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Hello,

Has it been proven that two elliptic curves defined over $\mathbb{Q}$ having the same rank necessarily have the same analytic rank (order of annulation of the L-functions associated to thoses curves in $z=1$) or is it still out of reach? What are the latest related results? Thanks in advance.

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    $\begingroup$ Nothing like that it know. Basically, the only things we know are the Gross-Zagier-Kolyvagin results for analytic rank $\leq 1$ and parity results. There is an MO thread concerning the latter: mathoverflow.net/questions/71609/the-parity-conjecture $\endgroup$
    – Alex B.
    Commented Jun 13, 2012 at 18:20
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    $\begingroup$ It is well-known that there exists an elliptic curve of analytic rank 0 and algebraic rank 0. Parity conjectures are now theorems in many cases. But if one could rule out the existence of an elliptic curve with analytic rank 2 and algebraic rank 0 this would be a major breakthrough. So your more general question seems to me to be hopelessly out of reach. $\endgroup$
    – Kevin Buzzard
    Commented Jun 13, 2012 at 21:06
  • $\begingroup$ Dear Alex, why don't you post your comment as an answer (as it is definitely such ;-)) ? $\endgroup$
    – Filippo Alberto Edoardo
    Commented Jun 14, 2012 at 3:41
  • $\begingroup$ Dear Filippo, done. $\endgroup$
    – Alex B.
    Commented Jun 14, 2012 at 6:47

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In general, it does not seem to be any easier to compare two elliptic curves than to say something about each individual curve. In other words, the only results in the direction that you are asking about are the usual results in the direction of the BSD: the Gross-Zagier-Kolyvagin results on analytic rank $\leq 1$, and parity results (which are conditional on finiteness of sha). So e.g. if you have two elliptic curves, both of algebraic rank 1, then the only thing you know unconditionally is that their analytic rank is not 0. If you also assume finiteness of sha, then you know that they both have odd analytic rank. That's it! They could have analytic ranks 3 and 5, respectively, for all you know.

There is an exception to this. If there is a special reason for the algebraic ranks to be equal, namely if you know that the curves are isogenous, then of course you know that both algebraic and analytic ranks are the same (the whole $L$-functions are the same). Much less trivially, you even know that the conjectural leading coeffiecient of the $L$-functions at $s=1$ is the same for both curves, provided the Tate-Shafarevich group of one of them is finite. This is a theorem of Cassels.

P.S. By the way, if such a comparison theorem existed, then proving the BSD for all curves of given algebraic rank would be reduced to proving it for one such curve. This might still be hard though. I don't know of a way of proving that the order of vanishing of an $L$-function, which you can only compute approximately, is exactly what it appears to be.

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  • $\begingroup$ Indeed, numerical computation only proves an upper bound for the analytic rank. Currently there is no elliptic curve known for which the analytic rank is provably $\geq 4$. $\endgroup$
    – François Brunault
    Commented Jun 14, 2012 at 11:47
  • $\begingroup$ In some sense you do not expect a lot of elliptic curves with large ranks if you believe the results of Bhargava and Shankar. So finding these curves(if they exist) will be very hard indeed. $\endgroup$
    – Arijit
    Commented Feb 12, 2013 at 15:28

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