III Differential Geometry

6Integration

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.