A Rapid Introduction to Differential Topology

Paul D. Mitchener

June 17, 2013

Contents

1 Higher Dimensional Calculus 21.1 The Total Derivative . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Differentiation of Paths . . . . . . . . . . . . . . . . . . . . . . . 41.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Manifolds 72.1 Topological Manifolds . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Smooth Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 Paracompactness . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 Boundaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 The Tangent Bundle 123.1 Vector Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.2 Tangent Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

4 Critical and Regular Values 174.1 Submersions, Immersions, and Embeddings . . . . . . . . . . . . 174.2 Sard’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.3 The Brouwer Fixed Point Theorem . . . . . . . . . . . . . . . . . 24

5 Vector and Tensor Fields 255.1 Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255.2 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275.3 Integral Curves and Flows . . . . . . . . . . . . . . . . . . . . . . 285.4 Tensors and Duals . . . . . . . . . . . . . . . . . . . . . . . . . . 325.5 Tensor Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

6 Riemannian Geometry 376.1 Riemannian Metrics . . . . . . . . . . . . . . . . . . . . . . . . . 376.2 Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406.3 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446.4 Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 516.5 Normal Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

1

7 Some Notions from Algebraic Topology 557.1 Homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557.2 Categories and Functors . . . . . . . . . . . . . . . . . . . . . . . 577.3 Some Homological Algebra . . . . . . . . . . . . . . . . . . . . . 60

8 De Rham Cohomology 648.1 Differential Forms . . . . . . . . . . . . . . . . . . . . . . . . . . 648.2 The de Rham Complex . . . . . . . . . . . . . . . . . . . . . . . 668.3 De Rham Cohomology . . . . . . . . . . . . . . . . . . . . . . . . 68

9 Orientation and Duality 739.1 Orientations on Vector Bundles . . . . . . . . . . . . . . . . . . . 739.2 Orientations of Manifolds . . . . . . . . . . . . . . . . . . . . . . 759.3 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 779.4 Poincare Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

1 Higher Dimensional Calculus

1.1 The Total Derivative

In order to talk about smooth manifolds, we need to be able to talk aboutsmooth maps and differentation for functions from an open subset of Rm to Rn.First, let us define the norm of a vector in Rn by the formula

‖(x1, . . . , xn)‖ =√

(x1)2 + · · ·+ (xn)2.

The Euclidean metric on Rn is defined by the formula d(x, y) = ‖x − y‖.This metric defines the topology used in the previous section when talking aboutopen sets and homeomorphisms.

Recall that a function f :R → R is differentiable at a point x ∈ R if thederivative

f ′(x) = limh→0

f(x+ h)− f(x)

h

exists.We can rewrite this equation

f(x+ h) = f(x) + f ′(x)h+ |h|r(h) limr→0

r(h) = 0

Definition 1.1 Let U ⊆ Rm be an open subset. We say that a functionϕ:U → Rn is differentiable at a point x ∈ U if there is a bounded linearmap (Dϕ)x:Rm → Rn, and a function r:V → Rn for some open set V ⊆ Rmcontaining 0 such that

ϕ(x+ v) = ϕ(x) + (Dϕ)xv + ‖v‖r(v)

for all v ∈ V , where limv→0 r(v) = 0.The above linear map (Dϕ)x:V → W is called the total differential of the

function ϕ at the point x.

We refer to a function ϕ:U → Rn as differentiable if it is differentiable atevery point of the domain U . The following is an exercise.

2

Proposition 1.2 Any differentiable function ϕ:U → Rn is continuous.

It is straightforward to prove that the total differential at a point, if itexists, is unique. Further, if we have two functions ϕ1, ϕ2:U → Rn that areboth differentiable at a point x ∈ U , then for any α1, α2 ∈ R, the functiony 7→ α1ϕ1(y) + α2ϕ2(y) is also differentialble at x, with

D(α1ϕ1 + α2ϕ2)x = α1(Dϕ1)x + α2(Dϕ2)x

Example 1.3 Let T :V → W be a bounded linear map. Then, for any pointx ∈ V and v ∈ V , we have

T (x+ v) = T (x) + Tv = T (x) + Tv + ‖v‖ · 0

It follows that the map T is differentiable, with total derivative T at anypoint x ∈ V .

The following is similarly easy to check.

Example 1.4 Let c:R → W be a constant map, where R ⊆ V is an open set,and V and W are normed vector spaces. Then the map c is differentiable, andthe total derivative is 0.

The following result is known as the chain rule. We omit the proof.

Theorem 1.5 Let U , V , and W be normed vector spaces. Let R ⊆ U andS ⊆ V be open sets, and let ϕ:R→ S and ψ:S →W be functions.

Suppose that the function ϕ is differentiable at the point x ∈ R, and thefunction ψ is differentiable at the point ϕ(x) ∈ S. Then the composite ψϕ:R→W is differentiable at the point x ∈ R, with total derivative

D(ψ ϕ)x = (Dψ)ψ(x) (Dϕ)x

2

Definition 1.6 Let U ⊆ Rm be open. We call a map ϕ:U → Rn a smooth ifit is infinitely differentiable, ie: the function Dϕ:Rm → Rn is differentiable, thefunction D(Dϕ):Rm → Rn is differentiable, the function D(D(Dϕ)):Rm → Rnis differentiable, and so on.

More generally, let X ⊆ Rm. We call a map f :X → Rn smooth if there is anopen neighbourhood U ⊇ X and a smooth map F :U → Rn such that F |U = f .

Given subsets X ⊆ Rm and Y ⊆ Rn, we call a smooth map ϕ:X → Y adiffeomorphism if there is a smooth inverse ϕ−1:Y → X.

Let U, V ⊆ Rn be open. Note that if ϕ:U → V is a diffeomorphism, foreach x ∈ Rn, by the chain rule we have that D(ϕ ϕ−1)x = D(id)x = I, whereI:Rn → Rn is the identity. Thus, for any diffeomorphism, the total derivative(Dϕ)x is invertible at each point x ∈ U .

The following result is called the local diffeomorphism theorem or inversefunction theorem, and provides a partial converse.

Theorem 1.7 Let ϕ:U → V be a smooth map between open subsets of Rn. Letx ∈ U , and suppose (Dϕ)x is invertible. Then there is an open set U ′ 3 x suchthat ϕ|U ′ :U ′ → ϕ[U ′] is a diffeomorphism. 2

3

1.2 Differentiation of Paths

A path in Rn is simply a continuous function γ: [a, b]→ Rn.

Definition 1.8 We say that the path γ is differentiable at the point t ∈ [a, b]if the derivative

γ′(t) = limh→0

γ(t+ h)− γ(t)

h

exists.We call a path γ smooth if it is infinitely differentiable, that is to say gamma′

is differentiable, (γ′)′ is differentiable, and so on.

We can rewrite the above equation

γ(t+ h) = γ(t) + γ′(t)h+ |h|r(h) limh→0

r(h) = 0

Thus, when t ∈ (a, b), the above is related to the total derivative by theformula

(Dγ)t(h) = γ′(t)h

Let V = Rn. Write γ(t) = (γ1(t), . . . , γn(t). Then it is straightforward tocheck that γ is differentiable at t if and only if each γi is differentiable at t, and

γ′(t) = ((γ1)′(t), . . . , (γn)′(t))

It follows that a path γ is constant if and only if γ′(t) = 0 for all t ∈ [a, b].This, along with the link with the total derivative, is the key to proving thefollowing.

Theorem 1.9 Let U ⊆ Rm be a connected open subset. Then a functionϕ:U → Rn is constant if and only if it is differentiable at each point x ∈ U ,with (Dϕ)x = 0.

Example 1.10 Define a path γ: [0, 2π]→ R2 by the formula

γ(t) = (cos t, sin t)

Then γ is differentiable, and γ′(t) = (− sin t, cos t).

In the above example, note that γ(0) = γ(2π) = (1, 0). So γ(2π)− γ(0) = 0.However, γ′(t) 6= 0 for all t, so there is no value c ∈ [0, 2π] such that

γ(2π)− γ(0) = (2π − 0)γ′(c)

The nearest analogy of the mean value theorem in higher dimensions is thefollowing result, which we call the mean value inequality.

Theorem 1.11 Let γ: [a, b]→ Rn be a differentiable function. Then there existsc ∈ [a, b] such that

‖γ(b)− γ(a)‖ ≤ (b− a)‖γ′(c)‖

2

4

1.3 Partial Derivatives

Let U ⊆ Rm be an open subset, and let f :R→ R be a map. Let (x1, . . . , xm) becoordinates in Rm. Then the partial derivative, ∂f

∂xj is defined by differentiatingthe function f with respect to the variable xj while treating the other variablesas constants.

More generally, if we have a function f :U → Rn, we can write

f(x1, . . . , xm) = (f1(x1, . . . , xm), . . . , fn(x1, . . . , xm))

and form the partial derivatives ∂fi

∂xj .Under many circumstances, we can swap the order of taking partial deriva-

tives. The following is sometimes known as Schwarz’s theorem.

Theorem 1.12 Let U ⊆ Rm be an open subset, and let (x1, . . . , xm) be coordi-nates in the space Rm. let i, j ∈ 0, . . . ,m, and let f :R → R be a map suchthat the second partial derivatives

∂2f

∂xi∂xj(x)

∂2f

∂xj∂xi(x)

exist and are continuous for all x ∈ R.Then

∂2f

∂xi∂xj=

∂2f

∂xj∂xi

2

Partial derivatives are useful for computations, but depend upon a choice ofbasis, and are therefore less useful theoretically.

Example 1.13 Define a function f :R2 → R by the formula

f(x, y) =xy3

x2 + y6(x, y) 6= (0, 0)

and f(0, 0) = 0.Then the partial derivatives ∂f

∂x (0, 0) and ∂f∂y (0, 0) exist. However, the func-

tion f is not continuous, and hence not differentiable, at the point (0, 0).

The following is simply a case of manipulating the definitions.

Proposition 1.14 Let U ⊆ Rm be an open subset, and let f :U → Rn bedifferentiable at the point x ∈ R. Let e1, . . . , em and e′1, . . . , e′n be thestandard bases of the spaces Rm and Rn respectively.

Then we have

(Df)x(ej) =∂f1

∂xj(x)e′1 + · · ·+ ∂fn

∂xj(x)e′n

2

5

Hence the total derivative (Df)x has matrix(∂fi

∂xj (x))

with respect to the

standard bases. This matrix is sometimes called the Jacobian matrix of f .Let y = f(x), and suppose we have a map g: f [U ] → Rp that is differen-

tiable at the point y. Abusing notation somewhat, let us write (f1, . . . , fn) todenote coordinates in the space Rn. Then the total derivative (Dg)y has ma-

trix(∂gi

∂fj (y))

, and by the chain rule the matrix of the derivative D(g f)x is

obtained by matrix multiplication of the above two matrices. In other words,the chain rule can be written

∂(g f)i

∂xj=

n∑k=1

∂gi

∂fk∂fk

∂xj.

Theorem 1.15 let U ⊆ Rm be an open subset, and let ϕ:U →Rn. Let (x1, . . . , xm) be coordinates in Rm, write ϕ(x1, . . . , xm) =(ϕ1(x1, . . . , xm), . . . ϕn(x1, . . . , xm)), and let x ∈ U .

Suppose that the partial derivatives of the function ϕ all exist on U , and arecontinuous at the point x. Then the total derivative (Dϕ)x exists, and is given

by the matrix(∂ϕi

∂xj

)with respect to the standard bases of the vector spaces Rm

and Rn. 2

In particular, note that ϕ is smooth if and only if all partial derivatives ofϕ of all orders exist.

Example 1.16 Define a map f :R2 → R2 by the formula

f(x, y) = (ex+y, x2 + y2)

Write f1 = ex+y and f2 = x2 + y2. Then

∂f1

∂x= ex+y ∂f1

∂y= ex+y

and∂f2

∂x= 2x

∂f2

∂y= 2y

These partial derivatives are all continuous. Hence the total differential isthe linear transformation defined by the matrix

(Df)(x,y) =

(ex+y ex+y

2x 2y

)The following is an immediate consequence of theorem 1.15 and the local

diffeomorphism theorem.

Corollary 1.17 Let f :U → V be a map between subsets of Rn. Let x ∈ U , and

suppose that the matrix of partial derivatives(∂fi

∂xj

)exists and is continuous,

and has non-zero determinant at the point x ∈ U . Then there is an open setU ′ 3 x such that ϕ|U ′ :U ′ → ϕ[U ′] is a diffeomorphism. 2

6

2 Manifolds

2.1 Topological Manifolds

A manifold of dimension n is a nice topological space that is locally homeomor-phic to Rn. More precisely, we have the following.

Definition 2.1 An n-dimensional manifold is a subspace M ⊆ RN such thatfor every point x ∈ X there are open sets V 3 x and U ⊆ Rn together with ahomeomorphism φ:U → V .

In section ??, we shall see that the condition that M ⊆ RN is not reallyneeded, and any compact metric space that is locally homeomorphic to Rn ishomeomorphic to a subspace of RN for some N , and so is a manifold.

Definition 2.2 Let M be a topological space. A triple, (U, V, φ), where U ⊆Rn, V ⊆ M are open subsets, and φ:U → V is a homeomorphism is called achart. A collection of charts (φλ, Uλ, Vλ) | λ ∈ Λ such that

⋃λ∈Λ Vλ = M is

called an atlas for M .

By definition, a subset of RN is a manifold if and only if it has an atlas.

Example 2.3 Euclidean space Rn is a manifold of dimension n. There is anatlas consisting of a single chart, (Rn,Rn, id).

Example 2.4 The sphere

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

is a manifold of dimension n. Observe that for any point x ∈ Sn the spaceSn\x is homeomorphic to Rn. Consequently there is an atlas containing twocharts.

Proposition 2.5 Let M be an m-dimensional manifold, and let M ′ be an n-dimensional manifold. Then the Cartesian product, M ×M ′ is an (m + n)-dimensional manifold.

Proof: Let M ⊆ RN and M ′ ⊆ RN ′ . Then M ×M ′ ⊆ RN+N ′ . Let

(Uλ, Vλ, φλ) | λ ∈ Λ

and(U ′µ, V ′µ, φ′µ) | µ ∈ Λ′

be atlases for the manifolds M and M ′ respectively.Each product Uλ×U ′µ is an open subset of Euclidean space Rm×Rn = Rm+n.

Hence the product M ×M ′ is an (m+ n)-dimensional manifold, with atlas

(Uλ × U ′µ, Vλ × V ′µ, (φλ, φ′µ)) | (λ, µ) ∈ Λ× Λ′

2

Example 2.6 The torus, T 2 is defined to be the product S1×S1 of two circles.

7

2.2 Smooth Manifolds

Definition 2.7 We call a subspace M ⊆ RN a smooth manifold of dimensionn if for every point x ∈ X there are open sets V 3 x and U ⊆ Rn together witha diffeomorphism φ:U → V .

The triple, (U, V, φ), where U ⊆ Rn, V ⊆ M are open subsets, andφ:U → V is a homeomorphism is called a smooth chart. A collection of charts(φλ, Uλ, Vλ) | λ ∈ Λ such that

⋃λ∈Λ Vλ = M is called a smooth atlas for M .

By definition, a subset of RN is a smooth manifold if and only if it has asmooth atlas.

All of the examples in the previous section are in fact smooth manifolds. Wewill only look at one such in detail for now.

Example 2.8 The sphere

S2 = (x, y, z) ∈ R3 | x2 + y2 + z2 = 1

is a smooth manifold of dimension 2. To see this, let V +z = (x, y, z) ∈

S2 |z > 0, V −z = (x, y, z) ∈ S2 |z < 0, and define V ±x , V ±y similarly. ThenU±x , U±y , U±z is an open cover of S2.

Let U = (x, y) ∈ R2 | x2 +y2 < 1, and define a diffeomorphism ϕ:U → V +z

by

ϕ(x, y) = (x, y,√

1− x2 − y2).

Diffeomorphisms U → V −z , U → V ±x , and U → V ±y are defined similarly. Sowe have a smooth atlas for S2.

Smooth maps between smooth manifolds are important in differential geom-etry. We usually check the smoothness of a map in terms of the ’coordinates’defined by the smooth charts. Specifically, by definition we have the followingobvious result.

Proposition 2.9 Let M and M ′ be manifolds, with smooth atlases(φλ, Uλ, Vλ) | λ ∈ Λ and (φµ, Uµ, Vµ) | µ ∈M respectively.

A map f :M → M ′ is smooth if and only if each map between open subsetsof Euclidean space φ−1

µ f φλ:φλf−1[Vµ]∩Uλ→ Uµ∩ φ−1

µ f [Vλ] is smooth. 2

2.3 Paracompactness

Paracompactness is a technical property of metric spaces that is useful in joiningfunctions defined locally to form a globally defined function. This idea is usefulwhen it comes to generalising constructions on Rn to constructions on manifoldsthrough charts.

Definition 2.10 Let X be a topological space, with open covers U and V. Theopen cover U is said to be a refinement of the open cover V if every open setU ∈ U is contained in a set belonging to the collection V.

The open cover U is said to be locally finite if each point x ∈ X has an openneighbourhood that is contained in only finitely many open sets of the cover U .

The space X is said to be paracompact if it is Hausdorff, and every opencover has a locally finite refinement.

8

Clearly any compact topological space is paracompact. The proof of thefollowing is rather technical, however; we will not go into details here.

Theorem 2.11 Any metric space is paracompact. 2

Paracompactness lets us define nice collections of maps called partitions ofunity.

Definition 2.12 Let X be a topological space. Then the support, supp(f), ofa function f :X → R is the closure of the set of points x ∈ X such that f(x) 6= 0.

Definition 2.13 Let Uλ | λ ∈ Λ be a locally finite open cover of a space X.Then a collection of functions

ρλ:X → [0, 1] | λ ∈ Λ

is said to be a partition of unity subordinate to the cover Uλ |λ ∈ Λ if:

• supp(ρλ) ⊆ Uλ

• For every point x ∈ X: ∑λ∈Λ

ρλ(x) = 1

Our second major technical result on paracompactness is the following.

Theorem 2.14 Let U be a locally finite open cover of a paracompact space X.Then there is a partition of unity on M subordinate to U . 2

If M is a smooth manifold, we refer to a partition of unity on M consistingof smooth functions as a smooth partition of unity. The above theorem can berefined for smooth manifolds. Specifically, if U is a locally finite open cover of asmooth manifold, then there is a partition of unity on M subordinate to U .

We now come to the promised result saying that the condition that M ⊆ RNis not needed in a manifold. Actually, we will only prove this in the compactcase.

Theorem 2.15 Let M be a compact space for where for every point x ∈ X thereare open sets V 3 x and U ⊆ Rn together with a homeomorphism φ:U → V .Then M is homeomorphic to a subset of some Euclidean space RN .

Proof: As a compact space, M is paracompact. By compactness and para-compactness we can find a finite atlas

(Ui, Vi, φi) | i = 1, . . . , k

and maps ρi:M → [0, 1] such that supp(ρi) ⊆ Vi and∑i ρi(x) = 1 for every

point x ∈M .The product U1 × · · · ×Uk is a subset of some Euclidean space RN . We can

thus define a map φ:M → RN by the formula

φ(x) = (ρ1(x)φ−11 (x), . . . , ρk(x)φ−1

k (x))

Now the map φ is an injective continuous map from a compact space to aHausdorff space, and therefore a homeomorphism onto its image. 2

9

Example 2.16 Let us define an equivalence relation, ∼, on the sphere Sn byidentifying opposite points. Then real projective space, RPn, is defined to bethe quotient

RPn = Sn/ ∼The space RPn is a manifold of dimension n. Since the space RPn is compact,

by the above we need only construct an atlas.Recall from example 2.8 that the sphere Sn is an n-dimensional manifold.

Any point x ∈ Sn has an open neighbourhood that does not contain the oppositepoint, −x. Consequently we can form an atlas

(Uλ, Vλ, φλ) | λ ∈ Λ

for the sphere Sn such that if x ∈ Vλ then −x 6∈ Vλ.It is now easy to see how to proceed; let q:Sn → RPn be the quotient map.

Then we can define an atlas

(Uλ, q[Vλ], q φλ) | λ ∈ Λ

for the space RPn.

An alternative description of projective space RPn is as the set of one-dimensional subspaces of the vector space Rn+1. Such a subspace determines apair of opposite points on the sphere Sn.

2.4 Boundaries

Definition 2.17 We define the upper half-space, Rn+, to be the space:

(x1, . . . , xn) ∈ Rn | xn ≥ 0

An n-dimensional manifold with boundary is a subspace M ⊆ RN such thatfor every point x ∈ X there are open sets V 3 x and U ⊆ Rn+ together with ahomeomorphism φ:U → V .

An n-dimensional smooth manifold with boundary is a subspace M ⊆ RNsuch that for every point x ∈ X there are open sets V 3 x and U ⊆ Rn+ togetherwith a diffeomorphism φ:U → V .

We will sometimes write just ‘manifold’ when we really mean ‘manifold withboundary’. When indulging in such an abuse of terminogy, a manifold in thesense defined in the previous section (ie: a manifold with no boundary) is termeda closed manifold.

A triple (U, V, φ) in the above definition is again called a chart, and a col-lection of coordinate charts covering the space M is again called an atlas.

Definition 2.18 Let M be an n-dimensional manifold with boundary. Thenwe define the boundary, ∂M , to be the set of all points x ∈ M that contain noneighbourhood homeomorphic to an open subset of the space Rn.

The complement of the boundary, M0 = M\∂M , is called the interior.

Thus, if x ∈ ∂M , and we have a chart φ:U → V , where x ∈ V , thenφ(x1, . . . , xn−1, 0) = x.

A manifold with boundary, M , is not a closed manifold unless ∂M = ∅.However, the interior M0 is a closed manifold, with the same dimension as thatof the bounded manifold M .

10

Example 2.19 The half space, Rn+, is an n-dimensional manifold with bound-ary. The boundary is the space

(x1, . . . , xn−1, 0) | xi ∈ R

which is just a copy of the Euclidean space Rn−1.

Example 2.20 The closed disk,

Dn

= (x1, . . . , xn) ∈ Rn |n∑i=1

(xi)2 ≤ 1

is an n-dimensional manifold with boundary. The boundary is the (n−1)-sphere,Sn−1.

The above two examples suggest the following result. Thr proof is straight-forward.

Proposition 2.21 Let M be an n-dimensional manifold with boundary. Thenthe boundary, ∂M , is an (n− 1)-dimensional closed manifold. If M is smooth,then so is ∂M . 2

Proposition 2.22 Let f :M → N be a homeomorphism between manifolds withboundary. Then f [∂M ] = ∂N .

Proof: Let the manifold M (and so N) have dimension n. Let x ∈ ∂M .Then the point x has no neighbourhood homeomorphic to an open subset ofthe space Rn. Since the map f is a homeomorphism, the point f(x) ∈ N hasno neighbourhood homeomorphic to an open subset of the space Rn, and sof(x) ∈ ∂N .

Repeating the above argument with the homeomorphism f−1 tells us thatf [∂M ] = ∂N . 2

The corresponding result holds for diffeomorphisms between smooth mani-folds.

Our next result lets us ”glue together” manifolds along boundaries, and buildmany more examples. It is geometrically quite clear.

Proposition 2.23 Let M+ and M− be n-dimensional manifolds with boundary.Suppose we have a homeomorphism f : ∂M+ → ∂M−. Then the union

M+ ∪f M− =M+

∐M−

x ∼ f(x)

is an n-dimensional manifold.Further, if M+ and M− are smooth manifolds, and the map f is a diffeo-

morphsim, then the manifold M+ ∪f M− is also smooth. 2

Example 2.24 Let M be a manifold. If we remove a small open neighbourhoodhomeomorphic to the open disk, we obtain a manifold with boundary. Theboundary is homeomorphic to the sphere, Sn.

The connected sum, M]N , of two manifolds M and N is defined by removingsmall open disks from each and joining together along the common boundaryaccording to the above proposition.

The join of k tori is sometimes termed the k-holed torus.

11

3 The Tangent Bundle

3.1 Vector Bundles

A vector bundle of dimension n over X is a space that locally resembles theproduct of X with an n-dimensional vector space. The precise definition isperhaps clearest when broken up into two parts, introducing some other relevantnotions along the way.

Definition 3.1 A pre-Rn-bundle over X is a space E equipped with a con-tinuous projection map p:E → X such that for each point x ∈ X the fibreEx = p−1(x) is an n-dimensional real vector space, with the expected topology.

A pre-Rn-bundle is also termed a real pre-vector bundle of dimension n.

Example 3.2 For any space X, X × Rn is a Rn-bundle, with projection mapdefined by the formula p(x, v) = v where x ∈ X and v ∈ Rn. We call it theproduct bundle over X with dimension n.

Definition 3.3 Let E and F be pre-vector bundles over a space X. A bundlemap is a continuous map φ:E → F such that we have a commutative diagram

Eφ //

@@@

@@@@

F

~~~~~~

~~~

X

with the projection maps and the maps of fibres φx:Ex → Fx are all linear. Abundle isomorphism is a bundle map whose inverse is also a bundle map.

We are now ready for our main definition.

Definition 3.4 An Rn-bundle over a space X is a pre-Rn-bundle such that foreach point x ∈ X there is an open neighbourhood U such that the restrictionEU = p−1[U ] can be equipped with a bundle isomorphism U × Rn → EU .

An Rn-bundle is also termed a real vector bundle of dimension n. It issimilarly possible to define Cn-bundles, or complex vector bundles.

A pair (U,ψ) where U is an open subset of X and ψ:U×Rn → EU is a bundleisomorphism is called a local trivialisation. An atlas of local trivialisations for anRn-bundle E is a collection of local trivialisations (Uλ, ψλ) | λ ∈ Λ such that⋃λ∈Λ Uλ = X. By definition any Rn-bundle has an atlas of local trivialisations.

Such an atlas completely determines the structure of the bundle.A one-dimensional vector bundle is sometimes referred to as a line bundle.

Definition 3.5 Let M be a smooth manifold. Then a vector bundle E overM is termed a smooth vector bundle if the space E is a smooth manifold, theprojection map p:E →M is smooth, and E has an atlas of local trivialisationsconsisting of diffeomorphisms.

12

Example 3.6 Let RPn be real projective space. We can view the space RPnas the space of one-dimensional subspaces of the vector space Rn. We definethe canonical line bundle, E, over the space RPn to be the subspace of theCartesian product RPn × Rn+1 consisting of pairs (V, v) where V ⊆ Rn+1 isa one-dimensional subspace, and v ∈ V . The structure map p:En → RPn isdefined by the formula p(V, v) = V .

The following result provides a useful way of checking when a bundle mapis an isomorphism. The proof depends on the fact that any vector bundle islocally trivial.

Proposition 3.7 Let φ:E → F be a bundle map. Then the map ψ is a bundleisomorphism if and only if each map of fibres ψ:Ex → Fx is an isomorphism ofvector spaces.

Proof: By definition each map of fibres φx:Ex → Fx is a vector space isomor-phism if the map φ is a bundle isomorphism.

To prove the converse, suppose initially that E = X ×Rn and F = X ×Rn.Then we have a collection of vector space isomorphisms φx: x×Rn → x×Rn.The isomorphisms Tx:Rn → Rn defined by the formula

ψ(x, v) = (x, Tx(v))

depend continuously on the point x ∈ X. Hence the inverses, T−1x , also depend

continuously on the point x ∈ X and we have a bundle map φ−1:E → F definedby the formula

φ−1(x, v) = (x, T−1x (v))

In the general case, consider atlases of local trivialisations (Uλ, ψEλ ) | λ ∈ Λand (Uλ, ψFλ ) | λ ∈ Λ for the bundles E and F respectively. Then for eachelement λ ∈ Λ we have a bundle map

φλ:Uλ × Rn → Uλ × Rn

defined by forming the composition (ψFλ )φ(ψEλ )−1. By the above calculation,

this bundle map has an inverse, φ−1λ .

We can therefore define an inverse to the bundle map φ by writing

φ−1(v) = (ψEλ )φ−1λ (ψFλ )−1(v)

whenever v ∈ FUλ . 2

The following result is proved similarly.

Proposition 3.8 Let φ:E → F be a smooth bundle map. Then the map ψ is asmooth bundle isomorphism if and only if each map of fibres ψ:Ex → Fx is anisomorphism of vector spaces. 2

Definition 3.9 Let E be an Rn-bundle over a space X. Then the bundle E iscalled trivial if it is isomorphic to the product bundle X × Rn.

We can make a similar definition in the smooth case.

13

Example 3.10 Consider the canonical line bundle, E1, over the projective lineRP1 defined in example 3.6. Let q:S1 → RP1 be the quotient map identifyingopposite points on the circle, S1. Then the bundle E1 is the set of points of theform

(q(cos θ, sin θ), t(cos θ, sin θ))

where θ ∈ [0, π] and t ∈ R.The quotient map q identifies the opposite ends of the strip [0, π]×R under

correspondence(0, t) 7→ (π,−t)

Thus the space E1 is a Mobius band. In particular, the space E1 is nothomeomorphic to the product RP1 × R (this is visually quite clear, thoughto prove it rigorously, we need to introduce the concept of orientation) so thebundle E1 is not trivial.

3.2 Tangent Space

Definition 3.11 Let M ⊆ RN be a smooth manifold of dimension n. Then wedefine the tangent space

TxM = γ′(0) ∈ RN‖γ: (−δ, δ)→M smooth , γ(0) = x.

We define the tangent bundle

TM =⋃x∈Mx × TxM ⊆M × RN .

We have a smooth surjection p:TM → M given by the obvious formulap(x, v) = x, with fibres p−1(x) ∼= TxM . We claim that TM is a real vectorbundle of dimension n.

Proposition 3.12 Pick a chart (U, V, φ) for M where x ∈ V . Set φ−1(x) =(x1, . . . , xn), and

γi(t) = (x1, . . . , xi−1, xi + t, xi+1, . . . , xn).

Set∂

∂xi= (φ γi)′(0).

Then ∂

∂x1, . . . ,

∂

∂xn

is a basis for TxM .

Proof: Let v ∈ TxM . Then v = γ′(0) for some smooth path γ: (−δ, δ) → Vwith γ(0) = x. Let q = φ−1(x).

Define y: (−δ, δ)→ U by y = φ−1 γ. Write

y(t) = (y1(t), . . . , yn(t))

so yi(0) = xi. Set αi = (yi)′(0). Then

y′(0) = α1e1 + · · ·+ αnen

14

where e1, . . . , en is the standard basis for Rn.Now γ = φ y, so by the chain rule

v = γ′(0) = (Dφ)q(y′(0)) = α1(Dφ)qe1 + · · ·+ αn(Dφ)1en

Recallγi(t) = (x1, . . . , xi−1, xi + t, xi+1, . . . , xn).

so∂

∂xi= (φ γi)′(0) = (Dφ)q(γ

i)′(0) = (Dφ)qei.

Hence

v = α1 ∂

∂x1+ · · ·+ αn

∂

∂xn

and ∂

∂x1, . . . ,

∂

∂xn

spans TxM .

Now, suppose

β1 ∂

∂x1+ · · ·+ βn

∂

∂xn= 0

where βi ∈ R.By the above

∂

∂xi= (φ γi)′(0) = (Dφ)qei.

so our equation tells us that

(Dφ)q(β1e1 + · · ·+ βnen) = 0

By the chain rule,(Dφ−1)x(Dφ)q = id

Rn

so, applying the linear map (Dφ−1)x, we have

β1e1 + · · ·+ βnen = 0.

The standard basis e1, . . . , en is certainly linearly independent. Henceβi = 0 for all i, and the set

∂

∂x1, . . . ,

∂

∂xn

is also linearly independent. 2

It follows that TM is a smooth pre-Rn-bundle over M .

Proposition 3.13 The manifold TM is a smooth Rn-bundle over M .

15

Proof: Let x ∈ M . Let (U, V, φ) be a chart, where x ∈ V . Writeφ(x1, . . . , xn) = x. Then we can define a local trivalisation ψ:V ×Rn → (TM)Vby

φ(x, α1, . . . , αn) = (x, α1∂

∂x1+ · · ·+ αn

∂

∂xn).

2

Let M be a smooth manifold, and x ∈ M . Suppose we have smooth charts(U, V, φ) and (U , V , φ) such that x = φ(x1, . . . , xn) = φ(y1, . . . , yn). Then we

have two possible bases

∂∂x1 , . . . ,

∂∂xn

and

∂∂y1 , . . . ,

∂∂yn

for TxM . They are

related by the following change of coordinate rule.

Proposition 3.14

∂

∂xj=

n∑i=1

∂yi∂xj

∂

∂yi.

2

The proof is a straightforward application of the chain rule.

3.3 Differentiation

Before talking about differentials of smooth maps between manifolds, we needto extend the notion of a bundle map.

Definition 3.15 Let f :X → Y be a continuous map. Let E be a vector bundleover the space X, and let F be a vector bundle over the space Y . Then a bundlemap covering the map f is a map ψ:E → F fitting into a commutative diagram

Eψ //

F

X

f // Y

such that each map of fibres ψx:Ex → Ff(x) is a homomorphism of vectorspaces.

We can similarly define smooth bundle maps covering a smooth map.

Definition 3.16 Let f :X → Y be a continuous map. Let E be a vector bundleover the space Y with structure map p. Then we define the pullback, f?(E), tobe the subspace of the Cartesian product E ×X consisting of pairs (v, x) suchthat p(v) = f(x). The structure map is defined by the formula p?(v, x) = x.

It is straightforward to show that the above pullback is a vector bundle overthe space X. It has fibres

f?(E)x = Ef(x)

Proposition 3.17 Suppose we have a bundle map ψ:E → F covering a mapf :X → Y such that each map of fibres ψx:Ex → Ff(x) is an isomorphism. Thenthe vector bundle E is isomorphic to the pullback f?[F ].

16

Proof: Let v ∈ Ex. Then φ(v) ∈ Ff(x) = f?(F )x. We can therefore definea bundle map ψ′:E → f?(F ) by writing ψ′(v) = φ(v). The map ψ′ is anisomorphism as it is an isomorphism on each fibre. 2

The following is straightforward to prove.

Lemma 3.18 Let M and N be manifolds, and let f :M → N be a smooth map.Let x ∈M . Then we have a well-defined linear map (Df)x:TxM → TxN givenby the formula

(Df)x(γ′(0)) = (f γ)′(0).

Let (U, V, φ) is a smooth chart for M where x ∈ V , and x = φ(x1, . . . , xm),and let (U , V , φ) be a smooth chart for N where f(x) = φ(y1, . . . , yn). Then

∂

∂xj=

m∑i=1

∂yi∂xj

∂

∂yi.

2

Proposition 3.19 Let f :M → N be a smooth map. Then we have a bundlemap Df :TM → TN covering f defined by the formula

Df(x, v) = (f(x), (Df)x(v)) x ∈M, v ∈ TxM.

2

The definition of smooth manifold along with the chain rule and the localdiffeomorphism theorem for open subsets of Rn immediately gives us a general-isation of the local diffeomorphism theorem for manifolds.

Theorem 3.20 Let M and N be smooth manifolds, and let f :M → N be asmooth map. Let x ∈ M , and suppose (Dϕ)x:TxM → Tf(x)N is invertible.Then there is an open set U 3 x such that f |U :U → f [U ] is a diffeomorphism.2

4 Critical and Regular Values

4.1 Submersions, Immersions, and Embeddings

Definition 4.1 Let f :M → N be a smooth map between manifolds. Then wecall a point x ∈ M a critical point if the differential (Df)x:TxM → Tf(x)N isnot surjective.

Let M be an m-dimensional manifold, and N an n-dimensional manifold.Then (Df)x:TxM → Tf(x)N is a linear map from an m-dimensional vectorspace to an n-dimensional vector space. Thus, if m < n, every x ∈ M is acritical point of f .

The notion is more meaningful if m ≥ n.

17

Example 4.2 Let f :M → R be a smooth map. Then a point x ∈ M is acritical point if and only if the induced map f :TxM → Tf(x)R is zero. In termsof local coordinates (x1, . . . , xn) this means that:

∂f

∂x1= 0 · · · ∂f

∂xn= 0

Definition 4.3 A map f :M → N without critical points is called a submersion.

Example 4.4 The inclusion i:U → M of an open subset in a manifold is asubmersion.

Example 4.5 Let m > n. The projection p:Rm → Rn, defined by the formula

p(x1, . . . , xm) = (x1, . . . , xn)

is a submersion.

Proposition 4.6 Let f :M → N be a smooth map. Suppose a point x ∈ M isnot a critical point. Then there is an open neighbourhood U 3 x such that therestriction f |U :U → N is a submersion.

Proof: Since the point x ∈ M is not critical, the linear transformation(Df)x:TxM → Tf(x)M has the greatest possible rank. The rank of a matrix isthe size of the largest square submatrix with non-zero determinant.

Picking out charts in a neighbourhood of the space x and f(x), and looking atthe matrix of the transformation (Df)x with respect to the bases ∂

∂x1 , . . . ,∂

∂xm and ∂

∂y1 , . . . ,∂∂yn for TxM and Tf(x)M arising from these charts, the result

follows once we note that the determinant function is continuous. 2

Corollary 4.7 The set of critical points of a smooth map is closed. 2

Definition 4.8 An immersion is a smooth map f :M → N such that for everypoint x ∈M the differential (Df)x:TxM → Tf(x)N is injective.

Example 4.9 The map f :R→ R2 defined by the formula

f(x) = (x, |x|)

is not an immersion.

In general, immersions can exhibit some fairly strange behaviour. In orderto cut down on such behaviour we make the following definition.

Definition 4.10 An immersion f :M → N is called an embedding if the mapf :M → f [M ] is a homeomorphism onto its image.

We call a subset M ⊆ N an embedded submanifold if the inclusion M → N isan embedding. If M ⊆ N is an embedded submanifold, we consider the tangentspace TxM to be a subset of the tangent space TxN .

18

Example 4.11 The inclusion i:U →M of an open subset in a manifold is anembedding.

Example 4.12 Let m < n. The map i:Rm → Rn, defined by the formula

p(x1, . . . , xm) = (x1, . . . , xm, 0, . . . , 0)

is an embedding.

Example 4.13 Let M ⊆ RN be a smooth manifold. Then the inclusion M →RN is an embedding.

The following result is proved in the same way as proposition 4.6

Proposition 4.14 Let f :M → N be a smooth map. Suppose we have a pointx ∈ M such that the differential (Df)x:TxM → Tf(x)M is injective. Thenthere is an open neighbourhood U 3 x such that the restriction f |U :U → N is aimmersion. 2

Theorem 4.15 Let f :M → N be a smooth map, and suppose that the differ-ential Df :TxM → Tf(x)N is injective. Then there are smooth charts (U, V, φ)

and (U , V , φ) where x ∈ V and f(x) ∈ V such that

φ−1fφ(x1, . . . , xm) = (x1, . . . , xm, 0, . . . , 0)

Proof: Pick smooth charts (U1, V1, φ1) and (U2, V2, φ2) where x ∈ V1 andf(x) ∈ V2. Write

φ−12 fφ1(x1, . . . , xm) = (y1, . . . , yn)

Since the differential (Df)x:TxM → Tf(x)N is injective, we can assumewithout loss of generality (by permuting the coordinates if necessary) that thedeterminant of the matrix

det

(∂yi∂xj

)mi,j=1

is non-zero at the point x.Hence, by the local diffeomorphism theorem, the map (x1, . . . , xm) 7→

(y1, . . . , ym) is a diffeomorphism in some neighbourhood of the point φ−11 (x).

We therefore have a smooth chart (U3, V3, φ3) where x ∈ V3, and

φ−12 fφ3(y1, . . . , ym) = (y1, . . . , ym, zm+1, . . . , zn)

Now, let us define a map g in a neighbourhood of the point f(x) by theformula

φ−13 gφ2(y1, . . . , yn) = (y1, . . . , ym, ym+1 − zm+1, . . . , yn − zn)

Then certainly

φ−13 g f φ3(y1, . . . , ym) = (y1, . . . , ym, 0, . . . , 0)

We need to check that the map g is a diffeomorphism in a neighbourhood ofthe point f(x). Observe that the differential D(φ−1

3 gφ2) has determinant

det

(1 0? −1

)6= 0

19

at the point φ−12 f(x). Therefore, by the local diffeomorphism theorem, the map

g is indeed a diffeomorphism in a neighbourhood of the point f(x), and we aredone. 2

The following result is proved similarly.

Theorem 4.16 Let f :M → N be a smooth map, and suppose that a pointx ∈ M is not critical. Then there are smooth charts (U, V, φ) and (U , V , φ)where x ∈ V and f(x) ∈ V such that

φ−1fφ(x1, . . . , xm) = (x1, . . . , xn)

2

Corollary 4.17 Suppose that the manifolds M and N are closed. Let f :M →N be a smooth map, and let y ∈ N be a regular value. Then the inverse imagef−1(y) is an embedded submanifold of M of dimension m− n.

Proof: If y is not in the image of f , the statement is vacuously true. Otherwise,choose a point x ∈ f−1(y). Then by the above, we have smooth charts (U, V, φ)and (U , V , φ) such that x ∈ V , y ∈ V , and

φ−1fφ(x1, . . . , xm) = (x1, . . . , xn)

Without loss of generality, suppose that y = φ(0, . . . , 0). Then the set V ∩f−1(y) is the set of images under φ of points that take the form

(0, . . . , 0, xn+1, . . . , xm)

It follows that the inverse image f−1(y) is a manifold of dimension m − n.It is clear that the inclusion f−1(y) →M is an embedding. 2

A similar result applies to manifolds with boundary.

Corollary 4.18 Let M and N be smooth manifolds with boundary. Let f :M →N be a smooth map, and let y ∈ N be a regular value. Then the inverse imagef−1(y) is an embedded submanifold of M , with dimension m− n and boundaryf−1(y) ∩ ∂M . 2

We now have a handy way to define manifolds by looking at smooth mapsf :Rm → Rn.

Example 4.19 Define f :Rn+1 → R by the formula

f(x1, . . . , xn) =∑i

(xi)2

Then f is a smooth map, and 1 is a regular value. Therefore the sphere

Sn−1 = f−1(1)

is an n-dimensional embedded submanifold of Rn.

20

4.2 Sard’s Theorem

Let f :M → N be a smooth map. Sard’s theorem is the statement that ‘almostall’ points of the manifold N are regular values. It is one of the most importanttools in differential topology. In order to frame a more precise statement of thetheorem, we need a small amount of measure theory.

Recall that we define the volume of a cuboid

Q = [a1, b1]× [a2, b2]× · · · × [an, bn] ⊆ Rn

to be the product

V olume(Q) =

n∏i=1

(bi − ai)

A subset S ⊆ Rn is said to have measure zero if for every ε > 0 we can finda sequence of cuboids (Qk)∞k=1 such that:

• S ⊆⋃∞k=1Qk

•∑k V olume(Qk) < ε

Example 4.20 Any countable set has measure zero. A countable union of setsof measure zero has measure zero.

Example 4.21 The hyperplane Rm−1 × 0 has measure zero as a subset ofthe space Rm.

The following is straightforward; we will need it later on.

Proposition 4.22 Consider a subset K ⊆ Rm. Suppose that for every realnumber t ∈ R the intersection with the hyperplace Rm−1×t has measure zeroas a subset of the hyperplane Rm−1 × t. Then the set K has measure zero asa subset of Rm. 2

Definition 4.23 Let M be a smooth manifold of dimension n. Then a subsetS ⊆M is said to have measure zero if there is a countable set of charts

(Uλ, Vλ, φλ) | λ ∈ Λ

such that S ⊆⋃λ∈Λ Vλ, and each set φ−1

λ (Vλ ∩ S) has measure zero in Rn.

The following topological fact sometimes makes it easier to prove that a sethas measure zero.

Proposition 4.24 Let M be a smooth manifold. Then M has a countablesmooth atlas. 2

We are now ready to state Sard’s theorem. Although we state it for closedmanifolds, Sard’s theorem is also valid for manifolds with boundary.

Theorem 4.25 Let M and N be smooth closed manifolds. Let f :M → N bea smooth map. Then the set of critical values is a set of measure zero in themanifold N .

21

In particular, a randomly chosen point in the manifold N will be a regularvalue of the map f . Of course, we consider a point of the manifold N that doesnot belong to the image of the map f to be a regular value.

By corollary 4.17, if the manifold M has dimension m, and the manifold Nhas dimension n, for a randomly chosen point y ∈ N , the inverse image f−1(y)is an embedded submanifold of dimension m− n.

Most of the hard work in the proof of Sard’s theorem is contained in thefollowing special case.

Theorem 4.26 Let U ⊆ Rm be an open subset. Let f :U → Rn be a smoothmap. Then the set of critical values of f has measure zero.

Proof of Sard’s theorem: Let f :M → N be a smooth map. By proposition4.24 we can find smooth countable atlases

(Uλ, Vλ, φλ) | λ ∈ Λ

and(Uµ, Vµ, φµ) | µ ∈M

for the manifolds M and N respectively.The composites φ−1

µ f φλ are all smooth, and a point x ∈M is a critical

point if and only if the inverse image φ−1λ (x) is a critical point of the composite

φ−1µ f φλ for some λ ∈ Λ and µ ∈ M . By theorem 4.26 the set of critical

values of the composite φλ f has measure zero.The result now follows from the definition of a subset of a manifold being of

measure zero. 2

We now come to the proof of our special case. Let U ⊆ Rm be an opensubset, and let f :U → Rn be a smooth map. Let us write C to denote the setof critical points of f , and Ci to denote the set of points x ∈ U such that allpartial derivatives of order less than or equal to i vanish.

The proof of theorem 4.26 is then divided into three lemmas.

Lemma 4.27 The image f [C\C1] is a subset of measure zero in Rn.

Lemma 4.28 The image f [Ci\Ci+1] is a subset of measure zero in Rn.

Lemma 4.29 The image f [Cm] is a subset of measure zero in Rn.

Proof of lemma 4.27: We work by induction on the dimension, m, of thedomain. The lemma is clearly true when m = 0.

Consider a point x ∈ C\D. Since the differential of the function f does notvanish at the point x, there is a projection p:Rn → R such that the compositep f has non-zero differential at x. Without loss of generality, let us assumethat p(x1, . . . , xn) = xn.

The point x has an open neighbourhood, V , where the differential of thecomposite p f does not vanish. Thus every point t ∈ R is a regular value ofthe composite p f |V , and the inverse image

Vt = (p f)−1(t) ∩ V

22

is a smooth submanifold of dimension m− 1.By working in local coordinates, let us assume that Vt ⊆ Rm−1. Suppose the

conclusion of the lemma is true in dimension m− 1. Then for each real numbert ∈ R the set of critical values of the function f |Vt where the differential doesnot vanish is a set of measure zero in Rm−1 × t.

Observe that a point x ∈ Vt is a critical value of the function f if and onlyif it is a critical value of the restriction f |Vt . Thus for every point t ∈ R theintersection

f [(C\D) ∩ V ] ∩ (Rm−1 × t)

is a set of measure zero in Rm−1 × t. The desired result now follows fromproposition 4.22 2

Proof of lemma 4.28: Again we work by induction on the dimension m. Theresult certainly holds when m = 0.

Choose a point x ∈ U\Ci+1. Let g be an order i derivative of the functionf such that the differential of g at x is non-zero. Then we can find an openneighbourhood of the point x in which the derivative of g is non-vanishing. Thusthe set U\Ci+1 is open.

Let us consider the restriction g|U\Ci+1. By construction, the point 0 is a

regular value of the function g|U\Ci+1. Let V = (g|U\Ei+1

)−1(0). Then V is anembedded submanifold of the space U of dimension m− 1.

By construction, g(x) = 0 for all points x ∈ Ci\Ci+1. Therefore the setEi\Ei+1 is a subset of a manifold of dimension m− 1. Looking at local coordi-nates, we can assume that the set Ci\Ci+1 is a subset of Rm−1.

Thus if the desired result holds in dimension m−1, it also holds in dimensionm. This completes the process of induction. 2

Proof of lemma 4.29: Let x ∈ U . Then we can write

f(x) = (f1(x), . . . , fn(x))

Let E be the set of points x ∈ U of the function f1 such that all partialderivatives of order less than or equal to m vanish at the point x. Then:

f [Cm] ⊆ f1[E]× Rn−1

It therefore suffices to show that the set f1[E] has measure zero in R. Weproceed by proving that the image f [Em ∩ Q] has measure zero for any m-dimensional cube Q ⊆ U .

Let s be the length of a side of the cube Q, and choose a positive integerk ∈ N. Then we can partition the cube Q into km smaller cubes, each with sidesof length s/k. The diameter of one of these smaller cubes is s

√m/k.

Consider a point x0 ∈ Em∩Q. Let Q0 be one of the smaller cubes containingthe point x0. Observe that the set of partial derivatives of the function f oforder at most m + 1 is bounded on the cube Q.1 Hence by Taylor’s theoremthere is a constant C > 0 such that:

|f1(x)− f1(x0)| ≤ C‖x− x0‖m+1

1Because the partial derivatives are all continuous and the cube is compact.

23

whenever x ∈ Q. In particular, if x ∈ Q0 then:

|f1(x)− f1(x0)| ≤ C(s√m

k

)m+1

Thus the image f1(Q0) is contained in an interval of length A/km+1 where Ais some fixed constant. The cube Q is made up of km cubes of the same size asthe cube Q0. Therefore the image f [Q∩En] is contained in a union of intervalsof length at most:

kmA

km+1=A

k

By increasing the natural number k, the length A/k can be made arbitrarilysmall. It follows that the set f1[Q ∩ E1] has measure zero. 2

4.3 The Brouwer Fixed Point Theorem

The following result is called Hirsch’s theorem after it’s discoverer.

Theorem 4.30 Let M be a smooth manifold with boundary, ∂M . Then thereis no smooth map f :M → ∂M such that f(x) = x for all x ∈ ∂M .

Proof: Suppose that f :M → ∂M is a smooth map where f(x) = x for allx ∈ ∂M . We seek a contradiction.

By Sard’s theorem, the set of critical values of f in ∂M has measure zero,so we have at least one regular value y ∈ ∂M . Then f |∂M is the identity map,so y is certainly also a regular value of f |∂M . If M has dimension n, then ∂Mhas dimension n − 1, so by corollary 4.18, f−1(y) is an embedded manifold ofdimension 1, with boundary

f−1(y) ∩ ∂M = y.

But f−1(y) is also compact. The only compact one-dimensional manifoldswith boundary are disjoint unions of closed line segments and circles. Thus theboundary of f−1(y) must have an even number of points. This is a contradiction.2

Now consider the disk

Dn+1 = (x0, x1, . . . , xn) ∈ Rn+1 | (x0)2 + (x1)2 + · · ·+ (xn)2 ≤ 1

This is an (n+ 1)-dimensional manifold with boundary

Sn = (x0, x1, . . . , xn) ∈ Rn+1 | (x0)2 + (x1)2 + · · ·+ (xn)2 = 1

Lemma 4.31 Let g:Dn+1 → Dn+1 be a smooth map. Then there is a pointx ∈ Dn+1 such that g(x) = x.

Proof: Suppose g(x) 6= x for all x ∈ Dn+1. Define a smooth map f :Dn+1 →Sn = ∂Dn+1 by setting f(x) to be the point on Sn obtained by following a linefrom g(x) through x.

Then f(x) = x if x ∈ Sn, which does not exist by the above. We concludethat g(x) = x for some x ∈ Dn+1. 2

24

A point x ∈ Dn+1 such that g(x) = x is called a fixed point of g.The Weierstrass approximation theorem asserts that given a continuous func-

tion f : [a, b]→ R, and a number ε > 0, there is a polynomial p: [a, b]→ R suchthat |p(x)− f(x)| < ε for all x ∈ [a, b].

Note in particular that polynomials are smooth. A careful argument usingthe Weierstrass aproximation theorem, combined with the above lemma, yieldsthe following result.

Theorem 4.32 (The Brouwer approximation theorem) Let g:Dn+1 →Dn+1 be a continuous map. Then there is a point x ∈ Dn+1 such that g(x) = x.2

5 Vector and Tensor Fields

5.1 Sections

Definition 5.1 A section of a vector bundle E over a space X is a continuousmap u:M → E such that u(x) ∈ Ex for all points x ∈M .

A set of sections u1, . . . , un is said to be linearly independent if and onlyif the set of vectors u1(x), . . . , un(x) is linearly independent for all pointsx ∈M .

We write Γ(E) for the set of sections on a vector bundle E. If E is a smoothvector bundle, we write Γ∞(E) to denote the set of smooth sections.

Proposition 5.2 Let E be an Rn-bundle over a space X. Then the bundle Eis trivial if and only if we can find a set u1, . . . , un of linearly independentsections.

Proof: Suppose that the bundle E is trivial. Then there is a bundle iso-morphism φ:E → X × Rn. Let e1, . . . , en be the standard basis for the spaceRn. Then we can define a set of linearly independent sections u1, . . . , un bywriting

ui(x) = ψ−1(x, ei)

Conversely, suppose we have a set of linearly independent sectionsu1, . . . , un. Then we can define a bundle map φ:E →M ⊗Rn by the formula

φ(α1u1(x) + · · ·+ αnun(x)) = (x, α1e1 + · · ·+ αnen)

The map φ is an isomorphism of vector bundles by proposition 3.7. 2

In particular, a line bundle is trivial if and only if it has a nowhere-vanishingsection. The above result has a smooth analogue.

Proposition 5.3 Let E be a smooth Rn-bundle over a manifold M . Then thebundle E is trivial if and only if we can find a set u1, . . . , un of linearlyindependent smooth sections. 2

25

Example 5.4 Recall that we define projective space, RPn by identifying op-posite points on the sphere Sn, or alternatively as the set of one-dimensionalsubspaces of the vector space Rn+1, since such a subspace determines a pair ofopposite points on the sphere Sn. In example 3.6 we defined the canonical linebundle, E, over RPn to be the space

(v, V ) | V ∈ RPn, v ∈ V

where the fibre at V ∈ RPn is the one-dimensional space V itself.The canonical line bundle, En, over projective space RPn defined in example

3.6 is non-trivial. To see this, we appeal to the above result.Suppose we have a section s:RPn → E. Let q:Sn → RPn be the map

associating a point on the sphere with the one-dimensional subspace of Rn+1

containing that point, and consider the composition sq:Sn → E. Then a pointx ∈ Sn is mapped to the pair

(q(x), t(x)x) ∈ E

for some continuous function t:RPn → R such that t(−x) = t(x).By the intermediate value theorem there must be a point x ∈ S1 such that

t(x) = 0. Thus there must be a point at which the section s is equal to zero,and therefore the bundle E is non-trivial.

We also need to consider vector bundles equipped with extra structure.

Definition 5.5 Let E be a vector bundle over a space X. Then a metric onE consists of an inner product on each fibre Ex such that for any two sectionsu, v ∈ Γ(E) the map

x 7→ 〈s(x), t(x)〉is continuous.

If E is a smooth vector bundle over a manifold M , a metric 〈−,−〉 is saidto be smooth if for any two smooth sections u, v ∈ Γ∞(E) the map

x 7→ 〈s(x), t(x)〉

is smooth.

Theorem 5.6 Let E be a smooth vector bundle over a manifold M . Then thebundle E can be equipped with a smooth metric.

Proof: To begin, observe that by theorem ?? we can find a smooth locallyfinite atlas of local trivialisations (Uλ, ψλ) | λ ∈ Λ. We can define a metric onthe restriction EUλ by writing

〈e1, e2〉λ = 〈v1, v2〉

where ψλ(ei) = (x, vi) for points ei ∈ Ex and vi ∈ Rn, and the inner product〈v1, v2〉 is the standard inner product of the vectors v1 and v2 in the space Rn.

Choose a smooth partition of unity ρλ | λ ∈ Λ subordinate to the the opencover Uλ | λ ∈ Λ. We can define a metric on the bundle E by writing:

〈e1, e2〉 =∑λ∈Λ

ρλ(x)〈e1, e2〉λ

for all points e1, e2 ∈ Ex. 2

26

5.2 Vector Fields

Definition 5.7 A vector field on a manifold M is a smooth section of thetangent bundle TM .

By proposition 5.2, if M is an n-dimensional manifold, the tangent bundleTM is trivial if and only if there are n linearly independent smooth vector fields.A manifold with a trivial tangent bundle is called parallelisable.

Example 5.8 The sphere, S3, can be viewed as the set of unit quarternions:

S3 = w ∈ H | |w| = 1

We have a smooth surjection f :R3 → S3 defined by the formula

f(θ, φ, ψ) = (cos θ + i sin θ)(cosφ+ j sinφ)(cosψ + k sinψ)

There are three linearly independent vector fields on the sphere S3 definedby the formulae:

X(f(θ, φ, ψ)) = f?(∂∂θ )

Y (f(θ, φ, ψ)) = f?(∂∂φ )

Z(f(θ, φ, ψ)) = f?(∂∂ψ )

Therefore the sphere S3 is parallelisable.

A similar argument, left as an exercise, can be used to show that the sphereS7 is parallelisable. However, it is a remarkable result due to Adams that S1,S3 and S7 are the only spheres which are parallelisable.

Recall that the tangent space TxM is defined to be the set of all derivativesγ′(0) where γ: (−δ, δ) → M is a smooth function with γ(0) = x. The followingtherefore makes sense.

Definition 5.9 Let X ∈ Γ∞(TM) be a vector field, and let f ∈ C∞(M) be assmooth function. Then we define the derivative of f along X to be the smoothfunction defined by writing

X(f)(x) = (f γ)′(0) γ: (−δ, δ)→M, γ(0) = x, γ′(0) = X(x).

Let (U, V, φ) be a chart. For x ∈M , write φ(x1, . . . , xn) = x and

X(x) = X1 ∂

∂x1+ · · ·+Xn ∂

∂xn

where X1, . . . , Xn are smooth real-valued functions.Then by the chain rule

X(f) = X1 ∂f φ∂x1

+ · · ·+Xn ∂f φ∂xn

By definition, a vector field X ∈ Γ∞(TM) is determined by the operator

f 7→ X(f)

on smooth maps f ∈ C∞(M).

27

Definition 5.10 We define the Lie bracket, [X,Y ], of vector fields X and Y bythe formula

[X,Y ](f) = X(Y (f))− Y (X(f))

where f ∈ C∞(M).

Proposition 5.11 The Lie bracket, [X,Y ], is a vector field.

Proof: With respect to some chart φ, let us write:

X =∑i

Xi ∂

∂xiY =

∑j

Y j∂

∂xj

Then:

X(Y (f)) =∑i,j

Xi ∂

∂xi

(Y j

∂f φ∂xj

)=∑i,j

XiY j∂2f φ∂xi∂xj

+Xi ∂Yj

∂xi∂f φ∂xj

Similarly:

Y (X(f)) =∑i,j

XiY j∂2f φ∂xj∂xi

+ Y i∂Xj

∂xi∂f φ∂xj

Thus we have the formula:

[X,Y ] =∑i,j

(Xi ∂Y

j

∂xi− Y i ∂X

j

∂xi

)∂

∂xj

which defines a vector field. 2

5.3 Integral Curves and Flows

A flow is a certain type of continuous deformation of a manifold.

Definition 5.12 Let M be a smooth manifold. A flow on M is a smooth mapθ:R×M →M such that:

• θ(0, x) = x for all x ∈M

• θ(s+ t, x) = θ(s, θ(t, x)) for all s, t ∈ R and x ∈M

The second of the above conditions can be written

θ(s+ t,−) = θ(s,−) θ(t,−)

Note that the above definition tells us that each map θ(s,−):M → M is adiffeomorphism, with inverse θ(−s,−). For this reason a flow on a manifold Mis sometimes called a one parameter group of diffeomorphisms.

Example 5.13 For any manifold M we can define the constant flow θ:R×M →M by writing

θ(t, x) = x

28

Example 5.14 We can define a flow θ:R× R→ R by the formula

θ(s, x) = s+ x

Most of the interesting examples of flows come from looking at vector fields.

Definition 5.15 The tangent field of a flow θ on a manifold M is the vectorfield X defined by writing

X(f)(x) = limh→0

[f(θ(h, x))− f(x)

h

]where f ∈ C∞(M) and x ∈M .

Example 5.16 The tangent field of a constant flow is the zero vector field.

Example 5.17 Consider the flow θ:R× R→ R given by the formula

θ(s, x) = s+ x

The tangent vector field is defined by writing

X(x) =d

dx

Let us say that a vector field is compactly supported if it is zero outside ofsome compact set.

Theorem 5.18 Let X be a compactly supported vector field on a closed mani-fold M . Then there is a unique flow on M with tangent field X.

Proof: Consider a smooth curve c:R→M . We can define the velocity vectordc/dt ∈ Tc(t)M by the formula

dc

dt= limh→0

[f(c(t+ h))− f(c(t))

h

]Suppose that θ is a flow with tangent field X. Then for each point x ∈ M

we can write:

dθ(t, x)

dt(f) = lim

h→0

[f(θ(t+ h, x))− f(θ(t, x))

h

]= X(f)(θ(t, x))

since θ(t+ h, x) = θ(h, θ(t, x)).Thus we have the differential equation

dθ(t, x)

dt= X(θ(t, x)) θ(0, x) = x

With respect to a chart (U, V φ), let us write x = φ(x1, . . . , xn), X = X1 ∂∂x1 +

· · ·+Xn ∂∂xn , and θ(t, x) = φ(θ1, . . . , θn). Then we have a system of equations

dθi(t, x1, . . . , xn)

dt= Xi φ(θ1, . . . , θn)

29

with initial conditions θi(0, x1, . . . , xn) = (x1, . . . , xn).It is a well-known fact from the theory of ordinary differential equations that

a system of the above form has a unique solution in some neighbourhood thatdepends smoothly on the initial conditions.Thus, given a point y ∈M , there isa neighbourhood Uy and a real number εy > 0 such that the equation

dθ(t, x)

dt= X(θ(t, x)) θ(0, x) = x

has a unique solution for x ∈ Uy and |t| < εy.Now, let the vector field X vanish outside of some compact set K. We can

find points y1, . . . , yn ∈ K such that

K ⊆n⋃i=1

Uyi

Let ε = infεy1 , . . . , εyn. Then the equation

dθ(t, x)

dt= X(θ(t, x)) θ(0, x) = x

has a unique smooth solution for x ∈ K and |t| < ε.For a point x 6∈ K the vector X(x) is zero, so the function θ(t, x) = x is a

solution. Thus we have a smooth map θ: (−ε, ε)×M →M satisfying the givendifferential equation.

Suppose that |s| < ε, |t| < ε, and |s+ t| < ε. Then:

dθ(s+ t, x)

ds= X(θ(s+ t, x))

dθ(x, θ(t, x))

ds= X(θ(s, θ(t, x)))

At s = 0:

dθ(s+ t, x)

ds

∣∣∣∣s=0

= X(θ(t, x))dθ(x, θ(t, x))

ds

∣∣∣∣s=0

= X(θ(t, x))

Thus the solutions to the above two differential equations are the same andwe have the formula

θ(s+ t, x) = θ(s, θ(t, x))

All that remains is to extend the solution θ(t, x) to hold when |t| ≥ ε. Lett ∈ R. Write

t =kε

2+ r; k ∈ Z, |r| < ε

2

If k ≥ 0 define

θ(t,−) = θ(ε

2,−) · · · θ(ε

2,−)︸ ︷︷ ︸

k times

θ(r,−)

and if k < 0 define

θ(t,−) = θ(−ε2,−) · · · θ(−ε

2,−)︸ ︷︷ ︸

−k times

θ(r,−)

Then we have defined a smooth function θ:R×M →M such that:

30

• θ(0, x) = x for all x ∈M

• θ(s+ t, x) = θ(s, θ(t, x)) for all s, t ∈ R and x ∈M

•X(f)(x) = lim

h→0

[f(θ(h, x))− f(x)

h

]as required. 2

Note that in the above we need the conditions that the manifold M is closedand the vector field X is compactly supported for the result to be true. Withoutthese conditions, the result is not true.

Consider a smooth map c:R → M . In the course of the above proof weintroduced the velocity vector

dc

dt∈ Tc(t)M

Let (U, V, φ) be a chart, and write

c(t) = φ(x1(t), . . . , xn(t))

anddc

dt=dx1

dt

∂

∂x1+ · · ·+ dxx

dt

∂

∂xn

Definition 5.19 Let X be a vector field on a manifold M . Then an integralcurve for X is a curve c:R→M such that

dc

dt= X(c(t))

for all t ∈ R.

Proposition 5.20 Let X be a compactly supported vector field on a closed man-ifold M . Consider a point x ∈ M . Then there is a unique integral curvec:R→M for the field X such that c(0) = x.

Proof: By theorem 5.18 there is a unique flow θ:R ×M → M with tangentvector field X. We can define an integral vector field X by the formula

c(t) = θ(t, x)

Uniqueness of the integral curve c follows locally by the theory of ordinarydifferential equations. Global uniqueness therefore holds. 2

If θ:R×M →M is a flow on M , a flux line for θ is a curve of the form

t 7→ θ(t, x)

for some fixed point x ∈M . Flux lines for a flow are the same thing as integralcurves for the tangent vector field of a flow.

We conclude with a characterisation of integral curves.

31

Theorem 5.21 Let c:R → M be an integral curve. Then one of the followingthree possibilities hold:

• The map c is constant.

• The map c is an embedding of R in M .

• The map c is an immersion and there is a positive number p > 0 such thatc(t) = c(t+ p) for all t ∈ R

Proof: Suppose there is a point t0 ∈ R such that dc/dt0 = 0. Write c(t0) = x0.Then the curve c(t) satisfies the differential equation

dc

dt= X(c(t)) c(t0) = x0

The condition dc/dt0 = 0 implies that the map c must be constant, and wehave the first of the above possibilities.

If the map c is not constant, the above argument tells us that the mapc:R → M is an immersion. If the map c is injective, it is also an embedding,and we have the second of the above possibilities.

Thus if neither the first nor the second of the above possibilities hold, themap c is an immersion, and we can find points t0 ∈ R and p > 0 such thatc(t0) = c(t0 + p). The map t 7→ c(t+ p) thus satisfies the differential equation

dc(t+ p)

dt= X(c(t+ p)) c(t0 + p) = c(t0)

which is the same equation as that used to define the integral curve c.Hence c(t) = c(t+ p) for all points t ∈ R, and we are done. 2

5.4 Tensors and Duals

Let V and W be real vector spaces. Suppose that V and W are finite-dimensional, with bases e1, . . . , em and f1, . . . , fn respectively. Then thetensor product V ⊗ W is the vector space with a basis made up of symbols,which we write ei ⊗ fj , where i = 1, . . . ,m, j = 1, . . . , n.

Thus, the space V ⊗W has dimension mn. Given vectors

v = α1e1 + · · ·+ αnen, w = β1f1 + · · ·+ βnfn

we define the elementary tensor

v ⊗ w =

m,n∑i,j=1

αiβjei ⊗ fj .

It is easy to check that these elementary tensors satisfy the relations:

• α(v ⊗ w) = (αv)⊗ w = v ⊗ (αw) where α ∈ R, v ∈ V , w ∈W .

• (v + v′)⊗ w = v ⊗ w + v′ ⊗ w where v, v′ ∈ V and w ∈W .

• v ⊗ (w + w′) = v ⊗ w + v ⊗ w′.

32

We have a natural bilinear map σ:V ×W → V ⊗W defined by the formulaσ(v, w) = v ⊗ w. The following result is called the universal property of thetensor product.

Theorem 5.22 Let T :V ×W → A be a bilinear map, where A is another vectorspace. Then there is a unique linear map T :V ⊗W → A such that T = T σ.

Proof: Define T :V ⊗W → A by the formula

T

m,n∑i,j=1

γijei ⊗ fj

=

m,n∑i,j=1

T (ei, ej).

Let

v =

m∑i=1

αiei w =

n∑j=1

βjfj

Then

T (v ⊗ w) =

m,n∑i,j=1

αiβjT (ei, ej) = T (v, w)

so T = T σ.Suppose T ′:V ⊗W → A is a linear map where T = T ′σ. Then

T ′(ei ⊗ fj) = T (ei, fj).

Thus T ′ agrees with T on each basis element ei ⊗ fj . By linearity, we have

T ′ = T , and uniqueness follows. 2

Corollary 5.23 Suppose we have a vector space X equipped with a bilinear mapτ :V ×W → X where for any bilinear map T :V ×W → A there is a uniquelinear map T :X → A such that T = T τ .

Then X ≡ V ⊗W . 2

Definition 5.24 Let E and F be vector bundles over a spaceX. Then we definethe vector bundle E ⊗ F to be the union of the fibres (E ⊗ F )x = Ex ⊗ Fx.

The topology is defined by defining an atlas of local trivialisations as follows;

let (Uλ, ψ(E)λ ) | λ ∈ Λ and (Uλ, ψ(F )

λ ) | λ ∈ Λ be atlases of local trivialisa-tions for the bundles E and F respectively. Then we define local trivialisationshλ: (E ⊗ F )Uλ → Uλ × (Rm ⊗ Rn) by the formula

hλ(e⊗ f) = ψ(E)λ (e)⊗ ψ(F )

λ (f)

where x ∈ Uλ and v ∈ Rm.

If E and F are smooth vector bundles, then so is E ⊗ F .We now turn to the second of our constructions. Let V be a real vector

space. We define the dual space V ∗, to be the set Hom(V,R) of all linear mapsV → R. This set is a vector space, with addition and scalar multiplicationdefined by writing

(f + g)(v) = f(v) + g(v) f, g ∈ V ∗, v ∈ V

33

and(αf)(v) = αf(v) α ∈ R, f ∈ V

respectively.Note that given a linear map T :V → W , we have a dual linear map

T ∗:W ∗ → V ∗ defined by writing T (f)(v) = f(T (v)) where f ∈ W ∗ and v ∈ V .If the map T is an isomorphism, then so is the dual map T ∗.

Suppose that V is finite-dimensional, with basis e1, . . . , en. Define e∗i ∈ V ∗by

e∗i (α1e1 + · · ·+ αnen) = αi.

Then it is easy to check that e∗1, . . . , e∗n is a basis for V ∗. We call it thedual basis.

Although this construction proves that V and V ∗ are isomorphic when V isfinite-dimensional, the isomorphism depends on the chosen basis.

Proposition 5.25 Let V be finite-dimensional. Then we have a basis-independent isomorphism τ :V → (V ∗)∗.

Proof: We can define a linear map τ :V → (V ∗)∗ by

τ(v)(f) = f(v) f ∈ V ∗, v ∈ V.

If τ(v) = 0, then f(v) = 0 for all f ∈ Hom(V,R). But this clearly impliesv = 0. So τ is injective. By the above

dim(V ) = dim(V ∗) = dim((V ∗)∗)

so the linear injection τ is an isomorphism. 2

Definition 5.26 Let E and F be vector bundles over a space X. Then wedefine the vector bundle E∗ to be the union of the fibres (Ex)∗.

The topology is defined by defining an atlas of local trivialisations as follows;let (Uλ, ψλ) | λ ∈ Λ be an atlas of local trivialisations for the bundle E. Thenwe define local trivialisations hλ:E∗Uλ → Uλ × (Rn)∗ by the formula

hλ(f) = (ψ∗λ)−1(f) f ∈ E∗x, x ∈ Uλ

If E is a smooth vector bundle, then so is E∗.Now, let V and W be real vector spaces, and let Hom(V,W ) be the set of

all linear maps V → W . This set is a vector space, with addition and scalarmultiplication defined by writing

(S + T )(v) = S(v) + T (v) S, T ∈ Hom(V,W ), v ∈ V

and(αT )(v) = αT (v) α ∈ R, T ∈ Hom(V,W )

respectively.If dim(V ) = m and dim(W ) = n, then dimHom(V,W ) = mn.

Lemma 5.27 Let V and W be finite-dimensional real vector spaces. Then wehave an isomorphism Hom(V,W ) ∼= V ∗ ⊗W .

34

Proof: Define γ:V ∗ ⊗W → Hom(V,W ) by the formula

γ(f ⊗ w)(v) = f(v)w f ∈ V ∗, v ∈ V, w ∈W.

Then γ is a linear map. If γ(f ⊗ w) = 0, then f(v)w = 0 for all v ∈ V .This means either f = 0 or w = 0; in either case, f ⊗ w = 0, and the map γ isinjective.

Now, dim(V ∗ ⊗W ) = dim(V ) dim(W ) = dimHom(V,W ) so the map γ isalso surjective, and we are done. 2

This suggests the following.

Definition 5.28 Let E and F be vector bundles over a space X. Then wedefine Hom(E,F ) to be the set of bundle maps from E to F . We have fibresHom(E,F )x = Hom(Ex, Fx) ≡ (Ex)∗ ⊗ Fx.

The topology on Hom(E,F ) is defined in such a way as the natural bijectionHom(E,F )→ E∗ ⊗ F is a bundle isomorphism.

We can of course also define the topology on Hom(E,F ) through an atlasof local trivialisations, in the same way as for duals and tensor products. If Eand F are smooth vector bundles, then so is Hom(E,F ).

5.5 Tensor Fields

A tensor field is a generalisation of a vector field to several variables. Tensorfields frequently arise in differential geometry.

Definition 5.29 Let M be a smooth manifold. Then we define the cotangentbundle of M to be the dual of the tangent bundle. A section of the cotangentbundle is called a covector field.

We denote the cotangent bundle of the manifold M by the symbol T ?M .The fibre at the point x ∈M is called the space of cotangent vectors and denotedby the symbol T ?xM .

Let (U, V, φ) be a chart, and φ(x1, . . . , xn) = x ∈ V . Then the tangent spaceT ?xM has a basis

∂

∂x1, · · · , ∂

∂xn

The cotangent space has a dual basis,

dx1, . . . , dxn

where

dxi(

∂

∂xj

)=

1 i = j0 i 6= j

Let (U , V , φ) be another chart, where φ(x1, . . . , xn) = xThen the cotangent vector

α1dx1 + · · ·+ αndx

n

becomesα1dx

1 + · · ·+ αndxn

35

where

αj =∑i

αi∂xi

∂xj

by the rule for changing coordinates for tangent vectors.

Example 5.30 Let f :M → R be a smooth function. Then we define a covectorfield df ∈ Γ(T ?M) by writing in local coordinates:

df =∂f

∂x1dx1 + · · ·+ ∂f

∂xndxn

Looking at bases in local coordinates yields the following result.

Proposition 5.31 Every covector field has the form df for some smooth func-tion f . 2

Definition 5.32 A tensor field of type (p, q) is a section of the tensor product:

TM ⊗ · · · ⊗ TM︸ ︷︷ ︸p times

⊗T ?M ⊗ · · · ⊗ T ?M︸ ︷︷ ︸q times

As special cases, note that a vector field is a tensor of type (1, 0), and acovector field is a tensor of type (0, 1). A tensor field of type (0, 0) is simply asmooth function f :M → R. We write Γpq(M) to denote the set of tensor fieldsof type (p, q).

Let (U, V, φ) be a chart, and let (x1, . . . , xn) ∈ U . Then a tensor, A, of type(p, q) takes the form:∑

i1,...,ip,j1,...,jq

Ai1···ipj1···jq

∂

∂xi1⊗ · · · ⊗ ∂

∂xip⊗ dxj1 ⊗ · · · ⊗ dxjq

If we pick a different chart (U , V , φ) where ˜phi(x1, . . . , xn) = φ(x1, . . . , xn),the tensor A is expressed by the formula∑

i1,...,ip,j1,...,jq

Ai1···ipj1···jq

∂

∂xi1⊗ · · · ⊗ ∂

∂xip⊗ dxj1 ⊗ · · · ⊗ dxjq

where:

Ai1···ipj1···jq =

∑k1,...,kp,l1,...,lq

Ak1···kpl1···lq

∂xi1

∂xk1· · · ∂x

ip

∂xkp∂xl1

∂xj1· · · ∂x

lq

∂xjq

Classically, a tensor field of type (p, q) is defined to be a set of functions

Ai1···ipj1···jq on the coordinates defined by the charts that transforms according to

the above rule. The modern approach in mathematics is to avoid such messyimplicit descriptions of tensors wherever possible.

However, if such a definition is forced upon us, it is worth bearing in mindthat the above formula is not quite as nasty as it at first appears. Note thatany summation is over pairs of indices, one higher and one lower.2 The Einstein

2Actually, we have tried to keep to this convention throughout these notes.

36

summation convention omits the summation sign from such expressions; weassume that any pair of indices, one higher and one lower, is summed over.According to this convention, the above expression takes the form

Ai1···ipj1···jq = A

k1···kpl1···lq

∂xi1

∂xk1· · · ∂x

ip

∂xkp∂xl1

∂xj1· · · ∂x

lq

∂xjq

Actually, we will not use the Einstein summation convention in these notes,but it is worth bearing in mind for calculations done in private. Many otherreferences, especially in mathematical physics, do use the Einstein summationconvention, so we need to at least be aware of its existence.

The following result is extremely helpful when defining tensors, especiallywhen we seek to avoid a definition based on local coordinates.

Example 5.33 The Kronecker delta function, δ, is the tensor of type (1, 1)defined by δ(x) = 1 ∈ Hom(TxM,TxM) for all x ∈M .

In coordinates (x1, . . . , xn) from some chart it takes the form

δ =∑i

∂

∂xi⊗ dxi

or

δ =∑i,j

δij∂

∂xi⊗ dxj

where

δij =

1 i = j0 i 6= j

By lemma 5.27, a tensor field of type (p, q) is the same thing as a smoothsection of the bundle

Hom

TM ⊗ · · · ⊗ TM︸ ︷︷ ︸q times

, TM ⊗ · · · ⊗ TM︸ ︷︷ ︸p times

Example 5.34 Suppose that we have a metric on the tangent bundle TM (seedefinition 5.5). Then we have a tensor g of type (0, 2) defined by the formula

g(X ⊗ Y )(x) = 〈X(x), Y (x)〉

With respect to coordinates (x1, . . . , xn) defined by some chart it takes theform

g =∑i,j

gijdxi ⊗ dxj

where gij = 〈∂/∂xi, ∂/∂xj〉.

6 Riemannian Geometry

6.1 Riemannian Metrics

Definition 6.1 A Riemannian metric on a smooth manifold M is a metricon the tangent bundle TM . A smooth manifold equipped with a Riemannianmetric is called a Riemannian manifold.

37

Thus a Riemannian manifold has a collection of inner products〈−,−〉:TxM ⊗ TxM → R such that the map:

x 7→ 〈u(x), v(x)〉

is smooth for all smooth sections u, v ∈ Γ∞(TM).By theorem 5.6 any smooth manifold can be equipped with a Riemannian

metric.As we discussed in example 5.34, a metric on a manifold M defines a tensor

field g, of type (2, 0), by the formula

g(X,Y )(x) = 〈X(x), Y (x)〉

We also have a converse. Let g: Γ∞(TM)⊗C∞(M) Γ∞(TM)→ Γ∞(TM) bea tensor field of type (0, 2). Call g symmetric if g(X,Y ) = g(Y,X) for all vectorfields X and Y , and positive definite if g(X,X)(x) > 0 for any vector field Xsuch that X(x) 6= 0. We have the following obvious result.

Proposition 6.2 Let g be a symmetric and non-degenerate tensor field of type(0, 2) on a manifold M . Then we can define a Riemannian metric on the man-ifold M by the formula

〈X(x), Y (x)〉 = g(X,Y )(x)

2

We also refer to the tensor field g as the metric on M . For a chart (U, V, φ)where (x1, . . . , xn) ∈ U , let us write

gij = 〈 ∂∂xi

,∂

∂xj〉

Then for vector fields

X =∑i

Xi ∂

∂xiY =

∑j

Y j∂

∂xj

we have the inner product

〈X,Y 〉 =∑i,j

gijXiY j

Proposition 6.3 The matrix of smooth functions (gij) is invertible.

Proof: The desired result is now a statement in elementary linear algebraabout finite-dimensional inner product spaces; if V is a finite-dimensional innerproduct space, with basis e1, . . . , en, then the matrix with coefficients Aijdefined by the formula

Aij = 〈ei, ej〉is invertible.

Now apply this to the matrix (gij(x)). 2

We write (gij) to denote the inverse of the matrix of smooth functions (gij).It is itself a matrix of smooth functions. The following is straightforward.

38

Proposition 6.4 We can define a tensor, g−1, of type (2, 0) by writing

g−1 =∑i,j

gij∂

∂xi⊗ ∂

∂xj

with respect to a chart. 2

Example 6.5 Consider Euclidean space, Rn, with the usual coordinate system(x1, . . . , xn). We can define a metric on Rn by writing

〈X,Y 〉 =∑i

XiY i

for vector fields

X =∑i

Xi ∂

∂xiY =

∑j

Y j∂

∂xj

The following result follows from the definition of an immersion.

Proposition 6.6 Let i:M → N be an immersion, and suppose we have a met-ric 〈−,−〉N on the manifold N . Then we have a metric on the manifold Mdefined by the formula

〈X,Y 〉M = 〈f?(X), f?(Y )〉N

2

In particular, a manifold embedded in Euclidean space inherits a metricbased on the embedding.

Definition 6.7 Let M and N be Riemannian manifolds. Then a smooth mapf :M → N is called an isometry if:

〈f?(X), f?(Y )〉M = 〈X,Y 〉N

for all vector fields X and Y over the manifold M .

Isometries between Riemannian manifolds are the ‘distance-preserving’maps. We call two Riemannian manifolds M and N isometric if there is adiffeomorphism f :M → N that is also an isometry.

For example, if i:M → N is an immersion, N is a Riemannian manifold, andwe define a metric on the manifold M as in proposition 6.6, then the immersioni is an isometry. Conversely, we have the following obvious result.

Proposition 6.8 Let i:M → N be an isometry between Riemannian manifolds.Then i is an immersion. 2

39

6.2 Connections

Let M be a smooth manifold, and let X ∈ Γ∞(TM) be a vector field. Recallfrom definition 5.9 that for any smooth function f ∈ C∞(M) we can define asmooth function X(f) ∈ Γ(TM) by writing

X(f)(x) = (f γ)′(0) γ: (−δ, δ)→M, γ(0) = x, γ′(0) = X(x).

A connection on a smooth vector bundle E over M is a piece of structurethat enables us to perform a similar operation of differentiation of sections ofthe bundle E.

Definition 6.9 A connection on a smooth vector bundle E is a bilinear map∇: Γ∞(TM)× Γ∞(E)→ Γ∞(E), written (X,u) 7→ ∇Xu, such that:

• ∇fXu = f∇Xu

• ∇X(fu) = X(f)u+ f∇Xu

for all vector fields X ∈ Γ∞(TM), smooth sections u ∈ Γ∞(E), and smoothfunctions f ∈ C∞(M).

For most of this section we concentrate on connections on the tangent bundle,TM .

Example 6.10 Consider Euclidean space, Rn, with the usual coordinate sys-tem (x1, . . . , xn). We can define a connection on TRn by writing

∇XY =∑i,j

Xi ∂Yj

∂xi∂

∂xj

for vector fields

X =∑i

Xi ∂

∂xiY =

∑j

Y j∂

∂xj

In what follows, we will need to do some calculations in terms of a chart(U, V, φ) and (x1, . . . , xn) ∈ U . To streamline the process, let us make theabbreviations:

∂i =∂

∂xi∇i = ∇∂i

Definition 6.11 We define the Christoffel symbols, Γkij , of a connection ∇ onthe tangent bundle TM by writing:

∇i∂j =∑k

Γkij∂k

Note that despite our notation, the Christoffel symbols do not define tensorfields.

Proposition 6.12 The Christoffel symbols completely determine the connec-tion ∇.

40

Proof: Let X and Y be vector fields. Write:

X =∑i

Xi∂i Y =∑j

Xj∂j

Then:∇XY =

∑iX

i∇iY=

∑i,j X

i((∂iYj)∂j + Y j∇i∂j)

=∑i,j,kX

i((∂iYj)∂j + Y jΓkij∂k)

and we are done. 2

Recall that for vector fields X and Y we define the Lie bracket [X,Y ] by theformula:

[X,Y ](f) = X(Y (f))− Y (X(f))

where f ∈ C∞(M).

Definition 6.13 Let ∇ be a connection on TM . Then we call the connection∇ symmetric if the formula:

∇XY −∇YX = [X,Y ]

is satisfied.

It is not hard to check that the connection ∇ is symmetric if and only if theChristoffel symbols satisfy the identity

Γkij = Γkji

Definition 6.14 Let M be a Riemannian manifold. Then a connection ∇ onTM is said to be compatible with the metric if the formula:

X(〈Y, Z〉) = 〈∇XY, Z〉+ 〈Y,∇XZ〉

is satisfied.

Actually the above definition makes sense for any bundle, E, equipped witha connection and a metric.

Theorem 6.15 The Fundamental Theorem of Riemannian GeometryLet M be a Riemannian manifold. Then there is a unique symmetric connectionon TM that is compatable with the metric.

Proof: Let ∇ be a symmetric connection on the manifold M that is compat-able with the metric. Consider the Christoffel symbols Γkij . The compatabilityconditions give us an equation:

∂igjk =∑l

〈∇i∂j , ∂k〉+ 〈∂j ,∇i∂k〉 =∑l

Γlijglk + Γlikglj

for every permutation of the set of indices i, j, k. By symmetry of the con-nection we can solve the above set of equations to yield the identity:∑

l

Γlijglk =1

2(∂igjk + ∂jgik − ∂kgij)

41

and so:

Γlij =1

2

∑k

(∂igjk + ∂jgik − ∂kgij)gkl

Thus the Christoffel symbols, and therefore the connection, are determinedby the metric. Conversely, if we define the Christoffel symbols by the aboveformula, we obtain a connection that is symmetric and compatable with themetric. 2

The connection defined in the above theorem is called the Levi-Cevita con-nection of the Riemannian manifold M .

Example 6.16 Recall from example 6.10 that we can define a connection onEuclidean space, Rn by writing

∇XY =∑i,j

Xi(∂iYj)∂j

for vector fieldsX =

∑i

Xi∂i Y =∑j

Y j∂j

with respect to the standard coordinates (x1, . . . , xn).Observe that:

∇XY −∇YX =∑i,j

(Xi∂iYj − Y i∂iXj)∂j

which is just an expression for the Lie bracket [X,Y ] (see the proof of proposition5.11). Thus the given connection is symmetric.

Let 〈−,−〉 be the metric defined on Rn in example 6.5. Consider vectorfields

X =∑i

Xi∂i Y =∑j

Y j∂j Z =∑k

Zk∂k

Then:

X(〈Y,Z〉) =∑i,j

((Xi∂iYj)Zj∂j + Y j(Xi∂iZ

j)∂j) = 〈∇XY, Z〉+ 〈Y,∇XZ〉

so the connection is compatible with the metric.Hence the given connection is the Levi-Cevita connection for Euclidean space

equipped with the standard metric.

Let c:R → M be a smooth curve on a manifold M . A vector field along c,Y , consists of a tangent vector Yt ∈ Tc(t)(M) for each point t ∈ R such that forany smooth function f ∈ C∞(M) the map

t 7→ Yt(f)

is smooth.

Example 6.17 The velocity vector field, dc/dt, of the curve c is a vector fieldalong c.

42

Proposition 6.18 Let M be a smooth manifold equipped with a connection ∇.Let Y be a vector field along a smooth curve c:R→M . Then there is a uniquevector field DY/Dt characterised by the following properties.

• If Y and Z are both vector fields along c then

D(Y + Z)

Dt=DY

Dt+DZ

Dt

• For a smooth function f :R→ R and a vector field Y along c we have theequation

D(fY )

Dt=df

dtY + f

DY

Dt

• Let X be a vector field on M such that X(c(t)) = Yt. Then:

∇dc/dtX =DY

Dt

Proof: Let (U, V, φ) be a chart. Write c(t) = φ(x1(t), . . . , xn(t)) and

Y =∑i

Y i∂i

for some smooth functions Y i defined on an open subset of R. The statedconditions tell us that:

DY

Dt=∑i

(dY i

dt∂i + Y i∇dc/dt∂i

)so

DY

Dt=∑k

dY kdt

+∑i,j

Γkijdxi

dtY j

∂k

by definition of the Christoffel symbols Γkij .Conversely, the above formula defines a vector field DY/Dt along c that has

the required properties. 2

Definition 6.19 We call the vector field DY/Dt the covariant derivative ofY . The vector field V along c is called a parallel vector field if the covariantderivative is zero.

The usual methods of establishing the existence of unique solutions to firstorder differential equations give us the following.

Proposition 6.20 Let c:R → M be a smooth curve. Let Y0 ∈ Tc(0)M . Thenthere is a unique parallel vector field Yt along c that is equal to the vector Y0

when t = 0. 2

We say that the vector Yt ∈ Tc(t)M is obtained from Y0 by parallel transportalong c.

43

Definition 6.21 Let M be a Riemannian manifold equipped with some con-nection. Let us say that parallel transprort preserves inner products if for anycurve c and any two parallel vector fields Y and Z along c the product 〈Yt, Zt〉is constant.

The following is straightforward.

Proposition 6.22 The connection on the manifold M is compatible with themetric if and only if parallel translation preserves inner products. 2

We end this section by noting that we can formulate a two-dimensionalversion of proposition 6.18; the proof is virtually identical.

Definition 6.23 Let M be a smooth manifold equipped with a connection ∇.Let σ:R2 →M be a smooth map (we could call σ a smooth surface in M).

A vector field along σ, Y , consists of a tangent vector Yx,y ∈ Tσ(x,y)(M) foreach point (x, y) ∈ R2 such that for any smooth function f ∈ C∞(M) the map

(x, y) 7→ Y(x, y)(f)

is smooth.

Proposition 6.24 There are unique vector fields DY/∂x and DY/∂y charac-terised by the following properties.

• If Y and Z are vector fields along σ then

D(Y + Z)

∂x=DY

∂x+DZ

∂x

D(Y + Z)

∂y=DY

∂y+DZ

∂y

• For a smooth function f :R2 → R and a vector field V along σ we havethe equations

D(fY )

∂x=∂f

∂xY + f

DY

∂x

D(fY )

∂y=∂f

∂yY + f

DY

∂y

• Let X be a vector field on M such that X(σ(x, y)) = Yx,y. Then:

∇∂c∂xX =DY

∂x∇∂c∂yX =

DY

∂y

2

6.3 Geodesics

Let M be a connected Riemannian manifold, equipped with the Levi-Cevitaconnection.

Definition 6.25 Let I ⊆ R be some interval. A smooth curve γ: I → M iscalled a geodesic if the acceleration vector field D/Dt(γ′(t)) is zero for all t ∈ I.

44

Note that if γ is a geodesic, by definition the velocity vector field γ′(t) isparallel. Observe that:

d

dt〈γ′(t), γ′(t)〉 = 2〈 D

Dt(γ′(t)) ,

dγ

dt〉 = 0

Hence the norm

‖γ′(t)‖ = 〈dγdt,dγ

dt〉 12

is constant.

Definition 6.26 Let c: [a, b]→M be a piecewise-smooth curve. We define thearclength of c by the formula

Length(c) =

∫ b

a

‖c′(t)‖ dt

The following result is obvious.

Proposition 6.27 Let c: [a, b] → M be a piecewise-smooth curve, and letf :M → N be an isometry between Riemannian manifolds. Then:

Length(c) = Length(F c)

2

We will show in this section that geodesics are precisely those curves whichlocally minimise arclength. It is easy to see that if c′ is a reparametrisation ofa curve c, then Length(c′) = Length(c).

Consider a chart (U, V, φ) where (x1, . . . , xn) ∈ U . Write γ(t) =φ(x1(t), . . . , xn(t)). Then, as in the proof of proposition 6.18

dY

dt=∑k

DY kdt

+∑i,j

Γki,jdxi

dtY j

∂k

for any vector field Y along γ. In particular, the geodesic equation becomes thesystem of equations

d2xk

dt2+∑i,j

Γki,jdxi

dt

dxj

dt

Proposition 6.28 Let x0 ∈ M . Then there is an open neighbourhood U 3 x0

and a real number ε > 0 such that for each point x ∈ U and tangent vectorV ∈ TxM such that ‖V ‖ < ε there is a unique geodesic γV : (−2, 2) → M suchthat

γV (0) = xdγVdt

(0) = V

Moreover, the function(V, t) 7→ γV (t)

is smooth.

45

Proof: By choosing a suitable chart. the statement almost follows immediatelyfrom the theory of existence and uniqueness of solutions to ordinary differentialequations. To be precise, there is an open neighbourhood U 3 x0 and realnumbers ε1, ε2 > 0 such that for every tangent vector V ∈ TxM with ‖V ‖ < ε1

there is a unique geodesic γV : (−2ε2, 2ε2)→M such that

γV (0) = xdγVdt

(0) = V

Further, the required smoothness condition holds.Let ε = ε1ε2. If ‖V ‖ < ε and |t| < 2 note that ‖V/ε2‖ < ε1 and |ε2t| < 2ε2.

Thus we can define our geodesic by the formula

γV (t) γV/ε1(ε2t)

2

Definition 6.29 Let x ∈ M and let V ∈ TxM . Suppose there exists a uniquegeodesic γV : [0, 1]→M such that

γV (0) = xdγVdt

(0) = V

Then we define the exponential function of the vector V by writing

expx(V ) = γV (1)

We also write exp(x, V ) = γV (1). Consider a point x ∈M , and let 0x ∈ TxMbe the zero vector. Proposition 6.28 tells us that there is an open neighbourhoodU 3 0x in the manifold TM such that we have a well-defined smooth exponentialfunction

exp:U →M

Given a tangent vector V ∈ TxM we can describe the unique geodesic γVsuch that

γV (0) = xdγVdt

(0) = V

by the equationγV (t) = expx(tV )

for all sufficiently small t.

Proposition 6.30 The arclength of the geodesic

γV (t) = expx(tV ) t ∈ [0, 1]

is equal to ‖V ‖.

Proof: The path γV is the unique geodesic such that

γV (0) = xdγVdt

(0) = V

Since γV is a geodesic, the norm ‖γ′V (t)‖ is constant. Hence ‖γ′V (t)‖ = Vfor all points t ∈ [0, 1]. The desired result follows by definition of arclength. 2

46

Theorem 6.31 For each point x0 ∈ M there is a neighbourhood W 3 x and areal number ε > 0 such that:

• Any two points of W are joined by a unique geodesic with arclength lessthan ε.

• Lett 7→ expx(tV ) t ∈ [0, 1]

be the unique geodesic of arclength less than ε joining two points x, y ∈W .Then the map M×M → TM defined by the formula (x, y) 7→ V is smooth.

• For each point x ∈W the map

expx:D(0, ε)→M

maps the open ε-ball in TxM diffeomorphically onto an open set Ux ⊇W .

Proof: Let 0x ∈ Tx0M be the zero tangent vector to x0. By proposition 6.28we have a well-defined smooth map

F :U0 →M ×M ; F (V ) = (x, expx(V ))

for all points V ∈ TxM belonging to some neighbourhood, U0, of the point0x0∈ TM .Local coordinates (x1, . . . , xn) on the manifold M give us local coordinates

(x1, . . . , xn, α1, . . . , αn) on the tangent bundle TM ; the coordinates αi are de-fined by writing a given tangent vector in the form

α1∂1 + · · ·+ αn∂n

Let us write (x1, . . . , xn, y1, . . . , yn) to denote the local coordinates on theproduct manifold M ×M induced by the local coordinated (x1, . . . , xn) on M .Observe that:

F?

(∂

∂xi

)=

∂

∂xi+

∂

∂yiF?

(∂

∂αi

)=

∂

∂yi

at the point (x0, 0).

Thus the differential of the map F at the point (x0, 0) has the form

(I I0 I

)and so is invertible. By the inverse function theorem the map F is a diffeomor-phism from some neighbourhood V ′ 3 0x0

onto a neighbourhood V ′′ 3 (x0, x0).Let U ′ be a neighbourhood of the point x0 ∈ M such that V ′ contains all

tangent vectors of the form V ∈ TxM where x ∈ U ′ and ‖V ‖ < ε. Let W 3 x0

be a neighbourhood such that F [U ′] ⊇W ×W .Then by definition of the map F the required properties of the neighbour-

hood W hold. 2

We now introduce some terminology.

Definition 6.32 Consider a point x ∈M . The largest real number R > 0 suchthat the exponential map

expx:D(0, R)→M

47

maps the open r-ball in TxM diffeomorphically onto an open set Ux ⊆ M iscalled the injectivity radius of M at the point x. If ε ≤ R, the image N(x, ε) =expx[D(0, ε)] is termed a geodesic neighbourhood of the point x, of radius ε.

Thus, theorem 6.31 tells us that every point x ∈M has a non-zero injectivityradius, and so contains geodesic neighbourhoods.

It is possible for the injectivity radius to be infinite; for example, at everypoint x ∈ Rn the injectivity radius is infinite.

Theorem 6.33 Let x ∈M , and let U be a geodesic neighbourhood of x of radiusε. Let γ: [0, 1] → M be a geodesic of length less than ε joining the point x toanother point y ∈ U . Let ω: [0, 1]→M be a piecewise smooth path from x to y.Then Length(γ) ≤ Length(ω).

Further, equality holds if and only if the path ω is a reparametrisation of thepath γ.

Thus, at least within geodesic neighbourhoods, geodesics are the uniquearclength-minimising paths between points.

Before we prove the above theorem, we need to do some technical work. Thefirst of the technical results we need is sometimes called Gauss’ lemma.

Lemma 6.34 Let U be a geodesic neighbourhood of a point x ∈ M . Then thegeodesics through the point x in the set U are all perpendicular to the hypersur-faces

expx(V ) | ‖V ‖ = constant

Proof: Let ε be the radius of the given geodesic neighbourhood U . Lett 7→ V (t) be a curve in the tangent space TxM such that ‖V (t)‖ = 1. Supposethat 0 < r0 < ε. We must prove that the correponding curves

t 7→ expx(r0V (t)) r 7→ expx(rV (t0))

are orthogonal.Hence for the parametrised surface defined by the equation

f(r, t) = expx(rV (t)) 0 ≤ r < ε

we must prove that

〈∂f∂r,∂f

∂t〉 = 0

for all r and t. A simple calculation yields the formula:

∂

∂r〈∂f∂r,∂f

∂t〉 = 〈D

∂r

(∂f

∂r

),∂f

∂t〉+ 〈∂f

∂r,D

∂r

(∂f

∂t

)〉

Since the curves r 7→ expx(rV (t)) are geodesics, the first expression on theright is zero. The second expression is equal to:

〈∂f∂r,D

∂t

(∂f

∂r

)〉 =

1

2

∂

∂t〈∂f∂r,∂f

∂r〉 = 0

since the norm ‖∂f/∂r‖ = ‖V (t)‖ is constant.

48

Thus the quantity 〈∂f/∂r, ∂f/∂t〉 does not depend on r. But for r = 0:

f(0, t) = expx(0) = x

so ∂f/∂t(0, t) = 0. Therefore

〈∂f∂r,∂f

∂t〉 = 0

for all points r ∈ [0, ε) and t ∈ R and we are done. 2

The hypersurface

S(x,C) = expx(V ) | ‖V ‖ = C

is called a spherical shell around x of radius C.Consider a piecewise-smooth curve ω: [a, b] → U\x, where I is a geodesic

neighbourhood of the point x ∈M with radius ε. Each point ω(t) can be writtenuniquely in the form

expx(r(t)V (t)) 0 < r(t) < ε ‖V (t)‖ = 1

Lemma 6.35 We have the inequality

Length(ω) ≥ |r(b)− r(a)|

Equality holds if and only if the function r(t) is monotone and the functionV (t) is constant.

Proof: Let f(r, t) = expx(rV (t)). Then ω(t) = f(r(t), t). We have the formula

dω

dt=∂f

∂r

dr

dt+∂f

∂t

By Gauss’ lemma, the vectors (∂f/∂r)(dr/dt) and ∂f/∂t are orthogonal.Since ‖∂f∂r‖ = 1, we know that∥∥∥∥dωdt

∥∥∥∥2

=

∣∣∣∣drdt∣∣∣∣2 +

∥∥∥∥∂f∂t∥∥∥∥2

where equality holds only when ∂f/∂t = 0, ie: when the dV/dt = 0. Hence:

Length(ω) =

∫ b

a

∥∥∥∥dωdt∥∥∥∥ dt ≥

∫ b

a

|r′(t)| dt ≥ |r(b)− r(a)|

and equality holds if and only if the function r(t) is monotone and the functionV (t) is constant. 2

Armed with the above lemma, it is now quite straightforward to prove the-orem 6.33.

Proof of theorem 6.33: Consider a piecewise smooth path, ω, from the pointx ∈ M to a point y = expx(rV ) ∈ U , where 0 < r ≤ ε and ‖V ‖ = 1. Then forany real number δ ∈ (0, r) the path ω must contain a path joining the spherical

49

sphells S(x, δ) and S(x, r) and lying between these shells. By the above lemma,the length of this piece of the path ω is at least r − δ.

Hence, letting δ approach zero, we see that

Length(ω) ≥ r = Length(γ)

If the path ω is a reparametrisation of the path γ, then the lengths of thetwo curves are certainly equal. If such is not the case, the conditions stated inthe above lemma ensure that the above inequality is strict. 2

Let us say that a path ω: [0, l]→M is parametrised by arclength if:

s =

∫ s

0

∥∥∥∥dωdt∥∥∥∥ dt

for any point s ∈ [0, l].Theorem 6.33 has the following easy corollary.

Corollary 6.36 Let γ: [0, l]→M be a path parametrised by arclength. Supposethat for every path ω: [0, l] → M with ω(0) = γ(0) and ω(l) = γ(l) we have theinequality

Length(γ) ≤ Length(ω)

Then the path γ is a geodesic.

Proof: The desired statement follows locally from theorem 6.33. But thecondition for a path to be a geodesic is a purely local condition so the resultfollows. 2

Example 6.37 Consider the space Rn equipped with the standard metric,and coordinates (x1, . . . , xn). Let γ:R → Rn be a geodesic. Write γ(t) =(x1(t), . . . , xn(t)).

The Christoffel symbols, Γkij are then all equal to zero, and the geodesicequation for the curve γ becomes:

d2xi

dt2= 0

Thus geodesics in Rn are straight lines. It follows that the shortest pathbetween any two points of Rn is a straight line.

Example 6.38 Consider the circle, Sn, with the metric it inherits as the subsetof Rn consisting of all vectors of length 1. A great circle in Sn is then theintersection of a two-dimensional subspace of Rn with Sn.

The set of geodesics in Sn is the set of segments of great circles. To see thisfact, let R:Sn → Sn be the map defined by reflection in some two-dimensionalsubspace, E, of Rn. Then R is an isometry, with fixed point set the great circleC = Sn ∩ E.

Choose points x, y ∈ C, and let C ′ be the image of the unique geodesic ofminimum length between them. Since the map R is an isometry, it follows thatf(C ′) is also the unique geodesic of minimal length between x and y. ThusC ′ = f(C ′), and so C ′ ⊆ C.

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Thus any given geodesic in Sn is a segment of a great circle. Since any twopoints of Sn lie on some great circle, the above argument tells us that everysegment of a great circle is a geodesic.

Locally, any two points x, y ∈ Sn that are sufficiently close together arejoined by a unique geodesic of minimum length. However, by going the “wrongway” around the sphere, there is another geodesic joining them. In fact, we candefine still more geodesics from x to y by winding around the sphere more thanonce.

If x and y are opposite points of Sn, any semicircle joining x to y is a geodesicof minimal lenght. Thus, there is no geodesics of minimal length joining thepoints x and y.

Definition 6.39 We call a subset U ⊆M geodesically convex if any two pointsx, y ∈ U are joined by a unique geodesic lying entirely in U .

Theorem 6.40 Any point x ∈M has a geodesically convex neighbourhood. 2

6.4 Completeness

Let M be a connected closed Riemannian manifold. We can define a metric onM (in the sense of topology, rather than in the sense of Riemannian geometry)by the formula

d(x, y) = infLength(γ) | γ is a piecewise-smooth curve from x to y

The results of the previous section ensure that this metric does define theusual topology on the manifold M . In this section we compare completeness ofM as a metric space with the following notion.

Definition 6.41 We call M geodesically complete if the exponential map isalways defined.

Equivalently, the manifold M is geodesically complete if a given geodesicsegment γ: I →M can always be extended to a geodesic γ:R→M .

Theorem 6.42 (The Hopf-Rinow Theorem) Any two points in a geodesi-cally complete manifold can be joined by a geodesic of minimum length.

Proof: Let M be geodesically complete. Consider two points x, y ∈ M , andlet d(x, y) = r. We can of course assume that r > 0.

Choose a geodesic neighbourhoodN(x,R), and a positive real number δ < R.Since the spherical shell S(x, δ) is compact, we can find a point y0 = expx(δV ) ∈S(x, δ) such that d(y0, y) ≤ d(x, y) for all s ∈ S(x, δ). We claim that

expx(rV ) = y

Thus the geodesic segment

γ(t) = expx(tV ) t ∈ [0, r]

51

is a geodesic from x to y of minimum length. Of course, we need geodesiccompleteness at this final step to show that the above geodesic is defined forr > R.

To prove our claim, we will show that

d(γ(t), y) = r − t t ∈ [δ, r]

Taking t = r we complete the proof.Observe that every curve from x to y must intersect the shell S(x, δ). There-

fore:d(x, y) = supd(x, s) + d(s, y) | s ∈ S(x, δ) = δ + d(y0, y)

Thus d(γ(δ), y) = r − δ. Let

t0 = supt ∈ [δ, r] | d(γ(t), y) = r − t

By continuity, we have the formula d(γ(t0), y) = r− t0. Suppose T0 < r. LetR′ be the injectivity radius of M at γ(t0), and let δ′ < infR′, r − t0. Sincethe spherical shell S(γ(t0), δ′) is compact, we can find a point y′0 ∈ S(γ(t0), δ′)such that d(y′0, y) ≤ d(x, y) for all s ∈ S′. Observe that every path from γ(t0)to y must intersect the shell S(γ(t0), δ′). Therefore

d(γ(t0), y) = infd(γ(t0, s) + d(s, y) | s ∈ S(γ(t0), δ′) = δ′ + d(y′0, y)

Thus d(y′0, y) = (r− t0)− δ′. We claim that y′0 = γ(t0 + δ′). By the triangleinequality for the metric d:

d(x, y′0) ≥ d(x, y)− d(y′0, y) = t0 + δ′

But we can obtain a path of length t0 + δ′ from x to y′0 by following thegeodesic γ from x to γ(t0), then the minimal geodesic from γ(t0) to y′0. Bycorollary 6.36 this join is itself a geodesic, and so must coincide with the geodesicγ.

Hence γ(t0 + δ′) = y′0 and we have the equation

d(γ(t0 + δ′), y) = d(y′0, y) = r − (t0 + δ′)

If t0 < r the above equation contradicts the definition

t0 = supt ∈ [δ, r] | d(γ(t), y) = r − t

Therefore t0 = r and we are done. 2

Note that the above theorem tells us nothing about the uniqueness of min-imal geodesics. For example, the sphere Sn is geodesically complete, but thereis more than one geodesic of minimum length joining two opposite points of Sn.

Corollary 6.43 A connected closed Riemannian manifold M is geodisicallycomplete if and only if it is complete as a metric space.

Proof: Let M be complete as a metric space. Consider a geodesic γ: (a, b)→M . Let (tn) be a sequence of points in the interval (a, b) converging to b. Thenthe sequence (γ(tn)) is a Cauchy sequence in M , and so converges to a point

52

x ∈ M . Defining γ(b) = x we have a geodesic γ: (a, b] → M . By proposition6.28 we can extend the geodesic γ past b. Similarly, we can define the geodesicγ past a.

Thus we have an extended geodesic γ: (a′, b′)→M where a′ < a and b′ > b.But we can repeat the above process, so by a least upper bound argument, weobtain a geodesic γ:R→M .

Conversely, let M be geodesically complete. Let B ⊆ M be a boundedsubset, of diameter d. Let x ∈ B. Then by the above theorem we can find areal number R > 0 such that the exponential map

expx:D(0, R)→M

maps the closed disk D(0, R) ⊆ TxM onto a compact subset of M containingB. Hence the closure B is compact, which means that M is a complete metricspace. 2

We therefore do not, at least in the connected case, have to distinguish be-tween geodesically complete Riemannian manifolds, and Riemannian manifoldsthat are complete as metric spaces. We refer simply to complete Riemannianmanifolds, where we could be using either definition of completeness.

Corollary 6.44 Let M be a connected closed compact Riemannian manifold.Then any two points in M are joined by a minimal geodesic.

Proof: Since the manifold M is compact, it must be complete. The result nowfollows immediately from the above corollary and Hopf-Rinow theorem. 2

6.5 Normal Bundles

Let M be a Riemannian manifold. Let N be an embedded submanifold. Then Ninherits a metric from the manifold M . The tangent bundle, TN , is a subbundleof the tangent bundle TM . We define the normal bundle, T⊥N , to be the bundleover N with fibres

T⊥x N = V ∈ TxM | 〈V,W 〉 = 0 for all W ∈ TxN

Since TN is a sub-bundle of TM , we can form the quotient TM/TN , which isthe bundle with fibres TxM/TxN . The bundles T⊥N and TM/TN have atlasesof local trivialisations coming from the atlas of trivialisations for the tangentbundle TM (in the same way as the topological structure of dual bundles isdefined).

The following result is straightforward.

Proposition 6.45 The quotient bundle TM/TN and normal bundle T⊥N aresmoothly isomorphic. 2

Example 6.46 Consider the sphere, Sn, embedded in Rn+1 in the usual way.Then the normal bundle T⊥Rn is one-dimensional. It is trivial since we candefine a nowhere-vanishing section u:Sn → T⊥Sn by the formula u(x) = x,where we identify the tangent space TxRn with Rn.

53

Definition 6.47 Let N be an embedded submanifold of a smooth manifold M .Then a tubular neighbourhood of N in M is a vector bundle p:U → N , where Uis an open neighbourhood of N in M .

Before we prove the existence of tubular neighbourhoods, we need a topo-logical lemma.

Lemma 6.48 Let X be a compact metric space, and let X0 ⊆ X be a closedsubset. Let f :X → Y be a local homeomorphism such that f |X0

is injective.Then there is a real number ε > 0 such that the map f is injective when restrictedto the open neighbourhood

N(X0, ε) = x ∈ X | d(x,X0) < ε

Proof: Let

C = (x0, y0) ∈ X ×X | x 6= y, f(x) = f(y)

The set C is closed because the map f is a local homeomorphism. Considerthe map g:C → R defined by the formula

g(x, y) = d(x,X0) + d(y,X0)

Since the set C, as a closed subset of a compact metric space, is compact,we can find a real numbet ε > 0 such that g(x, y) ≥ 2ε for all (x, y) ∈ C. Itfollows that the map f is injective on the ε-neighbourhood of X0. 2

Theorem 6.49 (The Tubular Neighbourhood Theorem) Let N be acompact embedded submanifold of a manifold M . Then there is a tubular neigh-bourhood of N in M which is equivalent to the normal bundle of N in M .

Proof: By theorem 5.6 we can choose a Riemannian metric 〈−,−〉 for M ,with corresponding norm ‖ − ‖ and metric d. For ε > 0, let us write

Eε = V ∈ T⊥x N | x ∈ N, ‖V ‖ < ε

andUε = x ∈M | d(x,N) < ε

By theorem 6.31 and compactness of the manifold N the exponential map(x, V ) 7→ exp(x, V ) is defined on Eε for some ε > 0. We claim that the expo-nential map is a diffeomorphism provided ε is sufficiently small. The theoremclearly follows from this claim.

Let V ⊆ Eε be the set of regular points for the smooth map exp. Letus consider N to be the set of zero tangent vectors in the space Eε. Thencertainly V ⊇ N , and V1 = V ∩ E1 is compact. Since the exponential functionis injective on N ⊆ V1, by the above lemma the exponential function is injectiveon Eε provided ε is sufficiently small.

Certainly exp[Eε] ⊆ Uε. We need to prove that exp[Eε] = Uε. Choosey ∈ Uε. Since N is compact, there is a point x0 ∈ N such that d(x0, y) ≤ d(x, y)for all x ∈ N . Let γ: [0, 1]→ N be the unique geodesic of length less that ε suchthat γ(0) = x0 and γ(1) = y.

54

The fact that x0 is the nearest point on N to the point y means that thetangent vector γ′(0) ∈ Tx0

M is normal to N . Hence

y = exp(x0, γ′(0)) γ′(0) ∈ Eε

and we are done. 2

In fact the tubular neighbourhood theorem is also true for non-compactsubmanifolds. However, we will not give the proof here.

It is interesting to see that some fairly deep facts in Riemannian geometryare used to determine the existence of tubular neighbourhoods— a statementwhich does not itself even mention Riemannian geometry. The same is true ofthe following result, which has a similar proof.

Theorem 6.50 (Collared Neighbourhood Theorem) Let M be a mani-fold with compact boundary ∂M . Then ∂M has a neighbourhood in M thatis diffeomorphic to ∂M × [0, 1). 2

7 Some Notions from Algebraic Topology

7.1 Homotopy

When looking at topological spaces, it is often easy to see when two spaces arehomeomorphic; we simply construct a homeomorphism! For example the openinterval (−1, 1) and the entire real line R are homeomorphic; we can define ahomeomorphism f : (−1, 1)→ R by the formula

f(x) =x

1− |x|

In general a far more difficult problem is to prove that two spaces are nothomeomorphic. A homeomorphism-invariant is a property of a space that ispreserved by the relation of homeomorphism. For example, compactness andconnectedness are homeomorphism-invariants. Hence the intervals [0, 1] and(0, 1) are not homeomorphic because the space [0, 1] is compact whereas thespace (0, 1) is not. The spaces [0, 3] and [0, 1] ∪ [2, 3] are not homeomorphicbecause the space [0, 3] is connected whereas the space [0, 1] ∪ [2, 3] is not.

The goal of algebraic topology is to associate invariants to topological spacesthat enable us to tell whether or not they are homeomorphic.

Generally the notion of homeomorphism is rather restrictive; in algebraictopology it is common to work with a more general equivalence relation.

Definition 7.1 Let f, g:X → Y be continuous maps. Then we say the maps fand g are homotopic if there is a continuous map F :X × [0, 1] → Y such thatF (−, 0) = f and F (−, 1) = g.

The map F in the above definition is sometimes termed a homotopy betweenthe map f and the map g.

Proposition 7.2 The notion of homotopy is an equivalence relation on the setof all continuous maps from the space X to the space Y .

55

Proof: It is obvious that the relation of homotopy is reflexive and symmetric.We prove transitivity.

Let F be a homotopy between a map f :X → Y and a map g:X → Y . LetG be a homotopy between the map g:X → Y and a map h:X → Y . Then wecan define a homotopy F ?G:X × [0, 1]→ Y between the maps f and h by theformula

F ? G(x, t) =

F (x, 2t) t ≤ 1

2G(x, 2t− 1) t ≥ 1

2

2

The set of all maps g:X → Y that are homotopic to a map f :X → Y isthus called the homotopy class of the map f .

Definition 7.3 A continuous map f :X → Y is said to be a homotopy equiva-lence if there is a continuous map g:Y → X such that the composites gf andfg are homotopic to the identity maps 1X and 1Y respectively.

SpacesX and Y are said to be homotopy-equivalent if there exists a homotopyequivalence f :X → Y . It is easy to see that homeomorphic spaces must behomotopy equivalent.

Example 7.4 Euclidean space Rn is homotopy-equivalent to a single point 0.To see this fact, define maps f :Rn → 0 and g: 0 → Rn by the formulaef(x) = 0 and g(0) = 0 respectively. The composite f g is the identity map10. We can define a homotopy between the composite g f and the identitymap 1Rn by the formula

F (v, tv) = tv; t ∈ [0, 1], v ∈ Rn

A space that is homotopy-equivalent to a single point is called contractible.A space is contractible if and only if there is a homotopy between the constantmap onto a single point in that space and the identity map.

Definition 7.5 We define π0(X) to be the set of path-components of a topo-logical space.

Let us write 〈x〉 to denote the path-component of a point x ∈ X. Then acontinuous map f :X → Y induces a map f?:π0(X)→ π0(Y ) by the formula

f?〈x〉 = 〈f(x)〉

If maps f, g:X → Y are homotopic, the induced maps f?, g?:π0(X)→ π0(Y )are equal. It follows that if a map f :X → Y is a homotopy-equivalence, theinduced map f?:π0(X)→ π0(Y ) is a bijection.

Definition 7.6 Let f, g:M → N be smooth maps. Then we say the maps fand g are smoothly homotopic if there is a smooth map F :X × [0, 1]→ Y suchthat F (−, 0) = f and F (−, 1) = g.

As in definition 7.3 we can talk about a smooth map f :M → N being asmooth homotopy equivalence.

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Theorem 7.7 Let f :M → N be a continuous map betweem smooth compactmanifolds. Then f is homotopic to a smooth map.

Proof: In this proof, we need a small amount of functional analysis. Let X bea compact Hausdorff topological space, and let C(X) be the set of continuousfunctions f :X → R. The set C(X) is a normed vector space; the operations ofaddition and scalar multiplication are defined pointwise, and the norm is definedby the formula

‖f‖ = sup|f(x)| | x ∈ XLet us call a subset A ⊆ C(X) a subalgebra if for any two functions f, g ∈ A

the product, fg, defined pointwise, also belongs to the set A. We say that thealgebra A separates points of X if for any two points x, y ∈ X with x 6= y wecan find a function f ∈ A such that f(x) = 0 but f(y) 6= 0.

Theorem [The Stone-Weierstrass Theorem] Let A be a subalgebra of C(X)that contains the constant functions and separates points. Then A is a densesubset of C(X). 2

Note that the Stone-Weierstrass theorem implies that the set of smoothfunctions of M , C∞(M), is a dense subset of the normed space C(M). It is thisfact that we need here.

Suppose N ⊆ RN . Write

f(x) = (f1(x), . . . , fN (x)) x ∈M

Then for any real number ε > 0 we can find smooth functions gi:M → Rsuch that |fi(x)−gi(x)| < ε for all x ∈M . Hence, by the tubular neighbourhoodtheorem, there is a smooth map g:M → RN such that that the image g[M ] liesin some tubular neighbourhood, U , of N in RN .

According to example ??, the structure map p:U → N is a smoothhomotopy-equivalence. We therefore obtain a smooth map

p g:M → N

that is homotopic to f . 2

The above argument is the one used to fill in the details in the proof ofthe Brouwer fixed point theorem when we passed from consideration of smoothmaps to continuous maps.

Corollary 7.8 Let f, g:M → N be smooth maps between compact manifolds.Suppose that the maps f and g are homotopic. Then they are smoothly homo-topic. 2

7.2 Categories and Functors

Algebraic topology is frequently best expressed within an abstract languagecalled category theory.

Definition 7.9 A category, C, consists of a collection Ob(C) of objects and aset Hom(A,B) of morphisms for any two objects A,B ∈ Ob(C) such that thefollowing axioms are satisfied:

57

• For all morphisms f ∈ Hom(A,B) and g ∈ Hom(B,C) there is assigneda morphism gf ∈ Hom(A,C), called the composition of f and g.

• Composition is associative, that is to say the formula (hg)f = h(gf) holdsfor all morphisms f ∈ Hom(A,B), g ∈ Hom(B,C), and h ∈ Hom(C,D).

• For all objects A ∈ Ob(A) there is a morphism 1A ∈ Hom(A,A), calledthe identity at A, such that 1Ag = g and f1A = f for all morphismsf ∈ Hom(A,B) and g ∈ Hom(B,A).

For instance we may form the category of sets. The collection of objectsis the collection of all sets. For sets A and B the morphism set Hom(A,B)consists of all maps from A to B. This example indicates that the collection ofobjects in a given category is not in general a set.

Other examples include the category of all abelian groups and group ho-momorphisms, the category of all real vector spaces and linear maps, and thecategory of all topological spaces and continuous maps. Indeed we generallythink of a category as a collection of certain algebraic objects and a morphismset as a collection of homomorphisms between these algebraic objects. Becauseof this idea we sometimes use the notation f :A → B to mean that f is anelement of the morphism set Hom(A,B). We call a morphism f :A → B aninvertible morphism or an isomorphism if there is a morphism g:B → A suchthat gf = 1A and fg = 1B .

Two more examples that are relevant in these notes are the category ofsmooth manifolds and smooth maps, and the category of vector bundles andbundle maps over a given topological space.

The morphisms of a category need not be actual maps. We can see this bymentioning another example of importance to us in these notes: the categoryof topological spaces and homotopy classes of continuous maps. We call thiscategory the homotopy category.

The objects of a given category are not always sets. To see this, let P be apartially ordered set. Then P is a category; the objects are the elements of theset P and there is precisely one morphism in the set Hom(i, j) whenever i ≤ j.To take another example, let G be a group. Then G is a category with just oneobject; the morphisms from this object to itself are simply the elements of G.

Definition 7.10 Let C and D be categories. Then a functor F : C → D is aprocedure that assigns an object F (A) ∈ Ob(D) to each object A ∈ Ob(C) anda morphism f? ∈ Hom(F (A), F (B)) to each morphism f ∈ Hom(A,B). Theinduced morphisms f? must satisfy the formulae:

(fg)? = f?g? (1A)? = 1F (A)

Actually, a functor as defined above is usually called a covariant functor.There is a dual notion of a contravariant functor; a contravariant functor F : C →D is a procedure that assigns an object F (A) ∈ Ob(D) to each object A ∈ Ob(C)and a morphism f? ∈ Hom(F (B), F (A)) to each morphism f ∈ Hom(A,B)such that (fg)? = g?f? and (1A)? = 1F (A).

A trivial example of a (covariant) functor is the identity functor 1: C → C ona category C, which takes any object or morphism to itself.

58

Example 7.11 Let V be a real vector space. Then the assignment V 7→ V ?

taking V to its dual space is a contravariant functor from the category of realvector spaces to itself. Given a linear map T :V → W we define the inducedmap T ?:W ? → V ? by the formula

T ?(w)(v) = w(Tv)

Example 7.12 The assignment X 7→ π0(X), taking a topological space X toits set of path components is a covariant functor from the category of topologicalspaces to the category of sets.

If we write 〈x〉 to denote the path-component of a point x ∈ X. Then acontinuous map f :X → Y induces a map f?:π0(X)→ π0(Y ) by the formula

f?〈x〉 = 〈f(x)〉

The above example can also be considered a functor from the homotopycategory to the category of sets. Generalising, we can state a more formaldefinition of what we mean by a homotopy-invariant; it is any functor from thehomotopy category to some other category.

Definition 7.13 Let F,G: C → D be functors. Then a natural transformationg:F → G consists of a morphism gA ∈ Hom(F (A), G(A)) for each object A ∈Ob(C) such that G(f)gA = gBF (f) for any morphism f ∈ Hom(A,B).

We call a natural transformation g:F → G a natural isomorphism if themorphisms gA ∈ Hom(F (A), G(A)) are isomorphisms for all A.

We can express the condition G(f)gA = gBF (f) by saying that we have acommutative diagram

F (A)gA //

F (f)

G(A)

F (f)

F (B)

gB // G(B)

More complicated commutative diagrams are fairly common in category the-ory and algebraic topology, and have the obvious meaning. Proofs involvingcommutative diagrams are sometimes called diagram chases.

Example 7.14 Let V2 be the category in which the objects are pairs, (V1, V2)of vector spaces, and the morphisms are pairs (α1, α2) of linear transformations.Define functors T1 and T2 from the category V2 to the category of vector spacesand linear transformations by the formulae

T1(V1, V2) = V1 ⊗ V2 T1(α1, α2) = α1 ⊗ α2

andT2(V1, V2) = V2 ⊗ V1 T2(α1, α2) = α2 ⊗ α1

Then we can define a natural isomorphism g:T1 → T2 by the formula

g(V1,V2)(u1 ⊗ u2) = u2 ⊗ u1

59

Example 7.15 Let V be a vector space. Then we have an isomorphismαV :V → (V ?)? from the space V to its second dual space defined by the formula

αV (v)(f) = f(v)

for all vectors v ∈ V and functionals f ∈ V ?. For any linear map T :V →W weknow that αWT = (T ?)?αV so we have a commutative diagram

VαV→ (V ?)?

↓ ↓W

αW→ (W ?)?

Thus the collection of maps α:V → (V ?)? is a natural transformation be-tween the identity functor on the category of vector spaces and the functortaking a vector space to its second dual space.

On the category of finite-dimensional vector spaces, the natural transforma-tion α is a natural isomorphism.

7.3 Some Homological Algebra

Definition 7.16 A cochain complex, A•, is a sequence of abelian groups andhomomorphisms

· · · → Ap−1 dp−1→ Apdp→ Ap+1 → · · ·

such that dpdp−1 = 0 for all integers k ∈ Z

The homomorphisms dp are called differentials. We frequently simplify no-tation and write d for each differential. The defining condition for a sequenceto be a cochain complex is then written d2 = 0.

Example 7.17 The de Rham complex of a smooth manifold is a cochain com-plex.

Notice that for any cochain complex we have the inclusion im(d:Ap−1 →Ap) ⊆ ker(d:Ap → Ap+1). The cochain complex A• is called exact at p ifim(d:Ap−1 → Ap) = ker(d:Ap → Ap+1). The cochain complex A• is calledexact if it is exact at all integers p ∈ Z. We sometimes refer to an exact cochaincomplex as an exact sequence.

Definition 7.18 A short exact sequence is an exact complex of the form

0→ Aα→ B

β→ C → 0

The above short exact sequence

0→ Aα→ B

β→ C → 0

is called split exact if the group B can be written as a direct sum α[A]⊕C ′ andthe restriction β|C′ :C ′ → C is an isomorphism.

The proof of the following is a ’diagram chase’ and is more interesting towork out yourself than read.

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Proposition 7.19 5-lemmaSuppose we have a commutative diagram

A → B → C → D → E↓ ↓ ↓ ↓ ↓A′ → B′ → C ′ → D′ → E′

in which the rows are exact sequences and all of the vertical maps except forthe central one are isomorphisms. Then the central vertical map is also anisomorphism. 2

The proof of the following result is another diagram chase.

Proposition 7.20 Suppose we have a commutative diagram

→ Api→ Bp

j→ Cpk→ Ap+1 →

α ↓ β ↓ ↓ γ ↓ α→ A′p

i′→ B′pj′→ C ′p

k′→ A′p+1 →

in which the rows are exact and the vertical maps are all isomorphisms. Thenthere is an exact sequence

−→ Ap(α,−i)−→ A′p ⊕Bp i

′+β−→ B′pkγ−1j′−→ Ap−1 −→

2

The notion of the cohomology of a cochain complex is a measure of how farit is from being exact.

Definition 7.21 Let A• be a cochain complex. Then we define the cohomologygroups by the formula

Hp(A•) =ker(d:Ap → Ap+1)

im(d:Ap−1 → Ap)

Observe that a cochain complex is exact if and only if its cohomology groupsare all trivial.

Definition 7.22 A morphism of cochain complexes, α:A• → B•, is a collectionof homomorphisms α:Ap → Bp such that αd = dα.

We can form the category of all cochain complexes and morphisms. Simi-larly we can form the category of all short exact sequences and morphisms. Amorphism of cochain complexes is sometimes called a chain map.

Example 7.23 A smooth map f :M → N induces a chain map f?: Ω•(N) →Ω•(M) between the associated de Rham complexes.

Proposition 7.24 The cohomology groups are covariant functors from the cat-egory of cochain complexes to the category of abelian groups.

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Proof: Let α:A• → B• be a morphism of cochain complexes. Then for anyelement x ∈ Ap:

dx = 0 ⇒ αdx = 0 ⇒ d(αx) = 0x = dy ⇒ αx = α(dy) ⇒ αx = d(αy)

Thus α[ker(d:Ap → Ap+1)] ⊆ ker(d:Bp → Bp+1) and α[im(d:Ap−1 →Ap)] ⊆ im(d:Bp−1 → Bp) and we have a well-defined induced map

α?:Hp(A•)→ Hp(B•)

It is easy to see that with such induced maps the assignment A• 7→ Hp(A•)is a covariant functor. 2

We can similarly define morphisms of chain complexes. The analogue of theabove result holds for homology theories.

Definition 7.25 Let α, β:A• → B• be morphisms of cochain complexes. Thena chain homotopy in between α and β is a collection of homomorphisms T :Ap →Bp+1 such that:

α− β = Td± dT

We call morphisms α and β chain homotopic if there exists a chain homotopybetween them.

Proposition 7.26 Let α, β:A• → B• be chain homotopic morphisms ofcochain complexes. Then the induced maps α?, β?:H

p(A•)→ Hp(B•) are equal.

Proof: Let T be a chain homotopy between the morphisms α and β. Letx ∈ Ap and suppose that dx = 0. Then:

α(x)− β(x) = T (dx)± dT (x) = d(±T (x)) ∈ im(d:Bp+1 → Bp)

Hence α?([x])− β?([x]) = 0 for all elements [x] ∈ Hp(A•). 2

We call a morphism α:A• → B• a chain homotopy equivalence if thereis a morphism β:B• → A• such that the compositions αβ and βα are chainhomotopic to the idendities 1B• and 1A• respectively. By the above result achain homotopy equivalence of cochain complexes induces an isomorphism ofcohomology groups.

A cochain complex A• is called chain contractible if the identity and zeromorphisms are chain homotopic. The above result tells us that the homologygroups of a chain contractible cochain complex are all trivial.

Definition 7.27 A short exact sequence of cochain complexes is a sequence ofthe form

0→ A•α→ B•

β→ C• → 0

such that each sequence of abelian groups

0→ Apα→ Bp

β→ Cp → 0

is a short exact sequence.

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We can form the category of all short exact sequences of cochain complexes;the morphisms are commutative diagrams

0 → A• → B• → C• → 0↓ ↓ ↓

0 → A′• → B′• → C ′• → 0

in which the vertical maps are morphisms of cochain complexes.

Theorem 7.28 Snake LemmaLet

0→ A•α→ B•

β→ C• → 0

be a short exact sequence of cochain complexes. Then there are natural maps∂:Hp(C•)→ Hp+1(A•) such that we have a long exact sequence

· · · → Hp(A•)α?→ Hp(B•)

β?→ Hp(C•)∂→ Hp+1(A•)→ · · ·

Proof: We have a commutative diagram

↓ ↓ ↓0 → Ap−1 α→ Bp−1 β→ Cp−1 → 0

↓ ↓ ↓0 → Ap

α→ Bpβ→ Cp → 0

↓ ↓ ↓0 → Ap+1 α→ Bp+1 β→ Cp+1 → 0

↓ ↓ ↓

in which the rows are exact and the vertical maps are the differentials, d.Let x ∈ Cp and suppose that dx = 0. Then x = βy and dβy = 0 so βdy = 0.

Hence dy = αz for some element z ∈ Ap+1. The claim is that we can define anatural homorphism ∂:Hp(Cp)→ Hp(Ap+1) by writing ∂([x]) = [z] where

x = βy dy = αz

The above argument shows that a suitable element z ∈ Ap+1 exists. Further,αdz = dαz = d2y = 0 so dz = 0 since the homomorphism α is injective. If wecan also write x = βy′ and dy′ = αz′ then:

β(y − y′) = 0 ⇒ y − y′ = αw⇒ dy − dy′ = dαw = αdw⇒ α(z − z′) = αdw⇒ z − z′ = dw

so the elements [z], [z′] ∈ Hp+1(A•) are equal. Hence the homomorphism ∂ iswell-defined.

It is left as an exercise to show that the homomorphism ∂ is natural andthat the sequence of homology groups

· · · → Hp(A•)α?→ Hp(B•)

β?→ Hp(C•)∂→ Hp+1(A•)→ · · ·

63

is exact. The proof consists of a number of ‘diagram chases’ similar to the above.2

The notion of a split exact sequence can sometimes be useful in homologicalcomputations.

Proposition 7.29 A split exact sequence

0→ A•α→ B•

β→ C• → 0

of cochain complexes induces short exact sequences

0→ Hp(A•)α?→ Hp(B•)

β?→ Hp(C•)→ 0

of cohomology groups. 2

8 De Rham Cohomology

8.1 Differential Forms

Let M be a smooth manifold. Consider the space Γ0p(M) of tensor fields of type

(0, p). We have a subspace, N , generated by tensor fields of the form

α1 ⊗ · · · ⊗ αp + ασ(1) ⊗ · · · ⊗ ασ(p)

where αi is a covector field and σ is an odd permutation of the set 1, . . . , p.

Definition 8.1 We define the set of differential forms of order p to be thequotient

Ωp(M) = Γ0p(M)/N

A differential form of order p is sometimes referred to as a p-form.Observe that a differential form of order 0 is just a function f ∈ C∞(M). A

differential form of order 1 is the same thing as a covector field.We define the support of a p-form ω to be the closure of the set of points

x ∈ M such that ω(x) 6= 0. In parts of the theory it is important to considercompactly supported differential forms. We write Ωpc(M) to denote the set ofdifferential forms of order p with compact support.

Of course, on a compact manifold every differential form has compact sup-port.

Definition 8.2 We define the exterior product

∧: Ωp(M)× Ωq(M)→ Ωp+q(M)

by the formula[α] ∧ [β] = [α⊗ β]

where α ∈ Γ0p(M) and β ∈ Γ0

p(M).

64

Note that the product of a differential form with a compactly supporteddifferential form is compactly supported.

Now, observe that any differential p-form can be written as a sum of p-formsof the type

α1 ∧ · · · ∧ αp

where αi is a covector field. The definition of the space Ωp(M) means that wehave the identity

α1 ∧ · · · ∧ αp = sgn(σ)ασ(1) ∧ · · · ∧ ασ(p)

whenever σ is a permutation of the set 1, . . . , p. The following result is easilydeduced.

Proposition 8.3 Consider differential forms ω ∈ Ωp(M) and η ∈ Ωq(M).Then we have the identity

ω ∧ η = (−1)pqη ∧ ω

2

Corollary 8.4 Consider a differential form ω ∈ Ωp(M) where p is odd. Then:

ω ∧ ω = 0

2

Proposition 8.5 Let ω ∈ Ωp(M). Then in coordinates (x1, . . . , xn) defined bysome chart we can write

ω =∑

i1<···<ip

fi1,...,ipdxi1 ∧ · · · ∧ dxip

where fi1,...,ip is a smooth function.

Proof: Let us writeω =

∑j

αj1 ∧ · · · ∧ αjp

where αjk is a covector field. In coordinates (x1, . . . , xn) defined by a chart wecan certainly write

αjk = gjk1 dx1 + · · ·+ gjkn dx

n

where gjk is a smooth function.Hence

ω =∑i,j

(gj11 dx1 + · · ·+ gj1n dx

n) ∧ · · · ∧ (gjp1 dx

1 + · · ·+ gjpn dxn)

Bilinearity of the exterior product enables us to write:

ω =∑

j1,...,jp

hj1,...,jpdxj1 ∧ · · · ∧ dxjp

65

But the expression dxj1 ∧ · · · ∧ dxjp is either zero (if dxjk = dxjl for somek 6= l) or else takes the form

±dxi1 ∧ · · · ∧ dxip

where i1 < · · · < ip. Hence:

ω =∑

i1<···<ip

fi1,...,ipdxi1 ∧ · · · ∧ dxip

as required. 2

A slightly more condensed notation can be convenient when dealing withdifferential forms and coordinates. Let us write I to denote a p-tuple (i1, . . . , ip),and dxI to denote the product dxi1 ∧ · · · ∧ dxip . Then for coordinates definedby some chart a differential form ω ∈ Ωp(M) can be written

ω =∑I

fIdxI

Corollary 8.6 Let M be a smooth manifold of dimension n. then Ωp(M) = 0if p > n.

Proof: In coordinates (x1, . . . , xn) there are no expressions of the form

dxi1 ∧ · · · ∧ dxip

where i1 < · · · < ip. 2

8.2 The de Rham Complex

Recall that for a smooth function f ∈ C∞(M) we can define a covector field bythe formula

df =∑i

∂f

∂xidxi

in coordinates defined by a chart. We can extend this definition.

Definition 8.7 Consider a differential form ω ∈ Ωp(M). Write in coordinatesdefined by a chart

ω =∑I

fIdxI

Then we define a (p+ 1)-form dω ∈ Ωp+1(M) by the formula

dω =∑I

dfI ∧ dxI

The operator d: Ωp(M)→ Ωp+1(M) is called the (de Rham) differential. Oneway to prove that the differential is well-defined is by brute force and lemma9.15— a calculation with the chain rule shows that the form dω defined as aboveis independent of any choice of chart. Here we describe a more elegant method.

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Proposition 8.8 Let (U, V, φ) be a chart. Then the differential d: Ωp(U) →Ωp+1(U) defined by the above formula has the following properties:

• d(ω + η) = dω + dη for all ω, η ∈ Ωp(U).

• d(ω ∧ η) = dω ∧ η + (−1)pω ∧ dη for all ω ∈ Ωp(U) and η ∈ Ωq(U).

• d(dω) = 0 for all ω ∈ Ωp(U).

Proof: The first of these formulae is obvious. To prove the second, observethat the first formula means we can assume that

ω = fdxI η = gdxJ

where f and g are smooth function. In this case:

d(ω ∧ η) = d(fg) ∧ dxI ∧ dxJ= gdf ∧ dxI ∧ dxJ + fdg ∧ dxI ∧ dxJ= dω ∧ η + (−1)pfdxI ∧ (dg ∧ dxJ)= dω ∧ η + ω ∧ dη

To prove the last formula, we again consider the case ω = fdxI so that

dω =∑i

∂f

∂xidxi ∧ dxI

and

d(dω) =∑i,j

∂2f

∂xjxidxj ∧ dxi ∧ dxI =

∑i<j

∂2f

∂xjxi(dxj ∧ dxi + dxi ∧ dxj)∧ dxI = 0

since dxi ∧ dxi = 0 and dxi ∧ dxj = −dxj ∧ dxi 2

The last of the above formulae is frequently abbrebriated by writing ‘d2 = 0’.

Proposition 8.9 Let (U, V, φ) be a coordinate chart on M with local coordi-nates (x1, . . . , xn). Suppose we have an operator d′: Ωp(U)→ Ωp+1(U) with thefollowing properties.

• d′(ω + η) = d′ω + d′η for all ω, η ∈ Ωp(U).

• d′(ω ∧ η) = d′ω ∧ η + (−1)pω ∧ d′η for all ω ∈ Ωp(U) and η ∈ Ωq(U).

• d′(d′ω) = 0 for all ω ∈ Ωp(U).

• d′f = df when f is a smooth function.

Then d′ = d.

Proof: The first property means that it suffices to show that d′(fdxI) =d(fdxI) when f is a smooth function on U . The second and last propertiesyield the formula

d′(fxI) = d′f ∧ dxI + fd′(dxI) = df ∧ dxI + f ∧ d′(dxI)

67

With the first property, we can write

dxI = dxi1 ∧ · · · ∧ dxik = d′xi1 ∧ · · · ∧ d′xik

We know that d′(d′xil) = 0 for each il. It follows by induction using thethird formula that d′(dxI) = 0 and we are done. 2

Corollary 8.10 Let M be a smooth manifold. Then there is a unique operatord: Ωp(M)→ Ωp+1(M) such that:

• d(ω + η) = dω + dη for all ω, η ∈ Ωp(U).

• d(ω ∧ η) = dω ∧ η + (−1)pω ∧ dη for all ω ∈ Ωp(U) and η ∈ Ωq(U).

• d(dω) = 0 for all ω ∈ Ωp(U).

• Let f ∈ C∞(M). Then

df =∑i

∂f

∂xidxi

in terms of local coordinates x1, . . . , xn.

Proof: For each chart (U, V, phi) we have a unique operator dU : Ωp(U) →Ωp+1(U) satisfying the above properties. We can therefore define a suitableoperator d: Ωp(M)→ Ωp+1(M) by writing

dω(x) = dU (ω|U )(x)

if x ∈ U . 2

Because of proposition 8.9 the operator d is defined in terms of local coor-dinates by our original formula

dω =∑I

dfI ∧ dxI

whereω =

∑I

fIdxI

Definition 8.11 The sequence of groups and homomorphisms

0→ Ω0(M)d→ Ω1(M)

d→ Ω2(M)d→ · · ·

is called the de Rham complex of the manifold M .

8.3 De Rham Cohomology

Definition 8.12 The cohomology groups of the de Rham complex:

Hp(M) =ker(d: Ωp(M)→ Ωp+1(M))

im(d: Ωp−1(M)→ Ωp(M))

are called the de Rham cohomology groups of the smooth manifold M .

68

Example 8.13

Hp(point) =

R p = 00 p 6= 0

Example 8.14 Let M be any connected manifold. Suppose we have a smoothfunction f ∈ C∞(M) such that df = 0. Then, in terms of local coordinates(x1, . . . , xn), we know that ∂f/∂xi = 0. Hence the function f is constant.

By definition of de Rham cohomology, it follows that H0(M) ∼= R.

Example 8.15 Let M and N be smooth manifolds. Then Hp(M∐N) =

Hp(M)⊕Hp(N) for all p.In particular, using the above example, for any smooth manifold M , we have

an isomorphism H0(M) ∼= Rπ0(M).

To work out the de Rham cohomology groups of more complicated examples,and to see why they are interesting, we need to develop the general theory.

Proposition 8.16 Let f :M → N be a smooth map of manifolds. Thenthere is a contravariantly functorial induced chain map of cochain complexesf?: Ω?(N)→ Ω?(M).

Proof: Let ω ∈ Ωp(N). Choose local coordinates (x1, . . . , xm) on the manifoldM and write f(x1, . . . , xn) = (y1, . . . , yn). Write:

ω =∑

i1<···<ip

gi1,...,ipdxi1 ∧ · · · ∧ dxip

Then we can define a differential form f?(ω) ∈ Ωp(M) by the formula:

f?ω =∑

i1<···<ipj1<···<jp

gi1,...,ip f∂

∂xi1dyj

1

· · · partialxik

∂yjkdyj1 ∧ · · · ∧ dyjk

It is easy to check that f?(dω) = d(f?ω). Hence the map f? is a chain map.Functoriality is also easy to check. 2

We thus obtain contravariantly functorial induced maps of de Rham coho-mology groups:

f?:Hp(N)→ Hp(M)

Lemma 8.17 Let M be a smooth manifold. Define maps s:M → M × R andπ:M × R→M by the formulae:

s(x) = (x, 0) π(x, t) = x

respectively. Then the induced maps s?:Hp(M × R) → Hp(M) andπ?:Hp(M)→ Hp(M × R) are isomorphisms.

Proof: Observe that πs = 1M so the composition s? π? is the identity map.We aim to construct a chain homotopy between the map π? s?: Ωp(M ×R)→Ωp(M × R) and the identiy.

69

Let ω ∈ Ωp(M × R). Then we can write:

ω = f(x, t)π?(η) + g(x, t)π?(θ) ∧ dt

where η ∈ Ωp(M) and θ ∈ Ωp−1(M). Define a form Kω ∈ Ωp−1(M × R) bywriting:

Kω(x, t) = π?(θ)(x, t)

∫ t

0

g(x, s) ds

A fairly long but perfectly straightforward calculation tells us that:

ω − π? s?ω = (−1)p−1(dKω −Kdω)

so we have the desired chain homotopy. 2

Theorem 8.18 Let f, g:M → N be smoothly homotopic maps. Then the in-duced maps f?, g?:Hp(N)→ Hp(M) are equal.

Proof: We have a map F :M × R → N such that F (−, t) = f whenevert ≤ 0 and F (−, t) = g whenever t ≥ 1. Define maps s0, s1:M →M × R by theformulaee:

s0(x) = (x, 0) s1(x) = (x, 1)

By the above lemma the projection π:M ×R→M induces an isomorphismπ?:Hp(M)→ Hp(M ×R). The maps s?0 and s?1 are both inverses of the isomor-phism π? and therefore equal. Hence the maps f? = s?0 F ? and g? = s?1 F ?are equal. 2

Corollary 8.19 The assignment of homotopy groups, M 7→ Hp(M), is a con-travariant functor from the category of smooth manifolds and continuous mapsto the category of abelian groups. The functorially induced maps agree with thosedefined above in the smooth case.

Further, if f, g:M → N are homotopic maps, then the induced mapsf?, g?:Hp(N)→ Hp(M) are equal.

Proof: Let f :M → N be a smooth map. By theorem 7.7 there is a smoothmap, f :M → N that is homotopic to f . We try to define our induced mapf?:Hp(N)→ Hp(M) by the formula f? = f?.

Suppose that f ′ is another smooth map that is homotopic to f . Then themaps f ′ and f are homotopic. It follows by corollary 7.8 that the maps f ′ andf are smoothly homotopic, and so f ′? = f?, and the map f? is well-defined.

The proof that homotopic maps f, g:M → N induce the same homomor-phisms f?, g?:Hp(N)→ Hp(M) is proved similarly. 2

The following consequence of homotopy-invariance is sometimes called thePoincare lemma.

Corollary 8.20

Hp(Rn) =

R p = 00 p 6= 0

2

70

Lemma 8.21 Let M be a smooth manifold, and let U and V be open subman-ifolds such that M = U ∪ V . Let i:U ∩ V → U , j:U ∩ V → V , k:U →M , andl:V →M be the various inclusion maps. Then we have a short exact sequenceof cochain complexes:

0 −→ Ω?(M)(k?,l?)−→ Ω?(U)⊕ Ω?(V )

i?−j?−→ Ω?(U ∩ V ) −→ 0

Proof: The only remotely awkward part in proving exactness of the givensequence is proving that the map i? − j? is surjective.

Let ω ∈ Ωp(U ∩ V ). Let ρU , ρV be a partition of unity subordinate to theopen cover U, V . Then we have forms ρV ω ∈ Ωp(U) and ρUω ∈ Ωp(V ). Wecan write:

so the map i? − j? is surjective as required. 2

By the snake lemma, the above result implies the following theorem.

Theorem 8.22 There is a natural map ∂:Hp(U ∩ V ) → Hp+1(M) such thatwe have a long exact sequence:

−→ Hp(M)(k?,l?)−→ Hp(U)⊕Hp(V )

i?−j?−→ Hp(U ∩ V )∂−→ Hp+1(M) −→

2

The above long exact sequence is called the Mayer-Vietoris sequence in deRham cohomology. It is an extremely useful tool for calculations.

Example 8.23 The sphere, Sn+1, can be expressed as a union Sn+1 = U ∪V where the sets U and V are contractible, and the intersection, U ∩ V , ishomotopy-equivalent to the sphere Sn.

The Mayer-Vietoris sequence thus takes the following form

−→ Hp(Sn+1)(k?,l?)−→ Hp(+)⊕Hp(+)

i?−j?−→ Hp(Sn)∂−→ Hp+1(Sn+1) −→

where + is the one point space.Let p > 0. Then Hp(+) = 0 and Hp+1(+) = 0 so the map ∂:Hp(Sn) →

Hp+1(Sn+1) is an isomorphism.If n > 0 and p = 0, the above exact sequence takes the form

0 −→ R α−→ R⊕ R β−→ R ∂−→ H1(Sn+1) −→ 0

where the maps α and β are defined by the formulae

α(x) = (x, x) β(x, y) = x− y

respectively. The image of the map β is the entire set R, so by exactness, themap ∂ is zero, and the group H1(Sn+1) is zero. Since we have an isomorphismHp(Sn) ∼= Hp+1(Sn+1), we know that Hp(Sn) = 0 if n > p, and p > 0.

The ‘sphere’ S0 is just a set containing two points. Thus, if n = 0 and p > 0the above exact sequence becomes

0 −→ Hp+1(S1) −→ 0 −→ 0

71

and, using the isomorphism Hp(Sn) ∼= Hp+1(Sn+1), see that Hp(Sn) = 0 ifp > n.

Finally, if n = 0 and p = 0 we have the sequence

0 −→ R α−→ R⊕ R β−→ R⊕ R ∂−→ H1(Sn+1) −→ 0

where the maps α and β are defined by the formulae

α(x) = (x, x) β(x, y) = (x− y, x− y)

respectively. The map β has image

(x, x) ∈ R⊕ R

So we have a short exact sequence

0 −→ R γ−→ R⊕ R ∂−→ H1(Sn+1) −→ 0

where γ(x) = (x, x).We can define a homomorphism γ′:R⊕R→ R by the formula γ′(x, y) = x.

Then γ′γ = 1R so by proposition ?? the groups R ⊕ R and R ⊕ H1(S1) areisomorphic. It follows that the group H1(S1) is isomorphic to R. Using theisomorphism Hp(Sn) ∼= Hp+1(Sn+1), we see that Hn(Sn) = R if n > 0.

To summarise:

Hp(Sn) =

R p = n, p = 00 otherwise

The above example tells us that the spheres Sm and Sn are not homotopy-equivalent, and in particular not homeoemorphic if m 6= n. Therefore the spacesSm\+ and Sn\+ obtained by deleting a single point from these spheres arenot homeomorphic. But the spaces Sm\+ and Sn\+ are homeomorphicto the spaces Rm and Rm respectively, and so the spaces Rm and Rm are nothomeomorphic.

We will see some other fundamental geometric results proved through similarcomputations in the exercises and following chapters.

We now turn our attention to compactly supported differential forms. Webegin by noting that the differential of a compactly supported form is compactlysupported so we have a cochain complex:

0→ Ω0c(M)

d→ Ω1c(M)

d→ Ω2c(M)

d→ · · ·

The cohomology groups of this complex are the compactly supported de Rhamcohomology groups, Hp

c (M). Similar calculations to those performed above yieldthe following results:

Proposition 8.24

Hpc (Rn) =

R p = n0 p 6= n

2

72

Compactly supported cohomology also posesses a ‘wrong way functoriality’property. To be specific, if M is a manifold and U ⊆M is an open submanifold,then any compactly supported form ω ∈ Ωpc(U) can be extended (by zero) to acompactly supported form i(ω) ∈ Ωpc(M). The inclusion map i:U → M thusinduces a map of homology groups i?:H

pc (U)→ Hp

c (M).Mayer-Vietoris sequences also function in the compactly supported setting.

Theorem 8.25 Let M be a smooth manifold, and let U and V be open sub-manifolds such that M = U ∪ V . Let i:U ∩ V → U , j:U ∩ V → V , k:U →M ,and l:V → M be the various inclusion maps. Then there is a natural map∂:Hp

c (U ∩ V )→ Hp+1c (M) such that we have a long exact sequence:

−→ Hpc (U ∩ V )

(i?,j?)−→ Hpc (U)⊕Hp

c (V )k?−l?−→ Hp

c (M)∂−→ Hp−1

c (U ∩ V ) −→

2

9 Orientation and Duality

9.1 Orientations on Vector Bundles

Let e1, . . . , en and e′1, . . . , e′n be bases for a real vector space V . We canform an invertible linear transformation T :V → V by writing Tei = e′i. Thereis a matrix, A, associated to T ; the entries aij are defined by the formula

e′i =∑j

aijej

Since the linear transformation T is invertible, the matrix A has non-zerodeterminant.

Definition 9.1 The bases e1, . . . , en and e′1, . . . , e′n are said to have thesame orientation if the determinant det(A) is positive. Otherwise, they are saidto have the opposite orientation.

Thus an orientation of a vector space is a certain equivalence class of bases.A vector space has precisely two possible orientations. An oriented vector spaceis simply a vector space equipped with a choice of orientation; we say a basis inthe relevant equivalence class is one that defines the orientation.

Definition 9.2 Let V and V ′ be oriented vector spaces of the same dimension.Let e1, . . . , en be a basis for the vector space V defining the given orientation.Then an invertible linear transformation T :V → V ′ is said to be orientation-preserving if the basis Te1, . . . , T en defines the orientation given for the vectorspace V ′.

It is easy to check that the above definition does not depend on the exactchoice of basis e1, . . . , en with the given orientation. An invertible linear trans-formation that is not orientation-preserving is termed orientation-reversing.

The standard orientation for the vector space Rn is that defined by thestandard basis

(1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, . . . , 0, 1)

73

Definition 9.3 Let E be a real vector bundle over a space X. An orientationfor E is a choice of orientation for each vector space Ex such that if U ⊆ X isa connected open set, and ψ:EU → U ×Rn is a bundle isomorphism, the lineartransformations of fibres

ψ:Ex → x × Rn

are either all orientation-preserving or orientation-reversing.

A real vector bundle equipped with an orientation is called oriented. A realvector bundle that can be equipped with an orientation is called orientable.

Example 9.4 Any trivial real vector bundle is orientable.

The following result is an exercise in understanding the definition.

Proposition 9.5 Let E be a real orientable vector bundle. Then E can beequipped with precisely two possible orientations. 2

Definition 9.3 is perhaps the best definition to work with when proving factsabout oriented vector bundles or trying to prove that a given vector bundle isnot oriented. However, when trying to define orientations on vector bundles, areformulation is useful.

Proposition 9.6 Let E be a real vector bundle over a space X equipped withan orientation on each fibre Ex. Suppose we have an atlas of local trivialisations

(Uλ, ψλ) | λ ∈ Λ

such that for each λ ∈ Λ the linear transformations

ψλ:Ex → x × Rn

are either all orientation-preserving or orientation-reversing.Then the bundle E is oriented. 2

Example 9.7 Let E be a complex vector bundle over a space X, that is to saya vector bundle where the fibres are isomorphic to Cn for some n. There is anunderlying real vector bundle ER; the fibre (ER)x is defined by considering thefibre Ex as a real rather than a complex vector space. We claim that the bundleER is orientable.

To see this, let (Uλ, ψλ) | λ ∈ Λ be an atlas of local trivialisations for thecomplex vector bundle E. Let e1, . . . , en be a basis for the complex vectorspace Ex. Then the real vector space (ER)x has a basis

e1, ie1, e2, ie2, . . . , en, ien

Let us equip the space (ER)x with the orientation defined by the above basis.Let v1, . . . , v2n be the standard basis for the real vector space R2n, and letw1, . . . , wn be the standard basis for the complex vector space Cn. We havean isomorphism T :Cn → R2n of real vector spaces defined by writing

Twj = v2j−1 T (iwj) = v2j

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The real vector bundle ER has an atlas of local trivialisations (Uλ, T ψλ) | λ ∈ Λ. It is easy to check that the maps

T ψλ: (ER)x → x × R2n

are all orientation-preserving.

The argument used in the above example tells us not only that the real vectorbundle ER is orientable, but gives us a canonical orientation. This observationwill be important in later chapters when we explore the theory of characteristicclasses.

Example 9.8 Consider the Mobius band, E, as a one-dimensional real vectorbundle over the circle S1. Consider the space

A =[0, 1]× −1, 1

∼

where [0, t] ∼ [1,−t].Suppose that the Mobius band can be equipped with an orientation. Then

for each point x there is a vector s(x) ∈ A∩Ex of norm one such that the basiss(x) has the given orientation.

According to the definition of an orientation, we have constructed a nowhere-zero section s:S1 → E. But according to proposition 5.2 this would prove thatthe Mobius band is trivial, which is certainly not the case.

Thus the Mobius band is not orientable.

9.2 Orientations of Manifolds

Definition 9.9 Let M be a smooth manifold. Then M is termed orientable ifthe tangent bundle TM is orientable.

An orientation of an orientable manifold is a choice of orientation for thetangent bundle TM .

Example 9.10 Any manifold with trivial tangent bundle is orientable.

Manifolds with trivial tangent bundles are called parallelisable.

Example 9.11 The Mobius band is not an orientable manifold. The argumentto prove this is similar to that of example 9.8.

Let f :M → N be a diffeomorphism. Then the differential (Df)x:TxM →Tf(x)N is an isomorphism. If M and N are oriented manifolds, the diffeomor-phism f is termed orientation-preserving if the isomorphisms (Df)x:TxM →Tf(x)N are all orientation-preserving, and orientation-reversing if the isomor-phisms Df :TxM → Tf(x)N are all orientation-reversing.

Definition 9.12 Let M be a manifold. A smooth atlas (Uλ, Vλ, φλ) forM is said to be oriented if the diffeomorphisms (φ−1

µ φλ)|φ−1λ [Vλ∩Vµ] are all

orientation-preserving.

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A smooth manifold is said to be oriented if it is equipped with an orientedatlas. Disentangling the relevant definitions gives us the following result.

Proposition 9.13 An orientation on a smooth manifold determines, and isdetermined by, a choice of oriented atlas. 2

The above lets us easily see that, for example, the sphere Sn is orientable.Note that our work on orientations all also makes sense for manifolds with

boundary. Further, we have the following.

Proposition 9.14 Let M be an oriented manifold, with boundary ∂M . Thenthe boundary has an orientation induced from that of M . 2

Orientation can be reformulated in terms of differential forms. To prove thereformulation is valid, we need the following lemma. The proof is a straightfor-ward calculation based on the definition of a determinant.

Lemma 9.15 Let (U, V, φ) and (U , V , φ) be charts, and let φ(x1, . . . , xn) =φ(x1, . . . , xn)

Let J be the Jacobian determinant

J = det

(∂xi

∂xj

)Then

dx1 ∧ · · · ∧ dxn = Jdx1 ∧ · · · ∧ dxn

2

Proposition 9.16 Let M be a smooth n-dimensional manifold. Then the man-ifold M is orientable if and only if there is a differential form ω ∈ Ωn(M) suchthat ω(x) 6= 0 for all x ∈M .

Proof: Suppose the manifold M is orientable. Then we have an orientedlocally finite atlas

(Uλ.Vλ, φλ) | λ ∈ Λ

Let φλ(x1λ, . . . , x

nλ) = φµ(x1

µ, . . . , xnµ). Then the determinants

det

(∂xiλ∂xjµ

)

are all positive.Choose a smooth partition of unity, ρλ | λ ∈ Λ, subordinate to the locally

finite cover Uλ | λ ∈ Λ. Define

ω =∑λ∈Λ

ρλdx1λ ∧ · · · ∧ dxnλ

By lemma 9.15 and positivity of the determinants arising form coordinatechanges, it follows that ω(x) 6= 0 for every point x ∈M .

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Conversely, suppose we have a nowhere-zero differential form ω ∈ Ωn(M).Then for each point x ∈M we have local coordinates (x1, . . . , xn) such that

ω = fdx1 ∧ · · · ∧ dxn

for some non-vanishing smooth function f . By permuting two of the coordinates(xi) if necessary, we can assume that the function f is strictly positive.

If (x1, . . . , xn) is another set of such local coordinates, we can write

ω = fdx1 ∧ · · · ∧ dxn

where the function f is also strictly positive. Let J be the Jacobian determinantdet(∂xi/∂xj). Then by lemma 9.15, J = f/f . Hence the determinant J ispositive.

Thus the collection of local coordinate systems we have defined gives us anoriented atlas for M . 2

Definition 9.17 Let M be a Riemannian manifold equipped with some ori-ented atlas. In local coordinates (x1, . . . , xn) let us write

gij = 〈 ∂∂xi

,∂

∂xj〉

and let |g| denote the determinant of the matrix (gij). Then we define thevolume form on M by the formula

vol =√|g|dx1 ∧ · · · ∧ dxn

The volume form does not depend on a choice of local coordinates.

9.3 Integration

Suppose that M is a smooth n-dimensional manifold, equipped with an ori-ented atlas. Let [M ] denote the orientation defined by the atlas. Consider acompactly supported differential form ω ∈ Ωnc (M). Let (U, V, φ) be a chart, and(x1, . . . xn) ∈ U . Then we can write

ω(x) = f(x)dx1 ∧ · · · ∧ dxn

when x ∈ V .

Definition 9.18 If ω(x) = 0 for x 6∈ V , we define the integral:∫[M ]

ω =

∫f(φ(x1, . . . , xn)) dx1 . . . dxn

Suppose we have another chart with the same orientation, (U , V , φ), andhave φ(x1, . . . , xn) = φ(x1, . . . , xn). Then the determinant J(x1, . . . , xn) =det(∂xi/∂xj) is positive, and we have the formula

fdx1 ∧ · · · ∧ dxn = Jfdx1 ∧ · · · ∧ dxn

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by lemma 9.15. But by the change of variables formula for integration:∫fφ(x1, . . . , xn) dx1 · · · dxn =

∫J(x1, . . . xn)fφ(x1, . . . , xn) dx1 · · · dxn

so the integral of the differential form ω does not depend on a choice of localcoordinates. More generally, the following definition makes sense.

Definition 9.19 Let M be an n-dimensional manifold, with oriented locallyfinite atlas (Uλ, Vλ, φλ) | λ ∈ Λ. Let ρλ | λ ∈ Λ be a partition of unitysubordinate to the open cover Uλ | λ ∈ Λ, and let [M ] be the orientationdefined by the atlas. Let ω ∈ Ωnc (M) be a compactly supported n-form. Thenwe define the integral: ∫

[M ]

ω =∑λ∈Λ

∫ρλω

It is easy to check that the above definition does not depend on choice ofappropriate oriented atlas or on the partition of unity chosen.

Lemma 9.20 Let ω ∈ Ωn−1c (Rn) be a compactly supported (n−1)-form. Then:∫

Rndω = 0

Proof: Without loss of generality, assume that ω = fdx2 ∧ · · · ∧ dxn. Then:

dω =∂f

∂x1dx1 ∧ · · · ∧ dxn

By Fubini’s theorem:∫Rndω =

∫Rn−1

(∫ ∞−∞

∂f

∂x1dx1

)dx2 . . . dxn

But ∫ ∞−∞

∂f

∂x1dx1 = 0

by the fundamental theorem of calculus and the result follows. 2

A similar argument gives us the following result for integrals over half-space.

Lemma 9.21 Let ω ∈ Ωn−1c (Rn+) be a compactly supported (n−1)-form. Then:∫

Rndω =

∫Rn−1

ω

2

Theorem 9.22 Stokes’ theoremLet M be a smooth oriented bounded manifold of dimension n. Let ω ∈ Ωn−1

c (M)be a differential (n− 1)-form. Then:∫

[M ]

dω =

∫[∂M ]

ω

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Proof: Let (Uλ, Vλ, φλ) be an oriented atlas. Let ρλ | λ ∈ Λ be a partitionof unity subordinate to the open cover Uλ | λ ∈ Λ. Then:∫

[M ]

dω =∑λ∈Λ

∫[M ]

ρλdω

Without loss of generality, assume that the open set Vλ is diffeomorphic toeither Rn or Rn+. If the set Vλ is diffeomorphic to the space Rn then:∫

[M ]

ρλdω = 0

by lemma 9.20. If the set Vλ is diffeomorphic to the space Rn+ then:∫[M ]

ρλdω =

∫[∂M ]

ρλω

by lemma 9.21. Hence: ∑λ∈Λ

∫[M ]

ρλdω =∑λ∈Λ

∫[∂M ]

ρλω

and we are done. 2

Corollary 9.23 Let M be a smooth oriented manifold without boundary. Thenwe have a well-defined linear map∫

[M ]

:Hnc (M)→ R

defined by the formula ∫[M ]

[ω] =

∫[M ]

ω.

Proof: Let η ∈ Ωn−1c (M). Then by Stokes’ theorem∫

[M ]

dη =

∫∅η = 0.

Hence, if [ω] = [ω′], then ω − ω′ = dη, so∫[M ]

ω =

∫[M ]

ω′.

The result now follows. 2

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9.4 Poincare Duality

Let M be a smooth manifold. Let ω ∈ Ωp(M) and η ∈ Ωq(N). Then we havethe exterior product ω ∧ η ∈ Ωp+q(M). By proposition 8.10 we know that

d(ω ∧ η) = dω ∧ η + (−1)pω ∧ dη.

This gives us the following result.

Proposition 9.24 We have well-defined linear maps

∪:Hp(M)×Hq(M)→ Hp+q(M)

and∪:Hp(M)×Hq

c (M)→ Hp+qc (M)

both defined by the formula

[ω] ∪ [η] = [ω ∧ η].

2

We call the element [ω]∪ [η] the cup product of [ω] and [η]. The cup productis easily seen to be associative.

Suppose M is n-dimensional. Then by the above we have a bilinear map

〈−,−〉:Hp(M)×Hn−pc (M)→ R

defined by the formula

〈[ω], [η]〉 =

∫[M ]

ω ∧ η.

Definition 9.25 We say a bilinear map B:V ×W → R is non-degenerate ifB(v, w) = 0 for all v ∈ V implies w = 0.

It is equivalent to the above to say that B is non-degenerate if B(v, w) = 0for all w ∈ W implies v = 0. Given a bilinear map B, we have a linear mapA:V →W ∗ defined by the formula

A(v)(w) = B(v, w).

The bilinear map B is non-degenerate if and only if the linear map A is anisomorphism.

Poincare duality is the assertion that the bilinear map 〈−,−〉:Hp(M) ×Hn−pc (M)→ R is non-degenerate.

The key part of the proof is the following lemma.

Lemma 9.26 Let M = U ∪ V , where U and V are open. Then the followingdiagram is commutative up to a + or − sign:

−→ Hp(M)(k?,l?)−→ Hp(U)⊕Hp(V )

i?−j?−→ Hp(U ∩ V )∂−→ Hp+1(M) −→

⊗ ⊗ ⊗ ⊗←− Hp

c (M)k?−l?←− Hp

c (U)⊕Hpc (V )

(i?,j?)←− Hpc (U ∩ V )

∂←− Hp−1c (M) −→

↓ ↓ ↓ ↓R R R R

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Here, the top rows are Mayer-Vietoris sequences in de Rham and compactlysupported de Rham cohomology respectively, and the vertical arrows are definedby the map 〈−,−〉.

Proof: Let η ∈ Ωp(M), ξ1 ∈ Ωn−pc (U), ξ2 ∈ Ωn−pc (V ). Then

〈η, (k∗ − l∗)(ξ1, ξ2)〉 = 〈η, ξ1〉+ 〈η, ξ2〉

and〈(k∗, l∗(η), (ξ1, ξ2)〉 = 〈(η, η), (ξ1, ξ2)〉 = 〈η, ξ1〉+ 〈η, ξ2〉

so the first square commutes. The second square commutes similarly.Let η ∈ Ωp(U ∩ V ) and ξ ∈ Ωn−p−1

x (M). Then

〈η, dξ〉 =

∫[M ]

η ∧ dξ.

Recall from corollary 8.10 that

d(η ∧ ξ) = dη ∧ ξ + (−1)pη ∧ dξ

and by Stokes’ theorem ∫[M ]

d(η ∧ ξ) = 0.

Hence ∫[M ]

η ∧ dξ = ±∫

[M ]

dη ∧ ξ = 〈dη, ξ〉

and we are done. 2

Definition 9.27 Let M be a smooth manifold. An open cover Uλ | λ ∈ Λis called a good cover of all non-empty finite intersections Uλ1 ∩ · · · ∩ Uλn arediffeomorphic to Rn.

The proof of the following is straightforward.

Proposition 9.28 Every manifold (without boundary) has a good cover, andevery compact manifold has a finite good cover.

We say a manifold is of finite type if it has a finite good cover.

Theorem 9.29 (Poincare Duality) Let M be an oriented manifold of finitetype. Then the bilinear map

〈−,−〉:Hp(M)×Hn−pc (M)→ R

defined by the formula

〈[ω], [η]〉 =

∫[M ]

ω ∧ η

is non-degenerate.

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Proof: Recall, from the Poincare lemmas

Hp(Rn) ≡

R p = 00 otherwise

Hpc (Rn) ≡

R p = n0 otherwise

Suppose p = 0. Then η ∈ Ω0(Rn) means η ∈ C(Rn), and ξ ∈ Ωnc (Rn) meansξ = gdx1 · · · dxn where g ∈ Cc(Rn). So

〈[η], [ξ]〉 =

∫Rnη(x1, . . . , xn)g(x1, . . . , xn) dx1 · · · dxn

Suppose 〈[η], [ξ]〉 = 0 for all ξ ∈ Ωnc (Rn). Then by the fact that the aboveintegral defines an inner product on Rn, it follows that g = 0, and so η = 0.Therefore the bilinear map 〈−,−〉:H0(M)×Hn

c (M)→ R is non-degenerate.Similarly, the product 〈−,−〉:Hn(Rn) × H0

c (Rn) → R is non-degenerate.If p 6= 0, n, then Hp(Rn) = 0 and Hn−p

c (Rn) = 0, so the bilinear map〈−,−〉:Hp(Rn)×Hn−p

c (Rn)→ R has to be non-degenerate.Now, pick a finite good cover U1, . . . , Um of M . Then by the above, the

pairings〈−,−〉:Hp(Ui)×Hn−p

c (Ui)→ R

and〈−,−〉:Hp(Ui ∩ Uj)×Hn−p

c (Ui ∩ Uj)→ R

are non-degenerate, meaning Hp(Ui) ≡ Hn−pc (Ui)

∗ and Hp(Ui ∩ Uj) ≡Hn−pc (Ui ∩ Uj)∗.

By lemma 9.26, we have a commutative diagram

Hp−1(Ui)⊕Hp−1(Uj) → Hp−1(Ui ∩ Uj) → Hp(UicupUj) → Hp(Ui)⊕Hp(Uj) → Hp(Ui ∩ Uj)↓ ↓ ↓ ↓

Hp−1c (Ui)

∗ ⊕Hp−1c (Uj)

∗ → Hp−1c (Ui ∩ Uj)∗ → Hp

c (UicupUj)∗ → Hp

c (Ui)∗ ⊕Hp

c (Uj)∗ → Hp

c (Ui ∩ Uj)∗

where the left two and right two vertical maps are isomorphisms. By the fivelemma, the maps Hp(Ui ∪ Uj)→ Hn−p

c (Ui ∪ Uj)∗ are all isomorphisms.Repeating this process, we see that for every finite union X = Ur1∪· · ·∪Urs ,

the map Hp(X)→ Hpc (X)∗ is an isomorphism. In particular, as M = U1∪· · ·∪

Um, we have that the map Hp(M) → Hn−pc (M)∗ is an isomorphism, ie: that

the bilinear map〈−,−〉:Hp(M)×Hn−p

c (M)→ R

is non-degenerate. 2

Corollary 9.30 Let M be a compact orientable manifold. Then

Hp(M) ∼= Mn−p(M)∗.

2

In particular, it follows that Hn(M) = R when M is an n-dimensionalcompact orientable manifold. This, together with the fact that Hp(M) = 0if p > n tells us that the dimension of a compact orientable manifold is ahomotopy-invariant.

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