WEBVTT 00:00:00.000 --> 00:00:01.952 [SQUEAKING] 00:00:01.952 --> 00:00:03.404 [PAPER RUSTLING] 00:00:03.904 --> 00:00:04.880 [CLICKING] 00:00:18.080 --> 00:00:20.960 YUFEI ZHAO: Last time we started talking about Roth's theorem, 00:00:20.960 --> 00:00:26.420 and we showed a Fourier analytic proof of Roth's theorem 00:00:26.420 --> 00:00:28.760 in the finite field model. 00:00:28.760 --> 00:00:32.770 So Roth's theorem in F3 to the N. 00:00:32.770 --> 00:00:37.670 And I want to today show you how to modify that proof 00:00:37.670 --> 00:00:39.650 to work in integers. 00:00:39.650 --> 00:00:46.130 And this will be basically Roth's original proof 00:00:46.130 --> 00:00:46.820 of his theorem. 00:00:56.440 --> 00:00:57.410 OK. 00:00:57.410 --> 00:01:00.860 So what we'll prove today is the statement 00:01:00.860 --> 00:01:10.580 that the size of the largest 3AP-free subset of 1 through N 00:01:10.580 --> 00:01:21.080 is, at most, N divided by log log N. OK, 00:01:21.080 --> 00:01:22.880 so we'll prove a bound of this form. 00:01:26.450 --> 00:01:29.600 The strategy of this proof will be very similar to the one 00:01:29.600 --> 00:01:30.920 that we had from last time. 00:01:30.920 --> 00:01:32.837 So let me review for you what is the strategy. 00:01:44.680 --> 00:01:49.530 So from last time, the proof had three main steps. 00:01:49.530 --> 00:01:52.920 In the first step, we observed that if you 00:01:52.920 --> 00:02:02.170 are in the 3AP-free set then there exists a large Fourier 00:02:02.170 --> 00:02:03.274 coefficient. 00:02:11.810 --> 00:02:13.730 From this Fourier coefficient, we 00:02:13.730 --> 00:02:22.370 were able to extract a large subspace where 00:02:22.370 --> 00:02:25.220 there is a density increment. 00:02:25.220 --> 00:02:28.910 I want to modify that strategy so that we 00:02:28.910 --> 00:02:31.580 can work in the integers. 00:02:31.580 --> 00:02:34.190 Unlike an F2 to the N, where things 00:02:34.190 --> 00:02:37.610 were fairly nice and clean, because you have subspaces, 00:02:37.610 --> 00:02:39.530 you can take a Fourier coefficient, 00:02:39.530 --> 00:02:41.900 pass it down to a subspace. 00:02:41.900 --> 00:02:43.760 There is no subspaces, right? 00:02:43.760 --> 00:02:46.980 There are no subspaces in the integers. 00:02:46.980 --> 00:02:49.020 So we have to do something slightly different, 00:02:49.020 --> 00:02:51.120 but in the same spirit. 00:02:51.120 --> 00:02:53.577 So you'll find a large Fourier coefficient. 00:03:00.400 --> 00:03:10.370 And we will find that there is density increment when 00:03:10.370 --> 00:03:14.410 you restrict not to subspaces, but what 00:03:14.410 --> 00:03:17.335 could play the role of subspaces when it comes to the integers? 00:03:19.900 --> 00:03:23.050 So I want something which looks like a smaller 00:03:23.050 --> 00:03:26.260 version of the original space. 00:03:26.260 --> 00:03:29.280 So instead of it being integers, if we 00:03:29.280 --> 00:03:33.780 restrict to a subprogression, so to a smaller arithmetic 00:03:33.780 --> 00:03:34.970 progression. 00:03:34.970 --> 00:03:38.040 I will show that you can restrict to a subprogression 00:03:38.040 --> 00:03:40.697 where you can obtain density increment. 00:03:45.840 --> 00:03:50.070 So we'll restrict integers to something smaller. 00:03:50.070 --> 00:04:04.680 And then, same as last time, we can iterate this increment 00:04:04.680 --> 00:04:08.490 to obtain the conclusion that you 00:04:08.490 --> 00:04:14.545 have an upper bound on the size of this 3AP-free set. 00:04:14.545 --> 00:04:15.670 OK, so that's the strategy. 00:04:15.670 --> 00:04:18.310 So you see the same strategy as the one we did last time, 00:04:18.310 --> 00:04:20.829 and many of the ingredients will have parallels, 00:04:20.829 --> 00:04:23.830 but the execution will be slightly different, 00:04:23.830 --> 00:04:26.750 especially in the second step where, 00:04:26.750 --> 00:04:29.440 because we no longer have sub spaces, which 00:04:29.440 --> 00:04:30.880 are nice and clean, so that's why 00:04:30.880 --> 00:04:33.070 we started with a finite fuel model, just 00:04:33.070 --> 00:04:35.530 to show how things work in a slightly easier setting. 00:04:35.530 --> 00:04:39.670 And today, we'll see how to do a same kind of strategy 00:04:39.670 --> 00:04:42.890 here, where there is going to be a bit more work. 00:04:42.890 --> 00:04:45.290 Not too much more, but a bit more work. 00:04:45.290 --> 00:04:47.791 OK, so before we start, any questions? 00:04:53.200 --> 00:04:54.920 All right. 00:04:54.920 --> 00:04:57.290 So last time I used the proof Roth's theorem 00:04:57.290 --> 00:04:59.245 as an excuse to introduce Fourier analysis. 00:04:59.245 --> 00:05:01.620 And we're going to see basically the same kind of Fourier 00:05:01.620 --> 00:05:05.420 analysis, but it's going to take on a slightly different form, 00:05:05.420 --> 00:05:07.940 because we're not working at F3 to the N. We're 00:05:07.940 --> 00:05:10.490 working inside the integers. 00:05:10.490 --> 00:05:13.870 And there's a general theory of Fourier analysis on the group, 00:05:13.870 --> 00:05:16.472 on a billion groups. 00:05:16.472 --> 00:05:18.680 I don't want to go into that theory, because that's-- 00:05:18.680 --> 00:05:20.442 I want to focus on the specific case, 00:05:20.442 --> 00:05:22.400 but the point is that given the billion groups, 00:05:22.400 --> 00:05:27.220 you always have a dual group of characters. 00:05:27.220 --> 00:05:29.800 And they play the role of Fourier transform. 00:05:29.800 --> 00:05:33.790 Specifically in the case of Z, we 00:05:33.790 --> 00:05:35.890 have the following Fourier transform. 00:05:41.390 --> 00:05:51.110 So the dual group of Z turns out to be the torus. 00:05:51.110 --> 00:05:54.540 So real numbers mod one. 00:05:54.540 --> 00:06:03.450 And the Fourier transform is defined 00:06:03.450 --> 00:06:14.910 as follows, starting with a function on the integers, OK. 00:06:14.910 --> 00:06:17.010 If you'd like, let's say it's finitely supported, 00:06:17.010 --> 00:06:18.730 just to make our lives a bit easier. 00:06:18.730 --> 00:06:20.680 Don't have to deal with technicalities. 00:06:20.680 --> 00:06:23.820 But in general, the following formula holds. 00:06:23.820 --> 00:06:32.280 We have this Fourier transform defined 00:06:32.280 --> 00:06:39.620 by setting f hat of theta to be the following sum. 00:06:44.922 --> 00:06:50.030 OK, where this e is actually somewhat standard notation, 00:06:50.030 --> 00:06:52.480 additive combinatorics. 00:06:52.480 --> 00:06:56.400 It's e to the 2 pi i t, all right? 00:06:56.400 --> 00:07:02.350 So it goes a fraction, t, around the complex unit circle. 00:07:02.350 --> 00:07:02.850 OK. 00:07:02.850 --> 00:07:05.507 So that's the Fourier transform on the integers. 00:07:05.507 --> 00:07:08.090 OK, so you might have seen this before under a different name. 00:07:08.090 --> 00:07:09.770 This is usually called Fourier series. 00:07:13.790 --> 00:07:14.290 All right. 00:07:14.290 --> 00:07:16.332 You know, the notation may be slightly different. 00:07:16.332 --> 00:07:19.050 OK, so that's what we'll see today. 00:07:19.050 --> 00:07:22.760 And this Fourier transform plays the same role as the Fourier 00:07:22.760 --> 00:07:29.390 transform from last time, which was on the group F3 to the N. 00:07:29.390 --> 00:07:33.062 And just us in-- 00:07:33.062 --> 00:07:37.020 so last time, we had a number of important identities, 00:07:37.020 --> 00:07:39.900 and we'll have the same kinds of identities here. 00:07:39.900 --> 00:07:41.970 So let me remind you what they are. 00:07:41.970 --> 00:07:43.810 And the proofs are all basically the same, 00:07:43.810 --> 00:07:45.102 so I won't show you the proofs. 00:07:54.130 --> 00:08:01.540 f hat of 0 is simply the sum of f over the domain. 00:08:05.220 --> 00:08:16.470 We have this Plancherel Parseval identity, which tells us 00:08:16.470 --> 00:08:22.620 that if you look at the inner product 00:08:22.620 --> 00:08:27.610 by linear form in the physical space, 00:08:27.610 --> 00:08:33.450 it equals to the inner product in the Fourier space. 00:08:42.620 --> 00:08:43.120 OK. 00:08:43.120 --> 00:08:45.250 So in the physical space now, you sum. 00:08:45.250 --> 00:08:47.350 In the frequency space, you take integral 00:08:47.350 --> 00:08:51.960 over the torus, or the circle, in this case. 00:08:51.960 --> 00:08:54.570 It's a one-dimensional torus. 00:08:54.570 --> 00:08:58.470 There is also the Fourier inversion formula, 00:08:58.470 --> 00:09:05.350 which now says that f of x is equal to f hat of theta. 00:09:05.350 --> 00:09:09.780 E of x theta, you integrate theta from 0 to 1. 00:09:09.780 --> 00:09:13.830 Again, on the torus, on the circle. 00:09:13.830 --> 00:09:18.060 And third-- and finally, there was this identity last time 00:09:18.060 --> 00:09:20.940 that related three-term arithmetic progressions 00:09:20.940 --> 00:09:22.680 to the Fourier transform, OK? 00:09:22.680 --> 00:09:26.530 So this last one was slightly not as-- 00:09:26.530 --> 00:09:29.550 I mean, it's not as standard as the first several, which are 00:09:29.550 --> 00:09:31.560 standard Fourier identities. 00:09:31.560 --> 00:09:34.360 But this one will be useful to us. 00:09:34.360 --> 00:09:38.280 So the identity relating the Fourier transform, the 3AP, now 00:09:38.280 --> 00:09:39.643 has the following form. 00:09:39.643 --> 00:09:47.098 OK, so if we define lambda of f, g, 00:09:47.098 --> 00:09:55.790 and h to be the following sum, which 00:09:55.790 --> 00:10:02.600 sums over all 3APs in the integers, 00:10:02.600 --> 00:10:11.470 then one can write this expression 00:10:11.470 --> 00:10:15.970 in terms of the Fourier transform as follows. 00:10:26.750 --> 00:10:27.720 OK. 00:10:27.720 --> 00:10:28.220 All right. 00:10:28.220 --> 00:10:31.940 So comparing this formula to the one that we saw from last time, 00:10:31.940 --> 00:10:34.490 it's the same formula, where different domains, 00:10:34.490 --> 00:10:35.990 where you're summing or integrating, 00:10:35.990 --> 00:10:37.160 but it's the same formula. 00:10:37.160 --> 00:10:39.000 And the proof is the same. 00:10:39.000 --> 00:10:40.070 So go look at the proof. 00:10:40.070 --> 00:10:40.903 It's the same proof. 00:10:43.710 --> 00:10:44.210 OK. 00:10:44.210 --> 00:10:47.810 So these are the key Fourier things that we'll use. 00:10:47.810 --> 00:10:50.980 And then we'll try to follow on those with-- 00:10:50.980 --> 00:10:55.170 the same as the proof as last time, and see where we can get. 00:10:55.170 --> 00:10:57.180 So let me introduce one more notation. 00:10:57.180 --> 00:11:04.490 So I'll write lambda sub 3 of f to be lambda of f, f, f, 00:11:04.490 --> 00:11:05.030 three times. 00:11:07.509 --> 00:11:08.009 OK. 00:11:08.009 --> 00:11:09.980 So at this point, if you understood the lecture 00:11:09.980 --> 00:11:13.220 from last time, none of anything I've said so far 00:11:13.220 --> 00:11:14.450 should be surprising. 00:11:14.450 --> 00:11:16.280 We are working integers, so we should 00:11:16.280 --> 00:11:19.170 look at the corresponding Fourier transform in integers. 00:11:19.170 --> 00:11:21.680 And if you follow your notes, this is all the things 00:11:21.680 --> 00:11:23.660 that we're going to use. 00:11:23.660 --> 00:11:25.970 OK, so what was one of the first things 00:11:25.970 --> 00:11:29.270 we mentioned regarding the Fourier transform 00:11:29.270 --> 00:11:30.830 from last time after this point? 00:11:40.250 --> 00:11:40.750 OK. 00:11:40.750 --> 00:11:42.075 AUDIENCE: The counting lemma. 00:11:42.075 --> 00:11:42.700 YUFEI ZHAO: OK. 00:11:42.700 --> 00:11:44.260 So let's do a counting lemma. 00:11:44.260 --> 00:11:49.780 So what should the counting lemma say? 00:11:49.780 --> 00:11:52.700 Well, the spirit of the counting lemma is that if you have two 00:11:52.700 --> 00:11:56.360 functions that are close to each other-- and now, "close" 00:11:56.360 --> 00:11:58.940 means close in Fourier-- 00:11:58.940 --> 00:12:03.910 then their corresponding number of 3APs should be similar, OK? 00:12:03.910 --> 00:12:06.150 So that's what we want to say. 00:12:06.150 --> 00:12:14.740 And indeed, right, so the counting lemma for us 00:12:14.740 --> 00:12:25.510 will say that if f and g are functions on Z, and-- 00:12:28.420 --> 00:12:47.375 such that their L2 norms are both bounded by this M. OK, 00:12:47.375 --> 00:12:51.940 the sum of the squared absolute value entries are both bounded. 00:12:51.940 --> 00:13:03.710 Then the difference of their 3AP counts 00:13:03.710 --> 00:13:08.570 should not be so different from each other if f 00:13:08.570 --> 00:13:10.790 and g are close in Fourier, OK? 00:13:10.790 --> 00:13:21.990 And that means that if all the Fourier coefficients of f 00:13:21.990 --> 00:13:28.270 minus g are small, then lambda 3ff, 00:13:28.270 --> 00:13:32.290 which considers 3AP counts in f, is close to that of g. 00:13:35.685 --> 00:13:38.110 OK. 00:13:38.110 --> 00:13:42.744 Same kind of 3AP counting lemma from last time. 00:13:42.744 --> 00:13:45.706 OK, so let's prove it, OK? 00:13:51.520 --> 00:13:53.620 As with the counting lemma proofs 00:13:53.620 --> 00:13:56.350 you've seen several times already in this course, 00:13:56.350 --> 00:14:00.070 we will prove it by first writing this difference 00:14:00.070 --> 00:14:01.450 as a telescoping sum. 00:14:07.420 --> 00:14:12.920 The first term being f minus g f f, 00:14:12.920 --> 00:14:25.045 and then g of f minus g f and lambda g, g, f, f minus g. 00:14:25.045 --> 00:14:27.790 OK, and we will like to show that each of these terms 00:14:27.790 --> 00:14:34.273 is small if f minus g has small Fourier coefficients. 00:14:39.690 --> 00:14:40.190 OK. 00:14:40.190 --> 00:14:41.900 So let's bound the first term. 00:14:50.077 --> 00:14:55.340 OK, so let me bound this first term using the 3AP identity, 00:14:55.340 --> 00:14:57.950 relating 3AP to Fourier coefficients, 00:14:57.950 --> 00:15:05.150 we can write this lambda as the following integral 00:15:05.150 --> 00:15:07.118 over Fourier coefficients. 00:15:24.070 --> 00:15:25.730 And now, let me-- 00:15:25.730 --> 00:15:27.550 OK, so what was the trick last time? 00:15:27.550 --> 00:15:32.600 So we said let's pull out one of these guys 00:15:32.600 --> 00:15:37.245 and then use triangle inequality on the remaining factors. 00:15:37.245 --> 00:15:38.150 OK, so we'll do that. 00:15:56.410 --> 00:15:58.852 So far, so good. 00:15:58.852 --> 00:16:02.295 And now you see this integral. 00:16:02.295 --> 00:16:03.773 Apply Cauchy-Schwartz. 00:16:18.090 --> 00:16:20.890 OK, so apply Cauchy-Schwartz to the first factor, 00:16:20.890 --> 00:16:25.533 you got this l2 sum, this l2 integral. 00:16:25.533 --> 00:16:26.950 And then you apply Cauchy-Schwartz 00:16:26.950 --> 00:16:28.295 to the second factor. 00:16:34.633 --> 00:16:35.550 You get that integral. 00:16:38.826 --> 00:16:40.920 OK, now what do we do? 00:16:44.912 --> 00:16:46.409 Yep? 00:16:46.409 --> 00:16:49.980 AUDIENCE: [INAUDIBLE] 00:16:49.980 --> 00:16:50.980 YUFEI ZHAO: OK, so yeah. 00:16:50.980 --> 00:16:54.700 So you see an l2 of Fourier, the first automatic reaction 00:16:54.700 --> 00:16:56.890 should be to use a Plancherel or Parseval, OK? 00:16:56.890 --> 00:17:04.589 So apply Plancherel identity to each of these factors. 00:17:04.589 --> 00:17:13.520 We find that each of those factors 00:17:13.520 --> 00:17:20.040 is equal to this l2 sum in the physical space. 00:17:23.750 --> 00:17:25.609 OK, so this square root, the same thing. 00:17:25.609 --> 00:17:27.980 Square root again. 00:17:27.980 --> 00:17:28.520 OK. 00:17:28.520 --> 00:17:34.750 And then we find that because there 00:17:34.750 --> 00:17:36.720 was an hypothesis-- in the hypothesis, 00:17:36.720 --> 00:17:40.430 there was a bound M on this sum of squares. 00:17:43.610 --> 00:17:46.090 You have that down there. 00:17:46.090 --> 00:17:48.080 And similarly, with the other two terms. 00:18:01.526 --> 00:18:02.050 OK. 00:18:02.050 --> 00:18:03.511 So that proves the counting lemma. 00:18:07.359 --> 00:18:08.802 Question? 00:18:08.802 --> 00:18:12.760 AUDIENCE: Last time, the term on the right-hand side 00:18:12.760 --> 00:18:16.525 was the maximum over non-zero frequency? 00:18:16.525 --> 00:18:17.730 YUFEI ZHAO: OK. 00:18:17.730 --> 00:18:18.230 OK. 00:18:18.230 --> 00:18:22.230 So the question is, last time we had a counting lemma that 00:18:22.230 --> 00:18:24.120 looked slightly different. 00:18:24.120 --> 00:18:26.730 But I claimed they're all really the same counting lemma. 00:18:26.730 --> 00:18:28.320 They're all the same proofs. 00:18:28.320 --> 00:18:30.810 If you run this proof, it won't work. 00:18:30.810 --> 00:18:33.150 If you take what we did last time, 00:18:33.150 --> 00:18:35.260 it's the same kind of proofs. 00:18:35.260 --> 00:18:36.760 So last time we had a counting lemma 00:18:36.760 --> 00:18:44.120 where we had the same f, f, f essentially. 00:18:44.120 --> 00:18:48.660 We know have-- I allow you to essentially take three 00:18:48.660 --> 00:18:49.640 different things, and-- 00:18:49.640 --> 00:18:52.980 OK, so both-- in both cases, you're 00:18:52.980 --> 00:18:54.750 running through this calculation, 00:18:54.750 --> 00:18:57.378 but they look slightly different. 00:18:57.378 --> 00:19:02.555 AUDIENCE: [INAUDIBLE] 00:19:02.555 --> 00:19:03.430 YUFEI ZHAO: So, yeah. 00:19:03.430 --> 00:19:04.435 So I agree. 00:19:04.435 --> 00:19:05.810 It doesn't look exactly the same, 00:19:05.810 --> 00:19:07.520 but if you think about what's involved in the proof, 00:19:07.520 --> 00:19:08.520 they're the same proofs. 00:19:11.232 --> 00:19:12.630 OK. 00:19:12.630 --> 00:19:16.170 Any more questions? 00:19:16.170 --> 00:19:19.060 All right. 00:19:19.060 --> 00:19:20.670 So now, we have this counting lemma, 00:19:20.670 --> 00:19:25.660 so let's start our proof of Roth's theorem in the integers. 00:19:25.660 --> 00:19:27.820 As with last time, there will be three steps, 00:19:27.820 --> 00:19:32.175 as mentioned up there, OK? 00:19:32.175 --> 00:19:37.590 In the first step, let us show that if you are 3AP-free, 00:19:37.590 --> 00:19:40.130 then we can obtain a density-- 00:19:40.130 --> 00:19:42.198 a large Fourier coefficient. 00:19:56.640 --> 00:19:59.200 Yeah, so in this course, this counting lemma, 00:19:59.200 --> 00:20:01.450 we actually solved this-- 00:20:01.450 --> 00:20:03.970 basically this kind of proof for the first time when we 00:20:03.970 --> 00:20:07.000 discussed graph counting lemma, back in the chapter 00:20:07.000 --> 00:20:09.810 on Szemerédi's regularity lemma. 00:20:09.810 --> 00:20:12.430 And sure, they all look literally-- 00:20:12.430 --> 00:20:14.740 not exactly the same, but they're all really 00:20:14.740 --> 00:20:16.400 the same kind of proofs, right? 00:20:16.400 --> 00:20:17.050 So I want-- 00:20:17.050 --> 00:20:19.540 I'm showing you the same thing in many different guises. 00:20:19.540 --> 00:20:20.873 But they're all the same proofs. 00:20:23.690 --> 00:20:28.540 So if you are a set that is 3AP-free-- 00:20:37.630 --> 00:20:41.480 and as with last time, I'm going to call 00:20:41.480 --> 00:20:46.550 alpha the density of A now inside this progression, 00:20:46.550 --> 00:20:49.820 this length N progression. 00:20:49.820 --> 00:20:54.560 And suppose N is large enough. 00:20:59.460 --> 00:21:07.580 OK, so the conclusion now is that there exists some theta 00:21:07.580 --> 00:21:19.170 such that if you look at this sum over here 00:21:19.170 --> 00:21:23.930 as a sum over both integers-- 00:21:23.930 --> 00:21:26.680 actually, let me do the sum only from 1 00:21:26.680 --> 00:21:35.040 to uppercase N. Claim that-- 00:21:35.040 --> 00:21:39.510 OK, so-- so it's saying what this title says. 00:21:39.510 --> 00:21:43.205 If you are 3AP-free and this N is large 00:21:43.205 --> 00:21:45.580 enough relative to the density, you think of this density 00:21:45.580 --> 00:21:49.990 alpha is a constant, then I can find a large Fourier 00:21:49.990 --> 00:21:50.860 coefficient. 00:21:50.860 --> 00:21:54.010 Now, there's a small difference, and this 00:21:54.010 --> 00:21:56.140 is related to what you were asking earlier, 00:21:56.140 --> 00:21:59.710 between how we set things up now versus what happened last time. 00:21:59.710 --> 00:22:05.530 So last time, we just looked for a Fourier coefficient 00:22:05.530 --> 00:22:10.080 corresponding to a non-zero r. 00:22:10.080 --> 00:22:13.340 Now, I'm not restricting non-zero, 00:22:13.340 --> 00:22:15.700 but I don't start with an indicator function. 00:22:15.700 --> 00:22:19.280 I start with the demeaned indicator function. 00:22:19.280 --> 00:22:24.340 I take out the mean so that the zeroth coefficient, 00:22:24.340 --> 00:22:28.450 so to speak, which corresponds to the mean, is already 0. 00:22:28.450 --> 00:22:32.790 So you don't get to use that for your coefficient. 00:22:32.790 --> 00:22:34.840 So if you didn't do this, if you just 00:22:34.840 --> 00:22:36.420 tried to do this last time, I mean, 00:22:36.420 --> 00:22:38.045 you can also do exactly the same setup. 00:22:38.045 --> 00:22:40.180 But if you don't demean it, then-- 00:22:40.180 --> 00:22:42.400 if you don't have this term, then this statement 00:22:42.400 --> 00:22:46.960 is trivially true, because I can take theta equal to 0, OK? 00:22:46.960 --> 00:22:48.300 But I don't want. 00:22:48.300 --> 00:22:52.900 I want an actual significant Fourier improvement. 00:22:52.900 --> 00:22:53.930 So I take-- 00:22:53.930 --> 00:22:58.540 I do this demean, and then I consider its Fourier 00:22:58.540 --> 00:22:59.583 coefficient. 00:23:02.260 --> 00:23:02.760 OK. 00:23:02.760 --> 00:23:06.220 Any questions about the statement? 00:23:06.220 --> 00:23:09.810 Yeah, so this demeaning is really important, right? 00:23:09.810 --> 00:23:12.530 So that's something that's a very common technique whenever 00:23:12.530 --> 00:23:14.140 you do these kind of analysis. 00:23:14.140 --> 00:23:16.510 So make sure you're-- 00:23:16.510 --> 00:23:19.940 so that you're-- yeah, so you're looking at functions with mean 00:23:19.940 --> 00:23:22.777 0. 00:23:22.777 --> 00:23:23.610 Let's see the proof. 00:23:28.190 --> 00:23:33.290 We have the following information 00:23:33.290 --> 00:23:39.670 about all 3AP counts in A. Because A is 3AP-free, OK, 00:23:39.670 --> 00:23:43.320 so what is the value of lambda sub 3 of the indicator of A? 00:23:46.100 --> 00:23:49.440 Lambda of 3, if you look at the expression, 00:23:49.440 --> 00:23:55.770 it basically sums over all 3APs, but A has no 3APs, 00:23:55.770 --> 00:23:58.500 except for the trivial ones. 00:23:58.500 --> 00:24:01.020 So we'll only consider the trivial 3APs, 00:24:01.020 --> 00:24:06.210 which has size exactly the size of A, which 00:24:06.210 --> 00:24:10.820 is alpha N from trivial 3APs. 00:24:14.870 --> 00:24:20.030 On the other hand, what do we know about lambda 3 00:24:20.030 --> 00:24:25.256 of this interval from 1 to N? 00:24:25.256 --> 00:24:28.410 OK, so how many 3APs are there? 00:24:28.410 --> 00:24:33.640 OK, so roughly, it's going to be about N squared over 2. 00:24:33.640 --> 00:24:38.470 And in fact, it will be at least N squared over 2, 00:24:38.470 --> 00:24:41.860 because to generate a 3AP, I just 00:24:41.860 --> 00:24:49.070 have to pick a first term and a third term, 00:24:49.070 --> 00:24:51.760 and I'm OK as long as they're the same parity. 00:24:58.195 --> 00:25:00.410 And then you have a 3AP. 00:25:00.410 --> 00:25:04.220 So the same parity cuts you down by half, 00:25:04.220 --> 00:25:11.190 so you have at least N squared over 2 3APs from 1 through N. 00:25:11.190 --> 00:25:14.930 So now, let's look at how to apply the counting 00:25:14.930 --> 00:25:16.480 lemma all to the setting. 00:25:16.480 --> 00:25:18.420 So we have the counting lemma up there, 00:25:18.420 --> 00:25:19.930 where I now want to apply it-- 00:25:22.600 --> 00:25:28.960 so apply counting to, on one hand, 00:25:28.960 --> 00:25:34.780 the indicator function of A so we get the count 3APs in A, 00:25:34.780 --> 00:25:43.770 but also compared to the normalized indicator 00:25:43.770 --> 00:25:45.252 on the interval. 00:25:45.252 --> 00:25:46.940 OK, so maybe this is a good point for me 00:25:46.940 --> 00:25:50.425 to pause and remind you that the spirit of this whole proof 00:25:50.425 --> 00:25:54.950 is understanding structure versus pseudorandomness, OK? 00:25:54.950 --> 00:25:57.680 So as was the case last time. 00:25:57.680 --> 00:26:03.620 So we want to understand, in what ways is A pseudorandom? 00:26:03.620 --> 00:26:06.020 And here, "pseudorandom," just as with last time, 00:26:06.020 --> 00:26:08.540 means having small Fourier coefficients, 00:26:08.540 --> 00:26:11.856 being Fourier uniform. 00:26:11.856 --> 00:26:16.850 If A is pseudorandom, which here, means f and g 00:26:16.850 --> 00:26:18.830 are close to each other. 00:26:18.830 --> 00:26:20.420 That's what being pseudorandom means, 00:26:20.420 --> 00:26:21.920 then the counting lemma will tell us 00:26:21.920 --> 00:26:25.790 that f and g should have similar AP counts. 00:26:25.790 --> 00:26:31.190 But A has basically no AP count, so they should not 00:26:31.190 --> 00:26:35.970 be close to each other. 00:26:35.970 --> 00:26:38.610 So that's the strategy, to show that A is not pseudorandom 00:26:38.610 --> 00:26:41.490 in this sense, and thereby extracting a large Fourier 00:26:41.490 --> 00:26:43.240 coefficient. 00:26:43.240 --> 00:26:47.490 So we apply counting to these two functions, 00:26:47.490 --> 00:26:49.350 and we obtain that. 00:26:54.860 --> 00:26:55.360 OK. 00:26:55.360 --> 00:27:03.000 So this quantity, which corresponds to lambda 3 of g, 00:27:03.000 --> 00:27:17.300 minus alpha N. So these were lambda 3 of g, lambda 3 of F. 00:27:17.300 --> 00:27:18.760 So it is upper-bounded. 00:27:18.760 --> 00:27:23.897 The difference is up rebounded by the-- 00:27:28.010 --> 00:27:29.760 using the counting lemma, we find 00:27:29.760 --> 00:27:31.500 that their difference is upper-bounded 00:27:31.500 --> 00:27:32.670 by the following quantity. 00:27:43.760 --> 00:27:46.850 Namely, you look at the difference between f and g 00:27:46.850 --> 00:27:52.082 and evaluate its maximum Fourier coefficient. 00:27:52.082 --> 00:27:53.010 OK. 00:27:53.010 --> 00:27:58.760 So if A is pseudorandom, meaning that Fourier uniform-- 00:27:58.760 --> 00:28:04.840 this l infinity norm is small, then 00:28:04.840 --> 00:28:10.010 I should expect lots and lots of 3APs in A, 00:28:10.010 --> 00:28:12.390 but because that is not the case, 00:28:12.390 --> 00:28:14.510 we should be able to conclude that there is 00:28:14.510 --> 00:28:16.745 some large Fourier coefficient. 00:28:20.000 --> 00:28:21.090 All right, so thus-- 00:28:33.131 --> 00:28:36.720 so rearranging the equation above, we have that-- 00:28:44.030 --> 00:28:45.270 so this should be a square. 00:28:51.480 --> 00:28:52.100 OK. 00:28:52.100 --> 00:28:54.260 So we have this expression here. 00:28:54.260 --> 00:28:56.690 And now we are-- 00:28:56.690 --> 00:29:03.320 OK, so let me simplify this expression slightly. 00:29:03.320 --> 00:29:07.550 And now we're using that N is sufficiently large, OK? 00:29:07.550 --> 00:29:12.290 So we're using N is sufficiently large. 00:29:12.290 --> 00:29:23.202 So this quantity is at least a tenth of alpha squared N. 00:29:23.202 --> 00:29:24.910 OK, and that's the conclusion, all right? 00:29:24.910 --> 00:29:28.937 So that's the conclusion of this step here. 00:29:28.937 --> 00:29:29.770 What does this mean? 00:29:29.770 --> 00:29:33.600 This means there exists some theta 00:29:33.600 --> 00:29:37.050 so that the Fourier coefficient at theta 00:29:37.050 --> 00:29:38.750 is at least the claimed quantity. 00:29:42.530 --> 00:29:43.834 Any questions? 00:29:51.420 --> 00:29:52.050 All right. 00:29:52.050 --> 00:29:53.960 So that finishes step 1. 00:29:53.960 --> 00:29:57.720 So now let me go on step 2. 00:29:57.720 --> 00:30:02.120 In step 2, we wish to show that if you have a large Fourier 00:30:02.120 --> 00:30:11.956 coefficient, then one can obtain a density increment. 00:30:21.200 --> 00:30:26.720 So last time, we were working in a finite field vector space. 00:30:26.720 --> 00:30:30.470 A Fourier coefficient, OK, so which is a dual vector, 00:30:30.470 --> 00:30:34.020 corresponds to some hyperplane. 00:30:34.020 --> 00:30:37.610 And having a large Fourier coefficient 00:30:37.610 --> 00:30:40.380 then implies that the density of A 00:30:40.380 --> 00:30:43.140 on the co-sets of those hyperplanes 00:30:43.140 --> 00:30:46.720 must be not all close to each other. 00:30:46.720 --> 00:30:48.570 All right, so one of the hyperplanes 00:30:48.570 --> 00:30:53.380 must have significantly higher density than the rest. 00:30:53.380 --> 00:30:55.300 OK, so we want to do something similar here, 00:30:55.300 --> 00:30:57.580 except we run into this technical difficulty 00:30:57.580 --> 00:31:01.250 where there are no subspaces anymore. 00:31:01.250 --> 00:31:05.243 So the Fourier character, namely corresponding to this theta, 00:31:05.243 --> 00:31:06.160 is just a real number. 00:31:06.160 --> 00:31:07.510 It doesn't divide up your space. 00:31:07.510 --> 00:31:09.070 It doesn't divide up your 1 through N 00:31:09.070 --> 00:31:11.050 very nicely into sub chunks. 00:31:11.050 --> 00:31:15.820 But we still want to use this theta to chop up 1 through N 00:31:15.820 --> 00:31:20.860 into smaller spaces so that we can iterate and do 00:31:20.860 --> 00:31:24.730 density increment. 00:31:24.730 --> 00:31:25.910 All right. 00:31:25.910 --> 00:31:28.920 So let's see what we can do. 00:31:28.920 --> 00:31:36.980 So given this theta, what we would like to do 00:31:36.980 --> 00:31:50.260 is to partition this 1 through N into subprogressions. 00:31:53.564 --> 00:32:07.030 OK, so chop up 1 through N into sub APs such that 00:32:07.030 --> 00:32:12.230 if you evaluate for-- so this theta is fixed. 00:32:12.230 --> 00:32:17.990 So on each sub AP, this function here 00:32:17.990 --> 00:32:28.000 is roughly constant on each of your parts. 00:32:31.150 --> 00:32:34.000 Last time, we had this Fourier character, 00:32:34.000 --> 00:32:37.120 and then we chopped it up using these three hyperplanes. 00:32:37.120 --> 00:32:40.390 And each hyperplane, the Fourier character 00:32:40.390 --> 00:32:43.078 is literally constant, OK? 00:32:43.078 --> 00:32:46.670 So you have-- and so that's what we work with. 00:32:46.670 --> 00:32:49.450 And now, you cannot get them to be exactly constant, 00:32:49.450 --> 00:32:53.020 but the next best thing we can hope for is to get this Fourier 00:32:53.020 --> 00:32:56.097 character to be roughly constant. 00:32:56.097 --> 00:32:58.430 OK, so we're going to do some positioning that allows us 00:32:58.430 --> 00:33:00.860 to achieve this characteristic. 00:33:00.860 --> 00:33:03.740 And let me give you some intuition 00:33:03.740 --> 00:33:04.910 about why this is true. 00:33:04.910 --> 00:33:07.570 And this is not exactly a surprising fact. 00:33:07.570 --> 00:33:10.040 The intuition is just that if you 00:33:10.040 --> 00:33:11.900 look at what this function behaves like-- 00:33:17.444 --> 00:33:19.950 all right, so what's going on here? 00:33:19.950 --> 00:33:27.600 You are on the unit circle, and you are jumping by theta. 00:33:31.288 --> 00:33:42.080 OK, so you just keep jumping by theta and so on. 00:33:42.080 --> 00:33:49.370 And I want to show that I can sharp up my progression 00:33:49.370 --> 00:33:57.690 into a bunch of almost periodic pieces, where in each part, 00:33:57.690 --> 00:34:01.370 I'm staying inside a small arc. 00:34:01.370 --> 00:34:06.560 So in the extreme case of this where it is very easy to see 00:34:06.560 --> 00:34:18.159 is if x is some rational number, a over b, with b fairly small, 00:34:18.159 --> 00:34:22.010 then we can-- 00:34:22.010 --> 00:34:34.190 so then, this character is actually constant on APs 00:34:34.190 --> 00:34:35.659 with common difference b. 00:34:43.063 --> 00:34:43.563 Yep? 00:34:43.563 --> 00:34:46.708 AUDIENCE: Is theta supposed to be [INAUDIBLE]?? 00:34:46.708 --> 00:34:48.875 YUFEI ZHAO: Ah, so theta, yes. so theta-- thank you. 00:34:48.875 --> 00:34:50.270 So theta 2 pi. 00:34:54.174 --> 00:34:57.370 AUDIENCE: Like, is x equal to your theta? 00:34:57.370 --> 00:34:58.078 YUFEI ZHAO: Yeah. 00:34:58.078 --> 00:34:59.054 Thank you. 00:34:59.054 --> 00:35:00.520 So theta equals-- yeah. 00:35:00.520 --> 00:35:06.180 So if theta is some rational with some small denominator-- 00:35:06.180 --> 00:35:13.420 so then you are literally jumping in periodic steps 00:35:13.420 --> 00:35:15.690 on the unit circle. 00:35:15.690 --> 00:35:21.230 So if you partition N according to the exact same periods, 00:35:21.230 --> 00:35:26.300 you have that this character is exactly constant in each 00:35:26.300 --> 00:35:28.190 of your progressions. 00:35:28.190 --> 00:35:32.210 Now, in general, the theta you get out of that proof 00:35:32.210 --> 00:35:35.400 might not have this very nice form, 00:35:35.400 --> 00:35:38.340 but we can at least approximately 00:35:38.340 --> 00:35:39.840 achieve the desired effect. 00:35:42.630 --> 00:35:43.130 OK. 00:35:46.890 --> 00:35:47.700 Any questions? 00:36:15.560 --> 00:36:16.260 OK. 00:36:16.260 --> 00:36:20.100 So to achieve approximately the desired effect, what we'll do 00:36:20.100 --> 00:36:25.370 is to find something so that b times theta 00:36:25.370 --> 00:36:29.485 is not quite an integer, but very close to an integer. 00:36:29.485 --> 00:36:31.610 OK, so this, probably many of you have seen before. 00:36:31.610 --> 00:36:37.470 It's a classic pigeonhole-type result. 00:36:37.470 --> 00:36:41.510 It's usually attributed to Dirichlet. 00:36:45.020 --> 00:36:54.990 So if you have theta, a real number, and a delta, 00:36:54.990 --> 00:37:00.300 kind of a tolerance, then there exists 00:37:00.300 --> 00:37:09.350 a positive integer d at most 1 over delta such 00:37:09.350 --> 00:37:19.030 that d times theta is very close to an integer. 00:37:19.030 --> 00:37:23.720 OK, so this norm here is distance 00:37:23.720 --> 00:37:25.850 to the closest integer. 00:37:33.215 --> 00:37:38.390 All right, so the proof is by pigeonhole principle. 00:37:38.390 --> 00:37:44.560 So if we let N be 1 over delta rounded down 00:37:44.560 --> 00:37:51.970 and consider the numbers 0, theta, 2 theta, 3 theta, and so 00:37:51.970 --> 00:37:55.630 on, to N theta-- 00:37:55.630 --> 00:38:08.040 so by pigeonhole, there exists i theta and j theta, so 00:38:08.040 --> 00:38:12.660 two different terms of the sequence such 00:38:12.660 --> 00:38:20.700 that they differ by less than-- at most 00:38:20.700 --> 00:38:22.590 delta in their fractional parts. 00:38:30.260 --> 00:38:36.920 OK, so now take d to be difference between i and j. 00:38:36.920 --> 00:38:37.820 OK, and that works. 00:38:48.060 --> 00:38:48.560 OK. 00:38:48.560 --> 00:38:52.720 So even though you don't have exactly rational, 00:38:52.720 --> 00:38:55.210 you have approximately rational. 00:38:55.210 --> 00:38:56.200 So this is a-- 00:38:56.200 --> 00:38:58.970 it's a simple rational approximation statement. 00:38:58.970 --> 00:39:00.890 And using this rational approximation, 00:39:00.890 --> 00:39:05.660 we can now try to do the intuition here, 00:39:05.660 --> 00:39:10.490 pretending that we're working with rational numbers, indeed. 00:39:16.442 --> 00:39:20.670 OK, so if we take eta between 0 and 1 00:39:20.670 --> 00:39:34.850 and theta irrational and suppose N is large enough-- 00:39:34.850 --> 00:39:37.090 OK, so here, C means there exists 00:39:37.090 --> 00:39:40.490 some sufficiently large-- 00:39:40.490 --> 00:39:45.420 some constant C such that the statement is true, OK? 00:39:45.420 --> 00:39:49.420 So suppose you think a million here. 00:39:49.420 --> 00:39:51.160 That should be fine. 00:39:51.160 --> 00:39:55.740 So then there exists-- 00:39:55.740 --> 00:40:12.330 so then one can partition 1 through N into sub-APs, 00:40:12.330 --> 00:40:14.960 which we'll call P i. 00:40:14.960 --> 00:40:24.250 And each having length between cube root of N 00:40:24.250 --> 00:40:33.110 and twice the cube root of N such that this character 00:40:33.110 --> 00:40:35.700 that we want to stay roughly constant indeed 00:40:35.700 --> 00:40:39.550 does not change very much. 00:40:39.550 --> 00:40:47.610 If you look at two terms in the same AP, in the sub-AP, 00:40:47.610 --> 00:41:00.670 then the value of this character on each P sub i 00:41:00.670 --> 00:41:03.080 is roughly the same. 00:41:03.080 --> 00:41:08.920 So they don't vary by more than eta on each P i. 00:41:08.920 --> 00:41:16.350 So here, we're partitioning this 1 through N into a sub-A piece 00:41:16.350 --> 00:41:20.355 so that this guy here stays roughly constant. 00:41:25.940 --> 00:41:26.440 OK. 00:41:26.440 --> 00:41:29.470 Any questions? 00:41:29.470 --> 00:41:29.970 All right. 00:41:29.970 --> 00:41:33.728 So think about how you might prove this. 00:41:33.728 --> 00:41:34.770 Let's take a quick break. 00:41:37.580 --> 00:41:41.160 So you see, we are basically following the same strategy 00:41:41.160 --> 00:41:44.770 as the proof from last time, but this second step, 00:41:44.770 --> 00:41:47.370 which we're on right now, needs to be somewhat modified 00:41:47.370 --> 00:41:52.230 because you cannot cut this space up into pieces where 00:41:52.230 --> 00:41:54.330 your character is constant. 00:41:54.330 --> 00:41:58.140 Well, if they're roughly constant then we're go to go, 00:41:58.140 --> 00:42:01.300 so that's what we're doing now. 00:42:01.300 --> 00:42:03.000 So let's prove the statement up there. 00:42:15.350 --> 00:42:15.860 All right. 00:42:15.860 --> 00:42:18.950 So let's prove this statement over here. 00:42:18.950 --> 00:42:37.600 So using Dirichlet's lemma, we find that there exists some d. 00:42:37.600 --> 00:42:39.615 OK, so I'll write down some number for now. 00:42:39.615 --> 00:42:40.490 Don't worry about it. 00:42:40.490 --> 00:42:45.090 It will come up shortly why I write this specific quantity. 00:42:47.670 --> 00:43:06.330 So there exists some d which is not too big, such that d theta 00:43:06.330 --> 00:43:09.830 is very close to an integer. 00:43:09.830 --> 00:43:12.540 So now, I'm literally applying Dirichlet's lemma. 00:43:16.050 --> 00:43:17.680 OK. 00:43:17.680 --> 00:43:24.826 So given such d-- 00:43:24.826 --> 00:43:27.640 so how big is this d? 00:43:27.640 --> 00:43:32.740 You see that because I assumed that N is sufficiently large, 00:43:32.740 --> 00:43:48.800 if we choose that C large enough, d is at most root N. So 00:43:48.800 --> 00:43:53.570 given such d, which is at most root N, 00:43:53.570 --> 00:44:04.180 you can partition 1 through N into subprogressions 00:44:04.180 --> 00:44:09.580 with common difference d. 00:44:09.580 --> 00:44:14.550 Essentially, look at-- let's do classes mod d. 00:44:14.550 --> 00:44:19.130 So they're all going to have length by basically N over d. 00:44:19.130 --> 00:44:25.140 And I chop them up a little bit further to get-- 00:44:25.140 --> 00:44:33.390 so a piece of length between cube root of N 00:44:33.390 --> 00:44:40.990 and twice cube root of N. 00:44:40.990 --> 00:44:41.490 OK. 00:44:41.490 --> 00:44:44.520 So I'm going to make sure that all of my APs 00:44:44.520 --> 00:44:48.000 are roughly the same length. 00:44:48.000 --> 00:44:56.090 And now, inside each subprogression-- 00:45:00.790 --> 00:45:06.610 let me call this subprogression P prime, subprogression P, 00:45:06.610 --> 00:45:09.430 let's look at how much this character value can 00:45:09.430 --> 00:45:13.119 vary inside this progression. 00:45:19.346 --> 00:45:20.304 All right. 00:45:26.010 --> 00:45:29.420 OK, so how much can this vary? 00:45:29.420 --> 00:45:36.330 Well, because theta is such that d times theta is very close 00:45:36.330 --> 00:45:41.778 to an integer and the length of each progression is not too 00:45:41.778 --> 00:45:42.570 large-- so here's-- 00:45:42.570 --> 00:45:44.710 I want some control on the length. 00:45:44.710 --> 00:45:49.020 So we find that the maximum variation 00:45:49.020 --> 00:45:56.695 is, at most, the size of P, the length of P, times-- 00:45:59.580 --> 00:46:04.530 so this-- that difference over there. 00:46:04.530 --> 00:46:08.920 So all of these are exponential, so I can shift them. 00:46:08.920 --> 00:46:19.995 Well, the length of P is at most twice cube root of N. And-- 00:46:19.995 --> 00:46:22.160 OK, so what is this quantity? 00:46:22.160 --> 00:46:25.470 So the point is that if this fractional part here 00:46:25.470 --> 00:46:29.080 is very close to an integer, then e to that, 00:46:29.080 --> 00:46:31.560 e to the 2 pi times that i times some number should 00:46:31.560 --> 00:46:36.630 be very close to 1, because what is happening here? 00:46:36.630 --> 00:46:42.050 This is the distance between those two points 00:46:42.050 --> 00:46:45.230 on the circle, which is at most bounded 00:46:45.230 --> 00:46:46.490 by the length of the arc. 00:46:57.460 --> 00:47:02.530 OK, so cord length, at most of the arc. 00:47:02.530 --> 00:47:05.560 So now, you put everything here together, 00:47:05.560 --> 00:47:09.230 and apply the bound that we got on d theta. 00:47:09.230 --> 00:47:11.410 So this is the reason for choosing that weird number 00:47:11.410 --> 00:47:13.960 up there. 00:47:13.960 --> 00:47:30.050 We find that the variation within each progression 00:47:30.050 --> 00:47:31.903 is at most eta, right? 00:47:31.903 --> 00:47:33.320 So the variation of this character 00:47:33.320 --> 00:47:36.175 within each progression is not very large, OK? 00:47:36.175 --> 00:47:37.050 And that's the claim. 00:47:39.830 --> 00:47:41.780 Any questions? 00:47:41.780 --> 00:47:45.920 All right, so this is the analogous claim to the one 00:47:45.920 --> 00:47:46.780 that we had-- 00:47:46.780 --> 00:47:48.500 the one that we used last time, where 00:47:48.500 --> 00:47:51.950 we said that the character is constant on each coset 00:47:51.950 --> 00:47:52.915 of the hyperplane. 00:47:52.915 --> 00:47:55.430 They're not exactly constant, but almost good enough. 00:48:05.610 --> 00:48:06.300 All right. 00:48:06.300 --> 00:48:10.020 So the goal of step 2 is to show an energy-- 00:48:10.020 --> 00:48:13.020 show a density increment, that if you have a large Fourier 00:48:13.020 --> 00:48:15.180 coefficient, then we want to claim 00:48:15.180 --> 00:48:18.720 that the density goes up significantly 00:48:18.720 --> 00:48:21.940 on some subprogression. 00:48:21.940 --> 00:48:24.910 And the next part, the next lemma, 00:48:24.910 --> 00:48:27.280 will get us to that goal. 00:48:27.280 --> 00:48:29.920 And this part is very similar to the one 00:48:29.920 --> 00:48:34.600 that we saw from last time, but with this new partition 00:48:34.600 --> 00:48:35.980 in mind, like I said. 00:48:35.980 --> 00:48:47.860 If you have A that is 3AP-free with density alpha, 00:48:47.860 --> 00:48:57.990 and N is large enough, then there 00:48:57.990 --> 00:49:08.980 exists some subprogression P. So by subprogression, 00:49:08.980 --> 00:49:11.560 I just mean that I'm starting with original progression 00:49:11.560 --> 00:49:15.340 1 through N, and I'm zooming into some subprogression, 00:49:15.340 --> 00:49:22.160 with the length of P fairly long, so the length of P 00:49:22.160 --> 00:49:29.430 is at least cube root of N, and such that A, 00:49:29.430 --> 00:49:32.760 when restricted to this subprogression, 00:49:32.760 --> 00:49:36.295 has a density increment. 00:49:41.995 --> 00:49:44.950 OK, so originally, the density of A is alpha, 00:49:44.950 --> 00:49:47.300 so we're zooming into some subprogression P, 00:49:47.300 --> 00:49:48.980 which is a pretty long subprogression, 00:49:48.980 --> 00:49:50.870 where the density goes up significantly 00:49:50.870 --> 00:49:53.360 from A to essentially A-- 00:49:53.360 --> 00:49:56.330 from alpha to roughly alpha plus alpha squared. 00:50:00.240 --> 00:50:00.740 OK. 00:50:07.120 --> 00:50:10.660 So we start with A, a 3AP-free set. 00:50:10.660 --> 00:50:25.110 So from step 1, there exists some theta with large-- 00:50:25.110 --> 00:50:27.932 so that corresponds to a large Fourier coefficient. 00:50:31.790 --> 00:50:46.070 So this sum here is large. 00:50:48.782 --> 00:50:54.710 OK, and now we use-- 00:50:54.710 --> 00:51:00.510 OK, so-- so step 1 obtains us, you know, this consequence. 00:51:00.510 --> 00:51:14.780 And from this theta, now we apply the lemma up there to-- 00:51:14.780 --> 00:51:16.760 so we apply lemma with, let's say, 00:51:16.760 --> 00:51:21.350 eta being alpha squared over 30. 00:51:21.350 --> 00:51:23.690 OK, so the exact constants are not so important. 00:51:23.690 --> 00:51:31.010 But when we apply the lemma to partition, 00:51:31.010 --> 00:51:37.720 and into a bunch of subprogressions, 00:51:37.720 --> 00:51:40.570 which we'll call P1 through Pk. 00:51:43.462 --> 00:51:48.080 And each of these progressions have 00:51:48.080 --> 00:51:59.750 length between cube root of N and twice cube root of N. 00:51:59.750 --> 00:52:04.070 And I want to understand what happens to the density of A 00:52:04.070 --> 00:52:07.680 when restricted to these progressions. 00:52:07.680 --> 00:52:14.030 So starting with this inequality over here, 00:52:14.030 --> 00:52:17.150 which suggests to us that there must be some deviation. 00:52:30.191 --> 00:52:34.700 OK, so starting with what we saw. 00:52:34.700 --> 00:52:40.950 And now, inside each progression this e x theta 00:52:40.950 --> 00:52:44.050 is roughly constant. 00:52:44.050 --> 00:52:46.760 So if you pretend them as actually constant, 00:52:46.760 --> 00:52:54.230 I can break up the sum, depending on where the x's lie. 00:52:54.230 --> 00:52:57.530 So i from 1 to k. 00:52:57.530 --> 00:53:01.492 And let me sum inside each progression. 00:53:13.560 --> 00:53:17.610 So by triangle inequality, I can upper bound the first sum 00:53:17.610 --> 00:53:23.540 by where I now cut the sum into progression by progression. 00:53:23.540 --> 00:53:29.520 And on each progression, this character is roughly constant. 00:53:29.520 --> 00:53:34.760 So let me take out the maximum possible deviations 00:53:34.760 --> 00:53:36.440 from them being constant. 00:53:36.440 --> 00:53:48.525 So upper bound-- again, you'll find that we can essentially 00:53:48.525 --> 00:53:49.025 pretend-- 00:53:52.190 --> 00:53:55.410 all right, so if each exponential 00:53:55.410 --> 00:53:57.450 is constant on each subprogression, 00:53:57.450 --> 00:54:00.750 then I might as well just have this sum here. 00:54:00.750 --> 00:54:03.510 But I lose a little bit, because it's not exactly constant. 00:54:03.510 --> 00:54:04.680 It's almost constant. 00:54:04.680 --> 00:54:06.140 So I loose a little bit. 00:54:06.140 --> 00:54:11.350 And that little bit is this eta. 00:54:11.350 --> 00:54:13.440 So you lose that little bit of eta. 00:54:13.440 --> 00:54:18.960 And so on each progression, P i, you 00:54:18.960 --> 00:54:22.830 lose at most something that's essentially of alpha squared 00:54:22.830 --> 00:54:24.265 times the length of P i. 00:54:27.240 --> 00:54:29.160 OK. 00:54:29.160 --> 00:54:32.880 Now, you see, I've chosen the error parameter 00:54:32.880 --> 00:54:37.260 so that everything I've lost is not 00:54:37.260 --> 00:54:41.250 so much more than the initial bound I began with. 00:54:41.250 --> 00:54:47.100 So in particular, we see that even 00:54:47.100 --> 00:54:49.800 if we had pretended that the characters were constant, 00:54:49.800 --> 00:54:54.090 on each progression we would have still obtained some lower 00:54:54.090 --> 00:54:56.880 bound of the total deviation. 00:55:08.680 --> 00:55:09.860 OK. 00:55:09.860 --> 00:55:15.130 And what is this quantity over here? 00:55:15.130 --> 00:55:18.880 Oh, you see, I'm restricting each sum 00:55:18.880 --> 00:55:23.590 to each subprogression, but the sum 00:55:23.590 --> 00:55:26.440 here, even though it's the sum, but it's really 00:55:26.440 --> 00:55:31.990 counting how many elements of A are in that progression. 00:55:31.990 --> 00:55:35.710 So this sum over here is the same thing. 00:55:35.710 --> 00:55:38.543 OK, so let me write it in a new board. 00:55:43.480 --> 00:55:45.213 Oh, we don't need step 1 anymore. 00:55:49.077 --> 00:55:51.270 All right. 00:55:51.270 --> 00:55:52.440 So what we have-- 00:56:01.386 --> 00:56:05.431 OK, so left-hand side over there is this quantity here, 00:56:05.431 --> 00:56:07.690 all right? 00:56:07.690 --> 00:56:15.890 We see that the right-hand side, even though you have that sum, 00:56:15.890 --> 00:56:19.010 it is really just counting how many elements of A 00:56:19.010 --> 00:56:23.360 are in each progression versus how many you should expect 00:56:23.360 --> 00:56:28.750 based on the overall density of A. OK, 00:56:28.750 --> 00:56:32.220 so that should look similar to what we got last time. 00:56:32.220 --> 00:56:35.280 I know the intuition should be that, well, 00:56:35.280 --> 00:56:42.210 if the average deviation is large, then one of them, 00:56:42.210 --> 00:56:47.447 one of these terms, should have the density increment. 00:56:51.190 --> 00:56:53.950 If you try to do the next step somewhat naively, 00:56:53.950 --> 00:56:56.840 you run into an issue, because it could be-- 00:56:56.840 --> 00:56:59.290 now, here you have k terms. 00:56:59.290 --> 00:57:07.270 It could be that you have all the densities except for one 00:57:07.270 --> 00:57:12.640 going up only slightly, and one density dropping dramatically, 00:57:12.640 --> 00:57:15.580 in which case you may not have a significant density 00:57:15.580 --> 00:57:18.210 increment, all right? 00:57:18.210 --> 00:57:20.840 So we want to show that on some progression 00:57:20.840 --> 00:57:24.350 the density increases significantly. 00:57:24.350 --> 00:57:26.870 So far, from this inequality, we just 00:57:26.870 --> 00:57:29.720 know that there is some subprogression where 00:57:29.720 --> 00:57:32.450 the density changes significantly. 00:57:32.450 --> 00:57:34.850 But of course, the overall density, the average density, 00:57:34.850 --> 00:57:36.120 should remain constant. 00:57:36.120 --> 00:57:38.950 So if some goes up, others must go down. 00:57:38.950 --> 00:57:41.180 But if you just try to do an averaging argument, 00:57:41.180 --> 00:57:42.890 you have to be careful, OK? 00:57:42.890 --> 00:57:45.290 So there was a trick last time, which we didn't really 00:57:45.290 --> 00:57:49.910 need last time, but now, it's much more useful, where 00:57:49.910 --> 00:57:52.220 I want to show that if this holds, 00:57:52.220 --> 00:57:55.850 then some P i sees a large energy-- 00:57:55.850 --> 00:57:58.520 sees a large density increment. 00:57:58.520 --> 00:58:08.990 And to do that, let me rewrite the sum as the following, 00:58:08.990 --> 00:58:11.660 so I keep the same expression. 00:58:14.880 --> 00:58:18.980 And I add a term, which is the same thing, 00:58:18.980 --> 00:58:20.560 but without the absolute value. 00:58:29.434 --> 00:58:35.000 OK, so you see these guys, they total to 0, 00:58:35.000 --> 00:58:40.350 so adding that term doesn't change my expression. 00:58:40.350 --> 00:58:45.460 But now, the summand is always non-negative. 00:58:45.460 --> 00:58:49.380 So it's either 0 or twice this number, 00:58:49.380 --> 00:58:51.500 depending on the sign of that number. 00:58:54.610 --> 00:58:55.110 OK. 00:58:55.110 --> 00:58:57.600 So comparing left-hand side and right-hand side, 00:58:57.600 --> 00:59:00.790 we see that there must be some i. 00:59:00.790 --> 00:59:08.730 So hence, there exists some eye such that the left-hand side-- 00:59:13.410 --> 00:59:15.270 the i-th term on the left hand side 00:59:15.270 --> 00:59:18.272 is less than or equal to the i-th term 00:59:18.272 --> 00:59:19.230 on the right-hand side. 00:59:24.940 --> 00:59:28.640 And in particular, that term should be positive, 00:59:28.640 --> 00:59:31.460 so it implies-- 00:59:31.460 --> 00:59:35.740 OK, so how can you get this inequality? 00:59:35.740 --> 00:59:40.540 It implies simply that the restriction of a to this P i 00:59:40.540 --> 00:59:50.810 is at least alpha plus alpha squared over 40 times P i. 00:59:50.810 --> 00:59:54.320 So this claim here just says that on the i-th progression, 00:59:54.320 --> 00:59:58.570 there's a significant energy increment. 00:59:58.570 --> 01:00:02.342 If it's more decrement, that term would have been 0. 01:00:02.342 --> 01:00:04.610 So remember that. 01:00:16.040 --> 01:00:17.280 OK. 01:00:17.280 --> 01:00:20.610 So this achieves what we were looking for in step 2, 01:00:20.610 --> 01:00:24.720 namely to find that there is a density increment 01:00:24.720 --> 01:00:28.041 on some long subprogression. 01:00:28.041 --> 01:00:30.630 OK, so now we can go to step 3, which is basically 01:00:30.630 --> 01:00:35.460 the same as what we saw last time, where now we 01:00:35.460 --> 01:00:38.929 want to iterate this density increment. 01:00:47.911 --> 01:00:55.900 OK, so it's basically the same argument as last time, 01:00:55.900 --> 01:01:00.300 but you start with density alpha, 01:01:00.300 --> 01:01:04.140 and each step in the iteration, the density 01:01:04.140 --> 01:01:05.600 goes up quite a bit. 01:01:09.740 --> 01:01:14.030 And we want to control the total number of steps, 01:01:14.030 --> 01:01:20.704 knowing that the final density is at most 1, always. 01:01:20.704 --> 01:01:23.370 OK. 01:01:23.370 --> 01:01:25.020 So how many steps can you take? 01:01:25.020 --> 01:01:28.080 Right, so this was the same argument that we saw last time. 01:01:28.080 --> 01:01:38.450 We see that starting with alpha, alpha 0 being alpha, 01:01:38.450 --> 01:01:41.810 it doubles after a certain number of steps, right? 01:01:41.810 --> 01:01:45.590 So we double after-- 01:01:45.590 --> 01:01:48.710 OK, so how many steps do you need? 01:01:48.710 --> 01:01:53.430 Well, I want to get from alpha to 2 alpha. 01:01:53.430 --> 01:02:00.960 So I need at most alpha over 40 steps. 01:02:00.960 --> 01:02:03.770 OK, so last time I was slightly sloppy. 01:02:03.770 --> 01:02:09.630 And so there's basically a floor, upper floor down-- 01:02:09.630 --> 01:02:11.060 rounding up or down situation. 01:02:11.060 --> 01:02:13.620 But I should add a plus 1. 01:02:13.620 --> 01:02:14.120 Yeah. 01:02:14.120 --> 01:02:16.120 AUDIENCE: Shouldn't it be 40 over alpha? 01:02:16.120 --> 01:02:17.340 YUFEI ZHAO: 40-- thank you. 01:02:17.340 --> 01:02:19.930 40 over alpha, yeah. 01:02:19.930 --> 01:02:25.080 So you double after at most that many steps. 01:02:25.080 --> 01:02:31.560 And then now, you add density at least 2 alpha. 01:02:31.560 --> 01:02:39.880 So we double after at most 20 over alpha steps, and so on. 01:02:39.880 --> 01:02:44.690 And we double at most-- 01:02:44.690 --> 01:02:51.690 well, basically, log sub 2 of 1 over alpha times. 01:02:55.658 --> 01:02:58.600 OK, so anyway, putting everything together, 01:02:58.600 --> 01:03:04.350 we see that the total number of steps 01:03:04.350 --> 01:03:08.790 is at most on the order of 1 over alpha. 01:03:12.340 --> 01:03:15.490 When you stop, you can only stop for one reason, 01:03:15.490 --> 01:03:18.210 because in the-- 01:03:18.210 --> 01:03:18.710 yeah. 01:03:18.710 --> 01:03:19.880 So, yeah. 01:03:19.880 --> 01:03:24.790 So in step 1, remember, the iteration 01:03:24.790 --> 01:03:28.762 said that the process terminates. 01:03:28.762 --> 01:03:37.589 The process can always go on, and then terminates 01:03:37.589 --> 01:03:43.170 if the length-- 01:03:43.170 --> 01:03:45.590 so you're now at step i, so let N i be 01:03:45.590 --> 01:03:54.580 the length of the progression, that step i is at most C times 01:03:54.580 --> 01:03:56.990 alpha i to the minus 12th. 01:03:56.990 --> 01:04:01.180 So we have a-- 01:04:01.180 --> 01:04:04.980 right, so provided that N is large enough, 01:04:04.980 --> 01:04:07.020 you can always pass to a subprogression. 01:04:07.020 --> 01:04:09.540 And here, when you pass to subprogression, of course, 01:04:09.540 --> 01:04:12.030 you can re-label that subprogression. 01:04:12.030 --> 01:04:15.140 And it's now, you know, 1 through N i. 01:04:15.140 --> 01:04:16.553 Right, so I can-- it's-- 01:04:16.553 --> 01:04:17.970 all the progressions are basically 01:04:17.970 --> 01:04:21.373 the same as the first set of positive integers. 01:04:21.373 --> 01:04:23.850 I'm sorry, prefix of the positive integers. 01:04:23.850 --> 01:04:26.590 So when we stop a step i, you must 01:04:26.590 --> 01:04:31.290 have N sub i being at most this quantity over here, which 01:04:31.290 --> 01:04:36.510 is at most C times the initial density raised to this minus 01:04:36.510 --> 01:04:37.450 12th. 01:04:37.450 --> 01:04:49.050 So therefore, the initial length N of the space is bounded by-- 01:04:49.050 --> 01:04:55.595 well, each time we went down by a cube root at most. 01:04:55.595 --> 01:04:57.560 Right, so the fine-- 01:04:57.560 --> 01:05:00.980 if you stop a step i, then the initial length 01:05:00.980 --> 01:05:07.670 is at most N sub i to the 3 times 3 to the power of i 01:05:07.670 --> 01:05:10.922 each time you're doing a cube root. 01:05:10.922 --> 01:05:13.135 OK, so you put everything together. 01:05:20.280 --> 01:05:22.360 At most that many iterations when you stop 01:05:22.360 --> 01:05:24.781 the length is at most this. 01:05:24.781 --> 01:05:26.680 So you put them together, and then you 01:05:26.680 --> 01:05:37.320 find that the N must be at most double exponential in 1 01:05:37.320 --> 01:05:40.140 over the density. 01:05:40.140 --> 01:05:47.670 In other words, the density is at most 1 01:05:47.670 --> 01:05:55.665 over log log N, which is what we claimed in Roth's theorem, 01:05:55.665 --> 01:05:57.060 so what we claimed up there. 01:06:01.460 --> 01:06:01.960 OK. 01:06:01.960 --> 01:06:03.228 So that finishes the proof. 01:06:12.210 --> 01:06:13.307 Any questions? 01:06:18.780 --> 01:06:22.230 So the message here is that it's the same proof as last time, 01:06:22.230 --> 01:06:24.660 but we need to do a bit more work. 01:06:24.660 --> 01:06:26.220 And none of this work is difficult, 01:06:26.220 --> 01:06:27.780 but there are more technical. 01:06:27.780 --> 01:06:29.580 And that's often the theme that you 01:06:29.580 --> 01:06:31.500 see in additive combinatorics. 01:06:31.500 --> 01:06:34.650 This is part of the reason why the finite field model is 01:06:34.650 --> 01:06:38.130 a really nice playground, because there, things 01:06:38.130 --> 01:06:41.850 tend to be often cleaner, but the idea's often similar, 01:06:41.850 --> 01:06:43.140 or the same ideas. 01:06:43.140 --> 01:06:43.890 Not always. 01:06:43.890 --> 01:06:46.242 Next lecture, we'll see one technique 01:06:46.242 --> 01:06:47.700 where there's a dramatic difference 01:06:47.700 --> 01:06:50.940 between the finite field vector space and over the integers. 01:06:50.940 --> 01:06:53.170 But for many things in additive combinatorics, 01:06:53.170 --> 01:06:54.720 the finite field vector space is just 01:06:54.720 --> 01:06:58.172 a nicer place to be in to try all your ideas and techniques. 01:07:04.440 --> 01:07:09.090 Let me comment on some analogies between these two approaches 01:07:09.090 --> 01:07:10.200 and compare the bounds. 01:07:13.090 --> 01:07:15.280 So on one hand-- 01:07:15.280 --> 01:07:20.130 OK, so we saw last time this proof in F3 to the N, 01:07:20.130 --> 01:07:26.770 and now in the integers inside this interval of length 01:07:26.770 --> 01:07:33.280 N. So let me write uppercase N in both cases 01:07:33.280 --> 01:07:38.205 to be the size of the overall ambient space. 01:07:38.205 --> 01:07:43.310 OK, so what kind of bounds do we get in both situations? 01:07:43.310 --> 01:07:46.050 So last time, for-- 01:07:46.050 --> 01:07:51.720 in F3 to the N, we got a bound which 01:07:51.720 --> 01:08:06.220 is of the order N over log N, whereas today, the bound is 01:08:06.220 --> 01:08:08.560 somewhat worse. 01:08:08.560 --> 01:08:09.920 It's a little bit worse. 01:08:09.920 --> 01:08:13.690 Now we lose an extra log. 01:08:13.690 --> 01:08:16.810 So where do we lose an extra log in this argument? 01:08:16.810 --> 01:08:19.060 So where does these two argument-- 01:08:19.060 --> 01:08:22.980 where do these two arguments differ, quantitatively? 01:08:26.240 --> 01:08:26.740 Yep? 01:08:26.740 --> 01:08:29.569 AUDIENCE: When you're dividing by 3 versus [INAUDIBLE]?? 01:08:29.569 --> 01:08:31.560 YUFEI ZHAO: OK, so you're dividing by-- 01:08:31.560 --> 01:08:41.960 so here, in each iteration, over here, 01:08:41.960 --> 01:08:46.399 your size of the iteration-- 01:08:46.399 --> 01:08:48.300 I mean, each iteration the size of the space 01:08:48.300 --> 01:08:53.279 goes down by a factor of 3, whereas over here, it 01:08:53.279 --> 01:08:56.979 could go down by a cube root. 01:08:56.979 --> 01:08:58.399 And that's precisely right. 01:08:58.399 --> 01:09:02.420 So that explains for this extra log in the balance. 01:09:05.630 --> 01:09:08.490 So while this is a great analogy, 01:09:08.490 --> 01:09:10.760 it's not a perfect analogy. 01:09:10.760 --> 01:09:13.279 You see there is this divergence here between the two 01:09:13.279 --> 01:09:14.790 situations. 01:09:14.790 --> 01:09:16.430 And so then you might ask, is there 01:09:16.430 --> 01:09:21.240 some way to avoid the loss, this extra log factor 01:09:21.240 --> 01:09:22.950 loss over here? 01:09:22.950 --> 01:09:25.260 Is there some way to carry out the strategy 01:09:25.260 --> 01:09:28.689 that we did last time in a way that 01:09:28.689 --> 01:09:31.300 is much more faithful to that strategy of passing down 01:09:31.300 --> 01:09:33.060 to subspaces. 01:09:33.060 --> 01:09:36.090 So here, we pass to progressions. 01:09:36.090 --> 01:09:40.649 And because we have to do this extra pigeonhole-type argument, 01:09:40.649 --> 01:09:41.970 it was somewhat lost-- 01:09:41.970 --> 01:09:48.770 we lost a power, which translated into this extra log. 01:09:48.770 --> 01:09:51.910 So it turns out there is some way to do this. 01:09:51.910 --> 01:09:53.600 So let me just briefly mention what's 01:09:53.600 --> 01:09:55.640 the idea that is involved, all right? 01:09:55.640 --> 01:09:59.930 So last time, we went down from-- 01:09:59.930 --> 01:10:02.660 so the main objects that we were passing 01:10:02.660 --> 01:10:05.180 would start with a vector space and pass down 01:10:05.180 --> 01:10:08.310 to a subspace, which is also a vector space, right? 01:10:08.310 --> 01:10:17.490 So you can define subspaces in F3 to the N by the following. 01:10:17.490 --> 01:10:24.310 So I can start with some set of characters U, and I define-- 01:10:24.310 --> 01:10:26.630 some set of characters S, and I define 01:10:26.630 --> 01:10:40.822 U sub S to be basically the orthogonal complement of S. 01:10:40.822 --> 01:10:42.972 OK, so this is a subspace. 01:10:42.972 --> 01:10:44.430 And these were the kind of subspace 01:10:44.430 --> 01:10:48.168 that we saw last time, because the S's or the R's that 01:10:48.168 --> 01:10:50.460 came out of the proof last time, every time we saw one, 01:10:50.460 --> 01:10:51.390 we threw it in. 01:10:51.390 --> 01:10:56.330 We cut down to a smaller subspace, and we repeat. 01:10:56.330 --> 01:10:59.018 But the progressions, they don't really look like this. 01:10:59.018 --> 01:11:00.560 So the question is, is there some way 01:11:00.560 --> 01:11:02.935 to do this argument so that you end up with progressions, 01:11:02.935 --> 01:11:04.010 and looked like that? 01:11:04.010 --> 01:11:06.300 And it turns out there is a way. 01:11:06.300 --> 01:11:08.150 And there are these objects, which 01:11:08.150 --> 01:11:13.215 we'll see more later in this course, called Bohr sets. 01:11:13.215 --> 01:11:18.030 OK, so they were used by Bourgain 01:11:18.030 --> 01:11:23.220 to mimic this Machoulin argument that we saw last time more 01:11:23.220 --> 01:11:26.010 faithfully into the integers, where 01:11:26.010 --> 01:11:30.840 we're going to come up with some set of integers that resemble-- 01:11:30.840 --> 01:11:34.950 much more closely resemble this notion of subspaces 01:11:34.950 --> 01:11:37.290 in the finite field setting. 01:11:37.290 --> 01:11:41.020 And for this, it's much easier to work inside a group. 01:11:41.020 --> 01:11:42.900 So instead of working in the integers, 01:11:42.900 --> 01:11:45.110 let's work inside and Z mod nZ. 01:11:45.110 --> 01:11:47.580 So we can do Fourier transform in Z mod nZ, so 01:11:47.580 --> 01:11:50.040 the discrete Fourier analysis here. 01:11:50.040 --> 01:11:52.335 So in Z mod nZ, we define-- 01:11:55.510 --> 01:11:59.260 so given a S, let's define this Bohr 01:11:59.260 --> 01:12:14.080 set to be the set of elements of Z mod nZ such 01:12:14.080 --> 01:12:20.470 that if you look at what really is supposed to resemble 01:12:20.470 --> 01:12:26.280 this thing over here, OK? 01:12:26.280 --> 01:12:34.280 If this quantity is small for all S-- 01:12:34.280 --> 01:12:36.930 OK, so we put that element into this Bohr set. 01:12:39.670 --> 01:12:44.870 OK, so these sets, they function much more like subspaces. 01:12:44.870 --> 01:12:48.890 So there are the analog of subspaces 01:12:48.890 --> 01:12:53.420 inside Z mod nZ, which, you know, n is prime, 01:12:53.420 --> 01:12:54.430 has no subgroups. 01:12:54.430 --> 01:12:56.990 It has no natural subspace structure. 01:12:56.990 --> 01:12:58.880 But by looking at these Bohr sets, 01:12:58.880 --> 01:13:03.020 they provide a natural way to set up 01:13:03.020 --> 01:13:05.330 this argument so that you can-- 01:13:05.330 --> 01:13:07.910 but with much more technicalities, 01:13:07.910 --> 01:13:13.630 repeat these kind of arguments more similar to last time, 01:13:13.630 --> 01:13:17.220 but passing not to subspaces but to Bohr sets. 01:13:17.220 --> 01:13:22.270 And then with quite a bit of extra work, 01:13:22.270 --> 01:13:32.370 one can obtain bounds of the quantity N over a poly log N. 01:13:32.370 --> 01:13:36.510 So the current best bound I mentioned last time 01:13:36.510 --> 01:13:40.080 is of this type, which is through further refinements 01:13:40.080 --> 01:13:41.347 of this technique. 01:13:50.610 --> 01:13:52.230 The last thing I want to mention today 01:13:52.230 --> 01:13:56.040 is, so far we've been talking about 3APs. 01:13:56.040 --> 01:13:59.600 So what about four term arithmetic progressions? 01:13:59.600 --> 01:14:04.998 OK, do any of the things that we talk about here work for 4APs? 01:14:04.998 --> 01:14:07.290 And there's an analogy to be made here compared to what 01:14:07.290 --> 01:14:09.085 we discussed with graphs. 01:14:09.085 --> 01:14:11.770 So in graphs, we had a triangle counting lemma and a triangle 01:14:11.770 --> 01:14:12.930 removal lemma. 01:14:12.930 --> 01:14:15.210 And then we said that to prove 4APs, 01:14:15.210 --> 01:14:18.390 we would need the hypergraph version, the simplex removal 01:14:18.390 --> 01:14:20.580 lemma, hypergraph regularity lemma. 01:14:20.580 --> 01:14:22.500 And that was much more difficult. 01:14:22.500 --> 01:14:24.170 And that analogy carries through, 01:14:24.170 --> 01:14:27.040 and the same kind of difficulties that come up. 01:14:27.040 --> 01:14:30.700 So it can be done, but you need something more. 01:14:30.700 --> 01:14:34.590 And the main message I want you to take away 01:14:34.590 --> 01:14:42.940 is that 4APs, while we had a counting lemma that 01:14:42.940 --> 01:14:45.980 says that the Fourier coefficients, 01:14:45.980 --> 01:14:54.540 so the Fourier transform, controls 3AP counts, 01:14:54.540 --> 01:14:59.100 it turns out the same is not true for 4APs. 01:14:59.100 --> 01:15:05.200 So the Fourier does not control 4AP counts. 01:15:10.410 --> 01:15:12.340 Let me give you some-- 01:15:12.340 --> 01:15:12.840 OK. 01:15:12.840 --> 01:15:14.910 So in fact, in the homework for this week, 01:15:14.910 --> 01:15:19.870 there's a specific example of a set where it has uniformly 01:15:19.870 --> 01:15:21.490 small Fourier coefficients. 01:15:21.490 --> 01:15:22.990 But that's the wrong number of 4APs. 01:15:27.130 --> 01:15:28.500 So the following-- it is true-- 01:15:28.500 --> 01:15:32.610 OK, so it is true that you have Szemerédi's term in-- 01:15:32.610 --> 01:15:34.770 let's just talk about the finite field setting, 01:15:34.770 --> 01:15:36.840 where things are a bit easier to discuss. 01:15:36.840 --> 01:15:42.510 So it is true that the size of the biggest subset of F5 01:15:42.510 --> 01:15:45.180 to the N is a tiny fraction-- it's a little, one 01:15:45.180 --> 01:15:46.530 fraction of the entire space. 01:15:46.530 --> 01:15:50.460 OK, I use F5 here, because if I set F3, 01:15:50.460 --> 01:15:53.690 it doesn't make sense to talk about 4APs. 01:15:53.690 --> 01:15:58.990 So F5, but it doesn't really matter which specific field. 01:15:58.990 --> 01:16:03.550 So you can prove this using hypergraph removal, same proof, 01:16:03.550 --> 01:16:07.330 verbatim, that we saw earlier, if you have hypergraph removal. 01:16:07.330 --> 01:16:10.610 But if you want to try to prove it using Fourier analysis, 01:16:10.610 --> 01:16:15.640 well, it doesn't work quite using the same strategy. 01:16:15.640 --> 01:16:17.590 But in fact, there is a modification 01:16:17.590 --> 01:16:20.800 that would allow you to make it work. 01:16:20.800 --> 01:16:23.670 But you need an extension of Fourier analysis. 01:16:23.670 --> 01:16:30.790 And it is known as higher order Fourier analysis, which 01:16:30.790 --> 01:16:35.470 was an important development in modern additive combinatorics 01:16:35.470 --> 01:16:38.590 that initially arose in Gowers' work 01:16:38.590 --> 01:16:41.190 where he gave a new proof of similarities theorem. 01:16:41.190 --> 01:16:42.880 So Gowers didn't work in this setting. 01:16:42.880 --> 01:16:44.020 He worked in integers. 01:16:44.020 --> 01:16:47.140 But many of the ideas originated from his paper, 01:16:47.140 --> 01:16:49.120 and then subsequently developed by a lot 01:16:49.120 --> 01:16:51.250 of people in various settings. 01:16:51.250 --> 01:16:53.877 I just want to give you one specific statement, what 01:16:53.877 --> 01:16:55.710 this high-order Fourier analysis looks like. 01:16:55.710 --> 01:16:58.450 So it's a fancy term, and the statements often 01:16:58.450 --> 01:17:00.070 get very technical. 01:17:00.070 --> 01:17:04.140 But I just want to give you one concrete thing to take away. 01:17:04.140 --> 01:17:05.790 All right, so for a Fourier piece, 01:17:05.790 --> 01:17:08.220 higher-order Fourier analysis, roughly-- 01:17:08.220 --> 01:17:12.484 OK, so it also goes by the name quadratic Fourier analysis. 01:17:16.436 --> 01:17:22.370 OK, so let me give you a very specific instance 01:17:22.370 --> 01:17:23.570 of the theorem. 01:17:27.090 --> 01:17:31.710 And this can be sometimes called an inverse theorem 01:17:31.710 --> 01:17:34.272 for quadratic Fourier analysis. 01:17:40.790 --> 01:17:48.700 OK, so for every delta, there exists some c such 01:17:48.700 --> 01:17:51.490 that the following is true. 01:17:51.490 --> 01:18:03.390 If A is a subset of a F5 to the N with density alpha and such 01:18:03.390 --> 01:18:05.132 that it's-- 01:18:05.132 --> 01:18:07.750 OK, so now lambda sub 4, so this is 01:18:07.750 --> 01:18:10.380 the 4AP density, so similar to 3AP, 01:18:10.380 --> 01:18:11.710 but now you write four terms. 01:18:11.710 --> 01:18:18.630 The 4AP density of A differs from alpha to the fourth 01:18:18.630 --> 01:18:21.542 by a significant amount. 01:18:21.542 --> 01:18:25.050 OK, so for 3APs, then we said that now A has a large Fourier 01:18:25.050 --> 01:18:26.750 coefficient, right? 01:18:29.700 --> 01:18:33.640 So for-- OK. 01:18:33.640 --> 01:18:37.530 For 4APs, that may not be true, but the following 01:18:37.530 --> 01:18:38.375 is true, right? 01:18:38.375 --> 01:18:48.668 So then there exists a non-zero quadratic polynomial. 01:18:55.500 --> 01:19:08.620 F and N variables over F5 such that the indicator 01:19:08.620 --> 01:19:13.280 function of A correlates with this quadratic exponential 01:19:13.280 --> 01:19:13.780 face. 01:19:23.200 --> 01:19:24.833 So Fourier analysis, the conclusion 01:19:24.833 --> 01:19:26.250 that we got from counting lemma is 01:19:26.250 --> 01:19:28.440 that you have some linear function 01:19:28.440 --> 01:19:33.650 F, such that this quantity is large, 01:19:33.650 --> 01:19:35.630 this large Fourier coefficient. 01:19:35.630 --> 01:19:37.980 OK, so that is not true 4APs. 01:19:37.980 --> 01:19:39.870 But what is true is that now you can 01:19:39.870 --> 01:19:44.860 look at quadratic exponential faces, and then it is true. 01:19:44.860 --> 01:19:48.720 So that's the content of higher order Fourier. 01:19:48.720 --> 01:19:51.660 I mean, that's the example of higher-order Fourier analysis. 01:19:51.660 --> 01:19:56.010 And you can imagine with this type of result, 01:19:56.010 --> 01:19:58.200 and with quite a bit more work, you 01:19:58.200 --> 01:20:01.560 can try to follow a similar density increment 01:20:01.560 --> 01:20:07.110 strategy to prove similarities term for 4APs.