 Thank you. First of all, thank you very much for inviting me to speak at this seminar, which is so largely attended by so many number theorists around the world and hopefully not only number theorists. So today I'm going to talk about approximation by algebraic numbers and the main question for today would be to understand how well algebraic numbers of a bounded degree are distributed or more exactly given a real number of psi, we want to understand how small can we make the difference psi minus alpha, where alpha runs through algebraic numbers of bounded degrees or degree of alpha is at most n and bounded height of alpha is at most h. And here in this talk by height we mean the naïve height of the minimal polynomial of alpha and primitive minimal polynomial of alpha. So here's the naïve height of p alpha, basically the maximum absolute value of its coefficients. Right. Okay, let's firstly start with some heuristics. What can we expect from this difference? So we know that there are around h to the power n plus one different polynomials of height at most h. Okay. And then with a bit of work we can derive that there are approximately the same amount of algebraic numbers with bounded height and with bounded degree. So if we assume that all of them are equidistributed, are uniformly distributed, then we expect something like psi minus alpha to be smaller than h of alpha to the power minus n minus one, probably some constant. And that should happen for infinitely many algebraic numbers. So this is our expectation from very brief, very rough heuristics. Okay. If you look a bit deeper into the discussion, we will see that this heuristics does not always work. For example, it does not work for algebraic numbers. For example, let's take an algebraic number psi of degree d at most n, then the classical Luegel's inequality gives us that psi minus alpha is always at least h of alpha to the power minus d, which is, as you can see, this expression h of alpha to the power minus d is much bigger than h of alpha to the power minus n minus one. In some sense, it shows that algebraic numbers do not like each other. If we have an algebraic number of small height, then the neighbors will be quite far apart from it. Okay. And because of this issue, to be on the safe side, we usually assume that psi is transcendental. So we don't can see the algebraic numbers of psi. Okay. Right. Now let's make our problem more formal. Let's define the first, the finite exponent for today, the quantity w star of psi, which is a supremum of all numbers, positive numbers, double star, such that the following inequality, psi minus alpha is less than h to the power minus w star minus one, has solutions in algebraic numbers, alpha of bounded degree, degree at most n and bounded height, h of alpha at most h, for arbitrarily large values of h. So it should not necessarily have solutions for all large values of h, but it should have solutions for some sequence of h, which tends to infinity. Okay. Then if we translate our heuristics into this language, then we should get that the w n star of psi, it should be at most, at least n, at least n. And that's actually something posed quite long ago that was firstly posed by Verzing in 1961, and now it's called the conjecture Verzing. So the conjecture is that for any transcendental number of psi, w and star of psi is at least n. Right. Now, before we go into some proofs and some ideas, let's have a look what is already known about this conjecture. First of all, this conjecture is known for small values as well. For n equals one, that's basically the Dirichlet theorem, the classical Dirichlet theorem. So basically we have psi minus p over q is always smaller than one over q squared. So yeah, so w one star of psi is always at least one. So that's basically Dirichlet theorem. So nothing difficult, but when n becomes bigger, the problem becomes much harder. For n equals two, this conjecture is also verified by Davenport and Schmidt in 1967. However, for higher values, so then nothing is known. So for the case and the case n bigger than two is still open. Now, let's have a look what bounds, what estimates for w and star of psi we know for the moment. And his initial paper version in 1961 proved that w and star of psi is at least n plus one over two. So it is at least something. Well, it's quite far from n, but still it's at least n plus one over two. In 1969, Springjuke verified the conjecture for almost all real nonmask psi. So basically it means that if we have counter examples to this conjecture, the Lebeck's measure of those counter examples should be zero. So the set of counter examples is very small. Then later, Bernick and Tyschenko made some improvements to the law of bounds of raising. Namely, in 1993, they showed that w and star of psi is at least n over two plus two minus something which tends to zero as n tends to infinity. So they have precise formula, but it doesn't matter. So it's quite technical. But what does matter that we still have n over two here. So asymptotics is still n over two, but we win a little bit with a constant. And just for the reference, the number three star of psi in their work is at least two to two point five, two and a half. Actually not too good. I think the number three star is two point five exactly, but yeah, but it's not always rational. Later Tyschenko improved this bound again a little bit in 2007. He showed that w and star of psi is at least n over two plus three minus something which tends to zero. And again, for n equals three, that low bound gives us two points 73 and something. And yeah, if you look at all this law of bounds, you can see that we always have n over two. So the asymptotics in all this law of bounds is n over two. And for a long time, nothing better was known. So no better estimates or asymptotic estimate than n over two were known. Okay, I will introduce our result mine and later where we beat this asymptotics. But for the moment, let's not do this. Let's leave it to the end of this talk. Right. Now I want to talk a little bit more about the ideas how to, what can be done? How can we attack this problem? How can we estimate the distance between the given number and given rational number? Usually it is relatively hard to to study the distribution of algebraic numbers themselves. And it is much easier to study the polynomials, the minimal polynomials. So we will introduce several more, several more different time exponents, which will be used later. So the first exponent describes how small can we make the values of polynomials at a given point? So that's namely wn of xi, or respectively, w hat n of xi is defined as the supremum of all w's, real w's, such that the following system of inequalities. So basically the main one is this, the value of the polynomial at xi is at most x to the power minus w. And all the coefficients of that polynomial accept possibly a zero are between one and x, capital X. So if this system of inequalities has solutions in integers a zero and one and so on, a n for arbitrarily large x, that gives us wn of xi. And if we require the system of inequalities to be true for all large x, then that gives us wn hat of xi. Right. And one, and we will need a dual version of this, of this exponents. So here we basically consider the dual body to do the previous one. And yeah, and we define the exponent lambda n of xi and respectively lambda hat n of xi as a supremum of all lambdas, such that the following inequalities x zero times xi to the power i minus xi, all of those are small or are not bigger than x to the power minus lambda. And x zero is between one and x. And again, if that happens for some arbitrarily large values of x, that gives us lambda n of xi. And if, if we require it to be true for all large values of x, that gives us lambda hat n of xi. Cool. So now we have five different definite exponents. Right. And now let's, let's, let's see what can be done with all of them. So we start with some basic ideas. Right. Right. The most basic idea is if, if alpha is, is very close to xi, that implies that immediately implies that the value of minimal polynomial of alpha at this point xi is very small is very close to zero. Right. And well, that's not, that's not something hard. Right. We can estimate the absolute value of pf of xi as xi minus alpha times the derivative of this, of this polynomial at some point between xi and alpha. That's, that's just the basic calculus. Right. And yeah, that can be estimated as h of alpha times xi minus alpha. Right. And if we write it down in terms of different exponents, that, that gives us this, this inequality. Wn of xi is always at least Wn star of xi. That's of course good, but we need low bounds for Wn star, not upper bounds. And so we need, we need to get, we want to get the inverse implication if, if the value of polynomial is small, does it imply that xi and alpha are close to each other? And that inverse implication is only true if somehow we can guarantee that the, the derivative of that polynomial is, is large. Otherwise we cannot guarantee that. So if you, if you have some play that then the inverse is also true. But for, for arbitrary polynomial we can't, there is no way for us to guarantee this. And then for in all, in all the methods, in all the, in the other works of various mathematicians who try to attack this problem, the biggest challenge was to construct or to show the existence of one or several polynomials p, which are very small at point xi and satisfy some other, some other conditions such that, well, that guarantees that one of the roots of that polynomial of one of those polynomials is close enough to xi. So that's like the biggest challenge. And that's just a very brief and vague idea what, what we want to achieve. So Dimitri, we have a question in the audience. Could you repeat the definition of Wn of xi? Wn. Wn of xi. It's here. So it's, it measures how, how small we can make the value of a polynomial. Is it, is it okay or I can, I can repeat. Is it okay? I just don't know who asked the question. Yes, that's good enough. Thank you. Where did we stop? Somewhere here, right? Yeah, right. So now let's, let's think about how can we make one polynomial small at, at a given real point xi. That's, that's not a hard problem, right? That's just a consequence of classical Minkowski's theorem. So for any, for an arbitrary capital H bigger than zero, bigger than zero, there exists a polynomial p such that its value with integer coefficients such that its value at xi is less than h to the power minus m and all its coefficients, the absolute value of all its coefficients are at most h. That's just a classical Minkowski's theorem, a convex boundary Minkowski's theorem, right? And in terms of, in terms of the, the afro tonic spoilers, that means that Wn hat of xi is at least m and also Wn of xi, which is also, which is always at least Wn hat of xi is also at least m. We can do the same for dual, dual body. We can apply Minkowski's theorem for dual body and show that the following system of inequalities, a zero xi to the power i minus ai, absolute value less than capital H to the power minus one over m for all i's between one and m and a zero is at most h, that this system of inequalities also has a non-zero integer solution in values a zero and one and so on a m for all values of h and that gives us that lambda and hat of xi is always at least one over m and lambda n of xi is at least lambda hat of xi. So at least if we, if we just want to estimate the value of polynomial or if you want to estimate lambdas or like for, for dual body, that's not, that's not a hard problem. The hardest problem is then to estimate or to somehow derive the information about Wn star. Okay, now I want to discuss several ideas of, from the, from the words of Wrensing and then other mathematicians. Actually, in his paper, Wrensing has many, had many ideas, many relations between different different exponents. I will discuss only two of them, which will in some way be used later. So the, the first idea is, as, as I also already mentioned in the previous slide, we can always get sequence of polynomials, p of xi, which are smaller than h to the power minus Wn of xi. Yeah, that's just by definition, oh, plus epsilon, plus epsilon, that's just by definition of double of xi, of xi. And by, by some easy tricks, we can, we can guarantee that p of xi is irreducible. And then by, by working with this discriminant, we can, we can derive, we can derive that it's closest, it's, it's closest root to xi. It has, it's, for it's closest root to xi, xi minus alpha is always at most the value of the polynomial times the height of the polynomial to the power n minus two. So here I skip all the details, I just, I just provide the idea, right? Actually, this, this number alpha is not necessarily real, it's, it's a complex number, right? And then from, from this inequality, p of xi is less than h of p to the power minus W, we derive that xi minus alpha is smaller than h of p to the power minus W plus n minus two, right? After that, we use some tricks to replace a complex number by a real number. So if, if, if alpha is, is not real, then we have two conjugate roots, which are very close to xi, and then we can replace p of xi by its derivative and its derivative again has a, has a root which is very close to xi. So it's some trick, some trick, which, which ensures that we can find a real algebraic number alpha with all, with all the same properties. So xi minus alpha is smaller than h of p to the power minus W plus n minus two. Yeah. And finally, the conclusion from, from, from this idea is that we have W and star of xi is at least W, Wn of xi minus n plus one. In principle, this, this low bound usually does not give us a good low bound, right? For example, if, well, we, we, we know that Wn of xi is at least n, and if it is n, then we just get Wn star of xi is at least one, which is, which is rather weak low bound. But if Wn of xi appears to be very high, then it could give some theories about what is reasonable. And also this method shows what can we do, what, what result we can derive if we can see that only one polynomial, we, which is small at xi. So possibly this, this, this factor can be improved, but not too much. Now the next idea, again, YouTubersing was instead considering one polynomial, which is small at xi. He considered two polynomials. It's one polynomial p1 comes from the definition of Wn of xi. So we can always find such a polynomial that p1 of xi is smaller than h to the power minus Wn plus epsilon. And then by reducing h a little bit, a little bit, a little bit, and applying Minkowski theorem, we can find another polynomial p2 of xi, which is, which is also small at xi, right? p2 of xi is at most h to the power minus m. And we can, we can guarantee, again, by doing some tricks that they are co-prime. They don't share any, any root. And after we have two polynomials of this form, and considering some careful consideration of the resultant, we, we can, we can derive that they, there exists the alpha of their product, or basically the root of one of those polynomials, which gives the following lower bound for Wn star of xi. So I don't want to give all the details, because they're technical, and I just want to give an idea. We get Wn star of xi is at least the minimum of these two numbers, and, and Wn of xi plus one over two. And, yeah, and, and this gives our lower bound n plus one over two, and that is the result of verzing. Okay. Yeah, and the conclusion from, from here is that Wn star of xi is at least, yeah, n plus one over two. Right. Now, since verzing, various people tried to make some improvements into this idea. The first improvement I want to discuss is due to Peugeot and Schleschitz. That's in, that was written in 2016, but the, the improvement in self is not too deep, I would say. So if you, if you look closer to, to the initial idea of, of, of verzing, we can find two prime polynomials. And for the first one, we have the same estimators before P1 of xi is at most h to the minus double n plus epsilon. That's because of the definition of double n of xi of xi but for the second polynomial, we can, we can do a little bit better. We can guarantee that it's smaller than h to the power minus W hat of n, n plus epsilon. And that's due to the definition of double n hat. So if double n hat of xi is n, then we don't have any improvement. But in some cases we can, we can, yeah, if double n hat is bigger, strictly bigger than n, we do have an improvement. And then again, if we apply all the same, all the same arguments of verzing, we have an improved low bound for double n star of xi in terms of double n hat. So it's, it's now bigger than the minimum of these two numbers. W hat n of xi and W hat n of xi minus n plus what we had before, double n xi plus one over two, right? Or this can be written as the following formula. So I probably have it, oh no, I don't have it. Okay, so it's three halves double hat and xi minus n plus one half, right? And yeah, and later we will use this low bound. So let's, let's have it here. So also Bernick and Titian in 1993 and 2007 also use similar ideas, which I'm not going to reproduce. I just want to say that they chose polynomials p i of x in more delicate way and considered various cases for the absolute values and for absolute values of their derivatives and that, well, and all of those helped them to reach, to reach this, this low bound, which is still, yeah, which is still n over two plus plus something bounded. So compared to the initial bound of verzing, we improve by basically two and a half, by the constant two and a half. Okay, so that's these other ideas of verzing. Now I also want to, to talk about the ideas of Darin Potenshmid. The first idea of Darin Potenshmid, so they also tried to, to, to attack verzing, verzing conjunction. And yeah, the first idea was to construct polynomials p such that p of xi is much smaller than the derivative p dash of xi. And this idea worked for n equals two. So they, they, they proved the following theorem that L and M be two nearly linearly independent linear forms of three variables. Then, and actually they can be arbitrary linearly dependent forms. Then they exist infinitely many non-zero integer vectors a such that absolute value of L of a is at most with some absolute constant, then the absolute value of M of a times the length of a to the power minus three. Now, if we, if it's in substitute in place of L of a, we substitute the value of a polynomial, which is, which can be considered as a linear form in terms of its coefficients. And for M of a, we consider, we take the value of its, its derivative at xi. Then this inequality basically implies that w two star of xi is at least two. Okay. And of course, after, after you get this theorem, it's tempting to, to generalize it to not be true again. And of course, Darwin potentially tried to do that. And unfortunately, for, unfortunately, they, they, they realized that it cannot be generalized to high velocity. So basically, they proved the following result in the next year, in 1968, they proved that for any n, at least three, one can construct linear forms L of a M of a, such that the inverse inequality is true. L of a is at least with some absolute constant M of a absolute value of M of a times the length of a to the power minus three. So yeah, so for this, this result, the required result is not, is not true anymore for arbitrary linear forms L and M. Of course, maybe if one uses the specific linear forms and we have specific linear forms, one is a failure polynomial, another one is a value of derivative maybe if we use specific linear forms, maybe something can be done, but the, the, the general result is not true. And yeah, and, and, and, and since then this idea, well, I, I, I, I, at least I didn't, I didn't see that anyone pursued this idea any further. Right. That they have another idea, idea two. They, instead of dealing with polynomials to P of psi themselves, they looked at the dual case. So they consider two convex body dual to each other. So the first one is, is, is this, so it consists of points of real points such that x zero plus psi x one plus and so on, plus psi to the power n x n is at most y to the power minus n. And, yeah, and all x i's are at most y. That's the first, the first convex body and it's dual, dual to that convex body is, is this one. So that's the, the set of all points x and plus one dimensional points, points x such that x zero times psi to the power i minus x i is at most y to the power minus one over n and the absolute value of x zero is at most y. So we have these two dual convex bodies and what is good for them about them, we have, we can relate the successive minima. We have the smaller duality theorem. So if we, for us, two successive minima are important. The first successive minima for the first body convex body lambda one of y and n plus first successive minima for the dual, for the dual convex body for c star of y, which is lambda n plus one star of y. And by model duality theorem, they are related by the following equation. So that product is, is bounded from above and from below by some absolute constants. So it's comparable with one. And what does it give us? So what we can do by the definition of lambda and head of psi, the following system of inequalities. So a zero psi to the power i minus a i smaller than x to the power minus lambda and head minus some epsilon slightly less than x to the power minus lambda and head and a zero is less than x. This system does not have solutions for some arbitrary large values of x and that gives us the upper bound, right? The upper bound for lambda one of, sorry, lower bound for lambda one of y. So from there, we can derive the lower bound on lambda one of y for some values, for some large values of y. And then that will, through this duality theorem, that will give us some upper bounds on the lambda n plus one star of y for some values of y. And now, now we need to, we need to understand what's, what does this mean? It means that we have n plus one linearly independent polynomials, p i, such that their values are rather small. So we can, we can derive, yes, their values p i of psi all, for all of those, for all of those independent polynomials, the values are at most h to the power minus one over lambda head n of psi and the heights are at most stage. So we have n plus one linearly independent polynomials, p. And the, yeah, are there any questions? And after that, because we have so many linearly independent polynomials, we can carefully choose the linear combination so that the value of the linear combination is still small. So p of psi is still smaller than h to the power minus one over lambda head of psi. But the, the derivative is rather high, p dashed of psi is at least h. And the, the conclusion from, from this method is that w and star of psi is always at least one over lambda head, lambda n head of psi. So that gives us the low bound for w and star of psi in terms of lambda n head of psi. And I think by, by the time, Darin Potenshmid found this, this relation, it looked, this way, it looked quite promising. And even now, and even now, if we find some good upper bound for lambda n head of psi, that will give us some good low bound for, and maybe some improvement for w and star of psi. But, but at the moment, we, we know quite little about lambda and head of psi. And that's, that's actually another interesting open question about this, the front-end exponents. We actually do not know, do not even know if n is at least three other transcendental numbers, psi, psi, that lambda hat n of psi is strictly bigger than one over n. We do not know. And of course, if we, if the answer to this question is no, then we solve for this in conjunction. Yeah, if there are no, if, if, if lambda n hat of psi is always at most one over n, then w and star of psi is always at least ten, right? But the conjecture is that the answer is yes. The conjecture is that the answer is yes. But as I said, not no one knows. No one knows for sure, right? Up to, up to now, the best general upper bound for lambda n hat of psi is due to Laurent. I'm afraid I don't remember in which year he got it, but he got that lambda n hat of psi is at most the n over two upper integer, upper, upper, upper integer value of n over two to the power minus one. And if we, if we, if we take this, this lower, lower bound and substitute it there, we still get w and star of psi is at least n over two plus something bounded. So it's, it's not an improvement, right? Of, of known results. Right. And now, now I think it's time to introduce our, what time do I have? Our, our result, Schleschitz from 2019. So we actually managed to beat the asymptotics n over two. So we show that if n is at least four and psi is transcendental, then w and star of psi is always bigger than n divided by square root of three, which is, of course, bigger than n over two asymptotically. And moreover, if we, if you consider the, the, the, the ratios w star of psi divided by n, then the limit super this ratios is even bigger. It's, it's at least some delta, which is approximately zero point 64 or eight and so on and so forth. And here you can see the formula for this delta, but you don't have to remember it. I don't, I don't remember it. It's, it comes from the proof and it's very technical. So it's, it's, it's some, it's some delta which, which is bigger than one over two three. Right. Now I'm, I will, I will, I will give some, a little insight into the proof, but very little insight, because firstly I don't have time and it becomes too technical, but they're just, they're just very general ideas behind the proof. Right. So we actually use a combination of the previous bounds and clever, clever links between them. So if, if lambda n hat of psi is, is small, then we use a Davenport and Schmidt estimate, which is let me write it down for, for you to remember, which is w n star of psi is at least one over lambda n hat psi. So if lambda n hat of psi is small, we use this one, but if it is not small, we get, we get an additional information about, yeah, about, about psi. So in that case, we, we get a reasonable lower estimate on one, one of the, of these different exponents, w n hat of psi, w n minus one hat of psi, and so w n minus m hat of psi. And after that, after we get this reasonable lower bound, whatever it is, we use a Bijou-Schleich's estimate, which is, let me remind you, here, which is this w n star of psi is at least three halves w hat of psi minus n plus one half, right. Yeah, so here, here, what we do, we define the sequence of best approximations to psi, psi squared, and so on, psi to the n. Basically, we take the, the, the minimal possible vectors x i, such that this value, which is a maximum of, of x i zero psi to the j minus x i j, which minimizes that maximum, right. And then from, from the definition of lambda n hat, we get this estimate. L of x i is always at most h of x i plus one to the power minus lambda n hat of psi. So that's a standard estimate. It's not hard, but yeah, I'm, I'm, I'm leaving all the details, I'm skipping all the details, right. And after that, we use the following crucial lemma, which is, and the ideas of this lemma come from the work of Laurent. So if, if lambda n hat is, is relatively big, yeah. Then, and you remember, we have, we have vector, minimal vector x i zero x i one up to x i n. Yeah. Then from, if, from, from this vector, we can choose the, the set, the, the, the collection of, of smaller vectors x i zero x i one and so on x i n minus m, then x i one x i two up to x i n minus m plus one, and so on. So each time we kind of shift, right. We take this one, then we take the next one and so on and so forth, right. So all of those are linearly independent, right. And as soon as we know that all of those are linearly independent, we, we know, we can estimate the, yeah, then we, we can translate this, this statement that the, the following system of inequalities has m plus one linearly independent solutions for all large x. And that gives us a, so it has a, I think it's, it gives us a lower, is it lower bound or upper bound? So it gives us the estimate for m plus first successive minimum for, for some value of y and actually for all large values of y of a dual body, right. And since we have, we have so many small values, it gives us upper, upper bound. So upper bound, upper bound on, on, on this, on this m plus first successive minimum. And if we have an upper bound or this successive minimum, we can, from there we can use the Minkowski second theory about successive minimum and get a lower bound. So we get lower bound for the, for the biggest successive minimum from there. And we get some, yeah, so we, we get, that gives us the lower bound for, for this successive minimum. And that gives, that is the exact formula, but it doesn't, you don't have to remember it. And from there, we use a maleduity theorem again to get an upper bound for, for the first successive minimum of, of, of, of, of C of y. So we, you know, we now come to, from the dual body to the initial body. Yeah. And after, after we do all of this, we get lower bound for w hat and minus m of psi, because it's related with the first successive minimum. Right. And, and the rest is just, just, just quite straightforward. So from there, after, after we have lower bound for, for w and minus m hat, we derive lower bound. So after we have lower bound for this, we derive lower bound for w and star of psi, which, which appears to be some complicated expression of lambda and hat of psi. Right. And, and, and then we combine two things together. So we have this, this formula for small values, lambda and hat, this formula for large values of lambda and hat. And that, that's, that becomes the problem from, from analysis. Right. We need to optimize, optimize this, this right hand side. And we optimize it and we optimize it firstly. We optimize it by choosing the right, the right value of lambda and hat. So we want to minimize this expression over all lambda and hat of psi. And after that, we maximize it by, by them. That's, that's something technical, but rather straightforward. Yeah. And, and, and that gives us, finally, in the end, that gives us the, the result that lambda and star of psi is, is bigger than n over root of three. Yeah. I hope, I hope that gave you at least some idea. I'm, I, I, I don't think I was clear enough to explain all the details, but at least I hope that you, you got to grasp some ideas for, from the, from this book. Yeah. And yeah, thank you very much for, for your attention.