So, thank you very much, and thank you, Maxim, for existing.

So, yeah, important words, I was the thesis supervisor of Maxim. I was just in Moscow a couple of weeks ago and met a certain person to whom I mentioned. She mentioned Koncevich, and I said, yes, I know him. In fact, I was his thesis supervisor.

Ha ha, she was telling everyone. This guy thinks he was the advisor of Maxim Koncevich. So, it's true that, for instance, we have another man in the front row who much more reasonably considers that he was the advisor, but he wasn't, he was only the teacher. He only taught Maxim mathematics.

But Maxim's PhD was not obtained in Moscow, but in Bond, and I was his thesis advisor. And I even suggested the problem, not that I understood it, even the problem, let alone the answer, but this was in 1991 when Maxim had just arrived, I think 1991. He had arrived fairly recently in Bond.

I had met him before in Moscow in 87, so he was already a lot older than when you first knew him, but he was still extremely young, and, of course, I was completely overwhelmed. And if my memory is right, we invited you to Bond once before, and then you came later as a second visit, the second time not as a visitor.

And, well, Professor Hertzberg said to him, you know, if you don't have a PhD, we can only pay you like a student. So, you need a PhD. He didn't have, I mean, not the doctor, he didn't have even the candidate, he didn't have a PhD. So, he needed a PhD, and so he needed a supervisor, so I was assigned. And then the Abweitstagen happened,

and Atiyah gave the, as always, the opening talk. No, no, it was before, it was 19. That was 19, yeah, I wasn't sure which year. And so, Atiyah gave, as always, the opening talk, and he explained Witten's conjecture, and at the end, I brought, you know, it's a long row, I brought Maxime down to introduce to Atiyah,

who had been my first thesis supervisor, and said, Michael, I would like to introduce to you my student, Maxime, who will probably be, I imagine, will be able to prove the conjecture you just told us. And so, Atiyah looked at me like he's completely mad. You know, you come with this, you know, kid,

and tell him that he's, this guy can probably, you know, prove the conjecture, you know, Witten, Atiyah, I mean, we're not talking high school mathematics here. And so, he looked at me like, you know, I've known you a long time, Todd, but now you've really gone over the tent. And then, I think, two days later, there's the famous boat trip on the Rhine part of the Abweitstagen, and the entire,

my memory at least, the entire three hours going, and the entire hour and a half coming back, because, you know, it's downhill. It goes much faster. Atiyah and Maxime were together discussing, and Maxime showed the proof. Well, it wasn't complete. I think it was not completely finished until the end of the week, but the idea of the proof, two days after hearing of the conjecture,

he had solved it, essentially. And so, of course, it was considered sufficient in Bond for a PhD, and it was considered sufficient by the IMU for a Fields Medal. And so, strangely enough, it's a long thesis, yeah, I mean,

but it was, yeah, 25 pages. He hadn't yet read, so he didn't know how to make it shorter. Now he would write it in three pages, and nobody would understand. So, I've been very disappointed since then.

Sorry? Oh, that's another good, if I start telling about the qualifying exam, I'll never get my talk. But, so I'll just say that I've been very disappointed. I haven't had a single student since who was that good. And it's not going to happen. Okay, so he's also a wonderful friend

and a wonderful person, and I'm not going to say anything more, but the thoughts are there. So, I'm supposed to give a talk, which means I have to take, roll up my sleeves. So, the title of my talk had three ingredients. One of them I understand more or less,

which is partitions, and one is quasi-modular forms, which I understand very well, and one is Ziegel-Wich constants, which I don't really understand at all, but since that's the part that led to this work, the only part that's specifically connected

with something Maxime has done, I decided I have to include it, but luckily we had a beautiful explanation yesterday by Anton Zorich, who's still there, and so I can skip trying to explain things that I only half understand. So, since that was explained, I won't say anything, but I won't say something first about the first two ingredients and then how it all comes together.

So, the last part, this is essentially going to be about, well, first of all, some old work, but in particular something I did many years ago with Masanova Kaneko, and then a wonderful generalization, which my main point is to popularize,

because maybe some of you don't know it, and also it's a beautiful theorem of Bloch and Okunkov of a few years ago, and I found a very, very simple proof, so I'll be able to essentially give you a complete proof of their theorem, and this part is joint work, well, with Daibei Chen and Martin Müller, except that I actually haven't,

I'm not even sure I've ever met Chen, and the joint work is Chen, Müller, and Müller and me, there's a part on Siegel-Wietch and a part on modern forms, and you need them both, but they're disjoint, and the common link is Müller, and just for people who were at the talk yesterday and saw the very unusual spelling of Müller, this is the standard one, the one that we saw on the slide yesterday

is based, I think, on the French champagne, mouer et chandon, but it's just, it's too much. Okay. So, okay, so I want to start with partitions because they're certainly the easiest. Whoops, this thing doesn't go up as high as one expects.

So, a partition is a partition of a number, and the notation will be lambda is in n, and I'll use p, the script p for the set of all partitions, so I won't specify the n, you could put p n, and the usual way of writing a partition

would be lambda one, lambda two, and you can either stop at the last positive one or you can just take a sequence of integers decreasing to zero and only find out the many, and then to say that it's a partition of n means that the sum of the lambda i is n. So, that's what a partition is,

and of course, when you do, you know, what I'm always saying that people always laugh that I always say this, but the only person who really lives by this precept apart from me is Maxine. Whenever you see numbers in life, a sequence of numbers, and it happens, then you should always make the generating function

and take it seriously, and it almost always helps you. Sometimes it doesn't. So, here, if you take p of n to be the number of partitions of a given number n, then, of course, what you should do is make the partition function, the generating function, n from zero to infinity,

p of nq to the n, and so this was noticed already very long ago by Euler that this is equal to, well, I'll have a notation that I'll use. Sometimes it's convenient just so I don't have to write so much. The infinite product, one minus q times one minus q squared and so on,

which is called the infinite q Paul Humber symbol. Sometimes this is just for convenience that I write that, and Euler, it's very trivial, of course, to prove that. Now, Euler noticed this fact, but Euler was then interested in computing the partitions, let's say, up to 50, and so to do that, he said, well, if I can compute the generating function

of the reciprocal, so q infinity itself, this product, then that should be easier, and then I can always invert a power series, and so he found, well, this famous result that this is the sum where every coefficient of q to the n is zero, most of them are zero, and otherwise plus or minus one, so this is the sum minus one to the n,

q to the three n squared plus n over two, and then if you know from high school that when you see a quadratic function, you should complete it to a square, to a perfect square, then you immediately see that actually the natural function is the product of this thing with q to the 1 24th because then you've completed the square,

so it now becomes six n plus one squared. Think about modular forms, you know that if you have q to a perfect square and you sum over a lattice, maybe with some congruence conditions like n is congruent to either one or seven, modular or 12, and you take the alternating sign, this thing is a modular form,

and of course it has a name, it's called eta, and so this is a modular form, in this case of weight a half, and it's the dedicated eta function, and again, I assume that people do know what a modular form is, although I'll very briefly

remind you, so you take a variable tau, which should now be in the upper half plane, this is meant to be a German H, so the upper half plane is the complex numbers with positive imaginary part, and then one always writes, so this I won't repeat, q is e to the two pi tau, and therefore if tau is in the upper half plane, q is in the open unit disk,

and to say that it's a modular form in general means that if you take any matrix, a, b, c, d, and s, l, two, z, then well here there's the 24th rule of unity, so there's a little factor, but you have c, d plus d to a certain power called the weight, which here is a half times eight of tau, and this is what

Dedekind discovered. So in other words, if we start with partitions, we immediately see that we're led to these classical modular forms, which have this transformation law. So now let me introduce the second ingredient of my title before putting them together at all. So that's quasi-modular forms. Maybe I'll go here.

So modular forms, if I take m star of s, l, two, z, so all modular forms on s, l, two, z, I won't write the group each time to m star, then it's very well known that there are Eisenstein series, e4, e6, e8, and so on,

which are modular forms of that weight, and it's very convenient to introduce shorthand notations due to Raman, and actually there's also an e2, but it's not a modular form, so it's not in this ring. So we have Eisenstein series p, q, and r. I won't define the general one,

but I can define these three since it's so easy. p is one minus the sum from one to infinity, n times q to the n over one minus q to the n. The q is one plus 240 times the same sum with n cubed. This is the same, and r is one minus 504 times the same sum with n5.

And then the remarkable thing is that these functions, q and r, are modular forms of weight four and six, so they transform exactly like this. Now there's no funny factor, just with c tau plus d to the four and c tau plus d to the sixth. And of course it's very well known that you have this relation, and the eta that I had before, there are two relations with eta,

and this, one of which is that the 24th power of eta is q cubed minus r squared, well divided by 1728, and also that the derivative of eta divided by eta, so the derivative is just the derivative, but it's very convenient to divide by two pi i

so that we don't keep getting formulas full of two pi i's I want to work over q. And so this in terms of the q is just q d by dq, and then you see very easily that the derivative, d eta by eta, is exactly 124th times e2, which I'm calling p. So you already have a connection between what I told you here,

which is that this dedicated eta function, which is a modular form, is related to the modular forms q and r, but there's a bigger ring called quasimodular forms, the word was introduced, in fact, in this paper of Kaneko and myself that I mentioned about 10 years ago,

quasimodular forms, you can define it for any group, I won't bother with, but the answer here is it happens to be a free algebra on three generators which are exactly this p, q, and r. So since that's not my main object, let's just take this as a definition,

it's certainly included because you change it's an original ring. Thank you. For me, I won't work over q all the time just to keep life simple. Thank you. So now you have something very nice. We just had this derivative, and you see that the derivative of eta, well, eta isn't in this ring, but eta to the 24th is,

and that would just multiply this logarithmic drift by 24. So you see that the logarithm, the derivative d of eta to the 24th is in this ring, but in fact, the d acts on this ring, and let me tell you, so the famous formula is very easy to prove that Ramanujan found, and just for completeness,

I'll write them on the board. So the derivative of p is p squared minus q over 12, the derivative of q is pq minus r over three, and the derivative of r is pr minus q squared over two. So since it's a derivation, it's enough to do the generators,

so it acts on the ring, and the last comment I want to make in this generality about quasi-modular forms is that we have three operators. So d sends f to, as I said, qd by dq of f, and this will be in mk plus two.

If, let's say that I have an f in mk tilde, so quasi-modular form of weight k, the derivative will be a quasi-modular form of weight k plus two. Then I have the weight operator, which is the Hamiltonian, if you prefer. It just multiplies a form by its weight, so of course it doesn't change the weight, but I have a third operator.

This is meant to be a German d, for those of you who like this Utilin script, and this is, well, it's the derivative, 12 times the derivative with respect to e2 of f, so this will be a quasi-modular form of weight k minus two, and the derivative means if you,

in this definition, there's no problem, since I haven't defined f in an intrinsic way as a function of tau, you can do it by transformation properly, but simply it's a polynomial in modular forms, in a polynomial in p with coefficients that are modular. If you do that, you're just differentiated, so d by dp, the notation I've been using.

So the reason I want these three operators, well, they'll come later, so trivially the commutator of w and d is 2d, because w is the weight and d raises the weight by two, and for the same reason the commutator with d is minus 2d, and then finally the commutator of d and d is w, and so this is an SL2,

which is acting on quasi-modular forms, so that's kind of, and the primitive part, the kernel of German d is exactly the modular form, so that's how to see modular forms in quasi-modular. So we'll see explicit examples in a minute. So now I want to say what this has to do, well, how these things first came together,

at least for me. So the notion of quasi-modular form, it came from the physicists, well, several also wrote, but the main person was Deschraff, who had the idea many years ago of looking at mirror symmetry in dimension one. So mirror symmetry, which you heard of many lectures here, you map curves, holomorphic curves, complex curves, into an algebraic variety,

but you fix the homology class that the fundamental classes mapped onto, and you count them in some suitable way, and in string theory and in other parts, one especially wants to do this for Calabi-Yau's, so you want to count the curves sitting in the Calabi-Yau, and so Calabi-Yau in dimension one,

mirror symmetry in dimension one, would be that I, well, you can't really speak of putting curves inside a curve if there's not room, but you can still map a curve of genus G to a curve which is a Calabi-Yau, but that means that it's a torus, it's genus one,

because that's a one-dimensional Calabi-Yau, and now saying that you fix where the fundamental class goes means that you fix the degree, so the degree is some fixed number n, and so you can ask the question, how many coverings are there, let's say topological, because then you can pull back the complex structure, how many coverings are there, appropriately counted, so we have sort of n of,

well, I won't give it a name because I'm going to introduce the notation in a second, it'll be confusing. Now, we want to count how many coverings, and as usual, you count things up twice more when you count them with multiplicity,

which is one of the number of automorphisms. I don't want to explain it exactly, but this will be certainly a rational number. It'll be a finite sum if you do it right, I'll come in a second, but I want to specify the ramification, and the ramification is this, I assume that it's generic ramification, so that means that any critical, if I have a critical value down here,

then there's a unique critical point above it, so at any point, the preimage is either n sheets or it's n minus one and exactly two cross. Okay, so that's the nature of the ramification, and so we want to count this number, and we want to make a generating function, so let me define fg of q to be the sum over all such,

well, I take n from one to infinity. Of course, n can't be one if the genus is not one because you can't map degree one, so the first coefficient will be zero. It's this number, number of coverings with all the things I said, generic branching and counted with one of automorphisms, and then again, you make a generating function,

q to the degree of the covering, and the prediction of, in particular, was that this one was supposed to be a modular form, a quasi-modular form, and in fact, of weight six g minus six, and there are two parts. One, you have to find a way to compute it,

what those numbers are. You don't want to go to your computer and say, please draw pictures of Riemann surfaces, so you need a combinatorial description. That comes from group theory, and he gave, and then he had also a proof of this, a proof in quotes using ideas from quantum field theory and path integrals, but it wasn't a proof. In any way, this is certainly a theorem,

and an explicit concrete proof was given in this very short paper by Kaneko and myself, which appeared in the same volume as Descraft's paper where he explains the whole thing. So this was, I don't know, 10 or 12 years ago. So this is the theorem. Let me give you an example just so that you see something concrete. So g equals one is, and actually this is only if g is at least two.

For g equals one, there's a slight exception because Taurus is an infinite group of other morphs. By translation, it somehow messes up the counting. So for instance, f2 explicitly is 10p cubed. It has to be of weight six because g is two,

and in weight six, we have only three, a base of three elements, p cubed, pq, and I have no idea why I wrote this this way, five p cubed minus three pq plus two r divided by 5,000, it's not 5,000, but it's 51,840.

Up to a constant. The second line has a constant term. The line before it doesn't have a constant term. I have no idea what you mean. The q to the zero. Well, okay, the coefficient of q to the zero is equal to the coefficient of q to the one, and they're both equal to zero.

So you can start with zero, one, or two. Those coefficients are zero. What you wrote doesn't have constant. It certainly does. Oh, thank you. That's all I was trying to say. Thank you very much. We have somebody in the audience who can subtract five minus three plus two. Thank you very much. I thought you were complaining about the first line because the second line. It's much better.

What's more, you're even right that you know with what I had, there would have been horrible denominators, but this thing actually is integer coefficients, which wouldn't have been true at all for the other, but that's an accident. In no other genus do you get integers. You always get a bounded denominator. Okay, so that's the theorem, and so this is the theorem of, well, that's the formal theorem of Kanneh-Kunni with the formal proof,

but it was the discovery of Dijkgraaf, and also of the physicist Rud. Well, I should have maybe also written. Now comes the amazing general, but let me say the combinatorics. As I said, there are two steps. So two steps. One, give a formula, a computable formula that you can put on the computer, so to speak, for FG, and then, which Rud Dijkgraaf had already done,

and the other is then show the quasi-modularity. So let me tell you the first part first, and I go back to partitions. If you have actually for any finite group, but let's now think of SN. So this is meant to be a German S. As you all know, irreducible representations of SN

correspond to Young diagrams, and therefore also to, so I remind you that any partition, for instance, the partition three plus one can be represented pictorially by a Young diagram or Young tableau or whatever it's called, so you have N squares, which on the upper left-hand corner, the way that I do it is the lower right-hand quadrant.

So that corresponds to unique irreducible representation, and now let me define an invariant new T of lambda. You could do this where T is any composite class in SN. You could do it for any finite group. It's not, it's proportional to, if you take the pi, if you have any element of the group,

for any finite group, if you have an irreducible representation and a conjugacy class, you of course have the character and the well-known character table by evaluating or by taking the trace of pi of T or pi whatever the element is, but you have something else, which is the action. You take all elements, so this is really the conjugacy class,

but I didn't write. You take all elements in the symmetric group, which are in the conjugacy class. Here it's the transposition. So T is the transposition. You interchange two of the two numbers, one and two. The conjugacy class consists of all transpositions of I and J, and you see that that's relevant for our problem because here if I do monodromy, I number these points,

and then when you go around and come back, you interchange two for the monodromy of the singularity you went around. So it's reasonable that this plays a role, but now that's why we need T. But for any element of the group, for any conjugacy class of the group, you take the sum, and then the sum of the corresponding matrices is a scalar by Shor's Lemma because it's an irreducible representation,

so it's a number, and so this is an integer. And it's actually, this is a very trivial formula that these two numbers are proportional. You have to multiply. It's true for any group and any conjugacy class. So if I have a finite group, a conjugacy class, and an irreducible representation, then chi P of the conjugacy class is equal. Well, I'll do it the other way.

So nu C, the way I'm writing it, of pi, is equal to chi P of psi multiplied by the size of the conjugacy class and divided by chi P of one, which is the dimension of pi. So completely trivial. Just take the trace of this element. The trace of a scalar is the scalar multiplied by the dimension of the space, but the trace of a sum is the sum of the trace, so they all have the same trace.

You just get the trace multiplied by the size of the conjugacy class. But these are integers, and these are integers, but the ones I want are nu C of P. So the answer is, let me give you a little picture. So if I take n equals four, then you have the following diagrams.

And the value, this is the picture of nu, and then the value of this, sorry, this is the lambda, which corresponds to a pi, and then the nu T of lambda will be six, two, zero,

minus two and minus six. It's easy to see that it'll change sign when you flip the Young Diagram, which means the conjugate representation. So now we associate to this now a polynomial H four of U, which will, in this particular case, be U to the sixth plus U squared plus one plus U to the minus two

plus U to the minus six. H n of U is just the sum over all partitions of n of U to the power of this funny invariant, U, what am I calling it? U T of lambda. Okay? So that's that. And then what you show is the following.

If you take H n, now I hope I can do this, U sum over all partitions, which again, you shouldn't really even ever put the n. You just sum right from the beginning. I could have just said you take U to the power of nu T of lambda

times Q to the size of the partition. So if you take this sum, then this is a power series in Q, of course, but also in X, and this will exactly correspond the coefficient of Q to the n times X to the K, or X to the K maybe over K factorial,

if I remember correctly, will exactly correspond to coverings, n-fold coverings, generically branched, as I said, n-fold coverings of T two with K branch points downstairs. Now, if I have a Riemann surface of genus G, then the, and here's a genus Taurus,

then you find trivially that because the order characteristic downstairs is zero, that in the Riemann Hurwitz formula, n doesn't play any role, and the K, the number of, so I have K branch points, then K will be independent of n. It'll simply be two G minus two.

That's because downstairs we have order characteristic zero, so I could have used K instead of G, but the point is these coverings may not be connected, and so that will correspond to that. Now, if you take the log of this, then this will correspond to connected coverings, and so that's the generating function we want, but, of course, taking the log won't change the quasi-modularity, so actually in the proof, one shows that the coefficients of this thing

are quasi-modeled forms, and then when you take the log, well, they're still quasi-modeled forms. The log of the power series has coefficients in the same ring as the original power series, so this is the way that you compute it, so I hope this is clear, so you write down this sum. I wrote down explicitly how to define this new T and an example for n equals four,

how it looks. You take these polynomials, then you take the log of this power series, and now the coefficient of Q to the n of X to the K over K factor over K is two G minus two will be exactly the rational number we're talking about. So now comes the theorem of Bloch and Okunkov. So first, let me introduce an abstract ring.

This is not quite, also the notations are not theirs, so if you know their paper, you still have to listen, unfortunately, because I've changed every single notation slightly. These QK are related to their PK but shifted by one and divided by factorial, but these are more convenient that they don't have a Q1. You'll see in a moment why they might have felt they didn't need a Q1,

but it's very, very useful. So let R be the abstract polynomial algebra in infinitely many variables, where QK has weight K. So of course, the dimension of this space in a given weight K is simply the number of partitions of K, and there's a nice Fox-based representation. I'm not going to talk about that at all, but it's explicit,

but they didn't implicit, but I'll tell. And now this will act on, I mean, if I have an F in R and a partition of some number, then I can evaluate F of lambda. So what I'm saying is that there exists specific functions, Q1 of lambda,

Q2 of lambda, Q3 of lambda, and it continues, but the board, you know, the margin is too small to contain it. So we have a series of specific functions and they're very explicit. Q1 of lambda is a function you all know very well. It's the function that assigns to every partition zero. That's why Bloch and Okunkov didn't include Q1, but I wanted my abstract ring. Q2 of lambda is the size of the partition,

so the simplest imaginable invariant, except that it doesn't work unless you correct it by subtracting a 24th. And Q3 of lambda is exactly the new T of lambda that I told you. Okay, so Q3 is the function that we needed. And so their generalization will say that,

I mean, we knew, I knew, and in fact, it's written in their paper that I suggested that such a form might be true, but I'd forgotten and was trying to wonder how they could possibly have discovered this. But anyway, I knew that there were higher functions, but I saw no way to prove the cause of modularity. So the statement is going to be that not just if you do this procedure that I just did with,

if you multiply this out, then you'll find that in this series, of course, since this is the sum, what was it, E to the U, but I've replaced E by, U by E to the X. So this becomes E to the new T of lambda X. And so the coefficient of lambda to the K is of course simply the sum over all partitions

of the very special function, U T to the N to the K, which will be an even integer later, two G minus two times, sorry, new T of lambda times Q to the lambda. Or, yeah, it says, sorry, it's essentially that. So the thing we're talking about is this,

and so they had the idea, given that new T is just the third of a list of infinitely many interesting functions, why not take any polynomial in these special functions which I'll define in a minute, and then their theorem is going to be that you'll always get another form, but let me first introduce the notation I want. So I define the Q bracket

of a function. Let's say that I have any function on the set of all partitions. As I say, I'll always work over Q just because everything, all the examples are rational. So if you have any function, arbitrary, no convergence, then the Q bracket is the Q average. We call it the Q bracket, but maybe Q average is better. And you do what you always do

in statistical physics. In statistical physics, you have a large ensemble, you have a lot of states, and you have an observable, which in this case is F, so the states would be partitions, and you have an observable and you want to know the expectation values in quantum theory. So what will you get when you evaluate F of lambda? So what you'd like to do is take F of lambda,

the sum over all partitions, and divide by the sum of one. But of course, that doesn't make sense. Both sums are infinite. So what you do in statistical physics is, of course, you put in a factor in statistical physics, it would be E to the minus one, I forget where the K is, I think downstairs, one, sorry, the energy, sorry, it would be E to the minus one

over Boltzmann's constant times the temperature, and then you would take if the state is energy eight, then you would weight with that power. So you would typically have a constant, Q, which sort of depends on temperature, and you weight. So here we'll have a constant Q, which is just called Q, but you can think of it as Q gets bigger and smaller, it's getting hotter and colder.

And so this is the definition, and then, of course, downstairs you do the same. So this is the Q average. But of course, what it is, is a Q series. I mean, a power series, there are a lot of Qs, that's a capital Q for rational numbers, it's a power series in Q. So we associate to every function, and it's a formal power series, I don't care about convergence at all.

Although, actually, all the ones I write down are convergent for Q less than one, because they're, in fact, they're all convergent. But now, if you remember that this denominator, that's what I started with. If you just sum over all partitions, Q to the size of lambda, that's the same as P of N, Q, N. So that's this, so I could also write

this thing, as I take the sum over all lambda in P, and now I simply take F of lambda, Q to the lambda, and now I have to divide by this denominator, but the denominator is the reciprocal of this, so I have to multiply by eta, which is, of course, a modular form,

but then eta, remember, was not quite right. There was a Q to the 124, so I have to actually wait. Now, with Q to the lambda minus 24, and you see that's nice, because that lambda minus 24 is exactly our Q2. So the result I was talking about before would be the special case, Q3 of lambda to some power, which is 2G minus 2,

let's say. Well, I pretend you didn't have to take the log, and then Q to the power Q2 of lambda. Okay, so it's completely in terms of this ring, and so now I can give the theorem of Bloch and Okunkov, which is, I don't know, 2003, if I remember correctly, anyway, about 10 years old, and it says that this

Q bracket is a modular, a quasi-modular form for every F in this ring, which they call the ring of shifted symmetric polynomials. So that's an absolutely wonderful theorem, but it turns out to also have a wonderfully simple proof. When did I start? When should I stop,

and so on? Of course, the remarks about Maxime don't count, right? That's... Sorry? Left? Or since the beginning? Aye, aye, aye. It's all about positive and negative things. Well, then, I really don't want to skip my sketch of the proof

because it's really a beautiful proof, very, very, very simple. Let me say briefly what these Q3 of Lambda are. So you start with the partition. Let's take a very simple example, and you are going to... Let me say how I define QK of Lambda. QK of Lambda, well, you take what are called

Frobenius coordinates. So you take the longest diagonal in this Young Diagram, so even more complicated, as far as you can go down the diagonal, it's the Hirsch index. If you think of this as a researcher, who has one article with three citations, Maxime is more, but if you have one article with three citations and two with one citation, that's all that you ever wrote, then absolute value of Lambda

is the number of citations, and the Hirsch index is the smallest i for which your i-th paper is i-citations. In the literature, that would be this diagonal. So in general, if I had a more complicated thing, then you would have this diagonal, and so that divides up the Young Diagram

into this piece, this piece, and then the pieces below. So you'd have lengths here, which would be 7 halves, 1 half, and then you would also have minus a half, in this case, and minus 5 halves. So you would associate to see a finite set of,

this will be contained, I'll call it f, for fermions, which is z shifted by a half. And then this finite set, well, it'll live equally many, it'll be a balanced fermionic set. So this, it's balanced in the sense that it's a finite subset and it's equally many positive and negative things

because that's the length of this diagonal. And of course, the sum. So if c lambda, if I take the sum of the absolute value of c in the c lambda, then that's obviously the total area of the Young Diagram, so that's the size for a partition. But now the qk is the sum of,

now this is a shift and I think it's q to the k minus one. That's right, so I have to take c to the k minus one, but I have to put a sine. Sine of c times c to the k minus one. So you see if k is two, then sine of c times c is just the absolute value of c, so I get this. But then you have to divide by k minus one factorial

and you have to subtract a constant, which is the coefficient, the kth coefficient of x over two over sinh x over two or maybe sine. So I've forgotten the sine. So in particular if k is two, you see this is one. This is sine of c times c, so it's the absolute value, so you'll get lambda and the thing you have to add is minus the 24th

and so that's the official definition of qk of lambda. So that's now I've defined to you, so at least you now know the statement of the theorem because that's a complete definition. But of course it looks very artificial and there's a much nicer way of doing it, which is also in their paper. This one doesn't work.

This one works. The way that you do it is you're going to associate to lambda the set x lambda, which will be simply you take the elements of your partition and you shift them all down. That's why it's called shifted symmetric. You shift the jth entry down

by j minus a half, okay? And so this is going to be in a set that I'll call x fermionic zero, which is a subset of x fermionic. So these are sets in the fermions, finite sets,

sorry, not elements, but finite sets such that it's bounded above and co-bounded below. So if you can think of this like Dirac's sea of electrons, you start with the vacuum where all of the negative states, minus a half, minus three halves, and so on are all occupied.

And then you have a new state, which is not the vacuum state, where some of the positive energy states, so the energy here is a half integer, and some of the positive states are occupied, but only finitely many. And there are some holes on the left-hand part. And balanced would be that it's balanced in the same sense that we just had that there are as many holes

on the left. So if I make one hole, I take this point and move it here and of course, you apply some operator like that, then you'll get a balanced set and that's XF0, where there are as many positive elements as negative holes. And so in fact, you can see very easily that XF lambda plus is C lambda plus and X lambda minus

is all negative fermions minus C lambda minus. So that's why, you know, that's a correspondence. So now if you do this, oh, I was lucky and now I need that dangerous implement. Where is it?

Okay. So I'll pull this thing down. So, okay. So now what I do is I associate to any X, in particular X lambda,

I associate to X a formal power series, well it's actually a half power series, just T to the X. So this will be, so in my example it would be T to the, now I've covered So not the example, also for you, so it would be T to the, where has it gone?

It was seven halves, I think it was the biggest, T to the seven halves plus T to the one half plus now T, but now it's not the things that you see, but remember it was the holes in this case. So in this particular case, it would be, well it's the lambda J minus a half and in fact the next one would be T to the minus three halves plus T to the minus

five halves and so on. So in our example, and you see that after a while, since you've everything, this whole thing will be, it'll have a Laurent part, shifted by T to the one half plus and then the tail after a certain point will be the same as it would be for the vacuum which is this. So although this is an infinite sum, it becomes a rational function and so then I can define

now capital WX, let's say Z is W little X of E to the Z and once I put in this rational function that makes sense and this will be exactly the sum K from zero to infinity, QK of lambda, well QK of X but then QK of lambda will be QK of the associated

X lambda in the way that I just defined and then here it's simply Z to the K minus one. You see that the Q zero for any set X will always be one because there, it's always a finite set plus this thing which in the new language is Z over two over same Z over two and of course that has a simple pole

with res U one. So the one over Z term will always be one so for any such bounded set but Q one of lambda, Q one of X if you do the calculation will be the excess of the number of positive occupied things minus negative unoccupied so Q one of X equals zero if and only if

X is in this set fermionic. So as I said they didn't use the Q one because on the partitions which correspond to the balanced ones you don't see it but it's actually there. Okay and now I'm going to give you a nice theorem which and I'll try to sketch the proof because it's so

once you've set everything up it's literally one line and then that theorem very, very easily like in two or three lines that I might skip implies the Bloch-Kuhn-Kauff theorem. So that's the next thing I want to show you but I want to get the formulas right. So I want to first define a derivation from this ring to itself

which is the derivation that sends since it's a derivation I only have to say what it does on generators it sends Q K to the previous one and of course Q zero I should say it's one it's not one of the generators and obviously Q one goes to zero. Okay so that's the generator and I also need the classical theta series which depends on Q which is the sum over all fermions

which remember are just half integers of minus one to the integer part of mu times E to the power mu Z times Q to the power mu squared over two. So this is the Jacobi theta function. Okay and it's very very easy to see

it's well known I mean Jacobi found that the this is an odd function that's trivial and because it's an odd function it has a power series with odd powers and so I can pull out the derivative but that's what Jacobi found by his triple product formula that's the cube of eta is the constant the leading term and then it'll be one plus

H two Z squared plus H four Z to the fourth it's a power series expansion and each H K I could write down a couple is a quasi modular form of weight K that's the only thing I'll actually need about modular forms is that and this is a one line proof maybe since it's one line I'll write down the line

and then you have it well you I mean you have to prove the proof which is another one line but the the thing that you do in one line is that four N times N plus one times H N if N is at least two is H times the derivative of H N minus two plus P remember our P

the special quasi modular form times H N minus two that's a very very simple form that'll prove and you start with H zero is one and so since I already told you the differentiation raises the weight by two and preserves quasi modularity these H's are completely explicit quasi modular forms so those are my two ingredients is this derivation and the theta series and now the theorem

is the following that for all F which are in the sub ring of Q which is actually the sub ring they used where now I don't include R one so the sub ring of sorry it's a polynomial so this is the sub ring of R where Q one does not appear then you have the following statement

that theta you see this is a power series and so you can apply a power series to a derivation and apply it to an element and that makes sense because since this derivation reduces the degree by one on any polynomial large powers of D will kill it and so I can evaluate a whole power series and the claim is

that the Q bracket is simply zero so now if you think what it means the Q bracket I can divide by eight of tau cubed and by Z if I want not by Z so this tells me that if F is D of G where G is in the sub ring

Q two and so on then if I take this expansion then I'll find that zero is equal to the Q bracket of F plus H two times the Q bracket of D squared F plus dot dot dot and so by induction since this thing is lower weight I've already shown that it's quasi-modular H two is quasi-modular

so by induction this gives the quasi-modularity and the only slight problem is that not every F is in the image but it's very easy to show that every F of positive weight is the sum of D of something in this sub ring plus something which is divisible by Q one and for that the Q bracket is zero so in one line once you have this theorem you find an inductive form that it's completely programmable

you can make a table that you can actually not only prove but also evaluate the Q brackets you prove the modularity and evaluate the Q brackets as polynomials in P Q and R directly you never have to write out their Q expansions as we did for the first two years for our computations we had huge sums over partitions and then we could only get 40 terms

of the series because there are a lot of partitions when you go beyond 40 or 50 but now it's completely algebraic so let me sketch the proof of this theorem and then I want to say a little bit about the Ziegel-Wiech part but I'll just okay so the idea is this if you have a set X in X F and you also have

a number in F which remember is a half integer then I can simply translate X you just shove it let's say this is three halves you translate it by three halves and this will be the set X Z which is defined exactly the same way as X F was which has probably disappeared so these are subsets now of the integers which are bounded above and co-bounded below

in other words their complement is bounded below so you have this rather trivial change and so now I claim well every element X and Z so here you have an X which is a subset of Z which is bounded above unbounded below then it has a shift the number of positive things I mean the it turns out that there's a unique nu that you can shift it by

so under this map actually I want to write it the other way I think I can say given any fermionic number nu so half integer and any partition lambda then I simply map nu lambda to the set X lambda that we had before which was one of these

things for the fermions and then you shift it by nu and that no longer will satisfy the boundedness condition but given this X then Q1 of X it's very easy to see is going to be exactly nu and therefore if you give me an X in this set you evaluate Q1 and that tells you what nu you need and that tells you

which nu to subtract and then when you subtract it you get a bounds fermionic set and you interpret that as a Young diagram and so you have this very nice projection which actually is certainly known and is used in other forms in particular in connection with the free fermion free fermionic field but I don't want to talk about that and it plays no role. So now under this correspondence

you have a very simple lemma so the lemma is simply this if I have again an X in this set of subsets bounded above and co-bounded below of XZ and here I have a pair consisting of a half integer and a partition

then and if I have any f in the Bloch-Kuhn-Kauf ring then if I want to evaluate f of X it's very easy I take e to the power this nu this half integer times my derivation and apply it to f

that remember was a well-defined thing and then well it's that for lambda but that means on X lambda well sorry the way I've done it just on lambda okay let me prove this because it's really completely stupid since X is equal to nu plus X lambda that means that w

small w x of t which remember was the sum over all x in my set bounded above and co-bounded below of t to the x well you've just shifted so this is obviously just t to the w times the corresponding thing for the partition that's completely obvious so that means that when I go to the capital w which was the generating function remember qk of

lambda's qk in this case of x z to the k minus one this was the same thing where you change t to e to the z well then this will be e to the nu z times w well times the same thing qk of lambda z to the k minus one so now if you compare

then you see that qk of lambda or qk of x will be qk of the partition I won't write the lambda plus nu times qk minus one plus nu squared over two times qk minus two and so on but then remember our derivation d sends qk to qk minus one qk minus one qk minus two so this is that's this

in the case of qk but then if you have a derivation then the exponential of it is an automorphism and the qk are ring generators so you get this formula for free and in particular you have two special cases that already well one of them already mentioned q1 of x is the nu and q2 of x is the q2 of the partition which remember

was the size of the partition minus the 24th there's no contribution from q1 because it's zero for partitions but then there's a contribution from q0 so nu squared over two so you have these two formulas and now let me prove the theorem which is above so now proof of the theorem

so I'm going to write I want to take the q bracket of e of theta of this derivation applied to f f remember was in the subring not involving q1 I want the q bracket of that and I might as well divide it by eight of tau because I want to show it's zero so it's

equally zero if I divide by eight of tau so by what I've already told you that's the sum of all over all partitions of q to the power the size of the lambda minus the partition but now I have to insert theta of delta right so therefore I have to put here theta of

delta which by the definition of the theta series which is still there is the sum so this is all partitions this is all fermionic numbers half integers I have minus one to the integer part of mu times e to the nu times the derivation because I'm involved in my thing here that's the next term then I've e to the

u squared over q to the u squared over two so that comes here okay and then I have to take this and apply it to f at lambda right that's the recipe that's what the left hand side is but now by the lemma the lemma said that there was a bijection between pairs

consisting of a partition and a half integer shift and under this lemma well this thing is just q2 of x because that was the the second formula written here the size of lambda minus the 24th plus nu squared over 2 similarly e to the u delta f of lambda well that's just f of x

f of x and finally minus one to the integer part of lambda is minus one to the power integer part of q1 of x because q1 of x is lambda so now I've written in terms of lambda and now I promised you a one line proof and I'm almost at the end of the line but it's still going to work this is zero and the reason that that's zero is

completely obvious you have an involution an involution which is really stupid you map x to x star which is the stupidest thing you can think of if your set which is now a subset of z if it contains zero you take zero out and if it doesn't contain zero you put

zero in it's a pretty silly involution but if you think of this involution then our w x of t which was the sum of t to the x well you just add or subtract it so when you do this you're going to simply add or subtract one and so if you remember that the coefficient of z to the k

minus one and this is qk says qk of x star will be equal to qk of x for all k different from one but q1 of x star will differ from q1 of x by plus or minus one well it'll differ by one and one direction to the other so if you just apply the involution this sum is zero because the terms just cancel in pairs

every time you have x and you go to x star f of x only involves q2, q3 but those things don't change so this is invariant q2 is also at least q2 so it's also invariant but this q1 goes up by one so

proof of a theorem that immediately applies Bloch-Kunckoff now my time is I hope still positive very small can I apply a scaling of a minute or okay I wanted to talk briefly about

the Siegel-Wich constant thing so maybe I'll just say then in words how it works that's what started this whole thing and it's a very very complicated investigation with Mueller it's been going on for three years using this Bloch-Kunckoff theorem so basically the numbers that I'm not going to then repeat the forms we had them all in this lecture I spent till three in the

morning last night with my colleague on the phone well that was only half an hour but then myself preparing that I would be able to explain it correctly and I won't need to but we had a formula from the paper of well Maxime and collaborators I'll remind you it was a formula that said that if you're looking at the stratum 1 1 1 1 a flat surface sorry any

m1 up to mn the stratum of flat surfaces so a Riemann surface with a differential form with zeros of order m1 up to mn then there's a Lyapunov story and the sum of these positive Lyapunov exponents Maxime had somehow I think it was mostly him or all him I'm not at all in this field had seen that this has everything to do with intersection numbers

and moduli spaces that's one of the things that goes in but here there's an explicit formula which if I remember correctly is like this and then plus well there was a factor pi squared over 3 but I want to renormalize it so the thing called C area of the well of the particular surface you're looking at and C area is the thing that Anton explained

in his talk when you have these cylinders then you take you count them you count the number of geodesics up to a certain length but you count them according to weighting them by the area of the cylinder which states so the original Ziegler-Wich you didn't put in that area factor but it turns out this had much better properties and then you do

something with the asymptotics of that but the final upshot when you take the average in some sense or the generic point at the stratum is that your relation between one thing you want this number which is very simple and for the principal stratum which is the only one well no we have results in principle for all stratum but we've worked out in great detail for 1 1 1 this is just a simple number and this is

another number you want and because of the relation with modalized spaces these things are related to the volumes of modalized spaces or various strata in the space of flat surfaces so they can be expressed that was done well in particular by the work of Chen and Muller you can express that all in these

Bloch-Kunckhoff terms and therefore show that the generating functions of various counting functions like this so you have to go to Horvitz space that brings to a towards but with more generic ramification you can translate the problem of evaluating these

numbers into a problem of counting of averaging functions over partitions of these q brackets which we know are q modular and in particular these numbers like c area which the prediction was that they're always rational because they're supposed to be intersections in the in some modular space and in many cases it was known I mean that's Eskenmeister and Zorich and so on there are

formulas but it's very very hard to get your hands on either the volumes or the areas or it they also Chen and Muller found that you should also do other counts so here we're counting the cylinders by the by the width times the height of the cylinder because it's

the area but it turns out it's also a good idea to put powers I forget exactly the formula so they're weighted things so they make higher siegel-veach constant cp which are also interesting c minus one would be the c area but the others are perfectly good invariants also and it turns out that the c area are exactly given by the limiting behavior as

tau tends to zero of the quasi-modular form associated to a very particular element of the Bloch-Kuhn-Kopf ring and so you can answer all of these questions completely if you can calculate some rather difficult generating functions in this Bloch-Kuhn-Kopf thing and really calculate these things so we have

many many results on that and I was going to say something about them but maybe it's just hopeless maybe I'll just mention one thing because that's at least still on the level of partitions so purely combinatorial so what Chen and Muller found is that to compute this cp and

For any P, including P equals minus one, so the area case, but also these weighted, here you have to normalize the whole area of the surface to be one. What they found is that those CP correspond in some way to TP and then Q3, Q3, so I already told you for the generic ramification, if you have a bunch of ramification points,

K of them, but they're all generic, then that corresponds to the Q3. But now, if you multiply this thing, K times would actually have to take the so-called cumulants, it's an alternating sum by inclusion-exclusion, then this thing corresponds to CP. And TP is a function of, so this is the Q average, so remember the Q average is defined for every function.

It's only quasi-multitour when you're in this special ring, but it's defined. So they found that the CPs were in terms of TP, and TP is defined as full as it's one over N factorial, lambda is a partition of N, but then it's a huge sum over all the other partitions of N of the size of the CommsD class

times the square of the character, times SP, where SP of a thing is very simple, it's just the moment, so the sum lambda J to the P. So you have to take this kind of transform using the character table of the original thing into this. I think the form is right, maybe the two is wrong. But then, so this is lambda one to the P

up to the last one to the P. And then they did a lot of calculation. You get completely weird numbers, and they asked me to see if I could recognize them. And so you start, if you do a computer calculation, the first thing you'd find is that when you do this N, then T, they especially want T minus one. T minus one of N is a very beautiful number. It's the Nth partial sum of zeta of two.

And so that generalized that TP of lambda, so this is the theorem that I found on the computer, and Miller then proved, it's not that hard to prove. It was somehow, it took a while for us to understand that that was the statement, that TP is just the sum over all elements of the N diagram of the hook length

of S to the power P minus one. So the hook length of a thing is what you would think. You have a thing, and you go as far as you can to the top and the bottom, and you add that up. That's called the hook, and the number of squares is the hook length. So for this very simple diagram, the hook lengths are one, two, three, up to N. And so if P is minus one, if P is minus one, then you get the sum one over N squared.

So this is a non-trivial thing. But then it turned out that T, so another theorem, less obvious, although alts are not very deep, is that if P is greater than or equal to one, so non-minus one and odd, then these TPs belong to the Bloch-Kuhn-Kauffring. They are polynomials. So this TP is just a function. You can evaluate it for any real number, P,

but if P is a strictly positive odd number, then that function is a shifted symmetric polynomial. And therefore, it's bracket, and what's more, it's bracket when you multiply it by anything in the ring, is a quasi-modular form.

So if F is in the ring, and P is positive and odd, then by Bloch-Kuhn-Kauff plus this fact, DP times F, Q will be in the ring of quasi-modular forms. And then what we found with a huge amount of computer calculation by looking at hundreds of examples into very, very complicated formula, we still haven't completely proved,

although we've proved many cases, that this is the sum of certain differential operators, rho ij, which are very, well, the first, if i or j is less than five, we got explicit formulas by the computer, and then we finally guessed the general thing. You have certain differential operators on the ring F, and then you apply those universal operators, and here you take the i-th derivative

of the Eisenstein series, well, essentially the e before renormalized, of way to j plus p plus one. So this is still conjectural, but we've proved in very, very many cases, and the form of this is such that if it's true for positive p, it's also true for p equals minus one with a very small multiplication,

and using that, you get exact formulas for the volumes of all of these strata and asymptotic formulas, and also for the area thing, so for the Siegel-Wiech counting, but some of the results are now unconditional, but the most general in particular for the area counting is still conditional on proving this thing,

which we've only proved for a lot of special cases, but not in general. So there's this very, very interesting algebraic and combinatorial structure on this ring with these tps, and that very amazingly, even though they're defined, well, in two different ways, completely differently, but they turn out to be polynomials in the qk

and therefore to act on the ring, and so there's a very rich algebraic and combinatorial structure. So as said in the introduction, I'm somebody who likes formulas with no content, and to twist a quote that we heard from Maxim when he was a young man, I'm a specialist in special problems.

Okay, so thank you, and thank you, Maxim. For the great talk which you presented with exponential increase in speed. Sorry, sorry. Let me just interrupt again for you. In fact, it's some kind of system which is rational.

It's not obviously called this part of algebraic geometry. Part is some kind of different measure and it might have a proof that it's a priori that's strongly rational. No, but when you take the average or something, I mean for the generic. And that's one for the g. No, but I said that, that it's not, but that you have an expectation that it's connected because of modulized spaces.

But here, well this is certainly rational. Nobody can take that away from me. In fact, it's n over eight in our case. And this thing is rational because we have a formula for it. Well, that was already done by Eskin, Maser, and Zorich. But we have, actually I was going to write down the generating function we find for the volumes. It's, I'll just say it in words.

Take the Horvitz Zeta function at three halves, but shift it by half. So you take one of, well, or even at one half. So you take one over the square root of n, well it's slightly divergent, summed over n is a half, three halves, five halves. Sorry, shifted. So one over the square root of x plus the square root of x plus a half, x plus three halves, x plus five halves.

Expand that as a power series in x. It does an asymptotic expansion. It starts square root of x. So call that y. So y is a power series starting with square root of x. And now you invert that. Then x is a Laurent series starting y squared. And the coefficients of that inverse power series of the shift and Horvitz Zeta function are essentially the volume of the corresponding strata.

That's the formula. And for that, you can get asymptotic formulas not that easily, but we can. I was going to write them down, but I ran out of time. I suggest you start by giving them.

Thank you.