WEBVTT 00:00:17.865 --> 00:00:18.490 YUFEI ZHAO: OK. 00:00:18.490 --> 00:00:20.950 So let's get started. 00:00:20.950 --> 00:00:23.170 So we spent quite a bit of time with graph theory 00:00:23.170 --> 00:00:25.010 in the first part of this course, 00:00:25.010 --> 00:00:27.760 and today I want to move beyond that. 00:00:27.760 --> 00:00:31.630 So we're going to talk about more central topics 00:00:31.630 --> 00:00:33.730 and additive combinatorics, starting 00:00:33.730 --> 00:00:37.180 with the Fourier analytic proof of Roth's theorem. 00:00:48.760 --> 00:00:51.340 We discussed Roth's theorem, and we gave a proof, 00:00:51.340 --> 00:00:54.760 during the course, using similarities, graph regularity 00:00:54.760 --> 00:00:59.220 lemma, as well as the triangle removal lemma. 00:00:59.220 --> 00:01:01.810 Today, I want to show you a different approach 00:01:01.810 --> 00:01:05.540 to proving Roth's theorem that goes through Fourier analysis. 00:01:05.540 --> 00:01:07.240 So this is a very important proof, 00:01:07.240 --> 00:01:13.330 and it's one of the main tools in additive combinatorics. 00:01:13.330 --> 00:01:15.520 Let me remind you what Roth's theorem says. 00:01:15.520 --> 00:01:24.710 So Roth proved, in 1953, that if we 00:01:24.710 --> 00:01:36.140 write our sub-3 of interval n to be the maximum size of a 3AP3 00:01:36.140 --> 00:01:52.820 subset of 1 through n, then Roth showed that our 3 of n 00:01:52.820 --> 00:01:58.280 is little O of n. 00:01:58.280 --> 00:02:01.790 So in other words, if you have a positive density 00:02:01.790 --> 00:02:06.560 subset of the integers, then it must 00:02:06.560 --> 00:02:09.150 contain a three-term arithmetic progression then. 00:02:09.150 --> 00:02:13.740 So what I said is equivalent to the statement here. 00:02:13.740 --> 00:02:16.831 So previously, we gave a proof using regularity. 00:02:23.000 --> 00:02:25.500 Actually the regularity approach of Szemerédi was only found 00:02:25.500 --> 00:02:28.530 the '70s so Roth's original proof was through Fourier 00:02:28.530 --> 00:02:29.460 analysis. 00:02:29.460 --> 00:02:32.520 And we'll see that tomorrow. 00:02:32.520 --> 00:02:35.843 Today, we'll see a toy version of this proof. 00:02:35.843 --> 00:02:37.260 But it's not really a toy version. 00:02:37.260 --> 00:02:40.430 It has the same ideas, but in a slightly easier setting 00:02:40.430 --> 00:02:42.330 that has fewer technicalities. 00:02:44.553 --> 00:02:46.220 But before showing you that, let me just 00:02:46.220 --> 00:02:48.320 discuss a bit of history around Roth's theorem. 00:02:51.230 --> 00:02:55.600 We will show, next time, the next lecture, 00:02:55.600 --> 00:02:59.405 we'll show the bound. 00:02:59.405 --> 00:03:02.717 Also by regularity, we get some bound, which is little O then, 00:03:02.717 --> 00:03:04.550 but because of the use of regularity lemmas, 00:03:04.550 --> 00:03:06.880 it's pretty poor dependence. 00:03:06.880 --> 00:03:11.370 We got something like n over log star n. 00:03:11.370 --> 00:03:16.680 Next lecture, and basically Roth's original proof, 00:03:16.680 --> 00:03:20.340 gives you a bound which is n over log-log n. 00:03:20.340 --> 00:03:23.100 So it's a much more reasonable bound. 00:03:23.100 --> 00:03:29.880 The current best upper bound known 00:03:29.880 --> 00:03:43.000 has the form, essentially, n over log n raised to 1 00:03:43.000 --> 00:03:47.470 plus little 1, roughly n over log n. 00:03:47.470 --> 00:03:49.870 We do not know, or even have great guesses on, 00:03:49.870 --> 00:03:51.840 what the answer should be. 00:03:51.840 --> 00:03:57.250 So the best lower bound, and this is a construction 00:03:57.250 --> 00:04:02.620 that we saw earlier in the course due to Behrend 00:04:02.620 --> 00:04:10.360 is of the form n over E to the c root log n. 00:04:10.360 --> 00:04:12.550 It seems it may be very difficult to improve 00:04:12.550 --> 00:04:16.070 this upper bound without some genuine new ideas. 00:04:16.070 --> 00:04:18.940 On the other hand, there is some evidence 00:04:18.940 --> 00:04:20.529 that the lower bound might be closer 00:04:20.529 --> 00:04:24.617 to the truth in that there are variants of the Roth problem 00:04:24.617 --> 00:04:27.200 for which we know that the lower bound is basically the truth. 00:04:32.390 --> 00:04:34.670 What I want to do today is look at a variant 00:04:34.670 --> 00:04:37.760 of this problem in what's called a finite field model. 00:04:47.520 --> 00:04:49.480 And that basically just means we're 00:04:49.480 --> 00:04:52.870 going to be looking at Roth's theorem, not in the integers, 00:04:52.870 --> 00:04:56.170 but in some finite field vector space, specifically F3 00:04:56.170 --> 00:04:57.200 to the n. 00:04:57.200 --> 00:05:00.110 So we're going to define our sub-3 of this F3 00:05:00.110 --> 00:05:06.100 to the n to be the maximum size of the 3AP3 00:05:06.100 --> 00:05:13.700 subset of this finite field vector space. 00:05:16.588 --> 00:05:18.380 So the finite field model is really useful. 00:05:18.380 --> 00:05:22.140 We're going to see this again later in the course as well. 00:05:22.140 --> 00:05:24.140 Many of the ideas and techniques that 00:05:24.140 --> 00:05:28.940 work for the real problem, so to speak, many of those techniques 00:05:28.940 --> 00:05:31.040 also work in the finite field model, 00:05:31.040 --> 00:05:34.320 but they are technically simpler to execute. 00:05:34.320 --> 00:05:36.470 So this is often-- 00:05:36.470 --> 00:05:39.980 you view it as a sandbox, a playing ground, for testing out 00:05:39.980 --> 00:05:41.295 many of the ideas. 00:05:41.295 --> 00:05:42.920 And once you have those ideas, then you 00:05:42.920 --> 00:05:46.430 can see if you can bring them to the integer setting. 00:05:46.430 --> 00:05:48.560 And this is a very successful program, 00:05:48.560 --> 00:05:52.130 and we'll see one aspect of what happens when we do this. 00:05:55.220 --> 00:05:59.450 For this specific problem of Roth's theorem, in F3 to the n, 00:05:59.450 --> 00:06:00.950 there are some nice interpretations 00:06:00.950 --> 00:06:02.660 of what this problem means. 00:06:02.660 --> 00:06:05.750 So here's a pretty easy fact that for-- 00:06:05.750 --> 00:06:13.460 so n F3 to the n for three elements, x, y, z, 00:06:13.460 --> 00:06:17.370 the following interpretation, so what it means to be a 3AP, 00:06:17.370 --> 00:06:18.680 are equivalent. 00:06:18.680 --> 00:06:23.060 So x. y, z form a 3AP. 00:06:27.540 --> 00:06:31.680 So 3AP means the y is x plus D. Z is x plus 2D. 00:06:34.640 --> 00:06:38.560 Equivalently, they satisfy this equation, x minus 2y plus z 00:06:38.560 --> 00:06:40.430 equals zero. 00:06:40.430 --> 00:06:40.930 OK. 00:06:40.930 --> 00:06:43.630 In F3, minus 2 is plus 1. 00:06:43.630 --> 00:06:47.770 So it's the same as this even nicer looking equation, 00:06:47.770 --> 00:06:49.350 x plus y plus z equals 2. 00:06:53.588 --> 00:06:55.880 It also turns out to be the same as saying that x, y, z 00:06:55.880 --> 00:07:00.880 would lie on a line. 00:07:00.880 --> 00:07:01.760 But they are aligned. 00:07:06.780 --> 00:07:10.050 So the line has three points in F3. 00:07:10.050 --> 00:07:13.650 And if you look at the coordinate for every i, 00:07:13.650 --> 00:07:28.820 the i-th coordinate of x, y, z are all distinct or all equal. 00:07:32.187 --> 00:07:33.730 So easy to check all these things 00:07:33.730 --> 00:07:35.280 are equivalent to each other. 00:07:35.280 --> 00:07:37.990 And the last one is a nice interpretation 00:07:37.990 --> 00:07:43.460 in terms of a game that many of you know, Set. 00:07:43.460 --> 00:07:45.580 So in the game Set, you have a bunch of cards. 00:07:45.580 --> 00:07:47.740 There have some number of properties, n properties, 00:07:47.740 --> 00:07:50.560 like color, the number of symbols, the shape. 00:07:50.560 --> 00:07:53.560 And you want to form a set being three 00:07:53.560 --> 00:07:56.030 cards such that every property, they're 00:07:56.030 --> 00:07:58.120 all same or all different. 00:07:58.120 --> 00:08:02.777 So that's exactly this model over here. 00:08:02.777 --> 00:08:04.360 So what can we say about this problem? 00:08:04.360 --> 00:08:06.520 What's the size of the maximum subset 00:08:06.520 --> 00:08:09.760 of F3 to the n-th without 3-AP. 00:08:09.760 --> 00:08:11.740 If you look at the proof that we did earlier 00:08:11.740 --> 00:08:14.440 in this course, the one using triangle removal lemma, 00:08:14.440 --> 00:08:16.522 you see the proof works verbatim. 00:08:16.522 --> 00:08:17.980 Previously, we worked over Z mod n. 00:08:17.980 --> 00:08:21.388 Now, you work over a different group, same proof. 00:08:21.388 --> 00:08:25.090 So triangle removal lemma, it tells you 00:08:25.090 --> 00:08:34.490 that this r3 is always little o of the size of the space. 00:08:34.490 --> 00:08:35.860 But we would like to do better. 00:08:35.860 --> 00:08:37.710 So this gives you something like log star. 00:08:37.710 --> 00:08:39.659 It's not very good dependence. 00:08:39.659 --> 00:08:41.820 So we would like to do better. 00:08:41.820 --> 00:08:45.165 So what we will show today, so this theorem 00:08:45.165 --> 00:08:47.020 is attributed to Meshulam. 00:08:51.010 --> 00:08:54.350 So in this case, the order of history is somewhat reversed. 00:08:54.350 --> 00:08:57.010 So we'll see the finite field toy model, 00:08:57.010 --> 00:09:00.020 but it historically actually came afterwards. 00:09:00.020 --> 00:09:03.950 But you'll see that the Fourier analytic proof that we'll 00:09:03.950 --> 00:09:08.510 see today, it's basically the same proof in the two settings. 00:09:08.510 --> 00:09:19.730 So F is r3, we will prove a bound, well, just like that. 00:09:19.730 --> 00:09:23.990 So much better than what you get from the regularity method. 00:09:27.470 --> 00:09:28.495 In terms of-- 00:09:28.495 --> 00:09:30.620 OK, so let me tell you a bit more about the history 00:09:30.620 --> 00:09:32.078 of this problem in terms of what we 00:09:32.078 --> 00:09:35.300 know in terms of upper bounds and lower bounds. 00:09:40.770 --> 00:09:42.920 So let me say more about F3. 00:09:45.920 --> 00:09:49.200 So what's the best that you might hope for? 00:09:49.200 --> 00:09:58.350 So the best lower bound is due to E del. 00:09:58.350 --> 00:10:01.890 And it's some construction, some very specific construction, 00:10:01.890 --> 00:10:04.590 which gives you a bound that's something 00:10:04.590 --> 00:10:08.870 like 2.21 to the n-th. 00:10:08.870 --> 00:10:11.470 And the upper bound is 3 to the n-th-- 00:10:11.470 --> 00:10:13.920 on the-- 3 minus little 1 to the n-th. 00:10:13.920 --> 00:10:25.750 So for a long time, it was open whether the answer 00:10:25.750 --> 00:10:29.080 should be basically roughly like 3 00:10:29.080 --> 00:10:33.690 to the n-th or some constant less than 3 to the n-th. 00:10:33.690 --> 00:10:36.960 And improvements on the upper bound were very slow, 00:10:36.960 --> 00:10:40.170 and/or some very difficult works that nudge that down 00:10:40.170 --> 00:10:42.120 below just a little bit. 00:10:42.120 --> 00:10:44.940 And then a couple years ago, a few years ago, there 00:10:44.940 --> 00:10:48.210 was this incredible breakthrough, 00:10:48.210 --> 00:10:52.770 where in this paper that was just a couple pages long, 00:10:52.770 --> 00:10:56.160 they managed to significantly improve the upper bound 00:10:56.160 --> 00:11:07.030 to basically 2.76 to the n-th. 00:11:07.030 --> 00:11:09.100 So this was an incredible breakthrough that 00:11:09.100 --> 00:11:11.020 happened just a few years ago. 00:11:11.020 --> 00:11:14.330 And we'll talk about this proof in a couple of lectures. 00:11:14.330 --> 00:11:17.710 It turns out this proof, which uses what's now 00:11:17.710 --> 00:11:19.300 called the polynomial method-- 00:11:22.030 --> 00:11:24.950 so not Fourier analytic, but a different method-- 00:11:24.950 --> 00:11:28.100 unfortunately does not seem to generalize 00:11:28.100 --> 00:11:30.580 to the original Roth's theorem. 00:11:30.580 --> 00:11:32.510 In fact, you shouldn't expect it to generalize 00:11:32.510 --> 00:11:35.610 in a straightforward way, because up there you 00:11:35.610 --> 00:11:38.630 we know that you do not have a power saving, whereas here you 00:11:38.630 --> 00:11:39.780 have a power saving. 00:11:39.780 --> 00:11:43.360 So the exponent goes down. 00:11:43.360 --> 00:11:45.690 OK, so this is roughly the history of this theorem. 00:11:49.150 --> 00:11:50.356 Any questions? 00:11:53.272 --> 00:11:55.216 AUDIENCE: Do we have to have [INAUDIBLE]?? 00:11:55.216 --> 00:11:56.070 YUFEI ZHAO: Yeah. 00:11:56.070 --> 00:12:01.860 So I'll-- So I can tell you this is known as a Croot-Lev-Pach. 00:12:01.860 --> 00:12:05.050 So I'll say more about it in a couple lectures. 00:12:05.050 --> 00:12:09.460 But this for F3 is due to Ellenberg and Gijswijt. 00:12:16.560 --> 00:12:18.780 So I'll tell you more about it in a couple lectures. 00:12:21.900 --> 00:12:25.860 What I want to focus on today is the Fourier analytic nature 00:12:25.860 --> 00:12:28.800 of the proof that gives you this bound up there, 00:12:28.800 --> 00:12:31.080 3 to the n over n. 00:12:31.080 --> 00:12:33.395 And it may seem like a completely different topic 00:12:33.395 --> 00:12:34.770 compared to what we've been doing 00:12:34.770 --> 00:12:37.650 so far in the course, which is more about graph theory. 00:12:37.650 --> 00:12:39.870 But I want you to think about what 00:12:39.870 --> 00:12:43.590 are the relationships between what we'll see today 00:12:43.590 --> 00:12:45.280 and what we've seen so far. 00:12:45.280 --> 00:12:46.970 And there are lots of connections. 00:12:46.970 --> 00:12:48.720 So even though the proof may superficially 00:12:48.720 --> 00:12:51.120 look quite different, many of these ideas 00:12:51.120 --> 00:12:54.570 about quasirandomness versus structure will come up. 00:12:54.570 --> 00:12:57.510 And I want to present the proof in a way that highlights 00:12:57.510 --> 00:13:00.580 the similarities between what we did previously 00:13:00.580 --> 00:13:02.320 and this Fourier analytic proof. 00:13:04.980 --> 00:13:07.790 So let's talk about the strategy. 00:13:10.390 --> 00:13:15.220 In the proof of the Szemerédi graph regularity lemma, 00:13:15.220 --> 00:13:17.920 we had the strategy that we called the energy increment 00:13:17.920 --> 00:13:19.590 strategy. 00:13:19.590 --> 00:13:22.012 So you start-- you want to find a good partition. 00:13:22.012 --> 00:13:23.220 You start doing partitioning. 00:13:23.220 --> 00:13:26.580 And you keep track of this thing called the energy-- 00:13:26.580 --> 00:13:30.700 must go up at every step, cannot go up forever, 00:13:30.700 --> 00:13:34.220 so has a bounded number of steps. 00:13:34.220 --> 00:13:38.140 This strategy for Roth's theorem is also an important strategy. 00:13:38.140 --> 00:13:40.270 It's a variant of energy increment, 00:13:40.270 --> 00:13:42.256 but now density increment. 00:13:53.930 --> 00:13:56.240 So we start with a set-- 00:13:56.240 --> 00:14:00.560 A subset of F3 to the n-th, and we 00:14:00.560 --> 00:14:02.420 would like to understand something 00:14:02.420 --> 00:14:07.160 about its structure versus pseudorandomness 00:14:07.160 --> 00:14:10.040 in a way that is similar to when we discussed 00:14:10.040 --> 00:14:13.670 the similar issue for graphs. 00:14:13.670 --> 00:14:16.280 In particular, there will be this dichotomy 00:14:16.280 --> 00:14:24.460 that if A is in some sense pseudorandom-- 00:14:24.460 --> 00:14:28.020 so earlier, we saw what it means for a graph to be pseudorandom. 00:14:28.020 --> 00:14:30.750 So now, what does it mean for a subset of F3 00:14:30.750 --> 00:14:33.030 to the n-th to be pseudorandom. 00:14:33.030 --> 00:14:34.440 So we'll address that today. 00:14:34.440 --> 00:14:37.010 If A pseudorandom-- 00:14:37.010 --> 00:14:43.470 OK, so the short answer is that it is Fourier uniform-- 00:14:43.470 --> 00:14:48.120 in other words, all Fourier coefficients small. 00:14:51.220 --> 00:14:53.470 That's what pseudorandom will refer to. 00:14:53.470 --> 00:14:56.720 So then there is a counting lemma. 00:15:01.720 --> 00:15:03.760 And the counting lemma will in particular 00:15:03.760 --> 00:15:07.160 imply that A has lots of 3-APs. 00:15:13.700 --> 00:15:16.880 So then you find your 3-AP. 00:15:16.880 --> 00:15:19.410 If this is not the case, then-- 00:15:19.410 --> 00:15:21.890 so what's the opposite of Fourier uniform-- 00:15:21.890 --> 00:15:25.890 is that A now has some large Fourier coefficient. 00:15:36.190 --> 00:15:39.250 And what we'll do is to use this Fourier 00:15:39.250 --> 00:15:47.080 coefficient to extract some codimension 1 affine subspace-- 00:15:51.960 --> 00:15:57.420 it's also called a hyperplane-- 00:15:57.420 --> 00:16:06.470 where the density of A goes up significantly, 00:16:06.470 --> 00:16:09.126 if you restrict to that sub hyperplane. 00:16:14.090 --> 00:16:15.500 And you can repeat this process. 00:16:15.500 --> 00:16:18.920 Now, restrict to this hyperplane and ask yourself 00:16:18.920 --> 00:16:20.660 the same question. 00:16:20.660 --> 00:16:25.220 Is A, when restricted to this hyperplane, pseudorandom? 00:16:25.220 --> 00:16:27.830 In which case, we find APs. 00:16:27.830 --> 00:16:31.230 Or is A restricted to this hyperplane, 00:16:31.230 --> 00:16:33.420 does it have a large Fourier coefficient? 00:16:33.420 --> 00:16:34.920 In which case, we restrict further. 00:16:38.400 --> 00:16:42.970 And each time you iterate, you obtain a density increment. 00:16:42.970 --> 00:16:50.130 And the density increment cannot go on forever, 00:16:50.130 --> 00:16:53.400 because your total density is at most 1. 00:16:57.890 --> 00:16:59.860 So the number of steps must be bounded. 00:17:02.590 --> 00:17:04.290 So that's the strategy. 00:17:04.290 --> 00:17:06.750 So this should remind you somewhat of the energy 00:17:06.750 --> 00:17:10.210 increment strategy from Szemerédi's regularity lemma, 00:17:10.210 --> 00:17:12.210 although there are some fundamental differences. 00:17:12.210 --> 00:17:13.790 We're not doing partitionings. 00:17:16.603 --> 00:17:18.020 Any questions about this strategy? 00:17:21.470 --> 00:17:23.810 OK. 00:17:23.810 --> 00:17:26.099 I want to tell you about Fourier analysis. 00:17:34.740 --> 00:17:37.460 So probably, all of you have seen some version 00:17:37.460 --> 00:17:40.410 of Fourier analysis, maybe in your calculus class 00:17:40.410 --> 00:17:43.140 with Fourier series and whatnot. 00:17:43.140 --> 00:17:45.690 So you play with formulas, and solve some differential 00:17:45.690 --> 00:17:46.930 equations. 00:17:46.930 --> 00:17:49.830 So I want to give you more than just a bunch 00:17:49.830 --> 00:17:53.110 of ways about handling Fourier coefficients, 00:17:53.110 --> 00:17:55.600 a way to think about Fourier analysis. 00:17:55.600 --> 00:17:58.920 So think of this as a crash course about Fourier analysis 00:17:58.920 --> 00:18:01.200 from the perspective of combinatorics. 00:18:04.002 --> 00:18:05.460 And Fourier analysis, I think, it's 00:18:05.460 --> 00:18:08.160 much easier if you work in a finite group, 00:18:08.160 --> 00:18:11.880 in a finite abelian group, which is what we're doing here. 00:18:11.880 --> 00:18:13.690 Many of the technicalities go away. 00:18:13.690 --> 00:18:17.960 So we'll be looking specifically at Fourier analysis in F3 00:18:17.960 --> 00:18:21.840 to the n-th, although the 3 can be any prime. 00:18:21.840 --> 00:18:24.270 So it's really the same. 00:18:24.270 --> 00:18:29.700 So the main actors in Fourier analysis 00:18:29.700 --> 00:18:31.293 are the Fourier characters. 00:18:35.370 --> 00:18:39.840 The Fourier characters are denoted gamma sub r. 00:18:39.840 --> 00:18:42.420 And they're characters on the group, 00:18:42.420 --> 00:18:45.030 meaning that they are maps which-- 00:18:45.030 --> 00:18:47.850 so they turn out, happen to be homomorphisms for the 00:18:47.850 --> 00:18:50.520 multiplicative group under-- 00:18:50.520 --> 00:18:53.760 so C under multiplication. 00:18:53.760 --> 00:19:02.850 And they're indexed by r, which also elements of F3 00:19:02.850 --> 00:19:03.510 to the n-th. 00:19:03.510 --> 00:19:05.670 So I'm going to be fairly concrete here. 00:19:05.670 --> 00:19:07.520 There are ways to do this more abstractly. 00:19:07.520 --> 00:19:08.880 But I'll be fairly concrete. 00:19:08.880 --> 00:19:12.670 So it's defined by gamma sub r evaluated 00:19:12.670 --> 00:19:19.470 on x equals to omega raised to r dot product x, where 00:19:19.470 --> 00:19:24.450 here omega is a third root of unity 00:19:24.450 --> 00:19:28.060 and the dot is a dot product. 00:19:33.050 --> 00:19:33.550 So-- 00:19:42.660 --> 00:19:45.090 So that's the definition of the Fourier transform-- 00:19:45.090 --> 00:19:47.400 sorry, that's the definition of the Fourier characters. 00:19:47.400 --> 00:19:49.140 And once you have the Fourier characters, 00:19:49.140 --> 00:19:56.830 you can have this Fourier transform, just defined 00:19:56.830 --> 00:19:58.010 as follows. 00:19:58.010 --> 00:20:00.086 If you start with a function-- 00:20:03.814 --> 00:20:07.540 let's say, a complex-valued function on your space-- 00:20:07.540 --> 00:20:13.550 then I define the Fourier transform 00:20:13.550 --> 00:20:20.080 to be another function, like that, defined 00:20:20.080 --> 00:20:22.180 by the following formula. 00:20:34.240 --> 00:20:37.178 So that's the formula for the Fourier transform. 00:20:40.770 --> 00:20:44.220 It is basically the inner product 00:20:44.220 --> 00:20:47.520 between F and the Fourier character. 00:20:47.520 --> 00:20:50.220 So let me make the comment here, I 00:20:50.220 --> 00:20:52.440 think this is actually a pretty important comment, 00:20:52.440 --> 00:20:54.003 about the normalization. 00:20:58.350 --> 00:21:03.390 Now, when you first learn Fourier transforms, usually 00:21:03.390 --> 00:21:07.350 in the reals, there are all these questions about what 00:21:07.350 --> 00:21:08.730 number to put in the exponent. 00:21:08.730 --> 00:21:09.660 Is it 2 pi? 00:21:09.660 --> 00:21:10.690 Is it root 2 pi? 00:21:10.690 --> 00:21:12.000 Is it some other thing? 00:21:12.000 --> 00:21:17.770 And somehow, one answer is better than the others. 00:21:17.770 --> 00:21:19.873 And the same thing is true here in groups. 00:21:19.873 --> 00:21:21.790 So we'll stick with the following convention-- 00:21:21.790 --> 00:21:23.957 and I want all of you to stick with this convention, 00:21:23.957 --> 00:21:26.320 otherwise, we'll confuse ourselves to no end-- 00:21:26.320 --> 00:21:27.680 is that for a finite group-- 00:21:34.940 --> 00:21:36.930 actually, let me, start of a board. 00:21:36.930 --> 00:21:43.788 So the convention is that, in a finite group, 00:21:43.788 --> 00:21:46.310 the Fourier transform is defined-- 00:21:46.310 --> 00:21:49.650 and more generally, anything you do in the physical space, 00:21:49.650 --> 00:21:56.320 we always use the averaging measure. 00:21:59.590 --> 00:22:02.660 Don't sum, always average in the physical space. 00:22:02.660 --> 00:22:10.040 And in the frequency space, always use sums, 00:22:10.040 --> 00:22:12.860 use the counting measure. 00:22:12.860 --> 00:22:14.458 Keep this in mind. 00:22:14.458 --> 00:22:16.250 Any of these questions about normalization, 00:22:16.250 --> 00:22:17.667 if you stick with this convention, 00:22:17.667 --> 00:22:19.545 things will become much easier. 00:22:19.545 --> 00:22:21.670 So there won't be any of these questions about when 00:22:21.670 --> 00:22:23.390 you take the inverse Fourier transform, 00:22:23.390 --> 00:22:25.750 do I put an extra factor in front or not, 00:22:25.750 --> 00:22:27.691 if you stick with this correct convention. 00:22:30.520 --> 00:22:33.580 So with that convention in mind, what the Fourier 00:22:33.580 --> 00:22:40.510 transform really is is inner product between F and a Fourier 00:22:40.510 --> 00:22:41.478 character. 00:22:49.810 --> 00:22:51.670 There are some important properties 00:22:51.670 --> 00:22:52.760 of the Fourier transform. 00:22:52.760 --> 00:22:54.760 So let me go through a few of the key properties 00:22:54.760 --> 00:22:55.540 that we'll need. 00:23:04.940 --> 00:23:07.970 The first one is pretty easy. 00:23:07.970 --> 00:23:14.250 What is the meaning of the 0-th Fourier coefficient? 00:23:14.250 --> 00:23:24.020 You plug it in, and you see that it is just the average of F. 00:23:24.020 --> 00:23:27.800 So 0-th coefficient is the average of F. 00:23:27.800 --> 00:23:32.270 The second fact goes under one of two names, 00:23:32.270 --> 00:23:34.870 and they're often used interchangeably-- 00:23:34.870 --> 00:23:36.110 Plancherel or Parseval. 00:23:41.340 --> 00:23:48.560 And it says that if you look at the inner product 00:23:48.560 --> 00:23:57.080 in the physical space, then this product 00:23:57.080 --> 00:24:02.216 is preserved, if you take the Fourier transform. 00:24:05.617 --> 00:24:07.700 But now, of course, you're in the frequency space, 00:24:07.700 --> 00:24:10.070 so you should sum instead of doing the inner product. 00:24:15.180 --> 00:24:20.490 So this identity can be proved in a fairly straightforward way 00:24:20.490 --> 00:24:24.360 by plugging in what the definition is for the Fourier 00:24:24.360 --> 00:24:25.418 transform. 00:24:25.418 --> 00:24:26.960 This is a straightforward computation 00:24:26.960 --> 00:24:27.850 I'm not going to do on the board, 00:24:27.850 --> 00:24:29.970 but I highly encourage you to actually do at home, 00:24:29.970 --> 00:24:33.685 just to do it once to make sure you understand how it goes. 00:24:33.685 --> 00:24:35.310 But there is also a more conceptual way 00:24:35.310 --> 00:24:38.410 to understand this identity. 00:24:38.410 --> 00:24:41.490 And that's because-- now, this is also 00:24:41.490 --> 00:24:44.670 important to understand what the Fourier transform is. 00:24:44.670 --> 00:24:46.980 It's not just some magical formula somebody wrote down, 00:24:46.980 --> 00:24:50.080 like this is a very natural operation. 00:24:50.080 --> 00:24:59.280 It's because the characters, the set of characters, 00:24:59.280 --> 00:25:02.620 is an orthonormal basis. 00:25:02.620 --> 00:25:10.815 So the Fourier characters form an orthonormal basis. 00:25:15.750 --> 00:25:18.420 As a result, what the Fourier transform 00:25:18.420 --> 00:25:22.310 is is a unitary change of basis. 00:25:35.510 --> 00:25:36.320 You can check. 00:25:36.320 --> 00:25:37.737 It's very straightforward to check 00:25:37.737 --> 00:25:39.800 that the Fourier characters, indeed, form 00:25:39.800 --> 00:25:45.210 a orthonormal basis, because-- 00:25:45.210 --> 00:25:51.650 well, you can evaluate the inner product between two Fourier 00:25:51.650 --> 00:25:53.480 characters. 00:25:53.480 --> 00:25:55.730 So remember, in the physical space, 00:25:55.730 --> 00:25:57.786 we're always doing averaging. 00:26:01.050 --> 00:26:06.240 And so now, I'll just write down first 00:26:06.240 --> 00:26:08.310 what I mean by the inner product. 00:26:11.100 --> 00:26:12.920 So that's the inner product. 00:26:12.920 --> 00:26:21.977 And by the definition of the Fourier character, 00:26:21.977 --> 00:26:22.560 you have that. 00:26:25.410 --> 00:26:29.010 So think about what this expectation is-- 00:26:29.010 --> 00:26:37.470 unless r equals to s, in which case, this expectation is 1. 00:26:37.470 --> 00:26:39.240 Unless that is the case, you always 00:26:39.240 --> 00:26:43.360 have some coordinate of x in the exponent. 00:26:43.360 --> 00:26:46.230 So as you average over all possibilities, 00:26:46.230 --> 00:26:47.270 they average out to 0. 00:26:58.020 --> 00:27:01.050 So this calculation shows you that the Fourier characters 00:27:01.050 --> 00:27:02.880 form a orthonormal basis. 00:27:02.880 --> 00:27:05.910 And a basic fact you know from linear algebra 00:27:05.910 --> 00:27:07.470 is that if you do a change of basis, 00:27:07.470 --> 00:27:10.510 if you do a unitary change of basis, 00:27:10.510 --> 00:27:13.640 then inner product is preserved. 00:27:13.640 --> 00:27:15.430 It's like a rotation. 00:27:15.430 --> 00:27:18.150 It's the same-- so you're not changing the inner product. 00:27:18.150 --> 00:27:19.660 So the inner product is preserved 00:27:19.660 --> 00:27:21.640 under this change of basis. 00:27:21.640 --> 00:27:24.280 And that's why Plancherel is true. 00:27:27.780 --> 00:27:30.870 Another important thing is what's 00:27:30.870 --> 00:27:34.684 known as the Fourier inversion formula. 00:27:34.684 --> 00:27:39.720 The Fourier transform tells you how to go from a function 00:27:39.720 --> 00:27:42.730 to the Fourier transform. 00:27:42.730 --> 00:27:46.020 Well, now, if you are given the Fourier transform, 00:27:46.020 --> 00:27:48.090 how do you go back? 00:27:48.090 --> 00:27:50.040 There's a formula which tells you 00:27:50.040 --> 00:28:01.350 that you can go back by the following formula there. 00:28:01.350 --> 00:28:04.370 So that's the Fourier inversion formula. 00:28:04.370 --> 00:28:07.110 It allows you to do this inversion. 00:28:07.110 --> 00:28:09.540 And again, it's one of these formulas where 00:28:09.540 --> 00:28:12.990 I encourage you to try it out yourself by plugging 00:28:12.990 --> 00:28:16.770 in the formula and expanding. 00:28:16.770 --> 00:28:19.660 And it's pretty easy to check. 00:28:19.660 --> 00:28:22.090 It's much easier in the finite field setting, by the way. 00:28:22.090 --> 00:28:25.630 So if you use the usual Fourier transform on the real line, 00:28:25.630 --> 00:28:27.070 there are some technicalities even 00:28:27.070 --> 00:28:28.690 to prove the Fourier inversion. 00:28:28.690 --> 00:28:31.070 But in finite groups, it's almost trivial. 00:28:31.070 --> 00:28:33.170 You expand, and then you'll see. 00:28:33.170 --> 00:28:35.025 So it's very easy to prove. 00:28:35.025 --> 00:28:37.150 But you can also see this Fourier inversion formula 00:28:37.150 --> 00:28:39.040 more conceptually, because you're 00:28:39.040 --> 00:28:41.550 in a unitary change of basis. 00:28:41.550 --> 00:28:45.300 So to go back, well, think about what it means in linear algebra 00:28:45.300 --> 00:28:48.990 to revert a unitary transformation. 00:28:48.990 --> 00:28:52.530 You simply multiply the coefficients 00:28:52.530 --> 00:28:56.490 with the coordinates. 00:28:56.490 --> 00:29:00.335 Orthogonal, orthonormal change of basis. 00:29:00.335 --> 00:29:06.430 Finally, Fourier transform behaves while 00:29:06.430 --> 00:29:07.900 under convolution. 00:29:07.900 --> 00:29:15.430 So by convolution, we define the convolution 00:29:15.430 --> 00:29:19.320 of two functions, f and g, using the following formula. 00:29:27.510 --> 00:29:38.090 And so then the claim is that the Fourier transform behaves 00:29:38.090 --> 00:29:39.800 very well under convolution. 00:29:39.800 --> 00:29:44.250 It's basically multiplicative under convolution. 00:29:44.250 --> 00:29:45.860 So what this means is, if I put in-- 00:29:50.490 --> 00:29:54.670 so it's pointwise true everywhere. 00:29:54.670 --> 00:29:59.590 Again, very easy proof, because I just evaluate the left hand 00:29:59.590 --> 00:30:04.850 side, see what-- 00:30:04.850 --> 00:30:07.730 plug in the formula for the Fourier transform. 00:30:07.730 --> 00:30:16.170 And I find it's that. 00:30:16.170 --> 00:30:19.268 And now, I plug in the formula for convolution. 00:30:40.450 --> 00:30:45.140 So now, you can do a change of variables. 00:30:45.140 --> 00:30:48.865 And then you-- it's not hard to see. 00:30:48.865 --> 00:30:50.740 You eventually end up at the right hand side. 00:30:54.760 --> 00:30:56.260 So these are some of the properties. 00:30:56.260 --> 00:30:58.658 So there are important properties 00:30:58.658 --> 00:30:59.700 of the Fourier transform. 00:30:59.700 --> 00:31:01.743 So this is something that, whenever 00:31:01.743 --> 00:31:03.160 you learn about Fourier transform, 00:31:03.160 --> 00:31:06.720 you always see these few properties. 00:31:06.720 --> 00:31:08.230 And so we'll use them. 00:31:08.230 --> 00:31:10.690 But we'll also need another property 00:31:10.690 --> 00:31:14.440 that is specific to the analysis of 3-term arithmetic 00:31:14.440 --> 00:31:15.410 progressions. 00:31:17.990 --> 00:31:22.165 So what does Fourier transform have to do with 3-APs? 00:31:22.165 --> 00:31:24.040 So we want to use it to prove Roth's theorem. 00:31:24.040 --> 00:31:27.400 So we better have some tool that allows 00:31:27.400 --> 00:31:29.330 us to analyze the number 3-APs. 00:31:29.330 --> 00:31:45.250 And here is a key identity relating Fourier with 3-APs. 00:31:45.250 --> 00:31:56.410 And it's that, if you have three functions, 00:31:56.410 --> 00:32:02.920 then the following quantity, which relates the number 00:32:02.920 --> 00:32:06.840 of 3-APs-- 00:32:14.880 --> 00:32:18.720 So this function basically counts the number 3-APs, 00:32:18.720 --> 00:32:21.500 if your f, g, and h are indicator functions of a set. 00:32:24.240 --> 00:32:27.960 I want to express this formula in terms of the Fourier 00:32:27.960 --> 00:32:29.835 transforms of these functions. 00:32:32.550 --> 00:32:35.880 The formula turns out to be fairly simple, 00:32:35.880 --> 00:32:45.970 that it is simply that. 00:32:45.970 --> 00:32:50.190 So it's a single sum over the r's of f hat 00:32:50.190 --> 00:32:55.240 of r, g hat of minus 2r, and h hat of r. 00:32:55.240 --> 00:32:57.880 You might wonder why I put a minus 2 here, because minus 2 00:32:57.880 --> 00:33:00.520 is r, and it looks certainly much nicer 00:33:00.520 --> 00:33:02.212 with just r in there. 00:33:02.212 --> 00:33:05.080 And that is true. 00:33:05.080 --> 00:33:08.170 This formula as written is true for over any group. 00:33:13.940 --> 00:33:16.340 And our proof will show it. 00:33:16.340 --> 00:33:18.560 So it's not really about F3 at all, but any group. 00:33:25.300 --> 00:33:29.880 So let me prove this for you in a couple of different ways. 00:33:29.880 --> 00:33:39.520 So the first proof is basically a straightforward no 00:33:39.520 --> 00:33:46.240 thinking involved proof, as in we apply these formula 00:33:46.240 --> 00:33:48.250 for using either Fourier inversion 00:33:48.250 --> 00:33:53.820 or the inverse Fourier transform, and plug it in, 00:33:53.820 --> 00:33:56.545 and expand, and check. 00:33:56.545 --> 00:33:58.170 So it's worth doing at least this once. 00:33:58.170 --> 00:34:01.980 So let's do this together at least once. 00:34:01.980 --> 00:34:04.420 But this something that is a fairly straightforward 00:34:04.420 --> 00:34:05.580 computation. 00:34:05.580 --> 00:34:10.900 The left hand side can be expanded 00:34:10.900 --> 00:34:12.340 using Fourier inversion. 00:34:14.920 --> 00:34:21.389 so r1 f hat r1 omega to the minus r1 00:34:21.389 --> 00:34:32.280 dot x, and sum over r2 g hat r2 omega to the minus r2 00:34:32.280 --> 00:34:38.199 dot x plus y, and then finally sum 00:34:38.199 --> 00:34:47.550 over r3 h hat of r3 omega to the minus r3 dot x plus 2y. 00:34:50.989 --> 00:34:54.770 So I'm using Fourier inversion, replace f, g, and h 00:34:54.770 --> 00:34:58.800 by their Fourier transforms. 00:34:58.800 --> 00:35:02.280 Oh, sorry, there should be-- 00:35:02.280 --> 00:35:03.800 yeah, so no minus. 00:35:06.960 --> 00:35:10.620 So now, we exchange sums and expectations, 00:35:10.620 --> 00:35:18.265 do a switch in the order of summation, so r1, r2, r3 and f 00:35:18.265 --> 00:35:27.050 hat of r1 g hat of r2 h hat of r3. 00:35:27.050 --> 00:35:33.860 And you have this expectation over x and y and omega 00:35:33.860 --> 00:35:37.970 of x dot r1 plus r2 plus r3. 00:35:40.610 --> 00:35:45.340 In fact, I can even write the x and y separately, 00:35:45.340 --> 00:35:57.180 y omega to the y dot r2 plus 2r3. 00:35:57.180 --> 00:35:59.540 So just rearranging. 00:35:59.540 --> 00:36:01.570 And now, you see that as you take 00:36:01.570 --> 00:36:09.110 expectation over x, this expectation is equal to 1, 00:36:09.110 --> 00:36:19.130 if r1 plus r2 plus r3 is equal to 0, and 0 otherwise. 00:36:19.130 --> 00:36:23.930 And likewise, the third expectation is either 1 or 0, 00:36:23.930 --> 00:36:28.610 depending on the sums of r2 and r3. 00:36:33.650 --> 00:36:36.760 So the only terms that remain, after you take out 00:36:36.760 --> 00:36:43.920 these 0's, are cases where both these two equations 00:36:43.920 --> 00:36:44.730 are satisfied. 00:36:47.350 --> 00:36:55.150 And then you see that the only remaining terms are basically 00:36:55.150 --> 00:37:02.410 the ones given in the sum on the right hand side. 00:37:05.410 --> 00:37:06.000 OK? 00:37:06.000 --> 00:37:07.930 So that's the proof. 00:37:07.930 --> 00:37:10.363 Pretty straightforward, you plug in Fourier inversion. 00:37:13.540 --> 00:37:17.073 I want to show you a different proof that 00:37:17.073 --> 00:37:19.240 hopefully will be more familiar and more conceptual. 00:37:19.240 --> 00:37:21.657 Now, it doesn't involve carrying through this calculation, 00:37:21.657 --> 00:37:26.380 even though this is not at all hard calculation. 00:37:26.380 --> 00:37:31.610 But first, let me rewrite the formula up there. 00:37:31.610 --> 00:37:37.000 So in F3, it will be convenient, and so 00:37:37.000 --> 00:37:39.190 the formula is actually slightly easier 00:37:39.190 --> 00:37:42.700 to interpreting F3, in F3, the identity 00:37:42.700 --> 00:37:47.110 says that, if you look at the quantity-- 00:37:52.460 --> 00:37:53.020 I need. 00:37:58.732 --> 00:38:04.293 So let me give you a second proof that works just in F3, 00:38:04.293 --> 00:38:06.210 but you can modify it to work in other groups. 00:38:06.210 --> 00:38:08.950 But in F3, it's particularly nice. 00:38:08.950 --> 00:38:14.240 The left hand side, you see, the left hand side, 00:38:14.240 --> 00:38:21.980 I can rewrite it as the following form, 00:38:21.980 --> 00:38:24.840 where I sum over-- 00:38:24.840 --> 00:38:29.430 well, I take expectation over all triples x, y, z that sum 00:38:29.430 --> 00:38:29.930 to 0. 00:38:32.600 --> 00:38:36.470 Because a 3-AP is the same as three elements, 00:38:36.470 --> 00:38:38.720 three points in the vector space summing to 0. 00:38:41.830 --> 00:38:46.000 But now, you see that this quantity 00:38:46.000 --> 00:38:54.310 is the same as the convolution evaluated at 0-- 00:38:57.640 --> 00:39:00.610 so if you extend the definition of convolution to more than one 00:39:00.610 --> 00:39:03.630 func-- more than two functions. 00:39:03.630 --> 00:39:05.540 But now, we apply Fourier inversion. 00:39:09.650 --> 00:39:11.540 And we find that-- 00:39:21.780 --> 00:39:22.660 OK. 00:39:22.660 --> 00:39:26.940 So by Fourier inversion, you have that. 00:39:26.940 --> 00:39:30.700 But now, by the identity that relates the Fourier transform 00:39:30.700 --> 00:39:41.560 and inversion, you have that. 00:39:41.560 --> 00:39:45.410 And that's the proof, because minus 2 r is the same as r. 00:39:48.470 --> 00:39:52.100 So it's shorter, because we're using some properties here 00:39:52.100 --> 00:39:54.310 about convolution and-- 00:39:54.310 --> 00:39:56.229 yeah, so about the convolution. 00:40:03.720 --> 00:40:05.580 That formula up there is, of course, 00:40:05.580 --> 00:40:07.530 related to counting 3-APs. 00:40:07.530 --> 00:40:22.080 Because if f, g, and h are all indicators of some set, 00:40:22.080 --> 00:40:25.710 then the left hand side is the same 00:40:25.710 --> 00:40:35.535 as basically the number of triples of elements in A 00:40:35.535 --> 00:40:38.940 whose sum is equal to 0. 00:40:42.210 --> 00:40:46.680 And the right hand side is the sum 00:40:46.680 --> 00:40:52.184 of the third power of the Fourier coefficients. 00:40:56.360 --> 00:40:59.980 And this formula should look somewhat familiar, 00:40:59.980 --> 00:41:03.940 because we also used this kind of formula 00:41:03.940 --> 00:41:07.610 back when we discussed spectral graph theory. 00:41:07.610 --> 00:41:11.350 And remember, the third moment of the eigenvalues 00:41:11.350 --> 00:41:16.330 is the trace of the third power, which counts closed 00:41:16.330 --> 00:41:18.880 walks in Cayley graph. 00:41:18.880 --> 00:41:21.100 So this is actually the same formula. 00:41:21.100 --> 00:41:26.210 So in the case if A is symmetric, 00:41:26.210 --> 00:41:34.460 let's say, then this is the same as the formula 00:41:34.460 --> 00:41:47.530 that counts closed walks of length three 00:41:47.530 --> 00:41:48.660 in the Cayley graph. 00:41:53.945 --> 00:41:55.820 The point of this comment is just to tell you 00:41:55.820 --> 00:41:57.440 that Fourier transform is somehow 00:41:57.440 --> 00:42:02.160 it's not this brand new concept that we've never seen before. 00:42:02.160 --> 00:42:04.400 It is intimately tied to many of the things 00:42:04.400 --> 00:42:08.866 that we have seen earlier in this course but in disguise. 00:42:08.866 --> 00:42:11.390 So it is related to the spectral graph theory 00:42:11.390 --> 00:42:13.640 that we discussed at length earlier in this course. 00:42:21.290 --> 00:42:23.080 Now that we have the Fourier transform, 00:42:23.080 --> 00:42:28.420 I want to develop some machinery to prove 00:42:28.420 --> 00:42:32.930 Roth's theorem following the strategy up there. 00:42:32.930 --> 00:42:34.100 So let's take a quick break. 00:42:34.100 --> 00:42:36.392 And then when we come back, we'll prove Roth's theorem. 00:42:40.570 --> 00:42:41.500 Any questions so far? 00:42:45.204 --> 00:42:50.596 AUDIENCE: So for this one, you said-- 00:42:50.596 --> 00:42:53.150 is it like we only used the fact that it's 00:42:53.150 --> 00:42:57.255 F3 to the n-th at the end of the proof, like in that last step? 00:42:57.255 --> 00:42:57.880 YUFEI ZHAO: OK. 00:42:57.880 --> 00:43:01.040 So question is, where do we use that in F3 to the n-th? 00:43:01.040 --> 00:43:04.970 This formula here holds in every finite abelian group, 00:43:04.970 --> 00:43:07.190 if you use the correct definition of Fourier 00:43:07.190 --> 00:43:10.240 transform with the averaging normalization. 00:43:10.240 --> 00:43:15.020 So in the other formula where you replace minus 2 by 1, 00:43:15.020 --> 00:43:17.890 that requires F3. 00:43:17.890 --> 00:43:18.850 But you can-- 00:43:18.850 --> 00:43:22.280 I mean, you can follow the proof and come up 00:43:22.280 --> 00:43:27.080 with a similar formula for every equation. 00:43:27.080 --> 00:43:28.580 So there's a general principle here, 00:43:28.580 --> 00:43:30.600 which I'll discuss more at length 00:43:30.600 --> 00:43:34.120 in a bit, that for patterns that are 00:43:34.120 --> 00:43:36.760 governed by a single equation-- 00:43:36.760 --> 00:43:41.260 in this case, 3-APs, x minus 2y plus z equal to 0-- 00:43:41.260 --> 00:43:43.900 patterns that can be governed by a single equation 00:43:43.900 --> 00:43:46.196 can be controlled by a Fourier transform. 00:43:49.770 --> 00:43:53.210 So let's begin our proof, the Fourier analytic proof 00:43:53.210 --> 00:43:55.636 of Roth's theorem in F3 to the n-th. 00:43:55.636 --> 00:43:57.695 AUDIENCE: So at the end, you said 00:43:57.695 --> 00:43:59.445 it was going to be connected with counting 00:43:59.445 --> 00:44:00.362 the [INAUDIBLE] graph. 00:44:00.362 --> 00:44:03.045 Does this mean that the Fourier transform 00:44:03.045 --> 00:44:06.982 of the indicator of A, those are exactly the eigenvalues 00:44:06.982 --> 00:44:08.304 of the Cayley graph? 00:44:08.304 --> 00:44:09.387 Or is it like [INAUDIBLE]? 00:44:09.387 --> 00:44:12.050 YUFEI ZHAO: OK, so you're asking about the final step, where 00:44:12.050 --> 00:44:13.505 we're talking about-- 00:44:13.505 --> 00:44:15.380 so I mentioned that there was this connection 00:44:15.380 --> 00:44:18.170 between counting walks in graphs and spectral graph theory. 00:44:18.170 --> 00:44:27.230 So you can check that, if you have a subset A of an abelian 00:44:27.230 --> 00:44:31.840 group, then the Fourier transforms of A 00:44:31.840 --> 00:44:39.180 are exactly the eigenvalues of the Cayley graph. 00:44:44.090 --> 00:44:46.350 AUDIENCE: So then I guess, have we 00:44:46.350 --> 00:44:48.810 done anything so far that could have been done 00:44:48.810 --> 00:44:50.950 in a spectral way yet? 00:44:50.950 --> 00:44:54.190 Well, I guess, where is the Fourier analysis better 00:44:54.190 --> 00:44:55.555 than the spectral [INAUDIBLE]? 00:44:55.555 --> 00:44:56.280 YUFEI ZHAO: OK. 00:44:56.280 --> 00:44:57.950 Question, where is the Fourier analysis 00:44:57.950 --> 00:44:59.130 better than the spectral posts? 00:44:59.130 --> 00:45:00.170 Well, let's see the proof first. 00:45:00.170 --> 00:45:01.520 And then you'll see, yeah. 00:45:01.520 --> 00:45:03.470 So there's no graphs anymore. 00:45:03.470 --> 00:45:05.535 So we're going to work inside F3 to the n-th. 00:45:08.270 --> 00:45:10.528 But just like the proof of regularity in counting, 00:45:10.528 --> 00:45:12.070 we're going to have a counting lemma. 00:45:12.070 --> 00:45:14.240 So all of these are analytic. 00:45:14.240 --> 00:45:19.243 And at this point, they should be very familiar to you. 00:45:19.243 --> 00:45:20.660 They may come in a different form. 00:45:20.660 --> 00:45:22.790 They may be dressed in different clothing. 00:45:22.790 --> 00:45:24.738 But it's still a counting lemma. 00:45:24.738 --> 00:45:25.280 So let's see. 00:45:34.360 --> 00:45:36.190 The counting lemma, in this case, 00:45:36.190 --> 00:45:44.170 says that if you are in the setting of A F3 to the n-th-- 00:45:44.170 --> 00:45:47.470 and I'm going to throughout write 00:45:47.470 --> 00:45:56.460 the density of A as alpha, then let me write, 00:45:56.460 --> 00:46:06.420 let me define this lambda 3 of A to be the function which 00:46:06.420 --> 00:46:17.470 basically counts 3-APs in A but with the averaging 00:46:17.470 --> 00:46:18.762 normalization. 00:46:21.660 --> 00:46:23.590 So this is-- we saw this earlier. 00:46:26.550 --> 00:46:33.550 So the counting lemma says that this normalized number of 3-APs 00:46:33.550 --> 00:46:34.900 in A-- 00:46:34.900 --> 00:46:37.970 so including trivial 3-APs; that's why this is a nice 00:46:37.970 --> 00:46:39.650 analytic expression-- 00:46:39.650 --> 00:46:46.490 differs from what you might guess based on density alone. 00:46:46.490 --> 00:46:50.960 This difference should be small if all the nonzero Fourier 00:46:50.960 --> 00:46:53.920 coefficients of A are small. 00:47:04.190 --> 00:47:07.330 So in this strategy, I said that if-- so the counting 00:47:07.330 --> 00:47:10.330 lemma tells you, if A is Fourier uniform, 00:47:10.330 --> 00:47:11.550 then it is pseudorandom. 00:47:11.550 --> 00:47:13.630 And this is where it comes in. 00:47:13.630 --> 00:47:16.940 If A has all small Fourier coefficients, 00:47:16.940 --> 00:47:19.890 then you have a counting lemma, which tells you 00:47:19.890 --> 00:47:21.810 that the counts of A should not be 00:47:21.810 --> 00:47:25.980 so different from the guess based on density. 00:47:34.290 --> 00:47:36.880 So the proof is very short. 00:47:36.880 --> 00:47:41.310 It's based on the identity that we saw earlier. 00:47:41.310 --> 00:47:47.040 The 3-AP count of A by the identity earlier 00:47:47.040 --> 00:47:51.890 is simply the third power of the Fourier transforms. 00:47:55.180 --> 00:47:58.585 And all of these calculations should be reminiscent, 00:47:58.585 --> 00:48:00.460 because we've done these kind of calculations 00:48:00.460 --> 00:48:03.460 in some form or another earlier in this course. 00:48:03.460 --> 00:48:06.370 So we're going to separate out the main term 00:48:06.370 --> 00:48:07.690 and subsequent terms. 00:48:07.690 --> 00:48:13.750 So the main term is the one corresponding to r equals to 0. 00:48:13.750 --> 00:48:15.970 So that's the density. 00:48:15.970 --> 00:48:18.640 And all the other terms I'm going 00:48:18.640 --> 00:48:24.310 to lump together into this sum. 00:48:24.310 --> 00:48:33.610 So we now know that the difference 00:48:33.610 --> 00:48:36.850 we're trying to bound is upper bounded 00:48:36.850 --> 00:48:47.850 by the third moment of the absolute values of the Fourier 00:48:47.850 --> 00:48:50.730 transform. 00:48:50.730 --> 00:48:53.040 And I want to upper bound this quantity here, 00:48:53.040 --> 00:48:58.320 assuming that all of these Fourier coefficients are small. 00:48:58.320 --> 00:49:00.540 We've also done this kind of calculations before. 00:49:00.540 --> 00:49:01.957 So where have we seen this before? 00:49:06.310 --> 00:49:08.240 We saw this calculation earlier in the class 00:49:08.240 --> 00:49:09.700 with the 3 replaced by a 4. 00:49:15.640 --> 00:49:18.160 So in counting four cycles in a proof equivalent 00:49:18.160 --> 00:49:21.940 of quasirandomness, we said that, if all the eigenvalues 00:49:21.940 --> 00:49:24.010 other than the top one are small, then 00:49:24.010 --> 00:49:26.980 you can count four cycles. 00:49:26.980 --> 00:49:28.320 It's the same proof. 00:49:28.320 --> 00:49:30.370 And remember, in that proof, there 00:49:30.370 --> 00:49:34.510 was a important trick, where you do not uniformly 00:49:34.510 --> 00:49:38.470 bound each term by the max, because then you lose. 00:49:38.470 --> 00:49:40.900 You lose by an extra factor of n that you don't want. 00:49:40.900 --> 00:49:45.100 So you only take out one factor. 00:49:45.100 --> 00:49:47.260 So you take out one factor. 00:49:53.840 --> 00:49:56.900 And you keep the rest in there. 00:49:56.900 --> 00:49:59.980 In fact, I can be more generous and even 00:49:59.980 --> 00:50:02.860 throw the r equal to 0 term back in. 00:50:06.460 --> 00:50:10.950 And now, by Plancherel-- 00:50:10.950 --> 00:50:15.290 so by Plancherel/Parseval, this here 00:50:15.290 --> 00:50:22.740 is equal to the expectation of the indicator 00:50:22.740 --> 00:50:23.910 function of A squared. 00:50:23.910 --> 00:50:25.775 So you take this-- 00:50:25.775 --> 00:50:30.170 you go back to physical space, and that's simply the density. 00:50:30.170 --> 00:50:33.710 So then that proves the theorem. 00:50:39.175 --> 00:50:41.550 So the moral of the counting lemma is the same as the one 00:50:41.550 --> 00:50:43.550 that we've seen before when we discussed graphs. 00:50:43.550 --> 00:50:46.440 If you're in pseudorandom, then you have good counting. 00:50:46.440 --> 00:50:49.890 And here, pseudorandom means having small Fourier 00:50:49.890 --> 00:50:55.040 coefficients, uniformly small Fourier coefficients. 00:50:55.040 --> 00:50:57.890 So now, let's begin the proof of Roth's theorem. 00:50:57.890 --> 00:51:01.858 The Roth's theorem proof will have three steps. 00:51:01.858 --> 00:51:08.455 The first step, we will observe that if you're 00:51:08.455 --> 00:51:14.700 in the 3-AP-free set, then there exists a large Fourier 00:51:14.700 --> 00:51:15.731 coefficient. 00:51:26.050 --> 00:51:28.990 Throughout, I'm going to use uppercase N to denote 00:51:28.990 --> 00:51:32.620 the size of the ambient group. 00:51:32.620 --> 00:51:36.560 And specifically, we will prove the following. 00:51:36.560 --> 00:51:41.320 And also throughout, A is a subset of F3 to the n-th 00:51:41.320 --> 00:51:45.075 and with density alpha. 00:51:45.075 --> 00:51:47.420 So I'll keep this convention throughout this proof. 00:51:50.380 --> 00:52:05.230 We will show that if A is 3-AP-free and N is at least 2 00:52:05.230 --> 00:52:09.030 to the alpha to the minus 2-- 00:52:09.030 --> 00:52:11.680 so N is at least somewhat large-- 00:52:11.680 --> 00:52:16.750 then there exists a nonzero r such 00:52:16.750 --> 00:52:23.230 that the r-th Fourier coefficient is at least alpha 00:52:23.230 --> 00:52:24.520 squared over 2. 00:52:27.450 --> 00:52:30.900 If you're 3-AP-free, free then provided 00:52:30.900 --> 00:52:34.290 that you're working in a large enough ambient space, 00:52:34.290 --> 00:52:37.763 you always have some large Fourier coefficient. 00:52:40.601 --> 00:52:43.720 So the proof is essentially-- well, 00:52:43.720 --> 00:52:50.213 this claim is essentially a corollary of counting lemma, 00:52:50.213 --> 00:52:51.130 by the counting lemma. 00:52:54.820 --> 00:52:57.860 And using the fact that in a 3-AP-free 00:52:57.860 --> 00:53:02.611 set, what is the quantity lambda 3 of A? 00:53:02.611 --> 00:53:08.170 Up there, you only have the trivial 3-APs present. 00:53:08.170 --> 00:53:11.590 So this quantity lambda must then 00:53:11.590 --> 00:53:17.770 be the size of A divided by N squared or alpha over N, which 00:53:17.770 --> 00:53:22.100 are precisely counting the trivial 3-APs. 00:53:26.040 --> 00:53:28.220 So by the counting lemma, then we 00:53:28.220 --> 00:53:32.540 have that the upper bound on the right hand 00:53:32.540 --> 00:53:43.120 side, which we now write alpha max over nonzero r's, is 00:53:43.120 --> 00:53:49.840 at least alpha cubed minus the lambda 3 term, which 00:53:49.840 --> 00:53:55.370 should be alpha over N. So provided 00:53:55.370 --> 00:53:57.380 that N is large enough-- 00:53:57.380 --> 00:53:58.890 big N is large enough-- 00:53:58.890 --> 00:54:02.450 then the trivial 3-APs should not contribute very much. 00:54:02.450 --> 00:54:06.770 So I can lower bound the right hand side by, let's say, 00:54:06.770 --> 00:54:07.570 alpha 3-- 00:54:07.570 --> 00:54:10.020 alpha cubed over 2. 00:54:10.020 --> 00:54:13.945 So then you deduce the conclusion. 00:54:13.945 --> 00:54:16.070 I want you to think about how this proof is related 00:54:16.070 --> 00:54:19.270 to Szemerédi's graph regularity lemma. 00:54:19.270 --> 00:54:21.020 The analogy will break down at some point. 00:54:21.020 --> 00:54:23.780 But we've seen this step before as well. 00:54:23.780 --> 00:54:30.050 From lack of 3-APs, you extract some useful information 00:54:30.050 --> 00:54:32.980 and from this will extract some structure. 00:54:32.980 --> 00:54:34.430 And the structure here-- 00:54:34.430 --> 00:54:36.590 and this is where the proof now diverges 00:54:36.590 --> 00:54:40.190 from that of regularity-- 00:54:40.190 --> 00:54:46.540 having a large Fourier coefficient 00:54:46.540 --> 00:54:56.946 will now imply a density increment on a hyperplane. 00:55:04.730 --> 00:55:06.950 Specifically, if you have-- 00:55:11.390 --> 00:55:15.070 so keeping the same convention as before, 00:55:15.070 --> 00:55:21.100 if the Fourier coefficient of A at r 00:55:21.100 --> 00:55:29.310 is at least delta for some nonzero r, 00:55:29.310 --> 00:55:42.160 then A has density at least alpha plus delta over 2 00:55:42.160 --> 00:55:44.460 when restricted to a hyperplane. 00:55:55.280 --> 00:55:57.640 So if you have a large Fourier coefficient, 00:55:57.640 --> 00:56:00.730 then I can pass down to a smaller part of the space 00:56:00.730 --> 00:56:03.518 where the density of A goes up significantly. 00:56:10.210 --> 00:56:11.980 To see why this is true, let's go back 00:56:11.980 --> 00:56:15.910 to the definition of the Fourier coefficient, the Fourier 00:56:15.910 --> 00:56:16.510 transform. 00:56:19.120 --> 00:56:25.040 So recall that Fourier transform is given by the following 00:56:25.040 --> 00:56:32.030 formula, where I'm looking at this expectation over points 00:56:32.030 --> 00:56:39.610 in F3 to the n-th together with indicator of A multiplied 00:56:39.610 --> 00:56:43.840 by this Fourier character. 00:56:43.840 --> 00:56:49.840 And you see that this function here, it 00:56:49.840 --> 00:57:01.090 is constant on cosets of the hyperplane defined 00:57:01.090 --> 00:57:06.930 by the orthogonal complement of r. 00:57:10.400 --> 00:57:14.550 So the value of this dot product is this constant 00:57:14.550 --> 00:57:17.780 on the three hyperplanes. 00:57:17.780 --> 00:57:22.130 So I can rewrite this expectation simply 00:57:22.130 --> 00:57:28.010 as 1/3 of alpha 0 plus alpha 1 omega 00:57:28.010 --> 00:57:37.220 plus alpha 2 omega squared, where alpha 0, alpha 1, alpha 2 00:57:37.220 --> 00:57:48.451 are the densities of A on the three cosets of r perp. 00:57:52.139 --> 00:57:56.697 So group this expectation into these three hyperplanes. 00:57:59.320 --> 00:58:05.300 So now, you see that if this guy is large, 00:58:05.300 --> 00:58:12.570 then you should expect that alpha 0, alpha 1, and alpha 2 00:58:12.570 --> 00:58:15.550 are not all too close to each other. 00:58:15.550 --> 00:58:19.260 So if they were all equal to each other, you should get 0. 00:58:19.260 --> 00:58:22.380 But you should not expect them to be too close to each other. 00:58:22.380 --> 00:58:27.150 In particular, we would want to say that one of them goes up-- 00:58:27.150 --> 00:58:29.520 is much bigger than alpha. 00:58:29.520 --> 00:58:32.730 So one of these, they must be much bigger than alpha. 00:58:32.730 --> 00:58:34.230 That's an elementary inequality. 00:58:34.230 --> 00:58:36.522 This is something that I'm sure I give you five minutes 00:58:36.522 --> 00:58:37.320 you can figure out. 00:58:37.320 --> 00:58:41.170 But let me show you a small trick to show this. 00:58:41.170 --> 00:58:46.200 And the reason for this trick is because in next lecture, when 00:58:46.200 --> 00:58:48.900 we look at Roth's theorem over the integers, 00:58:48.900 --> 00:58:50.220 we'll need this extra trick. 00:58:55.290 --> 00:58:56.850 And the trick here is this. 00:58:56.850 --> 00:59:01.560 We now know that because of the hypothesis, 00:59:01.560 --> 00:59:08.430 3 delta is lower bound to the absolute value of alpha 0 00:59:08.430 --> 00:59:12.240 plus alpha 1 omega plus alpha 2 omega squared. 00:59:16.010 --> 00:59:21.680 OK, so note here that the average of the three alphas 00:59:21.680 --> 00:59:28.840 is equal to the original alpha by definition of density. 00:59:28.840 --> 00:59:35.430 So this inner sum I can read write like that. 00:59:44.620 --> 00:59:46.900 So the sum of the three groups of unity add up to 0. 00:59:49.410 --> 01:00:02.260 And now, I apply triangle inequality 01:00:02.260 --> 01:00:06.220 to extract the terms. 01:00:06.220 --> 01:00:07.820 So now, you should already deduce 01:00:07.820 --> 01:00:12.080 that one of the alphas i's has to be significantly 01:00:12.080 --> 01:00:14.157 larger than alpha. 01:00:14.157 --> 01:00:15.740 And has to be significantly different, 01:00:15.740 --> 01:00:16.730 but there are only three times. 01:00:16.730 --> 01:00:18.620 One of them has to be significantly larger. 01:00:18.620 --> 01:00:21.590 But let me do this one extra trick, which 01:00:21.590 --> 01:00:24.740 we'll need next time, which is that let 01:00:24.740 --> 01:00:34.850 me add an extra term like that, which sums out to 0. 01:00:34.850 --> 01:00:39.894 But now, you see that each summand is always nonnegative. 01:00:45.700 --> 01:00:49.120 So one of the-- 01:00:52.050 --> 01:00:58.830 so there exists some j such that delta lower abounds 01:00:58.830 --> 01:01:01.290 the j-th summand. 01:01:01.290 --> 01:01:06.960 And if you look at what that means, then alpha lower 01:01:06.960 --> 01:01:11.370 bounds the j-th summand. 01:01:11.370 --> 01:01:15.588 So in particular, this sum here should be nonnegative. 01:01:21.480 --> 01:01:22.480 So you have that. 01:01:22.480 --> 01:01:22.980 Good. 01:01:26.390 --> 01:01:30.413 So we obtained a density increment of this hyperplane. 01:01:34.760 --> 01:01:42.130 And finally, I want to iterate this density increment. 01:01:42.130 --> 01:01:44.749 So I want to iterate this density increment-- 01:02:01.400 --> 01:02:07.520 so summary so far is that we have-- 01:02:07.520 --> 01:02:20.520 so if A is 3-AP-free with density alpha and N at least 2 01:02:20.520 --> 01:02:26.300 to the alpha to the minus 2, then A 01:02:26.300 --> 01:02:35.510 has density at least alpha plus alpha squared over 4 01:02:35.510 --> 01:02:37.175 on some hyperplane. 01:02:40.430 --> 01:02:44.740 So you combine step one and step two, we obtain this conclusion. 01:02:48.340 --> 01:02:54.970 Well, I can now repeat this operation. 01:02:54.970 --> 01:03:09.890 So I can repeat by restricting A to this hyperplane. 01:03:09.890 --> 01:03:13.590 If A is originally 3-AP-free, I restrict it to a hyperplane, 01:03:13.590 --> 01:03:15.460 it's still 3-AP-free. 01:03:15.460 --> 01:03:17.790 So I can keep going. 01:03:17.790 --> 01:03:20.400 I can keep going provided that my space is still 01:03:20.400 --> 01:03:27.500 large enough, because I still need this lower bound on F. 01:03:27.500 --> 01:03:29.640 So don't forget this one here. 01:03:29.640 --> 01:03:40.340 So I can keep going as long as N is-- 01:03:44.960 --> 01:03:47.110 so I'm using N sub j to denote what 01:03:47.110 --> 01:03:50.240 happens after the j-th step. 01:03:50.240 --> 01:03:53.840 I can keep going as long as this is still satisfied. 01:03:53.840 --> 01:03:56.540 But of course, you cannot keep on going forever, 01:03:56.540 --> 01:04:00.030 because the density is bounded. 01:04:00.030 --> 01:04:01.970 So density cannot exceed 1. 01:04:01.970 --> 01:04:06.420 So these two will give you a bound on the total dimension. 01:04:06.420 --> 01:04:08.750 So let's work this out. 01:04:08.750 --> 01:04:18.140 So let alpha i denote the density 01:04:18.140 --> 01:04:25.770 after step i in this iteration. 01:04:25.770 --> 01:04:35.910 And we see from over here that you start with density alpha, 01:04:35.910 --> 01:04:44.940 and each step you go up by an increment, which is basically 01:04:44.940 --> 01:04:46.260 what-- 01:04:46.260 --> 01:04:48.720 so you go up by some increment. 01:04:48.720 --> 01:04:51.570 And you want to know, if you start with alpha, 01:04:51.570 --> 01:04:57.690 how many steps at most can you take before you blow out 1. 01:05:00.440 --> 01:05:02.530 So can you give me some bound? 01:05:06.640 --> 01:05:08.160 So what's the maximum-- 01:05:08.160 --> 01:05:09.590 at most, how many steps? 01:05:17.160 --> 01:05:24.184 So we know that the density cannot exceed 1. 01:05:24.184 --> 01:05:27.523 AUDIENCE: 4 by alpha squared. 01:05:27.523 --> 01:05:31.920 YUFEI ZHAO: So you see that you have at most 4 01:05:31.920 --> 01:05:38.850 over alpha squared steps, because density is at most 1. 01:05:38.850 --> 01:05:44.710 And if you plug this in, you get something which 01:05:44.710 --> 01:05:46.810 is not quite what I stated. 01:05:46.810 --> 01:05:50.920 So it turns out that if you plug this in, 01:05:50.920 --> 01:05:53.620 you find that alpha is-- 01:05:53.620 --> 01:05:57.730 you find that the size of A is at most 3 01:05:57.730 --> 01:05:59.660 to the n-th over square root n. 01:06:03.210 --> 01:06:04.945 So let me do a little bit better, 01:06:04.945 --> 01:06:09.730 so then simply seeing that this term here is 01:06:09.730 --> 01:06:13.260 at least alpha squared over 4. 01:06:13.260 --> 01:06:15.970 And the point is that when you increment, 01:06:15.970 --> 01:06:19.040 you increment faster and faster. 01:06:19.040 --> 01:06:21.680 So I can use that to give a better 01:06:21.680 --> 01:06:24.420 bound on the number of steps. 01:06:24.420 --> 01:06:26.840 And here's the way to see it. 01:06:26.840 --> 01:06:29.810 So let me-- we can do better. 01:06:33.600 --> 01:06:39.540 So starting at alpha, I then now ask, how many steps do you 01:06:39.540 --> 01:06:42.594 need to take before it doubles? 01:06:42.594 --> 01:06:44.426 AUDIENCE: [INAUDIBLE] 01:06:44.426 --> 01:06:48.610 YUFEI ZHAO: It goes up by alpha squared over 4. 01:06:48.610 --> 01:06:57.700 So it doubles after at most 4 over alpha steps. 01:07:01.280 --> 01:07:08.860 And at which point this alpha, new alpha, 01:07:08.860 --> 01:07:10.930 becomes at least the original-- 01:07:10.930 --> 01:07:13.250 twice the original alpha? 01:07:13.250 --> 01:07:14.740 But now, you keep going. 01:07:14.740 --> 01:07:18.560 How many times does it take to double again? 01:07:18.560 --> 01:07:21.030 2 over alpha, because the alpha became twice as much. 01:07:26.760 --> 01:07:33.400 So it doubles again after at most 2 over alpha steps. 01:07:33.400 --> 01:07:34.670 And then you keep going. 01:07:34.670 --> 01:07:38.390 The next iteration is 1 over alpha. 01:07:38.390 --> 01:07:52.700 So in total-- so you see that we must stop after at most 8 01:07:52.700 --> 01:07:56.000 over alpha steps. 01:07:58.570 --> 01:08:01.250 So the number of times it doubles, 01:08:01.250 --> 01:08:04.830 actually it decreases by at least half each time. 01:08:11.053 --> 01:08:11.970 So now, we know that-- 01:08:14.860 --> 01:08:17.500 we see that the-- 01:08:17.500 --> 01:08:19.609 so you keep on going. 01:08:19.609 --> 01:08:23.170 So you must stop after at most 8 over alpha steps. 01:08:23.170 --> 01:08:26.859 What is the final density when you have to stop, 01:08:26.859 --> 01:08:28.430 because when are you forced to stop? 01:08:28.430 --> 01:08:32.399 You are forced to stop if you run out of space. 01:08:32.399 --> 01:08:34.930 So you're forced to stop when you run out of space. 01:08:34.930 --> 01:08:46.290 So if the process terminates after m steps-- 01:08:46.290 --> 01:09:04.305 so we're at density alpha m, so then the final subspace 01:09:04.305 --> 01:09:16.100 has size less than 2 to the alpha m raised to minus 2, 01:09:16.100 --> 01:09:19.130 which is-- 01:09:19.130 --> 01:09:22.830 So now, I use this bound alpha. 01:09:25.490 --> 01:09:32.189 So the initial N is upper bounded by what? 01:09:32.189 --> 01:09:34.250 So how many steps did you take? 01:09:34.250 --> 01:09:36.990 You took at most 8 over alpha steps. 01:09:36.990 --> 01:09:41.760 Each of those steps, you pass down to codimension what? 01:09:41.760 --> 01:09:43.979 You lose a dimension for each step. 01:09:43.979 --> 01:09:47.050 And the final subspace has at least this-- 01:09:47.050 --> 01:09:51.490 has at most that much space. 01:09:51.490 --> 01:10:01.300 So the final dimension is this, basically log 1 over alpha. 01:10:01.300 --> 01:10:07.650 So put them together, we see that the size of the space 01:10:07.650 --> 01:10:11.322 originally is at most 1 over alpha. 01:10:11.322 --> 01:10:11.822 Yeah. 01:10:11.822 --> 01:10:14.466 AUDIENCE: Should this be a lower case n? 01:10:14.466 --> 01:10:15.383 YUFEI ZHAO: Thank you. 01:10:15.383 --> 01:10:16.210 Yeah. 01:10:16.210 --> 01:10:19.499 This should be a lower case n, so the dimension. 01:10:19.499 --> 01:10:19.999 Good. 01:10:22.780 --> 01:10:30.540 And OK, so then that's the conclusion, that the density 01:10:30.540 --> 01:10:36.361 alpha is big O of 1 over n. 01:10:41.360 --> 01:10:44.990 That proves the main theorem for today, 01:10:44.990 --> 01:10:48.940 so Roth's theorem over F3 to the n-th. 01:10:48.940 --> 01:10:51.540 So we went through this Fourier analytic proof. 01:10:51.540 --> 01:10:57.290 Next lecture, we will see the same proof again but done 01:10:57.290 --> 01:11:00.800 in the integers for interval. 01:11:00.800 --> 01:11:05.460 And there, there are some difficulties 01:11:05.460 --> 01:11:08.340 that we don't see over here. 01:11:08.340 --> 01:11:11.940 Because in the finite field space, 01:11:11.940 --> 01:11:15.120 in the finite field model, there's 01:11:15.120 --> 01:11:18.490 this very nice idea of looking at subspaces, 01:11:18.490 --> 01:11:21.410 so looking at hyperplanes. 01:11:21.410 --> 01:11:25.820 Each Fourier coefficient gets you down to one dimension less. 01:11:25.820 --> 01:11:29.730 But when you're working in the integers, 01:11:29.730 --> 01:11:32.310 there are no subspaces you can use. 01:11:34.860 --> 01:11:38.040 So we'll be looking at ways to get around 01:11:38.040 --> 01:11:39.685 the lack of subspaces. 01:11:39.685 --> 01:11:41.310 And this is why I said in the beginning 01:11:41.310 --> 01:11:43.790 that the finite field model is often 01:11:43.790 --> 01:11:50.940 a very good playground for additive combinatorics type 01:11:50.940 --> 01:11:53.940 techniques, especially Fourier analytic techniques. 01:11:53.940 --> 01:11:55.830 Because in the additive-- 01:11:55.830 --> 01:11:57.450 in all of these techniques, they just 01:11:57.450 --> 01:11:59.050 come out to be much cleaner. 01:11:59.050 --> 01:12:01.560 If you're working in a finite field setting, 01:12:01.560 --> 01:12:03.480 you have nice subspaces, you have Fourier 01:12:03.480 --> 01:12:05.160 transform in a very clean way. 01:12:05.160 --> 01:12:08.130 The Fourier transform always takes, in this case, 01:12:08.130 --> 01:12:10.220 one of three values. 01:12:10.220 --> 01:12:11.220 Everything's very clean. 01:12:11.220 --> 01:12:12.300 Everything's very simple. 01:12:12.300 --> 01:12:13.490 And you get to see the idea here. 01:12:13.490 --> 01:12:15.615 You get to see the sense of the increment argument. 01:12:18.210 --> 01:12:21.600 But once you understand those ideas 01:12:21.600 --> 01:12:26.320 and you're willing to do more work, then oftentimes, 01:12:26.320 --> 01:12:29.590 you can bring those ideas to other settings, 01:12:29.590 --> 01:12:32.830 to other abelian groups, to the integers, 01:12:32.830 --> 01:12:37.300 for instance, but with more work in the-- 01:12:37.300 --> 01:12:41.800 there are some extra ingredients that you need to use. 01:12:41.800 --> 01:12:45.260 I mentioned that there was a bound-- 01:12:45.260 --> 01:12:46.330 OK, so initially-- 01:12:49.812 --> 01:12:51.020 So next time, we'll see that. 01:12:51.020 --> 01:12:56.000 Next time, we'll see what happens over the integers. 01:12:56.000 --> 01:12:58.490 Any questions? 01:12:58.490 --> 01:12:59.187 Yes. 01:12:59.187 --> 01:13:03.300 AUDIENCE: [INAUDIBLE] 01:13:03.300 --> 01:13:04.250 YUFEI ZHAO: OK, great. 01:13:04.250 --> 01:13:07.850 So question is why the process must stop after at most 8 01:13:07.850 --> 01:13:08.840 over alpha steps? 01:13:12.430 --> 01:13:15.700 So you know that the density doubles after this many steps, 01:13:15.700 --> 01:13:17.380 doubles again after that many steps. 01:13:17.380 --> 01:13:20.710 So eventually, if it keeps on doubling, 01:13:20.710 --> 01:13:24.360 it cannot keep on doubling forever. 01:13:24.360 --> 01:13:26.570 So this process cannot keep on doubling forever. 01:13:26.570 --> 01:13:28.820 So it must stop-- 01:13:28.820 --> 01:13:39.560 so cannot double more than log base 2 of 1 over alpha times. 01:13:42.980 --> 01:13:45.440 And that point, you have to stop. 01:13:45.440 --> 01:13:47.270 So how many steps have you taken? 01:13:47.270 --> 01:13:49.420 Well, you sum this geometric series. 01:13:49.420 --> 01:13:55.940 So this-- and the next thing is that you 01:13:55.940 --> 01:13:57.840 sum this geometric series. 01:13:57.840 --> 01:14:01.310 And that geometric series sums to 8 over alpha. 01:14:10.230 --> 01:14:10.730 Great. 01:14:10.730 --> 01:14:12.280 So let's finish here. 01:14:12.280 --> 01:14:16.680 So next time, we'll see Roth's proof of Roth's theorem.