6Integration
III Differential Geometry
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
H
n
= {(x
1
, · · · , x
n
) ∈ R
n
: x
n
≥ 0}.
A chart-with-boundary on a set
M
is a bijection
ϕ
:
U → ϕ
(
U
) for some
U ⊆ M
such that
ϕ
(
U
)
⊆ H
n
is open. Note that this image may or may not hit the
boundary of H
n
. So a “normal” chart is also a chart with boundary.
An atlas-with-boundary on
M
is a cover by charts-with-boundary (
U
α
, ϕ
α
)
such that the transition maps
ϕ
β
◦ ϕ
−1
α
: ϕ
α
(U
α
∩ U
β
) → ϕ
β
(U
α
∩ U
β
)
are smooth (in the usual sense) for all α, β.
A manifold-with-boundary is a set
M
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)
.
If
M
is a manifold with boundary and
p ∈ M
,
then we say
p
is a boundary point if
ϕ
(
p
)
∈ ∂H
n
for some (hence any) chart-
with-boundary (
U, ϕ
) containing
p
. 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
M
be a manifold with boundary. Then
Int
(
M
) and
∂M
are
naturally manifolds, with
dim ∂M = dim Int M − 1.
Example.
The solid ball
B
1
(0)
is a manifold with boundary, whose interior is
B
1
(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
,
1]
2
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
1
p
, · · · ,
∂
∂x
n
p
is a basis for T
p
M. In particular, dim T
p
M = n.
Proof.
Since this is a local thing, it suffices to prove it for
M
=
H
n
. We write
C
∞
(
H, R
) for the functions
f
:
H
n
→ R
n
that extend smoothly to an open
neighbourhood of H
n
. We fix a ∈ ∂H
n
. Then by definition, we have
T
a
H
n
= Der
a
(C
∞
(H
n
, R)).
We let i
∗
: T
a
H
n
→ T
a
R
n
be given by
i
∗
(X)(g) = X(g|
H
n
)
We claim that
i
∗
is an isomorphism. For injectivity, suppose
i
∗
(
X
) = 0. If
f ∈ C
∞
(
H
n
), then
f
extends to a smooth
g
on some neighbourhood
U
of
H
n
.
Then
X(f) = X(g|
H
n
) = i
∗
(X)(g) = 0.
So X(f) = 0 for all f. Then X = 0. So i
∗
is injective.
To see surjectivity, let Y ∈ T
a
R
n
, and let X ∈ T
a
H
n
be defined by
X(f) = Y (g),
where
g ∈ C
∞
(
H
n
, R
) is any extension of
f
to
U
. To see this is well-defined, we
let
Y =
n
X
i=1
α
i
∂
∂x
i
a
.
Then
Y (g) =
n
X
i=1
α
i
∂g
∂x
i
(a),
which only depends on
g|
H
n
, i.e.
f
. So
X
is a well-defined element of
T
a
H
n
, and
i
∗
(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
T
p
∂M ⊆ T
p
M. If X
p
∈ T
p
M, then in a chart, we can write
X
p
=
n
X
i=1
a
i
∂
∂x
i
,
where
a
i
∈ R
and
∂
∂x
1
, · · · ,
∂
∂x
n−1
are a basis for
T
p
∂M
. We say
X
p
is outward
pointing if a
n
< 0, and inward pointing if a
n
> 0.
Definition
(Induced orientation)
.
Let
M
be an oriented manifold with boundary.
We say a basis
e
1
, · · · , e
n−1
is an oriented basis for
T
p
∂M
if (
X
p
, e
1
, · · · , e
n−1
)
is an oriented basis for
T
p
M
, where
X
p
is any outward pointing element in
T
p
M
.
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
∼
=
R
n−1
(x
1
, · · · , x
n−1
, 0) 7→ (x
1
, · · · , x
n−1
).
So
−
∂
∂x
n
∂H
n
is an outward pointing vector. So we know
x
1
, · · · , x
n−1
is an oriented chart for
∂H
n
iff
−
∂
∂x
n
,
∂
∂x
1
, · · · ,
∂
∂x
n−1
is oriented, which is true iff n is even.
Example.
If
n
= 1, say
M
= [
a, b
]
⊆ R
with
a < b
, then
{a, b}
, then
T
p
∂M
=
{
0
}
. So an orientation of
∂M
is a choice of numbers
±
1 attached to each point.
The convention is that if
M
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
M
be an oriented manifold with boundary of
dimension n. Then if ω ∈ Ω
n−1
(M) has compact support, then
Z
M
dω =
Z
∂M
ω.
In particular, if M has no boundary, then
Z
M
dω = 0
Note that this makes sense. d
ω
is an
n
-form on
M
, so we can integrate it.
On the right hand side, what we really doing is integrating the restriction of
ω
to
∂M
, i.e. the (
n −
1)-form
i
∗
ω
, where
i
:
∂M → M
is the inclusion, so that
i
∗
ω ∈ Ω
n−1
(∂M).
Note that if
M
= [
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
n
. Then we have
ω =
n
X
i=1
ω
i
dx
1
∧ · · · ∧
d
dx
i
∧ · · · ∧ dx
n
,
where ω
i
is compactly supported, and the hat denotes omission. So we have
dω =
X
i
dω
i
∧ dx
1
∧ · · · ∧
d
dx
i
∧ · · · ∧ dx
n
=
X
i
∂ω
i
∂x
i
dx
i
∧ dx
1
∧ · · · ∧
d
dx
i
∧ · · · ∧ dx
n
=
X
i
(−1)
i−1
∂ω
i
∂x
i
dx
1
∧ · · · ∧ dx
i
∧ · · · ∧ dx
n
Let’s say
supp(ω) = {x
j
∈ [−R, R] : j = 1, · · · , n − 1; x
n
∈ [0, R]} = A.
Then suppose i 6= n. Then we have
Z
H
n
∂ω
i
∂x
i
dx
1
∧ · · · ∧ dx
i
∧ · · · ∧ dx
n
=
Z
A
∂ω
i
∂x
i
dx
1
· · · dx
n
=
Z
R
−R
Z
R
−R
· · ·
Z
R
−R
Z
R
0
∂ω
i
∂x
i
dx
1
· · · dx
n
By Fubini’s theorem, we can integrate this in any order. We integrate with
respect to dx
i
first. So this is
= ±
Z
R
−R
· · ·
Z
R
−R
Z
R
0
Z
R
−R
∂ω
i
∂x
i
dx
i
!
dx
1
· · ·
d
dx
i
· · · dx
n
By the fundamental theorem of calculus, the inner integral is
ω(x
1
, · · · , x
i−1
, R, x
i+1
, · · · , x
n
)−ω(x
1
, · · · , x
i−1
, −R, x
i+1
, · · · , x
n
) = 0−0 = 0.
So the integral vanishes. So we are only left with the i = n term. So we have
Z
H
n
dω = (−1)
n−1
Z
A
∂ω
n
∂x
n
dx
1
· · · dx
n
= (−1)
n−1
Z
R
−R
· · ·
Z
R
−R
Z
R
0
∂ω
n
∂x
n
dx
n
!
dx
1
· · · dx
n−1
Now that integral is just
ω
n
(x
1
, · · · , x
n−1
, R) − ω
n
(x
1
, · · · , x
n−1
, 0) = −ω
n
(x
1
, · · · , x
n−1
, 0).
So this becomes
= (−1)
n
Z
R
−R
· · ·
Z
R
−R
ω
n
(x
1
, · · · , x
n−1
, 0) dx
1
· · · dx
n−1
.
Next we see that
i
∗
ω = ω
n
dx
1
∧ · · · ∧ dx
n−1
,
as i
∗
(dx
n
) = 0. So we have
Z
∂H
n
i
∗
ω = ±
Z
A∩∂H
n
ω(x
1
, · · · , x
n−1
, 0) dx
1
· · · dx
n
.
Here the sign is a plus iff
x
1
, · · · , x
n−1
are an oriented coordinate for
∂H
n
, i.e.
n is even. So this is
Z
∂H
n
ω = (−1)
n
Z
R
−R
· · ·
Z
R
−R
ω
n
(x
1
, · · · , x
n−1
, 0) dx
1
· · · dx
n−1
=
Z
H
n
dω.
Now for a general manifold
M
, suppose first that
ω ∈
Ω
n−1
(
M
) is compactly
supported in a single oriented chart (
U, ϕ
). Then the result is true by working
in local coordinates. More explicitly, we have
Z
M
dω =
Z
H
n
(ϕ
−1
)
∗
dω =
Z
H
n
d((ϕ
−1
)
∗
ω) =
Z
∂H
n
(ϕ
−1
)
∗
ω =
Z
∂M
ω.
Finally, for a general
ω
, we just cover
M
by oriented charts (
U, ϕ
α
), and use a
partition of unity χ
α
subordinate to {U
α
}. So we have
ω =
X
χ
α
ω.
Then
dω =
X
(dχ
α
)ω +
X
χ
α
dω = d
X
χ
α
ω +
X
χ
α
dω =
X
χ
α
dω,
using the fact that
P
χ
α
is constant, hence its derivative vanishes. So we have
Z
M
dω =
X
α
Z
M
χ
α
dω =
X
α
Z
∂M
χ
α
ω =
Z
∂M
ω.
Then all the things likes Green’s theorem and divergence theorem follow from
this.
Example.
Let
M
be a manifold without boundary with a symplectic form
ω ∈
Ω
2
(
M
) that is closed and positive definite. Then by basic Linear algebra we
know
Z
M
ω
n
6= 0.
Since
ω
is closed, it is an element [
ω
]
∈ H
2
dR
(
M
). Does this vanish? If
ω
= d
τ
,
then we have
d(τ ∧ ω ∧ · · · ∧ ω) = ω
n
.
So we have
Z
M
ω
n
=
Z
M
d(τ ∧ ω ∧ · · · ∧ ω) = 0
by Stokes’ theorem. This is a contradiction. So [ω] is non-zero in H
2
dR
(M).