7De Rham's theorem*

III Differential Geometry



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
H
p
dR
(M)
=
H
p
(M, R),
where
H
p
(
M, R
) is the singular cohomology of
M
, and this is in fact an iso-
morphism of rings, where
H
p
dR
(
M
) has the product given by the wedge, and
H
p
(M, R) has the cup product.
Recall that singular cohomology is defined as follow:
Definition
(Singular
p
-complex)
.
Let
M
be a manifold. Then a singular
p
-
simplex is a continuous map
σ : ∆
p
M,
where
p
=
(
p
X
i=0
t
i
e
i
:
X
t
I
= 1
)
R
n+1
.
We define
C
p
(M) = {formal sums
X
a
i
σ
i
: a
i
R, σ
i
a singular p simplex}.
We define
C
p
(m) = {formal sums
X
a
i
σ
i
: a
i
R, σ
i
a smooth singular p simplex}.
Definition (Boundary map). The boundary map
: C
p
(M) C
p1
(M)
is the linear map such that if σ : ∆
p
M is a p simplex, then
σ =
X
(1)
i
σ F
i,p
,
where
F
i,p
maps
p1
affine linearly to the face of
p
opposite the
i
th vertex.
We similarly have
: C
p
(M) C
p1
(M).
We can then define singular homology
Definition (Singular homology). The singular homology of M is
H
p
(M, R) =
ker : C
p
(M) C
p1
(M)
im : C
p+1
(M) C
p
(M)
.
The smooth singular homology is the same thing with
C
p
(
M
) replaced with
C
p
(M).
H
p
has the same properties as
H
p
, 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
p
(M) C
p
(M),
which obviously commutes with , giving
i
: H
p
(M) H
p
(M).
Theorem. The map i
: H
p
(M) H
p
(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
F
:
M N
between manifolds is
homotopic to a smooth one. But this is painful to prove.
(ii)
What we really care about is maps
σ
: ∆
p
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)
As
H
p
and
H
p
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
R
, we can cheat and define singular cohomology
in a simple way:
Definition
(Singular cohomology)
.
The singular cohomology of
M
is defined as
H
p
(M, R) = Hom(H
p
(M, R), R).
Similarly, the smooth singular cohomology is
H
p
(M, R) = Hom(H
p
(M, R), R).
This is a bad definition in general! It just happens to work for singular
cohomology with coefficients in
R
, and we are not too bothered to write dowm
the proper definition.
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
p
(
M
) with d
ω
= 0. Suppose that
σ
: ∆
p
M is smooth with σ = 0. We can then define
I([ω]) =
Z
p
σ
ω R.
Note that we have not defined what
R
p
means, because
p
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
×
p
= ∆
p
\ {codimension 2 faces}.
Now
ω|
p
does not have compact support, but has the property that it is the
restriction of a (unique)
p
-form on
p
, so in particular it is bounded. So the
integral is finite.
Now in general, if τ =
P
a
i
σ
i
C
p
(M), we define
I([ω])(τ) =
X
a
i
Z
p
σ
i
ω R.
Now Stokes theorem tell us
Z
σ
ω =
Z
σ
dω.
So we have
Lemma. I is a well-defined map H
p
dR
(M) H
p
(M, R).
Proof. If [ω] = [ω
0
], then ω ω
0
= dα. Then let σ H
p
(M, R). Then
Z
σ
(ω ω
0
) =
Z
σ
dα =
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
σσ
0
ω =
Z
β
ω =
Z
β
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
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 Mayer-Vietoris sequence
commute because they are induced by maps of manifolds.
Proof. Trace through the definitions.
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 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
{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 Mayer-Vietoris 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 )
II
I
I
II
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.
If
U
1
, · · · , U
k
is a finite de Rham cover of
U
1
· · · U
k
=
N
, then
M is de Rham.
Proof. By induction on k.
Proposition. The disjoint union of de Rham spaces is de Rham.
Proof. Let A
i
be de Rham. Then we have
H
p
dR
a
A
i
=
Y
H
p
dR
(A
i
)
=
Y
H
p
(A
i
)
=
H
p
a
A
i
.
Lemma. Let M be a manifold. If it has a de Rham basis, then it is de Rham.
Proof sketch.
Let
f
:
M R
be an “exhaustion function”, i.e.
f
1
([
−∞, c
]) for
all c R. This is guaranteed to exist for any manifold. We let
A
m
= {q M : f(q) [m, m + 1]}.
We let
A
0
m
=
q M : f(q)
m
1
2
, m +
3
2

.
Given any
q A
m
, there is some
U
α(q)
A
0
m
in the de Rham basis containing
q
.
As
A
m
is compact, we can cover it by a finite number of such
U
α
i
, with each
U
α
i
A
0
m
. Let
B
m
= U
α
1
· · · U
α
r
.
Since
B
m
has a finite de Rham cover, so it is de Rham. Observe that if
B
m
B
˜m
6= , then
˜
M {m, m 1, m + 1}. We let
U =
[
m even
B
m
, V =
[
m odd
B
m
.
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.
If
U R
n
is open, then it is de Rham, since there is a basis of convex
sets {U
α
} (e.g. take open balls). So they form a de Rham basis.
Finally,
M
has a basis of subsets diffeomorphic to open subsets of
R
n
. So it
is de Rham.