1 00:00:17,963 --> 00:00:19,380 YUFEI ZHAO: All right, today we're 2 00:00:19,380 --> 00:00:22,850 going to start a new topic an additive combinatorics. 3 00:00:22,850 --> 00:00:25,260 And this is a fairly central topic 4 00:00:25,260 --> 00:00:28,740 having to do with the structure of set addition. 5 00:00:43,480 --> 00:00:44,980 So the main players that we're going 6 00:00:44,980 --> 00:00:49,810 to be seeing in this chapter have to do with, 7 00:00:49,810 --> 00:00:56,410 if you start with a subset of some obedient group 8 00:00:56,410 --> 00:00:59,980 under addition-- 9 00:00:59,980 --> 00:01:01,150 not necessarily finite. 10 00:01:01,150 --> 00:01:03,580 So the obedient group that I'm going 11 00:01:03,580 --> 00:01:06,580 to keep in mind, the ones that will come up generally 12 00:01:06,580 --> 00:01:11,410 are integers, z mod n or the finite field model. 13 00:01:15,260 --> 00:01:20,750 We're going to be looking at objects such as a sum set, so 14 00:01:20,750 --> 00:01:26,000 a plus b, meaning the set of elements that can be written 15 00:01:26,000 --> 00:01:29,180 as a sum, where you take one element 16 00:01:29,180 --> 00:01:34,510 from a and another from b. 17 00:01:34,510 --> 00:01:39,340 Likewise, you can also have a minus b defined similarly, 18 00:01:39,340 --> 00:01:41,080 now taking a minus b. 19 00:01:45,550 --> 00:01:48,430 We can iterate this operation. 20 00:01:48,430 --> 00:01:54,760 So kA, so 2A, 3A, 4A, for instance, 21 00:01:54,760 --> 00:02:05,520 means I add A to itself k times, not 22 00:02:05,520 --> 00:02:11,330 to be confused with a dilation, which we'll denote by k dot A. 23 00:02:11,330 --> 00:02:16,370 So this is notation for multiplying every element of A 24 00:02:16,370 --> 00:02:17,540 by the number k. 25 00:02:24,360 --> 00:02:28,120 So given a subset of integers I can do these operations 26 00:02:28,120 --> 00:02:29,200 to the set. 27 00:02:29,200 --> 00:02:32,230 And I want to ask, how does the size of the set 28 00:02:32,230 --> 00:02:37,750 change when I do these operations? 29 00:02:37,750 --> 00:02:43,120 For example, what is the largest or the smallest? 30 00:02:43,120 --> 00:02:59,590 So how large or small can A plus A be for a given set size, A? 31 00:02:59,590 --> 00:03:04,070 So if I allow you to use 10 elements, 32 00:03:04,070 --> 00:03:07,290 how can you make A plus A as big as possible? 33 00:03:07,290 --> 00:03:10,160 And how can you make it as small as possible? 34 00:03:10,160 --> 00:03:12,500 So this is not a hard question. 35 00:03:12,500 --> 00:03:14,280 How can you make it as big as possible? 36 00:03:18,700 --> 00:03:21,080 So what's the maximum size A plus A 37 00:03:21,080 --> 00:03:24,430 can be as a function of A? 38 00:03:24,430 --> 00:03:27,120 Well, I'm looking at pairwise sums, 39 00:03:27,120 --> 00:03:30,020 so if there are no collisions between different pairwise 40 00:03:30,020 --> 00:03:32,720 sums, this is as large as possible. 41 00:03:32,720 --> 00:03:36,200 And then it's not hard to see that the maximum possible is 42 00:03:36,200 --> 00:03:38,790 the size of A plus 1, choose 2. 43 00:03:41,370 --> 00:03:52,200 So since at most, this many pairs 44 00:03:52,200 --> 00:03:56,475 and space possible if all sums are distinct. 45 00:04:00,610 --> 00:04:07,280 So for example, in integers, you can take 1, 2, 2 squared, 46 00:04:07,280 --> 00:04:07,780 and so on. 47 00:04:11,126 --> 00:04:15,040 So that will give you the span. 48 00:04:15,040 --> 00:04:17,480 The minimum possible is also not too hard. 49 00:04:20,910 --> 00:04:23,230 We're allowed to work in a general obedient group. 50 00:04:23,230 --> 00:04:29,770 So in that case, the minimum could be just the size of A. 51 00:04:29,770 --> 00:04:32,860 The size is always at least the size of A. 52 00:04:32,860 --> 00:04:39,080 And this is tight if A is a subgroup. 53 00:04:39,080 --> 00:04:42,780 If you have a subgroup, then it's closed under addition. 54 00:04:42,780 --> 00:04:48,030 So the set does not expand under addition. 55 00:04:48,030 --> 00:04:52,520 In the integers, you don't have any finite subgroups. 56 00:04:52,520 --> 00:04:55,940 So if I give you k integers, what's the smallest, 57 00:04:55,940 --> 00:04:57,564 the sum set can be? 58 00:04:57,564 --> 00:04:58,670 AUDIENCE: 2k minus 1. 59 00:04:58,670 --> 00:04:59,920 YUFEI ZHAO: 2k minus 1, right? 60 00:04:59,920 --> 00:05:02,628 So the example is when you have an arithmetic progression. 61 00:05:02,628 --> 00:05:07,750 So in integers, the minimum is 2k minus 1. 62 00:05:10,440 --> 00:05:18,360 And it's achieved for an arithmetic progression. 63 00:05:18,360 --> 00:05:21,810 So let me just give you the one-line proof why you always 64 00:05:21,810 --> 00:05:24,000 have at least this many elements, 65 00:05:24,000 --> 00:05:31,530 is if A has elements sorted like this, 66 00:05:31,530 --> 00:05:40,020 then the following elements are distinct in the sum set. 67 00:05:40,020 --> 00:05:41,940 So you start with A plus A. And then 68 00:05:41,940 --> 00:05:47,850 you move A1 plus A2, A1 plus A3, and so on, to A1 plus Ak. 69 00:05:47,850 --> 00:05:49,815 And then you move the first element forward. 70 00:05:56,310 --> 00:06:04,810 OK, so here you already see 2k minus 1 distinct elements 71 00:06:04,810 --> 00:06:10,420 in A plus A. 72 00:06:10,420 --> 00:06:13,060 OK, so these are fairly simple examples, 73 00:06:13,060 --> 00:06:14,350 fairly simple questions. 74 00:06:14,350 --> 00:06:18,280 So now let's get to some more interesting questions, which 75 00:06:18,280 --> 00:06:22,840 is, what can you say about a set if you know 76 00:06:22,840 --> 00:06:25,390 that it has small doubling? 77 00:06:25,390 --> 00:06:28,030 If it doesn't expand by very much, 78 00:06:28,030 --> 00:06:30,510 what can you tell me about the set? 79 00:06:30,510 --> 00:06:33,330 And for that, let me define the notion of a doubling constant. 80 00:06:38,960 --> 00:06:43,670 So the doubling constant of A is defined 81 00:06:43,670 --> 00:06:45,980 to be the number which we often denote 82 00:06:45,980 --> 00:06:54,730 by k, the number obtained by dividing the size of A plus A 83 00:06:54,730 --> 00:06:59,075 by the size of A. And we would like to understand-- 84 00:06:59,075 --> 00:07:01,340 and this is the main question that's 85 00:07:01,340 --> 00:07:07,160 addressed in the upcoming lectures is, 86 00:07:07,160 --> 00:07:22,405 what is the structure of a set with bounded doubling constant? 87 00:07:29,960 --> 00:07:33,070 So for instance, think of k as fixed. 88 00:07:35,960 --> 00:07:37,400 Let's say k is 100. 89 00:07:37,400 --> 00:07:41,730 If you know a set has doubling constant, at most, 100, 90 00:07:41,730 --> 00:07:45,570 what can you tell me about the structure of the set? 91 00:07:45,570 --> 00:07:47,310 So that's the main question. 92 00:07:47,310 --> 00:07:50,370 Let me show you in a second a few examples of sets 93 00:07:50,370 --> 00:07:53,600 the have bounded doubling constant. 94 00:07:53,600 --> 00:07:56,280 So that's easy to check that those examples indeed 95 00:07:56,280 --> 00:07:58,570 have bounded doubling constant. 96 00:07:58,570 --> 00:08:00,330 And what this question amounts to 97 00:08:00,330 --> 00:08:04,250 is what is often known as an inverse question. 98 00:08:04,250 --> 00:08:10,320 So it's an inverse problem that asks 99 00:08:10,320 --> 00:08:13,070 you to describe in reverse-- 100 00:08:13,070 --> 00:08:15,750 so it's easy to check in the upcoming examples 101 00:08:15,750 --> 00:08:20,600 that all of those examples have bounded doubling constant. 102 00:08:20,600 --> 00:08:22,500 And what we want to say is, in reverse, 103 00:08:22,500 --> 00:08:25,740 that if a set has bounded doubling constant, 104 00:08:25,740 --> 00:08:31,160 then it must in some sense look like one of our examples. 105 00:08:31,160 --> 00:08:32,820 It's the harder inverse question. 106 00:08:35,530 --> 00:08:37,929 OK, so let me give you some examples of sets 107 00:08:37,929 --> 00:08:39,744 with small doubling constant. 108 00:08:49,650 --> 00:08:51,900 One example we already saw earlier 109 00:08:51,900 --> 00:08:56,600 is that if you have an arithmetic progression. 110 00:08:56,600 --> 00:08:58,400 If you have an arithmetic progression, 111 00:08:58,400 --> 00:09:01,430 then the size of A plus A is always 112 00:09:01,430 --> 00:09:06,085 2 times the size of A minus 1. 113 00:09:06,085 --> 00:09:11,340 So the doubling constantly is always, at most, 2. 114 00:09:11,340 --> 00:09:12,460 That's pretty small. 115 00:09:12,460 --> 00:09:16,580 That's as small as you can get in arithmetic progressions is 116 00:09:16,580 --> 00:09:17,330 in the integers. 117 00:09:20,460 --> 00:09:23,370 But if you start with an arithmetic progression 118 00:09:23,370 --> 00:09:27,810 and now I take just a subset of the elements 119 00:09:27,810 --> 00:09:33,580 of this progression, so if I take AP, 120 00:09:33,580 --> 00:09:38,650 and if I cross out a few elements, just a small number 121 00:09:38,650 --> 00:09:41,350 of elements from this progression, or even 122 00:09:41,350 --> 00:09:44,320 cross out most, but keeping a constant fraction 123 00:09:44,320 --> 00:09:48,850 of elements still remaining, I claim 124 00:09:48,850 --> 00:09:52,200 that's still a pretty good set. 125 00:09:52,200 --> 00:10:01,290 So if A can be embedded inside an AP 126 00:10:01,290 --> 00:10:06,690 such that the AP has size no more a constant factor 127 00:10:06,690 --> 00:10:15,880 and more than that of A, then the size of A plus A is, 128 00:10:15,880 --> 00:10:16,700 at most-- 129 00:10:16,700 --> 00:10:22,240 so we bound it by the size of P plus P, which is, at most, 2P. 130 00:10:25,160 --> 00:10:30,250 So the doubling constant of A is, at most, 2C. 131 00:10:30,250 --> 00:10:35,960 So if you have a set which is at least 1/10 fraction of an AP, 132 00:10:35,960 --> 00:10:40,630 then you are doubling constant at most, 20, bounded. 133 00:10:40,630 --> 00:10:42,920 So this is another class of examples. 134 00:10:42,920 --> 00:10:45,950 So it's kind of a modification, some alteration 135 00:10:45,950 --> 00:10:48,827 of the arithmetic progression. 136 00:10:52,170 --> 00:10:56,220 Another more substantial generalization of APs 137 00:10:56,220 --> 00:11:01,790 is that of a two-dimensional arithmetic progression. 138 00:11:01,790 --> 00:11:03,600 So you think of an arithmetic progression 139 00:11:03,600 --> 00:11:08,420 as equally spaced points on a line. 140 00:11:08,420 --> 00:11:16,600 But you can extend this in multiple dimensions, 141 00:11:16,600 --> 00:11:18,710 so like a grid. 142 00:11:18,710 --> 00:11:21,560 So this is a two-dimensional arithmetic progression, 143 00:11:21,560 --> 00:11:24,330 but I still want to work inside the integers. 144 00:11:24,330 --> 00:11:30,080 So what we are going to do is project this picture 145 00:11:30,080 --> 00:11:31,025 onto the integers. 146 00:11:33,880 --> 00:11:36,700 So that's a two-dimensional arithmetic progression. 147 00:11:36,700 --> 00:11:45,240 And specifically, we have a set of the form, 148 00:11:45,240 --> 00:11:52,300 so x0 is the starting point, plus l1 of x1-- 149 00:11:52,300 --> 00:11:57,830 l1 times x1, and l2 2 times x2, where 150 00:11:57,830 --> 00:12:04,070 the little l's are integers, non-negative 151 00:12:04,070 --> 00:12:19,140 integers up to big L. 152 00:12:19,140 --> 00:12:23,110 So that's a two-dimensional arithmetic progression. 153 00:12:23,110 --> 00:12:28,740 So the picture that you can have in mind is, on the number line, 154 00:12:28,740 --> 00:12:35,470 we can get, write down first an AP 155 00:12:35,470 --> 00:12:40,490 and then a few more points like that so 156 00:12:40,490 --> 00:12:42,980 that you can have a two-dimensional arithmetic 157 00:12:42,980 --> 00:12:43,911 progression. 158 00:12:48,630 --> 00:12:53,910 We say that this set, this two-dimensional arithmetic 159 00:12:53,910 --> 00:13:05,635 progression is proper if all terms are distinct. 160 00:13:12,570 --> 00:13:17,356 And if that's the case, then I can write A plus A 161 00:13:17,356 --> 00:13:19,290 in a very similar format. 162 00:13:19,290 --> 00:13:26,780 So A plus A contains elements still of the same form, 163 00:13:26,780 --> 00:13:32,420 but now the indices go up to 2L minus 1. 164 00:13:39,120 --> 00:13:43,130 So you see that A plus A has size, at most, 4 times 165 00:13:43,130 --> 00:13:47,600 the original set, A. Also easy to see from this blue picture 166 00:13:47,600 --> 00:13:48,360 up there-- 167 00:13:48,360 --> 00:13:49,880 you expand that grid. 168 00:13:49,880 --> 00:13:52,300 It goes to, at most, 4 times the size. 169 00:13:52,300 --> 00:13:53,055 Yes? 170 00:13:53,055 --> 00:13:56,520 AUDIENCE: [INAUDIBLE] 171 00:13:56,520 --> 00:13:58,840 YUFEI ZHAO: So the question is, should it be? 172 00:13:58,840 --> 00:14:04,430 AUDIENCE: [INAUDIBLE] 173 00:14:04,430 --> 00:14:05,600 YUFEI ZHAO: 2x0? 174 00:14:05,600 --> 00:14:08,480 AUDIENCE: [INAUDIBLE] 175 00:14:08,480 --> 00:14:09,710 YUFEI ZHAO: What do you mean? 176 00:14:09,710 --> 00:14:11,485 AUDIENCE: 2x0 plus l1 x0 plus 1? 177 00:14:11,485 --> 00:14:13,360 YUFEI ZHAO: Ah, thank you, so 2x0, thank you. 178 00:14:13,360 --> 00:14:15,220 Yeah, 2x0, great. 179 00:14:18,070 --> 00:14:21,400 OK, so that's the size. 180 00:14:21,400 --> 00:14:24,130 And of course, you can generalize this example 181 00:14:24,130 --> 00:14:27,220 of a fairly straightforward way to d dimensional arithmetic 182 00:14:27,220 --> 00:14:28,150 progressions. 183 00:14:28,150 --> 00:14:30,010 And we call those things generalized 184 00:14:30,010 --> 00:14:33,200 arithmetic progressions. 185 00:14:33,200 --> 00:14:46,270 So a Generalized Arithmetic Progression, 186 00:14:46,270 --> 00:14:50,250 which we will abbreviate by the letters GAP, 187 00:14:50,250 --> 00:14:56,440 is a set of numbers of the form as above, 188 00:14:56,440 --> 00:15:09,720 except now you have d different directions and indices, 189 00:15:09,720 --> 00:15:12,270 are also straightforward generalizations 190 00:15:12,270 --> 00:15:14,440 of what was earlier. 191 00:15:14,440 --> 00:15:17,340 So this is the notion of a generalized arithmetic 192 00:15:17,340 --> 00:15:18,070 progression. 193 00:15:18,070 --> 00:15:20,220 So think about projection of a d dimensional grid 194 00:15:20,220 --> 00:15:20,970 onto the integers. 195 00:15:23,640 --> 00:15:30,650 And for GAPs, we say that it's proper 196 00:15:30,650 --> 00:15:32,015 if all the terms are distinct. 197 00:15:38,940 --> 00:15:47,310 We call d the dimension of the GAP. 198 00:15:47,310 --> 00:15:51,150 And for a GAP, whether it's proper or not, 199 00:15:51,150 --> 00:15:56,925 we call the size to be the product of the lengths. 200 00:16:01,550 --> 00:16:05,290 And this is potentially larger. 201 00:16:05,290 --> 00:16:12,220 So this is larger than the number of distinct elements 202 00:16:12,220 --> 00:16:15,790 if it's not proper. 203 00:16:15,790 --> 00:16:18,040 So when I refer to the size of a GAP-- 204 00:16:18,040 --> 00:16:20,260 so I view the GAP more than just as a set, 205 00:16:20,260 --> 00:16:22,570 but also with the data of the initial point 206 00:16:22,570 --> 00:16:24,050 and the directions. 207 00:16:24,050 --> 00:16:25,570 If I talk about the size, I'm always 208 00:16:25,570 --> 00:16:27,630 referring to this quantity over here. 209 00:16:32,360 --> 00:16:33,790 Great. 210 00:16:33,790 --> 00:16:39,910 So you see, if you take a GAP or a fraction of a GAP, then, 211 00:16:39,910 --> 00:16:43,420 as with earlier examples, you have small doubling. 212 00:16:45,930 --> 00:16:59,310 So if P is a proper GAP, of dimension d, 213 00:16:59,310 --> 00:17:05,370 then P plus P is, at most, 2 raised 214 00:17:05,370 --> 00:17:11,400 to power d times the size of P. And furthermore, 215 00:17:11,400 --> 00:17:22,319 if A is an arbitrary subset of P and such that A has size-- 216 00:17:22,319 --> 00:17:25,400 such that the GAP has size, at most, 217 00:17:25,400 --> 00:17:28,250 a constant fraction bigger than that of A, 218 00:17:28,250 --> 00:17:32,430 then A has small doubling as well. 219 00:17:45,120 --> 00:17:47,600 So all of these are examples of constructions 220 00:17:47,600 --> 00:17:51,230 of sets where, for some fixed constant, the doubling 221 00:17:51,230 --> 00:17:56,900 constant, we can find a family of sets 222 00:17:56,900 --> 00:17:59,760 with doubling constant bounded by that number. 223 00:18:02,646 --> 00:18:07,150 And the natural question though is, are these all the examples? 224 00:18:07,150 --> 00:18:10,810 So have we missed some important family 225 00:18:10,810 --> 00:18:14,700 of constructions not covered by any of these examples? 226 00:18:14,700 --> 00:18:16,450 And so that's the kind of inverse question 227 00:18:16,450 --> 00:18:17,717 I was referring to earlier. 228 00:18:17,717 --> 00:18:20,050 So all of these examples, easy to check that they indeed 229 00:18:20,050 --> 00:18:21,258 have small doubling constant. 230 00:18:23,680 --> 00:18:24,990 Can you go in reverse? 231 00:18:24,990 --> 00:18:27,880 So can you ask the inverse question, 232 00:18:27,880 --> 00:18:30,100 if a set has small doubling constant, 233 00:18:30,100 --> 00:18:34,610 must it look like, in some sense, one of these sets? 234 00:18:34,610 --> 00:18:38,770 It turns out this is not such an easy problem. 235 00:18:38,770 --> 00:18:45,130 And there is a central result in additive combinatorics known 236 00:18:45,130 --> 00:18:51,310 as Freiman's theorem which gives a positive answer 237 00:18:51,310 --> 00:18:53,850 to that question. 238 00:18:53,850 --> 00:18:58,300 So Freiman's theorem is now considered a central result 239 00:18:58,300 --> 00:19:00,040 in additive combinatorics. 240 00:19:00,040 --> 00:19:02,230 And it completely describes, in some sense, 241 00:19:02,230 --> 00:19:05,380 the sets that have small doubling. 242 00:19:05,380 --> 00:19:07,730 And let me write down the statement. 243 00:19:07,730 --> 00:19:20,500 So if A is a subset of Z and has bounded doubling, 244 00:19:20,500 --> 00:19:40,310 then A is contained in a GAP of bounded dimension and size 245 00:19:40,310 --> 00:19:46,160 bounded by some constant times the size of the set. 246 00:19:52,710 --> 00:19:54,085 This is a really important result 247 00:19:54,085 --> 00:19:55,730 in additive combinatorics. 248 00:19:59,170 --> 00:20:04,480 The title of this chapter, "Structure of Set Addition," 249 00:20:04,480 --> 00:20:06,970 Freiman's theorem tells us something about the structure 250 00:20:06,970 --> 00:20:09,220 of a set with small doubling. 251 00:20:11,740 --> 00:20:14,020 The next few lecturers are going to be occupied 252 00:20:14,020 --> 00:20:15,550 with proving this theorem. 253 00:20:15,550 --> 00:20:18,090 So this theorem will have-- 254 00:20:18,090 --> 00:20:20,290 its proof is involved and probably the most involved 255 00:20:20,290 --> 00:20:22,330 proof that we have in this course. 256 00:20:22,330 --> 00:20:25,570 And the proof will take the next several lectures. 257 00:20:25,570 --> 00:20:27,610 And we'll see a lot of different ingredients, 258 00:20:27,610 --> 00:20:29,377 a lot of really nice tools. 259 00:20:29,377 --> 00:20:31,210 Fourier analysis will come up at some point, 260 00:20:31,210 --> 00:20:34,630 but also other tools like the geometry of numbers and also 261 00:20:34,630 --> 00:20:37,935 some more classical additive combinatorics ideas. 262 00:20:37,935 --> 00:20:39,310 But before starting on a proof, I 263 00:20:39,310 --> 00:20:42,940 want to offer a few remarks and historical remarks to just 264 00:20:42,940 --> 00:20:46,280 give you some more context about Freiman's theorem, 265 00:20:46,280 --> 00:20:47,905 but first, a few mathematical comments. 266 00:20:50,980 --> 00:20:53,290 In this conclusions of Freiman's theorem, 267 00:20:53,290 --> 00:20:55,150 I didn't mention properness. 268 00:20:55,150 --> 00:20:57,220 And that's mostly a matter of convenience. 269 00:20:57,220 --> 00:21:01,150 So you can, in fact, make the conclusion proper 270 00:21:01,150 --> 00:21:05,200 as well at the cost of increasing the number somewhat, 271 00:21:05,200 --> 00:21:08,110 but still constants depending only on k-- 272 00:21:11,890 --> 00:21:21,540 can guarantee properness as well. 273 00:21:26,130 --> 00:21:31,240 So there is an extra step involved which we'll not cover, 274 00:21:31,240 --> 00:21:32,670 because it's not entirely trivial, 275 00:21:32,670 --> 00:21:33,795 but it's also not too hard. 276 00:21:37,260 --> 00:21:41,160 Freiman's original proof, so it's named after Freiman. 277 00:21:41,160 --> 00:21:43,310 So he proved that in the '60s. 278 00:21:43,310 --> 00:21:46,320 But at that time, the proof was considered rather obscure. 279 00:21:46,320 --> 00:21:50,220 It actually did not get the attention and the recognition 280 00:21:50,220 --> 00:21:52,740 that it deserved until much later. 281 00:21:52,740 --> 00:21:55,530 So this was kind of a forgotten result, a forgotten proof 282 00:21:55,530 --> 00:22:00,700 for a very long time until quite a bit later when Ruzsa-- 283 00:22:00,700 --> 00:22:03,150 Ruzsa's name will come up many times in this chapter. 284 00:22:03,150 --> 00:22:06,290 Ruzsa came and gave a different proof of Freiman's theorem, 285 00:22:06,290 --> 00:22:07,915 and significantly cleaned up the proof, 286 00:22:07,915 --> 00:22:09,750 and offered many new ideas. 287 00:22:09,750 --> 00:22:12,390 So much of what we'll see today are results 288 00:22:12,390 --> 00:22:14,190 that we now attribute to Ruzsa. 289 00:22:14,190 --> 00:22:16,570 And theorem sometimes is also called 290 00:22:16,570 --> 00:22:17,930 the Freiman-Ruzsa theorem. 291 00:22:25,650 --> 00:22:27,740 But this result was really brought into-- 292 00:22:30,360 --> 00:22:33,520 brought as a highlight of additive combinatorics 293 00:22:33,520 --> 00:22:36,540 in the work of Gowers when he proved, 294 00:22:36,540 --> 00:22:40,000 that gave his new proof of Szemerédi's theorem, 295 00:22:40,000 --> 00:22:41,620 giving much better bounds. 296 00:22:41,620 --> 00:22:46,630 So he had to use quite a bit of serious additive combinatorics. 297 00:22:46,630 --> 00:22:50,410 And many of the ideas that went into Gowers' proof 298 00:22:50,410 --> 00:22:53,290 of Szemerédi's theorem came from this line of work, 299 00:22:53,290 --> 00:22:55,170 Freiman and Ruzsa. 300 00:22:55,170 --> 00:22:58,780 So and their work was, again, brought into prominence 301 00:22:58,780 --> 00:23:02,050 as a result of Gowers' Fields-Medal-winning work 302 00:23:02,050 --> 00:23:03,580 on Szemerédi's theorem. 303 00:23:06,590 --> 00:23:08,610 So this is some of the history. 304 00:23:08,610 --> 00:23:10,720 And now Freiman's theorem is considered 305 00:23:10,720 --> 00:23:13,118 a central result in the area. 306 00:23:13,118 --> 00:23:14,660 You can see, it's a beautiful result. 307 00:23:14,660 --> 00:23:17,730 And it's also quite a deep result. 308 00:23:17,730 --> 00:23:20,060 Let me mention a few things about bounds. 309 00:23:20,060 --> 00:23:25,380 So what do we know about this d of k and f of k? 310 00:23:25,380 --> 00:23:27,400 But first, an example-- 311 00:23:27,400 --> 00:23:32,460 so if the set A is dissociated in the sense of having 312 00:23:32,460 --> 00:23:40,460 no arithmetic structure, no coincidental sums 313 00:23:40,460 --> 00:23:44,960 colliding, so for example, if a of this form, 314 00:23:44,960 --> 00:23:47,360 then you see that-- 315 00:23:47,360 --> 00:23:51,800 also and we saw the size of A plus A, so A plus 1 choose 2. 316 00:23:51,800 --> 00:23:57,790 So in this case, the doubling constant 317 00:23:57,790 --> 00:24:03,385 is the size of A plus 1 divided by 2, so roughly 318 00:24:03,385 --> 00:24:07,260 on the same order as the size of A. 319 00:24:07,260 --> 00:24:11,010 But what do you need to take in Freiman's theorem 320 00:24:11,010 --> 00:24:13,820 for d and for f? 321 00:24:13,820 --> 00:24:17,140 So how can I embed this A in generalized arithmetic 322 00:24:17,140 --> 00:24:19,366 progression? 323 00:24:19,366 --> 00:24:23,990 See, there is not a great way to do it. 324 00:24:23,990 --> 00:24:26,005 So I want to keep the size small. 325 00:24:26,005 --> 00:24:27,660 There is not a great way to do it. 326 00:24:27,660 --> 00:24:31,700 So one way to do it is to use one direction 327 00:24:31,700 --> 00:24:35,170 for each of these elements. 328 00:24:35,170 --> 00:24:39,857 So contained in GAP-- 329 00:24:39,857 --> 00:24:41,440 now of course, there is always a trade 330 00:24:41,440 --> 00:24:43,447 off between dimension and size. 331 00:24:43,447 --> 00:24:44,530 But usually, not a great-- 332 00:24:44,530 --> 00:24:46,820 I mean, it's not such an important trade off. 333 00:24:46,820 --> 00:24:49,780 So but certainly it's contained in the GAP 334 00:24:49,780 --> 00:24:55,180 of dimension size of A minus 1 and size 335 00:24:55,180 --> 00:25:00,370 2 to the size of A minus 1, by thinking about A as a cube. 336 00:25:03,380 --> 00:25:06,330 And so you convince yourself that you basically cannot do 337 00:25:06,330 --> 00:25:06,830 much better. 338 00:25:10,400 --> 00:25:24,710 So the best possible bound that we can hope to prove 339 00:25:24,710 --> 00:25:33,820 is of the form d being, at most, linear in k, 340 00:25:33,820 --> 00:25:38,420 and f being, at most, exponential in k. 341 00:25:41,680 --> 00:25:43,510 So you see already, the bounds, that you 342 00:25:43,510 --> 00:25:45,130 have to lose some things. 343 00:25:45,130 --> 00:25:45,710 Yes? 344 00:25:45,710 --> 00:25:47,710 AUDIENCE: Why can't we just make the dimension 1 345 00:25:47,710 --> 00:25:50,295 and just let our arithmetic progression be 1 through 2 346 00:25:50,295 --> 00:25:51,647 to the sine of A minus 1? 347 00:25:51,647 --> 00:25:53,730 YUFEI ZHAO: OK, great, so that's a great question. 348 00:25:53,730 --> 00:25:56,600 So why can't we just make dimension 1 349 00:25:56,600 --> 00:25:59,000 and have the entire thing be as part 350 00:25:59,000 --> 00:26:01,170 of a single linear arithmetic progression? 351 00:26:01,170 --> 00:26:04,490 So you can do that, but then I can cook up other examples 352 00:26:04,490 --> 00:26:08,110 where I blow up this cube. 353 00:26:08,110 --> 00:26:10,080 So I ask you to think about how to do that. 354 00:26:10,080 --> 00:26:12,150 So you can try to blow up this cube 355 00:26:12,150 --> 00:26:14,960 so that you really do need the dimension 356 00:26:14,960 --> 00:26:21,492 to not be constant, so exercise. 357 00:26:21,492 --> 00:26:24,730 So the best result is not quite this claim. 358 00:26:24,730 --> 00:26:26,080 So this is still open. 359 00:26:26,080 --> 00:26:32,110 So the best result so far is due to Tom Sanders, whose name 360 00:26:32,110 --> 00:26:34,720 came up earlier, as he has basically the best bound 361 00:26:34,720 --> 00:26:35,752 on Roth's theorem. 362 00:26:35,752 --> 00:26:37,210 And you know, many of these results 363 00:26:37,210 --> 00:26:40,160 are all related to each other. 364 00:26:40,160 --> 00:26:51,860 So Sanders has-- so he showed that Freiman's theorem is 365 00:26:51,860 --> 00:27:00,090 true with d being, so basically k, 366 00:27:00,090 --> 00:27:04,600 but you lose a poly log factor. 367 00:27:04,600 --> 00:27:07,610 I think the big O is maybe 3, or 4, something like that, 368 00:27:07,610 --> 00:27:09,390 so not substantial. 369 00:27:09,390 --> 00:27:14,670 And then f of k is also basically exponential, 370 00:27:14,670 --> 00:27:16,880 but you lose a poly log factor in the exponent. 371 00:27:21,870 --> 00:27:25,020 Just a minor note about how to read this notation-- 372 00:27:25,020 --> 00:27:28,500 so I mean, it's written slightly sloppily as log k 373 00:27:28,500 --> 00:27:29,880 raised to big O of 1. 374 00:27:29,880 --> 00:27:33,750 You should think k as constant, but somewhat big, 375 00:27:33,750 --> 00:27:37,200 because if k were 2, this notation actually 376 00:27:37,200 --> 00:27:38,560 doesn't make sense. 377 00:27:38,560 --> 00:27:41,883 So just think of chaos, as at least 3 378 00:27:41,883 --> 00:27:43,050 when you read that notation. 379 00:27:45,750 --> 00:27:49,550 All right, so we will prove Freiman's theorem. 380 00:27:49,550 --> 00:27:51,580 So this bound will show a worse bound. 381 00:27:51,580 --> 00:27:53,737 It actually will be basically exponentially worse, 382 00:27:53,737 --> 00:27:54,820 but it will be a constant. 383 00:27:54,820 --> 00:27:57,370 So it will be just a function of k. 384 00:27:57,370 --> 00:28:00,800 And that will take us the next several lectures. 385 00:28:00,800 --> 00:28:05,260 So we'll begin by developing some tools that are, 386 00:28:05,260 --> 00:28:07,070 I think, of interest individually. 387 00:28:07,070 --> 00:28:09,308 And they can all be used for other things. 388 00:28:09,308 --> 00:28:10,850 So we'll develop some tools that will 389 00:28:10,850 --> 00:28:15,380 help us to show, eventually lead us to Freiman's theorem. 390 00:28:15,380 --> 00:28:18,020 And I'll try to structure this proof in such a way 391 00:28:18,020 --> 00:28:21,110 that there are several goal posts that 392 00:28:21,110 --> 00:28:22,580 are also interesting. 393 00:28:22,580 --> 00:28:25,960 So in particular, just as what we did with Roth's theorem, 394 00:28:25,960 --> 00:28:28,640 we'll begin by proving a finite field 395 00:28:28,640 --> 00:28:32,270 analog of Freiman's theorem. 396 00:28:32,270 --> 00:28:36,860 So what would that mean, a finite field analog? 397 00:28:36,860 --> 00:28:39,810 So what would a problem like this mean in F2 to the n? 398 00:28:52,680 --> 00:29:00,140 So in F2 to the n, so this is a finite field analog. 399 00:29:00,140 --> 00:29:05,060 If A plus A is small-- 400 00:29:10,005 --> 00:29:13,400 so I'm not trying to ask an inverse question. 401 00:29:13,400 --> 00:29:17,050 But what are examples of sets in F2 to the n that 402 00:29:17,050 --> 00:29:20,345 have small doubling? 403 00:29:20,345 --> 00:29:21,728 AUDIENCE: 2 to the n. 404 00:29:21,728 --> 00:29:25,167 YUFEI ZHAO: So 2 to the n, so you can take the entire space. 405 00:29:25,167 --> 00:29:27,000 Any other examples that have small doubling? 406 00:29:33,503 --> 00:29:34,920 AUDIENCE: You can take a subspace. 407 00:29:34,920 --> 00:29:36,712 YUFEI ZHAO: Exactly, I can take a subspace. 408 00:29:36,712 --> 00:29:38,780 So a subspace, well, it doesn't grow. 409 00:29:38,780 --> 00:29:43,460 So A plus A is the same as A. All right, so 410 00:29:43,460 --> 00:29:47,960 and also, as before, you can take a subset of a subspace. 411 00:29:47,960 --> 00:29:50,480 So then the analog of Freiman's theorem 412 00:29:50,480 --> 00:30:06,950 will say that A is contained in a subspace of size, at most, 413 00:30:06,950 --> 00:30:13,930 a constant times the size of A. 414 00:30:13,930 --> 00:30:18,950 So this is the analog of Freiman's theorem in F2. 415 00:30:18,950 --> 00:30:22,040 And so we'll see, so this will be much easier 416 00:30:22,040 --> 00:30:24,040 than the general result about Freiman's theorem, 417 00:30:24,040 --> 00:30:26,190 but it will involve a subset of F2. 418 00:30:26,190 --> 00:30:28,060 And we'll see this theorem first. 419 00:30:28,060 --> 00:30:30,760 So we'll prove that next lecture. 420 00:30:30,760 --> 00:30:33,400 Of course, this is much easier in many ways, 421 00:30:33,400 --> 00:30:35,470 because here, unlike before, I don't even 422 00:30:35,470 --> 00:30:37,700 have to think about what subspace to take. 423 00:30:37,700 --> 00:30:43,900 I can just take the subspace generated by the elements of A. 424 00:30:43,900 --> 00:30:49,084 All right, Any questions so far? 425 00:30:49,084 --> 00:30:49,584 Yes? 426 00:30:49,584 --> 00:30:52,853 AUDIENCE: Is the f of k here still exponential in k? 427 00:30:52,853 --> 00:30:54,770 YUFEI ZHAO: OK, so the question, is the f of k 428 00:30:54,770 --> 00:30:56,910 here still exponential in k? 429 00:30:56,910 --> 00:30:58,870 So the answer is, yes. 430 00:30:58,870 --> 00:31:01,560 And the construction is if you take A to be a basis. 431 00:31:05,790 --> 00:31:10,020 OK, so let's start with some techniques and some proofs. 432 00:31:23,830 --> 00:31:27,900 So in this chapter, many things are named after Ruzsa. 433 00:31:27,900 --> 00:31:30,960 And at some point, it becomes slightly confusing 434 00:31:30,960 --> 00:31:33,353 which ones are not named after Ruzsa. 435 00:31:33,353 --> 00:31:35,270 But the first thing will be named after Ruzsa. 436 00:31:35,270 --> 00:31:37,341 So it's a Ruzsa Triangle Inequality. 437 00:31:45,197 --> 00:31:48,520 All right, the Ruzsa Triangle Inequality 438 00:31:48,520 --> 00:31:52,390 tells us that, if A, B, and C-- 439 00:31:52,390 --> 00:31:54,500 so unless otherwise I tell you so, 440 00:31:54,500 --> 00:31:57,310 and I'll try to remind you each time, but basically, 441 00:31:57,310 --> 00:32:01,240 we're always going to be looking finite sets 442 00:32:01,240 --> 00:32:09,940 in an arbitrary obedient group always with an under addition-- 443 00:32:09,940 --> 00:32:15,790 then one has the inequality on their sizes of different sets. 444 00:32:22,750 --> 00:32:25,520 The size of A times the size of B minus C 445 00:32:25,520 --> 00:32:28,800 is upper bounded by the size of A minus B times the size of A 446 00:32:28,800 --> 00:32:36,500 minus C. So that's the Ruzsa Triangle Inequality. 447 00:32:36,500 --> 00:32:39,490 Let me show you the proof. 448 00:32:39,490 --> 00:32:51,300 We will construct an injection from A cross B minus C 449 00:32:51,300 --> 00:32:57,420 to A minus B cross A minus C. Of course, 450 00:32:57,420 --> 00:32:59,620 if you can exhibit such an injection, 451 00:32:59,620 --> 00:33:03,170 then you prove the desired inequality. 452 00:33:03,170 --> 00:33:11,640 To obtain this injection, we start with an element a, d. 453 00:33:11,640 --> 00:33:17,400 And for this a, d, so for each d, let me pick-- 454 00:33:20,410 --> 00:33:24,850 so if d is an element of B minus C, 455 00:33:24,850 --> 00:33:30,450 let us pick arbitrarily but stick with those choices 456 00:33:30,450 --> 00:33:41,550 a b of d in the set B and a c of d in the set C such 457 00:33:41,550 --> 00:33:47,180 that d equals to b of d minus c of d. 458 00:33:47,180 --> 00:33:51,660 So because d is the set B minus C, 459 00:33:51,660 --> 00:33:54,720 it can be represented as a difference from one element 460 00:33:54,720 --> 00:33:55,340 from each set. 461 00:33:55,340 --> 00:33:58,210 So it may be represented in many ways. 462 00:33:58,210 --> 00:34:00,740 But from the start, you pick a way to represent it. 463 00:34:00,740 --> 00:34:02,290 And you stick with that choice. 464 00:34:02,290 --> 00:34:05,598 And you label that function b of d and c of d. 465 00:34:08,670 --> 00:34:21,570 Now I map a, d to the element a minus b of d and a minus 466 00:34:21,570 --> 00:34:22,170 c of d. 467 00:34:28,070 --> 00:34:30,230 So this is a map. 468 00:34:30,230 --> 00:34:32,104 I want to show that it is injective. 469 00:34:35,830 --> 00:34:37,296 Why is it injective? 470 00:34:41,760 --> 00:34:43,650 Well, to show something is injective, 471 00:34:43,650 --> 00:34:46,080 I just need to show that I can recover where I came 472 00:34:46,080 --> 00:34:49,810 from if I tell you the image. 473 00:34:49,810 --> 00:34:54,850 So I can recover a and d from these two numbers. 474 00:34:54,850 --> 00:34:58,930 So if-- sorry, new board. 475 00:35:05,040 --> 00:35:07,470 OK, so well you basically can think 476 00:35:07,470 --> 00:35:12,320 about how you can recover a and d from the image elements. 477 00:35:21,540 --> 00:35:24,140 So if the image-- 478 00:35:24,140 --> 00:35:27,440 so I label that map phi. 479 00:35:27,440 --> 00:35:28,790 So that's phi up there. 480 00:35:28,790 --> 00:35:42,520 So if the image is given, then I can recover d. 481 00:35:42,520 --> 00:35:44,380 So how can we recover the element d? 482 00:35:50,810 --> 00:35:53,090 So you subtract these two numbers. 483 00:35:53,090 --> 00:35:55,860 So d is minus x. 484 00:35:55,860 --> 00:35:58,940 And once you recover d, you can also then take 485 00:35:58,940 --> 00:36:00,230 a look at the first element. 486 00:36:00,230 --> 00:36:01,680 And you can recover a. 487 00:36:06,030 --> 00:36:06,980 So now you know d. 488 00:36:06,980 --> 00:36:10,490 I can now recover a. 489 00:36:10,490 --> 00:36:13,420 OK, so then this is-- you can check this is an injection. 490 00:36:13,420 --> 00:36:19,269 And that proves the Ruzsa Triangle Inequality. 491 00:36:19,269 --> 00:36:24,468 OK, so it's short, but it's tricky. 492 00:36:24,468 --> 00:36:26,380 It's tricky. 493 00:36:26,380 --> 00:36:29,320 OK, so why is this called Ruzsa's Triangle Inequality? 494 00:36:29,320 --> 00:36:31,770 Where is the triangle in this? 495 00:36:31,770 --> 00:36:36,090 The reason that it's given that name is that you can write 496 00:36:36,090 --> 00:36:38,360 the inequality as follows. 497 00:36:38,360 --> 00:36:45,070 Suppose we use rho A, B to denote this quantity obtained 498 00:36:45,070 --> 00:36:49,630 by taking the log of the size of A minus B divided 499 00:36:49,630 --> 00:36:52,510 by the square root of the product 500 00:36:52,510 --> 00:36:59,470 of their individual sizes, then the inequality 501 00:36:59,470 --> 00:37:05,860 says that the rho of B, C is, at most, 502 00:37:05,860 --> 00:37:13,100 rho of A, B plus rho of A, C, which looks like a triangle 503 00:37:13,100 --> 00:37:14,590 inequality. 504 00:37:14,590 --> 00:37:16,910 So that's why it's called Ruzsa's Triangle Inequality, 505 00:37:16,910 --> 00:37:17,720 because this is-- 506 00:37:17,720 --> 00:37:20,660 don't take it too seriously, because this is not a distance. 507 00:37:24,960 --> 00:37:30,560 So rho of A, A is not equal to 0. 508 00:37:30,560 --> 00:37:33,540 But it certainly has the form of a triangle inequality, hence 509 00:37:33,540 --> 00:37:34,040 the name. 510 00:37:37,560 --> 00:37:39,780 How should you think of Ruzsa's triangle inequality? 511 00:37:39,780 --> 00:37:40,900 So in this chapter, there's going 512 00:37:40,900 --> 00:37:42,480 to be a lot of symbol pushing around. 513 00:37:42,480 --> 00:37:45,310 And it's easy to get lost and buried in all of these symbols. 514 00:37:45,310 --> 00:37:48,390 And I want to tell you about how you 515 00:37:48,390 --> 00:37:52,430 might think about what's the point of Ruzsa's Triangle 516 00:37:52,430 --> 00:37:52,930 Inequality. 517 00:37:52,930 --> 00:37:55,120 How would you use it? 518 00:37:55,120 --> 00:38:02,230 And the idea is that if you have a set with small doubling, 519 00:38:02,230 --> 00:38:04,780 we want to use Ruzsa's triangle inequality 520 00:38:04,780 --> 00:38:08,998 and other tools to control its further doublings. 521 00:38:11,810 --> 00:38:14,760 So in particular, if-- 522 00:38:14,760 --> 00:38:19,790 so I'll say, applications. 523 00:38:19,790 --> 00:38:27,755 So suppose you knew that 2A minus 2A 524 00:38:27,755 --> 00:38:31,040 is size, at most, k times A. So this is 525 00:38:31,040 --> 00:38:34,790 a stronger hypothesis than just A has small doublings. 526 00:38:34,790 --> 00:38:37,400 Even if you iterate it several times, 527 00:38:37,400 --> 00:38:42,156 you still have size, at most, constant times A. 528 00:38:42,156 --> 00:38:45,250 I would like to start from this hypothesis 529 00:38:45,250 --> 00:38:49,120 and control further iterations, further subsets 530 00:38:49,120 --> 00:38:51,700 of A. And Ruzsa's Triangle Inequality 531 00:38:51,700 --> 00:39:00,820 allows us to do it, because by the Ruzsa's Triangle 532 00:39:00,820 --> 00:39:09,630 Inequality, setting B and C to be 2A minus A, 533 00:39:09,630 --> 00:39:18,330 we find that 3A minus 3A is, at most, 2A minus 2A squared 534 00:39:18,330 --> 00:39:22,950 over A, the size of A. So plug it in. 535 00:39:22,950 --> 00:39:25,680 This is what you get. 536 00:39:25,680 --> 00:39:29,810 So if the size of 2A plus 2A is, at most, k times the size of A, 537 00:39:29,810 --> 00:39:35,500 then the size of 3A times 3A is-- 538 00:39:35,500 --> 00:39:38,550 blows up by a factor, at most, k squared. 539 00:39:38,550 --> 00:39:40,650 So it controls further doublings. 540 00:39:40,650 --> 00:39:42,368 And of course, we can iterate. 541 00:39:46,220 --> 00:39:53,180 If we know set B and C to be 3A minus 2A, 542 00:39:53,180 --> 00:39:57,890 then what we get is 5A minus 5A is, at most, 543 00:39:57,890 --> 00:40:02,930 a size of 3A minus 3A square divided by the size of A. 544 00:40:02,930 --> 00:40:08,760 And so now you have a bound which is k to the 4 times A. 545 00:40:08,760 --> 00:40:11,464 And you can continue. 546 00:40:11,464 --> 00:40:12,922 You can continue. 547 00:40:15,672 --> 00:40:17,880 OK, so this is all a consequence of Ruzsa's triangle. 548 00:40:17,880 --> 00:40:19,920 So starting with this hypothesis, 549 00:40:19,920 --> 00:40:25,160 now I get to control all the further doublings, 550 00:40:25,160 --> 00:40:26,850 the further subset iterations. 551 00:40:26,850 --> 00:40:29,620 I call them doublings, but they're no longer doubles, 552 00:40:29,620 --> 00:40:32,580 but further subsets. 553 00:40:32,580 --> 00:40:34,650 But this is a stronger hypothesis than the one 554 00:40:34,650 --> 00:40:37,790 that we start with in Freiman's theorem, 555 00:40:37,790 --> 00:40:42,300 because if you have that, then this 2A minus 2A 556 00:40:42,300 --> 00:40:45,640 is at least as large as the size of 2A. 557 00:40:45,640 --> 00:40:50,350 So can we start with just doubling constant 558 00:40:50,350 --> 00:40:55,410 and then obtain bounds on the iterations? 559 00:40:55,410 --> 00:40:56,960 | it turns out you can. 560 00:40:56,960 --> 00:41:00,270 It will require another theorem. 561 00:41:07,130 --> 00:41:10,753 So this theorem is called Plunnecke inequality. 562 00:41:14,070 --> 00:41:16,790 But actually, these days, in literature, 563 00:41:16,790 --> 00:41:21,560 it's often referred to as Plunnecke-Ruzsa inequality. 564 00:41:21,560 --> 00:41:23,330 So Plunnecke initially proved it. 565 00:41:23,330 --> 00:41:25,130 But nobody understood his proof. 566 00:41:25,130 --> 00:41:26,718 And Ruzsa gave a better proof. 567 00:41:26,718 --> 00:41:29,010 And actually, recently, there was an even better proof. 568 00:41:29,010 --> 00:41:31,688 And that's the one I will show you. 569 00:41:31,688 --> 00:41:34,680 So Plunnecke-Ruzsa inequality tells us 570 00:41:34,680 --> 00:41:45,680 that if A is subset of some obedient group, 571 00:41:45,680 --> 00:41:54,340 and has doubling constant, at most, k, 572 00:41:54,340 --> 00:42:03,930 then for all non-negative integers m and n, 573 00:42:03,930 --> 00:42:10,440 the size of mA minus nA is, at most, 574 00:42:10,440 --> 00:42:17,230 k to the m plus n times the size of A. 575 00:42:17,230 --> 00:42:22,490 So if you have bounded doubling, then the further iterations, 576 00:42:22,490 --> 00:42:27,520 the further subset iterations are also controlling size. 577 00:42:27,520 --> 00:42:31,480 I want you to think of polynomial transformations in k 578 00:42:31,480 --> 00:42:33,320 as negligible. 579 00:42:33,320 --> 00:42:36,340 So don't worry about that we're raising things here. 580 00:42:36,340 --> 00:42:37,210 k is constant. 581 00:42:37,210 --> 00:42:39,640 You should think of m and n as constant. 582 00:42:39,640 --> 00:42:42,015 So I'm changing k to some other constant. 583 00:42:42,015 --> 00:42:44,210 And in fact, I'm only changing it by a polynomial. 584 00:42:44,210 --> 00:42:46,150 So this is, like, almost no change at all. 585 00:42:50,810 --> 00:42:52,600 So this is tricky. 586 00:42:52,600 --> 00:42:56,210 So we'll do it after a short break. 587 00:42:56,210 --> 00:42:59,150 All right, let's prove Plunnecke's inequality, 588 00:42:59,150 --> 00:43:00,290 Plunnecke-Ruzsa inequality. 589 00:43:00,290 --> 00:43:05,770 So the history of Plunnecke's inequality 590 00:43:05,770 --> 00:43:08,160 has some similarities with Freiman's theorem. 591 00:43:08,160 --> 00:43:09,760 So Plunnecke initially proved it, 592 00:43:09,760 --> 00:43:13,390 but his proof was hard to understand 593 00:43:13,390 --> 00:43:16,330 and was sort of left not understood for a long time 594 00:43:16,330 --> 00:43:20,230 until others like Plun and Ruzsa came in 595 00:43:20,230 --> 00:43:22,690 and really simplified the proof. 596 00:43:22,690 --> 00:43:26,320 But even then, the proof was not so easy. 597 00:43:26,320 --> 00:43:30,260 And if I were teaching this course about 10 years ago, 598 00:43:30,260 --> 00:43:32,440 I would have just skipped this proof, maybe sketched 599 00:43:32,440 --> 00:43:34,398 some ideas, but I would have skipped the proof. 600 00:43:34,398 --> 00:43:36,595 And the proof, actually, it's a beautiful proof, 601 00:43:36,595 --> 00:43:38,860 but it uses some serious graph theory. 602 00:43:38,860 --> 00:43:41,658 It uses Menger's theorem about flows. 603 00:43:41,658 --> 00:43:42,700 You construct some graph. 604 00:43:42,700 --> 00:43:45,132 And then you try to understand its flows. 605 00:43:45,132 --> 00:43:46,090 It's very pretty stuff. 606 00:43:46,090 --> 00:43:50,170 And I do encourage you to look it up. 607 00:43:50,170 --> 00:43:53,560 And then about eight years ago, Petridis 608 00:43:53,560 --> 00:44:03,290 found a proof, so a proof by Petridis, 609 00:44:03,290 --> 00:44:06,450 who was a PhD student that Tim Gowers at the time. 610 00:44:06,450 --> 00:44:10,200 And that was surprisingly short, and beautiful, 611 00:44:10,200 --> 00:44:13,650 and kind of surprised everyone that such a short proof exists, 612 00:44:13,650 --> 00:44:16,200 given that this theorem sat in that state 613 00:44:16,200 --> 00:44:17,450 for such a long time. 614 00:44:17,450 --> 00:44:20,340 And it's a pretty central step in the proof 615 00:44:20,340 --> 00:44:23,500 of Freiman's theorem. 616 00:44:23,500 --> 00:44:27,600 We'll prove Plunnecke-Ruzsa via a slightly more general 617 00:44:27,600 --> 00:44:28,690 statement. 618 00:44:28,690 --> 00:44:32,393 So you see, it generalizes the earlier statement. 619 00:44:32,393 --> 00:44:33,810 Instead of having one set, it will 620 00:44:33,810 --> 00:44:36,930 be convenient to have two different sets. 621 00:44:36,930 --> 00:44:46,500 So let A and B be subsets of some obedient group, as usual. 622 00:44:50,350 --> 00:44:59,490 If size of A plus B is, at most, k times the size of A, 623 00:44:59,490 --> 00:45:08,490 then mB minus nB has size, at most, 624 00:45:08,490 --> 00:45:13,650 k to the m plus n times the size of A 625 00:45:13,650 --> 00:45:19,028 for all non-negative integers m and n. 626 00:45:19,028 --> 00:45:21,070 So instead of having one set, so I have two sets, 627 00:45:21,070 --> 00:45:24,580 A and B. Of course, then you derive the earlier statement 628 00:45:24,580 --> 00:45:27,040 setting A and B to equal. 629 00:45:27,040 --> 00:45:29,314 So we'll prove this more general statement. 630 00:45:34,640 --> 00:45:37,400 The proof uses a key lemma. 631 00:45:40,390 --> 00:45:54,330 And the key lemma says that if a subset x of A is non-empty 632 00:45:54,330 --> 00:45:55,430 and-- 633 00:45:55,430 --> 00:46:05,720 so if x is a non-empty subset of A 634 00:46:05,720 --> 00:46:21,500 that minimizes the ratio x plus B divided by size of x, 635 00:46:21,500 --> 00:46:32,160 and let k prime be this ratio, this minimum ratio, 636 00:46:32,160 --> 00:46:40,780 then so the conclusion says that x plus B plus C 637 00:46:40,780 --> 00:46:48,200 has size, at most, k prime times the size of x plus C for all 638 00:46:48,200 --> 00:46:53,230 sets C. 639 00:46:53,230 --> 00:46:54,510 So that's the statement. 640 00:46:54,510 --> 00:46:57,576 I'll explain how you should think about the statement. 641 00:47:00,770 --> 00:47:05,420 These ratios which you see in both hypotheses, 642 00:47:05,420 --> 00:47:10,210 how you should think about them is that there is this graph. 643 00:47:10,210 --> 00:47:14,840 Let's say it's the group bipartite graph with the group 644 00:47:14,840 --> 00:47:17,070 elements on both sides. 645 00:47:17,070 --> 00:47:22,430 And the graph has edges, the bipartite graph, 646 00:47:22,430 --> 00:47:30,500 where the edges are from each vertex a drawn edge 647 00:47:30,500 --> 00:47:35,400 for each element of B. So I expand by B. 648 00:47:35,400 --> 00:47:37,920 So if you have this graph and you 649 00:47:37,920 --> 00:47:45,280 start with some A on the left, then its neighbors on the right 650 00:47:45,280 --> 00:47:56,910 will be A plus B. And those ratios up there 651 00:47:56,910 --> 00:47:59,790 are the expansion ratios. 652 00:47:59,790 --> 00:48:05,975 so quantities like this, they are expansion ratios. 653 00:48:08,520 --> 00:48:11,070 You start with some set on the left 654 00:48:11,070 --> 00:48:13,890 and see by a what fraction does it expand if you 655 00:48:13,890 --> 00:48:16,730 look at the neighborhood. 656 00:48:16,730 --> 00:48:20,010 So let's read the statement of the key lemma. 657 00:48:20,010 --> 00:48:23,630 It says, if you have a set x-- 658 00:48:23,630 --> 00:48:25,420 I look, so I have a set A. And I'm 659 00:48:25,420 --> 00:48:31,420 choosing a subset of A that minimizes the expansion ratio, 660 00:48:31,420 --> 00:48:33,260 so choose a non-empty subset that 661 00:48:33,260 --> 00:48:36,080 minimizes the expansion ratio. 662 00:48:36,080 --> 00:48:41,950 And if this minimum expense ratio is k prime, then, 663 00:48:41,950 --> 00:48:53,550 so x minimizes expansion ratio and expense ratios k prime, 664 00:48:53,550 --> 00:49:04,660 then x plus C also has expansion ratio, at most, k prime 665 00:49:04,660 --> 00:49:05,160 as well. 666 00:49:08,560 --> 00:49:09,771 So that's the statement. 667 00:49:12,480 --> 00:49:15,060 I mentioned earlier that the previous proofs of this theorem 668 00:49:15,060 --> 00:49:18,750 went through some graph theory and Menger's theorem, 669 00:49:18,750 --> 00:49:20,530 that type of graph theory. 670 00:49:20,530 --> 00:49:23,695 You can kind of see where it might come in. 671 00:49:23,695 --> 00:49:24,820 We're not going to do that. 672 00:49:24,820 --> 00:49:26,100 We're going to stick with additive combinatorics. 673 00:49:26,100 --> 00:49:28,620 We're going to stick with playing with sums, 674 00:49:28,620 --> 00:49:30,870 playing with additive combinatorics. 675 00:49:30,870 --> 00:49:34,390 So let's see how we can prove the statement up there, 676 00:49:34,390 --> 00:49:35,620 so using the key lemma. 677 00:49:39,960 --> 00:49:50,580 So assuming key lemma, so let's prove the statement, 678 00:49:50,580 --> 00:49:52,610 the theorem up there. 679 00:49:52,610 --> 00:50:01,520 So take a non-empty subset x of A-- 680 00:50:01,520 --> 00:50:12,640 sorry, so x subset of A that minimizes the ratio x plus B 681 00:50:12,640 --> 00:50:16,600 divided by x. 682 00:50:16,600 --> 00:50:20,140 And let k prime be this minimum ratio. 683 00:50:25,710 --> 00:50:30,390 Note that k prime is, at most, k, 684 00:50:30,390 --> 00:50:33,810 because if you plug in x equals the k, you get-- 685 00:50:33,810 --> 00:50:36,980 if you plug in x equals to A, you get k. 686 00:50:36,980 --> 00:50:40,870 But I'm choosing x to be possibly even lower. 687 00:50:40,870 --> 00:50:42,510 So k prime is, at most, k. 688 00:50:50,738 --> 00:51:03,980 Now, applying the lemma, so applying the key lemma with C 689 00:51:03,980 --> 00:51:14,294 equals to B, we find that x plus 2B, so C, plug in B, 690 00:51:14,294 --> 00:51:23,780 x plus 2B has size, at most, k times size of x plus B. 691 00:51:23,780 --> 00:51:28,490 But the size of x plus B is, at most, k times the size of A. 692 00:51:28,490 --> 00:51:35,430 So we get k squared, so k times the size of x, at most, k 693 00:51:35,430 --> 00:51:37,860 squared x. 694 00:51:37,860 --> 00:51:41,430 So we're already in good shape. 695 00:51:41,430 --> 00:51:44,220 If you iterate expansion twice-- so 696 00:51:44,220 --> 00:51:46,170 I imagine there is several chains 697 00:51:46,170 --> 00:51:47,976 of these bipartite graphs. 698 00:51:47,976 --> 00:51:50,250 If you iterate this expansion twice, 699 00:51:50,250 --> 00:51:52,410 you still do not blow up by too much. 700 00:51:54,970 --> 00:51:58,610 So we can iterate further, so apply the lemma 701 00:51:58,610 --> 00:52:05,460 with C being now 2B, and then later 3B, and so on. 702 00:52:05,460 --> 00:52:13,530 So you find that x plus nB has size, 703 00:52:13,530 --> 00:52:18,320 at most, k raised to power n times the size of x 704 00:52:18,320 --> 00:52:21,430 for all non-negative integers, n. 705 00:52:30,930 --> 00:52:32,600 What do we want to control? 706 00:52:32,600 --> 00:52:39,810 So we want to prove a bound on the size of mB minus nB. 707 00:52:39,810 --> 00:52:43,721 Take a look at the statement of Ruzsa Triangle Inequality. 708 00:52:50,790 --> 00:52:53,970 Applying Ruzsa Triangle Inequality, 709 00:52:53,970 --> 00:53:02,800 we find that if we want to control mB minus nB, 710 00:53:02,800 --> 00:53:13,390 we can upper bound it by x plus mB x plus nB divided 711 00:53:13,390 --> 00:53:15,370 by the size of x. 712 00:53:18,160 --> 00:53:22,630 Because each of these two factors in the numerator 713 00:53:22,630 --> 00:53:26,080 are small expansions of x, now we 714 00:53:26,080 --> 00:53:30,430 can upper bound the whole expression by k to the m 715 00:53:30,430 --> 00:53:35,790 plus n times the size of x. 716 00:53:35,790 --> 00:53:43,950 And because x is a subset of A, we can do one more upper bound 717 00:53:43,950 --> 00:53:46,410 and obtain the bound that we are looking for. 718 00:53:51,400 --> 00:53:53,020 OK, so that proves the key lemma. 719 00:53:56,764 --> 00:53:57,700 It's OK? 720 00:53:57,700 --> 00:53:59,580 AUDIENCE: [INAUDIBLE] 721 00:53:59,580 --> 00:54:02,170 YUFEI ZHAO: Sorry, that proves the theorem, 722 00:54:02,170 --> 00:54:03,587 assuming the key lemma. 723 00:54:03,587 --> 00:54:05,170 Thank you, that's what I meant to say. 724 00:54:05,170 --> 00:54:06,680 Yeah, so that proves the theorem, 725 00:54:06,680 --> 00:54:07,638 assuming the key lemma. 726 00:54:07,638 --> 00:54:10,170 So now we do prove the key lemma. 727 00:54:10,170 --> 00:54:12,010 Great, we need to prove the key lemma. 728 00:54:12,010 --> 00:54:15,000 And so Petridis' proof of the key lemma, 729 00:54:15,000 --> 00:54:19,300 it's quite surprising, in that it uses induction. 730 00:54:19,300 --> 00:54:21,430 And basically, we have not used induction 731 00:54:21,430 --> 00:54:24,460 in this course ever since the first or maybe 732 00:54:24,460 --> 00:54:27,430 the second lecture, and for good reason. 733 00:54:27,430 --> 00:54:29,518 So everything in this course is fairly analytic. 734 00:54:29,518 --> 00:54:31,060 You know, you have these Roth bounds. 735 00:54:31,060 --> 00:54:36,010 And putting one extra vertex often doesn't really help. 736 00:54:36,010 --> 00:54:54,257 OK, so here, we're going to use induction on the size of C. OK, 737 00:54:54,257 --> 00:54:56,590 I just want to emphasize again that the use of induction 738 00:54:56,590 --> 00:54:59,190 here was surprising. 739 00:54:59,190 --> 00:55:02,320 So if the base case-- 740 00:55:02,320 --> 00:55:05,290 always check the base case-- 741 00:55:05,290 --> 00:55:10,780 when C is 1, then plus C is a translation. 742 00:55:15,070 --> 00:55:18,630 So this shifts the set over. 743 00:55:18,630 --> 00:55:27,070 And so you can see that if you do plus C and minus 1, 744 00:55:27,070 --> 00:55:33,040 you raise the plus C. And the conclusion follows basically 745 00:55:33,040 --> 00:55:35,830 from the hypothesis. 746 00:55:35,830 --> 00:55:40,840 So in this case, x plus B plus C is 747 00:55:40,840 --> 00:55:49,990 equal to x plus B, which is, at most, k prime times the size x, 748 00:55:49,990 --> 00:55:52,990 by definition of-- 749 00:55:52,990 --> 00:55:56,890 so this actually is equal to the size of k. 750 00:56:03,030 --> 00:56:04,120 The base case is easy. 751 00:56:06,642 --> 00:56:07,850 Now we do the induction step. 752 00:56:13,660 --> 00:56:20,080 So let's assume that the size of C is bigger than 1 753 00:56:20,080 --> 00:56:25,510 and C is C prime plus an additional element, which 754 00:56:25,510 --> 00:56:26,380 we'll call gamma. 755 00:56:29,680 --> 00:56:35,410 So let's see this expression, x plus B plus C, 756 00:56:35,410 --> 00:56:39,760 by separating it according to if its contribution came 757 00:56:39,760 --> 00:56:44,870 from C prime or not. 758 00:56:44,870 --> 00:56:47,460 The contributions that came from C prime, 759 00:56:47,460 --> 00:56:49,760 I can write it like that. 760 00:56:52,973 --> 00:56:54,890 And then there are other contributions, namely 761 00:56:54,890 --> 00:56:56,980 those that came from this extra element. 762 00:57:04,480 --> 00:57:08,960 But I may have some redundancies in doing this. 763 00:57:08,960 --> 00:57:11,760 So I may have some redundancies coming from the fact 764 00:57:11,760 --> 00:57:14,430 that some of the elements in this set 765 00:57:14,430 --> 00:57:18,100 might have already appeared earlier. 766 00:57:18,100 --> 00:57:27,300 So let me take out those elements 767 00:57:27,300 --> 00:57:37,230 by taking out elements where it already appeared earlier. 768 00:57:37,230 --> 00:57:42,330 So this means I'm looking at the set z 769 00:57:42,330 --> 00:57:51,150 being elements of x such that x plus B plus gamma is already 770 00:57:51,150 --> 00:57:56,940 a subset of x plus B plus C prime. 771 00:57:59,570 --> 00:58:02,600 So the stuff in yellow, I can safely discard, 772 00:58:02,600 --> 00:58:04,330 because it already appeared earlier. 773 00:58:11,490 --> 00:58:16,250 So because of the definition of z, 774 00:58:16,250 --> 00:58:24,020 we see that z plus B plus gamma appears 775 00:58:24,020 --> 00:58:27,950 in x plus B plus C prime. 776 00:58:27,950 --> 00:58:31,940 So that union is valid. 777 00:58:31,940 --> 00:58:34,790 Now, z is a subset of x. 778 00:58:39,500 --> 00:58:48,410 So the expansion ratio for z is at least k prime, 779 00:58:48,410 --> 00:58:52,695 because we chose x to minimize this expansion ratio. 780 00:59:07,530 --> 00:59:13,830 We would like to understand how big x plus B plus C. 781 00:59:13,830 --> 00:59:19,310 So let's evaluate the cardinality of that expression 782 00:59:19,310 --> 00:59:21,260 up there. 783 00:59:21,260 --> 00:59:23,790 The cardinality I can upper bound 784 00:59:23,790 --> 00:59:27,960 by the union of these sum of the sizes of the components. 785 00:59:39,420 --> 00:59:45,100 So up there, so I just do a union bound on that expression 786 00:59:45,100 --> 00:59:45,600 up there. 787 00:59:50,490 --> 00:59:53,310 And now you see z is a subset of x. 788 00:59:53,310 --> 00:59:56,400 So I can split this expression up even further. 789 01:00:09,657 --> 01:00:15,930 All right, now let's use the induction hypothesis. 790 01:00:15,930 --> 01:00:17,700 So we have some expression involving 791 01:00:17,700 --> 01:00:21,790 x plus B plus C prime. 792 01:00:21,790 --> 01:00:28,170 So now we apply induction hypothesis over here 793 01:00:28,170 --> 01:00:32,260 to this expression that has plus C prime. 794 01:00:32,260 --> 01:00:34,510 And we obtain an upper bound which 795 01:00:34,510 --> 01:00:38,650 is k prime x plus C prime. 796 01:00:43,370 --> 01:00:45,510 And the two expressions on the right, 797 01:00:45,510 --> 01:00:50,850 well, one of them here is, by definition, 798 01:00:50,850 --> 01:00:56,610 coming from the expansion ratio of x. 799 01:00:56,610 --> 01:01:00,300 And then the other, we gave a bound just now. 800 01:01:12,060 --> 01:01:13,770 OK, so we're almost there. 801 01:01:13,770 --> 01:01:17,520 So we are trying to upper bound the size of this quantity. 802 01:01:17,520 --> 01:01:22,690 So we decomposed it into pieces according to its contribution 803 01:01:22,690 --> 01:01:24,240 coming from this extra element. 804 01:01:24,240 --> 01:01:28,770 And we analyzed these pieces individually. 805 01:01:28,770 --> 01:01:32,630 But now I want to understand the right-hand side, 806 01:01:32,630 --> 01:01:37,750 so x plus C. So let's try to understand the right-hand side. 807 01:01:41,140 --> 01:01:46,510 See, the x plus C, I can likewise write it as earlier 808 01:01:46,510 --> 01:01:52,510 by decomposing it into contributions from C prime 809 01:01:52,510 --> 01:01:54,460 and those from the extra element. 810 01:02:01,490 --> 01:02:06,670 And as earlier, we can take out contributions that were already 811 01:02:06,670 --> 01:02:13,820 appearing earlier, which we now recall W plus gamma, where 812 01:02:13,820 --> 01:02:24,970 W is the set of elements in x, such that x plus gamma 813 01:02:24,970 --> 01:02:30,100 is already contained in x plus C prime. 814 01:02:30,100 --> 01:02:33,988 So this part was already included earlier. 815 01:02:33,988 --> 01:02:35,530 We don't need to include it any more. 816 01:02:39,234 --> 01:02:43,480 A couple of observations that were different from earlier-- 817 01:02:43,480 --> 01:02:46,890 now this union I claim and say disjoined union. 818 01:02:52,518 --> 01:02:55,250 So this union is a disjoined union. 819 01:02:55,250 --> 01:02:58,270 So there is actually no more overlaps. 820 01:03:02,780 --> 01:03:07,730 And furthermore, W is contained in the set z from earlier. 821 01:03:17,410 --> 01:03:18,778 Any questions? 822 01:03:24,634 --> 01:03:30,990 All right, therefore, the size of x plus C 823 01:03:30,990 --> 01:03:35,550 is equal to, because this is a disjoined union, x 824 01:03:35,550 --> 01:03:46,590 plus C prime plus the size of x minus the size of W, 825 01:03:46,590 --> 01:03:51,030 and which is-- 826 01:03:51,030 --> 01:03:54,700 so W, because W is contained in z, 827 01:03:54,700 --> 01:04:01,510 is x plus C prime plus the size of x minus the size of z. 828 01:04:07,160 --> 01:04:09,240 Now you compare these two expressions. 829 01:04:12,660 --> 01:04:13,920 And that proves the key lemma. 830 01:04:21,255 --> 01:04:23,211 OK? 831 01:04:23,211 --> 01:04:24,300 That's it. 832 01:04:24,300 --> 01:04:24,800 Yeah? 833 01:04:24,800 --> 01:04:26,425 AUDIENCE: Can you explain one more time 834 01:04:26,425 --> 01:04:27,620 why it's a disjoined union? 835 01:04:27,620 --> 01:04:30,453 YUFEI ZHAO: OK, great, so why is this a disjoined union? 836 01:04:30,453 --> 01:04:34,500 Now, I have the set here. 837 01:04:34,500 --> 01:04:40,200 So I'm looking at this x plus gamma. 838 01:04:43,760 --> 01:04:46,790 So think about, let's say, gamma equals to 0. 839 01:04:46,790 --> 01:04:51,910 So we translate, think about if gamma equals to 0. 840 01:04:51,910 --> 01:05:00,890 So I include x, but if some element of x was already here, 841 01:05:00,890 --> 01:05:03,340 I take it out. 842 01:05:03,340 --> 01:05:09,400 So here is x plus C prime. 843 01:05:09,400 --> 01:05:13,270 And let's say this set is x. 844 01:05:13,270 --> 01:05:16,850 This W then would there be their intersection. 845 01:05:16,850 --> 01:05:20,520 So now x minus W is just this set. 846 01:05:20,520 --> 01:05:22,448 So it's a disjoined union. 847 01:05:22,448 --> 01:05:24,740 So the points are, here, you're adding single elements, 848 01:05:24,740 --> 01:05:26,282 where there, you're adding some sets. 849 01:05:26,282 --> 01:05:30,350 So you cannot necessarily take a whole partition, necessarily. 850 01:05:30,350 --> 01:05:31,560 But here it's OK. 851 01:05:36,500 --> 01:05:40,410 It's tricky. 852 01:05:40,410 --> 01:05:41,288 Yeah, it's tricky. 853 01:05:41,288 --> 01:05:43,580 And you know, this took a long time for people to find. 854 01:05:43,580 --> 01:05:48,090 It was found about eight years ago. 855 01:05:48,090 --> 01:05:51,930 And yeah, it was surprising when this proof was discovered. 856 01:05:51,930 --> 01:05:54,090 People did not expect that this proof existed. 857 01:05:59,270 --> 01:06:00,990 And it's also tricky to get right. 858 01:06:00,990 --> 01:06:02,900 So the details-- I do it slowly. 859 01:06:02,900 --> 01:06:05,060 But the execution, like, the order 860 01:06:05,060 --> 01:06:07,850 that you take the minimalities is important. 861 01:06:07,850 --> 01:06:11,760 It's easy to mess up this proof. 862 01:06:11,760 --> 01:06:13,190 OK, any questions? 863 01:06:15,880 --> 01:06:20,290 Let me show you, just as an aside, an application 864 01:06:20,290 --> 01:06:23,520 of this key lemma. 865 01:06:23,520 --> 01:06:26,460 So earlier we saw Ruzsa's Triangle Inequality. 866 01:06:26,460 --> 01:06:30,000 And you may wonder, what if you replace the minus signs 867 01:06:30,000 --> 01:06:31,620 in the theorem by plus signs? 868 01:06:34,530 --> 01:06:36,370 I mean, if you replace the right-hand side, 869 01:06:36,370 --> 01:06:38,800 the two pluses by minuses, the same proof works. 870 01:06:41,340 --> 01:06:44,650 But if you replace all the minus signs by plus signs, you see, 871 01:06:44,650 --> 01:06:47,860 the proof doesn't work anymore. 872 01:06:47,860 --> 01:06:50,280 Just give yourself a moment to convince yourself that. 873 01:06:50,280 --> 01:06:53,305 If you just replace all the minus signs by plus signs, 874 01:06:53,305 --> 01:06:55,180 it doesn't work anymore, but it's still true. 875 01:06:59,540 --> 01:07:01,070 So this is more of an aside. 876 01:07:01,070 --> 01:07:02,290 We will not use it. 877 01:07:02,290 --> 01:07:03,320 But it's nice. 878 01:07:03,320 --> 01:07:04,190 It's fun. 879 01:07:04,190 --> 01:07:13,610 So we have the inequality A B plus C bounded by A plus B A 880 01:07:13,610 --> 01:07:17,477 plus C. 881 01:07:17,477 --> 01:07:19,200 So hopefully you've convince yourself 882 01:07:19,200 --> 01:07:21,540 that if you follow our notes with the previous proof, 883 01:07:21,540 --> 01:07:23,520 you are you're not going to get it. 884 01:07:23,520 --> 01:07:26,340 You're not going to prove this this way. 885 01:07:26,340 --> 01:07:27,010 It's still true. 886 01:07:27,010 --> 01:07:28,140 So how can we prove it? 887 01:07:28,140 --> 01:07:31,050 So we are going to use the key lemma. 888 01:07:31,050 --> 01:07:38,970 So first, the statement is trivial if A is empty. 889 01:07:38,970 --> 01:07:40,470 So let's assume that's not the case. 890 01:07:43,290 --> 01:07:53,870 Let x be a subset of A that minimizes 891 01:07:53,870 --> 01:08:02,330 the expression or the expansion ratio x plus B 892 01:08:02,330 --> 01:08:07,490 divided by x as in the key lemma. 893 01:08:07,490 --> 01:08:15,220 So let k denote the quantity A plus B over A, 894 01:08:15,220 --> 01:08:21,680 so the expansion ratio for A, and k prime 895 01:08:21,680 --> 01:08:25,720 be the expansion ratio for x. 896 01:08:25,720 --> 01:08:29,420 So the quantities came up earlier. 897 01:08:29,420 --> 01:08:36,560 k prime is, at most, value of k, because of our choice of x. 898 01:08:36,560 --> 01:08:47,069 So the key lemma gives x B plus C-- 899 01:08:47,069 --> 01:08:50,120 OK, and this, it's really amazing what's happening. 900 01:08:50,120 --> 01:08:55,100 It seems like we're just going to throw in some extra stuff. 901 01:08:55,100 --> 01:08:57,890 So I'm going to upper bound it by x plus B 902 01:08:57,890 --> 01:09:01,870 plus C. I'm just going to throw in some extra stuff. 903 01:09:01,870 --> 01:09:07,540 And then by the lemma, I can upper bound this expression 904 01:09:07,540 --> 01:09:15,540 by k prime times the size of x plus C. 905 01:09:15,540 --> 01:09:18,359 So that's what the lemma gives you. 906 01:09:18,359 --> 01:09:22,200 And because x is a subset of A, we 907 01:09:22,200 --> 01:09:31,220 can upper bound it by the size of A plus C. And now k prime 908 01:09:31,220 --> 01:09:34,380 is, at most, the size of k. 909 01:09:34,380 --> 01:09:36,520 k prime is, at most, k. 910 01:09:36,520 --> 01:09:38,529 So you have that. 911 01:09:38,529 --> 01:09:40,444 But now look at what the definition of k is. 912 01:09:48,139 --> 01:09:49,115 And that's it. 913 01:09:55,480 --> 01:09:58,300 So that's how you can prove this harder 914 01:09:58,300 --> 01:10:03,800 version of Ruzsa's Triangle Inequality, Yes, question? 915 01:10:03,800 --> 01:10:06,110 AUDIENCE: Are there equality cases for this? 916 01:10:06,110 --> 01:10:07,860 YUFEI ZHAO: All right, question, are there 917 01:10:07,860 --> 01:10:09,130 equality cases for this? 918 01:10:09,130 --> 01:10:11,640 Yes, so I mean, if you're in a subgroup, 919 01:10:11,640 --> 01:10:14,850 then all things are equal, although, 920 01:10:14,850 --> 01:10:18,480 if A, B, and C are all the same subgroup 921 01:10:18,480 --> 01:10:20,310 of some finite obedient group. 922 01:10:20,310 --> 01:10:22,615 AUDIENCE: What if you're working in the integers? 923 01:10:22,615 --> 01:10:23,580 YUFEI ZHAO: Great, yeah, so the question 924 01:10:23,580 --> 01:10:25,205 is, what if you're working in integers? 925 01:10:27,980 --> 01:10:29,010 That's a good question. 926 01:10:29,010 --> 01:10:31,550 I mean, you can suddenly get expansion ratio of two 927 01:10:31,550 --> 01:10:33,340 if you have-- 928 01:10:33,340 --> 01:10:33,840 no, OK. 929 01:10:36,650 --> 01:10:39,090 Right, yeah, so that's a good question. 930 01:10:39,090 --> 01:10:41,430 Can you get equality cases? 931 01:10:41,430 --> 01:10:50,310 If you set A, B, and C to be sets of very different sizes, 932 01:10:50,310 --> 01:10:51,600 AP is a very different set. 933 01:10:51,600 --> 01:10:51,860 Yes? 934 01:10:51,860 --> 01:10:54,193 AUDIENCE: If you set B being true for the single element 935 01:10:54,193 --> 01:10:58,305 and B and C to just be sets that have full extension 936 01:10:58,305 --> 01:10:59,670 so that B plus C is not [? B. ?] 937 01:10:59,670 --> 01:11:01,280 YUFEI ZHAO: Great, yeah, so you take 938 01:11:01,280 --> 01:11:04,670 A to be a single-element set, then 939 01:11:04,670 --> 01:11:07,790 it could be that B plus C is the same as the size of B 940 01:11:07,790 --> 01:11:11,973 times the size of C if B and C have no additive interactions. 941 01:11:16,603 --> 01:11:17,530 Yeah? 942 01:11:17,530 --> 01:11:19,113 AUDIENCE: Are there other known proofs 943 01:11:19,113 --> 01:11:21,002 of this that are less involved? 944 01:11:21,002 --> 01:11:23,210 YUFEI ZHAO: OK, are there other known proofs of this? 945 01:11:23,210 --> 01:11:24,020 I don't know. 946 01:11:24,020 --> 01:11:25,625 I'm not aware of other proofs. 947 01:11:25,625 --> 01:11:27,540 It would be nice to find a different proof. 948 01:11:31,570 --> 01:11:33,768 More questions? 949 01:11:33,768 --> 01:11:34,268 Yeah? 950 01:11:34,268 --> 01:11:36,260 AUDIENCE: How did come up with this? 951 01:11:36,260 --> 01:11:37,968 YUFEI ZHAO: How did he come up with this? 952 01:11:37,968 --> 01:11:39,680 You know, Petridis did a very long PhD. 953 01:11:39,680 --> 01:11:42,520 He spent, I think, seven or eight years in his PhD. 954 01:11:42,520 --> 01:11:44,270 And he eventually came up with this proof. 955 01:11:46,568 --> 01:11:48,610 So he must have thought a lot about this problem. 956 01:11:53,417 --> 01:11:55,000 But the already proofs are still nice. 957 01:11:55,000 --> 01:11:57,280 The earlier proofs, I think they are worth looking at. 958 01:11:57,280 --> 01:12:00,700 They are looking at expansion ratios in graphs. 959 01:12:00,700 --> 01:12:03,690 So you take a sequence of graphs, multi-partite graphs. 960 01:12:03,690 --> 01:12:05,310 And you think about expansion. 961 01:12:05,310 --> 01:12:06,960 And you think about flows. 962 01:12:06,960 --> 01:12:12,310 It's, again, not easy at all, but maybe more motivated 963 01:12:12,310 --> 01:12:16,960 if you're used to think about expansions and flows in graphs. 964 01:12:16,960 --> 01:12:21,440 And this one really distills the core ideas of that proof, 965 01:12:21,440 --> 01:12:25,040 but looks something you can teach in half a lecture, 966 01:12:25,040 --> 01:12:28,332 whereas before this proof came about, 967 01:12:28,332 --> 01:12:30,290 I could have taught the proof, but most likely, 968 01:12:30,290 --> 01:12:31,546 I would have just skipped it. 969 01:12:35,830 --> 01:12:40,180 To just give you a sense of what's coming up ahead, so 970 01:12:40,180 --> 01:12:45,070 going forward, the first thing we'll do in the next lecture 971 01:12:45,070 --> 01:12:46,240 is we'll show-- 972 01:12:46,240 --> 01:12:50,890 we'll see the proof of the Freiman's theorem 973 01:12:50,890 --> 01:12:55,750 in the finite field setting, so in F2 to the n. 974 01:12:55,750 --> 01:12:58,645 There is one more thing, one more very quick lemma called 975 01:12:58,645 --> 01:13:00,520 the covering lemma, Ruzsa are Covering Lemma, 976 01:13:00,520 --> 01:13:01,522 that I will tell you. 977 01:13:01,522 --> 01:13:02,980 And then once we have that, then we 978 01:13:02,980 --> 01:13:07,620 can prove Freiman's theorem in the finite field setting. 979 01:13:07,620 --> 01:13:09,540 But then moving on to the integers, 980 01:13:09,540 --> 01:13:16,330 we'll need to understand how to think about the integers. 981 01:13:16,330 --> 01:13:21,150 Well, if you start with a subset of integers, they could, 982 01:13:21,150 --> 01:13:23,130 even if you have a small number of elements, 983 01:13:23,130 --> 01:13:25,970 they could be spread out, really, all over the place. 984 01:13:25,970 --> 01:13:28,950 But because you only care about the additive structure 985 01:13:28,950 --> 01:13:32,400 within the integers, you can try to model 986 01:13:32,400 --> 01:13:34,650 that very spread-out set of integers 987 01:13:34,650 --> 01:13:36,722 to something that is very compact. 988 01:13:36,722 --> 01:13:38,430 So there is something called the modeling 989 01:13:38,430 --> 01:13:42,638 lemma, Ruzsa's Modeling Lemma, that we'll see next time. 990 01:13:42,638 --> 01:13:44,430 And that will play a pretty important role. 991 01:13:47,760 --> 01:13:49,920 Before finishing off, I also want 992 01:13:49,920 --> 01:13:54,150 to mention that Freiman in his work, so he had this result. 993 01:13:54,150 --> 01:13:58,290 And he also wrote a book I think called The Structural Theory 994 01:13:58,290 --> 01:14:01,980 of Set Addition, or something like that, that 995 01:14:01,980 --> 01:14:04,350 emphasized this connection. 996 01:14:04,350 --> 01:14:07,200 He tried to draw this analogy sort of comparing 997 01:14:07,200 --> 01:14:12,390 additive combinatorics to geometry in the sense of cline, 998 01:14:12,390 --> 01:14:14,625 where in order to understand sets, 999 01:14:14,625 --> 01:14:15,750 you don't think about sets. 1000 01:14:15,750 --> 01:14:17,910 You think about maps between sets, 1001 01:14:17,910 --> 01:14:20,940 which was kind of an obscure idea at the time. 1002 01:14:20,940 --> 01:14:23,120 But we'll see next lecture that this actually 1003 01:14:23,120 --> 01:14:25,650 is a very powerful, it's a very influential idea 1004 01:14:25,650 --> 01:14:28,620 to really think about a sets of integers 1005 01:14:28,620 --> 01:14:30,410 under transformations that only preserve 1006 01:14:30,410 --> 01:14:32,800 their additive structure. 1007 01:14:32,800 --> 01:14:35,210 So we'll see this next time.