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
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)
p1
(M) by
(i(X)ψ)(Y
1
, ··· , Y
p1
) = ψ(X, Y
1
, ··· , Y
p1
).
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
)(
αβ
)
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
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
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).