5Differential forms and de Rham cohomology
III Differential Geometry
5.3 Homological algebra and Mayer-Vietoris theorem
The main theorem we will have for computing de Rham cohomology will be the
Mayer-Vietoris theorem. Proving this involves quite a lot of setting up and hard
work. In particular, we need to define some notions from homological algebra to
even state Mayer-Vietoris theorem.
The actual proof will be divided into two parts. The first part is a purely
algebraic result known as the snake lemma, and the second part is a differential-
geometric part that proves that we satisfy the hypothesis of the snake lemma.
We will not prove the snake lemma, whose proof can be found in standard
algebraic topology texts (perhaps with arrows the wrong way round).
We start with some definitions.
Definition
(Cochain complex and exact sequence)
.
A sequence of vector spaces
and linear maps
· · · V
p−1
V
p
V
p+1
· · ·
d
p−1
d
p
is a cochain complex if d
p
◦
d
p−1
= 0 for all
p ∈ Z
. Usually we have
V
p
= 0 for
p <
0 and we do not write them out. Keeping these negative degree
V
p
rather
than throwing them away completely helps us state our theorems more nicely,
so that we don’t have to view
V
0
as a special case when we state our theorems.
It is exact at p if ker d
p
= im d
p−1
, and exact if it is exact at every p.
There are, of course, chain complexes as well, but we will not need them for
this course.
Example. The de Rham complex
Ω
0
(M) Ω
1
(M) Ω
2
(M) · · ·
d d
is a cochain complex as d
2
= 0. It is exact at p iff H
p
dR
(M) = {0}.
Example.
If we have an exact sequence such that
dim V
p
< ∞
for all
p
and
are zero for all but finitely many p, then
X
p
(−1)
p
dim V
p
= 0.
Definition (Cohomology). Let
V
·
= · · · V
p−1
V
p
V
p+1
· · ·
d
p−1
d
p
be a cochain complex. The cohomology of V
·
at p is given by
H
p
(V
·
) =
ker d
p
im d
p−1
.
Example.
The cohomology of the de Rham complex is the de Rham cohomology.
We can define maps between cochain complexes:
Definition
(Cochain map)
.
Let
V
·
and
W
·
be cochain complexes. A cochain
map
V
·
→ W
·
is a collection of maps
f
p
:
V
p
→ W
p
such that the following
diagram commutes for all p:
V
p
W
p
V
p+1
W
p+1
f
p
d
p
d
p
f
p+1
Proposition.
A cochain map induces a well-defined homomorphism on the
cohomology groups.
Definition
(Short exact sequence)
.
A short exact sequence is an exact sequence
of the form
0 V
1
V
2
V
3
0
α
β
.
This implies that
α
is injective,
β
is surjective, and
im
(
α
) =
ker
(
β
). By the
rank-nullity theorem, we know
dim V
2
= rank(β) + null(β) = dim V
3
+ dim V
1
.
We can now state the main technical lemma, which we shall not prove.
Theorem
(Snake lemma)
.
Suppose we have a short exact sequence of complexes
0 A
·
B
·
C
·
0
i
q
,
i.e. the i, q are cochain maps and we have a short exact sequence
0 A
p
B
p
C
p
0
i
p
q
p
,
for each p.
Then there are maps
δ : H
p
(C
·
) → H
p+1
(A
·
)
such that there is a long exact sequence
· · · H
p
(A
·
) H
p
(B
·
) H
p
(C
·
)
H
p+1
(A
·
) H
p+1
(B
·
) H
p+1
(C
·
) · · ·
i
∗
q
∗
δ
i
∗
q
∗
.
Using this, we can prove the Mayer-Vietoris theorem.
Theorem
(Mayer-Vietoris theorem)
.
Let
M
be a manifold, and
M
=
U ∪ V
,
where U, V are open. We denote the inclusion maps as follows:
U ∩ V U
V M
i
1
i
2
j
1
j
2
Then there exists a natural linear map
δ : H
p
dR
(U ∩ V ) → H
p+1
dR
(M)
such that the following sequence is exact:
H
p
dR
(M) H
p
dR
(U) ⊕ H
p
dR
(V ) H
p
dR
(U ∩ V )
H
p+1
dR
(M) H
p+1
dR
(U) ⊕ H
p+1
dR
(V ) · · ·
j
∗
1
⊕j
∗
2
i
∗
1
−i
∗
2
δ
j
∗
1
⊕j
∗
2
i
∗
1
−i
∗
2
Before we prove the theorem, we do a simple example.
Example. Consider M = S
1
. We can cut the circle up:
U
V
Here we have
S
1
= {(x, y) : x
2
+ y
2
= 1}
U = S
1
∩ {y > −ε}
V = S
1
∩ {y < ε}.
As
U, V
are diffeomorphic to intervals, hence contractible, and
U ∩V
is diffeomor-
phic to the disjoint union of two intervals, we know their de Rham cohomology.
0 H
0
dR
(S
1
) H
0
dR
(U) ⊕ H
0
dR
(V ) H
0
dR
(U ∩ V )
H
1
dR
(S
1
) H
1
dR
(U) ⊕ H
1
dR
(V ) · · ·
We can fill in the things we already know to get
0 R R ⊕ R R ⊕ R
H
1
dR
(S
1
) 0 · · ·
By adding the degrees alternatingly, we know that
dim H
1
dR
(S
1
) = 1.
So
H
1
dR
(S
1
)
∼
=
R.
Now we prove Mayer-Vietoris.
Proof of Mayer-Vietoris.
By the snake lemma, it suffices to prove that the
following sequence is exact for all p:
0 Ω
p
(U ∪ V ) Ω
p
(U) ⊕ Ω
p
(V ) Ω
p
(U ∩ V ) 0
j
∗
1
⊕j
∗
2
i
∗
1
−i
∗
2
It is clear that the two maps compose to 0, and the first map is injective. By
counting dimensions, it suffices to show that i
∗
1
− i
∗
2
is surjective.
Indeed, let
{ϕ
U
, ϕ
V
}
be partitions of unity subordinate to
{U, V }
. Let
ω ∈ Ω
p
(U ∩ V ). We set σ
U
∈ Ω
p
(U) to be
σ
U
=
(
ϕ
V
ω on U ∩ V
0 on U \ supp ϕ
V
.
Similarly, we define σ
V
∈ Ω
p
(V ) by
σ
V
=
(
−ϕ
U
ω on U ∩ V
0 on V \ supp ϕ
U
.
Then we have
i
∗
1
σ
U
− i
∗
2
σ
V
= (ϕ
V
ω + ϕ
U
ω)|
U∩V
= ω.
So i
∗
1
− i
∗
2
is surjective.