this is a very vague question, but I'm also glad about vague answers...

One knows that for an abelian variety over the complex numbers, one has a canonical exact sequence

$0\rightarrow \Omega^1(X) \rightarrow H^1_{DR}(X) \rightarrow H^1(X,\mathcal O_X) \rightarrow 0$.

On the other hand there is a canonical group isomorphism

$Ext(X,G_a) \simeq H^1(X,\mathcal O_X)$,

which I will call the iso (+). Here on the left we have the group of extensions of $X$ by the additive group $G_a$, i.e. in our case just the complex numbers considered as algebraic group.

Know I would like to have a group $H$ which has the following properties:

1) It induces an iso $H \simeq H^1_{DR}(X)$

2) It projects on $Ext(X,G_a)$ naturally

3) The properties in 1) and 2) are compatible with the iso (+).

Does such a group exist? How do you get it?

Does this have to do with biextensions?


  • $\begingroup$ There must be more to your conditions. I can just define the 'natural' projection $H^1_{DR}(X)\to Ext(X,G_a)$ to be the composition of the canonical projection onto $H^1(X,\mathcal{O}_X)$ followed by the canonical isomorphism with $Ext(X,G_a)$. What's not natural about this? $\endgroup$ – Keerthi Madapusi Pera Sep 29 '11 at 12:48

Perhaps what you are looking for is the universal vectorial (aka additive) extension of an abelian variety; this has been discussed by Rosenlicht, Mazur-Messing and been applied to height pairings by Coleman and others. This universal object is an algebraic group whose Lie algebra is the first de Rham cohomology of the abelian variety.

see here

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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