4Hodge theory on Riemannian manifolds

III Riemannian Geometry 4.4 Introduction to Bochner’s method
How can we apply the Hodge decomposition theorem? The Hodge decomposition
theorem tells us the de Rham cohomology group is the kernel of the Laplace–
Beltrami operator ∆. So if we want to show, say,
H
1
dR
(
M
) = 0, then we want to
show that α 6= 0 for all non-zero α
1
(M). The strategy is to show that
hhα, αii 6= 0
for all
α 6
= 0. Then we have shown that
H
1
dR
(
M
) = 0. In fact, we will show that
this inner product is positive. To do so, the general strategy is to introduce an
, and then write
∆ = T
T + C
for some operator C. We will choose T cleverly such that C is very simple.
Now if we can find a manifold such that C is always positive, then since
hhT
T α, σii = hhTα, T αii 0,
it follows that is always positive, and so H
1
dR
(M) = 0.
Our choice of
T
will be the covariant derivative
itself. We can formulate
this more generally. Suppose we have the following data:
A Riemannian manifold M.
A vector bundle E M.
An inner product h on E.
A connection =
E
: Ω
0
(E)
1
(E) on E.
We are eventually going to take
E
=
T
M
, but we can still proceed in the
general setting for a while.
E
)
: Ω
1
(E)
0
(E) is defined by the relation
Z
M
h∇α, βi
E,g
ω
g
=
Z
M
hα,
βi
E
ω
g
for all
α
0
(
E
) and
β
1
(
E
). Since
h
is non-degenerate, this defines
uniquely.
Definition (Covariant Laplacian). The covariant Laplacian is
: Γ(E) Γ(E)
We are now going to focus on the case
E
=
T
M
. It is helpful to have the
following explicit formula for
, which we shall leave as an exercise:
As mentioned, the objective is to understand
. The theorem is that
this difference is given by the Ricci curvature.
This can’t be quite right, because the Ricci curvature is a bilinear form on
T M
2
, but
is a linear endomorphism
1
(
M
)
1
(
M
). Thus, we need
to define an alternative version of the Ricci curvature by “raising indices”. In
coordinates, we consider g
jk
Ric
ik
We can also define this
Ric
ik
without resorting to coordinates. Recall that
given an
α
1
(
M
), we defined
X
α
Vect
(
M
) to be the unique field such that
α(z) = g(X
α
, Z)
for all Z Vect(M). Then given α
1
(M), we define Ric(α)
1
(M) by
Ric(α)(X) = Ric(X, X
α
).
With this notation, the theorem is
Theorem (Bochner–Weitzenock formula). On an oriented Riemannian mani-
fold, we have
∆ =
+ Ric .
Before we move on to the proof of this formula, we first give an application.
Corollary. Let (M, g) be a compact connected oriented manifold. Then
If Ric(g) > 0 at each point, then H
1
dR
(M) = 0.
If Ric(g) 0 at each point, then b
1
(M) = dim H
1
dR
(M) n.
If Ric(g) 0 at each point, and b
1
(M) = n, then g is flat.
Proof. By Bochner–Weitzenock, we have
hhα, αii = hh∇
α, αii +
Z
M
Ric(α, α) ω
g
= k∇αk
2
2
+
Z
M
Ric(α, α) ω
g
.
Suppose
Ric >
0. If
α 6
= 0, then the RHS is strictly positive. So the
left-hand side is non-zero. So α 6= 0. So H
1
M
=
H
1
dR
(M) = 0.
Suppose
α
is such that
α
= 0. Then the above formula forces
α
= 0.
So if we know
α
(
x
) for some fixed
x M
, then we know the value of
α
everywhere by parallel transport. Thus
α
is determined by the initial
condition
α
(
x
), Thus there are
n
=
dim T
x
M
linearly independent such
α.
If
b
1
(
M
) =
n
, then we can pick a basis
α
1
, ··· , α
n
of
H
1
M
. Then as above,
these are parallel 1-forms. Then we can pick a dual basis
X
1
, ··· , X
n
Vect
(
M
). We claim they are also parallel, i.e.
X
i
= 0. To prove this, we
note that
hα
j
, X
i
i + h∇α
j
, X
i
i = ∇hα
j
, X
i
i.
But
hα
j
, X
i
i
is constantly 0 or 1 depending on
i
and
j
, So the RHS vanishes.
Similarly, the second term on the left vanishes. Since the
α
j
span, we know
we must have X
i
= 0.
Now we have
R(X
i
, X
j
)X
k
= (
[X
i
,X
j
]
[X
i
,
X
j
])X
k
= 0,
Since this is valid for all
i, j, k
, we know
R
vanishes at each point. So we
are done.
Bochner–Weitzenock can be exploited in a number of similar situations.
In the third part of the theorem, we haven’t actually proved the optimal
statement. We can deduce more than the flatness of the metric, but requires
some slightly advanced topology. We will provide a sketch proof of the theorem,
Proposition. In the case of (iii), M is in fact isometric to a flat torus.
Proof sketch. We fix p M and consider the map M R
n
given by
x 7→
Z
x
p
α
i
i=1,···,n
R
n
,
where the
α
i
are as in the previous proof. The integral is taken along any path
from
p
to
x
, and this is not well-defined. But by Stokes’ theorem, and the fact
that dα
i
= 0, this only depends on the homotopy class of the path.
In fact,
R
x
p
depends only on
γ H
1
(
M
), which is finitely generated. Thus,
R
x
p
α
i
is a well-defined map to
S
1
=
R
i
Z
for some
λ
i
6
= 0. Therefore we obtain
a map
M
(
S
1
)
n
=
T
n
. Moreover, a bit of inspection shows this is a local
diffeomorphism. But since the spaces involved are compact, it follows by some
topology arguments that it must be a covering map. But again by compactness,
this is a finite covering map. So M must be a torus. So we are done.
We only proved this for 1-forms, but this is in fact fact valid for forms of any
degree. To do so, we consider E =
V
p
T
M, and then we have a map
: Ω
0
M
(E)
1
M
(E),
and this has a formal adjoint
: Ω
1
M
(E)
0
M
(E).
Now if α
p
(M), then it can be shown that
α =
α + R(α),
where
R
is a linear map
p
(
M
)
p
(
M
) depending on the curvature. Then
by the same proof, it follows that if
R >
0 at all points, then
H
k
(
M
) = 0 for all
k = 1, ··· , n 1.
If
R
0 only, which in particular is the case if the space is flat, then we have
b
k
(M)
n
k
= dim
V
k
T
M,
and moreover α = 0 iff α = 0.
Proof of Bochner–Weitzenock
We now move on to actually prove Bochner–Weitzenock. We first produce an
explicit formula for
, and hence
.
Proposition.
Let
e
1
, ··· , e
n
be an orthonormal frame field, and
β
1
(
T
M
).
Then we have
β =
n
X
i=1
i(e
i
)
e
i
β.
Proof. Let α
0
(T
M). Then by definition, we have
h∇α, βi =
n
X
i=1
h∇
e
i
α, β(e
i
)i.
Consider the 1-form given by
θ(Y ) = hα, β(Y )i.
Then we have
divX
θ
=
n
X
i=1
h∇
e
i
X
θ
, e
i
i
=
n
X
i=1
e
i
hX
θ
, e
i
i hX
θ
,
e
i
e
i
i
=
n
X
i=1
e
i
hα, β(e
i
)i hα, β(
e
i
e
i
)i
=
n
X
i=1
h∇
e
i
α, β(e
i
)i + hα,
e
i
(β(e
i
))i hα, β(
e
i
e
i
)i
=
n
X
i=1
h∇
e
i
α, β(e
i
)i + hα, (
e
i
β)(e
i
)i.
So by the divergence theorem, we have
Z
M
h∇α, βi ω
g
=
Z
M
n
X
i=1
hα, (
e
i
β)(e
i
)i ω
g
.
So the result follows.
Corollary. For a local orthonormal frame field e
1
, ··· , e
n
, we have
α =
n
X
i=1
e
i
e
i
α.
We next want to figure out more explicit expressions for d
δ
and
δ
d. To make
our lives much easier, we will pick a normal frame field:
Definition
(Normal frame field)
.
A local orthonormal frame
{e
k
}
field is normal
at p if further
e
k
|
p
= 0
for all k.
It is a fact that normal frame fields exist. From now on, we will fix a point
p M
, and assume that
{e
k
}
is a normal orthonormal frame field at
p
. Thus, the
formulae we derive are only valid at
p
, but this is fine, because
p
was arbitrary.
The first term dδ is relatively straightforward.
Lemma. Let α
1
(M), X Vect(M ). Then
hdδα, Xi =
n
X
i=1
h∇
X
e
i
α, e
i
i.
Proof.
hdδα, Xi = X(δα)
=
n
X
i=1
Xh∇
e
i
α, e
i
i
=
n
X
i=1
h∇
X
e
i
α, e
i
i.
This takes care of one half of for the other half, we need a bit more work.
Recall that we previously found a formula for
δ
. We now re-express the formula
in terms of this local orthonormal frame field.
Lemma. For any 2-form β, we have
(δβ)(X) =
n
X
k=1
e
k
(β(e
k
, X)) + β(e
k
,
e
k
X).
Proof.
(δβ)(X) =
n
X
k=1
(
e
k
β)(e
k
, X)
=
n
X
k=1
e
k
(β(e
k
, X)) + β(
e
k
e
k
, X) + β(e
k
,
e
k
X)
=
n
X
k=1
e
k
(β(e
k
, X)) + β(e
k
,
e
k
X).
Since we want to understand
δ
d
α
for
α
a 1-form, we want to find a decent
formula for dα.
Lemma. For any 1-form α and vector fields X, Y , we have
dα(X, Y ) = h∇
X
α, Y i h∇
Y
α, Xi.
Proof. Since the connection is torsion-free, we have
[X, Y ] =
X
Y
Y
X.
So we obtain
dα(X, Y ) = Xhα, Y i Y hα, Xi hα, [X, Y ]i
= h∇
X
α, Y i h∇
Y
α, Xi.
Finally, we can put these together to get
Lemma. For any 1-form α and vector field X, we have
hδdα, Xi =
n
X
k=1
h∇
e
k
e
k
α, Xi +
n
X
k=1
h∇
e
k
X
α, e
k
i
n
X
k=1
h∇
e
k
X
α, e
k
i.
Proof.
hδdα, Xi =
n
X
k=1
h
e
k
(dα(e
k
, X)) + dα(e
k
,
e
k
X)
i
=
n
X
k=1
h
e
k
(h∇
e
k
α, Xi h∇
X
α, e
k
i)
+ h∇
e
k
α,
e
k
Xi h∇
e
k
X
α, e
k
i
i
=
n
X
k=1
h
h∇
e
k
e
k
α, Xi h∇
e
k
α,
e
k
Xi + h∇
e
k
X
α, e
k
i)
+ h∇
e
k
α,
e
k
Xi h∇
e
k
X
α, e
k
i
i
=
n
X
k=1
h∇
e
k
e
k
α, Xi +
n
X
k=1
h∇
e
k
X
α, e
k
i
n
X
k=1
h∇
e
k
X
α, e
k
i.
What does this get us? The first term on the right is exactly the
term
we wanted. If we add dδα to this, then we get
n
X
k=1
h([
e
k
,
X
]
e
k
X
)α, e
k
i.
We notice that
[e
k
, X] =
e
k
X
X
e
k
=
e
k
X.
So we can alternatively write the above as
n
X
k=1
h([
e
k
,
X
]
[e
k
,X]
)α, e
k
i.
The differential operator on the left looks just like the Ricci curvature. Recall
that
R(X, Y ) =
[X,Y ]
[
X
,
Y
].
Lemma
(Ricci identity)
.
Let
M
be any Riemannian manifold, and
X, Y, Z
Vect(M) and α
1
(M). Then
h([
X
,
Y
]
[X,Y ]
)α, Zi = hα, R(X, Y )Zi.
Proof. We note that
h∇
[X,Y ]
α, Zi+hα,
[X,Y ]
Zi = [X, Y ]hα, Zi = h[
X
,
Y
]α, Zi+hα, [
X
,
Y
]Zi.
The second equality follows from writing [
X, Y
] =
XY Y X
. We then rearrange
and use that R(X, Y ) =
[X,Y ]
[
X
,
Y
].
Corollary. For any 1-form α and vector field X, we have
hα, Xi = h∇
α, Xi + Ric(α)(X).
This is the theorem we wanted.
Proof. We have found that
hα, Xi = h∇
α, Xi +
n
X
i=1
hα, R(e
k
, X)e
k
i.
We have
n
X
i=1
hα, R(e
k
, X)e
k
i =
n
X
i=1
g(X
α
, R(e
k
, X)e
k
) = Ric(X
α
, X) = Ric(α)(X).
So we are done.