Suppose is a compact complex manifold of dimension and is a holomorphic vector bundle of rank over . From the Dolbeault isomorphism, we know that the cohomology group is isomorphic to the set of all -closed smooth -valued -forms on modulo the set of all -closed smooth -valued -forms on . Though this gives us a more practical way of computing by using differential forms, it would be more convenient for computational purposes if a cohomology class is represented by a *unique* differential form rather than an equivalence class of differential forms.

When we have a quotient of a finite-dimensional vector space by a vector subspace, instead of a looking at an element in the quotient space, one way is to consider the orthogonal complement of the vector subspace provided we have an inner product in the ambient space. So every element in the quotient space is represented by a unique element in the orthogonal complement of the subspace instead of an equivalence class in the original ambient space. This is the same as using the element with minimum length in its equivalence class to represent the equivalence class.

In our case, the vector space of all -closed smooth -valued -forms on , in general, is an infinite-dimensional vector space. Suppose we give a Hermitian metric and also give a Hermitian metric along its fibers. So there is a natural inner product on the vector space over all -closed smooth -valued -forms on . If we use this method of orthogonal complement, or equivalently, the method of finding an element with minimum length in its equivalence class, we run into the trouble of closedness of the subspace, or equivalently, the convergence of a sequence of elements in an equivalence class minimizing the length function. We know that the vector space of all -closed smooth -valued -forms on is not complete in the natural inner product. In order to talk about closedness of subspaces or the convergence of a minimizing sequence of elements, we should first consider the vector space of all -closed smooth -valued -forms on to a Hilbert space. So we should consider the set of all -closed -valued -forms on .

- The first question that arises in considering the set of all -closed -valued -forms on is the definition of the operator on -valued -forms on .
- The second question is the closedness of the image of after it is defined.

Let denote the set of all -valued -forms on , and let be the set of all -valued -forms on . The operator from to is defined on the dense subset of . In order to make the image of closed, we should define it on as large a set as possible. So we consider the closure of the graph of in the product space of and . Given an element of and an element of , the pair is in the graph if and only if there exists a sequence converging in the norm of to so that converges in the norm of to . We have to check that this closure is the graph of a map. The trouble is that for the closure of the graph, we may have two distinct elements of given as the image of the same element of . By taking the difference of these two distinct elements of , we can assume that there exists a sequence converging in the norm of to 0 so that converges in the norm of to some nonzero element of . Take an arbitrary element of . We have the process of integration by parts. So we get

where means the inner product in and is a first-order linear partial differential operator. Letting , we conclude that

for any , contradicting that is nonzero. Hence, we know that the closure of the graph of is also a graph. This argument works in general for differential operators because we have integration by parts.

There is another way to extend the definition of from . For any , the expression makes sense when differentiation is done in the sense of distributions. So in general, would be a current. If this current can be represented by an element of , then we say that belongs to the domain of and define as . The earlier extension done by using the closure of the graph of is known as the strong extension. At first, it seems that the weak extension may have a bigger domain than the strong extension. It turns out that the two extensions are the same. This is known as the *Friedrichs Extension Lemma *(K.O. Friedrichs, “The identity of weak and strong extensions of differential operators”. *Trans. Amer. Math. Soc.* 55 (1944), 132-151). First, one observes that by using a partition of unity, we can assume that all forms involved are supported in a single coordinate chart. Then in that coordinate chart, one uses smoothing by convolution with a cutoff function. Suppose

is a first-order differential operator with smooth coefficients. Let be a nonnegative function supported on the unit open ball of . Let

Suppose is an function. Then

in norm as . We assume that is taken in the sense of distributions in . Then

in norm as . It suffices to show that

in norm as . This is clearly true when belongs to the dense subset of smooth functions. So it suffices to show that

is bounded in norm when belongs to a set bounded in norm. The zero-order part of clearly has bounded contribution. So we can assume, without loss of generality, that . Then

Clearly, the second term on the right-hand side is bounded. So we can drop it. We have

In the last integral,

because . We have

Hence,

and the norm of

is bounded by

which is bounded if the norm of is bounded.

We want to show that the image of

is closed, and the kernel of

quotiented by the image of

is isomorphic to the set of all -closed smooth -valued -forms on . First, we handle the question of the closedness of the image of

This means that if we have so that converges to some in , then we want to show that for some . Let . If we are able to solve the equation so that or a subsequence of it converges to some in , then we are done.

We are going to consider the operator

First, we would like to discuss the motivation for this operator . We look at the equation , where is in and . We would like to solve it so that has a good bound when is bounded. The equation is equivalent to for all in , i.e. . Now, is the unknown. To know is the same as knowing the linear functional . Now, we know this linear functional is , and is known. The set of all is a subset of . If we can extend this known linear functional from the set to a bounded linear functional on all of , then we know whose norm is bounded by that of the bounded linear functional. Such an extension is possible if we have an estimate

We write so that and . Since , we have . From , we have and . Hence,

is equivalent to

We can assume for that inequality that . By the Schwarz Inequality, it suffices to have an estimate

Since to have the inequality that

This inequality can be rewritten as

This shows that the operator is closely related to the solution . If we do have the inequality

then our preceding discussion shows that the equation admits a solution with

We do not expect in general to have the inequality

because this inequality would imply that vanishes. However, we have a related inequality that can serve our purpose of proving the closedness of the image of . This related inequality is *Gårding’s Inequality*. It is says that

where the norm means a norm that is equivalent to the sum of the -norm of and the norm of the first derivative of .

Before we prove Gårding’s Inequality, let us look at its consequences. Rewrite Gårding’s Inequality in the form

We know that the operator admits an inverse whose norm is , because the operator is nonnegative. Replacing by in Gårding’s Inequality, we get

Since

it follows that

This means that the map from to is continuous, where is the Hilbert space of all -valued -forms whose first derivatives are also . Since by the *Rellich-Kondrachov Theorem* the inclusion map

is a compact operator, it follows that the map from to is a compact operator. A compact operator means that it maps a bounded set of the domain space to a relatively compact subset of the target space. We will prove the Rellich-Kondrachov Theorem later. By the spectral theorem for self-adjoint compact operators, we know that the eigenvalues of the operator from to are in , the only possible limit point of is 0, the eigenspace for each is finite-dimensional, and the eigenfunctions of span . The eigenvalues are limited to , because is positive and its norm is . An equation

is equivalent to

So we conclude that the eigenvalues of are in , the eigenspace for each is finite-dimensional, and the eigenfunction of span . If 0 is an eigenvalue, we call it . By allowing some positive eigenvalues () to be counted more than once, we can assume that the eigenspace for each positive eigenvalue is only -dimensional, and the totality of the eigenfunction for () is an orthonormal basis of the orthonal complement of in . We know that is finite-dimensional. On , we can define the inverse of by

We extend to all of by defining to be zero on . We call *Green’s operator*. The reason for the name is that is the inverse of the Laplace operator . Let be the orthogonal projection from onto . Then we have the identity

Now, we are ready to prove the closedness of . We go back to our earlier notation of in so that . Suppose with . We have

By applying to both sides, we get

Taking the inner product with , we obtain

Hence,

This shows that though in general we can not solve the equation because may not vanish, we can still always solve the equation

From

and Gårding’s Inequality, we conclude that

because

We go back to our earlier notation of in so that . Since , we have

An element of belongs to if and only if

because

This implies every element of is perpendicular to both and . Since

belongs to , it follows from that . Hence, . Since is bounded independent of , by the Rellich-Kondrachov Theorem, we can select a subsequence of converging to some in . Then , because

Hence, the image of

Now, we want to show that the cohomology group can be calculated by using -valued forms instead of -valued forms. For this we need a regularity result for the equation in . More precisely, we want to show that if the norm , which is equivalent to the sum of all the norms of derivatives of of order is finite, then is finite. Note that we assume already that the norms of both and are finite. First, we observe that if and and for some , then . The case is given right away by Gårding’s Inequality. For the general case, we take any vector field and apply the argument to and use induction on . Now, we come back to the equation in . Again, we look at first the case . By Gårding’s Inequality,

Likewise,

Hence, by the above observation, is finite. For the case of a general , again we take any vector field and apply the argument to and use induction on . In these arguments, we have been quite sloppy above justifying such things as integration by parts even though our forms are not smooth. The rigorous way to justify such arguments is to smooth out the forms first *in the graph norm* of the first-order operators by using the result of Friedrichs and get estimates on the approximating smooth forms using constants dependent only on our original nonsmooth form and then take limits at the end. A form is in if and only if it can be approximated by smooth forms in the norm of .

Now that we have the regularity result for , we consider the map from the space

to the space

We want to show that is an isomorphism. First, we show that it is surjective. Take with . We have

As we argued before, when we apply to both sides and take the inner product with , we conclude that

From and the regularity of , we know that is smooth. Since differs from the -closed smooth form by the -exact form , we conclude that is surjective. Now, we look at the injectivity of . Suppose and for some . Take the decomposition

It is clear that the three spaces , , are mutually orthogonal, because

From

we conclude that

So

Since is smooth, by the regularity of and

we know that is smooth and is smooth. Since is the of the smooth form , we conclude that is injective. From the above argument, we also see that both spaces

and

are isomorphic to . An element of is called a *harmonic form*. So we have proved that the cohomology group is isomorphic to the space of all harmonic -valued -forms. Since is finite-dimensional from the spectral theorem, we know that is always finite-dimensional for a compact manifold .