UPDATE: Greg Kuperberg writes the same things with fewer typos here.

This blog post is meant for me to work through some things I want to present in class tomorrow, so it won’t have as much background as I usually try to include.

Let be a compact complex manifold with a metric. Then we have operators , , and . These go from -forms to , , and -forms respectively. They obey

, , , ,

and .

The exterior derivative is , the Hodge dual formula is . We define three Laplacians: , and .

So, in general,

.

When is Kähler, we have . This identity is actually made up out of the following identities, which strike me as more fundamental:

and .

.

The two quantities in the third equation are (by definition) and ; we’ll denote them both by . So takes -forms to -forms.

Also, commutes with all of , , and , this is an easy consequence of and (exercise!).

This is all standard. The rest is something that I haven’t seen written down, but strikes me as making the theory much simpler.

Results on elliptic operators tell us that there is a discrete sequence of eigenvalues so that breaks up as a direct sum of eigenspaces for , which each finite dimensional.

The precise meaning of the direct sum is a little tricky: The completion of is the -direct sum of these spaces and, more generally, the same is true for all of the Sobolev norms. But you don't need to understand that to understand the rest.

Since etcetera commute with , they take each -eigenspace to itself. So, for each , we get finite dimensional vector spaces and maps , , and between them satisfying , and

.

When , all for maps are ; you just have a diamond of vector spaces with no maps.

Now, what about ? There is an obvious way to satisfy the above equations: Put a one dimensional vector space in positions , , and . Let the two nontrivial maps be , the two nontrivial maps also be , and the nontrivial and maps be . Call this configuration a *square* and denote it .

Then here is a nice algebraic lemma: Conditions , and imply that is a direct sum of various ‘s. (I put this as Problem 6 on a recent problem set and got good solutions from most of my students.)

So, what the Hodge decomposition tells us is that looks like a direct sum of (a) terms where all four derivatives act by , call them -terms, and (b) various squares. In particular, even once we forget the Kähler structure, and thus can only talk about and , we still get a nontrivial consequence: The double complex is a direct sum of (a) -terms, where and act by , and (b) squares of one dimensional vector spaces where the nontrivial and maps are isomorphisms.

Notice how easy this makes the standard results.

The deRham cohomology, , is the cohomology of

The Dolbeault cohomology, is the coholomogy of

.

Hodge decomposition says that .

Well, that’s clear enough. Cohomology distributes over direct sum. The terms on which and both act by contribute one dimension to each. The square terms are exact in both settings.

Similarly, the lemma states that, if is -closed and in the image of then it is in the image of . Proof: Decompose into these summands. If is -closed then the component of on each summand is closed. The -closed elements are (a) the -terms, (b’) the bottom elements of squares (the element in degree in ) and (b”) the difference between the two middle terms of the square (the difference between the guy in degree and the guy in degree ). The image of can’t contain any terms of type (a) or (b”), so we see that is a sum of terms of type (b’), which are, indeed, in the image of .

~~Of course, this is essentially the statement that the spectral sequence degenerates at . But I really think authors should write it out directly. It makes everything else much more obvious.~~

This statement is related to the statement that the spectral sequence degenerates at , but stronger, see Greg Kuperberg’s MathOverflow post.

There’s a typo in “Dolbeault”…

Thanks!

(Actually I had to first make a search on Numdam to make sure it was Dolbeault and not Dolbeaut…)

I wrote about this topic in a MathOverflow post last year. In brief, any Dolbeault complex decomposes into dots, squares, and zigzags. The Hodge theorem implies that the Hodge spectral sequence degenerates at (I thought — you say ?). But the theorem is stronger than that, because the spectral sequence only detects even-length zigzags. The Hodge theorem actually says that there are no zigzags at all.

Sorry, should be $E_1$. The reason that I was thinking $E_2$ was that I was thinking of the Dolbeault complex as the $E_1$ page for the hypercohomology of the holomorphic deRham complex.

I’ll edit.