Okay, so it's my pleasure to introduce Don Seige's third lecture of the series and He will speak today on quasi modular forms and counting problems in topology and group theory So Thank you so Today will be a somewhat different subject from the first two well in fact every day is Somewhere different subject from every other day the common theme of course being that modular forms are always very much present I Wanted before I start for today to make a small Correction it is something I said on the first day. It wasn't wrong, but it was maybe Confusing and any way that will be a nice you can relax because you've seen this but on the first day I talked about applications of Modular forms in classical number theory like Diphantine equations or other fields and among the many many problems that I mentioned there must have been I Think ten Applications one was the very classical problem given the number n is it a sum of two squares and I gave a complete Answer and even a formula for the number of pairs a and b such that a squared plus b squared equals a given number n and that formula I can even remind you of the formula the number of Solutions that this is the number of pairs a b in z times e Such that the equation is true This was given my explicit and very simple formula you sum over all devices of D such that D is odd Minus one to the D minus one over two and then I did similar formula which I won't write up again for a sum of four squares But then I talked later about Sylvester's problem, which is to decide Whether a given numbers is sum of two cubes and I gave answers in two special cases one was twice a prime That's a theorem of such a one was a single prime. That's a theorem of Fernando and myself By quite a few years ago But then it struck me afterwards that I left out one important thing here Implicitly when I write this I mean that a and b are integers as I wrote here and Here they're integers if you ask simply is n a sum of two squares Then it's not quite obvious but true that if you have some n which can be rented represented as a sum of two squares even using rational numbers Then you can also find a solution using integers So if you just want to know is there any Representation it doesn't matter here whether you say it over q or over z and here it also doesn't matter because the answer is always Yes in both cases, but if you ask for the number of representation, of course here It's the number of integral representations, so I implicitly meant integral, but here I forgot to say it here It's absolutely important that you have to take a and b you have to allow rational numbers So if you only take integers, it's kind of a bit random They're only a handful for given n a handful of possibilities Well, the fan isn't too big and it seems to have no special structure. There seems to no good interest So when I gave theorems that said that certain numbers like for instance every prime of the form four modular nines every prime There's a 4k plus 9 so every prime which is congruent for multiple nine implies this is I Think this is now actually known due to a theorem of elkies that he did or did not ever finally publish Then it's true that this almost true, but only if you allow rational squares So if I take an example if I take 13, which is 9 plus 4 then according to this theorem It should be a sum of cubes, but if you think of the cubes, they're 1 8 27 and even if you allow negative cubes, which you should Then you you see that there's no way to get 13 because 27 is too big and 1 and 8 to add upright So it certainly doesn't work with integer cubes and in this case You have instead 7 3rds cubed plus 2 3rds cubed So in other words 343 Divided by 27 plus 8 divided by 27 is 351 over 27 which is indeed 13 But I just wanted to correct in case I gave a wrong impression and while I'm on the subject I should say that this equation and I mentioned at the other time And that's why Modular forms can be used if you fix p like 13 and think of this as an equation in two unknowns Let's call them x and y x cube plus y cubic plus p you could draw a graph that will be an elliptic curve And an elliptic curve is a group and so if you have one solution you have twice it in three times So you get infinitely many for free This was noticed already by Dave Funtas in the third century in some very special case Of course, he didn't know the the theory and Fermat in other special cases So in this case for instance as well as this you could also take 2513 over a thousand and five cubed and minus 1388 over a thousand and five cubed and I'm sure that all of you can immediately see that that thing is again 13 So I just wanted to mention that first while it's very beautiful that if you have a solution you have infinitely many But in a rather tricky way and quite big and secondly, I wanted not to Give the impression of having said something that isn't true that we know when a prime or any number is a sum of integral cubes So that's not part of today's lecture. That's part of the first lecture that I didn't say so today The main hero of today's lecture well apart from multiple forms, of course will be partitions of Numbers and so actually pretty much everything I say will be connected with partitions That's on the number theory side of the combinatorial side Then on the multiple forms side It's a particular function called eight of towel that I think I did not introduce the first day I'm beginning to forget what I've said and what I happened because after the lecture. There's a question Question answer session with the students very lively at least so far. I hope it continues and there There were many questions and I told but I told a little about partitions then but I'll have to repeat that and I think I'd I know I introduced a detail there But I think only there and then there will also be a topological side so the Part of the story will have to do with topology and this will have to do with coverings of surfaces But that part I'll be relatively brief So I want to start with partitions because it's something completely elementary This is not even number theory although partitions have very interesting number of theoretical properties It's pure combinatorics and so if I have a number n Say a positive integer then you speak of a partition event Well, there are many many ways to write it But let me first say in words, but it is let's say n is five and the idea is you have five objects And you divide them up into groups So for instance you might it might be five students in a class and you sense three of them into one room to work On a problem and one into another room and one into a third room to work on different problems But what we don't number the students we don't identify they don't have names So of typical partitions I divide up five into a group of three a group of one and a group of one So if you think of how many ways there are to do that well, you could either not divide the students at all Or you could divide them into two groups in which case the groups would be of four and one or three and two Or there could be three groups in which case it would have to be three plus one plus one or two plus two plus one Or it could be four groups then it has to be two plus one plus one plus one or three five groups in which case you would have this and So if you count one two three four five six seven you see that there are seven partitions So each of these so this is a partition of five But as I said the order doesn't matter. I don't number anything I could also put one plus three plus one It's gonna sit the same partition So if you want if you want a formal definition you could always as I did here put them in decreasing order So lambda is in the set of partitions of n or some people write lambda and then this symbol Lambda partitions n Means that lambda is a decreasing sequence Like here it would be three comma one comma one. So the sum of all the lambdas Is equal to n and they're all bigger than equal to zero And you can either stop with the finite number or if it makes you happy and it will make me happy a little later You can also add zeros to infinity. It still adds up to five. So that's a formal definition But this is the sort of Really what it is. It's just a way of dividing up n into positive pieces where the ordering doesn't count And then you have p of n, which is the number of partitions So in particular p of five. I just did that example. I wrote them all out and there were seven of them So p of five is seven So if you make a table, oh, I now realize that I Should have copied out a nice table somewhere, but I didn't if you make a table Then one has only one partition namely one two is two partitions one plus one or two Three has three partitions one plus one plus one or one plus two or two plus one if you prefer decreasing or three Four has five partitions. You can find them easily five. I just ate a seven six has If I remember correctly it continues eleven fifteen and twenty two I think are the next few I didn't specially prepare and there are various reasons that it's intelligent to say that zero has One partition namely the empty set is the only way you can do it, but there is one way namely the empty set Okay, you might argue about that, but for the formulas, it's better to set p of zero equal to one Okay, so this problem goes back to Euler who anyways my big Hero who invented sort of everything in modern mathematics in the in the first form you can see in order practically everything and So in particular he invented partitions and he also invented a wonderful thing called generating functions which Absolutely everybody and certainly every mathematics student. I think also everybody doing Physics or anything that uses mathematics should be aware many many are it's very well known But surprisingly many people don't think of generating functions that they're also called generating series as A key tool in the arsenal and it should be one of the best tools that Anybody who uses mathematics has it's a very very simple and very powerful idea And the idea which also goes back to Euler in fact in exactly this context is this if you have a sequence of numbers And you want to understand their property? So these have many many properties. I'll give you an example at random If you could ask the asymptotic Prince properties, which in this case is known These things grow less than exponential bit more than a polynomial They grow like this e to the square root of n times a constant But it's a strange constant pi times the square root of two-thirds Divided by four times n times. It's by memory I'll write three, but it could be two. I simply forgot to look it up It's one or the other if you take on your computer n equals a hundred you can find out immediately which They're very different except you won't know what p of n is but I'll tell you in a minute how to find it So that would be a typical property. So I'm just saying properties So it might be an asymptotic property or it might be a congruence property like the one that Ramanujan found you see that p of 4 is 5 p of 9 is 30 and They're both divisible by 4 and when he found is that whenever n is of the form 5j plus 4 Then p of this is divisible by 5 So you might want some property like a congruence and asymptotic property of many many others that I could think of and What Euler found and that what we know today is that a very very good way to understand the properties of any sequence of Numbers is to make the generating function. Actually I used it yesterday very strongly because at the upper e numbers a 0 is 1 a 1 is 5 a 2 is 73 But then I merely turned them into a power series some a and t to the end is 1 plus 5 t plus 73 t squared And that didn't seem to help much except that the recursion became a differential equation But then suddenly this thing Had the wonderful property that if t was a special modular function that this sum became a modular form So I just wanted to emphasize I should have said it already yesterday that it's very natural if you have a sequence to look at the corresponding Generating function you usually use x sometimes you might want a generating function with x to the n over n factorial There are many such case, but here we don't so it's p of n over x to the n and then Euler found and this Maybe I'll explain because you can almost do it in your head. He found that this factors as an infinite product Okay So it is this infinite product or in a math more modern mathematical notation It's the product m from 1 to infinity of 1 over 1 minus x to the m So this is very easy. He did not spend years proving this Namely the reason is this is pure thought if you multiply this out then the first term the right-hand side One over 1 minus x is a geometric series And if x is small and in a power series you always think of the variables being small even some sense infinitesimally small then 1 over 1 minus x is just the sum of 1 plus x to the i for all i So here we have the sum x to the a But then for the same reason 1 over 1 minus x squared is 1 plus x squared plus x the fourth and son So it's the sum x to the 2b and Then similarly 1 plus x cube plus x to the 6 would be the 1 over 1 minus x cube So this would be the sum x to the 2c and now the beauty of generating functions that when you just multiply this out Even if they're infinitely many factors. It doesn't matter Well, this is the sum over all possibilities a and b and c and so on all at least zero and the x one It will now be a plus 2b plus 3c and so on So if I multiply out the right-hand side and ask for the coefficient of x to the n I'm asking it how many ways can I write n as a sum of so many ones a ones? So many twos b twos There are many threes C threes and so on but then it's clear that simply the way of writing n as a sum of positive integers where you put them in some order And so it is the partitions of n and so I've just proved this identity of Euler, which he proved exactly the same way Okay, so he studied that he'd have had this wonderful idea of using generating function. He found this factorization a little similar but much simpler than well That's also very simple his famous Euler product for the so-called Riemann Zeta function But then he wanted to compute these numbers and he computed quite a few But very very quickly the way that I did it here just in my head or with you I said okay if you've divided five into three pieces It's 3 1 1 or 2 2 1 but try dividing a hundred into 57 pieces And you'll find you can't do it in your head reefed on paper because there are trillions and trillions of solutions These numbers grow very quickly as I told you less than exponentially But still if let's say n is a hundred two hundred over three is sixty seven squared of sixty seven is eight and a bit Times three in a bit. So this is about eight e to the thirty so about ten to the ten You're talking about a ten digit number So you can't just list the partitions and he wanted to actually compute them and he wrote down a formula that later The famous combinatorialist major McMahon actually used to compute by hand of course Up to three hundred and he got them all right. So and there are two hundred I think but they're absolutely gigantic numbers So what Euler did is he said This was a very natural question to ask in combinatorics. This was the answer But if you just look at this function Directly without knowing where it came from you would say what a funny thing to look at why are you taking the reciprocal of everything? Why not look at the actual product 1 minus x to the n which is 1 minus x times 1 minus x squared So he said that would be a much more natural thing to look at and Then if you know the coefficient of this power series Well, this power series multiplied by that power series gives one because that's the reciprocal And so if you have some answer for this, let's call them q of n although. It's a very poor name Then you'll live here q of let's call it something else q of l x to the l and that's one Well, that means the first coefficient p of 0 times q of 0 will be 1 so this is one and that's also one That's not very interesting, but all the higher coefficients are 0 So you'll get p of n plus p of n minus 1 times q of 1 up to dot dot dot up to q of n And this will add up to 0 so if you know the cues you get a recursion for the piece And if you if you're making a table well, then you can very very quickly get p of n if you know it's predecessors So he said if we're lucky and we can figure out a closed form of these coefficients Then we can find the partitions inductively so he did this and of course there were no computers He multiplied it out and I want to emphasize this 50 terms I mean now I can do 5,000 in one second in Paris on the computer, but Euler Obviously by hand in the 18th century did this to 50 terms by hand He did not consider it was beneath his dignity to do a big calculation if he was trying to understand how something worked And what he found was very very surprising and very beautiful He found that if you do this he went all the way to 50 I've just gone up to 22 That all of the coefficients are 0 1 or minus 1 so it starts 1 minus 1 minus 1 But the next two coefficients are 0 it's 0 times x cubed 0 times x to the fourth Then there's a 1 0 1 and the pattern was also there's a 1 then 1 minus 1 then 2 zeros then 1 Sorry 2 minus 1's minus 1 minus 1 with no space then 1 0 1 with 1 0 between them then 8 9 10 11 4 zeros and then minus 1 0 0 minus 1 so the pattern was very easy to see Namely the gaps between the were two zeros four zero six zeros and the numbers were two ones with no nothing between two ones with Zero between two ones with two zeros between and so on and then alternating sign So if you work out what that says then you find that his what he discovered experimentally can be written as the sum K for minus infinity to infinity Minus 1 to the K times Q to the 3 K squared plus K over 2 and these are the so-called Pantagonal numbers, so he called that the these numbers have nothing to do with those numbers, so they're both correct These numbers are the exponents of this series and these numbers are the coefficients of this series So that's I'm glad for the question because in fact That's why I was not sure of my memory because I remembered 11 15 22. It's a little confusing that there's 15 and a 22 in both and a 7 and a 5 and that I've no idea if there's any reason for that I think not but I've noticed it before so just to say it again. Let me call this Euler series capital P of x and Let me call this one Q of x But of course Q of x because of orders identity is 1 over P of x so the product P of x times Q of x is simply 1 and so to write it out in all its glory I should have done that P of x has these numbers as its coefficients So every exponent occurs not there are no gaps. There are no zeros but The coefficients are the numbers that you saw here 11 x to the 6th and so on Whereas Q of x which is 1 over P of x Starts completely differently 1 minus x minus x squared plus 0 x cubed plus 0 x to the fourth plus 1 x to the fifth Plus 0 x to the 6th minus plus 1 times x to the seventh and so on So those numbers are the exponents and the other co co fish are always 1 and the other co fish are simply 0 Whereas these numbers for the coefficient of the exponents to account every value and now indeed you see if I multiply Coefficient of x is 1 minus 1 is 0 coefficient x squared 2 Minus 1 minus 1 is 0 3 minus 2 minus 1 plus 0 is 0 5 Minus 3 minus 2 plus 0 plus 0 and so on the product of them is 1 So this is what Euler found now if Euler had been a genius, but of course Euler wasn't genius Then he would have now at this point said, aha, I should do two things Sorry, oh, sorry. I'm going ahead. I'm a genius and Euler wasn't that's the problem You'll see in a moment why I called it Q. Sorry. Thank you. That's an x So had Euler been a genius, which of course I say Euler certainly was then he would have said what Euler should have thought He should have thought two things and he missed them both. It was really very shameful one I need An x to the 1 24th Now why is that well, that's very simple because when you when you go to school and you know Euler had gone to school Of course quite a while ago Then you learn about completing the square if you have to solve the quadratic equation x squared plus bx plus e equals 0 Then what you do is you say well, it starts Let's pretend it's a square So then the thing I'm squaring would have to have a b over 2a and then unfortunately It's not quite right. So then I've subtract b minus b squared over 4a But you write it first by completing the square So we all learned that in school to solve quadratic equation if you see a quadratic expression Which is a linear term you shift the variable a little bit to kill it So here if I shift k by a sixth to the left k minus a sixth Then k squared will start will be replaced by k squared In the positive direction k squared plus k over 3. So this becomes that so if I rewrite 6k plus 1 as for instance m Then you see that m squared is 36 k squared plus 30 say a plus 12k plus 1 and So therefore m squared minus 1 Divided by 24 as it happens is 3k squared plus k over 2 and So what Euler should have done is looked at that and said ah ha 3k squared plus k over 2 is dying for an x to the 124 so my power series q of x shouldn't have been q of x at all It should have been x to the 1 24th and then my form would be the sum over all m of the form 6k plus 1 so m is now a I Can write it the 6z plus 1 it's 6 times integer plus 1 I could put m congruent to 1 modulo 6 if you like that and then I would have minus 1 to the k Which is now minus 1 to the m minus 1 over 6 times x to the m squared over 24 that's what Euler should have done and then secondly He should have said but this you really couldn't expect him to and I jumped the gun slightly of course x shouldn't be called x It should be called q because of course. This is a multivariable It's the e to the 2 pi I tell of multiple forms, but that was only to come You know 150 years later so that when he couldn't really have known that x should be called q But he could have seen the x to the 1 24th and I'm showing you that it comes in a natural way And if you remember in the first and the second lecture I think I had some theta series like the sum q to the n squared and various others I did theta 2 and a theta 3 where you shifted n by bit But every theta series has the form q to the length squared of a vector So it's never shifted by anything, but there may be congruence conditions So you may shift the vector, but the exponent should always be a perfect square I mean it's the square of the length so here. This is very natural and so if we do what he should have done Namely called it q, but he couldn't know that and if we also Know as we now know we should do we should think of this as a function of Q of towel where q is as always e to the 2 pi at all Then if I take this inverse power series that I've been talking about and then it should have a name And it shouldn't be capital Q anymore. In fact, it should be Eta. This is the famous Dedekind Eta function Which was not quite invented by Eta sort of by Dedekind certainly not by Eta It was named by him and he found its main properties But it was based he wrote this the paper in I think around 1867 a Riemann Who lived from 1826 till he was 40 1866 and when he died Dedekind was given the job of editing his unedited papers that he'd left with you know have finished Papers and notes and so on so various fragments and he wrote one paper on a fragment of Riemann where he found Essentially this function and worked out many things that Riemann hadn't completely worked out Of course Dedekind puts himself an extremely high-level mathematician, but not quite of the level of genius of Riemann And he proved what in a modern point of view is that it's a modern form because let's look at this function What Dedekind proved is that if you invert this towel and here it's important of the towel not the Q If you invert towel then the function only changes by a very simple function factory, which is the square root of towel over I and Since it's also clear that if you replace towel by towel plus one then you get just z to 24 Times 8 of towel where I use this kind of standard notation z to n is the standard n through the unity So if you draw the circle you take an angle 2 pi over n That's one and that's a to n So this is obvious for the function equation because if you change towel by one you don't change Q But you have a different 24th root e to the 2 pi i times towel has changed by factor e to the 2 pi i over 24 Which is this zeta so these two things tell you it's a modular form of way to half because here It's tau to the half except that it's a little strange And I actually never defined what multiple forms of way to half are but in particular if I now define delta to be a to the 24 that's the one that I did show you and That was the reason that we needed that funny 24th It's exactly to get rid of this zeta 24 now You simply have delta of towel plus one is equal to delta of towel and Delta of minus one over towel if you take the 24th power you get simply tell the 12th And so this exactly tells you as I told you on the first day that this is a modular form of way 12 So that's the dedicated a to function and the connection now is that the reciprocal of the dedicated a to function Is the generating function of partitions and you also see that Oiler is lucky because he wanted this recursion and as I wrote out the recursion. I think I raced it It's here, but I raced most of it. You had many terms That zero is equal to p of n Plus p of n minus one times q of n which is here minus But now we're very lucky because not only we have a closed formula for q But it's incredibly simple every q is either one or minus one or zero and most of them are zero So actually the recursion takes on the very simple form if I put p of n on the left that p of n is the sum of the two Predecessors, that's why it starts out like the Fibonacci numbers because each Fibonacci numbers the sum of its two predecessors But then this next that would be these two and Then this one kicks in so now it's minus p of n minus five minus p of n minus seven Plus p of n minus 12 etc So therefore, you know, you get that's very very quick if n is a hundred They're only approximately the squared of a hundred so about ten terms on the right and the coefficient are plus or minus one So it's extremely easy and as I said it was done by hand up to 200 by General McMain with no mistakes Yes, please well, I'm telling what Oiler did and the connection with multiple forms I'm not saying there aren't other ways when I'm trying to tell the connection to multiple forms. Yeah You're absolutely right. Yes, you're absolutely right. There is another way But yeah, but it's much harder to add up 10,000 numbers with a positive sign than a hundred numbers with plus or minus Because you put the positive at the left and the neg from the right. It's all positive signs and you do one subtraction It's not a big deal. You're completely right So let me translate the comment also you didn't have a microphone the others here when I went through the ones for five I sort of mentally group them into Dividing five up into one pieces two pieces three pieces four pieces or five pieces So if you make PJ of n the number of ways to do it There's an easier recursion and indeed Fernando will tell you that you know recently I had to program P of n not just the number but also the list of partitions That's if you want the complete list that's much faster And that it this doesn't work, but this is much faster But anyway, I'm trying to tell the connection with multiple forms than that isn't one So it is no I'm not saying there are no other ways to study partitions There are many books about partitions and of course there's much to be said. I don't know what you mean both formulas are true It's neither form is obvious. They're both have to be proved and when you prove them then of course they agree I don't understand the question is it obvious? It's it's not obvious that you get the same number, but both their theorems are true So there are many many recursions for any given set of numbers or there potentially there's no reason that the A number is a number the formula for the number is not unique You can write the number according to many many formulas So I'm just describing the one that that is relevant for my story I should go faster because I'm not going to I mean not faster speaking but faster mathematically because I'm not getting anywhere So let me say a little bit more about this function It's it's a wonderful function I mean yesterday when I talked about the lambda function I gave you two identities, but now I don't have them written down But one of them you might even remember because it's a little surprise We had this theta series, which is the sum Q to the n squared that I used the first day for talking about sums of two And four squares and then very remarkably this thing also has a product expansion Which is as follows. It's 8 of 2 tau to the fifth divided by 8 of tau squared times 8 of 4 tau squared So So in other words, this is the product overall m from 1 to infinity of 1 minus q to the 2m to the fifth divided by 1 minus q to the m squared times 1 minus q to the 4m squared Okay, so that's not obvious it follows from an identity of Jacobi So all of these are very classical things and so one can ask the question and I asked that many years ago If I have some product of adus some product 8 of d tau to some exponent Let's say, you know new new of D or N of D Of course, there should only be finite the many D So here I'll put d goes to infinity, but I could also put you know D goes up to some some finite value It's a finite product, but the exponents can be positive or negative like here We have a 5 we have minus 2's but sometimes this is Holomorphic in other words, it looks like I've divided by something But actually the quotient is this and it's a perfectly good holomorphic multiple forms there more form There are no poles so you can ask a question When is this holomorphic So I made a conjecture actually I made two conjectures and I'll write down the statement just for fun as a theorem So both theorems were proved This is many years ago 15 years ago by a student of mine, Merzmann in extremely complicated ways the harder theorem It's like an 60 page double induction It's never been published except in his thesis because it was just too complicated and then much more recent student who's sitting in the audience Show more part of Charger who's right there in his thesis. He did other things, but in particular he found a much Nicer it's based in some sense some of many of the ideas came from Merzmann that there are many new ones It's much simpler anyway, but I mean if I'm really rigorous I should say just this because he proved the theorem and the theorem is if The weight is a half Then there's a finite list And I'd conjecture that on the base of computer experiments. I found exactly 14 and just for fun Let me write them down because it's not that long But I'll use in a brief yet Where I'll just put the powers of you know the the multiple and the exponent So I would write this as just two to the fifth one to the fifth four to the fifth So I don't have to write all of that then the There's only one that involves one. There's there are two That involves one and two that's one squared over two That means eight a tau squared over eight of two tau and two squared over one There are three that involve four up up to four devices of four of which the third is the one that we just saw So that the fourth is this two to the sorry It would be if I wrote it correctly two to the fifth over one squared four squared Then there are four that involved the device of six one squared six over two times three two squared three Over one time six two three squared over one time six and one six squared over two times three There's some logic in them as you can probably even see and then there are four more that involve devices of 12 There's no particular point to writing them down if you're taking notes But they're of course in the one two three 12 to the fifth and the theorem that bears when proved is that if you have weighed a half Which simply means that this that there there's one more eight upstairs than downstairs So five is one more than two plus two because the weight of each eight is a half Then this is the complete list up to a stupid reskating of course I can replace tau by six tau and then there's a second theorem that I won't write down if the weight is One or three halves or any fixed half integer Then it's also a finite list if you only look at the irreducible ones in some sense I don't want to write it down carefully and that's the main theorem the hard one that Also, but I try you found a much simpler proof of So this is very pretty and in this connection. I should briefly mention a wonderful theorem of Sarenstark, so it's 25 or 30 years ago by now and the theorem is that all multiple forms of weight a half which I haven't defined rigorously But I've given you various examples. They're always theta series so we just saw that for the problem the a to function the a to function is this and then it's written as a Sum q to a square well square over 24 So it is a theta series and the Sarenstark theorem is that they always are so as an example if I take this one Let's just take this one So if I write this one out in case you've forgotten the notation it means eight of tau eight of four tau Eight of six tau to the fifth over eight of two three and six I think squared But this is quite a complicated expression But it has weight one plus one plus five is seven minus two plus two plus two is six So does there's one a to function net which is way to half So by the Sarenstark theorem this has to be a sum and actually it's given by exactly the same Formula as the one of Euler namely n is congruent to one mother of six and you've queued the m squared over 24 And the only thing that changes is for order. This was m minus one over two and now it's m squared minus one over 16 So just as an example of the power of the theory such an identity is quite amazing But not only you can prove it with modern forms, but you can predict it You know a priori that this thing has to have such an identity Well, I'm not even close to my actual subject of this lecture But I wanted to take a certain amount of time and in fact I want to still stay far from the subject by lecture and give one more example Because it's the famous Rodgers-Ramanutian identities which many Many people consider the most beautiful formulas in all the mathematics and if you just want a beautiful looking formula It's not one you can show to your friends if they're not mathematicians It's you know much more complicated looking than e to the two pi i equals one But it's it's a much much more interesting formula actually there are three formulas one is a pair So I'll write the pair first these are the so-called Rodgers there Actually, we discovered by Rodgers and then about 12 years later I think by independently by Ramanutian with a different proof And then he learned about Rodgers and they met and they wrote a third paper with the third proof the joint paper So the Rodgers-Ramanutian and then there are two of them and I they would have used X But I'll use Q right away You take in the numerator Q to the n squared and the denominator 1 minus Q up to 1 minus Q to the n Okay, let me just call this so I have a name for it G of Q I won't need it till in a few minutes So if I write this out It's 1 plus Q over 1 minus Q plus Q to the fourth over 1 minus Q to the Q times 1 minus Q squared and so on So of course you're going to expand each of these as a geometric series So this one starts 1 plus Q plus Q squared plus Q Q plus Q to the fourth and so on to infinity But then I get another Q to the fourth and so you can compute as many terms as you want and they grow very rapidly More or less like the partitions also e to the squared of n times a constant and then there's a pair and The other one is exactly the same In all respects I write the same thing but you change the exponent from n squared to n squared plus n So that's now 1 plus Q squared over 1 minus Q plus Q to the sixth over 1 minus Q times 1 minus Q squared and so on So that's equal to 1 plus Q squared plus Q Q plus Q to the fourth and so on And now the identity this is just you know computing the first few terms the wonderful identity Maybe I'll write in a different color to make it Stand out more The identity is that this thing well any power series that starts with 1 you can write as a product 1 minus Q to some power and Then 1 minus Q squared to some other power you can always expand the power series starting with 1 as a product 1 minus Q to the end Some power that's something you can see easily just by induction at each stage. You've adjusted the first n exponents to match the first n coefficient and then the next one You there's a unique way to make it fit So when you do this you find something very remarkable it starts 1 minus Q the next term is 1 minus Q to the fourth then 1 minus Q to the Sixth 1 minus Q to the ninth and so on you never get anything in the numerator and You never get any power of 1 minus Q to the I different from the first or the zero So every power is zero or minus one and the powers you get these numbers you quickly see One and four then six and nine is five plus one and four. These are simply the numbers Which are congruent to either one or four mother of five? So this is considered By many people certainly also by me very surprising and very beautiful Because even though it's you would say why look at this There's a reason and I should mention that these two functions and the whole story I'm about to tell this is used for instance rational conformal field theory for Conformal field theory for the rational conformal field theory with central charge 22 over 5 I mean this comes up in many place actually mathematics and physics But even without motivation what's surprising is that here you write something very uniform There's no five inside and suddenly there's a five and similarly the other one age if you do it It's one over and now what you get is the other ones not the multiples of five, but the other non-multiple. So two three seven eight and so on In other words you get the product n is congruent to two or three mod five Of the same thing So you notice each of these is just like orders form of the partitions But it's only this so this actually means that if you expand this out these coefficients like these two is the number of these coefficients are the number of partitions of n into parts which are quadratic residues of five and quadratic non-residues of five if you want the Sort of number theorists way of saying it and then there's a final formula Which is that's really the most beautiful of all if you divide one of these by the other So I assume most of you have seen the Legendre symbol But here I can just say it the Legendre symbol n over five is plus one if n is congruent to one or four Model of five. It's minus one if n is congruent to two or three mod of five and It's zero if n is congruent to zero model of five in other words divisible by five So that's the famous Legendre symbol in this case. So by what I've already written h over q By what I've already written is the product over all numbers of one minus q to the n to the Legendre symbol and now You don't you can't deduce directly from that but from the proof of these you find the world is reminiscent continued fraction Which is really a thing of beauty Because again, there are the fives on the left, but there are no fives on the right and you get something is it plus or minus now forgotten It should be plus of course You get it. I won't explain the continued fractions or if you've never seen them But even if you've never seen them if you expand just what I wrote Q to the cubed over one add one q squared over that add one q over that add one and take the receipt little Agree with this to four terms and that's what I or even more I think and that's what I mean in general So there's absolutely a wonderful continued fraction and this is the formula that many people say would be there Candidate for the single most astounding formula that you can just write down with no explanation if you know the Legendre symbol in all the mathematics Okay, that's not my topic for today actually at all Except that since my topic of the whole course is Multiforms I should say and that's absolutely crucial that of course these things are multiforms, but as usual Justice Euler, you know, you shouldn't you should use tell rather than q and Also, there's a power missing and it turns out to be here the minus one sixtieth power and here the 11 sixtieth power So if I take the same function of q but multiply by this rational power of q and call that a function of q Then I find that if I take the vector it's now a vector value to form something I haven't mentioned if I take the vector of g and h at tau plus one. Well, that's completely trivial because If you change tau to tau plus one you don't change q But you change q to the 160th by what I called z to 60 before so you'll get z to 60 to the minus one And here for the same reason I'll get z to 60 to the 11th So this is trivial and there's no meaning if I put any other power series at random This would be true that alone is a bore the second formula alone is also It's already surprising But it's the combination that makes it a multiform because I'm going to give you a formula Now for g of minus one over tau h. Sorry g of minus one over tau h of minus one over tau And if you remember the transformations tau goes to tau plus one and tau goes to minus one over tau Give you the whole multiple groups or anything that transforms Under those two is actually a multiform on the whole multiform group sl2z But here it's vector valued and the actual form that I can never remember So it's a matrix of course no longer diagonal. It's two over the square root of five I think I can remember the rest sine two pi over five Sine pi over five minus sine pi over five. No plus sine pi over five And here minus sine two pi over five Times g of tau h of tau So in other words this vector valued thing under all transformations of the multiple group Transforms into a multiple of itself, but the multiple is now a matrix So that's to show you a little more again of the riches of multiple forms and that there are more kinds than I've been telling you Here's vector valued and there are many many many other kinds of multiple forms Yes, sorry. What are you talking? Q equals one is not allowed. I remind you it's not allowed I remind you that q is e to the two pi i tau and that imaginary part of tau is positive Of course, you can take the limiting value. It's true if you take the If you take the that's true, okay, I'll make the comment It's a perfectly nice when you cannot take q equals into one here because that means tau is zero It's not in the upper half plane. So these functions make no sense in fact They're both infinite, but they both diverge the same way. So g of q g of one is infinite And h of one is also infinite and you're not allowed to take q equals one It's forbidden my q is almost strictly less than one because it's e to the two pi i tau and tau in the upper half plane However, it is true that g and h diverge the same way So if I put back which I've now erased what this thing was it was h of q over g of q and indeed that limit does have a limit And this of course you're quite right if I take q equals one This is the well-known continue fraction of one plus square to five over two and that indeed tells you that there is a Five floating around somewhere so it slightly reduces the surprise actually. That's why it was not going to mention that I thought of it, but now you've spoiled it by pointing out that the five actually is not quite as mysterious as it looks because here You see it but just to say that on the molecular level you cannot take q equals one, but it's a limiting value You can okay now I want to get back to my story and I've used up almost an hour and haven't come to the thing I meant to talk about in this third hour because well it was it's not an accident I planned I decided this morning to talk talk more about partitions first because it's more Elementary and the lecture some of them have been getting a little high-flown and also it's a lot of fun And it's very beautiful, and if you haven't seen it, it's worth having seen so now I want to come to my other subject so using a to remember yesterday I defined if I have a modular form or Any function of time to find f prime of tau is the usual derivative But for convenience I divided by 2 pi i so in other words prime is Usual derivative divided by 2 pi i or equivalently it's differentiating q and multiplied by q that at the advantage that it's kept rational coefficients, so q to the n prime Is simply n times q to the n that's very convenient And then I showed you that if you take eta prime and divide by eta and multiply by 24 Then just from the definition as an infinite product I won't even write it again. What you get is the sum Some of the divisive n q to the n I'm not going to go through this again What you get is this function e tau which is is an Eisenstein series, and it's very like e4 and e6 But it's not quite multiple and so what I told you yesterday, and I'm not going to go into any detail So we also remember e4 and e6 so in general e k is 1 plus a certain constant I don't know gamma k. Let's call it times the sum n from 1 to infinity Some of the divisive n to the power k minus 1 q to the n and gamma 2 is minus 24 as I just wrote gamma 4 is 240 Gamma 6 is minus 504 gamma 8 is 480 and so on so it's just some numbers that whose definition plays no role here But they're explicit numbers So then what we showed that what I discussed the first day is that if you take m star Which is the ring of all modular forms Then this is simply polynomials in e4 and e6. That's not obvious, but it's not a very deep theorem It's it's a one-page proof It's on you know the like the third page on my book of the beginning of the theory But still you have to prove it, but then this is in a larger ring, which is called quasi-modern forms I'll abbreviate q mf Although I also like to use q for quantum-modern forms. This is quasi Maybe I'll put small q mf for quasi-modern forms and this one well e2 I'm not going to give you the complete definition because it plays no real rule and in this case e2 satisfies that definition It forms a ring so every polynomial in e2 e4 and e6 is quasi-modern if you know the formal definition and then it's a theorem that they give the ring as you can just Take this as a definition so for our purposes Modern for quasi-modern just means a polynomial in those but it's a silly definition And there's a good one like the definition of a modular form by some kind of transformation properties We take me five minutes to explain. I don't want to it's not that enlightening and I'll skip it Okay, so now we have that and now I want to tell you a complete different story And that's my the story that I meant to tell in this lecture and I'll tell it now in in a curtailed form so in Physics and in mathematics more precisely physics is very big and a tiny part of it In fact many physicists say a part that's disjoint from it So it's not quite a part of it because not all physicists even like it But there's a thing called string theory in physics and there's a thing called algebraic geometry in mathematics and Whether or not this is physics in the sense of being testable in the lab sort of remains to be seen But in any case they're both wonderful Theories and in both of them you have slightly different versions of So of a thing called mirror symmetry Mirror symmetry I'm going to say I won't say really anything about the mirror symmetry part But it's connected with something else that certainly Almost everyone will have seen these words Gromov-Witton invariance And so I won't say just a word about what they want to do in these things Very very very briefly, but it's full of multiple forms sometimes and unfortunately in other cases One can't make multiple forms help and those are the hard cases But roughly this is the situation x is a very special kind of a variety of some dimension Well every variety of some dimensionals. I'm just gonna call it n, but it's what's called a Calabi all Variety the definition is very simple begin. I'll skip it because it plays no role So for string theory They're interested especially the case n equals 3, but I can mention that if n is 2 then these would be for instance What are called k3 surfaces? Those are the most interesting and if n is 1 this exactly elliptic curves, which I already talked about Again, don't feel you have to understand why it's the same because I haven't given any definitions But what you do in Gromov-Witton theory is you count that what they say is you count holomorphic curves In in x so that means you have a curve C So C is a holomorphic curve Meaning that it's a one-dimensional variety over the complex numbers So therefore it's a two-dimensional variety Over the integer over r so it's a surface, but it's got this complex structure So it's called the Riemann surface and as I'm sure you all know such a surface has a genus like this would be genus 2 Which is sort of the number of holes so the genus could be zero you have a sphere one You have a torus two you have this and so on and then you look at maps from C into x So see if this is a holomorphic map because we want holomorphic curves again I won't use any of this so if for if for some of you some of these words are completely meaningless Just don't worry about it. Let them wash over your head and then I'll come to the to the content Which is you know beautiful formulas. So if you have such a map. Well, then it also induces a map In homology and the homology in h2 of a curve is there's a thing called the fundamental class Which is unique so you will get some class in the homology of x Now again, even if you don't know every word here But h2 of xz whatever it is It's computable and so it's some lattice it might be z plus z and so you can pick some random vector in it Some vector that you like and if you know what these words mean you can ask How many holomorphic maps will have from C with this given psi as the image and so if I fix the genus I'm being very sloppy and what I'm saying is definitely not correct as it stands But roughly you get a number which could be rational for various reasons But you get a number quote number Which is the number of curves of a given genus G whose image is a given homology class and so you get a bunch of numbers and as you can imagine what you want to do is to make a Generating function so you have some variable like a capital Q which generalizes our Q to the power xi in some reasonable sense And you make a sum of xi and sometimes even a sum of a G with some other variable So that's what you want to do in Grom of Witten theory by the way mirror symmetry There's a word symmetry says that with one calabi all Actually with one family of calabi all says another calabi all actually another family and that the Grom of Witten Calculation on the one side which are very hard are the same as the Picard Fuchs differential equation on the other and its periods And that's just what I was talking about today And in lucky cases you're back in multiple forms and you can compute things and so multiple forms play a very big role in Doing this because of this mirror symmetry, but I won't talk about that so now the physicists day graph actually several others so Rudd and Douglas were also involved, but the main person was Robert day graph and he had the idea it's completely pointless for physics and it's kind of silly, but Let's embed curves not into something three-dimensional or even two-dimensional, but in something one dimensional So let's take the case when n is one and And ask what happens well if n is one you can't really embed a higher genus curve Into a lower genus curves So here's a curve of some genus g bigger than one like here two and downstairs You have a Calabi a one-fold but I already said a Calabi a one-fold soliptic curve We just means this curve of genus I won't call it g because that's g but genus equal to one and now we have a map and to say the fundamental class Here is just the whole thing and here the fundamental class generates Homology and so the only possible is that this goes to multiple of this so the only possibility for XI is There's XI some multiple n so this is here's my curve and here's a torus That it could be some multiple of the torus so therefore in that case will have n of gn It's simply the number of holomorphic maps Well, I'll just put maps and not I'm being so sloppy anyway that nothing I say will make it right so it's just a For you know free free association It's the number of maps correctly counted and there are lots of little providers from a surface See well, I've called this maybe I'll call the surface of genus g to a torus of Given degree and so I'm fixing g. Well, I've given so this is serves a given genus. So I fix that and Also given degree n because as I said this homology class whether or not you know about homology But in this case, it's very easy saying where the homology class goes is just saying the degree and the degree I remind you just means that if you take a generic point here There will be exactly and let's say n is for then above this point There'll be four points. So if I take a little neighborhood you usually draw this a sheet Well, I could take a little two-dimensional neighborhood then above that there will be typically four disjoint sheets However, they won't always be disjoint there will always be points where some of the sheets cross They're called ramification points So if I draw this schematically because it's too much work to draw these each time like this Then the sheets are just four lines The simplest kind of ramification would be only two of them to come together and the others wander off happily This is called generic ramification So over a bad point where some sheets meet only one pair meets in one point in other words the pre-image of a point Which here had four this point down here Had four pre-images But this point here has only three pre-images, but it's never less than three So it only goes down by one that's called generic and it's important for what I'm doing So I want but since I'm not giving any details It doesn't matter at all really, but I Didn't want to lie too completely so believe me there's a way to make this completely rigorous and Actually, there's a slightly easier number Well, I'll say the theorem now and now they could have thought of that and he found a way to compute them and Compute it a lot. He gave a closed formula in terms of the group theory of the symmetric group how to compute them and he discovered a property which Again, the fifth is somehow said they do I mean they called it a theorem in the sense They just said it's true and it is true But nothing that they wrote as far as any mathematician can see is even close to to approve It's just true because of some reasons from quantum field theory various actions and gauges and so it should be true But so I'll put theorem, but I hesitate to call it. So I'll put daycraft and company who was Rudd and Douglas and The other company was Kaneco and myself You gave it very simple I'm not saying it was a great breakthrough the great breakthrough was to discover it But still we gave an actual down-to-earth proof and the theorem is very very nice I make a function as you you'll certainly guess by now a generating function because we now know that's what you should always do So this remember is the degree So if you want you could say this is the sum over all maps from a given surface SG to T Generically ramified You just take them all correctly counted and you take q to the degree of the map and So that would be means that if you fix the degree to be n you're counting how many there are So this is the generating function counting that and you think of that as usual not as a function of q But as a function of tau though the faces which is called fg if q and then say something false So that's a lot of the form and the theorem is that this is always not quite a multidorm But it's a quasi-multidorm form in other words. It's a polynomial in e2 e4 and e6. That's really I think so That's the topology Coming in let me give you one example, which is completely misleading F to of tau I May be slightly lying, but let's say I'm not Because I just realized there's a change of notation, but I think it's correct It should be and I've no idea why I doubled everything. It's 5 e2 of tau cubed Minus 3 e2 of tau See, it's sorry. I didn't tell you it's actually a multidorm of a weight that you know And it's a famous number in theory for amount surfaces 6g minus 6 that I mentioned of the modular space So this fg is a quasi-multidorm and it's way to 6g minus 6 this for g at least two So if g is 2 6g minus 6 is then 6 and it weighed 6 well since the theorem was that quasi-multidorms are Polynomers in e2 e4 and e6. They're only three That's the partitions of three e2 cubed e2 e4 and e6. So it has to be a linear combination and Sorry, let me finish writing the formula before and then otherwise. I'll make a mistake and then I'll listen to the question And then the denominator is five one eight four zero which looks very rational But when you multiply this out, it starts with zero is it has to because you couldn't have a map of degree Zero and it also has no q-term because if you had a map of degree one They would be isomorphic, but they have different genera, but then the next coefficients are q squared So there's one map of degree two eight q cubed 30 q to the fourth 80 q to the fifth and before I answer the question. I just want to say one word so this Expansion you looks like it is all integer coefficients and indeed it does not just the first six It is all integer coefficients even though here. There's this big denominator fifty one thousand eight hundred forty Because there are congruences, but that's a fluke if you take f3 or any fg beyond genus two It will still have a rational denominator here a denominator But also here you'll get rational numbers because when you count these numbers there's some Technical thing you get rational numbers. So right now. What was the question? Who was it? And what? Okay, okay, so yeah, it was just I stopped in the middle of writing. Okay, so Yeah, I didn't see that was you sorry Okay, so That's the the theorem and as I say Kaneko and I proved it by an extremely ugly proof I mean it's only two pages And it was a nice paper because we introduced the word the function had been around for many years But we introduced the word quasi-model form and a formal definition that included them in their basic properties And now it's turned out very useful. It's used everywhere So that paper gets quoted a lot just because of the name and not because of the theorem But the theorem itself. It's a clever proof, but just a clever proof It's two pages very computational no enlightenment, but then there was a wonderful Generalization a few years later by bloch and O'Kuncoff and reading the introduction of their paper I discovered that I suggested that this theorem should be true and they proved it I couldn't remember at all and couldn't imagine how they even guessed it, but apparently I guess I don't remember I remember the Conversation with bloch but not guessing the statement. So there's a much bigger class big class of Combinatorial things all of which are quasi-model. Well, I'm calling them now quasi-model reforms So I have I do have a few minutes left I'm going to make an attempt to give you a little bit of the flavor of the bloch O'Kuncoff theorem and I'm doing a big piece of research with somebody. Well, he was in Bonn in Frankfurt mountain Miller connected with The well connected dynamical systems, but that's his part flat surfaces That's also his part Siegel Beach Constance many things in the theory of multi-ly and it requires a huge Generalization of this bloch O'Kuncoff theorem that we haven't yet got we know the statement finally after two years of work But it's still conjectural. We've only proved some cases But as part of that I studied the bloch O'Kuncoff formula and then I found a very simple proof And so last year to inaugurate being at the ICTP I gave my first course here at CISA actually jointly with Fernando Villegas and The theorem was exactly explaining the theorem I now want to tell you but of course with all the details with the proof Now I just want to give it a sketch. So this theorem to me is a little bit associated with the ICTP And actually the first calculations that I did for the project with Miller were also done on an earlier list with the ICTP and with Fernando's help. So it's got a definite TST no kind of a flavor in my it for me not for you, of course Just have to take my word for it Okay, so let me give you an idea and it's really a beautiful idea and before I come to the actual statement of the theorem First I'm going to give you the the framework. This is the idea of Bloch O'Kuncoff and all physicists among you will recognize immediately the nature of what I'm doing so Let me let me take P is the set of all partitions of All numbers So a partition is just lambda which I told you before is lambda one lambda two where the lambda eyes are Decreasing and go to zero and only find that the many are non-zero Those are lambda one greater than or equal to lambda two Etc down to zero. So every partition is a partition of some number the number it's a partition of is called You know size of lambda. So this is simply the sum of the lambdas It is finite and then as I said before I also write lambdas in Pn or some people write lambda bar And but I'll just use Pn. So this is just the union of P zero P one disjoint union of the various P ads and now let F But I see F is very poor because I've been consistently using F for Multi-forms, I'll use capital F and hope I remember let F from P to Q Be any function anything that you like So you just assign in some way that pleases you to every partition a Rational number now sometimes you may have made a stupid assignment The theorem will say that for a very large class of subjects But by no means all something nice will happen for the moment for what I'm going to do You could take any function and now I'm going to define a formal power series But it will have rational coefficients, but it's only a formal power series So again, if you haven't if you don't know that notion It means you write down and generating functions some AI Q to the I and you don't worry if there's any Q at all for which Convergence is just a way to Write down the coefficients, but you can add and even multiply formal power series So they form a very convenient ring and you don't care about convergence And so this is called the Q bracket or Q average and what you're doing as I say this the form of this Form then I'll say a few words about that is something very Familiar to physicists you take what you would like to do. It's supposed to be the average value So you want to take an average. So if you take an average you have if you have six numbers You know three seven well, let's have two three numbers three seven two The average is you add them up and you divide by how many there are But if there is the many then if you add them up you'll get infinity and if you add one plus one that many times You also get infinity. So you have to do something. So what I would like to do is take f of lambda For all partitions divided by the number of all partitions, which is the sum of one But obviously that's nonsense because there's actually many partitions and f of lambda will unless you chose something very funny Will not usually converge but now You put in a weighting factor and it's going to be this variable Q Which is just a formal variable for the moment There's no e to the two-by-tile here and here you put the n where Where that's Q to n but then since it should be an average here have to do the same thing with Q to the lambda and So this is called the Q average of the Q bracket and it's absolutely a familiar construction from statistical physics This would be you have a statistical system So you have a large collection of states here even infinite the piece would be the states and then to each state You have a certain Quantity that you want to measure and observable. Let's say in the quantum theoretic lounge that would be f of the state It's whatever it is But you want to take all states so you won't take the sum of all of them But that won't converge so you put a formula and our Q is as usual going to be e to the two-by-i tau But if I think of two-by-i tau to be the inverse temperature in statistical physics actually Inverse temperature times Boltzmann's constant Then that would be the typical weight that you put that if you think of lambda as the energy of The state then typically you would take you know eat the energy divided by Boltzmann's constant times the temperature So here that would become Q to the lambda and the point is that because this Q is small It's infinitesimally or in the cases we care about this will actually converge So though I said formal in all the case. I'll write down will actually convert So Q just has to be less than one in absolute value Then this Q to the energy is a weighting factor which tells you what the temperature is so the the higher or the lower the Temperatures the nearer Q is to zero or one and the more terms you have to take so in one direction kind of becomes rigid in the other There's a lot of Statistical motion, but you always want to divide by the thing you would have if the observable were just one So you want to normalize and that denominator. I won't use the word because it's totally confusing But that of course is what this is would always call the partition function and denote by capital Z So this is a completely familiar construction now You can forget those words because I'm not of course doing statistics and this is the definition so if you If you think of it then if you think of this denominator Well, how's the numerator the numerator? I could also write as the sum and now I take simply the partitions of a given number n f p n f of lambda And then Q to the end so that's the same thing I've just written it in a different way and similarly downstairs. I would have n from 0 to 10 Not just the number of partitions p of n Q to the n But we already saw that this is 8 of tau inverse times Q to the 1 24th So what the whole thing means is that if I multiply 8 of tau times this cube bracket Then I can also write this as the sum over all partitions F of lambda and now it's Q to the power lambda minus a 24th And the 24th comes from the so this isn't the theorem. This is just a restate So that's the definition except that I forgot I'm using capital F Somebody would have told me Okay, so that's the definition of this cube bracket and now Let me give an example so example There exists this is from the the calculation that I told you of Decroff and maybe Douglas. There is a specifics function I could tell anyone who wants to know in a question, but for now I just want to skip it It's a function of partition. It's actually z valued such that if I take the I'll first lie and then I'll correct it if I simply take all partitions of a given number n and I take this function nu of lambda to the power some even number So since it's even let me call it 2g minus 2 where g could be a genus Then this will be exactly The number n of gn that I talked about before Actually, it's not quite it's a slightly different number n tilde if you remember what n of gn was it was coverings For instance, if you know it was coverings of this by connected remount surface of a given genus Which means that the Euler characteristic here is 2 minus 2g But you could also disconnected things where you might have several surfaces above it So the coverings aren't always connected But you can take the union of this and if the sum of these Euler characters is 2 minus 2g for some g Then you get a new number and that's n tilde and there's a very easy passage from one to the other So the theorem I wrote down before which has gotten erased, but I want to put it back I know it's still here just the word theorem got erased the previous theorem was that the sum The generating function n of gn the number of ramified coverings of degree n q to the n was a multitude of form It's completely equivalent because of the easy passage from one to the other Just believe me you can equally prove it and it's three minutes to go from one theorem to the other You can do it with this so that theorem is this and so if you think what that means I can now restate this theorem. This is nothing else then Wait, something is slightly wrong it's not exactly like that, but Let me get it right. Let me just say what the correct theorem is so what I said is slightly wrong because of this denominator Very slight technical thing that I don't want to write down because it's just a distraction Let me write down what the theorem is so the theorem above this theorem is Equivalent to a new for so the theorem of daycraft and company and can echo myself the rigorous theorem is equivalent to the following theorem let G Be a positive number bigger than bigger than equal to two Then I take this function new that I told you about here that I haven't given you the definition to the power 2g minus 2 That's a specific function on partitions which associates the partition you the number new of lambda the power 2g minus 2 and then you see this sum here is exactly what was written here and so this whole generating function just what I want and so this Final theorem is that the cube bracket of this is a quasi-modular form of genus 6g minus 6 So that's completely equivalent to the theorem that we had But in a complete stated in a completely different way And it's not completely equivalent because daycraft first had to do a reduction to prove this statement That this sum of things that parts elementary using known things but coverings of surfaces and the connection with group theory But it takes some work, but we use that so we actually didn't prove directed the topology thing We used his formula so we actually proved the theorem in this form But we didn't know this beautiful formulation with this cube rack with this average value so now we have that and so when blog and a concoct actually blog and I had a long conversation about this theorem and I remember saying that it had to generalize, but I thought we didn't guess the generalization He found it later, but in their paper. It's written that actually I Guessed some some statement, but the theorem is very explicit, but I can't completely give it to you I think because I'm running out of time So I'll just say a few words about the nature of the proof I was going to so blog and a kunkhoff define a Series of functions and I'll give you an idea if I've timed us afterwards a series of functions Q Well, you could even have q0 of a partition would always be one but q1 of lambda q2 of lambda Q3 of lambda well, and you can imagine I goes on q4 and so on the reason I picked out these three is that these three I can tell you in closed format. They are this is exactly the function Well, I didn't introduce yet. It's the function called zero So q1 of lambda seems rather pointless, but I need it for my statement of the theorem q1 of lambda is always zero Q2 of lambda is n if lambda is a partition of n, but it's not quite There's a minus 1 over 24th, which we just saw in the theorem that I in the statement that I stupidly erased that the q bracket of Any function f times was equal to 1 over eta times the sum over partitions f of lambda and then we'd hear q to the power lambda minus the 24th So we already saw that combination lambda minus the 24th and finally q3 of lamb is exactly the function that I just called new of lambda So you see that Okay, and now they do they introduce a ring which is simply the ring Generated well as an abstract ring is just generated by infinitely many variables of weight 1 2 3 and so on But as if you can think of anything any element of this ring is a polynomial in the qi's and so you can evaluate it on Lambda by taking the same polynomial in qi of lambda it becomes a function. So any function here will have a cube racket and Then their big theorem is Locke-Augumkhoff of which then I found a very very simple proof and presented in this course This thing is always a quasi-modular form and to be more precise if f has total weight k So q1 is weight 1 q2 is weight 2 q3 is weight 3 and so on if the total weight is f is k Then its cube racket will have the same weight will be quasi-modular of that weight So you see this theorem is a very special case because new as is written here is q3 So this is the same as q3 to the power 2g minus 2 and so its weight is indeed 6g minus 6 So that's the absolutely wonderful generalization and The only thing that I owe you is to at least tell you what the functions are I was going to give a mini mini sketch of The key fact that you use for the at the end of the proof I give is one line, but of course like all the one line proofs There's a lot of preparation for that line So I can't quite give it to either the line or the preparation, but I'll tell you roughly with the qi's are So the last part of the lecture it's five more minutes You can forget multiple forms you can forget everything actually you can stop listing of course It's completely independent, but it's kind of cute. It's how you describe a partition So a partition you can do in in many different ways and the one I want to describe So it's called Frobenius coordinates, but I have a nice numerical example here So I'll write it. I'll take it so that I have it right So let's take the example the partition fifth n is 15 and the partition of lambda so lambda will be 15 It's the partition 5541 Okay, that's my partition my example and just to give you an example of the values Here's of this particular. I told you that we have this qi of lambda So q1 is always 0 q2 already told you the size of it 15 minus of 24th So if you're good at mental arithmetic, you'll believe me that that's 359 over 24 the third one is a little simpler. It's 10 the fourth one is way worse It's 176407 over 5,760 But then the next one is rather easy again There's a reason actually for the madness, but It doesn't matter just to say that this is completely explicit So when you make a partition what you do often is you represent it by what's called the young diagram So the first five you make a row of five squares the second five you make another row of five squares The next four you make a row of four squares the one you make a row of one square So you turn it into what's called the young diagram, which is an array of squares starting in the upper left in English-speaking countries, but in the in a different corner in France. So this is the English format and then Okay, so that's how such a partition would look and now we're going to introduce what's called Frobenius coordinates. So every such partition lambda will correspond to Frobenius coordinates a wonderful mathematician If you want to know all about his discoveries as Fernando Villegas because he's a big fan and his read many of his papers And this will consist of a triple. Well, it's an infant. It's a big triple There'll be an integer R. That's at least one and then there'll be a sequence strictly decreasing sequence of non-negative integers of length of that length I don't really need the art because you can read it off But it's and then a second strictly decreasing sequence independent of the first time they can be for the same Of course they feel like The same length so and conversely given any R and to decreasing sequence of interest at the same length You know it's two sets of positive integers of the same cardinality You can make a partition and here I can show you how he goes here are in this example would I can't draw a diagonal But you take the longest diagonal So this is the I like to tell this I told the joke in my course. It's my discovery. This is the H index I hope some of you are too young to have encountered this monstrosity called the H index This is the idea that you can judge a scientist. You can translate the scientist into a single number how good he is Called the H index like you know 73 that he's a wonderful scientist Three he's a terrible scientist and so the H index is you count how many not just how many papers He's written because he could read a right a lot and nobody reads them and not the total number of citations because you could Have written one very popular title You know multiple forms and sex and then nobody read as others So the H index you list his paper in decreasing order of citations So his best paper has five citations the next is five the next is four and then you look where they're diagonal How far it goes so the H index is three it means I've written at least three papers that at least three people have read It's 73 I've written at least 73 papers that at least 73 people have read It's not completely stupid, but it's pretty stupid. Anyway, this is the H index It's the length of if that's the publications Five citations five four and one then the H indexes are and now If I look at this then I have these diagonal squares Well, they're always there and then I've divided up the rest into four and three and one and here three and one So in this case the the for Venus coordinates would be three and then these are the lengths of the arms a for arm So before and three and one Which is strictly decreasing and positive Greater than or equal to zero I said and the then they're B Which are the bind of the legs it'd be three plus one plus zero. So in this case, this would be the Frobenius coordinates And I can actually now give you the formula for qi Will then be the sum I from one to well, it's actually one over k minus one factorial Times the sum I from one to r and then it's a I but you don't actually want to use air Plus a half you want to take those four plus this half. So this is the enhanced Frobenius coordinate So it should really be four and a half three and a half one and a half and here three and a half one and a half And a half and then you see it just adds up even nicer So I take a plus a half and then I take the moment Except that since the a eyes are above the diagonal the B eyes are below I take the alternating moment like that and the I should have been k So I guess this is I minus one factorial and that's not quite right Then you have to add a constant but the constant is zero for odd numbers to beat the zero is beta one is zero for instance So let's do it if I is One No This is also I minus one now we can do it if I is one this is one and that's also one So you have as many ones here as ones here because there are as many Arms as there are legs and so you get zeros. That's why q1 is zero now. Let's do q2 This is one factor here. It's the sum of the a plus a half. So that's the sum of these volumes minus I made another mistake Here you have to take these as negative numbers and nevertheless subtract So now depending whether when I was one this was one minus one But if I is to this is a plus a half minus minus B plus a half So it's the sum of these areas plus the sum of these areas. So that's the total area So q2 would be n which is the size of the partitions Except I have to add this beta 2 and beta 1 is 0 but beta 2 is not the odd ones aren't they're the coefficients of some easy Function it doesn't matter and so it's actually one minus a 24th. That's the formula that I showed you But then q3 again would be this would be one half times the sum A from 1 to R and then it would be now a I plus a half squared so it's a squared plus a I plus a quarter minus Be I squared minus be I minus a quarter so the quarter goes away and then you see that you're in luck Because a squared plus is even and so is this and so this is still an integer And then it's a theorem which means nothing to you because I happen to find you but the function you I told Well, that is new but there's an intrinsic definition. That's the formula. So the new function is this The functional lambda minus a 24th is That for I equals 2 and the function 0 is that so indeed the three special cases I gave you our special case of this uniform formula And now at least I've given you a complete statement of the block of kung fu formula and the one line proof is Very pretty, but you aren't going to get to see it unless there are questions. So thank you for today Don't be shy Apparently for the first time I was completely clear. I explained everything I hope that some students will say that they will not be shy and will have questions even if it's fewer than the other days The question sessions with the students on the first two days were absolutely wonderful very good questions And it was really at least for me fun and I was I hope that something will be coming back today So, okay, thanks


So there are a lot of things you can do to a person or their property. They will get you sued and sued good. But there are also several defenses you can raise in court. To start with, you could claim that the person consented or agreed to the action that hurt them, no matter what it might have been. It even counts as consent if they never come right out and tell you to do it, but they act in such a way that a reasonable person would assume that they have consented. You can't claim the person consented if they were obviously not capable of giving consent. There is a special exception though in many places where you can claim consent to help an injured unconscious person. It's not clear why someone would sue the person trying to help them, but it happens. You can also get in trouble for doing more than the person consented to unless there's some emergency that justifies it. While someone can give consent even if their consent is based on a mistake, the consent won't count if the mistake was caused by or known by the person getting sued. A famous case of this was where a woman invited her doctor to her house to help her give birth and the doctor brought some random dude with him who wasn't a doctor. The woman said it was okay because the second guy was a doctor and the actual doctor didn't correct her, so her consent didn't get him out of being sued. Consent also doesn't count if it's given under threats. Another way to get out of being sued is to claim self-defense. You can harm someone and not get sued for it if you are using reasonable force to keep someone from harming you or from touching you in an offensive way. Even if you're in a situation where force can be used for self-defense, that doesn't mean you can use as much force as you want to. The amount of force you use has to be reasonable. That means you can't use deadly force unless you're being threatened with death or serious bodily harm. You can even use force in self-defense if you think you're in danger and you're wrong as long as it was reasonable to think so. You are not, however, allowed to use force after the danger has passed. Words by themselves almost never justify the use of force in self-defense no matter how hurtful they are. You also can't use force in self-defense to avoid some distant future harm. You can use force to defend someone else without getting sued provided that they have the right to use force in their own defense. Some places will let you off though if you help someone who it turns out didn't deserve it. You can use force to defend property without being sued as long as the force is reasonable. Deadly force to preserve property is pretty much never reasonable. There's also the defense of necessity. It means when you're allowed to damage someone else's property to keep something much worse from happening. Not quite. Better. Moving on. Necessity is nice because then you don't have to pay damages or go to court. If you cause damage to save the public or a large group of people, then you don't even have to pay for whatever damage you caused. If you only save yourself, then you still have to pay for the damage you caused though you won't be liable for anything extra like punitive damages. And there you have it.


Hi everyone, Joe for jazby's casebrakes.com with a one box pick your team break number one from a fresh case of 2020 tops definitive baseball From if there's the fresh case right there big. Thanks all of these folks for getting into the action right here. I Don't know who got last spot mojo. I know who got last spot mojo in this randomizer here Oh, it was the the brave spot. All right. There you go. Thanks Chris. All right Now let's figure out which See which box we're gonna do so let's slide this out as Is right there So there's one two. That's a two One two and three will roll the dime. I like that one and then we'll go one two three four five six Two one two, so we're gonna be doing box one and we'll save the other boxes for for next time We can do that. I think we should be able to we could be able to do those tonight If you want to ladies and gentlemen Shallow box here. Does that mean redemption? We're just thinner hits. All right. Good luck ladies and gents first one is Miguel Sunno two-color patch 11 out of 15 twins Steve Locke with the twins and you can look up that number when you get this Steve and Go to mlb.com slash authentication check out exactly what date that jersey came from and you can see if you played a game that day It's pretty good same with the Reds Aristides Aquino to 29.  So that goes to Reds seven That'll be for Rory. There you go Rory So you can look that up mlb.com slash authentication, which I think is an excellent feature Hopefully it'll extend to more sports more products. There's no one Ryan Four out of ten Astros edition that'll go to New York Nice no one Ryan Then we've got a Hank Aaron. Wow Two out of three on that Hank Aaron going to Chris co and the Braves Nice Chris All aboard the Big Head Express 21-time all-star selectee we got 11 Nope 21 out of 35 mark McGuire and will Clark just one off of will Clark's jersey number The Bay Area teams right here A's are a number block team A's one is Adam Wilson the Giants Also a number block team no randomizer Adam Wilson There you go And there's Chris code. Thanks Chris Congrats Jersey and autograph Jim Tomei Philly's edition 38 out of 50. That'll be for Steve Locke in the Philadelphia Phillies Boba Shet nice 34 out of 50 on that Boba Shet That'll be for Richard Zitterman and the bluebirds The A's are playing the Blue Jays on TV right now and a nice one Soto Four out of ten three color patch and autograph We pulled the MLB logo man of his last night and here's another one for Scott Goodman and the Nationals That is a nice one box break ladies and gentlemen Thanks for watching. Thanks for breaking this. We as you can see we got a couple more to go So keep chasing on jazby's case breaks.com. I'm Joe. I'll see you next time. Bye. Bye


What's going on everybody? Got Kyle and Donna here with late sports cards and we are ripping 2022-23 top two AFA Champions League Sapphire five box number four, random team style. Did a random milestone for the breakout lands with PSG down to Oliver with Salzburg, Leipzig and Leverkusen. Can always find those results on our website guys late in sports cards. Got a results tab at the very bottom. Good luck, good luck here. Box one. Yeah, Ramars, I'm down man. Haven't seen one yet. Haven't seen one. This is about to be like box eight I've opened so far. I've only opened a couple for personals and then those first five today. Haven't opened a bunch but I know they're relatively hard to hit for sure, but we should see one here soon. Raheem Sterling, Thomas Moeller with Charleston and Wonder Kids is Carvalho for Liverpool. Rookie Bruno Iglesias for Real. Bruno Fernandez, Manu. Kubo and Refractor. Variation is Jude Bellingham. Nice one there, Dorman, the nice young English player. Jude Bellingham, the new Real player too. Nice variation there. Frankie De Jong. We've got Rookie Noah Sasa, Jared Bowen, and David Silva is going to be for associate dad there. Orange to 50. I believe he just retired, David Silva. Long time man, man city player. Patrick Sheik. Gonzalo Ramos and Mendes, Future Stars PSG. Aslani. Got a Sulamana Future Stars. Tomiasu and Fatahou, Wonder Kids. There's Holland for Man City. We've got Mario Lizzetti, Cherokee, AC Milan, Casamiro, and Giovanni Reina. Hey, thank you Stefano grabbing a spot in Almanac, guys. If you guys are interested in Almanac this weekend, it's Sunday and Monday night this weekend. Definitely check it out soon here, guys. First five sold out very quickly. Second five are coming in hot. Sunday and Monday? Yeah, Sunday and Monday. Luca, Graven Burch, Oblak, and Yellow is going to be number 299, Emil Smith Rowe for Arsenal. We've got a gold. Rookie Santiago Jimenez. Got Gorna Duov, Noah Lange, and our gold is number 250. Kai Havertz on these Sapphire selections. I believe these are about one per box on these. Or no, we didn't see any in the last five. Never mind. One per case or so maybe. Sapphire selections, Kai on the yellow to 50 or gold. Nice one there, Chelsea. Look, he got moved, too. I don't think he's on Chelsea anymore. I think he's on Arsenal now, right? The Ready Guys box, too. One per case, gotcha, Mitchell. I think maybe when I opened the last couple, I got a little lucky and maybe hit two of them in a couple boxes when I did a couple of personal. Might have been why I thought it was a little more common, but there's a Rodriguez rookie. Stanisic, we got Zlatan, and Cesco Wunderkidz, Benjamin Sesco, Valmark Rookie, Kohn, Antonio Nusa, and Leo Messi. He's going to be our variation. Nice one there, PSG on the Messi variation. Bergewein, Morata, Rashford, and Sobaslye. Number 250 there for Leipzig, newest Liverpool midfielder, Dominic Sobaslye. Got a rookie, Fabio Blanco-Barça, Degrassa, and Michiabu, rookie there for PSG. Doku, and this one is Amir Dedic, Salzburg. It's a lot of UPS track and just kidding. KDB, Tami Abraham, Bakayoko, and Nkunku for Leipzig. Tebow Kortua, Sadio Mane, Ari Kane, and rookie is Wunderkidz Fatahou. That one is green number to 70 for Sporting there on the green insert. Wunderkidz, Valentino Carboni, and another one, Cody Gakpo, Saka, and this one is Sapphire Selections of Hungman's Zone this time. Tottenham, got a second one out of this half case. Nice one there, Tottenham on the zone. Maybe if there's a numbered one, maybe you get more than one. Sometimes that does happen with the case hit type stuff. Alrighty guys, box three. Benzema, Future Stars is Sesco, Bellingham, and Wunderkidz Zalouski, for Roma, Lewandowski, rookie Alfie Devine, there's Zidane Iqbal, and Holland is gonna be our variation. Nice one there, Man City. The Holland variation, Matisse Tell, Morata, and this one is Takafusa Kubo for Sociedad number to 50, on the Orange, Rookie Enzo, Maida, Becker, and Pablo Marine, Rookie for Sociedad. They are hiding, come on now, Barca. Phil Foden, Olivia Giroux, Skarls Rookie, and Ali Cho Wunderkidz, a lot of Sociedad coming out of this box. Hey, thank you, Terry, taking a spot, Luminance, and John. John grabbing a spot in some Almanac and some Ram Divisions of NT. It's like John grabbed one in all the Almanacs there. Appreciate you, John. Yeah, guys, not gonna be available much longer. I will go ahead and say that for Almanac, second half, six through 10, guys. One through five sold out this morning quickly. There's a Garabi Future Stars PSG, another selections. There's Raphinha, Rookie Barrios, Rookie Douay, and this one is Kai Havertz on these Sapphire selections. We're Chelsea's second Kai. Purple. We've got Koku Rookie, Jadon Sancho, Tyron Buchanan, and our purple is going to be Ecare Bravo Future Stars for Real. That one is 11 out of 15. Nice little number there, Real. Ecare Bravo Future Stars. Two left, two left. Good luck. Hey, thank you, Kurt grabbing a spot in Almanac. Like I'm saying, guys, I don't think it's going to be lasting much longer. Every couple minutes a spot or two comes in. I think he posted them right about, right around five when we went live here. It's Pulisic, Fatih, Diketliere, and Eileen Jr. on the Rookie Juventus. Rookie Ali Dou for Frankfurt. Got Matisse Albin Rookie, Marquinhos, and our variation is going to be Jamal Moussiala for Barcelona. There have been some nice variations, some good players. Moussiala there that time for Byron. Hey, thank you, Josh, taking a spot in the Legacy of Luminance Mixer. Rookie Julian Alvarez, Durami, VVD, and this one is Green to 70, Palo Di Bala for Roma. The Legacy of Luminances. It's a big one. Benzema, Bragi, Weston McKinney, and Marquinhos, PSG, Kaku, Inter, Sesco Future Stars, there's Jude Bellingham, and Wonder Kids to Luski for Roma. Hey, thank you, and taking a spot in the NT and College iMac Mixer. There's Levendosky for Barca. Levendosky, we got Rookie Devine, Iqbal, and Marita. Rookie for Sporting. Fernandez Rookie, Matisse Tel, Maita, and this one is Bakayoko, number 250. Nice one there, Wonder Kids for Einhoven. Johann Bakayoko, Orange to 50. Come on, let's see an auto. Let's see an auto this last box. Kaku, Foden, Becker, and Pablo Marine Rookie. Associate up. Box five, got Rodriguez, Smosala, Marco Royce, and Sesco, Wonder Kids, Salzburg, Stanisic, Zlatan, there's a Philip Cohn, and Holland again on the variation. Second one here is variation, Erling Holland, Man City. Not bad at all. Second variation for Man City. Burgvine, Nusa, and this one is Harry Kane on the green. 41 to 70 for Tottenham. Iron's New Striker, Harry Kane. Chukwameka, Marata, there's Rashford, and Bitshiabu, Rookie for PSG. Say Rookie Fabio Blanco, Barcelona, Degrassa, Dooku, and Amar Dedich, Wonder Kids Salzburg. Hey, thank you, Jake taking a spot in Luminance, and Philip taking a spot in NT. Took the Jets and the Raiders. Thank you, Philip. Jake taking Eagles in some Luminance. There you go, Donnie. Appreciate you guys grabbing spots. Rookie is Gabriel Veron, KDB, Tommy Abraham, and Asugo. Rookie for sporting. Ooh, that's red. That's red. Tebow Courtois. Come on now. Bakayoko in Kumku. And red is Rookie Mohamed Alicho for Sociedad. Nice one there. Good Rookie there, Sociedad. That is Noah. Nice. Two of five on the Rookie Alicho. We'll mag that for you, Noah. Super low numbered on the Rookie. Very nice one there. And last pack here, guys. Saudi Omane. Got Harry Kane, Gakpo, and Phil Foden on our Sapphire selections for Man City. Alrighty there, guys. We'll do a quick recap. Purple, that's 15. Ecare Bravo for Ray Almadrin. We had green, number 270. We got Harry Kane, DiBala, and Fatahou. We had Orange to 50. Bakayoko, Rookie Einhoven. Got Kubo to 50. Silva Sly. Yellow to 99. We had Emil Smith Rowe and 50. Kai Havertz selections. And we had a mag to five. We had Rookie Mohamed Alicho. Nice one there. Two of five on the Rookie for Sociedad. Alrighty there, guys. That'll do it for UEFA Champions League Sapphire. Five box number four. Get some baseball going next.