1 00:00:18,502 --> 00:00:20,460 YUFEI ZHAO: The goal for the next few lecturers 2 00:00:20,460 --> 00:00:23,905 is to prove Freiman's theorem, which we discussed last time. 3 00:00:23,905 --> 00:00:27,410 And so we started with this tool that we proved 4 00:00:27,410 --> 00:00:29,820 called the Plunnecke-Ruzsa inequality, 5 00:00:29,820 --> 00:00:33,540 which tells us that if you have a set A now in an arbitrary 6 00:00:33,540 --> 00:00:37,890 abelian group, and if A has controlled doubling, 7 00:00:37,890 --> 00:00:40,530 it has bounded doubling, then all 8 00:00:40,530 --> 00:00:47,160 the further iterated sumsets have bounded growth, as well. 9 00:00:47,160 --> 00:00:49,130 So this is what we proved last time. 10 00:00:49,130 --> 00:00:51,900 And so we're going to be using the Plunnecke-Ruzsa inequality 11 00:00:51,900 --> 00:00:53,970 many times. 12 00:00:53,970 --> 00:00:56,910 But there are some other tools I need to tell you about. 13 00:01:00,250 --> 00:01:03,830 So the next tool is known as Ruzsa covering lemma. 14 00:01:12,090 --> 00:01:12,660 All right. 15 00:01:12,660 --> 00:01:14,285 So let me first give you the statement, 16 00:01:14,285 --> 00:01:17,222 and then I explain some intuition. 17 00:01:17,222 --> 00:01:19,680 I think the technique is more important than the statement. 18 00:01:19,680 --> 00:01:22,560 But in any case, here's what it says. 19 00:01:22,560 --> 00:01:28,010 If you have X and B-- 20 00:01:28,010 --> 00:01:30,480 they're subsets of some arbitrary abelian group-- 21 00:01:38,370 --> 00:01:46,500 if you have the inequality that X plus B is at most K times 22 00:01:46,500 --> 00:01:49,320 the size of B-- so the size of X plus B is at most K times 23 00:01:49,320 --> 00:01:50,610 the size of B-- 24 00:01:50,610 --> 00:02:03,870 then there exists some subset T of X, 25 00:02:03,870 --> 00:02:08,430 with the size of T being, at most, K, 26 00:02:08,430 --> 00:02:20,250 such that X is contained in T plus B minus B. So 27 00:02:20,250 --> 00:02:22,200 that's the statement of Ruzsa covering lemma. 28 00:02:22,200 --> 00:02:25,710 So let me play to explain what it is about. 29 00:02:25,710 --> 00:02:32,250 So the idea here is that, if you're in a situation where 30 00:02:32,250 --> 00:02:34,860 if it looks like-- 31 00:02:37,780 --> 00:02:40,770 so you should think of B as a ball. 32 00:02:40,770 --> 00:02:44,180 So if it looks like X plus B might 33 00:02:44,180 --> 00:02:58,260 be coverable by K translates of B-- 34 00:02:58,260 --> 00:03:04,080 so here, you're supposed to think of B like a ball 35 00:03:04,080 --> 00:03:05,520 in the metric space-- 36 00:03:05,520 --> 00:03:13,020 then X is actually coverable. 37 00:03:16,490 --> 00:03:19,120 So if it looks like, meaning just by size 38 00:03:19,120 --> 00:03:24,320 alone, by size info alone-- 39 00:03:24,320 --> 00:03:26,800 so just based on the size, if it looks 40 00:03:26,800 --> 00:03:31,390 like X plus B might be coverable by K different translates of B, 41 00:03:31,390 --> 00:03:34,750 then actually X is coverable, but you 42 00:03:34,750 --> 00:03:45,960 have to use slightly larger balls by K copies of B minus B. 43 00:03:45,960 --> 00:03:49,050 So you should think of B minus B as slightly larger balls 44 00:03:49,050 --> 00:03:50,090 than B itself. 45 00:03:50,090 --> 00:03:52,710 So if B were an actual ball in Euclidean space, 46 00:03:52,710 --> 00:03:55,200 then B minus B is the same ball with twice the radius. 47 00:04:03,160 --> 00:04:07,530 And so Ruzsa covering lemma is a really important tool. 48 00:04:07,530 --> 00:04:09,000 The proof is not very long. 49 00:04:09,000 --> 00:04:12,010 And it's important to understand the idea of this proof. 50 00:04:12,010 --> 00:04:14,820 So this is a proof not just in additive combinatorics 51 00:04:14,820 --> 00:04:17,160 but something that happens-- 52 00:04:17,160 --> 00:04:19,550 it's a standard idea in analysis. 53 00:04:19,550 --> 00:04:22,560 So it's very important to understand this idea. 54 00:04:22,560 --> 00:04:24,350 And the key idea-- 55 00:04:24,350 --> 00:04:26,850 here, I think the proof is more important than the statement 56 00:04:26,850 --> 00:04:28,110 up there-- 57 00:04:28,110 --> 00:04:36,840 the key idea here is that if you want to produce a covering, 58 00:04:36,840 --> 00:04:40,635 one way to produce a covering is to take a maximal packing. 59 00:04:45,960 --> 00:04:55,640 A maximal packing using with balls, 60 00:04:55,640 --> 00:05:07,260 for instance, implies a covering with balls twice as large. 61 00:05:11,940 --> 00:05:14,910 So let me illustrate that with a picture. 62 00:05:14,910 --> 00:05:20,920 Suppose you have some space that I want to cover. 63 00:05:20,920 --> 00:05:24,290 But you get to use, let's say, unit balls. 64 00:05:24,290 --> 00:05:27,590 So how can I make sure I can cover 65 00:05:27,590 --> 00:05:29,000 the space using unit balls? 66 00:05:29,000 --> 00:05:32,060 And I don't want to use too many unit balls. 67 00:05:32,060 --> 00:05:45,260 So what you can do is, let me start by a maximal set of unit 68 00:05:45,260 --> 00:05:47,250 balls, so with centers-- 69 00:05:47,250 --> 00:05:53,060 so it's now half unit balls, so the radius is 1/2. 70 00:05:58,450 --> 00:06:03,170 So I put in as many as I can so that I cannot put in any more. 71 00:06:03,170 --> 00:06:05,080 So that's what maximal means-- 72 00:06:05,080 --> 00:06:07,360 "mal" here doesn't mean the maximum number. 73 00:06:07,360 --> 00:06:11,350 Although, if you put in the maximum number, that's also OK. 74 00:06:11,350 --> 00:06:14,560 But maximal means that I just cannot put in any more balls 75 00:06:14,560 --> 00:06:16,810 that are not overlapping. 76 00:06:16,810 --> 00:06:19,600 If I have this configuration, now what I do 77 00:06:19,600 --> 00:06:22,390 is I double the radii of all the balls. 78 00:06:28,710 --> 00:06:31,178 So whoever takes today's notes will have a fun time drawing 79 00:06:31,178 --> 00:06:31,720 this picture. 80 00:06:36,310 --> 00:06:38,935 So this has to be a covering of the original space. 81 00:06:41,670 --> 00:06:44,520 Because if you had missed some point, 82 00:06:44,520 --> 00:06:47,500 I could have put a blue ball in. 83 00:06:47,500 --> 00:06:48,161 Yes? 84 00:06:48,161 --> 00:06:51,065 AUDIENCE: What if a space has some narrow portion? 85 00:06:51,065 --> 00:06:53,690 YUFEI ZHAO: Question-- what if a space has some narrow portion? 86 00:06:53,690 --> 00:06:54,990 It doesn't matter. 87 00:06:54,990 --> 00:06:58,310 If you formulate this correctly-- 88 00:06:58,310 --> 00:07:02,770 if you miss some point-- so here, 89 00:07:02,770 --> 00:07:06,635 it depends on how you formulate this covering. 90 00:07:06,635 --> 00:07:15,130 The point is that, if you take a maximal set of points, 91 00:07:15,130 --> 00:07:18,730 when you expand, you have to cover the whole space. 92 00:07:18,730 --> 00:07:20,920 Because if you missed some point-- 93 00:07:20,920 --> 00:07:23,655 so imagine if you had-- 94 00:07:23,655 --> 00:07:25,030 for example, suppose you missed-- 95 00:07:28,700 --> 00:07:31,490 suppose you had missed some point over here. 96 00:07:31,490 --> 00:07:34,510 Then I could have put an extra ball in. 97 00:07:34,510 --> 00:07:38,070 So if you had missed some point over here, 98 00:07:38,070 --> 00:07:42,730 that means that I should have been able to put in that ball 99 00:07:42,730 --> 00:07:45,030 there initially. 100 00:07:45,030 --> 00:07:48,930 So this is a very simple idea, but a very powerful idea. 101 00:07:48,930 --> 00:07:52,200 And it comes up all the time in analysis and geometry. 102 00:07:52,200 --> 00:07:54,930 And it also comes up here. 103 00:07:54,930 --> 00:07:56,640 So let's do the actual proof. 104 00:08:01,210 --> 00:08:19,160 So let me let T be a subset of X be a maximal subset of X, 105 00:08:19,160 --> 00:08:25,520 such that the sets little t plus B are 106 00:08:25,520 --> 00:08:35,659 disjoint for all elements little t of a set big T. 107 00:08:35,659 --> 00:08:37,299 So it's like this picture. 108 00:08:37,299 --> 00:08:41,350 I pick a subset of X so that if I center balls 109 00:08:41,350 --> 00:08:45,700 around these t's, then these translates of B's are 110 00:08:45,700 --> 00:08:46,700 destroyed. 111 00:08:46,700 --> 00:08:50,990 Then you put in a maximal such set. 112 00:08:50,990 --> 00:08:58,040 So due to the disjointness, we find 113 00:08:58,040 --> 00:09:04,990 that the product of the sizes of t and B 114 00:09:04,990 --> 00:09:10,985 equals to the size of the sum, because there's no overlaps. 115 00:09:13,780 --> 00:09:19,130 But t plus B has size at most X plus B, 116 00:09:19,130 --> 00:09:23,750 because T is a subset of X. But also, from the assumption 117 00:09:23,750 --> 00:09:28,310 we knew that X is-- 118 00:09:28,310 --> 00:09:35,870 size of x plus B is upper bounded by the size B times K. 119 00:09:35,870 --> 00:09:40,790 So in particular, we get that the size of T 120 00:09:40,790 --> 00:09:47,480 is at most K. So in other words, over here, 121 00:09:47,480 --> 00:09:50,810 the number of blue balls you can control by simply the volume. 122 00:09:53,890 --> 00:10:13,350 Now, since T is maximal, we have that for every little x, 123 00:10:13,350 --> 00:10:24,960 there exists some little t such that the translate of B 124 00:10:24,960 --> 00:10:28,680 given by x intersects one of my children translates. 125 00:10:31,520 --> 00:10:34,360 Because if this were not true, then I 126 00:10:34,360 --> 00:10:41,000 could have put in an extra translate of B. 127 00:10:41,000 --> 00:10:48,800 So in other words, there exist two elements, b and b prime, 128 00:10:48,800 --> 00:10:54,110 such that t plus b equals to x plus b prime. 129 00:10:54,110 --> 00:11:09,040 And hence, x lies in little t plus B minus B, which 130 00:11:09,040 --> 00:11:20,600 implies that the set X lies in T plus B minus B. OK. 131 00:11:20,600 --> 00:11:23,120 So that's the Ruzsa covering lemma. 132 00:11:23,120 --> 00:11:25,520 So it's an execution of this idea 133 00:11:25,520 --> 00:11:30,080 I mentioned earlier that a maximal packing implies 134 00:11:30,080 --> 00:11:30,830 a good covering. 135 00:11:33,460 --> 00:11:34,464 Any questions? 136 00:11:36,960 --> 00:11:37,460 Right. 137 00:11:37,460 --> 00:11:41,390 So now we have this tool, we can prove an easier version 138 00:11:41,390 --> 00:11:45,230 of Freiman's theorem where, instead of working in integers, 139 00:11:45,230 --> 00:11:47,430 we're going to work in the finite field model. 140 00:11:47,430 --> 00:11:51,088 So usually, it's good to start with finite field models. 141 00:11:51,088 --> 00:11:52,130 Things are a bit cleaner. 142 00:11:52,130 --> 00:11:54,500 And in this case, it actually only requires 143 00:11:54,500 --> 00:11:57,290 a subset of the tools that we need for the full theorem. 144 00:12:00,880 --> 00:12:02,975 So instead of working in finite field model, 145 00:12:02,975 --> 00:12:05,350 we're going to be working in something just slightly more 146 00:12:05,350 --> 00:12:05,850 general. 147 00:12:05,850 --> 00:12:06,850 But it's the same proof. 148 00:12:10,270 --> 00:12:16,580 So we're going to be working in a group of bounded exponent. 149 00:12:16,580 --> 00:12:26,420 Freiman's theorem in groups of bounded exponent. 150 00:12:26,420 --> 00:12:32,850 And the word "exponent" in group theory means the following-- 151 00:12:32,850 --> 00:12:41,440 so the exponent of an abelian group 152 00:12:41,440 --> 00:12:52,820 is the smallest positive integer, if it exists. 153 00:12:57,490 --> 00:13:04,420 So the smallest positive integer r so that rx 154 00:13:04,420 --> 00:13:08,430 equals to 0 for all x in the group. 155 00:13:18,300 --> 00:13:22,580 So for example, if you are working Fp to the n, 156 00:13:22,580 --> 00:13:24,680 then the exponent is p. 157 00:13:24,680 --> 00:13:28,560 So every element has-- 158 00:13:28,560 --> 00:13:34,710 if you add to itself r times, then it vanishes the element. 159 00:13:37,920 --> 00:13:39,948 The word "exponent" comes from-- 160 00:13:39,948 --> 00:13:41,490 I mean, you can define the same thing 161 00:13:41,490 --> 00:13:43,032 for non-abelian groups, in which case 162 00:13:43,032 --> 00:13:46,830 you should write this expression using the exponent. 163 00:13:46,830 --> 00:13:49,892 Instead of addition, you have multiplication. 164 00:13:49,892 --> 00:13:51,350 So that's why it's called exponent. 165 00:13:51,350 --> 00:13:54,680 So the name is stuck, even though we're still working 166 00:13:54,680 --> 00:13:55,680 on the additive setting. 167 00:13:58,470 --> 00:14:03,450 We're going to write this, so the angle 168 00:14:03,450 --> 00:14:09,340 brackets, to mean the subgroup generated 169 00:14:09,340 --> 00:14:15,010 by the subset A, where A is a subset of the group elements. 170 00:14:15,010 --> 00:14:20,140 And so the exponent of a group, then, 171 00:14:20,140 --> 00:14:22,870 is equal to the maximum number of-- 172 00:14:26,170 --> 00:14:28,330 so if you pick a group element, look 173 00:14:28,330 --> 00:14:31,630 at how many group elements does it generate. 174 00:14:31,630 --> 00:14:32,980 So it's equal to the max. 175 00:14:38,280 --> 00:14:39,148 All right. 176 00:14:39,148 --> 00:14:40,940 I mean, the example you can have in mind is 177 00:14:40,940 --> 00:14:44,390 F2 to the n, which has exponent 2. 178 00:14:44,390 --> 00:14:46,830 So in general, we're going to be looking 179 00:14:46,830 --> 00:14:49,740 at a special case in a group with bounded exponent. 180 00:14:52,350 --> 00:14:54,480 And Freiman's theorem, in this case, 181 00:14:54,480 --> 00:15:01,260 is due to Ruzsa, who showed that if you 182 00:15:01,260 --> 00:15:21,470 have a finite subset in an abelian group with exponent r-- 183 00:15:21,470 --> 00:15:22,730 so a finite exponent-- 184 00:15:25,890 --> 00:15:33,422 if A has doubling constant at most K, then 185 00:15:33,422 --> 00:15:34,380 what do we want to say? 186 00:15:34,380 --> 00:15:36,338 We want to say, just like in Freiman's theorem, 187 00:15:36,338 --> 00:15:38,370 that if A has bounded doubling, then 188 00:15:38,370 --> 00:15:42,770 it is a large portion of some structured set. 189 00:15:42,770 --> 00:15:46,922 And here "structured set" means subgroup. 190 00:15:46,922 --> 00:15:49,130 Well, if you're going to be looking at subgroups that 191 00:15:49,130 --> 00:15:52,610 contain A, you might as well look at the subgroup generated 192 00:15:52,610 --> 00:15:58,220 by A. So the claim, then, is that the subgroup generated 193 00:15:58,220 --> 00:16:04,280 by A is only a constant factor bigger than A itself. 194 00:16:12,610 --> 00:16:14,410 So let me say it again. 195 00:16:14,410 --> 00:16:17,890 If you have a set A in a group of bounded exponent, 196 00:16:17,890 --> 00:16:21,970 and A has bounded doubling, then conclusion 197 00:16:21,970 --> 00:16:27,080 is that A is a large proportion of some subgroup. 198 00:16:27,080 --> 00:16:28,910 Conversely-- we saw last time-- 199 00:16:28,910 --> 00:16:31,400 if you take a subgroup, it has doubling constant 1, 200 00:16:31,400 --> 00:16:34,100 so if you take a positive proportional constant 201 00:16:34,100 --> 00:16:37,750 proportion of the subgroup, it also has bounded doubling. 202 00:16:37,750 --> 00:16:40,480 And this statement is in some sense 203 00:16:40,480 --> 00:16:44,570 the converse of that observation. 204 00:16:44,570 --> 00:16:47,240 This bound here is not optimal. 205 00:16:47,240 --> 00:16:48,920 I'll make some comments about what 206 00:16:48,920 --> 00:16:50,073 we know about these bounds. 207 00:16:50,073 --> 00:16:52,115 But for now, just view this number as a constant. 208 00:16:54,750 --> 00:16:56,540 Any questions about the statement? 209 00:16:59,830 --> 00:17:01,240 All right. 210 00:17:01,240 --> 00:17:07,349 So let's prove this Ruzsa's theorem, giving you 211 00:17:07,349 --> 00:17:10,740 Freiman's theorem in groups with bounded exponent. 212 00:17:10,740 --> 00:17:13,470 So we're going to be applying the tools we've seen so far, 213 00:17:13,470 --> 00:17:15,530 starting with Plunnecke-Ruzsa. 214 00:17:21,650 --> 00:17:25,770 So by Plunnecke-Ruzsa inequality, we find that-- 215 00:17:25,770 --> 00:17:27,770 so then I'm going to write down some expression. 216 00:17:27,770 --> 00:17:29,270 You may wonder, why this expression? 217 00:17:29,270 --> 00:17:33,190 Because we're going to apply covering lemma in a second. 218 00:17:33,190 --> 00:17:39,930 So this set by Plunnecke-Ruzsa, its size is bounded, 219 00:17:39,930 --> 00:17:43,560 because the set can also be written as 3A minus A, 220 00:17:43,560 --> 00:17:49,620 so its size is bounded by K to the fourth times the size of A. 221 00:17:49,620 --> 00:17:53,112 And so Plunnecke-Ruzsa is a very nice tool to have. 222 00:17:53,112 --> 00:17:55,320 It basically tells you, if you have bounded doubling, 223 00:17:55,320 --> 00:17:57,630 then all the other iterated sums are essentially 224 00:17:57,630 --> 00:17:58,848 controlled in size. 225 00:17:58,848 --> 00:18:00,140 And then we're using that here. 226 00:18:03,280 --> 00:18:03,780 OK. 227 00:18:03,780 --> 00:18:05,447 So now we're in the setting where we can 228 00:18:05,447 --> 00:18:08,520 apply the Ruzsa covering lemma. 229 00:18:08,520 --> 00:18:14,642 So by covering lemma, we're going 230 00:18:14,642 --> 00:18:15,850 to apply the covering lemma-- 231 00:18:15,850 --> 00:18:17,990 so using the notation earlier-- 232 00:18:17,990 --> 00:18:29,430 with X equal to 2A minus A and B equal to A. By covering lemma, 233 00:18:29,430 --> 00:18:39,858 there exists some T being a subset of 2A minus A, with T 234 00:18:39,858 --> 00:18:42,450 not too large-- 235 00:18:42,450 --> 00:18:46,580 because of our earlier estimate-- 236 00:18:46,580 --> 00:18:58,910 and such that 2A minus A is contained in T plus A minus A. 237 00:18:58,910 --> 00:19:01,650 So it's easy to get lost in the details. 238 00:19:01,650 --> 00:19:04,830 But what's happening here is that I start with A, 239 00:19:04,830 --> 00:19:10,620 and I'm looking at how big these iterated growing 240 00:19:10,620 --> 00:19:12,880 sumsets can be. 241 00:19:12,880 --> 00:19:15,780 And if I keep on doing this operation, 242 00:19:15,780 --> 00:19:18,150 like if I just apply Plunnecke-Ruzsa, 243 00:19:18,150 --> 00:19:22,380 I can't really control the iterated sums 244 00:19:22,380 --> 00:19:23,880 by an absolute bound. 245 00:19:23,880 --> 00:19:28,930 This bound will keep growing if I take bigger iterations. 246 00:19:28,930 --> 00:19:30,660 But Ruzsa covering lemma gives me 247 00:19:30,660 --> 00:19:34,530 another way to control the bounded iterations. 248 00:19:34,530 --> 00:19:36,800 It says there is some bounded set 249 00:19:36,800 --> 00:19:42,880 T such that this iterated sum is nicely controlled. 250 00:19:42,880 --> 00:19:46,410 So let me iterate this down even further. 251 00:19:46,410 --> 00:19:48,410 We're going to iterate the containment 252 00:19:48,410 --> 00:19:58,360 by adding A to both sides. 253 00:19:58,360 --> 00:20:05,440 And we obtain that 3A minus A is contained in T plus 2A 254 00:20:05,440 --> 00:20:14,370 minus A. But now, 2A minus A was containing T plus A minus A. 255 00:20:14,370 --> 00:20:21,340 So we get 2T plus A minus A. 256 00:20:21,340 --> 00:20:27,150 So now we've gained quite a bit, because we 257 00:20:27,150 --> 00:20:31,350 can make this iteration go up, but not 258 00:20:31,350 --> 00:20:36,270 at the cost of iterating A but at the cost of iterating T. 259 00:20:36,270 --> 00:20:39,060 But T is a bounded size. 260 00:20:39,060 --> 00:20:44,720 T is bounded size, so T will have very nice control. 261 00:20:44,720 --> 00:20:48,210 We'll be able to very nicely control the iterations of T. 262 00:20:48,210 --> 00:20:57,570 So if we keep going, we find that for all integer n, 263 00:20:57,570 --> 00:21:02,780 positive integer n, n plus 1 A minus A 264 00:21:02,780 --> 00:21:11,070 is contained in nT plus A minus A. But for every integer n, 265 00:21:11,070 --> 00:21:14,640 the iterated sums of T are contained 266 00:21:14,640 --> 00:21:27,740 in the subgroup generated by T. So therefore, if we take n 267 00:21:27,740 --> 00:21:32,270 to be as large as you want, see that the left-hand side 268 00:21:32,270 --> 00:21:36,670 eventually becomes the subgroup generated by A, 269 00:21:36,670 --> 00:21:39,710 and the right hand side does not depend on n. 270 00:21:42,590 --> 00:21:46,640 So you have this containment over here. 271 00:21:46,640 --> 00:21:50,580 We would like to estimate how large the subgroup generated 272 00:21:50,580 --> 00:21:52,450 by A is. 273 00:21:52,450 --> 00:21:55,100 So we look at that formula, and we 274 00:21:55,100 --> 00:22:00,320 see that the size of the subgroup generated by T-- 275 00:22:00,320 --> 00:22:01,940 and here is where we're going to use 276 00:22:01,940 --> 00:22:04,760 the fact that the assumption that a group has bounded 277 00:22:04,760 --> 00:22:05,840 exponent. 278 00:22:05,840 --> 00:22:07,420 So think F2 to the n, if you will. 279 00:22:07,420 --> 00:22:11,840 So in F2 to the n, if I give you a set T, 280 00:22:11,840 --> 00:22:15,350 what's the subspace spanned by T? 281 00:22:15,350 --> 00:22:18,390 What's the maximum possible size? 282 00:22:18,390 --> 00:22:22,520 It's, at most, 2 raised to the number. 283 00:22:22,520 --> 00:22:25,630 So in general, you also have that the subgroup 284 00:22:25,630 --> 00:22:31,060 generated by T has size at most r raised to the size of T 285 00:22:31,060 --> 00:22:36,130 in a group of exponent r, which we can control, 286 00:22:36,130 --> 00:22:39,370 because T has bounded size. 287 00:22:42,740 --> 00:22:48,650 And the second term, A minus A, so we also 288 00:22:48,650 --> 00:22:51,192 know how to control that by Plunnecke-Ruzsa. 289 00:23:00,850 --> 00:23:07,660 Therefore, putting these two together, 290 00:23:07,660 --> 00:23:16,080 we see that the size up here is at most r to the K 291 00:23:16,080 --> 00:23:22,002 to the 4 times K squared size of A, which 292 00:23:22,002 --> 00:23:23,210 is the bound that we claimed. 293 00:23:28,320 --> 00:23:29,425 Any questions? 294 00:23:32,040 --> 00:23:36,640 So the trick here is to only use Plunnecke-Ruzsa somehow 295 00:23:36,640 --> 00:23:38,110 a bounded number of times. 296 00:23:38,110 --> 00:23:43,000 If you use it too many times, this bound blows up. 297 00:23:43,000 --> 00:23:45,250 But you use it only a small number of times, 298 00:23:45,250 --> 00:23:49,590 and then you use the Ruzsa covering lemma 299 00:23:49,590 --> 00:23:53,923 so that you get this bounded set T that I can iterate. 300 00:23:57,630 --> 00:24:00,090 Let me make some comments about the bounds that 301 00:24:00,090 --> 00:24:01,330 come out of this proof. 302 00:24:01,330 --> 00:24:05,850 So you get this very clean bound, following this proof. 303 00:24:05,850 --> 00:24:07,870 But what about examples? 304 00:24:07,870 --> 00:24:09,390 So what kinds of examples can you 305 00:24:09,390 --> 00:24:13,560 think of where you have a set of bounded doubling, 306 00:24:13,560 --> 00:24:17,370 but the set generated, the group generated by that set, 307 00:24:17,370 --> 00:24:20,100 is potentially large? 308 00:24:20,100 --> 00:24:34,570 So even in F2 to the n, so if A is an independent set-- 309 00:24:38,146 --> 00:24:41,360 so a basis or a subset of a basis, for instance-- 310 00:24:41,360 --> 00:24:47,820 then K-- so A is independent set, so all the pairwise sums 311 00:24:47,820 --> 00:24:54,346 are distinct-- so K is about the size of A/2. 312 00:24:54,346 --> 00:24:56,370 That's the doubling constant. 313 00:24:56,370 --> 00:25:02,710 Whereas, the group generated by A 314 00:25:02,710 --> 00:25:11,720 has size 2 to the size of A, which is around 2 to the 2K 315 00:25:11,720 --> 00:25:19,500 times the size of A. OK. 316 00:25:19,500 --> 00:25:24,210 So you see that you do need some exponential blow-up from K 317 00:25:24,210 --> 00:25:27,810 to this constant over here. 318 00:25:27,810 --> 00:25:29,910 And turns out, that's more or less correct. 319 00:25:29,910 --> 00:25:41,540 So the optimal constant for F2 to the n 320 00:25:41,540 --> 00:25:44,060 is now known very precisely. 321 00:25:47,800 --> 00:25:53,310 And so if you give me a real value of K, then I can tell you 322 00:25:53,310 --> 00:25:54,960 there are some recent results that 323 00:25:54,960 --> 00:25:57,690 tells you exactly what is the optimal constant you 324 00:25:57,690 --> 00:26:00,210 can put in front of the A. So very precise. 325 00:26:00,210 --> 00:26:09,700 But asymptotically, it looks like 2 to the 2K divided by. 326 00:26:09,700 --> 00:26:15,680 K So that's what it looks like. 327 00:26:15,680 --> 00:26:19,650 So this example is basically correct. 328 00:26:19,650 --> 00:26:26,400 For general r, we expect a similar phenomenon. 329 00:26:26,400 --> 00:26:31,635 So Ruzsa conjectured that-- 330 00:26:31,635 --> 00:26:35,790 in the hypothesis of the theorem, 331 00:26:35,790 --> 00:26:37,680 the constant you can take is only 332 00:26:37,680 --> 00:26:42,930 exponential in K, the r to the some constant C times K. 333 00:26:42,930 --> 00:26:48,840 And it has been verified for some values of r, 334 00:26:48,840 --> 00:26:49,750 but not in general-- 335 00:26:53,430 --> 00:26:55,215 for some r, for example primes. 336 00:27:02,625 --> 00:27:03,590 OK. 337 00:27:03,590 --> 00:27:04,583 Any questions? 338 00:27:09,650 --> 00:27:10,150 All right. 339 00:27:10,150 --> 00:27:13,420 So this is a good milestone. 340 00:27:13,420 --> 00:27:15,415 So we've developed some tools, and we 341 00:27:15,415 --> 00:27:18,940 were able to prove a easier version of Freiman's theorem 342 00:27:18,940 --> 00:27:22,630 in a group of bounded exponent. 343 00:27:22,630 --> 00:27:24,670 And you can ask yourself, does this proof 344 00:27:24,670 --> 00:27:27,000 work in the integers? 345 00:27:27,000 --> 00:27:28,473 And well, literally no. 346 00:27:28,473 --> 00:27:29,890 Because if you look at this proof, 347 00:27:29,890 --> 00:27:34,660 this set here is infinite, unlike in the finite field 348 00:27:34,660 --> 00:27:35,760 setting. 349 00:27:35,760 --> 00:27:38,860 In the integers, well, that's not very good. 350 00:27:38,860 --> 00:27:42,910 So the strategy of Freiman's theorem, 351 00:27:42,910 --> 00:27:46,630 the proof of Freiman's theorem, is to start with the integers, 352 00:27:46,630 --> 00:27:50,080 and then try to, not work in the integers, 353 00:27:50,080 --> 00:27:52,840 but try to work in a smaller group. 354 00:27:52,840 --> 00:27:55,690 Even though you start in a maybe very spread 355 00:27:55,690 --> 00:27:59,260 out set of integers, I want to work in a much smaller group 356 00:27:59,260 --> 00:28:03,390 so that I can control things within that group. 357 00:28:03,390 --> 00:28:05,460 And this is an idea called modeling. 358 00:28:05,460 --> 00:28:07,400 So I have a big set and want to model it 359 00:28:07,400 --> 00:28:11,480 by something in a small group. 360 00:28:11,480 --> 00:28:12,890 So we're going to see this idea. 361 00:28:12,890 --> 00:28:16,040 But to understand what does it mean 362 00:28:16,040 --> 00:28:18,470 to have a good model for a set in the sense 363 00:28:18,470 --> 00:28:20,240 of additive combinatorics, I need 364 00:28:20,240 --> 00:28:22,910 to introduce the notion of Freiman homomorphisms. 365 00:28:40,120 --> 00:28:44,120 So one of the central philosophies across mathematics 366 00:28:44,120 --> 00:28:46,610 is that if you want to study objects, 367 00:28:46,610 --> 00:28:49,280 you should try to understand maps between objects 368 00:28:49,280 --> 00:28:54,650 and understand properties that are preserved under those maps. 369 00:28:54,650 --> 00:28:56,480 So if you want to study groups, I 370 00:28:56,480 --> 00:28:58,740 don't really care how you label your group elements-- 371 00:28:58,740 --> 00:29:01,040 by 1, 2, 3, or A, B, C. What I care 372 00:29:01,040 --> 00:29:02,690 about is the relationships. 373 00:29:02,690 --> 00:29:04,830 And those are the data that I care about. 374 00:29:04,830 --> 00:29:06,680 And then, of course, then you have 375 00:29:06,680 --> 00:29:08,720 concepts like group homomorphisms, 376 00:29:08,720 --> 00:29:13,950 group isomorphisms, that preserve all the relevant data. 377 00:29:13,950 --> 00:29:15,630 Similarly in any other area-- 378 00:29:15,630 --> 00:29:17,240 in geometry you have manifolds. 379 00:29:17,240 --> 00:29:20,810 You understand not specifically how they embed into space 380 00:29:20,810 --> 00:29:22,760 but what are the intrinsic properties. 381 00:29:22,760 --> 00:29:24,290 So we would like to understand what 382 00:29:24,290 --> 00:29:28,280 are the intrinsic properties of a subset of an abelian group 383 00:29:28,280 --> 00:29:29,960 that we care about for the purpose 384 00:29:29,960 --> 00:29:32,360 of additive combinatorics, and specifically 385 00:29:32,360 --> 00:29:34,860 for Friedman's theorem. 386 00:29:34,860 --> 00:29:38,390 And what we care about is what kinds of additive relationships 387 00:29:38,390 --> 00:29:40,160 are preserved. 388 00:29:40,160 --> 00:29:45,120 And Freiman's homomorphisms capture that notion. 389 00:29:45,120 --> 00:29:46,880 So roughly speaking, we would like 390 00:29:46,880 --> 00:29:52,810 to understand maps between sets in possibly different groups-- 391 00:29:57,500 --> 00:29:59,720 in possibly different abelian groups-- 392 00:30:03,390 --> 00:30:06,897 that only partially preserve additive structure. 393 00:30:17,850 --> 00:30:21,840 So here's a definition. 394 00:30:21,840 --> 00:30:32,700 Suppose we have A and B, and they're 395 00:30:32,700 --> 00:30:37,530 subsets in possibly different abelian groups. 396 00:30:40,180 --> 00:30:42,740 Could be the same, but possibly different. 397 00:30:42,740 --> 00:30:45,970 And everything's written under addition, as usual. 398 00:30:45,970 --> 00:30:55,500 So we say that a map phi from A to B 399 00:30:55,500 --> 00:31:03,010 is a Freiman s-homomorphism. 400 00:31:08,610 --> 00:31:09,620 So that's the term-- 401 00:31:09,620 --> 00:31:13,750 Freiman s-homomorphism, sometimes also Freiman 402 00:31:13,750 --> 00:31:20,380 homomorphisms of order s, so equivalently I can call it 403 00:31:20,380 --> 00:31:22,000 that, as well-- 404 00:31:22,000 --> 00:31:24,400 if the following holds. 405 00:31:24,400 --> 00:31:30,540 If we have the equation phi of A plus dot, 406 00:31:30,540 --> 00:31:37,870 dot, dot plus phi of a sub s equal to phi of a prime 1 407 00:31:37,870 --> 00:31:49,160 plus dot, dot, dot plus phi of a prime s, whenever 408 00:31:49,160 --> 00:31:54,750 a through as, a prime through as prime, so 409 00:31:54,750 --> 00:32:04,770 a1 prime through as prime, satisfy the equation 410 00:32:04,770 --> 00:32:10,650 a1 plus dot, dot, dot plus as equal to a1 prime plus dot, 411 00:32:10,650 --> 00:32:13,010 dot, dot plus as prime. 412 00:32:16,160 --> 00:32:17,330 OK. 413 00:32:17,330 --> 00:32:22,680 So that's the definition of a Freiman s-homomorphism. 414 00:32:22,680 --> 00:32:27,110 It should remind you of the definition of a group 415 00:32:27,110 --> 00:32:32,360 homomorphism, which completely preserves additive structure, 416 00:32:32,360 --> 00:32:35,030 let's say, between abelian groups. 417 00:32:35,030 --> 00:32:36,740 And for Freiman homomorphisms, I'm 418 00:32:36,740 --> 00:32:40,292 only asking you to partially preserve additive structure. 419 00:32:44,210 --> 00:32:46,670 So the point here is that if I only care about, 420 00:32:46,670 --> 00:32:49,070 let's say, pairwise sums, if that's 421 00:32:49,070 --> 00:32:52,720 the only data I care about, then Freiman homomorphisms 422 00:32:52,720 --> 00:32:53,660 preserve that data. 423 00:33:01,830 --> 00:33:03,090 To give you some-- 424 00:33:03,090 --> 00:33:04,580 OK, so one more thing. 425 00:33:08,720 --> 00:33:19,180 If phi from A to B is, furthermore, a bijection, 426 00:33:19,180 --> 00:33:32,800 and both phi and phi inverse are Freiman s-homomorphisms, 427 00:33:32,800 --> 00:33:39,940 then we say that phi is a Freiman s-isomorphism. 428 00:33:48,850 --> 00:33:52,000 So it's not enough just to be a bijection, but it's a bijection 429 00:33:52,000 --> 00:33:54,710 and both the forward and the inverse 430 00:33:54,710 --> 00:33:59,970 maps are Freiman homomorphisms. 431 00:33:59,970 --> 00:34:03,240 So these are the definitions we're going to use. 432 00:34:03,240 --> 00:34:04,490 Let me give you some examples. 433 00:34:09,510 --> 00:34:17,385 So every group homomorphism is a Freiman homomorphism 434 00:34:17,385 --> 00:34:18,010 of every order. 435 00:34:24,677 --> 00:34:28,320 So group homomorphisms preserve all additive structure, 436 00:34:28,320 --> 00:34:30,880 and Freiman homomorphisms only partially 437 00:34:30,880 --> 00:34:34,500 preserve additive structure. 438 00:34:34,500 --> 00:34:40,560 A composition-- so if phi 1 and phi 2 439 00:34:40,560 --> 00:34:49,020 are Freiman s-homomorphisms, then phi 1 composed with phi 2 440 00:34:49,020 --> 00:34:53,764 is a Freiman s-homomorphism. 441 00:34:53,764 --> 00:34:56,900 So compositions preserve this property. 442 00:34:56,900 --> 00:35:01,190 And likewise, instead of homomorphisms, 443 00:35:01,190 --> 00:35:06,110 if you have isomorphisms, then that's also true, as well. 444 00:35:06,110 --> 00:35:09,920 So these are straightforward things to check. 445 00:35:09,920 --> 00:35:14,230 So a concrete example that shows you a difference between group 446 00:35:14,230 --> 00:35:17,760 homomorphisms and Freiman homomorphisms 447 00:35:17,760 --> 00:35:23,720 is, suppose you take an arbitrary map 448 00:35:23,720 --> 00:35:29,270 phi from a set that has no additive structure. 449 00:35:29,270 --> 00:35:35,570 So it's a four-element set, has no additive structure. 450 00:35:35,570 --> 00:35:39,590 And I map it to the integers, claim 451 00:35:39,590 --> 00:35:43,850 that this is a Freiman 2-homomorphism. 452 00:35:49,790 --> 00:35:50,620 So you can check. 453 00:35:50,620 --> 00:35:54,010 So whenever this is satisfied, but that's 454 00:35:54,010 --> 00:35:56,730 never non-trivially satisfied. 455 00:35:56,730 --> 00:36:00,078 So an arbitrary map here is a Freiman 2-homomorphism. 456 00:36:03,070 --> 00:36:08,560 And if furthermore-- so if you have, 457 00:36:08,560 --> 00:36:13,670 let's say, bijection between two sets, 458 00:36:13,670 --> 00:36:27,770 both having no additive structure, if it's a bijection, 459 00:36:27,770 --> 00:36:32,390 it's a Freiman isomorphism of here, order 2. 460 00:36:36,960 --> 00:36:38,460 Let me give you a few more examples. 461 00:36:53,450 --> 00:36:58,190 When you look at homomorphisms between finite groups, 462 00:36:58,190 --> 00:37:01,670 so you know that if you have a homomorphism 463 00:37:01,670 --> 00:37:08,470 and it's also a bijection, then it's an isomorphism. 464 00:37:08,470 --> 00:37:13,450 But that's not true for this notion of homomorphisms. 465 00:37:25,190 --> 00:37:34,110 So the natural embedding that sends the Boolean cube 466 00:37:34,110 --> 00:37:39,590 to the Boolean cube viewed as Z mod 2 to the n. 467 00:37:45,990 --> 00:37:47,700 So what's happening here? 468 00:37:50,790 --> 00:37:53,600 This is a part of a group homomorphism. 469 00:37:53,600 --> 00:37:56,870 And so if you look at Z to the n, and I do mod 2, 470 00:37:56,870 --> 00:38:00,650 and I restrict to this Boolean cube, 471 00:38:00,650 --> 00:38:03,950 that's the group homomorphism If I view this 472 00:38:03,950 --> 00:38:05,870 as a subset of a bigger group. 473 00:38:05,870 --> 00:38:11,150 So it is a Freiman homomorphisms of every order. 474 00:38:15,350 --> 00:38:17,510 And it's bijective. 475 00:38:17,510 --> 00:38:22,730 But it is not a Freiman 2-isomorphism. 476 00:38:27,570 --> 00:38:30,240 Because you have additive relations here 477 00:38:30,240 --> 00:38:32,890 that are not present over here. 478 00:38:32,890 --> 00:38:36,000 So if you read the definition, the inverse map, 479 00:38:36,000 --> 00:38:37,950 there are some additive relations here 480 00:38:37,950 --> 00:38:39,660 that are not preserved if you pull back. 481 00:38:45,310 --> 00:38:46,960 Here's another example that will be 482 00:38:46,960 --> 00:38:49,330 more relevant to our subsequent discussion. 483 00:38:53,270 --> 00:39:01,170 So the mod N map, which sends Z to Z mod N, so this is a group 484 00:39:01,170 --> 00:39:04,060 homomorphisms. 485 00:39:04,060 --> 00:39:09,580 So hence, it's a Freiman homomorphism of every order. 486 00:39:09,580 --> 00:39:11,840 But it's not-- 487 00:39:11,840 --> 00:39:20,050 OK, so if you look at this map, and even if I restrict to here, 488 00:39:20,050 --> 00:39:26,720 so it's not a Freiman isomorphism just like earlier. 489 00:39:26,720 --> 00:39:39,560 However-- let me go back to Z. So 490 00:39:39,560 --> 00:39:51,000 if A is a subset of integers with diameter less than N/s, 491 00:39:51,000 --> 00:40:05,900 then this map mod N maps A Freiman s-isomorphically 492 00:40:05,900 --> 00:40:06,770 onto its image. 493 00:40:12,810 --> 00:40:16,310 So even though mod N restricted to 1 through N 494 00:40:16,310 --> 00:40:21,170 is not a Freiman isomorphism of order 2, 495 00:40:21,170 --> 00:40:26,130 if I restrict to a subset that's, let's say, 496 00:40:26,130 --> 00:40:29,130 contained in some small interval, 497 00:40:29,130 --> 00:40:34,340 then all the additive structures are preserved. 498 00:40:34,340 --> 00:40:35,420 So let me show you why. 499 00:40:35,420 --> 00:40:37,910 So this is not too hard once you get your head 500 00:40:37,910 --> 00:40:40,740 around the definition. 501 00:40:40,740 --> 00:40:52,250 So indeed, if you have group elements a1 through as, 502 00:40:52,250 --> 00:41:02,777 a prime 1 through a prime s, and if they satisfy the equation-- 503 00:41:13,990 --> 00:41:15,490 so if they satisfy this equations, 504 00:41:15,490 --> 00:41:20,410 so we're trying to verify that it is a Freiman s-isomorphism, 505 00:41:20,410 --> 00:41:24,350 namely the inverse of this map is a Freiman s-homomorphism, 506 00:41:24,350 --> 00:41:27,920 so if they satisfy this equation, 507 00:41:27,920 --> 00:41:32,050 so this is satisfying this additive relation in the image, 508 00:41:32,050 --> 00:41:40,560 in Z mod N, then note that the left-hand side-- 509 00:41:40,560 --> 00:41:43,410 so all of these A's are contained in a small interval, 510 00:41:43,410 --> 00:41:46,690 because the diameter of the set is less than N/s. 511 00:41:46,690 --> 00:41:52,950 So if you look at how big a1 minus a1 prime can be it's, 512 00:41:52,950 --> 00:41:56,820 at most-- or, it's less than N/s in size. 513 00:41:56,820 --> 00:41:58,970 So the left-hand side, in absolute value, 514 00:41:58,970 --> 00:41:59,910 viewed as an integer-- 515 00:42:03,660 --> 00:42:12,510 so the left-hand side is less than N in absolute value, 516 00:42:12,510 --> 00:42:18,940 since the diameter of a is less than N/s. 517 00:42:18,940 --> 00:42:22,945 So you have some number here, which is strictly less than N 518 00:42:22,945 --> 00:42:27,850 in absolute value, and it's 0 mod N, 519 00:42:27,850 --> 00:42:35,230 so it must be actually equal to 0 as a number as an integer. 520 00:42:35,230 --> 00:42:38,790 So this verifies that the additive relations 521 00:42:38,790 --> 00:42:43,590 up to s-wise sums are preserved under the mod N map, 522 00:42:43,590 --> 00:42:46,590 if you restrict to a small interval. 523 00:42:54,715 --> 00:42:55,590 Any questions so far? 524 00:42:58,200 --> 00:43:00,970 So in additive combinatorics, we are 525 00:43:00,970 --> 00:43:05,980 trying to understand properties, specific additive properties. 526 00:43:05,980 --> 00:43:08,560 And the notion of Freiman homomorphisms 527 00:43:08,560 --> 00:43:12,490 Freiman isomorphism capture what specific properties we 528 00:43:12,490 --> 00:43:14,530 need to study and what are the maps that 529 00:43:14,530 --> 00:43:17,620 preserve those properties. 530 00:43:17,620 --> 00:43:21,280 And the next thing we will do is to understand this model lemma, 531 00:43:21,280 --> 00:43:22,990 the modeling lemma, that tells us 532 00:43:22,990 --> 00:43:27,620 that if you start with a set A with small doubling, initially 533 00:43:27,620 --> 00:43:30,550 A may be very much spread out in the integers-- it may 534 00:43:30,550 --> 00:43:33,140 have very large elements, very small elements, very spread 535 00:43:33,140 --> 00:43:33,640 out. 536 00:43:33,640 --> 00:43:36,040 But if A small doubling properties, 537 00:43:36,040 --> 00:43:42,880 then I can model A inside a small group, such 538 00:43:42,880 --> 00:43:46,860 that all the relevant data, namely 539 00:43:46,860 --> 00:43:49,170 relative to these Freiman homomorphisms, 540 00:43:49,170 --> 00:43:50,780 are preserved under this model. 541 00:43:53,550 --> 00:43:55,200 So let's move on to the modeling lemma. 542 00:44:06,180 --> 00:44:09,540 The main message of the modeling lemma 543 00:44:09,540 --> 00:44:26,440 is that if A has small doubling, then A can be modeled-- 544 00:44:26,440 --> 00:44:30,700 and here, that means being Freiman isomorphic-- 545 00:44:30,700 --> 00:44:37,400 to a subset of a small group. 546 00:44:43,990 --> 00:44:49,100 So first as a warm-up, let's work in the finite field model, 547 00:44:49,100 --> 00:44:51,610 just to see what such a result looks like. 548 00:44:51,610 --> 00:44:53,420 And it contains most of the ideas, 549 00:44:53,420 --> 00:44:54,710 but it's much more clean. 550 00:44:54,710 --> 00:44:57,230 It's much cleaner than in the integers. 551 00:44:57,230 --> 00:45:06,382 So in the finite field model, specifically F2 to the n, 552 00:45:06,382 --> 00:45:07,340 what do we want to say? 553 00:45:11,720 --> 00:45:16,340 Suppose you have A, a subset of F2 to the n, 554 00:45:16,340 --> 00:45:23,000 and suppose that m is some number such that 2 to the m 555 00:45:23,000 --> 00:45:28,580 is at least as large as sA minus sA. 556 00:45:28,580 --> 00:45:31,040 So remember, from Plunnecke-Ruzsa, 557 00:45:31,040 --> 00:45:34,380 you know that if A plus A is small, 558 00:45:34,380 --> 00:45:35,990 then this iterated sum is small. 559 00:45:38,500 --> 00:45:42,780 So suppose we have these parameters and sets. 560 00:45:42,780 --> 00:45:50,770 The conclusion is that A is Freiman 561 00:45:50,770 --> 00:46:06,900 s-isomorphic to some subset of F2 raised to m. 562 00:46:06,900 --> 00:46:10,650 So initially, A is in a potentially very large vector 563 00:46:10,650 --> 00:46:13,960 space, or it could be all over the place. 564 00:46:13,960 --> 00:46:16,650 And what we are trying to say here 565 00:46:16,650 --> 00:46:31,740 is that if A has small doubling, then by Plunnecke-Ruzsa, sA 566 00:46:31,740 --> 00:46:36,130 minus sA has size not too much bigger than A itself-- only 567 00:46:36,130 --> 00:46:38,640 bounded times the size of A itself. 568 00:46:38,640 --> 00:46:42,840 So I can take an m so that the size of this group 569 00:46:42,840 --> 00:46:45,600 is only a constant factor in larger 570 00:46:45,600 --> 00:46:46,950 than the size of A itself. 571 00:46:50,780 --> 00:46:52,460 So we're in a pretty small group. 572 00:46:52,460 --> 00:46:56,620 So we are able to model A, even though initially it 573 00:46:56,620 --> 00:46:58,430 sits inside a pretty large abelian group, 574 00:46:58,430 --> 00:47:01,270 by some subset in a pretty small group. 575 00:47:03,593 --> 00:47:05,510 So let's see how to prove this modeling lemma. 576 00:47:10,450 --> 00:47:18,290 So the finite field setting is much easier to-- 577 00:47:18,290 --> 00:47:21,200 it's not too hard to deal with, because we 578 00:47:21,200 --> 00:47:22,940 can look at linear maps. 579 00:47:22,940 --> 00:47:33,960 So the following are equivalent for linear maps phi, 580 00:47:33,960 --> 00:47:37,610 so for group homomorphisms. 581 00:47:37,610 --> 00:47:43,240 So phi is a Freiman s-isomorphism 582 00:47:43,240 --> 00:47:49,770 when restricted to A. So here, phi 583 00:47:49,770 --> 00:47:56,450 I'm going to let it be a linear map from F2 to the n to F2 584 00:47:56,450 --> 00:47:58,030 to the m. 585 00:47:58,030 --> 00:48:02,710 The following are equivalent. 586 00:48:02,710 --> 00:48:07,020 So we would like phi to be a Freiman s-isomorphism when 587 00:48:07,020 --> 00:48:10,830 restricted to A. Because this means that when I restrict to A 588 00:48:10,830 --> 00:48:14,670 and I restrict to its image, it Freiman isomorphically maps 589 00:48:14,670 --> 00:48:17,312 onto the image. 590 00:48:17,312 --> 00:48:18,270 So what does that mean? 591 00:48:22,060 --> 00:48:23,830 So phi is already a homomorphism. 592 00:48:23,830 --> 00:48:27,310 So it's automatically a Freiman s-homomorphism. 593 00:48:27,310 --> 00:48:30,490 For it to be an s-isomorphism, it just 594 00:48:30,490 --> 00:48:33,610 means that there are no additional linear relations 595 00:48:33,610 --> 00:48:38,660 in the image that were not present earlier, which means 596 00:48:38,660 --> 00:48:45,390 that phi is injective on sA. 597 00:48:48,600 --> 00:48:51,490 So let's just think about the definition. 598 00:48:51,490 --> 00:48:54,715 And in the definition, if you know additionally 599 00:48:54,715 --> 00:48:59,230 that phi is a homomorphism, everything's much cleaner. 600 00:48:59,230 --> 00:49:02,740 It is also equivalent to that phi 601 00:49:02,740 --> 00:49:10,180 of x is non-zero for every non-zero element 602 00:49:10,180 --> 00:49:13,210 x of sA minus sA. 603 00:49:17,470 --> 00:49:19,780 So this is a very clean characterization 604 00:49:19,780 --> 00:49:22,810 of what it means to be in Freiman s-isomorphism 605 00:49:22,810 --> 00:49:25,810 when you are in an abelian group and you have 606 00:49:25,810 --> 00:49:28,750 linear maps of homomorphisms. 607 00:49:28,750 --> 00:49:33,520 So if we start by taking phi to be 608 00:49:33,520 --> 00:49:41,560 a uniformly random linear map-- 609 00:49:46,920 --> 00:49:49,420 so for example, you pick a basis, 610 00:49:49,420 --> 00:49:52,420 and you send each basis element to a uniformly random element-- 611 00:49:55,720 --> 00:50:04,920 then we find that if 2m is at least A at minus sA, then-- 612 00:50:04,920 --> 00:50:08,150 so let me call these properties 1, 2, 3-- 613 00:50:08,150 --> 00:50:17,010 so then the probability that 3 is satisfied is positive. 614 00:50:17,010 --> 00:50:22,610 Because each element of sA minus sA-- 615 00:50:22,610 --> 00:50:28,060 I can also ignore 0-- so each non-zero element of sA minus sA 616 00:50:28,060 --> 00:50:32,260 violates this property with probability exactly 2 617 00:50:32,260 --> 00:50:36,930 to the minus m, everything is uniform. 618 00:50:36,930 --> 00:50:39,820 So if there are very few elements, 619 00:50:39,820 --> 00:50:44,620 and the space is large enough, then the third bullet 620 00:50:44,620 --> 00:50:47,360 is satisfied with positive probability. 621 00:50:47,360 --> 00:50:50,260 So you get Freiman s-isomorphism. 622 00:50:53,640 --> 00:50:54,680 Any questions? 623 00:50:57,440 --> 00:51:00,890 To get this model, in this case in the finite field setting, 624 00:51:00,890 --> 00:51:01,610 it's not so hard. 625 00:51:01,610 --> 00:51:03,650 So you kind of project the whole set, 626 00:51:03,650 --> 00:51:07,370 even though initially it might have or be very spread out 627 00:51:07,370 --> 00:51:09,170 into a lot of dimensions. 628 00:51:09,170 --> 00:51:15,040 You project it down to a small dimensional subspace randomly, 629 00:51:15,040 --> 00:51:17,050 and that works. 630 00:51:17,050 --> 00:51:18,860 So then with high probability, it 631 00:51:18,860 --> 00:51:21,360 preserves all the additive structures that you want, 632 00:51:21,360 --> 00:51:24,890 provided that you have small doubling. 633 00:51:24,890 --> 00:51:34,210 Now, let's look at what happens in Z. 634 00:51:34,210 --> 00:51:38,690 So in Z, things are a bit more involved. 635 00:51:38,690 --> 00:51:41,180 But the ideas-- actually, a lot of the ideas-- 636 00:51:41,180 --> 00:51:43,490 come from this proof, as well. 637 00:51:43,490 --> 00:51:48,260 So Ruzsa's modeling lemma tells us 638 00:51:48,260 --> 00:51:53,880 that if you have a set of integers-- 639 00:51:53,880 --> 00:51:55,650 always a finite set-- 640 00:51:55,650 --> 00:52:11,790 and integers s and N are such that N is at least sA minus sA, 641 00:52:11,790 --> 00:52:13,770 then so it turns out you might not 642 00:52:13,770 --> 00:52:16,350 be able to model the whole set A. 643 00:52:16,350 --> 00:52:20,660 But it will be good enough for us to model a large fraction. 644 00:52:20,660 --> 00:52:28,000 So then there exists an A prime subset of A, 645 00:52:28,000 --> 00:52:36,930 with A prime being at least an s fraction of the original set. 646 00:52:36,930 --> 00:52:48,130 And A prime is Freiman s-isomorphic to a subset 647 00:52:48,130 --> 00:52:55,940 of Z mod N. 648 00:52:55,940 --> 00:53:03,490 So same message as before, with an extra ingredient 649 00:53:03,490 --> 00:53:05,795 that we did not see before. 650 00:53:05,795 --> 00:53:07,170 But the point is that if you have 651 00:53:07,170 --> 00:53:11,320 a set A with controlled doubling, then well, now 652 00:53:11,320 --> 00:53:14,830 you can take a large fraction of A that 653 00:53:14,830 --> 00:53:20,260 is Freiman isomorphism to a subset of a small group. 654 00:53:20,260 --> 00:53:22,700 This is small, because we only need 655 00:53:22,700 --> 00:53:28,660 n to exceed sA minus sA, which are only constant factor more 656 00:53:28,660 --> 00:53:31,414 than the size of A. Yeah? 657 00:53:31,414 --> 00:53:33,850 AUDIENCE: Is s greater than 2 comma N? 658 00:53:33,850 --> 00:53:34,600 YUFEI ZHAO: Sorry. 659 00:53:34,600 --> 00:53:40,960 S greater than 2 separate and some integer. 660 00:53:40,960 --> 00:53:43,240 Thank you. 661 00:53:43,240 --> 00:53:45,930 So in our application, s will be a constant. 662 00:53:45,930 --> 00:53:48,690 s will be 8. 663 00:53:48,690 --> 00:53:50,895 So think of s as some specific constant. 664 00:53:54,210 --> 00:53:55,382 Any questions? 665 00:53:57,980 --> 00:54:01,910 So let's prove this modeling lemma. 666 00:54:01,910 --> 00:54:07,930 We want to try to do some kind of random map. 667 00:54:07,930 --> 00:54:12,460 But it's not clear how to start doing a random map 668 00:54:12,460 --> 00:54:15,740 if you just start in the integers. 669 00:54:15,740 --> 00:54:19,480 So what we want to do is first place ourself 670 00:54:19,480 --> 00:54:27,520 in some group where we can consider random automorphisms. 671 00:54:27,520 --> 00:54:29,800 So we start by-- 672 00:54:29,800 --> 00:54:35,860 perhaps we're very wastefully choosing a prime q bigger 673 00:54:35,860 --> 00:54:40,240 than the maximum possible sA minus sA. 674 00:54:43,840 --> 00:54:47,080 And so just choose a large enough prime. 675 00:54:47,080 --> 00:54:48,610 I don't care how large you pick. 676 00:54:48,610 --> 00:54:51,520 q can be very, very large. 677 00:54:51,520 --> 00:54:52,570 Pick a prime. 678 00:54:52,570 --> 00:54:57,520 And now I work inside Z mod q. 679 00:54:57,520 --> 00:55:00,230 I noticed that if you make q large enough, 680 00:55:00,230 --> 00:55:03,130 then A sits inside Z mod q Freiman 681 00:55:03,130 --> 00:55:05,820 isomorphically, or s-isomorphically. 682 00:55:05,820 --> 00:55:08,350 So just pick q large enough so that you don't 683 00:55:08,350 --> 00:55:11,580 have to worry about any issues. 684 00:55:11,580 --> 00:55:13,870 So Yeah. 685 00:55:13,870 --> 00:55:20,810 So the mod q map from A to Z mod q-- 686 00:55:20,810 --> 00:55:26,790 so this is Freiman s-isomorphic-- 687 00:55:26,790 --> 00:55:28,320 onto its image. 688 00:55:32,400 --> 00:55:36,968 So let's now consider a sequence of maps. 689 00:55:36,968 --> 00:55:39,010 And we're going to denote the sequence like this. 690 00:55:39,010 --> 00:55:43,550 So we start with Z. That's where A originally sits. 691 00:55:43,550 --> 00:55:51,060 And maps to Z mod q-- so that was the first map that we saw. 692 00:55:51,060 --> 00:55:54,150 And now we want to do a random automorphism, kind 693 00:55:54,150 --> 00:55:55,770 of like the random map earlier. 694 00:55:55,770 --> 00:56:00,680 But in Z mod q, there are lots of nice random automorphisms, 695 00:56:00,680 --> 00:56:06,940 namely multiplication by some non-zero element. 696 00:56:06,940 --> 00:56:12,460 And finally, we can consider the representative map, 697 00:56:12,460 --> 00:56:16,150 where every element of Z mod q, I can associate 698 00:56:16,150 --> 00:56:21,080 to it a positive integer from 1 to q 699 00:56:21,080 --> 00:56:22,820 which agrees with the Z mod q. 700 00:56:22,820 --> 00:56:26,890 So the final step is not a group homomorphism. 701 00:56:26,890 --> 00:56:29,870 So we need to be more careful. 702 00:56:29,870 --> 00:56:32,760 So let me denote by phi this entire map. 703 00:56:32,760 --> 00:56:34,950 So from the beginning to the end, this composition 704 00:56:34,950 --> 00:56:38,000 I'll denote by phi. 705 00:56:38,000 --> 00:56:44,340 And lambda here is some element between 1 and q minus 1. 706 00:56:47,000 --> 00:56:49,540 Now, remember what we said earlier, 707 00:56:49,540 --> 00:56:52,930 that this map, this final map here, 708 00:56:52,930 --> 00:56:56,680 might not be a Freiman homomorphism, because there 709 00:56:56,680 --> 00:57:01,630 are some additive relations here that are not preserved over 710 00:57:01,630 --> 00:57:02,650 here. 711 00:57:02,650 --> 00:57:07,360 But if I restrict myself to a small interval, 712 00:57:07,360 --> 00:57:10,300 then it is a Freiman homomorphism 713 00:57:10,300 --> 00:57:13,290 if we restrict to that interval. 714 00:57:13,290 --> 00:57:15,060 If you restrict yourself to an interval, 715 00:57:15,060 --> 00:57:18,330 you cannot have extra relations over here, 716 00:57:18,330 --> 00:57:19,395 because they cannot-- 717 00:57:19,395 --> 00:57:22,470 the interval is small enough, you can't wrap around. 718 00:57:22,470 --> 00:57:27,330 So let's consider restrictions to small intervals. 719 00:57:27,330 --> 00:57:29,190 I start with A over here. 720 00:57:29,190 --> 00:57:32,640 So I want to restrict myself to some interval 721 00:57:32,640 --> 00:57:38,670 so that I still have lots of A in that restriction. 722 00:57:38,670 --> 00:57:41,100 And you can do this by pigeonhole. 723 00:57:41,100 --> 00:57:46,710 So by pigeonhole, for every lambda 724 00:57:46,710 --> 00:57:52,670 there exists some interval we'll denote 725 00:57:52,670 --> 00:57:58,560 by I sub lambda inside q. 726 00:57:58,560 --> 00:58:05,250 So the length of this interval will be at most q/s, 727 00:58:05,250 --> 00:58:10,940 such that if I look at the restriction of this interval, 728 00:58:10,940 --> 00:58:16,550 I pull it all the way back to the beginning then I still get 729 00:58:16,550 --> 00:58:20,950 a lot of elements of A. So A sub lambda, namely 730 00:58:20,950 --> 00:58:29,570 the elements of A whose map gets sent to this interval, 731 00:58:29,570 --> 00:58:33,820 has at least A/s elements. 732 00:58:37,180 --> 00:58:41,780 So for instance, you can chop up q into s different intervals. 733 00:58:41,780 --> 00:58:51,712 So one of them will have lots of elements that came from A. 734 00:58:51,712 --> 00:58:54,310 And this is why in the end we only get 735 00:58:54,310 --> 00:58:56,410 a large subset of the A. So we're 736 00:58:56,410 --> 00:58:58,240 going to forget about everything else 737 00:58:58,240 --> 00:59:01,078 and focus our attention on the set here. 738 00:59:05,760 --> 00:59:09,990 So thus, by our earlier discussion 739 00:59:09,990 --> 00:59:16,020 having to do with the final map being a Freiman s-homomorphism 740 00:59:16,020 --> 00:59:19,440 when you're working inside a short interval, 741 00:59:19,440 --> 00:59:25,380 we see that phi, if you restrict to this A sub lambda, 742 00:59:25,380 --> 00:59:30,240 is a Freiman s-homomorphism. 743 00:59:32,760 --> 00:59:34,710 Each step is a Freiman homomorphism, 744 00:59:34,710 --> 00:59:36,540 because it's a group homomorphism. 745 00:59:36,540 --> 00:59:39,900 And the final step in the restriction 746 00:59:39,900 --> 00:59:43,440 is also a Freiman s-homomorphism, 747 00:59:43,440 --> 00:59:49,710 because what we said about working inside short intervals. 748 00:59:49,710 --> 00:59:51,050 So this part is very good. 749 00:59:53,870 --> 00:59:54,470 All right. 750 00:59:54,470 --> 00:59:57,000 So now let me consider one more composition. 751 00:59:57,000 --> 00:59:58,520 So at the end of the day, we would 752 00:59:58,520 --> 01:00:08,150 like to model this A lambda inside some small cyclic group. 753 01:00:08,150 --> 01:00:11,390 So far, we don't have small cyclic groups here. 754 01:00:11,390 --> 01:00:13,970 But I'm going to manufacture a small cyclic group. 755 01:00:13,970 --> 01:00:20,780 So we're going to consider the map where, first we 756 01:00:20,780 --> 01:00:23,960 take our phi all the way until the end, 757 01:00:23,960 --> 01:00:28,850 and now you take mod m map. 758 01:00:28,850 --> 01:00:31,450 So if I don't write anything, if it goes to Z mod m, 759 01:00:31,450 --> 01:00:34,230 it means the mod m map. 760 01:00:34,230 --> 01:00:39,470 So let me consider psi, which is the composition of these two 761 01:00:39,470 --> 01:00:39,970 maps. 762 01:00:44,910 --> 01:00:46,600 All right. 763 01:00:46,600 --> 01:00:52,720 So we would like to say that you can choose this lambda so 764 01:00:52,720 --> 01:00:59,020 that this A sub lambda gets mapped Freiman s-isomorphically 765 01:00:59,020 --> 01:01:02,480 until to the end. 766 01:01:02,480 --> 01:01:04,130 So far, everything looks pretty good. 767 01:01:04,130 --> 01:01:07,420 So you have Freiman s-homomorphism, 768 01:01:07,420 --> 01:01:09,640 and you have a group homomorphism. 769 01:01:09,640 --> 01:01:12,231 So the whole thing is a Freiman s-homomorphism. 770 01:01:14,940 --> 01:01:20,250 So psi is restricted to A sub lambda, 771 01:01:20,250 --> 01:01:23,226 is a Freiman s-homomorphism. 772 01:01:28,280 --> 01:01:31,130 But now the thing that we really want to check 773 01:01:31,130 --> 01:01:37,160 is if there are some relationships that 774 01:01:37,160 --> 01:01:42,950 are present at the end in Z mod m that 775 01:01:42,950 --> 01:01:44,770 were not present earlier. 776 01:01:49,790 --> 01:01:52,470 And so we need to check that-- 777 01:01:52,470 --> 01:02:03,320 we claim that if psi does not map A sub lambda 778 01:02:03,320 --> 01:02:06,530 Freiman isomorphically, then something 779 01:02:06,530 --> 01:02:08,010 has to have gone wrong. 780 01:02:08,010 --> 01:02:12,070 So if it does not map A sub lambda Freiman 781 01:02:12,070 --> 01:02:19,430 isomorphically onto its image, then 782 01:02:19,430 --> 01:02:21,560 what could have gone wrong? 783 01:02:21,560 --> 01:02:26,630 Claim that there must be some d which 784 01:02:26,630 --> 01:02:34,070 depends on lambda in sA minus sA, 785 01:02:34,070 --> 01:02:45,055 and d not 0, such that phi of d is 0 mod m. 786 01:02:49,510 --> 01:02:50,560 So we'll prove this. 787 01:02:50,560 --> 01:02:53,000 But like before, it's a very similar idea 788 01:02:53,000 --> 01:02:54,530 to what's happening earlier. 789 01:02:54,530 --> 01:02:56,990 The idea is that if you have-- 790 01:02:59,600 --> 01:03:02,190 we want to show that there are no additional additive 791 01:03:02,190 --> 01:03:05,710 relations in the image. 792 01:03:05,710 --> 01:03:07,370 So we would like-- 793 01:03:07,370 --> 01:03:09,840 so if it's a Freiman isomorphism, 794 01:03:09,840 --> 01:03:12,330 then there has to be some accidental collisions. 795 01:03:12,330 --> 01:03:18,230 And that accidental collision has to be witnessed by some d. 796 01:03:18,230 --> 01:03:19,780 So this requires some checking. 797 01:03:24,130 --> 01:03:33,360 So suppose-- indeed, suppose the hypothesis-- 798 01:03:33,360 --> 01:03:38,010 suppose that the psi does not map A sub lambda Freiman 799 01:03:38,010 --> 01:03:41,130 s-isomorphically onto its image. 800 01:03:41,130 --> 01:03:51,720 Then there exists a1 through as, a1 prime through as prime, 801 01:03:51,720 --> 01:04:03,740 in A lambda such that they do not have additive relation, 802 01:04:03,740 --> 01:04:08,700 but their images do have this additive relation. 803 01:04:12,900 --> 01:04:14,740 Their images all the way until the end 804 01:04:14,740 --> 01:04:16,510 having the additive relation means 805 01:04:16,510 --> 01:04:21,610 that phi has this additive relation mod m. 806 01:04:35,030 --> 01:04:35,530 OK. 807 01:04:35,530 --> 01:04:36,322 So how can this be? 808 01:04:39,490 --> 01:04:49,740 Recall that since the image-- 809 01:04:49,740 --> 01:04:52,610 so all of these elements-- lie inside some short interval. 810 01:04:58,526 --> 01:05:05,400 The interval has length less than 2 minus s. 811 01:05:05,400 --> 01:05:09,270 So we saw this argument, very similar argument earlier, 812 01:05:09,270 --> 01:05:11,220 before the break. 813 01:05:11,220 --> 01:05:13,830 Because everything lies in the short interval, 814 01:05:13,830 --> 01:05:18,150 we see that this difference between the left- 815 01:05:18,150 --> 01:05:31,440 and the right-hand sides, this difference 816 01:05:31,440 --> 01:05:33,670 is strictly less than q. 817 01:05:39,610 --> 01:05:49,020 Now, by switching the a's and a primes if necessary, 818 01:05:49,020 --> 01:05:58,120 we may assume that this difference is non-negative. 819 01:05:58,120 --> 01:06:00,460 Otherwise, it's just a labeling issue. 820 01:06:00,460 --> 01:06:01,590 Otherwise, I relabel them. 821 01:06:05,280 --> 01:06:07,560 So then this here-- 822 01:06:07,560 --> 01:06:11,510 so what's inside this expression-- 823 01:06:11,510 --> 01:06:16,442 we call this expression inside the absolute value, 824 01:06:16,442 --> 01:06:17,520 we call it star. 825 01:06:22,110 --> 01:06:29,450 So star is some number between 0 and strictly-- so at least 0 826 01:06:29,450 --> 01:06:30,770 and strictly less than q. 827 01:06:39,560 --> 01:06:40,510 Right. 828 01:06:40,510 --> 01:06:57,950 So if we set d to be this expression, so 829 01:06:57,950 --> 01:07:03,580 the difference between these two sums, on one hand this d here-- 830 01:07:06,350 --> 01:07:09,150 sorry, that's what I want to say. 831 01:07:09,150 --> 01:07:10,180 Suppose you don't have-- 832 01:07:16,420 --> 01:07:20,640 if you are not mapping Freiman isomorphically onto the image, 833 01:07:20,640 --> 01:07:24,300 then I can exhibit some witness for that non-isomorphism, 834 01:07:24,300 --> 01:07:27,780 meaning a bunch of elements that do not 835 01:07:27,780 --> 01:07:30,600 have additive relations in the domain. 836 01:07:30,600 --> 01:07:33,970 But I do have additive relation in image. 837 01:07:33,970 --> 01:07:40,290 So if we set this d, then it's some element of sA minus sA. 838 01:07:40,290 --> 01:07:44,760 And it's non-zero, because we assume that d is non-zero. 839 01:07:50,860 --> 01:08:06,950 And so then, what can we say about phi of d? 840 01:08:06,950 --> 01:08:16,850 So phi of d, I claim, must be this expression 841 01:08:16,850 --> 01:08:23,399 over here, the difference of the corresponding sums 842 01:08:23,399 --> 01:08:24,920 in the image. 843 01:08:24,920 --> 01:08:35,569 Because the two sides are congruent mod q-- 844 01:08:39,890 --> 01:08:42,560 two sides are congruent mod q. 845 01:08:42,560 --> 01:08:54,680 And furthermore, they are in the interval from 0 846 01:08:54,680 --> 01:08:56,558 to strictly less than q. 847 01:09:09,529 --> 01:09:12,300 So this is a slightly subtle argument. 848 01:09:12,300 --> 01:09:14,430 But the idea is all very simple. 849 01:09:14,430 --> 01:09:18,043 Just have to keep track of the relationships 850 01:09:18,043 --> 01:09:19,960 between what's happening in the domain, what's 851 01:09:19,960 --> 01:09:20,918 happening in the image. 852 01:09:23,295 --> 01:09:24,670 Somehow, I think the finite field 853 01:09:24,670 --> 01:09:27,910 case is quite illustrative of-- 854 01:09:27,910 --> 01:09:30,062 there, what goes wrong is similar to what 855 01:09:30,062 --> 01:09:30,729 goes wrong here. 856 01:09:30,729 --> 01:09:32,979 Except here, you have to keep track a bit more things. 857 01:09:35,840 --> 01:09:36,340 OK. 858 01:09:36,340 --> 01:09:42,430 So consequently, thus phi of d is 859 01:09:42,430 --> 01:09:50,987 congruent to 0 mod m, which is what we're looking for. 860 01:09:50,987 --> 01:09:52,029 So that proves the claim. 861 01:09:54,570 --> 01:09:55,584 Any questions? 862 01:09:58,902 --> 01:09:59,850 Yeah? 863 01:09:59,850 --> 01:10:01,225 AUDIENCE: [INAUDIBLE] 864 01:10:01,225 --> 01:10:01,850 YUFEI ZHAO: OK. 865 01:10:01,850 --> 01:10:03,060 So this part? 866 01:10:03,060 --> 01:10:04,160 All right. 867 01:10:04,160 --> 01:10:07,760 So we set d to be this expression. 868 01:10:07,760 --> 01:10:10,070 I claim this equality. 869 01:10:10,070 --> 01:10:11,950 So why is it true? 870 01:10:11,950 --> 01:10:18,600 First, the left-hand side and the right-hand side, they 871 01:10:18,600 --> 01:10:22,625 are congruent to each other mod q. 872 01:10:22,625 --> 01:10:23,250 So why is that? 873 01:10:31,186 --> 01:10:33,412 AUDIENCE: [INAUDIBLE] 874 01:10:33,412 --> 01:10:34,162 YUFEI ZHAO: Sorry? 875 01:10:41,106 --> 01:10:43,090 Yeah. 876 01:10:43,090 --> 01:10:47,140 AUDIENCE: [INAUDIBLE] every part of phi should preserve-- 877 01:10:47,140 --> 01:10:49,740 the first two parts are group homomorphisms. 878 01:10:49,740 --> 01:10:51,580 And then we just take [INAUDIBLE] mod q. 879 01:10:51,580 --> 01:10:52,413 YUFEI ZHAO: Exactly. 880 01:10:52,413 --> 01:10:54,330 If you look at phi-- 881 01:10:54,330 --> 01:11:00,180 where was it?-- up there, you see that everything preserves. 882 01:11:00,180 --> 01:11:04,040 Even though the very last step is not a group homomorphism, 883 01:11:04,040 --> 01:11:07,260 mod q is preserved. 884 01:11:07,260 --> 01:11:10,210 So even though the last step is not group homomorphism, 885 01:11:10,210 --> 01:11:12,012 it is all mod q. 886 01:11:12,012 --> 01:11:13,720 So if you're looking at mod q, everything 887 01:11:13,720 --> 01:11:15,348 is group homomorphism. 888 01:11:15,348 --> 01:11:16,140 So here we're good. 889 01:11:18,930 --> 01:11:21,577 Both are in this interval. 890 01:11:21,577 --> 01:11:23,160 The right-hand side is in the interval 891 01:11:23,160 --> 01:11:26,280 because of our assumption about short-- 892 01:11:26,280 --> 01:11:29,220 everything living inside a short interval. 893 01:11:29,220 --> 01:11:30,990 And the left-hand side is by definition. 894 01:11:30,990 --> 01:11:40,730 Because the image of phi is in that interval, 895 01:11:40,730 --> 01:11:43,040 especially if given that d is not equal to 0. 896 01:11:47,950 --> 01:11:50,160 Is that OK? 897 01:11:50,160 --> 01:11:52,870 So it's not hard, but it's a bit confusing. 898 01:11:52,870 --> 01:11:56,080 So think about it. 899 01:11:56,080 --> 01:11:56,580 All right. 900 01:11:56,580 --> 01:11:57,900 So we're almost done. 901 01:11:57,900 --> 01:12:03,540 So we're almost done proving the Ruzsa modeling 902 01:12:03,540 --> 01:12:06,375 lemma in Z mod m. 903 01:12:06,375 --> 01:12:07,950 So let me finish off the proof. 904 01:12:25,400 --> 01:12:31,850 So for each non-zero d in this iterated sumset, basically, 905 01:12:31,850 --> 01:12:35,970 we would like to pick a lambda so that that map up there does 906 01:12:35,970 --> 01:12:38,940 what we want to do. 907 01:12:38,940 --> 01:12:40,650 If it doesn't do what we want to do, 908 01:12:40,650 --> 01:12:45,060 then it is witnessed by some d, like this. 909 01:12:45,060 --> 01:12:48,980 Those are the bad lambda. 910 01:12:48,980 --> 01:12:53,250 So if there exists a d, then this lambda is bad. 911 01:12:53,250 --> 01:12:56,850 So for each d that potentially witnessed 912 01:12:56,850 --> 01:13:09,950 some bad lambda, the number of bad lambda, i.e., 913 01:13:09,950 --> 01:13:19,086 such that phi of d is congruent to 0 mod 914 01:13:19,086 --> 01:13:25,820 m, so here we're no longer even thinking 915 01:13:25,820 --> 01:13:27,740 about group homomorphisms anymore 916 01:13:27,740 --> 01:13:29,410 or the Freiman homomorphisms. 917 01:13:29,410 --> 01:13:33,460 It's just a question of, if I give you a non-zero integer, 918 01:13:33,460 --> 01:13:40,540 how many lambdas are there so that phi of d 919 01:13:40,540 --> 01:13:42,290 is divisible by m? 920 01:13:45,300 --> 01:13:50,790 Remember that we picked q large enough so that, initially, you 921 01:13:50,790 --> 01:13:52,905 are sitting very much inside-- 922 01:13:55,680 --> 01:13:57,750 everything's really between 0 and q. 923 01:13:57,750 --> 01:14:05,590 So this dot lambda up there lacks uniformity. 924 01:14:05,590 --> 01:14:08,220 So the number of such bad lambdas 925 01:14:08,220 --> 01:14:17,790 is exactly the number of elements in this interval that 926 01:14:17,790 --> 01:14:22,608 are divisible by m. 927 01:14:28,240 --> 01:14:30,340 So everything's more or less a bijection 928 01:14:30,340 --> 01:14:31,840 if you restrict to the right places. 929 01:14:35,800 --> 01:14:42,120 And the number of such elements is, at most, q minus 1 over m. 930 01:14:42,120 --> 01:14:49,320 So therefore, the total number of bad lambdas 931 01:14:49,320 --> 01:14:54,570 is, at most, for each element d of sA 932 01:14:54,570 --> 01:14:58,740 minus sA, a non-zero element. 933 01:14:58,740 --> 01:15:06,520 We have, at most, q minus 1 over m bad lambdas. 934 01:15:06,520 --> 01:15:10,290 So the total number of bad lambdas is strictly less than q 935 01:15:10,290 --> 01:15:11,860 minus 1. 936 01:15:11,860 --> 01:15:19,080 So there exists some lambda such that psi, 937 01:15:19,080 --> 01:15:30,100 when restricted to A sub lambda, maps Freiman s-isomorphically 938 01:15:30,100 --> 01:15:31,440 onto the image. 939 01:15:43,010 --> 01:15:45,340 Somehow, I think it's really the same kind of proof 940 01:15:45,340 --> 01:15:46,840 as the one in the finite field case, 941 01:15:46,840 --> 01:15:49,560 except you have this extra wrinkle about restricting 942 01:15:49,560 --> 01:15:53,130 too short diameter intervals, to short intervals. 943 01:15:53,130 --> 01:15:54,390 But the idea is very similar. 944 01:15:57,490 --> 01:15:57,990 OK. 945 01:15:57,990 --> 01:16:03,000 So that's the Freiman model lemma in the integers. 946 01:16:03,000 --> 01:16:07,350 And let me summarize what we know so far. 947 01:16:07,350 --> 01:16:09,150 And so that will give you a sense 948 01:16:09,150 --> 01:16:13,210 of where we're going in the proof of Freiman's theorem. 949 01:16:13,210 --> 01:16:18,160 So what we know so far is that if you 950 01:16:18,160 --> 01:16:22,570 have a subset of integers, a finite subset, 951 01:16:22,570 --> 01:16:27,760 such that A plus A is size A times K at most, 952 01:16:27,760 --> 01:16:36,900 doubling constant at most K, then there exists some prime N 953 01:16:36,900 --> 01:16:44,368 at most 2K to the 16th times the size of A 954 01:16:44,368 --> 01:16:50,830 and some subset A prime of A such 955 01:16:50,830 --> 01:17:04,977 that A prime is Freiman 8-isomorphic to a subset of Z 956 01:17:04,977 --> 01:17:05,540 mod NZ. 957 01:17:09,698 --> 01:17:13,970 So it follows from two things we've seen so far. 958 01:17:13,970 --> 01:17:21,950 Because by the Plunnecke-Ruzsa inequality 8A minus 8A 959 01:17:21,950 --> 01:17:27,146 is, at most, K to the 16 times A. 960 01:17:27,146 --> 01:17:34,130 And now we can choose a prime N between K 961 01:17:34,130 --> 01:17:38,950 to the 16 and 2 times K to the 16 962 01:17:38,950 --> 01:17:41,180 and apply the modeling lemma. 963 01:17:50,450 --> 01:17:52,010 So that's where we are at. 964 01:17:52,010 --> 01:17:56,980 So you start with a set of integers with small doubling. 965 01:17:56,980 --> 01:18:02,470 Then we can conclude that by keeping a large fraction of A, 966 01:18:02,470 --> 01:18:04,960 keeping-- 967 01:18:04,960 --> 01:18:07,730 I forgot to-- so very important. 968 01:18:07,730 --> 01:18:10,810 So there exist some A, which is a large fraction. 969 01:18:15,320 --> 01:18:19,550 Keeping a large fraction of A, I can model this large subset 970 01:18:19,550 --> 01:18:25,430 of A by some subset of a cyclic group, 971 01:18:25,430 --> 01:18:27,860 where the size of the cyclic group 972 01:18:27,860 --> 01:18:33,550 is only a constant times more than the size of A. 973 01:18:33,550 --> 01:18:41,360 So now, we are going to work inside a cyclic group 974 01:18:41,360 --> 01:18:46,640 and working with a set inside a cyclic group that's 975 01:18:46,640 --> 01:18:50,044 a constant size, a constant fraction of the cyclic group. 976 01:18:54,000 --> 01:18:54,940 Question is why 8? 977 01:18:54,940 --> 01:18:56,630 So that will come up later. 978 01:18:56,630 --> 01:18:59,250 So basically, you need to choose some numbers 979 01:18:59,250 --> 01:19:03,320 so that you want to preserve the structure of GAPs. 980 01:19:03,320 --> 01:19:06,260 So that will come up later. 981 01:19:06,260 --> 01:19:10,010 And now we're inside a cyclic group. 982 01:19:10,010 --> 01:19:13,668 And you have a constant fraction of a cyclic group. 983 01:19:13,668 --> 01:19:14,960 Where have we seen this before? 984 01:19:20,030 --> 01:19:24,065 So when we proved Roth's theorem, that was the setting. 985 01:19:24,065 --> 01:19:26,340 So in cyclic loop, you have a constant fraction 986 01:19:26,340 --> 01:19:29,270 of cyclic group, and you can do Fourier analysis. 987 01:19:29,270 --> 01:19:32,340 Initially, you could not do Fourier analysis 988 01:19:32,340 --> 01:19:34,340 starting with Freiman's theorem, because the set 989 01:19:34,340 --> 01:19:35,870 may be very, very spread out. 990 01:19:35,870 --> 01:19:39,410 But now, you are a large fraction of a cyclic group. 991 01:19:39,410 --> 01:19:42,830 So we're going to do Fourier analysis next time 992 01:19:42,830 --> 01:19:47,780 to show that such a set must contain lots of structure, 993 01:19:47,780 --> 01:19:51,890 just from the fact that it is large. 994 01:19:51,890 --> 01:19:54,260 So that will be the next step. 995 01:19:54,260 --> 01:19:54,760 Good. 996 01:19:54,760 --> 01:19:57,150 Happy Thanksgiving.