Hodge decomposition

We discussed in the previous post that we can understand the moduli of Elliptic curves by looking at the moduli of rank two lattices. That is, we consider them as \mathbb{C}/\Lambda_\tau for some element \tau in the upper half-plane. It followed that

\mathcal{M}_{1,1} \cong \mathbb{H}/SL_2\mathbb{Z}

is the moduli space of elliptic curves. The notation \mathcal{M}_{1,1} will be discussed further at a later point; in brief, it is the moduli space of 1-marked genus 1 curves: Elliptic curves, that is.

However, there is another way of discussing the moduli of Elliptic curves. If we instead consider an Elliptic curve as an abstract 1-dimensional complex manifold, then its genus can be defined as

g = \dim H^1(E, \mathcal{O}_E) = \dim H^0(E, K_E)

i.e. we can define the genus to be the dimension of the space of holomorphic 1-forms. Since this is the same as the topological genus, we know this to be one.

Now, we further know that all elliptic curves are homeomorphic. In particular, for every elliptic curve E we have that their cohomology groups are isomorphic, and in particular they all have H^1(E, \mathbb{Z}) isomorphic to the fixed lattice

\Lambda = \displaystyle \Big(\mathbb{Z}^2, \langle\, , \rangle\Big)

where the intersection form can be written as the matrix

\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}

We will use this to provide a new look at their moduli.

Let us first note that \Lambda has an almost-complex structure: that is, there is a linear endomorphism I : \Lambda \to \Lambda which satisfies

  • I^2 = -Id_\Lambda
  • \langle Iu, Iv \rangle = \langle u, v \rangle i.e. the intersection form is preserved.

Exercise: What is this endomorphism?

We can then look at the complexification \Lambda_\mathbb{C} = \Lambda \otimes \mathbb{C}. This splits into \pm i-eigenspaces, and so we write

\Lambda_\mathbb{C} \cong \Lambda^{1,0}\oplus \Lambda^{0,1}

where \Lambda^{1,0} is the i-eigenspace (and vice versa).

Lastly, since \Lambda has an almost-complex structure, it follows that \Lambda_\mathbb{R} = \Lambda \otimes \mathbb{R} can be given the structure of a complex vector space, and moreover, there is thus a complex-linear isomorphism \Lambda_\mathbb{R} \cong \Lambda^{1,0}.

So we now return to using this idea to describe the moduli of elliptic curves. In brief, the decomposition of \Lambda into \Lambda^{1,0} \oplus \Lambda^{0,1} is called the Hodge decomposition. This decomposition will also provide a way to discuss the moduli of Elliptic curves.

The idea is as follows. We define a lattice-polarized Elliptic curve to be an elliptic curve E together with a choice of isomorphism

\phi : H^1(E, \mathbb{Z}) \to \Lambda

which we know exists because Elliptic curves are homeomorphic. This of course yields a complexified map \phi_\mathbb{C} : H^1(E, \mathbb{C}) \to \Lambda_\mathbb{C}.

We also have an integration map

\iota : H^0(E, K_E) \to Hom\big(H_1(E, \mathbb{Z}), \mathbb{C}\big) \cong H^1(E, \mathbb{C})

given by mapping an element \omega \in H^0(E, K_E) to the map

\gamma \mapsto \int_\gamma \omega

The composition of these two maps is then a non-zero map \iota \circ \phi_\mathbb{C} : H^0(E, K_E) \to \Lambda_\mathbb{C}. Since this map is \mathbb{C}-linear, it follows that its image must be \Lambda^{1,0}.

There are now a few things that we can say about this map. First of all due to the intersection form on \Lambda_\mathbb{C}, it follows that for any \lambda \in \Lambda^{1,0} that

\mathfrak{Im}\,\langle \lambda, \overline\lambda\rangle > 0

Exercise: Check this!

It follows of course that for any \omega \in H^0(E, K_E) that \langle\iota \circ \phi_\mathbb{C} \omega , \overline{\iota \circ \phi_\mathbb{C} \omega }\rangle has positive imaginary part. The punchline is now the following.

Consider \mathbb{P}\Lambda_\mathbb{C} (we consider the projectivisation since we only care about the image of the line \mathbb{C}\omega). Then the collection of those lines [\omega] satisfying

\mathfrak{Im}\,\langle \omega , \overline\omega\rangle > 0

form a moduli space for lattice-polarized Elliptic curves. Moreover, if we quotient out by the natural action of SL_2\mathbb{Z} then we recover the usual moduli space of Elliptic curves.

Another way of saying this is that the complex structure on the Elliptic curve determines (and is determined by) its Hodge decomposition on H^1(E, \mathbb{C}).

In further posts, we will consider how to extend these ideas to higher genus curves, and even to higher dimensional analogs of Elliptic curves, namely Abelian Varieties.

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About charlesflorian

Mathematician. Climber. Living in Copenhagen.
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2 Responses to Hodge decomposition

  1. Atsushi says:

    Well written. Look forward to the next post.

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