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Lecture Notes

Geometry I

Boris Springborn

Winter Semester 2018/19

TU Berlin

—draft version of January 28, 2019—

These notes were prepared by Alexander Bobenko, Thilo Rörig andBoris Springborn with the help of Alina Hinzmann and Jan Techter.

If you find a mistake, we would be greatful if you told us by email toBoris Springborn <[email protected]>.

Contents

1 Introduction 1

2 Spherical geometry 52.1 Great circles and shortest curves . . . . . . . . . . . . . . . . . . . . . . . 52.2 Hemispheres and digons . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3 Spherical triangles and their polar triangles . . . . . . . . . . . . . . . . . 102.4 Side lengths, angles, and area of a spherical triangle . . . . . . . . . . . . . 122.5 The spherical cosine rules . . . . . . . . . . . . . . . . . . . . . . . . . . 162.6 The sine theorem, half-angle formulas, and half-side formulas . . . . . . . 192.7 Napier’s rule and pentagramma mirificum . . . . . . . . . . . . . . . . . . 202.8 Stereographic projection . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3 Linear algebra refresher: Bilinear and quadratic forms, indefinite scalar prod-ucts and orthogonal transformations 293.1 Bilinear and quadratic forms . . . . . . . . . . . . . . . . . . . . . . . . . 293.2 Diagonalizing a bilinear form with the generalized Gram–Schmidt algorithm 323.3 Completing the squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.4 Normalization in complex and real vector spaces . . . . . . . . . . . . . . 353.5 Scalar products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.6 Orthogonal transformations . . . . . . . . . . . . . . . . . . . . . . . . . . 403.7 A closer look at Lorentz vector spaces . . . . . . . . . . . . . . . . . . . . 41

4 Hyperbolic geometry 454.1 Hyperbolic space (in the hyperboloid model) . . . . . . . . . . . . . . . . . 464.2 “One-dimensional hyperbolic space” . . . . . . . . . . . . . . . . . . . . . 474.3 Lines in hyperbolic space . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.4 Two-dimensional hyperbolic space . . . . . . . . . . . . . . . . . . . . . . 504.5 Hyperbolic triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.6 The Klein model of the hyperbolic plane . . . . . . . . . . . . . . . . . . . 544.7 Angle of parallelism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.8 More distance formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.9 Hyperbolic “trilaterals” . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

i

ii Contents

4.10 Intersections of H2 ⊂ R2,1 with planes . . . . . . . . . . . . . . . . . . . . 584.11 The Poincaré disk model of the hyperbolic plane . . . . . . . . . . . . . . . 584.12 From Klein model to Poincaré disk model via the hemisphere model . . . . 604.13 The Poincaré half-plane model . . . . . . . . . . . . . . . . . . . . . . . . 614.14 Two examples for length calculations in the half-plane model (and some

remarks) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.15 Calculating hyperbolic areas in the half-plane model . . . . . . . . . . . . 624.16 Concluding remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

5 Projective geometry 695.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 695.2 Projective spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 725.3 Desargues’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805.4 Projective transformations . . . . . . . . . . . . . . . . . . . . . . . . . . 835.5 The cross-ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 905.6 Complete quadrilateral and quadrangle . . . . . . . . . . . . . . . . . . . . 945.7 The fundamental theorem of real projective geometry . . . . . . . . . . . . 1035.8 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

6 Conic sections 1116.1 Conic sections – The Euclidean point of view . . . . . . . . . . . . . . . . 1116.2 Conics – The projective point of view . . . . . . . . . . . . . . . . . . . . 1166.3 Pencils of conics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1216.4 Rational parametrizations of conics . . . . . . . . . . . . . . . . . . . . . 1236.5 The pole-polar relationship, the dual conic and Brianchon’s theorem . . . . 1266.6 Confocal conics and elliptic billiard . . . . . . . . . . . . . . . . . . . . . 131

7 Quadrics 1377.1 Quadrics. The Euclidean point of view. . . . . . . . . . . . . . . . . . . . 1377.2 Quadrics. The projective point of view . . . . . . . . . . . . . . . . . . . . 1407.3 Polarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1457.4 The synthetic approach to projective geometry . . . . . . . . . . . . . . . . 146

8 The Plücker quadric 151

9 Möbius geometry 153Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

10 Lie geometry 155

Chapter 1

Introduction

What is geometry?For now, let us answer this question in the form of a short historic overview of the

subject. Geometry is, after all, something that people have been doing for a very longtime, and the answers to the above question have changed considerably over the centuries.The following brief history of geometry will be incomplete, inaccurate (true history ismuch more complicated) and biased (we will ignore what happened in India or China, forexample). It is a shortened, smoothed out version of history that is meant only as a roughexplanation of how the material that will be covered in this course came into being.

The word geometry comes from the Greek word γεωμετρία which is a composite of thewords for earth and measure. Geometry began as the science of measuring the earth, orsurveying, and it began ∼2000 BC in Egypt and Mesopotamia (Babylon, in today’s Iraq).These were among the first great civilizations and they depended on agriculture along therivers Nile in Egypt, and Tigris and Euphrates in Mesopotamia. These rivers would period-ically inundate and fertilize the surrounding land, which made periodic surveying necessaryto delimit the fields. The science of geometry developed from this, with applications alsoin construction and astronomy. The Egyptians and Babylonians could compute areas andvolumes of simple geometric figures, they had some approximations for π, and they alreadyknew Pythagoras’ theorem. Strangely though, no records of general theorems or proofshave survived from this period. Egyptian papyri and Babylonian clay tables with theircuneiform script contain only worked exercises. Maybe they did not state general theorems,maybe they just did not write them down, or maybe they did but these documents did notsurvive. Basically we have no idea how they conducted their research.

This changed with the period of Greek geometry (Thales∼600 BC to Euclid∼300 BC).They clearly stated general theorems for which they gave proofs. That is, they deduced morecomplicated statements from simpler ones by logical reasoning. Let us look at a concreteexample:

Theorem (Inscribed angle theorem). The inscribed angle α is half the central anglethat subtends the same chord (see Figure 1.1).

1

2 Introduction

Proof. Consider the case where the triangleABC contains the centerO of the inscribedcircle as shown in Figure 1.2. Let 2α = ∠(COB), and let α′, δ, ε be the angles shownin the figure. Since the angle sum ine each of the triangles OPC and ABC is π, wehave

α′ =π

2− α and 2α′ + 2δ + 2ε = π,

and this impliesδ + ε = α.

Similar arguments hold in the other cases. (Find them (the cases and the arguments).)

This proof uses explicitly the theorem about the angle sum in a triangle, and implicitlya large number of other theorems, for example about isosceles triangles and the congruenceof triangles. How can one be shure that the proofs are free of circular reasoning? It mustbe possible to put all theorems in a linear order so that each theorem is proved using onlytheorems that have already been proved. By necessity, such a list must begin with a few(as few as possible) hopefully very simple statements that are accepted without any logicalproof.

In Euclid’s Elements, most or even all of what was known about geometry at the time ispresented in this form. It begins with a few definitions and postulates (today we say axioms)from which all theorems are deduced one by one. These postulates were simple statementslike “there is a unique straight line through any two points” and “any two lines in a planeintersect in a unique point or they are parallel”. But one of the postulates was consideredmore complicated and less obvious than the others, the parallel postulate: “Given a lineand a point not on the line, there is a unique parallel to the line through the point.” Forcenturies to come, people tried to prove this one postulate using the other, simpler ones, sothat it could be eliminated from the unproved postulates.

One way people tried to prove the parallel postulate was to assume instead that there

2α

αα

α

α

Figure 1.1: Inscribed angle theorem

A

O

PBC

δ

δ

ε

ε

α α

α′ α′

Figure 1.2: Proof

3

was more than one parallel through a given point and derive a contradiction. Some strangetheorems could be deduced from this alternative parallel postulate. For example it impliedthat there was an upper bound for the area of triangles. This lead some people to believethat such “absurd” consequences were a proof of the parallel postulate. But for a proofby contradiction, it is not sufficent for the consequences of the assumption to be “absurd”.They have to contradict other postulates or or proved statements.

In fact, the alternative parallel postulate would lead to no honest logical contradictionsbetween any proved statements. This finally lead to the realization (by Lobachevsky1829 and Bólyai 1831) that the alternative parallel postulate did not contradict the otherpostulates. Instead it leads to a logically equally valid version of geometry which is nowcalled hyperbolic geometry. Later it was realized that one may also assume that thereare no parallels (the other postulates also have to be changed a little for this), and theresulting geometry is called elliptic geometry. This is simply the geometry on the sphere,where pairs of opposite points are considered as one point, and lines are great circles. Bothhyperbolic and elliptic geometry are called non-Euclidean geometries, because their axiomsare different from Euclid’s.

Another important development in geometry was the introduction of coordinates byDescartes and Fermat in the first half of the 17th century. One could then describe geometricfigures and prove theorems using numbers. This way of doing geometry was called analyticgeometry, as opposed to the old way beginning with geometric axioms, which was calledsynthetic geometry. By the late 19th/early 20th century, it was proved that both approachesare in fact equivalent: One can either start with axioms for numbers and use them todefine the objects of geometry, or one can start with axioms of geometry and definenumbers geometrically, one gets the same theorems. This was not proved earlier becausethe the axiomatic treatment of geometry (and mathematics in general) had not been rigorousenough. The first axiomatic treatment of euclidean geometry that satisfies modern standardsis due to David Hilbert. At the same time, Veblen and Young axiomatized projectivegeometry (see below). It turns out that projective geometry is easier to axiomatize and onecan derive euclidean geometry by adding additional structure to it.

So form a purely logical perspective, there is no clear boundary between geometryon the one side, and algebra and analysis on the other. One can start with an axiomatictreatment of numbers, vector spaces, etc., and derive geometric theorems analytically.Or one can start with geometric axioms and construct number systems geometrically.Logically both approaches are the same, but there is a big psychological difference. Wethink differntly about geometry because geometric thinking allows us to use our spaceintuition in mathematics. To put this as a slogan: “Geometry is a branch of mathematicsthat stimulates part of the brain that other areas cannot reach.”

But there is more than euclidean, hyperbolic and elliptic geometry. The study ofthe rules of perspective in painting (da Vinci & Dürer ∼1500) lead to the developmentof projective geometry (Poncelet, 1822), dealing with the question: Which properties ofgeometric figures do not change under projections? For example, straight lines remainstraight lines, but parallel straight lines do not remain parallel.

4 Introduction

Another type of geometry is Möbius geometry. Planar Möbius geometry deals withthe properties of figures that remain unchanged under transformations that map all circlesto circles. These transformations include the euclidean motions (translations, rotations,reflections, and glide reflections) but also reflections in circles.

There are more types of geometry, like Lie geometry and Laguerre geometry, which wewill consider, but about which nothing shall be said at this point. Except that they all havetheir reason and purpose. In fact, one of the main points we would like to make with notesis that all of these different types of geometry are useful even if you are only interested ineuclidean geometry. If you shun the world of non-euclidean geometries, you are deprivingyourself of some very powerful toolboxes.

It turns out that there is a unified view of all these different geometries, illuminatingtheir relationships and Klein’s Erlangen Program (1871) provided a systematic treatmentof all these different kinds of geometry and their interrelationships. It also provided acomprehensive and maybe surprising answer to the question: What is geometry? We willcome back to this.

Chapter 2

Spherical geometry

In this chapter we introduce some basic notions of spherical geometry. Section 2.1 isconcerned with great circles and shortest curves in the sphere. Section 2.8 treats stereo-graphic projection. These discussions will be valid in any dimension. The other sectionsare concerned with spherical trigonometry, i.e., with two-dimensional spherical geometry.Nevertheless, we favor methods that generalize easily to higher dimensions (and other ge-ometries). For example, we derive the cosine laws using the Gram-Matrix method insteadof using the cross-product of R3.

This chapter also serves as a preparation for the following Chapter 4 on hyperbolicgeometry. Many calculations presented here carry over to the hyperbolic case with onlyminor changes.

Since we may take the sphere’s radius as our unit of length measurement, it is sufficientto consider geometry in the n-dimensional unit sphere

Sn = {x ∈ Rn+1 | |x| = 1} ⊂ Rn+1, (2.1)

where |x| =√〈x, x〉, and 〈x, y〉 =

∑xiyi is the standard Euclidean scalar product.

2.1 Great circles and shortest curvesIn this section, we show that the shortest curves in the sphere are arcs of great circles(Theorem 2.1.2). This allows us to define the spherical metric (Definition 2.1.4) and toconclude that the isometries of the sphere are orthogonal transformations (Corollary 2.1.5).

Definition 2.1.1. A great circle in Sn is the intersection Sn ∩ E of the n-dimensionalsphere Sn and a two-dimensional vector subspace E ⊂ Rn+1.

A nonempty intersection of Sn with an affine 2-plane that does not contain theorigin is called a small circle.

A circle in the sphere is either a great circle or a small circle. Points are considered asdegenerate circles.

5

6 Spherical geometry

For any pair of points x, y ∈ Sn that are different, x 6= y, and not opposite, x 6= −y,there is a unique great circle through x and y: it is the intersection of Sn with the 2-dimensional subspace spanned by x and y. If x = ±y, there is a (n−1)-dimensional familyof great circles through x and y.

In R3, two-dimensional subspaces are determined by a unit normal vector:

E = {x ∈ R3 | 〈x, n〉 = 0} ⊂ R3, |n| = 1. (2.2)

For the 2-dimensional sphere S2, this establishes a 2-to-1 correspondence between pointsand great circles. The points±n ∈ S2 that correspond to a great circle S2∩E are called thepoles of the great circle. Any two great circles in S2 intersect in two diametrically oppositepoints.

Theorem 2.1.2. (i) A piecewise continuously differentiable curve connecting two dif-ferent points x, y ∈ Sn is shortest among such curves if and only if it is the shorter arcof a great circle through x and y. (The great circle and the shorter arc are uniquelydetermined unless y = −x.)

(ii) The length of a shorter arc of a great circle connecting x and y is

d(x, y) := arccos〈x, y〉. (2.3)

Of course, d(x, y) is just the angle between the unit vectors x and y. Statement (ii) ofthe theorem essentially says that the definition of angles in terms of the scalar product andthe inverse cosine function is consistent with the definition in terms of lengths of circulararcs.

Before we embark on the proof of Theorem 2.1.2, see if you can deduce the followingconsequences:

Corollary 2.1.3. The function d : Sn → R defined by (2.3) is a metric on Sn. Thismeans d has the following properties: For all x, y, z ∈ Sn• d(x, y) ≥ 0, and d(x, y) = 0 if and only if x = y, (d is positive definite)• d(x, y) = d(y, x), (d is symmetric)• d(x, y) + d(y, z) ≥ d(x, z). (triangle inequality)

Definition 2.1.4. The metric (2.3) is called the intrinsic or spherical metric on Sn.

Corollary 2.1.5. Let f : Sn → Sn be any function. Then the following statements areequivalent:(i) The function f is an isometry with respect to the spherical metric d, i.e.,

d(f(x), f(y)) = d(x, y) for all x, y ∈ Sn. (2.4)

2.1 Great circles and shortest curves 7

(ii) The function f is the restriction to Sn of an orthogonal transformation of Rn+1

(with respect to the standard euclidean scalar product, see Section 3.6).

The rest of this section is devoted to the proof of Theorem 2.1.2. We will use theabbreviation

α := d(x, y) so cosα = 〈x, y〉.We will rely only on standard methods from the analysis of functions of several variables.The proof can be made shorter and conceptually clearer with the help of tools from theanalysis on manifolds, which are beyond the scope of this book. The initiated reader willrecognize the vector field v as the gradient tangent vector field of the function f defined byequation (2.1) on the submanifold Sn ⊂ Rn.

It is a good exercise to retro-engineer the proof of Theorem 2.1.2 to the setting ofeuclidean geometry:

Exercise 2.1.6. Prove that the shortest piecewise continuously differentiable curve inRn connecting two points x, y ∈ Rn is the straight line segment from x to y.

update proofto work fory = −x

update proofto work fory = −xProof of Theorem 2.1.2. (ii) Assume first that the distinct points x and y are not op-

posite. Then the unique great circle through x and y is the intersection of Sn withthe two-dimensional subspace E spanned by x and y. To parameterize the shorter arcfrom x to y, apply Gram-Schmidt orthonormalization to x, y to obtain an orthonormalbasis of E. The result is

e1 = x,

e2 =1

|y − 〈y, x〉x|(y − 〈y, x〉x

)

=1

sinα

(y − (cosα)x

),

The shorter arc from x to y is parameterized by the curve

γ : [0, α]→ Sn, γ(s) = (cos s) e1 + (sin s) e2, (2.5)

and since |γ′| = 1,

length(γ) =

∫ α

0

|γ′| dt = α.

Even if the points x and y = −x are opposite, equation (2.5) provides a parameter-ization of a great circular arc connecting them

This proves statement (ii) of the theorem.


(i) To prove the first statement, let

η : [a, b]→ Sn

be a piecewise continuously differentiable curve with η(a) = x and η(b) = y. We haveto show that

length(η) ≥ d(x, y),

and equality holds only if η is a monotonic reparameterization of γ. To simplify theargument, we will assume that η is C1. We leave it to the reader to complete theargument for curves that are only piecewise C1.

We want to consider d(x, y) as a function of y, so we define

f(p) := arccos 〈x, p〉.

This not only well defined for p ∈ Sn, but for all p ∈ Rd+1 satisfying

− 1 ≤ 〈x, p〉 ≤ 1. (2.6)

So we may consider f as a function defined and continuous in the closed region inRd+1 bounded by the parallel planes tangent to the sphere Sn in the points ±x. It isdifferentiable in the interior of this region, where the inequalities (2.6) hold strictly.There, the gradient is

grad f(p) = − 1√1− 〈x, p〉2

x.

What we are really interested in is the tangential component, let us call it v, of therestriction of grad f to Sn, i.e.,

v : Sn \ {x,−x} → Rd+1, v(p) = grad f(p)− 〈grad f(p), p〉 p.

A direct calculation shows that |v| = 1. Finally, we obtain

d(x, y) = f(y)− f(x)

=

∫ b

a

〈grad f(η(t)), η′(t)〉 dt (2.7)

=

∫ b

a

〈v(η(t)), η′(t)〉 dt (2.8)

≤∫ b

a

|η′(t)| dt = length(η). (2.9)

Equality (2.7) is essentially the fundamental theorem of calculus for line integrals.We should better assume that η(t) = x only for t = a and η(t) 6= −x for all t ∈ [a, b].

2.2 Hemispheres and digons 9

We may do this without loss of generality. (Can you see why?) Then the integrand isat least well defined for t ∈ (a, b]. So

f(η(a+ ε))− f(η(b)) =

∫ b

a+ε

〈grad f(η(t)), η′(t)〉 dt

for all small enough ε > 0. By the continuity of f and η we obtain (2.7) in the limit asε→ 0, where we interpret the integral over [a, b] as an improper integral.

Equality (2.8) holds because

〈grad f(η), η′〉 = 〈v(η), η′〉.

Indeed, the difference of v(η) − grad f(η) is in the radial direction η, and the radialcomponent of η′ is zero. To see this last point, take the derivative of 〈η, η〉 = 1 toobtain 〈η′, η〉 = 0.

The inequality (2.9) follows from the Cauchy-Schwarz inequality and |v| = 1.Equality holds if and only if η′(t) is a nonnegative multiple of v(η(t)) for all t, i.e., if

η′ = uv(η)

for some nonnegative C1-function u : [a, b] → R. A short calculation shows that γsatisfies

γ′ = v(γ).

Finally, use the Picard–Lindelöf theorem on the unique solvability of differentialequations with given initial conditions to see that (2.9) holds with equality if and onlyif η is a monotonic reparametrization of γ.

This completes the proof of statement (i), at least for C1-curves (with the extensionto piecewise-C1-curves left to the reader).

2.2 Hemispheres and digonsBefore we treat spherical triangles, let us consider an even simpler class of sphericalpolygons: digons, i.e., polygons with two vertices and two sides. First, a hemisphere is theintersection of S2 with a half-space

H = {x ∈ R3 | 〈x, n〉 ≥ 0}, |n| = 1.

A hemisphere is bounded by a great circle. One of its poles, n, lies inside the hemisphere,the other, −n, outside.

A spherical digon is the intersection of two hemispheres that are neither equal noropposite (see Figure 2.1). The interior angle α and the exterior angle α of the digon satisfy

α + α = π, cos α = 〈m,n〉, cosα = −〈m,n〉.


Figure 2.1: Spherical digon

The exterior angle is the spherical distance between the poles m and n. The area of thedigon is 2α.

Remark 2.2.1. One could alternatively define a spherical digon as a spherical regionbounded by two great half circles. Then the interior angle α could exceed π. Theformula for the area would still hold.

2.3 Spherical triangles and their polar trianglesA spherical triangle is the intersection of three hemispheres (see Figure 2.2). To avoiddegenerate cases, we require that their poles are not all contained in one great circle.Equivalently, a spherical triangle is the intersection of S2 with a triangular cone

C = {x ∈ R3 | 〈A′, x〉 ≥ 0, 〈B′, x〉 ≥ 0, 〈C ′, x〉 ≥ 0}, (2.10)

where A′, B′, C ′ ∈ S2 are linearly independent and hence form a basis of R3. The pointsA′, B′, C ′ are called the poles of the spherical triangle.

Remark 2.3.1. Spherical triangles according to this definition are sometimes calledEulerian spherical triangles (after Leonhard Euler, who wrote a series of articles onspherical geomery). There are other perfectly sensible definitions of spherical triangles.For example one could define a spherical triangle as a region in the sphere that is boundedby three great circular arcs. Such a spherical triangle is either a Eulerian triangle, the

2.3 Spherical triangles and their polar triangles 11

Figure 2.2: Spherical triangle

outside of a Eulerian triangle, the union of a Eulerian triangle and a hemisphere, or thedifference of a hemisphere and a Eulerian triangle.

We have defined spherical triangles in terms of hemispheres. But a spherical triangle isalso determined by its vertices A, B, C. This follows from the following proposition:

Proposition 2.3.2. (i) Suppose A,B,C ∈ R3 are linearly independent unit vectors inR3. Then the triangular cone of positive linear combinations

{λA+ µB + νC ∈ R3 | λ, µ, ν ≥ 0} (2.11)

is equal to the cone C defined by (2.10) if the following system of equations andinequalities holds:

〈A′, A〉 > 0, 〈A′, B〉 = 0, 〈A′, C〉 = 0,

〈B′, A〉 = 0, 〈B′, B〉 > 0, 〈B′, C〉 = 0,

〈C ′, A〉 = 0, 〈C ′, B〉 = 0, 〈C ′, C〉 > 0.

(2.12)

(ii) For every triple A, B, C of linearly independent unit vectors there is a uniquetriple A′, B′, C ′ of unit vectors satisfying (2.12), and it is linearly independent. Con-versely, by the symmetry of the equations, A′, B′, C ′ determine A, B, C in the sameway.


Exercise 2.3.3. Prove Proposition 2.3.2. Hint: If the system (2.12) is satisfied, then thebasis of (R3)∗ dual to the basis A, B, C of R3 consists of the linear forms 1

〈A′,A〉〈A′, · 〉,. . .

The points A′, B′, C ′ are the poles of the spherical triangle with vertices A, B, C. Thepoint A′, say, is the pole of the great circle through B and C that is on the same side as thepoint A.

The spherical triangle with verticesA′, B′, C ′ is called the polar triangle of the sphericaltriangle with vertices A,B,C.

The poles of the polar triangle are the vertices of the original triangle:

A′′ = A, B′′ = B, C ′′ = C.

So the polar triangle of the polar triangle is the original triangle.

2.4 Side lengths, angles, and area of a spherical triangleThe side lengths a, b, c and exterior angles α, β, γ of a spherical traingle with vertices A,B, C and poles A′, B′, C ′ as shown in Figure 2.2 are determined by

cos a = 〈B,C〉, cos α = 〈B′, C ′〉, etc. (2.13)

The interior angles α, β, γ are

α = π − α, etc.

The side lengths of the polar triangle are the exterior angles of the original triangle and viceversa. This follows immediately from equations (2.13).

Theorem 2.4.1 (Area of a spherical triangle). The area of a spherical triangle withinterior angles α, β, γ is α + β + γ − π. In particular, the sum of interior angles of aspherical triangle is greater than π.

Proof. The three great circles through A and B, B and C, C and A divide the sphereinto eight triangles. Two of them are the triangleA,B,C and the triangle−A,−B,−C.They have the same area because they are symmetric with respect to the origin. Theother six triangles all have a side in common with either triangle A,B,C or triangle−A,−B,−C. Together with one or the other they form six digons, two with angleα, two with angle β, two with angle γ. Altogether, these six digons cover each of the

2.4 Side lengths, angles, and area of a spherical triangle 13

triangles A,B,C and −A,−B,−C three times, and the rest of the sphere once. So

4α + 4β + 4γ = sum of areas of the six digons= area(S2) + 2 area(4ABC) + 2 area(4(−A)(−B)(−C))

= 4π + 4 area(4A,B,C).

What are the conditions for three positive numbers to be the side lengths of a sphericaltriangle? Do the side lengths determine the triangle up to congruence?

Theorem2.4.2 (Existence and uniqueness for given lengths). Leta, b, c be real numbers.(i) There exists a spherical triangle with side lengths a, b, c if and only if the

following inequalities are satisfied:

− a+ b+ c > 0, a− b+ c > 0, a+ b− c > 0, a+ b+ c < 2π. (2.14)

(Note that these inequalities imply 0 < a < π, 0 < b < π, 0 < c < π.)(ii) If it exists, the spherical triangle with side lengths a, b, c is unique up to an

orthogonal transformation of R3.

Remark 2.4.3. It is strangely subtle to show that the inequalities (2.14) are sufficientfor the existence of a spherical triangle. (Note that all four inequalities are necessary.So any argument for existence that does not require all four inequalities is false.) Theargument given below is nicely intuitive but involves some hand waving. In Section 2.5,we will introduce a useful toolbox, the Grammatrix method, for proving statements likeTheorem 2.4.2 using only standard arguments from linear algebra (see Remark 2.5.3).

Figure 2.3: Constructing a spherical triangle with given side lengths


Proof of Theorem 2.4.2. (i) “⇒”: Suppose there is a spherical triangleABC with sidelengths a, b, c. Then first three inequalities of (2.14) follow directly from the triangleinequality for the spherical metric d (see Corollary 2.1.3 (ii)). The fourth inequalityalso follows from the triangle inequality for d, but slightly less directly:

d(C,A) < d(C,−B) + d(−B,A).

implies

a+ b+ c = d(B,C) + d(C,A) + d(A,B)

< d(B,C) + d(C,−B)︸︷︷︸π

+ d(−B,A) + d(A,B)︸︷︷︸π

= 2π.

(Alternatively, one obtains the fourth inequality of (2.14) by applying Theorem 2.4.1to the polar triangle.)

“⇐”: Suppose a, b, c satisfy (2.14). We construct a spherical triangle with sidelengths a, b, c as follows (see Figure 2.3): Start with two points A, B on the sphere atdistance c. If the spherical circle aroundAwith radius b and the spherical circle aroundB with radius a intersect in two points, we may take one of these points to be the pointC. Then the triangle with vertices A, B, C has side lengths a, b, c.

It remains to show that the circles intersect in two points. To this end, consider thepoints X1, X2 and Y1, Y2 in which the two circles intersect the great circle through Aand B, as shown in the figure. The points X1 and X2 divide the great circle into twosegments, one containing A, the other not. The circles around A and B intersect intwo points if Y1 is contained in segment containing A and Y2 is contained in the othersegment.

Finally, note that Y1 is contained in segment containing A if

−b < c− a < b

and Y2 is contained in the segment not containing A if

b < c+ a < 2π − b.

(ii) Suppose A, B, C and A, B, C are the vertices of spherical triangles so thatcorresponding sides have the same length. Then corresponding scalar products areequal:

〈A,A〉 = 〈A, A〉, 〈A,B〉 = 〈A, B〉, 〈A,C〉 = 〈A, C〉,〈B,B〉 = 〈B, B〉, 〈B,C〉 = 〈B, C〉, 〈C,C〉 = 〈C, C〉.

(2.15)

2.4 Side lengths, angles, and area of a spherical triangle 15

Since (A,B,C) and (A, B, C) are bases of R3, there is a unique invertible linear mapT : R3 → R3 with T (A) = A, T (B) = B, T (B) = B. The equalities (2.15) implythat 〈T (x), T (y)〉 = 〈x, y〉 for all x, y ∈ R3. Hence T is orthogonal.

Applying Theorem 2.4.2 to the polar triangle, we obtain an analogous existence anduniqueness theorem for spherical triangles with given angles:

Theorem 2.4.4 (Existence and uniqueness for given angles). Let α, β, γ be realnumbers.

(i) There exists a spherical triangle with exterior angles α, β, γ if and only if thefollowing inequalities are satisfied:

−α+ β+ γ > 0, α− β+ γ > 0, α+ β− γ > 0, α+ β+ γ < 2π, (2.16)

(ii) If it exists, the spherical triangle with exterior angles α, β, γ is unique up to anorthogonal transformation of R3.

Consider the subset D ⊂ R3 defined by

D ={(

x1x2x3

)∈ R3

∣∣∣ −x1 + x2 + x3 > 0, x1 − x2 + x3 > 0,

x1 + x2 − x3 > 0, x1 + x2 + x3 < 2π}. (2.17)

Theorems 2.4.2 and 2.4.4 say there is a spherical triangle with side lengths a, b, c and aspherical triangle with exterior angles α, β, γ if and only if

(abc

)∈ D and

( αβγ

)∈ D,

respectively. By the uniqueness statements there is a bijective map

D → D, (a, b, c) 7→ (α, β, γ) (2.18)

mapping side lengths to corresponding exterior angles. The spherical cosine theorems(Theorem 2.5.1) provide explicit formulas for this map and its inverse.

Exercise 2.4.5. Show that D is the interior of the tetrahedron with vertices(

000

),(ππ0

),(π0π

),(

0ππ

). (2.19)

The lengths-to-angles-map (2.18) has the following peculiar property: If the side lengthsapproach a face of D, then the corresponding exterior angles approach the opposite vertexof D. If the exterior angles approach a face of D, then the corresponding side lengthsapproach the opposite vertex of D.


2.5 The spherical cosine rulesThe spherical cosine rules provide explicit formulas for the angles of a spherical triangle asfunctions of the side lengths, and vice versa.

Theorem 2.5.1 (Spherical cosine rules). Side lengths and exterior angles of a sphericaltriangle satisfy the equations

cos α =− cos a+ cos b cos c

sin b sin c, (2.20)

cos a =− cos α + cos β cos γ

sin β sin γ, (2.21)

and four more equations obtained by simultaneously permutating a, b, c and α, β, γ.

Written in terms of the interior anglesα, β, γ, the cosine rules (2.20) and (2.21) are

cosα =cos a− cos b cos c

sin b sin c, (2.22)

cos a =cosα + cos β cos γ

sin β sin γ. (2.23)

They are usually written in this form, which obscures the complete symmetry between theequations.

Exercise 2.5.2. Prove equation (2.20) using standard identities involving the scalar andcross products in R3. (Hint: Essentially, one only needs to expand the scalar productof two cross products using the formula for a triple cross product.)

The purpose of the following proof is to introduce the powerful and generalGrammatrixmethod. For example, the Gram matrix method• leads to a analytic proofs of the existence and uniqueness Theorems 2.4.2 and 2.4.4 (seeRemark 2.5.3).

• is straightforward to generalize to higher dimensions, for example to obtain (complicated)formulas relating the side lenghts and dihedral angles of spherical tetrahedra.

• will also show its versatility in the context of hyperbolic geometry.

The Gram matrix for a family v1, . . . vn of vectors vi ∈ Rn is the symmetric matrix(〈vi, vj〉) ∈ Rn×n of pairwise scalar products.

2.5 The spherical cosine rules 17

Proof of Theorem 2.5.1 (by the Gram matrix method). Weprove the cosine rule (2.20).The cosine rule (2.21) could be proved in the same way, but it also follows immediatelyby applying the cosine rule (2.20) to the polar triangle.

Let V = (A B C) ∈ R3×3 be the matrix whose columns are the vertices of thespherical triangle. Then the Gram matrix for A,B,C is

G = V tV =

〈A,A〉〈A,B〉〈A,C〉〈B,A〉〈B,B〉〈B,C〉〈C,A〉〈C,B〉〈C,C〉

=

1 cos c cos bcos c 1 cos acos b cos a 1

. (2.24)

Note for later that detG = (detV )2 > 0.Similarly, letW = (A′ B′ C ′) be the matrix of the poles. Their Gram matrix is

G′ = W tW =

〈A′, A′〉〈A′, B′〉〈A′, C ′〉〈B′, A′〉〈B′, B′〉〈B′, C ′〉〈C ′, A′〉〈C ′, B′〉〈C ′, C ′〉

=

1 cos γ cos βcos γ 1 cos α

cos β cos α 1

.

Now the product

W tV =

〈A′, A〉〈A′, B〉〈A′, C〉〈B′, A〉〈B′, B〉〈B′, C〉〈C ′, A〉〈C ′, B〉〈C ′, C〉

=

D11 0 00 D22 00 0 D33

=: D

is a diagonal matrix with positive entries by (2.12). We obtainW t = DV −1 and henceW = (V t)−1D, and finally a relation between the Gram matrices G and G′:

G′ = DV −1(V t)−1D = D(V tV )−1D = DG−1D. (2.25)

Apply Sarrus’ rule to find that the inverse of G is

G−1 =1

detG

sin2 a − cos c+ cos a cos b − cos b+ cos c cos a− cos c+ cos a cos b sin2 b − cos a+ cos b cos c− cos b+ cos c cos a − cos a+ cos b cos c sin2 c

.

Substitute this into (2.25) and consider first the diagonal elements. One finds

1 = D211

1

detGsin2 a,

so

D11 =

√detG

sin a,


and similarly

D22 =

√detG

sin b, D33 =

√detG

sin c.

Now consider for example the matrix elements (2, 3) in equation (2.25):

cos α = D221

detG(− cos a+ cos b cos c) D33.

This is the cosine rule (2.20).

Remark 2.5.3. The Gram matrix method reduces the existence and uniqueness The-orem 2.4.2 to linear algebra. To see that the inequalities (2.24) are sufficient for theexistence of a spherical triangle, which is unique up to concgruence, follow these steps:

1. For a, b, c ∈ R, letG =

(1 cos c cos b

cos c 1 cos acos b cos a 1

).

Then the principal minors of G are

1, 1− cos2 c = sin2 c,

and

detG = 1 + 2 cos a cos b cos c− cos2 a− cos2 b− cos2 c

= 4 sin(−a+ b+ c

2

)sin(a− b+ c

2

)sin(a+ b− c

2

)sin(a+ b+ c

2

).

(2.26)

(To derive the miraculous last equality above, start by applying the trigonometricidentity

2 sinx sin y = cos(x− y)− cos(x+ y)

twice to the right hand side.) So by Sylvester’s criterion, G is positive definite if a,b, c satisfy the inequalities (2.14).

2. For any positive definite symmetricmatrixG ∈ R3, there is a unique positive definitesymmetric matrix S ∈ R3 satisfying S2 = G.

3. A matrix V ∈ R3 satisfies G = V tV if and only if V = RS, for some orthogonalmatrix R ∈ O(3).

To see that the inequalities (2.14) are also necessary for the existence of a sphericaltriangle, note that G = V tV is positive definite if V is invertible. Adapt the argumentin step 1 to show that if a, b, c ∈ (0, π), they satisfy (2.14).

2.6 The sine theorem, half-angle formulas, and half-side formulas 19

2.6 The sine theorem, half-angle formulas, and half-sideformulas

All relations involving the angles and side lengths of a spherical triangle can in principlebe derived purely analytically and without any recourse to geometry from the side inequali-ties (2.14) and the side cosine rule (2.20) or, alternatively, from the angle inequaltities (2.16)and the angle cosine rule (2.21). This is a consequence of the Existence and UniquenessTheorems 2.4.2 and 2.4.4 for sphercial triangles with given side lengths or angles. In thissection, we show how to derive some of the most important relations in this way: thespherical sine theorem and the spherical half-angle theorems.

Theorem 2.6.1 (Sine theorem, half-angle and half-side formulas). Side lengths a, b, cand interior angles α, β, γ of a spherical triangle satisfy the sine theorem

sin a

sinα=

sin b

sin β=

sin c

sin γ, (2.27)

the half-angle formulas

cosα

2=

√sin(−a+b+c

2) sin(a+b+c

2)

sin b sin c, (2.28)

sinα

2=

√sin(a−b+c

2) sin(a+b−c

2)

sin b sin c, (2.29)

tanα

2=

√sin(a−b+c

2) sin(a+b−c

2)

sin(−a+b+c2

) sin(a+b+c2

), (2.30)

and the half-side formlas

cosa

2=

√cos(α−β+γ

2) cos(α+β−γ

2)


sina

2=

√

−cos(−α+β+γ2

) cos(α+β+γ2

)


tana

2=

√−cos(−α+β+γ

2) cos(α+β+γ

2)

cos(α−β+γ2

) cos(α+β−γ2

). (2.33)


Exercise 2.6.2. (a) Give a proof of the euclidean sine theorem without looking it up.(Hint: Use the Inscribed Angle Theorem mentioned in Chapter 1 to express eachside length in terms of the opposite angle and the circumradius.)

(b) Derive the half-angle theorems for euclidean triangles that are analogous to (2.28)–(2.30) from the euclidean cosine rule.

(c) Derive the cosine rule (2.21) purely analytically and without recourse to geometryfrom the cosine rule (2.20). (This is hard. A solution can be found in [Stu93].)

Proof of Theorem 2.6.1. Using

cosα = 2 cos2 α

2− 1 = 1− 2 sin2 α

2

and other trigonometric identities, one easily obtains the equations formulas (2.28)and (2.29) from the cosine rule (2.22) for the interior angle. The half-angle for-mula (2.30) for the tangent follows by taking quotients. The half-side formulas (2.31)–(2.33) can be obtained similarly from the cosine rule (2.23), or by applying the half-angleformulas to the polar triangle.

To prove the sine theorem, note that

sinα = 2 sinα

2cos

α

2=

P

sin b sin c,

where

P = 2

√sin(−a+ b+ c

2

)sin(a− b+ c

2

)sin(a+ b− c

2

)sin(a+ b+ c

2

). (2.34)

SoP = sinα sin b sin c. (2.35)

But since the right hand side of (2.34) is symmetric in a, b, c, one also has

P = sinα sin b sin c = sin β sin c sin a = sin γ sin a sin b.

Divide by sin a sin b sin c to obtain the sine theorem (2.27).

2.7 Napier’s rule and pentagramma mirificumNapier’s rule provides ten formulas relating the angles and side lengths of a right-angledspherical triangle. In principle, one could derive all ten equations (2.36)–(2.40) fromthe spherical cosine rule. But there is an easier way involving an amazing geometricconstruction called the pentagramma mirificum.

2.7 Napier’s rule and pentagramma mirificum 21

Figure 2.4: Right-angled spherical tri-angle and the “parts”of Napier’s rule

Figure 2.5: Pentagramma miri-ficum

Figure 2.6: Angles and sidesin the next triangle

Theorem 2.7.1 (Napier’s rule). Consider a right-angled spherical triangle with γ = π2

as shown in Figure 2.4 (top), and let a = π2− a, b = π

2− b. Then

cos c = sin b sin a = cotα cot β, (2.36)cos β = sinα sin b = cot c cot a, (2.37)cos a = sin c sinα = cot β cot b, (2.38)cos b = sin β sin c = cot a cotα, (2.39)cosα = sin a sin β = cot b cot c. (2.40)

In short: “The cosine of any part is equal to the product of sines of opposite parts andto the product of cotangents of adjacent parts,” where the “parts” are a, b, α, c, β, inthis cyclic order, as shown in Figure 2.4 (bottom).

Proof. Equations (2.36) follow directly from the cosine rules

cos γ =cos c− cos a cos b

sin a sin band cos c =

cos γ + cosα cos β

sinα sin β,

with cos γ = 0. To prove the remaining equations, assume first that

a, b, c, α, β <π

2. (2.41)


Now draw the two great circles that have A and B as poles (see Figure 2.5). Togetherwith the extended sides of the original triangle, they form 4 other right angled triangles.The five triangles form a right-angled pentagram called the pentagramma mirificum. Inits center there is a spherical pentagon each vertex of which is the pole of the oppositeside. Contemplate Figure 2.6 to see that

a1 =π

2− c, b1 =

π

2− β, α1 =

π

2− a, c1 =

π

2− b, β1 = α,

so thata1 = c, b1 = β, α1 = a, c1 = b, β1 = α.

It follows that the five right-angled triangles in the pentagramma mirificum have thesame parts in all five cyclic permutations. This proves the other equations of Napier’srule under the assumption (2.41).

Now suppose that a side length a, b, c or an angle α, β is strictly greater than π2.

It can be shown that, first, for one of the neighbor triangles into which the sphere isseparated by the same great circles, a, b, c, α, β < π

2, and, second, if Napier’s rule holds

for one of the neighbor triangles, it holds for all of them.The remaining cases where one side length or is equal to π

2consist of doubly or

triply right-angled triangles for which Napier’s rule can easily be checked.

Concept of neighbor triangles needs to be explained, with figure. Maybe in an exer-cise in Sec. 2.5

2.8 Stereographic projectionIn this section we discuss a peculiar map from the sphere to the plane, and we prove twostriking geometric properties. In Section 4.11 on the Poincaré disk model of the hyperbolicplane, we will project from a hyperboloid instead of a sphere. That is the reason whythe proofs in this section are purely analytical and rather brutally straightforward: Thesearguments can easily be adapted to the less familiar setting of hyperbolic geometry.

Let us first consider the two-dimensional case. Project the unit sphere S2 ⊂ R3 fromthe “north pole” e3 =

(001

)to the plane x3 = 0.

stereographic projectionMissingfigure

2.8 Stereographic projection 23

This map σ : S2 \ {e3} → R2 is called stereographic projection. One easily derives thefollowing equations for σ and its inverse:

σ

x1

x2

x3

=

1

1− x3

(x1

x2

), (2.42)

σ−1

(u1

u2

)=

1

u21 + u2

2 + 1

2u1

2u2

u21 + u2

2 − 1

. (2.43)

Theorem 2.8.1 (Stereographic projection preserves circles). The stereographic projec-tion σ maps the circles in S2 that contain the north pole e3 to lines in R2 and all othercircles in S2 to circles in R2. All circles and lines in R2 are images of circles in S2.

Remark 2.8.2. The fact that circles through e3 aremapped to lines is geometrically clear:A circle through e3 is the intersection of S2 with a plane through e3. So all projectionrays lie in this plane, and the circle is mapped to the line in which it intersects the imageplane x3 = 0.

A nice geometric proof of Theorem 2.8.1 is presented in [HCV52]. The following proofis analytic and rather brutally straightforward. The advantage is that it generalizes easily tohigher dimensions and to the setting of hyperbolic geometry (see Section 4.11).

Proof of Theorem 2.8.1. A circle in S2 is the intersection of S2 with a plane

E = {x ∈ R3 | 〈x, n〉 = d}, where |n| = 1, 0 ≤ d < 1.

The circle contains e3 precisely if

d = 〈e3, n〉 = n3.

A point u ∈ R2 in the image of the circle if and only if σ−1(u) ∈ E, that is, if and onlyif

d =1

u21 + u2

2 + 1

(2u1n1 + 2u2n2 + (u2

1 + u22 − 1)n3

).

Collecting powers of uk, we obtain the equivalent equation

0 = (n3 − d)(u21 + u2

2) + 2u1n1 + 2u2n2 − (n3 + d).

If n3 = d, this is the equation for a line. Otherwise, divide by n3 − d, complete the


squares, and use |n|2 = 1 to obtain

(u1 +

n1

n3 − d)2

+(u2 +

n2

n3 − d)2

− 1− d2

(n3 − d)2= 0.

This is the equation for a circle with center c = − 1n3−d

(n1n2

)and radius r =

√1−d2|n3−d| .

To show (without further calculations) that every circle and every line in R2 is theimage of a circle in S2, you can make an argument using the fact that three points inR2 uniquely determine a line or circle through them and three points in S2 uniquelydetermine a circle through them.

Theorem 2.8.3. Stereographic projection is conformal, that is, it preserves angles: Iftwo curves in S2 \ {e3} intersect at some angle, then their image curves in R2 intersectat the same angle.

Curves γ, η and their imagesMissingfigure

Proof of Theorem 2.8.3. Let

γ, η : (−ε, ε)→ S2 \ {e3}

be two differentiable curves with

γ(0) = η(0) =: p,

and letγ = σ ◦ γ, η = σ ◦ η

be their images curves in R2. Let

v = γ′(0), w = η′(0), v = γ′(0), w = η′(0),

and letp = γ(0) = η(0) = σ(p).


The intersection angles α and α between γ, η and γ, η, are determined by

cos α =〈v, w〉√〈v, v〉〈w, w〉

, cosα =〈v, w〉√〈v, v〉〈w,w〉

.

We want to show that α = α. In fact, we will show that

〈v, w〉 = λ 〈v, w〉, 〈v, v〉 = λ 〈v, v〉, 〈w, w〉 = λ 〈w,w〉,

withλ =

4

(〈p, p〉+ 1)2,

and this implies cos α = cosα, and hence α = α.First we derive equations for v, w in terms of v, w. By equation (2.43) we have

γ =1

〈γ, γ〉+ 1

2γ1

2γ2

〈γ, γ〉 − 1,

so

γ′ =1

〈γ, γ〉+ 1

2γ′12γ′2

2〈γ, γ′〉

− 2〈γ, γ′〉

(〈γ, γ〉+ 1)2

2γ1

2γ2

〈γ, γ〉 − 1,

=2

〈γ, γ〉+ 1

γ′1γ′2〈γ, γ′〉

− 〈γ, γ′〉

〈γ, γ〉+ 1

2γ1

2γ2

〈γ, γ〉 − 1

,

and hence

v = γ′(0) =2

〈p, p〉+ 1

v1

v2

〈p, v〉

− 〈p, v〉

〈p, p〉+ 1

2p1

2p2

〈p, p〉 − 1

.

In the same way one gets

w =2

〈p, p〉+ 1

w1

w2

〈p, w〉

− 〈p, w〉

〈p, p〉+ 1

2p1

2p2

〈p, p〉 − 1

,


so

〈v, w〉 =4

(〈p, p〉+ 1)2

(〈v, w〉+ 〈p, v〉〈p, w〉

− 〈p, w〉〈p, p〉+ 1

(2〈p, v〉+ 〈p, v〉

(〈p, p〉 − 1

)︸︷︷︸

〈p,v〉(〈p,p〉+1)

)

− 〈p, v〉〈p, p〉+ 1

(2〈p, w〉+ 〈p, w〉

(〈p, p〉 − 1

)︸︷︷︸

〈p,w〉(〈p,p〉+1)

)

+〈p, v〉〈p, w〉(〈p, p〉+ 1)2

(4〈p, p〉+

(〈p, p〉 − 1

)2

︸︷︷︸(〈p,p〉+1)2

))

=4

(〈p, p〉+ 1)2〈v, w〉,

and similarly for 〈v, v〉 and 〈w, w〉.

All of this works also for the n-dimensional sphere Sn ∈ Rn+1. The equations for thestereographic projection and its inverse are

σ : Sn \ {en+1} −→ Rn,

σ(x) =1

1− xn+1

x1

x2...xn

, (2.44)

σ−1(u) =1

〈u, u〉+ 1

2u1...2un

〈u, u〉 − 1

. (2.45)

An (n− 1)-dimensional sphere (or hypersphere) in Sn is a nonempty intersection of Snwith an n-dimensional affine subspace (or hyperplane) in Rn+1. Stereographic projectionmaps hyperspheres in Sn to hyperspheres and hyperplanes in Rn, and it is conformal. It isstraightforward to adapt the proofs of Theorems 2.8.1 and 2.8.3 to the higher dimensionalcase. (In the case of n = 1, stereographic projection from the circle to the line, the twotheorems are trivial.)

An (n−k)-dimensional sphere is Sn is the intersection of k hyperspheres. This impliesthat stereographic projection maps spheres of any dimension in Sn to spheres and affinehyperplanes of the same dimension in Rn. In particular, stereographic projection mapscircles (one-dimensional spheres) to circles and lines in Rn.


Add advanced topics section on spherical geometry and integrable systems andmaybe statistical physicsPetrera/Suris https://arxiv.org/pdf/1208.3625Onsager (uses hyperbolic)Shechtman (see refs therein) https://arxiv.org/abs/1106.3633

Chapter 3

Linear algebra refresher: Bilinear andquadratic forms, indefinite scalarproducts and orthogonaltransformations

In Chapter 4, we will define n-dimensional hyperbolic space as

Hn ={x ∈ Rn+1

∣∣ 〈x, x〉 = −1, xn+1 > 0},

where 〈·, ·〉 denotes the Lorentz scalar product

〈x, y〉 = x1y1 + l · · ·+ xnyn − xn+1yn+1.

Lengths of curves in Hn and angles between them are measured using this Lorentz scalarproduct instead of the usual euclidean one. Bilinear and quadratic forms will also play afundamental role in the treatment of conic sections and general quadrics (Chapter 7). Andwe will need more general indefinite scalar products for Möbius geometry and Lie geometry(Chapters 9 and 10). So it is probably a good idea to collect some relevant material fromlinear algebra in this chapter.

3.1 Bilinear and quadratic formsLet V be an n-dimensional vector space over a field F . We will primarily be interested inthe cases F = R and F = C, but the basic definitions make sense for an arbitrary field F .

A bilinear form on V is a function

b : V × V → F,

which is linear in each argument.

29

30 Linear algebra refresher

The matrix of a bilinear form b with respect a basis (v1, . . . , vn) of V is the matrix

B = (bij) ∈ F n×n, bij = b(vi, vj).

If x, y ∈ F n are the coordinate vectors for v, w ∈ V with respect to this basis, i.e.,

v =n∑

i=1

xivi, w =n∑

i=1

yivi,

then

b(v, w) = xtBy =n∑

i,j=1

Bijxiyj.

If (v1, . . . , vn) is another basis of V , then the coordinate transformation matrix is

T = (tij) ∈ F n×n so that vj =∑

i

tijvi. (3.1)

The matrix of the bilinear form b with respect to the new basis is

B = T tBT. (3.2)

Note that the transformation rule for the matrix A of a linear map V → V is different:

A = T−1AT. (3.3)

In the absence of a scalar product, there is no canonical way to identify bilinear forms withendomorphisms (see, however, Remark 3.1.2).

A bilinear form b on V is symmetric if

b(v, w) = b(w, v) for all v, w ∈ V.

This is the case if the matrix B of the bilinear form with respect to one (and hence any)basis is symmetric, i.e., Bt = B.

A function q : V → F is a quadratic form on V if there exists a symmetric bilinearform b on V such that

q(v) = b(v, v) for all v ∈ V. (3.4)

If q is a quadratic form and the characteristic of the field F is not 2, then the symmetricbilinear form b satisfying (3.4) is uniquely determined. This follows from the polarizationidentity for symmetric biliner forms,

2b(v, w) = b(v + w, v + w)− b(v, v)− v(w,w). (3.5)

Symmetric bilinear forms and quadratic forms are therefore in one-to-one correspondence,unless the characteristic is 2. In coordinates, quadratic forms are homogeneous polynomialsof degree 2.

3.1 Bilinear and quadratic forms 31

Example 3.1.1. (a) The quadratic form

x21 + x1x2 + x2

2 on R2

comes from the symmetric bilinear form

x1y1 +1

2x1y2 +

1

2x2y1 + x2y2.

(b) On the 2-dimensional vector spaceF22 over the fieldF2 = Z/2Zwith two elements,

the non-vanishing symmetric bilinear form

x1y2 + x2y1

represents the quadratic form 0.

The kernel of a symmetric bilinear form b on V is

ker b = {v ∈ V | b(v, w) = 0 for all w ∈ V }.

The kernel of a bilinear form is a linear subspace of V . If the characteristic is not 2, thekernel of a quadratic form is defined as the kernel of the corresponding bilinear form.

Remark 3.1.2. A bilinear form b on V can be interpreted as a linear map

V → V ∗, v 7→ b(v, ·)

into the dual vector space V ∗. The kernel of the bilinear form b is the kernel of this linearmap. Also, the transformation rule (3.1) for bilinear forms is just the transformationrule for linear maps from V to V ∗, provided that the dual bases are chosen for V ∗.

A symmetric bilinear form b is called

degeneratenondegenerate

}if

{ker b 6= {0}ker b = {0} .

A symmetric bilinear form is degenerate if and only if its matrix with respect to one (henceany) basis has determinant 0.

If the characteristic is not 2, a quadratic form is called degenerate or nondegenerate ifthe corresponding bilinear form is degenerate or nondegenerate.

If the base field F is R (or any other ordered field) one can further differentiate bilinearand quadratic forms depending on the signs of their values. On a vector space V over an


ordered field, a bilinear form b (and the corresponding quadratic form) is called

positive semidefinitepositive definite

indefinite

if

b(v, v) ≥ 0 for all v ∈ V .b(v, v) > 0 for all v ∈ V \ {0}.there are v, w ∈ V with b(v, v) > 0 and b(w,w) < 0.

The terms negative semidefinite and negative definite are defined analogously. A realnondegenerate symmetric bilinear form is either positive definite, negative definite, orindefinite.

3.2 Diagonalizing a bilinear form with the generalizedGram–Schmidt algorithm

For any finite dimensional euclidean vector space, the well-known Gram–Schmidt orthog-onalization procedure is an algorithm to construct an orhonormal basis. It provides analgorithmic proof that an orthonormal basis exists. In this section, we explain how theGram–Schmidt algorithm can be generalized to diagonalize arbitrary bilinear forms. Inparticular, this implies that any symmetric bilinear form can be diagonalized. More pre-cisely:

Theorem 3.2.1 (Any symmetric bilinear form can be diagonalized). Let V be an n-dimensional vector space over an arbitrary field F , and let b be a symmetric bilinearform on V . Then there exists a basis (v1, . . . , vn) of V such that

b(vi, vj) = 0 if i 6= j. (3.6)

The basis vectors vi with b(vi, vi) = 0 form a basis of ker b.

If a basis (v1, . . . , vn) satisfies (3.6), then the matrix of b with respect to this basis isdiagonal. That is why such a basis is said to diagonalize the bilinear form b. The rank ofthe bilinear form b is the number r = n− dim ker b of nonzero entries on the diagonal. Itis independent of the particular choice of diagonalizing basis. In fact, the rank of a bilinearform is the rank of the corrsponding linear map V 7→ V ∗ (see Remark 3.1.2).

Algorithm 3.2.1 is a version of the ususal Gram-Schmidt orthogonalization procedurewhich is known as the modified or stabilized Gram-Schmidt algorithm. Given a positive (ornegative) definite bilinear form on a real vector space V , and a basis (v1, . . . , vn) of V , itconstructs a basis that diagonalizes b. Note that the basis vectors are not normalized. Wewill generalize this algorithm to diagonalize forms on vector spaces over arbitrary fields,and whether or not the base vectors can be normalized depends on the existence of squareroots in the base field. We will discuss normalization in the cases of complex and realvector spaces in Section 3.4.

3.3 Completing the squares 33

Input: a basis (v1, . . . , vn) and a definite bilinear form bfor i = 1 to n− 1 dofor j = i+ 1 to n do

vj ← vj −b(vi, vj)

b(vi, vi)vi

end forend for

Output: a diagonalizing basis (v1, . . . , vn)

Algorithm 3.2.1: The modified Gram–Schmidt algorithm without normalization. In eachiteration of the outer for-loop, we subtract the component in the direction of vi from thebasis vectors vi+1, . . . , vn. (In the classical version of Gram–Schmidt, the for-loop over j isthe outer loop and the inner loop over i runs from 1 to j − 1.)

If b is not positive or negative definite, Algorithm 3.2.1 may fail because for some i inthe outer for-loop, it may happen that b(vi, vi) = 0. This can be repaired as follows. Let usfirst consider the case that b is nondegenerate. Then if b(vi, vi) = 0 for some i in the outerfor-loop, there are two possibilities :(i) There is a k ∈ {i + 1, . . . , n} for which b(vk, vk) 6= 0. If this is the case, simply

swap vi and vk and continue.(ii) b(vk, vk) = 0 for all k ∈ {i + 1, . . . , n}. Because we have assumed that b is nonde-

generate, there must be a k ∈ {i+ 1, . . . , n} such that b(vi, vk) 6= 0. This implies

b(vi ± vk, vi ± vk) = ±2b(vi, vk) 6= 0.

So we may replace vi and vk with vi + vk and vi − vk and continue.Adding these modifications to Algorithm 3.2.1 produces Algorithm 3.2.2, the generalizedGram–Schmidt algorithm.

It remains to consider the case that b is degenerate. In this case, let U be any comple-mentary subspace of ker b, so that V is the direct sum

V = U ⊕ ker b.

Then the restriction b|U of the bilinear form b to the subspace U is a non-degenerate bilinearform on U . Apply the modified Gram–Schmidt algorithm to a basis of U to obtain adiagonalizing basis of U . Extend it to a diagonalizing basis of V by appending a basis ofker b.

3.3 Completing the squaresNow consider the quadratic form q corresponding to a symmetric bilinear form b. Written incoordinates u1, . . . , un with respect to a basis (v1, . . . , vn) that diagonalizes b, the quadratic


Input: a basis (v1, . . . , vn) and a symmetric bilinear form bfor i = 1 to n− 1 doif b(vi, vi) = 0 thenif there is a k ∈ {i+ 1, . . . n} such that b(vk, vk) 6= 0 thenswap vi and vk

elsefind k ∈ {i+ 1, . . . n} such that b(vi, vk) 6= 0replace vi and vk by vi + vk and vi − vk

end ifend iffor j = i+ 1 to n do

vj ← vj −b(vi, vj)

b(vi, vi)vi

end forend for

Output: a diagonalizing basis (v1, . . . , vn)

Algorithm 3.2.2: The generalized Gram–Schmidt orthogonalization algorithm

form q has the form

q =n∑

i=1

λiu2i , where λi = b(vi, vi).

So when we diagonalize a symmetric bilinear form, we apply a linear change of coordi-nates that transforms the quadratic form into a “sum of squares”. Instead of applying thegeneralized Gram–Schmidt algorithm to the bilinear form, it is sometimes easier to findsuch a change of coordinates by completing the squares. Let us explain this by way of twoexamples:

Example 3.3.1. Suppose in some coordinates x1, x2, x3, the quadratic form is

x12 + x2

2 + 2x32 + x1x2 + x1x3 + x2x3.

First, consider the terms involving x1 and complete the squares, then do the same with

3.4 Normalization in complex and real vector spaces 35

the remaining terms involving x2 and x3:

x12 + x2

2 + 2x32 + x1x2 + x1x3 + x2x3

=(x1 + 12x2 + 1

2x3)2 + 3

4x2

2 + 74x3

2 + 12x2x3

=(x1 + 12x2 + 1

2x3︸︷︷︸

u1

)2 + 34(x2 + 1

3x3︸︷︷︸

u2

)2 + 53x3︸︷︷︸u3

2

= u12 + 3

4u2

2 + 53u3

2.

It may happen that there is not square to complete:

Example 3.3.2. Suppose in some coordinates the quadratic form is

x1x2 + x2x3.

Introduce new coordinates y1 = x1 +x2 and y2 = x1−x2 and write the quadratic formin terms of y1, y2, x3:

x1x2 + x2x3 = 14(x1 + x2︸︷︷︸

y1

)2 − 14(x1 − x2︸︷︷︸

y2

)2 + x2x3

= 14y1

2 + 14y2

2 + 12(y1 − y2)x3.

Now you can continue by completing the squares.

3.4 Normalization in complex and real vector spacesSuppose (v1, . . . , vn) is a basis of V that diagonalizes the symmetric bilinear form b and let

λi = b(vi, vi)

be the diagonal entries in the matrix of b with respect to this basis.

3.4.1 The complex caseIf V is a complex vector space, we can make each nonzero entry λi ∈ C equal to 1 bydividing the basis vector vi by one of the two square roots of λi. Thus we obtain thefollowing theorem:

Theorem 3.4.1. For any symmetric bilinear form b with rank r on an n-dimensionalcomplex vector space V , there is a basis (v1, . . . , vn) of V such that

b(vi, vj) = 0 if i 6= j


and

b(vi, vi) =

{1 if i = j and 1 ≤ i ≤ r

0 if i = j and r < i ≤ n.

As a consequence, the rank determines a complex symmetric bilinear form up to linearautomorphisms of V . More precisely:

Corollary 3.4.2. Let b and b be symmetric bilinear forms on a finite dimensionalcomplex vector space V . Then the following statements are equivalent:(i) There is a linear autormorphism F : V → V such that for all v, w ∈ V

b(v, w) = b(F (v), F (w)).

(ii) The forms b and b have the same rank.

3.4.2 The real caseIf V is a real vector space, we can make each nonzero entry λi ∈ R equal to 1 or −1 bydividing vi by

√|λi|. Thus we obtain the following theorem:

Theorem 3.4.3. For any symmetric bilinear form b on an n-dimensional real vectorspace V , there is a basis (v1, . . . , vn) of V such that

b(vi, vj) = 0 if i 6= j (3.7)

andb(vi, vi) ∈ {1,−1, 0}. (3.8)

The numbers of ones, minus-ones and zeros are invariants of the form:

Theorem 3.4.4 (Sylvester’s law of inertia). Let (v1, . . . , vn) be a basis of V thatsatisfies (3.7) and (3.8), and let

i+ = #{k ∈ {1, . . . , n}

∣∣ b(vk, vk) > 0},

i− = #{k ∈ {1, . . . , n}

∣∣ b(vk, vk) < 0},

i0 = #{k ∈ {1, . . . , n}

∣∣ b(vk, vk) = 0}.

Then the numbers i+, i−, i0 depend only on the bilinear form b and not on the particularchoice of basis satisfying (3.7) and (3.8).

3.5 Scalar products 37

Exercise 3.4.5. To prove Theorem 3.4.4, note that i0 = dim ker b and show that

i+ = max{dimU | U ⊆ V is a subspace on which b is positive definite}, (3.9)i− = max{dimU | U ⊆ V is a subspace on which b is negative definite}. (3.10)

The numbers i+ and i− are called the positive index and the negative index of b. Thetriple (i+, i−, i0), or if i0 = 0 also the pair (i+, i−), is called the signature of b.

Historical Remark 3.4.6. The peculiar name of Theorem 3.4.4 is due to J. J. Sylvester,who saw an analogy with the preservation of mass:

[This is] a law to which my view of the physical meaning of quantity ofmatter inclines me, upon the ground of analogy, to give the name of theLaw of Inertia for Quadratic Forms, as expressing the fact of the existenceof an invariable number inseparably attached to such forms. [Syl52]

3.5 Scalar productsA [bilinear] scalar product on a real or complex vector space V is a non-degeneratesymmetric bilinear form. Scalar products are often written 〈·, ·〉 or (·, ·). A vector spaceequipped with a scalar product (or, more precisely, a pair (V, 〈·, ·〉) consisting of a Vectorspace V and a scalar product 〈·, ·〉 on V ) is called an inner product space. Vectors v, w inan inner product space are called orthogonal if 〈v, w〉 = 0.

Remark 3.5.1. This notion of scalar products is useful in geometry. In linear algebratextbooks, scalar products are usually defined differently:• For real vector spaces, they are defined as positive definite symmetric bilinear forms.Our definition is more general. Every positive definite form is nondegenerate, but anondegenerate form may be indefinite.

• For complex vector spaces, scalar products are usually defined as positive definitehermitian forms. They are not bilinear but sesquilinear. Hermitian scalar productsare also important in geometry, but we do not want to discuss them right now.

3.5.1 Bilinear complex scalar products

Example 3.5.2. On the standard complex vector spaceCn, the canonical bilinear scalar


product is

〈x, y〉 =n∑

j=1

xjyj. (3.11)

A basis (v1, . . . , vn) of a complex vector spacewith scalar product is called orthonormal,if

〈vj, vk〉 =

{0 if j 6= k

1 if j = k.(3.12)

By Theorem 3.4.1, every complex vector space equipped with a scalar product has anorthonormal basis. By an appropriate choice of basis, any n-dimensional complex vectorspace with a scalar product can be identified with the standard space Cn, equipped with thecanonical scalar product (3.11).

3.5.2 Real scalar products

Example 3.5.3. On the standard real vector space Rp+q, the canonical scalar productwith signature (p, q) is

〈x, y〉 =

p∑

j=1

xjyj −p+q∑

j=p+1

xjyj. (3.13)

A basis (v1, . . . , vn) of a real vector space with a scalar product with signature (p, q) iscalled orthonormal, if

〈vj, vk〉 =

0 if j 6= k

1 if j = k ∈ {1, . . . , p}−1 if j = k ∈ {p+ 1, . . . , p+ q}.

(3.14)

By Theorem 3.4.3, every real vector space with scalar product has an orthonormal basis.The (p + q)-dimensional standard real vector space Rp+q equipped with the canonical

scalar product (3.13) with signature (p, q) is denoted by Rp,q. By an appropriate choiceof basis, any real vector space with a scalar product with signature (p, q) can be identifiedwith Rp,q.

A euclidean scalar product is a scalar product that is positive definite, i.e., one withsignature (n, 0). A vector space equipped with a euclidean scalar product is called aeuclidean vector space. In a euclidean vector space, the euclidean norm or length of avector v is

|v| =√〈v, v〉, (3.15)

3.5 Scalar products 39

and the angle between two nonzero vectors v, w is

∠(v, w) = arccos

(〈v, w〉√〈v, v〉〈w,w〉

). (3.16)

A Lorentz scalar product is an indefinite scalar product with signature (n − 1, 1), i.e.,with negative index 1. A vector space equipped with a Lorentz scalar product is called aLorentz vector space. We will discuss Lorentz vector spaces in Section 3.7. They play animportant role in hyperbolic geometry (see Chapter 4).

Remark 3.5.4. In Einstein’s special theory of relativity, spacetime is a four-dimensionalLorentz vector space. (More precisely, it is the affine space based on a four-dimensionalLorentz vector space.) The following terminology is motivated by Einstein’s specialtheory of relativity.

A vector v in a vector space with indefinite scalar product is called

space-like if 〈v, v〉 > 0,

time-like if 〈v, v〉 < 0,

light-like or isotropic if 〈v, v〉 = 0 and v 6= 0.

The set of lightlike vectors is called the light cone.

The signature of a scalar product can be determined with the Gram–Schmidt orthogo-nalization algorithm (see Section 3.2). With some luck, the signature of a scalar product canalso be read off from the sequence of principal minors with respect to some basis:

Theorem 3.5.5. Let V be a real vector space with scalar product 〈·, ·〉, let (v1, . . . , vn)be a basis of V , let B = (〈vi, vj〉) ∈ Rn×n be the matrix of the scalar product withrespect to his basis, and let (a0, . . . , an) be the finite sequence of principal minors:

a0 = 1, a1 = B11, . . . ak = det

B11 · · · B1k... ...

Bk1 · · · Bkk

, . . . an = detB.

Suppose that ak 6= 0 for all k ∈ {1, . . . , n}. Then the negative index of the scalarproduct 〈·, ·〉 is equal to the number of sign changes in the sequence (a0, . . . , an).


Corollary 3.5.6 (Sylvester’s criterion). The scalar product is euclidean if and only ifall principal minors are positive.

Exercise 3.5.7. Prove Theorem 3.5.5.

3.6 Orthogonal transformationsLet V be a finite dimensional real or complex vector space equipped with a [bilinear] scalarproduct 〈·, ·〉.

An orthogonal transformation on V is a linear map F : V → V that satisfies

〈Fv, Fw〉 = 〈v, w〉 for all v, w ∈ V. (3.17)

Because the scalar product is nondegenerate, this implies immediately that F is injective.Since V was assumed to be finite dimensional, it follows that F is also surjective, henceinvertible. (For infinite dimensional vector spaces, the requirement that an orthogonaltransformation is invertible is added to the definition.)

The orthogonal transformations of a inner product space (V, 〈·, ·〉) form a group, theorthogonal group of (V, 〈·, ·〉), denoted by O(V, 〈·, ·〉).

Let (v1, . . . , vn) be a basis of V and let B and A be the matrices of the scalar productand an endomorphism F with respect to this basis, respectively. Then F is orthogonal ifand only if

ATBA = B, (3.18)

or, equivalently, if A is invertible and

A−1 = B−1ATB. (3.19)

Equation (3.18) implies that the determinant of an orthogonal transformation is 1 or −1.The orthogonal transformations with determinant 1 form a subgroup, the special orthogonalgroup, denoted by SO(V, 〈·, ·〉).

The orthogonal and special orthogonal groups ofCn with standard scalar product (3.11)are denoted by O(n,C) and SO(n,C). The orthogonal and special orthogonal groups ofRp,q are denoted by O(p, q) and SO(p, q). In the euclidean case (q = 0) one also writesO(p) or O(p,R).

A (special) orthogonal matrix is the matrix of a (special) orthogonal transformationwith respect to an orthonormal basis of V . The columns of an orthogonal matrix form anorthonormal basis of Cn or Rp,q, respectively. The columns of a special orthogonal matrixform a positively oriented orthonormal basis.

3.7 A closer look at Lorentz vector spaces 41

Figure 3.1: The Lorentz vector spaces R1,1 (left) and R2,1 (right). The two vectors shownin R1,1 form an orthonormal basis.

The groups of orthogonal matrices are denoted by the same symbols:

O(n,C) = {A ∈ Cn×n |ATA = In},O(p, q) = {A ∈ Rn×n |AT Ip,qA = Ip,q},

SO(n,C) = {A ∈ O(n,C) | detA = 1},SO(p, q) = {A ∈ O(p, q) | detA = 1},

where In ∈ Rn×n ⊂ Cn×n denotes the n × n identity matrix and Ip,q denotes the blockmatrix

Ip,q =

(Ip 00 −Iq

).

3.7 A closer look at Lorentz vector spacesIn this Section, we will take a closer look at Lorentz vector spaces. The motivation istwofold. First and foremost, we need to gather some background material for the treatmentof hyperbolic geometry in Chapter 4. More generally, we need to develop some geometricintuition for vector spaces with indefinite scalar product. The particular case of signature(n, 1) is a good place to start. The following fact is special to this signature:

Proposition 3.7.1. In a Lorentz vector space, any non-zero vector that is orthogonal toa timelike vector is spacelike.

Exercise 3.7.2. Prove Proposition 3.7.1. (Use Theorem 3.4.4.)


Since (by choosing an orthonormal basis) any Lorentz vector space can be identifiedwith Rn,1 for some n ≥ 1, we only need to consider these standard spaces. To build upsome geometric intuition, we will first loot at the two- and three-dimensional cases. Thenwe will discuss the orthogonal transformations of Lorentz vector spaces.

3.7.1 Two-dimensional Lorentz spaceThe two-dimensional Lorentz vector space R1,1 is the standard vector space R2 equippedwith the indefinite scalar product

〈x, y〉 = x1y1 − x2y2, (3.20)

The corresponding quadratic form is

〈x, x〉 = x21 − x2

2,

and〈x, x〉 = 0 ⇐⇒ x1 = x2 or x1 = −x2.

The lightlike vectors lie on the two angle bisectors (in the euclidean sense) of the coordinateaxes. They divide the R1,1-plane into four regions (see Figure 3.1, left). Two regionscontain the time-like vectors and two regions contain space-like vectors. A nonzero vectorx is spacelike, lighlike or timelike depending on the slope of the line Rx generated by x:

A nonzero vector x ∈ R1,1 is

spacelikelightliketimelike

if

|x1| > |x2||x1| = |x2||x1| < |x2|

.

The spacelike unit vectors (with 〈x, x〉 = 1) form a hyperbola in the spacelike region. Thetimelike unit vectors (with 〈x, x〉 = −1) form a hyperbola in the timelike region. The anglebisectors of the coordinate axes are the asymptotes of both hyperbolas.

Two nonzero vectors x and y are orthogonal if

0 = 〈x, y〉 = x1y1 − x2y2 = det

(x1 y2

x2 y1

),

which is the case if and only if the lines Rx and Ry are symmetric (in the euclidean sense)with respect to the one angle bisector of the coordinate axes (and therefore also the other).

3.7.2 Three-dimensional Lorentz spaceThe three-dimensional Lorentz vector space R2,1 is the standard space R3 equipped withthe scalar product

〈x, y〉 = x1y1 + x2y2 − x3y3. (3.21)

3.7 A closer look at Lorentz vector spaces 43

The corresponding quadratic form is

〈x, x〉2,1 = x21 + x2

2 − x23

and〈x, x〉 = 0 ⇐⇒ x3 = ±

√x2

1 + x22.

The light cone is a right circular double cone (see Figure 3.1, rigth). The spacelike vectorsfill the connected region outside the cone. The timelike vectors fill the region inside thedouble cone, which consists of two connected components.

The spacelike unit vectors form a one-sheeted hyperboloid outside the light cone.The timelike unit vectors form a two-sheeted hyperboloid inside the light cone. Bothhyperboloids are asymptotic to the light cone.

Exercise 3.7.3. For a nonzero vector v ∈ R2,1 consider the Lorentz-orthogonal com-plement

v⊥ = {x ∈ R2,1 | 〈v, x〉 = 0 }.This is a two-dimensional subspace of R3. Describe it in terms of the euclideangeometry in R3. Don’t forget to consider also the case that v is lightlike.

3.7.3 Orthogonal transformations of a Lorentz vector spaceFor any n ≥ 1, the region of timelike vectors in (n + 1)-dimensional Lorentz space Rn,1

has two connected components separated by the light cone. For all x in one component oftimelike vectors, the last coordinate is positive, xn+1 > 0, whereas for all x in the othercomponent, the last coordinate is negative, xn+1 < 0.

An orthogonal transformation T ∈ O(n, 1) of Rn,1 either maps each connected com-ponent of timelike vectors onto itself, or it interchanges the two components. Equivalently,the transformation T either maps each sheet of the hyperboloid 〈x, x〉 = −1 of timelikeunit vectors onto itself, or it interchanges the sheets.

An orthogonal transformation of Rn,1 that maps each connected componenent of time-like vectors onto itself is called orthochronous or future preserving. (These terms are ofcourse taken from the special theory of relativity.)

The orthochronous transformations form a subgroup of the orthogonal group O(n, 1)with the tongue-twisting name orthochronous orthogonal group,

O+(n, 1) = {T ∈ O(n, 1) | 〈Ten+1, en+1〉 < 0}. (3.22)

(Note that the last component of Ten+1 is positive if and only if the Lorentz scalar product〈Ten+1, en+1〉 is negative.)

The orthochronous orthogonal transformations with determinant 1 form a subgroupcalled the special orthochronous orthogonal group,

SO+(n, 1) = O+(n, 1) ∩ SO(n, 1).


In all, the orthogonal group O(n, 1) has 4 connected components, each consisting ofthe transformations with determinant either +1 or−1 and either fixing or interchanging thecomponents timelike vectors.

Remark 3.7.4. InEinstein’s special theory of relativity, spacetime is the four-dimensionalLorentz vector spaceR3,1. Orthogonal transformations ofR3,1 are called Lorentz trans-formations, and the orthogonal group O(3, 1) is called the Lorentz group. Lorentztransformations describe coordinate transformations between inertial coordinate sys-tems of different observers, which may be moving at constant speed relative to eachother.

Chapter 4

Hyperbolic geometry

Just as spherical geometry is geometry in the sphere, hyperbolic geometry is geometry inhyperbolic space. But while there is hardly any confusion what the n-dimensional sphereis, it is less clear how to define n-dimensional hyperbolic space.

Usingmore advanced concepts fromdifferential geometry, one can definen-dimensionalhyperbolic space as the simply connectedn-dimensionalRiemannianmanifoldwith constantcurvature −1. Conceptionally, this is a satisfactory and clean definition. But this approachto hyperbolic geometry is fairly abstract and requires some heavy machinery that is beyondthe scope of this book. Analogously, one could define the n-dimensional sphere as thesimply connected n-dimensional Riemannian manifold with constant curvature 1. Fromthis point of view, the so-called round sphere Sn ⊂ Rn+1 defined by equation (2.1) is justa concrete model for this abstract concept. But this concrete model is incredibly useful ifone wants to study spherical geometry.

While there is one obvious model for spherical geometry (the round sphere Sn), it isless obvious what the best model for hyperbolic geometry is. Different models have distinctadvantages and disadvantages. We will introduce the most important models of hyperbolicspace in this chapter.

Any of these models could be used to define hyperbolic space. To fix ideas, we willdefine hyperbolic space in terms of the hyperboloid model (see Section 4.1). In this model,the analogy between spherical and hyperbolic geometry is most obvious. For example,we will treat hyperbolic lines, the hyperbolic metric, hyperbolic isometries, and hyperbolictrigonometry in much the same way as we treated great circles, the spherical metric, andspherical isometries in Chapter 2.

The Beltrami-Klein ball model (see Section 4.6) and the conformal Poincaré ball model(see Section 4.11) are better suited to visualize hyperbolic space. Some calculations areeasiest in the conformal half-space model (see Section 4.14). The hemisphere model (seeSection 4.12) makes it easier to translate between the other models.

45

46 Hyperbolic geometry

4.1 Hyperbolic space (in the hyperboloid model)We define n-dimensional hyperbolic spaceHn as the upper sheet of the two sheeted hyper-boloid of timelike unit vectors in the (n+1)-dimensional Lorentz vector spaceRn,1,

Hn = {x ∈ Rn,1 | 〈x, x〉 = −1, xn+1 > 0}, (4.1)

where

〈x, y〉 = x1y1 + . . .+ xnyn − xn+1yn+1 (4.2)

is the Lorentz scalar product of Rn,1 (see Sections 3.5–3.7). Most importantly:

Lengths of curves in Hn ⊂ Rn,1 and angles between curves are measured with theLorentz scalar product (4.2) ofRn,1 and not with the canonical euclidean scalar productof Rn+1.

So the hyperbolic length of a curve γ : [t0, t1]→ Hn is

length(γ) =

∫ t1

t0

√〈γ′(t), γ′(t)〉 dt (4.3)

=

∫ t1

t0

√γ′ 21 (t) + . . .+ γ′ 2n (t)− γ′ 2n+1(t) dt.

The hyperbolic angle α ∈ [0, π] between two curves

γ, η : (−ε, ε)→ Hn

intersecting at t = 0 in the point

p = γ(0) = η(0) ∈ Hn

with non-zero velocitiesv = γ′(0), w = η′(0)

is determined by

cosα =〈v, w〉√〈v, v〉〈w.w〉

(4.4)

=v1w1 + . . .+ vnwn − vn+1wn+1√

(v21 + . . .+ v2

n − v2n+1)(w2

1 + . . .+ w2n − w2

n+1)

Why is all this well defined? More precisely:Question 1. In equation (4.3), why is

〈γ′(t), γ′(t)〉 ≥ 0,

so that the square root is a positive real number?

4.2 “One-dimensional hyperbolic space” 47

Question 2. In equation (4.4), why is

− 1 ≤ 〈v, w〉√〈v, v〉〈w.w〉

≤ 1, (4.5)

so the angle α is well defined as the inverse cosine of this expression?To answer question 1, note that γ(t) is a timelike unit vector for all t ∈ [t0, t1],

〈γ(t), γ(t)〉 = −1.

Taking derivatives on both sides of this equation, we obtain

〈γ(t), γ′(t)〉 = 0.

So for all t ∈ [t0, t1], the velocity vector γ′(t) is orthogonal to the timelike vector γ(t). Itfollows that γ′(t) is spacelike if it is not the zero vector (see Proposition 3.7.1).

To answer question 2, note that by the same reasoning, the velocity vectors v and w areboth contained in the orthogonal complement of the vector p:

v, w ∈ p⊥ = {x ∈ Rn,1 | 〈p, x〉 = 0 }.

But because p is timelike, the restriction 〈·, ·〉|p⊥ of the Lorentz scalar product to the subspacep⊥ is positive definite. This follows from Sylvester’s Law of Inertia (Theorem 3.4.4). So therestriction of the Lorentz scalar product is a euclidean scalar product on p⊥. But then theinequalities (4.5) follow immediately from the Cauchy-Schwartz inequality for euclideanscalar products.

4.2 “One-dimensional hyperbolic space”The space H1 obtained by setting n = 1 in equation (4.1) is the upper branch of thehyperbola

x12 − x2

2 = −1 (4.6)

shown in Figure 4.1. In this section, we calculate hyperbolic lengths along this hyperbola.Geomerically, this is not very interesting. This “one-dimensional hyperbolic space” isisometric to the one-dimensional euclidean space R1. So there is no difference betweenhyperbolic and euclidean geometry in one dimension. Nervertheless, the calculation will beuseful in the next section, in which we discuss hyperbolic lines in n-dimensional hyperbolicspace.

Just like the unit circle can be parametrized with the cosine and sine functions, the upperbranch of the hyperbola (4.6) can be parametrized with the hyperbolic cosine and sine:

γ : R→ H1 ⊂ R1,1, γ(s) =

(sinh scosh s

). (4.7)


x2

x1

x12 − x22 = −1

H1

Figure 4.1: The hyperbola of unit timelike vectors in R1,1

This is due to the hyperbolic version of the trigonometric Pythagoras’ theorem:

cosh2 s− sinh2 s = 1. (4.8)

As in the spherical case, this parametrization has unit speed:

γ′(s) =

(cosh ssinh s

)and 〈γ′(s), γ′(s)〉 = cosh2 s− sinh2 s = 1.

So the length of the image of a parameter intervall [s1, s2] equals the length of the interval:

length(γ|[s1,s2]) =

∫ s2

s1

√〈γ′(s), γ′(s)〉 = s2 − s1. (4.9)

On the other hand, using the addition theorem for the hyperbolic cosine,

cosh(x+ y) = cosh x cosh y + sinhx sinh, (4.10)

we obtain the equation

〈γ(s1), γ(s2)〉 = sinh s1 sinh s2 − cosh s1 cosh s2 = − cosh(s1 − s2). (4.11)

Equations (4.9) and (4.11) imply the following result:The hyperbolic distance d(p1, p2) of two points p1, p2 ∈ H1 satisfies the equation

cosh d(p1, p2) = −〈p1, p2〉. (4.12)

4.3 Lines in hyperbolic space

Definition 4.3.1. A hyperbolic line in n-dimensional hyperbolic space Hn is a non-empty intersection of Hn with a 2-dimensional linear subspace U of Rn,1.

4.3 Lines in hyperbolic space 49

Proposition 4.3.2. If U is a 2-dimensional linear subspace of Rn,1 with U ∩Hn 6= ∅,then the restriction 〈·, ·〉|U has signature (1, 1).

Proof. By assumption, there is a u ∈ U with 〈u, u〉 = −1. Extend u to a basis u, v ofU , and let v = v + 〈u, v〉u. Then u, v is a basis of U with 〈u, v〉 = 0. Because anynon-zero vector which is orthogonal to a timelike vector is spacelike, 〈v, v〉 > 0. So〈·, ·〉|U has signature (1, 1).

By an appropriate choice of basis, any 2-dimensional U intersecting Hn can thereforebe identified with R1,1. The hyperbolic line U ∩Hn is thus identified with 1-dimensionalhyperbolic space H1 ∈ R1,1.

Remark 4.3.3. In the same way, any non-empty intersection of Hn with a (k + 1)-dimensional subspace can be identified with Hk.

For two points p1, p2 ∈ Hn, there is a unique hyperbolic line containing them:span(p1, p2) ∩Hn.

Theorem 4.3.4. The shortest piecewise continuously differentiable curve connectingtwo points p1, p2 ∈ Hn is the hyperbolic line segment between them. Its hyperboliclength is

d(p1, p2) = arcosh(−〈p1, p2〉).

This can be proved in the same way as we proved the corresponding theorem for thesphere.


4.4 Two-dimensional hyperbolic spaceThe hyperbolic plane H2 is one component of the hy-perboloid of two sheets x1

2 + x22 − x3

2 = −1 in R2,1.Any 2-dimensional subspaceU ofR2,1 which intersectsH2 is

U = {x ∈ R2,1 | 〈x, n〉 = 0}for some n ∈ R2,1 with 〈n, n〉 = 1. The vector −nwould give the same subspace, but up to sign, the unitnormal n of U is unique. Thus, the spacelike unitvectors in R2,1 are in 2-to-1 correspondence with thehyperbolic lines in H2. They are in 1-to-1 correspon-dence with the hyperbolic half-planes

{x ∈ R2,1 | 〈x, n〉 ≥ 0} ∩H2.

Proposition 4.4.1. Let n1, n2 ∈ R2,1 with 〈n1, n1〉 = 〈n2, n2〉 = 1, and let l1, l2 be thecorresponding hyperbolic lines, li = {x ∈ H2 | 〈x, ni〉 = 0}. Assume n1 6= ±n2, sothat l1 and l2 are different lines. Then the following statements are equivalent.(i) The lines l1 and l2 intersect.(ii) The restriction of 〈·, ·〉 to span(n1, n2) has signature (2, 0).(iii) |〈n1, n2〉| < 1.

Proof. (i)⇒(ii): If l1 ∩ l2 6= ∅ then there is an x ∈ H2 with 〈x, n1〉 = 〈x, n2〉 = 0.So x⊥ = span(n1, n2), and (ii) follows because any non-zero vector orthogonal to atimelike vector is spacelike.

(ii)⇒(i): If the restriction of the scalar product to span(n1, n2) has signature (2, 0),then its restriction to the orthogonal complement span(n1, n2)⊥ must have signature(0, 1), so the complement intersects H2.

(ii)⇔(iii): The vectors n1, n2 form a basis of span(n1, n2). In this basis, the matrixof the restriction of 〈·, ·〉 is

B =

(〈n1, n1〉〈n1, n2〉〈n2, n1〉〈n2, n2〉

)=

(1 〈n1, n2〉

〈n2, n1〉 1

).

So B11 = 1 and detB = 1− 〈n1, n2〉2. The equivalence of (ii) and (iii) follows fromlast lecture’s signature theorem (and the fact that the matrix of a non-degenerate bilinearform has non-zero determinant).

Now suppose the hyperbolic lines l1, l2 intersect in x ∈ H2, and let hi = {y ∈ H2 |〈y, ni〉 ≥ 0} be half-planes bounded by l1, l2. Since the Lorentz scalar product of R2,1 is

4.5 Hyperbolic triangles 51

Euclidean on the subspace x⊥ = span(n1, n2), we can measure angles between vectors inx⊥ in the usual way.

The exterior angle α of the half-planes h1, h2 at x is determined by

cos α = 〈n1, n2〉.

The interior angle α is π − α, so

− cosα = 〈n1, n2〉.

In particular, l1 and l2 intersect orthogonally if 〈n1, n2〉 = 0.

Remark 4.4.2. A hyperbolic rotation (ofH2) with center x ∈ H2 is a map T ∈ O(2, 1)with T (x) = x and which is a Euclidean rotation on x⊥. The exterior angle betweenthe two half-spaces h1, h2 is the angle of the hyperbolic rotation mapping one to theother.

Proposition 4.4.3. (i) If x ∈ H2 and l is a hyperbolic line, then there is a uniquehyperbolic line through x which intersects l orthogonally.

(ii) If l1, l2 are two hyperbolic lines with unit normals n1, n2 such that |〈n1, n2〉| > 1(so the lines do not intersect), then there is a unique line l3 intersecting both l1 and l2orthogonally.

Proof. Exercise.

4.5 Hyperbolic trianglesLet A,B,C ∈ H2 be three points in the hyperbolic plane. Assume that they do not all lieon one hyperbolic line (this is equivalent to assuming A,B,C to be linearly independent).The hyperbolic triangle with vertices A,B,C is the intersection of H2 ⊂ R2,1 with the setof non-negative linear combinations

{λA+ µB + νC | λ, µ, ν ∈ R≥0}.

The side lengths a = d(B,C), b = d(C,A), c = d(A,B) satisfy

− cosh a = 〈B,C〉, − cosh b = 〈C,A〉, − cosh c = 〈A,B〉.

Let A′, B′, C ′ be the spacelike unit vectors such that the half-plane bounded by the linethrough B,C and containing A is

hA′ = {x ∈ H2 | 〈A′, x〉 ≥ 0},


and analogously for B′ and C ′. Then the hyperbolic triangle with vertices A,B,C is alsothe intersection hA′ ∩ hB′ ∩ hC′ . The interior angles α, β, γ at A,B,C satisfy

− cosα = 〈B′, C ′〉, − cos β = 〈C ′, A′〉, − cos γ = 〈A′, B′〉.

Theorem 4.5.1. The side lengths and interior angles of a hyperbolic triangle satisfy

cosα =− cosh a+ cosh b cosh c

sinh b sinh c, (hyperbolic side cosine theorem)

cosh a =cosα + cos β cos γ

sin β sin γ. (hyperbolic angle cosine theorem)

sketch. This can be proved in the same way as we proved the spherical cosine theoremsusing the Gram matrices G,G′ of A,B,C and A′, B′, C ′. Only now the scalar productis

〈x, y〉 = xtEy with E =(

1 0 00 1 00 0 −1

).

So if V = (A B C) andW = (A′ B′ C ′), then

G = V tEV =( −1 − cosh c − cosh b− cosh c −1 − cosh a− cosh b − cosh a −1

), G′ = W tEW =

( 1 − cos γ − cosβ− cos γ 1 − cosα− cosβ − cosα 1

),

andD = W tEV is a diagonal matrix with positive elements on the diagonal. Continueas in the spherical case . . .

Remark 4.5.2. The spherical cosine theorems for a sphere of radiusR (instead of 1) are

cosα =cos a

R− cos b

Rcos c

R

sin bR

sin cR

, cosa

R=

cosα + cos β cos γ

sin β sin γ.

One gets the hyperbolic cosine theorems by setting R = i. That’s why it is sometimessaid that hyperbolic geometry is the geometry on a sphere with imaginary radius.

From the hyperbolic cosine theorems, one can derive

sinh a

sinα=

sinh b

sin β=

sinh c

sin γ(hyperbolic sine theorem)

in the same way in which we derived the spherical sine theorem from the spherical cosinetheorems. One can also derive hyperbolic versions of the half-angle and half-side theorems,and other formulas of spherical trigonometry.

4.5 Hyperbolic triangles 53

Theorem 4.5.3. (i) A hyperbolic triangle with side lengths a, b, c ∈ R>0 exists if andonly if the triangle inequalities are satisfied. (ii) A hyperbolic triangle with anglesα, β, γ ∈ (0, π) exists if and only if

α + β + γ < π.

Remark 4.5.4. The positive number π− (α+β+γ) is in fact the area of the hyperbolictriangle (see Theorem 4.15.1).

sketch. 1. Show that a symmetric (3 × 3)-matrix G is the Gram matrix of 3 linearlyindependent vectors in R2,1 if and only if the bilinear form xtGy has signature (2, 1).

2. Consider G =( −1 − cosh c − cosh b− cosh c −1 − cosh a− cosh b − cosh a −1

). Because

G11 = −1 < 0, det(G11 G12G21 G22

)= − sinh2 c < 0,

the signature of G is (2, 1) if and only if detG < 0 (by the theorem from Lecture 6).3. Show the remarkable identity

detG = −4 sinh(−a+b+c

2

)sinh

(a−b+c

2

)sinh

(a+b−c

2

)sinh

(a+b+c

2

),

and use it to prove part (i) of the theorem.Part (ii) can be shown in the same way by considering G′ =

( 1 − cos γ − cosβ− cos γ 1 − cosα− cosβ − cosα 1

)

and using the identity

detG′ = −4 cos(−a+b+c

2

)cos(a−b+c

2

)cos(a+b−c

2

)cos(a+b+c

2

).


4.6 The Klein model of the hyperbolic planeProject H2 ⊂ R2,1 to the plane x3 = 1 with0 ∈ R2,1 as center of projection. This is cen-tral projection, because 0 is the center of thehyperboloid of which H2 is one sheet. A point( x1x2x3

)∈ H2 is mapped to 1

x3

( x1x2x3

), and all of H2

is mapped to the inside of the unit circle in theimage plane x3 = 1. Hyperbolic lines are mappedto secants of this unit circle. If we forget about thex3-coordinate, we get an image of the hyperbolicplane which is called the Klein model. Thus, theKlein model of the hyperbolic plane is the unitdisk

From W. P. Thurston, Three-dimensional geometry and Topology.

D2 = {u ∈ R2 | u21 + u2

2 < 1},where

( x1x2x3

)∈ H2 corresponds to 1

x3

(x1x2

)∈ D2, and, inversely,

(u1u2

)∈ D2 corresponds to

1√1−u21−u22

( u1u21

)∈ H2. Angles and lengths in are measured in H2.

For two hyperbolic lines l1, l2 with unit normals n1, n2 ∈ R2,1, there are three possibil-ities:

(1) |〈n1, n2〉| < 1.

The lines intersect.

(2) |〈n1, n2〉| = 1.

The lines do not intersectand their images in the Kleinmodel intersect on the unit cir-cle.

(3) |〈n1, n2〉| > 1.

The lines do not intersectand their images in the Kleinmodel intersect outside theunit circle.

In cases 2 and 3 the lines do not intersect, thus they are parallel. However, to distinguishthe two cases, parallel is sometimes used to mean only lines in case 2, and lines in case 3are then called ultra-parallel.

4.7 Angle of parallelismIf l ⊂ H2 is a hyperbolic line and x ∈ H2 is a point noton l, then there exist two parallels (in the narrow sense)to l through x and infinitely many ultra-parallels. Theangle α between one of the parallels and the perpen-dicular through x is called the angle of parallelism. Itdepends only on the distance b from x to l. In fact,α = 2 arctan e−b (exercise).

b α

x

`

4.8 More distance formulas 55

4.8 More distance formulas

Let x ∈ H2 and let l = {y ∈ H2 | 〈y, n〉 = 0} be a hyperbolic line.

Proposition 4.8.1. The point on l that is closest to x is the intersection xp of l with theline lp through x that is perpendicular to l.

Proof. The case x ∈ l is trivial, so assume x 6∈ l. Let γ : R → l ⊂ H2 be aparameterization of l with unit speed. So 〈γ, γ〉 = −1, 〈γ, n〉 = 0, and 〈γ′, γ′〉 = 1.Let f(s) = d(x, γ(s)) = arcosh(−〈x, γ(s)〉). Convince yourself that f(s) → ∞ ass→ ±∞. So f must attain aminimum, say at s = s0. Then f ′(s0) = 0, and this implies〈x, γ′(s0)〉 = 0. Let xp = γ(s0), np = γ′(s0), and let lp = {y ∈ H2 | 〈y, np〉 = 0}.Then• xp is the point on l closest to x,• 〈x, np〉 = 0 so x ∈ lp.• 〈xp, np〉 = 0 because 〈γ, γ′〉 = 0, so xp ∈ lp,• 〈n, np〉 = 0 because 〈γ′, n〉 = 0, so l ⊥ lp.


Proposition 4.8.2. The distance d(x, l) from x to l satisfies |〈x, n〉| = sinh d(x, l).

Proof. Let V = (x xp n np) and let E be the diagonal matrix with 1, 1,−1 on thediagonal. Then

0 = detV tEV = det

( 〈x,x〉〈x,xp〉〈x,n〉〈x,np〉〈xp,x〉〈xp,xp〉〈xp,n〉〈xp,np〉〈n,x〉〈n,xp〉〈n,n〉〈n,np〉〈np,x〉〈np,xp〉〈np,n〉〈np,np〉

)= det

(−1 − cosh d(xp,x) 〈x,n〉 0

− cosh d(xp,x) −1 0 0〈n,x〉 0 1 0

0 0 0 1

)

= 1 + 〈x, n〉2 − cosh2 d(xp, x) = − sinh2 d(xp, x) + 〈x, n〉2

and d(xp, x) = d(x, l) by the previous proposition.

Of course the sign of 〈x, n〉 depends on whether or not x is contained in the half-plane〈x, n〉 ≥ 0.

Proposition 4.8.3. The distance d(l1, l2) between two lines li = {y ∈ H2 | 〈y, ni〉 = 0}with |〈n1, n2〉| > 1 satisfies |〈n1, n2〉| = cosh d(l1, l2).

sketch. This can be proved in the same way as the previous proposition, but lettingV = (x n1 n2 np), where lp = {y ∈ H2 | 〈y, np〉 = 0} is the common perpendicularto l1, l2, and x = l1 ∩ lp.

The sign of 〈n1, n2〉 depends on whether or not one of the half-planes 〈y, ni〉 ≥ 0contains the other.

Summary. Let x1, x2 ∈ H2, let n, n1, n2 be unit spacelike vectors and let l, l1, l2 be thecorresponding lines. The two sides of each line are marked + and − according to the signof the scalar product of points in H2 with the chosen normal on that side.

x1x2

〈x1, x2〉 = − cosh d(x1, x2)

x2

`

+ −

x1

〈x1, n〉 = sinh d(x1, l)〈x2, n〉 = − sinh d(x2, l)

`2`1

− ++ −

〈n1, n2〉 = cosh d(l1, l2)

4.9 Hyperbolic “trilaterals” 57

`2

`1

−+

+−

αα

〈n1, n2〉 = cos α = − cosα

`2`1

− −+ +

`2`1

+ +− −

〈n1, n2〉 = − cosh d(l1, l2)

4.9 Hyperbolic “trilaterals”

Let us define a hyperbolic trilateral as a non-empty intersection of three half-planes, ofwhich none is contained in another. If we only consider the generic cases where pairs ofboundary lines intersect or are ultra-parallel then there are four types of trilaterals accordingto the number of vertices. In the figures, the common perpendiculars of ultra-parallel linesare drawn in red.

(1) (2) (3)

(4)

b

a cb′

c′a′

The first case is the case of triangles. In the other cases, one can also derive trigonometricformulas in the same way as we did for triangles. Of particular interest are trilaterals oftype (4), which correspond to right-angled hexagons. For those one obtains the cosine andsine theorems

cosh a′ =cosh a+ cosh b cosh c

sinh b sinh cand

sinh a′

sinh a=

sinh b′

sinh b=

sinh c′

sinh c.


4.10 Intersections of H2 ⊂ R2,1 with planesAny plane in R2,1 is determined by an equation of the form 〈x, v〉 = b withv ∈ R2,1 \ {0} and b ∈ R. There are three types of planes according towhether v is spacelike, timelike or lightlike.• If 〈v, v〉 < 0, we may assume that 〈v, v〉 = −1 and v3 > 0, so v ∈ H2.A plane of this type which intersects H2 is of the form 〈v, x〉 = − cosh r.The intersection is therefore a circle with center v and radius r.• If 〈v, v〉 > 0, we may assume that 〈v, v〉 = 1 and b > 0. A plane ofthis type always intersects H2 and is of the form 〈v, x〉 = sinh r. Theintersection with H2 is therefore a curve of constant distance from the line〈v, x〉 = 0. (For r = 0, it is the line itself.)• If 〈v, v〉 = 0, a non-empty intersection with H2 is called a horocircle.

4.11 The Poincaré disk model of the hyperbolic planeProject H2 to the plane x3 = 0 with e3 =

(00−1

)as center

of projection. This is stereographic projection of H2. ItmapsH2 to the unit disk of the plane x3 = 0. Analytically,it is the map σH2 : H2 → D2,

σH2

x1

x2

x3

=

1

x3 + 1

(x1

x2

),

σH2−1

(u1

u2

)=

1

1− u12 − u2

2

2u1

2u2

1 + u12 + u2

2

.

Theorem 4.11.1. Stereographic projection ofσH2 : H2 → D2 maps intersections ofH2 ⊂ R2,1

with planes in R2,1 to intersections of D2 withcircles and lines.In particular, hyperbolic lines are mapped to in-tersections ofD2 with circles and lines that inter-sect the unit circle ∂D2 orthogonally.

This can be shown by a calculation like in the case ofS2.

7 lines, 2 circles, 1 horocircle(left). Curves of constant dis-tance form a line (right).

Theorem 4.11.2. Consider two curves γ, η : (−ε, ε) → H2 in the hyperbolic plane

4.11 The Poincaré disk model of the hyperbolic plane 59

with γ(0) = η(0) = p. Let γ = σH2 ◦ γ and η = σH2 ◦ η be their images in D2 understereographic projection, and let v = γ′(0), w = η′(0), v = γ′(0), w = η′(0). Then

〈v, w〉R2,1 =4

(1− p12 − p2

2)2〈v, w〉R2 ,

where 〈·, ·〉R2,1 is the Lorentz scalar product ofR2,1 and 〈·, ·〉R2 is the standard Euclideanscalar product of R2.

This, too, can be shown by a calculation like in the case of S2.Hence σH2 is conformal in the sense that curves in H2 intersectingat some angle are mapped to curves inside the unit circle of theEuclidean plane intersecting at the same angle.One can measure hyperbolic lengths and angles directly in D2 byusing the variable scalar product

gp(v, w) =4

(1− p12 − p2

2)2〈v, w〉R2 . (∗)

For example, the hyperbolic length of a curve in H2 ⊂ R2,1 whichσH2 maps to γ : [t1, t2]→ D2 is

∫ t2

t1

√gγ(t)(γ′(t), γ′(t)) dt.

In the image of H2 under stereographic projection, lengths appear scaled down by thevariable factor 1

2(1− p1

2− p22). The image inD2 of an object in the hyperbolic plane gets

smaller and smaller as it moves towards the boundary circle ∂D2.A Riemannian metric on an open set U ⊆ Rn is a variable Euclidean scalar product.

More precisely, it is a C∞ map g : U × Rn × Rn → R, (p, v, w) 7→ gp(v, w), such that foreach p ∈ U , gp(·, ·) is a Euclidean scalar product on Rn. (Thus, gp(v, w) = vtG(p)w witha matrix G depending on p ∈ U .) One can then measure lengths of curves and angles in Uusing the Riemannian metric, and this is called Riemannian geometry.

A Riemannian metric g on U ⊆ Rn is called conformal if

gp(u, v) = λ(p)〈u, v〉Rn

for some function λ : U → R>0. If this is the case, angles measured using g are equal tothe Euclidean angles measured using 〈·, ·〉Rn .

Thus, gp as defined by equation (∗) is a conformal Riemannian metric on D2. The unitdisk D2 with this Riemannian metric is called the Poincaré disk model of the hyperbolicplane.


4.12 From Klein model to Poincaré disk model via thehemisphere model

We have encountered two ways to map H2 ∈ R2,1 to the unit disk. Central projectiongives the Klein model and stereographic projection gives the Poincaré disk model. Thecomposition

D2 central−−−−−→projection

H2 stereographic−−−−−−→projection

D2

is a peculiar self-map of the unit disk D2, which maps secants of D2 to circles orthogonalto the boundary ∂D2.

Proposition 4.12.1. The same map D2 → D2 is alsothe result of the following construction: First, projectD2 ⊂ R2 ⊂ R3 orthogonally down to the lower hemi-sphere of S2 ⊂ R3. Then project stereographicallyback to D2.

From Prasolov & Tikhomirov,Geometry.

Proof. Let x =(x1x2x3

)∈ H2. Via central projection, this corresponds to 1

x3

(x1x2

)in the

Klein model. Orthogonally projecting down to the lower hemisphere of S2 gives

x1/x3

x2/x3

−√

1− (x1x3

)2 − (x2x3

)2

=

1

x3

x1

x2

−√x3

2 − x12 − x2

2

=

1

x3

x1

x2

−1

Projecting this point back to the unit disk (by stereographic projection of S2) results in

σ(

1x3

(x1x2−1

))=

1

1 + 1x3

( x1x3x2x3

)=

1

x3 + 1

(x1

x2

),

which is the same as σH2(x).

The calculation in the proof shows that(x1x2x3

)7→ 1

x3

(x1x2−1

)maps H2 to the lower

hemisphere of S2. Equally(x1x2x3

)7→ 1

x3

(x1x21

)maps H2 to the upper hemisphere of S2.

These images of H2 are called the (lower and upper) hemisphere models. The hemispheremodels are also conformal, and circles, horospheres, lines, and curves of constant distancefrom lines are represented by (parts of) circles in S2.

4.13 The Poincaré half-plane model 61

4.13 The Poincaré half-plane modelOne obtains the Poincaré half-plane model of H2 by pro-jecting the upper hemisphere model stereographically froma point on the equator, say from e1 =

(100

)to the plane

x1 = 0. This maps the equator (minus e1) to the x2-axisand the upper hemisphere to the upper half-plane of the thex2, x3-plane which we identify with the upper half-plane

H2+ =

{(u1u2

) ∣∣u2 > 0}⊂ R2. From W.P. Thurston, Three-Dimensional Geometry

and Topology.

Since stereographic projection is conformal and maps circles to circles and lines, hy-perbolic lines are represented in H2

+ by half circles meeting the u1-axis orthogonally andvertical lines. You can show by a direct calculation that hyperbolic lengths and angles canbe measured in H2

+ by using the Riemannian metric

gp(v, w) =1

u22〈v, w〉R2 .

The half-plane H2+ with this Riemannian metric is called the Poincaré half-plane model

of the hyperbolic plane. In H2+, hyperbolic lengths appear scaled by the variable factor u2

which is the Euclidean distance to the boundary.

4.14 Two examples for length calculations in the half-planemodel (and some remarks)

(1) Consider the curve γ : [0, α]→ H2+, γ(t) = r

(sin tcos t

). Its hyperbolic length is

∫ α

0

√gγ(t)(γ′(t), γ′(t)) dt =

∫ α

0

√1

r2 cos2 t

⟨r

(cos t− sin t

), r

(cos t− sin t

)⟩dt

=

∫ α

0

1

cos tdt =

1

2log

1 + sinα

1− sinα= log tan

(α2

+π

4

). (Check it out!)


(2) Consider the curve η : [t1, t2]→ H2+, η(t) =

(0t

). Its hyperbolic length is

∫ t2

t1

√gη(t)(η′(t), η′(t)) dt =

∫ t2

t1

√1

t2

⟨(01

),

(01

)⟩dt =

∫ t2

t1

1

tdt

= log t2 − log t1 = logt2t1.

Note that in the first example, the length does not depend on r, and in the secondexample, the length depends only on the quotient t2/t1. In fact, scaling transformations(u1u2

)7→ λ

(u1u2

)(with λ > 0) of the upper half-plane represent isometries of the hyperbolic

plane. For example, scaling by the factor 2 makes all objects in the upper half-plane looktwice as large. At the same time, all distances from the boundary also double. In effect,hyperbolic lengths stay the same.

Horizontal translations(u1u2

)7→(u1+cu2

)of the upper half-plane also represent isometries

of H2, and so do reflections on vertical lines.

4.15 Calculating hyperbolic areas in the half-plane modelIf Euclidean lengths in the upper half-plane model have to be scaled by a variable factor of1x2

to get the hyperbolic length, then Euclidean area has to be scaled by the factor ( 1x2

)2. Sothe hyperbolic area of a region R ⊂ H2

+ is

area(R) =

∫

R

1

x22dx1 dx2 .

Theorem 4.15.1. The area of a hyperbolic triangle with interior angles α, β, γ is

π − α− β − γ.

Remark 4.15.2. In Lecture 8, I sketched a purely analytic prooffor the fact that the angle sum in a hyperbolic triangle is alwaysless than π. A more visual argument can be made using thePoincaré disk model and moving one vertex of the triangle to thecenter of the disk.

4.15 Calculating hyperbolic areas in the half-plane model 63

αβ

γ

A

X1

X2

To prove the theorem, consider first a hyperbolic trian-gle T (α, β, 0) with one vertex “at infinity”. (Strictlyspeaking, this is not a triangle but what we called a tri-lateral.) The figure on the right shows such a trianglein the Klein model, in the Poincaré disk model, and inthe Poincaré half-plane (where the infinite vertex wasused to project from the hemisphere model).

area(T (α, β, 0)) =

∫

T (α,β,0)

1

u22du1 du2

=

∫ cosβ

u1=cos(π−α)

(∫ ∞

u2=√

1−u12

1

u22du2

)du1

=

∫ cosβ

u1=cos(π−α)

([− 1

u2

]∞√

1−u12

)du1

=

∫ cosβ

u1=cos(π−α)

1√1− u1

2du1

=[− arccosu1

]cosβ

cos(π−α)= π − α− β.

For β →∞ one obtains the area of a triangle with two vertices “at infinity”:

area(T (α, 0, 0)) = π − α.

Nowwe can calculate the area for a triangleT (α, β, γ)with anglesα, β, γ:

area(T (α, β, γ)) = area(T (α, 0, 0))− area(T (β1, γ1, 0))− area(T (β2, 0, 0)

= (π − α)− (π − β1 − γ1)− (π − β2)

= π − α− (π − β1 − β2)− (π − γ1) = π − α− β − γ.

This proves the theorem.


4.16 Concluding remarks(1) All the models we have discussed exist also for higher dimensional hyperbolic space.

(2) We have defined hyperbolic space as one sheet of a hyperboloid and then derived theother models from it. Actually, any metric space isometric to our Hn is called hyperbolicspace and Hn. The hyperboloid is just a model like the others, called the hyperboloidmodel.

ProblemsProblem 4.1. LetG be the orthogonal groupO(1, 1), acting on the spaceR(1,1) with scalarproduct 〈x, y〉 = x1y1 − x2y2 for the vectors x = (x1, x2) and y = (y1, y2).

• Show that if 〈v, v〉 = 1 then v = (±cosh(t), sinh(t)) for some t ∈ R, and if 〈v, v〉 = −1then v = (sinh(t),±cosh(t)) for some t ∈ R.

• Show that every matrix in O(1, 1) is of the form(

cosh(t) sinh(t)sinh(t) cosh(t)

),

(− cosh(t) sinh(t)sinh(t) − cosh(t)

),(

cosh(t) − sinh(t)sinh(t) − cosh(t)

), or

(− cosh(t) sinh(t)− sinh(t) cosh(t)

).

• Show that if R(t) :=

(cosh(t) sinh(t)sinh(t) cosh(t)

), then R(s)R(t) = R(s + t). (The operation

on the left hand side is matrix multiplication).

Problem 4.2. Let g ⊂ R2,1 be a space-like line generated by n ∈ R2,1 with 〈n, n〉 = 1 andU = g⊥ the orthogonal complement of g with respect to the Lorentz product. Further, letL = {x ∈ R2,1 | 〈x, x〉 = 0} be the light-cone.1. Show that U ∩ L consists of two lines l1 and l2.

2. Show that the span of l1 and g (resp. l2 and g) is a tangent plane to L. Make a sketch inthe plane x3 = 1.

3. Draw an arbitrary hyperbolic line l in the Klein model of the hyperbolic plane. Can youconstruct the corresponding point n?

Problem 4.3. Consider the hyperboloid model H2 = {x ∈ R2,1 | 〈x, x〉2,1 = −1, x3 > 0}of the hyperbolic plane. Define the circle of radius r around a point p ∈ H2 as

Cp,r := {x ⊂ H2 | d(p, x) = r},where d denotes the hyperbolic distance. Show that Cp,r is the intersection of H2 with aplane and draw a sketch. What is the equation of the plane?

What is the length of a hyperbolic circle with radius r? Hint: Apply a suitablenormalization to the picture before calculating the length.

4.16 Concluding remarks 65

Problem 4.4. Consider the Lorentz vector space Rn,1 (n ≥ 2), i.e. the real vector spaceRn+1 with scalar product 〈x, y〉 =

∑ni=1 xiyi − xn+1yn+1. The corresponding quadric Q

is given by the zero-set of the quadratic form q(x) := 〈x, x〉. In the lecture we defined theprojective model of hyperbolic space Hn

pr = {[x] ∈ RPn | q(x) < 0}. A point P ∈ RPn

is inside the quadric, if every line through P intersects the quadric in exactly two differentpoints. (The points of the quadric itself are not inside, since the tangents intersect thequadric in only one point.)

Show: P = [x] ∈ RPn is insideQ if and only if q(x) < 0. In other words the projectivemodel of hyperbolic space corresponds to the inside of the quadric.

Problem 4.5. Let p, q ∈ Hn.

1. Show that 〈p, q〉 ≤ −1 with equality if and only if p = q.

2. The line connecting p and q isHn ∩ span(p, q). Show that the length of the line segmentconnecting p and q equals arcosh(−〈p, q〉) = d(p, q).

Problem 4.6. Let O = (0, 0) and P = (x, 0) with x > 0 be points in the Klein disk modelof the hyperbolic plane, and assume d(O,P ) = t. Show that x = tanh(t). Show that thecorresponding formula in the Poincare model is x = tanh( t

2).

Problem 4.7. DefineM :=

−1 −2 2−2 −1 2−2 −2 3

. Show:

1. M ∈ O(2, 1).

2. A = (1, 1, 1), B = (1, 0, 1), C = (0, 1, 1) are eigenvectors of M corresponding toeigenvalues (−1, 1, 1), resp.

3. 〈A,B〉 = 〈A,C〉 = 0.

4. M fixes the hyperbolic line l := A⊥ pointwise.

Finally, describe in words and make a sketch of the action ofM on the Klein model of thehyperbolic plane.

Problem 4.8. A line l in H2 can be described as the intersection of H2 with the orthogonalcomplement of a space-like unit vector n, i.e., l = H2∩{x ∈ R2,1 | 〈x, n〉 = 0} for n ∈ R2,1

with 〈n, n〉 = 1. Two lines determined by normals n1 and n2 are orthogonal if 〈n1, n2〉 = 0.

1. Given a point x and a line l in H2 not containing x. Show that there exists a unique lineg in H2 containing x and orthogonal to l.

2. Let l1 and l2 be hyperbolic lines with normals n1 and n2. Show that if |〈n1, n2〉| > 1,then there exists a unique hyperbolic line l3 such that l1 ⊥ l3 and l2 ⊥ l3.


Problem 4.9. 1. Let b(·, ·) be a symmetric bilinear form on Rn. A map f : Rn → Rn iscalled orthogonal with respect to b if b(f(x), f(y)) = b(x, y)∀x, y ∈ Rn.Let M be the matrix representation of a linear map on Rn with respect to a basis B.When doesM describe an orthogonal transformation with respect to b? How does thisrelate to the description of the orthogonal group O(n) = {M ∈ Rn×n |M tM = I}?What is the analogous matrix description of the orthogonal group O(n, 1) with respectto the Lorentz product?Now, consider the Klein model H2

Klein of the hyperbolic plane and show the following.

2. Let l ⊂ H2Klein be a line and P ∈ l. For any line g ⊂ H2

Klein and Q ∈ g there exists ahyperbolic motion f such that f(l) = g and f(P ) = Q.

3. Let l ⊂ H2Klein be a line and P1, P2 ∈ l. For any line g ⊂ H2

Klein and Q1, Q2 ∈ gthere exists a hyperbolic motion f such that f(l) = g and f(Pi) = Qi if and only ifdpr(P1, P2) = dpr(Q1, Q2).

Problem 4.10. Let ABC be a right-angled hyperbolic triangle with right-angle at vertexC, with side-lengths (a, b, c) and interior angles (α, β, π

2).

1. Show that tanh(a) = sinh(b) tan(α).

2. Show that lima→∞ α = tan−1( 1sinh(b)

) = 2 tan−1(e−b)

Problem 4.11. Let l, l1, l2 ⊂ H2 be three lines in the hyperbolic plane with normalsn, n1,n2 satisfying |〈n1, n2〉| > 1. Further let x ∈ H2 be a point that does not lie on l and〈n, n〉 = 〈n1, n1〉 = 〈n2, n2〉 = 1 and 〈x, x〉 = 〈x1, x1〉 = 〈x2, x2〉 = −1. Show:a) The distance d(x, l) of x to l satisfies |〈n, x〉| = sinh d(x, l).b) The distance d(l1, l2) between the lines l1 and l2 satisfies |〈n1, n2〉| = cosh d(l1, l2). It

is equal to d(x1, x2) where x1 = l1 ∩ l3 and x2 = l2 ∩ l3 and l3 is the unique hyperbolicline orthogonal to l1 and l2.

Problem 4.12. Show that the distance between two points A and B in the Poincaré discmodel is given by the following formula:

d(A,B) = log cr(B,X,A, Y ),

whereX and Y are the points of intersection of the unique geodesic through A and B withthe unit circle.

Problem 4.13. Show that for each triple (α, β, γ) with α + β + γ < π there exists ahyperbolic triangle with these angles.

Problem4.14. Forwhichn ∈ N do there exist right-angled regularn-gons in the hyperbolicplane? What is the corresponding sidelength depending on n? (You may assume that foreach n all the polygons are congruent)

4.16 Concluding remarks 67

Problem 4.15. Show that there are circles in the hyperbolic plane which cannot be inscribedinto a triangle. What is the maximal radius of an incircle?

Problem 4.16. How many disjoint hyperbolic halfplanes are there in a hyperbolic plane?

Problem 4.17. An ideal triangle in the hyperbolic plane is a triangle with all three verticeson the boundary of the hyperbolic plane. Show that all ideal triangles are congruent.

Problem 4.18. For a line l in the hyperbolic plane and δ > 0, denote

E ={p ∈ H2 | d(p, l) = δ

}

the set of equi-distant points to l.

1. Show that in the hyperboloid model H2, the set E is obtained as intersection of H2 withtwo affine planes.

2. Draw a sketch of l andE in the Klein model and prove thatE and l meet at the boundaryof the Klein disc, i.e. at common points at infinity.

Problem 4.19. Let P = (x, y) be a point in the Poincare model of the hyperbolic planeand p be the corresponding point in the Klein disk model. Show that p = 2

1+x2+y2P . What

is the corresponding formula for P in terms of p?

Problem 4.20. Consider the Poincaré disc model in the complex plane. Show that

f(z) = eiϕz − z0

1− zz0

,

with z0 ∈ C with |z0| < 1 and ϕ ∈ [0, 2π) is a hyperbolic motion.Optionally (4 Extra pts): Show that all orientation preserving hyperbolic motions of

the Poincaré disc are of this form. (Hint: You already know the isometries in the halfplanemodel.)

Problem 4.21. In the Poincaré half-plane modelH , a hyperbolic line p is represented by anarc of the circleSp orthogonal to the boundary ∂H . Show that the (hyperbolic) reflection in p

is given by the inversion in the circleSp.

H

x

x′

Spp

〈p, p〉 = 1, x 7→ x′ = x− 2〈p, x〉p

geometry i - tu berlingeometry , as opposed to the old way beginning with geometric axioms, which...

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