a practical introduction to diﬀerential forms alexia e ...schulz/chapter4.pdf · a practical...

A Practical Introduction to

Differential Forms

Alexia E. Schulz

and

William C. Schulz

October 3, 2016

Transgalactic Publishing CompanyFlagstaff, Vienna, Cosmopolis

ii

c© 2012 by Alexia E. Schulz and William C. Schulz

Every creator painfully experiences the chasm between his inner vision andits ultimate expression. Isaac Bashevis Singer

iii

Dedicated to our parents, children, and cats

Contents

1 Applications to Differential Geometry 1

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 A Little History . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Embedded n-manifolds in Euclidean (n+ 1)-space . . . . . . . . 5

1.3.1 Connection and Curvature Forms . . . . . . . . . . . . . . 51.3.2 Curves and Geodesics . . . . . . . . . . . . . . . . . . . . 171.3.3 Special Case; Surfaces in R3 . . . . . . . . . . . . . . . . . 20

1.4 Some Tensors and the Proof of the Gauss-Bonnet theorem . . . . 301.4.1 Tensors and their algebra . . . . . . . . . . . . . . . . . . 301.4.2 Tensor Calculus . . . . . . . . . . . . . . . . . . . . . . . . 311.4.3 Raising and Lowering Indices . . . . . . . . . . . . . . . . 341.4.4 Epsilon tensors . . . . . . . . . . . . . . . . . . . . . . . . 351.4.5 Epsilon tensors and Dual tensors in Two Dimensions . . . 361.4.6 The Riemann Curvature Tensor in Two Dimenions . . . . 37

1.5 General Manifolds and Connections . . . . . . . . . . . . . . . . . 401.6 Parallel Displacement Along Curves . . . . . . . . . . . . . . . . 441.7 A little about Lie Groups and Lie Algebras . . . . . . . . . . . . 461.8 Frame Bundles and Principle Bundles . . . . . . . . . . . . . . . 53

1.8.1 Group Actions . . . . . . . . . . . . . . . . . . . . . . . . 541.8.2 Principal Bundles . . . . . . . . . . . . . . . . . . . . . . 54

1.9 Affine Connections . . . . . . . . . . . . . . . . . . . . . . . . . . 551.10 Riemannian Geometry . . . . . . . . . . . . . . . . . . . . . . . . 57

v

vi CONTENTS

Chapter 1

Applications to Differential

Geometry

1

2 CHAPTER 1. APPLICATIONS TO DIFFERENTIAL GEOMETRY

The Great Way is not difficultIf only you do not pick and choose.Neither love nor hateAnd you will clearly understandBe off by a hairAnd you are as far apart as heaven and earth

Seng Can(c. 590 CE)

1.1 Introduction

In this Chapter we wish to give an introduction to the ways differential formscan be used in Differential Geometry. There has been considereable neglect ofthis usage, and we discuss this below. There are many fine books on differentialgeometry and we do not want to plow this field again, so here we will concentrateon the important ideas that lie behind the use of differential forms. We willneed to introduce the concepts of vector bundle and frame bundles but theseare natural concepts, at least for the bundle of natural frames that comes fromthe coordinate system. We will mostly stick with frame bundles, but the theoryextends with almost no change to any sort of bundle.

In classical differential geometry (which has certainly not been replaced bymore modern treatments, although progress is being made), the principal com-puational tool was tensor analysis. Tensor analysis is fine as a computationaltool but poor as an aid to understanding what is really going on. For example,the concept vector is defined by saying an array of numbers (of the proper size)is a vector if it transforms in a certain way under coordinate change. This isnot too illuminating. Then, when objects are differentiated, certain surprisingquantities, the affine connections, are introduced so that the derivatives are notcoordinate dependent. All of this can be made quite clear and straightforwardwith a little help from modern mathematical concepts.

Our basic goal is to explain the sentence

A connection is a Lie Algebra valued differential form

Once you understand what this really means, the whole of connections andcurvature become accessible. This is important because curvature is the ba-sic notion of differential geometry and one which has a fairly clear emotionalmeaning. When you control curvature, you control it all.

We do not want to suggest that tensors and their algebra and calculus areobsolete; we want to complement them by the use of differential forms, whichoften drastically simplify calculations. Tensors retain their importance as toolsin places where differential forms are not appropriate, and even when differentialforms are used tensors can be used a practical calculational adjuncts. Theoccasional tendency to regard tensors as obsolete is just picking and choosing.One needs both.

We will begin by developing some classical results for an n-manifold embed-ded in (n + 1)-dimensional Euclidean space. From this we get a feeling of how

1.2. A LITTLE HISTORY 3

the theory should go and we easily introduce connection coefficients and theRiemann curvature tensor, as well as the differential form equivalents. Thenwe move over to abstract manifolds and show how to construct analogs of theconnections and curvature there, which is relatively easy. In fact, one of thereasons for doing this chapter is to show just how easy it is.

Next we will apply the previous material to 2-manifolds in 3-space and ab-stract 2-manifolds. Here we will introduce the most important invariant of a2-manifold, the Gaussian Curvature and develop some formulas for computingit.

We will be staying almost entirely with local differential geometry but afterthe above material we will have a short chapter on the Gauss-Bonnet theoremwhich is one of the greatest theorems in mathematics, and the jumping off pointfor a vast amount of modern mathematics. The generalizations to n-manifolds,begun by Chern1, uses tools which too advanced for us to discuss in a book ofthis nature, unfortunately, but the 2-manifold case is extremely impressive onits own.

We will develop the theory in a naive way first, so that you get to know theactors. At the end we will introduce the Lie Groups and Algebras to put thedevelopment in a modern context. These sections are a little more difficult andmay be omitted by less enthusiastic students.

1.2 A Little History

When C. F. Gauss got the job of Professor of Astronomy at Gottingen U. hefound out that it came with certain extra duties, including being responsiblefor surveying the Kingdom of Hannover. Gauss was not by nature a hiker andcamper, and thus found it useful to think about the theory of surfaces partiallyto minimize the amount of actual measurement in the great outdoors that wouldbe needed. He wrote a classic book on the subject, and this essentially began thesubject as an independent discipline. Euler and others had already made someprogress in applying Calculus to surfaces but Gauss pointed out the importantconcepts and proved many important theorems, thus creating a systematic bodyof knowledge from which new researchers could move forward sytematically.Another reason for Gauss’s interest was non-Euclidean geometry, which couldbe done in a not totally satifying way on a curve called the pseudosphere, anddifferential geometry was helpful in these investigations.

One of Gauss’s greatest theorems was that the Gaussian curvature dependedonly on the first fundamental form, or as we would say now the coefficiants gij ofthe inner product that the surface inherited from the surrounding, or ambient,3-dimensional Euclidean space. Gauss realized that the gij could be determinedby measurements on the surface itself and thus did not depend on how thesurfaces was embedded in the Euclidean 3-space, and thus was an invariant of

1Professor Chern had the misfortune of transliterating his name in a Romanization ofChinese which failed to catch on. The r in Chern is a tone mark, indicating a rising tone; itis not pronounced at all. Chern thus rhymes with gun.


the surface itself. Riemann, with his usual surprising insight, reinterpreted thisresult to mean that you could have “surfaces” of any dimension, or as we wouldsay, manifolds, and you could have curvature provided you gave the manifoldan inner product at every point (continuously varying of course). For Gauss,the use of the surrounding three dimensional space of the surface was to providethis inner product, but for Riemann there was no need of any surrounding spacesince each point had a Riemann-given inner product. Riemann showed in hisfamous paper that this was enough to define curvature2. We also mention theprobability that Riemann discovered his curvature tensor as an outgrowth ofhis researches on the integrability of first order partial differential equations, inwhich he was an expert.

Riemann’s work opened up the field of Manifold theory, which was thencarried on by the German Christoffel (1829-1900) and by Italians, who met Rie-mann on his trips south to a climate that was better for his health. Levi-Civitaand others developed the tensor methods that allowed ordinary mathematicians,physicists, and engineers into this new world, in which eventually Cartan andEinstein became enthusiastic explorers.

Riemann presented his ideas on what became Riemannian Geometry in alecture in 1854. Gauss was in the audience and, a very rare occurance, he wasvery impressed by Riemann’s ideas. For unclear reasons, perhaps because hewanted to add more detail to the manuscript, Riemann had not published themanuscript when he died in 1866. Riemann was a very diffident man and did notmake friends easily, but he was fortunate that the one friend he did make wasRichard Dedekind, who arranged for the publication of the lecture in 1868, themost important paper in geometry in a thousand years. By 1869 Chistoffel hadalready introduced the Christoffel symbols and showed how to get the RiemannCurvature Tensor from them. An important book about Riemann’s lecture andits influence has recently been published by Jurgen Jost.

However, the tensor methods introduced by the Italians, while natural forsome people were very difficult for others, and there was a desire to have a moreimmediate approach to the material that was less off putting than the “forests ofIndices” typical of tensor analysis. There are various methods, but differentialforms, along with a bit of Lie Groups and Lie Algebras, are one entree into thisworld which is a little less offputting. This is what we are going to introduceyou to. It is very easy in differential geometry to lose ones way and becomedistracted by all the beautiful objects to either side of the path. We are notgoing to do that (very much). We are going to march straight along the pathas suggested by Seng Can’s poem, and then it is not difficult. The methodwas pioneered by Elie Cartan in the first half of the 20th Century. However,enthusiasm has been less than one might anticipate, and one reason for thischapter has been to show how it can be done easily.

2It is possible that Gauss was actually aware of much of this, but Gauss disliked straininghis contemporaries with radical new ideas. Riemann had no such reluctance.

1.3. EMBEDDED N -MANIFOLDS IN EUCLIDEAN (N + 1)-SPACE 5

1.3 Embedded n-manifolds in Euclidean (n + 1)-space

1.3.1 Connection and Curvature Forms

We begin with an n-manifold embedded in an n − 1-dimensional Euclindianspace. We are interested in a theory which will describe the local properties ofthe manifold at a point. Specifically, we would like to be able to take derivativesof vector fields and to obtain expressions for the curvature. The manifold will bedescribed locally by the coordinates u1, . . . un and the Euclidean space will haveglobal coordinates x1, . . . xn, xn+1. The x-coordinates of points on the manifoldwill thus be functions of the u coordinates

x1 = x1(u1, . . . un), . . . , xn+1 = xn+1(u1, . . . un)

We can put these x-coordinates together into a position vector R for the surface,

R = 〈x1(u1, . . . un), . . . , xn+1(u1, . . . un)〉

We assume the coordinates u1, . . . un are independent which means that the(n+ 1)× n matrix

∂x1

∂u1 . . . ∂x1

∂un

. . . . . . . . .∂xn+1

∂u1 . . . ∂xn+1

∂un

has rank n.

We can now form the vectors

ei =∂R

∂ui

These vectors are linearly independent in view of the requirement on the rankof the matrix above. We can then form the vector n1 by using the base vectorsi1, . . . , in+1 of the Euclidean (n+ 1)-space:

n1 =

∣

∣

∣

∣

∣

∣

∂x1

∂u1 . . . ∂x1

∂un i1. . . . . . . . . . . .

∂xn+1

∂u1 . . . ∂xn+1

∂un in+1

∣

∣

∣

∣

∣

∣

Now n1 is perpedicular to each of the ei because the inner product of ei can bewritten as a determinant

∣

∣

∣

∣

∣

∣

∂x1

∂u1 . . . ∂x1

∂un∂x1

∂ui

. . . . . . . . . . . .∂xn+1

∂u1 . . . ∂xn+1

∂un∂xn+1

∂ui

∣

∣

∣

∣

∣

∣

which, since it has a repeated column, is equal to 0. We now set

n =n1

||n1||


and this is the unit normal vector for the manifold. Because of the position ofthe basis vectors in the definition of n1 the vectors

e1, . . . , en,n

form a basis for Rn+1 which is whose orientation is the same as that of i1, . . . , in+1,as is easily seen (check for n = 1 or n = 2). The n = 1 case will convince you thatthe column of base vectors needs to go at the end rather than at the beginning,where we put it for ordinary vector algebra.

If we don’t like the way the vector n is pointing, for example if we wantthe exterior normal for a closed surface, it is only necessary to renumber theparameters ui.

The vectors ei are tangent vectors to the coordinate lines for the coordinatesui. This is quite possibly familiar to you from advanced calculus courses. Thevectors e1, . . . , en form a basis of the tangent space Tp(M) at each p ∈M .

Although it is not of much use to us in our circumstances, we note that ifM happens to be (even locally) the zero set of some function f, then we can geta normal vector n1 by using the gradient

n1 = ∇f =( ∂f

∂x1, . . . ,

∂f

∂xn+1

)

Try it with the 2-sphere in 3-space with f(x, y, z) = x2 + y2 + z2.We wish to take the derivative of tangent vectors v = eiv

i and to do this weneed first to take the derivative of ei. Since e1, . . . , en,n is a basis of Rn+1 wehave

∂ei

∂uj= ekΓk

ij + nbij

for some coefficients Γkij and bij . For reasons to be explained later the Γk

ij arecalled connection coefficients. There is a symmetry here of the indices

∂ei

∂uj=

∂

∂uj

∂R

∂ui

=∂

∂ui

∂R

∂uj

=∂ej

∂ui

soekΓk

ij + nbij = ekΓkji + nbji

which gives the symmetry conditions

Γkij = Γk

ji bij = bji

At this point we must introduce an idea of great importance. The basisek used to prove the last equation is derived from the position vector R ofthe manifold via ek = ∂R

∂uk and this particular choice of a basis of the tangentspace T (M) is called the natural basis. Naturally there is a natural basis for


each coordinate system. However, for many purposes limitation to the use ofonly natural bases is too restrictive. We will now widen the applicability of ourtheory by removing this restriction. Henceforth, when it is not explicitly statedthat the basis is natural, we will assume it is general. This means that at eachpoint p ∈M the set of vectors e1(p), e2(p), . . . , en(p) will be assumed to be abasis for the tangent space Tp(M) at p. Naturally we assume that the ek(p) aresmooth vector functions of the coordinates ui. If el is the natural basis and ek

is our general basis there will be functions αkl on the manifold so that ek = el α

kl

but with a little luck we will not need to use this. The matrix (αkl ) is of course

invertible.Notice, and this is important, that the symmetry Γk

ij = Γkji is true for the

natural basis but there is no reason to think it is true for a general basis. Hencewhen using this symmetry be sure you are using a natural basis or that you haveproved the symmetry in whatever situation you are in by some other means.Remember symmetry of the Γk

ij is a very special circumstance and not something

to be expected in general3. However, the equation

dei = ∂ei

∂uj duj = ekΓk

ij duj + nbij du

j

remains just as true for general bases as for natural bases.

Now some philosophy. Quantities in differential geometry are of two types.intrinsic quantities are quantities that depend only on the surface itself and noton how it happens to lie in Rn+1 and non-intrinsic quantities. We consider thegij to be intrinsic, which sounds odd since as we have set things up it looks likeM inherits the gij from the surrounding Rn+1. The way to understand this isto think of the surface as having an inner product at each point and then beingmapped into Rn+1 in such a way that the inner product it came with and theinner product it inherits from the surrounding Rn+1 coincide. After you read thenext section this may seem clearer. For a non-intrinsic quantity, the unit normalvector n is very representative, and the bij are also naturally non-intrinsic. Wewill eventually show that the connection coefficients Γk

ij are intrinsic but thisis by no means obvious. The standard way to show something is intrinsic is toshow it depends on the ur-intrinsic quantities gij , the coefficients of the innerproduct, which, I emphasize again, are thought of as provided with the surface.

We now want to change the game in an essential way. We are not interestedvery much in the non-intrinsic aspects of the situation, so we would like thederivative of a tangent vector v to be again a tangent vector, and the differentialof a tangent vector to be an element of T ∗(M). These things are not true asthings stand, and we must modify the definitions to make them true. We dothis by creating an alternative to dv by orthogonally projecting dv onto T ∗(M),which means along n. Because we saw this coming we set things up so this is a

3Einstein played with the idea of finding a place for the electromagnetic field in an asym-metry of the Γk

ij and thus having a unified field theory for gravity and electromagnetics. Itdidn’t work, and we would not expect it to now because of the existence of strong force fieldsand weak force fields


trivial task; the covariant differential and covariant derivative are

Dei = Dei

∂uj duj = ekΓk

ij duj

Dei

∂uj= ekΓk

ij

We are not too interested in the second equation, but the first is critical for allthat follows.

We need the operator D in a couple of other circumstances, and we define it

Df = d f for functions f

D(eiω) = ei dω + (Dei) ∧ ω for differential forms ω

(We are gliding over some technicalities here with tensor products ⊗ but itwould just complicate the notation for very small payoff.) We emphasize thatkeeping the order consistent is very important here. We can now, using theabove formula, find the formula for the differential (in the new sense,) of avector

Dv = D(eivi) = ei dv

i + (Dei) vi

= ei

∂vi

∂ujduj + ekΓk

ij duj vi

= ek

(∂vk

∂uj+ Γk

ij vi)

duj

= ek vk|j du

j

where

vk|j =

∂vk

∂uj+ Γk

ij vi

is the classical notation for the covariant derivative in coordinate form. Thiswill help orient you if you have previous experience with differential geometryor general relativity.

Now that we have Dv what could be more natural than to take anothercoveriant differential. It is cosmically significant that while d2 = 0, this isnot generally true for D and it turns out that D2 churns up one of the mostimportant things in differential geometry.

D2ei = D(Dei)

= D(

ekΓkijdu

j)

= ek d(Γkij du

j) + (Dek)Γkij du

j

= eℓ

∂Γℓij

∂umdum ∧ duj + eℓΓ

ℓkmdu

m ∧ Γkij du

j

= eℓ

(∂Γℓij

∂um+ Γℓ

kmΓkij

)

dum ∧ duj


Now I rewrite the previous expression interchanging the summation indices mand j to get

D2ei = eℓ

(∂Γℓim

∂uj+ Γℓ

kjΓkim

)

duj ∧ dum

= −eℓ

(∂Γℓim

∂uj+ Γℓ

kjΓkim

)

dum ∧ duj

The reason for this sleight of hand is that since D2ei is a vector valued 2-formand we want it’s coefficients to be skew symmetric in m and j. Now we havetwo different expressions for D2ei and we can get still another expression byadding the two and dividing by 2. The result is then

D2ei =1

2eℓ

(∂Γℓij

∂um− ∂Γℓ

im

∂uj+ Γℓ

kmΓkij − Γℓ

kjΓkim

)

dum ∧ duj

= eℓ

1

2Ω ℓ

i

whereΩ ℓ

i = R ℓi mj du

m ∧ duj

is the curvature 2-form and

R ℓi mj =

∂Γℓij

∂um− ∂Γℓ

im

∂uj+ Γℓ

kmΓkij − Γℓ

kjΓkim

is the famous Riemann Curvature Tensor. Note that it is skew symmetric in mand j. Here is is again with more standard letters4.

R ij kl =

∂Γijl

∂uk−∂Γi

jk

∂ul+ Γi

mkΓmjl − Γi

mlΓmjk

Now there is nothing wrong with this derivation and we have presented it inthis way to make sure you see how it works. However, it is relatively inefficient,with far more writing than necessary. We now want to present it in a moreconvenient form which takes far less writing. Recall that

Dei = ekΓkij du

j

= ek ωk

i

where (ω ki ) is the n× n matrix of 1-forms

(ω ki ) = (Γk

ij duj)

Writing things this way, and remembering the old formula

d(ω ∧ η) = (dω) ∧ η + (−1)deg(η)ω ∧ dη4The sign of the Riemann Curature Tensor is not a matter of universal agreement, and

some books define it to be the negative of the definition given here. It is wise always tocheck how the reference you are using defines it. The position of the upper index also varies.Somewhat oddly, Riemann does not seem to have written it down anywhere.


which now becomes critically important, we have

D2ei = D(ekωk

i )

= ek dωk

i +D(ek) ∧ ω ki

= el dωl

i + el ωl

k ∧ ω ki

= el

(

dω li + ω l

k ∧ ω ki

)

Pretending that we haven’t already defined Ω it would be natural at this pointto define

Ω li = dω l

i + ω lk ∧ ω k

i

Let’s check that this is the same as our previous result.

Ω li = d(Γl

ij duj) + (Γl

km dum) ∧ (Γkij du

j)

=( ∂Γl

ij

∂um+ Γl

kmΓkij

)

dum ∧ duj

which is exactly the previous result. Notice however how much more elegantand efficient the new method is.

But if you REALLY want elegance, watch this. We set

~e = (e1, . . . , en) row vector

ω = (ω li ) n× n matrix of 1-forms

Ω = (Ω li ) n× n matrix of 2-forms

Then

D~e = ~eω

D2~e = D(~eω)

= ~e dω + ~eω ∧ ω= ~e (dω + ω ∧ ω)

= ~eΩ

and we see, with implied matrix multiplication

Ω = dω + ω ∧ ω

Note that ω∧ω 6= 0 because it is not multiplication of 1-forms but multiplicationof matrices of 1-forms, and so need not be 0.

Here’s another example; set

~v =

v1

...vn


and then

Dv = D(e~v)

= e d~v + (eω)~v)

= e(

d~v + ω ~v)

If we decode d~v + ω ~v we get for the ith component

dvi + Γijkv

j duk =( ∂vi

∂uk+ Γi

jkvj)

duk

= vi|k, du

k

which is as it should be.For the fun of it we should find dΩ. We use dω = Ω− ω ∧ ω.

dΩ = d(dω + ω ∧ ω)

= 0 + (dω) ∧ ω + (−1)deg(ω)ω ∧ dω= (Ω− ω ∧ ω) ∧ ω − ω ∧ (Ω− ω ∧ ω)

= Ω ∧ ω − ω ∧ Ω

= [Ω, ω]

where we adopt the usual notation [A,B] = AB−BA and we keep in mind thatthe AB and BA are matrix multiplications.

Mixed Covariant Derivatives

We want to find a connection between mixed covariant derivatives and the Rie-mann Curvature Tensor which is a little surprising and very important. Recallthat for ordinary derivatives we have ∂2f/(∂ui∂uj) = ∂2f/(∂uj∂ui) but this isnot true for covariant derivatives and we would like to know just how much itfails. This is a simple calculation. Recall the covariant differential

Dv = eivi|kdu

k where vi|k =

∂vi

∂uk+ Γi

jkvj

and we have the covariant partial derivative notation

Dv

∂uk= ei v

i|k = ei

(

∂vi

∂uk+ Γi

jkvj

)

Now we take another covariant derivative

D2v

∂ul∂uk=

D

∂ul(eiv

i|k) = ei

∂vi|k

∂ul+

(

D

∂ulei

)

vi|k

= em

∂vm|k

∂ul+ (emΓm

il ) vi|k


= em

(

∂vm|k

∂ul+ Γm

il vi|k

)

= em

(

∂2vm

∂ul∂uk+∂Γm

ik

∂ulvi + Γm

ik

∂vi

∂ul+ Γm

il

(

∂vi

∂uk+ Γi

jkvj

))

= em

(

∂2vm

∂ul∂uk+

(

∂Γmjk

∂ul+ Γm

il Γijk

)

vj +

(

Γmik

∂vi

∂ul+ Γm

il

∂vi

∂uk

))

From this we easily get, leaving out the terms symmetric in k and l,

D2v

∂ul∂uk− D2v

∂uk∂ul= em

(

∂Γmjk

∂ul−∂Γm

jl

∂uk+ Γm

il Γijk − Γm

ikΓijl

)

vj

= emRm

j lk vj

which can also be written

vm|k|l − vm

|l|k = R mj lk v

j

It is interesting and important that the final formula has no derivatives of thevi in it. This is related to the fact that R m

j lk is a tensor.Warning: The placement of indices in the Riemann Curvature Tensor R m

j lk

varies from book to book. Hence you should not expect, on consulting a differenttext, to find the indices lined up as we have done it here. This is annoying butunavoidable. One can usually see a book’s convention by checking the definitionof R m

j lk in terms of the Christoffel symbols.

Computation of the Γijk

In this section it is critical that Γkij = Γk

ji. This can be arranged by using anatural basis. The formulas given here are not valid for a general basis unlessit happens that the Christoffel symbols are symmetric: Γk

ij = Γkji.

Recall that∂ei

∂uj= ek Γk

ij + n bij

You may have wondered just how in a particular case you compute Γkij and we

are going to work that out now. It is easy but intricate. The first thing is totake the inner product with el to get

( ∂ei

∂uj, el

)

= (ek Γkij , el) + bij(n, el)

= gkl Γkij + 0

= Γij; l

where for convenience we have introduced the abbreviation

Γij; l = gkl Γkij


We use this formula to coupute ∂gij/∂uk as follows. The second and third

formulas are derived by applying the cyclic permuation i → j, j → k, k → itwice. From

gij = (ei, ej)

we get

∂gij

∂uk=

( ∂ei

∂uk, ej

)

+(

ei,∂ej

∂uk

)

= Γik; j + Γjk; i

∂gjk

∂ui= Γji; k + Γki; j

∂gki

∂uj= Γkj; i + Γij; k

Now we add the last two equations and subtract the one before them. Up tothis point all we have done is valid for a general basis. However, at this point werequire the symmetry of the Christoffel symbols: Γk

ij = Γkij which we would have,

for example, for a natural basis. This transfers immediately to Γjk;i = Γjk;i sowe get

∂gjk

∂ui+∂gki

∂uj− ∂gij

∂uk= Γji; k + Γki; j + Γkj; i + Γij; k − Γik; j − Γjk; i

= Γji; k + Γij; k + Γki; j − Γik; j + Γkj; i − Γjk; i

= 2Γij; k

Then we have

gkmΓij; k =1

2gkm

(∂gjk

∂ui+∂gki

∂uj− ∂gij

∂uk

)

gkmgklΓlij =

gkm

2

(∂gjk

∂ui+∂gki

∂uj− ∂gij

∂uk

)

δml Γl

ij =gkm

2

(∂gjk

∂ui+∂gki

∂uj− ∂gij

∂uk

)

∗∗∗ Γmij =

gkm

2

(∂gjk

∂ui+∂gki

∂uj− ∂gij

∂uk

)

Natural Basis ∗∗∗

which is the required formula for the Γijk in terms of the gij . This formula was

derived from the ei = ∂R∂ui using gij = (ei, ej). It is critical that the coordinates

ui in the formula and the gij are connected in this way, so don’t go using thisformula with gij that come from some other basis, or from somewhere else.

There is another formula for the ΓIij which can be useful when many of the

gij are 0, for example when the coordinates are orthogonal. Recalling the firstformula, we derive

∂ei

∂uj= ekΓk

ij + nbij


(el,∂ei

∂uj) = gklΓ

kij

glm(el,∂ei

∂uj) = Γk

ij = δmk Γk

iglmgklj= Γm

ij

so

Γkij = gkl(el,

∂ei

∂uj )

Now we wish to derive an interesting formula which will be important whenwe get to Riemannian Geometry. We return at this point to a general basis;the material used in the derivation does not depend on the symmetry of theChristoffel symbols. We want a formula for d(v,w) for v,w ∈ Tp(M). Ourapproach is very direct.

∂

∂uk(v,w) =

∂

∂uk

(

gijviwj)

=(∂gij

∂uk

)

viwj + gij

( ∂vi

∂uk

)

wj + gijvi(∂wj

∂uk

)

= (Γik;j + Γjk;i) viwj + gij

( ∂vi

∂uk

)

wj + gijvi(∂wj

∂uk

)

= glj

(

Γlik v

iwj +∂vl

∂uk

)

wj + gilvi(

Γljk v

iwj +∂wl

∂uk

)

=(Dv

∂uk, w)

+(

v,Dw

∂uk

)

d(v,w) =∂

∂uk

(

gijviwj)

duk

=(Dv

∂uk, w)

duk +(

v,Dw

∂uk

)

duk

=(Dv

∂ukduk, w

)

+(

v,Dw

∂ukduk

)

= (Dv, w) + (v, Dw)

Some additional formulas

This subsection deals with the derivation of some formulas which we need forproving the Theorema Egregium of Gauss. Once again we are deriving formulas

for general bases. These formulas have various uses but are not as important asthe previous material. Essentially they show that the Christoffel symbols Γi

jk

are not completely independent from the bij and the theorema egregium comesout of the dependence. We are going to prove these formulas for n-manifoldsembedded in Rn+1 although we will get the most use of them for the case of2-manifolds embedded in R3.

There are a few preliminaries to the proof. First we consider what we candeduce from Aij du

i ∧ duj = 0. Most emphatically this does not mean thatAij = 0, because the objects dui ∧ duj are not linearly independent. We can


get a linearly independent set by taking only the objects dui ∧ duj with i < j.Let us look at how this works.

ω = Aij dui ∧ duj

= Aji duj ∧ dui Interchanging i and j

= −Aji dui ∧ duj

Adding the two expressions for ω we have

2ω = (Aij −Aji) dui ∧ duj

Notice now that Aij − Aji is antisymmetric in i and j, as is dui ∧ duj . Hence,switching i and j, we have

(Aji −Aji) duj ∧ dui = (Aij −Aji) du

i ∧ duj

since the two minus signs cancel. Thus if we take only i < j terms we will haveonly half as many terms as the full i and j and thus

2ω = 2∑

i<j

(Aij −Aji) dui ∧ duj

ω =∑

i<j

(Aij −Aji) dui ∧ duj

This is often called antisymmetrizing the coefficients. Now because the dui∧duj

with i < j form a linearly independent set, we can say that

If ω = Aij dui ∧ duj = 0 then Aij −Aji = 0

We will use this several times in what follows. Clearly this could be generalizedto higher order forms but it would look quite complicated.

Next we need formulas for dn = ∂ n∂uj du

j . Since (n,n) = 1, we have

( ∂n

∂uj, n)

= 0

so ∂ n∂ui is a linear combination of the ek. Since (n, ej) = 0 we have

( ∂ n

∂uj, ei

)

= −(

n,∂ei

∂uj

)

= −bij

If ∂ n∂uj = ekc

kj then we have

(ekckj , ei) = −bijgkic

kj = −bij

gligkickj = −gli bij

def= −blj

clj = δlkc

kj = −blj


so we have∂ n

∂uj= −ekb

kj = −ekg

kibij

Recall that when we calculated D2ei it churned up the curvature tensor. Weare now going do the same thing with d2ei which will naturally be 0. We recallthat

dei = ekΓkij du

j + nbij duj

We recall that d(eiω) = ei ∧ dω + (dei) ∧ ω, so the calculation runs

0 = d2ei = d(ekΓkij du

j + n bij) duj

= ekd(Γkij du

j) + (dek)Γkij du

j + n d(bij duj) + (dn) ∧ bij duj

= ek

(∂Γkij

∂um

)

dum ∧ duj +(

elΓlkmdu

m + nbkmdum)

∧ Γkij du

j

+ n∂bij∂um

dum ∧ duj +∂n

∂umdum ∧ bij duj

= el

(∂Γlij

∂um+ Γl

kmΓkij

)

dum ∧ duj + n(

bkmΓkij +

∂bij∂um

)

dum ∧ duj

− el blmbij du

m ∧ duj

= el

(∂Γlij

∂um+ Γl

kmΓkij − blmbij

)

dum ∧ duj + n(

bkmΓkij +

∂bij∂um

)

dum ∧ duj

From this we see

0 =(∂Γl

ij

∂um+ Γl

kmΓkij − blmbij

)

dum ∧ duj

0 =(

bkmΓkij +

∂bij∂um

)

dum ∧ duj

Remembering the need to antisymmetrize the coefficients which we discusstedearlier we have

∂Γlij

∂um− ∂Γl

im

∂uj+ Γl

kjΓkim − Γl

kjΓkim − blmbij + bljbim = 0

∂bij∂um

− ∂bim∂uj

+ bkmΓkij − bkjΓ

kim = 0

The second equations are the equations of Codazzi-Mainardi. The first equa-tions, when rewritten as

R li mj + blmbij − bljbim = 0 Equations of Gauss

are called the equations of Gauss. Both equations show that the connectioncoefficients Γi

jk and the bij are not independent. We will use the equations of

Gauss to prove Gauss’s theorema egregium in the section on 2-manifolds in R3.


1.3.2 Curves and Geodesics

It has been a while since we used the fact that the n-manifold M was embeddedin Rn+1. For this section we will again make use of this.

The idea is to investigate some of the manifold’s local properties by meansof curves going through the point of interest. For this to be effective we needto know a bit about curves. Sadly, we must limit ourselves to only the mostessential material.

Let C(s) : (a, b) → Rn be a C∞ function from an interval (a, b) ⊆ R toRn. We assume that the parameter s is arc length as this makes things sim-pler without restricting generality very much, although it precludes kinematicinterpretations. Also for simplicity we will assume that C is one to one, so thecurve does not have double points, and we further assume that C′(s) is never0, so the curve does not have cusps and does not reverse direction. If we wereinterested in curves themselves these would be very restrictive conditions but weare interested only in curves on manifolds and for our purposes the conditionscause no problems.

We define the unit tangent vector T as

T =dC

ds

We know T is a unit vector because the parameter is arc length. We now takeanother derivative to form N = dT

ds. Because (T,T) = 1 we have

(dT

ds,T) + (T,

dT

ds) = 0

(dT

ds,T) = 0

dT

ds⊥ T

We now make another assumption and this one will not always be satisfied, butunder normal conditions it will be. We assume

dT

ds6= 0

The only way this can fail over an interval is if the unit tangent vector is constantand the curve is a straight line for that interval. However, it is certainly possible

that dTds

= 0 at the occasional isolated point, and our analysis will fail at thatpoint. The assumption that the curve C(s) is C∞ precludes some of the badthings that can happen. We will talk a bit more about this later.

Since, as we have assumed, dTds6= 0, we can form the unit principal normal

vector N by

N =dTds

||dTds||

and we will also define the curvature of the curve κ at a point as

κ =∣

∣

∣

∣

∣

∣

dT

ds

∣

∣

∣

∣

∣

∣ > 0


(There is a technical carp here; this works only for curves an ambient space ofdimension ≥ 3; for curves in R2 we have a slightly different situation and κ maybe negative. However, this case does not occur for our work.) Trivially we have

dT

ds= κN

Now we suppose that the curve lies in the manifold M , in which case the unittangent vector T will lie in the tangent space to the manifold. However, thereis no reason to think that κN will be in the tangent space. We now resolve κNinto a component parallel to n, the normal to the manifold, and a componentin the tangent space. This idea turns out to be of staggering import.

Here is the explanation. The curvature of a curve on a surface comes fromtwo imputs. There is a component that is due the curvature of the manifolditself and there is a component caused by the curve curving in the manifold.To illustrate the second idea, imagine that you are lost in a barren countrysidewith many hills and valleys and you wish to travel as straight as possible insome direction. You have three sticks of length, say, two meters. Stick one inthe ground. You walk in the direction you wish to go for a way, and, with thefirst stick still visible you stick a second stick in the ground. You then walk awhile further, keeping both sticks in sight, and sighting along the third stickyou adjust your position so that the first two sticks are in line with your presentposition, and then stick the third stick in the ground. Now go get the firststick, and repeat the process above keeping in line with the second and thirdsticks, and put the stick in the ground. Repeating this process over and over,you will trace a path that is as straight as possible in a hilly world5. Such aspath is called a geodesic. When you travel on a geodesic, you are not curvingin the surface, and the normal N to your path points the same direction as thenormal to the surface (or opposite to it). If you deviate from the path indicatedby the sticks, then you are curving in the surface and the principle normal toyour path will deviate from the normal to the surface. An example on the earthof a geodesic (no curving in the surface) is a great circle. Now take a smallercircle tangent to a great circle on the earth and you see that the second circleis curved on the earth’s surface. Think about the normal vectors to the curvesin each case.

Now we would like to make all this mathematical and we have the equipmentto do this; we discussed it in the paragraph above about resolving κN. ResolvingκN into its components perpendicular and parallel to the tangent space, we have

dT

ds= κN = κg s + κn n

(Recall n is the unit normal vector to the surface.) Here s is a unit vector inthe tangent space, κg is the geodesic curvature and κn is the normal curvature.

Def A curve in the manifold is a geodesic ⇐⇒ κg = 0.

5And will travel three times as far.


To clarify a bit, it is true that a geodesic minimizes the distance betweennearby points on it; any other path between the two points is longer than thegeodesic path, but that is NOT the definition. It is a theorem. We will prove itin the section on Riemannia Geometry. Note that the definition is quite closeto the intuitive discussion involving the sticks.

Now we must find formulas for κg and κn. This is fairly easy. The points ofthe curve, being on the surface, have surface coordinates u1(s), . . . , un(s). Thuswe have

C(s) = R(ui(s))

T =dC

ds=

∂R

∂ui

dui

ds= ei

dui

ds

κN =dT

ds= ei

d2ui

ds+∂ei

∂uj

duj

ds

dui

ds

= ek

d2uk

ds+ (ekΓk

ij + nbij)duj

ds

dui

ds

sκg + nκn = κN = ek

(d2uk

ds2+ Γk

ij

dui

ds

duj

ds

)

+ n bijdui

ds

duk

ds

from which we get immediately

sκg = ek

(d2uk

ds2+ Γk

ij

dui

ds

duj

ds

)

κn = bijdui

ds

duk

ds

Since s is a unit vector we have

(kg s, kg s) =(

ek

(d2uk

ds2+ Γk

ij

dui

ds

duj

ds

)

, el

(d2ul

ds2+ Γl

rs

dur

ds

dus

ds

)

)

κ2g = gkl

(d2uk

ds2+ Γk

ij

dui

ds

duj

ds

)(d2ul

ds2+ Γl

rs

dur

ds

dus

ds

)

and we have the formula for κg. Notice that κg depends on the curve and theΓi

jk which means that κg is intrinsic and does not depend on the embedding. Itis more difficult to get s but we have no real need for it so we won’t bother.

Now look at κn. Note that the formula for κn uses only the bij which havenothing to do with the curve C, and the components of the vector T, that is

T = eidui

ds. Hence any two curves C1(s) and C2(s) which go through p ∈ M

in the same direction have the same normal curvature. This means that thenormal curvature κn is a function only of the manifold and the direction inthe tangent space, and it is thus a property of the manifold itself and does notdepend on the curve.

With regard to κg we have the following important theorem. Since the ei

are linearly independent,

Theorem A curve C(s) = R(ui(s)) in the manifold is a geodesic if and only if


it satifies the (nonlinear) system of ordinary differential equations

d2uk

ds2+ Γk

ij

dui

ds

duj

ds= 0

(Here the parameter s is arc length.)

Note that by the usual existence theorems for differential equations there isa unique geodesic stating at a point p ∈M and going in a chosen direction.

Geodesics of of great importance and there are many books devoted to them.An example of a geodesic question is, if you pick a direction at a point on atorus and follow it, will it eventually come back on itself or will it wind aroundforever. Mostly it winds forever, but if you choose the direction carefully itcloses up. On a general manifold, the existence of closed geodesics is a matterof great interest.

From the physics point of view, geodesics are extremely important becausein four dimensional space time light travels in geodesics.

But we must move on.

1.3.3 Special Case; Surfaces in R3

It is worth looking at the special case n = 2 because it has interesting fea-tures, one of the most important of which is the Gaussian Curvature. GaussianCurvature is not only extremely interesting in itself but also is the jumping offplace to a great deal of modern work on manifolds. This was pioneered by S. S.Chern.

Gaussian Curvature

In this section we will take a classical approach to the Gaussian Curvature. Webegin by recalling that the normal curvature of a surface depends only on thesurface. If we sit at a point p ∈M and rotate a unit tangent vector v around thepoint the value of the normal curvature will change and because we are dealingwith a continuous function κn will have a maximum κmax and a minimum κmin.

To be explicit, we wish to maximize and minimize κn(v) under the constraintthat ||v|| = gijv

ivj = 1. This is a problem tailored to the method of LagrangeMultipliers (refer to your old Calculus textbook) whereby we form the auxiliaryfunction

F (vi) = κn(vi)− λ(gijvivj − 1)

and seek to maximize or minimize F . (Physicists and applied mathematicianstake note; the value of λ in problems with Lagrange Multipliers is often ofinterest, for no obvious reason.) We recall that

κn(vi) = bijvivj i, j = 1, 2

We now take the partial derivatives of F with respect to the variables vi and λand set them equal to 0.

∂F

∂ui= bi,jv

j − λgi,jvj = 0 i, j = 1, 2


∂F

∂λ= −(gijv

ivj − 1) = 0

Note that the last equation is simply the constraint, as is always the case. Thefirst two equations are a 2×2 system of homogeneous linear equations, and thushave a solution if and only if the coefficient determinant is 0.

det

(

b11 − λg11 b12 − λg12b21 − λg21 b22 − λg22

)

= 0

Suppose now that we find a solution λ to this equation. Then F will be a max ora min (by the nature of the problem we should have one of each) and moreoverthe vi corresponding to that λ will satisfy ||v|| = gijv

ivj = 1. Since there is amax and a min, we have found them and moreover the max and min of F arethe max and min of

F (vi) = κn(vi)− λ(gijvivj − 1) = κn(vi)− λ · 0

and are thus κmax and κmin. Thus it comes down to solving det(bij −λgij) = 0.This is not particularly challenging. We have

(b11 − λg11)(b22 − λg22)− (b12 − λg12)(b21 − λg21) = 0

(g11g22 − g12g21)λ2 − (g11b22 + g22b11 − g12b21 − g21b12)λ+ b11b22 − b12b21 = 0

Recalling that g = det(gij) and b = det(bij) and that gij = gji and bij = bji wecan rewrite this is

gλ2 − (g11b22 + g22b11 − 2g12b21)λ+ b = 0

and the two solutions of this equation are κmax and κmin. As it happens weare not so interested in these individual values; we want their average and theirproduct. Recall that if α1 and α2 are the two roots of a quadratic equationa(x − α1)(x − α1) = ax2 + bx + c = 0 then a(α1 + α2) = −b and a(α1α2) = cby comparing the coefficients of powers of x. Hence we have the equations

H =1

2(κmax + κmin) =

g11b22 + g22b11 − 2g12b212g

Mean Curvature

K = κmax · κmin =b

gGaussian Curvature

The Mean Curvature has its enthusiasts. For example, if we have a closed curvein 3-space, we might wonder what is the surface of minimum area bounded bythe closed curve. You can find a model by making the closed curve out of wireand dipping it in soapy water. The soap film will (usually) be the surface ofminimum area. This minimal suface has 0 mean curvature, so this is a necessarycondition for a minimal surface. Study of the mean curvature was initiated bySophie Germain in her work on elasticity theory.


The Gaussian Curvature is much more important for the following reason.The formula given, K = b/g, certainlty makes it look like K depends on theembedding of the surface in 3-space. However, this is an illusion. In fact,

Theorem (Gauss’s Theorma Egregium6) The Gaussian Curvature depend onlyon the (gij) and is thus an intrinsic invariant of the surface.

This means, to restate, that the Gaussian curvature does not depend on howthe surface is embedded in 3-space.

To make it clearer what this means, let Φ be a one to one map between twosurfaces M1 and M2. Let u1 and u2 be local coordinates around p ∈M1. Thenu1 and u2 can also be used as coordinates around Φ(p) on M2, and a metric canbe put on M2 by using the (gij) at Φ(p) ∈ M2. To be explicit, we take (v, w)at Φ(p) (for v, w ∈ TΦ(p)) to be (Φ∗(v),Φ∗(w)) on M1. Now suppose that M2

has a metric already defined on it. Then Φ is an isometry if and only if, whenusing the ui as coordinates on both surfaces, the inner products coincide, whichmeans that they have the same (gij) and thus, by the Theorma Egregium, thesame K. Hence it is not possible to have an accurate flat map of the earth(which would preserve the (gij)) because the earth has K = 1/R2 and the flatmap has K = 0 where R is the radius of the earth.

However, we can map a portion of a plane onto a cylinder of any radius assuggested by the fact that plane and cylinder both have K = 0.

The proof of the Theorema Egregium is not at all difficult because we havepreviously set everything up for it. Recall that K = det (bij)/ det (gij). Recallalso the formula of Gauss

R li mj − blmbij + bljbim = 0

We manipulate this equation a bit

R li mj = blmbij − bljbim

gklRl

i mj = gklblmbij − gklb

ljbim = bijbkm − bimbkj

Now we cleverly substitute particular values for the i and j; we let i = j = 1and k = m = 2 and we get

g2ℓRℓ

1 21 = b11b22 − b12b21 = det (bij)

K =det (bij)

det (gij)=

g2ℓ

det (gij)R ℓ

1 21

Since R ℓ1 21 depends only on the Γi

jk and the Γijk depend only on the gij , K

ultimately depends only on the gij and thus is intrinsic; the Theorema Egregiumis proved.

Example: The Sphere

We think it might be worthwhile to give an example of all of this, so we willpresent the results for the sphere of radius a. The conscientious student will

6Egregium is Latin for distinguished, excellent, admirable


work out the details and compare her results with the summary we give. Acouple of things will be worked out in more detail.

The standard parametrization of the sphere of radiua a is to use colatitudeφ = π/2− latitude and longitude θ (in that order). Then the parametrizationis

R = 〈a sinφ cos θ, a sinφ sin θ, a cosφ〉Then

e1 =∂R

∂u1=

∂R

∂φ= 〈a cosφ cos θ, a cosφ sin θ,−a sinφ〉

e2 =∂R

∂u2=

∂R

∂θ= 〈−a sinφ sin θ, a sinφ cos θ, 0〉

Using the standard inner product on R3 we have

(gij) =

(

a2 00 a2 sin2 φ

)

Using the cross product we have, dividing out the length,

n = 〈sinφ cos θ, sinφ sin θ, cosφ〉

To find the Christoffel symbols we use the formula

Γkij = gkl(el,

∂ei

∂uj)

We will need∂e2

∂u1= 〈−a cosφ sin θ, a sinφ sin θ, 0〉

so for example

Γ221 = g2ℓ(el,

∂e2

∂u1)) = g22(e2,

∂e2

∂u1))

=1

a2 sin2 φ

(

a2 sinφ cosφ)

=cosφ

sinφ= cotφ

After a page or two of calculation, we find

ω =

(

0 − sinφ cosφdθcotφdθ cotφdφ

)

Now to find the curvature form we just use

Ω = dω + ω ∧ ω

We have

dω =

(

0 (− cos2 φ+ sin2 φ) dφdθ−csc2φdφdθ 0

)


and

ω ∧ ω =

(


)

∧(


)

=

(

0 cos2 φdφdθcot2φdφdθ 0

)

so (remembering that −csc2φ = −1− cot2φ)

Ω = dω + ω ∧ ω =

(

0 sin2 φ−1 0

)

dφdθ

from which we read off

R 11 12 = 0 R 1

2 12 = sin2 φR 2

1 12 = −1 R 22 12 = 0

Now we can compute the Gaussian Curvature

K =g2j

gR j

1 21 =g22gR 2

1 21 = −g22gR 2

1 12 = −a2 sin2 φ

a4 sin2 φ(−1) =

1

a2

You can also get K by noting that κmax = κmin = 1/a but our method wasmore enlightening.

Lastly we will show that a great circle on the sphere is a geodesic. Recallthat a curve is a geodesic if and only if it satisfies the geodesic equation

d2uk

ds2+ Γk

ij

dui

ds

duj

ds= 0

Since all great circles are pretty much the same, we can take a particular greatcircle, say the equator, and check it for being a geodesic. The equator has thecoordinates u1 = φ = π/2 and u2 = θ which runs from 0 to 2π. For k = 1 inthe geodesic equation the only possible nonzero term in the equation is wheni = j = 2. But Γ2

22 = 0 since ω 22 = cotφdφ and has no dθ term. Thus we need

only worry about k = 2. Note ds = adθ so pulling out the 1/a2 the equationbecomes, remembering u1 = constant,

d2u2

dθ2+ Γ1

22

du2

dθ

du2

dθ= 0

But ω 12 = − sinφ cosφdθ so Γ1

22 = − sinφ cosφ = − sinπ/2 cosπ/2 = 0 Hencethe equator, and thus all great circles, satisfies the geodesic equation and thusall great circles are geodesics.

It would have been easy to reason geometrically, since the the unit normal toa great circle is just −n (where n is the normal to the sphere) and thus κg = 0but we wanted to see how this could be done analytically.


The Gauss-Bonnet theorem

The Gauss-Bonnet theorem is one of the most important theorems in mathe-matics, partly because it unifies a lot of geometry by proving many elementaryformulas in a coherant way and partly because it was the jumping off pointfor such advanced topics as Chern cohomology classes and following this roadhas lead to vast generalizations of our good old surface theory. Here differen-tial geometry intersects the algebraic topology of manifolds in highly non-trivialways.

The Gauss-Bonnet theorem comes in several forms and we will state differentlooking forms of it. The first might be called the Polygonal form.

Theorem (Gauuss-Bonnet I) Let M be a 2-manifold and S ⊆ M be a simplyconnected subset of M whose boundary is a smooth curve except for a finitenumber of corners with exterior angles αi. Then

∫

S

K dS +

∫

∂S

κg ds+∑

i

αi = 2π

Here K is the Gaussian curvature of S, κg is the geodesic curvature of ∂S,dS is the element of surface area, and ds is the element of arc length.

The second form might be called the Topological form. We need the followingideas. The genus g of a closed (no boundary) 2-manifold M is the number ofholes. For example a sphere has genus g = 0, a 2-torus has g = 1 and a pretzel(skin only) has g = 3. If a manifold is cut into regions by “polygons””, andthere are V vertices, E edges and F faces then the Euler-Poincare characteristicof M is

χ(M) = V − E + F

and it is known that

χ(M) = V − E + F = 2(1− g)

The topological form of the Gauss-Bonnet theorem is then

Theorem (Gauss-Bonnet II) Let M be an oriented 2-manifold of genus g. Then∫

M

K dS = 4π(1− g) = 2πχ(M)

Since the proof of this theorem is a little harder than most results in thisbook and we don’t want you to miss the applications since they are so beautiful,we have decided that we will first present the applications and then do the proof.

App 1. Surface area of a sphere. Use GBII and the fact that the Gaussiancurvature is K = 1/a2 to get

∫

M

K dS = 4π(1− g)∫

M

1

a2dS = 4π(1− 0)


1

a2

∫

M

dS = 4π

surface area of sphere = 4πa2

App 2 Surface area of Lune, with opening angle ψ. The lune is bounded bygreat circles that have κg = 0 and the exterior angles are π−ψ so from GBI wehave

∫

S

K dS +

∫

∂S

κg ds+∑

i

αi = 2π

∫

S

1

a2dS +

∫

∂S

0 ds+ π − ψ + π − ψ = 2π

surface area of lune = 2a2ψ

App 3 Area of a Geodesic Triangle on a Sphere. The “straight lines” on asphere are the geodesics which are great circles. Hence the closest thing to atriangle on a sphere is a the area bounded by three great circles. Three greatcircles actually bound a number of areas, but the area is selected by choosingthe angles. An angle on a sphere between two curves are defined as the anglebetween the tangent lines to the curves where the curves meet. Suppose theinterior angles of the triangle are α, β and γ. Then the exterior angles for thetriangle are π − α, π − β, and π − γ. Once again, for a sphere of radius a wehave the Gaussian Curvature K = 1/a2 and κg = 0 since we are dealing withgeodesics, so

∫

S

K dS +

∫

∂S

κg ds+∑

i

αi = 2π

∫

S

1

a2dS +

∫

∂S

0 ds+ π − α+ π − β + π − γ = 2π

1

a2S = 2π + α+ β + γ − 3π

surface area of tiangle = a2(α+ β + γ − π)

Triangles on Spheres always have an angle sum in excess of π in contrast totriangles on the plane, where the angle sum is exactly π. The amount by whichthe angle sum exceeds π is called the excess and what we have shown is the thatthe area = a2× (the excess).

App 4 Geodesic Polygon on a sphere. Let the interior angles be βi; i = 1, . . . , nso the exterior angles are π − βi; i = 1, . . . , n. Then, as above,

∫

S

1

a2dS +

∫

∂S

0 ds+

n∑

i=1

(π − βi) = 2π

1

a2· Area + nπ −

n∑

i=1

βi = 2π


Area = a2(n∑

i=1

βi − (n− 2)π)

App 5 You can only understand this one if you know something about Lobachevskygeometry, also called hyperbolic geometry. The Lobachevsky plane has a con-stant a analogous to the radius for a sphere and now the Gaussian curvature isK = −1/a2. Negative Gaussian Curvature is characteristic of saddle-like sur-faces and one model of part of the Lobachevski plane is the pseudosphere whichindeed is saddle like at every point. In Lobachevsky geometry the angle sum isalways less than π and there is thus a deficit which is π − (α+ β + γ). Almostprecisely the same calculation as in App 3 tells us that the area of a geodesictriangle is a2× (the deficit). Geodesics in Lobachevsky geometry can meet oneanother with angle 0. A geodesic triangle in which this happens will then havethe maximum possible area for a geodesic triangle which will be πa2.

App 6 Geodesic Curvature of a parallel. A parallel on a sphere is a curvedefined by φ = constant. (This is a ludicrously bad name because these curvesare not great circles. Since any two great circles clearly meet each other, thereare no “parallel lines” on a sphere. However the name parallel is traditional andwe are stuck with it.) The equator φ = π/2 is a geodesic, but the parallels withφ < π/2 are not geodesics, and thus have positive geodesic curvature κg whichwe are now going to compute analytically with the help of the Gauss Bonnettheorem. Let φ0 determine the parallel. The area of the cap above the parallelis

∫

cap

a2 sinφdφdθ =

∫ 2π

0

∫ φ0

0

a2 sinφdφdθ = 2πa2(1 − cosφ0)

We then have, making use of the obvious κg = constant,

∫

S

K dS +

∫

∂S

κg ds+∑

i

αi = 2π

1

a22πa2(1 − cosφ0) + 2πa sinφκg + 0 = 2π

− cosφ0 + a sinφ0 κg = 0

κg =1

acotφ0

Proof of the Gauss Bonnet Theorem

We first prove the theorem for a 2-manifold (surface) and a region S ⊆ Mwhose boundary ∂S is a smooth curve (we are putting off the corners till later)and which is contained in a single coordinate patch. It will be important to usthat the general basis we use is orthonormal. To arrange this let u1, u2 be anoriented coordinate system for M and set e1 = ∂R

∂u1 , where R is the positionvector of M in R3. Let g11 = (e1, e1) and set

e1 =1√g11

e1


Thus e1 is a unit vector. Let e2 be a unit vector in Tp(M) which at each pointis orthogonal to e1 and in a direction so that e1, e2 has the same orienta-tion as M . We have now created a orthonormal general basis of Tp(M). Theorthogonality will have an effect on the curvature form ω k

i . We have

(ei, ej) = δij

(Dei, ej) + (ei, Dej) = Dδij = 0

(ekωk

i , ej) + (ei, elωl

j ) = 0

ω ki δkj + ω l

j δil = 0

ω ji + ω i

j = 0

ω ji = −ω j

i

Now we are now ready to launch into the proof, which consists of 3 parts.In the first, we effect a cosmetic change on

∫

∂Sκg ds to get the integral of a

certain form η so the integral becomes∫

∂Sη. The second part is to apply stokes

theorem, which is very quick. The third part is more cosmetic shifting on dη sothat it comes out what we need for the Gauss Bonnet theorem. None of this isdeep; it’s just a bit tricky. Also, the only substantive step is the use of Stokestheorem; the rest is cosmetic shifting.

For the points along the curve ∂S we must use another basis. The directionof integration for ds is the one determined on ∂S by the orientation on M andinherited by S. The first vector is the unit tangent vector T to the curve directedin the direction of the orientation and the second is the unit vector we called s

in the section on curves but which we are now going to call b to avoid variableclash. We arrange b so that it is in the tangent space Tp(M) for p ∈ ∂S andis perpendicular to T. The direction of b is determined so that when we goaround the curve ∂S according to its orientation, T,b has the same orientationas M . Then we have

DT

ds= bκg

Db

ds= −Tκg

Note the coefficients are skew symmetric as we showed they had to be for anorthonormal basis. On the other hand we have

e1 = T cos θ − b sin θ

e2 = T sin θ + b cos θ

so that

De1

ds= (−T sin θ − b cos θ)

dθ

ds+DT

dscos θ − Db

dssin θ

= −e2dθ

ds+ bκg cos θ − (−Tκg) sin θ


= −e2dθ

ds+ e2κg

= e2

(

κg −dθ

ds

)

The reader may wish to practise by calculating De2

dsand she should come out

withDe2

ds= e1

(

− κg +dθ

ds

)

Note skew symmetry. (We do not actually need this result.) Now we have

De1 = eiωi

1 = e1ω1

1 + e2ω2

1 = e2ω2

1

since ω 11 = 0 due to skew symmetry and we have

ω 21 =

(

κg −dθ

ds

)

ds

= κgds− dθ

We have finished part 1; we have derived the differential form η to which we willapply Stokes theorem and it is ω 2

1 = κgds− dθ. This is of course not obvious.

The next step is to apply Stokes theorem. Before we do this let us observethe following.

ω 2k ∧ ω k

1 = ω 21 ∧ ω 1

1 + ω 22 ∧ ω k

2 = 0

because, due to skew symmetry, ω 11 = ω 2

2 = 0. Next note that

Ω 21 = dω 2

1 + ω 2k ∧ ω k

1

dω 21 = Ω 2

1 − ω 2k ∧ ω k

1

and so we have∫

∂S

(κg − dθ) =

∫

∂S

ω 21

=

∫

S

dω 21 the substantive step

=

∫

S

Ω 21 − ω 2

k ∧ ω k1

=

∫

S

Ω 21

since, as we showed, the second term in the integrand is 0.The general formula for the Gaussian Curvature K is

K =g2ℓ

det(gij)R ℓ

1 21


1.4 Some Tensors and the Proof of the Gauss-

Bonnet theorem

1.4.1 Tensors and their algebra

The reader may have wondered how tensors fit into the structure that we havebeen building, and this is a convenient place to discuss this. We will discusstensors in a very superficial way, and in addition we will discuss them in a non-traditional way. Lore on Tensors is vast and for readers who want to pursuethem in greater depth we recommend [9] for the pure theory and [?] and [?]for more applications oriented treatments. For those who wish to sail theirrowboats through index storms across the tensor sea, there is [?].

The most difficult thing to learn about tensors is that there is almost nothingto learn. Let V be a vector space with dual V ∗. For our purposes these areV = T (M) and V ∗ = T ∗(M) but the concepts have applications elsewhere.Letv1, . . . , vk ∈ V and λ1, . . . , λℓ ∈ V ∗. A tensor product is a formal product(where α is a scaler)

αv1 ⊗ . . .⊗ vk ⊗ λ1 ⊗ . . .⊗ λℓ

This product is multilinear and the scalers may move around freely. That is

α v1 ⊗ v2 ⊗ . . .⊗ vk ⊗ λ1 ⊗ . . .⊗ λℓ + β v1 ⊗ v2 ⊗ . . .⊗ vk ⊗ λ1 ⊗ . . .⊗ λℓ

= v1 ⊗ (α v2 + β v2)⊗ . . .⊗ vk ⊗ λ1 ⊗ . . .⊗ λℓ

and similarly with any other slot. For convenience we put λs from V ∗ at theend so, for example

(v1 ⊗ v2 ⊗ λ1 ⊗ λ2 ⊗ λ3) ⊗ (v3 ⊗ v4 ⊗ v5 ⊗ λ4 ⊗ λ5)

= v1 ⊗ v2 ⊗ v3 ⊗ v4 ⊗ v5 ⊗ λ1 ⊗ λ2 ⊗ λ3 ⊗ λ4 ⊗ λ5

We reiterate, because it seems hard to learn, that v1 ⊗ v2 is not some vector orsome other previously known object; it is a formal product which can mutateunder bilinerity and is associativity but that is all. It is remarkable that sucha construction is so useful, although the use often requires additional structure.Tensor analysis itself does not have powerful theorems like Stokes’ theorem fordifferential forms.

Naturally if e1, . . . en is a basis for V with dual basis e1, . . . en for V ∗ we can,using bilinearity, express all tensors as sums of products of the form

α e1 ⊗ . . .⊗ ek ⊗ e1 ⊗ . . .⊗ eℓ

Thus we could write such a sum as

ti1...ik

j1...jℓei1 ⊗ . . .⊗ eik

⊗ ej1 ⊗ . . .⊗ ejℓ

with the Einstein summation convention in force for the indices. In practice kand ℓ are stable in a calcultion.

This finishes our introduction to the algebra of tensors. Next we will bringon tensor calculus, which takes place on manifolds.

1.4. SOME TENSORS AND THE PROOF OF THE GAUSS-BONNET THEOREM31

1.4.2 Tensor Calculus

You cannot actually do any tensor calculus unless you have an affine connection.With an affine connection you can do quite a lot, but the full power of tensorcalculus only comes when you have an inner product on the tangent space of themanifold which means you have a (pseudo-)Riemannian metric. We will dealhere with only the basic stuff which involves the affine connection. Essentiallythis means that we have coefficients Γi

jk so that we have

Dej = ei Γijkdu

k

Dei

∂uk= ej Γi

jk

Affine connections are the subject of the next section but if you can just be-lieve the formulas temporarily we can deal with tensors and a few other minormatters.

It is very important that you realize that

Γijk is ∗NOT∗ a tensor

It does not have the correct transformation formula for a tensor when you changethe coordinates. We discuss this later in this section and much more extensivelyin the next section.

Next we want a formula for Dei

∂uk . Let us regard the formula ei(ej) = δij , or

using a more symmetric notation 〈ei, ej〉 = δij , as a kind of product, and then

let us insist that Leibniz’ formula for products works. Then we have, settingDei

∂uk = γijke

j for some constants γijk

Dδij

∂uk=

⟨Dei

∂uk, ej

⟩

+⟨

ei,Dej

∂uk

⟩

0 = 〈γimke

m, ej〉+ 〈ei, eℓΓℓjk〉

= γimkδ

mj + Γℓ

jkδiℓ

= γijk + Γi

jk

γijk = −Γi

jk

Dei

∂uk= −Γi

jk ej

Now let us take the derivative of some random tensor

tijei ⊗ ej

using, as usual, Leibniz’ rule. We get

(D tij ei ⊗ ej)

∂uk=

∂tij∂uk

ei ⊗ ej + tijDei

∂uk⊗ ej + tij ei ⊗

Dej

∂uk

=∂tij∂uk

ei ⊗ ej + tij Γℓikeℓ ⊗ ej − tij Γj

ℓkei ⊗ eℓ


=∂tij∂uk

ei ⊗ ej + tℓj Γiℓk ei ⊗ ej − tiℓΓℓ

jk ei ⊗ ej

=( ∂tij∂uk

+ Γiℓk t

ℓj − Γℓ

jk tiℓ

)

ei ⊗ ej

The object in parentheses is the covariant derivative of the tensor. In tensornotation we call tij the tensor, i a contravariant index and j a covariant index.(We recall contravariant means changes like the basis vectors and contravariantmeans changes like the coefficients of a vector7.) Then to take the covariantderivative of the tensor you have to add terms involving the connection coeffi-cients for each contravariant index (upstairs index) and subtract terms involvingthe connection coefficients for each coveriant index (downstairs index.) Commonnotations are

∇k tij = vi

j|k =∂tij∂uk

+ Γiℓk t

ℓj − Γℓ

jk tiℓ

Note

∇k vi = vi

|k =∂vi

∂uk+ Γi

ℓk vℓ

which is our old familiar coveriant derivative so that our new stuff is a general-ization of the material we got from differential forms to a wider class of objects.We note that the formulas we have derived are very general and do not dependon Γi

jk = Γikj .

Another possible way to look at it is

(D tij ei ⊗ ej)

∂uk= (∇k t

ij) ei ⊗ ej = vi

j|k ei ⊗ ej

As a useful example of this, let’s find the coveriant derivative to metric tensorgij

∇k gij =∂gij

∂uk− Γm

ik gmj − Γmjk gim

=∂gij

∂uk− Γik;j − Γjk;i

= 0

as we found when we were finding a formula for the Γijk. Note also

∇kg =∂g

∂gij

∇kgij = 0

Now it is time to figure out why some indices are upstairs (contravariant)and some are downstairs (covariant) and what happens to a tensor’s coefficientswhen the basis is changed. We will handle this for natural bases as this istraditional. We will also handle it for a simple example and leave it to the

7Some people feel that the names should be reversed, which would add more to a subjectalready awash in confusion. Anyway a good case can be made the the current choice is correct


reader to guess the general form, which is a trivial genralization. For yourconvenience we will rederive some things right here.

Recall that ei is just a convenient notation for the tangent vector ∂/∂ui andei is just convenient for dui, which forms the dual basis to ∂/∂ui. Hence we see

ei =∂

∂ui=

∂uj

∂ui

∂

∂uj=

∂uj

∂uiej contravariant change

ei = dui =∂ui

∂ujduj =

∂ui

∂ujej covariant change

Now how does the tensor tijk change. Well it is short for T = tijkei ⊗ ej ⊗ ek soit must change like

T = tijkei ⊗ ej ⊗ ek = tijk

∂ur

∂uier ⊗

∂uj

∂uses ⊗ ∂uk

∂utet

= tijk

∂ur

∂ui

∂uj

∂us

∂uk

∂uter ⊗ es ⊗ et

SinceT = trst er ⊗ es ⊗ et

we have

trst(um) = tijk(up(um))

∂ur

∂ui

∂uj

∂us

∂uk

∂ut

where we have slipped in the change of variables in the arguments of of tijk aswell.

This is the tensor change law. In olden times a tensor was defined as anindexed quantity that changed in this manner; the mantra was a tensor is an

indexed quantity that changes like a tensor, that is according to the above model.Mathematicians (in contrast to physicists) are no longer completely comfortablewith this approach and so we have taken a different approach, which may clarifythings for some people.

Now the big question is, is vj

|k a tensor; does the“deriviative index” k change

properly? We will do this example that shows that this is indeed the case. Youcan use the same technioques to show that it works for any tensor. For conve-nience, it helps to modify the change of basis law for the connection coefficientsa bit. We will derive this modification from scratch and the formulas may comein handy for other things.

ekΓkij =

Dei

∂uj=

D

∂uj

(

en

∂un

∂ui

)

=D

∂um(en)

∂um

∂uj

∂un

∂ui+ en

∂2un

∂uj∂ui

eℓΓℓij − en

∂2un

∂uj∂ui= epΓ

pmn

∂um

∂uj

∂un

∂ui

eℓΓℓij − eℓ

∂uℓ

∂un

∂2un

∂uj∂ui= eℓ

∂uℓ

∂upΓp

mn

∂um

∂uj

∂un

∂ui

Γℓij −

∂uℓ

∂un

∂2un

∂uj∂ui=

∂uℓ

∂upΓp

mn

∂um

∂uj

∂un

∂ui(

Γℓij −

( ∂uℓ

∂un

∂2un

∂uj∂ui

)

)

∂up

∂uℓ

∂uj

∂um

∂ui

∂un= Γp

mn


Note that the first term on the left is a standard tensorial change; it is the secondterm we are interested in. It compensates for things that appear because thebasis frame is moving. Keep your eye on how it works. We now show the vj

|k

changes in a tensorial manner. This is straightforward but tedius, as is commonwith tensor calculations.

vℓ|m =

∂vℓ

∂vm+ Γℓ

nmvn

=∂

∂ui

(

vj ∂uℓ

∂uj

) ∂ui

∂um+

(

Γkij −

(∂uk

∂up

∂2up

∂uj∂ui

)

)

∂uℓ

∂uk

∂uj

∂um

∂ui

∂unvn

=∂vj

∂ui

∂uℓ

∂uj

∂ui

∂um+ vj ∂2uℓ

∂ui∂uj

∂ui

∂um+ Γk

ij

∂uℓ

∂uk

∂uj

∂umvi − ∂uℓ

∂uk

∂uk

∂up

∂2up

∂ui∂uj

∂uj

∂umvi

=(∂vj

∂ui+ Γj

kivk)∂uℓ

∂uj

∂ui

∂um+ vj ∂2uℓ

∂ui∂uj

∂ui

∂um− ∂2uℓ

∂ui∂uj

∂uj

∂umvi

= vj

|i

∂uℓ

∂uj

∂ui

∂um

which is the proper tensor change law. All such covariant derivatives of tensorscan be proved to be tensors with exactly the same computation. We havepresented this so that the reader can observe how the second derivative term inthe change of variables formula just cancels the second derivative term whichshows up in the derivative ∂/∂ui(vj ∂uℓ/∂uj)∂ui/∂um.

1.4.3 Raising and Lowering Indices

In the presence of an inner product it is possible to shift covariant and con-travariant indices in a tensor. It is usually necessary to be quite careful aboutkeeping the indices in the same vertical strip as we move them up and down,and for clarity a dot may be used as a placeholder. For example, the RiemannCurvature Tensor may be written with the dot as

R ij kℓ = R i

j · kℓ

where the dot makes it clear that i is above the second slot.The raising and lowering is done with the gij and gij as follows.

Rj i kl = gimR mj · kℓ

The reason for doing this is that the modified tensor may have more congenialsymmetries, or be easier to calculate, or a number of other reasons. Rememberthat this is just convenience; there is no real content here. Also note that wecan always undo this

R mj · kℓ = gmiRj i kl

Another example is the lowering of an index on a contrvariant vector to geta covariant vector.

vj = gijvi


Remember that g is symmetric, e.g. gij = gji, so the order of the indices doesn’tmatter. When this is done we then have the inner product written as

(v, w) = gij viwj = vj w

j

and this can be convenient.

1.4.4 Epsilon tensors

That the Kronecker delta is a tensor is a trivial matter and we leave it to thereader. The next tensor of somewhat similar type is the epsilon tensor, whichwe will discuss after some introductory remarks. Epsilon tensors were used toform dual tensors before the knowledge of the ∗-operator became common 8.They are a rather clumsy tool to do the ∗-operator with, and the techniquesused earlier in the book are much more convenient. However, in low dimensionsthey do have their uses.

Let (i1, i2, . . . , in) be a permutation of (1, 2, . . . , n) which we think ofa as afunction f(1) = i1, f(2) = i2, . . . , f(n) = in and symbolize by

σ =

(

1 2 , . . . , ni1 i2 , . . . , in

)

We symbolize the sign of this permutation by sgn(i1, i2, . . . , in). Now recall thatthe determinant of an n× n matrix aij is a sum over all permutations σ of 1 ton

det(aij) =∑

σ

sgn(σ) a1i1 · · · aninor

det(aij) =∑

σ

sgn(σ) ai11 · · · ainn

and similarly for (aij) and (aij).

We remind the reader that we assume that on each Tangent space Tp(M)we have an inner product (v, w)p) which is smooth in p. For an embedded man-ifold this comes from the embedding space, but, as we will learn later, it canbe supplied to the manifold by fiat, and this is called a (pseudo-)Riemannianmanifold. It is pseudo-Riemannian if the inner product is non-degenerate butnot positive definite, and we have consideredsuch cases previously, but will notconsider them at present for simplicity. There is not much change from Rieman-nian to pseudo-Riemannian in the formalism. Of course there is a big changein the geometry.

For any basis e1, . . . , en we have a matrix (gij) for the inner product givenby gij = (ei, ej) with inverse matrix (gij) = (gij)

−1. Now recall Grassmann’stheorem

(e1 ∧ . . . ∧ en, e1 ∧ . . . ∧ en) = det(

(ei, ej))

= det(gij) = g

8Although the work had already been done by Grassmann in the AUSDEHNUNGSLEHREof 1861.


Now we will look at the situation when we change from one natural basis toanother. We have

ei =∂

∂uiei =

∂

∂ui

ei =∂

∂ui=

∂uj

∂ui

∂

∂uj=

∂uj

∂uiej

e1 ∧ . . . ∧ en = det(∂uj

∂ui

)

e1 ∧ . . . ∧ en

Taking the inner product of each side with itself we get

g = det(∂uj

∂ui

)2

g

√

g

g= det

(∂uj

∂ui

)

√

g

g= det

( ∂ui

∂uj

)

Now we want a tensor that acts rather like sgn but sgn itself does not quitetransform like a tensor. It must be multiplied by a function as follows

εi1...in=√g sgn(i1 . . . in) εi1...in =

sgn(i1 . . . in)√g

Now we will check the tensor character of ǫi1...in .

εj1,...jn∂ui1

∂uj1· · · ∂u

in

∂ujn=

1√g

det

∂ui1

∂u1 . . . ∂ui1

∂un

· · · · · · · · ·∂uin

∂u1 . . . ∂uin

∂un

=1√g

sgn(i1 · · · in) det

∂u1

∂u1 . . . ∂u1

∂un

· · · · · · · · ·∂un

∂u1 . . . ∂un

∂un

=1√g

sgn(i1 · · · in)

√

g

g=

1√g

sgn(i1 . . . in)

= εi1...in

1.4.5 Epsilon tensors and Dual tensors in Two Dimensions

We can use the epsilon tensors to construct dual tensors although the processis clumsy. However, in two dimensions it works rather well. Let v = ei v

i Then∗v is defined by the usual relation

w ∧ ∗v = (w, v)Ω0

for all w ∈ T (M) where Ω0 = e1 ∧ e2/||e1 ∧ e2|| = e1 ∧ e2/√g. We claim

∗w = ej (ε·jk vk) = ej (gki ε

ij vk)


This can be done by a calculation but since the dimension is 2 it is interestingto look at what happens in detail; to see the gears whirr and clank. To do thiswe first determine ε·jk = gik ε

kj .

ε·11 = g1kεk1 = −g12

1√g

ε·12 = g2kεk1 = −g22

1√g

ε·21 = g1kεk2 = g11

1√g

ε·22 = g2kεk2 = g21

1√g

Now we compute ∗v

∗v = ej ε·ji v

k

= e1 (ε·11 v1 + ε·12 v

2) + e1 (ε·21 v1 + ε·22 v

2)

=1√g

[

e1 (−g12 v1 − g22 v2) + e2 (g11 v1 + g21 v

2)]

Next we calculate

w ∧ ∗v = (e1w1 + e2w

2) ∧ 1√g

[

e1 (−g12 v1 − g22 v2) + e2 (g11 v1 + g21 v

2)]

=[

w1(g11 v1 + g21 v

2)− w2(−g12 v1 − g22 v2)]e1 ∧ e2√

g

=(

gij viwj)

Ω0 = (v, w)Ω0

which proves our assertion ∗v = ej (gki εij vk).

Now recall that

∗ ∗ v = (−1)deg(v)(dim(V )−deg(v)) = (−1)1(2−1)v = −v

This relation is also easily proved by using the above formula for ∗v twice. Wesuggest you do this so you can see how the factor 1/

√g once again saves the

day. Using this we can show || ∗ v|| = ||v|| and that v, ∗v form an orientedorthonormal basis for V = Tp(M). Indeed we have

(∗v, ∗v)Ω0 = ∗v ∧ ∗ ∗ v = ∗v ∧ (−v) = v ∧ ∗v = (v, v)Ω0

So, Λ2(V ) being one-dimensional, we have (∗v, ∗v) = (v, v). Also note theformula v ∧ ∗v = (v, v)Ω0 which, since (v, v) > 0, tells us that v, ∗v is orientedlike e1, e2.

Finally,(v, ∗v)Ω0 = v ∧ ∗ ∗ v = v ∧ (−v) = −v ∧ v = 0

so v is perpedicular to ∗v.

1.4.6 The Riemann Curvature Tensor in Two Dimenions

We will now look at some aspects of the Riemann curvature tensor for a twodimensional manifold embedded in 3-space. We will see that much informa-tion can be extracted from the symmetries of the coefficients. We recall some


previous equations to remind the reader who the actors are. First, recall that

∂ei

∂uj= ekΓk

ij + bij n

where e1, e2 is a basis of Tp(M) and n is the unit normal vector. The Riemanncurvature tensor we remember is

R ℓi mj =

∂Γℓij

∂um− ∂Γℓ

im

∂uj+ Γℓ

kmΓkij − Γℓ

kjΓkim

The first equation following is the equation of Gauss and it will be critical forus. In the second equation we lower the contravariant index.

R li mj = blmbij − bljbim

Rikmj = gklRl

i mj = gklblmbij − gklb

ljbim = bijbkm − bimbkj

The second equation allows us to easily determine some symmetries of the Rie-mann curvature tensor. It is not available in general Riemannian Geometry sothere the symmetries are a little harder to prove. First note, interchangeing thefinal two indices j and m, that

Rikjm = bimbkj − bijbkm = −(bijbkm − bimbkj) = −Rikmj

so Rikmj is antisymmetric in the last two indices. (We already knew this fromearlier theory.) Next we swap the first two indices i and k to get

Rkimj = bkjbim − bkmbij = −(bijbkm − bimbkj) = −Rikmj

so Rikmj is also antisymmetric in the first two indices. Then of course

Rkimj = −Rikmj = Rikjm

Now we are going to swap the first pair of indices ik with the second pair ofindices jm

Rmjik = bmkbji − bmibjk = bijbkm − bimbkj = Rikmj

So swapping the first pair and the second pair leaves Rikmj invariant. Now weknow the value of one set of indices from the the theorem egregium of Gauss.We have

R1221 = b11b22 − b12b21 = det(bij)

But we know that

K =det(bij)

det(gij)=R1221

g

so

R1221 = gK


Using the antibymmetries found above we can now write down the complete setof values of Rikmj . They are

R1111 = 0 R1112 = 0 R1121 = 0 R1122 = 0R1211 = 0 R1212 = −gK R1221 = gK R1222 = 0R2111 = 0 R2112 = gK R2121 − gK R2122 = 0R2211 = 0 R2212 = 0 R2221 = 0 R2222 = 0

Because of the antisymmetries it is typical of the Riemann curvature tensor tohave a large number of zero components. The fully covariant Riemann curvaturetensor is simpler than the original tensor. For example

R 21 21 = g2ℓR1ℓ 21 = g21R11 21 + g22R1221 = g22gk

We will need the formula (∇k∇j − ∇j∇k) vi = R im kj v

m. Indeed this is howthe Gaussian Curvature K enters into the Gauss-Bonnet theorem. To get a littletensor practise we will derive the formula with tensor methods.

∇jvi =

∂vi

∂uj+ Γi

mjvm = vi

|j

∇k vi|j =

∂

∂ukvi

|j + Γinkv

n|j − Γℓ

jkvi|ℓ

=∂

∂uk

( ∂vi

∂uj+ Γi

mjvm)

+ Γink

(∂vn

∂uj+ Γn

mjvm)

− Γℓjk

( ∂vi

∂uℓ+ Γi

mℓvm)

=∂2vi

∂uk∂uj+∂Γi

mj

∂ukvm + Γi

mj

∂vm

∂uk+ Γi

nk

∂vn

∂uj− Γℓ

jk

∂vi

∂uℓ

+(

ΓinkΓn

mj − ΓℓjkΓi

mℓ

)

vm

Now we swap j and k and hope a lot of stuff cancels out.

∇j vi|k =

∂2vi

∂uj∂uk+∂Γi

mk

∂ujvm + Γi

mk

∂vm

∂uj+ Γi

nj

∂vn

∂uk− Γℓ

kj

∂vi

∂uℓ

+(

ΓinjΓ

nmk − Γℓ

kjΓimℓ

)

vm

The terms involving derivatives of vi cancel out in pairs, as do two of the termsmultipling vm leaving us with

(∇k∇j −∇j∇k) vi =(∂Γi

mj

∂uk− ∂Γi

mk

∂uj+ Γi

nkΓnmj − Γi

njΓnmk

)

vm

= R im kj v

m = R im kj v

m

This can also be written, where we abbreviate vi|j|k as vi

|jk, (and note the

order)vi

|jk − vi|kj = R i

m kj vm

The advantage of the differential form method for deriving these things is thatoften terms that cancel out in the tensor derivation never appear at all in thedifferential form derivation.


1.5 General Manifolds and Connections

In this short section we introduce the idea of Connections on the tangent bundleof an arbitrary differentiable manifold. This is easy. A connection on the tangentbundle is often called an affine connection. It is trivial to put connections onany vector bundle on a manifold, and we will eventually talk about this.

A manifold is covered by a set of coordinate patches. We choose one of thesepatches with coordinates u1, . . . , un. Recall that in this case the basis vectors ei

are defined to be the differential operators ∂∂ui but we will continue to call them

ei so things will look familiar. In a few places we have to dredge up ei = ∂∂ui

but not often.We now introduce on this coordinate patch a Covariant Differential D by

settingDei = ej ω

ji = ej Γj

ik duk

Now you ask, what are the conditions on the Γjik for this to be a connection

and the answer is none whatever. You pick any Γjik on this patch and you have

a connection on this patch. You may choose the connection according to somemathematical or physical situation or by pure whimsey. However, once youhave chosen it for this patch it will migrate to the overlaps of other patcheswith this patch, and your choice of connection on other patches is constrainedby this. This is very important. Let us now look at the overlap of our patch witha nearby patch which has coordinates u1, . . . , un. This is one of those momentswhen we need to remember what ei actually is. We have

ej =∂

∂uj=

∂ui

∂uj

∂

∂ui= ei

∂ui

∂uj

(e1, . . . , en) = (e1, . . . , en)( ∂ui

∂uj

)

~e = ~eC

where C is the n× n matrix

C =( ∂ui

∂uj

)

Quantities that change in this way are called coveriant vectors, have low indicesand are written as rows.

Consider now a vector v = eivi. We can write this as

v = eivi = (e1, . . . , en)

v1

...vn

= ~e~v

To change bases, we compute as follows

~e ~v = v = ~e~v

~eC ~v = ~e~v

1.5. GENERAL MANIFOLDS AND CONNECTIONS 41

C ~v = ~v∂ui

∂ujvj = vi

Quantities that change in this way are called contravariant vectors, have highindices, and are written as columns.

Now we return to the change in the connection coefficients due to the changeof coordinates. We are so focused on change of coordinates for the reasonsdiscussed above and we also want our equations not to depend on particularcoordinate choices, but to be valid whatever coordinates we use. The system isset up to take care of all this by itself. We have

D~e = ~e ω D~e = ~eω ~e = ~eC

Thus we have

~e = ~eC

D~e = D(~eC)

~e ω = ~e dC +D(~e)C

~eC ω = ~e dC + ~eωC

C ω = dC + ωC

so we haveω = C−1 dC + C−1 ωC

This is an important formula and the way this change of variable is often de-scribed in advanced books. Make sure you understand it. If you have hadexperience with tensors, which we have not emphasized in the book, note thatthe formula is made up of two parts, the second of which is a standard tenso-rial formula. The first part, is not tensorial, and is what keeps the Christoffelsymbols from being tensors.

Now we just have to decode this, noting that C−1 = (∂uj

∂ui ) and that

dC = d(∂uii

∂uj

)

=( ∂2ui

∂uj∂ukduk

)

so the above formula decodes as

ω rs =

∂ur

∂ui

∂2ui

∂uj∂ukduk +

∂ur

∂uiω i

j

∂uj

∂us

which is nice to know. However, we would also like to go down another leveland see how the formula is expressed in terms of the Christoffel symbols Γi

jk.This is intricate but easy. We replace the ω r

s by their formulas in terms of theChristoffel symbols to get

Γrsmdu

m =∂ur

∂ui

∂2ui

∂uj∂ukduk +

∂ur

∂uiΓi

jkduk ∂u

j

∂us

=∂ur

∂ui

∂2ui

∂uj∂uk

∂uk

∂umdum +

∂ur

∂uiΓi

jk

∂uk

∂umdum ∂uj

∂us

Γrsm =

∂ur

∂ui

∂2ui

∂uj∂uk

∂uk

∂um+∂ur

∂uiΓi

jk

∂uj

∂us

∂uk

∂um


In differential geometry books this is often presented in the form

Γrsm =

∂2ui

∂uj∂uk

∂ur

∂ui

∂uk

∂um+ Γi

jk

∂ur

∂ui

∂uj

∂us

∂uk

∂um

but the first form is preferable since it reflects the matrix multiplications fromwhich the formula comes.

We have now got the coefficients of the connection on the Tangent Bundle onthe overlap of the original patch with each neighbouring patch. Next, on eachsuch patch, we construct Γi

jk arbitrarily except for the critical constraint that

on the overlaps it must coincide with the transform of the original Γijk to the

new coordinates. Care must of course be taken where more than two coordinatepatches overlap, but it turns out that because of the form of the transformationthis is automatically taken care of. However annoying in practise there is noproblem in principle. However, this is a very crude method of introducing aglobal connection on the manifold and we will soon see a much better method.However, the better method requires a digression through a bit of Lie Groupsand Lie Algebras.

The basic idea of a connection is to allow us to take the derivative of sectionsof a vector bundle. In our case the bundle is the tangent bundle. Thus weneed a machine into which we insert a section, and then a tangent vector, andthe outcome is the directional derivative of the section in the direction of thetangent vector. This amounts to having an operator ∇ so that for a sections : U → T(M) (think of U as a coordinate patch and for p ∈ U , s(p) ∈ Tp(M)is a tangent vector at p.) Then ∇s is a function which eats tangent vectorsX ∈ Tp(M) and spits out a real number; the directional derivative of s in thedirectionX . Thus∇s(X) is a real number giving the change of s in the directionX . This is often written ∇Xs. Now the only question is how to compute ∇s(X),

Let us now discuss why what has occurred to you won’t work. The naturalthing to want to do is to set

∇s(X) = limt→0

s(p+ tX)− s(p)t

WRONG!!

This has two defects. First, p+ tX makes no sense on a general manifold. Onemight attempt to get around this by using a flow, but that won’t work eitherbecause s(p+ tX) and s(p) are in different tangent spaces so how are we goingto subtract them? Nothing like this is going to work directly in this context.The notion of parallel transfer was invented to overcome this problem, but thisis just another way of introducing a connection.

We remind the reader that there is a function π from T (M) to M whichsends a tangent vector Y ∈ Tp(M) to p ; we have π(Y ) = p. Then a section scan be defined as a function on an open set U ⊆M to T (M) for which

π(s(p)) = p π s = I

The set of sections of U is denoted by Γ(T (M))(U) 9. We will usually omit the

9This has no relation to the use of Γ in Γijk

as connection coefficients, but hopefully (the

standard wish) no confusion will arise.

1.5. GENERAL MANIFOLDS AND CONNECTIONS 43

(U) since it is always there. Keep this in mind.With this equipment we can now say that ∇ is a function

∇ : Γ(T (M))→ Γ(T (M))⊗ Γ(T ∗(M))

See if you can decode this based on what was said above.A manifold with a connection on the Tangent Bundle (referred to as an

affine connection) is called an Affine Manifold or (older term) an Affinely relatedmanifold. As we said before, given a differential manifold we may put an affineconnection on it patch by patch being careful that the coefficients transformproperly on the overlaps. This having been done, almost all the remaining workcoincides with what we found in the embedded case, which I will now recall.

On a coordinate patch we have coordinates u1, . . . , un and a local basis oftangent vectors of the Tangent Bundle

e1 =∂

∂u1. . . en =

∂

∂un

Because we have been given an affine connection we can then write

Dei = ekΓkij du

j

This was the basic formula from which we derived all the other formulas. Weset

ω ki = Γk

ij duj ω = (ω k

i )

and then we have

Ω = dω + ω ∧ ω

with

Ω = (R ji kℓ du

k ∧ duℓ)

with the components R ji kℓ of the Riemann Curvature tensor. The same calcu-

lations as in the embedded case give us just as before

R ij kℓ =

∂Γijℓ

∂uk−∂Γi

jk

∂uℓ+ Γi

mkΓmjℓ − Γi

mℓΓmjk

We can then calculate, for s a local section over an open subset of the coordinatepatch

Ds = D(eisi) = eids

i +Deisi

= ekdsk + ekΓk

ijsi duj

= ek(dsk + Γkijs

i duj)

= ek

(∂sk

∂uj+ Γk

ijsi)

duj

= eksk|j du

j


where

sk|j =

∂sk

∂uj+ Γk

ijsi

and for a tangent vector X ∈ T (M) the directional derivative of s in the Xdirection is

DXs = Ds(X) = eksk|j du

j(

eℓXℓ) = eks

k|jδ

jℓX

ℓ = eksk|jX

j

and we have now solved the problem of taking derivatives of sections on anAffine Manifold. Thus our D may serve for the ∇ discussed above, and forsections s we can use Ds for ∇s. We introduced ∇ because in many books it isused instead of D and we thought it might be useful for the reader to have seenit in this context should she meet it elsewhere.

To complete our study of affine connections we will need to get a deeperinsight into what a connection is globally and this requires a look at paral-lel displacement along a curve and a small excursion into Lie Groups and LieAlgerbras.

1.6 Parallel Displacement Along Curves

Although we have not previously mentioned parallel displacement, it is an im-portant idea and indeed the whole theory can be based on it. This is a matterof personal preference. We will need a little of the theory to put connectionsinto a more abstract setting. Fortunately there is not much difficulty with thissubject and it is rather interesting.

The idea of parallel displacement of a vector along a curve C is that, to theextent possible, the vector (in the tangent bundle) does not change as we movealong the curve. This is of course an impossible requirement because the variousvectors lie in different tangent spaces; if ui(t) describes the curve and w(t) thevectors along the curve then w(t) ∈ Tp(t)M and so vectors at different pointscannot be directly compared. However, we already have equipment which willallow us to express this idea conveniently with the covariant derivative.

Let U ⊆ M be a coordinate patch and C a curve in the patch given byu1(t), . . . , un(t). Let p(t) be the point on the manifold given by ui(t) and w(t) ∈Tp(t)M be a vector at each point of the curve. Let ei = ∂

∂ui be the natural basis.Since we do not want w(t) to change, the obvious requirement is that

Dw

dt= 0

We may write this out in coordinates as

Dw

dt=

D

dt(eiw

i) = ei

dwi

dt+Dei

∂uj

duj

dtwi

= ek

dwk

dt+ ekΓk

ijwi du

j

dt

= ek

(dwk

dt+ Γk

ijwi du

j

dt

)

1.6. PARALLEL DISPLACEMENT ALONG CURVES 45

Thus for the vector field w along C we have

Def The vector field w(t) is parallel along C ⇐⇒

dwk

dt+ Γk

ijwi du

j

dt= 0

An amusing question is to seek the condition that the tangent vector to

the curve is itself parallel along its own curve. In this case wi = dui

dtand the

condition becomesd2ui

dt2+ Γk

ij

dui

dt

duj

dt= 0

which you will recall is the equation which says that C is a geodesic. That is,a geodesic C is characterized by its tangent vector being parallel along itself.This is good evidence that the notion of parallel displacement is an importantidea, though we will not make a lot of use for it in this book.

Now let us look at this in a slightly different way. Consider a tangent vectorw ∈ TpM to the manifold at the point p = p(0) of C. We would like tomanufacture a vector field w(t) ∈ Tp(t)M along the curve C which is parallelalong C and for which w(0) = w. This is called parallely transporting the vector

w along C or parallely translating the vector w along C and each w(t) is calleda parallel transport of w along C. Now we ask, is it possible always to do this.For it to happen, we must satisfy the differential equation an initial condition

dwk

dt+ Γk

ijwi du

j

dt= 0 wi(0) = wi

Since this is a linear system of differential equations with C∞ coefficients thereis always a unique solution. Thus parallel transport is always possible for anycurve C, and it is easy to shift coordinates as we go across coordinate patchboundaries.

Now here is a critical fact. Suppose we have two curves from point p to pointq on the manifold and we transport a tangent vector w at p along both curves.Will they coincide? No, in general they will not coincide. If they do coincidefor any two curves it indicates a flatness in the manifold in an area containingp and q. It is then no big jump to guess that the failure of the transports tocoincide is related to the Riemann Curvature Tensor. Sadly we will not go intothis in any further detail.

Notice that parallel transport from p to q sets up an isomorphism (becausethe transport equation is linear) between TpM and TqM . This seems to be theorigin of the term ”connection” since the connection connects the two tangentspaces. However, because the isomorphism depends on the choice of curve thisis not as exciting as it looks at first glance.

Looked at another way, if we take two curves C1 and C2 from p to q we canalso think of w as translated back to p by going around the combined curveC1 − C2. This will then give an automorphism of TpM into itself, and thus amapping from curves C to AUT(Tp(M)), the space of automorphisms of Tp(M).This is another object of study.


In some treatments of differential geometry the concept of parallel displace-ment is given center stage and other concepts are then derived from it. Onereason is because it can be easily visualized, and this is helpful. For exam-ple, on a two dimensional manifold embedded in three space and with a curveparametrized by arc length, parallel translation is easily visualized; startingwith a vector w at p and a curve C with points p(t) find the angle between wand the tangent vector to the curve C at p(0) and maintain this same angle asyou move along C. The result will be the parallel transport of w along C. Wewill discuss this further when we get to Riemannian geometry where the notionof angle is again available. (The need for the angle is why I had to return tothe embedded case.) Our need for parallel transport is to produce certain LieGroups from which we will get certain Lie Algebras which will then be used tojustify the mantra a connection is a Lie Algebra valued differential form.

1.7 A little about Lie Groups and Lie Algebras

We will need a short introduction to Lie Groups and Lie Algebras in order to putconnections into a more abstract setting. This is a vast and beautiful subjectbut we can only give the shortest possible introduction here. Fortunately onlysome very basic concepts are necessary for our purposes. At the end of thesection we will give a short bibliography for those who wish to learn more aboutthis central subject of modern mathematics.

Lie Groups were among the first crossover abstract structures which meansstructures combining two fundamental structures, in this case Groups and Dif-ferentiable Manifolds. Crossover structures have been a great source of inter-esting mathematics.

Lie Groups are defined as follows

Def G is a Lie Group ⇐⇒

1. G is a group

2. G is a (separable) differentiable manifold

3. The operations (a, b)→ ab and a→ a−1 are C∞ functions.

This marriage of group and manifold theory leads to a theory of great sub-tlety and beauty and occasional surprising difficulty for some of its importanttheorems.

While there is a beautiful theory of abstract Lie Groups, it is a little moredifficult than the special case of Lie Groups whose elements are matrices. A LieGroup whose elements are n× n matrices is a subgroup of the group GL(n,R)of invertible n× n matrices. There are also complex Lie Groups like GL(n,C)whose entries are complex rather than real numbers but we will not be usingthem. There are also infinite dimensional Lie Groups and this is an area ofactive research which we will not even glance at.

1.7. A LITTLE ABOUT LIE GROUPS AND LIE ALGEBRAS 47

Our favorite two example of Lie Groups are GL(n,R) and SO(n,R). Wedefine SO(n,R) by

Def A ∈ SO(n,R) ⇐⇒

1. AA⊤ = I

2. detA = +1

SO(n,R) is called the special orthogonal group and it is the second conditionthat makes it special. The orthogonal group O(n,R) is defined by only the firstcondition AA⊤ = I.

A lie Group is connected if and only if it is connected as a topological space.O(n,R) is not connected; it has two componenents, one with detA = 1 and theother, a coset, with detA = −1. (It is not completely obvious that the formeris connected.) The component of the identity is always itself a Lie Group anda subgroup of the original group with discrete factor group. In the present casethe component of the identity is SO(n,R). A Lie subgroup H of a Lie GroupG is an algebraic subgroup whose inclusion map ι : H → G is an immersion,which means dι is onto. We do NOT want to require that that the topologyof H is the same as the relative topology of H as a subset of G, as this wouldexclude many interesting and useful examples. This is one of the subtleties ofLie theory.

Now we are going to introduce the Lie Algebra of G. This turns out to beTI(G) but we will be more explicit about it. Let A(t) be a C∞ path in G wheret ∈ [a, b] where a < 0 < b and A(0) = I. Then X = A′(0) will be a tangentvector at I to G and the set of all such tangent vectors will be the Lie Algebrag of G. (It is customary to use the lower case fraktur letters for Lie Algebras.)

It is very important that the tangent space TI(G) is not just a vector space;it has a product structure on it. For matrix Lie Algerbras this is given by

[X,Y ] = XY − Y X

It is called the Lie Bracket and naturally we must show that if X,Y ∈ g then[X,Y ] ∈ g which is not obvious. It is easy to prove the following two identitiesfor Matrix Lie Algebras by direct computation.

[X,Y ] = −[Y,X ] antisymmetry

[X, [Y, Z]] + [Y, [Z,X ]] + [Z, [X.Y ]] = 0 the Jacobi Identity

We now prove that, for matrix Lie Groups G, [X,Y ] ∈ g. To do this (and forpurposes of general education) we note that G acts on g as we now explain. for

h, g ∈ G, we have ghg−1 ∈ G. Now let Y ∈ g and Y = dh(t)dt

∣

∣

∣

t=0for some path

h(t) in G where h(0) = I. Then gh(t)g−1 is a path in G and thus

gY g−1 =d

dt(gh(t)g−1)

∣

∣

∣

∣

t=0

∈ g


This is important enough to get a special name

Def Ad(g)(Y ) = gY g−1

Ad(g) is an endomorphism10 of g which is vector space. Thus

Ad : G→ End(g)

where End(V ) is the ring of all endomorphism of V . A homomorphism of agroup G into the ring of endomorphisms of a vector space is called a representa-

tion. By choosing a basis for V we may take the matrix representations of theendomorphisms and we get matrix representation. Since g is a vector space wemay choose a basis of g and then, in the usual way, Ad(G) will be representedin this basis as a matrix and so we have a homomorphism of G into a set ofmatrices. Such a homomorphism is referred to as a matrix representation of G.Ad: G → End(g) is called the Left Regular Represetation of G and the samename is used for the corresponding matrix representation.

Now we want to show that for X,Y ∈ g we have [X,Y ] = XY − Y X ∈ g.To do this, we take g(t) a path in G with X = d

dtg(t)

∣

∣

t=0and g(0) = I. Then

we have

Ad(g(t))Y = g(t)Y g(t)−1

d

dt

(

Ad(g(t))Y)

=d

dt(g(t)Y g(t)−1)

=dg(t)

dtY g(t)−1) + g(t)Y g(t)

dg(t)−1

dt

=dg(t)

dtY g(t)−1) + g(t)Y g(t)(−1)g(t)−1 dg(t)

dtg(t)−1

Taking the value for t = 0 we have

d

dt

(

Ad(g(t))Y)

∣

∣

∣

∣

t=0

= XY − Y X

Finally, we showed before that Ad(g(t))Y = g(t)Y g(t)−1 is in g and therefore

d

dt

(

Ad(g(t))Y)

= limu→0

1

u

(

Ad(g(t+ u))Y −Ad(g(t))Y)

will be in g since a vector space is closed under limits, and thus its value at t = 0which is XY − Y X = d

dt

(

Ad(g(t))Y)∣

∣

t=0is in g. This proof works for matrix

Lie Groups; it requires some slight modification for abstract Lie Groups.

For clarity, and because it is really all we need, let us specialize to GL(n,R).Then GL(n,R) is inside the linear space M(n,R) of n × n matrices and thetangent vector A′(0) will be in in M(n,R) also, which is handy. We will showthat gl(n,R) =M(n,R). We may now imitate the construction for SO(n,R) (or

10An endomorphism is a homomorphism of a vector space into itself.


any other real matrix Lie Group) to get its Lie Algebra so(n,R) but we will dothis later.

First, it is obvious that gl(n,R) ⊆ M(n,R). We must show the opposite. Todo this we must introduce the exponential map exp: g → G. This can be donefor abstract groups too but is easier for matrix groups. We may put a “natural”inner product, and thus a norm, on M(n,R) quite easily by, for X,Y ∈M(n,R)

(X,Y ) =∑

ij

XijYij

||X ||2 = (X,X) =∑

ij

X2ij

Then we can define the exponential map exp : M(n,R) → G(n,R) by, forX ∈M(n,R),

exp(X) =

∞∑

j=0

1

j!Xj

Then one proves the convergence of this series much as one proves it in real orcomplex variables using the Weierstrass M test which can easily be shown towork in these circumstance. It is a little trickier to prove

det(exp(X)) = exp(Tr(X))

where Tr(X) =∑

i Xii is the trace of X . One proves it for diagonal matrices,which is easy, and then proves it for all matrices similar to diagonal matrices,which is also easy, and then since diagonable matrices are dense in M(n,R)the result follows by continuity. You would think this would be easier but it isoften the case with Lie theory that things that should be easy are harder thanexpected. Note that

det(exp(X)) 6= 0 so exp(X) ∈ GL(n,R)

Now given X ∈ M(n,R) we may form a path in M(n,R) by

A(t) = exp(tX) =∞∑

j=0

1

j!(tX)j =

∞∑

j=1

tj

j!Xj

Note A(0) = I. Then, imitating the standard trickery for differentiating powerseries, we have

dA(t)

dt

∣

∣

∣

t=0= exp(tX) ·X |t=0 = IX = X

and since X is the tangent vector to a curve at the origin we have X ∈ gl(n,R)and thus

gl(n,R) = M(n,R)

as we promised.


We can try to imitate this construction for any matrix Lie Group but oftenthere are easier ways. Now note that, just as with power series in complexvariables, we may prove that

exp(tX) · exp(uX) = exp((t+ u)X)

which tells us that A(t) = exp(tX) is a lot better than a path; it is a one-dimensional subgroup of GL(n,R). Each tangent vector X at I has one of theseone dimensional subgroups to which it is tangent, so that an alternate definitionof the Lie Algebra would be the set of one dimensional subgroups of G. Notealso that A : R→ GL(n,R) is a homomorphism, so that X is not just a tangentvector to a curve but a tangent vector to a curve which is a homomorphism.

Caution In general exp(X)·exp(Y ) 6= exp(X+Y ). However, just as in complexvariables it is not difficult to prove that if X and Y commute then indeedexp(X) · exp(Y ) = exp(X + Y ).

These results or analogous ones can be proved for abstract Lie Groups (LieGroups which are not groups of matrices) but it takes a little more effort.

We now want to find the Lie Algebra so(n,R) of SO(n,R), which is char-acterized by AA⊤ = I and det(A) = +1. Let X ∈ TI(SO(n,R)) and select acurve A(t) with with A(0) = I and

X =dA(t)

dt

∣

∣

∣

t=0

Thend

dt

(

A(t)A⊤(t))

=d

dtI = 0

so

dA(t)

dtA⊤(t) +A(t)

dA⊤(t)

dt= 0

dA(t)

dt

∣

∣

∣

t=0A⊤(0) +A(0)

dA⊤(t)

dt

∣

∣

∣

t=0= 0

XI + IX⊤ = 0

X +X⊤ = 0

Thus so(n,R) ⊆ skew symmetric n× n matrices.On the other hand, let X be a skew symmetric n× n matrix. Then

exp(tX)(exp(tX))⊤ = exp(tX) exp(tX⊤) = exp(tX) exp(−tX)

= exp(tX) exp(tX)−1 = I

This shows that exp(tX) ∈ SO(n,R) so

X =d

dtexp(tX)

∣

∣

∣

t=0∈ so(n,R)


and we have shown

so(n,R) = skew symmetric n× n matrices

Now we wish to do an example so we can see this all in action. The grouphere is SO(2,R) and we will take X ∈ so(2,R) where X is the skew symmetricmatrix

X =

(

0 −11 0

)

Then we have

X2 =

(

−1 00 −1

)

X3 =

(

0 1−1 0

)

X4 =

(

1 00 1

)

X5 =

(

0 −11 0

)

etc

Then we set

A(t) = exp(tX) =

(

1 00 1

)

+ t

(

0 −11 0

)

+t2

2!

(

−1 00 −1

)

+t3

3!

(

0 1−1 0

)

+t4

4!

(

1 00 1

)

+t5

5!

(

0 −11 0

)

=

(

1− t2

2! + t4

4! + . . . −t+ t3

3! − t5

5! + . . .

t− t3

3! + t5

5! + . . . 1− t2

2! + t4

4! + . . .

)

=

(

cos t − sin tsin t cos t

)

Thus A(t) ∈ SO(2,R) is the one parameter group of rotations in the (x, y)-plane.Note

dA

dt

∣

∣

∣

∣

t=0

=

(

− sin t − cos tcos t − sin t

)∣

∣

∣

∣

t=0

=

(

0 −11 0

)

= X

which is just as it should be. In physics X is called the infinitesimal generatorof the Lie group of rotations. It should be clear that in 3-spacethe infinitesimalgenerator of the group of the one parameter group of rotations around the zaxis is

Xz =

0 −1 01 0 00 0 0

From this you should be able to figure out the rotation matrix Tz(t) for anglet around the z axis and then by symmetries the rotations Tx(t) and Ty(t) byyourself. Then you can work out the commutators [Xz, Xx], [Xx, Xy], [Xy, Xz].These are of great interest in Quantum Mechanics.


The Lie Algebra has a great deal of information about its Lie Group, butif two Lie Groups are alike around I their Lie Algebras will be the same. Forexample, the unit circle is a Lie Group with elements eit|t ∈ (−π, π] and R is aLie Group under addition. Both have Lie Algebras isomorphic to R with bracketidentically 0, so the Lie Algebra cannot determine the Lie Group. However, ifwe restrict ourselves to simply connected Lie Groups, then it is the case thatevery finite dimensional Lie Algebra does determine a unique simply connectedLie Group. This theorem is very difficult for people outside Lie studies.

For persons wishing to pursue this most beautiful and important part ofmodern mathmatics we give now a short annotated bibliography of the booksthat we found helpful. They are arranged more or less in order of difficulty (inour opinion).

SHORT BIBLIOGRAPHY OF LIE GROUPS AND LIE ALGEBRAS

These books cover only finite dimensional Lie theory.

1. Wu, Loring Tu. AN INTRODUCTION TO MANIFOLDS. Springer, NewYork, 2011. Chapter 4 of this book has a very gentle introduction to LieGroups and Lie Algebras much more complete than that in the presentbook.

2. Baker, Andrew, MATRIX GROUPS, AN INTRODUCTION TO LIE GROUPTHEORY, Springer, London 2002. An introduction to matrix Lie Groupswith many examples and lots of motivation. A fine introduction to thesubject including many subjects important for applications.

3. Knopp, Anthony W. LIE GROUPS BEYOND AN INTRODUCTION 2ndediton. Birkhauser, Boston 2002. Despite the name, the 2nd edition hasa initial chapter which is an introduction to Matrix Lie Groups. Thebook goes on from there to develop the abstract theory of Lie Groups ina way suitable for mathematical studies. Knopp explains things well andthe book covers the classical theory. If you read this book you will be abeginning professional in the subject.

4. Bump, Daniel LIE GROUPS, Springer, New York, 2004. Covers the classicalmaterial in a somewhat shorter treatment than 3.

5. Duistermaat, J.J. & Kolk, J.A.C. LIE GROUPS, Springer, New York, 2000.This is a modern treatment of Lie Groups and Lie Algebras meant for themathematically mature reader. It is not for the timid or untrained, butwe found it rewarding to look at their treatment when we already hadsome familiarity with the material.

6. Chevelley, Claude, THEORY OF LIE GROUPS, Princeton 1946. An impor-tant classic in the field by a master. The reader must accustom herselfto some terminology that is no longer current, but this is still a wonder-ful source. We highly recommend this book for readers who want to get

1.8. FRAME BUNDLES AND PRINCIPLE BUNDLES 53

the flavor of Lie Groups quickly. Very self contained book which developssome functional analysis for use in the Peter Weyl theorem. Cognoscentiwill recognize material here in the definition of manifolds which later wasincorporated into sheaf theory, which is equivalent to but looks differentfrom the treatment of manifolds in this book.

1.8 Frame Bundles and Principle Bundles

This is a natural place to introduce the Frame Bundle of a Differentiable Man-ifold. The frame bundle is the fundamental example of a Principle Bundle andso, with a little extra work, we can include this concept in our repertoire. Prin-ciple bundles have some importance in advanced areas of quantum mechanics soit is worth the slight extra effort to understand how frame bundles are examplesof principle bundles. However, since this topic is pretty advanced we will notbe using it much in the material that follows, so if you prefer you can just skimthe abstract part. But the easy stuff about frame bundles it would be good tounderstand completely.

Recall the concept of Tangent Bundle of a manifold. At each point of themanifold Tp(M) is an n-dimensional vector space and and we can, in a C∞ way,choose a local basis of sections of Tp(M). This means, for some U ⊆M (whichmay or may not be a coordinate patch) we have at each point p ∈ U a basis

σ(p) = (e1(p), . . . , en(p))

of Tp(M), where the ei(p) are C∞ sections. The basis σ(p) of Tp(M) is calleda frame. We can always come up with a natural basis σ = (∂/∂u1, . . . , ∂/∂un)but we do not want to emphasize the natural basis at all in the section. It isnot possible in general to come up with a σ for the entire manifold; in fact itmay not be possible to come up with a single non-zero section of the manifold,but this is a local book and such things are global results covered in books onDifferential Topology.

The frame bundle resembles the tangent bundle in certain respects, butinstead of a vector space over each p ∈ U we have the set of frames σ(p) forTp(M). This is just a set, but by requiring σ(p) to be C∞ we have already putsome additional structure on the frame bundle. We want more.

Recall that, given a frame σ(p) at each point we get an isomorphism ofTp(M) onto Rn which takes v = eiv

i ∈ Tp(M) to (v1, . . . , vn). Recall that π :T (U)→M is given by π(v) = p were v ∈ Tp(U). We thus have a commutativetriangle

v ←→ ((v1, . . . , vn), p)π ց ւ π2

p

We wish to do a similar thing for the frame bundle, giving coordinates of a sort,but this is rendered tricky by certain rules that principle bundles have. We willlook at this soon.


1.8.1 Group Actions

An Automorphism of a set S is just a one to one onto mapping (a bijection ) ofS to S. The set of all automorphisms of S is denoted by Aut(S).

A right action of a group G on a set S is a mapping of G into Aut(S) denotedgenerally by (s, g)→ sg subject to the following law:

(sg)h = s(gh) for s ∈ S and g.h ∈ G

Naturally if the set S and the group G have topologies it is assumed that(s, g)→ sg is C∞ and will will not talk much explicitly about these topologicalmatters, assuming the reader can just insert them as necessary.

The action is transitive if and only if for any two elements s1, s2 ∈ S thereexists an element g ∈ G for which s1g = s2. The action is free if and only if, iffor any s ∈ S we have sg = s then g = e, e being the identity of G. It is easyto prove the if the auction is both transitive and free then the g taking s1 to s2is unique.

For transitive free actions an interesting (and possibly slightly confusing)thing happens. Select s0 ∈ S arbitrarily. Then there is a one to one cor-respoindence between S and G; given s ∈ S there is a unique gs for whichs = s0gs. Hence S is bijective to the group G but since the choice of s0 (whichcorresponds to e in the bijection) is arbitrary S lacks an identifiable identity, notto mention the binary operation of a group. We express this whole situation,S, G and the transitive and free action of G on S by saying that

S is a G-torsor

Our important and nearly unique example is the frame bundle. Let U ⊆M bea subset of M over which a local basis of sections

σ0 = (e1, . . . , en)

of Tp(U) can be defined. Then G =GL(n,R) acts on the frame bundle by

σ = σ0g = (e1, . . . , en)(gij) where g ∈ G

where if σ = (f1, . . . , fn) then fj = eigij . This action is easily seen to be

transitive and free. Hence the frame bundle is (locally) a GL(n,R)-torsor, butthere is no distinguished section σ0 which is the typical torsor situation.

1.8.2 Principal Bundles

We now return to the Frame bundle although we will set up the notation so thatit looks very like the notation in an abstract treatment. The principal bundlehere will be the bundle of frames where we are working over U ⊆ M wherewe can set up a basis of local sections σ(p) = (e1(p)), . . . , en(p)) where p ∈ Uand (e1(p)), . . . , en(p)) is a basis of Tp(U). The definition of Priniple Bundlerequires us to come up with a mapping Φ : π−1[U ]→ G×U which is one to one

1.9. AFFINE CONNECTIONS 55

and onto (a bijection), C∞, and which satisfies the following condition which isoften called equivariance. That is

if Φ(σ) = (g, p)

then Φ(σh) = (gh, p)

(The σh were defined in the last section for the frame bundle but it worksthe same for any bundle with a transitive free action.) The mappings Φ arecorrelated with distinguished sections σ in the follwing manner.

→ Given a Φ, define a section σ0(p) by

σ0(p) = Φ−1(e, p)

← Given a section σ0(p), for any σ we can find a g(p) in GL(n,R) for whichσ(p) = σ0(p)g(p). Then we set

Φ(σ(p)) = (g, p)

1.9 Affine Connections

To make my point more clearly in this section I want to begin with a littlereview. Let us start with two bases

σ = (e1, . . . , en) σ = (e1, . . . , en)

and let them be connected by the matrix C = (hij) so that

(e1, . . . , en) = (e1, . . . , en)(hij) σ = σC

Then we know that if v = e1v1 + · · · + env

n = e1v1 + · · · + env

n that we willhave

v1

...vn

= C−1

v1

...vn

Now it is possible to define a vector as a column whose entries change underchange σ = σC in exactly this way. This analogous to defining a duck byDef If it quacks like a duck it’s a duck11.Note that having feathers or having n entries is not enough; it must have theproper quack or transformation behaviour under basis change.

Now this kind of definition, singling out some characteristic property anddefining the concept as something having it, is not in itself bad, and is used inphysics productively and often. Vectors are things that transform like vectors,tensors are things that transform like tensors, etc. However, in mathematics we

11I am indepted to Anthony Zee for this example


like to have things defined by having them belong to sets. This may sometimesbe just pyschological quibbling, but that’s the way we like to do it, and if thesets have structures this can be genuinely illuminating. So that’s what we willattempt to do for Affine Connections.

In a somewhat similar way we can define an affine connection ω on a manifoldlike this. In each coordinate patch U with coordinates (u1, . . . , un) a affineconnection is represented by an n × n matrix of first order differential forms,and thus looks like

ω is represented locally by(

Γijk du

k)

where the Γijk are called the connection coefficients. Now for the tricky part.

On the overlap of coordinate patches, the ω representatives must change in aspecified way. Using the letter ω itself for the representative on the first patch,let us derive the transition rule more time. As usual we have ei = ∂/∂uj andei = ∂/∂ui. They are related by

(∂

∂u1, . . . ,

∂

∂un) = (

∂

∂u1, . . . ,

∂

∂un)( ∂ui

∂uj

)

(e1, . . . , en) = (e1, . . . , en)C

with C = (∂ui/∂uj). Then, using the abbreviation e = (e1, . . . , en) etc we have

from our old formulas

De = eω

D(eC) = eCω

e dC + (De)C = eCω

e dC + eω C = eCω

dC + ωC = Cω

C−1 dC + C−1 ωC = ω

This is the formula which must be satisfied between the representatives ω andω on the overlap of the patches U ∩ U . We leave it to the reader to show that onthe overlap of three or more coordinate patches the requirement is consistent.

Another point of importance is that in deriving the change of basis formulafor ω I never used the fact that the bases involved originally were natural bases.This formula is good for any change of basis.

Now we must recall Euler’s method for solving a first order differential equa-tion. The equation with initial condition is

dy

dt= f(y, t) y(t0) = y0

The important point for us is that at t = t0 the value of dydt

tells us which wayto travel from the initial point. We now choose a small ∆t and compute ∆y =dydt|t=t0∆t. We will get an approximation y1 = y0 + ∆y. Iterating the process

1.10. RIEMANNIAN GEOMETRY 57

we will get a sequence of values yj(tj) = yj−1(tj−1) + dydt

∣

∣

t=tj−1∆t. Connecting

the dots in the sequence of points (tj , yj(tj)) we get an approximation of thesolution with points (t, y(t)). The smaller the ∆t the better the approximation.Again, the value of dy

dt

∣

∣

t=tj−1tells us which way to go to get to the next point

in the approximation.

The domain (t1, t2) of the solution with t1 < t0 < t2) is not very predictablefor general f(y, t) but for linear funcitons f(y, t) = Ay the solution is defined forall t. The situation with systems where y is a column vector and A ∈ GL(n,R)is similar and the solution exists for all t.

We now apply this to the following situation. Let C be a curve in the n-dimensional manifold M given by (u1(t), . . . , un(t)). Let v(t) be the tangentvector to C(t) with coordinates (u′1(t), . . . , u′n(t)). Let σ0(t) be a curve in theframe bundle of M lying over C so that π(σ0(t)) = C(t).

We assume that we have an affine connection given in our coordinate patchby ω. Now consider the system of equations on the curve C

dgji (t)

dt= ω(v(t)) = −Γj

ikduk(v(t)) = −Γj

ik

duk

dtgj

i (0) = δji

This system will have a solution and for small t continuity guarantees thatdet(gj

i (t)) 6= 0. (Actually, a little extra work would give us this for all t butwe don’t need it. A local solution will do.) Our solution can thus be regardeda curve in the Lie Group GL(n,R) and thus it’s derivative at t = 0, that is

−(

Γjik

duk

dt

)∣

∣

t=0, is an element of the Lie algebra of GL(n,R), which is gl(n,R).

Thus we can say that ω is a gl(n,R)−valued one-form. The entire process canbe generalized to other groups than GL(n,R) and other principle bundles. Laterwe will show that for Riemannian geometry the natural group is SL(n,R) andthe connection is thus an sl(n,R)−valued one-form which is to say that ω isskew symmetric: ωi

j = ωji .

The curious reader may wonder what the geometric significance of gji (t) is

and it is very interesting.

1.10 Riemannian Geometry

In a previous section we studied manifolds which are embedded in a Eucli-den space of one higher dimension. This gave us, among other things, a nor-mal vector and an inner product. Riemann realized from his knowledge ofGauss’s work on intrinsic geometry (for example, the Theorema Egregium),that the embedding was unnecessary. Whether Riemann foresaw Einstein’s useof non-embedded manifolds (perhaps the Universe as a spacial 3-manifold) isnot known; various people have suspected this was one of his motivations, buthe left no solid clue behind. Riemann realized that the basic tool one needed todevelop a geometry familiar enough to work with but much more general wasthe inner product.


Def A Riemannian Manifold is a finite dimensional differentiable manifoldM (C∞ for us) for which each point p ∈ M has an inner product on Tp(M)associated with it. We also require that if v, w are C∞ vector fields over someopen set of M then (v, w) is a C∞ function on the open set.

It is easy to see that this just means the coefficients (gij) are C∞ functionson M . The inner product lets us measure lengths and angles for vectors inTp(M).

We will introduce here an idea which has a lot of uses later. For any open setU a section of T (M) over U is a function v : U → T (M) for which v(p) ∈ Tp(M).Thus v(p) gives a vector in the tangent space at p for each p ∈ U . This is oftencalled a tangent vector field on U , which is the older terminology. Section ofthe tangent bundle and tangent vector field are thus synonyms.

Now it should be clear that locally (that is, in a neighborhood U of any pointp ∈ M) we can choose n vector fields which are linearly independent at eachpoint of U . It probably won’t be possible to do this globally, but no matter; thisis mostly a local book. We call the n vector fields e1, . . . , en. We could even usethe Gram-Schmidt orthogonalization procedure to get (locally) an orthonormalbasis which might, or might not, be a good idea. Note we can do this becausethe inner product lets us measure lengths and angles.

Our tour of embedded manifolds has taught us some things. We are goingto need a covariant derivative which we assume works like it did before; we needone forms on the manifold so that for a local basis of sections of the Tangentbundle e1, . . . , en w have a matrix (ω j

i ) so that

Dei = ej ωj

i

As before the ω ji = Γj

ik duk are called the connection 1-forms. Since all we have

to begin with are the (gij), we must somehow get the connection 1-forms fromthe (gij). Once we have these we have curvature and geodesics.

We have to start somewhere, and one of quite a large number of ways tostart is to require that a formula we previously derived for embedded manifoldscontinues to hold. It is quite natural to require the following analogue of Leibniz’rule:

d(v, w) = (Dv,w) + (v,Dw)

We would certainly like this to be true, and perhaps it will narrow down thechoice of ω j

i . Applying this to ei and ej , we have

Dei = ekωk

i = ekΓkim dum

Dej = elωk

j = elΓljn du

n

d(ei, ej) = (Dei, ej) +D(ei, Dej)

dgij = (ekΓkim dum, ej) + (ei, elΓ

ljn du

n)

= gljΓlim dum + gilΓ

ljn du

n

∂gij

∂ukduk =

(

gljΓlik + gilΓ

ljk

)

duk

1.10. RIEMANNIAN GEOMETRY 59

Using the linear independence of the dum and cyclically permuting the i j k wehave

∂gij

∂uk= gljΓ

lik + gilΓ

ljk

∂gjk

∂ui= glkΓl

ji + gjlΓlki

∂gki

∂uj= gliΓ

lkj + gklΓ

lij

This almost does it. If we add the second and third equations and subtract thefirst we have

∂gjk

∂ui+∂gki

∂uj− ∂gij

∂uk= glkΓl

ji + gjlΓlki + gliΓ

lkj + gklΓ

lij − gljΓ

lik − gilΓ

ljk

The second and fifth terms on the right, and the third and sixth terms, wouldcancel if Γl

ki = Γlik. Hence, besides the analogue of Leibniz rule we started with

it is also necessary to assume the connection coefficients Γlik are symmetric in

the lower two indices. We recall this was true in the embedded case and so itseems fairly natural to assume it here also. With this assumption the termscancel and the surviving first and fourth terms double up and we are left with

∂gjk

∂ui+∂gki

∂uj− ∂gij

∂uk= 2gklΓ

lij

1

2gmk

(

∂gjk

∂ui+∂gki

∂uj− ∂gij

∂uk

)

= gmkgklΓlij = δm

l Γlij

giving

Γmij =

1

2gmk

(

∂gjk

∂ui+∂gki

∂uj− ∂gij

∂uk

)

as the only possible choice of connection coefficients which are symmetric andsatisfies the analogue of Leibniz rule.

Now that we have ω ij we may define geodesics as curves parametrized by

arc length which satisfy the geodesic equation

d2uk

ds2+ Γk

ij

dui

ds

duj

ds= 0

We may study curvature by defining the Curvature Form Ω as

Ω = dω + ω ∧ ω

where ω = (ω ij ) is the matrix of curvature forms. This cherns up the Riemann

Curvature Tensor exactly as in the embedded case.

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[18] Zee, Anthony, EINSTEIN GRAVITY IN A NUTSHELL, PrincetonU.P, Princeton New Jersey, 2013

a practical introduction to diﬀerential forms alexia e ...schulz/chapter4.pdf · a practical...

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