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Fix a positive integer $g$. What positive integer $N$ can be the conductor of a $g$-dimensional abelian variety over $\mathbb Q$ ?

For example, as there is no abelian varieties over $\mathbb Z$, $N$ can not be $1$. And for elliptic curves, $N$ must be no less than $11$.

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2 Answers 2

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You seem to be asking about which small integers cannot occur as the conductor $N$ of an abelian variety over $\mathbb Q$ of dimension $g$, but possibly you'll also be interested in other constraints on the value of $N$. It's relatively easy to see that for primes $p>2g+1$, one has $\operatorname{ord}_p(N)\le2g$. For $p\le2g+1$, there are also upper bounds for $\operatorname{ord}_p(N)$, but they're more complicated. See for example

  • Lockhart-Rosen-Silverman, An upper bound for the conductor of an abelian variety. J. Algebraic Geom. 2 (1993), no. 4, 569–601 (MR1227469).
  • Brumer--Kramer, The conductor of an abelian variety. Compositio Math. 92 (1994), no. 2, 227–248 (MR1283229).
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  • $\begingroup$ Thank you! I want to know possible value of $N$, is your bound for valuation of $N$ the optimal bound? $\endgroup$
    – Zhiyu
    Commented Feb 26, 2019 at 3:34
  • $\begingroup$ The Brumer-Kramer bounds on ord$_p(N)$ are optimal, if I remember correctly, but you should check their paper. $\endgroup$
    – Joe Silverman
    Commented Feb 26, 2019 at 12:27
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There is work by Brumer and Kramer, Paramodular abelian varieties of odd conductor, on the possible conductors. They show for example that if $A$ is a semistable abelian surface over $\mathbb{Q}$ of odd non-square conductor $N$, then $N \geq 249$ (and there is an explicit abelian surface with this conductor). They also give tables of possible odd conductors $N \leq 1000$.

The Langlands philosophy predicts that abelian varieties of dimension $g$ over $\mathbb{Q}$ should (roughly) correspond to automorphic forms on $\mathrm{GSp}_{2g}/\mathbb{Q}$. There is even a precise conjecture, see Section 8 in the article of Brumer and Kramer. If you assume this conjecture, then you are lead to investigate the space of such automorphic forms, and for what levels the space is non-trivial. This gives only a necessary condition, because the field of Hecke eigenvalues may be larger than $\mathbb{Q}$. There has been work by Poor and Yuen, Paramodular cusp forms where they classify such automorphic forms. Their results support the above conjecture.

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    $\begingroup$ .. and I will add that on the Galois side, lists of genus two curves with small conductors have been produced and included in the lmfdb : e.g. lmfdb.org/Genus2Curve/Q/?cond=1-499 $\endgroup$
    – Chris Wuthrich
    Commented Feb 26, 2019 at 11:08
  • $\begingroup$ Thank you, are there some good sufficient conditions on $N$ to imply it's a conductor? $\endgroup$
    – Zhiyu
    Commented Feb 26, 2019 at 19:59
  • $\begingroup$ @zzy I think this is a difficult problem even for elliptic curves. I think Neumann has classified the E/Q with prime conductor and a rational 2-torsion point. But in general I'm not aware of any algorithm other than computing the whole space of modular symbols of level N. $\endgroup$
    – François Brunault
    Commented Feb 26, 2019 at 21:19
  • $\begingroup$ @FrançoisBrunault I see...For something weaker, can we find an elliptic curve over $\mathbb Q$ with prescribed places of bad reduction? $\endgroup$
    – Zhiyu
    Commented Feb 26, 2019 at 23:35
  • $\begingroup$ @zzy Shafarevich's theorem guarantees that there are only finitely many isomorphism classes for a fixed set of primes. There is an algorithm due to Cremona and Lingham to determine them, see homepages.warwick.ac.uk/~masgaj/papers/egros.pdf $\endgroup$
    – François Brunault
    Commented Feb 27, 2019 at 9:20

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