## Is it possible for the repeated doubling of a non torsion point of an elliptic curve tstays bounded in the affine plane ?

Let P=(x1,y1) be a non torsion point on an elliptic curve y^2=x^3+Ax+B. Let (xn,yn)=P^{2^n}. xn,yn are rationals with heights growing rapidly. Can {xn} {yn} stays bounded ?

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From the discussion above it looks like the answer is yes (EDIT: if you allow real numbers; the OP was unclear, perhaps they wanted a rational point, in which case I'm uncertain. Does anybody know anything about the binary expansion of complex numbers with rational Weierstrass p-values?). Let the origin of your group be the point at infinity in the curve in $\mathbb{RP}^2$, and pick a topological group isomorphism of $S^1$ to the component of the identity to $S^1\cong \mathbb{R}/\mathbb{Z}$. The doublings of a point are given by truncating off the first $m$ digits of the base 2 expansion of your point. Thus the doublings of a point stay bounded if and only if the length of a consecutive string of 0's and 1's in this expansion is bounded above (there are plenty of irrational numbers with this property).

There's a similar answer for putting the origin somewhere else: you can never allow too much of the beginning of the expansion of the point at infinity to show up in the expansion of your point.

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P seems to be rational. – S. Carnahan Apr 5 2010 at 19:33

EDIT: this answer is wrong. I misread the question as looking at the group generated by P, not the points obtained by repeated doubling. I would be OK if the subset of S^1 generated by taking a non-torsion point and repeatedly doubling came arbitrarily close to the origin---but it may not, as the comments below show. As I write, this question is still open. If a correct answer appears I might well delete this one.

Kevin, repeated doubling of a point doesn't produce the subgroup generated by that point. What is needed is a theorem to the effect that if $\xi$ is irrational then the set of $2^n\xi$ in $\mathbb{R}/\mathbb{Z}$ meets every neighbourhood of the origin in $\mathbb{R}/\mathbb{Z}$. I'm sure this is true but can't see an immediate proof. – Robin Chapman Mar 28 2010 at 10:32
However, $2^n \xi$ doesn't need to meet every neighborhood of the origin if you only assume $\xi$ is irrational. That $\xi$ is irrational just means the "digits" of the binary expansion aren't preperiodic, not that $\xi$ is normal base 2. – Douglas Zare Mar 28 2010 at 11:22
In that case should the elliptic logarithm'' of a rational point have this property then the answer to the original question would be `yes'. I don't think this can be the case but proving it suddenly looks like hard work :-( – Robin Chapman Mar 28 2010 at 11:31