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Let $E$ be an elliptic curve over $\mathbb{Q}$. It is known from Gross and Zagier that if $\textrm{rank}_{\textrm{an}}(E) \leq 1$, then

$$\textrm{rank}(E) \geq \textrm{rank}_{\textrm{an}}(E).$$

Instead, suppose I know that $\textrm{rank}(E) \leq 1$, are there any (unconditional) inequalities that relate rank and analytic rank of $E$?

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    $\begingroup$ This is tangential, but I believe by results of Kolyvagin (and the Modularity Theorem) one has equality in the displayed equation. [Added: perhaps my comment is even more tangential, as it occurs to me it could well be this is phrased like this on purpose for symmetry.] $\endgroup$
    – user9072
    Commented Mar 6, 2013 at 21:47
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    $\begingroup$ In the function field case, there is a known inequality, but it is the other way around than yours: the analytic rank is an upper bound to the algebraic rank. Iwasawa theory tries to mimic this for number fields, but all we currently know is that the $p$-adic analytic rank is an upper bound to the alebraic rank. Here the $p$-adic analytic rank is the order of vanishing of the $p$-adic L-function. $\endgroup$
    – Chris Wuthrich
    Commented Mar 7, 2013 at 9:13
  • $\begingroup$ @Chris: In the function field case, assuming finiteness of the primary part of the Tate-Shafarevich group, Kato and Trihan have proved the equality. Do you mean that the inequality rank is at most analytic rank always hold? $\endgroup$
    – ACL
    Commented Mar 7, 2013 at 9:51
  • $\begingroup$ @ACL: Yes I do. It can already be deduced from Tate's Boubaki talk on the geometric analogue of BSD. $\endgroup$
    – Chris Wuthrich
    Commented Mar 7, 2013 at 11:17

2 Answers 2

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This is not really the right question. I think it's easier to just list the two things that are known unconditionally and then you can figure out if this answers your question.

1) If analytic rank is zero then algebraic rank is zero.

2) If analytic rank is 1 then algebraic rank is 1.

Proof: see quid's comment.

One now deduces easily that if the algebraic rank is at most 1 then the analytic rank is at least the algebraic rank.

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The short answer to your question is no.

One of the central open problems related to BSD rank conjecture is to find a way to canonically construct points of infinite order on elliptic curves with analytic rank at least 2. This is difficult because it requires passing from analytic properties of the $L$-function to algebraic points on the elliptic curve.

For analytic rank 1, Heegner points do the trick but they turn out to be torsion points when the analytic rank is higher.

There is no theorem proven that says that it can never happen that the analytic rank is 2 and the algebraic rank is 0 (of course if BSD is true, this cannot happen).

That being said, if you have a particular elliptic curve and you know the algebraic rank is 0, after a finite amount of computation you should be able to show that the analytic rank is 0, i.e. you should be able to prove the BSD rank conjecture for this particular curve.

If you have a particular elliptic curve and you know that the analytic rank is 2, you probably proved this by observing three things:

  1. The analytic rank is even by the sign of the functional equation.
  2. The analytic rank is $\leq 2$ by some calculation with the L-function.
  3. The algebraic rank is not equal to 0, because you found a point of infinite order, so the analytic rank cannot be 0.

So you conclude the analytic rank must be 2.

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    $\begingroup$ You can (and usually do) prove $L(E,1) = 0$ without finding a single point of infinite order, because $L(E,1)$ is a priori a rational multiple of the real period with bounded denominator. $\endgroup$
    – Noam D. Elkies
    Commented Mar 6, 2013 at 23:34

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