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Manifolds. Lecture Notes - Fall 2017

Ioan Marcut,

1

Contents

Manifolds. Lecture Notes - Fall 2017 1

Some words about this course 6

Lecture 1. 91.1. Brief review of notions from Topology and Analysis 91.2. Manifolds 121.3. Exercises 17

Lecture 2. 212.1. Smooth maps 212.2. How many smooth structures? 222.3. Ck-manifolds 232.4. Embedded manifolds in Rn 242.5. Exercises 25

Lecture 3. 273.1. The topology is determined by the atlas 273.2. The real projective space 283.3. Smooth manifolds as quotients of group actions 303.4. Manifolds of dimensions 1 and 2 323.5. Exercises 33

Lecture 4. 374.1. Bump functions 374.2. Partitions of unity 374.3. Proof of Theorem 4.2.4 394.4. Corollaries of Theorem 4.2.4 414.5. Exercises 42

Lecture 5. 435.1. The tangent space 435.2. Notations for tangent vectors 455.3. The commutative algebra C∞(M) 465.4. Tangent vectors as derivations 465.5. Exercises 49

Lecture 6. 536.1. The inverse function theorem 536.2. Immersions and submersions 53

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4 IOAN MARCUT, , MANIFOLDS

6.3. Embedded submanifolds 556.4. Preimages of submanifolds 576.5. Exercises 59

Lecture 7. 617.1. Immersed submanifolds 617.2. A weak version of Whitney’s Embedding Theorem 657.3. Exercises 66

Lecture 8. 698.1. The tangent bundle 698.2. Vector bundles 708.3. Sections of vector bundles 718.4. Frames 728.5. Vector fields as sections of the tangent bundle 738.6. Exercises 75

Lecture 9. 779.1. Vector fields as derivations 779.2. The Lie bracket 789.3. Related vector fields 799.4. Change of coordinates rule for vector fields 819.5. Integral curves of a vector field 819.6. Exercises 83

Lecture 10. 8510.1. The flow of a vector field 8510.2. Complete vector fields 8710.3. The flow as the exponential of the Lie derivative 8810.4. The Lie bracket as infinitesimal commutator of flows 8910.5. Exercises 90

Lecture 11. 9311.1. Lie groups 9311.2. Exercises 97

Lecture 12. 10112.1. The exterior algebra 10112.2. The cotangent bundle 10712.3. The exterior powers of the cotangent bundle 10712.4. Differential forms 10812.5. The pullback of differential forms 10912.6. Exercises 110

Lecture 13. 11313.1. The exterior derivative 11313.2. De Rham cohomology 11613.3. Cartan calculus 12113.4. The algebra of graded derivations 12213.5. Exercises 123

IOAN MARCUT, , MANIFOLDS 5

Lecture 14. 12714.1. Volume elements 12714.2. The density bundle 12814.3. Densities 12914.4. Integration of densities 13014.5. Orientations 13214.6. Exercises 133

Lecture 15. 13715.1. Manifolds with boundary 13715.2. Integration of differential forms on oriented submanifolds 14115.3. Stokes’ Theorem 14315.4. Exercises 145

Bibliography 147

Exercises and assignments 149Problems for exercise classes 149Homework assignments 149


Some words about this course

The goal of this course is to introduce and to study basic properties of smoothmanifolds. The easiest way to picture smooth manifolds is as generalizations ofsmooth curves (1-dimensional) and surfaces (2-dimensional) in R3 to higher dimen-sions (and in Rn, instead of R3). However, the fact that these spaces lie in R3

is merely a tool for us to picture them; the axioms of a smooth manifold do notrefer to an ambient space (surprisingly, these axioms recover precisely the intuitivenotion of a manifold in Rn; this is the content of Whitney’s Embedding Theorem,to be discussed in the lectures).

Manifolds are the mathematical objects that are used to model the abstractshapes of “physical spaces”. A d-dimensional manifold is a topological space thatlocally looks like Rd. For example, the surface of the Earth looks locally flat, likea piece of the plane, but globally its topology is that of a sphere. The Universe ismodeled by a 3-dimensional manifold because locally it looks like a piece of R3, butits global topology might be more complicated. Space-time is a 4-dimensional man-ifold. The space of possible positions of a ball rolling on a plane is a 5-dimensionalmanifold.

Smooth manifolds are not just topological spaces; they are endowed with aso-called differentiable structure, which allows one to build classical multi-variablecalculus on smooth manifolds. In particular, one defines smooth maps, derivatives,integration etc. This forces one to treat many standard constructions from calculusin a rather abstract way, which yields a more conceptual understanding of suchoperations. On the other hand, using manifolds to model physical systems, thedifferentiable structure allows one to write the laws of Physics, which are usuallyexpressed as differential equations.

Smooth manifolds are used as the “background space” in many branches ofmodern geometry: Riemannian geometry = manifold + smooth distance, complexgeometry = holomorphic manifolds, symplectic geometry = manifold + symplecticstructure, Lie groups = manifold + smooth group multiplication etc.

So, let’s begin!

SOME WORDS ABOUT THIS COURSE 7

Some typographical conventions

• When a section is colored in blue, it indicates that the section was notcovered in the lectures.

• The symbol ♣ next to the statement of a theorem indicates that the proofwill not be given in this course, mostly because it is too involved or toolong.

LECTURE 1

1.1. Brief review of notions from Topology and Analysis

Did I use to know all these things? Was the summer so long?

We start by recalling some notions from Analysis and Topology, which will beused throughout the course.

Some Topology

Definition 1.1.1. A topology on a set X is a set T of subsets of X, whoseelements are called open subsets, satisfying the axioms:

∅, X ∈ T , U, V ∈ T ⇒ U ∩ V ∈ T , (Ui ∈ T , ∀i ∈ I)⇒⋃i∈I

Ui ∈ T .

The pair (X, T ) is called a topological space.

Definition 1.1.2. Let Y be a subset of a topological space (X, T ). The inducedtopology on Y , also called the subspace topology, is given by:

T |Y := U ∩ Y : U ∈ T .

Definition 1.1.3. Let (X, TX) and (Y, TY ) be topological spaces. A function

f : X −→ Y

is called continuous if f−1(U) is open in X for every open set U in Y .The function f is called a homeomorphism if it is continuous, bijective and

its inverse f−1 : Y → X is also continuous.

Definition 1.1.4. A topological space (X, T ) is called Hausdorff if every twodistinct points can be separated by two open sets; in other words, for every x, y ∈ X,with x 6= y, there exist U, V ∈ T such that x ∈ U , y ∈ V and U ∩ V = ∅.

Definition 1.1.5. A topological space (X, T ) is called second countable if thereexists a countable family of open sets Uii∈N with the following property: for everyopen set U there exists an index set I ⊂ N such that U =

⋃i∈I Ui. In other words,

the topology has a countable basis.

Next, we recall the definition of the Euclidean topology on

Rn = x = (x1, . . . , xn) : x1, . . . , xn ∈ R.

9


Definition 1.1.6. (1) The open ball in Rn of radius ε > 0 and center x0 ∈ Rnis the set

Bε(x0) = x ∈ Rn : |x− x0| < ε,where | · | is the usual Euclidean length:

|x− y| :=√

(x1 − y1)2 + . . .+ (xn − yn)2.

(2) A subset U ⊂ Rn is called open if for every x0 ∈ U there exists a positivenumber ε > 0 such that Bε(x0) ⊂ U . The collection of all such sets forms theso-called Euclidean topology on Rn.

Smooth functions

Let U ⊂ Rn be an open set. A function f : U → Rm is called differentiable at apoint x ∈ U , if there exists a linear map

dxf : Rn −→ Rm

such that:

lim|h|→0

|f(x+ h)− f(x)− dxf(h)||h|

= 0.

The map dxf is called the differential of f at x. If dxf exists, then it is alsouniquely determined, since it can be described as the directional derivative of f :

dxf(v) = limε→0

f(x+ εv)− f(x)

ε, v ∈ Rn.

Denote the components of f by f = (f1, . . . , fm). The matrix of dxf in the standardbases of Rn and Rm is given by the Jacobian matrix:

dxf =

∂f1

∂x1∂f1

∂x2 . . . ∂f1

∂xn∂f2

∂x1∂f2

∂x2 . . . ∂f2

∂xn

. . . . . . . . . . . .∂fm

∂x1∂fm

∂x2 . . . ∂fm

∂xn

.

Recall also the chain rule for the differential:

Proposition 1.1.7 (The chain rule). Consider two maps f : U → Rm, g : V → U ,where U ⊂ Rl and V ⊂ Rn are open sets. If g is differentiable at x ∈ V and fis differentiable at g(x) ∈ U then f g is differentiable at x, and its differential isgiven by

dx(f g) = (dg(x)f) dxg : Rn −→ Rm.

In terms of partial derivatives the chain rule takes the form:

∂(f g)

∂xi(x) =

l∑j=1

∂f

∂yj(g(x))

∂gj

∂xi(x), for all 1 ≤ i ≤ n.

The first, and most general class of maps are continuous maps; these are calledalso functions of class zero, or just C0-functions.

Denote the space of all linear maps from Rn to Rm by Lin(Rn,Rm). Writingthe linear maps in the standard bases, this space can be identified with the space ofn×m-matrices, hence with Rn×m. Using this identification, we endow Lin(Rn,Rm)with the Euclidean topology on Rn×m.

LECTURE 1. 11

Definition 1.1.8. Let U ⊂ Rn be an open set. A map f : U → Rm is said tobe a C1-function if f is differentiable at every x ∈ U , and its differential is acontinuous function from U to Lin(Rn,Rm):

df : U → Lin(Rn,Rm), x 7→ dxf ∈ Lin(Rn,Rm).

Functions of class Ck are defined inductively. We say that f : U → Rm isa Ck-function, for k ≥ 2, if it is differentiable and its differential is a Ck−1-function.

Let us recall some standard properties of Ck-functions:

Proposition 1.1.9. Let U ⊂ Rn be an open set, and let

f = (f1, . . . , fm) : U −→ Rm

be a function.

(1) If f is a Ck-function then it is also a Ck−1-function.(2) The function f is of class Ck if and only if its partial derivatives of order ≤ k:

∂lf i

∂xj1∂xj2 . . . ∂xjl=

∂

∂xj1

(∂

∂xj2

(. . .

∂f i

∂xjl. . .

)): U → R,

for 1 ≤ i ≤ m, 0 ≤ l ≤ k, 1 ≤ j1, . . . , jl ≤ n

exist and are continuous on U .(3) If f is of class Ck then its partial derivatives “commute”, meaning that:

∂lf i

∂xj1∂xj2 . . . ∂xjl=

∂lf i

∂xjσ(1)∂xjσ(2) . . . ∂xjσ(l),

for 1 ≤ i ≤ m, 1 ≤ l ≤ k, 1 ≤ j1, . . . , jl ≤ n, and every permutation σ of theset 1, . . . , l.

We will mostly work with the following type of maps:

Definition 1.1.10. A smooth function, also called a C∞-function, is a functionf : U → Rm, where U is an open subset of Rn, which is of class Ck for all k ≥ 0.Equivalently, f is smooth if and only if its partial derivatives of all orders exist andare continuous.

The following lemma, which will be repeatedly used throughout this course,illustrates the richness of the class of smooth functions.

Lemma 1.1.11. Fix a point a ∈ Rm, and positive numbers 0 < ε < δ. There existsa smooth map χ : Rm → R such that

• 0 ≤ χ(x) ≤ 1, for all x ∈ Rm,• χ(x) = 1, for all x ∈ Bε(a),• χ(x) = 0, for all x /∈ Bδ(a).

The functions constructed in the lemma are usually called bump-functions(think about their graph in Rm+1). In Exercise 1.2 we show how to construct suchfunctions and how to prove the lemma.


Analytic functions and Borel’s Lemma

For completeness, we recall also the class of analytic functions. Let f : U → Rmbe a smooth function defined on an open set U ⊂ Rn. The formal Taylor seriesof f at a point a ∈ U is given by:

P af (T1, . . . , Tn) :=∑

i1,...,in≥0

1

i1! . . . in!

∂i1+...+inf

(∂x1)i1 . . . (∂xn)in(a)T i11 . . . T inn .

This expression belongs to the vector space of formal power series in the Ti’s withcoefficients in Rm, denoted:

Rm[[T1, . . . , Tn]] :=

∑i1,...,in≥0

ci1,...,inTi11 . . . T inn : ci1,...,in ∈ Rm

.

Definition 1.1.12. The function f is called analytic around a if there exists ε > 0such that for all x ∈ Bε(a), the Taylor series evaluated on T = x − a convergesabsolutely to f(x):

f(x) =∑

i1,...,in≥0

1

i1! . . . in!

∂i1+...+inf

(∂x1)i1 . . . (∂xn)in(a)(x1 − a1)i1 . . . (xn − an)in .

Finally, f is called analytic on U , or of class Cω, if it is analytic around everypoint in U .

In this course we will develop the theory of smooth manifolds starting fromthe class of smooth functions. There are theories of manifolds which are devel-oped based on other classes of functions. There are two main reasons why smoothfunctions are usually preferred: first, smooth functions are closed under severaloperations (derivatives of smooth functions are smooth, solutions to ODE’s withsmooth coefficients are smooth etc.), and second, the class of smooth function isvery rich and very flexible (especially when compared to analytic functions; seeExercise 1.3).

Also to show the contrast with analytic functions, we state the following (for aproof, solve Exercises 1.4 and 1.5):

Theorem ♣ 1.1.13 (Borel’s Lemma). For any formal power series

Q(T1, . . . , Tn) :=∑

i1,...,in≥0

ci1,...,inTi11 . . . T inn ∈ R[[T1, . . . , Tn]],

there exists a smooth function f : Rn → R which has Q as its Taylor series at 0.

1.2. Manifolds

Topological manifolds

Definition 1.2.1. An m-dimensional topological manifold is a Hausdorff, sec-ond countable topological space, for which every point has an open neighborhoodhomeomorphic to an open set in Rm.

The last condition means that a topological manifold looks locally like Rm.The following terminology is used:

LECTURE 1. 13

Definition 1.2.2. Let M be a topological space. An m-dimensional chart (U,ϕ)

on M consists of a homeomorphism ϕ : U → U from an open set U ⊂ M to an

open set U ⊂ Rm. The components of ϕ are denoted by

ϕ = (x1ϕ, . . . , x

mϕ ), xiϕ : U −→ R, for 1 ≤ i ≤ m,

and are called the local coordinates on M corresponding to the chart (U,ϕ). The

inverse map ϕ−1 : U → U is called a local parameterization of M .

Non-mathematically, an atlas is a book with charts. Here is an atlas of Earth:

Definition 1.2.3. Let M be a topological space. An m-dimensional topologicalatlas on M (or C0-atlas on M) consists of a collection of m-dimensional chartsA = (Uα, ϕα)α∈I covering M , i.e. M = ∪α∈IUα.

With this terminology, note that a Hausdorff, second countable topologicalspace is an m-dimensional topological manifold if and only if it admits an m-dimensional topological atlas.

Smooth manifolds

Let us recall:

Definition 1.2.4. A map f : U → V between open sets U, V ⊂ Rm is called adiffeomorphism if f is bijective and both f and f−1 are smooth.

Note that (as in the case of homeomorphisms) the condition that f−1 be smoothis not automatically satisfied. The standard example of a smooth bijection whichis not a diffeomorphism is

f : R −→ R, f(t) = t3.

In a local chart a topological manifold is described as an open piece of Rm. Todevelop analysis on manifolds, one needs to introduce derivatives and integration offunctions, notions which, locally in charts, should coincide with those from multi-variable calculus. However, a function that is differentiable in one chart might failto be differentiable in different chart! To circumvent this, on a smooth manifoldone only works with mutually compatible charts.

Definition 1.2.5. Let M be a topological space. Let (U,ϕ) and (V, ψ) be two m-dimensional charts on M . The map

ϕ ψ−1 : ψ(U ∩ V ) −→ ϕ(U ∩ V )


is a called the change of coordinates map or the transition map between thetwo charts. The two charts are said to be compatible if the transition map is adiffeomorphism.

Definition 1.2.6. An m-dimensional topological atlas A on an m-dimensionaltopological manifold M is said to be an m-dimensional smooth atlas (or C∞-atlas) if every two charts in A are compatible.

Here are some simple examples of C∞-atlases:

Example 1.2.7. (1) On Rm there is a smooth atlas with only one chart

A = (Rm, idRm).Another smooth atlas is the collection of all diffeomorphisms between opensubsets of Rm:

B = (U,ϕ) : U, V ⊂ Rm are open and ϕ : U → V is a diffeomorphism.In fact, B consists of all charts compatible with (Rm, idRm).

(2) Consider the m-dimensional sphere with the induced topology from Rm+1:

Sm =

(x0, x1, . . . , xm) : (x0)2 + (x1)2 + . . .+ (xm)2 = 1⊂ Rm+1.

We construct an atlas on Sm with only two charts. The “south pole” and the“north pole” of Sm are the points

s := (0, 0, . . . , 0,−1) ∈ Sm, resp. n := (0, 0, . . . , 0, 1) ∈ Sm.The stereographic projection through the north pole is the map

πn : Sm\n −→ Rm

which sends a point p ∈ Sm, p 6= n, to the point q = πn(p) which is theintersection of the plane xm = 0 and the line through n and p.

n

Sm

p

qxm = 0

Explicitly, for p = (x0, . . . , xm), the point q must satisfy q = (y, 0) = (1−t)n+tpfor some t ∈ R. The last coordinate gives 1 − t + txm = 0, hence t = 1

1−xm ,and so,

πn(x0, . . . , xm) =1

1− xm(x0, . . . , xm−1).

Similarly, we have the stereographic projection through the south pole:

πs : Sm\s → Rm, πs(x0, . . . , xm) =

1

1 + xm(x0, . . . , xm−1).

We claim that the following is a smooth atlas on Sm:

A = (Sm\n, πn), (Sm\s, πs).

LECTURE 1. 15

First we show that the stereographic projections are homeomorphisms. Notethat the formula defining πn makes sense on the open set U = Rm+1\xm = 1,and that it defines a continuous map U → Rm. Thus πn is the restriction toU ∩ Sm of a continuous map and, using the definition of the induced topology,this implies that πn is continuous. The same argument implies that πs iscontinuous. Finally, let us determine the inverse of πn. We have that p =π−1n (y) is the point at the intersection of Sm and the line through q = (y, 0)

and n, and which is not n. So p = t(y, 0) + (1 − t)n, with t ∈ R. Writing|p|2 = 1, we obtain: t2|y|2 + (1 − t)2 = 1; or t2(|y|2 + 1) = 2t. The solutiont = 0 corresponds to n; thus t = 2

1+|y|2 . Using this, we obtain that the inverse

of πn is given by:

π−1n : Rm −→ Sm\n,

π−1n (y1, y2, . . . , ym) =

(2y1

|y|2 + 1,

2y2

|y|2 + 1, . . . ,

2ym

|y|2 + 1,|y|2 − 1

|y|2 + 1

).

All functions appearing in this expression are continuous, and this proves thatπn is a homeomorphism. A similar calculation shows that also πs is a homeo-morphism.

Finally, we need to check that the charts are C∞-compatible. Since πn(s) =πs(n) = 0, the transition map is defined as follows:

πs π−1n : Rm\0 −→ Rm\0,

and using the formulas above, we obtain that this map is given by:

πs π−1n (y) = πs

(2y

|y|2 + 1,|y|2 − 1

|y|2 + 1

)=

1

1 + |y|2−1|y|2+1

2y

|y|2 + 1=

y

|y|2.

This map is clearly smooth. Note that the map satisfies (πs π−1n )2 = idRm\0.

This shows that it equals its own inverse, hence it is a diffeomorphism.

We introduce a relation on atlases:

Definition 1.2.8. Let M be an m-dimensional topological manifold. Consider thefollowing relation on m-dimensional C∞-atlases on M :

A1 ∼ A2def.⇐⇒ A1 ∪ A2 is a C∞ − atlas.

As expected:

Proposition 1.2.9. The relation ∼ is an equivalence relation on the set of m-dimensional C∞-atlases of M .

Proof. Reflexivity and symmetry of the relation ∼ are obvious. We will check thattransitivity holds. Consider three m-dimensional C∞-atlases on M : A1, A2 andA3 such that A1 ∼ A2 and A2 ∼ A3. To check that A1 ∼ A3 we need to showthat each pair of charts (U1, ϕ1) ∈ A1 and (U3, ϕ3) ∈ A3 are compatible, i.e. thatthe map

(1) ϕ1 ϕ−13 : ϕ3(U1 ∩ U3) −→ ϕ1(U1 ∩ U3)

is a diffeomorphism. This map is clearly a bijection, with inverse

(2) ϕ3 ϕ−11 : ϕ1(U1 ∩ U3) −→ ϕ3(U1 ∩ U3).


So, it suffices to prove that the maps (1) and (2) are smooth. Let p ∈ U1 ∩ U3.Since A2 is an atlas, there exists a chart (U2, ϕ2) ∈ A2 such that p ∈ U2. SinceA1 ∼ A2 and A2 ∼ A3 is follows that the following maps are diffeomorphisms:

ϕ1 ϕ−12 : ϕ2(U1 ∩ U2) −→ ϕ1(U1 ∩ U2),

ϕ2 ϕ−13 : ϕ3(U2 ∩ U3) −→ ϕ2(U2 ∩ U3).

In particular, the restriction of their composition is a diffeomorphism:

ϕ1 ϕ−13 : ϕ3(U1 ∩ U2 ∩ U3) −→ ϕ1(U1 ∩ U2 ∩ U3).

Hence, the map (1) and its inverse (2) are smooth when restricted to the openneighborhoods ϕ3(U1 ∩ U2 ∩ U3) and ϕ1(U1 ∩ U2 ∩ U3), respectively, of ϕ3(p) andϕ1(p), respectively. Since p was chosen arbitrary in U1 ∩U3, it follows that (1) and(2) are smooth everywhere. This finishes the proof.

Next, we define:

Definition 1.2.10. An m-dimensional C∞-atlas A on the topological manifold Mis said to be maximal if

A ∼ B =⇒ B ⊂ Afor any m-dimensional C∞-atlas B on M .

We have that:

Proposition 1.2.11. Any equivalence class of smooth atlases has a unique maximalrepresentative. The maximal C∞-atlas equivalent to the C∞-atlas A is given by:

Amax = (U,ϕ) : A ∪ (U,ϕ) is a C∞-atlas .

Proof. Let A be an m-dimensional C∞-atlas on M . It is clear that there existsat most one maximal smooth atlas which is equivalent to A, because if there weretwo, A1

max and A2max, then by transitivity of the relation ∼ we would have that

A1max ∼ A2

max, and since both atlases are maximal, this implies that

A1max ⊂ A2

max ⊂ A1max;

thus A1max = A2

max.To prove existence of a maximal C∞-atlas equivalent to A, let Amax be the set

constructed in the statement. We check that Amax is indeed a C∞-atlas. First,since A ⊂ Amax, it follows that the open sets in Amax cover M . Second, we checkthat each pair of charts (U,ϕ) and (V, ψ) in Amax are compatible. Clearly A ∼A∪ (U,ϕ) and A ∼ A∪ (V, ψ), therefore, by transitivity of ∼, A∪ (U,ϕ) ∼A ∪ (V, ψ), hence A ∪ (U,ϕ) ∪ (V, ψ) is a C∞-atlas. This implies that thecharts (U,ϕ) and (V, ψ) are compatible, and proves that Amax is indeed a C∞-atlas.

Finally, we prove that Amax is maximal. Note first that A ∼ Amax. Let B bea C∞-atlas such that B ∼ Amax. Then, by transitivity of ∼, B ∼ A, which impliesthat every chart in B is compatible with every chart in A; hence A∪(U,ϕ) is anatlas for every (U,ϕ) ∈ B. The definition of Amax implies now that B ⊂ Amax.

Definition 1.2.12. An m-dimensional differentiable structure (or smoothstructure) on a topological manifold M is an equivalence class of m-dimensionalC∞-atlases on M . Equivalently, by Proposition 1.2.11, a differentiable structure isthe same as a maximal atlas on M .

We are ready to give the general definition of a smooth manifold:

LECTURE 1. 17

Definition 1.2.13. An m-dimensional smooth manifold is a topological manifoldendowed with an m-dimensional differentiable structure.

Example 1.2.14. The topological space Rm and Sm are smooth manifolds whenendowed with the differentiable structure corresponding to the C∞-atlases con-structed in Example 1.2.7. Note that B Example 1.2.7 (1) is the maximal atlas onRm. The obtained differentiable structures on Rm and Sm are called the standarddifferentiable structures. A differentiable structure which is not equivalent to thestandard one is called an exotic differentiable structure. These do exist in certaindimensions; see Section 2.3 for more details.

Remark 1.2.15. The axiom of being second countable insures that manifolds arenot “too big”. Let us mention that there are examples of Hausdorff topologicalspaces, endowed with a smooth atlas, but which are not second countable. Aneasy example is the disjoint union of an uncountable collection of manifolds of thesame dimension; e.g. an uncountable collection of points with the discrete topologyis not second countable, and has a 0-dimensional atlas. There are also connectedexamples, but these are more difficult to describe; a famous one is the Long Line(see e.g. [17, 18]).

1.3. Exercises

Exercise 1.1. (a) Show that Rn is Hausdorff and second countable.(b) Show that if X is a Hausdorff topological space and Y ⊂ X is a subset, then

the induced topology on Y is Hausdorff.(c) Show that if X is a second countable topological space and Y ⊂ X is a subset,

then the induced topology on Y is second countable.

Exercise 1.2. (a) Prove that the following function is smooth but is not analytic

f : R −→ R, f(x) =

0, x ≤ 0;e−1/x, x > 0.

(b) Show that g : R → R, g(x) := f(x)f(1 − x) is a smooth function which ispositive on (0, 1) and zero elsewhere. Show also that the function h : R→ R,

h(x) :=

∫ x0g(y)dy∫ 1

0g(y)dy

is a smooth and it satisfies: h(x) = 0 for x < 0, 0 < h(x) < 1 for 0 < x < 1,and h(x) = 1 for x > 1.

(c) Use (a) and (b) to prove Lemma 1.1.11.

Exercise 1.3. Prove that there are no analytic functions χ : Rm → R with theproperties from Lemma 1.1.11.

The following exercise is used to prove Borel’s Lemma:

Exercise 1.4. Let Ckb (Rn) denote the set of Ck-function f : Rn → R which haveall partial derivatives up to order k bounded. On this space we define the so-calledCk-norm:

‖f‖k := sup

∣∣ ∂i1+...+inf

(∂x1)i1 . . . (∂xn)in(x)∣∣ : x ∈ Rn, i1 + . . .+ in ≤ k

.

Let C∞b (Rn) denote the set of smooth maps with all partial derivatives bounded.


(a) Prove that (Ckb (Rn), ‖ · ‖k) is a Banach space, i.e. prove that: every sequenceof functions in Ckb (Rn) which is Cauchy with respect to the norm ‖ · ‖k is infact convergent to an element of Ckb (Rn).

(b) Prove that C∞b (Rn) endowed with the family of norms ‖ · ‖kk≥0 is a Frechetspace, i.e. prove that: every sequence of functions in C∞b (Rn) which is Cauchywith respect to all the norms ‖ · ‖kk≥0 is in fact convergent with respect toall the norms ‖ · ‖kk≥0 to an element in C∞b (Rn).

In the next exercise you are asked to prove Borel’s Lemma; note that it usesthe previous exercise.

Exercise 1.5. Consider a formal power series

Q(T1, . . . , Tn) :=∑

i1,...,in≥0

ci1,...,inTi11 . . . T inn ∈ R[[T1, . . . , Tn]].

Let χ : Rn → R be a smooth function such that χ(x) = 1, for x ∈ B1(0), andχ(x) = 0 for x /∈ B2(0) (by Lemma 1.1.11, such functions exist).

For k ≥ 0, define the following function depending on a positive number εk > 0:

fk(x) :=∑

i1+...+in=k

χ

(x

εk

)· ci1,...,in(x1)i1 . . . (xn)in .

(a) Prove that fk ∈ C∞b (Rn).(b) For k ≥ 1, prove that one can choose εk > 0 such that

‖fk‖k−1 < 2−k.

(c) Let εk > 0 be as in (b). Show that the series:∑k≥0

fk

converges to a smooth function f ∈ C∞b (Rn).(d) Prove that the Taylor series of f at 0 is Q.

Exercise 1.6. Consider the map

f : R −→ R, f(t) = t3.

(a) Show that f is a smooth homeomorphism of R.(b) Show that, there are no open neighborhoods U and V of 0 in R such that

f |U : U → V is a diffeomorphism.(c) On the topological manifold R consider the atlases

A1 = (R, idR) and A2 = (R, f).Is A1 ∼ A2?

Exercise 1.7. On the unit sphere Sm ⊂ Rm+1 consider the following collection of2(m+ 1) maps:

B = (U εi , ϕεi) : 0 ≤ i ≤ m, ε = ±1,U εi = (x0, . . . , xm) ∈ Sm : ε · xi > 0,ϕεi(x

0, . . . , xm) = (x0, . . . , xi, . . . , xm),

where the circumflex indicates that the underlying term has been deleted.

(a) Prove that the pairs (U εi , ϕεi) are indeed charts, and prove that B is a smooth

atlas on Sm.

LECTURE 1. 19

(b) Prove that the atlas A from Example 1.2.7 (2) and the atlas B define the samedifferentiable structure on Sm.

LECTURE 2

2.1. Smooth maps

We introduce smooth maps between smooth manifolds. First, let us define:

Definition 2.1.1. A smooth chart on a smooth manifold M is a chart (U,ϕ)which belongs to the maximal C∞-atlas of M .

Definition 2.1.2. Let M and N be smooth manifolds of dimension m and n,respectively. A continuous map f : M → N is said to be smooth (or C∞), if forevery smooth chart (U,ϕ) on M and every smooth chart (V, ψ) on N , the map:

ψ f ϕ−1 : ϕ(U ∩ f−1(V )) −→ ψ(V )

is smooth. The map ψ f ϕ−1, which is defined between open sets in Rm and Rnis called the local representation of f in the charts (U,ϕ) and (V, ψ).

The space of smooth maps from M to N is denoted by C∞(M,N). If N = R(as a smooth manifold) we simply write C∞(M) := C∞(M,R).

The condition that f be continuous in the previous definition is necessary toinsure that U ∩ f−1(V ) is open.

Remark 2.1.3. Smoothness can be checked on an open cover. Namley, a contin-uous map f : M → N is smooth iff for each p ∈ M there exists a smooth chart(U,ϕ) on M , with p ∈ U , and a smooth chart (V, ψ) on N with f(p) ∈ V such thatthe local representation of f in these charts is smooth. This follows because anytwo local representative around p and f(p) are related by composing on both sideswith local diffeomorphisms.

Composition of smooth maps is smooth:

Proposition 2.1.4. Let f : M → N and g : N → P be smooth maps betweensmooth manifolds. Then g f : M → P is a smooth map.

Next, we extend also the notion of a diffeomorphism:

Definition 2.1.5. A map f : M → N between smooth manifolds M and N iscalled a diffeomorphism if f is bijective and both f and f−1 are smooth. If thereexists a diffeomorphism f : M → N , then the manifolds M and N are said to bediffeomorphic.

21


Let us mention here that there is a product operation on manifolds:

Proposition 2.1.6. Let M and N be smooth manifolds of dimensions m, respec-tively n. Then the product M ×N endowed with the product topology is a smoothm+ n-dimensional manifold. A smooth atlas on M ×N is

AM×N := (U × V, ϕ× ψ) : (U,ϕ) ∈ AM , (V, ψ) ∈ AN ,

where AM is a smooth atlas on M and AN is a smooth atlas on N . Moreover, theprojection maps

pr1 : M ×N →M, pr1(x, y) = x, pr2 : M ×N → N, pr2(x, y) = y.

are smooth.

2.2. How many smooth structures?

This section is mostly for the general knowledge of the reader, and will not be usedin further lectures. The results stated here are not proven in this course, as theyare beyond the scope of these lectures; for more details, see the references in theWikipedia article [16].

Let us state one of the problems which has been of central interest since thebeginning of the theory of smooth manifolds:

Let M be a topological manifold. Can M be made into a smooth manifold? Ifso, in how many non-equivalent ways?

Let us be a bit more precise about the second part of the questions. Assume thatM is a smooth manifold in two different ways, corresponding to the differentiablestructures D1 and D2. We regard these differentiable structures are equivalent, ifthere exists a diffeomorphism relating the two structure:

ϕ : (M,D1) −→ (M,D2).

Thus, the second part of the questions asks how many non-equivalent such differ-entiable structures exist.

For small dimension, we have existence and uniqueness:

Theorem ♣ 2.2.1. For m ≤ 3, every m-dimensional topological manifold has asmooth structure, and any two such smooth structures are equivalent (for m = 1, 2Rado, 1920s, for m = 3 Moise, 1953, [9]).

However, in larger dimension:

Theorem ♣ 2.2.2. There exist compact topological manifolds of dimension ≥ 4which do not admit a differentiable structure (Kervaire 1960, [6] first example indimension 10; nowadays many examples are known even of dimension 4).

Finally, also uniqueness fails in dimension m ≥ 4; there are topological man-ifolds which admit non-equivalent smooth structures. The non-standard ones areusually called exotic. The first examples of non-equivalent smooth structures on atopological manifold were Milnor’s exotic 7-spheres:

Theorem ♣ 2.2.3. The topological manifold S7 admits exactly 28 non-equivalentsmooth structures (Milnor 1956, [7]).

LECTURE 2. 23

However, it is not known whether S4 has exotic smooth structures! Here is atable with the number of exotic structure for spheres of dim ≤ 20:

dimension 1 2 3 4 5 6 7 8 9 10C∞ − str.’s 1 1 1 ? 1 1 28 2 8 6

dimension 11 12 13 14 15 16 17 18 19 20C∞ − str.’s 992 1 3 2 16256 2 16 16 523264 24

Even more surprisingly are the exotic R4’s:

Theorem ♣ 2.2.4. For m 6= 4, the topological manifold Rm admits a uniquesmooth structure (Stallings, 1962 [12]). The topological manifold R4 admits un-countably many non-equivalent smooth structures (Taubes 1987, [13]).

2.3. Ck-manifolds

Throughout this section we fix k which is either an integer k ≥ 1, or k ∈ ∞, ω.We briefly introduce the notion of Ck-manifolds.

All notions discussed so far can be extended directly to the Ck-setting, bysimply replacing the requirement that a map be smooth with that of being Ck.

For example, a Ck-diffeomorphism is a map f : U → V between open setsU, V ⊂ Rm which is bijective and both f and f−1 are Ck-maps. Note the following:

Lemma 2.3.1. If f : U → V is a C1-diffeomorphism and a Ck-map, then f is aCk-diffeomorphism.

Proof. We need to check that the inverse of f , denoted by g : V → U , is also ofclass Ck. For k = ω, we omit the proof (one should write the power series expansionof the equation f g = id, and determine a term-by-term formula for g, and thenshow that this is convergent). Let k 6= ω. Differentiating the relation f g = id, bythe chain rule, we obtain that

dg(x)f dxg = id.

This gives that the differential of g is the inverse of the differential of f :

dxg = (dg(x)f)−1.

This relation shows that if g is Cl and f is Cl+1 then g is Cl+1 (the right hand sideis a composition of Cl-maps, and the smooth (!) operation of taking the inverse ofa matrix). An argument by induction, for 1 ≤ l ≤ k, proves the statement.

One introduces Ck-compatible charts, Ck-atlases, Ck-differentiable structures,and finally, Ck-manifolds, all in perfect analogy with the smooth setting.

Let l, k ∈ 0, 1, 2, . . . ,∞, ω l < k (where ∞ < ω). A Ck-differentiable struc-ture Dk on M is said to extend a given Cl-differentiable structure Dl on M if everyCk-atlas representing Dk, when regarded as a Cl-atlas it represents Dl. Two exten-sions Dk1 and Dk2 of Dl are said to be equivalent, if there exists a Ck-diffeomorphismrelating them:

ϕ : (M,Dk1 ) −→ (M,Dk2 ).

The following theorem says that Cl-differentiable structures, with 0 < l < ∞,are essentially the same as smooth structures (or even analytic structures).


Theorem ♣ 2.3.2. If 1 ≤ l < k, with l, k ∈ 1, 2, . . . ,∞, ω, then any Cl-differentiable structure can be extended to a Ck-differentiable structure, and anytwo extensions are equivalent (Whitney, 1936 [15]).

Whitney’s result also explains why the problem discussed in the previous sectionis only interesting for topological manifolds.

2.4. Embedded manifolds in Rn

In this section we introduce embedded manifolds in Rn. These are subsets of Rnwhich, locally, can be straighten out smoothly to look like an m-dimensional linearspace. This gives a large class of examples of smooth manifolds, and in fact, Whit-ney’s theorem (which will be discussed later on) states that any smooth manifoldcan be regarded as an embedded manifold in some Rn.

Definition 2.4.1. A subset M ⊂ Rn is called an m-dimensional embedded man-ifold in Rn if around every point in M there exists an open set U ⊂ Rn and thereexists a diffeomorphism f : U → V , where V ⊂ Rn is open, such that:

f(M ∩ U) = (Rm × 0) ∩ V.

We will call a diffeomorphism (U, f) as above a chart adapted to M .

Example 2.4.2. Let us check that the circle

S1 ⊂ R2, S1 = (x, y) ∈ R2 : x2 + y2 = 1

is a 1-dimensional embedded manifold in R2. Consider the map

f(x, y) := (x, x2 + y2 − 1).

Note that f is a diffeomorphism between the open sets

U := (x, y) ∈ R2 : y > 0, V := (u, v) ∈ R2 : v > u2 − 1,

with inverse given by

f−1 : V −→ U, f−1(u, v) = (u,√v + 1− u2).

Note that f(x, y) ∈ R× 0 if and only if (x, y) ∈ S1; thus, f satisfies:

f(S1 ∩ U) = (R× 0) ∩ V.

The open set U covers the upper half of the circle. On the open sets U1 := −U ,U2 := (x, y) : x > 0 and U3 := −U2, similar diffeomorphisms can be constructed:

f1 := f |U1 , f2 := g|U2 , f3 := g|U3 , where g(x, y) := (y, x2 + y2 − 1).

We will use the following notations for the canonical projections and inclusions,respectively, between Rm and Rn, with m ≤ n:

prnm : Rn −→ Rm, prnm(x1, . . . , xn) := (x1, . . . , xm)

imn : Rm −→ Rn, imn (x1, . . . , xm) := (x1, . . . , xm, 0, . . . , 0).

Proposition 2.4.3. Let M be an m-dimensional embedded manifold in Rn. ThenM , endowed with the induced topology, is a smooth m-dimensional manifold withdifferentiable structure represented by the atlas

A = (M ∩ U,prnm f |M∩U ) : (U, f) is a chart adapted to M.

LECTURE 2. 25

Proof. According to Exercise 1.1, the induced topology is Hausdorff and secondcountable. We need to check that A is a smooth atlas on M . By definition, M iscovered by adapted charts. Let (U, f) be an adapted chart, and denote V := f(U)and ϕ := prnm f |M∩U . First, note that M ∩ U is open in the induced topologyon M , and the map ϕ is continuous for the induced topology, because it is therestriction of a continuous map on U . We have that

ϕ(M ∩ U) = prnm f(M ∩ U) = prnm ((Rm × 0) ∩ V ) ⊂ Rm.Since prnm restricts to a homeomorphism between Rm × 0 (with the inducedtopology) and Rm, we have that ϕ(M ∩ U) is open in Rm. Finally, the map ϕ is ahomeomorphism onto its image, because its inverse is given by as a composition ofcontinuous maps:

ϕ−1 : ϕ(M ∩ U) −→M ∩ U, ϕ−1 = f−1 imn |ϕ(M∩U).

This proves that (M ∩ U,ϕ) is an m-dimensional chart on M . Thus, M is atopological m-dimensional manifold.

To show that A is a smooth atlas, we need to check that any two charts areC∞-compatible. Let f : U → V and f ′ : U ′ → V ′ be two adapted charts, inducingthe following charts in A:

(M ∩ U,ϕ := prnm f |M∩U ), (M ∩ U ′, ϕ′ := prnm f ′|M∩U ′).The corresponding transition map is smooth since it is given by:

ϕ′ ϕ−1 : ϕ(M ∩ U ∩ U ′) −→ ϕ′(M ∩ U ∩ U ′), prnm f ′ f−1 imn .

By interchanging the roles of f and f ′ we see that its inverse is also smooth. Henceany two charts in A are C∞-compatible.

The converse of this result also holds:

Theorem ♣ 2.4.4 (Whitney’s Embedding Theorem). Any m-dimensional smoothmanifold, for m ≥ 1, is diffeomorphic to an embedded manifold in R2m, which is aclosed subset.

2.5. Exercises

Exercise 2.1. Prove Proposition 2.1.6.

Exercise 2.2. Let M be an embedded manifold in Rn. Show that the inclusionmap M → Rn, x 7→ x, is smooth.

Exercise 2.3. (a) Prove that Sm is an embedded manifold in Rm+1 (as in Defini-tion 2.4.1).

(b) Prove that the smooth structure on Sm resulting from Proposition 2.4.3 co-incides with the smooth structure on Sm coming from the smooth atlas inExample 1.2.7.

LECTURE 3

3.1. The topology is determined by the atlas

It is important to notice that an atlas determines the topology uniquely. Thisremark allows one to describe smooth manifolds by giving a collection of charts ona set. We state this in the following proposition.

Proposition 3.1.1. Let M be a set and let A be a collection of maps

A = (Uα, ϕα) : α ∈ I, Uα ⊂M, ϕα : Uα −→ Rm,

which satisfy the following conditions:

(1) ∪α∈IUα = M ;(2) for each α ∈ I, ϕα is injective;(3) for each α, β ∈ I, ϕβ(Uα ∩ Uβ) is open in Rm;

in particular, for α = β, ϕα(Uα) is open in Rm;(4) for each α, β ∈ I, the map ϕα ϕ−1

β : ϕβ(Uα ∩ Uβ) → ϕα(Uα ∩ Uβ) is adiffeomorphism.

There exists a unique topology TA on M such that, for all α ∈ I, the pair (Uα, ϕα)is a chart on M . Moreover,

• If I is at most countable, then TA is second countable.• If for every two distinct points p, q ∈M there exist α, β ∈ I and open setsOα and Oβ in Rm such that

p ∈ ϕ−1α (Oα), q ∈ ϕ−1

β (Oβ), ϕ−1α (Oα) ∩ ϕ−1

β (Oβ) = ∅,

then TA is Hausdorff.• If TA is Hausdorff and second countable, then M has an m-dimensional

differentiable structure for which A is a C∞-atlas.

Proof. First, we show that there is at most one topology TA on M for which A isan atlas: since Uαα∈I is an open cover of M w.r.t. TA, we have that

V ∈ TA ⇐⇒(∀ α ∈ I : V ∩ Uα ∈ TA

);

and since ϕα : Uα → ϕα(Uα) is a homeomorphism, we have that

V ∈ TA ⇐⇒(∀ α ∈ I : ϕα(V ∩ Uα) is open in Rm).

27


The above defines TA uniquely. Next we show that TA is indeed a topology. Clearly,∅ ∈ TA, and by (3), M ∈ TA. Let V1, V2 ∈ TA. For every α ∈ I, we have that

ϕα(Uα ∩ V1 ∩ V2) = ϕα ((Uα ∩ V1) ∩ (Uα ∩ V2))(2)= ϕα(Uα ∩ V1) ∩ ϕα(Uα ∩ V2),

which shows that ϕα(Uα∩V1∩V2) is open. This implies that V1∩V2 ∈ TA. Finally,consider an arbitrary family Vλλ∈Λ ⊂ TA. Then, for each α ∈ I,

ϕα (Uα ∩ (∪λ∈ΛVλ)) = ϕα (∪λ∈ΛUα ∩ Vλ) = ∪λ∈Λϕα(Uα ∩ Vλ),

which shows that ϕα (Uα ∩ (∪λ∈ΛVλ)) is open. Hence, ∪λ∈ΛVλ ∈ TA. We concludethat TA is a topology.

Let us see that (Uα, ϕα) is indeed a chart on M for the topology TA. By (3),Uα is open. Next, we show that ϕα : Uα → ϕα(Uα) is a homeomorphism. For anyopen set O ⊂ ϕα(Uα), we have that ϕβ(Uβ ∩ϕ−1

α (O)) = ϕβ ϕ−1α (O) which is open

because, by (4), ϕβ ϕ−1α is a homeomorphism. Since β is arbitrary, ϕ−1

α (O) ∈ TA.So, ϕα is continuous. By definition of TA, we have that also ϕ−1

α : ϕα(Uα) → Mis continuous. Thus, the maps ϕα : Uα → ϕα(Uα) are homeomorphisms, i.e. chartson M for the topology TA.

Assuming that I is at most countable, we prove that TA is second countable.Let Okk∈N be a countable basis for the topology of Rm, i.e. every open set U ⊂ Rmcan be represented as U = ∪k∈JUOk, for a subset JU ⊂ N. We claim that the familyof open sets ϕ−1

α (Ok)α∈I,k∈N is a basis for TA. Since I is at most countable, theindex set I × N is countable. Let V ∈ TA be an open set. First, decomposeV = ∪α∈I(V ∩Uα). Since ϕα(V ∩Uα) is open in Rm, there exists a subset Jα ⊂ Nsuch that ϕα(V ∩ Uα) = ∪k∈JαOk. Thus we have that

V = ∪α∈IV ∩ Uα = ∪α∈I ∪k∈Jα ϕ−1α (Ok).

This shows that TA is second countable.The criterion for being Hausdorff is clear: the points p and q are separated by

the open sets ϕ−1α (Oα) and ϕ−1

β (Oβ).

By (4) the charts inA are pairwise compatible, which makes the final conclusionobvious.

3.2. The real projective space

Proposition 3.1.1 allows us to describe the structure of a smooth manifold on a setin terms of a collection of local maps to Rm satisfying certain conditions. Here isan important example:

Example 3.2.1. The m-dimensional real projective space is defined as the setof all lines through the origin in Rm+1:

Pm(R) := l : l ⊂ Rm+1 is a line with 0 ∈ l.

Note first that a non-zero vector in Rm+1 belongs to a unique line in Pm(R), andtwo vectors lie on the same line Pm(R) iff one is a scalar multiple of the other. Thisallows us to identify

Pm(R) =(Rm+1\0

)/ ∼,

where ∼ is the equivalence relation on Rm+1\0 given by

v ∼ w ⇐⇒ v = tw, for some 0 6= t ∈ R.

LECTURE 3. 29

We denote the equivalence class of v = (x0, x1, . . . , xm) 6= 0 by:

[v] = [x0 : x1 : . . . : xm] =

(tx0, tx1, . . . , txm) : 0 6= t ∈ R.

Consider the following m+ 1 subsets of Pm(R):

Ui :=

[x0 : x1 : . . . : xm] : xi 6= 0⊂ Pm(R), 0 ≤ i ≤ m.

and the maps:ϕi : Ui −→ Rm,

ϕi([x0 : . . . : xi−1 : xi : xi+1 : . . . : xm]) :=

(x0

xi, . . . ,

xi−1

xi,xi+1

xi, . . . ,

xm

xi

).

Note that the pair (Ui, ϕi) is well-defined: if (x0, . . . , xm) ∼ (y0, . . . , ym) then

yj = txj , for some t 6= 0. So, xi 6= 0 iff yi 6= 0, and xj

xi = txj

txi = yj

yi .

We claim that Pm(R) is a smooth m-dimensional manifold with C∞-atlas

A = (Ui, ϕi) : 0 ≤ i ≤ m.We verify the condition (1)-(4) in Proposition 3.1.1. Since each point in Rm+1\0has at least one non-zero coordinate, it follows that each point in Pm(R) lies in oneof the sets Ui, hence condition (1) is satisfied

Pm(R) =⋃

0≤i≤m

Ui.

Condition (2) follows since ϕi : Ui → Rm is a bijection with inverse

ϕ−1i : Rm −→ Ui, ϕ−1

i (y1, . . . , ym) = [y1 : . . . : yi : 1 : yi+1 : . . . : ym].

For (3), note that

ϕi(Ui ∩ Uj) =

y ∈ Rm : yj+1 6= 0, if i > jy ∈ Rm : yj 6= 0, if i < jRm, if i = j

Finally, (4) follows by calculating the transition maps explicitly; for i < j:

ϕj ϕ−1i : y ∈ Rm : yj 6= 0 −→ y ∈ Rm : yi+1 6= 0

ϕj ϕ−1i (y1, . . . , ym) =

(y1

yj, . . . ,

yi

yj,

1

yj,yi+1

yj, . . . ,

yj−1

yj,yj+1

yj. . . ,

ym

yj

),

andϕi ϕ−1

j : y ∈ Rm : yi+1 6= 0 −→ y ∈ Rm : yj 6= 0

ϕi ϕ−1j (y1, . . . , ym) =

(y1

yi+1, . . . ,

yi

yi+1,yi+2

yi+1, . . . ,

yj

yi+1,

1

yi+1,yj+1

yi+1, . . . ,

ym

yi+1

).

By Proposition 3.1.1, Pm(R) has a unique topology for which A is an m-dimensional atlas. Since the atlas is finite, this topology is second countable.

It remains to check that the topology is Hausdorff. Consider two distinctpoints p = [p0 : . . . : pm], q = [q0 : . . . : qm] ∈ Pm(R). Assume first that there exists0 ≤ i ≤ m such that p, q ∈ Ui. If Op and Oq are two open sets in Rm separating

ϕi(p) and ϕi(q), then p and q can be separated by the open sets ϕ−1i (Op) and

ϕ−1i (Oq). Assume now that there is no 0 ≤ i ≤ m such that Ui contains both p and

q. By (1) there exist 0 ≤ i, j ≤ m such that p ∈ Ui and q ∈ Uj . Since p /∈ Uj andq /∈ Ui, we have that

pi 6= 0, pj = 0, qi = 0, qj 6= 0.


We claim that the following two sets are open and separate p and q:

U(p) := [x0 : . . . : xm] : |xi| > |xj |, U(q) := [x0 : . . . : xm] : |xi| < |xj |.

Note first that these sets are well-defined. Clearly, p ∈ U(p), q ∈ U(q) and U(p) ∩U(q) = ∅. Next, note that U(p) ⊂ Ui and that ϕi(U(p)) is open in Rm. Since ϕiis a homeomorphism, we conclude that U(p) is open. Similarly, U(q) is open. Thisproves that Pm(R) is indeed Hausdorff.

3.3. Smooth manifolds as quotients of group actions

New examples of manifolds can be constructed by taking the quotient of a manifoldunder a group action. We begin with some definitions.

Definition 3.3.1. Consider an action of a group G on a smooth manifold M :

a : G×M −→M, a(g, x) = g · x.

(1) The action is called free if, for all x ∈ M , the equality g · x = x withg ∈ G implies that g = e, where e denotes the identity element in G.

(2) The action is called smooth (resp. continuous) if, for each g ∈ G, theaction by g is smooth (resp. continuous):

ag : M −→M, x 7→ g · x

(3) A continuous action of G on M is called proper if every two points x, y ∈M have open neighborhoods Ux and Uy such that there are only finitelymany g ∈ G for which Ux ∩ g · Uy 6= ∅.

(4) The quotient space M/G is defined as the set of orbits

M/G :=

[x] : x ∈M, [x] := G · x.

This set comes with a canonical projection

π : M −→M/G, π(x) = [x].

(5) The quotient topology on M/G is defined by

U ⊂M/G is open ⇐⇒ π−1(U) ⊂M is open.

In other words, the quotient topology is the largest topology (in the sensethat it has the most open sets) for which the projection π is still continuous.

In the following result we construct a smooth manifold structure on the quotientspace of a smooth, free and proper action.

Theorem 3.3.2. Consider a smooth, free and proper action of group G on a smoothm-dimensional manifold M . Then M/G has a unique m-dimensional differentiablestructure for which the canonical projection π : M → M/G is a local diffeomor-phism.

In the statement we have used the following concept:

Definition 3.3.3. We say that a smooth map f : M → N is a local diffeomor-phism if each p ∈ M has an open neighborhood U ⊂ M such that f(U) ⊂ N isopen and f |U : U → f(U) is a diffeomorphism.

LECTURE 3. 31

Proof of Theorem 3.3.2. We consider the quotient topology on M/G. First notethat π is an open map, i.e. π(V ) is open M/G for every open set V ⊂ M . Thisholds because π−1(π(V )) is the union of the open sets g · V , g ∈ G.

To see that M/G is second countable, let Ukk∈N be a countable basis for thetopology of M . Then the collection π(Uk)k∈N is a basis for the topology of M/G(since π is an open map, these sets are indeed open). Let V ⊂ M/G be any openset. Then π−1(V ) = ∪k∈IUk, for some subset I ⊂ N, and therefore V = ∪k∈Iπ(Uk).

Next, we show that M/G is Hausdorff. Let [x], [y] ∈M/G with [x] 6= [y]. Sincethe action is proper, the points have neighborhoods Ux and Uy, respectively, suchthat Ux and g · Uy intersect only for a finite number of elements g ∈ G; call theseelements g1, . . . , gk. Since [x] 6= [y] it follows that giy 6= x, for 1 ≤ i ≤ k; thus, sinceM is Hausdorff, we can find open neighborhoods Ui of giy and U ′x ⊂ Ux of x suchthat Ui ∩ U ′x = ∅. Let U ′y :=

(∩ki=1g

−1i Ui

)∩ Uy. Then the open neighborhoods U ′x

and U ′y of x and y, respectively, have the property that U ′x ∩ gU ′y = ∅ for all g ∈ G.This implies that gU ′x∩hU ′y = ∅ for all g, h ∈ G; or equivalently, π(U ′x)∩π(U ′y) = ∅.Since π is an open map, the sets π(U ′x) and π(U ′y) are disjoint open neighborhoodsof [x], respectively of [y]. Thus M/G is Hausdorff.

Next, we construct charts on M/G. Consider [x] ∈ M/G. Since the action isproper, there is an open neighborhood U of x such that g · U intersects U only fora finite number of elements g ∈ G; call these elements g0 := e, g1, . . . , gk. Since theaction is free, gi · x 6= x, for 1 ≤ i ≤ k; thus we can find open neighborhoods Ui ofgi · x and U ′ ⊂ U of x such that Ui ∩ U ′ = ∅. Let V :=

(∩ki=1g

−1i Ui

)∩ U ′. This

open neighborhood of x satisfies V ∩ g · V = ∅, for all g 6= e. This implies thatπ|V : V → π(V ) is injective. By shrinking V , we may assume that it is the domainof a chart (V, ϕ) on M . We define the following chart on M/G:

ϕ (π|V )−1 : π(V ) −→ Rm.

Since π is open, it follows that π|V : V → π(V ) is a homeomorphism; hence(π(V ), ϕ (π|V )−1) is indeed a chart on M/G. Consider the collection of all chartsobtained as above:

A :=

(π(V ), ϕ (π|V )−1) : (V, ϕ) is a smooth chart on M s.t. π|V is injective.

Above we have constructed a chart around each [x] ∈ M/G; thus the open sets inA cover M/G. Finally, we need to prove that the transition maps are smooth. Let(V, ϕ) and (W,ψ) be two smooth charts on M such that π|V and π|W are injective.We need to show that the following map is a diffeomorphism:

ϕ (π|V )−1 (π|W ) ψ−1 : ψ(V ∩ π−1(π(W ))) −→ ϕ(π−1(π(V )) ∩W ).

Let p ∈ π(V ) ∩ π(W ), and let x ∈ V and y ∈ W such that [x] = p = [y]. We showthat the transition map is smooth in a neighborhood of ψ(y). Since [x] = [y], thereexists g ∈ G such that x = g · y. Let U := g−1 · V ∩W . On U , we claim that thefollowing equality holds:

(π|V )−1 π|U = ag|U : U −→ V.

By composing on both sides with π|V , the equality clearly holds; and so the claimfollows by injectivity of π|V . Hence, on ψ(U), the transition map is a compositionof diffeomorphisms

ϕ ag ψ−1 : ψ(U) −→ ϕ(g · U);


therefore it is itself a diffeomorphism. This finishes the proof that A is a smoothatlas, and so M/G is a smooth m-dimensional manifold.

Finally, let us show that π : M →M/G is a local diffeomorphism. As we haveseen above, around any point in M we can find a chart (V, ϕ) such that π|V isinjective. The local representation of π in the charts (V, ϕ) and (π(V ), ϕ (π|V )−1)is simply given by the identity map of the set ϕ(V ). This implies that π is smooth,and π|V is a diffeomorphism; hence π is a local diffeomorphism.

Denote by T the quotient topology and by D the differentiable structure con-structed above. Next, we show that D and T are unique such that π is a localdiffeomorphism. Consider a second topology T ′ and a second differentiable struc-ture D′ on M/G such that π is a local diffeomorphism.

First we show that T = T ′. Since π : M →M/G is continuous, it follows thatT ′ ⊂ T . Since local diffeomorphisms are open maps, π(U) ∈ T ′ for any open setU in M . Thus, if V ∈ T then π−1(V ) is open in M , hence V = π(π−1(V )) ∈ T ′.This proves that T ⊂ T ′; hence the two topologies coincide.

Since π is a local diffeomorphism for D′, every point has an open neighborhoodV such that π(V ) is open and π|V : V → π(V ) is a diffeomorphism. By shrinkingV , we may assume that it is the domain of a smooth chart (V, ϕ) on M . Hence,ϕ (π|V )−1 : π(V ) → Rm is a diffeomorphism for D′, and so (π(V ), ϕ (π|V )−1)is a smooth chart for D′. The collection of such charts covers M/G, hence it is anatlas A′ representing D′. On the other hand, A′ ⊂ A, and so D = D′.

3.4. Manifolds of dimensions 1 and 2

Without providing proofs, in this section we discuss some classification results formanifolds in dimension one and two.

Theorem ♣ 3.4.1. Any connected 1-dimensional manifold is diffeomorphic toeither R or S1.

Two-dimensional smooth manifolds, also called surfaces, are more compli-cated. For example every open subset U of R2 is a surface; just think about thecases U = R2\Z2, or U is the complement of the Cantor set.

Nevertheless, compact surfaces have a simple classification; we introduce thesebasic examples in a less-rigorous fashion.

The genus-g surface, denoted by Σg, is a surface which looks like the one inthe picture, where g ≥ 0 stands for the number of holes:

Σg

For g = 0, we obtain the 2-sphere S2 and for g = 1 the 2-torus T 2:

Σ0∼= S2 Σ1

∼= T 2

LECTURE 3. 33

Place Σg in R3 such that it is symmetric with respect to the three coordinateplanes; here is the picture for Σ3:

x

y

z

Σ3

In this position, the group Z2 = 0, 1 acts freely on Σg:

0 · (x, y, z) = (x, y, z), 1 · (x, y, z) = (−x,−y,−z).The quotient space Σg/Z2 inherits therefore a smooth structure. For g = 0, S2/Z2

is the projective space (see Exercise 3.7), and for g = 1, T 2/Z2 is the Klein bottle(described also in Exercise 3.9).

We are ready to state the classification theorem for compact surfaces:

Theorem ♣ 3.4.2. Any compact connected 2-dimensional manifold is diffeomor-phic to either Σg or Σg/Z2, for some g ≥ 0.

The manifolds in the list Σg, Σg/Z2, for g ≥ 0, are non-diffeomorphic. Let usmention that, the surfaces Σg are the orientable ones, and the surfaces Σg/Z2 arethe non-orientable ones; the notion of orientable manifold will be introduced in alater lecture.

3.5. Exercises

In the following exercise we construct a non-example of a manifold; a non-Hausdorffsecond countable topological space with a smooth 1-dimensional atlas.

Exercise 3.1 (Two intervals glued along an open subinterval). Consider the set:

M = (−1, 1) ∪ [2, 3).

Let U = (−1, 1) and V = (−1, 0) ∪ [2, 3). On these subsets of M define the maps

ϕ : U → R, ϕ(t) = t, and

ψ : V → R, ψ(t) =

t, t ∈ (−1, 0)t− 2, t ∈ [2, 3)

Show that A = (U,ϕ), (V, ψ) satisfies the conditions (1)-(4) of Proposition 3.1.1.Show that the topology TA on M is non-Hausdorff.

Exercise 3.2. Let L be the set of all lines in R2. For a, b, c ∈ R, with a 6= 0 orb 6= 0, denote by l(a, b, c) the line given by the equation

ax+ by + c = 0.

Let U ⊂ L be the set of lines l(a, b, c) with a 6= 0, and let V ⊂ L be the set of linesl(a, b, c) with b 6= 0. Define the maps

ϕ : U → R2, ϕ(l(a, b, c)) = (b/a, c/a), ψ : V → R2, ψ(l(a, b, c)) = (a/b, c/b).


Prove that L is a smooth 2-dimensional manifold with atlas A = (U,ϕ), (V, ψ).

Exercise 3.3. Let S1 denote the unit circle in C

S1 := z = x+ iy : |z| =√x2 + y2 = 1 ⊂ C.

Let M be the collection of subsets of S1 with 2 elements:

M := z, w : z, w ∈ S1, z 6= w.

(Pay attention to the fact that z, w = w, z!!)(a) Consider the set

V := (α, β) : 0 < α < β < 2π ⊂ R2,

and for k ∈ 0, 1, 2 consider the map ψk : V →M given by

ψk(α, β) := ei(α+ 2πk3 ), ei(β+ 2πk

3 ).

Prove that ψk is injective.(b) Let Uk := ψk(V ), and let ϕk := ψ−1

k : Uk → V . Prove that M has the structureof a smooth 2-dimensional manifold, for which

A := (Uk, ϕk) : k ∈ 0, 1, 2

is a C∞-atlas. (Recall that eiα = cos(α) + i sin(α)).

Exercise 3.4. Show that the manifolds L and M constructed in Exercise 3.2 and3.3, respectively, are diffeomorphic. Note that both are a Mobius band:

The following exercise generalizes the construction of the projective space.

Exercise 3.5. The Grassmannian Gr(k, n) is the set of all k-dimensional lin-ear subspaces of Rn. We construct a smooth manifold structure on Gr(k, n) ofdimension k · (n− k).

Let ei : 1 ≤ i ≤ n denote the standard basis of Rn. Consider a subset Iof 1, 2, . . . , n with k elements. Denote the elements of I and of its complementincreasingly as follows:

I = i1 < i2 < . . . < ik, 1, . . . , n\I := j1 < j2 < . . . < jn−k.

For each such subset I we construct the following map

χI : Rk×(n−k) −→ Gr(k, n)

LECTURE 3. 35

χI (A) = Span⟨eiu +

n−k∑v=1

au,vejv : 1 ≤ u ≤ k⟩, for A = (au,v)1≤u≤k,1≤v≤n−k,

where Span denotes the linear span of a subset of Rn.

(a) Prove that χI is injective.(b) Let UI := χI(Rk×(n−k)) and let ϕI := χ−1

I : UI → Rk×(n−k).Prove that Gr(k, n) is a smooth k × (n− k)-dimensional manifold with atlas:

A = (UI , ϕI) : I = 1 ≤ i1 < i2 < . . . < ik ≤ n.

(c) Prove that the manifolds Gr(k, n) and Gr(n− k, n) are diffeomorphic.

The following example is from [14].

Exercise 3.6. A continuous action of a group G on a manifold M is called wan-dering if every point has a neighborhood U such that gU ∩ U = ∅, for all g 6= e.

Consider the action of (Z,+) on M := R2\0 given by

n · (x, y) = (2nx, 2−ny).

(a) Prove that this action is wandering, but not proper.(b) Consider the quotient topology on M/Z. Prove that M/Z is second countable,

but not Hausdorff.(c) Prove that M/Z has a smooth 2-dimensional atlas, i.e. a collection of charts A

satisfying conditions (1)-(4) from Proposition 3.1.1.

In the following exercise we construct the projective plane as a quotient of thesphere.

Exercise 3.7. Consider the following action of Z2 = 0, 1 on Sm:

0 · x = x, 1 · x = −x.

Prove that the action is smooth, proper and free. Prove that Sm/Z2 is diffeomorphicto Pm(R).

In the following exercise we construct the m-dimensional torus.

Exercise 3.8. Consider the action of the group (Zm,+) on Rm given by:

(k1, . . . , km) · (x1, . . . , xm) := (k1 + x1, . . . , km + xm).

Prove that the action is smooth, proper and free. The quotient manifold

Tm := Rm/Zm = x+ Zm : x ∈ Rm

is called the m-dimensional torus. Prove that Tm is diffeomorphic to the productof m-copies on S1 (see Proposition 2.1.6 for the construction of the product).

By using actions on R2, in the following exercise we construct several 2-dimensionalmanifolds: the cylinder, the Mobius band, the torus and the Klein bottle.

Exercise 3.9. Consider the following two diffeomorphisms of R2:

a, b : R2 −→ R2, a(x, y) = (x, y + 1), b(x, y) = (x+ 1/2,−y).

Let Diff(R2) be the group of diffeomorphisms of R2. Let Ga, Gb, Ga,b2 and Ga,b,respectively, denote the subgroups of Diff(R2) generated by the sets a, b, a, b2and a, b, respectively.


(a) Prove that the following relation holds: ba = a−1b. Using this relation, showthat every element in Ga,b can be represented as anbm, for unique n,m ∈ Z.Prove that the action of all four groups Ga, Gb, Ga,b2 and Ga,b on R2 is smooth,free and proper.

(b) Prove that the quotient R2/Ga is diffeomorphic to the cylinder S1 × R.(c) Prove that R2/Ga,b2 is diffeomorphic to the 2-torus T 2 (from Exercise 3.8).(d) Prove that R2/Gb is diffeomorphic to the manifold from Exercise 3.2. (A man-

ifold diffeomorphic to R2/Gb is called a Mobius band; see also Exercise 3.4.)(e) Prove that Ga,b2 is an index two subgroup of Ga,b, i.e. Ga,b/Ga,b2 ∼= Z2. Using

this, define a smooth, free and proper action of Z2 on T 2 such that T 2/Z2 isdiffeomorphic to R2/Ga,b. (A manifold diffeomorphic to R2/Ga,b is called aKlein bottle.)

Exercise 3.10. Let a : G×M →M be a proper smooth action, and assume thatM is compact. Prove that G is finite.

LECTURE 4

4.1. Bump functions

Recall the following topological notion:

Definition 4.1.1. Let X be a topological space, and let χ : X → R be a function.The support of χ is defined as the closure of the set where χ is not zero:

supp(χ) := x ∈ X : χ(x) 6= 0.

Lemma 1.1.11 shows that around any point in Rn we can find a smooth bumpfunction, i.e. a function with support in a fixed neighborhood of the point, andwhich is equal to the constant function 1 on a smaller neighborhood. This can beeasily generalized to manifolds.

Lemma 4.1.2. Let M be a smooth manifold, let p ∈M and let U ⊂M be an openneighborhood of p. Then there exists a smooth function χ : M → [0, 1] such thatsupp(χ) ⊂ U and χ|V = 1, where V is some neighborhood of p.

Proof. Let (W,ϕ) be a chart with p ∈ W ⊂ U , and let ε > 0 be such that theclosed ball Bε around ϕ(p) is included in ϕ(W ). By Lemma 1.1.11, there exists asmooth function θ : Rm → [0, 1] such that θ|Bε/2 = 1 and supp(θ) ⊂ Bε. Define

χ : M → [0, 1], χ|W = θ ϕ, χ|M\ϕ−1(Bε)= 0.

Note that χ is well-defined: on the overlap of the sets W and M\ϕ−1(Bε), we have:

θ ϕ|W\ϕ−1(Bε)= 0.

Since χ is smooth on these two open sets that cover M , it follows that χ is smoothon M . The conclusion holds with V := ϕ−1(Bε/2).

4.2. Partitions of unity

Partitions of unity play an important technical role in the theory of smooth mani-folds, because they allow to build global objects by gluing together locally definedobjects. In this section we prove the existence of partitions of unity.

We start by introducing the necessary terminology.

Definition 4.2.1. Let X be a topological space.

37


(1) An open cover of a subset Y ⊂ X is a family of open subsets of X

U = Uii∈I , such that Y ⊂ ∪i∈IUi,where I is an index set.

(2) A subcover of an open cover U of Y is a subfamily V ⊂ U which is still acover of Y .

(3) A subset Y ⊂ X is called compact, if any open cover of Y has a finite subcover.(4) A refinement of an open cover U of Y is a second open cover V of Y such

that for every V ∈ V there exists U ∈ U such that V ⊂ U .(5) A family S = Sii∈I of subsets of X is called locally finite if every x ∈ X

has an open neighborhood V which intersects only finitely many elements of S.(6) A family of functions ρi : X → Ri∈I is called locally finite if the family of

supports supp(ρi)i∈I is locally finite.

The following will be often used:

Lemma 4.2.2. Let M be a smooth manifold, and consider a locally finite familyof smooth functions on M , ρi ∈ C∞(M)i∈I . Since at every point x ∈ M thereare only a finite number of i ∈ I such that ρi(x) 6= 0, we can define:

ρ =∑i∈I

ρi.

The function ρ is smooth.

Proof. Let x ∈M . Since the family is locally finite, x has an open neighborhood Uwhich intersects only the supports of a finite number of functions, say ρi1 , . . . , ρin .Therefore, on U we have that:

ρ|U =

n∑k=1

ρik |U .

The sum of a finite number of smooth functions is smooth, hence ρ|U is smooth. So,every point has a neighborhood on which ρ is smooth; thus ρ is smooth on M .

In order to define the function ρ in the lemma above, it suffices that at everypoint only a finite number of functions are non-zero. However, this condition doesnot insure smoothness of the resulting function; in Exercise 4.1, you are asked toprovide an example.

The main notion of this section is the following:

Definition 4.2.3. A partition of unity on a smooth manifold M is a locallyfinite family of smooth functions

ρii∈I , ρi ∈ C∞(M), for i ∈ I,such that, for all x ∈M , 0 ≤ ρi(x) ≤ 1 and∑

i∈Iρi = 1.

The partition of unity ρii∈I is said to be subordinate to an open coverU = Uii∈I if supp(ρi) ⊂ Ui for all i ∈ I.

The following is the main result about partitions of unity:

Theorem 4.2.4. Any open cover of a smooth manifold has a partition of unitysubordinate to it.

LECTURE 4. 39

4.3. Proof of Theorem 4.2.4

The proof is divided into three main steps. First, we show that it suffices to provethe result for a refinement of the cover (Lemma 4.3.1); next, we show that opencovers of manifolds admit locally finite refinements whose elements have compactclosures (Lemma 4.3.4; this result implies that manifolds are paracompact, seeRemark 4.3.5); and finally, we prove Theorem 4.2.4 for open covers with theseproperties.

Lemma 4.3.1. Let V be a refinement of the open cover U . If V has a partition ofunity subordinate to it, then so does U .

Proof. Denote V = Vii∈I and U = Ujj∈J . The fact that V is a refinementof U implies that there exists a function f : I → J (axiom of choice) such thatVi ⊂ Uf(i). Let ρii∈I be a partition of unity subordinate to V. Since the familyof supports is locally finite, the following functions are smooth: χj :=

∑i∈f−1(j) ρi

(if f−1(j) = ∅, we set χj = 0). We claim that χjj∈J is a partition of unitysubordinate to U . First, note that:

(*) supp(χj) ⊂⋃

i∈f−1(j)

supp(ρi) ⊂⋃

i∈f−1(j)

Vi ⊂ Uj .

Next, we show next that the family χjj∈J is locally finite. Let p ∈ M , andconsider O an open neighborhood of p such that O intersects only a finite numberof the sets supp(ρi); denote these indexes by i1, . . . , in ∈ I. If q ∈ supp(χj), thenby (*), q ∈ supp(ρi) for some i satisfying f(i) = j. Therefore, O intersects only thesets supp(χj), for j = f(i1), . . . , f(ik). Finally, it is clear that:∑

j∈Jχj =

∑j∈J

∑i∈f−1(j)

ρi =∑i∈I

ρi = 1.

The following two results are used in the proof of Lemma 4.3.4.

Lemma 4.3.2. The topology of a manifold has a countable basis all of whose ele-ments have compact closures.

Proof. Let B be a countable basis for M , and let Bc be the collection of sets in Bwith compact closure. Clearly, Bc is countable. We prove that Bc is a basis. LetO ⊂M be an open set. For p ∈ O, let Kp ⊂ O be a compact neighborhood (e.g. letKp be the preimage by a chart around p of a small closed ball). Since B is a basis,there exists Up ∈ B such that p ∈ Up ⊂ int(Kp) ⊂ Op. In particular: Up ∈ Bc andUp ⊂ O. Thus, we can write O = ∪p∈OUp, which shows that Bc is a basis.

Lemma 4.3.3. Any manifold M has an open cover Gk∞k=1 such that, for k ≥ 1,

Gk is compact and Gk ⊂ Gk+1.

Proof. By Lemma 4.3.2, there exists a countable basis B = Bnn≥1 for the topol-

ogy of M such that Bn is compact for all n ≥ 1. Let Om := ∪mn=1Bn. Note thatOm ⊂ Om+1, Om = ∪mn=1Bn is compact, and that ∪m≥1Om = M . Therefore,

for any m ≥ 1 there is a smallest integer f(m) > m such that Om ⊂ Of(m).Let m1 := 1, and define inductively for k ≥ 2, mk := f(mk−1). The sequenceGk := Omk satisfies all requirements.


Lemma 4.3.4. Any open cover U of a manifold M has a refinement V which is atmost countable, locally finite, and whose elements have compact closures.

Proof. Denote U = Uii∈I . Consider a sequence of open sets Gk∞k=1 as inLemma 4.3.3, and denote G−1 = G0 = ∅. For each k ≥ 1, the family

Ui ∩ (Gk+1\Gk−2)i∈Iis an open cover of the compact set Gk\Gk−1, therefore it has a finite subcover Vk.Let V := ∪k≥1Vk. We prove that V has the required properties. Clearly, V is an

open cover because it covers Gk\Gk−1, and it is a refinement of U because eachelement is contained in some Ui. Since, for each k ≥ 1, Vk contains only a finitenumber of open sets, V is at most countable. Every set in Vk has compact closurebecause it is contained in the compact Gk+1. Finally, let p ∈M and l ≥ 1 be suchthat p ∈ Gl. Then Gl intersects at most elements in Vk with k ≤ l+ 1. This showsthat V is locally finite.

Let us make a general remark about what we have proven so far.

Remark 4.3.5. A simple analysis of the proofs of the previous three Lemmas(4.3.2,4.3.3,4.3.4) shows that we have used only the following topological propertiesof a manifold:

• second countable;• Hausdorff (in particular, compact sets are closed);• locally compact (i.e. every point has a compact neighborhood).

A topological space for which every open cover has a locally finite refinement iscalled paracompact. Thus we have proven that: Any topological space which issecond countable, Hausdorff and locally compact is paracompact.

Finally, we show that open covers which have the properties from Lemma 4.3.4admit partitions of unity, and therefore we complete the proof of Theorem 4.2.4.

Lemma 4.3.6. Let V be a locally finite open cover of M whose elements havecompact closures. Then there exists a partition of unity subordinate to V.

Proof. Let V ∈ V. By Lemma 4.1.2, for each p ∈ V there is a smooth functionψVp : M → [0, 1] such that ψVp (p) = 1 and supp(ψVp ) ⊂ V . Denote WV

p := x ∈M :

ψVp (x) > 0. Then the collection W := WVp : p ∈ V ∈ V is an open cover of M

which is a refinement of V. By Lemma 4.3.4 W has a locally finite refinement O.So, for O ∈ O there is a smooth function ψO : M → [0, 1] and V ∈ V such that

ψO|O > 0 and supp(ψO) ⊂ V.We show that an element V ∈ V intersects only finitely many elements in O. SinceO is locally finite, V can be covered by open sets which intersect only finitely manyelements in O; and by compactness of V , we may extract a finite subcover. Let Ube the union of the elements of this subcover. Then V ⊂ U and U intersects onlyfinitely many elements in O.

In particular, for any V ∈ V the set

OV := O ∈ O : supp(ψO) ⊂ V is finite. Consider the family of smooth maps

ψV :=∑O∈OV

ψO : M → [0,∞), V ∈ V.

LECTURE 4. 41

Because it is subordinate to V, it follows that the family is locally finite; and bythe properties of O, at every point in M at least one of these functions is positive.Therefore ψ :=

∑V ∈V ψ

V > 0. The family of smooth functions ρV := ψV /ψ, V ∈ Vis a partition of unity subordinate to V.

4.4. Corollaries of Theorem 4.2.4

For covers with two elements, Theorem 4.2.4 gives a smooth Urysohn’s lemma:

Corollary 4.4.1. Let A,B ⊂M be two disjoint closed sets. There exists a smoothfunction f : M → [0, 1] such that f |A = 0 and f |B = 1.

In particular, M is a normal topological space.

Proof. Consider the open cover M\A,M\B. If f, g is a partition of unitysubordinated to it, then f satisfies the condition from the statement. Finally, Mis normal, because A and B can be separated by the open sets x : f(x) < 1/2and x : f(x) > 1/2, respectively.

Theorem 4.2.4 can be used to extend smooth functions defined on closed em-bedded submanifolds in Rn.

Corollary 4.4.2. Let M ⊂ Rn be an embedded manifold in Rn which is closed as

a subset. For every f ∈ C∞(M) there exists f ∈ C∞(Rn) such that f |M = f .

Proof. Recall that around every point in M there is a diffeomorphism ϕ : U → V ,with V,U ⊂ Rn open, such that

ϕ(M ∩ U) = (Rm × 0) ∩ V.

We cover M by a family of such charts (Ui, ϕi)i∈I . Consider the following opencover: Rn\M∪Uii∈I , and let ρ∪ρii∈I be a subordinate partition of unity.For each i ∈ I, consider the following local extension of f :

fi := f (ϕ−1i prnm ϕi) ∈ C∞(Ui),

and consider also the function:

gi ∈ C∞(M), gi(x) :=

ρi(x)fi, x ∈ Ui0, x /∈ supp(ρi)

The family gii∈I is locally finite, and it is easy to see that the sum f :=∑i∈I gi

satisfies f |M = f .

Recall the following notion:

Definition 4.4.3. A continuous function f : X → Y between topological spaces Xand Y is called proper, if for any compact subset K ⊂ Y , we have that f−1(K) ⊂ Xis compact.

Corollary 4.4.4. On any manifold M there exists a proper smooth function h :M → [0,∞).

Proof. Let ρkk≥1 be a partition of unity subordinated to a locally finite, at most

countable cover Vkk≥1, with V k compact (as in Lemma 4.3.4). Then the functionh :=

∑k≥1 kρk is a positive proper function.


4.5. Exercises

Exercise 4.1. Construct a sequence of functions ρk ∈ C∞(R), k = 1, 2, . . ., suchthat at every x ∈ R, ρk(x) 6= 0 only for a finite number of k’s, but such that thefunction

∑k≥1 ρk is not smooth.

Exercise 4.2. Let U = Uii∈I and V = Vjj∈J be two open covers of M . Showthat W := Ui ∩ Vj(i,j)∈I×J is an open cover of M . If ρii∈I is a partition ofunity subordinate to U and χjj∈J is a partition of unity subordinate to V, provethat ρiχj(i,j)∈I×J is a partition of unity subordinate to W.

The locally finite cover found in Lemma 4.3.4 is at most countable. However,this fact is automatic.

Exercise 4.3. Let U be a locally finite open cover of a manifold M .

(a) If M is compact, prove that U has a finite number of elements.(b) In general, prove that U is at most countable.

Hint: choose a second cover V, as in Lemma 4.3.4, and show that every elementin V hits only a finite number of elements of U .

Exercise 4.4. Let ρii∈I be a locally finite family of smooth functions on M .

(a) If M is compact, prove that only a finite number of functions are not identicallyzero.

(b) In general, prove that the set of functions which are not identically zero is atmost countable.

LECTURE 5

Remark. For simplicity, from now on we will usually suppress the adjective smooth,by implicitly assuming smoothness of the objects involved: by manifold we meansmooth manifold, by chart we mean smooth chart, etc.

5.1. The tangent space

Intuitively, the tangent space of a smooth manifold M at a point p, is a vectorspace denoted by TpM of dimension m = dim(M), which represents all velocitiesof curves passing through p.

Example 5.1.1. For example, the tangent space to the 2-sphere S2 at a pointp ∈ S2 can be identified with the set of vectors perpendicular to p: TpS

2 ∼= v ∈R3 : v · p = 0. Namely, any smooth curve γ : (−ε, ε) → S2 (with γ(0) = p)satisfies γ(t) · γ(t) = 1, hence taking the derivative, γ′(t) · γ(t) = 0; in particularv · p = 0, where v = γ′(0). On the other hand, given a local parameterizationf : O ⊂ R2 → S2 (i.e. the inverse of a local chart) with f(q) = p, we can describethe tangent space as the image of the differential of f :

TpS2 ∼= (dqf)(R2).

Example 5.1.2. The tangent space of Rm at p ∈ Rm can be canonically identifiedTpRm ∼= Rm: any smooth curve γ : (−ε, ε) → Rm with γ(0) = p induces thetangent vector γ′(0) ∈ Rm. Conversely, any v ∈ Rm is the speed of some curvethrough p; for example, γ(t) = p + tv. We think about TpRm as being “vectorsbased at p”. Let us see what happens when we change coordinates on Rm, i.e.consider a diffeomorphism χ : U ∼−→ V , between open sets U, V ⊂ Rm. Then anysmooth curve γ : (−ε, ε)→ U , with γ(0) = p, can be transported to a smooth curveχ γ : (−ε, ε)→ V , and their speeds are related by the chain rule:

(χ γ)′(0) = (dpχ)γ′(0).

Thus, the differential of the change of coordinate matrix gives a natural identifica-tion between the tangent spaces.

On a general manifold, a fixed chart induces a model for the tangent space (i.e.just Rm), and we use the differential of the change of coordinate maps to identifythese various models.

43


Definition 5.1.3. Let M be a smooth manifold of dimension m. A tangent vectorat p ∈ M is a map v which associates to each chart (U,ϕ) around p an elementvϕ ∈ Rm satisfying the following rule

vψ = dχ(p)(ψ χ−1)(vχ),

for any two charts (O,χ) and (V, ψ) around p. The set of all tangent vectors at pis called the tangent space of M at p and is denoted by TpM .

Since the differential of the change of coordinates map is a linear map, it followsthat TpM is a vector space with operations

(λv)ϕ = λvϕ, (v + w)ϕ = vϕ + wϕ, ∀v, w ∈ TpM, λ ∈ R.

We have that:

Lemma 5.1.4. The tangent space of M at p is a vector space of dimension m =dim(M). Moreover, for any chart (U,ϕ) around p, the map

TpM → Rm, v 7→ vϕ

is a linear isomorphism.

Proof. That the map is linear is clear. The kernel of the map is zero: if vϕ = 0then, by linearity of the differential, vψ = dϕ(p)(ψϕ−1)(vϕ) = 0 for any other chart(V, ψ), thus v = 0. Thus the map is injective. To show surjectivity, let u ∈ Rm,and define v so that

vψ := dϕ(p)(ψ ϕ−1)(u),

for a chart (V, ψ). That v is indeed a tangent vector follows from the chain rule:for any two charts (V, ψ) and (O,χ), we have

vψ = dϕ(p)(ψ ϕ−1)(u) = dϕ(p)(ψ χ−1 χ ϕ−1)(u) =

= dχ(p)(ψ χ−1) dϕ(p)(χ ϕ−1)(u) = dχ(p)(ψ χ−1)(vχ).

Clearly, vϕ = u; thus the assignment is indeed onto.

Using the standard chart (Rm, id) on Rm, the lemma above allows us to identify

(∗) TpRm → Rm, v 7→ vid.

5.1.5. The differential of a smooth map

Next, we extend the notion of the differential of a smooth map. We will use thesame notation as for maps between open sets in Euclidean space; the standardnotion is obtained by using the identification (∗) above.

Definition 5.1.6. Let f : M → N be a smooth map. The differential of f atp ∈M is the linear map

dpf : TpM → Tf(p)N(dpf(v)

)ψ

= dϕ(p)(ψ f ϕ−1)(vϕ),

where (U,ϕ) is a chart around p and (V, ψ) is a chart around f(p).

To show that the above assignment ψ 7→ (dpf(v))ψ yields indeed a tangentvector, one needs to show that it transforms accordingly to the rule from Definition5.1.3; however, this is a straightforward application of the chain rule, which we omit(as in the proof of Lemma 5.1.4).

LECTURE 5. 45

With the canonical identification of the tangent space of Rm, note that thedifferential of a chart (U,ϕ) around p ∈M becomes

dpϕ : TpM → Tϕ(p)Rm, dpϕ(v) = vϕ.

The chain rule also extends to the setting of smooth manifolds, and its proofis again a straightforward application of the ‘classical’ chain rule:

Lemma 5.1.7. Let f : M → N and g : N → P be smooth maps between smoothmanifolds. The differentials satisfy the chain rule:

dp(g f) = df(p)g dpf, ∀ p ∈M.

5.2. Notations for tangent vectors

Let M be a smooth manifold and consider a chart (U,ϕ) around p ∈ M inducinglocal coordinates

xiϕ : U → R, ϕ = (x1ϕ, . . . , x

mϕ ).

Let e1, . . . , em be the standard basis of Rm. The corresponding basis of TpM willbe denoted by

∂

∂x1ϕ

∣∣∣p,

∂

∂x2ϕ

∣∣∣p, . . . ,

∂

∂xmϕ

∣∣∣p∈ TpM,

( ∂

∂xiϕ

∣∣∣p

)ϕ

= dpϕ( ∂

∂xiϕ

∣∣∣p

)= ei, 1 ≤ i ≤ m.

To explain the notation, let f ∈ C∞(M) be a smooth function on M . Then itsdifferential is given in this standard basis by:

dpf( ∂

∂xiϕ

∣∣∣p

)= dϕ(p)(f ϕ−1)(ei) =

∂f ϕ−1

∂xi(ϕ(p));

in other words the differential of f on the standard basis is given by the partialderivatives of the local expression f ϕ−1 of f .

More generally, let f : M → N be a smooth map, and consider a smooth chart(V, ψ) around q = f(p), with local coordinates ψ = (y1

ψ, . . . , ynψ).The matrix of dpf

in the bases ∂

∂xiϕ

∣∣∣p

: 1 ≤ i ≤ m

and

∂

∂yjψ

∣∣∣q

: 1 ≤ j ≤ n

of TpM and TqN , respectively, is the Jacobian matrix of the local representation

ψ f ϕ−1(x) = (f1(x), . . . , fn(x))

of f in these charts:

[dpf

]=

∂f1

∂x1∂f1

∂x2 . . . ∂f1

∂xm∂f2

∂x1∂f2

∂x2 . . . ∂f2

∂xm

. . . . . . . . . . . .∂fn

∂x1∂fn

∂x2 . . . ∂fn

∂xm

(ϕ(p));

dpf( ∂

∂xiϕ

∣∣∣p

)=

n∑j=1

∂f j

∂xi(ϕ(p))

∂

∂yjψ

∣∣∣q.


5.3. The commutative algebra C∞(M)

The “cleanest” way to introduce the tangent space is by using the commutativeR-algebra of smooth functions:

(C∞(M),+, ·), (f + g)(x) := f(x) + g(x), (f · g)(x) := f(x) · g(x).

Note that smooth maps induce algebra homomorphisms:

Proposition 5.3.1. Any smooth map ϕ : M → N induces a homomorphism ofalgebras, called the pullback along ϕ:

ϕ∗ : C∞(N)→ C∞(M), ϕ∗(f) = f ϕ.

The more surprising result is the converse of this Proposition, which impliesthat the commutative algebra C∞(M) encodes algebraically the entire informationabout the manifold M :

Theorem ♣ 5.3.2. Any algebra homomorphism s : C∞(N) → C∞(M) is of theform s = ϕ∗ for a unique smooth map ϕ : M → N .

In particular, this result implies that two manifolds are diffeomorphic if andonly if their commutative algebras of smooth functions are isomorphic. A proof ofTheorem 5.3.2 is outlined in the exercises at the end of this lecture.

5.4. Tangent vectors as derivations

Recall that on Rm the differential of a function f : U ⊂ Rm → Rn is defined usingdirectional derivatives:

dpf(v) =d

dtf(p+ tv)

∣∣t=0

.

We introduce the analog terminology on manifolds:

Definition 5.4.1. Let M be a smooth manifold. The derivative along v ∈ TpMis the map

Dv : C∞(U)→ R, Dv(f) := dpf(v).

The following holds:

Lemma 5.4.2. For all f, g ∈ C∞(M), and v ∈ TpM we have that:

Dv(f · g) = f(p)Dv(g) +Dv(f)g(p).

Proof. Using a chart, it suffices to prove this on open sets in Rm. Here it followsfrom the usual Leibniz rule (α · β)′ = α · β′ + α′ · β applied to the functionsα(t) = f(p+ tv) and β(t) = g(p+ tv) at t = 0.

Definition 5.4.3. A derivation of C∞(M) at p ∈M is a linear map

D : C∞(M) −→ R

satisfying the derivation rule:

D(f · g) = f(p)D(g) +D(f)g(p), ∀ f, g ∈ C∞(M)

The space of derivations of C∞(M) at p will be denoted by Derp(C∞(M)).

LECTURE 5. 47

Note that the space of derivations is a vector space: for all λ, µ ∈ R andD1, D2 ∈ Derp(C

∞(M)),

λD1 + µD2 : C∞(M)→ R, (λD1 + µD2)(f) := λD1(f) + µD2(f)

is also derivation of C∞(M) at p.The following result shows that all derivations come from derivatives along

tangent vectors.

Theorem 5.4.4. Let M be a smooth manifold, and let p ∈M . The map

TpM → Derp(C∞(M)), v 7→ Dv


To prove the theorem, we first show that derivations of C∞(M) at p are local,in the sense that they depend only on the behavior of the functions around p. Thefollowing terminology will be therefore rather useful:

Definition 5.4.5. Let M be a manifold and p ∈M . Two functions f, g ∈ C∞(M)are said to have the same germ at p, if there is an open neighborhood W of p suchthat f |W = g|W .

Lemma 5.4.6. Let f, g ∈ C∞(M) have the same germ at p ∈ M , then for anyD ∈ Derp(C

∞(M)), we have that D(f) = D(g).

Proof. Let h := f − g; then h|W = 0, for some neighborhood W of p. We showthat D(h) = 0. Consider a bump function χ ∈ C∞(M) such that χ(p) = 1 andsupp(χ) ⊂W . Then, χh = 0. Therefore:

0 = D(0) = D(χh) = h(p)D(χ) + χ(p)D(h) = D(h).

Lemma 5.4.7. Let U ⊂M be an open neighborhood of p. Then the map

I : Derp(C∞(U))→ Derp(C

∞(M)), I(D)(g) := D(g|U )


Proof. Let χ ∈ C∞(M) be a bump function such that χ|W = 1, on some neigh-borhood W ⊂ U of p and such that supp(χ) ⊂ U . We define the map

E : C∞(U)→ C∞(M), E(f)|U := χf, E(f)|M\supp(χ) := 0.

Note that f and E(f)|U = χf have the same germ at p. Consider the map

R : Derp(C∞(M))→ Derp(C

∞(U)), R(D′)(f) := D′(E(f)).

Since E(fg) and E(f)E(g) have the same germ at p, R(D′) is indeed a derivation:

R(D′)(fg) =D′(E(fg)) = D′(E(f)E(g)) = E(f)(p)D′(E(g))

+ E(g)(p)D′(E(f)) = f(p)R(D′)(g) + g(p)R(D′)(f).

Note that I R(D′)(f) = D′(E(f)|U ) = D′(f), hence I R(D′) = D′; and similarly,that R I(D)(g) = D(E(g|U)) = D(g), hence R I(D) = D. We obtain that I isinvertible with I−1 = R.

For the proof, we will need the following version of Taylor’s theorem:


Lemma 5.4.8. Let B ⊂ Rm be a convex open neighborhood of 0. For any f ∈C∞(B) there exist smooth functions gi ∈ C∞(B), 1 ≤ i ≤ m, such that

f(x) = f(0) +

m∑i=1

gi(x)xi,

and moreover, gi(0) = ∂f∂xi (0).

Proof. Fix x ∈ B, and consider the function h : [0, 1]→ R, h(t) = f(tx). Clearly,h is smooth, therefore the fundamental theorem of calculus applies:

h(1)− h(0) =

∫ 1

0

h′(t)dt.

This equation gives:

f(x)− f(0) =

∫ 1

0

d

dtf(tx1, . . . , txm)dt =

∫ 1

0

m∑i=1

∂f

∂xi(tx)xidt =

m∑i=1

gi(x)xi,

where gi(x) =∫ 1

0∂f∂xi

(tx)dt. Clearly, gi ∈ C∞(B) and gi(0) = ∂f∂xi

(0).

Proof of Theorem 5.4.4. Consider a chart (U,ϕ) on M around p, with coordi-nates ϕ = (x1

ϕ, . . . , xmϕ ) such that ϕ(p) = 0 and ϕ(U) = B is convex.

By Lemma 5.4.7, it suffices to show that the map

TpM → Derp(C∞(U)), v 7→ Dv

is an isomorphism.Let v ∈ TpM and denote vϕ = (v1, . . . , vm). By the formulas in Section 5.2,

Dv(xiϕ) = dpx

iϕ(v) =

m∑j=1

vj∂xi

∂xj(0) = vi;

thus, the map is injective.Let D ∈ Derp(C

∞(U)). We show that D vanishes on constant functions. Since

D(1) = D(1 · 1) = 1 ·D(1) +D(1) · 1 = 2 ·D(1),

we obtain that D(1) = 0. By linearity, D(c) = D(c · 1) = c ·D(1) = 0, for all c ∈ R.Let v ∈ TpM be the vector with vϕ = (v1, . . . , vm) ∈ Rm, where vi := D(xiϕ).

We show that D = Dv. Let f ∈ C∞(U). Applying Lemma 5.4.8, we find functionsgi ∈ C∞(U), for 1 ≤ i ≤ m, such that

f = f(p) +

m∑i=1

xiϕgi, gi(p) = dpf( ∂

∂xiϕ

∣∣∣p

).

Applying D, we obtain:

D(f) = D(f(p)) +

m∑i=1

D(xiϕgi) =

= 0 +

m∑i=1

(0 ·D(gi) +D(xiϕ) · gi(p)

)=

=

m∑i=1

vidpf( ∂

∂xiϕ

∣∣∣p

)= dpf(v) = Dv(f).

Thus, D = Dv.

LECTURE 5. 49

5.5. Exercises

Exercise 5.1. Let M ⊂ Rn be a manifold embedded in Rn, as in Definition 2.4.1,and let p ∈M . Let Vp ⊂ Rn consist of all vectors γ′(0) ∈ Rn, where γ : (−ε, ε)→Mis a smooth curve such that γ(0) = p. Show that Vp is a linear subspace of Rn, andthat there is a linear isomorphism Vp

∼−→ TpM , such that γ′(0) ∈ Vp corresponds

to d0γ( ddt |t=0) ∈ TpM .

The following exercise gives an alternative definition of the tangent space:

Exercise 5.2. Let M be a manifold and let p ∈ M . Denote by Ip ⊂ C∞(M) theideal of functions vanishing at p:

Ip = f ∈ C∞(M) : f(p) = 0.Denote by I2

p the square of this ideal, i.e. I2p consists of finite sums of the form:

f =

k∑j=1

gjhj , gj , hj ∈ Ip, 1 ≤ j ≤ k.

(a) Prove that Ip/I2p has dimension m = dim(M).

Hint: use Lemma 5.4.8 and Lemma 4.1.2.(b) Show that dpf(v) = 0 for all f ∈ I2

p and v ∈ TpM .(c) Prove that the following map is a linear isomorphism:

E : TpM → (Ip/I2p)∗, E(v)(f + I2

p) := dpf(v), f ∈ Ip.

We introduce some terminology for the following exercises.

Definition 5.5.1. An algebra over R is a real vector space A endowed with anassociative, bilinear multiplication · : A×A→ A, and with a unit 1A ∈ A; i.e.

(f · g) · h = f · (g · h), 1A · f = f = f · 1A,

(λf + µg) · h = λ(f · h) + µ(g · h), f · (λg + µh) = λ(f · g) + µ(f · h),

for all f, g, h ∈ A and all λ, µ ∈ R. The algebra A is called commutative iff · g = g · f for all f, g ∈ A.

A homomorphism between the R-algebras A and B is a linear map s : A→ B,which preserves multiplication and units:

s(f · g) = s(f) · s(g) and s(1A) = 1B .

Definition 5.5.2. Let A and R be commutative algebras over R, and let s : A→ Rbe an algebra homomorphism. A derivation at s of A is a linear map

D : A −→ R s.t. D(u · v) = s(u)D(v) +D(u)s(v), for all u, v ∈ A.We denote the space of all derivations at s by:

Ders(A,R).

Definition 5.5.3. Let A be a commutative algebra over R. A character of A isan algebra homomorphism χ : A→ R.

Definition 5.5.4. For a commutative algebra R, let R[X] denote the polynomialalgebra with coefficients in R, i.e. each element in p ∈ R[X] can be written as afinite sum:

p = u0 + u1X + . . .+ ukXk,


for unique u0, . . . , uk ∈ R; the algebraic operations are defined as usually for poly-nomials. Consider the following algebra:

R[ε] := R[X]/(X2),

where (X2) ⊂ R[X] is the ideal generated by X2, i.e. each element p ∈ R[ε] can beuniquely written as

p = u0 + u1ε,

where ε := X + (X2) ∈ R[X]/(X2); and the multiplication is such that ε2 = 0:

(u0 + u1ε) · (v0 + v1ε) = u0v0 + (u0v1 + u1v0)ε.

Exercise 5.3. (a) Let s : A → R be an algebra homomorphism, and let D ∈Ders(A,R). Prove that the map

s+ εD : A −→ R[ε], (s+ εD)(u) := s(u) + εD(u)

is an algebra homomorphism.(b) Prove that any algebra homomorphism σ : A → R[ε] is of the form described

at (a).

Exercise 5.4. Let s : A→ R be a homomorphism of algebras.

(a) Prove that a homomorphism of algebras t : B → A induces a linear mapbetween the derivation spaces, defined as follows:

t∗s : Ders(A,R) −→ Derst(B,R), t∗s(D) := D t.

(b) If t is surjective, prove that t∗s is injective.(c) Prove that the following “chain rule” holds: if t : B → A and r : C → B are

algebra homomorphisms, then

(t r)∗s = r∗st t∗s : Ders(A,R) −→ Derstr(C,R).

Exercise 5.5. Let X be a topological space, and let C(X) be the algebra ofcontinuous maps X → R. Consider a subalgebra A ⊂ C(X), with unit 1A theconstant map 1A(p) = 1, and which satisfies the condition:

(*) (f ∈ A, f(x) 6= 0, ∀ x ∈ X) =⇒ 1/f ∈ A.

We denote by A∨ the set of characters. For χ ∈ A∨, let Iχ denote the kernel of χ:

Iχ := f ∈ A : χ(f) = 0.

(a) Let p ∈ X. Prove that the following map gives a character of A:

χp : A −→ R, χp(f) := f(p).

We denote Ip := Iχp .(b) For χ ∈ A∨, and p ∈ X, prove that Ip ⊂ Iχ, implies χ = χp.(c) Prove that if f ∈ Iχ, then there exists p ∈ X such that f(p) = 0.(d) Let χ ∈ A∨ be a general character, and assume that χ 6= χp for every p ∈ X.

Prove that for any point p ∈ X there exists f ∈ Iχ such that f(p) > 0.(e) Prove that if X is compact, then every χ ∈ A∨ is of the form χ = χp for some

p ∈ X.(f) Assume that there exists f ∈ A such that f−1(λ) ⊂ X is compact for any

λ ∈ R. Show that every χ ∈ A∨ is of the form χ = χp for some p ∈ X.

LECTURE 5. 51

Exercise 5.6. Prove that for any algebra homomorphism χ : C∞(M) → R (i.e.any character of C∞(M)) there exists a unique point p ∈M such that:

χ(f) = f(p), for all f ∈ C∞(M).

Hint: use Exercise 5.5 and Corollary 4.4.4.

Exercise 5.7. Let M be a manifold. Show that there is a one-to-one correspon-dence between tangent vectors on M and algebra homomorphisms C∞(M)→ R[ε].Hint: use Exercises 5.3 and 5.6.

The following exercise proves Theorem 5.3.2.

Exercise 5.8. Let s : C∞(N)→ C∞(M) be an algebra homomorphism.

(a) Prove that there exists a unique function ϕ : M → N such that s(f) = f ϕfor all f ∈ C∞(N). Hint: use Exercise 5.6.

(b) Prove that for any smooth function f ∈ C∞(N), the set

Uf := p ∈ N : f(p) 6= 0is open, and that these sets forms a basis for the topology of N .Hint: use Lemma 4.1.2.

(c) Prove that ϕ is continuous.(d) Prove that ϕ is smooth.

LECTURE 6

6.1. The inverse function theorem

The inverse function theorem gives a very simple condition to check that a map islocally a diffeomorphism, namely, it suffices that its differential be invertible. Fora proof of this classical result, see for example [11]. This can be directly extendedto smooth manifolds:

Theorem 6.1.1 (The inverse function theorem). Let f : M → N be a smooth mapand let p ∈ M . If dpf : TpM → Tf(p)N is a linear isomorphism, then there existsan open neighborhood O ⊂ M of p, such that f(O) is open and f restricts to adiffeomorphism:

f |O : O ∼−→ f(O).

Proof. Let (U,ϕ) be a chart around p and (V, ψ) be a chart around f(p). Byshrinking U , we may assume that f(U) ⊂ V . By the chain rule, we have that

dϕ(p)(ψ f ϕ−1) = df(p)ψ dpf (dpϕ)−1.

Since the charts are diffeomorphisms, their differentials are linear isomorphisms;therefore, the right hand side is the composition of linear isomorphisms. We obtainthat also dϕ(p)(ψf ϕ−1) is a linear isomorphism. The map ψf ϕ−1 is a smoothmap between open sets in Rm, and its Jacobian matrix at ϕ(p) coincides with itsdifferential, thus it is invertible. By the standard inverse function theorem (see e.g.[11]), there exists an open neighborhood W of ϕ(p) such that ψ f ϕ−1(W ) isopen and ψ f ϕ−1|W : W ∼−→ ψ f ϕ−1(W ) is a diffeomorphism. Since thecharts are diffeomorphisms, this implies that conclusion with O := ϕ−1(W ).

6.2. Immersions and submersions

The following two classes of maps play an important role:

Definition 6.2.1. Let f : M → N be a smooth map, and let p ∈M .

(1) f is called an immersion at p if dpf : TpM → Tf(p)N is injective. If thisholds at all points in M , then f is called an immersion.

(2) f is called a submersion at p if dpf : TpM → Tf(p)N is surjective. If thisholds at all points in M , then f is called a submersion.

53


The most basic example of an immersion is the linear inclusion:

ikl : Rk −→ Rl, ikl (x1, . . . , xk) := (x1, . . . , xk, 0, . . . , 0), for k ≤ l,

and the most basic example of a submersion is the linear projection

prlk : Rl −→ Rk, prlk(x1, . . . , xl) := (x1, . . . , xk), for k ≤ l.

The following two theorems say that locally, up to a change of coordinates,these are the only examples:

Theorem 6.2.2 (The local immersion theorem). Let f : M → N be a smooth map.If f is an immersion at p ∈ M , then there are charts (U,ϕ) and (V, ψ) around pand f(p), respectively, such that f(U) ⊂ V and, in these coordinates, f becomesthe map imn ; i.e.

ψ f ϕ−1 = imn |ϕ(U) : ϕ(U) −→ Rn,ψ f ϕ−1(x1, . . . , xm) = (x1, . . . , xm, 0, . . . , 0).

Theorem 6.2.3 (The local submersion theorem). Let f : M → N be a smoothmap. If f is a submersion at p ∈M , then there are charts (U,ϕ) and (V, ψ) aroundp and f(p), respectively, such that f(U) ⊂ V and, in these coordinates, f becomesthe map prmn ; i.e.

ψ f ϕ−1 = prmn |ϕ(U) : ϕ(U) −→ Rn,

ψ f ϕ−1(x1, . . . , xm) = (x1, . . . , xn).

The proofs of these two results are completely analogous, therefore, we onlygive the details for the second result; the proof of the local immersion theorem isleft as an exercise.

Proof of the local submersion theorem. Let (U , ϕ) and (V, ψ) be two charts

around p and f(p), respectively, such that f(U) ⊂ V . Denote the local expressionof f by h := ψ f ϕ−1. We are going to “complete h to a diffeomorphism”.By applying Exercise 6.1 to the linear map A = dϕ(p)h (which, by the chain rule,

is surjective), we obtain a linear map B : Rm → Rm−n such that (dϕ(p)h,B) :Rm → Rm is a linear isomorphism. This map is precisely the differential at ϕ(p)

of the map (h,B) : ϕ(U) → Rm. Therefore, by the inverse function theorem,

there is a neighborhood O ⊂ ϕ(U) of ϕ(p) such that W := (h,B)(O) is open and(h,B) : O ∼−→W is a diffeomorphism; denote its inverse by g := (h,B)−1 : W ∼−→ O.Then we have that (h(g(x)), B(g(x))) = x for all x ∈ W , hence h g = prmn |W .Equivalently, ψ f ϕ−1 = prmn |W , where

ϕ := g−1 ϕ = (h,B) ϕ : U −→ Rm, U := ϕ−1(O);

hence, the charts (U,ϕ) and (V, ψ) satisfy the conditions from the statement.

Remark 6.2.4. In the local submersion theorem, the chart on the codomain (V, ψ)can be chosen arbitrary; this follows from the proof. Similarly, in the local immer-sion theorem the chart (U,ϕ) on the domain can be chosen arbitrary.

LECTURE 6. 55

6.3. Embedded submanifolds

The following generalizes the notion of embedded manifolds in Rn from Section 2.4:

Definition 6.3.1. A subset S ⊂ M of a smooth m-dimensional manifold M iscalled a k-dimensional embedded submanifold if for every point p ∈ S thereexists a chart (U,ϕ) around p, such that

(*) ϕ(S ∩ U) =(Rk × 0

)∩ ϕ(U) ⊂ Rm.

A chart (U,ϕ) with these property is called a chart adapted to S. The codimen-sion of S is the number m− k = dim(M)− dim(S).

Next, we show that embedded submanifolds are indeed smooth manifolds:

Theorem 6.3.2. Let S ⊂ M be a k-dimensional embedded submanifold. Then S,endowed with the subset topology, has a k-dimensional differentiable structure, withatlas obtained by restricting adapted charts:

AS = (U ∩ S, prmk ϕ|U∩S) : (U,ϕ) is a chart adapted to S.Moreover, the inclusion map i : S →M , i(x) = x is a smooth immersion.

Proof. The proof that S, endowed with the subset topology, has a k-dimensionaldifferentiable structure for which AS is a smooth atlas can be proven exactly as inthe case of embedded manifolds in Rn (see Proposition 2.4.3.

Next, since S has the induced topology, it follows that i is continuous. Let(U,ϕ) be an adapted chart around p ∈ S. Then, note that in the pair of charts(U ∩S, prmk ϕ|U∩S) and (U,ϕ) the local expression of i is the canonical immersionikm : Rk → Rm. This proves that i is a smooth immersion.

In Definition 6.3.1 above, denote that the coordinates by ϕ = (x1ϕ, . . . , x

mϕ ).

Then condition (*) means that S is locally described by the l = m− k equations:

S ∩ U = q ∈ U : xk+1ϕ (q) = 0, xk+2

ϕ (q) = 0, . . . , xmϕ (q) = 0.The converse also holds, and turns out to be a very useful criterion: any subsetwhich is locally described as the zeroes of by l independent equations is an embeddedsubmanifold of codimension l (here, independent means that the equations form asubmersion):

Proposition 6.3.3. A subset S ⊂ M is an embedded submanifold of codimensionl if and only if for every p ∈ S there exists an open set p ∈ U ⊂ M and smoothmap h : U → Rl which is a submersion at p, such that S ∩ U = h−1(0).

Proof. Assume first that S is an embedded submanifold of codimension l. Letp ∈ S and let (U,ϕ) be a chart adapted to S with p ∈ U . Define

h : U −→ Rl, h := prml ϕ,where prml : Rm → Rl denotes the projection onto the last l coordinates. Then his a submersion and S ∩ U = h−1(0).

Conversely, assume that S ⊂M is a subset such that, around any p ∈ S, we canfind an open set U and a smooth map h : U → Rl which is a submersion at p andsatisfies U ∩S = h−1(0). We apply the local submersion theorem to h. By Remark6.2.4, we do not need to change the coordinates on Rl. Thus, after shrinking U , wemay assume that there is a chart (U,ϕ) on M such that

h ϕ−1(x1, . . . , xm) = (xk+1, . . . , xm), where k = m− l;


or equivalently, in the usual notation for coordinates ϕ = (x1ϕ, . . . , x

mϕ ), we have:

h(q) = (xk+1ϕ (q), . . . , xmϕ (q)), for q ∈ U.

Thus, for q ∈ U : h(q) = 0 iff ϕ(q) ∈ Rk×0, and since U ∩S = h−1(0), we obtain:ϕ(S ∩ U) = (Rk × 0) ∩ ϕ(U). This concludes the proof.

Remark 6.3.4. Consider an embedded submanifold S ⊂ M . By Theorem 6.3.2the inclusion map i : S → M is a smooth immersion. Therefore if p ∈ S the mapdpi : TpS → TpM is injective, and so it sends TpS isomorphically onto a linearsubspace of TpM . We will identify these spaces:

TpS = dpi(TpS) ⊂ TpM.

Assume now that h : U → Rl is a local submersion around p such that S = h−1(0).We claim that

(*) TpS = ker(dph) = v ∈ TpM : dph(v) = 0.

Since h i(U ∩ S) = 0, it follows that dph dpi(TpS) = 0. Hence TpS ⊂ ker(dph).The other inclusion follows because dph is surjective, and so these vector spaceshave the same dimension:

dim ker(dph) = dim TpM − dim im(dph) = dim M − codim S = dim S = dim TpS.

Let us introduce some standard terminology:

Definition 6.3.5. A regular value of a smooth map f : M → N is a point q ∈ Nsuch that f is a submersion at every point in f−1(q). A point q ∈ N which is nota regular value is called a critical value for f .

Corollary 6.3.6 (Preimage Theorem). Let f : M → N be a smooth map. If q ∈ Nis a regular value f , then f−1(q) is an embedded submanifold of M with

codim(f−1(q)) = dim(N), Tp(f−1(q)) = ker(dpf), for p ∈ f−1(q).

Proof. Take a chart (U,ϕ) around p with ϕ(q) = 0. Then h := ϕ f : V → Rn isa submersion, and f−1(q) = h−1(0). Proposition 6.3.3 implies the result.

Let us also mention a deep result of mathematical analysis, which states thatmost points are regular values (note that points that are not in the image are alwaysregular values). Without introducing all notions involved, we state this result below(for details, the reader is encouraged to consult [8] or [4]).

Theorem ♣ 6.3.7 (Sard’s Theorem). The set of critical values of a smooth mapf : M → N has Lebesgue measure zero. In particular, the set of regular values of fis dense in N .

Remark 6.3.8. If dimM < dimN , then every point in M is a critical point off . In this situation, Sard’s Theorem says that f(M) has measure zero in N ; andtherefore N\f(M) is dense in N . In particular, one cannot have a surjective mapsmooth f : M → N if dimM < dimN .

For this property, smoothness is essential. In the continuous setting, there arethe co-called space-filling curves (e.g. Peano’s curve [21]). For example, one canprove that any connected, nonempty manifold is the image of a continuous curve!!

LECTURE 6. 57

6.4. Preimages of submanifolds

The following extends the notion of a regular value to submanifolds.

Definition 6.4.1. A smooth map f : M → N is said to be transverse to anembedded submanifold Q ⊂ N , if

Tf(p)N = Tf(p)Q+ dpf(TpM), ∀ p ∈ f−1(Q).

Remark 6.4.2. Note that the sum in the decomposition is not assumed to bea direct sum, i.e. the vector spaces Tf(p)Q and dpf(TpM) may have a nontrivialintersection. For example, a submersion is transverse to any embedded submanifoldof the codomain.

Theorem 6.4.3. Let f : M → N be a smooth map that is transverse to theembedded submanifold Q ⊂ N . Then P := f−1(Q) is an embedded submanifold ofM with codim P = codim Q and

TpP = (dpf)−1

(Tf(p)f−1(Q)), ∀p ∈ P.

Before giving the proof, let us consider an example.

Example 6.4.4. Consider the orthogonal projection from the 2-sphere to R2,

f : S2 −→ R2, f(x, y, z) := (x, y).

Note that f is a submersion outside of the equatorial circle, which we denote by Z.Consider two embedded submanifold Q1 and Q2 in R2 which are indicated on thepicture below:

p

f(p)

S2

R2

Q1

f−1(Q1)

Z

f(Z)

Q2

f−1(Q2)

p0

f(p0)

p1

f(p1)

p2

f(p2)

Note that f is not transverse to Q1, because at the point of intersection f(p0) off(Z) and Q1, we have that Tf(p0)Q1 = dp0f(Tp0S

2). Note also that f−1(Q1) is not


an embedded submanifold of S2, because it has a singularity which looks like theintersection of two lines.

On the other hand, f is transverse to Q2; outside of Z this is clear because fis a submersion, and at the two points p1, p2 ∈ Z ∩ f−1(Q2) note that

dpif(TpiS2) + Tf(pi)(Q2) = Tf(pi)R

2, i = 1, 2.

As predicted by the Theorem 6.4.3, f−1(Q2) is an embedded submanifold of S2; infact it is diffeomorphic to a circle.

Proof of Theorem 6.4.3. We check that P satisfies the condition in Proposition6.3.3. Consider p ∈ P , and denote q := f(p) ∈ Q. Since Q is an embeddedsubmanifold there exists an open neighborhood U of q and a smooth map h : U →Rl, where l = codimQ, such that h is a submersion at q and h−1(0) = U ∩Q. Onthe open neighborhood f−1(U) of p, consider the smooth map hf : f−1(U)→ Rl.We show that this map satisfies the conditions from Proposition 6.3.3. Note that

(h f)−1(0) = f−1(h−1(0)) = f−1(U ∩Q) = f−1(U) ∩ f−1(Q) = f−1(U) ∩ P.By Remark 6.3.4, TqQ = ker dqh; hence dqh(TqQ) = 0. Using this, that h is asubmersion at q, that f is transverse to Q, and the chain rule, we obtain:

T0Rl = dqh (TqN) = dqh (TqQ+ dpf(TpM)) = dqh dpf(TpM) = dp(h f)(TpM).

Thus hf is a submersion at p. So Proposition 6.3.3 implies that P is an embeddedmanifold of codim P = l = codim Q. Finally, again by (*), we have that

TpP = ker dp(h f) = ker (dqh dpf) = (dpf)−1(ker dqh) = (dpf−1)(TqQ).

Remark 6.4.5. Note that the empty set ∅ is a smooth manifold of any dimensiond ≥ 0: it has a (unique) topology; it is locally homeomorphic to the empty set in Rd(which is open); and the only transition map is the id∅, which is smooth, when weregard ∅ ⊂ Rd (prove this “by contradiction”!). In the setting of Theorem 6.4.3, itcould happen that f(M)∩Q = ∅; in this case f is transverse to Q and f−1(Q) = ∅.

Theorem 6.4.3 can be used to intersect submanifolds. First, we need the fol-lowing definition:

Definition 6.4.6. Two embedded submanifold R,S ⊂ M are said to intersecttransversely, if for every p ∈ R ∩ S we have that

TpM = TpR+ TpS.

This condition is equivalent to the inclusion i : R →M being transverse to S.Note also that

i−1(S) = R ∩ S, (dpi)−1(TpR) = TpR ∩ TpS, p ∈ R ∩ S.

Hence, Theorem 6.4.3 implies the following:

Corollary 6.4.7. The intersection R∩S of two embedded submanifolds R,S ⊂Mwhich intersect transversally is an embedded submanifold with

codim R ∩ S = codim R+ codim S;

moreover, for each p ∈ R ∩ S we have that

Tp(R ∩ S) = TpR ∩ TpS.

LECTURE 6. 59

6.5. Exercises

Exercise 6.1. Let k ≤ n, and let A : Rn → Rk be a surjective linear map, and letC : Rk → Rn be an injective linear map.

(a) Prove that there exists a linear map B : Rn → Rn−k such that the map

(A,B) : Rn → Rk × Rn−k ∼= Rn, (A,B)(x) := (A(x), B(x)),

is a linear isomorphism.(b) Prove that there exists a linear isomorphism G : Rn → Rn such that

A G = prnk .

(c) Prove that there exists a linear map D : Rn−k → Rn such that the map

C +D : Rk × Rn−k ∼= Rn → Rn, (C +D)(y, z) := C(y) +D(z),

is a linear isomorphism.(d) Prove that there exists a linear isomorphism H : Rn → Rn such that

H C = ikn.

Exercise 6.2. Let a ∈ R. Show that the following map is an immersion:

fa : R −→ S1 × S1, fa(t) := (eit, eita).

Make a sketch of fa(R) for a = 0, 1/2, 1,√

2, 3. For which a ∈ R is fa injective?

Exercise 6.3. Let f : P2(R)→ R3 be the map defined by

f([x, y, z]) =1

x2 + y2 + z2(yz, xz, xy).

Show that f is smooth and show that it only fails to be an immersion at 6 points.Make a sketch of the image of f .

Being an immersion/submersion is an open condition:

Exercise 6.4. If a smooth map f : M → N is an immersion (resp. a submersion)at p ∈ M prove that there is an open neighborhood V ⊂ M of p such that f |V isan immersion (resp. a submersion).Do not use the local immersion/submersion theorem!

Exercise 6.5. Prove the local immersion theorem.

Exercise 6.6 (Alternative definition of immersions and submersions). Let f : M →N be a smooth map, and let p ∈M . Prove the following:

(a) f is an immersion at p iff it has a local left inverse, i.e. there exist open neigh-borhoods U ⊂ M of p and V ⊂ N of f(p) and a smooth map σ : V → U suchthat f(U) ⊂ V and

σ(f(x)) = x, for all x ∈ U.(b) f is a submersion at p iff it has a local right inverse, i.e. there exist open

neighborhoods U ⊂ M of p and V ⊂ N of f(p) and a smooth map j : V → Usuch that j(f(p)) = p and

f(j(y)) = y, for all y ∈ V.

Exercise 6.7. Let f : M → N be a submersion. Prove that f is an open map (i.e.for every open set U ⊂M , f(U) is open in N).


Exercise 6.8. Let f : M → N be a smooth map, and let K ⊂ M be a compactsubset. If f |K : K → N is injective, and dpf : TpM → Tf(p)N is a linear isomor-phism for every p ∈ K, prove that there is an open set U such that K ⊂ U , f(U)is open in N and f : U → f(U) is a diffeomorphism.

Exercise 6.9. Use the Preimage Theorem (Corollary 6.3.6) to prove that Sn is anembedded submanifold of Rn+1.

Exercise 6.10. Let S1b ⊂ R3 be the circle in the plane z = 0 of radius b > 0 and

with center at the origin. The 2-dimensional torus of radii 0 < a < b, denoted byT 2a,b, is the set of points in R3 at distance a from S1

b . Use the Preimage Theorem

(Corollary 6.3.6) to prove that T 2a,b is a 2-dimensional manifold (Find an explicit

equation describing it! )

Exercise 6.11. Let M(n) denote the space of all real n × n matrices. Let S(n)denote the space of symmetric matrices:

S(n) := B ∈M(n) : B = Bt.Prove that I is a regular value of the map

f : M(n) −→ S(n), f(A) = AAt.

Using the Preimage Theorem (Corollary 6.3.6), prove that the set of orthogonalmatrices

O(n) := A ∈M(n) : AAt = I,is a compact embedded submanifold of M(n) of dimension n(n− 1)/2.

Hint: For compactness, show that O(n) ⊂ Sn2−1.

LECTURE 7

7.1. Immersed submanifolds

There are two ways to describe submanifolds: either by giving (local) equationsthat define them, or by giving a parameterization. As discussed in the previous lec-ture, the first class corresponds to embedded submanifolds. The second class, calledimmersed submanifolds, is more general; it incorporates several natural “subman-ifolds” which are not embedded, but play an important role in various geometricsettings: Lie subgroups of Lie groups, orbits of Lie group actions, leaves of folia-tions, etc.

Definition 7.1.1. Let M be a smooth manifold. An immersed k-dimensionalsubmanifold of M is a subset S ⊂M endowed with a topology and a k-dimensionaldifferentiable structure such that the inclusion map i : S →M , i(x) = x is a smoothimmersion.

Here are some obvious examples:

Example 7.1.2. (1) An embedded submanifold S ⊂ M with the differentiablestructure from Theorem 6.3.2 is also an immersed submanifold.

(2) If f : N → M is an injective immersion then S := f(N) is an immersedsubmanifold, with differentiable structure and topology such that f : N → S isa diffeomorphism (this condition defines a unique topology and differentiablestructure on S!). In fact, every immersed submanifold S is of this type: justtake S = N , f = i.

Remark 7.1.3. It is important to note that:

• The topology of an immersed submanifold is not necessarily the inducedtopology (!)

• There can be several immersed submanifolds with the same underlyingset S ⊂M , i.e. there can exist several differentiable structures on the setS, such that S is an immersed submanifold (!)

Here are some examples illustrating these remarks:

Example 7.1.4. Denote the six curves represented below by Ca, Cb, . . . , Cf :

61


a. Topologist’s sine curve b. Figure 8 c. Shade of a trefoil knot

d. Dutch figure 8 e. Alien creature f. Non-analytic bifurcation

All these curves are immersed submanifolds in R2! Here are some of their properties:

a. The red line segment that Ca approaches is not part of Ca. An injective immer-sion that parameterizes Ca (or a curve resembling Ca) is:

j : (0,∞) −→ R2, j(t) = (t, sin(1/t)).

Note that Ca is in fact an embedded submanifold of R2: this follows by Propo-sition 6.3.3, because Ca is the 0-set of the submersion:

h : (0,∞)× R −→ R, h(x, y) = y − sin(1/x).

The differentiable structure on Ca coming from j coincides with the differentiablestructure of Ca regarded as an embedded submanifold; in fact, we will prove that,in general, on an embedded submanifold there is a unique differentiable structurewhich makes it into an immersed submanifold; namely, the one constructed inTheorem 6.3.2. Note that Ca is the only embedded submanifold among the 6examples.

b. The curve Cb is the image of the immersion:

j : R −→ R2, j(t) = (sin(2t), sin(t)).

Note that j is injective on either of the two intervals I1 := (−π, π) and I2 :=(0, 2π); and that j(I1) = Cb = j(I2). Therefore, each of the maps

j1 := j|I1 : I1 −→ Cb, and j2 := j|I2 : I2 −→ Cb

endows Cb with the structure of an immersed submanifold. Note that the re-sulting topologies are different! In fact, these are the only two differentiablestructures on Cb making Cb an immersed submanifold. Note that (0, 0) = j1(0)

LECTURE 7. 63

and (0, 0) = j2(π). At this point, the tangent spaces corresponding to the dif-ferent smooth structures are different:

T(0,0)Cb = d0j1(T0R) =⟨

2∂

∂x

∣∣(0,0)

+∂

∂y

∣∣(0,0)

⟩T(0,0)Cb = dπj2(TπR) =

⟨2∂

∂x

∣∣(0,0)− ∂

∂y

∣∣(0,0)

⟩.

c. The curve Cc has 23 = 8 differentiable structures; at each intersection point thereare 2 options how to separate the curve. Also, the 8 topologies are different!

d. The curve Cd has a unique differentiable structure as an immersed submanifold;but note that it is not an embedded submanifold!

e. The curve Ce is a closed version of the curve Ca; the behavior at the two “ends”is similar to that of the Topologist’s sine curve. This curve has a unique dif-ferentiable structure as an immersed submanifold, but note that it is not anembedded submanifold!

f. The curve Cf represents two curves that are “infinitely tangent”. There are twodifferentiable structure on Cf which make it into an immersed submanifold. Forexample, as a model one can take the following two decompositions of Cf :

Cf = (x, 0) : x ∈ R ∪ (x, e−1/x) : x > 0;

Cf = (x, 0) : x > 0 ∪ (x, f(x)) : x ∈ R,where f is the smooth function: f(x) = 0, if x ≤ 0 and f(x) = e−1/x, if x > 0,which give different decompositions of Cf into connected components. Note thatboth smooth structures have the same tangent space at (0, 0):

T(0,0)Cf =⟨ ∂

∂x

∣∣(0,0)

⟩.

The following is the main tool for comparing different immersed submanifoldswith the same underlying subset.

Lemma 7.1.5. Let f : N → M be an injective immersion. Consider a smoothmap ϕ : X → M such that ϕ(X) ⊂ f(N), and denote by ϕ : X → N the uniquemap satisfying f ϕ = ϕ. If ϕ is continuous then it is smooth.

Xϕ

""ϕN

f // M

Proof. Let x ∈ X, and denote y := ϕ(x) ∈ N and z := f(y) = ϕ(x) ∈ M . Sincef is an immersion, by Exercise 6.6, there exists a smooth map σ : V → U , whereU ⊂ N is an open neighborhood of y and V ⊂ M is an open neighborhood ofz such that f(U) ⊂ V and σ f |U = idU. Since ϕ is continuous, we have thatW := ϕ−1(U) ⊂ X is an open neighborhood of x.

x ∈Wϕ

((ϕ

y ∈ Uf //

V 3 zσ

oo

Note that on W , the map ϕ is a composition of smooth functions:

ϕ|W = σ ϕ|W ,


hence ϕ is smooth on W . So, around every point x ∈ X we have found an openneighborhood on which ϕ is smooth; this implies that ϕ is smooth.

Corollary 7.1.6. Let fi : Ni → M , i = 1, 2, be two injective immersions frommanifolds of the same dimension dim(N1) = dim(N2) which have the same imagef1(N1) = f2(N2) = S. If the map f−1

2 f1 : N1 → N2 is continuous, then it isa diffeomorphism. Hence f1 and f2 induce the same structure of a k-dimensionalimmersed submanifold on S (recall Example 7.1.2 (2)).

Proof. By Lemma 7.1.5, χ := f−12 f1 is smooth. Since f1 = f2 χ, it follows

that χ is an immersion between manifolds of the same dimension, hence it is a localdiffeomorphism. Since χ is a bijection, it is a diffeomorphism.

Remark 7.1.7. The condition dim(N1) = dim(N2) in Corollary 7.1.6 superfluous.Namely, since χ : N1 → N2 is onto, it follows by Sard’s Theorem (see Remark 6.3.8)that dim(N1) ≥ dim(N2), and the equality follows because χ is an immersion.

As a consequence of the previous result, we obtain:

Corollary 7.1.8. The differentiable structure on an embedded k-dimensional sub-manifold S ⊂ M constructed in Theorem 6.3.2 is the unique k-dimensional differ-entiable structure on S for which the inclusion map is an immersion.

Proof. Let f : N →M be an injective immersion with f(N) = S, which induces asecond k-dimensional differentiable structure on S. As an embedded submanifold,S has the induced topology from M ; hence f : N → S is continuous, and soby Corollary 7.1.6, f is a diffeomorphism. Thus, the two differentiable structurecoincide.

Let us introduce the following notion:

Definition 7.1.9. (1) Let X and Y be topological spaces. An injective continuousmap f : X → Y is called a topological embedding if f : X → f(X) is ahomeomorphism, where f(X) is endowed with the induced topology from Y .

(2) Let N and M be smooth manifolds. A smooth map f : N → M is called asmooth embedding if f is an injective immersion, and f : N → f(N) is ahomeomorphism, where f(N) is endowed with the induced topology.

Embedded submanifolds are precisely the image of smooth embeddings:

Proposition 7.1.10. Let S ⊂M be a subset. The following are equivalent:

(1) S is an embedded k-dimensional submanifold.(2) S is an immersed k-dimensional submanifold, whose topology is the induced

topology from M (see Definition 1.1.2).(3) There exists a k-dimensional manifold N and a smooth embedding f : N →M

such that S = f(N).

Proof. (1)⇒ (2) follows because the topology on an embedded submanifold is theinduced topology. (2) ⇒ (3) follows with N := S, f := i : S → M . Finally, weprove (3) ⇒ (1). Let f : N → M be a smooth embedding. Note the followingconsequence of the Local Immersion Theorem 6.2.2: every point in N has an openneighborhood U ⊂ N such that f(U) is an embedded k-dimensional submanifold.Since f is a topological embedding, f(U) = V ∩ f(N), for some open set V ⊂ M .We conclude that every point in S := f(N) has an open neighborhood V ⊂ M

LECTURE 7. 65

such that S ∩ V is a k-dimensional embedded submanifold. This implies that S isa k-dimensional embedded submanifold.

Next, recall that a continuous map is called proper if the preimages of compactsubsets are compact (see Definition 4.4.3). We note the following topological result:

Lemma 7.1.11. Let f : X → Y be a continuous map. Assume that f is properand injective and that Y is Hausdorff and locally compact (i.e. every point has acompact neighborhood). Then f is a topological embedding and f(X) is closed.

Proof. We show that f is a closed map, i.e. the image of any closed set is closed.Let C ⊂ X be a closed set, and let y ∈ Y \f(C). Since Y is locally compact, yhas an open neighborhood V with V compact. Since f is proper, E := f−1(V ) iscompact in X. Thus, f(E) is compact. Since Y is Hausdorff, f(E) is closed. LetW = V \f(E). One easily checks that W is a neighborhood of y disjoint from f(C),as desired. In particular, f(X) ⊂ Y is closed. Therefore, f : X → f(X) is a closedcontinuous bijection; which implies that it is a homeomorphism.

Finally, Lemma 7.1.11 and Proposition 7.1.10 give:

Corollary 7.1.12. The image of a proper injective immersion is a closed embeddedsubmanifold. In particular, the image of an injective immersion from a compactmanifold is an embedded submanifold.

7.2. A weak version of Whitney’s Embedding Theorem

As an application of the existence of partitions of unity, and of Corollary 7.1.12, weprove of a weak version of Whitney’s Embedding Theorem 2.4.4.

Theorem 7.2.1. A manifold which admits a finite atlas can be embedded in Rn,for large enough n. In particular, a compact manifold can be embedded in some Rn.

Proof. Consider a finite atlas (Ui, ϕi)ki=1 on a manifold M . Let ρiki=1 be apartition of unity subordinated to the cover Uiki=1. For each i, define the function

ϕi : M → Rm, ϕi(x) =

0, x /∈ Ui;ρi(x)ϕi(x), x ∈ Ui.

Since ϕi is smooth on the open sets Ui and M\supp(ρi), it is smooth on M . Let

f : M → Rm(k+1), f(x) = (ρ1(x), . . . , ρk(x), ϕ1(x), . . . , ϕk(x)).

We prove that f is an injective immersion. Let x, y ∈ M such that f(x) = f(y).Since

∑i ρi(x) = 1, there is i such that ρi(x) > 0. Since f(x) = f(y), we have that

ρi(x) = ρi(y) > 0 and ϕi(x) = ϕi(y). In particular, x, y ∈ Ui, and therefore:

ϕi(x) =ϕi(x)

ρi(x)=ϕi(y)

ρi(y)= ϕi(y).

Since ϕi is injective, we conclude that x = y. To see that f is an immersion,consider x ∈M and v ∈ TxM such that dxf(v) = 0. This implies that dxρi(v) = 0and dxϕi(v) = 0. Let i be such that ρi(x) > 0. Then x ∈ Ui, and note that

dxϕi(v) = dx(ρiϕi)(v) = dxρi(v)ϕi(x) + ρi(x)dxϕi(v).

Thus, we obtain that dxϕi(v) = 0. Since ϕi is a diffeomorphism, v = 0.By Corollary 7.1.12, if M is compact then f is an embedding. In the general

case consider a proper smooth function h : M → R, which exists by Corollary 4.4.4.


Then the map (f, h) : M → Rm(k+1)+1 is a proper injective immersion, thus, byCorollary 7.1.12 an embedding.

7.3. Exercises

Exercise 7.1. If X is compact and Z is Hausdorff, prove that any continuousbijection f : X → Z is a homeomorphism. Do not use Lemma 7.1.11!

Definition 7.3.1. An immersed submanifold S ⊂M is called an initial subman-ifold, if for every smooth map ϕ : X →M , such that ϕ(X) ⊂ S, we have that theinduced map ϕ : X → S, ϕ(x) = ϕ(x) is smooth.

Xϕ

""ϕS

i // M

Exercise 7.2. Prove that embedded submanifolds are initial. Hint: Use Lemma7.1.5.

Exercise 7.3. Prove that an initial submanifold has a unique differentiable struc-ture such that the inclusion map is an immersion. Hint: Use Corollary 7.1.6.

Exercise 7.4. (a) Construct a smooth map χ : (−ε, 1 + ε)→ [0, 1] such that

χ|(−ε,0] = 0, χ|[1,1+ε) = 1, χ(0, 1) = (0, 1),

and χ : (0, 1)→ (0, 1) is a diffeomorphism.(b) Let M be a manifold and consider two smooth curves γ1, γ2 : [0, 1] → M such

that γ1(1) = γ2(0). Show that the following curve is smooth:

γ : [0, 2] −→M, γ(t) =

γ1(χ(t)), 0 ≤ t ≤ 1γ2(χ(t− 1)), 1 ≤ t ≤ 2

Recall that a map from a closed interval f : [a, b]→ M is smooth iff there is a

smooth map f : (a− ε, b+ ε)→M , for some ε > 0, such that f |[a,b] = f .

(c) Consider a subset in R2 which looks like the letter T :

S = (0, t) : t ∈ R ∪ (t, 0) : t ∈ [0,∞).Prove that S is an immersed submanifold of R2.

(d) Using item (b) prove that S is not an initial submanifold of R2.

Exercise 7.5. Consider the six curves from Example 7.1.4. Show that Cb, Cc, Cd,and Cf are not initial submanifold. Explain why Ca and Ce are initial manifolds(for Ce, you do not need to give a rigorous proof). Show that Ce is not an embeddedsubmanifold (note that Ce is a compact subset).

Exercise 7.6. Let M be a smooth manifold, and let X ⊂M be a subset which isat most countable. Prove that X is an initial submanifold of M . Prove that X isan embedded submanifold iff it is discrete, i.e. for every point x ∈ X has an openneighborhood U ⊂M such that X ∩ U = x.

Exercise 7.7. For a ∈ R\Q, consider the map from Exercise 6.2:

fa : R −→ S1 × S1, fa(t) := (eit, eita).

Show that fa(R) is an initial submanifold of the 2-torus S1 × S1.

LECTURE 7. 67

For the following exercises, we consider another class of submanifolds. Thisnotion is not standard; however, it seems useful because it is less general thaninitial submanifold, but it includes the interesting examples of leaves of foliations,Lie subgroups and orbits of smooth actions of Lie groups.

Definition 7.3.2. A subset S ⊂M is called a leaf-like submanifold of codimen-sion l, if for every p ∈ S there exists a submersion f : U → Rl, where U is anopen neighborhood of p in M , and there exists a subset Λ ⊂ Rl which is at mostcountable such that

U ∩ S = f−1(Λ).

Exercise 7.8. Prove that the initial submanifold from Exercise 7.7 is a leaf-likesubmanifold of codimension one.

Exercise 7.9. Prove that an embedded submanifold S ⊂M of codimension l is aleaf-like submanifold of codimension l.

Exercise 7.10. Show that the initial submanifold Ce from Example 7.1.4 is not aleaf-like submanifold.

Exercise 7.11. Prove that any leaf-like submanifold is an initial submanifold.Don’t forget second countability!

Exercise 7.12. Prove that a connected leaf-like submanifold S ⊂ M which isclosed (as a subset of M) is also embedded.

The following exercise represents the converse of Corollary 4.4.2, and gives analgebraic description of closed embedded submanifolds (recall also Theorem 5.3.2).

Exercise 7.13. Let ϕ : S →M be a smooth map such that the pullback map

ϕ∗ : C∞(M) −→ C∞(S), ϕ∗(f) = f ϕ,is surjective. The goal of this exercise is to show that ϕ is a closed embedding, andso, by Proposition 7.1.10 and Corollary 7.1.12, ϕ(S) ⊂ M is a closed embeddedsubmanifold and ϕ : S → ϕ(S) is a diffeomorphism. Prove this by showing thatthe following hold:

(a) ϕ is injective.(b) ϕ is an immersion (argue “by contradiction”: if it is not an immersion prove

that there is x ∈ S and a vector v ∈ TxS such that v 6= 0 but v(g) = 0 for allg ∈ C∞(S)).

(c) ϕ is an embedding (show first that open sets of the form f−1(0,∞), f ∈ C∞(S)form a basis for the topology of S; then show that the image via ϕ of such opensets is open in ϕ(S)).

(d) ϕ(S) is closed.(e) Show that ϕ(S) is a closed embedded submanifold (apply Corollary 7.1.12).

LECTURE 8

8.1. The tangent bundle

Definition 8.1.1. The tangent bundle of a smooth manifold M is the set of alltangent vectors:

TM :=⊔p∈M

TpM.

The canonical projection is the map

π : TM −→M, v ∈ TpM =⇒ π(v) = p.

Theorem 8.1.2. Let M be a smooth manifold of dimension m. The tangent bundleTM has a unique 2m-dimensional differentiable structure such that, for any chart(U,ϕ) on M , the pair (TU,dϕ) is a smooth chart on TM , where

dϕ : TU −→ Tϕ(U) = ϕ(U)× Rm, v ∈ TpM 7→ (ϕ(p),dpϕ(v)).

Moreover, the canonical projection π : TM →M is a surjective submersion.

Proof. Let A = (Uα, ϕα)α∈I be a maximal atlas on M . We will apply Propo-

sition 3.1.1 to the collection A := (TUα,dϕα)α∈I . Note that: (1) the collectionTUαα∈I covers TM ; (2) the maps dϕα : TUα → ϕα(Uα)×Rm are bijections; (3)

dϕα(TUα ∩ TUβ) = ϕα(Uα ∩ Uβ)× Rm

are open sets; (4) the transition maps of A are smooth because they are the differ-entials of the transition maps of A:

dϕβ (dϕα)−1 = d(ϕβ ϕ−1α ) : ϕα(Uα ∩ Uβ)× Rm −→ ϕβ(Uα ∩ Uβ)× Rm.

So the conditions (1)-(4) from Proposition 3.1.1 are satisfied, therefore TM has aunique topology for which the maps (TUα,dϕα) are (topological) charts. For thistopology, the map π : TM → M is continuous: for any open set V ⊂ M , we havethat π−1(V ) is open:

π−1(V ) =⋃α∈I

π−1(V ∩ Uα) =⋃α∈I

(dϕα)−1 (ϕα(Uα ∩ V )× Rm) .

To show that the topology is second countable, by Lemma 4.3.4, we can considera countable refinement Vkk≥1 of the cover Uαα∈I , i.e. Vk ⊂ Uik for some ik ∈ I.If Bll≥0 is a countable basis of Rm ×Rm, then π−1(Vk) ∩ (dϕik)−1(Bl)k,l≥1 isa countable basis for the topology on TM .

69


Next, we show that the topology is Hausdorff. Let v, w ∈ TM , with v 6= w. Ifπ(v) = π(w), consider a chart (Uα, ϕα) such that π(v) ∈ Uα. Then v, w ∈ TUα.Since dϕα : TUα → ϕα(U) × Rm is a homeomorphism, and the second space isHausdorff, v and w can be separated by open sets in TUα. If π(v) 6= π(w), letV,W ⊂ M be open sets separating π(v) and π(w). Since π is continuous, π−1(V )and π−1(W ) are open sets separating v and w.

Proposition 3.1.1 implies now that TM has a unique 2m-dimensional differen-

tiable structure such that A is a smooth atlas. Finally, we check that π : TM →Mis a submersion. Note that in the pair of charts (TUα,dϕα) and (Uα, ϕα), π becomesthe canonical projection:

ϕα π (dϕα)−1 = pr2mm : ϕα(U)× Rm −→ ϕα(U).

Thus π is locally a submersion; hence it is a submersion.

8.2. Vector bundles

Note that the tangent bundle TM of a manifold M has more structure: it is acollection of vector spaces TpM depending smoothly on p ∈ M . Such structuresarise in many natural constructions, therefore we introduce the following concept:

Definition 8.2.1. A smooth vector bundle of rank r consists of a surjectivesubmersion π : E →M satisfying

• For each p ∈M , the fibre of π over p, denoted by

Ep := π−1(p),

is endowed with the structure of a real vector space of dimension r.• Around every point in M , there is an open set U and a diffeomorphism

Φ : π−1(U) −→ U × Rr,

such that the following diagram commutes:

π−1(U)Φ //

π

U × Rr

pr1

UidU // U

pr1 Φ = π,

where pr1(x, v) = x, and such that for every p ∈ U , the restriction

pr2 Φ|Ep : Ep −→ Rr


Here are some examples:

Example 8.2.2. The trivial vector bundle of rank r over a manifold M is

pr1 : M × Rr −→M,

where the vector space structure on p × Rr is the obvious one.

Example 8.2.3. The tangent bundle TM of a manifold M is a vector bundle ofrank(TM) = dim(M).

LECTURE 8. 71

Example 8.2.4. The group (±1, ·) acts on S1 × R via

±1 · (z, x) = (±z,±x).

The action is free, and the quotient E := (S1 × R)/±1 is diffeomorphic to theMobius band. Then E is a vector bundle over S1 with surjective submersion:

π : E −→ S1, π([z, x]) = z2.

For w ∈ S1, fix z ∈ S1 one of the two solutions of the equation z2 = w. The vectorspace structure on the fibre Ew = [z, x] : x ∈ R is given by:

a[z, x] + b[z, y] = [z, ax+ by], a, b ∈ R, [z, x], [z, y] ∈ Ew.

Next, we introduce vector bundle morphisms:

Definition 8.2.5. A morphism between two vector bundles πE : E → M andπF : F → N is a pair of smooth maps Φ : E → F and ϕ : M → N such that for allp ∈M we have Φ(Ep) ⊂ Fϕ(p) and that the induced map

Φ|Ep : Ep −→ Fϕ(p)

is linear. If Φ and ϕ are diffeomorphisms, then (Φ, ϕ) is said to be a vector bundleisomorphism, and the two bundles are said to be isomorphic.

Remark 8.2.6. Note that the condition Φ(Ep) ⊂ Fϕ(p) in the definition above isequivalent to the commutativity of the following:

EΦ //

πE

F

πF

Mϕ // N

πF Φ = ϕ πE

Example 8.2.7. If ϕ : M → N is a smooth map, then (dϕ,ϕ), with df : TM →TN , is a vector bundle morphism.

Definition 8.2.8. The restriction of a vector bundle π : E → M to an open setU ⊂M is the vector bundle over U

E|U := π−1(U), π|E|U : E|U −→ U.

Definition 8.2.9. A vector bundle isomorphic to a trivial vector bundle is calledtrivializable.

Remark 8.2.10. Every vector bundle π : E → M is locally trivializable. Thisholds by definition, since M can be covered by oven subsets U such that thereexists a vector bundle isomorphism Φ : E|U ∼−→ U × Rr covering idU . Such a localisomorphism to the trivial bundle is called a local trivialization.

8.3. Sections of vector bundles

Definition 8.3.1. A smooth section of a vector bundle π : E → M is a smoothmap σ : M → E such that π σ = idM ; in other words: σ(p) ∈ Ep, for all p ∈M .

We denote the space of all sections of π : E →M by Γ(E).

Note that every vector bundle π : E → M has the zero-section 0 ∈ Γ(E)which sends a point p ∈M to the origin of the vector space Ep: p 7→ 0p ∈ Ep.


The space of sections Γ(E) forms a module over the commutative algebra ofsmooth functions C∞(M), with the point-wise defined operations:

+ : Γ(E)× Γ(E) −→ Γ(E), (σ1 + σ2)(p) := σ1(p) + σ2(p) ∈ Ep,· : C∞(M)× Γ(E) −→ Γ(E), (f · σ)(p) := f(p)σ(p) ∈ Ep,

where, on the right hand side, the vector space operations on Ep are used. Thezero-section plays the role of the zero-element of Γ(E).

Consider the case of a trivial bundle M × Rr → M . Its sections are smoothmaps of the form:

σ : M −→M × Rr, σ(p) = (p, (f1(p), . . . , fr(p))) ∈M × Rr.

Thus, we have a one-to-one correspondence between Γ(M × Rr) and C∞(M ;Rr).Using the structure of a C∞(M)-module of Γ(M × Rr), we can decompose anysection σ uniquely as

σ = f1σ1 + . . .+ frσr,

where, for 1 ≤ i ≤ r, σi ∈ Γ(M × Rr) represents the section

σi(p) := (p, (0, . . . , 1, . . . , 0)) ∈M × Rr,

with 1 is on the i-th position. In other words:

Γ(M × Rr) is a free C∞(M)-module with basis σ1, . . . , σr.

Since a general vector bundle π : E →M is locally trivializable, there is a coverof M by open sets U such that Γ(E|U ) is a free C∞(U)-module of rank r.

8.4. Frames

Definition 8.4.1. Let π : E →M be a vector bundle. A local frame over U ⊂Mis a smooth map

γ : E|U −→ Rr,such that, for all p ∈ U , γ|Ep : Ep → Rr is a linear isomorphism.

Lemma 8.4.2. Let π : E →M be a vector bundle, and let U ⊂M be an open set.There is a one-to-one correspondence between the following objects:

(1) local trivializations Φ : E|U ∼−→ U × Rr;(2) local frames γ : E|U → Rr;(3) sections σ1, . . . , σr ∈ Γ(E|U ) such that, for every p ∈ U , σ1(p), . . . , σr(p) forms

a basis of Ep.

This correspondence is such that, given a local frame γ, the corresponding localtrivialization is Φ := π × γ : E|U → U × Rr, and the corresponding set of sectionsσ1, . . ., σr is given by σi(p) := γ−1

p (ei), where e1, . . . , er is the standard basis of Rr.

Proof. By passing to the restriction E|U , we may assume that U = M .First, we explain the equivalence between (a) and (b). Consider a trivialization

Φ : E ∼−→ M × Rr. Then, by the definition of Φ, we have that Φ = π × γ, whereγ : E → Rr is a frame. Conversely, let γ : E → Rr be a frame on E over M . Weneed to show that Φ = π × γ : E → M × Rr is a diffeomorphism. Since Φ is abijection, and the manifolds have the same dimension, invoking the inverse functiontheorem, it suffices to show that π × γ is an immersion. Let v ∈ TeE be such that

0 = de (π × γ) (v) = (deπ(v),deγ(v)).

LECTURE 8. 73

By Corollary 6.3.6, ker(deπ) = Te(Eπ(e)). Therefore, v ∈ Te(Eπ(e)), and so 0 =

deγ(v) = de(γ|Eπ(e)

)(v). Since γ|Eπ(e)

is a diffeomorphism, we have that v = 0.

We explain now the equivalence between (b) and (c). Let γ : E → Rr be aframe on E. By the first part, Φ = π × γ : E → M × Rr is a diffeomorphism,therefore the maps:

σi : M → E, σi(p) := Φ−1(p, (0, . . . , 1, . . . , 0))

are smooth, and therefore σi ∈ Γ(E). Now, for all p ∈M , γp(σi(p)) = ei, thereforeσ1(p), . . ., σr(p) form a basis of Ep. Conversely, consider σ1, . . ., σr ∈ Γ(E) whichform a point-wise basis of E. Consider the map

Ψ : M × Rr −→ E, Ψ(p, (a1, . . . , ar)) := a1σ1(p) + . . .+ arσr(p).

Clearly, Ψ is a vector bundle morphism. An argument similar to the proof ofthe first part shows that Ψ is also a diffeomorphism; hence Ψ is a vector bundleisomorphism. The corresponding frame is the map γ : E → Rr determined byΨ−1 = π × γ.

The Lemma implies that:

Corollary 8.4.3. For a vector bundle π : E →M , the following are equivalent:

(1) E is trivializable;(2) E admits a global frame γ : E → Rr;(3) there are smooth sections σ1, . . . , σr ∈ Γ(E) such that, for all p ∈ M , their

values at p form a basis of Ep.

Definition 8.4.4. Let π : E →M be a vector bundle of rank r. Let γ : E|U → Rrbe a local frame over the open set U ⊂ M , and let σ1, . . . , σr ∈ Γ(E|U ) denote thecorresponding sections which, at every p ∈ U , form a basis σ1(p), . . . , σr(p) of Ep.

The local representation of a section σ ∈ Γ(E) in this local frame is thedecomposition of σ|U ∈ Γ(E|U ) as a linear combination

σ|U = f1σ1 + . . .+ frσr,

with coefficients fi ∈ C∞(U). The coefficients are determined by

(f1, . . . , fr) = γ σ|U : U −→ Rr.

Let us mention that vector bundles are determined by their module of smoothsections; in fact, one can characterize algebraically which modules arise from vectorbundles. Without introducing the necessary terminology, we state the following:

Theorem ♣ 8.4.5 (Serre-Swan). Let M be a smooth manifold. The correspon-dence E 7→ Γ(E) is a one-to-one correspondence between vector bundles over M(up to isomorphism) and finitely generated projective modules over C∞(M) (up toisomorphism).

8.5. Vector fields as sections of the tangent bundle

Definition 8.5.1. A vector field on a smooth manifold M is a smooth section ofthe tangent bundle π : TM →M ; in other words it is a smooth map

X : M −→ TM such that Xp ∈ TpM, for all p ∈M.

We use the notation Xp rather than X(p) for the value of X at p ∈M .


The space of all smooth vector fields is denoted by

X(M) := Γ(TM).

Let (U,ϕ) be a chart on M inducing coordinates ϕ = (x1ϕ, . . . , x

mϕ ), with xiϕ ∈

C∞(U). The chart induces a canonical frame on TM over U :

∂

∂x1ϕ

, . . . ,∂

∂xmϕ∈ X(U).

The local representation of a vector field X ∈ X(M) is given by:

X|U =

m∑i=1

Xi ∂

∂xiϕ, Xi ∈ C∞(U),

where the coefficients are smooth functions on U .The notion of trivializable specializes to:

Definition 8.5.2. If the tangent bundle of a manifold is trivializable, then themanifold is said to be parallelizable.

Example 8.5.3. Clearly, Rm is parallelizable. A global frame on TRm is given bythe vector fields ∂

∂xi , with 1 ≤ i ≤ m.

Example 8.5.4. The circle S1 is also parallelizable: if θ ∈ R/(2πZ) denotes the“angle coordinate” on S1, then a global frame is given by the vector field

∂

∂θ∈ X(S1).

Even though the θ makes sense as a coordinate only on sets of the form S1\eiβ,the vector field ∂

∂θ makes sense globally, as it can be defined as the derivative ofrotation:

∂

∂θ

∣∣eiα

:=d

dθeiθ∣∣θ=α

.

Example 8.5.5. Products of parallelizable manifolds are parallelizable. In par-ticular the m-dimensional torus Tm = (S1)m is parallelizable. Using the anglecoordinates (θ1, . . . , θm) ∈ (R/(2πZ))m, we obtain the global frame

∂

∂θ1, . . . ,

∂

∂θm∈ X(Tm).

The 2-dimensional sphere S2 is not parallelizable. This is a consequence of thefollowing result (see Corollary 8.4.3):

Theorem ♣ 8.5.6 (The Hairy Ball Theorem). Every smooth vector field X ∈X(S2) has at least one zero, i.e. a point p ∈ S2 such that Xp = 0.

Here is an interesting application of this result. Thinking about the surface ofthe Earth as a manifold diffeomorphic to S2, and assuming that the velocity of thewind can be represented by a smooth vector field X ∈ X(S2), this result impliesthat there always exists a point on Earth where the wind is not blowing (most likelynot in this part of the World, but you can check it out here: [22]).

It is known precisely which spheres are parallelizable:

Theorem ♣ 8.5.7. The only parallelizable spheres are S0, S1, S3 and S7 (i.e.S2c−1, with c = 0, 1, 2, 3).

LECTURE 8. 75

You can check that this is a consequence of the following classical theorem:

Theorem ♣ 8.5.8. On the sphere Sn the maximum number k = k(n) of vectorfields X1, . . . , Xk such that at every p ∈ Sn the vectors X1,p, . . . , Xk,p ∈ TpSn arelinearly independent is given as follows: write n+ 1 = (2a+ 1)24b+c, with a, b ∈ Nand c ∈ 0, 1, 2, 3, then k(n) = 8b+ 2c − 1.

If n is even then k(n) = 0, so every vector field on Sn has a zero. In general,note that k(n) is quite small compared to n; in particular, k(n) ≤ 2 log2(n+ 1) + 1.

8.6. Exercises

Exercise 8.1. Prove that TS1 is isomorphic to the trivial vector bundle S1 × R.

Exercise 8.2. Let f : M → N be a smooth map. Prove that df : TM → TN is asmooth vector bundle morphism.

Exercise 8.3. Let M be a smooth manifold. Prove that there exists a vector fieldV on the manifold TM (i.e. V ∈ X(TM) = Γ

(T (TM)

)) satisfying:

dvπ(Vv) = v, ∀ v ∈ TM.

Hint: construct V locally, and use a partition of unity.

Exercise 8.4. Recall that the m-dimensional real projective space is the space ofall lines through the origin in Rm+1, with the manifold structure from Example3.2.1:

Pm(R) := l : l ⊂ Rm+1 is a line with 0 ∈ l.The tautological line bundle over Pm(R) is the rank 1 vector bundle π : L(m)→Pm(R) whose fiber over a line l is the line l:

L(m) := (l, v) : l ∈ Pm(R), v ∈ l.(a) Sketch a picture of L(1).(b) Show that π : L(m)→ Pm(R) is indeed a vector bundle of rank one, such that

the following map is morphism of vector bundles:

L(m)Φ //

π

Pm(R)× Rm+1

pr1Pm(R)

id // Pm(R)

, Φ(l, v) := (l, v) ∈ Pm(R)× Rm+1.

(c) Sketch a picture of the maps above for m = 1.(d) Prove that L(1) is not trivializable (use Corollary 8.4.3).(e) Prove that L(m) is not trivializable, by using the morphism of vector bundles:

L(1)Ψ //

π

L(m)pr1

P1(R) // Pm(R)

,

Ψ(l, v) := (l × 0, (v, 0)) ∈ Pm(R)× Rm+1.

The following exercise gives a proof of the Hairy Ball Theorem 8.5.6. Any suchproof requires some knowledge of algebraic topology. We will give a proof usingonly the winding number of a curve, which should be familiar from a standardcourse in one-variable complex analysis.


Definition 8.6.1. Let γ : S1 → C∗ be a smooth closed curve, where C∗ := C\0.The winding number of γ is defined by the formula:

W (γ) :=1

2πi

∮γ

dz

z∈ Z.

This number is an integer, and counts how many of times γ goes around 0 ∈ C inthe positive direction, minus the number of times it goes around 0 in the negativedirection.

Note that the winding number is a homotopy invariant: if γt : S1 → C∗ is afamily of closed curves depending continuously on t ∈ [0, 1], then W (γ0) = W (γ1).This is because W (γt) has integer values, and depends continuously on t; thusW (γt) must be constant.

Exercise 8.5. For a vector field V = a(x, y) ∂∂x + b(x, y) ∂∂y ∈ X(R2), we denote by

γV : S1 → C the closed curve

γV (cos(θ), sin(θ)) := a(cos(θ), sin(θ)) + ib(cos(θ), sin(θ)) ∈ C.(a) Let V ∈ X(R2) be a vector field which is nowhere zero, i.e. V(x,y) 6= 0 for all

(x, y) ∈ R2. Construct a continuous family of closed curves γt : S1 → C∗, witht ∈ [0, 1], such that γ1 = γV and γ0 = λ, where λ ∈ C∗ is a constant. Concludethat W (γV ) = 0 (Hint: look at the vector field V on circles of radius t).

(b) Consider the stereographic atlas on S2 with two charts from Example 1.2.7.Let X ∈ X(S2) be a vector field, and consider its representatives in the thesetwo charts:

V s = a(x, y)∂

∂x+ b(x, y)

∂

∂y∈ X(R2),

V n = c(x, y)∂

∂x+ d(x, y)

∂

∂y∈ X(R2).

Show that, at (x, y) 6= (0, 0), these local representatives are related by:

c(x, y) = (y2 − x2)a

(x

x2 + y2,

y

x2 + y2

)− 2xy · b

(x

x2 + y2,

y

x2 + y2

),

d(x, y) = −2xy · a(

x

x2 + y2,

y

x2 + y2

)+ (x2 − y2)b

(x

x2 + y2,

y

x2 + y2

).

Conclude the following relation between the corresponding curves:

γV n = e2iθ(−γV s).(c) Prove that every vector field X ∈ X(S2) has at least one zero. Hint: assume

that a nowhere vanishing vector field X exists. By applying (a) to V s, and then(b), show that γV n is homotopic to the curve e2iθ. Deduce that W (γV n) = 2,which contradicts (a).

LECTURE 9

9.1. Vector fields as derivations

Vector fields have a natural action on smooth functions:

Definition 9.1.1. The derivative along a vector field X ∈ X(M) of a smoothfunction f ∈ C∞(M) is the function X(f) ∈ C∞(M), M 3 p 7→ Xp(f).

The fact that X(f) is indeed a smooth function can be checked locally. Repre-sent X in a local chart (U,ϕ = (x1

ϕ, . . . , xmϕ )) on M ,

X|U =

m∑i=1

Xi ∂

∂xiϕ, Xi ∈ C∞(U),

then we have the explicit formula:

X(f)|U =

m∑i=1

Xi ∂

∂xiϕ(f |U ) =

m∑i=1

Xi ∂f ϕ−1

∂xi ϕ

which shows that X(f)|U is smooth; and since (U,ϕ) is arbitrary, X(f) is smooth.The resulting operation f 7→ X(f) on C∞(M) has the following property:

Definition 9.1.2. A derivation of C∞(M) is a linear map

D : C∞(M) −→ C∞(M)

satisfying the rule

D(f · g) = f ·D(g) +D(f) · g, for all f, g ∈ C∞(M).

Proposition 9.1.3. Every derivation D of C∞(M) is the derivative along a uniquevector field X ∈ X(M).

Proof. Let X ∈ X(M). Because Xp is a derivation for all p ∈M :

Xp(f · g) = f(p)Xp(g) +Xp(f)g(p),

we obtain that the derivative along X is indeed a derivation of C∞(M):

X(f · g) = f ·X(g) +X(f) · g.Assume that X,Y ∈ X(M) induces the same derivation, i.e. X(f) = Y (f) for allf ∈ C∞(M). Hence Xp(f) = Yp(f), for all f ∈ C∞(M) and p ∈ M , and soXp = Yp for all p ∈M . Thus X = Y , which shows the uniqueness assertion.

77


Finally, consider a derivation D of C∞(M). Evaluating the equality D(fg) =fD(g) +D(f)g at a point p ∈M , we obtain that the linear map

Xp : C∞(M)→ R, Xp(f) := D(f)(p)

is a derivation at p. Thus, we obtain a map X : M → TM , with Xp ∈ TpM , whichsatisfies that, for all f ∈ C∞(M), the map M 3 p 7→ Xp(f) is smooth (becausethis map is D(f)). We show that this implies smoothness of X. Consider a chart(U,ϕ = (x1

ϕ, . . . , xmϕ )) around a point p ∈ M . Note that, even if X is not smooth,

we can still decompose X|U in the corresponding frame:

X|U =

m∑i=1

Xi ∂

∂xiϕ,

just that the coefficients Xi : U → R might not be smooth. For 1 ≤ i ≤ m,let xiϕ ∈ C∞(M) have the same germ as xiϕ at p, i.e. xiϕ|V = xiϕ|V for someneighborhood V ⊂ U of p. Then

X(xiϕ)|V = X|V (xiϕ|V ) = Xi|V .

Since X(xiϕ) ∈ C∞(M), we have shown that Xi|V is smooth, hence X|V : V → TMis smooth. So X is smooth because it is smooth around any p ∈M . This concludesthe proof, because D and X act the same on C∞(M).

From now on, we will identify the space of vector fields with the space ofderivations of C∞(M).

9.2. The Lie bracket

Let X,Y ∈ X(M) be two vector fields, which we regard as derivations:

X,Y : C∞(M) −→ C∞(M).

In general, their composition is not a derivation. The failure for this to hold is

(X Y )(f · g)− (X Y )(f) · g − f · (X Y )(g) = X(f) · Y (g) + Y (f) ·X(g).

Note that the right hand side is symmetric in X and Y . Therefore, if we subtractfrom this the same equation with X and Y interchanged, we obtain:

(X Y − Y X)(f · g) = (X Y − Y X)(f) · g + f · (X Y − Y X)(g).

Thus, X Y −Y X is again a derivation of C∞(M), and so, by Proposition 9.1.3,again a vector field.

Definition 9.2.1. The Lie bracket is the following operation on vector fields:

[·, ·] : X(M)× X(M)→ X(M), [X,Y ] := X Y − Y X.The Lie bracket [X,Y ] is also called the commutator of X and Y (because itmeasures whether the derivations X and Y commute, i.e. whether X Y = Y X).

The Lie bracket has the following properties (the proof is left to the reader):

Proposition 9.2.2. The Lie bracket

(1) is skew-symmetric:[X,Y ] = −[Y,X];

(2) is R-bilinear:[X, aY + bZ] = a[X,Y ] + b[X,Z];

LECTURE 9. 79

(3) satisfies the Jacobi identity:

[X, [Y,Z]] + [Y, [Z,X]] + [Z, [X,Y ]] = 0;

(4) satisfies the Leibniz rule:

[X, fY ] = f [X,Y ] +X(f)Y ;

for all X,Y, Z ∈ X(M), a, b ∈ R and f ∈ C∞(M).

Remark 9.2.3. A vector space V endowed with an operation [·, ·] : V × V → Vsatisfying (1), (2) and (3) is called a Lie algebra. This notion will be discussed inLecture 11.

Let us write the Lie bracket [X,Y ] of two vector fields X,Y ∈ X(M) in a localchart (U,ϕ = (x1

ϕ, . . . , xmϕ )) on M . Denote

X|U =

m∑i=1

Xi ∂

∂xiϕ, Y |U =

m∑i=1

Y i∂

∂xiϕ.

Then, for any f ∈ C∞(U) we have that:

X Y (f)|U = X( m∑j=1

Y j∂f

∂xjϕ

)=

m∑i=1

m∑j=1

Xi ∂

∂xiϕ

(Y j

∂f

∂xjϕ

)=

=( m∑i=1

m∑j=1

Xi ∂Yj

∂xiϕ

∂

∂xjϕ

)(f) +

m∑i=1

m∑j=1

XiY j∂2f

∂xiϕ∂xjϕ

.

Note that there are second order derivatives of f appearing, and this is the reason

why XY is not a vector field. However, since ∂2

∂xiϕ∂xjϕ

= ∂2

∂xjϕ∂xiϕ, these terms cancel

in the commutator [X,Y ]|U . We obtain the local expression of the Lie bracket:

[X,Y ]|U =

m∑j=1

(m∑i=1

Xi ∂Yj

∂xiϕ− Y i ∂X

j

∂xiϕ

)∂

∂xjϕ.

9.3. Related vector fields

Vector fields cannot be transported via smooth maps (only via diffeomorphisms).However, the following relation makes sense in general:

Definition 9.3.1. Let ϕ : M → N be a smooth map. A vector field X ∈ X(M) issaid to be ϕ-related to a vector field Y ∈ X(N) if, for all p ∈M ,

dpϕ(Xp) = Yϕ(p).

Recall that a smooth map ϕ : M → N induces a pullback map on smoothfunctions, denoted:

ϕ∗ : C∞(N)→ C∞(M), ϕ∗(f) := f ϕ.

Here is an alternative description of related vector fields in terms of the pullback:

Lemma 9.3.2. Let ϕ : M → N be a smooth map. Then X ∈ X(M) if ϕ-related toY ∈ X(N) iff

X(ϕ∗(f)) = ϕ∗(Y (f)), ∀ f ∈ C∞(N).


Proof. At p ∈M , the left-hand-side reads:

X(ϕ∗(f))(p) = Xp(f ϕ) = (dpϕ(Xp)) (f),

and the right-hand-side:

ϕ∗(Y (f))(p) = Yϕ(p)(f).

The conclusion follows by noting that two vectors in Tϕ(p)N are equal iff they actthe same on smooth functions.

The following is a very useful result:

Lemma 9.3.3. Let ϕ : M → N be a smooth map, and let X1, X2 ∈ X(M) andY 1, Y 2 ∈ X(N) be four vector fields. If X1 is ϕ-related to Y 1 and X2 is ϕ-relatedto Y 2, then also [X1, X2] ∈ X(M) is ϕ-related to [Y 1, Y 2] ∈ X(N).

Proof. Using Lemma 9.3.2, this result is straightforward:

ϕ∗([Y 1, Y 2](f)) = ϕ∗(Y 1 Y 2(f)

)− ϕ∗

(Y 2 Y 1(f)

)=

= X1(ϕ∗(Y 2(f))

)−X2

(ϕ∗(Y 1(f))

)=

= X1 X2(ϕ∗f)−X2 X1(ϕ∗f) =

= [X1, X2](ϕ∗f).

Vector fields can always be transported along diffeomorphisms.

Definition 9.3.4. The pushforward of a vector field X ∈ X(M) via a diffeomor-phism ϕ : M ∼−→ N is the vector field

ϕ∗X = dϕ X ϕ−1 ∈ X(N),

more explicitly:

(ϕ∗X)q := dpϕ(Xp), where p = ϕ−1(q).

Note that ϕ∗X is the unique vector field on N such that X and ϕ∗X are ϕ-related.The pullback of a vector field Y ∈ X(N) via the diffeomorphism ϕ : M ∼−→ N

is the vector field

ϕ∗Y := (ϕ−1)∗Y = (dϕ)−1 Y ϕ ∈ X(M).

Note that ϕ∗Y is the unique vector field on M such that ϕ∗Y and Y are ϕ-related.

Example 9.3.5. Let ι : M → N denote the inclusion of the embedded submanifoldM of N . Then X ∈ X(M) is ι-related to Y ∈ X(N) iff

Yp = Xp ∈ TpM ⊂ TpN, ∀ p ∈MFor example, consider ι : S1 → R2, and the following vector field on R2

Y := x∂

∂y− y ∂

∂x∈ X(R2).

First, note that Y is tangent to S1 because, for (u, v) ∈ S1, we have that

T(u,v)S1 := ker(d(u,v)(x

2 + y2 − 1)) =a∂

∂x

∣∣(u,v)

+ b∂

∂y

∣∣(u,v)

: ua+ bv = 0.

Consider the “angle coordinate” θ ∈ R/(2πZ) on S1 (see Example 8.5.4),

θ 7→ (x(θ), y(θ)) := (cos(θ), sin(θ)).

LECTURE 9. 81

We claim that Y is ι-related to X := ∂∂θ , i.e. Y |S1 = ∂

∂θ . We will use Lemma 9.3.2.

For any f ∈ C∞(R2), we have that

Y (f) = x∂f

∂y(x, y)− y ∂f

∂x(x, y),

and so

ι∗(Y (f)) = cos(θ)∂f

∂y(cos(θ), sin(θ))− sin(θ)

∂f

∂x(cos(θ), sin(θ)).

On the other hand, by the chain rule, we have that:

X(ι∗f) =∂

∂θ(f(cos(θ), sin(θ))) =

=∂f

∂x(cos(θ), sin(θ))

∂ cos(θ)

∂θ+∂f

∂y(cos(θ), sin(θ))

∂ sin(θ)

∂θ=

= cos(θ)∂f

∂y(cos(θ), sin(θ))− sin(θ)

∂f

∂x(cos(θ), sin(θ)).

9.4. Change of coordinates rule for vector fields

Let X be a smooth vector field on M , and let (U,ϕ = (x1ϕ, . . . , x

mϕ )) be a chart.

Denote the local expression of X by:

X|U =

m∑i=1

Xi ∂

∂xiϕ, Xi ∈ C∞(U).

Consider a second chart (V, ψ = (y1ψ, . . . , y

mψ )) on M , and denote the corre-

sponding local expression of X by

X|V =

m∑i=1

Y i∂

∂yiψ, Y i ∈ C∞(V ).

Thus, on U ∩ V , we have that:m∑i=1

Xi ∂

∂xiϕ=

m∑i=1

Y i∂

∂yiψ.

Applying this equality to the coordinate function yjψ, we obtain

Y j =

m∑i=1

Xi∂yjψ∂xiϕ

.

This expression represents the rule of change of coordinates for vector fields, i.e. itgives the local expression of X in the chart (V, ψ) in terms of the local expressionof X in the chart (U,ϕ).

9.5. Integral curves of a vector field

Let I ⊂ R be an open interval, and consider a smooth curve γ : I → M . Recallthat the derivative of γ is the following curve in TM :

dγ

dt(t) := dtγ

( ∂∂t

)∈ Tγ(t)M.

We will consider curves whose derivatives are given by a fixed vector field:


Definition 9.5.1. Let X ∈ X(M) be a vector field. An integral curve of X is asmooth curve γ : I →M , where I ⊂ R is an open interval, which satisfies:

dγ

dt(t) = Xγ(t), for all t ∈ I.

The integral curve γ is said to start at a point p ∈M , if 0 ∈ I and γ(0) = p.

Let (U,ϕ = (x1ϕ, . . . , x

mϕ )) be a chart on M . Denote the local expression of a

vector field X ∈ X(M) in this chart by:

X|U =

m∑i=1

Xi ∂

∂xiϕ, Xi ∈ C∞(U).

We will write locally what it means for a curve γ : I → U to be an integral curveof X. Denote the local expression of γ in this chart by:

ϕ γ(t) = (γ1ϕ(t), . . . , γmϕ (t)), t ∈ I.

Then, as explained in Subsection 5.2, we have that

dγ

dt(t) = dtγ

( ∂∂t

)=

m∑i=1

dγiϕdt

(t)∂

∂xiϕ

∣∣γ(t)

.

Therefore, the condition that γ be an integral curve of X is equivalent to thefollowing system of ordinary differential equations (ODE’s)

dγiϕdt

(t) = Xi ϕ−1(γ1ϕ(t), . . . , γmϕ (t)), for 1 ≤ i ≤ m,

on curves (γ1ϕ, . . . , γ

mϕ ) : I → ϕ(U). We write this system in a more compact way:

dz

dt(t) = F (z(t)), z : I → V,

where V = ϕ(U) ⊂ Rm and

z = (γ1ϕ, . . . , γ

mϕ ) : I → V, F = (X1, . . . , Xm) ϕ−1 : V → Rm.

Such systems are studied using standard techniques from the theory of ODE’s. Wewill use the following result (e.g. [1, Appendix C]):

Theorem 9.5.2. Let F : V → Rm be a smooth map, where V ⊂ Rm is an openset. For some p ∈ V , consider the ordinary differential equation

dz

dt(t) = F (z(t)), z(0) = p.

Every point in V has an open neighborhood W and there exists a smooth map

y : (−ε, ε)×W −→ V,

for some ε > 0, such that the curve t 7→ y(t, p) is a solution of the ODE, i.e.

dy

dt(t, p) = F (y(t, p)), y(0, p) = p.

Moreover, the solution is unique, in the sense that if p ∈ W and t 7→ y(t, p) is asecond solution defined for t around 0, then y(t, p) = y(t, p) for t around 0.

LECTURE 9. 83

9.6. Exercises

Exercise 9.1. Consider the polar coordinates (U, r, θ) on R2, where

U = R2\((−∞, 0]× 0

), x = r cos(θ), y = r sin(θ), (r, θ) ∈ (0,∞)× (−π, π).

Write in polar coordinates the expression of an arbitrary vector field on R2:

X = a(x, y)∂

∂x+ b(x, y)

∂

∂y∈ X(R2).

Exercise 9.2. Let ϕ : M → N be a local diffeomorphism (recall Definition 3.3.3).Define the pullback ϕ∗X of a vector field X ∈ X(N) along ϕ (your definition shouldgeneralize Definition 9.3.4). Show that this operation preserves the Lie bracket:

ϕ∗[X,Y ] = [ϕ∗X,ϕ∗Y ], for all X,Y ∈ X(N).

Exercise 9.3. Calculate the Lie bracket of the vector fields:

X = sin(x+ y)∂

∂x+ exy

∂

∂y, Y = y2 ∂

∂x+ x2 ∂

∂y.

Exercise 9.4. Consider the map

ϕ : R2 −→ R2, ϕ(x, y) := (ex cos(y), ex sin(y)),

i.e. ϕ is the complex exponential written in real coordinates.

(a) Find two vector fields X,Y ∈ X(R2) such that X is ϕ-related to ∂∂x and Y is

ϕ-related to ∂∂y . Are these vector fields uniquely determined by this condition?

Calculate [X,Y ].(b) Find two vector fields U, V ∈ X(R2) such that ∂

∂x is ϕ-related to U and ∂∂y is

ϕ-related to V . Are these vector fields uniquely determined by this condition?Calculate [U, V ].

Exercise 9.5. Let ϕ : M → N and ψ : N → P be smooth maps. If X ∈ X(M)is ϕ-related to Y ∈ X(N) and if Y is ψ-related to Z ∈ X(P ) show that X isψ ϕ-related to Z.

Exercise 9.6. Let M ⊂ N be an embedded submanifold. A vector field X ∈ X(N)is said to be tangent to M , if Xp ∈ TpM for all p ∈M . If X ∈ X(N) and Y ∈ X(N)are tangent to M show that also [X,Y ] ∈ X(N) is tangent to M .

Exercise 9.7. (a) Let X ∈ X(M) be a vector field. Show that a smooth curveγ : I →M is an integral curve of X iff ∂

∂t ∈ X(I) is γ-related to X.(b) Let ψ : M → N be a smooth map, and assume that X ∈ X(M) is ψ-related to

Y ∈ X(N). If γ : I →M is an integral curve of X, prove that ψ γ : I → N isan integral curve of Y .

Exercise 9.8. Calculate the integral curves of the following vector fields on R2

X =∂

∂x+

∂

∂y, Y = y

∂

∂x− x ∂

∂y, Z = x2 ∂

∂x+ y

∂

∂y.

LECTURE 10

10.1. The flow of a vector field

Let us fix a vector field X on a smooth manifold M . In this section we will describethe integral curves of X. Most results discussed here are in essence consequencesof Theorem 9.5.2. First let us remark:

Lemma 10.1.1. If γ : (a, b) → M is an integral curve of X then, for any s ∈ R,also σ : (a− s, b− s)→M , σ(t) := γ(t+ s) is an integral curve of X.

Proof. Let τs(t) := t+ s. Then σ = γ τs, and

d(γ τs)dt

(t) =dt(γ τs)( ∂∂t

∣∣t

)= (dt+sγ) dt(τs)

( ∂∂t

∣∣t

)=

=(dt+sγ) ( ∂∂t

∣∣t+s

)=

dγ

dt(t+ s) = Xγτs(t).

Lemma 10.1.2. For any p ∈M there exists an integral curve

γp : Ip →M

of X starting at p which is maximal in the sense that, any other integral curveγ : I →M of X starting at p satisfies

I ⊂ Ip and γ = γp|I .

Proof. Consider two integral curves starting at p, denoted γi : Ii → M , i = 1, 2.By continuity of the curves, the set

J := t ∈ I1 ∩ I2 : γ1(t) = γ2(t)

is closed in I1∩I2. Let s ∈ J . Note that σi(t) := γi(t+s), i = 1, 2, are two integralcurves which start at q := γi(s); hence, in a chart around q, they are solutions tothe same ODE with the same initial condition. Thus, by the uniqueness assertionin Theorem 9.5.2, we have that σ1 = σ2 in a neighborhood of 0; in other wordsJ contains a neighborhood of s. We have shown that J is both open and closedin the open, non-empty inteval open I1 ∩ I2, and since 0 ∈ J , this implies thatJ = I1 ∩ I2. In other words any two integral curves starting at p coincide on theintersection of their domains.

85


By Theorem 9.5.2, there is at least one integral curve γ : I →M starting at p.Let Ip be the union of all intervals I for which there is an integral curve γ : I →Mstarting at p. By the above, for any t ∈ Ip all integral curves starting at p anddefined at t ∈ Ip take the same value at t; we define γp(t) to be this common value.Clearly, γp : Ip →M satisfies the maximality property.

Note that the maximality property of the integral curve γp : Ip → M impliesits uniqueness. This allows us to define:

Definition 10.1.3. Let X be a vector field on a manifold M . The maximaldomain (or just the domain) of the flow of X is the set

D(X) := (t, p) : p ∈M, t ∈ Ip ⊂ R×M,

and the maximal flow (or just the flow) of X is the map

φX : D(X)→M, (t, p) 7→ φtX(p) := γp(t).

For some t ∈ R, the flow at time t is the map

φtX : Dt(X)→M, p 7→ φtX(p),

where its domain is given by

Dt(X) := p ∈M : t ∈ Ip ⊂ M.

Lemma 10.1.4. For s ∈ Ip, we have that t ∈ IφsX(p) iff t+ s ∈ Ip; and moreover,

φtX(φsX(p)) = φt+sX (p) ∀ t ∈ IφsX(p).

Proof. Denote q := φsX(p) and τs(t) := t + s. Then γp τs : τ−s(Ip) → M is anintegral curve starting at q, therefore τ−s(Ip) ⊂ Iq and γp τs = γq|τ−s(Ip); i.e.

φt+sX (p) = φtX(φsX(p)), ∀ t ∈ τ−s(Ip).

Similarly, γq τ−s : τs(Iq)→M is an integral curve starting at p, hence τs(Iq) ⊂ Ip.We obtain that Iq = τ−s(Ip) , and this concludes the proof.

We have that

Theorem 10.1.5. Let X be a vector field on M . Then the domain D(X) of X isopen in R×M and the flow of X is a smooth map

φX : D(X)→M, (t, p) 7→ φtX(p).

Proof. Theorem 9.5.2 implies that for every point p ∈ M there exists an openneighborhood W ⊂ M and there exists ε > 0 such that (−ε, ε) ×W ⊂ D(X), andφX is smooth on (−ε, ε)×W .

Let (t, p) ∈ D(X). Denote the image of the integral curve from p and φtX(p) by

C := φsX(p) : s ∈ [0, t] (or s ∈ [t, 0]).

Since C is compact, the argument above implies that there is some open set C ⊂W ⊂ M and some ε > 0 such that (−ε, ε) × W ⊂ D(X) and φX is smooth on

(−ε, ε)×W . Let n > 0 be such that t/n ∈ (−ε, ε); then φt/nX : W →M is smooth.

Consider the open sets Wini=0 defined inductively by

Wn := W, Wi :=(φt/nX |W

)−1

(Wi+1), 0 ≤ i ≤ n− 1.

LECTURE 10. 87

Since C ⊂W , it follows, by induction on j = 0, . . . , n and by Lemma 10.1.4 that

φtn−jnX (p) ∈Wn−j ;

therefore, W0 is not empty, because it contains p. Also, directly by the definitionof the sets Wi and by Lemma 10.1.4, it follows that

φt inX (W0) ⊂Wi.

In particular, we obtain that t ×W0 ⊂ D(X). Moreover, φtX is smooth on W0,because it is given by the composition of smooth maps:

φtX |W0= (φ

t/nX |Wn

) . . . (φt/nX |W0

) : W0 −→M.

On the other hand, since φX is smooth on (−ε, ε)×W (in particular continuous),there is an open set p ∈ V ⊂W and there is 0 < δ < ε such that φX((−δ, δ)×V ) ⊂W0. Hence, (t − δ, t + δ) × V ⊂ D(X) is an open neighborhood of (t, p) on whichφX is smooth, since it is given by:

φsX |V = (φtX |W0) (φs−tX |V ), s ∈ (t− δ, t+ δ).

This finishes the proof.

The following is a rather direct consequence of Theorem 10.1.5 and Lemma10.1.4; the proof is left to the reader.

Corollary 10.1.6. For t ∈ R, Dt(X) is open in M and φtX is a diffeomorphismbetween φtX : Dt(X) ∼−→ D−t(X) with inverse φ−tX : D−t(X) ∼−→ Dt(X).

10.2. Complete vector fields

Definition 10.2.1. A vector field X ∈ X(M) is called complete, if its integralcurves are defined at all times; in other words, if D(X) = R×M .

Given a complete vector field X ∈ X(M), for each t ∈ R, φtX is a diffeomorphismof M . Moreover, this assignment satisfies

φsX φtX = φs+tX ;

in other words, we obtain a group homomorphism:

(R,+) −→ (Diff(M), ), t 7→ φtX .

Equivalently, one can think about the flow of a complete vector field as a smoothaction of the group (R,+) on M :

φX : R×M −→M, (t, p) 7→ φtX(p).

The family φtXt∈R is also called a one-parameter group of diffeomor-phisms, in the sense that it is a one-dimensional subgroup of the infinite-dimensionalgroup of diffeomorphisms.

The following result is useful for checking completeness:

Lemma 10.2.2. Let X ∈ X(M) be a vector field. If there is ε > 0 such that(−ε, ε)×M ⊂ D(X), then X is complete.

Proof. Let p ∈M , and let t ∈ Ip. Then (−ε, ε) ⊂ IφtX(p) and so, by Lemma 10.1.4,

(t− ε, t+ ε) ⊂ Ip. Hence, Ip = R.

A similar argument implies the following behavior of integral curves:


Lemma 10.2.3. Let X ∈ X(M) be a vector field, and let p ∈ M . If there is acompact set K ⊂ M such that φtX(p) ∈ K for all t ∈ Ip, then Ip = R (in otherwords, an integral curve is either complete, or it leaves every compact subset).

Proof. Since K is compact, there is ε > 0 such that (−ε, ε)×K ⊂ D(X). So, forq ∈ K, (−ε, ε) ⊂ Iq. Thus if t ∈ Ip, then (−ε, ε) ⊂ IφtX(p), which by Lemma 10.1.4

implies that (t− ε, t+ ε) ⊂ Ip. We conclude that Ip = R.

Either of the two lemmas, implies the following:

Corollary 10.2.4. Any vector field on a compact manifold is complete. Moregenerally, any compactly supported vector field on a manifold is complete.

10.3. The flow as the exponential of the Lie derivative

Definition 10.3.1. Let X ∈ X(M) be a vector field. The Lie derivative along Xof a function f ∈ C∞(M) is the smooth function

LXf :=d

dt

∣∣t=0

(φtX)∗f ∈ C∞(M).

The Lie derivative along X of a vector field Y ∈ X(M) is the vector field

LXY :=d

dt

∣∣t=0

(φtX)∗Y ∈ X(M).

Remark 10.3.2. Let us be more precise about the definitions above. First, recallthat the flow of X at time t is defined as a map φtX : Dt(X)→M and that Dt(X)is an open subset of M . Thus, for f ∈ C∞(M) and Y ∈ X(M), we have

(φtX)∗f = f φtX ∈ C∞(Dt(X)) and (φtX)∗Y = (dφtX)−1(YφtX ) ∈ X(Dt(X)).

Let p ∈ M . The maps Ip 3 t 7→ ((φtX)∗f)(p) and Ip 3 t 7→ ((φtX)∗Y )p are smoothmaps into the vector spaces R and TpM , respectively. Therefore their derivativesat t = 0 ∈ Ip are well-defined, and these give (LXf)(p) and (LXY )p, respectively.

The following shows that the operators (φtX)∗ play the role of the exponentialof the operators tLX .

Proposition 10.3.3. Let X ∈ X(M). For f ∈ C∞(M) and Y ∈ X(M), we have:

d

dt(φtX)∗f = (φtX)∗ X(f) = X (φtX)∗f ∈ C∞(Dt(X)),

d

dt(φtX)∗Y = (φtX)∗[X,Y ] = [X, (φtX)∗Y ] ∈ X(Dt(X)).

In particular,

LX(f) = X(f), LXY = [X,Y ].

Proof. The first equality is equivalent to

d

dtf(φtX(p)) = XφtX(p)(f), for all p ∈ Dt(X),

which holds because φtX(p) is an integral curve of X. Using that φt+sX = φtX φsX ,and applying the first equality for s = 0, we obtain the second equality:

d

dt(φtX)∗f =

d

ds

∣∣s=0

(φt+sX )∗f =d

ds

∣∣s=0

(φsX)∗((φtX)∗f) = X((φtX)∗f).

LECTURE 10. 89

Next, note that, for any g ∈ C∞(M) and Z ∈ X(M), we have that:

(*) ((φtX)∗Z)g =((dφ−tX ) ZφtX

)g = ZφtX (g φ−tX ) = (φtX)∗ Z (φ−tX )∗g.

Taking the derivative of this equality, for Z = Y , and using the first part, we obtain:

d

dt((φtX)∗Y )g =

d

dt

((φtX)∗ Y (φ−tX )∗g

)=

= (φtX)∗ (X Y − Y X) (φ−tX )∗g =

= (φtX)∗ [X,Y ] (φ−tX )∗g = ((φtX)∗[X,Y ])g,

where in the last step we have used again (∗) for Z = [X,Y ]. Since, g is arbitrary, weobtain the third equality d

dt (φtX)∗Y = (φtX)∗[X,Y ]. Since [X,X] = 0, this implies

that (φtX)∗X = X. Next, note that the vector fields (φtX)∗X = X, (φtX)∗Y and(φtX)∗[X,Y ] are φtX -related to the vector fields X, Y and [X,Y ], respectively. Thus,by Lemma 9.3.3, and by the fact that φtX is a local diffeomorphism, (φtX)∗[X,Y ] =[X, (φtX)∗Y ]. This implies also the last equality.

10.4. The Lie bracket as infinitesimal commutator of flows

The Lie bracket [X,Y ] of two vector fields X,Y ∈ X(M) can be regarded as theinfinitesimal obstruction for the flows of the vector fields to commute. Aroundevery point in M we can find an open set W and ε > 0 such that the expression:

σp(t, s) := φ−sY φ−tX φ

sY φtX(p) ∈M,

is defined for (t, s, p) ∈ (−ε, ε) × (−ε, ε) × W . If X and Y are complete, thenσ(t, s) ∈ Diff(M) is the group-theoretic commutator of the diffeomorphisms φtXand φsY . Geometrically, σp(t, s) is obtained as follows: try to construct a “curvedparallelogram” starting at p with one pair of sides pq and nm in the direction ofX and of “length” t, and the other pair of sides qm and np′ parallel to Y and of“length” s, as in the picture below:

p

q

m

n

p′

The black and red curves represent the flow lines of X and Y , respectively, and

q = φtX(p), m = φsX(q), n = φ−tX (m), p′ = φ−sX (n) = σp(t, s).

In general, the curved parallelogram does not close up, i.e. p′ = σp(t, s) 6= p.


On the other hand, since σ(t, 0) = idM , the derivative t 7→ ∂∂sσp(t, 0) is a

smooth vector field on M , which is given by:

d

dsσp(t, s)

∣∣s=0

=d

dsφ−sY φ

−tX φ

sY φtX(p)

∣∣s=0

=

= −Yp + (dφ−tX )(YφtX(p)) =

=((φtX)∗Y − Y

)p.

Applying Proposition 10.3.3, we obtain the following description of the Lie bracket:

Proposition 10.4.1. With the notation from above, we have that:

d

dt

∣∣t=0

d

ds

∣∣s=0

σ(t, s) = [X,Y ].

10.5. Exercises

Exercise 10.1. Let ψ : M → N be a smooth map. Consider X ∈ X(M) which isψ-related to Y ∈ X(N). Prove that their flows are ψ-related in the following sense:for every p ∈M , Ip ⊂ Iψ(p), and for all t ∈ Ip we have that

ψ(φtX(p)) = φtY (ψ(p)).

In other words (idR × ψ)(D(X)) ⊂ D(Y ), and the following diagram commutes:

D(X)idR×ψ //

φX

D(Y )

φY

Mψ // N

Exercise 10.2. Let X ∈ X(M). For λ ∈ R, prove that the flow of λX is given by:

φtλX(p) = φλtX (p),

and moreover that

(t, p) ∈ D(λX) ⇐⇒ (λt, p) ∈ D(X).

Exercise 10.3. Let X be a vector field on M . Let p ∈ M be such that Xp = 0.Prove that Ip = R and that φtX(p) = p for all t ∈ R.

Exercise 10.4. On the m-dimensional sphere

Sm = (x1, . . . , xm+1) ∈ Rm+1 :

m+1∑i=1

(xi)2 = 1,

consider the stereographical atlas:

A = (U−, ϕ−), (U+, ϕ−),

where U± = Sm\(0, . . . , 0,±1) and ϕ± : U±∼−→ Rm induce the coordinates

ϕ± = (y1±, . . . , y

m± ):

yi± :=xi

1∓ xm+1, 1 ≤ i ≤ m.

LECTURE 10. 91

(a) Show that there exists a smooth vector field X on Sm which in the coordinatesyi+ is given by

X|U+=

m∑i=1

yi+∂

∂yi+∈ X(Rm),

and write the expression of X in the coordinates yi−.(b) Determine the flow of X in both charts.

(c) Prove that there exists a vector field X ∈ X(Rm+1) satisfying

Xp = Xp, ∀ p ∈ Sm.

Exercise 10.5. (a) For any vector field X = f(x) ∂∂x ∈ X(R), prove that the vector

field sin(x)f(x) ∂∂x ∈ X(R) is complete (Hint: use Lemma 10.2.3).

(b) Prove that any vector field on R can be written as the sum of two completevector fields on R (Hint: use part (a)).

Exercise 10.6. Let X =∑mi=1X

i(x) ∂∂xi ∈ X(Rm) be a bounded vector field, i.e.

there is a constant C > 0 such that |Xi(x)| < C for all x ∈ Rm and all 1 ≤ i ≤ m.Prove that X is complete (Hint: first, show that its flow at time t can displace apoint only by a distance of C

√m|t|; second, use an argument as in Lemma 10.2.3).

Exercise 10.7. For f ∈ C∞(M) and X ∈ X(M), prove that the Taylor series of(φtX)∗f at t = 0 is given by:

exp(TLX)f :=∑n≥0

Tn

n!L nX(f) ∈ C∞(M)[[T ]],

where C∞(M)[[T ]] denotes the ring of formal power series in T with coefficients inC∞(M).

Exercise 10.8. Let A be a real vector space endowed with a bilinear map

∗ : A×A −→ A.

We are not assuming any other axiom for the operation ∗ (like associativity, com-mutativity, Jacobi identity etc.).

An automorphism of (A, ∗) is a linear isomorphism g : A→ A satisfying

g(a ∗ b) = g(a) ∗ g(b).

Denote the space of all automorphisms by Aut(A, ∗).A derivation of (A, ∗) is a linear map D : A→ A satisfying:

D(a ∗ b) = D(a) ∗ b+ a ∗D(b), for all a, b ∈ A.

Denote the space of all derivations by Der(A, ∗).(a) Show that (Aut(A, ∗), ) is a group.(b) Show that, if D1, D2 ∈ Der(A, ∗), then

[D1, D2] := D1 D2 −D2 D1 ∈ Der(A, ∗).

(c) Prove that the operation [·, ·] on Der(A, ∗) satisfies the condition (1),(2) and(3) of Proposition 9.2.1; i.e. (Der(A, ∗), [·, ·]) is a Lie algebra.


For (d) and (e), assume A to be finite dimensional1 Hence, we can talk aboutsmooth curves g : (−ε, ε)→ LinR(A,A), where LinR(A,A) is the (finite dimensional)vector space of linear maps from A to A. Identifying Tg(t)LinR(A,A) ∼= LinR(A,A),

we regard dgdt (t) ∈ LinR(A,A).

(d) Consider a smooth curve g : (−ε, ε)→ LinR(A,A), such that g(0) = idA and

g(t) ∈ Aut(A, ∗), for all t ∈ (−ε, ε).Prove that dg

dt (0) ∈ Der(A, ∗).(e) Let D ∈ Der(A, ∗). Show that the following series converges absolutely to an

automorphism of (A, ∗)

exp(D) =∑n≥0

1

n!Dn ∈ Aut(A, ∗).

For (f) and (g) assume (A, ∗) to be associative and commutative, however,A can be infinite dimensional.

(f) For a ∈ A and D ∈ Der(A, ∗), prove that a ∗ D ∈ Der(A, ∗), where a ∗ D isdefined in the obvious way:

(a ∗D)(b) := a ∗ (D(b)).

(g) Prove that the following Leibniz rule holds:

[D1, a ∗D2] = D1(a) ∗D2 + a ∗ [D1, D2].

1This assumption simplifies the discussion; however, the results hold also in an infinite dimensionalsetting, e.g. for bounded operators on a Banach space.

LECTURE 11

11.1. Lie groups

Definition 11.1.1. A Lie group is a manifold G endowed with a smooth groupstructure, i.e. the multiplication and the inversion are smooth maps:

µ : G×G −→ G, (g, h) 7→ g · h,

ι : G −→ G, g 7→ g−1.

A Lie subgroup of G is a subgroup H ⊂ G which is an immersed submanifold.

Examples 11.1.2. (1) Every at most countable group is a 0-dimensional Lie group.(2) The additive group (Rn,+) is an n-dimensional Lie group.(3) The multiplicative group (C∗, ·) is a 2-dimensional Lie group. The following

are 1-dimensional subgroups R∗ ⊂ C∗ and S1 ⊂ C∗.(4) The general linear group. The group GL(n) of linear automorphisms of

Rn is an n2-dimensional Lie group. Its smooth structure comes by regardingGL(n) as an open subset in the n2-dimensional vector space M(n) of linearmaps from Rn to Rn.

(5) The special linear group. The group SL(n) is the subgroup of GL(n) con-sists of linear automorphisms of Rn which have determinant 1.

(6) The orthogonal group. The group O(n) of orthogonal matrices (i.e. A ∈M(n) such that AAt = I) is a compact Lie group of dimension n(n− 1)/2 (seeExercise 6.11).

(7) There are complex versions of the groups above: GL(n,C) the group of in-vertible complex matrices, with subgroup SL(n,C) consisting of matices withdeterminant 1; O(n,C) the group of complex matrices satisfying AAt = I.

(8) The product G1 × . . . × Gk of Lie groups is a Lie group. In particular, thek-torus T k := S1 × . . .× S1 is a k-dimensional compact Lie group.

(9) If G is any Lie group, then the connected component of the identity, denotedby G ⊂ G, is a Lie subgroup (which is both open and closed). In fact, G isa normal subgroup.

(10) The group GL(n) consists of matrices with positive determinant.(11) The special orthogonal group is the Lie group SO(n) := O(n).(12) If G is a Lie group, then TG is a Lie group. Its structure maps are the differ-

entials of the structure maps of G.

93


Definition 11.1.3. The left translation by g ∈ G is the diffeomorphism of G

λg : G −→ G, λg(h) := gh.

The right translation by g ∈ G is the diffeomorphism of G

ρg : G −→ G, ρg(h) := hg.

Using left translations one obtains the following:

Proposition 11.1.4. Every Lie group is a parallelizable manifold.

Proof. Using left translations, we construct a global frame on TG:

θL : TG −→ TeG, θL(X) := dgλg−1(X), X ∈ TgG.By Corollary 8.4.3, the existence of θL1 implies that G is parallelizable.

Definition 11.1.5. Let G be a Lie group. A vector field X ∈ X(G) is calledleft invariant if X is λg-related to X, for all g ∈ G. We denote the space ofleft invariant vector fields by XL(G). Similarly, one defines the space of rightinvariant vector fields XR(G).

Explicitly, the condition that X ∈ X(G) is left invariant can be written as:

dhλg(Xh) = Xgh, for all g, h ∈ G.Letting h = e, this formula implies that X is determined by its value at the identity:

Xg = deλg(Xe).

Conversely, the left invariant extension of v ∈ TeG is the vector field:

vL ∈ X(G), vLg := deλg(v).

That vL is left invariant follows by the chain rule:

dhλg(vLh ) = dhλg deλh(v) = de(λg λh)(v) = deλgh(v) = vLgh.

We conclude with the following:

Proposition 11.1.6. (1) The following map is a linear isomorphism:

XL(G) ∼−→ TeG, X 7→ Xe, with inverse v 7→ vL.

(2) If X,Y ∈ XL(G), then [X,Y ] ∈ XL(G).

Proof. Part (1) was discussed above; part (2) follows from Lemma 9.3.3.

Definition 11.1.7. A Lie algebra is a vector space g endowed with an operation:

[·, ·] : g× g −→ g,

called the Lie bracket, which

(1) is skew-symmetric:

[u, v] = −[v, u], u, v ∈ g,

(2) is R-bilinear:

[u, av + bw] = a[u, v] + b[u,w], a, b ∈ R, u, v, w ∈ g,

(3) and satisfies the Jacobi identity:

[u, [v, w]] + [v, [w, u]] + [w, [u, v]] = 0, u, v, w ∈ g.

1The map θL is called the left Maurer-Cartan form of G.

LECTURE 11. 95

Definition 11.1.8. The Lie algebra of a Lie group G is the vector space

g := TeG

with Lie bracket:

[·, ·] : g× g→ g, [v, w] := [vL, wL]e.

By Proposition 9.2.1, (g, [·, ·]) satisfies indeed the axioms of a Lie algebra.

Remarks 11.1.9. (1) The Lie algebra g of a Lie group G plays the role of a“linearized version” of G; it encodes infinitesimal information about G. In fact,if G is simply connected, it can be fully recovered as a Lie group from thestructure of g (see for example [2]).

(2) In general Lie groups are denoted by capital Latin letters: G, H, K, L, P ,Q, etc., while the corresponding Lie algebras are denoted by the correspondingsmall Fraktur letters (also called Gothic letters): g, h, k, l, p, q etc. (in LaTeXuse \mathfrak..).

(3) One can define the Lie algebra of G also using right invariant vector fields;nevertheless, one obtains an isomorphic object. The push-forward along theinversion map gives an isomorphism between left invariant and right invariantvector fields. Note however that, if both Lie algebras are identified with TeG(by evaluating vector fields at e), one obtains opposite Lie brackets on TeG,because the differential of the inversion map at e is −IdTeG.

Proposition 11.1.10. Left invariant vector fields are complete. Moreover,

φtX(gh) = gφtX(h), for all X ∈ XL(G), g, h ∈ G, t ∈ R.

Proof. Let X ∈ XL(G) and let g, h ∈ G. As in Lecture 10, let Ih denote themaximal interval on which the flow line starting at h is defined. For t ∈ Ih, wehave:

d

dt

(gφtX(h)

)= dφtX(h)λg

( d

dtφtX(h)

)= dφtX(h)λg

(XφtX(h)

)= XgφtX(h),

where we have used that X is left invariant. This shows that

Ih 3 t 7→ gφtX(h)

is an integral curve of X. Since the curve starts at gh, by Lemma 10.1.2, weconclude that Ih ⊂ Igh and that gφtX(h) = φtX(gh). In particular, Ie ⊂ Ig for allg ∈ G, and so, by Lemma 10.2.2, X is complete.

The Proposition above allows us to define:

Definition 11.1.11. The exponential map of a Lie group G with Lie algebra gis the map:

exp : g −→ G, v 7→ φ1vL(e).

Exercise 11.1 guides you through a possible proof that the exponential map isindeed smooth. In fact, the exponential map is a local diffeomorphism around 0.

Proposition 11.1.12. The exponential map exp : g→ G satisfies:

d0 exp = Idg,

where we identify T0g = g = TeG. Hence, exp is a diffeomorphism between aneighborhood of 0 in g and a neighborhood of e in G.


Proof. The result follows from the formula φtsvL(g) = φstvL(g) of Exercise 10.2:

d0 exp(v) =d

dε

∣∣ε=0

exp(εv) =d

dε

∣∣ε=0

φ1εvL(e) =

d

dε

∣∣ε=0

φεvL(e) = vLe = v.

Next, we discuss maps between Lie groups and Lie algebras

Definition 11.1.13. (1) A Lie group homomorphism is a smooth group ho-momorphism

F : (G, ·) −→ (H, ·).(2) A homomorphism of Lie algebras is a linear map

f : (g, [·, ·]) −→ (h, [·, ·])

satisfying

f([u, v]) = [f(u), f(v)], for all u, v ∈ g.

Lie group homomorphisms induce Lie algebra homomorphisms:

Proposition 11.1.14. Let (G, ·) and (H, ·) be two Lie groups with correspondingLie algebras (g, [·, ·]) and (h, [·, ·]). Consider a Lie group homomorphism:

F : (G, ·) −→ (H, ·).

Since F (e) = e, define:

f := deF : g = TeG −→ h = TeH.

Then f is a homomorphism of Lie algebras, and the following diagram commutes:

GF // H

gf //

expOO

h

expOO

i.e. F (exp(v)) = exp(f(v)) for all v ∈ g.

Proof. For v ∈ g, we show that vL and f(v)L are F -related. Since F is a Lie grouphomomorphism, we have F λg = λF (g) F . Therefore,

dgF (vLg ) = dgF deλg(v) = de(F λg)(v) = de(λF (g) F )(v) =

= deλF (g) deF (v) = deλF (g)(f(v)) = f(v)LF (g).

So, for any w ∈ g, wL and f(w)L are also F -related, and so, by Lemma 9.3.3,[vL, wL] is F -related to [f(v)L, f(w)L]. Evaluating this identity at the unit, impliesthat f is a Lie algebra homomorphism:

[f(v), f(w)] = [f(v)L, f(w)L]e = deF ([vL, wL]e) = f([v, w]).

Finally, using again that vL is F -related to f(v)L, and Exercise 10.1, we obtain thecommutativity of the diagram:

F (exp(v)) = F (φ1vL(e)) = φ1

f(v)L(F (e)) = exp(f(v)).

LECTURE 11. 97

11.2. Exercises

Exercise 11.1. Let G be a Lie group with Lie algebra g. Denote by vL ∈ XL(G)the left invariant extension of a vector v ∈ g.

(a) Show that the following formula defines a smooth vector field on G× g:

ξ ∈ X(G× g), ξ(g,v) := (vLg , 0) ∈ TgG⊕ Tvg ∼= T(g,v)(G× g).

(b) Prove that the flow of ξ is given by

φtξ(g, v) = (φtvL(g), v),

and conclude that ξ is complete.(c) Use (b) to prove that the exponential map exp : g→ G is smooth.

Exercise 11.2. Prove that left invariant vector fields commute with right invariantvector fields:

X ∈ XL(G), Y ∈ XR(G) =⇒ [X,Y ] = 0.

Exercise 11.3. For k ≥ 1, denote by T k the k-dimensional torus, i.e.

T k = (S1)k = (z1, . . . , zk) ∈ Ck : |z1| = . . . = |zk| = 1.

(a) Prove that left invariant vector fields on the k-torus are the same as rightinvariant vector fields:

XL(T k) = XR(T k).

Moreover, if θi, 1 ≤ i ≤ k denote the “angle coordinates” on T k (see Example8.5.5), prove that the following is a basis of XL(T k):

∂

∂θ1, . . . ,

∂

∂θk∈ X(T k).

(b) Calculate the exponential map:

exp : TeTk −→ T k.

Hint: use the basis from (a) for TeTk.

(c) Consider an n×m matrix with integer coefficients A = ai,j ∈ Z1≤i≤n,1≤j≤m.Show that the map

FA : (Tm, ·) −→ (Tn, ·),FA(z1, . . . , zm) = (z

a1,11 . . . za1,mm , . . . , z

an,11 . . . zan,mm )

is a Lie group homomorphism.(d) Show that any Lie group homomorphism F : Tm → Tn is of the form described

in (c). Hint: use Proposition 11.1.14.

Exercise 11.4. Consider the Lie group GL(n) of invertible n×n-matrices. Denotethe space of all n × n-matrices by gl(n). Since GL(n) is an open subset of gl(n),we identify:

TgGL(n) = gl(n), for all g ∈ GL(n).

(a) Prove that, with the above identification, the differential of left translation issimply given by matrix multiplication:

dhλg : ThGL(n) = gl(n) −→ TghGL(n) = gl(n),

dhλg(v) = g · v.


(b) Prove that the left invariant extension of v ∈ TIGL(n) = gl(n) is the vectorfield

vL ∈ XL(GL(n)), vLg = g · v ∈ TgGL(n) = gl(n).

(c) Prove that the flow of the left invariant extension vL ∈ XL(GL(n)) of v ∈TIGL(n) = gl(n) is given by

φtvL(g) = g · etv,

where ew =∑n≥0

wn

n! is the matrix exponential.

(d) Using Proposition 10.4.1 and (c) prove that the Lie bracket on gl(n) is thecommutator of matrices:

[v, w] = v · w − w · v, v, w ∈ gl(n).

Exercise 11.5. The quaternions are the real 4-dimensional associative algebraH with basis 1, i, j, k, where the multiplication is determined by the rules:

i2 = j2 = k2 = −1, ij = k, jk = i, ki = j, ji = −k, kj = −i, ik = −j.One can also represent H as the space of all 2× 2-matrices with complex entries ofthe form: (

z w−w z

), z, w ∈ C.

The correspondence between the two descriptions is given by

i←→(

i 00 −i

), j ←→

(0 1−1 0

), k ←→

(0 ii 0

).

(a) Prove that every element in H∗ = H\0 is invertible, and conclude the H∗ isa 4-dimensional Lie group.

(b) Prove that the map following map is a Lie group homomorphism:

F : (H∗, ·) −→ (R>0, ·), F (a+ bi+ cj + dk) = a2 + b2 + c2 + d2.

(c) Prove that G := F−1(1) is a Lie subgroup of H∗ which as a manifold is diffeo-morphic to S3.

(d) Identifying T1H∗ = H, prove that the exponential map of H∗ is given by theusual exponential series:

exp : H −→ H∗, A 7→∑n≥0

An

n!.

(e) Prove that the Lie algebra of G is given by

g = T1G = bi+ cj + dk : b, c, d ∈ R.(f) Consider the set of “square roots of the unity” in H:

Σ := X ∈ H : X2 = −1.Prove that Σ ⊂ g, and that Σ is diffeomorphic to S2.

(g) Prove that every element in A ∈ H can be decomposed as

A = λ+ µX, λ, µ ∈ R, X ∈ Σ.

(h) In the decomposition A = λ+ µX from (g), prove that A ∈ G iff λ2 + µ2 = 1.(i) In the decomposition A = λ+ µX from (g), prove that

exp(A) = eλ(cos(µ) + sin(µ)X).

LECTURE 11. 99

(j) On g we consider the metric such that Σ is the unit sphere around the origin,and denote by Br ⊂ g the open ball of radius r > 0. Using the previous items,prove the following about the exponential map exp : g→ G:

(i) exp is a diffeomorphism between Bπ and G\−1;(ii) exp sends the spheres of radii 2kπ to 1 ∈ G, and the spheres of radii

(2k + 1)π to −1 ∈ G;(iii) exp restricts to a diffeomorphism between B(k+1)π\Bkπ and G\1,−1.

(k) Show that every flow line of a left invariant vector field on G is a circle.

LECTURE 12

12.1. The exterior algebra

There are (at least) two important algebras associated to any vector space V . First,the algebra S(V ∗) of polynomial functions on V , which is also called the symmetricalgebra of V ∗. One can construct S(V ∗) as the algebra generated by the elementsof V ∗ subject to the relations that these generators commute (in fact this givesa presentation of S(V ∗) in the realm of unital associative algebras). The secondalgebra, which will be constructed in this section, is called the exterior algebra ofV ∗ and is denoted by

∧V ∗. The algebra

∧V ∗ is the algebra generated by elements

of V ∗ where these generators are subject to the condition that they anti-commute,i.e. αβ = −βα.

Definition 12.1.1. An alternating k-form on a vector space V is a map

ω :V × . . .× V︸︷︷︸ −→ R

k × V

which is multilinear

ω(v1, . . . ,vi−1, av + bw, vi+1, . . . , vk) =

= aω(v1, . . . , vi−1, v, vi+1, . . . , vk) + bω(v1, . . . , vi−1, w, vi+1, . . . , vk),

and satisfies

ω(v1, . . . , vi, vi+1, . . . , vk) = −ω(v1, . . . , vi+1, vi, . . . , vk).

The space of alternating k-forms on V is denoted by∧k

V ∗ and is called the

k-th exterior power of V ∗. We have that∧1

V ∗ = V ∗, and we define∧0

V ∗ := R.

Fixing v ∈ V to be the first argument of ω ∈∧k

V ∗, we obtain an alternating

k − 1 form, denoted by ιvω ∈∧k−1

V ∗,

(ιvω)(v1, . . . , vk−1) := ω(v, v1, . . . , vk−1).

The operation ω 7→ ιvω is called the interior product by v, or the contraction byv. For k = 1 and η ∈ V ∗, we have that ιvη = η(v). For k = 0 and a ∈ R =

∧0V ∗,

we set ιva := 0.

101


Remark 12.1.2. The symmetric group Sk consists of permutations of the set1, . . . , k. Recall that Sk is generated by the transpositions i↔ (i+ 1). If σ ∈ Sk

can be written as a product of p transpositions, then the parity of p is independentof this decomposition, and the sign of σ is defined as sign(σ) = (−1)p. This impliesthat an alternating k-from ω satisfies

ω(vσ(1), . . . , vσ(k)) = sign(σ)ω(v1 . . . , vk),

for all σ ∈ Sk. In particular, ω returns 0 whenever two entries of coincide:

ω(. . . , v, . . . , v, . . .) = 0.

Exercise 12.2 gives a more general version of this last observation.

The tensor product of α ∈∧k

V ∗ and β ∈∧l

V ∗ is the multilinear map:

α⊗ β :V × . . .× V︸︷︷︸ −→ R,

(k + l) × V

(α⊗ β)(v1, . . . , vk,w1, . . . , wl) := α(v1, . . . , vk)β(w1, . . . , wl).

Note that α ⊗ β is not alternating: if we permute the vi’s among themselves,or the wj ’s among themselves, the sign does change correspondingly, however, ifthe vi’s and the wj ’s get mixed up, one obtains a completely different outcome.For example, if k = l = 1, then α(v)β(w) and α(w)β(v) can be quite different.Nevertheless, note that their difference is alternating:

(α⊗ β − β ⊗ α)(v, w) = α(v)β(w)− α(w)β(v).

This operation can be extended to all degrees k, l ≥ 1, by considering the sum (withsigns) of all possible ways the vi’s and the wj ’s can mix. To make this precise weintroduce the following notion:

Definition 12.1.3. A (k, l)-shuffle is a permutation σ ∈ Sk+l which is increasingon the first k integers and on the last l integers:

σ(1) < σ(2) < . . . < σ(k), σ(k + 1) < σ(k + 2) < . . . < σ(k + l).

We denote the set of all (k, l)-shuffles by S(k, l).

A (k, l)-shuffle for k + l = 52.

Definition 12.1.4. The wedge product (or exterior product) is the followingoperation on alternating forms:

∧ :

k∧V ∗ ×

l∧V ∗ −→

k+l∧V ∗,

LECTURE 12. 103

(α ∧ β)(v1, . . . , vk+l) =∑

σ∈S(k,l)

sign(σ)α(vσ(1), . . . , vσ(k))β(vσ(k+1), . . . , vσ(k+l)).

For a ∈ R =∧0

V ∗ and α ∈∧k

V ∗ note that:

a ∧ α = aα = α ∧ a.

It is not obvious from the definition that α∧β is alternating; this will be shownthis in the proof of the following:

Theorem 12.1.5. The wedge product has the following properties:

(1) the interior product with any v ∈ V is a graded derivation:

ιv(α ∧ β) = (ιvα) ∧ β + (−1)kα ∧ (ιvβ);

(2) the wedge product is graded commutative:

α ∧ β = (−1)klβ ∧ α,

(3) the wedge product is associative:

(α ∧ β) ∧ γ = α ∧ (β ∧ γ),

for all α ∈∧k

V ∗, β ∈∧l

V ∗ and γ ∈∧m

V ∗.Moreover, if n = dim(V ∗) and e1, . . . , en is a basis of V ∗, then the set

ei1 ∧ ei2 ∧ . . . ∧ eik : 1 ≤ i1 < i2 < . . . < ik ≤ n

forms a basis of∧k

V ∗. Hence

dim( k∧

V ∗)

=

(nk

)= n!

k!(n−k)! , 0 ≤ k ≤ n0, n < k.

Proof. First we prove (1). Note that the set of (k, l)-shuffles admits the disjointdecomposition S(k, l) = S′(k, l) t S′′(k, l), where

S′(k, l) = σ ∈ S(k, l) : σ(1) = 1 S′′(k, l) = σ ∈ S(k, l) : σ(k + 1) = 1.

Therefore,

(ιv1(α ∧ β))(v2, . . . , vk+l) = (α ∧ β)(v1, . . . , vk+l) =

=∑

σ∈S(k,l)

sign(σ)α(vσ(1), . . . , vσ(k))β(vσ(k+1), . . . , vσ(k+l)) =

=∑

σ′∈S′(k,l)

sign(σ′)(ιv1α)(vσ′(2), . . . , vσ′(k))β(vσ′(k+1), . . . , vσ′(k+l))+

+∑

σ′′∈S′′(k,l)

sign(σ′′)α(vσ′′(1), . . . , vσ′′(k))(ιv1β)(vσ′′(k+2), . . . , vσ′′(k+l)).

Next, note that there are one-to-one correspondences

θ ∈ S(k − 1, l)↔ σ′ ∈ S′(k, l) η ∈ S(k, l − 1)↔ σ′′ ∈ S′′(k, l)

θ(i) = σ′(i+ 1)− 1 η(i) =

σ′′(i)− 1, 1 ≤ i ≤ kσ′′(i+ 1)− 1, k + 1 ≤ i ≤ k + l − 1


and moreover, that sign(σ′) = sign(θ) and sign(σ′′) = (−1)ksign(η). Using thisfact, the above formula becomes:

(ιv1(α ∧ β))(v2, . . . , vk+l) =

=∑

θ∈S(k−1,l)

sign(θ)(ιv1α)(vθ(1)+1, . . . , vθ(k)+1)β(vθ(k+1)+1, . . . , vθ(k+l)+1)+

(−1)k∑

η∈S(k,l−1)

sign(η)α(vη(1)+1, . . . , vη(k)+1)(ιv1β)(vη(k+1)+1, . . . , vη(k+l−1)+1) =

= (ιv1α ∧ β)(v2, . . . , vk+l) + (−1)k(α ∧ ιv1β)(v2, . . . , vk+l).

This proves (1).

Next, we prove by induction on n := k + l ≥ 0 that, for all α ∈∧k

V ∗

and β ∈∧l

V ∗, α ∧ β is alternating. Clearly, this holds for n = 0. Assuming the

induction hypothesis for n−1, it follows that (ιv1α)∧β+(−1)kα∧(ιv1β) ∈∧n−1

V ∗.Hence, if we switch vi and vi+1 in (α∧β)(v1, . . . , vk+l), for 1 < i < n−1, the resultchanges the sign. For v1 and v2, we need to check that

ιv1ιv2(α ∧ β) = −ιv2ιv1(α ∧ β).

For this, we apply the derivation rule twice, and use that α and β are alternating:

ιv2ιv1(α ∧ β) =ιv2((ιv1α) ∧ β + (−1)kα ∧ (ιv1β)

)=

=(ιv2ιv1α) ∧ β + (−1)k−1(ιv1α) ∧ (ιv2β)+

+ (−1)k(ιv2α) ∧ (ιv1β) + α ∧ (ιv2ιv1β) =

=− (ιv1ιv2α) ∧ β − (−1)k(ιv1α) ∧ (ιv2β)+

− (−1)k−1(ιv2α) ∧ (ιv1β)− α ∧ (ιv1ιv2β) =

=− ιv1ιv2(α ∧ β),

where in the last step we have performed first two steps backwards, with v1 andv2 interchanged. This finishes the inductive argument, and proves that the wedgeproduct of alternating forms is indeed alternating.

Properties (2) and (3) are proven similarly, by induction on n = k + l andn = k + l + m, respectively, and by using the derivation rule. We will prove only(2), and leave it to the reader to check (3). Clearly, (2) holds for n = k + l = 0.

Assume that it holds for n− 1. Let α ∈∧k

V ∗ and β ∈∧l

V ∗. For v ∈ V , we have

ιv(α ∧ β) = (ιvα) ∧ β + (−1)kα ∧ (ιvβ) =

= (−1)(k−1)lβ ∧ (ιvα) + (−1)k+k(l−1)(ιvβ) ∧ α =

= (−1)kl((−1)lβ ∧ (ιvα) + (ιvβ) ∧ α

)=

= (−1)klιv(β ∧ α),

where we have used the induction hypothesis in the second line. Since this holdsfor all v, (2) follows.

For the last part, let e1, . . . , en ∈ V be the dual basis, i.e. these elements aredefined by ei(ej) = δij , where δij = 1 if i = j and δij = 0 if i 6= j. By repeatedlyapplying the derivation rule, one can easily show that, for 1 ≤ i1 < . . . < ik ≤ nand 1 ≤ j1 < . . . < jk ≤ n, the following holds:(

ei1 ∧ . . . ∧ eik)

(ej1 , . . . , ejk) = δi1j1 . . . δikjk,

LECTURE 12. 105

in other words the result is zero unless ip = jp, for all 1 ≤ p ≤ k, and then theresult is 1.

Consider now a linear combination:

ω =∑

i1<...<ik

ai1...ikei1 ∧ . . . ∧ eik ∈

k∧V ∗.

Then for 1 ≤ j1 < . . . < jk ≤ n, we have that ω(ej1 , . . . , ejk) = aj1...jk . Thus ifω = 0, then all coefficients aj1...jk = 0. This shows that the k-forms ei1 ∧ . . . ∧ eikare linearly independent.

To show that these forms span∧k

V ∗, let ω ∈∧k

V ∗. Consider the k-form:

η := ω −∑

i1<...<ik

ω(ei1 , . . . , eik)ei1 ∧ . . . ∧ eik ∈k∧V ∗.

Then η(ej1 , . . . , ejk) = 0 for all 1 ≤ j1 < . . . < jk ≤ n. Since η is alternating,this implies that, η(el1 , . . . , elk) = 0, for all l1, . . ., lk ∈ 1, . . . , n (see Remark12.1.2). Finally, consider k vectors v1, . . . , vk ∈ V . By writing these in the basisvi =

∑nl=1 v

liel, and using that η is multilinear, we obtain:

η(v1, . . . , vk) =

n∑l1=1

. . .

n∑lk=1

vl11 . . . vlkk η(el1 , . . . , elk) = 0.

Thus η = 0, and therefore:

ω =∑

i1<...<ik

ω(ei1 , . . . , eik)ei1 ∧ . . . ∧ eik .

We conclude that the set ei1 ∧ . . . ∧ eik : 1 ≤ i1 < . . . < ik ≤ n forms a basis of∧kV ∗. Note that sequences of the form 1 ≤ i1 < . . . < ik ≤ n are in one-to-one

correspondence with subset I = i1, . . . , ik of 1, . . . , n with exactly k elements;

this implies that dim(∧k

V ∗) =(nk

).

As a consequence of the graded commutativity, we obtain:

Corollary 12.1.6. If k is odd and ω ∈∧k

V ∗ then ω2 = ω ∧ ω = 0.

Putting the exterior powers of V ∗ together, one obtains the exterior algebra:

Definition 12.1.7. Let V be a vector space. The exterior algebra of V ∗, denotedby (

∧V ∗,+,∧), is the direct sum of the exterior products of V ∗∧

V ∗ :=⊕k≥0

k∧V ∗,

endowed with the bilinear extension of the wedge product

∧ :∧V ∗ ×

∧V ∗ −→

∧V ∗.

The exterior algebra is an associative unital algebra of dim(∧V ∗) = 2dim(V ).

Remark 12.1.8. The unit of∧V ∗ is just 1 ∈ R =

∧0V ∗. The “bilinear extension

of the wedge product” is meant in the obvious way: if α, β ∈∧V ∗ decompose as

α = α0 + α1 + . . .+ αk, αi ∈i∧V ∗,


β = β0 + β1 + . . .+ βl, βj ∈j∧V ∗,

then their wedge product is given by:

α ∧ β = γ0 + γ1 + . . .+ γk+l, γs :=∑i+j=s

αi ∧ βj ∈s∧V ∗.

Definition 12.1.9. Let V and W be vector space, and let A : V → W be a linearmap. The pullback map induced by A is the algebra homomorphism

A∗ :(∧

W ∗,+,∧)−→

(∧V ∗,+,∧

),

defined by

A∗ :

k∧W ∗ −→

k∧V ∗, (A∗ω)(v1, . . . , vk) := ω(Av1, . . . , Avk),

for k ≥ 1, and for a ∈∧0

W ∗ = R, A∗a = a ∈∧0

V ∗ = R.

It can be easily checked that A∗ is indeed an algebra homomorphism:

A∗(α ∧ β) = (A∗α) ∧ (A∗β).

The top exterior power of V ∗, i.e. the 1-dimensional vector space∧dim(V )

V ∗,is also called the determinant of V ∗, because of the following:

Proposition 12.1.10. For any linear map A : V → V and any ω ∈∧dim(V )

V ∗:

A∗(ω) = det(A)ω;

explicitly, for v1, . . . , vn ∈ V , where n = dim(V ),

ω(Av1, . . . , Avn) = det(A)ω(v1, . . . , vn).

Remark 12.1.11. Note that (k, l)-shuffles are in one-to-one correspondence withpairs of subsets (A,B) of the set 1, . . . , k + l, such that:

1, . . . , k + l = A ∪B, |A| = k, |B| = l.

The element σ ∈ S(k, l) corresponds to the pair

A = σ(1), . . . , σ(k), B = σ(k + 1), . . . , σ(k + l).

Conversely, given such a pair (A,B), one first orders the elements of the two sets:

A = a1 < a2 < . . . < ak, B = b1 < b2 < . . . < bl,

and then the corresponding σ is given by:

σ(i) := ai, 1 ≤ i ≤ k; σ(k + j) := bj , 1 ≤ j ≤ l.

In particular, we have that

|S(k, l)| =(k + l

k

)=

(k + l)!

k! l!.

LECTURE 12. 107

12.2. The cotangent bundle

Definition 12.2.1. Let M be a manifold. The cotangent space at p ∈M , denotedby T ∗pM , is defined as the linear dual of the tangent space TpM ; i.e. T ∗pM consistsof all linear maps α : TpM → R. Elements of T ∗pM are called cotangent vectors.All the cotangent spaces together form the cotangent bundle of M :

T ∗M :=⊔p∈M

T ∗pM.

Example 12.2.2. Let f ∈ C∞(M) be a smooth function. The differential of f atp ∈M is a cotangent vector dpf ∈ T ∗pM at p

dpf : TpM −→ R, dpf(v) := v(f),

where, as usually, we identify Tf(p)R = R.

Let (U,ϕ = (x1, . . . , xm)) be a coordinates chart on M1. At p ∈ U , we have:

dpxi( ∂

∂xj∣∣p

)=∂xi

∂xj(p) = δij =

1, if i = j,0, if i 6= j.

This shows that dpx1, . . . ,dpx

m is a basis of the cotangent space T ∗pM dual to the

basis ∂∂x1 |p, . . . , ∂

∂xm |p of TpM .

Theorem 12.2.3. The cotangent bundle π : T ∗M →M is a vector bundle over Mof rank equal to m = dim(M). The vector bundle structure is uniquely determinedby the condition that, for every chart (U,ϕ = (x1, . . . , xm)) on M , the differentialsof the coordinate functions:

dx1, . . . ,dxm : U −→ T ∗M |U , p 7→ dpxi ∈ T ∗pM,

form a local frame on T ∗M .

The above result can be proven exactly like Theorem 8.1.2, therefore we omitthe proof. However, it is instructive to write down the transition functions fromone frame to another. Given a second set of coordinates (V, ψ = (y1, . . . , ym)), forp ∈ U ∩ V , we have that:

dxi|p =m∑j=1

∂xi

∂yj(p)dyj |p,

which can be easily seen by applying to both sides the basis vectors ∂∂yk

∣∣p.

12.3. The exterior powers of the cotangent bundle

Next, we consider the exterior powers of the cotangent spaces.

Definition 12.3.1. Let M be a manifold, and let p ∈ M . The k-th exterior

power of the cotangent space at p is denote by∧k

T ∗pM . The collection of all thesevector spaces is the k-th exterior power of the cotangent bundle:

k∧T ∗M :=

⊔p∈M

k∧T ∗pM.

1From this lecture on, we use simplified notation: instead of xiϕ we denote the coordinates by xi.


Consider a chart (U,ϕ = (x1, . . . , xm)) on M . By Theorem 12.1.5, for each

p ∈ U , a basis of∧k

T ∗pM is given by the elements:

dxi1 ∧ . . . ∧ dxik |p, 1 ≤ i1 < . . . < ik ≤ m.Generalizing the case k = 1 of the cotangent bundle, when choosing a second chart(V, ψ = (y1, . . . , ym)), on U ∩ V the frame changes as follows:

dxi1 ∧ . . . ∧ dxik =∑l1...lk

∂xi1

∂yl1. . .

∂xik

∂ylkdyl1 ∧ . . . ∧ dylk ,

where the sum runs over all 1 ≤ l1, . . . , lk ≤ m. After ordering the indices l1, . . . , lkincreasingly, and dropping the terms where some indexes are repeated (are thereforeare zero), we can rewrite:

dxi1 ∧ . . . ∧ dxik =∑

j1<...<jk

ci1,...,ikj1,...,jkdyj1 ∧ . . . ∧ dyjk ,

where the coefficients are given explicitly by

ci1,...,ikj1,...,jk=∑σ∈Sk

sign(σ)∂xi1

∂yjσ(1). . .

∂xik

∂yjσ(k)=∂(xi1 . . . xik)

∂(yj1 . . . yjk),

where the notation on the right is the standard notation for the minor consisting of

the rows i1, . . . , ik and columns j1, . . . , jk in the Jacobian matrix ( ∂xi

∂yj )mi,j=1 of the

change of coordinates map ψ ϕ−1, i.e.

∂(xi1 . . . xik)

∂(yj1 . . . yjk)= det

((∂xia∂yjb

)ka,b=1

).

The precise formula is not that important; what matters is that these coefficientsare smooth, because this allows us to apply the same argument as in the proof ofTheorem 8.1.2, and conclude:

Theorem 12.3.2. The k-th exterior power of the cotangent bundle

π :

k∧T ∗M −→M

is a vector bundle over M of rank equal to(mk

), where m = dim(M). A chart

(U,ϕ = (x1, . . . , xm)) on M induces a local frame on∧k

T ∗M over U

dxi1 ∧ . . . ∧ dxik , 1 ≤ i1 < . . . < ik ≤ m;

moreover, these frames define the vector bundle structure uniquely.

12.4. Differential forms

Definition 12.4.1. A differential form of degree k (or a k-form) on M is a

section∧k

T ∗M . The space of differential k-forms is denoted by

Ωk(M) := Γ( k∧

T ∗M).

For k = 0, Ω0(M) = C∞(M). The direct sum of these vector spaces is denoted by

Ω(M) :=⊕k≥0

Ωk(M).

LECTURE 12. 109

A differential form on M is an element

ω ∈ Ω(M), ω = ω0 + . . .+ ωm, ωi ∈ Ωi(M),

where m = dim(M).

The pointwise wedge product induces the structure of an associative unitalalgebra on the space of differential forms:

(Ω(M),+,∧), (α ∧ β)p := αp ∧ βp,

which is graded commutative:

α ∧ β = (−1)klβ ∧ α, α ∈ Ωk(M), β ∈ Ωl(M).

The interior product with a vector field X ∈ X(M) is the linear operator:

ιX : Ωk(M) −→ Ωk−1(M), (ιXω)p := ιXp(ωp),

which is a graded derivation for the wedge-product:

ιX(α ∧ β) = (ιXα) ∧ β + (−1)kα ∧ (ιXβ),

for α ∈ Ωk(M) and β ∈ Ω(M).Given a local chart (U,ϕ) with coordinates ϕ = (x1, . . . , xm), a differential

k-form ω ∈ Ωk(M) can be written as

ω|U =∑

i1<...<ik

ωi1...ikdxi1 ∧ . . . ∧ dxik ,

with smooth coefficients ωi1...ik ∈ C∞(U). The notation where the indexes are notordered is also useful:

ω|U =∑i1...ik

1

k!ωi1...ikdxi1 ∧ . . . ∧ dxik ,

where the coefficients are defined to be fully skew-symmetric:

ωiσ(1)...iσ(k) = sign(σ)ωi1...ik , for all σ ∈ Sk.

In this notation the interior by a vector fields X|U =∑mi=1X

i ∂∂xi is given by

(ιXω)|U =∑i1...ik

1

(k − 1)!Xi1ωi1...ikdxi2 ∧ . . . ∧ dxik .

12.5. The pullback of differential forms

Differential forms can be pulled back via any smooth maps:

Definition 12.5.1. The pullback along a smooth map f : M → N is defined by:

f∗ : Ωk(N) −→ Ωk(M), (f∗ω)p := (dpf)∗ωf(p),

that is, for v1, . . . , vk ∈ TpM :

(f∗ω)p(v1, . . . , vk) = ωf(p)(dpf(v1), . . . ,dpf(vk)).

The pullback satisfies the following:


Proposition 12.5.2. The pullback along a smooth map f : M → N is an algebrahomomorphism:

f∗ : (Ω(N),+,∧) −→ (Ω(M),+,∧) ,

i.e. for all α, β ∈ Ω(N) and all a, b ∈ R it satisfies:

f∗(α ∧ β) = f∗(α) ∧ f∗(β),

f∗(aα+ bβ) = af∗(α) + bf∗(β).

Moreover, if g : N → P is also a smooth map, then f∗ g∗ = (g f)∗.

12.6. Exercises

Exercise 12.1. Prove the following formulas for the wedge product:

(α ∧ β)(v1, . . . , vk+l) =1

k!l!

∑σ∈Sk+l

sign(σ)α(vσ(1), . . . , vσ(k))β(vσ(k+1), . . . , vσ(k+l)).

Exercise 12.2. Let v1, . . . , vp be p linearly dependent vectors in V . Show that for

any alternating form ω ∈∧k

V ∗, we have that

ιv1 . . . ιvpω = 0.

Exercise 12.3. Let α1, . . . , αk ∈ V ∗, and v1, . . . , vk ∈ V . Prove that:

(α1 ∧ . . . ∧ αk)(v1, . . . , vk) = det (αi(vj))ki,j=1 .

Exercise 12.4. Prove Proposition 12.1.10.

Exercise 12.5. Denote the product of k-copies of V by V ×k. For a k × k-matrixR = rji ki,j=1, consider the following map:

RV : V ×k −→ V ×k, RV (v1, . . . , vk) :=( k∑j=1

rj1vj , . . . ,

k∑j=1

rjkvj

).

Prove that a map ω : V ×k → R is an alternating k-form if and only if, for everyk × k-matrix R:

ω RV = det(R) · ω.

Exercise 12.6. The rank of a 2-form ω ∈∧2

V ∗ is the largest even integer2r such that ωr 6= 0. Prove that ω has rank 2r if and only if V ∗ has a basise1, . . . , er, f1, . . . , fr, g1, . . . , gc such that

ω = e1 ∧ f1 + . . .+ er ∧ fr.

Exercise 12.7. Let V be a vector space of dimension n. Let v ∈ V and ξ ∈ V ∗such that ξ(v) = 1. Consider the operators: interior product by v

ιv :∧V ∗ −→

∧V ∗, ω 7→ ιvω,

and exterior products by ξ

eξ :∧V ∗ −→

∧V ∗, ω 7→ ξ ∧ ω.

(a) Prove that these operators satisfy:

eξ ιv + ιv eξ = Id∧V ∗ .

(b) For ω ∈∧V ∗, prove ιvω = 0 if and only if there α ∈

∧V ∗ such that ω = ιvα.

(c) For ω ∈∧V ∗, prove ξ∧ω = 0 if and only if there β ∈

∧V ∗ such that ω = ξ∧β.

LECTURE 12. 111

Exercise 12.8. Let V be a vector space of dimension n. An element α ∈∧V ∗

is called invertible if there exists β ∈∧V ∗ such that α ∧ β = 1. Prove that an

element

α = α0 + α1 + . . .+ αn ∈∧V ∗, αi ∈

i∧V ∗

is invertible iff α0 6= 0 (Hint: think about the geometric series 11−q = 1+q+q2+. . .).

Exercise 12.9. Consider the differential forms, the vector field and the map below:

α = cos(y)dx+ dz − e−w2

dw ∈ Ω1(R4),

β = x2dx ∧ dy + w7dw ∧ dz ∈ Ω2(R4),

γ = w2dz + dz ∧ dy ∧ (dw + ydw ∧ dx) ∈ Ω(R4),

X = x∂

∂y− y ∂

∂z+ y3 ∂

∂w,

f : R2 → R4, f(u, v) = (u2, euv, u− v4, sin(v)).

(a) Calculate the nine exterior products:

α ∧ α α ∧ β α ∧ γβ ∧ α β ∧ β β ∧ γγ ∧ α γ ∧ β γ ∧ γ

.

(b) Calculate the three interior products: ιXα, ιXβ, ιXγ.(c) Calculate the pullback along f of the three forms: f∗(α), f∗(β), f∗(γ).(d) Construct, or prove that it is impossible to do so, two one-forms η, θ ∈ Ω1(R4)

such that β = η ∧ θ.

LECTURE 13

13.1. The exterior derivative

Recall that, with our usual identification TtR ∼= R, the differential at p ∈ M of asmooth map a ∈ C∞(M) can be viewed as a covector

dpa ∈ T ∗pM, dpa(v) = v(a), v ∈ TpM.

Moreover, the assignment p 7→ dpa defines a one-form da ∈ Ω1(M). Globally, weobtain a linear maps from functions to one-forms:

d : C∞(M) −→ Ω1(M), a 7→ da.

In local coordinates (U,ϕ = (x1, . . . , xm)) the differential is given by

da =

m∑i=1

∂a

∂xidxi.

The exterior derivative is an operator which extends the differential of functionsto forms of higher degree, and is uniquely determined by the properties below.

Theorem 13.1.1. For any manifold M there is a unique linear map

d : Ωk(M) −→ Ωk+1(M), k ≥ 0

satisfying the following properties:

(1) on zero-forms Ω0(M) = C∞(M) it is the differential of functions;(2) it squares to zero: d2α = 0, for all α ∈ Ω(M);(3) it is a graded derivation for the wedge product, i.e.

d(α ∧ β) = dα ∧ β + (−1)kα ∧ dβ,

for all α ∈ Ωk(M) and all β ∈ Ω(M).

Moreover, the following holds:

(4) if f : M → N is a smooth map, then df∗(α) = f∗(dα) for all α ∈ Ω(N).

Definition 13.1.2. The operator d : Ω(M)→ Ω(M) from Theorem 13.1.1 is calledthe exterior derivative.

Remark 13.1.3. It is important to note that (2), (3) and (4) are essentially equiv-alent to the following properties of smooth functions (this will be seen in the proof):

(2) partial derivatives commute;

113


(3) the Leibniz rule;(4) the chain rule.

We first prove Theorem 13.1.1 for open subsets of the Euclidean space.

Lemma 13.1.4. Theorem 13.1.1 holds for open subsets M ⊂ Rm and N ⊂ Rn.

Proof. Let us deduce a formula for d by using properties (1)-(3); this argumentwill imply uniqueness. Let α ∈ Ωk(M). To calculate dα we decompose α in thestandard basis:

(*) α =∑

i1<...<ik

ai1...ikdxi1 ∧ . . . ∧ dxik , ai1...ik ∈ C∞(M).

Using linearity of d and the derivation rule (3), we obtain

dα =∑

i1<...<ik

dai1...ik ∧ dxi1 ∧ . . . ∧ dxik + ai1...ik ∧ d(dxi1 ∧ . . . ∧ dxik

),

where by (1) the first term is determined since dai1...ik is the usual differential ofthe smooth map ai1...ik . For the second term, by applying (3), we obtain:

d(dxi1 ∧ . . . ∧ dxik

)= d

(dxi1

)∧ dxi2 ∧ . . . ∧ dxik − dxi1 ∧ d

(dxi2 ∧ . . . ∧ dxik

).

Since d extends the usual derivation (1), and since d2 = 0 (2), the first term is zero.By repeatedly applying this argument, the entire expression is zero.

We conclude that there is a unique linear operator d : Ω(M)→ Ω(M) satisfyingthe conditions (1)-(3), and that on a form α ∈ Ωk(M) as in (*) it acts by:

dα =∑

i1<...<ik

dai1...ik ∧ dxi1 ∧ . . . ∧ dxik =(**)

=∑

i1<...<ik

∑i

∂ai1...ik∂xi

dxi ∧ dxi1 ∧ . . . ∧ dxik .

Next, we check that this linear operator satisfies indeed the properties fromTheorem 13.1.1. Property (1) clearly holds.

We check (2)-(4) first on smooth functions. For a ∈ C∞(M), we have that:

d2a = d

(n∑i=1

∂a

∂xidxi

)=

n∑i=1

d

(∂a

∂xi

)∧ dxi =

n∑i=1

n∑j=1

∂2a

∂xi∂xjdxj ∧ dxi =

=

n∑i=1

∂2a

(∂xi)2dxi ∧ dxi +

∑i<j

∂2a

∂xi∂xj(dxj ∧ dxi + dxi ∧ dxj

)= 0,

where we have used that the partial derivatives commute, and the following alge-braic properties of the exterior algebra:

dxi ∧ dxi = 0 and dxi ∧ dxj = −dxj ∧ dxi.

The derivation rule (3), for a, b ∈ C∞(M), is immediate:

d(ab)(v) = v(ab) = v(a)b+ av(b) = (da · b+ a · db)(v), ∀ v ∈ TM.

Similarly, for f : N →M and a ∈ C∞(M), (4) holds by the chain rule:

f∗(da)(v) = da(df(v)) = d(a f)(v) = d(f∗a)(v), ∀ v ∈ TN.Next, we show that (3) holds in all degrees. Since the wedge product is bilinear,

and d is linear, it suffices to show (3) for elements of the form α = adxI ∈ Ωk(M)

LECTURE 13. 115

and β = bdxJ ∈ Ωl(M), where a, b ∈ C∞(M) and, for I = i1 < . . . < ik andJ = j1 < . . . < jl, we have denoted:

dxI := dxi1 ∧ . . . ∧ dxik , dxJ := dxj1 ∧ . . . ∧ dxjl ,

Note that d(dxI ∧ dxJ) = 0. Using (3) for functions, we obtain (3) in all degrees:

d(α ∧ β) = d(abdxI ∧ dxJ) = d(ab) ∧ dxI ∧ dxJ = ((da)b+ a(db)) ∧ dxI ∧ dxJ =

= da ∧ dxI ∧ bdxJ + (−1)kadxI ∧ db ∧ dxJ = dα ∧ β + (−1)kα ∧ dβ.

Again, by linearity of d, it suffices to prove (2) for α = adxI . Using (3) and(2) for functions, we obtain (2) in general:

d2(α) = d(da ∧ dxI

)= (d2a) ∧ dxI − da ∧ d(dxI) = 0.

By linearity of d and of the pullback map f∗, it suffices to check (4) for α =adxI . By (4) for functions, we have that

f∗(dxI) = df∗(xi1) ∧ . . . ∧ df∗(xik),

and therefore, applying (2) and (3) repeatedly,

df∗(dxI) =

k∑p=1

(−1)p+1df∗(xi1) ∧ . . . ∧ d2f∗(xip) ∧ . . . ∧ df∗(xik) = 0.

Using this, and again (4) for functions, we obtain that (4) holds in general:

df∗(α) = d(f∗(a)f∗(dxI)

)= d(f∗a) ∧ f∗(dxI) + f∗(a)df∗(dxI) =

= f∗(da) ∧ f∗(dxI) = f∗(da ∧ dxI) = f∗(dα).

Next, we prove existence of the operator d for any manifold:

Lemma 13.1.5. For any manifold M there exists an operator d : Ω(M)→ Ω(M)with the properties (1)-(4) from Theorem 13.1.1.

Proof. For α ∈ Ω(M), let dα be such that, for any chart (U,ϕ) on M ,

(dα)|U =(ϕ∗ dR (ϕ−1)∗

)(α|U ),

where on the right-hand-side we have denoted by dR the operator constructed inLemma 13.1.4 for forms on ϕ(U) ⊂ Rm. For this definition we need to check that,given a second chart (V, ψ) we obtain the same value for dα on U ∩ V . Considerthe change of coordinates diffeomorphism between open sets in Rm:

f : ψ(U ∩ V ) ∼−→ ϕ(U ∩ V ), f = ϕ ψ−1.

By Lemma 13.1.4, we have that the operator dR satisfies property (4); therefore:

dR f∗((ϕ−1)∗(α|U∩V )

)= f∗ dR ((ϕ−1)∗(α|U∩V )

).

Writing f∗ = (ψ−1)∗ ϕ∗, and composing both sides with ψ∗ on the left, we obtain:(ψ∗ dR (ψ−1)∗

)(α|U∩V ) =

(ϕ∗ dR (ϕ−1)∗

)(α|U∩V ).

In other words, the two definition of dα coincide on U ∩ V . We conclude that d iswell-defined. Clearly, it suffices to check properties (1)-(4) locally; and these followfrom the respective properties of dR which were proven in Lemma 13.1.4. Thisconcludes the proof.


Next, we show that any operator satisfying (3) is local.

Lemma 13.1.6. Any linear operator D : Ω(M) → Ω(M) satisfying (3) is local,i.e. Dα|U depends only on α|U for any α ∈ Ω(M) and any open set U ⊂M .

Proof. We need to show that, if α1, α2 ∈ Ω(M) satisfy α1|U = α2|U , then alsoD(α1)|U = D(α2)|U . Since D is linear, by taking the difference of these forms,it suffices to show that α|U = 0 implies that Dα|U = 0. Let p ∈ U , and letχ ∈ C∞(M) be a smooth function such that χ(p) = 1 and supp(χ) ⊂ U (this existsby Lemma 4.1.2). Then, χα = 0. Applying D, we obtain:

0 = D(χα) = Dχ ∧ α+ χDα.

Evaluating at p, and using that αp = 0 and χ(p) = 1, we obtain that (Dα)p = 0.

The following concludes the proof of Theorem 13.1.1

Lemma 13.1.7. The operator d : Ω(M)→ Ω(M) constructed in Lemma 13.1.5 isthe unique linear map satisfying (1)-(3).

Proof. Assume D : Ω(M) → Ω(M) is a second linear map satisfying (1)-(3). Asin the proof of Lemma 13.1.4, (1)-(3) imply that D = d on forms of the formα = a0da1 ∧ . . .∧dak, where ai ∈ C∞(M). Consider now a general α ∈ Ω(M). Letp ∈M and choose a chart (U, (x1, . . . , xm)) around p, and write

α|U =∑

i1<...<ik

ai1...ikdxi1 ∧ . . . ∧ dxik .

By Lemma 4.1.2, there are function xi, ai1...ik ∈ C∞(M) and an open set p ∈ V ⊂ Usuch that xi|V = xi|V and ai1...ik |V = ai1...ik |V . Consider the form:

α :=∑

i1<...<ik

ai1...ikdxi1 ∧ . . . ∧ dxik .

By the above, dα = Dα. Since α|V = α|V , Lemma 13.1.6 gives:

(Dα)|V = (Dα)|V = (dα)|V = (dα)|V .Thus, Dα and dα are equal in a neighborhood of p. We conclude that D = d.

13.2. De Rham cohomology

Differential forms can be used to extract topological information about manifoldsthrough the so-called de Rham cohomology.

Definition 13.2.1.

• A k-form ω ∈ Ωk(M) satisfying dω = 0 is called closed.• A k-form ω ∈ Ωk(M) for which there exists α ∈ Ωk−1(M) such that

dα = ω is called exact.

Since d2 = 0, every exact form is closed.

Definition 13.2.2.

• The k-th de Rham cohomology group of M is the quotient of the spaceof closed forms modulo the space of exact forms:

HkdR(M) :=

closed k-forms

exact k-forms=

ker(d : Ωk(M)→ Ωk+1(M))

im(d : Ωk−1(M)→ Ωk(M)).

LECTURE 13. 117

• The de Rham cohomology of M is the direct sum of all these spaces:

HdR(M) :=⊕k≥0

HkdR(M) =

closed forms

exact forms=

ker d

im d.

• The cohomology class of a closed k-form ω is the equivalence class:

[ω] = ω + dΩk−1(M) ∈ HkdR(M).

• The k-th Betti number of M is the dimension of the k-th de Rhamcohomology group of M (provided this vector space is finite dimensional):

bk(M) := dim(HkdR(M)

).

Remark 13.2.3. The general algebraic framework in which cohomology groupsare considered is that of a chain complex (C•, δ), i.e. a sequence of vector spacesCkk∈Z together with linear maps δk : Ck → Ck+1 satisfying δk+1 δk = 0. Thecohomology of the chain complex are the quotient spaces:

Hk(C•, δ) :=ker(δk)

im(δk−1).

Remark 13.2.4. We have that H0dR(M) is the space of function a ∈ C∞(M)

satisfying da = 0. In a chart, this condition is equivalent to all partial derivativesof a being zero. Therefore, da = 0 is equivalent to a being a locally constantfunction. Such functions are uniquely determined by their values on the connectedcomponents of M ; thus, H0

dR(M) = Maps(π0(M),R) where π0(M) denotes the setof connected components of M . So, if finite, b0(M) counts connected componentsof M .

Remark 13.2.5. Let M be a connected manifold of dimension m. The top degreecohomology has the following interpretation: if M is compact and orientable (to bedefined in the following lecture), then Hm

dR(M) = R, otherwise, HmdR(M) = 0.

Example 13.2.6. Since S1 is connected, we have that H0dR(S1) ∼= R. Let 0 ≤ θ ≤

2π be the “angle coordinate” on S1. Every one-forms on S1 can be written as

f(θ)dθ ∈ Ω1(S1), f ∈ C∞(S1).

The notation dθ is a bit misleading, because dθ is not an exact 1-form (θ is not asmooth function on S1); in fact, let us prove that [dθ] is a basis of H1

dR(S1), i.e.

H1dR(S1) = R[dθ] 6= 0.

Clearly, every one-form is closed. The exact one-forms have the form

f(θ)dθ = d(g(θ)) =∂g

∂θ(θ)dθ,

for some g(θ) ∈ C∞(S1). Note that an exact one-form integrates to 0 when viewedas a smooth function on [0, 2π]∫ 2π

0

f(τ)dτ = g(2π)− g(0) = 0.

Conversely, given f ∈ C∞(S1), we regard f as a 2π-periodic function on R. Then

g(x) :=

∫ x

0

f(τ)dτ,


defines a smooth function g : R→ R, which satisfies

g(x+ 2π) = g(x) +

∫ 2π

0

f(τ)dτ.

Thus, g ∈ C∞(S1) iff∫ 2π

0f(τ)dτ = 0; and so, f(θ)dθ is exact iff

∫ 2π

0f(τ)dτ = 0.

In particular, [dθ] 6= 0 in H1(S1). One the other hand, for f(θ)dθ ∈ Ω1(S1), we

have that (f(θ)− c2π )dθ has integral 0 for c =

∫ 2π

0f(τ)dτ , and so [f(θ)dθ] = c

2π [dθ].We conclude that:

H1(S1) = R[dθ] 6= 0.

Remark 13.2.7. Assume that M is connected, and let p ∈ M . It can be proventhat H1

dR(M) is isomorphic to the vector space of group homomorphisms from thefundamental group (π1(M,p), ∗) to (R,+):

H1dR(M) ∼= Hom(π1(M,p);R).

Even if the reader is not familiar with the notion of fundamental group, it is im-portant to note that this notion is purely topological; it uses only the topology ofM and not its differentiable structure. In fact, this is true in all degrees; namely,even though the groups HdR(M) are constructed using smooth forms, they encodeinformation only about the topology of M . We state the precise result below:

Theorem ♣ 13.2.8 (De Rham). We have that HkdR(M) ∼= Hom(Hk(M,Z);R)1.

Next, we give some important properties of de Rham cohomology.

Proposition 13.2.9. The wedge product on differential forms Ω(M) induces aproduct on de Rham cohomology HdR(M):

[α] ∧ [β] = [α ∧ β],

which turns (HdR(M),+,∧) into an associative algebra.

Proof. If α and β are closed forms, using that d is a derivation for the wedgeproduct (Theorem 13.1.1 (3)), we have that α ∧ β is closed; hence the class [α ∧ β]exists. Next, we show that [α ∧ β] depends only on [α], [β] ∈ HdR(M), and not onthe particular representatives α, β ∈ Ω(M). Any other representatives are of theform α′ = α+ dη and β′ = β + dθ. Again by the derivation rule, we have that:

α′ ∧ β′ = α ∧ β′ + d(η ∧ β′) = α ∧ β + d((−1)kα ∧ θ + η ∧ β′),

where k is the degree of α. Thus [α′ ∧ β′] = [α ∧ β]. This proves that the productis well-defined. Clearly, associativity and the other algebraic properties (unital,graded commutativity) are inherited from the wedge product on Ω(M).

Proposition 13.2.10. The pullback of differential forms along any smooth mapf : M → N induces an algebra homomorphism in de Rham cohomology

f∗ : HdR(N) −→ HdR(M), f∗[ω] := [f∗ω].

Moreover, if g : N → P is a second smooth map, then

f∗ g∗ = (g f)∗ : HdR(P ) −→ HdR(M).

1The group Hk(M,Z) is called the k-th singular homology of M ; it will not be introduced in thiscourse, but it is important to note that it is defined purely in topological terms

LECTURE 13. 119

Proof. By item (4) of Theorem 13.1.1, we have that f∗ : Ω(N) → Ω(M) sendsclosed forms to closed forms, and exact forms to exact forms; thus it induces indeeda map f∗ : HdR(N)→ HdR(M). The fact that f∗ is an algebra homomorphism ondifferential forms, implies that it is an algebra homomorphism in cohomology, andsimilarly, the equality f∗ g∗ = (g f)∗ holds already at the level of forms.

An immediate consequence of the Proposition, is that de Rham cohomology isan invariant of a manifold:

Corollary 13.2.11. If M and N are diffeomorphic manifolds, then HdR(M) andHdR(N) are isomorphic algebras.

Theorem 13.2.12 (Homotopy invariance of de Rham cohomology). Let ft : M →N be a family of smooth maps depending smoothly on t ∈ [0, 1], then

f∗0 = f∗1 : HdR(N) −→ HdR(M).

Proof. We can assume that ft is defined for all t ∈ R (first, extend the family to asmall open interval containing [0, 1], and then use a diffeomorphism of this intervalto R which fixes [0, 1]). The family can be regarded as a smooth map

f : R×M −→ N.

Differential forms on R×M can be uniquely decomposed as

dt ∧ αt + βt ∈ Ω(R×M),

where αt, βt are smooth families of differential forms on M . We denote the exteriorderivative of forms on R×M by dtot, and the exterior derivatives on M and N asusually by d. In the above decomposition, dtot becomes:

dtot (dt ∧ αt + βt) = −dt ∧ dαt + dt ∧ d

dtβt + dβt.

Note that the pullback of ω ∈ Ωk(N) by f decomposes as follows:

f∗(ω) = dt ∧Ht(ω) + f∗t (ω),

where Ht(ω) ∈ Ωk−1(M) is a smooth family of k−1-forms on M . Next, we calculateboth sides of the equality dtotf

∗(ω) = f∗(dω) (Theorem 13.1.1 (4)):

dtot

(dt ∧Ht(ω) + f∗t (ω)

)= dt ∧

(− dHt(ω) +

d

dtf∗t (ω)

)+ df∗t (ω)

f∗(dω) = dt ∧Ht(dω) + f∗t (dω).

Using that also f∗t commutes with the exterior derivative, and the uniqueness ofthe decomposition of differential forms on R×M , we obtain:

d

dtf∗t (ω) = Ht(dω) + dHt(ω).

Integrating from 0 to 1:

(*) f∗1 (ω)− f∗0 (ω) = h d(ω) + d h(ω),

where h denotes the following linear operator:

h : Ωk(N) −→ Ωk−1(M), h(η) :=

∫ 1

0

Ht(η)dt.


This operator is called a homotopy operator, and relation (*) is all what is neededto conclude the proof. Namely, if [ω] ∈ Hk

dR(M), then dω = 0, and so

f∗1 ([ω]) = [f∗1 (ω)] = [f∗0 (ω) + d h(ω)] = f∗0 ([ω]).

Thus f1 and f0 induce the same map in cohomology.

Remark 13.2.13. The operator Ht : Ωk(N)→ Ωk−1(M) can be given explicitly:

Ht(ω) = f∗t (ι ddt ft

ω).

Therefore, the homotopy operators are given by

h(ω) =

∫ 1

0

f∗t (ι ddt ft

ω)dt.

Homotopy invariance can be used to calculate the cohomology of Rn.

Corollary 13.2.14 (The Poincare Lemma). We have that

HkdR(Rn) =

0, k 6= 0;R, k = 0.

Proof. Consider the family of maps ft : Rn → Rn, ft(x) := tx. Then f1 = id,and so f∗1 [ω] = [ω] for all [ω] ∈ Hk

dR(Rn). On the other hand, f0(x) = 0 for allx ∈ Rn, and so f∗0 [ω] = 0 for all [ω] ∈ Hk

dR(Rn) and k > 0, and f∗0 (c) = c, forc ∈ R = H0(Rn). Applying Theorem 13.2.12, we obtain the result.

More generally, we have that:

Corollary 13.2.15. Let π : E → M be a vector bundle. Then π∗ : HdR(M) →HdR(E) is a linear isomorphism.

Example 13.2.16. We show that

H1dR(Sn) = 0, n ≥ 2.

Consider a closed one-form α ∈ Ω1(Sn). We will prove that α is exact. DecomposeSn = UN ∪ US , where UN is the complement of the north pole, and US is thecomplement of the south pole. Since UN ∼= Rn we have that [α|UN ] = 0 ∈ H1

dR(UN ).Therefore, there exists fN ∈ C∞(UN ) such that dfN = α|UN . Similarly, we findfS ∈ C∞(US) such that dfS = α|US . On the sphere without the two poles, we havethat

d(fN |UN∩US − fS |UN∩US ) = α|UN∩US − α|UN∩US = 0.

This implies that fN |UN∩US−fS |UN∩US is a constant function c ∈ R (this is becauseUN∩US is connected; here we are using that n ≥ 2). Therefore, the following definesa smooth function on Sn:

f : Sn −→ R, f |UN := fN − c, and f |US = fS ,

which clearly satisfies df = α.

The general structure of the cohomology of spheres is given below (the proofis not difficult; see e.g. [3]):

Theorem ♣ 13.2.17. The cohomology of spheres is given by:

HkdR(Sn) =

0, 0 < k < n;R, k = 0, n.

LECTURE 13. 121

13.3. Cartan calculus

In this section we introduce the Lie derivative of a differential form, and we proveseveral relations between this operator, the exterior derivative and the interiorproduct. This entire algebraic machinery, including the definition of the exteriorderivative, was developed by Elie Cartan, and therefore it is commonly called Car-tan calculus.

Definition 13.3.1. The Lie derivative of differential forms along a vector fieldX ∈ X(M) is the operator

LX : Ωk(M) −→ Ωk(M)

LXα :=d

dt(φtX)∗(α)

∣∣t=0

= limt→0

1

t((φtX)∗(α)− α).

Theorem 13.3.2. The following relations hold:

(1) LX(α ∧ β) = LXα ∧ β + α ∧LXβ(2) LX d = d LX

(3) LX ιY − ιY LX = ι[X,Y ]

(4) LX = d ιX + ιX d (Cartan’s magic formula)(5) LX LY −LY LX = L[X,Y ]

for all X,Y ∈ X(M) and all α, β ∈ Ω(M).

Proof. Since the pullback by the flow of X preserves the wedge product, we obtain:

(φtX)∗(α ∧ β) = (φtX)∗(α) ∧ (φtX)∗(β).

Taking the derivative at t = 0, and using that φ0X = id, we obtain (1).

For (2), we use that the pullback commutes with the exterior derivative:

(φtX)∗(dα) = d(φtX)∗(α),

and take the derivative at t = 0.Formula (3) is a version of (1) for the interior product instead of the exterior

product. Namely, directly from the definitions of the pullback, it follows that

(φtX)∗(ιY α) = ι(φtX)∗(Y )(φtX)∗(α).

Taking the derivative at t = 0, we obtain that:

LX(ιY α) = ιLXY α+ ιY (LXα).

Using that LXY = [X,Y ] (see Proposition 10.3.3), we obtain (3).Denote the operator in (4) by DX := d ιX + ιX d. By using the derivation

rule for d and ιX , one easily checks that DX satisfies the same derivation rule asLX (this follows also from the comments in the previous subsection)

DX(α ∧ β) = DXα ∧ β + α ∧DXβ.

Using that d2 = 0, we obtain that DX commutes with d:

DX d = (d ιX + ιX d) d = d ιX d = d (d ιX + ιX d) = d DX .

Next, note that DX and LX coincide on functions a ∈ C∞(M):

DX(a) = (d ιX + ιX d)(a) = ιXda = LX(a).

Finally, since both operators LX and DX are local and linear, it suffices that showthat they coincide on forms of the form a0da1 ∧ . . .∧ dak, with ai ∈ C∞(M). This


follows by applying the derivation rule, that both commute with d, and that bothact the same on functions.

Formula (5) follows by using (2)-(4):

L[X,Y ] = d ι[X,Y ] + ι[X,Y ] d =

= d (LX ιY − ιY LX) + (LX ιY − ιY LX) d =

= LX d ιY − d ιY LX + LX ιY d− ιY d LX =

= LX (d ιY + ιY d)− (d ιY + ιY d) LX =

= LX LY −LY LX .

A differential k-form ω ∈ Ωk(M) can be viewed also as a C∞(M)-multilineark-form on the C∞(M)-module X(M) with values in C∞(M):

ω :X(M)× . . .× X(M)︸︷︷︸ −→ C∞(M)

k × X(M)

ω(X1, . . . , Xk)(p) := ωp((X1)p, . . . , (Xk)p),

for X1, . . . , Xk ∈ X(M). This allows for a coordinate-free definition of the exteriorderivative:

Theorem 13.3.3. The exterior derivative satisfies:

dω(X0, . . . , Xk) =

k∑i=0

(−1)iLXi(ω(X0, . . . , Xi, . . . , Xk))

+∑

0≤i<j≤k

(−1)i+jω([Xi, Xj ], X0, . . . , Xi, . . . , Xj , . . . , Xk),

for all ω ∈ Ωk(M) and all X0, . . . , Xk ∈ X(M), where the notation Xi means thatthe vector field Xi is being skipped.

The proof is left as an exercise. In low degrees the formula gives:

da(X0) =LX0(a),

dθ(X0, X1) =LX0(θ(X1))−LX1

(θ(X0))− θ([X0, X1]),

dω(X0, X1, X2) =LX0(ω(X1, X2))−LX1

(ω(X0, X2)) + LX2ω(X0, X1)

− ω([X0, X1], X2) + ω([X0, X2], X1)− ω([X1, X2], X0),

for a ∈ C∞(M), θ ∈ Ω1(M) and ω ∈ Ω2(M).

13.4. The algebra of graded derivations

We explain the algebraic framework in which the relations of Theorem 13.3.2 fitbest. A graded algebra is an algebra (A,+, ·) which decomposes as a direct suminto vector space:

A =⊕k∈Z

Ak,

and satisfiesa · b ∈ Ak+l, for all a ∈ Ak, b ∈ Al.

LECTURE 13. 123

For this discussion, it is not important whether A is unital, associative, or (graded)commutative. A derivation of degree p ∈ Z of A is a linear map D : A → A,satisfying D(Ak) ⊂ Ak+p, and the derivation rule:

D(a · b) = D(a) · b+ (−1)kpa ·D(b), for all a ∈ Ak, b ∈ A.Let us denote by Derp(A) the space of all derivations of degree p of A. The gradedcommutator of D1 ∈ Derp(A) and D2 ∈ Derq(A) is the map

[D1, D2]c := D1 D2 − (−1)pqD2 D1 : A −→ A.

It can be easily checked that [D1, D2]c is a derivation of degree p+ q. Consider

Der(A) :=⊕p∈Z

Derp(A).

This vector space endowed with the bilinear extension of the commutator is thegraded algebra of derivations of A:

(Der(A),+, [·, ·]c).The commutator satisfies the following relations:

[D1, D2]c = −(−1)pq[D2, D1]c,

[D1, [D2, D3]c]c = [[D1, D2]c, D3]c + (−1)pq[D2, [D1, D3]c]c,

for all D1 ∈ Derp(A), D2 ∈ Derq(A) and D3 ∈ Der(A). These relations are calledgraded commutativity and the graded Jacobi identity, respectively. Notethat the second relation is equivalent to the operator [D1, ·]c being a derivation ofdegree p of Der(A), i.e. [D1, ·]c ∈ Derp(Der(A)). A graded algebra which satisfiesthese two axioms is called a graded Lie algebra.

In our case, A = Ω(M), with multiplication the usual wedge product. In thelanguage of graded algebras, the derivation rule for the interior product and theexterior derivative become:

ιX ∈ Der−1(Ω(M)) and d ∈ Der1(Ω(M)).

Note that we have the relations d2 = 0 and ιX ιY = −ιY ιX can be expressedusing the graded commutator:

[d, d]c = 2d2 = 0 and [ιX , ιY ]c = ιX ιY + ιY ιX = 0.

Moreover, Theorem 13.3.2 can be restated as follows:

(1) LX ∈ Der0(Ω(M))(2) [LX ,d]c = 0(3) [LX , ιY ]c = ι[X,Y ]

(4) LX = [ιX ,d]c(5) [LX ,LY ]c = L[X,Y ]

13.5. Exercises

The exterior derivative of a function a ∈ C∞(M) can be calculated as follows:

da(v) =d

dta(γ(t))

∣∣t=0

,

where t 7→ γ(t) is a curve on M , with dγdt (0) = v. The following exercise extends

this result to higher degrees:


Exercise 13.1. Let v1, . . . , vk ∈ TpM be k-vectors. A small parallelotope withdirections v1, . . . , vk is a smooth map (t1, . . . , tk) 7→ γ(t1, . . . , tk) defined on an openneighborhood of 0 in Rk such that

γ(0, . . . , 0) = p,∂γ

∂ti(0, . . . , 0) = vi.

Prove that for any k − 1-form ω ∈ Ωk−1(M), the exterior derivative is given by:

dω(v1, . . . , vk) =

k∑i=1

(−1)i−1 ∂

∂tiω( ∂γ∂t1

, . . . ,∂γ

∂ti, . . . ,

∂γ

∂tk

)∣∣(t1,...,tk)=(0,...,0)

.

Hint: Calculate dγ∗(ω), and use item (4) from Theorem 13.1.1.

Exercise 13.2. Let U ⊂ R3 be an open set. Recall the standard operators fromCalculus: the gradient, the curl and the divergence:

grad = ∇ : C∞(U) −→ X(U),

curl = ∇× : X(U) −→ X(U),

div = ∇· : X(U) −→ C∞(U),

which for f ∈ C∞(U) and F = Fx∂∂x + Fy

∂∂y + Fz

∂∂z ∈ X(U) are given by:

∇f =∂f

∂x

∂

∂x+∂f

∂y

∂

∂y+∂f

∂z

∂

∂z,

∇× F =(∂Fz∂y− ∂Fy

∂z

) ∂∂x

+(∂Fx∂z− ∂Fz

∂x

) ∂∂y

+(∂Fy∂x− ∂Fx

∂y

) ∂∂z,

∇ · F =∂Fx∂x

+∂Fy∂y

+∂Fz∂z

.

Denote by 〈·, ·〉 the standard dot-product on R3 = TpU .

(a) Show that the following give C∞(U)-linear isomorphisms:

A : X(U) ∼−→ Ω1(U), A(F ) := 〈F, ·〉B : X(U) ∼−→ Ω2(U), B(F ) := ιF (dx ∧ dy ∧ dz)

C : C∞(U) ∼−→ Ω3(U), C(f) := fdx ∧ dy ∧ dz.

(b) Prove that the following diagram commutes:

C∞(U)∇ //

id

X(U)∇× //

A

X(U)∇· //

B

C∞(U)

CC∞(U)

d // Ω1(U)d // Ω2(U)

d // Ω3(U)

.

(c) Deduce the classical relations:

curl(grad(f)) = 0 and div(curl(F )) = 0,

and show that under the map A conservative vector fields correspond to exact1-forms, irrotational vector fields correspond to closed 1-forms; and under themap B divergence free vector fields correspond to closed 2-forms. (Hint: opena calculus book.)

LECTURE 13. 125

Exercise 13.3. (a) For ω ∈ Ωk(M) and X0, . . . , Xk ∈ X(M), prove that:

LX0(ω(X1, . . . , Xk)) = LX0

(ω)(X1, . . . , Xk) +

k∑i=1

ω(X1, . . . , [X0, Xi], . . . , Xk).

Hint: either directly from the definitions, or use item (3) from Theorem 13.3.2.(b) Prove Theorem 13.3.3.

Hint: argue by induction; use (a) and item (4) from Theorem 13.3.2.

Exercise 13.4. Let U ⊂ C be an open set, and consider a smooth function

f : U → C, f(z) = u(z) + iv(z), u(z), v(z) ∈ R.Consider the one-forms θ1, θ2 ∈ Ω1(U):

θ1(x, y) := v(x+ iy)dx+ u(x+ iy)dy, θ2(x, y) := u(x+ iy)dx− v(x+ iy)dy.

Prove that f is a holomorphic map if and only if both θ1 and θ2 are closed.

Exercise 13.5. Prove Corollary 13.2.15.

Exercise 13.6. (a) For 1 ≤ k ≤ n, given a closed k-form on Rn:

ω =∑

i1<...<ik

ωi1,...,ik(x1, . . . , xn)dxi1 ∧ . . . ∧ dxik ∈ Ωk(Rn)

give an explicit formula for a k − 1-form α ∈ Ωk−1(Rn) such that dα = ω.(Hint: extract a formula from the proofs of Theorem 13.2.12 and of Corollary13.2.14, and then check directly that it works.)

(b) Prove Corollary 13.2.14 directly, by using (a).

Exercise 13.7. Calculate the de Rham cohomology of the Mobius band.

LECTURE 14

In this lecture we construct the density bundle of a smooth manifold. Thedensity bundle and smooth densities give a precise geometric meaning to the notionof “volume element” dx1 . . . dxn which appears when writing integrals∫

f(x1, . . . , xn)dx1 . . . dxn.

We will show that smooth densities, i.e. smooth section of the density bundle, canbe canonically integrated. The discussion below follows [10].

14.1. Volume elements

Let us review some elementary facts about volumes of n-dimensional parallelotope,i.e. the n-dimensional generalization of the notions of: line segment, parallelogram,parallelepiped. Let P ⊂ Rn be an n-dimensional parallelotope which has a corner inthe origin of Rn, and is spanned by n linearly independent vectors v1, . . . , vn ∈ Rn:

P = t1v1 + . . .+ tnvn : 0 ≤ ti ≤ 1.

Note that, up to a permutation, P determines the basis (v1, . . . , vn). Writing vj =(a1j , . . . , a

nj ), we parameterize P by the standard n-cube [0, 1]n

A : [0, 1]n ∼−→ P, (t1, . . . , tn) 7→ (x1, . . . , xn), xi =

n∑j=1

aijtj .

The volume of P is simply the absolute value of the determinant of A:

Vol(P ) =

∫P

dx1 . . . dxn =

∫[0,1]n

|det(A)|dt1 . . . dtn = |det(A)|,

where we used the change of coordinate formula for multiple integrals. More gen-erally, the same argument applies to the following situation: if Q is any othern-dimensional parallelotope spanned by w1, . . . , wn, and B : Rn → Rn is linearisomorphism satisfying B(vi) = wi, for 1 ≤ i ≤ n (i.e. B is the transition matrixbetween the two bases) then we have that:

Vol(Q) = |det(B)|Vol(P ).

The following is the abstract notion of a volume of parallelotopes:

127


Definition 14.1.1. Let V be an n-dimensional vector space. A volume elementon V is a map

µ :V × . . .× V︸︷︷︸ −→ R

n × V

such that, for all linear maps A : V → V and all v1, . . . , vn ∈ V :

µ(Av1, . . . , Avn) = |det(A)|µ(v1, . . . , vn).

We denote the space of volume elements on V by Ve(V ).

Up to a constant, there is a unique non-zero volume element:

Lemma 14.1.2. Let e1, . . . , en be a basis of the vector space V . Then there is aunique volume element µ0 ∈ Ve(V ) such that µ0(e1, . . . , en) = 1. Moreover, for anyother µ ∈ Ve(V ) there is a unique c ∈ R such that µ = cµ0; in other words thevolume elements on V form a 1-dimensional vector space with basis µ0.

Proof. Let v1, . . . , vn ∈ V , and define µ0(v1, . . . , vn) = |det(A)|, where A : V → Vis the linear map determined by Aei = vi, for 1 ≤ i ≤ n. Note that µ0 is indeed avolume element: for any linear map B : V → V , we have that

µ0(Bv1, . . . , Bvn) = µ0(BAe1, . . . , BAen) = |det(BA)| == |det(B)||det(A)| = |det(B)|µ0(v1, . . . , vn).

Consider an arbitrary µ ∈ Ve(V ), and denote c := µ(e1, . . . , en). Let v1, . . . , vn ∈V , and let A : V → V be such that Aei = vi, for 1 ≤ i ≤ n. Then, we have that

µ(v1, . . . , vn) = |det(A)|µ(e1, . . . , en) = cµ0(v1, . . . , vn).

Hence µ = cµ0. This shows that Ve(V ) is indeed 1-dimensional, and also that µ0

is the unique volume element such that µ0(e1, . . . , en) = 1.

14.2. The density bundle

Let M be a manifold. For p ∈M , denote the space of volume elements on TpM by

Dp(M) := Ve(TpM).

The density bundle of M is the union of all these vector spaces:

D(M) :=⊔p∈M

Dp(M).

Let (U,ϕ = (x1, . . . , xm)) be a chart around a point p ∈ M . Lemma 14.1.2 showsthat there exists a unique volume element

|dx1 ∧ . . . ∧ dxm|p ∈ Dp(M)

such that

|dx1 ∧ . . . ∧ dxm|p( ∂

∂x1|p, . . . ,

∂

∂xm|p)

= 1,

and moreover, that |dx1 ∧ . . . ∧ dxm|p forms a basis of Dp(M).Consider a second chart (V, ψ = (y1, . . . , ym)), with p ∈ U ∩ V . Then,

∂

∂yi∣∣p

=

m∑j=1

∂xj

∂yi(p)

∂

∂xj∣∣p,

LECTURE 14. 129

and so, by Lemma 14.1.2, this implies that:

|dx1 ∧ . . . ∧ dxm|p( ∂

∂y1

∣∣p, . . . ,

∂

∂ym∣∣p

)=

∣∣∣∣∂(x1, . . . , xm)

∂(y1, . . . , ym)(p)

∣∣∣∣ ,where we have denoted by ∂(x1,...,xm)

∂(y1,...,ym) the determinant of the Jacobian matrix of

ϕ ψ−1, i.e.

∂(x1, . . . , xm)

∂(y1, . . . , ym)(p) = det(dpϕ (dpψ)−1) = det

(∂xj∂yi

(p)).

Thus we have obtained the following change of coordinate formula:

|dx1 ∧ . . . ∧ dxm|p =

∣∣∣∣∂(x1, . . . , xm)

∂(y1, . . . , ym)(p)

∣∣∣∣ · |dy1 ∧ . . . ∧ dym|p.

Using this, the result below can be proven exactly like Theorem 8.1.2:

Theorem 14.2.1. The density bundle π : D(M) → M is a smooth vector bundleof rank 1. The vector bundle structure is uniquely determined by the condition that,for every chart (U,ϕ = (x1, . . . , xm)) on M , |dx1 ∧ . . . ∧ dxm| : U → D(M) is asmooth section.

14.3. Densities

Definition 14.3.1. The space of sections of the density bundle will be denoted by

D(M) := Γ(D(M)).

Sections of the density bundle are called smooth densities (or 1-densities, or justdensities) on M . A density µ ∈ D(M) is said to be positive, if µp(v1, . . . , vm) > 0for every p ∈M and every basis v1, . . . , vm of TpM .

Given a chart (U,ϕ = (x1, . . . , xm)) on M , the standard volume element definesa positive density over U

|dx1 ∧ . . . ∧ dxm| ∈ Γ(D(M)|U ) = D(U).

Any other density µ ∈ D(U) over U is of the form

µp = f(p)|dx1 ∧ . . . ∧ dxm|p,

for a unique smooth map f ∈ C∞(U).Next, we show that D(M) is a trivializable vector bundle (however, the trivi-

alization is not canonical).

Proposition 14.3.2. On any manifold M there exists a positive density µ ∈ D(M).In particular, the density bundle D(M) is trivializable.

Proof. Consider an atlas on M , (Ui, ϕi)i∈I , and let ρii∈I be a subordinatepartition of unity. Let µi = |dx1

ϕi ∧ . . . ∧ dxmϕi | ∈ D(Ui) be the local densitycorresponding to the chart ϕi. It is easy to check that µ :=

∑i∈I ρiµi is a positive

density on M .


14.4. Integration of densities

We define the pullback of densities under diffeomorphisms

Definition 14.4.1. The pullback of a smooth density µ ∈ D(N) along a localdiffeomorphism ϕ : M → N is the smooth density ϕ∗(µ) ∈ D(M) defined by:

ϕ∗(µ)p(v1, . . . , vm) := µϕ(p)(dpϕ(v1), . . . ,dpϕ(vm)).

Let us write the pullback of densities in local coordinates. Let U ⊂ Rm be anopen set, and let µ ∈ D(U) be a density on U . We can write:

µ = f(x1, . . . , xm)|dx1 ∧ . . . ∧ dxm|, f ∈ C∞(U).

Consider a diffeomorphism ϕ : V ∼−→ U , where V ⊂ Rm is another open set. Then,

ϕ∗(µ) = f(ϕ(y1, . . . , ym))

∣∣∣∣det

(∂ϕi

∂yj

)∣∣∣∣ · |dy1 ∧ . . . ∧ dym|.

Note that this is precisely the change of variables formula in multiple integrals!More precisely, let us recall the following:

Theorem 14.4.2 (Substitution rule for multiple integrals). Let U, V ⊂ Rm beopen sets, and let ϕ : V → U be a diffeomorphism. If f ∈ C∞(U) is a compactlysupported function on U , then we have that∫

U

f(x1, . . . , xm)dx1 . . . dxm =

∫V

f(ϕ(y1, . . . , ym))

∣∣∣∣det

(∂ϕi

∂yj

)∣∣∣∣dy1 . . . dym.

Let Dc(M) denote the space of compactly supported densities on M . Com-pactly supported densities can be integrated canonically:

Theorem 14.4.3. For each manifold M , there is a linear map∫M

: Dc(M) −→ R,

which is uniquely determined by the following conditions:

(1) It extends the usual Riemann/Lebesgue integration of compactly supported func-tions on Rm: if µ ∈ Dc(Rm) is a compactly supported density, then∫

Rmµ :=

∫Rm

f(x1, . . . , xm)dx1 . . . dxm,

where µ = f(x1, . . . , xm)|dx1 ∧ . . . ∧ dxm|.1(2) It is local in the sense that: if µ ∈ Dc(M) has support inside the open U , then∫

U

µ =

∫M

µ.

(3) It is invariant under diffeomorphisms: if ϕ : M ∼−→ N is diffeomorphism, andµ ∈ Dc(N), then ∫

N

µ =

∫M

ϕ∗(µ).

1This property looks quite tautological. However, note that the right hand side of the equation is

the integral of the smooth compactly supported function f on U (the Riemann/Lebesgue integral),

and dx1 . . . dxm plays a formal notational role (the “volume element” which one often forgets towrite), whereas µ = f(x1, . . . , xm)|dx1 ∧ . . . ∧ dxm| is an “actual volume element”, as defined in

Definition 14.1.1.

LECTURE 14. 131

We will make the following identification: if U ⊂ M is an open set, we willidentify Dc(U) with elements in Dc(M) with support in U . This is used implicitlyin writing condition (2), and will be used throughout the proof.

Proof. We will construct the integration map in several steps:Step 1. Let U ⊂ Rm be an open set, and let µ = f |dx1 ∧ . . . ∧ dxm| ∈ Dc(U)be a compactly supported density on U . Conditions (1) and (2) force us to define∫Uµ =

∫Rm fdx1 . . . dxm.

Step 2. Consider two open sets U, V ⊂ Rm, and a diffeomorphism θ : V ∼−→ U . Letµ ∈ Dc(U) be a compactly supported density on U . Then θ∗(µ) is a compactlysupported density on V , and Theorem 14.4.2 can be restated as:∫

U

µ =

∫V

θ∗(µ).

This shows that (3) holds for open subsets of Rm.Step 3. Let µ ∈ Dc(M), and assume that there is a coordinate chart (U,ϕ) on Msuch that supp(µ) ⊂ U . Then we define:∫

M

µ =

∫ϕ(U)

(ϕ−1)∗(µ).

Note that this definition is forced by conditions (1)-(3). We prove that this def-inition does not depend on the chart. Consider a second chart (V, ψ) such thatsupp(µ) ⊂ V . Then, by applying Step 2 to the change of coordinates diffeomor-phism θ = ϕ ψ−1 : ψ(U ∩ V )→ ϕ(U ∩ V ), we obtain∫

ϕ(U∩V )

(ϕ−1)∗(µ) =

∫ψ(U∩V )

θ∗(ϕ−1)∗(µ) =

∫ψ(U∩V )

(ψ−1)∗(µ).

Step 4. Consider a cover of M by coordinate charts (Ui, ϕi)i∈I , and let ρii∈Ibe a partition of unity subordinate to it. For µ ∈ Dc(M), define∫

M

µ :=∑i∈I

∫M

ρiµ.

Since ρiµ is supported in the domain Ui of the chart ϕi, it follows that∫Mρiµ is

well-defined; since the support of µ is compact, and the family ρii∈I is locallyfinite, only for a finite number of i ∈ I we have that ρiµ 6= 0, therefore the sumabove is finite. Thus,

∫Mµ is well-defined. Moreover, the definition is forced by

linearity of the integral and by uniqueness of the integral on the domains of charts(see Step 3).Step 5. Next, we prove that the definition is independent on the atlas and onthe partition of unity. Consider a second atlas (Vj , ψj)j∈J , with subordinatepartition of unity σjj∈J . For each i ∈ I, we have ρiµ =

∑j∈J ρiσjµ is a finite

sum of densities supported in Ui. Since∫M

: Dc(Ui)→ R is linear,∫M

ρiµ =∑j∈J

∫M

ρiσjµ.

This implies that: ∑i∈I

∫M

ρiµ =∑i∈I

∑j∈J

∫M

ρiσjµ.


Since all sums involved are finite, we can change the summation order on theright hand side. Therefore, by applying the same argument with the two atlasesinterchanged, we obtain: ∑

i∈I

∫M

ρiµ =∑j∈J

∫M

σjµ.

This shows that the integral is independent on the choice of the atlas.We have proven existence of the integral, and we have proven that the condi-

tions (1)-(3) imply uniqueness of the integral. We leave it to the reader to checkthat conditions (1)-(3) are indeed satisfied for the integral we have constructed.

14.5. Orientations

Definition 14.5.1. Let V be an n-dimensional vector space V , and let B(V ) denotethe space of all bases of V . An orientation on V is a map

o : B(V ) −→ −1, 1,

such that, for all linear isomorphisms A : V ∼−→ A, it satisfies:

o(Av1, . . . , Avn) = sign(det(A))o(v1, . . . , vn).

We denote by Or(V ) the set of orientations on V . Given an orientation o on V abasis v1, . . . , vn of V is said to be a positive basis for o (resp. negative basis), ifo(v1, . . . , vn) = 1 (resp. o(v1, . . . , vn) = −1).

Clearly, every vector space has precisely two orientations. This is true even fora 0-dimensional vector space, which has a unique basis, namely the empty set ∅.The two orientations are o(∅) = 1 and o(∅) = −1; we simply write o = 1 or o = −1.

The standard orientation on Rn is the orientation for which the standardbasis e1, . . . , en is positive. In general, a basis v1, . . . , vn of Rn is positive withrespect to the standard orientation iff the matrix A = (vji )1≤i,j≤n has positive de-terminant, where we have denoted vi = (v1

i , . . . , vni ).

Let M be a manifold. Putting together the orientations of all tangent spacesTpM one obtains the orientation bundle of M :

O(M) :=⊔p∈M

Or(TpM).

Now O(M) comes with a canonical projection π : O(M)→M . For p ∈M , π−1(p)is composed of the two orientations of TpM . The orientation bundle is a smoothmanifold and π : O(M) → M is a local diffeomorphism. To a chart (U,ϕ) on Mwe associate two charts on O(M):

(U+, ϕ+) and (U−, ϕ−).

Let ϕ = (x1, . . . , xm). For each p ∈ U consider the orientation on TpM denoted by

oϕp ∈ Op(M) = Or(TpM)

which is determined by the condition that

∂

∂x1

∣∣p, . . . ,

∂

∂xm∣∣p

LECTURE 14. 133

forms a positive basis. Then oϕ : U → O(M)|U is a section of π, i.e. π oϕ = idU .Let −oϕ : U → O(M)|U be the opposite orientation. The charts are given by:

U+ := oϕ(U), ϕ+ := ϕ π|U+ ,

U− := (−oϕ)(U) = −U+, ϕ− := ϕ π|U− .

We leave it to the reader to check that O(M), endowed with these family of charts,is a smooth manifold of the same dimension as M .

Example 14.5.2. The orientation bundle of the Mobius band is a cylinder; youcan see this here: [19].

Definition 14.5.3. An orientation on a manifold M is a smooth choice of ori-entations on all the tangent spaces of M . In other words, an orientation of M isa smooth section of the orientation bundle, i.e. a smooth map o : M → O(M) suchthat π o = idM .

A manifold M is said to be orientable if it admits an orientation.An oriented manifold is a manifold M with a fixed orientation o.If (M1, o1) and (M2, o2) are two oriented manifolds, then a local diffeomorphism

ϕ : M1 →M2 is said to be orientation preserving, if for every p ∈M1 and everypositive basis v1, . . . , vm of TpM1, we have that dpϕ(v1), . . . ,dpϕ(vm) is a positivebasis of Tϕ(p)M2.

Proposition 14.5.4. A connected orientable manifold has two orientations.

Proof. Since M is orientable, there exists an orientation o : M → O(M) on M .Now, also −o : M → O(M) is an orientation on M . Thus there are at least twoorientations. In general, given any other orientation, o1 : M → O(M), we can lookat the quotient map o/o1 : M → 1,−1, which returns 1 if the orientations agree,and −1 if they disagree. This map is smooth; and therefore it is locally constant.Since M is connected, o/o1 is constant. Therefore, either o1 = o or o1 = −o.

14.6. Exercises

Exercise 14.1. (a) Let (V, 〈·, ·〉) be an n-dimensional vector space together withan inner-product. Prove that the following expression defines a positive volumeelement on V :

µ(v1, . . . , vn) :=√

det (〈vi, vj〉i,j).

(b) Let M ⊂ Rn be an embedded submanifold. For each p ∈M , we regard TpM asa linear subspace of TpRn = Rn. The restriction of the standard dot-product onRn gives an inner-product on TpM . Show that the following defines a smoothdensity on M :

µMp (v1, . . . , vm) :=√

det (〈vi, vj〉i,j),

where v1, . . . , vm is a basis of TpM .

(c) With the notation from (b), prove that µRn = |dx1 ∧ . . . ∧ dxn|.(d) For n = 3, show that the construction from (b) recovers the usual notions of

length of a curve and area of a surface, more precisely:• Let γ : (a, b) → R3 be a smooth embedding, and let C be the image ofγ. Prove that µC = ds, where, at p = γ(t), dsp denotes the “arc-length


element”,∣∣∣dγdt (t)

∣∣∣dt. Conclude that∫CµC =

∫ b

a

∣∣∣∣dγdt (t)

∣∣∣∣dt,assuming that both sides are bounded (note that the right hand side isthe length of C).• Let U ⊂ R2 be an open set, and let σ : U → R3 be a smooth embedding

parameterizing the curve S := σ(U). Prove that µSp = dSp, where dSp isthe “area form”:

dSp :=

∣∣∣∣∂σ∂u (u, v)× ∂σ

∂v(u, v)

∣∣∣∣ dudv, at p = σ(u, v).

Conclude that the following holds:∫SµS =

∫U

∣∣∣∣∂σ∂u (u, v)× ∂σ

∂v(u, v)

∣∣∣∣ dudv,

assuming that both sides are bounded (note that the right hand side isthe area of S).

Exercise 14.2. Let G be a Lie group with Lie algebra g := TeG. A smooth densityµ ∈ D(G) is called left invariant, if λ∗g(µ) = µ for all g ∈ G (recall that λg denotesleft translation by g). Let DL(G) denote the space of left invariant densities.

(a) Prove that evaluation at the identity

DL(G) −→ Ve(g), µ 7→ µ|pis a linear isomorphism.

(b) If G is compact, prove that there exists a unique left invariant density µG ∈DL(G) such that

∫GµG = 1 (The density µG is called the Haar measure of

G).(c) For g ∈ G and µ ∈ DL(G), prove that ρ∗g(µ) ∈ DL(G) (recall that ρg denotes

right translation by g). Moreover, show that ρ∗g(µ) = u(g)µ, where u(g) ∈R\0 does not depend on µ 6= 0. Prove that the map u : G → R\0,g 7→ u(g) is a Lie group homomorphism.

(d) If G is compact, prove that u(g) = 1, for all g ∈ G.(e) Consider the Lie group G = Aff(1) consisting of affine transformation of the real

line R, i.e. the elements of Aff(1) are maps f : R→ R of the form f(x) = ax+b,with a 6= 0, and the group structure is the composition of maps. Calculate themap u : Aff(1)→ R\0.

Exercise 14.3. Prove that a manifold M is orientable if and only if it admitsan atlas (Ua, ϕa)a∈I for which all transition maps ϕa (ϕb)

−1 : ϕb(Ua ∩ Ub) →ϕa(Ua ∩ Ub), for a, b ∈ I, have positive Jacobian determinant:

det

(∂xia

∂xjb

)> 0,

where ϕa = (x1a, . . . , x

ma ).

Exercise 14.4. Let M be a connected manifold. Prove that M is orientable iffO(M) is disconnected.

LECTURE 14. 135

Exercise 14.5. Let (M,o) be a connected oriented manifold, and let ϕ : M → N bea local diffeomorphism. Assume that ϕ is not injective, and consider p, q ∈M , withp 6= q, such that ϕ(p) = ϕ(q) =: z. Consider the orientation oz ∈ Or(TzN) for whichdpϕ : (TpM, op)→ (TzN, oz) is orientation preserving (i.e. it sends positive bases topositive bases). If dqϕ : (TqM, oq)→ (TzN, oz) is not orientation preserving, provethat N is not orientable.

Exercise 14.6. (a) Show that the n-sphere Sn is orientable.(b) Let τ : Sn → Sn be the antipodal map, τ(x) = −x. Prove that τ is orientation

preserving iff n is odd.(c) Show that Pn(R) is orientable iff n is odd.(d) If n is even, prove that O(Pn(R)) is diffeomorphic to Sn.

LECTURE 15

15.1. Manifolds with boundary

The simplest example of a manifold with boundary is the upper half-space

Hn := (x1, . . . , xn) ∈ Rn : xn ≥ 0.

The boundary of Hn is the n− 1-dimensional space:

∂Hn := Rn−1 × 0 ⊂ Hn,

and its complement is called the interior of Hn:

int(Hn) := Rn−1 × (0,∞) ⊂ Hn.

Clearly, these coincide with the topological notions of boundary and interior assubsets in Rn, but soon we will introduce the more general class of manifoldswith boundary, which also are made of an “interior” and a “boundary” but which,intrinsically, are not subsets of a larger space.

The boundary and interior of an open set U ⊂ Hn are defined as follows:

∂U := U ∩ ∂Hn int(U) := U ∩ int(Hn).

Note that already in this case, these sets do not coincide with the topological notionsof boundary and interior.

A map f : U → Rm defined on an open subset U ⊂ Hn is called smooth, if

there exists a smooth map f : U → Rm, where U is an open set in Rn containing

U such that f |U = f . If U is already open as a subset of Rn, this is equivalent tothe usual smoothness condition on f . The differential of f at p ∈ U is defined as

dpf := dpf : Rn −→ Rm,

where f is an extension of f . The differential does not depend on the chosen

extension of f : at p ∈ int(U) this is clear; at p ∈ ∂U , this is because df iscontinuous and int(U) is dense in U , and therefore:

dpf = limq→p

q∈int(U)

dq f = limq→p

q∈int(U)

dqf,

and clearly the right hand side is independent of f .

137


Clearly the composition of smooth maps between open subsets of upper half-spaces is again smooth (its extension is the composition of the extensions), and theusual chain rule holds as well (it follows by the chain rule for the extensions).

The notion of a diffeomorphism generalizes accordingly: if U, V ⊂ Hn areopen subsets, then f : U → V is a diffeomorphism if and only if f is smooth,bijective, and f−1 : V → U is also smooth.

The boundary and the interior are preserved under diffeomorphisms:

Lemma 15.1.1. Let U, V ⊂ Hn be open sets, and let f : U → V be a diffeomor-phism. Then

f(∂U) = ∂V f(int(U)) = int(V ).

Proof. Note that f |int(U) : int(U) → Rn is an injective local diffeomorphism;hence f(int(U)) is an open set (as a subset of Rn). This implies that f(int(U)) ⊂int(V ). The same argument for f−1 implies that f−1(int(V )) ⊂ int(U). Since f is abijection, we conclude that f(int(U)) = int(V ). But then f restricts to a bijectionbetween the complements of these sets; thus, f(∂U) = ∂V .

Remark 15.1.2. In fact any homeomorphism f : U → V , where U, V ⊂ Hn areopen sets, satisfies the conclusions of the Lemma: f(∂U) = ∂V and f(int(U)) =int(V ). To show this, one needs to find a topological property which differentiatespoints on the boundary from interior points. For example, a point p ∈ ∂U has abasis of neighborhoods which consists of contractible open sets O such that O\pis also contractible. This is not true for a point p ∈ int(U); actually, if O is anyopen neighborhood of p ∈ int(U) then O\p then is not contractible (a small n−1-sphere around p in O in a nontrivial element of Hn−1(O;Z), because its image underthe map Hn−1(O;Z)→ Hn−1(Rn\p;Z) ∼= Z is the generator of this group).

In the usual definition of a smooth manifold, if one considers charts with valuesin Hm, one obtains the concept of a manifolds with boundary. Let us give a briefoutline of this construction.

First, we consider atlases with values in Hm:

Definition 15.1.3. Let M be a topological space. A C∞-atlas with values inHm on M is a family A := (Uα, ϕα)α∈I , where

• Uαα∈I is an open cover of M ;• ϕα : Uα → Hm are maps such that ϕα(Uα) are open sets in Hm, andϕα : Uα → ϕα(Uα) are homeomorphisms,

such that, for all α, β ∈ I, the transition map

ϕα ϕ−1β : ϕβ(Uα ∩ Uβ)→ ϕα(Uα ∩ Uβ)

is a diffeomorphism between open sets in Hm.Two C∞-atlases A and B with values in Hm on M are said to be equivalent,

if A ∪ B is again such a C∞-atlas with values in Hm.

Definition 15.1.4. A smooth m-dimensional manifold with boundary is aHausdorff, second countable topological space M endowed with an equivalence classof C∞-atlases with values in Hm.

Note that a usual m-dimensional manifold can be regarded also as a manifoldwith boundary, by choosing an atlas with values in int(Hm) ∼= Rm.

LECTURE 15. 139

Definition 15.1.5. Let M be a manifold with boundary. A boundary point ofM is a point p ∈ M such that there exists a chart ϕ : U → Hm around p withϕ(p) ∈ ∂Hm. By Lemma 15.1.1, the transition functions preserve ∂Hm; therefore,if p is a boundary point, then every chart around p sends p to a point on ∂Hm. Theset of boundary points of M is called the boundary of M is denote by ∂M .

An interior point of M is a point p ∈M which is not a boundary point, i.e.for every chart ϕ : U → Hm we have that ϕ(p) ∈ int(Hm) (which is equivalent tothe existence of at least one such chart). The set of interior points of M is calledthe interior of M and is denoted by int(M).

Thus, any manifold with boundary M has a partition into two subsets M =∂M t int(M). Note that int(M) is open in M and so ∂M is closed in M . Thesesubsets are manifolds (without boundary):

Proposition 15.1.6. Let M be an m-dimensional manifold with boundary. Thenint(M) and ∂M are smooth manifolds without boundary of dimensions m and m−1,respectively. Their differentiable structures are such that, if ϕ : U → Hm is a charton M , then

ϕ|int(U) : int(U) −→ int(Hm) and ϕ|∂U : ∂U −→ Rm−1

are smooth charts on int(M) and ∂M , respectively, where we have denoted int(U) :=U ∩ int(M) and ∂U := U ∩ ∂M , and we have used the identification

∂Hm ∼= Rm−1, (x1, . . . , xm−1, 0) 7→ (x1, . . . , xm−1).

Example 15.1.7. The intervals [0, 1], (0, 1], [0, 1) and (0, 1) are manifolds withboundary; all have interior (0, 1), and their respective boundaries are:

∂[0, 1] = 0, 1, ∂(0, 1] = 1, ∂[0, 1) = 0, ∂(0, 1) = ∅.

Example 15.1.8. The closed ball in Rn

Bn := (x1, . . . , xn) : (x1)2 + . . .+ (xn)2 ≤ 1is a manifold with boundary the n − 1-dimensional sphere and interior the openball in Rn:

∂Bn = Sn−1, int(Bn) = Bn.

This can be checked by applying Exercise 15.1 to the map

f : Rn −→ R, f(x1, . . . , xn) = 1− ((x1)2 + . . .+ (xn)2).

All the construction we have performed for manifold without boundary gener-alize to manifolds with boundary. We briefly enumerate what one needs to keeptrack of. Let M and N be manifolds with boundary. Then

• The notion of a smooth map generalizes accordingly: f : M → N issmooth iff it is continuous and

ϕ f ψ−1 : ψ(U ∩ f−1(V )) −→ Hn

is smooth for every pair of charts chart ψ : U → Hm and ϕ : V → Hn onM and N , respectively.

• The tangent space TpM at a point p is defined as for usual manifolds. Evenat p ∈ ∂M , one has that dim(TpM) = m. The tangent bundle π : TM →M is a 2m-dimensional manifold with boundary ∂TM = π−1(∂M). Notethat ∂TM 6= T∂M : dim(∂TM) = 2m−1, but dim(T∂M) = 2m−2. Thedifferential of smooth maps generalizes accordingly.


• A vector bundle π : E → M is defined exactly as in Definition 8.2.1,just that E is also a manifold with boundary ∂E = π−1(∂M).

• The density bundle D(M), the orientation bundle O(M), the cotangent

bundle T ∗M and its exterior powers∧k

T ∗M are constructed exactly as inthe case of manifold without boundary (note that all these were obtainedby applying a linear algebra construction to TpM for all p ∈M).

• The exterior derivative d : Ω(M) → Ω(M) is defined in the same way,and has the usual properties. One can consider the associated de Rhamcohomology HdR(M) (one can show that HdR(M) ∼= HdR(int(M))).

• There is a canonical integration of compactly supported densities definedas for usual manifolds

∫M

: Dc(M)→ R.

Vectors along the boundary, not tangent to the boundary, are of two types:

Definition 15.1.9. Let p ∈ ∂M . A vector v ∈ TpM such that v /∈ Tp∂M is saidto point inwards if there exists a smooth curve

γ : [0, ε) −→M such that γ(0) = p andd

dtγ(t)

∣∣t=0

= v,

and v is said to point outwards, if −v points inwards.

In other words, the hyperplane Tp(∂M) divides TpM into two components:vectors pointing inwards and vectors pointing outwards. Note that, for p ∈ ∂M ,and v ∈ TpM precisely one of the following situations holds: v points inwards,v points outwards, or v ∈ Tp(∂M). In a chart, ϕ : U → Hm around p, withdpϕ(v) = (v1, . . . , vm) ∈ Tϕ(p)Rm ∼= Rm this can be checked as follows:

• v points inwards iff vm > 0 iff dpϕ(v) ∈ int(Hm);• v points outwards iff vm < 0 iff dpϕ(v) /∈ Hm;• v is tangent to ∂M iff vm = 0 iff dpϕ(v) ∈ ∂Hm.

In other words, the one-dimensional vector space TpM/Tp∂M , which consists of“directions transverse to ∂M”, comes with a canonical orientation. This impliesthat an orientable manifold with boundary has an orientable boundary, as thefollowing construction shows:

Definition 15.1.10. Let (M,o) be an oriented manifold with boundary. Theboundary orientation ∂o is the orientation on the manifold ∂M defined on thebasis v1, . . . , vm−1 of Tp∂M by:

∂o(v1, . . . , vm−1) = o(v, v1, . . . , vm−1),

where v ∈ TpM is any outwards pointing vector, and m = dim(M).

Example 15.1.11. Consider the manifold with boundary [0, 1]. Consider the ori-entation o on [0, 1], such that ∂

∂t ∈ Tt[0, 1] is a positive basis for all t ∈ [0, 1].Outwards pointing vectors at the boundary ∂[0, 1] = 0 ∪ 1 are given by:

− ∂

∂t

∣∣t=0∈ T0[0, 1] and

∂

∂t

∣∣t=1∈ T1[0, 1].

Recall that an orientation of a 0-dimensional vector space is the same as the choiceof a sign (see the comments after Definition 14.5.1). The boundary orientation ∂oconsists of the following two signs:

(∂o)0 = o(− ∂

∂t

∣∣t=0

)= −1 and (∂o)1 = o

( ∂∂t

∣∣t=1

)= +1.

LECTURE 15. 141

The following result describes a manifold in a neighborhood of its boundary.

Theorem 15.1.12. (Collar Neighborhood Theorem) Let M be a manifold withboundary. There exists a neighborhood U ⊂ M of ∂M and a diffeomorphism ϕ :∂M × [0, 1) ∼−→ U , such that ϕ(x, 0) = x for all x ∈M .

For the proof of the above, we will use:

Lemma 15.1.13. Let M be a manifold with boundary. There exists a vector fieldX ∈ X(M) such that Xp points inwards, for all p ∈ ∂M .

Proof. Let A = (Ui, ϕi)i∈I be a smooth atlas on M with valued Hm. In eachchart, consider the inwards pointing vector field Xi := (ϕi)

∗( ∂∂xm ) ∈ X(Ui). Let

ρii∈I be a partition of unity subordinated to the cover Uii∈I . Define X :=∑i∈I ρiXi ∈ X(M). Since each Xi is pointing inwards, it follows that also X

points inwards (note that pointing inwards is a convex condition).

We prove Theorem 15.1.12 only in the case of compact boundary:

Proof of Theorem 15.1.12 for compact ∂M . Let X ∈ X(M) be a vector fieldpointing inwards. Because X points inwards, there is a neighborhood U ⊂ [0,∞)×M of 0 ×M on which the flow is defined:

φX : U −→M, (t, p) 7→ φtX(p).

Consider the restriction of the flow to the boundary of M :

ψ : U ∩([0,∞)× ∂M

)−→M, ψ(t, p) := φtX(p).

For any p ∈ ∂M , we have that ψ(0, p) = p, hence d(0,p)ψ(0, v) = v for any v ∈Tp∂M . On the other hand, since the curve t 7→ ψ(t, p) is a flow line of X,

d(0,p)ψ( ∂∂t, 0)

= Xp.

Since TpM = Tp∂M ⊕RXp, we conclude that d(0,p)ψ is a linear isomorphism. TheInverse Function Theorem 6.1.1, implies that ψ is a diffeomorphism in a neigh-borhood of p. On the other hand, since ∂M is compact and ψ(0, p) = p for allp ∈ ∂M , Exercise 6.8 implies that ψ is a diffeomorphism between a neighborhoodV ⊂ U of 0 × ∂M and a neighborhood O ⊂ M of ∂M . On the other hand,since ∂M is compact, there exists ε > 0 such that [0, ε)× ∂M ⊂ V . Finally, defineϕ : [0, 1)× ∂M →M , ϕ(t, p) = ψ(εt, p).

15.2. Integration of differential forms on oriented submanifolds

The top power of V ∗ is closely related to volume elements on V . Proposition 12.1.10implies the following:

Lemma 15.2.1. Let V be a vector space of dimension n. An orientation o ∈ Or(V )induces a linear isomorphism:

n∧V ∗ ∼−→ Ve(V ), ω 7→ oω,

(oω)(v1, . . . , vn) := o(v1, . . . , vn)ω(v1, . . . , vn),

for all (v1, . . . , vn) ∈ B(V ).


Consider an oriented manifold (M,o) of dimensionm. The above lemma impliesthat the orientation o gives a vector bundle isomorphism

m∧T ∗M ∼−→ D(M), ω 7→ oω.

Definition 15.2.2. Let (M, o) be an oriented manifold of dimension m. The in-tegral of a compactly supported top-form ω ∈ Ωmc (M) on M is defined as theintegral of the compactly supported density oω ∈ Dc(M):∫

(M,o)

ω :=

∫M

oω.

Remark 15.2.3. If the orientation changes, the sign of the integral changes:∫(M,−o)

ω = −∫

(M,o)

ω.

In notation, if the orientation o on M is fixed, it is common to omit the orienta-tion under the orientation under the integral symbol; we also adopt this convention,and write ∫

M

ω instead of

∫(M,o)

ω.

Integration of forms is invariant under orientation preserving diffeomorphisms:

Proposition 15.2.4. Let M and N be connected oriented manifolds, and let ϕ :M ∼−→ N be a diffeomorphism. Then, for all compactly supported top-forms ω ∈Ωmc (N), we have that ∫

M

ϕ∗(ω) = ε

∫N

ω,

where ε = 1 if ϕ is orientation preserving, and ε = −1 if ϕ is orientation reversing.

Proof. Let oM and oN be the respective orientations. We have that followingequality of densities: ϕ∗(oNω) = εoMϕ

∗(ω). Using that integration of densities isinvariant under diffeomorphisms (see Theorem 14.4.3), we obtain the result:∫

(M,oM )

ϕ∗(ω) =

∫M

oMϕ∗(ω) = ε

∫M

ϕ∗(oNω) = ε

∫N

oNω = ε

∫(N,oN )

ω.

Forms of lower degree can be integrated on oriented submanifolds:

Definition 15.2.5. Let M be an m-dimensional manifold. Consider an embeddedsubmanifold N ⊂ M of dimension n, which is oriented, and consider an n-formη ∈ Ωn(M). If supp(η) ∩N is compact, we define the integral of η over N by:∫

N

η :=

∫(N,o)

i∗(η),

where o is the orientation on N , and i : N →M denotes the inclusion map.

Note that supp(i∗(η)) = supp(η) ∩ N ; thus the compactness assumption isequivalent to i∗(η) being compactly supported.

LECTURE 15. 143

15.3. Stokes’ Theorem

Stokes’ Theorem is the higher-dimensional version of the Fundamental Theorem ofCalculus:

Theorem 15.3.1 (Stokes’ Theorem). Let (M, o) be an oriented manifold withboundary, and consider on ∂M the induced orientation ∂o. Then, for any com-pactly supported m− 1-form ω ∈ Ωm−1

c (M), where m = dim(M), we have that:∫(M,o)

dω =

∫(∂M,∂o)

ω.

Proof. We first prove the result for Hm endowed with the standard orientationost. Consider ω ∈ Ωm−1

c (Hm). Then ω can be written uniquely as the sum:

ω =

m∑i=1

fi(x)dx1 ∧ . . . ∧ dxi ∧ . . . ∧ dxm,

where the coefficients fi ∈ C∞c (Hm) are compactly supported function on Hm, and

dxi denotes that dxi is missing in the product. The differential of ω is given by:

dω =

(m∑i=1

(−1)i−1 ∂fi∂xi

(x)

)dx1 ∧ . . . ∧ dxm.

Note that, under the standard orientation on Hm, we have that dx1 ∧ . . . ∧ dxm

corresponds to the density |dx1 ∧ . . . ∧ dxm|, and therefore:∫Hm

dω =

m∑i=1

(−1)i−1

∫Hm

∂fi∂xi

(x)dx1 . . . dxm.

Choosing a large enough number N > 0 such that

supp(ω) ⊂ [−N,N ]m−1 × [0, N ].

Consider a term in the sum above with 1 ≤ i < m. By interchanging the order ofintegration, we first calculate the integral with respect to dxi:∫ N

−N

∂fi∂xi

(x)dxi = fi(x1, . . . , xi−1, N, xi+1, . . . , xm)

− fi(x1, . . . , xi−1,−N, xi+1, . . . , xm) = 0,

where we have used the Fundamental Theorem of Calculus and that supp(fi) ⊂[−N,N ]m−1 × [0, N ]. The term with i = m gives:∫ N

0

∂fm∂xm

(x)dxm = fm(x1, . . . , xm−1, N)− fm(x1, . . . , xm−1, 0) =

= −fm(x1, . . . , xm−1, 0).

We conclude that

(*)

∫Hm

dω = (−1)m∫Rm−1

fm(x1, . . . , xm−1, 0)dx1 . . . dxm−1.

On the other hand, if i : ∂Hm → Hm denotes the inclusion, we have that:

i∗(ω) = fm(x1, . . . , xm−1, 0)dx1 ∧ . . . ∧ dxm−1.


Note that the vector − ∂∂xm is pointing outwards, therefore, the boundary orienta-

tion on ∂Hm is such that

(∂ost)( ∂

∂x1, . . . ,

∂

∂xm−1

)= ost

(− ∂

∂xm,∂

∂x1, . . . ,

∂

∂xm−1

)= (−1)m.

We obtain that

(∂ost)dx1 ∧ . . . ∧ dxm−1 = (−1)m|dx1 ∧ . . . ∧ dxm−1|,

and therefore∫∂Hm

ω =

∫∂Hm

(∂ost) i∗(ω) = (−1)m

∫Rm−1

fm(x1, . . . , xm−1, 0)dx1 . . . dxm−1.

This and (∗) prove Stokes’ Theorem for Hm with the standard orientation.Consider an oriented manifold (M,o) with boundary. Let ω ∈ Ωm−1

c (M).Assume first that the support of ω lies inside the domain of a coordinate chartϕ : U → Hm, such that U and ∂U := U ∩∂M are connected. By Proposition 15.2.4and Stokes’ Theorem for Hm, we obtain that∫M

dω =

∫U

dω = ε

∫ϕ(U)

(ϕ−1)∗(dω) = ε

∫ϕ(U)

d(ϕ−1)∗(ω) = ε

∫Hm

d(ϕ−1)∗(ω) =

= ε

∫∂Hm

(ϕ−1)∗(ω) = ε

∫∂ϕ(U)

(ϕ−1)∗(ω),

where ε = 1 if ϕ : U → ϕ(U) is orientation preserving, and otherwise ε = −1. If∂ϕ(U) = ∅, then the last term is zero, but also U ∩ ∂M = ∅, and therefore also∫∂M

ω = 0. On the other hand, if ∂ϕ(U) 6= ∅, then note that ϕ : U → ϕ(U) isorientation preserving if and only if ϕ|∂U : ∂U → ∂ϕ(U) is orientation preserving;therefore, applying Proposition 15.2.4 to ϕ|∂U , we obtain that

ε

∫∂ϕ(U)

(ϕ−1)∗(ω) =

∫∂U

ω =

∫∂M

ω.

In both cases, we obtain that∫M

dω =∫∂M

ω.

In general, using partitions of unity, we can decompose any ω ∈ Ωm−1c (M) as a

finite sum ω = ω1 + . . .+ωl, with ωi ∈ Ωm−1c (M) having support inside the domain

of a coordinate chart with the properties from above. Applying Stokes’ Theorem toeach ωi, and summing up the resulting equalities, we obtain that Stokes’ Theoremholds for ω.

Example 15.3.2. Consider the closed interval [a, b] with the standard orientation.Stokes’ Theorem in this case gives∫ b

a

dF =

∫a,b

F, for all F ∈ Ω0([a, b]) = C∞([a, b]).

The left hand side is just∫ baF ′(t)dt, while the right hand side gives oaF (a) +

obF (b) = F (b) − F (a), where oa, ob ∈ −1, 1 are the induced orientations on theboundary (see example 15.1.11). Thus, in this case, Stokes’ Theorem is equivalentto the Fundamental Theorem of Calculus:∫ b

a

f(t)dt = F (b)− F (a), where f(t) = F ′(t).

For manifolds without boundary, Stokes’ Theorem gives:

LECTURE 15. 145

Corollary 15.3.3. Let M be a compact oriented manifold without boundary ofdimension m. Then, for any m− 1 form ω ∈ Ωm−1(M), we have that

∫M

dω = 0.

Equivalently, if µ ∈ Ωm(M) satisfies∫Mµ 6= 0, then its cohomology class [µ] ∈

HmdR(M) is non-trivial [µ] 6= 0.

We have that:

Corollary 15.3.4. Let M be a compact oriented manifold without boundary ofdimension m. Then Hm

dR(M) 6= 0.

Proof. Let p ∈M , and consider an oriented chart (U,ϕ = (x1, . . . , xm)), around p.Let χ : M → [0, 1] be smooth function with compact support in U , and such thatχ(p) > 0. Then µ = χ · dx1 ∧ . . . ∧ dxm satisfies

∫Mµ > 0, thus, by the previous

Corollary, [µ] 6= 0 in HmdR(M).

15.4. Exercises

Exercise 15.1. Let M be a manifold without boundary, and let f : M → R be asmooth function. If 0 ∈ R is a regular value of f , prove that N := f−1([0,∞)) canbe given the structure of a manifold with boundary ∂N = f−1(0) such that theinclusion N →M is smooth. Hint: use the local submersion Theorem 6.2.3.

Exercise 15.2. Show that Stokes’ Theorem implies the following classical resultsfrom Calculus:

(a) Green’s Theorem. Let Ω ⊂ R2 be an open set bounded by the simple closedsmooth curve C. For every smooth vector field F = (f1, f2) on R2, we havethat: ∮

Cf1dx+ f2dy =

∫∫Ω

(∂f2

∂x− ∂f1

∂y

)dxdy.

(b) Divergence Theorem. Let Ω ⊂ R3 be a bounded open set with smooth boundary.Then for every smooth vector field F on R3, we have that:∫∫∫

D

divF dv =

∫∫∂D

F · n dA.

(c) Classical Stokes’ Theorem. Let S ⊂ R3 be a smooth bounded surface withsmooth boundary. Then for every smooth vector field F = (f1, f2, f3) on R3,we have that: ∮

∂S

f1dx+ f2dy + f3dz =

∫∫S

curl(F) · n dA.

Use Exercise 13.2 and a Calculus book (or Wikipedia) to make sense of all terms.

Exercise 15.3. Let U ⊂ C be an open set, and let f : U → C be a holomorphicfunction. Consider a closed disk D ⊂ U with center a and boundary C. ProveCauchy’s integral formula:

f(a) =1

2πi

∮C

f(z)

z − adz.

Hint: Use Exercise 13.4 and Stokes’ Theorem.

BIBLIOGRAPHY

1. L. Conlon, Differentiable manifolds. Reprint of the 2001 second edition. ModernBirkhauser Classics. Birkhauser Boston, Inc., Boston, MA, 2008.

2. J. J. Duistermaat, J. Kolk, Lie groups, Universitext, Springer-Verlag Berlin,2000.

3. R. Loja Fernandes, Differentiable Manifolds I, http://www.math.illinois.edu/~ruiloja/Math518/notes.pdf.

4. V. Guillemin, A. Pollack, Differential topology. Prentice-Hall, Inc., EnglewoodCliffs, N.J., 1974.

5. N. Hitchin, Differentiable manifold, lecture notes, http://people.

maths.ox.ac.uk/hitchin/hitchinnotes/Differentiable_manifolds/

manifolds2012.pdf

6. M. Kervaire, A manifold which does not admit any differentiable structure,Coment. Math. Helv., (1960) 34, 257–270.

7. J.W. Milnor, On manifolds homeomorphic to the 7-sphere, Ann. of Math. (2)64 (1956): 399–405.

8. J.W. Milnor, Topology from the differentiable viewpoint, Princeton UniversityPress, Princeton, NJ, 1997.

9. E. Moise, Affine structures in 3-manifolds. V. The triangulation theorem andHauptvermutung Ann. of Math. (2) 56 (1952), 96–114.

10. L. Nicolaescu, Lectures on the Geometry of Manifolds, http://www3.nd.edu/

~lnicolae/Lectures.pdf

11. W. Rudin, Principles of mathematical analysis. Third edition. InternationalSeries in Pure and Applied Mathematics. McGraw-Hill Book Co., New York-Auckland-Dusseldorf, 1976.

12. J. Stallings, The piecewise-linear structure of Euclidean space, Proc. CambridgePhilos. Soc. 58 (3), 1962, 481–488.

13. C.H. Taubes, Gauge theory on asymptotically periodic 4-manifolds, J. of Diff.Geom. 25 (3), (1987), 363–430.

14. W. Thurston, Three-dimensional geometry and topology. Vol. 1. Edited bySilvio Levy. Princeton Mathematical Series, 35. Princeton University Press,Princeton, NJ, 1997.

15. H. Whitney, Differentiable manifolds, Ann. of Math. (2) 37 (1936), no. 3, 645–680.

16. https://en.wikipedia.org/wiki/Differential_structure

147

http://www.math.illinois.edu/~ruiloja/Math518/notes.pdf

http://www.math.illinois.edu/~ruiloja/Math518/notes.pdf

http://people.maths.ox.ac.uk/hitchin/hitchinnotes/Differentiable_manifolds/manifolds2012.pdf



http://www3.nd.edu/~lnicolae/Lectures.pdf

http://www3.nd.edu/~lnicolae/Lectures.pdf

https://en.wikipedia.org/wiki/Differential_structure


17. https://en.wikipedia.org/wiki/Long_line_(topology)

18. http://pages.uoregon.edu/koch/math431/LongLine.pdf

19. https://commons.wikimedia.org/wiki/File:Orientation_cover_of_

Mobius_strip.webm

20. https://en.wikipedia.org/wiki/Prfer_manifold

21. https://en.wikipedia.org/wiki/Space-filling_curve

22. http://earth.nullschool.net/

https://en.wikipedia.org/wiki/Long_line_(topology)

http://pages.uoregon.edu/koch/math431/LongLine.pdf

https://commons.wikimedia.org/wiki/File:Orientation_cover_of_Mobius_strip.webm

https://commons.wikimedia.org/wiki/File:Orientation_cover_of_Mobius_strip.webm

https://en.wikipedia.org/wiki/Prüfer_ manifold

https://en.wikipedia.org/wiki/Space-filling_curve

http://earth.nullschool.net/

EXERCISES AND ASSIGNMENTS

Problems for exercise classes

• Exercise class 11/9: 1.1, 1.2, 1.6, 1.7.• Exercise class 18/9: 2.1, 2.2, 2.3.• Exercise class 25/9: 3.2, 3.9.• Exercise class 2/10: 4.1, 4.3.• Exercise class 9/10: 5.1, 5.3.• Exercise class 16/10: 6.3, 6.6.• Exercise class 23/10: 6.5, 6.11.• Exercise class 13/11: 8.1, 8.2, 8.3.• Exercise class 20/11: 9.3, 9.7, 9.8.• Exercise class 27/11: 10.2,10.3,10.4.• Exercise class 4/12: 11.3.• Exercise class 11/12: 12.3,12.9.• Exercise class 18/12: 13.4, 13.5, 13.6.• Exercise class 8/1: 14.1, 14.6.• Exercise class 15/1: 15.2.

Homework assignments

• Homework for 15/9: 1.2, 1.7.• Homework for 22/9: 2.3.• Homework for 29/9: 3.9.• Homework for 6/10: 4.3.• Homework for 13/10: 5.1.• Homework for 20/10: 6.3, 6.6.• Homework for 27/10: 6.5, 6.11.• Homework for 17/11: 8.1, 8.2, 8.3.• Homework for 24/11: 9.3, 9.7, 9.8.• Homework for 1/12: 10.2,10.3,10.4.• Homework for 8/12: 11.3.• Homework for 15/12:12.3,12.9.• Homework for 22/12: 13.4, 13.5, 13.6.• Homework for 12/1: 14.1, 14.6.• Homework for 19/1: 15.2.

149

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