## The Hodge Theorem

Previously, we talked a bit about the category of Hodge structures, and did some basic constructions.  However, I’d claimed that this was algebraic geometry (at least, in the categories on the post) so today, we’ll talk about a LOT of Hodge structure that arise in nature.  Everything I say is true in more generality (for compact Kähler manifolds, in fact) but for now, let $X$ be a smooth projective variety (as always, over $\mathbb{C}$.

I’m going to be VERY sketchy on the details today, they’re largely analytic and though I can use them a bit, I won’t be able to do them justice.  So now, we’ll do the local case first, and by local, I mean local in the analytic topology.

Let $T_X$ be the tangent bundle, and then tensor with $\mathbb{C}$ to get the complexified tangent bundle $T_X^\mathbb{C}$.  This decomposes into a holomorphic and an antiholomorphic part, which are complex conjugates.  Call sections of the holomorphic part type (1,0) and the anti type (0,1).

This decomposition is a wonderful thing! It induces decompositions of all tensors on $X$, and in particular, on $k$-forms, which are of type $(p,q)$ if they are a linear combination (over the $C^\infty$ functions) of forms consisting of $p$ terms $dz_i$ and $q$ terms of the form $d\bar{z}_j$.

Now, let’s say $H^{p,q}\subset H^{p+q}(X)$ is the set of classes where there is a $(p,q)$ form representing it.  We’ll need to show that these are disjoint (except zero, of course) and their direct sum is everything to get a Hodge decomposition.

To check that every class is represented by a type $(p,q)$ form, and that the $(p,q)$ don’t overlap, the trick is to apply Harmonic Analysis (which I only have the barest outlines of at the moment).  The trick is that everything is represented by a unique(!) harmonic form, and you just need to check the Harmonic forms.

Then, a computation shows that the Laplacian is a degree (0,0) operator, so it will take a $(p,q)$ form to a $(p,q)$ form.  This tells us that the $(p,q)$ components of a harmonic form are still harmonic, and then we have the Hodge decomposition!

So this gets us a Hodge structure of weight $k$ on $H^k(X,\mathbb{C})$ for all $k$ and all $X$ complex, smooth and projective.  In general, we don’t get Hodge structures on varieties failing either smoothness or projectivity.  There, we’ll get what’s called a mixed Hodge structure.  But for now, a couple of rather concrete examples:

Example 1: on a curve $C$, the Hodge structure on $H^0(C)$ is uninteresting, there just aren’t many Hodge structure of weight 0, much less with a one dimensional vector space.  On $H^2(C)$, we also have a one dimensional space, and there are no holomorphic 2-forms on a curve, so we have $H^2(C)=H^{1,1}$.  Now, $H^1(C)=H^{1,0}\oplus H^{0,1}$, and the two spaces are conjugate, so have equal dimension. Thus, $\dim H^{1,0}=h^{1,0}=g$, the genus, and is the space of holomorphic forms, isomorphic to $H^0(C,\Omega_C^1)$.

More generally, we’ll have $H^{p,q}(X)\cong H^q(X,\Omega^p_X)$, which will be a very useful fact, in general, and will be implicit in many arguments.

Example 2: Look at $\mathbb{P}^n$.  For $k$ odd, we have no cohomology, and so things are dull.  For $k=2\ell$, however, we have a one dimensional space.  Now, as anything off center must give rise to its conjugate, it must be $H^{2\ell}=H^{\ell,\ell}$.

And here’s a quick application of Hodge structures, and the related constructions from last time.  Let $E$ be an elliptic curve.  Using the universal coefficients theorem, we have a Hodge structure of weight -1 on $H_1(E,\mathbb{Z})$.  Next, we construct the intermediate Jacobian, which is going to be $H^{0,-1}/H_1(E,\mathbb{Z})$.  Finally, we note that $H^{0,-1}$ and $H^{-1,0}$ are dual (exercise!) and that $H^{-1,0}$ and $H^{1,0}$ are also dual (these two facts hold in general) so, $H^{0,-1}$ and $H^{1,0}$ are naturally isomorphic, and we get a map we get a map $H_1(E,\mathbb{Z})\to H^{1,0}$, which is injective, and the quotient is a complex torus with complex dimension one.  Now, every complex manifold of dimension one is an algebraic curve, so we have an elliptic curve $E'$.  But in fact, this is our original curve! Why? Because an elliptic curve is determined by its lattice in $\mathbb{C}$ and we can recover the lattice by integrating against the form $dz$ on $E$.

This is an example of a Torelli theorem, which lets us reconstruct a variety in a certain class from it’s Hodge theory.  In fact, it’s a special case of THE Torelli Theorem, which is the following:

Theorem: Let $\mathcal{M}_g$ be the moduli space of curves of genus $g$ and let $\mathcal{A}_g$ the moduli space of principally polarized Abelian varieties of dimension $g$.  The Jacobian map $C\mapsto H^{1,0}/H_1(C,\mathbb{Z})$ is injective, and away from the hyperellitpic locus of $\mathcal{M}_g$, it is an embedding.

This is a very nice theorem.  It says that a curve’s Hodge structure determines it (the injectivity) and that the only thing that stops $\mathcal{M}_g$ from being a nice subvariety of $\mathcal{A}_g$ is that the singularities don’t match up on the hyperelliptic locus.  Now, the converse, finding conditions on $\mathcal{A}_g$ that cut out the image of this map, is a famous problem called the Schottky problem.

## About Charles Siegel

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
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### 2 Responses to The Hodge Theorem

1. Steven Sam says:

In the last post, you defined $({\rm H}^{p,q})^{\vee}$ to be ${\rm H}^{-p,-q}$, so shouldn’t it be that ${\rm H}^{-1,0}$ and ${\rm H}^{1,0}$ are naturally dual and not naturally isomorphic? It looks like we want to say that ${\rm H}^{1,0}$ and ${\rm H}^{0,-1}$ are naturally isomorphic via the double dual?

Also, what does ${\rm H}^{0,-1}/{\rm H}_1(E,\mathbf{Z})$ mean? If I understand correctly, ${\rm H}^{0,-1} \subseteq {\rm H}_1(E, {\bf C})$, and the natural embedding of ${\rm H}_1(E, {\bf Z})$ doesn’t sit in either ${\rm H}^{0,-1}$ nor ${\rm H}^{-1,0}$?

2. For the first part of your comment, yes, that’s what’s going on, sadly, the above didn’t go through much editing before being posted. I’ll clarify that part tonight.

For the second, if you project into one side or the other, the map is still injective, because everything had to be sent to a pair $(h,\bar{h})$, so $H_1(E,\mathbb{Z})$ lies in both $H^{0,-1}$ and $H^{-1,0}$.