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

Ω

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 Ω

−1

(M) = 0.

As before, the compactness of M is essential, and cannot be dropped.

Corollary. We have orthogonal decompositions

Ω

p

(M) = H

p

⊕ dδΩ

p

(M) ⊕ δdΩ

p

(M)

= H

p

⊕ dΩ

p−1

(M) ⊕ δΩ

p+1

(M).

Proof. Now note that for an α, β, we have

hhdδα, δdβii

g

= hhddδα, dβii

g

= 0.

So

dδΩ

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

p−1

(M) = d(H

p−1

⊕ ∆Ω

p−1

(M)) = d(δdΩ

p−1

(M)) ⊆ dδΩ

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 ∈ Ω

p

(M) be such that dα = 0. We write

α = α

1

+ dα

2

+ δα

3

∈ H

p

⊕ dΩ

p−1

(M) ⊕ δΩ

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 ∈ Ω

p

(M) : dξ = 0, [ξ] = a}.

This is an infinite dimensional affine space.

We now ask ourselves — which

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 ∈ Ω

p−1

(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

∈

Ω

n

(

M

) satisfies

kα

n

k <

C and k∆α

n

k < C for all n, then α

n

contains a Cauchy subsequence.

This is almost like saying Ω

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 ∈ Ω

p

(M), we have

hω, ∆ϕi = h∆ω, 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

h∆ω, 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 ∈ Ω

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 Ω

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 ∈ 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

cares about what

`

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 Ω

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

Ω

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 ∈ Ω

p

(M). Then we have

hα, ∆βi = h∆α, 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 ∈ Ω

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) ≤ Ck∆ηk

for all

η ∈

Ω

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 ≤ Ck∆ηk

for every η ∈ Ω

p

(M).

Suppose not. Then we can find a sequence

η

k

∈

(

H

p

)

⊥

such that

kη

k

k

= 1

and k∆η

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 ∈

Ω

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→∞

h∆η

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 ∈ Ω

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.

But this is bad. Since

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 Ω

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

k∈Z

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

).