 So maybe what I will do is just, first of all, recall some of what I said in the first two lectures. So we talked about nil sequences and Gower's norms. And at the end of last time I stated the inverse theorem for the Gower's norms. So let me just state that again. Suppose that f is a function from the first n integers to c, bounded everywhere by 1. And suppose that its Gower's norm, its Keith Gower's norm, is at least delta. Then the conclusion is that f correlates with a nil sequence. Then there is a nil sequence, chi of n, which is an automatic function phi of a polynomial sequence, such that the average of f of n times chi of n is at least delta primed. And here, so where delta primed is bigger than 0. And various complexity quantities connected with chi are bounded. So there is the basis B for the Lie algebra of G, such that this notion of complexity that I defined before is bounded and the smoothness of phi is bounded. And actually there was one thing I forgot last time, also the dimension of G is bounded. So the way to think about this is to not worry too much about the last three lines. So all of this, I mean it's important to do things rigorously, but you should just think bounded complexity. So I said last time that this is a very difficult theorem to prove, but that the converse is somewhat easier. So today I'm going to talk a little bit about the converse, which explains somehow at least a bit why nil sequences are important. So today we'll talk about the converse. And the crucial fact here is, so the crucial observation is that nil sequences behave a little bit like polynomials. So under favorable circumstances, and I'll explain what that means, the derivative of a nil sequence is a nil sequence of degree one lower, of step one lower. So the derivative delta sub h chi of n, which is chi of n, chi of n plus h bar is a nil sequence of step of class s minus one, if chi is a nil sequence of class s. So in this respect nil sequences, they behave like polynomial phases. So for example, so this is a familiar fact for certain special cases, namely the polynomial phases. So if I take chi of n to be e to the 2 pi i alpha n squared, for example, such as, I mentioned before that this is a nil sequence. So this is a nil sequence of class two, in which the underlying group G is just the reals. The lattice gamma is the integers. The automorphic function, well the polynomial p of n is alpha n squared and then phi of x is e of x, which let me remind you is e to the 2 pi i times x. So it is a nil sequence of class two. What's its derivative in the sense that I just defined there? Delta sub h chi of n is e of alpha n squared minus n plus h squared. And the point is that for fixed h, let me just make it clear that this is for fixed h. So the derivative is e to the minus 2 alpha n h times a constant, c h. And that's a linear nil sequence. So this is a class one nil sequence for fixed h. In fact, this is just an additive character. In fact, an additive character. So what I claim is true is that there's a complete generalization of this observation to arbitrary nil sequences. Maybe I'll just mention another example. I'll leave this as a not very easy exercise. So exercise, and this is definitely a bit tricky. Reminds yourself that so if chi of n is e of alpha n times brackets beta n, then the same is true. So delta sub h chi of n is a class one nil sequence. I mentioned this particular nil sequence before. It comes from the Heisenberg group. So recall that chi comes from Heisenberg. It was actually not quite a genuine nil sequence because the phi was discontinuous. So phi was discontinuous, but this doesn't actually matter here. So this isn't important. So that's already quite a tricky exercise. Any questions so far? So to explain why this is so, I need to say a little bit more about these polynomial sequences. So let me state formally a proposition. So I'm going to just describe various equivalent forms of what it means to be a polynomial sequence. So let g be a g-bullet, be a pre-filtration, and suppose that p from z to g is just some map, then the following are equivalent. So first of all is that p is a polynomial sequence in the sense that I defined before. So that is to say that p is a product, p of n is a product of g i to the p i n, where g i is in g of the degree of p i. So that's the definition we saw before. Next is that the derivatives of p behave a bit like a polynomial. So delta sub h1 delta sub hk p of n takes values in g sub k for all integers k and all choices of h1 up to hk and n in the integers. And I want to mention, so I don't think I'm going to have time to prove this. I mentioned this before and I said that this is quite a non-trivial fact. So the object satisfying one are obviously a group, but the same is far less clear for object satisfying two. So I'm not going to, I don't think I'm going to have time to give the proof in full, but I do want to mention, I mean there are two other equivalent properties morning which are useful inside the proof. But it relies, well it relies really on one other, on a further equivalent property, which is three, that for every n and h the 2 to the k tuple p of n plus omega h has omega ranges over 0, 1 to the k. So that takes values in something called the Host-Krah cube group. Takes values in a certain subgroup of g to the 0, 1 to the k called the Host-Krah cube group, hk sub k of g bullet. And I'll at least tell you what it is. So this is the group generated by all elements of the form. I'll write g to the square omega. So this is the element that's defined by g to the omega sub epsilon is equal to g if epsilon is contained in omega, so dominated by omega, and is equal to the identity otherwise. And g lies in a certain element of this filtration. So it's quite difficult to explain why this is important. Maybe if I draw a picture, this will make a certain amount more sense. Good job, I have thin wrists. So you can, the Host-Krah cube group of dimension three, you can think of it as eight tuples in the group. So one thing I could do is I could have g, g, g, g, g, g and g. So this is the element that I'd call g to the 1, 1, 1. So it's just everything on the, every vertex of the cube is g. And here, so for any g in g0, so that's just any g whatsoever. So all of those elements are in the Host-Krah cube group. But I can also have things like this. So I could have g primed, g primed and then everything else is the identity. So this would be g primed to the 1, 0, 0. And I'm allowed this for any g primed in g2. So this is a certain group and somehow the combinatorics of polynomial sequences and nilpotent groups is intimately related to the combinatorics of these cubes and how they multiply and commutate together. As I say, it's, yeah, it would take me a whole lecture to develop this. But I do actually have full printed notes on this by the way. If anyone wants them, just send me an email. And I'll give you the, actually I've got notes for all three of the lectures that I've given this week. Anyway, so there's a proposition about different equivalences of what it means to be a polynomial sequence. Okay, so now I'm going to explain a little bit this. So the statement that if I have a nil sequence and conditions are favorable, then its derivative is a nil sequence of class 1 less. So favorable turns out to mean that the automorphic function phi has just a certain additional invariance. So favorable means that phi is additionally invariant, which we call having a vertical frequency. So what this means is, so if chi from the last element, the last non-trivial element of the filtration is a homomorphism onto the multiplicative complexes, character, continuous homomorphism. And if it annihilates, if it annihilates the lattice, so annihilating the lattice intersect g sub s, then we say it's a vertical character and phi has chi as a vertical frequency. Precisely if phi of x gs is equal to chi of gs phi of x for all x in the group and for all central elements, well for all elements gs in big gs. So maybe I should remark explicitly, so nb g sub s is in the center of g and that's because I've got a filtration and so the commutator of g with gs is supposed to be contained within gs plus 1, but that is by assumption trivial because I've got a filtration of class s. So think of this as an automorphic function phi that has some additional variance in the center and I gave two examples at the top and those both do have a vertical frequency. So both of the examples do have a vertical frequency. So furthermore you can kind of do a Fourier expansion just in the direction of this central group gs to decompose an arbitrary automorphic function phi into automorphic functions having a vertical frequency. So by a kind of vertical Fourier expansion we can write phi is as a sum over psi of phi psi where phi psi has psi as a vertical frequency and in fact I may come back to this and mention it again later. So for this to be useful we need to know some additional things about this. So we'll need to know how the complexity of this phi psi depends on that of phi and we'd also need to have some rapid decay, but these are the same all the ideas are the same as when you decompose a smooth function into its Fourier modes and the proofs are also the same. But for now let's just think that this is just an additional invariance that it's not very difficult to attain in any given case. Any questions at all before I carry on? Alright so let me try and explain then why the derivative of a nil sequence with a vertical character is a nil sequence of lower class. So let's look at the derivative. There is not much difficulty in interpreting the derivative delta sub h chi of n which is chi of n chi of n plus h bar as a nil sequence. So in fact it's equal to phi times phi bar of p of n p of n plus h. So you can think of it as a nil sequence on g cross g. So on g cross g with filtration in which g cross g sub i is g i cross g i and well I suppose it's not completely trivial. So indeed for fixed h pn plus h is also a polynomial sequence adapted to the same filtration on g bullet because it is well it's the derivative delta sub h p of n inverse times p of n and p of n is a polynomial sequence adapted to g and so is the derivative by part 2 and hence the product is also a polynomial sequence. So here I've really used the proposition. So at least the derivative is a nil sequence but unfortunately I've not proved anything like this observation here because the nil sequence is more complicated. The situation is more complicated than the one I started with. I've simply taken a direct product of two groups of class s. So unfortunately g cross g is a bigger group so also with class so that's unfortunate. But you can make some small modifications in such a way that this actually lives inside a smaller group. So by making some small modifications we can ensure we can pass to a smaller group. So I claim in particular that in fact the sequence p of n p of n plus h I can multiply that through by a constant which is pretty harmless. So p of 0 p of h inverse just check I've got that correct. Oh no sorry before I do that let me not do it yeah. So I'm actually going to first pass to a smaller filtration before I pass to a smaller group. So in fact this already the pair p of n p of n plus h already lies inside a smaller filtration. So I claim so let's write so I'll write p of n p of n plus h is a of n inverse times b of n where so a of n is going to be just the identity and then the derivative of p at n and b of n will be just p of n p of n. So that's straightforward. And both of those sequences lie in much smaller groups. So the the kth derivative that delta h1 delta hkb of n takes values in let's write a j and i takes values in gi diagonal which is just a set of all pairs gg for g in gi. So that's obvious b is in the diagonal of g so its derivatives are all in the diagonal and the derivative of a well the derivatives of a the i derivative of a is essentially an i plus first derivative of p. So that takes values in in g in the identity cross g sub i plus 1 being an i plus first derivative of p. So by the product property so therefore the product or that is to say the sequence p of n p of n plus h has its i th derivative lying in well the group that you get by taking the group generated by those two groups which I will call gi cross sub gi plus 1 gi and that's just precisely the set of all pairs gg primed with gg primed in gi and and these two things being equivalent mod gi plus 1. So it turns out that that sequence of groups that I've defined gives you a filtration on the product gi cross gi and it's a finer filtration than merely the product filtration. So this gives well it's not a filtration yet it's actually a pre filtration it gives a pre filtration on gi cross gi so that's an exercise it's really just a simple group theory exercise you just need to check that the commutator of elements there does what you think it does has the appropriate filtration property. So to make a filtration I need to to get a filtration there's a little trick we just multiply through by a constant so we consider instead the sequence so I'll instead consider the sequence p of n p of n plus h this is what I started writing before times p of 0 p of h inverse so because it's just the previous sequence times a constant the derivatives are essentially the same so this also has derivatives so it also has i th derivatives taking values in gi cross gi plus 1 gi but now the sequence itself takes values in the first so the sequence itself will take values in g1 cross sub g2 g1 which is just the same thing as g cross sub g2 g that's not difficult to prove let me just write it down this sort of half an exercise half a half an actual proof site it wasn't as quite as straightforward as I thought so to prove this so what one does is observe that the following pn plus h times pn inverse is equal to sorry ph inverse is equal to the derivative n p of h inverse times the derivative n p of 0 times p of n p of 0 inverse so that's just an identity but then this so delta n ph inverse delta n p of 0 differs from up to a commutator I can swap the order of these so it differs from delta n p of 0 delta n p of h inverse by a commutator which lies in certainly lies in g2 and this is in fact the derivative delta h delta n p of 0 it's a second derivative which also lies in g2 so I bet you there's a simpler I mean that's not very hard but there's probably a simpler way to see that so I'll have to think about that anyway so now we have a polynomial sequence which lies inside a filtration on this group here maybe I should remind you what I'm trying to do here and so remember what I'm trying to do is understand why the derivative of a nil sequence is a nil sequence of lower class and what I've done so far is interpret the derivative of a nil sequence as a nil sequence on this group here now you can probably guess that I'm not finished because I haven't actually used the fact that phi has a vertical character so if you remember that I talked about this notion of vertical character I've not made any use of that yet actually so that group unfortunately still has class s so unfortunately g cross g2 g still has class s and in fact the the s the s the that's extremely hard to say I know that th is supposed to be hard to say for French people so I don't know if s must be very hard to say yes so it still has class s the s term in the filtration is well it's gs cross gs plus one gs and that's just the diagonal of gs but it turns out that the vertical frequency of phi is precisely what lets us mod out by that diagonal and hence pass to a class s minus one group so however phi has a vertical frequency psi and so well phi cross phi bar of gs gs x y if you just compute away is going to be phi of g of s x phi of g of s y bar and that will be psi of gs phi of x psi bar of gs phi of y bar and of course that's just phi times phi bar of x y so phi times phi bar descends to a function phi times phi bar descends to a function on on a group that I will call the square of g so g square which is defined to be g cross g2 g mod by the center by the diagonal of the center and this is which has class s minus one so here's an exercise that you if you if you want to check your understanding of these concepts when g is the Heisenberg group what is the square so exercise if g is the Heisenberg 1 1 1 r r then the square is isomorphic to r cubed the Heisenberg group of course is not isomorphic to r cubed okay well there's more that needs to be said in this discussion but I don't think I'm going to say it I'm going to just state a final theorem what more needs to be said well one problem is that so actually I mean there's a slight inaccuracy here when I multiplied on the right by this constant here p of naught p of h inverse I have to correspondingly modify the automorphic function phi cross phi bar by turning it in fact to me just briefly sketch this I'd in fact need to instead consider phi cross phi bar of x y times p naught pH inverse times so maybe I should have said that but it makes the notation worse but the problem is that no one said that these p naught and pH are in any way nicely bounded and the there's no right invariance of the Sobolev norms that I use last time so what this means is if p of naught and p of h are very very big then actually the smoothness norms of this phi cross phi bar could blow up hugely and that's problematic you want them to be bounded so you need to do some extra tricks essentially conjugating by elements of the lattice to make these elements smaller again but then things start becoming even harder to understand so I'm just going to say that there are some further things that you need to do and then state of final theorem so theorem let chi of n equals phi of p of n be a nil sequence of class s with a vertical frequency psi then it's derivative so the derivative delta h chi of n which is chi of n chi of n plus h bar may be interpreted as that can be interpreted as a nil sequence phi sub h box p sub h box of n on g box which is defined to be this group so that's a group of filtered group of class s minus one with the filtration I mentioned before and furthermore with more care with some additional arguments one may ensure one may bound the smoothness norms of phi h box in terms of those of phi nicely in terms of those of phi so of course there's a more precise statement that needs to be made there but that's that's the basic idea does anyone want any further elaboration of any of that I'm guessing no there's a lot more that can be said but if you if you really want to see it I have I have all these notes on it giving the precise precise bounds for all of those but what I want to explain to you now then is why this fact so this is saying that nil sequences behave in many ways like polynomials why this implies that they are obstructions to what we call Gower's uniformity so in other words if you correlate with a nil sequence then you have a large Gower's norm so that's the converse to the inverse theorem for the Gower's norms so in fact well it's useful to introduce something called the Gower's dual norm so we introduce the dual norms and the dual norms so the dual norm of a function psi uk and star is by definition the supremum over all f with just with bounded u2 norm bounded l2 norm no sorry supremum over all f with bounded Gower's norm uk n equals one of the inner product of f with psi so this is actually this is just the standard construction of the dual of a norm this is this construction works for any norm in fact and it gives another norm so the statement I the converse to the inverse theorem for the Gower's norm is the statement that nil sequences have a small dual norm so that's the theorem so the theorem is we have if chi is a nil sequence that the dual norm of class s the dual s plus one norm is bounded by a suitable smoothness norm of phi I forget I worked out at some point how many you need the number of derivatives you need depends on it's not terribly important but the number of derivatives you need does depend on the dimension of the group G and also on s I think 2 to the s times the dimension of G is going to be enough so this statement is true well you also need to take account of the complexity of G mod gamma so this will depend on on s and the complexity of the nil sequence but the basic idea is that a fixed nil sequence has a bounded Gower's dual norm and therefore so that is nil sequences obstruct Gower's uniformity or to spell it out even more correlation with the nil sequence implies large Gower's norm and the way the numerology works is that if the nil sequence has class s then the Gower's norm is the s plus one norm so sometimes it's difficult to remember the numerology here so just to remind you when I was interested in four term progression four term arithmetic progressions I needed to talk about the Gower's three norm u3 norm and to understand the Gower's u3 norm I care about nil sequences of class 2 which are just the step up from a billion and to work with three term arithmetic progressions it's the Gower's second norm that's important and that is related to nil sequences of class 1 which is just a billion foyer analysis so for the proof well I'm first going to observe a different formula for the Gower's norm and actually this is quite a suggestive formula I think this is what tells you that the Gower's norm has something to do with derivatives so first of all observe that the Gower's norm Gower's k norm is equal to to the power 2 to the k is equal to the average over h1 hk and x of the derivative where delta sub h f of x is f of x f of x plus h bar so that's nothing but a restatement of the definition but it makes it sort of a bit clear that derivatives have some role to play and hence you can use this to work inductively with the Gower's norms so the Gower's u k plus one norm to the power 2 to the k plus one is the same thing as the average over h of the Gower's u k norm of the derivative to the power 2 k so that's a useful fact so it turns out that this fact implies a corresponding fact about the dual norms and it's that fact that we will use to prove this theorem and that's really very natural because we understand the derivatives I mean then we can prove the theorem by induction on s so on the dual side I claim that so my claim is that the Gower's dual norm u k plus one of n star is bounded by the maximum the supremum over h of the dual norm of the derivative to the power one half okay so what's how do we prove that well let f be arbitrary then the inner product of f with psi is the same thing as you get by taking the inner product the inner product of f with psi squared it's the same as taking the average now I should say that there are there are a couple of technical issues here that I'm overlooking what I write is correct if I replace this one up to n by the group z and then all these averages are over the whole group if I'm not taking averages over the whole group I have to be a tiny bit careful but let me ignore that so this this is just an equality just expanded out and then this is going to be bounded by the average over h of the dual norm of psi times the the Gower's norm of f just by definition of the dual norm and this will be bounded by the supremum over h of the dual norm times the average over h of this Gower's norm of the derivatives of f and then I can take Holder's inequality let's me raise that to the power 2 to the k so that's at most the supremum over h of the dual norm of the derivative times the average of this to the 2 to the k to the 1 over 2 to the k and that's Holder's inequality and that is basically the end so this equals sup over h times the Gower's UK plus one norm of f so that's a proof of the statement that I wrote at the top let me see if I can just get that back so let's call that something so this is a proof of star so I've basically already told you how that proves the theorem so we prove the theorem by induction I'll just sketch the proof but before you go and apply the induction you first need to expand phi into vertical characters so expand phi as a sum of vertical of automorphic functions with a vertical character as psi as a vertical character and then so just by the norm property the dual norm of this is at most the sum over psi of the dual norms of those then imply star sorry I don't mean Gower's norms of phi I mean the Gower's norms of phi so here chi sub psi of n is equal to phi sub psi of p of n so now apply star but by induction and induction on s so by induction on the class s and the fact that chi sub psi has class s minus one we're done so that's that's what there is to it vertical Fourier expansion into automorphic functions with this additional vertical invariance by the center then the observation that the derivative of a nil sequence of class s is of class s minus one and then this fact so here's an exercise so if you didn't follow any of what I said about groups and filtrations and so on without any of that you can convince yourself that the dual norm of e to the 2 pi i alpha n squared in the u3 dual norm is bounded and the thing is that that's already a vertical frequency so you don't need to do a Fourier expansion and so all you need to do there is apply star and then induction okay so it's almost 12 o'clock what I'm going to do is just tell you I don't know if anyone's still going to be here next week but if anyone anyone is maybe I can help you choose which day to come so I'll try and make us in as much as I can three lectures that are independent of one another and relatively independent of what I said so far so next week so lecture four I'm going to talk about distributional properties properties of nil sequences so this has something somewhere halfway between kind of traditional uniform distribution problems in analytic number theory and questions in a gothic theory so the basic question I'll be asking is is p of n for n less than or equal to n close to equidistributed in gamma mod g so this is really a question about equal distribution on a homogeneous space so it's kind of uniform distribution theory meets a gothic theory and then in the fifth lecture well I'm going to then talk about the two applications that I mentioned at the start so in the fifth lecture I'll talk a little bit about samarades theorem and related issues and then in the sixth lecture I will talk about some aspects of linear equations in primes now these will all use so this to talk about distributional properties one does need to use facts about polynomial sequences again but I'll just I will recall what I needed and then to talk about either of these two issues one needs to use the Gower's norms and also things from here and from what I've said this week other than that I think I will stop there but let me just remind you that I have here well they're a bit uncorrected at the moment but as long as it's a private thing if you just email me I will send you my uncorrected notes so this means that if you want to read over the weekend everything I've said with all the details from the first three lectures then you will be able to do so so send me an email and my email is ben.greenatmaths.ox.ac.uk very important to note that it's Ben and not Benjamin because there is somebody called Benjamin Green at Oxford and that's a different person to me he does not have these notes okay so that's it for this week here there should be a square there that's an interesting question it's quite unclear clear to me whether that's true or not I'm I would think not and in any case I don't see why it would be easier in the free and open group it's not even clear to me the sense in which an arbitrary filtration can be seen as a quotient coming from the free and open group because these filtrations can be a lot fatter than just the lower central series there can be very sort of weak filtration yes now well all of those things are true and in fact in that sense you don't need the notion of polynomial sequence or arbitrary filtration at all because a polynomial nil sequence can in fact be written as just a linear nil sequence phi of g to the end adapted to and this is all automatically adapted to the lower central series filtration on some group so why did I not do that well the point is that when it comes to think about distributional properties you can't guarantee that this is nicely distributed and if you it's very important to have a theory in which you can get a nicely distributed nil sequence and to allow that you have to work in this bigger category of polynomial sequences so it it's not enough to interpret your nil sequence as living somewhere else you need to you need all of the information about the distribution to move to that new setting as well and it generally wouldn't 230 on Monday yes