# Almost Northcott properties for heights of abelian varieties

Let $h$ be a function on the moduli space of abelian varieties of dimension $g$ over $\overline{\mathbf{Q}}$.

Let $K$ be a number field and let $g\geq 2$ be an integer. Fix a real number $C$. Does the finiteness of the set $$\{A/\overline{\mathbf{Q}}: \dim A = g, h(A) \leq C \}/\{\overline{\mathbf{Q}}-\mathrm{isomorphism}\}$$ imply the finiteness of the set $$\{A/K: \dim A=g, h(A_{\overline{\mathbf{Q}}}) \leq C, A/K \ \textrm{ has semi-stable reduction}\}/\{K-\mathrm{isomorphism}\}$$

In simple words, suppose you have a function on the moduli space of abelian varieties over $\overline{\mathbf{Q}}$ with the Northcott property. Then, can one deduce an almost Northcott property for abelian varieties over a fixed number field?

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It might help if you could suggest what your function h should look like. For example, let's take $g=1$. Then I guess you could use $h(E) = h(j(E))+[\mathbb{Q}(j(E)):\mathbb{Q}]$, where $h(j(E))$ is the usual Weil height. Then bounding $h(E)$ bounds both $h(j(E))$ and the degree of the field of definition, so you'll get finiteness. You second set will also be finite. Or you could use the height $h(A)$ defined by Faltings, in which case again I think the answer to your question is yes. (Although I worry a little bit about polarizations, maybe it would be better to restrict to principally polarized $A$.) Anyway, do I understand correctly that assuming that the first set is finite, then you're asking if, for a fixed $A/K$, there are only finitely many twists $B/K$ of $A/K$ that are semistable and have bounded $h(B)$? That seems likely to be true for any reasonable function $h$ having the initial property, but maybe you can cook up a strange $h$ that won't work.

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Actually, the function I'm interested in is only defined for curves (i.e., Jacobians of curves) and I actually only want to restrict to this set. I forgot about the difficulty with polarizations, and therefore thought this generalization could be true without assuming polarizations. In any case, restricting to curves we don't encounter this difficulty. I agree with your last sentence. I guess if we take h to be the Faltings height or the theta height, then what I'm asking for is true if bounded Faltings height (or theta height) means that the degree of the field of definition is bounded. True? –  Hafez Oct 27 '11 at 8:03
Also, as you made clear to me, I'm asking if, for a fixed A/K with bounded h(A_Qbar), there are only finitely many twists B/K of A/K that are semi-stable. (This is slightly different than how you phrased it because every twist B/K of A/K has the same h as A. That is, h(B) = h(A).) –  Hafez Oct 27 '11 at 8:11

Your function $h$ seems irrelevant. Your question is equivalent to the following:

Given semistable $A/K$ of dimension $g$, are there finitely many semi-stable $B/K$ which are isomorphic to $A$ after base change to $\bar{K}$? I believe the answer is yes. To prove this, note that $B$ must have good reduction everywhere $A$ has potentially good reduction, so everywhere away from some finite set $S$ of places. Now $(B\times B^{\vee})^4$ is a principally polarized abelian variety with good reduction outside a finite set $S$ of places. There are finitely many of these by Faltings theorem. Since $B$ is a direct summand of $(B\times B^{\vee})^4$ its corresponding Tate module is a summand of the tate module of $(B\times B^{\vee})^4$ and by semisimplicity (Falting's theorem) there are finitely many isomorphism classes of these. Hence there are finitely many possibilities for the isogeny class of $B$ , and hence finitely many possibilities for $B$ by , once again, Falting's theorem.

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Nice, but seems like you're using pretty heavy machinery. If you fix $A$ with good reduction at $v$ and twist by a cocycle $\xi$ to get $B=A^\xi$, and if $\xi$ is ramified at $v$, then $B$ will have bad reduction at $v$ by Neron-Ogg-Shafarevich. So more-or-less, you only need to worry about twists that are unramified outside of $S$. I suspect, but haven't worked it out, that from there one will really only need the finiteness of class number and finite generation of the unit group. But I like your proof, it's very slick. –  Joe Silverman Nov 15 '13 at 23:52
@JoeSilverman: Thanks! I tried making the argument you describe work, but had a little trouble. The getting $\xi$ unramified part is fine, but the automorphism group of $A$, since its without a polarization, can be huge. For instance, if $A=ExE$ then you get $\mathrm{GL}_2(\mathbb{Z})$ and you need to show that cohomology of $G$ with those coefficients is finite. This is true but it was not so obvious to me how to see it directly, even if $G$ is finite, and of course we need to handle the scarier $G_{K,S}$. So i found this hard, and if you could make it work I'd love to see it! –  jacob Nov 16 '13 at 0:13
As I said, I didn't work out the details. Sounds as if you did, and that there are some serious complications. Still, somehow it feels as if there should be a proof that doesn't require all those tools from Faltings' Inventionnes paper. But it seems that one may at least have to do some challenging Galois cohomology. –  Joe Silverman Nov 16 '13 at 1:52
@jason: This is "circular" (but OK): that an abelian variety over a field admits only finitely many direct factors (up to isomorphism as abelian varieties) is used by Faltings to reduce to the semistable case in his own paper (via Zarhin too). So that finiteness is a fact on its own (valid over any field); it is a consequence of a deep finiteness theorem of Borel and Harish-Chandra on arithmetic groups (Theorem 9.11 of Borel's book on arithmetic groups) via arguments with idempotents in endomorphism rings and with unit groups of orders in finite-dimensional semisimple $\mathbf{Q}$-algebras. –  user27920 Nov 1 '14 at 15:10
@jacob: Also, though it is weaker than the question, the finiteness question when comparing isomorphism classes of abelian varieties over $K$ and over a finite extension $K'$ (for any field $K$!) is used in Faltings' proof as well (for $K$ a number field) as part of the reduction to treating just the semistable case. It will be very surprising if one could bypass the appeal to Borel & Harish-Chandra in these matters. –  user27920 Nov 1 '14 at 15:11