Question

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?

Answer 1

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.