 Hi, thank you for inviting me here. It's a great group of people. And I'd like to say got it, I guess. And I'd like to tell you something in the interface of dynamics and number theory. So I'll basically start with number theory. And then gradually you'll see more and more dynamical stuff. So what, what I should say here is that it is the main result is joined with Andreas Strombergson and Xu Cheng Yu. Time allows I'll also talk about results with an aura grow. Some extensions. Maybe somewhere then. And let me explain. So there are many strange words. Why do I have shrinking targets. Which you mentioned in the spaces and also what is the last year. So let me start with the last year in the front end approximation. This is the number. Well, basically that's that's the theorem. What I'm doing here is taking the supreme norm. Why is the matrix Q is a vector. Actually, I think this is not correct. Q is the end. And, and then you can approximate so this theorem says that you can basically approximate integer values of the system of linear forms with integers with a good precision and they like to raise norms to approximate to appropriate powers and then the formulas are look very, very nice. So, for example, if you specialize to just approximation of real numbers, you have the familiar inequality y minus p or q less than one over t q and q is less than t, or maybe maybe you can just write like this. It's simpler q y minus p. Okay, so that's that's what you want to solve and you can always do this. And there is a corollary also proved by Dirichlet that you can compare the discrepancy with the denominator itself so this is just familiar inequality. So this is y minus p over q, so it's in one over q squared. Okay, and then, and then so called metric number theory deals with various improvements enhancements, and so on of these inequalities what can be approximated better what can be done. And in particular, the main question is what happens if in this inequality systems one and two the right hand side is replaced by a faster decreasing function of T so I'm talking about this right hand side here, or this right hand side here. What happens for typical in particular typical with respect to the bed measure. So, most of what I'm talking about will be about results for typical wise, although it's not. There are other things that can be said. And oops, doesn't work for me here. Okay, and what I what I was trying to say is that it is actually quite well studied in the setting of number two, so called asymptotic approximation, but not so well for one so the so the setting improving. This setting is basically some new stuff and people were paying attention to it recently much more. And I'll talk about the history as well. Okay, so let me first talk about the setting to quickly. So, I want to take the function, some decreasing function which I'll call the approximating function psi. And then w mn of psi will be the set of psi approximately matrices so those wise for which you have infinitely many solutions of this inequality. So now instead of one over Q to the n, you have this function and then there is King hinge and Sierra. And the advantage of having this powers here is that the condition and hinges serum is uniform and doesn't depend on the dimensions which is what they really really like. Okay, so you have this convergence or divergence. And, in fact, monotonicity of psi is only needed when in dimension one, and the rest is true anyway. So this is related to differential injection. And also, what did what you can see is that the condition is stable with respect to replace psi by constant right so this series converges if you multiply by two still converges and so on. And, for example, I want to pay attention to this. So the specific function one over T and call it psi one. So that from directly follows that functions that matrices are proximal with with this function is the whole thing. And then by hinge and serum if you multiply it by C it will still have full measure. And by the way, the complement to the union of all these guys is a set of badly approximate matrices. So, so this is what so what I'm trying to say is that the complement is still in some sense big, but it has measure zero. And one other remark that the theory comes from the fact that it is a limb soup said what we're talking about so in fact why it belongs to this set if and only if this system has a non trivial integer solution for infinitely many natural numbers T. Okay, so, and this system is rather easy right so it's basically some kind of so we're talking about unions of neighborhoods of affine subspaces. Okay, so we can we can write to write something right in such a way. Again, if we restrict and then equal to one then we're talking about various intervals, taking the unions of intervals. And then to prove something like this you need to have some kind of quasi independence estimates for these unions and various methods to do this coming from geometry or arithmetic. So, it turned out to be not such a difficult problem. And then, let me switch to improving number one. And, well, to basically to improve the original degree of theorem, not its corollary, I need to consider not leave soup set but limit set. So this is my main definition. For today, I'll define the set dmn of psi to be the set of psi directly improvable matrices. What does it mean. It means that instead of T here, I have some right hand side, sorry, instead of one over T. So, before we had psi one of T which is one over T. And, well, then it's clear that you, it's too good to be true to wanted for all T but I want to demand that this has a solution for all large enough values of T. So the theorem says that if you put psi one here you have everything. Okay, so why is it an interesting setup. So the interest comes from the paper of them important Schmidt in 1969, who showed this very surprising sharp drop well at least for me when I first learned about it it was very surprising. You replace, you replace it by 0.99 over T and suddenly the set of solutions. The set of why is for which you can solve this has has measure zero. So in other words, what they proved. Well, again, let me maybe write it in the case in one dimensional case. So if C is less than one. And for almost every Y real numbers, real number exists sequence to K going to infinity, such that you write the Y Q minus P less than C over T K and then Q is less than T K. No, non zero integer solution. Okay, so if you put one things are very different, but just a little bit less than one. Suddenly everything changes. Okay, so this sounds rather, rather strange and I'll try to explain why this is happening. So but first let me pose some questions. So now that this we have this result, you can just see and try to find the right condition for this psi. So, and this will be kind of between one over T. And C over T where C is less than one. So maybe there is some kind of log factor or some other power, which distinguishes measure zero from full measure so that that's a natural question and I think we first tried to. So it is actually answered for M and then equal to one in my paper with Nick Wadley 2018 and the method was using continued fractions. So you can express in one dimension is things in continued fractions and figure it out and then if minimum of M and then is bigger than one, it's still not done. What I'll talk about is some partial result, which is quite close to full result but still not full result. So, so this seems to be much more difficult and much more sensitive problem. So I ask, why is it the case, because there should be some duality between lip soup and limit sets right so I was before I was talking about some kind of a case which were unions of neighborhoods of a fine subspaces. We end up talking about Lim soup set of their complements and this complements are much more complicated basically. So there are complements of those neighborhoods and what to do with this is not, is not clear so okay so there is a, there's a question about continued fractions maybe I'll answer later if you don't mind so I can actually just trained respond to chat because I happen to see it so so so you can write that why belongs to the set. The man. Si, if and only if some conditional continued fractions happens. Okay, and this is and. Yeah, and what, what is an efficient condition I'll also tell later because I'll try to state the general theorem, where I'll have this condition plus something else and this is something else will not need so I'll answer this question. A bit later. Okay, so now let me try to get to the main result. So, so the crucial critical case for this to have zero measure notice when when this happens when. So your function psi is one over team minus something right and this something. Now this something has to go to zero, because if it doesn't go to zero that I am in in very important Schmidt situation. Okay, so, so you can we can work with psi, or sometimes it makes more sense to work with this fight, which you can of course express in terms of psi. And so my condition will be involving this function fine. And what I, what I want to do is I want to assume both these functions to be non increasing. Okay, it's some kind of technical condition maybe it's not so important. But let me do it like this. And the case of the function having positive limit corresponds to the world for Schmidt situation with C less than one let me shut the door and I'll be right there. Okay, now it's quieter. So, so here's a theorem, and it will answer the question I got in chat it's kind of very long. But at least it has the condition. So it's joined with Andreas and should change. Take arbitrary and then you take this function psi and this phi gives you the discrepancy the difference from one over T. And I, we have two constants, capital D and lambda D, which depend quadratic from on D where D is M plus M. And now we have certain convergence divergence condition. So if this series converges. Well, I am also calling it one, even though now this one means different thing maybe, maybe I should call it a red one. Something like this. If this series converges then this set is a fully back measure. So we have phi to this power times log one or a fine to some other power. And we want to want this condition to be optimal. And then where it comes from. And then it diverges. Unfortunately, we're not able to prove that the measure is zero. We need a very strange condition, which is very hard to read so I'm not even going to read it. The limit of certain ratios will, if the limit for certain ratios is going to zero, then the divergence case can be can be can be treated. Well, we really believe that this condition shouldn't be here but for now it is here so let me give some examples. So this condition will be called to in the red square. And by the way, our result involving continued fractions was gave exactly exactly this condition. And of course, we have that lambda two is equal to. Okay, so, so, so, so easy D is equal to two. Right, so, so we have side D. We have a D is equal to one. Lambda D is also equal to one. Okay, and, and the condition was just the sum. Okay, so let me show some examples. For example, what if you have this log discrepancy we have one over T minus C times log T to some power over T and then you can check that the critical exponent is one over capital D. Critical exponent tau, and there is no problem with condition to it is always satisfied. Okay, so the conclusion is that you have for this family of functions, you have full measure if tau is bigger than one over ID, in one dimensional case it was bigger than one, and of zero measure otherwise. And if you want to fine tune this a little bit for example, you keep this critical exponent one over ID, and then you put log log T to some factor, then your situation becomes trickier. So the divergence of this one is equivalent to saying that tau is less than lambda D plus one over ID, but sometimes we do the we get this to sometimes and sometimes we don't. So we can only prove that that this condition grantees full measure and this condition grantees zero. Okay, so it's a bit annoying, but at least I'll I'll try to really say why where this anonymous comes from. Okay, so anyway, so the bottom line is that we have this series that decides what happens in this in this problem. So fine tuning of. All right, so now, I guess it's time to explain why dynamics is involved so usually the history of using dynamics and number theory is not great. Sometimes it's great but more often than not you do something by dynamics and then number theories come and say well all right it's very easy you just do this and that and you can also be fine. So as far as this problem is goes goes I, I don't really see how to do it yet. So it's, it is one of the domains where you can do something using dynamics and classical methods. I don't show anything close but anyway, let me go through the dynamical approach which actually can be extracted from them important Schmidt because the way they proved it can be translated to dynamical terms. There was a correspondence between approximation dynamics was spelled by Danny, and it goes. Now it, I mean, usually goes by the name Danny correspondence, which we was the term we invented in the paper with Margulis in 1998 and then with what labor adapted it to the directly improvement situation and also I should mention my work with Barack wise 2008 where we also utilize this correspondence. So, so let me try to. So what I want to do today is try to explain how dynamics enters the picture, and I want to take the easier case. I want to take the easier explanation so my function for now will be constant times psi one so it is and see is less than one so this is the Davenport Schmidt situation. And I want to see what happens after we translate the setup into the language of flaws in the space of lattices. So why let us as well. So if let me fix the C fix the matrix why and then you have this definitely different improvement condition. So let me write it down. For all large enough tea. This system has a non trivial integer solution. Okay, that's, that's basically, basically the definition and let me draw a picture corresponding to this definition it means that you have this R M here. And then you have a very long and thin rectangle, which so one side is C over T and other side is T. And then you have this objects, which are the vectors with coordinates, why Q minus P and Q. Maybe it's something like this for all integer Q and somehow there must be an integer point here. Okay, so that's what what this condition says. Okay, and then you can tweak it a little bit and the way you tweak it is that you take this rectangle and turn it into a box. So, equivalently with some with a little bit of computation you can do the following again you have these two axis. And now you just squeeze this rectangle so that it's area is still the same. No area this D dimensional volume. Okay. Well, so you have M coordinates with okay so. So the way the way this exponents are worked out guarantees that see the T cancels, and you and you only have this constant term here, and you can make it to be the same now you have a cube and what you are doing. So this is to is multiply the first value by something big and the second value by something small. So you're applying a diagonal matrix. And again, you have some kind of So now you're talking about applying this diagonal matrix to this and again, you want an integer vector here. That's the condition. Yeah, so we are. And so basically I moved from this setup to so called geometry of numbers studying ledges. And let me give this notation so so lambda sub y will be exactly this latest. So this is the set of all vectors of this form will be. I'll call lambda sub y. Okay, and a sub s will be this diagonal matrix. And then what what we showed is the following that you have this directly approval condition, even only for all large enough as this lattice as a non zero vector in the bowl. With radius C and power m over D. Okay, so this ball with respect to the supreme norm. Okay, and you can restate it as follows. It means that AS lambda y is not in certain set K sub see the power m over D, where I'm using this notation case of our to denote the set of lattices which have trivial intersection with a ball of radials are. Okay, and here we are in the situation of certain certain nice compact subsets in the space of lattices. So I can maybe draw a picture of this space XD. It's actually SLDR module SL. The space of any module lattices. And then you have certain set case of our right here. And then you start with lambda y. Maybe it's somewhere here. So, apply a sub s. And then you don't want to enter this KR for large enough as so this area is forbidden completely. Okay, that's, that's a dynamical interpretation of the improvement which is essentially between the lines of them and put Smith. It's important to remark that if R is less than one, this sets actually have none empty interior. And if also if, if R is bigger than one, then this KR is empty. Okay, so R equal to one is some kind of critical parameter. Okay, because what happens with K equal to one is this cube of radials one. And then by Minkowski's lemma where you absolutely must have intersection of any unimoderal lattice with this cube but if you replace if you remove, if you decrease the radials then it's not going to sometimes happen and it's not going to happen for all lattices and you have some set of lattices for which it doesn't happen. This is the green set case of part. Okay. So, any questions. Maybe I should stop and ask if I do something here. Anyway, so now, the conclusion is that this explains how David Palsch meets theorem can actually be derived from the rigidity of the action of this group as, and this is a case of fixed targets. My topic is shrinking targets but for now I'm not drinking anything I'm just keeping the same target. Let's figure it out. Okay, so we have this improvement property with the function given by C times psi one, and this happens if and only if the trajectory eventually avoid some neighborhoods of this set. Okay, so case of one, I mean what is case of one case of one is the set of lattices with this property that have vectors on the boundary of this cube but no vectors inside. So this is the set, and this set is called the critical locus of the supreme norm so these lattices are called critical and what picture again again this this will be the space xd I'm drawing it like this because it's not compact, but it has finite measure so this part which goes to infinity the cusp is kind of thin, I mean it's thick but in terms of the measure is thin. And then somewhere you have this set case of one, and here you have k sub r. Okay, and again if your starting point is in such a way that the trajectory does not reach it this set at all so this is not allowed. And then, then we have this directly improvement business with this function C one and this is very strange if you believe that this trajectory is connected should actually go anywhere so you have the notion of organicity of this action, and organicity tells you that almost every why will actually not have this property. Okay, and by the way, this set k one plays an important role in our proof. So the structure of the set is given by the theorem of higher, it was actually conjectured by minkowski and proved by higher. So this set is what you think it is. Namely, you have, maybe I should draw a picture here let's let me take D equal to two D is equal to two. Of course you have the latest Z two which belongs to K one right it's in the interior, but also you can share it's in the vertical direction and also share in horizontal direction separately. Okay, and so what you what you'll see is the union of the orbits of this upper triangle and lower triangular groups. So, in the space of ledges. You can. This is space X two space X two. Then this, this guy is the union of two close to her cycles. This is K one, or maybe number theorists prefer to understand this space as a union tension bundle to the modular surface. This is also great. Because here you have this her cycle one her cycle consists of vectors pointing upwards and another consists of vectors pointing downwards. Okay, so that's, that's, but topologically, it is just the union of these two circles and for for the bigger than two. It's kind of a mess you have a lot of so so when D is equal, let's say D equal to equal to three. You have a lot of three dimensional dry, which six of them actually which are somehow glued together in a very strange way and that's, this is the, this is this is the set. Okay, and this is, again, this is K one. And if you want to have K sub R when R is less than one unit to have some kind of a neighborhood with this set, but not just the absolute neighborhood. Very complicated neighborhood. Okay, but anyway, for the purpose of proving them and put Smith's theorem and doesn't matter we just have some open set and every almost every orbit should enter. And then, then you derive everything. So, now the argument goes similarly for functions which decay. I guess I have to speed up a little bit. So, so it is. And then, if, if the fun, if this five goes to zero, it turns out that you actually talking about targets which shrink. So you have this K one, which is somewhere here. And then you have R of s here K R of s which becomes sure smaller and smaller. And you need to have this condition. This is not in case so it's about the rate of shrinking rate of approximating this set, you said, K, K one. Okay, so, so that's, that's how this strictly improvement is reduced to shrinking targets theory in dynamical system that's basically what we did with Nick Wadley. And then you can make it even nicer if you discretize. So, you have this yellow set and you make a say Brown set out of it by moving it in the AS direction by time one, and then you have this discrete condition. A j, j is a natural number. And this becomes this set becomes a big J. And for large enough, J, you don't want to get there. Okay, so, in other words, take here this should be a compliment. So why isn't the compliment if and only if a J applied to lambda y belongs to this be J from it into many J. Okay, so, so we're reduced it to basically to some kind of barrel can tell you and, and of course, but I'll come to the limit says that the right condition and the sum of measures of the J's is if the sum of measures is finite if almost surely have finally many solutions, if the sum of measures is infinite and something else, then the number of solutions is infinite. And this something else, basically some kind of quasi independence properties of this family. Okay, so now I have to rush a little bit. I guess, at least what is left are several steps. We need to come come come up with a with a measure I mean what is really the measure of this set. And new is now the hard measure SLDR invariant probability measure on the space of lattices. So we need to come to compute it or to estimate for small enough are not sorry for our are close to one. And then you want to also have this BJ, which is this brown set, which you get from KR by extending it in in one direction. Actually, I have to say that I put this equivalence in quotation marks, because it's kind of, it's a little bit approximate you need to squeeze it between two sets. And in this approximation actually using the monotonicity property of the function. Okay, so, so then, then you need to estimate this, this measure. And then, you need to establish some kind of quasi dependence of pre images. And also, another step is to go from almost every lattice lambda to almost every lambda of the form. So, so again, for the first part. Remember that in the D is equal to two, we are talking about a set of the form, some kind of neighborhood of these two circles, and it's even more complicated in high dimensions. And now you need to move it a little bit in the transfer so the action and also do some work. So this, this is actually the bulk of the paper. Okay, so this one is work is done is done here. And so let me show you the main measure estimate. Remember that we had this exponents capital D and lambda D. And then the measure of K sub R is a SR goes to one is approximately one minus R to some power times log of one minus R to some other power. So, with Schuchin you actually did the case D equal to two before. Okay. So, and then as a corollary we can actually look at this shifts the cylinder and the only difference is that we go from this power to this power. Okay, so just simply because you lose one dimension so so it's, but it also requires requires a lot of work. And then for parts three and four. Here is where dynamics is important. So before we only had the geometry, and basically, I want to use the fact that you start with this set of so I want to do three and four, the same time, because it makes sense. Alternatively, you can take care of three. So remember that three was some kind of quasi independence conditions. And then you have a affiliated version but in fact, modern technique allows you to have the affiliated version right away. So what you what you do is, is the following. Let me try to draw a picture don't have much time so maybe I'll just use pictures. So, let me start with some kind of the set of the form, let us solve the problem with some why. When you apply a some s and the way it is set up shows you that in fact you get. So you this this manifold gets expanded and it actually becomes equidistributed. In the sense that if you fix some, some open set you, then the proportion will be approximately the measure of you. Okay, and in fact, there is an exponential rate of mixing and this gives rise to exponential rate of equity distribution with some error bounds so we first spelled it out. Well, the first was spelled out in the thesis of Margulis about genetic flows and also flows, but then we did it for flows of this type, driving from the exponential mixing of the action. And in fact, we need the refined version so called effective double equity distribution which is the paper of Michael Bjorkland and Alex Rodnik. So, so there are some strong dynamical, dynamical results, which make it possible to try at least to extract this was independence conditions, but the bad news is that these results. So applicable to smooth functions. And so to apply this machinery, this sets BJ, which you start with, need to be approximated by smooth functions and the coefficients and the mixing estimates depend on the size of their derivatives and the size blows up as J goes to infinity because they this sets become thinner and thinner. Well, I don't have time to convince you that if you take care of this factor, appearing of big derivatives you actually end up having this very strange technical condition which I don't even want to mention I'll mention it anyway. It's somewhere somewhere here. The point is that if you, in addition, assume this, this convergence you get, you are able to use this mixing mixing technique. Okay. So, I guess, I have to stop now let me, let me just mention. So I left this page blank just in case I have one minute so maybe I'll have one minute to mention the extensions and one is so called approximation with weights. Again, I don't want to spend much time on it but weights basically means that you treat various components of Q and various linear forms differently. So you have a vector alpha and vector beta, and then you do absolutely the same but your dynamical system is now as depends on this alpha and beta and this is just e to minus alpha one s up to e to minus alpha m s and here you have beta one s beta m s. Okay, so all these guys were the same. But in fact, you can make them a little bit different and still prove more or less the same result because this Bjerkel and Gerdnik condition actually takes care of even this, this type of expansion. Okay, and finally, you can also use other norms. Okay, so everything I did was for the supremum norm, and then we ended up having this very strange critical laws. You can choose another norm, you can still save, you can still state the number theoretical problem. For example, you can take Euclidean norm, and then you have a different setup and you can still ask the same questions about the improvements of the theorem and this is work I did with an aura grow that I don't really have time to mention. For example, when the norm is Euclidean and when m and then are equal to one, we have a necessary and sufficient condition for this set to have measured zero full measure without any extra technical assumptions. And here we're lucky to be able to derive it from some dynamical systems results from some current. Okay, anyway, I guess I'm over time already so let me stop here and write a big thank you for your attention and invitation.