Martin's Blog

Universal vector extensions of abelian varieties

Posted by Martin Orr on Friday, 09 May 2014 at 18:25

In the last post we showed that extensions of an abelian variety A by the additive group \mathbb{G}_a are classified by a vector space \operatorname{Ext}^1(A, \mathbb{G}_a) which is canonically isomorphic to T_0(A^\vee). In this post I will show that there is a so-called universal vector extension of A, that is, a vector extension E_0 of A such that every vector extension of A can be obtained in a unique way as a pushout of E_0. The vector group part of E_0 is \operatorname{Ext}^1(A, \mathbb{G}_a)^\vee, and an ingredient which we require from last time is that this is finite-dimensional.

Even if you are not interested in vector extensions of abelian varieties for their own sake (and I am not), they still have interesting properties. For example, I will show that they provide examples of algebraic varieties over the complex numbers which are non-isomorphic as algebraic varieties but which are isomorphic as complex manifolds.

Definition of universal vector extensions

As last time, we will only consider fields k of characteristic zero but most of the post remains valid without this assumption.

Let A be an abelian variety over k. The universal vector extension of A is an extension E_0 of A by a vector group V_0, such that for each vector extension E of A there exist unique morphisms of algebraic groups f and F making the following diagram commute:  \usepackage{xypic} \xymatrix{
   0   \ar[r]
 & V_0 \ar[r]  \ar[d]^f
 & E_0 \ar[r]  \ar[d]^F
 & A   \ar[r]  \ar[d]^{\id}
 & 0
\\ 0   \ar[r]
 & V   \ar[r]
 & E   \ar[r]
 & A   \ar[r]
 & 0
}

As defined in the previous post, this says that E is the pushout of E_0 by f \colon V_0 \to V.

By the standard universal property argument, if the universal vector extension of A exists, then it is unique up to unique isomorphism of vector extensions.

Construction of universal vector extension

We shall construct the universal vector extension, using the vector group \operatorname{Ext}^1(A, \mathbb{G}_a)^\vee. Last time we showed that this vector space is canonically isomorphic to T_0(A^\vee)^\vee, which is finite-dimensional.

Choose a basis x_1, \dotsc, x_n for \operatorname{Ext}^1(A, \mathbb{G}_a) and let E_1, \dotsc, E_n denote the corresponding extensions of A by \mathbb{G}_a. We can form the fibre product extension  0 \to \mathbb{G}_a^n \to E_1 \times_A \dotsb \times_A E_n \to A \to 0.

The choice of basis x_1, \dotsc, x_n induces an isomorphism u \colon \operatorname{Ext}^1(A, \mathbb{G}_a)^\vee \to \mathbb{G}_a^n given by  u(\alpha) = (\alpha(x_1), \dotsc, \alpha(x_n)). We define E_0 to be the extension  0 \to \operatorname{Ext}^1(A, \mathbb{G}_a)^\vee \to E_0 \to A \to 0 obtained by pushing out the above fibre product extension by u^{-1}. One can show by hand that the extension E_0 is independent of the choice of basis, or one can deduce this after proving the universal property.

Now we shall prove that E_0 has the required universal property. Every extension E of A by an arbitrary vector group V is isomorphic to a fibre product of extensions of A by \mathbb{G}_a (via an isomorphism between V and \mathbb{G}_a^m) so it suffices to prove that the universal property holds for extensions of A by \mathbb{G}_a.

So suppose that E is an extension of A by \mathbb{G}_a, and let x be the associated element of \operatorname{Ext}^1(A, \mathbb{G}_a). Define a morphism  f \colon \operatorname{Ext}^1(A, \mathbb{G}_a)^\vee \to \mathbb{G}_a by evaluating at x. Considering the definition of the vector space structure on \operatorname{Ext}^1(A, \mathbb{G}_a) shows that E is the pushout of E_0 by f, as required.

It remains to prove that f is unique. Suppose we had some other morphism  g \colon \operatorname{Ext}^1(A, \mathbb{G}_a)^\vee \to \mathbb{G}_a such that E is the pushout of E_0 by g. Since  \operatorname{Ext}^1(A, \mathbb{G}_a)^{\vee\vee} \cong \operatorname{Ext}^1(A, \mathbb{G}_a), g must be evaluation at some other point y \in \operatorname{Ext}^1(A, \mathbb{G}_a). But then E is the extension corresponding to y, and the definition of \operatorname{Ext}^1(A, \mathbb{G}_a) implies that x = y.

Universal extensions over the complex numbers

Let us consider what universal vector extensions look like over \mathbb{C}.

As remarked above, the vector space part of the universal vector extension is T_0(A^\vee)^\vee, the vector-space dual of the tangent space to dual abelian variety. Until recently, I imagined that this was naturally isomorphic to T_0(A), but in fact this is false (of course they are isomorphic because they are vector spaces of the same dimension, but there is no natural isomorphism).

For example, if the ground field is \mathbb{C} and V = T_0(A), then T_0(A^\vee)^\vee is the complex conjugate space \bar{V}, that is, the vector space with the same underlying set but where the action of \mathbb{C}^\times is replaced by its complex conjugate.

To construct the universal vector extension complex analytically, take an abelian variety A over \mathbb{C}, and let \Lambda be a lattice in V = T_0(A) such that A(\mathbb{C}) = V/\Lambda. Then the universal vector extension of A is  0 \to \bar{V} \to (V \oplus \bar{V}) / \{ (\lambda, \lambda) \mid \lambda \in \Lambda \} \to V/\Lambda \to 0.

It turns out that the universal vector extensions of different abelian varieties of the same dimension are all isomorphic to each other as complex Lie groups. To prove this, choose a \mathbb{Z}-basis \{ \lambda_1, \dotsc, \lambda_{2g} \} for \Lambda. The positivity of a Riemann form for A implies that (\lambda_i, \lambda_i) form a basis for V \oplus \bar{V} as a complex vector space, and hence  (V \oplus \bar{V}) / \{ (\lambda, \lambda) \mid \lambda \in \Lambda \} is isomorphic to (\mathbb{C}/\mathbb{Z})^{2g} as a complex Lie group, for every abelian variety A.

On the other hand, the universal vector extensions E_0, E'_0 of non-isomorphic abelian varieties A, A' are not isomorphic as algebraic varieties. To prove this, we will use the well-known fact (Proposition 3.9 of Milne's notes) that there every morphism of varieties from \mathbb{G}_a into an abelian variety is constant. If E_0 and E'_0 are isomorphic, we can consider the square  \usepackage{xypic} \xymatrix{
   E_0   \ar[r]  \ar[d]
 & A             \ar@{.>}[d]
\\ E'_0  \ar[r]
 & A'
} The fact we just mentioned implies that the composition E_0 \to A' is constant on each fibre of E_0 \to A, and so there exists a morphism A \to A' making the square commute. By also doing the same in the other direction, we get a morphism A' \to A and conclude that A and A' are isomorphic.

Hence we have constructed examples of smooth complex algebraic varieties which are not isomorphic, but for which the underlying complex manifolds are isomorphic. Note that this contrasts with projective varieties, such as abelian varieties themselves, where Chow's theorem implies that if two smooth projective varieties over \mathbb{C} are isomorphic as complex manifolds then they are isomorphic as algebraic varieties.

It is interesting to note that there is another algebraic group with the same underlying complex Lie group (\mathbb{C}/\mathbb{Z})^{2g} as the universal vector extension of an abelian variety of dimension g, namely \mathbb{G}_m^{2g}. This follows from the exact sequence  0 \rightarrow \mathbb{Z} \rightarrow \mathbb{C} \overset{\exp(2\pi i -)}{\longrightarrow} \mathbb{G}_m(\mathbb{C}). The algebraic group \mathbb{G}_m^{2g} is of course affine, while universal vector extensions of abelian varieties are not affine because any quotient of an affine algebraic group is also affine.

Tags abelian-varieties, alg-geom, maths

Trackbacks

  1. The Hodge filtration and universal vector extensions From Martin's Blog

    We will begin this post by looking at the isomorphism between the Hodge filtration H^{0,-1}(A) \subset H_1(A, \mathbb{C}) of a complex abelian variety A and the natural filtration ... on the tangent space to the universal vector extension of A. ...

Comments

  1. he said on Monday, 02 January 2017 at 21:26 :

    'pushing out the above fibre product extension by ''u^{-1}'''?

  2. Martin said on Friday, 06 January 2017 at 15:20 :

    Fixed. Thank you very much for the correction.

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