Are ranks bounded?

Important update, 23 Jul:  I missed one very important thing about Bjorn’s talk:  it was about joint work with a bunch of other people, including one of my own former Ph.D. students, whom I left out of the original post!  Serious apologies.  I have modified the post to include everyone and make it clear that Bjorn was talking about a multiperson project.  There are also some inaccuracies in my second-hand description of the mathematics, which I will probably deal with by writing a new post later rather than fixing this one.

I was only able to get to two days of the arithmetic statistics workshop in Montreal, but it was really enjoyable!  And a pleasure to see that so many strong students are interested in working on this family of problems.

I arrived to late to hear Bjorn Poonen’s talk, where he made kind of a splash talking about joint work by Derek Garton, Jennifer Park, John Voight, Melanie Matchett Wood, and himself, offering some heuristic evidence that the Mordell-Weil ranks of elliptic curves over Q are bounded above.  I remember Andrew Granville suggesting eight or nine years ago that this might be the case.  At the time, it was an idea so far from conventional wisdom that it came across as a bit cheeky!  (Or maybe that’s just because Andrew often comes across as a bit cheeky…)

Why did we think there were elliptic curves of arbitrarily large rank over Q?  I suppose because we knew of no reason there shouldn’t be.  Is that a good reason?  It might be instructive to compare with the question of bounds for rational points on genus 2 curves.  We know by Faltings that |X(Q)| is finite for any genus 2 curve X, just as we know by Mordell-Weil that the rank of E(Q) is finite for any elliptic curve E.  But is there some absolute upper bound for |X(Q)|?  When I was in grad school, Lucia Caporaso, Joe Harris, and Barry Mazur proved a remarkable theorem:  that if Lang’s conjecture were true, there was some constant B such that |X(Q)| was at most B for every genus 2 curve X.  (And the same for any value of 2…)

Did this make people feel like |X(Q)| was uniformly bounded?  No!  That was considered ridiculous!  The Caporaso-Harris-Mazur theorem was thought of as evidence against Lang’s conjecture.  The three authors went around Harvard telling all the grad students about the theorem, saying — you guys are smart, go construct sequences of genus 2 curves with growing numbers of points, and boom, you’ve disproved Lang’s conjecture!

But none of us could.

And nobody could generate sequences of elliptic curves with unbounded ranks, either!  People have constructed lots of cool examples over the years.  But Noam Elkies’s elliptic curve with rank 28 has stood as champion for almost a decade now.  We may be approaching a Noamsymptote.

Now here’s the new stuff.  (This is based on my chats with others about his talk, since I didn’t see it.  Please correct/refine in comments.)  The BKLPR heuristics propose a very rich conjectural description of the distribution of Selmer, Shafarevich, and Mordell-Weil groups of a random elliptic curve over Q.  In particular:  the p-adic Selmer group of E should be modeled by the intersection between two randomly chosen maximal isotropic subspaces in a large orthogonal space over Z_p.

So it seems natural to model the actual Mordell-Weil group as the intersection between two random maximal isotropic lattices in a large orthogonal space over Z!

But now there’s a problem with “random.”  The space of maximal p-adic isotropics is a nice compact p-adic manifold with a natural probability distribution on it.  And in this probability distribution, there is zero probability that the intersection of two isotropics — the “Mordell-Weil rank” — is greater than 1.  Which is the answer we’re supposed to get!  But for the present problem, the claim that “0% of elliptic curves have rank greater than 1” isn’t good enough.  We don’t want 0%.  We want 0.

What’s more, what can it mean to talk about a random isotropic lattice now?  There are a countably infinite set of such things, with no natural distribution on them to call “uniform.”

So here’s what to do.  You can count all pairs of isotropic subspaces in a “box” — say, just count those generated by vectors with entries at most B, or better, count all subspaces of height at most B.  There are finitely many of them.  And let p(r,B) be the probability that two of these subspaces, chosen uniformly at random from the finitely many choices, intersect in a lattice of rank r.  As B goes to infinity, we ought to expect p(r,B) to go to 0.

Similarly, let P(r,X) be the probability that an elliptic curve of conductor at most X has rank r.

We would like a heuristic to say that p = P!  But this is meaningless without a way of “matching” B and X.  They finesse this in a very clever way.  We already have random matrix predictions that tell us that P(2,X) is supposed to be on order X^{-1/24}.  So you “tune” B to match X by letting B be whatever power of X makes p(2,B) ~ B^{-1/24}!

(Note:  I originally had an exponent of -1/4 above, but that was for quadratic twist families.  Michael Rubinstein pointed out that I should have been quoting this paper of Mark Watkins instead.  Interesting — I’m very used to casually saying “quadratic twist families are the same as the general family,” and I think that’s true for questions about measures, as in BKLPR, but there’s no reason they have to behave the same way w/r/t more refined questions like this one!  This big difference in exponents makes me wonder — should we expect that 100% of elliptic curves should have no quadratic twist with rank greater than, I dunno, 5? Update:  I am told the answer is yes; more later on this, I hope.)

And what you find is that, having done so, you get that p(r,B) varies as a negative power of B, and for r big enough, p(r,B) is smaller than B^{5/6}, which is supposed to be the number of elliptic curves of conductor at most B, and so for r this big, you find yourself predicting no elliptic curves of that rank at all.  Or finitely many, better to say.  After all, we know by work of Ulmer that there are elliptic curves over F_q(t) of arbitrarily large Mordell-Weil rank, and the BKLPR heuristics work just as well in this case — but the Ulmer examples will be very sparse.  Maybe a better way to read what the new heuristic says is that, for sufficiently large r, the number of E/Q with rank at least r and conductor at most X grows more slowly than any power of X.

All aspects of what I’ve said here are oversimplified and no doubt some are wrong!  But it’s very exciting to see conventional wisdom on such a fundamental question begin to shift, so I wanted to record the moment here.





Tagged , , , , , ,

6 thoughts on “Are ranks bounded?

  1. Jason Starr says:

    Wasn’t Viehweg also part of that story?

  2. JenJL says:

    Thanks for this post, Jordan! That is very cool, and I was not yet aware of the results, nor the definition of Noamsymptote. Does Poonen (or Granville) give a reason for not expecting a sparse set of curves with unbounded rank as in the function field case?

  3. JSE says:

    I’m not sure either of them feel the heuristics are strong enough to say confidently there should be no such examples. You’d have to ask them! I think what one can say for sure is that the methods used by Ulmer to construct those examples really have no analogue in the number field case. E.G. you might find a curve over E/F_q(t) which acquires very large rank over F_q(t^{1/1000000}). But the latter field is isomorphic to F_q(t)! So you’ve found a curve of high rank over a rational function field. In the number field case, you can’t have arbitrarily high-degree extensions of Q which are isomorphic to Q. To say the least.

  4. […] articles:  Jordan Ellenberg describes progress on understanding the rank of elliptic curves:  Are Ranks Unbounded? and Cathy O’Neil produced a neat little python notebook to walk people through RSA’s […]

  5. […] interesting feature of the heuristics of Garton, Park, Poonen, Wood, Voigt, discussed here previously: they predict there are fewer elliptic curves of rank 3 than there are of rank 2.  Is this what we […]

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: