4
$\begingroup$

Question

Given an abelian variety $V$ and an integer $n$, is there a natural abelian category with a natural object $X$ and natural coefficients $F$ so that $V\simeq H^n (X,F)$?

Motivation

Studying abelian varieties is awesome. Studying objects in long exact sequences is awesome. How do (somewhat forcefully) combine these two? I mean without taking cohomology of the variety like everyone else does...

Possible answers

The abelian variety is a $G$-module, where $G=Gal(\bar{k}/k)$, $k$ the field over which the variety is defined. So, maybe there is an interesting $G$-module that answers the above? The cases of abelian varieties over number fields and finite fields are the most interesting, so $G$ is assumed to be interesting as well (i.e. not trivial).

Maybe it arises as the $n$-th cohomology of some interesting sheaf of some interesting related variety?

$\endgroup$

2 Answers 2

2
$\begingroup$

Let $\mathcal{A}/S$ be an Abelian scheme. Then the dual Abelian scheme is given by $R^1\pi_*\mathcal{A}$, if I remember correctly. Also, $\mathcal{A}^\vee(V) = \mathrm{Ext}^1_V(\mathcal{A},\mathbf{G}_m)$.

$\endgroup$
1
  • $\begingroup$ Yes, this is correct (it is discussed in Milne's article from the Storr's volume on Arithmetic Geometry). I did not think about fppf sheaves when I posted my "answer" below. My bad. $\endgroup$ Jan 27, 2010 at 14:37
1
$\begingroup$

You probably want the isomorphism above to respect some additional structures; otherwise if we view $V(\bar k)$ (assuming $V$ is defined over a field $k$) as just an abelian group, then take any finitely generated group $G$ (say the trivial group) and endow it with the trivial action on $V(\bar k)$, we will have $H^0(G, V(\bar k)) = V(\bar k)^G = V(\bar k)$. I don't think this is what you want.

Perhaps you would like to have some Galois action on the cohomology as well?

Putting some additional restriction on $X$, say, demanding it to be a geometric object (like a scheme) would help, too.

$\endgroup$
2
  • $\begingroup$ I'm not sure I understand your final two comments, both ideas are in the question. The question is - can it be done? $\endgroup$ Jan 26, 2010 at 8:34
  • $\begingroup$ My initial concern was that cohomology is usually a functor $$H^i \colon \mathbf{K}(\mathcal{A}) \to \mathcal{A}$$ from the derived category $\mathbf{K}(\mathcal{A})$ to an abelian category $\mathcal{A}$. Now the cohomological groups I have encountered are ususally modules, which, unlike abelian varieties, are not geometric objects. With that said, I looked up Milne's article on Abelian Varieties from the Storrs volume, apparently you can identify $$V^{\vee} \simeq \mathit{Ext}^1(V, \mathbf{G}_m)$$ where the is $\mathit{Ext}$ taken in the cateogry of fppf sheaves. Hopefully this helps. $\endgroup$ Jan 27, 2010 at 14:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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