III Differential Geometry (Full)

Part III — Differential Geometry

Based on lectures by J. A. Ross

Notes taken by Dexter Chua

Michaelmas 2016

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

This course is intended as an introduction to modern differential geometry. It can be

taken with a view to further studies in Geometry and Topology and should also be

suitable as a supplementary course if your main interests are, for instance in Analysis

or Mathematical Physics. A tentative syllabus is as follows.

•

Local Analysis and Differential Manifolds. Definition and examples of manifolds,

smo oth maps. Tangent vectors and vector fields, tangent bundle. Geometric

consequences of the implicit function theorem, submanifolds. Lie Groups.

•

Vector Bundles. Structure group. The example of Hopf bundle. Bundle mor-

phisms and automorphisms. Exterior algebra of differential forms. Tensors.

Symplectic forms. Orientability of manifolds. Partitions of unity and integration

on manifolds, Stokes Theorem; de Rham cohomology. Lie derivative of tensors.

Connections on vector bundles and covariant derivatives: covariant exterior

derivative, curvature. Bianchi identity.

•

Riemannian Geometry. Connections on the tangent bundle, torsion. Bianchi’s

identities for torsion free connections. Riemannian metrics, Levi-Civita con-

nection, Christoffel symbols, geodesics. Riemannian curvature tensor and its

symmetries, second Bianchi identity, sectional curvatures.

Pre-requisites

An essential pre-requisite is a working knowledge of linear algebra (including bilinear

forms) and multivariate calculus (e.g. differentiation and Taylor’s theorem in several

variables). Exposure to some of the ideas of classical differential geometry might also

be useful.

Contents

0 Introduction

1 Manifolds

1.1 Manifolds

1.2 Smooth functions and derivatives

1.3 Bump functions and partitions of unity

1.4 Submanifolds

2 Vector fields

2.1 The tangent bundle

2.2 Flows

2.3 Lie derivative

3 Lie groups

4 Vector bundles

4.1 Tensors

4.2 Vector bundles

5 Differential forms and de Rham cohomology

5.1 Differential forms

5.2 De Rham cohomology

5.3 Homological algebra and Mayer-Vietoris theorem

6 Integration

6.1 Orientation

6.2 Integration

6.3 Stokes Theorem

7 De Rham’s theorem*

8 Connections

8.1 Basic properties of connections

8.2 Geodesics and parallel transport

8.3 Riemannian connections

8.4 Curvature

0 Introduction

In differential geometry, the main object of study is a manifold. The motivation

is as follows — from IA, we know well how to do calculus on

. We can talk

about continuity, differentiable functions, derivatives etc. happily ever after.

However, sometimes, we want to do calculus on things other than

. Say,

we live on a sphere, namely the Earth. Does it make sense to “do calculus” on a

sphere? Surely it does.

The key insight is that these notions of differentiability, derivatives etc. are

local properties. To know if a function is differentiable at a point

, we only need

to know how the function behaves near

, and similarly such local information

tells us how to compute derivatives. The reason we can do calculus on a sphere

is because the sphere looks locally like

. Therefore, we can make sense of

calculus on a sphere.

Thus, what we want to do is to study calculus on things that look locally like

, and these are known as manifolds. Most of the time, our definitions from

usual calculus on

transfer directly to manifolds. However, sometimes the

global properties of our manifold will give us some new exciting things.

In fact, we’ve already seen such things when we did IA Vector Calculus. If

we have a vector field

→ R

whose curl vanishes everywhere, then we know

it is the gradient of some function. However, if we consider such a vector field

\ {

}

instead, then this is no longer true! Here the global topology of the

space gives rise to interesting phenomena we do not see at a local level.

When doing differential geometry, it is important to keep in mind that

what we’ve learnt in vector calculus is actually a mess.

has a lot of special

properties. Apart from being a topological space, it is also canonically a vector

space, and in fact an inner product space. When we did vector calculus, these

extra structure allowed us conflate many different concepts together. However,

when we pass on to manifolds, we no longer have these identifications, and we

have to be more careful.

1 Manifolds

1.1 Manifolds

As mentioned in the introduction, manifolds are spaces that look locally like

This local identification with R

is done via a chart.

Many sources start off with a topological space and then add extra structure

to it, but we will be different and start with a bare set.

Definition

(Chart)

A chart (

U, ϕ

) on a set

is a bijection

U → ϕ

(

)

⊆ R

where U ⊆ M and ϕ(U) is open.

A chart (U, ϕ) is centered at p for p ∈ U if ϕ(p) = 0.

Note that we do not require

to be open in

, or

to be a homeomorphism,

because these concepts do not make sense!

is just a set, not a topological

space.

ϕ(p)

With a chart, we can talk about things like continuity, differentiability by

identifying U with ϕ(U):

Definition

(Smooth function)

Let (

U, ϕ

) be a chart on

and

M → R

We say

is smooth or

∞

p ∈ U

f ◦ ϕ

−1

(

)

→ R

is smooth at

(

) in

the usual sense.

⊇ ϕ(U) U R

−1

ϕ(p)

f ◦ ϕ

−1

We can define all other notions such as continuity, differentiability, twice differ-

entiability etc. similarly.

This definition has a problem that some points might not be in the chart, and

we don’t know how to determine if a function is, say, smooth at the point. The

solution is easy — we just take many charts that together cover

. However,

we have the problem that a function might be smooth at a point relative to some

chart, but not relative to some other chart. The solution is to require the charts

to be compatible in some sense.

Definition

(Atlas)

An atlas on a set

is a collection of charts

{

(

, ϕ

)

}

M such that

(i) M =

(ii)

For all

α, β

, we have

(

∩U

) is open in

, and the transition function

◦ ϕ

−1

: ϕ

∩ U

) → ϕ

∩ U

)

is smooth (in the usual sense).

−1

Lemma.

If (

, ϕ

) and (

, ϕ

) are charts in some atlas, and

M → R

then

f ◦ ϕ

−1

is smooth at

(

) if and only if

f ◦ ϕ

−1

is smooth at

(

) for all

p ∈ U

∩ U

Proof. We have

f ◦ ϕ

−1

= f ◦ ϕ

−1

◦ (ϕ

◦ ϕ

−1

So we know that if we have an atlas on a set, then the notion of smoothness

does not depend on the chart.

Example. Consider the sphere

= {(x

, x

) :

= 1} ⊆ R

We let

= S

∩ {x

> 0}, U

−

= S

∩ {x

< 0}, · · ·

We then let

: U

→ R

, x

) 7→ (x

, x

It is easy to show that this gives a bijection to the open disk in

. We similarly

define the other ϕ

. These then give us an atlas of S

Definition

(Equivalent atlases)

Two atlases

and

are equivalent if

∪A

is an atlas.

Then equivalent atlases determine the same smoothness, continuity etc.

information.

Definition

(Differentiable structure)

A differentiable structure on

is a choice

of equivalence class of atlases.

We want to define a manifold to be a set with a differentiable structure. How-

ever, it turns out we can find some really horrendous sets that have differential

structures.

Example.

Consider the line with two origins given by taking

R × {

} ∪ R × {

}

and then quotienting by

(x, 0) ∼ (x, 1) for x 6= 0.

Then the inclusions of the two copies of R gives us an atlas of the space.

The problem with this space is that it is not Hausdorff, which is bad. However,

that is not actually true, because

is not a topological space, so it doesn’t

make sense to ask if it is Hausdorff. So we want to define a topology on

, and

then impose some topological conditions on our manifolds.

It turns out the smooth structure already gives us a topology:

Exercise.

An atlas determines a topology on

by saying

V ⊆ M

is open iff

(

U ∩ V

) is open in

for all charts (

U, ϕ

) in the atlas. Equivalent atlases give

the same topology.

We now get to the definition of a manifold.

Definition

(Manifold)

A manifold is a set

with a choice of differentiable

structure whose topology is

(i)

Hausdorff, i.e. for all

x, y ∈ M

, there are open neighbourhoods

, U

⊆ M

with x ∈ U

, y ∈ U

and U

∩ U

= ∅.

(ii)

Second countable, i.e. there exists a countable collection (

)

n∈N

of open

sets in

such that for all

V ⊆ M

open, and

p ∈ V

, there is some

such

that p ∈ U

⊆ V .

The second countability condition is a rather technical condition that we

wouldn’t really use much. This, for example, excludes the long line.

Note that we will often refer to a manifold simply as

, where the differen-

tiable structure is understood from context. By a chart on

, we mean one in

some atlas in the equivalence class of atlases.

Definition

(Local coordinates)

Let

be a manifold, and

U → ϕ

(

) a

chart of M. We can write

ϕ = (x

, · · · , x

)

where each x

: U → R. We call these the local coordinates.

So a point p ∈ U can be represented by local coordinates

(p), · · · , x

(p)) ∈ R

By abuse of notation, if

M → R

, we confuse

and

f ◦ ϕ

−1

(

)

→ R

So we write f(x

, · · · , x

) to mean f(p), where ϕ(p) = (x

, · · · , x

) ∈ ϕ(U).

U M R

ϕ(U)

Of course, we can similarly define

, C

, · · ·

manifolds, or analytic manifolds.

We can also model manifolds on other spaces, e.g.

, where we get complex

manifolds, or on infinite-dimensional spaces.

Example.

(i)

Generalizing the example of the sphere, the

-dimensional sphere

{(x

, · · · , x

) ∈ R

n+1

= 1} is a manifold.

(ii)

is open in

, then the inclusion map

M → R

given by

(

) =

is a chart forming an atlas. So

is a manifold. In particular,

a manifold, with its “standard” differentiable structure. We will always

assume R

is given this structure, unless otherwise specified.

(iii) M

(

n, n

), the set of all

n × n

matrices is also a manifold, by the usual

bijection with

. Then

⊆ M

(

n, n

) is open, and thus also a manifold.

(iv)

The set

, the set of one-dimensional subspaces of

n+1

is a manifold.

We can define charts as follows: we let

to be the lines spanned by a

vector of the form (v

, v

, · · · , v

i−1

, 1, v

i+1

, · · · , v

) ∈ R

n+1

We define the map

→ R

∼

{x ∈ R

n+1

= 1

}

that sends

(

) = (

, · · · ,

, · · · , v

), where

is spanned by (

, · · · ,

, · · · , v

). It

is an easy exercise to show that this defines a chart.

Note that when we defined a chart, we talked about charts as maps

U → R

We did not mention whether

is fixed, or whether it is allowed to vary. It turns

out it cannot vary, as long as the space is connected.

Lemma.

Let

be a manifold, and

→ R

and

→ R

be charts.

If U

∩ U

6= ∅, then n = m.

Proof. We know

−1

: ϕ

∩ U

) → ϕ

∩ U

)

is a smooth map with inverse ϕ

−1

. So the derivative

D(ϕ

−1

)(ϕ

(p)) : R

→ R

is a linear isomorphism, whenever p ∈ U

∩ U

. So n = m.

Definition

(Dimension)

p ∈ M

, we say

has dimension

if for one

(thus all) charts

U → R

with

p ∈ U

, we have

. We say

has

dimension n if it has dimension n at all points.

1.2 Smooth functions and derivatives

From now on,

and

will be manifolds. As usual, we would like to talk about

maps between manifolds. What does it mean for such a map to be smooth? In

the case of a function

M → R

, we had to check it on each chart of

. Now that

we have functions M → N, we need to check it on charts of both N and M.

Definition

(Smooth function)

A function

M → N

is smooth at a point

p ∈ M

if there are charts (

U, ϕ

) for

and (

V, ξ

) for

with

p ∈ U

and

(

)

∈ V

such that ξ ◦ f ◦ ϕ

−1

: ϕ(U) → ξ(V ) is smooth at ϕ(p).

A function is smooth if it is smooth at all points p ∈ M .

A diffeomorphism is a smooth f with a smooth inverse.

We write

∞

(

M, N

) for the space of smooth maps

M → N

. We write

∞

(

) for

∞

(

M, R

), and this has the additional structure of an algebra, i.e.

a vector space with multiplication.

ξ ◦ f ◦ ϕ

−1

Equivalently,

is smooth at

ξ ◦ f ◦ ϕ

−1

is smooth at

(

) for any such charts

(U, ϕ) and (V, ξ).

Example.

Let

U → R

be a chart. Then

U → ϕ

(

) is a diffeomorphism.

Definition

(Curve)

A curve is a smooth map

I → M

, where

is a non-empty

open interval.

To discuss derivatives, we first look at the case where

U ⊆ R

is open.

Suppose

U → R

is smooth. If

p ∈ U

and

v ∈ R

, recall that the directional

derivative is defined by

Df|

(v) = lim

t→0

f(p + tv) − f (p)

If v = e

= (0, · · · , 0, 1, 0, · · · , 0), then we write

Df|

) =

∂f

∂x



Also, we know Df|

: R

→ R is a linear map (by definition of smooth).

Note that here

and

are both vectors, but they play different roles —

is an element in the domain

, while

is an arbitrary vector in

. Even if

is enormous, by taking a small enough

, we find that

will eventually be

inside U .

If we have a general manifold, we can still talk about the

. However, we

don’t have anything that plays the role of a vector. Our first goal is to define

the tangent space to a manifold that captures where the “directions” live.

An obvious way to do so would be to use a curve. Suppose

I → M

is a

curve, with

(0) =

p ∈ U ⊆ M

, and

U → R

is smooth. We can then take the

derivative of f along γ as before. We let

X(f) =



t=0

f(γ(t)).

It is an exercise to see that

∞

(

)

→ R

is a linear map, and it satisfies the

Leibniz rule

X(fg) = f(p)X(g) + g(p)X(f).

We denote

˙γ

(0). We might think of defining the tangent space as curves

up to some equivalence relation, but if we do this, there is no obvious vector

space on it. The trick is to instead define a vector by the derivative

induces.

This then has an obvious vector space structure.

Definition

(Derivation)

A derivation on an open subset

U ⊆ M

p ∈ U

is a

linear map X : C

∞

(U) → R satisfying the Leibniz rule

X(fg) = f(p)X(g) + g(p)X(f).

Definition

(Tangent space)

Let

p ∈ U ⊆ M

, where

is open. The tangent

space of M at p is the vector space

M = { derivations on U at p } ≡ Der

∞

(U)).

The subscript p tells us the point at which we are taking the tangent space.

Why is this the “right” definition? There are two things we would want to

be true:

(i) The definition doesn’t actually depend on U.

(ii) This definition agrees with the usual definition of tangent vectors in R

We will do the first part at the end by bump functions, and will do the second

part now. Note that it follows from the second part that every tangent vector

comes from the derivative of a path, because this is certainly true for the usual

definition of tangent vectors in

(take a straight line), and this is a completely

local problem.

Example. Let U ⊆ R

be open, and let p ∈ U . Then we have tangent vectors

∂

∂x



∈ T

, i = 1, . . . , n.

These correspond to the canonical basis vectors in R

Lemma.

∂

∂x



, · · · ,

∂

∂x



is a basis of T

. So these are all the derivations.

The idea of the proof is to show that a derivation can only depend on the

first order derivatives of a function, and all possibilities will be covered by the

∂

∂x

Proof. Independence is clear as

∂x

= δ

We need to show spanning. For notational convenience, we wlog take

= 0. Let

X ∈ T

We first show that if

g ∈ C

∞

(

) is the constant function

= 1, then

X(g) = 0. Indeed, we have

X(g) = X(g

) = g(0)X(g) + X(g)g(0) = 2X(g).

Thus, if

is any constant function, say,

, then

(

) =

(

) =

(

). So the

derivative of any constant function vanishes.

In general, let f ∈ C

∞

(U). By Taylor’s theorem, we have

f(x

, · · · , x

) = f(0) +

i=1

∂f

∂x



+ ε,

where ε is a sum of terms of the form x

h with h ∈ C

∞

(U).

We set λ

= X(x

) ∈ R. We first claim that X(ε) = 0. Indeed, we have

X(x

h) = x

(0)X(x

h) + (x

h)(0)X(x

) = 0.

So we have

X(f) =

i=1

∂f

∂x



So we have

X =

i=1

∂

∂x



Given this definition of a tangent vector, we have a rather silly and tautological

definition of the derivative of a smooth function.

Definition

(Derivative)

Suppose

F ∈ C

∞

(

M, N

), say

(

) =

. We define

DF |

: T

M → T

N by

DF |

(X)(g) = X(g ◦ F )

for X ∈ T

M and g ∈ C

∞

(V ) with q ∈ V ⊆ N .

This is a linear map called the derivative of F at p.

M N

g◦F

With a silly definition of a derivative comes a silly definition of the chain

rule.

Proposition

(Chain rule)

Let

M, N, P

be manifolds, and

F ∈ C

∞

(

M, N

G ∈ C

∞

(N, P ), and p ∈ M, q = F (p). Then we have

D(G ◦ F )|

= DG|

◦ DF |

Proof. Let h ∈ C

∞

(P ) and X ∈ T

M. We have

DG|

(DF |

(X))(h) = DF |

(X)(h ◦ G) = X(h ◦ G ◦ F ) = D(G ◦ F )|

(X)(h).

Note that this does not provide a new, easy proof of the chain rule. Indeed,

to come this far into the course, we have used the actual chain rule something

like ten thousand times.

Corollary.

is a diffeomorphism, then D

F |

is a linear isomorphism, and

(DF |

)

−1

= D(F

−1

F (p)

In the special case where the domain is

, there is a canonical choice of

tangent vector at each point, namely 1.

Definition

(Derivative)

Let

R → M

be a smooth function. Then we write

dγ

(t) = ˙γ(t) = Dγ|

(1).

We now go back to understanding what

is if

p ∈ M

. We let

p ∈ U

where (

U, ϕ

) is a chart. Then if

(

), the map D

ϕ|

M → T

is a

linear isomorphism.

Definition (

∂

∂x

). Given a chart ϕ : U → R

with ϕ = (x

, · · · , x

), we define

∂

∂x



= (Dϕ|

)

−1

∂

∂x



ϕ(p)

∈ T

∂

∂x



, · · · ,

∂

∂x



is a basis for T

Recall that if

U → R

is smooth, then we can write

(

, · · · , x

). Then

we have

∂

∂x



(f) =

∂f

∂x



ϕ(p)

So we have a consistent notation.

Now, how does this basis change when we change coordinates? Suppose we

also have coordinates

, · · · , y

near

given by some other chart. We then have

∂

∂y



∈ T

M. So we have

∂

∂y



j=1

∂

∂x



for some

. To figure out what they are, we apply them to the function

. So

we have

∂

∂y



) =

∂x

∂y

(p) = α

So we obtain

∂

∂y



j=1

∂x

∂y

(p)

∂

∂x



This is the usual change-of-coordinate formula!

Now let

F ∈ C

∞

(

M, N

), (

U, ϕ

) be a chart on

containing

with coordinates

, · · · , x

, and (

V, ξ

) a chart on

containing

(

) with coordinates

, · · · , y

. By abuse of notation, we confuse

and

ξ ◦ F ◦ ϕ

−1

. So we write

F = (F

, · · · , F

) with F

= F

, · · · , x

) : U → R.

As before, we have a basis

∂

∂x



, · · · ,

∂

∂x



for T

∂

∂y



, · · · ,

∂

∂y



for T

Lemma. We have

DF |

∂

∂x



j=1

∂F

∂x

(p)

∂

∂y



In other words, DF |

has matrix representation



∂F

∂x

(p)



Proof. We let

DF |

∂

∂x



j=1

∂

∂y



for some λ

. We apply this to the local function y

to obtain





j=1

∂

∂y







)

= DF

∂

∂x



)

∂

∂x



◦ F )

∂

∂x



)

∂F

∂x

(p).

Example.

Let

∞

(

) where

U ⊆ M

is an open set containing

. Then

M → T

f(p)

∼

is a linear map. So D

is an element in the dual

space (

)

∗

, called the differential of

, and is denoted d

. Then we

have

df|

(X) = X(f).

(this can, e.g. be checked in local coordinates)

1.3 Bump functions and partitions of unity

Recall that there is one thing we swept under the carpet — to define the tangent

space, we needed to pick an open set

. Ways to deal with this can be found in

the example sheet, but there are two general approaches — one is to talk about

germs of functions, where we consider all open neighbourhoods, and identify two

functions if they agree on some open neighbourhood of the point. The other way

is to realize that we can “extend” any function on

U ⊆ M

to a function on the

whole of M, using bump functions.

In general, we want a function that looks like this:

Lemma.

Suppose

W ⊆ M

is a coordinate chart with

p ∈ W

. Then there is an

open neighbourhood

such that

V ⊆ W

and an

X ∈ C

∞

(

M, R

) such that

X = 1 on V and X = 0 on M \ W .

Proof.

Suppose we have coordinates

, · · · , x

. We wlog suppose these

are defined for all |x| < 3.

We define α, β, γ : R → R by

α(t) =

(

−t

−2

t > 0

0 t ≤ 0

We now let

β(t) =

α(t)

α(t) + α(1 − t)

Then we let

γ(t) = β(t + 2)β(2 − t).

Finally, we let

X(x

, · · · , x

) = γ(x

) · · · γ(x

on W . We let

V = {x : |x

| < 1}.

Extending

to be identically 0 on

M \ W

to get the desired smooth function

(up to some constant).

Lemma.

Let

p ∈ W ⊆ U

and

W, U

open. Let

, f

∈ C

∞

(

) be such that

= f

on W . If X ∈ Der

∞

(U)), then we have X(f

) = X(f

)

Proof.

Set

− f

. We can wlog assume that

is a coordinate chart. We

pick a bump function

χ ∈ C

∞

(

) that vanishes outside

. Then

χh

= 0. Then

we have

0 = X(χh) = χ(p)X(h) + h(p)X(χ) = X(h) + 0 = X(f

) − X(f

While we’re doing boring technical work, we might as well do the other one,

known as a partition of unity. The idea is as follows — suppose we want to

construct a global structure on our manifold, say a (smoothly varying) inner

product for each tangent space

. We know how to do this if

because there is a canonical choice of inner product at each point in

. We

somehow want to patch all of these together.

In general, there are two ways we can do the patching. The easy case is that

not only is there a choice on

, but there is a unique choice. In this case, just

doing it on each chart suffices, because they must agree on the intersection by

uniqueness.

However, this is obviously not the case for us, because a vector space can

have many distinct inner products. So we need some way to add them up.

Definition

(Partition of unity)

Let

}

be an open cover of a manifold

. A

partition of unity subordinate to

}

is a collection

∈ C

∞

(

M, R

) such that

(i) 0 ≤ ϕ

≤ 1

(ii) supp(ϕ

) ⊆ U

(iii) For all p ∈ M , all but finitely many ϕ

(p) are zero.

(iv)

= 1.

Note that by (iii), the final sum is actually a finite sum, so we don’t have to

worry about convergence issues.

Now if we have such a partition of unity, we can pick an inner product on

each

, say

(

· , ·

), and then we can define an inner product on the whole

space by

q(v

, w

) =

(p)q

, w

where

, w

∈ T

are tangent vectors. Note that this makes sense. While

each

is not defined everywhere, we know

(

) is non-zero only when

defined at p, and we are also only taking a finite sum.

The important result is the following:

Theorem.

Given any

}

open cover, there exists a partition of unity subor-

dinate to {U

Proof.

We will only do the case where

is compact. Given

p ∈ M

, there exists

a coordinate chart

p ∈ V

and

(

) such that

⊆ U

α(p)

. We pick a bump

function

∈ C

∞

(

M, R

) such that

= 1 on a neighbourhood

⊆ V

Then supp(χ

) ⊆ U

α(p)

Now by compactness, there are some

, · · · , p

such that

is covered by

∪ · · · ∪ W

. Now let

˜ϕ

i:α(p

)=α

Then by construction, we have

supp( ˜ϕ

) ⊆ U

Also, by construction, we know

˜ϕ

> 0. Finally, we let

˜ϕ

The general proof will need the fact that the space is second-countable.

We will actually not need this until quite later on in the course, but we might

as well do all the boring technical bits all together.

1.4 Submanifolds

You have a manifold, and a subset of it is a manifold, so you call it a submanifold.

Definition

(Embedded submanifold)

Let

be a manifold with

dim M

and

be a submanifold of

. We say

is an embedded submanifold if for all

p ∈ S

, there are coordinates

, · · · , x

on some chart

U ⊆ M

containing

such

that

S ∩ U = {x

k+1

= x

k+2

= · · · = x

= 0}

for some k. Such coordinates are known as slice coordinates for S.

This is a rather technical condition, rather than “a subset that is also a

manifold under the inherited smooth structure”. The two definitions are indeed

equivalent, but picking this formulation makes it easier to prove things about it.

Lemma.

is an embedded submanifold of

, then there exists a unique

differential structure on

such that the inclusion map

S → M

is smooth and

S inherits the subspace topology.

Proof.

Basically if

, · · · , x

is a slice chart for

, then

, · · · , x

will be

coordinates on S.

More precisely, let π : R

→ R

be the projection map

π(x

, · · · , x

) = (x

, · · · , x

Given a slice chart (

U, ϕ

) for

, consider

˜ϕ

S ∩U → R

˜ϕ

π ◦ϕ

. This

is smooth and bijective, and is so a chart on

. These cover

by assumption.

So we only have to check that the transition functions are smooth.

Given another slice chart (

V, ξ

) for

, we let

π ◦ ξ

, and check that

ξ ◦ ˜ϕ

−1

= π ◦ ξ ◦ ϕ

−1

◦ j,

where j : R

→ R

is given by j(x

, · · · , x

) = (x

, · · · , x

, 0, · · · , 0).

From this characterization, by looking at local charts, it is clear that

has

the subspace topology. It is then easy to see that the embedded submanifold is

Hausdorff and second-countable, since these properties are preserved by taking

subspaces.

We can also check easily that

S → M

is smooth, and this is the only

differential structure with this property.

It is also obvious from the slice charts that:

Proposition.

Let

be an embedded submanifold. Then the derivative of the

inclusion map ι : S → M is injective.

Sometimes, we like to think of a subobject not as a subset, but as the inclusion

map

S → M

instead. However, when we are doing topology, there is this

funny problem that a continuous bijection need not be a homeomorphism. So if

we define submanifolds via inclusions maps, we get a weaker notion known as an

immersed submanifold.

Definition

(Immersed submanifold)

Let

S, M

be manifolds, and

S → M

be a smooth injective map with D

ι|

S → T

injective for all

p ∈ S

. Then

we call (

ι, S

) an immersed submanifold. By abuse of notation, we identify

and

ι(S).

Example.

If we map

into

via the following figure of eight (where the arrow

heads denote the “end points” of

), then this gives an immersed submanifold

that is not an embedded submanifold.

Example.

Consider the line

, and define the map

R → T

(

) =

αx

, where

is some irrational number. Then this map gives an immersed

submanifold of

, but is not an embedded submanifold, since

certainly does

not have the subspace topology from T

How do we construct submanifolds? The definition is rather difficult to work

with. It is not immediately clear whether

= {x ∈ R

n+1

: |x| ≤ 1} ⊆ R

n+1

is an embedded submanifold, even though it feels like it should be.

More generally, if

M, N

are manifolds,

F ∈ C

∞

(

M, N

) and

c ∈ N

, under

what circumstances will

−1

(

) be an embedded submanifold of

? The answer

is that c has to be a regular value.

Definition

(Regular value)

Let

F ∈ C

∞

(

M, N

) and

c ∈ N

. Let

−1

(

We say

is a regular value if for all

p ∈ S

, the map D

F |

M → T

surjective.

Proposition.

Let

F ∈ C

∞

(

M, N

), and let

c ∈ N

. Suppose

is a regular value.

Then S = F

−1

Proof.

We let

dim M

and

dim N

. Notice that for the map D

to be

surjective, we must have n ≥ m.

Let

p ∈ S

, so

(

) =

. We want to find a slice coordinate for

near

Since the problem is local, by restricting to local coordinate charts, we may wlog

assume N = R

, M = R

and c = p = 0.

Thus, we have a smooth map

→ R

with surjective derivative at 0.

Then the derivative is



∂F

∂x





i=1,...,n; j=1,...,m

which by assumption has rank

. We reorder the

so that the first

columns

are independent. Then the m × m matrix

R =



∂F

∂x





i,j=1,...,m

is non-singular. We consider the map

α(x

, · · · , x

) = (F

, · · · , F

, x

m+1

, · · · , x

We then obtain

Dα|



R ∗

0 I



and this is non-singular. By the inverse function theorem,

is a local diffeomor-

phism. So there is an open

W ⊆ R

containing 0 such that

α|

W → α

(

) is

smooth with smooth inverse. We claim that α is a slice chart of S in R

Since it is a smooth diffeomorphism, it is certainly a chart. Moreover, by

construction, the points in

are exactly those whose image under

have the

first m coordinates vanish. So this is the desired slice chart.

Example.

We want to show that

is a manifold. Let

n+1

→ R

defined by

F (x

, · · · , x

) =

Then F

−1

(1) = S

. We find that

DF |

= 2(x

, · · · , x

) 6= 0

when p ∈ S

. So S

is a manifold.

Example.

Consider the orthogonal group. We let

∼

be the space of

all n × n matrices with the obvious smooth structure. We define

N = {A ∈ M

: A

= A}.

Since this is a linear subspace, it is also a manifold. We define

F : M

→ N

A 7→ AA

Then we have

O(n) = F

−1

(I) = {A : AA

= I}.

We compute the derivative by looking at

F (A + H) = (A + H)(A + H)

= AA

+ HA

+ AH

+ HH

So we have

DF |

(H) = HA

+ AH

Now if A ∈ O(n), then we have

DF |

(HA) = HAA

+ AA

= H + H

for any

. Since every symmetric matrix is of the form

, we know

DF |

: T

→ T

F (A)

N is surjective. So O(n) is a manifold.

2 Vector fields

2.1 The tangent bundle

Recall that we had the notion of a tangent vector. If we have a curve

I → M

then we would like to think that the derivative

˙γ

“varies smoothly” with time.

However, we cannot really do that yet, since for different

, the value of

˙γ

lies in

different vector spaces, and there isn’t a way of comparing them.

More generally, given a “vector field”

p 7→ v

∈ T

for each

p ∈ M

, how

do we ask if this is a smooth function?

One way to solve this is to pick local coordinates

, · · · , x

U ⊆ M

. We

can then write

(p)

∂

∂x



Since

(

)

∈ R

, we can say

varies smoothly if the functions

(

) are smooth.

We then proceed to check that this does not depend on coordinates etc.

However, there is a more direct approach. We simply turn

T M =

[

p∈M

into a manifold. There is then a natural map

T M → M

sending

∈ T

for each

p ∈ M

, and this is smooth. We can then define the smoothness of

f using the usual notion of smoothness of maps between manifolds.

Assuming that we have successfully constructed a sensible

T M

, we can define:

Definition

(Vector field)

A vector field on some

U ⊆ M

is a smooth map

X : U → T M such that for all p ∈ U, we have

X(p) ∈ T

In other words, we have π ◦ X = id.

Definition

(

Vect

(

))

Let

Vect

(

) denote the set of all vector fields on

. Let

X, Y ∈ Vect(U ), and f ∈ C

∞

(U). Then we can define

(X + Y )(p) = X(p) + Y (p), (fX)(p) = f(p)X(p).

Then we have X + Y, fX ∈ Vect(U). So Vect(U) is a C

∞

(U) module.

Moreover, if V ⊆ U ⊆ M and X ∈ Vect(U), then X|

∈ Vect(V ).

Conversely, if

}

is a cover of

, and

∈ Vect

(

) are such that they

agree on intersections, then they patch together to give an element of

Vect

(

So we say that Vect is a sheaf of C

∞

(M) modules.

Now we properly define the manifold structure on the tangent bundle.

Definition (Tangent bundle). Let M be a manifold, and

T M =

[

p∈M

There is a natural projection map π : T M → M sending v

∈ T

M to p.

Let

, · · · , x

be coordinates on a chart (

U, ϕ

). Then for any

p ∈ U

and

∈ T

M, there are some α

, · · · , α

∈ R such that

i=1

∂

∂x



This gives a bijection

−1

(U) → ϕ(U) × R

7→ (x

(p), · · · , x

(p), α

, · · · , α

These charts make

T M

into a manifold of dimension 2

dim M

, called the tangent

bundle of M.

Lemma. The charts actually make T M into a manifold.

Proof. If (V, ξ) is another chart on M with coordinates y

, · · · , y

, then

∂

∂x



j=1

∂y

∂x

(p)

∂

∂y



So we have

ξ ◦ ˜ϕ

−1

: ϕ(U ∩ V ) × R

→ ξ(U ∩ V ) × R

given by

ξ ◦ ˜ϕ

−1

, · · · , x

, α

, · · · , α

) =

, · · · , y

i=1

∂y

∂x

, · · · ,

i=1

∂y

∂x

and is smooth (and in fact fiberwise linear).

It is easy to check that

T M

is Hausdorff and second countable as

is.

There are a few remarks to make about this.

(i) The projection map π : T M → M is smooth.

(ii) If U ⊆ M is open, recall that

Vect(U) = {smooth X : U → T M | X(p) ∈ T

M for all p ∈ U }.

We write

for

(

). Now suppose further that

is a coordinate chart,

then we can write any function

U → T M

such that

∈ T

(uniquely) as

i=1

(p)

∂

∂x



Then X is smooth iff all α

are smooth.

(iii)

F ∈ C

∞

(

M, N

), then D

T M → T N

given by D

(

) = D

F |

(

) is

smooth. This is nice, since we can easily talk about higher derivatives, by

taking the derivative of the derivative map.

(iv)

F ∈ C

∞

(

M, N

) and

is a vector field on

, then we cannot obtain a

vector field on

by D

(

), since

might not be injective. If

(

) =

F (p

), we need not have DF (X(p

)) = DF (X(p

)).

However, there is a weaker notion of being F -related.

Definition

(

-related)

Let

M, N

be manifolds, and

X ∈ Vect

(

Y ∈

Vect(N) and F ∈ C

∞

(M, N). We say they are F -related if

= DF |

)

for all

p ∈ M

and

(

) =

. In other words, if the following diagram commutes:

T M T N

M N

So what does

Vect

(

) look like? Recall that a vector is defined to be a

derivation. So perhaps a vector field is also a derivation of some sort.

Definition

(

Der

(

∞

(

)))

Let

Der

(

∞

(

)) be the set of all

-linear maps

X : C

∞

(M) → C

∞

(M) that satisfy

X (fg) = fX (g) + X (f )g.

This is an R-vector space, and in fact a C

∞

(M) module.

Given X ∈ Vect(M), we get a derivation X ∈ Der(C

∞

(M)) by setting

X (f)(p) = X

(f).

It is an exercise to show that

(

) is smooth and satisfies the Leibniz rule.

Similar to the case of vectors, we want to show that all derivations come from

vector fields.

Lemma. The map X 7→ X is an R-linear isomorphism

Γ : Vect(M ) → Der(C

∞

(M)).

Proof. Suppose that α is a derivation. If p ∈ M, we define

(f) = α(f)(p)

for all f ∈ C

∞

(M). This is certainly a linear map, and we have

(fg) = α(fg)(p) = (fα(g) + gα(f))(p) = f(p)X

(g) + g(p)X

(f).

∈ T

. We just need to check that the map

M → T M

sending

p 7→ X

is smooth. Locally on M, we have coordinates x

, · · · , x

, and we can write

i=1

(p)

∂

∂x



We want to show that α

: U → R are smooth.

We pick a bump function

that is identically 1 near

, with

supp ϕ ⊆ U

Consider the function ϕx

∈ C

∞

(M). We can then consider

α(ϕx

)(p) = X

(ϕx

ϕx

is just

near

, by properties of derivations, we know this is just equal

to α

. So we have

α(ϕx

) = α

So α

is smooth.

From now on, we confuse

and

, i.e. we think of any

X ∈ Vect

(

) as a

derivation of C

∞

(M).

Note that the product of two vector fields (i.e. the composition of derivations)

is not a vector field. We can compute

XY (fg) = X(Y (f g))

= X(fY (g) + gY (f))

= X(f)Y (g) + f XY (g) + X(g)Y (f) + gXY (f).

So this is not a derivation, because we have the cross terms

(

)

(

). However,

what we do have is that XY − Y X is a derivation.

Definition

(Lie bracket)

X, Y ∈ Vect

(

), then the Lie bracket [

X, Y

] is

(the vector field corresponding to) the derivation XY − Y X ∈ Vect(M).

So Vect(M ) becomes what is known as a Lie algebra.

Definition

(Lie algebra)

A Lie algebra is a vector space

with a bracket

[ · , · ] : V × V → V such that

(i) [ · , · ] is bilinear.

(ii) [ · , · ] is antisymmetric, i.e. [X, Y ] = −[Y, X].

(iii) The Jacobi identity holds

[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y ]] = 0.

It is a (painful) exercise to show that the Lie bracket does satisfy the Jacobi

identity.

The definition of the Lie algebra might seem a bit weird. Later it will come

up in many different guises and hopefully it might become more clear.

2.2 Flows

What can we do with vector fields? In physics, we can imagine a manifold as all

of space, and perhaps a vector field specifies the velocity a particle should have

at that point. Now if you actually drop a particle into that space, the particle

will move according to the velocity specified. So the vector field generates a flow

of the particle. These trajectories are known as integral curves.

Definition

(Integral curve)

Let

X ∈ Vect

(

). An integral curve of

is a

smooth γ : I → M such that I is an open interval in R and

˙γ(t) = X

γ(t)

Example. Take M = R

, and let

X = x

∂

∂y

− y

∂

∂x

The field looks like this:

We would expect the integral curves to be circles. Indeed, suppose

I → R

an integral curve. Write γ = (γ

, γ

). Then the definition requires

(t)

∂

∂x

+ γ

(t)

∂

∂y

= γ

(t)

∂

∂y

− γ

(t)

∂

∂x

So the equation is

(t) = −γ

(t)

(t) = γ

(t).

For example, if our starting point is p = (1, 0), then we have

(t) = cos t, γ

(t) = sin t.

We see that to find an integral curve, all we are doing is just solving ordinary

differential equations. We know that all ODEs have smooth and unique solutions,

and they have all the nice properties we can hope for. So we are going to get nice

corresponding results for integral solutions. However, sometimes funny things

happen.

Example. Take M = R, and

X = x

Then if γ is an integral curve, it must satisfy:

(t) = γ(t)

This means that the solution is of the form

γ(t) =

C − t

for C a constant. For example, if we want γ(0) =

, then we have

γ(t) =

2 − t

The solution to this ODE is defined only for

t <

2, so we can only have

I = (−∞, 2) at best.

We are going to prove that integral curves always exist. To do so, we need

to borrow some powerful theorems from ODE theory:

Theorem

(Fundamental theorem on ODEs)

Let

U ⊆ R

be open and

U →

smooth. Pick t

∈ R.

Consider the ODE

˙γ

(t) = α

(γ(t))

) = c

where c = (c

, · · · , c

) ∈ R

Then there exists an open interval

containing

and an open

⊆ U

such

that for every

c ∈ U

, there is a smooth solution

I → U

satisfying the ODE.

Moreover, any two solutions agree on a common domain, and the function

Θ : I × U

→ U defined by Θ(t, c) = γ

(t) is smooth (in both variables).

Theorem

(Existence of integral curves)

Let

X ∈ Vect

(

) and

p ∈ M

. Then

there exists some open interval

I ⊆ R

with 0

∈ I

and an integral curve

I → M

for X with γ(0) = p.

Moreover, if

˜γ

I → M

is another integral curve for

, and

˜γ

(0) =

, then

˜γ = γ on I ∩

Proof.

Pick local coordinates for

centered at

in an open neighbourhood

So locally we write

X =

i=1

∂

∂x

where α

∈ C

∞

(U). We want to find γ = (γ

, · · · , γ

) : I → U such that

i=1

(t)

∂

∂x



γ(t)

i=1

(γ(t))

∂

∂x



γ(t)

, γ

(0) = 0.

Since the

∂

∂x

form a basis, this is equivalent to saying

(t) = α

(γ(t)), γ

(0) = 0

for all i and t ∈ I.

By the general theory of ordinary differential equations, there is an interval

I and a solution γ, and any two solutions agree on their common domain.

However, we need to do a bit more for uniqueness, since all we know is that

there is a unique integral curve lying in this particular chart. It might be that

there are integral curves that do wild things when they leave the chart.

So suppose

I → M

and

˜γ

I → M

are both integral curves passing

through the same point, i.e. γ(0) = ˜γ(0) = p.

We let

J = {t ∈ I ∩

I : γ(t) = ˜γ(t)}.

This is non-empty since 0

∈ J

, and

is closed since

and

˜γ

are continuous. To

show it is all of I ∩

I, we only have to show it is open, since I ∩

I is connected.

So let

∈ J

, and consider

(

). Then

and

˜γ

are integral curves of

passing through

. So by the first part, they agree on some neighbourhood of

So J is open. So done.

Definition

(Maximal integral curve)

Let

p ∈ M

, and

X ∈ Vect

(

). Let

the union of all I such that there is an integral curve γ : I → M with γ(0) = p.

Then there exists a unique integral curve

→ M

, known as the maximal

integral curve.

Note that I

does depend on the point.

Example. Consider the vector field

X =

∂

∂x

\ {

}

. Then for any point

= (

x, y

), if

y 6

= 0, we have

, but if

= 0 and

x <

0, then

= (

−∞, −x

). Similarly, if

= 0 and

x >

0, then

= (−x, ∞).

Definition

(Complete vector field)

A vector field is complete if

for all

p ∈ M .

Given a complete vector field, we obtain a flow map as follows:

Theorem.

Let

be a manifold and

a complete vector field on

. Define

: R × M → M by

(p) = γ

(t),

where

is the maximal integral curve of

through

with

(0) =

. Then Θ

is a function smooth in p and t, and

= id, Θ

◦ Θ

= Θ

s+t

Proof.

This follows from uniqueness of integral curves and smooth dependence

on initial conditions of ODEs.

In particular, since Θ

◦ Θ

−t

= Θ

= id, we know

−1

= Θ

−t

So Θ

is a diffeomorphism.

More algebraically, if we write

Diff

(

) for the diffeomorphisms

M → M

then

R → Diff(M )

t 7→ Θ

is a homomorphism of groups. We call this a one-parameter subgroup of diffeo-

morphisms.

What happens when we relax the completeness assumption? Everything is

essentially the same whenever things are defined, but we have to take care of

the domains of definition.

Theorem. Let M be a manifold, and X ∈ Vect(M). Define

D = {(t, p) ∈ R × M : t ∈ I

In other words, this is the set of all (t, p) such that γ

(t) exists. We set

(p) = Θ(t, p) = γ

(t)

for all (t, p) ∈ D. Then

(i) D is open and Θ : D → M is smooth

(ii) Θ(0, p) = p for all p ∈ M .

(iii)

If (

t, p

)

∈ D

and (

Θ(

s, p

))

∈ D

, then (

t, p

)

∈ D

and Θ(

Θ(

s, p

)) =

Θ(t + s, p).

(iv) For any t ∈ R , the set M

: {p ∈ M : (t, p) ∈ D} is open in M, and

: M

→ M

−t

is a diffeomorphism with inverse Θ

−t

This is really annoying. We now prove the following useful result that saves

us from worrying about these problems in nice cases:

Proposition.

Let

be a compact manifold. Then any

X ∈ Vect

(

) is

complete.

Proof. Recall that

D = {(t, p) : Θ

(p) is defined}

is open. So given

p ∈ M

, there is some open neighbourhood

U ⊆ M

and an

ε >

0 such that (

−ε, ε

)

× U ⊆ D

. By compactness, we can find finitely many

such U that cover M, and find a small ε such that (−ε, ε) × M ⊆ D.

In other words, we know Θ

(

) exists and

p ∈ M

and

|t| < ε

. Also, we

know Θ

◦

= Θ

t+s

whenever

|t|, |s| < ε

, and in particular Θ

t+s

is defined. So

= (Θ

)

is defined for all N and |t| < ε, so Θ

is defined for all t.

2.3 Lie derivative

We now want to look at the concept of a Lie derivative. If we have a function

defined on all of

, and we have a vector field

, then we might want to

ask what the derivative of

in the direction of

is at each point. If

is a

real-valued function, then this is by definition

(

). If

is more complicated,

then this wouldn’t work, but we can still differentiate things along

using the

flows.

Notation. Let F : M → M be a diffeomorphism, and g ∈ C

∞

(M). We write

∗

g = g ◦ F ∈ C

∞

(M).

We now define the Lie derivative of a function, i.e. the derivative of a function

in the direction of a vector field

. Of course, we can obtain this by just

applying X(f ), but we want to make a definition that we can generalize.

Definition

(Lie derivative of a function)

Let

be a complete vector field, and

Θ be its flow. We define the Lie derivative of g along X by

(g) =



t=0

∗

Here this is defined pointwise, i.e. for all p ∈ M, we define

(g)(p) =



t=0

∗

(g)(p).

Lemma. L

(g) = X(g). In particular, L

(g) ∈ C

∞

(M, R).

Proof.

(g)(p) =



t=0

∗

(g)(p)



t=0

g(Θ

(p))

= dg|

(X(p))

= X(g)(p).

So this is quite boring. However, we can do something more exciting by

differentiating vector fields.

Notation.

Let

Y ∈ Vect

(

), and

M → M

be a diffeomorphism. Then

−1

F (p)

: T

F (p)

M → T

M. So we can write

∗

(Y )|

= DF

−1

F (p)

) ∈ T

Then F

∗

(Y ) ∈ Vect(M ). If g ∈ C

∞

(M), then

∗

(Y )|

(g) = Y

F (p)

(g ◦ F

−1

Alternatively, we have

∗

(Y )|

(g ◦ F ) = Y

F (p)

(g).

Removing the p’s, we have

∗

(Y )(g ◦ F ) = (Y (g)) ◦ F.

Definition

(Lie derivative of a vector field)

Let

X ∈ Vect

(

) be complete,

and

Y ∈ Vect

(

) be a vector field. Then the Lie derivative is given pointwise

(Y ) =



t=0

∗

(Y ).

Lemma. We have

Y = [X, Y ].

Proof. Let g ∈ C

∞

(M, R). Then we have

∗

(Y )(g ◦ Θ

) = Y (g) ◦ Θ

We now look at

∗

(Y )(g) − Y (g)

∗

(Y )(g) − Θ

∗

(Y )(g ◦ Θ

)

| {z }

Y (g) ◦ Θ

− Y (g)

| {z }

We have

lim

t→0

= L

(Y (g)) = XY (g)

by the previous lemma, and we have

lim

t→0

= lim

t→0

(Θ

∗

(Y ))



g − g ◦ Θ



= Y (−L

(g)) = −Y X(g).

Corollary. Let X, Y ∈ Vect(M) and f ∈ C

∞

(M, R). Then

(i) L

(fY ) = L

(f)Y + f L

Y = X(f )Y + f L

(ii) L

Y = −L

(iii) L

[Y, Z] = [L

Y, Z] + [Y, L

Z].

Proof. Immediate from the properties of the Lie bracket.

3 Lie groups

We now have a short digression to Lie groups. Lie groups are manifolds with

a group structure. They have an extraordinary amount of symmetry, since

multiplication with any element of the group induces a diffeomorphism of the Lie

group, and this action of the Lie group on itself is free and transitive. Effectively,

this means that any two points on the Lie group, as a manifold, are “the same”.

As a consequence, a lot of the study of a Lie group reduces to studying

an infinitesimal neighbourhood of the identity, which in turn tells us about

infinitesimal neighbourhoods of all points on the manifold. This is known as the

Lie algebra.

We are not going to go deep into the theory of Lie groups, as our main focus

is on differential geometry. However, we will state a few key results about Lie

groups.

Definition

(Lie group)

A Lie group is a manifold

with a group structure

such that multiplication

G × G → G

and inverse

G → G

are smooth

maps.

Example. GL

(R) and GL

Example. M

(R) under addition is also a Lie group.

Example. O(n) is a Lie group.

Notation.

Let

be a Lie group and

g ∈ G

. We write

G → G

for the

diffeomorphism

(h) = gh.

This innocent-seeming translation map is what makes Lie groups nice. Given

any local information near an element

, we can transfer it to local information

near

by applying the diffeomorphism

−1

. In particular, the diffeomorphism

G → G

induces a linear isomorphism D

G → T

, so we have a

canonical identification of all tangent spaces.

Definition

(Left invariant vector field)

Let

X ∈ Vect

(

) be a vector field.

This is left invariant if

) = X

for all g, h ∈ G.

We write Vect

(G) for the collection of all left invariant vector fields.

Using the fact that for a diffeomorphism F , we have

∗

[X, Y ] = [F

∗

X, F

∗

Y ],

it follows that Vect

(G) is a Lie subalgebra of Vect(G).

If we have a left invariant vector field, then we obtain a tangent vector at

the identity. On the other hand, if we have a tangent vector at the identity, the

definition of a left invariant vector field tells us how we can extend this to a left

invariant vector field. One would expect this to give us an isomorphism between

and

Vect

(

), but we have to be slightly more careful and check that the

induced vector field is indeed a vector field.

Lemma. Given ξ ∈ T

G, we let

= DL

(ξ) ∈ T

(G).

Then the map

G → Vect

(

) by

ξ 7→ X

is an isomorphism of vector spaces.

Proof.

The inverse is given by

X 7→ X|

. The only thing to check is that

actually is a left invariant vector field. The left invariant part follows from

) = DL

(DL

(ξ)) = DL

(ξ) = X

To check that

is smooth, suppose

f ∈ C

∞

(

U, R

), where

is open and

contains e. We let γ : (−ε, ε) → U be smooth with ˙γ(0) = ξ. So

= DL

(ξ)(f) = ξ(f ◦ L

) =



t=0

(f ◦ L

◦ γ)

But as (

t, g

)

7→ f ◦ L

◦ γ

(

) is smooth, it follows that

is smooth. So

∈ Vect

(G).

Thus, instead of talking about

Vect

(

), we talk about

, because it

seems less scary. This isomorphism gives T

G the structure of a Lie algebra.

Definition

(Lie algebra of a Lie group)

Let

be a Lie group. The Lie algebra

is the Lie algebra

whose Lie bracket is induced by that of the

isomorphism with Vect

(G). So

[ξ, η] = [X

, X

We also write Lie(G) for g.

In general, if a Lie group is written in some capital letter, say

, then the

Lie algebra is written in the same letter but in lower case fraktur.

Note that dim g = dim G is finite.

Lemma. Let G be an abelian Lie group. Then the bracket of g vanishes.

Example.

For any vector space

and

v ∈ V

, we have

∼

. So

as a

Lie group has Lie algebra

itself. The commutator vanishes because the group

is commutative.

Example.

Note that

(

) is an open subset of

, so it is a manifold.

It is then a Lie group under multiplication. Then we have

(R) = Lie(GL

(R)) = T

(R) = T

∼

If A, B ∈ GL

(R), then

(B) = AB.

(H) = AH

as L

is linear.

We claim that under the identification, if ξ, η ∈ gl

(R) = M

, then

[ξ, η] = ξη − ηξ.

Indeed, on G, we have global coordinates U

: GL

(R) → R where

(A) = A

where A = (A

) ∈ GL

(R).

Under this chart, we have

= L

(ξ) =

i,j

(Aξ)

∂

∂U



i,j,k

∂

∂U



So we have

i,j,k

∂

∂U

So we have

, X

] =





i,j,k

∂

∂U

p,r,q

∂

∂U





We now use the fact that

∂

∂U

= δ

We then expand

, X

] =

i,j,k,r

− U

)

∂

∂U

So we have

, X

] = X

ξη−ηξ

Definition

(Lie group homomorphisms)

Let

G, H

be Lie groups. A Lie group

homomorphism is a smooth map that is also a homomorphism.

Definition

(Lie algebra homomorphism)

Let

g, h

be Lie algebras. Then a Lie

algebra homomorphism is a linear map β : g → h such that

β[ξ, η] = [β(ξ), β(η)]

for all ξ, η ∈ g.

Proposition.

Let

be a Lie group and

ξ ∈ g

. Then the integral curve

for

through

e ∈ G

exists for all time, and

R → G

is a Lie group homomorphism.

The idea is that once we have a small integral curve, we can use the Lie

group structure to copy the curve to patch together a long integral curve.

Proof.

Let

I → G

be a maximal integral curve of

, say (

−ε, ε

)

∈ I

. We fix

a t

with |t

| < ε. Consider g

= γ(t

We let

˜γ(t) = L

(γ(t))

for |t| < ε.

We claim that ˜γ is an integral curve of X

with ˜γ(0) = g

. Indeed, we have

˜γ|

γ(t) = DL

˙γ(t) = DL

γ(t)

= X

·γ(t)

= X

˜γ(t)

By patching these together, we know (

− ε, t

)

⊆ I

. Since we have a fixed

ε that works for all t

, it follows that I = R.

The fact that this is a Lie group homomorphism follows from general proper-

ties of flow maps.

Example. Let G = GL

. If ξ ∈ gl

, we set

k≥0

We set

(

) =

tξ

. We observe that this is in

since

tξ

has an inverse

−tξ

(alternatively, det(e

tξ

) = e

tr(tξ)

6= 0). Then

(t) =

= e

tξ

ξ = L

tξ

ξ = L

F (t)

ξ.

Also, F (0) = I. So F (t) is an integral curve.

Definition

(Exponential map)

The exponential map of a Lie group

exp : g → G given by

exp(ξ) = γ

(1),

where γ

is the integral curve of X

through e ∈ G.

So in the case of G = GL

, the exponential map is the exponential map.

Proposition.

(i) exp is a smooth map.

(ii)

(

) =

exp

(

tξ

), then

R → G

is a Lie group homomorphism and

DF |





= ξ.

(iii) The derivative

D exp : T

∼

g → T

∼

is the identity map.

(iv) exp

is a local diffeomorphism around 0

∈ g

, i.e. there exists an open

U ⊆ g

containing 0 such that exp : U → exp(U ) is a diffeomorphism.

(v) exp

is natural, i.e. if

G → H

is a Lie group homomorphism, then the

diagram

g G

h H

exp

Df|

exp

commutes.

Proof.

(i) This is the smoothness of ODEs with respect to parameters

(ii) Exercise.

(iii) If ξ ∈ g, we let σ(t) = tξ. So ˙σ(0) = ξ ∈ T

∼

g. So

D exp |

(ξ) = D exp |

( ˙σ(0)) =



t=0

exp(σ(t)) =



t=0

exp(tξ) = X

= ξ.

(iv) Follows from above by inverse function theorem.

(v) Exercise.

Definition

(Lie subgroup)

A Lie subgroup of

is a subgroup

with a smooth

structure on H making H an immersed submanifold.

Certainly, if H ⊆ G is a Lie subgroup, then h ⊆ g is a Lie subalgebra.

Theorem.

h ⊆ g

is a subalgebra, then there exists a unique connected Lie

subgroup H ⊆ G such that Lie(H) = h.

Theorem.

Let

be a finite-dimensional Lie algebra. Then there exists a (unique)

simply-connected Lie group G with Lie algebra g.

Theorem.

Let

G, H

be Lie groups with

simply connected. Then every Lie

algebra homomorphism g → h lifts to a Lie group homomorphism G → H.

4 Vector bundles

Recall that we had the tangent bundle of a manifold. The tangent bundle gives

us a vector space at each point in space, namely the tangent space. In general,

a vector bundle is a vector space attached to each point in our manifold (in a

smoothly-varying way), which is what we are going to study in this chapter.

Before we start, we have a look at tensor products. These will provide us a

way of constructing new vector spaces from old ones.

4.1 Tensors

The tensor product is a very important concept in Linear Algebra. It is something

that is taught in no undergraduate courses and assumed knowledge in all graduate

courses. For the benefit of the students, we will give a brief introduction to

tensor products.

A motivation for tensors comes from the study of bilinear maps. A bilinear

map is a function that takes in two vectors and returns a number, and this is

linear in both variables. An example is the inner product, and another example

is the volume form, which tells us the volume of a parallelepiped spanned by the

two vectors.

Definition

(Bilinear map)

Let

U, V, W

be vector spaces. We define

Bilin

(

V ×

W, U ) to be the functions V × W → U that are bilinear, i.e.

α(λ

+ λ

, w) = λ

α(v

, w) + λ

α(v

, w)

α(v, λ

+ λ

) = λ

α(v, w

) + λ

α(v, w

It is important that a bilinear map is not a linear map. This is bad. We

spent so much time studying linear maps, and we now have to go back to our

linear algebra book and rewrite everything to talk about bilinear maps as well.

But bilinear maps are not enough. We want to do them for multi-linear maps!

But linear maps were already complicated enough, so this must be much worse.

We want to die.

Tensors are a trick to turn the study of bilinear maps to linear maps (from a

different space).

Definition

(Tensor product)

A tensor product of two vector spaces

V, W

a vector space

V ⊗ W

and a bilinear map

V × W → V ⊗ W

such that a

bilinear map from

V × W

is “the same as” a linear map from

V ⊗ W

. More

precisely, given any bilinear map

V × W → U

, we can find a unique linear

map ˜α : V ⊗ W → U such that the following diagram commutes:

V × W

V ⊗ W U

˜α

So we have

Bilin(V × W, U)

∼

Hom(V ⊗ W, U).

Given

v ∈ V

and

w ∈ W

, we obtain

(

v, w

)

∈ V ⊗ W

, called the tensor product

of v and w, written v ⊗ w.

We say V ⊗ W represents bilinear maps from V × W .

It is important to note that not all elements of

V ⊗ W

are of the form

v ⊗ w

Now the key thing we want to prove is the existence and uniqueness of tensor

products.

Lemma.

Tensor products exist (and are unique up to isomorphism) for all pairs

of finite-dimensional vector spaces.

Proof.

We can construct

V ⊗ W

Bilin

(

V × W, R

)

∗

. The verification is left as

an exercise on the example sheet.

We now write down some basic properties of tensor products.

Proposition.

Given maps

V → W

and

→ W

, we obtain a map

f ⊗ g : V ⊗ V

→ W ⊗ W

given by the bilinear map

(f ⊗ g)(v, w) = f(v) ⊗ g(w).

Lemma. Given v, v

∈ V and w, w

∈ W and λ

∈ R, we have

(λ

+ λ

) ⊗ w = λ

⊗ w) + λ

⊗ w)

v ⊗ (λ

+ λ

) = λ

(v ⊗ w

) + λ

(v ⊗ w

Proof. Immediate from the definition of bilinear map.

Lemma. If v

, · · · , v

is a basis for V , and w

, · · · , w

is a basis for W , then

⊗ w

: i = 1, · · · , n; j = 1, · · · , m}

is a basis for V ⊗ W . In particular, dim V ⊗ W = dim V × dim W .

Proof.

We have

V ⊗ W

Bilin

(

V × W, R

)

∗

. We let

V × W → R

be given





= a

Then

∈ Bilin

(

V × W, R

), and (

⊗ w

) are dual to

. So it suffices to

show that

are a basis. It is clear that they are independent, and any bilinear

map can be written as

α =

where

= α(v

, w

So done.

Proposition. For any vector spaces V, W, U , we have (natural) isomorphisms

(i) V ⊗ W

∼

W ⊗ V

(ii) (V ⊗ W ) ⊗ U

∼

V ⊗ (W ⊗ U )

(iii) (V ⊗ W )

∗

∼

∗

⊗ W

∗

Definition

(Covariant tensor)

A covariant tensor of rank

is an element

α ∈ V

∗

⊗ · · · ⊗ V

∗

| {z }

k times

i.e. α is a multilinear map V × · · · × V → R.

Example. A covariant 1-tensor is an α ∈ V

∗

, i.e. a linear map α : V → R.

A covariant 2-tensor is a

β ∈ V

∗

⊗ V

∗

, i.e. a bilinear map

V × V → R

, e.g.

an inner product.

Example.

α, β ∈ V

∗

, then

α ⊗ β ∈ V

∗

⊗ V

∗

is the covariant 2-tensor given

(α ⊗ b)(v, w) = α(v)β(w).

More generally, if

is a rank

tensor and

is a rank



tensor, then

α ⊗ β

is a

rank k +  tensor.

Definition (Tensor). A tensor of type (k, ) is an element in

(V ) = V

∗

⊗ · · · ⊗ V

∗

| {z }

k times

⊗ V ⊗ · · · ⊗ V

| {z }

` times

We are interested in alternating bilinear maps, i.e.

(

v, w

) =

−α

(

w, v

), or

equivalently, α(v, v) = 0 (if the characteristic is not 2).

Definition (Exterior product). Consider

T (V ) =

k≥0

⊗k

as an algebra (with multiplication given by the tensor product) (with

⊗0

We let

(

) be the ideal (as algebras!) generated by

{v ⊗ v

v ∈ V } ⊆ T

(

We define

Λ(V ) = T (V )/I(V ),

with a projection map

(

)

→

Λ(

). This is known as the exterior algebra.

We let

(V ) = π(V

⊗k

the k-th exterior product of V .

We write a ∧ b for π(α ⊗ β).

The idea is that Λ

is the dual of the space of alternating multilinear maps

V × V → R.

Lemma.

(i) If α ∈ Λ

V and β ∈ Λ

V , then α ∧ β = (−1)

β ∧ α.

(ii) If dim V = n and p > n, then we have

dim Λ

V = 1, dim Λ

V = 1, Λ

V = {0}.

(iii) The multilinear map det : V × · · · × V → R spans Λ

V .

(iv) If v

, · · · , v

is a basis for V , then

∧ · · · ∧ v

: i

< · · · < i

}

is a basis for Λ

V .

Proof.

(i) We clearly have v ∧ v = 0. So

v ∧ w = −w ∧ v

Then

∧ · · · ∧ v

) ∧ (w

∧ · · · ∧ w

) = (−1)

∧ · · · ∧ w

∧ v

∧ · · · ∧ v

since we have pq swaps. Since

∧ · · · ∧ v

: i

, · · · , i

∈ {1, · · · , n}} ⊆ Λ

spans Λ

(by the corresponding result for tensor products), the result

follows from linearity.

(ii) Exercise.

(iii) The det map is non-zero. So it follows from the above.

(iv) We know that

∧ · · · ∧ v

: i

, · · · , i

∈ {1, · · · , n}} ⊆ Λ

spans, but they are not independent since there is a lot of redundancy (e.g.

∧ v

−v

∧ v

). By requiring

< · · · < i

, then we obtain a unique

copy for combination.

To check independence, we write

= (

, · · · , i

) and let

∧· · · ∧v

Then suppose

= 0

for

∈ R

. For each

, we let

be the multi-index

{

, · · · , n} \ I

. So

if I 6= I

, then v

∧ v

= 0. So wedging with v

gives

∧ v

= a

∧ v

= 0.

So a

= 0. So done by (ii).

V → W

is a linear map, then we get an induced linear map Λ

V → Λ

W in the obvious way, making the following diagram commute:

⊗p

V Λ

⊗p

π π

More concretely, we have

F (v

∧ · · · ∧ v

) = F (v

) ∧ · · · ∧ F (v

Lemma.

Let

V → V

be a linear map. Then Λ

: Λ

V →

multiplication by det F .

Proof.

Let

, · · · , v

be a basis. Then Λ

is spanned by

∧ · · · ∧ v

. So we

have

(Λ

F )(v

∧ · · · ∧ v

) = λ v

∧ · · · ∧ v

for some λ. Write

F (v

) =

for some A

∈ R, i.e. A is the matrix representation of F . Then we have

(Λ

F )(v

∧ · · · ∧ v

) =





∧ · · · ∧





If we expand the thing on the right, a lot of things die. The only things that

live are those where we get each of

once in the wedges in some order. Then

this becomes

σ∈S

ε(σ)(A

σ(1),1

· · · A

σ(n),n

∧ · · · ∧ v

= det(F ) v

∧ · · · ∧ v

where

(

) is the sign of the permutation, which comes from rearranging the

to the right order.

4.2 Vector bundles

Our aim is to consider spaces

M ⊗ T

M, . . . ,

etc as

varies, i.e. con-

struct a “tensor bundle” for these tensor products, similar to how we constructed

the tangent bundle. Thus, we need to come up with a general notion of vector

bundle.

Definition

(Vector bundle)

A vector bundle of rank

is a smooth

manifold E with a smooth π : E → M such that

(i) For each p ∈ M, the fiber π

−1

(p) = E

is an r-dimensional vector space,

(ii)

For all

p ∈ M

, there is an open

U ⊆ M

containing

and a diffeomorphism

t : E|

= π

−1

(U) → U × R

such that

U × R

commutes, and the induced map

→ {q} × R

is a linear isomorphism

for all q ∈ U.

We call

a trivialization of

over

; call

the total space; call

the

base space; and call

the projection. Also, for each

q ∈ M

, the vector

space E

= π

−1

({q}) is called the fiber over q.

Note that the vector space structure on

is part of the data of a vector bundle.

Alternatively,

can be given by collections of smooth maps

, · · · , s

U → E

with the property that for each

q ∈ U

, the vectors

(

)

, · · · , s

(

) form a basis

for E

. Indeed, given such s

, · · · , s

, we can define t by

t(v

) = (q, α

, · · · , α

where v

∈ E

and the α

are chosen such that

i=1

(q).

The s

, · · · , s

are known as a frame for E over U.

Example

(Tangent bundle)

The bundle

T M → M

is a vector bundle. Given

any point

, find some coordinate charts around

with coordinates

, · · · , x

Then we get a frame

∂

∂x

, giving trivializations of

T M

over

. So

T M

is a vector

bundle.

Definition

(Section)

A (smooth) section of a vector bundle

E → M

over some

open

U ⊆ M

is a smooth

U → E

such that

(

)

∈ E

for all

p ∈ U

, that is

π ◦ s = id. We write C

∞

(U, E) for the set of smooth sections of E over U.

Example. Vect(M ) = C

∞

(M, T M ).

Definition

(Transition function)

Suppose that

→ U

× R

and

: E|

→ U

× R

are trivializations of E. Then

◦ t

−1

: (U

∩ U

) × R

→ (U

∩ U

) × R

is fiberwise linear, i.e.

◦ t

−1

(q, v) = (q, ϕ

αβ

(q)v),

where ϕ

αβ

(q) is in GL

(R).

In fact,

αβ

∩ U

→ GL

(

) is smooth. Then

αβ

is known as the

transition function from β to α.

Proposition.

We have the following equalities whenever everything is defined:

(i) ϕ

αα

= id

(ii) ϕ

αβ

= ϕ

−1

βα

(iii) ϕ

αβ

βγ

= ϕ

αγ

, where ϕ

αβ

βγ

is pointwise matrix multiplication.

These are known as the cocycle conditions.

We now consider general constructions that allow us to construct new vector

bundles from old ones.

Proposition

(Vector bundle construction)

Suppose that for each

p ∈ M

, we

have a vector space E

. We set

E =

[

We let

E → M

be given by

(

) =

for

∈ E

. Suppose there is an open

cover {U

} of open sets of M such that for each α, we have maps

: E|

= π

−1

) → U

× R

over

that induce fiberwise linear isomorphisms. Suppose the transition

functions

αβ

are smooth. Then there exists a unique smooth structure on

making π : E → M a vector bundle such that the t

are trivializations for E.

Proof. The same as the case for the tangent bundle.

In particular, we can use this to perform the following constructions:

Definition

(Direct sum of vector bundles)

Let

be vector bundles on

. Suppose

∼

× R

is a trivialization for

over

, and

∼

× R

˜r

is a trivialization for

E over U

We let

αβ

be transition functions for

}

and

˜ϕ

αβ

be transition functions

for {

Define

E ⊕

E =

[

⊕

and define

: (E ⊕

E)|

= E|

⊕

→ U

× (R

⊕ R

˜r

) = U

× R

r+˜r

be the fiberwise direct sum of the two trivializations. Then

clearly gives a

linear isomorphism (E ⊕

∼

r+˜r

, and the transition function for T

◦ T

−1

= ϕ

αβ

⊕ ˜ϕ

αβ

which is clearly smooth. So this makes E ⊕

E into a vector bundle.

In terms of frames, if

, · · · , s

}

is a frame for

and

{˜s

, · · · , ˜s

˜r

}

is a frame

for

E over some U ⊆ M , then

⊕ 0, 0 ⊕ ˜s

: i = 1, · · · , r; j = 1, · · · , ˜r}

is a frame for E ⊕

Definition

(Tensor product of vector bundles)

Given two vector bundles

over M, we can construct E ⊗

E similarly with fibers (E ⊗

E)|

= E|

⊗

Similarly, we can construct the alternating product of vector bundles Λ

Finally, we have the dual vector bundle.

Definition

(Dual vector bundle)

Given a vector bundle

E → M

, we define the

dual vector bundle by

∗

[

p∈M

)

∗

Suppose again that

→ U

× R

is a local trivialization. Taking the

dual of this map gives

∗

: U

× (R

)

∗

→ E|

∗

since taking the dual reverses the direction of the map. We pick an isomorphism

(

)

∗

→ R

once and for all, and then reverse the above isomorphism to get a

map

∗

→ U

× R

This gives a local trivialization.

, · · · , s

}

is a frame for

over

, then

∗

, · · · , s

∗

}

is a frame for

∗

over U, where {s

∗

(p), · · · , s

∗

(p)} is a dual basis to {s

(p), · · · , s

(p)}.

Definition (Cotangent bundle). The cotangent bundle of a manifold M is

∗

M = (T M)

∗

In local coordinate charts, we have a frame

∂

∂x

, · · · ,

∂

∂x

T M

over

. The

dual frame is written as dx

, · · · , dx

. In other words, we have

∈ (T

∗

and

∂

∂x



= δ

Recall the previously, given a function

f ∈ C

∞

(

U, R

), we defined d

as the

differential of f given by

df|

= Df|

: T

M → T

f(p)

∼

Thinking of x

as a function on a coordinate chart U, we have

∂

∂x



∂

∂x

) = δ

for all i, j. So the two definitions of dx

agree.

We can now take powers of this to get more interesting things.

Definition

(

-form)

-form on a manifold

over

is a smooth section of

∗

M, i.e. an element in C

∞

(U, Λ

∗

M).

Example. A 1-form is an element of T

∗

M. It is locally of the form

+ · · · + α

for some smooth functions α

, · · · , α

Similarly, if ω is a p-form, then locally, it is of the form

ω =

where I = (i

, · · · , i

) with i

< · · · < i

, and dx

= dx

∧ · · · ∧ dx

It is important to note that these representations only work locally.

Definition (Tensors on manifolds). Let M be a manifold. We define

M = T

∗

M ⊗ · · · ⊗ T

∗

| {z }

k times

⊗ T M ⊗ · · · ⊗ T M

| {z }

` times

A tensor of type (k, ) is an element of

∞

(M, T

M).

The convention when k =  = 0 is to set T

M = M × R.

In local coordinates, we can write a (k, ) tensor ω as

ω =

,...,j

,...,i

⊗ · · · ⊗ dx

⊗

∂

∂x

⊗ · · · ⊗

∂

∂x

where the α are smooth functions.

Example. A tensor of type (0, 1) is a vector field.

A tensor of type (1, 0) is a 1-form.

A tensor of type (0, 0) is a real-valued function.

Definition

(Riemannian metric)

A Riemannian metric on

is a (2

0)-tensor

such that for all

, the bilinear map

M × T

M → R

is symmetric and

positive definite, i.e. an inner product.

Given such a g and v

∈ T

M, we write kv

k for

, v

Using these, we can work with things like length:

Definition (Length of curve). Let γ : I → M be a curve. The length of γ is

(γ) =

k ˙γ(t)k dt.

Finally, we will talk about morphisms between vector bundles.

Definition

(Vector bundle morphisms)

Let

E → M

and

→ M

be vector

bundles. A bundle morphism from

is a pair of smooth maps (

E →

, f : M → M

) such that the following diagram commutes:

E E

M M

i.e. such that F

: E

→ E

f(p)

is linear for each p.

Example.

Let

T M

and

T M

. If

M → M

is smooth, then (D

f, f

)

is a bundle morphism.

Definition

(Bundle morphism over

)

Given two bundles

E, E

over the same

base

, a bundle morphism over

is a bundle morphism

E → E

of the form

(F, id

Example.

Given a Riemannian metric

, we get a bundle morphism

T M →

∗

M over M by

v 7→ F (v) = g(v, −).

Since each

(

v, −

) is an isomorphism, we have a canonical bundle isomorphism

T M

∼

∗

Note that the isomorphism between

T M

and

∗

requires the existence of

a Riemannian metric.

5 Differential forms and de Rham cohomology

5.1 Differential forms

We are now going to restrict our focus to a special kind of tensors, known as

differential forms. Recall that in

(as a vector space), an alternating

-linear

map tells us the signed volume of the parallelepiped spanned by

vectors.

In general, a differential

-form is an alternating

-linear map on the tangent

space at each point, so it tells us the volume of an “infinitesimal

-dimensional

parallelepiped”.

In fact, we will later see than on an (oriented)

-dimensional manifold, we

can integrate a

-form on the manifold to obtain the “volume” of the manifold.

Definition (Differential form). We write

Ω

(M) = C

∞

(M, Λ

∗

M) = {p-forms on M }.

An element of Ω

(M) is known as a differential p-form.

In particular, we have

Ω

(M) = C

∞

(M, R).

In local coordinates x

, · · · , x

on U we can write ω ∈ Ω

(M) as

ω =

<...<i

,...,i

∧ · · · ∧ dx

for some smooth functions ω

,...,i

We are usually lazy and just write

ω =

Example. A 0-form is a smooth function.

Example.

A 1-form is a section of

∗

. If

ω ∈

Ω

(

) and

X ∈ Vect

(

then ω(X) ∈ C

∞

(M, R).

For example, if f is a smooth function on M , then df ∈ Ω

(M) with

df(X) = X(f)

for all X ∈ Vect(M).

Locally, we can write

df =

i=1

To work out what the a

’s are, we just hit this with the

∂

∂x

. So we have

= df



∂

∂x



∂f

∂x

So we have

df =

i=1

∂f

∂x

This is essentially just the gradient of a function!

Example. If dim M = n, and ω ∈ Ω

(M), then locally we can write

ω = g dx

∧ · · · ∧ dx

for some smooth function

. This is an alternating form that assigns a real

number to n tangent vectors. So it measures volume!

If y

, · · · , y

is any other coordinates, then

∂x

∂y

So we have

ω = g det



∂x

∂y



i,j

∧ · · · ∧ dy

Now a motivating question is this — given an

ω ∈

Ω

(

), can we find some

f ∈ Ω

(M) such that ω = df?

More concretely, let U ⊆ R

be open, and let x, y be the coordinates. Let

ω = a dx + b dy

If we have w = df for some f, then we have

a =

∂f

∂x

, b =

∂f

∂y

So the symmetry of partial derivatives tells us that

∂a

∂y

∂b

∂x

. (∗)

So this equation (∗) is a necessary condition to solve ω = df . Is it sufficient?

To begin with, we want to find a better way to express (

∗

) without resorting

to local coordinates, and it turns out this construction will be very useful later

on.

Theorem (Exterior derivative). There exists a unique linear map

d = d

M,p

: Ω

(M) → Ω

p+1

(M)

such that

(i) On Ω

(M) this is as previously defined, i.e.

df(X) = X(f) for all X ∈ Vect(M).

(ii) We have

d ◦ d = 0 : Ω

(M) → Ω

p+2

(M).

(iii) It satisfies the Leibniz rule

d(ω ∧ σ) = dω ∧ σ + (−1)

ω ∧ dσ.

It follows from these assumptions that

(iv)

d acts locally, i.e. if

ω, ω

∈

Ω

(

) satisfy

ω|

for some

U ⊆ M

open, then dω|

= dω

(v) We have

d(ω|

) = (dω)|

for all U ⊆ M.

What do the three rules tell us? The first rule tells us this is a generalization

of what we previously had. The second rule will turn out to be a fancy way of

saying partial derivatives commute. The final Leibniz rule tells us this d is some

sort of derivative.

Example. If we have

ω = a dx + b dy,

then we have

dω = da ∧ dx + a d(dx) + db ∧ dy + b d(dy)

= da ∧ dx + db ∧ dy



∂a

∂x

dx +

∂a

∂y



∧ dx +



∂b

∂x

dx +

∂b

∂y



∧ dy



∂b

∂x

−

∂a

∂y



dx ∧ dy.

So the condition (∗) says dω = 0.

We now rephrase our motivating question — if ω ∈ Ω

(M) satisfies dω = 0,

can we find some

such that

= d

for some

f ∈

Ω

(

)? Now this has the

obvious generalization — given any

-form

, if d

= 0, can we find some

such that ω = dσ?

Example.

, we have coordinates

x, y, z

. We have seen that for

f ∈

Ω

(

we have

df =

∂f

∂x

dx +

∂f

∂y

dy +

∂f

∂z

dz.

Now if

ω = P dx + Q dy + R dz ∈ Ω

then we have

d(P dx) = dP ∧ dx + P ddx



∂P

∂x

dx +

∂P

∂y

dy +

∂P

∂z



∧ dx

= −

∂P

∂y

dx ∧ dy −

∂P

∂z

dx ∧ dz.

So we have

dω =



∂Q

∂x

−

∂P

∂y



dx ∧ dy +



∂R

∂x

−

∂P

∂z



dx ∧ dz +



∂R

∂y

−

∂Q

∂z



dy ∧ dz.

This is just the curl! So d

= 0 just says that curl ◦ grad = 0.

Proof.

The above computations suggest that in local coordinates, the axioms

already tell use completely how d works. So we just work locally and see that

they match up globally.

Suppose

is covered by a single chart with coordinates

, · · · , x

. We

define d : Ω

(M) → Ω

(M) as required by (i). For p > 0, we define





<...<i

,...,i

∧ · · · ∧ dx





dω

,...,i

∧ dx

∧ · · · ∧ dx

Then (i) is clear. For (iii), we suppose

ω = f dx

∈ Ω

(M)

σ = g dx

∈ Ω

(M).

We then have

d(ω ∧ σ) = d(fg dx

∧ dx

)

= d(fg) ∧ dx

∧ dx

= g df ∧ dx

∧ dx

+ f dg ∧ dx

∧ dx

= g df ∧ dx

∧ dx

+ f(−1)

∧ (dg ∧ dx

)

= (dω) ∧ σ + (−1)

ω ∧ dσ.

So done. Finally, for (ii), if f ∈ Ω

(M), then

f = d

∂f

∂x

i,j

∂

∂x

∧ dx

= 0,

since partial derivatives commute. Then for general forms, we have

ω = d





= d



dω

∧ dx



= d



dω

∧ dx

∧ · · · ∧ dx



= 0

using Leibniz rule. So this works.

Certainly this has the extra properties. To claim uniqueness, if

∂

: Ω

(

)

→

Ω

p+1

(M) satisfies the above properties, then

∂ω = ∂





∂ω

∧ dx

+ ω

∧ ∂dx

dω

∧ dx

using the fact that ∂ = d on Ω

(M) and induction.

Finally, if

is covered by charts, we can define d : Ω

(

)

→

Ω

p+1

(

) by

defining it to be the d above on any single chart. Then uniqueness implies this is

well-defined. This gives existence of d, but doesn’t immediately give uniqueness,

since we only proved local uniqueness.

So suppose

∂

: Ω

(

)

→

Ω

p+1

(

) again satisfies the three properties. We

claim that

∂

is local. We let

ω, ω

∈

Ω

(

) be such that

ω|

for some

U ⊆ M

open. Let

x ∈ U

, and pick a bump function

χ ∈ C

∞

(

) such that

χ ≡ 1 on some neighbourhood W of x, and supp(χ) ⊆ U . Then we have

χ · (ω − ω

) = 0.

We then apply ∂ to get

0 = ∂(χ · (ω − ω

)) = dχ ∧ (ω − ω

) + χ(∂ω − ∂ω

But χ ≡ 1 on W . So dχ vanishes on W . So we must have

∂ω|

− ∂ω

= 0.

So ∂ω = ∂ω

on W .

Finally, to show that

∂

, if

ω ∈

Ω

(

), we take the same

as before,

and then on x, we have

∂ω = ∂





= ∂χ

+ χ

∂ω

∧ dx

= χ

dω

∧ dx

= dω.

So we get uniqueness. Since x was arbitrary, we have ∂ = d.

One useful example of a differential form is a symplectic form.

Definition

(Non-degenerate form)

A 2-form

ω ∈

Ω

(

) is non-degenerate if

ω(X

, X

) = 0 implies X

= 0.

As in the case of an inner product, such an

gives us an isomorphism

M → T

∗

M by

α(X

)(Y

) = ω(X

, Y

Definition

(Symplectic form)

A symplectic form is a non-degenerate 2-form

such that dω = 0.

Why did we work with covectors rather than vectors when defining differential

forms? It happens that differential forms have nicer properties. If we have some

F ∈ C

∞

(M, N) and g ∈ Ω

(N) = C

∞

(N, R), then we can form the pullback

∗

g = g ◦ F ∈ Ω

(M).

More generally, for x ∈ M , we have a map

DF |

: T

M → T

F (x)

This does not allow us to pushforward a vector field on

to a vector field of

N, as the map F might not be injective. However, we can use its dual

(DF |

)

∗

: T

∗

F (x)

N → T

∗

to pull forms back.

Definition

(Pullback of differential form)

Let

ω ∈

Ω

(

) and

F ∈ C

∞

(

M, N

We define the pullback of ω along F to be

∗

ω|

= Λ

(DF |

)

∗

(ω|

F (x)

In other words, for v

, · · · , v

∈ T

M, we have

∗

ω|

)(v

, · · · , v

) = ω|

F (x)

(DF |

), · · · , DF |

)).

Lemma. Let F ∈ C

∞

(M, N). Let F

∗

be the associated pullback map. Then

(i) F

∗

is a linear map Ω

(N) → Ω

(M).

(ii) F

∗

(ω ∧ σ) = F

∗

ω ∧ F

∗

σ.

(iii) If G ∈ C

∞

(N, P ), then (G ◦ F )

∗

= F

∗

◦ G

∗

(iv) We have dF

∗

= F

∗

Proof.

All but (iv) are clear. We first check that this holds for 0 forms. If

g ∈ Ω

(N), then we have

∗

dg)|

(v) = dg|

F (x)

(DF |

(v))

= DF |

(v)(g)

= v(g ◦ F )

= d(g ◦ F )(v)

= d(F

∗

g)(v).

So we are done.

Then the general result follows from (i) and (ii). Indeed, in local coordinates

, · · · , y

, if

ω =

,...,i

∧ · · · ∧ dy

then we have

∗

ω =

∗

,...,i

)(F

∗

∧ · · · ∧ dy

Then we have

∗

ω = F

∗

dω =

∗

dω

,...,i

)(F

∗

∧ · · · ∧ dy

5.2 De Rham cohomology

We now get to answer our original motivating question — given an ω ∈ Ω

(M)

with dω = 0, does it follow that there is some σ ∈ Ω

p−1

(M) such that ω = dσ?

The answer is “not necessarily”. In fact, the extent to which this fails tells

us something interesting about the topology of the manifold. We are going to

define certain vector spaces

(

) for each

, such that this vanishes if and

only if all

forms

with d

= 0 are of the form d

. Afterwards, we will come

up with techniques to compute this

(

), and then we can show that certain

spaces have vanishing H

(M).

We start with some definitions.

Definition (Closed form). A p-form ω ∈ Ω

(M) is closed if dω = 0.

Definition

(Exact form)

-form

ω ∈

Ω

(

) is exact if there is some

σ ∈

Ω

p−1

(M) such that ω = dσ.

We know that every exact form is closed. However, in general, not every

closed form is exact. The extent to which this fails is given by the de Rham

cohomology.

Definition

(de Rham cohomology)

The

th de Rham cohomology is given by

the R-vector space

(M) =

ker d : Ω

(M) → Ω

p+1

(M)

im d : Ω

p−1

(M) → Ω

(M)

closed forms

exact forms

In particular, we have

(M) = ker d : Ω

(M) → Ω

(M).

We could tautologically say that if d

= 0, then

is exact iff it vanishes

(

). But this is as useful as saying “Let

be the set of solutions to

this differential equation. Then the differential equation has a solution iff

is non-empty”. So we want to study the properties of

and find ways of

computing them.

Proposition.

(i) Let M have k connected components. Then

(M) = R

(ii) If p > dim M, then H

(M) = 0.

(iii)

F ∈ C

∞

(

M, N

), then this induces a map

∗

(

)

→ H

(

)

given by

∗

[ω] = [F

∗

ω].

(iv) (F ◦ G)

∗

= G

∗

◦ F

∗

(v)

M → N

is a diffeomorphism, then

∗

(

)

→ H

(

) is an

isomorphism.

Proof.

(i) We have

(M) = {f ∈ C

∞

(M, R) : df = 0}

= {locally constant functions f}

= R

number of connected components

(ii) If p > dim M, then all p-forms are trivial.

(iii)

We first show that

∗

indeed represents some member of

(

). Let

[ω] ∈ H

(N). Then dω = 0. So

d(F

∗

ω) = F

∗

(dω) = 0.

So [F

∗

ω] ∈ H

(M). So this map makes sense.

To see it is well-defined, if [

] = [

], then

ω − ω

= d

for some

. So

∗

ω − F

∗

= d(F

∗

σ). So [F

∗

ω] = [F

∗

(iv) Follows from the corresponding fact for pullback of differential forms.

(v) If F

−1

is an inverse to F , then (F

−1

)

∗

is an inverse to F

∗

by above.

It turns out that de Rham cohomology satisfies a stronger property of being

homotopy invariant. To make sense of that, we need to define what it means to

be homotopy invariant.

Definition

(Smooth homotopy)

Let

, F

M → N

be smooth maps. A

smooth homotopy from

is a smooth map

: [0

× M → N

such that

(x) = F (0, x), F

(x) = F (1, x).

If such a map exists, we say F

and F

are homotopic.

Note that here

is defined on [0

× M

, which is not a manifold. So we

need to be slightly annoying and say that

is smooth if it can be extended to a

smooth function I × M → N for I ⊇ [0, 1] open.

We can now state what it means for the de Rham cohomology to be homotopy

invariant.

Theorem

(Homotopy invariance)

Let

, F

be homotopic maps. Then

∗

: H

(N) → H

(M).

Proof. Let F : [0, 1] × M → N be the homotopy, and

(x) = F (t, x).

We denote the exterior derivative on

by d

(and similarly d

), and that on

[0, 1] × M by d.

Let

ω ∈

Ω

(

) be such that d

= 0. We let

be the coordinate on [0

1].

We write

∗

ω = σ + dt ∧ γ,

where σ = σ(t) ∈ Ω

(M) and γ = γ(t) ∈ Ω

p−1

(M). We claim that

σ(t) = F

∗

ω.

Indeed, we let ι : {t} × M → [0, 1] × M be the inclusion. Then we have

∗

ω|

{t}×M

= (F ◦ ι)

∗

ω = ι

∗

= ι

∗

(σ + dt ∧ γ)

= ι

∗

σ + ι

∗

dt ∧ ι

∗

= ι

∗

σ,

using the fact that ι

∗

dt = 0. As d

ω = 0, we have

0 = F

∗

= dF

∗

= d(σ + dt ∧ γ)

= d

(σ) + (−1)

∂σ

∂t

∧ dt + dt ∧ d

= d

σ + (−1)

∂σ

∂t

∧ dt + (−1)

p−1

γ ∧ dt.

Looking at the dt components, we have

∂σ

∂t

= d

γ.

So we have

∗

ω − F

∗

ω = σ(1) − σ(0) =

∂σ

∂t

dt =

γ dt = d

γ(t) dt.

So we know that

∗

ω] = [F

∗

ω].

So done.

Example.

Suppose

U ⊆ R

is an open “star-shaped” subset, i.e. there is some

∈ U such that for any x ∈ U and t ∈ [0, 1], we have

tx + (1 − t)x

∈ U.

We define F

: U → U by

(x) = tx + (1 − t)x

Then

is a smooth homotopy from the identity map to

, the constant map

. We clearly have

∗

being the identity map, and

∗

is the zero map on

(U) for all p ≥ 1. So we have

(U) =

(

0 p ≥ 1

R p = 0

Corollary

(Poincar´e lemma)

Let

U ⊆ R

be open and star-shaped. Suppose

ω ∈

Ω

(

) is such that d

= 0. Then there is some

σ ∈

Ω

p−1

(

) such that

ω = dσ.

Proof. H

(U) = 0 for p ≥ 1.

More generally, we have the following notion.

Definition

(Smooth homotopy equivalence)

We say two manifolds

M, N

are

smoothly homotopy equivalent if there are smooth maps

M → N

and

G : N → M such that both F ◦ G and G ◦ F are homotopic to the identity.

Corollary.

and

are smoothly homotopy equivalent, then

(

)

∼

(N).

Note that by approximation, it can be shown that if

and

are homotopy

equivalent as topological spaces (i.e. the same definition where we drop the word

“smooth”), then they are in fact smoothly homotopy equivalent. So the de Rham

cohomology depends only on the homotopy type of the underlying topological

space.

5.3 Homological algebra and Mayer-Vietoris theorem

The main theorem we will have for computing de Rham cohomology will be the

Mayer-Vietoris theorem. Proving this involves quite a lot of setting up and hard

work. In particular, we need to define some notions from homological algebra to

even state Mayer-Vietoris theorem.

The actual proof will be divided into two parts. The first part is a purely

algebraic result known as the snake lemma, and the second part is a differential-

geometric part that proves that we satisfy the hypothesis of the snake lemma.

We will not prove the snake lemma, whose proof can be found in standard

algebraic topology texts (perhaps with arrows the wrong way round).

We start with some definitions.

Definition

(Cochain complex and exact sequence)

A sequence of vector spaces

and linear maps

· · · V

p−1

p+1

· · ·

p−1

is a cochain complex if d

◦

p−1

= 0 for all

p ∈ Z

. Usually we have

= 0 for

p <

0 and we do not write them out. Keeping these negative degree

rather

than throwing them away completely helps us state our theorems more nicely,

so that we don’t have to view

as a special case when we state our theorems.

It is exact at p if ker d

= im d

p−1

, and exact if it is exact at every p.

There are, of course, chain complexes as well, but we will not need them for

this course.

Example. The de Rham complex

Ω

(M) Ω

(M) · · ·

d d

is a cochain complex as d

= 0. It is exact at p iff H

(M) = {0}.

Example.

If we have an exact sequence such that

dim V

< ∞

for all

and

are zero for all but finitely many p, then

(−1)

dim V

= 0.

Definition (Cohomology). Let

= · · · V

p−1

p+1

· · ·

p−1

be a cochain complex. The cohomology of V

at p is given by

) =

ker d

im d

p−1

Example.

The cohomology of the de Rham complex is the de Rham cohomology.

We can define maps between cochain complexes:

Definition

(Cochain map)

Let

and

be cochain complexes. A cochain

map

→ W

is a collection of maps

→ W

such that the following

diagram commutes for all p:

p+1

Proposition.

A cochain map induces a well-defined homomorphism on the

cohomology groups.

Definition

(Short exact sequence)

A short exact sequence is an exact sequence

of the form

0 V

This implies that

is injective,

is surjective, and

(

) =

ker

(

). By the

rank-nullity theorem, we know

dim V

= rank(β) + null(β) = dim V

+ dim V

We can now state the main technical lemma, which we shall not prove.

Theorem

(Snake lemma)

Suppose we have a short exact sequence of complexes

0 A

i.e. the i, q are cochain maps and we have a short exact sequence

0 A

for each p.

Then there are maps

δ : H

) → H

p+1

)

such that there is a long exact sequence

· · · H

) H

)

p+1

) H

p+1

) H

p+1

) · · ·

∗

Using this, we can prove the Mayer-Vietoris theorem.

Theorem

(Mayer-Vietoris theorem)

Let

be a manifold, and

U ∪ V

where U, V are open. We denote the inclusion maps as follows:

U ∩ V U

V M

Then there exists a natural linear map

δ : H

(U ∩ V ) → H

p+1

(M)

such that the following sequence is exact:

(M) H

(U) ⊕ H

(V ) H

(U ∩ V )

p+1

(M) H

p+1

(U) ⊕ H

p+1

(V ) · · ·

∗

⊕j

∗

−i

∗

⊕j

∗

−i

∗

Before we prove the theorem, we do a simple example.

Example. Consider M = S

. We can cut the circle up:

Here we have

= {(x, y) : x

+ y

= 1}

U = S

∩ {y > −ε}

V = S

∩ {y < ε}.

U, V

are diffeomorphic to intervals, hence contractible, and

U ∩V

is diffeomor-

phic to the disjoint union of two intervals, we know their de Rham cohomology.

0 H

) H

(U) ⊕ H

(V ) H

(U ∩ V )

) H

(U) ⊕ H

(V ) · · ·

We can fill in the things we already know to get

0 R R ⊕ R R ⊕ R

) 0 · · ·

By adding the degrees alternatingly, we know that

dim H

) = 1.

)

∼

Now we prove Mayer-Vietoris.

Proof of Mayer-Vietoris.

By the snake lemma, it suffices to prove that the

following sequence is exact for all p:

0 Ω

(U ∪ V ) Ω

(U) ⊕ Ω

(V ) Ω

(U ∩ V ) 0

∗

⊕j

∗

−i

∗

It is clear that the two maps compose to 0, and the first map is injective. By

counting dimensions, it suffices to show that i

∗

− i

∗

is surjective.

Indeed, let

{ϕ

, ϕ

}

be partitions of unity subordinate to

{U, V }

. Let

ω ∈ Ω

(U ∩ V ). We set σ

∈ Ω

(U) to be

(

ω on U ∩ V

0 on U \ supp ϕ

Similarly, we define σ

∈ Ω

(V ) by

(

−ϕ

ω on U ∩ V

0 on V \ supp ϕ

Then we have

∗

− i

∗

= (ϕ

ω + ϕ

ω)|

U∩V

= ω.

So i

∗

− i

∗

is surjective.

6 Integration

As promised, we will be able to integrate differential forms on manifolds. However,

there is a slight catch. We said that differential forms give us the signed volume of

an infinitesimal parallelepiped, and we can integrate these infinitesimal volumes

up to get the whole volume of the manifold. However, there is no canonical

choice of the sign of the volume, so we do not, in general, get a well-defined

volume.

In order to fix this issue, our manifold needs to have an orientation.

6.1 Orientation

We start with the notion of an orientation of a vector space. After we have one,

we can define an orientation of a manifold to be a smooth choice of orientation

for each tangent space.

Informally, an orientation on a vector space

is a choice of a collection of

ordered bases that we declare to be “oriented”. If (

, · · · , e

) is an oriented

basis, then changing the sign of one of the

changes orientation, while scaling

by a positive multiple does not. Similarly, swapping two elements in the basis

will induce a change in orientation.

To encode this information, we come up with some alternating form

ω ∈

(

∗

). We can then say a basis

, · · · , e

is oriented if

(

, · · · , e

) is positive.

Definition

(Orientation of vector space)

Let

be a vector space with

dim V

. An orientation is an equivalence class of elements

ω ∈

(

∗

), where we say

ω ∼ ω

iff ω = λω

for some λ > 0. A basis (e

, · · · , e

) is oriented if

ω(e

, · · · , e

) > 0.

By convention, if V = {0}, an orientation is just a choice of number in {±1}.

Suppose we have picked an oriented basis

, · · · , e

. If we have any other

basis ˜e

, · · · , ˜e

, we write

˜e

Then we have

ω(˜e

, · · · , ˜e

) = det B ω(e

, · · · , e

So ˜e

, · · · , ˜e

is oriented iff det B > 0.

We now generalize this to manifolds, where we try to orient the tangent

bundle smoothly.

Definition

(Orientation of a manifold)

An orientation of a manifold

defined to be an equivalence class of elements

ω ∈

Ω

(

) that are nowhere

vanishing, under the equivalence relation

ω ∼ ω

if there is some smooth

M → R

such that ω = fω

Definition

(Orientable manifold)

A manifold is orientable if it has some

orientation.

is a connected, orientable manifold, then it has precisely two possible

orientations.

Definition

(Oriented manifold)

An oriented manifold is a manifold with a

choice of orientation.

Definition

(Oriented coordinates)

Let

be an oriented manifold. We say

coordinates x

, · · · , x

on a chart U are oriented coordinates if

∂

∂x



, · · · ,

∂

∂x



is an oriented basis for T

M for all p ∈ U .

Note that we can always find enough oriented coordinates. Given any

connected chart, either the chart is oriented, or

−x

, · · · , x

is oriented. So any

oriented M is covered by oriented charts.

Now by the previous discussion, we know that if

, · · · , x

and

, · · · , y

are oriented charts, then the transition maps for the tangent space all have

positive determinant.

Example. R

is always assumed to have the standard orientation given by

∧ · · · ∧ dx

Definition

(Orientation-preserving diffeomorphism)

Let

M, N

be oriented

manifolds, and

F ∈ C

∞

(

M, N

) be a diffeomorphism. We say

preserves

orientation if D

F |

M → T

F (p)

takes an oriented basis to an oriented

basis.

Alternatively, this says the pullback of the orientation on

is the orientation

on M (up to equivalence).

6.2 Integration

The idea is that to define integration, we fist understand how we can integrate

on R

, and then patch them up using partitions of unity.

We are going to allow ourselves to integrate on rather general domains.

Definition

(Domain of integration)

Let

D ⊆ R

. We say

is a domain of

integration if D is bounded and ∂D has measure zero.

Since

can be an arbitrary subset, we define an

-form on

to be some

ω ∈ Ω

(U) for some open U containing D.

Definition

(Integration on

)

Let

be a compact domain of integration,

and

ω = f dx

∧ · · · ∧ dx

be an n-form on D. Then we define

ω =

f(x

, · · · , x

) dx

· · · dx

In general, let U ⊆ R

and let ω ∈ Ω

) have compact support. We define

ω =

for some D ⊆ U containing supp ω.

Note that we do not directly say we integrate it on

supp ω

, since

supp ω

need

not have a nice boundary.

Now if we want to integrate on a manifold, we need to patch things up, and

to do so, we need to know how these things behave when we change coordinates.

Definition

(Smooth function)

Let

D ⊆ R

and

D → R

. We say

smooth if it is a restriction of some smooth function

U → R

where

U ⊇ D

Lemma.

Let

D → E

be a smooth map between domains of integration in

, and assume that

F |

D →

is an orientation-preserving diffeomorphism.

Then

ω =

∗

ω.

This is exactly what we want.

Proof.

Suppose we have coordinates

, · · · , x

and

, · · · , y

. Write

ω = f dy

∧ · · · ∧ dy

Then we have

ω =

f dy

· · · dy

(f ◦ F ) | det DF | dx

· · · dx

(f ◦ F ) det DF dx

· · · dx

∗

ω.

Here we used the fact that

| det

F |

det

because

is orientation-preserving.

We can now define integration over manifolds.

Definition

(Integration on manifolds)

Let

be an oriented manifold. Let

ω ∈

Ω

(

). Suppose that

supp

(

) is a compact subset of some oriented chart

(U, ϕ). We set

ω =

ϕ(U)

(ϕ

−1

)

∗

ω.

By the previous lemma, this does not depend on the oriented chart (U, ϕ).

ω ∈

Ω

(

) is a general form with compact support, we do the following:

cover the support by finitely many oriented charts

}

α=1,...,m

. Let

{χ

}

be a

partition of unity subordinate to {U

}. We then set

ω =

ω.

It is clear that we have

Lemma.

This is well-defined, i.e. it is independent of cover and partition of

unity.

We will not bother to go through the technicalities of proving this properly.

Note that it is possible to define this for non-smooth forms, or not-everywhere-

defined form, or with non-compact support etc, but we will not do it here.

Theoretically, our definition is perfectly fine and easy to work with. However,

it is absolutely useless for computations, and there is no hope you can evaluate

that directly.

Now how would we normally integrate things? In IA Vector Calculus, we

probably did something like this — if we want to integrate something over a

sphere, we cut the sphere up into the Northern and Southern hemisphere. We

have coordinates for each of the hemispheres, so we integrate each hemisphere

separately, and then add the result up.

This is all well, except we have actually missed out the equator in this

process. But that doesn’t really matter, because the equator has measure zero,

and doesn’t contribute to the integral.

We now try to formalize our above approach. The below definition is not

standard:

Definition

(Parametrization)

Let

be either an oriented manifold of dimen-

sion

, or a domain of integration in

. By a parametrization of

we mean a

decomposition

M = S

∪ · · · ∪ S

with smooth maps

→ S

, where

is a compact domain of integration,

such that

(i) F

→

is an orientation-preserving diffeomorphism

(ii) ∂S

has measure zero (if

is a manifold, this means

(

∂S

∩ U

) for all

charts (U, ϕ)).

(iii) For i 6= j, S

intersects S

only in their common boundary.

Theorem.

Given a parametrization

}

and an

ω ∈

Ω

(

) with compact

support, we have

ω =

∗

ω.

Proof.

By using partitions of unity, we may consider the case where

has

support in a single chart, and thus we may wlog assume we are working on

and then the result is obvious.

There is a problem — in all our lives, we’ve been integrating functions, not

forms. If we have a function f : R → R, then we can take the integral

f dx.

Now of course, we are not actually integrating

. We are integrating the

differential form

. Why we seem like we are integrating functions is because

we have a background form d

. So if we have a manifold

with a “background”

n-form ω ∈ Ω

(M), then we can integrate f ∈ C

∞

(M, R) by

fω.

In general, a manifold does not come with such a canonical background form.

However, in some cases, it does.

Lemma.

Let

be an oriented manifold, and

a Riemannian metric on

Then there is a unique

ω ∈

Ω

(

) such that for all

, if

, · · · , e

is an oriented

orthonormal basis of T

M, then

ω(e

, · · · , e

) = 1.

We call this the Riemannian volume form, written dV

Note that d

is a notation. It is not the exterior derivative of some mysterious

object V

Proof.

Uniqueness is clear, since if

is another, then

λω

for some

, and

evaluating on an orthonormal basis shows that λ = 1.

To see existence, let

be any nowhere vanishing

-form giving the orientation

. On a small set

, pick a frame

, · · · , s

for

T M|

and apply the Gram-

Schmidt process to obtain an orthonormal frame

, · · · , e

, which we may wlog

assume is oriented. Then we set

f = σ(e

, · · · , e

which is non-vanishing because σ is nowhere vanishing. Then set

ω =

This proves existence locally, and can be patched together globally by uniqueness.

6.3 Stokes Theorem

Recall from, say, IA Vector Calculus that Stokes’ theorem relates an integral on

a manifold to a integral on its boundary. However, our manifolds do not have

boundaries! So we can’t talk about Stokes’ theorem! So we now want to define

what it means to be a manifold with boundary.

Definition (Manifold with boundary). Let

= {(x

, · · · , x

) ∈ R

: x

≥ 0}.

A chart-with-boundary on a set

is a bijection

U → ϕ

(

) for some

U ⊆ M

such that

(

)

⊆ H

is open. Note that this image may or may not hit the

boundary of H

. So a “normal” chart is also a chart with boundary.

An atlas-with-boundary on

is a cover by charts-with-boundary (

, ϕ

)

such that the transition maps

◦ ϕ

−1

: ϕ

∩ U

) → ϕ

∩ U

)

are smooth (in the usual sense) for all α, β.

A manifold-with-boundary is a set

with an (equivalence class of) atlas

with boundary whose induced topology is Hausdorff and second-countable.

Note that a manifold with boundary is not a manifold, but a manifold is a

manifold with boundary. We will often be lazy and drop the “with boundary”

descriptions.

Definition

(Boundary point)

is a manifold with boundary and

p ∈ M

then we say

is a boundary point if

(

)

∈ ∂H

for some (hence any) chart-

with-boundary (

U, ϕ

) containing

. We let

∂M

be the set of boundary points

and Int(M) = M \ ∂M.

Note that these are not the topological notions of boundary and interior.

Proposition.

Let

be a manifold with boundary. Then

Int

(

) and

∂M

are

naturally manifolds, with

dim ∂M = dim Int M − 1.

Example.

The solid ball

(0)

is a manifold with boundary, whose interior is

(0) and boundary is S

n−1

Note that the product of manifolds with boundary is not a manifold with

boundary. For example, the interval [0

1] is a manifold with boundary, but [0

has corners. This is bad. We can develop the theory of manifolds with corners,

but that is more subtle. We will not talk about them.

Everything we did for manifolds can be done for manifolds with boundary,

e.g. smooth functions, tangent spaces, tangent bundles etc. Note in particular

the definition of the tangent space as derivations still works word-for-word.

Lemma.

Let

p ∈ ∂M

, say

p ∈ U ⊆ M

where (

U, ϕ

) is a chart (with boundary).

Then

∂

∂x



, · · · ,

∂

∂x



is a basis for T

M. In particular, dim T

M = n.

Proof.

Since this is a local thing, it suffices to prove it for

. We write

∞

(

H, R

) for the functions

→ R

that extend smoothly to an open

neighbourhood of H

. We fix a ∈ ∂H

. Then by definition, we have

= Der

∞

, R)).

We let i

∗

: T

→ T

be given by

∗

(X)(g) = X(g|

)

We claim that

∗

is an isomorphism. For injectivity, suppose

∗

(

) = 0. If

f ∈ C

∞

(

), then

extends to a smooth

on some neighbourhood

Then

X(f) = X(g|

) = i

∗

(X)(g) = 0.

So X(f ) = 0 for all f. Then X = 0. So i

∗

is injective.

To see surjectivity, let Y ∈ T

, and let X ∈ T

be defined by

X(f) = Y (g),

where

g ∈ C

∞

(

, R

) is any extension of

. To see this is well-defined, we

let

Y =

i=1

∂

∂x



Then

Y (g) =

i=1

∂g

∂x

(a),

which only depends on

, i.e.

. So

is a well-defined element of

, and

∗

(X) = Y by construction. So done.

Now we want to see how orientations behave. We can define them in exactly

the same way as manifolds, and everything works. However, something interesting

happens. If a manifold with boundary has an orientation, this naturally induces

an orientation of the boundary.

Definition

(Outward/Inward pointing)

Let

p ∈ ∂M

. We then have an inclusion

∂M ⊆ T

M. If X

∈ T

M, then in a chart, we can write

i=1

∂

∂x

where

∈ R

and

∂

∂x

, · · · ,

∂

∂x

n−1

are a basis for

∂M

. We say

is outward

pointing if a

< 0, and inward pointing if a

> 0.

Definition

(Induced orientation)

Let

be an oriented manifold with boundary.

We say a basis

, · · · , e

n−1

is an oriented basis for

∂M

if (

, e

, · · · , e

n−1

)

is an oriented basis for

, where

is any outward pointing element in

This orientation is known as the induced orientation.

It is an exercise to see that these notions are all well-defined and do not

depend on the basis.

Example. We have an isomorphism

∂H

∼

n−1

, · · · , x

n−1

, 0) 7→ (x

, · · · , x

n−1

−

∂

∂x



∂H

is an outward pointing vector. So we know

, · · · , x

n−1

is an oriented chart for

∂H

iff

−

∂

∂x

∂

∂x

, · · · ,

∂

∂x

n−1

is oriented, which is true iff n is even.

Example.

= 1, say

= [

a, b

]

⊆ R

with

a < b

, then

{a, b}

, then

∂M

{

}

. So an orientation of

∂M

is a choice of numbers

1 attached to each point.

The convention is that if

is in the standard orientation induced by

M ⊆ R

then the orientation is obtained by giving +1 to b and −1 to a.

Finally, we get to Stokes’ theorem.

Theorem

(Stokes’ theorem)

Let

be an oriented manifold with boundary of

dimension n. Then if ω ∈ Ω

n−1

(M) has compact support, then

dω =

∂M

ω.

In particular, if M has no boundary, then

dω = 0

Note that this makes sense. d

is an

-form on

, so we can integrate it.

On the right hand side, what we really doing is integrating the restriction of

∂M

, i.e. the (

n −

1)-form

∗

, where

∂M → M

is the inclusion, so that

∗

ω ∈ Ω

n−1

(∂M).

Note that if

= [

a, b

], then this is just the usual fundamental theorem of

calculus.

The hard part of the proof is keeping track of the signs.

Proof. We first do the case where M = H

. Then we have

ω =

i=1

∧ · · · ∧

∧ · · · ∧ dx

where ω

is compactly supported, and the hat denotes omission. So we have

dω =

dω

∧ dx

∧ · · · ∧

∧ · · · ∧ dx

∂ω

∂x

∧ dx

∧ · · · ∧

∧ · · · ∧ dx

(−1)

i−1

∂ω

∂x

∧ · · · ∧ dx

Let’s say

supp(ω) = {x

∈ [−R, R] : j = 1, · · · , n − 1; x

∈ [0, R]} = A.

Then suppose i 6= n. Then we have

∂ω

∂x

∧ · · · ∧ dx

∂ω

∂x

· · · dx

−R

· · ·

−R

∂ω

∂x

· · · dx

By Fubini’s theorem, we can integrate this in any order. We integrate with

respect to dx

first. So this is

= ±

−R

· · ·

−R

∂ω

∂x

· · ·

· · · dx

By the fundamental theorem of calculus, the inner integral is

ω(x

, · · · , x

i−1

, R, x

i+1

, · · · , x

)−ω(x

, · · · , x

i−1

, −R, x

i+1

, · · · , x

) = 0−0 = 0.

So the integral vanishes. So we are only left with the i = n term. So we have

dω = (−1)

n−1

∂ω

∂x

· · · dx

= (−1)

n−1

−R

· · ·

−R

∂ω

∂x

· · · dx

n−1

Now that integral is just

, · · · , x

n−1

, R) − ω

, · · · , x

n−1

, 0) = −ω

, · · · , x

n−1

, 0).

So this becomes

= (−1)

−R

· · ·

−R

, · · · , x

n−1

, 0) dx

· · · dx

n−1

Next we see that

∗

ω = ω

∧ · · · ∧ dx

n−1

as i

∗

(dx

) = 0. So we have

∂H

∗

ω = ±

A∩∂H

ω(x

, · · · , x

n−1

, 0) dx

· · · dx

Here the sign is a plus iff

, · · · , x

n−1

are an oriented coordinate for

∂H

, i.e.

n is even. So this is

∂H

ω = (−1)

−R

· · ·

−R

, · · · , x

n−1

, 0) dx

· · · dx

n−1

dω.

Now for a general manifold

, suppose first that

ω ∈

Ω

n−1

(

) is compactly

supported in a single oriented chart (

U, ϕ

). Then the result is true by working

in local coordinates. More explicitly, we have

dω =

(ϕ

−1

)

∗

dω =

d((ϕ

−1

)

∗

ω) =

∂H

(ϕ

−1

)

∗

ω =

∂M

ω.

Finally, for a general

, we just cover

by oriented charts (

U, ϕ

), and use a

partition of unity χ

subordinate to {U

}. So we have

ω =

ω.

Then

dω =

(dχ

)ω +

dω = d





ω +

dω =

dω,

using the fact that

is constant, hence its derivative vanishes. So we have

dω =

∂M

ω =

∂M

ω.

Then all the things likes Green’s theorem and divergence theorem follow from

this.

Example.

Let

be a manifold without boundary with a symplectic form

ω ∈

Ω

(

) that is closed and positive definite. Then by basic Linear algebra we

know

6= 0.

Since

is closed, it is an element [

]

∈ H

(

). Does this vanish? If

= d

then we have

d(τ ∧ ω ∧ · · · ∧ ω) = ω

So we have

d(τ ∧ ω ∧ · · · ∧ ω) = 0

by Stokes’ theorem. This is a contradiction. So [ω] is non-zero in H

(M).

7 De Rham’s theorem*

In the whole section, M will be a compact manifold.

Theorem (de Rham’s theorem). There exists a natural isomorphism

(M)

∼

(M, R),

where

(

M, R

) is the singular cohomology of

, and this is in fact an iso-

morphism of rings, where

(

) has the product given by the wedge, and

(M, R) has the cup product.

Recall that singular cohomology is defined as follow:

Definition

(Singular

-complex)

Let

be a manifold. Then a singular

simplex is a continuous map

σ : ∆

→ M,

where

∆

(

i=0

= 1

)

⊆ R

n+1

We define

(M) = {formal sums

: a

∈ R, σ

a singular p simplex}.

We define

∞

(m) = {formal sums

: a

∈ R, σ

a smooth singular p simplex}.

Definition (Boundary map). The boundary map

∂ : C

(M) → C

p−1

(M)

is the linear map such that if σ : ∆

→ M is a p simplex, then

∂σ =

(−1)

σ ◦ F

i,p

where

i,p

maps ∆

p−1

affine linearly to the face of ∆

opposite the

th vertex.

We similarly have

∂ : C

∞

(M) → C

∞

p−1

(M).

We can then define singular homology

Definition (Singular homology). The singular homology of M is

(M, R) =

ker ∂ : C

(M) → C

p−1

(M)

im ∂ : C

p+1

(M) → C

(M)

The smooth singular homology is the same thing with

(

) replaced with

∞

(M).

∞

has the same properties as

, e.g. functoriality, (smooth) homotopy

invariance, Mayer-Vietoris etc with no change in proof.

Any smooth p-simplex σ is also continuous, giving a natural inclusion

i : C

∞

(M) → C

(M),

which obviously commutes with ∂, giving

∗

: H

∞

(M) → H

(M).

Theorem. The map i

∗

: H

∞

(M) → H

(M) is an isomorphism.

There are three ways we can prove this. We will give the ideas for these

proofs:

(i)

We can show that any continuous map

M → N

between manifolds is

homotopic to a smooth one. But this is painful to prove.

(ii)

What we really care about is maps

: ∆

→ M

, and we can barycentrically

subdivide the simplex so that it only lies in a single chart, and then it is

easy to do smooth approximation.

(iii)

and

∞

have enough properties in common, in particular they

both have Mayer-Vietoris and agree on convex sets, this implies they are

the same. We will not spell out this proof, because we are going to do this

to prove that de Rham cohomology is the same as singular cohomology

Since we are working with

, we can cheat and define singular cohomology

in a simple way:

Definition

(Singular cohomology)

The singular cohomology of

is defined as

(M, R) = Hom(H

(M, R), R).

Similarly, the smooth singular cohomology is

∞

(M, R) = Hom(H

∞

(M, R), R).

This is a bad definition in general! It just happens to work for singular

cohomology with coefficients in

, and we are not too bothered to write dowm

the proper definition.

Our goal is now to describe an isomorphism

(M)

∼

∞

(M, R).

The map itself is given as follows:

Suppose [

]

∈ H

(

), so

ω ∈

Ω

(

) with d

= 0. Suppose that

: ∆

p →

M is smooth with ∂σ = 0. We can then define

I([ω]) =

∆

∗

ω ∈ R.

Note that we have not defined what

∆

means, because ∆

is not even a

manifold with boundary — it has corners. We can develop an analogous theory

of integration on manifolds with corners, but we can also be lazy, and just

integrate over

∆

= ∆

\ {codimension 2 faces}.

Now

ω|

∆

∗

does not have compact support, but has the property that it is the

restriction of a (unique)

-form on ∆

, so in particular it is bounded. So the

integral is finite.

Now in general, if τ =

∈ C

(M), we define

I([ω])(τ) =

∆

∗

ω ∈ R.

Now Stokes theorem tell us

∂σ

ω =

dω.

So we have

Lemma. I is a well-defined map H

(M) → H

∞

(M, R).

Proof. If [ω] = [ω

], then ω − ω

= dα. Then let σ ∈ H

∞

(M, R). Then

(ω − ω

) =

dα =

∂σ

α = 0,

since ∂σ = 0.

On the other hand, if [

] = [

], then

σ − σ

∂β

for some

. Then we have

σ−σ

ω =

∂β

ω =

dω = 0.

So this is well-defined.

Lemma. I

is functorial and commutes with the boundary map of Mayer-Vietoris.

In other words, if F : M → N is smooth, then the diagram

(M) H

(N)

∞

(M) H

∞

(N)

∗

I I

∗

And if M = U ∪ V and U, V are open, then the diagram

(U ∩ V ) H

p+1

(U ∪ V )

∞

(U ∩ V, R) H

(U ∪ V, R)

I I

also commutes. Note that the other parts of the Mayer-Vietoris sequence

commute because they are induced by maps of manifolds.

Proof. Trace through the definitions.

Proposition. Let U ⊆ R

is convex, then

U : H

(U) → H

∞

(U, R)

is an isomorphism for all p.

Proof.

p >

0, then both sides vanish. Otherwise, we check manually that

I : H

(U) → H

∞

(U, R) is an isomorphism.

These two are enough to prove that the two cohomologies agree — we can

cover any manifold by convex subsets of

, and then use Mayer-Vietoris to

patch them up.

We make the following definition:

Definition (de Rham).

(i) We say a manifold M is de Rham if I is an isomorphism.

(ii)

We say an open cover

}

is de Rham if

∩ · · · ∩ U

is de Rham

for all α

, · · · , α

(iii)

A de Rham basis is a de Rham cover that is a basis for the topology on

Our plan is to inductively show that everything is indeed de Rham.

We have already seen that if

U ⊆ R

is convex, then it is de Rham, and a

countable disjoint union of de Rham manifolds is de Rham.

The key proposition is the following:

Proposition.

Suppose

{U, V }

is a de Rham cover of

U ∪ V

. Then

U ∪ V

is de

Rham.

Proof.

We use the five lemma! We write the Mayer-Vietoris sequence that is

impossible to fit within the margins:

(U) ⊕ H

(V ) H

(U ∪ V ) H

p+1

(U ∩ V ) H

(U) ⊕ H

p+1

(V ) H

p+1

(U ∪ V )

∞

(U) ⊕ H

∞

(V ) H

∞

(U ∪ V ) H

p+1

∞

(U ∩ V ) H

∞

(U) ⊕ H

p+1

∞

(V ) H

p+1

∞

(U ∪ V )

I⊕I

This huge thing commutes, and all but the middle map are isomorphisms. So by

the five lemma, the middle map is also an isomorphism. So done.

Corollary.

, · · · , U

is a finite de Rham cover of

∪ · · · ∪ U

, then

M is de Rham.

Proof. By induction on k.

Proposition. The disjoint union of de Rham spaces is de Rham.

Proof. Let A

be de Rham. Then we have





∼

)

∼

∞

)

∼

∞





Lemma. Let M be a manifold. If it has a de Rham basis, then it is de Rham.

Proof sketch.

Let

M → R

be an “exhaustion function”, i.e.

−1

([

−∞, c

]) for

all c ∈ R. This is guaranteed to exist for any manifold. We let

= {q ∈ M : f(q) ∈ [m, m + 1]}.

We let



q ∈ M : f(q) ∈



m −

, m +



Given any

q ∈ A

, there is some

α(q)

⊆ A

in the de Rham basis containing

is compact, we can cover it by a finite number of such

, with each

⊆ A

. Let

= U

∪ · · · ∪ U

Since

has a finite de Rham cover, so it is de Rham. Observe that if

∩ B

˜m

6= ∅, then

M ∈ {m, m − 1, m + 1}. We let

U =

[

m even

, V =

[

m odd

Then this is a countable union of de Rham spaces, and is thus de Rham. Similarly,

U ∩ V is de Rham. So M = U ∪ V is de Rham.

Theorem. Any manifold has a de Rham basis.

Proof.

U ⊆ R

is open, then it is de Rham, since there is a basis of convex

sets {U

} (e.g. take open balls). So they form a de Rham basis.

Finally,

has a basis of subsets diffeomorphic to open subsets of

. So it

is de Rham.

8 Connections

8.1 Basic properties of connections

Imagine we are moving in a manifold

along a path

I → M

. We already

know what “velocity” means. We simply have to take the derivative of the path

(and pick the canonical tangent vector 1

∈ T

) to obtain a path

I → T M

Can we make sense of our acceleration? We can certainly iterate the procedure,

treating

T M

as just any other manifold, and obtain a path

I → T T M

. But

this is not too satisfactory, because

T T M

is a rather complicated thing. We

would want to use the extra structure we know about

T M

, namely each fiber is

a vector space, to obtain something nicer, perhaps an acceleration that again

lives in T M.

We could try the naive definition

= lim

h→0

γ(t + h) − γ(t)

but this doesn’t make sense, because

(

) and

(

) live in different vector

spaces.

The problem is solved by the notion of a connection. There are (at least) two

ways we can think of a connection — on the one hand, it is a specification of

how we can take derivatives of sections, so by definition this solves our problem.

On the other hand, we can view it as telling us how to compare infinitesimally

close vectors. Here we will define it the first way.

Notation. Let E be a vector bundle on M . Then we write

Ω

(E) = Ω

(E ⊗ Λ

∗

M)).

So an element in Ω

(E) takes in p tangent vectors and outputs a vector in E.

Definition

(Connection)

Let

be a vector bundle on

. A connection on

is a linear map

: Ω

(E) → Ω

(E)

such that

(fs) = df ⊗ s + f d

for all f ∈ C

∞

(M) and s ∈ Ω

(E).

A connection on T M is called a linear or Koszul connection.

Given a connection d

on a vector bundle, we can use it to take derivatives

of sections. Let

s ∈

Ω

(

) be a section of

, and

X ∈ Vect

(

). We want to

use the connection to define the derivative of

in the direction of

. This is

easy. We define ∇

: Ω

(E) → Ω

(E) by

∇

(s) = hd

(s), Xi ∈ Ω

(E),

where the brackets denote applying d

(

) :

T M → E

. Often we just call

∇

the connection.

Proposition.

For any

∇

is linear in

over

, and linear in

over

∞

(

Moreover,

∇

(fs) = f∇

(s) + X(f)s

for f ∈ C

∞

(M) and s ∈ Ω

(E).

This doesn’t really solve our problem, though. The above lets us differentiate

sections of the whole bundle

along an everywhere-defined vector field. However,

what we want is to differentiate a path in

along a vector field defined on that

path only.

Definition

(Vector field along curve)

Let

I → M

be a curve. A vector field

along γ is a smooth V : I → T M such that

V (t) ∈ T

γ(t)

for all t ∈ I. We write

J(γ) = {vector fields along γ}.

The next thing we want to prove is that we can indeed differentiate these

things.

Lemma.

Given a linear connection

∇

and a path

I → M

, there exists a

unique map D

: J(γ) → J(γ) such that

(i) D

(fV ) =

fV + f D

V for all f ∈ C

∞

(I)

(ii)

is an open neighbourhood of

(

) and

is a vector field on

such

that

V |

γ(t)

= V

for all t ∈ I, then

(V )|

= ∇

˙γ(0)

V .

We call D

the covariant derivative along γ.

In general, when we have some notion on

that involves derivatives and

we want to transfer to general manifolds with connection, all we do is to replace

the usual derivative with the covariant derivative, and we will usually get the

right generalization, because this is the only way we can differentiate things on

a manifold.

Before we prove the lemma, we need to prove something about the locality

of connections:

Lemma.

Given a connection

∇

and vector fields

X, Y ∈ Vect

(

), the quantity

∇

Y |

depends only on the values of Y near p and the value of X at p.

Proof. It is clear from definition that this only depends on the value of X at p.

To show that it only depends on the values of

near

, by linearity, we just

have to show that if

= 0 in a neighbourhood

, then

∇

Y |

= 0. To do

so, we pick a bump function

that is identically 1 near

, then

supp

(

)

⊆ U

Then χY = 0. So we have

0 = ∇

(χY ) = χ∇

(Y ) + X(χ)Y.

Evaluating at

, we find that

(

)

vanishes since

is constant near

. So

∇

(Y ) = 0.

We now prove the existence and uniqueness of the covariant derivative.

Proof of previous lemma. We first prove uniqueness.

By a similar bump function argument, we know that D

V |

depends only

on values of V (t) near t

. Suppose that locally on a chart, we have

V (t) =

(t)

∂

∂x



γ(t)

for some V

: I → R. Then we must have

V |

(t)

∂

∂x



γ(t

)

)∇

˙γ(t

)

∂

∂x

by the Leibniz rule and the second property. But every term above is uniquely

determined. So it follows that D

V must be given by this formula.

To show existence, note that the above formula works locally, and then they

patch because of uniqueness.

Proposition. Any vector bundle admits a connection.

Proof.

Cover

such that

is trivial. This is easy to do locally, and

then we can patch them up with partitions of unity.

Note that a connection is not a tensor, since it is not linear over

∞

(

However, if d

and

are connections, then

−

)(fs) = df ⊗ s + f d

s − (df ⊗ s + f

S) = f(d

−

)(s).

So the difference is linear. Recall from sheet 2 that if

E, E

are vector bundles

and

α : Ω

(E) → Ω

)

is a map such that

α(fs) = fα(s)

for all

s ∈

Ω

(

) and

f ∈ C

∞

(

), then there exists a unique bundle morphism

ξ : E → E

such that

α(s)|

= ξ(s(p)).

Applying this to

= d

−

: Ω

(

)

→

Ω

(

) = Ω

(

E ⊗ T

∗

), we know

there is a unique bundle map

ξ : E → E ⊗ T

∗

such that

(s)|

+ ξ(s(p)).

So we can think of d

−

as a bundle morphism

E → E ⊗ T

∗

In other words, we have

−

∈ Ω

(E ⊗ E

∗

⊗ T

∗

M) = Ω

(End(E)).

The conclusion is that the set of all connections on

is an affine space modelled

on Ω

(End(E)).

Induced connections

In many cases, having a connection on a vector bundle allows us to differentiate

many more things. Here we will note a few.

Proposition.

The map d

extends uniquely to d

: Ω

(

)

→

Ω

p+1

(

) such

that d

is linear and

(w ⊗ s) = dω ⊗ s + (−1)

ω ∧ d

for

s ∈

Ω

(

) and

ω ∈

Ω

(

). Here

ω ∧

means we take the wedge on the

form part of d

s. More generally, we have a wedge product

Ω

(M) × Ω

(E) → Ω

p+q

(E)

(α, β ⊗ s) 7→ (α ∧ β) ⊗ s.

More generally, the extension satisfies

(ω ∧ ξ) = dω ∧ ξ + (−1)

ω ∧ d

ξ,

where ξ ∈ Ω

(E) and ω ∈ Ω

(M).

Proof.

The formula given already uniquely specifies the extension, since every

form is locally a sum of things of the form

ω ⊗ s

. To see this is well-defined, we

need to check that

((fω) ⊗ s) = d

(ω ⊗ (fs)),

and this follows from just writing the terms out using the Leibniz rule. The

second part follows similarly by writing things out for ξ = η ⊗ s.

Definition

(Induced connection on tensor product)

Let

E, F

be vector bundles

with connections d

respectively. The induced connection is the connection

E⊗F

on E ⊗ F given by

E⊗F

(s ⊗ t) = d

s ⊗ t + s ⊗ d

for s ∈ Ω

(E) and t ∈ Ω

(F ), and then extending linearly.

Definition

(Induced connection on dual bundle)

Let

be a vector bundle

with connection d

. Then there is an induced connection d

∗

given by

requiring

dhs, ξi = hd

s, ξi + hs, d

∗

ξi,

for

s ∈

Ω

(

) and

ξ ∈

Ω

(

∗

). Here

h · , · i

denotes the natural pairing Ω

(

)

Ω

∗

) → C

∞

(M, R).

So once we have a connection on

, we have an induced connection on all

tensor products of it.

Christoffel symbols

We also have a local description of the connection, known as the Christoffel

symbols.

Say we have a frame

, · · · , e

for

over

U ⊆ M

. Then any section

s ∈ Ω

(E|

) is uniquely of the form

s = s

where

∈ C

∞

(

U, R

) and we have implicit summation over repeated indices (as

we will in the whole section).

Given a connection d

, we write

= Θ

⊗ e

where Θ

∈ Ω

(U). Then we have

s = d

= ds

⊗ e

+ s

= (ds

+ Θ

) ⊗ e

We can write s = (s

, · · · , s

). Then we have

s = ds + Θs,

where the final multiplication is matrix multiplication.

It is common to write

= d + Θ,

where Θ is a matrix of 1-forms. It is a good idea to just view this just as a

formal equation, rather than something that actually makes mathematical sense.

Now in particular, if we have a linear connection

∇

T M

and coordinates

, · · · , x

U ⊆ M

, then we have a frame for

T M|

given by

∂

, · · · , ∂

. So

we again have

∂

= Θ

⊗ ∂

where Θ

∈

Ω

(

). But we don’t just have a frame for the tangent bundle, but

also the cotangent bundle. So in these coordinates, we can write

= Γ

where Γ

∈ C

∞

(U). These Γ

are known as the Christoffel symbols.

In this notation, we have

∇

∂

= hd

∂

, ∂

= hΓ

⊗ ∂

, ∂

= Γ

∂

8.2 Geodesics and parallel transport

One thing we can do with a connection is to define a geodesic as a path with

“no acceleration”.

Definition

(Geodesic)

Let

be a manifold with a linear connection

∇

. We

say that γ : I → M is a geodesic if

˙γ(t) = 0.

A natural question to ask is if geodesics exist. This is a local problem, so we

work in local coordinates. We try to come up with some ordinary differential

equations that uniquely specify a geodesic, and then existence and uniqueness

will be immediate. If we have a vector field

V ∈ J

(

), we can write it locally as

V = V

∂

then we have

V =

∂

+ V

∇

˙γ(t

)

∂

We now want to write this in terms of Christoffel symbols. We put

(γ

, · · · , γ

). Then using the chain rule, we have

V =

∂

+ V

˙γ

∇

∂

= (

+ V

˙γ

)∂

Recall that γ is a geodesic if D

˙γ = 0 on I. This holds iff we locally have

¨γ

+ ˙γ

˙γ

= 0.

As this is just a second-order ODE in

, we get unique solutions locally given

initial conditions.

Theorem.

Let

∇

be a linear connection on

, and let

W ∈ T

. Then there

exists a geodesic γ : (−ε, ε) → M for some ε > 0 such that

˙γ(0) = W.

Any two such geodesics agree on their common domain.

More generally, we can talk about parallel vector fields.

Definition

(Parallel vector field)

Let

∇

be a linear connection on

, and

I → M

be a path. We say a vector field

V ∈ J

(

) along

is parallel if

V (t) ≡ 0 for all t ∈ I.

What does this say? If we think of D

as just the usual derivative, this tells

us that the vector field

is “constant” along

(of course it is not literally

constant, since each V (t) lives in a different vector space).

Example. A path γ is a geodesic iff ˙γ is parallel.

The important result is the following:

Lemma

(Parallel transport)

Let

∈ I

and

ξ ∈ T

γ(t

)

. Then there exists a

unique parallel vector field

V ∈ J

(

) such that

(

) =

. We call

the parallel

transport of ξ along γ.

Proof.

Suppose first that

(

)

⊆ U

for some coordinate chart

with coordinates

, · · · , x

. Then V ∈ J(γ) is parallel iff D

V = 0. We put

V =

(t)

∂

∂x

Then we need

+ V

˙γ

= 0.

This is a first-order linear ODE in

with initial condition given by

(

) =

which has a unique solution.

The general result then follows by patching, since by compactness, the image

of γ can be covered by finitely many charts.

Given this, we can define a map given by parallel transport:

Definition

(Parallel transport)

Let

I → M

be a curve. For

, t

, we define

the parallel transport map

: T

γ(t

)

M → T

γ(t

)

given by ξ 7→ V

It is easy to see that this is indeed a linear map, since the equations for

parallel transport are linear, and this has an inverse

given by the inverse

path. So the connection ∇ “connects” T

γ(t

)

M and T

γ(t

)

Note that this connection depends on the curve

chosen! This problem is in

general unfixable. Later, we will see that there is a special kind of connections

known as flat connections such that the parallel transport map only depends on

the homotopy class of the curve, which is an improvement.

8.3 Riemannian connections

Now suppose our manifold

has a Riemannian metric

. It is then natural to

ask if there is a “natural” connection compatible with g.

The requirement of “compatibility” in some sense says the product rule is

satisfied by the connection. Note that saying this does require the existence of a

metric, since we need one to talk about the product of two vectors.

Definition

(Metric connection)

A linear connection

∇

is compatible with

(or

is a metric connection) if for all X, Y, Z ∈ Vect(M ),

∇

g(Y, Z) = g(∇

Y, Z) + g(Y, ∇

Z).

Note that the first term is just X(g(Y, Z)).

We should view this formula as specifying that the product rule holds.

We can alternatively formulate this in terms of the covariant derivative.

Lemma.

Let

∇

be a connection. Then

∇

is compatible with

if and only if for

all γ : I → M and V, W ∈ J(γ), we have

g(V (t), W (t)) = g(D

V (t), W (t)) + g(V (t), D

W (t)). (∗)

Proof. Write it out explicitly in local coordinates.

We have some immediate corollaries, where the connection is always assumed

to be compatible.

Corollary.

V, W

are parallel along

, then

(

)

, W

(

)) is constant with

respect to t.

Corollary. If γ is a geodesic, then | ˙γ| is constant.

Corollary. Parallel transport is an isometry.

In general, on a Riemannian manifold, there can be many metric conditions.

To ensure that it is actually unique, we need to introduce a new constraint,

known as the torsion. The definition itself is pretty confusing, but we will chat

about it afterwards to explain why this is a useful definition.

Definition

(Torsion of linear connection)

Let

∇

be a linear connection on

The torsion of ∇ is defined by

τ(X, Y ) = ∇

Y − ∇

X − [X, Y ]

for X, Y ∈ Vect(M).

Definition

(Symmetric/torsion free connection)

A linear connection is sym-

metric or torsion-free if τ (X, Y ) = 0 for all X, Y .

Proposition. τ is a tensor of type (2, 1).

Proof. We have

τ(fX, Y ) = ∇

Y − ∇

(fX) − [fX, Y ]

= f∇

Y − Y (f)X − f ∇

X − f XY + Y (fX)

= f(∇

Y − ∇

X − [X, Y ])

= fτ(X, Y ).

So it is linear.

We also have τ (X, Y ) = −τ (Y, X) by inspection.

What does being symmetric tell us? Consider the Christoffel symbols in

some coordinate system x

, · · · , x

. We then have



∂

∂x

∂

∂x



= 0.

So we have



∂

∂x

∂

∂x



= ∇

∂

− ∇

∂

= Γ

∂

− Γ

∂

So we know a connection is symmetric iff the Christoffel symbol is symmetric,

i.e.

= Γ

Now the theorem is this:

Theorem.

Let

be a manifold with Riemannian metric

. Then there exists

a unique torsion-free linear connection ∇ compatible with g.

The actual proof is unenlightening.

Proof. In local coordinates, we write

g =

⊗ dx

Then the connection is explicitly given by

(∂

+ ∂

− ∂

where g

is the inverse of g

We then check that it works.

Definition

(Riemannian/Levi-Civita connection)

The unique torsion-free met-

ric connection on a Riemannian manifold is called the Riemannian connection

or Levi-Civita connection.

Example.

Consider the really boring manifold

with the usual metric. We

also know that T R

∼

→ R

is trivial, so we can give a trivial connection



∂

∂x



= df ⊗

∂

∂x

In the ∇ notation, we have

∇



∂

∂x



= X(f)

∂

∂x

It is easy to see that this is a connection, and also that it is compatible with the

metric. So this is the Riemannian connection on R

This is not too exciting.

Example.

Suppose

M ⊆ R

is an embedded submanifold. This gives us a

Riemannian metric on M by pulling back

g = φ

∗

on M .

We also get a connection on

as follows: suppose

X, Y ∈ Vect

(

). Locally,

we know X, Y extend to vector fields

Y on R

. We set

∇

Y = π(

∇

Y ),

where π is the orthogonal projection T

) → T

It is an exercise to check that this is a torsion-free metric connection on M.

It is a (difficult) theorem by Nash that every manifold can be embedded in

such that the metric is the induced metric. So all metrics arise this way.

8.4 Curvature

The final topic to discuss is the curvature of a connection. We all know that

is flat, while S

is curved. How do we make this precise?

We can consider doing some parallel transports on

along the loop coun-

terclockwise:

We see that after the parallel transport around the loop, we get a different vector.

It is in this sense that the connection on S

is not flat.

Thus, what we want is the following notion:

Definition

(Parallel vector field)

We say a vector field

V ∈ Vect

(

) is parallel

if V is parallel along any curve in M.

Example.

, we can pick

ξ ∈ T

∼

. Then setting

(

) =

ξ ∈ T

∼

gives a parallel vector field with V (0) = ξ.

However, we cannot find a non-trivial parallel vector field on S

This motivates the question — given a manifold

and a

ξ ∈ T

non-zero,

does there exist a parallel vector field

on some neighbourhood of

with

V (p) = ξ?

Naively, we would try to construct it as follows. Say

dim M

= 2 with

coordinates

x, y

. Put

= (0

0). Then we can transport

along the line

= 0

}

to define

(

0). Then for each

, we parallel transport

(

α,

0) along

α}

So this determines V (x, y).

Now if we want this to work, then

has to be parallel along any curve, and

in particular for lines

β}

for

β 6

= 0. If we stare at it long enough, we figure

out a necessary condition is

∇

∂

∂x

∇

∂

∂x

= ∇

∂

∂x

∇

∂

∂x

So the failure of these to commute tells us the curvature. This definition in fact

works for any vector bundle.

The actual definition we will state will be slightly funny, but we will soon

show afterwards that this is what we think it is.

Definition

(Curvature)

The curvature of a connection d

: Ω

(

)

→

Ω

(

) is

the map

= d

◦ d

: Ω

(E) → Ω

(E).

Lemma. F

is a tensor. In particular, F

∈ Ω

(End(E)).

Proof.

We have to show that

is linear over

∞

(

). We let

f ∈ C

∞

(

) and

s ∈ Ω

(E). Then we have

(fs) = d

(fs)

= d

(df ⊗ s + f d

= d

f ⊗ s − df ∧ d

s + df ∧ d

s + fd

= fF

(s)

How do we think about this? Given X, Y ∈ Vect(M), consider

(X, Y ) : Ω

(E) → Ω

(E)

(X, Y )(s) = (F

(s))(X, Y )

Lemma. We have

(X, Y )(s) = ∇

∇

s − ∇

∇

s − ∇

[X,Y ]

In other words, we have

(X, Y ) = [∇

, ∇

] − ∇

[X,Y ]

This is what we were talking about, except that we have an extra term

∇

[X,Y ]

which vanishes in our previous motivating case, since

∂

∂x

and

∂

∂x

commute in

general.

Proof. We claim that if µ ∈ Ω

(E), then we have

µ)(X, Y ) = ∇

(µ(Y )) − ∇

(µ(X)) − µ([X, Y ]).

To see this, we let µ = ω ⊗ s, where ω ∈ Ω

(M) and s ∈ Ω

(E). Then we have

µ = dω ⊗ s − ω ∧ d

So we know

µ)(X, Y ) = dω(X, Y ) ⊗ s − (ω ∧ d

s)(X, Y )

By a result in the example sheet, this is equal to

= (Xω(Y ) − Y ω(X) − ω([X, Y ])) ⊗ s

− ω(X)∇

(s) + ω(Y )∇

(s)

= Xω(Y ) ⊗ s + ω(Y )∇

− (Y ω(X) ⊗ s + ω(X)∇

s) − ω([X, Y ]) ⊗ s

Then the claim follows, since

µ([X, Y ]) = ω([X, Y ]) ⊗ s

∇

(µ(Y )) = ∇

(ω(Y )s)

= Xω(Y ) ⊗ s + ω(Y )∇

Now to prove the lemma, we have

s)(X, Y ) = d

s)(X, Y )

= ∇

((d

s)(Y )) − ∇

((d

s)(X)) − (d

s)([X, Y ])

= ∇

∇

s − ∇

∇

s − ∇

[X,Y ]

Definition (Flat connection). A connection d

is flat if F

= 0.

Specializing to the Riemannian case, we have

Definition

(Curvature of metric)

Let (

M, g

) be a Riemannian manifold with

metric

. The curvature of

is the curvature of the Levi-Civita connection,

denoted by

∈ Ω

(End(T M)) = Ω

(Λ

∗

M ⊗ T M ⊗ T

∗

M).

Definition (Flat metric). A Riemannian manifold (M, g) is flat if F

= 0.

Since flatness is a local property, it is clear that if a manifold is locally

isometric to

, then it is flat. What we want to prove is the converse — if

you are flat, then you are locally isometric to

. For completeness, let’s define

what an isometry is.

Definition

(Isometry)

Let (

M, g

) and (

N, g

) be Riemannian manifolds. We

say G ∈ C

∞

(M, N) is an isometry if G is a diffeomorphism and G

∗

= g, i.e.

DG|

: T

M → T

G(p)

is an isometry for all p ∈ M.

Definition

(Locally isometric)

A manifold

is locally isometric to

if for

all

p ∈ M

, there is a neighbourhood

and a

V ⊆ N

and an isometry

G : U → V .

Example.

The flat torus obtained by the quotient of

is locally isometric

to R

, but is not diffeomorphic (since it is not even homeomorphic).

Our goal is to prove the following result.

Theorem.

Let

be a manifold with Riemannian metric

.Then

is flat iff

it is locally isometric to R

One direction is obvious. Since flatness is a local property, we know that if

M is locally isometric to R

, then it is flat.

To prove the remaining of the theorem, we need some preparation.

Proposition.

Let

dim M

and

U ⊆ M

open. Let

, · · · , V

∈ Vect

(

) be

such that

(i)

For all

p ∈ U

, we know

(

)

, · · · , V

(

) is a basis for

, i.e. the

are

a frame.

(ii) [V

, V

] = 0, i.e. the V

form a frame that pairwise commutes.

Then for all

p ∈ U

, there exists coordinates

, · · · , x

on a chart

p ∈ U

such

that

∂

∂x

Suppose that

is a Riemannian metric on

and the

are orthonormal in

M. Then the map defined above is an isometry.

Proof.

We fix

p ∈ U

. Let Θ

be the flow of

. From example sheet 2, we know

that since the Lie brackets vanish, the Θ

commute.

Recall that (Θ

)

(

) =

(

), where

is the maximal integral curve of

through q. Consider

α(t

, · · · , t

) = (Θ

)

◦ (Θ

n−1

)

n−1

◦ · · · ◦ (Θ

)

(p).

Now since each of Θ

is defined on some small neighbourhood of

, so if we just

move a bit in each direction, we know that

will be defined for (

, · · · , t

)

∈

B = {|t

| < ε} for some small ε.

Our next claim is that

Dα



∂

∂t



= V

whenever this is defined. Indeed, for t ∈ B and f ∈ C

∞

(M, R). Then we have

Dα



∂

∂t





(f) =

∂

∂t



f(α(t

, · · · , t

))

∂

∂t



f((Θ

)

◦ (Θ

)

◦ · · · ◦

(Θ

)

◦ · · · ◦ (Θ

)

(p))

= V

α(t)

(f).

So done. In particular, we have

Dα|



∂

∂t





= V

(p),

and this is a basis for

. So D

α|

→ T

is an isomorphism. By the

inverse function theorem, this is a local diffeomorphism, and in this chart, the

claim tells us that

∂

∂x

The second part with a Riemannian metric is clear.

We now actually prove the theorem

Proof of theorem.

Let (

M, g

) be a flat manifold. We fix

p ∈ M

. We let

, · · · , x

be coordinates centered at

, say defined for

| <

1. We need to

construct orthonormal vector fields. To do this, we pick an orthonormal basis at

a point, and parallel transport it around.

We let

, · · · , e

be an orthonormal basis for

. We construct vector

fields

, · · · , E

∈ Vect

(

) by parallel transport. We first parallel transport

along (

, · · · ,

0) which defines

(

, · · · ,

0), then parallel transport along

the

direction to cover all

(

, x

, · · · ,

0) etc, until we define on all of

By construction, we have

∇

= 0 (∗)

on {x

k+1

= · · · = x

= 0}.

We will show that the

}

are orthonormal and [

, E

] = 0 for all

i, j

. We

claim that each E

is parallel, i.e. for any curve γ, we have

= 0.

It is sufficient to prove that

∇

= 0

for all i, j.

By induction on k, we show

∇

= 0

for

j ≤ k

k+1

· · ·

= 0

}

. The statement for

= 1 is already given

by (∗). We assume the statement for k, so

∇

= 0 (A)

for

j ≤ k

and

k+1

· · ·

= 0

}

. For

+ 1, we know that

∇

k+1

= 0

k+2

· · ·

= 0

}

by (

∗

). So the only problem we have is for

and

k+2

= · · · = x

= 0}.

By flatness of the Levi-Civita connection, we have

[∇

k+1

, ∇

] = ∇

[∂

k+1

,∂

]

= 0.

So we know

∇

k+1

∇

= ∇

∇

k+1

= 0 (B)

k+2

· · ·

= 0

}

. Now at

k+1

= 0 , we know

∇

vanishes. So it

follows from parallel transport that ∇

vanishes on {x

k+2

= · · · = x

= 0}.

As the Levi-Civita connection is compatible with

, we know that parallel

transport is an isometry. So the inner product product

(

, E

) =

(

, e

) =

So this gives an orthonormal frame at all points.

Finally, since the torsion vanishes, we know

, E

] = ∇

− ∇

= 0,

as the E

are parallel. So we are done by the proposition.

What does the curvature mean when it is non-zero? There are many answers

to this, and we will only give one.

Definition

(Holonomy)

Consider a piecewise smooth curve

: [0

→ M

with

(0) =

(1) =

. Say we have a linear connection

∇

. Then we have a notion of

parallel transport along γ.

The holonomy of ∇ around γ: is the map

H : T

M → T

given by

H(ξ) = V (1),

where V is the parallel transport of ξ along γ.

Example.

∇

is compatible with a Riemannian metric

, then

is an isometry.

Example.

Consider

with the usual connection. Then if

ξ ∈ T

, then

H(ξ) = ξ for any such path γ. So the holonomy is trivial.

Example.

Say (

M, g

) is flat, and

p ∈ M

. We have seen that there exists a

neighbourhood of

such that (

U, g|

) is isometric to

. So if

([0

1])

∈ U

then H = id.

The curvature measures the extent to which this does not happen. Suppose

we have coordinates x

, · · · , x

on some (M, g). Consider γ as follows:

(s, 0, · · · , 0)

(s, t, · · · , 0)(0, t, · · · , 0)

Then we can Taylor expand to find

H = id +F

∂

∂x



∂

∂x



st + O(s

t, st