Hodge theory on compact hermitian manifolds - 2

In this post, we introduce some basic tools in functional analysis concerning differential operators and equations. The proof of the Hodge theorem will be presented on the basis of three key theorems, which will be proven in the next post.

Sobolev spaces

Recall our assumption that is a compact hermitian manifold and is a holomorphic hermitian vector bundle on . The hermitian structure enables us to define an inner product on the space of sections, which further induces a norm on .

However, as we are concerned with the derivatives of elements in , this norm is not satisfactory. The derivatives of sections are closely related to the connections on . Denote the metric connection on by . Then we define the -norm, where , of -valued forms on by the following formula:

where is the inner product on the fiber and is the volumn form of induced by the hermitian metric. It is clear that each is indeed a norm on and that is exactly itself.

Although this norm has a concise expression and is clearly globally well-defined, it is not so convenient in specific computations. Therefore we also introduce the following equivalent norms to simply our work.

Suppose is an open subset. For a smooth function on and a multi-index , we define Now consider the compact complex manifold . We may take an open cover of such that on each we have a holomorphic coordinate map and a holomorphic frame of . Suppose is a smooth partition of unity subordinate to the open cover . Then for each we can express locally as We see that each is a smooth function on . Since is identified with an open subset of through , we can consider for each . Then we define where is the Lebesgue measure on .

By comparing the order of differentiation, it can be verified that the norms and are equivalent to each other.

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Note that is not complete with respect to the norm , it is not a choice to do analysis on . Thus we need to consider the completion of with respect to these norms. The completion of with respect to is denoted by , usually called the Sobolev space. These Sobolev spaces are actually Hilbert spaces with respect to some inner product. It worth mentioning that the inner product on is induced from the inner product on and will be denoted by the same notation.

It is not hard to see that consists of those forms of on . Precisely, an element can be written locally as where . Using this description of and the equivalence of and , we can see that can be identified with the subspace of consisting of such that for all . We could also see that for each , if , then . Meanwhile, if holds for each , then .

For the sake of techinal convenience, the Sobolev spaces with are introduced. First let us consider the Sobolev spaces on an open subset . When , we define to be given by The norm on is defined as If we let then we can see that is exactly the completion of with respect to the norm . For each , the partial derivatives is defined for any , with .

If we want to consider the derivative of with , it is necessary to figure out what means for . For , we define to be the dual space of , i.e., the space of bounded linear functional on . The norm on is given by Since is identified with its dual by where this above definition of is suitable for .

Noting that when , we can view as a subspace of in a natural way. Analogously, we want to cannonically embed into . For , the corresponding element in is obtained by the restriction of to . Since indeed gives an element in . Thus we obtain a sequence of inclusion:

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Denote the space of smooth functions on with compact support by . For each and , the integral by parts yields that This inspires us to generalized the concept of derivatives in the following way. Suppose and . We write if for each , we have where the pairing is given by the inner product on if , and if .

Propostion. Suppose is an open subset. Then for each integer and multi-index with , defines a bounded operator with .

Proof: The assertion is direct if as we have for and is dense in .

Then suppose that and . We need to find for each some such that . Consider the linear funcctional given by . Since , we can find a multi-index such that for each and . Then for each , we have and hence Thus defines a bounded linear functional on . By Hahn-Banach theorem, can be extended to a bounded linear functional on without changing its norm, i.e., we obtain with and .

Now we consider the case . Similarly for define by . We see that The same argument as the previous case is valid and hence the conclusion is shown.

The above constructions can be done for in a corresponding way. The space with is defined to be the dual space of , and defined in this way agrees with the original definition. The seqeuence of inclusion also holds for , that is, for any integer . We see that , and give operators from to , while yields an operator from to .

Key theorems

Now we can state our three key theorems before the proof of the Hodge theorem. From now on, we also assume that is a compact hermitia manifold and is a holomorphic hermitian vector bundle on .

Theorem A (Gårding’s inequality). There exist positive constants such that for each , we have

Theorem B. If satisfies that , then .

Theorem C. Suppose is a sequence in such that is bounded. Then there exists a subsequence of which is a Cauchy sequence with respect to , i.e., converges in .

The proof of these theorems will not be given in this post. Instead, we will apply them in the proof of the Hodge theorem.

The first assertion of the Hodge theorem follows from Theorem A and Theorem C.

Proof of the first assertion of the Hodge theorem: Assume that is infinite dimensional. Then we can take a countable orthonormal list of elements in , with respect to the inner product on . By Theorem A we obtain showing the boundedness of with respect to . Thus Theorem C implies that contains a subsequence that is a Cauchy sequence with respect to . However, the orthonormal property of suggests that for any , which yields a contradiction.

As the Hodge theorem is concerned with instead of , we need a regularity lemma for similar to Theorem B.

Lemma. If satisfies that , then .

Proof: As , we have . Note that , it is clear that , and hence Therefore which implies by Theorem B. This is exactly , which again by Theorem B suggests that .

Write instead of for short. Denote the orthogonal complement of in by . As is finite dimensional, we have the direct sum decomposition .

Lemma. There exists a positive constant such that for each , we have .

Proof: Assume that the assertion is not true. Then there exists a sequence in such that for each and . By Theorem C, passing to a subsequence if necessary, we may assume that converges to with respect to . Note that we have

This exactly means that is zero as an element in . Clearly the zero functional belongs to , and then it follows from Theorem B that .

On the one hand, as is a closed subspace of with respect to the norm , we have . On the other hand, we have showing that . Thus , and hence . However, Theorem A yields that a contradiction.

Noting that and that and are both differential operators of order , we see that can be bounded by some constant multiple of for . Thus if we let then and are equivlent norms on .

Lemma. The image of is contained in , and the restriction of to gives a bijective operator .

Proof: For each and , we have implying that . If there is some such that , then and hence , showing the injectivity of .

Now we turn to the surjectivity. Fix a and we have to find some such that . Let be the closure of in with respect to . Then the regularity lemma for implies that it suffices to show that there exists such that in the sense that for any .

Define an inner product on by Then the norm induced by is exactly , which is equivalent to on . As is the completion of with respect to , the inner product naturally extends to . Consider the linear functional on given by Since defines a bounded linear functional on with respect to .

Hahn-Banach theorem implies that extends to a linear functional with the same operator norm, and then the Riesz representation theorem yields such that for each . In particular, we have Meanwhile, since , we have Together we obtain the desired relation between and , and hence the assertion is proven.

Now we can prove the second assertion of the Hodge theorem.

Proof of the second assertion of the Hodge theorem: Define by

First we show that is a compact operator. Suppose is a sequence in with for each . We need to show that there is a subsequence of such that is a Cauchy sequence with respect to . By Theorem C, it suffices to show that is bounded with respect to .

For each , we have implying that Suppose with and for each . Then showing the boundedness of in .

Next we show the commutatvity of with and . Note that and the image of and are both contained in . For , there exists such that . Then For , we have and , implying that Together we see that

Last we show that . For each we have , showing that which directly prove our assertion.