 Okay, so today I'm going to be talking about distribution or equidistribution of nil sequences. I'll remind you just quickly what a nil sequence is shortly. But really I'm going to be talking about generalizations of a very, very well-known fact. So generalizations and quantitative variants of the following very well-known fact, which is that if theta is a real number, then the sequence of values theta, 2 theta, 3 theta, etc., is equidistributed mod 1 if and only if theta is irrational. So I'm going to be generalizing that to polynomial sequences on arbitrary quotients g mod gamma. Here I've just got r modulo z. And I'm also going to be interested not in the whole sequence up to infinity, but in finite segments of it. And all of that with a mind to applications next time and the time after. So let me just remind you again what this thing called a nil sequence is. And recall, we have a simply connected nil-potent Lie group g, which has on it a proper filtration g-bullet. And that's a collection of, it's a collection of closed Lie subgroups of g, which satisfy this commutator inclusion property that I've mentioned before. And then I define this notion of a polynomial sequence that we defined in several different ways, which I didn't show to be equivalent, but I stated it. The notion of polynomial sequence p of n, p from z to g. So I won't go over that again. I did it last time. And you don't really need to know what the general definition is this time. So the basic question I'm going to be interested in is, when is, let me take a lattice gamma, let gamma be a lattice. If I have a finite set of values of that polynomial, so when is p of n from n equals one to big n close to equal distributed in the quotient gamma mod g? And what does this even mean? So that's the question I'm going to be interested in. Well to motivate this, maybe I'll give you a few examples. I've already had one. So let me first of all say that the example I put up there actually generalizes, this is perhaps a less well-known fact, it generalizes to arbitrary polynomial sequences. So g equals r and gamma equals z. And if p is just any polynomial. So I think I briefly mentioned last time how to make the traditional notion of a polynomial compatible with this slightly fancy notion to do with filtrations. If the polynomial has degree d, you just take the first d elements of the filtration to equal the whole group, which is r in this case. So if p is any polynomial, then p of n is the whole sequence, is equidistributed. If and only if the polynomial has at least one irrational coefficient, other than the constant term, other than the constant term. So this is not so trivial to prove. I may say a few words about it later. So the proof uses vials inequality, which I will mention later. So vials inequality will tell you that if the polynomial is not equidistributed, then the lead coefficient is irrational. And then you'd have to iterate down to show that all of the other coefficients are irrational as well. So uses vials inequality and an iteration. And there are theorems in a Goddick theory. So a theorem of Leon Green, no relation of minus should point out. And this is about equidistribution of flows on nilpotent groups. So if g is a nilpotent, so a simply connected nilpotent lead group. And if p is a linear sequence, so p of n is a to the n for some a. So that qualifies as a polynomial sequence with the definition I've given. Then the conclusion is, so p of n, n equals 1 to infinity, is equidistributed. What does equidistributed mean? Well, I'll be going over that shortly. But let's just say for the moment, with respect to the Haar measure, so with respect to Haar measure on gamma mod g. Well, if and only if the only way that a sequence can fail to be equidistributed is if it fails to be equidistributed in an abelian projection of g. So if and only if the abelian projection of p fails to be equidistributed. So what I mean by that is that you have g and you have a natural projection to g mod its commutator. And the sequence p projects there. So this will be isomorphic to just a Euclidean torus. And when you project p, it will be isomorphic to a certain rotation. So the p projects to a map. Pi composes with p. So this induces a map on a torus. So let me give an example with the Heisenberg group. So if and only it is equidistributed, if and only if the abelian projection is equidistributed, exactly. Thank you. I think it's best if I give an example. So example, if g is the Heisenberg with the usual lattice, and if a is just some element, so if a is 1, 1, 1 alpha beta gamma, then the flow that's induced on the two-dimensional torus by that is just rotation by alpha and beta. So the induced map on the torus, r mod z squared, is given by p tilde of n is alpha n beta n. So you have Leon Green's theorem is the non-trivial fact that the sequence a to the n in the nilpatent group is equidistributed if and only if this sequence is equidistributed in the torus, which by Kroniker's theorem is the same as saying that alpha and beta are independent over the rationals. So alpha, beta and 1 are independent over the rationals. By Kroniker's theorem, this is equidistributed if and only if 1 alpha and beta are linearly independent over q. So that's a complete criterion for when a nil rotation on the Heisenberg group is equidistributed. So I'm really going to be interested in quantitative variants of this, and even to say what I mean, I should give a bit more discussion. So are there any questions so far? So quantitative equidistribution. If you want to make things quantitative, you have to quantify the notion of close. So let me introduce a parameter delta greater than or equal to zero be a parameter. Well, if somebody asked you how should I define equidistributed on the circle, probably the first thing I would say is it's a sequence that spends the right portion of time in each interval. It's slightly problematic to use that as a definition, because what should be the analog of interval in a more general group? So we don't think of it quite like that, but we think of the interval, the characteristic function of the interval as a function, as it is, and then equal distribution is saying that averages of our sequence along a function are close to the integral of that function. So we say that a sequence p of n, so it will always be for us a polynomial sequence taking values, well, the sequence p of n, is delta equidistributed in gamma mod g if, so when you look at the average of this sequence weighted by any function on, so this phi will be a function on g mod gamma, and then you compare that to the integral with respect to harm measure. Then that's at most delta times an appropriate norm of the function. I'll tell you which norm in a moment we're here for every, let's say for every smooth, automorphic function on g. So here we put the harm measure on gamma mod g induced from the unique harm measure on g, and it normalized in such a way that the total volume of this quotient is one. So the measure of the whole space is one. And these harm measures are not too difficult to understand in the Heisenberg group, it's actually just dx dy dz. So in Heisenberg the integral is simply with respect to dx dy dz. So the harm measure is just the product of the harm measures in the three coordinates. So you won't lose much by thinking about that case or even just the circle case for now. It doesn't matter hugely which norm you take, I mean if you change the norm it will change the notion that you're dealing with, but in relatively minor ways. So for technical convenience it's actually useful to have our norm to be one of those Sobolev norms that I mentioned last time. So we'll take the norm. So you just take it to be a sufficiently high Sobolev norm, and I think I found that the Sobolev norm with 2 to the s times the dimension of g derivatives was enough. But that's really a technical matter. In fact you could just look at first derivatives of phi and every function with bounded first derivatives is basically Lipschitz, or it is Lipschitz, and it can easily be approximated by much smoother functions. So that's really just a technical point. Anyway the basic idea here is simply that you're comparing space averages and time averages and that's how you decide whether a sequence is close to equidistributed. Questions on this? So a crucial result on, well I suppose the main result that I want to talk about is a more or less complete statement about when a polynomial sequence on g mod gamma is equidistributed in this sense. And it generalizes really the examples that I showed you above. So the main result, a generalization of the above examples to, well first of all, two polynomial sequences on gamma mod g and with this quantitative notion of distribution. Now maybe I should say, well I should say, so Leon Green's theorem is valid for any simply connected nilperton group and for any linear sequence on that group. If you instead want a polynomial sequence, that's a more general result. But it turns out that actually a polynomial sequence can be lifted to a linear sequence on a much bigger group. So actually you get the polynomial equal distribution result for free from the linear one. But for various reasons that's not, that argument fails you in this quantitative setting. So I thought I'd mention it but I don't want to place any emphasis on it. So what I do want to do is state precisely this main result and then I'll sketch some ideas of the proof. So here is a reasonably precise statement and it seems to be a bit of a theme in these lectures that the most I can hope to do with precision is state results. I guess that's, sometimes things are difficult. So theorem, let the notation be as above. Let's suppose that p fails to be delta equidistributed from 1 up to n. Well then there's an abelian explanation for that and I'll write down what it is. There is a horizontal character. So I'll introduce a few terms that I will explain later. A horizontal character eta from g to the reals, mod z, such that what we call the smoothness norm of the abelianization of p is bounded. So this bounded by delta to the minus a constant where the constant depends on the group. So I have a lot of things here to explain. But if you don't wish to take too much notice of my explanations, this is an abelian obstruction to a sequence being equidistributed. So here pi is projection from g to g mod, actually, sorry, with the way I've defined it here I don't need the pi. This will automatically be a, so I just have to tell you what a horizontal character is. So here horizontal character just means it's a homomorphism. So it will automatically annihilate the commutator subgroup g brackets g and it also is supposed to annihilate gamma and the smoothness norm. So thus eta composed with p of n, well it's going to be an r mod z valued polynomial. So it's a polynomial map. Let's call it f from the integers to r mod z. And so I can just write it in the usual way that I write polynomials. So this will have the form f of n is alpha naught plus alpha 1n plus up to alpha d n to the d. And this smoothness norm is basically telling you how close to being rational all of those coefficients are. So to say that, so the statement and that f in smoothness norm c infinity of n is at most m means, well it means that those coefficients are very close to integers. It means that alpha j in r mod z, which is the distance from alpha j to an integer, is at most m over n to the j for every j. So that's a lot to take in. Let me tell you at least how it specializes down to just even some abelian settings. Before I do that, I said that this smoothness norm is telling you that those alpha j's are close to being rational. Actually it's telling you that they're close to being integers. To somehow, that's enough because you can always multiply eta through by an appropriate integer to make things that are close to rationals close to integers. Let me give an example. An example just with g equals r and gamma equals z. The sequence theta n squared from n equals one to n fails to be delta equal distributed if and only if theta is very, very close to a rational. So if and only if there is a q in the integers such that the distance from q theta to the nearest integer is bounded by delta to the minus constant over n squared. Where is the q coming from? Well multiplication by an integer q is the same thing as one of these horizontal characters. Let me just write that. So eta is multiplication by q. So this is a known result. This follows from vials inequality, but this is not the usual way of phrasing vials inequality. Actually it's the way that I prefer to, when I'm lecturing on vials inequality, I prefer to state it this way. I find the usual way a little bit confusing. I don't know if anyone's familiar with the statement of the inequality. It's a statement about certain exponential sums with polynomial phases, but it's really saying precisely this. So a polynomial sequence and when it's equal distributed. So does anyone notice something that's lacking in that statement? Well, sorry? It's bound on q. As stated, it has no content whatsoever, right? Because every theta by just by Dirichlet's pigeonhole principle, for any theta you can find a q, it might have to be very big, such that theta times q is close to an integer. So I additionally need a quantification to give this content. We also need to place a bound on the complexity of this horizontal character eta, that a certain complexity quantity, so the complexity of eta. And as before, this has got to be measured with respect to a basis on the Lie algebra, curly b, is bounded. So I don't want to go over again what the basis was and how you define exactly this notion of complexity. But in this particular example, it just degenerates to the size of q. So in the example just given, it asserts a bound on q, mod q. Also q should not be zero, otherwise it's again completely trivial. So it asserts a bound on the size of q, in this case, well, the size of q is again bounded by delta to some power. So to state the theorem precisely, I would need to include also some bound on the complexity of this horizontal character. So let me just write explicitly what the contents of this theorem is in one phrase. So p of n can only fail to be equidistributed for a billion reasons. Probably all of a sudden the power of delta on the left hand side delta to the k equals 100. Power of delta where? On the left hand side of the equivalence because you can replace delta by delta. Ah, yeah, yeah. Let me just, okay, that's a good point. The notion of equivalence in this sort of discrete setting is always a bit, you don't quite get back to where you started from. So there is an if and only if, but if you go that way and then back, delta will have become delta to some bigger power. So let me just phrase it like that for now. Here's the basic idea. A polynomial sequence, the only way it fails to be equidistributed is for a billion reasons. Now I want to say a little bit about the proof, obviously not very much because it's a, this is a really difficult theorem to prove, but the proof uses a trick, well it's more than a trick, a technique that is the same technique that's used in the proof of vials inequality and that's called the van der corporate lemma. So the main idea of the proof. So following vile, and this is not, so this follows vile which is a paper in analytic number theory. It's not the same thing that Leon Green does in his paper, but by following vile we will, we use what's called the van der corporate lemma. So the van der corporate lemma asserts a relation between a sequence being equidistributed, well the original van der corporate lemma, so the original version of which, asserted a link between the distribution of a sequence, the uniform distribution of a sequence un mod 1 and its derivatives, which are un plus h minus un. Specifically if all of the derivatives are equidistributed then so is the sequence. So that would actually immediately give you qualitative equidistribution results about theta n squared, because the derivatives will then be linear sequences and then you can just apply the known results on linear sequences. Here is our take on van der corporate inequality, which is very standard. Suppose you have just any bounded sequence, so suppose that a n, n equals 1 to n, are complex numbers of norm at most 1 and suppose that their average is not close to 0. So suppose the expected value of a of n is at least delta, then the conclusion is that many of the derivatives have the same property. So I'm going to state a precise version even though I'm certainly not going to prove a precise version. So suppose that big h, slightly technical condition, big h lies between, big h is neither incredibly tiny nor quite as big as n, then we have that the average value of many of the derivatives is at least delta squared over 4 for many values of h. Remember how many? The 4 and the 32 clearly don't matter, so for at least delta squared over 32 times h values of h. Sometimes I find it less confusing, I don't know if you agree to put in explicit constants rather than unspecified absolute constants when they're not too big. So the statement is if you've got some bias in a sequence then also many of its derivatives are biased. Now let me sketch the proof. Well the idea is that if h is small and that's going to be guaranteed by this inequality then averaging the average of a n is very close to the average of a n shifted by h. I mean they're averaging essentially the same thing except you've got some slight edge effects. So many of those averages are well, so then it's basically the Cauchy-Schwarz inequality. So you can choose just some phases, so let's choose some unit complex numbers, zh expected value n less than or equal to n of a n plus h is bigger than delta over 2 let's say for an appropriate phase zh just to make it real. So if you average that over h, so the average over h less than h and the average over n less than or equal to n of z of h a of n plus h is also at least delta over 2 and now you apply Cauchy-Schwarz. So by Cauchy the average over n of the average over h a n plus h squared is at least delta squared over 4 and if you expand that out so that implies that the average over h1 and h2 less than or equal to h of the average over n less than or equal to n of a n plus h1 a n plus h2 bar is at least delta squared over 4. Well you can see that with a bit of messing around you'll get the statement that I wanted. Every average on the inside here is basically like the average in the statement of van der Korpen's inequality so this is roughly the expected value over n less than or equal to n of a n a n plus h1 minus h2 bar and then you just have to count how many times each h1 minus h2 occurs so etc. But that's the basic idea is just one application of Cauchy-Schwarz after you've done a shift. Questions? The other key idea is also a key idea a very famous idea that is due to Vile which is that to understand equal distribution you could you should expand into Fourier series. So the other crucial idea is Vile's idea of proving equal distribution by expanding into Fourier series or put another way we had a definition of equal distribution which sadly I've erased but the definition was about an arbitrary function phi and you just need to check that definition on a basis of well l2 of gamma mod g so check the definition so check the definition of equal distribution only for phi lying in a basis of l2. Now of course it's not quite enough to just check things if you're dealing with a quantitative questions you can't just check things for a basis and then hope that the same statement will hold when you add basis elements together but as long as your way of decomposing into these basis elements is suitably effective which it will be if the functions are highly smooth then that's more or less the case. So what Vile did famously in the case of r mod z it's enough to check it so it's enough to handle phi of t being e to the 2 pi i mt for m in the integers and as I said you do lose a little bit in the delta so this causes some slight degradation some slight change in delta but just a power. So I'll go through that in the case of null sequences in just a second but let me explain how it works already in the case of say theta n squared. So if you want to test the distribution of theta n squared on the circle it's enough to test it for e to the 2 pi i theta n squared and if that average happened to be large then you could apply the van der koper inequality and get that many derivatives of that have to have large average but a derivative of e to the 2 pi i theta n squared is just a linear phase function which is just a geometric series so it's quite easy to evaluate that. So what's the appropriate variant of that of this Fourier expansion in gamma mod g g mod gamma where g is a nilpotent group well of course that's a that's a very deep question in general but actually it turns out we can get away with a fairly simple answer to that we're not going to look at a sophisticated way of decomposing this space it's actually just enough to do a Fourier expansion in the vertical components so in the general case a similar trick works to test equal distribution for an arbitrary phi you decompose it into automorphic functions with a vertical frequency so to test equal distribution against an arbitrary phi in c infinity of gamma mod g it's enough to handle the case when phi has a vertical frequency psi which let me remind you means that phi of if you twist x by an element of the last group in the filtration then it has a natural transformation property under psi and the reason for that simply is that there is um you can do a Fourier expansion in the direction of this group g so as I mentioned last time uh phi can be expanded as a sum of automorphic functions with a vertical frequency and of course all of this would have to be done quantitatively but I shan't bore you with the details of that so suppose you've done that and suppose that you have now a function that additionally has a vertical frequency so suppose that psi is a non-trivial frequency then that that means just as with the non-trivial exponentials on the circle the integral of this function with respect to the harm measure because of the vertical oscillation it just cancels out and you get zero so what does equal what does failure of equal distribution mean in this case so if p of n from n equals one to n fails to be delta equal distributed with respect to uh this particular automorphic function with a vertical frequency then it just means that the average is at least delta well actually at least delta times the smoothness norm of this function but I'll suppress that so this is something that we can apply the van der corporate lemma to and by van der corporate there are many h such that the same average but now with an additional term of phi of p of n plus h is at least some constant times delta squared now I don't know if anybody remembers what happened last week but you can see that something very similar to what happened in the proof of vials inequality has happened here so for vials inequality van der corporate took a quadratic phase function and by differentiating it made it linear and what is the derivative of a nil sequence while I spent a whole hour explaining that that is a nil sequence of class one less so we can turn this into an induction by this device of taking derivatives so by the main discussion by the discussion of um I think it was mainly in lecture three uh the object here phi psi p of n phi psi p of n plus h bar for fixed h so is for fixed h a nil sequence of class at most s minus one so what this means is that you can imply you can hope to prove the main theorem by induction on the class so this offers the possibility of proceeding by induction on s so this is very special to the case of these nil flows one can consider flows on g mod gamma in other cases for example where g is sl2r um gammas a subgroup but the same sort of trick just wouldn't work I mean you'd get this would not be a simpler object in fact it would be a worse object sitting on g cross g but in the nil potent case we do have this simplification so that's the main idea of the proof but you still have to actually carry out the induction so carrying out this induction is uh rather difficult and I just want to show you just give you a glimpse of what happens in the heisenberg case so you can see the way in which it's difficult um so let me actually revert completely to the notation of last time so let's set chi of n to be this nil sequence and then in the notation I had last time delta sub h of chi is precisely the derivative so last time I said that this derivative is a nil sequence of class s minus one well actually let me let me specialize also to the case of the heisenberg just for illustration so g is the heisenberg and um and the polynomial is just a to the n for some n linear so I said last time last time we saw that the derivative delta h chi of n can be thought of as a nil sequence of class one and I wrote down precisely what those actually I wrote down the general form so here p sub h delta of n is p of n plus h so it's p of n p of n plus h times p of naught p of h inverse and that's actually in the linear case it's just the same it's just the diagonal flow a to the n a to the n in the linear case if you calculate very easy calculation and the automorphic function phi phi sub h box of n was phi times phi bar of x y p naught p h which in my case is phi of x times phi of y times a to the h and here we come across a serious issue which I alluded to last time so I'm trying to prove a statement about equal distribution by induction but the statement I'm trying to prove is quantitative it cares about how smooth my functions phi are and how lipschitz they are but unfortunately this a to the h here could be huge I said nothing about how a is bounded this could be an absolutely enormous element in the group g so there's no guarantee that this automorphic function has any nice control at all so a could be huge a to the h could be huge so no control on phi h square so the induction won't work unless you do something to get around that issue and what you need to do is to insert a conjugation by elements of the lattice gamma so to get around this we can conjugate by elements of gamma cross gamma so here and maybe I'll just show you how to modify this to do that so you can put a you can conjugate this by an arbitrary gamma gamma h and this has the effect of adjusting the automorphic function in here so put some maybe to explain this I did a computation last time explaining why the derivative of a null sequence could be written in this form it turned out that there's extra flexibility inside there that I didn't make use of you can do this conjugation in any way you like and you have a different representation of the derivative in this form and so this will then become delta h I can take gamma naught equals zero gamma naught is the identity so I get to multiply a to the h by any element of the lattice that I like on the left and by doing that I can make it small so by choosing choose gamma sub h so that gamma sub h a to the h lies in the fundamental domain so in particular we'll have all of its matrix entries bounded by one and so when you do that you do then recover control on this automorphic function here unfortunately you've made the problem more complicated so you had to do this to make get yourself a derivative function that's got bounded smoothness norms but you've made the polynomial the derivative of the polynomial more complicated so p sub h is now while it's rotation by a by the diagonal a a conjugated so it's delta sub h gamma sub h a gamma sub h inverse to the n so now you can imply the inductive hypothesis this is just a rotation on a torus it's a rotation on a torus so I introduced this torus last time it's what you get by applying the constructions I showed you last time on the Heisenberg and it's a three-dimensional Euclidean torus and so by Kroniker's theorem we have essentially the coordinates of this vector a and the conjugate of a fail to be independent over q so the coordinates of a gamma h a gamma h inverse well reduced to lie in this torus so reduced mod g2 diagonal fail to be independent over q for many values of h so it's an abelian problem but it's a rather complicated one so you have to somehow go from that assertion to saying something about the coordinates of a itself it's a difficult matter to go from this to a statement about a but it can be done um I think it's well I'm not going to go into any more details the the the paper in which this is done is difficult but at least you're reassured that you kind of know it has to work if the theorem that we're aiming for about distribution of nil sequences is correct and that's because this van der korp is almost an if and only if so it's not you haven't really lost anything apart from a few powers of delta in going through all of this machinery the expansion into Fourier series again it's more or less an if and only if so this is a it's kind of a weird sort of integration problem you've taken derivatives and you have to deduce it was easy to take derivatives but the statement you ended up with when you did that is quite difficult to integrate back again and I should say that it's the same problem with the proof of the inverse conjectures for the Gauss norms I showed you last time why why nil sequences are obstructions to Gauss uniformity by induction so I said if a function correlates with a nil sequence chi many of its derivatives correlate with derivatives and then it I was done by induction but to get the converse statement is much harder so many of the derivatives of f correlate with a nil sequence but you have to then deduce from that that f itself correlates with a nil sequence so it's a similar sort of issue that's difficult there this is not to use it it is all over the place actually yes for in the case of the u4 norm and higher this is used everywhere I mean I didn't use the inverse conjecture at any point here this is purely an almost an algebraic fact really I've said nothing about Gauss norms for the u4 norm and higher it's needed this is really this is somehow key to the entire theory of higher order Fourier analysis and nil sequences it's also needed crucially in showing that well tomorrow I'll be talking a bit about the primes and there you need to show that the Möbius function is orthogonal to these nil sequences and again this theorem is a crucial ingredient there so it's somehow quite fundamental um well just to finish I want to say two things about how this is applied so why I haven't really said anything about why I should care about equal distribution of nil sequences so why do I care about this particular notion equal distribution of polynomial sequences on gamma mod g so there are two reasons for that and the first one is that essentially every sequence is equidistributed so every p of n is almost equidistributed I'll say a little more about what that means in a moment um but it means that p an arbitrary polynomial sequence can be decomposed into some very benign pieces a periodic piece and a very smooth piece plus an equidistributed piece so somehow the equidistributed case is pretty general and then secondly if p of n is equidistributed then we can we can count various linear configurations so if I have time I'll give you an example so eg the average of I can count say four term progressions involving nil sequences let me make this r p of n plus two r phi of p of n plus three r and it will be I get a very algebraic answer it's an integral of phi to the four over a certain subgroup called the whole potresco group h p four of g modulo so this where h p four of g which is a subgroup of g cross g cross g cross g is a certain group explicit group called the whole potresco group the whole potresco four group um and actually that this generalizes there's nothing special about arithmetic progressions there's a corresponding result for any linear forms which is called the leadman group um let me tell you what the whole potresco group is because we've already seen it in action eg when g is so when g is r but with the filtration so I'm going to put a degree across two filtration on g then the whole potresco four group h p four of g is it's a subgroup of r four and it's the set of all topples x naught x one x two x three such that x naught minus three x one plus three x two plus x three is equal to zero and we've seen that before um because I've already we know an example of a polynomial sequence here would just be theta n squared um but the example I showed you before was where theta was root two so let's give that example and we noticed before that there's a constraint amongst four term progressions uh in this function so I mentioned that when I was showing you that traditional Fourier analysis is not enough to handle four term progressions and what this theorem is saying is that essentially that's the only constraint so it's saying that the four term progressions or inside this function equal distribute over the set of all four topples satisfying this constraint so um in a sense so n squared root two minus n three n plus d squared root two plus three n plus two d squared root two minus n plus three d squared root two equal zero is the only constraint on four term progressions for ap's in p of n so these equal distributed sequences are very nice actually I lied to you slightly uh that's false you actually need a slightly more elaborate notion a stronger notion of equal distributed called irrational it's quite technical and I I don't feel inclined to say what it is here in many cases for example for for these linear sequences on the Heisenberg group the two notions coincide so to finish today I'm going to explain what I mean by this statement what do I mean by every polynomial sequence is almost equal distributed uh well this is basically a generalization of what's called the Hardy-Littlewood method so Hardy and Littlewood were concerned with well things like polynomial phases on the circle when they were thinking about wearing's problem and they decomposed those phases or the coefficients of those phases into major and minor arcs and um so in other words they basically said every every numbers either essentially rational or it's not rational it's irrational and that's essentially what's generalized by this assertion here so let me explain point one is uh really a generalization of the Hardy-Littlewood idea of decomposing into major and minor arcs and in fact the heart this Hardy-Littlewood idea can be seen as the special case where g is the circle a g is r of this so a more precise statement so given an arbitrary polynomial sequence I can write it as a product so I can write p as epsilon times p prime times gamma prime this is a point wise product of polynomial sequences uh where epsilon epsilon just varies very slowly smooth so that means something like uh the I mean some on on the group measured on the group in some way it's just doesn't vary very much so this means essentially that you can always use some standard tricks in analytic number theory like partial summation to just ignore it and then gamma is rational and that means that um basically that gamma is periodic modulate the lattice for some small q and again in practice that means you can get rid of gamma just by or just by splitting into sub progressions of common difference q and then finally p dashed is equal distributed is highly equal distributed somewhere um not necessarily in g I mean the p could just be trivial p could be identically equal to it just could be the identity always and then obviously you're going to have a hard time saying that anything related to it is equal distributed on g but p primed of n is highly equal distributed somewhere on some g primed mod gamma with g primed contained in g a closed connected subgroup and as with all of these statements a proper quantification of this is quite technical um you need to say what you want by highly equidistributed and you get to choose it to an extent at the at a certain cost um and then you get to control the complexity of the g primed etc etc so there is a much more precise statement that needs to be made but this is the this is what's true in spirit so in the example I mentioned where p is constant this decomposition is trivial so epsilon is also constant gamma is constant and p primed is constant and it's highly equidistributed in where g primed is identically one the identity um and then the other extreme would be where p is already equidistributed and then you could take epsilon and gamma to be trivial so just to finish let me explain how is this proven and this point is why this is the reason that we have to deal with polynomial sequences rather than merely linear ones so the way this is proven is by induction on downwards induction on the dimension so if p is already equidistributed then you've got nothing to do stop otherwise the main theorem I've been talking about today tells you that there's an abelian explanation for that fact and what this means is that p is basically near it sits inside a proper subgroup so you then get to induct downwards on dimension and that will eventually stop so the proof is by downwards induction on dimension using the equidistribution theorem and let me give you an example of what can happen when you do that so e g again with the Heisenberg suppose that you take p of n to be a to the n a linear sequence where a is one one one theta one zero and here theta is of size about n to the minus three halves so you can compute what a to the n is a to the n is one theta n one one n and then a half theta times n times n minus one now a n a to the n n equals one to n is not close to equidistributed and we know by the theorem that there must be an abelian explanation for that but in this case it's staring us in the face because the x-coordinate never really leaves zero so it's visibly not it not close to equidistributed because theta of n is stuck near very near to zero and so if I follow through this idea of in downwards induction on the dimension let's restrict the whole setting to the group where x is zero but then a to the n is no longer a linear sequence it's then a polynomial sequence with those entries so this can turn a linear sequence so p of n turns into a polynomial sequence upon restriction to x equals zero so in fact it's a polynomial sequence on an abelian torus in that case so what this point shows is that you have to work in this category of polynomial sequences in order to be able to make any of this work just the linear sequences don't have enough closure they don't have this it's essentially the group property of polynomial sequences again I mean a to the n times b to the n is not of the form c to the n in the not abelian world okay so that's what I'm going to say about that there are two lectures left and they will only tangentially rely on this and not on the details of this I think next time I'll talk about semi-radius theorem cases of that and some variants of that and then last lecture I will talk about applications to the primes but that's all for today thanks um questions at all yeah yes and you to do that you need a two variable version of the equal distribution theorem so instead of just a sequence index by n you need a sequence index by n and d and actually so the paper in which these results were proven is this paper quantitative behavior of polynomial orbits on no manifolds annals of mass 2000 and well it says 2012 but this paper was written in 2006 and recently we discovered that this was wrong the proof of that multi parameter version so it's now being corrected but it was it was wrong for six years actually um hence actually all of the in a sense the proof of the inverse conjectures was I mean it was wrong but fixable but yeah that's what's needed here