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This question is a follow up to an earlier question of mine on enumerating elliptic curves of a given conductor.

I've heard people say that studying higher dimensional varieties via explicit defining equations often leads to hopelessly unmanageable complexity. As apparent evidence of this, on page 65 of Cassels' and Flynn's book titled Prolegomena to a Middlebrow Arithmetic of Curves of Genus $2$, the authors state that the defining equations that they find for Jacobian varieties of genus $2$ curves consist of $72$ quadratic equations in $\mathbb P^{15}$. People say that rather than studying higher dimensional algebraic varieties as solution sets to explicit equations, one typically studies such varieties in a more abstract and geometric way.

This makes sense. Yet I wonder how one can get look at concrete examples without defining equations. I know that there are some varieties such as moduli spaces which provide examples. But suppose, say, you want to prove that there are finitely abelian varieties of an arbitrary fixed dimension $d$ over a fixed arbitrary number field $K$ with a fixed conductor $N$ without looking at the automorphic side of things.

Is there a (conjectural) method of proving finiteness without writing down explicit defining equations?

[Edit: As Barinder Banwait points out, this follows immediately from the Shafarevich conjecture, which was proved by Faltings.]

Pushing the envelop further,

Is there a (conjectural) algorithm for enumerating these objects?

Pushing the envelop still further,

Suppose beyond enumerating such varieties, you want to determine, e.g., how many of them have surjective (mod $5$) Galois representation (say, attached to $H^1$) - is there a conjectural algorithm for doing so?

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Jonah Sinick
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    The answer to your first question is YES; it is known as "Shafarevich conjecture for abelian varieties" (though it's not a conjecture anymore). See this blog post of Martin Orr for an overview: http://www.martinorr.name/blog/2011/09/19/finiteness-theorems-for-abelian-varieties – Barinder Banwait Sep 18 '12 at 18:46
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    Warning: Even for Jacobians $J$ of curves $C$ of genus $g \geq 2$, it can happen that $C$ has good reduction at a prime $p$ but $C$ does not. For $g=2$ the list of curves over ${\bf Q}$ with good reduction outside $\lbrace 2 \rbrace$ is known, but not so the list of Jacobians with good reduction outside $\lbrace 2 \rbrace$. Yes, it follows from Faltings that this list is finite, but the proof is not effective (though I think it does yield an upper bound on the size of the list, at least in principle). – Noam D. Elkies Sep 18 '12 at 19:02
  • @ Barinder Banwait - yes, of course - I'm sheepish for having missed this, as it was already mentioned in the answers to my last question! – Jonah Sinick Sep 18 '12 at 19:33
  • @ Noam D. Elkies - I was asking about abelian varieties rather than curves and so don't understand the relevance of your comment - am I missing something? – Jonah Sinick Sep 18 '12 at 19:36
  • In the case $d=2$, every abelian surface is isogenous to a Jacobian or to a product of elliptic curves, so we are reduced to the case of Jacobians. But as Noam D. Elkies shows, this problem is strictly more difficult than that of curves. – François Brunault Sep 18 '12 at 20:36
  • @Noam: Typo in your comment, it should say "it can happen that $J$ has good reduction but $C$ does not". Anyway, on a more serious note, the last time I thought about Faltings' proof, I convinced myself that it doesn't even give an effective method for finding an upper bound for the number of abelian varieties having good reduction outside of $S$. And hence it does not give an effective bound for $#C(K)$, where $C$ is a curve of genus $g\ge2$ and $K$ is a number field. However, Vojta's proof does give an effective (huge) bound for $#C(K)$. – Joe Silverman Sep 18 '12 at 20:46
  • @Jonah: You might run across a list of the genus-2 curves with good reduction outside {2} (it's somewhat surprisingly large), so I was warning that it doesn't answer you question even for principally polarized abelian surfaces. (@ François Brunault: every principally polarized surface.) @Joe: Yes, a typo; can't correct it on a comment :-( $\phantom+$ OK, I wasn't sure about Faltings' first proof; I think his second proof (which is much the same as Vojta's) does give an effective bound, which isn't all that huge if you know the rank of the Jacobian and exclude points of small height. – Noam D. Elkies Sep 18 '12 at 21:24
  • @Noam: If you exclude points of small height, then conjecturally there aren't any points left in $C(K)$, so I think the correct bound would be zero. The provable bounds are certainly much larger than that. I'm not quite sure what you mean by Faltings' second proof. Does the Vojta/Faltings proof of Mordell-Lang for subvarieties of abelian varieties give the Shafarevich conjecture? There are also the transcendence theory proofs of some versions of Tate's isogeny conjecture (proven by Faltings). I'm not sure to what extent they're strong enough to apply to the Shafarevich conjecture. – Joe Silverman Sep 18 '12 at 22:26
  • Sorry for being unclear: indeed I meant only Faltings' second proof of Mordell, which as far as I know is not known or expected to imply Shafarevich, and is thus relevant only to the size of $C(K)$, not abelian varieties with fixed conductor which was the original question. (And yes, the provable bounds on the number of large points are much larger than zero, but the combined bound on the number of all rational points might be not too far from the truth, at least for some $C$.) – Noam D. Elkies 0 secs ago – Noam D. Elkies Sep 18 '12 at 23:18
  • @Noam : Right, I was thinking over $\mathbf{C}$, here one may need to extend the base field. – François Brunault Sep 21 '12 at 14:41

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