Can an abelian variety be represented as the cohomology of some other object?

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...

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?

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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, $\mathcal{A}^\vee(V) = \mathrm{Ext}^1_V(\mathcal{A},\mathbf{G}_m)$.

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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. – Johnson Jia Jan 27 '10 at 14:37

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.

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I'm not sure I understand your final two comments, both ideas are in the question. The question is - can it be done? – Dror Speiser Jan 26 '10 at 8:34
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. – Johnson Jia Jan 27 '10 at 14:32