1 00:00:00,000 --> 00:00:01,952 [SQUEAKING] 2 00:00:01,952 --> 00:00:03,404 [PAPER RUSTLING] 3 00:00:03,904 --> 00:00:04,880 [CLICKING] 4 00:00:18,080 --> 00:00:20,960 YUFEI ZHAO: Last time we started talking about Roth's theorem, 5 00:00:20,960 --> 00:00:26,420 and we showed a Fourier analytic proof of Roth's theorem 6 00:00:26,420 --> 00:00:28,760 in the finite field model. 7 00:00:28,760 --> 00:00:32,770 So Roth's theorem in F3 to the N. 8 00:00:32,770 --> 00:00:37,670 And I want to today show you how to modify that proof 9 00:00:37,670 --> 00:00:39,650 to work in integers. 10 00:00:39,650 --> 00:00:46,130 And this will be basically Roth's original proof 11 00:00:46,130 --> 00:00:46,820 of his theorem. 12 00:00:56,440 --> 00:00:57,410 OK. 13 00:00:57,410 --> 00:01:00,860 So what we'll prove today is the statement 14 00:01:00,860 --> 00:01:10,580 that the size of the largest 3AP-free subset of 1 through N 15 00:01:10,580 --> 00:01:21,080 is, at most, N divided by log log N. OK, 16 00:01:21,080 --> 00:01:22,880 so we'll prove a bound of this form. 17 00:01:26,450 --> 00:01:29,600 The strategy of this proof will be very similar to the one 18 00:01:29,600 --> 00:01:30,920 that we had from last time. 19 00:01:30,920 --> 00:01:32,837 So let me review for you what is the strategy. 20 00:01:44,680 --> 00:01:49,530 So from last time, the proof had three main steps. 21 00:01:49,530 --> 00:01:52,920 In the first step, we observed that if you 22 00:01:52,920 --> 00:02:02,170 are in the 3AP-free set then there exists a large Fourier 23 00:02:02,170 --> 00:02:03,274 coefficient. 24 00:02:11,810 --> 00:02:13,730 From this Fourier coefficient, we 25 00:02:13,730 --> 00:02:22,370 were able to extract a large subspace where 26 00:02:22,370 --> 00:02:25,220 there is a density increment. 27 00:02:25,220 --> 00:02:28,910 I want to modify that strategy so that we 28 00:02:28,910 --> 00:02:31,580 can work in the integers. 29 00:02:31,580 --> 00:02:34,190 Unlike an F2 to the N, where things 30 00:02:34,190 --> 00:02:37,610 were fairly nice and clean, because you have subspaces, 31 00:02:37,610 --> 00:02:39,530 you can take a Fourier coefficient, 32 00:02:39,530 --> 00:02:41,900 pass it down to a subspace. 33 00:02:41,900 --> 00:02:43,760 There is no subspaces, right? 34 00:02:43,760 --> 00:02:46,980 There are no subspaces in the integers. 35 00:02:46,980 --> 00:02:49,020 So we have to do something slightly different, 36 00:02:49,020 --> 00:02:51,120 but in the same spirit. 37 00:02:51,120 --> 00:02:53,577 So you'll find a large Fourier coefficient. 38 00:03:00,400 --> 00:03:10,370 And we will find that there is density increment when 39 00:03:10,370 --> 00:03:14,410 you restrict not to subspaces, but what 40 00:03:14,410 --> 00:03:17,335 could play the role of subspaces when it comes to the integers? 41 00:03:19,900 --> 00:03:23,050 So I want something which looks like a smaller 42 00:03:23,050 --> 00:03:26,260 version of the original space. 43 00:03:26,260 --> 00:03:29,280 So instead of it being integers, if we 44 00:03:29,280 --> 00:03:33,780 restrict to a subprogression, so to a smaller arithmetic 45 00:03:33,780 --> 00:03:34,970 progression. 46 00:03:34,970 --> 00:03:38,040 I will show that you can restrict to a subprogression 47 00:03:38,040 --> 00:03:40,697 where you can obtain density increment. 48 00:03:45,840 --> 00:03:50,070 So we'll restrict integers to something smaller. 49 00:03:50,070 --> 00:04:04,680 And then, same as last time, we can iterate this increment 50 00:04:04,680 --> 00:04:08,490 to obtain the conclusion that you 51 00:04:08,490 --> 00:04:14,545 have an upper bound on the size of this 3AP-free set. 52 00:04:14,545 --> 00:04:15,670 OK, so that's the strategy. 53 00:04:15,670 --> 00:04:18,310 So you see the same strategy as the one we did last time, 54 00:04:18,310 --> 00:04:20,829 and many of the ingredients will have parallels, 55 00:04:20,829 --> 00:04:23,830 but the execution will be slightly different, 56 00:04:23,830 --> 00:04:26,750 especially in the second step where, 57 00:04:26,750 --> 00:04:29,440 because we no longer have sub spaces, which 58 00:04:29,440 --> 00:04:30,880 are nice and clean, so that's why 59 00:04:30,880 --> 00:04:33,070 we started with a finite fuel model, just 60 00:04:33,070 --> 00:04:35,530 to show how things work in a slightly easier setting. 61 00:04:35,530 --> 00:04:39,670 And today, we'll see how to do a same kind of strategy 62 00:04:39,670 --> 00:04:42,890 here, where there is going to be a bit more work. 63 00:04:42,890 --> 00:04:45,290 Not too much more, but a bit more work. 64 00:04:45,290 --> 00:04:47,791 OK, so before we start, any questions? 65 00:04:53,200 --> 00:04:54,920 All right. 66 00:04:54,920 --> 00:04:57,290 So last time I used the proof Roth's theorem 67 00:04:57,290 --> 00:04:59,245 as an excuse to introduce Fourier analysis. 68 00:04:59,245 --> 00:05:01,620 And we're going to see basically the same kind of Fourier 69 00:05:01,620 --> 00:05:05,420 analysis, but it's going to take on a slightly different form, 70 00:05:05,420 --> 00:05:07,940 because we're not working at F3 to the N. We're 71 00:05:07,940 --> 00:05:10,490 working inside the integers. 72 00:05:10,490 --> 00:05:13,870 And there's a general theory of Fourier analysis on the group, 73 00:05:13,870 --> 00:05:16,472 on a billion groups. 74 00:05:16,472 --> 00:05:18,680 I don't want to go into that theory, because that's-- 75 00:05:18,680 --> 00:05:20,442 I want to focus on the specific case, 76 00:05:20,442 --> 00:05:22,400 but the point is that given the billion groups, 77 00:05:22,400 --> 00:05:27,220 you always have a dual group of characters. 78 00:05:27,220 --> 00:05:29,800 And they play the role of Fourier transform. 79 00:05:29,800 --> 00:05:33,790 Specifically in the case of Z, we 80 00:05:33,790 --> 00:05:35,890 have the following Fourier transform. 81 00:05:41,390 --> 00:05:51,110 So the dual group of Z turns out to be the torus. 82 00:05:51,110 --> 00:05:54,540 So real numbers mod one. 83 00:05:54,540 --> 00:06:03,450 And the Fourier transform is defined 84 00:06:03,450 --> 00:06:14,910 as follows, starting with a function on the integers, OK. 85 00:06:14,910 --> 00:06:17,010 If you'd like, let's say it's finitely supported, 86 00:06:17,010 --> 00:06:18,730 just to make our lives a bit easier. 87 00:06:18,730 --> 00:06:20,680 Don't have to deal with technicalities. 88 00:06:20,680 --> 00:06:23,820 But in general, the following formula holds. 89 00:06:23,820 --> 00:06:32,280 We have this Fourier transform defined 90 00:06:32,280 --> 00:06:39,620 by setting f hat of theta to be the following sum. 91 00:06:44,922 --> 00:06:50,030 OK, where this e is actually somewhat standard notation, 92 00:06:50,030 --> 00:06:52,480 additive combinatorics. 93 00:06:52,480 --> 00:06:56,400 It's e to the 2 pi i t, all right? 94 00:06:56,400 --> 00:07:02,350 So it goes a fraction, t, around the complex unit circle. 95 00:07:02,350 --> 00:07:02,850 OK. 96 00:07:02,850 --> 00:07:05,507 So that's the Fourier transform on the integers. 97 00:07:05,507 --> 00:07:08,090 OK, so you might have seen this before under a different name. 98 00:07:08,090 --> 00:07:09,770 This is usually called Fourier series. 99 00:07:13,790 --> 00:07:14,290 All right. 100 00:07:14,290 --> 00:07:16,332 You know, the notation may be slightly different. 101 00:07:16,332 --> 00:07:19,050 OK, so that's what we'll see today. 102 00:07:19,050 --> 00:07:22,760 And this Fourier transform plays the same role as the Fourier 103 00:07:22,760 --> 00:07:29,390 transform from last time, which was on the group F3 to the N. 104 00:07:29,390 --> 00:07:33,062 And just us in-- 105 00:07:33,062 --> 00:07:37,020 so last time, we had a number of important identities, 106 00:07:37,020 --> 00:07:39,900 and we'll have the same kinds of identities here. 107 00:07:39,900 --> 00:07:41,970 So let me remind you what they are. 108 00:07:41,970 --> 00:07:43,810 And the proofs are all basically the same, 109 00:07:43,810 --> 00:07:45,102 so I won't show you the proofs. 110 00:07:54,130 --> 00:08:01,540 f hat of 0 is simply the sum of f over the domain. 111 00:08:05,220 --> 00:08:16,470 We have this Plancherel Parseval identity, which tells us 112 00:08:16,470 --> 00:08:22,620 that if you look at the inner product 113 00:08:22,620 --> 00:08:27,610 by linear form in the physical space, 114 00:08:27,610 --> 00:08:33,450 it equals to the inner product in the Fourier space. 115 00:08:42,620 --> 00:08:43,120 OK. 116 00:08:43,120 --> 00:08:45,250 So in the physical space now, you sum. 117 00:08:45,250 --> 00:08:47,350 In the frequency space, you take integral 118 00:08:47,350 --> 00:08:51,960 over the torus, or the circle, in this case. 119 00:08:51,960 --> 00:08:54,570 It's a one-dimensional torus. 120 00:08:54,570 --> 00:08:58,470 There is also the Fourier inversion formula, 121 00:08:58,470 --> 00:09:05,350 which now says that f of x is equal to f hat of theta. 122 00:09:05,350 --> 00:09:09,780 E of x theta, you integrate theta from 0 to 1. 123 00:09:09,780 --> 00:09:13,830 Again, on the torus, on the circle. 124 00:09:13,830 --> 00:09:18,060 And third-- and finally, there was this identity last time 125 00:09:18,060 --> 00:09:20,940 that related three-term arithmetic progressions 126 00:09:20,940 --> 00:09:22,680 to the Fourier transform, OK? 127 00:09:22,680 --> 00:09:26,530 So this last one was slightly not as-- 128 00:09:26,530 --> 00:09:29,550 I mean, it's not as standard as the first several, which are 129 00:09:29,550 --> 00:09:31,560 standard Fourier identities. 130 00:09:31,560 --> 00:09:34,360 But this one will be useful to us. 131 00:09:34,360 --> 00:09:38,280 So the identity relating the Fourier transform, the 3AP, now 132 00:09:38,280 --> 00:09:39,643 has the following form. 133 00:09:39,643 --> 00:09:47,098 OK, so if we define lambda of f, g, 134 00:09:47,098 --> 00:09:55,790 and h to be the following sum, which 135 00:09:55,790 --> 00:10:02,600 sums over all 3APs in the integers, 136 00:10:02,600 --> 00:10:11,470 then one can write this expression 137 00:10:11,470 --> 00:10:15,970 in terms of the Fourier transform as follows. 138 00:10:26,750 --> 00:10:27,720 OK. 139 00:10:27,720 --> 00:10:28,220 All right. 140 00:10:28,220 --> 00:10:31,940 So comparing this formula to the one that we saw from last time, 141 00:10:31,940 --> 00:10:34,490 it's the same formula, where different domains, 142 00:10:34,490 --> 00:10:35,990 where you're summing or integrating, 143 00:10:35,990 --> 00:10:37,160 but it's the same formula. 144 00:10:37,160 --> 00:10:39,000 And the proof is the same. 145 00:10:39,000 --> 00:10:40,070 So go look at the proof. 146 00:10:40,070 --> 00:10:40,903 It's the same proof. 147 00:10:43,710 --> 00:10:44,210 OK. 148 00:10:44,210 --> 00:10:47,810 So these are the key Fourier things that we'll use. 149 00:10:47,810 --> 00:10:50,980 And then we'll try to follow on those with-- 150 00:10:50,980 --> 00:10:55,170 the same as the proof as last time, and see where we can get. 151 00:10:55,170 --> 00:10:57,180 So let me introduce one more notation. 152 00:10:57,180 --> 00:11:04,490 So I'll write lambda sub 3 of f to be lambda of f, f, f, 153 00:11:04,490 --> 00:11:05,030 three times. 154 00:11:07,509 --> 00:11:08,009 OK. 155 00:11:08,009 --> 00:11:09,980 So at this point, if you understood the lecture 156 00:11:09,980 --> 00:11:13,220 from last time, none of anything I've said so far 157 00:11:13,220 --> 00:11:14,450 should be surprising. 158 00:11:14,450 --> 00:11:16,280 We are working integers, so we should 159 00:11:16,280 --> 00:11:19,170 look at the corresponding Fourier transform in integers. 160 00:11:19,170 --> 00:11:21,680 And if you follow your notes, this is all the things 161 00:11:21,680 --> 00:11:23,660 that we're going to use. 162 00:11:23,660 --> 00:11:25,970 OK, so what was one of the first things 163 00:11:25,970 --> 00:11:29,270 we mentioned regarding the Fourier transform 164 00:11:29,270 --> 00:11:30,830 from last time after this point? 165 00:11:40,250 --> 00:11:40,750 OK. 166 00:11:40,750 --> 00:11:42,075 AUDIENCE: The counting lemma. 167 00:11:42,075 --> 00:11:42,700 YUFEI ZHAO: OK. 168 00:11:42,700 --> 00:11:44,260 So let's do a counting lemma. 169 00:11:44,260 --> 00:11:49,780 So what should the counting lemma say? 170 00:11:49,780 --> 00:11:52,700 Well, the spirit of the counting lemma is that if you have two 171 00:11:52,700 --> 00:11:56,360 functions that are close to each other-- and now, "close" 172 00:11:56,360 --> 00:11:58,940 means close in Fourier-- 173 00:11:58,940 --> 00:12:03,910 then their corresponding number of 3APs should be similar, OK? 174 00:12:03,910 --> 00:12:06,150 So that's what we want to say. 175 00:12:06,150 --> 00:12:14,740 And indeed, right, so the counting lemma for us 176 00:12:14,740 --> 00:12:25,510 will say that if f and g are functions on Z, and-- 177 00:12:28,420 --> 00:12:47,375 such that their L2 norms are both bounded by this M. OK, 178 00:12:47,375 --> 00:12:51,940 the sum of the squared absolute value entries are both bounded. 179 00:12:51,940 --> 00:13:03,710 Then the difference of their 3AP counts 180 00:13:03,710 --> 00:13:08,570 should not be so different from each other if f 181 00:13:08,570 --> 00:13:10,790 and g are close in Fourier, OK? 182 00:13:10,790 --> 00:13:21,990 And that means that if all the Fourier coefficients of f 183 00:13:21,990 --> 00:13:28,270 minus g are small, then lambda 3ff, 184 00:13:28,270 --> 00:13:32,290 which considers 3AP counts in f, is close to that of g. 185 00:13:35,685 --> 00:13:38,110 OK. 186 00:13:38,110 --> 00:13:42,744 Same kind of 3AP counting lemma from last time. 187 00:13:42,744 --> 00:13:45,706 OK, so let's prove it, OK? 188 00:13:51,520 --> 00:13:53,620 As with the counting lemma proofs 189 00:13:53,620 --> 00:13:56,350 you've seen several times already in this course, 190 00:13:56,350 --> 00:14:00,070 we will prove it by first writing this difference 191 00:14:00,070 --> 00:14:01,450 as a telescoping sum. 192 00:14:07,420 --> 00:14:12,920 The first term being f minus g f f, 193 00:14:12,920 --> 00:14:25,045 and then g of f minus g f and lambda g, g, f, f minus g. 194 00:14:25,045 --> 00:14:27,790 OK, and we will like to show that each of these terms 195 00:14:27,790 --> 00:14:34,273 is small if f minus g has small Fourier coefficients. 196 00:14:39,690 --> 00:14:40,190 OK. 197 00:14:40,190 --> 00:14:41,900 So let's bound the first term. 198 00:14:50,077 --> 00:14:55,340 OK, so let me bound this first term using the 3AP identity, 199 00:14:55,340 --> 00:14:57,950 relating 3AP to Fourier coefficients, 200 00:14:57,950 --> 00:15:05,150 we can write this lambda as the following integral 201 00:15:05,150 --> 00:15:07,118 over Fourier coefficients. 202 00:15:24,070 --> 00:15:25,730 And now, let me-- 203 00:15:25,730 --> 00:15:27,550 OK, so what was the trick last time? 204 00:15:27,550 --> 00:15:32,600 So we said let's pull out one of these guys 205 00:15:32,600 --> 00:15:37,245 and then use triangle inequality on the remaining factors. 206 00:15:37,245 --> 00:15:38,150 OK, so we'll do that. 207 00:15:56,410 --> 00:15:58,852 So far, so good. 208 00:15:58,852 --> 00:16:02,295 And now you see this integral. 209 00:16:02,295 --> 00:16:03,773 Apply Cauchy-Schwartz. 210 00:16:18,090 --> 00:16:20,890 OK, so apply Cauchy-Schwartz to the first factor, 211 00:16:20,890 --> 00:16:25,533 you got this l2 sum, this l2 integral. 212 00:16:25,533 --> 00:16:26,950 And then you apply Cauchy-Schwartz 213 00:16:26,950 --> 00:16:28,295 to the second factor. 214 00:16:34,633 --> 00:16:35,550 You get that integral. 215 00:16:38,826 --> 00:16:40,920 OK, now what do we do? 216 00:16:44,912 --> 00:16:46,409 Yep? 217 00:16:46,409 --> 00:16:49,980 AUDIENCE: [INAUDIBLE] 218 00:16:49,980 --> 00:16:50,980 YUFEI ZHAO: OK, so yeah. 219 00:16:50,980 --> 00:16:54,700 So you see an l2 of Fourier, the first automatic reaction 220 00:16:54,700 --> 00:16:56,890 should be to use a Plancherel or Parseval, OK? 221 00:16:56,890 --> 00:17:04,589 So apply Plancherel identity to each of these factors. 222 00:17:04,589 --> 00:17:13,520 We find that each of those factors 223 00:17:13,520 --> 00:17:20,040 is equal to this l2 sum in the physical space. 224 00:17:23,750 --> 00:17:25,609 OK, so this square root, the same thing. 225 00:17:25,609 --> 00:17:27,980 Square root again. 226 00:17:27,980 --> 00:17:28,520 OK. 227 00:17:28,520 --> 00:17:34,750 And then we find that because there 228 00:17:34,750 --> 00:17:36,720 was an hypothesis-- in the hypothesis, 229 00:17:36,720 --> 00:17:40,430 there was a bound M on this sum of squares. 230 00:17:43,610 --> 00:17:46,090 You have that down there. 231 00:17:46,090 --> 00:17:48,080 And similarly, with the other two terms. 232 00:18:01,526 --> 00:18:02,050 OK. 233 00:18:02,050 --> 00:18:03,511 So that proves the counting lemma. 234 00:18:07,359 --> 00:18:08,802 Question? 235 00:18:08,802 --> 00:18:12,760 AUDIENCE: Last time, the term on the right-hand side 236 00:18:12,760 --> 00:18:16,525 was the maximum over non-zero frequency? 237 00:18:16,525 --> 00:18:17,730 YUFEI ZHAO: OK. 238 00:18:17,730 --> 00:18:18,230 OK. 239 00:18:18,230 --> 00:18:22,230 So the question is, last time we had a counting lemma that 240 00:18:22,230 --> 00:18:24,120 looked slightly different. 241 00:18:24,120 --> 00:18:26,730 But I claimed they're all really the same counting lemma. 242 00:18:26,730 --> 00:18:28,320 They're all the same proofs. 243 00:18:28,320 --> 00:18:30,810 If you run this proof, it won't work. 244 00:18:30,810 --> 00:18:33,150 If you take what we did last time, 245 00:18:33,150 --> 00:18:35,260 it's the same kind of proofs. 246 00:18:35,260 --> 00:18:36,760 So last time we had a counting lemma 247 00:18:36,760 --> 00:18:44,120 where we had the same f, f, f essentially. 248 00:18:44,120 --> 00:18:48,660 We know have-- I allow you to essentially take three 249 00:18:48,660 --> 00:18:49,640 different things, and-- 250 00:18:49,640 --> 00:18:52,980 OK, so both-- in both cases, you're 251 00:18:52,980 --> 00:18:54,750 running through this calculation, 252 00:18:54,750 --> 00:18:57,378 but they look slightly different. 253 00:18:57,378 --> 00:19:02,555 AUDIENCE: [INAUDIBLE] 254 00:19:02,555 --> 00:19:03,430 YUFEI ZHAO: So, yeah. 255 00:19:03,430 --> 00:19:04,435 So I agree. 256 00:19:04,435 --> 00:19:05,810 It doesn't look exactly the same, 257 00:19:05,810 --> 00:19:07,520 but if you think about what's involved in the proof, 258 00:19:07,520 --> 00:19:08,520 they're the same proofs. 259 00:19:11,232 --> 00:19:12,630 OK. 260 00:19:12,630 --> 00:19:16,170 Any more questions? 261 00:19:16,170 --> 00:19:19,060 All right. 262 00:19:19,060 --> 00:19:20,670 So now, we have this counting lemma, 263 00:19:20,670 --> 00:19:25,660 so let's start our proof of Roth's theorem in the integers. 264 00:19:25,660 --> 00:19:27,820 As with last time, there will be three steps, 265 00:19:27,820 --> 00:19:32,175 as mentioned up there, OK? 266 00:19:32,175 --> 00:19:37,590 In the first step, let us show that if you are 3AP-free, 267 00:19:37,590 --> 00:19:40,130 then we can obtain a density-- 268 00:19:40,130 --> 00:19:42,198 a large Fourier coefficient. 269 00:19:56,640 --> 00:19:59,200 Yeah, so in this course, this counting lemma, 270 00:19:59,200 --> 00:20:01,450 we actually solved this-- 271 00:20:01,450 --> 00:20:03,970 basically this kind of proof for the first time when we 272 00:20:03,970 --> 00:20:07,000 discussed graph counting lemma, back in the chapter 273 00:20:07,000 --> 00:20:09,810 on Szemerédi's regularity lemma. 274 00:20:09,810 --> 00:20:12,430 And sure, they all look literally-- 275 00:20:12,430 --> 00:20:14,740 not exactly the same, but they're all really 276 00:20:14,740 --> 00:20:16,400 the same kind of proofs, right? 277 00:20:16,400 --> 00:20:17,050 So I want-- 278 00:20:17,050 --> 00:20:19,540 I'm showing you the same thing in many different guises. 279 00:20:19,540 --> 00:20:20,873 But they're all the same proofs. 280 00:20:23,690 --> 00:20:28,540 So if you are a set that is 3AP-free-- 281 00:20:37,630 --> 00:20:41,480 and as with last time, I'm going to call 282 00:20:41,480 --> 00:20:46,550 alpha the density of A now inside this progression, 283 00:20:46,550 --> 00:20:49,820 this length N progression. 284 00:20:49,820 --> 00:20:54,560 And suppose N is large enough. 285 00:20:59,460 --> 00:21:07,580 OK, so the conclusion now is that there exists some theta 286 00:21:07,580 --> 00:21:19,170 such that if you look at this sum over here 287 00:21:19,170 --> 00:21:23,930 as a sum over both integers-- 288 00:21:23,930 --> 00:21:26,680 actually, let me do the sum only from 1 289 00:21:26,680 --> 00:21:35,040 to uppercase N. Claim that-- 290 00:21:35,040 --> 00:21:39,510 OK, so-- so it's saying what this title says. 291 00:21:39,510 --> 00:21:43,205 If you are 3AP-free and this N is large 292 00:21:43,205 --> 00:21:45,580 enough relative to the density, you think of this density 293 00:21:45,580 --> 00:21:49,990 alpha is a constant, then I can find a large Fourier 294 00:21:49,990 --> 00:21:50,860 coefficient. 295 00:21:50,860 --> 00:21:54,010 Now, there's a small difference, and this 296 00:21:54,010 --> 00:21:56,140 is related to what you were asking earlier, 297 00:21:56,140 --> 00:21:59,710 between how we set things up now versus what happened last time. 298 00:21:59,710 --> 00:22:05,530 So last time, we just looked for a Fourier coefficient 299 00:22:05,530 --> 00:22:10,080 corresponding to a non-zero r. 300 00:22:10,080 --> 00:22:13,340 Now, I'm not restricting non-zero, 301 00:22:13,340 --> 00:22:15,700 but I don't start with an indicator function. 302 00:22:15,700 --> 00:22:19,280 I start with the demeaned indicator function. 303 00:22:19,280 --> 00:22:24,340 I take out the mean so that the zeroth coefficient, 304 00:22:24,340 --> 00:22:28,450 so to speak, which corresponds to the mean, is already 0. 305 00:22:28,450 --> 00:22:32,790 So you don't get to use that for your coefficient. 306 00:22:32,790 --> 00:22:34,840 So if you didn't do this, if you just 307 00:22:34,840 --> 00:22:36,420 tried to do this last time, I mean, 308 00:22:36,420 --> 00:22:38,045 you can also do exactly the same setup. 309 00:22:38,045 --> 00:22:40,180 But if you don't demean it, then-- 310 00:22:40,180 --> 00:22:42,400 if you don't have this term, then this statement 311 00:22:42,400 --> 00:22:46,960 is trivially true, because I can take theta equal to 0, OK? 312 00:22:46,960 --> 00:22:48,300 But I don't want. 313 00:22:48,300 --> 00:22:52,900 I want an actual significant Fourier improvement. 314 00:22:52,900 --> 00:22:53,930 So I take-- 315 00:22:53,930 --> 00:22:58,540 I do this demean, and then I consider its Fourier 316 00:22:58,540 --> 00:22:59,583 coefficient. 317 00:23:02,260 --> 00:23:02,760 OK. 318 00:23:02,760 --> 00:23:06,220 Any questions about the statement? 319 00:23:06,220 --> 00:23:09,810 Yeah, so this demeaning is really important, right? 320 00:23:09,810 --> 00:23:12,530 So that's something that's a very common technique whenever 321 00:23:12,530 --> 00:23:14,140 you do these kind of analysis. 322 00:23:14,140 --> 00:23:16,510 So make sure you're-- 323 00:23:16,510 --> 00:23:19,940 so that you're-- yeah, so you're looking at functions with mean 324 00:23:19,940 --> 00:23:22,777 0. 325 00:23:22,777 --> 00:23:23,610 Let's see the proof. 326 00:23:28,190 --> 00:23:33,290 We have the following information 327 00:23:33,290 --> 00:23:39,670 about all 3AP counts in A. Because A is 3AP-free, OK, 328 00:23:39,670 --> 00:23:43,320 so what is the value of lambda sub 3 of the indicator of A? 329 00:23:46,100 --> 00:23:49,440 Lambda of 3, if you look at the expression, 330 00:23:49,440 --> 00:23:55,770 it basically sums over all 3APs, but A has no 3APs, 331 00:23:55,770 --> 00:23:58,500 except for the trivial ones. 332 00:23:58,500 --> 00:24:01,020 So we'll only consider the trivial 3APs, 333 00:24:01,020 --> 00:24:06,210 which has size exactly the size of A, which 334 00:24:06,210 --> 00:24:10,820 is alpha N from trivial 3APs. 335 00:24:14,870 --> 00:24:20,030 On the other hand, what do we know about lambda 3 336 00:24:20,030 --> 00:24:25,256 of this interval from 1 to N? 337 00:24:25,256 --> 00:24:28,410 OK, so how many 3APs are there? 338 00:24:28,410 --> 00:24:33,640 OK, so roughly, it's going to be about N squared over 2. 339 00:24:33,640 --> 00:24:38,470 And in fact, it will be at least N squared over 2, 340 00:24:38,470 --> 00:24:41,860 because to generate a 3AP, I just 341 00:24:41,860 --> 00:24:49,070 have to pick a first term and a third term, 342 00:24:49,070 --> 00:24:51,760 and I'm OK as long as they're the same parity. 343 00:24:58,195 --> 00:25:00,410 And then you have a 3AP. 344 00:25:00,410 --> 00:25:04,220 So the same parity cuts you down by half, 345 00:25:04,220 --> 00:25:11,190 so you have at least N squared over 2 3APs from 1 through N. 346 00:25:11,190 --> 00:25:14,930 So now, let's look at how to apply the counting 347 00:25:14,930 --> 00:25:16,480 lemma all to the setting. 348 00:25:16,480 --> 00:25:18,420 So we have the counting lemma up there, 349 00:25:18,420 --> 00:25:19,930 where I now want to apply it-- 350 00:25:22,600 --> 00:25:28,960 so apply counting to, on one hand, 351 00:25:28,960 --> 00:25:34,780 the indicator function of A so we get the count 3APs in A, 352 00:25:34,780 --> 00:25:43,770 but also compared to the normalized indicator 353 00:25:43,770 --> 00:25:45,252 on the interval. 354 00:25:45,252 --> 00:25:46,940 OK, so maybe this is a good point for me 355 00:25:46,940 --> 00:25:50,425 to pause and remind you that the spirit of this whole proof 356 00:25:50,425 --> 00:25:54,950 is understanding structure versus pseudorandomness, OK? 357 00:25:54,950 --> 00:25:57,680 So as was the case last time. 358 00:25:57,680 --> 00:26:03,620 So we want to understand, in what ways is A pseudorandom? 359 00:26:03,620 --> 00:26:06,020 And here, "pseudorandom," just as with last time, 360 00:26:06,020 --> 00:26:08,540 means having small Fourier coefficients, 361 00:26:08,540 --> 00:26:11,856 being Fourier uniform. 362 00:26:11,856 --> 00:26:16,850 If A is pseudorandom, which here, means f and g 363 00:26:16,850 --> 00:26:18,830 are close to each other. 364 00:26:18,830 --> 00:26:20,420 That's what being pseudorandom means, 365 00:26:20,420 --> 00:26:21,920 then the counting lemma will tell us 366 00:26:21,920 --> 00:26:25,790 that f and g should have similar AP counts. 367 00:26:25,790 --> 00:26:31,190 But A has basically no AP count, so they should not 368 00:26:31,190 --> 00:26:35,970 be close to each other. 369 00:26:35,970 --> 00:26:38,610 So that's the strategy, to show that A is not pseudorandom 370 00:26:38,610 --> 00:26:41,490 in this sense, and thereby extracting a large Fourier 371 00:26:41,490 --> 00:26:43,240 coefficient. 372 00:26:43,240 --> 00:26:47,490 So we apply counting to these two functions, 373 00:26:47,490 --> 00:26:49,350 and we obtain that. 374 00:26:54,860 --> 00:26:55,360 OK. 375 00:26:55,360 --> 00:27:03,000 So this quantity, which corresponds to lambda 3 of g, 376 00:27:03,000 --> 00:27:17,300 minus alpha N. So these were lambda 3 of g, lambda 3 of F. 377 00:27:17,300 --> 00:27:18,760 So it is upper-bounded. 378 00:27:18,760 --> 00:27:23,897 The difference is up rebounded by the-- 379 00:27:28,010 --> 00:27:29,760 using the counting lemma, we find 380 00:27:29,760 --> 00:27:31,500 that their difference is upper-bounded 381 00:27:31,500 --> 00:27:32,670 by the following quantity. 382 00:27:43,760 --> 00:27:46,850 Namely, you look at the difference between f and g 383 00:27:46,850 --> 00:27:52,082 and evaluate its maximum Fourier coefficient. 384 00:27:52,082 --> 00:27:53,010 OK. 385 00:27:53,010 --> 00:27:58,760 So if A is pseudorandom, meaning that Fourier uniform-- 386 00:27:58,760 --> 00:28:04,840 this l infinity norm is small, then 387 00:28:04,840 --> 00:28:10,010 I should expect lots and lots of 3APs in A, 388 00:28:10,010 --> 00:28:12,390 but because that is not the case, 389 00:28:12,390 --> 00:28:14,510 we should be able to conclude that there is 390 00:28:14,510 --> 00:28:16,745 some large Fourier coefficient. 391 00:28:20,000 --> 00:28:21,090 All right, so thus-- 392 00:28:33,131 --> 00:28:36,720 so rearranging the equation above, we have that-- 393 00:28:44,030 --> 00:28:45,270 so this should be a square. 394 00:28:51,480 --> 00:28:52,100 OK. 395 00:28:52,100 --> 00:28:54,260 So we have this expression here. 396 00:28:54,260 --> 00:28:56,690 And now we are-- 397 00:28:56,690 --> 00:29:03,320 OK, so let me simplify this expression slightly. 398 00:29:03,320 --> 00:29:07,550 And now we're using that N is sufficiently large, OK? 399 00:29:07,550 --> 00:29:12,290 So we're using N is sufficiently large. 400 00:29:12,290 --> 00:29:23,202 So this quantity is at least a tenth of alpha squared N. 401 00:29:23,202 --> 00:29:24,910 OK, and that's the conclusion, all right? 402 00:29:24,910 --> 00:29:28,937 So that's the conclusion of this step here. 403 00:29:28,937 --> 00:29:29,770 What does this mean? 404 00:29:29,770 --> 00:29:33,600 This means there exists some theta 405 00:29:33,600 --> 00:29:37,050 so that the Fourier coefficient at theta 406 00:29:37,050 --> 00:29:38,750 is at least the claimed quantity. 407 00:29:42,530 --> 00:29:43,834 Any questions? 408 00:29:51,420 --> 00:29:52,050 All right. 409 00:29:52,050 --> 00:29:53,960 So that finishes step 1. 410 00:29:53,960 --> 00:29:57,720 So now let me go on step 2. 411 00:29:57,720 --> 00:30:02,120 In step 2, we wish to show that if you have a large Fourier 412 00:30:02,120 --> 00:30:11,956 coefficient, then one can obtain a density increment. 413 00:30:21,200 --> 00:30:26,720 So last time, we were working in a finite field vector space. 414 00:30:26,720 --> 00:30:30,470 A Fourier coefficient, OK, so which is a dual vector, 415 00:30:30,470 --> 00:30:34,020 corresponds to some hyperplane. 416 00:30:34,020 --> 00:30:37,610 And having a large Fourier coefficient 417 00:30:37,610 --> 00:30:40,380 then implies that the density of A 418 00:30:40,380 --> 00:30:43,140 on the co-sets of those hyperplanes 419 00:30:43,140 --> 00:30:46,720 must be not all close to each other. 420 00:30:46,720 --> 00:30:48,570 All right, so one of the hyperplanes 421 00:30:48,570 --> 00:30:53,380 must have significantly higher density than the rest. 422 00:30:53,380 --> 00:30:55,300 OK, so we want to do something similar here, 423 00:30:55,300 --> 00:30:57,580 except we run into this technical difficulty 424 00:30:57,580 --> 00:31:01,250 where there are no subspaces anymore. 425 00:31:01,250 --> 00:31:05,243 So the Fourier character, namely corresponding to this theta, 426 00:31:05,243 --> 00:31:06,160 is just a real number. 427 00:31:06,160 --> 00:31:07,510 It doesn't divide up your space. 428 00:31:07,510 --> 00:31:09,070 It doesn't divide up your 1 through N 429 00:31:09,070 --> 00:31:11,050 very nicely into sub chunks. 430 00:31:11,050 --> 00:31:15,820 But we still want to use this theta to chop up 1 through N 431 00:31:15,820 --> 00:31:20,860 into smaller spaces so that we can iterate and do 432 00:31:20,860 --> 00:31:24,730 density increment. 433 00:31:24,730 --> 00:31:25,910 All right. 434 00:31:25,910 --> 00:31:28,920 So let's see what we can do. 435 00:31:28,920 --> 00:31:36,980 So given this theta, what we would like to do 436 00:31:36,980 --> 00:31:50,260 is to partition this 1 through N into subprogressions. 437 00:31:53,564 --> 00:32:07,030 OK, so chop up 1 through N into sub APs such that 438 00:32:07,030 --> 00:32:12,230 if you evaluate for-- so this theta is fixed. 439 00:32:12,230 --> 00:32:17,990 So on each sub AP, this function here 440 00:32:17,990 --> 00:32:28,000 is roughly constant on each of your parts. 441 00:32:31,150 --> 00:32:34,000 Last time, we had this Fourier character, 442 00:32:34,000 --> 00:32:37,120 and then we chopped it up using these three hyperplanes. 443 00:32:37,120 --> 00:32:40,390 And each hyperplane, the Fourier character 444 00:32:40,390 --> 00:32:43,078 is literally constant, OK? 445 00:32:43,078 --> 00:32:46,670 So you have-- and so that's what we work with. 446 00:32:46,670 --> 00:32:49,450 And now, you cannot get them to be exactly constant, 447 00:32:49,450 --> 00:32:53,020 but the next best thing we can hope for is to get this Fourier 448 00:32:53,020 --> 00:32:56,097 character to be roughly constant. 449 00:32:56,097 --> 00:32:58,430 OK, so we're going to do some positioning that allows us 450 00:32:58,430 --> 00:33:00,860 to achieve this characteristic. 451 00:33:00,860 --> 00:33:03,740 And let me give you some intuition 452 00:33:03,740 --> 00:33:04,910 about why this is true. 453 00:33:04,910 --> 00:33:07,570 And this is not exactly a surprising fact. 454 00:33:07,570 --> 00:33:10,040 The intuition is just that if you 455 00:33:10,040 --> 00:33:11,900 look at what this function behaves like-- 456 00:33:17,444 --> 00:33:19,950 all right, so what's going on here? 457 00:33:19,950 --> 00:33:27,600 You are on the unit circle, and you are jumping by theta. 458 00:33:31,288 --> 00:33:42,080 OK, so you just keep jumping by theta and so on. 459 00:33:42,080 --> 00:33:49,370 And I want to show that I can sharp up my progression 460 00:33:49,370 --> 00:33:57,690 into a bunch of almost periodic pieces, where in each part, 461 00:33:57,690 --> 00:34:01,370 I'm staying inside a small arc. 462 00:34:01,370 --> 00:34:06,560 So in the extreme case of this where it is very easy to see 463 00:34:06,560 --> 00:34:18,159 is if x is some rational number, a over b, with b fairly small, 464 00:34:18,159 --> 00:34:22,010 then we can-- 465 00:34:22,010 --> 00:34:34,190 so then, this character is actually constant on APs 466 00:34:34,190 --> 00:34:35,659 with common difference b. 467 00:34:43,063 --> 00:34:43,563 Yep? 468 00:34:43,563 --> 00:34:46,708 AUDIENCE: Is theta supposed to be [INAUDIBLE]?? 469 00:34:46,708 --> 00:34:48,875 YUFEI ZHAO: Ah, so theta, yes. so theta-- thank you. 470 00:34:48,875 --> 00:34:50,270 So theta 2 pi. 471 00:34:54,174 --> 00:34:57,370 AUDIENCE: Like, is x equal to your theta? 472 00:34:57,370 --> 00:34:58,078 YUFEI ZHAO: Yeah. 473 00:34:58,078 --> 00:34:59,054 Thank you. 474 00:34:59,054 --> 00:35:00,520 So theta equals-- yeah. 475 00:35:00,520 --> 00:35:06,180 So if theta is some rational with some small denominator-- 476 00:35:06,180 --> 00:35:13,420 so then you are literally jumping in periodic steps 477 00:35:13,420 --> 00:35:15,690 on the unit circle. 478 00:35:15,690 --> 00:35:21,230 So if you partition N according to the exact same periods, 479 00:35:21,230 --> 00:35:26,300 you have that this character is exactly constant in each 480 00:35:26,300 --> 00:35:28,190 of your progressions. 481 00:35:28,190 --> 00:35:32,210 Now, in general, the theta you get out of that proof 482 00:35:32,210 --> 00:35:35,400 might not have this very nice form, 483 00:35:35,400 --> 00:35:38,340 but we can at least approximately 484 00:35:38,340 --> 00:35:39,840 achieve the desired effect. 485 00:35:42,630 --> 00:35:43,130 OK. 486 00:35:46,890 --> 00:35:47,700 Any questions? 487 00:36:15,560 --> 00:36:16,260 OK. 488 00:36:16,260 --> 00:36:20,100 So to achieve approximately the desired effect, what we'll do 489 00:36:20,100 --> 00:36:25,370 is to find something so that b times theta 490 00:36:25,370 --> 00:36:29,485 is not quite an integer, but very close to an integer. 491 00:36:29,485 --> 00:36:31,610 OK, so this, probably many of you have seen before. 492 00:36:31,610 --> 00:36:37,470 It's a classic pigeonhole-type result. 493 00:36:37,470 --> 00:36:41,510 It's usually attributed to Dirichlet. 494 00:36:45,020 --> 00:36:54,990 So if you have theta, a real number, and a delta, 495 00:36:54,990 --> 00:37:00,300 kind of a tolerance, then there exists 496 00:37:00,300 --> 00:37:09,350 a positive integer d at most 1 over delta such 497 00:37:09,350 --> 00:37:19,030 that d times theta is very close to an integer. 498 00:37:19,030 --> 00:37:23,720 OK, so this norm here is distance 499 00:37:23,720 --> 00:37:25,850 to the closest integer. 500 00:37:33,215 --> 00:37:38,390 All right, so the proof is by pigeonhole principle. 501 00:37:38,390 --> 00:37:44,560 So if we let N be 1 over delta rounded down 502 00:37:44,560 --> 00:37:51,970 and consider the numbers 0, theta, 2 theta, 3 theta, and so 503 00:37:51,970 --> 00:37:55,630 on, to N theta-- 504 00:37:55,630 --> 00:38:08,040 so by pigeonhole, there exists i theta and j theta, so 505 00:38:08,040 --> 00:38:12,660 two different terms of the sequence such 506 00:38:12,660 --> 00:38:20,700 that they differ by less than-- at most 507 00:38:20,700 --> 00:38:22,590 delta in their fractional parts. 508 00:38:30,260 --> 00:38:36,920 OK, so now take d to be difference between i and j. 509 00:38:36,920 --> 00:38:37,820 OK, and that works. 510 00:38:48,060 --> 00:38:48,560 OK. 511 00:38:48,560 --> 00:38:52,720 So even though you don't have exactly rational, 512 00:38:52,720 --> 00:38:55,210 you have approximately rational. 513 00:38:55,210 --> 00:38:56,200 So this is a-- 514 00:38:56,200 --> 00:38:58,970 it's a simple rational approximation statement. 515 00:38:58,970 --> 00:39:00,890 And using this rational approximation, 516 00:39:00,890 --> 00:39:05,660 we can now try to do the intuition here, 517 00:39:05,660 --> 00:39:10,490 pretending that we're working with rational numbers, indeed. 518 00:39:16,442 --> 00:39:20,670 OK, so if we take eta between 0 and 1 519 00:39:20,670 --> 00:39:34,850 and theta irrational and suppose N is large enough-- 520 00:39:34,850 --> 00:39:37,090 OK, so here, C means there exists 521 00:39:37,090 --> 00:39:40,490 some sufficiently large-- 522 00:39:40,490 --> 00:39:45,420 some constant C such that the statement is true, OK? 523 00:39:45,420 --> 00:39:49,420 So suppose you think a million here. 524 00:39:49,420 --> 00:39:51,160 That should be fine. 525 00:39:51,160 --> 00:39:55,740 So then there exists-- 526 00:39:55,740 --> 00:40:12,330 so then one can partition 1 through N into sub-APs, 527 00:40:12,330 --> 00:40:14,960 which we'll call P i. 528 00:40:14,960 --> 00:40:24,250 And each having length between cube root of N 529 00:40:24,250 --> 00:40:33,110 and twice the cube root of N such that this character 530 00:40:33,110 --> 00:40:35,700 that we want to stay roughly constant indeed 531 00:40:35,700 --> 00:40:39,550 does not change very much. 532 00:40:39,550 --> 00:40:47,610 If you look at two terms in the same AP, in the sub-AP, 533 00:40:47,610 --> 00:41:00,670 then the value of this character on each P sub i 534 00:41:00,670 --> 00:41:03,080 is roughly the same. 535 00:41:03,080 --> 00:41:08,920 So they don't vary by more than eta on each P i. 536 00:41:08,920 --> 00:41:16,350 So here, we're partitioning this 1 through N into a sub-A piece 537 00:41:16,350 --> 00:41:20,355 so that this guy here stays roughly constant. 538 00:41:25,940 --> 00:41:26,440 OK. 539 00:41:26,440 --> 00:41:29,470 Any questions? 540 00:41:29,470 --> 00:41:29,970 All right. 541 00:41:29,970 --> 00:41:33,728 So think about how you might prove this. 542 00:41:33,728 --> 00:41:34,770 Let's take a quick break. 543 00:41:37,580 --> 00:41:41,160 So you see, we are basically following the same strategy 544 00:41:41,160 --> 00:41:44,770 as the proof from last time, but this second step, 545 00:41:44,770 --> 00:41:47,370 which we're on right now, needs to be somewhat modified 546 00:41:47,370 --> 00:41:52,230 because you cannot cut this space up into pieces where 547 00:41:52,230 --> 00:41:54,330 your character is constant. 548 00:41:54,330 --> 00:41:58,140 Well, if they're roughly constant then we're go to go, 549 00:41:58,140 --> 00:42:01,300 so that's what we're doing now. 550 00:42:01,300 --> 00:42:03,000 So let's prove the statement up there. 551 00:42:15,350 --> 00:42:15,860 All right. 552 00:42:15,860 --> 00:42:18,950 So let's prove this statement over here. 553 00:42:18,950 --> 00:42:37,600 So using Dirichlet's lemma, we find that there exists some d. 554 00:42:37,600 --> 00:42:39,615 OK, so I'll write down some number for now. 555 00:42:39,615 --> 00:42:40,490 Don't worry about it. 556 00:42:40,490 --> 00:42:45,090 It will come up shortly why I write this specific quantity. 557 00:42:47,670 --> 00:43:06,330 So there exists some d which is not too big, such that d theta 558 00:43:06,330 --> 00:43:09,830 is very close to an integer. 559 00:43:09,830 --> 00:43:12,540 So now, I'm literally applying Dirichlet's lemma. 560 00:43:16,050 --> 00:43:17,680 OK. 561 00:43:17,680 --> 00:43:24,826 So given such d-- 562 00:43:24,826 --> 00:43:27,640 so how big is this d? 563 00:43:27,640 --> 00:43:32,740 You see that because I assumed that N is sufficiently large, 564 00:43:32,740 --> 00:43:48,800 if we choose that C large enough, d is at most root N. So 565 00:43:48,800 --> 00:43:53,570 given such d, which is at most root N, 566 00:43:53,570 --> 00:44:04,180 you can partition 1 through N into subprogressions 567 00:44:04,180 --> 00:44:09,580 with common difference d. 568 00:44:09,580 --> 00:44:14,550 Essentially, look at-- let's do classes mod d. 569 00:44:14,550 --> 00:44:19,130 So they're all going to have length by basically N over d. 570 00:44:19,130 --> 00:44:25,140 And I chop them up a little bit further to get-- 571 00:44:25,140 --> 00:44:33,390 so a piece of length between cube root of N 572 00:44:33,390 --> 00:44:40,990 and twice cube root of N. 573 00:44:40,990 --> 00:44:41,490 OK. 574 00:44:41,490 --> 00:44:44,520 So I'm going to make sure that all of my APs 575 00:44:44,520 --> 00:44:48,000 are roughly the same length. 576 00:44:48,000 --> 00:44:56,090 And now, inside each subprogression-- 577 00:45:00,790 --> 00:45:06,610 let me call this subprogression P prime, subprogression P, 578 00:45:06,610 --> 00:45:09,430 let's look at how much this character value can 579 00:45:09,430 --> 00:45:13,119 vary inside this progression. 580 00:45:19,346 --> 00:45:20,304 All right. 581 00:45:26,010 --> 00:45:29,420 OK, so how much can this vary? 582 00:45:29,420 --> 00:45:36,330 Well, because theta is such that d times theta is very close 583 00:45:36,330 --> 00:45:41,778 to an integer and the length of each progression is not too 584 00:45:41,778 --> 00:45:42,570 large-- so here's-- 585 00:45:42,570 --> 00:45:44,710 I want some control on the length. 586 00:45:44,710 --> 00:45:49,020 So we find that the maximum variation 587 00:45:49,020 --> 00:45:56,695 is, at most, the size of P, the length of P, times-- 588 00:45:59,580 --> 00:46:04,530 so this-- that difference over there. 589 00:46:04,530 --> 00:46:08,920 So all of these are exponential, so I can shift them. 590 00:46:08,920 --> 00:46:19,995 Well, the length of P is at most twice cube root of N. And-- 591 00:46:19,995 --> 00:46:22,160 OK, so what is this quantity? 592 00:46:22,160 --> 00:46:25,470 So the point is that if this fractional part here 593 00:46:25,470 --> 00:46:29,080 is very close to an integer, then e to that, 594 00:46:29,080 --> 00:46:31,560 e to the 2 pi times that i times some number should 595 00:46:31,560 --> 00:46:36,630 be very close to 1, because what is happening here? 596 00:46:36,630 --> 00:46:42,050 This is the distance between those two points 597 00:46:42,050 --> 00:46:45,230 on the circle, which is at most bounded 598 00:46:45,230 --> 00:46:46,490 by the length of the arc. 599 00:46:57,460 --> 00:47:02,530 OK, so cord length, at most of the arc. 600 00:47:02,530 --> 00:47:05,560 So now, you put everything here together, 601 00:47:05,560 --> 00:47:09,230 and apply the bound that we got on d theta. 602 00:47:09,230 --> 00:47:11,410 So this is the reason for choosing that weird number 603 00:47:11,410 --> 00:47:13,960 up there. 604 00:47:13,960 --> 00:47:30,050 We find that the variation within each progression 605 00:47:30,050 --> 00:47:31,903 is at most eta, right? 606 00:47:31,903 --> 00:47:33,320 So the variation of this character 607 00:47:33,320 --> 00:47:36,175 within each progression is not very large, OK? 608 00:47:36,175 --> 00:47:37,050 And that's the claim. 609 00:47:39,830 --> 00:47:41,780 Any questions? 610 00:47:41,780 --> 00:47:45,920 All right, so this is the analogous claim to the one 611 00:47:45,920 --> 00:47:46,780 that we had-- 612 00:47:46,780 --> 00:47:48,500 the one that we used last time, where 613 00:47:48,500 --> 00:47:51,950 we said that the character is constant on each coset 614 00:47:51,950 --> 00:47:52,915 of the hyperplane. 615 00:47:52,915 --> 00:47:55,430 They're not exactly constant, but almost good enough. 616 00:48:05,610 --> 00:48:06,300 All right. 617 00:48:06,300 --> 00:48:10,020 So the goal of step 2 is to show an energy-- 618 00:48:10,020 --> 00:48:13,020 show a density increment, that if you have a large Fourier 619 00:48:13,020 --> 00:48:15,180 coefficient, then we want to claim 620 00:48:15,180 --> 00:48:18,720 that the density goes up significantly 621 00:48:18,720 --> 00:48:21,940 on some subprogression. 622 00:48:21,940 --> 00:48:24,910 And the next part, the next lemma, 623 00:48:24,910 --> 00:48:27,280 will get us to that goal. 624 00:48:27,280 --> 00:48:29,920 And this part is very similar to the one 625 00:48:29,920 --> 00:48:34,600 that we saw from last time, but with this new partition 626 00:48:34,600 --> 00:48:35,980 in mind, like I said. 627 00:48:35,980 --> 00:48:47,860 If you have A that is 3AP-free with density alpha, 628 00:48:47,860 --> 00:48:57,990 and N is large enough, then there 629 00:48:57,990 --> 00:49:08,980 exists some subprogression P. So by subprogression, 630 00:49:08,980 --> 00:49:11,560 I just mean that I'm starting with original progression 631 00:49:11,560 --> 00:49:15,340 1 through N, and I'm zooming into some subprogression, 632 00:49:15,340 --> 00:49:22,160 with the length of P fairly long, so the length of P 633 00:49:22,160 --> 00:49:29,430 is at least cube root of N, and such that A, 634 00:49:29,430 --> 00:49:32,760 when restricted to this subprogression, 635 00:49:32,760 --> 00:49:36,295 has a density increment. 636 00:49:41,995 --> 00:49:44,950 OK, so originally, the density of A is alpha, 637 00:49:44,950 --> 00:49:47,300 so we're zooming into some subprogression P, 638 00:49:47,300 --> 00:49:48,980 which is a pretty long subprogression, 639 00:49:48,980 --> 00:49:50,870 where the density goes up significantly 640 00:49:50,870 --> 00:49:53,360 from A to essentially A-- 641 00:49:53,360 --> 00:49:56,330 from alpha to roughly alpha plus alpha squared. 642 00:50:00,240 --> 00:50:00,740 OK. 643 00:50:07,120 --> 00:50:10,660 So we start with A, a 3AP-free set. 644 00:50:10,660 --> 00:50:25,110 So from step 1, there exists some theta with large-- 645 00:50:25,110 --> 00:50:27,932 so that corresponds to a large Fourier coefficient. 646 00:50:31,790 --> 00:50:46,070 So this sum here is large. 647 00:50:48,782 --> 00:50:54,710 OK, and now we use-- 648 00:50:54,710 --> 00:51:00,510 OK, so-- so step 1 obtains us, you know, this consequence. 649 00:51:00,510 --> 00:51:14,780 And from this theta, now we apply the lemma up there to-- 650 00:51:14,780 --> 00:51:16,760 so we apply lemma with, let's say, 651 00:51:16,760 --> 00:51:21,350 eta being alpha squared over 30. 652 00:51:21,350 --> 00:51:23,690 OK, so the exact constants are not so important. 653 00:51:23,690 --> 00:51:31,010 But when we apply the lemma to partition, 654 00:51:31,010 --> 00:51:37,720 and into a bunch of subprogressions, 655 00:51:37,720 --> 00:51:40,570 which we'll call P1 through Pk. 656 00:51:43,462 --> 00:51:48,080 And each of these progressions have 657 00:51:48,080 --> 00:51:59,750 length between cube root of N and twice cube root of N. 658 00:51:59,750 --> 00:52:04,070 And I want to understand what happens to the density of A 659 00:52:04,070 --> 00:52:07,680 when restricted to these progressions. 660 00:52:07,680 --> 00:52:14,030 So starting with this inequality over here, 661 00:52:14,030 --> 00:52:17,150 which suggests to us that there must be some deviation. 662 00:52:30,191 --> 00:52:34,700 OK, so starting with what we saw. 663 00:52:34,700 --> 00:52:40,950 And now, inside each progression this e x theta 664 00:52:40,950 --> 00:52:44,050 is roughly constant. 665 00:52:44,050 --> 00:52:46,760 So if you pretend them as actually constant, 666 00:52:46,760 --> 00:52:54,230 I can break up the sum, depending on where the x's lie. 667 00:52:54,230 --> 00:52:57,530 So i from 1 to k. 668 00:52:57,530 --> 00:53:01,492 And let me sum inside each progression. 669 00:53:13,560 --> 00:53:17,610 So by triangle inequality, I can upper bound the first sum 670 00:53:17,610 --> 00:53:23,540 by where I now cut the sum into progression by progression. 671 00:53:23,540 --> 00:53:29,520 And on each progression, this character is roughly constant. 672 00:53:29,520 --> 00:53:34,760 So let me take out the maximum possible deviations 673 00:53:34,760 --> 00:53:36,440 from them being constant. 674 00:53:36,440 --> 00:53:48,525 So upper bound-- again, you'll find that we can essentially 675 00:53:48,525 --> 00:53:49,025 pretend-- 676 00:53:52,190 --> 00:53:55,410 all right, so if each exponential 677 00:53:55,410 --> 00:53:57,450 is constant on each subprogression, 678 00:53:57,450 --> 00:54:00,750 then I might as well just have this sum here. 679 00:54:00,750 --> 00:54:03,510 But I lose a little bit, because it's not exactly constant. 680 00:54:03,510 --> 00:54:04,680 It's almost constant. 681 00:54:04,680 --> 00:54:06,140 So I loose a little bit. 682 00:54:06,140 --> 00:54:11,350 And that little bit is this eta. 683 00:54:11,350 --> 00:54:13,440 So you lose that little bit of eta. 684 00:54:13,440 --> 00:54:18,960 And so on each progression, P i, you 685 00:54:18,960 --> 00:54:22,830 lose at most something that's essentially of alpha squared 686 00:54:22,830 --> 00:54:24,265 times the length of P i. 687 00:54:27,240 --> 00:54:29,160 OK. 688 00:54:29,160 --> 00:54:32,880 Now, you see, I've chosen the error parameter 689 00:54:32,880 --> 00:54:37,260 so that everything I've lost is not 690 00:54:37,260 --> 00:54:41,250 so much more than the initial bound I began with. 691 00:54:41,250 --> 00:54:47,100 So in particular, we see that even 692 00:54:47,100 --> 00:54:49,800 if we had pretended that the characters were constant, 693 00:54:49,800 --> 00:54:54,090 on each progression we would have still obtained some lower 694 00:54:54,090 --> 00:54:56,880 bound of the total deviation. 695 00:55:08,680 --> 00:55:09,860 OK. 696 00:55:09,860 --> 00:55:15,130 And what is this quantity over here? 697 00:55:15,130 --> 00:55:18,880 Oh, you see, I'm restricting each sum 698 00:55:18,880 --> 00:55:23,590 to each subprogression, but the sum 699 00:55:23,590 --> 00:55:26,440 here, even though it's the sum, but it's really 700 00:55:26,440 --> 00:55:31,990 counting how many elements of A are in that progression. 701 00:55:31,990 --> 00:55:35,710 So this sum over here is the same thing. 702 00:55:35,710 --> 00:55:38,543 OK, so let me write it in a new board. 703 00:55:43,480 --> 00:55:45,213 Oh, we don't need step 1 anymore. 704 00:55:49,077 --> 00:55:51,270 All right. 705 00:55:51,270 --> 00:55:52,440 So what we have-- 706 00:56:01,386 --> 00:56:05,431 OK, so left-hand side over there is this quantity here, 707 00:56:05,431 --> 00:56:07,690 all right? 708 00:56:07,690 --> 00:56:15,890 We see that the right-hand side, even though you have that sum, 709 00:56:15,890 --> 00:56:19,010 it is really just counting how many elements of A 710 00:56:19,010 --> 00:56:23,360 are in each progression versus how many you should expect 711 00:56:23,360 --> 00:56:28,750 based on the overall density of A. OK, 712 00:56:28,750 --> 00:56:32,220 so that should look similar to what we got last time. 713 00:56:32,220 --> 00:56:35,280 I know the intuition should be that, well, 714 00:56:35,280 --> 00:56:42,210 if the average deviation is large, then one of them, 715 00:56:42,210 --> 00:56:47,447 one of these terms, should have the density increment. 716 00:56:51,190 --> 00:56:53,950 If you try to do the next step somewhat naively, 717 00:56:53,950 --> 00:56:56,840 you run into an issue, because it could be-- 718 00:56:56,840 --> 00:56:59,290 now, here you have k terms. 719 00:56:59,290 --> 00:57:07,270 It could be that you have all the densities except for one 720 00:57:07,270 --> 00:57:12,640 going up only slightly, and one density dropping dramatically, 721 00:57:12,640 --> 00:57:15,580 in which case you may not have a significant density 722 00:57:15,580 --> 00:57:18,210 increment, all right? 723 00:57:18,210 --> 00:57:20,840 So we want to show that on some progression 724 00:57:20,840 --> 00:57:24,350 the density increases significantly. 725 00:57:24,350 --> 00:57:26,870 So far, from this inequality, we just 726 00:57:26,870 --> 00:57:29,720 know that there is some subprogression where 727 00:57:29,720 --> 00:57:32,450 the density changes significantly. 728 00:57:32,450 --> 00:57:34,850 But of course, the overall density, the average density, 729 00:57:34,850 --> 00:57:36,120 should remain constant. 730 00:57:36,120 --> 00:57:38,950 So if some goes up, others must go down. 731 00:57:38,950 --> 00:57:41,180 But if you just try to do an averaging argument, 732 00:57:41,180 --> 00:57:42,890 you have to be careful, OK? 733 00:57:42,890 --> 00:57:45,290 So there was a trick last time, which we didn't really 734 00:57:45,290 --> 00:57:49,910 need last time, but now, it's much more useful, where 735 00:57:49,910 --> 00:57:52,220 I want to show that if this holds, 736 00:57:52,220 --> 00:57:55,850 then some P i sees a large energy-- 737 00:57:55,850 --> 00:57:58,520 sees a large density increment. 738 00:57:58,520 --> 00:58:08,990 And to do that, let me rewrite the sum as the following, 739 00:58:08,990 --> 00:58:11,660 so I keep the same expression. 740 00:58:14,880 --> 00:58:18,980 And I add a term, which is the same thing, 741 00:58:18,980 --> 00:58:20,560 but without the absolute value. 742 00:58:29,434 --> 00:58:35,000 OK, so you see these guys, they total to 0, 743 00:58:35,000 --> 00:58:40,350 so adding that term doesn't change my expression. 744 00:58:40,350 --> 00:58:45,460 But now, the summand is always non-negative. 745 00:58:45,460 --> 00:58:49,380 So it's either 0 or twice this number, 746 00:58:49,380 --> 00:58:51,500 depending on the sign of that number. 747 00:58:54,610 --> 00:58:55,110 OK. 748 00:58:55,110 --> 00:58:57,600 So comparing left-hand side and right-hand side, 749 00:58:57,600 --> 00:59:00,790 we see that there must be some i. 750 00:59:00,790 --> 00:59:08,730 So hence, there exists some eye such that the left-hand side-- 751 00:59:13,410 --> 00:59:15,270 the i-th term on the left hand side 752 00:59:15,270 --> 00:59:18,272 is less than or equal to the i-th term 753 00:59:18,272 --> 00:59:19,230 on the right-hand side. 754 00:59:24,940 --> 00:59:28,640 And in particular, that term should be positive, 755 00:59:28,640 --> 00:59:31,460 so it implies-- 756 00:59:31,460 --> 00:59:35,740 OK, so how can you get this inequality? 757 00:59:35,740 --> 00:59:40,540 It implies simply that the restriction of a to this P i 758 00:59:40,540 --> 00:59:50,810 is at least alpha plus alpha squared over 40 times P i. 759 00:59:50,810 --> 00:59:54,320 So this claim here just says that on the i-th progression, 760 00:59:54,320 --> 00:59:58,570 there's a significant energy increment. 761 00:59:58,570 --> 01:00:02,342 If it's more decrement, that term would have been 0. 762 01:00:02,342 --> 01:00:04,610 So remember that. 763 01:00:16,040 --> 01:00:17,280 OK. 764 01:00:17,280 --> 01:00:20,610 So this achieves what we were looking for in step 2, 765 01:00:20,610 --> 01:00:24,720 namely to find that there is a density increment 766 01:00:24,720 --> 01:00:28,041 on some long subprogression. 767 01:00:28,041 --> 01:00:30,630 OK, so now we can go to step 3, which is basically 768 01:00:30,630 --> 01:00:35,460 the same as what we saw last time, where now we 769 01:00:35,460 --> 01:00:38,929 want to iterate this density increment. 770 01:00:47,911 --> 01:00:55,900 OK, so it's basically the same argument as last time, 771 01:00:55,900 --> 01:01:00,300 but you start with density alpha, 772 01:01:00,300 --> 01:01:04,140 and each step in the iteration, the density 773 01:01:04,140 --> 01:01:05,600 goes up quite a bit. 774 01:01:09,740 --> 01:01:14,030 And we want to control the total number of steps, 775 01:01:14,030 --> 01:01:20,704 knowing that the final density is at most 1, always. 776 01:01:20,704 --> 01:01:23,370 OK. 777 01:01:23,370 --> 01:01:25,020 So how many steps can you take? 778 01:01:25,020 --> 01:01:28,080 Right, so this was the same argument that we saw last time. 779 01:01:28,080 --> 01:01:38,450 We see that starting with alpha, alpha 0 being alpha, 780 01:01:38,450 --> 01:01:41,810 it doubles after a certain number of steps, right? 781 01:01:41,810 --> 01:01:45,590 So we double after-- 782 01:01:45,590 --> 01:01:48,710 OK, so how many steps do you need? 783 01:01:48,710 --> 01:01:53,430 Well, I want to get from alpha to 2 alpha. 784 01:01:53,430 --> 01:02:00,960 So I need at most alpha over 40 steps. 785 01:02:00,960 --> 01:02:03,770 OK, so last time I was slightly sloppy. 786 01:02:03,770 --> 01:02:09,630 And so there's basically a floor, upper floor down-- 787 01:02:09,630 --> 01:02:11,060 rounding up or down situation. 788 01:02:11,060 --> 01:02:13,620 But I should add a plus 1. 789 01:02:13,620 --> 01:02:14,120 Yeah. 790 01:02:14,120 --> 01:02:16,120 AUDIENCE: Shouldn't it be 40 over alpha? 791 01:02:16,120 --> 01:02:17,340 YUFEI ZHAO: 40-- thank you. 792 01:02:17,340 --> 01:02:19,930 40 over alpha, yeah. 793 01:02:19,930 --> 01:02:25,080 So you double after at most that many steps. 794 01:02:25,080 --> 01:02:31,560 And then now, you add density at least 2 alpha. 795 01:02:31,560 --> 01:02:39,880 So we double after at most 20 over alpha steps, and so on. 796 01:02:39,880 --> 01:02:44,690 And we double at most-- 797 01:02:44,690 --> 01:02:51,690 well, basically, log sub 2 of 1 over alpha times. 798 01:02:55,658 --> 01:02:58,600 OK, so anyway, putting everything together, 799 01:02:58,600 --> 01:03:04,350 we see that the total number of steps 800 01:03:04,350 --> 01:03:08,790 is at most on the order of 1 over alpha. 801 01:03:12,340 --> 01:03:15,490 When you stop, you can only stop for one reason, 802 01:03:15,490 --> 01:03:18,210 because in the-- 803 01:03:18,210 --> 01:03:18,710 yeah. 804 01:03:18,710 --> 01:03:19,880 So, yeah. 805 01:03:19,880 --> 01:03:24,790 So in step 1, remember, the iteration 806 01:03:24,790 --> 01:03:28,762 said that the process terminates. 807 01:03:28,762 --> 01:03:37,589 The process can always go on, and then terminates 808 01:03:37,589 --> 01:03:43,170 if the length-- 809 01:03:43,170 --> 01:03:45,590 so you're now at step i, so let N i be 810 01:03:45,590 --> 01:03:54,580 the length of the progression, that step i is at most C times 811 01:03:54,580 --> 01:03:56,990 alpha i to the minus 12th. 812 01:03:56,990 --> 01:04:01,180 So we have a-- 813 01:04:01,180 --> 01:04:04,980 right, so provided that N is large enough, 814 01:04:04,980 --> 01:04:07,020 you can always pass to a subprogression. 815 01:04:07,020 --> 01:04:09,540 And here, when you pass to subprogression, of course, 816 01:04:09,540 --> 01:04:12,030 you can re-label that subprogression. 817 01:04:12,030 --> 01:04:15,140 And it's now, you know, 1 through N i. 818 01:04:15,140 --> 01:04:16,553 Right, so I can-- it's-- 819 01:04:16,553 --> 01:04:17,970 all the progressions are basically 820 01:04:17,970 --> 01:04:21,373 the same as the first set of positive integers. 821 01:04:21,373 --> 01:04:23,850 I'm sorry, prefix of the positive integers. 822 01:04:23,850 --> 01:04:26,590 So when we stop a step i, you must 823 01:04:26,590 --> 01:04:31,290 have N sub i being at most this quantity over here, which 824 01:04:31,290 --> 01:04:36,510 is at most C times the initial density raised to this minus 825 01:04:36,510 --> 01:04:37,450 12th. 826 01:04:37,450 --> 01:04:49,050 So therefore, the initial length N of the space is bounded by-- 827 01:04:49,050 --> 01:04:55,595 well, each time we went down by a cube root at most. 828 01:04:55,595 --> 01:04:57,560 Right, so the fine-- 829 01:04:57,560 --> 01:05:00,980 if you stop a step i, then the initial length 830 01:05:00,980 --> 01:05:07,670 is at most N sub i to the 3 times 3 to the power of i 831 01:05:07,670 --> 01:05:10,922 each time you're doing a cube root. 832 01:05:10,922 --> 01:05:13,135 OK, so you put everything together. 833 01:05:20,280 --> 01:05:22,360 At most that many iterations when you stop 834 01:05:22,360 --> 01:05:24,781 the length is at most this. 835 01:05:24,781 --> 01:05:26,680 So you put them together, and then you 836 01:05:26,680 --> 01:05:37,320 find that the N must be at most double exponential in 1 837 01:05:37,320 --> 01:05:40,140 over the density. 838 01:05:40,140 --> 01:05:47,670 In other words, the density is at most 1 839 01:05:47,670 --> 01:05:55,665 over log log N, which is what we claimed in Roth's theorem, 840 01:05:55,665 --> 01:05:57,060 so what we claimed up there. 841 01:06:01,460 --> 01:06:01,960 OK. 842 01:06:01,960 --> 01:06:03,228 So that finishes the proof. 843 01:06:12,210 --> 01:06:13,307 Any questions? 844 01:06:18,780 --> 01:06:22,230 So the message here is that it's the same proof as last time, 845 01:06:22,230 --> 01:06:24,660 but we need to do a bit more work. 846 01:06:24,660 --> 01:06:26,220 And none of this work is difficult, 847 01:06:26,220 --> 01:06:27,780 but there are more technical. 848 01:06:27,780 --> 01:06:29,580 And that's often the theme that you 849 01:06:29,580 --> 01:06:31,500 see in additive combinatorics. 850 01:06:31,500 --> 01:06:34,650 This is part of the reason why the finite field model is 851 01:06:34,650 --> 01:06:38,130 a really nice playground, because there, things 852 01:06:38,130 --> 01:06:41,850 tend to be often cleaner, but the idea's often similar, 853 01:06:41,850 --> 01:06:43,140 or the same ideas. 854 01:06:43,140 --> 01:06:43,890 Not always. 855 01:06:43,890 --> 01:06:46,242 Next lecture, we'll see one technique 856 01:06:46,242 --> 01:06:47,700 where there's a dramatic difference 857 01:06:47,700 --> 01:06:50,940 between the finite field vector space and over the integers. 858 01:06:50,940 --> 01:06:53,170 But for many things in additive combinatorics, 859 01:06:53,170 --> 01:06:54,720 the finite field vector space is just 860 01:06:54,720 --> 01:06:58,172 a nicer place to be in to try all your ideas and techniques. 861 01:07:04,440 --> 01:07:09,090 Let me comment on some analogies between these two approaches 862 01:07:09,090 --> 01:07:10,200 and compare the bounds. 863 01:07:13,090 --> 01:07:15,280 So on one hand-- 864 01:07:15,280 --> 01:07:20,130 OK, so we saw last time this proof in F3 to the N, 865 01:07:20,130 --> 01:07:26,770 and now in the integers inside this interval of length 866 01:07:26,770 --> 01:07:33,280 N. So let me write uppercase N in both cases 867 01:07:33,280 --> 01:07:38,205 to be the size of the overall ambient space. 868 01:07:38,205 --> 01:07:43,310 OK, so what kind of bounds do we get in both situations? 869 01:07:43,310 --> 01:07:46,050 So last time, for-- 870 01:07:46,050 --> 01:07:51,720 in F3 to the N, we got a bound which 871 01:07:51,720 --> 01:08:06,220 is of the order N over log N, whereas today, the bound is 872 01:08:06,220 --> 01:08:08,560 somewhat worse. 873 01:08:08,560 --> 01:08:09,920 It's a little bit worse. 874 01:08:09,920 --> 01:08:13,690 Now we lose an extra log. 875 01:08:13,690 --> 01:08:16,810 So where do we lose an extra log in this argument? 876 01:08:16,810 --> 01:08:19,060 So where does these two argument-- 877 01:08:19,060 --> 01:08:22,980 where do these two arguments differ, quantitatively? 878 01:08:26,240 --> 01:08:26,740 Yep? 879 01:08:26,740 --> 01:08:29,569 AUDIENCE: When you're dividing by 3 versus [INAUDIBLE]?? 880 01:08:29,569 --> 01:08:31,560 YUFEI ZHAO: OK, so you're dividing by-- 881 01:08:31,560 --> 01:08:41,960 so here, in each iteration, over here, 882 01:08:41,960 --> 01:08:46,399 your size of the iteration-- 883 01:08:46,399 --> 01:08:48,300 I mean, each iteration the size of the space 884 01:08:48,300 --> 01:08:53,279 goes down by a factor of 3, whereas over here, it 885 01:08:53,279 --> 01:08:56,979 could go down by a cube root. 886 01:08:56,979 --> 01:08:58,399 And that's precisely right. 887 01:08:58,399 --> 01:09:02,420 So that explains for this extra log in the balance. 888 01:09:05,630 --> 01:09:08,490 So while this is a great analogy, 889 01:09:08,490 --> 01:09:10,760 it's not a perfect analogy. 890 01:09:10,760 --> 01:09:13,279 You see there is this divergence here between the two 891 01:09:13,279 --> 01:09:14,790 situations. 892 01:09:14,790 --> 01:09:16,430 And so then you might ask, is there 893 01:09:16,430 --> 01:09:21,240 some way to avoid the loss, this extra log factor 894 01:09:21,240 --> 01:09:22,950 loss over here? 895 01:09:22,950 --> 01:09:25,260 Is there some way to carry out the strategy 896 01:09:25,260 --> 01:09:28,689 that we did last time in a way that 897 01:09:28,689 --> 01:09:31,300 is much more faithful to that strategy of passing down 898 01:09:31,300 --> 01:09:33,060 to subspaces. 899 01:09:33,060 --> 01:09:36,090 So here, we pass to progressions. 900 01:09:36,090 --> 01:09:40,649 And because we have to do this extra pigeonhole-type argument, 901 01:09:40,649 --> 01:09:41,970 it was somewhat lost-- 902 01:09:41,970 --> 01:09:48,770 we lost a power, which translated into this extra log. 903 01:09:48,770 --> 01:09:51,910 So it turns out there is some way to do this. 904 01:09:51,910 --> 01:09:53,600 So let me just briefly mention what's 905 01:09:53,600 --> 01:09:55,640 the idea that is involved, all right? 906 01:09:55,640 --> 01:09:59,930 So last time, we went down from-- 907 01:09:59,930 --> 01:10:02,660 so the main objects that we were passing 908 01:10:02,660 --> 01:10:05,180 would start with a vector space and pass down 909 01:10:05,180 --> 01:10:08,310 to a subspace, which is also a vector space, right? 910 01:10:08,310 --> 01:10:17,490 So you can define subspaces in F3 to the N by the following. 911 01:10:17,490 --> 01:10:24,310 So I can start with some set of characters U, and I define-- 912 01:10:24,310 --> 01:10:26,630 some set of characters S, and I define 913 01:10:26,630 --> 01:10:40,822 U sub S to be basically the orthogonal complement of S. 914 01:10:40,822 --> 01:10:42,972 OK, so this is a subspace. 915 01:10:42,972 --> 01:10:44,430 And these were the kind of subspace 916 01:10:44,430 --> 01:10:48,168 that we saw last time, because the S's or the R's that 917 01:10:48,168 --> 01:10:50,460 came out of the proof last time, every time we saw one, 918 01:10:50,460 --> 01:10:51,390 we threw it in. 919 01:10:51,390 --> 01:10:56,330 We cut down to a smaller subspace, and we repeat. 920 01:10:56,330 --> 01:10:59,018 But the progressions, they don't really look like this. 921 01:10:59,018 --> 01:11:00,560 So the question is, is there some way 922 01:11:00,560 --> 01:11:02,935 to do this argument so that you end up with progressions, 923 01:11:02,935 --> 01:11:04,010 and looked like that? 924 01:11:04,010 --> 01:11:06,300 And it turns out there is a way. 925 01:11:06,300 --> 01:11:08,150 And there are these objects, which 926 01:11:08,150 --> 01:11:13,215 we'll see more later in this course, called Bohr sets. 927 01:11:13,215 --> 01:11:18,030 OK, so they were used by Bourgain 928 01:11:18,030 --> 01:11:23,220 to mimic this Machoulin argument that we saw last time more 929 01:11:23,220 --> 01:11:26,010 faithfully into the integers, where 930 01:11:26,010 --> 01:11:30,840 we're going to come up with some set of integers that resemble-- 931 01:11:30,840 --> 01:11:34,950 much more closely resemble this notion of subspaces 932 01:11:34,950 --> 01:11:37,290 in the finite field setting. 933 01:11:37,290 --> 01:11:41,020 And for this, it's much easier to work inside a group. 934 01:11:41,020 --> 01:11:42,900 So instead of working in the integers, 935 01:11:42,900 --> 01:11:45,110 let's work inside and Z mod nZ. 936 01:11:45,110 --> 01:11:47,580 So we can do Fourier transform in Z mod nZ, so 937 01:11:47,580 --> 01:11:50,040 the discrete Fourier analysis here. 938 01:11:50,040 --> 01:11:52,335 So in Z mod nZ, we define-- 939 01:11:55,510 --> 01:11:59,260 so given a S, let's define this Bohr 940 01:11:59,260 --> 01:12:14,080 set to be the set of elements of Z mod nZ such 941 01:12:14,080 --> 01:12:20,470 that if you look at what really is supposed to resemble 942 01:12:20,470 --> 01:12:26,280 this thing over here, OK? 943 01:12:26,280 --> 01:12:34,280 If this quantity is small for all S-- 944 01:12:34,280 --> 01:12:36,930 OK, so we put that element into this Bohr set. 945 01:12:39,670 --> 01:12:44,870 OK, so these sets, they function much more like subspaces. 946 01:12:44,870 --> 01:12:48,890 So there are the analog of subspaces 947 01:12:48,890 --> 01:12:53,420 inside Z mod nZ, which, you know, n is prime, 948 01:12:53,420 --> 01:12:54,430 has no subgroups. 949 01:12:54,430 --> 01:12:56,990 It has no natural subspace structure. 950 01:12:56,990 --> 01:12:58,880 But by looking at these Bohr sets, 951 01:12:58,880 --> 01:13:03,020 they provide a natural way to set up 952 01:13:03,020 --> 01:13:05,330 this argument so that you can-- 953 01:13:05,330 --> 01:13:07,910 but with much more technicalities, 954 01:13:07,910 --> 01:13:13,630 repeat these kind of arguments more similar to last time, 955 01:13:13,630 --> 01:13:17,220 but passing not to subspaces but to Bohr sets. 956 01:13:17,220 --> 01:13:22,270 And then with quite a bit of extra work, 957 01:13:22,270 --> 01:13:32,370 one can obtain bounds of the quantity N over a poly log N. 958 01:13:32,370 --> 01:13:36,510 So the current best bound I mentioned last time 959 01:13:36,510 --> 01:13:40,080 is of this type, which is through further refinements 960 01:13:40,080 --> 01:13:41,347 of this technique. 961 01:13:50,610 --> 01:13:52,230 The last thing I want to mention today 962 01:13:52,230 --> 01:13:56,040 is, so far we've been talking about 3APs. 963 01:13:56,040 --> 01:13:59,600 So what about four term arithmetic progressions? 964 01:13:59,600 --> 01:14:04,998 OK, do any of the things that we talk about here work for 4APs? 965 01:14:04,998 --> 01:14:07,290 And there's an analogy to be made here compared to what 966 01:14:07,290 --> 01:14:09,085 we discussed with graphs. 967 01:14:09,085 --> 01:14:11,770 So in graphs, we had a triangle counting lemma and a triangle 968 01:14:11,770 --> 01:14:12,930 removal lemma. 969 01:14:12,930 --> 01:14:15,210 And then we said that to prove 4APs, 970 01:14:15,210 --> 01:14:18,390 we would need the hypergraph version, the simplex removal 971 01:14:18,390 --> 01:14:20,580 lemma, hypergraph regularity lemma. 972 01:14:20,580 --> 01:14:22,500 And that was much more difficult. 973 01:14:22,500 --> 01:14:24,170 And that analogy carries through, 974 01:14:24,170 --> 01:14:27,040 and the same kind of difficulties that come up. 975 01:14:27,040 --> 01:14:30,700 So it can be done, but you need something more. 976 01:14:30,700 --> 01:14:34,590 And the main message I want you to take away 977 01:14:34,590 --> 01:14:42,940 is that 4APs, while we had a counting lemma that 978 01:14:42,940 --> 01:14:45,980 says that the Fourier coefficients, 979 01:14:45,980 --> 01:14:54,540 so the Fourier transform, controls 3AP counts, 980 01:14:54,540 --> 01:14:59,100 it turns out the same is not true for 4APs. 981 01:14:59,100 --> 01:15:05,200 So the Fourier does not control 4AP counts. 982 01:15:10,410 --> 01:15:12,340 Let me give you some-- 983 01:15:12,340 --> 01:15:12,840 OK. 984 01:15:12,840 --> 01:15:14,910 So in fact, in the homework for this week, 985 01:15:14,910 --> 01:15:19,870 there's a specific example of a set where it has uniformly 986 01:15:19,870 --> 01:15:21,490 small Fourier coefficients. 987 01:15:21,490 --> 01:15:22,990 But that's the wrong number of 4APs. 988 01:15:27,130 --> 01:15:28,500 So the following-- it is true-- 989 01:15:28,500 --> 01:15:32,610 OK, so it is true that you have Szemerédi's term in-- 990 01:15:32,610 --> 01:15:34,770 let's just talk about the finite field setting, 991 01:15:34,770 --> 01:15:36,840 where things are a bit easier to discuss. 992 01:15:36,840 --> 01:15:42,510 So it is true that the size of the biggest subset of F5 993 01:15:42,510 --> 01:15:45,180 to the N is a tiny fraction-- it's a little, one 994 01:15:45,180 --> 01:15:46,530 fraction of the entire space. 995 01:15:46,530 --> 01:15:50,460 OK, I use F5 here, because if I set F3, 996 01:15:50,460 --> 01:15:53,690 it doesn't make sense to talk about 4APs. 997 01:15:53,690 --> 01:15:58,990 So F5, but it doesn't really matter which specific field. 998 01:15:58,990 --> 01:16:03,550 So you can prove this using hypergraph removal, same proof, 999 01:16:03,550 --> 01:16:07,330 verbatim, that we saw earlier, if you have hypergraph removal. 1000 01:16:07,330 --> 01:16:10,610 But if you want to try to prove it using Fourier analysis, 1001 01:16:10,610 --> 01:16:15,640 well, it doesn't work quite using the same strategy. 1002 01:16:15,640 --> 01:16:17,590 But in fact, there is a modification 1003 01:16:17,590 --> 01:16:20,800 that would allow you to make it work. 1004 01:16:20,800 --> 01:16:23,670 But you need an extension of Fourier analysis. 1005 01:16:23,670 --> 01:16:30,790 And it is known as higher order Fourier analysis, which 1006 01:16:30,790 --> 01:16:35,470 was an important development in modern additive combinatorics 1007 01:16:35,470 --> 01:16:38,590 that initially arose in Gowers' work 1008 01:16:38,590 --> 01:16:41,190 where he gave a new proof of similarities theorem. 1009 01:16:41,190 --> 01:16:42,880 So Gowers didn't work in this setting. 1010 01:16:42,880 --> 01:16:44,020 He worked in integers. 1011 01:16:44,020 --> 01:16:47,140 But many of the ideas originated from his paper, 1012 01:16:47,140 --> 01:16:49,120 and then subsequently developed by a lot 1013 01:16:49,120 --> 01:16:51,250 of people in various settings. 1014 01:16:51,250 --> 01:16:53,877 I just want to give you one specific statement, what 1015 01:16:53,877 --> 01:16:55,710 this high-order Fourier analysis looks like. 1016 01:16:55,710 --> 01:16:58,450 So it's a fancy term, and the statements often 1017 01:16:58,450 --> 01:17:00,070 get very technical. 1018 01:17:00,070 --> 01:17:04,140 But I just want to give you one concrete thing to take away. 1019 01:17:04,140 --> 01:17:05,790 All right, so for a Fourier piece, 1020 01:17:05,790 --> 01:17:08,220 higher-order Fourier analysis, roughly-- 1021 01:17:08,220 --> 01:17:12,484 OK, so it also goes by the name quadratic Fourier analysis. 1022 01:17:16,436 --> 01:17:22,370 OK, so let me give you a very specific instance 1023 01:17:22,370 --> 01:17:23,570 of the theorem. 1024 01:17:27,090 --> 01:17:31,710 And this can be sometimes called an inverse theorem 1025 01:17:31,710 --> 01:17:34,272 for quadratic Fourier analysis. 1026 01:17:40,790 --> 01:17:48,700 OK, so for every delta, there exists some c such 1027 01:17:48,700 --> 01:17:51,490 that the following is true. 1028 01:17:51,490 --> 01:18:03,390 If A is a subset of a F5 to the N with density alpha and such 1029 01:18:03,390 --> 01:18:05,132 that it's-- 1030 01:18:05,132 --> 01:18:07,750 OK, so now lambda sub 4, so this is 1031 01:18:07,750 --> 01:18:10,380 the 4AP density, so similar to 3AP, 1032 01:18:10,380 --> 01:18:11,710 but now you write four terms. 1033 01:18:11,710 --> 01:18:18,630 The 4AP density of A differs from alpha to the fourth 1034 01:18:18,630 --> 01:18:21,542 by a significant amount. 1035 01:18:21,542 --> 01:18:25,050 OK, so for 3APs, then we said that now A has a large Fourier 1036 01:18:25,050 --> 01:18:26,750 coefficient, right? 1037 01:18:29,700 --> 01:18:33,640 So for-- OK. 1038 01:18:33,640 --> 01:18:37,530 For 4APs, that may not be true, but the following 1039 01:18:37,530 --> 01:18:38,375 is true, right? 1040 01:18:38,375 --> 01:18:48,668 So then there exists a non-zero quadratic polynomial. 1041 01:18:55,500 --> 01:19:08,620 F and N variables over F5 such that the indicator 1042 01:19:08,620 --> 01:19:13,280 function of A correlates with this quadratic exponential 1043 01:19:13,280 --> 01:19:13,780 face. 1044 01:19:23,200 --> 01:19:24,833 So Fourier analysis, the conclusion 1045 01:19:24,833 --> 01:19:26,250 that we got from counting lemma is 1046 01:19:26,250 --> 01:19:28,440 that you have some linear function 1047 01:19:28,440 --> 01:19:33,650 F, such that this quantity is large, 1048 01:19:33,650 --> 01:19:35,630 this large Fourier coefficient. 1049 01:19:35,630 --> 01:19:37,980 OK, so that is not true 4APs. 1050 01:19:37,980 --> 01:19:39,870 But what is true is that now you can 1051 01:19:39,870 --> 01:19:44,860 look at quadratic exponential faces, and then it is true. 1052 01:19:44,860 --> 01:19:48,720 So that's the content of higher order Fourier. 1053 01:19:48,720 --> 01:19:51,660 I mean, that's the example of higher-order Fourier analysis. 1054 01:19:51,660 --> 01:19:56,010 And you can imagine with this type of result, 1055 01:19:56,010 --> 01:19:58,200 and with quite a bit more work, you 1056 01:19:58,200 --> 01:20:01,560 can try to follow a similar density increment 1057 01:20:01,560 --> 01:20:07,110 strategy to prove similarities term for 4APs.