introduction to differential · pdf filechapter 1 introduction 1.1 some history in the words...

Introduction to Differential Geometry

Lecture Notes for MAT367

Contents

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Some history . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 The concept of manifolds: Informal discussion . . . . . . . . . . . . . . . . . . 31.3 Manifolds in Euclidean space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Intrinsic descriptions of manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.5 Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.1 Atlases and charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Definition of manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3 Examples of Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.3.1 Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.3.2 Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.3.3 Real projective spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.3.4 Complex projective spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.3.5 Grassmannians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.3.6 Complex Grassmannians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.4 Oriented manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.5 Open subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.6 Compact subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.7 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.7.1 Countability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.7.2 Equivalence relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3 Smooth maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.1 Smooth functions on manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.2 Smooth maps between manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.2.1 Diffeomorphisms of manifolds . . . . . . . . . . . . . . . . . . . . . . . . . 433.3 Examples of smooth maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.3.1 Products, diagonal maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.3.2 The diffeomorphism RP1 ∼= S1 . . . . . . . . . . . . . . . . . . . . . . . . . 45

-3

-2 Contents

3.3.3 The diffeomorphism CP1 ∼= S2 . . . . . . . . . . . . . . . . . . . . . . . . . 463.3.4 Maps to and from projective space . . . . . . . . . . . . . . . . . . . . . . 473.3.5 The quotient map S2n+1→ CPn . . . . . . . . . . . . . . . . . . . . . . . . 48

3.4 Submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.5 Smooth maps of maximal rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

3.5.1 The rank of a smooth map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.5.2 Local diffeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.5.3 Level sets, submersions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583.5.4 Example: The Steiner surface . . . . . . . . . . . . . . . . . . . . . . . . . . 623.5.5 Immersions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

3.6 Appendix: Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

4 The tangent bundle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.1 Tangent spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.2 Tangent map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

4.2.1 Definition of the tangent map, basic properties . . . . . . . . . . . . 764.2.2 Coordinate description of the tangent map . . . . . . . . . . . . . . . 784.2.3 Tangent spaces of submanifolds . . . . . . . . . . . . . . . . . . . . . . . . 804.2.4 Example: Steiner’s surface revisited . . . . . . . . . . . . . . . . . . . . . 84

4.3 The tangent bundle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

5 Vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 875.1 Vector fields as derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 875.2 Vector fields as sections of the tangent bundle . . . . . . . . . . . . . . . . . . . 895.3 Lie brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 905.4 Related vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 945.5 Flows of vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 965.6 Geometric interpretation of the Lie bracket . . . . . . . . . . . . . . . . . . . . . 1045.7 Frobenius theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1075.8 Appendix: Derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

6 Differential forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1136.1 Review: Differential forms on Rm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1136.2 Dual spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1156.3 Cotangent spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1176.4 1-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1186.5 Pull-backs of function and 1-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1206.6 Integration of 1-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1226.7 2-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1236.8 k-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

6.8.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1246.8.2 Wedge product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1256.8.3 Exterior differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

6.9 Lie derivatives and contractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1286.9.1 Pull-backs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

Contents -1

6.10 Integration of differential forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1326.11 Integration over oriented submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . 1336.12 Stokes’ theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1336.13 Volume forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

A Topology of manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141A.1 Topological notions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141A.2 Manifolds are second countable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142A.3 Manifolds are paracompact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142A.4 Partitions of unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

B Vector bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147B.1 Tangent bundle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

B.1.1 Vector bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148B.1.2 Tangent bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150B.1.3 Some constructions with vector bundles . . . . . . . . . . . . . . . . . 151

B.2 Dual bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

Chapter 1Introduction

1.1 Some history

In the words of S.S. Chern, ”the fundamental objects of study in differential geome-try are manifolds.” 1 Roughly, an n-dimensional manifold is a mathematical objectthat “locally” looks like Rn. The theory of manifolds has a long and complicatedhistory. For centuries, manifolds have been studied as subsets of Euclidean space,given for example as level sets of equations. The term ‘manifold’ goes back to the1851 thesis of Bernhard Riemann, “Grundlagen fur eine allgemeine Theorie derFunctionen einer veranderlichen complexen Grosse” (“foundations for a generaltheory of functions of a complex variable”) and his 1854 habilitation address “Uberdie Hypothesen, welche der Geometrie zugrunde liegen” (“on the hypotheses un-derlying geometry”).

2 However, in neither reference Riemann makes an attempt to give a precise defi-nition of the concept. This was done subsequently by many authors, including Rie-

1 Page 332 of Chern, Chen, Lam: Lectures on Differential Geometry, World Scientific2http://en.wikipedia.org/wiki/Bernhard_Riemann

1

http://en.wikipedia.org/wiki/Bernhard_Riemann

2 1 Introduction

mann himself. 3 Henri Poincare in his 1895 work analysis situs, introduces the ideaof a manifold atlas. 4

The first rigorous axiomatic definition of manifolds was given by Veblen and White-head only in 1931.

We will see below that the concept of a manifold is really not all that compli-cated; and in hindsight it may come as a bit of a surprise that it took so long toevolve. Quite possibly, one reason is that for quite a while, the concept as suchwas mainly regarded as just a change of perspective (away from level sets in Eu-clidean spaces, towards the ‘intrinsic’ notion of manifolds). Albert Einstein’s theoryof General Relativity from 1916 gave a major boost to this new point of view; In histheory, space-time was regarded as a 4-dimensional ‘curved’ manifold with no dis-tinguished coordinates (not even a distinguished separation into ‘space’ and ‘time’);a local observer may want to introduce local xyzt coordinates to perform measure-ments, but all physically meaningful quantities must admit formulations that arecoordinate-free. At the same time, it would seem unnatural to try to embed the 4-dimensional curved space-time continuum into some higher-dimensional flat space,in the absence of any physical significance for the additional dimensions. Someyears later, gauge theory once again emphasized coordinate-free formulations, andprovided physics motivations for more elaborate constructions such as fiber bundlesand connections.

Since the late 1940s and early 1950s, differential geometry and the theory ofmanifolds has developed with breathtaking speed. It has become part of the ba-sic education of any mathematician or theoretical physicist, and with applicationsin other areas of science such as engineering or economics. There are many sub-branches, for example complex geometry, Riemannian geometry, or symplectic ge-ometry, which further subdivide into sub-sub-branches.

3 See e.g. the article by Scholz http://www.maths.ed.ac.uk/ aar/papers/scholz.pdf for the long listof names involved.4http://en.wikipedia.org/wiki/Henri_Poincare

http://en.wikipedia.org/wiki/Henri_Poincare

1.2 The concept of manifolds: Informal discussion 3

1.2 The concept of manifolds: Informal discussion

To repeat, an n-dimensional manifold is something that “locally” looks like Rn. Theprototype of a manifold is the surface of planet earth:

It is (roughly) a 2-dimensional sphere, but we use local charts to depict it as subsetsof 2-dimensional Euclidean spaces. 5

To describe the entire planet, one uses an atlas with a collection of such charts, suchthat every point on the planet is depicted in at least one such chart.

This idea will be used to give an ‘intrinsic’ definition of manifolds, as essentiallya collection of charts glued together in a consistent way. One can then try to de-velop analysis on such manifolds – for example, develop a theory of integration anddifferentiation, consider ordinary and partial differential equations on manifolds, byworking in charts; the task is then to understand the ‘change of coordinates’ as oneleaves the domain of one chart and enters the domain of another.

5 Note that such a chart will always give a somewhat ‘distorted’ picture of the planet; the distanceson the sphere are never quite correct, and either the areas or the angles (or both) are wrong. Forexample, in the standard maps of the world, Canada always appears somewhat bigger than it reallyis. (Even more so Greenland, of course.)

4 1 Introduction

1.3 Manifolds in Euclidean space

In multivariable calculus, you will have encountered manifolds as solution sets ofequations. For example, the solution set of an equation of the form f (x,y,z) = ain R3 defines a ‘smooth’ hypersurface S ⊆ R3 provided the gradient of f is non-vanishing at all points of S. We call such a value of f a regular value, and henceS = f−1(a) a regular level set. Similarly, the joint solution set C of two equations

f (x,y,z) = a, g(x,y,z) = b

defines a smooth curve in R3, provided (a,b) is a regular value of ( f ,g) in the sensethat the gradients of f and g are linearly independent at all points of C. A familiarexample of a manifold is the 2-dimensional sphere S2, conveniently described as alevel surface inside R3:

S2 = (x,y,z) ∈ R3| x2 + y2 + z2 = 1.

There are many ways of introducing local coordinates on the 2-sphere: For exam-ple, one can use spherical polar coordinates, cylindrical coordinates, stereographicprojection, or orthogonal projections onto the coordinate planes. We will discusssome of these coordinates below. More generally, one has the n-dimensional sphereSn inside Rn+1,

Sn = (x0, . . . ,xn) ∈ Rn+1| (x0)2 + . . .+(xn)2 = 1.

The 0-sphere S0 consists of two points, the 1-sphere S1 is the unit circle. Anotherexample is the 2-torus, T 2. It is often depicted as a surface of revolution: Given realnumbers r,R with 0 < r < R, take a circle of radius r in the x− z plane, with centerat (R,0), and rotate about the z-axis.

The resulting surface6 is given by an equation,

T 2 = (x,y,z)|(√

x2 + y2−R)2

+ z2 = r2. (1.1)

Not all surfaces can be realized as ‘embedded’ in R3; for non-orientable surfacesone needs to allow for self-intersections. This type of realization is referred to as an

6http://calculus.seas.upenn.edu/?n=Main.CentroidsAndCentersOfMass.

http://calculus.seas.upenn.edu/?n=Main.CentroidsAndCentersOfMass

1.4 Intrinsic descriptions of manifolds 5

immersion: We don’t allow edges or corners, but we do allow that different parts ofthe surface pass through each other. An example is the Klein bottle7

The Klein bottle is an example of a non-orientable surface: It has only one side. (Infact, the Klein bottle contains a Mobius band – see exercises.) It is not possible torepresent it as a regular level set f−1(0) of a function f : For any such surface onehas one side where f is positive, and another side where f is negative.

1.4 Intrinsic descriptions of manifolds

In this course, we will mostly avoid concrete embeddings of manifolds into any RN .Here, the term ‘embedding’ is used in an intuitive sense, for example as the real-ization as the level set of some equations. (Later, we will give a precise definition.)There are a number of reasons for why we prefer developing an ‘intrinsic’ theory ofmanifolds.

1. Embeddings of simple manifolds in Euclidean space can look quite complicated.The following one-dimensional manifold8

is intrinsically, ‘as a manifold’, just a closed curve, that is, a circle. The problemof distinguishing embeddings of a circle into R3 is one of the goals of knot theory,a deep and difficult area of mathematics.

2. Such complications disappear if one goes to higher dimensions. For example, theabove knot (and indeed any knot in R3) can be disentangled inside R4 (with R3

viewed as a subspace). Thus, in R4 they become unknots.3. The intrinsic description is sometimes much simpler to deal with than the extrin-

sic one. For instance, the equation describing the torus T 2 ⊆R3 is not especially

7http://www.map.mpim-bonn.mpg.de/2-manifolds

8http://math201s09.wdfiles.com/local--files/medina-knot/alternating.jpg

http://www.map.mpim-bonn.mpg.de/2-manifolds

http://math201s09.wdfiles.com/local--files/medina-knot/alternating.jpg

6 1 Introduction

simple or beautiful. But once we introduce the following parametrization of thetorus

x = (R+ r cosϕ)cosθ , y = (R+ r cosϕ)sinθ , z = r sinϕ,

where θ ,ϕ are determined up to multiples of 2π , we recognize that T 2 is simplya product:

T 2 = S1×S1. (1.2)

That is, T 2 consists of ordered pairs of points on the circle, with the two factorscorresponding to θ ,ϕ . In contrast to (1.1), there is no distinction between ‘small’circle (of radius r) and ‘large circle’ (of radius R). The new description suggestsan embedding of T 2 into R4 which is ‘nicer’ then the one in R3. (But does ithelp?)

4. Often, there is no natural choice of an embedding of a given manifold inside RN ,at least not in terms of concrete equations. For instance, while the triple torus 9

is easily pictured in 3-space R3, it is hard to describe it concretely as the level setof an equation.

5. While many examples of manifolds arise naturally as level sets of equations insome Euclidean space, there are also many examples for which the initial con-struction is different. For example, the set M whose elements are all affine linesin R2 (that is, straight lines that need not go through the origin) is naturally a2-dimensional manifold. But some thought is required to realize it as a surface inR3.

1.5 Surfaces

Let us briefly give a very informal discussion of surfaces. A surface is the samething as a 2-dimensional manifold. We have already encountered some examples:The sphere, the torus, the double torus, triple torus, and so on:

9http://commons.wikimedia.org/wiki/File:Triple_torus_illustration.png

http://commons.wikimedia.org/wiki/File:Triple_torus_illustration.png

1.5 Surfaces 7

All of these are ‘orientable’ surfaces, which essentially means that they have twosides which you might paint in two different colors. It turns out that these are allorientable surfaces, if we consider the surfaces ‘intrinsically’ and only consider sur-faces that are compact in the sense that they don’t go off to infinity and do nothave a boundary (thus excluding a cylinder, for example). For instance, each of thefollowing drawings depicts a double torus:

We also have one example of a non-orientable surface: The Klein bottle. More ex-amples are obtained by attaching handles (just like we can think of the torus, doubletorus and so on as a sphere with handles attached).

Are these all the non-orientable surfaces? In fact, the answer is no. We have missedwhat is in some sense the simplest non-orientable surface. Ironically, it is the surfacewhich is hardest to visualize in 3-space. This surface is called the projective planeor projective space, and is denoted RP2. One can define RP2 as the set of all lines(i.e., 1-dimensional subspaces) in R3. It should be clear that this is a 2-dimensionalmanifold, since it takes 2 parameters to specify such a line. We can label such linesby their points of intersection with S2, hence we can also think of RP2 as the setof antipodal (i.e., opposite) points on S2. In other words, it is obtained from S2

by identifying antipodal points. To get a better idea of how RP2 looks like, let ussubdivide the sphere S2 into two parts:

(i) points having distance ≤ ε from the equator,(ii) points having distance ≥ ε from the equator.

8 1 Introduction

If we perform the antipodal identification for (i), we obtain a Mobius strip. If weperform antipodal identification for (ii), we obtain a 2-dimensional disk (think of itas the points of (ii) lying in the upper hemisphere). Hence, RP2 can also be regardedas gluing the boundary of a Mobius strip to the boundary of a disk:

Now, the question arises: Is it possible to realize RP2 smoothly as a surface insideR3, possibly with self-intersections (similar to the Klein bottle)? Simple attempts ofjoining the boundary circle of the Mobius strip with the boundary of the disk willalways create sharp edges or corners – try it. Around 1900, David Hilbert posedthis problem to his student Werner Boy, who discovered that the answer is yes. Thefollowing picture of Boy’s surface was created by Paul Nylander. 10

There are some nice videos illustrating the construction of the surface: See in par-ticular

https://www.youtube.com/watch?v=9gRx66xKXek

and

www.indiana.edu/˜minimal/archive/NonOrientable/NonOrientable/Bryant-anim/web/

While these pictures are very beautiful, it certainly makes the projective space ap-pear more complicated than it actually is. If one is only interested in RP2 itself,

10http://mathforum.org/mathimages/index.php/Boy’s_Surface

https://www.youtube.com/watch?v=9gRx66xKXek

www.indiana.edu/~minimal/archive/NonOrientable/NonOrientable/Bryant-anim/web/

www.indiana.edu/~minimal/archive/NonOrientable/NonOrientable/Bryant-anim/web/

http://mathforum.org/mathimages/index.php/Boy's_Surface

1.5 Surfaces 9

rather than its realization as a surface in R3, it is much simpler to work with thedefinition (as a sphere with antipodal identification).

Going back to the classification of surfaces: It turns out that all closed, connectedsurfaces are obtained from either the 2-sphere S2, the Klein bottle, or RP2, by at-taching handles.

Remark 1.1. Another operation for surfaces, generalizing the procedure of ‘attach-ing handles’, is the connected sum. Given two surfaces Σ1 and Σ2, remove smalldisks around given points p1 ∈ Σ1 and p2 ∈ Σ2, to create two surfaces with bound-ary circles. Then glue-in a cylinder connecting the two boundary circles, withoutcreating edges. The resulting surface is denoted

Σ1]Σ2.

For example, the connected sum Σ]T 2 is Σ with a handle attached. You may wantto think about the following questions: What is the connected sum of two RP2’s?And what is the connected sum of RP2 with a Klein bottle? Both must be in the listof 2-dimensional surfaces given above.

Chapter 2Manifolds

It is one of the goals of these lectures to develop the theory of manifolds in intrinsicterms, although we may occasionally use immersions or embeddings into Euclideanspace in order to illustrate concepts. In physics terminology, we will formulate thetheory of manifolds in terms that are ‘manifestly coordinate-free’.

2.1 Atlases and charts

As we mentioned above, the basic feature of manifolds is the existence of ‘localcoordinates’. The transition from one set of coordinates to another should be smooth.We recall the following notions from multivariable calculus.

Definition 2.1. Let U ⊆ Rm and V ⊆ Rn be open subsets. A map F : U → V iscalled smooth if it is infinitely differentiable. The set of smooth functions from U toV is denoted C∞(U,V ). The map F is called a diffeomorphism from U to V if it isinvertible, and the inverse map F−1 : V →U is again smooth.

Example 2.1. The exponential map exp : R→R, x 7→ exp(x) = ex is smooth. It maybe regarded as a map onto R>0 = y|y > 0, and is a diffeomorphism

exp : R→ R>0

with inverse exp−1 = log (the natural logarithm). Similarly,

tan : x ∈ R|−π/2 < x < π/2→ R

is a diffeomorphism, with inverse arctan.

Definition 2.2. For a smooth map F ∈C∞(U,V ) between open subsets U ⊆Rm andV ⊆ Rn, and any x ∈ U , one defines the Jacobian matrix DF(x) to be the n×m-matrix of partial derivatives

(DF(x))ij =

∂F i

∂x j

11

12 2 Manifolds

Its determinant is called the Jacobian matrix of F at x.

The inverse function theorem states that F is a diffeomorphism if and only if it isinvertible, and for all x ∈U , the Jacobian matrix DF(x) is invertible. (That is, onedoes not actually have to check smoothness of the inverse map!)

The following definition formalizes the concept of introducing local coordinates.

Definition 2.3 (Charts). Let M be a set.

1. An m-dimensional (coordinate) chart (U,ϕ) on M is a subset U ⊆ M togetherwith a map ϕ : U →Rm, such that ϕ(U)⊆Rm is open and ϕ is a bijection fromU to ϕ(U).

2. Two charts (U,ϕ) and (V,ψ) are called compatible if the subsets ϕ(U ∩V ) andψ(U ∩V ) are open, and the transition map

ψ ϕ−1 : ϕ(U ∩V )→ ψ(U ∩V )

is a diffeomorphism.

As a special case, charts with U ∩V = /0 are always compatible.

Question: Is compatibility of charts an equivalence relation? (See the appendix tothis section for a reminder of equivalence relations.)

Let (U,ϕ) be a coordinate chart. Given a point p ∈ U , and writing ϕ(p) =(u1, . . . ,um), we say that the ui are the coordinates of p in the given chart. (Letting pvary, these become real-valued functions p 7→ ui(p).) The transition maps ψ ϕ−1

amount to a change of coordinates. Here is a picture1 of a ‘coordinate change’:

Definition 2.4 (Atlas). Let M be a set. An m-dimensional atlas on M is a collectionof coordinate charts A = (Uα ,ϕα) such that

1. The Uα cover all of M, i.e.,⋃

α Uα = M.

1http://en.wikipedia.org/wiki/Differentiable_manifold

http://en.wikipedia.org/wiki/Differentiable_manifold

2.1 Atlases and charts 13

2. For all indices α,β , the charts (Uα ,ϕα) and (Uβ ,ϕβ ) are compatible.

Example 2.2 (An atlas on the 2-sphere). Let S2 ⊆ R3 be the unit sphere, consistingof all (x,y,z) ∈ R3 satisfying the equation x2 + y2 + z2 = 1. We shall define an atlaswith two charts (U+,ϕ+) and (U−,ϕ−). Let n = (0,0,1) be the north pole, let s =(0,0,−1) be the south pole, and put

U+ = S2−s, U− = S2−n.

Regard R2 as the coordinate subspace of R3 on which z = 0. Let

ϕ+ : U+→ R2, p 7→ ϕ+(p)

be stereographic projection from the south pole. That is, ϕ+(p) is the unique pointof intersection of R2 with the affine line passing through p and s. Similarly,

ϕ− : U−→ R2, p 7→ ϕ−(p)

is stereographic projection from the north pole, where ϕ−(p) is the unique point ofintersection of R2 with the affine line passing through p and n. A picture of ϕ−, withp′ = ϕ−(p) (the picture uses capital letters): 2

A calculation shows that for p = (x,y,z),

ϕ+(x,y,z) =( x

1+ z,

y1+ z

), ϕ−(x,y,z) =

( x1− z

,y

1− z

).

Exercise Verify these formulas.

Both ϕ± : U±→R2 are bijections onto R2. Let us verify this in detail for the mapϕ+. Given (u,v) we may solve the equation (u,v) = ϕ+(x,y,z), using the conditionthat x2 + y2 + z2 = 1 and z >−1. One has

u2 + v2 =x2 + y2

(1+ z)2 =1− z2

(1+ z)2 =(1− z)(1+ z)

(1+ z)2 =1− z1+ z

,

from which one obtains

z =1− (u2 + v2)

1+(u2 + v2),

2http://en.wikipedia.org/wiki/User:Mgnbar/Hemispherical_projection

http://en.wikipedia.org/wiki/User:Mgnbar/Hemispherical_projection

14 2 Manifolds

and since x = u(1+ z), y = v(1+ z) one obtains

ϕ−1+ (u,v) =

( 2u1+(u2 + v2)

,2v

1+(u2 + v2),

1− (u2 + v2)

1+(u2 + v2)

).

For the map ϕ−, we obtain by a similar calculation

ϕ−1− (u,v) =

( 2u1+(u2 + v2)

,2v

1+(u2 + v2),(u2 + v2)−11+(u2 + v2)

).

(Actually, it is also clear from the geometry that ϕ−1+ ,ϕ−1

− only differ by the sign ofthe z-coordinate.) Note that ϕ+(U+∩U−) =R2\(0,0). The transition map on theoverlap of the two charts is

(ϕ− ϕ−1+ )(u,v) =

( uu2 + v2 ,

vu2 + v2

)which is smooth on R2\(0,0) as required. ut

Here is another simple, but less familiar example where one has an atlas with twocharts.

Example 2.3 (Affine lines in R2). A line in a vector space E is the same as a 1-dimensional subspace. By an affine line, we mean a subset l ⊆ E, such that the setof differences v−w| v,w ∈ l is a 1-dimensional subspace. Put differently, l isobtained by adding a fixed vector v0 to all elements of a 1-dimensional subspace.In plain terms, an affine line is simply a straight line that does not necessarily passthrough the origin.

Let M be a set of affine lines in R2. Let U ⊆ M be the subset of lines that arenot vertical, and V ⊆M the lines that are not horizontal. Any l ∈U is given by anequation of the form

y = mx+b,

where m is the slope and b is the y-intercept. The map ϕ : U→R2 taking l to (m,b)is a bijection. On the other hand, lines in V are given by equations of the form

x = ny+ c,

and we also have the map ψ : V →R2 taking such l to (n,c). The intersection U ∩Vare lines l that are neither vertical nor horizontal. Hence, ϕ(U ∩V ) is the set of all(m,b) such that m 6= 0, and similarly ψ(U ∩V ) is the set of all (n,c) such that n 6= 0.To describe the transition map ψ ϕ−1 : ϕ(U ∩V )→ψ(U ∩V ), we need to express(n,c) in terms of (m,b). Solving y = mx+b for x we obtain

x =1m

y− bm.

Thus, n = 1m and c =− b

m , which shows that the transition map is

2.1 Atlases and charts 15

(ψ ϕ−1)(m,b) = (

1m, − b

m).

Note that this is smooth; similarly, ϕ ψ−1 is smooth; hence U,V define an 2-dimensional atlas on M.

Question: What is the resulting surface?

As a first approximation, we may take an m-dimensional manifold to be a setwith an m-dimensional atlas. This is almost the right definition, but we will makea few adjustments. A first criticism is that we may not want any particular atlasas part of the definition: For example, the 2-sphere with the atlas given by stereo-graphic projections onto the x− y-plane, and the 2-sphere with the atlas given bystereographic projections onto the y−z-plane, should be one and the same manifoldS2. To resolve this problem, we will use the following notion.

Definition 2.5. Suppose A = (Uα ,ϕα) is an m-dimensional atlas on M, and let(U,ϕ) be another chart. Then (U,ϕ) is said to be compatible with A if it is com-patible with all charts (Uα ,ϕα) of A .

Example 2.4. On the 2-sphere S2, we had constructed the atlas

A = (U+,ϕ+), (U−,ϕ−)

given by stereographic projection. Consider on the chart (V,ψ), with domain V theset of all (x,y,z) ∈ S2 such that y < 0, with ψ(x,y,z) = (x,z). To check that it iscompatible (U+,ϕ+), note that U+∩V =V , and

ϕ+(U+∩V ) = (u,v)| v < 0, ψ(U+∩V ) = (x,z)| x2 + z2 < 1

Expressing the coordinates (u,v) = ψ(x,y,z) in terms of x,z and vice versa, we find

(x,z) = (ψ ϕ−1+ )(u,v) =

( 2u1+(u2 + v2)

,1− (u2 + v2)

1+(u2 + v2)

)(u,v) = (ϕ+ ψ

−1)(x,z) =( x

1+ z, −√

1− (x2 + z2)

1+ z

)Both maps are smooth, proving that (V,ψ) is compatible with (U+,ϕ+).

Note that (U,ϕ) is compatible with the atlas A = (Uα ,ϕα) if and only if theunion A ∪(U,ϕ) is again an atlas on M. This suggests defining a bigger atlas, byusing all charts that are compatible with the given atlas. In order for this to work,we need that the new charts are also compatible not only with the charts of A , butalso with each other.

Lemma 2.1. Let A = (Uα ,ϕα) be a given atlas on the set M. If two charts(U,ϕ), (V,ψ) are compatible with A , then they are also compatible with eachother.

16 2 Manifolds

Proof. For every chart Uα , the sets ϕα(U ∩Uα) and ϕα(V ∩Uα) are open, hencetheir intersection is open. Since the map ϕα is injective, this intersection is

ϕα(U ∩Uα)∩ϕα(V ∩Uα) = ϕα(U ∩V ∩Uα),

see the exercise below. Since ϕ ϕ−1α : ϕα(U ∩Uα)→ ϕ(U ∩Uα) is a diffeomor-

phism, it follows that

ϕ(U ∩V ∩Uα) = (ϕ ϕ−1α )(ϕα(U ∩V ∩Uα)

)is open. Taking the union over all α , we see that

ϕ(U ∩V ) =⋃α

ϕ(U ∩V ∩Uα)

is open. A similar argument applies to ψ(U ∩V ). The transition map ψ ϕ−1 :ϕ(U ∩V )→ψ(U ∩V ) is smooth since for all α , its restriction to ϕ(U ∩V ∩Uα) is acomposition of two smooth maps ϕα ϕ−1 : ϕ(U∩V ∩Uα)−→ϕα(U∩V ∩Uα) andψ ϕ−1

α : ϕα(U ∩V ∩Uα)−→ψ(U ∩V ∩Uα). Likewise, the composition ϕ ψ−1 :ψ(U ∩V )→ ϕ(U ∩V ) is smooth. ut

Exercise: Show that if f : X → Y is an injective map between sets, and A,B ⊆ Xare two subsets, then

f (A∩B) = f (A)∩ f (B). (2.1)

Show that (2.1) is not true, in general, if f is not injective.

Theorem 2.1. Given an atlas A = (Uα ,ϕα) on M, let A be the collection of allcharts (U,ϕ) that are compatible with A . Then A is itself an atlas on M, containingA . In fact, A is the largest atlas containing A .

Proof. Note first that A contains A , since the set of charts compatible with A

contains the charts from the atlas A itself. In particular, the charts in A cover M.By the Lemma, any two charts in A are compatible. Hence A is an atlas. If (U,ϕ)

is a chart compatible with all charts in A , then in particular it is compatible with allcharts in A ; hence (U,ϕ) ∈ A by the definition of A . This shows that A cannotbe extended to a larger atlas.

Definition 2.6. An atlas A is called maximal if it is not properly contained in anylarger atlas. Given an arbitrary atlas A , one calls A (as in Theorem 2.1) the maximalatlas determined by A .

Remark 2.1. Although we will not need it, let us briefly discuss the notion of equiv-alence of atlases. (For background on equivalence relations, see the appendix tothis chapter, Section 2.7.2.) Two atlases A = (Uα ,ϕα) and A ′ = (U ′α ,ϕ ′α) arecalled equivalent if every chart of A is compatible with every chart in A ′. For ex-ample, the atlas on the 2-sphere given by the two stereographic projections to the

2.2 Definition of manifold 17

x− y-plane is equivalent to the atlas A ′ given by the two stereographic projectionsto the y− z-plane. Using Lemma 2.1, one sees that equivalence of atlases is indeedan equivalence relation. (In fact, two atlases are equivalent if and only if their unionis an atlas.) Furthermore, two atlases are equivalent if and only if they are containedin the same maximal atlas. That is, any maximal atlas determines an equivalenceclass of atlases, and vice versa.

2.2 Definition of manifold

As a next approximation towards definition of manifolds, we can take an m-dimensional manifold to be a set M together with an m-dimensional maximal atlas.This is already quite close to what we want, but for technical reasons we would liketo impose two further conditions.

First of all, we insist that M can be covered by countably many coordinate charts.In most of our examples, M is in fact covered by finitely many coordinate charts.This countability condition is used for various arguments involving a proof by in-duction.

Example 2.5. A simple example that is not countable: Let M = R, with A =(Uα ,ϕα) the 0-dimensional maximal (!) atlas, where each Uα consists of a singlepoint, and ϕα : Uα → 0 is the unique map to R0 = 0. Compatibility of chartsis obvious. But M cannot be covered by countably many of these charts. Thus, wewill not allow to consider R as a zero-dimensional manifold.

Secondly, we would like to avoid the following type of example.

Example 2.6. Let X be a disjoint union of two copies of the real line R. We denotethe two copies by R×1 and R×−1, just so that we can tell them apart. Definean equivalence relation on X generated by

(x,1)∼ (x′,−1) ⇔ x′ = x < 0,

and let M = X/ ∼ the set of equivalence classes. That is, we ‘glue’ the two reallines along their negative real axes (taking care that no glue gets on the origins ofthe axes). Here is a (not very successful) attempt to sketch the resulting space:

18 2 Manifolds

As a set, M is a disjoint union of R<0 with two copies of R≥0. Let π : X →M bethe quotient map, and let

U = π(R×1), V = π(R×−1)

the images of the two real lines. The projection map X→R, (x,±1) 7→ x is constanton equivalence classes, hence it descends to a map f : M→ R; let ϕ : U → R bethe restriction of f to U and ψ : V → R the restriction to V . Then

ϕ(U) = ψ(V ) = R, ϕ(U ∩V ) = ψ(U ∩V ) = R<0,

and the transition map is the identity map. Hence, A = (U,ϕ), (V,ψ) is an atlasfor M. A strange feature of M with this atlas is that the points

p = ϕ−1(0), q = ψ

−1(0)

are ‘arbitrarily close’, in the sense that if I,J ⊆ R are any open subsets containing0, the intersection of their pre-images is non-empty:

ϕ−1(I)∩ψ

−1(J) 6= /0.

Yet, p 6= q! There is no really satisfactory way of drawing M (our picture above isinadequate), since it cannot be realized as a submanifold of any Rn.

Since such a behaviour is inconsistent with the idea of a manifold that ‘locally lookslike Rn’, we shall insist that for two distinct points p,q∈M, there are always disjointcoordinate charts containing the two points. This is called the Hausdorff condition,after Felix Hausdorff (1868-1942). 3

3http://en.wikipedia.org/wiki/Felix_Hausdorff

http://en.wikipedia.org/wiki/Felix_Hausdorff

2.2 Definition of manifold 19

Definition 2.7. An m-dimensional manifold is a set M, together with a maximalatlas A = (Uα ,ϕα) with the following properties:

1. (Countability condition) M is covered by countably many coordinate charts inA . That is, there are indices α1,α2, . . . with

M =⋃

i

Uαi .

2. (Hausdorff condition) For any two distinct points p,q ∈M there are coordinatecharts (Uα ,ϕα) and (Uβ ,ϕβ ) in A such that p ∈Uα , q ∈Uβ , with

Uα ∩Uβ = /0.

The charts (U,ϕ) ∈A are called (coordinate) charts on the manifold M.

Before giving examples, let us note the following useful fact concerning the Haus-dorff condition:

Lemma 2.2. Let M be a set with a maximal atlas A = (Uα ,ϕα), and supposep,q ∈ M are distinct points contained in a single coordinate chart (U,ϕ) ∈ A .Then we can find indices α,β such that p ∈Uα , q ∈Uβ , with Uα ∩Uβ = /0.

Proof. We begin with the following remark, which we leave as an exercise. Suppose(U,ϕ) is a chart, with image U = ϕ(U) ⊆ Rm. Let V ⊆ U be a subset such thatV = ϕ(V )⊆ U is open, and let ψ = ϕ|V be the restriction of ϕ . Then (V,ψ) is againa chart, and is compatible with (U,ϕ). If (U,ϕ) is a chart from an atlast A , then(V,ψ) is compatible with that atlas.

Now let (U,ϕ) be as in the Lemma. Since

p = ϕ(p), q = ϕ(q)

are distinct points in U ⊆ Rm, we can choose disjoint open subsets Uα and Uβ ⊆ Ucontaining p = ϕ(p) and q = ϕ(q), respectively.4 Let Uα , Uβ ⊆U be their preim-ages, and take ϕα = ϕ|Uα

, ϕβ = ϕ|Uβ. Then (Uα ,ϕα) and (Uβ ,ϕβ ) are charts in A ,

4 For instance, take these subsets to be the elements in U of distance less than ||p− q||/2 from pand q, respectively.

20 2 Manifolds

with disjoint chart domains, and by construction we have that p ∈Uα and q ∈Uβ .ut

Example 2.7. Consider the 2-sphere S2 with the atlas given by the two coordinatecharts (U+,ϕ+) and (U−,ϕ−). This atlas extends uniquely to a maximal atlas. Thecountability condition is satisfied, since S2 is already covered by two charts. TheHausdorff condition is satisfied as well: Given distinct points p,q ∈ S2, if both arecontained in U+ or both in U−, we can apply the Lemma. The only case to beconsidered is thus if one point (say p) is the north pole and the other (say q) thesouth pole. But here we can construct Uα ,Uβ by replacing U+ and U− with the openupper hemisphere and open lower hemisphere, respectively. Alternatively, we canuse the chart given by stereographic projection to the x− z plane, noting that this isalso in the maximal atlas.

Remark 2.2. As we explained above, the Hausdorff condition rules out some strangeexamples that don’t quite fit our idea of a space that is locally like Rn. Nevertheless,the so-called non-Hausdorff manifolds (with non-Hausdorff more properly callednot necessarily Hausdorff ) do arise in some important applications. Much of thetheory can be developed without the Hausdorff property, but there are some com-plications: For instance, the initial value problems for vector fields need not haveunique solutions, in general.

Remark 2.3 (Charts taking values in ‘abstract’ vector spaces). In the definition ofan m-dimensional manifold M , rather than letting the charts (Uα ,ϕα) take values inRm we could just as well let them take values in m-dimensional real vector spacesEα :

ϕα : Uα → Eα .

Transition functions are defined as before, except they now take an open subset ofEβ to an open subset of Eα . The choice of basis identifies Eα = Rm, and takes usback to the original definition.

As far as the definition of manifolds is concerned, nothing has been gained byadding this level of abstraction. However, it often happens that the Eα ’s are given tous ‘naturally’. For example, if M is a surface inside R3, one would typically use x−ycoordinates, or x− z coordinates, or y− z coordinates on appropriate chart domains.It can then be useful to regard the x− y plane, x− z plane, and y− z plane as thetarget space of the coordinate maps, and for notational reasons it may be convenientto not associate them with a single R2.

2.3 Examples of Manifolds

We will now discuss some basic examples of manifolds. In each case, the manifoldstructure is given by a finite atlas; hence the countability property is immediate. Wewill not spend too much time on verifying the Hausdorff property; while it may bedone ‘by hand’, we will later have some better ways of doing this.

2.3 Examples of Manifolds 21

2.3.1 Spheres

The construction of an atlas for the 2-sphere S2, by stereographic projection, alsoworks for the n-sphere

Sn = (x0, . . . ,xn)| (x0)2 + . . .+(xn)2 = 1.

Let U± be the subsets obtained by removing (∓1,0, . . . ,0). Stereographic projectiondefines bijections ϕ± : U±→ Rn, where ϕ±(x0, x1, . . . , xn) = (u1, . . . ,un) with

ui =xi

1± x0 .

For the transition function one finds (writing u = (u1, . . . ,un))

(ϕ− ϕ−1+ )(u) =

u||u||2

.

We leave it as an exercise to check the details. An equivalent atlas, with 2n+ 2charts, is given by the subsets U+

0 , . . . ,U+n ,U−0 , . . . ,U−n where

U+j = x ∈ Sn| x j > 0, U−j = x ∈ Sn| x j < 0

for j = 0, . . . ,n, with ϕ±j : U±j → Rn the projection to the j-th coordinate plane (in

other words, omitting the j-th component x j):

ϕ±j (x

0, . . . ,xn) = (x0, . . . ,xi−1,xi+1, . . . ,xn).

2.3.2 Products

Given manifolds M,M′ of dimensions m,m′, with atlases (Uα ,ϕα) and (U ′β,ϕ ′

β),

the cartesian product M ×M′ is a manifold of dimension m + m′. An atlas isgiven by the product charts Uα ×U ′

βwith the product maps ϕα ×ϕ ′

β: (x,x′) 7→

(ϕα(x),ϕ ′β (x′)). For example, the 2-torus T 2 = S1×S1 becomes a manifold in this

way, and likewise for the n-torus

T n = S1×·· ·×S1.

22 2 Manifolds

2.3.3 Real projective spaces

The n-dimensional projective space RPn, also denoted RPn, is the set of all linesl⊆ Rn+1. It may also be regarded as a quotient space5

RPn = (Rn+1\0)/∼

for the equivalence relation

x∼ x′⇔∃λ ∈ R\0 : x′ = λx.

Indeed, any x∈Rn+1\0 determines a line, and two points x,x′ determine the sameline if and only if they agree up to a non-zero scalar multiple. The equivalence classof x = (x0, . . . ,xn) under this relation is commonly denoted

[x] = (x0 : . . . : xn).

RPn has a standard atlas

A = (U0,ϕ0), . . . ,(Un,ϕn)

defined as follows. For j = 0, . . . ,n, let

U j = (x0 : . . . : xn) ∈ RPn| x j 6= 0

be the set for which the j-th coordinate is non-zero, and put

ϕ j : U j→ Rn, (x0 : . . . : xn) 7→ (x0

x j , . . . ,x j−1

x j ,x j+1

x j , . . . ,xn

x j ).

This is well-defined, since the quotients do not change when all xi are multipliedby a fixed scalar. Put differently, given an element [x] ∈ RPn for which the j-thcomponent x j is non-zero, we first rescale the representative x to make the j-thcomponent equal to 1, and then use the remaining components as our coordinates.As a random example (with n = 2),

ϕ1(7 : 3 : 2) = ϕ1(7

3: 1 :

23)=(7

3,

23).

From this description, it is immediate that ϕ j is a bijection from U j onto Rn, withinverse map

ϕ−1j (u1, . . . ,un) = (u1 : . . . : u j : 1 : u j+1 : . . . : un).

Geometrically, viewing RPn as the set of lines in Rn+1, the subset U j ⊆ RPn

consists of those lines l which intersect the affine hyperplane

5 See the appendix to this chapter for some background on quotient spaces.


H j = x ∈ Rn+1| x j = 1,

and the map ϕ j takes such a line l to its unique point of intersection l∩H j, followedby the identification H j ∼= Rn (dropping the coordinate x j = 1).

Let us verify that A is indeed an atlas. Clearly, the domains U j cover RPn, sinceany element [x] ∈ RPn has at least one of its components non-zero. For i 6= j, theintersection Ui ∩U j consists of elements x with the property that both componentsxi, x j are non-zero. There are two cases:Case 1: 0≤ i < j ≤ n. We have

ϕi ϕ−1j (u1, . . . ,un) = (

u1

ui+1 , . . . ,ui

ui+1 ,ui+2

ui+1 , . . . ,u j

ui+1 ,1

ui+1 ,u j+1

ui+1 , . . . ,un

ui+1 ),

defined on ϕ j(Ui∩U j) = u ∈ Rn| ui+1 6= 0.Case 2: If 0≤ j < i≤ n. We have 6

ϕi ϕ−1j (u1, . . . ,un) = (

u1

ui , . . . ,u j

ui ,1ui ,

u j+1

ui , . . . ,ui−1

ui ,ui+1

ui , . . . ,un

ui ),

defined on ϕ j(Ui∩U j) = u ∈ Rn| ui 6= 0.In both cases, we see that ϕi ϕ

−1j is smooth. To complete the proof that this atlas

(or the unique maximal atlas containing it) defines a manifold structure, it remainsto check the Hausdorff property.

This can be done with the help of Lemma 2.2, but we postpone the proof sincewe will soon have a simple argument in terms of smooth functions. See Proposition3.1 below.

Remark 2.4. In low dimensions, we have that RP0 is just a point, while RP1 is acircle.

Remark 2.5. Geometrically, Ui consists of all lines in Rn+1 meeting the affine hyper-plane Hi, hence its complement consists of all lines that are parallel to Hi, i.e., thelines in the coordinate subspace defined by xi = 0. The set of such lines is RPn−1.In other words, the complement of Ui in RPn is identified with RPn−1.

Thus, as sets, RPn is a disjoint union

RPn = RntRPn−1,

where Rn is identified (by the coordinate map ϕi) with the open subset Un, andRPn−1 with its complement. Inductively, we obtain a decomposition

RPn = RntRn−1t·· ·tRtR0,

where R0 = 0. At this stage, it is simply a decomposition into subsets; later it willbe recognized as a decomposition into submanifolds.

6 If i = n, the last entry in the following expression is un−1/un.

24 2 Manifolds

Exercise: Find an identification of the space of rotations in R3 with the 3-dimensionalprojective space RP3. Hint: Associate to any v ∈R3 a rotation, as follows: If v = 0,take the trivial rotation, if v 6= 0, take the rotation by an angle ||v|| around the ori-ented axis determined by v. Note that for ||v|| = π , the vectors v and −v determinethe same rotation.

2.3.4 Complex projective spaces

Similar to the real projective space, one can define a complex projective space CPn

as the set of complex 1-dimensional subspaces of Cn+1. We identify C with R2, thusCn+1 with R2n+2. Thus

CPn = (Cn+1\0)/∼

where the equivalence relation is z∼ z′ if and only if there exists a complex λ withz′= λ z. (Note that the scalar λ is then unique, and is non-zero.) Alternatively, lettingS2n+1 ⊆ Cn+1 = R2n+2 be the ‘unit sphere’ consisting of complex vectors of length||z||= 1, we have

CPn = S2n+1/∼,

where z′ ∼ z if and only if there exists a complex number λ with z′ = λ z. (Note thatthe scalar λ is then unique, and has absolute value 1.) One defines charts (U j,ϕ j)similar to those for the real projective space:

U j = (z0 : . . . : zn)| z j 6= 0, ϕ j : U j→ Cn = R2n,

ϕ j(z0 : . . . : zn) =( z0

z j , . . . ,z j−1

z j ,z j+1

z j , . . . ,zn

z j

).

The transition maps between charts are given by similar formulas as for RPn (just re-place x with z); they are smooth maps between open subsets of Cn =R2n. Thus CPn

is a smooth manifold of dimension 2n. 7 Similar to RPn there is a decomposition

CPn = CntCn−1t·· ·tCtC0.

2.3.5 Grassmannians

The set Gr(k,n) of all k-dimensional subspaces of Rn is called the Grassmannian ofk-planes in Rn. (Named after Hermann Grassmann (1809-1877).) 8

7 The transition maps are not only smooth but even holomorphic, making CPn into an example ofa complex manifold (of complex dimension n).8http://en.wikipedia.org/wiki/Hermann_Grassmann

http://en.wikipedia.org/wiki/Hermann_Grassmann


As a special case, Gr(1,n) = RPn−1.We will show that for general k, the Grassmannian is a manifold of dimension

dim(Gr(k,n)) = k(n− k).

An atlas for Gr(k,n) may be constructed as follows. The idea is to present linearsubspaces of dimension k as graphs of linear maps from Rk to Rn−k. Here Rk isviewed as the coordinate subspace corresponding to a choice of k components fromx = (x1, . . . ,xn) ∈ Rn, and Rn−k the coordinate subspace for the remaining coordi-nates. To make it precise, we introduce some notation. For any subset I ⊆ 1, . . . ,nof the set of indices, let

I′ = 1, . . . ,n\I

be its complement. Let RI ⊆ Rn be the coordinate subspace

RI = x ∈ Rn| xi = 0 for all i ∈ I′.

If I has cardinality |I|= k, then RI ∈ Gr(k,n). Note that RI′ = (RI)⊥. Let

UI = E ∈ Gr(k,n)|E ∩RI′ = 0.

Each E ∈UI is described as the graph of a unique linear map AI : RI → RI′ ,that is,

E = y+AI(y)|y ∈ RI.

This gives a bijection

26 2 Manifolds

ϕI : UI → L(RI ,RI′), E 7→ ϕI(E) = AI ,

where L(F,F ′) denotes the space of linear maps from a vector space F to a vectorspace F ′. Note L(RI ,RI′) ∼= Rk(n−k), because the bases of RI and RI′ identify thespace of linear maps with (n−k)×k-matrices, which in turn is just Rk(n−k) by listingthe matrix entries. In terms of AI , the subspace E ∈UI is the range of the injectivelinear map (

1AI

): RI → RI⊕RI′ ∼= Rn (2.2)

where we write elements of Rn as column vectors.To check that the charts are compatible, suppose E ∈UI ∩UJ , and let AI and AJ

be the linear maps describing E in the two charts. We have to show that the map

ϕJ ϕ−1I : L(RI ,RI′)→ L(RJ ,RJ′), AI = ϕI(E) 7→ AJ = ϕJ(E)

is smooth. By assumption, E is described as the range of (2.2) and also as the rangeof a similar map for J. Here we are using the identifications RI ⊕RI′ ∼= Rn andRJ⊕RJ′ ∼= Rn. It is convenient to describe everything in terms of RJ⊕RJ′ . Let(

a bc d

): RI⊕RI′ → RJ⊕RJ′

be the matrix corresponding to the identification RI ⊕RI′ → Rn followed by theinverse of RJ ⊕RJ′ → Rn. For example, c is the inclusion RI → Rn as the corre-sponding coordinate subspace, followed by projection to the coordinate subspaceRJ′ . 9 We then get the condition that the injective linear maps(

a bc d

)(1AI

): RI → RJ⊕RJ′ ,

(1

AJ

): RJ → RJ⊕RJ′

have the same range. In other words, there is an isomorphism S : RI→RJ such that(a bc d

)(1AI

)=

(1

AJ

)S

as maps RI → RJ⊕RJ′ . We obtain(a+bAIc+dAI

)=

(S

AJS

)Using the first row of this equation to eliminate the second row of this equation, weobtain the formula

AJ = (c+dAI)(a+bAI)−1.

9 Put differently, the matrix is the permutation matrix ‘renumbering’ the coordinates of Rn.


The dependence of the right hand side on the matrix entries of AI is smooth, byCramer’s formula for the inverse matrix. It follows that the collection of all ϕI :UI → Rk(n−k) defines on Gr(k,n) the structure of a manifold of dimension k(n− k).The number of charts of this atlas equals the number of subsets I ⊆1, . . . ,n of car-dinality k, that is, it is equal to

(nk

). (The Hausdorff property may be checked similar

to that for RPn. Alternatively, given distinct E1,E2 ∈ Gr(k,n), choose a subspaceF ∈ Gr(k,n) such that F⊥ has zero intersection with both E1,E2. (Such a subspacealways exists.) One can then define a chart (U,ϕ), where U is the set of subspaces Etransverse to F⊥, and ϕ realizes any such map as the graph of a linear map F→ F⊥.Thus ϕ : U→ L(F,F⊥). As above, we can check that this is compatible with all thecharts (UI ,ϕI). Since both E1,E2 are in this chart U , we are done by Lemma 2.2.)

Remark 2.6. As already mentioned, Gr(1,n) = RPn−1. One can check that our sys-tem of charts in this case is the standard atlas for RPn−1.

Exercise: This is a preparation for the following remark. Recall that a linear mapΠ : Rn→ Rn is an orthogonal projection onto some subspace E ⊆ Rn if Π(x) = xfor x ∈ E and Π(x) = 0 for x ∈ E⊥. Show that a square matrix P ∈MatR(n) is thematrix of an orthogonal projection if and only if it has the properties

P> = P, PP = P,

where the superscript > indicates ‘transpose’. What is the matrix of the orthogonalprojection onto E⊥?

Remark 2.7. For any k-dimensional subspace E ⊆Rn, let Π E : Rn→Rn be the lin-ear map given by orthogonal projection onto E, and let PE ∈MatR(n) be its matrix.By the exercise,

P>E = PE , PEPE = PE ,

Conversely, any square matrix P with the properties P>=P, PP=P with rank(P) =k is the orthogonal projection onto a subspace Px| x∈Rn⊆Rn. This identifies theGrassmannian Gr(k,n) with the set of orthogonal projections of rank k. In summary,we have an inclusion

Gr(k,n) →MatR(n)∼= Rn2, E 7→ PE .

By construction, this inclusion take values in the subspace SymR(n)∼= Rn(n+1)/2 ofsymmetric n×n-matrices.

Remark 2.8. For all k, there is an identification Gr(k,n) ∼= Gr(n− k,n) (taking ak-dimensional subspace to the orthogonal subspace).

Remark 2.9. Similar to RP2 = S2/ ∼, the quotient modulo antipodal identification,one can also consider

M = (S2×S2)/∼

28 2 Manifolds

the quotient space by the equivalence relation

(x,x′)∼ (−x,−x′).

It turns out that this manifold M is the same as Gr(2,4), where ‘the same’ is meantin the sense that there is a bijection of sets identifying the atlases.

2.3.6 Complex Grassmannians

Similar to the case of projective spaces, one can also consider the complex Grass-mannian GrC(k,n) of complex k-dimensional subspaces of Cn. It is a manifold ofdimension 2k(n−k), which can also be regarded as a complex manifold of complexdimension k(n− k).

2.4 Oriented manifolds

The compatibility condition between charts (U,ϕ) and (V,ψ) on a set M is that themap ψ ϕ−1 : ϕ(U ∩V )→ ψ(U ∩V ) is a diffeomorphism. In particular, the Jaco-bian matrix D(ψ ϕ−1) of the transition map is invertible, and hence has non-zerodeterminant. If the determinant is > 0 everywhere, then we say (U,ϕ),(V,ψ) areoriented-compatible. An oriented atlas on M is an atlas such that any two of itscharts are oriented-compatible; a maximal oriented atlas is one that contains everychart that is oriented-compatible with all charts in this atlas. An oriented manifoldis a set with a maximal oriented atlas, satisfying the Hausdorff and countability con-ditions as in definition 2.7. A manifold is called orientable if it admits an orientedatlas.

The notion of an orientation on a manifold will become crucial later, since inte-gration of differential forms over manifolds is only defined if the manifold is ori-ented.

Example 2.8. The spheres Sn are orientable. To see this, consider the atlas withthe two charts (U+,ϕ+) and (U−,ϕ−), given by stereographic projections. (Sec-tion 2.3.1.) Here ϕ−(U+ ∩U−) = ϕ+(U+ ∩U−) = Rn\0, with transition mapϕ− ϕ

−1+ (u) = u/||u||2. The Jacobian matrix D(ϕ− ϕ

−1+ )(u) has entries(

D(ϕ− ϕ−1+ )(u)

)i j=

∂

∂u j

( ui

||u||2)=

1||u||2

δi j−2uiu j

||u||4. (2.3)

2.5 Open subsets 29

Its determinant is −||u||−2n (see exercise below).10 Hence, the given atlas is not anoriented atlas. But this is easily remedied: Simply compose one of the charts, sayU−, with the map (u1,u2, . . . ,un) 7→ (−u1,u2, . . . ,un); then with the resulting newcoordinate map ϕ− the atlas (U+,ϕ+),(U−, ϕ−) will be an oriented atlas.

Exercise: Calculate the determinant of the matrix with entries (2.3). Hint: Checkthat u is an eigenvector of the matrix, as is any vector orthogonal to u. Alternatively,use that the Jacobian determinant must be invariant under rotations in u-space.

Example 2.9. One can show that RPn is orientable if and only if n is odd or n = 0.More generally, Gr(k,n) is orientable if and only if n is even or n = 1. The complexprojective spaces CPn and complex Grassmannians GrC(k,n) are all orientable. Thisfollows because the transition maps for their standard charts, as maps between opensubsets of Cm, are actually complex-holomorphic, and this implies that as real maps,their Jacobian has positive determinant. See the following exercise.

Exercise: Let A∈MatC(n) be a complex n×n-matrix, and AR ∈MatR(2n) the samematrix regarded as a real-linear transformation of R2n ∼= Cn. Show that

detR(AR) = |detC(A)|2.

You may want to start with the case n = 1, and next consider the case that A is uppertriangular.

2.5 Open subsets

Let M be a set equipped with an m-dimensional maximal atlas A = (Uα ,ϕα).

Definition 2.8. A subset U ⊆M is open if and only if for all charts (Uα ,ϕα) ∈ Athe set ϕα(U ∩Uα) is open.

To check that a subset U is open, it is not actually necessary to verify this conditionfor all charts. As the following proposition shows, it is enough to check for any col-lection of charts whose union contains U . In particular, we may take A in definition2.8 to be any atlas, not necessarily a maximal atlas.

Proposition 2.1. Given U ⊆M, let B ⊆A be any collection of charts whose unioncontains U. Then U is open if and only if for all charts (Uβ ,ϕβ ) from B, the setsϕβ (U ∩Uβ ) are open.

10 Actually, to decide the sign of the determinant, one does not have to compute the determinanteverywhere. If n > 1, since Rn\0, it suffices to compute the determinant at just one point, e.g.u = (1,0, . . . ,0).

30 2 Manifolds

Proof. In what follows, we reserve the index β to indicate charts (Uβ ,ϕβ ) from B.Suppose ϕβ (U ∩Uβ ) is open for all such β . Let (Uα ,ϕα) be a given chart in themaximal atlas A . We have that

ϕα(U ∩Uα) =⋃β

ϕα(U ∩Uα ∩Uβ )

=⋃β

(ϕα ϕ−1β

)(ϕβ (U ∩Uα ∩Uβ )

)=⋃β

(ϕα ϕ−1β

)(ϕβ (Uα ∩Uβ )∩ϕβ (U ∩Uβ )

).

Since B ⊆A , all ϕβ (Uα ∩Uβ ) are open. Hence the intersection with ϕβ (U ∩Uβ )

is open, and so is the pre-image under the diffeomorphism ϕα ϕ−1β

. Finally, we usethat a union of open sets is again open. This proves the ‘if’ part; the ‘only if’ part isobvious. ut

If A is an atlas on M, and U ⊆M is open, then U inherits an atlas by restriction:

AU = (U ∩Uα ,ϕα |U∩Uα).

Exercise: Verify that if A is a maximal atlas, then so is AU , and if this maximalatlas A satisfies the countability and Hausdorff properties, then so does AU .

This then proves:

Proposition 2.2. An open subset of a manifold is again a manifold.

The collection of open sets of M with respect to an atlas has properties similar tothose for Rn:

Proposition 2.3. Let M be a set with an m-dimensional maximal atlas. The collec-tion of all open subsets of M has the following properties:

• /0,M are open.• The intersection U ∩U ′ of any two open sets U,U ′ is again open.• The union

⋃i Ui of an arbitrary collection Ui, i ∈ I of open sets is again open.

Proof. All of these properties follow from similar properties of open subsets in Rm.For instance, if U,U ′ are open, then

ϕα((U ∩U ′)∩Uα) = ϕα(U ∩Uα)∩ϕα(U ′∩Uα)

is an intersection of open subsets of Rm, hence it is open and therefore U ∩U ′ isopen. ut

These properties mean, by definition, that the collection of open subsets of M definea topology on M. This allows us to adopt various notions from topology:

2.6 Compact subsets 31

1. A subset A⊆M is called closed if its complement M\A is open.2. M is called connected if the only subsets A ⊆ M that are both closed and open

are A = /0 and A = M.3. If U is an open subset and p ∈U , then U is called an open neighborhood of p.

More generally, if A ⊆U is a subset contained in M, then U is called an openneighborhood of A.

The Hausdorff condition in the definition of manifolds can now be restated as thecondition that any two distinct points p,q in M have disjoint open neighborhoods.(It is not necessary to take them to be domains of coordinate charts.)

It is immediate from the definition that domains of coordinate charts are open.Indeed, this gives an alternative way of defining the open sets:

Exercise: Let M be a set with a maximal atlas. Show that a subset U ⊆M is open ifand only if it is either empty, or is a union U =

⋃i∈I Ui where the Ui are domains of

coordinate charts.

2.6 Compact subsets

Another important concept from topology that we will need is the notion of com-pactness. Recall (e.g. Munkres, Chapter 1 § 4) that a subset A⊆Rm is compact if ithas the following property: For every collection Uα of open subsets of Rm whoseunion contains A, the set A is already covered by finitely many subsets from thatcollection. One then proves the important result (see Munkres, Theorems 4.2 and4.9)

Theorem 2.2 (Heine-Borel). A subset A⊆ Rm is compact if and only if it is closedand bounded.

While ‘closed and bounded’ is a simpler characterization of compactness to workwith, it does not directly generalize to manifolds (or other topological spaces), whilethe original definition does:

Definition 2.9. Let M be a manifold.11 A subset A ⊆ M is compact if it has thefollowing property: For every collection Uα of open subsets of M whose unioncontains A, the set A is already covered by finitely many subsets from that collection.

In short, A⊆M is compact if every open cover admits a finite subcover.

Proposition 2.4. If A⊆M is contained in the domain of a coordinate chart (U,ϕ),then A is compact in M if and only if ϕ(A) is compact in Rn.

Proof. Suppose ϕ(A) is compact. Let Uα be an open cover of A. Taking intersec-tions with U , it is still an open cover (since A⊆U). Hence

11 More generally, the same definition is used for arbitrary topological spaces – e.g., sets with anatlas.

32 2 Manifolds

A⊆⋃α

(U ∩Uα),

and thereforeϕ(A)⊆

⋃α

ϕ(U ∩Uα).

Since ϕ(A) is compact, there are indices α1, . . . ,αN such that

ϕ(A)⊆ ϕ(U ∩Uα1)∪ . . .∪ϕ(U ∩UαN ).

But thenA⊆ (U ∩Uα1)∪ . . .∪ (U ∩UαN )⊆Uα1 ∪ . . .∪UαN .

The converse is proved similarly. ut

Exercise: Complete the proof, by working out the details for the other direction.

The proposition is useful, since we can check compactness of ϕ(A) by using theHeine-Borel criterion. For more general subsets of M, we can often decide compact-ness by combining this result with the following:

Proposition 2.5. If A1, . . . ,Ak ⊆ M is a finite collection of compact subsets, thentheir union A = A1∪ . . .∪Ak is again compact.

Proof. If Uα is an open cover of A, then in particular it is an open cover of eachof the sets A1, . . . ,Ak. For each Ai, we can choose a finite subcover. The collectionof all Uα ’s such that appear in at least one of these subcovers, for i = 1, . . .k are thena finite subcover for A.

Example 2.10. Let M = Sn. The closed upper hemisphere x ∈ Sn| x0 ≥ 0 is com-pact, because is contained in the coordinate chart (U+,ϕ+) for stereographic projec-tion, and its image under ϕ+ is the closed and bounded subset u ∈ Rn| ||u|| ≤ 1.Likewise the closed lower hemisphere is compact, and hence Sn itself (as the unionof upper and lower hemispheres) is compact.

Example 2.11. Let (Ui,ϕi)| i = 0, . . . ,n be the standard atlas for RPn. Let

Ai = (x0 : . . . : xn) ∈ RPn| ||x||2 ≤ (n+1)x2i .

Then Ai ⊆Ui (since necessarily xi 6= 0 for elements of Ai). Furthermore,⋃n

i=0 Ai =RPn: Indeed, given any (x0 : · · · : xn) ∈ RPn, let i be an index for which |xi| is maxi-mal. Then ||x||2 ≤ (n+1)x2

i (since the right hand side is obtained from the left handside by replacing each (x j)2 with (xi)2 ≥ (x j)2)), hence (x0 : · · · : xn) ∈ Ai. Finally,one checks that ϕi(Ai) ⊆ Rn is a closed ball of radius

√n+1, and in particular is

compact.

In a similar way, one can prove compactness of CPn, Gr(k,n), GrC(k,n). How-ever, soon we will have a simpler way of verifying compactness, by showing thatthey are closed and bounded subsets of RN for a suitable N.

2.7 Appendix 33

Proposition 2.6. Let M be a set with a maximal atlas. If A ⊆ M is compact, andC ⊆M is closed, then A∩C is compact.

Proof. Let Uα be an open cover of A∩C. Together with the open subset M\C,these cover A. Since A is compact, there are finitely many indices α1, . . . ,αN with

A⊆ (M\C)∪Uα1 ∪ . . .∪UαN .

Hence A∩C ⊆Uα1 ∪ . . .∪UαN . ut

The following fact uses the Hausdorff property (and holds in fact for any Haus-dorff topological space).

Proposition 2.7. If M is a manifold, then every compact subset A⊆M is closed.

Proof. Suppose A ⊆ M is compact. Let p ∈ M\A be given. For any q ∈ A, thereare disjoint open neighborhoods Vq of q and Uq of p. The collection of all Vq forq ∈ A are an open cover of A, hence there exists a finite subcover Vq1 , . . . ,Vqk . Theintersection U =Uq1 ∩ . . .∩Uqk is an open subset of M with p ∈M and not meetingVq1 ∪ . . .∪Vqk , hence not meeting A. We have thus shown that every p ∈ M\A hasan open neighborhood U ⊆M\A. The union over all such open neighborhoods forall p ∈M\A is all of M\A, which hence is open. It follows that A is closed. ut

Exercise: Let M be the non-Hausdorff manifold from Example 2.6. Find a compactsubset A⊆M that is not closed.

2.7 Appendix

2.7.1 Countability

A set X is countable if it is either finite (possibly empty), or there exists a bijectivemap f : N→ X . We list some basic facts about countable sets:

• N,Z,Q are countable, R is not countable.• If X1,X2 are countable, then the cartesian product X1×X2 is countable.• If X is countable, then any subset of X is countable.• If X is countable, and f : X → Y is surjective, then Y is countable.• If (Xi)i∈I are countable sets, indexed by a countable set I, then the (disjoint) unionti∈IXi is countable.

2.7.2 Equivalence relations

We will make extensive use of equivalence relations; hence it may be good to reviewthis briefly. A relation from a set X to a set Y is simply a subset

34 2 Manifolds

R⊆ Y ×X .

We write x ∼R y if and only if (y,x) ∈ R. When R is understood, we write x ∼ y. IfY = X we speak of a relation on X .

Example 2.12. Any map f : X → Y defines a relation, given by its graph Gr f =( f (x),x)|x ∈ X. In this sense relations are generalizations of maps; for example,they are often used to describe ‘multi-valued’ maps.

Remark 2.10. Given another relation S ⊆ Z×Y , one defines a composition S R ⊆Z×X , where

SR = (z,x)| ∃y ∈ Y : (z,y) ∈ S, (y,x) ∈ R.

Our conventions are set up in such a way that if f : X → Y and g : Y → Z are twomaps, then Grg f = Grg Gr f .

Example 2.13. On the set X = R we have relations ≥,>,<, ≤,=. But there is alsothe relation defined by the condition x∼ x′⇔ x′− x ∈ Z, and many others.

A relation ∼ on a set X is called an equivalence relation if it has the followingproperties,

1. Reflexivity: x∼ x for all x ∈ X ,2. Symmetry: x∼ y⇒ y∼ x,3. Transitivity: x∼ y, y∼ z⇒ x∼ z.

Given an equivalence relation, we define the equivalence class of x ∈ X to be thesubset

[x] = y ∈ X | x∼ y.

Note that X is a disjoint union of its equivalence classes. We denote by X/ ∼ theset of equivalence classes. That is, all the elements of a given equivalence class arelumped together and represent a single element of X/ ∼. One defines the quotientmap

q : X → X/∼, x 7→ [x].

By definition, the quotient map is surjective.

Remark 2.11. There are two other useful ways to think of equivalence relations:

• An equivalence relation R on X amounts to a decomposition X = ti∈IXi as adisjoint union of subsets. Given R, one takes Xi to be the equivalence classes;given the decomposition, one defines R = (y,x) ∈ X×X |∃i ∈ I : x,y ∈ Xi.

• An equivalence relation amounts to a surjective map q : X → Y . Indeed, givenR one takes Y := X/∼ with q the quotient map; conversely, given q one definesR = (y,x) ∈ X×X | q(x) = q(y).

Remark 2.12. Often, we will not write out the entire equivalence relation. For exam-ple, if we say “the equivalence relation on S2 given by x∼−x”, then it is understoodthat we also have x ∼ x, since reflexivity holds for any equivalence relation. Sim-ilarly, when we say “the equivalence relation on R generated by x ∼ x+ 1”, it is

2.7 Appendix 35

understood that we also have x∼ x+2 (by transitivity: x∼ x+1∼ x+2) as well asx∼ x−1 (by symmetry), hence x∼ x+k for all k ∈Z. (Any relation R0 ⊆ X×X ex-tends to a unique smallest equivalence relation R; one says that R is the equivalencerelation generated by R0.)

Example 2.14. Consider the equivalence relation on S2 given by

(x,y,z)∼ (−x,−y,−z).

The equivalence classes are pairs of antipodal points; they are in 1-1 correspondencewith lines in R3. That is, the quotient space S2/∼ is naturally identified with RP2.

Example 2.15. The quotient space R/∼ for the equivalence relation x∼ x+1 on Ris naturally identified with S1. If we think of S1 as a subset of R, the quotient mapis given by t 7→ (cos(2πt),sin(2πt)).

Example 2.16. Similarly, the quotient space for the equivalence relation on R2 givenby (x,y)∼ (x+ k,y+ l) for k, l ∈ Z is the 2-torus T 2.

Example 2.17. Let E be a k-dimensional real vector space. Given two ordered bases(e1, . . . ,ek) and (e′1, . . . ,e

′k), there is a unique invertible linear transformation A :

E → E with A(ei) = e′i. The two ordered bases are called equivalent if det(A) > 0.One checks that equivalence of bases is an equivalence relation. There are exactlytwo equivalence classes; the choice of an equivalence class is called an orientationon E. For example, Rn has a standard orientation defined by the standard basis(e1, . . . ,en). The opposite orientation is defined, for example, by (−e1,e2, . . . ,en).A permutation of the standard basis vectors defines the standard orientation if andonly if the permutation is even.

Chapter 3Smooth maps

3.1 Smooth functions on manifolds

A real-valued function on an open subset U ⊆ Rn is called smooth if it is infinitelydifferentiable. The notion of smooth functions on open subsets of Euclidean spacescarries over to manifolds: A function is smooth if its expression in local coordinatesis smooth.

Definition 3.1. A function f : M→ R on a manifold M is called smooth if for allcharts (U,ϕ) the function

f ϕ−1 : ϕ(U)→ R

is smooth. The set of smooth functions on M is denoted C∞(M).

Remarks 3.1. 1. Since transition maps are diffeomorphisms, it suffices to check thecondition for the charts from any given atlas A = (Uα ,ϕα), which need notbe the maximal atlas. Indeed, if the condition on f holds for all charts from theatlas A , and if (U,ϕ) is another chart compatible with A , then the functions

f ϕ−1∣∣

ϕ(U∩Uα )= ( f ϕ

−1α ) (ϕα ϕ

−1) : ϕ(U ∩Uα)→ R

are smooth, and since the open sets ϕ(U ∩Uα) cover ϕ(U) this implies smooth-ness of f ϕ−1.

2. Given an open subset U ⊆ M, we say that a function f is smooth on U if itsrestriction f |U is smooth. (Here we are using that U itself is a manifold.) Givenp ∈M, we say that f is smooth at p if it is smooth on some open neighborhoodof p.

Example 3.1. The ‘height function’

f : S2→ R, (x,y,z) 7→ z

is smooth. In fact, we see that for any smooth function h ∈C∞(R3) (for example thecoordinate functions), the restriction f = h|S2 is again smooth. This may be checked

37

38 3 Smooth maps

using the atlas with 6 charts given by projection to coordinate planes: E.g., in thechart U = (x,y,z)| z > 0 with ϕ(x,y,z) = (x,y), we have

( f ϕ−1)(x,y) = h

(x,y,

√1− (x2 + y2)

)which is smooth on ϕ(U) = (x,y)| x2+y2 < 1. (The argument for the other chartsin this atlas is similar.)

Of course, if h is not smooth, it might still happen that its restriction to S2 issmooth. On the other hand, the map f : S2→R, (x,y,z) 7→

√1− z2 is smooth only

on S2\(0,0,1),(0,0,−1). To analyse the situation near the north pole, use thecoordinate chart (U,ϕ) as above. In these coordinates, z =

√1− (x2 + y2), hence√

1− z2 =√

x2 + y2 which is not smooth near (x,y) = (0,0).

Example 3.2. Letπ : Rn+1\0→ RPn

be the quotient map. Given f : RPn→ R, the function

f = f π : Rn+1\0→ R

satisfies f (λx) = f (x) for λ 6= 0; conversely any f with this property descends toa function f on the projective space. We claim that f is smooth if and only if f issmooth. To see this, note that in the standard coordinate chart (Ui,ϕi) for RPn, thefunction f ϕ

−1i may be written as the smooth map

Ui→ Rn+1\0, (u1, . . . ,un) 7→ (u1, . . . ,ui,1,ui+1, . . . ,un)

followed by f . Hence, if f is smooth then so is f . (The converse is similar.) As aspecial case, we see that for all 0≤ j ≤ k ≤ n the functions

f : RPn→ R, (x0 : . . . : xn) 7→ x jxk

||x||2(3.1)

are well-defined and smooth. By a similar argument, the functions

f : CPn→ C, (z0 : . . . : zn) 7→ z jzk

||z||2(3.2)

(where the bar denotes complex conjugation) are well-defined and smooth, in thesense that both the real and imaginary part are smooth.

Lemma 3.1. Smooth functions f ∈ C∞(M) are continuous: For every open subsetJ ⊆ R, the pre-image f−1(J)⊆M is open.

Proof. We have to show that for every (U,ϕ), the set ϕ(U ∩ f−1(J))⊆Rm is open.But this subset coincides with the pre-image of J under the map f ϕ−1 : ϕ(U)→R, which is a smooth function on an open subset of Rm, and these are (by definition)continuous.

3.1 Smooth functions on manifolds 39

Exercise Show that f : M→ R continuous (i.e., the pre-image of any open subsetJ ⊆ R under f is open) if and only if for all charts (U,ϕ) the function f ϕ−1 iscontinuous.

From the properties of smooth functions on Rm, one immediately gets the fol-lowing properties of smooth functions on manifolds M:

• If f ,g ∈C∞(M) and λ ,µ ∈ R, then λ f +µg ∈C∞(M).• If f ,g ∈C∞(M), then f g ∈C∞(M).• 1 ∈C∞(M) (where 1 denotes the constant function p 7→ 1).

These properties say that C∞(M) is an algebra with unit 1. (See the appendix tothis chapter for some background information on algebras.) Below, we will developmany of the concepts of manifolds in terms of this algebra of smooth functions.

Suppose M is any set with a maximal atlas (Uα ,ϕα). The definition of C∞(M)does not use the Hausdorff or countability conditions; hence it makes sense in thismore general context. We may use functions to check the Hausdorff property:

Proposition 3.1. Suppose M is any set with a maximal atlas, and p 6= q are twopoints in M. Then the following are equivalent:

(i) There are open subsets U,V ⊆M with p ∈U, q ∈V, U ∩V = /0,(ii) There exists f ∈C∞(M) with f (p) 6= f (q).

Proof. “(i)⇒ (ii)”. Suppose (i) holds. As explained in Section 2.5, we may takeU,V to be the domains of coordinate charts (U,ϕ) and (V,ψ) around p,q. Chooseε > 0 such that the closed ε-ball

Bε(ϕ(p)) =

x ∈ Rm∣∣ ||x−ϕ(p)|| ≤ ε

is contained in ϕ(U); let A ⊆ U be its pre-image under ϕ . Let χ ∈ C∞(Rm) be a‘bump function’ centered at ϕ(p), with χ(ϕ(p)) = 1 and χ(x) = 0 for ||x−ϕ(p)|| ≥ε . (For the existence of such a function see Munkres, Lemma 16.1, or Lemma A.2in the appendix.)

The function f : M→ R such that f = χ ϕ on U and f = 0 on M\A is smooth,and satisfies f (p) = 1, f (q) = 0.

“(ii)⇐ (i)”. Suppose (ii) holds. Let δ = | f (q)− f (p)|/2, and put

40 3 Smooth maps

U = x ∈M| | f (x)− f (p)|< δ, (3.3)V = x ∈M| | f (x)− f (q)|< δ (3.4)

By the Lemma 3.1, U,V are open, and clearly p ∈U, q ∈V , U ∩V = /0.

A direct consequence of this result is:

Corollary 3.1 (Criterion for Haudorff condition). A set M with an atlas satisfiesthe Hausdorff condition if and only if for any two distinct points p,q ∈ M, thereexists a smooth function f ∈C∞(M) with f (p) 6= f (q). In particular, if there existsa smooth injective map F : M→ RN , then M is Hausdorff.

Remark 3.2. One may replace ‘smooth’ with continuous in Proposition 3.1 andCorollary 3.1.

Example 3.3 (Projective spaces). Write vectors x ∈ Rn+1 as column vectors, hencex> is the corresponding row vector. The matrix product xx> is a square matrix withentries x jxk. The map

RPn→MatR(n+1), (x0 : . . . : xn) 7→ x x>

||x||2(3.5)

is a smooth; indeed, its matrix components are the functions (3.1). For any given(x0 : . . . : xn) ∈ RPn, at least one of these components is non-zero. IdentifyingMatR(n + 1) ∼= RN , where N = (n + 1)2, this gives the desired smooth injectivemap from projective space into RN ; hence the criterion applies, and the Hausdorffcondition follows. For the complex projective space, one similarly has a smooth andinjective map

CPn→MatC(n+1), (z0 : . . . : zn) 7→ z z†

||x||2(3.6)

(where z† = z> is the conjugate transpose of the complex column vector z) intoMatC(n+1) = RN with N = 2(n+1)2.

Exercise: Verify that the map

Gr(k,n)→MatR(n), E 7→ PE , (3.7)

taking a subspace E to the matrix of the orthogonal projection onto E, is smoothand injective, hence Gr(k,n) is Hausdorff. Discuss a similar map for the complexGrassmannian GrC(k,n).

In the opposite direction, the criterion tells us that for a set M with an atlas, if theHausdorff condition does not hold then no smooth injective map into RN exists.

Example 3.4. Consider the non-Hausdorff manifold M from Example 2.6. Here,there are two points p,q that do not admit disjoint open neighborhoods. We seedirectly that any smooth function on M must take on the same values at p and q:With the coordinate charts (U,ϕ),(V,ψ) in that example,

3.2 Smooth maps between manifolds 41

f (p) = f (ϕ−1(0)) = limt→0−

f (ϕ−1(t)) = limt→0−

f (ψ−1(t)) = f (ψ−1(0)) = f (q),

since ϕ−1(t) = ψ−1(t) for t < 0.

3.2 Smooth maps between manifolds

The notion of smooth maps from M to R generalizes to smooth maps between man-ifolds.

Definition 3.2. A map F : M→ N between manifolds is smooth at p ∈M if thereare coordinate charts (U,ϕ) around p and (V,ψ) around F(p) such that F(U)⊆Vand such that the composition

ψ F ϕ−1 : ϕ(U)→ ψ(V )

is smooth. The function F is called a smooth map from M to N if it is smooth at allp ∈M.

Remarks 3.3. 1. 1. The condition for smoothness at p does not depend on the choiceof charts: Given a different choice of charts (U ′,ϕ ′) and (V ′,ψ ′) with F(U ′) ⊆V ′, we have

ψ′ F (ϕ ′)−1 = (ψ ′ ψ

−1) (ψ F (ϕ)−1) (ϕ (ϕ ′)−1)

on ϕ ′(U ∩U ′).2. To check smoothness of F , it suffices to take any atlas (Uα ,ϕα) of M with

the property that F(Uα) ⊆ Vα for some chart (Vα ,ψα) of N, and then checksmoothness of the maps

42 3 Smooth maps

ψα F ϕ−1α : ϕα(Uα)→ ψα(Vα).

3. Smooth maps M→ R are the same thing as smooth functions on M:

C∞(M,R) =C∞(M).

Smooth functions γ : J → M from an open interval J ⊆ R to M are called(smooth) curves in M. Note that the image of a smooth curve need not looksmooth. For instance, the image of γ : R→R2, t 7→ (t2, t3) has a ‘cusp singular-ity’ at (0,0).

Example 3.5. a) Consider the map F : RP1→ RP1 given by

(t : 1) 7→ (et2: 1) for t ∈ R, (1 : 0) 7→ (1 : 0).

In the chart U1, we have F(U1) ⊆U1, and ϕ1 F ϕ−11 (t) = et2

, which is smooth.It remains to check smoothness at the point p = (1 : 0). Since F(p) = p, we willverify this using the chart U0 around p. We have, for u 6= 0

F(ϕ−10 (u)) = F(1 : u) = F(

1u

: 1) = (e1

u2 : 1) = (1 : e−1

u2 ),

Hence ϕ0 F ϕ−10 is the map

u 7→ e−1

u2 for u 6= 0, 0 7→ 0.

As is well-known, this map is smooth even at u = 0.b) The same calculation applies for the map F : CP1→ CP1, given by the same

formulas. However, the conclusion is different: The map

z 7→ e−1z2 for u 6= 0, 0 7→ 0.

is not smooth (or even continuous) at z = 0. (For a non-zero complex number a,consider the limit of this function for z = sa as s→ 0. If a = 1, the limit is 0. If

3.2 Smooth maps between manifolds 43

z = 1+ i, the absolute value is always one, and the limit doesn’t exist. If a = i, thelimit is ∞.)

Proposition 3.2. Suppose F1 : M1→M2 and F2 : M2→M3 are smooth maps. Thenthe composition

F2 F1 : M1→M3

is smooth.

Proof. Given p ∈M1, choose charts (U1,ϕ1) around p, (U2,ϕ2) around F1(p), and(U3,ϕ3) around F2(F1(p)), with F2(U2) ⊆ U3 and F1(U1) ⊆ U2. (This is alwayspossible – see exercise below.) Then F2(F1(U2))⊆U3, and we have:

ϕ3 (F2 F1)ϕ−11 = (ϕ3 F2 ϕ

−12 ) (ϕ2 F1 ϕ

−11 ),

a composition of smooth maps between open subsets of Euclidean spaces. ut

Exercise: Suppose F ∈C∞(M,N).

1. Let (U,ϕ) be a coordinate chart for M and (V,ψ) a coordinate chart for N, withF(U)⊆V . Show that for all open subsets W ⊆ N the set U ∩F−1(W ) is open.Hint: Show that ϕ(U ∩ F−1(W )) is the pre-image of the open set ψ(V ∩W )under the smooth map ψ F ϕ−1 : ϕ(U)→ ψ(V ).

2. Show that F is continuous: For every open W ⊆ N the pre-image F−1(W ) isopen.

3. Given p ∈ M and any chart (V,ψ) around F(p), show that there exists a chart(U,ϕ) around p such that F(U)⊆V .Hint: Start with any open chart (U1,ϕ1) around p, and replace U1 with U =U1∩F−1(V ).

Exercise: Using the previous exercise, show that smooth maps F ∈ C∞(M,N) arecontinuous: The pre-images of every open subsets of N is open in M.

3.2.1 Diffeomorphisms of manifolds

Definition 3.3. A smooth map F : M→ N is called a diffeomorphism if it is invert-ible, with a smooth inverse F−1 : N→M. Manifolds M,N are called diffeomorphicif there exists a diffeomorphism from M to N.

In other words, a diffeomorphism of manifolds is a bijection of the underlying setsthat identifies the maximal atlases of the manifolds. Manifolds that are diffeomor-phic are therefore considered ‘the same manifolds’.

Similarly, a continuous map F : M→N is called a homeomorphism if it is invert-ible, with a continuous inverse. Manifolds that are homeomorphic are considered‘the same topologically’. Since every smooth map is continuous, every diffeomor-phism is a homeomorphism.

44 3 Smooth maps

Example 3.6. By definition, every coordinate chart (U,ϕ) on a manifold M gives adiffeomorphism ϕ : U → ϕ(U) onto an open subset of Rm.

Example 3.7. The standard example of a homeomorphism of smooth manifolds thatis not a diffeomorphism is the map

R→ R, x 7→ x3.

Indeed, this map is smooth and invertible, but the inverse map y 7→ y13 is not smooth.

Example 3.8. Give a manifold M, with maximal atlas A , then any homeomorphismF : M→M can be used to define a new atlas A ′ on M, with charts (U ′,ϕ ′) ∈A ′

obtained from charts (U,ϕ) ∈ A as U ′ = F(U), ϕ ′ = ϕ F−1. One can verify(please do) that A ′ = A if and only if F is a diffeomorphism. Thus, if F is ahomeomorphism of M which is not a diffeomorphism, then F defines a new atlasA ′ 6= A .

However, the new manifold structure on M is not genuinely different from theold one. Indeed, while F : M→M is not a diffeomorphism relative to the atlas Aon the domain M and target M, it does define a diffeomorphism if we use the atlasA on the domain and the atlas A ′ on the target. Hence, even though A and A ′ aredifferent atlases, the resulting manifold structures are still diffeomorphic.

Remark 3.4. In the introduction, we explained (without proof) the classification of1-dimensional and 2-dimensional connected compact manifolds up to diffeomor-phism. This classification coincides with their classification up to homeomorphism.This means, for example, that for any maximal atlas A ′ on S2 which induces thesame system of open subsets as the standard maximal atlas A , there exists a home-omorphism F : S2→ S2 taking A to A ′, in the sense that (U,ϕ) ∈A if and only if(U ′,ϕ ′) ∈A ′, where U ′ = F(U) and ϕ ′ F = ϕ . In higher dimensions, it becomesmuch more complicated: .

It is quite possible for two manifolds to be homeomorphic but not diffeomor-phic (unlike example 3.8). The first example of ‘exotic’ manifold structures wasdiscovered by John Milnor in 1956, who found that the 7-sphere S7 admits manifoldstructures that are not diffeomorphic to the standard manifold structure, but inducethe standard topology. Kervaire and Milnor in 1963, proved that there are exactly28 distinct manifold structures on S7, and in fact classified all manifold structureson all spheres Sn with the exception of the case n = 4. For example, they showedthat S3,S5,S6 do not admit exotic (i.e., non-standard) manifold structures, while S15

has 16256 different manifold structures. For S4 the existence of exotic manifoldstructures is an open problem; this is known as the smooth Poincare conjecture.

Around 1982, Michael Freedman (using results of Simon Donaldson) discoveredthe existence of exotic manifold structures on R4; later Clifford Taubes showed thatthere are uncountably many such. For Rn with n 6= 4, it is known that there are noexotic manifold structures on Rn.

3.3 Examples of smooth maps 45

3.3 Examples of smooth maps

3.3.1 Products, diagonal maps

a) If M,N are manifolds, then the projection maps

prM : M×N→M, prN : M×N→ N

are smooth. (This follows immediately by taking product charts Uα ×Vβ .)b) The diagonal inclusion

∆M : M→M×M

is smooth. (In a coordinate chart (U,ϕ) around p and the chart (U ×U,ϕ × ϕ)around (p, p), the map is the restriction to ϕ(U) ⊆ Rn of the diagonal inclusionRn→ Rn×Rn.)

c) Suppose F : M → N and F ′ : M′ → N′ are smooth maps. Then the directproduct

F×F ′ : M×M′→ N×N′

is smooth. This follows from the analogous statement for smooth maps on opensubsets of Euclidean spaces.

3.3.2 The diffeomorphism RP1 ∼= S1

We have stated before that RP1 ∼= S1. To obtain an explicit diffeomorphism, weconstruct a bijection identifying the standard atlas for RP1 with (essentially) thestandard atlas for S1. Recall that the atlas for RP1 is given by

U1 = (u : 1)| u ∈ R, ϕ1(u : 1) = u,

U0 = (1 : u)| u ∈ R, ϕ0(1 : u) = u

with ϕi(Ui) = R and ϕ0(U0∩U1) = ϕ1(U0∩U1) = R\0, with the transition mapϕ1 ϕ

−10 : u 7→ u−1. Similarly, the atlas for S1 is

U+ = (x,y) ∈ S1| y 6=−1 ϕ+(x,y) =x

1+ y,

U− = (x,y) ∈ S1| y 6=+1 ϕ−(x,y) =x

1− y.

again with ϕ±(U±) = R, ϕ±(U+∩U−) = R\0, and transition map u 7→ u−1.Hence, there is a well-defined diffeomorphism F : RP1 → S1 which identifies

the chart (U−,ϕ−) with (U1,ϕ1) and (U+,ϕ+) with (U0,ϕ0), in the sense that both

46 3 Smooth maps

ϕ− F ϕ−11 : R→ R, ϕ+ F ϕ

−10 : R→ R

are the identity idR. Namely, the restriction of F to U1 is FU1 = ϕ−1− ϕ1 : U1→U−,

the restriction to U0 is F |U0 = ϕ−1+ ϕ0 : U0 → U+. The inverse map G = F−1 :

S1 → RP(1) is similarly given by ϕ−10 ϕ+ over U+ and by ϕ

−11 ϕ− over U−. A

calculation gives

F : RP1→ S1, (w0 : w1) 7→ 1||w||2

(2w1w0, (w0)2− (w1)2);

with inverse.

G(x,y) = (1+ y : x), y 6=−1,G(x,y) = (x : 1− y), y 6= 1

(note that the two expressions agree if−1 < y < 1). For example, to get the formulafor G(x,y) for y 6=−1, i.e. (x,y) ∈U+, we calculate as follows:

ϕ−10 ϕ+(x,y) = ϕ

−10

( x1+ y

)=(

1 :x

1+ y

)= (1+ y : x).

Exercise: Work out the details of the calculation of F(w0 : w1).

3.3.3 The diffeomorphism CP1 ∼= S2

By a similar reasoning, we find CP1 ∼= S2. For S2 we use the atlas given by stereo-graphic projection.

U+ = (x,y,z) ∈ S2| z 6=−1 ϕ+(x,y,z) =1

1+ z(x,y),

U− = (x,y,z) ∈ S2| z 6=+1 ϕ−(x,y,z) =1

1− z(x,y).

The transition map is u 7→ u||u||2 , for u = (u1,u2). Regarding u as a complex number

u = u1 + iu2, the norm ||u|| is just the absolute value of u, and the transition mapbecomes

u 7→ u|u|2

=1u.

Note that it is not quite the same as the transition map for the standard atlas of CP1,which is given by u 7→ u−1. We obtain a unique diffeomorphism F : CP1 → S2

such that ϕ+ F ϕ−10 is the identity, while ϕ− F ϕ

−11 is complex conjugation. A

calculation shows that this map is


F(w0 : w1) =1

|w0|2 + |w1|2(

2Re(w1w0), 2Im(w1w0), |w0|2−|w1|2)

;

the inverse map G = F−1 : S2→ CP(1) is

G(x,y,z) = (1+ z : x+ iy), z 6=−1,G(x,y,z) = (x− iy : 1− z), z 6= 1

(note that the two expressions agree if −1 < z < 1).

Exercise: Work out the details of the calculation.

3.3.4 Maps to and from projective space

The quotient map

π : Rn+1\0→ RPn, x = (x0, . . . ,xn) 7→ (x0 : . . . : xn)

is smooth, as one verifies by checking in the standard atlas for RPn. Indeed, on theopen subset where xi 6= 0, we have π(x) ∈Ui, and

(ϕi π)(x0, . . . ,xn) = (x0

xi , . . . ,xi−1

xi ,xi+1

xi , . . . ,xn

xi ).

which is a smooth function on the open set of x’s for which xi 6= 0.Given a map F : RPn→ N to a manifold N, let F = F π : Rn+1\0 → N be

its composition with the projection map π : Rn+1\0→ RPn. That is,

F(x0, . . . ,xn) = F(x0 : . . . : xn).

Note that F(λx0 : . . . : λxn) = F(x0, . . . ,xn) for all non-zero λ ; conversely, everymap F with this property descends to a map F on projective space. We claim thatthe map F is smooth if and only the corresponding map F is smooth. One directionis clear: If F is smooth, then F = F π is a composition of smooth maps. For theother direction, assuming that F is smooth, note that for the standard chart (U j,ϕ j),the maps

(F ϕ−1j )(u1, . . . ,un) = F(u1, . . . ,ui,1,ui+1, . . . ,un),

are smooth.An analogous argument applies to the complex projective space CPn, taking the

xi to be complex numbers zi. That is, the quotient map π : Cn+1\0 → CPn issmooth, and a map F : CPn → N is smooth if and only if the corresponding mapF : Cn+1\0→ N is smooth.

As an application, we can see that the map

CP1→ CP2, (z0 : z1) 7→ ((z0)2 : (z1)2 : z0z1)

48 3 Smooth maps

is smooth, starting with the (obvious) fact that the lifted map

C2\0→ C3\0, (z0,z1) 7→ ((z0)2,(z1)2,z0z1)

is smooth.

3.3.5 The quotient map S2n+1→ CPn

As we explained above, the quotient map q : Cn+1\0→CPn is smooth. Since anyclass [z] = (z0 : . . . : zn) has a representative with |z0|2 + . . .+ |zn|2 = 1, and |zi|2 =(xi)2 +(yi)2 for zi = xi +

√−1yi, we may also regard CPn as a set of equivalence

classes in the unit sphere S2n+1 ⊆ R2n+2 = Cn+1. The resulting quotient map

π : S2n+1→ CPn

is again smooth, because it can be written as a composition of two smooth maps

π = q ι

where ι : S2n+1→ R2n+2\0= Cn+1\0 is the inclusion map.For any p ∈ CPn, the corresponding fiber π−1(p) ⊆ S2n+1 is diffeomorphic to a

circle S1 (which we may regard as complex numbers of absolute value 1). Indeed,given any point (z0, . . . ,zn) ∈ π−1(p) in the fiber, the other points are obtained as(λ z0, . . . ,λ zn) where |λ |= 1.

In other words, we can think of

S2n+1 =⋃

p∈CPn

π−1(p)

as a union of circles, parametrized by the points of CPn. This is an example ofwhat differential geometers call a fiber bundle or fibration. We won’t give a formaldefinition here, but let us try to ‘visualize’ the fibration for the important case n = 1.

Identifying CP1 ∼= S2 as above, the map π becomes a smooth map

π : S3→ S2

with fibers diffeomorphic to S1. This map appears in many contexts; it is called theHopf fibration (after Heinz Hopf (1894-1971)).


Let S ∈ S3 be the ‘south pole’, and N ∈ S3 the ‘north pole’. We have that S3−S ∼=R3 by stereographic projection. The set π−1(π(S))−S projects to a straight line(think of it as a circle with ‘infinite radius’). The fiber π−1(N) is a circle that goesaround the straight line. If Z ⊆ S2 is a circle at a given ‘latitude’, then π−1(Z) isis a 2-torus. For Z close to N this 2-torus is very thin, while for Z approaching thesouth pole S the radius goes to infinity. Each such 2-torus is itself a union of circlesπ−1(p), p ∈ Z. Those circles are neither the usual ‘vertical’ or ‘horizontal’ circlesof a 2-torus in R3, but instead are ‘tilted’. In fact, each such circle is a ‘perfect ge-ometric circle’ obtained as the intersection of its 2-torus with a carefully positionedaffine 2-plane.

Moreover, any two of the circles π−1(p) are linked:

The full picture looks as follows:1

1http://perso-math.univ-mlv.fr/users/kloeckner.benoit/images.html

http://perso-math.univ-mlv.fr/users/kloeckner.benoit/images.html

50 3 Smooth maps

A calculation shows that over the charts U+,U− (from stereographic projection), theHopf fibration is just a product. That is, one has

π−1(U+)∼=U+×S1, π

−1(U−)∼=U−×S1.

In particular, the pre-image of the closed upper hemisphere is a solid 2-torus D2×S1

(with D2 = z ∈C| |z| ≤ 1 the unit disk), geometrically depicted as a 2-torus in R3

together with its interior.2 We hence see that the S3 may be obtained by gluing twosolid 2-tori along their boundaries S1×S1.

3.4 Submanifolds

Let M be a manifold of dimension m. We will define a k-dimensional submanifoldS ⊆ M to be a subset that looks locally like Rk ⊆ Rm (which we take to be thecoordinate subspace defined by xk+1 = · · ·= xm = 0.

Definition 3.4. A subset S ⊆M is called a submanifold of dimension k ≤ m, if forall p ∈ S there exists a coordinate chart (U,ϕ) around p such that

ϕ(U ∩S) = ϕ(U)∩Rk.

Charts (U,ϕ) of M with this property are called submanifold charts for S.

Remark 3.5. 1. A chart (U,ϕ) such that U ∩S = /0 and ϕ(U)∩Rk = /0 is considereda submanifold chart.

2. We stress that the existence of submanifold charts is only required for points pthat lie in S. For example, the half-open line S = (0,∞) is a submanifold of R.There does not exist a submanifold chart containing 0, but this is not a problemsince 0 6∈ S.

Strictly speaking, a submanifold chart for S is not a chart for S, but is a chart for Mwhich is adapted to S. On the other hand, submanifold charts restrict to charts for S,and this may be used to construct an atlas for S:

Proposition 3.3. Suppose S is a submanifold of M. Then S is a k-dimensional man-ifold in its own right, with atlas consisting of all charts (U ∩ S,ϕ|U∩S) such that(U,ϕ) is a submanifold chart.

Proof. Let (U,ϕ) and (V,ψ) be two submanifold charts for S. We have to show thatthe charts (U ∩S,ϕ|U∩S) and (V ∩S,ψ|V∩S) are compatible. The map

ψ|V∩S ϕ|−1U∩S : ϕ(U ∩V )∩Rk→ ψ(U ∩V )∩Rk

2 A solid torus is an example of a ”manifold with boundary”, a concept we haven’t properly dis-cussed yet.

3.4 Submanifolds 51

is smooth, because it is the restriction of ψ ϕ−1 : ϕ(U ∩V )→ψ(U ∩V ) to the co-ordinate subspace Rk. Likewise its inverse map is smooth. The Hausdorff conditionfollows because any two distinct points p,q ∈ S, one can take disjoint submanifoldcharts around p,q. (Just take any submanifold charts, and intersect with the domainsof disjoint charts around p,q.)

The proof that S admits a countable atlas is unfortunately a bit technical3. Weuse the following

Fact: Every open subset of Rm is a union of rational ε-balls Bε (x), ε > 0. Here, ‘rational’means that both the center of the ball and its radius are rational: x ∈Qn, ε ∈Q.

(We leave this as an exercise.) Our goal is to construct a countable collection ofsubmanifold charts covering S. (The atlas for S itself is then obtained by restriction.)Start with any countable atlas (Uα ,ϕα) for M. Given p ∈ S∩Uα , we can choosea submanifold chart (V,ψ) containing p. Using the above fact, we can choose arational ε-ball with

ϕ(p) ∈ Bε(x)⊆ ϕα(Uα ∩V ).

This shows that the subsets of the form ϕ−1α (Bε(x)), with Bε(x) ⊆ ϕα(Uα) a ra-

tional ε-ball such that ϕ−1α (Bε(x)) is contained in some submanifold chart, cover

all of S. Take these to be the domains of a charts (Vβ ,ψβ ), where Vβ is one of theϕ−1

α (Bε(x)), and ψβ is the restriction of the coordinate maps of a submanifold chartcontaining ϕ−1

α (Bε(x)). Then (Vβ ,ψβ ) is a countable collection of submanifoldcharts covering S. (Recall that a countable union of countable sets is again count-able.) ut

Example 3.9 (Open subsets). The m-dimensional submanifolds of an m-dimensionalmanifold are exactly the open subsets.

Example 3.10 (Spheres). Let Sn = x∈Rn+1| ||x||2 = 1. Write x = (x0, . . . ,xn), andregard

Sk ⊆ Sn

for k < n as the subset where the last n− k coordinates are zero. These are subman-ifolds: The charts (U±,ϕ±) for Sn given by stereographic projection

ϕ±(x0, . . . ,xn) =1

1± x0 (x1, . . . ,xn)

are submanifold charts. In fact, the charts U±i , given by the condition that ±xi > 0,with ϕ

±i the projection to the remaining coordinates, are submanifold charts as well.

Example 3.11 (Projective spaces). For k < n, regard

RPk ⊆ RPn

as the subset of all (x0 : . . . : xn) for which xk+1 = . . .= xn = 0. These are submani-folds, with the standard charts (Ui,ϕi) for RPn as submanifold charts. (Note that the

3 You are welcome to ignore the following proof.

52 3 Smooth maps

charts Uk+1, . . . ,Un don’t meet RPk, but this does not cause a problem.) In fact, theresulting charts for RPk obtained by restricting these submanifold charts, are justthe standard charts of RPk. Similarly,

CPk ⊆ CPn

are submanifolds, and for n < n′ we have Gr(k,n)⊆ Gr(k,n′) as a submanifold.

Proposition 3.4. Let F : M→N be a smooth map between manifolds of dimensionsm and n. Then

graph(F) = (F(p), p)| p ∈M ⊆ N×M

is a submanifold of N×M, of dimension equal to the dimension of M.

Proof. Given p ∈M, choose charts (U,ϕ) around p and (V,ψ) around F(p), withF(U)⊆V , and let W =V ×U . We claim that (W,κ) with

κ(q, p) = (ϕ(p), ψ(q)−ψ(F(p))) (3.8)

is a submanifold chart for graph(F) ⊆ N ×M. Note that this is indeed a chart ofN×M, because it is obtained from the product chart (V ×U,ψ×ϕ) by compositionwith the diffeomorphism ψ(V )×ϕ(U)→ ϕ(U)×ψ(V ), (v,u) 7→ (u,v), followedby the diffeomorphism

ϕ(U)×ψ(V )→ κ(W ), (u,v) 7→ (u,v− F(u)). (3.9)

where F = ψ F ϕ−1. (The map (3.9) is smooth and injective, and its Jacobian hasdeterminant one, hence is invertible everywhere.) Furthermore, the second compo-nent in (3.8) vanishes if and only if F(p) = q. That is,

κ(W ∩graph(F)) = κ(W )∩Rm

as required. ut

This result has the following consequence: If a subset of a manifold, S⊆M, can belocally described as the graph of a smooth map, then S is a submanifold. In moredetail, suppose that S can be covered by open sets U , such that for each U there is adiffeomorphism U→ P×Q taking S∩U to the graph of a smooth map Q→ P, thenS is a submanifold.

Example 3.12. The 2-torus S = f−1(0)⊆ R3, where

f (x,y,z) = (√

x2 + y2−R)2 + z2− r2

is a submanifold of R3, since it can locally be expressed as the graph of a functionof x,y, or of y,z, or of x,z. For example, on the subset where z > 0, it is the graph ofthe smooth function on the annulus (x,y)| (R− r)2 < x2 +y2 < (R+ r)2, given as

H(x,y) =√

r2− (√

x2 + y2−R)2.

3.4 Submanifolds 53

This function is obtained by solving the equation f (x,y,z) for z. Similarly, on eachof the four components of the subset where x2 + y2 6= R2 and x 6= 0 (respectively,y 6= 0), one can solve the equation f (x,y,z) = 0 uniquely for x (respectively, for y),expressing S as the graph of a smooth function of y and z (respectively, of x and z).

Exercise: Work out the formula for the F(y,z) on the subset where x2+y2 < R2 andx > 0.

Example 3.13. More generally, suppose S ⊆ R3 is given as a level set S = f−1(0)for a smooth map f ∈C∞(R3). (Actually, we only need f to be defined and smoothon an open neighborhood of S.) Let p ∈ S, and suppose

∂ f∂x

∣∣∣p6= 0.

By the implicit function theorem from multivariable calculus, there is an open neigh-borhood U ⊆ R3 of p on which the equation f (x,y,z) = 0 can be uniquely solvedfor x. That is,

S∩U = (x,y,z) ∈U | x = F(y,z)

for a smooth function F , defined on a suitable open subset of R2. This shows that Sis a submanifold near p, and in fact we may use y,z as coordinates near p. Similararguments apply for ∂ f

∂y |p 6= 0 or ∂ f∂ z |p 6= 0. Hence, if the gradient

∇ f = (∂ f∂x

,∂ f∂y

,∂ f∂ z

)

is non-vanishing at all points p∈ S= f−1(0), then S is a 2-dimensional submanifold.Of course, there is nothing special about 2-dimensional submanifolds of R3, andbelow, we will put this discussion in a more general framework.

As we saw, submanifolds S of manifolds M are themselves manifolds. They comewith an inclusion map

i : S→M, p 7→ p,

taking any point of S to the same point but viewed as a point of M. Unsurprisingly,we have:

Proposition 3.5. The inclusion map i : S→M is smooth.

Proof. Given p ∈ S ⊆ M, let (U,ϕ) be a submanifold chart around p ∈ M, and(U ∩S,ϕ|U∩S) the corresponding chart around p ∈ S. The composition

ϕ i (ϕ|U∩S)−1 : ϕ(U ∩S)→ ϕ(U)

is simply the inclusion map from ϕ(U)∩Rk to ϕ(U), which is obviously smooth.ut

54 3 Smooth maps

This shows in particular that if F ∈C∞(M,N) is a smooth map, then its restrictionF |S : S→ N is again smooth. Indeed, F |S = F i is a composition of smooth maps.This is useful in practice, because in such cases there is no need to verify smoothnessin local coordinates of S! For example, the map S2→R, (x,y,z) 7→ z is smooth sinceit is the restriction of a smooth map R3→R to the submanifold S2. A related result,which we leave as an exercise, is the following:

Exercise. Let S ⊆M be a submanifold, with inclusion map i, and let F : Q→ S bea map from another manifold Q. Then F is smooth if and only if iF is smooth. (Inother words, if and only if F is smooth as a map into M.)

For the following proposition, recall that a subset U of a manifold is open ifand only if for all p ∈ U , and any coordinate chart (V,ψ) around p, the subsetψ(U ∩V )⊆ Rm is open. (This does not depend on the choice of chart.)

Proposition 3.6. Suppose S is a submanifold of M. Then the open subsets of S forits manifold structure are exactly those of the form U ∩S, where U is an open subsetof M.

In other words, the topology of S as a manifold coincides with the ‘subspace topol-ogy’ as a subset of the manifold M.

Proof. We have to show:

U ′ ⊆ S is open ⇔ U ′ =U ∩S where U ⊆M is open.

“⇒”. Suppose U ⊆M is open, and let U ′ =U∩S. For any submanifold chart (V,ψ),with corresponding chart (V ∩S,ψ|V∩S) for S, we have that

ψ((V ∩S)∩U ′) = ψ(V ∩S∩U) = ψ(U)∩ψ(V )∩Rk

is the intersection of the open set ψ(U)∩ψ(V ) ⊆ Rn with the subspace Rk, henceis open in Rk. Since submanifold charts cover all of S, this shows that U ′ is open.

“⇐” Suppose U ′ ⊆ S is open in S. Define

U =⋃V

ψ−1(ψ(U ′∩V )×Rm−k)⊆M,

where the union is over any collection of submanifold charts (V,ψ) that cover all ofS. Since U ′ is open in S, so is U ′∩V ≡U ′∩(V ∩S). Hence ψ(U ′∩V ) =ψ(U ′∩(V ∩S)) is open in Rk, and its cartesian product with Rm−k is open in Rm. The pre-imageψ−1(ψ(U ′∩V )×Rm−k) is thus open in V , hence also in M, and the union over allsuch sets is open in M. Since U ∩S =U ′ (see exercise below) we are done. ut

Exercise: Fill in the last detail of this proof: Check that U ∩S =U ′.

Remark 3.6. As a consequence, if a manifold M can be realized realized as a sub-manifold M ⊆ Rn, then M is compact with respect to its manifold topology if andonly if it is compact as a subset of Rn, if and only if it is a closed and bounded subset

3.5 Smooth maps of maximal rank 55

of Rn. This can be used to give quick proofs of the facts that the real or complexprojective spaces, as well as the real or complex Grassmannians, are all compact.

Remark 3.7. Sometimes, the result can be used to show that certain subsets are notsubmanifolds. Consider for example the subset

S = (x,y) ∈ R2| xy = 0 ⊆ R2

given as the union of the coordinate axes. If S were a 1-dimensional submanifold,then there would exist an open neighborhood U ′ of p = (0,0) in S which is dif-feomorphic to an open interval. But for any open subset U ⊆ R2 containing p, theintersection U ′ =U ∩S cannot possibly be an open interval, since (U ∩S)\p hasat least four components, while removing a point from an open interval gives onlytwo components.

3.5 Smooth maps of maximal rank

Let F ∈C∞(M,N) be a smooth map. Then the fibers (level sets)

F−1(q) = x ∈M| F(x) = q

for q ∈ N need not be submanifolds, in general. Similarly, the image F(M) ⊆ Nneed not be a submanifold – even if we allow self-intersections. (More precisely,there may be points p such that the image F(U) ⊆ N of any open neighborhood Uof p is never a submanifold.) Here are some counter-examples:

1. The fibers f−1(c) of the map f (x,y) = xy are hyperbolas for c 6= 0, but f−1(0) isthe union of coordinate axes. What makes this possible is that the gradient of fis zero at the origin.

2. As we mentioned earlier, the image of the smooth map

γ : R→ R2, γ(t) = (t2, t3)

does not look smooth near (0,0) (and replacing R by an open interval around 0does not help). 4 What makes this is possible is that the velocity γ(t) vanishes fort = 0: the curve described by γ ‘comes to a halt’ at t = 0, and then turns around.

In both cases, the problems arise at points where the map does not have maximalrank. After reviewing the notion of rank of a map from multivariable calculus, wewill generalize to manifolds.

4 It is not a submanifold, although we haven’t proved it (yet).

56 3 Smooth maps

3.5.1 The rank of a smooth map

The following discussion will involve some notions from multivariable calculus. LetU ⊆ Rm and V ⊆ Rn be open subsets, and F ∈C∞(U,V ) a smooth map.

Definition 3.5. The derivative of F at p ∈U is the linear map

DpF : Rm→ Rn, v 7→ ddt

∣∣∣t=0

F(p+ tv).

The rank of F at p is the rank of this linear map:

rankp(F) = rank(DpF).

(Recall that the rank of a linear map is the dimension of its range.) Equivalently,DpF is the n×m matrix of partial derivatives (DpF)i

j =∂F i

∂x j

∣∣∣p:

DpF =

∂F1

∂x1

∣∣p

∂F1

∂x2

∣∣p · · ·

∂F1

∂xm

∣∣p

∂F2

∂x1

∣∣p

∂F2

∂x2

∣∣p · · ·

∂F2

∂xm

∣∣p

· · · · · · · · · · · ·∂Fn

∂x1

∣∣p

∂Fn

∂x2

∣∣p · · ·

∂Fn

∂xm

∣∣p

and the rank of F at p is the rank of this matrix (i.e., the number of linearly inde-pendent rows, or equivalently the number of linearly independent columns). Noterankp(F)≤min(m,n). By the chain rule for differentiation, the derivative of a com-position of two smooth maps satisfies

Dp(F ′ F) = DF(p)(F′)Dp(F). (3.10)

In particular, if F ′ is a diffeomorphism then rankp(F ′ F) = rankp(F), and if F is adiffeomorphism then rankp(F ′ F) = rankF(p)(F ′).

Definition 3.6. Let F ∈C∞(M,N) be a smooth map between manifolds, and p ∈M.The rank of F at p ∈M is defined as

rankp(F) = rankϕ(p)(ψ F ϕ−1)

for any two coordinate charts (U,ϕ) around p and (V,ψ) around F(p) such thatF(U)⊆V .

By (3.10), this is well-defined: if we use different charts (U ′,ϕ ′) and (V ′,ψ ′), thenthe rank of

ψ′ F (ϕ ′)−1 = (ψ ′ ψ

−1) (ψ F ϕ−1) (ϕ (ϕ ′)−1)

at ϕ ′(p) equals that of ψ F ϕ−1 at ϕ(p), since the two maps are related by dif-feomorphisms.


The following discussion will focus on maps of maximal rank. We have that

rankp(F)≤min(dimM, dimN)

for all p ∈ M; the map F is said to have maximal rank at p if rankp(F) =min(dimM, dimN). A point p ∈ M is called a critical point for F if rankp(F) <min(dimM, dimN).

3.5.2 Local diffeomorphisms

In this section we will consider the case dimM = dimN. Our ‘workhorse theorem’from multivariable calculus is going to be the following fact.

Theorem 3.1 (Inverse Function Theorem for Rm). Let F ∈C∞(U,V ) be a smoothmap between open subsets of Rm, and suppose that the derivative DpF at p ∈U isinvertible. Then there exists an open neighborhood U1 ⊆U of p such that F restrictsto a diffeomorphism U1→ F(U1).

Remark 3.8. The theorem tells us that for a smooth bijection, a sufficient conditionfor smoothness of the inverse map is that the differential (i.e., the first derivative) isinvertible everywhere. It is good to see, in just one dimensions, how this is possible.Given an invertible smooth function y = f (x), with inverse x = g(y), and usingddy =

dxdy

ddx , we have

g′(y) =1

f ′(x),

g′′(y) = =− f ′′(x)f ′(x)3 ,

g′′′(y) = =− f ′′′(x)

f ′(x)4 +3f ′′(x)2

f ′(x)5 ,

and so on; only powers of f ′(x) appear in the denominator.

Theorem 3.2 (Inverse function theorem for manifolds). Let F ∈ C∞(M,N) be asmooth map between manifolds of the same dimension m = n. If p ∈M is such thatrankp(F) = m, then there exists an open neighborhood U ⊆ M of p such that Frestricts to a diffeomorphism U → F(U).

Proof. Choose charts (U,ϕ) around p and (V,ψ) around F(p) such that F(U)⊆V .The map

F = ψ F ϕ−1 : U := ϕ(U)→ V := ψ(V )

has rank m at ϕ(p). Hence, by the inverse function theorem for Rm, after replac-ing U with a smaller open neighborhood of ϕ(p) (equivalently, replacing U with

58 3 Smooth maps

a smaller open neighborhood of p) the map F becomes a diffeomorphism from Uonto F(U) = ψ(F(U)). It then follows that

F = ψ−1 F ϕ : U →V

is a diffeomorphism U → F(U). ut

A smooth map F ∈ C∞(M,N) is called a local diffeomorphism if dimM = dimN,and F has maximal rank everywhere. By the theorem, this is equivalent to the con-dition that every point p has an open neighborhood U such that F restricts to adiffeomorphism U → F(U). It depends on the map in question which of these twoconditions is easier to verify.

Example 3.14. The quotient map π : Sn→ RPn is a local diffeomorphism. Indeed,one can see (using suitable coordinates) that π restricts to diffeomorphisms fromeach U±j = x ∈ Sn| ± x j > 0 to the standard chart U j.

Example 3.15. The map R→ S1, t 7→ (cos(2πt), sin(2πt)) is a local diffeomor-phism. (Exercise.)

Example 3.16. Let M be a manifold with a countable open cover Uα, and let

Q =⊔α

Uα

be the disjoint union. Then the map π : Q→M, given on Uα ⊆ Q by the inclusioninto M, is a local diffeomorphism. Since π is surjective, it determines an equivalencerelation on Q, with π as the quotient map and M = Q/∼.

We leave it as an exercise to show that if the Uα ’s are the domains of coordinatecharts, then Q is diffeomorphic to an open subset of Rm. This then shows that anymanifold is realized as a quotient of an open subset of Rm, in such a way that thequotient map is a local diffeomorphism.

3.5.3 Level sets, submersions

The inverse function theorem is closely related to the implicit function theorem, andone may be obtained as a consequence of the other. (We have chosen to take theinverse function theorem as our starting point.)

Proposition 3.7. Suppose F ∈C∞(U,V ) is a smooth map between open subsets U ⊆Rm and V ⊆ Rn, and suppose p ∈U is such that the derivative DpF is surjective.Then there exists an open neighborhood U1 ⊆ U of p and a diffeomorphism κ :U1→ κ(U1)⊆ Rm such that

(F κ−1)(u1, . . . ,um) = (um−n+1, . . . ,um)

for all u = (u1, . . . ,um) ∈ κ(U1).


Thus, in suitable coordinates F is given by a projection onto the last n coordinates.

Although it belongs to multivariable calculus, let us recall how to get this result fromthe inverse function theorem.

Proof. The idea is to extend F to a map between open subsets of Rm, and then applythe inverse function theorem.

By assumption, the derivative DpF has rank equal to n. Hence it has n linearlyindependent columns. By re-indexing the coordinates of Rm (this permutation isitself a change of coordinates) we may assume that these are the last n columns.That is, writing

DpF =(

C,D)

where C is the n× (m− n)-matrix formed by the first m− n columns and D then×n-matrix formed by the last n columns, the square matrix D is invertible. Writeelements x ∈Rm in the form x = (x′,x′′) where x′ are the first m−n coordinates andx′′ the last n coordinates. Let

G : U → Rm, x = (x′,x′′) 7→ (x′,F(x)).

Then the derivative DpG has block form

DpG =

(Im−n 0

C D

),

(where Im−n is the square (m− n)× (m− n) matrix), and is hence is invertible.Hence, by the inverse function theorem there exists a smaller open neighborhoodU1 of p such that G restricts to a diffeomorphism κ : U1→ κ(U1)⊆ Rm. We have,

Gκ−1(u′,u′′) = (u′,u′′)

for all (u′,u′′) ∈ κ(U1). Since F is just G followed by projection to the x′′ compo-nent, we conclude

F κ−1(u′,u′′) = u′′.

ut

60 3 Smooth maps

Again, this result has a version for manifolds:

Theorem 3.3. Let F ∈C∞(M,N) be a smooth map between manifolds of dimensionsm ≥ n, and suppose p ∈M is such that rankp(F) = n. Then there exist coordinatecharts (U,ϕ) around p and (V,ψ) around F(p), with F(U)⊆V , such that

(ψ F ϕ−1)(u′,u′′) = u′′

for all u = (u′,u′′) ∈ ϕ(U). In particular, for all q ∈V the intersection

F−1(q)∩U

is a submanifold of dimension m−n.

Proof. Start with coordinate charts (U,ϕ) around p and (V,ψ) around F(p) suchthat F(U)⊆V . Apply Proposition 3.7 to the map F = ψ F ϕ−1 : ϕ(U)→ψ(V ),to define a smaller neighborhood ϕ(U1)⊆ ϕ(U) and change of coordinates κ so thatF κ−1(u′,u′′) = u′′. After renaming (U1,κ ϕ|U1) as (U,ϕ) we have the desiredcharts for F . The last part of the Theorem follows since (U,ϕ) becomes a subman-ifold chart for F−1(q)∩U (after shifting ϕ by ψ(q) ∈ Rn). ut

Definition 3.7. Let F ∈ C∞(M,N). A point q ∈ N is called a regular value of F ∈C∞(M,N) if for all x ∈ F−1(q), one has rankx(F) = dimN. It is called a singularvalue if it is not a regular value.

Note that regular values are only possible if dimN ≤ dimM. Note also that all pointsof N that are not in the image of the map F are considered regular values. We mayrestate Theorem 3.3 as follows:

Theorem 3.4 (Regular Value Theorem). For any regular value q ∈ N of a smoothmap F ∈C∞(M,N), the level set S = F−1(q) is a submanifold of dimension

dimS = dimM−dimN.

Example 3.17. The n-sphere Sn may be defined as the level set F−1(1) of the func-tion F ∈C∞(Rn+1,R) given by

F(x0, . . . ,xn) = (x0)2 + . . .+(xn)2.

The derivative of F is the 1× (n+1)-matrix of partial derivatives, that is, the gradi-ent ∇F :

DpF = (2x0, . . . ,2xn).

For x 6= 0 this has maximal rank. Note that any nonzero real number q is a regularvalue since 0 6∈ F−1(q). Hence all the level sets F−1(q) for q 6= 0 are submanifolds.

Example 3.18. Let 0 < r < R. Then

F(x,y,z) = (√

x2 + y2−R)2 + z2

has r2 as a regular value, with corresponding level set the 2-torus.


Example 3.19. The orthogonal group O(n) is the group of matrices A ∈ MatR(n)satisfying A> = A−1. We claim that O(n) is a submanifold of MatR(n). To see this,consider the map

F : MatR(n)→ SymR(n), A 7→ A>A,

where SymR(n) ⊆MatR(n) denotes the subspace of symmetric matrices. We wantto show that the identity matrix I is a regular value of F . We compute the differentialDAF : MatR(n)→ SymR(n) using the definition5

(DAF)(X) =ddt

∣∣∣t=0

F(A+ tX)

=ddt

∣∣∣t=0

((A>+ tX>)(A+ tX))

= A>X +X>A.

To see that this is surjective, for A ∈ F−1(I), we need to show that for any Y ∈SymR(n) there exists a solution of

A>X +X>A = Y.

Using A>A = F(A) = I we see that X = 12 AY is a solution. We conclude that I is a

regular value, and hence that O(n) = F−1(I) is a submanifold. Its dimension is

dimO(n) = dimMatR(n)−dimSymR(n) = n2− 12

n(n+1) =12

n(n−1).

Note that it was important here to regard F as a map to SymR(n); for F viewed as amap to MatR(n) the identity would not be a regular value.

Definition 3.8. A smooth map F ∈C∞(M,N) is a submersion if rankp(F) = dimNfor all p ∈M.

Thus, for a submersion all level sets F−1(q) are submanifolds.

Example 3.20. Local diffeomorphisms are submersions; here the level sets F−1(q)are discrete points, i.e. 0-dimensional manifolds.

Example 3.21. Recall that CPn can be regarded as a quotient of S2n+1. Using charts,one can check that the quotient map π : S2n+1 → CPn is a submersion. Hence itsfibers π−1(q) are 1-dimensional submanifolds. Indeed, as discussed before thesefibers are circles. As a special case, the Hopf fibration S3→ S2 is a submersion. Asa special case, the Hopf fibration S3→ S2 is a submersion.

Remark 3.9. (For those who are familiar with quaternions.) Let H = C2 = R4 bethe quaternionic numbers. The unit quaternions are a 3-sphere S3. Generalizing the

5 Note that it would have been confusing to work with the description of DAF as a matrix of partialderivatives.

62 3 Smooth maps

definition of RPn and CPn, there are also quaternionic projective spaces, HPn. Theseare quotients of the unit sphere inside Hn+1, hence one obtains submersions

S4n+3→HPn;

the fibers of this submersion are diffeomorphic to S3. For n = 1, one can show thatHP1 = S4, hence one obtains a submersion

π : S7→ S4

with fibers diffeomorphic to S3.

3.5.4 Example: The Steiner surface

In this section, we will give more lengthy examples, investigating the smoothnessof level sets. 6

Example 3.22 (Steiner’s surface). Let S⊆ R3 be the solution set of

y2z2 + x2z2 + x2y2 = xyz.

in R3. Is this a smooth surface in R3? (We use surface as another term for 2-dimensional manifold; by a surface in M we mean a 2-dimensional submanifold.)Actually, we can easily see that it’s not. If we take one of x,y,z equal to 0, then theequation holds if and only if one of the other two coordinates is 0. Hence, the inter-section of S with the set where xyz = 0 (the union of the coordinate hyperplanes) isthe union of the three coordinate axes.

Hence, let us rephrase the question: Letting U ⊆R3 be the subset where xyz 6= 0,is S∩U is surface? To investigate the problem, consider the function

f (x,y,z) = y2z2 + x2z2 + x2y2− xyz.

The differential (which in this case is the same as the gradient) is the 1×3-matrix

D(x,y,z) f =(

2x(y2 + z2)− yz 2y(z2 + x2)− zx 2z(x2 + y2)− xy)

This vanishes if and only if all three entries are zero. Vanishing of the first entrygives, after dividing by 2xy2z2, the condition

1z2 +

1y2 =

12xyz

;

we get similar conditions after cyclic permutation of x,y,z. Thus we have

6 We won’t cover this example in class, for lack of time, but you’re encouraged to read it.


1z2 +

1y2 =

1x2 +

1z2 =

1y2 +

1x2 =

12xyz

,

with a unique solution x = y = z = 14 . Thus, D(x,y,z) f has maximal rank (i.e., it is

nonzero) except at this point. But this point doesn’t lie on S. We conclude that S∩Uis a submanifold. How does it look like? It turns out that there is a nice answer. First,let’s divide the equation for S∩U by xyz. The equation takes on the form

xyz(1x2 +

1y2 +

1z2 ) = 1. (3.11)

The solution set of (3.11) is contaned in the set of all (x,y,z) such that xyz > 0. Onthis subset, we introduce new variables

α =

√xyzx

, β =

√xyzy

, γ =

√xyzz

;

the old variables x,y,z are recovered as

x = βγ, y = αγ, z = αβ .

In terms of α,β ,γ , Equation (3.11) becomes the equation α2 +β 2 + γ2 = 1. Actu-ally, it is even better to consider the corresponding points

(α : β : γ) = (1x

:1y

:1z) ∈ RP2,

because we could take either square root of xyz (changing the sign of all α,β ,γdoesn’t affect x,y,z). We conclude that the map U → RP2, (x,y,z) 7→ ( 1

x : 1y : 1

z )restricts to a diffeomorphism from S∩U onto

RP2\(α : β : γ)| αβγ = 0.

The image of the map

RP2→ R3, (α : β : γ) 7→ 1|α|2 + |β |2 + |γ|2

(βγ,αβ ,αγ).

is called Steiner’s surface, even though it is not a submanifold (not even an immersedsubmanifold). Here is a picture: 7

7 Source: http://upload.wikimedia.org/wikipedia/commons/7/7a/RomanSurfaceFrontalView.PNG

http://upload.wikimedia.org/wikipedia/commons/7/7a/RomanSurfaceFrontalView.PNG

http://upload.wikimedia.org/wikipedia/commons/7/7a/RomanSurfaceFrontalView.PNG

64 3 Smooth maps

Note that the subset of RP2 defined by αβγ = 0 is a union of three RP1 ∼= S1,each of which maps into a coordinate axis (but not the entire coordinate axis).For example, the circle defined by α = 0 maps to the set of all (0,0,z) with− 1

2 ≤ z≤ 12 . In any case, S is the Steiner surface together with the three coordinate

axes. See http://www.math.rutgers.edu/courses/535/535-f02/pictures/romancon.jpg for a very nice picture.

Example 3.23. Let S⊆ R4 be the solution set of

y2x2 + x2z2 + x2y2 = xyz, y2x2 +2x2z2 +3x2y2 = xyzw.

Again, this cannot quite be a surface because it contains the coordinate axes forx,y,z. Closer investigation shows that S is the union of the three coordinate axes,together with the image of an injective map

RP2→ R4, (α : β : γ) 7→ 1α2 +β 2 + γ2 (βγ,αβ ,αγ,α2 +2β

2 +3γ2).

It turns out (see Section 4.2.4 below) that the latter is a submanifold, which realizesRP2 as a surface in R4.

3.5.5 Immersions

We next consider maps F : M→ N of maximal rank between manifolds of dimen-sions m≤ n. Once again, such a map can be put into a ‘normal form’: By choosingsuitable coordinates it becomes linear.

Proposition 3.8. Suppose F ∈C∞(U,V ) is a smooth map between open subsets U ⊆Rm and V ⊆Rn, and suppose p∈U is such that the derivative DpF is injective. Thenthere exist smaller neighborhoods U1⊆U of p and V1⊆V of F(p), with F(U1)⊆V1,and a diffeomorphism χ : V1→ χ(V1), such that

(χ F)(u) = (u,0) ∈ Rm×Rn−m

http://www.math.rutgers.edu/courses/535/535-f02/pictures/romancon.jpg

http://www.math.rutgers.edu/courses/535/535-f02/pictures/romancon.jpg


Proof. Since DpF is injective, it has m linearly independent rows. By re-indexingthe rows (which amounts to a change of coordinates on V ), we may assume thatthese are the first m rows.

That is, writing

DpF =

(AC

)where A is the m×m-matrix formed by the first m rows and C is the (n−m)×m-matrix formed by the last n−m rows, the square matrix A is invertible. Consider themap

H : U×Rn−m→ Rn, (x,y) 7→ F(x)+(0,y)

Then

D(p,0)H =

(A 0C In−m

)is invertible. Hence, by the inverse function theorem for Rn, H is a diffeomorphismfrom some neighborhood of (p,0) in U ×Rn−m onto some neighborhood V1 ofH(p,0) = F(p), which we may take to be contained in V . Let

χ : V1→ χ(V1)⊆U×Rn−m

be the inverse; thus(χ H)(x,y) = (x,y)

for all (x,y) ∈ χ(V1). Replace U with the smaller open neighborhood

U1 = F−1(V1)∩U

of p. Then F(U1)⊆V1, and

(χ F)(u) = (χ H)(u,0) = (u,0)

66 3 Smooth maps

for all u ∈U1. ut

The manifolds version reads as follows:

Theorem 3.5. Let F ∈C∞(M,N) be a smooth map between manifolds of dimensionsm ≤ n, and p ∈ M a point with rankp(F) = m. Then there are coordinate charts(U,ϕ) around p and (V,ψ) around F(p) such that F(U)⊆V and

(ψ F ϕ−1)(u) = (u,0).

In particular, F(U)⊆ N is a submanifold of dimension m.

Proof. Once again, this is proved by introducing charts around p, F(p) to reduceto a map between open subsets of Rm, Rn, and then use the multivariable version ofthe result to obtain a change of coordinates, putting the map into normal form. ut

Definition 3.9. A smooth map F : M→N is an immersion if rankp(F) = dimM forall p ∈M.

Example 3.24. Let J ⊆ R be an open interval. A smooth map γ : J → M is alsocalled a smooth curve. We see that the image of γ is an immersed submanifold,provided that rankp(γ) = 1 for all p ∈ M. In local coordinates (U,ϕ), this meansthat d

dt (ϕ γ)(t) 6= 0 for all t with γ(t) ∈U . For example, the curve γ(t) = (t2, t3)fails to have this property at t = 0.

Example 3.25 (Figure eight). The map

γ : R→ R2, t 7→(

sin(t),sin(2t))

is an immersion; the image is a figure eight.

(Indeed, for all t ∈ R we have Dtγ ≡ γ(t) 6= 0.)

Example 3.26 (Immersion of the Klein bottle). The Klein bottle admits a ‘figureeight’ immersion into R3, obtained by taking the figure eight in the x− z-plane,moving in the x-direction by R > 1, and then rotating about the z-axis while at thesame time rotating the figure eight, so that after a full turn ϕ 7→ ϕ + 2π the figureeight has performed a half turn. 8

8 Picture source: http://en.wikipedia.org/wiki/Klein_bottle

http://en.wikipedia.org/wiki/Klein_bottle


We can regard this procedure as a composition of the following maps:

F1 : (t,ϕ) 7→ (sin(t),sin(2t),ϕ) = (u,v,ϕ),

F2 : (u,v,ϕ) 7→(

ucos(ϕ

2)+ vsin(

ϕ

2), vcos(

ϕ

2)−usin(

ϕ

2),ϕ)= (a,b,ϕ)

F3 :(a,b,ϕ) 7→ ((a+R)cosϕ, (a+R)sinϕ, b

)= (x,y,z).

Here F1 is the figure eight in the u− v-plane (with ϕ just a bystander). F2 rotatesthe u− v-plane as it moves in the direction of ϕ , by an angle of ϕ/2; thus ϕ = 2π

corresponds to a half-turn. The map F3 takes this family of rotating u,v-planes, andwraps it around the circle in the x−y-plane of radius R, with ϕ now playing the roleof the angular coordinate.

The resulting map F = F3 F2 F1 : R2 → R3 is given by F(t,ϕ) = (x,y,z),where with

x =(

R+ cos(ϕ

2)sin(t)+ sin(

ϕ

2)sin(2t)

)cosϕ,

y =(

R+ cos(ϕ

2)sin(t)+ sin(

ϕ

2)sin(2t)

)sinϕ,

z = cos(ϕ

2)sin(2t)− sin(

ϕ

2)sin(t)

is an immersion. To verify that this is an immersion, it would be cumbersome towork out the Jacobian matrix directly. It is much easier to use that F is obtained asa composition F = F3 F2 F1 of the three maps considered above, where F1 is animmersion, F2 is a diffeomorphism, and F3 is a local diffeomorphism from the opensubset where |a|< R onto its image.

Since the right hand side of the equation for F does not change under the trans-formations

(t,ϕ) 7→ (t +2π,ϕ), (t,ϕ) 7→ (−t,ϕ +2π),

this descends to an immersion of the Klein bottle. It is straightforward to checkthat this immersion of the Klein bottle is injective, except over the ‘central circle’corresponding to t = 0, where it is 2-to-1.

Note that under the above construction, any point of the figure eight creates acircle after two‘full turns’, ϕ 7→ ϕ +4π . The complement of the circle generated bythe point t = π/2 consists of two subsets of the Klein bottle, generated by the parts

68 3 Smooth maps

of the figure eight defined by −π/2 < t < π/2, and by π/2 < t < 3π/2. Each ofthese is a ‘curled-up’ immersion of an open Mobius strip. (Remember, it is possibleto remove a circle from the Klein bottle to create two Mobius strips!) The point t = 0also creates a circle; its complement is the subset of the Klein bottle generated by0< t < 2π . (Remember, it is possible to remove a circle from a Klein bottle to createone Mobius strip.) We can also remove one copy of the figure eight itself; then the‘rotation’ no longer matters and the complement is an open cylinder. (Remember, itis possible to remove a circle from a Klein bottle to create a cylinder.)

Example 3.27. Let M be a manifold, and S⊆M a k-dimensional submanifold. Thenthe inclusion map ι : S→M, x 7→ x is an immersion. Indeed, if (V,ψ) is a subman-ifold chart for S, with p ∈U =V ∩S, ϕ = ψ|V∩S then

(ψ F ϕ−1)(u) = (u,0),

which shows that

rankp(F) = rankϕ(p)(ψ F ϕ−1) = k.

By an embedding, we will mean an immersion given as the inclusion map for asubmanifold. Not every injective immersion is an embedding; the following picturegives a counter-example:

In practice, showing that an injective smooth map is an immersion tends to be easierthan proving that its image is a submanifold. Fortunately, for compact manifolds wehave the following fact:

Theorem 3.6. If M is a compact manifold, then every injective immersion F : M→N is an embedding as a submanifold S = F(M).

Proof. Let p ∈ M be given. By Theorem 3.5, we can find charts (U,ϕ) around pand (V,ψ) around F(p), with F(U) ⊆ V , such that F = ψ F ϕ−1 is in normalform: i.e., F(u) = (u,0). We would like to take (V,ψ) as a submanifold chart forS = F(M), but this may not work yet since F(M)∩V = S∩V may be strictly largerthan F(U)∩V . Note however that A := M\U is compact, hence its image F(A) iscompact, and therefore closed (here we are using that N is Haudorff). Since F isinjective, we have that p 6∈ F(A). Replace V with the smaller open neighborhoodV1 =V\(V ∩F(A)). Then (V1,ψ|V1) is the desired submanifold chart.

Remark 3.10. Some authors refer to injective immersions ι : S→ M as ‘subman-ifolds’ (thus, a submanifold is taken to be a map rather than a subset). To clarify,

3.6 Appendix: Algebras 69

‘our’ submanifolds are sometimes called ‘embedded submanifolds’ or ‘regular sub-manifolds’.

Example 3.28. Let A,B,C be distinct real numbers. We will leave it as a homeworkproblem to verify that the map

F : RP2→ R4, (α : β : γ) 7→ (βγ,αγ,αβ , Aα2 +Bβ

2 +Cγ2),

where we use representatives (α,β ,γ) such that α2 +β 2 + γ2 = 1, is an injectiveimmersion. Hence, by Theorem 3.6, it is an embedding of RP2 as a submanifold ofR4.

To summarize the outcome from the last few sections: If F ∈C∞(M,N) has max-imal rank near p ∈M, then one can always choose local coordinates around p andaround F(p) such that the coordinate expression of F becomes a linear map of maxi-mal rank. (This simple statement contains the inverse and implicit function theoremsfrom multivariable calculus are special cases.)

Remark 3.11. This generalizes further to maps of constant rank. In fact, if rankp(F)is independent of p on some open subset U , then for all p ∈ U one can choosecoordinates in which F becomes linear.

3.6 Appendix: Algebras

An algebra (over the field R of real numbers) is a vector space A , together with amultiplication (product) A ×A →A , (a,b) 7→ ab such that

1. The multiplication is associative: That is, for all a,b,c ∈A

(ab)c = a(bc).

2. The multiplication map is linear in both arguments: That is,

(λ1a1 +λ2a2)b = λ1(a1b)+λ2(a2b),

a(µ1b1 +µ2b2) = µ1(ab1)+µ2(ab2),

70 3 Smooth maps

for all a,a1,a2,b,b1,b2 ∈A and all scalars λ1,λ2,µ1,µ2 ∈ R.

The algebra is called commutative if ab = ba for all a,b ∈ A . A unital algebra isan algebra A with a distinguished element 1A ∈A (called the unit), with

1A a = a = a1A

for all a ∈A .

Remark 3.12. One can also consider non-associative product operations on vectorspaces, most importantly one has the class of Lie algebras. If there is risk of confu-sion with these or other concepts, we may refer to associative algebras.

For example, the space C of complex numbers (regarded as a real vector spaceR2) is a unital, commutative algebra. A more sophisticated example is the alge-bra H ∼= R4 of quaternions, which is a unital non-commutative algebra. (Recallthat elements of H are expressions x + iu + jv + kw, where i, j,k have productsi2 = j2 = k2 = −1, i j = k = − ji, jk = i = −k j, ki = j = −ik.) For any n, thespace MatR(n) of n× n matrices, with product the matrix multiplication, is a non-commutative unital algebra. One can also consider matrices with coefficients in C,or in fact with coefficients in any given algebra. For any set X , the space of func-tions f : X → R is a unital commutative algebra, where the product is given bypointwise multiplication. Given a topological space X , one has the algebra C(X) ofcontinuous R-valued functions. A homomorphism of algebras Φ : A → A ′ is alinear map preserving products: Φ(ab) = Φ(a)Φ(b). (For a homomorphism of uni-tal algebras, one asks in addition that Φ(1A ) = 1A ′ .) It is called an isomorphismof algebras if Φ is invertible. For the special case A ′ = A , these are also calledalgebra automorphisms of A . Note that the algebra automorphisms form a groupunder composition.

Example 3.29. Consider R2 as an algebra, with product coming from the identifica-tion R2 =C. The complex conjugation z 7→ z defines an automorphism Φ : R2→R2

of this algebra.

Example 3.30. The algebra H of quaternions has an automorphism given by cyclicpermutation of the three imaginary units. That is, Φ(x+ iu+ jv+ kw) = x+ ju+kv+ iw

Example 3.31. Let A = MatR(n) the algebra of n×n-matrices. If U ∈A is invert-ible, then X 7→Φ(X) =UXU−1 is an algebra automorphism.

Example 3.32. Suppose A is a unital algebra. Let A × be the set of invertible ele-ments, that is, elements u ∈ A for which there exists v ∈ A with uv = vu = 1A .Given u, such v is necessarily unique (write v = u−1), and the map A → A , a 7→uau−1 is an algebra automorphism. Such automorphisms are called ‘inner’.

Chapter 4The tangent bundle

4.1 Tangent spaces

For embedded submanifolds M⊆Rn, the tangent space TpM at p∈M can be definedas the set of all velocity vectors v = γ(0), where γ : J→M is a smooth curve withγ(0) = p; here J ⊆ R is an open interval around 0.

It turns out (not entirely obvious!) that TpM becomes a vector subspace of Rn.(Warning: In pictures we tend to draw the tangent space as an affine subspace, wherethe origin has been moved to p.)

Example 4.1. Consider the sphere Sn ⊆Rn+1, given as the set of x such that ||x||2 =1. A curve γ(t) lies in Sn if and only if ||γ(t)|| = 1. Taking the derivative of theequation γ(t) · γ(t) = 1 at t = 0, we obtain (after dividing by 2, and using γ(0) = p)

p · γ(0) = 0.

That is, TpM consists of vectors v ∈ Rn+1 that are orthogonal to p ∈ R3\0. It isnot hard to see that every such vector v is of the form γ(0),1 hence that

TpSn = (Rp)⊥,

the hyperplane orthogonal to the line through p.

1 Given v, take γ(t) = (p+ tv)/||p+ tv||.

71

72 4 The tangent bundle

To extend this idea to general manifolds, note that the vector v = γ(0) defines a“directional derivative” C∞(M)→ R:

v : f 7→ ddt |t=0 f (γ(t)).

For a general manifold, we will define TpM as a set of directional derivatives.

Definition 4.1 (Tangent spaces – first definition). Let M be a manifold, p ∈ M.The tangent space TpM is the set of all linear maps v : C∞(M)→ R of the form

v( f ) = ddt |t=0 f (γ(t))

for some smooth curve γ ∈C∞(J,M) with γ(0) = p.

The elements v ∈ TpM are called the tangent vectors to M at p.The following local coordinate description makes it clear that TpM is a linear

subspace of the vector space L(C∞(M),R) of linear maps C∞(M)→R, of dimensionequal to the dimension of M.

Theorem 4.1. Let (U,ϕ) be a coordinate chart around p. A linear map v : C∞(M)→R is in TpM if and only if it has the form,

v( f ) =m

∑i=1

ai ∂ ( f ϕ−1)

∂ui

∣∣∣u=ϕ(p)

for some a = (a1, . . . ,am) ∈ Rm.

Proof. Given a linear map v of this form, let γ : R→ ϕ(U) be a curve with γ(t) =ϕ(p)+ ta for |t| sufficiently small. Let γ = ϕ−1 γ . Then

ddt

∣∣∣t=0

f (γ(t)) =ddt

∣∣∣t=0

( f ϕ−1)(ϕ(p)+ ta)

=m

∑i=1

ai ∂ ( f ϕ−1)

∂ui

∣∣∣u=ϕ(p)

,

by the chain rule. Conversely, given any curve γ with γ(0) = p, let γ = ϕ γ be thecorresponding curve in ϕ(U) (defined for small |t|). Then γ(0) = ϕ(p), and

ddt

∣∣∣t=0

f (γ(t)) =ddt

∣∣∣t=0

( f ϕ−1)(γ(t))

=m

∑i=1

ai ∂ ( f ϕ−1)

∂ui |u=γ(p),

where a = dγ

dt

∣∣∣t=0

. ut

We can use this result as an alternative definition of the tangent space, namely:

4.1 Tangent spaces 73

Definition 4.2 (Tangent spaces – second definition). Let (U,ϕ) be a chart aroundp. The tangent space TpM is the set of all linear maps v : C∞(M)→ R of the form

v( f ) =m

∑i=1

ai ∂ ( f ϕ−1)

∂ui

∣∣∣u=ϕ(p)

(4.1)

for some a = (a1, . . . ,am) ∈ Rm.

Remark 4.1. From this version of the definition, it is immediate that TpM is an m-dimensional vector space. It is not immediately obvious from this second definitionthat TpM is independent of the choice of coordinate chart, but this follows from theequivalence with the first definition. Alternatively, one may check directly that thesubspace of L(C∞(M),R) characterized by (4.1) does not depend on the chart, bystudying the effect of a change of coordinates.

According to (4.1), any choice of coordinate chart (U,ϕ) around p defines a vectorspace isomorphism TpM ∼= Rm, taking v to a = (a1, . . . ,am). In particular, we seethat if U ⊆ Rm is an open subset, and p ∈U , then TpU is the subspace of the spaceof linear maps C∞(M)→R spanned by the partial derivatives at p. That is, TpU hasa basis

∂

∂x1 |p, . . . ,∂

∂xm |p

identifying TpU ≡ Rm. Given

v = ∑ai ∂

∂xi |p

the coefficients ai are obtained by applying v to the coordinate functions x1, . . . ,xm :U → R, that is, ai = v(xi).

We now describe yet another approach to tangent spaces which again charac-terizes “directional derivatives” in a coordinate-free way, but without reference tocurves γ . Note first that every tangent vector satisfies the product rule, also calledthe Leibniz rule:

Lemma 4.1. Let v ∈ TpM be a tangent vector at p ∈M. Then

v( f g) = f (p)v(g)+ v( f )g(p) (4.2)

for all f ,g ∈C∞(M).

Proof. Letting v be represented by a curve γ , this follows from

ddt

∣∣∣t=0

(f(γ(t)

)g(γ(t)

))= f (p)

( ddt

∣∣∣t=0

g(γ(t)

))+( d

dt

∣∣∣t=0

f(γ(t)

))g(p).

ut

Alternatively, in local coordinates it is just the product rule for partial derivatives. Itturns out that the product rule completely characterizes tangent vectors:


Theorem 4.2. A linear map v : C∞(M)→ R defines an element of TpM if and onlyif it satisfies the product rule (4.2).

The proof of this result will require the following fact from multivariable calculus:

Lemma 4.2 (Hadamard Lemma). Let U = BR(0)⊆ Rm be an open ball of radiusR > 0 and h ∈ C∞(U) a smooth function. Then there exist smooth functions hi ∈C∞(U) with

h(u) = h(0)+m

∑i=1

uihi(u)

for all u ∈U. Here hi(0) = ∂h∂ui (0).

Proof. Let hi be the functions defined for u = (u1, . . . ,um) ∈U by

hi(u) =

1ui

(h(u1, . . . ,ui,0, . . . ,0)−h(u1, . . . ,ui−1,0,0, . . . ,0)

)if ui 6= 0

∂h∂ui (u1, . . . ,ui−1,0,0, . . . ,0) if ui = 0

Using Taylor’s formula with remainder, one sees that these functions are smooth.2 If all ui 6= 0, then the sum ∑

mi=1 uihi(u) is a telescoping sum, equal to h(u)−h(0).

By continuity, this result extends to all u. Finally, evaluating the derivative

∂h∂ui = hi(u)+∑

kuk ∂hk

∂ui

at u = 0, we see that ∂h∂ui

∣∣u=0 = hi(0). ut

Proof (Theorem 4.2). Let v : C∞(M)→ R be a linear map satisfying the productrule (4.2).

Step 1: v vanishes on constants.

By the product rule, applied to the constant function 1 = 1 ·1, we have v(1) = 0.Thus v vanishes on constants.

Step 2: If f1 = f2 on some open neighborhood U of p, then v( f1) = v( f2).

Equivalently, letting f = f1− f2, we show that v( f ) = 0 if f = 0 on U . Choosea ‘bump function’ χ ∈C∞(M) with χ(p) = 1, with χ|M\U = 0. Then f χ = 0. Theproduct rule tells us that

0 = v( f χ) = v( f )χ(p)+ v(χ) f (p) = v( f ).

Step 3: If f (p) = g(p) = 0, then v( f g) = 0.

2 It is a well-known fact from calculus (proved e.g. by using Taylor’s theorem with remainder) thatif f is a smooth function of a real variable x, then the function g, defined as g(x) = x−1( f (x)− f (0))for x 6= 0 and g(0) = f ′(0), is smooth.

4.1 Tangent spaces 75

This is immediate from the product rule.Step 4: Let (U,ϕ) be a chart around p, with image U = ϕ(U). Then there is

unique linear map v : C∞(U)→ R such that v( f ) = v( f ) whenever f agrees withf ϕ−1 on some neighborhood of p.

Given f , we can always find a function f such that f agrees with f ϕ−1 onsome neighborhood of p. Given another such function f ′, it follows from Step 2that v( f ) = v( f ′).

Step 5: In a chart (U,ϕ) around p, the map v : C∞(M)→R is of the form (4.1).

Since the condition (4.1) does not depend on the choice of chart around p, wemay assume that p = ϕ(p) = 0, and that U is an open ball of some radius R > 0around 0. Define v as in Step 4. Since v satisfies the product rule on C∞(M), themap v satisfies the product rule on C∞(U). Given f ∈ C∞(M), consider the Taylorexpansion of the coordinate expression f = f ϕ−1 near u = 0:

f (u) = f (0)+∑i

ui ∂ f∂ui

∣∣∣u=0

+ r(u)

The remainder term r is a smooth function that vanishes at u = 0 together with itsfirst derivatives. By Lemma 4.2, it can be written in the form r(u) =∑i uiri(u) whereri are smooth functions that vanish at 0. Let us now apply v to the formula for f .Since v vanishes on products of functions vanishing at 0 (by Step 3), we have thatv(r) = 0. Since it also vanishes on constants (by Step 1), we obtain

v( f ) = v( f ) = ∑i

ai ∂ f∂ui

∣∣∣u=0

,

where we put ai = v(ui).

To summarize, we have the following alternative definition of tangent spaces:

Definition 4.3 (Tangent spaces – third definition). The tangent space TpM is thespace of linear maps C∞(M)→ R satisfying the product rule,

v( f g) = f (p)v(g)+ v( f )g(p)

for all f ,g ∈C∞(M).

At first sight, this characterization may seem a bit less intuitive then the defini-tion as directional derivatives along curves. But it has the advantage of being lessredundant – a tangent vector may be represented by many curves. Also, as in the co-ordinate definition it is immediate that TpM is a linear subspace of the vector spaceL(C∞(M),R). One may still want to use local charts, however, to prove that thisvector subspace has dimension equal to the dimension of M.

The following remark gives yet another characterization of the tangent space.Please read it only if you like it abstract – otherwise skip this!


Remark 4.2 (A fourth definition). There is a fourth definition of TpM, as follows.For any p ∈M, let C∞

p (M) denotes the subspace of functions vanishing at p, and letC∞

p (M)2 consist of finite sums ∑i fi gi where fi,gi ∈C∞p (M). We have a direct sum

decompositionC∞(M) = R⊕C∞

p (M),

where R is regarded as the constant functions. Since any tangent vector v : C∞(M)→R vanishes on constants, v is effectively a map v : C∞

p (M)→R. By the product rule,v vanishes on the subspace C∞

p (M)2 ⊆ C∞p (M). Thus v descends to a linear map

C∞p (M)/C∞

p (M)2 → R, i.e. an element of the dual space (C∞p (M)/C∞

p (M)2)∗. Themap

TpM→ (C∞p (M)/C∞

p (M)2)∗

just defined is an isomorphism, and can therefore be used as a definition of TpM.This may appear very fancy on first sight, but really just says that a tangent vectoris a linear functional on C∞(M) that vanishes on constants and depends only onthe first order Taylor expansion of the function at p. Furthermore, this viewpointlends itself to generalizations which are relevant to algebraic geometry and non-commutative geometry: The ‘vanishing ideals’ C∞

p (M) are the maximal ideals inthe algebra of smooth functions, with C∞

p (M)2 their second power (in the sense ofproducts of ideals). Thus, for any maximal ideal I in a commutative algebra Aone may regard (I /I 2)∗ as a ‘tangent space’.

After this lengthy discussion of tangent spaces, observe that the velocity vectorsof curves are naturally elements of the tangent space. Indeed, let J ⊆ R be an openinterval, and γ ∈ C∞(J,M) a smooth curve. Then for any t0 ∈ J, the tangent (orvelocity) vector

γ(t0) ∈ Tγ(t0)M.

at time t0 is given in terms of its action on functions by

(γ(t0))( f ) =ddt

∣∣∣t=t0

f (γ(t))

We will also use the notation dγ

dt (t0) or dγ

dt |t0 to denote the velocity vector.

4.2 Tangent map

4.2.1 Definition of the tangent map, basic properties

For smooth maps F ∈ C∞(U,V ) between open subsets U ⊆ Rm and V ⊆ Rn ofEuclidean spaces, and any given p∈U , we considered the derivative to be the linearmap

DpF : Rm→ Rn, a 7→ ddt

∣∣∣t=0

F(p+ ta).

4.2 Tangent map 77

The following definition generalizes the derivative to smooth maps between mani-folds.

Definition 4.4. Let M,N be manifolds and F ∈C∞(M,N). For any p∈M, we definethe tangent map to be the linear map

TpF : TpM→ TF(p)N

given by (TpF(v)

)(g) = v(gF)

for v ∈ TpM and g ∈C∞(N).

We leave it as an exercise to check that the right hand side does indeed define atangent vector:

Exercise: Show that for all v ∈ TpM, the map g 7→ v(gF) satisfies the product ruleat q = F(p), hence defines an element of TqN.

Proposition 4.1. If v ∈ TpM is represented by a curve γ : J→M, then (TpF)(v) isrepresented by the curve F γ .

Proof. For g ∈C∞(N),

TpF(v)(g) = v(gF) =ddt

∣∣∣t=0

(gF)(γ(t)) =ddt

∣∣∣t=0

g((F γ)(t)

).

This shows that TpF(v) is represented by F γ : R→ N.

Remark 4.3 (Pull-backs, push-forwards). For smooth maps F ∈C∞(M,N), one canconsider various ‘pull-backs’ of objects on N to objects on M, and ‘push-forwards’of objects on M to objects on N. Pull-backs are generally denoted by F∗, push-forwards by F∗. For example, functions on N pull back

g ∈C∞(N) F∗g = gF ∈C∞(M).

Curves push on M forward:

γ : J→M F∗γ = F γ : J→ N.

Tangent vectors to M also push forward,

v ∈ TpM F∗(v) = (TpF)(v).

The definition of the tangent map can be phrased in these terms as (F∗v)(g) =v(F∗g). Note also that if v is represented by the curve γ , then F∗v is representedby the curve F∗γ .

Proposition 4.2 (Chain rule). Let M,N,Q be manifolds. Under composition ofmaps F ∈C∞(M,N) and F ′ ∈C∞(N,Q),


Tp(F ′ F) = TF(p)F′ TpF.

Proof. Let v ∈ TpM be represented by a curve γ . Then both Tp(F ′ F)(v) andTF(p)F ′(TpF(v)) are represented by the curve F ′ (F γ) = (F ′ F) γ . ut

Exercise: a) Show that the tangent map of the identity map idM : M→M at p ∈Mis the identity map on the tangent space:

Tp idM = idTpM.

b) Show that if F ∈C∞(M,N) is a diffeomorphism, then TpF is a linear isomor-phism, with inverse

(TpF)−1 = (TF(p)F−1).

4.2.2 Coordinate description of the tangent map

To get a better understanding of the tangent map, let us first consider the spacialcase that F ∈C∞(U,V ) is a smooth map between open subsets U ⊆Rm and V ⊆Rn.For p ∈U , the tangent space TpU is canonically identified with Rm, using the basis

∂

∂x1

∣∣∣p, . . . ,

∂

∂xm

∣∣∣p∈ TpU

of the tangent space. Similarly, TF(p)V ∼=Rn, using the basis given by partial deriva-tives ∂

∂y j |F(p). Using this identifications, the tangent map becomes a linear mapTpF : Rm → Rn, i.e. it is given by an n×m-matrix. This matrix is exactly theJacobian:

Proposition 4.3. Let F ∈C∞(U,V ) is a smooth map between open subsets U ⊆ Rm

and V ⊆Rn. For all p∈M, the tangent map TpF is just the derivative (i.e., Jacobianmatrix) DpF of F at p.

Proof. For g ∈C∞(V ), we calculate((TpF)

( ∂

∂xi

∣∣∣p

))(g) =

∂

∂xi

∣∣∣p(gF)

=n

∑j=1

∂g∂y j

∣∣∣F(p)

∂F j

∂xi

∣∣∣p

=( n

∑j=1

∂F j

∂xi

∣∣∣p

∂

∂y j

∣∣∣F(p)

)(g).

This shows

(TpF)( ∂

∂xi

∣∣∣p

)=

n

∑j=1

∂F j

∂xi

∣∣∣p

∂

∂y j

∣∣∣F(p)

.

4.2 Tangent map 79

Hence, in terms of the given bases of TpU and TF(p)V , the matrix of the linear map

TpF has entries ∂F j

∂xi

∣∣∣p.

Remark 4.4. For F ∈ C∞(U,V ), it is common to write y = F(x), and accordinglywrite ( ∂y j

∂xi )i, j for the Jacobian. In these terms, the derivative reads as

TpF(

∂

∂xi

∣∣∣p

)= ∑

j

∂y j

∂xi

∣∣∣p

∂

∂y j

∣∣∣F(p)

.

This formula is often used for explicit calculations.

For a general smooth map F ∈ C∞(M,N), we obtain a similar description oncewe pick coordinate charts. Given p ∈M, choose charts (U,ϕ) around p and (V,ψ)around F(p), with F(U)⊆V . Let U = ϕ(U), V = ψ(V ), and put

F = ψ F ϕ−1 : U → V .

Since the coordinate map ϕ : U → Rm is a diffeomorphism onto U , It gives anisomorphism

Tpϕ : TpU → Tϕ(p)U = Rm.

Similarly, TF(p)ψ gives an isomorphism of TF(p)V with Rn. Note also that sinceU ⊆M is open, we have that TpU = TpM. We obtain,

Tϕ(p)F = TF(p)ψ TpF (Tpϕ)−1.

which may be depicted in a commutative diagram

RmDϕ(p)F

// Rn

TpM = TpU

∼=Tpϕ

OO

TpF// TF(p)V = TF(p)N

∼= TF(p)ψ

OO

Now that we have recognized TpF as the derivative expressed in a coordinate-free way, we may liberate some of our earlier definitions from coordinates:

Definition 4.5. Let F ∈C∞(M,N).

• The rank of F at p ∈M, denoted rankp(F), is the rank of the linear map TpF .• F has maximal rank at p if rankp(F) = min(dimM,dimN).• F is a submersion if TpF is surjective for all p ∈M,• F is an immersion if TpF is injective for all p ∈M,• F is a local diffeomorphism if TpF is an isomorphism for all p ∈M.• p ∈M is a critical point of F is TpF does not have maximal rank at p.• q∈N is a regular value of F if TpF is surjective for all p∈ F−1(q) (in particular,

if q 6∈ F(M)).


• q ∈ N is a singular value if it is not a regular value.

Exercise: Using this new definitions, show that the compositions of two submer-sions is again a submersion, and that the composition of two immersions is an im-mersion.

4.2.3 Tangent spaces of submanifolds

Suppose S ⊆M is a submanifold, and p ∈ S. Then the tangent space TpS is canon-ically identified as a subspace of TpM. Indeed, since the inclusion i : S →M is animmersion, the tangent map is an injective linear map,

Tpi : TpS→ TpM,

and we identify TpS with the subspace given as the image of this map. (Hopefully,the identifications are not getting too confusing: S gets identified with i(S) ⊆ M,hence also p ∈ S with its image i(p) in M, and TpS gets identified with (Tpi)(TpS)⊆TpM.) As a special case, we see that whenever M is realized as a submanifold of Rn,then its tangent spaces TpM may be viewed as subspaces of TpRn = Rn.

Proposition 4.4. Let F ∈ C∞(M,N) be a smooth map, having q ∈ N as a regularvalue, and let S = F−1(q). For all p ∈ S,

TpS = ker(TpF),

as subspaces of TpM.

Proof. Let m = dimM, n = dimN. Since TpF is surjective, its kernel has dimensionm− n. By the normal form for submersions, this is also the dimension of S, henceof TpS. It is therefore enough to show that TpS⊆ ker(TpF). Letting i : S→M be heinclusion, we have to show that

TpF Tpi = Tp(F i)

is the zero map. But F i is a constant map, taking all points of S to the constantvalue q ∈ N. The tangent map to a constant map is just zero. (See below.) HenceTp(F i) = 0. ut

Exercise: Suppose that F ∈ C∞(M,N) is a constant map, that is, F(M) = q forsome element q ∈ N. Show that TpF = 0 for all p ∈ M. (Hint: Use the definitionof TpF , and observe that for g ∈ C∞(N) the pull-back F∗g = g F is a constantfunction.)

As a special case, we can describe the tangent spaces to level sets:

4.2 Tangent map 81

Corollary 4.1. Suppose V ⊆ Rn is open, and q ∈ Rk is a regular value of F ∈C∞(M,Rk), defining an embedded submanifold M = F−1(q). For all p ∈ M, thetangent space TpM ⊆ TpRn = Rn is given as

TpM = ker(TpF)≡ ker(DpF).

Example 4.2. Let F : Rn+1→R be the map F(x) = x ·x = (x0)2+ . . .+(xn)2. Then,for all p ∈ F−1(1) = Sn,

(DpF)(a) =ddt

∣∣∣t=0

F(p+ ta) =ddt

∣∣∣t=0

(p+ ta) · (p+ ta) = 2p ·a,

henceTpSn = a ∈ Rn+1| a · p = 0= span(p)⊥.

As another typical application, suppose that S ⊆ M is a submanifold, and f ∈C∞(S) is a smooth function given as the restriction f = h|S of a smooth functionh ∈C∞(M). Consider the problem of finding the critical points p ∈ S of f , that is,

Crit( f ) = p ∈ S| Tp f = 0.

Letting i : S→M be the inclusion, we have f = h|S = h i, hence Tp f = TphTpi.It follows that Tp f = 0 if and only if Tph vanishes on the range of Tpi, that is on TpS:

Crit( f ) = p ∈ S| TpS⊆ ker(Tph).

If M =Rm, then Tph is just the Jacobian Dph, whose kernel is sometimes rather easyto compute – in any case this approach tends to be much faster than a calculation incharts. Here is a concrete example:

Example 4.3. Problem. Find the critical points of

f : S2→ R, f (x,y,z) = xy.

Solution. Following the strategy outlined above, we write f = h i with h(x,y,z) =xy. For p = (x,y,z) we have

Tph = Dph = (y x 0),

as a linear map R3→ R. There are two cases: Case 1. Dph = 0, i.e. x = y = 0. Thismeans p = (0,0,±1). In this case ker(Tph) = R3, which of course contains TpS.Thus both

(0,0,±1) (4.3)

are critical points of f .

Case 2. Dph 6= 0. Then


ker(Tph) = span

0

01

,

x−y0

.

This contains TpS if and only if it is equal to TpS. To check whether the two basisvectors are in TpS, we just have to check their dot products with p = (x,y,z). Thisgives the conditions z = 0 and x2− y2 = 0, which together with x2 + y2 + z2 = 1leads to the four critical points

(± 1√2, ± 1√

2, 0). (4.4)

In summary, the function F has six critical points, corresponding to two sign choicesin (4.3) and four sign choices in (4.4). ut

As another application of the same idea, you should try to prove:

Exercise: Let S⊆R3 be a surface. Show that p ∈ S is a critical point of the functionf ∈C∞(S) given by f (x,y,z) = z, if and only if TpS is the x-y-plane.

Example 4.4. Problem. Show that the equations

x2 + y = 0, x2 + y2 + z3 +w4 + y = 1

define a two dimensional submanifold S of R4, and find the equation of the tangentspace at the point (x0,y0,z0,w0) = (−1,−1,−1,−1).

Solution. Let F ∈C∞(R4,R2) be the function

F(x,y,z,w) = (x2 + y, x2 + y2 + z3 +w4 + y).

The Jacobian matrix is (2x 1 0 02x 2y+1 3z2 4w3

).

Since the first row is non-zero, this has rank 2 unless the second row is a scalarmultiple of the first row. This is the case if either x 6= 0 and y = z = w, or x = 0 andz = w = 0. In particular, x = 0 or y = 0. But if such a point (x,y,z,w) also satisfiesthe first equation for S, that is x2 + y = 0, we see that x,y must both be zero. Thisonly leaves the point (0,0,0,0), which however does not solve the second equationx2+y2+z3+w4+y= 1. This shows that S is a submanifold of dimension 4−2= 2.At (−1,−1,−1,−1) the Jacobian matrix becomes(

−2 1 0 0−2 −1 3 −4

);

so the equation of the tangent space TpS = ker(DpF) reads as

4.2 Tangent map 83

(−2 1 0 0−2 −1 3 −4

)xyzw

=

(00

);

that is,−2x+ y = 0, −2x− y+3z−4w = 0.

Example 4.5. We had discussed various matrix Lie groups G as examples of mani-folds. By definition, these are submanifolds G ⊆MatR(n), consisting of invertiblematrices with the properties

A,B ∈ G⇒ AB ∈ G, A ∈ G⇒ A−1 ∈ G.

The tangent space to the identity (group unit) for such matrix Lie groups G turns outto be important; it is commonly denoted by lower case fracture letters:

g= TIG.

Some concrete examples:

1. The matrix Lie group

GL(n,R) = A ∈MatR(n)| det(A) 6= 0

of all invertible matrices is an open subset of MatR(n), hence

gl(n,R) = MatR(n)

is the entire space of matrices.2. For the group O(n), consisting of matrices with F(A) := A>A = I, we has com-

puted TAF(X) = X>A+AX>. For A = I, the kernel of this map is

o(n) = X ∈MatR(n)| X> =−X.

3. For the group SL(n,R) = A ∈ MatR(n)| det(A) = 1, given as the level setF−1(1) of the function det : MatR(n)→ R, we calculate

DAF(X)=ddt

∣∣∣t=0

F(A+tX)=ddt

∣∣∣t=0

det(A+tX)=ddt

∣∣∣t=0

det(I+tA−1X)= tr(A−1X),

where tr : MatR(n)→ R is the trace (sum of diagonal entries). (See exercisebelow.) Hence

sl(n,R) = X ∈MatR(n)| tr(X) = 0.

Exercise: Show that for every X ∈MatR(n),

ddt

∣∣t=0 det(I + tX) = tr(X).


(Hint: Use that every matrix is conjugate (i.e., similar) to an upper triangular matrix,and that both determinant and trace are unchanged under conjugation (i.e., similaritytransformation).)

4.2.4 Example: Steiner’s surface revisited

As we discussed in Section 3.5.4, Steiner’s ‘Roman surface’ is the image of the map

RP2→ R3, (x : y : z) 7→ 1x2 + y2 + z2 (yz, xz, xy).

(We changed notation from α,β ,γ to x,y,z.) At what points p ∈ RP2 does this maphave maximal rank (so that the map is an immersion on an open neighborhoodof p?). To investigate this question, one can express the map in local charts, andcompute the resulting Jacobian matrix. However, while this approach is perfectlyfine, the resulting expressions will become rather complicated. A simpler approachis to consider the composition with the local diffeomorphism π : S2→ RP2, givenas

S2→ R3, (x,y,z) 7→ (yz, xz, xy).

In turn, this map is the restriction F |S2 of the map

F : R3→ R3, (x,y,z) 7→ (yz, xz, xy).

We have Tp(F |S2) = TpF |TpS2 , hence ker(Tp(F |S2)) = ker(TpF)∩TpS2. But TpF =

DpF for p = (x,y,z) is given by the Jacobian matrix

DpF =

0 z yz 0 xy x 0

.

This has determinant det(DpF) = 2xyz, hence its kernel is zero unless x = 0 or y = 0or z = 0. If x = 0, thus p = (0,y,z), the matrix simplifies to

DpF =

0 z yz 0 0y 0 0

,

which (unless both y and z are zero as well) has a 1-dimensional kernel spanned bycolumn vectors of the form (0,−y,z)>. Such a vector is tangent to S2 if and onlyif its dot product with p = (0,y,z) is zero, that is, y2 = z2. Since p ∈ S2 this meansp = (0,± 1√

2,± 1√

2). Similarly if y = 0, or if z = 0. We have thus shown: The map

F |S2 has maximal rank at all points of S2, except at the following twelve points:

4.3 The tangent bundle 85

(0,± 1√2,± 1√

2), (± 1√

2,0,± 1√

2), (± 1√

2,± 1√

2,0).

The kernel of Tp(F |S2) at (0,± 1√2,± 1√

2) is the 1-dimensional space spanned by

(0,± 1√2,∓ 1√

2) (sign change in the last entry), and similarly for the points where

y = 0 or z = 0. We conclude that the map RP2→ R3 defining Steiner’s surface hasexactly six points where it fails to be an immersion, and we have computed thekernel of the tangent map at those points.

4.3 The tangent bundle

Proposition 4.5. For any manifold M of dimension m, the tangent bundle

T M =⊔

p∈M

TpM

(disjoint union of vector spaces) is a manifold of dimension 2m. The map

π : T M→M

taking v ∈ TpM to the base point p, is a smooth submersion, with fibers the tangentspaces.

Proof. The idea is simple: Take charts for M, and use the tangent map to get chartsfor T M. For any open subset U of M, we have

TU =⊔

p∈U

TpM = π−1(U).

(Note TpU = TpM.) Every chart (U,ϕ) for M, with ϕ : U→Rm, gives vector spaceisomorphisms

Tpϕ : TpM→ Tϕ(p)Rm = Rm

for all p ∈U . The collection of all maps Tpϕ for p ∈U gives a bijection,

T ϕ : TU → ϕ(U)×Rm, v 7→ (ϕ(p),(Tpϕ)(v))

for v ∈ TpU ⊆ TU . The image of these bijections are the open subsets subsets

(T ϕ)(TU) = ϕ(U)×Rm ⊆ R2m,

hence they define charts. We take the collection of all such charts as an atlas for T M:


TUT ϕ

//

π

ϕ(U)×Rm

(u,v)7→u

Uϕ

// ϕ(U)

We need to check that the transition maps are smooth. If (V,ψ) is another coordinatechart with U ∩V 6= /0, the transition map for TU ∩TV = T (U ∩V ) = π−1(U ∩V ) isgiven by,

T ψ (T ϕ)−1 : ϕ(U ∩V )×Rm→ ψ(U ∩V )×Rm. (4.5)

But Tpψ (Tpϕ)−1 = Tϕ(p)(ψ ϕ−1) is just the derivative (Jacobian matrix) for thechange of coordinates ψ ϕ−1; hence (4.5) is given by

(x,a) 7→((ψ ϕ

−1)(x), Dx(ψ ϕ−1)(a)

)Since the Jacobian matrix depends smoothly on x, this is a smooth map. This showsthat any atlas A = (Uα ,ϕα) for M defines an atlas (TUα ,T ϕα) for T M. TakingA to be countable the atlas for T M is countable. The Hausdorff property is easilychecked as well. ut

Proposition 4.6. For any smooth map F ∈C∞(M,N), the map

T F : T M→ T N

given on TpM as the tangent maps TpF : TpM→ TF(p)N, is a smooth map.

Proof. Given p ∈M, choose charts (U,ϕ) around p and (V,ψ) around F(p), withF(U) ⊆ V . Then (TU,T ϕ) and (TV,T ψ) are charts for T M and T N, respectively,with T F(TU)⊆ TV . Let F = ψ F ϕ−1 : ϕ(U)→ ψ(V ). The map

T F = T ψ T F (T ϕ)−1 : ϕ(U)×Rm→ ψ(V )×Rn

is given by(x,a) 7→

((F)(x), Dx(F)(a)

).

It is smooth, by smooth dependence of the differential DxF on the base point. Con-sequently, T F is smooth, ut

Chapter 5Vector fields

5.1 Vector fields as derivations

A vector field on a manifold may be regarded as a family of tangent vectors Xp ∈TpM for p ∈M, depending smoothly on the base points p ∈M. One way of makingprecise what is meant by ‘depending smoothly’ is the following.

Definition 5.1 (Vector fields – first definition). A collection of tangent vectorsXp, p∈M defines a vector field X ∈X∈M if and only if for all functions f ∈C∞(M)the function p 7→ Xp( f ) is smooth. The space of all vector fields on M is denotedX(M).

We hence obtain a linear map X : C∞(M)→C∞(M) such that

X( f )|p = Xp( f ). (5.1)

Since each Xp satisfy the product rule (at p), it follows that X itself satisfies a productrule. We can use this as an alternative definition:

Definition 5.2 (Vector fields – second definition). A vector field on M is a linearmap

X : C∞(M)→C∞(M)

satisfying the product rule,

X( f g) = X( f )g+ f X(g) (5.2)

for f ,g ∈C∞(M).

Remark 5.1. The condition (5.2) says that X is a derivation of the algebra C∞(M)of smooth functions. More generally, a derivation of an algebra A is a linear mapD : A→ A such that

D(a1a2) = D(a1) a2 +a1 D(a2).

(Appendix 5.8 reviews some facts about derivations.)

87

88 5 Vector fields

We can also express the smoothness of the tangent vectors Xp in terms of coordi-nate charts (U,ϕ). Recall that for any p ∈U , and all f ∈C∞(M), the tangent vectorXp is expressed as

Xp( f ) =m

∑i=1

ai ∂

∂ui

∣∣∣u=ϕ(p)

( f ϕ−1).

The vector a = (a1, . . . ,am) ∈ Rm represents Xp in the chart; i.e., (Tpϕ)(Xp) = aunder the identification Tϕ(p)ϕ(U) = Rm. As p varies in U , the vector a becomes afunction of p ∈U , or equivalently of u = ϕ(p).

Proposition 5.1. The collection of tangent vectors Xp, p ∈M define a vector field ifand only if for all charts (U,ϕ), the functions ai : ϕ(U)→ R defined by

Xϕ−1(u)( f ) =m

∑i=1

ai(u)∂

∂ui ( f ϕ−1),

are smooth.

Proof. If the ai are smooth functions, then for every f ∈C∞(M) the function X( f )ϕ−1 : ϕ(U)→ R is smooth, and hence X( f )|U is smooth. Since this is true for allcharts, it follows that X( f ) is smooth. Conversely, if X is a vector field, and p ∈Msome point in a coordinate chart (U,ϕ), and i ∈ 1, . . . ,m a given index, choosef ∈C∞(M) such that f (ϕ−1(u)) = ui. Then X( f )ϕ−1 = ai(u), which shows thatthe ai are smooth.

Exercise: In the proof, we used that for any coordinate chart (U,ϕ) around p, onecan choose f ∈C∞(M) such that f ϕ−1 : ϕ(U)→ R coincides with u j near ϕ(p).Write out the details in the construction of such a function f , using a choice of‘bump function’.

In particular, we see that vector fields on open subsets U ⊆ Rm are of the form

X = ∑i

ai ∂

∂xi

where ai ∈C∞(U). Under a diffeomorphism F : U →V, x 7→ y = F(x), the coordi-nate vector fields transform with the Jacobian

T F(∂

∂xi ) = ∑j

∂F j

∂xi

∣∣∣x=F−1(y)

∂

∂y j

Informally, this ‘change of coordinates’ is often written

∂

∂xi = ∑j

∂y j

∂xi∂

∂y j .

5.2 Vector fields as sections of the tangent bundle 89

Here one thinks of the xi and y j as coordinates on the same set, and doesn’t worryabout writing coordinate maps, and one uses the (somewhat sloppy, but convenient)notation y = y(x) instead of y = F(x).)

Example 5.1. Problem. Express the coordinate vector fields ∂

∂x ,∂

∂y in polar coordi-nates, given by

x = r cosθ , y = r sinθ

(valid for r > 0 and −π < θ < π).Solution. We have

∂

∂ r=

∂x∂ r

∂

∂x+

∂y∂ r

∂

∂y= cosθ

∂

∂x+ sinθ

∂

∂y

and similarly

∂

∂θ=

∂x∂θ

∂

∂x+

∂y∂θ

∂

∂y=−r sinθ

∂

∂x+ r cosθ

∂

∂y.

The matrix of coefficients is of course the Jacobian. Inverting this matrix(cosθ sinθ

−r sinθ r cosθ

)−1

=1r

(r cosθ −sinθ

r sinθ cosθ

)(in other words, solving the equations for ∂

∂x and ∂

∂y ) we obtain

∂

∂x= cosθ

∂

∂ r− 1

rsinθ

∂

∂θ,

∂

∂y= sinθ

∂

∂ r+

1r

cosθ∂

∂θ.

5.2 Vector fields as sections of the tangent bundle

The ‘best’ way of describing the smoothness of p 7→ Xp is that it is literally a smoothmap into the tangent bundle.

Definition 5.3 (Vector fields – third definition). A vector field on M is a smoothmap X ∈C∞(M,T M) such that π X is the identity.

It is common practice to use the same symbol X both as a linear map from smoothfunctions to smooth functions, or as a map into the tangent bundle. Thus

X : M→ T M, X : C∞(M)→C∞(M)

coexist. But if it gets too confusing, one uses a symbol

90 5 Vector fields

LX : C∞(M)→C∞(M)

for the interpretation as a derivation; here the L stands for ‘Lie derivative’ (namedafter Sophus Lie). Both viewpoints are useful and important, and both have theiradvantages and disadvantages. For instance, from (A) it is immediate that vectorfields on M restrict to open subsets U ⊆M; this map

X(M)→ X(U), X 7→ X |U

may seem a little awkward from viewpoint (B) since C∞(U) is not a subspace ofC∞(M). (There is a restriction map C∞(M)→ C∞(U), but no natural map in theother direction.) On the other hand, (B) gives the Lie bracket operation discussedbelow, which seems unexpected from viewpoint (A).

5.3 Lie brackets

Let M be a manifold. Given vector fields X ,Y : C∞(M)→C∞(M), the compositionX Y is not a vector field: For example, if X = Y = ∂

∂x as vector fields on R, then

X Y = ∂ 2

∂x2 is a second order derivative, which is not a vector field (it does notsatisfy the Leibnitz rule). However, the commutator turns out to be a vector field:

Theorem 5.1. For any two vector fields X ,Y ∈ X(M) (regarded as derivations), thecommutator

[X ,Y ] := X Y −Y X : C∞(M)→C∞(M)

is again a vector field.

Proof. To check that [X ,Y ] is a vector field, we verify the derivation property, bydirect calculation. We have that

(X Y )( f1 f2) = X(Y ( f1) f2 + f1Y ( f2)

)= X(Y ( f1)) f2 + f1 X(Y ( f2))+X( f1)Y ( f2)+Y ( f1)X( f2);

subtracting a similar expression with 1 and 2 interchanged, some terms cancel, andwe obtain

[X ,Y ]( f1 f2) = X(Y ( f1)) f2 + f1 X(Y ( f2))−X(Y ( f2)) f1 + f2 X(Y ( f1))

= [X ,Y ]( f1) f2 + f1 [X ,Y ]( f2)

as required.

Remark 5.2. A similar calculation applies to derivations of algebras in general: Thecommutator of two derivations is again a vector field.

Definition 5.4. The vector field

5.3 Lie brackets 91

[X ,Y ] := X Y −Y X

is called the Lie bracket of X ,Y ∈ X(M).

It is instructive to see how this works in local coordinates. For open subsets U ⊆Rm,if

X =m

∑i=1

ai ∂

∂xi , Y =m

∑i=1

bi ∂

∂xi ,

with coefficient functions ai,bi ∈ C∞(U), the composition X Y is a second orderdifferential operators on functions f ∈C∞(U):

X Y =m

∑i=1

m

∑j=1

a j ∂bi

∂x j∂

∂xi +m

∑i=1

m

∑j=1

aib j ∂ 2

∂xi∂x j

Subtracting a similar expression for Y X , the terms involving second derivativescancel, and we obtain

[X ,Y ] =m

∑i=1

m

∑j=1

(a j ∂bi

∂x j −b j ∂ai

∂x j

)∂

∂xi .

(This calculation applies to general manifolds, by taking local coordinates.) Thesignificance of the Lie bracket will become clear later. At this stage, let us givesome examples.

Example 5.2. Consider the following two vector fields on R2,

X =∂

∂x, Y = (1+ x2)

∂

∂y.

We have

X Y = (1+ x2)∂ 2

∂x∂y+2x

∂

∂y, Y X = (1+ x2)

∂ 2

∂y∂x.

Both a second order differential operators. Taking the difference, the second orderderivatives cancel, due to the equality of mixed partials. We obtain

[X ,Y ] = 2x∂

∂y.

Note that the vector fields X ,Y are linearly independent everywhere. Is it possible tointroduce coordinates (u,v) = ϕ(x,y), such that in the new coordinates, these vectorfields are the coordinate vector fields ∂

∂u ,∂

∂v ? The answer is no: the coordinatevector fields have zero Lie bracket

[∂

∂u,

∂

∂v] = 0,

92 5 Vector fields

(they ‘commute’), but [X ,Y ] 6= 0.

Example 5.3. Consider the following two vector fields on R2, on the open subsetwhere xy > 0,

X =xy

∂

∂x+

∂

∂y, Y = 2

√xy

∂

∂x

Indicating second order derivatives by dots we have

X Y =(x

y

√yx+

√xy

)∂

∂x+ . . .= 2

√xy

∂

∂x+ . . .

Y X = 2√

xy

∂

∂x+ . . . .

Thus, [X ,Y ] = X Y −Y X = 0. Can one introduce coordinates u,v in which thesevector fields become the coordinate vector fields? This time, the answer is yes: De-fine a change of coordinates u,v by putting x = uv2, y = u. Then

∂

∂u=

∂x∂u

∂

∂x+

∂y∂u

∂

∂y= v2 ∂

∂x+

∂

∂y=

xy

∂

∂x+

∂

∂y= X ,

∂

∂v=

∂x∂v

∂

∂x+

∂y∂v

∂

∂y= 2uv

∂

∂x= 2√

xy∂

∂x= Y,

Note: When calculating Lie brackets X Y −Y X of vector fields X ,Y in localcoordinates, it is not necessary to work out the second order derivatives – we knowin advance that these are going to cancel out! This is why we indicated second orderderivatives by “. . .” in the calculation above.

Example 5.4. Consider the same problem for the vector fields

X = x∂

∂y− y

∂

∂x, Y = x

∂

∂x+ y

∂

∂y.

This time, we may verify that [X ,Y ] = 0. Introduce polar coordinates,

x = r cosθ , y = r sinθ .

(this is a well-defined coordinate chart for r > 0 and −π < θ < π). We have 1

∂

∂ r=

∂x∂ r

∂

∂x+

∂y∂ r

∂

∂y=

1r

Y

and

1 In the following, we are using somewhat sloppy notation. Given (θ ,r) = ϕ(x,y), we should moreproperly write ϕ∗X ,ϕ∗Y for the vector fields in the new coordinates.

5.3 Lie brackets 93

∂

∂θ=

∂x∂θ

∂

∂x+

∂y∂θ

∂

∂y= X

Hence X = ∂

∂θ, Y = r ∂

∂ r . To get this into the desired form, we make another changeof coordinates ρ = f (r) in such a way that Y becomes ∂

∂ρ. Since

∂

∂ r=

∂ρ

∂ r∂

∂ρ= f ′(r)

∂

∂ρ

we want f ′(r) = 1r , thus f (r) = ln(r). So, r = eρ . Hence, the desired change of

coordinates isx = eρ cosθ , y = eρ sinθ .

Let S ⊆M be a submanifold. A vector field X ∈ X(M) is called tangent to S if forall p ∈ S, the tangent vector Xp lies in TpS⊆ TpM. (Thus X restricts to a vector fieldX |S ∈ X(S).)

Example 5.5. The three vector fields

X = y∂

∂ z− z

∂

∂y, Y = z

∂

∂x− x

∂

∂ z, Z = x

∂

∂y− y

∂

∂x

on R3 are tangent to the 2-sphere S2: For example, under the identification TpR3 =R3, for p = (x,y,z), each of

Xp = (0,−z,y), Yp = (z,0,−x), Zp = (−y,x,0)

have zero dot product with p. The bracket of any two of X ,Y,Z is again tangent; infact we have

[X ,Y ] = Z, [Y,Z] = X , [Z,X ] = Y.

More generally, we have:

Proposition 5.2. If two vector fields X ,Y ∈ X(M) are tangent to a submanifold S⊆M, then their Lie bracket is again tangent to S.

Proposition 5.2 can be proved by using the coordinate expressions of X ,Y in sub-manifold charts. But we will postpone the proof for now since there is a muchshorter, coordinate-independent proof, see the next section.

Example 5.6. Consider the vector fields on R3,

X =∂

∂x, Y =

∂

∂y+ x

∂

∂ z.

We have

[X ,Y ] =∂

∂ z;

94 5 Vector fields

hence Xp, Yp, Zp are a basis of R3 for all p ∈ R3. In particular, there cannot exist asurface S⊆ R3 such that both X and Y are tangent to S.

5.4 Related vector fields

Definition 5.5. Let F ∈ C∞(M,N) be a smooth map. Vector fields X ∈ X(M) andY ∈ X(N) are called F-related, written as

X ∼F Y,

ifTpF(Xp) = YF(p)

for all p ∈M.

Example 5.7. If F is a diffeomorphism, then X ∼F Y if and only if Y = F∗X . Inparticular, if N = M, then an equation X ∼F X means that X is invariant under F .

Example 5.8. Let S⊆M be an embedded submanifold and i : S→M the inclusion.Let X ∈ X(S) and Y ∈ X(M). Then

X ∼i Y

if and only if Y is tangent to S, with X as its restriction. In particular,

0∼i Y

if and only if Y vanishes along the submanifold S.

Example 5.9. If F : M → N is a submersion, and X ∈ X(M), then X ∼F 0 if andonly if X is tangent to the fibers of F .

Example 5.10. Let π : Sn→ RPn be the quotient map. Then X ∼π Y if and only ifthe vector field X is invariant under the transformation F : Sn→ Sn, x 7→ −x (thatis, T F X = X F , and with Y the induced vector field on the quotient.

The F-relation of vector fields also has a simple interpretation in terms of the‘differential operator’ picture.

Proposition 5.3. One has X ∼F Y if and only if for all g ∈C∞(N),

X(gF) = Y (g)F.

In terms of the pull-back notation, with F∗g = g F for g ∈ C∞(N), this meansX F∗ = F∗ Y :

5.4 Related vector fields 95

C∞(M)X// C∞(M)

C∞(N)

F∗

OO

Y// C∞(N)

F∗

OO

Proof. The condition X(gF) = Y (g)F says that

(TpF(Xp))(g) = YF(p)(g)

for all p ∈M. ut

The key fact concerning related vector fields is the following.

Theorem 5.2. Let F ∈C∞(M,N) For vector fields X1,X2 ∈X(M) and Y1,Y2 ∈X(M),we have

X1 ∼F Y1, X2 ∼F Y2⇒ [X1,X2]∼F [Y1,Y2].

Proof. Using the differential operator picture, we have that

[X1,X2](gF) = X1(X2(gF))−X2(X1(gF))

= X1(Y2(g)F)−X2(Y1(g)F)

= Y1(Y2(g))F−Y2(Y1(g))F

= [Y1,Y2](g)F.

ut

Example 5.11. If two vector fields Y1,Y2 are tangent to a submanifold S ⊆ M thentheir Lie bracket [Y1,Y2] is again tangent to S, and the Lie bracket of their restrictionis the restriction of the Lie brackets. Indeed, letting Xi be the restrictions, we have

X1 ∼i Y1, X2 ∼i Y2 ⇒ [X1,X2]∼i [Y1,Y2].

Similarly, if Y1 is tangent to S and Y2 vanishes along S, then the Lie bracket vanishesalong S. This follows from the above by putting X2 = 0, since [X1,0] = 0.

Exercise. Explore the consequences of Proposition 5.3 for the other examples ofrelated vector fields given above.

Exercise. Show that in the description of vector fields as sections of the tangentbundle, two vector fields X ∈ X(M), Y ∈ X(M) are F-related if and only if thefollowing diagram commutes:

T M T F // T N

MF//

X

OO

N

Y

OO

96 5 Vector fields

5.5 Flows of vector fields

For any curve γ : J→M, with J ⊆ R an open interval, and any t ∈ J, the velocityvector

γ(t)≡ dγ

dt∈ Tγ(t)M

is defined as the tangent vector, given in terms of its action on functions as

(γ(t))( f ) =ddt

f (γ(t)).

(The dot signifies a t-derivative.) The curve representing this tangent vector for agiven t, in the sense of our earlier definition, is the shifted curve τ 7→ γ(t + τ).Equivalently, one may think of the velocity vector as the image of ∂

∂ t |t ∈ TtJ ∼= Runder the tangent map Ttγ:

γ(t) = (Ttγ)(∂

∂ t|t).

Definition 5.6. Suppose X ∈ X(M) is a vector field on a manifold M. A smoothcurve γ ∈C∞(J,M), where J ⊆R is an open interval, is called a solution curve to Xif

γ(t) = Xγ(t) (5.3)

for all t ∈ J.

Geometrically, Equation (5.3) means that at any given time t, the value of X at γ(t)agrees with the velocity vector to γ at t.

Equivalently, in terms of related vector fields,

∂

∂ t∼γ X .

Consider first the case that M =U ⊆ Rm. Here curves γ(t) are of the form

5.5 Flows of vector fields 97

γ(t) = x(t) = (x1(t), . . . ,xm(t)),

hence

γ(t)( f ) =ddt

f (x(t)) =m

∑i=1

dxi

dt∂ f∂xi (x(t)).

That is

γ(t) =m

∑i=1

dxi

dt∂

∂xi

∣∣∣x(t)

.

On the other hand, the vector field has the form X = ∑mi=1 ai(x) ∂

∂xi . Hence (5.3)becomes the system of first order ordinary differential equations,

dxi

dt= ai(x(t)), i = 1, . . . ,m. (5.4)

Example 5.12. The solution curves of the coordinate vector field ∂

∂x j are of the form

xi(t) = xi0, i 6= j, x j(t) = x j

0 + t.

More generally, if a = (a1, . . . ,am) is a constant function of x (so that X = ∑ai ∂

∂xi isthe constant vector field, the solution curves are affine lines,

x(t) = x0 + ta.

Example 5.13. Consider the vector field on R2,

X =−y∂

∂x+ x

∂

∂y.

The corresponding differential equation is x = −y, y = x. Its solutions are γ(t) =(x(t),y(t)), where

x(t) = x0 cos(t)− y0 sin(t), y(t) = y0 cos(t)+ x0 sin(t),

for any given (x0,y0) ∈ R2.

Example 5.14. Consider the following vector field on Rm,

X =m

∑i=1

xi ∂

∂xi .

98 5 Vector fields

The corresponding differential equation is

xi = xi(t),

with solution xi(t) = et xi0, for i = 1, . . . ,m. That is,

x(t) = et x0.

One of the main results from the theory of ODE’s says that for any given initialcondition x(0) = x0, a solution to the system (5.4) exists and is (essentially) unique:

Theorem 5.3 (Existence and uniqueness theorem for ODE’s). Let U ⊆Rm be anopen subset, and a ∈ C∞(U,Rm). For any given x0 ∈U, there is an open intervalJx0 ⊆ R around 0, and a solution x : Jx0 →U of the ODE

dxi

dt= ai(x(t)), i = 1, . . . ,m

with initial condition x(0) = x0, and which is maximal in the sense that any othersolution to this initial value problem is obtained by restriction to some subintervalof Jx0 .

Thus, Jx0 is the maximal open interval on which the solution is defined. The solutiondepends smoothly on initial conditions, in the following sense. For any given x0, letΦ(t,x0) be the solution x(t) of the initial value problem with initial condition x0,that is,

Φ(0,x0) = x0,ddt

Φ(t,x0) = a(Φ(t,x0)).

Theorem 5.4 (Dependence on initial conditions for ODE’s). For a ∈C∞(U,Rm)as above, the set

J = (t,x) ∈ R×U | t ∈ Jx.

is an open neighborhood of 0×U in R×U, and the map

Φ : J →U, (t,x) 7→Φ(t,x)

is smooth.

In general, the interval Jx0 may be strictly smaller than R, because a solution mightescape to infinity in finite time.

Examples 5.15. 1. Consider the ODE

x = 1

on U = (0,1) ⊆ R. Thus a(x) = 1. The solution curves with initial conditionx0 ∈U are x(t) = x0 + t, defined for −x0 < t < 1− x0. Thus Jx0 = (−x0,1− x0),and


J = (t,x)|x ∈ (0,1), t + x ∈ (0,1), Φ(t,x) = t + x.

2. Conside the ODEx = x2

on U = R. Here the solution curves escape to infinity in finite time. The initialvalue problem has solutions

x(t) =x0

1− tx0,

with domain of definition

Jx0 = t ∈ R| tx0 < 1.

The set J = (t,x)|tx < 1 is the region between the two branches of the hy-perbola tx = 1, and Φ(t,x) = x

1−tx .3. A similar example, which we leave as an exercise, is

x = 1+ x2.

For a general vector field X ∈ X(M) on manifolds, Equation (5.3) becomes (5.4)after introduction of local coordinates. In detail: Let (U,ϕ) be a coordinate chart. Inthe chart, X becomes the vector field

ϕ∗(X) =m

∑j=1

a j(u)∂

∂u j

and ϕ(γ(t)) = u(t) withui = ai(u(t)).

If a = (a1, . . . ,am) : ϕ(U)→ Rm corresponds to X in a local chart (U,ϕ), thenany solution curve x : J → ϕ(U) for a defines a solution curve γ(t) = ϕ−1(x(t))for X . The existence and uniqueness theorem for ODE’s extends to manifolds, asfollows:

Theorem 5.5 (Solutions of vector fields on manifolds). Let X ∈X(M) be a vectorfield on a manifold M. For any given p ∈ M, there is an open interval Jp ⊆ Raround 0, and a solution γ : Jp→M of the initial value problem

γ(t) = Xγ(t), γ(0) = p, (5.5)

which is maximal in the sense that any other solution of the initial value problem isobtained by restriction to a subinterval. The set

J = (t, p) ∈ R×M| t ∈Jp

is an open neighborhood of 0×M, and the map

Φ : J →M, (t, p) 7→Φ(t, p)

100 5 Vector fields

such that γ(t) = Φ(t, p) solves the initial value problem (5.5), is smooth.

Proof. Existence and uniqueness of solutions for small times t follows from theexistence and uniqueness theorem for ODE’s, by considering the vector field inlocal charts. To prove uniqueness even for large times t, let γ : J→M be a maximalsolution of (5.5) (i.e., a solution that cannot be extended to a larger open interval),and let γ1 : J1→M be another solution of the same initial value problem, but withγ1(t) 6= γ(t) for some t ∈ J, t > 0. (There is a similar discussion if the solution isdifferent for some t < 0). Then we can define

b = inft ∈ J| t > 0, γ1(t) 6= γ(t).

By the uniqueness for small t, we have b > 0. We will get a contradiction in both ofthe following cases:

Case 1: γ1(b) = γ(b) =: q. Then both λ1(s) = γ1(b+ s) and λ (s) = γ(b+ s) aresolutions to the initial value problem

λ (0) = q, λ (s) = Xλ (s);

hence they have to agree for small |s|, and consequently γ1(t),γ(t) have to agree fort close to b. This contradicts the definition of b.

Case 2: γ1(b) 6= γ(b). Using the Hausdorff property of M, we can choose disjointopen neighborhoods U of γ(b) and U1 of γ(b1). For t = b−ε with ε > 0 sufficientlysmall, γ(t) ∈U while γ1(t) ∈U1. But this is impossible since γ(t) = γ1(t) for 0 ≤t < b.

The result for ODE’s about the smooth dependence on initial conditions shows,by taking local coordinate charts, that J contains an open neighborhood of 0×M, on which Φ is given by a smooth map. The fact that J itself is open, and themap Φ is smooth everywhere, follows by the ‘flow property’ to be discussed below.(We will omit the details of this part of the proof.) ut

Note that the uniqueness part uses the Hausdorff property in the definition ofmanifolds. Indeed, the uniqueness part may fail for non-Hausdorff manifolds.

Example 5.16. A counter-example is the non-Hausdorff manifold

Y = (R×1)∪ (R×−1)/∼,

where ∼ glues two copies of the real line along the strictly negative real axis. LetU± denote the charts obtained as images of R×±1. Let X be the vector field onY , given by ∂

∂x in both charts. It is well-defined, since the transition map is just theidentity map. Then γ+(t) = π(t,1) and γ−(t) = π(t,−1) are both solution curves,and they agree for negative t but not for positive t.

Given a vector field X , the map Φ : J → M is called the flow of X . For anygiven p, the curve γ(t) = Φ(t, p) is a solution curve. But one can also fix t andconsider the time-t flow,


Φt(p)≡Φ(t, p).

It is a smooth map Φt : Ut →M, defined on the open subset

Ut = p ∈M| (t, p) ∈J .

Note that Φ0 = idM .Intuitively, Φt(p) is obtained from the initial point p ∈M by flowing for time t

along the vector field X . One expects that first flowing for time t, and then flowingfor time s, should be the same as flowing for time t+s. Indeed one has the followingflow property.

Theorem 5.6 (Flow property). Let X ∈X(M), with flow Φ : J →M. Let (t2, p)∈J , and t1 ∈ R. Then

(t1,Φt2(p)) ∈J ⇔ (t1 + t2, p) ∈J ,

and one hasΦt1(Φt2(p)) = Φt1+t2(p).

Proof. Given t2 ∈ Jp, we consider both sides as functions of t1 = t. Write q=Φt2(p).We claim that both

t 7→Φt(Φt2(p)), t 7→Φt+t2(p)

are maximal solution curves of X , for the same initial condition q. This is clear forthe first curve, and follows for the second curve by the calculation, for f ∈C∞(M),

ddt

f (Φt+t2(p)) =dds

∣∣∣s=t+t2

Φs(p) = XΦs(p)( f )∣∣∣s=t+t2

= XΦt+t2 (p)( f ).

Hence, the two curves must coincide. The domain of definition of t 7→ Φt+t2(p) isthe interval Jp, shifted by t2. Hence, t1 ∈ JΦ(t2,p) if and only if t1 + t2 ∈ Jp. ut

We see in particular that for any t, the map Φt : Ut → M is a diffeomorphismonto its image Φt(Ut) =U−t , with inverse Φ−t .

Example 5.17. Let us illustrate the flow property for various vector fields on R. Theflow property is evident for ∂

∂x with flow Φt(x) = x+ t, as well as for x ∂

∂x , withflow Φt(x) = et x. The vector field x2 ∂

∂x has flow Φt(x) = x/(1− tx), defined for1− tx < 1. We can explicitly verify the flow property:

Φt1(Φt2(x)) =Φt2(x)

1− t1Φt2(x)=

x1−t2x

1− t1 x1−t2x

=x

1− (t1 + t2)x= Φt1+t2(x).

Let X be a vector field, and J = J X be the domain of definition for the flowΦ = ΦX .

Definition 5.7. A vector field X ∈ X(M) is called complete if J X = R×M.

Thus X is complete if and only if all solution curves exist for all time.

102 5 Vector fields

Example 5.18. The vector field x ∂

∂x on M = R is complete, but x2 ∂

∂x is incomplete.

A vector field may fail to be complete if a solution curve escapes to infinity infinite time. This suggests that a vector fields X that vanishes outside a compact setmust be complete, because the solution curves are ‘trapped’ and cannot escape toinfinity.

Proposition 5.4. If X ∈X(M) is a vector field that has compact support, in the sensethat X |M−A = 0 for some compact subset A, then X is complete. In particular, everyvector field on a compact manifold is complete.

Proof. By the uniqueness theorem for solution curves γ , and since X vanishes out-side A, if γ(t0) ∈M−A for some t0, then γ(t) = γ(t0) for all t. Hence, if a solutioncurve γ : J → M has γ(0) ∈ A, then γ(t) ∈ A for all t. Let Uε ⊆ M be the set ofall p such that the solution curve γ with initial condition γ(0) = p exists for |t|< ε

(that is, (−ε,ε)⊆ Jp). By smooth dependence on initial conditions, Uε is open. Thecollection of all Uε with ε > 0 covers A, since every solution curve exists for suf-ficiently small time. Since A is compact, there exists a finite subcover Uε1 , . . . ,Uεk .Let ε be the smallest of ε1, . . . ,εk. Then Uεi ⊆Uε , for all i, and hence A⊆Uε . Hence,for any p ∈ A we have (−ε,ε)⊆ Jp, that is any solution curve γ(t) starting in A ex-ists for times |t|< ε . But γ(−ε/2),γ(ε/2) ∈ A, hence the solution curve starting atthose points again exist for times < ε . This shows (−3ε/2,3ε/2)⊆ Jp. Continuingin this way, we find that (−ε−Nε/2,ε +Nε/2)⊆ Jp for all N, thus Jp = R for allp∈ A. For points p∈M−A, it is clear anyhow that Jp =R, since the solution curvesare constant. ut

Theorem 5.7. If X is a complete vector field, the flow Φt defines a 1-parametergroup of diffeomorphisms. That is, each Φt is a diffeomorphism and

Φ0 = idM, Φt1 Φt2 = Φt1+t2 .

Conversely, if Φt is a 1-parameter group of diffeomorphisms such that the map(t, p) 7→Φt(p) is smooth, the equation

Xp( f ) =ddt

∣∣∣t=0

f (Φt(p))

defines a complete vector field X on M, with flow Φt .

Proof. It remains to show the second statement. Given Φt , the linear map

C∞(M)→C∞(M), f 7→ ddt

∣∣∣t=0

f (Φt(p))

satisfies the product rule, hence it is a vector field X . Given p ∈M the curve γ(t) =Φt(p) is an integral curve of X since

ddt

Φt(p) =dds

∣∣∣s=0

Φt+s(p) =dds

∣∣∣s=0

Φs(Φt(p)) = XΦt (p).


ut

Remark 5.3. In terms of pull-backs, the relation between the vector field and its flowreads as

ddt

Φ∗t ( f ) = Φ

∗t

dds

∣∣∣s=0

Φ∗s ( f ) = Φ

∗t X( f ).

This identityddt

Φ∗t = Φ

∗t X

as linear maps C∞(M)→C∞(M) may be viewed as the definition of the flow.

Example 5.19. Given A ∈MatR(m) let

Φt : Rm→ Rm, x 7→ etAx =( ∞

∑j=0

t j

j!A j)

x

(using the exponential map of matrices). Since e(t1+t2)A = et1Aet2A, and since (t,x) 7→etAx is a smooth map, Φt defines a flow. What is the corresponding vector field X?For any function f ∈C∞(Rm) we calculate,

X( f )(x) =ddt

∣∣∣t=0

f (etAx)

= ∑j

∂ f∂x j (Ax) j

= ∑i j

Aijxi ∂ f

∂x j

showing that

X = ∑i j

Aijxi ∂

∂x j .

2

As a special case, taking A to be the identity matrix, we recover the Euler vectorfield X = ∑i xi ∂

∂xi , and its flow Φt(x) = etx.

Example 5.20. Let X be a complete vector field, with flow Φt . For each t ∈ R, thetangent map T Φt : T M→ T M has the flow property,

T Φt1 T Φt2 = T (Φt1 Φt2) = T (Φt1+t2),

2 Here we wrote the matrix entries for the i-th row and j-th column as Aij rather than Ai j . That is,

one standard basis vectors ei ∈ Rm (written as column vectors), we have A(ei) = ∑ j Aije j , hence

for x = ∑xiei we getAx = ∑

i jAi

jxie j

from which we read off (Ax) j = ∑i Aijxi.

104 5 Vector fields

and the map R×T M→ T M,(t,v) 7→Φt(v) is smooth (since it is just the restrictionof the map T Φ : T (R×M)→ T M to the submanifold R×T M). Hence, T Φt is aflow on T M, and therefore corresponds to a complete vector field X ∈X(T M). Thisis called the tangent lift of X .

Proposition 5.5. Let F ∈ C∞(M,N), and X ∈ X(M), Y ∈ X(N) complete vectorfields, with flows ΦX

t , ΦYt .

X ∼F Y ⇔ F ΦXt = Φ

Yt F for all t.

3

In short, vector fields are F-related if and only if their flows are F-related.

Proof. Suppose F ΦXt = ΦY

t F for all t. For g ∈ C∞(N), and p ∈ M, taking at-derivative of

g(F(ΦXt (p))) = g(ΦY

t (F(p)))

at t = 0 on both sides, we get(TpF(Xp)

)(g) = YF(p)(g)

i.e. TpF(Xp) =YF(p). Hence X ∼F Y . Conversely, suppose X ∼F Y . As we had seen,if γ : J→M is a solution curve for X , with initial condition γ(0) = p then F γ :J→M is a solution curve for Y , with initial condition F(p). That is, F(ΦX

t (p)) =ΦY

t (F(p)), or F ΦXt = ΦY

t F . ut

5.6 Geometric interpretation of the Lie bracket

For any smooth map F ∈C∞(M,N) we defined the pull-back

F∗ : C∞(N)→C∞(M), g 7→ gF.

If F is a diffeomorphism, then we can also pull back vector fields:

F∗ : X(N)→ X(M), Y 7→ F∗Y,

by the condition (F∗Y )(F∗g) = F∗(Y (g)) for all functions g. That is, F∗Y ∼F Y , orin more detail

(F∗Y )p = (TpF)−1 YF(p).

By Theorem 5.2, we have F∗[X ,Y ] = [F∗X ,F∗Y ].Any complete vector field X ∈ X(M) with flow Φt gives rise to a families of

pull-back maps

3 This generalizes to possibly incomplete vector fields: The vector fields are related if and only ifF Φ = Φ (idR×F). But for simplicity, we only consider the complete case.

5.6 Geometric interpretation of the Lie bracket 105

Φ∗t : C∞(M)→C∞(M), Φ

∗t : X(M)→ X(M).

The Lie derivative of a function f with respect to X is the function

LX ( f ) =ddt

∣∣∣t=0

Φ∗t f ;

thus LX ( f ) = X( f ). The Lie derivative measures how f changes in the direction ofX . Similarly, for a vector field Y one defines the Lie derivative LX (Y ) by

LX (Y ) =ddt

∣∣∣t=0

Φ∗t Y ∈ X(M).

The definition of Lie derivative also works for incomplete vector fields, since thedefinition only involves derivatives at t = 0. The Lie derivative measures how Ychanges in the direction of X . Note that

(Φ∗t Y )p = (TpΦ−1t ) YΦt (p);

that is, we use the inverse to the tangent map of the flow of X to move YΦt (p) to p. IfY were invariant under the flow of X , this would agree with Yp; hence (Φ∗t Y )p−Ypmeasures how Y fails to be Φt -invariant. LXY is the infinitesimal version of this. Aswe will see below, the infinitesimal version actually implies the global version.

Theorem 5.8. For any X ,Y ∈ X(M), the Lie derivative LXY is just the Lie bracket:

LX (Y ) = [X ,Y ].

Proof. Let Φt = ΦXt be the flow of X . For all f ∈C∞(M) we obtain, by taking the

t-derivative at t = 0 of both sides of

Φ∗t (Y ( f )) = (Φ∗t Y )(Φ∗t f ),

that

X(Y ( f )) =( d

dt

∣∣∣t=0

Φ∗t Y)( f )+Y

( ddt

∣∣∣t=0

Φ∗t f)= (LXY )( f )+Y (X( f )).

That is, LXY = X Y −Y X = [X ,Y ]. ut

Thus, the Lie bracket [X ,Y ] measures ‘infinitesimally’ how the vector field Ychanges along the flow of X . Note that in particular, LXY is skew-symmetric inX and Y – this is not obvious from the definition.

One can also interpret the Lie bracket as measuring how the flows of X and Y failto commute.

Theorem 5.9. Let X ,Y be complete vector fields, with flows Φt ,Ψs. Then

[X ,Y ] = 0⇔ Φ∗t Y = Y for all t

⇔Ψ∗

s X = X for all s

106 5 Vector fields

⇔ Φt Ψs =Ψs Φt for all s, t.

Proof. The calculation

ddt(Φt)

∗Y = (Φt)∗LXY = (Φt)

∗[X ,Y ]

shows that Φ∗t Y is independent of t if and only if [X ,Y ] = 0. Since [Y,X ] =−[X ,Y ],interchanging the roles of X ,Y this is also equivalent to Ψ ∗s X being independent ofs. The property Φ∗t Y =Y means that Y is Φt -related to itself, hence it takes the flowof Ψs to itself, that is

Φt Ψs =Ψs Φt .

Conversely, if this equation holds then Φ∗t (Ψ∗

s f ) = Ψ ∗s (Φ∗t f ) for all f ∈ C∞(M).

Differentiating with respect to s at s = 0, we obtain

Φ∗t (Y ( f )) = Y (Φ∗t f ).

Hence Φ∗t (Y ) = Y . Differentiating with respect to t at t = 0, we get that [X ,Y ] = 0.ut

Example 5.21. If X = ∂

∂y as a vector field on R2, then [X ,Y ] = 0 if and only if Y isinvariant under translation in the y-direction.

Example 5.22. The vector fields X = x ∂

∂y − y ∂

∂x and Y = x ∂

∂x + y ∂

∂y commute. Thisis verified by direct calculation but can also be ‘seen’ in the following picture

The flow of X is rotations around the origin, but Y is invariant under rotations.Likewise, the flow of Y is by dilations away from the origin,, but X is invariantunder dilations.

Aside from being skew-symmetric [X ,Y ] = −[Y,X ], the Lie bracket of vectorfields satisfies the important Jacobi identity.

Proposition 5.6. The Lie bracket of vector fields satisfies the Jacobi identity

[X , [Y,Z]]+ [Y, [Z,X ]]+ [Z, [X ,Y ]] = 0.

Proof. This may be proved ‘by hand’, expanding the definition of the Lie bracket[X ,Y ] = X Y −Y X . Each summand gives rise to 4 terms, hence there are alto-gether 12 terms. Each of the 3! = 6 orderings of X ,Y,Z appears twice, with opposite

5.7 Frobenius theorem 107

signs. For example, the term Y Z X appears with coefficient−1 in [X , [Y,Z]], withcoefficient +1 in [Y, [Z,X ]], with coefficient 0 in [Z, [X ,Y ]]. ut

The identity may be equivalently stated as

[LX ,LY ]Z = L[X ,Y ]Z,

or also as a ‘derivation property’

LX [Y,Z] = [LXY,Z]+ [Y,LX Z].

This last form gives an ‘explanation’ of the Jacobi identity, as the derivative at t = 0of the identity

Φ∗t [Y,Z] = [Φ∗t Y,Φ∗t Z],

where Φt is the flow of X .

5.7 Frobenius theorem

We saw that for any vector field X ∈ X(M), there are solution curves through anygiven point p ∈M. The image of this curve is an (immersed) submanifold to whichX is everywhere tangent. One might similarly ‘integral surfaces’ for pairs of vectorfields, and ‘integral submanifolds’ for collections of vector fields.

Suppose X1, . . . ,Xr are vector fields on the manifold M, such that the tangentvectors X1|p, . . . ,Xr|p ∈ TpM are linearly independent for all p∈M. A r-dimensionalsubmanifold S ⊆M is called an integral submanifold if the vector fields X1, . . . ,Xrare all tangent to S.

Suppose that there exists an integral submanifold S through any given point p ∈M. Then each Lie bracket [Xi,X j]|p ∈ TpS, and hence is a linear combination ofX1|p, . . . ,Xr|p. It follows that

[Xi,X j] =r

∑k=1

cki jXk (5.6)

for certain (smooth) functions cki j.

A bit more generally, consider a sub-bundle E ⊆ T M of rank r. Such a subbundleis called involutive if the Lie bracket of any two sections of E is again a section ofE. For vector fields Xi as above, the pointwise spans

Ep = spanX1|p, . . . ,Xr|p

define a subbundle with this property. Indeed, given X =∑mi=1 aiXi and Y =∑

mi=1 biXi

with functions ai,bi, the condition (5.6) guarantees that E is involutive. Given anyrank r subbundle E ⊆ T M (not necessarily involutive), a submanifold S ⊆ M is

108 5 Vector fields

called an integral submanifold if Ep = TpS for all p ∈ S. The following result is dueto F. G. Frobenius. 4

Theorem 5.10 (Frobenius theorem). Let E ⊆ T M be a subbundle of rank r. Thefollowing are equivalent:

1. There exists an integral submanifold through every p ∈M.2. E is involutive.

In fact, if E is involutive, then it is possible to find a coordinate chart (U,ϕ) nearany given p, in such a way that the subbundle (T ϕ)(E|U ) ⊆ T ϕ(U) is spanned bythe first r ≤ m coordinate vector fields

∂

∂u1 , . . . ,∂

∂ur .

Proof. The statement is local; hence, by choosing coordinates we may assume Mis an open subset U ⊆ Rm, with p = (0,0). In particular, the tangent spaces are allidentified with Rm. By re-indexing the coordinates, we may assume that Ep ∩ (0⊕Rm−r) = 0. It is convenient to denote the first r coordinates by x1, . . . ,xr and theremaining coordinates by y1, . . . ,ym−r. Thus Ep projects isomorphically onto thecoordinate subspace spanned by x1, . . . ,xr. Taking U smaller if necessary, we mayassume that this remains true for all points in U . Then E is spanned by vector fieldsof the form

Xi =∂

∂xi +m−r

∑j=1

a ji (x,y)

∂

∂y j .

We claim that the Xi commute. Indeed, since [Xi,X j] takes values in E, it is of theform ∑k ck

i jXk for some functions cki j. By comparing the coefficients in front of ∂

∂xk ,we see that ck

i j = 0. (Indeed, [Xi,X j] is a linear combination of vector fields in they-direction.) Thus

[Xi,X j] = 0.

Since the Xi commute, also their flows Φi,ti commute:

Φi,ti Φ j,t j = Φ j,t j Φi,ti .

4http://upload.wikimedia.org/wikipedia/en/c/c9/Ferdinand_Georg_Frobenius.jpg

http://upload.wikimedia.org/wikipedia/en/c/c9/Ferdinand_Georg_Frobenius.jpg

5.7 Frobenius theorem 109

Note also that Φi,ti(x1, . . . ,xr,∗) = (x1, . . . ,xi + ti, . . . ,xr,∗). (This follows because

Xi is related to ∂

∂xi under projection (x,y) 7→ x.) We define a change of coordinatesby the equation

(x,y) = κ(u,v) := Φ1,u1 · · · Φr,ur(0,v).

(The Jacobian for this change of variables is invertible at (0,0). Indeed, note thatκ(0,v) = (0,v), while κ(u,0) = (u,∗) where ∗ indicates some function. This impliesthat the Jacobian matrix at (0,0) is upper triangular with 1’s along the diagonal,hence that its determinant is 1.) In these new coordinated the flow of the Xi is simplyaddition of t i in the i-th entry. This means that Xi =

∂

∂ui . Each subspace consistingof elements (u,v) with v = const is an integral submanifold. [MORE DETAILSNEEDED] ut

Thus, for any involutive subbundle E ⊆ T M, then any p ∈M has an open neigh-borhood U with a nice decomposition into r-dimensional submanifolds.

One calls such a decomposition (or sometimes the involutive subbundle E itself)a (local) foliation.

Example 5.23. Let Φ : M→ N be a submersion. Then the subbundle E ⊆ T M withfibers

Ep = ker(TpΦ)⊆ TpM

is an involutive subbundle of rank dimM−dimN. Every fiber Φ−1(q) is an integralsubmanifold.

Example 5.24. Consider the vector fields on R3,

X = (y− z)∂

∂x, Y =

∂

∂y+

∂

∂ z.

Away from y = z these are linearly independent. Since [X ,Y ] = 0, the Frobeniustheorem tells us there are integral submanifolds. Indeed, for any given C 6= 0 onehas the integral submanifold given by the equation y− z =C.

Remark 5.4. A foliation gives a decomposition into submanifolds on a neighbor-hood of any given point. Globally, the integral submanifolds are often only im-mersed submanifolds, given by immersions i : S→ M with (Tpi)(TpS) = Ep forall p ∈ S. The problem is already present for the foliation defined by a single non-vanishing vector field X = X1: It may happen that a solution curve γ through p getsarbitrarily close to p for large t; hence one cannot get a submanifold chart at p unlessone restricts the domain of γ .

110 5 Vector fields

5.8 Appendix: Derivations

A vector field on a manifold can be regarded as a derivation of the algebra of smoothfunctions.

Let us quickly recall the notion of a derivation.

Definition 5.8. A derivation of an algebra A is a linear map D : A →A satisfyingthe product rule

D(a1a2) = D(a1)a2 +a1D(a2).

Remarks 5.5. 1. If dimA < ∞, a derivation is an infinitesimal automorphism of analgebra. Indeed, let U : R→ End(A), t 7→Ut be a smooth curve with U0 = I,such that each Ut is an algebra automorphism. Consider the Taylor expansion,

Ut = I + tD+ . . .

hereD =

ddt

∣∣∣t=0

Ut

is the velocity vector at t = 0. By taking the derivative of the condition

Ut(a1a2) =Ut(a1)Ut(a2)

at t = 0, we get the derivation property for D. Conversely, if D is a derivation,then

Ut = exp(tD) =∞

∑n=0

tn

n!Dn

(using the exponential of a matrix) is a well-defined curve of algebra automor-phisms. We leave it as an exercise to check the automorphism property; it in-volves proving the property

Dn(a1a2) = ∑k

(nk

)Dk(a1) Dn−k(a2)

for all a1,a2 ∈ A.If A has infinite dimensions, one may still want to think of derivations D as in-finitesimal automorphisms, even though the discussion will run into technicalproblems. (For instance, the exponential map of infinite rank endomorphisms isnot well-defined in general.)

2. Any given x ∈ A defines a derivation

D(a) = [x,a] := xa−ax.

(Exercise: Verify that this is a derivation.) These are called inner derivations. IfA is commutative (for example A = C∞(M)) the inner derivations are all trivial.At the other extreme, for the matrix algebra A = MatR(n), one may show thatevery derivation is inner.

5.8 Appendix: Derivations 111

3. If A is a unital algebra, with unit 1A, then D(1A) = 0 for all derivations D. (Thisfollows by applying the defining property of derivations to 1A = 1A1A.)

4. Given two derivations D1,D2 of an algebra A, their commutator

[D1,D2] = D1D2−D2D1

is again a derivation. Indeed, if a,b ∈ A then

D1D2(ab) = D1(D2(a)b+aD2(b)

)= (D1D2)(a)b+a(D1D2)(b)+D1(a)D2(b)+D2(a)D1(b).

Subtracting a similar expression with 1,2 interchanged, one obtains the deriva-tion property of [D1,D2].

5. If the algebra A is commutative, then the space of derivations is a ‘left-moduleover A’. That is, if D is a derivation and x ∈ A then a 7→ (xD)(a) := xD(a) isagain a derivation:

(xD)(ab) = x(D(ab)) = x(D(a)b+a(D(b)) = (xD)(a)b+a(xD)(b),

where we used xa = ax.

Chapter 6Differential forms

6.1 Review: Differential forms on Rm

A differential k-form on an open subset U ⊆ Rm is an expression of the form

ω = ∑i1···ik

ωi1...ik dxi1 ∧·· ·∧dxik

where ωi1...ik ∈C∞(U) are functions, and the indices are numbers

1≤ i1 < · · ·< ik ≤ m.

Let Ω k(U) be the vector space consisting of such expressions, with the pointwiseaddition. It is convenient to introduce a short hand notation I = i1, . . . , ik for theindex set, and write ω = ∑I ωIdxI with

ωI = ωi1...ik , dxI = dxi1 ∧·· ·∧dxik .

Since a k-form is determined by these functions ωI , and since there are m!k!(m−k)! ways

of picking k-element subsets from 1, . . . ,m, the space Ω k(U) can be identifiedwith vector-valued smooth functions,

Ωk(U) =C∞(U, R

m!k!(m−k)! ).

The dxI are just formal expressions; at this stage they don’t have any particularmeaning. They are used, however, to define an associative product operation

Ωk(U)×Ω

l(U)→Ωk+l(U)

by the ‘rule of computation’

dxi∧dx j =−dx j ∧dxi

113

114 6 Differential forms

for all i, j; in particular dxi ∧ dxi = 0. In turn, using the product structure we maydefine the exterior differential

d : Ωk(U)→Ω

k+1(U), d(∑

IωIdxI

)=

m

∑i=1

∑I

∂ωI

∂xi dxi∧dxI . (6.1)

The key property of the exterior differential is the following fact:

Proposition 6.1. The exterior differential satisfies

dd = 0,

i.e. ddω = 0 for all ω .

Proof. By definition,

ddω =m

∑j=1

m

∑i=1

∑I

∂ 2ωI

∂x j∂xi dx j ∧dxi∧dxI ,

which vanishes by equality of mixed partials ∂ωI∂xi∂x j =

∂ωI∂x j∂xi . (We have dxi∧dx j =

−dx j ∧dxi, but the coefficients in front of dxi∧dx j and dx j ∧dxi are the same.) ut

Example 6.1. Consider forms on R3.

• The differential of a function f ∈Ω 0(R3) is a 1-form

d f =∂ f∂x

dx+∂ f∂y

dy+∂ f∂ z

dz,

with components the gradient

grad f = ∇ f .

• A 1-form ω ∈Ω 1(R3) is an expression

ω = f dx+gdy+hdz

with functions f ,g,h. The differential is

dω =(

∂g∂x− ∂ f

∂y

)dx∧dy+

(∂h∂y− ∂g

∂ z

)dy∧dz+

(∂ f∂ z− ∂h

∂x

)dz∧dx.

Thinking of the coefficients of ω as the components of a function F = ( f ,g,h) :U → R3, we see that the coefficients of dω give the curl of F ,

curl(F) = ∇×F.

• Finally, any 2-form ω ∈Ω 2(R3) may be written

ω = a dy∧dz+b dz∧dx+ c dx∧dy,

6.2 Dual spaces 115

with A = (a,b,c) : U → R3. We obtain

dω = (∂a∂x

+∂b∂y

+∂c∂ z

) dx ∧dy ∧dz;

the coefficient is the divergence

div(A) = ∇ ·A

The usual properties

curl(grad( f )) = 0, div(curl(F)) = 0

are both special cases of dd = 0.

The support supp(ω) ⊆U of a differential form is the smallest closed subset suchthat ω vanishes on U\supp(ω). Suppose ω ∈Ω m(U) is a compactly supported formof the top degree k = m. Such a differential form is an expression

ω = f dx1∧·· ·∧dxm

where f ∈C∞(U) is a compactly supported function. One defines the integral of ω

to be the usual Riemann integral:∫U

ω =∫Rm

f (x1, . . . ,xm)dx1 · · ·dxm. (6.2)

Note that we can regard ω as a form on all of Rm, due to the compact supportcondition.

Our aim is now to define differential forms on manifolds, beginning with 1-forms. Even though 1-forms on U ⊆ Rm are identified with functions U → Rm,they should not be regarded as vector fields, since their transformation propertiesunder coordinate changes are different. In fact, while vector fields are sections ofthe tangent bundle, the 1-forms are sections of its dual, the cotangent bundle. Wewill thus begin with a review of dual spaces in general.

6.2 Dual spaces

For any real vector space E, we denote by E∗ = L(E,R) its dual space, consistingof all linear maps α : E→R. We will assume that E is finite-dimensional. Then thedual space is also finite-dimensional, and dimE∗ = dimE. 1 It is common to writethe value of α ∈ E∗ on v ∈ E as a pairing, using the bracket notation:2

1 For possibly infinite-dimensional vector spaces, the dual space E∗ is not isomorphic to E, ingeneral.2 In physics, one also uses the Dirac bra-ket notation 〈α| v〉 := α(v); here α = 〈α| is the ‘bra’ andv = |v〉 is the ‘ket’.


〈α,v〉 := α(v).

Let e1, . . . ,er be a basis of E. Any element of E∗ is determined by its values onthese basis vectors. For i = 1, . . . ,r, let ei ∈ E∗ (with upper indices) be the linearfunctional such that

〈ei, e j〉= δij =

0 if i 6= j,1 if i = j.

The elements e1, . . . ,er are a basis of E∗; this is called the dual basis. The elementα ∈ E∗ is described in terms of the dual bases as

α =r

∑j=1

α j e j, α j = 〈α,e j〉.

Similarly, for vectors v ∈ E we have

v =r

∑i=1

viei, vi = 〈ei,v〉.

Notice the placement of indices: In a given summation over i, j, . . ., upper indicesare always paired with lower indices.

Remark 6.1. As a special case, for Rr with its standard basis, we have a canonicalidentification (Rr)∗ =Rr. For more general E with dimE <∞, there is no canonicalisomorphism between E and E∗ unless more structure is given.

Given a linear map R : E→ F between vector spaces, one defines the dual map

R∗ : F∗→ E∗

(note the direction), by setting

〈R∗β , v〉= 〈β ,R(v)〉

for β ∈ F∗ and v ∈ E. This satisfies (R∗)∗ = R, and under the composition of linearmaps,

(R1 R2)∗ = R∗2 R∗1.

In terms of basis e1, . . . ,er of E and f1, . . . , fs of F , and the corresponding dual bases(with upper indices), a linear map R : E→ F is given by the matrix with entries

Rij = 〈 f j, R(ei)〉,

while R∗ is described by the transpose of this matrix (the roles of i and j are re-versed). Namely,3

3 In bra-ket notation, we have Rij = 〈 f j |R |ei〉, and

|Rei〉= R|ei〉= ∑j| f j〉〈 f j |R |ei〉, 〈R∗( f j)|= 〈( f j)|R = 〈 f j |R |ei〉〈ei|

6.3 Cotangent spaces 117

R(ei) =s

∑j=1

Rij f j, R∗( f j) =

r

∑i=1

Rij f i.

Thus,(R∗) j

i = Rij.

6.3 Cotangent spaces

Definition 6.1. The dual of the tangent space TpM of a manifold M is called thecotangent space at p, denoted

T ∗p M = (TpM)∗.

Elements of T ∗p M are called cotangent vectors, or simply covectors. Given a smoothmap F ∈C∞(M,N), and any p ∈M we have the cotangent map

T ∗p F = (TpF)∗ : T ∗F(p)N→ T ∗p M

defined as the dual to the tangent map.

Thus, a co(tangent) vector at p is a linear functional on the tangent space, as-signing to each tangent vector at p a number. The very definition of the tangentspace suggests one such functional: Every function f ∈C∞(M) defines a linear map,TpM→ R, v 7→ v( f ). This linear functional is denoted (d f )p ∈ T ∗p M.4

Definition 6.2. Let f ∈C∞(M) and p ∈M. The covector

(d f )p ∈ T ∗p M, 〈(d f )p,v〉= v( f ).

is called the differential of f at p.

Lemma 6.1. For F ∈C∞(M,N) and g ∈C∞(N),

d(F∗g)p = T ∗p F((dg)F(p)).

Proof. Check on tangent vectors v ∈ TpM,

〈T ∗p F((dg)F(p)), v〉 = 〈(dg)F(p)), (TpF)(v)〉= ((TpF)(v))(g)

= v(F∗g)

= 〈d(F∗g)p, v〉.

ut

4 Note that this is actually the same as the tangent map Tp f : TpM→ Tf (p)R= R.


Consider an open subset U ⊆ Rm, with coordinates x1, . . . ,xm. Here TpU ∼= Rm,with basis

∂

∂x1

∣∣∣p, . . . ,

∂

∂xm

∣∣∣p∈ TpU (6.3)

The basis of the dual space T ∗p U , dual to the basis (6.3), is given by the differentialsof the coordinate functions:

(dx1)p, . . . , (dxm)p ∈ T ∗p U.

Indeed, ⟨(dxi)p,

∂

∂x j

∣∣∣p

⟩=

∂

∂x j

∣∣∣p(xi) = δ

ij

as required. For f ∈C∞(M), the coefficients of (d f )p = ∑i〈(d f )p, ei〉ei are deter-mined as ⟨

(d f )p,∂

∂x j

∣∣∣p

⟩=

∂

∂x j

∣∣∣p( f ) =

∂ f∂x j

∣∣∣p.

Thus,

(d f )p =m

∑i=1

∂ f∂xi

∣∣∣p(dxi)p.

Let U ⊆ Rm and V ⊆ Rn be open, with coordinates x1, . . . ,xm and y1, . . . ,yn. ForF ∈C∞(U,V ), the tangent map is described by the Jacobian matrix, with entries

(DpF)ij =

∂F j

∂xi (p)

for i = 1, . . . ,m, j = 1, . . . ,n. We have:

(TpF)(∂

∂xi

∣∣∣p) =

n

∑j=1

(DpF)ij ∂

∂y j

∣∣∣F(p)

,

hence dually

(TpF)∗(dy j)F(p) =m

∑i=1

(DpF)ij (dxi)p. (6.4)

Thought of as matrices, the coefficients of the cotangent map are the transpose ofthe coefficients of the tangent map.

6.4 1-forms

Similar to the definition of vector fields, one can define co-vector fields, more com-monly known as 1-forms: Collections of covectors αp ∈ T ∗p M depending smoothlyon the base point. One approach of making precise the smooth dependence on thebase point is to endow the cotangent bundle

6.4 1-forms 119

T ∗M =⋃p

T ∗p M.

(disjoint union of all cotangent spaces), and require that the map p 7→ αp issmooth. The construction of charts on T ∗M is similar to that for the tangent bun-dle: Charts (U,ϕ) of M give cotangent charts (T ∗U,T ∗ϕ−1) of T ∗M, using thefact that T ∗(ϕ(U)) = ϕ(U)×Rm canonically (since ϕ(U) is an open subset ofRm). Here T ∗ϕ−1 : T ∗U → T ∗ϕ(U) is the union of inverses of all cotangent mapsT ∗p ϕ : T ∗

ϕ(p)ϕ(U)→ T ∗p U . A second approach is observe that in local coordinates,1-forms are given by expressions ∑i fidxi, and smoothness should mean that thecoefficient functions are smooth.

We will use the following (equivalent) approach.

Definition 6.3. A 1-form on M is a linear map

α : X(M)→C∞(M), X 7→ α(X) = 〈α, X〉,

which is C∞(M)-linear in the sense that

α( f X) = f α(X)

for all f ∈C∞(M), X ∈ X(M). The space of 1-forms is denoted Ω 1(M).

Let us verify that a 1-form can be regarded as a collection of covectors:

Lemma 6.2. Let α ∈ Ω 1(M) be a 1-form, and p ∈M. Then there is a unique cov-ector αp ∈ T ∗p M such that

α(X)p = αp(Xp)

for all X ∈ X(M).

(We indicate the value of the function α(X) at p by a subscript, just like we did forvector fields.)

Proof. We have to show that α(X)p depends only on the value of X at p. By consid-ering the difference of vector fields having the same value at p, it is enough to showthat if Xp = 0, then α(X)p = 0. But any vector field vanishing at p can be writtenas a finite sum X = ∑i fiYi where fi ∈ C∞(M) vanish at p. 5 By C∞-linearity, thisimplies that

α(X) = α(∑i

fiYi) = ∑i

fiα(Yi)

vanishes at p. ut

The first example of a 1-form is described in the following definition.

Definition 6.4. The exterior differential of a function f ∈C∞(M) is the 1-form

d f ∈Ω1(M),

5 For example, using local coordinates, we can take the Yi to correspond to ∂

∂xi near p, and the fito the coefficient functions.


defined in terms of its pairings with vector fields X ∈ X(M) as 〈d f , X〉= X( f ).

Clearly, d f is the 1-form defined by the family of covectors (d f )p. Note that criticalpoints of f may be described in terms of this 1-form: p ∈M is a critical point of fif and only if (d f )p = 0.

Similar to vector fields, 1-forms can be multiplied by functions; hence one hasmore general examples of 1-forms as finite sums,

α = ∑i

fi dgi

where fi,gi ∈C∞(M).Let us examine what the 1-forms are for open subsets U ⊆Rm. Given α ∈Ω 1(U),

we have

α =m

∑i=1

αi dxi

with coefficient functions αi =⟨α, ∂

∂xi

⟩∈C∞(U). (Indeed, the right hand side takes

on the correct values at any p ∈ U , and is uniquely determined by those values.)General vector fields on U may be written

X =m

∑j=1

X j ∂

∂x j

(to match the notation for 1-forms, we write the coefficients as X i rather than ai, aswe did in the past), where the coefficient functions are recovered as X j = 〈dx j, X〉.The pairing of the 1-form α with the vector field X is then

〈α, X〉=m

∑i=1

αiX i.

Lemma 6.3. Let α : p 7→ αp ∈ T ∗p M be a collection of covectors. Then α defines a1-form, with

α(X)p = αp(Xp)

for p ∈M, if and only if for all charts (U,ϕ), the coefficient functions for α in thechart are smooth.

Proof. This is similar to the discussion for vector fields, and is left as an exercise.

6.5 Pull-backs of function and 1-forms

Recall again that for any manifold M, the vector space C∞(M) of smooth functions isan algebra, with product the pointwise multiplication. Any smooth map F : M→M′

between manifolds defined an algebra homomorphism, called the pull-back

6.5 Pull-backs of function and 1-forms 121

F∗ : C∞(M′)→C∞(M), f 7→ F∗( f ) := f F.

The fact that this preserves products is the following simple calculation:

(F∗( f )F∗(g))(p) = f (F(p))g(F(p)) = ( f g)(F(p)) = F∗( f g)(p).

Given another smooth map F ′ : M′→M′′ we have

(F ′ F)∗ = F∗ (F ′)∗

(note the ordering).Let F ∈ C∞(M,N) be a smooth map. Recall that for vector fields, there is no

general ‘push-forward’ or ‘pull-back’ operation, unless F is a diffeomorphism. For1-forms the situation is better. Indeed, for any p ∈M one has the dual to the tangentmap

T ∗p F = (TpF)∗ : T ∗F(p)N→ T ∗p M.

For a 1-form β ∈Ω 1(N), we can therefore define

(F∗β )p := (T ∗p F)(βF(p)).

Lemma 6.4. The collection of co-vectors (F∗β )p ∈ T ∗p M depends smoothly on p,defining a 1-form F∗β ∈Ω 1(M).

Proof. By working on local coordinates, we may assume that M is an open subsetU ⊆ Rm, and N is an open subset V ⊆ Rn. Write

β =n

∑j=1

β j(y)dy j.

By (6.4), the pull-back of β is given by

F∗β =m

∑i=1

( n

∑j=1

β j(F(x))∂F j

∂xi

)dxi.

In particular, the coefficients are smooth. ut

The Lemma shows that we have a well-defined pull-back map

F∗ : Ω1(N)→Ω

1(M), β 7→ F∗β .

Under composition of two maps, (F1 F2)∗ = F∗2 F∗1 . The pull-back of forms is

related to the pull-back of functions, g 7→ F∗g = gF :

Proposition 6.2. For g ∈C∞(N),

F∗(dg) = d(F∗g).


Proof. We have to show (F∗(dg))p = (d(F∗g))p for all p ∈ M. But this is justLemma 6.1. ut

Remark 6.2. Recall once again that while F ∈ C∞(M,N) induces a tangent mapT F ∈ C∞(T M,T N), there is no natural push-forward operation for vector fields.By contrast, for cotangent bundles there is no naturally induced map from T ∗N toT ∗M (or the other way), yet there is a natural pull-back operation for 1-forms!

In the case of vector fields, rather than working with ‘F∗(X)’ one has the notion ofrelated vector fields, X ∼F Y . For any related vector fields X ∼F Y , and β ∈Ω 1(N),we then have that

(F∗β )(X) = F∗(β (Y )).

Indeed, at any given p ∈M this just becomes the definition of the pullback map.

6.6 Integration of 1-forms

Given a curve γ : J→M in a manifold, and any 1-form α ∈Ω 1(M), we can considerthe pull-back γ∗α ∈Ω 1(J). By the description of 1-forms on R, this is of the form

γ∗α = f (t)dt

for a smooth function f ∈C∞(J).To discuss integration, it is convenient to work with closed intervals rather than

open intervals. Let [a,b] ⊆ R be a closed interval. A map γ : [a,b]→ M into amanifold will be called smooth if it extends to a smooth map from an open intervalcontaining [a,b]. We will call such a map a smooth path.

Definition 6.5. Given a smooth path γ : [a,b]→ M, we define the integral of a 1-form α ∈Ω 1(M) along γ as ∫

γ

α =∫ b

aγ∗α.

The fundamental theorem of calculus has the following consequence for manifolds.It is a special case of Stokes’ theorem.

Proposition 6.3. Let γ : [a,b]→M be a smooth path, with γ(a) = p, γ(b) = q. Forany f ∈C∞(M), we have ∫

γ

d f = f (q)− f (p).

In particular, the integral of d f depends only on the end points of the path, ratherthan the path itself.

Proof. We have

γ∗d f = dγ

∗ f = d( f γ) =∂ ( f γ)

∂ tdt.

6.7 2-forms 123

Integrating from a to b, we obtain, by the fundamental theorem of calculus, f (γ(b))−f (γ(a)). ut

A 1-form α ∈ Ω 1(M) such that α = d f for some function f ∈ C∞(M) is calledexact.

Example 6.2. Consider the 1-form

α = y2exdx+2yexdy ∈Ω(R2).

Problem: Find the integral of α along the path

γ : [0,1]→M, t 7→ (sin(πt/2), t3).

Solution: Observe that the 1-form α is exact:

α = d(y2ex)= d f

with f (x,y) = y2ex. The path has end points γ(0) = (0,0) and γ(1) = (1,1). Hence,∫γ

α = f (γ(1))− f (γ(0)) = e.

Note that the integral over, say, α = y2exdx would be much harder.

Remark 6.3. The proposition gives a necessary condition for exactness: The integralof α along paths should depend only on the end points. This condition is also suf-ficient, since we can define f on the connected components of M, by fixing a basepoint p0 on each such component, and putting f (p) =

∫γ

α for any path from p0 top.

If M is an open subset U ⊆ Rm, so that α = ∑i αidxi, then α = d f means thatαi =

∂ f∂xi . A necessary condition is the equality of partial derivatives,

∂αi

∂x j =∂α j

∂xi ,

In multivariable calculus one learns that this condition is also sufficient, providedU is simply connected (e.g., convex). Using the exterior differential of forms inΩ 1(U), this condition becomes dα = 0. To obtain a coordinate-free version of thecondition, we need higher order forms.

6.7 2-forms

To get a feeling for higher degree forms, and constructions with higher forms, wefirst discuss 2-forms.


Definition 6.6. A 2-form on M is a C∞(M)-bilinear skew-symmetric map

α : X(M)×X(M)→C∞(M), (X ,Y ) 7→ α(X ,Y ).

Here skew-symmetry means that α(X ,Y ) = −α(Y,X) for all vector fields X ,Y ,while C∞(M)-bilinearity means

α( f X ,Y ) = f α(X ,Y ) = α(X , fY )

for f ∈ C∞(M), as well as α(X ′+X ′′,Y ) = α(X ′,Y )+α(X ′′,Y ), and similarly inthe second argument. (Actually, by skew-symmetry it suffices to require C∞(M)-linearity in the first argument.) By the same argument as for 1-forms, the valueα(X ,Y )p depends only on the values Xp,Yp. Also, if α is a 2-form then so is f α forany smooth function f .

First examples of 2-forms are obtained from 1-forms: Let α,β ∈ Ω 1(M). Thenwe define a wedge product α ∧β ∈Ω 2(M), as follows:

(α ∧β )(X ,Y ) = α(X)β (Y )−α(Y )β (X).

This is well-defined, since the right hand side is skew-symmetric and bi-linear in Xand Y .

For an open subset U ⊆ Rm, a 2-form ω ∈ Ω 2(U) is uniquely determined by itsvalues on coordinate vector fields. By skew-symmetry the functions

ωi j = ω

(∂

∂xi ,∂

∂x j

)satisfy ωi j = −ω ji; hence it suffices to know these functions for i < j. As a conse-quence, we see that the most general 2-form on U is

ω =12

m

∑i, j=1

ωi jdxi∧dx j = ∑i< j

ωi jdxi∧dx j.

6.8 k-forms

We now generalize to forms of arbitrary degree.

6.8.1 Definition

Definition 6.7. Let k be a non-negative integer. A k-form on M is a C∞(M)-multilinear,skew-symmetric map

6.8 k-forms 125

α : X(M)×·· ·×X(M)︸︷︷︸k times

→C∞(M).

The space of k-forms is denoted Ω k(M); in particular Ω 0(M) =C∞(M).

Here, skew-symmetry means that α(X1, . . . ,Xk) changes sign under exchange of anytwo of its elements. For example, α(X1,X2,X3, . . .) =−α(X2,X1,X3, . . .). More gen-erally, if Sk is the group of permutations of 1, . . . ,k, and sign(s) is the sign of apermutation s ∈Sk (+1 for an even permutation, −1 for an odd permutation) then

α(Xs(1), . . . ,Xs(k)) = sign(s)α(X1, . . . ,Xk).

The C∞(M)-multilinearity means C∞(M)-linearity in each argument, similar to thecondition for 2-forms. It implies, in particular, α is local in the sense that the value ofα(X1, . . . ,Xk) at any given p ∈M depends only on the values X1|p, . . . ,Xk|p ∈ TpM.One thus obtains a skew-symmetric multilinear form

αp : TpM×·· ·×TpM→ R,

for all p ∈M.If α1, . . . ,αk are 1-forms, then one obtains a k-form α =: α1∧ . . .∧αk by ‘wedge

product’.

(α1∧ . . .∧αk)(X1, . . . ,Xk) = ∑s∈Sk

sign(s)α1(Xs(1)) · · ·αk(Xs(k)).

(More general wedge products will be discussed below.) Here, the signed summa-tion over the permutation group guarantees that the result is skew-symmetric.

Using C∞-multilinearity, a k-form on U ⊆ Rm is uniquely determined by its val-ues on coordinate vector fields ∂

∂x1 , . . . ,∂

∂xm , i.e. by the functions

αi1...ik = α

(∂

∂xi1, . . . ,

∂

∂xik

).

Moreover, by skew-symmetry we only need to consider ordered index sets I =i1, . . . , ik ⊆ 1, . . . ,m, that is, i1 < .. . < ik. Using the wedge product notation,we obtain

α = ∑i1<···<ik

αi1...ik dxi1 ∧·· ·dxik .

6.8.2 Wedge product

We next turn to the definition of a wedge product of forms α ∈ Ω k(M) and β ∈Ω l(M). A permutation s ∈Sk+l is called a k, l shuffle if it satisfies

s(1)< .. . < s(k), s(k+1)< .. . < s(k+ l).


Definition 6.8. The wedge product of α ∈Ω k(M), β ∈Ω l(M) is the element

α ∧β ∈Ωk+l(M)

given as

(α ∧β )(X1, . . . ,Xk+l) = ∑sign(s)α(Xs(1), . . . ,Xs(k)) β (Xs(k+1), . . . ,Xs(k+l))

where the sum is over all k, l-shuffles.

Example 6.3. For α,β ∈Ω 2(M),

(α ∧β )(X ,Y,Z,W ) = α(X ,Y )β (Z,W )−α(X ,Z)β (Y,W )

+α(X ,W )β (Y,Z)+α(Y,Z)β (X ,W )

−α(Y,W )α(X ,Z)+α(Z,W )β (X ,Y ).

ut

The wedge product is graded commutative: If α ∈Ω k(M) and β ∈Ω l(M) then

α ∧β = (−1)klβ ∧α.

Furthermore, it is associative:

Lemma 6.5. Given αi ∈Ωki(M) we have

(α1∧α2)∧α3 = α1∧ (α2∧α3)

Proof. For both sides, the evaluation on X1, . . . ,Xk with k = k1 + k2 + k3, is a signedsum over all k1,k2,k3-shuffles (it should be clear how this is defined).

So, we may in fact drop the parentheses when writing wedge products.

6.8.3 Exterior differential

Recall that we defined the exterior differential on functions by the formula

(d f )(X) = X( f ). (6.5)

we will now extend this definition to all forms.

Theorem 6.1. There is a unique collection of linear maps d : Ω k(M)→Ω k+1(M),extending the map (6.5) for k = 0, such that d(d f ) = 0 and satisfying the gradedproduct rule,

d(α ∧β ) = dα ∧β +(−1)kα ∧dβ (6.6)

for α ∈Ω k(M) and β ∈Ω l(M). This exterior differential satisfies dd = 0.

6.8 k-forms 127

Proof. Suppose first that such an exterior differential is given. Then d is local, in thesense that for any open subset U ⊆M the restriction (dα)|U depends only on α|U ,or equivalently (dα)|U = 0 when α|U = 0. Indeed, if this is the case and p ∈U , wemay choose f ∈C∞(M) = Ω 0(M) such that f vanishes on M\U and f |p = 1. Thenf α = 0, hence the product rule (6.6) gives

0 = d( f α) = d f ∧α + f dα.

Evaluating at p we obtain (dα)p = 0 as claimed. Using locality, we may thus workin local coordinates. If α ∈Ω 1(M) is locally given by

α = ∑i1<···<ik

αi1···ik dxi1 ∧·· ·∧dxik ,

then the product rule together with ddxi = 0 forces us to define

dα = ∑i1<···<ik

dαi1···ik ∧dxi1 ∧·· ·∧dxik =m

∑l=1

∑i1<···<ik

∂αi1···ik∂xl dxl ∧dxi1 ∧·· ·∧dxik .

Conversely, we may use this explicit formula (cf. (6.1)) to define dα|U for a coor-dinate chart domain U ; by uniqueness the local definitions on overlas of coordinatechart domains agree. Proposition 6.1 shows that (ddα)|U = 0, hence it also holdsglobally. ut

Definition 6.9. A k-form ω ∈ Ω k(M) is called exact if ω = dα for some α ∈Ω k−1(M). It is called closed if dω = 0.

Since dd = 0, the exact k-forms are a subspace of the space of closed k-forms. Forthe case of 1-forms, we had seen that the integral

∫γ

α of an exact 1-form α = d falong a smooth path γ : [a,b]→ M is given by the difference of the values at theend points; a necessary condition for α to be exact is that it is closed. An exampleof a 1-form that is closed but not exact is

α =ydx− xdy

x2 + y2 ∈Ω1(R2\0).

Remark 6.4. The quotient space (closed k-forms modulo exact k-forms) is a vectorspace called the k-th (de Rham) cohomology

Hk(M) =α ∈Ω k(M)| α is closed α ∈Ω k(M)| α is exact

.

It turns out that whenever M is compact (and often also if M is non-compact), Hk(M)is a finite-dimensional vector space. The dimension of this vector space

bk(M) = dimHk(M)


is called the k-th Betti number of M; these numbers are important invariants ofM which one can use to distinguish non-diffeomorphic manifolds. For example, ifM = CPn one can show that

bk(CPn) = 1 for k = 0,2, . . . ,2n

and bk(CPn) = 0 otherwise. For M = SN the Betti numbers are

bk(Sn) = 1 for k = 0,n

while bk(Sn) = 0 for all other k. Hence CPn cannot be diffeomorphic to S2n unlessn = 1.

6.9 Lie derivatives and contractions

Given a vector field X , and a k-form α ∈Ω k(M), we can define a k−1-form

ιX α ∈Ωk−1(M)

by contraction: Thinking of α as a multi-linear form, one simply puts X into thefirst slot:

(ιX α)(X1, . . . ,Xk−1) = α(X ,X1, . . . ,Xk−1).

Contractions have the following compatibility with the wedge product, similar tothat for the exterior differential:

ιX (α ∧β ) = ιX α ∧β +(−1)kα ∧ ιX β , (6.7)

for α ∈ Ω k(M),β ∈ Ω l(M), which one verifies by evaluating both sides on vectorfields. Another important operator on forms is the Lie derivative:

Theorem 6.2. Given a vector field X, there is a unique collection of linear mapsLX : Ω k(M)→Ω k(M), such that

LX ( f ) = X( f ), LX (d f ) = dX( f ),

and satisfying the product rule,

LX (α ∧β ) = LX α ∧β +α ∧LX β (6.8)

for α ∈Ω k(M) and β ∈Ω l(M).

Proof. As in the case of the exterior differential, we can use the product rule to showthat LX is local: (LX α)|U depends only on α|U and X |U . Since any differential formis a sum of wedge products of 1-forms, LX is uniquely determined by its action onfunctions and differential of functions. This proves uniqueness. For existence, wegive the following formula:

6.9 Lie derivatives and contractions 129

LX = d ιX + ιX d.

On functions, this gives the correct result since

LX f = ιX d f = X( f ),

and also on differentials of functions since

LX d f = dιX d f = dLX f = 0.

ut

To summarize, we have introduced three operators

d : Ωk(M)→Ω

k+1(M), LX : Ωk(M)→Ω

k(M), ιX : Ωk(M)→Ω

k−1(M).

These have the following compatibilities with the wedge product: For α ∈ Ω k(M)and β ∈Ω l(M) one has

d(α ∧β ) = (dα)∧β +(−1)kα ∧dβ ,

LX (α ∧β ) = (LX α)∧β +α ∧LX β ,

ιX (α ∧β ) = (ιX α)∧β +(−1)kα ∧ ιX β .

One says that LX is an even derivation relative to the wedge product, whereas d, ιXare odd derivations. They also satisfy important relations among each other:

dd = 0LX LY −LY LX = L[X ,Y ]

ιX ιY + ιY ιX = 0dLX −LX d = 0

LX ιY − ιY LX = ι[X ,Y ]

ιX d+d ιX = LX .

Again, the signs are determined by the even/odd parity of these operators; oneshould think of the left hand side as ‘graded’ commutators, where a plus sign ap-pears whenever two entries are odd. Writing [·, ·] for the graded commutators (withthe agreement that the commutator of two odd operators has a sign built in) the iden-tities becomes [d,d] = 0, [LX ,LY ] = L[X ,Y ], [ιX , ιY ] = 0, [d,LX ] = 0, [LX , ιY ] = ι[X ,Y ]and [d, ιX ] = LX .

This collection of identities is referred to as the Cartan calculus, after Elie Cartan(1861-1951), and in particular the last identity (which certainly is the most intrigu-ing) is called the Cartan formula. Basic contributions to the theory of differentialforms were made by his son Henri Cartan (1906-1980), who also wrote a textbookon the subject.


Exercise: Prove these identities. (Note that some have already been established.)Hint: First check that the left hand side satisfies a (graded) product rule with respectto wedge product. It therefore suffices to check that both sides agree on functions fand differentials of functions d f .

As an illustration of the Cartan identities, let us prove the following formula forthe exterior differential of a 1-form α ∈Ω 1(M):

(dα)(X ,Y ) = LX (α(Y ))−LY (α(X))−α([X ,Y ]).

(In the Cartan Calculus, we prefer to wrote LX f instead of X( f ) since expressionssuch as X(α(Y )) would look too confusing.) The calculation goes as follows:

(dα)(X ,Y ) = ιY ιX dα

= ιY LX α− ιY dιX α

= LX ιY α− ι[X ,Y ]α−LY ιX α +dιY ιX α

= LX (α(Y ))−LY (α(X))−α([X ,Y ]).

In the last step we used that ιY ιX α = 0, because α is a 1-form.

Exercise: Prove a similar formula for the exterior differential of a 2-form, and tryto generalize to arbitrary k-forms.

6.9.1 Pull-backs

Similar to the pull-back of functions (0-forms) and 1-forms, we have a pull-backoperation for k-forms,

F∗ : Ωk(N)→Ω

k(M)

for any smooth map between manifolds, F ∈C∞(M,N). Its evaluation at any p ∈Mis given by

(F∗β )p(v1, . . . ,vk) = βF(p)(TpF(v1), . . . ,TpF(vk)).

The pull-back map satisfies d(F∗β ) = F∗dβ , and for a wedge product of forms,

F∗(β1∧β2) = F∗β1∧F∗β2.

In local coordinates, if F : U → V is a smooth map between open subsets of Rm

and Rn, with coordinates x,y, the pull-back just amounts to ‘putting y = F(x)’.

Example 6.4. If F : R3→ R2 is given by (u,v) = F(x,y,z) = (y2z, x) then

F∗(du∧dv) = d(y2z)∧dx = y2dz∧dx+2yzdy∧dx.

The next example is very important, hence we state it as a proposition. It is the‘key fact’ toward the definition of an integral.

6.9 Lie derivatives and contractions 131

Proposition 6.4. Let U ⊆ Rm with coordinates xi, and V ⊆ Rn with coordinates y j.Suppose m = j, and F ∈C∞(U,V ). Then

F∗(dy1∧·· ·∧dyn) = J dx1∧·· ·∧dxn

where J(x) is the determinant of the Jacobian matrix,

J(x) = det(

∂F i

∂x j

)n

i, j=1.

Proof.

F∗β = dF1∧·· ·∧dFn

= ∑i1...in

∂F1

∂xi1· · · ∂Fn

∂xindxi1 ∧·· ·∧dxin

= ∑s∈Sn

∂F1

∂xs(1) · · ·∂Fn

∂xs(n)dxs(1)∧·· ·∧dxs(n)

= ∑s∈Sn

sign(s)∂F1

∂xs(1) · · ·∂Fn

∂xs(n)dx1∧·· ·∧dxn

= J dx1∧·· ·∧dxn,

Here we noted that the wedge product dxi1 ∧ ·· ·∧dxin is zero unless i1, . . . , in are apermutation of 1, . . . ,n. ut

One may regard this result as giving a new, ‘better’ definition of the Jacobian deter-minant.

Remark 6.5. The Lie derivative LX α of a differential form with respect to a vectorfield X has an important interpretation in terms of the flow Φt of X . Assuming forsimplicity that X is complete (so that Φt is a globally defined diffeomorphism), onehas the formula

LX α =ddt

∣∣∣t=0

Φ∗t α.

(If X is incomplete, the flow Φt is defined only locally, but the definition still works.)To prove this identity, it suffices to check that the right hand side satisfies a productrule with respect to the wedge product of forms, and that it takes on the correctvalues on functions and on differentials of functions. The formula shows that LXmeasures to what extent α is invariant under the flow of X .


6.10 Integration of differential forms

Differential forms of top degree can be integrated over oriented manifolds. Let M bean oriented manifold of dimension m, and ω ∈Ω m(M). Let supp(ω) be the supportof ω . 6

If supp(ω) is contained in an oriented coordinate chart (U,ϕ), then one defines∫M

ω =∫Rm

f (x)dx1 · · ·dxm

where f ∈C∞(Rm) is the function, with supp( f )⊆ ϕ(U), determined from

(ϕ−1)∗ω = f dx1∧·· ·∧dxm.

This definition does not depend on the choice of oriented chart. Indeed, suppose(V,ψ) is another oriented chart with supp(ω)⊆V , and write

(ψ−1)∗ω = g dy1∧·· ·∧dym.

where we write y1, . . . ,ym for the coordinates on V . Letting F = ψ ϕ−1 be thechange of coordinates y = F(x), Proposition 6.4 says that

F∗(dy1∧·· ·∧dym) = J(x)dx1∧·· ·∧dxm,

where J(x) = det(DF(x)) is the determinant of the Jacobian matrix of F at x. Hence,f (x) = g(F(x))J(x), and we obtain∫

ψ(U)g(y)dy1 · · ·ym =

∫ϕ(U)

g(F(x))J(x)dx1 · · ·dxm =∫

ϕ(U)f (x)dx1 · · ·dxm,

as required.

Remark 6.6. Here we used the change-of-variables formula from multivariable cal-culus. It was very important that the charts are oriented, so that J > 0 everywhere.Indeed, for general changes of variables, the change-of-variables formula involves|J| rather than J itself.

If ω is not necessarily supported in a single oriented chart, we proceed as follows.Let (Ui,ϕi), i = 1, . . . ,r be a finite collection of oriented charts covering supp(ω).Together with U0 = M\supp(ω) this is an open cover of M.

Lemma 6.6. Given a finite open cover of a manifold there exists a partition of unitysubordinate to the cover, i.e. functions χi ∈C∞(M) with supp(χi)⊆Ui and ∑

ri=0 χi =

1.

6 The support of a form is defined similar to the support of a function, or support of a vector field.For any differential form α ∈ Ω k(M), we define the support supp(α) to be the smallest closedsubset of M outside of which α is zero. (Equivalently, it is the closure of the subset over which α

is non-zero.)

6.12 Stokes’ theorem 133

Indeed, partitions of unity exists for any open cover, not only finite ones. A proof isgiven in the appendix on ‘topology of manifolds’.

Let χ0, . . . ,χr be a partition of unity subordinate to this cover. We define∫M

ω =r

∑i=1

∫M

χiω

where the summands are defined as above, since χiω is supported in Ui for i≥ 1. (Wedidn’t include the term for i = 0, since χ0ω = 0.) We have to check that this is well-defined, independent of the choices. Thus, let (Vj,ψ j) for j = 1, . . . ,s be anothercollection of oriented coordinate charts covering supp(ω), put V0 = M− supp(ω),and let σ0, . . . ,σs a corresponding partition of unity subordinate to the cover by theVi’s.

Then the Ui∩Vj form an open cover, with the collection of χiσ j as a partition ofunity. We obtain

s

∑j=1

∫M

σ jω =s

∑j=1

∫M(

r

∑i=1

χi)σ jω =s

∑j=1

r

∑i=1

∫M

σ jχiω.

This is the same as the corresponding expression for ∑ri=1∫

M χiω .

6.11 Integration over oriented submanifolds

Let M be a manifold, not necessarily oriented, and S is a k-dimensional orientedsubmanifold, with inclusion i : S→M. We define the integral over S, of any k-formω ∈Ω k(M) such that S∩ supp(ω) is compact, as follows:∫

Sω =

∫S

i∗ω.

Of course, this definition works equally well for any smooth map from S into M.For example, the integral of compactly supported 1-forms along arbitrary paths γ :R→ M is defined. Note also that M itself does not have to be oriented, it sufficesthat S is oriented.

6.12 Stokes’ theorem

Let M be an m-dimensional oriented manifold.

Definition 6.10. A region with (smooth) boundary in M is a closed subset D ⊆ Mwith the following property: There exists a smooth function f ∈C∞(M,R) such that0 is a regular value of f , and


D = p ∈M| f (p)≤ 0.

We do not consider f itself as part of the definition of D, only the existence of fis required. The interior of a region with boundary, given as the largest open subsetcontained in D, is int(D) = p ∈M| f (p)< 0, and the boundary itself is

∂D = p ∈M| f (p) = 0,

a codimension 1 submanifold (i.e., hypersurface) in M.

Example 6.5. The region with bounday defined by the function f ∈C∞(R2), givenby f (x,y) = x2 + y2−1, is the unit disk D⊆ R2; its boundary is the unit circle.

Example 6.6. Recall that for 0 < r < R, zero is a regular value of the function on R3,

f (x,y,z) = z2 +(√

x2 + y2−R)2− r2.

The corresponding region with boundary D ⊆ R3 is the solid torus, its boundary isthe torus.

Recall that we are considering D inside an oriented manifold M. The boundary∂D may be covered by oriented submanifold charts (U,ϕ), in such a way that ∂Dis given in the chart by the condition x1 = 0, and D by the condition x1 ≤ 0: 7

ϕ(U ∩D) = ϕ(U)∩x ∈ Rm| x1 ≤ 0.

(Indeed, given an oriented submanifold chart for which D lies on the side wherex1 ≥ 0, one obtains a region chart by composing with the orientation-preserving co-ordinate change (x1, . . . ,xm) 7→ (−x1,−x2,x3 . . . ,xm).) We call oriented submanifoldcharts of this kind ‘region charts’.8

Lemma 6.7. The restriction of the region charts to ∂D form an oriented atlas for∂D.

Proof. Let (U,ϕ) and (V,ψ) be two region charts, defining coordinates x1, . . . ,xm

and y1, . . . ,ym, and let F = ψ ϕ−1 : ϕ(U ∩V )→ ψ(U ∩V ), x 7→ y = F(x). Itrestricts to a map

F1 : x ∈ ϕ(U ∩V )| x1 = 0→ y ∈ ψ(U ∩V )|y1 = 0.

Since y1 > 0 if and only if x1 > 0, the change of coordinates satisfies

∂y1

∂x1

∣∣∣x1=0

> 0,∂y1

∂x j

∣∣∣x1=0

= 0, for j > 0.

7 Note that while we originally defined submanifold charts in such a way that the last m− k coor-dinates are zero on S, here we require that the first coordinate be zero. It doesn’t matter, since onecan simply reorder coordinates, but works better for our description of the ‘induced orientation’.8 This is not a standard name.


Hence, the Jacobian matrix DF(x)|x1=0 has a positive (1,1) entry, and all otherentries in the first row equal to zero. Using expansion of the determinant across thefirst row, it follows that

det(DF(0,x2, . . . ,xm)) =∂y1

∂x1

∣∣∣x1=0

det(DF ′(x2, . . . ,xm)).

which shows that det(DF ′)> 0.

In particular, ∂D is again an oriented manifold. To repeat: If x1, . . . ,xm are localcoordinates near p∈ ∂D, compatible with the orientation and such that D lies on theside x1≤ 0, then x2, . . . ,xm are local coordinates on ∂D. This convention of ‘inducedorientation’ is arranged in such a way that the Stokes’ theorem holds without extrasigns.

For an m-form ω such that supp(ω)∩D is compact, the integral∫D

ω

is defined similar to the case of D = M: One covers D∩ supp(ω) by finitely manysubmanifold charts (Ui,ϕi) with respect to ∂D (this includes charts that are entirelyin the interior of D), and puts ∫

Dω = ∑

∫D∩Ui

χiω

where the χi are supported in Ui and satisfy ∑i χi over D∩ supp(ω). By the sameargument as for D = M, this definition of the integral is independent of the choicemade.

Theorem 6.3 (Stokes’ theorem). Let M be an oriented manifold of dimension m,and D ⊆ M a region with smooth boundary ∂D. Let α ∈ Ω m−1(M) be a form ofdegree m−1, such that supp(α)∩D is compact. Then∫

Ddα =

∫∂D

α.

As explained above, the right hand side means∫

∂D i∗α , where i : ∂D → M is theinclusion map.

Proof. We will see that Stokes’ theorem is just a coordinate-free version of thefundamental theorem of calculus. Let (Ui,ϕi) for i = 1, . . . ,r be a finite collectionof region charts covering supp(α)∩D. Let χ1, . . . ,χr ∈ C∞(M) be functions withχi ≥ 0, supp(χi) ⊆ Ui, and such that χ1 + . . .+ χr is equal to 1 on supp(α)∩D.(E.g., we may take U1, . . . ,Ur together with U0 = M\supp(ω) as an open covering,and take the χ0, . . . ,χr ∈C∞(M) to be a partition of unity subordinate to this cover.)Since ∫

Ddα =

r

∑i=1

∫D

d(χiα),∫

∂Dα =

r

∑i=1

∫∂D

χiα,


it suffices to consider the case that α is supported in a region chart.Using the corresponding coordinates, it hence suffices to prove Stokes’ theorem

for the case that α ∈Ω m−1(Rm) is a compactly supported form in Rm, and D = x∈Rm|x1 ≤ 0. That is, α has the form

α =m

∑i=1

fi dx1∧·· · dxi∧·· ·∧dxm,

with compactly supported fi where the hat means that the corresponding factor is tobe omitted. Only the i = 1 term contributes to the integral over ∂D = Rm−1, and∫

Rm−1α =

∫f1(0,x2, . . . ,xm) dx2 · · ·dxm.

On the other hand,

dα =( m

∑i=1

(−1)i+1 ∂ fi

∂xi

)dx1∧·· ·∧dxm

Let us integrate each summand over the region D given by x1 ≤ 0. For i > 1, wehave ∫

∞

∞

· · ·∫

∞

−∞

∫ 0

−∞

∂ fi

∂xi(x1, . . . ,xm)dx1 · · ·dxm = 0

where we used Fubini’s theorem to carry out the xi-integration first, and applied thefundamental theorem of calculus to the xi-integration (keeping the other variablesfixed, the integrand is the derivative of a compactly supported function). It remainsto consider the case i= 1. Here we have, again by applying the fundamental theoremof calculus, ∫

Ddα =

∫∞

∞

· · ·∫

∞

−∞

∫ 0

−∞

∂ f1

∂x1(x1, . . . ,xm)dx1 · · ·dxm

=∫

∞

∞

· · ·∫

∞

−∞

fm(0,x2, . . . ,xm)dx2 · · ·dxm =∫

∂Dα

utAs a special case, we have

Corollary 6.1. Let α ∈ Ω m−1(M) be a compactly supported form on the orientedmanifold M. Then ∫

Mdα = 0.

Note that it does not suffice that dα has compact support. For example, if f (t) is afunction with f (t) = 0 for t < 0 and f (t) = 1 for t > 0, then d f has compact support,but

∫R d f = 1.

A typical application of Stokes’ theorem shows that for a closed form ω ∈Ω k(M), the integral of ω over an oriented compact submanifold does not changewith smooth deformations of the submanifold.


Theorem 6.4. Let ω ∈Ω k(M) be a closed form on a manifold M, and S a compact,oriented manifold of dimension k. Let F ∈C∞(R×S,M) be a smooth map, thoughtof as a smooth family of maps

Ft = F(t, ·) : S→M.

Then the integrals ∫S

F∗t ω

do not depend on t.

If Ft is an embedding, then this is the integral of ω over the submanifold Ft(S)⊆M.

Proof. Let a < b, and consider the domain D = [a,b]×S⊆R×S. The boundary ∂Dhas two components, both diffeomorphic to S. At t = b the orientation is the givenorientation on S, while at t = a we get the opposite orientation. Hence,

0 =∫

DF∗dω =

∫D

dF∗ω =∫

∂DF∗ω =

∫S

F∗b ω−∫

SF∗a ω.

Hence∫

S F∗b ω =∫

S F∗a ω . ut

Remark 6.7. Note that if one member of this family of maps, say the map F1, takesvalues in a k− 1-dimensional submanifold (for instance, if F1 is a constant map),then F∗1 ω = 0. (Indeed, the assumption means that F1 = j F ′1, where j is the inclu-sion of a k− 1-submanifold and F ′1 takes values in that submanifold. But j∗ω = 0for degree reasons.) It then follows that

∫S F∗t ω = 0 for all t.

Given a smooth map ϕ : S→ M, one refers to a smooth map F : R× S→ Mwith F0 = ϕ as an ‘smooth deformation’ (or ‘isotopy’) of ϕ . We say that ϕ canbe smoothly deformed into ϕ ′ if there exists a smooth isotopy F with ϕ = F0 andϕ ′ = F1. The theorem shows that if S is oriented, and if there is a closed formω ∈Ω k(M) with ∫

Sϕ∗ω 6=

∫S(ϕ ′)∗ω

then ϕ cannot be smoothly deformed into ϕ ′. This observation has many applica-tions; here are some of them. 9

Example 6.7. Suppose ϕ : S→M is a smooth map, where S is oriented of dimensionk, and ω ∈Ω k(M) is closed. If

∫S ϕ∗ω 6= 0, then ϕ cannot smoothly be ‘deformed’

into a map taking values in a lower-dimensional submanifold. (In particular it cannotbe deformed into a constant map.) Indeed, if ϕ ′ takes values in a lower-dimensionalsubmanifold, then ϕ ′ = jϕ ′1 where j is the inclusion of that submanifold. But then

9 You may wonder if it is still possible to find a continuous deformation, rather than smooth. Itturns out that it doesn’t help: Results from differential topology show that two smooth maps canbe smoothly deformed into each other if and only if they can be continuously deformed into eachother.


j∗ω = 0, hence (ϕ ′)∗ω = 0. For instance, the inclusion ϕ : S2 → M = R3\0cannot be smoothly deformed inside M so that ϕ ′ would take values in R2\0 ⊆R3\0.

Example 6.8 (Winding number). Let ω ∈Ω 2(R2\0) be the 1-form

ω =1

x2 + y2 (xdy− ydx)

In polar coordinates x = r cosθ , y = r sinθ , one has that ω = dθ . Using this factone sees that ω is closed (but not exact, since θ is not a globally defined functionon R2\0.) Hence, if

γ : S1→ R2\0

is any smooth map (a ‘loop’), then the integral∫S1

γ∗ω

does not change under deformations (isotopies) of the loop. In particular, γ cannotbe deformed into a constant map, unless the integral is zero. The number

w(γ) =1

2π

∫S1

γ∗ω

is the winding number of γ . (One can show that this is always an integer, and thattwo loops can be deformed into each other if and only if they have the same windingnumber.)

Example 6.9 (Linking number). Let f ,g : S1 → R3 be two smooth maps whoseimages don’t intersect, that is, with f (z) 6= g(w) for all z,w ∈ S1 (we regard S1 asthe unit circle in C). Define a new map

F : S1×S1→ S2, (z,w) 7→ f (z)−g(w)|| f (z)−g(w)||

.

On S2, we have a 2-form ω of total integral 4π . It is the pullback of

xdy∧dz− ydx∧dz+ zdx∧dy ∈Ω2(R3)

to the 2-sphere. The integral

L( f ,g) =1

4π

∫S1×S1

F∗ω

is called the linking number of f and g. (One can show that this is always an integer.)Note that if it is possible to deform one of the loops, say f , into a constant loopthrough loops that are always disjoint from g, then the linking number is zero. In hiscase, we consider f ,g as ‘unlinked’.

6.13 Volume forms 139

6.13 Volume forms

A top degree differential form Γ ∈ Ω m(M) is called a volume form if it is non-vanishing everywhere: Γp 6= 0 for all p ∈M. In a local coordinate chart (U,ϕ), thismeans that

(ϕ−1)∗Γ = f dx1∧·· ·∧dxm

where f (x) 6= 0 for all x ∈ ϕ(U).

Example 6.10. The Euclidean space Rn has a standard volume form Γ0 = dx1∧·· ·∧dxn. Suppose S ⊆ Rn is a submanifold of dimension n−1, and X a vector field thatis nowhere tangent to S. Let i : S→ Rn be the inclusion. Then

Γ := i∗(ιXΓ0

)∈Ω

n−1(S)

is a volume form. For instance, if S is given as a level set f−1(0), where 0 is a regularvalue of f , then the gradient vector field

n

∑i=1

∂ f∂xi

∂

∂xi

has this property.

Exercise: Verify the claim that Γ := i∗(ιXΓ0

)is a volume form.

Example 6.11. Let i : Sn → Rn+1 be the inclusion of the standard n-sphere. LetX = ∑

ni=0 xi ∂

∂xi . Then

ιX (dx0∧·· ·∧dxn) =n

∑i=0

(−1)ixidx1∧·· ·dxi−1∧dxi+1∧·· ·∧dxn

pulls back to a volume form on Sn.

Lemma 6.8. A volume form Γ ∈Ω m(M) determines an orientation on M, by takingas the oriented charts those charts (U,ϕ) such that (ϕ−1)∗Γ = f dx1 ∧ ·· · ∧ dxm

with f > 0 everywhere on Φ(U).

Proof. We have to check that the condition is consistent. Suppose (U,ϕ) and (V,ψ)are two charts, where (ϕ−1)∗Γ = f dx1∧·· ·∧dxm and (ψ−1)∗Γ = g dy1∧·· ·∧dym

with f > 0 and g > 0. If U ∩V is non-empty, let F = ψ ϕ−1 : ϕ(U)→ ψ(V ) bethe transition function. Then

F∗(ψ−1)∗Γ |U∩V = (ϕ−1)∗Γ |U∩V ,

henceg(F(x)) J(x) dx1∧·· ·∧dxm = f (x) dx1∧·· ·∧dxm.


where J is that Jacobian determinant of the transition map F = ψ ϕ−1. Hencef = J (gF) on ϕ(U ∩V ). Since f > 0 and g > 0, it follows that J > 0. Hence thetwo charts are oriented compatible. ut

Theorem 6.5. A manifold M is orientable if and only if it admits a volume form.In this case, any two volume forms compatible with the orientation differ by aneverywhere positive smooth function:

Γ′ = fΓ , f > 0.

Proof. As we saw above, any volume form determines an orientation. Conversely,if M is an oriented manifold, there exists a volume form compatible with the orien-tation: Let (Uα ,ϕα) be an oriented atlas on M. Then each

Γα = ϕ∗α(dx1∧·· ·∧dxm) ∈Ω

m(Uα)

is a volume form on Uα ; on overlaps Uα ∩Uβ these are related by the Jacobiandeterminants of the transition functions, which are strictly positive functions. Letχα be a locally finite partition of unity subordinate to the cover Uα, see Ap-pendix A.4. The forms χαΓα have compact support in Uα , hence they extend byzero to global forms on M (somewhat imprecisely, we use the same notation for thisextension). The sum

Γ = ∑α

χαΓα ∈Ωm(M)

is a well-defined volume form. Indeed, near any point p at least one of the summandsis non-zero; and if other summands in this sum are non-zero, they differ by a positivefunction.

For a compact manifold M with a given volume form Γ ∈ Ω m(M), one can definethe volume of M,

vol(M) =∫

MΓ .

Here the orientation used in the definition of the integral is taken to be the orientationgiven by Γ . Thus vol(M)> 0.

Note that volume forms are always closed, for degree reasons (since Ω m+1(M) =0). But on a compact manifold, they cannot be exact:

Theorem 6.6. Let M be a compact manifold with a volume form Γ ∈Ω m(M). ThenΓ cannot be exact.

Proof. We have vol(M) =∫

M Γ > 0. But if Γ were exact, then Stokes’ theoremwould give

∫M Γ = 0.

Of course, the compactness of M is essential here: For instance, dx is an exact vol-ume form on R.

Appendix ATopology of manifolds

A.1 Topological notions

A topological space is a set X together with a collection of subsets U ⊆ X calledopen subsets, with the following properties:

• /0,X are open.• If U,U ′ are open then U ∩U ′ is open.• For any collection Ui of open subsets, the union

⋃i Ui is open.

The collection of open subsets is called the topology of X . In the third condition, theindex set need not be finite, or even countable.

The space Rm has a standard topology given by the usual open subsets. Likewise,the open subsets of a manifold M define a topology on M. For any set X , one hasthe trivial topology where the only open subsets are /0 and X , and the discrete topol-ogy where every subset is considered open. An open neighborhood of a point p isan open subset containing it. A topological space is called Hausdorff of any twodistinct points have disjoint open neighborhoods.

Let X be a topological space. Then any subset A ⊆ X has a subspace topology,with open sets the collection of all intersections U ∩A such that U ⊆ X is open.Given a surjective map q : X → Y , the space Y inherits a quotient topology, whoseopen sets are all V ⊆Y such that the pre-image q−1(V ) = x∈ X | q(x)∈V is open.

A subset A is closed if its complement X\A is open. Dual to the statements foropen sets, one has

• /0,X are closed.• If A,A′ are closed then A∪A′ is closed.• For any collection Ai of open subsets, the intersection

⋂i Ai is closed.

For any subset A, denote by A its closure, given as the smallest closed subset con-taining A.

141

142 A Topology of manifolds

A.2 Manifolds are second countable

A basis for the topology on X is a collection B = Uα of open subsets of X suchthat every U is a union from sets from B.

Example A.1. The collection of all open subsets of a topological space X is a basisof the topology.

Example A.2. Let X = Rn. Then the collection of all open balls Bε(x), with ε > 0and x ∈ Rn, is a basis for the topology on Rn.

A topological space is said to be second countable if its topology has a countablebasis.

Proposition A.1. Rn is second countable.

Proof. A countable basis is given by the collection of all rational balls, by whichwe mean ε-balls Bε(x) such that x ∈ Qm and ε ∈ Q>0. To check it is a basis, letU ⊆ Rm be open, and p ∈U . Choose ε ∈Q>0 such that B2ε(p)⊆U . There exists arational point x ∈Qn with ||x− p||< ε . This then satisfies p ∈ Bε(x)⊆U . Since pwas arbitrary, this proves the claim.

The same reasoning shows that for any open subset U ⊆Rm, the rational ε-balls thatare contained in U form a basis of the topology of U .

Proposition A.2. Manifolds are second countable.

Proof. Given a manifold M, let A = (Uα ,ϕα) be a countable atlas. Then the setof all ϕ−1

α (Bε(x)), where Bε(x) is a rational ball contained in ϕα(Uα), is a countablebasis for the topology of M. Indeed, any open subset U is a countable union overall U ∩Uα , and each of these intersections is a countable union over all ϕ−1

α (Bε(x))such that Bε(x) is a rational ε-ball contained in U ∩Uα . ut

A.3 Manifolds are paracompact

A collection Uα of open subsets of X is called an open covering of A ⊆ X ifA ⊆

⋃α Uα . Consider the case A = X . A refinement of an open cover Uα of X is

an open cover Vβ of X such that each Vβ is contained in some Uα . It is a subcoverif each Vβ ’s is equal to some Uα .

A topological space X is called compact if every open cover of X has a finitesubcover. A topological space is called paracompact if every open cover Uα hasa locally finite refinement Vβ: that is, every point has an open neighborhood meet-ing only finitely many Vβ ’s.

Proposition A.3. Manifolds are paracompact.

We will need the following auxiliary result.

A.4 Partitions of unity 143

Lemma A.1. For any manifold M, there exists a sequence of open subsets W1,W2, . . .of M such that ⋃

Wi = M,

and such that each Wi has compact closure with Wi ⊆Wi+1.

Proof. Start with a a countable open cover O1,O2, . . . of M such that each Oi hascompact closure Oi. (We saw in the proof of Proposition A.2 how to construct sucha cover, by taking pre-images of ε-balls in coordinate charts.) Replacing Oi withO1∪·· ·∪Oi we may assume O1 ⊆O2 ⊆ ·· ·. For each i, the covering of the compactset Oi by the collection of all O j’s admits a finite subcover. Since the sequence ofO j’s is nested, this just means Oi is contained in O j for j sufficiently large. Wecan thus define W1,W2, . . . as a subsequence Wi = O j(i), starting with W1 = O1, andinductively letting j(i) for i > 1 be the smallest index j(i) such that W i−1 ⊆ O j(i).

Proof (Proof or Proposition A.3). Let Uα be an open cover of M. Let Wi be asequence of open sets as in the lemma. For every i, the compact subset W i+1\Wi iscontained in the open set Wi+2\W i−1, hence it is covered by the collection of opensets

(Wi+2\W i−1)∩Uα . (A.1)

By compactness, it is already covered by finitely many of the subsets (A.1). LetV (i) be this finite collection, and V =

⋃∞i=1 V (i) = Vβ the resulting countable

open cover of M.Note that by construction, if Vβ ∈ V (i), then Vβ ∩Wi−1 = /0. That is, a given Wi

meets only Vβ ’s from V (k) with k ≤ i. Since these are finitely many Vβ ’s, it followsthat the cover V = Vβ is locally finite. ut

Remark A.1. (Cf. Lang, page 35.) One can strengthen the result a bit, as follows:Given a cover Uα, we can find a refinement to a cover Vβ such that each Vβ isthe domain of a coordinate chart (Vβ ,ψβ ), with the following extra properties, forsome 0 < r < R:

(i) ψβ (Vβ ) = BR(0), and(ii) M is already covered by the smaller subsets V ′

β= ψ

−1β

(Br(0)).

To prove this, we change the second half of the proof as follows: For each p ∈W i+1\Wi choose a coordinate chart (Vp,ψp) such that ψp(p) = 0, ψp(Vp) = BR(0),and Vp ⊆ (Wi+2\W i−1)∩Uα . Let V ′p ⊆ Vp be the pre-image of Br(0). The V ′p coverW i+1\Wi; let V (i) be a finite subcover and proceed as before. This remark is usefulfor the construction of partitions of unity.

A.4 Partitions of unity

We will need the following result from multivariable calculus.

144 A Topology of manifolds

Lemma A.2 (Bump functions). For all 0 < r < R, there exists a function f ∈C∞(Rm), with f (x) = 0 for ||x|| ≥ R and f (x) = 1 for ||x|| ≤ r.

Proof. It suffices to prove the existence of a function g ∈C∞(R) such that g(t) = 0for t ≥ R and g(t) = 1 for t ≤ r: Given such g we may take f (x) = g(||x||). Toconstruct g, recall that the function

h(t) = 0 if t ≤ 0, h(t) = exp(−1/t) if x > 0

is smooth even at t = 0. The function h(t− r)+ h(R− t) is strictly positive every-where, since for t ≥ r the first summand is positive and for t ≥ R the second sum-mand is positive. Furthermore, it agrees with h(t− r) for t ≥ R. Hence the functiong ∈C∞(R) given as

g(t) = 1− h(t− r)h(t− r)+h(R− t)

is 1 for t ≤ r, and 0 for t ≤ R. ut

The support supp( f ) of a function f on M is the smallest closed subset such that fvanishes on M\supp( f ). Equivalently, p ∈M\supp( f ) if and only if f vanishes onsome open neighborhood of p. In the Lemma above, we can take f to have supportin BR(0) – simply apply the Lemma to 0 < r < R′ := 1

2 (R+ r).

Definition A.1. A partition of unity subordinate to an open cover Uα of a mani-fold M is a collection of smooth functions χα ∈C∞(M), with 0≤ χα ≤ 1, such thatsupp(χα)⊆Uα , and

∑α

χα = 1.

Proposition A.4. For any open cover Uα of a manifold, there exists a partitionof unity χα subordinate to that cover. One can take this partition of unity to belocally finite: That is, for any p ∈M there is an open neighborhood U meeting thesupport of only finitely many χα ’s.

Proof. Let Vβ be a locally finite refinement of the cover Uα , given by coordinatecharts of the kind described in Remark A.1, and let V ′

β⊆ Vβ be as described there.

Since the images of V ′β⊆ Vβ are Br(0) ⊆ BR(0), we can use Lemma A.2 to define

a function fβ ∈C∞(M) with fβ (p) = 1 for p ∈V ′p and supp( fβ )⊆Vβ . Since the V ′β

are already a cover, the sum ∑β fβ is strictly positive everywhere.For each index β , pick an index α such that Vβ ⊆ Uα . This defines a map d :

β 7→ d(β ) between the indexing sets. The functions

χα =∑β∈d−1(α) fβ

∑γ fγ

.

give the desired partition of unity. ut

An important application of partitions of unity is the following result, a weakversion of the Whitney embedding theorem.

A.4 Partitions of unity 145

Theorem A.1. Let M be a manifold admitting a finite atlas with r charts. Then thereis an embedding of M as a submanifold of Rr(m+1).

Proof. Let (Ui,ϕi), i = 1, . . . ,r be a finite atlas for M, and χ1, . . . ,χr a partition ofunity subordinate to the cover by coordinate charts. Then the products χiϕi : Ui→Rm extend by zero to smooth functions ψi : M→ Rm. The map

F : M→ Rr(m+1), p 7→ (ψ1(p), . . . ,ψr(p),χ1(p) . . . ,χr(p))

is the desired embedding. Indeed, F is injective: if F(p) = F(q), choose i withχi(p)> 0. Then χi(q) = χi(p)> 0, hence both p,q ∈Ui, and the condition ψi(p) =ψi(q) gives ϕi(p) = ϕi(q), hence p = q. Similarly TpF is injective: For v ∈ TpM inthe kernel of TpF , choose i such that χi(p) > 0, thus v ∈ TpUi. Then v being in thekernel of Tpψi and of Tpχi implies that it is in the kernel of Tpϕi, hence v = 0 sinceϕi is a diffeomorphism. This shows that we get an injective immersion, we leaveit as an exercise to verify that the image is a submanifold (e.g., by constructingsubmanifold charts).

The theorem applies in particular to all compact manifolds. Actually, one can showthat all manifolds admit a finite atlas; for a proof see e.g. Greub-Halperin-Vanstone,connections, curvature and cohomology, volume I. Hence, every manifold can berealized as a submanifold.

Appendix BVector bundles

B.1 Tangent bundle

Let M be a manifold of dimension m. The disjoint union over all the tangent spaces

T M =⋃

p∈M

TpM

is called the tangent bundle of M. It comes with a projection map

π : T M→M

taking a tangent vector v ∈ TpM to its base point p, and with an inclusion map

i : M→ T M

taking p ∈ M to the zero vector in TpM ⊆ T M. We will show that T M is itself amanifold of dimension 2m, in such a way that π and i are smooth maps.

Example B.1. Suppose M is given as a submanifold M ⊆ Rn. Then each tangentspace TpM is realized as a subspace of Rn, and

T M = (p,v) ∈ Rn×Rn| p ∈M, v ∈ TpM.

We will show that this subset is a submanifold of dimension 2m. (The dimension isto be expected: p varies in an m-dimensional manifold, and once p is fixed then vvaries in an m-dimensional vector space.) For instance, if M = S1, and using coor-dinates x,y on the first copy of R2 and a,b on the second copy, then

T S1 = (x,y,r,s) ∈ R4| x2 + y2 = 1, xr+ ys = 0;

one can check directly that (1,0) is a regular value of the function Φ(x,y,r,s) =(x2 + y2, xr+ ys).

147

148 B Vector bundles

The tangent bundle of a general manifold M is a special case of the more generalconcept of a vector bundle, which we will first define.

B.1.1 Vector bundles

Let Ep be k-dimensional vector spaces indexed by the points p ∈ M of an m-dimensional manifold M, and

E =⋃

p∈M

Ep

their disjoint union. We denote by π : E → M the projection and i : M → E theinclusion of zeroes.

Definition B.1. E is called a vector bundle of rank r over M if for each p ∈M thereare charts (U,ϕ) around p and (U , ϕ) around i(p), with U = π−1(U), such that ϕ

restricts to vector space isomorphisms

Ep = π−1(p)→ϕ(p)×Rr ∼= Rr

for all p ∈ M. One calls E the total space and M the base of the vector bundle.Charts (U , ϕ) are called vector bundle charts.

The vector bundle charts may be pictured in terms of a diagram,

E ⊇ Uϕ

//

π

ϕ(U)×Rr

(u,v)7→u

M ⊇Uϕ

// ϕ(U)

The key condition is that ϕ restricts to vector space isomorphisms fiberwise. Thus,just like a manifold M looks locally like an open subset of Rm, a vector bundle looksover M locally like a product of an open subset of Rm with the vector space Rr.

Proposition B.1. For any vector bundle E over M, the projection π : E → M is asmooth submersion, while the inclusion i : M→ E is an embedding as a submani-fold.

Proof. In vector bundle charts, the maps π and i are just the obvious projectionϕ(U)×Rr → ϕ(U) and inclusion ϕ(U)→ ϕ(U)×Rr, which obviously are sub-mersions and embeddings respectively. ut

Note that the vector bundle charts (U , ϕ) are submanifold charts for i(M)⊆ E. Wewill identify M with its image i(M); it is called the zero section of the vector bundle.

Example B.2 (Trivial bundles). The trivial vector bundle over M is the direct productM×Rr. Charts for M directly give vector bundle charts for M×Rr.

B.1 Tangent bundle 149

Example B.3 (The infinite Mobius strip). View M = S1 as a quotient R/ ∼ for theequivalence relation x∼ x+1. Let E = (R×R)/∼ be the quotient under the equiv-alence relation (x,y)∼ (x+1,−y), with the natural map

π : E→M, [(x,y)] 7→ [x].

Then E is a rank 1 vector bundle (a line bundle) over S1. Its total space is an infiniteMobius strip.

Example B.4 (Vector bundles over the Grassmannians). For any p ∈ Gr(k,n), letEp ⊆ Rn be the k-dimensional subspace that it represents. Then

E = ∪p∈Gr(k,n)Ep

is a vector bundle over Gr(k,n), called the tautological vector bundle. Recall ourconstruction of charts (UI ,ϕI) for the Grassmannian, where UI is the set of all psuch that the projection map ΠI : Rn → RI restricts to an isomorphism Ep → RI ,and

ϕI : UI → L(RI ,RI′),

takes p ∈UI to the linear map A having Ep as its graph. Let

ϕI : π−1(UI)→ L(RI ,RI′)×RI

be the map given on the fiber Ep by

ϕI(v) = (ϕI(p),ΠI(v)).

The ϕI are vector bundle charts for E (once we identify L(RI ,RI′) = Rk(n−k) andRI = Rk).

There is another natural vector bundle E ′ over Gr(k,n), with fiber E ′p := E⊥p theorthogonal complement of Ep. In terms of the identification Gr(k,n) = Gr(n−k,n),E ′ is the tautological vector bundle over Gr(n− k,n).

Remark B.1. Note that we did not worry about the ‘Hausdorff property’ for the totalspace. We leave it as an exercise to show that for any (possibly non-Hausdorff)vector bundle E→M, the Hausdorff property for the total space E follows from theHausdorff property of the base M.

Example B.5. As a special case, we obtain the tautological line bundle E and thehyperplane bundle E ′ over RPn = Gr(1,n+1). The tautological line bundle is non-trivial: i.e., there do not exist global trivializations E → RPn×R. In the case n = 1the line bundle over RP1 ∼= S1 is the ‘infinite Mobius strip’ considered above.

Definition B.2. A vector bundle map (also called vector bundle morphism) fromE →M to F → N is a smooth map Φ : E → F of the total spaces, together with asmooth map Φ : M→ N of the base manifolds, such that Φ restricts to linear maps

Φ : Ep→ FΦ(p).


If Φ is a diffeomorphisms, then it is called a vector bundle isomorphism. An iso-morphism E→M×Rr with a trivial vector bundle is called a trivialization of M.

Vector bundle maps are pictured by commutative diagrams:

EΦ

//

F

MΦ

// N

Example B.6. Any vector bundle chart (U , ϕ) defines a vector bundle map ϕ :E|U := π−1(U)→ ϕ(U)×Rr.

Example B.7. The tautological line bundle E over RP1 is the simplest example of avector bundle that does not admit a global trivialization. (For example, one can notethat if one removes the zero section from E, then E −M stays connected; on theother hand (M×R)−M falls into two components.)

B.1.2 Tangent bundles

We return to the discussion of tangent bundles of manifolds.

Proposition B.2. For any manifold M of dimension m, the tangent bundle

T M = ∪p∈MTpM

(disjoint union of vector spaces) has the structure of a rank m vector bundle overM. Here π : T M→M takes v ∈ TpM to the base point p.

Proof. Recall that any chart (U,ϕ) for M gives vector space isomorphisms

Tpϕ : TpM = TpU → Tϕ(p)ϕ(U) = Rm

for all p ∈U . Let TU = ∪p∈U TpM = π−1(U). The collection of maps Tpϕ gives abijection,

T ϕ : TU → ϕ(U)×Rm.

We take the collection of (U , ϕ) = (TU,T ϕ) as vector bundle charts for T M:

T M ⊇ TUT ϕ

//

π

ϕ(U)×Rm

(u,v)7→u

M ⊇Uϕ

// ϕ(U)


We need to check that the transition maps are smooth. If (V,ψ) is another coordinatechart with U ∩V 6= /0, the transition map for TU ∩TV = T (U ∩V ) = π−1(U ∩V ) isgiven by,

T ψ (T ϕ)−1 : ϕ(U ∩V )×Rm→ ψ(U ∩V )×Rm.

But Tpψ (Tpϕ)−1 = Tϕ(p)(ψ ϕ−1) is just the derivative (Jacobian matrix) for thechange of coordinates ψ ϕ−1; hence this map is given by

ϕ(U ∩V )×Rm→ ψ(U ∩V )×Rm, (x,a) 7→ ((ψ ϕ−1)(x), Dx(ψ ϕ

−1)(a))

Since the Jacobian matrix depends smoothly on x, this is a smooth map. ut

Proposition B.3. For any smooth map Φ ∈C∞(M,N), the map

T Φ : T M→ T N

given on TpΦ as the tangent maps TpΦ : TpM→ TΦ(p)N, is a vector bundle map.

Proof. Given p ∈M, choose charts (U,ϕ) around p and (V,ψ) around Φ(p), withΦ(U) ⊆ V . As explained above, these give vector bundle charts (TU,T ϕ) and(TV,T ψ). Let Φ = ψ Φ ϕ−1 : ϕ(U)→ ψ(V ). The map

T Φ = T ψ T Φ (T ϕ)−1 : T ϕ(TU)→ T ψ(TV )

is smooth, since by smooth dependence of the differential DxΦ on the base point.Consequently, T Φ is smooth, ut

B.1.3 Some constructions with vector bundles

There are several natural constructions producing new vector bundles out of givenvector bundles.

1. If E1 → M1 and E2 → M2 are vector bundles of rank r1,r2, then the cartesianproduct E1×E2 is a vector bundle over M1×M2, of rank r1 + r2.

2. Let π : E→M be a given vector bundle.

Proposition B.4. Given a submanifold S⊆M, the restriction E|S := π−1(S) is avector bundle over S, in such a way that the inclusion map E|S → E is a vectorbundle map (and also an embedding as a submanifold).

Proof. Given p ∈ S ⊆M, let (U , ϕ) be a vector bundle chart for E, with under-lying chart (U,ϕ) containing p. Let (U ′,ϕ ′) be a submanifold chart for S at p.Replacing U,U ′ with their intersection, we may assume U ′ = U . Let ϕ ′ be thecomposition of ϕ : π−1(U)→ ϕ(U)×Rr with the map

(ϕ ′ ϕ−1)× id : ϕ(U)×Rr→ ϕ

′(U)×Rr.


Then ϕ ′ takes π−1(S) to (Rl ∩ϕ ′(U))×Rr; hence it is a vector bundle chart andalso a submanifold chart. ut

More generally, suppose Φ ∈C∞(M,N) is a smooth map between manifolds, andπ : E→ N is a vector bundle. Then the pull-back bundle

Φ∗E := ∪p∈MEΦ(p)

is a vector bundle over M. One way to prove this is as follows: Consider theembedding of M as the submanifold of N×M given as the graph of Φ :

M ∼= gr(Φ) = (Φ(x),x)| x ∈M.

The vector bundle E→ N defines a vector bundle over N×M, by cartesian prod-uct with the zero bundle 0M →M. We have

Φ∗E ∼= (E×0M)|graph(Φ).

3. Let E,E ′ be two vector bundles over M. Then the direct sum (also called Whitneysum)

E⊕E ′ := ∪p∈MEp⊕E ′p

is again a vector bundle over M. One way to see this is to regard E ⊕ E ′ asthe pull-back of the cartesian product E×E ′ under the diagonal inclusion M→M×M, x 7→ (x,x).

4. Suppose π : E→M is a vector bundle of rank r, and E ′⊆E is a vector subbundleof rank r′ ≤ r. That is, E ′ is a submanifold of E, and is itself a vector bundleover M, with the map π ′ : E ′→M given by restriction of π . In particular, eachE ′p = π ′−1(p) a vector subspace of Ep. The quotient bundle

E/E ′ :=⋃p

Ep/E ′p

is again a vector bundle over M.5. For any vector bundle E→M, the dual bundle

E∗ = ∪p∈ME∗p

(where E∗p = L(Ep,R) is the dual space to Ep) is again a vector bundle.

Example B.8. Given a manifold M with a submanifold S, one calls T M|S the tangentbundle of M along S. It contains T S as a subbundle; the normal bundle of S in M isdefined as a quotient bundle νS = T M|S/T S with fibers,

(νS)p = TpM/TpS.

Example B.9. The dual of the tangent bundle T M is called the cotangent bundle, andis denoted T ∗M. Given a submanifold S ⊆M, one can consider the set of covectors


α ∈ T ∗p M for p ∈ S that annihilate TpS, that is, 〈α,v〉 = 0 for all v ∈ TpS. This is avector bundle called the conormal bundle ν∗S of S. The notation is justified, since itis the dual bundle to νS.

Example B.10. The direct sum of the two natural bundles E,E ′ over the Grass-mannian Gr(k,n) has fibers Ep ⊕ E ′p = Rn, hence E ⊕ E ′ is the trivial bundleGr(k,n)×Rn.

Definition B.3. A smooth section of a vector bundle π : E → M is a smooth mapσ : M → E with the property π σ = idM . The space of smooth sections of E isdenoted Γ ∞(M,E), or simply Γ ∞(E).

Thus, a section is a family of vectors σp ∈ Ep depending smoothly on p.

Examples B.11. 1. Every vector bundle has a distinguished section, the zero section

p 7→ σp = 0,

where 0 is the zero vector in the fiber Ep. One usually denotes the zero sectionitself by 0.

2. For a trivial bundle M×Rr, a section is the same thing as a smooth function fromM to Rr:

Γ∞(M,M×Rr) =C∞(M,Rr).

Indeed, any such function f : M→ Rr defines a section σ(p) = (p, f (p)); con-versely, any section σ : M → E = M×Rr defines a function by compositionwith the projection M×Rr→ Rr.In particular, if κ : E|U →U ×Rr is a local trivialization of a vector bundle Eover an open subset U , then a section σ ∈ Γ ∞(E) restricts to a smooth functionψ σ |U : U → Rr.

3. Let π : E→M be a rank r vector bundle. A frame for E over U ⊆M is a collectionof sections σ1, . . . ,σr of EU , such that (σ j)p are linearly independent at eachpoint p ∈U . Any frame over U defines a local trivialization ψ : EU →U ×Rr,given in terms of its inverse map ψ−1(p,a) = ∑ j a j(σ j)p. Conversely, each localtrivialization gives rise to a frame.

The space Γ ∞(M,E) is a vector space under pointwise addition:

(σ1 +σ2)p = (σ1)p +(σ2)p.

Moreover, it is a module over the algebra C∞(M), under multiplication1: ( f σ)p =fpσp.

1 Here and from now on, we will often write fp or f |p for the value f (p).


B.2 Dual bundles

More generally, if E →M is a vector bundle of rank r, then we can define its dualE∗→M with fibers E∗p = (Ep)

∗.

Proposition B.5. The dual E∗ of a vector bundle E is itself a vector bundle.

Proof. For any open subset U ⊆M, write EU =⋃

p∈U Ep for the restriction of E toU , and similarly E∗U =

⋃p∈U Ep. Recall that a vector bundle chart for E is given by

a chart (U,ϕ) for M together with a chart for E, of the form (EU , ϕ), where

ϕ : EU → ϕ(U)×Rr

is a fiberwise linear isomorphism. Taking the inverse of the dual of the vector spaceisomorphism Ep→ Rr, we get maps E∗p→ (Rr)∗, hence

(ϕ∗)−1 : E∗U → ϕ(U)× (Rr)∗.

Identifying (Rr)∗ ∼= Rr, these serve as vector bundle charts (E∗U , (ϕ∗)−1) for E∗.

ut

Given smooth sections σ ∈ Γ ∞(M,E) and τ ∈ Γ ∞(M,E∗), one can take the pairingto define a function

〈τ,σ〉 ∈C∞(M), 〈τ,σ〉(p) = 〈τp, σp〉.

This pairing is C∞(M)-linear in both entries: That is,

〈τ, f σ〉= f 〈τ, σ〉= 〈 f τ, σ〉

for all τ ∈Γ ∞(M,E∗), σ ∈Γ ∞(M,E), f ∈C∞(M). We can use pairings with smoothsections of E to characterize the smooth sections of E∗.

Proposition B.6. 1. A family of elements τp ∈E∗p for p∈M defines a smooth sectionof E∗ if and only if for all σ ∈ Γ ∞(M,E), the function

M→ R, p 7→ 〈τp,σp〉

is smooth.2. The space of sections of the dual bundle is identified with the space of C∞(M)-

linear mapsτ : Γ

∞(M,E)→C∞(M), σ 7→ 〈τ,σ〉.

Here C∞(M)-linear means that 〈τ, f σ〉= f 〈τ, σ〉 for all functions f .

Proof. The proof uses local bundle charts and bump functions. Details are left as anexercise. 2

2 It is similar to the fact, proved earlier, that a collection of tangent vectors Xp ∈ TpM defines asmooth vector field if and only if for any f ∈C∞(M) the map 7→ Xp( f ) is smooth.

B.2 Dual bundles 155

Remark B.2. A slightly more precise version of the second part of this propositionis as follows: Regard E = Γ ∞(M,E) as a module over the algebra A = C∞(M) ofsmooth functions, and likewise for E ∗ = Γ ∞(M,E∗). The space HomA(E ,A) ofA =C∞(M)-linear maps E → A is again an A-module. There is a natural A-modulemap

E ∗→ HomA(E ,A)

defined by the pairing of sections. This map is an isomorphism.

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