Bjorn Poonen, Heuristics for the arithmetic of elliptic curves

RP: We’re very happy this semester to start with a theme of rational points on elliptic curves, and if you go to the website, you can see the lineup of speakers we have for this fall, at least the first half of the fall, and I hope that everything is going well for you all in wherever you are. So today, I’m very happy to introduce Bjorn Poonen Who will be speaking about heuristics for the arithmetic of elliptic curves Bjorn, do I have your permission to record this lecture? BP: Yes, please go ahead. RP: Okay. Thank you BP: Okay, thank you everybody. So I’m going to talk about heuristics that try to predict whether the ranks of elliptic curves over Q are bounded or not. So let me start by saying that what I’m going to talk about is joint work with a lot of people in a lot of papers so I’ve listed all the authors here, some of whom were in multiple papers Um, yeah. All right. Let me get started So, an elliptic curve over Q is given by an equation like this, except strictly speaking it’s supposed to be a projective curve.

01:20 - So you take this affine curve in the plane and you take its closure in P^2. And in order for it to be an elliptic curve it needs to be a smooth curve, which amounts to saying that the discriminant of this polynomial should be non-zero, and if you calculate the discriminant it’s some constant times this. Anyway, it’s essentially a cubic curve that’s given by an equation like this where these coefficients are rational numbers. Okay, who cares? So I mean why should you care about these curves at all? So yeah, who cares? Well I care because first of all, they’re they’re the simplest varieties – I mean from the point of view of number theory, they’re the simplest varieties whose rational points are not fully understood. So rational point just means a pair of rational numbers x and y that satisfy the equation together with the point of infinity And also these are the simplest examples of algebraic groups – of projective algebraic groups of positive dimension and so they’re really like the first testing ground The first testing ground where I think interesting things start happening.

02:28 - All right, and it’s because it’s a group variety, because it’s an algebraic grouping the rational points also form a group and the group law is commutative so you get an abelian group. Okay, so… All right, and it’s an old theorem of Mordell that was conjectured even earlier that this abelian group is always finitely generated So, okay. So Mordell proved that in 1922 and if you know the structure theorem for finitely generated abelian groups, it must be in this form. It’ll have the direct sum of a power of the Z and then a finite abelian group. And um, okay and this finite abelian group, that’s called the torsion subgroup, and Mazur worked out all the possibilities for this group, so he proved that the only possibilities are these, I guess Yeah, 15 groups So it’s either cyclic of order up to 12 and not Z/11Z, or else it’s almost cyclic it’s got an extra factor of Z/2Z cross an even cyclic group.

And then there are, yeah, so 03:45 - Okay, so there’s no mystery. I mean it’s pretty easy if somebody gives you an elliptic curve to figure out what the torsion subgroup is but the mystery is what is the rank I mean In fact, it’s not even known whether there’s an algorithm to compute the rank in general but, well, you can still ask questions about it. For example, what can you say about that rank? And, so this question actually goes back even before it was proved that that the group was finally generated Even before it was known that “the rank” made sense. So Poincaré asked in 1901 He essentially asked what are the possibilities? What are the possible values for this rank? and in particular as part of this question you can ask implicit in this is the question of whether the rank is uniformly bounded, so bounded as you vary the elliptic curve. And a lot of people have thought about this question, so…

04:43 - Initially it seemed that people thought the answer would be yes. So, Néron in 1950 wrote he thought probably it’s yes, and Honda also made a conjecture along those lines. And then, and Cassels said oh, I think you’re – all these earlier authors are wrong I believe it’s actually unbounded, and then I don’t know. I’m not sure if these are, all these people were sort of independently thinking about this or if they just said “oh, well Cassels said it, it must be true.” So Right. But anyway, so they all, everybody started saying “oh, probably, it’s a folklore conjecture that it’s unbounded” and so on and so on.

05:25 - I mean, I think some of these people actually did have other good reasons to believe it was unbounded I mean there were some other reasons. I mean people started using computers to investigate the question. They started finding elliptic curves of higher and higher rank and like every year it seemed that there would be a new record. And so it seemed to be going up and also it was shown in the function field case that the rank could be arbitrarily large. On the other hand, more recently, a couple of authors had not just guesses, but they actually had some heuristic reasons for believing that the rank should be bounded.

06:10 - So yeah, so Rubin and Silverberg had some argument based on lattices and they actually an argument that suggested in a family of quadratic twists the rank should be bounded by eight, but they didn’t really believe themselves because I mean because they knew that the rank is actually not bounded by eight. So I mean, whatever their heuristic was it, I mean there at least there’s something fishy about it, but… Granville had a different heuristic based on counting integral solutions to polynomial equations that, well, he originally did it for quadratic twists – a family of quadratic twists too and then Watkins adapted Granville’s argument to the family of all elliptic curves and they got an upper bound. They said they got an upper bound of 21. And what I’m going to do in this talk is present a different heuristic which tries to not just study ranks in isolation but actually try to study a complete package where you look at ranks and all sorts of other things, all sorts of other invariants attached to an elliptic curve, and the idea is maybe you can find a simple model that predicts everything together. And if you’re predicting many things at once then you can try to corroborate it with theorems about the various pieces and so you have more ways to check it and so you can get more confidence maybe that these predictions are correct.

07:55 - And mysteriously it turns out that our heuristic also leads to a prediction of 21 as as an upper bound well, not a strict upper bound but I mean In fact, there are some elliptic curves known that have rank greater than 21, but the prediction that comes out of this heuristic is that there are only finitely many elliptic curves that exceed this bound of 21. Okay So, now I should be a little bit careful about what I mean when I say for all but finitely many E because if you just look at the equations each elliptic curve up to isomorphism appears infinitely often because you can always do a change of variable on your equation and get something that’s really just the same elliptic curve. So in order to take that into account, let me mention that You can always put the equation of an elliptic curve in some sort of canonical form where you scale the variables in an appropriate way to make it so that the rational numbers A and B are actually integers and you can also just scale it exactly just just right to get rid of any extraneous factors in these integers A and B So whenever you have a prime such that p^4 divides A and p^6 divides B you can do a change of variables to eliminate it. I mean it’s kind of like taking a gcd or something But once you take care of that then it turns out that you get a unique representative for each isomorphism class and so I’m going to let this fancy E be the set of all these representatives so now this is a set where each elliptic curve up to isomorphism appears only once and so now whenever I say something like all but finitely many E, what I really mean is I’m only looking at these representatives, and I mean for all but finitely many of those elliptic curves And the other nice thing about putting an elliptic curve in this sort of simplest form is that you can define some notion of height which measures the complexity of this elliptic curve. and it’s just something that’s based on the size of the coefficients A and B and because the discriminant is some combination of these quantities, 4A^3 and 27B^2, you can… you see it’s…

it’s sort of, well, sort of natural to define a height this way. This is some sort of naive height. There’s some more sophisticated versions of this but this is going to be sufficient for what we’re trying to do and it’s more or less the same as other heights. So, okay. So, right, so then you can look at if you want to sort of order these elliptic curves you can order them by height, and what that means is you’ll look at for any given value of H you can look at all the elliptic curves of height less than or equal to that and that will be some finite set because once you bound H you’re you’re giving an upper bound on these integers A and B and you can actually count how many, how finite it is because once you fix H then if these two numbers are supposed to be bounded by H, that means A can go up to something around H^(¹⁄₃) and B can go up to something around H^(¹⁄₂) And, well, you can’t just choose any integers up to that range because there’s this condition, but you can check that when you sieve out the pairs (A,B) that violate this then you you’re only sieving out a positive fraction – some fraction between zero and one, so you still end up with roughly the same number. So if you ignore these constants, then you find that there’re H^(⁵⁄₆) elliptic curves of that height. Okay, so remember this ⁵⁄₆ for later. All right, so Now let me talk about…

so I mentioned, 12:00 - okay, so going back two slides, I mentioned that we’re going to model not just ranks, because that’s hard to do just by themselves, but we’re going to model it together with these other invariants. So I want to tell you a little bit about what these other things are. At least I’ll tell you what a Selmer group is. Okay, so… in order to understand this definition you have to know a little bit about Galois cohomology. So what you do is fix an integer n at least two, and then what we’re going to try to do is to figure out what the rank of E is by looking at this quotient because if E(Q) for example looks like Z^3, then this quotient should have a copy of (Z/nZ)^3 in it. And yeah, so if you can bound this group then you can also bound the rank.

And so the question is 12:54 - to what extent can you take a rational point and divide it by two? So what’s the obstruction to dividing a rational point by two? Okay, so if you’re trying to divide a… a point over Q bar instead of over Q if you’re trying to divide a Q bar point by two you’d always be able to do it because you can geometrically the multiplication by n map or I guess I’d say two – but the multiplication by n map is surjective On geometric points – on Q bar points, and so you can always divide it So you have a surjective map here and you can define the kernel to be this But now we want to look at if you actually start with a rational point can you divide it and get a rational point? And then there’s going to be an obstruction coming in Galois cohomology, so to see it you take Galois cohomology of this sequence and you get a long exact sequence coming out where here the first three terms are just the Galois invariants of these things so by Galois theory you just get the rational n-torsion points – you get E(Q) going to E(Q) and then the next term is the H1 of this group which I’ve written here and then it keeps going. And if you want to – you get a long exact sequence But you can cut it off at this point By replacing this group here by the co-kernel of this map, which I’ve written here. So if you do that then you get this well, you get the beginning of an exact sequence that has the group that we’re interested in and then that embeds into this cohomology group. And remember, we’re trying to prove that this group is finite – or we’re trying to bound this in order to bound the rank, so wouldn’t it be nice if this were just some finite group? And then we could – maybe we could figure out the size of this thing, and then that would give you a bound on the rank, but unfortunately this group is always infinite, so it’s not very helpful by itself.

14:57 - Okay, well I mean it’s infinite but maybe I mean this thing is eventually supposed to be finite. Maybe you can figure out what the image of this map is with this sort of global co-boundary map. And so if you can bound the image of this, maybe that’ll give you a bound on the rank. So okay, we can try to figure out what’s the image of this map, but that’s really – that problem is – that’s a hard problem in general. I mean things over Q and number fields those tend to be hard questions.

So instead 15:31 - we can look at a corresponding local question. Because there’s nothing in the formation of this sequence – there’s nothing special really about Q. You could do this for other fields at least other fields where the characteristic does not divide n. So in particular you could do this for the field of p-adic numbers and you’ll get a corresponding local sequence and it’s much easier to figure out what the image of a sequence like this is. It’s much easier to understand the Q_p points on E because you can use tools like Hensel’s lemma to understand that.

16:08 - So you can really compute this image of this map and moreover this thing here turns out to be finite, so it’s actually quite easy to figure out what this image is for any given prime p. In fact, you can even do this for all primes together. You can figure out sort of the – You can take the product over all primes including the infinite prime, by which I mean Q_p is the real numbers, and you can essentially determine what this is locally. You have a good understanding of what this image is here. So here’s the idea now, so we want to know here, in this group, what are the things that actually are in the global image? But that’s too hard to test.

So what we’re going to do is, 16:56 - we’re going to look at this map, and we’re going to ask, okay, we can’t test whether it’s globally in the image, but let’s check whether, when we map down here, is it locally in the image? And so that gives you a condition because of the commutativity of this diagram, and so you can define the Selmer group – the n-Selmer group to be a subgroup of this group. Those are the ones – they’re the classes c belonging here such that when you map them down they’re in the image of this local map. And that’s necessary in order for that c to have a chance of coming from here. All right, so So this group that these are the classes – maybe they’re not in the global image, but they at least satisfy this weaker condition of being in the local image, so that’s going to be an upper bound for the global image. So it’s an upper bound for this group here.

17:56 - And not only that, it turns out that because these local conditions are so easy to understand you can actually compute this group. There’s actually an algorithm to compute this group in principle for any elliptic curve and any n, And it turns out to be a finite group as well. So once you prove this is a finite group, that proves that this is finite, and that’s how you can get an upper bound on the rank. All right, so just now the question is well how good an upper bound is it. It turns out that by studying this sequence – Okay, let me go back one slide again.

18:37 - By studying this and by looking at the next term in these long exact sequences, you can measure the error. You can measure how good an upper bound this is by comparing it to how different this is from the actual from the actual group it’s trying to bound. And yeah, by the way there’s some people who like to define the Selmer group, instead of doing it in this square, they define it as the kernel of some diagonal map. I never liked that very much because I I’m like why look at why look at that diagonal map and what’s the motivation for this? Somehow I find this this kind of story more compelling, so I always like to define the Selmer group this way. Okay anyway, so the difference between the group we were trying to bound and then the Selmer group, the difference is measured by the n-torsion of some other group that’s defined – I mean, maybe I won’t define it here, but some other group defined using cohomology that also comes from the exact sequence and the sequences on the diagrams on the previous page.

19:47 - Okay, so this – as I said before, this group here is finite and computable, so if you were able to figure out which of these elements map to zero here, then you would know what this group was here. Unfortunately that seems to be a hard question; nobody knows how to do that in general. Another thing you can do is you can do this not just for one n – I mean, maybe it’s hard to do it for one n – but maybe you can combine the information for many n. If you do it for n that’s a product of prime powers then this sort of all breaks down. You can use the Chinese Remainder Theorem and break it down into prime power cases.

So you might as well take n to be a prime power and then 20:35 - you can do it not just for one prime power, but you can take a direct limit of all these groups as you start increasing the power, and you get more and more refined information as you increase the exponent. And so you might hope “oh, maybe if you take the limit then you’ll get precise information.” So you can do this and you get – so here if you rewrite this as E(Q), and then tensored with (1/n)Z/Z Then when we take the direct limit and – well it was n and being used as powers – you get E(Q) tensored with this group. Yeah, so I mean this group here, what it looks like is, it’s a power of Q_p/Z_p. It’s a finite power of Q_p/Z_p. And these other groups, they’re also kind of similar, they’re going to be groups that are called Z_p modules of cofinite type.

21:36 - I mean, their Pontryagin duals look like our finitely generated Z_p modules. So they all have the structure like (a finite group) direct sum (a finite power of Q_p/Z_p). And then the only question is, how many copies of Q_p/Z_p are there in there? That’s going to be the rank. So conjecturally, actually this group here is finite. So this group is supposed to have zero copies of Q_p/Z_p. And so if you could count the number of copies of Q_p/Z_p here, that should be the sharp answer, that should be what the rank is. So yeah, we’d like to actually understand what all these things look like, and now what we’re going to do is we’re going to start varying the elliptic curve and see What does this look like? What’s – if you choose random elliptic curves, or you look at all the elliptic curves up to height H, what’s the distribution of these groups here, as well as the individual Selmer groups. All right, so the general principle here that was guiding us is that if – for example, suppose you just look at the p-Selmer group for one prime p. We know that that’s a finite group but it varies as you vary the elliptic curve E so you get a collection of finite groups. Now, what do you think the distribution of that – in fact, it’s not just a finite group it’s a finite F_p-vector space – the p-Selmer group is just an F_p- vector space.

23:12 - So what do you expect the distribution of a finite-dimensional vector space to be? Well, we don’t know but there were some theorems in the literature about what happens for p=2, and there’s a theorem of Heath-Brown from the 1990s that had a prediction for what the distribution of the dimension of that F_2 vector space should be. So Eric Rains and I looked at that and we said: Okay, come on, there can’t be that many ways of generating a random F_2 vector space. I mean, it probably just comes from some random linear algebra construction, like you choose you choose n generators and you mod up by some number of relations, something like that. So we just tried all kinds of, all kinds of linear algebra constructions, and eventually we came upon this conjecture. So let me explain what this conjecture is. We found a particular linear algebra construction that produced a distribution that matched what Heath-Brown had discovered.

24:16 - So just by matching the formula – I mean, at that point, we didn’t have any idea why this construction worked. Okay, anyway I’ve been talking too much. Let me actually tell you – show you what this linear algebra construction is. So you start with a vector space – a big vector space, so n is going to be large – an F_p vector space of dimension 2n, but not just a vector space. I want to make it a quadratic space, which means it’s a vector space equipped with a quadratic form, a homogeneous polynomial of degree two that exists on this vector space. So it’s a quadratic form in 2n variables, and this is the simplest non-degenerate quadratic form there is, which is just this – it’s called the hyperbolic quadratic form, where you just take the sum of the products of the…

25:08 - It’s kind of like a dot product except I’m viewing it as a function of one argument, one vector of length 2n. Okay. So anyway, that’s a quadratic form, and then you can define a notion of, You can call a subspace maximal isotropic if first of all So isotropic means that the quadratic form is zero everywhere on it, and then maximal means that among the ones where the quadratic form is zero, it should be of maximal dimension, which is equivalent also to saying that the set of vectors that are perpendicular to that subspace under the associated bilinear pairing, that should just be exactly the vector space itself. I mean normally this ⟂ (“perp”) has the complementary dimension inside this 2n dimensional vector space. So this actually, in the presence of this condition, this is really just the same thing as saying that Z is of dimension n inside this – [unclear] Sorry, it’s a vector space whose dimension is half the dimension of the whole space. Okay, so that’s a maximal isotropic subspace, and well, I mean everything here is finite because we have a fixed n and we have a fixed prime p and so, there’re only going to be finitely many spaces all together, so there’re only finally many of these Z’s. Okay, so here’s what you do. You take two of them.

You take two maximal isotropic subspaces, Z and W, and you intersect them. So, I’m intersecting two n-dimensional subspaces in a 2n-dimensional space, so the intersection could be zero, and, well, it could be anything from zero up to n, I guess, but you take the dimension of that intersection, and if the Z and W are random, then you’ll get some random – you’ll get some random variable. This dimension will be some random variable. And you can look at what happens to this random variable, what its distribution is on the non-negative integers, in the limit where the ambient space becomes larger and larger. And it turns out that the limit exists, so there’s gonna be a certain probability. If you look at the probability, for example, that this dimension is zero, that tends to a limit as n goes to infinity, and same for the probability that this dimension equals one, and so on.

27:38 - So then you get a distribution that now is supported on all non- negative integers. It somehow decays rapidly as… And then the conjecture that we made is that that is what the p-Selmer group of an elliptic curve looks like: the dimension of that as you vary the elliptic curve among all elliptic curves over Q. It behaves like this linear algebra construction And now, at this point, there are lots of reasons to believe this conjecture. So first of all, the original – well, this is kind of where it came from, is this result of Heath-Brown that said that this is actually the correct answer, at least, well, maybe not if you look at the family of all elliptic curves, but if you look at the – if you restrict to certain quadratic twist families, then you get exactly this distribution, at least for p=2. But I mean, to say that that’s evidence for this conjecture is kind of cheating, because we because we used this, because we modeled our – I mean we came upon this conjecture by starting with what Heath- Brown’s theorem was, and said Oh, what linear algebra construction could get this? Oh look, now it’s compatible with that! Big surprise, right? A bunch of other people have worked on extending Heath-Brown’s method to other families of quadratic twists, so, okay, so there’s that evidence.

29:05 - Also, after we found this linear algebra construction, we actually proved that the Selmer group actually is an intersection of two maximal isotropic subgroups, so there’s something you can define in the arithmetic of an elliptic curve where you can actually see some isotropic subgroups showing up. I mean it turns out that they’re actually maximal isotropic subgroups in infinite-dimensional vector space, which you you equip with some topology to make it some locally compact, some locally compact F_p vector space, but anyway, I mean that actually that also kind of fits with this because, I mean if it’s happening if it’s really the intersection of two infinite-dimensional vector spaces, then it makes sense that you should get the right answer when you let n go to infinity here. So RP: There’s a question from Allison Miller about just whether there’s a simple explanation for why the distribution converges to a limit as n goes to infinity? BP: Oh, that’s a good question. Yeah, I don’t know if there’s a simple explanation. I mean we sort of – I mean, what we did, we sort of calculated it and I just checked that it did. Yeah, I don’t know. I don’t know if there’s a simple explanation.

I mean 30:23 - there is some relationship between the situation when you take n and you take n+1 – there’s a way to go from a quadratic space to a smaller one, so if you have a quadratic space and you choose a one-dimensional isotropic subspace, let’s call it X, then you can take X^⟂, which is of dimension 2n-1. You can take X^⟂/X and then that gives you, if you started with a non-degenerate quadratic space you’ll get X^⟂/X will be another non-degenerate quadratic space. So you can do some comparisons – things like this to match up, but I think the way – My memory is the way we did it is just by checking, by just actually figuring out what the distribution was and just checking that it converged to a limit. AM: Okay thanks. BP: So, sorry, were there other questions here? There’s a… Okay, it’s a question for the future, all right. [inaudible] Your time will come.

31:34 - All right, so those are two reasons – well one cheating reason and one actual reason – to believe this, and also another reason, that this conjecture is compatible, well, I mean, there’s not a general theorem about this yet, but it’s compatible with some consequences of this theorem. So if you knew this complete distribution, you would in particular be able to predict what the average size of the, of the Selmer group was, and as far as we know that’s compatible – I mean that’s compatible with all the theorems that have been proved. So first there was, Johan de Jong proved a result in the function field case for the 3-Selmer group, for the average number of elements in the 3-Selmer group, and for a global function field, and then Bhargava and Shankar proved a series of results for elliptic curves over Q for the 2-Selmer group and the 3-Selmer group and for 4-Selmer group and 5-Selmer group and in all cases, this is compatible, well, I guess maybe not for 4 because 4 is not a prime but anyway, at least for as much as it’s compatible – It’s as compatible as it could be with this. Everything looks right. And recently there’s some further evidence for this. If you take the function field version, but then you take what’s called a “large q limit” version of it where, it’s not quite where you… Yeah.

Anyway, you bound the height first and then you let the q go to infinity 33:05 - and then you try to look at the distribution of these things, and that is actually, then it’s proved that this distribution is correct for that variant of the problem. Okay, so I hope you believe this conjecture. I mean if, well, if you don’t believe this you’re definitely not going to believe what I tell you later. So now, I and a bunch of my co-authors that were listed on the first slide, we tried to extend this – so we had this model for this p-Selmer group, but now we want to adapt it to get a model for the higher Selmer groups, for the p^e Selmer group, and then if you can do that for all e, you can put them together and take a direct limit and hopefully get a model for the distribution of the p^∞-Selmer group, which gives you more refined information and maybe gives you more information about the rank. So we succeeded in doing that, we just guessed, well, after a couple of false starts, we sort of guessed what the…

34:10 - what a reasonable conjecture could be of some linear algebra construction that would give a distribution that is likely to be the distribution of the p^e-Selmer group, and then we actually put it all together to make just a conjecture for the p^∞-Selmer group, and in fact, we found a simple model that would not just predict that, but also predict all the terms in the sequence. So we have a – so we found a simple linear algebra construction that would… where you just choose some random matrices over Z_p and you do some co-kernels and kernels and so on, and you get a random sequence of Z_p modules and we conjecture that that should be, the distribution of that should be the distribution of what comes from elliptic curves. And so in particular, from this, well, even more you can actually, you can actually look at the distribution of this Ш (“sha”) of this Shafarevich-Tate group, and it looks like you get a different distribution for each – if you condition on the rank of the elliptic curve. I mean, it’s almost as if I mean, people who’ve studied the Shafarevich-Tate group know this, but it’s almost as if the Shafarevich-Tate group is a different group for each value of the rank.

35:32 - And so for each value of the rank it has its own distribution, its p-primary part is p-power torsion in this group. And yeah, so we predicted a distribution for what that should be. You might ask, well, maybe that doesn’t make any sense. I mean, suppose r is 100. Maybe there aren’t any rank 100 elliptic curves in this set. Then what does it mean to talk about the distribution? And well, I don’t have a good answer for that, so don’t worry about it.

36:07 - Yeah, and what this dist – so there’s some linear algebra construction that gives you this, and it amounts to taking a random matrix over Z_p, with respect to Haar measure, that’s alternating, which means that it’s equal to the negative of its transpose, so it’s like a skew-symmetric matrix, that’s what this little alt means. So that’s, okay, so you can choose a matrix in there at random, and then you can look at its cokernel, viewed as a map from Z_p^n to Z_p^n, just as a linear transformation of Z_p modules. And well, Z_p is not a field. So the cokernel is a, well, it’s a Z_p module, but it could have a torsion subgroup. And so you take the torsion subgroup of that, and that gives you some finite abelian p-group, and we conjecture that the distribution of that is the distribution of this p^∞ torsion here. Oh, I forgot to say one thing. So we don’t just take any random matrix in here, but if you want the distribution of this for rank r elliptic curves, then you should take the A’s in here such that the rank of the kernel is r Okay, and again, there are many reasons to believe all these conjectures on this page.

37:36 - So first of all, there are some prior conjectures by Christophe Delaunay and Jouhet, starting in 2000, that made predictions about some sort of Cohen- Lenstra heuristic kind of things, adapted to predicting the distribution of these groups, and our conjecture matches theirs in the cases where they both apply, and also recently there is, so Alex Smith proved a version, well he proved that this distribution is correct for p=2. Well, not for the family of all elliptic curves but for, for many families of quarterback twists. Okay, so I hope you all believe this. Oh, so the question is what is n? I guess you’re talking about this n here. Yeah. Sorry, I So n is going to be it’s, yeah, it’s similar to what happened on the previous slide. You’re supposed to take this distribution for each n And then you let n go to infinity and you actually have to let n go to infinity through integers of the parity depending on what this r is, but yeah, just otherwise you’ll get this co – yeah, to… In order to have it… Yeah, but anyway, so yeah that’s… n is going to always go to infinity at some point.

39:04 - All right, so now let me – okay, sorry, let me now go to the – I mean, oh so – Yeah, so let me just – I want to focus on this one thing here. So you notice that we are modeling these groups for each value of the rank, and the rank comes in in this way. The rank – So we condition, to get this distribution for the rank r elliptic curves we condition on this, on the rank of this kernel being r. But now you can sort of turn things around: suppose you want to – you don’t want to condition on the rank. You don’t want to condition on – you don’t want to fix attention to rank r elliptic curves.

Suppose instead 39:43 - you want to look at all elliptic curves and figure out what what’s the distribution of the rank? So you can turn it around and say okay, then maybe you’re supposed to just choose A at random and just see what kind of ranks of these kernels comes out, and maybe that should be the distribution of the ranks. And I mean, that would be the simplest possible explanation for why it’s showing up this way, in this here and in these conjectures, and, well, I mean, you don’t necessarily want to do this for p-adic matrices, because if you do it for p-adic matrices, then this condition about the rank being a given number, that’s going to be some kind of hyper surface or some sort of sub-variety inside the space of… inside the moduli space for these matrices and the chance that you land on a particular hypersurface is always going to be zero if it’s lower dimensional, so you won’t see higher rank if you do this. If you do it this way, it means the probability of higher rank will be zero, which actually is what is predicted for – I mean beyond these parity things, it is predicted actually that, that the probability if you choose a random elliptic curve, the rank is as small as possible given – I mean, given the parity constraint. So it’s predicted that 100 percent of elliptic curves have rank zero or one asymptotically.

41:08 - But we really wanna know about the zero percent that have higher rank. And so to analyze those we need a more refined model where we don’t just take Z_p coefficients, but we take Z. We take matrices that are with Z entries, and Z entries of a particular size, and so there’s actually some chance that, you know, some determinant will actually be exactly zero. And so this is what we’re going to do here. So now let me tell you now the, the model, well our model at least, for how to model the rank of an elliptic curve.

41:43 - And the model is going to depend on the height of the elliptic curve. So the chance, the chance of having a higher rank will depend on how big the coefficients of E are. So here, we’re going to choose – there’re going to be some functions that are going to go to infinity but go to infinity at a controlled rate depending on the height, and we’re going to choose – we’re going to do n by n matrices for n that’s large but whose size is again controlled by the height of an elliptic curve. And we want this n to be random parity. That’s because we want the parity of the rank of the elliptic curve to be equally distributed. And, okay, so, so now once you’ve chosen this n, you choose a random n by n matrix with integer entries, but you also control the size of these integers in this matrix.

42:41 - It’s an alternating matrix and the entries of this matrix, the integer entries, are bounded by this, this function that’s growing at a controlled rate. So, I mean, there’re only finally many matrices that satisfy this bound and you just choose this uniformly at random. And if you wanted to model many elliptic curves, you would just do this independently for each elliptic curve. Okay. So now you have this random matrix, random integer matrix, and now based on the previous slide you take the co-kernel of that matrix viewed as a linear transformation from Z^n to Z^n, and so that co-kernel will be some finitely generated abelian group, and you take its torsion. And then that’s predicted to be the model for the Shafarevich-Tate group, and on the other hand if you take the kernel of the matrix, that should be the model for the rank.

43:34 - Okay, so these are really random variables, these two things. I mean I’ve given them notations to make it look as if they’re kind of related to these things but these things are defined using just this – they’re just pure linear algebra constructions. These do not actually have anything to do with the elliptic curve, well, except that their construction depends on the height of the curve, but this really doesn’t have anything to do with the arithmetic of elliptic curves, except we conjecture that it does. Okay, so I’ll call this like the “pseudo-sha” of this E and this is the “pseudo-rank” of this elliptic curve E. And so we predict that these model – the distribution of these things is going to be like the distribution of the actual things that come in – the actual things in the arithmetic of the elliptic curves.

44:27 - Okay, so that’s the model, and the only thing I haven’t told you about this model is what these – how quickly do these functions grow? Like what is this controlled rate of growth? How do I calibrate the model by choosing those functions? And the way we did that is by using some things that we know about the arithmetic of elliptic curves. So it turns out that under standard conjectures like the Riemann Hypothesis for the L-functions of the elliptic curves, things like this, and the Birch and Swinnerton-Dyer conjecture, you can actually figure out what the average size of this group is. And so now you can try to set these parameters so the average size of this comes out to be the same thing. And it turns out in order to do that you need – you need a term – you can do the calculation, it finds out that you need this exponent, this thing to be roughly H^(¹⁄₁₂). Okay, so yeah, so as I said this calibration ensures that this model is giving you the right answers for the average size of these of these two groups, at least as E varies over rank zero curves.

45:45 - Okay, so now once you have that now the model is completely specified. I’ve told you what these functions – well, I guess it doesn’t really matter which functions you do that have this combined. It’s really only this combination that matters. But yeah, so you choose some growing functions that satisfy this and that and now you have this well-defined model, and now you can see, okay, what does this model predict? So if you define these random variables this way, what, for example, what is the distribution of these numbers? of these pseudo ranks? And so, yeah, and that just proving theorems in linear algebra, so that’s actually something you can do without knowing the hard – without answering the hard questions about the arithmetic of elliptic curves. And so we did that. We found that with probability one, – so I mean, you know, it must be true because it holds with probability one – So – this is a theorem also. Look, it says theorem. So there’s not a heuristic, except we’re not proving theorems about actual ranks of elliptic curves, we’re proving theorems about pseudo ranks, which are really, yeah, which…

46:52 - So, the theorem is that, well, you can count how many elliptic curves have each value of the pseudo rank. So if you start with all the elliptic curves, then as I mentioned before up to height H you have H^(⁵⁄₆) of those. And H^(⁵⁄₆), ⁵⁄₆ is the same as ²⁰⁄₂₄. And it turns out that rank zero and one, they appear most – most of the, the pseudo ranks are equal to zero or one. Most of the time it’s zero or one So each one is like half of H^(⁵⁄₆), or some constant – half of whatever the constant is. But then, after it becomes – it’s zero percent that have higher rank, but you can quantify how much of that – how…

47:39 - I mean it’s asymptotically zero percent, but you can quantify how quick – what the rate of growth is for having, for having pseudo rank at least two. In fact each time you ask for one higher pseudo rank, the fraction of elliptic curves that has that pseudo rank goes down by a factor of H^(¹⁄₂₄). So we started with H^(²⁰⁄₂₄), If you want pseudo rank at least 2, then it’s only going to be H^(¹⁹⁄₂₄), which is zero percent asymptotically compared to all the elliptic curves. And then there are even fewer that have pseudo rank at least three, and it keeps going like this. So this is like a theorem. And then eventually, okay, you get down to something where it’s H^0 or, well, We actually – our analysis is not fine enough to figure out whether this is plus or minus here, so this, we’re not quite sure whether this is H, like, it could be like log H or something or it might be like 1/log H, we don’t know.

Anyway, but then, once you go beyond that then, 48:43 - conjecturally it should be finite. In fact, we prove that there’re only finitely many – with probability one, there’re only finitely many elliptic curves with pseudo rank greater than 21. So, okay. So that’s a theorem. That’s – this is a theorem in linear algebra, and then the heuristic is that… So this is the theorem, but then the heuristic is that these pseudo ranks are actually, have something to do with actual ranks. So the heuristic is that whatever happens to these pseudo ranks is the same thing that happens to the actual ranks.

49:15 - So, based on this, it’s reasonable – this suggests that maybe it’s true that the ranks, yeah, the actual ranks have the same behavior. Or at least, I mean, yeah, and by the way, this implies that the pseudo ranks are uniformly bounded, because if there’re only finitely many E that have pseudo rank exceeding 21, you can just look at those finitely many E, and look at, well, and just take the maximum of their ranks, and that will be the overall upper bound. So it might be a little bit more than 21, but it’ll still be a single upper bound. And so, yeah, so I don’t know if I actually believe all this. I mean we made quite a few assumptions going along the way.

50:01 - but what I do, I mean, what I do believe is that even if 21 is not quite the right answer, I mean that I do believe that there should be some uniform upper bound on the rank, because I mean in order for this, for the rank to be unbounded, I mean, this heuristic would have to fail in a very strong way. It would have to, I mean, it would have to be not even close. yeah, so I do believe, yeah, so anyway, so I do believe, I do believe it’s bounded. I mean, there are other reasons why this could fail. There might be some, I mean as Noam Elkies and other people have pointed out, there could be some algebraic families over the curves that for somehow causal reasons have higher rank than you would expect.

50:48 - But even so I kind of – I still kind of believe that even if that does happen, I don’t believe that things are going to go so wrong that it will make the rank unbounded. But yeah. And by the way, you can also try to compare these predictions with reality. I mean, and so, well, what’s known, I mean, Elkies found infinitely many elliptic curves of rank at least 19 which is getting pretty – I mean, we predict that there should be infinitely many up to rank, well, if you believe this thing, there should be infinitely many of rank 20 and maybe also infinitely many of rank 21, so it’s getting kind of dangerously close, but so far, so far our predictions agree with reality and he also found one elliptic curve of rank at least 28, I mean, but that’s also compatible because we don’t we don’t say that 21 is the absolute upper bound, we just say that beyond there, there should be only finitely many. So, so far, everything looks sort of consistent. Now, as further evidence you can look at not just the family of all elliptic curves, but you can look at, you can sort of stratify it according to the torsion subgroup.

52:01 - For example, if you look at elliptic curves whose torsion sub group is Z/6Z, then you don’t have H^(⁵⁄₆) curves anymore, you have, you have only H^(¹⁄₆) curves up to height H, and so you have fewer tries, you have fewer chances to get high rank. And you can redo the analysis and you will get a different, a smaller upper bound for the critical value, beyond which you have finitely many exceptions – elliptic curves of higher rank than that. And then you compare that with the reality about what, what’s actually – what, yeah, so the reality of what, of what is actually – what’s the largest known rank, or the largest known infinite family of elliptic curves with this subgroup. So, yeah, so this 19 here means that there is an infinite family of elliptic curves that has a trivial torsion. There are infinitely many elliptic curves over Q with trivial torsion subgroup and of rank at least 19. And well so far, it looks good.

I mean the reality 53:13 - is – seems to be matching us now. I think – I’m not actually sure if this table is completely up to date. I think, maybe Noam Elkies will talk about this later, some of these numbers might have been improved in the last few years, in the last year or two, but I think it’s – I don’t know. Maybe it’s still true that I think it’s still true that there’s not quite – there’s no – well, I think it’s still true that they haven’t quite contradicted any of these bounds we predict. And not only that, but they look I mean, it looks – oops.

Sorry 53:47 - So it looks – I mean, not only are we getting a bound that looks, that is compatible with reality, but it looks like the bounds are actually close to what, what people have actually been able – the examples that people have really been able to discover in each case. Because I mean, not only are our upper bounds actually the upper bounds for what’s known but they’re actually – they seem to go in, parallel, so it seems like our predictions have some, there’s some connect – I mean, it’s not completely disconnected with the reality of what people are finding when they actually try to search for these elliptic curves with given torsion subgroup. RP: Oh hey, Bjorn There are a couple questions in the chat. A few from Ross Patterson and Grant Mulnar about the significance of the numbers 21 and 24, and then one from Alice Silverberg about whether you think it’s more likely that the bounds aren’t sharp or that there are infinite families that reach the bounds that haven’t been found yet. BP: Okay. Yeah, I see all these questions. Yeah. Okay. So let me start with the significance of 21. Well, it comes from the previous – let’s see, sorry.

54:58 - It comes from this, this previous, this previous slide here. You start with H^(⁵⁄₆) being the total number of elliptic curves and you have H^(²⁰⁄₂₄), but that’s the total number of elliptic curves, and then, well, I mean each time you impose one higher rank beyond one, the rank – the number goes down by H^(¹⁄₂₄) and so you have to do 20 steps and then you’re – and then If you do 20 steps starting at 1, that gives you 21. Now, okay, that just raises the question why is it H^(¹⁄₂₄)? So, where does the 24 come from, and I think it’s the, well, I think it’s the same 24 that appears in, you know, in a lot of other math, you know, a lot of other math, I mean the 24 comes from the 12. And it’s I mean it’s related to the fact that the delta is a modular form of weight 12 and all that and, I mean, so yeah, I guess it comes from the – so going back here, so, let’s see. I mean it goes back to this calculation of the average size of the sha using the, using the Birch and Swinnerton-Dyer conjecture to analyze, analyze this.

It has to do with – actually it boils down eventually to the, 56:29 - to the size of the real period of an elliptic curve of height H, and you can show that the the size of the real period is something like H^(-¹⁄₁₂). And then if you plug that into the Birch and Swinnerton-Dyer conjecture then that has to be compensated by the size of sha in order for the Birch and Swinnerton-Dyer conjecture to hold. And so that’s kind of where it comes from. So it comes from some calculus calculation of just what do you get when you do the integral of this elliptic integral that computes the real period for when you have an elliptic curve of height H. So, yeah, so that’s what it is. So, okay. So, okay now, let’s see. So I guess the next question is… RP: Maybe from Alice about BP: Okay, let me just get – So, Brad Brock pointed out that yeah, the first step didn’t drop. So if I go to this, this thing it’s actually – actually you can see it coming out from linear algebra.

57:34 - I mean, so maybe it’s a little bit surprising, the first step doesn’t drop. So the graph of this exponent goes, you know, it goes dun and then it goes doooown. So that little kink at the beginning is actually there. I mean, and that’s what you also would expect from the elliptic, from the arithmetic of elliptic curves. So that’s another reason why we sort of think that there might be something right about this. But then, yeah, then it goes down, and so that’s where the 21 comes from. Um, okay Ross Patterson, okay I think I explained that. So Alice Silverberg’s question “what do I think it’s more likely that our bounds aren’t sharp or that there are infinite families that reach all their bounds, but we haven’t found them yet.” I don’t know, I think, I mean, it wouldn’t surprise me too much if I go to this page, I mean there are a lot of predictions here, and some of them are actually very close so I wouldn’t surprise me if somebody someday found an example that violated one of these. So maybe, you know, maybe here maybe there’s – maybe you can actually improve this by one and find a family. Maybe there is one.

58:47 - So, but, yeah, but I think there are already some examples where there’re infinitely many, there’re infinite families that, yeah. Well, I mean there are some of these cases where there are these infinite families that obtained the bound. So anyway, maybe I should defer this to people who’ve actually been doing these searches and, yeah. NE: If I may, all of the numbers on the right don’t exploit parity. So, if you have two numbers that are equal, that means that you’d guess that about half the time, just for parity reason, unless it’s some weird counter-example to equidistribution of parity, that you actually will get, half those five, they’re actually sixes and half, and half the threes are actually fours.

These are not – 59:45 - these are not from my families, they go back about 12 years to [inaudible] and so forth BP: Uh, okay, so in some sense, maybe we already expect that some of these bounds are going to be exceeded. All right, okay, so I guess that’s what Armand Brumer was also asking about. So, okay. Armand Brumer: May I just add one thing? BP: Sure. AM: In experiment, where one did assume, or one imposed, let’s say one or two or more, rational points then the rank did jump. It was sometimes the imposed one and sometimes one more. That’s why – that’s what I meant by my question. You know, I’m surprised that the rank 2 and rank 3 are not roughly the same distribution, expected distribution, because half of the rank – if you’ve imposed two, the rank two, you will get rank three half the time, is what it seems like. BP: Well, I think that, well, if you first divide by parity – if you first divide your elliptic curves by parity, and you ask, okay, for the even rank – for the even parity ones, how many of them have rank at least two? It seems that it’s easier for that to happen than it is for an odd rank to have – for an odd parity elliptic curve to have rank at least three. At least that’s what our model predicts, and I think that’s what you people tend to see in practice as well, but I think some people – I mean at least the computations I know of, I think, tend to, tend to agree with this I think. So I think I do believe that at least. Oh, yeah, sorry, so I guess, so I guess I’ll – let me continue going through these.

I’m not sure if I answered your question, but, 01:51 - at least that’s what I think is happening. And let me – okay, let me consider, there are a couple more questions. My talk is basically done by the way, but let me just, I mean – but let me try to answer the rest of these questions. “So the bounds are in terms of height. Can we predict the size of the coefficients of the finitely many exceptions?” That’s a good question. I mean, the model I think would predict that when there are exceptions, then they should tend to have not so large coefficients, so that’s maybe a failing of the model, given that there’s a curve of rank 28 that has big coefficients.

02:29 - So, yeah, I don’t know what to, I don’t know what to say about that. I mean, because the prediction – the model says that the larger the coefficients become, the less and less likely it is to be an exception – to have that high rank. Yeah, so. Okay, Naomi Sweeting says “what are the two maximal isotropic subgroups that intersect to Selm_p.” So, okay, maybe I’ll defer that to later. I mean it, it comes in the cohomology – I can show you in terms of the, I’ll show you maybe later, but let me try to So, okay.

Oh, and then you have a second question about where in the model is the torsion sub group? 03:22 - So the only way we use the torsion subgroup is to look at how many elliptic curves there are of height up to H, so if… I mean most – if you look at elliptic curves of – with torsion sub-group Z/3Z, then the coefficient, the A and B, they satisfy some algebraic condition that cuts down the number of elliptic curves up to height H, and so you get this smaller exponent of H, and it’s just from the fact that you have fewer elliptic curves to try that you have fewer chances to get a higher rank elliptic curve. And so that’s the only way in which the torsion subgroup comes in in the model. Okay, so Will Sawin? WS: I’m just commenting in response to questions that people had asked. I don’t have questions. BP: Okay, you’re just answering questions about… WS: Trying to. BP:…

impose rank two, 04:18 - BP: Right, you could first, right, you could first take all the elliptic curves of rank two, and then what I would predict there is that 100% of them will have rank 2. Or you… WS: If you take this… BP: You’re saying something different. Okay, you’re saying that if you algebraically start with two rational points, then parity will make half of them of rank three. WS: Yeah, so if you have a specific family of elliptic curves with rank two, then there will be this parity phenomenon that half of them have rank three, which is what Armand was mentioning. BP: Right. But for these calculations – for the calculation on the previous slide – Oops, sorry here I was thinking of just taking all elliptic curves, in which case I do expect that rank three is less likely. Great, so, okay. So let me just end with this summary slide just to say what I told you.

05:21 - So yeah, so we got, in order to estimate what’s happening with ranks, we’ve bounded – we’ve sort of made a model for this complete package, and then you can corroborate it by using theorems about all the different aspects of the package, and then the model predicts that the pseudo ranks are – all but finitely many of the pseudo ranks are bound by 21. So we suggest – we believe that the ranks are, are uniformly bounded based on that. Okay, and then there are these things that I didn’t have time to talk about, that you can try to do similar heuristics for other number fields or function fields, or you can also try this for abelian varieties, and I think, I think there should be similar theorems for these two – or not theorems, but you can – it seems, it seems reasonable to predict similar things, at least to guess reasonable things, but I mean there’s – the evidence is maybe more shaky for those but, but I still believe it so, Okay. All right. Thank you all for your, yeah. Thank you all. I enjoyed hearing all the questions and everything so yeah. RP: Thanks, Bjorn [applause] Let’s see are there some more questions Oh, um, and there was Fabien’s original question.

BP: Oh, yeah, so, let’s see, so let me scroll back in the chat 06:59 - Okay, so could you point to where the heuristics fail for function fields? Well, sort of, Yeah, so, well, let’s go here, so there’s the next slide. Okay, so I mean, yeah, I mean – by the way you should all feel free to leave. I mean this is like me cheating to give a longer talk than I’m allowed to give, so. Anyway, so let’s take a global field, well maybe it’s a function field, and, you can again talk about this fancy E, except now it’s going to be all isomorphism class representatives for elliptic curves over K, and because we want to exclude finitely many exceptions, instead of taking the maximum of the ranks we want to look at sort of the lim sup of the ranks, which is the number above which there are only finitely many elliptic curves of that higher rank. So, our heuristic predicts that this lim sup for Q should be either 20 or 21, I don’t know whether it’s 20 or 21 because I don’t actually – in the case of 21, the heuristic was not fine enough, it was not refined enough to tell whether there’re infinitely many or finitely many at 21, because like H^o(1), that could be positive – the exponent positive or negative so, it might be only finitely many of rank 21.

08:23 - Yeah, so you can run the heuristic. I mean, it works pretty much the same way for a global field, and it gives exactly the same, the same estimate, so, so if you believe that, then you would get that the lim sup should be again 20 or 21 for every global field. And this is very, uh, this does not match well with reality, so that’s, yeah, so that’s maybe, unfortunate thing, but what I can do is I can tell you, well, here’s a reason why you shouldn’t expect it to compare well with reality. Okay. So first of all, here’s a way in which it doesn’t match reality. First of all, for the global function field case, the actual lim sup is infinity, because the rank is actually unbounded, and even for number fields you can find examples of special number fields over which – so we can prove, actually we prove in this paper that there is a number field, for example, where there are, there are infinitely many elliptic curves of rank at least a million.

09:33 - So we can actually construct, unconditionally, a number field that, that has infinitely many elliptic curves of rank over a million. So that’s another evidence that says that this B_K is not going to be 20 or 21, but all the constructions of these – okay, okay, here’s some specific number fields that we used for this theorem, for these theorems. But, to reconcile this with our heuristic, there’s these elliptic curves of high rank that have been constructed in the literature by Shafarevich and Tate and then also by by Omar and then by us – they’re all special in that they, these are elliptic curves that – I mean nominally, they’re defined over K, but they actually come from base change from a much smaller field, and so, you can – instead of, yeah, so, so maybe, yeah, so maybe, I mean those elliptic curves that come, that come from a smaller field, I mean, there’re arithmetic reasons why they should have higher rank. So the way we propose to, to correct our heuristic is just to exclude – not allow the elliptic curves that come from a smaller field. And if you exclude those, then it’s conceivable that there is – that there’s again a uniform upper bound.

11:01 - There’s no counter example known to that statement, even for function fields. And so that’s what made me believe. I mean, it’s still not true that 21 is exactly the number, because there is a special – there are a few special families, like there’s this example of Shioda that has gives you infinitely many of rank at least 68. But, yeah, but okay, but outside of that special family it’s possible, yeah. But even so it seems reasonable to guess that maybe it’s uniformly bounded once you exclude these curves coming from lower, lower fields. Okay. RP: Thank you so much, Bjorn. This was a wonderful talk. And next time we’re going to have Noam Elkies speak on September 15th, so I hope you can all join with us then. Okay, goodbye! BP: There’s something…

RP: Oh there’s more? 12:01 - BP: I don’t know if I should keep answering questions, but… RP: Oh yeah, definitely, if people can stick around. BP: Okay, and the question is would the generalization to the abelian varieties be only for self-dual ones? So, I guess, maybe, I mean principally polarized maybe so, um, well I mean, actually, I have a slide about this. Yes, I did come well prepared. So, yes. So you can ask, is there a bound on the rank that depends only on – of course you have to – the bound’s gonna have to depend on the dimension, because you can just take a elliptic curve of rank 28 and just take E x E x E x E and you can make the, the rank big that way. Also, you could start with – you could just take one abelian variety and just let the number field grow, so the bound is also going to have to depend on the degree.

12:46 - But you could ask, is there a bound that depends only on the degree and the, and the dimension of the abelian variety? And well, I mean There’s no – you don’t actually have to restrict to principally polarized ones, because you can use Zarhin’s trick to reduce to the principally polarized case. By doing that and also by doing restriction of scalars you can reduce to the family of just abelian varieties of a fixed dimension over Q, and principally polarized abelian varieties over Q. And then you can define a height in some naive way again and you can prove that the number of abelian varieties of a given height is bounded by some polynomial in H, maybe not H^(⁵⁄₆) anymore, but it’s H to some fixed number, and so if you have a pseudo rank that decays by a factor of H to the some fraction, for each – every time you ask for a one higher rank then again you would expect the pseudo rank should be bounded with probability one, and so, So yeah, so my guess is yes that this should be true. Okay, there’s another paper, okay, so I mean, I was mentioning a paper by Richard Griffon, I probably should have mentioned that as well. So yeah, okay. “Where does our heuristic to get rank 21 use Q?” Unfortunately, our heuristic doesn’t use Q.

The heuristic also predicts rank 21 for other number fields, too. So it’s a – so the Shioda thing is really, is really in contradiction with our heuristic. But on the other hand, I guess, I don’t know. I can kind of like wave my hands and explain away saying “oh, it’s just a special family maybe if you exclude these finitely” -- I mean there are the others conjectures, like Manin’s conjectures about integral, rational points about how – you have to always exclude some sub-varieties and stuff like that. So maybe you’re supposed to do the same thing here.

So I don’t know 14:50 - Yeah, but technically it does violate our heuristic, so. WS: So, um, along the lines of Manin’s conjecture, do you think, like, so in Manin’s conjecture doesn’t those tell you how many rational points there are? Or at least in the modern form it tells you that they’re distributed – how they’re distributed p-adically for each prime p. And so, do you have a guess about how like elliptic curves over Q of rank seven are distributed p-adically in the moduli space of elliptic curves? BP: well, I haven’t tried – no, I don’t have a prediction, but I think that, I think that our analysis could be used to get that, because I mean it’s, partly, well, I mean, let’s see. Actually I’m not sure. But I mean part of the model, at least the calibration part of it, comes from looking at the Birch and Swinnerton-Dyer conjecture, and you sort of, so – I mean one way to think about it is you take the Birch and Swinnerton-Dyer conjecture and you solve for the square root of sha because square root of sha is supposed to be an integer. And if you want to measure the probability, and if – sort of, if the rank seven sha is zero, I mean, that means that your elliptic curve actually has higher rank if you know what I mean by rank zero sha.

16:28 - So if you pretend that your elliptic curve has rank seven and you look at the Birch and Swinnerton-Dyer conjecture for an elliptic curve of rank seven and then you solve for the square root of sha you get some formula involving the seventh derivative of the L-series and then that formula, it’s going to be an integer, and if that is zero, that means that the L-function, the seventh derivative of the L-function was actually zero so that means you were actually higher rank. You were rank eight or higher. And so, you see, so then the question is, you take that square root of sha, and you ask what’s the chance that it’s exactly zero? And you can estimate that based on what the range in which the square root of sha is, where it lies, and that, you can, you can use that as sort of a guess as to how often it is that your family – the rank seven elliptic curves actually have higher rank. But that probably will change – I mean that range in which it lies will change depending on things like the Tamagawa factors at p, which are going to depend on your elliptic curve, how it looks p-adically. and so you could incorporate that into these heuristics I think, and maybe you could use that to, to get some kind of idea – answer to your question, but as far as I know nobody’s tried that, but maybe you could do that. I mean even for rank two, I think it’d be interesting to do that. I mean, it’s definitely true.

I think, I mean 18:01 - that if you impose p-adic conditions on your elliptic curves, you can make it more likely to have higher rank. In fact that was Mestre’s idea back in the 80s for finding higher rank elliptic curves. He took – he chose elliptic curves that had the maximal number of points mod 2, the maximum number of points mod 3, and the maximum number of points mod 5, and then just by imposing these conditions, it turned out by searching within those families it already made it very likely to have high rank elliptic curves. And so what he’s doing effectively is he’s like, He’s pushing – he’s like putting his hand on one side of the scale, of the Birch and Swinnerton-Dyer scale, and like making it more like, pushing it to make it more likely that the sha, the square root of sha, is going to vanish. The square root of whatever rank sha is going to vanish. .