4Hodge theory on Riemannian manifolds

III Riemannian Geometry

4.3 Divergence

In ordinary multi-variable calculus, we had the notion of the divergence. This

makes sense in general as well. Given any X ∈ Vect(M), we have

∇X ∈ Γ(T M ⊗ T

∗

M) = Γ(End T M).

Now we can take the trace of this, because the trace doesn’t depend on the

choice of the basis.

Definition (Divergence). The divergence of a vector field X ∈ Vect(M ) is

divX = tr(∇X).

It is not hard to see that this extends the familiar definition of the divergence.

Indeed, by definition of trace, for any local frame field {e

i

}, we have

divX =

n

X

i=1

g(∇

e

i

X, e

i

).

It is straightforward to prove from definition that

Proposition.

div(fX) = tr(∇(fX)) = fdivX + hdf, Xi.

The key result about divergence is the following:

Theorem.

Let

θ ∈

Ω

1

(

M

), and let

X

θ

∈ Vect

(

M

) be such that

hθ, V i

=

g(X

θ

, V ) for all V ∈ T M. Then

δθ = −divX

θ

.

So the divergence isn’t actually a new operator. However, we have some

rather more explicit formulas for the divergence, and this helps us understand

δ

better.

To prove this, we need a series of lemmas.

Lemma. In local coordinates, for any p-form ψ, we have

dψ =

n

X

k=1

dx

k

∧ ∇

k

ψ.

Proof.

We fix a point

x ∈ M

, and we may wlog we work in normal coordinates

at x. By linearity and change of coordinates, we wlog

ψ = f dx

1

∧ ··· ∧ dx

p

.

Now the left hand side is just

dψ =

n

X

k=p+1

∂f

∂x

k

dx

k

∧ dx

1

∧ ··· ∧ dx

p

.

But this is also what the RHS is, because ∇

k

= ∂

k

at p.

To prove this, we first need a lemma, which is also useful on its own right.

Definition

(Interior product)

.

Let

X ∈ Vect

(

M

). We define the interior

product i(X) : Ω

p

(M) → Ω

p−1

(M) by

(i(X)ψ)(Y

1

, ··· , Y

p−1

) = ψ(X, Y

1

, ··· , Y

p−1

).

This is sometimes written as i(X)ψ = Xyψ.

Lemma. We have

(divX) ω

g

= d(i(X) ω

g

),

for all X ∈ Vect(M).

Proof. Now by unwrapping the definition of i(X), we see that

∇

Y

(i(X)ψ) = i(∇

Y

X)ψ + i(X)∇

Y

ψ.

From example sheet 3, we know that ∇ω

g

= 0. So it follows that

∇

Y

(i(X) ω

g

) = i(∇

Y

X) ω

g

.

Therefore we obtain

d(i(X)ω

g

)

=

n

X

k=1

dx

k

∧ ∇

k

(i(X)ω

g

)

=

n

X

k=1

dx

k

∧ i(∇

k

X)ω

g

=

n

X

k=1

dx

k

∧ i(∇

k

X)(

p

|g|dx

1

∧ ··· ∧ dx

n

)

= dx

k

(∇

k

X) ω

g

= (divX) ω

g

.

Note that this requires us to think carefully how wedge products work (

i

(

X

)(

α∧β

)

is not just α(X)β, or else α ∧ β would not be anti-symmetric).

Corollary (Divergence theorem). For any vector field X, we have

Z

M

div(X) ω

g

=

Z

M

d(i(X) ω

g

) = 0.

We can now prove the theorem.

Theorem.

Let

θ ∈

Ω

1

(

M

), and let

X

θ

∈ Vect

(

M

) be such that

hθ, V i

=

g(X

θ

, V ) for all V ∈ T M. Then

δθ = −divX

θ

.

Proof.

By the formal adjoint property of

δ

, we know that for any

f ∈ C

∞

(

M

),

we have

Z

M

g(df, θ) ω

g

=

Z

M

fδθ ω

g

.

So we want to show that

Z

M

g(df, θ) ω

g

= −

Z

M

fdivX

ω

ω

g

.

But by the product rule, we have

Z

M

div(fX

θ

) ω

g

=

Z

M

g(df, θ) ω

g

+

Z

M

fdivX

θ

ω

g

.

So the result follows by the divergence theorem.

We can now use this to produce some really explicit formulae for what

δ

is,

which will be very useful next section.

Corollary. If θ is a 1-form, and {e

k

} is a local orthonormal frame field, then

δθ = −

n

X

k=1

i(e

k

)∇

e

i

θ = −

n

X

k=1

h∇

e

k

θ, e

k

i.

Proof. We note that

e

i

hθ, e

i

i = h∇

e

i

θ, e

i

i + hθ, ∇

e

i

e

i

i

e

i

g(X

θ

, e

i

) = g(∇

e

i

X

θ

, e

i

) + g(X

θ

, ∇

e

i

e

i

).

By definition of X

θ

, this implies that

h∇

e

i

θ, e

i

i = g(∇

e

i

X

θ

, e

i

).

So we obtain

δθ = −divX

θ

= −

n

X

i=1

g(∇

e

i

X

θ

, e

i

) = −

n

X

k=1

h∇

e

i

θ, e

i

i,

We will assume a version for 2-forms (the general result is again on the third

example sheet):

Proposition. If β ∈ Ω

2

(M), then

(δβ)(Y ) = −

n

X

k=1

(∇

e

k

β)(e

k

, Y ).

In other words,

δβ = −

n

X

k=1

i(e

k

)(∇

e

k

β).