1 00:00:17,865 --> 00:00:18,490 YUFEI ZHAO: OK. 2 00:00:18,490 --> 00:00:20,950 So let's get started. 3 00:00:20,950 --> 00:00:23,170 So we spent quite a bit of time with graph theory 4 00:00:23,170 --> 00:00:25,010 in the first part of this course, 5 00:00:25,010 --> 00:00:27,760 and today I want to move beyond that. 6 00:00:27,760 --> 00:00:31,630 So we're going to talk about more central topics 7 00:00:31,630 --> 00:00:33,730 and additive combinatorics, starting 8 00:00:33,730 --> 00:00:37,180 with the Fourier analytic proof of Roth's theorem. 9 00:00:48,760 --> 00:00:51,340 We discussed Roth's theorem, and we gave a proof, 10 00:00:51,340 --> 00:00:54,760 during the course, using similarities, graph regularity 11 00:00:54,760 --> 00:00:59,220 lemma, as well as the triangle removal lemma. 12 00:00:59,220 --> 00:01:01,810 Today, I want to show you a different approach 13 00:01:01,810 --> 00:01:05,540 to proving Roth's theorem that goes through Fourier analysis. 14 00:01:05,540 --> 00:01:07,240 So this is a very important proof, 15 00:01:07,240 --> 00:01:13,330 and it's one of the main tools in additive combinatorics. 16 00:01:13,330 --> 00:01:15,520 Let me remind you what Roth's theorem says. 17 00:01:15,520 --> 00:01:24,710 So Roth proved, in 1953, that if we 18 00:01:24,710 --> 00:01:36,140 write our sub-3 of interval n to be the maximum size of a 3AP3 19 00:01:36,140 --> 00:01:52,820 subset of 1 through n, then Roth showed that our 3 of n 20 00:01:52,820 --> 00:01:58,280 is little O of n. 21 00:01:58,280 --> 00:02:01,790 So in other words, if you have a positive density 22 00:02:01,790 --> 00:02:06,560 subset of the integers, then it must 23 00:02:06,560 --> 00:02:09,150 contain a three-term arithmetic progression then. 24 00:02:09,150 --> 00:02:13,740 So what I said is equivalent to the statement here. 25 00:02:13,740 --> 00:02:16,831 So previously, we gave a proof using regularity. 26 00:02:23,000 --> 00:02:25,500 Actually the regularity approach of Szemerédi was only found 27 00:02:25,500 --> 00:02:28,530 the '70s so Roth's original proof was through Fourier 28 00:02:28,530 --> 00:02:29,460 analysis. 29 00:02:29,460 --> 00:02:32,520 And we'll see that tomorrow. 30 00:02:32,520 --> 00:02:35,843 Today, we'll see a toy version of this proof. 31 00:02:35,843 --> 00:02:37,260 But it's not really a toy version. 32 00:02:37,260 --> 00:02:40,430 It has the same ideas, but in a slightly easier setting 33 00:02:40,430 --> 00:02:42,330 that has fewer technicalities. 34 00:02:44,553 --> 00:02:46,220 But before showing you that, let me just 35 00:02:46,220 --> 00:02:48,320 discuss a bit of history around Roth's theorem. 36 00:02:51,230 --> 00:02:55,600 We will show, next time, the next lecture, 37 00:02:55,600 --> 00:02:59,405 we'll show the bound. 38 00:02:59,405 --> 00:03:02,717 Also by regularity, we get some bound, which is little O then, 39 00:03:02,717 --> 00:03:04,550 but because of the use of regularity lemmas, 40 00:03:04,550 --> 00:03:06,880 it's pretty poor dependence. 41 00:03:06,880 --> 00:03:11,370 We got something like n over log star n. 42 00:03:11,370 --> 00:03:16,680 Next lecture, and basically Roth's original proof, 43 00:03:16,680 --> 00:03:20,340 gives you a bound which is n over log-log n. 44 00:03:20,340 --> 00:03:23,100 So it's a much more reasonable bound. 45 00:03:23,100 --> 00:03:29,880 The current best upper bound known 46 00:03:29,880 --> 00:03:43,000 has the form, essentially, n over log n raised to 1 47 00:03:43,000 --> 00:03:47,470 plus little 1, roughly n over log n. 48 00:03:47,470 --> 00:03:49,870 We do not know, or even have great guesses on, 49 00:03:49,870 --> 00:03:51,840 what the answer should be. 50 00:03:51,840 --> 00:03:57,250 So the best lower bound, and this is a construction 51 00:03:57,250 --> 00:04:02,620 that we saw earlier in the course due to Behrend 52 00:04:02,620 --> 00:04:10,360 is of the form n over E to the c root log n. 53 00:04:10,360 --> 00:04:12,550 It seems it may be very difficult to improve 54 00:04:12,550 --> 00:04:16,070 this upper bound without some genuine new ideas. 55 00:04:16,070 --> 00:04:18,940 On the other hand, there is some evidence 56 00:04:18,940 --> 00:04:20,529 that the lower bound might be closer 57 00:04:20,529 --> 00:04:24,617 to the truth in that there are variants of the Roth problem 58 00:04:24,617 --> 00:04:27,200 for which we know that the lower bound is basically the truth. 59 00:04:32,390 --> 00:04:34,670 What I want to do today is look at a variant 60 00:04:34,670 --> 00:04:37,760 of this problem in what's called a finite field model. 61 00:04:47,520 --> 00:04:49,480 And that basically just means we're 62 00:04:49,480 --> 00:04:52,870 going to be looking at Roth's theorem, not in the integers, 63 00:04:52,870 --> 00:04:56,170 but in some finite field vector space, specifically F3 64 00:04:56,170 --> 00:04:57,200 to the n. 65 00:04:57,200 --> 00:05:00,110 So we're going to define our sub-3 of this F3 66 00:05:00,110 --> 00:05:06,100 to the n to be the maximum size of the 3AP3 67 00:05:06,100 --> 00:05:13,700 subset of this finite field vector space. 68 00:05:16,588 --> 00:05:18,380 So the finite field model is really useful. 69 00:05:18,380 --> 00:05:22,140 We're going to see this again later in the course as well. 70 00:05:22,140 --> 00:05:24,140 Many of the ideas and techniques that 71 00:05:24,140 --> 00:05:28,940 work for the real problem, so to speak, many of those techniques 72 00:05:28,940 --> 00:05:31,040 also work in the finite field model, 73 00:05:31,040 --> 00:05:34,320 but they are technically simpler to execute. 74 00:05:34,320 --> 00:05:36,470 So this is often-- 75 00:05:36,470 --> 00:05:39,980 you view it as a sandbox, a playing ground, for testing out 76 00:05:39,980 --> 00:05:41,295 many of the ideas. 77 00:05:41,295 --> 00:05:42,920 And once you have those ideas, then you 78 00:05:42,920 --> 00:05:46,430 can see if you can bring them to the integer setting. 79 00:05:46,430 --> 00:05:48,560 And this is a very successful program, 80 00:05:48,560 --> 00:05:52,130 and we'll see one aspect of what happens when we do this. 81 00:05:55,220 --> 00:05:59,450 For this specific problem of Roth's theorem, in F3 to the n, 82 00:05:59,450 --> 00:06:00,950 there are some nice interpretations 83 00:06:00,950 --> 00:06:02,660 of what this problem means. 84 00:06:02,660 --> 00:06:05,750 So here's a pretty easy fact that for-- 85 00:06:05,750 --> 00:06:13,460 so n F3 to the n for three elements, x, y, z, 86 00:06:13,460 --> 00:06:17,370 the following interpretation, so what it means to be a 3AP, 87 00:06:17,370 --> 00:06:18,680 are equivalent. 88 00:06:18,680 --> 00:06:23,060 So x. y, z form a 3AP. 89 00:06:27,540 --> 00:06:31,680 So 3AP means the y is x plus D. Z is x plus 2D. 90 00:06:34,640 --> 00:06:38,560 Equivalently, they satisfy this equation, x minus 2y plus z 91 00:06:38,560 --> 00:06:40,430 equals zero. 92 00:06:40,430 --> 00:06:40,930 OK. 93 00:06:40,930 --> 00:06:43,630 In F3, minus 2 is plus 1. 94 00:06:43,630 --> 00:06:47,770 So it's the same as this even nicer looking equation, 95 00:06:47,770 --> 00:06:49,350 x plus y plus z equals 2. 96 00:06:53,588 --> 00:06:55,880 It also turns out to be the same as saying that x, y, z 97 00:06:55,880 --> 00:07:00,880 would lie on a line. 98 00:07:00,880 --> 00:07:01,760 But they are aligned. 99 00:07:06,780 --> 00:07:10,050 So the line has three points in F3. 100 00:07:10,050 --> 00:07:13,650 And if you look at the coordinate for every i, 101 00:07:13,650 --> 00:07:28,820 the i-th coordinate of x, y, z are all distinct or all equal. 102 00:07:32,187 --> 00:07:33,730 So easy to check all these things 103 00:07:33,730 --> 00:07:35,280 are equivalent to each other. 104 00:07:35,280 --> 00:07:37,990 And the last one is a nice interpretation 105 00:07:37,990 --> 00:07:43,460 in terms of a game that many of you know, Set. 106 00:07:43,460 --> 00:07:45,580 So in the game Set, you have a bunch of cards. 107 00:07:45,580 --> 00:07:47,740 There have some number of properties, n properties, 108 00:07:47,740 --> 00:07:50,560 like color, the number of symbols, the shape. 109 00:07:50,560 --> 00:07:53,560 And you want to form a set being three 110 00:07:53,560 --> 00:07:56,030 cards such that every property, they're 111 00:07:56,030 --> 00:07:58,120 all same or all different. 112 00:07:58,120 --> 00:08:02,777 So that's exactly this model over here. 113 00:08:02,777 --> 00:08:04,360 So what can we say about this problem? 114 00:08:04,360 --> 00:08:06,520 What's the size of the maximum subset 115 00:08:06,520 --> 00:08:09,760 of F3 to the n-th without 3-AP. 116 00:08:09,760 --> 00:08:11,740 If you look at the proof that we did earlier 117 00:08:11,740 --> 00:08:14,440 in this course, the one using triangle removal lemma, 118 00:08:14,440 --> 00:08:16,522 you see the proof works verbatim. 119 00:08:16,522 --> 00:08:17,980 Previously, we worked over Z mod n. 120 00:08:17,980 --> 00:08:21,388 Now, you work over a different group, same proof. 121 00:08:21,388 --> 00:08:25,090 So triangle removal lemma, it tells you 122 00:08:25,090 --> 00:08:34,490 that this r3 is always little o of the size of the space. 123 00:08:34,490 --> 00:08:35,860 But we would like to do better. 124 00:08:35,860 --> 00:08:37,710 So this gives you something like log star. 125 00:08:37,710 --> 00:08:39,659 It's not very good dependence. 126 00:08:39,659 --> 00:08:41,820 So we would like to do better. 127 00:08:41,820 --> 00:08:45,165 So what we will show today, so this theorem 128 00:08:45,165 --> 00:08:47,020 is attributed to Meshulam. 129 00:08:51,010 --> 00:08:54,350 So in this case, the order of history is somewhat reversed. 130 00:08:54,350 --> 00:08:57,010 So we'll see the finite field toy model, 131 00:08:57,010 --> 00:09:00,020 but it historically actually came afterwards. 132 00:09:00,020 --> 00:09:03,950 But you'll see that the Fourier analytic proof that we'll 133 00:09:03,950 --> 00:09:08,510 see today, it's basically the same proof in the two settings. 134 00:09:08,510 --> 00:09:19,730 So F is r3, we will prove a bound, well, just like that. 135 00:09:19,730 --> 00:09:23,990 So much better than what you get from the regularity method. 136 00:09:27,470 --> 00:09:28,495 In terms of-- 137 00:09:28,495 --> 00:09:30,620 OK, so let me tell you a bit more about the history 138 00:09:30,620 --> 00:09:32,078 of this problem in terms of what we 139 00:09:32,078 --> 00:09:35,300 know in terms of upper bounds and lower bounds. 140 00:09:40,770 --> 00:09:42,920 So let me say more about F3. 141 00:09:45,920 --> 00:09:49,200 So what's the best that you might hope for? 142 00:09:49,200 --> 00:09:58,350 So the best lower bound is due to E del. 143 00:09:58,350 --> 00:10:01,890 And it's some construction, some very specific construction, 144 00:10:01,890 --> 00:10:04,590 which gives you a bound that's something 145 00:10:04,590 --> 00:10:08,870 like 2.21 to the n-th. 146 00:10:08,870 --> 00:10:11,470 And the upper bound is 3 to the n-th-- 147 00:10:11,470 --> 00:10:13,920 on the-- 3 minus little 1 to the n-th. 148 00:10:13,920 --> 00:10:25,750 So for a long time, it was open whether the answer 149 00:10:25,750 --> 00:10:29,080 should be basically roughly like 3 150 00:10:29,080 --> 00:10:33,690 to the n-th or some constant less than 3 to the n-th. 151 00:10:33,690 --> 00:10:36,960 And improvements on the upper bound were very slow, 152 00:10:36,960 --> 00:10:40,170 and/or some very difficult works that nudge that down 153 00:10:40,170 --> 00:10:42,120 below just a little bit. 154 00:10:42,120 --> 00:10:44,940 And then a couple years ago, a few years ago, there 155 00:10:44,940 --> 00:10:48,210 was this incredible breakthrough, 156 00:10:48,210 --> 00:10:52,770 where in this paper that was just a couple pages long, 157 00:10:52,770 --> 00:10:56,160 they managed to significantly improve the upper bound 158 00:10:56,160 --> 00:11:07,030 to basically 2.76 to the n-th. 159 00:11:07,030 --> 00:11:09,100 So this was an incredible breakthrough that 160 00:11:09,100 --> 00:11:11,020 happened just a few years ago. 161 00:11:11,020 --> 00:11:14,330 And we'll talk about this proof in a couple of lectures. 162 00:11:14,330 --> 00:11:17,710 It turns out this proof, which uses what's now 163 00:11:17,710 --> 00:11:19,300 called the polynomial method-- 164 00:11:22,030 --> 00:11:24,950 so not Fourier analytic, but a different method-- 165 00:11:24,950 --> 00:11:28,100 unfortunately does not seem to generalize 166 00:11:28,100 --> 00:11:30,580 to the original Roth's theorem. 167 00:11:30,580 --> 00:11:32,510 In fact, you shouldn't expect it to generalize 168 00:11:32,510 --> 00:11:35,610 in a straightforward way, because up there you 169 00:11:35,610 --> 00:11:38,630 we know that you do not have a power saving, whereas here you 170 00:11:38,630 --> 00:11:39,780 have a power saving. 171 00:11:39,780 --> 00:11:43,360 So the exponent goes down. 172 00:11:43,360 --> 00:11:45,690 OK, so this is roughly the history of this theorem. 173 00:11:49,150 --> 00:11:50,356 Any questions? 174 00:11:53,272 --> 00:11:55,216 AUDIENCE: Do we have to have [INAUDIBLE]?? 175 00:11:55,216 --> 00:11:56,070 YUFEI ZHAO: Yeah. 176 00:11:56,070 --> 00:12:01,860 So I'll-- So I can tell you this is known as a Croot-Lev-Pach. 177 00:12:01,860 --> 00:12:05,050 So I'll say more about it in a couple lectures. 178 00:12:05,050 --> 00:12:09,460 But this for F3 is due to Ellenberg and Gijswijt. 179 00:12:16,560 --> 00:12:18,780 So I'll tell you more about it in a couple lectures. 180 00:12:21,900 --> 00:12:25,860 What I want to focus on today is the Fourier analytic nature 181 00:12:25,860 --> 00:12:28,800 of the proof that gives you this bound up there, 182 00:12:28,800 --> 00:12:31,080 3 to the n over n. 183 00:12:31,080 --> 00:12:33,395 And it may seem like a completely different topic 184 00:12:33,395 --> 00:12:34,770 compared to what we've been doing 185 00:12:34,770 --> 00:12:37,650 so far in the course, which is more about graph theory. 186 00:12:37,650 --> 00:12:39,870 But I want you to think about what 187 00:12:39,870 --> 00:12:43,590 are the relationships between what we'll see today 188 00:12:43,590 --> 00:12:45,280 and what we've seen so far. 189 00:12:45,280 --> 00:12:46,970 And there are lots of connections. 190 00:12:46,970 --> 00:12:48,720 So even though the proof may superficially 191 00:12:48,720 --> 00:12:51,120 look quite different, many of these ideas 192 00:12:51,120 --> 00:12:54,570 about quasirandomness versus structure will come up. 193 00:12:54,570 --> 00:12:57,510 And I want to present the proof in a way that highlights 194 00:12:57,510 --> 00:13:00,580 the similarities between what we did previously 195 00:13:00,580 --> 00:13:02,320 and this Fourier analytic proof. 196 00:13:04,980 --> 00:13:07,790 So let's talk about the strategy. 197 00:13:10,390 --> 00:13:15,220 In the proof of the Szemerédi graph regularity lemma, 198 00:13:15,220 --> 00:13:17,920 we had the strategy that we called the energy increment 199 00:13:17,920 --> 00:13:19,590 strategy. 200 00:13:19,590 --> 00:13:22,012 So you start-- you want to find a good partition. 201 00:13:22,012 --> 00:13:23,220 You start doing partitioning. 202 00:13:23,220 --> 00:13:26,580 And you keep track of this thing called the energy-- 203 00:13:26,580 --> 00:13:30,700 must go up at every step, cannot go up forever, 204 00:13:30,700 --> 00:13:34,220 so has a bounded number of steps. 205 00:13:34,220 --> 00:13:38,140 This strategy for Roth's theorem is also an important strategy. 206 00:13:38,140 --> 00:13:40,270 It's a variant of energy increment, 207 00:13:40,270 --> 00:13:42,256 but now density increment. 208 00:13:53,930 --> 00:13:56,240 So we start with a set-- 209 00:13:56,240 --> 00:14:00,560 A subset of F3 to the n-th, and we 210 00:14:00,560 --> 00:14:02,420 would like to understand something 211 00:14:02,420 --> 00:14:07,160 about its structure versus pseudorandomness 212 00:14:07,160 --> 00:14:10,040 in a way that is similar to when we discussed 213 00:14:10,040 --> 00:14:13,670 the similar issue for graphs. 214 00:14:13,670 --> 00:14:16,280 In particular, there will be this dichotomy 215 00:14:16,280 --> 00:14:24,460 that if A is in some sense pseudorandom-- 216 00:14:24,460 --> 00:14:28,020 so earlier, we saw what it means for a graph to be pseudorandom. 217 00:14:28,020 --> 00:14:30,750 So now, what does it mean for a subset of F3 218 00:14:30,750 --> 00:14:33,030 to the n-th to be pseudorandom. 219 00:14:33,030 --> 00:14:34,440 So we'll address that today. 220 00:14:34,440 --> 00:14:37,010 If A pseudorandom-- 221 00:14:37,010 --> 00:14:43,470 OK, so the short answer is that it is Fourier uniform-- 222 00:14:43,470 --> 00:14:48,120 in other words, all Fourier coefficients small. 223 00:14:51,220 --> 00:14:53,470 That's what pseudorandom will refer to. 224 00:14:53,470 --> 00:14:56,720 So then there is a counting lemma. 225 00:15:01,720 --> 00:15:03,760 And the counting lemma will in particular 226 00:15:03,760 --> 00:15:07,160 imply that A has lots of 3-APs. 227 00:15:13,700 --> 00:15:16,880 So then you find your 3-AP. 228 00:15:16,880 --> 00:15:19,410 If this is not the case, then-- 229 00:15:19,410 --> 00:15:21,890 so what's the opposite of Fourier uniform-- 230 00:15:21,890 --> 00:15:25,890 is that A now has some large Fourier coefficient. 231 00:15:36,190 --> 00:15:39,250 And what we'll do is to use this Fourier 232 00:15:39,250 --> 00:15:47,080 coefficient to extract some codimension 1 affine subspace-- 233 00:15:51,960 --> 00:15:57,420 it's also called a hyperplane-- 234 00:15:57,420 --> 00:16:06,470 where the density of A goes up significantly, 235 00:16:06,470 --> 00:16:09,126 if you restrict to that sub hyperplane. 236 00:16:14,090 --> 00:16:15,500 And you can repeat this process. 237 00:16:15,500 --> 00:16:18,920 Now, restrict to this hyperplane and ask yourself 238 00:16:18,920 --> 00:16:20,660 the same question. 239 00:16:20,660 --> 00:16:25,220 Is A, when restricted to this hyperplane, pseudorandom? 240 00:16:25,220 --> 00:16:27,830 In which case, we find APs. 241 00:16:27,830 --> 00:16:31,230 Or is A restricted to this hyperplane, 242 00:16:31,230 --> 00:16:33,420 does it have a large Fourier coefficient? 243 00:16:33,420 --> 00:16:34,920 In which case, we restrict further. 244 00:16:38,400 --> 00:16:42,970 And each time you iterate, you obtain a density increment. 245 00:16:42,970 --> 00:16:50,130 And the density increment cannot go on forever, 246 00:16:50,130 --> 00:16:53,400 because your total density is at most 1. 247 00:16:57,890 --> 00:16:59,860 So the number of steps must be bounded. 248 00:17:02,590 --> 00:17:04,290 So that's the strategy. 249 00:17:04,290 --> 00:17:06,750 So this should remind you somewhat of the energy 250 00:17:06,750 --> 00:17:10,210 increment strategy from Szemerédi's regularity lemma, 251 00:17:10,210 --> 00:17:12,210 although there are some fundamental differences. 252 00:17:12,210 --> 00:17:13,790 We're not doing partitionings. 253 00:17:16,603 --> 00:17:18,020 Any questions about this strategy? 254 00:17:21,470 --> 00:17:23,810 OK. 255 00:17:23,810 --> 00:17:26,099 I want to tell you about Fourier analysis. 256 00:17:34,740 --> 00:17:37,460 So probably, all of you have seen some version 257 00:17:37,460 --> 00:17:40,410 of Fourier analysis, maybe in your calculus class 258 00:17:40,410 --> 00:17:43,140 with Fourier series and whatnot. 259 00:17:43,140 --> 00:17:45,690 So you play with formulas, and solve some differential 260 00:17:45,690 --> 00:17:46,930 equations. 261 00:17:46,930 --> 00:17:49,830 So I want to give you more than just a bunch 262 00:17:49,830 --> 00:17:53,110 of ways about handling Fourier coefficients, 263 00:17:53,110 --> 00:17:55,600 a way to think about Fourier analysis. 264 00:17:55,600 --> 00:17:58,920 So think of this as a crash course about Fourier analysis 265 00:17:58,920 --> 00:18:01,200 from the perspective of combinatorics. 266 00:18:04,002 --> 00:18:05,460 And Fourier analysis, I think, it's 267 00:18:05,460 --> 00:18:08,160 much easier if you work in a finite group, 268 00:18:08,160 --> 00:18:11,880 in a finite abelian group, which is what we're doing here. 269 00:18:11,880 --> 00:18:13,690 Many of the technicalities go away. 270 00:18:13,690 --> 00:18:17,960 So we'll be looking specifically at Fourier analysis in F3 271 00:18:17,960 --> 00:18:21,840 to the n-th, although the 3 can be any prime. 272 00:18:21,840 --> 00:18:24,270 So it's really the same. 273 00:18:24,270 --> 00:18:29,700 So the main actors in Fourier analysis 274 00:18:29,700 --> 00:18:31,293 are the Fourier characters. 275 00:18:35,370 --> 00:18:39,840 The Fourier characters are denoted gamma sub r. 276 00:18:39,840 --> 00:18:42,420 And they're characters on the group, 277 00:18:42,420 --> 00:18:45,030 meaning that they are maps which-- 278 00:18:45,030 --> 00:18:47,850 so they turn out, happen to be homomorphisms for the 279 00:18:47,850 --> 00:18:50,520 multiplicative group under-- 280 00:18:50,520 --> 00:18:53,760 so C under multiplication. 281 00:18:53,760 --> 00:19:02,850 And they're indexed by r, which also elements of F3 282 00:19:02,850 --> 00:19:03,510 to the n-th. 283 00:19:03,510 --> 00:19:05,670 So I'm going to be fairly concrete here. 284 00:19:05,670 --> 00:19:07,520 There are ways to do this more abstractly. 285 00:19:07,520 --> 00:19:08,880 But I'll be fairly concrete. 286 00:19:08,880 --> 00:19:12,670 So it's defined by gamma sub r evaluated 287 00:19:12,670 --> 00:19:19,470 on x equals to omega raised to r dot product x, where 288 00:19:19,470 --> 00:19:24,450 here omega is a third root of unity 289 00:19:24,450 --> 00:19:28,060 and the dot is a dot product. 290 00:19:33,050 --> 00:19:33,550 So-- 291 00:19:42,660 --> 00:19:45,090 So that's the definition of the Fourier transform-- 292 00:19:45,090 --> 00:19:47,400 sorry, that's the definition of the Fourier characters. 293 00:19:47,400 --> 00:19:49,140 And once you have the Fourier characters, 294 00:19:49,140 --> 00:19:56,830 you can have this Fourier transform, just defined 295 00:19:56,830 --> 00:19:58,010 as follows. 296 00:19:58,010 --> 00:20:00,086 If you start with a function-- 297 00:20:03,814 --> 00:20:07,540 let's say, a complex-valued function on your space-- 298 00:20:07,540 --> 00:20:13,550 then I define the Fourier transform 299 00:20:13,550 --> 00:20:20,080 to be another function, like that, defined 300 00:20:20,080 --> 00:20:22,180 by the following formula. 301 00:20:34,240 --> 00:20:37,178 So that's the formula for the Fourier transform. 302 00:20:40,770 --> 00:20:44,220 It is basically the inner product 303 00:20:44,220 --> 00:20:47,520 between F and the Fourier character. 304 00:20:47,520 --> 00:20:50,220 So let me make the comment here, I 305 00:20:50,220 --> 00:20:52,440 think this is actually a pretty important comment, 306 00:20:52,440 --> 00:20:54,003 about the normalization. 307 00:20:58,350 --> 00:21:03,390 Now, when you first learn Fourier transforms, usually 308 00:21:03,390 --> 00:21:07,350 in the reals, there are all these questions about what 309 00:21:07,350 --> 00:21:08,730 number to put in the exponent. 310 00:21:08,730 --> 00:21:09,660 Is it 2 pi? 311 00:21:09,660 --> 00:21:10,690 Is it root 2 pi? 312 00:21:10,690 --> 00:21:12,000 Is it some other thing? 313 00:21:12,000 --> 00:21:17,770 And somehow, one answer is better than the others. 314 00:21:17,770 --> 00:21:19,873 And the same thing is true here in groups. 315 00:21:19,873 --> 00:21:21,790 So we'll stick with the following convention-- 316 00:21:21,790 --> 00:21:23,957 and I want all of you to stick with this convention, 317 00:21:23,957 --> 00:21:26,320 otherwise, we'll confuse ourselves to no end-- 318 00:21:26,320 --> 00:21:27,680 is that for a finite group-- 319 00:21:34,940 --> 00:21:36,930 actually, let me, start of a board. 320 00:21:36,930 --> 00:21:43,788 So the convention is that, in a finite group, 321 00:21:43,788 --> 00:21:46,310 the Fourier transform is defined-- 322 00:21:46,310 --> 00:21:49,650 and more generally, anything you do in the physical space, 323 00:21:49,650 --> 00:21:56,320 we always use the averaging measure. 324 00:21:59,590 --> 00:22:02,660 Don't sum, always average in the physical space. 325 00:22:02,660 --> 00:22:10,040 And in the frequency space, always use sums, 326 00:22:10,040 --> 00:22:12,860 use the counting measure. 327 00:22:12,860 --> 00:22:14,458 Keep this in mind. 328 00:22:14,458 --> 00:22:16,250 Any of these questions about normalization, 329 00:22:16,250 --> 00:22:17,667 if you stick with this convention, 330 00:22:17,667 --> 00:22:19,545 things will become much easier. 331 00:22:19,545 --> 00:22:21,670 So there won't be any of these questions about when 332 00:22:21,670 --> 00:22:23,390 you take the inverse Fourier transform, 333 00:22:23,390 --> 00:22:25,750 do I put an extra factor in front or not, 334 00:22:25,750 --> 00:22:27,691 if you stick with this correct convention. 335 00:22:30,520 --> 00:22:33,580 So with that convention in mind, what the Fourier 336 00:22:33,580 --> 00:22:40,510 transform really is is inner product between F and a Fourier 337 00:22:40,510 --> 00:22:41,478 character. 338 00:22:49,810 --> 00:22:51,670 There are some important properties 339 00:22:51,670 --> 00:22:52,760 of the Fourier transform. 340 00:22:52,760 --> 00:22:54,760 So let me go through a few of the key properties 341 00:22:54,760 --> 00:22:55,540 that we'll need. 342 00:23:04,940 --> 00:23:07,970 The first one is pretty easy. 343 00:23:07,970 --> 00:23:14,250 What is the meaning of the 0-th Fourier coefficient? 344 00:23:14,250 --> 00:23:24,020 You plug it in, and you see that it is just the average of F. 345 00:23:24,020 --> 00:23:27,800 So 0-th coefficient is the average of F. 346 00:23:27,800 --> 00:23:32,270 The second fact goes under one of two names, 347 00:23:32,270 --> 00:23:34,870 and they're often used interchangeably-- 348 00:23:34,870 --> 00:23:36,110 Plancherel or Parseval. 349 00:23:41,340 --> 00:23:48,560 And it says that if you look at the inner product 350 00:23:48,560 --> 00:23:57,080 in the physical space, then this product 351 00:23:57,080 --> 00:24:02,216 is preserved, if you take the Fourier transform. 352 00:24:05,617 --> 00:24:07,700 But now, of course, you're in the frequency space, 353 00:24:07,700 --> 00:24:10,070 so you should sum instead of doing the inner product. 354 00:24:15,180 --> 00:24:20,490 So this identity can be proved in a fairly straightforward way 355 00:24:20,490 --> 00:24:24,360 by plugging in what the definition is for the Fourier 356 00:24:24,360 --> 00:24:25,418 transform. 357 00:24:25,418 --> 00:24:26,960 This is a straightforward computation 358 00:24:26,960 --> 00:24:27,850 I'm not going to do on the board, 359 00:24:27,850 --> 00:24:29,970 but I highly encourage you to actually do at home, 360 00:24:29,970 --> 00:24:33,685 just to do it once to make sure you understand how it goes. 361 00:24:33,685 --> 00:24:35,310 But there is also a more conceptual way 362 00:24:35,310 --> 00:24:38,410 to understand this identity. 363 00:24:38,410 --> 00:24:41,490 And that's because-- now, this is also 364 00:24:41,490 --> 00:24:44,670 important to understand what the Fourier transform is. 365 00:24:44,670 --> 00:24:46,980 It's not just some magical formula somebody wrote down, 366 00:24:46,980 --> 00:24:50,080 like this is a very natural operation. 367 00:24:50,080 --> 00:24:59,280 It's because the characters, the set of characters, 368 00:24:59,280 --> 00:25:02,620 is an orthonormal basis. 369 00:25:02,620 --> 00:25:10,815 So the Fourier characters form an orthonormal basis. 370 00:25:15,750 --> 00:25:18,420 As a result, what the Fourier transform 371 00:25:18,420 --> 00:25:22,310 is is a unitary change of basis. 372 00:25:35,510 --> 00:25:36,320 You can check. 373 00:25:36,320 --> 00:25:37,737 It's very straightforward to check 374 00:25:37,737 --> 00:25:39,800 that the Fourier characters, indeed, form 375 00:25:39,800 --> 00:25:45,210 a orthonormal basis, because-- 376 00:25:45,210 --> 00:25:51,650 well, you can evaluate the inner product between two Fourier 377 00:25:51,650 --> 00:25:53,480 characters. 378 00:25:53,480 --> 00:25:55,730 So remember, in the physical space, 379 00:25:55,730 --> 00:25:57,786 we're always doing averaging. 380 00:26:01,050 --> 00:26:06,240 And so now, I'll just write down first 381 00:26:06,240 --> 00:26:08,310 what I mean by the inner product. 382 00:26:11,100 --> 00:26:12,920 So that's the inner product. 383 00:26:12,920 --> 00:26:21,977 And by the definition of the Fourier character, 384 00:26:21,977 --> 00:26:22,560 you have that. 385 00:26:25,410 --> 00:26:29,010 So think about what this expectation is-- 386 00:26:29,010 --> 00:26:37,470 unless r equals to s, in which case, this expectation is 1. 387 00:26:37,470 --> 00:26:39,240 Unless that is the case, you always 388 00:26:39,240 --> 00:26:43,360 have some coordinate of x in the exponent. 389 00:26:43,360 --> 00:26:46,230 So as you average over all possibilities, 390 00:26:46,230 --> 00:26:47,270 they average out to 0. 391 00:26:58,020 --> 00:27:01,050 So this calculation shows you that the Fourier characters 392 00:27:01,050 --> 00:27:02,880 form a orthonormal basis. 393 00:27:02,880 --> 00:27:05,910 And a basic fact you know from linear algebra 394 00:27:05,910 --> 00:27:07,470 is that if you do a change of basis, 395 00:27:07,470 --> 00:27:10,510 if you do a unitary change of basis, 396 00:27:10,510 --> 00:27:13,640 then inner product is preserved. 397 00:27:13,640 --> 00:27:15,430 It's like a rotation. 398 00:27:15,430 --> 00:27:18,150 It's the same-- so you're not changing the inner product. 399 00:27:18,150 --> 00:27:19,660 So the inner product is preserved 400 00:27:19,660 --> 00:27:21,640 under this change of basis. 401 00:27:21,640 --> 00:27:24,280 And that's why Plancherel is true. 402 00:27:27,780 --> 00:27:30,870 Another important thing is what's 403 00:27:30,870 --> 00:27:34,684 known as the Fourier inversion formula. 404 00:27:34,684 --> 00:27:39,720 The Fourier transform tells you how to go from a function 405 00:27:39,720 --> 00:27:42,730 to the Fourier transform. 406 00:27:42,730 --> 00:27:46,020 Well, now, if you are given the Fourier transform, 407 00:27:46,020 --> 00:27:48,090 how do you go back? 408 00:27:48,090 --> 00:27:50,040 There's a formula which tells you 409 00:27:50,040 --> 00:28:01,350 that you can go back by the following formula there. 410 00:28:01,350 --> 00:28:04,370 So that's the Fourier inversion formula. 411 00:28:04,370 --> 00:28:07,110 It allows you to do this inversion. 412 00:28:07,110 --> 00:28:09,540 And again, it's one of these formulas where 413 00:28:09,540 --> 00:28:12,990 I encourage you to try it out yourself by plugging 414 00:28:12,990 --> 00:28:16,770 in the formula and expanding. 415 00:28:16,770 --> 00:28:19,660 And it's pretty easy to check. 416 00:28:19,660 --> 00:28:22,090 It's much easier in the finite field setting, by the way. 417 00:28:22,090 --> 00:28:25,630 So if you use the usual Fourier transform on the real line, 418 00:28:25,630 --> 00:28:27,070 there are some technicalities even 419 00:28:27,070 --> 00:28:28,690 to prove the Fourier inversion. 420 00:28:28,690 --> 00:28:31,070 But in finite groups, it's almost trivial. 421 00:28:31,070 --> 00:28:33,170 You expand, and then you'll see. 422 00:28:33,170 --> 00:28:35,025 So it's very easy to prove. 423 00:28:35,025 --> 00:28:37,150 But you can also see this Fourier inversion formula 424 00:28:37,150 --> 00:28:39,040 more conceptually, because you're 425 00:28:39,040 --> 00:28:41,550 in a unitary change of basis. 426 00:28:41,550 --> 00:28:45,300 So to go back, well, think about what it means in linear algebra 427 00:28:45,300 --> 00:28:48,990 to revert a unitary transformation. 428 00:28:48,990 --> 00:28:52,530 You simply multiply the coefficients 429 00:28:52,530 --> 00:28:56,490 with the coordinates. 430 00:28:56,490 --> 00:29:00,335 Orthogonal, orthonormal change of basis. 431 00:29:00,335 --> 00:29:06,430 Finally, Fourier transform behaves while 432 00:29:06,430 --> 00:29:07,900 under convolution. 433 00:29:07,900 --> 00:29:15,430 So by convolution, we define the convolution 434 00:29:15,430 --> 00:29:19,320 of two functions, f and g, using the following formula. 435 00:29:27,510 --> 00:29:38,090 And so then the claim is that the Fourier transform behaves 436 00:29:38,090 --> 00:29:39,800 very well under convolution. 437 00:29:39,800 --> 00:29:44,250 It's basically multiplicative under convolution. 438 00:29:44,250 --> 00:29:45,860 So what this means is, if I put in-- 439 00:29:50,490 --> 00:29:54,670 so it's pointwise true everywhere. 440 00:29:54,670 --> 00:29:59,590 Again, very easy proof, because I just evaluate the left hand 441 00:29:59,590 --> 00:30:04,850 side, see what-- 442 00:30:04,850 --> 00:30:07,730 plug in the formula for the Fourier transform. 443 00:30:07,730 --> 00:30:16,170 And I find it's that. 444 00:30:16,170 --> 00:30:19,268 And now, I plug in the formula for convolution. 445 00:30:40,450 --> 00:30:45,140 So now, you can do a change of variables. 446 00:30:45,140 --> 00:30:48,865 And then you-- it's not hard to see. 447 00:30:48,865 --> 00:30:50,740 You eventually end up at the right hand side. 448 00:30:54,760 --> 00:30:56,260 So these are some of the properties. 449 00:30:56,260 --> 00:30:58,658 So there are important properties 450 00:30:58,658 --> 00:30:59,700 of the Fourier transform. 451 00:30:59,700 --> 00:31:01,743 So this is something that, whenever 452 00:31:01,743 --> 00:31:03,160 you learn about Fourier transform, 453 00:31:03,160 --> 00:31:06,720 you always see these few properties. 454 00:31:06,720 --> 00:31:08,230 And so we'll use them. 455 00:31:08,230 --> 00:31:10,690 But we'll also need another property 456 00:31:10,690 --> 00:31:14,440 that is specific to the analysis of 3-term arithmetic 457 00:31:14,440 --> 00:31:15,410 progressions. 458 00:31:17,990 --> 00:31:22,165 So what does Fourier transform have to do with 3-APs? 459 00:31:22,165 --> 00:31:24,040 So we want to use it to prove Roth's theorem. 460 00:31:24,040 --> 00:31:27,400 So we better have some tool that allows 461 00:31:27,400 --> 00:31:29,330 us to analyze the number 3-APs. 462 00:31:29,330 --> 00:31:45,250 And here is a key identity relating Fourier with 3-APs. 463 00:31:45,250 --> 00:31:56,410 And it's that, if you have three functions, 464 00:31:56,410 --> 00:32:02,920 then the following quantity, which relates the number 465 00:32:02,920 --> 00:32:06,840 of 3-APs-- 466 00:32:14,880 --> 00:32:18,720 So this function basically counts the number 3-APs, 467 00:32:18,720 --> 00:32:21,500 if your f, g, and h are indicator functions of a set. 468 00:32:24,240 --> 00:32:27,960 I want to express this formula in terms of the Fourier 469 00:32:27,960 --> 00:32:29,835 transforms of these functions. 470 00:32:32,550 --> 00:32:35,880 The formula turns out to be fairly simple, 471 00:32:35,880 --> 00:32:45,970 that it is simply that. 472 00:32:45,970 --> 00:32:50,190 So it's a single sum over the r's of f hat 473 00:32:50,190 --> 00:32:55,240 of r, g hat of minus 2r, and h hat of r. 474 00:32:55,240 --> 00:32:57,880 You might wonder why I put a minus 2 here, because minus 2 475 00:32:57,880 --> 00:33:00,520 is r, and it looks certainly much nicer 476 00:33:00,520 --> 00:33:02,212 with just r in there. 477 00:33:02,212 --> 00:33:05,080 And that is true. 478 00:33:05,080 --> 00:33:08,170 This formula as written is true for over any group. 479 00:33:13,940 --> 00:33:16,340 And our proof will show it. 480 00:33:16,340 --> 00:33:18,560 So it's not really about F3 at all, but any group. 481 00:33:25,300 --> 00:33:29,880 So let me prove this for you in a couple of different ways. 482 00:33:29,880 --> 00:33:39,520 So the first proof is basically a straightforward no 483 00:33:39,520 --> 00:33:46,240 thinking involved proof, as in we apply these formula 484 00:33:46,240 --> 00:33:48,250 for using either Fourier inversion 485 00:33:48,250 --> 00:33:53,820 or the inverse Fourier transform, and plug it in, 486 00:33:53,820 --> 00:33:56,545 and expand, and check. 487 00:33:56,545 --> 00:33:58,170 So it's worth doing at least this once. 488 00:33:58,170 --> 00:34:01,980 So let's do this together at least once. 489 00:34:01,980 --> 00:34:04,420 But this something that is a fairly straightforward 490 00:34:04,420 --> 00:34:05,580 computation. 491 00:34:05,580 --> 00:34:10,900 The left hand side can be expanded 492 00:34:10,900 --> 00:34:12,340 using Fourier inversion. 493 00:34:14,920 --> 00:34:21,389 so r1 f hat r1 omega to the minus r1 494 00:34:21,389 --> 00:34:32,280 dot x, and sum over r2 g hat r2 omega to the minus r2 495 00:34:32,280 --> 00:34:38,199 dot x plus y, and then finally sum 496 00:34:38,199 --> 00:34:47,550 over r3 h hat of r3 omega to the minus r3 dot x plus 2y. 497 00:34:50,989 --> 00:34:54,770 So I'm using Fourier inversion, replace f, g, and h 498 00:34:54,770 --> 00:34:58,800 by their Fourier transforms. 499 00:34:58,800 --> 00:35:02,280 Oh, sorry, there should be-- 500 00:35:02,280 --> 00:35:03,800 yeah, so no minus. 501 00:35:06,960 --> 00:35:10,620 So now, we exchange sums and expectations, 502 00:35:10,620 --> 00:35:18,265 do a switch in the order of summation, so r1, r2, r3 and f 503 00:35:18,265 --> 00:35:27,050 hat of r1 g hat of r2 h hat of r3. 504 00:35:27,050 --> 00:35:33,860 And you have this expectation over x and y and omega 505 00:35:33,860 --> 00:35:37,970 of x dot r1 plus r2 plus r3. 506 00:35:40,610 --> 00:35:45,340 In fact, I can even write the x and y separately, 507 00:35:45,340 --> 00:35:57,180 y omega to the y dot r2 plus 2r3. 508 00:35:57,180 --> 00:35:59,540 So just rearranging. 509 00:35:59,540 --> 00:36:01,570 And now, you see that as you take 510 00:36:01,570 --> 00:36:09,110 expectation over x, this expectation is equal to 1, 511 00:36:09,110 --> 00:36:19,130 if r1 plus r2 plus r3 is equal to 0, and 0 otherwise. 512 00:36:19,130 --> 00:36:23,930 And likewise, the third expectation is either 1 or 0, 513 00:36:23,930 --> 00:36:28,610 depending on the sums of r2 and r3. 514 00:36:33,650 --> 00:36:36,760 So the only terms that remain, after you take out 515 00:36:36,760 --> 00:36:43,920 these 0's, are cases where both these two equations 516 00:36:43,920 --> 00:36:44,730 are satisfied. 517 00:36:47,350 --> 00:36:55,150 And then you see that the only remaining terms are basically 518 00:36:55,150 --> 00:37:02,410 the ones given in the sum on the right hand side. 519 00:37:05,410 --> 00:37:06,000 OK? 520 00:37:06,000 --> 00:37:07,930 So that's the proof. 521 00:37:07,930 --> 00:37:10,363 Pretty straightforward, you plug in Fourier inversion. 522 00:37:13,540 --> 00:37:17,073 I want to show you a different proof that 523 00:37:17,073 --> 00:37:19,240 hopefully will be more familiar and more conceptual. 524 00:37:19,240 --> 00:37:21,657 Now, it doesn't involve carrying through this calculation, 525 00:37:21,657 --> 00:37:26,380 even though this is not at all hard calculation. 526 00:37:26,380 --> 00:37:31,610 But first, let me rewrite the formula up there. 527 00:37:31,610 --> 00:37:37,000 So in F3, it will be convenient, and so 528 00:37:37,000 --> 00:37:39,190 the formula is actually slightly easier 529 00:37:39,190 --> 00:37:42,700 to interpreting F3, in F3, the identity 530 00:37:42,700 --> 00:37:47,110 says that, if you look at the quantity-- 531 00:37:52,460 --> 00:37:53,020 I need. 532 00:37:58,732 --> 00:38:04,293 So let me give you a second proof that works just in F3, 533 00:38:04,293 --> 00:38:06,210 but you can modify it to work in other groups. 534 00:38:06,210 --> 00:38:08,950 But in F3, it's particularly nice. 535 00:38:08,950 --> 00:38:14,240 The left hand side, you see, the left hand side, 536 00:38:14,240 --> 00:38:21,980 I can rewrite it as the following form, 537 00:38:21,980 --> 00:38:24,840 where I sum over-- 538 00:38:24,840 --> 00:38:29,430 well, I take expectation over all triples x, y, z that sum 539 00:38:29,430 --> 00:38:29,930 to 0. 540 00:38:32,600 --> 00:38:36,470 Because a 3-AP is the same as three elements, 541 00:38:36,470 --> 00:38:38,720 three points in the vector space summing to 0. 542 00:38:41,830 --> 00:38:46,000 But now, you see that this quantity 543 00:38:46,000 --> 00:38:54,310 is the same as the convolution evaluated at 0-- 544 00:38:57,640 --> 00:39:00,610 so if you extend the definition of convolution to more than one 545 00:39:00,610 --> 00:39:03,630 func-- more than two functions. 546 00:39:03,630 --> 00:39:05,540 But now, we apply Fourier inversion. 547 00:39:09,650 --> 00:39:11,540 And we find that-- 548 00:39:21,780 --> 00:39:22,660 OK. 549 00:39:22,660 --> 00:39:26,940 So by Fourier inversion, you have that. 550 00:39:26,940 --> 00:39:30,700 But now, by the identity that relates the Fourier transform 551 00:39:30,700 --> 00:39:41,560 and inversion, you have that. 552 00:39:41,560 --> 00:39:45,410 And that's the proof, because minus 2 r is the same as r. 553 00:39:48,470 --> 00:39:52,100 So it's shorter, because we're using some properties here 554 00:39:52,100 --> 00:39:54,310 about convolution and-- 555 00:39:54,310 --> 00:39:56,229 yeah, so about the convolution. 556 00:40:03,720 --> 00:40:05,580 That formula up there is, of course, 557 00:40:05,580 --> 00:40:07,530 related to counting 3-APs. 558 00:40:07,530 --> 00:40:22,080 Because if f, g, and h are all indicators of some set, 559 00:40:22,080 --> 00:40:25,710 then the left hand side is the same 560 00:40:25,710 --> 00:40:35,535 as basically the number of triples of elements in A 561 00:40:35,535 --> 00:40:38,940 whose sum is equal to 0. 562 00:40:42,210 --> 00:40:46,680 And the right hand side is the sum 563 00:40:46,680 --> 00:40:52,184 of the third power of the Fourier coefficients. 564 00:40:56,360 --> 00:40:59,980 And this formula should look somewhat familiar, 565 00:40:59,980 --> 00:41:03,940 because we also used this kind of formula 566 00:41:03,940 --> 00:41:07,610 back when we discussed spectral graph theory. 567 00:41:07,610 --> 00:41:11,350 And remember, the third moment of the eigenvalues 568 00:41:11,350 --> 00:41:16,330 is the trace of the third power, which counts closed 569 00:41:16,330 --> 00:41:18,880 walks in Cayley graph. 570 00:41:18,880 --> 00:41:21,100 So this is actually the same formula. 571 00:41:21,100 --> 00:41:26,210 So in the case if A is symmetric, 572 00:41:26,210 --> 00:41:34,460 let's say, then this is the same as the formula 573 00:41:34,460 --> 00:41:47,530 that counts closed walks of length three 574 00:41:47,530 --> 00:41:48,660 in the Cayley graph. 575 00:41:53,945 --> 00:41:55,820 The point of this comment is just to tell you 576 00:41:55,820 --> 00:41:57,440 that Fourier transform is somehow 577 00:41:57,440 --> 00:42:02,160 it's not this brand new concept that we've never seen before. 578 00:42:02,160 --> 00:42:04,400 It is intimately tied to many of the things 579 00:42:04,400 --> 00:42:08,866 that we have seen earlier in this course but in disguise. 580 00:42:08,866 --> 00:42:11,390 So it is related to the spectral graph theory 581 00:42:11,390 --> 00:42:13,640 that we discussed at length earlier in this course. 582 00:42:21,290 --> 00:42:23,080 Now that we have the Fourier transform, 583 00:42:23,080 --> 00:42:28,420 I want to develop some machinery to prove 584 00:42:28,420 --> 00:42:32,930 Roth's theorem following the strategy up there. 585 00:42:32,930 --> 00:42:34,100 So let's take a quick break. 586 00:42:34,100 --> 00:42:36,392 And then when we come back, we'll prove Roth's theorem. 587 00:42:40,570 --> 00:42:41,500 Any questions so far? 588 00:42:45,204 --> 00:42:50,596 AUDIENCE: So for this one, you said-- 589 00:42:50,596 --> 00:42:53,150 is it like we only used the fact that it's 590 00:42:53,150 --> 00:42:57,255 F3 to the n-th at the end of the proof, like in that last step? 591 00:42:57,255 --> 00:42:57,880 YUFEI ZHAO: OK. 592 00:42:57,880 --> 00:43:01,040 So question is, where do we use that in F3 to the n-th? 593 00:43:01,040 --> 00:43:04,970 This formula here holds in every finite abelian group, 594 00:43:04,970 --> 00:43:07,190 if you use the correct definition of Fourier 595 00:43:07,190 --> 00:43:10,240 transform with the averaging normalization. 596 00:43:10,240 --> 00:43:15,020 So in the other formula where you replace minus 2 by 1, 597 00:43:15,020 --> 00:43:17,890 that requires F3. 598 00:43:17,890 --> 00:43:18,850 But you can-- 599 00:43:18,850 --> 00:43:22,280 I mean, you can follow the proof and come up 600 00:43:22,280 --> 00:43:27,080 with a similar formula for every equation. 601 00:43:27,080 --> 00:43:28,580 So there's a general principle here, 602 00:43:28,580 --> 00:43:30,600 which I'll discuss more at length 603 00:43:30,600 --> 00:43:34,120 in a bit, that for patterns that are 604 00:43:34,120 --> 00:43:36,760 governed by a single equation-- 605 00:43:36,760 --> 00:43:41,260 in this case, 3-APs, x minus 2y plus z equal to 0-- 606 00:43:41,260 --> 00:43:43,900 patterns that can be governed by a single equation 607 00:43:43,900 --> 00:43:46,196 can be controlled by a Fourier transform. 608 00:43:49,770 --> 00:43:53,210 So let's begin our proof, the Fourier analytic proof 609 00:43:53,210 --> 00:43:55,636 of Roth's theorem in F3 to the n-th. 610 00:43:55,636 --> 00:43:57,695 AUDIENCE: So at the end, you said 611 00:43:57,695 --> 00:43:59,445 it was going to be connected with counting 612 00:43:59,445 --> 00:44:00,362 the [INAUDIBLE] graph. 613 00:44:00,362 --> 00:44:03,045 Does this mean that the Fourier transform 614 00:44:03,045 --> 00:44:06,982 of the indicator of A, those are exactly the eigenvalues 615 00:44:06,982 --> 00:44:08,304 of the Cayley graph? 616 00:44:08,304 --> 00:44:09,387 Or is it like [INAUDIBLE]? 617 00:44:09,387 --> 00:44:12,050 YUFEI ZHAO: OK, so you're asking about the final step, where 618 00:44:12,050 --> 00:44:13,505 we're talking about-- 619 00:44:13,505 --> 00:44:15,380 so I mentioned that there was this connection 620 00:44:15,380 --> 00:44:18,170 between counting walks in graphs and spectral graph theory. 621 00:44:18,170 --> 00:44:27,230 So you can check that, if you have a subset A of an abelian 622 00:44:27,230 --> 00:44:31,840 group, then the Fourier transforms of A 623 00:44:31,840 --> 00:44:39,180 are exactly the eigenvalues of the Cayley graph. 624 00:44:44,090 --> 00:44:46,350 AUDIENCE: So then I guess, have we 625 00:44:46,350 --> 00:44:48,810 done anything so far that could have been done 626 00:44:48,810 --> 00:44:50,950 in a spectral way yet? 627 00:44:50,950 --> 00:44:54,190 Well, I guess, where is the Fourier analysis better 628 00:44:54,190 --> 00:44:55,555 than the spectral [INAUDIBLE]? 629 00:44:55,555 --> 00:44:56,280 YUFEI ZHAO: OK. 630 00:44:56,280 --> 00:44:57,950 Question, where is the Fourier analysis 631 00:44:57,950 --> 00:44:59,130 better than the spectral posts? 632 00:44:59,130 --> 00:45:00,170 Well, let's see the proof first. 633 00:45:00,170 --> 00:45:01,520 And then you'll see, yeah. 634 00:45:01,520 --> 00:45:03,470 So there's no graphs anymore. 635 00:45:03,470 --> 00:45:05,535 So we're going to work inside F3 to the n-th. 636 00:45:08,270 --> 00:45:10,528 But just like the proof of regularity in counting, 637 00:45:10,528 --> 00:45:12,070 we're going to have a counting lemma. 638 00:45:12,070 --> 00:45:14,240 So all of these are analytic. 639 00:45:14,240 --> 00:45:19,243 And at this point, they should be very familiar to you. 640 00:45:19,243 --> 00:45:20,660 They may come in a different form. 641 00:45:20,660 --> 00:45:22,790 They may be dressed in different clothing. 642 00:45:22,790 --> 00:45:24,738 But it's still a counting lemma. 643 00:45:24,738 --> 00:45:25,280 So let's see. 644 00:45:34,360 --> 00:45:36,190 The counting lemma, in this case, 645 00:45:36,190 --> 00:45:44,170 says that if you are in the setting of A F3 to the n-th-- 646 00:45:44,170 --> 00:45:47,470 and I'm going to throughout write 647 00:45:47,470 --> 00:45:56,460 the density of A as alpha, then let me write, 648 00:45:56,460 --> 00:46:06,420 let me define this lambda 3 of A to be the function which 649 00:46:06,420 --> 00:46:17,470 basically counts 3-APs in A but with the averaging 650 00:46:17,470 --> 00:46:18,762 normalization. 651 00:46:21,660 --> 00:46:23,590 So this is-- we saw this earlier. 652 00:46:26,550 --> 00:46:33,550 So the counting lemma says that this normalized number of 3-APs 653 00:46:33,550 --> 00:46:34,900 in A-- 654 00:46:34,900 --> 00:46:37,970 so including trivial 3-APs; that's why this is a nice 655 00:46:37,970 --> 00:46:39,650 analytic expression-- 656 00:46:39,650 --> 00:46:46,490 differs from what you might guess based on density alone. 657 00:46:46,490 --> 00:46:50,960 This difference should be small if all the nonzero Fourier 658 00:46:50,960 --> 00:46:53,920 coefficients of A are small. 659 00:47:04,190 --> 00:47:07,330 So in this strategy, I said that if-- so the counting 660 00:47:07,330 --> 00:47:10,330 lemma tells you, if A is Fourier uniform, 661 00:47:10,330 --> 00:47:11,550 then it is pseudorandom. 662 00:47:11,550 --> 00:47:13,630 And this is where it comes in. 663 00:47:13,630 --> 00:47:16,940 If A has all small Fourier coefficients, 664 00:47:16,940 --> 00:47:19,890 then you have a counting lemma, which tells you 665 00:47:19,890 --> 00:47:21,810 that the counts of A should not be 666 00:47:21,810 --> 00:47:25,980 so different from the guess based on density. 667 00:47:34,290 --> 00:47:36,880 So the proof is very short. 668 00:47:36,880 --> 00:47:41,310 It's based on the identity that we saw earlier. 669 00:47:41,310 --> 00:47:47,040 The 3-AP count of A by the identity earlier 670 00:47:47,040 --> 00:47:51,890 is simply the third power of the Fourier transforms. 671 00:47:55,180 --> 00:47:58,585 And all of these calculations should be reminiscent, 672 00:47:58,585 --> 00:48:00,460 because we've done these kind of calculations 673 00:48:00,460 --> 00:48:03,460 in some form or another earlier in this course. 674 00:48:03,460 --> 00:48:06,370 So we're going to separate out the main term 675 00:48:06,370 --> 00:48:07,690 and subsequent terms. 676 00:48:07,690 --> 00:48:13,750 So the main term is the one corresponding to r equals to 0. 677 00:48:13,750 --> 00:48:15,970 So that's the density. 678 00:48:15,970 --> 00:48:18,640 And all the other terms I'm going 679 00:48:18,640 --> 00:48:24,310 to lump together into this sum. 680 00:48:24,310 --> 00:48:33,610 So we now know that the difference 681 00:48:33,610 --> 00:48:36,850 we're trying to bound is upper bounded 682 00:48:36,850 --> 00:48:47,850 by the third moment of the absolute values of the Fourier 683 00:48:47,850 --> 00:48:50,730 transform. 684 00:48:50,730 --> 00:48:53,040 And I want to upper bound this quantity here, 685 00:48:53,040 --> 00:48:58,320 assuming that all of these Fourier coefficients are small. 686 00:48:58,320 --> 00:49:00,540 We've also done this kind of calculations before. 687 00:49:00,540 --> 00:49:01,957 So where have we seen this before? 688 00:49:06,310 --> 00:49:08,240 We saw this calculation earlier in the class 689 00:49:08,240 --> 00:49:09,700 with the 3 replaced by a 4. 690 00:49:15,640 --> 00:49:18,160 So in counting four cycles in a proof equivalent 691 00:49:18,160 --> 00:49:21,940 of quasirandomness, we said that, if all the eigenvalues 692 00:49:21,940 --> 00:49:24,010 other than the top one are small, then 693 00:49:24,010 --> 00:49:26,980 you can count four cycles. 694 00:49:26,980 --> 00:49:28,320 It's the same proof. 695 00:49:28,320 --> 00:49:30,370 And remember, in that proof, there 696 00:49:30,370 --> 00:49:34,510 was a important trick, where you do not uniformly 697 00:49:34,510 --> 00:49:38,470 bound each term by the max, because then you lose. 698 00:49:38,470 --> 00:49:40,900 You lose by an extra factor of n that you don't want. 699 00:49:40,900 --> 00:49:45,100 So you only take out one factor. 700 00:49:45,100 --> 00:49:47,260 So you take out one factor. 701 00:49:53,840 --> 00:49:56,900 And you keep the rest in there. 702 00:49:56,900 --> 00:49:59,980 In fact, I can be more generous and even 703 00:49:59,980 --> 00:50:02,860 throw the r equal to 0 term back in. 704 00:50:06,460 --> 00:50:10,950 And now, by Plancherel-- 705 00:50:10,950 --> 00:50:15,290 so by Plancherel/Parseval, this here 706 00:50:15,290 --> 00:50:22,740 is equal to the expectation of the indicator 707 00:50:22,740 --> 00:50:23,910 function of A squared. 708 00:50:23,910 --> 00:50:25,775 So you take this-- 709 00:50:25,775 --> 00:50:30,170 you go back to physical space, and that's simply the density. 710 00:50:30,170 --> 00:50:33,710 So then that proves the theorem. 711 00:50:39,175 --> 00:50:41,550 So the moral of the counting lemma is the same as the one 712 00:50:41,550 --> 00:50:43,550 that we've seen before when we discussed graphs. 713 00:50:43,550 --> 00:50:46,440 If you're in pseudorandom, then you have good counting. 714 00:50:46,440 --> 00:50:49,890 And here, pseudorandom means having small Fourier 715 00:50:49,890 --> 00:50:55,040 coefficients, uniformly small Fourier coefficients. 716 00:50:55,040 --> 00:50:57,890 So now, let's begin the proof of Roth's theorem. 717 00:50:57,890 --> 00:51:01,858 The Roth's theorem proof will have three steps. 718 00:51:01,858 --> 00:51:08,455 The first step, we will observe that if you're 719 00:51:08,455 --> 00:51:14,700 in the 3-AP-free set, then there exists a large Fourier 720 00:51:14,700 --> 00:51:15,731 coefficient. 721 00:51:26,050 --> 00:51:28,990 Throughout, I'm going to use uppercase N to denote 722 00:51:28,990 --> 00:51:32,620 the size of the ambient group. 723 00:51:32,620 --> 00:51:36,560 And specifically, we will prove the following. 724 00:51:36,560 --> 00:51:41,320 And also throughout, A is a subset of F3 to the n-th 725 00:51:41,320 --> 00:51:45,075 and with density alpha. 726 00:51:45,075 --> 00:51:47,420 So I'll keep this convention throughout this proof. 727 00:51:50,380 --> 00:52:05,230 We will show that if A is 3-AP-free and N is at least 2 728 00:52:05,230 --> 00:52:09,030 to the alpha to the minus 2-- 729 00:52:09,030 --> 00:52:11,680 so N is at least somewhat large-- 730 00:52:11,680 --> 00:52:16,750 then there exists a nonzero r such 731 00:52:16,750 --> 00:52:23,230 that the r-th Fourier coefficient is at least alpha 732 00:52:23,230 --> 00:52:24,520 squared over 2. 733 00:52:27,450 --> 00:52:30,900 If you're 3-AP-free, free then provided 734 00:52:30,900 --> 00:52:34,290 that you're working in a large enough ambient space, 735 00:52:34,290 --> 00:52:37,763 you always have some large Fourier coefficient. 736 00:52:40,601 --> 00:52:43,720 So the proof is essentially-- well, 737 00:52:43,720 --> 00:52:50,213 this claim is essentially a corollary of counting lemma, 738 00:52:50,213 --> 00:52:51,130 by the counting lemma. 739 00:52:54,820 --> 00:52:57,860 And using the fact that in a 3-AP-free 740 00:52:57,860 --> 00:53:02,611 set, what is the quantity lambda 3 of A? 741 00:53:02,611 --> 00:53:08,170 Up there, you only have the trivial 3-APs present. 742 00:53:08,170 --> 00:53:11,590 So this quantity lambda must then 743 00:53:11,590 --> 00:53:17,770 be the size of A divided by N squared or alpha over N, which 744 00:53:17,770 --> 00:53:22,100 are precisely counting the trivial 3-APs. 745 00:53:26,040 --> 00:53:28,220 So by the counting lemma, then we 746 00:53:28,220 --> 00:53:32,540 have that the upper bound on the right hand 747 00:53:32,540 --> 00:53:43,120 side, which we now write alpha max over nonzero r's, is 748 00:53:43,120 --> 00:53:49,840 at least alpha cubed minus the lambda 3 term, which 749 00:53:49,840 --> 00:53:55,370 should be alpha over N. So provided 750 00:53:55,370 --> 00:53:57,380 that N is large enough-- 751 00:53:57,380 --> 00:53:58,890 big N is large enough-- 752 00:53:58,890 --> 00:54:02,450 then the trivial 3-APs should not contribute very much. 753 00:54:02,450 --> 00:54:06,770 So I can lower bound the right hand side by, let's say, 754 00:54:06,770 --> 00:54:07,570 alpha 3-- 755 00:54:07,570 --> 00:54:10,020 alpha cubed over 2. 756 00:54:10,020 --> 00:54:13,945 So then you deduce the conclusion. 757 00:54:13,945 --> 00:54:16,070 I want you to think about how this proof is related 758 00:54:16,070 --> 00:54:19,270 to Szemerédi's graph regularity lemma. 759 00:54:19,270 --> 00:54:21,020 The analogy will break down at some point. 760 00:54:21,020 --> 00:54:23,780 But we've seen this step before as well. 761 00:54:23,780 --> 00:54:30,050 From lack of 3-APs, you extract some useful information 762 00:54:30,050 --> 00:54:32,980 and from this will extract some structure. 763 00:54:32,980 --> 00:54:34,430 And the structure here-- 764 00:54:34,430 --> 00:54:36,590 and this is where the proof now diverges 765 00:54:36,590 --> 00:54:40,190 from that of regularity-- 766 00:54:40,190 --> 00:54:46,540 having a large Fourier coefficient 767 00:54:46,540 --> 00:54:56,946 will now imply a density increment on a hyperplane. 768 00:55:04,730 --> 00:55:06,950 Specifically, if you have-- 769 00:55:11,390 --> 00:55:15,070 so keeping the same convention as before, 770 00:55:15,070 --> 00:55:21,100 if the Fourier coefficient of A at r 771 00:55:21,100 --> 00:55:29,310 is at least delta for some nonzero r, 772 00:55:29,310 --> 00:55:42,160 then A has density at least alpha plus delta over 2 773 00:55:42,160 --> 00:55:44,460 when restricted to a hyperplane. 774 00:55:55,280 --> 00:55:57,640 So if you have a large Fourier coefficient, 775 00:55:57,640 --> 00:56:00,730 then I can pass down to a smaller part of the space 776 00:56:00,730 --> 00:56:03,518 where the density of A goes up significantly. 777 00:56:10,210 --> 00:56:11,980 To see why this is true, let's go back 778 00:56:11,980 --> 00:56:15,910 to the definition of the Fourier coefficient, the Fourier 779 00:56:15,910 --> 00:56:16,510 transform. 780 00:56:19,120 --> 00:56:25,040 So recall that Fourier transform is given by the following 781 00:56:25,040 --> 00:56:32,030 formula, where I'm looking at this expectation over points 782 00:56:32,030 --> 00:56:39,610 in F3 to the n-th together with indicator of A multiplied 783 00:56:39,610 --> 00:56:43,840 by this Fourier character. 784 00:56:43,840 --> 00:56:49,840 And you see that this function here, it 785 00:56:49,840 --> 00:57:01,090 is constant on cosets of the hyperplane defined 786 00:57:01,090 --> 00:57:06,930 by the orthogonal complement of r. 787 00:57:10,400 --> 00:57:14,550 So the value of this dot product is this constant 788 00:57:14,550 --> 00:57:17,780 on the three hyperplanes. 789 00:57:17,780 --> 00:57:22,130 So I can rewrite this expectation simply 790 00:57:22,130 --> 00:57:28,010 as 1/3 of alpha 0 plus alpha 1 omega 791 00:57:28,010 --> 00:57:37,220 plus alpha 2 omega squared, where alpha 0, alpha 1, alpha 2 792 00:57:37,220 --> 00:57:48,451 are the densities of A on the three cosets of r perp. 793 00:57:52,139 --> 00:57:56,697 So group this expectation into these three hyperplanes. 794 00:57:59,320 --> 00:58:05,300 So now, you see that if this guy is large, 795 00:58:05,300 --> 00:58:12,570 then you should expect that alpha 0, alpha 1, and alpha 2 796 00:58:12,570 --> 00:58:15,550 are not all too close to each other. 797 00:58:15,550 --> 00:58:19,260 So if they were all equal to each other, you should get 0. 798 00:58:19,260 --> 00:58:22,380 But you should not expect them to be too close to each other. 799 00:58:22,380 --> 00:58:27,150 In particular, we would want to say that one of them goes up-- 800 00:58:27,150 --> 00:58:29,520 is much bigger than alpha. 801 00:58:29,520 --> 00:58:32,730 So one of these, they must be much bigger than alpha. 802 00:58:32,730 --> 00:58:34,230 That's an elementary inequality. 803 00:58:34,230 --> 00:58:36,522 This is something that I'm sure I give you five minutes 804 00:58:36,522 --> 00:58:37,320 you can figure out. 805 00:58:37,320 --> 00:58:41,170 But let me show you a small trick to show this. 806 00:58:41,170 --> 00:58:46,200 And the reason for this trick is because in next lecture, when 807 00:58:46,200 --> 00:58:48,900 we look at Roth's theorem over the integers, 808 00:58:48,900 --> 00:58:50,220 we'll need this extra trick. 809 00:58:55,290 --> 00:58:56,850 And the trick here is this. 810 00:58:56,850 --> 00:59:01,560 We now know that because of the hypothesis, 811 00:59:01,560 --> 00:59:08,430 3 delta is lower bound to the absolute value of alpha 0 812 00:59:08,430 --> 00:59:12,240 plus alpha 1 omega plus alpha 2 omega squared. 813 00:59:16,010 --> 00:59:21,680 OK, so note here that the average of the three alphas 814 00:59:21,680 --> 00:59:28,840 is equal to the original alpha by definition of density. 815 00:59:28,840 --> 00:59:35,430 So this inner sum I can read write like that. 816 00:59:44,620 --> 00:59:46,900 So the sum of the three groups of unity add up to 0. 817 00:59:49,410 --> 01:00:02,260 And now, I apply triangle inequality 818 01:00:02,260 --> 01:00:06,220 to extract the terms. 819 01:00:06,220 --> 01:00:07,820 So now, you should already deduce 820 01:00:07,820 --> 01:00:12,080 that one of the alphas i's has to be significantly 821 01:00:12,080 --> 01:00:14,157 larger than alpha. 822 01:00:14,157 --> 01:00:15,740 And has to be significantly different, 823 01:00:15,740 --> 01:00:16,730 but there are only three times. 824 01:00:16,730 --> 01:00:18,620 One of them has to be significantly larger. 825 01:00:18,620 --> 01:00:21,590 But let me do this one extra trick, which 826 01:00:21,590 --> 01:00:24,740 we'll need next time, which is that let 827 01:00:24,740 --> 01:00:34,850 me add an extra term like that, which sums out to 0. 828 01:00:34,850 --> 01:00:39,894 But now, you see that each summand is always nonnegative. 829 01:00:45,700 --> 01:00:49,120 So one of the-- 830 01:00:52,050 --> 01:00:58,830 so there exists some j such that delta lower abounds 831 01:00:58,830 --> 01:01:01,290 the j-th summand. 832 01:01:01,290 --> 01:01:06,960 And if you look at what that means, then alpha lower 833 01:01:06,960 --> 01:01:11,370 bounds the j-th summand. 834 01:01:11,370 --> 01:01:15,588 So in particular, this sum here should be nonnegative. 835 01:01:21,480 --> 01:01:22,480 So you have that. 836 01:01:22,480 --> 01:01:22,980 Good. 837 01:01:26,390 --> 01:01:30,413 So we obtained a density increment of this hyperplane. 838 01:01:34,760 --> 01:01:42,130 And finally, I want to iterate this density increment. 839 01:01:42,130 --> 01:01:44,749 So I want to iterate this density increment-- 840 01:02:01,400 --> 01:02:07,520 so summary so far is that we have-- 841 01:02:07,520 --> 01:02:20,520 so if A is 3-AP-free with density alpha and N at least 2 842 01:02:20,520 --> 01:02:26,300 to the alpha to the minus 2, then A 843 01:02:26,300 --> 01:02:35,510 has density at least alpha plus alpha squared over 4 844 01:02:35,510 --> 01:02:37,175 on some hyperplane. 845 01:02:40,430 --> 01:02:44,740 So you combine step one and step two, we obtain this conclusion. 846 01:02:48,340 --> 01:02:54,970 Well, I can now repeat this operation. 847 01:02:54,970 --> 01:03:09,890 So I can repeat by restricting A to this hyperplane. 848 01:03:09,890 --> 01:03:13,590 If A is originally 3-AP-free, I restrict it to a hyperplane, 849 01:03:13,590 --> 01:03:15,460 it's still 3-AP-free. 850 01:03:15,460 --> 01:03:17,790 So I can keep going. 851 01:03:17,790 --> 01:03:20,400 I can keep going provided that my space is still 852 01:03:20,400 --> 01:03:27,500 large enough, because I still need this lower bound on F. 853 01:03:27,500 --> 01:03:29,640 So don't forget this one here. 854 01:03:29,640 --> 01:03:40,340 So I can keep going as long as N is-- 855 01:03:44,960 --> 01:03:47,110 so I'm using N sub j to denote what 856 01:03:47,110 --> 01:03:50,240 happens after the j-th step. 857 01:03:50,240 --> 01:03:53,840 I can keep going as long as this is still satisfied. 858 01:03:53,840 --> 01:03:56,540 But of course, you cannot keep on going forever, 859 01:03:56,540 --> 01:04:00,030 because the density is bounded. 860 01:04:00,030 --> 01:04:01,970 So density cannot exceed 1. 861 01:04:01,970 --> 01:04:06,420 So these two will give you a bound on the total dimension. 862 01:04:06,420 --> 01:04:08,750 So let's work this out. 863 01:04:08,750 --> 01:04:18,140 So let alpha i denote the density 864 01:04:18,140 --> 01:04:25,770 after step i in this iteration. 865 01:04:25,770 --> 01:04:35,910 And we see from over here that you start with density alpha, 866 01:04:35,910 --> 01:04:44,940 and each step you go up by an increment, which is basically 867 01:04:44,940 --> 01:04:46,260 what-- 868 01:04:46,260 --> 01:04:48,720 so you go up by some increment. 869 01:04:48,720 --> 01:04:51,570 And you want to know, if you start with alpha, 870 01:04:51,570 --> 01:04:57,690 how many steps at most can you take before you blow out 1. 871 01:05:00,440 --> 01:05:02,530 So can you give me some bound? 872 01:05:06,640 --> 01:05:08,160 So what's the maximum-- 873 01:05:08,160 --> 01:05:09,590 at most, how many steps? 874 01:05:17,160 --> 01:05:24,184 So we know that the density cannot exceed 1. 875 01:05:24,184 --> 01:05:27,523 AUDIENCE: 4 by alpha squared. 876 01:05:27,523 --> 01:05:31,920 YUFEI ZHAO: So you see that you have at most 4 877 01:05:31,920 --> 01:05:38,850 over alpha squared steps, because density is at most 1. 878 01:05:38,850 --> 01:05:44,710 And if you plug this in, you get something which 879 01:05:44,710 --> 01:05:46,810 is not quite what I stated. 880 01:05:46,810 --> 01:05:50,920 So it turns out that if you plug this in, 881 01:05:50,920 --> 01:05:53,620 you find that alpha is-- 882 01:05:53,620 --> 01:05:57,730 you find that the size of A is at most 3 883 01:05:57,730 --> 01:05:59,660 to the n-th over square root n. 884 01:06:03,210 --> 01:06:04,945 So let me do a little bit better, 885 01:06:04,945 --> 01:06:09,730 so then simply seeing that this term here is 886 01:06:09,730 --> 01:06:13,260 at least alpha squared over 4. 887 01:06:13,260 --> 01:06:15,970 And the point is that when you increment, 888 01:06:15,970 --> 01:06:19,040 you increment faster and faster. 889 01:06:19,040 --> 01:06:21,680 So I can use that to give a better 890 01:06:21,680 --> 01:06:24,420 bound on the number of steps. 891 01:06:24,420 --> 01:06:26,840 And here's the way to see it. 892 01:06:26,840 --> 01:06:29,810 So let me-- we can do better. 893 01:06:33,600 --> 01:06:39,540 So starting at alpha, I then now ask, how many steps do you 894 01:06:39,540 --> 01:06:42,594 need to take before it doubles? 895 01:06:42,594 --> 01:06:44,426 AUDIENCE: [INAUDIBLE] 896 01:06:44,426 --> 01:06:48,610 YUFEI ZHAO: It goes up by alpha squared over 4. 897 01:06:48,610 --> 01:06:57,700 So it doubles after at most 4 over alpha steps. 898 01:07:01,280 --> 01:07:08,860 And at which point this alpha, new alpha, 899 01:07:08,860 --> 01:07:10,930 becomes at least the original-- 900 01:07:10,930 --> 01:07:13,250 twice the original alpha? 901 01:07:13,250 --> 01:07:14,740 But now, you keep going. 902 01:07:14,740 --> 01:07:18,560 How many times does it take to double again? 903 01:07:18,560 --> 01:07:21,030 2 over alpha, because the alpha became twice as much. 904 01:07:26,760 --> 01:07:33,400 So it doubles again after at most 2 over alpha steps. 905 01:07:33,400 --> 01:07:34,670 And then you keep going. 906 01:07:34,670 --> 01:07:38,390 The next iteration is 1 over alpha. 907 01:07:38,390 --> 01:07:52,700 So in total-- so you see that we must stop after at most 8 908 01:07:52,700 --> 01:07:56,000 over alpha steps. 909 01:07:58,570 --> 01:08:01,250 So the number of times it doubles, 910 01:08:01,250 --> 01:08:04,830 actually it decreases by at least half each time. 911 01:08:11,053 --> 01:08:11,970 So now, we know that-- 912 01:08:14,860 --> 01:08:17,500 we see that the-- 913 01:08:17,500 --> 01:08:19,609 so you keep on going. 914 01:08:19,609 --> 01:08:23,170 So you must stop after at most 8 over alpha steps. 915 01:08:23,170 --> 01:08:26,859 What is the final density when you have to stop, 916 01:08:26,859 --> 01:08:28,430 because when are you forced to stop? 917 01:08:28,430 --> 01:08:32,399 You are forced to stop if you run out of space. 918 01:08:32,399 --> 01:08:34,930 So you're forced to stop when you run out of space. 919 01:08:34,930 --> 01:08:46,290 So if the process terminates after m steps-- 920 01:08:46,290 --> 01:09:04,305 so we're at density alpha m, so then the final subspace 921 01:09:04,305 --> 01:09:16,100 has size less than 2 to the alpha m raised to minus 2, 922 01:09:16,100 --> 01:09:19,130 which is-- 923 01:09:19,130 --> 01:09:22,830 So now, I use this bound alpha. 924 01:09:25,490 --> 01:09:32,189 So the initial N is upper bounded by what? 925 01:09:32,189 --> 01:09:34,250 So how many steps did you take? 926 01:09:34,250 --> 01:09:36,990 You took at most 8 over alpha steps. 927 01:09:36,990 --> 01:09:41,760 Each of those steps, you pass down to codimension what? 928 01:09:41,760 --> 01:09:43,979 You lose a dimension for each step. 929 01:09:43,979 --> 01:09:47,050 And the final subspace has at least this-- 930 01:09:47,050 --> 01:09:51,490 has at most that much space. 931 01:09:51,490 --> 01:10:01,300 So the final dimension is this, basically log 1 over alpha. 932 01:10:01,300 --> 01:10:07,650 So put them together, we see that the size of the space 933 01:10:07,650 --> 01:10:11,322 originally is at most 1 over alpha. 934 01:10:11,322 --> 01:10:11,822 Yeah. 935 01:10:11,822 --> 01:10:14,466 AUDIENCE: Should this be a lower case n? 936 01:10:14,466 --> 01:10:15,383 YUFEI ZHAO: Thank you. 937 01:10:15,383 --> 01:10:16,210 Yeah. 938 01:10:16,210 --> 01:10:19,499 This should be a lower case n, so the dimension. 939 01:10:19,499 --> 01:10:19,999 Good. 940 01:10:22,780 --> 01:10:30,540 And OK, so then that's the conclusion, that the density 941 01:10:30,540 --> 01:10:36,361 alpha is big O of 1 over n. 942 01:10:41,360 --> 01:10:44,990 That proves the main theorem for today, 943 01:10:44,990 --> 01:10:48,940 so Roth's theorem over F3 to the n-th. 944 01:10:48,940 --> 01:10:51,540 So we went through this Fourier analytic proof. 945 01:10:51,540 --> 01:10:57,290 Next lecture, we will see the same proof again but done 946 01:10:57,290 --> 01:11:00,800 in the integers for interval. 947 01:11:00,800 --> 01:11:05,460 And there, there are some difficulties 948 01:11:05,460 --> 01:11:08,340 that we don't see over here. 949 01:11:08,340 --> 01:11:11,940 Because in the finite field space, 950 01:11:11,940 --> 01:11:15,120 in the finite field model, there's 951 01:11:15,120 --> 01:11:18,490 this very nice idea of looking at subspaces, 952 01:11:18,490 --> 01:11:21,410 so looking at hyperplanes. 953 01:11:21,410 --> 01:11:25,820 Each Fourier coefficient gets you down to one dimension less. 954 01:11:25,820 --> 01:11:29,730 But when you're working in the integers, 955 01:11:29,730 --> 01:11:32,310 there are no subspaces you can use. 956 01:11:34,860 --> 01:11:38,040 So we'll be looking at ways to get around 957 01:11:38,040 --> 01:11:39,685 the lack of subspaces. 958 01:11:39,685 --> 01:11:41,310 And this is why I said in the beginning 959 01:11:41,310 --> 01:11:43,790 that the finite field model is often 960 01:11:43,790 --> 01:11:50,940 a very good playground for additive combinatorics type 961 01:11:50,940 --> 01:11:53,940 techniques, especially Fourier analytic techniques. 962 01:11:53,940 --> 01:11:55,830 Because in the additive-- 963 01:11:55,830 --> 01:11:57,450 in all of these techniques, they just 964 01:11:57,450 --> 01:11:59,050 come out to be much cleaner. 965 01:11:59,050 --> 01:12:01,560 If you're working in a finite field setting, 966 01:12:01,560 --> 01:12:03,480 you have nice subspaces, you have Fourier 967 01:12:03,480 --> 01:12:05,160 transform in a very clean way. 968 01:12:05,160 --> 01:12:08,130 The Fourier transform always takes, in this case, 969 01:12:08,130 --> 01:12:10,220 one of three values. 970 01:12:10,220 --> 01:12:11,220 Everything's very clean. 971 01:12:11,220 --> 01:12:12,300 Everything's very simple. 972 01:12:12,300 --> 01:12:13,490 And you get to see the idea here. 973 01:12:13,490 --> 01:12:15,615 You get to see the sense of the increment argument. 974 01:12:18,210 --> 01:12:21,600 But once you understand those ideas 975 01:12:21,600 --> 01:12:26,320 and you're willing to do more work, then oftentimes, 976 01:12:26,320 --> 01:12:29,590 you can bring those ideas to other settings, 977 01:12:29,590 --> 01:12:32,830 to other abelian groups, to the integers, 978 01:12:32,830 --> 01:12:37,300 for instance, but with more work in the-- 979 01:12:37,300 --> 01:12:41,800 there are some extra ingredients that you need to use. 980 01:12:41,800 --> 01:12:45,260 I mentioned that there was a bound-- 981 01:12:45,260 --> 01:12:46,330 OK, so initially-- 982 01:12:49,812 --> 01:12:51,020 So next time, we'll see that. 983 01:12:51,020 --> 01:12:56,000 Next time, we'll see what happens over the integers. 984 01:12:56,000 --> 01:12:58,490 Any questions? 985 01:12:58,490 --> 01:12:59,187 Yes. 986 01:12:59,187 --> 01:13:03,300 AUDIENCE: [INAUDIBLE] 987 01:13:03,300 --> 01:13:04,250 YUFEI ZHAO: OK, great. 988 01:13:04,250 --> 01:13:07,850 So question is why the process must stop after at most 8 989 01:13:07,850 --> 01:13:08,840 over alpha steps? 990 01:13:12,430 --> 01:13:15,700 So you know that the density doubles after this many steps, 991 01:13:15,700 --> 01:13:17,380 doubles again after that many steps. 992 01:13:17,380 --> 01:13:20,710 So eventually, if it keeps on doubling, 993 01:13:20,710 --> 01:13:24,360 it cannot keep on doubling forever. 994 01:13:24,360 --> 01:13:26,570 So this process cannot keep on doubling forever. 995 01:13:26,570 --> 01:13:28,820 So it must stop-- 996 01:13:28,820 --> 01:13:39,560 so cannot double more than log base 2 of 1 over alpha times. 997 01:13:42,980 --> 01:13:45,440 And that point, you have to stop. 998 01:13:45,440 --> 01:13:47,270 So how many steps have you taken? 999 01:13:47,270 --> 01:13:49,420 Well, you sum this geometric series. 1000 01:13:49,420 --> 01:13:55,940 So this-- and the next thing is that you 1001 01:13:55,940 --> 01:13:57,840 sum this geometric series. 1002 01:13:57,840 --> 01:14:01,310 And that geometric series sums to 8 over alpha. 1003 01:14:10,230 --> 01:14:10,730 Great. 1004 01:14:10,730 --> 01:14:12,280 So let's finish here. 1005 01:14:12,280 --> 01:14:16,680 So next time, we'll see Roth's proof of Roth's theorem.