# Pic Zero and the Poincare Bundle

For any scheme , is the group of line bundles on . In the case of an abelian variety , there is a particularly nice subgroup .

If then,

- is trivial, in particular
- There is a surjection with finite kernel (an isogeney)
- If is a line bundle on s.t. there exists with then for all .

### Some big results

The above claims follow pretty easily from two pretty big results:

Thm: (See-saw priciple) Let be a complete variety, and a line bundle on with trivial restriction on all the fibers: . Then for some line bundle on .

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pf(sketch): Set . The significant part of the proof is an application of cohomology with base change (see page 20-21 or J.Milne’s abel. var. notes). The condition on the fibers of gives that is a locally free sheaf and in particular invertible. Now there is always a natural map

Set . For any open set , one can compute

Using and using that for a complete variety there are only constant global section (see post) I conclude is an isomorphism. Its an isomorphism on fibers, and in particular an iso for any point , i.e. for every the map on stalks is an iso after modding out by the maximal ideal and a Nakayama’s lemma argument shows it was already an isomorphism. So is an isomorphism (since it is on stalks). QED.

Thm (of the cube): Let complete and another variety and a line bundle on . If there exists such that the restriction of to is trivial, then is trivial.

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The proof is not too bad. A rough summary: first show it holds when is a complete curve (this uses that the Jacobian of a curve is a certain moduli space and the rigidity lemma).

For the general case, let be the line bundle on . Use a complete curve to connect any to the distinguished (see this post). So have

$latex C \times Y \times Z \xrightarrow{i} X \times Y

\times Z \xrightarrow{p} Z \times Z$

By the case of a curve, is trivial. This shows for every the restriction of to is trivial, i.e. trivial on the fibers of . By the see-saw theorem it follows , but which is trivial, hence is trivial.

### Corollaries

Let denote the three projections . Then

Cor1: For any line bundle on A, the line bundle

is trivial on .

Cor2: For morphisms the line bundle

on is trivial.

pf: , where .QED.

Cor3. For all and a line bundle there is an isomorphism

In particular, is a homomorphism.

pf: apply the previous corollary to , where sends everything to . The last statement follows by multiplying the above by QED.

Given a line bundle , define by . The following properties are readily checked:

For the last statement it suffices to show :

Finally, define . Set . Note that by 3 above, lands in . Now here are some more properties that are not that hard to prove

- iff is trivial (see-saw principle)
- because and use divisibility.
- For , if is finite, then is surjective. (uses Leray Spectral sequence)

For the last bullet point in the introduction, namely If is a line bundle on s.t. there exists with then for all . The trick is to consider the augmented on and use the theorem of the cube.

### Dual Variety and the Poincare Bundle

From the discussion above, for such that is finite, the scheme can be constructed. Its called the dual abelian variety. There is a line bundle on called the Poincare bundle that has the following property. For every , , and is trivial. Furthermore, it is universal for this: if is a line bundle on such that and trivial on the identity on A, then there is a morphism such that . The most popular way to construct is to give decent data for on and the morphism , see page 78 of Mumford Abelian varieties for details.

Its not hard to check in the case of on , the morphism should in fact be . I probably should have said this earlier, but the good to choose for are the ample line bundles, any ample one will do.

## About this entry

You’re currently reading “Pic Zero and the Poincare Bundle,” an entry on Math Meandering

- Published:
- June 23, 2009 / 10:18 pm

- Category:
- Teleman

- Tags:
- theorem of the cube

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