Our goal is now to describe an isomorphism
H
p
dR
(
M
)
∼
=
H
p
∞
(
M,
R
)
.
The map itself is given as follows:
Suppose [
w
]
∈
H
p
dR
(
M
), so
ω
∈
Ω
p
(
M
) with d
ω
= 0. Suppose that
σ
: Δ
p
→
M
is smooth with
∂σ
= 0. We can then define
I
([
ω
]) =
Z
Δ
p
σ
*
ω
∈
R
.
Note that we have not defined what
R
Δ
p
means, because Δ
p
is not even a
manifold with boundary — it has corners. We can develop an analogous theory
68
7
De Rham’s theorem*
III Differential Geometry
of integration on manifolds with corners, but we can also be lazy, and just
integrate over
Δ
×
p
= Δ
p
\ {
codimension 2 faces
}
.
Now
ω

Δ
*
p
does not have compact support, but has the property that it is the
restriction of a (unique)
p
form on Δ
p
, so in particular it is bounded. So the
integral is finite.
Now in general, if
τ
=
∑
a
i
σ
i
∈
C
p
(
M
), we define
I
([
ω
])(
τ
) =
X
a
i
Z
Δ
p
σ
*
i
ω
∈
R
.
Now
Stokes theorem
tell us
Z
∂σ
ω
=
Z
σ
d
ω.
So we have
Lemma.
I
is a welldefined map
H
p
dR
(
M
)
→
H
p
∞
(
M,
R
).
Proof.
If [
ω
] = [
ω
0
], then
ω

ω
0
= d
α
. Then let
σ
∈
H
p
∞
(
M,
R
). Then
Z
σ
(
ω

ω
0
) =
Z
σ
d
α
=
Z
∂σ
α
= 0
,
since
∂σ
= 0.
On the other hand, if [
σ
] = [
σ
0
], then
σ

σ
=
∂β
for some
β
. Then we have
Z
σ

σ
0
ω
=
Z
∂β
ω
=
Z
β
d
ω
= 0
.
So this is welldefined.
Lemma.
I
is functorial and commutes with the boundary map of MayerVietoris.
In other words, if
F
:
M
→
N
is smooth, then the diagram
H
p
dR
(
M
)
H
p
dR
(
N
)
H
p
∞
(
M
)
H
p
∞
(
N
)
F
*
I
I
F
*
.
And if
M
=
U
∪
V
and
U, V
are open, then the diagram
H
p
dR
(
U
∩
V
)
H
p
+1
dR
(
U
∪
V
)
H
p
∞
(
U
∩
V,
R
)
H
p
(
U
∪
V,
R
)
δ
I
I
δ
also commutes.
Note that the other parts of the MayerVietoris sequence
commute because they are induced by maps of manifolds.
Proof.
Trace through the definitions.
69
7
De Rham’s theorem*
III Differential Geometry
Proposition.
Let
U
⊆
R
n
is convex, then
U
:
H
p
dR
(
U
)
→
H
p
∞
(
U,
R
)
is an isomorphism for all
p
.
Proof.
If
p >
0, then both sides vanish. Otherwise, we check manually that
I
:
H
0
dR
(
U
)
→
H
0
∞
(
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
R
n
, and then use MayerVietoris 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
{
U
α
}
of
M
is
de Rham
if
U
α
1
∩ · · · ∩
U
α
p
is de Rham
for all
α
1
,
· · ·
, α
p
.
(iii)
A
de Rham basis
is a de Rham cover that is a basis for the topology on
M
.
Our plan is to inductively show that everything is indeed de Rham.
We have already seen that if
U
⊆
R
n
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 MayerVietoris sequence that is
impossible to fit within the margins:
H
p
dR
(
U
)
⊕
H
p
dR
(
V
)
H
p
dR
(
U
∪
V
)
H
p
+1
dR
(
U
∩
V
)
H
p
dR
(
U
)
⊕
H
p
+1
dR
(
V
)
H
p
+1
dR
(
U
∪
V
)
H
p
∞
(
U
)
⊕
H
p
∞
(
V
)
H
p
∞
(
U
∪
V
)
H
p
+1
∞
(
U
∩
V
)
H
p
∞
(
U
)
⊕
H
p
+1
∞
(
V
)
H
p
+1
∞
(
U
∪
V
)
I
⊕
I
I
I
I
⊕
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.
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 Fall '19
 Derivative, Manifold