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Considering elliptic curves over Q with positive discriminant and rank>0, are there any results or proposed heuristics regarding the fraction of generators that are located on the identity vs non-identity component? How does this vary with rank?

The family of curves I've been examining (associated with $a/(b+c) + b/(a+c) + c/(a+b) = N$) have the following distributions for N even:

Rank 1: 313 have generator on identity, 601 on non-identity
Rank 2: 107 have 2 on identity, 153 have 1, 16 have 0
Rank 3: 3 have 3 on identity, 16 have 2, 3 have 1, 0 have 0

For N odd, all of the generators appear on the identity component.

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  • $\begingroup$ I think there is a result saying that there is no non-trivial rational point for $N$ odd. Also, I don't understand what you mean by "have 2 on identity", etc. It seems to me that it only makes sense to say whether there is a generator on the non-identity component: if $P, Q$ is a set of generators, then $P, P + Q$ also. This is equivalent to asking whether $E(\mathbb{Q})$ is contained in the identity component. $\endgroup$
    – WhatsUp
    Commented Nov 30, 2018 at 13:02
  • $\begingroup$ Did you try to see where the generators are located for, say, the family of all elliptic curves of rank 1, positive discriminant and trivial torsion? $\endgroup$ Commented Nov 30, 2018 at 13:10
  • $\begingroup$ Of course, $E(\mathbb{Q})_{tors}$ should a priori be contained in the identity component, for this to make sense. $\endgroup$
    – WhatsUp
    Commented Nov 30, 2018 at 13:10
  • $\begingroup$ Sorry, now I remember that the result for $N$ odd is that this equation doesn't have solutions in positive integers, which is equivalent to saying that all rational points are located on the identity component. $\endgroup$
    – WhatsUp
    Commented Nov 30, 2018 at 13:13
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    $\begingroup$ You are asking if $E(\mathbb{Q})/2E(\mathbb{Q})\to E(\mathbb{R})/2E(\mathbb{R})$ is surjective. Maybe one can say something about the image of the $2$-Selmer group instead; there might be reasons that other local conditions impose the same as the real condition. Instead for the actual rational points, I doubt it is easy to say much. $\endgroup$ Commented Nov 30, 2018 at 13:35

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I gathered some statistics using the Cremona tables. There are 565803 elliptic curves with rank 1, trivial torsion group, positive discriminant (which means that the group of real points has two connected components) and conductor $\leq 4 \times 10^5$ (the current limit of the tables). Among them 239106 curves have the generator of $E(\mathbf{Q})$ located on the identity component.

Here is the histogram representing the location of the generator $P$ on the identity component for these curves. More precisely, this is the distribution of the (unique) real number $z_P \in (0,1/2)$ such that the Mordell-Weil group $E(\mathbf{Q})$ is generated by the class of $z_P \Omega_E^+$, where $\Omega_E^+$ is the real period of $E$. We can note the behaviour near $z=0,1/4,1/3,1/2$ which might be explained by the fact that the torsion points somehow repel the point of infinite order.

enter image description here

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