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It is an easy consequence of the Serre open image theorem that for the torsion point count on elliptic curves, the following possibilities arise.

  1. If $E/\bar{\mathbb{Q}}$ is an elliptic curve without CM, then the number of torsion points $x \in E$ with $[\mathbb{Q}(x):\mathbb{Q}] \leq d$ is $\asymp d^{3/2}$ as $d \to \infty$.

  2. If $E/\bar{\mathbb{Q}}$ is an elliptic curve with CM, then this number is $\asymp d^2$ as $d \to \infty$.

This appears on page 44 of Serre's book, Lectures on Mordell-Weil. As stated there, it is easy to show that, for a $g$-dimensional abelian variety, the corresponding count is bounded by $\leq O(d^{N})$ with $N = N(g) < \infty$.

Should we expect, for $g$-dimensional abelian varieties, the count to be always asymptotic to $d^{\alpha}$ for a finite set of exponents $\alpha$ depending on $g$? Are there any clues as to the spectrum of those exponents?

Second, I wanted to ask about any results on the uniform torsion count problem giving bounds that are uniform in $E$ but still polynomial in $d$. (Thus I do not ask about Merel's uniform boundedness theorem and its relatives). For example, restricting to $g$-dimensional abelian varieties over $\bar{\mathbb{Z}}$ (that is, with integral moduli, or with everywhere potentially good reduction), or at least to the abelian schemes over the spectra of rings of integers of number fields of bounded degree, it is clear a priori that there is a uniform exponential bound in $d$. Should we expect this uniform bound to be strengthened to a polynomial bound, or even (for elliptic curves) to $Cd^{2+\epsilon}$? What about the generalization where we count the small algebraic points, say those of (canonical) height $< 1/d$?

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You mean to count only torsion points in 1. and 2. –  Felipe Voloch Feb 8 '13 at 17:17
    
Yes, sorry for the omission: I will edit. However, I am particularly interested, more generally, in counting the algebraic points of degree $d$ and height $< 1/d$. As for those of degree $d$ and height $< 1$, this is perhaps a much more delicate problem; I wonder if there are any non-trivial bounds available. –  Vesselin Dimitrov Feb 8 '13 at 17:23
    
There are various Lehmer-type lower bound (Masser and other). For CM-Abelian varieties, there is an almost-optimal result by David and Hindry ($h(p)\gg d(p)^{-1/g-\varepsilon}$). –  ACL Feb 8 '13 at 22:59

1 Answer 1

up vote 9 down vote accepted

As far as I know, we expect that the image of the Galois group in $GL_{2g}(\mathbb A_\mathbb Q)$ is open in $G(\mathbb A_\mathbb Q)$ for $G$ the monodromy group of the Galois representation (which is expected to be independent of $l$. Then the asymptotic for this function clearly depends only on the connected component of the identity of this monodromy group, which is expected to be the Mumford-Tate group.

(1) For each possible Mumford-Tate group $G$, is the number of points in $(\mathbb Q/\mathbb Z)^{2g}$ fixed by an index $d$ subgroup in $G(\mathbb A_\mathbb Q)$ asymptotic to $d^\alpha$ for some $\alpha$?

(2) Are there finitely many possible Mumford-Tate groups?

The first one is not too hard to see. The index of the subgroup fixing an $n$-torsion point is the size of the orbit of that point, which is approximately the number of $\mathbb Z/n$ points of some algebraic variety, which is approximately polynomial in $n$. Then we can sum over all points $n$-torsion for $n\leq N$, another polynomial. If there are different sorts of orbits of different dimensions, then we can just sum over the different sorts of orbits and still get something of the form $d^\alpha$ up to some constant error.

The second one is clear since the Mumford-Tate group should be reductive, and thus there is a nice classification.

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@Will: The Mumford-Tate group should be reductive, and it is never semisimple. For example, the Mumford-Tate group of an abelian variety of CM-type is a torus. –  Mikhail Borovoi Feb 18 '13 at 19:19
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The Mumford-Tate group of a point is semisimple! More seriously, thanks for the correction. –  Will Sawin Feb 18 '13 at 21:06

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