Martin's Blog

Dual abelian varieties and line bundles

Posted by Martin Orr on Monday, 09 May 2011 at 14:30

The definition I gave last time of dual abelian varieties was very much dependent on complex analytic methods. In this post I will explain how dual varieties can be interpreted geometrically: the points of A^\vee correspond to a certain group of line bundles on A. We construct a single line bundle \mathcal{P} on the product A \times A^\vee, the Poincaré bundle, such that all line bundles on A arise as restrictions of \mathcal{P}, and show that the pair (A^\vee, \mathcal{P}) satisfies a universal property.

Homologically equivalent line bundles

Let A be a complex abelian variety and \Lambda = H_1(A). Recall that \operatorname{Pic}(A) denotes the group of isomorphism classes of line bundles on A. We shall define an equivalence relation, called homological equivalence, on \operatorname{Pic}(A).

Recall that line bundles on A are classified by the Appell-Humbert theorem. This will play a vital role throughout this post.

Appell-Humbert Theorem. Every line bundle on a complex abelian variety A is isomorphic to \mathcal{L}(H, \alpha) for exactly one pair (H, \alpha) of a Hermitian form H on \Lambda_\mathbb{R} and a semi-character \alpha : \Lambda_{\mathbb{Z}} \to S^1.

We say that two line bundles are homologically equivalent if their Hermitian forms H are the same.

In order to explain the terminology, we should interpret this via homology: because abelian varieties are non-singular varieties, line bundles biject with linear equivalence classes of Weil divisors. A Weil divisor is a linear combination of subvarieties of codimension 1. Now a subvariety of complex codimension 1 is a real manifold of dimension 2g-2 which gives us a homology class in H_{2g-2}(A, \mathbb{Z}).

This gives a homomorphism \operatorname{Pic}(A) \to H_{2g-2}(A, \mathbb{Z}). If you compose this with Poincaré duality H_{2g-2}(A, \mathbb{Z}) \cong H^2(A, \mathbb{Z}) then you get the cohomology class which corresponds to the Hermitian form H. So two line bundles are homologically equivalent iff they have the same homology class in H_{2g-2}(A, \mathbb{Z}).

The dual abelian variety and \operatorname{Pic}^0(A)

Write \operatorname{Pic}^0(A) for the group of isomorphism classes of line bundles on A homologically equivalent to 0. By using Appell-Humbert again, we shall show that \operatorname{Pic}^0(A) is isomorphic to the group of complex points of A^\vee.

By Appell-Humbert, line bundles homologically equivalent to 0 correspond to semi-characters \Lambda_\mathbb{Z} \to S^1 with respect to the zero Hermitian form. The latter are just group homomorphisms \Lambda_\mathbb{Z} \to S^1 so we get  \operatorname{Pic}^0(A) \cong \operatorname{Hom}(\Lambda_{\mathbb{Z}}, S^1).

We defined the dual abelian variety A^\vee to be the complex torus \Lambda_\mathbb{R}^\vee / \Lambda_\mathbb{Z}^\vee coming from the H_1-dual Hodge structure \Lambda^\vee.

Since \Lambda_\mathbb{Z} is a free \mathbb{Z}-module, the functor \operatorname{Hom}_{\mathbb{Z}}(\Lambda_{\mathbb{Z}}, -) is exact. Applying it to the short exact sequence  0 \to \mathbb{Z} \to \mathbb{R} \overset{\exp(2 \pi i -)}\longrightarrow S^1 \to 0 gives a short exact sequence  0 \to \Lambda_\mathbb{Z}^\vee \to \Lambda_\mathbb{R}^\vee \to \operatorname{Hom}(\Lambda_{\mathbb{Z}}, S^1) \to 0.

So we get an isomorphism of groups  A^\vee(\mathbb{C}) \cong \Lambda_\mathbb{R}^\vee / \Lambda_\mathbb{Z}^\vee
   \cong \operatorname{Hom}(\Lambda_{\mathbb{Z}}, S^1) \cong \operatorname{Pic}^0(A), with a point f \in \Lambda_\mathbb{R}^\vee corresponding to the line bundle \mathcal{L}_f = \mathcal{L}(0, \exp(2 \pi i f(-))).

The Poincaré bundle

We would like to show that the above group isomorphism A^\vee(\mathbb{C}) \to \operatorname{Pic}^0(A) is geometrically well-behaved. The right way to interpret "geometrically well-behaved" turns out to be that the line bundles \mathcal{L}_f on A fit together (as we vary f \in A^\vee) into a single line bundle \mathcal{P} on A \times A^\vee.

We define a Poincaré bundle for A to be a line bundle \mathcal{P} on A \times A^\vee satisfying

  1. \mathcal{P}_{|A \times \{ f \}} \cong \mathcal{L}_f for all f \in A^\vee;
  2. \mathcal{P}_{|\{ 0 \} \times A^\vee} is trivial.

Condition 1 is the important condition; condition 2 is just a normalisation to ensure that the Poincaré bundle is unique.

We construct \mathcal{P} by using the Appell-Humbert theorem again: we need to give a Hermitian form H on \Lambda \times \Lambda^\vee (or equivalently a Hodge symplectic form E on \Lambda \times \Lambda^\vee) and a semi-character \alpha : \Lambda_\mathbb{Z} \times \Lambda_\mathbb{Z} \to S^1 and then we take \mathcal{P} = \mathcal{L}(H, \alpha).

We know that \mathcal{P}_{|A \times \{ 0 \}} \cong 0. Subject to this condition, we can do a calculation which shows that  \mathcal{P}_{|A \times \{ f \}} \cong \mathcal{L}(0, \delta_f)
   \text{ where } \delta_f(\lambda) = \exp(2 \pi i E((0, f), (\lambda, 0))).

Hence in order to satisfy condition 1, we must have  E((0, f), (\lambda, 0)) = f(\lambda). Combined with condition 2, this forces  E((v_1, f_1), (v_2, f_2)) = f_1(v_2) - f_2(v_1),  \alpha((v, f)) = (-1)^{f(v)}.

This shows that there is indeed a unique Poincaré bundle \mathcal{P} on A \times A^\vee.

Universal property of the Poincaré bundle

We finish by proving that the Poincaré bundle satisfies a universal property. In this proposition, you should regard \mathcal{M} as being "a family of line bundles on A parameterised by T".

This is not just a universal property of \mathcal{P}, but of the pair (A^\vee, \mathcal{P}): this pair is universal among "families of line bundles on A (homologically equivalent to 0) parameterised by normal varieties and trivial at 0 \in A".

As always, the universal pair (A^\vee, \mathcal{P}) is unique up to unique isomorphism, and so we could use this property to define the dual abelian variety. This is exactly how we will soon define the dual abelian variety over fields other than \mathbb{C}.

Proposition. Let T be a normal variety over \mathbb{C} 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 varieties \psi : T \to A^\vee such that (1 \times \psi)^* \mathcal{P} \cong \mathcal{M}.

Sketch proof. Set theoretically, \psi must be defined by: \psi(t) is the point in A^\vee corresponding to \mathcal{M}_{|A \times \{ t \}} \in \operatorname{Pic}^0(A). We need to show that this is a morphism of varieties.

Let \Gamma be the graph of \psi in T \times A^\vee.

Let \mathcal{N} be the line bundle p_{12}^* \mathcal{M} \otimes p_{13}^* \mathcal{P}^{-1} on A \times T \times A^\vee. Then  \Gamma = \{ (t, f) \mid \mathcal{N}_{|A \times \{t, f\}} \text{ is trivial} \} and hence is closed in T \times A^\vee.

The projection p_1: \Gamma \to T is bijective, so by Zariski's Main Theorem, it is an isomorphism of varieties. Hence \psi : T \overset{p_1^{-1}}\longrightarrow \Gamma \overset{p_2}\longrightarrow A^\vee is a morphism of varieties.

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

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