# 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.

2. 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.
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.