Modularity of the q-Pochhammer symbol and application
Drappeau, Sary
This talk will report on a work with S. Bettin (University of Genova) in which we obtained exact modularity relations for the q-Pochhammer symbol, which is a finite version of the Dedekind eta function. We will overview some of their useful aspects and applications, in particular to the value distribution of a certain knot invariants, the Kashaev invariants, constructed with q-Pochhammer symbols.
 Thank you. Thanks for the opportunity to talk about this topic during this week. So the main object of this topic is something called QPore Hammer Symbols. So first I have to introduce what it is. So the Po-Harmor Symbol is just a short name for something which is a convenient way to denote things in combinatorics and binomial coefficient. So I'm not at all an expert on combinatorics, but I need to introduce a name because that's an important. So the quantity which will be interesting for us is not exactly this one, but a Q variant, Q analog, which is the following one. So A and Q are complex numbers. And then the notation in combinatorics is the following one. There won't be much combinatorics in our work. It would be a good complex analysis and this kind of thing. But let's see. So this is the quantities. So there are many identities and generalization of binomial coefficients, which makes sense when you replace formally the Po-Harmor Symbol by Q Po-Harmor Symbols. Some formulas remain true. More precisely, what the things which would be recurrent in this talk is this one. When you set A equals to Q, so then we forget about the A. So Qn will be when we set A equals to Q. So it's the product of the one, the powers 1 minus Q to 1 minus Q to the n. So here n is an integer greater than or equal to 0. If n is 0, this product is empty. It's equal to 1. Okay. And Q can be anything. So why are people interested in this? So there are many reasons. I will quote the four connected ones. So it appears in partition-related appears. Okay. My handwriting usually is bad. No, it's even worse. It appears if you country the word, please, then if it appears in partition-related generating series. So I will quote some. So of course, there is a partition function. The number of ways to write an integer as a sum of positive integer. So it's denoted P of n. And this generating series is nothing else then. So it's easy to see if you group all the components of the sum together. It's an integrated product. And this is just you can denote formally by, so it's just Qn, but then you take the limit as n tends to infinity. So here Q has to be of modulus less than 1 for this formula to make sense. And this product is useful because so for instance, you can obtain a hardier management. Okay. The theorem on the asymptotic formula for P of n by exploiting the polar behavior of this product on the unit circle. There are many works in finding identities. Works. I'm finding identities between two series. It's a big business. I will quote one, which is called the Rogers Manusen identity. Which is the following. So if you take this summation, it's Q to the n squared divided by this symbol that we had before, then it's equal to the product resembling also interving the Q-PoH symbol, but this one extended to infinity. So now I need to use the other notation with the A. It's a product like this one, but restricted with congruence condition on R mod Q, mod 5. Q to the 4, Q to the 5. Okay. Another interesting is that it's connected to the connected to a particular module of form. So it's a function eta defined on the upper half plane. Okay. This is z of imaginary part, because in zero, which is so E of z over 24 times the power symbol extended to infinity. And E of z here is the one periodic e to the 2 pi i z, the one periodic exponential function. So saying that this function is a modular form, yes. So imaginary part of that is positive. So this number is of modules strictly less than one. So this makes sense. And saying that it's a modular form for this specific function means that it satisfies some identities. So because of course this is one invariant, when you shift z by one, you will just have a factor coming out, which is E of 1 over 24 times eta of z. And eta of minus 1 over z is z over i to the one half. So z over i has real part positive. So you take the principle determination of the logarithm times eta of z. So, okay. So closer to what I discussed today, there has been another line of research that was devoted to study sine products. So this is understanding the size as a complex number. The size and distribution of values of the programmer symbol when the modulus of q is one. Okay, so when this happens, of course, we have that this thing has modulus between 0 and 2. And note that, okay, so I use this. And, okay, so I recall the definition. This is the product from r from 1 to n of 1 minus q to the r. So if you try to imagine this as some kind of free man sum for function, you can, we noticed that the integral between 0 and 1, if you imagine that all these powers at q is a number, a typical number of the unit circle, then all these powers will become uniformly distributed. When you take the logarithm of this, you notice that the modulus, the nodule I tend to compensate each other. So because this holds, there is a lot of consolation you expect that for typical q, at least some consolation happened in this product. So the size of this product relates to the secondary terms in this human sum approximation. So this is a fact that makes this problem, tends to make this problem somewhat difficult. So I mentioned some names. So there is our works of Erdos and Seckeres, like the 50s, I think, followed by work of Södler, the 60s. So for instance, one of their theorems, the theorem of Södler, I think, is the fact that if you fix an n and you compute the maximum of this symbol on the circle, then this is exponential. There is a constant such that this relation holds. So it's n to some power times an exponential growing. So k is some constant. So I mentioned some further names. So Lubinski has studied this problem. And for instance, one of the theorems that he proved is that for a typical q on the for the bag almost all q on the unit circle, the size is actually much less than this. It's not exponentially than it's o of log n to the one plus epsilon for any epsilon. And for q, just one, q almost everywhere for the bag measure on the unit circle. One further work, which is, so this is 1990. Some further very recent work, very nice theorems. This has been revived very recently by Krebsstadt, several others, so Krebsstadt. Sorry, I apologize for this. A captain book, which are in the Norwegian, I think, and I'm going to learn from Linz. And then some two or three works with two or three of all three of these people, then followed by further work, joint is asked this late now. Christopher is now a new plastic now. And the theory of Paulos from Gratz and Benz border. So for deterministic, they studied deterministic q, fixed q, deterministic is a bit over. And especially for q equals the exponential. So on the unit circle with the angle, which is quadratic. A quadratic number. So for instance, what a theorem of Krebsstadt, a captain book and name is the following from 2019. This, the cube for a symbol at the golden ratio is bounded below by one and upper bounded by n. This was open. It's a question which was asked by at least means keep it from the fore. Okay, so when we started to study this function, so before we knew about so many of these works, we started, we were studying a paper by Zagier. That's the four, the force motivation. And in this paper, I will put this reference as I get 2010, he has a paper which is called quantum module forms in which he studies many functions. And there is a very funny function in there, which is related to not theory. So we don't know anything about not theory, but the function is not very complete. Well, it is not very complicated to write down. So we say it is also the cupola machine board is also a basic component in constructing quantum invariant quantum net invariant. And also not only for that. So Jones and for us, my name, which is associated with this in not series, the Kashi of invariance. So I won't try to draw a knot because this is not theory people. I would say miserably, but for also a knot is a diagram. Okay, with some crossing, it's a knot in three dimensions that you can represent it in two dimensions. It depends on some skills which I don't have here. And to a knot, okay, define the knot. When you draw it, you can, when you look at the diagram and the crossing and which brand, which part of the knot goes on to another, and you look at some combinatorics. And from this, you can construct some invariant. So Kashi of associates a function, which so we state two examples to show you what it looks like to make it a bit more concrete. So five one and five two are some knots. I can try to, maybe I can show them to you, but after the talk, because I need to find them on the, I didn't think of throwing, I'm sorry about this, which depends on the variable Q. And so for instance, in one of the most simple cases, the function is the following, with some from n from zero to infinity, the Kui per Haamer symbol modulus squared. Okay, so this is just, we take the product for any one equals one, the product is empty, in other words, you take all the, all the successive one minus Q to the i, modulus squared. Okay, so if Q is modulus then less than one, this of course diverges. If Q is modulus bigger than one, it also diverges, but when Q is modulus exactly one, the interesting case for us would be when Q is exponential of a rational X. So it's a root of unity. So when Q is a root of unity eventually this product will vanish. And so there's some extent for a trivial reason. So this, I will state another one because I will need it. It's to illustrate a later statement. So this one is a bit more complicated to write down, but it's to show you that sometimes the sounds are not always positive. Okay, Q bar n times modulus Qm squared. So now it's a double sum. Okay, sorry. There's no X there, it's capital N. What is capital N? It's the denominator, it's the order of Q as a root of unity. Okay, so okay, maybe strictly, otherwise there is a definition drop. All right, so why was Kachev interested in this invariance is that it's easy to compute given the knot and when you look at some asymptotic expansion of this knot, they will tell you information about the knot itself, about the geometry of the knot. And this is something that interests not theory. But the reason why we're interested in this function is that, well, it's a very simple looking function to involve the polymer symbol. And there are many things that we still don't quite know about this function. So here's a statement. It's called the volume conjecture, but it's actually a theorem in, it's not a theorem, still. It's a theorem in some cases. And the cases that I'll talk about today, it's a theorem. But for generic knot, it's still wide open and not theory people are very, it's an important conjecture, wide open conjecture. So when you go, when you look at the first truth of unity, and you let the order tend to infinity, then they expect that for any such invariant constructed from the knot, and especially the two that I wrote above, there will be an expansion like this. So it will be exponential in n with a CK is some geometric invariant, you think the knot and divided by 2 pi. So it's some constant. Which has a geometric meaning that I want to discuss today. Okay. But then, so this doesn't tell you what happens for other roots of unity. And this is one of the first question which motivated us. So very happy to try to understand this. So you let an irrational be given, it will be between zero and one. Maybe this. And, okay, I will write what will come here afterwards. But what we want to know is what's the size of J for E of X for other kind of rationales. And we can answer this question under, so a generalization of this theorem, which is a theorem of, I wish to take the names afterwards. Understand and Hansen, in the case of four one, we can generalize this statement for other rational functions. So this simple looking function from before, when you evaluate it as a at other rational, it will take the following form. Of course, there is a specific constant here, plus little of one times the quantity which will depend very simply on the on X, simply, yes and no, it's the continuous fraction expansion of X. And so this, if you use a classical notation for the continuous fraction expansion, R is the length. And this asymptotic formula is as the mean value of the AI tends to infinity. If you give you a sequence of rational numbers with their, the mean value of their coefficients standing to infinity, you will get an asymptotic formula for the log of the J invariant, for this case, for the not just not. And so we are very lucky here because this is typically the case. If you take one over N, it will, you obtain back this statement, but this in fact holds for many other rational numbers and actually almost all in a certain sense. That's a corollary. As Q tends to infinity for proportion one plus little of one of rationals. X with denominator, I was okay, I can, okay, I won't be using Q for the denominator, I need to stop doing this. Of denominator less than Q. So you, when you average over all rationals less than, denominator less than Q, when you pick one at random, then for almost all of them, you have the asymptotic formula that this thing is exponential of, so what I wrote before, but it turns out that, that for almost all rational, there is a simple low of large numbers. Well, okay, simple to state, but to prove it, you need to work a bit harder. So we obtained a low of large numbers with the sum of the continuous fraction digits, and from this you can immediately deduce. From this end of previous theorem, you can immediately deduce this statement. So it's a form of low of large numbers for the log of the J invariant introsional numbers. So you can ask about this condition, is it necessary? What? And we're so very happy because Christophe and Benz, Christophe Eisleiter and Benz Borda earlier this year, they obtained a very nice statement, which I'm going to state now. So the condition, this condition is actually necessary because of the following theorem of that, so you take one example of a number of which, a very simple example, the simplest example of a number for which the average doesn't go to infinity is when you take all the a i equal to one. And so if you fix for any n x being the rational with all the numbers equal to one, and I will say more about what they can do and what else they can do. So this is a question of two B-Shibonacci numbers. And for these numbers, so this depends on n, then the J function grows exponentially exactly like before, but now the constant is different. So this is the same sum, the sum of the a i, but now the constant is not the same, the constant c is 1.1 something, at least, well numerically, they show that numerically it seems that the function is like this. Maybe they have some proof that the first digits are this, but maybe Benz cannot surface around, but they show that this constant is different than the one from the almost all results above. So the result is much more general and in particular it works for all quadratic rational. So when the expansion is periodic, they can obtain this formula and the constant is not always the same as before. More general. X with quadratic. So now I'll explain a bit why all of a sudden we were talking about this pochameric symbol and then we got into a constant attraction expansion. So before this, maybe I can talk for a few instance and ask if there are any questions. There seems to be one. No, that was on the rule. Okay. There are no questions. There's no question at the moment in the chat. So maybe someone will arise questions. Okay. Seems not the case. So please. Okay, continue. So why is the continued fraction expansion relevant here? Okay. Continue traction expansion. So that's because of the conjecture by the game who observed numerically the following. So you let H of X be, so I will state it only for the variant that I mentioned before, H of X be the log J. See everything happens inside the log. So let's take the log. Leg log J of that function from before at E of X minus log J, same thing as before at E of 1 over X. Then this function, which is defined on Q, is extends to A to A. A nice function of X in R star. So what I mean by nice is, well, not very clearly, the best way I can illustrate this is by showing you a picture. So this is the newspaper, quantum radioforms on 2018. It has lots of 17 pages, but still has a lot of stuff. And he has the following two graphics. So this is a log of the function that we had, plotted at rational numbers. You can see the size grows at the denominator. And then it doesn't look so regular. But if you take the difference between the difference that I just wrote, you obtain this function, which is at least much nicer looking. It's not very clear what we conjecture, but by zooming in at a rational, we see that it has some jumps at rational. So certainly we don't expect it to be continuous. And at irrational numbers, so 1 over phi is here, the golden ratio. And you see, well, okay, it looks continuous. It's not so, doesn't look at all differentiable, but at least we can imagine that the function has a limit. The limit is close to 1.1. And this is related to the values that I showed you before. So let's see, sorry, this one. So let's see. h of X, it looks continuous near the golden ratio or the inverse of the golden ratio. And h at the golden ratio looks close to 1.1. And this explains, this would explain the theorem that I showed you before, the theorem AB. Okay, let me call this one the generic theorem. And secondly, let me get back to this function. You see, it goes to infinity. And it seems to go to infinity like 1 over X, it seems to go like the constant that I showed you before divided by X. As X goes to zero plus. And this would explain the generic theorem. So sorry, sorry for this. So it would explain the theorem that I called the generic theorem. Why? Well, this is because when you start with evaluating the function at a certain rational, when you replace X by 1 over X, because this is one periodic, you at each time you apply, you replace a continued fraction. Oops, sorry, I should take the log. You replace the continued fraction. So say this is X. I'll get h of X plus. So log J of 1 over X, but 1 over X, what do you do? 1 is just taking away the first coefficient for 1, 0 of E of 0, A2, AR. So if the first coefficient is big, it means that X is very small. And then this will look like C4, 1 times A1. And then when you iterate, you get C4, 1 times A2. Well, of course, these numbers have to be big for this to make sense. And we can only prove this when the average goes to infinity. Okay, yeah, maybe you can keep this section. Okay, so I have a little time to explain to you. Yeah, still have maybe 15 minutes. So I will set you what happens inside the, well, what is the main theorem that is driving everything. So we don't have this conjecture. If we had this conjecture, things would be much easier and also with this information. But we get around this conjecture. By proving two theorems, which are modularity looking, the modularity relation, much like the one I wrote before for eta, let me go back to it for an instant. Much like this one, but having to do with the Pochamer symbol. So the main theorem is that, so there are two. If I give you Q for gamma in SL2z, then the function which 2x and n associates E of x over 24. So this is in Q cross natural numbers. Well, actually cross. So let's see 0, denominator of x minus 1, because then it vanishes afterwards. Let me see. So E of x, so then, so this function, which is interesting for this value, x is rational and n varies between 0 and the denominator of x minus 1 satisfies, satisfies, satisfies a transformation formula. Which, maybe I can show you, show it to you if you want, but it will take some time. So I will concentrate first of this, of this informal description. It's a formula which relates x, which sends x to gamma x like modular iteration, and it sends n to, let's state it like this, yes, it sends n to the floor part of n divided by denominator of x times the denominator of gamma x. And this is a number between 0 and the denominator of gamma x. And this formula is, is exact. 0 term depends holomorphically on the following, on the, the fractional part of this number. So I have to explain to you what I mean by this. And yes, maybe I have time. It will take, I think, some five minutes more to explain the full statement of this theorem. So the first part, the formula that we have, I can show, can show it to you. Actually, I'm wondering, okay, it takes some notation, so more precisely. But the point, there is a point I want to make in this formula. And to make the point, I need to, to actually show the formula, but the formula is not very, not particularly good looking. So to state it, I need to, okay, x as a rational and gamma of x. Here is gamma of x. You can express it using nd on this, but first it will be h over k. Then for one less than r, less than k. If you let l to be the floor part that I wrote before, and lambda to be the fractional part that I wrote before, if you take the quotient between these two things, then this will be, this will be an expo, we can express this exactly as some of many terms. There is the first one which will be maybe ring a bell for people to study cotangent sum. So this is a dedicant sum plus pi i over four plus half log of k over d. I will explain each term. Plus k over q d times something, some function which is related to the logarithm evaluated at this number between zero and one plus nr term which depends on r, lambda, and d over k. The main point is that this term is holomorphic. It's a function which depends on l on lambda. It has an expression depending on lambda which makes it extendable to a holomorphic function. In lambda, so in this set, whatever, it's some other set, more general than this one. And this came as a surprise because all these are finite products. So maybe it doesn't come immediately as a surprise that such a formula exists because when you take out going to infinity, if you could, you would obtain the data kind, the eta function that I wrote before. And the eta function is a modular form. So you might expect something like this to exist. But this is not the infinite product. It's a finite product. And even by taking finite products, you can still, you have still a formula which is useful. It's useful for us. And this is relevant. It's relevant for us when the denominator of x is bounded. This is a case when we can control the error term. And there is a second important agreement that I want to mention in the previous theorem, which is a second relation formula. Now, okay, I won't state it, but there is a second second reciprocity formula. Okay, modularity, sorry, modularity formula, second modularity formula, which relates now it's a bit. There is another layer of complication coming from the fact that it doesn't relate x with gamma x or one over x is going to relate the gamma x bar, which is when x you write it as a irreducible fraction, x bar, which be each you inverse the numerator modulo the denominator. Of course, this depends. This is only defined modulo k. So we take fraction upon. There's a formula which relates this. And basically, it means that when you apply, it allows you to take away not the first coefficient when you is making UK algorithm, but the last coefficient. This transformation going from x to x bar is equivalent to reading the continuous fraction expansion in the reverse order. And there is a reciprocity formula, which relates these two for the Q-PoH symbol. And it's useful for us. So I'm going to lay a bit to make the point more clear. It's useful for us when the denominator of x goes to infinity. So it's actually depends on the way you stated it. But we do stated when which makes it useful when the denominator, the denominator stands to infinity. And this is precisely the complementary, the complementary case of the previous one. So if we could get rid of this condition, we would be getting very close to the guess conjecture. We cannot. We're not able to right now. But by proving this complementary modulity formula, which has a lot of disadvantages, but it has the one big advantage that it's useful as soon as the denominator goes to infinity. And by combining the two, we can successfully iterate the formula. And maybe I can finish with the last remark that in this formula, in the formula two, there are some peculiar objects that pop up, which are the following. It's a cotangent sum. So let me say, okay, n, k bar over h maybe. Oh, sorry. Okay, okay, sorry. I confused between the button for go back page and okay cotangent of pi n h bar over k. h inverse over k. This function is one periodic. So this makes sense. But you're okay. Times n over k. And if there was, if the sum was over all, residue, non-zero, residue, modulo k, this would be a cotangent sum. But in this formula, it turns out that very naturally, we didn't, we certainly didn't call for it, that this sum appears when r is restricted to be something else to the k. So it's a partial cotangent sum. And this, okay, we struggled for a long time with this kind of object. It has some reciprocity, but it's not, it's not exact. So, okay, there is something, there was something new going on in this formula that we didn't expect. So if you stumble on something like this, maybe take a look at this pair of paper or Loubinski's paper and let's save you some headaches. And so, yeah, okay, so you can say more about some elements of the proof if someone asked, but otherwise I stop here. Thank you for your attention. Yeah, Saris, thank you very much for this very nice Blackboard talk. So we could end this conference with a Blackboard talk. Of course, in auditorium A1 in Exier, we have much more Blackboards. We have three big ones and some smaller ones. Here is just one Blackboard, but it has one advantage compared to a real Blackboard one. We can go back to any Blackboard. So I have a small question. You introduced this geometric constants somewhere in the first third of the talk. Could you say some words on the geometric interpretation of that constant? Yes, so maybe for this I can find a picture of the knot. So I need to find a picture of the knot. Let me see if it's in the gift paper. Because these interpretation as knots would be really very instructive. Yeah, so let me get right back with the screenshot. I need to find a picture of the 4-1 knot. But if it takes too long, we should postpone it to discussion room or something like that. It's okay, I found it. It's actually on Wikipedia. It wasn't difficult. Okay, so let me see. Yes, so this is the figure 8 knot. This is the 4-1 knot that I was talking about. And when you consider the complement of the knot, so everywhere in the space except inside, it turns out there is a theorem of... Okay, I forgot his name. Okay, that's embarrassing. He's one of the biggest people in this thing. You can put a hyperbolic structure, a metric, on the complement of the space, which makes it a hyperbolic space and it has finite volume. And the constant is the volume of the space. Maybe it's you to make my... Sorry? Maybe it's you to make my own. I don't know, but it could be. No? Okay. Yeah, I don't remember something. I'm sorry. It's not... It's not... It's not... It's not embarrassing. Yes, okay, so he showed that you can decompose the complement space into a simpler hyperbolic space. And when you compute the volume, it's interesting, intrinsic quantity, and the constant is the volume of that. Okay, so thank you for this very nice explanation. Are there other questions? Short ones? Long ones are postponed to the discussion rooms. Yeah, Benze is typing a question. Maybe it will come soon. Yeah, here is... Can you read the question? Yeah, so is it conjecture that the guess function is continuous at every rational? And I believe so, yes. And it certainly looks so numerically. Of course, you have to... Of the numerical, so I can show you why. This would be a funny explanation. And there is a graph from this paper. So this is what you happen when you compare j of x with j of 1 over x. And this is what you happen when you compare j of x bar with j of 1 over x bar. So it looks much or less the same. But this function takes actually... It goes to plus and minus infinity at the neighborhood of any rational number. So it looks to be nice and continuous, but when you zoom in, you remark that it's not very nice at all. So you need to be aware of the conjecture, but I believe that h of x is continuous at every rational.
 Thank you. Thanks for the opportunity to talk about this topic during this week.  So the main object of this topic is something called QPore Hammer Symbols. So first I have  to introduce what it is. So the Po-Harmor Symbol is just a short name for something which  is a convenient way to denote things in combinatorics and binomial coefficient. So I'm not at all  an expert on combinatorics, but I need to introduce a name because that's an important.  So the quantity which will be interesting for us is not exactly this one, but a Q variant,  Q analog, which is the following one. So A and Q are complex numbers.  And then the notation in combinatorics is the following one. There won't be much combinatorics  in our work. It would be a good complex analysis and this kind of thing. But let's see. So this is  the quantities. So there are many identities and generalization of binomial coefficients,  which makes sense when you replace formally the Po-Harmor Symbol by Q Po-Harmor Symbols.  Some formulas remain true. More precisely, what the things which would be recurrent in this talk  is this one. When you set A equals to Q, so then we forget about the A. So Qn will be when we set  A equals to Q. So it's the product of the one, the powers 1 minus Q to 1 minus Q to the n.  So here n is an integer greater than or equal to 0. If n is 0, this product is empty. It's equal to 1.  Okay. And Q can be anything. So why are people interested in this? So there are many reasons.  I will quote the four connected ones. So it appears in partition-related appears. Okay.  My handwriting usually is bad. No, it's even worse. It appears if you country the word, please, then  if it appears in partition-related generating series.  So I will quote some. So of course, there is a partition function. The number of ways to write  an integer as a sum of positive integer. So it's denoted P of n. And this generating series  is nothing else then. So it's easy to see if you group all the components of the sum together.  It's an integrated product. And this is just you can denote formally by, so it's just Qn,  but then you take the limit as n tends to infinity. So here Q has to be of modulus less than 1 for  this formula to make sense. And this product is useful because so for instance, you can obtain a  hardier management. Okay. The theorem on the asymptotic formula for P of n by exploiting the  polar behavior of this product on the unit circle. There are many works in finding identities.  Works. I'm finding identities between two series. It's a big business.  I will quote one, which is called the Rogers Manusen identity.  Which is the following. So if you take this summation,  it's Q to the n squared divided by this symbol that we had before, then it's equal to the product  resembling also interving the Q-PoH symbol, but this one extended to infinity. So now I need  to use the other notation with the A. It's a product like this one, but restricted with congruence  condition on R mod Q, mod 5. Q to the 4, Q to the 5. Okay. Another interesting is that it's connected to  the connected to a particular module of form.  So it's a function eta defined on the upper half plane. Okay. This is z  of imaginary part, because in zero, which is so  E of z over 24 times the power symbol extended to infinity. And E of z here is the one periodic  e to the 2 pi i z, the one periodic exponential function. So saying that this function is a  modular form, yes. So imaginary part of that is positive. So this number is of modules  strictly less than one. So this makes sense. And saying that it's a modular form for this  specific function means that it satisfies some identities. So because of course this is one  invariant, when you shift z by one, you will just have a factor coming out, which is E of  1 over 24 times eta of z. And eta of minus 1 over z is z over i to the one half. So z over i has  real part positive. So you take the principle determination of the logarithm times eta of z.  So, okay. So closer to what I discussed today, there has been another line of research  that was devoted to study sine products. So this is understanding the size as a complex number.  The size and distribution of values of the programmer symbol when the modulus of q is one.  Okay, so when this happens, of course, we have that this thing has modulus between 0 and 2.  And note that, okay, so I use this. And, okay, so I recall the definition. This is the product  from r from 1 to n of 1 minus q to the r. So if you try to imagine this as some kind of  free man sum for function, you can, we noticed that the integral between 0 and 1, if you imagine  that all these powers at q is a number, a typical number of the unit circle, then all these powers  will become uniformly distributed. When you take the logarithm of this, you notice that the modulus,  the nodule I tend to compensate each other. So because this holds, there is a lot of  consolation you expect that for typical q, at least some consolation happened in this product.  So the size of this product relates to the secondary terms in this  human sum approximation. So this is a fact that makes this problem, tends to make this problem somewhat difficult.  So I mentioned some names. So there is our works of Erdos and Seckeres,  like the 50s, I think, followed by work of Södler, the 60s. So for instance, one of their theorems,  the theorem of Södler, I think, is the fact that if you fix an n and you compute the maximum  of this symbol on the circle, then this is exponential. There is a constant such that this relation holds.  So it's n to some power times an exponential growing. So k is some constant.  So I mentioned some further names. So Lubinski has studied this  problem. And for instance, one of the theorems that he proved is that for a typical q on the  for the bag almost all q on the unit circle, the size is actually much less than this.  It's not exponentially than it's o of log n to the one plus epsilon for any epsilon.  And for q, just one, q almost everywhere for the bag measure on the unit circle.  One further work, which is, so this is 1990.  Some further very recent work, very nice theorems. This has been revived very recently by Krebsstadt,  several others, so Krebsstadt. Sorry, I apologize for this. A captain book,  which are in the Norwegian, I think, and I'm going to learn from Linz.  And then some two or three works with two or three of all three of these people,  then followed by further work, joint is asked this late now. Christopher is now  a new plastic now. And the theory of Paulos  from Gratz and Benz border. So for deterministic, they studied deterministic q, fixed q, deterministic  is a bit over. And especially for q equals the exponential. So on the unit circle with the  angle, which is quadratic.  A quadratic number. So for instance, what a theorem of Krebsstadt,  a captain book and name is the following from 2019. This, the cube for a symbol at the golden  ratio is bounded below by one and upper bounded by n. This was open.  It's a question which was asked by at least means keep it from the fore.  Okay, so  when we started to study this function, so before we knew about so many of these works,  we started, we were studying a paper by Zagier.  That's the four, the force motivation. And in this paper,  I will put this reference as I get 2010, he has a paper which is called quantum module forms  in which he studies many functions. And there is a very funny function in there,  which is related to not theory. So we don't know anything about not theory, but the function is  not very complete. Well, it is not very complicated to write down. So we say it is also the cupola  machine board is also a basic component in constructing  quantum invariant quantum net invariant. And also not only for that.  So Jones and for us, my name, which is associated with this in not series, the Kashi  of invariance. So I won't try to draw a knot because this is not theory people.  I would say miserably, but for also a knot is a diagram.  Okay, with some crossing, it's a knot in three dimensions that you can represent it in two  dimensions. It depends on some skills which I don't have here. And to a knot,  okay, define the knot. When you draw it, you can, when you look at the diagram and the crossing  and which brand, which part of the knot goes on to another, and you look at some combinatorics.  And from this, you can construct some invariant. So Kashi of  associates a function, which so we state two examples to  show you what it looks like to make it a bit more concrete. So five one and five two are some knots.  I can try to, maybe I can show them to you, but after the talk, because I need to find them on  the, I didn't think of throwing, I'm sorry about this, which depends on the variable Q.  And so for instance, in one of the most simple cases, the function is the following,  with some from n from zero to infinity, the Kui per Haamer symbol modulus squared.  Okay, so this is just, we take the product for any one equals one, the product is empty,  in other words, you take all the, all the successive one minus Q to the i, modulus squared.  Okay, so if Q is modulus then less than one, this of course diverges. If Q is modulus bigger  than one, it also diverges, but when Q is modulus exactly one, the interesting case for us would  be when Q is exponential of a rational X. So it's a root of unity.  So when Q is a root of unity eventually this product will vanish. And so there's some  extent for a trivial reason. So this, I will state another one because  I will need it. It's to illustrate a later statement.  So this one is a bit more complicated to write down, but it's to show you that sometimes the  sounds are not always positive.  Okay, Q bar n times modulus Qm squared. So now it's a double sum.  Okay, sorry. There's no X there, it's capital N.  What is capital N? It's the denominator, it's the order of Q as a root of unity.  Okay, so okay, maybe strictly, otherwise there is a definition drop.  All right, so why was Kachev interested in this  invariance is that it's easy to compute given the knot and when you look at some  asymptotic expansion of this knot, they will tell you information about the knot itself,  about the geometry of the knot. And this is something that interests  not theory. But the reason why we're interested in this function is that, well,  it's a very simple looking function to involve the polymer symbol. And there are many things  that we still don't quite know about this function. So here's a statement. It's called the volume  conjecture, but it's actually a theorem in, it's not a theorem, still.  It's a theorem in some cases. And the cases that I'll talk about today, it's a theorem.  But for generic knot, it's still wide open and not theory people are very,  it's an important conjecture, wide open conjecture. So when you go, when you look at the first  truth of unity, and you let the order tend to infinity, then they expect that for any  such invariant constructed from the knot, and especially the two that I wrote above,  there will be an expansion like this. So it will be exponential in n with a CK is some  geometric invariant, you think the knot and divided by 2 pi. So it's some constant.  Which has a geometric meaning that I want to discuss today.  Okay. But then, so this doesn't tell you what happens for other roots of unity. And this is  one of the first question which motivated us. So very happy to try to understand this. So you  let an irrational be given, it will be between zero and one. Maybe this. And, okay, I will write  what will come here afterwards. But what we want to know is what's the size of J for E of X for  other kind of rationales. And we can answer this question under, so a generalization of this theorem,  which is a theorem of, I wish to take the names afterwards. Understand and Hansen,  in the case of four one, we can generalize this statement for other rational functions.  So this simple looking function from before, when you evaluate it as a  at other rational, it will take the following form.  Of course, there is a specific constant here, plus little of one times  the quantity which will depend very simply on the on X, simply, yes and no, it's the  continuous fraction expansion of X. And so this, if you use a classical notation for the  continuous fraction expansion, R is the length. And this asymptotic formula is as  the mean value of the AI tends to infinity. If you give you a sequence of rational numbers  with their, the mean value of their coefficients standing to infinity,  you will get an asymptotic formula for the log of the J invariant, for this case, for the not just  not. And so we are very lucky here because this is typically the case.  If you take one over N, it will, you obtain back this statement, but this in fact holds for many  other rational numbers and actually almost all in a certain sense. That's a corollary.  As Q tends to infinity for proportion  one plus little of one of rationals.  X with denominator, I was okay, I can, okay, I won't be using Q for the denominator, I need to  stop doing this. Of denominator less than Q. So you, when you average over all  rationals less than, denominator less than Q, when you pick one at random, then for almost all of them,  you have the asymptotic formula that this thing is exponential of, so what I wrote before, but it  turns out that, that for almost all rational, there is a simple low of large numbers. Well,  okay, simple to state, but to prove it, you need to work a bit harder.  So we obtained a low of large numbers with the sum of the continuous fraction digits,  and from this you can immediately deduce. From this end of previous theorem, you can immediately  deduce this statement. So it's a form of low of large numbers for the log of the J invariant  introsional numbers. So you can ask about this condition, is it necessary? What? And  we're so very happy because Christophe and Benz, Christophe Eisleiter and Benz Borda  earlier this year, they obtained a very nice statement, which I'm going to state now. So the condition,  this condition is actually necessary  because of the following theorem of  that, so you take one example of a number of which,  a very simple example, the simplest example of a number for which the average doesn't go to  infinity is when you take all the a i equal to one. And so if you fix for any n x being the  rational with all the numbers equal to one, and I will say more about what they can do  and what else they can do. So this is a question of two B-Shibonacci numbers.  And for these numbers, so this depends on n, then the J function  grows exponentially exactly like before, but now the constant is different.  So this is the same sum, the sum of the a i, but now the constant is not the same,  the constant c is 1.1 something, at least, well numerically,  they show that  numerically it seems that the function is like this. Maybe they have some proof that the first  digits are this, but maybe Benz cannot surface around, but they show that this constant is  different than the one from the almost all results above.  So the result is much more general and in particular it works for all quadratic rational.  So when the expansion is periodic, they can obtain this formula and the constant is not  always the same as before. More general.  X with quadratic.  So now I'll explain a bit why all of a sudden we were talking about this  pochameric symbol and then we got into a constant attraction expansion.  So before this, maybe I can talk for a few instance and ask if there are any questions.  There seems to be one.  No, that was on the rule. Okay. There are no questions.  There's no question at the moment in the chat. So maybe someone will arise questions.  Okay. Seems not the case. So please. Okay, continue. So why is the continued fraction  expansion relevant here?  Okay. Continue traction expansion.  So that's because of the conjecture by the game who observed numerically the following.  So you let H of X be, so I will state it only for the variant that I mentioned before,  H of X be the log J. See everything happens inside the log. So let's take the log.  Leg log J of that function from before at E of X minus log J, same thing as before at E of 1 over X.  Then this function, which is defined on Q, is extends to A to A.  A nice function of X in R star. So what I mean by nice is, well, not very clearly,  the best way I can illustrate this is by showing you a picture. So this is the newspaper,  quantum radioforms on 2018. It has lots of 17 pages, but still has a lot of stuff.  And he has the following two graphics. So this is a log of the function that we had,  plotted at rational numbers. You can see the size grows at the denominator.  And then it doesn't look so regular. But if you take the difference between the difference that  I just wrote, you obtain this function, which is at least much nicer looking. It's not very clear  what we conjecture, but by zooming in at a rational, we see that it has some jumps at  rational. So certainly we don't expect it to be continuous. And at irrational numbers,  so 1 over phi is here, the golden ratio. And you see, well, okay, it looks continuous.  It's not so, doesn't look at all differentiable, but at least we can imagine that the function  has a limit. The limit is close to 1.1. And this is related to the values that I showed you before.  So let's see, sorry, this one. So let's see. h of X, it looks continuous  near the golden ratio or the inverse of the golden ratio.  And h at the golden ratio looks close to 1.1. And this explains, this would explain  the theorem that I showed you before, the theorem AB.  Okay, let me call this one the generic theorem.  And secondly, let me get back to this function. You see, it goes to infinity.  And it seems to go to infinity like 1 over X, it seems to go like the constant that I  showed you before divided by X. As X goes to zero plus. And this would explain  the generic theorem. So sorry, sorry for this. So it would explain the theorem that I called  the generic theorem. Why? Well, this is because when you start with evaluating the function  at a certain rational, when you replace X by 1 over X, because this is one periodic,  you at each time you apply, you replace a continued fraction.  Oops, sorry, I should take the log. You replace the continued fraction.  So say this is X.  I'll get h of X plus. So log J of 1 over X, but 1 over X, what do you do? 1 is just taking away  the first coefficient for 1, 0 of E of 0, A2, AR. So if the first coefficient is big,  it means that X is very small. And then this will look like C4, 1 times A1.  And then when you iterate, you get C4, 1 times A2. Well, of course, these numbers have to be big  for this to make sense. And we can only prove this when the average goes to infinity.  Okay, yeah, maybe you can keep this section. Okay, so I have a little time to explain to you.  Yeah, still have maybe 15 minutes. So I will set you what happens inside the, well,  what is the main theorem that is driving everything. So we don't have this conjecture.  If we had this conjecture, things would be much easier and also with this information.  But we get around this conjecture.  By proving two theorems,  which are modularity looking,  the modularity relation, much like the one I wrote before for eta, let me go back to it for  an instant. Much like this one, but having to do with the Pochamer symbol.  So the main theorem is that, so there are two. If I give you  Q for gamma in SL2z, then the function which 2x and n associates E of x over 24. So this is in Q  cross natural numbers. Well, actually cross. So let's see 0, denominator of x minus 1,  because then it vanishes afterwards. Let me see. So E of x, so then, so this function,  which is interesting for this value, x is rational and n varies between 0 and the denominator of x  minus 1 satisfies, satisfies, satisfies a transformation formula.  Which, maybe I can show you, show it to you if you want, but it will take some time. So I will  concentrate first of this, of this informal description. It's a formula which relates x,  which sends x to gamma x like modular iteration, and it sends n to, let's state it like this,  yes, it sends n to the floor part of n divided by denominator of x times the denominator of gamma x.  And this is a number between 0 and the denominator of gamma x.  And this formula is, is exact.  0 term depends holomorphically  on the following, on the, the fractional part of this number. So I have to explain to you what I mean  by this. And yes, maybe I have time. It will take, I think, some five minutes more to explain  the full statement of this theorem. So the first part, the formula that we have, I can show,  can show it to you.  Actually, I'm wondering, okay, it takes some notation, so more precisely. But the point,  there is a point I want to make in this formula. And to make the point, I need to,  to actually show the formula, but the formula is not very,  not particularly good looking. So to state it, I need to, okay,  x as a rational and gamma of x. Here is gamma of x. You can express it using nd on this,  but first it will be h over k. Then for one less than r, less than k. If you let l to be  the floor part that I wrote before, and lambda to be the fractional part that I wrote before,  if you take the quotient between these two things,  then this will be, this will be an expo, we can express this exactly as some of many terms.  There is the first one which will be maybe ring a bell for people to study cotangent sum. So this  is a dedicant sum plus pi i over four plus half log of k over d. I will explain each term.  Plus k over q d times something, some function which is related to the logarithm  evaluated at this number between zero and one plus nr term which depends on r, lambda,  and d over k. The main point is that this term is holomorphic. It's a function which depends  on l on lambda. It has an expression depending on lambda which makes it extendable to a holomorphic  function. In lambda, so in this set, whatever, it's some other set, more general than this one.  And this came as a surprise because all these are finite products. So maybe it doesn't come  immediately as a surprise that such a formula exists because when you take out going to infinity,  if you could, you would obtain the data kind, the eta function that I wrote before.  And the eta function is a modular form. So you might expect something like this to exist.  But this is not the infinite product. It's a finite product. And even by taking finite products,  you can still, you have still a formula which is useful. It's useful for us.  And this is relevant. It's relevant for us  when the denominator of x is bounded. This is a case when we can control the error term.  And there is a second important agreement that I want to mention in the previous theorem,  which is a second relation formula. Now, okay, I won't state it, but there is a second  second reciprocity formula. Okay, modularity, sorry, modularity formula, second modularity  formula, which relates  now it's a bit. There is another layer of  complication coming from the fact that it doesn't relate x with gamma x or one over x is going to  relate the gamma x bar, which is when x you write it as a irreducible fraction, x bar, which be  each you inverse the numerator modulo the denominator.  Of course, this depends. This is only defined modulo k. So we take fraction upon.  There's a formula which relates this. And basically, it means that when you apply,  it allows you to take away not the first coefficient when you is making UK algorithm,  but the last coefficient. This transformation going from x to x bar is equivalent to reading the  continuous fraction expansion in the reverse order. And there is a reciprocity formula,  which relates these two for the Q-PoH symbol. And it's useful for us.  So I'm going to lay a bit to make the point more clear. It's useful for us when  the denominator of x goes to infinity. So it's actually depends on the way you  stated it. But we do stated when which makes it useful when the denominator,  the denominator stands to infinity. And this is precisely the complementary,  the complementary case of the previous one. So if we could get rid of this condition,  we would be getting very close to the guess conjecture. We cannot. We're not able to right now.  But by proving this complementary modulity formula, which has a lot of disadvantages,  but it has the one big advantage that it's useful as soon as the denominator goes to infinity.  And by combining the two, we can successfully iterate the formula.  And maybe I can finish with the last remark that in this formula, in the formula two,  there are some peculiar objects that pop up, which are the following. It's a cotangent sum.  So let me say, okay, n, k bar over h maybe.  Oh, sorry. Okay, okay, sorry. I confused between the button for go back page and  okay cotangent of pi n h bar over k.  h inverse over k. This function is one periodic. So this makes sense. But you're okay.  Times n over k. And if there was, if the sum was over all,  residue, non-zero, residue, modulo k, this would be a cotangent sum.  But in this formula, it turns out that very naturally, we didn't, we certainly didn't call for it,  that this sum appears when r is restricted to be something else to the k. So it's a partial  cotangent sum. And this, okay, we struggled for a long time with this kind of object.  It has some reciprocity, but it's not, it's not exact. So, okay, there is something,  there was something new going on in this formula that we didn't expect. So if you stumble on something  like this, maybe take a look at this pair of paper or Loubinski's paper and let's save you some headaches.  And so, yeah, okay, so you can say more about some elements of the proof if someone asked, but  otherwise I stop here. Thank you for your attention.  Yeah, Saris, thank you very much for this very nice Blackboard talk. So we could end this  conference with a Blackboard talk. Of course, in auditorium A1 in Exier, we have much more  Blackboards. We have three big ones and some smaller ones. Here is just one Blackboard,  but it has one advantage compared to a real Blackboard one. We can go back to any Blackboard.  So I have a small question. You introduced this geometric constants somewhere in the first third  of the talk. Could you say some words on the geometric interpretation of that constant?  Yes, so maybe for this I can find a picture of the knot. So I need to find a picture of the knot.  Let me see if it's in the gift paper. Because these interpretation as knots would be really  very instructive. Yeah, so let me get right back with the screenshot. I need to find a picture of  the 4-1 knot. But if it takes too long, we should postpone it to discussion room or something like  that. It's okay, I found it. It's actually on Wikipedia. It wasn't difficult. Okay, so let me see.  Yes, so this is the figure 8 knot. This is the 4-1 knot that I was talking about.  And when you consider the complement of the knot, so everywhere in the space except inside,  it turns out there is a theorem of...  Okay, I forgot his name. Okay, that's embarrassing. He's one of the biggest people in this thing.  You can put a hyperbolic structure, a metric, on the complement of the space,  which makes it a hyperbolic space and it has finite volume. And the constant is the volume  of the space. Maybe it's you to make my... Sorry? Maybe it's you to make my own. I don't know,  but it could be. No? Okay. Yeah, I don't remember something. I'm sorry. It's not...  It's not... It's not... It's not embarrassing. Yes, okay, so he showed that you can decompose  the complement space into a simpler hyperbolic space. And when you compute the volume, it's  interesting, intrinsic quantity, and the constant is the volume of that. Okay, so thank you for  this very nice explanation. Are there other questions? Short ones? Long ones are postponed  to the discussion rooms. Yeah, Benze is typing a question.  Maybe it will come soon. Yeah, here is... Can you read the question?  Yeah, so is it conjecture that the guess function is continuous at every rational?  And I believe so, yes. And it certainly looks so numerically. Of course, you have to...  Of the numerical, so I can show you why. This would be a funny explanation.  And there is a graph from this paper. So this is what you happen when you compare j of x with  j of 1 over x. And this is what you happen when you compare j of x bar with j of 1 over x bar.  So it looks much or less the same. But this function takes actually... It goes to plus and minus  infinity at the neighborhood of any rational number. So it looks to be nice and continuous,  but when you zoom in, you remark that it's not very nice at all. So you need to be aware of  the conjecture, but I believe that h of x is continuous at every rational.
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Thank you. Thanks for the opportunity to talk about this topic during this week. So the main object of this topic is something called QPore Hammer Symbols. So first I have to introduce what it is. So the Po-Harmor Symbol is just a short name for something which is a convenient way to denote things in combinatorics and binomial coefficient. So I'm not at all an expert on combinatorics, but I need to introduce a name because that's an important. So the quantity which will be interesting for us is not exactly this one, but a Q variant, Q analog, which is the following one. So A and Q are complex numbers. And then the notation in combinatorics is the following one. There won't be much combinatorics in our work. It would be a good complex analysis and this kind of thing. But let's see. So this is the quantities. So there are many identities and generalization of binomial coefficients, which makes sense when you replace formally the Po-Harmor Symbol by Q Po-Harmor Symbols. Some formulas remain true. More precisely, what the things which would be recurrent in this talk is this one. When you set A equals to Q, so then we forget about the A. So Qn will be when we set A equals to Q. So it's the product of the one, the powers 1 minus Q to 1 minus Q to the n. So here n is an integer greater than or equal to 0. If n is 0, this product is empty. It's equal to 1. Okay. And Q can be anything. So why are people interested in this? So there are many reasons. I will quote the four connected ones. So it appears in partition-related appears. Okay. My handwriting usually is bad. No, it's even worse. It appears if you country the word, please, then if it appears in partition-related generating series. So I will quote some. So of course, there is a partition function. The number of ways to write an integer as a sum of positive integer. So it's denoted P of n. And this generating series is nothing else then. So it's easy to see if you group all the components of the sum together. It's an integrated product. And this is just you can denote formally by, so it's just Qn, but then you take the limit as n tends to infinity. So here Q has to be of modulus less than 1 for this formula to make sense. And this product is useful because so for instance, you can obtain a hardier management. Okay. The theorem on the asymptotic formula for P of n by exploiting the polar behavior of this product on the unit circle. There are many works in finding identities. Works. I'm finding identities between two series. It's a big business. I will quote one, which is called the Rogers Manusen identity. Which is the following. So if you take this summation, it's Q to the n squared divided by this symbol that we had before, then it's equal to the product resembling also interving the Q-PoH symbol, but this one extended to infinity. So now I need to use the other notation with the A. It's a product like this one, but restricted with congruence condition on R mod Q, mod 5. Q to the 4, Q to the 5. Okay. Another interesting is that it's connected to the connected to a particular module of form. So it's a function eta defined on the upper half plane. Okay. This is z of imaginary part, because in zero, which is so E of z over 24 times the power symbol extended to infinity. And E of z here is the one periodic e to the 2 pi i z, the one periodic exponential function. So saying that this function is a modular form, yes. So imaginary part of that is positive. So this number is of modules strictly less than one. So this makes sense. And saying that it's a modular form for this specific function means that it satisfies some identities. So because of course this is one invariant, when you shift z by one, you will just have a factor coming out, which is E of 1 over 24 times eta of z. And eta of minus 1 over z is z over i to the one half. So z over i has real part positive. So you take the principle determination of the logarithm times eta of z. So, okay. So closer to what I discussed today, there has been another line of research that was devoted to study sine products. So this is understanding the size as a complex number. The size and distribution of values of the programmer symbol when the modulus of q is one. Okay, so when this happens, of course, we have that this thing has modulus between 0 and 2. And note that, okay, so I use this. And, okay, so I recall the definition. This is the product from r from 1 to n of 1 minus q to the r. So if you try to imagine this as some kind of free man sum for function, you can, we noticed that the integral between 0 and 1, if you imagine that all these powers at q is a number, a typical number of the unit circle, then all these powers will become uniformly distributed. When you take the logarithm of this, you notice that the modulus, the nodule I tend to compensate each other. So because this holds, there is a lot of consolation you expect that for typical q, at least some consolation happened in this product. So the size of this product relates to the secondary terms in this human sum approximation. So this is a fact that makes this problem, tends to make this problem somewhat difficult. So I mentioned some names. So there is our works of Erdos and Seckeres, like the 50s, I think, followed by work of Södler, the 60s. So for instance, one of their theorems, the theorem of Södler, I think, is the fact that if you fix an n and you compute the maximum of this symbol on the circle, then this is exponential. There is a constant such that this relation holds. So it's n to some power times an exponential growing. So k is some constant. So I mentioned some further names. So Lubinski has studied this problem. And for instance, one of the theorems that he proved is that for a typical q on the for the bag almost all q on the unit circle, the size is actually much less than this. It's not exponentially than it's o of log n to the one plus epsilon for any epsilon. And for q, just one, q almost everywhere for the bag measure on the unit circle. One further work, which is, so this is 1990. Some further very recent work, very nice theorems. This has been revived very recently by Krebsstadt, several others, so Krebsstadt. Sorry, I apologize for this. A captain book, which are in the Norwegian, I think, and I'm going to learn from Linz. And then some two or three works with two or three of all three of these people, then followed by further work, joint is asked this late now. Christopher is now a new plastic now. And the theory of Paulos from Gratz and Benz border. So for deterministic, they studied deterministic q, fixed q, deterministic is a bit over. And especially for q equals the exponential. So on the unit circle with the angle, which is quadratic. A quadratic number. So for instance, what a theorem of Krebsstadt, a captain book and name is the following from 2019. This, the cube for a symbol at the golden ratio is bounded below by one and upper bounded by n. This was open. It's a question which was asked by at least means keep it from the fore. Okay, so when we started to study this function, so before we knew about so many of these works, we started, we were studying a paper by Zagier. That's the four, the force motivation. And in this paper, I will put this reference as I get 2010, he has a paper which is called quantum module forms in which he studies many functions. And there is a very funny function in there, which is related to not theory. So we don't know anything about not theory, but the function is not very complete. Well, it is not very complicated to write down. So we say it is also the cupola machine board is also a basic component in constructing quantum invariant quantum net invariant. And also not only for that. So Jones and for us, my name, which is associated with this in not series, the Kashi of invariance. So I won't try to draw a knot because this is not theory people. I would say miserably, but for also a knot is a diagram. Okay, with some crossing, it's a knot in three dimensions that you can represent it in two dimensions. It depends on some skills which I don't have here. And to a knot, okay, define the knot. When you draw it, you can, when you look at the diagram and the crossing and which brand, which part of the knot goes on to another, and you look at some combinatorics. And from this, you can construct some invariant. So Kashi of associates a function, which so we state two examples to show you what it looks like to make it a bit more concrete. So five one and five two are some knots. I can try to, maybe I can show them to you, but after the talk, because I need to find them on the, I didn't think of throwing, I'm sorry about this, which depends on the variable Q. And so for instance, in one of the most simple cases, the function is the following, with some from n from zero to infinity, the Kui per Haamer symbol modulus squared. Okay, so this is just, we take the product for any one equals one, the product is empty, in other words, you take all the, all the successive one minus Q to the i, modulus squared. Okay, so if Q is modulus then less than one, this of course diverges. If Q is modulus bigger than one, it also diverges, but when Q is modulus exactly one, the interesting case for us would be when Q is exponential of a rational X. So it's a root of unity. So when Q is a root of unity eventually this product will vanish. And so there's some extent for a trivial reason. So this, I will state another one because I will need it. It's to illustrate a later statement. So this one is a bit more complicated to write down, but it's to show you that sometimes the sounds are not always positive. Okay, Q bar n times modulus Qm squared. So now it's a double sum. Okay, sorry. There's no X there, it's capital N. What is capital N? It's the denominator, it's the order of Q as a root of unity. Okay, so okay, maybe strictly, otherwise there is a definition drop. All right, so why was Kachev interested in this invariance is that it's easy to compute given the knot and when you look at some asymptotic expansion of this knot, they will tell you information about the knot itself, about the geometry of the knot. And this is something that interests not theory. But the reason why we're interested in this function is that, well, it's a very simple looking function to involve the polymer symbol. And there are many things that we still don't quite know about this function. So here's a statement. It's called the volume conjecture, but it's actually a theorem in, it's not a theorem, still. It's a theorem in some cases. And the cases that I'll talk about today, it's a theorem. But for generic knot, it's still wide open and not theory people are very, it's an important conjecture, wide open conjecture. So when you go, when you look at the first truth of unity, and you let the order tend to infinity, then they expect that for any such invariant constructed from the knot, and especially the two that I wrote above, there will be an expansion like this. So it will be exponential in n with a CK is some geometric invariant, you think the knot and divided by 2 pi. So it's some constant. Which has a geometric meaning that I want to discuss today. Okay. But then, so this doesn't tell you what happens for other roots of unity. And this is one of the first question which motivated us. So very happy to try to understand this. So you let an irrational be given, it will be between zero and one. Maybe this. And, okay, I will write what will come here afterwards. But what we want to know is what's the size of J for E of X for other kind of rationales. And we can answer this question under, so a generalization of this theorem, which is a theorem of, I wish to take the names afterwards. Understand and Hansen, in the case of four one, we can generalize this statement for other rational functions. So this simple looking function from before, when you evaluate it as a at other rational, it will take the following form. Of course, there is a specific constant here, plus little of one times the quantity which will depend very simply on the on X, simply, yes and no, it's the continuous fraction expansion of X. And so this, if you use a classical notation for the continuous fraction expansion, R is the length. And this asymptotic formula is as the mean value of the AI tends to infinity. If you give you a sequence of rational numbers with their, the mean value of their coefficients standing to infinity, you will get an asymptotic formula for the log of the J invariant, for this case, for the not just not. And so we are very lucky here because this is typically the case. If you take one over N, it will, you obtain back this statement, but this in fact holds for many other rational numbers and actually almost all in a certain sense. That's a corollary. As Q tends to infinity for proportion one plus little of one of rationals. X with denominator, I was okay, I can, okay, I won't be using Q for the denominator, I need to stop doing this. Of denominator less than Q. So you, when you average over all rationals less than, denominator less than Q, when you pick one at random, then for almost all of them, you have the asymptotic formula that this thing is exponential of, so what I wrote before, but it turns out that, that for almost all rational, there is a simple low of large numbers. Well, okay, simple to state, but to prove it, you need to work a bit harder. So we obtained a low of large numbers with the sum of the continuous fraction digits, and from this you can immediately deduce. From this end of previous theorem, you can immediately deduce this statement. So it's a form of low of large numbers for the log of the J invariant introsional numbers. So you can ask about this condition, is it necessary? What? And we're so very happy because Christophe and Benz, Christophe Eisleiter and Benz Borda earlier this year, they obtained a very nice statement, which I'm going to state now. So the condition, this condition is actually necessary because of the following theorem of that, so you take one example of a number of which, a very simple example, the simplest example of a number for which the average doesn't go to infinity is when you take all the a i equal to one. And so if you fix for any n x being the rational with all the numbers equal to one, and I will say more about what they can do and what else they can do. So this is a question of two B-Shibonacci numbers. And for these numbers, so this depends on n, then the J function grows exponentially exactly like before, but now the constant is different. So this is the same sum, the sum of the a i, but now the constant is not the same, the constant c is 1.1 something, at least, well numerically, they show that numerically it seems that the function is like this. Maybe they have some proof that the first digits are this, but maybe Benz cannot surface around, but they show that this constant is different than the one from the almost all results above. So the result is much more general and in particular it works for all quadratic rational. So when the expansion is periodic, they can obtain this formula and the constant is not always the same as before. More general. X with quadratic. So now I'll explain a bit why all of a sudden we were talking about this pochameric symbol and then we got into a constant attraction expansion. So before this, maybe I can talk for a few instance and ask if there are any questions. There seems to be one. No, that was on the rule. Okay. There are no questions. There's no question at the moment in the chat. So maybe someone will arise questions. Okay. Seems not the case. So please. Okay, continue. So why is the continued fraction expansion relevant here? Okay. Continue traction expansion. So that's because of the conjecture by the game who observed numerically the following. So you let H of X be, so I will state it only for the variant that I mentioned before, H of X be the log J. See everything happens inside the log. So let's take the log. Leg log J of that function from before at E of X minus log J, same thing as before at E of 1 over X. Then this function, which is defined on Q, is extends to A to A. A nice function of X in R star. So what I mean by nice is, well, not very clearly, the best way I can illustrate this is by showing you a picture. So this is the newspaper, quantum radioforms on 2018. It has lots of 17 pages, but still has a lot of stuff. And he has the following two graphics. So this is a log of the function that we had, plotted at rational numbers. You can see the size grows at the denominator. And then it doesn't look so regular. But if you take the difference between the difference that I just wrote, you obtain this function, which is at least much nicer looking. It's not very clear what we conjecture, but by zooming in at a rational, we see that it has some jumps at rational. So certainly we don't expect it to be continuous. And at irrational numbers, so 1 over phi is here, the golden ratio. And you see, well, okay, it looks continuous. It's not so, doesn't look at all differentiable, but at least we can imagine that the function has a limit. The limit is close to 1.1. And this is related to the values that I showed you before. So let's see, sorry, this one. So let's see. h of X, it looks continuous near the golden ratio or the inverse of the golden ratio. And h at the golden ratio looks close to 1.1. And this explains, this would explain the theorem that I showed you before, the theorem AB. Okay, let me call this one the generic theorem. And secondly, let me get back to this function. You see, it goes to infinity. And it seems to go to infinity like 1 over X, it seems to go like the constant that I showed you before divided by X. As X goes to zero plus. And this would explain the generic theorem. So sorry, sorry for this. So it would explain the theorem that I called the generic theorem. Why? Well, this is because when you start with evaluating the function at a certain rational, when you replace X by 1 over X, because this is one periodic, you at each time you apply, you replace a continued fraction. Oops, sorry, I should take the log. You replace the continued fraction. So say this is X. I'll get h of X plus. So log J of 1 over X, but 1 over X, what do you do? 1 is just taking away the first coefficient for 1, 0 of E of 0, A2, AR. So if the first coefficient is big, it means that X is very small. And then this will look like C4, 1 times A1. And then when you iterate, you get C4, 1 times A2. Well, of course, these numbers have to be big for this to make sense. And we can only prove this when the average goes to infinity. Okay, yeah, maybe you can keep this section. Okay, so I have a little time to explain to you. Yeah, still have maybe 15 minutes. So I will set you what happens inside the, well, what is the main theorem that is driving everything. So we don't have this conjecture. If we had this conjecture, things would be much easier and also with this information. But we get around this conjecture. By proving two theorems, which are modularity looking, the modularity relation, much like the one I wrote before for eta, let me go back to it for an instant. Much like this one, but having to do with the Pochamer symbol. So the main theorem is that, so there are two. If I give you Q for gamma in SL2z, then the function which 2x and n associates E of x over 24. So this is in Q cross natural numbers. Well, actually cross. So let's see 0, denominator of x minus 1, because then it vanishes afterwards. Let me see. So E of x, so then, so this function, which is interesting for this value, x is rational and n varies between 0 and the denominator of x minus 1 satisfies, satisfies, satisfies a transformation formula. Which, maybe I can show you, show it to you if you want, but it will take some time. So I will concentrate first of this, of this informal description. It's a formula which relates x, which sends x to gamma x like modular iteration, and it sends n to, let's state it like this, yes, it sends n to the floor part of n divided by denominator of x times the denominator of gamma x. And this is a number between 0 and the denominator of gamma x. And this formula is, is exact. 0 term depends holomorphically on the following, on the, the fractional part of this number. So I have to explain to you what I mean by this. And yes, maybe I have time. It will take, I think, some five minutes more to explain the full statement of this theorem. So the first part, the formula that we have, I can show, can show it to you. Actually, I'm wondering, okay, it takes some notation, so more precisely. But the point, there is a point I want to make in this formula. And to make the point, I need to, to actually show the formula, but the formula is not very, not particularly good looking. So to state it, I need to, okay, x as a rational and gamma of x. Here is gamma of x. You can express it using nd on this, but first it will be h over k. Then for one less than r, less than k. If you let l to be the floor part that I wrote before, and lambda to be the fractional part that I wrote before, if you take the quotient between these two things, then this will be, this will be an expo, we can express this exactly as some of many terms. There is the first one which will be maybe ring a bell for people to study cotangent sum. So this is a dedicant sum plus pi i over four plus half log of k over d. I will explain each term. Plus k over q d times something, some function which is related to the logarithm evaluated at this number between zero and one plus nr term which depends on r, lambda, and d over k. The main point is that this term is holomorphic. It's a function which depends on l on lambda. It has an expression depending on lambda which makes it extendable to a holomorphic function. In lambda, so in this set, whatever, it's some other set, more general than this one. And this came as a surprise because all these are finite products. So maybe it doesn't come immediately as a surprise that such a formula exists because when you take out going to infinity, if you could, you would obtain the data kind, the eta function that I wrote before. And the eta function is a modular form. So you might expect something like this to exist. But this is not the infinite product. It's a finite product. And even by taking finite products, you can still, you have still a formula which is useful. It's useful for us. And this is relevant. It's relevant for us when the denominator of x is bounded. This is a case when we can control the error term. And there is a second important agreement that I want to mention in the previous theorem, which is a second relation formula. Now, okay, I won't state it, but there is a second second reciprocity formula. Okay, modularity, sorry, modularity formula, second modularity formula, which relates now it's a bit. There is another layer of complication coming from the fact that it doesn't relate x with gamma x or one over x is going to relate the gamma x bar, which is when x you write it as a irreducible fraction, x bar, which be each you inverse the numerator modulo the denominator. Of course, this depends. This is only defined modulo k. So we take fraction upon. There's a formula which relates this. And basically, it means that when you apply, it allows you to take away not the first coefficient when you is making UK algorithm, but the last coefficient. This transformation going from x to x bar is equivalent to reading the continuous fraction expansion in the reverse order. And there is a reciprocity formula, which relates these two for the Q-PoH symbol. And it's useful for us. So I'm going to lay a bit to make the point more clear. It's useful for us when the denominator of x goes to infinity. So it's actually depends on the way you stated it. But we do stated when which makes it useful when the denominator, the denominator stands to infinity. And this is precisely the complementary, the complementary case of the previous one. So if we could get rid of this condition, we would be getting very close to the guess conjecture. We cannot. We're not able to right now. But by proving this complementary modulity formula, which has a lot of disadvantages, but it has the one big advantage that it's useful as soon as the denominator goes to infinity. And by combining the two, we can successfully iterate the formula. And maybe I can finish with the last remark that in this formula, in the formula two, there are some peculiar objects that pop up, which are the following. It's a cotangent sum. So let me say, okay, n, k bar over h maybe. Oh, sorry. Okay, okay, sorry. I confused between the button for go back page and okay cotangent of pi n h bar over k. h inverse over k. This function is one periodic. So this makes sense. But you're okay. Times n over k. And if there was, if the sum was over all, residue, non-zero, residue, modulo k, this would be a cotangent sum. But in this formula, it turns out that very naturally, we didn't, we certainly didn't call for it, that this sum appears when r is restricted to be something else to the k. So it's a partial cotangent sum. And this, okay, we struggled for a long time with this kind of object. It has some reciprocity, but it's not, it's not exact. So, okay, there is something, there was something new going on in this formula that we didn't expect. So if you stumble on something like this, maybe take a look at this pair of paper or Loubinski's paper and let's save you some headaches. And so, yeah, okay, so you can say more about some elements of the proof if someone asked, but otherwise I stop here. Thank you for your attention. Yeah, Saris, thank you very much for this very nice Blackboard talk. So we could end this conference with a Blackboard talk. Of course, in auditorium A1 in Exier, we have much more Blackboards. We have three big ones and some smaller ones. Here is just one Blackboard, but it has one advantage compared to a real Blackboard one. We can go back to any Blackboard. So I have a small question. You introduced this geometric constants somewhere in the first third of the talk. Could you say some words on the geometric interpretation of that constant? Yes, so maybe for this I can find a picture of the knot. So I need to find a picture of the knot. Let me see if it's in the gift paper. Because these interpretation as knots would be really very instructive. Yeah, so let me get right back with the screenshot. I need to find a picture of the 4-1 knot. But if it takes too long, we should postpone it to discussion room or something like that. It's okay, I found it. It's actually on Wikipedia. It wasn't difficult. Okay, so let me see. Yes, so this is the figure 8 knot. This is the 4-1 knot that I was talking about. And when you consider the complement of the knot, so everywhere in the space except inside, it turns out there is a theorem of... Okay, I forgot his name. Okay, that's embarrassing. He's one of the biggest people in this thing. You can put a hyperbolic structure, a metric, on the complement of the space, which makes it a hyperbolic space and it has finite volume. And the constant is the volume of the space. Maybe it's you to make my... Sorry? Maybe it's you to make my own. I don't know, but it could be. No? Okay. Yeah, I don't remember something. I'm sorry. It's not... It's not... It's not... It's not embarrassing. Yes, okay, so he showed that you can decompose the complement space into a simpler hyperbolic space. And when you compute the volume, it's interesting, intrinsic quantity, and the constant is the volume of that. Okay, so thank you for this very nice explanation. Are there other questions? Short ones? Long ones are postponed to the discussion rooms. Yeah, Benze is typing a question. Maybe it will come soon. Yeah, here is... Can you read the question? Yeah, so is it conjecture that the guess function is continuous at every rational? And I believe so, yes. And it certainly looks so numerically. Of course, you have to... Of the numerical, so I can show you why. This would be a funny explanation. And there is a graph from this paper. So this is what you happen when you compare j of x with j of 1 over x. And this is what you happen when you compare j of x bar with j of 1 over x bar. So it looks much or less the same. But this function takes actually... It goes to plus and minus infinity at the neighborhood of any rational number. So it looks to be nice and continuous, but when you zoom in, you remark that it's not very nice at all. So you need to be aware of the conjecture, but I believe that h of x is continuous at every rational.
size and distribution of values of the programmer This, the cube for a symbol at the golden ratio which part of the knot goes on to another, and numbers with their, the mean value of their harder. So we obtained a low of large numbers and in particular it works for all quadratic start with evaluating the function at a certain before. And the eta function is a modular form. which is useful. It's useful for us. And this is