Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

On an elliptic curve given by a degree three equation y^2 = x(x - 1)(x - λ), we can define the group law in the following way (cf. Hartshorne):

  1. We note that the map to its Jacobian given by $\mathcal{O}(p - p_0)$ for a fixed point $p_0$ is an isomorphism; ergo it inherits a group structure from the Jacobian.

  2. In fact, if we embed it into $\mathbb{P}^2$ via the linear system $|3p_0|$, then three colinear points have $p + q + r \sim 3p_0$ and so this is in fact the group law inherited from Pic0.

Is there an analogous way to do this for Abelian varieties? In Lange & Birkenhake they simply define an Abelian variety to be $\mathbb{C}^n$ modulo a lattice, and so it automatically comes with a group structure. Still, this seems unsatisfying in comparison to the way we can do so for elliptic curves.

That being said, the previous method doesn't seem to work for Abelian varieties; divisors no longer correspond to formal sums of points and so any comparison with Pic0 wouldn't obviously yield a group structure on the points of X.

To make matters worse, the map that I tend to think of which takes X to Pic0(X) is given by $p \mapsto t_p^*L \otimes L^{-1}$ for a given line bundle $L$ on X, where $t_p : X \to X$ is the map... defined by translation in X. So this map already requires the group structure on X to be defined.

So is there a way of defining the group law analogous to that of an elliptic curve?

NB: I do note that an elliptic curve can be defined as the zero locus of a cubic equation in $\mathbb{P}^2$, where I'm not sure how else we might define an Abelian variety other than as $\mathbb{C}^n$ modulo a lattice, and so perhaps the question is moot.

share|improve this question
4  
I think that the best definition of abelian variety is "a complete variety with algebraic group structure", so in this sense, the question $is$ moot –  Victor Protsak May 19 '10 at 19:02
    
Yeah, I was thinking that, but that just feels so unsatisfying. –  Simon Rose May 19 '10 at 19:19

3 Answers 3

up vote 5 down vote accepted

Any torsor $V$ under an abelian variety over any field $K$ is caonically isomorphic to its degree $1$ Albanese variety $\operatorname{Alb}^1(V)$, which is itself a torsor under the degree $0$ Albanese variety $\operatorname{Alb}^0(V)$. Note that the $\operatorname{Alb}^0$ of any smooth projective variety is a complete, geometrically connected group variety.

Assuming there exists a $K$-rational point $O$, one can subtract $O$ to obtain an isomorphism $\operatorname{Alb}^1(V) \stackrel{\sim}{\rightarrow} \operatorname{Alb}^0(V)$. Pulling back via the composite of these isomorphisms puts a group structure on $V$ depending only on the chosen base point $O$.

share|improve this answer
    
What do you mean by "degree 1 Albanese variety"? The definition I know of the Albanese variety of a (smooth or normal say) proper variety $X$ is as a universal map $X \to V$ to torsors over abelian varieties. Hence, in your case the Albanese map is the identity map (which is what you say) but I do not understand what you mean by degree $1$ and degree $0$. –  Torsten Ekedahl May 19 '10 at 19:53
    
For a smooth projective variety $V$, one has a total Albanese scheme $\operatorname{Alb}(V)$, whose points parameterize all zero-cycles on $V$. This comes with a degree map to $\mathbb{Z}$, and $\operatorname{Alb}^i(V)$ is the component consisting of zero-cycles of degree $i$. Each $\operatorname{Alb}^i(V)$ is -- in an evident way -- a torsor under $\operatorname{Alb}^0(V)$. –  Pete L. Clark May 19 '10 at 20:02
    
For some more details, see Sections 4.1 and 4.3 of math.uga.edu/~pete/wc2.pdf. (Nevertheless these concepts are not due to me.) –  Pete L. Clark May 19 '10 at 20:04
    
I understand, I was thinking about the possibility that you meant zero-cycles but was thrown off by the fact that rational equivalence doesn't work. If I remember correctly the equivalence relation you are talking about is "abelian equivalence", the equivalence relation generated by maps into torsors over abelian varieties. This gives the quite tautological relation with the definition I was referring to. Of course there is a formal definition of the degree $1$ part, in that a torsor over $A$ gives an extension $0\to A \to A' \to \mathbb Z \to 0$ with the torsor the inverse image of $1$. –  Torsten Ekedahl May 19 '10 at 20:53

In dimension 1, the situation is like this: Every smooth proper genus 1 curve with a "marked" rational point has a unique group structure such that the given rational point becomes the neutral element.

For abelian varieties it is still true that the group structure, the origin being fixed, is unique: an abelian variety is canonically isomorphic to its Albanese variety, and to construct these you don't need the group structure.

share|improve this answer
    
The canonicity of the structure of a group is easier than the Albanese stuff. Any morphism from an Abelian to another sending 0 to 0 is a morphism of groups. So if you thought you had two group structures on (A,0), well, you only have one and they coincide! –  ACL Nov 26 '10 at 19:33

For algebraically complete integrable systems Abelian varieties usually show up as follows: One has the preimage under n commuting integrals on a complex 2n-dim symplectic manifold. Then, for regular values, assuming that the integrals are proper, one has compact fibers whith n commuting holomorphic vectorfields (the symplectic gradients of the integrals) without zeros. This is an algebraic variety, and the group structure comes from its Lie algebra by flowing along the vectorfields.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.