Manjul Bhargava: What is the Birch-Swinnerton-Dyer Conjecture, and what is known about it?

well thank you very much it's a it's really a great honor to be here at the celebration of the Appel prize to Professor Wiles professor Wiles has had a great influence on mathematics and on mathematicians in general myself included I was a PhD student of his and well that was a that was a really wonderful experience and really fun experience for me as we all got to see this morning presser Wiles is not just a great mathematician but also a really great teacher he had the thing that always struck me the most was that he has a real feel for what's important in mathematics what are the most promising directions in mathematics and as a result of this incredible intuition that he has about the direction of the field he's had a great impact on on many central problems in mathematics of course burma's Last Theorem is the most famous one but actually many many central problems have seen progress in large part influenced by his work and so the problem I want to talk about today is the vergence Winterton dire conjecture which has become one of the most fundamental problems in mathematics and in particular in number theory it's the next frontier in the solving of equations as I'll explain in number theory and of course andrew has had a great impact on the progress that we've seen towards this conjecture although of course there's also much more to be done to reach the final resolution of this conjecture and so that's what I want to talk about today is first of all I want to describe in elementary terms what this conjecture is and then explain a little bit about what's known about it what Andrews influence has been and what remains to be done hopefully that'll attract many people into the area including many of the young people who are here yeah okay so the versed in spirit generic conjecture it's one of the clay millennium problems so in celebration of the new millennium in 2000 the clay mathematics Institute established a list of seven millennium prize problems for which they offer 1 million dollars each for for the solutions you don't have to solve all seven for the million dollars each one is worth and really knows and and the problems are a lot of computer science most basic problem P versus NP there's the Riemann hypothesis there's the Hodge conjecture there's the yang-mills existence and mass gap problem there's the navier-stokes problem there's the Bert since winter 10 Dyer conjecture and there's the Poincare conjecture so these are the seven millennium prize problems of a clay mathematics Institute of these seven so these were announced in 2001 has been solved so far which is the pranic query conjecture that was solved in 2002 only two years after they announced the problems which kind of surprised them I don't know how ready they were with that what the money yeah but luckily for them Perelman famously turned down the million dollars so they didn't have to worry about that actually the the money has now been given to the pond query Institute in Paris to fund positions there in celebration of the solution of the park Creek injector so each of these problems has has descriptions official descriptions written about them which you can find on the clay mathematics Institute webpage and in particular the bird since winners in dire conjecture the description has been written by by Andrew Wiles and so you're welcome to go and look at that after this lecture what I wanted to do in this lecture though is to describe the problem in elementary terms and really motivate where the problem came from and why it's there and and why mathematicians and number theorists in particular are so interested in its solution so that's what I that's what I hoped you do today so I'm gonna concentrate just on this one problem namely the bird since Winterton dire conjecture so number theory in large part is about solving equations in the whole numbers or in the rational numbers as we saw for example in fermat's last theorem we're interested in solving a certain equation namely X to the n plus y to the N equals Z to the N in whole numbers and that's been the basic problem of number theory since ancient times number theorists have been interested in finding integer solutions or rational solutions to polynomial equations so those may be equations in one variable like x squared minus 2 X plus 3 x squared plus y squared equals 100 or higher degree equations like y squared equals x q plus 2 X plus 3 or sometimes we're interested in solving equations in more variables x squared plus y squared plus 2z squared equals 7 so these are kind of the typical kinds of equations that arise in number theory and instead of defining whole number solutions number theorists in general are interested in finding rational number solutions to such equations rational numbers being ratios of whole numbers so the case of one variable equations turns out over the rational numbers is fairly easy to understand so an equation in one variable looks like a polynomial in one variable set equal to zero so it looks like a 0 X to the n plus a a 1 X to the n minus 1 plus da plus a N equals 0 where the AI are integers and we're interested in finding what are the rational number solutions to that equation and well this is something that we we all learned in high school it's called it's called the rational root theorem so the fundamental theorem of algebra says that there can be at most end solutions even over the complex numbers but to know which ones can be rational there we learned what it's called the rational root theorem which says that any rational solution to this equation has to be a factor of a n divided by a factor of a zero right a factor of the last coefficient divided by a factor the first coefficient and so you can that gives you an algorithm to solve such an equation in rational numbers you take all the factors of a N and you divide them by all the factors of a nut and you try each one and whichever ones work all you found all the solutions and that's I follows from the rational root theorem that's not the most efficient way to do it but at least it is an algorithm that will terminate usually the way we would go about it today is you'd use the fundamental theorem of algebra to numerically approximate those end solutions and then try to rationally approximate it that would be the quickest way to do it but in case there is a quick algorithm to decide what are the rational solutions to one variable equations so it's two variable polynomial equations where it really starts to get interesting and in fact unsolved one variable equations we know how to do very quickly and already with two variable equations a number of years today still don't know how to find all rational solutions to two variable polynomial equations but in small degrees we do know how to find all rational solutions so for example if you look at two variable equations of degree one those are not so difficult to solve so if you want to find all rational solutions to a two variable equation of degree one well that looks like y equals ax plus B a and B say are integers and you want to find all rational solutions in x and y such that y equals ax plus B that's pretty easy plug in any rational value of x and that will give you a rational value of y and we found all the rational solutions okay so that's the case of degree one so degree one is is it solved we know how to find all rational solutions to two variable equations of degree one what about degree two if you look at two variable polynomial equations of degree two it turns out that those finding the rational solutions to two variable equations i've degree two it's also pretty easy and just to give you an example because one examples actually tells you how to how to do all of them for example suppose you want to find all the rational solutions to x squared plus y squared equals one so these are all the rational coordinate points on the unit circle right so here's a picture of the unit circle and when number three was talked about rational points it just means a point on a graph of your polynomial equation whose coordinates are rational so when we're looking for rational solutions to x squared plus y squared equals one that's equivalent to finding rational points on the unit circle right so there are some obvious rational points on the unit circle for example minus one comma zero which is the leftmost point on the unit circle here well that's a rational point so that's a that's a rational solution 2x squared plus y squared equals one and the key observation to find all the other rational points on the unit circle is as follows is that if X comma Y is another rational point on the unit circle then if you connect so say it's over here X comma Y if you take the line that connects minus 1 0 with X comma Y if you take that line well that's going to have rational slope right because it's connecting its connecting two points with rational coordinates so that when you compete the slope that's also going to be rational what maybe the more surprising thing here is that the converse is also true that if you take a line of rational slope through minus one comma zero and look at the second point of intersection with the circle that will give you a rational point and so that would that will then give you a complete description of the rational points namely just take all the rational sloped lines through minus one comma zero look at the second point of intersection and that will give you all rational points so why is that the case that any rational sloped line through minus one comma zero will intersect the circle in another rational point ok well suppose we take a line of rational slope through minus 1 comma 0 the equation of this line will look like y equals s times X plus 1 right where s is the slope and if we want to find the second point of intersection of this line that has slope s with the unit circle well we can take that equation y equals s times X plus 1 and plug it into the equation x squared plus y squared equals 1 and solve and that'll give us the points of intersection well if you if you plug in equation one there into equation two well you get a quadratic equation in X and so you can solve that using the quadratic formula you'll find that one of them one of the solutions is going to be given by x equals minus one which of course had to happen because we know that one of the points of intersection is minus one comma zero so that of course had to be a root but the other root will give you the x coordinate of the second point of intersection and if you solve for the other root you find it's the negative of s squared minus one over s squared plus one and once you have the x value you can plug it into equation one write the equation of a line and you find that Y is given by 2 s over s squared plus one and you notice that if s was a rational number all right if s was the slope of rational number then X comma Y the second point of intersection is also going to be a rational number we have an explicit formula for it and so this point P equals x comma Y which if s is rational is also rational gives the other point of intersection with the unit circle and it's a rational point and so that show the other direction that if you take a point a line of rational slope through minus one comma zero it will intersect a second point on the circle that has rational coordinates and so this is a complete parametrization he was just plug in all rational values of s and you get all pairs X comma Y that are going to be rational points on the circle and so we found all rational solutions to x squared plus y squared equals one so the main theorem about rational points on the unit circle what we've proven is the following is that the rational points X comma Y in the unit circle x squared plus y squared equals 1 are naturally in one-to-one correspondence with the elements s the slopes s in the set of rational numbers together with infinity infinity also has to be considered a rational number here what's the second point of intersection when you when you take the line of slope infinity well it has a double intersection with minus 1 comma 0 so this infinity that's corresponding to minus 1 comma 0 and then all the other rational numbers that are given as slopes will correspond to all the other rational points on the unit circle and the correspondence is between rational slopes s and rational slopes s correspond to that explicit value of x comma y that we found earlier which is the second point of intersection of the line with the unit circle having slope s going through minus 1 comma 0 and you'll notice in this formula that if you plug in s equals infinity what do you get well the second point of intersection should be minus 1 comma 0 and of course when you plug in s equals infinity here well the first coordinate becomes minus 1 as you can see and the second coordinate when you plug in s equals infinity right as s approaches infinity that becomes 0 that's now so in plugging in s equals infinity gives the original point minus 1 comma 0 but all the other rational numbers give all the other rational points on the unit circle and we've so we've explicitly determined all rational points on the unit circle and so we found all rational points all rational solutions to x squared plus y squared equals 1 so we've completely solved quadratic and two variables over the rational numbers and there's nothing special that we used about the unit circle here we only use the fact that the unit circle its equation has degree 2 and so in fact we could solve any conic this way over the rational numbers by by the same method and so the fundamental theorem about rational points on conics is is as follows let f of XY be a quadratic polynomial in x and y with rational coefficients so in other words the equation of a conic then the set of rational points X Y on the conic is given by the same procedure that we just used you start with a rational point on the on the conic and then you take lines of rational slope through it and the second points of intersection will parameterize all the rational points now there's the problem with this argument which is that they're made there may not be a rational point to begin with and then you can't even get started and that actually happens sometimes if you look at here an example is x squared plus y squared equals 3 you will not be able to find a rational solution even to get started so the fundamental theorem of all points on the konnex says that the set of all rational points X comma Y on a conic f of X y equals 0 is either empty because you can't even get started with this procedure yeah but if it's not empty then it's in one-to-one correspondence with rational slopes together with infiniti the infiniti slope and if we're in case two above where there are some rational points then one can explicitly find all the rational points as follows you start with the base rational point which was minus one comma zero in the case when we were looking at x squared plus y squared equals one start with the base rational point if it exists if it doesn't exist where in the empty case if it does exist we start with it and we take all lines of rational slope through P and we compute the second point of intersection of each of these lines with rational slope and those will will give you all the rational points on mechanic and so that's how you can solve the general quadratic in two variables over the rational numbers and the further amazing thing which we haven't proved here is that turns out there is an effective way of determining whether we're in case one or case two okay so given a conic there is an effective method there's an algorithm that'll tell you whether the set of rational points is empty or whether there is a rational point to start with to get started on finding all the rational points and this is called the Haase Minkowski theorem it allows you to take a conic and decide whether it has a rational point or not in an algorithmic way and so this theorem is is entirely effective okay so that's that's the whole story about quadratics in two variables solving them over the rational numbers we can this you can go and do on any of your favorite cuttings okay so the next case naturally then to consider if we're talking about two variable equations is those of degree three okay suppose you have a two variable polynomial equation of degree three we've already solved the degrees one and two well it turns out if you have a two variable equation of degree three it can have zero solutions that happens it could have some positive finite number of solutions remember that wasn't possible in the degree two case right if you have a conic it is not possible to have four solutions that just doesn't never happens right because our argument showed that if we had one solution to start with we could take any line of rational slope and that would produce infinitely many Solutions but in degree three it turns out you can have zero solutions you can have four solutions or some positive finite number of solutions or you can have implementing rational solutions all three of these possibilities can happen for degree three equations in two variables and the embarrassing thing is it's already an unsolved problem to decide which of those cases we're in given a two variable equation of degree three it's an unsolved problem to decide whether say it has zero solutions or whether it has implementing solutions so the case of cubic equations in two variables is really the first case where we just don't know how to find all the rational solutions and this is exactly where the bertson's Winterton dire conjecture comes in if the bird since Winterton died our conjecture were true it would give an effective method to find all rational solutions to a cubic equation in two variables and in particular to determine whether it has finitely many solutions or infinitely many solutions and so this really is the next frontier this is a problem that mathematicians have been studying for for centuries and yet we still don't know how to solve it but at least we now have a conjectured method to solve it and that is what the birch and Swinton dire conjecture gives us so I want to describe what the birch and swenerton Derik conjecture says because if it's true it would give us a method to solve the very next case of equations that number theorists don't yet know how to solve so as was already mentioned in Andrews and Henry's lecture the the structure of rational points on cubic curves in other words the graphs of cubic equations and two variables the set of rational points has a very amazing structure so suppose we have a smooth cubic equation smooth just means that when you graph it there they're no kinks that it's smooth when you graph it and suppose so suppose we have an equation f of X y equals 0 of degree 3 f is a degree 3 polynomial that has a rational solution X naught comma y naught okay let's call that P then it's an elementary fact that one can always make a rational change of variable so that P is sent to infinity in a way so that the equation of the cubic curve just looks like Y squared equals x cubed plus ax plus B and if the f of X Y had rational coefficients then we can actually assume that a and B are now integers and the discriminant of this cubic polynomial is called the discriminant of this curve which is minus 4a cubed minus 27 B squared and the condition of this cubic curve being smooth is exactly that this discriminant is nonzero it turns out and if you want to see how this argument works of how you can always make such a change of variable you can look at this amazing book by Silverman and Tate Tate is of course one of the recent Babel Prize winners he's written a book with Silverman that's really accessible to a high school student about elliptic curves highly recommended as a first book to learn about elliptic curves ok so any cubic curve in two variables you can always if it has a rational point you can always transform it into something that looks like Y squared equals x cubed plus ax plus B and where we can assume that a and B are integers if the original equation had rational coefficients and an equation of this type is called an elliptic curve or an elliptic curve and short Weierstrass form and it's very convenient to use this canonical form so most of the time you'll see when you see elliptic curves will often be written in this form y squared equals x q plus a X plus B where a and B are constants and once you've written into this in this way it's actually easy to describe an amazing structure on the set of rational solutions to Y squared equals x cubed plus ax plus B that are possessed by by the rational points on this on this curve called in lipstick curve so first of all what is the what is the graph of y squared equals x q plus ax plus B look like well if you just think about it if you graph it in r2 in the real plane it turns out it tends to look like one of these two pictures so either it looks like a picture with two connected components like this this part is called the egg of the elliptic curve and this is the infinite part sometimes you only have the infinite part and the egg is empty and so the graph of an elliptic curve in r2 tends to look like one of these two pictures and it just depends on whether the cubic polynomial so we have y squared equals a cubic polynomial in x xu plus ax plus b if that cubic polynomial in X has three real roots then you'll see three points of intersection right through the x axis and that's why you get this extra component here but if your cubic equation only has one real root then the elliptic curve will only intersect the x axis and one in one point and the lists have this one infinite component so the graph tends to look like one of these two pictures depending on whether the cubic polynomial in X has one real root or three real roots okay so that's the picture to keep in mind when you're talking about elliptic curves looks like one of those two pictures and now we can describe what the group law on elliptic curves look like so if one has two rational points on a plain elliptic curve okay so y squared equals x q plus ax plus B and then the line connecting those two rational points so say you have two solutions to Y squared equals x cubed plus ax plus B then the line connecting those two rational points will always intersect the elliptic curve and a third point just like when we were talking about conics we started with one point we took a line through it it would always intersect in a second point in the cubic case you can start with two points you take the line through them it allow it the first two points were rational coordinated coordinated have rational coordinates then the third will also have coordinates that's Annie that's something that's easy easy to check these this has to do with the fact that if you when you're finding that third point of intersection you'll be solving the cubic equation just like we were solving a quadratic equation before we had one rational solution therefore the other had to be rational because the quadratic had rational coefficients in the same way here when we're trying to find that third point of intersection we start with two rational points take a line through it find the third point of intersection with your elliptic curve you're gonna be solving a cubic equation to find that third x coordinate that cubic equation will have rational coefficients the first two roots are rational therefore the third has to be rational and that's what allows you to prove that if you have two rational points P and Q on an elliptic curve and you take the line through them that third point of intersection will again be a rational point so this is what's amazing about cubic curves is it allows you to take two points that are rational and produce a third that's rational and there's a natural law of addition that you can define namely if you have two rational points on your elliptic curve two points that have rational coordinates P and Q you can define P plus Q as the point that you get by taking the line through P and Q taking the third point of intersection and then reflecting it across the x-axis because remember the graphs of the elliptic curves are symmetric about the x-axis so you can reflect it across the x-axis and that you can define that to be P plus Q and that gives you a way of taking two rational points in producing a third one if you have two rational points P and Q 2 points with rational coordinates then this point P plus Q will also have rational coordinates and why do we call it P plus Q well one checks that together with the point at infinity which is the point that you get when you follow the curve all the way up to infinity or equivalently all the way down to infinity they're the same point and you call that the point at infinity then this law of addition plus and dows the set of rational points on a with the structure of an abelian group where the identity of this abelian group is the point at infinity so B Lynn group just means that you can add points and every point has negative and a building just means that P plus Q is the same as Q plus P which of course you can see because line connecting P and Q is the same as the line connecting Q and P so clearly P plus Q is Q plus P the hardest thing about this law of addition to prove but what you can do in an elementary way is to show that P plus Q plus R is the same as P plus Q plus R that's not obvious right if you take P plus Q and add it to our why is that the same as P added to Q plus R that's not obvious at all but that's true with this addition law it has a natural explanation in terms of modern algebraic geometry but it can also be verified just elementary by playing with coordinates okay so this gives the law of addition on on rational points on an elliptic curve something that's very special for elliptic curves is that it's rational solutions have this natural addition structure where if you have two points you can add them and make a third point that also has rational coordinates and that's a really beautiful structure on elliptic curves so when you're studying rational points on an elliptic curve you don't have to just think of it as a set but you can think of as a set with this addition structure that gives it the structure of an abelian group and that's very special to cubic curves so cubic equations in two variables so if you have a rational elliptic curve he was given with rational coefficients this group with this abelian group of rational points on e it's usually denoted a of Q and one question that you might ask is can you always start with some finite set of rational points so that just by playing with this addition operation you can produce all the rational points okay and if you don't want to think of it in terms of the B Lynn group you can think of it can you always start with a finite set of rational points even if it hasn't fully many rational points your elliptic curve may have been flume any rational points but the question is can you always start with some finite set of rational points so that just by playing let's connect the dots procedure where you take two and connect them and find a third rational point and keep repeating that can you get to all rational points and that was one of the very first fundamental questions that came up about elliptic curves there may be implementing rational points but maybe you can always start with some finite set so if that's just by playing let's connect the dots game you can find all the rational points just by using the addition law and this is the celebrated theorem of more Dell more Dell theorem says that the group II of Q of rational points on e is finitely generated and what that means in elementary terms is that you can always start with some finite set of rational points that's that just by using this addition operation or just connecting the dots operation you can produce all the rational points so even if they're inflamed any rational points more Dells theorem says you can there's actually only finite amount of information there you can always start with some finite set of rational points to start with and then use this addition law to produce all the rational points kind of just like in the case of conics or even that when there were implementing rational points on x squared plus y squared equals 1 you could always start with one point and that was enough to to find all the rational points just by taking rational sloped lines through there the analog here is on a cubic curve you can always start with some finite set of rational points and then just by connecting pairs at a time you can actually produce all the rational points this is the amazing theorem of more Dell in terms of group theory since e of Q is a finitely generated group you can use the addition law to produce all the rational points from some finite starting set the fundamental theorem of abelian groups says that if you have a finitely generated being in group it always looks like the product of cyclic groups some number of infinite cyclic groups which look like the integers so Eve Q since it's finally generated fundamental theorem of abelian group says that if Q has to look like some product of copies of infinite cyclic groups namely the integers raised to this M there may be are copies times some finite abelian group called T which is also going to be a product of finite cyclic groups and so you can kind of see from this that the big part of the group is sort of measured by this R this R how many copies how many infinite cyclic groups do we have inside e that kind of measures the size of Eve Q now it's a theorem of Mazur that the group T this finite group here is bounded in size by 16 so it never gets very big so even in a stronger sense it is really this R that's measuring the size of this this set of solutions C of Q so our kind of measure is how big the group II of Q is and it's called the rank the rank of e is the R that comes out of tell measures the number of infinite cyclic groups that you're getting inside this group of rational points in elementary terms if you're not familiar with group Theory the rank of e just measures the number of points basically it's measuring the number of points that you need to start with on your elliptic curve so that every other rational point can be obtained just by connecting pairs and finally the third point of intersection and repeating that procedure so what does the minimum number of points that you need to start with to get all the information about all the rational points on the elliptic curve by just using the addition law yeah and that's the rank and buy more Dells theorem this number is always finite you can always start with some finite set of rational points so that all the rational points can be found I had the connect-the-dots procedure the addition law and the minimal number of points you need is essentially the rank ignoring this two part here okay so are this rank is really the fundamental invariant right that measures how big how many solutions you have R equals zero corresponds to the case right if I are zero here that corresponds to the case where a of Q is finite but once R is 1 or more than the number of rational points is infinite and then R measures how infinite right the bigger R is the more infinite the set of rational points so R is the fundamental invariant given an elliptic curve over the rational numbers given a cubic equation in two variables R is the fundamental invariant that measures how many rational points you have R equals 0 corresponds to finally many rational points R equals one or more corresponds to lay many rational points and most basic questions that you might ask about the basic invariant about cubic equations and two variables are unsolved so for example you can ask what is the maximum that the rank of an elliptic curve can be no one has any idea in fact no one even knows whether the rank can be unbounded is there a maximum we don't know it's a maximum it might go all the way the ranks Michael all the way up to infinity or maybe they they stop at around 30 I say 30 because the current record for the largest rank ever found is 28 found by no Amell keys in 2006 so he showed that the rank of this curve that he found is at least 28 well you can ask more statistical questions you can ask what is the expected size that is the average size of the rank as you vary across the loop two curves that's also unknown do most curves have rank 0 or 1 most people expect that most curves have rank 0 or one can one prove that even say 1% of all two curves have Frank 0 1 and and then of course the most fundamental question that I was asking the beginning Oh is there any algorithm to determine the rank of an elliptic curve that will provably terminate with the correct answer so given an elliptic curve y squared equals x people say X plus B can you design an algorithm that will you know you put in the elliptic curve and it spits out are the rank well it's exactly that last question given it looked a curve can you algorithmically determine the rank that's exactly the question that the bertson's Winterton dire conjecture addresses okay so I want to tell you how it addresses this question and it gives you it gives you a reasonable way actually to find what the rank is in a way that actually terminates if the conjecture is true so as is often the case in number theory to try and solve to solve an equation over the integers or the rational numbers the key idea is to look at the solutions mod P so you look at y squared equals x people say X plus B reduce it mod P and you count the number of solutions in in the integers mod P in other words just just in that finite field so in 1960 bertson's for Newton Dyer did some computations of ranks of elliptic curves and and they also did computations of the number of solutions mod P on the elliptic curve alright so what is the maximum number of solutions mod p to an elliptic curve well remember it's a two variable equation so x and y each will have about P possibilities so the maximum number of solutions is P squared and so they would loop over all P squared possibilities for x and y right and they would count the number of solutions but they had to write code to do these computations in extremely clever ways because this is phone this was in 1960 their computer had 4k of memory so even storing the elliptic curve and the prime number was already taking a lot of the memory and then their program was taking a lot of the memory and then whatever they had left they had to do the computations and so they had to design very very clever ways to do these computations and so actually a lot of the the breakthroughs that they made at the time which we've taken advantage of theoretically later on they made a lot of theoretical breakthroughs in order to do these computations because they had to fit their programs in such tiny amounts of memory so as I said if you're looking at y squared equals x q plus ax plus B mod P then the maximum number of solutions mod P is P squared because X has P possibilities and Y has P possibilities but actually the correct order of magnitude of how many points you'll actually have on this curve modulo P you expected to have about P points not p-squared points why is that well once you plug in a value of X mod P then how many then you can solve for y Y will hat Y squared equals this number of mod p that'll happen most two solutions so but how often how often you expect when you when you plug in a value mod p 4x cubed plus ax plus B you don't expect it to be a square all the time it'll be a square about half the time because half the numbers mod P are squares so when you plug in a number mod p for X half the time you'll get a square and then Y will take one of those two square roots as values so half the time that X Q plus ax plus B will be a square and half the time when that is a square Y will have two solutions so when you plug in the P possible values for X about half of them will be a square and then you'll get two solutions for Y so the number the expected number of solutions is P right because about P over two values of X will give you a square but then for each of those you have two solutions okay so you have about P solutions for X comma Y mod P right on this curve okay so you expect about P solutions and so if we use the same notation NP that say are used in his very first I have a lecture and the under used earlier today NP if you use that to denote the number of points mod p on this on his curve y squared equals x q plus ax plus B you expect NP to be about P and therefore you expect NP over P to be about one yeah as P varies you expect that to be about one most of the time but here was Bert since Winterton dyers incredible observation which is well what they said is if he has lots of rational points on it in other words if this rank R is very large then these points if you take those rational numbers and you reduce the mod P remember you can take a rational number and reduce it mod P right because you can it's a mod P as a finite field so you can divide so you have a rational point on elliptic curve you can reduce in mod P and that will give you a point mod P on the elliptic curve if there are lots of rational points and you reduce the mod P shouldn't be getting lots and lots of points mod P and so somehow NP should be pushed a little higher than P in those cases so bertson's winners and I are they hypothesized that if the rank of the elliptic curve a is large then on average one should notice a having more than P points the expected number of people in that case so if the rank is zero then maybe NP should be about P but if there are lots of rational points and the rank is large then maybe NP tends to be a little bit higher than P as you vary across prime numbers P so maybe you should be able to pick up this large rank by noticing lots of points mod P for for lots of Prime's P so their computations in this direction confirmed the suspicion in large part and it led them to the following really spectacular conjecture so there's the birch since we understand our conjecture says that if EES an elliptic curve y squared equals x q plus ax plus B and ours it's rank an NP denotes the number of points mod P on the on the elliptic curve which tends to be about P but might be higher when the rank is larger than what they what they conjectured is that if you take n P over P and you multiply it over all Prime's ok so n P over P remember we expect it to be about 1 but we're trying to test maybe it's a little bit higher when the rank is large what they can what the conjecture does if you take n P over P which tends to let be very close to 1 and you multiply n P over P over all Prime's less than or equal to X where X is some large number then as X grows this should also grow as C times log of X to the rank so this is how the rank so this is exactly what we were saying before if the rank is large but somehow these NP s over P's should be maybe a little bit bigger than one so that when you take the product it grows as log of X they take the product over all Prime's less than equal to X it actually grows in a way that depends on the rank and if the rank is zero then then the product of n P over P as for all P less than X as X gets big they said we remain bounded in other words these things really were like 1 but when R is big then maybe as you take the products of NP s over P it actually grows and it grows in a way that depends on the rank and this is this is the birch and Swinton dire conjecture the original format the Burton's mentioned our conjecture in a really very beautiful conjecture because it's relating the structure of rational points how many rational points you have we which is a widget an infinite set with the counts of these finite sets name your the set of solutions mod P and it's saying there's a correlation that if there are lots of rational points then the number of points mod P tends to be large and this very explicit concrete way so there's the bridge since we're in dire conjecture what is this constancy well they actually made a conjecture about that too so bertson's foreigner also gave an explicit expression for c in terms of various other invariants of the elliptic curve e and that's called the strong form of the conjecture and well this is exactly the if you can prove this conjecture you win a million dollars and actually I should mention that you only need to prove the weak form of the conjecture to get a million dollars in case that helps but there is also a formula for C but that's not part of you that's not part of them to claim millenium problem okay so that's it that's the that's the bertson's for engine data conjecture there's a modern formulation that has been hinted at today in the earlier lectures already now the modern formulation is in terms of what is called an l function of the elliptic curve fees an elliptic curve and NP denotes the number of points mod p on the on the elliptic curve then remember we said that NP tends to be about P so you can measure the deviation from that expected value of P or P plus 1 by setting a P to be P plus 1 minus NP so this number now tends to be fairly small and then you can define what's called the incomplete l function of e by just saying the product over all primes that don't divide the discriminant so basically product over all Prime's of this expression 1 over 1 minus a P times P to the minus s plus P to the 1 minus 2 s and there's a natural motivation for for why you define this but again it's just depending on the right Dale functions just depending on the count of the number of points mod P on the curve and you make this make the cell function and you think of it as a function of a complex variable s and you find that this product converges for all s with real part of s bigger than three-halves but a conjecture of hosta says that it should have an analytic continuation a continuation to the entire complex plane and if it did have a analytic continuation to the entire complex plane then it would make sense to talk about the value of this function at s equals 1 and when you plug in s equals 1 well then you notice that you're basically taking products of P over N P in this expression right and so that might have something to do with vergence winners and IRAs original conjecture and they in fact reformulated in terms of the value at s equals 1 which they didn't know existed because we don't know but we didn't know about the analytic continuation at that time but notice that the partial products of this function at s equals 1 are exactly when you plug in s equals 1 that's exactly looking at the reciprocals of the expressions product of n P over P and so the conjecture of right since winners and dire was about this analytic function the behavior at s equals 1 and what they conjectured is that the rank of EE should be equal to the order of vanishing of the ALF function that s equals 1 so in other words if you took a Taylor expansion around s equals 1 the first Taylor coefficient would look like C times s minus 1 to the R and that would be the first and then all the other terms would be higher order terms so just the order of vanishing remember where it looks like the reciprocal now so going to the rate of going to infinity has been replaced by the rate of going to zero and so how fast is vanishing the order to which is vanishing is now measuring this the growth of n P over P for P less than equal to X and so their conjectures just says that the rank of E is equal to the order of vanishing at s equals 1 of a particular analytic function it's quite an amazing conjectures relating the behavior of an analytic function of quite a funky analysis to something algebraic the structure of rational solutions to an equation quite an amazing conjecture especially because this al function was not even known to be defined at s equals 1 at the time one of the great contributions of what of Andrews modularity theorem so proof of the modularity theorem of Wiles and boy Conrad diamond Taylor is that it allows this modern formulation of the birch in swinish entire conjecture to make sense so a consequence of the work of wilds is that for any elliptic curve e the l function Eliav s does actually have an analytic continuation to the entire complex plane so this of course was many years after the conjecture of britain swenerton dire was made about the l function at s equals 1 theta knows defined there but the work of Wiles allows us to know that the function is actually defined at s equals 1 so the version swenerton dire conjecture and the modern formulation makes sense because of the work of Wiles and boy Conrad diamond Taylor and the proof of the modularity theorem and the reason that the L function has an analytic continuation is basically because because it's modular modular forms are analytic in the entire complex plane and therefore the elf function is to through a technique called the Mellon transform so because modular functions are defined on the entire complex plane so are the elf functions of these elliptic curves if you look to curves are modular and so we now know that the L function is defined at s equals 1 and so this formulation of the britons mentioned our conjecture makes sense because of that and and that's that's huge because to solve a if you want to solve the bridge since Winterson our conjecture you do want to know that it makes sense first and so this really helps a lot ok so let u be an elliptic curve and let r be its rank ok by the theorem of wiles and board conrad diamond taylor the modularity theorem we know now that therefore there exists in our prime such that the taylor expansion of this analytic function at s equals 1 is of the form some constant times s minus r to the r prime plus higher order terms what does analytic means it just means that you have a power series expansion at that point if it's analytic at s equals one it means you can do a Taylor expansion at s equals one and you get some leading term the first exponent you can call our prime and then you have higher exponents forever after that and this our prime is exactly what it means to talk about the order of vanishing at s equals one and that's called the analytic rank quantity our prime is called the analytic rank okay it's the exponent of the first term in the Taylor series at s equals one where we now know it's known to be defined by the modularity theorem of Wiles well okay so the quantity R prime is called the analytic rank of e and so the conjecture bertson's Winterton dier now makes sense and we can say the rank of e is equal to the analytic rank of e that's what the bertson's Winterton dyer conjecture says the rank is defined in terms of the rank of the group of solutions of rational points on this elliptic curve totally algebraic quantity our prime is the order of vanishing of this analytic function that's defined in terms of the point point counts mod P make a you make a generating function called the L function and you look at the order of vanishing a totally analytic construction that's called R Prime and the bertson's finish and our conjecture says that R equals R prime ones defined algebraically wants to find analytically and they're the same that's the kind of thing that really excites mathematicians the unity of the way you're defining these quantities in two completely different ways algebra and analysis and they're the same that's the that's the modern formulation of the bertson's winners and our conjecture in a way that is now known to make sense due to the work of Wiles okay so that's that's the Burton swenerton Dyer conjecture I explained it both in elementary terms as well as it's modern formulation both of which now makes sense it takes much less time to describe what is known about the conjecture but one of the very first theoretical results to where you know there are lots of there was lots of mounting computational evidence for the bird since we mentioned dire conjecture but one of the very first theoretical evidences of the Virgin swenerton dire conjecture was actually an in Andrews PhD thesis with his PhD advisor John Coates so those coats and Wiles in 1977 they gave the first theoretical evidence towards the bridge in swinish and our conjecture and what they showed is that if he's an elliptic curve of the form y squared equals x cubed plus ax or y squared equals x cubed plus B and these were actually exactly the kinds of equations that bertson's Winterton dire were doing their computations on to test their to to test out their conjecture they were doing computation of these kinds of curves and these are exactly the kinds of curves that coats and while studied in 1977 and what they showed is that for these kinds of curves y squared equals x people say x or y squared equals x people's be if our prime equals 0 that analytic order of vanishing of the l function is 0 then the rank is also 0 they're finitely many rational points that was that was the results of coats and Wiles their results actually applied to a more general class of elliptic curves not just those two kinds I wrote there and although those two kinds are the ones that it's been cited for the most and used for the most because that was part of the original bsd computations but more generally the quotes whilst theorem if you're familiar with what it means to have complex multiplication the Cotswolds theorem applies to any elliptic curve having complex multiplication by the ring of integers in an imaginary quadratic field having class number 1 but these two kinds of equations are typical examples of curves that have complex multiplication by the ring of integers of an imaginary quadratic field of class number 1 namely this has complex multiplication inside q joint I and this has complex multiplication inside Q join the cube root of unity ok but the result still applied to a very special class of elliptic curves but that that was the first indication that it was possible to make progress on the bridge since Winterton dire conjecture even if it's for a special class of elliptic curves and only in the case of our prime equals 0 it was the first theoretical evidence that said the verse since winners and dire conjecture is true for for infinitely many elliptic curves in this in the special case the the results of coats and Wiles was extended to general elliptic curves and also for the case of our prime equals one my remarkable works in the 1980s by Coley bagging and my gross and saggy what they showed is that for any elliptic curve if our prime equals 0 or 1 they have to assume that the elliptic curve is modular which they didn't know yet but that's about to happen a few years later so I'll let it slide if our prime equals 0 or 1 for an elliptic curve II Thunderbird since we mentioned dire conjecture is true for e in other words if our prime equals 0 then R equals 0 for our prime equals 1 then R equals 1 this was a great contribution of gross saggy and Koli vagon in the 80s and applies to any elliptic curve no restrictions except that it's modular which we knew a few years later by the work of Wiles their techniques other than using modular their techniques were quite different than those with coats and miles and the question has been for a while if our prime equals 0 or 1 then that implies R equals 0 1 what about the converse if R equals 0 can we conclude that our prime equals 0 actually we still don't know that but we know it in certain cases now and same thing if R equals 1 can we conclude that our prime equals 1 that converse is still not known but inspired by and taking further these ideas of coats and Wiles and gross Agia kohli buggin as well as works that came later Cocteau actually proved reproved the result of gross saggy and Kohli bagging in the case our prime equals 0 using methods that are very reminiscent of coats and Wiles so they found catio found an alternate proof and then there was all this amazing work of Mazur Wiles and Wiles on the Asama main conjecture and putting all these ideas together the recent work of Skinner and urbe and John just in 2013 they've been able to prove a converse of sorts although there's still lots of restrictions but they prove that if R equals 0 or 1 for an elliptic curve and E satisfies various technical conditions and I'll give you an example for some of you who know these terms on some of the terminology so one of their conditions is for some prime P greater than equal to 5 he should have peace on my rank 0 or 1 you should have certain kinds of reduction mod P their various other things that it would take a whole slide to say but I want to mention that these technical conditions are actually checkable conditions so given an elliptic curve you can actually check these conditions fairly easily and if those technical conditions are satisfied then they conclude that our prime equals 0 or our prime equals 1 so this kind of converse is now known under certain conditions that are very checkable so we kind of basically know BSD for these cases of rank 0 and 1 and analytic rank 0 and 1 under certain technical hypotheses now the question is so if R equals 0 or 1 we can check these technical hypotheses and often they turns out to be well they're not true when you prove with theorem like this with lots of technical hypotheses their first question is always does it apply to any lip to curves at all so do these listen to any elliptic curve satisfy these conditions that's the natural question to ask and that was a question that I got very interested in because I wanted to know ok this converse is it actually hold for any lipstick rips so so how do you what does it mean to does it how do you test whether it holds for many lipstick curves well recall that you can write any elliptic curve in the form y squared equals x q plus ax plus B so we can order we can call that EAB and as a and B vary that varies across all elliptic curves over Q and so what you can do is you can just order all elliptic curves right list them all in order of increasing a and B right just take bigger and bigger a and B and just list them all so make way to make that precise you define the height of e by the size of the coefficients of the defining equation which are a and B you can define the height of EAB is just to be the maximum of a and B or it's more natural to compare the size of a cubed and B squared so you can order all up the curve by what's called the height of e which is just the maximum for a cubed maximum of the absolute values of for a cubed and 27 B squared remember these are the two terms in the discriminant I've the elliptic curve but what this just means is that we're taking all up two curves and we're just listing them all in order of increasing sizes of a and B and then when we order all elliptic curves that way by what we call the height we order all elliptic curves in order of increasing height and then we can ask statistical questions relating to the rank okay what did how does the rank very does it get bigger and bigger as a and B get bigger we can ask what is the probability that B SD is satisfied as a and B the conjecture is that it should be satisfied by a hundred percent of curves and we can ask what proportions satisfy the conditions of skinner by and young so that we actually know that the Burstyn spinach and our conjecture is true so so these are kind of statistical questions that some of us have been asking lately and first question is do most elliptic curves have small rank and it's been conjectured that a hundred percent of elliptic curves have rank zero or one one hundred percent doesn't mean all it just means one hundred percent there could be zero percent that have higher rank and in a density sense it's just like Prime's are zero percent of the integers but they're still inflamed any of them okay so do most of Lipton curves have small rankle all the conjectures at 100% should have rank zero or one it wasn't known before whether more than zero percent of rank zero or one but we now know in recent work that most elliptic curves do have rank zero or one so that's good for that's good from the point of the VSD theorems that we have which are only about rank zero one so to know that there are lots of curves with rank 0 one is nice to know because that means we know BSD for some curves and so this is joint work with RL Schenker just in 2013 we proved that at least eighty-three percent of all elliptic curves at rank zero or one okay the conjecture is still that one hundred percent have zero rank zero one but at least we know that most actually do have ranks U or one so that's good so lots of the at least some of the BSD theorems those first conditions about the rank of being zero one are satisfied in fact our methods that we use to show that eighty three percent of elliptic curves have rank zero one actually establish some of the technical conditions that Skinner inner bra and jiang require in their converse Niram to deduce that the analytic rank is 0-1 and so a corollary of our of our proof that most elliptic curves have rank zero one is that a positive proportion of elliptic curves satisfy the burners and our conjecture this positive proportion we established was less than 1% but it was still positive so we were happy with that but that led us to to ask actually what proportion do all current results actually allow us to prove if we just hit every if we just hit every work on the lifted curve that's out there so including gross zaggy and Colavecchia and skinnier bronzong as well as other recent results that says the doc should sir brothers who proved results about the burst minute and dire conjecture modulo two if we throw everything at it what's the maximum proportion that we can prove satisfied the rich and dark injector and so in joint work with with Chris Skinner and way Jiang we just threw everything at it just to see how how how much for what proportion of elliptic curves do we understand the risk mentioned our conjecture and we didn't know more for more than 0 percent before but now we can say that the bursar mentioned ir conjecture is true for more than 66 percent of all elliptic curves so it's definitely true more often than not and and that's well that's that's a good feeling at least we feel like we know we understand it for for most elliptic curves a common question before you ask it this does not mean that we win six hundred sixty thousand dollars [Laughter] okay so what remains to be done that's basically I just summarized basically everything we know about the about the British mentioned dire conjecture what remains to be done well everything we talked about so far has been about curbs of rank zero or one all right Oh Inc that should be ranked at zero or one but of course rank zero one curves are conjectured to be a hundred percent of curves so to really prove the Britain's winters and direct conjecture for a hundred percent of curves well the technical the technical conditions in the theorem of Skinner van Jiang must still be removed we've shown that they're not actually that major that most curves actually do satisfy those conditions but to really get to a hundred percent and some of those conditions still must be removed and once that's accomplished this would likely mean that we understand BSC for a hundred percent of elliptic curves because 100 percent of elliptic curves are conjectured to have ranks U or one however it really is the remaining zero percent of elliptic curves having ranked at least two that has been causing mathematicians the greatest difficulty so even if it's zero percent of the elliptic curves it's still it's still the real sticking point in making further progress in BST understanding rank two and higher and it is known that they're inflamed any so even though they're rare there are implementing Liptak herbs having rank at least two and four such curves essentially nothing is known regarding BSD and so that's really what needs to be done next this is where the next big idea is needed is to understand ranks at least to understanding the personage and dire conjecture there so that's really where the next big ideas needed I think that's the next that's really the next area that really needs to be studied and just yeah it needs right now they're really no ideas that have have made any progress and so that's where the next big ideas needed also what I mentioned finally that there are many beautiful extensions of the Burt since winners and dire conjecture and all the results that I mentioned there many beautiful extensions of that too for these generalizations of the Bertrand are conjecture for something called the block Kanto conjecture which extends the births winter entire conjecture beyond cubic equations and so unless you think that this is just about cubic equations actually if you make progress the bird since winters and dire conjecture they're likely implications for this more general conjecture called the block cacio conjecture which applies to the situations beyond cubic equations they're also p-adic analogues of the vs. mentioned our conjecture instead of looking at the l function as a complex analytic function you can look at it as a P attic analytic function and their corresponding conjectures there the first one was due to a maser Tate and entitled bomb and there's parent guru and there's their V attic versions of the Barossa a formula due to Bertolini and Henry and Karthik for sundar and then finally they're generalizations over other number fields for example for totally real number fields there's been this amazing work of John and his school proving versions of gross Sagi a / totally real fields so it's not just the BSD conjecture but what it leads to that's become so central and in in the area of number theory so I just want to end by first of all congratulating Andrew for for the Apple prize we're very excited about it he's done so much already in in bringing us to where we are for the births per Newton dark conjecture but I also want to point out that there's so much more to do I hope people more people get interested in it especially lots of the end people here yeah thank you very much [Applause]

Channel: The Abel Prize

Views: 17,070

Rating: 4.9172416 out of 5

Keywords: Manjul Bhargava, Birch-Swinnerton-Dyer Conjecture, Birch-Swinnerton-Dyer, Conjecture, Swinnerton-Dyer, number theory, millennium prize problems

Id: _-feKGb6-gc

Length: 62min 23sec (3743 seconds)

Published: Tue Feb 04 2020