# Results and conjectures on bounds on degrees of isogenies

Dear MO Community,

given an isogeny between two abelian varieties $\varphi: A\rightarrow B$ (everything definied over a number field $K$), we can factor $\varphi$ through a multiplication-by-$n$-endomorphism on $A$ and a `remaining' isogeny $\psi:A\rightarrow B$.

What do we know or expect about an upper bound $N$ on the degree of $\psi$?

For elliptic curves over $\mathbf{Q}$ it is known that $N=163$ is such a bound. It is even known that precisely the 26 cases $N= 1, \ldots , 19, 21, 25, 27, 37, 43, 67, 163$ do occur. This is the classification of cyclic isogenies of elliptic curves over $\mathbf{Q}$ and is due to Mazur and Kenku (1978-1982).

I am wondering if for bounded dimension $d$ of the abelian varieties and bounded degree $n$ of the number fields one expects that there is such a bound $N(d,n)$ for the possible degrees of $\psi$?

Is such a bound expected? (Is it clear that such a bound exists in case one fixes the number field and only varies the abelian varieties up to some bounded dimension?)

A subquestion would be the existence of rational $N$-torsion. Here it is known that for elliptic curves over a number field $K$ the order of the $K$-rational torsion is bounded by a constant that only depends on $[K:\mathbf{Q}]$. This is due to Merel (1996). What do we expect or know about the torsion in higher dimensions?

Thanks a lot.

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As far as I know, the conclusion of Merel's theorem is believed to hold for abelian varieties up to bounded dimension. In fact, the bound on the $K$-rational torsion is supposed to be polynomial in $[K:\mathbb{Q}]$. (This is wide open even for elliptic curves; Merel's theorem gives an exponential bound.) –  Vesselin Dimitrov Feb 27 '13 at 13:57
@Vesselin: do you know a reference where it was suggested that the bound should be polynomial in the degree? –  Michael Zieve Feb 27 '13 at 14:06
@Michael: This is stated, for example, in Remark 2 in Marusia Rebolledo's expository article ("Merel's theorem on the boundedness of the torsion of elliptic curves.") I am sure I have seen it in at least one other paper, but right now I can't remember the reference. But the idea is that since for individual elliptic curves the count is proportional to $[K:\mathbb{Q}]^{3/2}$ and $[K:\mathbb{Q}]^2$ in the non-CM and CM cases, respectively, it is not too wild to ask whether this count is uniform. Thus, one can ask whether the count is in fact uniformly bounded by $C(r) [K:\mathbb{Q}]^{2+r}$. –  Vesselin Dimitrov Feb 27 '13 at 14:22
By the way, Hindry and Silverman have shown that for elliptic curves with integral modulus (everywhere good reduction), the number of torsion points rational over a given number field of degree $d$ is at most $\mathrm{const} \cdot d\log{d}$. This lends, perhaps, additional support to the expectation that the bound in Merel's theorem should be polynomial. (Note that the trivial bound provided by good reduction at $2$ is exponential in $d$.) –  Vesselin Dimitrov Feb 27 '13 at 16:06
Dear Vesselin, thanks a lot for presenting the expectations for the size of torsion. –  Stefan Keil Feb 28 '13 at 9:20
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Mein lieber Stefan,

I think that these sorts of questions are wide-open. As far as I know, even if you fix the number field, and vary over AVs up to some fixed dimension, then it is not clear that there exists a bound on the possible degrees of your $\psi$ (which I'm guessing is a cyclic isogeny). You in any case have to be careful of complex multiplications; for example, if you have an elliptic curve over $\mathbb{Q}(i)$ with CM by the maximal order, then you can create $\mathbb{Q}(i)$-rational prime-degree isogenies for every prime that splits in $\mathbb{Z}[i]$. But if you exclude these CM isogenies, then I think it is believed that such a bound on cyclic isogeny degrees for AVs exists.

But even pushing Mazur's isogeny theorem to number fields is fairly recent stuff; we can now say that, if $K$ is a number field not containing the Hilbert Class Field of an imaginary quadratic field, and GRH is true, then there are only finitely many prime-degree isogenies for elliptic curves over $K$; this goes back to Momose, but with some refinements by Agnès David; see also the work of Larson and Vaintrob. These bounds are, however, quite big. I once computed the bound for $\mathbb{Q}(\sqrt{5})$ by computing all the constants in David's paper "Caractère d'isogénie..." and got something like $10^{120}$!

Serre's open image theorem holds for abelian varieties $A/K$ of dimension 2,6 and odd, where $End(A) = \mathbb{Z}$; so there is a bound $C(A,K)$ such that, for all primes $l > C(A,K)$, the mod-$l$ representation is surjective. It is believed that this constant can be made independent of $A$. So, if you fix your base $K$, and vary over all AVs $A/K$ of fixed dimension $d$ (=2,6,or odd) which have no extra endomorphisms, then the prime-degree isogenies are bounded by a constant depending only on $K$. But as you well know, this uniformity conjecture is still open even for elliptic curves over $\mathbb{Q}$!

The paper "Expander Graphs, gonality, and variation of Galois Representations" by Ellenberg, Hall, and Kowalski has some interesting results about families of AVs; see theorems 4 and 7, for example. In particular, if you have a family of AVs over a number field $K$, then there is an absolute constant $C$ such that, if $l > C$, then almost all members of the family have "large" mod-$l$ image. And (morally) "large" image means you don't get isogenies or torsion. See their paper for more details.

Finally, regarding your subquestion about torsion: François gave a link to a paper of Clark and Xarles where they address the question of bounding torsion primes for certain classes of AVs. In particular, they prove that a "generalised Szpiro conjecture" implies uniform boundedness of torsion for all "Hilbert-Blumenthal" abelian varieties. But it seems that doing this for all abelian varieties of a fixed dimension is very hard.

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Dear Barinder, thanks a lot for this excellent answer! (Are there any expectations in case $\psi$ is not assumed to be cyclic?) –  Stefan Keil Mar 7 '13 at 12:05