How to think about Hodge decomposition

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 X be a compact complex manifold with a metric. Then we have operators \partial, \overline{\partial}, \partial^* and \overline{\partial}^*. These go from (p,q)-forms to (p+1,q), (p,q+1), (p-1,q) and (p,q+1)-forms respectively. They obey

\displaystyle{\partial^2=0}, \displaystyle{\overline{\partial}^2=0}, \displaystyle{(\partial^*)^2=0}, \displaystyle{(\overline{\partial}^*)^2=0},

\displaystyle{\partial \overline{\partial} = - \overline{\partial} \partial} and \displaystyle{\partial^* \overline{\partial}^* = - \overline{\partial}^* \partial^*}  (1).

The exterior derivative d is \partial + \overline{\partial}, the Hodge dual formula is d^* = \partial^* + \overline{\partial}^*. We define three Laplacians: \Delta_d = d d^* + d^* d, \Delta_{\partial} =  \partial \partial^* + \partial^* \partial and \Delta_{\overline{\partial}} =  \overline{\partial} \overline{\partial}^* + \overline{\partial}^* \overline{\partial}.

So, in general,

\displaystyle{ \Delta_d = (\partial+\overline{\partial}) (\partial^* + \overline{\partial}^*) + (\partial^* + \overline{\partial}^*) (\partial+\overline{\partial}) = }

\displaystyle{\Delta_{\partial} + \Delta_{\overline{\partial}} + (\partial \overline{\partial}^* + \overline{\partial}^* \partial) +  ( \overline{\partial} \partial^* +  \partial^* \overline{\partial})}.

When X is Kähler, we have \Delta_{\partial} = \Delta_{\overline{\partial}} = (1/2) \Delta_{d}. This identity is actually made up out of the following identities, which strike me as more fundamental:

\displaystyle{  \partial \overline{\partial}^* + \overline{\partial}^* \partial =0} and \displaystyle{  \overline{\partial} \partial^* +  \partial^* \overline{\partial}=0}   (2).

\displaystyle{  \partial \partial^* + \partial^* \partial = \overline{\partial} \overline{\partial}^* + \overline{\partial}^* \overline{\partial} }   (3).

The two quantities in the third equation are (by definition) \Delta_{\partial} and \Delta_{\overline{\partial}}; we’ll denote them both by \Delta. So \Delta takes (p,q)-forms to (p,q)-forms.

Also, \Delta commutes with all of \partial, \overline{\partial}, \partial^* and \overline{\partial}^*, this is an easy consequence of (1) and (3) (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 0=\lambda_0 < \lambda_1 < \lambda_2 < \cdots so that \Omega^{p,q}(X) breaks up as a direct sum \bigoplus  \Omega_{\lambda}^{p,q}(X) of eigenspaces for \Delta, which each \Omega^{p,q}_{\lambda} finite dimensional.

The precise meaning of the direct sum is a little tricky: The L^2 completion of \Omega^{p,q}(X) is the L^2-direct sum of these spaces and, more generally, the same is true for all of the (2,s) Sobolev norms. But you don't need to understand that to understand the rest.

Since \partial etcetera commute with \Delta, they take each \lambda-eigenspace to itself. So, for each \lambda, we get (n+1)^2 finite dimensional vector spaces \Omega^{p,q}_{\lambda} and maps \partial, \overline{\partial}, \partial^* and \overline{\partial}^* between them satisfying (1), (2) and

\displaystyle{  \partial \partial^* + \partial^* \partial = \overline{\partial} \overline{\partial}^* + \overline{\partial}^* \overline{\partial} = \lambda }   (3').

When \lambda=0, all for maps are 0; you just have a diamond of vector spaces with no maps.

Now, what about \lambda>0? There is an obvious way to satisfy the above equations: Put a one dimensional vector space in positions (p,q), (p+1,q), (p,q+1) and (p+1,q+1). Let the two nontrivial \partial maps be 1, the two nontrivial \overline{\partial} maps also be 1, and the nontrivial \partial^* and \overline{\partial}^* maps be \lambda. Call this configuration a square and denote it S^{p,q}_{\lambda}.

Then here is a nice algebraic lemma: Conditions (1), (2) and (3') imply that \Omega^{\bullet, \bullet}_{\lambda} is a direct sum of various S^{p,q}_{\lambda}‘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 \Omega^{\bullet, \bullet}(X) looks like a direct sum of (a) terms where all four derivatives act by 0, call them 0-terms, and (b) various squares. In particular, even once we forget the Kähler structure, and thus can only talk about \partial and \overline{\partial}, we still get a nontrivial consequence: The double complex \Omega^{\bullet, \bullet} is a direct sum of (a) 0-terms, where \partial and \overline{\partial} act by 0, and (b) 2 \times 2 squares of one dimensional vector spaces where the nontrivial \partial and \overline{\partial} maps are isomorphisms.

Notice how easy this makes the standard results.

The deRham cohomology, H^k, is the cohomology of

\displaystyle{ \cdots \to  \bigoplus_{p+q=k-1} \Omega^{p,q} \stackrel{\partial+\partial^*}{\longrightarrow} \bigoplus_{p+q=k} \Omega^{p,q} \stackrel{\partial+\partial^*}{\longrightarrow} \bigoplus_{p+q=k+1} \Omega^{p,q} \to \cdots }.

The Dolbeault cohomology, H^{p,q} is the coholomogy of

\displaystyle{ \cdots \to \Omega^{p,q-1} \stackrel{\overline{\partial}}{\longrightarrow} \Omega^{p,q}  \stackrel{\overline{\partial}}{\longrightarrow} \Omega^{p,q+1} \to \cdots}.

Hodge decomposition says that H^k \cong \bigoplus H^{p,q}.

Well, that’s clear enough. Cohomology distributes over direct sum. The terms on which \partial and \overline{\partial} both act by 0 contribute one dimension to each. The square terms are exact in both settings.

Similarly, the \partial \overline{\partial} lemma states that, if \alpha is d-closed and in the image of \partial then it is in the image of \partial \overline{\partial}. Proof: Decompose into these summands. If \alpha is d-closed then the component of \alpha on each summand is closed. The d-closed elements are (a) the 0-terms, (b’) the bottom elements of squares (the element in degree (p+1,q+1) in S^{p,q}) and (b”) the difference between the two middle terms of the square (the difference between the guy in degree (p,q+1) and the guy in degree (p+1,q)). The image of \partial can’t contain any terms of type (a) or (b”), so we see that \alpha is a sum of terms of type (b’), which are, indeed, in the image of \partial \overline{\partial}.

Of course, this is essentially the statement that the spectral sequence degenerates at E^2. 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 E^1, but stronger, see Greg Kuperberg’s MathOverflow post.


5 thoughts on “How to think about Hodge decomposition

  1. 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 E_1 (I thought — you say E_2?). 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.

  2. 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.

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