This, like the previous and probably the next few posts, is an attempt at understanding some of the ideas in Mumford’s *Abelian Varieties.*

Let be an abelian variety. Last time, I described a formula which allowed us to express the pull-back of a line bundle as

This was a special case of the so-called “theorem of the cube,” which allowed us to express, for three morphisms , the pull-back

in terms of the various pull-backs of partial sums. Namely, we had the formula

In this formula, we take and constant maps at , respectively. Let , etc. denote the translation maps on . Then are the relevant sums, and we have

We get the theorem of the square:

Theorem 1 (Theorem of the square)Let be an abelian variety, , and points. Then

In the rest of this post, I’ll describe some applications of this result, for instance the fact that abelian varieties are projective.

**1. Translation-invariant line bundles**

Let be an abelian variety. We will be interested in line bundles on which are *translation-invariant*, i.e. line bundles such that for all .

Definition 2The group of which are translation-invariant is .

In fact, turns out to be not just a group, but an abelian variety itself. This is far from obvious, but it turns out to be a corollary of the (non-obvious) fact that line bundles on are themselves parametrized by a scheme, called the **Picard scheme.** The collection of translation-invariant bundles turns out to form the connected component at zero of the Picard scheme.

What are some examples of translation-invariant line bundles?

Example 1Let be any line bundle. Then is translation-invariant; this is a consequence of the theorem of the square.

One of the next goals is to show that any element of can be obtained in the manner of the previous example.

Example 2Let have dimension one, so is an elliptic curve. Then the translation-invariant line bundles correspond to divisors of total degree zero. In fact, by the theorem of the square, we find that the line bundle associated to for any is in . The other direction will follow because any translation-invariant bundle is algebraically equivalent to zero, as we will see later, and consequently has degree zero.

Let be the multiplication map, and let . Then the line bundle has the property that restriction to any fiber is precisely . If we think of as a family of line bundles on , parametrized by , then to say that is translation-invariant is a certain trivialty of this family. This is expressed in the following result.

Proposition 3A line bundle belongs to if and only if

for the multiplication map.

In fact, this follows from a “rigidity lemma” on families of line bundles. If we consider the line bundle , then we have seen that it is trivial on each vertical fiber, and similarly on each horizontal fiber. But the rigidity lemma (which is proved in Mumford’s book) states precisely that such a line bundle is trivial. The converse is easy, by restricting to the vertical fibers.

A consequence of this fact is that

because is obtained by pulling back under the map

However, because is constant.

In particular, translation-invariant line bundles are very strongly non-ample: if were ample, then would be too (since is an isomorphism). But, and cannot both be ample unless is zero-dimensional. In fact, it turns out that ampleness is a polar opposite of translation-invariance: a line bundle is ample if and only if the set of such that is finite, as it turns out.

**2. A little intersection theory**

Let be an abelian variety, and let be a divisor on . We saw in the previous section that

where denotes linear equivalence, for any . In particular, we get

This means that if is any effective divisor on , then the linear system is base-point free: given , we can choose such that does not contain . In particular, defines a map for some . We’d like to choose to make this map an immersion, or at least finite. We can thus state:

**Observation:** On an abelian variety, there are a lot of line bundles generated by global sections (and thus a lot of maps to projective space.)

Suppose is any base-point free divisor, defining a map

We’d like to know when is a finite morphism. By Zariski’s main theorem, finiteness is equivalent to quasi-finiteness. Consequently, fails to be finite precisely when there exists an irreducible curve such that . This means that every divisor in the linear system either contains or fails to intersect . In the language of line bundles, any section of vanishing on a point of must vanish everywhere on . By changing if necessary, we can assume that .

**Observation**: The intersections are either empty or all of , for all . In other words, can intersect no translate of .

In fact, the intersection number for is constant, and must thus be identically zero. Alternatively, one considers the family of line bundles on the normalization of . When , this line bundle is trivial; it follows that for all , the degree of the line bundle is zero. Since is an effective divisor, this implies that for all such that intersection is not .

This is a pretty useful geometric argument: “wiggling” the divisor by a translation by can’t force it to intersect . Motivated by this, we can deduce something seemingly stronger.

**Observation:** If contracts the irreducible curve to a point and , then is invariant under translation by for .

In fact, if and , then the translation has the property that (because the intersection contains ). In particular, it contains by the previous observation: thus

which is what we wanted.

These two observations show that there are strong constraints on a base-point free divisor with the property that the induced map to is not finite: the last observation shows that has to be invariant by a proper curve’s worth of translations. We thus find:

Proposition 4Let be a divisor on contained in for an open affine. If is base-point free, then the map is finite.

*Proof:* In fact, we can assume . In this case, we find that the only translations which leave invariant have to be in (i.e., they have to send into something in ). This means that there can’t be a proper curve’s worth of translations leaving invariant , and by the previous observations, can’t contract a curve.

Finally, putting all this together, we may deduce:

Theorem 5An abelian variety is projective.

*Proof:* It suffices to show that admits a finite map for a suitable divisor , because the divisor is then the pull-back of the ample divisor on by a finite map and is thus itself ample. In order to do this, we need to find a divisor on without base points and which is contained in for an open affine. That’s easy: just start by picking an irreducible component of , and then take twice that for . The analysis at the beginning of the section shows that is base-point free.

In fact, using a refinement of these arguments, one can prove a characterization of ampleness:

Theorem 6A line bundle associated to an effective divisor is ample if and only if the set of all such that is finite.

April 16, 2012 at 4:58 am

If you repeat the typo: check the missing L^{-1} in the cube formula.

Do you have other ideas in mind for studying Mumford’s Abelian Varieties?

April 16, 2012 at 8:50 pm

I don’t think it’s a typo: the equation is just cut off. (WordPress margins are a lot shorter than those of a typical LaTeX document.)

I’m not sure — my near goal for the blog is pretty limited (e.g., to prove that is smooth and of the right dimension, and maybe Mordell-Weil a little further down the line). There’s a lot of fascinating stuff in Mumford’s book that I’m hoping to explore over the summer, though.