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.



Does the equality of ranks imply equality of analytic ranks?
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](https://mathoverflow.net/questions/71609/the-parity-conjecture) (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.


