5Differential forms and de Rham cohomology
III Differential Geometry
5.2 De Rham cohomology
We now get to answer our original motivating question — given an ω ∈ Ω
p
(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
H
p
dR
(
M
) for each
p
, such that this vanishes if and
only if all
p
forms
ω
with d
ω
= 0 are of the form d
θ
. Afterwards, we will come
up with techniques to compute this
H
p
dR
(
M
), and then we can show that certain
spaces have vanishing H
p
dR
(M).
We start with some definitions.
Definition (Closed form). A p-form ω ∈ Ω
p
(M) is closed if dω = 0.
Definition
(Exact form)
.
A
p
-form
ω ∈
Ω
p
(
M
) 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
p
th de Rham cohomology is given by
the R-vector space
H
p
dR
(M) =
ker d : Ω
p
(M) → Ω
p+1
(M)
im d : Ω
p−1
(M) → Ω
p
(M)
=
closed forms
exact forms
.
In particular, we have
H
0
dR
(M) = ker d : Ω
0
(M) → Ω
1
(M).
We could tautologically say that if d
ω
= 0, then
ω
is exact iff it vanishes
in
H
p
dR
(
M
). But this is as useful as saying “Let
S
be the set of solutions to
this differential equation. Then the differential equation has a solution iff
S
is non-empty”. So we want to study the properties of
H
p
dR
and find ways of
computing them.
Proposition.
(i) Let M have k connected components. Then
H
0
dR
(M) = R
k
.
(ii) If p > dim M, then H
p
dR
(M) = 0.
(iii)
If
F ∈ C
∞
(
M, N
), then this induces a map
F
∗
:
H
p
dR
(
N
)
→ H
p
dR
(
M
)
given by
F
∗
[ω] = [F
∗
ω].
(iv) (F ◦ G)
∗
= G
∗
◦ F
∗
.
(v)
If
F
:
M → N
is a diffeomorphism, then
F
∗
:
H
p
dR
(
N
)
→ H
p
dR
(
M
) is an
isomorphism.
Proof.
(i) We have
H
0
dR
(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
F
∗
ω
indeed represents some member of
H
p
dR
(
M
). Let
[ω] ∈ H
p
dR
(N). Then dω = 0. So
d(F
∗
ω) = F
∗
(dω) = 0.
So [F
∗
ω] ∈ H
p
dR
(M). So this map makes sense.
To see it is well-defined, if [
ω
] = [
ω
0
], then
ω − ω
0
= d
σ
for some
σ
. So
F
∗
ω − F
∗
ω
0
= d(F
∗
σ). So [F
∗
ω] = [F
∗
ω
0
].
(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
0
, F
1
:
M → N
be smooth maps. A
smooth homotopy from
F
0
to
F
1
is a smooth map
F
: [0
,
1]
× M → N
such that
F
0
(x) = F (0, x), F
1
(x) = F (1, x).
If such a map exists, we say F
0
and F
1
are homotopic.
Note that here
F
is defined on [0
,
1]
× M
, which is not a manifold. So we
need to be slightly annoying and say that
F
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
0
, F
1
be homotopic maps. Then
F
∗
0
=
F
∗
1
: H
p
dR
(N) → H
p
dR
(M).
Proof. Let F : [0, 1] × M → N be the homotopy, and
F
t
(x) = F (t, x).
We denote the exterior derivative on
M
by d
M
(and similarly d
N
), and that on
[0, 1] × M by d.
Let
ω ∈
Ω
p
(
N
) be such that d
N
ω
= 0. We let
t
be the coordinate on [0
,
1].
We write
F
∗
ω = σ + dt ∧ γ,
where σ = σ(t) ∈ Ω
p
(M) and γ = γ(t) ∈ Ω
p−1
(M). We claim that
σ(t) = F
∗
t
ω.
Indeed, we let ι : {t} × M → [0, 1] × M be the inclusion. Then we have
F
∗
t
ω|
{t}×M
= (F ◦ ι)
∗
ω = ι
∗
F
∗
ω
= ι
∗
(σ + dt ∧ γ)
= ι
∗
σ + ι
∗
dt ∧ ι
∗
γ
= ι
∗
σ,
using the fact that ι
∗
dt = 0. As d
N
ω = 0, we have
0 = F
∗
d
N
ω
= dF
∗
ω
= d(σ + dt ∧ γ)
= d
M
(σ) + (−1)
p
∂σ
∂t
∧ dt + dt ∧ d
M
γ
= d
M
σ + (−1)
p
∂σ
∂t
∧ dt + (−1)
p−1
d
M
γ ∧ dt.
Looking at the dt components, we have
∂σ
∂t
= d
M
γ.
So we have
F
∗
1
ω − F
∗
0
ω = σ(1) − σ(0) =
Z
1
0
∂σ
∂t
dt =
Z
1
0
d
M
γ dt = d
M
Z
1
0
γ(t) dt.
So we know that
[F
∗
1
ω] = [F
∗
0
ω].
So done.
Example.
Suppose
U ⊆ R
n
is an open “star-shaped” subset, i.e. there is some
x
0
∈ U such that for any x ∈ U and t ∈ [0, 1], we have
tx + (1 − t)x
0
∈ U.
x
0
x
We define F
t
: U → U by
F
t
(x) = tx + (1 − t)x
0
.
Then
F
is a smooth homotopy from the identity map to
F
0
, the constant map
to
x
0
. We clearly have
F
∗
1
being the identity map, and
F
∗
0
is the zero map on
H
p
dR
(U) for all p ≥ 1. So we have
H
p
dR
(U) =
(
0 p ≥ 1
R p = 0
.
Corollary
(Poincar´e lemma)
.
Let
U ⊆ R
n
be open and star-shaped. Suppose
ω ∈
Ω
p
(
U
) is such that d
ω
= 0. Then there is some
σ ∈
Ω
p−1
(
M
) such that
ω = dσ.
Proof. H
p
dR
(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
F
:
M → N
and
G : N → M such that both F ◦ G and G ◦ F are homotopic to the identity.
Corollary.
If
M
and
N
are smoothly homotopy equivalent, then
H
p
dR
(
M
)
∼
=
H
p
dR
(N).
Note that by approximation, it can be shown that if
M
and
N
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.