bijective map from the plane P to itself, and that h ◦ h isthe identity.

• Claim: h is a projective transformation.Indeed, a linear map inducing h can be found as follows. Let P = P (W ) whereW now is three-dimensional, L = P (U), A = [a]. Due to the condition A /∈ L,the space W decomposes as the direct sum of the two-dimensional subspace Uand the one-dimensional subspace Ra. Let ϕ : W → W be the linear reflectionassociated with this decomposition, that is, ϕ|U = idU and ϕ(a) = −a. Thenh = [ϕ]. For if x ∈ W decomposes as x = αa + βb with b ∈ U , then ϕ(x) =−αa+ βb, therefore

([a][b][x][ϕ(x)]

)= (β/α)

/(β/− α) = −1.

6.3 Geometry of the projective line

One-dimensional projective geometry is different from higher dimensions becauseof the lack of incidence structure. As it turns out, the structure of a projective lineis given by cross-ratio. The following two theorems clarify the picture.

• Theorem. Let L and L′ be projective lines, A,B,C ∈ L and A′, B′, C ′ ∈ L′ bearbitrary triples of distinct points in each. Then there exists a unique projectivetransformation L→ L′ taking A to A′, B to B′, and C to C ′.

Proof: We may assume L′ is contained in a projective plane P . Choose anotherline L′′ ⊂ P , different from L′, such that L′ and L′′ intersect at point A′. Choosea projective transformation f : L→ L′′ which takes A to A′, and put B′′ = f(B),C ′′ = f(C).Let O denote the point of intersection of lines B′B′′ and C ′C ′′, then centralprojection g within P from line L′′ onto line L′ with center at O keeps A′ fixed,

52 GABOR MOUSSONG

and maps B′′ to B′ and C ′′ to C ′. Therefore, the composition g ◦ f : L → L′ isas desired.Uniqueness follows from projective transformations preserving cross-ratio, andfrom the fact that the value of cross-ratio uniquely determines the fourth point.

Specializing to the case L = L′ we get that the group of projective transformationsof the line L act simply transitively on the set of ordered triples of distinct pointsof L.

• Theorem. A map between two projective lines is a projective transformation ifand only if it is bijective and preserves cross-ratio.

Proof: We already know that all projective transformations are bijective andpreserve cross-ratio, therefore we only have to prove the converse.Let F : L → L′ be a cross-ratio preserving bijective map between projectivelines L and L′. Choose three distinct points A, B, C ∈ L, and let A′, B′,C ′ denote their images under F , respectively. By the preceding theorem thereexists a projective transformation f : L → L′ with f(A) = A′, f(B) = B′, andf(C) = C ′. We claim that F = f (thereby showing that F is indeed a projectivetransformation). Let D ∈ L − {A,B,C} be an arbitrary further point of L,then using that both F and f preserve cross-ratio we have

(A′B′C ′F (D)

)=

(ABCD) =(A′B′C ′f(D)

), from which F (D) = f(D) follows.

The main reason for using homogeneous coordinates in projective geometry is thatgenerally we cannot handle points at infinity with ordinary coordinates. Thereis, however, one exception: when the space is one-dimensional. In the projectiveextension of an ordinary line there is only one point at infinity, and the symbol ∞may work well in calculations. Another distinctive feature of dimension one is thatinversive extension L+ = L∪{∞} of a Euclidean line L coincides with its projectiveextension L = L ∪ {∞L} (where now ∞ = ∞L). This relates projective geometryto inversive geometry. Now we want to make this relationship clear.

We use the one-dimensional version of the standard setup given in 5.2: we dealwith the standard projective line P 1 = P (R2) identified with R = R ∪ {∞} via[x1 : x2] ↔ x = x1/x2 and [1 : 0] ↔ ∞.

Let a projective transformation f = [M ] : P 1 → P 1 be given by a matrix M =(a bc d

)∈ GL(2,R). The map f as a function of the affine coordinate x is ex-

pressed as

f(x) =

[a bc d

] [x1x2

]=

[ax1 + bx2cx1 + dx2

]=ax1 + bx2cx1 + dx2

=ax+ b

cx+ d,

f(∞) =

[a bc d

] [10

]=

[ac

]=a

c.


(As usual, we may use some “rules of thumb” regarding ∞, which unite these twoformulas, and make the first one work also for x = −d/c when c 6= 0, see 3.5.)

As a corollary to these calculations we get the following theorem:

• Theorem. Projective transformations of the standard projective line P 1, afterthe usual identifications P 1 = R = R ∪ {∞} = R+, are precisely the Mobiustransformations of the real line.

What this practically means is that one-dimensional (real) projective geometry isthe same as one-dimensional inversive geometry.

We make an important remark on possible generalizations. Our treatment of pro-jective geometry, from Section 4.4 on, is based on the algebraic model. This model,and the whole mathematical apparatus of linear algebra behind it, works exactlythe same way over an arbitrary field instead of the reals. That is, we can use thesame definitions, and prove the same theorems when building the theory of projec-tive spaces and projective transformations over a fixed field F other than R. Thefollowing concepts go through basically without change:

– projective spaces and projective subspaces,

– homogeneous coordinates and their relation to affine coordinates in Fn,

– projective transformations, extending affinities, central projections,

– cross-ratio (taking its values in F), and its invariance,

– projective transformations of F = F ∪ {∞} and linear fractional maps over F.

There are only minor instances where the nature of the field plays a role. Such is, forexample, the question of separation of points, which is a typical feature of geometryover the reals – no surprise that it is related to an inequality in the ordered field R.In general, working over an arbitrary field, harmonic relation between pairs is notrelated to the concept of separation. One more fine point about harmonic relationis when dealing with harmonic pairs, we must assume that −1 6= 1 in the field, thatis, the characteristic of F be different from 2.

In geometric applications the most important field, besides the reals, is the complexfield C.

6.4 Complex projective line

We can repeat the calculation done in 6.3 now with complex numbers. The complexprojective line is P 1(C) = P (C2).

• Warning: although C2, as a set, is usually identified with R4, now P (C2) is verymuch different from P (R4), as the equivalence relation by which the quotient istaken, is different.

The standard setup [z1 : z2] ↔ z = z1/z2, [1 : 0] ↔ ∞ identifies P 1(C) withC = C ∪ {∞} = C+. As in the real case, invertible 2 × 2 complex matrices

M =

(a bc d

)induce projective transformations which, as functions of the affine

coordinate z, turn out to be linear fractional maps

f(z) =az + b

cz + d

54 GABOR MOUSSONG

where we use the “rule of thumb” regarding ∞. These maps, by 3.5, are preciselythe orientation preserving Mobius transformations in the plane. Therefore, we endup with the following theorem:

• Theorem. Projective transformations of the complex projective line P 1(C),after the standard identification P 1(C) = C+, are precisely the orientation pre-serving Mobius transformations of the inversive plane C+.

This theorem says that one-dimensional complex projective geometry is preciselythe orientation preserving inversive geometry of Euclidean plane.

If a projective transformation of the real line is orientation preserving, then itsPoincare extension (cf. 3.6) is the same transformation with the same real coeffi-cients but viewed as a complex projective transformation. Poincare extensions oforientation reversing transformations (such as inversions themselves) are not com-plex projective as they are orientation reversing.

6.5 Classification of one-dimensional projectivities

We now return to the real projective line, and classify its projective transformationsby the number of fixed points.

• Definition (elliptic, parabolic, hyperbolic). Let L be a projective line. Aprojective transformation f : L → L is called elliptic, parabolic, or hyperbolic,if the number of fixed points of f in L is 0, 1, or 2, respectively.

By the first theorem of 6.3 a projective transformation of L fixing three points isthe identity of L. Therefore, any non-identity projective transformation of L iseither elliptic, parabolic, or hyperbolic.

• Examples

– Let r denote a rotation of Euclidean plane E about an ordinary point through anangle which is not an integer multiple of π. The projective extension r restrictedto the ideal line ∞E is an elliptic projective transformation of ∞E .

– Projective extension of a nonzero translation of a Euclidean line is a parabolicprojective transformation of the extended line. Its single fixed point is the pointat infinity.

– If h is any homothety of ratio 6= 1 of a Euclidean line, then the projectiveextension h is a hyperbolic projective transformation of the extended line, fixingboth the center of h and the point at infinity.

Remark: One can visualize some projective transformations of P 1 = R using com-plex numbers and Poincare extensions. For t ∈ R let E(t), P (t), and H(t) denotethe following 2× 2 real matrices:

E(t) =

(cos t − sin tsin t cos t

), P (t) =

(1 0t 1

), H(t) =

(cosh t sinh tsinh t cosh t

).

It is easy to check that the projective transformations induced by these matricesare elliptic in the case of E(t), t /∈ Zπ, parabolic for P (t), t 6= 0, and hyperbolicfor H(t), t 6= 0. It is not hard to verify that every orientation preserving projectivetransformation of P 1 to itself, written in a suitable basis, is given by one of thesematrices.


The maps t 7→ E(t), P (t), H(t) are differentiable homomorphisms from the additivegroup R into GL(2,R), and define differentiable actions of R (“flows”) on theprojective line P 1.

It helps us understand these transformations better if we study their Poincareextensions in P 1(C). The orbits of these flows are curves in the complex planealong which the points move as the parameter t varies. One can show that orbitsof the flow E(t) are members of the elliptic (Apollonian) pencil of circles with null-circles at i and −i ∈ C. Under the flow P (t) points move along members of theparabolic pencil of circles tangent to the real axis at 0. Finally, the flow H(t) movesthe points along members of a hyperbolic pencil of circles intersecting at −1 and 1.

6.6 Exercises

1. Calculate the cross-ratio (ABCD) if the four points are given by homogeneous coordinates as

A[2 : −3 : 1], B[1 : 2 : 1], C[7 : 0 : 5], D[1 : −5 : 0].

2. Given three points A[2 : 3 : −2], B[1 : −1 : 2] and C[7 : 3 : 2] in the projective plane. Findhomogeneous coordinates for the point D so that (ABCD) = 9/8.

3. Let A,B,C,D be four distinct collinear points in Euclidean plane and O be a fifth point not onthe same line. Let a, b, c, d denote the lines OA,OB,OC,OD, respectively. Fix an orientationfor the lines a, b, c, d in an arbitrary way. Denote by (ab) the oriented angle (with sign) of

the rotation about O which takes the directed line a to the directed line b. Define the angles(ac), (cb), (ad), (db) similarly. Prove:

(ABCD) =sin(ac)

sin(cb)

/ sin(ad)

sin(db).

4. Let OABCD be a regular pentagon in Euclidean plane, and let A′, B′, C′ and D′ denote

the ideal points of the lines OA, OB, OC and OD, respectively. Show that the cross-ratio(A′B′C′D′) equals the “golden ratio” τ = (1+

√5)/2. (You may use that in a regular pentagon

diagonal/side equals τ . The identity τ2 = τ + 1 may also be useful.)

5. Prove that (ACBD) = 1− (ABCD).

6. Let A, B, C, D be four distinct collinear points, and put λ = (ABCD). Work out the effect of

all possible permutations of the four points on the value of the cross-ratio, like (ABDC) = 1/λ,(ACBD) = 1− λ, etc.

(Hint: Use that three transpositions can generate all permutations.)

7. Let the points A, B and C of Euclidean plane be given by Cartesian coordinates as

A = (3,−1), B = (−3,−4), C = (9, 2).

Find the Cartesian coordinates of the harmonic conjugate of C with respect to the points A,B.

56 GABOR MOUSSONG

8. Suppose a > 1. Find the harmonic conjugate of the point 1

ain the projective line R with

respect to the pair{

1

a−1, 1

a+1

}

.

9. Let A, B, C, D be four distinct collinear points. Show that the pairs {A,B} and {C,D}separate each other harmonically if and only if (ACBD) = (DACB).

10. Let A, B, C and D be four distinct points in a projective line. Prove that there exists a pair of

points that separate harmonically both A from B and C from D if and only if (ABCD) > 0.

11. Let P 2 be understood as extended Euclidean plane. What is the harmonic homology in P 2

determined by the pair (L,A), if(a) A is the ideal point in the direction orthogonal to the ordinary line L, or

(b) L is the ideal line?(Recognize projective extensions of well-known Euclidean transformations in both cases.)

12. Prove: any order two projective transformation (i.e., a projective involution) of the projectiveplane is a harmonic homology.

13. Suppose that a projective transformation f of a projective line L interchanges a pair of points,i. e., f(A) = B and f(B) = A for some A,B ∈ L, A 6= B. Prove that f ◦ f is the identity.

(Hint: Consider the cross-ratio (ABC f(C)), and use that (XY UV ) = (Y XV U).)

14. Prove that equivalence classes of the matrices

M =

(

0 11 0

)

and N =

(

−1 10 1

)

generate a subgroup of PGL(2,R), isomorphic to the dihedral group of order six. Write theelements of this group in fractional linear form. Explain relationship with effect of permutations

on cross-ratio.(Hint: Assuming (ABCD) = x, consider the possible values of the cross-ratio of the same fourpoints under all permutations of the four letters A, B, C, D (cf. 6). Recognize a homomorphism

S4 → D3.)

15. Prove that an element [M ] has order two in the group PGL(2,R) if and only if trM = 0.

(Here trM denotes the trace of the matrix M , that is, the sum of the diagonal elements.)

16. Show that any orientation reversing (i.e., induced by a matrix of negative determinant) pro-

jective transformation of the standard projective line R is hyperbolic.

17. Prove if a projective transformation f of the real projective line has order two in the projectivegroup of the line, then f cannot be parabolic.

18. Let f be a parabolic projective transformation of P 1 with fixed point F . For any point X ∈ P 1

different from F put X′ = f(X) and X′′ = f(X′). Compute the cross-ratio (FXX′X′′).

19. Let F1 and F2 be two distinct points in P 1. For any hyperbolic projective transformation f

with fixed points F1 and F2 define λf = (F1 F2 X f(X)). Check correctness of this definition,verify that λf◦g = λfλg , and conclude that the group of projective transformations fixing both

F1 and F2 is isomorphic to the multiplicative group of nonzero real numbers. Generalize tothe projective line over other fields.

20. Give an example of a projective transformation of the line P 1 = R (that is, a real linear

fractional map) with a single fixed point at 0 in R.

21. Find x ∈ R if (0 1x∞) = (∞ 0 1x) holds in P 1 = R.

7 Conics

7.1 Conics in Euclidean geometry (overview)

Remark: Here, as in 1.9, the term ‘conic’ will mean either an ellipse, a parabola,or a hyperbola. That is, degenerate quadratic curves will not be considered conics.


Rotate a line L in Euclidean 3-space about an axis which meets L at point A atangle α ∈ (0, π/2). The surface C swept by the rotating line is called a cone ofrevolution. Rotated copies of L are called the generators of the cone.

A conic section is obtained as C ∩P where P is any plane not containing the apexA of the cone. The type of the conic depends on the angle β between P and theaxis: it is an ellipse if α < β (including the case β = π/2 when the ellipse is actuallya circle), a parabola if α = β, and a hyperbola if α > β. (If P is parallel to theaxis, then β is understood as 0.) Note that the three cases are characterized byhow many generators are parallel to plane P : none in the ellipse case, one in theparabola case, and two in the hyperbola case.

Interior (exterior, respectively) of the conic K = C∩P is the part of plane P whichfalls in the interior (exterior, resp.) of the cone C. A line in P is tangent to K if ithas one point in common with K, and all the other points of the line are exteriorpoints of K. (Note that this last requirement about exterior points is essential: inthe case of a parabola or a hyperbola there are many lines that meet the conic at asingle point without being tangent.) At each point of K there is a unique tangentline. In general, if a point in the exterior of K is given, two tangent lines meet atthis point. Exceptional are all points of two lines, called asymptotes, in the case ofa hyperbola: there is only one tangent line drawn to the hyperbola from points ofthe asymptotes, and none from their point of intersection.

As discussed in 1.9, a quadratic equation in Cartesian coordinates x, y of Euclideanplane as variables typically defines either an ellipse, a parabola, or a hyperbola.Standard forms of these equations are:

x2

a2+y2

b2= 1 (ellipse), y2 = 2px (parabola),

x2

a2− y2

b2= 1 (hyperbola).

The parameters in these equations are directly related to the shape and size ofthese curves. In the case of the ellipse a and b are equal to the semiaxes. In the

58 GABOR MOUSSONG

case of the parabola the parameter p equals the distance of the focus from thedirectrix. In the case of the hyperbola a equals the real semiaxis, while b (calledthe imaginary semiaxis) is related to the asymptotes of the hyperbola: the slopesof the asymptotes are equal to ±b/a.

7.2 Conics in extended Euclidean plane

When a Euclidean plane P is extended with ideal points, there is a natural way toattach certain ideal points to conics lying in P . We discuss two equivalent ways offinding ideal points for conics.

Let K = C ∩ P be a conic in P where C is a cone of revolution in Euclidean3-space. Choose another plane Q, not through the apex A, and orthogonal to theaxis of C, and consider central projection of plane Q to plane P with center at A.This central projection takes the circle D = C ∩ Q to the conic K. Points of thiscircle are projected along the generators of the cone, interior points of the circleare projected to interior points of the conic, exterior to exterior.

Some points of the circle may get projected to ideal points of P . This happensprecisely when the corresponding generator of the cone is parallel to P . The numberof such generators is zero, one, or two if K is an ellipse, a parabola, or a hyperbola,respectively. If K is a parabola, then this generator of the cone is parallel to theaxis of symmetry of K, and if K is a hyperbola, then the two generators parallelto P are parallel to the asymptotes of K.

This suggests that in the extended plane P , ideal points are attached to conics bythe following rule:

– an ellipse has no ideal points;– a parabola has a single ideal point, namely, that of its axis of symmetry;– a hyperbola has two ideal points, namely, those of its asymptotes.

The Euclidean conic K together with its ideal points, is called the projective closureof K, and is denoted K. Ideal points of P are all exterior to K if K is an ellipse, allexcept one if K is a parabola. If K is a hyperbola, the two ideal points of K dividethe ideal line ∞P to two segments one of which is interior, the other is exteriorto K.

These agreements on ideal points simplify the picture when tangent lines are con-sidered. From now on, a line of P is called tangent to K if it has precisely onepoint in common with K. For example, the ideal line is tangent to a parabola(with the ideal point of the parabola as the point of tangency). The asymptotes ofa hyperbola are tangent to the hyperbola at its ideal points. Now it is true, without


exception, that at any exterior point precisely two tangent lines of the conic inter-sect. Actually, the number of tangents drawn to a conic from a particular point is0, 1, or 2, according as the point is interior, is on the conic, or is exterior.

An important consequence of extending conics with ideal points is that all conicsbecome projectively equivalent. Indeed, all are obtained as central projectionsof circles. Therefore, up to projective transformation, there is only one conic inprojective geometry. Whether it appears an ellipse, a parabola, or a hyperbola inEuclidean plane depends on the position of the ideal line relative to the conic.

Another method of defining ideal points for conics is through ideal solutions tohomogenized equations (as is done in 5.2 in the case of straight lines). Considerthe standard equations given in 7.1; after homogenization they become

x21a2

+x22b2

= x23 , x22 = 2p x1x3 ,x21a2

− x22b2

= x23 .

To find ideal points, set x3 = 0. In the case of the ellipse the only solution isx1 = x2 = x3 = 0 which defines no point in P 2. There is one solution [1 : 0 : 0] inthe parabola case, which precisely is the ideal point of the axis of the parabola. Inthe case of the hyperbola there are two solutions [a : b : 0] and [a : −b : 0], whichprecisely are the ideal points of the asymptotes.

7.3 Conics and quadratic forms

We consider conics in projective plane as defined by quadratic equations in homo-geneous coordinates. General form of a homogeneous quadratic equation in threevariables is

Ax21 +Bx1x2 + Cx22 +Dx1x3 + Ex2x3 + Fx23 = 0

where not all coefficients are zero. It is convenient to introduce the following nota-tion for the coefficients in the same equation using double subscripts:

a11x21 + 2a12x1x2 + a22x

22 + 2a13x1x3 + 2a23x2x3 + a33x

23 = 0 ,

that is, a11 = A, a12 = B/2, etc. The left hand side of this equation is called aquadratic form in variables x1, x2, x3. With vector notation x = (x1, x2, x3) wewrite q(x) for the quadratic form.

In general, a quadratic form in a real vector spaceW is a function q :W → R which,when expressed as a function of the coordinates of its vector variable, becomes ahomogeneous quadratic polynomial. If this is the case with respect to one particularbasis, then it is also homogeneous quadratic with respect to any basis (typicallywith different coefficients), as the change of basis results in homogeneous linearsubstitutions of coordinate variables.

To any three-variable quadratic form

q(x) = a11x21 + 2a12x1x2 + a22x

22 + 2a13x1x3 + 2a23x2x3 + a33x

23

we associate the symmetric 3 × 3 matrix Q =(aij

). It is easily checked that for

the quadratic form one has the concise formula

q(x) = x⊤Qx (x ∈ R3) .

60 GABOR MOUSSONG

We say that the quadratic form q is non-degenerate if detQ 6= 0. In order for aquadratic form to define a conic, it is necessary to be non-degenerate. In addition,in order to rule out equations like x21 + x22 + x23 = 0 which are non-degeneratebut have no solution in P 2, it is also necessary that the quadratic form take bothpositive and negative values. In this case the quadratic form is called indefinite.

Therefore, it follows from the classical theory of quadratic equations in two variablesx, y (already mentioned in 1.9) that the equation given by a three-variable quadraticform defines a conic in the projective plane if and only if the quadratic form is non-degenerate and indefinite.

A projective transformation f = [ϕ] where ϕ ∈ GL(W ), transforms the quadraticform q to another quadratic form q′ = q ◦ ϕ. If coordinates are used, that is,W = R3, the matrix of q is Q, and the matrix of ϕ is M ∈ GL(3,R), thenq′(x) = q(Mx) = (Mx)⊤Q(Mx) = x⊤(M⊤QM)x, showing that the matrix of q′ isM⊤QM . If q is non-degenerate and indefinite, so is q′, therefore we can concludethat projective transformations take conics to conics.

A change of basis in W also corresponds to a projective transformation in P 2. It isa standard fact from linear algebra that any quadratic form can be “diagonalized”,that is, by a suitable change of basis, its matrix can be made diagonal with diagonalelements equal to ±1 or 0. If the quadratic form is non-degenerate, 0 cannot occur,and if indefinite, both +1 and −1 must occur. Then, reordering variables andmultiplying through by −1 if necessary, we (again) obtain the result that all conicsare projectively equivalent to the one given by the equation

x21 + x22 − x23 = 0 .

Note that by converting to Cartesian coordinates this becomes the equation of theunit circle in R2.

7.4 Polarity

Any conic in projective plane gives rise to a beautiful correspondence between pointsand lines of the plane. This correspondence has numerous interesting applications,most notably in various models of hyperbolic geometry, to be discussed later.

For the remainder of this chapter let us fix a conic K ⊂ P 2 with equation q(x) = 0where, as before, q is a quadratic form in R3 with matrix Q, detQ 6= 0.

• Definition (conjugate points). Two points, A = [a] and B = [b], of P 2, arecalled conjugate with respect to K if a⊤Qb = 0.

Note that this relation really depends on the two points, and not on the particularchoice of representative vectors.Conjugacy of points is a symmetric relation: b⊤Qa = (b⊤Qa)⊤ = a⊤Q⊤b = a⊤Qbsince Q is a symmetric matrix.Since the equation of K is x⊤Qx = 0, a point is conjugate to itself if and only if itbelongs to K.

• Theorem(1) Given any point A ∈ P 2, the set of all points conjugate to A is a straight line

in P 2.(2) For any straight line L in P 2 there exists a unique point conjugate to all points

of L.


Proof: (1): A point B = [b] is conjugate to A = [a] if and only if b · Qa =b⊤Qa = 0, that is, if b ∈ (Qa)⊥. Here the vector u = Qa is nonzero sincedetQ 6= 0. Therefore, the set of conjugates to A is the projectivization of the2-dimensional linear subspace with normal vector u, that is, a line.

(2): We can repeat the proof of (1) in reverse direction. Assume L = P (U)where U ≤ R3 is a 2-dimensional linear subspace with normal vector u. Puta = Q−1u. Then a 6= 0 and b⊤Qa = b · u = 0 for all b ∈ U , that is, the pointA = [a] is conjugate to all points of L. Up to a scalar factor, u is the only vectororthogonal to all elements of U , therefore, A is the only point with the desiredproperty.

• Definition (polar, pole). The line given in part (1) of the theorem is calledthe polar line of A, and the point given in part (2) is called te pole of L withrespect to the conic K.

This correspondence, which assigns polars to points and poles to lines, is calledpolarity with respect to K. These assignments clearly are mutually inverse,therefore they are bijective maps between the set of points and the set of linesin P 2.

It is immediate from the definitions that polarity respects incidence: if the pointA belongs to line L, then the pole of L belongs to the polar of A. Indeed, A ∈ Limplies that the pole of L and A are conjugate points, so the pole of L is containedin the polar of A.

• Theorem

(1 ) If A ∈ K, then the polar of A is the line tangent to K at point A.

(1′) If a line L is tangent to K, then its pole is the point of tangency.

(2 ) If A is in the exterior of K, then the polar of A is the line connecting thepoints of tangency of the two tangent lines to K drawn from A.

(2′) If a line L intersects K at two distinct points B and C, then the pole of L isthe point where the tangents to K drawn at B and C intersect.

(3 ) If A is in the interior of K, then the polar of A lies in the exterior of K.

(3′) If a line L misses K altogether, then its pole is in the interior of K.

Proof: Note that claims (1′), (2′), (3′) follow directly from (1), (2), (3), respec-tively, as they merely restate them in different words. It suffices then to prove(1), (2), and (3).

(1): As A is self-conjugate, its polar L must pass through A. By way of con-tradiction, assume L has another point B 6= A in common with K. Then Bis conjugate to both A and itself, implying that L is the polar of B. This is acontradiction since L cannot have two distinct poles A and B.

62 GABOR MOUSSONG

(2): Let the two tangent lines through A have their point of tangency with K atB and C, respectively. Then by (1) (or by (1′)), A is conjugate to both B andC, therefore its polar must be line BC.

(3): If a line L is not in the exterior of K, then it is either tangent to K, orhas two points in common with K. Both these possibilities are exhausted by thecases (1′) and (2′), so L must lie in the exterior.

Our last theorem in projective geometry relates polarity to harmonic separation.This theorem has its most important application in hyperbolic geometry. Anothernice application is given in Exercise 12.

• Theorem. Let A, B be two distinct points of the conicK. Two further points Cand D of line AB are conjugate with respect to K if and only if (ABCD) = −1.

Proof: Let L denote the line AB, then {A,B} = L ∩K. For any point C ∈ L,C 6= A,B, there exists a unique D ∈ L conjugate to C, since the polar line ofC does not pass through C, so it intersects L at a unique point. Conjugacy ofpoints with respect to K is a symmetric relation, therefore it puts all points ofthe set L−{A,B} into pairs. Harmonic relation with respect to pair {A,B} alsoputs these points in pairs. We have to show that these two pairings are actuallythe same. To this end it suffices to show that C and D are conjugate whenever(ABCD) = −1, as the converse will automatically follow.

So assume (ABCD) = −1. Choose representative vectors for C and D: writeC = [c], D = [d], then write a representative for point A in the form a = λc+µd.Then, since (CDAB) = −1, point B is represented by vector b = λc−µd. Bothpoints A and B belong to the conic K with equation q(x) = 0, therefore

0 = q(λc+ µd) = (λc+ µd)⊤Q(λc+ µd) = λ2q(c) + 2λµc⊤Qd+ µ2q(d)

and 0 = q(λc− µd) = (λc− µd)⊤Q(λc− µd) = λ2q(c)− 2λµc⊤Qd+ µ2q(d),

from which λµc⊤Qd = 0 follows. Here λ, µ 6= 0, so c⊤Qd = 0, which meansthat C and D are conjugate.

7.5 Exercises

1. Consider the following equation in Euclidean plane written in Cartesian coordinates:

8x2 + 10xy − 3y2 − 8x+ 2y + 5 = 0

Check that this equation defines a non-degenerate conic. Find ideal points, and determine thetype of this conic (ellipse, parabola, or hyperbola).

2. Let q(x) = 0 be the homogeneous equation of a conic where q is a quadratic form with 3 × 3

symmetric matrix Q = (aij). Show that whether the conic appears as an ellipse, a parabola,or a hyperbola in Euclidean plane is determined by the sign of the upper left 2× 2 minor of Q(i. e., the sign of a11a22 − a2

12).


3. Let K denote a conic in P 2. Call two lines L and M conjugate with respect to K if L passes

through the pole of M . Show that this is a symmetric relation on the set of lines in P 2. Whichlines in P 2 are conjugate to themselves?

4. Let K be a conic, and A, B, C three distinct points in P 2. We say that these three pointsform a self-polar triangle with respect to K if all three pairs A and B, B and C, C and A areconjugate with respect to K. Check that in this case the three points cannot be collinear.

5. Prove that any self-polar triangle (cf. 4) has one vertex in the interior of the conic, and thatthe other two vertices are exterior.

6. Where is the pole of the ideal line of Euclidean plane with respect to (a) an ellipse, (b) a

parabola, (c) a hyperbola?

7. Where is the pole of (a) an asymptote, (b) the ideal line, (c) the imaginary axis, with respectto a hyperbola?

8. Show that the focus and directrix are pole and polar with respect to a parabola.

9. The polar line of a focus F with respect to an ellipse/hyperbola is called a directrix d of thecurve. (So every ellipse/hyperbola has two directrices.) Prove that the ratio of the distance

from d to the distance from F is constant along the ellipse/hyperbola.

10. Prove that (ABCD) = −1.

(Hint: Apply the last theorem of 7.4 to a suitably chosen line, then use the Pappus–Steiner

theorem.)

11. Draw a tangent line to a hyperbola. Prove that the point of tangency bisects the segment ofthe line between the asymptotes.

(Hint: Apply a suitable projective transformation to the diagram in Exercise 10.)

12. Prove that midpoints of parallel chords of a conic in Euclidean plane are all collinear.(Hint: Consider the polar line of the common ideal point.)

64 GABOR MOUSSONG

Part Three: Hyperbolic Geometry

Hyperbolic geometry historically grew out of questions about axioms. In thesenotes a different approach is taken: hyperbolic geometry is treated as a concretemathematical structure through some of its models. Main ingredients of thesemodels are: the structure of subspaces (lines, planes, etc.), the group of congruences,and the metric. For simplicity, the models are presented in dimension two. Mostof the definitions and theorems readily generalize to higher dimension.

8 The Projective Model

8.1 Incidence structure

• Definition (projective model). Let K be a conic in a projective plane P =P (W ), and let X denote its interior. Topologically X is an open disk.The underlying set of the model, that is, the set of points, is X. Hyperbolic linesin X are the sets l = L ∩X where L ⊂ P is any projective line with L ∩X 6= ∅.

Remarks: (1) The axiom of parallels clearly fails for hyperbolic lines. That is,several different lines pass through any point not in a line l without intersecting l.

(2) Hyperbolic lines are proper open intervals of projective lines. Therefore, thereexists a natural concept of order and separation inherited from the geometry of realaffine geometry. So without further ado, we can speak about straight segments,rays, half-planes, triangles, convexity, convex hulls, etc., in hyperbolic plane.

• Definition (Cayley–Klein model). Special choice of coordinates: ifW = R3,the quadratic form definingK is q(x) = x21+x

22−x23, and the plane P is identified

withR2 the usual way, thenX becomes the open unit disk about the origin inR2.In this case the projective model is called the Cayley–Klein model of hyperbolicplane.

It is clear that a suitable projective transformation takes any projective model tothe Cayley–Klein model. A projective transformation taking a model to another iscalled an isomorphism between the two models. All projective models of hyperbolicplane therefore are isomorphic to one another, in particular, to the Cayley–Kleinmodel.


8.2 Congruences

• Definition (congruence). A map f : X → X is called a congruence of themodel if there exists a projective transformation F : P → P which restricts tof . In other words, congruences are isomorphisms of the model onto itself. Twosubsets of X are called congruent if a suitable congruence takes one to the other.All congruences of the model form a group, denoted G(X), under compositionof maps. One may regard G(X) ≤ PGL(W ) as f determines F uniquely.

Remark: The subgroup G(X) of the projective group PGL(W ) consists preciselyof those projectivities F : P → P for which F (K) = K. Indeed, if F preserves K,then it must take interior points to interior points, and exterior to exterior. Thisis easiest to see by counting the number of tangents drawn from a given point.

• Definition (reflection). Let l ⊂ X be any hyperbolic line. Let L ⊂ P be theprojective line for which l = L ∩X, and let A denote the pole of L with respectto the conic K. Consider the harmonic homology h(L,A) : P → P . By the lasttheorem of 7.4, h(L,A) takes K to K, that is, defines a hyperbolic congruenceσl = h(L,A)|X : X → X. This congruence σl ∈ G(X) is called hyperbolicreflection of X in the hyperbolic line l.It is clear that σl is an element of order two in the group G(X), and that lconsists of its fixed points.

For example, if X is the Cayley–Klein model, and l is a diameter of the boundarycircle K, then σl simply is the restriction to X of a Euclidean orthogonal reflection.

Remark: If A ∈ X, then its polar line L is disjoint from K (and from X). Theharmonic homology h(L,A) restricts to a congruence of order two again, but now Ais the only point fixed in X. By analogy with Euclidean geometry this congruencemay be called a central symmetry (or half turn) of X with center at A.

In Euclidean plane two distinct straight lines are perpendicular to each other if andonly if reflection in one of them takes the other one to itself. This motivates thefollowing definition of orthogonality among hyperbolic lines.

• Definition (orthogonal lines). Let l = L ∩ X and m = M ∩ X be twohyperbolic lines in X. We say that l and m are orthogonal lines, l ⊥ m, if Lpasses through the pole of M .

Note that this relation is symmetric as it is equivalent to the poles of L and Mbeing conjugate points with respect to K. Clearly all congruences of X preserveorthogonality of lines, since they are induced by projective transformations whichrespect polarity. Note also that orthogonal hyperbolic lines necessarily intersect.

66 GABOR MOUSSONG

• Lemma. For any A,B ∈ X, A 6= B, there exists a hyperbolic reflection of Xwhich takes A to B and B to A.

Proof: We find the axis of the reflection through the following procedure. LetM denote the projective line through A and B, and let M intersect the conic Kat points U and V . Draw tangent lines to K at U and V which intersect at thepole Q of M . The projective lines QA and QB intersect both arcs into which Kis divided by line M . Choose the points of intersection with one of the arcs, andcall them R and S. Draw tangents at R and S, and let T denote their point ofintersection. Finally, let L be the projective line through Q and T .

Let Z denote the pole of L, then the harmonic homology h(L,Z) restricts to ahyperbolic reflection of X. We claim that h(L,Z)(A) = B. Since T is fixed underh(L,Z), and tangents are taken to tangents, h(L,Z) must take R to S and S toR. Then line QR is taken to line QS and vice versa. Line M passes throughpoint Z, so it is invariant under h(L,Z). Therefore, the point A of intersection ofM with line QR must be taken to the point of intersection of M with line QS,which is B.

Remark: By Euclidean analogy, the hyperbolic line constructed in the lemma maybe called the perpendicular bisector between points A and B. Perpendicularityclearly holds by the construction. After the definition of hyperbolic distance it willbe easy to see that this line indeed bisects the straight segment between A and B.

• Theorem. The group G(X) acts transitively on the set X, and the stabilizersof the points are isomorphic to the orthogonal group O(2).

Proof: Transitivity immediately follows from the preceding lemma. In order tofind the stabilizer we may assume that X is the Cayley–Klein model (i.e., K isthe unit circle in R2), and due to transitivity, it will suffice to find the stabilizerG(X)0 fixing the origin 0 ∈ R2.

All Euclidean symmetries of the model (after projective extension and restrictionto X) belong to G(X), and form the group O(2). Therefore clearly O(2) ≤G(X)0.

Conversely, assume that f(0) = 0 holds for some congruence f of the Cayley–Klein model. Let F ∈ PGL(3,R) be the projectivity inducing f . Now the unitcircle K is invariant under F , and F (0) = 0, therefore the polar line of point 0with respect to K, the ideal line, must also be invariant under F . This meansthat F is the projective extension of an affinity of the plane (which we also denoteby F ). This affinity keeps the origin fixed, so F : R2 → R2 is a linear map. Now


F (K) = K implies that F preserves the norm of vectors, which means that F isan orthogonal linear map. Therefore, G(X)0 ≤ O(2).

Remark: Note that there is a remarkable similarity in how transformations of thethree classical two-dimensional geometries (Euclidean, spherical, and hyperbolic)act: transitively with stabilizer O(2). This roughly means that these geometrieshave the same degree of homogeneity.

• Corollary(1) All lines of the projective model are congruent.(2) Hyperbolic reflections in lines form a generating set in the group of congru-

ences of the projective model.

Proof: (1): It suffices to show that for any line l of the Cayley–Klein model thereexists a congruence which takes l to one particular line of the model, say, theinterval (−1, 1) of the first coordinate axis. Such a map is easily found in twosteps: first choose a point A ∈ l and move it to the origin by a congruence, thenrotate the image of l about the origin through the necessary angle.

(2): We may assume again that X is the Cayley–Klein model. If a congruenceof X leaves the origin fixed, then, being given by an orthogonal 2× 2 matrix, itis either a reflection or a rotation, that is, a product of one or two hyperbolicreflections. If a congruence f takes the origin to some other point A, thenapplying this to σl ◦ f , where σl brings A back to the origin, we get that f itselfis a product of two or three hyperbolic reflections.

8.3 Distance

The key ingredient of the structure of any model of hyperbolic geometry is thecorrect definition of distance. Naturally, we expect lines to have infinite length, andcongruent figures to have equal size. This is achieved by the following definition ofdistance in the projective model.

• Definition (hyperbolic distance, dp). For any two points A,B ∈ X we definetheir hyperbolic distance, a real number denoted dp(A,B), as follows. If A = B,then dp(A,B) = 0. If A 6= B, then let U and V denote the points of intersectionof the projective line through A and B with the conic K, and put

dp(A,B) =1

2

∣∣ ln(ABUV )∣∣ .

Note that the cross-ratio (ABUV ) is positive as {A,B} and {U, V } are non-separating pairs.

• Theorem(1) The function dp is a metric on the set X.(2) The metric dp is invariant under the action of the group G(X) of congruences.

(In other words, all congruences of the model preserve the metric dp.)(3) All hyperbolic lines (equipped with the restriction of dp as metric) are iso-

metric to the real line R with its standard metric.

Proof: (1): Clearly dp ≥ 0, and since (ABUV ) can never equal 1, the hyperbolicdistance is always positive between distinct points. Symmetry of dp is alsoclear from the definition. The triangle inequality for dp is far less obvious. Its

68 GABOR MOUSSONG

easiest proof uses some trigonometric formulas, see 5.4 below. Those formulasare easiest to obtain using a different model, therefore we postpone the proof oftriangle inequality to Section 10.5.

(2): Congruences are projective transformations takingK to itself. As projectivetransformations take lines to lines, and preserve cross-ratio, the claim followsimmediately.

(3): Consider the distance dp restricted to a hyperbolic line l. By (2), we mayassume that X is the Cayley–Klein model, and that l is the interval (−1, 1) ofthe first coordinate axis. For a, b ∈ (−1, 1), a 6= b, by the distance formula wehave

dp(a, b) =1

2

∣∣ ln(−1 1 a b)∣∣ = 1

2

∣∣∣∣ ln(a+ 1

1− a

/b+ 1

1− b

) ∣∣∣∣ =1

2

∣∣∣∣ ln1 + a

1− a− ln

1 + b

1− b

∣∣∣∣ .

The function ϕ : (−1, 1) → R, defined by ϕ(x) = (1/2) ln((1 + x)/(1 − x)

),

is therefore a distance preserving map from the hyperbolic line l = (−1, 1) toR. It is easy to check that ϕ is onto (actually, its inverse is the function y 7→(e2y − 1)/(e2y + 1)). Thus, ϕ is an isometry between l and R.

Remarks: (1) It is known that the properties given in the theorem determine themetric uniquely up to a positive multiplicative constant. That is, the incidencestructure and the group of congruences determine hyperbolic distance up to thechoice of the unit of measurement.

(2) The reason for the presence of the constant factor 1/2 in the distance formulais that this choice of the unit of measurement makes the projective model isometricto other models to be defined later. The distance given by this formula (with theconstant 1/2) is called the “natural” distance in hyperbolic geometry.

(3) The distance formula was discovered by Cayley in 1859 in connection withquadratic equations, and it was Klein who realized in 1868 that this is what defines amodel for non-Euclidean geometry, thereby solving the several centuries old problemof parallels. (That is, whether the axiom of parallels is deductible from the restof Euclid’s axioms. The mere existence of this model answers this question in thenegative.)

8.4 Angle

• Definition (hyperbolic angle). Let l and m be two rays with a commonendpoint A in the projective model X. We measure the hyperbolic angle formedby l and m as follows. Choose an isomorphism f between X and the Cayley–Klein model such that f(A) = 0. The images f(l) and f(m) of l and m underthis isomorphism are two half-open straight segments emanating from the origin,i.e., two radii of the unit circle. The angle of l and m is then defined as the angleof f(l) and f(m).


For correctness of this definition let us choose another isomorphism g with g(A) =0, and check whether the same angle is obtained using g instead of f . By thetheorem of 8.2, the image under g of the pair of rays k,m differs from the imageunder f by an orthogonal transformation (namely, by g ◦ f−1), therefore, thetwo angles are equal.

It is clear that hyperbolic angle is invariant under congruences of the projectivemodel. By looking at polarity with respect to the unit circle in R2, we see thatlines we defined orthogonal in 8.2 indeed form angle π/2.

Three non-collinear points of the projective model span a hyperbolic triangle. An-gles of a triangle are formed by pairs of rays emanating from a vertex, and passingthrough the other two vertices. The next theorem, called the hyperbolic Law ofCosines, will be proved later in Section 10.5, using the apparatus of a differentmodel.

• Theorem Let a = dp(B,C), b = dp(C,A), and c = dp(A,B) denote the sides ofa triangle ABC in X, and let α be the angle at vertex A. Then

cosh a = cosh b cosh c− sinh b sinh c cosα .

• Corollary If A, B, and C are three non-collinear points in X, then dp(A,B) +dp(A,C) > dp(B,C).

Proof: We use the notations of the theorem. Because of α < π, we have cosα >−1. Then from the theorem cosh a < cosh b cosh c + sinh b sinh c = cosh(b + c)follows, which implies a < b+ c.

It follows that the triangle inequality holds for the distance function dp of theprojective model, that is, dp is indeed a metric on the set X. (For collinear triplesof points this follows from statement (3) of the theorem of 8.3.)

8.5 Ideal boundary

• Definition (points at infinity, ideal boundary). The conic K used in thedefinition of the projective model X is often called the ideal boundary of X, andis denoted as ∂X. Its elements are called the points at infinity of the model.The set ∂X = K is homeomorphic to a circle, and the set X = X ∪ ∂X, theclosure of X, is homeomorphic to a closed disk.

If l = L ∩ X is any hyperbolic line in X, then the set L ∩ ∂X consists of twopoints. These two points are called the points at infinity of the hyperbolic line l.

It is clear now that for any two distinct points (ordinary or at infinity) of X thereexists a unique hyperbolic line connecting them.

Remark: Isomorphisms between projective models (in particular, congruences ofone model) are defined on the closure X, and they are homeomorphisms. In con-trast, the hyperbolic metric of X cannot be extended to the ideal boundary: it canbe shown that no metric exists on ∂X which is invariant under the action of G(X),and is compatible with the topology of ∂X.

70 GABOR MOUSSONG

8.6 Exercises

1. Given a hyperbolic line l ⊂ X and a point A ∈ X, show that there exists a unique hyperbolicline in X through A which is orthogonal to l.

2. Prove or disprove: If l and m are two distinct hyperbolic lines in hyperbolic plane, then eitherl ∩m 6= ∅, or there exists a hyperbolic line orthogonal to both l and m.

3. Given two distinct hyperbolic lines l and m in hyperbolic plane, use the projective model toshow that there cannot exist more than one hyperbolic line orthogonal to both l and m.

4. Prove that all right angles in hyperbolic plane are congruent. More generally, let l,m and l′,m′

be two pairs of intersecting hyperbolic lines, and show that the two pairs are congruent if andonly if they intersect at equal angle.

5. Let l = L ∩ X and m = M ∩ X be two distinct hyperbolic lines in the projective modelof hyperbolic plane X such that the projective lines L and M meet on the boundary conic.Apply hyperbolic reflections in l and m to an arbitrary point A ∈ X to get points B and C,

respectively. Prove if A /∈ l ∪m, then A,B,C cannot be collinear.

6. Prove that two distinct hyperbolic reflections commute if and only if their axes are orthogonal.

(Hint: For the “if” part apply a hyperbolic congruence to take the point of intersection ofthe axes to the center of the Cayley–Klein model. For the “only if” part observe that two

transformations can commute only if the set of fixed points of one is invariant under theother.)

7. Suppose that a point A of the Cayley–Klein model has hyperbolic distance d from the origin.

Show that Euclidean distance of point A from the origin equals tanh d.

8. In the Cayley–Klein model an angle is formed by two rays one of which passes through the

origin. Show that the angle is a right angle (in the sense of the model) if and only if it appearsas a right angle in the ambient Euclidean plane.

9. Let α denote the angle of an equilateral hyperbolic triangle of side length a. Use the hyperbolicLaw of Cosines to prove that α → 0 as a → ∞.

9 The Poincare Models

The construction of our next model of hyperbolic plane is based on inversive geom-etry. This model was discovered by Poincare when working with complex functions.


• Definition (Poincare disk model). Let E denote a Euclidean plane, and letK ⊂ E be a circle with interior Y . The underlying set of the so-called Poincaredisk model is Y .


Hyperbolic lines in Y are subsets of the form l = C ∩ Y where C ⊂ E is anycircle or straight line such that C ⊥ K. In other words, a line in this model iseither an open circular arc orthogonally connecting two points of K, or an opendiameter.

The usual incidence properties of points and lines are less obvious now than inthe case of the projective model. Let us see, for instance, why there is a uniquehyperbolic line through any given pair of distinct points A, B ∈ Y . Any circle orline that meets K orthogonally is its own inverse with respect to K. Let thereforeA′ = σK(A) be the inverse of point A. Then A, A′, and B are three distinct pointsof the inversive plane E+, so there extists a unique circle or line C through them.As C passes through a pair of distinct inverse points, it intersects K orthogonally.

Actually, it turns out that the incidence structure of this model is isomorphic tothat of the projective model. Let us use the same circle K as a conic to build theprojective model on the underlying set X = Y in the projective extension of E. Wedefine the isomorphism Φ : X → Y as follows.

• Definition (map Φ). Embed E as a plane in Euclidean 3-space, and let S bea sphere with K as a great circle. Think of E as a horizontal plane, and K asthe equator of S. The map Φ is the following composition of two maps.

First, the disk X is projected vertically to the northern hemisphere of S, thenthis hemisphere is projected centrally (through a stereographic projection, see3.8), from the south pole as center, back to E. It is clear that the composition ofthe two projections is a bijective map from the interior of K to itself. (Actually,it is also defined on K, keeping all points of K fixed.) Hyperbolic lines of theprojective model (that is, chords ofK) are first taken to vertical semicircles, then,as a stereographic projection takes circles to circles or lines, and preserves angle,to hyperbolic lines of the Poincare disk model. Therefore, Φ is an isomorphismbetween the two incidence structures.

72 GABOR MOUSSONG

All further structure of the Poincare disk model is defined so that the map Φeventually is an isomorphism between the two models.

9.2 Congruences

• Definition (congruence, reflection). Congruences of the Poincare disk modelare obtained as Poincare extensions of Mobius transformations of the boundarycircle. That is, a congruence of Y is any map of the form f |Y : Y → Y wheref = PE

K (g) and g ∈ M(K). The group of congruences of this model is thereforeisomorphic to the Mobius group M(K).

We know that the group M(K) is generated by inversions of K with respectto pairs of points in K, and that Poincare extensions of these generators areprecisely the inversions and reflections of the plane in circles and lines meetingK ortogonally. Thus, the group of congruences of the model is generated bythese inversions and reflections, restricted to Y . These generators are calledhyperbolic reflections in this model. Axes of these reflections are just what wecalled hyperbolic lines in 9.1.

Recall that the group of congruences G(X) of the projective model of hyperbolicplane is generated by hyperbolic reflections, that is, harmonic homologies wherecenter and axis are pole and polar with respect to K. When K is a circle, note thatrestriction of such a harmonic homology to K is an inversion of K with respect toa pair of points, and all inversions of K are obtained this way. This means that,when restricted to circle K, the groups of congruences of the projective model andof the Poincare disk model have precisely the same transformations as generators.This implies that the two groups are isomorphic, and act exactly the same way onthe boundary circle K. Thus, we have proved the following theorem.

• Theorem. If K is the boundary circle of both the projective model X and thePoincare disk model Y , then G(X) ∼= M(K), and the actions of the two groupson K are identical.

Remark: Transformations of the two models are the same on the boundary, butthey are extended differently to the interior of the circle. It is not hard to seethat the relationship between the two types of transformations is given by themap Φ. A map f : X → X is a congruence of the projective model if and only ifΦ ◦ f ◦ Φ−1 : Y → Y is a congruence of the Poincare disk model.

9.3 Distance and angle

As we have an explicit bijection Φ : X → Y between the underlying sets, and wewant this map to be an isomorphism between the projective and Poincare models,we have no choice when defining the concepts of distance and angle in the latter.We simply copy these from X to Y via Φ. As it turns out, the result is particularlypleasing in the case of angles.

• Definition (hyperbolic distance, dc). For any two points A,B ∈ Y we definetheir hyperbolic distance in the Poincare disk model by the formula

dc(A,B) = dp(Φ−1(A),Φ−1(B)

)


where dp is distance in the projective model. (The subscript c indicates that dc isdistance in the “conformal model” which is a common name of various versionsof the Poincare model.)

Remark: It is possible to define the cross-ratio of four distinct points of a circle (orof any conic, for that matter). Using this version of cross-ratio, one has a formulasimilar to the Cayley distance formula: dc(A,B) =

∣∣ ln(ABUV )∣∣, where U and V

are the points where the circle or line containing the hyperbolic line through A andB meets K.

• Definition (hyperbolic angle). Let k and m be two hyperbolic rays with acommon endpoint in the Poincare disk model Y . The hyperbolic angle formedby k and m is defined as the hyperbolic angle of the rays Φ−1(k) and Φ−1(m)in the projective model.

This definition of angle clearly turns Φ into an angle preserving map between thetwo models. As a consequence, it follows that all congruences of the Poincare diskmodel preserve hyperbolic angle.

• Theorem. Let k and m be two rays, with common endpoint A, in the Poincaredisk model. Then the hyperbolic angle formed by k and m equals their apparentEuclidean angle, that is, the angle in E between the two tangent rays drawn tok and m at A.

Proof: If A is at the center of circle K, then Φ−1(k) = k and Φ−1(m) = m, sothe claim follows from the definition of angle in the Cayley–Klein model. If A isarbitrary, then a suitable congruence of the Poincare disk model takes A to thecenter. Congruences of the Poincare model preserve both hyperbolic angle andEuclidean angle, therefore the two angles are equal at A.

Remark: The property of the Poincare disk model stated in the theorem is usuallycalled “conformality” of the model. This simply means that all angles are repre-sented faithfully. What’s behind this visually appealing property is the fact that allcongruences of this model are angle-preserving (a.k.a. conformal) transformationsof the ambient inversive plane.

9.4 Poincare half-plane model

If an arbitrary Mobius transformation µ ∈ M(E) of the plane E is applied to thePoincare disk Y , then the whole structure of the disk model can be copied via µonto the new underlying set µ(Y ). This way a new conformal model is obtainedwhich clearly is isomorphic to the original Poincare disk model. We apply thisprinciple to construct a model whose underlying set is an open half-plane.

• Definition (Poincare half-plane model). Choose a point P ∈ K, and con-sider an inversion σ : E+ → E+ centered at P , with arbitrary radius. Thisinversion takes K to an inversive line L+ ⊂ E+, and Y to one of the openhalf-planes determined by L. Copy the structure of the Poincare disk by σonto the half-plane U = σ(Y ). This model, which is a conformal model by theangle-preserving nature of inversion, is called the Poincare half-plane model withunderlying set U .

Remark: Clearly, instead of inversion σ, we could have used any Mobius transfor-mation which takes a point of K to ∞.

74 GABOR MOUSSONG

Hyperbolic lines in the half-plane model are open semicircles and open rays in U ,meeting L orthogonally. Congruences are restrictions to U of Poincare extensionsof Mobius transformations of L+. The inversive line L+ is the ideal boundary ofthe Poincare half-plane model.

A special choice of coordinates defines the “canonical” version of the Poincare halfplane. For underlying set we choose the upper half-plane

U = {(x, y) ∈ R2 : y > 0}

of the coordinate plane R2. Ideal boundary of this model is the inversive line R+

(the first coordinate axis), group of congruences is the Mobius group M(R) of thereal line.

This model establishes a direct link between the geometry of hyperbolic plane,and the geometry of real and complex projective line. We know from 6.3 thatMobius transformations of R+ are the same as projective transformations of thereal projective line R = R+. Thus, the group of congruences of hyperbolic planeis isomorphic to the projective group PGL(2,R). If R2 is identified with thecomplex plane C the usual way, then this group acts on U through linear fractionaland semilinear fractional maps with real coefficients (cf. 3.6 and 6.3). The formulafor this action is the same as the formula in 3.6 for the Poincare extension: letf = [M ] ∈ PGL(2,R) be any projective transformation of the line where M =(a bc d

), then for z ∈ U

f(z) =

az + b

cz + dif detM > 0,

az + b

cz + dif detM < 0.

• Examples. Consider the matrices E(t), P (t), and H(t) given at the end of 6.5as examples of elliptic, parabolic, and hyperbolic, respectively, transformationsof real projective line. The flows defined by these matrices on U now are typicalexamples of congruences (rigid motions) of hyperbolic plane, viewed in the upperhalf-plane model. We shall discuss these examples again in 11.4.

9.5 Exercises

1. Show that no circle that represents a hyperbolic line in the Poincare disk model can contain

the center of the model in its interior.

2. We know (from the projective model) that in the hyperbolic plane one can drop a uniqueperpendicular from any given point to any given line. Verify this statement directly in thePoincare disk model.

(Hint: Use orthogonal pencils of circles.)


3. Let K ⊂ R2 be the unit circle about the origin (i.e., the circle x2 + y2 − 1 = 0). Consider the

Poincare disk model in K, and suppose that some hyperbolic line is represented by a circlewith equation x2 + y2 + ux+ vy + w = 0. Prove w = 1.

4. Let O denote the center of the boundary circle K of the Poincare disk model Y , and let A ∈ Ybe a point different from O. Let C be the circle such that hyperbolic reflection in the hyperbolic

line l = C ∩ Y maps O to A. (That is, let l be the hyperbolic perpendicular bisector of thehyperbolic line segment from O to A.) Show that the center of C is the inverse of A withrespect to circle K.

5. Consider the following four points in the Poincare upper half-plane model U ⊂ R2:

A = (−0.001, 0.001), B = (0.001, 0.001), A′ = (−1000, 1000), B′ = (1000, 1000) .

Find a congruence of the model that takes A to A′ and B to B′, thereby showing that the

hyperbolic distance from A to B equals the hyperbolic distance from A′ to B′.

6. Define a (doubly infinite) sequence of points in the upper half plane model by the formula

(a, 2nb), n ∈ Z,

where a, b are arbitrarily fixed real numbers, b > 0. Prove that these points are evenly spacedalong the hyperbolic line x = a, y > 0.(Hint: use congruences.)

7. Consider the circle with equation x2 + y2 − 4y + 3 = 0 in the upper half-plane model ofhyperbolic plane. We know that this circle represents a circle in hyperbolic plane. Find the

coordinates of its center (where the center is understood in hyperbolic sense).

8. Let A = (0, a) and B = (0, b) be two points in the Poincare upper half-plane model, 0 < a < b.

Find the equation of the semicircle that represents the perpendicular bisector of A and B.

9. Let A = (1, 6) and B = (3, 4) be two points in the Poincare upper half-plane. Find theequation (in terms of coordinates x, y in R2) of the circle that represents the hyperbolicperpendicular bisector of A and B.

10. In the upper half plane model U ⊂ R2 of hyperbolic plane consider the following two subsets:

G = { (x, y) ∈ U : y = 3 } , H = { (x, y) ∈ U : x2 + y2 − 4y = 0 } .Show that G and H are congruent.

11. Given two distinct points z, w of the upper half plane model U ⊂ C, find formulas (in terms ofz, w ∈ C) for the center and radius of the semicircle that represents the perpendicular bisectorof z and w.

10 The Hyperboloid Model

The last one of our models of hyperbolic plane comes with an extensive mathemat-ical formalism based on linear algebra, and is most usable to carry out calculations.A remarkable feature of this model is its resemblance to the geometry of the sphere.Most of the constructions and formulas have their counterpart in spherical geome-try.

10.1 Minkowski space and Lorentz transformations

• Definition (Minkowski space). A pair (W, q) is called a (3-dimensional)Minkowski space ifW is a real vector space with dimW = 3, and q :W → R is aquadratic form of type (2, 1), that is, when expressed as a function of coordinatesin a suitable (diagonalizing) basis, it becomes

q(x) = x21 + x22 − x23 .

76 GABOR MOUSSONG

Standard Minkowki space: R2,1 is R3 equipped with the quadratic form above.Any Minkowski space is isomorphic to this one.

The symmetric bilinear function associated with q is denoted by 〈 , 〉, that is, instandard coordinates

〈x,y〉 = x1y1 + x2y2 − x3y3 .

• Definition (q-orthogonality). Two vectors x and y of Minkowski space Ware called q-orthogonal (denoted x ⊥ y) if 〈x,y〉 = 0. For a linear subspaceV ≤ W , put V ⊥ = {x ∈ W : x ⊥ y for all y ∈ V }. The restricted quadraticform q|V is non-degenerate if and only if V ∩ V ⊥ = {0}.Terminology: x ∈ W is called spacelike, timelike, or lightlike if q(x) > 0, < 0,= 0, respectively. The “light cone” is the (double) cone with equation q(x) = 0.Spacelike vectors form the interior of the light cone. A 2-dimensional subspaceV ≤W is called spacelike, timelike, or lightlike, if q|V is definite, non-degenerateindefinite (i.e., of type (1, 1)), or degenerate, respectively. (Equivalently, if V =u⊥ where u 6= 0 is timelike, spacelike, lightlike, respectively.)

Remark: If (W, q) is a Minkowski space, and the projective plane P (W ) is consid-ered, the equation q(x) = 0 defines a conic K. Conjugacy of points with respect toK is then equivalent to q-orthogonality of representative vectors. A nonzero vectoris spacelike, timelike, or lightlike if and only if it represents an exterior point, aninterior point, or a point of K, respectively. A 2-dimensional subspace V ≤ W isspacelike, timelike, or lightlike if and only if the projective line P (V ) avoids K,intersects K at two points, or is tangent to K, respectively.

• Lemma (Reverse Cauchy–Schwarz inequality). Let x and y be timelikevectors in W . Then

|〈x,y〉| ≥√q(x)q(y) ,

where equality holds if and only if x and y are linearly dependent.

Proof: For linearly dependent vectors, say when y = λx, direct substitutionimmediately shows equality. If x and y are independent, then they span a two-dimensional subspace V . The matrix of the quadratic form q|V in the basisformed by x and y is

B =

(q(x) 〈x,y〉〈x,y〉 q(y)

).

Since q|V is non-degenerate and indefinite, detB < 0. This is just equivalent tothe strict version of the inequality.

• Definition (Lorentz transformation, Lorentz group). A linear map ϕ ∈GL(W ) is a Lorentz transformation if q is invariant under ϕ, that is, q ◦ ϕ = q.Equivalently, 〈ϕ(x), ϕ(y)〉 = 〈x,y〉 for all x,y ∈ V (i.e., ϕ is a q-orthogonalmap). A Lorentz transformation is called positive if it does not interchange thetwo nappes of the light cone.

The group O(q) formed by Lorentz transformations (i.e., the orthogonal group ofq) is called the Lorentz group of the Minkowski space. In standard coordinatesit is denoted as O(2, 1). The positive Lorentz transformations form a subgroupof index two (denoted O+(q) or O+(2, 1)).



• Definition (hyperboloid model). Consider the quadric surface in W = R2,1

given by the equation q(x) = −1, that is, by

x21 + x22 − x23 = −1

in standard coordinates. This is a hyperboloid of two sheets. Choose one sheet,say, the one with x3 > 0, and denote this sheet by Z. The underlying set ofthe model is Z. All subsets of the form V ∩ Z where V ≤ W is a timelike2-dimensional subspace, are called hyperbolic lines in Z.

• Definition (map Ψ). Let K denote the conic with equation q(x) = 0 in theprojective plane P (W ), and let X be the interior of K. Points of X are repre-sented by vectors x ∈W with q(x) < 0. We define the so-called projectivizationmap Ψ : Z → X, from the hyperboloid model to the projective model of hyper-bolic plane, by Ψ(x) = [x] (x ∈ Z). This map is bijective, and respects incidencestructure (i.e., takes hyperbolic lines to hyperbolic lines).

It is useful to consider the following “visual” interpretation of the map Ψ. As-sumeW = R2,1 and embed R2 in R2,1 as the affine plane x3 = 1. This identifiesK with the unit circle about the origin in R2, and X with the Cayley–Kleinmodel. Then Ψ simply is projection centered at the origin of R2,1.

78 GABOR MOUSSONG

10.3 Congruences

• Definition (congruence). Positive Lorentz transformations take the modelspace Z to itself, preserving the incidence structure. Their restrictions to Z arecalled the congruences of hyperboloid model. Thus, the group of congruences isO+(q).

• Theorem. Under the projectivization map Ψ : Z → X, congruences of thehyperboloid model get transformed precisely to the congruences of the projectivemodel. That is, the correspondence ϕ 7→ [ϕ]|X is a group isomorphism betweenthe positive Lorentz group O+(q) and the group G(X).

Proof: It is clear that the given correspondence is an injective homomorphism.Surjectivity follows from the fact that the quadratic equation of a conic is deter-mined up to a scalar factor by the conic (as a set of points).

• Example. We define reflections in lines of the hyperboloid model. Let l ⊂ Z bea hyperbolic line, l = V ∩Z where V ≤W is a timelike plane. Choose a nonzerovector u ∈ V ⊥, then u is spacelike. Normalize so that q(u) = 1, and define thelinear map σV :W →W as

σV (x) = x− 2〈u,x〉u .

One can verify by direct calculation that σV is a Lorentz transformation:

q(σV (x)

)= 〈x− 2 〈u,x〉u , x− 2 〈u,x〉u〉= 〈x,x〉 − 4〈u,x〉〈u,x〉+ 4〈u,x〉2〈u,u〉= q(x) .

As σV has fixed points within the light cone (for instance, all points of linel are fixed), it is a positive Lorentz transformation. Therefore, it restricts to acongruence σl = σV |Z . This congruence fixes l pointwise, and is called hyperbolicreflection in l.

Linear reflections in W projectivize to harmonic homologies in P (W ), thereforereflections in the two models correspond to each other under the projectivizationmap. That is, σΨ(l)

(Ψ(x)

)= Ψ

(σl(x)

)holds for all x ∈ Z.

10.4 Distance

• Definition (hyperbolic distance, dh). For x, y ∈ Z their hyperbolic distancein Z is defined as

dh(x,y) = cosh−1(− 〈x,y〉

).

For this formula to make sense −〈x,y〉 must be ≥ 1. The reason for this is thefollowing. If x is timelike (as is the case when x ∈ Z), then x⊥ is a spacelike plane.Therefore, the whole nappe of the light cone containing x is on the same side of thisplane as x. So, 〈x,y〉 < 0 for all x, y ∈ Z. The reverse Cauchy–Schwarz inequality

then implies −〈x,y〉 = |〈x,y〉| ≥√q(x)q(y) = 1.

It is clear that dh is non-negative (equal to zero only when x = y), symmetric, andinvariant under positive Lorentz transformations. Triangle inequality for dh willfollow as a consequence of the Law of Cosines below.


• Theorem. The projectivization map Ψ : Z → X is an isometry between thehyperboloid model and the projective model. That is,

dp([x], [y]

)= dh(x,y)

holds for all x, y ∈ Z.

Proof: Given x 6= y, one can always choose lightlike vectors u and v which forma basis for the 2-dimensional subspace spanned by x and y, such that x = u+v.Write y = λu+ µv, then

−1 = 〈u+ v,u+ v〉 = 2〈u,v〉 and − 1 = 〈λu+ µv, λu+ µv〉 = 2λµ〈u,v〉 ,

from which 〈u,v〉 = −1/2 and λµ = 1 follows. Put t = dh(x,y), then

cosh t = −〈x,y〉 = −〈u+ v, λu+ µv〉 = −(λ+ µ)〈u,v〉 = λ+ λ−1

2,

so either λ = et or λ = e−t. Now

dp([x], [y]

)=

1

2

∣∣ ln([u] [v] [x] [y]

)∣∣ = 1

2

∣∣∣∣ln(1

1

/λ

µ

)∣∣∣∣ =1

2| ln(λ2)| = | lnλ| = t .

10.5 Tangent vectors, parametric lines, angles

• Definition (tangent vector, tangent plane). A vector v ∈ R2,1 is calledtangent to Z at the point x ∈ Z if v ⊥ x ( ⇐⇒ 〈x,v〉 = 0). The tangent planeat x ∈ Z is the q-orthogonal complement TxZ = x⊥ ≤ R2,1, a 2-dimensionalspacelike linear subspace.

Remark: Just like in spherical geometry, by this definition TxZ is not actuallytangent to Z, but instead, it is a linear subspace through 0. It is not hard to seethat its translate x + TxZ really is tangent to Z at x: we know that 〈x,x〉 = −1,while 〈x,y〉 < −1 for all y ∈ Z, y 6= x, follows from the reverse Cauchy inequality.Therefore, the whole set Z (except for its single point x) lies on the side of theplane x+ TxZ = {y ∈W : 〈y,x〉 = −1} that does not contain the origin.

The restriction of the quadratic form q to any tangent plane is positive definite,therefore the tangent planes carry the structure of a Euclidean vector space. Inparticular, norms and angles of vectors in TxZ are defined as in Euclidean geometry,now using the inner product 〈 , 〉.• Definition (direction vector). Given a hyperbolic line l = V ∩Z and a point

x ∈ l, a direction vector for l at x is any nonzero vector in the one-dimensionalsubspace V ∩ TxZ.

Suppose u is a direction vector of unit norm for the hyperbolic line l ⊂ Z at x ∈ l.Then the formula

r(t) = (cosh t)x+ (sinh t)u

is a parametric representation of l as shown by the calculation

q(r(t)

)= 〈cosh tx+ sinh tu , cosh tx+ sinh tu〉= cosh2 t 〈x,x〉+ sinh2 t 〈u,u〉 = − cosh2 t+ sinh2 t = −1 .

80 GABOR MOUSSONG

This is arc length parameterization, i.e., for any t1, t2 we have dh(r(t1), r(t2)) =|t1 − t2|. This follows from the calculation cosh dh(r(t1), r(t2)) = 〈r(t1), r(t2)〉 =− cosh t1 cosh t2 + sinh t1 sinh t2 = − cosh(t1 − t2).

• Definition (hyperbolic angle). Given two rays in Z with a common endpointx, the angle between the rays is defined as the angle of the direction vectors atx. As TxZ is a Euclidean vector space, this angle is calculated using the innerproduct and the familiar inverse cosine formula.

This concept of angle clearly is invariant under positive Lorentz transformations.The projectivization map Φ : Z → X preserves this angle at the bottom point(0, 0, 1) ∈ R2,1 of the model. Then it follows from invariance that Φ is anglepreserving everywhere.

Now we turn to triangles in hyperbolic plane. The reader may note that the waywe prove our basic theorem in hyperbolic trigonometry (the Law of Cosines below)is almost word by word identical to our argument in spherical geometry given inSections 2.2–3.

Three points x, y, z ∈ Z determine a triangle if and only if they are linearlyindependent. The triangle ∆ with vertices at x,y, z can be defined just like in thecase of spherical triangles (cf. 2.2), namely:

∆ = Z∩ (convex hull of the three rays from the origin through x, y and z), orequivalently,∆ = Z∩ (intersection of the three half-spaces in R2,1 bounded by the linear2-planes spanned by two of x, y, z and containing the third as interior point).

Sides of ∆ are the pairwise connecting hyperbolic line segments between thevertices.

• Theorem (hyperbolic Law of Cosines). Let α, β, γ denote the angles atvertices x, y, z, respectively, of a hyperbolic triangle, and let a, b, c denote theside lengths opposite x, y, z, respectively. Then


Proof:

Choose unit tangent vectors u,v ∈ TxZ at the endpoint x to the sides from xto y and z, respectively.Parameterize side xy and substitute the parameter value c, and get the formulay = (cosh c)x+ (sinh c)u.Similarly, parameterize xz, substitute b, and get z = (cosh b)x+ (sinh b)v.Taking inner product, 〈y , z〉 = (cosh b)(cosh c) 〈x ,x〉 + (sinh b)(sinh c) 〈u ,v〉.Here 〈y , z〉 = − cosh a, 〈x ,x〉 = −1, 〈u ,v〉 = cosα, and the theorem follows.


• Corollary. In any hyperbolic triangle with sides a, b, c the strict triangleinequality a < b+ c holds.

Proof: In the Law of Cosines formula one has cosα > −1, so

cosh a < cosh b cosh c+ sinh b sinh c = cosh(b+ c) .

This implies the triangle inequality since the hyperbolic cosine function is strictlymonotone increasing.

It follows now that dh is a metric on the set Z. As dp([x], [y]) = dh(x,y) (for all x,y ∈ Z), and dc

(Φ(A),Φ(B)

)= dp(A,B) (for all A,B ∈ X), this finally proves that

the triangle inequality holds for the distances defined in both the projective modeland the Poincare models.

Remark: Recall that the maps Ψ : Z → X and Φ : X → Y , both definedthrough projections, are isomorphisms between the hyperboloid, Cayley–Klein, andPoincare disk models. There exists a similarly defined direct isomorphism betweenthe hyperboloid model Z and the Poincare disk model Y . As in 10.2, K ⊂ R2

is identified with the unit circle about origin in the plane x3 = 1. Now centralprojection with center at (−1, 0, 0) ∈ R2,1 maps Z onto the interior Y of K, andpreserves all structure. This map, as the reader may verify, happens to coincidewith the composition Φ ◦Ψ.

10.6 Exercises

1. Let a and b be two linearly independent vectors in Minkowski space. Show that the two-dimensional linear subspace spanned by a and b is spacelike, timelike, or lightlike if and onlyif 〈a,b〉2 < q(a)q(b), 〈a,b〉2 > q(a)q(b), or 〈a,b〉2 = q(a)q(b), respectively.

2. Show that a matrix M = (aij) defines a Lorentz transformation in R2,1 if and only if J =

M⊤JM where

J =

1 0 0

0 1 00 0 −1

.

Compare this with the (definite) orthogonality condition M⊤M = I given in Section 1.8.Check that the Lorentz transformation is positive if and only if a33 > 0.

3. Prove that the determinant of any Lorentz transformation is either 1 or −1.

4. Verify that all matrices

cosα − sinα 0sinα cosα 00 0 1

,

1− s2/2 −s −s2/2s 1 s

s2/2 s 1 + s2/2

,

cosh t sinh t 0sinh t cosh t 00 0 1

,

where α, s, t are real parameters, define positive Lorentz transformations in R2,1.

5. Let x,y ∈ Z. Using vectors in W , find a formula for the midpoint of the hyperbolic linesegment from x to y in Z.

(Hint: Make a guess using spherical analogy, then prove.)

6. If x,y ∈ Z are two distinct points, show that the vector y − x is always spacelike, and prove

the formula√

q(y − x) = 2 sinh(

dh(x,y)/2)

. (Note that the norm of y − x is greater thanthe hyperbolic distance between x and y.) Compare this with the elementary geometry fact

that the length of a chord of the unit sphere equals 2 sin(d/2) if the endpoints are sphericaldistance d apart.

82 GABOR MOUSSONG

11 The Hyperbolic Plane

We do not define “the” hyperbolic plane as one concrete mathematical object. Thesymbol H2 for hyperbolic plane denotes a mathematical structure only defined upto isomorphism. All the models discussed in the preceding sections are isomorphicstructures, therefore either one may be taken as a concrete realization of H2. Invarious questions concerning the geometry of H2, we may use the model which isbest suited to provide the answer.

In the construction of the projective and Poincare models a crucial role was playedby the ideal boundary of the model. As in 8.5, this ideal boundary is denoted∂H2, and its elements are called points at infinity of hyperbolic plane. The nat-ural structure of ∂H2 is one-dimensional inversive geometry in which the Mobiustransformations are induced by congruences of H2.

11.1 Hyperbolic parallelism

• Definition (parallel, ultraparallel lines). Let l,m ⊂ H2 be two distinctlines. Consider the projective model, and let L and M be the projective linescontaining l and m, respectively. The hyperbolic lines l and m are called inter-secting, parallel, or ultraparallel, according as the point of intersection L∩M isan interior point of the boundary conic of the model, belongs to the conic, or isexterior to the conic. (For convenience, two coincident lines may be called eitherintersecting, parallel, or ultraparallel.)

Equivalently, two hyperbolic lines with no point in common are parallel if theyshare a point at infinity, and are ultraparallel if they do not.

Remark: In our diagrams, if the ideal boundary is not explicitly shown, the factthat two (or more) lines are parallel and share a point at infinity is indicated by anarrow on each line involved.

In contrast with Euclidean geometry, parallelism of hyperbolic lines is not a tran-sitive relation. Since parallelism involves only one of the two ideal points of aline, it is straighforward to define parallelism of directed lines, or of half-lines, andthis relation is already an equivalence relation. Equivalence classes are in bijectivecorrespondence with points of ∂H2.

• Theorem. Two hyperbolic lines of H2 are ultraparallel if and only if thereexists a third line perpendicular to both. If the first two lines are distinct, thenthe third is uniquely determined.


Proof: Use the projective model, and let the hyperbolic lines l and m be sup-ported by projective lines L and M .

For a third line n, supported by projective line N , to be orthogonal to both land m, it is necessary and sufficient that N pass through the poles of L and M .Such a projective line N exists, and if the two lines are distinct, the poles arealso distinct, and N is unique. The question is whether it supports a hyperbolicline, i.e., whether it has nonempty intersection with the model. This is the case ifand only if its pole L∩M is exterior to the boundary conic, which, by definition,means that L and M are ultraparallel.

The fact whether two lines are intersecting, parallel, or ultraparallel, can be char-acterized by distances between their points. For the following calculations we usethe hyperboloid model Z. Consider a hyperbolic line l = V ∩ Z where V ≤ W isa two-dimensional timelike subspace in Minkowski space W . A normal vector forthe subspace V (i.e., any v ∈ W such that V = v⊥) will serve as a normal vectorfor line l.

• Lemma. Choose a unit normal vector v ∈ W for line l = V ∩ Z, that is, letV = v⊥, ‖v‖ = 1. Then the hyperbolic distance of any point x ∈ Z from line lequals

dh(x, l) = cosh−1(1 + 〈x,v〉2

).

Proof: It follows from the Law of Cosines that the point of l nearest to x isthe foot of the perpendicular dropped from x to l. From the reflection formulagiven in 7.3 this point is x − 〈x,v〉v. The distance of the two points is thencosh−1

(− 〈x,x− 〈x,v〉v〉

)= cosh−1

(1 + 〈x,v〉2

).

• Theorem. Let l and m be two lines of hyperbolic plane. Let a point movealong m indefinitely toward a point A of m at infinity. Then the distance of themoving point from line l converges to zero if A is also a point at infinity of l(that is, if l and m are parallel in the direction of A), otherwise it tends to +∞.

Proof: Parameterize line m in the hyperboloid model, as in 10.5, by the formula

r(t) = cosh tx+ sinh tu =1

2e−t(x− u) +

1

2et(x+ u)

where u ∈ TxZ is the direction vector of unit norm, for line m, pointing in thedirection of A, at an arbitrarily chosen point x ∈ l. Let the line l be given by aunit normal vector v. By the lemma, the distance of the moving point r(t) ∈ mfrom line l is

dh(r(t), l

)= cosh−1

(1 + 〈r(t),v〉2

)

= cosh−1

(1 +

1

4

(e−t〈x− u,v〉+ et〈x+ u,v〉

)2).

84 GABOR MOUSSONG

The limit of this expression, as t → +∞, is either 0 or +∞ according as thecoefficient of et is zero or not.

We claim that the vector x+u represents point A. First, it represents some pointat infinity as q(x+u) = q(x)+q(u) = −1+1 = 0. Second, r(t) → A as t→ +∞,since working in the projective model [r(t)] = [x+(sinh t/ cosh t)u] → [x+u] aslimt→+∞(sinh t/ cosh t) = 1.

Finally, if point A is a point at infinity of line l, then 〈x + u,v〉 = 0 and sodh(r(t), l) → 0. Otherwise dh(r(t), l) → +∞.

Remark: It follows that for distinct ultraparallel lines l and m the distance betweentheir points attains a positive minimum value. Clearly this minimum is the lengthof the segment of the common perpendicular line between l and m.

11.2 Pencils of lines

• Definition (pencil of lines). In a projective plane a pencil of lines is the setof all lines passing through a common point. We use the projective model todefine pencils of lines in hyperbolic plane: a set of lines in H2 = X is a pencilif it consists of all the hyperbolic lines l = L ∩X defined by projective lines Lwhich belong to a pencil in the ambient projective plane.

Depending on whether the common point of the projective pencil is inside theboundary conic, belongs to the conic, or outside the conic, we call the pencil ofhyperbolic lines intersecting, parallel, or ultraparallel. Any two distinct hyper-bolic lines are included in a unique pencil of lines in H2.

An intersecting pencil of hyperbolic lines consists of all lines through a commonpoint of H2, a parallel pencil is a collection of all hyperbolic lines that share acommon point at infinity, and a hyperbolic pencil is the set of all hyperbolic linesperpendicular to a given hyperbolic line.

The following statements describe pencils of lines in other models of hyperbolicplane. They immediately follow by applying the isomorphisms Φ and Ψ betweenthe models.

• Theorem

(1) If the boundary circle of the Poincare disk model intersects all members ofa pencil C of circles (or lines) of the ambient inversive plane orthogonally,then the part of C inside the boundary is a pencil of hyperbolic lines inthe model. It is intersecting, parallel, or ultraparallel, if and only if C is


hyperbolic, parabolic, or elliptic, respectively. All pencils of lines in the modelare obtained this way.

(2) Consider a pencil of planes in the Minkowski space W , that is, the systemof all 2-dimensional linear subspaces containing a fixed line L through origin.The hyperbolic lines of the hyperboloid model Z obtained as intersections ofthese planes with Z form a pencil of lines which is intersecting, parallel, orultraparallel according as the direction of line L is timelike, lightlike, or space-like, respectively. All pencils of lines in the hyperboloid model are obtainedthis way.

• Example. Consider any three distinct points in H2. We claim that the threeperpendicular bisectors, taken for each pair of the points, belong to one pencilof lines.Use the hyperboloid model, and let x, y, and z be the three points in Z. By thereflection formula of 10.3, the vectors z − y, x − z, and y − x serve as normalvectors for the three perpendicular bisectors. These three vectors are linearlydependent, so their q-orthogonal complements belong to a pencil of planes inW .

11.3 Cycles

In Euclidean plane there exist only two types of curves (namely, straight lines andcircles) which are homogeneous in the sense that planar congruences which mapthe curve to itself are capable of moving any point to any other point on the curve.In hyperbolic geometry there exist more types of homogeneous curves, commonlycalled cycles.

• Definition (corresponding points). Let P be a pencil of lines in H2. Twopoints, A,B ∈ H2, are called corresponding points with respect to P if B = σl(A)for some line l ∈ P. (Recall that σl denotes reflection in l.) We use the symbolA ≏P B for correspondence.

Clearly A ≏P B holds if and only if either A = B or the perpendicular bisectorof A and B belongs to P. The relation ≏P oviously is reflexive and symmetric.Due to the last example of 11.2, it is also transitive, therefore, it is an equivalencerelation. If P is an intersecting pencil, then the common point is an equivalenceclass in itself, but in all other cases the equivalence classes are infinite.

• Definition (cycle, horocycle, hypercycle). Equivalence classes defined byrelation ≏P , where P is any pencil of lines, are called cycles generated by P.Note that P is uniquely determined by any cycle; lines that belong to P arecalled the axes of the cycle.

If P is an intersecting pencil, then the cycles generated by P are circles (includinga null-circle) centered at the common point.

If a cycle C is generated by a parallel pencil, then C is called a horocycle.The common point at infinity of its axes is called the center of C. Horocyclesgenerated by the same parallel pencil are called concentric.

A cycle is called a hypercycle if it is generated by an ultraparallel pencil of lines.Let C be a hypercycle with axes in P, and let l be the line orthogonal to allmembers of P. Then l is invariant under all reflections in lines in P, thereforeeither C = l, or C consists of all points at a fixed positive distance from l in

86 GABOR MOUSSONG

one half-plane of l. This is why hypercycles are often called equidistant curves.The line l and the constant distance are called the base line and radius of C,respectively.

• Theorem. All horocycles are congruent. Two circles or two hypercycles arecongruent if and only if they have equal radius.

Proof: Two pencils of lines of the same type are clearly congruent. This impliesthe claim about circles and hypercycles. For horocycles it suffices to check thatconcentric horocycles are congruent. Choose a common axis for the two con-centric horocycles, and observe that the composition f of any two reflections inlines orthogonal to this axis maps the generating pencil to itself. Let one suchline pass through the intersection of one of the horocycles with the axis, and theother pass through the midpoint between the two intersections, then f takes thefirst horocycle to the other.

• Theorem. Any three distinct points of H2 belong to a unique cycle.

Proof: Existence follows from the last example of 11.2. All three perpendicularbisectors must belong to the pencil generating any cycle through the points, fromwhich uniqueness follows.

Remark: It follows that hyperbolic triangles do not necessarily have a circumscribedcircle. There always exists however a unique circumscribed cycle.

It is an interesting question what kind of curves represent cycles in various mod-els of hyperbolic plane. The answer in the case of the Poincare models and thehyperboloid model is as nice as possible.

• Theorem(1) In the Poincare disk model Y the cycles are precisely the nonempty subsets

of form C = D ∩ Y where D is a circle or line in the ambient inversive plane.The cycle C is a circle if D ⊂ Y , a horocycle if D is tangent internally to ∂Y ,and a hypercycle if D intersects ∂Y at a pair of distinct points.

(2) In the hyperboloid model Z the cycles are precisely the nonempty subsetsof form C = P ∩ Z where P is any two-dimensional affine plane (that is, atranslate of a two-dimensional linear subspace) in Minkowski space W . Thecycle C is a circle if the direction of P is spacelike, a horocycle if the directionof P is lightlike, and a hypercycle if the direction of P is timelike. The axesof two cycles belong to the same pencil of hyperbolic lines if and only if thecorresponding planes are parallel.

Proof: (1): Reflections are represented by inversions, and inverse of a circle orline orthogonal to the circle of inversion is itself. Therefore, the cycles generatedby a pencil P are represented by members of the pencil of circles orthogonal tothe pencil of circles that represents P.


(2): A pencil P of lines in the model corresponds to a pencil of planes in Wby the theorem in 11.2. Let L be the common line of this pencil of planes.Reflection in members of P are restrictions of q-orthogonal reflections of Wwhich all keep L pointwise fixed. For x,y ∈ Z the condition x ≏P y meansthat their q-orthogonal bisector plane contains L. By 10.3, the vector y− x is anormal vector to this plane, so x ≏P y is equivalent to y − x ⊥ L. Therefore,equivalence classes by the relation ≏P are intersections with Z of affine planes inW parallel to the two-dimensional subspace L⊥. We know that P is intersecting,parallel, or ultraparallel, according as Z is timelike, lightlike, or spacelike, thatis, as Z⊥ is spacelike, lightlike, or timelike, respectively.

11.4 Transformations

We know that the group of congruences of hyperbolic plane is isomorphic to theprojective group PGL(2,R). When viewed in the Poincare half-plane model, thisgroup acts through linear fractional or semilinear fractional maps, depending onthe sign of the determinant of the matrix (or equivalently, depending on whetherthe transformation is orientation preserving or orientation reversing).

We have already defined hyperbolic reflections in hyperbolic lines. Further typesof congruences can be defined as compositions of two reflections.

• Definition (rotation). Let l,m ⊂ H2 be two intersecting hyperbolic lines.The product σm ◦ σl is called a rotation of H2. When l 6= m (that is, when therotation is different from the identity map), the point l ∩m is called the centerof the rotation.

In the l 6= m case the pencil of lines generated by l and m is invariant under therotation, and so are the cycles generated by this pencil, that is, the circles centeredat the point l ∩m.

We know that in the group of congruences of H2 the stabilizators of points areisomorphic to the orthogonal group O(2). Rotations of H2 are precisely the trans-formations that belong to the SO(2) subgroups.

• Definition (translation). Let l,m ⊂ H2 be two ultraparallel hyperbolic lines.The product σm ◦ σl is called a translation. If l 6= m, let n denote the uniquehyperbolic line orthogonal to both l and m; in this case line n is called the axisof the translation (and the map is called translation along line n).

When l 6= m, the ultraparallel pencil generated by l and m is invariant under thetranslation, and so are the hypercycles, including line n, generated by this pencil.Nontrivial translations clearly have no fixed point.

• Theorem. Any nontrivial translation uniquely determines its axis.

Proof: Consider how the translation acts on the ideal boundary ∂H2. If ahyperbolic line is translated within itself, then both of its points at infinity arefixed. As the action of PGL(2,R) on ∂H2 is the same as its action on theprojective line, a nontrivial group element can have at most two fixed points in∂H2, see 6.5. This means that no other line can be translated within itself.

• Definition (horolation). Let l,m ⊂ H2 be two parallel hyperbolic lines. Theproduct σm ◦ σl is called a horolation.

88 GABOR MOUSSONG

If l 6= m, that is, the horolation is different from the identity, then it cannot havea fixed point as σl(A) = σm(A) implies l = m. The parallel pencil generated by land m is invariant under the horolation, and so are all the horocycles generated bythis pencil.

Remark: The name of this transformation is an abbreviated form of “horocyclictranslation”, and comes from this last property of horocycles being translatedwithin themselves.

Recall that a projective transformation, different from identity, of the real projectiveline is either elliptic, parabolic, or hyperbolic, that is, can have zero, one, or twofixed points. Working in the Poincare half-plane U , we can apply this to themap induced a congruence on ∂H2 = ∂U = R+ = R. Note that in this senserotations are elliptic, translations are hyperbolic, and horolations are parabolicmaps on the ideal boundary. The converse is also true: it can be shown thatany orientation preserving congruence, different from identity, of H2, is either arotation, a translation, or a horolation, according as the number of fixed points atinfinity is zero, two, or one, respectively.

• Examples. Consider again the matrices

E(t) =

(cos t − sin tsin t cos t

), P (t) =

(1 0t 1

), H(t) =

(cosh t sinh tsinh t cosh t

)

given at the end of 6.5 as examples of elliptic, parabolic, and hyperbolic trans-formations of the real projective line. The congruences of the upper half-planemodel U defined by these matrices are as follows: E(t) is rotation about centeri ∈ U , P (t) is horolation fixing the point 0 ∈ ∂U , H(t) is translation along thehyperbolic line connecting points −1 and 1. It can be shown that any orientationpreserving congruence of U is conjugate to one of these examples.

Observe that as t varies, points of hyperbolic plane move along cycles: alongcircles in the case of rotations, along horocycles in the case of horolations, andalong hypercycles in the case of translations. This shows that cycles can bedefined as orbits of so-called “one-parameter subgroups” of motions in hyperbolicplane.

11.5 Some trigonometric formulas

Recall the hyperbolic Law of Cosines proved in 10.5:


where a, b, c and α, β, γ denote the side lengths and opposite angles of a hyperbolictriangle. Now we derive some more trigonometric identities.


• Theorem (Hyperbolic Law of Sines)

sinα

sinh a=

sinβ

sinh b=

sin γ

sinh c

Proof: We express cosα from the Law of Cosines:

cosα =cosh b cosh c− cosh a

sinh b sinh c.

From this we get a formula for sin2 α. Dividing by sinh2 a, and applying theidentity cosh2 − sinh2 = 1 several times yields

sin2 α

sinh2 a=

2 cosh a cosh b cosh c− sinh2 a− sinh2 b− sinh2 c

sinh2 a sinh2 b sinh2 c.

Here the right hand side is invariant under permutations of symbols a, b, c.Therefore, it also equals sin2 β/ sinh2 b and sin2 γ/ sinh2 c . Taking square rootsthe Law of Sines follows.

• Theorem. In a right triangle with γ = π/2 the following identities hold:

(1) cosh c = cosh a cosh b,

(2) cosα = cosh a sinβ.

Proof: (1): Immediately follows from the Law of Cosines written for c.

(2): Write the Law of Cosines for a, and using (1), substitute cosh a cosh b forcosh c, and then write 1 + sinh2 b for cosh2 b:

cosh a = cosh a cosh2 b− sinh b sinh c cosα

= cosh a+ sinh2 b cosh a− sinh b sinh c cosα .

From this, cosα can be expressed, then the Law of Sines can be applied:

cosα =sinh b

sinh ccosh a =

sinβ

sin γcosh a .

Now the claim follows as sin γ = 1.

• Definition (angle of parallelism). Let x be a positive number. We define anangle Π(x), called the angle of parallelism for x, as follows. Choose a hyperbolicline l, and a point A at distance x from l. Drop a perpendicular from A to l,and let C denote its foot. Let the ray m connect A with a point at infinityof l. The angle between m and AC depends only on distance x, since theconfiguration which consists of l and A is uniquely determined up to congruenceby x, and is axially symmetric in line AC. This angle, denoted Π(x), is the angleof parallelism for x.

90 GABOR MOUSSONG

Rotate a ray with endpoint at A about A so that its angle from AC increase from0 to π/2. When α < Π(x), the ray meets the line l, and when α ≥ Π(x), itis disjoint from l. Therefore, the ray making angle equal to Π(x) separates non-intersecting rays from intersecting ones (itself being non-intersecting). This explainsthat Π(x) < π/2 since the line containing the ray and l are ultraparallel if both areorthogonal to AC.

• Theorem

sinΠ(x) =1

coshx.

Proof: Use notations of the preceding definition. Let U be the common pointat infinity of m and l. If B ∈ l is an arbitrary point between C and U , then wecan apply formula (2) of the preceding theorem to the right triangle ABC (withb = x):

cosβ = coshx sinα .

If the point B moves towards U indefinitely, then α → Π(x). On the otherhand, using a conformal model, parallelism of l and m means tangency at U ofthe circular arcs representing them, therefore β → 0 due to the angle preservingproperty of the model. Taking limits of both sides, we get 1 = coshx sinΠ(x).

It is intuitively clear from the definition of the angle of parallelism that to a largerdistance a smaller angle of parallelism must correspond. The theorem confirms this:it follows that the function Π is bijective and strictly monotone decreasing fromthe set of positive real numbers onto the interval (0, π/2).

Surjectivity of the function Π : (0,+∞) → (0π/2) means that in hyperbolic ge-ometry the angle at which a whole line is visible can be made arbitrarily smallwhen the line is viewed from sufficiently far away. This is a striking difference fromEuclidean geometry where a line, no matter how far away, always occupies one halfof the total field of vision.

• Theorem The sum of the three angles of any hyperbolic triangle is less than π.

Proof: Place the triangle ABC in the Poincare disk model so that vertex Acoincide with the center of the model. Now we compare the angles α, β, γ of thetriangle to the angles α′, β′, γ′, of the Euclidean triangle ABC defined in theplane containing the model.

Sides AB and AC are straight segments in the model, therefore α′ = α. SideBC in the model is part of a Euclidean circle D which intersects the boundarycircle orthogonally. By a lemma in 3.1 it follows that the power of point A withrespect to circle D is a positive number, therefore A is in the exterior of D. Thisimplies β′ > β and γ′ > γ, from which α+ β + γ < α′ + β′ + γ′ = π follows.


11.6 Area of triangles

The rigorous concept of area is technically harder to define in hyperbolic geometrythan in Euclidean. Therefore, we do not go into precise definitions, but insteadwe take it for granted that it is possible to measure the area of geometric figures.We also assume that area has the same fundamental properties as in Euclideangeometry:

– figures with nonempty interior have positive area,– bounded regions have finite area,– congruent figures have equal area, and– the area of the union of two non-overlapping figures equals the sum of the two

areas.

Just like for measuring distances, there exists a natural unit of measurement forarea in hyperbolic geometry. Squares don’t exist, so it is not easy to define whatexactly this natural unit is. We introduce it by way of the following theorem whichwe shall prove below.

• Theorem. The area bounded by three pairwise parallel lines in hyperbolic planeis finite.

Such an object is called an “ideal triangle” in hyperbolic geometry as it has its threevertices on the ideal boundary of the plane. Recall that the action of congruenceson ∂H2 is the same as the action of the projective group of a real projective line.Now we can apply the first theorem of 6.3 to conclude that all ideal triangles arecongruent. Therefore, all have the same area.

• Definition (asymptotic triangles). If A, B, C are three non-collinear pointsin H2 ∪∂H2, then using the Cayley–Klein model we can form the convex hull ∆of these three points in the ambient Euclidean plane. If all three points belongto H2, then ∆ is the triangle ABC in the usual sense. If not, then depending onwhether one, two, or all three of A, B, and C belong to the ideal boundary ∂H2,the set ∆ ∩ H2 is called a singly, doubly, or triply asymptotic triangle. (Notethat triply asymptotic triangles are just what we called ideal triangles above.)

More generally, we can define asymptotic convex polygons in H2 as convex hulls offinitely many points of H2 ∪ ∂H2, at least one of which is at infinity.

Proof of the theorem: It is straightforward to subdivide an ideal triangle intoa few (e.g., six) singly asymptotic triangles. Therefore, it suffices to show thatsingly asymptotic triangles have finite area. Let ABC be an arbitrary singlyasymptotic triangle with vertex C at infinity.Extend side AB beyondB through a finite positive distance to pointD. Draw theray DC. Move point D along this ray, and translate the whole plane, togetherwith the moving point D, along axis DC. As the plane translates, point Btraces out a hypercycle. This hypercycle intersects AC at a certain point A1.

92 GABOR MOUSSONG

(This follows from the fact that the distance of a variable point of AC fromline DC strictly monotone decreasingly converges to zero, see 11.1.) This pointcorresponds to a specific choice of the translation. Let t denote this translation,then t(B) = A1 and t(C) = C, therefore t moves the ray BC to the ray A1C.Put D1 = t(D), and let B1 be the point where BC meets A1D1.

Now we have another singly asymptotic triangle A1B1C with point D1 on theextension of side A1B1. Translation t will move B1 to some point A2 on the rayA1C, and we get points D2 and B2 analogously. Apply the inverse translationt−1 to the quadrilateral A1B1B2A2 to get BEE1B1, where E is on AD and E1

is on A1D1. These two quadrilaterals are therefore congruent.

We repeat the process indefinitely. The nth quadrilateral An−1Bn−1BnAn istranslated back in n − 1 steps (via the translation t−(n−1)) into ADD1A1, andso on. Eventually, the original triangle ABC is subdivided into a sequence ofquadrilaterals, and congruent (translated) copies of all these quadrilaterals arepacked with no overlaps into the bounded region ADD1A1. This proves that thearea of ABC is finite.

For any triangle ∆, ordinary or asymptotic, there exists an ideal triangle whichcontains ∆. Indeed, take any point in the interior of ∆, and draw the three raysfrom this point through the vertices. The points at infinity of these rays span anideal triangle that covers ∆. This implies that the area of no triangle can exceedthe common area of ideal triangles.

• Definition (natural unit of area). We say that the hyperbolic plane isequipped with the natural unit of measurement of area if the area of ideal trian-gles equals π.

This definition may not seem very natural at first, but the following results explainwhy this is the correct choice. From now on we assume that the natural unit ischosen for measuring areas in H2.

An angle of an asymptotic triangle is defined whenever the vertex of the angle isin H2 (i.e., is not at infinity). A doubly asymptotic triangle has only one angle.This angle may be any real number (strictly) between 0 and π. Note that this angledetermines the doubly asymptotic triangle uniquely up to congruence.

• Lemma. The area of a doubly asymptotic triangle with angle ϕ equals π − ϕ.

Proof: Let A(ϕ) denote the area in question.

If ϕ, ψ > 0, and ϕ + ψ < π, then two doubly asymptotic triangles with anglesϕ and ψ can be joined together along a boundary ray to form an asymptotic


quadrilateral with three vertices at infinity. This quadrilateral can also be ob-tained by joining together an ideal triangle and a doubly asymptotic trianglewith angle ϕ+ ψ. This yields the equation A(ϕ) +A(ψ) = A(ϕ+ ψ) + π, whichcan be rewritten as

π −A(ϕ+ ψ) =(π −A(ϕ)

)+(π −A(ψ)

).

This last equation means that the function π − A : (0, π) → (0, π) is additive.Therefore this fuction is linear, that is, A(ϕ) = π − cϕ with some constant c.Finally, joining together two doubly asymptotic triangles with angle π/2 we getan ideal triangle, so 2A(π/2) = π, from which c = 1 follows.

• Theorem. If the angles of a triangle in hyperbolic plane are α, β, and γ, thenfor the area A of the triangle we have

A = π − (α+ β + γ) .

Proof:

Let A, B, C be the vertices, and let P , Q, R denote the points at infinity ofthe rays drawn from A through B, from B through C, and from C through A,respectively. These rays subdivide the ideal triangle PQR into triangle ABC,and three doubly asymptotic triangles ARP , BPQ, and QCR, with angles π−α,π−β, and π−γ, respectively. Using the preceding lemma we get A+α+β+γ = π.

Remark: Compare this result with the Girard formula of spherical geometry.

11.7 Exercises

1. Prove that the configuration formed by two distinct, parallel, lines in hyperbolic plane is unique

up to congruence.

2. Suppose that two ultraparallel lines in H2 are distance d apart. Prove that cosh d = |〈u,v〉|where u,v ∈ W are unit normal vectors for the lines in the hyperboloid model.

94 GABOR MOUSSONG

3. Show that two distinct cycles in H2 cannot have more than two common points.

4. Show that the three Lorentz transformations given in Exercise 4 of 10.6 define a rotation, ahorolation, and a translation in the hyperboloid model of H2.

(Hint: Count fixed points in ∂H2.)

5. Prove that two hyperbolic translations can commute only if they have the same axis.(Hint: Consider the action on the ideal boundary, and apply the fact that for two commuting

transformations the set of fixed points of one is invariant under the other.)

6. Let a, b, c, α, β, γ denote the sides and angles of a hyperbolic triangle the usual way. Ifγ = π/2, prove tanα tanβ cosh c = 1.

7. Prove that a circle of radius r in hyperbolic plane has perimeter 2π sinh r.(Hint: Take an isosceles triangle with sides r, r, a, and with angle 2π/n opposite side a. Usethe hyperbolic Law of Cosines to get a formula for a in terms of r and n, then evaluate the

limit of na as n → ∞.)

8. Prove Bolyai’s Absolute Law of Sines which says that the formula

sinα

©a=

sinβ

©b=

sin γ

©c

holds for all triangles in all three types of geometry: Euclidean, spherical, and hyperbolic. Inthe formula a, b, c are the sides of the triangle, α, β, γ are the corresponding angles, and ©xdenotes the perimeter of the circle of radius x.

9. Show that the radius of the inscribed circle of any hyperbolic triangle is less than cosh−1(

2/√3)

.(Hint: The angle of parallelism for the radius must be greater than 60◦.)

10. Check that at least two of the theorems of this chapter imply that rectangles (i.e., quadrilaterals

with four right angles) cannot exist in hyperbolic geometry. In contrast, show that for anyinteger n ≥ 5 there exist n-sided polygons in H2 with all interior angles equal to π/2.(Hint: Use the projective model.)

11. For an n-sided polygon P in hyperbolic plane show that the sum of interior angles is less than(n− 2)π.

(Hint: subdivide into triangles.)

12. Define the defect δ(P ) of a polygon P in hyperbolic geometry by the formula

δ(P ) = (n− 2)π − (α1 + . . .+ αn) ,

where n is the number of sides of P , and α1, . . . , αn are the interior angles of P (cf. 11).Prove that the defect is an additive function, that is, if P is a non-overlapping union of two

polygons P1 and P2, then δ(P ) = δ(P1) + δ(P2). Deduce that δ(P ) equals the area of P (withrespect to the natural unit of measurement).

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