# Martin Orr's Blog

## Line bundles and morphisms to the dual variety

Posted by Martin Orr on Saturday, 28 May 2011 at 15:25

Over the complex numbers, the dual of an abelian variety is defined to have a Hodge structure dual to that of . Hence morphisms can be interpreted as bilinear forms on the Hodge structure of . Of particular importance are the morphisms corresponding to Hodge symplectic forms.

Last time we saw that can also be interpreted as a group of line bundles on . Today we will use this interpretation to define morphisms which turn out to be the same as those corresponding to Hodge symplectic forms. Then we generalise the definition of to base fields other than , which we will use next time in constructing dual abelian varieties over number fields.

### The morphisms over

Let be a complex abelian variety, and let be any line bundle on . Define the function by where is the "translation by " map.

By the Appell-Humbert Theorem, we may suppose that with a Riemann form on . Let . Then we can calculate and deduce

We can deduce several important facts from this calculation:

1. is a group homomorphism.
2. depends only on the Riemann form . In particular is identically iff .
3. The image of is contained in .

Furthermore, unpacking the isomorphism which we defined last time, we get that the map corresponding to is given by . This is a morphism of Hodge structures, so is a morphism of abelian varieties; more precisely, it is the morphism associated to the Hodge symplectic form .

### The maps over arbitrary fields

Now we define the maps for an abelian variety over an arbitrary field . We have not yet defined the dual abelian variety, so we cannot say that they are morphisms , but we can still define a function by

Now consider the numbered facts above. Fact 1 is still true: is a group homomorphism. Explicitly, this says that for all .

This result is called the Theorem of the Square. For general , this is a corollary of the Theorem of the Cube, whose proof involves the cohomology of line bundles on families of complete varieties. Alternatively, if is a number field, then we can deduce it from the fact that we already know it is true over . We just need the following lemma:

Lemma. Let be a complete variety over a field and a line bundle on . If is trivial on for some field containing , then is trivial on .

Proof. A line bundle on a complete variety is trivial iff both and have nonzero global sections. So it suffices to show that This follows from the fact that "cohomology commutes with flat base change" (Hartshorne III.9.3).

We will turn fact 2 around and use it as the definition of : is defined to be the group of isomorphism classes of line bundles on for which for all . (As often with varieties over non-algebraically closed fields, we need to use -points in this definition because there might not be enough -points; e.g. when is a number field can be finite.)

Finally, now that we have defined , fact 3 follows from the theorem of the square.

So for every line bundle on , we have defined a group homomorphism

### Nondegenerate line bundles

Define . This is a subgroup of . We can extend this to a closed subgroup scheme of . It turns out that the correct definition for the functor of points of this scheme for any -algebra is

We say that is nondegenerate if is finite. In the complex case, these are the line bundles whose corresponding symplectic form is nondegenerate.

Next time, we will need a nondegenerate line bundle for the construction of the dual abelian variety. The following proposition ensures that they always exist (every abelian variety has an ample line bundle because it is projective).

Proposition. An ample line bundle on an abelian variety is nondegenerate.

Proof. If is a number field, then we can deduce this from the fact that it is true over . Over , let be the Riemann form associated to . Some power is very ample, and the proof that every complex abelian variety is polarisable tells us that is positive definite. So is positive definite, and hence nondegenerate.

Over a general field, let be the connected component of the scheme containing the identity. You show that the restriction to of is both ample and trivial, which implies that .

This theorem also has a converse, which says that a line bundle on an abelian variety which is nondegenerate and has a nonzero global section is ample.

1. Dual varieties over general fields From Martin's Blog

Today we will construct dual abelian varieties over number fields. We use the universal property from two posts ago to define dual abelian varieties, then we give a simple construction inspired by the complex case. Proving that this construction s...

1. Barinder said on Monday, 20 June 2011 at 10:12 :

I guess that, next time, to construct the dual, you'll have to cook up a variety D and an invertible sheaf P such that the pair (D,P) satisfies the previous universal property. How hard is this? I seem to recall there being another approach that works in great generality, about representing picard functors by schemes or algebraic spaces. Are these two approaches basically the same?

Some typos:

1. Line 2 should be "dual to that of A", not A-dual.

2. Paragraph 2 Line 2: "Today we will use this..."

3. In the "Furthermore, unpacking..." paragraph, you're missing a dual on an A.

2. Martin Orr said on Monday, 20 June 2011 at 21:26 :

I will only give a sketch of the construction, hopefully later this week. Specifying what D and P are is not hard, though there is some technical stuff I will skip. Proving that it satisfies the universal property is harder. We already know that it works analytically over , and I will leave out algebraically closed fields of positive characteristic. I had thought that the method for algebraically closed fields could be extended to other fields but I realised today that this does not work. However it should be possible to deduce the general fields case from the algebraically closed case by Galois descent.

The universal property that I gave tells you what the morphisms are for any normal variety , and actually it works for any scheme - the restriction to normal varieties is just needed for that proof with Zariski's Main Theorem. So this tells you what the functor of points of is. In other words, the universal property is basically the same thing as the definition of the Picard functor. However the method of construction that I will give is specific to abelian varieties - the general representability method requires much more about schemes and sheaves.