There are a number of results in geometry which allow to conclude that a certain group vanishes or is bounded under hypotheses on the curvature. For instance, we have:

Theorem 1 If ${M}$ is a compact manifold of positive curvature, then ${H^1(M; \mathbb{R}) = 0}$.

Another such result is the Kodaira vanishing theorem, which enables one to show that certain cohomology groups of an ample line bundle on a smooth projective variety vanish in characteristic zero.

I’ve been trying to gain an understanding of such results, and it seems that there is a common technique in such arguments. The first strategy is to identify the desired cohomology group (e.g. ${H^p(M; \mathbb{R})}$) with the kernel of a Laplacian-type operator, by Hodge theory. The second step is to bound below the relevant Laplacian-type operator. In this post, I’d like to try to explain what’s going on, in a special case.

1. Hodge theory

It is a classical fact that on a smooth manifold ${M}$, one can write

$\displaystyle H^p(M; \mathbb{R}) = \frac{\mathrm{closed} \ p- \mathrm{forms}}{\mathrm{exact} \ p- \mathrm{forms}}$

In other words, one can use differential forms and the exterior derivative as a replacement for cochains, and one gets a representation of ${H^p(M; \mathbb{R})}$ in terms of cocycles (closed forms) modulo coboundaries (exact forms).

Hodge theory enables one to pick out a subspace ${V}$ of closed ${p}$-forms such that the natural map

$\displaystyle V \rightarrow H^p(M; \mathbb{R})$

is an isomorphism. The idea is that, if one puts a ${L^2}$ norm on the space of all ${p}$-forms, then each element in the quotient ${H^p(M; \mathbb{R})}$ can be represented by an element of smallest norm. In other words, to a cohomology class ${[\omega] \in H^p(M; \mathbb{R})}$, we can choose a representative closed form ${\eta}$ which is orthogonal to the subspace of exact forms (by which we are quotienting). That is, we require

$\displaystyle (\eta, d \nu) = 0, \quad \forall \nu \ \ \ \ \ (1)$

with respect to some inner product ${(\cdot, \cdot)}$ on the space of ${p}$-forms. Ideally, for every cohomology class ${[\omega]}$, we should be able to find a unique representative ${\eta}$ which is in the cohomology class ${[\omega]}$ and which satisfies the additional condition (1). As a result, we will want to take ${V}$ as the subspace of closed forms satisfying (1); it turns out that this is in fact isomorphic to ${H^p(M; \mathbb{R})}$, as desired.

If one has an oriented Riemannian manifold ${M}$, then one can introduce a metric on the space of all ${p}$-forms: namely, one introduces a metric ${(\cdot, \cdot)'}$ in ${\bigwedge^p T_{x, M}^*}$ for each ${x \in M}$, and then a metric on global sections by integrating ${(\cdot, \cdot)'}$. In this case, ${d}$ will have an adjoint operator ${d^*}$ that sends a ${q}$-form to a ${q+1}$-form. Then (1) can be rephrased as

$\displaystyle (d^* \eta , \nu) = 0, \quad \forall \nu.$

This is equivalent to saying that

$\displaystyle d^* \eta = 0.$

The main result is:

Theorem 2 (Hodge theorem) Every cohomology class ${[\omega] \in H^p(M; \mathbb{R})}$ has a unique representative ${\eta}$ such that ${d \eta = d^* \eta = 0}$. As a result, ${H^p(M; \mathbb{R})}$ is isomorphic to the space of forms such that ${d\eta = d^* \eta = 0}$.

In other words, every cohomology class has a representative of smallest length. In fact, it can be stated via a direct sum decomposition of the space ${\mathcal{A}^p(M)}$ of all ${p}$-forms on ${M}$:

$\displaystyle \mathcal{A}^p(M) = d(\mathcal{A}^{p-1}(M)) \oplus d^*(\mathcal{A}^{p+1}(M)) \oplus V, \ \ \ \ \ (2)$

where ${V}$ is the space of forms in the kernel of both ${d}$ and ${d^*}$.

Definition 3 A form ${\eta}$ is harmonic if ${d\eta = d^* \eta = 0}$.

Thus, the Hodge theorem states that the space of harmonic forms is isomorphic to the cohomology ${H^*(M; \mathbb{R})}$. We can give a simpler characterization of harmonic forms.

Definition 4 The Hodge Laplacian is given by ${\Delta = d d^* + dd^*}$ on forms.

The elements of ${\ker \Delta}$ are precisely the harmonic forms (this is a general fact, easily proved, about inner product spaces), and the Laplacian ${\Delta}$ is a nonnegative, self-adjoint operator on the space of all forms.

Let’s try to see why the Hodge theorem should be believable. As I said, ${\Delta}$ is a self-adjoint, nonnegative operator on forms. It is, moreover, elliptic, and so we should think of it as something like a Fredholm operator. For a Fredholm operator, we get a decomposition

$\displaystyle \mathcal{A}^p(M) = \mathrm{Ker} \Delta^* \oplus \mathrm{Im} \Delta$

but since ${\Delta}$ is self-adjoint, we have

$\displaystyle \mathcal{A}^p(M) = \mathrm{Ker} \Delta \oplus \mathrm{Im} \Delta.$

One part of the direct sum decomposition is the space ${\ker \Delta}$ of harmonic forms, as desired. Since ${\mathrm{Im}\Delta }$ is a subset of ${d (\mathcal{A}^{p-1}(M)) \oplus d^* (\mathcal{A}^{p+1}(M))}$ (note that the two spaces are orthogonal as ${d^2 = 0}$), we can conclude the decomposition (2). From this, the Hodge theorem follows.

The problem, of course, is that the space ${\mathcal{A}^p(M)}$ of smooth ${p}$-forms on ${M}$ is not a Hilbert space, and the talk of Fredholm operators was a bit imprecise. Fortunately, though, there is a detailed theory of Sobolev spaces which allows one to make the above ideas precise.

2. The exterior derivative from the connection

The idea of the Bochner technique is, in effect, “completion of squares.” Namely, we have seen that we can represent cohomology classes on a smooth, oriented Riemannian manifold ${M}$ via elements in ${\ker \Delta}$. The strategy is now to bound below the nonnegative operator ${\Delta}$.

In order to do so, we’ll need to construct more second-order, nonnegative operators on ${M}$. Namely, the theme is to show that

$\displaystyle \Delta = T^* T + C$

where ${T}$ is a first-order operator (so ${T^* T}$ is nonnegative), and ${C}$ is a lower-order term that can be computed explicitly. If we can bound ${C}$ from below and show that it is positive, it will follow ${\Delta}$ is positive, and we can thus show that the space of harmonic forms vanishes. This is the technique of completion of squares.

The relevant operator ${T}$ in this case is the covariant derivative ${\nabla}$: this sends a ${p}$-form to a ${T^*M}$-valued ${p}$-form. That is, if ${\left\{e_i\right\}}$ is a local orthonormal frame for the tangent bundle, with dual basis ${\{\eta_i\}}$, then we have

$\displaystyle \nabla \omega = \eta_i \otimes \nabla_{e_i} \omega$

where ${\nabla_{e_i}}$ is differentiation associated to the Levi-Civita connection on ${M}$. Here the Einstein summation convention is in effect.

One of the consequences of ${\nabla}$‘s being the Levi-Civita connection (in particular, torsion-free), is that ${\eta_i \wedge \nabla_{e_i} \omega}$ is the same as the usual exterior derivative ${d \omega}$: that is, covariant differentiation becomes exterior differentiation after antisymmetrizing. We can state this formally:

Proposition 5 Notation as above,

$\displaystyle d \omega = \eta_i \wedge \nabla_{e_i} \omega:\ \ \ \ \ (3)$

thus the exterior derivative can be obtained from the covariant derivative.

Proof: The strategy is to note that ${d\omega, \eta_i \wedge \nabla_{e_i} \omega}$ are tensorial, so we may prove the equality (3) desired by working in a local coordinate system that is particularly convenient. Fix a point ${r}$. To prove (3), we may assume that we have chosen local coordinates ${x^1, \dots, x^n}$ around ${r}$ such that ${\partial_i = \frac{\partial}{\partial x_i}}$ are orthonormal at ${r}$ and such that

$\displaystyle g(\partial_i, \partial_j) = \delta_{ij} + O(|x|^2).$

That is, we are choosing normal coordinates such that the metric is to order one the same as the euclidean metric. Equivalently, we are assuming that the Christoffel symbols ${\Gamma_{ij}^k}$ vanish at ${p}$.

At the point ${r}$, we can thus prove (3) by taking ${e_i = \partial_i, \eta_i = dx^i}$. If ${\omega}$ is, say, a form ${\sum f dx^{i_1} \wedge \dots \wedge dx^{i_p}}$, then we have

$\displaystyle d \omega = \sum_i \frac{\partial f}{\partial x_i} dx_i \wedge dx^{i_1} \dots \wedge dx^{i_p}$

while

$\displaystyle \nabla_{\partial_i} \omega = \frac{\partial f}{\partial x_i} dx^{i_1} \wedge \dots \wedge dx^{i_p} .$

So, ${\eta_i \wedge \nabla_{e_i} \omega = \sum \frac{\partial f}{\partial x_i} dx_i \wedge dx^{i_1} \dots dx^{i_p} }$, and the two are equal as desired. $\Box$

The above technique, though elementary, is pretty powerful in differential geometry: when one wants to prove that two invariantly defined objects are equal, it may be useful to prove it a point by working in a particularly convenient system of local coordinates at that point. This is the basis for the so-called “Kähler identities” that lead to the high degree of structure in the cohomology of a Kähler manifold.

By a similar technique, we can get a clean description of the adjoint ${d^*}$:

Proposition 6 Notation as above, for ${\iota(e_i)}$ the interior product,

$\displaystyle d^* \omega = -\iota(e_i) \nabla_{e_i} \omega.$

This proposition is a corollary of the previous one if we use the Hodge star operator.

Anyway, so ${\nabla}$ is pretty close to ${d}$, except for the lack of antisymmetrizing. This suggests we might look for a relation between

$\displaystyle \Delta = d^* d + d d^*, \quad \nabla^* \nabla.$

It will turn out that there is one; however, there is an additional term that comes up because the covariant derivative is not symmetric. This is related to the curvature.

3. The Weitzenbock formula

Let us start by studying the Laplacian ${\Delta}$. Let’s choose any orthonormal frame ${e_i}$ of vector fields, locally, with dual coframe field ${\eta_i}$. For a ${p}$-form ${\omega}$, we can use the formulas of the previous section to see that

$\displaystyle d^* d \omega = d^* ( \eta_i \wedge \nabla_{e_i} \omega ) = -\iota(e_j)\nabla_{e_j} (\eta_i \wedge \nabla_{e_i} \omega). \ \ \ \ \ (4)$

Similarly,

$\displaystyle d d^* \omega = d ( -\iota(e_j) \nabla_{e_j} \omega ) = \eta_i \wedge \nabla_{e_i} ( -\iota(e_j) \nabla_{e_j} \omega ). \ \ \ \ \ (5)$

Let us now suppose the ${e_i}$ are normal at a point ${r}$, i.e. ${\nabla e_i = 0}$ at ${r}$. If we work only at ${r}$, then we can simplify some of the terms. Namely, from the first equation, we get (at ${r}$):

$\displaystyle d^* d \omega = - \iota(e_j) \eta_i \wedge \nabla_{e_j} \nabla_{e_i} \omega.$

From the second equation, we get

$\displaystyle d d^* \omega = - \eta_i \wedge \iota(e_j) \nabla_{e_i} \nabla_{e_j} \omega.$

These two look almost the same, and there is indeed a fair bit of cancellation.

We shall use the following formula. Consider the operators ${a_i }$ on the exterior algebra given by wedging with ${\eta_i}$ and ${a_j^* = \iota(e_j)}$. Then the ${a_i, a_j^*}$ satisfy the Clifford relations

$\displaystyle a_i a_i^* + a_i^* a_i = 1, \quad a_i a_j^* + a_j^* a_i = 0 \ \text{for } j \neq i.$

In fact, the operators ${a_i + a_i^*}$ give an action of an appropriate Clifford algebra on the exterior algebra.

So, anyway, we have proved that, at ${r}$,

$\displaystyle \Delta \omega = - ( a_j a_i^* \nabla_{e_j} \nabla_{e_i } \omega + a_i^* a_j \nabla_{e_i} \nabla_{e_j} \omega ). \ \ \ \ \ (6)$

We can simplify this. When ${i = j}$, we get a term ${-\sum \nabla_{e_i} \nabla_{e_i} \omega}$. When ${i \neq j}$, we have the sum

$\displaystyle - \sum_{i \neq j } \left( (a_j a_i^* + a_i^* a_j) \nabla_{e_j}\nabla_{e_i} \omega + a_i^* a_j R(e_i, e_j) \omega \right) ,$

at least if we make the further simplifying assumption that ${[e_i, e_j] = 0}$ at ${r}$. By the commutator relations, this becomes

$\displaystyle \Delta \omega = -\sum_i \nabla_{e_i}\nabla_{e_i} \omega - a_i^* a_j R(e_i, e_j) \omega.$

However, let’s note that since we are in a good frame ${\left\{e_i\right\}}$ which has no covariant derivatives at ${r}$, we have that the adjoint to ${\nabla_{e_i}}$ is ${-\nabla_{e_i}}$ (since it just looks like differentiation in this frame), and consequently, we have proved that ${\Delta = \nabla^* \nabla - a_i^* a_j R(e_i, e_j) \omega}$.

Definition 7 The curvature operator ${C: \bigwedge^\bullet T^*M \rightarrow \bigwedge^\bullet T^* M }$ is the operator ${- a_i^* a_j R(e_i, e_j)}$ on the space of forms. Note that this is invariantly defined for an orthonormal frame.

As a result, we can state:

Theorem 8 (Weitzenbock formula) On an oriented Riemannian manifold, we have:

$\displaystyle \Delta = \nabla^* \nabla + C. \ \ \ \ \ (7)$

Again, we have proved the Weitzenbock formula purely in a very specific choice of local coordinates, but since the two operators are invariantly defined and global, that is not a problem.

4. Applications to geometry

With the “completion of squares” identity established in (7), we can make an observation.

Meta-observation: If ${C}$ is a nonnegative operator, then any harmonic form on the compact Riemannian manifold ${M}$ is parallel. If ${C}$ is positive, then there are no nontrivial harmonic forms.

In fact, we notice that the operator ${\nabla^* \nabla}$ is always nonnegative, and the elements in the kernel are precisely the parallel forms. So if ${\omega}$ is harmonic, then we have

$\displaystyle 0 = \nabla^* \nabla \omega + C \omega$

and consequently, taking inner products with ${\omega}$, we get:

$\displaystyle \left \lVert\nabla \omega \right \rVert^2 + (C \omega, \omega) = 0.$

As a result, if we can bound below ${C}$ (by zero), then we will get very strong results in the cohomology of the manifold.

Theorem 9 (Bochner) If ${M}$ has nonnegative (resp. positive) Ricci curvature, then the curvature operator ${C}$ is nonnegative (resp. positive) on the space of 1-forms. Consequently, under these hypotheses we have

$\displaystyle \dim H^1(M; \mathbb{R}) \leq \dim M$

resp.

$\displaystyle H^1(M; \mathbb{R}) = 0.$

To prove this result, we will need to study the curvature operator, which was defined in local coordinates as ${\omega \mapsto -a_i^* a_j R(e_i, e_j) \omega}$, on one-forms. (The notation ${a_i^*, a_j}$ is as in the previous section; ${\left\{e_i\right\}}$ is a local frame, not necessarily normal.) The key point is the following:

Proposition 10 The curvature operator on one-forms can be identified with the Ricci tensor.

Proof: The Riemann tensor is given in a local orthonormal frame ${\left\{e_i\right\}}$ via four indices ${R_{ijkl}}$ (which is defined as the component of ${e_l}$ in ${R(e_j, e_k) e_i)}$), and the Ricci tensor is a contraction of it,

$\displaystyle Rc_{\sigma \tau} = R_{k \sigma k \tau}.$

The curvature operator, as defined, takes in two tangent vectors and a form and spits out a form. We can thus get it by raising and lowering indices in the Riemann tensor (which does nothing since we are in an orthonormal frame)

$\displaystyle R(e_i, e_j) \eta^k = R_{ k i jl} \eta^l.$

The operator ${a_j}$ is wedging by ${e_j}$. Consequently, we can work out the action of the curvature operator:

$\displaystyle C \omega_k e^k = - a_i^* a_j R(e_i, e_j) \omega_k e^k = - R_{kijl} \omega_k (a_i^* a_j e^l ).$

But acting by ${a_j}$ multiplies by ${e_j}$, while acting by ${a_i^*}$ lops off a factor of ${e_i}$ if it exists and otherwise does nothing. So the only terms occur for ${j = i}$, and we have

$\displaystyle C \omega_k e^k = -R_{kiil}\omega_k e^l ,$

and this is the Ricci tensor acting on ${\omega}$. In fact, we have ${-R_{kiil} = R_{kili} = Rc_{ki}}$. $\Box$

Now it is easy to prove Bochner’s theorem. We have shown that if the Ricci curvature is nonnegative (resp. positive), then the operator ${C}$ on 1-forms is just the Ricci tensor, and consequently nonnegative (resp. positive). This means that a harmonic form is in the kernel of ${\nabla}$ (resp. is zero). So we’ve established the second part of the theorem. In the first case, we’ve shown that harmonic forms are parallel, and since a parallel form is determined by its value at a point, we get the upper bound on ${\dim H^1(M; \mathbb{R})}$.

Incidentally, the result in the case of positive curvature can be strenghtened: by a theorem of Myers and Bonnet, the fundamental group of such a manifold (compact, with positive Ricci curvature) is in fact finite.