 Moscow and he will tell us about Zaremba's conjecture and growth in groups. Please, Yelena. Thank you very much for the invitation. It's only for me to participate in this seminar. I want to talk about Zaremba conjecture and how this conjecture is connected with growth in groups. Let me first recall Zaremba conjecture. It's about continued fraction. We have irrational alpha, generally or rational, from 0 to 1 and we consider defining the continued fraction expansion. So the expression of this form has J are natural numbers. So Zaremba conjecture asks that given positive integer Q, if there exists A, which is comparing this Q, it's said that if we consider the continued fraction expansion for this rational number A over Q, so it's a finite continued fraction expansion, then there is A side that all as J, which are called partial quotients, are bounded by some constant and Zaremba suggested to take this constant equals 5. So once again, we have Q and we want to find the deunumerator side that A over Q has all partial quotients bounded by 5. So Q is fixed and A is a variety. So it's a strangely looking conjecture maybe, but let me say a few words about Zaremba's motivation and after that about mathematical motivation, why do we believe in this conjecture. So Zaremba's motivation has roots in numerical integration. We have a the dimensional function f on this cube from 0 to 1 with bounded variation. Yes, and we want to approximate its integral by a finite sum, a finite weighted sum. So a well-known Koksma-Glavka inequality says that this expression can be estimated as the variation times the discrepancy of this sequence x. The discrepancy is defined as usual. We take the frequency of x in any rectangle, r minus the volume of rectangle and after that we take the supremum over all rectangles. So you see if we want to approximate our integral by sum we have this quantity which depends on the function f only and the question now how to find x with relatively good upper bound for the discrepancy. So how optimally find this sequence x. There is a lower bound, another famous bound by Volga Schmidt who proved that for any set x there is lower bound for discrepancy. So for any set x it cannot be too random, cannot be perfectly uniformly distributed. So the limit of these discrepancies is log x over x and if on the other hand you take x ux be a random set then discrepancy will be much worse. It's just one over square root of x and there is a huge gap between low and upper bounds. So Zaremba suggested the following two-dimensional widen of two-dimensional torus. For simplicity I consider just two-dimensional torus, two-dimensional situation. So you have a finite number of points roughly speaking with the tangent equals a over q, a and q are parameters which we will choose later and he proved that for this sequence x if you consider the continuous fraction expansion of a over q and denote a to be the maximum of all partial quotients as j, then the discrepancy of this sequence is log q over q basically times this constant, this function actually because we don't know what is m. So you see that if m is constant then this upper bound coincides with Schminz's lower bound and thus the choice of this x is optimal. But to choose this x to be optimal you need to prove that there is a side that all partial quotients of its continuous fraction expansion is bounded by constant m. So that was the motivation of Zaremba. If you indeed prove that for any q there is a then you obtain an optimal upper bound for the discrepancy. So what do we know about m? There is a general result of Korobov who proved for q equals prime that there is always a side that a over q has the continuous fraction expansion side that as j not bounded by some growing function namely log of q. This result actually it's a nice result but I should say that a is typical here. So for almost all a if you choose a randomly then you obtain this logarithm. So for typical a we have logarithm and in Zaremba conjecture we need to have due with a set of zero measure. There's a problem that the difficulty of the Zaremba conjecture. There are further results here about the distribution of this maximum. For example, Rukavishnikov proved Korobov result for all q and moreover there is the law of large numbers for this maximum. For example, if you calculate the number of a side that this maximum is greater than t then the fraction of such j is bounded by log q over t. Okay so we have due with a set of zero measure. Let me remind you another beautiful result of Nieder Eiter who proved that Zaremba conjecture holds for special q. For q of the form to the to the n or three to the n with m equals four not five but four and four five to the n with m equals five. There are further results of another mathematicians and the proof is based on so-called the folding lemma. There are several variants of such results. Let me write just one of them. So if you have the continuous fractional expansion of a over q then you have such fraction and you have the denominator which are polynomials in your variables s1 ss. So Zaremba's conjecture is just a question about representation of your q as the value of polynomial. This polynomial is called continent. So continent s1 sj represents q and you want to find s1 sj bounded by m. Okay so the folding lemma variant of the folding lemma is the following. If you consider this almost palindromical sequence you see s1 s1 s2 s2 then this almost palindromical continuum can be calculated via a half of this continuum square times x plus one. It gives you the basis of induction. If you represent q say of the form to the end with this short sequence then by the folding lemma choosing I don't know x equals one I suppose you can you take you obtain a longer continuum and after that you are not very correct here but with the spirit of the proof. A huge progress in Zaremba's conjecture was made rather recently five six nine years ago by Bogankontarovich who proved that Zaremba's conjecture holds for almost all q. It means that if I take q from the segment from one to n and consider Zaremba's conjecture for such q then there is n minus this error term q side that Zaremba's conjecture holds for this q. Cfm is a constant and for simplicity if you take say m equals 50 then there is a positive proportion of q for which Zaremba's conjecture takes place. So this error term is very good enormous for example there are a lot of primes side that Zaremba's conjecture takes place. Then the result and after that this dependency this constant will be improved by several mathematicians like Frenkoff-Kann, Lang, Muggle-Winter and let me write just one result here and the best result in the system of this m. If m equals four then for all but smaller of n numbers q Zaremba's conjecture holds. That's the result of the Grikan maybe such result takes place even for m equals two. Yeah but it's close you see. So it was the motivation was it was the progress relative to Zaremba's conjecture but what about mathematical motivation of such things? Why do we believe in this conjecture? To do this let's consider even more general conjecture which is called Hensley conjecture. Take the following set of real numbers not finite continuous fraction not rational numbers but infinite set of real numbers sides that all partial portions of such sets abounded by m. Then and this result of Hensley you can calculate the household dimension of this set with very precise accuracy it's a nice result Hensley used functional analysis to obtain such asymptotics. For example if you take W2 so this is the household dimension of sets with partial portions abounded by 1 and 2 then the household dimension is a little bit greater than one-half 1.53. I should say that the dependence of Wm on m or the side sort was known by Henshin in 30s I think and actually for us this would be enough. Okay what about Hensley conjecture? Instead of this set this countertop set it's easy to see that this set has a countertype iteratively defined set. He considered a more general alphabet alpha finite alphabet so now my set has all partial portions from alphabet alpha not from numbers from 1 to m and he conjectured Hensley conjectured that if the household dimension of the corresponding counter set is greater than one-half then Zarembets conjecture takes place for sufficient large q. It means that for such q there is a which is compared with q side that all partial coefficients not bounded by m but belong to this alphabet alpha. I should say that literally speaking this conjecture is wrong. It doesn't help because of some local modular abstractions. It was pointed by Bugankontarovich but maybe this conjecture holds for this special alphabet from 1 to m or maybe modular these local abstractions. Okay so why why this condition on the host of dimension? The motivation of Hensley was the following. He considered this two-dimensional set of rational numbers so it's a finite set. We take rational u over v all is j bounded by m again or belong to to alphabet alpha it's not important. u and v are co-prime and u and v run from 1 to q. Then it's easy exercise size of f m can be recogulated via the host of dimension of the corresponding counter set. f m of q this high accuracy equals q to the 2 w m. Okay so it's a two-parametric set u and v arrive but we are interested in one-parametric set so we fixed p for simplicity I will suppose that q is a prime number. Okay so and below as well so a is from 1 to p minus 1 and we want to find a size that all partial quotients are bounded by m. If we believe in the uniform distribution in the denominators then we need to just divide this bound by the number of denominators so we need to divide p to the 2 w m by p. This is p to the 2 w m minus 1 and you see if oh sorry if you see if w m is strictly greater than one-half then this quantity tends to infinity and it gives some explanation why we think that Zarem's conjecture holds if we believe in some uniform distribution or denominators of some sets then Zarem's conjecture takes place. So the first results which will be obtained by grossing groups are the following. So let me recall Zarem's conjecture once again the asymptotic of this Zarem's set of numerators is the following and we proved with machete in north then the same upper bound takes place. So for any prime p and any positive epsilon there is m dependent on epsilon so all partial quotients are bounded by m and the same upper bound takes place with this error. Okay so of course we need in lower bounds but nevertheless some kind of uniform distribution takes place because it's the same upper bound. On the other hand there is a positive result of this direction for any p and any epsilon there is m side that if you take the continuous fraction expansion s1 sj you can bound all sj up to small interval in the middle. So you remove this interval of length epsilon s from the middle and you can prove that all the rest has partial quotient bounded by m. So up to this interval Zarem's conjecture takes place. Okay so if the first bunch of hours out and how we prove them I will talk about it later I will give more details but which tools we use. Of course the proof is based on a beautiful theorem of how you feel about growth in SL2. So we have a set which generates SL2 then either a cube equals SL2 or we have the following exponential group. So a cube if a to the power one plus some positive number c it's a very fast growth. So it's a nice result because it says that if my set in principle generates SL2 then it's generated SL2 very quickly because at each step at each multiplication from one to a cube we have these exponential growth so it means that basically you generate SL2 in logarithmic number of steps more or less. So if you generate SL2 you generate it's very quickly. Even more surprising that actually there is a result about uniform distribution here. Suppose that we have a set which not just generates SL2 of FP but it doesn't relate with any subgroup H of SL2 or even with all cassettes of subgroup. It means that intersection of fave with any cassette is bounded by a over k. k is a parameter which tends to infuse. Then actually you not just generate SL2 but you do it very uniformly. The number of solutions of this equation is just expectation right plus the error term. It's a consequence of Hilbert theorem plus the famous Frobenian theorem about non-trivial irreducible representation of SL2. We know that the dimension of such representation is bounded below like p minus 1 over 2. So it's a general conceptions that if you have growth in the sense of Therigod plus you have quasi-randomness by Gauss actually this definition was given before by Sarnak and Xui and before by Huxley I think even. So if you have quasi-randomness in the sense that the dimension of all irreducible presentation is comparable this size of group in this sense then you automatically have the uniform distribution. So that is our tools. One more result I need to say about it. So Hilbert theorem gives us some c and after that this c was calculated by Kowalski and recently rather recently we proved with Misha Rudnev that this c can be taken equals 1 over 20 and even we calculated the diameter of the corresponding paragraph. Okay. So these results used growth in group and for the second part I need in another results about growth in groups and after that I will give you the scheme of the proof. So the second bunch of our results corresponds to so-called modulus arambo conjecture. This theme was initiated by Hensley as well. So for a given prime p now we take a composite q side that q is divided by p so q is greater than p right but somehow q is comparable with p so it's bound by big old p over to the 30 and we want to find a side that a over q has all partial quotients are bound by m. So if you take q equals p then you obtain the arambo conjecture but now q just is divisible by p and q is comparable with p as well. So that is our result with machine written and moreover if you allow to replace the 30 by a larger constant then you can take this m equals even 2 like in Hensley conjecture. So there is an absolute constant c side that for any prime p if q is comparable with p in this sense so q is p to the c again q is divisible by p then you can find a side that all is j bounded by 2. So that's the modular form of the arambo conjecture and recently I proved the following result actually q can be taken very close to p q equals big old p 1 plus epsilon for any epsilon for any such q you can find a side that a over q has bounded partial quotient. So if you take epsilon equals 0 then you obtain the arambo conjecture and we are absolutely epsilon close to the arambo conjecture but for me I don't know how to replace this epsilon so I don't know completely. Of course we are not. Okay another thing that you can obtain modular form of Hensley conjecture so now alpha is a finite alphabet and the house of dimension of the corresponding counter set is greater than one-half plus delta then there is q which is comparable with p q is divided by p and side that all partial quotients of a over q belong to this alphabet alpha. So you see it is interesting that this general form of Hensley conjecture holds for modular Hensley conjecture is true but general Hensley conjecture is wrong. Aki. So what do we use here? I want to talk about this in a little bit more general context. I want to say a little bit about Chevrolet groups or groups of lead type by me they're small as the same. So these groups are naturally belong to naturally gives you the families of finite simple groups so by the classification of finite simple groups any finite simple group either the prime field or alternative group or belongs to the family of groups of lead type for us they will be Chevrolet groups or belongs to one of 27 finite groups which are called sporadic groups. So we are interested in this in this third point. Chevrolet groups are classified. There are classical groups, classical matrix groups SLN, unitary groups and so on. There are such type of groups and also Steinberg and Suzuki groups. I think I will skip the formal definition. Chevrolet groups are defined by the prime field. I think I will skip the definition. Let's consider the simplest and the main actual example for us. So SLN is a Chevrolet group. It contains the following Borel subgroup, the Borel subgroup of the upper triangle matrices. And we are interested not only in Borel subgroup, not only in the maximal solvable subgroup, but also in parabolic subgroups. These are groups by definition just containing a Borel subgroup. In SLN it's easy to describe such subgroups. Instead of alpha one, alpha two, lambda one, lambda two, lambda n, you take blocks, square blocks, c1, c2, and c something, ck. And of course, such matrices, such groups below Borel subgroups, and that is the full description of parabolic subgroups in SLN. So this s1 is to sN corresponds to the partition of the number n. So there are two to the n minus one of parabolic subgroups, and n minus one is called the rank of SLN. So what do we know about expansion in simple groups and in Chevrolet groups, in group of lead types? There is a famous conjecture of Bob who say that if a generalized the whole group, then again it generates it very quickly. There is n, which has a logarithmic size, type that a to the n equals g. So again, if a generates g, then it do it quickly. And for groups of lead type with finite rank, we know that conjecture of Bob takes place. So for any its result, which was unshaded by Hegoth after the Brea, Greentail, Piper sub, proved the following nice result. So any subset of g, which generates g, g is a group of lead type, either has the following growth or a cube equals g. And as a consequence, they obtain Bob's conjecture for such groups, but you see this expansion depends on rank, so actually it works just for lead groups of small rank. Aiki, so we are interested in another questions. I will talk about it later why we need in such formulation. We don't want to generate the whole group g. We want to take a subgroup h and we want to find non-empty intersection of the orbit of my set A with this subgroup. For example, I can take rather large subgroup h, like a parabola subgroup, parabola subgroup have very nice structure and I want to take a general a, actually, side that a to the n intersect h. There is a generalization of Frobeno's theorem, which belongs to Lanza's results, that for any Chevalier group, there is a lower bound for all irreducible, not irreducible representations. So this dimension is bounded below by q to the r, q is the size of my field, r is rank, and this bound is optimal and it's very natural. This quantity is just the index of my Chevalier group over the maximal parabola subgroup, maximal by size. All parabola subgroups are maximal, but I take a parabola subgroup of maximal size. So that is the sense of this q to the r and in particular the theorem says that any Chevalier group is a quasi random group in the sense of Gauss. So a simple corollary of Lanza's theorem is the following. Suppose I take any set in my Chevalier group of size, a little bit larger than the size of the maximal parabolic subgroup, a little bit larger. Then I generate the whole group and I generate it quickly. This number of n depends on delta, just one over delta, I think. Time some constant depending on the rank. So it's very natural. Of course, the maximal parabola subgroup doesn't generate the whole group, but if I take a little bit larger subgroup, then I generate it all. And I consider the following equation. I consider the intersection of a to the end of this parabola subgroup and I prove the following results. So let q be ordinal square, p is any parabolic subgroup, a is any set, and suppose that size of a is not g over q to the r, but g of q over q to the r minus 1. So basically I divide this period bound by q. Okay, so it's a weaker condition and I have a weaker consequence. So a to the end intersect p. My parabolic subgroup and the gain n can be estimated precisely. So this one is super, super important for me here. And that is the first result about expansion in Chevalier groups. Another result is the following. Suppose that we have a set a of the following form, x times p, p is a parabolic group, say, and x is any set. Then of course, if I multiply a by p from the right, I obtain the set a. If I multiply a from the left of this form, then again obtain a. But suppose that I multiply both from the left and from the right. Then for a set, it's very difficult to be the right, the union of right cassettes and left cassettes simultaneously. And there's the result. In any Chevalier group, and parabolic subgroup, the set either a p or p a must grow. I have this assumption, but it's a stupid assumption. I have much more assumption here. It's not important. So either a p or p a is large. It reminds a little bit the sum product phenomenon maximum of two expression grows. For example, if a is less than p, the maximum of a p and p a is a times square root of q. So it's it grows. So we are going to use such results in our study of SL2 or another word in our study of continuous fractions. So how gross is connected with continuous fraction? Very easy and very classical. So if I take two approximation p s over q s and correspond to the continuous fraction expansion, another continuous fraction expansion, then I can multiply such matrices and obtain the following matrices. The determinant of these matrices is plus minus one. So basically it belongs to SL2. So I want to start the following set. I take all products of such sort. I take sg abounded by m, because I consider the Ramos conjecture, all sg abounded by m. And support that q s bounded by p. So I take all products up to q s is less than p. q s is the largest number here. So all elements of these matrices are bounded by p. And so it's a fair multiplication in z, not in p. Okay, so I consider such set of matrices. You can calculate the size of this a by Hensley lemma, because such matrices corresponds to p s and q s. So size of a is p2 minus p to the 2 Wm. So it's a little bit less than p squared. So I have the following set of matrices. And to compare, let me recall that the size of SL2 is p cubed. And the size of the Borel subgroups and the Borel subgroup of the upper triangle matrices is p squared. So it's a little bit less than the size of Borel subgroup. That is our set of matrices. Okay, so again, I repeat the definition of my set of matrices, and suppose that I consider a to the n in fp, now the multiplication in fp, and suppose that I find q s minus one equals zero model p. Then by definition, this rational number has bounded partial quotients, they are bounded by m, right, by definition. And because of this element belongs to a to the n, I can estimate q s minus one. That is more or less p to the n, two p to the n. That's all. If I find this element in a to the n, which is zero model IP, then I solve my modulus of Rambo conjecture. That's all. You can reformulate this problem in a more, I would say, additive way. So you consider the standard Borel subgroup. It's the following subgroup. And you need to find a non-trivial intersection of a to the n with b, right, because you need to find zero here. It's a typical, I would say additive problem. You have rather random set A. I will talk about it later while it's random. And you have a structural set B, which corresponds in a mathematical language like progression or something like that. And you intersect the random set, a power front of set with the structural set B. Okay, so again, I repeat the formulation. And after that, I multiply this intersection by B from the left and from the right. I have BA here, AAB here. And that is equivalent to the original problem. And now you see non-computability helps us because we have BA, AAB. And we know by our growth result that either AB or BA must grow. And because size of A is more or less p squared, I break this p squared bearer. I have p to the 5 over 2 minus epsilon. After that, I break this bearer, this bearer, which is naturally given by the size of Borel subgroup. And after that, I apply Hewlett growth result or representation and so on and so forth. And that is how we prove this result with machine. So for model Hensley conjecture, we need in the classification of subgroups in SL2. So now the dimension is one half plus delta, so size of A not close to p squared, but close to p, a little bit larger than p. But thanks to classical Dixon classification theory, we know that all subgroups of SL2 either, roughly speaking, conjugates of the standard Borel subgroup or deferral groups or finite groups. So because size of A is greater than p, I don't care about these dihedral or finite groups. And my enemy is now just standard Borel subgroups. So if I want to find something in Hensley and Intercept Borel subgroup, I don't need to see on these small objects. So after that, we apply our second growth result. We know that if A has the following size, then the intersection of A to the N with any parabolic subgroup is non-empty with effective N. In our case, the problem subgroup is just a Borel subgroup. Size of A is this one. And you see, thanks to this lovely one, I beat this bound. And hence, I find a non-trivial intersection of A to the N with my Borel subgroup. So it gives us the following result. You see, I repeat again, so if household is greater than 1.5 plus data, then I find q which is driven by p and A over q has partial quotient belong to this alphabet. So you see, this condition in the sense of groups allows you to beat, allows to arise these small and chaotic groups here. So this condition gives you this. And finally, how to prove this result with p to the 1 plus epsilon? You need to analyze more carefully your set A. You need to analyze the intersection of A with subgroups. You need to prove more or less this upper bound for exponential sums, non-computative exponential sums, I mean the representations. And moreover, you need to analyze not only A but some useful subsets of A. It's a little bit tricky. So I just recall your result. If q is p to the 1 plus epsilon and q is divisible by p, then you can find A over q side that all is j, bounded by m. I think that's all. Thank you for your attention.