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Dual varieties over general fields

Posted by Martin Orr on Friday, 24 June 2011 at 17:26

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 satisfies the universal property is harder; in the case of number fields, we will use Galois descent to deduce it from the complex case which we already know analytically.

Definition of the dual abelian variety

We define the dual abelian variety A^\vee to be an abelian variety over k such that there exists a line bundle \mathcal{P} on A \times A^\vee satisfying the following universal property:

Let T be a normal k-variety and \mathcal{M} a line bundle on A \times T such that

  1. \mathcal{M}_{|A \times \{ t \}} \in \operatorname{Pic}^0(A) for all t \in T;
  2. \mathcal{M}_{|\{ 0 \} \times T} is trivial.

Then there is a unique morphism of k-schemes \psi : T \to A^\vee such that (1 \times \psi)^* \mathcal{P} \cong \mathcal{M}.

In fact, this property is satisfied for any k-scheme T. This tells us exactly what the morphisms T \to A^\vee, and so an equivalent statement would be:

The dual abelian variety A^\vee is the k-scheme whose functor of points is  A^\vee(T) = \{ \mathcal{M} \in \operatorname{Pic}^0(A \times T) \mid \mathcal{M}_{|\{0\} \times T} \text{ is trivial} \}.

Construction of dual abelian variety as a quotient

Last time we defined a homomorphism \phi_\mathcal{L} : A(k) \to \operatorname{Pic}^0(A) for each line bundle \mathcal{L} on A. Over the complex numbers, we know that if \mathcal{L} is ample then \phi_\mathcal{L} is an isogeny A \to A^\vee. Hence A^\vee is isomorphic to the quotient A/\operatorname{ker} \phi_\mathcal{L}.

We will copy this construction over an arbitrary field k: Choose an ample line bundle \mathcal{L} on A. Let A^\vee = A/K(\mathcal{L}), where K(\mathcal{L}) is the subgroup scheme of A defined last time whose k-points are \operatorname{ker} \phi_\mathcal{L}. Then we "just" have to specify a Poincaré bundle on A \times A^\vee and prove that it satisfies the universal property.

Before doing that, we need to specify what we mean by the quotient of group schemes A/K(\mathcal{L}). The theory of quotients of schemes by group actions is tricky, but fortunately we need only a very simple case, because we saw last time that K(\mathcal{L}) is finite. (Note that K(\mathcal{L}) is a finite group scheme, not just a finite group, but this does not really cause trouble.) The key result is the following:

Theorem. Let V be a k-variety and G a finite group k-scheme acting on V by regular morphisms. Suppose that every G(\bar{k})-orbit is contained in an open affine subset of V_{\bar{k}}. Then there exists a variety W and a finite regular morphism \pi : V \to W such that any G-invariant regular morphism f : V \to W' factorises uniquely as  \usepackage[matrix,arrow]{xy} \xymatrix@1{ V \ar[r]^\pi & W \ar@{-->}[r]^{f'} & W'. }

Sketch proof. The condition on orbits implies that it suffices to prove the theorem for affine varieties and glue. So suppose that V = \operatorname{Spec} A. Let B = (A \otimes_k \bar{k})^{G(\bar{k})} \cap A and W = \operatorname{Spec} B. Then proving that W has the required properties is a matter of commutative algebra.

The action of K(\mathcal{L}) on A (by translations) satisfies the conditions of this theorem, so the quotient variety A^\vee = A/K(\mathcal{L}) exists. It inherits a group structure from that of A. To say more about A^\vee, we need the following result on sheaves on quotient varieties.

Theorem. In the situation of the previous theorem, suppose also that the action of G on V is free. Then \mathcal{F} \mapsto \pi^*\mathcal{F} is an equivalence of categories  \{ \text{coherent sheaves on } W \} \to \{ \text{coherent sheaves on } V \text{ with } G\text{-action} \}. Here a sheaf with G-action means a sheaf on V together with isomorphisms \mathcal{F} \otimes R \cong g^* \mathcal{F} \otimes R for every g \in G(R) and every k-algebra R satisfying the obvious cocycle and functoriality conditions.

By the definition of K(\mathcal{L}), the line bundle \mathcal{L} itself has a K(\mathcal{L})-action, so it corresponds to a line bundle \mathcal{L'} on A^\vee, which can be proved to be ample. So A^\vee is an abelian variety.

Similarly, in order to construct the Poincaré bundle \mathcal{P} on A \times A^\vee, we need to find the bundle \Lambda(\mathcal{L}) = (1 \times \phi_\mathcal{L})^* \mathcal{P} on A \times A, then take its quotient by the action of K(\mathcal{L}) on the second A.

This bundle should have the property  \Lambda(\mathcal{L})_{|A \times \{x\}} \cong \phi_\mathcal{L}(x). The following bundle has this property:  \Lambda(\mathcal{L}) = m^* \mathcal{L} \otimes p_1^* \mathcal{L}^{-1} \otimes p_2^* \mathcal{L}^{-1} where m : A \times A \to A is the group law. Furthermore, this bundle has a canonical action of K(\mathcal{L}) compatible with the action on the second factor of A \times A, so it descends to a line bundle \mathcal{P} on A \times A^\vee as required.

Proof of universal property

The proof that the pair (A^\vee, \mathcal{P}) satisfies the universal property is too hard to include here. It can be found in Mumford's book (at least for algebraically closed fields). It is similar in outline to the proof we gave over the complex numbers, but involves hard cohomological arguments.

However, I am really interested in number fields. In this case, we can deduce the universal property for (A^\vee, \mathcal{P}) from the fact that we already know it holds over \mathbb{C}.

So suppose we are given a k-variety T and a line bundle \mathcal{M} on A \times T satisfying conditions 1 and 2 above. Extending scalars to \mathbb{C}, we know that there is a unique morphism  \psi : T_{\mathbb{C}} \to A^\vee_{\mathbb{C}} such that (1 \times \psi)^* \mathcal{P} \cong \mathcal{M}. We just have to show that \psi is defined over k.

But if \sigma \in \operatorname{Aut}(\mathbb{C}/k), then (1 \times {^\sigma \psi}) \mathcal{P} \cong {^\sigma \mathcal{M}} \cong \mathcal{M}, since both \mathcal{P} and \mathcal{M} are defined over k. So by the uniqueness of \psi we have {^\sigma \psi} = \psi.

Polarisations

We can show that the group homomorphisms \phi_\mathcal{L} : A(k) \to \operatorname{Pic}^0 A (for any line bundle \mathcal{L}) are actually morphisms of abelian varieties A \to A^\vee using the universal property: \phi_\mathcal{L} is the morphism associated with the line bundle \Lambda(\mathcal{L}) = m^* \mathcal{L} \otimes p_1^* \mathcal{L}^{-1} \otimes p_2^* \mathcal{L}^{-1}.

A polarisation of A is defined to be an isogeny \phi : A \to A^\vee such that, after extending scalars to \bar{k}, \phi = \phi_\mathcal{L} for some ample line bundle \mathcal{L} on A_{\bar{k}}. According to Milne, this is not quite the same as \phi being of the form \phi_\mathcal{L} for some ample line bundle on A itself.

Tags abelian-varieties, alg-geom, hodge, maths

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Comments

  1. Barinder said on Sunday, 07 August 2011 at 12:15 :

    In the non-principally polarised case, over a number field, how different can be the arithmetic between A and its dual? I think the torsion and ranks have to be the same, for example.

    Where do we "see" the dual abelian variety in number theory? Am I right in saying that it only matters when our abelian variety is not principally polarised? If so, what are some interesting arithmetic questions regarding non-principally polarised abelian varieties? Where do they arise in your work on André-Oort?

    By the way, I will come on the evening of the 8th, and my train leaves on the 10th at 17:13. Perhaps we can visit Versailles on the Saturday?

  2. Martin Orr said on Sunday, 14 August 2011 at 20:21 :

    The arithmetic is going to be pretty similar because the cohomology of the dual abelian variety is the dual of the cohomology of A. For example this applies to the Tate modules which implies that the torsion groups are isomorphic.

    The important thing about the dual abelian variety is that you need it to define the Weil pairing which I intend to discuss in my last post in the series. I am not really aware of anything interesting about non-principally polarised AVs - one often begins by reducing questions about them to something about ppAVs. For example we might use the following theorems:

    Theorem. If A has a polarisation whose degree is not divisible by the characteristic of the base field, then A is isogenous to a ppAV.

    Zarhin's Trick. If A is any abelian variety, then (A \times A^\vee)^4 is principally polarisable.

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