1 00:00:17,460 --> 00:00:21,180 YUFEI ZHAO: OK, we are still on our journey 2 00:00:21,180 --> 00:00:23,170 to proving Freiman theorem. 3 00:00:23,170 --> 00:00:23,670 Right? 4 00:00:23,670 --> 00:00:25,650 So we've been looking at some tools 5 00:00:25,650 --> 00:00:29,280 for analyzing sets of small doubling. 6 00:00:29,280 --> 00:00:33,270 And last time, we showed the following result, 7 00:00:33,270 --> 00:00:38,790 that if a has small doubling, then there exists a prime. 8 00:00:38,790 --> 00:00:41,700 It's not too much bigger than a, such 9 00:00:41,700 --> 00:00:46,560 that a big subset of a, of at least 1/8 proportion, 10 00:00:46,560 --> 00:00:51,840 a big subset of a is Freiman 8-isomorphic to a subset 11 00:00:51,840 --> 00:00:57,120 of this small cyclic group. 12 00:00:57,120 --> 00:01:01,710 So last time, we developed this tool called of modeling lemma, 13 00:01:01,710 --> 00:01:03,930 so Ruzsa's modeling lemma that allows 14 00:01:03,930 --> 00:01:09,030 us to pass from a set of small doubling, which could have 15 00:01:09,030 --> 00:01:11,940 elements very spread out in the integers, 16 00:01:11,940 --> 00:01:17,610 to something that is much more compact, a much tighter set. 17 00:01:17,610 --> 00:01:20,700 That's a subset, a positive proportionate 18 00:01:20,700 --> 00:01:23,880 subset of a small cyclic group. 19 00:01:23,880 --> 00:01:27,270 And remember, the last time we defined this notion of Freiman 20 00:01:27,270 --> 00:01:31,530 8-isomorphic, Freiman isomorphism, in this case, 21 00:01:31,530 --> 00:01:36,020 it just means that it preserves partially additive structure. 22 00:01:36,020 --> 00:01:38,310 It preserves additive structure when 23 00:01:38,310 --> 00:01:41,520 you look at most 8-wise sums. 24 00:01:44,250 --> 00:01:45,530 All right. 25 00:01:45,530 --> 00:01:48,570 Well, this is where we left off last time. 26 00:01:48,570 --> 00:01:50,410 If you start with small doubling, 27 00:01:50,410 --> 00:01:54,610 then I can model a big portion of this set 28 00:01:54,610 --> 00:02:00,850 by a large fraction of a small cyclic group. 29 00:02:00,850 --> 00:02:03,333 All right. 30 00:02:03,333 --> 00:02:04,750 So now, we're in the setting where 31 00:02:04,750 --> 00:02:10,370 we are looking at some space, a cyclic group, for instance. 32 00:02:10,370 --> 00:02:12,130 And now, we have positive proportion, 33 00:02:12,130 --> 00:02:15,370 a constant proportion subset of that group. 34 00:02:15,370 --> 00:02:19,810 And we would like to extract some additional additive 35 00:02:19,810 --> 00:02:23,690 structure from this large set. 36 00:02:23,690 --> 00:02:25,750 And that should remind you of things 37 00:02:25,750 --> 00:02:28,030 we've discussed when we talked about Roth's theorem. 38 00:02:28,030 --> 00:02:28,530 Right? 39 00:02:28,530 --> 00:02:30,580 So Roth's theorem also had this form. 40 00:02:30,580 --> 00:02:34,450 If you start with Z mod n or in the finite field setting, 41 00:02:34,450 --> 00:02:37,480 and you have a constant proportion of the space, 42 00:02:37,480 --> 00:02:41,110 then you must find a three-term arithmetic progression. 43 00:02:41,110 --> 00:02:43,222 In fact, you must find many three-APs. 44 00:02:43,222 --> 00:02:45,430 So we're going to do something very similar, at least 45 00:02:45,430 --> 00:02:46,710 in spirit, here. 46 00:02:46,710 --> 00:02:49,630 We're starting from this large proportion of some space. 47 00:02:49,630 --> 00:02:53,950 We're going to extract a very large additive structure, just 48 00:02:53,950 --> 00:02:55,480 from the size alone. 49 00:02:55,480 --> 00:03:00,280 So let me begin by motivating it with a question. 50 00:03:00,280 --> 00:03:04,210 And we're going to start, as we've done in the past, 51 00:03:04,210 --> 00:03:07,270 with a finite field model, where things are much easier to state 52 00:03:07,270 --> 00:03:10,060 and to analyze. 53 00:03:10,060 --> 00:03:14,470 The question is, suppose you have a set a, which 54 00:03:14,470 --> 00:03:17,740 is a subset of f2 to the m. 55 00:03:17,740 --> 00:03:25,630 And a is an alpha proportion of the space where you think 56 00:03:25,630 --> 00:03:27,520 of alpha as some constant. 57 00:03:32,150 --> 00:03:36,680 Question is, OK, suppose this is true. 58 00:03:36,680 --> 00:03:42,920 And must it be the case that a plus a the subset-- 59 00:03:42,920 --> 00:03:45,830 all right, so a itself, just because it's 60 00:03:45,830 --> 00:03:48,240 a large proportion of the space. 61 00:03:48,240 --> 00:03:51,320 So just because it's the 1% of the space, 62 00:03:51,320 --> 00:03:55,070 doesn't mean that it contains any large structures. 63 00:03:55,070 --> 00:03:58,610 It doesn't contain necessarily any large sub-spaces, 64 00:03:58,610 --> 00:04:04,050 because it could be a more random subset of f to the m. 65 00:04:04,050 --> 00:04:07,110 But there's a general principle in additive combinatorics 66 00:04:07,110 --> 00:04:10,070 or even analysis where if you start with a set that 67 00:04:10,070 --> 00:04:13,840 is quite large, and it might be a bit rough, a is a bit rough, 68 00:04:13,840 --> 00:04:16,350 it's all over the place. 69 00:04:16,350 --> 00:04:21,940 If you add a to itself, it smooths out the set. 70 00:04:21,940 --> 00:04:26,060 So a plus a is much smoother than a. 71 00:04:26,060 --> 00:04:35,120 And the question is, must A plus A contain a large subspace? 72 00:04:35,120 --> 00:04:39,200 And here, by "large," I mean the following. 73 00:04:39,200 --> 00:04:41,540 Because we're looking at constant proportions 74 00:04:41,540 --> 00:04:44,680 of the entire space, by "large," I 75 00:04:44,680 --> 00:04:49,170 would also want a constant proportion of the entire space. 76 00:04:49,170 --> 00:04:53,780 So does there exist some subspace 77 00:04:53,780 --> 00:04:56,930 of bounded codimension? 78 00:04:56,930 --> 00:04:59,350 So if alpha is a constant, I want 79 00:04:59,350 --> 00:05:10,020 a bounded codimensional subspace that lives inside A plus A. 80 00:05:10,020 --> 00:05:12,270 It turns out the answer is no. 81 00:05:12,270 --> 00:05:16,500 So there exists sets that, even though there are very large 82 00:05:16,500 --> 00:05:18,720 and you add it to itself, it still 83 00:05:18,720 --> 00:05:21,280 doesn't have large subspaces. 84 00:05:21,280 --> 00:05:22,530 So let me give you an example. 85 00:05:26,180 --> 00:05:28,390 And this construction is called a nivo set. 86 00:05:32,980 --> 00:05:41,110 So let's take A sub n to be the set of points in F2 to the n 87 00:05:41,110 --> 00:05:43,030 who's Hamming weight-- 88 00:05:43,030 --> 00:05:45,400 so Hamming weight here is just the number 89 00:05:45,400 --> 00:05:50,110 of 1's or it's a number of non-zero elements-- 90 00:05:50,110 --> 00:05:53,730 number of 1's in x or, in general, 91 00:05:53,730 --> 00:05:57,100 the number of non-zero elements among the coordinates 92 00:05:57,100 --> 00:06:01,110 of x, so the number of non-zero coordinates. 93 00:06:01,110 --> 00:06:05,620 And I want the Hamming weight to be less than this quantity 94 00:06:05,620 --> 00:06:08,170 here. 95 00:06:08,170 --> 00:06:10,990 So visually, what this looks like is 96 00:06:10,990 --> 00:06:14,890 I'm thinking of the Hamming cube placed 97 00:06:14,890 --> 00:06:16,840 so that the all-zero vector is here, 98 00:06:16,840 --> 00:06:18,550 the all-ones vector is up there. 99 00:06:18,550 --> 00:06:21,700 And then it's sorted by Hamming weight. 100 00:06:21,700 --> 00:06:23,920 So this is called a Boolean lattice. 101 00:06:23,920 --> 00:06:30,320 And I'm looking at all the elements, A, 102 00:06:30,320 --> 00:06:35,280 which are within a Hamming ball of the 0 vector. 103 00:06:35,280 --> 00:06:37,340 So this is the set. 104 00:06:37,340 --> 00:06:43,660 It's not too hard to calculate the size of the set, 105 00:06:43,660 --> 00:06:48,400 because I'm taking everything with Hamming weight 106 00:06:48,400 --> 00:06:53,500 less than this quantity over here by central limit theorem. 107 00:06:53,500 --> 00:06:59,720 The number of elements in the set is of the form 108 00:06:59,720 --> 00:07:02,380 a constant fraction of the entire space, 109 00:07:02,380 --> 00:07:09,430 where alpha is some constant if c is a constant. 110 00:07:12,330 --> 00:07:15,570 So it has the desired size. 111 00:07:15,570 --> 00:07:20,830 But also, A added to itself consists 112 00:07:20,830 --> 00:07:26,360 of points in the Boolean cube whose Hamming weight 113 00:07:26,360 --> 00:07:33,260 is at most n minus c root n. 114 00:07:33,260 --> 00:07:36,390 And I claim that this sumset does not 115 00:07:36,390 --> 00:07:54,080 contain any subspace of dimension larger than n minus c 116 00:07:54,080 --> 00:07:54,580 root n. 117 00:07:58,980 --> 00:08:00,400 So this is the final claim. 118 00:08:00,400 --> 00:08:03,960 It's something that's, again, one of these linear algebraic 119 00:08:03,960 --> 00:08:09,510 exercises that we've actually seen earlier when we discussed 120 00:08:09,510 --> 00:08:13,400 the proof of a cap set, right, so the polynomial method 121 00:08:13,400 --> 00:08:14,520 proof of cap set. 122 00:08:14,520 --> 00:08:18,960 If you have a subspace of dimension 123 00:08:18,960 --> 00:08:21,690 greater than some quantity, then you 124 00:08:21,690 --> 00:08:26,710 should be able to find a vector in that subspace whose 125 00:08:26,710 --> 00:08:29,593 support has size at least the dimension. 126 00:08:34,423 --> 00:08:35,872 OK. 127 00:08:35,872 --> 00:08:41,770 So you see, in particular, we do not 128 00:08:41,770 --> 00:08:52,270 have any bounded codimensional subspaces in this A plus A. 129 00:08:52,270 --> 00:08:54,257 So even though the philosophy is roughly right 130 00:08:54,257 --> 00:08:56,590 that if you start with a set A and you add it to itself, 131 00:08:56,590 --> 00:08:58,310 it smooths out the set, it should 132 00:08:58,310 --> 00:09:01,350 contain-- we expect it to contain some large structure. 133 00:09:01,350 --> 00:09:02,670 That's not quite true. 134 00:09:02,670 --> 00:09:05,930 But what turns out to be true, and this is the first result 135 00:09:05,930 --> 00:09:09,500 that I will show today, is that if you add A 136 00:09:09,500 --> 00:09:12,960 to itself a few more times, that, indeed, 137 00:09:12,960 --> 00:09:15,740 you can get large subspaces. 138 00:09:15,740 --> 00:09:19,940 And this is an important step in a proof of Freiman's theorem. 139 00:09:19,940 --> 00:09:22,010 And this step is known as Bogolyubov's lemma. 140 00:09:33,280 --> 00:09:37,450 So Bogolyubov's lemma, in the case of F2 to the n, 141 00:09:37,450 --> 00:09:40,150 says that if you have a subset A of F2 142 00:09:40,150 --> 00:09:52,840 to the n of fraction alpha of the space, then 2A minus 2A 143 00:09:52,840 --> 00:09:58,090 contains a bounded codimensional subspace, so a very large 144 00:09:58,090 --> 00:09:59,290 subspace. 145 00:09:59,290 --> 00:10:08,210 so 2A minus 2A contains a subspace of codimension 146 00:10:08,210 --> 00:10:11,780 less than 1 over alpha squared. 147 00:10:16,810 --> 00:10:19,700 Here, I write 2A minus 2A even though we're in F2. 148 00:10:19,700 --> 00:10:21,680 So this is the same as 4A. 149 00:10:21,680 --> 00:10:24,530 But in general-- and you'll see later on when we do it 150 00:10:24,530 --> 00:10:25,580 in integers-- 151 00:10:25,580 --> 00:10:28,000 2A minus 2A is the right expression to look at. 152 00:10:28,000 --> 00:10:31,570 So 2A minus 2A, something that works in every group. 153 00:10:31,570 --> 00:10:34,870 But for F2 to the n, it's the same as 4A. 154 00:10:37,440 --> 00:10:41,781 So the main philosophy here is that adding is smoothing. 155 00:10:47,560 --> 00:10:49,810 You start with a large subset of F2 to the n. 156 00:10:49,810 --> 00:10:51,430 It's large. 157 00:10:51,430 --> 00:10:53,020 Does it contain large structures? 158 00:10:53,020 --> 00:10:54,160 Not necessarily. 159 00:10:54,160 --> 00:10:57,700 But you add it to itself, and it smooths out the picture. 160 00:10:57,700 --> 00:11:00,180 So it has a rough spot. 161 00:11:00,180 --> 00:11:01,950 It smooths it out. 162 00:11:01,950 --> 00:11:03,780 And if you keep adding A to itself, 163 00:11:03,780 --> 00:11:05,470 it smooths it out even further. 164 00:11:05,470 --> 00:11:07,180 You add it to itself enough times, 165 00:11:07,180 --> 00:11:10,070 and then it will contain a large structure 166 00:11:10,070 --> 00:11:14,320 and just from the size of A alone. 167 00:11:14,320 --> 00:11:16,300 And there is a very similar idea, which 168 00:11:16,300 --> 00:11:18,370 comes up all over the place in analysis, 169 00:11:18,370 --> 00:11:22,710 is that convolutions are smoothing. 170 00:11:22,710 --> 00:11:27,740 So you start with some function that might be very rough. 171 00:11:27,740 --> 00:11:30,710 If you convolve it with itself and if you 172 00:11:30,710 --> 00:11:34,580 do it many more times, you get something 173 00:11:34,580 --> 00:11:37,080 that is much smoother. 174 00:11:37,080 --> 00:11:39,830 And in fact, adding and convolutions 175 00:11:39,830 --> 00:11:41,630 are almost the same things. 176 00:11:41,630 --> 00:11:43,950 And I'll explain that in a second. 177 00:11:43,950 --> 00:11:48,140 So this is an important idea to take away from all of this. 178 00:11:48,140 --> 00:11:52,040 So when we do these free analytic calculations-- 179 00:11:52,040 --> 00:11:54,380 so there will be some free analytic calculations-- 180 00:11:54,380 --> 00:11:56,213 the first time you see them, they might just 181 00:11:56,213 --> 00:11:57,720 seem like calculations. 182 00:11:57,720 --> 00:12:00,630 So you push the symbols around, and you get some inequalities. 183 00:12:00,630 --> 00:12:01,640 You get some answers. 184 00:12:01,640 --> 00:12:04,203 But that's no way to learn the subject. 185 00:12:04,203 --> 00:12:05,620 So you need to figure out, what is 186 00:12:05,620 --> 00:12:07,690 the intuition behind each step? 187 00:12:07,690 --> 00:12:10,280 Because when you need to work on it yourself, 188 00:12:10,280 --> 00:12:13,430 you're not just guessing the right symbols to put down. 189 00:12:13,430 --> 00:12:16,760 You have to understand the intuition, like why, 190 00:12:16,760 --> 00:12:20,020 intuitively, each inequality should be expected to hold? 191 00:12:20,020 --> 00:12:22,550 And this is an important idea that adding is smoothing 192 00:12:22,550 --> 00:12:25,128 and convolution is smoothing. 193 00:12:25,128 --> 00:12:26,000 All right. 194 00:12:26,000 --> 00:12:28,415 So let me remind you about convolutions. 195 00:12:34,000 --> 00:12:36,680 So recall, in a general abelian group, 196 00:12:36,680 --> 00:12:39,930 if I have two functions, f and g, on the group-- 197 00:12:39,930 --> 00:12:43,100 so complex value functions-- then the convolution 198 00:12:43,100 --> 00:12:45,990 is given by the following formula. 199 00:12:49,520 --> 00:12:53,550 So that's the convolution. 200 00:12:53,550 --> 00:12:56,130 And it behaves very well with respect 201 00:12:56,130 --> 00:12:58,350 to the Fourier transform. 202 00:12:58,350 --> 00:13:04,020 The Fourier transform turns convolutions 203 00:13:04,020 --> 00:13:05,910 into multiplications. 204 00:13:05,910 --> 00:13:12,600 So this means, point wise, I have that. 205 00:13:12,600 --> 00:13:17,240 Convolutions also relate to some sets, because if I-- 206 00:13:17,240 --> 00:13:19,260 and this is the interpretation of convolutions 207 00:13:19,260 --> 00:13:21,420 that I want you to keep in mind for the purpose 208 00:13:21,420 --> 00:13:24,060 of additive combinatorics. 209 00:13:24,060 --> 00:13:26,790 If you have two sets, A and B, then 210 00:13:26,790 --> 00:13:29,670 look at the convolution of their indicators. 211 00:13:33,090 --> 00:13:35,050 It has an interpretation. 212 00:13:35,050 --> 00:13:37,590 So if you read out what this value says, 213 00:13:37,590 --> 00:13:41,670 then what this comes out to is 1 divided 214 00:13:41,670 --> 00:13:48,450 by the size of the group over the number of pairs in A and B 215 00:13:48,450 --> 00:13:52,860 such that their sum is x. 216 00:13:52,860 --> 00:13:55,960 So up to normalization, convolution records 217 00:13:55,960 --> 00:13:58,540 some sets with multiplicities. 218 00:13:58,540 --> 00:14:01,380 So the convolution tells you how many ways 219 00:14:01,380 --> 00:14:05,130 are there to express x in terms of a sum of one element 220 00:14:05,130 --> 00:14:07,470 from A and another element from B. 221 00:14:07,470 --> 00:14:09,810 And in particular, this function here 222 00:14:09,810 --> 00:14:15,930 is supported on the sumset A plus B. 223 00:14:15,930 --> 00:14:19,440 So this is the way that convolutions and sumsets 224 00:14:19,440 --> 00:14:21,300 are intimately related to each other. 225 00:14:24,100 --> 00:14:26,550 So that's proof Bogolyubov's lemma. 226 00:14:33,930 --> 00:14:40,050 We're going to be looking at this sumset, which is related 227 00:14:40,050 --> 00:14:42,160 to the following convolution. 228 00:14:42,160 --> 00:14:51,910 So let f be this convolution of indicators, AA minus A minus 229 00:14:51,910 --> 00:14:53,285 A. Of course, in F2 to the n, you 230 00:14:53,285 --> 00:14:55,035 don't need to worry about the minus signs. 231 00:14:55,035 --> 00:14:57,710 But I'll keep them there for future reference. 232 00:14:57,710 --> 00:15:03,760 Here, by what we said earlier, the support of f 233 00:15:03,760 --> 00:15:06,130 is 2A minus 2A. 234 00:15:09,480 --> 00:15:12,960 It's not too hard to evaluate the Fourier transform of f, 235 00:15:12,960 --> 00:15:17,050 because Fourier transform plays very well with convolutions. 236 00:15:17,050 --> 00:15:21,390 So in this case, it is the Fourier transform of A squared, 237 00:15:21,390 --> 00:15:24,380 Fourier transform of minus A squared. 238 00:15:24,380 --> 00:15:27,360 The Fourier transform of minus A, if you look at the formula, 239 00:15:27,360 --> 00:15:28,540 is the conjugate. 240 00:15:28,540 --> 00:15:31,440 It's a complex conjugate of the Fourier transform of A. 241 00:15:31,440 --> 00:15:36,350 So again, in F2 to the n, they're actually the same. 242 00:15:36,350 --> 00:15:41,610 But in general, it's the complex conjugate. 243 00:15:41,610 --> 00:15:44,130 So we always have this formula here. 244 00:15:48,150 --> 00:15:50,910 So we always have that. 245 00:15:50,910 --> 00:15:56,410 So by the Fourier inversion formula, 246 00:15:56,410 --> 00:16:02,082 we can write f in terms of its Fourier transform. 247 00:16:07,164 --> 00:16:08,265 Here, I'm using that. 248 00:16:08,265 --> 00:16:09,140 We're now using that. 249 00:16:09,140 --> 00:16:11,780 We're in F2, so that's what the inverse Fourier 250 00:16:11,780 --> 00:16:14,090 transform looks like. 251 00:16:14,090 --> 00:16:22,780 And so we have the following formula 252 00:16:22,780 --> 00:16:30,130 for the value of f in terms of the Fourier transform 253 00:16:30,130 --> 00:16:33,050 of the original set. 254 00:16:35,820 --> 00:16:36,510 OK. 255 00:16:36,510 --> 00:16:43,650 We want to show that f, whose support is the 2A minus 2A 256 00:16:43,650 --> 00:16:44,770 we're interested in-- 257 00:16:44,770 --> 00:16:47,340 we want to show the support of f contains 258 00:16:47,340 --> 00:16:52,390 a large subspace, a small, codimensional subspace. 259 00:16:52,390 --> 00:17:02,250 So observe that if f has positive value, then-- 260 00:17:07,089 --> 00:17:12,010 if f of x is positive, then x lies in its support. 261 00:17:12,010 --> 00:17:15,760 So we just want to find a large subspace 262 00:17:15,760 --> 00:17:20,290 on which f is positive. 263 00:17:20,290 --> 00:17:24,250 But we can choose our subspace by looking 264 00:17:24,250 --> 00:17:29,120 at Fourier coefficients according to their size. 265 00:17:29,120 --> 00:17:41,640 So what we can do is let R be the set of, essentially, 266 00:17:41,640 --> 00:17:49,900 Fourier characters whose corresponding value 267 00:17:49,900 --> 00:17:52,360 in the Fourier transform is large. 268 00:17:55,130 --> 00:17:57,100 So it's at least alpha to the 3/2. 269 00:17:57,100 --> 00:18:01,000 And that value will come up later. 270 00:18:01,000 --> 00:18:03,480 So let's look at this R. And what 271 00:18:03,480 --> 00:18:05,550 we're going to do is we're going to look 272 00:18:05,550 --> 00:18:08,040 at the orthogonal complement of R 273 00:18:08,040 --> 00:18:16,830 and show that f is positive on the orthogonal complement of R. 274 00:18:16,830 --> 00:18:19,590 First, R is not too large. 275 00:18:19,590 --> 00:18:27,880 The size of R is, I claim, less than 1 over alpha squared. 276 00:18:27,880 --> 00:18:28,856 Why's that? 277 00:18:33,233 --> 00:18:34,900 So this is an important trick that we've 278 00:18:34,900 --> 00:18:36,280 seen a few times before. 279 00:18:38,810 --> 00:18:42,950 The number of large Fourier coefficients cannot be too 280 00:18:42,950 --> 00:18:51,320 large, because they're Parseval, which tells us that the sum 281 00:18:51,320 --> 00:18:59,550 of the squares of the Fourier coefficients is equal to the L2 282 00:18:59,550 --> 00:19:05,680 norm of the original function, which, in this case, 283 00:19:05,680 --> 00:19:09,480 is just the density of A. So that's alpha. 284 00:19:09,480 --> 00:19:13,450 So just looking at that, the number of large terms cannot be 285 00:19:13,450 --> 00:19:13,950 too many. 286 00:19:18,650 --> 00:19:20,590 OK. 287 00:19:20,590 --> 00:19:25,810 So we have this small set, R, on which f has large Fourier 288 00:19:25,810 --> 00:19:27,150 transform values. 289 00:19:31,390 --> 00:19:35,730 Now, let's look at f of x. 290 00:19:35,730 --> 00:19:36,800 So let's look at f of x. 291 00:19:36,800 --> 00:19:42,010 We want to find out, when can we control f of x 292 00:19:42,010 --> 00:19:45,850 to make sure it is positive? 293 00:19:45,850 --> 00:19:50,020 Well, for the values of r-- 294 00:19:50,020 --> 00:19:58,490 little r-- not in big R or 0, we see that the Fourier 295 00:19:58,490 --> 00:20:01,770 transform-- 296 00:20:01,770 --> 00:20:04,890 we would like to upper bound this quantity here so that this 297 00:20:04,890 --> 00:20:05,760 is negligible. 298 00:20:05,760 --> 00:20:07,195 This is a small term. 299 00:20:07,195 --> 00:20:08,820 Again, this is a computation that we've 300 00:20:08,820 --> 00:20:11,950 seen several times earlier in this course. 301 00:20:11,950 --> 00:20:13,840 All of these terms are small. 302 00:20:13,840 --> 00:20:16,750 So I want to show that the whole sum is small. 303 00:20:16,750 --> 00:20:20,050 I don't want to bound each term individually and then sum up 304 00:20:20,050 --> 00:20:21,670 all the possible contributions. 305 00:20:21,670 --> 00:20:22,780 That will be too big. 306 00:20:22,780 --> 00:20:25,240 But we've seen this trick before where we just take out 307 00:20:25,240 --> 00:20:27,725 some subset of the factors. 308 00:20:27,725 --> 00:20:31,150 So in particular, I'll take out two of the factors 309 00:20:31,150 --> 00:20:45,495 and get alpha cubed upper bound plus the remaining factors. 310 00:20:52,260 --> 00:20:58,900 And once again, use Parseval on this very last sum, 311 00:20:58,900 --> 00:21:02,700 keeping in mind that I'm throwing away 312 00:21:02,700 --> 00:21:05,670 some of the r's, including 0. 313 00:21:05,670 --> 00:21:08,509 So it will be a strict inequality. 314 00:21:14,365 --> 00:21:14,865 OK. 315 00:21:14,865 --> 00:21:15,365 Yeah. 316 00:21:15,365 --> 00:21:17,910 So this step should be reminiscent of very similar 317 00:21:17,910 --> 00:21:20,580 computations that we did in the proof of Roth's theorem. 318 00:21:23,510 --> 00:21:34,640 So if x lies in the orthogonal complement of uppercase R, then 319 00:21:34,640 --> 00:21:35,840 f of x-- 320 00:21:35,840 --> 00:21:40,040 well, let's evaluate f of x from the Fourier inversion formula. 321 00:21:40,040 --> 00:21:47,540 We have this. 322 00:21:52,600 --> 00:22:06,815 So I can now split the sum as the 0-th term, the large terms. 323 00:22:11,337 --> 00:22:14,660 Now, you see, for the large terms, 324 00:22:14,660 --> 00:22:17,750 because we're in the orthogonal complement of A, 325 00:22:17,750 --> 00:22:20,170 I can make sure that they all come with a positive sign. 326 00:22:24,510 --> 00:22:26,450 And finally, the small terms. 327 00:22:42,930 --> 00:22:49,080 And you see that the main term is alpha to the 4. 328 00:22:51,780 --> 00:22:56,980 This term is always non-negative. 329 00:22:56,980 --> 00:22:59,860 And the error terms, the small terms, 330 00:22:59,860 --> 00:23:05,960 are strictly less than alpha to the 4th in magnitude. 331 00:23:08,780 --> 00:23:14,070 So as a result, this whole sum is positive. 332 00:23:14,070 --> 00:23:14,570 Yeah. 333 00:23:14,570 --> 00:23:17,233 AUDIENCE: These 1's are also 1 sub A's, right? 334 00:23:17,233 --> 00:23:18,150 YUFEI ZHAO: Thank you. 335 00:23:18,150 --> 00:23:20,658 The 1's are 1 sub A's. 336 00:23:20,658 --> 00:23:21,158 Yeah. 337 00:23:24,470 --> 00:23:26,130 So this is the very similar philosophy 338 00:23:26,130 --> 00:23:28,920 to when we proved Roth's theorem. 339 00:23:28,920 --> 00:23:33,670 We look at a sum like this, so some trigonometric series, 340 00:23:33,670 --> 00:23:35,220 some Fourier series. 341 00:23:35,220 --> 00:23:38,460 And we decompose it into several terms 342 00:23:38,460 --> 00:23:43,500 based on how large their Fourier coefficients are. 343 00:23:43,500 --> 00:23:46,840 We can control the small ones using what essentially amounts 344 00:23:46,840 --> 00:23:54,070 to a counting lemma and show that the small ones cannot ever 345 00:23:54,070 --> 00:23:59,140 annihilate the large, dominant terms. 346 00:23:59,140 --> 00:24:03,540 So as a result, f of x is positive 347 00:24:03,540 --> 00:24:06,240 on the orthogonal complement of R. 348 00:24:06,240 --> 00:24:14,600 So thus R lies in the support of f, 349 00:24:14,600 --> 00:24:17,600 which is equal to 2A minus 2A. 350 00:24:21,300 --> 00:24:34,342 And furthermore, the codimension of R is at most-- 351 00:24:34,342 --> 00:24:36,050 so it could be some linear dependencies-- 352 00:24:36,050 --> 00:24:39,990 is at most the size of R, which is strictly less than 1 353 00:24:39,990 --> 00:24:40,850 over alpha squared. 354 00:24:45,020 --> 00:24:47,250 And that proves Bogolyubov's lemma. 355 00:24:47,250 --> 00:24:50,170 So if you have a large subset of F2 to the n, 356 00:24:50,170 --> 00:24:52,980 you add it to itself enough times 357 00:24:52,980 --> 00:24:55,030 so that it's a smoothing operation. 358 00:24:55,030 --> 00:24:57,790 And then eventually, you must find a large structure. 359 00:24:57,790 --> 00:25:00,340 And we only start by assuming the size of it. 360 00:25:00,340 --> 00:25:01,930 If it's just large enough, then we 361 00:25:01,930 --> 00:25:05,350 can find a large structure within this iterated sumset. 362 00:25:07,910 --> 00:25:08,954 Any questions? 363 00:25:12,720 --> 00:25:13,220 Yeah? 364 00:25:13,220 --> 00:25:17,462 AUDIENCE: Isn't R in support of R [INAUDIBLE]?? 365 00:25:20,240 --> 00:25:21,764 YUFEI ZHAO: Sorry, come again? 366 00:25:21,764 --> 00:25:24,740 AUDIENCE: You got that the orthogonal complement of R-- 367 00:25:24,740 --> 00:25:25,490 YUFEI ZHAO: Sorry. 368 00:25:25,490 --> 00:25:27,817 The orthogonal complement of R is in the support. 369 00:25:27,817 --> 00:25:28,571 Yeah. 370 00:25:28,571 --> 00:25:31,560 So R lives in the character space. 371 00:25:34,530 --> 00:25:35,430 OK, great. 372 00:25:35,430 --> 00:25:37,740 So this is the proof of Bogolyubov's lemma 373 00:25:37,740 --> 00:25:39,960 in the finite field setting, working 374 00:25:39,960 --> 00:25:43,590 in F2 to the n, which is fine. 375 00:25:43,590 --> 00:25:47,310 It's a useful setting as a playground for us to work in. 376 00:25:47,310 --> 00:25:48,990 But ultimately, we want to understand 377 00:25:48,990 --> 00:25:50,830 what happens in the integers. 378 00:25:50,830 --> 00:25:52,860 So if you look at where we left off last time, 379 00:25:52,860 --> 00:25:56,610 we started in the cyclic group, Z mod n. 380 00:25:56,610 --> 00:26:01,140 So we would like to know how to formulate a similar result 381 00:26:01,140 --> 00:26:07,980 but in the cyclic group where there are no more subspaces. 382 00:26:07,980 --> 00:26:09,822 We encountered a similar situation, 383 00:26:09,822 --> 00:26:11,280 although we didn't go into it, when 384 00:26:11,280 --> 00:26:12,990 we discussed Roth's theorem. 385 00:26:12,990 --> 00:26:15,150 In the first proof of Roth's theorem 386 00:26:15,150 --> 00:26:18,960 that we showed, in the first Fourier analytic proof 387 00:26:18,960 --> 00:26:20,910 in the finite field setting, the proof 388 00:26:20,910 --> 00:26:26,570 won by restricting to subspaces, to hyperplanes. 389 00:26:26,570 --> 00:26:30,530 And then we keep on iterating by restricting to hyperplanes. 390 00:26:30,530 --> 00:26:32,230 So you can stay in subspaces. 391 00:26:32,230 --> 00:26:37,110 And the finite field setting has lots of subspaces. 392 00:26:37,110 --> 00:26:40,640 And we said that to get that proof to work in the integers, 393 00:26:40,640 --> 00:26:43,400 we had to do something different. 394 00:26:43,400 --> 00:26:46,710 And we did something by restricting to intervals. 395 00:26:46,710 --> 00:26:48,460 But I also mentioned that, somehow, that's 396 00:26:48,460 --> 00:26:51,100 not the natural analog of subspaces. 397 00:26:51,100 --> 00:26:54,880 The natural analog of subspaces is something called a Bohr set. 398 00:26:54,880 --> 00:26:57,480 And so I want to explore this idea further now. 399 00:27:00,650 --> 00:27:05,160 So the natural analog of subspaces in Z mod n 400 00:27:05,160 --> 00:27:06,750 are these objects called Bohr sets. 401 00:27:09,820 --> 00:27:11,980 And they're defined as follows. 402 00:27:11,980 --> 00:27:19,840 So suppose you are given some R, a subset of Z mod n. 403 00:27:19,840 --> 00:27:24,310 We define a Bohr set, denoted like this, so 404 00:27:24,310 --> 00:27:34,220 Bohr of R and epsilon, to be the subset of Z mod n, 405 00:27:34,220 --> 00:27:40,150 so including elements x, such that rx 406 00:27:40,150 --> 00:27:44,330 is pretty close to a multiple of n. 407 00:27:44,330 --> 00:27:48,220 So here, we're looking at the R mod Z norm. 408 00:27:48,220 --> 00:27:52,200 So this is the distance to the closest integer such 409 00:27:52,200 --> 00:27:59,120 that this fraction is very close to an integer for all little r 410 00:27:59,120 --> 00:28:08,872 and big R. You see, this is the analog of subspaces, 411 00:28:08,872 --> 00:28:11,080 because in the finite field setting, the finite field 412 00:28:11,080 --> 00:28:16,010 vector space, even if I set epsilon to equal to 0 413 00:28:16,010 --> 00:28:18,800 and turn this into an inner product, 414 00:28:18,800 --> 00:28:23,360 then Bohr sets are exactly subspaces-- namely, 415 00:28:23,360 --> 00:28:27,730 the orthogonal complement of the set R. 416 00:28:27,730 --> 00:28:31,600 But now we're in the integers, where you don't have exact 0. 417 00:28:31,600 --> 00:28:35,320 But I just want that quantity, that norm, to be small enough. 418 00:28:40,020 --> 00:28:41,460 So let me give you some names. 419 00:28:41,460 --> 00:28:43,800 So given the Bohr set, which, technically 420 00:28:43,800 --> 00:28:45,720 speaking, is more than just the set 421 00:28:45,720 --> 00:28:49,950 itself but also includes the information of R and epsilon-- 422 00:28:49,950 --> 00:28:52,470 so it's the entire data written on the board-- 423 00:28:52,470 --> 00:28:58,530 we call the size of the R the dimension of the Bohr set 424 00:28:58,530 --> 00:28:59,880 and epsilon, the width. 425 00:29:04,690 --> 00:29:11,230 Bogolyubov's lemma for Z mod n now takes the following form. 426 00:29:20,960 --> 00:29:25,880 If you start with a subset A of Z mod n, and all I need to know 427 00:29:25,880 --> 00:29:32,450 is that A is a constant fraction of the cyclic group, 428 00:29:32,450 --> 00:29:35,870 then the iterated sumset 2A minus 2A 429 00:29:35,870 --> 00:29:51,250 contains some Bohr set Bohr R of 1/4 430 00:29:51,250 --> 00:29:56,740 with the size of R less than 1 over alpha squared. 431 00:30:01,400 --> 00:30:03,970 So earlier, we said that if you have a large subset of F2 432 00:30:03,970 --> 00:30:08,990 to the n, then 2A minus 2A contains a large subspace. 433 00:30:08,990 --> 00:30:11,930 And now we say that if A is a large subset 434 00:30:11,930 --> 00:30:14,450 of the cyclic group, then 2A minus 2A 435 00:30:14,450 --> 00:30:19,130 contains a large Bohr set of small dimension. 436 00:30:19,130 --> 00:30:22,790 And so this terminology may be slightly confusing. 437 00:30:22,790 --> 00:30:27,090 The dimension corresponds to codimension previously. 438 00:30:27,090 --> 00:30:33,020 So if you do this translation, this dimension-- 439 00:30:33,020 --> 00:30:35,078 I mean, if R were a set of independent vectors 440 00:30:35,078 --> 00:30:36,620 and you have 2 to the n, then that'll 441 00:30:36,620 --> 00:30:39,290 be the codimension of the corresponding subspace. 442 00:30:39,290 --> 00:30:43,770 But this is the terminology that we're stuck with. 443 00:30:43,770 --> 00:30:44,270 OK. 444 00:30:44,270 --> 00:30:46,060 Any questions about the statement? 445 00:30:50,770 --> 00:30:53,590 You see, even the bounds are exactly the same, 1 over alpha 446 00:30:53,590 --> 00:30:54,430 squared. 447 00:30:54,430 --> 00:30:56,320 And I mean, the proof is going to be 448 00:30:56,320 --> 00:30:57,970 pretty much exactly the same once you 449 00:30:57,970 --> 00:31:01,247 make the correct notational modifications. 450 00:31:01,247 --> 00:31:02,330 So we're going to do that. 451 00:31:02,330 --> 00:31:05,230 So I'm going to write on top of this earlier proof 452 00:31:05,230 --> 00:31:08,080 and show you what are the notational modifications so 453 00:31:08,080 --> 00:31:12,640 that you can get exactly the same result here 454 00:31:12,640 --> 00:31:14,760 but with Bohr sets instead of a subspace. 455 00:31:17,840 --> 00:31:20,690 The thing to keep in mind is that we have a somewhat 456 00:31:20,690 --> 00:31:23,210 different Fourier transform. 457 00:31:23,210 --> 00:31:28,580 So let me now use different colored chalk. 458 00:31:28,580 --> 00:31:40,870 So the Fourier transform of a function f from Z mod n, 459 00:31:40,870 --> 00:31:46,330 so a complex value, is a function 460 00:31:46,330 --> 00:31:55,570 also on Z mod n defined by f hat of r equal to expectation 461 00:31:55,570 --> 00:32:00,100 over x in Z mod n of f of x times omega 462 00:32:00,100 --> 00:32:07,690 to the minus rx, where omega is a primitive n root of unity. 463 00:32:17,950 --> 00:32:20,010 And you also had the Fourier inversion formula. 464 00:32:20,010 --> 00:32:21,500 It's what you expect. 465 00:32:21,500 --> 00:32:23,735 I won't bother writing it down. 466 00:32:23,735 --> 00:32:24,860 So we go back to the proof. 467 00:32:24,860 --> 00:32:27,550 And pretty much everything will read exactly the same. 468 00:32:27,550 --> 00:32:28,920 So f is still the same f. 469 00:32:32,180 --> 00:32:36,866 And the Fourier transform has the same property. 470 00:32:36,866 --> 00:32:40,880 So all of these nice properties of the Fourier transform hold. 471 00:32:40,880 --> 00:32:43,340 For inversion, it's basically the same 472 00:32:43,340 --> 00:32:48,210 except that the formula is slightly different. 473 00:32:48,210 --> 00:32:50,960 So instead of minus 1 to the r dot 474 00:32:50,960 --> 00:32:54,700 x, what we have now is omega to the rx. 475 00:32:58,310 --> 00:33:01,990 So here, we have omega to the rx. 476 00:33:05,530 --> 00:33:06,193 OK. 477 00:33:06,193 --> 00:33:08,660 Great. 478 00:33:08,660 --> 00:33:12,620 The next part is the same, where we define r. 479 00:33:12,620 --> 00:33:21,680 So now we define r to consist of elements of z mod n, whose 480 00:33:21,680 --> 00:33:24,190 Fourier transform is large. 481 00:33:24,190 --> 00:33:25,170 I can take out 0. 482 00:33:36,555 --> 00:33:39,433 OK. 483 00:33:39,433 --> 00:33:40,600 This part is still the same. 484 00:33:40,600 --> 00:33:42,540 It's the same calculation. 485 00:33:42,540 --> 00:33:44,340 Now, it's the very last part that needs 486 00:33:44,340 --> 00:33:46,320 to be just slightly changed. 487 00:33:49,890 --> 00:33:53,030 Where does the 1/4 come in? 488 00:33:53,030 --> 00:33:54,990 So where does this come in? 489 00:33:54,990 --> 00:34:06,120 So observe that if x is in the Bohr set with width 1/4, 490 00:34:06,120 --> 00:34:16,830 then rx divided by n is-- 491 00:34:16,830 --> 00:34:20,860 OK, so by definition, all of these fractions 492 00:34:20,860 --> 00:34:24,070 are within the 1/4 of an integer. 493 00:34:24,070 --> 00:34:29,320 And if you think about what happens on the unit circle, 494 00:34:29,320 --> 00:34:33,250 if you are within 1/4 of the integer, 495 00:34:33,250 --> 00:34:36,550 then that means the corresponding place on the unit 496 00:34:36,550 --> 00:34:44,900 circle is on the left half circle. 497 00:34:44,900 --> 00:34:54,630 So in particular, the cosine of 2rx over n is non-negative. 498 00:34:54,630 --> 00:35:00,110 So it has non-negative real part. 499 00:35:00,110 --> 00:35:02,980 So now we go back to this part of the proof, 500 00:35:02,980 --> 00:35:07,180 where we're applying Fourier inversion formula to f of x. 501 00:35:07,180 --> 00:35:09,430 So we had the Fourier inversion formula up there. 502 00:35:09,430 --> 00:35:16,160 But because f of x is real, it's really the cosine 503 00:35:16,160 --> 00:35:18,660 that should come in play. 504 00:35:21,795 --> 00:35:23,580 It should be a cosine. 505 00:35:23,580 --> 00:35:31,320 And now, for the next step, we have no negative sign here, 506 00:35:31,320 --> 00:35:34,510 because this step-- 507 00:35:34,510 --> 00:35:40,420 OK, let me just cross out this step over there. 508 00:35:40,420 --> 00:35:43,000 All of these terms, the terms that 509 00:35:43,000 --> 00:35:45,190 correspond to little r and big R, 510 00:35:45,190 --> 00:35:49,380 they have non-negative contribution. 511 00:35:53,080 --> 00:35:55,840 Whatever the contributions here, it's non-negative. 512 00:35:55,840 --> 00:35:59,480 So I cross out this term. 513 00:35:59,480 --> 00:36:01,580 All I'm left with is the main term, 514 00:36:01,580 --> 00:36:05,310 corresponding to the density, and the error term, 515 00:36:05,310 --> 00:36:08,540 so to speak, the minor terms, which is less than alpha 516 00:36:08,540 --> 00:36:12,315 to the 4th in absolute value. 517 00:36:12,315 --> 00:36:14,190 OK. 518 00:36:14,190 --> 00:36:17,883 So it's positive. 519 00:36:17,883 --> 00:36:19,050 So basically the same proof. 520 00:36:19,050 --> 00:36:21,210 Once you make the appropriate modifications, 521 00:36:21,210 --> 00:36:23,140 it's the same proof in Z mod n. 522 00:36:28,320 --> 00:36:29,460 OK, great. 523 00:36:29,460 --> 00:36:33,602 So this concludes our discussion of Bogolyubov's lemma. 524 00:36:33,602 --> 00:36:35,550 So it says that-- 525 00:36:35,550 --> 00:36:37,540 OK, so continuing our previous thread, 526 00:36:37,540 --> 00:36:42,670 we start with a subset of z mod n of constant proportion. 527 00:36:42,670 --> 00:36:49,820 Then 2A minus 2A necessarily contains a large Bohr set. 528 00:36:49,820 --> 00:36:53,285 And the next thing I want to do is to start with this Bohr set. 529 00:36:53,285 --> 00:36:54,910 So that's the definition of a Bohr set. 530 00:36:54,910 --> 00:36:56,100 But what does it look like? 531 00:36:56,100 --> 00:36:59,887 So it's a bit hard to imagine. 532 00:36:59,887 --> 00:37:00,970 So what does it look like? 533 00:37:00,970 --> 00:37:04,390 In the finite field setting, we know it's a subspace. 534 00:37:04,390 --> 00:37:07,510 But in the Z mod n setting, right now, it's 535 00:37:07,510 --> 00:37:09,550 just some subset of Z mod n. 536 00:37:09,550 --> 00:37:12,520 OK, so in the next step, we want to extract 537 00:37:12,520 --> 00:37:18,610 some geometric structure from this Bohr set. 538 00:37:18,610 --> 00:37:20,830 So we're going to show that this Bohr 539 00:37:20,830 --> 00:37:24,940 set will contain a large, generalized arithmetic 540 00:37:24,940 --> 00:37:25,802 progression. 541 00:37:28,460 --> 00:37:30,790 So you asked something earlier about-- 542 00:37:30,790 --> 00:37:34,510 something seems a bit fishy about the general strategy. 543 00:37:34,510 --> 00:37:38,500 Seems like our goal for proving-- 544 00:37:38,500 --> 00:37:40,630 we want to prove Freiman's theorem, which 545 00:37:40,630 --> 00:37:43,640 says that the conclusion is that A 546 00:37:43,640 --> 00:37:50,890 is contained in some GAP, some fairly compact additive 547 00:37:50,890 --> 00:37:52,770 structure. 548 00:37:52,770 --> 00:37:54,850 And we're already losing quite a bit. 549 00:37:54,850 --> 00:37:58,240 So we pass down to 1/8 of A. So it 550 00:37:58,240 --> 00:38:02,110 seems like even if you contain the rest, 551 00:38:02,110 --> 00:38:05,230 even if you can contain this fraction, this large fraction 552 00:38:05,230 --> 00:38:10,270 of A, what are you going to do about the rest of A? 553 00:38:10,270 --> 00:38:12,440 That's an unanswered question. 554 00:38:12,440 --> 00:38:14,440 A second unanswered question-- so right now, 555 00:38:14,440 --> 00:38:16,570 what I've told you, the strategy is 556 00:38:16,570 --> 00:38:25,280 we're going to find a large GAP inside 2A minus 2A, 557 00:38:25,280 --> 00:38:28,710 which is not quite the thing that we want to do. 558 00:38:28,710 --> 00:38:32,880 We want to contain A in a small GAP. 559 00:38:32,880 --> 00:38:35,050 But at least it's some progress, right? 560 00:38:35,050 --> 00:38:37,140 It's some progress to find some structure. 561 00:38:37,140 --> 00:38:38,880 I mean, the name of the game is to try 562 00:38:38,880 --> 00:38:40,890 to find additive structure. 563 00:38:40,890 --> 00:38:43,590 So in the theme of this whole semester 564 00:38:43,590 --> 00:38:46,770 course is trying to understand the dichotomy between structure 565 00:38:46,770 --> 00:38:48,050 and pseudorandomness. 566 00:38:48,050 --> 00:38:50,960 And when you have structure, let's use that structure. 567 00:38:50,960 --> 00:38:53,820 See if you can boost that structure. 568 00:38:53,820 --> 00:38:57,150 So there will be an additional argument, which 569 00:38:57,150 --> 00:38:59,880 I will show you at the beginning of next lecture 570 00:38:59,880 --> 00:39:04,380 at the conclusion of the proof of Freiman's theorem, which 571 00:39:04,380 --> 00:39:08,460 will allow you to start with the structure on a small part of A, 572 00:39:08,460 --> 00:39:10,950 but not too small-- it's a constant fraction of A-- 573 00:39:10,950 --> 00:39:15,516 and pass it up to the whole of A. 574 00:39:15,516 --> 00:39:17,670 And we've actually already seen a tool 575 00:39:17,670 --> 00:39:20,990 that allows us to do that. 576 00:39:20,990 --> 00:39:26,610 So I want to cover all of A. So last time, we 577 00:39:26,610 --> 00:39:28,530 did something called the covering lemma, 578 00:39:28,530 --> 00:39:30,600 Ruzsa covering lemma, that tells us 579 00:39:30,600 --> 00:39:33,585 that if you have some nice control on A 580 00:39:33,585 --> 00:39:38,160 and you can cover some part of A very well, 581 00:39:38,160 --> 00:39:42,110 then I can cover the entirety of A very well. 582 00:39:42,110 --> 00:39:45,460 So those tools will come in hand. 583 00:39:45,460 --> 00:39:47,100 I mean, so similar to actually how we 584 00:39:47,100 --> 00:39:50,357 proved Freiman's theorem in groups with bounded exponent. 585 00:39:50,357 --> 00:39:51,940 And so we're going to use the covering 586 00:39:51,940 --> 00:39:54,540 lemma to conclude the theorem. 587 00:39:57,290 --> 00:39:58,790 But now I want to get into the issue 588 00:39:58,790 --> 00:40:00,140 of the geometry of numbers. 589 00:40:08,900 --> 00:40:09,570 OK. 590 00:40:09,570 --> 00:40:11,580 I want to tell you some necessary tools 591 00:40:11,580 --> 00:40:19,632 that we'll need to find a large GAP inside 2A minus 2A. 592 00:40:19,632 --> 00:40:23,750 Now, it will seem like a bit of a digression, 593 00:40:23,750 --> 00:40:26,050 but we'll come back into additive combinatorics 594 00:40:26,050 --> 00:40:27,070 in a bit. 595 00:40:27,070 --> 00:40:30,155 So the geometry of numbers concerns the study of lattices. 596 00:40:32,910 --> 00:40:37,200 So it concerns the study of lattices and convex bodies. 597 00:40:41,370 --> 00:40:45,710 So this is a really important area of mathematics, especially 598 00:40:45,710 --> 00:40:49,220 about a century ago with mathematicians like Minkowski 599 00:40:49,220 --> 00:40:51,340 playing foundational roles in the subject. 600 00:40:51,340 --> 00:40:54,050 So number theorists were very interested in trying 601 00:40:54,050 --> 00:40:58,100 to understand how lattices behave. 602 00:40:58,100 --> 00:41:01,340 So I'll tell you some very classical results 603 00:41:01,340 --> 00:41:04,580 that we'll use for proving Freiman's theorem. 604 00:41:08,140 --> 00:41:10,270 So first, what is a lattice? 605 00:41:10,270 --> 00:41:16,870 So let me give you the following definition of a lattice in R 606 00:41:16,870 --> 00:41:18,390 to the d. 607 00:41:18,390 --> 00:41:26,910 It's a structure on a group, if you will, as an integer 608 00:41:26,910 --> 00:41:31,270 span of d independent vectors. 609 00:41:37,460 --> 00:41:43,080 So I start with v1 through vd vectors 610 00:41:43,080 --> 00:41:45,890 that are linearly independent. 611 00:41:45,890 --> 00:41:47,730 And I look at their integer span. 612 00:41:47,730 --> 00:41:50,710 I think this is best explained with a picture. 613 00:41:50,710 --> 00:41:55,790 So if I have a bunch of-- 614 00:41:58,900 --> 00:42:00,760 so here, I'm drawing a picture in R2. 615 00:42:06,350 --> 00:42:09,080 And this picture extends in all directions. 616 00:42:09,080 --> 00:42:15,460 If I start with two vectors, v1 and v2, linearly independent, 617 00:42:15,460 --> 00:42:20,080 and look at their integer span, so that's a lattice. 618 00:42:20,080 --> 00:42:21,703 So that's what a lattice is. 619 00:42:21,703 --> 00:42:23,870 You can come up with all sorts of fancy definitions, 620 00:42:23,870 --> 00:42:28,250 like a discrete subgroup of R to the n. 621 00:42:28,250 --> 00:42:29,540 But this is what it is. 622 00:42:32,740 --> 00:42:36,670 So just to emphasize this definition for a bit-- 623 00:42:36,670 --> 00:42:39,400 and also, one more definition that we'll need 624 00:42:39,400 --> 00:42:43,100 is the determinant of a lattice. 625 00:42:43,100 --> 00:42:46,530 So what's the determinant of a lattice? 626 00:42:46,530 --> 00:42:51,120 One way to define it is you look at these v's, and you 627 00:42:51,120 --> 00:42:59,630 construct a matrix with the v's as columns. 628 00:42:59,630 --> 00:43:05,070 And you evaluate the absolute value of this determinant. 629 00:43:05,070 --> 00:43:11,720 More visually, the determinant of a lattice is also equal 630 00:43:11,720 --> 00:43:24,100 to the volume of its fundamental parallelepiped, 631 00:43:24,100 --> 00:43:27,570 which is a parallelepiped-- well, 632 00:43:27,570 --> 00:43:30,200 in the two-dimensional case, it's a parallelogram-- 633 00:43:30,200 --> 00:43:37,060 which is spanned by v1 and v2 or these v's, although you 634 00:43:37,060 --> 00:43:38,780 have more choices, right? 635 00:43:38,780 --> 00:43:41,530 So you could have chosen a different set 636 00:43:41,530 --> 00:43:43,162 of generating vectors. 637 00:43:43,162 --> 00:43:45,370 For example, you could have chosen these two vectors, 638 00:43:45,370 --> 00:43:48,280 and they also generate the same lattice. 639 00:43:48,280 --> 00:43:50,170 And that's also a fundamental parallelepiped. 640 00:43:50,170 --> 00:43:51,628 And they will have the same volume. 641 00:43:54,120 --> 00:43:56,290 You can make some wrong choices, and then they 642 00:43:56,290 --> 00:43:57,760 will not have the right volume. 643 00:43:57,760 --> 00:44:08,630 So if you had chosen these two, so this 644 00:44:08,630 --> 00:44:13,936 is not a fundamental parallelepiped. 645 00:44:18,040 --> 00:44:19,160 Great. 646 00:44:19,160 --> 00:44:20,535 So let me give you some examples. 647 00:44:23,610 --> 00:44:31,330 The simplest lattice is just the integer lattice, Zd, 648 00:44:31,330 --> 00:44:34,150 which has determinant 1. 649 00:44:40,090 --> 00:44:41,860 If I'm in the complex plane, which 650 00:44:41,860 --> 00:44:50,790 is viewed as two-dimensional real plane, then if I take, 651 00:44:50,790 --> 00:44:59,710 let's say, omega being the 3rd root of unity, 652 00:44:59,710 --> 00:45:01,370 I have a triangular lattice. 653 00:45:08,240 --> 00:45:10,960 And the fundamental parallelepiped 654 00:45:10,960 --> 00:45:14,290 of this lattice, that's one example. 655 00:45:14,290 --> 00:45:16,910 And you can evaluate its determinant 656 00:45:16,910 --> 00:45:20,838 as the area of that parallelogram. 657 00:45:24,120 --> 00:45:28,540 If I take two nonlinearly independent vectors-- 658 00:45:28,540 --> 00:45:32,250 so for example, if I'm in one dimension 659 00:45:32,250 --> 00:45:35,970 and I look at the integer span of 1 and root 2, 660 00:45:35,970 --> 00:45:37,020 this is not a lattice. 661 00:45:45,732 --> 00:45:49,440 Now, the next definition will initially 662 00:45:49,440 --> 00:45:50,610 be slightly confusing. 663 00:45:50,610 --> 00:45:52,920 But I will explain it through an example 664 00:45:52,920 --> 00:45:56,940 or at least try to help you visualize what's going on. 665 00:45:56,940 --> 00:46:03,720 So if I give you a centrally symmetric convex body-- 666 00:46:03,720 --> 00:46:08,360 "centrally symmetric" means that k equals to minus k. 667 00:46:08,360 --> 00:46:14,780 So centrally symmetric convex body, OK. 668 00:46:14,780 --> 00:46:20,640 So here, centrally symmetric is x in k 669 00:46:20,640 --> 00:46:22,510 if and only if minus x is in k. 670 00:46:25,760 --> 00:46:30,610 And I'm in d dimensions. 671 00:46:30,610 --> 00:46:40,490 Let me define the i-th successive minimum 672 00:46:40,490 --> 00:46:44,500 to be lambda i. 673 00:46:44,500 --> 00:46:53,180 OK, so i-th successive minimum of k with respect to lambda 674 00:46:53,180 --> 00:47:04,450 to be the infimum of all non-negative lambda such that 675 00:47:04,450 --> 00:47:14,810 the dimension of the span of the intersection of lambda k and-- 676 00:47:14,810 --> 00:47:19,190 well, little o lambda k and the lattice-- 677 00:47:19,190 --> 00:47:24,772 has dimension at least i. 678 00:47:24,772 --> 00:47:25,700 OK. 679 00:47:25,700 --> 00:47:26,450 So let me explain. 680 00:47:33,380 --> 00:47:35,490 I start with a lattice. 681 00:47:35,490 --> 00:47:38,450 So I start with some lattice. 682 00:47:38,450 --> 00:47:40,280 And I have some convex body. 683 00:47:45,110 --> 00:47:46,450 So this is 0, let's say. 684 00:47:49,110 --> 00:47:55,040 So I have some convex body, a centrally symmetric convex 685 00:47:55,040 --> 00:47:56,578 body like that. 686 00:47:56,578 --> 00:47:58,120 I initially could be bigger, as well, 687 00:47:58,120 --> 00:48:02,650 but that's scale it so that it's quite small initially. 688 00:48:02,650 --> 00:48:11,740 And let's consider an animation where I look at lambda k 689 00:48:11,740 --> 00:48:17,710 where k lambda goes from 0 to infinity. 690 00:48:17,710 --> 00:48:18,580 This is k. 691 00:48:18,580 --> 00:48:22,590 So initially, lambda k is very, very small. 692 00:48:22,590 --> 00:48:24,450 And I imagine it growing. 693 00:48:24,450 --> 00:48:27,040 It gets bigger and bigger and bigger. 694 00:48:27,040 --> 00:48:29,500 So it gets bigger and bigger. 695 00:48:29,500 --> 00:48:32,230 And let's think about the first time 696 00:48:32,230 --> 00:48:36,760 that this growing body hits a lattice point, 697 00:48:36,760 --> 00:48:39,310 a non-zero lattice point. 698 00:48:39,310 --> 00:48:42,930 At that point, I freeze the animation. 699 00:48:42,930 --> 00:48:46,826 And I record this vector. 700 00:48:46,826 --> 00:48:58,280 I record this vector where I've hit a lattice point. 701 00:48:58,280 --> 00:48:59,790 And now I continue the animation. 702 00:48:59,790 --> 00:49:01,498 It's going to keep on growing and growing 703 00:49:01,498 --> 00:49:05,110 and growing until when I hit a vector in a direction 704 00:49:05,110 --> 00:49:07,560 I haven't seen before. 705 00:49:07,560 --> 00:49:08,980 So it's going to keep growing. 706 00:49:08,980 --> 00:49:14,950 And then the next time I hit a vector in a new direction, 707 00:49:14,950 --> 00:49:18,450 I stop the animation. 708 00:49:18,450 --> 00:49:21,540 And I look at the other vector. 709 00:49:21,540 --> 00:49:27,000 So I keep growing this ball until I hit new vectors, 710 00:49:27,000 --> 00:49:29,200 keep growing this convex body. 711 00:49:29,200 --> 00:49:41,920 So for example, if your initial convex body is very elongated, 712 00:49:41,920 --> 00:49:43,050 if that's your k-- 713 00:49:43,050 --> 00:49:44,820 so you keep growing, growing-- 714 00:49:44,820 --> 00:49:50,610 you might initially hit that vector. 715 00:49:50,610 --> 00:49:52,280 And then you keep on growing it. 716 00:49:52,280 --> 00:49:54,020 And the next vector you hit might still 717 00:49:54,020 --> 00:49:55,850 be in the same direction. 718 00:49:55,850 --> 00:49:57,700 But I don't count it. 719 00:49:57,700 --> 00:49:59,450 I don't stop the animation here, because I 720 00:49:59,450 --> 00:50:01,070 didn't see a new direction yet. 721 00:50:01,070 --> 00:50:05,450 I only stop the animation when I see a new direction. 722 00:50:05,450 --> 00:50:10,410 So I keep growing until I see a new direction. 723 00:50:10,410 --> 00:50:13,480 And I stop the animation there. 724 00:50:13,480 --> 00:50:17,410 So think about this growing body, and stop in every place 725 00:50:17,410 --> 00:50:22,000 when you see a new direction contained in your lambda k. 726 00:50:22,000 --> 00:50:26,000 And the places where you stop the animations, 727 00:50:26,000 --> 00:50:31,292 they're the successive minimum of k. 728 00:50:31,292 --> 00:50:32,276 Yeah? 729 00:50:32,276 --> 00:50:36,002 AUDIENCE: Is this defined if i is greater than d? 730 00:50:36,002 --> 00:50:38,210 YUFEI ZHAO: Is this defined when i is greater than d? 731 00:50:38,210 --> 00:50:38,710 No. 732 00:50:38,710 --> 00:50:44,880 So you only have exactly d successive minimum. 733 00:50:49,040 --> 00:50:52,180 Now, sometimes you might see two new directions 734 00:50:52,180 --> 00:50:53,620 at the same time. 735 00:50:53,620 --> 00:50:55,240 That's OK. 736 00:50:55,240 --> 00:50:58,393 But once you exhaust all d directions, 737 00:50:58,393 --> 00:51:00,560 then there's no more new directions you can explore. 738 00:51:03,230 --> 00:51:10,680 We also consider the vectors that you see. 739 00:51:10,680 --> 00:51:15,470 So let me also call these so that we can-- 740 00:51:15,470 --> 00:51:21,570 OK, so we can select these lattice vectors bi. 741 00:51:21,570 --> 00:51:26,140 I am going to use underscore to denote. 742 00:51:26,140 --> 00:51:29,650 So I'm going to use this underline to denote boldface. 743 00:51:29,650 --> 00:51:38,870 So it's a vector bi, which is in, basically, this. 744 00:51:38,870 --> 00:51:42,290 You should think of bi as the new vector that you see. 745 00:51:42,290 --> 00:51:53,240 And it will have the property such that b1 through bd 746 00:51:53,240 --> 00:52:03,210 form a basis of Rd. 747 00:52:03,210 --> 00:52:05,740 So I keep growing this convex body. 748 00:52:05,740 --> 00:52:09,050 When I see a vector in a new direction, I record lambda. 749 00:52:09,050 --> 00:52:11,402 And I record the vector bi. 750 00:52:11,402 --> 00:52:13,510 I keep on going, keep going, keep 751 00:52:13,510 --> 00:52:17,380 going until I exhaust all d directions. 752 00:52:17,380 --> 00:52:22,195 I call these b's the directional basis. 753 00:52:28,742 --> 00:52:29,710 OK. 754 00:52:29,710 --> 00:52:30,743 Any questions? 755 00:52:35,473 --> 00:52:37,850 All right. 756 00:52:37,850 --> 00:52:40,970 So the result from the geometry of numbers 757 00:52:40,970 --> 00:52:43,310 that we're going to need is something called 758 00:52:43,310 --> 00:52:44,870 Minkowski's second theorem. 759 00:52:56,150 --> 00:52:59,330 So Minkowski's second theorem says 760 00:52:59,330 --> 00:53:05,800 that if you have lambda, a lattice, 761 00:53:05,800 --> 00:53:15,580 in Rd and k, a centrally symmetric body, also in Rd, 762 00:53:15,580 --> 00:53:19,200 such that lambda 1 through lambda 763 00:53:19,200 --> 00:53:28,560 d are the successive minima of k with respect to lambda, 764 00:53:28,560 --> 00:53:33,580 then one has the inequality lambda 1, lambda 2. 765 00:53:33,580 --> 00:53:35,850 So the product of these successive minima 766 00:53:35,850 --> 00:53:40,710 times the volume of k is upper bounded by 2 767 00:53:40,710 --> 00:53:44,520 to the d times the determinant of lambda. 768 00:53:48,000 --> 00:53:54,160 For example, and here is a very easy case of this 769 00:53:54,160 --> 00:54:02,960 Minkowski's second theorem, if your k is an axis-aligned box-- 770 00:54:02,960 --> 00:54:12,240 namely, it is a box where the width in the i-th direction is 771 00:54:12,240 --> 00:54:14,220 2 over lambda i-- 772 00:54:22,380 --> 00:54:28,230 so then you see that the successive minima of this box 773 00:54:28,230 --> 00:54:31,880 are exactly the lambda i's. 774 00:54:31,880 --> 00:54:34,760 And you can check that for-- 775 00:54:34,760 --> 00:54:37,288 this inequality is actually an equality. 776 00:54:42,410 --> 00:54:42,910 OK. 777 00:54:42,910 --> 00:54:47,110 So actually, in this case, lambda, the lattice, 778 00:54:47,110 --> 00:54:49,150 is the integer lattice. 779 00:54:49,150 --> 00:54:52,150 Now, this is a pretty easy case of Minkowski's second theorem. 780 00:54:52,150 --> 00:54:55,900 But the general case, which we're not going to prove, 781 00:54:55,900 --> 00:55:00,030 is actually quite subtle. 782 00:55:00,030 --> 00:55:02,270 I mean, the proof itself is not so long. 783 00:55:02,270 --> 00:55:05,810 It's worth looking up and trying to see what the proof is about. 784 00:55:05,810 --> 00:55:08,095 But it's actually rather counterintuitive 785 00:55:08,095 --> 00:55:08,720 to think about. 786 00:55:08,720 --> 00:55:11,743 It's one of those theorems where you sit down 787 00:55:11,743 --> 00:55:12,910 for half an hour or an hour. 788 00:55:12,910 --> 00:55:13,520 You're trying to prove. 789 00:55:13,520 --> 00:55:15,560 You think you might have come up with a proof. 790 00:55:15,560 --> 00:55:18,520 And then on closer examination, it'll 791 00:55:18,520 --> 00:55:21,425 be very likely that you made some very subtle error. 792 00:55:21,425 --> 00:55:25,550 So it's not so easy to get all the details right. 793 00:55:25,550 --> 00:55:28,040 And we're going to skip the proof. 794 00:55:28,040 --> 00:55:29,936 But any questions about the statement? 795 00:55:34,796 --> 00:55:37,240 OK. 796 00:55:37,240 --> 00:55:39,220 We're going to use Minkowski's second theorem 797 00:55:39,220 --> 00:55:44,860 to show that a large Bohr set contains a large GAP. 798 00:55:58,760 --> 00:56:05,530 And specifically, we will prove that every Bohr 799 00:56:05,530 --> 00:56:21,910 set of dimension d and width epsilon-- 800 00:56:21,910 --> 00:56:25,000 epsilon is between 0 and 1-- 801 00:56:25,000 --> 00:56:42,680 in Z mod nZ contains a proper GAP with dimension at most d 802 00:56:42,680 --> 00:56:48,650 and size at least this quantity, which 803 00:56:48,650 --> 00:56:53,720 is epsilon divided by d raised to the power of d fraction 804 00:56:53,720 --> 00:56:54,820 of the cyclic group. 805 00:57:01,570 --> 00:57:05,100 So just to step back a bit and see 806 00:57:05,100 --> 00:57:07,530 where we're going, from everything 807 00:57:07,530 --> 00:57:11,670 that we've done earlier, we conclude that 2A minus 2A 808 00:57:11,670 --> 00:57:13,770 contains a large Bohr set. 809 00:57:13,770 --> 00:57:15,360 Here, epsilon is 1/4. 810 00:57:15,360 --> 00:57:17,160 So epsilon is a constant. 811 00:57:17,160 --> 00:57:21,620 And R is also going to be a constant. 812 00:57:21,620 --> 00:57:23,460 It's depending on the doubling constant. 813 00:57:26,050 --> 00:57:27,870 And this proposition will tell us 814 00:57:27,870 --> 00:57:31,080 that inside this 2A minus 2A, we will 815 00:57:31,080 --> 00:57:33,910 be able to find a very large, proper GAP. 816 00:57:33,910 --> 00:57:36,150 So "proper" means that in this generalized arithmetic 817 00:57:36,150 --> 00:57:38,860 progression, all the individual terms are distinct, 818 00:57:38,860 --> 00:57:40,360 or you don't have collisions. 819 00:57:40,360 --> 00:57:42,540 So you're going to find this proper GAP that 820 00:57:42,540 --> 00:57:47,550 is constant dimension and at least 821 00:57:47,550 --> 00:57:52,110 a constant fraction of the size of the group, so pretty 822 00:57:52,110 --> 00:57:53,000 large GAP. 823 00:58:01,490 --> 00:58:05,540 To find this GAP, we will set up a lattice 824 00:58:05,540 --> 00:58:07,430 and apply Minkowski's second theorem. 825 00:58:13,540 --> 00:58:19,330 Suppose the Bohr set is given by R 826 00:58:19,330 --> 00:58:21,450 where the individual elements, I'm 827 00:58:21,450 --> 00:58:26,340 going to denote by little r1 through little rd. 828 00:58:26,340 --> 00:58:33,960 And let uppercase lambda be a lattice explicitly given 829 00:58:33,960 --> 00:58:35,880 as follows. 830 00:58:35,880 --> 00:58:52,190 It consists of all points in Rd that are congruent mod 1 831 00:58:52,190 --> 00:59:05,840 to some integer multiple of the vector r1 over n, r2 over n, 832 00:59:05,840 --> 00:59:11,880 through rd over n, so congruent mod 1. 833 00:59:14,620 --> 00:59:18,910 So for example, in two dimensions, which is all 834 00:59:18,910 --> 00:59:27,950 I can draw on the board, if r1 and r2 are 1 and 3 835 00:59:27,950 --> 00:59:33,260 and n equals to 5, then basically, 836 00:59:33,260 --> 00:59:43,090 what we're going to have is a refinement of the integer 837 00:59:43,090 --> 00:59:50,960 lattice, where this box is going to be the integer lattice. 838 00:59:50,960 --> 00:59:54,320 And I'm going to tell you some additional lattice vectors. 839 00:59:54,320 --> 00:59:59,370 And here, it's going to repeat, or it's going to tile all over. 840 00:59:59,370 --> 01:00:04,390 So I start with 1, 3. 841 01:00:04,390 --> 01:00:06,520 And I look at multiples of it. 842 01:00:06,520 --> 01:00:08,930 But I mod 1. 843 01:00:08,930 --> 01:00:15,650 So I would end up with these points and then repeat it. 844 01:00:26,162 --> 01:00:27,120 And so you would have-- 845 01:00:32,180 --> 01:00:33,420 so that's the lattice. 846 01:00:33,420 --> 01:00:37,810 So you have this lattice, lambda. 847 01:00:37,810 --> 01:00:39,460 What is the volume? 848 01:00:39,460 --> 01:00:43,120 What is the determinant of this lattice? 849 01:00:43,120 --> 01:00:44,750 So the determinant of the lattice, 850 01:00:44,750 --> 01:00:50,900 remember, is the volume of its fundamental parallelepiped. 851 01:00:50,900 --> 01:00:58,480 So I claim that the determinant is exactly 1 over n. 852 01:00:58,480 --> 01:01:00,420 There are a few ways to see this. 853 01:01:00,420 --> 01:01:04,300 So one is that, originally, I had the integer 854 01:01:04,300 --> 01:01:06,170 lattice as determinant 1. 855 01:01:06,170 --> 01:01:07,870 And now I put-- 856 01:01:07,870 --> 01:01:09,940 instead of one point, I have endpoints 857 01:01:09,940 --> 01:01:12,070 in each original parallelepiped. 858 01:01:12,070 --> 01:01:16,030 So the determinant has to go down by a factor of n. 859 01:01:16,030 --> 01:01:20,890 Or you can construct an explicit fundamental parallelepiped 860 01:01:20,890 --> 01:01:21,790 like that. 861 01:01:21,790 --> 01:01:25,186 And then you use base times height. 862 01:01:25,186 --> 01:01:25,686 OK. 863 01:01:33,110 --> 01:01:36,750 We're going to apply Minkowski's second theorem. 864 01:01:36,750 --> 01:01:38,450 And I will need to tell you-- 865 01:01:38,450 --> 01:01:40,950 I don't need the definition of Bohr set up there. 866 01:01:40,950 --> 01:01:45,660 So I want to tell you what to use as the convex body. 867 01:01:57,760 --> 01:02:00,580 The convex body that we're going to use 868 01:02:00,580 --> 01:02:11,280 is k being this box of width 2 epsilon. 869 01:02:11,280 --> 01:02:12,780 So that's the lattice. 870 01:02:12,780 --> 01:02:14,990 That's the convex body. 871 01:02:14,990 --> 01:02:17,910 And we're going to apply Minkowski's second theorem. 872 01:02:17,910 --> 01:02:25,370 So let's let little lowercase lambda 1 through lambda d-- 873 01:02:25,370 --> 01:02:28,470 so n is d-- 874 01:02:28,470 --> 01:02:39,290 be the successive minima of k with respect to lambda 875 01:02:39,290 --> 01:02:47,800 and b be the directional vectors, 876 01:02:47,800 --> 01:02:52,750 the rational basis corresponding to those successive minima. 877 01:02:56,640 --> 01:03:03,830 I claim that the L-infinity norm of bj 878 01:03:03,830 --> 01:03:12,310 is at most lambda j epsilon for each j. 879 01:03:12,310 --> 01:03:16,080 And this is basically because of the definition. 880 01:03:16,080 --> 01:03:18,615 I mean, if you look at the definition of successive minima 881 01:03:18,615 --> 01:03:28,900 and directional basis, this is k. 882 01:03:28,900 --> 01:03:34,020 I grow k, grow it by a factor of lambda j. 883 01:03:34,020 --> 01:03:39,030 And that's the first point when I see b sub j. 884 01:03:39,030 --> 01:03:42,030 So every coordinate of b sub j has 885 01:03:42,030 --> 01:03:45,230 to be at most this quantity in absolute value. 886 01:03:50,470 --> 01:03:55,900 So now let me denote uppercase L sub j 887 01:03:55,900 --> 01:04:02,380 to be 1 over lambda jd rounded up. 888 01:04:07,510 --> 01:04:16,200 And I claim that if little l is less than big L-- 889 01:04:16,200 --> 01:04:26,710 lj, Lj-- then little lj-- 890 01:04:26,710 --> 01:04:33,070 OK, so if I dilate the bj vector by factor little l, 891 01:04:33,070 --> 01:04:38,980 so if I plug it in and just look at these two inequalities, 892 01:04:38,980 --> 01:04:42,280 I obtain an upper bound of epsilon 893 01:04:42,280 --> 01:04:50,345 over d on the L-infinity norm of lj bj, 894 01:04:50,345 --> 01:04:55,100 so just looking at this bound here and the size of lj. 895 01:04:58,650 --> 01:05:14,020 And if this holds for all j, then summing up 896 01:05:14,020 --> 01:05:16,240 all of these individual inequalities, 897 01:05:16,240 --> 01:05:28,770 we find that the sum of these lj bj's is at most 898 01:05:28,770 --> 01:05:30,700 epsilon in L-infinity norm. 899 01:05:47,040 --> 01:05:49,380 So the point here is that we want 900 01:05:49,380 --> 01:05:52,980 to find the GAP in this Bohr set. 901 01:05:52,980 --> 01:05:56,550 And how does one think of a Bohr set? 902 01:05:56,550 --> 01:05:59,730 So it's kind of hard to imagine, because the Bohr set 903 01:05:59,730 --> 01:06:01,890 is a subset of Z mod n. 904 01:06:01,890 --> 01:06:04,140 But the right way to think about a Bohr set 905 01:06:04,140 --> 01:06:09,810 is in a higher dimensional lift, because a Bohr set is defined 906 01:06:09,810 --> 01:06:18,220 by looking at these numbers for R different values, 907 01:06:18,220 --> 01:06:19,510 R different coordinates. 908 01:06:19,510 --> 01:06:24,540 So we think of each r as its own coordinate. 909 01:06:24,540 --> 01:06:28,630 So we think of there being capital uppercase 910 01:06:28,630 --> 01:06:31,690 R many coordinates. 911 01:06:31,690 --> 01:06:36,460 And we want to consider the set of x's 912 01:06:36,460 --> 01:06:41,510 so that all the coordinate values are small. 913 01:06:41,510 --> 01:06:43,800 So instead of considering a one-dimensional picture, 914 01:06:43,800 --> 01:06:46,008 as we do in the Bohr sets, we're considering a higher 915 01:06:46,008 --> 01:06:48,030 dimensional or d-dimensional picture 916 01:06:48,030 --> 01:06:50,760 and then eventually projecting what happens up 917 01:06:50,760 --> 01:06:53,125 there down to this Bohr set. 918 01:06:53,125 --> 01:06:54,750 So what does Minkowski's second theorem 919 01:06:54,750 --> 01:06:56,310 have to do with anything? 920 01:06:56,310 --> 01:07:02,200 Well, once you have this higher dimensional lattice, what we're 921 01:07:02,200 --> 01:07:09,120 going to do is find a large lattice parallelepiped, 922 01:07:09,120 --> 01:07:12,360 so a large structure inside this higher dimensional 923 01:07:12,360 --> 01:07:15,150 lattice, and then project it down 924 01:07:15,150 --> 01:07:19,370 onto one-dimensional Z mod n. 925 01:07:19,370 --> 01:07:21,340 So this is the process of-- 926 01:07:21,340 --> 01:07:26,000 so you already see some aspects of a GAP in here. 927 01:07:26,000 --> 01:07:28,730 So these guys, they're essentially 928 01:07:28,730 --> 01:07:33,960 the GAP that we're going to eventually wish to find. 929 01:07:33,960 --> 01:07:37,460 And right now, they live in this higher dimensional lattice. 930 01:07:37,460 --> 01:07:39,870 But we're going to pull them down to Z mod n. 931 01:07:44,480 --> 01:07:46,280 All right. 932 01:07:46,280 --> 01:07:50,530 Now, where do these b's come from? 933 01:07:50,530 --> 01:08:08,950 So each b sub j is congruent to some x sub j times this vector 934 01:08:08,950 --> 01:08:20,200 mod 1 where x sub j is an integer 935 01:08:20,200 --> 01:08:27,370 between 0 and uppercase N. 936 01:08:27,370 --> 01:08:34,960 So this inequality up here, star. 937 01:08:34,960 --> 01:08:41,859 So the i-th coordinate for star-- 938 01:08:41,859 --> 01:08:44,189 "coordinate" meaning this is an L-infinity bound, 939 01:08:44,189 --> 01:08:47,229 so the i-th coordinate is upper bounded by epsilon. 940 01:08:47,229 --> 01:08:50,340 But the i-th coordinate bound implies 941 01:08:50,340 --> 01:09:04,840 that if you look at this sum over here times Ri divided 942 01:09:04,840 --> 01:09:12,220 by N, this quantity, whatever it is, 943 01:09:12,220 --> 01:09:20,810 is very close to an integer for each i. 944 01:09:20,810 --> 01:09:24,100 So the i-th coordinates implies this inequality, 945 01:09:24,100 --> 01:09:25,354 and it's true for every i. 946 01:09:28,600 --> 01:09:36,510 Thus what we find is that the GAP, which you 947 01:09:36,510 --> 01:09:38,600 already see in this formula over here-- 948 01:09:38,600 --> 01:09:45,322 so the GAP is given like that. 949 01:09:50,640 --> 01:09:54,658 So this GAP is contained in the Bohr set. 950 01:10:04,130 --> 01:10:10,390 So we found a large structure in the lattice. 951 01:10:10,390 --> 01:10:13,340 But the lattice came from this construction, 952 01:10:13,340 --> 01:10:16,240 which was directly motivated by the Bohr set. 953 01:10:16,240 --> 01:10:18,990 So we find a large GAP in the Bohr set. 954 01:10:18,990 --> 01:10:24,100 Well, we haven't shown yet it is large or that it is proper. 955 01:10:24,100 --> 01:10:25,920 So we need to check those two things. 956 01:10:40,960 --> 01:10:45,460 To check that this GAP that we found is large, 957 01:10:45,460 --> 01:10:47,860 we're going to apply Minkowski's second theorem. 958 01:10:53,950 --> 01:10:58,215 Let's check GAP is large. 959 01:11:01,440 --> 01:11:11,000 So by Minkowski's second theorem, 960 01:11:11,000 --> 01:11:15,710 we find that the size of the GAP, which is, by definition, 961 01:11:15,710 --> 01:11:19,150 the product of these upper case L's-- 962 01:11:19,150 --> 01:11:22,090 so if you look at how the uppercase L's is defined, 963 01:11:22,090 --> 01:11:24,640 you see that this quantity is at least 1 964 01:11:24,640 --> 01:11:30,230 over the product of the successive minima times 965 01:11:30,230 --> 01:11:33,030 denominator d to the d. 966 01:11:33,030 --> 01:11:36,530 And now we apply Minkowski's second. 967 01:11:36,530 --> 01:11:42,250 And we find that this quantity is at least 968 01:11:42,250 --> 01:11:47,800 the volume of k divided by 2 to the d 969 01:11:47,800 --> 01:11:55,010 times the determinant of the lattice times d to the d. 970 01:11:55,010 --> 01:11:59,450 But we saw what is the determinant of the lattice. 971 01:11:59,450 --> 01:12:06,210 It is 1 over N. You have d to the d, 2 to the d. 972 01:12:06,210 --> 01:12:09,320 And the volume of k, well, k is just that box. 973 01:12:09,320 --> 01:12:13,830 So the volume of k is 2 epsilon raised to d. 974 01:12:13,830 --> 01:12:16,470 So putting everything together, we 975 01:12:16,470 --> 01:12:23,940 find that the size of this GAP is the claimed quantity. 976 01:12:23,940 --> 01:12:26,352 It's a constant fraction of the entire group. 977 01:12:29,672 --> 01:12:31,880 The second thing that we need to check is properness. 978 01:12:38,090 --> 01:12:40,510 So what does it mean to be proper? 979 01:12:40,510 --> 01:12:42,940 So we just want to know that you don't 980 01:12:42,940 --> 01:12:45,250 have two different ways of representing 981 01:12:45,250 --> 01:12:48,490 the same term in the GAP. 982 01:12:48,490 --> 01:12:58,480 So if I have the following congruence, 983 01:12:58,480 --> 01:13:07,180 so if this combination of the x's is 984 01:13:07,180 --> 01:13:21,700 congruent to a different combination of the x's where 985 01:13:21,700 --> 01:13:28,240 these little l's are between 1 and-- 986 01:13:28,240 --> 01:13:31,190 OK, so I want to show-- so to check that it's proper-- 987 01:13:31,190 --> 01:13:32,780 so we're in Z mod n-- 988 01:13:32,780 --> 01:13:35,360 we just need to check that if this holds, 989 01:13:35,360 --> 01:13:38,690 then all the corresponding little l's must 990 01:13:38,690 --> 01:13:42,530 be the same as their primes. 991 01:13:42,530 --> 01:13:49,240 Well, if it is true, then setting-- 992 01:13:49,240 --> 01:13:50,830 let's go back to the lattice-- 993 01:13:50,830 --> 01:13:59,720 setting the vector b to be a vector originally 994 01:13:59,720 --> 01:14:02,400 that corresponds to the difference of these two 995 01:14:02,400 --> 01:14:02,900 numbers-- 996 01:14:14,800 --> 01:14:17,580 so if we set b to be the difference of these two 997 01:14:17,580 --> 01:14:27,250 numbers, we find that, first of all, it lies in Z to the d, 998 01:14:27,250 --> 01:14:32,220 because these two numbers are congruent to each other mod n. 999 01:14:41,220 --> 01:14:56,140 And furthermore, the L-infinity norm of b is upper bounded by-- 1000 01:14:58,780 --> 01:15:03,550 I mean, each one of them has small l-infinity norm. 1001 01:15:03,550 --> 01:15:07,200 And this is some number that is bounded. 1002 01:15:07,200 --> 01:15:11,000 It's less than uppercase L. 1003 01:15:11,000 --> 01:15:15,730 So the whole thing, this whole sum, the L-infinity norm, 1004 01:15:15,730 --> 01:15:21,460 cannot be larger than this quantity over here, 1005 01:15:21,460 --> 01:15:26,190 where I essentially use the triangle inequality to analyze 1006 01:15:26,190 --> 01:15:27,500 this b term by term. 1007 01:15:30,680 --> 01:15:33,380 All of these numbers are very small, 1008 01:15:33,380 --> 01:15:42,520 because if you look at what we saw up there, so the size of b, 1009 01:15:42,520 --> 01:15:47,230 we see that this whole thing is at most epsilon. 1010 01:15:50,020 --> 01:15:53,370 And epsilon is strictly less than 1. 1011 01:15:53,370 --> 01:16:01,110 So you have some vector b, which is an integer vector, 1012 01:16:01,110 --> 01:16:10,400 such that all of its coordinates have L-infinity norms strictly 1013 01:16:10,400 --> 01:16:12,630 less than 1. 1014 01:16:12,630 --> 01:16:14,920 So that means that b is equal to 0. 1015 01:16:22,280 --> 01:16:23,590 So b is the 0 vector. 1016 01:16:27,292 --> 01:16:28,890 So b is a zero vector. 1017 01:16:34,180 --> 01:16:39,920 Thus this thing here equals to 0. 1018 01:16:39,920 --> 01:16:42,760 So this sum here equals to 0. 1019 01:16:42,760 --> 01:16:50,650 And since the bi's form a basis, we 1020 01:16:50,650 --> 01:16:55,480 find that the li's and l prime i's 1021 01:16:55,480 --> 01:16:58,371 are equal to each other for all i. 1022 01:17:01,670 --> 01:17:04,020 And this checks the properness of this GAP. 1023 01:17:08,520 --> 01:17:09,020 Yeah. 1024 01:17:09,020 --> 01:17:10,740 So this argument, it's not hard. 1025 01:17:10,740 --> 01:17:13,350 But you need to check the details. 1026 01:17:13,350 --> 01:17:17,640 So you need to wrap your mind around changing 1027 01:17:17,640 --> 01:17:20,280 from working in a higher dimensional lattice setting 1028 01:17:20,280 --> 01:17:23,520 to going back down to Z mod n. 1029 01:17:23,520 --> 01:17:26,340 And the main takeaway here is that the right way 1030 01:17:26,340 --> 01:17:29,780 to think about a Bohr set is to not stay in Z mod n 1031 01:17:29,780 --> 01:17:32,970 but to think about what happens in d-dimensional space 1032 01:17:32,970 --> 01:17:35,070 where d is the dimension of the Bohr set. 1033 01:17:39,520 --> 01:17:40,020 OK. 1034 01:17:40,020 --> 01:17:42,240 So now we have pretty much all the ingredients 1035 01:17:42,240 --> 01:17:45,240 that we need to prove Freiman's theorem. 1036 01:17:45,240 --> 01:17:48,190 And that's what we'll do at the beginning of next lecture. 1037 01:17:48,190 --> 01:17:50,550 We'll conclude the proof of Freiman's theorem. 1038 01:17:50,550 --> 01:17:53,505 And then I'll tell you also about an important conjecture 1039 01:17:53,505 --> 01:17:56,490 in additive combinatorics called a polynomial Freiman-Ruzsa 1040 01:17:56,490 --> 01:17:59,100 conjecture, which many people think 1041 01:17:59,100 --> 01:18:02,730 is the most important open conjecture 1042 01:18:02,730 --> 01:18:05,660 in additive combinatorics.