 Okay, so to gain some time, so welcome to the second of the series of the Salam Lectures in 2015 from Don Zaghiere. So, you've heard the first lecture yesterday, and it's the magic of modular forms. And every day has a different sub-subject that Don is writing at the moment for each of the days. So, yesterday was in general applications for arithmetic. Today is modular forms and application for differential equations and the irrationality of Zeta3. And then he will go on tomorrow, the next two days, on different applications of the modular forms. If I can... I remember if I can... I think tomorrow it will be about applications to black holes, if it is correct. Yes, that will be on Thursday. So, every day the good thing is that every lecture can be self-contained. So, in that sense, we can restart every day and try to follow up. Yes, it's the fourth lecture of application to string theory in black holes. So, let me just remind you that this series of lectures this year, we have been fortunate to get the funding from the CAFAS, which is the Kuwait Foundation for the Advancement of Science. So, now I'll let you with Don. The topics disappeared mysteriously from the website. So, many people asked me what I was going to do, and I said it was all written, but it has somehow gone into cyberspace. I've written the five titles. So, I start by saying that I'm very happy that many of the people who were here yesterday came back anyway, and that some people who weren't here yesterday came showing that ignorance can be bliss. So, remember the two most important things from yesterday? Well, now there were three most important things. One is that since I used the blackboard, people in back might want to come forward, because I shout, I do my best, but I won't use the computer. In fact, maybe somebody could even turn it off, and I just will use the blackboard and my voice. So, the two important things were that the letter tau is not z. And so, if I say z, please correct me. And the other, I'm sure nobody has forgotten, and you've all already done it, but by the... Okay, so today my theme, as you can read it here, MF is modular forms. Oh, I should have written it out at least once. The magic of modular forms is my general title, and to me the magic has both aspects. One is the one I tried to emphasize yesterday, that these very simple functions in a certain sense have two completely different aspects, which are not obviously the same, but that somehow makes them very rich. And also that it gives you kind of an automatic tool to prove many, many identities that you couldn't prove in a direct way. So, we saw many examples yesterday, and there will be more tomorrow, today. In some sense, that's today's theme. But the other magic you see illustrated in this list, which as I said yesterday is just a selection of five topics that I've happened to like because I've worked on them, but I could have easily found another five disjoint from this list and probably a third five disjoint from both lists. So, the magic is that these things that start as being a very, very specific class of functions invented by people like Poincare and Fuchs, and they look like very specific functions, but they come up in every domain of mathematics and more and more in recent years in every domain or many domains, at least of theoretical physics. So, in conformal field theory or many other, because there are many things that I won't mention here at all or just maybe a sentence in passing. So, today, the connection is between modular forms and differential equations. Differential equations is surely one of the most central themes in all of mathematics, probably historically, the most important part of mathematics as far as applications in physics goes. I mean, certainly Newtonian and the 19th and 19th century physics, but still in 20th century physics and, of course, in mathematics. And there, the amazing thing to me is the following. When modular forms were invented, which was, I mean, it was a very, very slow process, you can see traces of them in the paper of Gauss in the first half of the 19th century, but something that resembles what we know really started in the late 19th century and the people were, well, I mean, I don't know in any particular order, Weber, Frecke and Klein, who one often ends together because they wrote several books together on elliptic and on modular forms, of course, Poincaré, books, and various other people. Fuchs is, of course, also very famous specifically for differential equations. And the whole field in those years was part of the theory of linear differential equations and people studied it because they were interested in, you know, triangle groups and special monodromy groups of special differential equations. And then somehow that aspect of the theory disappeared out of the collective consciousness to the extent, I mean, I've often told this story, so friends of mine in the audience will have heard me say this before, but about, I don't know, maybe 20 years ago, maybe a little more, I gave a lecture in a very top European university, in fact, in Orsay, with maybe 80 or 100 people there, all very good number theorists. It was a specialized lecture for experts on multiple forms, and I simply asked, I said it's a basic fact that all multiple forms satisfy linear differential equations. Just out of curiosity, how many of you know that fact? Please raise your hand. And there were zero. And that included Serr in the audience that actually had written about it but in a very different language. I said, but surely Serr, you know this. No, he said that's new to me. I said, of course you know, but maybe you didn't recognize. But it so disappeared that what was originally the origin of the subject, the essence, people didn't, even the experts no longer knew it was even true. I mean, a few, but really very, very few. And in the last years, it's become much more popular in particular because in theoretical physics, in particular in string theory, where you look at a mirror symmetry, for instance, families of Calabals, they have associated differential equations, which are the so-called Picard-Fuchs differential equations. I think I hope I'll say something about. And those frequently turn out to be the same, frequently, unfortunately not, but frequently that the same is the differential equations coming from other forms. So that's one reason people have started getting interested again. And another I think was, in fact, this irrationality of, say, the three, which is a story I'll tell today. But to me, this was really incredible. If you think of the functions you learn about as a student, as an undergraduate. So you start with functions like, you know, e to the x log, well first, of course, polynomials. And then you have slowly more and more transcendental functions, maybe the gamma function. And then depending on whether you have an old-fashioned upbringing like me or a modern one, you may or may not see things like Bessel functions and meet polynomials and so on. And at least when I was a student or classically, certainly also in the 19th century, one distinguished two broad classes of functions, the good ones and the bad ones. The good ones would be these. These all satisfy a differential equation. So for me, a differential equation will mean, when I say just differential equation, I mean a linear differential equation with polynomial coefficients. So, well, this one is sort of really stupid because I guess x, y prime is one. And here there's a famous second order equation. Of course, the hypergeometric function, a, b, c, x, and many, many others. And if you, but so all of these functions have differential equations here, second order, here's second, here's second. And this one does not. So this was considered the next level of the hierarchy. So essentially all mathematicians distinguish a function that doesn't have a linear differential equation from one that does. As the basic classification of all functions. And then you have the specialists in the whole field of modular forms. And not a single one even knows that their functions all have a differential equation. So it's normally not the central point of view. They don't even know. I mean, it's somehow very, to me that's, I wanted to make that remark because it's somehow a little strange. So we won't see Bessel functions, but we will see the hypergeometric function. Well, these functions are somehow two, not the right sort. So that's one thing I wanted to say. So actually it's even better. There are two kinds. Every modular form normally satisfies the differential equation, but actually two very different kinds. And so I'm going to explain both kinds. So there are two main theorems, theorem a and theorem b. Maybe I can already write them. And then I'll actually want to prove both of them. So theorem a. So first of all, since not everyone was here yesterday and certainly people are here yesterday will be more confused than people who weren't here yesterday. So let me remind you very briefly that a modular form in the correct way of thinking of it, so to speak, a modular form f of tau f is a holomorphic function with some growth condition that I'm not going to write down again from h to c. So it's a complex value. Function h is the upper complex half plane, so that means imaginary part of tau. Tau is always the variable unless I forget any z. It's a function such that for all a, b, c, d in the group gamma, which in my lecture yesterday, but today it'll be less and less so, gamma is the group s l 2 z of 2 by 2 matrices with determinant 1 integer entries, a, b, c and d are integers and I'll put here for today or a subgroup of finite index. Yesterday was the first day I didn't want to emphasize that but already yesterday some of the examples I gave, I said actually, you know, this is not really on s l 2 z, it's on a subgroup of index 12. But today most of the examples will be on subgroups. So and then it satisfies this famous equation that for every such matrix a, b, c, d, but in the case of s l 2 z, you only need these two matrices usually called s and t because they generate the group so it's enough to check it for two matrices and then the famous equation is that the function is not simply invariant by gamma so it does not simply satisfy f of a tau plus b over c tau plus d equal to f of tau. That is also interesting, it's called a modular function but in general there's a formula of c tau plus d to the k and I'll put here, here k is an integer or sometimes a half integer is called the weight of f so it's a basic invariant so each modular form has a weight but there are many, many modular forms of course of the same weight, they form a vector space if you fix the group it's a finite dimensional vector space so that's called the weight and what else did I want to say? If k equals 0 in which case you have to relax the growth condition in some way then f is called a modular function rather than a modular form so a form is really in a fancy modern language a section of some line bundle over the modular curve and if it's the trivial line bundle it's just a complex valued function a form is a little fancier so that's the language of modular functions and modular forms just to remind you, please a k well it doesn't make sense so let's take this equation we have this c tau plus d to the k so if you have a complex number and if you're a mathematician, not a physicist then if you have a number which is an integer you know what z to the k is but if k is not an integer then what z to the k is if you're a physicist you probably do know because you choose some branch of some log and you cut the plane but that's no longer a function you can't cut up the plane so we simply don't have well-defined k powers for instance even I said that k is allowed it's a good question but I'm explaining even half I said very cavalierly for instance I gave the example the theta of tau q was always e to the 2 pi i tau and I had this very simple theta function the sum q to all square powers and this was some kind of a modular form I told you of way to half but there's actually a problem or I gave another example well I'll state with this for the moment what does it mean to have way to half well in particular it means in this case let me take the function theta 3 of tau which is the same thing but I divide here by 2 okay so that's just a reshift this is a modular form on a group called gamma of 2 I'll define it in a minute so there's no problem because q to the n squared over 2 makes sense because it's e to the pi i n squared tau and you can exponentiate any complex number but now when I write for instance this theta 3 of minus 1 over tau well it should satisfy tau to the 1 half but actually the problem is what does this mean I mean tau is a complex number it is 2 square roots one of them is the negative of the other it's not a well defined function so there is a way around it in the case of half integral weight there's an elaborate theory developed by Schumacher and other people and so you can make sense but a priori k has to be an integer just to make sense of this kth power and it really will give a problem some root like the standard root because we're in the upper half plane but then there won't be any such function because when you take the important thing is that this was a group action if I take f of gamma tau then it's invariant but if I take f of gamma 1 gamma 2 tau then I can think of this in two ways as f of gamma 1 of gamma 2 of tau or as f of gamma 1 times gamma 2 of tau because it's a group action if I do it this way I'll get something times f of tau gamma 2 but if I do it this way I'll get something times f of gamma 1 tau gamma 2 tau excuse me and that will be something else times f of tau so in either case I'll get a transformation but these things won't agree because in complex numbers it is not always true even if you pick the standard square root that the square root of a b is the square root of a times b but it is true that if k is an integer then a b to the k is a to the k b to the k so in other words the consistency of this equation essentially requires k to be an integer and the only reason I can allow a half integer it's a very tricky theory with co-cycles and the metaplectic group I don't want to talk about it there's a trick but basically you must have an integer I hope that's a good answer no that's what I just explained I spent 5 minutes I just spent 5 minutes to explain no we require this for every a, b, c, d and you can't choose all the brands because that's what I was explaining but of course you can choose a branch for each number at random but then this won't be true and therefore this consistency that's exactly my whole explanation f of gamma 1, gamma 2 is something times f of tau but you've chosen a branch but f of gamma 1, gamma 2 is also f of gamma 1 of gamma 2 of tau so it will be something times f of gamma 2 tau applying it again is something else times f of tau and there's no way, believe me there's no way to make them agree the space of forms will always be zero dimensional they will not exist sure but I don't want to get into that yeah but that's because I'm doing a global thing I don't want to get into a very technical discussion coming from conformal fields but here we have global fields they're defined in the whole upper half plane it's absolutely essential it's not defined in some small area and I'm simply telling you as a fact and please simply believe me 10 minutes to give you a complete proof this equation will not have any solution even if k is a half integer unless you do something very tricky and for fractional indices it simply won't work so simply you won't get any function this is remember for all a, b, c, d in the group and for all tau in the upper half plane not in a little neighbourhood of a point where you choose something it just won't work there aren't any functions so you can make the definition but you will get only the zero function it's really true I'm not trying to sweep over difficulty it simply won't work so that's why it's a very good question I mean of course it's why does k have to be an integer or sometimes a half integer so now let me say the two main theorems so we've already seen that a modular form is a function of tau but as we also saw yesterday if I look at these two things these two generators then one of them says f of minus one over tau if tau is an integer should be simply tau to the k f of tau but this one so this is this matrix and again if you have the two equations I'm writing suffice because these two matrices generate the group so we have two functional equations this is very simple it's just an involution so it's an even function under some involution the other is also very simple it's just a translation by one so each group is an abelian group this is the group Z this is the group of order two but together they give you this more complicated group but because of this fact as I explained yesterday any function which is a function of tau plus one you can also think of as a function of q which is either the two pi tau because by order's identity q only depends on tau up to translation by one so you can think of every modular form I'll never write f of q but I'll write f of tau but I'll certainly write it's Fourier expansion which we talked about a lot yesterday that's where the applications come from and that will be a power series so we're not expanding around tau equals zero we couldn't it's not a regular point but around q equals zero which in tau is the point at infinity so now there are two theorems two basic theorems one is the tau point of view it's very simple to state and as far as I know completely very simple to prove and as far as I know completely useless but it's a very nice theorem so every modular form each time every modular form of any weight satisfies I have one more abbreviation I'd like to make it's very very standard in mathematics and it's such a long phrase please forgive me if I write differential equations I'll write dE because it takes too long but it's pretty standard everybody knows you know ODE PDE so ordinary differential equations partial differential equations maybe LDE for linear I won't use anything but I will use DE so satisfies an LDE which I'll write out the linear sorry except I would write it if it were true but I won't write it because it's false a non-linear differential equation of and now comes the surprising thing so this is now with respect to the original variable with respect to tau so f a priori is a function of tau and now the surprising thing this equation always is the same order it's always order 3 never more also never less unless it's the constant so that's a very nice theorem and in fact it's even better if I had colored chalk which I do then I would insert a word which I will well I have to insert an N also because I'm writing in English an autonomous first because maybe not everyone knows the word let me remind you the terminology very briefly in case it's either non-standard or you're not familiar with it it's fairly standard so when I say so when I say a linear which I usually won't say a linear differential equation so I have a function y which is a function of x so this means or let's just call it f of x then this means that there's an equation a linear differential equation of order N I should define that too because I used the word it means that among the first capital N derivatives of f there's a linear relation whose coefficients are for me just polynomials they keep life simple although actually in the theorem B the more general statement you would allow algebraic functions rather than polynomials but I'll avoid that so linear differential equation means this in other words it's a combination of x it's polynomial at x of arbitrary degree but it's linear in f f prime up to the nth derivative a non-linear differential equation as in theorem A is simply of a polynomial in several variables again of order N you have a polynomial a fixed polynomial with constant coefficients just complex coefficients in n plus 1 variables and this function if you substitute excuse me you substitute in at x f and its derivatives so the 0 up to the nth derivative it's identically 0 as a function of x so this is a polynomial and it's called autonomous it's not very important but just it's a well-known word and it happens to be true for free in this theorem if the coefficients of this polynomial so I could also say polynomial in the derivatives whose coefficients like here are polynomials in x but if the coefficients are constant we simply have a polynomial relation among the derivatives with no x at all then it's even simpler well the typical equation would be I already mentioned e to the x satisfies y prime equals y that's linear of course but it's also autonomous but a Bessel function wouldn't you'd have some x y double prime equals something okay so those are just the words so that's theorem B and as I said as far as I know it's useless I don't know any direct consequence of this general theorem and the reason it's useless is because of this innocent looking word non linear differential equations have a very very rigid theory with monodromy and beautiful properties nonlinear there are wonderful parts of mathematics that are all about nonlinear equations but there's no theory that you can just throw your equation into and get an answer you have to study each class individually it can be very hard I mean think of Perlman's work and things like that I mean it's nonlinear equations are notoriously difficult and here maybe if some serious nonlinear differential equations people studied this maybe they could say things and maybe it's even happened in the early 20th century and has become forgotten I'll mention one nonlinear equation that was used but so far as I know you can't really make much hay out of this but theorem B is let me write it out now more carefully let F be a modular form of some weight K but now I do want the weight to be first of all an integer not a half integer and positive quickly positive in English positive being strictly bigger than zero weight K and let T be a modular function so remember a modular function is a modular form of weight zero you can ask does it have to be on the same group I can answer well it's automatic because each one is on some subgroup of finite index of SL2Z and then they're both modular on the intersection of those they're still a finite subgroup so for all the difference it makes on the same group okay and now then the claim is F satisfies and now it's much much better a linear this time differential equation but it's worse in two or three respects two respects first of all it's not autonomous that's not a big deal I'm trying to be totally boring if it were linear and autonomous it would just be some of the exponential functions and it certainly isn't linear equation but now the order is no longer always three the order is K plus one exactly K plus one so it now does depend on the weight but that's why K had to be an integer order three halves wouldn't make too much sense but now it's respect to T not to tau so this is quite remarkable and as I say that's the theorem we'll really use but I'm going to prove both or at least sketch the proof of both because they're both not too hard and very pretty and very enlightening I think so those are the two basic facts and it's the second one that was so important in the 19th century and got forgotten but I'm the first one got forgotten too but it deserved it okay so that those are the statements I want to put meet into this now by telling you why they're true and how it works and examples and then the beautiful application found by burgers to the irrational the of Z of three so I'll just as advertising for later in 1978 a peri who was a relatively unknown French mathematician became an extremely well known French mathematician immediate after that proved a problem that had been open Euler had already mentioned that Z of three of course is what people call for some reason the Riemann Zeta function that Euler invented 110 years earlier and it's simply the sum of the reciprocals of the integers to the power you see so in this case the cubes so Z of three is simply the number one plus an eighth plus a 27th which is approximately 1.20205 I think 6903 I forget and already Euler who found that Zeta of two and Zeta of four and Zeta of six could be given by closed formulas said he couldn't find a formula for Zeta three and he believed there wasn't one so it was in 1849 and in 1978 say 1749 what am I saying so it took you know 230 years until it was proved that this number maybe it can be expressed in terms of some other numbers but at least it's not rational but about one year later so around I think 79 Burgers it was a very mysterious proof very ad hoc nobody understood actually nobody even quite understood why it was true there was he gave such a bizarre presentation the proof that people doubted it and actually people asked me to prove the key identity he claimed and I did prove it and then everyone believed it but it was of course the identity this thing but anyway the proof was very mysterious even with the proof you didn't understand where the ideas came from and then Burgers found an interpretation of Apiris proof using this connection with modular forms and differential equations that's what I'll explain later but first I want to talk about theorems A and B so the first thing is if you talk about differential equations you have to differentiate so let's start with the modular form well luckily I don't have to write it again because it's still here we take this equation and we ask okay F is a modular form is F prime anything special so let's simply differentiate this so I applied D by D tau so by the chain rule when you differentiate F of A tau plus B over C tau plus D you get F prime of A tau plus B over C tau plus D but you have to multiply it by a factor which is of course the tau derivative of this thing so you have to exercise that the answer is just 1 over C tau plus D squared it's 1 because remember the determinant is 1 otherwise you would have the determinant in the numerator but here it's 1 so that's the left hand side and on the right hand side you know it's really easy if you just if you forget to differentiate the first term and apply the product rule the derivative of the product well you get C tau plus D to the K F prime of tau and if that were true then C tau plus D to the square to the right and it would become C tau plus D to the K plus 2 and so I would have if that were true the F prime would be in M K plus 2 but since it isn't true it's not but I want to emphasize that it's not a modular form but the weight of the modular form that it's not is exactly K plus 2 because you see this K plus 2 but unfortunately there's a slight correction term called an anomaly which is that when you differentiate the second term of course you get K times C times C tau plus D to the K minus 1 F of tau and so F prime no longer is a modular form because of this bothersome term now actually F is something and I'll talk about that it's still on the list of talks tomorrow I'll talk about quasi modular forms so I'm not going to define them today yet but they'll come up again tomorrow so this is what's called a quasi modular form I mean this is an example of what's called a quasi modular form that's a slightly wider class that allows even more applications and more liberty in applying the thing but for the moment you don't need a name this is just a fact and so what do you do with this well what you do with it is very nice you can so there are several things when mathematicians run into a difficulty it never stops them if a theorem is false you either change the definition so that you allow things to make it true so that's what I just did so modular forms do not have the property that the derivative of a modular form is a modular form it's false so we just say well I have a wider class called quasi modular forms and now it's okay because I just defined it that way and it turns out if you define it right the derivative of all quasi modular forms is still quasi modular you have a bigger ring you'll see it in a minute in an example another thing you do is you say we eliminate the difficulty so let's take another form which is some let's say G which is another weight L so we would have the same equation G of this C tau plus D to the L times G of tau and now let me multiply this equation the F prime equation by the G equation then I'll have an F prime G term and I'll have an F prime G term that's correct but then there'll be a correction term all of that would be fine because this is weight L if I just had that weight K plus 2 when I multiply them F prime times G would have weight K plus L plus 2 but it doesn't but the error will be K times C times C tau plus D to the K minus 1 that I just had times F of tau but I'm multiplying by the first equation there will also be a G of tau and there will be this so if you mentally multiply this F prime equation by the G equation the extra term that bothers you has to form K times C times C tau plus D to the K plus L minus 1 times F times G but this thing is completely symmetric if I interchange K and L because F and G is symmetric and K plus L is symmetric and so one of the equations will have an extra K times a certain correction term so I make what's called the rank and coin bracket I'm not going to explain why there are two names the papers were 25 years apart it wasn't a drawing paper at all I define the bracket rank and coin bracket of two forms to be F prime G but then I also could have done it the other way I could take F G prime and then remember the correction term here was K times sorry F G prime here it was L times the same to keep things in alphabetical order then I can put the K here and the L here under minus sign so now if I take this particular combination K which was the weight of F times F times G prime minus L which was the weight of G times F prime times G then those correction terms will go away and this thing now will be a modular form of weight K plus L plus 2 and that's very nice you can check easily it's an exercise that this thing is anti-symmetric it's twice the Jacobi identity so that M star which was already a ring now also becomes a Lie algebra and it's even graded but the grading is not K but K plus 2 you have to shift it by 2 because of this shift so suddenly and they are compatible with each other it's actually so called Poisson algebra and actually it has much more structure so called rank and coin algebra so this is you can do that and actually you can continue I'll only write the next one so there's an FG2 this would have been FG1 but I didn't bother to that would be K times K plus 1 times FG double prime minus 2 by normal coefficient 2 times K plus 1 times L plus 1 times F prime G prime plus L times L plus 1 F double time G and that would now be a modular form you can guess of course of weight K plus L plus 2 so we have these rank and coin brackets so that's one way you can get out of the difficulty I just mentioned that because it's important but I'll actually use it in the proof of theorem B later so I'll try to remember not to erase it maybe if I circle it in green that will remind me that it's meant to remain here for a few minutes ok so that's another thing you can do so as I say if you have a difficulty you either get out of it by changing the definition so the difficulty goes away saying modular forms are part of a more general class or you make combinations to eliminate the difficulty on this combination is a true modular form in the original sense but it's not just the product of F and G prime it's a more complicated bilinear form or higher bilinear forms ok so those are things you can do so now let me give you some examples and that will immediately lead to a proof of theorem 1 a theorem A that anyway the examples are very pretty I thought I said it I'm sure I said it anyway nobody can prove it now although I'm on video maybe you can prove it K plus L plus 4 thank you very much and the third one would be K plus L plus 6 I even said you can all guess and what you would have guessed would have been right but I wrote the wrong thing thank you ok so let me give some examples so remember the examples I gave yesterday the basic examples of modular forms I'll write them out in some detail well this one was 240 times the sum and then the nth coefficient was the sum m cubed m divides n but I called sigma 3 of n q to the n which you can write a little more efficiently this part is simply the sum m for bug to infinity and then you can do the other sum the multiples of m is just a geometric series so you can write it a little more efficiently it's a simple sum rather than a double sum and similarly E6 of tau was a similar thing m to the fifth q to the m over 1 minus q to the m and finally I had another one delta of tau which was something very strange looking which was the product 1 minus q to the n to the 24th and then that product not at all obviously this is a modular form of weight 4 this is a modular form of weight 6 and this is even much less obviously a modular form of weight 12 but those are all very classical theorems so we had that and we had a relation between the which is that 1728 which I'll remind you is 12 cubed it'll play a role later times delta of tau is E4 of tau cubed minus E6 of tau squared at least we're in the right bold part this is weight 12 that is weight 12 that is weight 12 so at least they have a chance to be linearly related so now let me if I were tall enough if I stand on tiptoe actually it's obvious how to define E2 of tau you just take the same definition instead of 5 and 3 I'm just going to put 1 and so it'll be simply m so there should be an m m type is q to the m over 1 minus q to the m or if you prefer the sum sigma 1 of n q to the m the sum of the devices for the first part and then there should be a constant and if you know the rule of formation of these so it works perfectly well in weight 2 so E2 is a perfectly well defined form but it's not a modular form that's why yesterday I didn't give it on the list of mod forms but it exists and it is the same as the function I called G2 but G2 which I won't repeat today was defined by non-absolutely convergent series and if you sum in the obvious order you get this but if you change the order you mess it up and that's why it's not modular so we have this E2 so a very very simple thing pretty much in our head well it depends on the quality of our heads of course so maybe I'll do it on the board I can't do it in my head but let me change this n to m to make the calculation easier let me look at the logarithmic derivative of delta so that would be the derivative of log delta I mean delta f prime over f for any function is the derivative of log and log delta well it starts with log q because here's delta so that will be q but the q prime over q is 1 because sorry not quite 1, q prime is e to the 2 pi i tau prime which is 2 pi i times q so actually very soon I'm going to abuse notation and prime will change its meaning and mean I'd simply divide by 2 pi to get rid of this but for the moment maybe I better not so the derivative normalized by 2 pi of delta prime I start with the derivative q prime over q but q prime up to this 2 pi i is q so it starts with 1 now the log of something to the 24th is 24th times the something so I get 24 and now the product turns into a something and now the derivative of the log of 1 minus q to the m is m times q to the m over 1 minus q to the m and I hope that's familiar to you because that's the function I wrote down 10 seconds ago that's e2 so in other words this e2 was not quite modular but for the same reason in fact it's very much an example unfortunately I just erased it but I had the equation for how f prime for a modular form transforms and the way f prime transforms maybe remember there was an extra term with c tau plus d to the k minus 1 but there was a c tau plus d squared so actually there's an extra term with c tau plus d to the k plus 1 that was how I'm sorry I'd erased it so if you divide by f of gamma tau which is the same just do with that if you divide then you see that f prime over f in this case we will find that delta prime over delta which we now know is e2 satisfies e2 of gamma tau c tau plus d squared but then you have a factor minus or plus minus I think 6i over c tau plus d plus if I put it like that time sorry 6i I'm not dividing so c times c tau plus d and first of all you have the formula that you would have if it were modular c tau plus d to the k e tau of 2 and now the correction term so I'm trying to calculate and think at the same time it would be 1 over 2 pi i with a minus sign but I'm multiplying by minus 24 sorry 1 over 2 pi times minus 24 is not quite 12 anyway believe me it's correct with the 6 c times c tau plus d so this is what you get if you simply apply the equation that I just well I erased it and then I wrote it here very illegibly the equation that we had for f prime you apply to delta divided by the original equation for delta and you'll get this transformation law just as an example let's take this just so you see it in all of its glory I'll put it here with some if you maybe can't read the special case of the matrix 0 minus 1 1 0 would be tau squared e2 of tau plus 6i tau over pi which means that the sign is wrong after all nobody can remember signs and so here if I specialize to e2 of i I find that e2 of i is minus e2 of i so I can put it 2 and then you get 6 over pi and so e2 of i is 3 over pi and indeed if you substitute i into this series it's exponentially convergent it converges to 3 over pi okay so we have this somewhat modified thing and this is typical of what tomorrow will be repeated and use these so called quasi-multidiforms so now we have something very pretty to remember the calculation that I did a few minutes ago if f is an ordinary multidiform then f of gamma tau is something times f of tau but then if I differentiate that I got f prime of gamma tau with something else this something with c tau plus d to the k this was c tau plus d to the k plus 2 I'm not going to write it all out it involves only f and if I differentiate again and again of course I'll keep having more terms so if you do this and now so that we don't go crazy please allow me to change my notation f prime of tau I'm going to change the definition is still the derivative but it's the derivative with respect to 2 pi i tau so it's 1 over 2 pi i times the usual derivative and if you like q it's q times the q derivative so that's very convenient because it means if f is the sum a n q to the n then f prime is simply the sum n a n q to the n if I didn't divide by 2 pi i all of the formulas would fill up with transcendental factors 2 pi i and it completely mess I want to be doing arithmetic so let me just redefine the meaning of prime so all formulas on the right just became wrong this is correct because I was very careful not to say what I was saying so now let's do a little calculation so I have theorem I won't call it theorem c because somehow it's completely independent theorem it's not really a theorem it's an easy easy observation but it was a wonderful discovery of Ramana John obviously many years ago since he died in 1920 theorem is that if you take the derivatives of e2 e4 or e6 then even though they're no longer modular forms they are still polynomials in e2 e4 and e6 actually they're quasi-multidiforms this word that I mentioned and in this case we know that all quasi-multiforms are polynomials in e2 e4 and e6 and the formulas are very simple so these are Ramana John's famous formulas the weight always has to go up by 2 so here the weight with 6 it has to become 8 2 plus 6 is 8 2 times 4 is 8 similarly here 2 plus this has to weight 6 so it could be e2 e4 or it could be e6 and you notice that in the formula that I just wrote and erased sum a n q to the n prime is the sum n a n q to the n you've always killed the constant term q to the 0 goes away so there can't be a constant since every e starts with 1 this starts with 1 then I have to take this multiple has to be 1 to make it have no constant terms so only the thing in front is a little mysterious and the values are half a third and a 12 everything else is the only thing it could be so those are the theorems now let me give you the proof e4 is a modular form so you write down the equation for e4 and for e4 prime is I did it on this board but you also write down the equation for e2 that I also had on the board and in fact still do somewhere for e2 it's over there and if you do that what you'll find is that e4 prime fails to be a modular form and so does e2 times e4 but a linear combination of them named exactly this combination is a modular form of way 6 and since this starts with 0 as a constant this starts with minus a third it therefore has to be minus the third e6 and the same for the other two so it's completely trivial using again the magic of modular forms if you suspect that identity there are clever proofs of this of course but you don't need a clever proof the modularity properties automatically tell you that it's true but it's not true for the other two cases so each one is a one line proof so we have this lovely fact and that means that the ring of quasi-multidiforms which is usually called m star tilde is in fact well it doesn't mean but it's true that that ring here is generated by e4 e2 and e6 and it's closed under differentiation closed under prime okay now let me prove theorem A let me make a lot of space for the proof it's not going to be very hard so proof of theorem A if f that's too easy to prove a theorem an important theorem in such a huge space let me strengthen the theorem I already strengthened by adding autonomous every moded form or quasi-multidiform satisfies a nonlinear differential equation it doesn't end but be modular we'll see an example for e2 in a minute so the answer is if let f be a quasi-multidiform well I just showed that since every quasi or I told you every quasi-multiform is a polynomial e2 e4 e6 but their derivatives are still in the ring and they generate the ring so the whole ring as I already wrote is closed under differentiation so if f is in there well then it's still in there after I write it the second time but so is its derivative so is its second derivative and so is its third derivative so if you have a multidiform or for that matter a quasi-multiform it makes no difference here then it and its first three derivatives are all quasi-multidiforms but this thing as I told you in fact that's how I just proved this is generated by only three objects so therefore an algebra generated by three things has transcendence to get most three here exactly three so any four objects are linearly dependent such that p of f f prime f double prime f triple prime vanishes identically it's simply a fact about an algebra with three generators that any four generators are polynomially dependent it's kind of a trivial algebraic fact so this theorem is not at all deep it's sort of obvious and you see why we need it exactly third order because I needed four derivatives because four is bigger than three and I have three generators e2 e4 e6 and in fact it doesn't work with fewer so now I want to concentrate on the theorem which is theorem b so theorem b also I'll have to sharpen in several ways first of all I should now admit that I lied well as it's written it's correct because I didn't say what I meant by linear differential equation later I did tell you what I meant I said linear differential equation to me means with polynomial coefficients and then this thing should be what's called a Hobbes module I suggest I just don't worry about it some of you know some if anybody wants to ask later certain modular functions are and certain aren't and all the ones I'll show you are so if you take a random modular function so for instance j that we had yesterday yesterday we had one interesting example of a modular function as opposed to modular form that was the famous j function defined as e4 cubed divided by delta so it started q inverse plus 744 plus 796,884 q and so on very famous function this is a Hobbes module Hobbes module is of course German as you might guess it means principle modular function so it generates the ring of all the field of all functions and j squared isn't but it is still a modular function so as I just said if you look at all modular functions for a given group they form a field and if that field is simply generated by one particular element like j then that's called a Hobbes module and in all the cases we'll look at that's the case otherwise the theorem is not false but the linear differential equation will have algebraic outfits rather than polynomials it's a small detail I just didn't want to actually lie but let's leave it for now ok so let's look first what this theorem says and why it's not nonsense because as I've written it the theorem is nonsense and somebody should have complained and particularly the students who are supposed to still be paying attention to what they hear when they hear something wrong they should complain because what do I mean when I say that f satisfies a linear differential with respect to t well I can only mean one thing and I do but it's slightly wrong unless you interpret it correctly which is what I'm coming to it means that you change the make a substitution you change the name of the independent variable from tau to t so what was originally small f of tau which was a modular form of some weight k now it becomes a function capital f not of tau but of t of tau but this is so for some for some f of t which might be you know the function power series and then the claim was that there's a linear differential operator l which annihilates f and l is some kind of a combination of pn of t some polynomial times d to the n over dt to the n where n goes from 0 to the order which was supposed to be k plus 1 so that was the claim but there's a big problem with this theorem namely I can't write this because if I replace tau by gamma tau which is some a tau plus b over c tau plus d for my group then f of tau as we very much know by now does not remain invariant it changes by this famous factor but t is a modular function so t of tau just goes to itself so now we have a function capital f which at the same value t of tau takes on two different values and I remind you professional mathematicians forget this amazingly often a function is something that assigns to every number one value that's what we mean by a function there are no many valued functions it's an abuse of terminology of course they exist but there shouldn't be called that ok so in other words this f cannot exist you cannot have that for all tau in the upper half plane that you have this simply because they're different tells with the same t but different values of f so that's certainly impossible but you can have this locally and by the way if I did things locally to come back to your question then of course I could take all the roots I want and we in fact sometimes do that so if you take a point in the upper half plane and you take a small neighborhood now that small neighborhood you will not find if it's not a fixed point you will not find two points that are equivalent under the group and so you won't run into this problem nothing will stop you locally from inverting t of tau and writing your function which was a function of tau as a function of t and then that function will satisfy differential equation but then by analytic continuation it still will if you move around the polynomials can't change so as you move the different neighborhoods you'll always have the same operator indeed this operator will work everywhere so the operator is well defined but f isn't but that's just what you want when you do linear differential equations you have monodromy if you have an nth order differential equation you have n solutions if you pick a basis n at a point and then you move around in a closed loop which goes around the singularity of the equation then you get also basis of n solutions but it's not the same n there's some n by n matrix and if you keep doing that for all the loops you get a group of matrices called the monodromy group well that's perfect because that's what we want we have a many valued function so it's just set up to be and in fact I'll reveal now the secret the formal proof there's I in my book the 123 that you thought I've already bought there are three different proofs given of theorem B the three together take up one page it's not that hard but it's such an important theorem I want to give different points of view so there are three proofs and one of them tells you what all the solutions are so the theorem says as I just wrote that there exists an operator of order k plus one which kills f so therefore the kernel of L and the things annihilated the solution space of this differential operator will contain f but we know that the solution space of an nth order differential operator has to be n dimensional so here we need k plus one functions and what are they well it's not hard to guess at all it's simply tau f of tau f of tau up to tau to the k f of tau and that's exactly why there are k plus one of them because you go and now you see this space is wonderful if I call this space I mean I'm not going to do the proof along these lines but I wanted to mention this if I just call this space v of course it depends on tau it's not a good notation locally it's a collection of functions but I don't mean this collection I mean the space spanned by that was a basis and you take all the new combinations so in other words it's equal to f of tau times polynomials in tau of degree at most k as a vector space so it has dimension k plus one but now you see that that space is very nice because if you think that f of a tau plus b over c tau plus d is equal to c tau plus d to the k times f of tau then if I call fi is tau to the i times f where i goes from 0 to k then you see that fi will be fi of tau multiplied by oh my gosh can I do this yes by a tau plus b over c tau plus d to the i but then of course this c tau plus d I can change k to k minus i and just kill it and now you see that c tau plus d to the k minus i times a tau plus b to the i is a polynomial of degree k so it's still in the space so in other words if fi is any function in this space then fi of gamma tau is also in the space there's no other morphe factor anymore it took care of itself what was previously the c tau plus d to the k is now taken care of by this collection and that's the basis of the actual natural proof of this theorem but I thought I would give you also a calculational proof at least in the case when k is one how you actually write down the differential equation and then I'll turn finally to examples well just to relieve the boredom where to make it less abstract I forgot to give the example of theorem one so I backtracked slightly to make an example of theorem one of theorem A, excuse me that was the non-linear I take, remember that the theorem wasn't just true for multiple forms but quasi-multidiforms so it's easiest to take e2 it's the smallest and it's the only one where the equation isn't it gets more and more complicated so if f is e2 and now I think I take my new prime then you just check using so I gave you the formula how to differentiate e2 but then you have e4 then if you differentiate again you'll need e6 but then you can differentiate as many times you want you'll never need more than e2 e4 and e6 so if you take f2 and differentiate three times you'll get some polynomial very easy to compute in those three variables if you take f and f double prime you'll also get a polynomial and if you take f prime squared and multiply by three halves you'll also get a polynomial and when you check you'll find that this equation is simply true using those so this is an immediate consequence of Ramanujan's formulas and this is actually a famous equation connected with with the Pandevez story and it's called Shaziz equations Shaziz equation and so it's studied in I don't want to go into that at all and I don't know much about it anyway so that was an example of the nonlinear thing which I didn't want to talk about now but I just remembered that I haven't given you an example so now let's go back and I want to prove this theorem now in an explicit way and then I won't give you the examples yeah it's a wonderful question the answer is in general no but there are two necessary so let me say louder because you didn't have a microphone the question is can you invert if somebody gives me a differential equation linear with some monodromy group can I somehow can I say that it's multidar the answer certainly not it almost never is I'll give examples in a second this is a super rare phenomenon among the multidary equations but the monodromy group must satisfy two properties which I'll come to a little later one of which is easier to check algebraically one isn't but if the equation came as a so called Picard-Fuchs equation which I've mentioned before for instance of Affirmative Calabios the second property is automatic and in that case it's conjectually always multidar but it's not proved so the answer is it's usually not multidar when it is we have criteria that we think are necessary insufficient they aren't necessarily easier to check but some of you can check them and we think we have the picture but that's a very very good question and I'll come to that a little I hope I don't know how I'm doing on time worse than yesterday so I thought today I had so little material I could go really slowly I hope you noticed but I'm going to run out of time nevertheless I do want to give the proof of the theorem I was going to give it at the end but I'm going to give it now right now so let's prove that this theorem be explicitly as I said the true reason is what I told you that the kernel is f of tau times all polynomials to create most k let me assume k is one just to simplify life a little so luckily I didn't erase my rank and cone things so let f be a multitude of form of weight 1 and t remember was a multitude of function I'm not going to write it all again because it's still here f is a multitude of form now of weight 1 t is a multitude of function so it is weight 0 now I look at t prime the derivative well 1 over 2 pi times the derivative that's a detail then that will be a multitude of form of weight 2 because if you think of the calculation I did at the beginning the derivative of a multitude of form of weight k failed to be multitude of weight k plus 2 but the failure was k times something so if k is 0 you're ok there is no failure so the derivative of a multitude of function is a true multitude of form not a quasi multitude so therefore that means that I can make the expression f and remember I had the bracket you don't have to remember because I carefully didn't erase it the bracket of two multitude of forms here's the definition but who cares is something whose weight is the sum of the weight so this is weight 1 this is weight 2 and then you have to add 2 so this form of weight 5 but I don't want to keep that I want to divide by f and by t prime squared let's call this a of tau then this has weighed 0 and similarly I can take f and itself now if I took the rank and bracket I would get 0 because this is anti symmetric but I can take the second rank and coin bracket that's why I wrote it ff2 and that will have weight 2 plus 2 plus not 2 but 4 so it is weight 8 no f at weight 1 1 plus 1 plus 4 is 6 it's much better because it's what I wanted and I divide for summary by 2 times f squared times t prime squared and I even put a minus sign and I call that b of tau and now this will also be a multitude of form of weight 0 simply because of the properties of the rank and coin bracket that I told you so this will be a very calculation proof quite sure it will give no insight the insight is what I told you before that the solutions are f of tau tau f of tau up to tau to the k this will be computational but it has the advantage of giving you the answer of giving you the differential equation explicitly so we have these two functions I hope this was clear up to now t is weight 0 therefore its derivative is weight 2 f is weight 1 by assumption and therefore these rank and coin brackets are weight 6 what was it 5 and 6 respectively the denominator also is weight 5 or 6 respectively so these quotients a of tau and b of tau have weight 0 but remember that I told you that the Haute module which is what we actually need has the property that any modular function of weight 0 is some rational function of t so all modular functions can be written as rational functions of t of tau so a and b a of t and b of t which are just functions of a single variable are simply rational functions quotients of polynomials very very easy ok and now I simply write out this I cannot do by heart and I don't think anybody could be expected to remember the calculation by heart if I now take the second derivative but not in tau but in t of f and then I take a of t times the first derivative in t but let me write it like this this is going to be my differential operator and I claim that this operator is going to kill my function and then I'm done I've given my second order differential equation because a and b I said I needed polynomials they're rational but then I just multiply this by common denominator so I want to show that this is 0 so let's just do it well when I differentiate in t that's the same as differentiating in tau but then dividing by the derivative of tau well sorry it should be prime so if I have f if I differentiate in t that's the same as differentiating in tau with the 2 pi i and then dividing by t also with the 2 pi i so it doesn't matter so that's the first derivative and now to differentiate a second time I do the same I put another prime and t prime then I just continue a I copy from this formula so if you look at the definition that it doesn't matter at all if you're taking notes don't bother if you want to see this calculation of course it's written in the book so here we'll have again the derivative in t is 1 over t prime f prime because it's not the tau derivative which is f prime with the t derivative and a is this factor that's the rank and cone bracket divided by f t prime you don't have to follow every detail I want you to see that it's a two line calculation that it's very elementary obviously it takes a little time to check every step so now finally I write b which is this expression I'm just copying it the 2 goes into the definition of ff2 I hope I didn't make mistake f and if you multiply this out you just find that it's identically 0 for any function f so that's the end of that story so you see that that proves completely explicitly if you give me an f I can actually work out a and b as multiple forms and then write them as rational functions of t that's my differential equation I'm going to make a one minute pause and clean sub-boards and then you'll have only examples no more theory now I don't need any of this all gone so I'm going to give you three examples two are very classed and very simple also very beautiful and the third will be the up every one so example one so gamma remember with sl2z in my notation it is a subgroup which is index 6 ah psl2z don't worry it's up to sign the matrices forget it gamma of 2 is the matrices abcd such that a and d are both odd and b and c are both even actually it's enough to say b and c are even and d have to be odd so in other words this is simply congruent to the identity matrix modulo 2 so if I'm really correct I should put here psl2z which means matrices up to sign because a tau plus b over c tau plus d to the negative so that's really the group but I don't want to be that pedantic anyway this is the subgroup and here we have the help module j of tau but here we have a help module called lambda of tau so it's an expressive function and let me write it out to give the example so this is a very classical function it's due to the genre this is the famous the genre form of elliptic curves I'm not sure if you gave the form as I now gave so that will remind you that I already had I wrote it before and took it away theta 3 of tau is the sum it was the q to the n squared we've already had but I changed tau to tau over 2 so this thing is some modulo form of way to half since I haven't really defined it just forgive me but if I squared which I will in a second it would have way one which I have defined this is on gamma of 2 and similarly there are three of these Jacobi theta series called 2, 3 and 4 1 for some reason isn't there there's a very good reason so if you simply relate the same thing q to the n squared over 2 but instead of shifting over 0 summing over 0 plus or minus 1 plus or minus 2 you sum over plus or minus a half plus or minus 3 halves etc that's called theta 2 and this is also a modulo form in the same group and actually there's a very important function that the students who came afterwards yesterday so because in answer to a question I wrote it out and I used to give it yesterday it's the Dedekind theta function and it's defined exactly like the delta but without the 24 or rather there is therefore 24 here so it's simply the 24th root of delta but it was delta was found by Ramon and John in 1916 this was found by Dedekind in 1868 so it's much older and it's actually a much simpler function this of course is way to half actually on gamma where you have to interpret this the right way so on the full group that's because this it weighed 12 and 12 divided by 24 it's a half so this is way to half and then each of these functions have a very nice representation in terms of theta function not that it matters but just to to make it really explicit so the other one is 2 times 8 of 2t squared over I probably made a mistake but this is what I wrote in my notes maybe it's correct now the claim is I'm going to apply the theorem my t is going to be lambda and my f is going to be not theta 3 because remember in the theorem the weight had to be an integer so the weight k had to be an integer I said that positive integer weight so I take the square of theta 3 that will of weight 1 and so the claim is that theta 3 this is an identity view I think to Jacobi I guess but I have no idea I didn't check it I expand it as a power series locally in lambda of tau then the coefficients are very simple you take the middle binomial coefficient 2n over n so 2n factorial over n factorial squared divided by 16 to the n and that's the coefficient that's the expansion so it's much much simpler than the q-expansion because the q-expansion of a multideform by delta has some unknown difficult arithmetic function like the Ramanujan-Tel function that depends on various things this is just a simple thing and in fact if I call this now as I was doing before f of lambda of tau so that f of t is the function I'll just write it again but without the lambda 1 over 16 to the n times 2n over n sorry this would be trivial 2n over n that would be a Newton's theorem that would be the square root it's 2n over n squared excuse me that would be a trivial theorem to binomial expansion if I take this function this has a name it's the wrong name it's called the Gal's hypergeometric function because Euler wrote huge papers about 100 years or 80 years earlier and Gal studied them and quoted Euler profusely but for some reason they got named after him anyway this is the has everyone seen the hypergeometric function or should I write it down it's terribly unimportant for what I'm doing this is a standard function that satisfies a second order linear differential equation which I certainly won't write down but I'll just give it the first coefficient is ab over c the next one is a times a plus 1 times b times b plus 1 over c times c plus 1 times 2 factorial and so on so this is called the Gal's it should be the order hypergeometric function and it satisfies the famous second order linear differential equation the second one is the Gal's squared is this hypergeometric equation a hypergeometric function which does satisfy a differential equation as promised in lambda so that's the first example completely classical example very very well known with many applications I don't need this anymore the second one is a little more complicated we take the first modular form on the full group as we saw yesterday as weight 4 and it's e4 up to a constant it's unique up to a constant e4 of tau so that will be my f and my t will be the j function but I prefer 1 over j of tau so if you remember j which I think I just erased but j started q inverse plus 744 and so on so if I take its reciprocal this starts q minus 744 q squared and so on it's integer coefficients but it starts with q so you can also write q as t plus 744t squared and the further coefficients are completely different it's an invertible expression you can write locally q is a power series in t or t is a power series in q and this e4 remember started 240q plus 2068 q squared and so on so I can certainly write this then also in terms of q so e is also I don't have them written down but 240t plus dot dot dot where t remembers 1 over j so I can write my e4 as some function of t which is 1 over j and now that should satisfy fifth order differential equation so if you're a reasonable person even if you're an expert in mathematics you'll say this is not going to be familiar I knew the Gauss hypergeometric I know Bessel functions in the second order nobody carries around fifth order equations in their head nobody actually knows by heart the solution of a fifth order equation but of course that's not true everybody does, namely this function is the same Gauss hypergeometric function that I just had the parameters now instead of being a half, a half and one there are 12, 5, 12 and 1 but it's similar to the fourth power of that so if you know a function you also know it's fourth power it's an easy fact that if a function satisfies a second order in your equation then its kth power satisfies a k plus first order in your equation and so this is the example and so it is a familiar function even though you might think that you don't know by heart any functions satisfying fifth order in your differential equations but in fact after the fourth is such a one so those were the easy example by the way the first one I said I don't know I think it's Jacobi this one I do know this is the famous identity of Frick M. Klein maybe Klein and Frick they wrote several papers and books and the order kept changing so who was the senior author ok so now I want to finally 10 minutes then I'm going to have to just tell you the facts without I was going to give you some of the proofs a little bit of the proof I want to tell you the apiary application because I would think that you came for except since the titles that got lost in the internet maybe you didn't know but anyway I promise to show you this proof using modularity that's a to 3 is not in q so let me first write what apiary did and it's an absolutely beautiful theorem which now has several proofs he gave one modulo that some things were a bit mysterious Berkers gave it actually two more and I think by now there are others so to find two sequences an for apiary and bn for I don't know for apiary well well before I give I'll give you a little table so that you get a feeling for how they look and if you're taking notes leave a little bit of space as I'll also do so the first one the an which are called the apiary numbers although he found both sequences well I know them by heart they start 1573 1445 the next one is 33001 but then you define the second sequence I'll write it down in a second and maybe some people can't see if I write any lower because of that thing so let me move translate to the left so I define them by recursion and the recursion is very beautiful or very ugly depending on chasing such things if you already know an and an minus one then an plus one is one over n plus one cubed times a somewhat complicated looking cubic polynomial which is 34n cubed plus 51n squared plus 27n plus 5 times an minus n cubed times an minus one and then bn is defined by exactly the same recursion so because this is a three-term recursion if I tell you two initial values then you get all the further values so if I start with one and five then I get these but if I start with zero and one I get some other numbers of which the first two I think are this but it might be 62931 and the next one I definitely don't remember and I need it let me drop the next one anyway for the moment here no that's yesterday's notes it's even worse I had a little table it's disappeared I'll have it in many of the papers here here 62 my memory is all wrong 62 531 over 216 the next one I'll write in a minute so if you look at these numbers so I now finish the set it's defined a and b and that's just a little table apply this recursion with the initial values as you see in the table one five and then the others take care of themselves okay then there are three statements so if you look at this let's say that a n and a n minus one were integers as you see here for the first few but then this right hand side is an integer but I have to divide by n plus one cubed so a priori a priori a n would be an integer but only with an integer divided by n factorial cubed can really be n for the same reason but in fact if you look at the table you see that the first five a n's are all integers and the first statement which he proved by giving an explicit formula and the explicit formula was certainly an integer but it was far from obvious that it satisfied the recursion and his proof was somehow didn't use non-standard things like a divergent series and infinite expressions and nobody could quite make sense out of it but I mean everyone knew it was right because he checked and he certainly had found it and it proved it so there's a mysterious that his proof is very ad hoc you write down an explicit formula which is obviously an integer that part is easy, well guessing it is not easy but then you have to prove the recursion and for that you need some kind of cleverness now the second statement is that b n well it's not an integer so you can see in the table and so far it looks as bad as it could be because two factorial is two two cubed is eight one is six, six cubed is two sixteen so it looks like we've gained nothing but in fact if you take the next number in the sequence it's really big one, one, four, two, four, six, nine, five but the denominator is only 1728 which I pointed out to you before and told you to remember is 12 cubed but four factorial is not 12, it's 24 so we've gained something and so the second statement is that b n is true but it goes to the denominator most n n which is not the product of the numbers from one up to n but the least come multiple which is much, much smaller the n factorial grows more than exponentially but the least come multiple grows like e to the n, that's a well known fact so therefore this denominator only grows like e cubed to the n so approximately 21 to the n so roughly the denominator of b n is something well it's less than 21 to the n but a n and b n themselves grow that's very easy to see like roughly 34 to the n because of the nature, because of that 34 so therefore that means that a n divided by b n divided by a n tends to its limit very quickly because both satisfy the same recursion so that part's very easy that it's quick but it's not obvious what the limit is part of Apiris theorem is that this limit is one sixth, say now three so now together this has a corollary and I was going to do it takes three or four minutes maybe afterwards if there's a question or in the time of the students I can go through if somebody wants to see but an easy and completely standard number theoretical argument tells you immediately if you have any numbers with these properties now you can forget the definition just these properties and call the limit whatever you want that limit is irrational automatically and the reason is very roughly any solution of the equation has to grow like either roughly 34 to the n or roughly 34 to the minus n because of the nature of the recursion so therefore because the limit b n over a n is a sixth, eight, or three if I take this difference it's much smaller than 34 to the n therefore it has to be roughly 34 to the minus n and that means that this rapidity the difference between this and this is 1 over 34 squared it's about 1100 to the n whereas the denominator is only 34 to the n times 21 to the n which is less than 700 and so this is converging a sequence of rational numbers converges to this limit and the difference is much smaller than the denominator the difference goes like 1 over 1100 to the n and the denominator is less than 1 over 700 to the n and that's enough to prove irrationality that's an almost trivial argument so the big puzzle was how do you prove these three things I won't say anything about those two it's slightly more technical but just a brief word about this so as I said the way that Appierre did it easy wrote down an explicit formula which is obviously an integer but it's very opposite it satisfies the recursion just for fun I'll tell you what it was but it doesn't help at all to see why the recursion is true you take the binomial coefficients square them multiplied by the square but this binomial coefficient that's an you can take the first three values so what I told you that's obviously an integer but the recursion is non-dominous so the question is why so he as I said gave this calculational proof rather ad hoc he explained where it came from there were of course ideas that led him to it but then Barker's found that there's something very beautiful let me define f of t to be the sum a and t to the n so this is the power series whose coefficients are these numbers 1, 5, 7, 3, 14, 45 and so on then the recursion give it a name I mean I won't use it because I'm not going to give you the details of the calculation here the recursion everybody knows who's ever written down linear differential equations already in university you learn that as exercises it's completely easy to translate the recursion with polynomial coefficients into a differential equation for the generating function with polynomial coefficients and vice versa so he tells you that L here's an explicit third order differential operator I want to write it down completely explicit you just write it down and that's just a translation of the recursion so so far we've done nothing this is standard so what we have to do is show that that's an equation and the nature of the equation is such it has a unique holomorphic solution at the origin unique power series and so that power series is just the a n satisfying this recursion but then as I said a priori they would have a huge denominator from the differential equation so the question is why if f is equal to 1 plus 0 into the power series with well it's automatically rational from the recursion complex coefficient starting with 1 and this L f equals 0 why does that imply that f has integer coefficients that was the whole mystery and now to come back to the question of how frequent this is a priori gave two similar proofs for zeta 3 and zeta 2 zeta 2 was not exciting because it was known by order 3 pi squared over 6 but it gave exactly similar proofs for zeta 2 it was a second order equation that equation belongs to a very natural 3 parameter family and so I once did an experiment I took 100 million values integer values of those 3 parameters and only 6 times did this integrality miracle happen up a reason 5 more and those with the 6 modular ones I could show they were the only modular ones so it happened but only 6 times up 100 million so that's a very rare phenomenon just to come back but anyway we're now in this check so the question is why is it true and so now I tell you well it's standard calculations once we've discovered it but it's an amazing discovery it was really one of the most beautiful pieces of number theory found so with Burgers Fritz Burgers who is my colleague from many years in Irtecht Burgers's discovery the s is correct of course Burgers's discoveries take the function so I'm going to take f of tau remember the a to function if you don't you remember at least that there was such a function it had weight of half so if I take its 7th power with 2 tau and its 7th power with 3 tau that is weight 7 7 halves plus 7 halves and I divide by a to tau to the 6th and a to of 6 tau sorry to the 5th so this is weight 5 halves and 5 halves so the total weight is 7 5s and 5 halves this is weight 2 on a group that happens to be called gamma 0 of 6 which means A B C D such that 6 divides C that group is called gamma 0 of 6 but it doesn't matter so that's a multiple form of weight 2 and T of tau is exactly well T of tau I know by heart you do the same but the numerator goes upstairs is the denominator the denominator is the numerator and now all powers are 6 instead of 5 and 7 so since 6 plus 6 equals 6 plus 6 that has weight 0 so indeed I'm in the situation of theorem B no longer weight 1 now it's weight 2 so this must satisfy a third order equation so if you compute the beginning of the expansion well it's very easy because remember eta well maybe you don't remember but I'll tell you again it was a product 1 minus q to the n which it has a very simple expansion as Euler found all coefficients are plus or minus 1 or 0 so it's very easy to take 8 of tau 8 of 2 tau you put 1 minus q squared and some you multiply it out and you get the beginning of both expansion it's no problem at all now the coefficients are very small because this is a holomorphic multiplier form they have just polynomial growth so they're 1, 5, 13, 23 the next one is 29 and so on and similarly T it's coefficient to grow a little faster but it's got completely explicit coefficients it's q minus 12 q squared plus 66 q cubed minus 220 q to the fourth and so on those are just standard q-expansions to stick in the expansion of eta but now as I said before since T is a power series in q with integer coefficients and leading coefficients q of course I can invert this q will therefore be T plus 12 T squared plus something or other and then I can substitute that into this and so this will become a new power series in T and guess what, well I'm sure you've already guessed what you see if you look at the first 10 coefficients are the up here re-numbers are exactly the coefficients at the expansion of a multiform of weight 2 in terms of multiform of weight 0 but then by theorem B we know that automatically any multiform of weight 2 in terms of any multiform of weight 0 is the solution of a differential equation of order 2 plus 1 3 so therefore you know that this satisfies the differential equation and it's explicit I showed you how to find it you do it and you find exactly the same equation and so they both have the same recursion the first 110,000 terms on the computer are the same but you only need two the first two coefficients are the same same recursion so they're the same and therefore everything's an integer because here I never left the world of interest I never divided by nq that gave a completely natural proof using multiforms a very very unexpected application and that's the one I wanted to end with today so thank you thank you very much BN was crucial for the proof which unfortunately I now erased then you write I'll answer but I want to show you but I erased it, I have to write it again up here is theorem I've got three parts I'm answering I'm saying it's clear that if you did the ratio of these things in the modular also you don't mean the rule in the proof you mean how do I get BN that's unfortunately another very good question which has to be a little evasive I'll tell you but using words that you haven't quite seen so it's the question indeed is the absolutely right question to make up here is proof work remember I needed three statements one was that AM is an integer AM was the first solution and I just showed you why that's true second I needed that BN was a near integer it's denominated with much smaller than it could have been in factorial but if this GCD which grows on the like 2.7 to the N and finally I needed that the ratio which is it's easy to see that it tends very rapidly to something and you can check it numerically but that it's 8 of 3 so the question is what is the so the modular meaning of A I gave you if I take this T and F which I just erased but you can still see part of it so it doesn't matter but F and T were some explicit products of A to functions then what I told you was that so T of tau was a modular function actually it's a HULT module not quite on gamma 0 of 6 but on a slightly bigger group that contains index 2 called gamma 0 store forget that but the statement was that if I took the sum AM T of tau to the N then this was the function I called F of tau which was a modular form in this case of way 2 on the same group which happens to be gamma 0 of 6 that's a detail important is that it's way 2 and so the question was what is the modern nature of the other function and so I can tell you it turns out it's the product of 2 functions one of which is called it is the same F and the other all called G T and so G of tau is again a modular form on gamma 0 of 6 and it's an Eisenstein series and well just for completeness I can even say what it is because it takes a line and then since I haven't defined Eisenstein series on subgroups the easiest just to define this one it is in fact 1 over 240 times E4 of tau minus 28 E4 of 2 tau plus 63 E4 of 3 tau minus 36 E4 of 6 tau and the next coefficient starts with 0 and then it goes on Q 14 Q squared 91 Q cubed and so on so then I called G but now this G tilde is what's called the Eichler integral G tilde tau is the same Q expansion it starts with Q but the next coefficient instead of being 14 it's 14 divided by 2 cubed and the next coefficient instead of being 91 is 91 over 3 cubed and the next coefficient instead of which I didn't even write being 179 it's 179 over 4 cubed in other words this is the function such that the third derivative of it is G remember my derivative just multiplies Q to the end by N so to integrate 3 times I differentiate by N cubed that is what's called an Eichler integral I would have to go much further into multiple forms than I did to prove the second part that it satisfies the right differential equation which is now in home it's the same operator in homogeneous but it does give the modular connection this thing has a completely multiple description it's F times G tilde G tilde is this Eichler integral now you immediately see if you believe that this part because if I go up to order 100 let's just take N to be 100 but if I go up to 100 G has integer coefficients so G tilde the coefficients are 1 over N cubed N going up to 100 so a common denominator for all of that is N of 100 Q the LCM so if I make any change of variables like between T and Q with integer coefficients I will still only have that GCD as a coefficient so it's obvious that this is true and finally to get the limit I said that the limit as N goes to infinity of BN over AN is the same as the limit when tau goes to some singular point which is a cost which I won't specify to take the nearest singularity which is roughly 1 over 34 so T goes to the limit which is roughly 1 over 34 it's actually quadratic irrationality and then you have to divide this by this but this is already this times something so it's just G tilde of tau but this singularity here in tau tilde just corresponds to G tilde of 0 and now knowing things about Eisenstein series tells you that this is 8 of 3 so indeed I would have said that if I did more time thank you very much for the kind question for letting me say it I have a simple question what does it what is it known about ZFN for all the values of N? well that is absolutely nothing to do with this series of lectures because although things are known to my great disappointment none of the higher proofs of any connection with multiple forms so we're in the wrong lecture room for that but I'll answer anyway because I understand basically very little if I'm very honest they proved some wonderful theorems but for instance certainly the two main things one would like to know well the first one was is 8 of 3 rational and that up here re-answered as I said 34, 36 years ago the next two questions are how about 8 of 5 is that rational and actually the question that up here we should have asked Euler didn't show that ZF2 is rational he showed that it's a rational multiple of pi squared nobody believes that these are both periods rational numbers are periods of weight 0 ZF3 is a period of weight 3 and symmetric conjecture that in all of mathematics no period of weight k can ever be a period of weight k prime nobody can prove it so it's not possible that this is rational but it certainly would be possible from that point of view for it to be a rational multiple of pi cubed because they're both at weight 3 so the actual natural question is if ZF3 rational multiple of pi cubed or to say it differently is ZF3 squared which is weight 6 equal up to q cross to ZF2 cubed which is also pi to the sixth well to 100,000 digits on the computer it's not true nobody can prove that so both of these things are still open after 36 years so if the three most basic questions up here re-answered one and the others remain open what was proved now by several people all of them I know is well the Rivo Arles should say maybe is the most made the most important contribution and maybe the second most is Zudelin but several people in various theorems what's known nothing is known when you divide by powers of pi but on this question of irrationality it's known maybe ZF5 is irrational is rational nobody believes it but it could be it's not disproved maybe ZF7 is maybe ZF9 is ZF11 is all that's open but they're not all rational so it's known that of these four numbers at least one is irrational and more generally it's known that if you take all Z to ends with an odd they generate an infinite dimensional vector space over q so they can't all be written and it's effective if you take the first 10,000 you can say that has at least dimension 1,000 over q but it's very disappointing because they're all meant to be irrational anyway frankly nobody cares except to make themselves famous but mathematically this is not a reasonable question this is the reasonable question are the Z to ends over pi to the n irrational and that's not even known for 3 so we know very little and the little we know is not worth talking because it's not 1 to the 4 thank you very much let's say a thing about them again