Lecture Notes on Differential Geometry

Jay Wilkins

August 19, 2008

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Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1 Manifolds and Smooth Functions 5

2 The Tangent and Cotangent Spaces 28

3 Mapping Theorems and Submanifolds 59

4 Vector Fields and the Tangent Bundle 60

5 Tensor Analysis on Manifolds 73

6 Riemannian Manifolds 91

7 A Brief Excursion into Lorentzian Manifolds 123

8 Connections and Covariant Differentiation 143

9 Geodesics, Normal Neighborhoods, and Hopf-Rinow 145

10 Curvature 146

Appendix 147

Bibliography 157

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Preface

This is a collection of lecture notes on differential geometry, focusing primarily onRiemannian geometry. They were compiled over a span of several years from varyingsources. Consequently, they are, at present, somewhat uneven. A large portion ofthe material was put together while doing the research for my Masters Degree atClemson University, under Dr. James Peterson. About half of the material came froma course in Riemannian geometry, taught by Dr. Conrad Plaut at the University ofTennessee. The rest of the material is pieced together from my own personal notes.I owe Dr. Peterson and Dr. Plaut a great deal for their support and help in mypursuit of this subject.

Since these are lecture notes and not a textbook, the reader is expected to haveknowledge of basic linear algebra, particularly of linear transformations between fi-nite dimensional vector spaces, change of basis, etc, as well as a strong grasp of realanalysis on Euclidean spaces. Specifically, the use of the differential calculus is ofprimary importance, including knowledge of the Inverse and Implicit Function The-orems and the Rank Theorem. While these results will be essential to the materialhere, their proofs will not. So the reader willing to take them on faith can find themin [Ba], [Bo], or [Sp,2]. In addition, these notes take a decidedly topological point ofview, so a certain degree of knowledge in that area will be assumed. The necessaryresults are not difficult or deep, and come primarily from the concepts of connected-ness, compactness, separation, and continuity. At some point, we will use the notionof paracompactness, but the results in this section will be given with proof, sincethey are not standard. The standard topological reference for this material will be[Mu]. I understand that this may deter some aspiring differential geometers who donot yet have the necessary background in topology. The fact, however, is that if onewishes to pursue this subject, the topology must be learned at some point, as it isan indispensable tool in geometry. Moreover, the prevalent use of topology makesthe proofs much more rigorous and formal.

These notes begin with the most basic fundamentals of topological manifolds,proceeding to smooth manifolds and their properties. After the definitions are pre-sented, several examples of smooth manifolds are given. This list, and this collectionof notes in general, is still growing, so the notes will be updated periodically. Myhope is to eventually include in these notes a catalog of useful and interesting andexamples of Riemannian manifolds. Local structures, such as the tangent space, arediscussed in great detail before any global results are pursued. In particular, I givedetailed and independent constructions of the tangent and cotangent spaces, show-ing, only afterward, that there is a natural duality between the two. I then proceedto global structures such as vector fields, the tangent bundle, tensor fields, metric

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tensors, etc. In particular, there are detailed constructions of the Riemannian andLorentzian metric tensors, as well as the standard proof that a Riemannian manifoldis an inner metric space.

After this metric space structure is developed, the more technical constructionsof differential geometry are presented, which, of course, leads into the concept ofcurvature. The study of curvature and how it affects various other properties of amanifold lies at the heart of modern differential geometry. These notes are still beingadded as time permits me.

As a final remark, I should inform the reader ahead of time that my choice ofpresentation, notation, and organization, as it is for all mathematicians, is influencedby the work of those mathematicians whose research and textbooks I have personallyfound to be most accessible. In particular, the fundamental definitions given hereof smooth manifolds, smooth mappings, the tangent and cotangent spaces, etc., areadapted from the work of S.S. Chern and William Boothby (see [Bo] and [Ch] in thebibliography). The later concepts and constructions, including geodesics, convexity,curvature, Jacobi fields, etc, are heavily influenced by Barret ONeills book [ON,2]and the course notes of Dr. Plaut. I have added my own touches in various places, andI have arranged the material according to my own taste, which, of course, may not bethe preference of others. Such is the nature, Im afraid, of the subject of DifferentialGeometry. I have, however, been careful to unify the notation and terminology. So,even though the material is drawn from different sources, it is presented here in thesenotes in a uniform way.

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1 Manifolds and Smooth Functions

We begin by introducing the formal concept of a manifold. Roughly speaking, amanifold is a topological space that is locally Euclidean. More formally, we have thefollowing.

Definition 1.1 A topological manifold is a second-countable Hausdorff space Msuch that for every p M , there is a neighborhood, U , of p, a natural number m, anda corresponding map U : U Rm with the property that U is a homeomorphismof U onto an open subset of Rm.

First, there are some remarks on notation and terminology. As we have done here,we will not reference the specific topology on M unless it is necessary. The existenceof the topology is the primary concern and will usually, but not always, be assumed.Second, as is common in topology, we will typically refer to a topological manifoldmerely as a manifold. We will introduce a different class of manifolds later on, so,occasionally, we will use the full name to distinguish topological manifolds fromothers.

The sets U are called coordinate neighborhoods, and their correspondingmappings U are called coordinate mappings. Each pair (U , U) is called acoordinate chart, or a coordinate system, and the collection of all such pairsis called an atlas. Note that the collection of coordinate neighborhoods forms anopen covering of M . We will occasionally use the word covering when referringto an atlas, even though technically the elements of an atlas are pairs consisting ofneighborhoods and corresponding mappings.

Two charts are equal if and only if their domains and mappings are identical.We will usually drop the subscript U from the map U , since we will almost alwaysreference a chart in the form (,U), from which it should be clear that is thecoordinate mapping corresponding to the coordinate neighborhood U . For brevity,we will often simply call the pair (,U) a chart around p or at p to denote the factthat (,U) is a coordinate chart with p U .

The reason for this terminology is that these charts can be used to establish localcoordinate systems around any point p M . If p M , then there is some chart(,U) such that p U and (U) is a neighborhood of (p) in Rm for some m. (Wetake neighborhood here to mean an open set, and not, as some define it, as a setcontaining an open set.) We can define the coordinates of q U to be the Euclideancoordinates of (q) (U). It is in this respect that we describe a manifold as locallyEuclidean. Note also that if (,U) is any chart around p, then the restriction of to any open subset, V , of U gives us a new coordinate chart, namely (|V , V ), since

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this restriction gives us a homeomorphism of V onto (V ) (U). More generally,if V is open in M and U V 6= , then U V is open in U and (|UV , U V ) is acoordinate chart.

The dimension of a manifold, M , at a point p M , is defined to be the dimen-sion of the Euclidean space containing (U), where (,U) is a chart such that p U .This definition brings us to our first instance of coordinate independence. The pointp may lie in several different coordinate neighborhoods, so in order to show thatdimension is a well-defined concept, we must show that each one is homeomorphic toa subset of the same Euclidean space. So, suppose there are two coordinate charts,(,U) and (, V ) such that p U V , and suppose (U) Rn and (V ) Rm.Then U V is a neighborhood of p on which we have two coordinate mappings, and . Moreover, (U V ) and (U V ) are open subsets of Rm and Rn, respec-tively. The mapping 1 : (U V ) (U V ) is a homeomorphism, as itis a composition of homeomorphisms. That is, 1 maps an open subset of Rmhomeomorphically to an open subset of Rn. Two open subsets of Euclidean spacescannot be homeomorphic unless the Euclidean spaces are of the same dimension.(This result is known as Invariance of Domain.) Hence, we must have m = n, andthe dimension of a manifold is well-defined, meaning that it does not depend on theparticular coordinate system chosen around a point. While local coordinate systemsare useful for many purposes, a fundamental goal of differential geometry is to deriveand express results that are independent of any particular coordinate system.

The dimension of a manifold is only defined locally, and a topological manifold,as we have defined it here, need not be of constant dimension. For example, thesubspace M = {(x, y) R2 : 0 < x < 1} is an open subset of R2. It is also amanifold, since given any point (x, y) M , we can let M , itself, be a coordinateneighborhood, and the coordinate map, then, is simply the inclusion map of Minto R2. This also shows that M is of constant dimension, since every point in Mhas a neighborhood homeomorphic to an open set in R2. Similarly, the subspaceN = M {(x, y) R2 : x = 2} is also a manifold, though not of constant dimension.Every point in M has a neighborhood homeomorphic to an open set in R2, whileevery point in {(x, y) R2 : x = 2} has a neighborhood (in the subspace topology)homeomorphic to an open interval in R.

In geometry, physics, and other applications, most manifolds are of constant di-mension, and those that are not, like N above, are usually pathological and onlyserve as counter-examples. Indeed, in this example, N was disconnected. A con-nected manifold, M , must be of constant dimension. To see this, note that we candefine f : M R by letting f(p) be the dimension of M at p. Since weve shown thedimension to be a well-defined concept, f is well-defined. Moreover, f is continuous,

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for if p M , there is some coordinate neighborhood, U , around p, and f is constant,and therefore continuous, on U . Since the coordinate neighborhoods cover M , f iscontinuous. It now follows that f(M) is a connected subspace of R. But f takesvalues only in N, and the only connected subspaces of N are the one-point sets. So,f , and the dimension of M , must be constant.

While we will not universally require that our manifolds be connected, we willassume from here on that they are of constant dimension. Consequently, we will usethe following notational conventions when referring to manifolds. When we denote amanifold by capital letters M , N , etc, this is meant to imply that these manifolds areof constant dimension m, n, etc respectively. Occasionally, it will be convenient touse the same capital letter to denote multiple manifolds. In this case, we will alwaysnumber them and denote the dimension with a superscript. For example, Mm1 andMn2 will denote manifolds of dimension m and n, respectively.

We will now prove a series of topological properties of manifolds which will beuseful in later arguments.

Lemma 1.1 A manifold, M , is locally compact.

Proof Since M is Hausdorff, it suffices to show that for any point, p M , and anyneighborhood, V , of p, there is a neighborhood, W , of p, so that W is compact andW V .

So, let p be any point in M , and let V be any neighborhood of p. There is acoordinate chart (,U) such that p U , and, as we have seen, by restricting to anappropriate intersection, we can assume that U V . Since (U) is an open subsetof Rm containing (p) = x, there is some > 0 such that B(x) (U). Now, B(x)is compact both in Rm and in the subspace (U). Thus, 1(B(x)) is compact inthe subspace U M . It is also compact in M , for if {G} is an open cover of1(B(x)) by sets open in M , then the collection {G U} is an open coverof 1(B(x)) by sets open in U . This latter collection has, then, a finite subcover,say {Gk U}sk=1, implying that the collection {Gk}sk=1 is a finite subcover from theoriginal collection.

Now, let W = 1(B(x)). Let W and cl(W ) denote the closures of W in Mand U , respectively. Note that the closure of B(x) in (U) is just B(x). Since isa homeomorphism, we have 1(B(x)) = cl(1(B(x))) = cl(W ). Thus, cl(W ) iscompact in M , and so is also closed in M . It follows that W cl(W ). Conversely,we have cl(W ) = W U , implying that cl(W ) W . Thus, cl(W ) = W , and W iscompact in M . Finally, since cl(W ) U V , the result follows.

QED

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Theorem 1.2 A manifold, M , is normal, separable, metrizable, and is a Lindelofspace. Furthermore, it is locally path connected (therefore locally connected), and Mis connected if and only if any two points in M can be joined by a continuous curve.That is, M is connected if and only if it is path connected.

Proof Every regular space with a countable basis is normal, while every locallycompact Hausdorff space is regular. Thus, by the previous lemma and the secondcountability hypothesis in Definition 1.1, M is normal.

Now, every second countable space is both separable and Lindelof, but we willprove these less obvious results directly. By hypothesis, there is a countable basis{Bn}n1 for M . For each n, let pn be a point in Bn. Define D = {pn : pn Un, n 1}. Given any nonempty open set U M , let q be in U . Then there is some basisset, Bn, such that q Bn U . It follows that pn U U D 6= . Hence, Dintersects every open set in M , implying that D is a countable dense subset of M .Thus, M is separable.

Next, let S be an open covering of M . For each p M , there is an open set,Up S such that p Up. So, there is a basis set, Bnp , such that p Bnp Up. Thatis, to each p, we associate some natural number, np. For example, we can choosethe smallest index n such that p Bn Up. Let I = {np : p M}. Then I iscountable. For each k I, choose a set Uk S such that Bk Uk. This is possibleby our construction. The collection {Uk : k I} covers M and is countable. Thus,S has a countable subcover, implying that M is a Lindelof space.

Since M is second countable and normal, Urysohns metrization theorem impliesthat M is metrizable.

Being locally Euclidean, every point in M has a neighborhood that his homeo-morphic to an open subset of Rm. The open sets in Rm are locally path connected,implying that M has this same property.

Finally, it is a standard topological result that connected and locally path con-nected implies path connected.

QED

Next, we prove that a manifold can be covered by a countable collection of com-pact sets. A space with this property is called -compact.

Theorem 1.3 A manifold, M , is -compact.

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Proof If p M , there is a basis set, Bk, such that p Bk. Applying the localcompactness, there is a neighborhood, V , of p such that V is compact and p V V Bk. There is a basis set, Bj, such that p Bj V . It follows that Bj V .Moreover, since V is compact, Bj is compact. Thus, given p M , we have found abasis element, Bj, such that p Bj and Bj is compact. Let Bc be the set of of theclosures of all such basis elements. Then Bc is a countable compact covering of M .

QED

Thus far, we have only worked with manifolds as topological spaces with certainspecial properties. In order to use calculus to study geometrical properties, we needto add one more condition to our definition of a manifold. As a preliminary remarkon terminology, recall that C describes a function or mapping that has continuousderivatives of all orders. We will, from this point on, use the words differentiableand smooth as synonyms for C, and we will interchange them without hesitation.Varying degrees of differentiability are required throughout differential geometry.Thus, to avoid having to continually state what degree is necessary, it will be assumedthat all constructions for which the description differentiable makes sense are, infact, C.

Definition 1.2 A differentiable (or smooth, or C) manifold is a topologicalmanifold, M , with an atlas, denoted here by A , that satisfies the following additionalproperties.

i) For any charts (,U) and (, V ) in A such that U V 6= , the map 1 :(U V ) (U V ) is C, meaning that its component functions havecontinuous partial derivatives of all orders. Such pairs of charts are said to beC-compatible.

ii) The atlas is maximal in the sense that any chart (,U) on M that is C

compatible with all charts in A is also a member of A . (We will assume thatcharts whose domains do not overlap are automatically C-compatible, so thatchecking this condition reduces to looking at charts that do overlap.)

Technically, we should refer to the pair (M,A ) as the smooth manifold, since theexistence of the atlas and its properties are the essential parts of this definition.However, as is common in mathematics, when no confusion will arise, we will omitexplicit mention of the atlas, A . Smooth manifolds will be the our primary objectsof study, so some explanatory remarks are in order.

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Remark 1.2.1 An atlas satisfying these properties is said to be a differentiable, orsmooth, structure on M , and the atlas, A , described above is called a maximalatlas to distinguish it from an atlas. An atlas, as defined after definition 1.1, maycontain only a few charts, perhaps finitely many, since it is only required to coverthe manifold. A maximal atlas will, in general, contain many more charts, and canbe considered as an extension of the atlas. (See below.)

Remark 1.2.2 Condition i in this definition has a familiar geometric interpretation.If two charts (,U) and (, V ) overlap (overlap meaning that the domains inter-sect), then the map 1 : (U V ) (U V ) is just a change of coordinatesfrom one coordinate system to another. The requirement here is that this transitionfrom one system to another be smooth. Think, for example, of the transition fromrectangular to cylindrical or spherical coordinates in R3.

Remark 1.2.3 Condition ii is not as transparent, but it turns out to be very useful.Most importantly for our purposes, it provides us with a great deal of freedom inchoosing our coordinate charts for specific purposes.

Given a topological manifold, there really are no conditions imposed on the atlasexcept that it does, in fact, cover the space. The atlas may, indeed, be quite sparse inthe sense that there may be just enough coordinate charts to cover the space. Thus,given a particular point p M , there will, in general, be little freedom in choosing acoordinate chart at p. The maximality condition in the previous definition is muchstronger, though. Given any point, p, in a smooth manifold, we will see that thereare always infinitely many distinct coordinate charts at p that are C-compatiblewith all of the charts in the maximal atlas. Some will occasionally be more usefulthan others, thus allowing us to choose, in a given situation, a coordinate systemthat suits our needs. For instance, given a point p M , we will always be able tochoose, if it is helpful, a chart (,U) at p such that (p) = 0 Rm. The drawback isthat this maximal atlas will, consequently, be uncountable, and it will be impossibleto write down all of the coordinate charts in the covering and check the compatibilityof each pair. So, one might think that showing that a space is a smooth manifoldis an impossible task. It turns out, though, that it is not necessary to check thecompatibility of each pair of charts. As the following theorem will show, all we needto do is find an atlas for M (not necessarily maximal) such that all of the charts inthat particular atlas are C-compatible. Then this next result does the rest of thework for us.

Theorem 1.4 Let M be a topological manifold with an atlas, {(,U), (, V ), . . . },such that any two charts in this atlas are C-compatible. Then there is a unique

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maximal atlas, A , on M containing these coordinate charts, thus making (M,A ) asmooth manifold.

Proof We define our differentiable structure, A , to be the collection of all charts onM that are C compatible with all charts in the given atlas. Then A is nonemptyby hypothesis and is an atlas for M . Suppose (,W ) and (, Z) are two charts in Asuch that W Z 6= . If at least one of them is in the original coordinate covering,then they are C-compatible by hypothesis. So, we assume that they are not. Themaps 1 and 1 are well defined homeomorphisms between open sets in Rm,so we need only show that these maps are smooth. Let x = (p) be an arbitrary pointof (W Z). Then there is some chart (,U) in the original coordinate covering suchthat p U . So, N = U Z W is a neighborhood of p and (N) is a neighborhoodof x. On (N), we have 1 = 1 1. But 1 and 1 are Cby hypothesis. Hence, their composition is also C on (N). Since p was arbitrary,it follows that for every point, q, in (W Z), there is a neighborhood of q on which 1 is C, implying that this map is C. A similar argument shows that 1is C on (W Z). Hence, condition (1) of the definition of smooth manifold issatisfied. As for property (2), it is satisfied by our definition of A .

QED

So, to prove that a topological space is a smooth manifold, we need only find an atlasin which all of the charts are C-compatible, and then apply this theorem to extendthat covering to a maximal atlas, thus giving us a smooth manifold. Whenever weconstruct a smooth manifold, we will usually just construct the initial atlas andshow that the charts are C-compatible. We will typically not even mention theapplication of this theorem, but it will be implicit in all such examples.

Example 1.1 Euclidean Space

Any Euclidean space, Rm, is a smooth manifold. The single chart (i,Rm), where iis the identity map i : Rm Rm, gives us an atlas, and this map is trivially C-compatible with itself. Applying Theorem 1.4 gives us a smooth structure on Rm.This is referred to as the canonical smooth structure on Rm.

Example 1.2 Finite Dimensional Vector Spaces

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If V is a vector space of dimension n, then there is an isomorphism : V Rn.This isomorphism allows us to define a topology on V making it homeomorphic toRn, and, thus, making a homeomorphism. The single chart (, V ) gives us an atlasjust as in the previous example. Theorem 1.4 extends this to a smooth structure onV .

Example 1.3 Graph of a smooth function f : U R, where U is open in R2

In multivariable calculus, the graph of a smooth function f : U R, viewed asa subset of R3, is known to induce a surface. As manifolds are meant to be gen-eralizations of surfaces, it stands to reason that surfaces should certainly be mani-folds. Suppose U is an open set in R2, and f : U R is a smooth function. LetM = {(x, y, f(x, y)) R3 : (x, y) U}. We will show that M is a manifold. First,since M is a subspace of a Euclidean space, M is Hausdorff and second countable.Define a map : M U by (x, y, f(x, y)) = (x, y). This map is clearly surjec-tive, and it is injective, since (x1, y1, f(x1, y1)) = (x2, y2, f(x2, y2)) x1 = x2and y1 = y2. This implies that f(x1, y1) = f(x2, y2). Moreover, is smooth as itscomponent functions are smooth. The inverse mapping, 1 : U M , is given by1(x, y) = (x, y, f(x, y)). Since f is smooth by hypothesis, so is 1. Thus, thesingle chart (,M) covers M and is C-compatible with itself. Hence, it can beextended to a smooth structure on M .

Example 1.4 S1: the 1-dimensional sphere, or circle

Let S1 = {(x, y) R2 : x2 + y2 = 1}. This is, of course, the unit circle in R2. Itis a subspace of R2, so it is Hausdorff and second countable. It is a curve, or a 1-dimensional surface, but it cannot be represented as the graph of a single function.Define the following mappings.

1. Let U1 = {(x, y) S1 : y > 0}. Define 1 : U1 (1, 1) by 1(x, y) = x.2. Let U2 = {(x, y) S1 : y < 0}. Define 2 : U2 (1, 1) by 2(x, y) = x.3. Let U3 = {(x, y) S1 : x > 0}. Define 3 : U3 (1, 1) by 3(x, y) = y.4. Let U4 = {(x, y) S1 : x < 0}. Define 4 : U4 (1, 1) by 4(x, y) = y.

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Noting that U1 can be represented as the set of all points in S1 of the form (x,

1 x2)

with x (1, 1), we see that 11(x) = (x,

1 x2). The inverses of the other mapsare defined similarly. We are simply projecting hemispheres of the circle onto an openinterval. Given their respective domains, the maps i, 1 i 4 are easily seen tobe homeomorphisms. The charts (i, Ui) are also all C

-compatible. To see this,consider the case 1 31 : 3(U1 U3) 1(U1 U3). U1 U3 is the set of allpoints, (x, y), on the circle such that x > 0 and y > 0. We can represent such a pointas (x,

1 x2) or (1 y2, y). So, if y is a point in 3(U1 U3), then 0 < y < 1,

and 31(y) = (

1 y2, y). Thus, 1 31(y) = 1(

1 y2, y) = 1 y2. Since

0 < y < 1, this map is C. The other possible compositions are computed in asimilar way, and are also C. So, S1, with the atlas induced by the charts (i, Ui),is a smooth manifold.

In a similar way, we can show that the set Sn = {(x1, ..., xn+1) Rn+1 : x12 + + xn+12 = 1}, called the n-dimensional sphere, is a smooth manifold. As in the1-dimensional case, we simply divide Sn into 2(n + 1) hemispheres (two for eachdirection or dimension in Rn+1) and project to the appropriate coordinates. Thiswill be referred to as the standard coordinate atlas on Sn. However, it is possible todefine, on any sphere, a smooth coordinate covering consisting of just two coordinatecharts. This is called the stereographic projection, and it has certain advantages overthe standard coordinate covering that will become apparent later on.

Example 1.5 Sn in stereographic coordinates

Let n and s denote the points (0, . . . , 0, 1) and (0, . . . , 0,1), respectively, on Sn Rn+1. These are the north and south poles of Sn. Given any point p Sn {n}, the line passing through p and n intersects the hyperplane xn+1 = 0, which ishomeomorphic to Rn, in exactly one point. We define a coordinate chart by sendingp to that point. Define : Sn {n} Rn by

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

1 xn+1 (x1, . . . , xn).

The inverse map is given by

1(y1, . . . , yn) =

(2y1

1 + y2 , . . . ,2yn

1 + y2 ,1 + y21 + y2

).

Similarly, we define : Sn {s} Rn by

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

1 + xn+1(x1, . . . , xn).

It is easy to check that and are both homeomorphisms onto Rn, thus making{(, Sn{n}), (, Sn{s})} an atlas for Sn. Another computation shows that thecompositions 1 and 1 are smooth, thus inducing a smooth structure onSn. Note that each chart covers all of Sn except for a single point.

Example 1.6 Open submanifolds of a manifold, M

Let M be a smooth manifold, and let O be an open subspace of M . Then O isboth Hausdorff and second countable. Suppose the maximal atlas of M is given by{(,U), (, V ), . . . }. Define an atlas on O by {(|UO, U O), (|V O, V O), . . . }.The restriction of a homeomorphism to an open subset of its domain is still a home-omorphism, so this covering clearly makes O a topological manifold. Moreover, if(,U) and (, V ) are two charts such that (U O) (V O) = U V O 6= ,then 1 : (U V O) (U V O) is C, as it is just the restriction ofthe C map 1 : (U V ) (U V ) to an open subset of (U V ). Thus,any two charts in this atlas on O are C-compatible, and we can conclude that O,with the induced maximal atlas, is a smooth manifold. We call it a smooth opensubmanifold, or just an open submanifold, of M . It follows that every coordi-nate neighborhood, U , is an open submanifold of M . Note that open submanifoldsnecessarily have the same dimension as the ambient, or parent, manifold. Thiswill not be true for more general submanifolds that will be discussed later.

Example 1.7 Product Manifolds

Suppose M and N are smooth manifolds. Then the products A B, where A andB are open in M and N , respectively, form a basis for the topology on M N . Thismakes M N Hausdorff and second countable. Given any point (p, q) M N ,there are charts (,U) on M and (, V ) on N at p and q, respectively. We form achart (, U V ) on M N by letting (r, s) = ((r), (s)) for (r, s) U V . Doingthis for each point in MN gives us an atlas, making MN a topological manifold.It is easy to check that any two of these charts which overlap are C-compatible. (Itjust reduces to compatibility of charts on M and N , individually.) Hence, this atlasextends to a smooth structure, making M N a smooth manifold. This is referredto as the standard product manifold structure. Whenever we form productmanifolds, it will be assumed, unless specifically stated otherwise, that this is thesmooth structure we place on them.

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The previous examples have all been either standard or trivial in their construction.The next example will show how manifolds can arise from more complicated andabstract structures. This example, itself, however, is considered standard. Manifoldsmay have a much more complex structure.

Example 1.8 Real Projective n-space, RP n

This example follows a construction given in Boothby. Let X = Rn+1 {0}. Definean equivalence relation, , on X as follows. For x, y X, x y if and only ifthere exists a real number, 6= 0, such that x = y. Denote the equivalence classcontaining x by [x]. Let Pn = {[x] : x X}. Then Pn is the quotient space X/ ,and we assume it is endowed with the usual quotient space topology. This meansthat O Pn is open if and only if pi1(O) is open in X, where pi : X Pn is theusual quotient map, taking x X to the equivalence class [x]. This automaticallymakes pi continuous. Note that, geometrically speaking, we can identify Pn with theset of lines in Rn+1 passing through the origin. We will show that Pn is a smoothn-dimensional manifold. The proof is direct, following our definition of a smoothmanifold.

Note that, for [x] = [(x1, x2, ..., xn+1)] Pn, if xi 6= 0 for some i = 1, 2, ..., n+ 1, then[(x1, ..., xi, ..., xn+1)] = [(x1/xi, ..., xi1/xi, 1, xi+1/xi, ..., xn+1/xi)]. Also, for [x] Pn,if any element (x1, ..., xn+1) [x] satisfies xi = 0, then every element of [x] has 0 ithcomponent. So, we can speak of an equivalence class [x] Pn as having a zero ornonzero ith component, [x]i.

Let U Pn be given. Since pi1(U) = [x]U [x], it follows that a set U Pnis open if and only if the union of all the equivalence classes in U is open in X.Define sets Ui, i = 1, ..., n + 1, by Ui = {[x] Pn : [x]i 6= 0}. If [x] Pn, thenat least one of the components of [x] must be nonzero. So, the sets Ui cover Pn.They are also open. Consider pi1(Ui) =

[x]Ui [x]. This is just the union of all the

equivalence classes that have a nonzero ith component. Let x = (x1, ..., xi, ..., xn+1)

be in pi1(Ui). Then xi 6= 0. Choose > 0 such that 0 < < |xi|2 . The set{y Rn+1 : x y < } = B(x) is an open ball centered at x. If y B(x), then|xi yi| x y |xi yi| < |xi|2 < yi xi < |xi|2 . It follows that we musthave yi 6= 0. Thus B(x) pi1(Ui), implying that pi1(Ui) is open in X. Hence,each set Ui is open. We will use this collection of sets to form the initial atlas.

For each i = 1, ..., n+ 1, define the map i : Ui Rn by

15

i([x]) = i

([(x1, ..., xi, ..., xn+1)

])=

(x1

xi,x2

xi, ...,

xi1

xi,xi+1

xi, ...,

xn+1

xi

)Note that i is well-defined, for if x = (x

1, ..., xn+1), and if y = (y1, ..., yn+1) is in [x],then y = x for some 6= 0. Thus,

(y1yi, ...,

yi1

yi,yi+1

yi, ...,

yn+1

yi

)=

(x1xi

, ...,xi1

xi,xi+1

xi, ...,

xn+1

xi

)=

(x1xi, ...,

xi1

xi,xi+1

xi, ...,

xn+1

xi

).

Now, if i([x]) = i([y]) for [x], [y] Ui, then for any elements x [x] and y [y],we have

(x1

xi, ...,

xi1

xi,xi+1

xi, ...,

xn+1

xi

)=

(y1

yi, ...,

yi1

yi,yi+1

yi, ...,

yn+1

yi

)

(x1, ..., xi1, xi+1, ..., xn+1) = xiyi(y1, ..., yi1, yi+1, ..., yn+1

) (x1, ..., xi1, xi, xi+1, ..., xn+1) = xi

yi(y1, ..., yi1, yi, yi+1, ..., yn+1

)from which it follows that x = x

i

yiy. So [x] = [y], and each i is injective. If

y Rn, consider the element in X given by y = (y1, ..., yi1, 1, yi, ..., yn). That is,we put a 1 in the ith position and shift the remaining components up one indexvalue. If i = n + 1, we simply attach a 1 on the end of y. Then [y] Ui andi([y]) = (y

1, ..., yi1, yi, ..., yn) = y. So, each i is also surjective.

To show that each i is continuous, it suffices to show that i1(B(y)) is open in Pn

for arbitrary and y Rn. Hence, we want to show that pi1(i1(B(y))) is open inX. But pi1(i1(B(y))) = (i pi)1(B(y)), so, if the map i pi : pi1(Ui) Rnis continuous, this will prove the result. Now, pi1(Ui) consists of all those points inX such that xi 6= 0. This set is open in X. If x is such a point, then pi(x) = [x] and

i(pi(x)) =

(x1

xi, ...,

xi1

xi,xi+1

xi, ...,

xn+1

xi

).

16

Hence, since xi never vanishes for any point x pi1(Ui), we see that each of thecomponent functions of this map is continuous. It follows that i pi is continuouson pi1(Ui), implying that i is continuous on Ui.

Since i is a bijection, it has a well-defined inverse. Explicitly, the inverse mappingi1 : Rn Ui satisfies

i1(y1, ..., yn) = [y1, ..., yi1, 1, yi, ..., yn].

To see that i1 is continuous, define a map f : Rn Rn+1 by f(y1, ..., yn) =

(y1, ..., yi1, 1, yi, ..., yn). This map is continuous and its range lies in pi1(Ui). More-over, i

1 = pi f . Thus, i1 is continuous, and each i is a homeomorphism.

To show that Pn is a manifold, we still must show that Pn is Hausdorff and secondcountable. These results are not trivial, as topological properties are not generallywell-preserved under the formation of quotient spaces.

First, we claim that pi is an open map. For t R, t 6= 0, consider t : X X definedby t(x) = tx. This is just a scaling map, so it is a homeomorphism with t

1 = 1t.

If U X is open, then t(U) is open in X also. Moreover, if y [U ] :=xU [x],

then y [x] for some x U , implying that y = tx for some t 6= 0. So, y t(U),and we have

[U ] tRt6=0

t(U).

Conversely, if y tR,t 6=0 t(U), then y t(U) for some t. So, y = tx for somex U , implying that y x y [U ]. Thus, we have

[U ] =tRt6=0

t(U)

Since each t(U) is open, [U ] is open in X. But it is easily seen that pi1(pi(U)) = [U ].

Thus, by the definition of the quotient topology, pi(U) is open in Pn.

This shows that Pn is second countable. If {Bn}n1 is a countable basis for X, then{pi(Bn)}n1 is a countable basis for Pn. Given any open set O Pn and any [x] O,pi1(O) is open inX and contains x. So, there is some Bn such that x Bn pi1(O),

17

implying that [x] = pi(x) pi(Bn) pi(pi1(O)) O. Therefore, {pi(Bn)} is, indeed,a basis.

To prove the Hausdorff property, we will digress momentarily to prove the followinglemma.

Lemma 1.5 Let be an equivalence relation on a topological space X, and assumethat the quotient map pi : X X/ is an open map. Then R = {(x, y) X X :x y} is a closed subset of X X (with the standard product topology) if and onlyif X/ is Haudsdorff.Proof First, suppose X/ is Hausdorff. Suppose (x, y) / R. Then x y, whichimplies that pi(x) 6= pi(y). Hence, there are disjoint open subsets, U and V , of X/ such that pi(x) U and pi(y) V . Then pi1(U) and pi1(V ) are open sets in Xcontaining x and y, respectively. So, the set pi1(U) pi1(V ) is an open set inX X that contains (x, y). Suppose this set intersects R. Then it must containsome element (x, y) such that x y. This implies that pi(x) = pi(y). But (x, y)is also in pi1(U) pi1(V ), implying that pi(x) U and pi(y) V . It follows thatU and V have a nonempty intersection, contradicting the fact that they are disjoint.Thus, the set pi1(U) pi1(V ) does not intersect R. Therefore, for any (x, y) Rc,there is an open set containing (x, y) that does not intersect R. So, Rc is open,implying that R is closed.

Conversely, suppose R is closed. Since any point in X/ is the image of someelement under the map pi, we can consider two elements of X/ as pi(x) and pi(y).Suppose pi(x) and pi(y) are two distinct points of X/ . Then we cannot have x y.Thus, the element (x, y) must be in Rc, which is open. Hence, there are open setsU and V in X such that (x, y) U V and the set U V does not intersect R.Consider the sets pi(U) and pi(V ) in X/ . If there was an element, say [z], in bothof these sets, then we would have x z and y z, implying that x y. Hence, thiscontradiction shows that pi(U) and pi(V ) are disjoint. The previous lemma showsthat they are open, and we have pi(x) pi(U) and pi(y) pi(V ). Since these twoelements were arbitrary, this shows that X/ is Hausdorff.

QED

We now apply this result to prove the Haudsorff condition. Define a function f :X X R by

f(x, y) = f(x1, ..., xn+1, y1, ..., yn+1) =n

i,j=1

(xiyj xjyi)2.

18

Then f is continuous on X X. We can also show that f(x, y) = 0 y = txfor some t 6= 0. If y = tx, then it is clear that f vanishes. Conversely, supposef(x, y) = 0. Then, viewing x and y as vectors in Rn+1, this implies

ni=1

nj=1

(xi)2(yj)2 2ni=1

nj=1

xiyixjyj +ni=1

nj=1

(xj)2(yj)2 = 0

( ni=1

(xi)2)( n

j=1

(yj)2) 2( ni=1

xiyi)( n

j=1

xjyj)

+( ni=1

(xi)2)( n

j=1

(yj)2)

= 0

x2y2 = x, y2 xy = x, y,

from which it follows that x and y must be linearly dependent. That is, y = tx forsome t 6= 0. Thus, f1(0) consists of all those pairs, (x, y), in X X such thaty = tx for some t 6= 0. Hence, f1(0) = R = {(x, y) : x y} is closed, andLemma 1.6 implies that Rn is Hausdorff. So, Pn is a manifold. All that remains is toshow that the maps i are C

-compatible. This is nothing more than an elaboratecomputation.

Suppose Ui Uj 6= and x i(Ui Uj) Rn. Then x = i([y]) for some[y] UiUj. This [y] must, then, have nonzero i and j components. We can assumewithout loss of generality that i < j. The map i will remove the i

th component of[y] and shift the jth component to the (j 1)st position. Hence, x has a nonzero(j 1)st component, and it follows that

i1(x) = [x1, ..., xi1, 1, xi, ..., xj1, ..., xn]

where the component xj1 is nonzero and in the jth position of the equivalence class.We then have

j(i1(x)) = j

([x1, ..., xi1, 1, xi, ..., xj1, ..., xn]

)=

(x1

xj1, ...,

xi1

xj1,

1

xj1,xi

xj1, ...,

xj2

xj1,xj

xj1, ...,

xn

xj1

).

This map is defined on (Ui Uj), which consists only of points whose (j 1)stcomponent is nonzero. It is, therefore, C. The indices i and j were arbitrary,

19

so this shows that the coordinate charts (i, Ui) are all C-compatible. So, Pn is

a smooth n-dimensional manifold. In the cases n = 1 and n = 2, we call Pn theProjective Line and the Projective Plane, respectively.

We should remark that the extensive effort that went in to showing the projectivespace was, in fact, a smooth manifold, is not always required. Indeed, in many cases,there are quicker and more indirect ways of showing that a space is a manifold. As weprogress and further develop our theory, these methods will arise naturally. However,there are occasions where the direct method is the only viable choice, so the readershould have some experience with it.

Smooth Functions and Mappings

We will now introduce the concepts of smooth, or differentiable, functions on asmooth manifold, M . As M is a topological space, we have a well-defined notionof what it means for a function to be continuous. However, the methods of thedifferential calculus are the backbone of differential geometry. So, we need somemeans of differentiating functions on M . It is the local Euclidean structure of amanifold that makes this possible.

Definition 1.3 Let f be a real-valued function defined on an open subset, W , ofa smooth manifold, M . We say that f is C on W if for every p W there isa coordinate chart (,U) at p such that the function f 1 : (U W ) R isC on (U W ). We call the function f 1 : (U W ) R the coordinaterepresentation of f on U W . We let F (M) denote the set of smooth functionson M .

Remark In this definition, it is important to remember that (U W ) is anopen subset of Rm, so we know from basic Euclidean calculus what it means forf 1 : (U W ) R to be differentiable.

Even though this definition uses local coordinate systems and requires the exis-tence of only one such chart at the point p, (when, as we know, there will be manycharts at any given point) the smoothness of a function is, in fact, a well-definedcoordinate-independent property. Given p W , if there is one chart (,U) thatsatisfies the above condition, then any chart at p will. For if (, V ) is another chartat p, then f 1 : (U V W ) R is C, and the map 1 : (U V W )

20

(UV W ) is C. Thus, by the chain rule in Euclidean space (note that we have notyet developed a chain rule on manifolds), the function f 1 = (f 1) (1)is C on (U V W ), a neighborhood of (p). Thus, the definition does notdepend on the particular coordinate representation of f .

Given this coordinate independence, a more compact way of saying all of thiswould be the following. A function f : W R, where W M is open, issmooth on W if and only if for any chart (,U) such that U W 6= , thefunction f 1 : (U W ) R is smooth. Moreover, as we should expect,differentiability of f at p implies continuity of f at p. We have f = (f 1) on(U W ). Being a homeomorphism, is continuous at p, and f 1 is continuousat (p), being differentiable there. Thus, smoothness implies continuity.

As a note on the algebraic structure of F (M), we should point out that it followsalmost directly from the definition that F (M) is a commutative algebra over R witha multiplicative identity.

This definition of smooth real-valued functions leads us to the definition of asmooth mapping between manifolds. As a point of terminology, we will reserve theterm function for real-valued functions on manifolds, while we will use the term mapto refer to mappings between general manifolds.

Definition 1.4 Let M and N be smooth manifolds, and let F : M N be amapping of M into N . We say that F is a smooth mapping if for every p M thereexist coordinate charts (,U) and (, V ) such that p U , F (p) V , F (U) V , andthe mapping F 1 : (U) (V ) is C. We call F 1 the coordinaterepresentation of F with respect to the charts (,U) and (, V ).

As with the previous definition, this definition of smoothness is coordinate inde-pendent and does not depend on the specific charts we choose. If (, U ) and(, V ) are two other coordinate systems around p and F (p), respectively, then F 1 = 1 F 1 1 on (U U ). Since the maps1, F 1, and 1 are C, so is F 1, by applying the Euclideanchain rule. Thus, F : M N is C if and only if every possible coordinaterepresentation of F is C. Moreover, it follows from this definition that smoothmaps are continuous.

Note, also, that we have not bothered to restrict F to an open subset of M inthis definition, as we did in the definition of a smooth function. Recalling Example1.5, an open subset of M is an open submanifold of M , and so is a manifold, itself.So, if we have a mapping F : W N , where W is an open subset of M , we canstill apply this definition by simply replacing M with W . Definition 1.4, therefore,includes all of the cases in which F may be defined on only some open subset of M .

21

There is a consequence of these definitions that we will use frequently and usuallywithout comment. Each coordinate mapping : U (U) is a smooth functionfrom the coordinate neighborhood (and manifold), U , to the open set (U) Rm.This just follows from the C compatibility of the charts in the smooth structureon M . If p is any point in U and (, V ) is any coordinate chart around p (possibly(,U) itself), then 1 : (U V ) (U V ) is C because the charts areC-compatible. Thus, is smooth. A similar argument shows that 1 is smoothas a map from the open set (U) onto U . Hence, the coordinate maps on a smoothmanifold are diffeomorphisms.

Definition 1.5 Let M and N be smooth manifolds. A map F : M N is adiffeomorphism if it has an inverse F1 : N M and if both F and F1 aresmooth. If such a map exists, we say that M and N are diffeomorphic

Just as homeomorphisms preserve all topological properties between two topologicalspaces, diffeomorphisms preserve all properties that can be expressed in terms of thetopological and smooth structures of manifolds. This concept is the foundation of thesubject of Differential Topology. Clearly every diffeomorphism is a homeomorphism.The converse is not true, as can be seen by the mapping t 7 t1/3 from R onto R.

We now prove a series of useful lemmas concerning smooth maps on manifolds.The first is a version of the chain rule. We will strengthen this result in the nextsection.

Lemma 1.6 Let F : M N and G : N L be smooth maps between smoothmanifolds. Then G F is smooth.Proof Let p M be given. Let q = F (p) N , and let r = G(q) L. Byhypothesis, there are charts (,U) and (, V ) at q and r, respectively, such thatG(U) V and G 1 : (U) (V ) is C. Choose a chart (,W ) at p,and, using the continuity of F , shrink W so that F (W ) U . Since F is smooth, F 1 : (W ) (U) is smooth. Thus, by the Euclidean chain rule, the map G F 1 = G 1 F 1 is C on (W ). Since we also have(G F )(W ) V , this shows that G F is smooth.

QED

Note that this version of the chain rule did not use the notion of the differentialof a map, which is the manner in which the Euclidean chain rule is usually stated.We are assuming the Euclidean chain rule in this case. Later on, we will definea corresponding notion of the differential of a map and prove a chain rule in thatcontext as well.

22

Lemma 1.7 Let M be a smooth manifold, and let W M be open. Suppose F :W Rm is a diffeomorphism from W onto some open subset of Rm. Then (F,W )is a coordinate chart on M .

Proof Being a diffeomorphism, F is necessarily a homeomorphism. So (F,W ) isa coordinate chart in the topological sense. Let (,U) be any chart in the maximalatlas on M such that UW 6= . Since F is smooth, F 1 : (UW ) F (UW )is also smooth by the previous lemma (or simply by the definition of a smooth mapon the open submanifold W ). Likewise, the differentiability of F1 implies that F1 : F (U W ) (U W ) is C. Since (,U) was arbitrary, it followsthat (F,W ) is C-compatible with every overlapping chart on M . Hence, (F,W ) iscontained in the maximal smooth structure on M .

QED

As an example of the utility of this Lemma, consider any chart (,U) on a smoothmanifold M , and let p be a fixed point in U . We know that is a diffeomorphismof U onto the open set (U) Rm, which contains (p). Let T : Rm Rm bethe translation map, T (x) = x (p). It is easy to verify that translations arediffeomorphisms of Euclidean spaces. Since the restriction of a diffeomorphism toan open subset of its domain is, again, a diffeomorphism, it follows that the mapT : U T ((U)) is a diffeomorphism. By the preceding lemma, we can concludethat (T ,U) is a coordinate chart on M . Moreover, we see that (T )(p) = 0.Thus, we may choose, whenever convenient, a chart that maps any given point of Mto 0 Rm.

In this same way, and utilizing the same lemma, we can always find charts withvery desirable properties. We can assume, for instance, that a given chart mapsits corresponding coordinate neighborhood to an open ball centered at any pointwe choose. We can also permute coordinates if we need to, since permutation ofcoordinates in Rm is a diffeomorphism.

Next, we prove two very important lemmas regarding the existence of certainkinds of smooth functions on manifolds. These are smooth analogues of well-knownresults in general topology concerning the existence of continuous functions on certaintopological spaces. (Compare, for example, to Urysohns Lemma.) Proving thetheorems for manifolds follows easily once one has established them on ordinaryEuclidean space, so we will prove those results first.

Lemma Suppose A,K Rm, with A closed, K compact, and A K = . Thenthere is a smooth function f : Rm [0, 1] such that f = 0 on A and f = 1 on K.

23

Proof The function h(t) = e1/t2, extended to (, 0] by setting h(t) = 0 for

t 0, is C. Define g : Rm R by

g(x) =h(2 x2)

h(2 x2) + h(x2 2/4) .

It is straightforward to check that g is C on Rm, positive on B(0), 1 on B/2(0), and0 outside B(0). Thus, for a Rm, g(x) = g(xa) is C on Rm, 1 on B/2(a), and 0outside B(a). Let B(ai), i = 1, . . . , k, be a finite collection of balls in Rm F suchthat K iB/2(ai). For each i, let gi be as above. Then the function f : Rm [0, 1]defined by

f(x) = 1 ki=1(1 gi(x))satisfies the desired conclusions.

QED

Corollary Suppose U Rm is open and f : U R is C. Let p be a point in U .There is a neighborhood, V , of p, with V U , and a C function g : Rm R suchthat g = f on V and g = 0 outside of U .

Proof Since Rm is regular and locally compact, choose neighborhoods V and W ofp such that p V V W W U with V compact. By the lemma, there is asmooth function h : Rm R such that h = 1 on V and h = 0 outside W . Define gby letting g = fh on U and letting g = 0 on RmU . Then g is smooth and satisfiesthe desired properties.

QED

Lemma 1.8 Let M be a smooth manifold. Suppose A M is closed, K M iscompact, and A K = . Then there is a smooth function, f : M [0, 1] such thatf = 1 on K and f = 0 on A.

Proof Given p K, let (,U) be a chart at p such that U A = , (p) = 0, and(U) = Br(0) Rm. We will first show that there exist neighborhoods of p, W andV , such that p W V V U , with V compact in M , and a smooth functiong : M [0, 1] such that g = 1 on W , g is strictly positive on V , and g = 0 outsideof V .

24

Choose < r, so that B/2(0) B(0) B(0) Br(0). Let W = 1(B/2(0))and V = 1(B(0)). From the Lemma above, there is a C function g : Rm [0, 1]such that g = 1 on B/2(0), g is positive on B(0), and g = 0 outside B(0).

It is clear that W V . The closure of B(0) in Br(0) is the same as the closureof B(0) in Rm. Thus, 1(B(0)) = clU(1(B(0))) = clU(V ), where clU denotesthe closure of a set in U . (We use an overhead bar to denote closure in the wholespace, M .) Moreover, B(0) is compact in Br(0), so

1(B(0)) is compact in U .That is, clU(V ) is compact in U . Following reasoning similar to that used in Lemma1.1, we find that clU(V ) is compact in M and that clU(V ) = V . So V is compact inM and V = clU(V ) U .

Define g : M [0, 1] by

g(q) =

{g((q)), q U0, q M U.

Note that q W g(q) = 1, q V g(q) > 0, and q U V g(q) = 0. Thus,g = 1 on W , g > 0 on V , and g = 0 outside V . On the open set U , g agrees withthe C function g . On the open set M V , g is identically 0. So g is smooth ontwo open sets which, together, cover all of M . Hence, g is smooth.

Now, for each p K, choose a chart (p, Up) so that Up A = , and let Wp, Vp,gp be the corresponding neighborhoods and function as determined above, so thatp Wp Vp Vp Up, gp maps M smoothly into [0, 1], gp = 1 on Wp, gp > 0 onVp, and gp = 0 oustide of Vp. The collection {Wp}pK covers K, so there is a finitecollection of points, {p1, . . . , pn}, such that K

ni=1Wpi . Define f : M [0, 1] by

f(q) = 1 ni=1(1 gpi(q)).Clearly f is smooth. If q M , then 0 gpi(q) 1 for each i, from which it followsthat 0 f(q) 1. So f does, indeed, map into [0, 1]. If q K, then q Wpi forsome i {1, . . . , n}, so 1 gpi(q) = 0 for this i and f(q) = 1. That is, f = 1 onK. Finally, suppose q A. Then q cannot be in pK Up since Up A = for allp K. So, q / Vpi for all i = 1, . . . , n, implying that gpi(q) = 0 for all such i. Thus,f(q) = 0, and we see that f = 0 on A. So, f is the desired function.

QED

Corollary 1.9 Let U M be open, and let f : U R be smooth. Let p be afixed point in U . Then there is a neighborhood, V , of p, with V U , and a smoothfunction g : M R such that g = f on V and g = 0 outside U .

25

Proof As in the corollary above, we use the local compactness and the regularityof M to obtain neighborhoods V and W of p such that V is compact and p V V W W U . Apply the previous Lemma (with K = V and A = M W ) toget a smooth function h : M [0, 1] such that h = 1 on V and h = 0 on M W .Define g : M [0, 1] by letting g = fh on U and letting g = 0 on M U . Thenq V V h(q) = 1 g(q) = f(q), so g = f on V . Clearly g is smooth on U .On M W , g = 0 because h vanishes outside of W . So g is smooth on M W .Since U and M W cover M , g is smooth on M .

QED

Corollary 1.10 Let M be a smooth manifold. Given any point p M and anyneighborhood, U , of p, there is a smooth function f : M [0, 1] (called a bumpfunction) such that f has compact support, supp(f) U , and f is identically 1 onsome neighborhood of p contained in U .

Proof Choose neighborhoods V and W of p such that p V V W W Uand W is compact (thus making V compact). Applying Lemma 1.8, we obtain asmooth function f : M [0, 1] such that f = 1 on V and f = 0 outside W . Thensupp(f) W , so f has compact support and supp(f) U . Moreover, f is identically1 on V .

QED

Of these results, Corollary 1.10 and the existence of bump functions will provemost useful.

As a final note on smooth functions and maps, we will consider more carefullytheir coordinate representations. In the following chapters, when we discuss tangentvectors and derivatives of maps, we will find a way to represent the derivatives ofthese smooth maps. For the present, however, we will give some simple examplesthat will be useful later on.

Let I be an interval in R (closed or open, perhaps infinite), and let : I Mbe a smooth map. This is a curve in the manifold M . If p (I) and (,U) is acoordinate system around p, then the coordinate representation of on I 1(U)is given by

( )(t) = (x1(t), x2(t), ..., xm(t))

26

where each xi is a real-valued coordinate function on I 1(U). That is, if pii isthe ith coordinate projection on Rm, then each xi is defined by xi = pii . Thesmoothness of these maps implies that each xi is smooth on the open set I 1(U).This coordinate representation is just a usual curve in Rm, and it has derivative((x1)(t), ..., (xm)(t)). We will see later that this is a coordinate representation ofthe tangent vector to the the curve .

Next, consider a point, p M , and suppose there are two charts, (,U) and(, V ), such that p UV . Then we have two different means of representing pointsnear p by local coordinates. Each coordinate mapping defines m coordinate functionsfrom M to R. That is, we can represent the map on U by (q) = (x1(q), ..., xm(q)),where each xi maps U into R. We refer to xi as the ith coordinate function of .If pii is the ith coordinate projection in Rm, then xi = pii . It is easily seen thatthe smoothness of implies the smoothness of the functions xi. Likewise, there aresmooth functions yj such that (q) = (y1(q), ..., ym(q)) for q V .

Now, a point q U V will have coordinates (q) (U V ) and (q) (U V ). Suppose we want to change from one coordinate system to another, sayfrom -coordinates to -coordinates. We need only apply to (q) the map 1,which will give us (q). Likewise, we change the other way by applying to (q) themap 1. In terms of the coordinate functions, these mappings indicate that wecan think of the functions xi as each being functions of the yjs, and vice versa. Inshort, we have, via the mapping 1, that

(y1, ..., ym) 7 (x1(y1, ..., ym), ..., xm(y1, ..., ym)),and, via the map 1, we have

(x1, ..., xm) 7(y1(x1, ..., xm), ..., ym(x1, ..., xm)).Moreover, since the manifold is smooth, the coordinate transformations are C.Thus, each of the functions xi(y1, . . . , ym) is smooth, as is each of the functionsyj(x1, . . . , xm).

As a final remark on notation, whenever we utilize a specific chart (,U) andwe need to specifically use the coordinate system, we will often simply write =(x1, . . . , xm) to indicate that the xi are the coordinate functions of .

27

2 The Tangent and Cotangent Spaces

2.1 The Tangent Space

The tangent space is the first additional structure we will add to our smoothmanifolds. The motivation for this concept is clear. In classical differential geometry,the notion of a tangent plane to a surface was a well-defined concept, and it could beidentified with a 2-dimensional subspace of R3, thus giving us the idea of the tangentspace. However, in that context, the tangent space at a particular point was definedby means of tangent vectors to curves in the surface passing through that point. Itis possible to extend this idea to manifolds, and we will do so. We will proceed,though, with a definition of the tangent space that is completely equivalent, thoughnot quite as familiar. For the remainder of this section, p is a fixed point in thesmooth manifold, M .

Let C(p) be the set of all real-valued functions that are defined and smooth onsome neighborhood of p. Thus,

C(p) = {f : Uf R Uf is a neighborhood of p, f is C on Uf}.

Note that we can always assume that a neighborhood, Uf , associated with a function,f , lies within a coordinate neighborhood.

Define a relation on C(p) by f g there exists some neighborhood V of pon which f = g. It is easy to see that this is an equivalence relation on C(p), andwe will denote the equivalence class containing f by f . Denote the quotient spaceC(p)/ by Fp.Lemma 2.1 Fp is a commutative algebra with a multiplicative identity

Proof For f , g Fp, define f + g to be the equivalence class of the functionf + g, where f + g is defined on Nf Ng. That is, we define f + g = f + g. Weneed to show that this is independent of the choice of representatives from f andg. Suppose f1, f2 f and g1, g2 g. Then there are neighborhoods, Nf and Ng,of p such that f1 = f2 on Nf and g1 = g2 on Ng. Then, for q Nf Ng, we havef1(q) = f2(q) and g1(q) = g2(q), implying that f1(q) + g1(q) = f2(q) + g2(q). So,

f1 + g1 f2 + g2 f1 + g1 = f2 + g2. Hence, the addition operation is well-defined.Thus, we have a well-defined notion of vector addition in Fp.

A similar argument shows that scalar multiplication is well-defined. The vectorspace properties are easily established from here. For example, the zero element, 0,is the set of all functions in C(p) that vanish in some neighborhood of p.

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Finally, define f g = f g. As with addition, this is a well-defined multiplicationoperation. Associativity and commutativity of multiplication follow easily, and dis-tributivity of multiplication over addition is also clear. The multiplicative identity isjust the set of all functions that are equal to the constant, 1, on some neighborhoodof p.

QED

The reason for defining the equivalence classes above is one of convenience. Wewill generally only be concerned with derivatives of smooth functions at p, and func-tions that are equal on some neighborhood of p will have the same derivatives. Thisbrings us to our definition of the tangent space at p.

Definition 2.1 A tangent vector at p is a map, Xp : Fp R such that, for anyf , g Fp and , R,

i) Xp(f + g) = Xp(f) + Xp(g) (linearity)

ii) Xp(f g) = f(p)Xp(g) +Xp(f)g(p) (Leibniz property).

Note that these are the defining characteristics of derivative operator. In general, amap on an algebra of functions satisfying properties i and ii is called a derivation.So, a tangent vector is just a derivation on Fp. The set of all tangent vectors at pis called the tangent space at p, and is denoted by TpM . Since a tangent vector is alocal object, we will often denote an element of TpM by Xp, to indicate that Xp is atangent vector at p.

Clearly, TpM is nonempty, as the zero derivation that maps all functions to 0 willbe in this set. However, if it is not clear that there are nontrivial derivations, it willbe as we progress and construct explicit examples. Indeed, we will construct a basisfor TpM with respect to a given coordinate chart.

First, however, note that TpM is made into a real vector space in the obviousway. For Xp and Yp in TpM , and any f Fp, R, we define

(Xp + Yp)(f) = Xp(f) + Yp(f)

(Xp)(f) = Xp(f).

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Of course, one must show that these definitions do, in fact, produce tangent vectors,but that is an easy exercise.

Note that the existence and linear structure of TpM is coordinate independent.We have not used any local coordinate system around p to construct anything so far.This shows that the tangent space and its linear structure are well-defined concepts,independent of any choice of coordinates. However, to construct a basis and representtangent vectors in explicit form, a coordinate system is necessary.

Let (,U) be a coordinate system around p. In Euclidean spaces, we can identifythe standard basis vectors with the partial derivative operators that give the direc-tional derivatives in the direction of the coordinate axes. So, it is natural for us toask whether we can construct a basis for our tangent space using the same idea. Infact, we can. For i = 1, 2, ...,m define mappings i : Fp R by

i(f) =

xi(f 1)

(p)

.

Note that this is a well-defined mapping, since any f1, f2 f are equal on someneighborhood of p, implying that f1 1 and f2 1 are equal on some neighbor-hood of (p). Hence, i just maps f to the i

th partial derivative of this coordinaterepresentation of f at (p).

Lemma 2.2 Each i is a tangent vector. That is, each i is a derivation on Fp.

Proof Let f and g be in Fp. Then

i(f + g) = i(f + g) =

xi((f + g) 1)

(p).

But (f + g) 1 = f 1 + g 1, so

i(f + g) =

xi(f 1 + g 1)

(p)

=

xi(f 1)

(p)+

xi(g 1)

(p)

= i(f) + i(g).

Now, if R, then we have

30

i(f) = i(f)

=

xi((f) 1)

(p)

=

xi((f 1))

(p)

=

xi(f 1)

(p)

= i(f).

Finally, we have

i(f g) = i(f g)

=

xi((fg) 1)

(p)

=

xi((f 1)(g 1))

(p)

= (f 1)((p)) xi

(g 1)(p)

+ (g 1)((p)) xi

(f 1)(p)

= f(p)i(g) + g(p)i(f).

Thus, each Ei is tangent vector.

QED

Before we continue in showing that the set {i : 1 i m} forms a basisfor TpM , we should say something about our coordinate representations. We havebeen using the partial derivative notation with respect to the variable xi. There isa specific meaning behind this. These are the coordinate functions representing .As in the previous section, if pii is the ith coordinate projection from Rm to R, then,for q U , xi(q) = pii((q)). Note that these coordinate functions are smooth real-valued functions on U . Our tangent vectors, i, then, are just the partial derivativesof f 1 with respect to these coordinate functions.

Now, we will prove that the tangent vectors {i} are linearly independent. Let0p denote the zero element of TpM . That is, 0p(f) = 0 for all f Fp. Suppose thereare scalars c1, ..., cm such that

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mi=1

cii = 0p

Consider the elementm

i=1 cii acting on xk, the equivalence class containing the kth

coordinate function. We have

mi=1

cii(xk) =

mi=1

ci

xi(xk 1)

(p)

=mi=1

ci

xi(pik 1)

(p)

=mi=1

ci

xipik(p)

=mi=1

ciki

= ck

= 0.

This shows that ck = 0. Applying this sum to each k, in turn, shows that c1 = c2 = = cm = 0. Hence, these tangent vectors are linearly independent. To prove thatthey span TpM , we need a preliminary result from multivariable calculus. This isjust a consequence of the multivariable mean value theorem.

Lemma 2.3 Let f be a real-valued function that is differentiable on some neighbor-hood, U , of a Rm, and suppose that U is such that the straight line from a to x, forany x U , is contained in U . Such a neighborhood is called star-shaped about a.Then there are functions gk : U R, for k = 1, 2, ...,m, such that

f(x) = f(a) +mk=1

(xk ak)gk(x) and gk(a) = fxk

a.

Proof For each x U , define hx : [0, 1] R by hx(t) = f(a + t(x a)). Then wehave

dhx(t)

dt=

mk=1

(xk ak)Dkf(a+ t(x a)).

32

By the fundamental theorem of calculus, we know that 10

hx(t)dt = hx(1) hx(0) = f(x) f(a),so it follows that

f(x) f(a) =mk=1

(xk ak) 1

0

Dkf(a+ t(x a))dt x U.

Letting gk(x) be the integral in the above expression, we have gk(a) = Dkf(a) and

f(x) = f(a) +mk=1

(xk ak)gk(x).

QED

To use this result, we must see how to translate it to the manifold. Suppose f isa C function defined on some neighborhood, V , of p. Assume, by shrinking V ifnecessary, that V U and that V is the image under 1 of an open ball at p (orany open set in (U) that is star-shaped about (p)). This implies that f 1 is aC function on (V ) (U). We can apply this lemma to conclude that there arefunctions gk : (V ) R, such that, for all x (V ),

(f 1)(x) = (f 1)((p)) +mk=1

(xk [(p)]k)gk(x)

where [(p)]k is the kth coordinate of (p) Rm and

gk((p)) =

xk(f 1)

(p).

Given q V , with x = (q), the above equality implies that

f(q) = f(p) +mk=1

((pik )(q) (pik )(p))gk((q)). (1)

Since this last equality holds on V , we see that f and the function on the right-hand side belong to the same equivalence class. Remember that p is fixed in thisdiscussion, so f(p) and (pik )(p) are constants in the right-hand side.

33

Lemma 2.4 The elements {i : 1 i m} span TpM .

Proof Let Xp be an arbitrary tangent vector, and let f be in Fp. By the represen-tation formula we just derived, there is some neighborhood, V , of p, contained in U ,and there are functions gk, k = 1, . . . ,m, such that (1) holds for all q V . It followsthat the function on the right-hand side of (1) represents the same equivalence classas f . Thus, we have

Xp(f) = Xp

(f(p) +

mk=1

[(pik ) (pik )(p)

](gk )

)

= Xp(f(p)

)+

mk=1

Xp

[((pik ) (pik )(p)

)(gk )

]= Xp

(f(p)

)+

mk=1

Xp

((pik ) (pik )(p)

)gk((p))

+mk=1

((pik )(p) (pik )(p)

)Xp(gk

)= Xp

(f(p)

)+

mk=1

Xp( (pik ) (pik )(p))gk((p)).

It is easy to see that a tangent vector (or any derivation) must map to 0 any elementof Fp that is constant on a neighborhood of p. Since p is fixed in this argument, weget

Xp(f) =mk=1

Xp(pik )k(f).

Since f Fp was arbitrary, we can write this as the operator equation

Xp =mk=1

Xp(xk)k,

and this shows that the set {i : 1 i m} spans TpM . Moreover, it also showsthat the components of a tangent vector, Xp, with respect to the basis {i} are thevalues obtained by Xp acting on the equivalence classes of the coordinate functionsrepresenting .

34

QED

As a consequence of this result, we see that the dimension of the tangent space,TpM , is the same as that of the manifold at p. It should be emphasized again,however, that the basis we have constructed depends upon the coordinate system(,U). If we had used another chart, (, V ), such that p V , then we would stillhave obtained a basis for an m-dimensional vector space, but that basis would, ingeneral, be different. We call the basis {i : 1 i m} obtained at p froma particular chart the standard coordinate frame , or simply the coordinateframe , induced by the coordinate system.

As another consequence of this theorem, it becomes clear that we can considertangent vectors as acting on individual functions instead of equivalence classes offunctions. This will make our notation a great deal more convenient. To see why thisfollows, suppose f g in C(p). Then, since f and g agree on some neighborhoodof p, we see from the representation of Xp in the coordinate frame that

Xp(f) =mk=1

Xp(xk)

xk(f 1)

(p)

=mk=1

Xp(xk)

xk(g 1)

(p)

= Xp(g).

In other words, the valueX(f) can be computed using any function in the equivalenceclass f . Thus, we can actually formally define Xp(f), for f C(p), to be thevalue Xp(f). The linearity and Leibniz properties still hold in this context since

Xp(f)+Xp(g) = Xp(f)+Xp(g) = Xp(f+g) = Xp( f + g) = Xp(f+g)and Xp(fg) = Xp(f g) = f(p)Xp(g) + g(p)Xp(f) = f(p)Xp(g) + g(p)Xp(f).

First, since we have shown that the action of a tangent vector on f is independentof the function, f , we choose from f , we will omit the tilde in our notation wheneverconvenient. For example, the operator equation we derived for the representation of atangent vector with respect to a particular coordinate system can be more succinctlywritten as

Xp =mk=1

Xp(xk)Ek,

where it is implied that xk actually represents the equivalence class xk. This alternatenotation will not be used exclusively. In particular, when we discuss the cotangentspace later on, we will use the tilde again, since referring to the entire equivalenceclass will be necessary. However, it should be clear from the context which particular

35

notation is appropriate. In general, all of the mappings and operators we define willbe independent of any particular element chosen from an equivalence class. So, therewill be no difference in the results regardless of which notation we use. This is reallyjust a matter of convenient notation, which is very useful in differential geometry.The technical nature of the field is such that ones results can quickly become miredin what Elie Cartan called the debauch of indices. Thus, expressions are simplified,both visually and mathematically, whenever possible.

Second, the following transformation theorem, along with its analogue in the caseof the cotangent space, will require that we take partial derivatives of the change ofvariable maps. Recall that if (,U) and (, V ) are two coordinate systems such thatp U V , then we can change from -coordinates to -coordinates via the map 1, as well as from -coordinates to -coordinates via the inverse map, 1.Each of these is a map from Rm into Rm. As before, we will denote the coordinatefunctions of by xi, meaning that, for p U , we have xi(p) = pii((p)), wherepii is the ith coordinate projection in Rm. Likewise, we will denote the coordinatefunctions of by yi. Hence, in forming the composition 1, we see that eachcoordinate map, xi, becomes a function of the coordinates y1, . . . , ym, since 1maps points described by -coordinates to points described by -coordinates. Inother words, we can represent this map in terms of its input and output coordinatesby the expression

1(y1, . . . , ym) = (x1(y1, . . . , ym), . . . , xm(y1, . . . , ym)).Similarly, in forming the composition 1, we see that each coordinate map,yi, becomes a function of the coordinates x1, . . . , xm, since 1 maps pointsdescribed by -coordinates to points described by -coordinates. That is, we obtaina representation of the form

1(x1, . . . , xm) = (y1(x1, . . . , xm), . . . , ym(x1, . . . , xm)).Now, suppose we wish to compute the ith partial derivative of the jth component ofthe map 1 at the point (p). The correct, coordinate-free notation, of course,would be to write this partial derivative as

Di( 1)j

(p),

where Di represents the general partial derivative operator with respect to the ith

variable of the function in question, and the superscript j indicates the jth componentfunction of the map 1. Note that this is equivalent to the classically motivatedexpression

36

(pij 1)

yi

(p)

,

since composing pij with 1 just gives us the jth component function of this map.The classical partial derivative operator, /yi, simply indicates that the derivativeis to be taken with respect to the ith variable. Since the composition operation isassociative, consider composing pij with in this expression, and leaving 1 as is.We know that pij is just xj. So, the previous expression is further equivalent to

(xj 1)yi

(p)

.

So, in a very real sense, we are simply computing the ith partial derivative, withrespect to the -coordinate system, of the jth -coordinate. These particular partialderivatives are often simply written in the abbreviated form

xj

yi

(p)

,

since the fact that we are composing xj with 1 is implied by the fact that we aretaking the partial derivative with respect to one of the -coordinates. The point ofall of this is to alleviate any confusion that might arise later on. This abbreviatedform is just a notational device. It is very elegant and convenient, but it shouldalways be taken to represent the formal derivative Di( 1)j|(p). In the exactsame fashion, we will use the abbreviated expression

yj

xi

|(p)

to represent the formal partial derivative Di( 1)j|(p), which is the ith partialderivative of the jth component of the map 1 at (p). With these conventionsin mind, we can finally present the following result.

Theorem 2.5 Suppose p U V , where (,U) and (, V ) are two coordinate sys-tems around p. Let = (x1, . . . , xm) and let = (y1, . . . , ym) denote the coordinatefunctions making up the coordinate representation of , and let {yi} denote the co-ordinates of . Let {Exi } and {Eyi } denote the coordinate frames induced by (,U)and (, V ), respectively, on TpM . Then

Exi =mk=1

yk

xi

(p)

Eyk and Eyj =

ml=1

xl

yj

(p)

Exl .

37

If Xp =iExi =

jEyj is a tangent vector at p, then

i =mj=1

jxi

yj

(p)

and j =mi=1

iyj

xi

(p)

.

Proof The proof is just an application of change of coordinates at p. By our previouslemma, we know that Exi =

k E

xi (y

k)Eyk . By our definition of Exi , and using our

abbreviated notation, we have

Exi (yk) =

xi(yk 1)

(p)

=(pik 1)

xi

(p)

= Di( 1)k

(p)

=yk

xi

(p)

.

It follows that

Exi =mk=1

yk

xi

(p)

Eyk ,

and the second formula follows similarly. Once those formulas are established, givenXp, we know that its i

th component, i, with respect to the basis {Exi }, is given byXp(x

i). So, we have

i = Xp(xi)

=mj=1

jEyj (xi)

=mj=1

jxi

yj

(p)

,

and this verifies the third formula. The last one follows in the same manner.

QED

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There is an interesting consequence of this last lemma. Recall from elementarylinear algebra that, if B1 = {e1, . . . , em} and B2 = {f1, . . . , fm} are two bases foran m-dimensional vector space, V , then we can easily construct a change of basismatrix, P1,2. Given any vector, v V , if [v]1 represents the vector of the componentsof v with respect to B1, then P1,2(v) will give us [v]2, the vector of the componentsof v with respect to B2. The ith column of P1,2 is given by [ei]2, the vector of thecomponents of ei with respect to B2. Likewise, the inverse of this matrix, denotedP2,1, maps [v]2 to [v]1.

Now, consider applying this idea to the results given in the previous lemma. If Xpis an arbitrary tangent vector in TpM , with representations Xp =

iExi =

jEyj ,

then we can transform between the two component vectors [Xp] = [1 m]T and

[Xp] = [1 m]T . According to the lemma, the matrix that maps [Xp] to [Xp]

is given by y1

x1 y1

xm...

. . ....

ym

x1 ym

xm

,where the partial derivatives are evaluated at (p). Thus, the change of basis matrixthat maps tangent vectors in span{Exi } to tangent vectors in span{Eyj } is just theJacobian of the change of coordinate mapping 1 : (U V ) (U V )evaluated at (p). Likewise, the change of basis matrix that maps tangent vectorsin span{Eyj } to tangent vectors in span{Exi } is the inverse of that matrix, whichis nothing more than the Jacobian of the change of coordinate mapping 1 :(UV ) (UV ) evaluated at (p). These Jacobians are often given in shorthandnotation by

D( 1)(p) and D( 1)(p).

3.2 The Cotangent Space

We have seen that the tangent space TpM has the same dimension as the manifold,which we assume to be finite. Hence, after choosing a basis, the tangent spaceis isomorphic to Rm. Consequently, it has a well-defined dual space of the samedimension. This dual space is called the cotangent space, and it is the set of all linearfunctionals on TpM . We denote the cotangent space by T

pM . It is a perplexing but

common feature of most differential geometry texts to leave the discussion of thecotangent space after just these few statements, if it is even mentioned at all. It is

39

instructive, however, to actually construct the cotangent space independently, as wedid the tangent space, and then produce a pairing between the two showing thatthey are dual spaces of each other. Given what we have constructed so far, we canshow that the cotangent space at a point p is actually a quotient space of Fp.

Let p be the set of all smooth curves in M through p. By a suitable translation,we can assume that each curve maps 0 R to p and is defined on an interval of theform (, ). So, precisely, we have

p = { : (, )M | (0) = p, is C}.Note that depends on in this definition. Now, define a map T : Fp p R by

T (f , ) =d

dt

(f )

t=0.

This map is well-defined, for if f1, f2 f , then f1 = f2 on some neighborhood, N , ofp, implying that f1 = f2 on N (, ). T is also linear in the first variable, sinceif f , g Fp and , R, then

T (f + g, ) = T ( f + g, )

=d

dt

((f + g) )

t=0

=d

dt

(f + g )

t=0

= d

dt

(f )

t=0+

d

dt

(g )

t=0

= T (f , ) + T (g, ).

Next, define a set Hp by

Hp = {f Fp : T (f , ) = 0 p}.Since T is linear in the first variable, it follows that Hp is a linear subspace of Fp.In fact, we can completely characterize those elements of Fp that are in Hp.

Lemma 2.6 Let f be in Fp. Then f Hp if and only if for any coordinate chart,(,U), with p U , we have

xi(f 1)

(p)= 0

40

for all i = 1, 2, ...,m. In short, f Hp if and only if all coordinate partial derivativesof f vanish at p.

Proof Let (,U) be any coordinate system around p, and let be any curve in p.We can write the coordinate representation of as

( )(t) = (x1(t), ..., xm(t)), t (, ).Then for any f f , we have

T (f , ) =d

dt

(f )

t=0

=d

dt

(f 1 )

t=0

=d

dt

(f 1)(x1(t), ..., xm(t))

t=0

=mi=1

xi(f 1)

(p)

dxi

dt

t=0. (2)

So, if all the partial derivatives of f 1 vanish at p, we must have T (f , ) = 0.Conversely, suppose f Hp, so that T (f , ) = 0 for all p. For each i =1, 2, ...,m, define a particular curve in p as follows. Let i : R Rm be given byi(t) = (((p))

1, ..., ((p))i1, ((p))i + t, ((p))i+1, ..., ((p))m), where ((p))k is thekth component of (p) Rm. Then define i : R M by i = 1 i. Eachcurve, i, is smooth and satisfies i(0) = p. Then the coordinate representation, i = (x1(t), ..., xm(t)), satisfies

dxk

dt

t=0

=

{1 k = i0 k 6= i .

Thus, applying this to equation (2), we see that

T (f , i) = 0 xi(f 1)

(p)= 0.

Applying this to each i = 1, 2, ...,m shows that xi

(f 1) = 0 at (p) for all i.Moreover, the coordinate system (,U) was arbitrary. So, this proves the result.

QED

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We are now ready to define the cotangent space. Recall that, for a vector space,V , and a subspace W V , the quotient space V/W is defined as follows. We definean equivalence relation on V by x y if and only if xy W . Then V/W is the setof all equivalence classes under this relation. It is a linear space with the operations[x] + [y] = [x+ y] and [x] = [x]. The zero element of V/W is the entire subspaceW . Using this concept, we have the following.

Definition 2.2 The quotient space Fp/Hp is called the cotangent space and isdenoted by T pM . The Hp equivalence class of f is denoted dfp, and is called acotangent vector at p.

So, we see right away that T pM is a linear space, as it is the quotient space definedby a subspace ofFp. Moreover, by the general structure of a quotient space, we know

that for any f , g Fp and R, we have dfp+dgp = d(f + g)p and dfp = d(f)p.As a remark on notation, we will usually denote a cotangent vector simply by

df , if it is clear from the context to which specific point we are referring. Since allof our discussions in this chapter are focused on the single point p M , we will usethis notation.

The definition of the cotangent space may not be as immediately transparentas that of the tangent space. A moment of thought, though, will convince us thattwo elements f , g Fp are in the same Hp equivalence class if and only if each ofthe coordinate partial derivatives of f , in terms of any coordinate system aroundp, equals the corresponding coordinate partial derivative of g. This characterizationdoes not, of course, depend on which functions f f and g g we might choose tocompute such partial derivatives. Hence, a cotangent vector, df , can be thought ofas a set of smooth functions defined in a neighborhood of p that all have identicalcoordinate partial derivatives at p.

Our next goal is to show that the dimension of T pM is m. To do this, we willproceed as in the case of the tangent space, choosing a local coordinate system andconstructing a basis. Our construction thus far has not made use of any coordinatesystem, so the existence and linear structure of T pM is well-defined and independentof any coordinate system around p.

To illustrate, first, the natural choice for a basis, consider the following simpleexample. Let M = R2, and let p = (0, 0). Suppose two functions, f and g, definedand smooth on respective neighborhoods of p, have the same coordinate partialderivatives at the origin. Then we would equate these functions as a cotangentvector. In essence, their gradients, f and g, are equal at p. Any other functionwhose gradient equalled these would also be identified with f and g. Hence, we canthink of the cotangent vector df as simply defining a vector in R2 emanating from the

42

origin, namely the vector defined by the gradient of any function identified with f .Conversely, given a vector in R2, is straightforward to construct a smooth functionhaving this vector as its gradient. Hence, we can actually identify the cotangentspace with R2. This space is spanned by the standard basis vectors e1 = (1, 0) ande2 = (0, 1). So, to think of these vectors as a basis for the cotangent space at p, weneed to consider what functions have these vectors as gradients. The simple choice,of course, is the set of coordinate functions f1(x, y) = x and f2(x, y) = y. That is,the coordinate functions are the natural choice for a basis for the cotangent space.This is the idea we will pursue.

Let (,U) be a coordinate system around p. Then we have local coordinates onU defined by (q) = (x1(q), ..., xm(q)) for q U . The functions xk : U R arethe coordinate functions on U . As before, we will also use the fact that each xi canbe obtained by composing the ith component projection on Rm, pii, with . Thatis, xi = pii . Each coordinate function xi is also smooth, so it is an element ofC(p). We want to show that {dxi : 1 i m} is a basis for T pM induced by thecoordinate system (,U).

We will begin by proving the following technical lemma, which will be useful inour basis construction.

Lemma 2.7 Suppose f1, ..., fk are in C(p), and let F (y1, ..., yk) be a C real-valued

function defined in a neighborhood of (f1(p), ..., fk(p)). Then f = F (f1, ..., fk) is inC(p) and

df =ki=1

F

fk

(f1(p), ..., fk(p)

)dfi.

Proof If Uj is the neighborhood of of p on which fj is smooth, then all the functionsfj are smooth on N = kj=1Uj. It follows that f = F (f1, ..., fk) is smooth on N . So,f C(p), and we can consider f Fp and df T pM . Let be any curve in p.Then, using the linearity of the map T in its first variable, we have

43

T (f , ) =d

dt

(f )

t=0

=d

dt

(F(f1 , ..., fk

))t=0

=ki=1

F

fi

(f1(p), ..., fk(p)

) ddt

(fi

)t=0

=ki=1

F

fi

(f1(p), ..., fk(p)

)T (fi, )

= T

(ki=1

F

fi

(f1(p), ..., fk(p)

)fi,

).

This implies that

T

(f

ki=1

F

fi

(f1(p), ..., fk(p)

)fi,

)= 0.

Since was arbitrary, this implies that

f ki=1

F

fi

(f1(p), ..., fk(p)

)fi Hp.

Applying the linear structure of the cotangent space, it follows that

df =ki=1

F

fi

(f1(p), ..., fk(p)

)dfi.

QED

Now, to show that the cotangent vectors dxi form a basis for T pM , we first needto know that they form a set of m distinct elements. That is, we want to knowthat if i 6= j, then dxi 6= dxj. This result is obvious in the Euclidean case, but itmay not be in an arbitrary coordinate system. So, suppose i and j are indices suchthat 1 i < j m. If xi = xj, then there is a neighborhood, N , of p on whichxi = xj. Thus, every point (y1, ..., ym) (N) has the same ith and jth coordinates,implying that there is an open ball, B((p)), such that every point in this ball,

44

including (p), has the same ith and jth coordinates. Consider, however, the pointy = (p) +

2ei, where ei is the i

th standard basis vector in Rm. Then this point isin the ball B((p)), but its i

th and jth coordinates are not the same, as we haveshifted one by a positive amount while leaving the other fixed. Hence we cannothave xi = xj when i 6= j. This shows that the elements {xi Fp : 1 i m} areall distinct.

Next, with indices i and j as before, suppose dxi = dxj. Then for any k = 1, ...,m,we have

xk(xi 1)

(p)=

xk(xj 1)

(p).

But xi = pii, so xi1 = pii, and likewise for xj. Thus, letting k = i, the previousequation implies that

xipii(p)

=

xipij(p)

,

from which it follows that 1 = 0, as i 6= j. The contradiction shows that, if i 6= j,then dxi 6= dxj. Hence, the set {dxi : 1 i m} does, indeed, contain m distinctelements. Thus, we can prove the following result.

Theorem 2.8 The set {dxi : 1 i m} forms a basis for T pM with respect to thecoordinate system (,U).

Proof The fact that this set spans T pM follows from Lemma 3.7. We let k = mand let the functions f1, ..., fm be the coordinate functions x

1, ..., xm. Then, for anyelement df T pM , we let F be f 1. Then the function g = F (x1, ..., xm) satisfiesg = f 1 = f . Hence, dg = df , and the lemma implies