Part IB - Geometry - SRCF · 2020-01-24 · Part IB | Geometry Based on lectures by A. G. Kovalev...

Part IB — Geometry

Based on lectures by A. G. KovalevNotes taken by Dexter Chua

Lent 2016

These notes are not endorsed by the lecturers, and I have modified them (oftensignificantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Parts of Analysis II will be found useful for this course.

Groups of rigid motions of Euclidean space. Rotation and reflection groups in two andthree dimensions. Lengths of curves. [2]

Spherical geometry: spherical lines, spherical triangles and the Gauss-Bonnet theorem.Stereographic projection and Mobius transformations. [3]

Triangulations of the sphere and the torus, Euler number. [1]

Riemannian metrics on open subsets of the plane. The hyperbolic plane. Poincaremodels and their metrics. The isometry group. Hyperbolic triangles and the Gauss-Bonnet theorem. The hyperboloid model. [4]

Embedded surfaces in R3. The first fundamental form. Length and area. Examples. [1]

Length and energy. Geodesics for general Riemannian metrics as stationary points ofthe energy. First variation of the energy and geodesics as solutions of the correspondingEuler-Lagrange equations. Geodesic polar coordinates (informal proof of existence).Surfaces of revolution. [2]

The second fundamental form and Gaussian curvature. For metrics of the form

du2 + G(u, v)dv2, expression of the curvature as√Guu/

√G. Abstract smooth surfaces

and isometries. Euler numbers and statement of Gauss-Bonnet theorem, examples and

applications. [3]

1

Contents IB Geometry

Contents

0 Introduction 3

1 Euclidean geometry 41.1 Isometries of the Euclidean plane . . . . . . . . . . . . . . . . . . 41.2 Curves in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Spherical geometry 122.1 Triangles on a sphere . . . . . . . . . . . . . . . . . . . . . . . . . 122.2 Mobius geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Triangulations and the Euler number 23

4 Hyperbolic geometry 274.1 Review of derivatives and chain rule . . . . . . . . . . . . . . . . 274.2 Riemannian metrics . . . . . . . . . . . . . . . . . . . . . . . . . 284.3 Two models for the hyperbolic plane . . . . . . . . . . . . . . . . 324.4 Geometry of the hyperbolic disk . . . . . . . . . . . . . . . . . . 364.5 Hyperbolic triangles . . . . . . . . . . . . . . . . . . . . . . . . . 394.6 Hyperboloid model . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5 Smooth embedded surfaces (in R3) 445.1 Smooth embedded surfaces . . . . . . . . . . . . . . . . . . . . . 445.2 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485.3 Surfaces of revolution . . . . . . . . . . . . . . . . . . . . . . . . 525.4 Gaussian curvature . . . . . . . . . . . . . . . . . . . . . . . . . . 54

6 Abstract smooth surfaces 60

2

0 Introduction IB Geometry

0 Introduction

In the very beginning, Euclid came up with the axioms of geometry, one ofwhich is the parallel postulate. This says that given any point P and a line `not containing P , there is a line through P that does not intersect `. Unlikethe other axioms Euclid had, this was not seen as “obvious”. For many years,geometers tried hard to prove this axiom from the others, but failed.

Eventually, people realized that this axiom cannot be proved from the others.There exists some other “geometries” in which the other axioms hold, but theparallel postulate fails. This was known as hyperbolic geometry. Together withEuclidean geometry and spherical geometry (which is the geometry of the surfaceof a sphere), these constitute the three classical geometries. We will study thesegeometries in detail, and see that they actually obey many similar properties,while being strikingly different in other respects.

That is not the end. There is no reason why we have to restrict ourselvesto these three types of geometry. In later parts of the course, we will massivelygeneralize the notions we began with and eventually define an abstract smoothsurface. This covers all three classical geometries, and many more!

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1 Euclidean geometry IB Geometry

1 Euclidean geometry

We are first going to look at Euclidean geometry. Roughly speaking, this is thegeometry of the familiar Rn under the usual inner product. There really isn’tmuch to say, since we are already quite familiar with Euclidean geometry. Wewill quickly look at isometries of Rn and curves in Rn. Afterwards, we will tryto develop analogous notions in other more complicated geometries.

1.1 Isometries of the Euclidean plane

The purpose of this section is to study maps on Rn that preserve distances, i.e.isometries of Rn. Before we begin, we define the notion of distance on Rn in theusual way.

Definition ((Standard) inner product). The (standard) inner product on Rn isdefined by

(x,y) = x · y =

n∑i=1

xiyi.

Definition (Euclidean Norm). The Euclidean norm of x ∈ Rn is

‖x‖ =√

(x,x).

This defines a metric on Rn by

d(x,y) = ‖x− y‖.

Note that the inner product and the norm both depend on our choice oforigin, but the distance does not. In general, we don’t like having a choice oforigin — choosing the origin is just to provide a (very) convenient way labellingpoints. The origin should not be a special point (in theory). In fancy language,we say we view Rn as an affine space instead of a vector space.

Definition (Isometry). A map f : Rn → Rn is an isometry of Rn if

d(f(x), f(y)) = d(x,y)

for all x,y ∈ Rn.

Note that f is not required to be linear. This is since we are viewing Rnas an affine space, and linearity only makes sense if we have a specified pointas the origin. Nevertheless, we will still view the linear isometries as “special”isometries, since they are more convenient to work with, despite not being specialfundamentally.

Our current objective is to classify all isometries of Rn. We start with thelinear isometries. Recall the following definition:

Definition (Orthogonal matrix). An n× n matrix A is orthogonal if AAT =ATA = I. The group of all orthogonal matrices is the orthogonal group O(n).

In general, for any matrix A and x,y ∈ Rn, we get

(Ax, Ay) = (Ax)T (Ay) = xTATAy = (x, ATAy).

4


So A is orthogonal if and only if (Ax, Ay) = (x,y) for all x,y ∈ Rn.Recall also that the inner product can be expressed in terms of the norm by

(x,y) =1

2(‖x + y‖2 − ‖x‖2 − ‖y‖2).

So if A preserves norm, then it preserves the inner product, and the converseis obviously true. So A is orthogonal if and only if ‖Ax‖ = ‖x‖ for all x ∈ Rn.Hence matrices are orthogonal if and only if they are isometries.

More generally, letf(x) = Ax + b.

Thend(f(x), f(y)) = ‖A(x− y)‖.

So any f of this form is an isometry if and only if A is orthogonal. This is nottoo surprising. What might not be expected is that all isometries are of thisform.

Theorem. Every isometry of f : Rn → Rn is of the form

f(x) = Ax + b.

for A orthogonal and b ∈ Rn.

Proof. Let f be an isometry. Let e1, · · · , en be the standard basis of Rn. Let

b = f(0), ai = f(ei)− b.

The idea is to construct our matrix A out of these ai. For A to be orthogonal,{ai} must be an orthonormal basis.

Indeed, we can compute

‖ai‖ = ‖f(ei)− f(0)‖ = d(f(ei), f(0)) = d(ei,0) = ‖ei‖ = 1.

For i 6= j, we have

(ai,aj) = −(ai,−aj)

= −1

2(‖ai − aj‖2 − ‖ai‖2 − ‖aj‖2)

= −1

2(‖f(ei)− f(ej)‖2 − 2)

= −1

2(‖ei − ej‖2 − 2)

= 0

So ai and aj are orthogonal. In other words, {ai} forms an orthonormal set. Itis an easy result that any orthogonal set must be linearly independent. Since wehave found n orthonormal vectors, they form an orthonormal basis.

Hence, the matrix A with columns given by the column vectors ai is anorthogonal matrix. We define a new isometry

g(x) = Ax + b.

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We want to show f = g. By construction, we know g(x) = f(x) is true forx = 0, e1, · · · , en.

We observe that g is invertible. In particular,

g−1(x) = A−1(x− b) = ATx−ATb.

Moreover, it is an isometry, since AT is orthogonal (or we can appeal to themore general fact that inverses of isometries are isometries).

We defineh = g−1 ◦ f.

Since it is a composition of isometries, it is also an isometry. Moreover, it fixesx = 0, e1, · · · , en.

It currently suffices to prove that h is the identity.Let x ∈ Rn, and expand it in the basis as

x =

n∑i=1

xiei.

Let

y = h(x) =

n∑i=1

yiei.

We can compute

d(x, ei)2 = (x− ei,x− ei) = ‖x‖2 + 1− 2xi

d(x,0)2 = ‖x‖2.

Similarly, we have

d(y, ei)2 = (y − ei,y − ei) = ‖y‖2 + 1− 2yi

d(y,0)2 = ‖y‖2.

Since h is an isometry and fixes 0, e1, · · · , en, and by definition h(x) = y, wemust have

d(x,0) = d(y,0), d(x, ei) = d(y, ei).

The first equality gives ‖x‖2 = ‖y‖2, and the others then imply xi = yi for all i.In other words, x = y = h(x). So h is the identity.

We now collect all our isometries into a group, for convenience.

Definition (Isometry group). The isometry group Isom(Rn) is the group of allisometries of Rn, which is a group by composition.

Example (Reflections in an affine hyperplane). Let H ⊆ Rn be an affinehyperplane given by

H = {x ∈ Rn : u · x = c},

where ‖u‖ = 1 and c ∈ R. This is just a natural generalization of a 2-dimensionalplane in R3. Note that unlike a vector subspace, it does not have to contain theorigin (since the origin is not a special point).

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Reflection in H, written RH , is the map

RH : Rn → Rn

x 7→ x− 2(x · u− c)u

It is an exercise in the example sheet to show that this is indeed an isometry.We now check this is indeed what we think a reflection should be. Note that

every point in Rn can be written as a + tu, where a ∈ H. Then the reflectionshould send this point to a− tu.

0

a

a− tu

H

a + tu

This is a routine check:

RH(a + tu) = (a + tu)− 2tu = a− tu.

In particular, we know RH fixes exactly the points of H.The converse is also true — any isometry S ∈ Isom(Rn) that fixes the points

in some affine hyperplane H is either the identity or RH .To show this, we first want to translate the plane such that it becomes a

vector subspace. Then we can use our linear algebra magic. For any a ∈ Rn, wecan define the translation by a as

Ta(x) = x + a.

This is clearly an isometry.We pick an arbitrary a ∈ H, and let R = T−aSTa ∈ Isom(Rn). Then R fixes

exactly H ′ = T−aH. Since 0 ∈ H ′, H ′ is a vector subspace. In particular, ifH = {x : x · u = c}, then by putting c = a · u, we find

H ′ = {x : x · u = 0}.

To understand R, we already know it fixes everything in H ′. So we want to seewhat it does to u. Note that since R is an isometry and fixes the origin, it is infact an orthogonal map. Hence for any x ∈ H ′, we get

(Ru,x) = (Ru, Rx) = (u,x) = 0.

So Ru is also perpendicular to H ′. Hence Ru = λu for some λ. Since R is anisometry, we have ‖Ru‖2 = 1. Hence |λ|2 = 1, and thus λ = ±1. So either

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λ = 1, and R = id; or λ = −1, and R = RH′ , as we already know for orthogonalmatrices.

It thus follow that S = idRn , or S is the reflection in H.Thus we find that each reflection RH is the (unique) isometry fixing H but

not idRn .

It is an exercise in the example sheet to show that every isometry of Rnis a composition of at most n + 1 reflections. If the isometry fixes 0, then nreflections will suffice.

Consider the subgroup of Isom(Rn) that fixes 0. By our general expressionfor the general isometry, we know this is the set {f(x) = Ax : AAT = I} ∼= O(n),the orthogonal group.

For each A ∈ O(n), we must have det(A)2 = 1. So detA = ±1. We use thisto define a further subgroup, the special orthogonal group.

Definition (Special orthogonal group). The special orthogonal group is thegroup

SO(n) = {A ∈ O(n) : detA = 1}.

We can look at these explicitly for low dimensions.

Example. Consider

A =

(a bc d

)∈ O(2)

Orthogonality then requires

a2 + c2 = b2 + d2 = 1, ab+ cd = 0.

Now we pick 0 ≤ θ, ϕ ≤ 2π such that

a = cos θ b = − sinϕ

c = sin θ d = cosϕ.

Then ab+ cd = 0 gives tan θ = tanϕ (if cos θ and cosϕ are zero, we formally saythese are both infinity). So either θ = ϕ or θ = ϕ± π. Thus we have

A =

(cos θ − sin θsin θ cos θ

)or A =

(cos θ sin θsin θ − cos θ

)respectively. In the first case, this is a rotation through θ about the origin. Thisis determinant 1, and hence A ∈ SO(2).

In the second case, this is a reflection in the line ` at angle θ2 to the x-axis.

Then detA = −1 and A 6∈ SO(2).So in two dimensions, the orthogonal matrices are either reflections or rota-

tions — those in SO(2) are rotations, and the others are reflections.

Before we can move on to three dimensions, we need to have the notion oforientation. We might intuitively know what an orientation is, but it is ratherdifficult to define orientation formally. The best we can do is to tell whethertwo given bases of a vector space have “the same orientation”. Thus, it wouldmake sense to define an orientation as an equivalence class of bases of “the sameorientation”. We formally define it as follows:

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Definition (Orientation). An orientation of a vector space is an equivalenceclass of bases — let v1, · · · ,vn and v′1, · · · ,v′n be two bases and A be the changeof basis matrix. We say the two bases are equivalent iff detA > 0. This is anequivalence relation on the bases, and the equivalence classes are the orientations.

Definition (Orientation-preserving isometry). An isometry f(x) = Ax + bis orientation-preserving if detA = 1. Otherwise, if detA = −1, we say it isorientation-reversing.

Example. We now want to look at O(3). First focus on the case where A ∈SO(3), i.e. detA = 1. Then we can compute

det(A− I) = det(AT − I) = det(A) det(AT − I) = det(I −A) = − det(A− I).

So det(A− I) = 0, i.e. +1 is an eigenvalue in R. So there is some v1 ∈ R3 suchthat Av1 = v1.

We set W = 〈v1〉⊥. Let w ∈W . Then we can compute

(Aw,v1) = (Aw, Av1) = (w,v1) = 0.

So Aw ∈W . In other words, W is fixed by A, and A|W : W →W is well-defined.Moreover, it is still orthogonal and has determinant 1. So it is a rotation of thetwo-dimensional vector space W .

We choose {v2,v3} an orthonormal basis of W . Then under the bases{v1,v2,v3}, A is represented by

A =

1 0 00 cos θ − sin θ0 sin θ cos θ

.

This is the most general orientation-preserving isometry of R3 that fixes theorigin.

How about the orientation-reversing ones? Suppose detA = −1. Thendet(−A) = 1. So in some orthonormal basis, we can express A as

−A =

1 0 00 cos θ − sin θ0 sin θ cos θ

.

So A takes the form

A =

−1 0 00 cosϕ − sinϕ0 sinϕ cosϕ

,

where ϕ = θ + π. This is a rotated reflection, i.e. we first do a reflection, thenrotation. In the special case where ϕ = 0, this is a pure reflection.

That’s all we’re going to talk about isometries.

1.2 Curves in Rn

Next we will briefly look at curves in Rn. This will be very brief, since curves inRn aren’t too interesting. The most interesting fact might be that the shortestcurve between two points is a straight line, but we aren’t even proving that,because it is left for the example sheet.

9


Definition (Curve). A curve Γ in Rn is a continuous map Γ : [a, b]→ Rn.

Here we can think of the curve as the trajectory of a particle moving throughtime. Our main objective of this section is to define the length of a curve. Wemight want to define the length as∫ b

a

‖Γ′(t)‖ dt,

as is familiar from, say, IA Vector Calculus. However, we can’t do this, since ourdefinition of a curve does not require Γ to be continuously differentiable. It ismerely required to be continuous. Hence we have to define the length in a moreroundabout way.

Similar to the definition of the Riemann integral, we consider a dissectionD = a = t0 < t1 < · · · < tN = b of [a, b], and set Pi = Γ(ti). We define

SD =∑i

‖−−−−→PiPi+1‖.

P0

P1 P2

PN−1

PN

Notice that if we add more points to the dissection, then SD will necessarilyincrease, by the triangle inequality. So it makes sense to define the length as thefollowing supremum:

Definition (Length of curve). The length of a curve Γ : [a, b]→ Rn is

` = supDSD,

if the supremum exists.

Alternatively, if we let

mesh(D) = maxi

(ti − ti−1),

then if ` exists, then we have

` = limmesh(D)→0

sD.

Note also that by definition, we can write

` = inf{˜ : ˜≥ SD for all D}.

The definition by itself isn’t too helpful, since there is no nice and easy way tocheck if the supremum exists. However, differentiability allows us to computethis easily in the expected way.

Proposition. If Γ is continuously differentiable (i.e. C1), then the length of Γis given by

length(Γ) =

∫ b

a

‖Γ′(t)‖ dt.

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The proof is just a careful check that the definition of the integral coincideswith the definition of length.

Proof. To simplify notation, we assume n = 3. However, the proof works for allpossible dimensions. We write

Γ(t) = (f1(t), f2(t), f3(t)).

For every s 6= t ∈ [a, b], the mean value theorem tells us

fi(t)− fi(s)t− s

= f ′i(ξi)

for some ξi ∈ (s, t), for all i = 1, 2, 3.Now note that f ′i are continuous on a closed, bounded interval, and hence

uniformly continuous. For all ε ∈ 0, there is some δ > 0 such that |t − s| < δimplies

|f ′i(ξi)− f ′(ξ)| <ε

3

for all ξ ∈ (s, t). Thus, for any ξ ∈ (s, t), we have∥∥∥∥Γ(t)− Γ(s)

t− s− Γ′(ξ)

∥∥∥∥ =

∥∥∥∥∥∥f ′1(ξ1)f ′2(ξ2)f ′3(ξ3)

−f ′1(ξ)f ′2(ξ)f ′3(ξ)

∥∥∥∥∥∥ ≤ ε

3+ε

3+ε

3= ε.

In other words,‖Γ(t)− Γ(s)− (t− s)Γ′(ξ)‖ ≤ ε(t− s).

We relabel t = ti, s = ti−1 and ξ = s+t2 .

Using the triangle inequality, we have

(ti − ti−1)

∥∥∥∥Γ′(ti + ti−1

2

)∥∥∥∥− ε(ti − ti−1) < ‖Γ(ti)− Γ(ti−1)‖

< (ti − ti−1)

∥∥∥∥Γ′(ti + ti−1

2

)∥∥∥∥+ ε(ti − ti−1).

Summing over all i, we obtain∑i

(ti − ti−1)

∥∥∥∥Γ′(ti + ti−1

2

)∥∥∥∥− ε(b− a) < SD

<∑i

(ti − ti−1)

∥∥∥∥Γ′(ti + ti−1

2

)∥∥∥∥+ ε(b− a),

which is valid whenever mesh(D) < δ.Since Γ′ is continuous, and hence integrable, we know∑

i

(ti − ti−1)

∥∥∥∥Γ′(ti + ti−1

2

)∥∥∥∥→ ∫ b

a

‖Γ′(t)‖ dt

as mesh(D)→ 0, and

length(Γ) = limmesh(D)→0

SD =

∫ b

a

‖Γ′(t)‖ dt.

This is all we are going to say about Euclidean space.

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2 Spherical geometry IB Geometry

2 Spherical geometry

The next thing we are going to study is the geometry on the surface of a sphere.This is a rather sensible thing to study, since it so happens that we all live onsomething (approximately) spherical. It turns out the geometry of the sphere isvery different from that of R2.

In this section, we will always think of S2 as a subset of R3 so that we canreuse what we know about R3.

Notation. We write S = S2 ⊆ R3 for the unit sphere. We write O = 0 for theorigin, which is the center of the sphere (and not on the sphere).

When we live on the sphere, we can no longer use regular lines in R3, sincethese do not lie fully on the sphere. Instead, we have a concept of a sphericalline, also known as a great circle.

Definition (Great circle). A great circle (in S2) is S2 ∩ (a plane through O).We also call these (spherical) lines.

We will also call these geodesics, which is a much more general term definedon any surface, and happens to be these great circles in S2.

In R3, we know that any three points that are not colinear determine aunique plane through them. Hence given any two non-antipodal points P,Q ∈ S,there exists a unique spherical line through P and Q.

Definition (Distance on a sphere). Given P,Q ∈ S, the distance d(P,Q) is theshorter of the two (spherical) line segments (i.e. arcs) PQ along the respectivegreat circle. When P and Q are antipodal, there are infinitely many line segmentsbetween them of the same length, and the distance is π.

Note that by the definition of the radian, d(P,Q) is the angle between−−→OP

and−−→OQ, which is also cos−1(P ·Q) (where P = OP , Q = OQ).

2.1 Triangles on a sphere

One main object of study is spherical triangles – they are defined just likeEuclidean triangles, with AB,AC,BC line segments on S of length < π. Therestriction of length is just merely for convenience.

A

B C

c b

a

α

β γ

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We will take advantage of the fact that the sphere sits in R3. We set

n1 =C×B

sin a

n2 =A×C

sin b

n3 =B×A

sin c.

These are unit normals to the planes OBC,OAC and OAB respectively. Theyare pointing out of the solid OABC.

The angles α, β, γ are the angles between the planes for the respective sides.Then 0 < α, β, γ < π. Note that the angle between n2 and n3 is π + α (not αitself — if α = 0, then the angle between the two normals is π). So

n2 · n3 = − cosα

n3 · n1 = − cosβ

n1 · n2 = − cos γ.

We have the following theorem:

Theorem (Spherical cosine rule).

sin a sin b cos γ = cos c− cos a cos b.

Proof. We use the fact from IA Vectors and Matrices that

(C×B) · (A×C) = (A ·C)(B ·C)− (C ·C)(B ·A),

which follows easily from the double-epsilon identity

εijkεimn = δjmδkn − δjnδkm.

In our case, since C ·C = 1, the right hand side is

(A ·C)(B ·C)− (B ·A).

Thus we have

− cos γ = n1 · n2

=C×B

sin a· A×C

sin b

=(A ·C)(B ·C)− (B ·A)

sin a sin b

=cos b cos a− cos c

sin a sin b.

Corollary (Pythagoras theorem). If γ = π2 , then

cos c = cos a cos b.

Analogously, we have a spherical sine rule.

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Theorem (Spherical sine rule).

sin a

sinα=

sin b

sinβ=

sin c

sin γ.

Proof. We use the fact that

(A×C)× (C×B) = (C · (B×A))C,

which we again are not bothered to prove again. The left hand side is

−(n1 × n2) sin a sin b

Since the angle between n1 and n2 is π + γ, we know n1 × n2 = C sin γ. Thusthe left hand side is

−C sin a sin b sin γ.

Thus we knowC · (A×B) = sin a sin b sin γ.

However, since the scalar triple product is cyclic, we know

C · (A×B) = A · (B×C).

In other words, we have

sin a sin b sin γ = sin b sin c sinα.

Thus we havesin γ

sin c=

sinα

sin a.

Similarly, we know this is equal to sin βsin b .

Recall that for small a, b, c, we know

sin a = a+O(a3).

Similarly,

cos a = 1− a2

2+O(a4).

As we take the limit a, b, c→ 0, the spherical sine and cosine rules become theusual Euclidean versions. For example, the cosine rule becomes

ab cos γ = 1− c2

2−(

1− a2

2

)(1− b2

2

)+O(‖(a, b, c)‖3).

Rearranging gives

c2 = a2 + b2 − 2ab cos γ +O(‖(a, b, c, )‖3).

The sine rule transforms similarly as well. This is what we would expect, sincemaking a, b, c small is equivalent to zooming into the surface of the sphere, andit looks more and more like flat space.

Note that if γ = π, it then follows that C is in the line segment given by AB.So c = a+ b. Otherwise, we get

cos c > cos a cos b− sin a sin b = cos(a+ b),

since cos γ < 1. Since cos is decreasing on [0, π], we know

c < a+ b.

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Corollary (Triangle inequality). For any P,Q,R ∈ S2, we have

d(P,Q) + d(Q,R) ≥ d(P,R),

with equality if and only if Q lies is in the line segment PR of shortest length.

Proof. The only case left to check is if d(P,R) = π, since we do not allow ourtriangles to have side length π. But in this case they are antipodal points, andany Q lies in a line through PR, and equality holds.

Thus, we find that (S2, d) is a metric space.On Rn, straight lines are curves that minimize distance. Since we are calling

spherical lines lines, we would expect them to minimize distance as well. This isin fact true.

Proposition. Given a curve Γ on S2 ⊆ R3 from P toQ, we have ` = length(Γ) ≥d(P,Q). Moreover, if ` = d(P,Q), then the image of Γ is a spherical line segmentPQ.

Proof. Let Γ : [0, 1]→ S and ` = length(Γ). Then for any dissection D of [0, 1],say 0 = t0 < · · · < tN = 1, write Pi = Γ(ti). We define

SD =∑i

d(Pi−1, Pi) > SD =∑i

|−−−−→Pi−1Pi|,

where the length in the right hand expression is the distance in Euclidean 3-space.Now suppose ` < d(P,Q). Then there is some ε > 0 such that `(1 + ε) <

d(P,Q).Recall from basic trigonometric that if θ > 0, then sin θ < θ. Also,

sin θ

θ→ 1 as θ → 0.

Thus we haveθ ≤ (1 + ε) sin θ.

for small θ. What we really want is the double of this:

2θ ≤ (1 + ε)2 sin θ.

This is useful since these lengths appear in the following diagram:

2 sin θ2θ

2θ

This means for P,Q sufficiently close, we have d(P,Q) ≤ (1 + ε)|−−→PQ|.

From Analysis II, we know Γ is uniformly continuous on [0, 1]. So we canchoose D such that

d(Pi−1, Pi) ≤ (1 + ε)|−−−−→Pi−1Pi|

for all i. So we know that for sufficiently fine D,

SD ≤ (1 + ε)SD < d(P,Q),

15


since SD → `. However, by the triangle inequality SD ≥ d(P,Q). This is acontradiction. Hence we must have ` ≥ d(P,Q).

Suppose now ` = d(P,Q) for some Γ : [0, 1] → S, ` = length(Γ). Then forevery t ∈ [0, 1], we have

d(P,Q) = ` = length Γ|[0,t] + length Γ|[t,1]

≥ d(P,Γ(t)) + d(Γ(t), Q)

≥ d(P,Q).

Hence we must have equality all along the way, i.e.

d(P,Q) = d(P,Γ(t)) + d(Γ(t), Q)

for all Γ(t).However, this is possible only if Γ(t) lies on the shorter spherical line segment

PQ, as we have previously proved. So done.

Note that if Γ is a curve of minimal length from P to Q, then Γ is aspherical line segment. Further, from the proof of this proposition, we knowlength Γ|[0,t] = d(P,Γ(t)) for all t. So the parametrisation of Γ is monotonic.Such a Γ is called a minimizing geodesic.

Finally, we get to an important theorem whose prove involves complicatedpictures. This is known as the Gauss-Bonnet theorem. The Gauss-Bonnettheorem is in fact a much more general theorem. However, here we will specializein the special case of the sphere. Later, when doing hyperbolic geometry, wewill prove the hyperbolic version of the Gauss-Bonnet theorem. Near the end ofthe course, when we have developed sufficient machinery, we would be able tostate the Gauss-Bonnet theorem in its full glory. However, we will not be ableto prove the general version.

Proposition (Gauss-Bonnet theorem for S2). If ∆ is a spherical triangle withangles α, β, γ, then

area(∆) = (α+ β + γ)− π.

Proof. We start with the concept of a double lune. A double lune with angle0 < α < π is two regions S cut out by two planes through a pair of antipodalpoints, where α is the angle between the two planes.

A

A′

16


It is not hard to show that the area of a double lune is 4α, since the area of thesphere is 4π.

Now note that our triangle ∆ = ABC is the intersection of 3 single lunes,with each of A,B,C as the pole (in fact we only need two, but it is moreconvenient to talk about 3).

A

A′

B

B′C

C ′

Therefore ∆ together with its antipodal partner ∆′ is a subset of each of the 3double lunes with areas 4α, 4β, 4γ. Also, the union of all the double lunes coverthe whole sphere, and overlap at exactly ∆ and ∆′. Thus

4(α+ β + γ) = 4π + 2(area(∆) + area(∆′)) = 4π + 4 area(∆).

This is easily generalized to arbitrary convex n-gons on S2 (with n ≥ 3).Suppose M is such a convex n-gon with interior angles α1, · · · , αn. Then wehave

area(M) =

n∑1

αi − (n− 2)π.

This follows directly from cutting the polygon up into the constituent triangles.This is very unlike Euclidean space. On R2, we always have α+ β + γ = π.

Not only is this false on S2, but by measuring the difference, we can tell thearea of the triangle. In fact, we can identify triangles up to congruence just byknowing the three angles.

2.2 Mobius geometry

It turns out it is convenient to identify the sphere S2 withe the extended complexplane C∞ = C ∪ {∞}. Then isometries of S2 will translate to nice class of mapsof C∞.

We first find a way to identify S2 with C∞. We use coordinates ζ ∈ C∞. Wedefine the stereographic projection π : S2 → C∞ by

NP

π(P )

17


π(P ) = (line PN) ∩ {z = 0},which is well defined except where P = N , in which case we define π(N) =∞.

To give an explicit formula for this, consider the cross-section through theplane ONP .

O

N

r P

zπ(P )

R

If P has coordinates (x, y), then we see that π(P ) will be a scalar multiple ofx + iy. To find this factor, we notice that we have two similar triangles, andhence obtain

r

R=

1− z1

.

Then we obtain the formula

π(x, y, z) =x+ iy

1− z.

If we do the projection from the South pole instead, we get a related formula.

Lemma. If π′ : S2 → C∞ denotes the stereographic projection from the SouthPole instead, then

π′(P ) =1

π(P ).

Proof. Let P (x, y, z). Then

π(x, y, z) =x+ iy

1− z.

Then we have

π′(x, y, z) =x+ iy

1 + z,

since we have just flipped the z axis around. So we have

π(P )π′(P ) =x2 + y2

1− z2= 1,

noting that we have x2 + y2 + z2 = 1 since we are on the unit sphere.

We can use this to infer that π′ ◦ π−1 : C∞ → C∞ takes ζ 7→ 1/ζ, which isthe inversion in the unit circle |ζ| = 1.

From IA Groups, we know Mobius transformations act on C∞ and form agroup G by composition. For each matrix

A =

(a bc d

)∈ GL(2,C),

18


we get a Mobius map C∞ → C∞ by

ζ 7→ aζ + b

cζ + d.

Moreover, composition of Mobius map is the same multiplication of matrices.This is not exactly a bijective map between G and GL(2,C). For any

λ ∈ C∗ = C \ {0}, we know λA defines the same map Mobius map as A.Conversely, if A1 and A2 gives the same Mobius map, then there is some λ1 6= 0such that A1 = λA2.

Hence, we haveG ∼= PGL(2,C) = GL(2,C)/C∗,

whereC∗ ∼= {λI : λ ∈ C∗}.

Instead of taking the whole GL(2,C) and quotienting out multiples of theidentities, we can instead start with SL(2,C). Again, A1, A2 ∈ SL(2,C) definethe same map if and only if A1 = λA2 for some λ. What are the possible valuesof λ? By definition of the special linear group, we must have

1 = det(λA) = λ2 detA = λ2.

So λ2 = ±1. So each Mobius map is represented by two matrices, A and −A,and we get

G ∼= PSL(2,C) = SL(2,C)/{±1}.

Now let’s think about the sphere. On S2, the rotations SO(3) act as isometries.In fact, the full isometry group of S2 is O(3) (the proof is on the first examplesheet). What we would like to show that rotations of S2 correspond to Mobiustransformations coming from the subgroup SU(2) ≤ GL(2,C).

Theorem. Via the stereographic projection, every rotation of S2 induces aMobius map defined by a matrix in SU(2) ⊆ GL(2,C), where

SU(2) =

{(a −bb a

): |a|2 + |b|2 = 1

}.

Proof.

(i) Consider the r(z, θ), the rotations about the z axis by θ. These correspondsto the Mobius map ζ 7→ eiθζ, which is given by the unitary matrix(

eiθ/2 00 e−iθ/2.

)(ii) Consider the rotation r(y, π2 ). This has the matrix 0 0 1

0 1 0−1 0 0

xyz

=

zy−x

.

This corresponds to the map

ζ =x+ iy

1− z7→ ζ ′ =

z + iy

1 + x

19


We want to show this is a Mobius map. To do so, we guess what theMobius map should be, and check it works. We can manually computethat −1 7→ ∞, 1 7→ 0, i 7→ i.

1 −1

i

0

∞

The only Mobius map that does this is

ζ ′ =ζ − 1

ζ + 1.

We now check:

ζ − 1

ζ + 1=x+ iy − 1 + z

x+ iy + 1− z

=x− 1 + z + iy

x+ 1− (z − iy)

=(z + iy)(x− 1 + z + iy)

(x+ 1)(z + iy)− (z2 + y2)

=(z + iy)(x− 1 + z + iy)

(x+ 1)(z + iy) + (x2 − 1)

=(z + iy)(x− 1 + z + iy)

(x+ 1)(z + iy + x− 1)

=z + iy

x+ 1.

So done. We finally have to write this in the form of an SU(2) matrix:

1√2

(1 −11 1

).

(iii) We claim that SO(3) is generated by r(y, π2

)and r(z, θ) for 0 ≤ θ < 2π.

To show this, we observe that r(x, ϕ) = r(y, π2 )r(z, ϕ)r(y,−π2 ). Note thatwe read the composition from right to left. You can convince yourself thisis true by taking a physical sphere and try rotating. To prove it formally,we can just multiply the matrices out.

20


Next, observe that for v ∈ S2 ⊆ R3, there are some angles ϕ,ψ such thatg = r(z, ψ)r(x, ϕ) maps v to x. We can do so by first picking r(x, ϕ) torotate v into the (x, y)-plane. Then we rotate about the z-axis to send itto x.

Then for any θ, we have r(v, θ) = g−1r(x, θ)g, and our claim follows bycomposition.

(iv) Thus, via the stereographic projection, every rotation of S2 correspondsto products of Mobius transformations of C∞ with matrices in SU(2).

The key of the proof is step (iii). Apart from enabling us to perform theproof, it exemplifies a useful technique in geometry — we know how to rotatearbitrary things in the z axis. When we want to rotate things about the x axisinstead, we first rotate the sphere to move the x axis to where the z axis usedto be, do those rotations, and then rotate it back. In general, we can use someisometries or rotations to move what we want to do to a convenient location.

Theorem. The group of rotations SO(3) acting on S2 corresponds preciselywith the subgroup PSU(2) = SU(2)/± 1 of Mobius transformations acting onC∞.

What this is in some sense a converse of the previous theorem. We are sayingthat for each Mobius map from SU(2), we can find some rotation of S2 thatinduces that Mobius map, and there is exactly one.

Proof. Let g ∈ PSU(2) be a Mobius transformation

g(z) =az + b

bz + a.

Suppose first that g(0) = 0. So b = 0. So aa = 1. Hence a = eiθ/2. Then gcorresponds to r(z, θ), as we have previously seen.

In general, let g(0) = w ∈ C∞. Let Q ∈ S2 be such that π(Q) = w. Choose arotation A ∈ SO(3) such that A(Q) = −z. Since A is a rotation, let α ∈ PSU(2)be the corresponding Mobius transformation. By construction we have α(w) = 0.Then the composition α ◦ g fixes zero. So it corresponds to some B = r(z, θ).We then see that g corresponds to A−1B ∈ SO(3). So done.

Again, we solved an easy case, where g(0) = 0, and then performed somerotations to transform any other case into this simple case.

We have now produced a 2-to-1 map

SU(2)→ PSU(2) ∼= SO(3).

If we treat these groups as topological spaces, this map does something funny.Suppose we start with a (non-closed) path from I to −I in SU(2). Applying

the map, we get a closed loop from I to I in SO(3).Hence, in SO(3), loops are behave slightly weirdly. If we go around this loop

in SO(3), we didn’t really get back to the same place. Instead, we have actuallymoved from I to −I in SU(2). It takes two full loops to actually get back to I.In physics, this corresponds to the idea of spinors.

21


We can also understand this topologically as follows: since SU(2) is defined bytwo complex points a, b ∈ C such that |a|2 + |b|2, we can view it as as three-sphereS3 in SO(3).

A nice property of S3 is it is simply connected, as in any loop in S3 can beshrunk to a point. In other words, given any loop in S3, we can pull and stretchit (continuously) until becomes the “loop” that just stays at a single point.

On the other hand, SO(3) is not simply connected. We have just constructeda loop in SO(3) by mapping the path from I to −I in SU(2). We cannot deformthis loop until it just sits at a single point, since if we lift it back up to SU(2), itstill has to move from I to −I.

The neat thing is that in some sense, S3 ∼= SU(2) is just “two copies” ofSO(3). By duplicating SO(3), we have produced SU(2), a simply connectedspace. Thus we say SU(2) is a universal cover of SO(3).

We’ve just been waffling about spaces and loops, and throwing around termswe haven’t defined properly. These vague notions will be made precise in theIID Algebraic Topology course, and we will then (maybe) see that SU(2) is auniversal cover of SO(3).

22

3 Triangulations and the Euler number IB Geometry

3 Triangulations and the Euler number

We shall now study the idea of triangulations and the Euler number. We aren’tgoing to do much with them in this course — we will not even prove that theEuler number is a well-defined number. However, we need Euler numbers in orderto state the full Gauss-Bonnet theorem at the end, and the idea of triangulationsis useful in the IID Algebraic Topology course for defining simplicial homology.More importantly, the discussion of these concepts is required by the schedules.Hence we will get some exposure to these concepts in this chapter.

It is convenient to have an example other than the sphere when discussingtriangulations and Euler numbers. So we introduce the torus

Definition ((Euclidean) torus). The (Euclidean) torus is the set R2/Z2 ofequivalence classes of (x, y) ∈ R2 under the equivalence relation

(x1, y1) ∼ (x2, y2)⇔ x1 − x2, y1 − y2 ∈ Z.

It is easy to see this is indeed an equivalence relation. Thus a point in Trepresented by (x, y) is a coset (x, y) + Z2 of the subgroup Z2 ≤ R2.

Of course, there are many ways to represent a point in the torus. However, forany closed square Q ⊆ R2 with side length 1, we can obtain T is by identifyingthe sides

Q

We can define a distance d for all P1, P2 ∈ T to be

d(P1, P2) = min{‖v1 − v2‖ : vi ∈ R2,vi + Z2 = Pi}.

It is not hard to show this definition makes (T, d) into a metric space. Thisallows us to talk about things like open sets and continuous functions. We willlater show that this is not just a metric space, but a smooth surface, after wehave defined what that means.

We write Q for the interior of Q. Then the natural map f : Q → T givenby v 7→ v + Z2 is a bijection onto some open set U ⊆ T . In fact, U is justT \ {two circles meeting in 1 point}, where the two circles are the boundary ofthe square Q.

Now given any point in the torus represented by P +Z2, we can find a squareQ such that P ∈ Q. Then f : Q→ T restricted to an open disk about P is anisometry (onto the image, a subset of R2). Thus we say d is a locally Euclideanmetric.

One can also think of the torus T as the surface of a doughnut, “embedded”in Euclidean space R3.

23


Given this, it is natural to define the distance between two points to be thelength of the shortest curve between them on the torus. However, this distancefunction is not the same as what we have here. So it is misleading to think ofour locally Euclidean torus as a “doughnut”.

With one more example in our toolkit, we can start doing what we reallywant to do. The idea of a triangulation is to cut a space X up into many smallertriangles, since we like triangles. However, we would like to first define what atriangle is.

Definition (Topological triangle). A topological triangle on X = S2 or T (orany metric space X) is a subset R ⊆ X equipped with a homeomorphism R→ ∆,where ∆ is a closed Euclidean triangle in R2.

Note that a spherical triangle is in fact a topological triangle, using the radialprojection to the plane R2 from the center of the sphere.

Definition (Topological triangulation). A topological triangulation τ of a metricspace X is a finite collection of topological triangles of X which cover X suchthat

(i) For every pair of triangles in τ , either they are disjoint, or they meet inexactly one edge, or meet at exactly one vertex.

(ii) Each edge belongs to exactly two triangles.

These notions are useful only if the space X is “two dimensional” — there isno way we can triangulate, say R3, or a line. We can generalize triangulation toallow higher dimensional “triangles”, namely topological tetrahedrons, and ingeneral, n-simplices, and make an analogous definition of triangulation. However,we will not bother ourselves with this.

Definition (Euler number). The Euler number of a triangulation e = e(X, τ) is

e = F − E + V,

where F is the number of triangles; E is the number of edges; and V is thenumber of vertices.

Note that each edge actually belongs to two triangles, but we will only countit once.

There is one important fact about triangulations from algebraic topology,which we will state without proof.

Theorem. The Euler number e is independent of the choice of triangulation.

So the Euler number e = e(X) is a property of the space X itself, not aparticular triangulation.

Example. Consider the following triangulation of the sphere:

24


This has 8 faces, 12 edges and 6 vertices. So e = 2.

Example. Consider the following triangulation of the torus. Be careful notto double count the edges and vertices at the sides, since the sides are gluedtogether.

This has 18 faces, 27 edges and 9 vertices. So e = 0.

In both cases, we did not cut up our space with funny, squiggly lines. Instead,we used “straight” lines. These triangles are known as geodesic triangles.

Definition (Geodesic triangle). A geodesic triangle is a triangle whose sidesare geodesics, i.e. paths of shortest distance between two points.

In particular, we used spherical triangles in S2 and Euclidean triangles in Q.Triangulations made of geodesic triangles are rather nice. They are so nice thatwe can actually prove something about them!

Proposition. For every geodesic triangulation of S2 (and respectively T ) hase = 2 (respectively, e = 0).

Of course, we know this is true for any triangulation, but it is difficult toprove that without algebraic topology.

Proof. For any triangulation τ , we denote the “faces” of ∆1, · · · ,∆F , and writeτi = αi + βi + γi for the sum of the interior angles of the triangles (withi = 1, · · · , F ).

Then we have ∑τi = 2πV,

since the total angle around each vertex is 2π. Also, each triangle has threeedges, and each edge is from two triangles. So 3F = 2E. We write this in amore convenient form:

F = 2E − 2F.

How we continue depends on whether we are on the sphere or the torus.

25


– For the sphere, Gauss-Bonnet for the sphere says the area of ∆i is τi − π.Since the area of the sphere is 4π, we know

4π =∑

area(∆i)

=∑

(τi − π)

= 2πV − Fπ= 2πV − (2E − 2F )π

= 2π(F − E + V ).

So F − E + V = 2.

– For the torus, we have τi = π for every face in Q. So

2πV =∑

τi = πF.

So2V = F = 2E − 2F.

So we get2(F − V + E) = 0,

as required.

Note that in the definition of triangulation, we decomposed X into topologicaltriangles. We can also use decompositions by topological polygons, but they areslightly more complicated, since we have to worry about convexity. However,apart from this, everything works out well. In particular, the previous propositionalso holds, and we have Euler’s formula for S2: V −E+F = 2 for any polygonaldecomposition of S2. This is not hard to prove, and is left as an exercise on theexample sheet.

26

4 Hyperbolic geometry IB Geometry

4 Hyperbolic geometry

At the beginning of the course, we studied Euclidean geometry, which was nothard, because we already knew about it. Later on, we studied spherical geometry.That also wasn’t too bad, because we can think of S2 concretely as a subset ofR3.

We are next going to study hyperbolic geometry. Historically, hyperbolicgeometry was created when people tried to prove Euclid’s parallel postulate (thatgiven a line ` and a point P 6∈ `, there exists a unique line `′ containing P thatdoes not intersect `). Instead of proving the parallel postulate, they managed tocreate a new geometry where this is false, and this is hyperbolic geometry.

Unfortunately, hyperbolic geometry is much more complicated, since wecannot directly visualize it as a subset of R3. Instead, we need to develop themachinery of a Riemannian metric in order to properly describe hyperbolicgeometry. In a nutshell, this allows us to take a subset of R2 and measuredistances in it in a funny way.

4.1 Review of derivatives and chain rule

We start by reviewing some facts about taking derivatives, and make explicitthe notation we will use.

Definition (Smooth function). Let U ⊆ Rn be open, and f = (f1, · · · , fm) :U → Rm. We say f is smooth (or C∞) if each fi has continuous partial derivativesof each order. In particular, a C∞ map is differentiable, with continuous first-order partial derivatives.

Definition (Derivative). The derivative for a function f : U → Rm at a pointa ∈ U is a linear map dfa : Rn → Rm (also written as Df(a) or f ′(a)) such that

limh→0

‖f(a+ h)− f(a)− dfa · h‖‖h‖

→ 0,

where h ∈ Rn.If m = 1, then dfa is expressed as

(∂f∂xa

(a), · · · , ∂f∂xn(a))

, and the linear map

is given by

(h1, · · · , hn) 7→n∑i=1

∂f

∂xi(a)hi,

i.e. the dot product. For a general m, this vector becomes a matrix. The Jacobianmatrix is

J(f)a =

(∂fi∂xj

(a)

),

with the linear map given by matrix multiplication, namely

h 7→ J(f)a · h.

Example. Recall that a holomorphic (analytic) function of complex variablesf : U ⊆ C→ C has a derivative f ′, defined by

lim|w|→0

|f(z + w)− f(z)− f ′(z)w||w|

→ 0

27


We let f ′(z) = a+ ib and w = h1 + ih2. Then we have

f ′(z)w = ah1 − bh2 + i(ah2 + bh1).

We identify R2 = C. Then f : U ⊆ R2 → R2 has a derivative dfz : R2 → R2

given by (a −bb a

).

We’re also going to use the chain rule quite a lot. So we shall write it outexplicitly.

Proposition (Chain rule). Let U ⊆ Rn and V ⊆ Rp. Let f : U → Rm andg : V → U be smooth. Then f ◦ g : V → Rm is smooth and has a derivative

d(f ◦ g)p = (df)g(p) ◦ (dg)p.

In terms of the Jacobian matrices, we get

J(f ◦ g)p = J(f)g(p)J(g)p.

4.2 Riemannian metrics

Finally, we get to the idea of a Riemannian metric. The basic idea of a Rieman-nian metric is not too unfamiliar. Presumably, we have all seen maps of theEarth, where we try to draw the spherical Earth on a piece of paper, i.e. a subsetof R2. However, this does not behave like R2. You cannot measure distances onEarth by placing a ruler on the map, since distances are distorted. Instead, youhave to find the coordinates of the points (e.g. the longitude and latitude), andthen plug them into some complicated formula. Similarly, straight lines on themap are not really straight (spherical) lines on Earth.

We really should not think of Earth a subset of R2. All we have done wasto “force” Earth to live in R2 to get a convenient way of depicting the Earth, aswell as a convenient system of labelling points (in many map projections, the xand y axes are the longitude and latitude).

This is the idea of a Riemannian metric. To describe some complicatedsurface, we take a subset U of R2, and define a new way of measuring distances,angles and areas on U . All these information are packed into an entity knownas the Riemannian metric.

Definition (Riemannian metric). We use coordinates (u, v) ∈ R2. We letV ⊆ R2 be open. Then a Riemannian metric on V is defined by giving C∞

functions E,F,G : V → R such that(E(P ) F (P )F (P ) G(P )

)is a positive definite definite matrix for all P ∈ V .

Alternatively, this is a smooth function that gives a 2× 2 symmetric positivedefinite matrix, i.e. inner product 〈 · , · 〉P , for each point in V . By definition, ife1, e2 are the standard basis, then

〈e1, e1〉P = E(P )

〈e1, e2〉P = F (P )

〈e2, e2〉P = G(P ).

28


Example. We can pick E = G = 1 and F = 0. Then this is just the standardEuclidean inner product.

As mentioned, we should not imagine V as a subset of R2. Instead, we shouldthink of it as an abstract two-dimensional surface, with some coordinate systemgiven by a subset of R2. However, this coordinate system is just a convenient wayof labelling points. They do not represent any notion of distance. For example,(0, 1) need not be closer to (0, 2) than to (7, 0). These are just abstract labels.

With this in mind, V does not have any intrinsic notion of distances, anglesand areas. However, we do want these notions. We can certainly write downthings like the difference of two points, or even the compute the derivative of afunction. However, these numbers you get are not meaningful, since we can easilyuse a different coordinate system (e.g. by scaling the axes) and get a differentnumber. They have to be interpreted with the Riemannian metric. This tellsus how to measure these things, via an inner product “that varies with space”.This variation in space is not an oddity arising from us not being able to makeup our minds. This is since we have “forced” our space to lie in R2. InsideV , going from (0, 1) to (0, 2) might be very different from going from (5, 5) to(6, 5), since coordinates don’t mean anything. Hence our inner product needsto measure “going from (0, 1) to (0, 2)” differently from “going from (5, 5) to(6, 5)”, and must vary with space.

We’ll soon come to defining how this inner product gives rise to the notionof distance and similar stuff. Before that, we want to understand what we canput into the inner product 〈 · , · 〉P . Obviously these would be vectors in R2, butwhere do these vectors come from? What are they supposed to represent?

The answer is “directions” (more formally, tangent vectors). For example,〈e1, e1〉P will tell us how far we actually are going if we move in the directionof e1 from P . Note that we say “move in the direction of e1”, not “move bye1”. We really should read this as “if we move by he1 for some small h, thenthe distance covered is h

√〈e1, e1〉P ”. This statement is to be interpreted along

the same lines as “if we vary x by some small h, then the value of f will varyby f ′(x)h”. Notice how the inner product allows us to translate a length in R2

(namely ‖he1‖eucl = h) into the actual length in V .What we needed for this is just the norm induced by the inner product. Since

what we have is the whole inner product, we in fact can define more interestingthings such as areas and angles. We will formalize these ideas very soon, aftergetting some more notation out of the way.

Often, instead of specifying the three functions separately, we write the metricas

E du2 + 2F du dv +G dv2.

This notation has some mathematical meaning. We can view the coordinatesas smooth functions u : V → R, v : U → R. Since they are smooth, they havederivatives. They are linear maps

duP : R2 → R dvP : R2 → R(h1, h2) 7→ h1 (h1, h2) 7→ h2.

These formula are valid for all P ∈ V . So we just write du and dv instead.Since they are maps R2 → R, we can view them as vectors in the dual space,

29


du,dv ∈ (R2)∗. Moreover, they form a basis for the dual space. In particular,they are the dual basis to the standard basis e1, e2 of R2.

Then we can consider du2,du dv and dv2 as bilinear forms on R2. Forexample,

du2(h,k) = du(h)du(k)

du dv(h,k) =1

2(du(h)dv(k) + du(k)dv(h))

dv2(h,k) = dv(h)dv(k)

These have matrices (1 00 0

),

(0 1

212 0

),

(0 00 1

)respectively. Then we indeed have

E du2 + 2F du dv +G dv2 =

(E FF G

).

We can now start talking about what this is good for. In standard Euclideanspace, we have a notion of length and area. A Riemannian metric also gives anotion of length and area.

Definition (Length). The length of a smooth curve γ = (γ1, γ2) : [0, 1]→ V isdefined as ∫ 1

0

(Eγ2

1 + 2F γ1γ2 +Gγ22) 1

2 dt,

where E = E(γ1(t), γ2(t)) etc. We can also write this as∫ 1

0

〈γ, γ〉12

γ(t) dt.

Definition (Area). The area of a region W ⊆ V is defined as∫W

(EG− F 2)12 du dv

when this integral exists.

In the area formula, what we are integrating is just the determinant of themetric. This is also known as the Gram determinant.

We define the distance between two points P and Q to be the infimum ofthe lengths of all curves from P to Q. It is an exercise on the second examplesheet to prove that this is indeed a metric.

Example. We will not do this in full detail — the details are to be filled in inthe third example sheet.

Let V = R2, and define the Riemannian metric by

4(du2 + dv2)

(1 + u2 + v2)2.

30


This looks somewhat arbitrary, but we shall see this actually makes sense byidentifying R2 with the sphere by the stereographic projection π : S2\{N} → R2.

For every point P ∈ S2, the tangent plane to S2 at P is given by {x ∈ R3 :

x ·−−→OP = 0}. Note that we translated it so that P is the origin, so that we can

view it as a vector space (points on the tangent plane are points “from P”).

Now given any two tangent vectors x1,x2 ⊥−−→OP , we can take the inner product

x1 · x2 in R3.We want to say this inner product is “the same as” the inner product provided

by the Riemannian metric on R2. We cannot just require

x1 · x2 = 〈x1,x2〉π(P ),

since this makes no sense at all. Apart from the obvious problem that x1,x2

have three components but the Riemannian metric takes in vectors of twocomponents, we know that x1 and x2 are vectors tangent to P ∈ S2, but toapply the Riemannian metric, we need the corresponding tangent vector atπ(P ) ∈ R2. To do so, we act by dπp. So what we want is

x1 · x2 = 〈dπP (x1),dπP (x2)〉π(P ).

Verification of this equality is left as an exercise on the third example sheet. Itis helpful to notice

π−1(u, v) =(2u, 2v, u2 + v2 − 1)

1 + u2 + v2.

In some sense, we say the surface S2 \ {N} is “isometric” to R2 via thestereographic projection π. We can define the notion of isometry between twoopen sets with Riemannian metrics in general.

Definition (Isometry). Let V, V ⊆ R2 be open sets endowed with Riemannianmetrics, denoted as 〈 · , · 〉P and 〈 · , · 〉∼Q for P ∈ V,Q ∈ V respectively.

A diffeomorphism (i.e. C∞ map with C∞ inverse) ϕ : V → V is an isometryif for every P ∈ V and x,y ∈ R2, we get

〈x,y〉P = 〈dϕP (x),dϕP (y)〉∼ϕ(P ).

Again, in the definition, x and y represent tangent vectors at P ∈ V , andon the right of the equality, we need to apply dϕP to get tangent vectors atϕ(P ) ∈ V .

How are we sure this indeed is the right definition? We, at the very least,would expect isometries to preserve lengths. Let’s see this is indeed the case. Ifγ : [0, 1]→ V is a C∞ curve, the composition γ = ϕ ◦ γ : [0, 1]→ V is a path inV . We let P = γ(t), and hence ϕ(P ) = γ(t). Then

〈γ′(t), γ′(t)〉∼γ(t) = 〈dϕP ◦ γ′(t),dϕP ◦ γ′(t)〉∼ϕ(P ) = 〈γ′(t), γ′(t)〉γ(t)=P .

Integrating, we obtain

length(γ) = length(γ) =

∫ 1

0

〈γ′(t), γ′(t)〉γ(t) dt.

31


4.3 Two models for the hyperbolic plane

That’s enough preparation. We can start talking about hyperbolic plane. Wewill in fact provide two models of the hyperbolic plane. Each model has its ownstrengths, and often proving something is significantly easier in one model thanthe other.

We start with the disk model.

Definition (Poincare disk model). The (Poincare) disk model for the hyperbolicplane is given by the unit disk

D ⊆ C ∼= R2, D = {ζ ∈ C : |ζ| < 1},

and a Riemannian metric on this disk given by

4(du2 + dv2)

(1− u2 − v2)2=

4|dζ|2

(1− |ζ|2)2, (∗)

where ζ = u+ iv.

Note that this is similar to our previous metric for the sphere, but we have1− u2 − v2 instead of 1 + u2 + v2.

To interpret the term |dζ|2, we can either formally set |dζ|2 = du2 + dv2, orinterpret it as the derivative dζ = du+ idv : C→ C.

We see that (∗) is a scaling of the standard Riemannian metric by a factordepending on the polar radius r = |ζ|2. The distances are scaled by 2

1−r2 , while

the areas are scaled by 4(1−r2)2 . Note, however, that the angles in the hyperbolic

disk are the same as that in R2. This is in general true for metrics that are justscaled versions of the Euclidean metric (exercise).

Alternatively, we can define the hyperbolic plane with the upper-half plane.

Definition (Upper half-plane). The upper half-plane is

H = {z ∈ C : Im(z) > 0}.

What is the appropriate Riemannian metric to put on the upper half plane?We know D bijects to H via the Mobius transformation

ϕ : ζ ∈ D 7→ i1 + ζ

1− ζ∈ H.

This bijection is in fact a conformal equivalence, as defined in IB ComplexAnalysis/Methods. The idea is to pick a metric on H such that this map isan isometry. Then H together with this Riemannian metric will be the upperhalf-plane model for the hyperbolic plane.

To avoid confusion, we reserve the letter z for points z ∈ H, with z = x+ iy,while we use ζ for points ζ ∈ D, and write ζ = u+ iv. Then we have

z = i1 + ζ

1− ζ, ζ =

z − iz + i

.

Instead of trying to convert the Riemannian metric on D to H, which would bea complete algebraic horror, we first try converting the Euclidean metric. TheEuclidean metric on R2 = C is given by

〈w1, w2〉 = Re(w1w2) =1

2(w1w2 + w1w2).

32


So if 〈 · , · 〉eucl is the Euclidean metric at ζ, then at z such that ζ = z−iz+i , we

require (by definition of isometry)

〈w, v〉z =

⟨dζ

dzw,

dζ

dzv

⟩eucl

=

∣∣∣∣dζdz

∣∣∣∣2 Re(wv) =

∣∣∣∣dζdz

∣∣∣∣2 (w1v1 + w2v2),

where w = w1 + iw2, v = v1 + iv2.Hence, on H, we obtain the Riemannian metric∣∣∣∣dζdz

∣∣∣∣2 (dx2 + dy2).

We can computedζ

dz=

1

z + i− z − i

(z + i)2=

2i

(z + i)2.

This is what we get if we started with a Euclidean metric. If we start with thehyperbolic metric on D, we get an additional scaling factor. We can do somecomputations to get

1− |ζ|2 = 1− |z − i|2

|z + i|2,

and hence1

1− |ζ|2=

|z + i|2

|z + i|2 − |z − i|2=|z + i|2

4 Im z.

Putting all these together, metric corresponding to 4|dζ|2(1−|ζ|2)2 on D is

4 · 4

|z + i|4·(|z + i|2

4 Im z

)2

· |dz|2 =|dz|2

(Im z)2=

dx2 + dy2

y2.

We now use all these ingredients to define the upper half-plane model.

Definition (Upper half-plane model). The upper half-plane model of the hy-perbolic plane is the upper half-plane H with the Riemannian metric

dx2 + dy2

y2.

The lengths on H are scaled (from the Euclidean one) by 1y , while the areas

are scaled by 1y2 . Again, the angles are the same.

Note that we did not have to go through so much mess in order to definethe sphere. This is since we can easily “embed” the surface of the sphere in R3.However, there is no easy surface in R3 that gives us the hyperbolic plane. Aswe don’t have an actual prototype, we need to rely on the more abstract data ofa Riemannian metric in order to work with hyperbolic geometry.

We are next going to study the geometry of H, We claim that the followinggroup of Mobius maps are isometries of H:

PSL(2,R) =

{z 7→ az + b

cz + d: a, b, c, d ∈ R, ad− bc = 1

}.

Note that the coefficients have to be real, not complex.

33


Proposition. The elements of PSL(2,R) are isometries of H, and this preservesthe lengths of curves.

Proof. It is easy to check that PSL(2,R) is generated by

(i) Translations z 7→ z + a for a ∈ R

(ii) Dilations z 7→ az for a > 0

(iii) The single map z 7→ − 1z .

So it suffices to show each of these preserves the metric |dz|2

y2 , where z = x+ iy.The first two are straightforward to see, by plugging it into formula and noticethe metric does not change.

We now look at the last one, given by z 7→ − 1z . The derivative at z is

f ′(z) =1

z2.

So we get

dz 7→ d

(−1

z

)=

dz

z2.

So ∣∣∣∣d(−1

z

)∣∣∣∣2 =|dz|2

|z|4.

We also have

Im

(−1

z

)= − 1

|z|2Im z =

Im z

|z|2.

So|d(−1/z)|2

Im(−1/z)2=

(|dz|2

|z4|

)/( (Im z)2

|z|4

)=|dz|2

(Im z)2.

So this is an isometry, as required.

Note that each z 7→ az + b with a > 0, b ∈ R is in PSL(2,R). Also, we canuse maps of this form to send any point to any other point. So PSL(2,R) actstransitively on H. Moreover, everything in PSL(2,R) fixes R ∪ {∞}.

Recall also that each Mobius transformation preserves circles and lines in thecomplex plane, as well as angles between circles/lines. In particular, consider theline L = iR, which meets R perpendicularly, and let g ∈ PSL(2,R). Then theimage is either a circle centered at a point in R, or a straight line perpendicularto R.

We let L+ = L ∩H = {it : t > 0}. Then g(L+) is either a vertical half-lineor a semi-circle that ends in R.

Definition (Hyperbolic lines). Hyperbolic lines in H are vertical half-lines orsemicircles ending in R.

We will now prove some lemmas to justify why we call these hyperbolic lines.

Lemma. Given any two distinct points z1, z2 ∈ H, there exists a uniquehyperbolic line through z1 and z2.

34


Proof. This is clear if Re z1 = Re z2 — we just pick the vertical half-line throughthem, and it is clear this is the only possible choice.

Otherwise, if Re z1 6= Re z2, then we can find the desired circle as follows:

R

z1

z2

It is also clear this is the only possible choice.

Lemma. PSL(2,R) acts transitively on the set of hyperbolic lines in H.

Proof. It suffices to show that for each hyperbolic line `, there is some g ∈PSL(2,R) such that g(`) = L+. This is clear when ` is a vertical half-line, sincewe can just apply a horizontal translation.

If it is a semicircle, suppose it has end-points s < t ∈ R. Then consider

g(z) =z − tz − s

.

This has determinant −s+t > 0. So g ∈ PSL(2,R). Then g(t) = 0 and g(s) =∞.Then we must have g(`) = L+, since g(`) is a hyperbolic line, and the onlyhyperbolic lines passing through ∞ are the vertical half-lines. So done.

Moreover, we can achieve g(s) = 0 and g(t) = ∞ by composing with − 1z .

Also, for any P ∈ ` not on the endpoints, we can construct a g such thatg(P ) = i ∈ L+, by composing with z 7→ az. So the isometries act transitively onpairs (`, P ), where ` is a hyperbolic line and P ∈ `.

Definition (Hyperbolic distance). For points z1, z2 ∈ H, the hyperbolic distanceρ(z1, z2) is the length of the segment [z1, z2] ⊆ ` of the hyperbolic line throughz1, z2 (parametrized monotonically).

Thus PSL(2,R) preserves hyperbolic distances. Similar to Euclidean spaceand the sphere, we show these lines minimize distance.

Proposition. If γ : [0, 1] → H is a piecewise C1-smooth curve with γ(0) =z1, γ(1) = z2, then length(γ) ≥ ρ(z1, z2), with equality iff γ is a monotonicparametrisation of [z1, z2] ⊆ `, where ` is the hyperbolic line through z1 and z2.

Proof. We pick an isometry g ∈ PSL(2,R) so that g(`) = L+. So without lossof generality, we assume z1 = iu and z2 = iv, with u < v ∈ R.

35


We decompose the path as γ(t) = x(t) + iy(t). Then we have

length(γ) =

∫ 1

0

1

y

√x2 + y2 dt

≥∫ 1

0

|y|y

dz

≥∣∣∣∣∫ 1

0

y

ydt

∣∣∣∣= [log y(t)]10

= log( vu

)This calculation also tells us that ρ(z1, z2) = log

(vu

). so length(γ) ≥ ρ(z1, z2)

with equality if and only if x(t) = 0 (hence γ ⊆ L+) and y ≥ 0 (hence monotonic).

Corollary (Triangle inequality). Given three points z1, z2, z3 ∈ H, we have

ρ(z1, z3) ≤ ρ(z1, z2) + ρ(z2, z3),

with equality if and only if z2 lies between z1 and z2.

Hence, (H, ρ) is a metric space.

4.4 Geometry of the hyperbolic disk

So far, we have worked with the upper half-plane model. This is since the upperhalf-plane model is more convenient for these calculations. However, sometimesthe disk model is more convenient. So we also want to understand that as well.

Recall that ζ ∈ D 7→ z = i 1+ζ1−ζ ∈ H is an isometry, with an (isometric) inverse

z ∈ H 7→ ζ = z−iz+i ∈ D. Moreover, since these are Mobius maps, circle-lines are

preserved, and angles between the lines are also preserved.Hence, immediately from previous work on H, we know

(i) PSL(2,R) ∼= {Mobius transformations sending D to itself} = G.

(ii) Hyperbolic lines in D are circle segments meeting |ζ| = 1 orthogonally,including diameters.

(iii) G acts transitively on hyperbolic lines in D (and also on pairs consistingof a line and a point on the line).

(iv) The length-minimizing geodesics on D are a segments of hyperbolic linesparametrized monotonically.

We write ρ for the (hyperbolic) distance in D.

Lemma. Let G be the set of isometries of the hyperbolic disk. Then

(i) Rotations z 7→ eiθz (for θ ∈ R) are elements of G.

(ii) If a ∈ D, then g(z) = z−a1−az is in G.

36


Proof.

(i) This is clearly an isometry, since this is a linear map, preserves |z| and|dz|, and hence also the metric

4|dz|2

(1− |z|2)2.

(ii) First, we need to check this indeed maps D to itself. To do this, we firstmake sure it sends {|z| = 1} to itself. If |z| = 1, then

|1− az| = |z(1− az)| = |z − a| = |z − a|.

So|g(z)| = 1.

Finally, it is easy to check g(a) = 0. By continuity, G must map D to itself.We can then show it is an isometry by plugging it into the formula.

It is an exercise on the second example sheet to show all g ∈ G is of the form

g(z) = eiθz − a1− az

or

g(z) = eiθz − a1− az

for some θ ∈ R and a ∈ D.We shall now use the disk model to do something useful. We start by coming

up with an explicit formula for distances in the hyperbolic plane.

Proposition. If 0 ≤ r < 1, then

ρ(0, reiθ) = 2 tanh−1 r.

In general, for z1, z2 ∈ D, we have

g(z1, z2) = 2 tanh−1

∣∣∣∣ z1 − z2

1− z1z2

∣∣∣∣ .Proof. By the lemma above, we can rotate the hyperbolic disk so that reiθ isrotated to r. So

ρ(0, reiθ) = ρ(0, r).

We can evaluate this by performing the integral

ρ(0, r) =

∫ r

0

2 dt

1− t2= 2 tanh−1 r.

For the general case, we apply the Mobius transformation

g(z) =z − z1

1− z1z.

Then we have

g(z1) = 0, g(z2) =z2 − z1

1− z1z2=

∣∣∣∣ z1 − z2

1− z1z2

∣∣∣∣ eiθ.So

ρ(z1, z2) = ρ(g(z1), g(z2)) = 2 tanh−1

∣∣∣∣ z1 − z2

1− z1z2

∣∣∣∣ .37


Again, we exploited the idea of performing the calculation in an easy case, andthen using isometries to move everything else to the easy case. In general, whenwe have a “distinguished” point in the hyperbolic plane, it is often convenient touse the disk model, move it to 0 by an isometry.

Proposition. For every point P and hyperbolic line `, with P 6∈ `, there is aunique line `′ with P ∈ `′ such that `′ meets ` orthogonally, say ` ∩ `′ = Q, andρ(P,Q) ≤ ρ(P, Q) for all Q ∈ `.

This is a familiar fact from Euclidean geometry. To prove this, we againapply the trick of letting P = 0.

Proof. wlog, assume P = 0 ∈ D. Note that a line in D (that is not a diameter)is a Euclidean circle. So it has a center, say C.

Since any line through P is a diameter, there is clearly only one line thatintersects ` perpendicularly (recall angles in D is the same as the Euclideanangle).

`′ P

D

C`Q

It is also clear that PQ minimizes the Euclidean distance between P and `.While this is not the same as the hyperbolic distance, since hyperbolic linesthrough P are diameters, having a larger hyperbolic distance is equivalent tohaving a higher Euclidean distance. So this indeed minimizes the distance.

How does reflection in hyperbolic lines work? This time, we work in theupper half-plane model, since we have a favorite line L+.

Lemma (Hyperbolic reflection). Suppose g is an isometry of the hyperbolichalf-plane H and g fixes every point in L+ = {iy : y ∈ R+}. Then G is eitherthe identity or g(z) = −z, i.e. it is a reflection in the vertical axis L+.

Observe we have already proved a similar result in Euclidean geometry, andthe spherical version was proven in the first example sheet.

Proof. For every P ∈ H \ L+, there is a unique line `′ containing P such that`′ ⊥ L+. Let Q = L+ ∩ `′.

L+

`′ PQ

38


We see `′ is a semicircle, and by definition of isometry, we must have

ρ(P,Q) = ρ(g(P ), Q).

Now note that g(`′) is also a line meeting L+ perpendicularly at Q, since gfixes L+ and preserves angles. So we must have g(`′) = `′. Then in particularg(P ) ∈ `′. So we must have g(P ) = P or g(P ) = P ′, where P ′ is the imageunder reflection in L+.

Now it suffices to prove that if g(P ) = P for any one P , then g(P ) must bethe identity (if g(P ) = P ′ for all P , then g must be given by g(z) = −z).

Now suppose g(P ) = P , and let A ∈ H+, where H+ = {z ∈ H : Re z > 0}.

L+

PP ′

AA′B

Now if g(A) 6= A, then g(A) = A′. Then ρ(A′, P ) = ρ(A,P ). But

ρ(A′, P ) = ρ(A′, B) + ρ(B,P ) = ρ(A,B) + ρ(B,P ) > ρ(A,P ),

by the triangle inequality, noting that B 6∈ (AP ). This is a contradiction. So gmust fix everything.

Definition (Hyperbolic reflection). The map R : z ∈ H 7→ −z ∈ H is the(hyperbolic) reflection in L+. More generally, given any hyperbolic line `, let Tbe the isometry that sends ` to L+. Then the (hyperbolic) reflection in ` is

R` = T−1RT

Again, we already know how to reflect in L+. So to reflect in another line `,we move our plane such that ` becomes L+, do the reflection, and move back.

By the previous proposition, R` is the unique isometry that is not identityand fixes `.

4.5 Hyperbolic triangles

Definition (Hyperbolic triangle). A hyperbolic triangle ABC is the regiondetermined by three hyperbolic line segments AB,BC and CA, including extremecases where some vertices A,B,C are allowed to be “at infinity”. More precisely,in the half-plane model, we allow them to lie in R ∪ {∞}; in the disk model weallow them to lie on the unit circle |z| = 1.

We see that if A is “at infinity”, then the angle at A must be zero.Recall for a region R ⊆ H, we can compute the area of R as

area(R) =

∫∫R

dx dy

y2.

Similar to the sphere, we have

39


Theorem (Gauss-Bonnet theorem for hyperbolic triangles). For each hyperbolictriangle ∆, say, ABC, with angles α, β, γ ≥ 0 (note that zero angle is possible),we have

area(∆) = π − (α+ β + γ).

Proof. First do the case where γ = 0, so C is “at infinity”. Recall that we liketo use the disk model if we have a distinguished point in the hyperbolic plane.If we have a distinguished point at infinity, it is often advantageous to use theupper half plane model, since ∞ is a distinguished point at infinity.

So we use the upper-half plane model, and wlog C =∞ (apply PSL(2,R))if necessary. Then AC and BC are vertical half-lines. So AB is the arc of asemi-circle. So AB is an arc of a semicircle.

C C

B

A

βπ − α

αβ

We use the transformation z 7→ z + a (with a ∈ R) to center the semi-circle at 0.We then apply z 7→ bz (with b > 0) to make the circle have radius 1. Thus wlogAB ⊆ {x2 + y2 = 1}.

Now we have

area(T ) =

∫ cos β

cos(π−α)

∫ ∞√

1−x2

1

y2dy dx

=

∫ cos β

cos(π−α)

1√1− x2

dx

= [− cos−1(x)]cos βcos(π−α)

= π − α− β,

as required.In general, we use H again, and we can arrange AC in a vertical half-line.

Also, we can move AB to x2 + y2 = 1, noting that this transformation keepsAC vertical.

B

Aα

Cγ

δ

β

40


We consider ∆1 = AB∞ and ∆2 = CB∞. Then we can immediately write

area(∆1) = π − α− (β + δ)

area(∆2) = π − δ − (π − γ) = γ − δ.

So we have

area(T ) = area(∆2)− area(∆1) = π − α− β − γ,

as required.

Similar to the spherical case, we have some hyperbolic sine and cosine rules.For example, we have

Theorem (Hyperbolic cosine rule). In a triangle with sides a, b, c and anglesα, β, γ, we have

cosh c = cosh a cosh b− sinh a sinh b cos γ.

Proof. See example sheet 2.

Recall that in S2, any two lines meet (in two points). In the Euclidean planeR2, any two lines meet (in one point) iff they are not parallel. Before we moveon to the hyperbolic case, we first make a definition.

Definition (Parallel lines). We use the disk model of the hyperbolic plane. Twohyperbolic lines are parallel iff they meet only at the boundary of the disk (at|z| = 1).

Definition (Ultraparallel lines). Two hyperbolic lines are ultraparallel if theydon’t meet anywhere in {|z| ≤ 1}.

In the Euclidean plane, we have the parallel axiom: given a line ` and P 6∈ `,there exists a unique line `′ containing P with ` ∩ `′ = ∅. This fails in both S2

and the hyperbolic plane — but for very different reasons! In S2, there are nosuch parallel lines. In the hyperbolic plane, there are many parallel lines. Thereis a more deep reason for why this is the case, which we will come to at the veryend of the course.

4.6 Hyperboloid model

Recall we said there is no way to view the hyperbolic plane as a subset of R3,and hence we need to mess with Riemannian metrics. However, it turns out wecan indeed embed the hyperbolic plane in R3, if we give R3 a different metric!

Definition (Lorentzian inner product). The Lorentzian inner product on R3

has the matrix 1 0 00 1 00 0 −1

41


This is less arbitrary as it seems. Recall from IB Linear Algebra that we canalways pick a basis where a non-degenerate symmetric bilinear form has diagonalmade of 1 and −1. If we further identify A and −A as the “same” symmetricbilinear form, then this is the only other possibility left.

Thus, we obtain the quadratic form given by

q(x) = 〈x,x〉 = x2 + y2 − z2.

We now define the 2-sheet hyperboloid as

{x ∈ R2 : q(x) = −1}.

This is given explicitly by the formula

x2 + y2 = z2 − 1.

We don’t actually need to two sheets. So we define

S+ = S ∩ {z > 0}.

We let π : S+ → D ⊆ C = R2 be the stereographic projection from (0, 0, -1) by

π(x, y, z) =x+ iy

1 + z= u+ iv.

P

π(P )

We put r2 = u2 + v2. Doing some calculations, we show that

(i) We always have r < 1, as promised.

(ii) The stereographic projection π is invertible with

σ(u, v) = π−1(u, v) =1

1− r2(2u, 2v, 1 + r2) ∈ S+.

(iii) The tangent plane to S+ at P is spanned by

σu =∂σ

σu, σv =

∂σ

∂v.

42


We can explicitly compute these to be

σu =2

(1− r2)2(1 + u2 − v2, 2uv, 2u),

σv =2

(1− r2)2(2uv, 1 + v2 − u2, 2v).

We restrict the inner product 〈 · , · 〉 to the span of σu, σv, and we get asymmetric bilinear form assigned to each u, v ∈ D given by

E du2 + 2F du dv +G dv2,

where

E = 〈σu, σu〉 =4

(1− r2)2,

F = 〈σu, σv〉 = 0,

G = 〈σv, σv〉 =4

(1− r2)2.

We have thus recovered the Poincare disk model of the hyperbolic plane.

43

5 Smooth embedded surfaces (in R3) IB Geometry

5 Smooth embedded surfaces (in R3)

5.1 Smooth embedded surfaces

So far, we have been studying some specific geometries, namely Euclidean, spheri-cal and hyperbolic geometry. From now on, we go towards greater generality, andstudy arbitrary surfaces. We will mostly work with surfaces that are smoothlyembedded as subsets of R3, we can develop notions parallel to those we have hadbefore, such as Riemannian metrics and lengths. At the very end of the course,we will move away from the needless restriction restriction that the surface isembedded in R3, and study surfaces that just are.

Definition (Smooth embedded surface). A set S ⊆ R3 is a (parametrized)smooth embedded surface if every point P ∈ S has an open neighbourhood U ⊆ S(with the subspace topology on S ⊆ R3) and a map σ : V → U from an openV ⊆ R2 to U such that if we write σ(u, v) = (x(u, v), y(u, v), z(u, v)), then

(i) σ is a homeomorphism (i.e. a bijection with continuous inverse)

(ii) σ is C∞ (smooth) on V (i.e. has continuous partial derivatives of all orders).

(iii) For all Q ∈ V , the partial derivatives σu(Q) and σv(Q) are linearly inde-pendent.

Recall that

σu(Q) =∂σ

∂u(Q) =

∂x∂u∂y∂u∂z∂u

(Q) = dσQ(e1),

where e1, e2 is the standard basis of R2. Similarly, we have

σv(Q) = dσQ(e2).

We define some terminology.

Definition (Smooth coordinates). We say (u, v) are smooth coordinates onU ⊆ S.

Definition (Tangent space). The subspace of R3 spanned by σu(Q), σv(Q) isthe tangent space TPS to S at P = σ(Q).

Definition (Smooth parametrisation). The function σ is a smooth parametrisa-tion of U ⊆ S.

Definition (Chart). The function σ−1 : U → V is a chart of U .

Proposition. Let σ : V → U and σ : V → U be two C∞ parametrisations of asurface. Then the homeomorphism

ϕ = σ−1 ◦ σ : V → V

is in fact a diffeomorphism.

This proposition says any two parametrizations of the same surface arecompatible.

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Proof. Since differentiability is a local property, it suffices to consider ϕ on somesmall neighbourhood of a point in V . Pick our favorite point (v0, u0) ∈ V . Weknow σ = σ(u, v) is differentiable. So it has a Jacobian matrixxu xv

yu yvzy zv

.

By definition, this matrix has rank two at each point. wlog, we assume the firsttwo rows are linearly independent. So

det

(xu xvyu yv

)6= 0

at (v0, u0) ∈ V . We define a new function

F (u, v) =

(x(u, v)y(u, v)

).

Now the inverse function theorem applies. So F has a local C∞ inverse, i.e.there are two open neighbourhoods (u0, v0) ∈ N and F (u0, v0) ∈ N ′ ⊆ R2 suchthat f : N → N ′ is a diffeomorphism.

Writing π : σ → N ′ for the projection π(x, y, z) = (x, y) we can put thesethings in a commutative diagram:

σ(N)

N N ′

π

F

σ .

We now let N = σ−1(σ(N)) and F = π ◦ σ, which is yet again smooth. Then wehave the following larger commutative diagram.

σ(N)

N N ′ N

π

F

σ

F

σ .

Then we have

ϕ = σ−1 ◦ σ = σ−1 ◦ π−1 ◦ π ◦ σ = F−1 ◦ F ,

which is smooth, since F−1 and F are. Hence ϕ is smooth everywhere. Bysymmetry of the argument, ϕ−1 is smooth as well. So this is a diffeomorphism.

A more practical result is the following:

Corollary. The tangent plane TQS is independent of parametrization.

Proof. We knowσ(u, v) = σ(ϕ1(u, v), ϕ2(u, v)).

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We can then compute the partial derivatives as

σu = ϕ1,uσu + ϕ2,uσv

σv = ϕ1,vσu + ϕ2,vσv

Here the transformation is related by the Jacobian matrix(ϕ1,u ϕ1,v

ϕ2,u ϕ2,v

)= J(ϕ).

This is invertible since ϕ is a diffeomorphism. So (σu, σv) and (σu, σv) aredifferent basis of the same two-dimensional vector space. So done.

Note that we have

σu × σv = det(J(ϕ))σu × σv.

So we can define

Definition (Unit normal). The unit normal to S at Q ∈ S is

N = NQ =σu × σv‖σu × σv‖

,

which is well-defined up to a sign.

Often, instead of a parametrization σ : V ⊆ R2 → U ⊆ S, we want thefunction the other way round. We call this a chart.

Definition (Chart). Let S ⊆ R3 be an embedded surface. The map θ = σ−1 :U ⊆ S → V ⊆ R2 is a chart.

Example. Let S2 ⊆ R3 be a sphere. The two stereographic projections from±e3 give two charts, whose domain together cover S2.

Similar to what we did to the sphere, given a chart θ : U → V ⊆ R2, we caninduce a Riemannian metric on V . We first get an inner product on the tangentspace as follows:

Definition (First fundamental form). If S ⊆ R3 is an embedded surface, theneach TQS for Q ∈ S has an inner product from R3, i.e. we have a family of innerproducts, one for each point. We call this family the first fundamental form.

This is a theoretical entity, and is more easily worked with when we havea chart. Suppose we have a parametrization σ : V → U ⊆ S, a,b ∈ R2, andP ∈ V . We can then define

〈a,b〉P = 〈dσP (a),dσP (b)〉R3 .

With respect to the standard basis e1, e2 ∈ R2, we can write the first fundamentalform as

E du2 + 2F du dv +G dv2,

where

E = 〈σu, σu〉 = 〈e1, e1〉PF = 〈σu, σv〉 = 〈e1, e2〉PG = 〈σv, σv〉 = 〈e2, e2〉P .

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Thus, this induces a Riemannian metric on V . This is also called the firstfundamental form corresponding to σ. This is what we do in practical examples.

We will assume the following property, which we are not bothered to prove.

Proposition. If we have two parametrizations related by σ = σ ◦ ϕ : V → U ,then ϕ : V → V is an isometry of Riemannian metrics (on V and V ).

Definition (Length and energy of curve). Given a smooth curve Γ : [a, b] →S ⊆ R3, the length of γ is

length(Γ) =

∫ b

a

‖Γ′(t)‖ dt.

The energy of the curve is

energy(Γ) =

∫ b

a

‖Γ′(t)‖2 dt.

We can think of the energy as something like the kinetic energy of a particlealong the path, except that we are missing the factor of 1

2m, because they areannoying.

How does this work with parametrizations? For the sake of simplicity, weassume Γ([a, b]) ⊆ U for some parametrization σ : V → U . Then we define thenew curve

γ = σ−1 ◦ Γ : [a, b]→ V.

This curve has two components, say γ = (γ1, γ2). Then we have

Γ′(t) = (dσ)γ(t)(γ1(t)e1 + γ2(t)e2) = γ1σu + γ2σv,

and thus‖Γ′(t)‖ = 〈γ, γ〉

12

P = (Eγ21 + 2F γ1γ2 +Gγ2

2)12 .

So we get

length Γ =

∫ b

a

(Eγ21 + 2F γ1γ2 +Gγ2

2)12 dt.

Similarly, the energy is given by

energy Γ =

∫ b

a

(Eγ21 + 2F γ1γ2 +Gγ2

2) dt.

This agrees with what we’ve had for Riemannian metrics.

Definition (Area). Given a smooth C∞ parametrization σ : V → U ⊆ S ⊆ R3,and a region T ⊆ U , we define the area of T to be

area(T ) =

∫θ(T )

√EG− F 2 du dv,

whenever the integral exists (where θ = σ−1 is a chart).

Proposition. The area of T is independent of the choice of parametrization.So it extends to more general subsets T ⊆ S, not necessarily living in the imageof a parametrization.

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Proof. Exercise!

Note that in examples, σ(V ) = U often is a dense set in S. For example, ifwe work with the sphere, we can easily parametrize everything but the poles. Inthat case, it suffices to use just one parametrization σ for area(S).

Note also that areas are invariant under isometries.

5.2 Geodesics

We now come to the important idea of a geodesic. We will first define these forRiemannian metrics, and then generalize it to general embedded surfaces.

Definition (Geodesic). Let V ⊆ R2u,v be open, and E du2 + 2F du dv +G dv2

be a Riemannian metric on V . We let

γ = (γ1, γ2) : [a, b]→ V

be a smooth curve. We say γ is a geodesic with respect to the Riemannian metricif it satisfies

d

dt(Eγ1 + F γ2) =

1

2(Euγ

21 + 2Fuγ1γ2 +Guγ

22)

d

dt(F γ1 +Gγ2) =

1

2(Evγ

21 + 2Fvγ1γ2 +Gvγ

22)

for all t ∈ [a, b]. These equations are known as the geodesic ODEs.

What exactly do these equations mean? We will soon show that these arecurves that minimize (more precisely, are stationary points of) energy. To do so,we need to come up with a way of describing what it means for γ to minimizeenergy among all possible curves.

Definition (Proper variation). Let γ : [a, b] → V be a smooth curve, and letγ(a) = p and γ(b) = q. A proper variation of γ is a C∞ map

h : [a, b]× (−ε, ε) ⊆ R2 → V

such thath(t, 0) = γ(t) for all t ∈ [a, b],

andh(a, τ) = p, h(b, τ) = q for all |τ | < ε,

and thatγτ = h( · , τ) : [a, b]→ V

is a C∞ curve for all fixed τ ∈ (−ε, ε).

Proposition. A smooth curve γ satisfies the geodesic ODEs if and only if γ isa stationary point of the energy function for all proper variation, i.e. if we definethe function

E(τ) = energy(γτ ) : (−ε, ε)→ R,

thendE

dτ

∣∣∣∣τ=0

= 0.

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Proof. We let γ(t) = (u(t), v(t)). Then we have

energy(γ) =

∫ b

a

(E(u, v)u2 + 2F (u, v)uv +G(u, v)v2) dt =

∫ b

a

I(u, v, u, v) dt.

We consider this as a function of four variables u, u, v, v, which are not necessarilyrelated to one another. From the calculus of variations, we know γ is stationaryif and only if

d

dt

(∂I

∂u

)=∂I

∂u,

d

dt

(∂I

∂v

)=∂I

∂v.

The first equation gives us

d

dt(2(Eu+ F v)) = Euu

2 + 2Fuuv +Guv2,

which is exactly the geodesic ODE. Similarly, the second equation gives the othergeodesic ODE. So done.

Since the definition of a geodesic involves the derivative only, which is a localproperty, we can easily generalize the definition to arbitrary embedded surfaces.

Definition (Geodesic on smooth embedded surface). Let S ⊆ R3 be an embed-ded surface. Let Γ : [a, b]→ S be a smooth curve in S, and suppose there is aparametrization σ : V → U ⊆ S such that im Γ ⊆ U . We let θ = σ−1 be thecorresponding chart.

We define a new curve in V by

γ = θ ◦ Γ : [a, b]→ V.

Then we say Γ is a geodesic on S if and only if γ is a geodesic with respect tothe induced Riemannian metric.

For a general Γ : [a, b]→ V , we say Γ is a geodesic if for each point t0 ∈ [a, b],there is a neighbourhood V of t0 such that im Γ|V lies in the domain of somechart, and Γ|V is a geodesic in the previous sense.

Corollary. If a curve Γ minimizes the energy among all curves from P = Γ(a)to Q = Γ(b), then Γ is a geodesic.

Proof. For any a1, a2 such that a ≤ a1 ≤ b1 ≤ b, we let Γ1 = Γ|[a1,b1]. Then Γ1

also minimizes the energy between a1 and b1 for all curves between Γ(a1) andΓ(b1).

If we picked a1, b1 such that Γ([a1, b1]) ⊆ U for some parametrized neigh-bourhood U , then Γ1 is a geodesic by the previous proposition. Since theparametrized neighbourhoods cover S, at each point t0 ∈ [a, b], we can find a1, b1such that Γ([a1, b1]) ⊆ U . So done.

This is good, but we can do better. To do so, we need a lemma.

Lemma. Let V ⊆ R2 be an open set with a Riemannian metric, and let P,Q ∈ V .Consider C∞ curves γ : [a, b]→ V such that γ(0) = P, γ(1) = Q. Then such a γwill minimize the energy (and therefore is a geodesic) if and only if γ minimizesthe length and has constant speed.

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This means being a geodesic is almost the same as minimizing length. It’sjust that to be a geodesic, we have to parametrize it carefully.

Proof. Recall the Cauchy-Schwartz inequality for continuous functions f, g ∈C[0, 1], which says(∫ 1

0

f(x)g(x) dx

)2

≤(∫ 1

0

f(x)2 dx

)(∫ 1

0

g(x)2 dx

),

with equality iff g = λf for some λ ∈ R, or f = 0, i.e. g and f are linearlydependent.

We now put f = 1 and g = ‖γ‖. Then Cauchy-Schwartz says

(length γ)2 ≤ energy(γ),

with equality if and only if γ is constant.From this, we see that a curve of minimal energy must have constant speed.

Then it follows that minimizing energy is the same as minimizing length if wemove at constant speed.

Is the converse true? Are all geodesics length minimizing? The answer is“almost”. We have to be careful with our conditions in order for it to be true.

Proposition. A curve Γ is a geodesic iff and only if it minimizes the energylocally, and this happens if it minimizes the length locally and has constantspeed.

Here minimizing a quantity locally means for every t ∈ [a, b], there is someε > 0 such that Γ|[t−ε,t+ε] minimizes the quantity.

We will not prove this. Local minimization is the best we can hope for, sincethe definition of a geodesic involves differentiation, and derivatives are localproperties.

Proposition. In fact, the geodesic ODEs imply ‖Γ′(t)‖ is constant.

We will also not prove this, but in the special case of the hyperbolic plane,we can check this directly. This is an exercise on the third example sheet.

A natural question to ask is that if we pick a point P and a tangent directiona, can we find a geodesic through P whose tangent vector at P is a?

In the geodesic equations, if we expand out the derivative, we can write theequation as (

E FF G

)(γ1

γ2

)= something.

Since the Riemannian metric is positive definite, we can invert the matrix andget an equation of the form(

γ1

γ2

)= H(γ1, γ2, γ1, ˙γ2)

for some function H. From the general theory of ODE’s in IB Analysis II, subjectto some sensible conditions, given any P = (u0, v0) ∈ V and a = (p0, q0) ∈ R2,there is a unique geodesic curve γ(t) defined for |t| < ε with γ(0) = P and

50


γ(0) = a. In other words, we can choose a point, and a direction, and then thereis a geodesic going that way.

Note that we need the restriction that γ is defined only for |t| < ε since wemight run off to the boundary in finite time. So we need not be able to define itfor all t ∈ R.

How is this result useful? We can use the uniqueness part to find geodesics.We can try to find some family of curves C that are length-minimizing. To provethat we have found all of them, we can show that given any point P ∈ V anddirection a, there is some curve in C through P with direction a.

Example. Consider the sphere S2. Recall that arcs of great circles are length-minimizing, at least locally. So these are indeed geodesics. Are these all? Weknow for any P ∈ S2 and any tangent direction, there exists a unique greatcircle through P in this direction. So there cannot be any other geodesics on S2,by uniqueness.

Similarly, we find that hyperbolic line are precisely all the geodesics on ahyperbolic plane.

We have defined these geodesics as solutions of certain ODEs. It is possibleto show that the solutions of these ODEs depend C∞-smoothly on the initialconditions. We shall use this to construct around each point P ∈ S in a surfacegeodesic polar coordinates. The idea is that to specify a point near P , we canjust say “go in direction θ, and then move along the corresponding geodesic fortime r”.

We can make this (slightly) more precise, and provide a quick sketch of howwe can do this formally. We let ψ : U → V be some chart with P ∈ U ⊆ S. Wewlog ψ(P ) = 0 ∈ V ⊆ R2. We denote by θ the polar angle (coordinate), definedon V \ {0}.

θ

Then for any given θ, there is a unique geodesic γθ : (−ε, ε) → V such thatγθ(0) = 0, and γθ(0) is the unit vector in the θ direction.

We defineσ(r, θ) = γθ(r)

whenever this is defined. It is possible to check that σ is C∞-smooth. Whilewe would like to say that σ gives us a parametrization, this is not exactly true,since we cannot define θ continuously. Instead, for each θ0, we define the region

Wθ0 = {(r, θ) : 0 < r < ε, θ0 < θ < θ0 + 2π} ⊆ R2.

Writing V0 for the image of Wθ0 under σ, the composition

Wθ0 V0 U0 ⊆ Sσ ψ−1

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is a valid parametrization. Thus σ−1 ◦ ψ is a valid chart.The image (r, θ) of this chart are the geodesic polar coordinates. We have

the following lemma:

Lemma (Gauss’ lemma). The geodesic circles {r = r0} ⊆W are orthogonal totheir radii, i.e. to γθ, and the Riemannian metric (first fundamental form) on Wis

dr2 +G(r, θ) dθ2.

This is why we like geodesic polar coordinates. Using these, we can put theRiemannian metric into a very simple form.

Of course, this is just a sketch of what really happens, and there are manyholes to fill in. For more details, go to IID Differential Geometry.

Definition (Atlas). An atlas is a collection of charts covering the whole surface.

The collection of all geodesic polars about all points give us an example.Other interesting atlases are left as an exercise on example sheet 3.

5.3 Surfaces of revolution

So far, we do not have many examples of surfaces. We now describe a nice wayof obtaining surfaces — we obtain a surface S by rotating a plane curve η arounda line `. We may wlog assume that coordinates a chosen so that ` is the z-axis,and η lies in the x− z plane.

More precisely, we let η : (a, b)→ R3, and write

η(u) = (f(u), 0, g(u)).

Note that it is possible that a = −∞ and/or b =∞.We require ‖η′(u)‖ = 1 for all u. This is sometimes known as parametrization

by arclength. We also require f(u) > 0 for all u > 0, or else things won’t makesense.

Finally, we require that η is a homeomorphism to its image. This is morethan requiring η to be injective. This is to eliminate things like

Then S is the image of the following map:

σ(u, v) = (f(u) cos v, f(u) sin v, g(u))

for a < u < b and 0 ≤ v ≤ 2π. This is not exactly a parametrization, since it isnot injective (v = 0 and v = 2π give the same points). To rectify this, for eachα ∈ R, we define

σα : (a, b)× (α, α+ 2π)→ S,

given by the same formula, and this is a homeomorphism onto the image. Theproof of this is left as an exercise for the reader.

Assuming this, we now show that this is indeed a parametrization. It isevidently smooth, since f and g both are. To show this is a parametrization,

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we need to show that the partial derivatives are linearly independent. We cancompute the partial derivatives and show that they are non-zero. We have

σu = (f ′ cos v, f ′ sin v, g′)

σv = (−f sin v, f cos v, 0).

We then compute the cross product as

σu × σv = (−fg′ cos v,−fg′ sin v, ff ′).

So we have‖σu × σv‖ = f2(g′2 + f ′2) = f2 6= 0.

Thus every σα is a valid parametrization, and S is a valid embedded surface.More generally, we can allow S to be covered by several families of parametriza-

tions of type σα, i.e. we can consider more than one curve or more than one axisof rotation. This allows us to obtain, say, S2 or the embedded torus (in the oldsense, we cannot view S2 as a surface of revolution in the obvious way, since wewill be missing the poles).

Definition (Parallels). On a surface of revolution, parallels are curves of theform

γ(t) = σ(u0, t) for fixed u0.

Meridians are curves of the form

γ(t) = σ(t, v0) for fixed v0.

These are generalizations of the notions of longitude and latitude (in someorder) on Earth.

In a general surface of revolution, we can compute the first fundamental formwith respect to σ as

E = ‖σu‖2 = f ′2 + g′2 = 1,

F = σu · σv = 0

G = ‖σv‖2 = f2.

So its first fundamental form is also of the simple form, like the geodesic polarcoordinates.

Putting these explicit expressions into the geodesic formula, we find that thegeodesic equations are

u = fdf

duv2

d

dt(f2v) = 0.

Proposition. We assume ‖γ‖ = 1, i.e. u2 + f2(u)v2 = 1.

(i) Every unit speed meridians is a geodesic.

(ii) A (unit speed) parallel will be a geodesic if and only if

df

du(u0) = 0,

i.e. u0 is a critical point for f .

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Proof.

(i) In a meridian, v = v0 is constant. So the second equation holds. Also, weknow ‖γ‖ = |u| = 1. So u = 0. So the first geodesic equation is satisfied.

(ii) Since o = ou, we know f(u0)2v2 = 1. So

v = ± 1

f(u0).

So the second equation holds. Since v and f are non-zero, the first equationis satisfied if and only if df

du = 0.

5.4 Gaussian curvature

We will next consider the notion of curvature. Intuitively, Euclidean space is“flat”, while the sphere is “curved”. In this section, we will define a quantityknown as the curvature that characterizes how curved a surface is.

The definition itself is not too intuitive. So what we will do is that we firststudy the curvature of curves, which is something we already know from, say, IAVector Calculus. Afterwards, we will make an analogous definition for surfaces.

Definition (Curvature of curve). We let η : [0, `]→ R2 be a curve parametrizedwith unit speed, i.e. ‖η′‖ = 1. The curvature κ at the point η(s) is determinedby

η′′ = κn,

where n is the unit normal, chosen so that κ is non-negative.

If f : [c, d]→ [0, `] is a smooth function and f ′(t) > 0 for all t, then we canreparametrize our curve to get

γ(t) = η(f(t)).

We can then find

γ(t) =df

dtη′(f(t)).

So we have

‖γ‖2 =

(df

dt

)2

.

We also have by definitionη′′(f(t)) = κn,

where κ is the curvature at γ(t).On the other hand, Taylor’s theorem tells us

γ(t+ ∆t)− γ(t) =

(df

dt

)η′(f(t))∆t

+1

2

[(d2f

dt2

)η′(f(t)) +

(df

dt

)2

η′′(f(t))

]+ higher order terms.

Now we know by assumption that

η′ · η′ = 1.

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Differentiating thus give sη′ · η′′ = 0.

Hence we getη′ · n = 0.

We now take the dot product of the Taylor expansion with n, killing off all theη′ terms. Then we get

(γ(t+ ∆t)− γ(t)) · n =1

2κ‖γ‖2(∆t)2 + · · · , (∗)

where κ is the curvature. This is the distance denoted below:

γ(t)

γ(t+ ∆t)

(γ(t+ ∆t)− γ(t)) · n

We can also compute

‖γ(t+ ∆t)− γ(t)‖2 = ‖γ‖2(∆t)2. (†)

So we find that 12κ is the ratio of the leading (quadratic) terms of (∗) and (†),

and is independent of the choice of parametrization.We now try to apply this thinking to embedded surfaces. We let σ : V →

U ⊆ S be a parametrization of a surface S (with V ⊆ R2 open). We applyTaylor’s theorem to σ to get

σ(u+ ∆u, v + ∆v)− σ(u, v) = σ(u, v) + σu∆u+ σv∆v

+1

2(σuu(∆u2) + 2σuv∆u∆v + σvv(∆v)2) + · · · .

We now measure the deviation from the tangent plane, i.e.

(σ(u+ ∆u, v + ∆v)− σ(u, v)) ·N =1

2(L(∆u)2 + 2M∆U∆v +N(∆v)2) + · · · ,

where

L = σuu ·N,

M = σuv ·N,

N = σvv ·N.

Note that N and N are different things. N is the unit normal, while N is theexpression given above.

We can also compute

‖σ(u+ ∆u, v + ∆v)− σ(u, v)‖2 = E(∆u)2 + 2F∆u∆v +G(∆v)2 + · · · .

We now define the second fundamental form as

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Definition (Second fundamental form). The second fundamental form on Vwith σ : V → U ⊆ S for S is

L du2 + 2M du dv +N dv2,

where

L = σuu ·NM = σuv ·NN = σvv ·N.

Definition (Gaussian curvature). The Gaussian curvature K of a surface of Sat P ∈ S is the ratio of the determinants of the two fundamental forms, i.e.

K =LN −M2

EG− F 2.

This is valid since the first fundamental form is positive-definite and in particularhas non-zero derivative.

We can imagine that K is a capital κ, but it looks just like a normal capitalK.

Note that K > 0 means the second fundamental form is definite (i.e. eitherpositive definite or negative definite). If K < 0, then the second fundamentalform is indefinite. If K = 0, then the second fundamental form is semi-definite(but not definite).

Example. Consider the unit sphere S2 ⊆ R3. This has K > 0 at each point.We can compute this directly, or we can, for the moment, pretend that M = 0.Then by symmetry, N = M . So K > 0.

On the other hand, we can imagine a Pringle crisp (also known as a hyperbolicparaboloid), and this has K < 0. More examples are left on the third examplesheet. For example we will see that the embedded torus in R3 has points atwhich K > 0, some where K < 0, and others where K = 0.

It can be deduced, similar to the curves, that K is independent of parametriza-tion.

Recall that around each point, we can get some nice coordinates where thefirst fundamental form looks simple. We might expect the second fundamentalform to look simple as well. That is indeed true, but we need to do somepreparation first.

Proposition. We let

N =σu × σv‖σu × σv‖

be our unit normal for a surface patch. Then at each point, we have

Nu = aσu + bσv,

Nv = cσu + dσv,

where

−(L MM N

)=

(a bc d

)(E FF G

).

In particular,K = ad− bc.

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Proof. Note thatN ·N = 1.

Differentiating givesN ·Nu = 0 = N ·Nv.

Since σu, σv and N for an orthogonal basis, at least there are some a, b, c, d suchthat

Nu = aσu + bσv

Nv = cσu + dσv.

By definition of σu, we haveN · σu = 0.

So differentiating givesNu · σu + N · σuu = 0.

So we knowNu · σu = −L.

Similarly, we find

Nu = σv = −M = Nv · σu, Nv · σv = −N.

We dot our original definition of Nu,Nv in terms of a, b, c, d with σu and σv toobtain

−L = aE + bF −M = aF + bG

−M = cE + dF −N = cF + dG.

Taking determinants, we get the formula for the curvature.

If we have nice coordinates on S, then we get a nice formula for the Gaussiancurvature K.

Theorem. Suppose for a parametrization σ : V → U ⊆ S ⊆ R3, the firstfundamental form is given by

du2 +G(u, v) dv2

for some G ∈ C∞(V ). Then the Gaussian curvature is given by

K =−(√G)uu√G

.

In particular, we do not need to compute the second fundamental form of thesurface.

This is purely a technical result.

Proof. We set

e = σu, f =σv√G.

Then e and f are unit and orthogonal. We also let N = e× f be a third unitvector orthogonal to e and f so that they form a basis of R3.

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Using the notation of the previous proposition, we have

Nu ×Nv = (aσu + bσv)× (cσu + dσv)

= (ad− bc)σu × σv= Kσu × σv= K√Ge× f

= K√GN.

Thus we know

K√G = (Nu ×Nv) ·N

= (Nu ×Nv) · (e× f)

= (Nu · e)(Nv · f)− (Nu · f)(Nv · e).

Since N · e = 0, we knowNu · e + N · eu = 0.

Hence to evaluate the expression above, it suffices to compute N · eu instead ofNu · e.

Since e · e = 1, we know

e · eu = 0 = e · ev.

So we can write

eu = αf + λ1N

ev = βf + λ2N.

Similarly, we have

fu = −αe + µ1N

fv = −βe + µ2N.

Our objective now is to find the coefficients µi, λi, and then

K√G = λ1µ2 − λ2µ1.

Since we know e · f = 0, differentiating gives

eu · f + e · fu = 0

ev · f + e · fv = 0.

Thus we getα = α, β = β.

But we have

α = eu · f = σuu ·σv√G

=

((σu · σv)u −

1

2(σu · σu)v

)1√G

= 0,

since σu · σv = 0, σu · σu = 1. So α vanishes.

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Also, we have

β = ev · f = σuv ·σv√G

=1

2

Gu√G

= (√G)u.

Finally, we can use our equations again to find

λ1µ1 − λ2µ1 = eu · fv − ev · fu= (e · fv)u − (e · fu)v

= −βu − (−α)u

= −(√G)uu.

So we haveK√G = −(

√G)uu,

as required. Phew.

Observe, for σ as in the previous theorem, K depends only on the firstfundamental from, not on the second fundamental form. When Gauss discoveredthis, he was so impressed that he called it the Theorema Egregium, which means

Corollary (Theorema Egregium). If S1 and S2 have locally isometric charts,then K is locally the same.

Proof. We know that this corollary is valid under the assumption of the previoustheorem, i.e. the existence of a parametrization σ of the surface S such that thefirst fundamental form is

du2 +G(u, v) dv2.

Suitable σ includes, for each point P ∈ S, the geodesic polars (ρ, θ). However,P itself is not in the chart, i.e. P 6∈ σ(U), and there is no guarantee that therewill be some geodesic polar that covers P . To solve this problem, we notice thatK is a C∞ function of S, and in particular continuous. So we can determine thecurvature at P as

K(P ) = limρ→0

K(ρ, σ).

So done.Note also that every surface of revolution has such a suitable parametrization,

as we have previously explicitly seen.

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6 Abstract smooth surfaces IB Geometry

6 Abstract smooth surfaces

While embedded surfaces are quite general surfaces, they do not include, say,the hyperbolic plane. We can generalize our notions by considering surfaces“without embedding in R3”. These are known as abstract surfaces.

Definition (Abstract smooth surface). An abstract smooth surface S is a metricspace (or Hausdorff (and second-countable) topological space) equipped withhomeomorphisms θi : Ui → Vi, where Ui ⊆ S and Vi ⊆ R2 are open sets suchthat

(i) S =⋃i Ui

(ii) For any i, j, the transition map

φij = θj ◦ θ−1i : θj(Ui ∩ Uj)→ θi(Ui ∩ Uj)

is a diffeomorphism. Note that θj(Ui ∩ Uj) and θi(Ui ∩ Uj) are opensets in R2. So it makes sense to talk about whether the function is adiffeomorphism.

Like for embedded surfaces, the maps θi are called charts, and the collectionof θi’s satisfying our conditions is an atlas etc.

Definition (Riemannian metric on abstract surface). A Riemannian metric onan abstract surface is given by Riemannian metrics on each Vi = θi(Ui) subject tothe compatibility condition that for all i, j, the transition map φij is an isometry,i.e.

〈dϕP (a),dϕP (b)〉ϕ(P ) = 〈a,b〉PNote that on the left, we are computing the Riemannian metric on Vi, while onthe left, we are computing it on Vj .

Then we can define lengths, areas, energies on an abstract surface S.It is clear that every embedded surface is an abstract surface, by forgetting

that it is embedded in R3.

Example. The three classical geometries are all abstract surfaces.

(i) The Euclidean space R2 with dx2 + dy2 is an abstract surface.

(ii) The sphere S2 ⊆ R2, being an embedded surface, is an abstract surfacewith metric

4(dx2 + dy2)

(1 + x2 + y2)2.

(iii) The hyperbolic disc D ⊆ R2 is an abstract surface with metric

4(dx2 + dy2)

(1− x2 − y2)2.

and this is isometric to the upper half plane H with metric

dx2 + dy2

y2

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Note that in the first and last example, it was sufficient to use just one chartto cover every point of the surface, but not for the sphere. Also, in the case of thehyperbolic plane, we can have many different charts, and they are compatible.

Finally, we notice that we really need the notion of abstract surface for thehyperbolic plane, since it cannot be realized as an embedded surface in R3. Theproof is not obvious at all, and is a theorem of Hilbert.

One important thing we can do is to study the curvature of surfaces.Given a P ∈ S, the Riemannian metric (on a chart) around P determinesa “reparametrization” by geodesics, similar to embedded surfaces. Then themetric takes the form

dρ2 +G(ρ, θ) dθ2.

We then define the curvature as

K =−(√G)ρρ√G

.

Note that for embedded surfaces, we obtained this formula as a theorem. Forabstract surfaces, we take this as a definition.

We can check how this works in some familiar examples.

Example.

(i) In R2, we use the usual polar coordinates (ρ, θ), and the metric becomes

dρ2 + ρ2 dθ2,

where x = ρ cos θ and y = ρ sin θ. So the curvature is

−(√G)ρρ√G

=−(ρ)ρρρ

= 0.

So the Euclidean space has zero curvature.

(ii) For the sphere S, we use the spherical coordinates, fixing the radius to be1. So we specify each point by

σ(ρ, θ) = (sin ρ cos θ, sin ρ sin θ, cos ρ).

Note that ρ is not really the radius in spherical coordinates, but just oneof the angle coordinates. We then have the metric

dρ2 + sin2 ρ dθ2.

Then we get √G = sin ρ,

and K = 1.

(iii) For the hyperbolic plane, we use the disk model D, and we first expressour original metric in polar coordinates of the Euclidean plane to get(

2

1− r2

)2

(dr2 + r2 dθ2).

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This is not geodesic polar coordinates, since r is given by the Euclideandistance, not hyperbolic distance. We will need to put

ρ = 2 tanh−1 r, dρ =2

1− r2dr.

Then we haver = tanh

ρ

2,

which gives4r2

(1− r2)2= sinh2 ρ.

So we finally get √G = sinh ρ,

withK = −1.

We see that the three classic geometries are characterized by having constant0, 1 and −1 curvatures.

We are almost able to state the Gauss-Bonnet theorem. Before that, we needthe notion of triangulations. We notice that our old definition makes sense for(compact) abstract surfaces S. So we just use the same definition. We thendefine the Euler number of an abstract surface as

e(S) = F − E + V,

as before. Assuming that the Euler number is independent of triangulations, weknow that this is invariant under homeomorphisms.

Theorem (Gauss-Bonnet theorem). If the sides of a triangle ABC ⊆ S aregeodesic segments, then∫

ABC

K dA = (α+ β + γ)− π,

where α, β, γ are the angles of the triangle, and dA is the “area element” givenby

dA =√EG− F 2 du dv,

on each domain U ⊆ S of a chart, with E,F,G as in the respective firstfundamental form.

Moreover, if S is a compact surface, then∫S

K dA = 2πe(S).

We will not prove this theorem, but we will make some remarks. Note thatwe can deduce the second part from the first part. The basic idea is to take atriangulation of S, and then use things like each edge belongs to two trianglesand each triangle has three edges.

This is a genuine generalization of what we previously had for the sphereand hyperbolic plane, as one can easily see.

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Using the Gauss-Bonnet theorem, we can define the curvature K(P ) for apoint P ∈ S alternatively by considering triangles containing P , and then takingthe limit

limarea→0

(α+ β + γ)− πarea

= K(P ).

Finally, we note how this relates to the problem of the parallel postulate we havementioned previously. The parallel postulate, in some form, states that given aline and a point not on it, there is a unique line through the point and parallelto the line. This holds in Euclidean geometry, but not hyperbolic and sphericalgeometry.

It is a fact that this is equivalent to the axiom that the angles of a trianglesum to π. Thus, the Gauss-Bonnet theorem tells us the parallel postulate iscaptured by the fact that the curvature of the Euclidean plane is zero everywhere.

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Part IB - Geometry - SRCF · 2020-01-24 · Part IB | Geometry Based on lectures by A. G. Kovalev...

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