The n-Category Café

Complex line bundles on a topological space $X$ are classified up to isomorphism by elements of $H^2(X,\mathbb{Z})$.

This becomes particularly nice when $X$ is a real torus. Let’s equip $X$ with the structure of an abelian Lie group. Then the universal cover $\tilde{X}$ is a finite-dimensional real vector space, and the kernel of the projection

$$p \colon \tilde{X} \to X$$

is a lattice

$$L = \ker p$$

in the vector space $\tilde{X}$ — that is, a free abelian group of rank equal to the dimension of that vector space. This lets us write

$$X = \tilde{X}/L$$

In this situation $H^2(X,\mathbb{Z})$ is isomorphic to the group of alternating bilinear maps

$$A \colon L \times L \to \mathbb{Z}$$

There’s nothing magic about the number $2$ and the word ‘bilinear’ here — the same kind of description works for $H^i(X, \mathbb{Z})$ for any $i$, using alternating multilinear maps.

So, complex line bundles on a real torus $X$ are classified up to isomorphism by alternating bilinear maps $A \colon L \times L \to \mathbb{Z}$. And I should be clear: we could use isomorphism of topological line bundles, or of smooth line bundles: it makes no difference here.

But now suppose $\tilde{X}$ is equipped with the structure of a complex vector space. This makes $X = \tilde{X}/L$ into a complex manifold, and indeed a compact abelian complex Lie group. We call $X$ a complex torus. In fact, every compact connected complex Lie group is a complex torus.

Complex tori are a lot subtler than real tori. All real tori of a given dimension are isomorphic. This is not true for complex tori! For example, complex tori of complex dimension 1 — hence real dimension 2, so they look like doughnuts — are called elliptic curves. There’s an interesting space of different isomorphism classes of elliptic curves, called the moduli space of elliptic curves. The story gets even more complicated in higher dimensions.

Now for the real point of this article. If $X$ is a complex torus, it’s an interesting question to classify holomorphic complex line bundles on $X$. How does it work?

This question breaks up nicely into two parts:

Question 1. Which alternating bilinear $A \colon L \times L \to \mathbb{Z}$ come from holomorphic complex line bundles?

Question 2. If $A \colon L \times L \to \mathbb{Z}$ comes from a holomorphic complex line bundle, how many different such bundles give this particular $A$?

A nice thing is that the answer to Question 2 turns out not to depend on the choice of $A$, as long as there’s any holomorphic line bundle at all giving that $A$.

Answer to Question 1. The answer to this question is simple, though it takes work to show it’s correct. Using the fact that every vector in $\tilde{X}$ is a real linear combination of vectors in the lattice $L \subseteq \tilde{X}$, you can show any alternating bilinear map

$$A \colon L \times L \to \mathbb{Z}$$

extends uniquely to an alternating real-bilinear map

$$\tilde{A} \colon \tilde{X} \times \tilde{X} \to \mathbb{R}$$

It then turns out that $A$ comes from a holomorphic complex line bundle if and only if

$$\tilde{A}(i v, i w) = \tilde{A}(v,w)$$

for all $v,w \in \tilde{X}$.

Answer to Question 2. This answer is also simple. Suppose $A$ comes from some holomorphic complex line bundle. Then the set of isomorphism classes of holomorphic complex line bundles giving this $A$ is $X^*$, the ‘dual’ of torus $X$.

What’s dual of a torus? In this game it’s another torus — not the usual Pontryagin dual of the torus. Remember, for any real torus $X$ we can write

$$X = \tilde{X}/L$$

The real vector space $\tilde{X}$ has a dual $\tilde{X}^*$, defined in the usual way. Sitting inside this dual vector space we have the so-called dual lattice of the lattice $L$, defined like this:

$$L^\ast = \{ f \in \tilde{X}^\ast : \; f(v) \in \mathbb{Z} \; for \; all \; v \in L \}$$

So, the quotient ${\tilde{X}}^\ast / L^\ast$ is a torus, called the dual torus of the torus $X$ and again denoted with a star:

$$X^\ast = {\tilde{X}}^\ast / L^\ast$$

Now if $X$ is a complex torus, the vector space $\tilde{X}$ is a complex vector space, and thus so is its dual ${\tilde{X}}^\ast$. Beware: ${\tilde{X}}^\ast$ is still defined in the usual way: it’s the real dual of the underlying real vector space of $\tilde{X}$. But it gets the structure of a complex vector space from that of $\tilde{X}$ — this is a bit tricky, and I’ll talk about it next time. So the dual torus $X^\ast$ is a complex torus if $X$ is.

People like to package up the answers to Question 1 and Question 2 into a single theorem, the Appell–Humbert theorem. And the usual statement of this theorem involves lots of extra jargon — which you have to learn if you want to study this subject:

• The set of isomorphism classes of holomorphic complex line bundles on a complex torus $X$ is called its Picard group, $\mathrm{Pic}(X)$. This is actually an abelian group, because you can tensor line bundles. In fact, it’s a complex Lie group.

• The identity component $\mathrm{Jac}(X)$ of the Picard group is called the Jacobian of $X$, or, as if there weren’t enough jargon already, the Picard variety of $X$. This is a complex torus, and an abelian variety when $X$ is.

• The quotient $\mathrm{Pic}(X)/\mathrm{Jac}(X)$ is called the Néron–Severi group of $X$ and denoted $\mathrm{NS}(X)$. This is a finitely generated free abelian group.

So we have a short exact sequence

$$0 \to \mathrm{Jac}(X) \to \mathrm{Pic}(X) \to \mathrm{NS}(X) \to 0$$

You should think of it this way:

• The Néron–Severi group $\mathrm{NS}(X)$ describes the ‘discrete degrees of freedom’ required to choose a holomorphic complex line bundle over $X$. That’s because it’s the group of connected components of $\mathrm{Pic}(X)$. It’s a finitely generated free abelian group.

• The Jacobian $\mathrm{Jac}(X)$ describes the ‘continuous degrees of freedom’ required to choose a holomorphic complex line bundle over $X$. That’s because it’s the identity component of $\mathrm{Pic}(X)$. It’s a complex torus.

But even better:

• $\mathrm{NS}(X)$ is naturally isomorphic to the subgroup of $H^2(X,\mathbb{Z})$ coming from holomorphic complex line bundles — which we described in our answer to Question 1.

• $\mathrm{Jac}(X)$ is naturally isomorphic to the dual torus $X^*$, which we described in our answer to Question 2.

Putting all these facts together, we get this:

Appell–Humbert Theorem. If $X$ is a complex torus, we have a short exact sequence

$$0 \to \mathrm{Jac}(X) \to \mathrm{Pic}(X) \to \mathrm{NS}(X) \to 0$$

where

• $\mathrm{NS}(X)$ is isomorphic to the abelian group whose elements are alternating bilinear forms $A \colon L \times L \to \mathbb{Z}$ whose real-linear extension $\tilde{A}$ obeys $\tilde{A}(i v, i w) = \tilde{A}(v,w)$. (The set of these becomes an abelian group under addition.)

and

• $\mathrm{Jac}(X)$ is isomorphic to the dual torus $X^\ast$.

The dimension of the complex torus $\mathrm{Jac}(X)$ depends only on the dimension of $X$, since it’s just the dual torus. But the rank of the free abelian group $\mathrm{NS}(X)$ depends on more than just the dimension of $X$. Indeed, a generic complex torus has no holomorphic complex line bundles on them except those that are topologically trival. For such complex, $\mathrm{NS}(X)$ has just one element! But in general $\mathrm{NS}(X)$ is a subgroup of $H^2(X,\mathbb{Z})$ whose rank depends on the complex structure of $X$, not just its dimension. It’s bigger for complex tori where the lattice $L \subseteq \tilde{X}$ gets along better with the complex structure on $\tilde{X}$, and these are the ones worth focusing on.

There’s a lot more to say, and I’m proud of myself for having not said it. Maybe I’ll say more later. I’m having tons of fun looking at examples of the theorems I just stated.