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4P9

An Introduction to Differential Geometry.

Michael D. Alder

November 29, 2008

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Contents

1 Introduction 9

2 Smooth Manifolds and Vector Fields 11

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2 Smooth Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.3 Smooth maps and tangent vectors . . . . . . . . . . . . . . . . 15

2.4 Notation: Vector Fields . . . . . . . . . . . . . . . . . . . . . . 27

2.5 Cotangent Bundles . . . . . . . . . . . . . . . . . . . . . . . . 31

2.6 The Tangent Functor . . . . . . . . . . . . . . . . . . . . . . . 32

2.6.1 The (non-existent) Cotangent Functor . . . . . . . . . 35

2.7 Autonomous Systems of ODEs . . . . . . . . . . . . . . . . . . 37

2.7.1 Systems of ODEs and Vector Fields . . . . . . . . . . . 37

2.7.2 Exponentiation ofThings . . . . . . . . . . . . . . . . 39

2.7.3 Solving Linear Autonomous Systems . . . . . . . . . . 40

2.7.4 Existence and Uniqueness . . . . . . . . . . . . . . . . 41

2.8 Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2.9 Lie Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.10 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3 Tensors and Tensor Fields 51

3.1 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.1.1 Natural and Unnatural Isomorphisms . . . . . . . . . . 51

3.1.2 Multilinearity . . . . . . . . . . . . . . . . . . . . . . . 54

3.1.3 Dimension of Tensor spaces . . . . . . . . . . . . . . . 57

3.1.4 The Tensor Algebra . . . . . . . . . . . . . . . . . . . . 62

3.2 Tensor Fields on a Manifold . . . . . . . . . . . . . . . . . . . 67

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4 CONTENTS

3.3 The Riemannian Metric Tensor . . . . . . . . . . . . . . . . . 72

3.3.1 What this means: Ancient History . . . . . . . . . . . 76

3.4 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

3.5 The Exterior Algebra . . . . . . . . . . . . . . . . . . . . . . . 91

3.6 The Exterior Calculus . . . . . . . . . . . . . . . . . . . . . . 96

3.7 Hodge Duality: The Hodge Operator . . . . . . . . . . . . . 102

3.7.1 The Riemannian Case . . . . . . . . . . . . . . . . . . 102

3.7.2 The SemiRiemannian Case . . . . . . . . . . . . . . . . 106

4 Some Elementary Physics 109

4.1 Three weird forces . . . . . . . . . . . . . . . . . . . . . . . . 109

4.2 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

4.2.1 Gradient Fields . . . . . . . . . . . . . . . . . . . . . . 116

4.2.2 What are Flux? . . . . . . . . . . . . . . . . . . . . . . 117

4.3 Maxwell and Faraday . . . . . . . . . . . . . . . . . . . . . . . 120

4.4 Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

4.4.1 The Idea of Invariance . . . . . . . . . . . . . . . . . . 126

4.4.2 The Lorentz Group . . . . . . . . . . . . . . . . . . . . 130

4.4.3 The Maxwell Equations . . . . . . . . . . . . . . . . . 1354.5 Saying it with Differential Forms . . . . . . . . . . . . . . . . 136

4.6 Lorentz Invariance . . . . . . . . . . . . . . . . . . . . . . . . 141

4.6.1 Special Relativity . . . . . . . . . . . . . . . . . . . . . 145

5 DeRham Cohomology: Counting holes 149

5.1 Cultural Anthropology . . . . . . . . . . . . . . . . . . . . . . 149

5.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

5.3 Infinite Variety . . . . . . . . . . . . . . . . . . . . . . . . . . 155

5.4 Gauge Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . 1565.5 Exact and Closed forms . . . . . . . . . . . . . . . . . . . . . 157

5.6 Homotopies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

5.7 Counting Holes . . . . . . . . . . . . . . . . . . . . . . . . . . 165

5.8 More Cultural Anthropology . . . . . . . . . . . . . . . . . . . 166

5.9 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

6 Lie Groups 169

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CONTENTS 5

6.1 Introduction and Motivation . . . . . . . . . . . . . . . . . . . 169

6.1.1 The rest of the course . . . . . . . . . . . . . . . . . . 169

6.2 Introduction to Lie Groups . . . . . . . . . . . . . . . . . . . . 169

6.3 Group Representations . . . . . . . . . . . . . . . . . . . . . . 173

6.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 173

6.3.2 Irreducible Representations . . . . . . . . . . . . . . . 176

6.3.3 Tensor Representations . . . . . . . . . . . . . . . . . . 177

6.3.4 Schurs Lemma . . . . . . . . . . . . . . . . . . . . . . 178

6.3.5 Representations of SU(2, C) . . . . . . . . . . . . . . . 179

6.3.6 Representations of SU(2) . . . . . . . . . . . . . . . . . 1826.4 Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

7 Fibre Bundles 185

7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

7.2 Principal Bundles . . . . . . . . . . . . . . . . . . . . . . . . . 188

7.3 The Endomorphism Bundle . . . . . . . . . . . . . . . . . . . 191

8 Connections 193

8.1 Fundamental Ideas . . . . . . . . . . . . . . . . . . . . . . . . 193

8.2 Back in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197

8.2.1 Covariant differentiation . . . . . . . . . . . . . . . . . 197

8.2.2 Curves and transporting vectors . . . . . . . . . . . . . 199

8.3 Covariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

8.4 Extensions to Tensor Fields on R2 . . . . . . . . . . . . . . . . 201

8.5 The Koszul Connection . . . . . . . . . . . . . . . . . . . . . . 203

8.6 Vector Potentials . . . . . . . . . . . . . . . . . . . . . . . . . 203

8.6.1 Tensor formulation . . . . . . . . . . . . . . . . . . . . 206

8.7 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . 207

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Preface

This is a first course in Differential Geometry. I follow a number of sources:first the text for the course, Baez and Muniains Gauge Fields, Knots andGravity, second the unique Michael Spivakss Comrehensive Introduction toDifferential Geometry, which is almost encyclopaedic and also readable, if attimes demanding. Third R.W.R Darlings Differential Forms and Connec-tions and finally the rather old fashioned Sternberg Lectures in DifferentialGeometry. I shall also make some allusions to Helgasons Differential Geom-etry, Lie Groups and Symmetric Spaces.

My aim is to cover some of the ideas with applications to theoretical physicsfrom the text book while covering basic ideas. I hope that the students willhave done my 3P0 course which introduces tensors and tensor fields, but itseems unsafe to count on it having been absorbed as thoroughly as desired.So some of the introductory material has been lifted from my 3P0 notes.

Mike Alder, February 2007

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Chapter 1

Introduction

This course is about Differential Geometry and the text book is really im-portant. You need your own copy unless you are sharing with a very goodfriend. It is particularly important if you want to see where the physicalreasons for studying this topic tie in with the mathematics. We shant getthrough the whole book but we shall have started on the journey.

I copied some of that from the introduction to 3P0. Its still true.

You can see what is in the course by reading the contents page. Not thatit will help; this is Mathematics, not one of those subjects where you learn

to say the right things without considering whether they are true or false oreven whether or not they mean anything.

There are many different reasons for studying this subject but one impor-tant one is that you will come to grips with some of the ideas that modernPhysics needs to make sense of the universe. You may be a physicist or amathematician or even an engineer (embryonically at least). The three sub-jects tend to attract slighly different kinds of people (only slightly different:compared with poets, pop-stars, princes, politicians and philosophers we arebarely different at all). Engineers tend to see the world in terms of facts andprotocols which they have to learn and which may or may not make much

sense, physicists see the world in terms of facts and theories, the theoriesbeing there to summarise and predict the facts. Mathematicians expect tosee reasons and logic and relatively few basic facts from which the otherscan be deduced. For a mathematician it has to make sense or it is definitelywrong. For a theoretical physicist it has to be elegant or it is definitely wrong.Because I am a mathematician, I have put in material which is left out ofthe text book where it is treated as some bunch of facts which you need toknow, whereas I want to show why these things are the case. Maybe I justhave a bad memory for facts but a good one for arguments. Whatever the

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10 CHAPTER 1. INTRODUCTION

reason, I am going to try to show you the essential beauty of the subject,

to get you to agree that it is amazingly cool, because this is ultimately whymathematicians do it. The fact that it is also very useful is not why we doit, it is why we get paid to do it. Though not much 1.

This is very tough stuff, so dont expect an easy time. On the other hand itwill be very exciting.

1This is a really bad subject to do if you want to get rich, or boss people about, but avery good one if you want to be happy and have lots of interesting and important thingsto think about.

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Chapter 2

Smooth Manifolds and Vector

Fields

2.1 Introduction

This chapter considers the machinery needed to say what we mean by asmooth manifold. We also look at vector fields on smooth manifolds andexplain what this has to do with systems of ordinary differential equations.

The first idea is that a curve (in the plane or in three space) is a one di-mensional object, a surface such as a sphere (the surface of a beach ball) ora torus (the surface of an American doughnut where they sell you a hole inthe middle) is a two dimensional object. And there ought reasonably to behigher dimensional variants of these things, as for example the n-sphere Sn

given bySn {x Rn+1 : x = 1}

It is also reasonable to look at smooth maps between manifolds. If you draw asmooth curve on a beach ball without stopping, which joins up to stop whereit starts and has the same final and initial velocity, then we could think of

this as a smooth map from S1 to S2. But we have a problem in dealingwith what we mean by a differentiable map in this case since neither S1 norS2 are Banach spaces, and Banach spaces are the setting for talking aboutdifferentiation, since they contain the linear algebra and distance notionsneeded to talk of linear approximations, which is what derivatives are.

We might be able to make sense of this if the sphere is sitting in R3, and thecircle in R2, in which case we have a way of defining smoothness extrinsically.But smoothness ought to make sense intrinsically, that is without referenceto some external space in which the manifold may or may not be sitting. An

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12 CHAPTER 2. SMOOTH MANIFOLDS AND VECTOR FIELDS

important reason for this is that we live in what looks to be a 3-manifold

called the physical universe, at least at some scales. Make it a 4-manifold ifyou want to throw in time. If the universe is sitting in some higher dimen-sional space, we cant know much about it, so it is slightly lunatic to believeit is there. Google branes for an alternative viewpoint. Also string theory forsome disturbing ideas. But these do not contradict the idea that if we livein a three dimensional physical universe it makes sense to talk about smoothmotion in it without having to postulate some inaccessible space external tothe universe.

So we seek to specify enough extra structure on a manifold so that we cantalk about smooth maps between them without reference to any space in

which they may be sitting. This will certainly be necessary if we are tosuppose that we live in a 3-manifold and want to talk about geodesics in it,curves of minimal length. We shall certainly want to do this if we are to talkof the path of a photon in our universe.

All this means generalising ideas about maps from Rn, or subsets of Rn, toRm, which involve differentiability. Which we understand. Or do we?

Recall that if U, V are open subsets of Rn and f : U V is a differentiablemap we have that at each point a U, the derivative of f at a is the linearmap Df(a) : Rn Rn which is represented in the standard basis by then

n matrix of partial derivatives:

[Dfa] =

D1f

1(a) D2f1(a) Dnf1(a)

D1f2(a) D2f

2(a) Dnf2(a)...

... ...D1f

n(a) D2fn(a) Dnfn(a)

=

f1

x1f1

x2 f1

xnf2

x1f2

x2 f2

xn...

...

...

fnx1

fnx2 fnxn

x=a

Usually I shant bother to distinguish between the linear map and its matrixrepresentation. You know how to compute this matrix if it should be abso-lutely necessary, and you should understand that the linear map is the linearpart of the best affine approximation to f at a. Note that I have used fi forthe n component functions off and xi for the components of a vector in Rn.I shall explain this notation later.

An important point about smooth curves needs to be considered:

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2.2. SMOOTH MANIFOLDS 13

Figure 2.1.1: A smooth curve.

Exercise 2.1.1. Figure 2.1.1 shows two line segments joined together. Thehorizontal one is the set of points in R2 with y = 0 and 0 x 1 and thevertical one is the set of points in R2 with x = 1 and 0 y 1Show that there is a continuous but non-differentiable function from [0, 2] toR2 which traces the curve formed by the two segments from the origin to thepoint (1, 1)T.

Show that there is a differentiable function from [0, 2] to R2 which does thesame job.

Exercise 2.1.2. Show that [1, 1] is the image of [1, 1] by a continuousbijection which is not differentiable.

The conclusion you should draw from this is that you cannot decide if a curveis smooth or not merely by looking at the image!

2.2 Smooth Manifolds

Definition 2.2.1. A chart on a topological space X is a homeomorphismfrom some open subset of X onto an open subset of Rn. I shall call theinverse of such a homeomorphism a local parametrisation.

We show a typical local parametrisation map from a rectangular neighbour-hood of the origin in R2 to a region on the surface in figure 2.2.1. A chartcan be used to give coordinates for points of the space, at least some of them.Different charts will, of course, give different coordinates in general to thosepoints on which the domains overlap.

Definition 2.2.2. Two charts on a space X, f : U Rn and g : V Rn aresmoothly compatible iff the maps fg1 and gf1 are infinitely differentiablewherever they are defined.

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Figure 2.2.1: A local coordinate map.

Figure 2.2.2: Two charts.

In other words, the composite map f g1 must have partial derivatives ofall orders at every point of the domain, and the same is true of the inversemap.

IfU and V have empty intersection then this holds vacuously. If they do havean intersection, then f g1 has domain and codomain some open subsetsof Rn and is certainly continuous. It makes sense to demand that this map

be smooth, that is, infinitely differentiable. The picture of figure 2.2.2 mayhelp.

Definition 2.2.3. A smooth atlas for a space X is a collection of smoothlycompatible charts such that every point of X is in the domain of at leastone chart. Such an atlas is maximal iff every possible (smoothly compatible)chart is in it.

Definition 2.2.4. A smooth n-manifold is a hausdorff topogical space to-gether with a maximal atlas of smoothly compatible charts. The atlas is said

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2.3. SMOOTH MAPS AND TANGENT VECTORS 15

to define a smooth differential structure on X.

The reason for wanting the atlas to be maximal is just so that anyone wan-dering in with a new local coordinate map cant cause us trouble. Either itis compatible with our atlas in which case we already have it, or it is not, inwhich case it may be part of a different differential structure for the manifold.

Exercise 2.2.1.

1. Show that S1 and S2 as usually defined are smooth manifolds.

2. Show that the flat torus obtained by gluing opposite edges of a squareis also a smooth manifold.

3. Show that Sn is a smooth manifold for any n Z+. Hint: use theImplicit Function Theorem.

(Generic hint: you dont need to have many charts. Enough to cover themanifold will do, then just add the instruction to fill up with all other possiblesmoothly compatible charts.)

Exercise 2.2.2. Construct a definition of an orientable manifold.

2.3 Smooth maps and tangent vectors

Now we have enough to say what it means for a map f : X Y to besmooth when X and Y are smooth manifolds:

Definition 2.3.1. A map f : X Y between smooth manifolds is differen-tiable when h1 f g is differentiable for all charts h on X and all charts gon Y belonging to the differential structures.

The diagram of figure 2.3.1 gives the idea.

We can define higher order differentiability in the same way and we can saythat a map f : X Y is smooth whenever all composites h1 f g aresmooth for all charts h on X and all charts g on Y

Exercise 2.3.1. Show that if f : X Y has composite h1 f g differ-entiable at some point a in X then it is differentiable in any other pair ofcharts containing a, f(a).

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Figure 2.3.1: A smooth map.

Figure 2.3.2: Some tangent curves in a manifold.

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Note that although we can say that f is differentiable, we cannot provide

a derivative, since this will generally be different in different charts. If wemove away from simple linear spaces we must pay the price: there is no longera best affine approximation because affine maps dont make sense betweenmanifolds in general.

We can however say when two maps from R into a manifold X are tangent.Let f, g : (1, 1) X be smooth maps into a manifold and without lossof generality let f(0) = g(0) = a X. Then we can say that f and g aretangent at a iff the derivative of f and the derivative of g are the same forany chart h : U Rn where f(0) = g(0) = a U. If they are the same inany one chart they must be the same in any other.

Exercise 2.3.2. Prove the last remark.

Exercise 2.3.3. Show that tangency is an equivalence relation on the set ofmaps from R to X, and that we can do the same thing with maps from Xto R.

We can take a tangency equivalence class of maps from R to the manifoldX, and regard it as an object in its own right. The picture 2.3.2 shows somemembers of a tangency equivalence class.

The curves can be thought of as the trajectories of moving points, and they

are all moving through the point a at the same speed, and in the samedirection, although we cannot give the direction a particular vector to specifyit, and the speed may also be different in different charts.

Definition 2.3.2. A tangency equivalence class at a point a in a manifoldX is called a tangent vector at a in X.

Remark 2.3.1. Watching the faces of students in class when giving thisdefinition is a real treat. The look of stark horror and incomprehensionis very encouraging, as it proves that some at least are listening. A smallamount of imagination, however, goes a long way to making this definition

quite reasonable.Suppose that the North pole has been cleared of snow and turned into askating rink for penguins1, and the North pole itself is marked by a flashingred light. There are two space craft hovering up there, call them A and B. Inspace craft A, an astronaut leans out and takes a photograph of the regionaround the north pole. Suppose for simplicity he is directly above the north

1It has been pointed out to me that there are no penguins at the North pole only atthe South pole. On the other hand there isnt a skating rink at the North pole either. Soif we are going to make a skating rink we might as well import the penguins.

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Figure 2.3.3: Penguins (imported from the antarctic) skating.

pole so his photograph, when enlarged is a disc as in the picture figure 2.3.4.Astronaut B is somewhere over Russia and he also takes a photograph ofwhat he can see.

Now each astronaut looks at his photograph and lays it out flat and enlarges

it to a nice size, and each marks on a coordinate grid using a ruler and pen,and so each has a chart of a bit of the polar regions, with the north polein the domain of each chart. If both put the origin in the centre, astronautA will have the flashing red light at the origin, and astronaut B will have anegative x coordinate for the red light if he puts his coordinates on the chartin the way suggested by the diagram. I regard the chart as both the bit ofthe earth the astronaut can see, and also the process of turning it into a flatpicture with a coordinate grid on it. Call them u and v for the maps and Uand V for the domains of the maps back on earth.

I claim it makes sense to talk of a penguin skating over the north pole as

having a velocity vector as it passes through the north pole. Each astronautcan plot the position of the green penguin in his chart, and each will agree ifthe curve is differentiable. Note that if g : (1, 1) S2 describes the greenpenguin then astronaut A will plot the green curve at the top of the pictureand will be able to give it a perfectly respectable velocity on his chart relativeto the cartesian coordinates marked on the chart. Similarly astronaut B cando the same. The problem is that they will have, usually, different estimatesor what the velocity is. If B is much higher up in space, his scale will be suchthat the penguins will seem to be moving more slowly, for instance.

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Figure 2.3.4: Two penguins skating under the watchful eyes of two astro-nauts.

Does this mean my claim that we can assign a meaningful velocity vector tothe penguin is just nonsense? No, for if b is the blue penguin, also skating

over the north pole at the same time as the green penguin (and mysteriouslynot knocking over the first penguin: maybe they are ghost penguins and canoccupy the same space), it certainly makes sense to say if they are travellingin the same direction at the same speed. A penguin cutting across the pathwould obviously be travelling in a different direction, and a really slow pen-guin would be slower for both astronauts, and the fast penguins would passthrough it. So I claim that the penguin velocity is a real thing which existsat the penguin level if not at the astronaut level. But if one astronaut saidthat the blue penguin and the green penguin had the same velocity at theinstant they went through the pole, the other astronaut would agree even

though disagreeing as to the actual value of the vector in both direction andspeed, these things being properties of the charts, not the penguins.

The reason this happens is that there are two things going on here. The actualvelocity at the north pole is a real thing, penguins are actually moving, andeither they pass through the north pole at the same time in the same directionat the same speed or they dont. But attempts by the two astronauts todescribe the penguin motion to each other with numbers involve inventingcoordinate systems which are bits of language. So the numerical value of anyvector is dependent on the language. But the fact that different languages

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agree on whether two penguins have the same velocity tells you that the

velocity is real. It exists independent of the coordinate system, provided thetwo coordinate systems are related by a diffeomorphism. So there are movingpenguins and there is language, and the penguins will have the same velocityat the pole or they wont, and this is true no matter what language you useto talk about it unless your language is really weird.

The problem then is to say what a velocity vector is given that any pair ofastronauts can disagree about the actual numbers. And the most elegantsolution is to say that it is what all the penguin trajectories, real and poten-tial, have in common. And what they have in common is that every observerwill agree that they pass through the north pole in the same direction at the

same speed. This is the tangency equivalence class.Note that I have assumed that all observers use synchronised clocks so theyall agree that the time at which the penguins hit the north pole is timezero. This doesnt have to be the case either. They will all agree on thesimultaneity of the events, whatever time they claim they occur. This isbecause two penguins either meet or they dont, and this is not a matter oflanguage but of fact.

The ghost-penguins are negotiable. Having a nice vivid picture of some sortis essential: you should be prepared to invent your own, but this time youmay borrow my penguins if they help. If I give you more definitions like this,

it is your job to supply the penguins, or whatever it takes.Exercise 2.3.4. Show that the claim that the two astronauts would agree iftwo penguins have the same velocity at the north pole is true provided thatu v1 and v u1 are both differentiable.Remark 2.3.2. There is, of course, a simpler way of defining tangent vectorson S2. It is usually viewed as a subspace of R3, so a curve on S2 is also acurve in R3 and we can define velocities on S2 as tangent vectors in R3 inthe sense of the derivatives of maps from (1, 1) to R3 which, for tangentvectors at a particular point, happen to lie in a plane in R3 which is tangentto S2 at that point. This certainly removes some tricky conceptual problemsbut at the expense of making tangent vectors extrinsic rather than intrinsicto the space. The whole thrust of the text book is to using intrinsic ideasfor the very good reason that we live in a 3-manifold and cannot form anyuseful idea of an embedding of it in some higher dimensional space.

The next proposition tells us that the set of tangency equivalence classes ata fixed point a in a manifold form a vector space, the tangent space at a.

Proposition 2.3.1. The set of tangent vectors at a point a of a smoothn-manifold X comprise a real vector space of dimension n.

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Figure 2.3.5: The sum of tangent vectors.

Proof:

We have to produce sensible rules for adding and scaling tangent vectors.Then we have to show that the result satisfies the axioms for a real vectorspace. Suppose we have a tangency equivalence class v and that v is anelement of it, that is a curve v : R X with v(0) = a and in any chartw : W Rn with a U there is some derivative of w v. Then we canscale the function v by a scalar k R to get v(kt) instead of v(t) for t Rand the derivative of w v will also be scaled by the factor k. This will bethe same scaling in any chart, so it makes sense to call this new function kv.This has its own tangency equivalence class, kv.

It would not make a difference if we had chosen another function v v,kv is a function tangent to kv since they both have the same derivative nomatter what chart we choose, although in different charts the derivatives willbe different but still equal to each other.

So we can say that kv exists and we have scaled the equivalence class.

If v and u are distinct tangency equivalence classes through the point aas in figure 2.3.5, we can take representative functions v, u : R

X with

u(0) = v(0) = a and composing with w : W Rn, a chart, we have twomaps, w u and w v from R into Rn. Such maps may be added: we takethe map w u + w v w(a). At t = 0 this passes through w(a) Rn. Theresulting curve in Rn can be mapped back into the manifold by w1, or atleast a bit of it in a neighbourhood of w(a) can be. This gives a sort of sumcurve ofu and v in the manifold, w1 (w u + w v w(a)). The tangencyclass of this sum curve is defined to be the sum of u and v. It is easy to seethat the tangency equivalence class does not depend on the choice of chart.(Although the sum curve does.)

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Nor does it depend on which representatives u of u and v of v which we

choose because they all have the same derivative. We may write u + v is thetangency equivalence class ofw1(w u) + (w v) w(a)) therefore, and wemay add tangency equivalence classes, otherwise known as tangent vectorsat a.

(If in doubt about this argument, say it with penguins.)

It is clear that the sum is associative and commutative and there is a zerowhich contains the constant function sending R to a. The rest of the axiomsfor a vector space are easily checked.

The claim that it has dimension n the same as the dimension of X is left asan exercise.

Exercise 2.3.5. Check all the axioms for a vector space. This kind of thingis called axiom bashing and is good for you.

The resulting vector space is called Ta(X) and is isomorphic to Rn when X

is an n-manifold. I want to emphasise an important point: there is in generalno particular or natural isomorphism between Ta(X) and R

n. If X = Rn,then I can get away with calling T0(R

n) by the slang name Rn, because inthis case there is an obvious basis for the tangent space at the origin, I havethe unit vectors along the axes. And by a simple translation I can carry Rn

to Rn and take the origin to any point a, and this translation will also takecurves through the origin (and hence vectors) to curves through a. So inthis rather special case, I do have a natural basis for Ta(X). But there is nonatural basis for Ta(S

2) for any a S2; the best I could do is to fudge one byusing the embedding in R3, but this is a property of the embedding, not ofS2. This loss of a natural basis, or if you prefer a natural isomorphism withRn, has important implications. It parallels the fact that there is no obviouschoice of a coordinate frame in the space we inhabit2.

Exercise 2.3.6. Prove the last statement. Hint: Do it for X = Rn first,

then observe that locally any X isRn

as near as dammit, and that tangencyis a very local kind of business.

There is one such tangent space Ta(X) for each point a X. There is, ingeneral, no particular isomorphism between any Ta(X) and R

n. If X = Rn

then there is (what?), but in general there is a huge choice and no way ofpicking any particular one.

2Although in earlier days, it was thought in some quarters that Jerusalem was a goodplace to put one. Where exactly in Jerusalem was not altogether clear.

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Figure 2.3.6: The simplest tangent bundle.

Exercise 2.3.7. Show that the tangent plane at the north pole to S2 asusually embedded in R3 can be mapped isomorphically to the tangent spaceas defined here. Is there an obvious isomorphism?

Exercise 2.3.8. Show that there is an isomorphism between Ta(X) andTb(X) for any two points a, b X.Examples 2.3.1.

1. The simplest case is where X = R. A tangent vector at the point 1can be thought of as the space of velocities of moving points as they gothrough 1; the chart consisting of the identity map does it all nicely.So we have a line of possible velocities attached to each point of R andthe tangent bundle is the collection of all the tangent spaces. We candraw it as R R where the first component is the space itself and thesecond is the space of velocities. I am making up the notation of Rfor the space of velocities, and you wont find it in the books, but itmakes sense and reduces confusion. Since R is isomorphic to R, R Rlooks an awful lot like R2. We think of the different tangent spaces Ra

attached to each a R and draw some of them as in figure 2.3.6The reason it is called a bundle is because it looks like a bundle of (red)tangent spaces. The tangent spaces are called the fibres of the bundle.The manifold to which the fibres are attached is called the base spaceof the fibre bundle.

2. Let the manifold X be S1. Again it makes sense to have curves in S1

which all pass through a point and have the same velocity vector at thatpoint. The different velocities again form a vector space R and there is

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Figure 2.3.7: The next simplest tangent bundle.

one attached to each point of S1. If we draw the possible tangents inthe plane, they intersect; this is a property of the space we are tryingto squash the tangent bundle into and if we turn them through a rightangle as in the last example we get the fibre bundle of figure 2.3.7

Again the fibres are all copies of a line and the bundle is pretty much the

same as S

1

R

. The red dot sitting over the black one represents a speedin the positive direction passing through the black point underneath it.

3. We have now run out of cases where we can draw the pictures, sinceR and S1 are the only one dimensional manifolds, and if we go to S2

we get a tangent bundle of dimension four. We can draw one tangentplane, but any more would usually intersect, and this is what happenswhen we try to embed a four dimensional space in R3. We can seehowever that there is a collection of planes, one for each point of S2

and they form a four dimensional space. It is useful to visualise at leasta part of the tangent space of S2 as a sphere in R3 with some bits of

tangent planes attached to it, as in figure 2.3.8, because it is better tohave a partial idea than stick entirely to the algebra, but you shouldbe aware of the limitations of the picture.

The two earlier examples came out to be simple cartesian products of thetangent space at any one point with the manifold. Such bundles are calledtrivial bundles. An example of a non-trivial fibre bundle is the Mobius bundleshown in figure 2.3.9

This has fibre an interval, say (1, 1), from R and base space S1. But it is

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Figure 2.3.8: A bit of the tangent bundle for S2.

Figure 2.3.9: A non-trivial fibre bundle.

not the cartesian product of the two.

Every tangent bundle has, however, a projection onto the base space, theunderlying manifold. We may write this as a vertical pair of spaces

TM

?

M

where M is the manifold, T M is the tangent bundle and is the projectionwhich sends a tangent vector to the point in the manifold to which it isattached.

Now we have described the tangent bundle as a union of all the tangentspaces Ta(M) for a M but that does not specify a topology on it. To dothat we say a subset U of T M is open iff the projection (U) is open in Mand the intersection ofU with any fibre is open in the fibre. Since the fibresare all real vector spaces we can give them the usual topology, obtained froman isomorphism with Rn.

Definition 2.3.3. The tangent bundle to a smooth manifold M is the setaM

Ta(M)

with the topology specified by saying U T M is open whenever (U) is

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open in M and for every a

(U), U

Ta(M) is open in Ta(M), where

Ta(M) has a topology induced by any isomorphism with Rn.

Note that this assumes that any two isomorphisms with Rn will induce thesame topology.

Exercise 2.3.9.

1. Show that a linear map from Rn to Rm is continuous iff it is continuousat the origin.

2. Show that any linear map from Rn to Rm is continuous.

3. Show that any isomorphism from Rn to itself is a homeomorphism.

Now for some formal definitions:

Definition 2.3.4. A fibre bundle is a quartet (E, B , F , ) where E is calledthe total space, B is called the base space, : E B is a continuous mapcalled the projection and for every b B, 1(b) is homeomorphic to F. Thespaces 1(b) are called the fibres of the bundle.

Definition 2.3.5. A fibre bundle is called locally trivial iff for every b Bthere is an open set U B containing b such that

1

(U) is homeomorphicto U F

The bundle B F is called a trivial bundle.Exercise 2.3.10. Describe clearly the trivial bundle with base space S2 andfibre S1 and give an example of a non-trivial bundle with the same base andfibre. Hint: you might find it easier if you specify some gluings.

Exercise 2.3.11. Show that the tangent bundle of a smooth manifold islocally trivial.

Exercise 2.3.12. Show there is a natural atlas on the tangent bundle whichmakes it a smooth manifold. Is the bundle projection smooth?

Note that for a locally trivial fibre bundle a topology on the bundle musthave as base the cartesian product of sets which are open in B (and overwhich the bundle is locally trivial) with open sets in the fibre.

Definition 2.3.6. A section of a fibre bundle E with projection to basespace B is a map s : B E such that s is the identity on B.

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2.4. NOTATION: VECTOR FIELDS 27

Definition 2.3.7. A vector field on a manifold M is a section of the tangent

bundle T M.

You should be able to see that this makes sense and we can talk aboutcontinuous, differentiable and smooth vector fields according as the section(which is after all a map) is continuous, differentiable or smooth.

Exercise 2.3.13.

1. Draw a vector field on R2 which is nice and easy and write it as asection of the tangent bundle.

2. Show that the tangent bundle for S2

is not trivial. Use the hairy balltheorem which says that any continuous vector field on S2 must haveat least one place where the vector is of length zero.

2.4 Notation: Vector Fields

On R2, I can write the tangent space as R2 R2 which is mildly useful forthinking about the meaning but not standard and not particularly useful forcomputations. I shall extend this to talking about the standard basis for R2

and call it (e1, e

2). A vector field on R2 is an assignment to each point ofR2

of a vector, and if it is a smooth vector field this vector changes smoothly aswe move around in R2. So there is a tangent vector, with two components,which both depends smoothly on x and y and hence is given by a pair offunctions P(x, y), Q(x, y). We might write the vector field as

P(x, y)e1 + Q(x, y)e2

but we dont. We write it as

P(x, y)

x+ Q(x, y)

y

This notation takes a bit of explaining.

If we have a smooth function f : R2 R and a smooth vector field on R2 wecan take the directional derivative of f at any point in the direction of thevector field at the point, and multiply it by the length of the vector. Thiswill give us a new smooth function on R2. This means that such a vectorfield can be thought of as an operator on the space of smooth functions fromR2 to R, which is usually written as C(R2). The constant vector field whichassigns the vector e1 to every point ofR

2 can easily be seen to be the operator

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/x and similarly the orthogonal constant vector field which assigns e2 is

the operator /y. This explains the notation for vector fields on R2 and byan obvious extension we can write a vector field v on Rn as

i[1:n]

vi(x)

xi

where each vi is a function from Rn to R and where I called v1 the functionP(x, y) and v2 the function Q(x, y) when n = 2.

Since the procedure for interpreting a vector field as an operator on C(R2)is local, a vector field on a manifold M is an operator on

C(M) although

there is an issue involved in choosing a basis for each tangent space Ta(M)if we wish to do calculations.

This gives two quite different ways of looking at a vector field on a smoothmanifold. We have the tangency equivalence classes which we may think ofas little arrows, each selected by a section of the tangent bundle. This isa quite straightforward transfer of ideas from Rn and should seem naturaland reasonable once you have come to terms with the problem of having tosay everything via charts. But the other way of thinking of a vector field asan operator on the space C(M) has some advantages. One of these is thatit makes sense without immediate reference to charts. Of course, we need

charts to say what it means for some map f : M R to be differentiable, butgiven that, we have a pleasant freedom from particular coordinate systems.Physicists are particularly interested in this, because the physical universedoes not come equipped with charts anymore than it has an origin and axessticking out of it. Recall penguins, and what they do, versus the language fortalking about them given by charts. Now we want to focus on the behaviourof the physical universe (penguins) and not be to distracted by the language(charts). So an invariant description, that is one which does not dependon choosing a particular language, is definitely more physical. Note OliverHeavisides remarks quoted at the top of chapter three of the text book.

On Rn

we can therefore write a vector field v as a map

v : C(Rn) C(Rn)

with vf the map

i[1:n] vi(x) f

xi.

This can be compressed into

v =

i[1:n]

vi

xi

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2.4. NOTATION: VECTOR FIELDS 29

An even more compact form is

v =

i[1:n]

vii

We can make this even terser by using the Einstein Summation Conventionwhich is that if an index is repeated as a superscript and a subscript then weautomatically sum over the possible values. This gives us

v = vi i

where you have to know what the space is in which we are working to knowhow many is there are. For some reason physicists prefer to use greek lettersas indices which means that you are likely to find expressions such as

v = v

instead. I fear that you will have to get used to this as the textbook iscommitted to it.

This leads to a new definition of a vector field on a smooth manifold M.

First we define C(M) to be the set of smooth maps from M to R. Thisis clearly a real vector space. It is certainly possible to add and scale thefunctions, and the rest is simple axiom bashing, as done in second year. It

is rather more than just a vector space, it is an algebra, which is to say it ispossible to multiply any pair of elements, f g being the function

a M, f g(a) = f(a)g(a)where the right hand side of the equality means we just multiply the two realnumbers f(a) and g(a). The multiplication is associative, commutative, andleft and right distributive over addition. In other words, it is a real vectorspace which is also a commutative ring, which is basically what we mean bya real algebra. You should write down the complete list of axioms for such athing, not relying on the text book too much.

Now we define a linear operator on such an algebra A by saying it is a map

v : A Awhich is linear, that is,

f, g A v(f + g) = vf + vgand

f A, t R v(tf) = tv(f)

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Such an operator is called a derivation if it also satisfies

f, g A v(f g) = fv(g) + gv(f)

which you will recognise as Leibnitz Rule for differentiating a product func-tion.

Exercise 2.4.1. Take M = R2 and any smooth vector field on it. Show thatit is a derivation.

Note that it makes sense to define a derivation over any real algebra andalgebraists indeed do exactly this. This is a long way from differentiating

functions, but it gives all the essential properties, and algebraists have ahabit of studying the properties without much caring where they came from.They have their uses. Algebraists, that is.

We can finally define a vector field on a manifold M as a derivation on the realalgebra C(M). Such a definition has advantages and disadvantages. Theobvious disadvantage is that it is so abstract it seems to have nothing to dowith the things we care about, but the advantage is that the abstraction hasremoved all the irrelevancies which get in the way of thinking about thingsand left the bare essentials. Any lingering suspicion that the geometric babyhas been thrown out with all that bathwater may be put to rest by checking

through the last exercise carefully, and by doing it with S1

instead of R2

:

Exercise 2.4.2. Take M = S1 and any smooth vector field v on it regardedas a section of the tangent bundle. Show that v is a derivation: take somesimple functions from S1 to R and operate on them by v. Confirm that allthe rules for a derivation are satisfied.

We also need to be able to go in the opposite direction: if v : C(M) C(M) is a derivation, then it must be able to be expressed as a vector fieldin the earlier sense.

Exercise 2.4.3. Do this onR2

at the origin. Suppose f :R2 R is asmooth map. Then we can write

f

xy

= f(0) +

f

x,

f

y

0

xy

+ ax2 + 2bxy + cy2

where a,b,c are second order partial derivatives off evaluated at some pointbetween the origin and (x, y)T (and hence, we have to admit, depend uponx and y). This is just the Taylor expansion with Lagrange form of theremainder in two dimensions.

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2.5. COTANGENT BUNDLES 31

Now apply v to f to get a new function g: then g(0) must be the limit of

g(x, y)T as (x, y)T 0, as g is certainly continuous, and show that sincev is linear, g(x, y)T must be the sum of the action of v on the above threeterms in a neighbourhood of the origin, that v takes the constant first termto zero, and that since v satisfies the Leibnitz condition, g(0) must be

f

x,

f

y

uv

for some vector (u, v)T R2. Finally show that if it works on R2 it mustwork on Rn and also on any smooth manifold.

We can now define Vect(M) or V(M) as the set of all vector fields on thesmooth manifold M.

Exercise 2.4.4. Show that Vect(M) (V(M) ) is a real vector space. Showthat it is a module over C(M), that is, it is like a vector space over C(M)except that C(M) is not a field but a ring.

Exercise 2.4.5. Show that V(M) as a module over C(Rn), is finite dimen-sional and has the obvious basis.

2.5 Cotangent Bundles

I mentioned earlier that we could do the business of equivalence classes ofmaps from the manifold to R in exactly the same way as we took maps fromR to the manifold. If we do this we get an exact parallel and a tangencyequivalence class of such maps at a point is called a cotangent or covectorat the point. Somewhat easier is to define the space of cotangents at a Xfor a smooth manifold X as the dual space of Ta(X). Recall that the dual(vector) space for a space V is the space V of linear maps from V to R. I

shall say more about this in the next chapter. We can do exactly the sameprocess of taking the union of all the Ta(X)

as we did for the tangent bundleand this gives us a slightly different object called the cotangent bundle. Ithas to be admitted that there is no difference between them as topologicalspaces. All the difference is in the algebra and it manifests itself stronglywhen we look to see what happens under maps between manifolds.

Exercise 2.5.1. I have given two different definitions of the cotangent space.Show they are equivalent.

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The same sort of considerations as worked for vector fields apply to covec-

tor fields or differential 1-forms as they are more commonly known. At eachpoint ofR2, we select an element ofR2, the cotangent space, which again hastwo components. I suppose we might call the standard basis for this space(e1, e

2) where e

1 is the linear map from R

2 to R which projects everythingonto the first component and e2 projects everything onto the second com-ponent. But we actually call them dx, dy to be loosely consistent with theclassical notation. So we interpret dx as the linear map which takes (x, y)T

to x where (x, y)T is a point in the tangent space Ta(R2) at some point a.

Similarly for dy. So a differential 1-form or covector field on R2 is written

P(x, y) dx + Q(x, y) dy

The generalisation to Rn is of coursei[1:n]

i(x) dxi (or idx

i using the Einstein summation convention)

and this, for smooth functions i, i [1 : n] represents a covector fieldor differential 1-form on Rn. The preference for letters towards the end ofthe Greek alphabet to denote differential forms is widespread so again youought to get used to it. The subscripts instead of superscripts for indicestells you something about the covariance or contravariance of the entities. I

shall explain this properly shortly.

If you wonder why on earth anybody bothers to distinguish between vectorfields and differential 1-forms, one answer is that it is natural to differentiatek-forms to get (k + 1)-forms for k N. This is what Stokes theorem is reallyall about. As you ought to have learnt in second year but probably didnt.

2.6 The Tangent Functor

Suppose f : X

Y is a differentiable map between manifolds. Then for the

case where X = Rn and Y = Rm there is a map between the tangent spacesat each point which takes the tangent space at a X to the tangent spaceat f(a) Y. To take a tangent vector va in the tangent space Ta(X) to onein the tangent space Tf(a)(Y) all we have to do is to operate on it by Df(a)which is by definition a linear map and has the right dimensions for domainand codomain. If we are prepared to choose a basis for Ta(X) and Tf(a)(Y)we could represent Df(a) by a matrix, and there is a perfectly sensible wayof choosing the same basis for tangent spaces over different points. All thismakes sense even if X and Y are just finite dimensional real vector spaces

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2.6. THE TANGENT FUNCTOR 33

without the extra structure of Rn. In fact it makes sense in arbitrary Banach

spaces.Of course, there is a slight problem of how to extend this to manifolds whichare not Banach spaces. Spheres and tori spring to mind.

If we take va, and recall that it is a tangency equivalence class of curvesv : (1, 1) X taking 0 to a then f v is a curve through f(a) and itspecifies a tangency class. Moreover if v is tangent to v at a then f v istangent to f v at f(a).Exercise 2.6.1. Most of this should have been a second year exercise butprobably wasnt. Do it now and all about tangent vectors and maps will be

clear. Well, clearer.

1. Let f : R2 R2 be defined byxy

uv

=

x2 + x + y + y2

1 + xy

Compute f on the set of points

t0

for t [0, 1]. Do this by choosing

ten points along the interval and evaluating f on them and plot themon a sheet of graph paper to obtain ten points which should lie on a

smooth curve. Do the same for points on the interval 0t for t [0, 1].

2. Calculate f

1/10

0

and f

0

1/10

if you havent already.

3. Calculate Df

10

.

4. Evaluate the above matrix on the tangent vector e1

5. Evaluate the above matrix on the tangent vector e2

6. Map the two tangent vectors obtained by the last two jobs on the samegraph.

7. Represent the tangent vector e1 by any curve c1 in the tangency equiv-alence class and compose with f. Differentiate to find a linear repre-sentative of T f(0, e1)

8. Repeat for a curve c2 representing e2

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9. Sketch the curves f

c1 and f

c2

10. Prove the claim that if v is tangent to v at a then f v is tangent tof v at f(a).

It follows that f induces a map T f which takes tangent vectors at a to tan-gent vectors at f(a). This process doesnt, on the face of things, involvedifferentiation. Nor does it involve charts. Of course it does involve differ-entiation, as the last series of exercises shows convincingly. And it is easy tosee that it goes through on charts for the usual reasons, which involve thechain rule.

In the case when we have a differentiable f : Rn

Rm the last exercises

should convince you we have at each point a X the diagram

Ta(X)

?

X

X

Tf(a)(Y)

?

Y

Y

-

-

Df(a)

f

This diagram commutes which means whichever way around you go you getthe same result. We can do this for every point a

X to get the commutative

diagram:

T X

?

X

X

T Y

?

Y

Y

-

-

T f

f

The process of taking a manifold and producing its tangent bundle is said to

be functorial because if we have two manifolds and a smooth map betweenthem the process gives a map between the bundles.

Instead of writing T f we often write f for the same map. This is moregeneral because it makes sense for some other vector bundles and not justthe tangent bundle.

Such a map between tangent bundles is said to be fibre preserving, sinceit takes anything in the fibre over a to the fibre over f(a). And we cangeneralise this to maps between any fibre bundles, so they are also calledbundle maps. If the fibre is a vector space we talk of vector bundles and we

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2.6. THE TANGENT FUNCTOR 35

require the bundle maps to be linear, so the map T f is also a vector bundle

map.Note that the map T f contains all the information about the derivative andalso tells you where things are, which the derivative (being only the linearpart of an affine map) does not. So this is actually cleaner and conceptuallysimpler than the usual description of the business of differentiation. Anotherway to put this, in the light of the last exercise, is that when you calculatelots of partial derivatives you are merely trying to calculate the linear part ofan affine map which specifies a tangency equivalence class, that is, a tangentvector.

We can usefully think ofT f as coming in two parts, since locally the tangent

space is simply a cartesian product of possible tangent vectors over a spacewith a part of the space. On the first part T f is simply f and on the secondpart, the fibres, it is Df, the derivative off. We can now choose to define thederivative of a smooth map this way. I have hankered after teaching calculusthis way in first year. It is actually easier, probably because you need toisolate the core ideas in order to generalise things and fronting up to the coreideas although demanding at first makes life a lot easier subsequently.

Note that the chain rule can now be formulated as

T(f g) = T f T g

Exercise 2.6.2. Confirm that the chain rule holds. This is also a part ofsaying that T is functorial.

Exercise 2.6.3. Guess what a functor is and what it is a map between.Confirm your guess by doing some googling. I warn against doing the googlingfirst.

Exercise 2.6.4. Take f : R+ R+ defined by x x2. Show this is adiffeomorphism. Let V be the vector field on R+ which has constant vectorsof length 1 at every point. Show that T f takes this into a new vector field on

R

+

, and say what the new vector field is. Regarding the two vector fields asdifferential equations, find both solutions.

2.6.1 The (non-existent) Cotangent Functor

Suppose we have f : X Y a smooth map between smooth manifolds,and we look to see what happens in the cotangent bundle. Thinking of acotangent at a X as a tangency equivalence class of maps from someneighbourhood of a to R, we see that the map between the fibres goes in

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the reverse direction. Given v : W

Y

R as a representative function in

the tangency equivalence class at f(a) (with f(a) W), f induces v f :f1(W) R on X which defines a cotangent vector at a. So we obtain thediagram:

Ta

X

?

X

X

Tf(a)Y

?

Y

Y

-

f

f

This makes T, a hypothetical induced map on the whole cotangent spacea mess, because it goes one way (left to right) on the space part and theopposite way (right to left) on the cotangent part. If f has a smooth inversewe can get around this, but it is not so neat. Incidentally:

Definition 2.6.1. A smooth map with a smooth inverse is called a (smooth)diffeomorphism

Exercise 2.6.5. How, if at all, can we relate the derivative of f to f whenX = Rn, Y = Rm?

Remark 2.6.1. In older books, a covector field is called a contravariantvector field and a vector field is called a covariant vector field. See forexample, Mackeys Theoretical Foundations of Quantum Mechanics. As weshall see later, a covariant vector field is a contravariant tensor field. Dontblame me for this.

This is all rather confusing on first encounter. Familiarity breeds acceptanceand the best way to become familiar with these ideas is to work them throughin very simple cases. So make up a set of exercises yourself in which you workwith particular simple maps between very simple manifolds (Rn and Rm forn, m small positive integers.) As a start:

Exercise 2.6.6. Let f : R R be given by f(x) = x2

. Put a = 2 andinvestigate what happens if we take (a) a tangent vector at 1 and (b) acotangent vector at 4.

Now try it for f : R2 R2 withxy

x2 + y2

xy

ans some suitable points for a and f(a). In this case you can convenientlyrepresent tangent vectors as columns and cotangents as rows.

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2.7. AUTONOMOUS SYSTEMS OF ODES 37

Exercise 2.6.7. Write out a lecture for first year students which describes

tangent vectors on R in a really simple way as possible velocities along theline, and hence define the tangent bundle R R. Define differentiation ofmaps from R to R in terms of bundle maps. Prove the chain rule as T(fg) =T f T g. Be prepared to answer any awful questions an intelligent studentmight ask.

Write out a lecture on ordinary differential equations in terms of sections ofthe tangent bundle. Set up and solve some easy ones in this notation.

Do you think this is easier or harder than the traditional way of doing it?Assume that since Mathematica can solve ODEs, the idea is not to trainstudents to jump through hoops but to get them to understand what they

are doing.

2.7 Autonomous Systems of ODEs

2.7.1 Systems of ODEs and Vector Fields

Consider the system of linear ordinary differential equations:

x = y x(0) = 1y = x y(0) = 0

We can write this as a two dimensional problem:xy

=

0 11 0

xy

or more succinctly:

x = Ax (2.7.1)

where A is the above matrix.The matrix A defines a vector field on R2 by taking the location x to thevector A(x). We are now used to the idea of a vector field on R2 both visuallyin terms of lots of little arrows stuck on the space (which can incidentally begenerated quickly and painlessly using Mathematica), and algebraically asa map from R2 to R2 sending locations to arrows (with their tails attachedto those locations).

Such a system of ordinary differential equations is called autonomous, mean-ing that the vector field specified by the system doesnt change in time.

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Figure 2.7.1: A vector field or system of ODEs in R2

Consequently we can either refer to an Autonomous System of Ordinary Dif-ferential Equations defined on an open set U Rn, or we can talk about aSmooth Vector Field on U. The second is much shorter and easier to think

about.If we draw the vector field in the above case, we get arrows which go aroundthe space in a positive direction as in figure 2.7.1

A solution to the system of differential equations, or an integral curve for thevector field is a map f : R R2, usually written

x(t)y(t)

with the property that x and y satisfy the given system of equations. Whatthis means is that we think of a point moving in R2 so that its velocity at

any point is just the vector attached to that point. So the solution curve hasto have the vector field tangent to it always.

It is possible to learn to solve autonomous systems of differential equationswithout ever understanding that they are all about vector fields which givethe velocity of a moving point, and that a solution is simply a function whichsays where the moving point is at any time, and which agrees with the givenvector field in what the velocity vector is. This is a pity.

In the above case, you can see by looking at the system what the solutionis: obviously the solution orbits are circles, and given the initial condition

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where at time t = 0 we start at the point (1, 0)T, the solution can be written

down asx = cos(t), y = sin(t)

and it is easy to verify that this works.

Exercise 2.7.1. Do it.

Obviously, solving initial value ODE problems for more complicated vectorfields isnt going to be so easy, and doing it in dimensions greater than threeby the look at it and think method also looks doomed. So it is desirableto have a general rule for getting out the solution. Fortunately this is easyenough for linear vector fields in principle, although the calculations can be

messy in preactice. But again, thats what computers are for.

2.7.2 Exponentiation ofThings

I did this in second year M213 but some of you may have missed out on it inwhich case here it is. Those of you who did it can read this rather quickly.

If you write down the usual series for the exponential function you get:

exp(x) = 1 + x +x2

2!+

x3

3!+

xn

n!+

Now think about this and ask yourself what x has to be for this to makesense. You are used to x being a real number, but it should be obvious thatit could equally well be a complex number. After all, what do you do withx? Answer, you have to be able to multiply it by itself lots of times, and youhave to be able to scale it by a real number, and you have to be able to addthe results of this. You also have to have an identity to represent x0. Oh, andyou need to be able to take limits of these things. So it will certainly workfor x a real or a complex number. But it also makes sense if x is a squarematrix. Or, with any system where the objects can be added and scaled and

multiplied by themselves. And have limits of sequences of these things.The name of a system of objects which can be added and scaled by realnumbers is a vector space, and a vector space where the vectors can alsobe multiplied is called an algebra. We can do exponentiation in any algebrawhich has a norm and a multiplicative identity. (And it would be a help if itwas complete in that norm, i.e. limits of cauchy sequences exist.) The squaren n matrices form such an algebra. We can also hope to take sequences ofthem and maybe have them converge to some matrix. So we can exponentiatesquare matrices.

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Exercise 2.7.2. Exponentiate the matrix A in equation 2.7.1. Now expo-

nentiate the matrix tA. Do you recognise the result?It should be obvious that we could, in principle, calculate the exponential ofa matrix to some number of terms, and if the infinite sum makes sense andthe sequence of partial sums converges, then we could always get some sort ofestimate of exp(A) for any matrix A by computing enough terms. We wouldhope that multiplying A by itselfn times would give some reasonable sort ofmatrix, and when we divided all the entries by n! we would get somethingpretty close to the zero matrix. If this happened for all the n past somepoint, then we could optimistically suppose that exp(A) was some matrixwhich we could at least get better and better approximations to, which after

all is exactly what we have with exp(x) for x a real number.

Exercise 2.7.3. Define the norm of an n n matrix A to be

A = supx=1

A(x)

as in an earlier problem, and show that A2 (A)2. Hence prove thatthe function exp is always defined for any n n matrix.Exercise 2.7.4. IfetA exp(tA) denotes a map from R to the space ofnnmatrices, show that its derivative is AetA.

There are other algebras where a bit of exponentiation makes sense, so beprepared for them.

2.7.3 Solving Linear Autonomous Systems

In principle this is now rather trivial:

Proposition 2.7.1. If x = Ax is an autonomous linear system of ODEswith x(0) = a, then

x = etAa

is the solution.

Proof:

Differentiating etA gives AetA by the last exercise and since exp0 = I theidentity matrix, the initial value x(0) = a is satisfied. So it is certainly asolution.

If this looks a bit like a miracle and in need of explanation, you are thinkingsensibly and merely need to do more of it. It may help to note that the

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exponential function is the unique function with slope at a point the same

as the value at the point, and that this leads to the general solution for thelinear ODE in dimension one, and that this goes over to higher dimensionswith no essential changes. In effect, the exponential function was invented tosolve all these cases. It actually goes deeper than this, see Vladimir Arnoldsbook Ordinary Differential Equations.

2.7.4 Existence and Uniqueness

Could you have two different solutions (or more)? No, not for linear systems,but this requires thought. Certainly the 1-dimensional ODE given by

x(t) = 3x2/3, x(0) = 0

has the solution x(t) = t3 but also the solution x(t) = 0 It also has infinitelymany other solutions. (Can you find some?) Of course this is not a linearODE, but it is clear that some sort of conditions will need to be imposedbefore we can look at vector fields which are not linear and expect them tohave solutions. Happily, there is a simple one which guarantees at least localexistence and uniqueness:

Theorem 2.7.1. If f : U

Rn

Rn is a continuously differentiable

vector field, then for any point a in U there is a neighbourhood W Uof a containing a solution to the system of equations x = f(x) with a asinitial value, and the solution is unique. Moreover, there is a continuouslydifferentiable map F : W J Rn for some interval J = (a, a) on 0 Rsuch that for all b in W, the map Fb : J Rn is the solution for initialvalue b at t = 0.

There is a proof in Hirsch and Smales Differential Equations, DynamicalSystems and Linear Algebra, pages 163 to 169.

There is a better proof in Arnolds book on page 213. It is actually the

same proof but much better explained. It is given for the general (non-autonomous) case. Both arguments use the contraction mapping theorem.You should read through it if you have not already done a proof in yourODEs course. Assuming you did one.

The results follow easily from a more basic result sometimes called TheStraightening Out Theorem (In Arnold The basic theorem of the theory ofordinary differential equations or the rectification theorem. See chapter 2).The theorem says that in a neighbourhood U of a point of Rn where the(continuously differentiable) vector field is non-zero, we can find a one-one

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differentiable map from U to W

Rn with a differentiable inverse, such that

the transformed vector field on W is uniform and constant.Given that we can do that, we could also make the vectors all have lengthone and lie along the x1 axis in Rn with a rotation and scaling. The systemof ODEs then would be, in this transformed region W, the rather boringsystem:

x1 = 1

x2 = 0...

xn = 0

with the solution

x1(t) = t + a1; x2(t) = a2; xn(t) = an

If you believe in the Straightening Out Theorem, then it is obvious that anycontinuously differentiable vector field has at any point where the vector fieldis non-zero a solution which is unique in some neighbourhood of the pointand which depends smoothly on the point. All we have to do is to map thestraight line boring solution(s) back by the differentiable inverse.

Exercise 2.7.5. Prove the last remark.

When the vector field is zero at a point, the solution is the constant functiontaking all of R to the point. So there is a unique solution here too.

Remark 2.7.1. You will find a proof of the straightening out theorem inArnold. I shant prove it in this course on the grounds that this isnt a courseon ODEs. At least, I dont think it is.

Remark 2.7.2. It should be obvious that although we have looked at sys-tems of ordinary differential equations on Rn, the fact that everything is

defined locally means that they ship over to any smooth manifold. If themanifold is compact then the completeness is guaranteed, and the solutioncan be found by doing everything in charts and piecing the bits together.

2.8 Flows

I rather slithered over one important point, which is the question of whetherwe always get a solution for all time, past and future. It is not hard to see

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2.8. FLOWS 43

that the vector field X(x) = x2, X(0) = 1 on R has a solution

x(t) =1

1 twhich goes off to infinity in finite time. From which we deduce that it isnot in general possible to ensure that there is a solution for all time, andthis explains the cautious statement of the last theorem. The best we canhope to do, the theorem tells us, for a smooth vector field at a point is tofind a neighbourhood of the point in which there is a parametrised curve,x(t) : t (a, a) where if we are lucky a will be and if we arent it willbe some possibly rather small positive number.

Definition 2.8.1. A vector field on U Rn is said to be complete if anysolution can be extended to the whole real line.

Exercise 2.8.1. Show that if a vector field has compact support then it iscomplete.

Exercise 2.8.2. Show that if U is the unit open ball in Rn centred on theorigin and X is a smooth vector field on U, then if X is complete, and ifProj(X(x), x) is the projection of X(x) on x, then

limx1 Proj(X(x), x) = 0

Remark 2.8.1. It should be obvious that there are not many physical situa-tions where things go belting off to infinity in finite time, and for that reasonI shall restrict myself from now on to complete vector fields. If I forget toput the word in, put it in yourself. Also put the word smooth in front ofthe term vector field whenever it occurs since I shall not consider any othersort.

The business of getting a solution is going to work not just for the point weselected as our starting point but also for neighbouring points provided we

dont go too far away. In the happy case where the vector field has solutionsfor all time, the space U on which the vector field is defined is decompos-able as a set of integral curves, since solutions cant intersect each other, orthemselves, although they can, of course, be closed loops. This statementfollows from the uniqueness of a solution. Hence we deduce that a vectorfield gives rise to what is called a foliation of the space into integral curves.You can, perhaps, guess that partial differential operators more complicatedthan vector fields will give rise to higher dimensional foliations, decomposingthe space into surfaces and other manifolds.

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Exercise 2.8.3. Describe the foliation of R2 by the vector field

y x

+ x

y

Recall that in second year (M213) we discussed the idea of groups acting onsets and came to the conclusion that they were conveniently seen as homo-morphisms from a group G into the group Aut(V) of maps from the set Vinto itself. Then a complete smooth vector field X on U Rn gives rise toan action of the group R on U as follows:

x : R U U(t, x0)

x(t)

where x(t) is the integral curve of X with x(0) = x0.

To prove this is indeed a group action, we need to show that x(0, x0) = x0for every x0 which follows immediately from my definition of x. (Since theadditive identity of R is 0.) We also need to show that

s, t R, x0 Rn, v(s, v(t, x0)) = v(s + t, x0)which merely means that if you travel for time t from x0 along the solutioncurve, and then go on for time s, this gives the same result as travelling fortime s + t from the starting point x0, which is, after all, what we expect a

solution curve to do.If we fix t and look to see what the group action does, it is a map from Rn

to itself. Well, we knew that. It is a truth that this map is always a smoothdiffeomorphism. The old fashioned way of saying this is that the solutionsdepend smoothly upon the initial conditions, but I much prefer the modernway of saying it. You should be able to see that all we are doing is takingeach point as input, and outputting the point it will get to after time t.

Proposition 2.8.1. For a complete smooth vector field X on U open inRn, for any t R, the map xt : U U, which sends x0 to x(t, x0) is adiffeomorphism of U

Proof:

The map xt certainly has an inverse, xt. And the theorem on existence ofsolutions to an ODE establishes that the map is continuously differentiablewhen X is. So ifX is smooth, so is xt.

Remark 2.8.2. The set of diffeomorphisms {xt : t R}, or in other wordsthe map x : R U U, is called in old fashioned books a one-parametergroup of diffeomorphisms. I shall simply say that the map x obtained fromthe vector field X is the flow of X.

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2.9. LIE BRACKETS 45

Remark 2.8.3. Given a flow x on U

Rn we can always recover the vector

field by simple taking any point, a and differentiating the map xa : R Uwhich sends t to x(t, a) at t = 0. This must give us the required vector fieldfrom which the flow can be derived. So there is a correspondence betweenflows and vector fields.

You now have four ways of thinking about vector fields. They are bunchesof arrows tacked onto a space; they are autonomous systems of ordinarydifferential equations. And they are also flows, obtained by solving the au-tonomous system. And last but not least they are operators on the algebra ofsmooth functions from the space to R. This demonstrates that vector fieldsare more interesting and complicated than you might have supposed.

I shall give one important feature of vector fields which arises from thismultiple perspective and which is much less obvious if you stick only tosystems of ordinary differential equations.

2.9 Lie Brackets

Writing, as is conventional in some areas, X and W for two vector fields inV(Rn) and bearing in mind that we can compose any such operators to getX W and W X (which we write XW and W X for short). In general theresult is a perfectly good operator but some calculations will rapidly convinceyou that XW is not, in general, a vector field operator but something muchnastier.

Example 2.9.1. Let V = y /x + x /y and W = x /x + y /yThen V W h is

xy 2h

x2 y h

x y2

2h

xy 0 + x2

2h

yx+ x

h

y+ xy

2h

y 2+ 0

and W V h =

xy 2hx2

+ 0 + x2 2h

yx+ x h

y y2 2h

xy h

x+ xy

2hy 2

+ 0

Neither of these look like a vector field operating on h. If however we takethe difference, V W W V we get some happy cancellation and wind up with

V W W V = (y x

+ x

y) (x

y y

x) = 0

which is a vector field although not a very interesting one.

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Exercise 2.9.1. Write down another pair of vector fields V, W on R2 and

compute V W W V. Check to see if you always get the zero vector field.What is it telling you about the vector fields when V W W V = 0? (Someintelligent conjectures would be of interest but only if supported by evidencenot used in framing the conjecture.)

Exercise 2.9.2. IfX = P(x, y)/x + Q(x, y)/y and W = R(x, y)/x +S(x, y)/y, calculate XW W X and verify that is is a vector field.Exercise 2.9.3. Compute XW W X for X, W V(Rn) and show it is avector field in V(Rn) Show that this also holds for V(U) for any open setU

Rn.

All this gives the following definition:

Definition 2.9.1. The Lie Bracket or Poisson Bracket of two vector fieldsX, W in V(U) for U Rn is written [X, W] and defined by

[X, W] XW W X

It is a multiplication on the vector space of Vector fields on U.

Exercise 2.9.4. Do some simple calculations preferably for U R1 and con-vince yourself that the Lie bracket multiplication is not in general associative

but does satisfy the Jacobi Identity:

X, Y , Z X(U), [X, [Y, Z]] + [Y[X, Z]] + [Z, [X, Y]] = 0

Exercise 2.9.5. Prove that the Jacobi Identity is always satisfied for VectorFields.

The Lie bracket almost makes the vector space of vector fields on U, anopen subset of Rn, into an algebra, which you will recall is merely a vectorspace where the vectors can be multiplied, to make a ring. Here the LieBracket operation fails to be associative in general, but a vector space with a

non-associative multiplication which satisfies the Jacobi Identity is, notwith-standing, called a Lie Algebra. There are others besides these and againalgebraists have gone to town on investigating abstract Lie Algebras. Well,we wouldnt like them to be at a loose end and hang around street corners3.

Exercise 2.9.6. Prove [X, (Y + Z)] = [X, Y] + [X, Z] and [(X + Y), Z] =[X, Z] + [Y, Z] Prove also that a R, [aX,Y] = a[X, Y] and [X,aY] =a[X, Y].

3Although theyd probably have an interesting line in graffiti.

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2.9. LIE BRACKETS 47

Remark 2.9.1. The above properties you will recognise as bilinearity.

Exercise 2.9.7. Investigate the relation between [hX,Y], [X,hY] and h[X, Y].

It should be apparent that although the calculations tend to be messy andprovide great scope for making errors, they are not essentially difficult. Anatural candidate for a good symbolic algebra package, you might say.

Exercise 2.9.8. Is there a multiplicative identity for the Lie Bracket oper-ation on vector spaces? That is, is there a vector field J such that for everyother vector field, X, [J, X] = X? (Hint: what is [J, J]?)

You might be interested in an area of applications of these ideas. If so readon.

It is easy to find the solution, h(x, y) = x2 + y2 to the PDE

y hx

+ xh

y= 0

Now this is one solution, and finding a single solution is very nice, but weusually want the general solution. In this particular case you can probablyguess it. But in general, if we have some linear partial differential operatorL acting on F, a suitable space of smooth functions, and if we want the setof all solutions of Lh = 0, then it will usually be a lot harder to find them.

This process is aided by the following idea: The set of solutions of L is goingto be a linear subspace ofF, by definition of the term linear operator. Callit F0. Now a symmetry of the solution space of the operator L, often calleda symmetry of the operator L, is some vector field operator X such that Xtakes F0 into itself, i.e. if whenever h is a solution to Lh = 0, so is Xh. If weknow the collection of all symmetry operators for L and we have a solution,then we can find all the other solutions. In trivial cases this will amountto no more than adding in arbitrary constant functions, but in non-trivialcases it will do a whole lot more than this. So it would be a good idea tobe able to find, for a given L, the set of all symmetries X for L. It is clearthat the Poisson-Lie bracket can be used for any pair of linear operators, notjust vector fields. The following observation goes some way to explaining ourinterest in them:

Proposition 2.9.1. If [L, X] = wL for some function w F, then X is asymmetry of L.

Proof:

We need to show that h F, L(Xh) = 0 NowLX XL = gL LX = gL + XL

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and

h F, (gL + XL)h = gLh + XLh = 0 + 0 = 0

Exercise 2.9.9. Prove the converse, that if X is a (vector space) symmetryof L, then [L, X] = gL for some g F.Exercise 2.9.10. What symmetry is involved in finding the general solutionto the equation

y hx

+ xh

y= 0

and how does it give the general solution?

Now it is possible to prove that the set of all vector space symmetries of anoperator L is itself a Lie Algbra. Which is one reason for wanting to knowmore about them.

Some students of PDEs want to know why it is that the standard partialdifferential equations all had their variables separable: does this happen forall possible PDEs and why does it work for these cases? The answer to thisquestion is rather long and may be found in Volume 4 of the Encylopediaof Mathematics and Its Applications, Symmetry and Separation of Variablesby Willard Miller. It has a lot to do with Lie Algebras.

It is now possible to state properly a significant problem.

Going back to the idea of flows, it makes sense to discover whether flowscommute. For a suitable pair of flows, x, y : R U Rn we can start offfrom a U and go by flow x for a time s and then by flow y for time t. Thiswill get us to some point in U, written naturally enough as yt xs(a). Or wecould go the other way around, first by y and then by x to get xs yt(a). Ifwe always wind up at the same point for any starting point and any pair oftimes s, t then we may say that the flows commute.

Then when the flows x, y correspond to the vector fields X, Y, we have thefollowing result: x and y commute iff [X, Y] = 0. You can see that this worksfor the case of the two vector fields V, W in Example 2.9.1.

At present we lack the machinery to prove this result economically, so I shallskip it until it is needed.

Remark 2.9.2. Again all this makes perfectly good sense on manifolds forthe usual reasons. The idea of thinking of a vector field on a manifold Mas a special kind of operator on C(M) ensures that we can compose themand add them and subtract them, so the Lie Bracket makes sense there too,and we can write down vector fields on manifolds via charts and find integralcurves for them and so foliate the manifold.

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2.10. CONCLUSION 49

Exercise 2.9.11. Demonstrate the truth of the last remark by doing some

of these things on S1 and, if you are feeling very brave, S2.

2.10 Conclusion

This has been a quick introduction to the ideas of smooth manifolds andvector fields on them. There are whole books dedicated to these ideas andyou will find some in the library. You will find some of these ideas coveredvery quickly in chapters two and three of the text book, which you shouldread and satisfy yourself that it is intelligible. You should be able to see why

definitions are as they are.

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Chapter 3

Tensors and Tensor Fields

This chapter deals with the machinery needed to talk about diferential geom-etry, although it only starts on actually doing so. It contains the informationin Chapter four of the text book and goes into the algebra in more detail.This is because we are doing it right, on account of being mathematiciansand therefore feeling uneasy about relying on our intuitions without beingable to check the logic. We also cover part of Chapter one of Part three,where the text book is decidedly scrappy, and part of Chapter five of Partone. I do things in a slightly different order. You should however read the

text book in conjunction with the notes and do the exercises.I also throw in a few remarks about the exterior calculus and Stokes Theo-rem, not because this is part of the course but because it is a part of everyeducated persons background in the twenty-first century. It has to be ad-mitted that there arent many educated people around, but then there neverhave been.

3.1 Tensors

3.1.1 Natural and Unnatural IsomorphismsLet V denote a real vector space of finite dimension, so V is isomorphic toRn for some n Z+. Then I can define the space of shifts of V which is acollection of maps from V to V by taking any v V and writing

v : V V, w V, w w + v

The set of all such maps I shall call V. The map v is the map that adds v toeverything. I can compose such maps, and it is immediate that uv = u + v.

51

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52 CHAPTER 3. TENSORS AND TENSOR FIELDS

Similarly, for every t

R, tu = tu where we scale maps in the usual way,(tf)(x) = t(f(x)).

This makes V a vector space and gives an

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