Richard Taylor: 2015 Breakthrough Prize in Mathematics Symposium

our third speaker will be Richard Taylor from the Institute for Advanced Study Sir Richard is one of the world's leading experts on a web of conjectures that were introduced in the 1960s and 1970s by Robert Langlands which have gone on to completely transform how number theorists think about their discipline and although the field is perhaps more widely respected and feared than understood among the broader mathematical community it really is motivated by some concrete questions and does have very tangible applications as well so it's a great honor to have Richard Taylor who's going to explain to us a bit about the Langlands program Thank You Bryan so I was asked to explain some big problem in mathematics to a general audience so I I'm making an attempt to do that now it may be that in fact I don't have a general audience I have a mathematical audience it's just one way I could fail or it may be that I have a general audience and I haven't simplified things sufficiently but anyway this is my attempt so as Brad said the Langlands program is has a long history it you can trace it back to the work of say Euler in the middle of the 18th century certainly to the work of Gauss but in the 50s 60s and 70s 1950s 60s and 70s it suddenly took on a new lease of life I guess about 1950 people thought might have thought that they had sort of got the end of these ideas the great achievements of class field theory and so on and then suddenly it took off again with a far more space than the original ideas throughout this long long history it's not it's had dramatic applications and maybe most striking of all is that is the sort of is the beauty of the way it brings together completely different areas of mathematics so I I don't think I'm able to convey the breadth of the program very successfully to a general audience so rather I'm going to take the approach of trying to illustrate it through some specific problems and hope that these specific examples will will give some insight into inter into the the general question so the way I want to present it is theory the theory of congruence is so some of my works been done for me because Jacob just introduced congruences we say that two numbers a and B are congruent modulo a third number P these are whole numbers integers if and we write a congruent to B modulo P if P divides the difference a minus B so in this talk P will often be a prime number which is why I call it P but at the moment it can be any integer so for example all even number and even number is exactly a number that's congruent to 0 modulo 2 and an odd number is exactly a number congruent to 1 modulo 2 two numbers of congruent modulo 10 exactly when the house share have the same last digit 1 is congruent to 13 modulo 12 I have put in another example there does anybody know why that example is relevant probably not it's the the Christian year is congruent the current Christian year is current congruent to the current Muslim year modulo 17 anyway the the so the last example in this slide is one congruent to 13 modulo 12 which is of course the basis of the of the way we tell time 1300 hours is the same as 1:00 p.m. and a good way to imagine congruence is if they are new to you is as a sort of clock there's no reason why a clock should have 12 hours it can have any number of hours so here's a clock with 7 hours and just as when you first to learn to do arithmetic you do it on the in elementary school you do it on an it's sort of infinite number line 4 congresses you just take that number line and wind it round a circle so it keeps repeating itself and you do arithmetic in exactly the same way so for instance if you wanted to add 4 & 5 you count round 4 get to 4 and then you count 5 more notches around this clock and you end up at 2 and I've Illustrated some are typical arithmetic operations here in arithmetic modulo seven in particular I've listed the squares modulo seven you'll see there zero one four and two the only we'll come back to this in a little bit so pay attention the if you run through the seven numbers modulo seven and look at the squares you only find zero one two and four I'll give you to do the calculation okay well why should one one care about congruence is well today what might give a number of answers for instance that they form the basis for the way information is moved about the web used on DVDs send by satellite links the way we encode vast amounts of information in a way that will protect against small errors and still allow you to to recover that information so it's crucial for from much modern technology but that's certainly not the reason it was originally invented I don't know when the theory of congruence is was first bended it was certainly known to the Greeks and the motivation as I say was certainly not that it was I'm guessing that well at least one possible motivation was the theory of Diophantus or it's an example of how a topic was originally invented from pure curiosity for its beauty but it does in fact find many many years later many centuries later practical applications so I'm guessing as I say that the application the one of the original motivations might have been the theory of da fantone equations that I'm going to use the theory of da find her equations in this talk as a way to illustrate as a repeated illustration so down find an equation is a polynomial equation which one's asked to solve in I'm going to say rational numbers fractions so something like a third is on a loud solution but something like the square root of two is not an allowed solution so this is a very old so I say very old topic it was certainly studied by the Greeks it's now for Diophantus who's a Greek living in Alexandria in the third century AD who wrote a massive tome on it on the subject which except for a small amount of work in India in the 6th and 7th centuries was really stood as was the up to date account of of the theory of define ten equations for about a thousand years after it was written so here is a typical example described all right angled triangles with hypotenuse one and the other two sides rational numbers so by Pythagoras theorem what we're looking for is rational numbers x and y such that sum of their squares is 1 what we don't know who first considered this question but there's a Babylonian cuneiform tablet nearly 4000 years old which lists a whole number of actually rather complicated solutions to this equation so in the second column in fact this just lists the ax I was solving the equation x squared plus y squared equals 1 this tablet lifts the possible values of x or a series of possible values of X for which there is a Y it doesn't actually tell you the Y but it's easy to find once you know the X the second column on this tablet gives the numerator of X and the third column gives the denominator and here are the same solutions to this equation in modern notation so for instance you see towards the bottom on the left hand side the famous three four five right-angled triangle is a right angle triangle with hypotenuse one and the shorter sides 3/5 and 4/5 but you get some much more complicated ones once we've done five digit numerators and denominators so it's curious to notice in fact that so the the the first of all these pairs is the one that's listed on the tablet the second the one that's not listed the number has the numerator has a high common fat factor with 60 or a power of 60 and of course the Babylonians worked number system work tube a 60 so to them all these Y's would have looks particularly simple numbers well we don't we don't know how they find this we don't know if they had a method or there was trial and error and we don't know why they found it was it curiosity was it some sort of trigonometric table as I say we don't know but we do know the Greeks knew how to solve this sort of problem of course x squared plus y squared equals 1 if I let X and y be real numbers describes a circle so we're asking for points on the unit circle whose coordinates are fractions or rational numbers and there's a very I'm not don't think I have time to describe it for you but there's a very elementary way to parameterize all such points in terms of rational numbers giving it well given any rational number T I draw the line through 1 0 with slope T and the second point of intersection with the circle will be a rational point and this gives a bijection between rational points on the circle and rational numbers T and in fact you see for instance that the rational solutions are dense on the unit circle that you can find them everywhere in any little interval okay let me vary this problem now slightly let me shrink the circle by a tiny amount unless look at solutions to x squared plus y squared equals nor point 9 9 9 9 9 9 now what do we do well you can start looking for solutions you can try x + y with small denominators nothing will seem to work and in fact there's a reason for that there are no rational points on this circle so geometrically this looks hardly distinguishable in the real world but in the world of die off and hand equations it's suddenly behaving entirely differently and the explanation for this is the theory of congruence --is the mathematician Hendrik Lenstra likes to say that a math taught without a proof is like a movie without a love scene he obviously thought neither was a very good idea so here is Maya the proof for this talk I'm going to prove this fact for you to illustrate one utility for congruences so I'm going to assume that there was a rational point XY on this cryogen x squared plus y squared equals 0.99 9 9 9 9 and I'm going to get a contradiction so if I there was a rational point I could write the x and y as fractions with integer whole numbers numerator and denominator and I could ensure that the denominator of just by multiplying up was always divisible by 1000 if I wanted and I can ensure that the corresponding a B and C have no common factor because if there was one I could divide it out and then substituting this into our equation in clearing denominators I'm asked to solve the equation a squared plus B squared is 999,999 C squared and you notice that if I factorize 999,999 there's the factorization and the key fact for us will be that seven divides this number exactly once now if this equation holds in the usual integers on the usual infinite number line it will hold modulo seven just as Jacob was explaining that you can prove things about the usual integers were initialized the rational numbers by generalizing to the ring of integers modulo seven so I'm gonna if this equation hold it also a codes modulo seven and I get the equation a squared plus B squared is 0 modulo 7 because 7 divided 999,999 and now I had one my previous slide a list of all the squares modulo seven they will want zero one two and four so I have an equation zero one or two or four plus another number of the same form is 0 modulo 7 is divisible by 7 and I can run through the sixteen possibilities in no time at all and only one of them gives zero if both a and B is 0 modulo seven then a squared plus B squared is 0 modulo 7 so I've proved if there is a solution 7 must divide a and 7 must divide be so 49 divides a-squared plus b-squared but 7 divided 999,999 only once so a nother 7 must divide C squared so 7 must divide C so I've got my contradiction they have a common factor which I could have factored out well so what have I done I've looked at the particular equation x squared plus y squared is congruent to 0 modulo 7 and the key fact I used where there's only had one zero one solution X and y both 0 modulo 7 but I can look at any of these equations modulo syphon I mean I can take two has two square roots modulo seven three and four one has three cube roots rodgero seven one two and four for instance two cubed is 8 which is 7 plus 1 4 cubed is 64 which is 63 plus one I can look at more complicated equations here's a cubic equation in two variables y squared plus y equals x cubed minus x squared it has nine solutions modulo seven which I list here again this is an example I will come back to okay well that's congruence is the Langdon's program is going to tell us something about congruence --is in terms of some very other very different sort of mathematics and the particular sort of question about congruence is it's going to answer is fix an equation or equations in one variable or several variables with integer coefficients let's say and let's look at the when I fix the equation let me look at the number of solutions modulo a varying prime number P and the question I'm going to ask is how does the number of solutions to this equation vary with P so as I said the this topic goes back to the middle of the 18th century when people looked at quadratic equations in one variable so here's a very simple example x squared congruent to 5 modulo P so I'm going to let P be a varying prime number I'm going to ask when I - arithmetic on the crock with P hours when is five a perfect square and I've lived listed the first 20 primes while excluding two and five which behave a bit differently because they occur in the equation and breech of the first 20 primes you see that there are either no solutions or there are two solutions in which case I've listed them and it's not actually very hard I think in this case to spot the pattern you think for a moment which Prime's for which Prime's are the no solutions well there's a prime for no solution exactly when the last digit of that Prime in the usual decimal notation is a 3 or a 5 there is no solution modulo 3 7 13 17 23 37 when the last digit of the prime P is a 1 or a 9 there is a solution so modulo 11 4 squared is 16 which is 5 and so on so there seems to be a very simple pattern and this is this in fact generalizes to quadratic equations there we go or quadratic equations in in one variable as was real as I say was realized in the middle of the 18th century first by Euler or as I discovered two nights ago as Keira Knightley says in the new movie the imitation game EULA and by legendre so i should apologize to lissandra for this portrait according to wikipedia this is the only extant portrait of lissandra in in most books you'll find a much more dignified picture of him but again according to wikipedia that's actually his cousin and what what they conjectured was proved at the very end of the 18th century by Gauss nineteen-year-old Gauss the for any whole number n the question of whether n has a square root modulo some prime number P only depends on P modulo 4 times n so this to somebody who hasn't seen it before may not appear very surprising you've got Congress you got congruence modulo P and then a congruence modulo four times n it's all the same sort of thing but I want to stress it really to me at least to me it is extremely surprising fact why should trying to find a square root of n on a clock with P hands have anything to do with where P is on a clock with 4n hands this this was supposed to be Gauss's favorite theorem various stories about it it's certainly true he came back to it again and again during his life he gave about eight proofs I think he gave six different proofs before anybody else found one and I and although I know proofs and I could give a proof to you none of them seem to have to really explain to me why it should be true many facts in mathematics once you can prove them sort of seems obvious it's the only proof you could give this is not such a theorem at least to my mind let's go so to illustrate it it's power let me take a particular example let me ask wherever seventeen has a square root modulo twenty million a hundred and forty two thousand and thirteen which I'll give you is a prime number well of course with a computer you could run through the twenty million possibilities fairly quickly but by hand I guess even Gauss who is a great calculator would have had trouble but using his theorem it's you can get the answer very quickly we know wherever seven Dean has a square root modulo 2014 2013 only depends on 2014 2013 modulo 4 times 1768 so when we divide by 68 the remainder is 5 so 17 has a square root modulo 2014 2013 if and only if it has a solution modulo 5 so we're asking whether 17 has a square root modulo 5 forever 2 has 17 is 2 modulo 5 15 plus 2 so we're asking whether 2 is a square modulo 5 and there are only 5 possibilities to try and if you run through the squares modulo 5 you get 0 squared is 0 1 squared is 1 2 squared is 4 3 squared is 9 which is 4 and 4 squared is 16 which is 1 so 2 is not a square modulo 5 so 17 was not a square modulo 2014 2013 okay well what about equations with more variables as I as I alluded to earlier at the beginning of my talk Gauss's example was a huge motivation to mathematicians throughout the 19th century and into the 20th century they try to generalize his theorem and it led to a class field theory one of the great achievements of the first half of the 20th century and it appeared about 1950 that somehow the subject had been closed that the it was a beautiful theory but it had somehow got to the limits of what how you could generalize Gauss's law of quadratic reciprocity but then in 1954 Martin eichler found a reciprocity law which didn't fit into this mold in all the work up to that time had basically been one variable equations and not any but one variable equations only those that were in some sense abelian technically the Galois group involved had to be an abelian group but then suddenly as I say Martin Eichler found an example which certainly wasn't a beam in it was too variable in fact it was a cubic equation in two variables of this form y squared some constant times y is X is a cubic equation in X where 8 these coefficients a B C and D are all fixed and he found such a law so let me describe this to you I'm going to write n P for the number of solutions and I'm going to take this example that I had a minute ago y squared plus y equals x cubed minus x squared and I said it had 9 solutions modulo 7 and I calculated it for a number of other small primes so if I it's about eight Prime's or something in the number of solutions to this cubic equation in two variables modulo each of them and the first thing you'll notice is that it doesn't seem to be like a quadratic equation in one variable the number of solutions seems to be growing with the primes and that's indeed the case Hass approved in 1933 but the number of solutions to this equation modulo a prime P was approximately p2 within an era that was bounded by twice the square root of P so the leading term was P and then there was a smaller error and one can ask the question so more natural then is to study the error rather than the leading term which is rather simple so we're going to study the difference between P and the number of solutions to this congruence modulo P so I've tabulated that and now I'm going to predict the same thing from a completely different point of view I'm going to consider this the following the infinite product that I've written there in an indeterminate Q so Q is just some indeterminate quantity I take Q times 1 minus Q squared and then 1 minus Q to the 11 all squared and then 1 minus Q squared all squared and 1 minus Q to the 22 or squared 1 minus Q cubes grad and so on of course the product goes on forever but if I'm only interested in the coefficient of Q to the 10 or something I only need to take finitely many terms in this infinite product so I can start multiplying it out and it takes quite a lot of work but there are the first 20 terms in this product and a to my mind another quite remarkable thing happens if I look at the coefficient of q squared it was minus two which was exactly the error in the point count modulo two if I look at the coefficient of Q to the seventh it's minus two which again is exactly the error in the point count modulo a number of points on this curve modulo seven so on the error and nineteen was exactly nineteen the coefficient was zero and you see that Q to the nineteen doesn't occur in this expansion and that's not a and a coincidence for the first few primes as I say mind trick Martine ichael approved in 1954 that this will be true for all Prime's P for this cubic equation in two variables and this opened the floodgates within a year or so people were beginning to believe this was a much more general phenomenon so taniyama suggested that in some sense the same thing should be true for any cubic equation in two variables shamora made it a bit more precise a couple of years later very a decade later really nailed it down precise form with the conjecture and I say there's some similar they proposed some similar algorithm for calculating the producing these number of counting the number of solutions to the cubic equation modulo P it does not involve in general an infinite product I chose this infinite product firstly because it was the case that Eichler studied and secondly because it was easy to explain but it that's not how it generalizes to generalize it similar means I should think of this as a function of Q this infinite product actually converges when Q is a complex number of modulus less than one so I can think of this as a function on the interior of the unit disk in the complex plane picture of the complex plane or the unit disk in the complex plane now the unit disk has certain symmetries it's a model for the so called hyperbolic plane in particular sl2 z the set of two-by-two matrices with integer entries and determinants one acts on it and this function the key thing about the function is not that it has this infinite product expansion but when I think of it as a function on the hyperbolic plane the India in the interior of the unit disk it's very symmetric this is a very symmetric function and that's the generalization this should be an algorithm to calculate the number of solutions to such a cubic congruence modular number P which involves somehow the group of two by two matrices with integer entries and its action on the so called hyperbolic plane so there's another picture of the hyperbolic plane to give you some idea of what it looks like all all those regions white and black regions in this picture of triangles so in this hyperbolic geometry some of these things that look like curved lines to us are in fact straight lines and all those triangles are the same size in this hyperbolic world so although to our eyes they look very tiny as they go out towards the boundary of the disk they should in the hyperbolic world space is spreading out very rapidly so they're to be considered the same size this sort of geometry also came up it was communicated to the artist Martin Escher Asia by a coxeter the mathematician coxeter and he used it in some of his woodcuts well we now know this is true it was proved beginning of this century following ideas introduced by Andrew Wiles and again it has daya fantail occations this case famously to Fermat's Last Theorem it's probably most of you know in the middle of the 17th century firma wrote in the margin of his copy of Diophantus his book on Diophantine equations that no cube could be written as the sum of two cubes no fourth power of the sum of two fourth powers and so on any power greater than two he claimed to have a proof that he couldn't fit in the margin and nobody could find a proof until Andrew Wiles at the end of the twentieth century so this was the theorem that the Wiles proved and how he proved it was exactly by a reciprocity law in fact a few years earlier God Frey and Ken ribert had established that if Fermat's theorem was wrong you could write down a cubic equation that did not satisfy the Shimura taniyama conjecture this reciprocity law for cubic equations in two variables so modulo some permuting of a b and c this this particular cubic equation will do and what Wiles did was he proved certain reciprocity laws so again it's a case that reciprocity laws are telling us things about Daiya fantail equations that couldn't be obtained by other means well it doesn't stop at cubic equations in two variables within a another decade or two bob Langlands had realized it should apply to any number of equations in any number of variables of any degree and the answer should again depend on symmetries in certain geometry so it was a connection on the one hand between solving polynomial equations modulo n this algebraic system of solving equations in certain rings and on the other hand some geometry symmetries in hide I'm at usually high dimensional geometries and non-traditional geometries where those space is curved so I can't even for sl3 so again there's an action of n by n matrices with integer entries and determinants one and we have to consider all n already fine equals three I have to consider a five dimensional geometry that I can't illustrate so instead I will talk about sl2 working over the gaussian integers Xia join the square root of minus there you get a three-dimensional geometry which I may be able to illustrate oops no maybe I can't ah here we go this was a movie made by people at the University of Minnesota which is moving through hyperbolic 3-space so you may notice various things first of all the space is tessellated by dodecahedra you can't tessellate normal space by dodecahedra ii again there's this phenomenon that as you move through things from the outside come towards you at a different rate the space is collapses towards you at a different rate than you would as if you were moving through usual Euclidean space so again it's symmetries in spaces like this which should determine the solubility of these equations and I'm today we know mrs. slide just tells us the generality in which which we now know these reciprocity laws as much greater generality we can look at varieties of any dimension there's still a very strong restriction it appears to be a very strong restriction called regularity a geometry will be horrified although somebody working in autumn Orphic forms would maybe say it was the generic case it's a sort of strange situation but anyway we we know much much more but on but we still those are awful lot we don't know for instance there's a sample of a particular higher degree equation and lots of variables for which we know a reciprocity law okay I want to finish with a final dive I'm running out of time I final add a fan tine question another old one that goes back to the 10th century the mathematician Alka Raji who worked in Baghdad posed the following question for which whole numbers n is there a right-angled triangle with area n all whose sides have fractional length so again this is problem we want to triangle area n sides XY and z are all rational numbers so we've got to find X Y given n we got to find x y&z with x squared plus y squared equals ed squared and then equals X Y over 2 so again it's not very clear how you would go about doing this I mean given that say you won you try say your n was one you want to write angle triangle area one you could try x y&z with small denominator see if you could find any examples you would find that you couldn't find any but what does that mean what we now believe we know how to answer this question instead of the original problem let me take this putative area of the triangle N and write it try and write it as the perfect square plus twice a perfect square plus eight times a perfect square and for simplicity let me concern is odd and square free just so do the other cases too well that's the easy thing to calculate because you the solutions UV and W have to be bounded by the square root of n so there are only finitely many possibilities you can easily run through and and then I'm going to count the ones for which W is even and the ones for which W is odd and write that a and even an a or not so two things that are easy to calculate given N and what we believe to be true is that there is a right-angled triangle with fractional side lengths and area n if and only if a and even is equal to a and on so this is a computable criterion now well we don't know this but we know half of it we know if a and even in a and order different there's no such triangle this was proved by Gerry tunnel dick gross Dawn's Agee and Victor khaliv argon in the 1980s and again their proof was based on the reciprocity or it was based on our tins reciprocity law from the first half of the 20th century in fact this isn't happens to be an abelian reciprocity law in sort of ubiquity of reciprocity laws endow fontaine equations the people involved so here are some examples for instance if N equals 1 a and even is to a an odd is zero so we know by this kolavari in serum that there's no such triangle that's not news in fact that was proved by a firma in about 1640 again n equals three there's no triangle of area three with sorry there's no triangle area three with rational sides but for N equals five and even is a and odd which is zero so there is a we believe but don't immediately know that there is a right-angled triangle with area five and rational sides and in fact that's not news either it was found by Fibonacci in about 1200 not terribly complicated has sides nine six forty six and forty one six but now if we let the numbers get a bit bigger just got to 100 look for triangles of area 150 550 759 and so on well we see that we the caller bargains theorem tells us there's no triangle of area 155 or 163 which wouldn't be so easy to determine any other way but we expect there are triangles of area 157 159 and so on Richard Feynman in his book you must be joking mr. Feynman said that when he was a graduate student in Princeton Ely used to like to tease the mathematicians he could say he could answer any problem they gave him he didn't promise to find a proof but by by calculating some examples or guesswork or psychology he could always answer their questions and he claimed that he was undefeated I wish somebody had asked him given him say in the numbers 153 for 165 and asked him for which such numbers there's a triangle with right and go triangle with rational sides in that area I don't think it would have been so easy without knowingly this theory and this reciprocity laws to figure such a thing out for instance in the case of 157 Dawn's Agee I found the simplest such triangle but the jury tunnels method predicted this very beautifully and very quickly ok I'll stop that so we have time for some questions yep in my research I don't use them at all although I mean that's that's my lack of ability as I said this morning they've certainly had an influence this this particular question is is the the fact that we know that answer it's a special case of the Bertrand S&R conjecture which I also mentioned this morning which arose through numerical computation to numerical work of brown birch and Peter swinish and our that I mean and hooped am I supposed to do here this this conjectured explanation for when there's a where a and odd equaling a an even predicts whether there's a right angle triangle as I say that's a that's an explicit version of the Bertrand it and our conjecture from a particular lifted curve but the way we but even to think about it in the right way to formulate it like that relies on on ideas that were inspired by my beautiful numerical computations any other any questions okay let's thank for it [Applause]

Channel: Breakthrough

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Keywords: Ricard Taylor, Institute For Advanced Study (College/University), Mathematics, Breakthrough Prize

Id: gGg-bmBm6oY

Length: 40min 11sec (2411 seconds)

Published: Thu Dec 04 2014