4Hodge theory on Riemannian manifolds

III Riemannian Geometry 4.2 Hodge decomposition theorem
We now work towards proving the Hodge decomposition theorem. This is a very
important and far-reaching result.
Theorem
(Hodge decomposition theorem)
.
Let (
M, g
) be a compact oriented
Riemannian manifold. Then
For all p = 0, ··· , dim M, we have dim H
p
< .
We have
p
(M) = H
p
∆Ω
p
(M).
Moreover, the direct sum is orthogonal with respect to the
L
2
inner product.
We also formally set
1
(M) = 0.
As before, the compactness of M is essential, and cannot be dropped.
Corollary. We have orthogonal decompositions
p
(M) = H
p
dδ
p
(M) δdΩ
p
(M)
= H
p
dΩ
p1
(M) δ
p+1
(M).
Proof. Now note that for an α, β, we have
hhdδα, δdβii
g
= hhddδα, dβii
g
= 0.
So
dδ
p
(M) δdΩ
p
(M)
is an orthogonal direct sum that clearly contains ∆Ω
p
(
M
). But each component
is also orthogonal to harmonic forms, because harmonic forms are closed and
co-closed. So the first decomposition follows.
To obtain the final decomposition, we simply note that
dΩ
p1
(M) = d(H
p1
∆Ω
p1
(M)) = d(δdΩ
p1
(M)) dδ
p
(M).
On the other hand, we certainly have the other inclusion. So the two terms are
equal. The other term follows similarly.
This theorem has a rather remarkable corollary.
Corollary.
Let (
M, g
) be a compact oriented Riemannian manifold. Then for
all
α H
p
dR
(
M
), there is a unique
α H
p
such that [
α
] =
a
. In other words,
the obvious map
H
p
H
p
dR
(M)
is an isomorphism.
This is remarkable. On the left hand side, we have
H
p
, which is a completely
analytic thing, defined by the Laplacian. On the other hand, the right hand
sides involves the de Rham cohomology, which is just a topological, and in fact
homotopy invariant.
Proof.
To see uniqueness, suppose
α
1
, α
2
H
p
are such that [
α
1
] = [
α
2
]
H
p
dR
(M). Then
α
1
α
2
= dβ
for some
β
. But the left hand side and right hand side live in different parts of
the Hodge decomposition. So they must be individually zero. Alternatively, we
can compute
kdβk
2
g
= hhdβ, α
1
α
2
ii
g
= hhβ, δα
1
δα
2
ii
g
= 0
since harmonic forms are co-closed.
To prove existence, let α
p
(M) be such that dα = 0. We write
α = α
1
+ dα
2
+ δα
3
H
p
dΩ
p1
(M) δ
p+1
(M).
Applying d gives us
0 = dα
1
+ d
2
α
2
+ dδα
3
.
We know d
α
1
= 0 since
α
1
is harmonic, and d
2
= 0. So we must have d
δα
3
= 0.
So
hhδα
3
, δα
3
ii
g
= hhα
3
, dδα
3
ii
g
= 0.
So δα
3
= 0. So [α] = [α
1
] and α has a representative in H
p
.
We can also heuristically justify why this is true. Suppose we are given some
de Rham cohomology class a H
p
dR
(M). We consider
B
a
= {ξ
p
(M) : dξ = 0, [ξ] = a}.
This is an infinite dimensional affine space.
α B
a
minimizes the
L
2
norm? We consider
the function
F
:
B
a
R
given by
F
(
α
) =
kαk
2
. Any minimizing
α
is an
extremum. So for any β
p1
(M), we have
d
dt
t=0
F (α + tdβ) = 0.
In other words, we have
0 =
d
dt
t=0
(kαk
2
+ 2thhα, dβii
g
+ t
2
kdβk
2
) = 2hhα, dβii
g
.
This is the same as saying
hhδα, βii
g
= 0.
So this implies
δα
= 0. But d
α
= 0 by assumption. So we find that
α H
p
. So
the result is at least believable.
The proof of the Hodge decomposition theorem involves some analysis, which
we are not bothered to do. Instead, we will just quote the appropriate results.
For convenience, we will use
h·, ·i
for the
L
2
inner product, and then
k · k
is
the L
2
norm.
The first theorem we quote is the following:
Theorem
(Compactness theorem)
.
If a sequence
α
n
n
(
M
) satisfies
kα
n
k <
C and kα
n
k < C for all n, then α
n
contains a Cauchy subsequence.
This is almost like saying
n
(
M
) is compact, but it isn’t, since it is not
complete. So the best thing we can say is that the subsequence is Cauchy.
Corollary. H
p
is finite-dimensional.
Proof.
Suppose not. Then by Gram–Schmidt, we can find an infinite orthonormal
sequence
e
n
such that
ke
n
k
= 1 and
k
e
n
k
= 0, and this certainly does not have
a Cauchy subsequence.
A large part of the proof is trying to solve the PDE
ω = α,
which we will need in order to carry out the decomposition. In analysis, one
useful idea is the notion of weak solutions. We notice that if
ω
is a solution,
then for any ϕ
p
(M), we have
hω, ϕi = hω, ϕi = hα, ϕi,
using that ∆ is self-adjoint. In other words, the linear form
`
=
hω, ·i
: Ω
p
(
M
)
R satisfies
`(∆ϕ) = hα, ϕi.
Conversely, if
hω, ·i
satisfies this equation, then
ω
must be a solution, since for
any β, we have
hω, βi = hω, βi = hα, βi.
Definition
(Weak solution)
.
A weak solution to the equation
ω
=
α
is a
linear functional ` : Ω
p
(M) R such that
(i) `(∆ϕ) = hα, ϕi for all ϕ
p
(M).
(ii) ` is bounded , i.e. there is some C such that |`(β)| < Ckβk for all β.
Now given a weak solution, we want to obtain a genuine solution. If
p
(
M
)
were a Hilbert space, then we are immediately done by the Riesz representation
theorem, but it isn’t. Thus, we need a theorem that gives us what we want.
Theorem
(Regularity theorem)
.
Every weak solution of
ω
=
α
is of the form
`(β) = hω, βi
for ω
p
(M).
Thus, we have reduced the problem to finding weak solutions. There is one
final piece of analysis we need to quote. The definition of a weak solution only
`
does to ∆Ω
p
(
M
). And it is easy to define what
`
should do
on ∆Ω
p
(M) we simply define
`(∆η) = hη, αi.
Of course, for this to work, it must be well-defined, but this is not necessarily
the case in general. We also have to check it is bounded. But suppose this
worked. Then the remaining job is to extend this to a bounded functional on all
of
p
(
M
) in whatever way we like. This relies on the following (relatively easy)
theorem from analysis:
Theorem
(Hahn–Banach theorem)
.
Let
L
be a normed vector space, and
L
0
be
a subspace. We let
f
:
L
0
R
be a bounded linear functional. Then
f
extends
to a bounded linear functional L R with the same bound.
We can now begin the proof.
Proof of Hodge decomposition theorem.
Since
H
p
is finite-dimensional, by basic
linear algebra, we can decompose
p
(M) = H
p
(H
p
)
.
Crucially, we know (H
p
)
is a closed subspace. What we want to show is that
(H
p
)
= ∆Ω
p
(M).
One inclusion is easy. Suppose α H
p
and β
p
(M). Then we have
hα, βi = hα, βi = 0.
So we know that
∆Ω
p
(M) (H
p
)
.
The other direction is the hard part. Suppose
α
(
H
p
)
. We may assume
α
is
non-zero. Since our PDE is a linear one, we may wlog kαk = 1.
By the regularity theorem, it suffices to prove that
ω
=
α
has a weak
solution. We define ` : ∆Ω
p
(M) R as follows: for each η
p
(M), we put
`(∆η) = hη, αi.
We check this is well-defined. Suppose
η
= ∆
ξ
. Then
η ξ H
p
, and we have
hη, αi hξ, αi = hη ξ, αi = 0
since α (H
p
)
.
We next want to show the boundedness property. We now claim that there
exists a positive C > 0 such that
`(∆η) Ckηk
for all
η
p
(
M
). To see this, we first note that by Cauchy–Schwartz, we have
|hα, ηi| kαk · kηk = kηk.
So it suffices to show that there is a C > 0 such that
kηk Ckηk
for every η
p
(M).
Suppose not. Then we can find a sequence
η
k
(
H
p
)
such that
kη
k
k
= 1
and kη
k
k 0.
But then
k
η
k
k
is certainly bounded. So by the compactness theorem, we
may wlog
η
k
is Cauchy. Then for any
ψ
p
(
M
), the sequence
hψ, η
k
i
is Cauchy,
by Cauchy–Schwartz, hence convergent.
We define a : Ω
p
(M) R by
a(ψ) = lim
k→∞
hψ, η
k
i.
Then we have
a(∆ψ) = lim
k→∞
hη
k
, ψi = lim
k→∞
hη
k
, ψi = 0.
So we know that
a
is a weak solution of
ξ
= 0. By the regularity theorem
again, we have
a(ψ) = hξ, ψi
for some ξ
p
(M). Then ξ H
p
.
We claim that
η
k
ξ
. Let
ε >
0, and pick
N
such that
n, m > N
implies
kη
n
η
m
k < ε. Then
kη
n
ξk
2
= hη
n
ξ, η
n
ξi |hη
m
ξ, η
n
ξi| + εkη
n
ξk.
Taking the limit as
m
, the first term vansihes, and this tells us
kη
n
ξk ε
.
So η
n
ξ.
η
k
(
H
p
)
, and (
H
p
)
is closed, we know
ξ
(
H
p
)
.
But also by assumption, we have
ξ H
p
. So
ξ
= 0. But we also know
kξk = lim kη
k
k = 1, whcih is a contradiction. So ` is bounded.
We then extend
`
to any bounded linear map on
p
(
M
). Then we are
done.
That was a correct proof, but we just pulled a bunch of theorems out of
nowhere, and non-analysts might not be sufficiently convinced. We now look
at an explicit example, namely the torus, and sketch a direct proof of Hodge
decomposition. In this case, what we needed for the proof reduces to the fact
Fourier series and Green’s functions work, which is IB Methods.
Consider
M
=
T
n
=
R
n
/
(2
πZ
)
n
, the
n
-torus with flat metric. This has local
coordinates (
x
1
, ··· , x
n
), induced from the Euclidean space. This is convenient
because
V
p
T
M
is trivialized by
{
d
x
i
1
···
d
x
i
p
}
. Moreover, the Laplacian is
just given by
∆(α dx
i
1
··· dx
i
p
) =
n
X
i=1
2
α
x
2
i
dx
i
1
··· dx
i
p
.
So to do Hodge decomposition, it suffices to consider the case
p
= 0, and we are
just looking at functions C
(T
n
), namely the 2π-periodic functions on R.
Here we will quote the fact that Fourier series work.
Fact.
Let
ϕ C
(
T
n
). Then it can be (uniquely) represented by a convergent
Fourier series
ϕ(x) =
X
kZ
n
ϕ
k
e
ik·x
,
where
k
and
x
are vectors, and
k · x
is the standard inner product, and this is
uniformly convergent in all derivatives. In fact, ϕ
k
can be given by
ϕ
k
=
1
(2π)
n
Z
T
n
ϕ(x)e
ik·x
dx.
Consider the inner product
hϕ, ψi = (2π)
n
X
¯ϕ
k
ψ
k
.
on `
2
, and define the subspace
H
=
(ϕ
k
) `
2
: ϕ
k
= o(|k|
m
) for all m Z
.
Then the map
F : C
(T
n
) `
2
ϕ 7→ (ϕ
k
).
is an isometric bijection onto H
.
So we have reduced our problem of working with functions on a torus to
working with these infinite series. This makes our calculations rather more
explicit.
The key property is that the Laplacian is given by
F(∆ϕ) = (−|k|
2
ϕ
k
).
In some sense, F “diagonalizes” the Laplacian. It is now clear that
H
0
= {ϕ C
(T
n
) : ϕ
k
= 0 for all k 6= 0}
(H
0
)
= {ϕ C
(T
n
) : ϕ
0
= 0}.
Moreover, since we can divide by
|k|
2
whenever
k
is non-zero, it follows that
(H
0
)
= ∆C
(T
n
).