4/6 Nilsequences
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Transkript: Englisch(automatisch erzeugt)
00:32
OK, so today I'm going to be talking about distribution, or equidistribution, of nil
00:45
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.
01:04
So generalizations and quantitative variance of the following very well-known fact, which
01:28
is that if theta is a real number, then the sequence of values theta, 2 theta, 3 theta, et cetera, is equidistributed mod 1 if and only if theta is irrational.
02:04
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.
02:24
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. So recall, we have a simply connected nilpotent Lie group G, which has on it a proper filtration,
03:07
G bullet. And that's a collection of closed Lie subgroups of G, which satisfy this commutator inclusion
03:25
property that I mentioned before.
03:42
And then I define this notion of a polynomial sequence. And 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.
04:19
So I won't go over that again.
04:21
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.
04:51
If I have a finite set of values of that polynomial, so when is P of N from N equals
05:00
one to big N close to equal distributed in the quotient gamma mod G? And what does this even mean?
05:34
So that's the question I'm going to be interested in. Well, to motivate this, maybe I'll give you a few examples.
05:43
I've already had one. So let me first of all say that the example I put up there actually generalizes.
06:01
This is perhaps a less well-known fact. It generalizes to arbitrary polynomial sequences. So if G equals R and gamma equals Z, and if P is just any polynomial.
06:23
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.
06:41
So if P is only polynomial, then P of N is the whole sequence, is equidistributed.
07:00
If and only if the polynomial has at least one irrational coefficient other than the constant term.
07:35
So this is not so trivial to prove. I may say a few words about it later.
07:40
So the proof uses Weyl's inequality, which I will mention later. So Weyl's 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
08:03
as well. So it uses Weyl's inequality and an iteration. And there are theorems in ergodic theory. So a theorem of Leon Green, no relation of minor, I should point out.
08:28
And this is about equidistribution of flows on nilpotent groups. So if G is nilpotent, so a simply connected nilpotent Lie group.
08:56
And if P is a linear sequence, so P of N is A to the N for some A.
09:07
So that is, qualifies as a polynomial sequence with the definition I've given. Then the conclusion is, so P of N, N equals one to infinity, is equidistributed.
09:31
What does equidistributed mean? Well, I'll be going over that shortly. But let's just say for the moment, with respect to the Ha measure, with respect to Ha measure on gamma mod G.
09:53
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.
10:01
So if and only if, the abelian projection of P fails to be equidistributed.
10:28
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.
10:49
So this will be isomorphic to just a Euclidean torus. And when you project P, it will be isomorphic to a certain rotation.
11:03
So the, projects to a map pi composed with P. So this induces a map, so on a torus.
11:32
So let me give an example with the Heisenberg group. Yes. I see the variety of the number of negations in the sentence.
11:42
So if and only, so it, it is equidistributed, if and only if the abelian projection is equidistributed. Thank you. I think it's best if I give an example.
12:02
So example, if G is the Heisenberg with the usual lattice,
12:22
and if A is just some element, so if A is one, one, one, alpha, beta, gamma, then the flow that's induced on the two-dimensional torus by that
12:42
is just rotation by alpha and beta. So the induced map on the torus, R mod Z squared,
13:02
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 nilpotent group is equidistributed if and only if this sequence is equidistributed in the torus.
13:25
Which by Kronecker's theorem is the same as saying that alpha and beta are independent over the rationals. So alpha, beta, and one are independent over the rationals. By Kronecker's theorem, this is equidistributed
13:52
if and only if one, alpha, and beta are linearly independent over Q.
14:06
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 variance of this.
14:23
And even to say what I mean, I should give a bit more discussion. So are there any questions so far?
14:58
So quantitative equidistribution.
15:05
If you want to make things quantitative, you have to quantify the notion of close. So let me introduce a parameter that delta greater than or equal to zero be a parameter.
15:23
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
15:42
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 characteristic function of the interval as a function as it is. And then equidistribution is saying that averages of our sequence along a function
16:01
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
16:20
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,
16:54
so this phi will be a function on g mod gamma, and then you compare that to the integral with respect to harm measure,
17:13
then that's at most delta times an appropriate norm of the function.
17:21
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
18:03
and normalised in such a way that the total volume of this quotient is one. So the measure of the whole space is one.
18:23
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.
18:43
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.
19:09
I mean, if you change the norm, it will change the notion that you're dealing with, but in relatively minor ways.
19:24
So for technical convenience, it's actually useful to have our norm to be one of those Sobolev norms that I mentioned last time.
19:43
So we'll take the norm. So you just take it to be a sufficiently high Sobolev norm. I think I found that the Sobolev norm with two to the s times the dimension of g derivatives was enough. But that's really a technical matter.
20:01
In fact, you could just look at first derivatives of phi. And every function with bounded first derivatives is basically Lipschitz. Well, it is Lipschitz. And it can easily be approximated by much smoother functions. So that's really just a technical point.
20:23
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?
20:46
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
21:00
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,
21:26
well, first of all, to polynomial sequences on gamma mod g
21:42
and with this quantitative notion of distribution.
22:00
Now, maybe I should say, well, I should say, so Léon Green's theorem is valid for any simply connected nilpotent group and for any linear sequence on that group. So 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.
22:23
So actually you get the polynomial equidistribution 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.
22:45
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.
23:06
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.
23:27
Let's suppose that p fails to be delta equidistributed from one up to n.
23:49
Well then there's an abelian explanation for that, and I'll write down what it is. There is a horizontal character,
24:05
so I'll introduce a few terms that I will explain later, a horizontal character eta from g to the reals of mod z, such that what we call the smoothness norm
24:28
of the abelianization of p is bounded.
24:47
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.
25:01
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.
25:28
Actually, sorry, I don't, with the way I've defined it here, I don't need the pi. This will automatically be a,
25:42
so I just have to tell you what a horizontal character is. So here, a horizontal character just means it's a homomorphism. So it will automatically annihilate the commutator subgroup
26:02
g brackets g, and it also is supposed to annihilate gamma and this smoothness norm. So thus, eta composed with p of n,
26:26
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.
26:42
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 nought plus alpha one n plus up to alpha d n to the d.
27:08
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
27:27
f in smoothness norm C infinity of n is at most m means, well, it means that those coefficients are very close to integers.
27:41
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.
28:07
So that's a lot to take in. Let me tell you at least how it specializes down to just even some abelian settings.
28:21
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. Somehow, that's enough because you can always multiply eta through by an appropriate integer to make things
28:42
that are close to rationals close to integers. So let me give an example. An example, just with g equals r and gamma equals z,
29:00
the sequence theta n squared from n equals one to n fails to be delta equal distributed if and only if
29:27
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
29:48
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
30:02
as one of these horizontal characters. Let me just write that. So eta is multiplication by q. So this is a known result.
30:20
This follows from Weyl's inequality. But this is not the usual way of phrasing Weyl's inequality. Actually, it's the way that I prefer to, when I'm lecturing on Weyl's inequality, I would 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
30:41
with polynomial phases. But it's really saying precisely this. So polynomial sequence and when it's equal distributed. So does anyone notice something
31:09
that's lacking in that statement?
31:21
Well, I, sorry? It's bounded on q. Yeah, as stated, it has no content whatsoever, right? Because every theta, 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.
31:43
So I additionally need a quantification to give this content. We also need to place a band on the complexity of this horizontal character eta
32:02
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.
32:25
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,
32:45
it asserts a band on q, mod q. Also, q should not be zero, otherwise it's again completely trivial.
33:03
So it asserts a band 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
33:21
some band on the complexity of this horizontal character. So let me just write explicitly what the content of this theorem is in one phrase. So p of n can only fail to be equidistributed
33:53
for a billion reasons.
34:02
Probably also need a power of delta on the left-hand side of delta to the k equidistant. 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.
34:25
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.
34:46
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 this is a really difficult theorem to prove.
35:21
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 Weyl's inequality and that's called the Van der Corpet lemma. So the main idea of the proof.
35:47
So following Weyl, and this is not... So this follows Weyl, which is a paper in analytic number theory. It's not the same thing that Leon Green does in his paper.
36:03
But by following Weyl, we use what's called the Van der Corpet lemma. So the Van der Corpet lemma asserts a relation between a sequence being equidistributed,
36:30
well, the original Van der Corpet lemma, so the original version of which,
36:46
asserted a link between the distribution of a sequence, the uniform distribution of a sequence un mod one
37:10
and its derivatives, which are un plus h minus un.
37:23
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.
37:41
And then you can just apply the known results on linear sequences. Here is our take on Van der Corpet inequality, which is very standard.
38:05
Suppose you have just any bounded sequence. So suppose that a n, n equals one to n, are complex numbers of norm at most one.
38:25
And suppose that their average is not close to zero. 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.
38:47
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,
39:03
big H is neither incredibly tiny nor quite as big as n. Then we have that the average value of many of the derivatives
39:26
is at least delta squared over four for many values of h. Remember how many?
39:42
The four 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
40:00
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. Let me sketch the proof.
40:23
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.
40:55
I mean, they're averaging essentially the same thing except you've got some slight edge effects.
41:03
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,
41:21
Z H expected value n less than or equal to n of a n plus H is bigger than delta over two, let's say, for an appropriate phase Z H, just to make it real.
42:29
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
42:41
of Z of H a of n plus H is also at least delta over two. And now you apply Cauchy-Schwarz. So by Cauchy, the average over n of the average over H
43:02
a n plus H squared is at least delta squared over four. And if you expand that out, so that implies that the average over H one and H two less than or equal to H of the average over n less than or equal to n of a n plus H one a n plus H two bar
43:26
is at least delta squared over four. 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
43:40
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 H one minus H two bar. And then you just have to count how many times each H one minus H two occurs, so et cetera.
44:07
But the basic idea is just one application of Cauchy-Schwarz after you've done a shift.
44:27
Questions? The other key idea is also a key idea, a very famous idea that is due to Weyl,
44:44
which is that to understand equal distribution, you should expand into Fourier series. So the other crucial idea is Weyl's idea
45:02
of proving equal distribution by expanding into Fourier series.
45:23
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, L two of gamma mod G.
45:43
So check the definition. So check the definition of equal distribution only for phi lying in a basis of L two.
46:14
Now, of course, it's not quite enough to just check things. If you're dealing with a quantitative question, you can't just check things for a basis
46:20
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 Weyl did famously in the case of R mod Z,
46:47
it's enough to check it. So it's enough to handle phi of T being E to the two pi I MT for M in the integers.
47:08
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.
47:28
So I'll go through that in the case of nil 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
47:42
of theta N squared on the circle, it's enough to test it for E to the two pi I theta N squared. And if that average happened to be large, then you could apply the van der Koppel inequality and get that many derivatives of that
48:01
have to have large average. But a derivative of E to the two 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.
48:37
So what's the appropriate variant of this Fourier expansion
48:43
in gamma mod G, G mod gamma, where G is a nilpotent group? Well, of course, that's a very deep question in general, but actually it turns out we can get away
49:01
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.
49:29
And to test equal distribution for an arbitrary phi, you decompose it into automorphic functions with a vertical frequency.
49:40
So to test equal distribution against an arbitrary phi in C infinity of gamma mod G, it's enough to handle the case
50:08
when phi has a vertical frequency psi,
50:23
which, let me remind you, means that phi of, if you twist x by an element of the last group in the filtration,
50:41
then it has a natural transformation property under psi. And the reason for that simply is that there is, you can do a Fourier expansion in the direction of this group G. So as I mentioned last time,
51:05
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,
51:21
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 psi is a non-trivial frequency,
51:53
then that means, just as with the non-trivial exponentials on the circle, the integral of this function with respect to the Haar measure,
52:10
because of the vertical oscillation, it just cancels out and you get zero. So what does failure of equal distribution mean in this case?
52:27
So if p of n from n equals one to n fails to be delta equal distributed with respect to this particular automorphic function
52:43
with a vertical frequency, then it just means that the average is at least delta.
53:04
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 Korpen lemma to. By Van der Korpen, there are many h
53:29
such that the same average,
53:45
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,
54:01
but you can see that something very similar to what happened in the proof of Weyl's inequality has happened here. So for Weyl's inequality, Van der Korpen took a quadratic phase function and by differentiating it, made it linear. And what is the derivative of a nil sequence? Well, I spent a whole hour explaining
54:20
that that is a nil sequence of class one less. So we can turn this into an induction by this device of taking derivatives.
54:54
So by the main discussion, by the discussion of, I think it was mainly in lecture three,
55:06
the object here, phi psi p of n, phi psi p of n plus h bar,
55:22
for fixed h, so is for fixed h, a nil sequence of class at most s minus one.
55:43
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.
56:10
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 SL two r,
56:22
and gamma is 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 nilpotent case, we do have this simplification.
56:42
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 rather difficult.
57:02
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.
57:32
So let me actually revert completely to the notation of last time. So let's set chi of n to be this nil sequence.
57:43
And then in the notation I had last time, delta sub h of chi is precisely the derivative.
58:02
So last time I said that this derivative is a nil sequence of class S minus one. Well, actually, let me specialize also to the case of the Heisenberg, just for illustration.
58:35
So g is the Heisenberg,
58:41
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
59:04
can be thought of as a nil sequence of class one.
59:21
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.
59:40
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.
01:00:00
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
01:00:21
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
01:00:44
But the statement I'm trying to prove is quantitative It cares about how smooth my functions fire 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
01:01:09
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
01:01:22
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
01:01:44
By elements of the lattice gamma so to get around this we can conjugate by elements of
01:02:04
gamma cross gamma So here 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
01:02:22
Gamma H and this has the effect of adjusting the automorphic function in here
01:02:42
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
01:03:00
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
01:03:38
Gamma sub H a to the H lies in the fundamental domain
01:03:49
so in particular we'll have all of its matrix entries banded by one and So when you do that you do then recover control on this automorphic function
01:04:00
Here
01:04:25
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
01:04:41
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
01:05:01
To the end so Now you can imply that inductive hypothesis this is just a rotation on a torus It's a rotation on a torus
01:05:26
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
01:05:41
so by Kronecker's theorem we have essentially the coordinates of this vector a and the conjugate of a
01:06:02
Fail to be independent over Q. So the coordinates of a gamma H a gamma H inverse
01:06:20
While reduced to lie in this torus so reduced mod G to 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
01:06:51
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
01:07:19
I think it's well, I'm not going to go into any more details the
01:07:24
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 de korpot is almost an if and only if
01:07:42
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 Kind of a weird sort of integration problem
01:08:03
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 it's the same problem with the proof of the inverse conjectures for the Gowers norms
01:08:21
I showed you last time why? Why nil sequences are obstructions to Gowers 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
01:08:44
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
01:09:06
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
01:09:24
I've said nothing about Gowers norms For the u4 norm and higher
01:09:43
From the It's needed this is really This is somehow key to the entire theory of higher order for your analysis and nil sequences. It's also needed crucially in showing that
01:10:02
Well tomorrow, I'll be Talking a bit about the primes And now you need to show that the Möbius function is orthogonal to these nil sequences And again, this theorem is a crucial ingredient there And it's somehow quite fundamental
01:10:21
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
01:10:40
notion equal distribution of polynomial sequences on gamma mod G so there are two reasons for that and
01:11:03
The First one is that essentially every sequence is equidistributed So every p of n is almost equal distributed. I'll say a little more about what that means in a moment
01:11:25
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
01:11:46
p of n is equidistributed Then we can We can count various linear configurations
01:12:09
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 2 R
01:12:37
Phi of p of n plus 3 R and it will be
01:12:47
I get a very algebraic answer. It's an integral of Phi to the 4 Over a certain subgroup called the whole Patresco group HP 4 of G
01:13:01
Modulo so this where HP 4 of G which is a subgroup of G cross G cross G cross G is a certain group Explicit group called the whole Patresco group the whole Patresco for group
01:13:37
And actually that this generalizes there's nothing special about arithmetic progressions. There's a
01:13:46
corresponding result for any linear forms Which is called the Liebman group? Um, let me tell you what the whole Patresco group is because we've already seen it in action Eg when G is so when G is R
01:14:04
But with the filtration So I'm going to put a degree a class to filtration on G
01:14:24
Then the whole Patresco for group HP 4 of G is It's a subgroup of R4 and it's the set of all topples x naught x 1 x 2 x 3 Such that x naught minus 3 x 1
01:14:43
Plus 3 x 2 plus x 3 is equal to 0 and we've seen that before Because I've already we know an example of a polynomial sequence here would just be
01:15:11
Theta n squared But the what the example I showed you before was where theta was root 2 So let's give that example and we noticed before that there's a constraint amongst four term progressions
01:15:29
in this function So I mentioned that when I was showing you that Traditional Fourier analysis is not enough to handle four term progressions
01:15:40
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 the this function equal distribute over the set of all four tuples satisfying this constraint, so
01:16:02
In a sense so n squared root 2 minus and 3 n plus d squared root 2 plus 3 n plus 2 d squared root 2 minus n Plus 3d squared root 2 equals 0 is the only constraint on four term progressions
01:16:28
for a piece in P of n So these equal distributed sequences are very nice
01:16:45
Actually, I lied to you slightly 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
01:17:03
Say what it is here in many cases, for example 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?
01:17:21
Polynomial sequence is almost equal distributed Well, this is basically a generalization of what's called the Hardy little wood method. So Hardy and little wood were concerned with well Things like polynomial phases on the circle when they were thinking about Waring's problem and they decomposed those phases
01:17:42
or the coefficients of those phases into major and minor arcs and So in other words, they basically said every every numbers either essentially rational or it's Not rational is irrational and that's essentially what's generalized by this assertion here
01:18:31
So let me explain point one Is really a generalization of the Hardy little wood idea of decomposing
01:18:58
into Major and minor arcs and in fact the heart this Hardy little wood idea can be seen as the special case
01:19:11
Where G is the circle G is R of this? So a more precise statement
01:19:23
So given an arbitrary Polynomial sequence I can write it as a product so I can write P as
01:19:50
Epsilon times P prime times gamma prime. This is a point wise product of Polynomial sequences
01:20:02
where epsilon Epsilon just varies very slowly smooth so that means something like The I mean some on the group measured on the group in some way
01:20:32
It 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
01:20:50
and that means that Basically that gamma is periodic
01:21:00
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
01:21:23
Equal-distributed is highly equal-distributed somewhere 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
01:21:47
You're going to have a hard time Saying that anything related to it is equal-distributed on G, but P primed event is highly equal-distributed somewhere on some G primed
01:22:01
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
01:22:22
You need to say what you want by highly equal-distributed and You get to choose it to an extent at the at a certain cost 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
01:22:41
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 equal-distributed in Where G primed is identically?
01:23:03
1 the identity And then the other extreme would be where P is already equal-distributed and then you could take epsilon and gamma to be trivial So just to finish
01:23:22
Let me explain how is this proven and This point is what 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
01:23:41
So if P is already equal-distributed, 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
01:24:02
So you then get to induct downwards on dimension and that will eventually stop. So the proof is by downwards induction on dimension
01:24:29
Using the equidistribution theorem and let me give you an example of what can happen when you do that so eg
01:24:47
Again with the Heisenberg suppose that you take P of n to be a to the n a linear sequence
01:25:00
Where a is 1 1 1 theta 1 0 and here theta is Of size about n to the minus 3 halves So you can compute what a to the n is a to the n is 1 theta n 1 1 n
01:25:28
And then a half theta Times n times n minus 1 Now a n a to the n n equals 1 to n is not
01:25:43
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
01:26:00
The x-coordinate never really leaves zero here 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
01:26:22
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 linearist sequence So P of n turns into a polynomial sequence upon restriction
01:26:51
To x equals zero. So in fact, it's a polynomial sequence on an abelian torus in that case
01:27:02
So what this point shows is that some 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 When a to the n times b to the n is not of the form c to the n in the non-abelian world
01:27:26
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 Semirides theorem cases of that and some variants of that and then last lecture
01:27:42
I will talk about applications to the primes But that's all for today. Thanks questions at all
01:28:10
Yeah Yes, and you to do that you need a two variable version of the equal distribution theorem. So instead of just a sequence indexed by n you need a sequence indexed by n and d and
01:28:24
actually, so the paper in which These results were proven is this paper quantitative behavior of polynomial orbits on nil manifolds annals of maths 2000 and when 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
01:28:44
So it's now being corrected But it was it was wrong for six years actually And 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