The biggest killer isn't always money being broke biggest killer in this world is loneliness It's a lot of science on this and loneliness is on the rise. So I'm traveling the world and Let me give you the simplest best life hack to make friends all around the world no matter where you go and Do it fast and you make the right? Vibe of friends, right your social circle you build a global social circle. So here's what I do I'm I'm In Europe right now, and you know, it's intimidating to make new friends My best friend is died, you know, it has me thinking a lot about Friendships and how they're hard to make quote-unquote, but that's a real limiting belief An Amish guy once told me, you know new friends are good But old friends are better and there's truth to that, right? but you need new friends because sometimes old friends are moving on or like I said my best friend died and Sadly, but here's the hack Because you got two parameters two constraints. You want to make new friends. You don't want it to be shallow global was great because you get all this like diverse kind of Mindset and you get all this eclectic would be the right word.  I found a good teacher You want to go with me or I'm doing a Muay Thai class or I'm doing it going to the gym You can use the gym. Jim's not quite as novel remember humans have these 25 cognitive biases That that attract to their the human brain's attention So humans have this reward system and novelties huge So when you say Brazilian jujitsu or you say Muay Thai or you say even boxing like not that many people do that Now if you're not into those sports, you can swap it out with whatever Kayaking I just like like boxing jujitsu because there's usually group classes It's not that expensive to bring people and it's super novel But again, you can adjust this a little bit, but try this way and like here's the thing It's a there's this concept of self-selecting groups and adverse selection, right people who say no to that Either they're busy, which is fine But if they're not in that mindset of self-development, you know self-actualization Then good riddance if they say no, but you'll find people will be like ah I'll try that and then you go into like a group class or I yesterday did one with like four or five people From all around Europe that I just have met in different places and we did a boxing class And it's crazy. It actually builds camaraderie faster There's this science that when humans do hard things together I don't know if you ever heard of this like best place you can take someone on a date is like a roller coaster Because they say this rush of dopamine When you're next to it, it's like an anchoring so they start associating you with this rush So like even a horror movie works because even though it's a negative Rush, right court is all all this stuff. It's still anchors to you as kind of like Oh, remember them around this person. It's emotion inducing. So That's why I like to do a tough class like jiu-jitsu people are gonna feel uncomfortable most I mean, I've done it for many years, but the average person didn't ever even thought of it But it's in our consciousness like the average person in the world wants to be more fit Okay, and like I said, you can do the gym. I just find the gym to not be so novel So it's like, oh, I'll come to the gym with me Half the world's especially if you're in Europe is already going to the gym Maybe in America or not many people are in shape. You can say, you know the gym But this starts to build a social circle shared activity in something That's a little bit tough that gives that emotional that psychological rush that they then anchor on you as like I want to run around this person. I like my emotions change remember no matter what you think humans are a Emotional being by the way anybody who tells you what whichever your friends tell you they're highly logical is probably your most emotional friend There's a good book on this called Descartes error by the way on a side note saying it's not even possible for the human brain to function Without emotion So you we're not able to be spock from Star Trek or robotic as much as we try it would actually in that book It's a scientific book Descartes error talks about how basically brain function was shut down So when you're traveling the world making friends, the reason this hack works is because one If that self-selecting group and the people will say no is kind of the adverse selection You don't even want right they're not on that mazel hierarchy of needs trying to get to the top of self-actualization By trying new things right so you're gonna get a person who's a little bit of a risk-taker Right you don't want friends that are totally risk-averse. I mean you want some But you know Homo sapiens our ancestors the ones who survived and evolved had a little bit lower risk aversion They were able to move before The storm or before the Ice Age hit you know the migration of humans across the globe Shows that those who are a little bit open they call it in the hexaco, which is kind of the best personality Test in the world better than 16 personalities or Myers-Briggs these 25 facets of human behavior one of them is openness to new experience So and and intelligence has been linked to openness to new experience So by bringing up stuff like you want to go to jujitsu or Muay Thai like since most people haven't done it It's kind of subconsciously. I shouldn't say subconscious.  So anyway, I hope this helps you, you know one thing I like about Scandinavia Scandinavia If you Google the least lonely people in the world, it's like Denmark Sweden, you know and the reason being I mean partly is it's a complex reason if you're an anthropologist And you study Viking culture even though we know Vikings as you know berserkers and this kind of aggressive plundering group, but really they were very Could almost say communal So if you go to a Scandinavian school, they foster this group mentality So I think the average Danish or Swedish person has like 12 friends the average American has like two Adult I'm talking about like people just like lose their friends So you have to work against this because as I said at the beginning of video all the money in the world Doesn't matter if you don't have anyone to enjoy it with Money best spent I look back and I've spent money on different things I'm kind of a mad scientist with life But like the expenses that I don't regret even if they were kind of extravagant are the ones like flying in my friends You know going on a global trip. So Grow your social circle. So yeah, the average Danish person. I think that's 12 friends. I'm average American. I was like two adult Like actual close friends. You see it. You go out in Scandinavia People are at big tables with friends for just like hours They're not just I got a restaurant On their phone getting a quick meal with maybe a buddy. They're like enjoying life So all money in the world doesn't matter if you don't have anybody to enjoy it with that's their science to that Like the actual marginal utility of wealth You can graph it with if you're in a bubble You know, imagine you're on Mars alone and you're the richest person. You have all this material stuff The famous billionaire the Greek billionaire. What was his name? I forget he married JFK's widow He said, um All the money in the world wouldn't matter if there wasn't women. Sorry, that's one take on the social thing But I mean friends So use this hack Make friends build your social circle. You should have at least 12 people in your phone That you talk to on a daily or weekly basis Unrelated to business or stuff just friends and not all of them shouldn't be family either Family is good. But you you know sometimes there's a friend that's Closer than a brother. That's a famous saying, you know, there's sometimes there's a friend that's closer than a brother Friendships if you study like dr. David buss is evolutionary psychology Families and friends have other social dynamics that sometimes make make them in conflict You know, leave a comment below Contributed conversation. What's a hack to make friends anywhere you go that I forgot about Contribute tell me what you agree or if you don't agree. What's something better than jujitsu? What's something better than kickboxing? By the way, I highly recommend you jump in my new 67 steps I made a new version of 67 steps. It's one of the most Widely purchased um self-development programs in history. I lost it in 2014 But I got a new version that has The four pillars health wealth love happiness 67 videos go through do a 67 day challenge to transform your life I've had hundreds of thousands of people go through the program. It's insane So and all this stuff that i'm talking about and I got lots of stuff that's even more in-depth than what i'm talking about here I'll see you in the 67 steps go here go to tidalopis.com Slash 67 steps podcast. Just the number six seven steps with a s steps podcast It'll redirect you to where you should go tidalopis.com 67 steps podcast I'll take you right get in the program try a 67 day challenge to transform your life I got stuff about health wealth love happiness all of these kind of Way more in-depth than i'm doing here. I mean a free podcast I wanted to put this out there because I like giving stuff away free, but for people it's not expensive either So go through it commit try something man life short.  That's the problem It's a good book on this now by daniel pink on the power of regret So avoid regret to the extent that you can so go to I'll put a link here or if you're listening Tai Lopez calm slash 67 steps podcast 67 steps podcast so the new version of my 67 steps If you're one of the whatever 50 or 100 thousand people who went through the paid 67 steps This is updated, you know seven eight years later with much more powerful stuff I still have the old stuff there too But um, yeah


Appalachian Wireless is transforming the way you upgrade your smart phone. Trade in, trade up and experience the benefits of Appalachian Wireless's trade-in program. Get your estimate on AppalachianWireless.com now. Better service and bigger savings, that's today's Appalachian Wireless. A Pye County couple is alive today, thanks to a Kentucky State trooper who risked his life to run into a burning home to save them. Just after 1 a.m. last Sunday morning, trooper Corey Charles was in the belfry area when he heard the call of a house fire just 5 minutes away. He was the first to arrive. The original caller was standing outside and I could see another female in the home because the door was open and she was back lit by the fire behind her. Started yelling at her to come out and she couldn't hear me. So I went in probably 10, 15 feet and grabbed her by the arm and got her out. But that's when things went from bad to worse. I asked her if anybody else was in the house and she said, yeah, my husband's asleep in the back farthest bedroom. The fire department hadn't arrived yet and trooper Charles made the decision to go back in. I crawled to stay under the smoke because like I said, it was two-thirds down and you couldn't see anything without your flashlight. So I shot my flashlight and the only thing I seen, once I did crawl through the hall and got to the back bedroom where she told me he was, was his feet on the bed. I screamed at him a couple times in between trying to breathe and he never acknowledged me so I just grabbed his feet and I jerked him off the bed and I drug him through the hall into the foyer and I stood him up and we went outside. As soon as they were all safe, the fire department arrived. They were able to save the exterior of the home but trooper Charles was the true hero that night. But he says he was just doing the job he signed up for. If I can, you know, I'd like to help anybody just like any other trooper at post nine would. You know, I didn't do anything that any other trooper here wouldn't have done the way I was just putting her out spot throughout time. I was glad I was there. In Pikeville, I'm Shannon Deskins, EKB News.


If the series are stand-alone, there are some people here who weren't here last week and probably some people not here who were here last week, but I won't talk to them. And so each lecture is meant to be self-sufficient, but of course they're linked. So if you go to all, you'll see more cross-connections. So in the first lecture, I mostly talked about the acosahedron and its geometry and also the representation theory associated a little bit to the group S5 and symmetries. And then in the second lecture, also in the first one, I talked about the other words that occur in my title, which is the Roger's Ramanujan Identities. But then in the second lecture, I explained about modular forms. It was meant to be a very quick introduction. But if you don't know modular forms to give you a feeling for the beauty why they're such beautiful objects and where they occur in various places in mathematics. So today I'm going to repeat some of that even more briefly, giving you the three kind of salient examples or types of examples. All of them will now come back into my story. And today the theme will be the links between modular forms. I mentioned it a little bit in, I think, this fifth of my five examples. So the examples which I'll write down in a second were Eisenstein series, product expansions like the Ada function, the Theta series, then modular forms coming from algebraic geometry and then very sporadic modular forms coming from hypergeometric functions, acute hypergeometric functions, which the Roger's Ramanujan Identities are an example. And so today I'm going to talk about this. And so it really should be part of the general culture, I think, of every mathematician that there is this notion of modular forms in number theory that there are also things that are not geometries, geometry, I'm sleepy, that there's a connection, a very deep connection, almost always conjectural, but checkable. And then in the last lecture, which is on Friday, which I will give, I wrote in the abstract that I would only give it if some people remained in the audience, which is the case A, and B, if I succeeded in solving the problem I wanted to solve. So I've solved two-thirds of that problem. I hope by Friday to have solved three-thirds, but I'll tell you the story on Friday anyway. So this will be lectured number three. This was more or less one and two. Well, one was already this. And one was, I call Ciedra Adler's Ramanujan II with this. And the last one will be an application, an example coming from mirror symmetry where all of these things come together. But today I might mention that at the very end of the lecture, but probably I won't get to it. So today I first want to repeat the part of last time's lectures where I explained in a few words what molecular functions are and molecular forms and gave examples of the three main constructions. So I said that I actually only know five constructions that give explicit molecular forms. I discussed all five. All five will come back today. So these, as I just said, are Eisenstein series, infinite products, product expansions. I'm not sure if the order was like this or if these two were reversed. Theta series. And here again, I don't know if they're reversed. These are Q-hypergeometric series, which is not a systematic way, but occasionally gives examples. And finally, the ones linked to algebraic geometry, which is the theme for today. So on the basis that I speak quickly and I therefore say everything twice, I'll repeat that part and also some people weren't there, but I'll do it both very differently and also, of course, much more briefly. So let me first remind you a multidorfer... So gamma will be a subgroup of SL2R, which is discrete and co-finite. That means that it's not too discrete. So SL2R is a huge two-dimensional group. If you took a very small group, like the trivial subgroup or group isomorphic to Z, then the quotient would still have infinite volume. There's a natural volume on SL2R. And what I want is that this thing is co-finelite volume. So if you go outside where you see pictures of the icosahedrons on, you can imagine we have a group, which in that case would be S3, divided by in that case a finite group, and then it's broken up into finite pieces, which are triangles in that case. But if it's an infinite group, if SL2R is a non-compact group and gamma is therefore an infinite group, you still might have pieces of finite volume. So the basic example, there were two... Well, several examples. One I'll just call gamma 1, which is the group SL2Z. So 2 by 2 matrices with integer coefficients and determinant 1. I won't write it. I think everyone knows what SL2 means. Determinant 1 and Z means ABCD are in 1. And then I have gamma 5, or more generally gamma of n. It's the set of matrices ABCD in SL2Z. But the ones that are congruent to the identity 1001 modulo 5. And you can imagine gamma of 7 is the same as 7. But I'll only use this with 5. This is called the principal congruent subgroup. So EG, and another example would be gamma 0 of, for instance, 11, which is the set of ABCD, where you only put a congruent's condition on C. It's congruent to 0 mod 11. And again, you can easily imagine the gamma 0 of 900 is the same thing with 900. And I gave an example with 900. And it'll come back on Friday after tomorrow. So that's the thing. And then the basic two definitions, well, we're a modular function and a modular form. I'll first remind a modular function is a holomorphic function f of a variable tau. I think I might have called it z, but let me call it tau. Tau's in the upper half plane, which means, as I drew before, the complex numbers with positive imaginary part, which is gamma invariant. So the group gamma acts. I mean, the whole group SL2 are acts by Mobius transformations. And so what I mean by gamma invariant is that f of this is equal to f of tau for all tau in h and all ABCD in my group. So an example of that was the Roder's Ramanujan function. I don't remember. Maybe somebody who takes notes can say it was either small r or little r. I've used both in my life, so I forget what it was. But anyway, I can give the example. And it'll be an example of an infinite product expansion because it's q to the power 1 5th. So these modular functions, all of my groups, contain either 1 1 0 1, that's, for instance, SL2z, or it's also in gamma 0 of n for any n. But it's not in gamma 5 because it's not congruent to the identity, but it's 5th power is congruent. But it's 5th power is in that group. So that means that if your group does contain 1 1 0 1, there's a Fourier expansion, f of tau is the sum a n q to the n, and q is always e to the 2 pi i tau. And that's exactly the expression e to the 2 pi tau that's invariant under tau goes to tau plus 1. But if your group is, for instance, gamma 5, then you have to send tau to tau plus 5, then you'll have fractional powers. And so there's this fractional powers q to the 1 5th, and then it's given by a product, n from 1 to infinity, 1 minus q to the n. And the power is the Legendre symbol, n over 5, which I remind you is 1 if n is congruent to plus or minus 1 mod 5, minus 1 if n is congruent to plus or minus 2 mod 5, and 0 if n is congruent to 0 mod 5. Every integer is congruent to one of those five things, 1 to the 5. So this is the power, and then it also, this is the wonderful continued fraction of Ramanujan that So amazed hardly when he first saw it and still can amaze all of us today, I think. So it's the infinite continued fraction, except for the q to the 1 5th that Ramanujan didn't put. He just had the identity like that. But we know that we need it to get a modular function. So this is the definition of the function, and the fact is that R, I'll call it capital R, does anyone remember if it was little R, big R? Okay, then it doesn't matter. I'm used to little R. So the claim is that little R of tau is the same as little R of gamma tau for all gamma in this group gamma 5. So this is what's called the Houtt module for the group gamma 5. If you think geometrically, which we're not really discussing, but it's behind everything I say, it means that R is invariant, and therefore it gives a map into C. But you can think of C as part of P1C, and then actually it's an isomorphism under P1 of C minus a finite number of points, and all those points are exactly the vertices, or the central points I forget now, of the dodecahedron. So there was a connection here with the dodecahedron that I don't want to go back to. So that was the first example. That's a modular function. But where things get more interesting is a modular form of weight K. K is usually integer, although it can be half inch, or even a fractional number, but I won't go into that, of weight K on gamma is the same. F is, again, holomorphic, and some growth condition at infinity that I don't want to repeat. But this time, instead of it being invariant, it's invariant up to a factor. And that factor is the Kth power of C tau plus D. And so that's called the weight, and that gives you a lot more freedom, allowing this K, and in particular there are now examples in a modular form. We also require that it has an expansion, some a and q to the n, and the a and sort of polynomial growth. That will never happen for a modular function because those things have to have poles, at least at infinity. So that's just a technical thing, but just to say that allowing the weight gives you the possibility to have much better behaved functions. And of course if you multiply two modular forms of weight K1 and K2, you get a form of weight K1 plus K2. So if you divide two forms of the same weight, then you'll get a modular function. And every modular function arises that way. So these are very intimately related. So I gave various examples, and the point was again that with a Fourier expansion, I'll just write n from 0 to infinity, but it might not be integer, there might be a denominator, although in my examples now, I think there won't be. So let me remind these three examples. So example one, or sorts of examples one, is F, is the Eisenstein series of weight one, and in that case, so that we can call this A and of F. In that case, it's characterized roughly over simply varying very slightly. It's a divisor sum. In the case of the full modular group, it was just a constant times the sum D to the K minus one, but in general there might be an epsilon of D where this is some periodic function. It's slightly more general, but let's pretend it's just like that. So I'm doing modular forms from the point of view of arithmetic. So what I care about is their coefficients. We want interesting number of theoretical coefficients like this, the simple things. This would, for instance, say that for a prime, this would just be epsilon of one plus epsilon of P, which would only depend on P multiple of something. So this is periodic with some period, let's say some n, times P to the K minus one. So in other words, these are very, very explicit, simple modular forms whose, for instance, their Pth coefficient is given by a very simple formula. Okay, then example two, I mean as a class of examples, was how did I do it? Theta infinite products. Well, we already had one here, but a more interesting example is A of tau is queued to the one 24th times the sum, sorry, times the product, one to infinity, one minus queued to the end of the 24th. And in this case, this is a model form and the full model group, because if you shift tau by one, that doesn't change queue, but it does change queue to the one 24th by Zeta of 24, where Zeta n was my standard n through divinity. And A of minus, this is trivial, but it's not trivial. There are many proofs, but it's not very, very hard, but it's certainly not trivial, is that A of tau is the square root of tau over I, sorry, A of minus one over tau is the square root of tau over I times A of tau. And since those two transformations, tau plus one and tau of minus one over tau generate all of SL2Z, it tells you that here you'll have some power, depending on ABCD, times the square root of C tau plus D, times A of tau for all matrices ABCD in SL2Z, and that means that this is a model form of weight one half. So here K is a half. By the way, when K is an integer, I'll write just for convenience that F is an element of MK of gamma. So I don't have to keep writing out all the words, so I tell you which group it's invariant under and in what weight. Okay, so those are two classes, and then I'll give identities in a second, but first one example. The third one was Theta series, and here A n of F is typically the number or maybe weighted. It could be weighted sometimes, but it's the weighted, it's the number of vectors V in L, which is some lattice of some dimension with scalar product, so lattice has a length, scalar product equal to n. So there are countably many vectors in the hexagonal lattice, then there are six vectors of length one, one of length zero, six of length one, a certain number of length two and so on. So you count that, and that's again a model form, and here the weight, if there's no weighting factor, if it's just this, the weight is equal to the rank of L over two. So those were my classes and the hybrid geometry will come in a second with Rodis Ramanujan, and the main thing that I emphasized is that there are identities, zillions of identities, and many deep facts of number theory become very easy to prove if you know about multiple forms, because one knows that this space is finite dimensional, and in principle it's computable, so although we don't know closed formulas so to speak for every gamut, every k, they're computable, they're computer programs, it's easy, so if k is 900 and gamma is 0 of 11, it's whatever it is, a 23 dimensional space, and you can tell your computer to give you a basis as Q expansion to as many coefficients as you want. So it means if you have two multiple forms and you believe they're equal, well you check that they're the same weight, otherwise they're not equal, and then you just look in to find that dimensional space and it's just a verification. So I gave many examples of that, I want to give, remind you very briefly I'll give slightly different examples today, because I don't want to repeat completely, so I'll keep the rod's function here, so let me give an example of an identity. Well, we had first of all the Eisenstein series on SL2z of every weight k, so there's a k in this where k is 4, 6, 8, e2 exists, but it's not quite a multiple form, and e4 for e4 well I don't have to give you the whole formula because I already wrote a and f, so e4 starts with 1, they all start with 1, it continues with 240, and here the epsilon of d which should be a periodic function, it's a simple periodic function, it's a constant 240, so a and f in this case is just 240 times the sum of the divisors to the power 3, because the weight here is 4, and similarly e6 is the same, they all start with 1, so e6 always starts with 1, and here epsilon d is minus 504, and for e8 it starts, so this is minus 504, and e8 starts with 480, so here k is 6, here k is 8, and here epsilon of d is a constant, 480, and then the identity which is trivial if you know just multiple forms, but not at all, easy to prove in an elementary way, there is a complicated elementary proof, is that the square of the power series e8 is e4 squared, which means that for any integer, the sum of the divisors of n to the power 7 can be expressed in terms of the sums of the cubes of the divisors of smaller numbers, so very surprising. Now with theoretical identity, but the point is not the identity, it's not that exciting, but that it's trivial to prove, the proof of this is simply both of these are in the one dimensional space, they both start with 1, we don't even need to observe that 2 times 240 is 480 it follows, and therefore they're equal. So that was the first identity, and the identity for e8 is actually already due to Euler, although he wrote it differently and I wrote it last time, sum, here the weighting factor will be minus 1 to the n minus 1 over 6, if n is 1 mod 6 and 0 otherwise, and the quadratic form, the lattice is just z with the length n squared over 24, and slightly rewritten this is Euler's formula for what's called the Dedekind's eta function, which was a Dedekind eta which Euler invented 150 years earlier, that's an exaggeration, 120 years earlier. Okay, so again, and again the proof of this today, he had after he discovered numerically, it took him several years to leave found a proof, it's not at all easy, but if we know multiple forms both sides are multiple forms of way to have and if a couple of coefficients agree, then they're already equal. So that was a typical example there and then okay, then an example from q-hybrid geometric functions were the two Rode's Ramanujan functions so here, g of q was the sum q to the power n squared over 1 minus q times 1 minus q to the n and there was the famous Euler product expansion of this, this is one of the Rode's Ramanujan identities, which I talked about sorry, this is not n from 1 to infinity, this is n positive and congruent to plus or minus 1 mod 5 and there was another one for h of q where you had plus or minus 2 mod 5 and here, these formulas tell you that it's multiple and so the identities in this case were that if you multiply g of q first you have to multiply by g of q by minus the 60th and h of q, so I remind you the other one is the same with n squared plus n over the same denominator and this is then the product n congruent to plus or minus 2 mod 5 but they aren't quite modular as they stand from the so-called Jacobian triple product event, you know that you have to include the power of q, q to the minus the 60th and q to the 11 over 60 and now if you multiply these two things by eta, these are way zero, these are modular functions on some group but if I multiply them by eta, these are modular forms and this is exact like others formula except that this is n q to the n squared over 40 instead of n squared over 12 and therefore n is 1 mod 10 instead of 1 mod 6 and the other one is exactly the same but n is 3 mod 10, so they're a pair they belong together but again, all of these identities which are totally amazing, if you multiply them out and do it on the computer, if you've never seen it before, it's amazing that it works these coefficients, almost every coefficient is zero, the coefficient of q to the m over 40 is zero and that's m is a perfect square, so when you multiply it out, you get lots and lots of zeros but all of these identities are obvious once you know that these things are modular forms which is easy with the product expansion, not at all obvious using that, so that's a series of different examples and I want to come soon to the ones coming from Algebraic Geometry, but first maybe I'll do something different because I promised it in the abstract I'll give a very beautiful application of this theory of modular forms which is Apery's proof of the irrationality of Zeta of 2 so Apery became very famous in I think 1987 roughly, I don't remember the exact year by proving that Zeta of 3 was irrational, which had been a conjecture while Euler had already asked whether it was formed or for Zeta of 3 Zeta of 3 remember is the sum 1 over n cubed Zeta of 2 which is the sum 1 over n squared Euler had already proved that it's pi squared over 6 and the irrationality of pi squared over 6 was proved in the mid 19th century, later in the 19th century even the transcendence so Apery gave a proof but a completely different proof for Zeta of 2 not using that it was pi, just using this and also for Zeta of 3 but the one that's relevant for this series of lectures is actually Zeta of 2, so I'll tell that one now, it's very pretty and then you'll see a completely different way that multireforms enter into into number theory so let me tell you Apery's proof so we define numbers a collection of numbers a n for the Apery numbers n greater than or equal to 0 by A0 equals 1 to start to get a positive life to normalize and then there's recursion and the recursion says that if you already know the numbers up to a n then you put here 11 n squared plus 11 n plus 3 times a n plus n squared times a n minus 1 and you divide by n plus 1 squared and that's a n so here I'll give you a little table and if you're taking nodes leave a little space there'll be a second line of the table in a second so if I start 0, 1, 2, 3, 4, 5 you might say don't I have to give you two values, two initial values because there are two terms but you see if n is 0 then this term is 0 squared times a minus 1 so it doesn't matter so it's completely well different term so since a1 is 1 if I take n to be 0 I have 11 plus 11 plus 3 which is where it's 3 times 1 plus 0 so it's 3 when you compute the next one it's 19 then it's 147 then it's 1251 the next one I don't remember by heart nor does it very much matter it's this but there's something remarkable and this is the first part so there are three remarkable facts but I'm only ready for the first one now all three were discovered or followed from the formulas that Appy gave and then from them and I'll show you that that's almost a one line proof if you believe these three facts then everything follows so one of them is that a n is almost an integer as you can see from this table but I mean I think you can see that this is quite surprising because if I compute a4 then I have to take 3 so if n is 3 this is 99 plus 33 is 132 plus 3 so it's 135 times a3 which is 147 and then since n is 3 it's plus 9 times 19 but now I have to divide by 4 squared and not every number is divisible by 4 squared but if you multiply this out if you work it out you find that it is divisible by 4 squared and the quotient is 1251 okay well that you can do by hand and you can let your computer do it up to 10,000 it takes about a minute and check that the first 10,000 are all integers but it's completely amazing because every time I compute an I divide by n squared so a priori we would have an would it denominator 1 over n factorial squared but in fact there's no denominator so that's the Appy-Ree cancellation miracle he gave a closed formula for which it follows that there's a better proof I'll come to in a minute now what Appy-Ree did in his proof he starts with the second solution b with the same recursion but now we start with 0 we start with two numbers so it's not true for n equals 0 because that was unique I start with 0 and 1 okay and now I run the recursion so now the Appy-Ree miracle doesn't happen it's 25 over 4 1741 over 36 6585 over 16 something over 36 I'll write it down okay so you get a sequence of numbers so it looks like there's no Appy-Ree miracle because there is a denominator and here you see the denominator is 4 which is 2 squared which is 2 factorial squared 36 is 6 squared which is 3 factorial squared but here the denominator is better actually that's a coincidence that's not a good example but if I take the next one I don't remember the numerator the denominator is 3600 that's 60 squared but 60 is not 5 factorial which is 120 which is the product 1 times 2 times 3 times 4 times 5 but instead this is the greatest the least common multiple of 1 2 3 4 5 so if you have a lot of numbers their product is always a multiple of all of them but it's usually way too big the least common multiple is much smaller and so here the theorem is that the denominator you need is only nn squared rather than n factorial squared where nn is the least common multiple of 1 up to n which of course divides n factorial but it's way smaller n factorial you know by Sterling grows roughly like n to the n or for e to the n but nn by Chebyshev grows well we actually know it's exactly like e to the n Chebyshev we tell you it's at most it's bounded by 2.9 to the n or 3 to the n but actually we know it grows rough like e to the n so n factorial is more than exponential growth and n is only exponential growth so that's also very surprising so there is cancellation but not as striking as for an and then finally if you take bn over an then the limit is pi squared over 30 that there is a limit is very easy because both of these sequences satisfy the same recursion and that tells you it's very very easy to check that the recursion implies that you have a constant and the same for bn with a different constant times pi to the power 5n divided by n as n goes to infinity where pi as before is 1 plus the square root of 5 over 2 and pi to the 5 which actually occurred in the previous lectures is 11 plus 5 squared of 5 over 2 it's the fifth power of the golden ratio so that's trivial simply because if you look at the growth the n squared I mean if you divide by n squared this is roughly 1, 11 and 1 and so this it grows like h to the n where h squared is equal to 11 h plus 1 that's a quadratic equation its solution is exactly 5 to the 5 that's trivial that's not a miracle and similarly but that's equally true for bn with a different constant so therefore bn over an has to tend to some limit and then bn minus that limit since now it satisfies the same recursion but it's got a smaller order of growth well there are only two solutions of this 5 to the 5th and 5 to the minus 5th so now it's a constant times h to the minus n over n so the first one an and bn grow like 11 to the n but their difference with a factor grows like 11 to the minus n so it's much much much smaller that's easy that's just from the recursion but what's not obvious is what is lambda and what he actually showed it's actually z of 2 divided by 5 but because of order we know that that's pi squared over 30 which is what I wrote before but his proof really gives z of 2 over 5 and he has an exactly similar sequence for z of 3 you have n plus 1 cubed here n cubed here and the polynomial of degree 3 here and then you get z of 3 well z of 3 over 6 so now I'm not going to explain those miracles for a second I will very soon but just to show how up he reads assertion of the irrational default is you see if I take bn minus pi squared over 30 or z of 2 over 5 times an well that's going to be very small this is going to be o of roughly 11.1 to the minus n by what I just said because any solution that's not the big solution but it's not an integer but if I multiply it by nn squared so I've already used part 3 but parts 1 and 2 say that nn squared is a common denominator of both n and bn so this is an integer plus an integer times 8 of 2 over 5 well now nn squared I already told you grows like to the 2n e squared is about 7 squared over 11 is about 2 thirds it's actually 0.666 slightly smaller I think 0.667 is I think even 0.666 will do it if I put 7 it's certainly smaller so this goes very rapidly to 0 that simply follows from these 3 statements but now you see we're done because if 8 of 2 over 5 were rational number it would have some fixed denominator then any integer combination of 1 and 8 would always have that denominator it can't therefore be going to 0 unless it's 0 but this isn't 0 that's very easy to see that it's not identically 0 and so that's up here is proof and you see it's a 1 line proof and he did the same for 8 of 3 but it's only a 1 line proof if you believe or if you know these 3 remarkable facts so up here is proof which was very mysterious he wrote down formulas that nobody really knew how he found them and from those explicit formulas you could know these but then Burkers Fritz Burkers that many people pronounce boykers but don't discovered a couple of years later I think within a year or two that there's a multirexplanation and it exactly corresponds to the functions we've been having here so let me take r of tau which is still on the board I think I've decided to call it little r of tau since nobody objected so let me give another example of a modular form so it's another example of these identities and you'll yarn you'll say who cares it's much more complicated than the others but then you'll see why I write this particular example so another example another another modular identity is f of tau is well it's not even closed form for f it's closed form for square so at the end I'll have to take a square root but you can believe me that you can take that square root oh sorry it's not I don't need parentheses it's just a square root okay so a to remember starts q to the 1 24th so this thing starts q to the 25 24 it's minus 1 24 that's q to the 1 r started q to the 1 fifth so it's also q to the 1 so the thing in the square it starts with 1 so I can take the square root that also starts with 1 and when you compute it well it's something it's a modular form 2 q cubed and so on and you can believe me that it's a modular form on the group that I call gamma 5 before actually it's even a better group but who cares so it's a modular form of weight 1 very boring but that it's a modular form of weight 1 it's clear a to his weight a half therefore this has weight 5 halves minus a half that's 2 this r with a modular function has weight 0 so this is weight 2 and I take the square root it's weight half of 2 so 1 but then the identity says that this is also 1 plus the sum remember if it's an Eisenstein series and so here it's the sum d divides n epsilon of d q to the n where epsilon of d is 3 it's 0 it's a little like the chronicle symbol it only depends on n mod 5 remember an Eisenstein series it was epsilon of d times d to the k minus 1 k was the weight 1 so there's no power and epsilon was periodic here it's periodic at period 5 if n is 1 mod 5 you get 3 but if n is minus 1 mod 5 you get minus 3 and if n is congruent to plus or minus 2 mod 5 then you get plus or minus 1 so you can check that the first couple of coefficients agree so this is a more complicated example of the same magic but it's a very boring I mean you would say who the hell cares why should I be interested in this particular form which looks even weird you have to show that this thing has a square root and here's another one fine it's a big deal it's another example but now comes the amazing thing I can take the function r of tau to the 5th power right r of tau is already a modular function when I take it to the 5th power it's still the 5th power so this one starts obviously q to the 1 5th times 1 minus q plus q squared I guess I can't do this on my head anyway you can compute this so since that one starts q to the 1 5th this will start with q so at the beginning this has therefore a proper q expansion and so I can write f of tau as a power series in r to the tau well it starts 1 plus 3 q but q is r to the 5th so it starts 3 r of tau to the 5th but then the 4q squared I have to change by 3 times 5 it is 15 15 plus 4 is 19 so the next coefficient will be 19 and if you continue then what you find is that the expansion of this modular form this is a modular form of weight 1 this is a modular function so weight 0 so it's not a rational function it's an infinite expansion but the power series expansion has as its coefficient exactly the upper e numbers 1 3 9 47 that's the wonderful thing that Burgers discovered I should write his name because if you aren't a number theorist you may not have heard of him but you don't know Dutch you probably can't guess how it spelled Fritz Burgers ok so we have this wonderful fact and the proof is actually easy it's a general theorem if you take any modular function and expand as a power series in another modular function then that power series is an algebraic function any two modular functions are algebraic functions of each other but if this is a modular form remember this was a modular form of weight 1 on some group which here was gamma 5 whereas this one even R that loaned its 5th power with a modular function of weight 0 not all the morphic infinity so the general theorem is if you take a modular form and you expand terms of multifunction it's an infinite power series it's not algebraic but it satisfies a linear differential equation that just means that the coefficient satisfies recursion and it's algorithmic so you compute the differential equation of this power series a and t to the n and that turns into a recurrence and then you're very happy it's exactly a previous recurrence so you can't there's no way to reverse engineer you cannot write down the multiform you have to guess it find it but once you have a multiform there is a differential equation you get a recursion and Burgers discovered that this particular example I don't need the identity it's enough to take just the Eisenstein series this particular example gives the differential equation and hence the recursion that gives the up your renumbers so now you see that this is now clear and believe me I'm not going to show you that but there's a multi-explanation but you see it's clear because this r of t obviously has coefficients it starts with 1 and they're all skewed to the 5th I don't even care but they start with 1 and they're all integers and this is integer coefficients this was also in z of q and so when I invert r it's still integer it's clear from this that the n's are integers and then when you work out the recursions you have a sequence of numbers that's already integral and satisfies the recursion then that does it for you and there is a multi-explanation a little more complicated that Burgers also gave for bn and for this limiting behavior so though that's a beautiful example of multiforms for a proof of z of 2 and it's kind of the way I talked about that in earlier lectures I think in the salam lectures here in the ICTP my life but it's a cute example anyway to see how multiforms can be used in unexpected ways like here transcendence theory and it fits very well because the key players here were exactly my relative Ramanujan but now I come to the actual theme of today's lecture I've used up already three quarters of my time which is the connection between but I'll probably not finish and continue it on Friday and then come to my main example in detail on Friday so the last examples are examples coming from algebraic geometry but before that let's take another example so let delta of tau be 8 of tau to the 24th so since 8 of tau was on SL2z of weight 12 of weight one half but it had a 24th root of unity when I multiplied by the 24th root of unity this is just an ordinary multiform on SL2z and so this of course is q times the product 1 minus q to the end of the 24th you can just multiply this out it's not difficult Ramanujan did it obviously by hand there were no computers yet to 30 terms and what he found I won't give all 30 effects so I don't even remember them to be very honest but I do remember the first 6 or 7 actually to be precise so Ramanujan computed 30 terms and now you get a quick lesson in how to be a genius unfortunately it's only very useful because it's hard to apply in practice but if you're a genius which Ramanujan certainly was first of all 99% perspiration 1% inspiration you take the trouble to compute 30 terms I don't know any mathematician today would do that without a computer Euler when he found the identity I wrote earlier for Eta had 50 and he had to multiply it all out then he discovered the thing and it took him 10 years to prove he was another genius so first you should be willing to do your homework and work a little bit but the other thing is you have to have your eyes open and what Ramanujan immediately noticed the kind of thing most of us would not notice is that these three coefficients minus 6048 is the product of minus 24 times 252 and similarly if you took a3 times a5 you got a15 so what he found is that if we call these numbers some a and they're usually called tau then but I'm using tau for my variable so for instance this would be tau of 6, the a of 6 would be minus 6048 it's the product a2 times a3 so this is what's called a multiplicative sequence and what one knows from elementary number theory is that what you should always do when you have a sequence of numbers with this multiplicative property this multiplicative does not mean that amn is always aman but it means it's only true if m and n are co-prime like 2 and 3 so a2 times a4 will not give the correct coefficient for a there's a correction that Ramanujan also found that it was proved the following year by Mordell who proved the whole thing so we have this property and whenever you have such a multiplicative function that means if you make for any L of sf so f is again a sum anq to the n there might be a constant in general here there isn't then here you take the corresponding Dirichlet series so you take the same coefficients an but instead of taking q to the n you divide by n to the s but of course you don't take a0 you don't want to divide by 0 so you just throw away a0 but here there is no a0 so here if I write this this will be 1 minus 24 plus 252 over 3 to the s but now it has an Euler problem because of this so it's the product of something 1 minus 24 over 2 to the s plus the coefficient of q to the 4th which was minus 1472 over 4 to the s and so on then the next term will be 1 plus 252 over 3 to the s and so on and then there will be a term for 5 so I can even put it etc so the statement that this is multiplicative which Ramanujan discovered experimentally in 1916 and Mordell proved in 1917 one of the fastest proofs of any major conjecture that I know of in mathematics is that this thing is an Euler product and actually Ramanujan gave more explanation and so you can write the product in a very nice way namely it turns out that this thing you can compute all the coefficients easily from this or for instance this factor is just 1 over 1 minus 24 over 2 to the s plus it's a power series in 2 to the s so I can take the reciprocal it's also a power series but that power series terminates and this term is 2048 which I'll recognize as 2 to the 11th and so if you take this power series in 2 to the minus s this one in 3 to the minus s this in 5 to the minus s if you take the reciprocal it's always only quadratic and the quadratic coefficient is always p to the k minus 1 so it's and this will be true actually here it's p to the 11 because the weight was 12 but in general it's p to the k minus 1 so this will be true and some there's for finitely many primes you have to change it slightly let me know here it's correct for delta so if f has this multiplicative property which we now call the HECI form because HECI showed that such forms not only exist that there are some examples which more delta already shown that delta was an example but HECI showed that for any nice group so for instance mk of gamma 0 of n it's spanned by the HECI form so the forms with this multiplicative property there aren't very many they're only found at the many and in the simplest case they're exactly a basis they're exactly as many as that I mentioned but they're always at least as many as I mentioned they span the space okay so there's a slight base about old forms and new forms I don't want to go into it so you have this wonderful multiplicative property now comes higher mathematics so the key player in all of modern number theory I would say of the 20th century is L functions with a multiplicative property that's everywhere when you see multiplicative things the idea is if you have two different parts of mathematics that give you such functions with similar properties then maybe there will be some overlap you'll have actual L functions as these are called which belong to both categories and then there should be a link between those two objects that's the meta idea and so here we just saw that multiplicative forms some multiplicative forms namely the HECI multiplicative forms some multiplicative forms give you L series I'll just say what I mean by that is a Dirichlet series say with integer coefficients and with an Euler product so it's a product over primes meaning that the coefficients are some multiplicative function but algebraic geometry also gives you such L series that's the point and I'll give you a couple of examples I gave one last time but now I'm doing it more systematically and so then you can ask and the Langmuir's program tells you it'll always be a case these are called and there's a generalization for more general things than just ordinary multiplicative forms but let's say automorphic L series and these are called geometric or usually they're called motivic but you could just say if you don't like such a fancy word geometrically motivated L series and I'll explain in a second how you get them and roughly the Langmuir's program says that in some huge generality all geometric L series these ones come from algebraic geometry come from some kind of a modular object but in special cases and we'll see in a few minutes when they are supposed to come from multiple forms that's almost always conjectural but for instance Wiles failed to get the Fields Medal because he was 40 and a half but would have got the Fields Medal and became one of the most famous mathematicians of the century by proving a very very very special case of the conjecture that certain motivic L series are multiple namely the ones from Liptic Curves which will come to in a minute so to prove this in general is very hard but the motivic philosophy even if you don't know what the words mean or if you don't like a theory that's largely based on conjectures still although there's a lot of proof theory but it predicts certain equalities you can check those numerically and sometimes you can prove them so it predicts connections between very different things so let me give you an example from algebraic geometry so example well now first let me do the general case and then give an example so how does it work so let's x be some algebraic variety but I'm doing arithmetic so I don't want it just with complex coefficients with coefficients defined by equations I won't write all the words defined by equations with coefficients in z or in q it doesn't matter you can always multiply through so it's given by some equations and the equations should have integer coefficients so that I can talk about number theory then I can make a wonderful object which is called the Haase-Ve zeta function so here s is a complex variable but I'm going to define it by an Euler product and the Euler product will only converge if real part of s is sufficiently large something like bigger than the dimension of x anyway this is a specific bound that converges but you have to start in some right half plane so it's defined first as a formal Dirichlet series but it converges on a half plane and it's got a very simple definition you sum over all prime powers so p is a prime and n is an integer and then you take you take your variety and you know that if p is a prime and n is an integer then p to the n is the order of a unique field up to isomorphism that's finite field of order p to the n for every prime power and no finite field of any order that's not a prime power so that's why here I want to restrict the prime powers and so this is just the set of solutions set of solutions so you know these are some equations so we have a bunch of equations x1 up to xm and then some polynomials with integer coefficients i goes from 1 to i you have a bunch of polynomial equations and a bunch of variables and a set of solutions in this finite field which makes sense it's easy if n is 1 you just count numbers multiple of p but you can do it for every p and that's a finite set so since it's finite you can take its cardinality and then here's the formula okay so in particular it has an Euler product because it's an exponent of a sum the sum over primes I could also write that is the product over p prime but the exponential function with just n which would be a simpler thing so let's take an example if x is a point then the number of points of x over fp to the n is of course just 1 it's just a point and so in that case we'll get x of the sum just over primes but then the sum 1 over np to the minus ns is of course just log of 1 over 1 minus p to the minus s so therefore that's the product over p of 1 over p to the minus s and that is as Euler discovered the Riemann's Aida function so in that case so this house of Aida's Aida function is not a trivial thing at all for instance we conjecture something like a Riemann hypothesis for it about this position of the zeros and poles but even in the case of a point that's a million dollar question nobody knows of course the Riemann hypothesis even for the Riemann's Aida function or Euler's Aida function if I take a slightly less trivial example it's still completely trivial take p1 so it's just given by no equations if you wish but two variables homogeneous so the number of points over any field is just p1 of that field so it's that field union of point at infinity and so therefore the cardinality is p to the n plus 1 and so here if you do the same calculation you'll get p to the n plus 1 so the 1 will give you what we had before the p to the n will give you the same with s shifted by 1 and you'll find that here z of xs is z of p1s is very simple but it's the product of the Riemann's Aida function times the Riemann's Aida function shifted by 1 so those are two very very simple examples but now what is known in general but what I now say is not something you can just sit down and prove for yourself it's a culmination of maybe 50 years of hard mathematics so the main names would be maybe vey, Grotendieg dwork and delinear we know if x has some dimension d it's a smooth let's say a smooth projective variety to keep everything then we know that this zeta function always splits in the following way it's the product there are new series ld of xs times l2 I won't keep writing the xs up to l2d and divided by l1, l3 up to l2d minus 1 well that's an empty statement if I don't tell you anything where each lj of s has another product of a very special form first of all it's not just a rational function of p to the s but it's the reciprocal of a polynomial depending for some p in p to the minus s so that's the form of the other product so in our examples here it's sort of stupid here because here the other product pp of t would just be 1 minus t that would give the Riemann's data function of 1 minus pt that would be the original one and here this would be l0 and this would be l2 so l2 l0 the polynomial between 1 minus t and l2 the polynomial between 1 minus pt in this stupid case so the first case is this but now there come the amazing properties pjp of t is normally a polynomial with integer coefficients but it starts with 1 sort of from the definition but the degree is at most bj of x is the jth betty number so it's the dimension if you like topology the dimension of the complex solution of the equation we're doing it in characteristic p we're counting over fields of p elements but I do it over c and then the dimension of this is 2d because it's a real variety 2d and then the smooth variety has coefficients well it doesn't matter it's the dimension of this model it's called the betty number so they're an intrinsic geometric invariance so that's the beginning of the geometric thing so this is the first statement that was maybe improved these are all part of what were originally the vague conjectures and then using work of growth and other things to work proved to be part of it in the last part so the first statement is that you have an Euler product which already is a special form it's one for a polynomial but that polynomial is integer coefficients and and this it also satisfies the symmetry it's up to rescaling it's palindromic but then the deep part is that all roots of pjp p to the power minus j over 2 so you see here this polynomial has the root 1 and j with 0 and here this is the polynomial p to the root p inverse so it's absolute values p to the minus 1 which is j over 2 since j is 2 so this was proved by Delinia this is the famous the vague conjecture of Delinia that was the main reason for his Fields Medal although he certainly deserved many Fields Medals for his other work but then there's a 3 but this is known and 3 is completely conjectural question mark question mark and there's even more there's a Riemann hypothesis which is even more conjectural because we don't even know it for a point so this lj satisfies a Riemann hypothesis I'm going to even write it down but this part says that lj of s extends holomorphically but it extends as a meromorphic function to all of c and there's a functional equation lj of s is equal up to a known I mean it's conjectural but we know what you should write there's a functional equation that sends j to an integer minus s s to an integer minus s and that integers all this j plus 1 okay so in the case of l0 then in the case of p1 l0 remember it was the Riemann's aida function and here since j is 0 s goes to 1 minus s 1 minus s but the next one is l2 of s is aida of s minus 1 and this has a function equation it's the same function equation under s goes to 3 minus s because j plus 1 here is 3 so this is completely unknown this is what was proved by Wiles if first of all if the dimension of x is 1 so it's already really special we're only talking about curves before I was talking about points these are only curves but on top of the curve is a genus and the genus also has to be 1 so it's like a double specialization and it's still one of the hardest theorems in mathematics and nobody can prove it for curves of genus 2 in general let alone for higher dimensional varieties but you can check it numerically Fernandu is an expert on this with the so-called hypergeometric let's check this kind of statement well I wanted to give a little more but I'll talk more about that next time and periods but I want to at least give a couple of examples so since we carefully took the time 230 to 4 so that I could run over time a little since we started weight I'll run over time a little bit so let me give some examples let me write another conjecture also conjectural if now I want to say something more this lj of s might be a product of different functions in general so u will belong to some set which depends on j and then there will be some lj of new s so often I mean it doesn't have to happen it may be that each lj of s is just something like in my example there l0 with z of s you can't decompose that l2 of s with z of s minus 1 but sometimes these individual pieces are the products of smaller pieces canonical pieces and these are then I'm not going to explain how you get them they're individually called motivic l-functions there's a description of how you can look for them and so on and the conjecture is if lj new and each of these pieces has a degree so remember this will also be product over p one over some polynomial depending on j and u and p of s polynomial will now be some bj new and of course the sum of all of these pieces will be the bedding number so the idea is if you think of the homology that's the dimension of a space a homology group it might be a sum of two or three subspaces or a thousand subspaces and then that's splitting of the homology for instance there might be a group that acts a finite group and then you split it up by eigenspaces then each piece all the same conjectures you make and these are automatically true but this one is of course automatically still the conjecture so the last part if lj has degree two so I told you that for a multidorm form you always this was for any multidorm which has this multiplicative property the l-series always is one over a polynomial some polynomial of course depending on p but that polynomial here is degree always two it's one minus a px plus p to the k minus one x squared so multidorm forms are always degree two and the conjecture the huge conjecture or one of these are all huge conjectures then this thing if you have a two dimensional piece which sometimes happens and I'll give you a couple of examples then lj of s is then it is the l function of a HECU form and of course we know what the weight is the weight has to be j plus one because what I didn't tell you here is that this thing for any cuss form actually for that you don't need multiplicative it has a function equation there's some gamma factor but you send s to k minus s where k is the weight but here we send s to j plus one and therefore j plus one should be the weight so that's the conjecture so let's look at two examples and then I'll stop for now and next time I'll repeat this last part a little bit and give the example I'm actually interested in which is the mirror quintic so example one well there will be a sub example if x is an elliptic curve over q so that means that x is given by an equation y squared is x cubed plus ax plus b where a and b are let's say integers well then in this case we know that the z function by what I told you will always be L0 which will always be actually true for every curve it'll be z of s so L0 is always z for every curve L2 is always z of s minus one just as it was for the trivial curve p1 but now we'll have the we can do this for any curve you have the part coming from the so this is L1 in my previous notation but L1 of xs here has degree so if it's any curve actually it's twice the genus so here it's 2 because the genus is 1 and remember the elliptic curve means that it looks like that so the middle homology group which is h1 this dimension 2 it's e plus z so here it's degree 2 and therefore the conjecture is and that's the theorem of well I'll just put wiles et al because Taylor and so on but there are several people in the full theorem the basic breakthrough was wiles the theorem is that this is true for elliptic curves in other words L1 of xs should be equal to L of fs where f is a modular form and it has to have weight 2 for the reason and it happens to be on some group gamma 0 of n and n is a number that you can predict it's called the conductor of the curve but that's not important it was already known that if it's true for any f the n has to be L1 that wasn't what was known by ribbit so it's true I can just write this so if you have an elliptic curve there will always be a cuss form which eats it up which produces it so I'll give that one example of that and then very briefly one more example that also gives a multi-form and the next time I'll tell the example that I actually want for this series of lectures so if I take that example so example I take the following curve y squared is 4x cubed so I've put a 4 it doesn't really matter minus 4x squared plus 1 so actually I could tell you here that remember this L1 of xs this you can show easily and that was why Taniyama had made his conjecture in 19 he didn't make a conjecture but he asked the question in 1955 if for elliptic curves so if L is an elliptic curve he noticed that this L of x1 the properties I said had already been proved but maybe for elliptic curves they had been but he noticed that it had the following form just as before the weight remember we're talking about L1 so the weight here is 1 sorry the power here is 1 the weight is 2 so it's 2 minus 1 so it has that form sometimes for a finite number of p's you have to change it slightly I don't want to worry right in a second and ap of x there's actually a formula for it if you take the definition I gave before you find rather easily I'll just do it here but it's true for any curve if p is different from 2 then ap of this elliptic curve is just minus the sum and you take all x modulo p and now you take this expression 4x cubed minus 4x squared plus 1 but x is modulo p and we want to know whether it's a square or not modulo p but for that you have to look at the genre symbol and if you do a one line computation you find that you exactly get the sum of the genre symbols so here I can make a little table in this case you can just compute this thing it's very easy and you can't do it for 2 but you can also do it for 2 but this formula doesn't make sense but the original definition makes sense so here are the primes up to 20 and the numbers are minus 2, 1, 4 sorry this one excuse me I shifted everything by 1 it starts minus 2 everything was shifted sorry because my table and my note starts at 3 1, minus 2, 1, 4 and minus 2 but anyway it's very easy to compute these numbers and by the way they're small it's true for any elliptic curve that's an older theorem that was known that these numbers are bounded by 2 squared of p so they grow rather slowly there's some integers positive or negative you know if the conjecture is right there should be a modular form well I think I actually gave it last time you take 8 of tau it's a modular form of way to half so it squares weight 1 8 of 11 tau also is way to half it square also is weight 1 so if I take this this will be a modular form of weight 2 on some group which is here gamma 0 of 11 because this was on SL2z I've been 11 and you can compute the Fourier expansion 1 squared minus q cubed plus 2q to the fourth minus plus q to the fifth and so on and you see that these coefficients are exactly what I wrote here minus 2 minus 1 1 so if I write this as a nq to the n a n of f then what I find is that for primes AP of the elliptic curve is the same as this coefficient so again that's a marvelous thing there's no elementary proof and C this is a completely elementary product I mean you don't have to know what the eta product is just write it out it's 1 minus q to the n squared times 1 minus q to the 11 n squared so you can just multiply this out quickly and you get the first few coefficients but that the pth coefficient of that expansion is the same as minus the sum of these of the genre symbols there's no really elementary proof of that so this is this marvelous thing but here we didn't have to talk about motives because as I drew the picture the for elliptic curve for any curve of g is g the first homology group is dimension 2g for elliptic curve g is 1 it's already a two dimensional piece so in that case there's only one motivic piece as I wrote so here well I wrote the formula for elliptic curve you'll have L0 is a to the s L2 is a to the s minus 1 L1 always will be a single piece like this but let me just give you one more example just very very briefly this is from my friend Ron Livne many years ago I remember how much we all liked this example when it was new which was before most of you were born so another example, Ron Livne he didn't have the accent yet then but later he decided to write it with an accent Livne 1987 you take x so it's going to be the set of points in projective 10 space sorry 9 space and there are two equates 9 space means there are 10 coordinates so first you require that the sum of the coordinates is 0 that's a homogeneous equation and it defines a subspace which is p8 but I've written it in a more symmetric way I could have just called it p8 and had 9 variables by you know throwing away one of the x's but now it's the Fermat surface also it's the Fermat cubic so the sum of the cubes of the 10 variables is 0 so that's two equations in 9 space that I mentioned of this is 7 and now one of the one of the pieces actually there's a shift but so for this x if you take one of the pieces I talked about here the homology is quite big so I forget which dimension you have to take but there's several pieces including several Riemann's a defunctions but one of the lg of x is l of fs where f it doesn't even matter what the coefficients are but I wrote them down somewhere except now I've lost and what piece of paper I did write it down maybe here I don't know where I put this example nor does it make any kind of difference but I did give the first few coefficients I wanted to write them out for you here we are except I can't read it ok so f f is a modular form all our members the weight is for the level of 10 and in my notes the expansion will again start with q plus something I have the first 5 or 6 coefficients to show you but there's some numbers some integers and I can't find on which side of this piece of paper I wrote it half an hour ago so I'll give myself another 3 seconds in case it turns up here it is so it's q plus 2 q it does have weight 4 and level 10 as I said it starts q plus 2 q squared minus 8 q cubed plus 4 q to the 4th plus 5 q to the 5th and so on so but it's a completely explicit form you can write it down in terms of the eta function you can write down every multiple form in terms of simple functions like ades and Eisenstein series and so there's a formula for this f but again the point is that if you count the number of solutions so it means for every prime we count 10 tuples of numbers mod of that prime such that their sum is 0 and the sum of the cubes is 0 but we count only non-zero tuples and up to a common scalar then that number will be essentially up to some simple transformation it will be the pth coefficient of this modular form so it's you know these are super super contractual statements in general as I said the easiest case of elliptic curves is a huge breakthrough we don't even know the case of curves of genus 2 so next time I'll tell a little more about this connection and the idea is in one direction it was shown by Eichler and Schimura 40 years ago if the weight is 2 and by DeLine if the weight is anything that if you have a modular form so this is equal this is known by Eichler Schimura and I'll explain that a little next time for weight 2 and for DeLine in general that this is the lj nu of xs for some k is of course as I said before k minus 1, j is k minus 1 for some explicit variety x so there's an actual variety x and you can show that the Hecke forms their l-series are geometry they are motility that was a huge breakthrough but in the other direction as I say it's completely conjectural but at least one way we know and in the case that I and then there's another connection that there's always that's part of the Groten-Diek philosophy there will then be a geometric correspondence some kind of a geometric map correspondence between the variety that you started with in this conjecture where you have the lj nu and the variety that Eichler Schimura and DeLine produce for the modular form those two varieties actually it's not to do with modular forms if you have any two varieties and some piece lj nu for one of them is also the piece for another there should be a geometric correspondence with a conjecture that's some way to state it and that's again a super-super conjecture completely unapproachable you can check it in examples and our example next time will tell us that that's happening for a famous variety called the mirror quintic which is a three-dimensional Calabi-Yau that the fist sits like and on the modular side it will be a modular form directly connected with the Eichler Schimura and the Ramanujan functions which is of course why I introduced them okay so thank you for listening and for coming and now you can have questions for at least 11 minutes I don't know if anyone needs the room in 11 minutes so I promised I would finish by 4 so does anyone dare ask a question yeah I don't want to go into your rationality at all that was a completely random application the theme of these lectures is that modular forms are beautiful things and of many unexpected applications which is transcendence you're absolutely right and in fact the proof that you're saying which is for instance something like pi log pi e and gamma of a quarter and something else at least three of them are algebraically independent that in fact uses the theory of modular forms but it used in a very different way I don't want to even think about transcendence but yes there is there are many proofs about transcendence field theory to my sorrow is a major field I find it a kind of strange thing to show that something isn't going on that the number that doesn't look algebraic you really show it's not algebraic so you know big surprise but people get very excited and who can blame them so many of those theorems I think alpha and schneider theorem have nothing to do with modular forms the transcendence of pi of e of e to the alpha where alpha is a log alpha where alpha is an algebraic number those theorems do not use modular forms there is one I mentioned that relates something like pi e to the pi and gamma of a quarter that those are the field that they generate is transcendence degree three so they're not only individually but they're even algebraically independent that's the result of the Schmofsky's and later from Nestorenko and that uses modular forms and quasi modular forms so there are other applications of modular forms to transcendence theory yes I don't know them well there are one two three of modular forms I mean it's not mine there are three authors but in my part which is the one of one two three there are many many applications of modular forms classical modular forms to other domains about many and there is a short section like half a page called transcendence and there I mentioned explicitly Schmofsky's theorem you can look and Nestorenko's theorem so you can see exactly what the statement is but of course I don't say anything about the proof I just give a reference here the general theorem is lj has degree well the whole lj has degree which is the bedding number it has powers I mean the absolute value of the roots is like p to the minus j over two now if you split it up into motific pieces then the bedding number will split up into lots of pieces and there's no general recipe for that but the conjecture here is if that maybe I raced already if one of the pieces has degree two that's what you need to have a modular form because a modular form I mean degree one by the way is trivial that's the Riemann's aida function that's considered trivial today but degree two are the ones coming from multiplicative forms and the conjecture and as I say it's proved by Deline after Eithler and Chimora that if you have a hacke form its l-series is motific it is a piece of the cosmology that means there's some variety somewhere and some cosmology group some j which has to then be k minus one and in the k minus first cosmology group you will find a two-dimensional subspace a natural subspace and if you take the corresponding l-series that I didn't explain what that means then you will get this that's actually a theorem and so I don't know which of the indices you're referring to but anyway this is degree two so whenever it's a modular form we're looking for a degree two piece of the cosmology it's because it's SL2 if we were doing GLN then it would be an n-dimensional piece but nobody knows how to write down anything arithmetic on GLN if n is not two so it's kind of pointless that statement I mean it's not pointless but it's very abstract there are no concrete examples except ones that come from GL2 so far as I know there are none yep yes that's a very good question so the question you can't hear I have a microphone Kevin the question is can I say more words philosophically about this motivic thing the answer is yes there was plan for this lecture it will take me five minutes and I've already gone over time so come on Friday and at the beginning I'll finish what should have been this lecture so the end of this I gave you a bit of a sketch is the following philosophy if some l-series occurs I already said that as a piece of lj of x and also of lj of y for a different variety it has to be the same j because the absolute value of the roots can't be different it's the same polynomial but if and it happens all the time in practice like this example of Ron Lifney you write down some crazy variety which Ron Lifney had this one but then for this particular multiform the proof by Delinio will give a completely different variety and we know that the same l-series will come there always because there's a physical correspondence between the two and I'll explain what that means next time and there's another thing which you might know or many of you know I wrote a survey paper many years ago with Komsaevich called periods and periods are numbers, real numbers or complex numbers that you associate in algebraic variety on your algebraic variety you take some cycle so something representing homology class and you integrate a differential form to run but that differential form has algebraic coefficients the numbers you get are almost always transcendental they're called periods and then you have the same conjectures for motives that's called the Hodge conjecture this was the tape conjecture if two different varieties have some periods in common then there should be a physical correspondence between them and their l-series should match all of those things should be equivalent so if there's this physical correspondence if they're geometrically related then some part of the l-series of one will be some part of the l-series of the other and some of the periods of one will be some of the periods of the other but the converse conjectures are called the Hodge conjecture and the tape conjecture these are very fancy versions of them and they're completely conjecturally general but you can check them and what I'll do next time is first explain this much more concretely and the example I'll have is the mirror quintic which I wrote down last time we'll write down again Wednesday in this dimensional well, complex three dimensions so D is three we're looking at J is three so the middle comology the full bedding number is 204 so L3 has an Euler problem with a 204 degree polynomial but that polynomial is a product for a lot of linear polynomials and one quadratic factor and so that quadratic factor by this general conjecture should come from a modern form is actually known the modern form is identified many years ago but on the level of periods you also expect the same period and that's a calculation, I did numerical calculation with Klam and Scheidegger that actually all of us have talked about at the ICTP last year some of you heard it and that worked out numerically and so the periods show up, the L-series agree and so if the Golden League philosophy is true there should be a physical correspondence between the variety predicted by the linear in the form of weight four and the original MiraQuintic of Candela's and company and that's the one I've been trying to construct and it's a function of three variables and I had a function of one variable a month ago that's when I announced this series of lectures and I was sure I could get the other two variables and until the day before yesterday I only had one variable two days ago I got the second variable so now I have two thirds of it and I hope by Friday I have the third variable last night I made some progress and I tell you that in this case the conjecture is really what it predicts is really true there is an actual map or a correspondence between these two varieties so I hope I'll be close to that by Friday so that's where we're going but I'll give more details next time now I think no more questions well two minutes quick question okay I think we can stop them