1 00:00:19,142 --> 00:00:20,600 PROFESSOR: So we've been discussing 2 00:00:20,600 --> 00:00:23,900 Szemeredi's regularity level for the past couple of lectures. 3 00:00:23,900 --> 00:00:25,580 And one of the theorems that we proved 4 00:00:25,580 --> 00:00:30,410 was Roth's theorem, which tells us how large can a subset of 1 5 00:00:30,410 --> 00:00:35,220 through N be if it has no 3-term arithmetic progressions. 6 00:00:35,220 --> 00:00:37,130 And I mentioned at the end of last time 7 00:00:37,130 --> 00:00:41,780 that we can construct fairly large subsets of 1 8 00:00:41,780 --> 00:00:44,990 through N without 3-term arithmetic progressions. 9 00:00:44,990 --> 00:00:48,740 I want to begin today's lecture with showing you that example, 10 00:00:48,740 --> 00:00:51,140 showing you that construction. 11 00:00:51,140 --> 00:00:55,250 But first, let's try to do something 12 00:00:55,250 --> 00:00:59,580 that is somewhat more naive, somewhat more straightforward, 13 00:00:59,580 --> 00:01:14,240 to try to just construct greedily 3-AP-free sets of 1 14 00:01:14,240 --> 00:01:15,830 through n. 15 00:01:15,830 --> 00:01:19,970 And recall from last time, we showed Roth's theorem 16 00:01:19,970 --> 00:01:27,740 that such a set must have size little o of N. 17 00:01:27,740 --> 00:01:29,570 So what's one thing you might do? 18 00:01:29,570 --> 00:01:33,533 Well, you can try to greedily construct such a set. 19 00:01:33,533 --> 00:01:35,450 It will just make our life a little bit easier 20 00:01:35,450 --> 00:01:37,740 if we start with 0 instead of 1. 21 00:01:37,740 --> 00:01:40,310 So you put one element in, and you 22 00:01:40,310 --> 00:01:42,200 keep putting in the next integer, 23 00:01:42,200 --> 00:01:46,500 as long as it doesn't create a 3-term arithmetic progression. 24 00:01:46,500 --> 00:01:47,820 So you keep doing this. 25 00:01:47,820 --> 00:01:51,140 Well, you can't put 2 in, so let's skip 2. 26 00:01:51,140 --> 00:01:53,654 So we go to the next one, 3. 27 00:01:53,654 --> 00:01:55,340 So 4 is OK. 28 00:01:57,870 --> 00:02:00,180 So we skip 5. 29 00:02:00,180 --> 00:02:05,260 We have to skip 6 as well because 0, 3, 6, that's a 3-AP. 30 00:02:05,260 --> 00:02:06,000 So we keep going. 31 00:02:06,000 --> 00:02:08,904 So what's the next number we can put in? 32 00:02:08,904 --> 00:02:11,257 AUDIENCE: 9. 33 00:02:11,257 --> 00:02:11,840 PROFESSOR: Up. 34 00:02:11,840 --> 00:02:12,610 Go to 9. 35 00:02:15,980 --> 00:02:19,010 Then the next one is 10. 36 00:02:19,010 --> 00:02:21,888 What's the next one we can put in? 37 00:02:21,888 --> 00:02:22,930 So we can play this game. 38 00:02:25,460 --> 00:02:30,490 Find out what is the next number that you can put in. 39 00:02:30,490 --> 00:02:32,360 Greedily, if you could put it in, 40 00:02:32,360 --> 00:02:37,460 put it in in a way that generates a subset of integers 41 00:02:37,460 --> 00:02:41,530 without 3-term arithmetic progressions. 42 00:02:41,530 --> 00:02:43,867 So this actually has a name, so this 43 00:02:43,867 --> 00:02:45,075 is called a Stanley sequence. 44 00:02:51,540 --> 00:02:55,530 And there's an easier way to see what the sequence is. 45 00:02:55,530 --> 00:03:01,930 Namely, if you write the sequence in base 3, 46 00:03:01,930 --> 00:03:09,758 you find that these numbers are 0; 0, 1; 1, 0; 1, 1. 47 00:03:12,370 --> 00:03:13,060 1, 0, 0. 48 00:03:13,060 --> 00:03:14,160 1, 0, 1. 49 00:03:14,160 --> 00:03:14,950 1, 1, 0. 50 00:03:14,950 --> 00:03:16,980 1, 1, 1. 51 00:03:16,980 --> 00:03:25,460 So these are just numbers whose base 3 representation 52 00:03:25,460 --> 00:03:30,050 consists of zeros and ones. 53 00:03:30,050 --> 00:03:32,600 So I'll leave it to you as an exercise 54 00:03:32,600 --> 00:03:34,830 to figure out why this is the case, if you generate 55 00:03:34,830 --> 00:03:38,570 the sequence greedily this way, this is exactly the sequence 56 00:03:38,570 --> 00:03:41,140 that you obtain. 57 00:03:41,140 --> 00:03:44,040 But once you know that, it's not too hard to find out 58 00:03:44,040 --> 00:03:46,300 how many numbers you generate. 59 00:03:49,240 --> 00:03:57,440 So up to-- suppose N is equal to 3 to the k. 60 00:03:57,440 --> 00:04:08,500 We get 2 to the k terms, which gives you 61 00:04:08,500 --> 00:04:12,450 N raised to the power of log base 3 of 2. 62 00:04:14,852 --> 00:04:17,310 So you can figure out what that number is, but some numbers 63 00:04:17,310 --> 00:04:18,279 strictly less than 1. 64 00:04:28,810 --> 00:04:30,540 And actually, for a very long time, 65 00:04:30,540 --> 00:04:32,830 people thought this was the best construction. 66 00:04:32,830 --> 00:04:35,620 So this construction was known even before Stanley, 67 00:04:35,620 --> 00:04:39,630 so it was something that is very natural to come up with. 68 00:04:39,630 --> 00:04:45,658 And it was somewhat of a surprise when in the '40s, 69 00:04:45,658 --> 00:04:47,200 it was discovered that you can create 70 00:04:47,200 --> 00:04:50,380 much larger subsets of the positive integers 71 00:04:50,380 --> 00:04:52,275 without arithmetic progressions. 72 00:04:52,275 --> 00:04:54,400 So that's the first thing I want to show you today. 73 00:04:58,470 --> 00:05:06,180 So these sets were first discovered by Salem and Spencer 74 00:05:06,180 --> 00:05:08,310 back in the '40s. 75 00:05:08,310 --> 00:05:17,080 And a few years later, Behrend gave a somewhat improved 76 00:05:17,080 --> 00:05:19,820 and simplified version of the Salem-Spencer construction. 77 00:05:19,820 --> 00:05:24,190 So these were all back in the '40s. 78 00:05:24,190 --> 00:05:26,160 So these days, we usually refer, at least 79 00:05:26,160 --> 00:05:27,910 in the additive combinatorics community, 80 00:05:27,910 --> 00:05:30,665 to this construction as Behrend's construction. 81 00:05:30,665 --> 00:05:33,040 And this, indeed, what I will show you is due to Behrend, 82 00:05:33,040 --> 00:05:36,150 but somehow, this Salem-Spencer name has been forgotten 83 00:05:36,150 --> 00:05:38,650 and I just wanted to point out that it 84 00:05:38,650 --> 00:05:45,640 was Salem and Spencer who first demonstrated that there exists 85 00:05:45,640 --> 00:05:50,590 a subset of 1 through N that is 3-AP-free 86 00:05:50,590 --> 00:05:56,710 and has size N to the 1 minus little o 1, that there's 87 00:05:56,710 --> 00:05:58,330 no power saving. 88 00:05:58,330 --> 00:06:01,100 So there exists examples with no power saving. 89 00:06:01,100 --> 00:06:04,240 And this is important because as we 90 00:06:04,240 --> 00:06:06,150 saw last time, the proof of Roth's theorem-- 91 00:06:06,150 --> 00:06:08,800 and we basically spent two full lectures 92 00:06:08,800 --> 00:06:11,020 proving Roth's theorem. 93 00:06:11,020 --> 00:06:13,010 And it's somewhat involved, right. 94 00:06:13,010 --> 00:06:15,340 It wasn't this one line of inequality 95 00:06:15,340 --> 00:06:18,620 you can just use to deduce the result. 96 00:06:18,620 --> 00:06:21,700 And that is part of the difficulty. 97 00:06:21,700 --> 00:06:24,640 So having such an example is indication 98 00:06:24,640 --> 00:06:28,350 that Roth's theorem should not be so easy. 99 00:06:28,350 --> 00:06:30,130 That's not a rigorous demonstration, 100 00:06:30,130 --> 00:06:32,910 but it's some indication of the difficulty of Roth's theorem. 101 00:06:36,430 --> 00:06:39,355 So let's see this construction due to Behrend. 102 00:06:47,890 --> 00:06:50,190 So let me write down the precise statement. 103 00:06:50,190 --> 00:06:56,760 There exists a constant, C, such that there exists a subset of 1 104 00:06:56,760 --> 00:07:07,070 through N that is 3-AP-free and has size at least N times e 105 00:07:07,070 --> 00:07:17,320 to the minus C root log N. 106 00:07:17,320 --> 00:07:19,870 Perhaps somewhat amazingly enough, this bound 107 00:07:19,870 --> 00:07:23,290 is still the current best known. 108 00:07:23,290 --> 00:07:25,920 We do not know any constructions that 109 00:07:25,920 --> 00:07:29,040 is essentially better than Behrend's construction 110 00:07:29,040 --> 00:07:30,870 from the '40s. 111 00:07:30,870 --> 00:07:33,180 There were some small recent improvements 112 00:07:33,180 --> 00:07:39,983 that clarified what the C could be, but in this form, 113 00:07:39,983 --> 00:07:41,650 we do not know anything better than what 114 00:07:41,650 --> 00:07:43,050 was known back in the '40s. 115 00:07:43,050 --> 00:07:46,020 And the construction, I will show yow-- 116 00:07:46,020 --> 00:07:48,387 hopefully you should believe that it's quite simple. 117 00:07:48,387 --> 00:07:50,220 So I will show you what the construction is. 118 00:07:50,220 --> 00:07:52,980 It's clever, but it's not complicated. 119 00:07:52,980 --> 00:07:56,350 And it's a really interesting direction to figure out 120 00:07:56,350 --> 00:08:00,330 is this really the best construction out there. 121 00:08:00,330 --> 00:08:01,080 Can you do better? 122 00:08:04,358 --> 00:08:04,900 So let's see. 123 00:08:08,590 --> 00:08:12,100 We're going to set some parameters to be decided later, 124 00:08:12,100 --> 00:08:14,200 m and d. 125 00:08:14,200 --> 00:08:22,510 Let me consider x to be this discrete box in d dimensions. 126 00:08:22,510 --> 00:08:27,850 So this is the box of lattice points in d dimensions, 1 127 00:08:27,850 --> 00:08:30,790 through m raised to d. 128 00:08:30,790 --> 00:08:34,210 And let me consider an intersection 129 00:08:34,210 --> 00:08:44,640 of this box by a sphere of radius root L. 130 00:08:44,640 --> 00:08:49,190 So namely, we look at points in this x, 131 00:08:49,190 --> 00:08:53,550 such that the sum of their squares 132 00:08:53,550 --> 00:08:59,890 is equal to exactly L. You take a bunch of these spheres, 133 00:08:59,890 --> 00:09:12,040 they partition your set x, so the smallest possible sum 134 00:09:12,040 --> 00:09:16,180 of squares, largest possible sum of squares. 135 00:09:16,180 --> 00:09:22,440 So in particular, there exists some L such that x of L 136 00:09:22,440 --> 00:09:33,250 is large, just by pigeonhole principle. 137 00:09:33,250 --> 00:09:36,790 You can probably do this step with a bit more finesse, 138 00:09:36,790 --> 00:09:40,132 but it's not going to change any of the asymptotics. 139 00:09:43,366 --> 00:09:46,340 The intuition here-- and we'll come back to this in a second-- 140 00:09:46,340 --> 00:09:49,950 is that xL lies on a sphere. 141 00:09:49,950 --> 00:09:51,480 So this is the set of lattice points 142 00:09:51,480 --> 00:09:53,070 that lies in a given sphere. 143 00:09:53,070 --> 00:09:57,180 And because you are looking at a sphere, it has no 3-term APs. 144 00:09:59,580 --> 00:10:01,080 So we're going to use that property, 145 00:10:01,080 --> 00:10:04,570 but right now it's not yet a subset of the integers. 146 00:10:04,570 --> 00:10:09,510 So what we're going to do is to take this section of a sphere 147 00:10:09,510 --> 00:10:14,460 and then project it to one dimension, 148 00:10:14,460 --> 00:10:17,770 so that it is now going to be a subset of integers. 149 00:10:17,770 --> 00:10:19,440 And we're going to do this in such a way 150 00:10:19,440 --> 00:10:25,104 that it does not affect the presence of 3-term APs. 151 00:10:25,104 --> 00:10:33,090 So let us map x to the integers by base expansion. 152 00:10:43,290 --> 00:10:47,660 So this is base 2m expansion. 153 00:10:57,580 --> 00:10:58,080 All right. 154 00:10:58,080 --> 00:11:01,356 So what's the point of this construction here so far? 155 00:11:01,356 --> 00:11:05,870 Well, you verified a couple of properties, the first being 156 00:11:05,870 --> 00:11:13,550 that this construction here is injective, this map. 157 00:11:13,550 --> 00:11:20,310 So if you call this map phi, this phi is injective. 158 00:11:20,310 --> 00:11:24,480 Well, it's base expansion, so it's injective. 159 00:11:24,480 --> 00:11:28,680 But also a somewhat stronger claim 160 00:11:28,680 --> 00:11:31,230 is that if you have three points-- 161 00:11:36,630 --> 00:11:40,050 so if you have three points in x, 162 00:11:40,050 --> 00:11:47,770 that map to a 3-AP in the integers, 163 00:11:47,770 --> 00:11:58,750 then the three points originally must be a 3-AP in x. 164 00:12:02,494 --> 00:12:04,020 Again, this is not a hard claim. 165 00:12:04,020 --> 00:12:05,670 Just think about it. 166 00:12:05,670 --> 00:12:08,833 Here, because we're using base 2m expansion 167 00:12:08,833 --> 00:12:10,750 and you're only allowed to use digits up to m, 168 00:12:10,750 --> 00:12:12,190 you don't have any wrap around effects. 169 00:12:12,190 --> 00:12:14,482 You don't have any carryovers when you do the addition. 170 00:12:17,760 --> 00:12:26,800 So combining these two observations, 171 00:12:26,800 --> 00:12:43,130 we find that the image of xL is a 3-AP-free set off 1 172 00:12:43,130 --> 00:12:48,430 through N. So what is N? 173 00:12:48,430 --> 00:12:53,430 I can take N to be, for instance, 2m raised to power d, 174 00:12:53,430 --> 00:12:57,710 so all the numbers up there are less than this quantity. 175 00:12:57,710 --> 00:13:04,400 And the size is the size of x sub 176 00:13:04,400 --> 00:13:11,512 L, which is N to the d divided by d m squared. 177 00:13:11,512 --> 00:13:13,720 And now you just need to find the appropriate choices 178 00:13:13,720 --> 00:13:19,430 of the parameters m and d to maximize the size of x sub L. 179 00:13:19,430 --> 00:13:23,440 And that's an exercise. 180 00:13:23,440 --> 00:13:29,690 So, for instance, if you take m to be e to the root log 181 00:13:29,690 --> 00:13:37,880 N and d to be root log N, then you 182 00:13:37,880 --> 00:13:42,980 find that the size here is the bound that we claim. 183 00:13:48,570 --> 00:13:54,920 And that finishes the construction of Behrend, 184 00:13:54,920 --> 00:13:57,960 giving you a fairly large subset of 1 185 00:13:57,960 --> 00:14:01,480 through N without 3-term arithmetic progressions. 186 00:14:01,480 --> 00:14:04,640 And the idea here is you look at a higher dimensional 187 00:14:04,640 --> 00:14:07,490 object, namely, a higher dimensional sphere which 188 00:14:07,490 --> 00:14:09,380 has this property of being 3-AP-free, 189 00:14:09,380 --> 00:14:13,185 and then you project it onto the integers. 190 00:14:13,185 --> 00:14:14,060 Any questions so far? 191 00:14:20,850 --> 00:14:21,610 OK. 192 00:14:21,610 --> 00:14:24,050 So we have some proofs, we have some examples. 193 00:14:24,050 --> 00:14:27,960 So now let's go on to variations of Roth's theorem. 194 00:14:27,960 --> 00:14:31,570 And I want to show you a higher dimensional 195 00:14:31,570 --> 00:14:33,750 version of Roth's theorem. 196 00:14:33,750 --> 00:14:35,500 So I mentioned in the very first lecture-- 197 00:14:35,500 --> 00:14:38,110 so you have this whole host of theorems and additive 198 00:14:38,110 --> 00:14:38,880 combinatorics-- 199 00:14:38,880 --> 00:14:42,880 Roth, Szemeredi, multi-dimensional Szemeredi. 200 00:14:42,880 --> 00:14:45,640 So this is, in some sense, the simplest example 201 00:14:45,640 --> 00:14:48,760 of multi-dimensional Szemeredi theorem. 202 00:14:48,760 --> 00:14:50,770 And this example is known as corners. 203 00:14:53,330 --> 00:14:54,370 So what's a corner? 204 00:14:54,370 --> 00:14:58,660 A corner is, well, we're working inside two dimensions, 205 00:14:58,660 --> 00:15:04,700 so a corner is simply three points, such that the two-- 206 00:15:04,700 --> 00:15:07,990 so they're positioned like that-- such that these two 207 00:15:07,990 --> 00:15:11,132 segments, they are parallel to the axes 208 00:15:11,132 --> 00:15:12,340 and they are the same length. 209 00:15:12,340 --> 00:15:14,890 So that's, by definition, what a corner is. 210 00:15:14,890 --> 00:15:16,810 And the question is, if you give me 211 00:15:16,810 --> 00:15:19,840 a subset of a grid that is corner-free, 212 00:15:19,840 --> 00:15:21,862 how large can this subset be? 213 00:15:25,478 --> 00:15:27,170 Here's a theorem. 214 00:15:27,170 --> 00:15:38,650 So if A is a subset of 1 through N with no corners, 215 00:15:38,650 --> 00:15:44,320 particular, no three points of the form, x comma y; 216 00:15:44,320 --> 00:15:48,680 x plus d comma y; and x comma y plus d, 217 00:15:48,680 --> 00:15:56,780 where d is some positive integer, then the size of A 218 00:15:56,780 --> 00:15:59,030 has to be little o of N squared. 219 00:16:02,840 --> 00:16:03,340 Question. 220 00:16:03,340 --> 00:16:04,965 AUDIENCE: Do we only care about corners 221 00:16:04,965 --> 00:16:06,260 oriented in that direction? 222 00:16:06,260 --> 00:16:07,120 PROFESSOR: OK, good question. 223 00:16:07,120 --> 00:16:08,495 So that's one of the first things 224 00:16:08,495 --> 00:16:10,150 we will address in this proof. 225 00:16:10,150 --> 00:16:12,880 So your question was, do we only care about corners oriented 226 00:16:12,880 --> 00:16:14,510 in the positive direction. 227 00:16:14,510 --> 00:16:18,300 So you can have a more relaxed version of this problem 228 00:16:18,300 --> 00:16:22,150 where you allow d to be negative as well. 229 00:16:22,150 --> 00:16:23,650 The first step in the proof is we'll 230 00:16:23,650 --> 00:16:26,806 see that that constraint doesn't actually matter. 231 00:16:26,806 --> 00:16:27,306 Yes. 232 00:16:32,790 --> 00:16:33,350 OK. 233 00:16:33,350 --> 00:16:33,850 Great. 234 00:16:37,360 --> 00:16:38,110 Let's get started. 235 00:16:40,640 --> 00:16:47,070 So as Michael mentioned, we do have this constraint in this 236 00:16:47,070 --> 00:16:48,950 over here that d is positive. 237 00:16:48,950 --> 00:16:52,680 And it's somewhat annoying because if you remember 238 00:16:52,680 --> 00:16:56,160 in our proof of Roth's theorem, they are positive, negative, 239 00:16:56,160 --> 00:16:57,810 they don't play a role. 240 00:16:57,810 --> 00:17:01,820 So let's try to find a way to get rid of this constraint 241 00:17:01,820 --> 00:17:05,650 so that we are in a more flexible situation. 242 00:17:05,650 --> 00:17:07,510 So first step is that we'll get rid 243 00:17:07,510 --> 00:17:14,190 of this d being positive requirement. 244 00:17:17,480 --> 00:17:20,119 So here's a trick. 245 00:17:20,119 --> 00:17:24,160 Let's consider the sumset A plus A. So sumset 246 00:17:24,160 --> 00:17:31,450 here means I'm looking at all pairwise sums as a set. 247 00:17:31,450 --> 00:17:35,860 So you don't keep track of duplicates, 248 00:17:35,860 --> 00:17:37,720 so you'll keep it as a set. 249 00:17:37,720 --> 00:17:40,930 And we're living inside this grid, 250 00:17:40,930 --> 00:17:43,600 but now somewhat wider in width. 251 00:17:49,930 --> 00:18:01,670 Then there exists an element in this domain of the sumset. 252 00:18:01,670 --> 00:18:03,350 By pigeonhole, that's represented 253 00:18:03,350 --> 00:18:05,900 in many different ways. 254 00:18:05,900 --> 00:18:11,870 So there exists a z represented as a plus b 255 00:18:11,870 --> 00:18:19,290 in at least size of A squared divided 256 00:18:19,290 --> 00:18:26,640 by size of 2N squared different ways, 257 00:18:26,640 --> 00:18:29,190 just because if you look at how many things come up 258 00:18:29,190 --> 00:18:34,086 in that representation, just use pigeonhole. 259 00:18:34,086 --> 00:18:50,033 And now let's take A prime to be A intersect z minus A. So 260 00:18:50,033 --> 00:18:50,950 what's happening here? 261 00:18:50,950 --> 00:18:55,870 So you have this set A. And basically what 262 00:18:55,870 --> 00:18:58,660 I want to do is look at the-- 263 00:18:58,660 --> 00:19:00,130 so suppose A looks like that. 264 00:19:00,130 --> 00:19:04,900 So look at minus A and then shift minus A 265 00:19:04,900 --> 00:19:10,420 over so that they intersect in as many elements as you can. 266 00:19:10,420 --> 00:19:12,400 And by pigeonhole, I can guarantee 267 00:19:12,400 --> 00:19:14,820 that their intersection is fairly large. 268 00:19:21,130 --> 00:19:24,570 So because the size of A prime, it's 269 00:19:24,570 --> 00:19:27,600 essentially the number of ways that z can be represented 270 00:19:27,600 --> 00:19:31,450 in the aforementioned manner. 271 00:19:31,450 --> 00:19:44,303 So it suffices to show that A prime is little o of N squared. 272 00:19:44,303 --> 00:19:45,970 If you show that, then you automatically 273 00:19:45,970 --> 00:19:48,920 show A is little o of N squared. 274 00:19:48,920 --> 00:19:51,820 But now A prime is symmetric. 275 00:19:51,820 --> 00:19:54,460 A prime is symmetric around 0 over 2. 276 00:19:59,290 --> 00:20:05,370 So this is centrally symmetric about z over 2, 277 00:20:05,370 --> 00:20:09,450 meaning that A prime equals to z minus A prime. 278 00:20:12,280 --> 00:20:13,900 And so you see now we've gotten rid 279 00:20:13,900 --> 00:20:19,090 of this d positive requirement because if A prime had 280 00:20:19,090 --> 00:20:23,390 a positive corner, then it has a negative corner and vice versa. 281 00:20:27,647 --> 00:20:35,520 So no corner in A prime with d positive 282 00:20:35,520 --> 00:20:44,520 implies that no corner with d negative. 283 00:20:44,520 --> 00:20:46,170 So now let's forget about A and A prime 284 00:20:46,170 --> 00:20:50,640 and just replace A by A prime and forget 285 00:20:50,640 --> 00:20:53,890 about this d positive condition. 286 00:20:56,650 --> 00:21:03,143 So let's forget about this part, but I do want d not equal to 0. 287 00:21:03,143 --> 00:21:05,310 Otherwise, you have trivial corners, and, of course, 288 00:21:05,310 --> 00:21:07,310 you always have trivial corners. 289 00:21:07,310 --> 00:21:07,810 All right. 290 00:21:15,120 --> 00:21:20,080 So let's remember how the proof of Roth's theorem went. 291 00:21:20,080 --> 00:21:22,610 And this relates to the very first lecture where 292 00:21:22,610 --> 00:21:26,480 I showed you this connection between additive combinatorics 293 00:21:26,480 --> 00:21:29,440 on one hand and graph theory on the other hand, 294 00:21:29,440 --> 00:21:31,930 or you take some arithmetic pattern 295 00:21:31,930 --> 00:21:34,840 and you try to encode it in a graph 296 00:21:34,840 --> 00:21:37,240 so that the patterns in your arithmetic 297 00:21:37,240 --> 00:21:43,160 set correspond to patterns in the graph. 298 00:21:43,160 --> 00:21:45,840 So we're going to do the same thing here. 299 00:21:45,840 --> 00:21:50,000 So we're going to encode the subset of the grid 300 00:21:50,000 --> 00:21:53,990 as a tripartite graph in such a way 301 00:21:53,990 --> 00:21:58,520 that the corners correspond to triangles in the graph. 302 00:21:58,520 --> 00:22:03,365 So I'll show you how to do this and it will be fairly simple. 303 00:22:03,365 --> 00:22:07,557 And in general, sometimes it takes a little bit of ingenuity 304 00:22:07,557 --> 00:22:09,890 to figure out what is the right way to set up the graph. 305 00:22:09,890 --> 00:22:12,920 So one of an upcoming homework problem 306 00:22:12,920 --> 00:22:14,630 will be for you to figure out how 307 00:22:14,630 --> 00:22:18,710 to set up a corresponding graph when the pattern is not 308 00:22:18,710 --> 00:22:20,690 a corner, but a square. 309 00:22:20,690 --> 00:22:24,260 If I add one extra point up there, how would you do that? 310 00:22:27,470 --> 00:22:29,690 So what does this graph look like? 311 00:22:29,690 --> 00:22:41,784 Let's build a tripartite graph where-- 312 00:22:41,784 --> 00:22:43,570 so this should be somewhat reminiscent 313 00:22:43,570 --> 00:22:48,120 of the proof of Roth's theorem-- 314 00:22:48,120 --> 00:22:56,240 where I give you three sets, x, y, and z; and x and y 315 00:22:56,240 --> 00:22:59,600 are both going to have N elements 316 00:22:59,600 --> 00:23:03,355 and z is now going to have 2N elements. 317 00:23:08,720 --> 00:23:13,000 x is supposed to enumerate or index 318 00:23:13,000 --> 00:23:19,680 all the vertical lines in your grid. 319 00:23:19,680 --> 00:23:30,130 So you have this grid of N by N. So x has size 4. 320 00:23:30,130 --> 00:23:33,160 Each vertex in x corresponds to a vertical line. 321 00:23:38,450 --> 00:23:48,350 y corresponds to the horizontal lines and z corresponds 322 00:23:48,350 --> 00:23:52,220 to these negatively sloped diagonal lines-- 323 00:23:52,220 --> 00:23:55,713 slope minus 1 lines. 324 00:23:55,713 --> 00:23:57,380 And of course you should only take lines 325 00:23:57,380 --> 00:24:01,250 that affect your N by N grid. 326 00:24:01,250 --> 00:24:03,440 So that's why there are this many of them 327 00:24:03,440 --> 00:24:06,080 for each direction. 328 00:24:06,080 --> 00:24:07,760 So what's the graph? 329 00:24:07,760 --> 00:24:21,685 I join two vertices if their corresponding lines 330 00:24:21,685 --> 00:24:39,055 meet in a point of A. So I might have-- 331 00:24:45,320 --> 00:24:50,060 this may be A, a point of A. So I 332 00:24:50,060 --> 00:24:53,570 put an edge between those two lines-- 333 00:24:53,570 --> 00:24:56,300 one for x, one for z-- 334 00:24:56,300 --> 00:25:02,075 because their intersection lies in A. And more explicitly-- 335 00:25:02,075 --> 00:25:03,450 I mean, that is pretty explicit-- 336 00:25:03,450 --> 00:25:06,000 and alternatively, you could also 337 00:25:06,000 --> 00:25:12,465 write this graph by telling us how to put 338 00:25:12,465 --> 00:25:15,540 in the edge between x and z. 339 00:25:15,540 --> 00:25:22,280 Namely, you do this if x comma z minus x lies in A; 340 00:25:22,280 --> 00:25:28,960 you put an edge between x and y if x comma y lies in A; 341 00:25:28,960 --> 00:25:31,710 and likewise, you put in the final edge 342 00:25:31,710 --> 00:25:37,680 if z minus y comma y lies in A. 343 00:25:37,680 --> 00:25:39,760 So two equivalent descriptions of the same graph. 344 00:25:44,050 --> 00:25:46,900 Any questions? 345 00:25:46,900 --> 00:25:47,498 All right. 346 00:25:47,498 --> 00:25:49,540 So the rest of the proof is more or less the same 347 00:25:49,540 --> 00:25:52,940 as the proof of Roth's theorem we saw last time. 348 00:25:52,940 --> 00:25:55,930 So we need to figure out how many edges are there. 349 00:26:00,710 --> 00:26:05,840 Well, every element of A gives you three edges. 350 00:26:10,510 --> 00:26:12,990 So that's one of the edges corresponding to that element. 351 00:26:12,990 --> 00:26:17,060 The other two pairs with two other edges. 352 00:26:17,060 --> 00:26:24,530 And most importantly, the triangles in this graph, 353 00:26:24,530 --> 00:26:28,435 I claim, correspond to corners. 354 00:26:31,890 --> 00:26:33,680 So what is a triangle? 355 00:26:33,680 --> 00:26:36,790 A triangle corresponds to a horizontal line, 356 00:26:36,790 --> 00:26:40,660 a vertical line, and a slope 1 minus line 357 00:26:40,660 --> 00:26:44,520 that pairwise intersect in elements of A. 358 00:26:44,520 --> 00:26:47,730 If you look at their intersections, that's a corner. 359 00:26:47,730 --> 00:26:51,550 And conversely, if you have a corner, you build a triangle. 360 00:26:51,550 --> 00:26:59,040 And because your graph, your set is corner-free, well, 361 00:26:59,040 --> 00:27:00,390 you don't have any triangles. 362 00:27:00,390 --> 00:27:02,340 Actually, no, that's not true. 363 00:27:02,340 --> 00:27:05,670 So like we saw in Roth's theorem, 364 00:27:05,670 --> 00:27:07,860 you have some triangles, but corresponding 365 00:27:07,860 --> 00:27:10,410 to trivial corners. 366 00:27:10,410 --> 00:27:18,880 So triangles correspond to trivial corners 367 00:27:18,880 --> 00:27:24,160 because your set A is corner-free. 368 00:27:24,160 --> 00:27:28,840 Trivial corners, meaning the same point with d close to 0, 369 00:27:28,840 --> 00:27:34,770 so it's not a genuine corner like that. 370 00:27:34,770 --> 00:27:39,050 And in particular, in this graph, 371 00:27:39,050 --> 00:27:46,515 every edge is in a unique triangle. 372 00:27:55,412 --> 00:27:56,870 And so we're in the same situation. 373 00:27:56,870 --> 00:28:02,900 By the corollary of triangle removal lemma, 374 00:28:02,900 --> 00:28:11,350 we find that the number of edges must 375 00:28:11,350 --> 00:28:17,060 be little o of the number of vertices. 376 00:28:20,930 --> 00:28:23,400 The number of edges must be subquadratic, 377 00:28:23,400 --> 00:28:27,380 so it must be little o of N squared. 378 00:28:27,380 --> 00:28:30,360 And so that implies that the size of A 379 00:28:30,360 --> 00:28:32,310 is little o of N squared. 380 00:28:35,256 --> 00:28:39,300 So that proves the Corners theorem. 381 00:28:39,300 --> 00:28:40,381 Any questions? 382 00:28:44,530 --> 00:28:46,270 So once you set up this graph, the rest 383 00:28:46,270 --> 00:28:47,540 is the same as Roth's theorem. 384 00:28:47,540 --> 00:28:49,840 So this connection between graph theory and additive 385 00:28:49,840 --> 00:28:52,210 combinatorics, well, you need to figure out 386 00:28:52,210 --> 00:28:53,390 how to set up this graph. 387 00:28:53,390 --> 00:28:57,760 And then sometimes it's clear, but sometimes you 388 00:28:57,760 --> 00:29:00,760 have to really think hard about how to do this. 389 00:29:04,000 --> 00:29:05,080 All right. 390 00:29:05,080 --> 00:29:08,680 What does corner have to do with Roth's theorem, 391 00:29:08,680 --> 00:29:12,415 other than that they have very similar looking proofs? 392 00:29:12,415 --> 00:29:14,290 Well, actually, you can deduce Roth's theorem 393 00:29:14,290 --> 00:29:15,550 from the Corners theorem. 394 00:29:18,990 --> 00:29:24,070 And to show you this precisely, so let me use r sub 3 of N 395 00:29:24,070 --> 00:29:37,950 to denote the size of the largest 3-AP-free set of 1 396 00:29:37,950 --> 00:29:41,520 through N. 397 00:29:41,520 --> 00:29:44,790 So this notation, the r sub 3 is actually fairly standard. 398 00:29:44,790 --> 00:29:47,443 The next one's not so standard, but let's just do that. 399 00:29:47,443 --> 00:29:49,110 So that's not an L, but that's a corner. 400 00:29:51,690 --> 00:29:58,410 So that is the size of the largest corner-free subset 401 00:29:58,410 --> 00:30:02,940 of N squared. 402 00:30:02,940 --> 00:30:06,890 So we gave bounds for both quantities, 403 00:30:06,890 --> 00:30:11,780 but they are actually related to each other through this fairly 404 00:30:11,780 --> 00:30:14,300 simple proposition that if you have 405 00:30:14,300 --> 00:30:16,370 an upper bound for corners, then you 406 00:30:16,370 --> 00:30:20,497 have an upper bound for Roth's theorem. 407 00:30:25,270 --> 00:30:39,430 Indeed, given A a 3-AP-free subset of 1 through N, 408 00:30:39,430 --> 00:30:42,720 let me build for you a corner-free subset 409 00:30:42,720 --> 00:30:47,430 of the grid that is a fairly large subset of that grid. 410 00:30:50,220 --> 00:31:04,220 I can form B by setting it to be the set of pairs 411 00:31:04,220 --> 00:31:09,930 inside this grid whose difference, x minus y, 412 00:31:09,930 --> 00:31:16,530 lies in A. So what does this look like? 413 00:31:23,910 --> 00:31:28,450 This is the grid of size 2N. 414 00:31:28,450 --> 00:31:37,590 And if I start with A that is 3-AP-free, what I can do-- 415 00:31:37,590 --> 00:31:38,880 so over here-- 416 00:31:38,880 --> 00:31:52,260 is look at the lines like that, putting all of those points. 417 00:31:52,260 --> 00:31:54,060 And you see that this set of points 418 00:31:54,060 --> 00:32:00,590 should not have any corners because if they have corners, 419 00:32:00,590 --> 00:32:02,920 then the corners would project down to a 3-AP. 420 00:32:06,740 --> 00:32:08,386 So B is corner-free. 421 00:32:16,970 --> 00:32:19,680 To recap, if we have upper bound on corners, 422 00:32:19,680 --> 00:32:22,630 we have upper bound on Roth's theorem. 423 00:32:22,630 --> 00:32:25,425 But you also know that if you have lower 424 00:32:25,425 --> 00:32:26,800 bound on Roth's theorem, then you 425 00:32:26,800 --> 00:32:28,420 have lower bound on corners. 426 00:32:28,420 --> 00:32:31,210 So the Behrend construction we saw at the beginning of today 427 00:32:31,210 --> 00:32:34,540 extends to, you know, through exactly this way to a fairly 428 00:32:34,540 --> 00:32:35,980 large corner-free subset. 429 00:32:35,980 --> 00:32:38,920 And that's more or less the best thing that we know how to do. 430 00:32:38,920 --> 00:32:41,760 In fact, there aren't that many constructions 431 00:32:41,760 --> 00:32:43,990 in additive combinatorics that's known. 432 00:32:43,990 --> 00:32:45,430 Almost everything that I know how 433 00:32:45,430 --> 00:32:47,830 to construct that's fairly large come 434 00:32:47,830 --> 00:32:50,160 from Behrend's construction or some variant 435 00:32:50,160 --> 00:32:51,850 of Behrend's construction. 436 00:32:51,850 --> 00:32:53,170 So it looks pretty simple. 437 00:32:53,170 --> 00:32:55,060 It comes from the '40s, yet we don't really 438 00:32:55,060 --> 00:32:59,110 have too many new ideas besides playing and massaging 439 00:32:59,110 --> 00:33:00,520 Behrend's construction. 440 00:33:04,360 --> 00:33:07,180 Let me tell you what is the best known upper bound 441 00:33:07,180 --> 00:33:08,680 on the Corners theorem. 442 00:33:14,440 --> 00:33:15,440 This is due to Shkredov. 443 00:33:18,430 --> 00:33:22,185 So the proof using triangle removal lemma, 444 00:33:22,185 --> 00:33:25,110 it goes to Szemeredi's regularity lemma. 445 00:33:25,110 --> 00:33:26,760 It gives you pretty horrible bounds, 446 00:33:26,760 --> 00:33:31,070 but using Fourier analytic methods-- 447 00:33:31,070 --> 00:33:33,010 so you see if you have upper bound on Roth, 448 00:33:33,010 --> 00:33:35,200 it doesn't give you an upper bound on corner, 449 00:33:35,200 --> 00:33:37,700 so you need to do something extra. 450 00:33:37,700 --> 00:33:43,330 And so the best known bound so far is of the form, 451 00:33:43,330 --> 00:33:47,825 N squared divided by polylog log N, 452 00:33:47,825 --> 00:33:50,305 so log log N raised to some small constant, 453 00:33:50,305 --> 00:33:53,748 C. Any questions? 454 00:34:02,050 --> 00:34:04,730 Last time, we discussed the triangle counting lemma 455 00:34:04,730 --> 00:34:06,070 and the triangle removal lemma. 456 00:34:08,870 --> 00:34:11,020 Well, it shouldn't be a surprise to you 457 00:34:11,020 --> 00:34:12,480 that if we can do it for triangles, 458 00:34:12,480 --> 00:34:15,929 we may be able to do it for other subgraphs. 459 00:34:15,929 --> 00:34:17,790 So that's the next thing I want to discuss-- 460 00:34:17,790 --> 00:34:22,380 how to generalize the techniques and results 461 00:34:22,380 --> 00:34:25,500 that we obtain for triangles to other graphs, 462 00:34:25,500 --> 00:34:27,090 and what are some of the implications 463 00:34:27,090 --> 00:34:30,000 if you combine it with Szemeredi's regularity lemma. 464 00:34:49,179 --> 00:34:56,236 So let's generalize the triangle counting lemma. 465 00:35:09,370 --> 00:35:11,650 So the strategy for the triangle counting lemma, 466 00:35:11,650 --> 00:35:15,110 let me remind you, was that we embedded the vertices one 467 00:35:15,110 --> 00:35:15,610 by one. 468 00:35:18,280 --> 00:35:20,890 Putting a vertex and a typical vertex here 469 00:35:20,890 --> 00:35:27,610 should have many neighbors to both vertex sets. 470 00:35:27,610 --> 00:35:32,980 So these two guys should have sizes typically roughly 471 00:35:32,980 --> 00:35:36,730 the same as a fraction of them corresponding 472 00:35:36,730 --> 00:35:38,920 to the edge density between the vertex parts. 473 00:35:41,610 --> 00:35:43,490 And if they're not too small, then 474 00:35:43,490 --> 00:35:46,670 from the epsilon regularity of these two sets, 475 00:35:46,670 --> 00:35:50,250 you can deduce the number of edges between them. 476 00:35:50,250 --> 00:35:54,210 So that was the strategy for the triangle counting lemma. 477 00:35:54,210 --> 00:35:58,580 So you can try to extend the same strategy for other graphs. 478 00:35:58,580 --> 00:36:01,410 So let me show you how this would have been done, 479 00:36:01,410 --> 00:36:04,520 but I don't want to give too many details because it 480 00:36:04,520 --> 00:36:07,420 does get somewhat hairy if you try to execute it. 481 00:36:11,300 --> 00:36:18,730 So the first strategy is to embed the vertices of H one 482 00:36:18,730 --> 00:36:19,230 at a time. 483 00:36:24,030 --> 00:36:29,100 So my H now is going to be, let's say, a K4. 484 00:36:32,770 --> 00:36:38,190 And I wish to embed this H in this setting 485 00:36:38,190 --> 00:36:41,940 where you have these four parts in the regularity partition, 486 00:36:41,940 --> 00:36:45,570 and they are pairwise epsilon regular with edge densities 487 00:36:45,570 --> 00:36:49,700 that are not too small. 488 00:36:49,700 --> 00:36:52,600 Well, what you can try to do, mimicking the strategy 489 00:36:52,600 --> 00:36:58,570 over there, is to first find a typical image 490 00:36:58,570 --> 00:36:59,875 for the top vertex. 491 00:37:04,760 --> 00:37:07,920 And a typical image over here, minus 492 00:37:07,920 --> 00:37:14,460 some small bad exceptions, will have many neighbors 493 00:37:14,460 --> 00:37:15,600 to each of the three parts. 494 00:37:19,160 --> 00:37:25,340 Next, I need to figure out where this vertex can go. 495 00:37:25,340 --> 00:37:31,020 I'm going to embed this vertex somewhere here. 496 00:37:31,020 --> 00:37:34,440 Again, a typical place for this vertex, 497 00:37:34,440 --> 00:37:37,440 modulo some small fraction, which I'm going to throw away. 498 00:37:37,440 --> 00:37:40,380 So now you see you need somewhat stronger hypotheses 499 00:37:40,380 --> 00:37:43,260 on the epsilon regularity, but they are still 500 00:37:43,260 --> 00:37:45,720 all polynomial dependent, so you just 501 00:37:45,720 --> 00:37:47,830 have to choose your parameters correctly. 502 00:37:47,830 --> 00:37:52,120 So this typical green vertex should have 503 00:37:52,120 --> 00:37:56,940 lots of neighbors over here. 504 00:37:59,740 --> 00:38:01,450 So you just keep embedding. 505 00:38:01,450 --> 00:38:03,280 So it's almost this greedy strategy. 506 00:38:03,280 --> 00:38:06,190 You keep embedding each vertex, but you 507 00:38:06,190 --> 00:38:10,930 have to guarantee that there are still lots of options left. 508 00:38:16,690 --> 00:38:20,170 So embed vertices one at a time. 509 00:38:20,170 --> 00:38:34,050 And I want to embed each vertex so that the yet to be embedded 510 00:38:34,050 --> 00:38:47,550 vertices have many choices left. 511 00:39:01,690 --> 00:39:05,490 So epsilon regularity guarantees that you can always do this. 512 00:39:05,490 --> 00:39:08,290 And you do this all the way until the end 513 00:39:08,290 --> 00:39:11,490 and you arrive at some statement. 514 00:39:11,490 --> 00:39:13,540 And depending on how you do this, 515 00:39:13,540 --> 00:39:15,340 the exact formulation of the statement 516 00:39:15,340 --> 00:39:16,798 will be somewhat different, but let 517 00:39:16,798 --> 00:39:19,120 me give you one statement which we will not prove, 518 00:39:19,120 --> 00:39:22,060 but you can deduce using this strategy. 519 00:39:22,060 --> 00:39:25,720 And this is known as a graph embedding lemma. 520 00:39:25,720 --> 00:39:28,420 And again, as I mentioned when I started 521 00:39:28,420 --> 00:39:29,855 discussing Szemeredi's regularity 522 00:39:29,855 --> 00:39:34,870 lemma, the exact statements, they are fairly robust 523 00:39:34,870 --> 00:39:39,520 and they are not as important as the spirit of the ideas. 524 00:39:39,520 --> 00:39:41,380 So if you have some application in mind, 525 00:39:41,380 --> 00:39:44,170 you might have to go into the proof, tweak a thing 526 00:39:44,170 --> 00:39:47,210 here and there, but you get what you want. 527 00:39:47,210 --> 00:39:50,240 So the graph embedding lemma says, for example, 528 00:39:50,240 --> 00:39:59,020 that if H is bipartite, and with maximum degree and most delta-- 529 00:39:59,020 --> 00:40:03,120 so the H maximum degrees at most delta-- 530 00:40:03,120 --> 00:40:11,930 and suppose you have vertex sets V1 through Vr, such 531 00:40:11,930 --> 00:40:21,970 that each vertex set is not too small; 532 00:40:21,970 --> 00:40:29,620 and if these vertex sets are pairwise epsilon 533 00:40:29,620 --> 00:40:36,550 regular and the density is not too small. 534 00:40:40,890 --> 00:40:44,440 So here I'm assuming some lower bound 535 00:40:44,440 --> 00:40:50,220 on the density which depends on your epsilon. 536 00:40:50,220 --> 00:41:02,910 Then the conclusion is that G contains a copy of H. 537 00:41:02,910 --> 00:41:06,090 I just want to give a few remarks on the statement 538 00:41:06,090 --> 00:41:06,900 of this theorem. 539 00:41:06,900 --> 00:41:10,000 Again, we will not discuss the proof. 540 00:41:10,000 --> 00:41:14,250 So what is this hypothesis on H being 541 00:41:14,250 --> 00:41:16,200 r-partite have to do with anything? 542 00:41:16,200 --> 00:41:19,620 So here, as an example, when r equals to 4, instead 543 00:41:19,620 --> 00:41:24,870 of this K4, maybe I also care about that graph over there. 544 00:41:29,520 --> 00:41:31,520 Well, maybe some more vertices, some more edges, 545 00:41:31,520 --> 00:41:32,840 but if it's 4-partite. 546 00:41:32,840 --> 00:41:37,640 And the point is that I want to embed the vertices in such 547 00:41:37,640 --> 00:41:42,290 a way that that top vertex goes to the part 548 00:41:42,290 --> 00:41:45,160 that it's supposed to go into. 549 00:41:45,160 --> 00:41:47,660 So I'm embedding this configuration 550 00:41:47,660 --> 00:41:52,400 in the same way that corresponds to the proper coloring of H. 551 00:41:52,400 --> 00:41:56,840 So if you do this, there's enough room still 552 00:41:56,840 --> 00:42:01,580 to go, as long as you have some lower bound on the edge 553 00:42:01,580 --> 00:42:06,240 density between the individual parts, 554 00:42:06,240 --> 00:42:10,170 and you only depend really on not the number of edges 555 00:42:10,170 --> 00:42:13,740 of H, but the maximum degree. 556 00:42:13,740 --> 00:42:18,640 Because if you look at how many times each vertex, 557 00:42:18,640 --> 00:42:22,170 its possibilities can be shrunk, it's in most delta times. 558 00:42:25,290 --> 00:42:26,993 Now, this graph embedding lemma-- 559 00:42:26,993 --> 00:42:28,410 so I give you this statement here, 560 00:42:28,410 --> 00:42:29,920 but it's a fairly robust statement. 561 00:42:29,920 --> 00:42:32,800 And if you want to get, for example, not just a single copy 562 00:42:32,800 --> 00:42:35,505 of H, if you want to get many copies of H, 563 00:42:35,505 --> 00:42:36,880 you can tweak the hypotheses, you 564 00:42:36,880 --> 00:42:38,920 can tweak the proofs somewhat to get you 565 00:42:38,920 --> 00:42:42,045 what you want, again, following what we did for the triangle 566 00:42:42,045 --> 00:42:42,670 counting lemma. 567 00:42:45,350 --> 00:42:46,670 Question. 568 00:42:46,670 --> 00:42:50,226 AUDIENCE: Is the bound on the H edge density between partitions 569 00:42:50,226 --> 00:42:51,040 correct? 570 00:42:51,040 --> 00:42:53,770 So if the maximum degree increases, 571 00:42:53,770 --> 00:42:55,240 the lower bound decreases? 572 00:42:55,240 --> 00:42:58,981 PROFESSOR: If maximum degree increases, this number goes up. 573 00:42:58,981 --> 00:42:59,963 AUDIENCE: Oh, OK. 574 00:43:08,810 --> 00:43:11,300 PROFESSOR: I want to show you a different way 575 00:43:11,300 --> 00:43:15,710 to do counting that does not go through this embedding vertices 576 00:43:15,710 --> 00:43:18,680 one by one, but instead we will try 577 00:43:18,680 --> 00:43:22,250 to analyze what happens if you take out an edge of H one 578 00:43:22,250 --> 00:43:24,190 by one. 579 00:43:24,190 --> 00:43:27,010 And that's an alternative approach which I like more. 580 00:43:27,010 --> 00:43:29,270 It's somewhat less intuitive if you are not 581 00:43:29,270 --> 00:43:31,880 used to thinking about it, but the execution 582 00:43:31,880 --> 00:43:33,350 works out to be much cleaner. 583 00:43:33,350 --> 00:43:37,460 And it's also in line with some of the techniques 584 00:43:37,460 --> 00:43:41,398 that we'll see later on when we discuss graph limits. 585 00:43:41,398 --> 00:43:42,440 Let's take a quick break. 586 00:43:45,620 --> 00:43:46,630 Any questions so far? 587 00:43:54,100 --> 00:43:58,880 We will see a second strategy for proving a graph counting 588 00:43:58,880 --> 00:44:00,650 lemma. 589 00:44:00,650 --> 00:44:06,100 And the second strategy is more analytic in nature, 590 00:44:06,100 --> 00:44:15,740 which is to embed, well, to analytically somehow we'll 591 00:44:15,740 --> 00:44:19,670 analyze what happens when we pick out one edge of H 592 00:44:19,670 --> 00:44:20,292 at a time. 593 00:44:26,635 --> 00:44:28,260 So let me give you the statement first. 594 00:44:38,050 --> 00:44:41,070 So the graph counting lemma says that if you 595 00:44:41,070 --> 00:44:51,030 have a graph H with vertex set elements of 1 through k-- 596 00:44:51,030 --> 00:44:55,110 I also have an epsilon parameter and I 597 00:44:55,110 --> 00:45:04,580 have a graph G and subsets V1 through Vk of G, 598 00:45:04,580 --> 00:45:11,900 such that Vi Vj is epsilon regular 599 00:45:11,900 --> 00:45:18,370 whenever ij is an edge of H. 600 00:45:18,370 --> 00:45:21,130 So here, the setup is slightly different from the picture 601 00:45:21,130 --> 00:45:22,910 I drew up there. 602 00:45:22,910 --> 00:45:28,770 So what's going on here is that you have, 603 00:45:28,770 --> 00:45:35,390 let's say, H being this graph. 604 00:45:35,390 --> 00:45:38,990 And suppose you are in a situation where 605 00:45:38,990 --> 00:45:41,180 I know that some of these-- 606 00:45:47,700 --> 00:45:53,370 so I know that five of these pairs are epsilon regular, 607 00:45:53,370 --> 00:45:58,200 and what I really want to do is embed this H 608 00:45:58,200 --> 00:45:59,933 into this configuration. 609 00:46:06,090 --> 00:46:08,920 So 1, 2; V1, V2; and so on. 610 00:46:11,800 --> 00:46:15,010 And I want to know how many ways can you embed this way. 611 00:46:18,690 --> 00:46:21,240 The conclusion is that the number 612 00:46:21,240 --> 00:46:36,900 of tuples, the number of such embeddings, such that Vi, Vj 613 00:46:36,900 --> 00:46:45,130 is an edge of G for all ij being the edge of H-- 614 00:46:45,130 --> 00:46:47,040 so exactly as showing that picture, 615 00:46:47,040 --> 00:46:52,390 the number of such embeddings, the little v's, is 616 00:46:52,390 --> 00:46:57,040 within a small error, which is the number of edges 617 00:46:57,040 --> 00:47:05,660 of H epsilon times the total, the product of these vertex 618 00:47:05,660 --> 00:47:16,810 set sizes of this number, which is what you would predict 619 00:47:16,810 --> 00:47:21,400 the number of embeddings to be if all of your bipartite graphs 620 00:47:21,400 --> 00:47:23,100 were actually random. 621 00:47:36,350 --> 00:47:38,530 So like in the triangle counting lemma, 622 00:47:38,530 --> 00:47:42,520 if you look at the edge densities in this configuration 623 00:47:42,520 --> 00:47:46,420 and predict how many copies of H you would get, 624 00:47:46,420 --> 00:47:48,400 that's the number you should write down. 625 00:47:48,400 --> 00:47:51,730 And this counting lemma tells you 626 00:47:51,730 --> 00:47:55,955 that the truth is not so far from the prediction. 627 00:48:04,510 --> 00:48:05,679 Any questions? 628 00:48:14,170 --> 00:48:16,000 So we will prove the graph counting 629 00:48:16,000 --> 00:48:21,430 lemma in an analytic manner. 630 00:48:21,430 --> 00:48:24,130 It helps to-- so it will be convenient 631 00:48:24,130 --> 00:48:28,090 for me to rephrase the result just a little bit 632 00:48:28,090 --> 00:48:29,940 in this probabilistic form. 633 00:48:29,940 --> 00:48:34,870 So it has the equivalent to show that if you 634 00:48:34,870 --> 00:48:42,325 have uniformly randomly chosen vertices, little v1 and big V1, 635 00:48:42,325 --> 00:48:45,770 and so on-- 636 00:48:45,770 --> 00:48:50,700 so little vk and big VK. 637 00:48:50,700 --> 00:48:53,770 So they're independent, uniformly, and random, 638 00:48:53,770 --> 00:48:59,000 then the probability-- 639 00:48:59,000 --> 00:49:02,830 so basically, I am putting down a potential image 640 00:49:02,830 --> 00:49:06,225 for each vertex of H and asking what's 641 00:49:06,225 --> 00:49:07,600 the probability that you actually 642 00:49:07,600 --> 00:49:08,980 have an embedding of H. 643 00:49:08,980 --> 00:49:12,070 So the probability that little vi, little vj 644 00:49:12,070 --> 00:49:18,910 is an actual edge of G for all ij being an edge of H, 645 00:49:18,910 --> 00:49:24,150 this number here, we're saying that it differs 646 00:49:24,150 --> 00:49:28,500 from the prediction, which is simply multiplying all the edge 647 00:49:28,500 --> 00:49:29,670 densities together. 648 00:49:36,070 --> 00:49:39,710 So the difference between the actual and the predicted values 649 00:49:39,710 --> 00:49:43,117 is fairly small. 650 00:49:43,117 --> 00:49:45,450 So I haven't done anything, just rephrasing the problem. 651 00:49:45,450 --> 00:49:48,104 Instead of counting, now we're looking at probabilities. 652 00:49:53,050 --> 00:49:56,920 As I mentioned, we'll take out one edge at a time. 653 00:49:56,920 --> 00:50:01,510 So relabeling if necessary, let's assume 654 00:50:01,510 --> 00:50:09,870 that 1, 2 is an edge of H. So now we 655 00:50:09,870 --> 00:50:18,080 will show the following plane, so I'll denote star. 656 00:50:18,080 --> 00:50:31,380 That, if you look at this quantity over here 657 00:50:31,380 --> 00:50:40,700 compared to if you take out just the edge density between V1 658 00:50:40,700 --> 00:50:47,670 and V2, but now you put in a similar quantity 659 00:50:47,670 --> 00:50:55,820 where I'm considering all of the edges of H, except for 1, 2. 660 00:51:11,750 --> 00:51:14,930 I claim that this difference is, at most, epsilon. 661 00:51:24,390 --> 00:51:26,070 So you can think of this quantity 662 00:51:26,070 --> 00:51:32,340 here as the same quantity as the green one, except not on H, 663 00:51:32,340 --> 00:51:34,990 but on H minus the edge 1, 2. 664 00:51:43,350 --> 00:51:50,190 To show this star claim, let us couple the two random processes 665 00:51:50,190 --> 00:51:51,860 choosing these little vi's. 666 00:52:07,990 --> 00:52:11,710 By that, I mean here you have the random little vi's and here 667 00:52:11,710 --> 00:52:14,590 you have the random little vi's, but use the same little vi's 668 00:52:14,590 --> 00:52:17,430 in both probabilities. 669 00:52:17,430 --> 00:52:21,720 So in both, there are two different random events, 670 00:52:21,720 --> 00:52:24,430 but you use the same little vi's, 671 00:52:24,430 --> 00:52:35,310 it suffices to show this inequality star with-- 672 00:52:37,900 --> 00:52:39,770 so what's this process, this random process? 673 00:52:39,770 --> 00:52:42,830 You pick V1, you pick V2, you pick V3, all of them 674 00:52:42,830 --> 00:52:45,690 independently uniformly at random. 675 00:52:45,690 --> 00:52:50,520 But if I show you the inequality under a further constraint 676 00:52:50,520 --> 00:52:55,540 of arbitrary V3 through VN, then that's even better. 677 00:52:58,540 --> 00:53:10,510 So with V3 through Vk, fixed arbitrary, 678 00:53:10,510 --> 00:53:18,420 and only little v1 and little v2 random. 679 00:53:27,028 --> 00:53:29,570 So you can phrase this in terms of conditional probabilities, 680 00:53:29,570 --> 00:53:31,450 if you like. 681 00:53:31,450 --> 00:53:33,950 You're comparing these two probabilities. 682 00:53:33,950 --> 00:53:38,660 Now I fix the V3's through Vk's, and I just 683 00:53:38,660 --> 00:53:40,850 let V1 and V2 be random. 684 00:53:40,850 --> 00:53:47,720 And if condition on V3 through Vk, you have this inequality, 685 00:53:47,720 --> 00:53:51,080 then letting V3 through Vk go, letting 686 00:53:51,080 --> 00:53:53,750 them be random as well, by triangle inequality, 687 00:53:53,750 --> 00:53:57,260 you obtain the bound that we're looking for. 688 00:54:02,060 --> 00:54:03,340 Any questions about this step? 689 00:54:06,220 --> 00:54:08,630 If you're confused about what's happening here, 690 00:54:08,630 --> 00:54:11,840 another way to bypass all the probability language 691 00:54:11,840 --> 00:54:14,720 is to go back to the counting language. 692 00:54:14,720 --> 00:54:17,260 So in other words, we're trying to count embeddings. 693 00:54:17,260 --> 00:54:23,590 And I'm asking if you arbitrarily fix V3 through Vk, 694 00:54:23,590 --> 00:54:26,600 how many different choices are there to V1 and V2 695 00:54:26,600 --> 00:54:27,850 in the two different settings. 696 00:54:32,600 --> 00:54:34,420 OK. 697 00:54:34,420 --> 00:54:47,950 So let A1 be the set of places where little v1 can go, 698 00:54:47,950 --> 00:54:51,880 if you already knew what V3 through Vk should be. 699 00:55:05,980 --> 00:55:06,510 OK. 700 00:55:06,510 --> 00:55:12,600 So you look at all the neighbors of V1, neighbors of 1 and H, 701 00:55:12,600 --> 00:55:14,650 except for 2. 702 00:55:14,650 --> 00:55:19,250 And I want to make sure that V1, Vi, as i 703 00:55:19,250 --> 00:55:24,820 ranges over all such neighbors, is indeed a valid H in G. 704 00:55:24,820 --> 00:55:26,410 I'll draw a picture in a second. 705 00:55:26,410 --> 00:55:36,870 And A2, likewise, is the same quantity, 706 00:55:36,870 --> 00:55:38,310 but with 2 instead of 1. 707 00:55:47,200 --> 00:55:51,640 So for example, if you're trying to embed a K4, what's 708 00:55:51,640 --> 00:56:00,260 happening here is that you have this V1, this V2, and somebody 709 00:56:00,260 --> 00:56:03,130 already arbitrarily fixed the locations 710 00:56:03,130 --> 00:56:06,120 where V3 and V4 are embedded. 711 00:56:06,120 --> 00:56:10,500 And you're asking, well, how many different choices are now 712 00:56:10,500 --> 00:56:12,960 left for little v1. 713 00:56:12,960 --> 00:56:21,830 It's the common neighborhood of V3 and V4. 714 00:56:21,830 --> 00:56:24,730 So that's for A1. 715 00:56:24,730 --> 00:56:32,150 And V3 and V4, also their common neighborhood in B2 is A2. 716 00:56:38,550 --> 00:56:39,050 OK. 717 00:56:39,050 --> 00:56:42,680 So with that notation, what is it that we're 718 00:56:42,680 --> 00:56:46,060 trying to show over here? 719 00:56:46,060 --> 00:56:48,590 So if you rewrite this inequality with the V3's 720 00:56:48,590 --> 00:56:56,880 through Vk's fixed, you find that what we're trying to show 721 00:56:56,880 --> 00:56:59,116 is the following. 722 00:57:03,880 --> 00:57:10,680 So we claim-- and this claim implies the star, 723 00:57:10,680 --> 00:57:16,040 that if you look at what the first term should be-- 724 00:57:16,040 --> 00:57:21,820 so this is the number of edges between A1 and A2 725 00:57:21,820 --> 00:57:26,590 as a fraction of the product of V1 and V2. 726 00:57:33,120 --> 00:57:36,990 And then the second guy here is if you use 727 00:57:36,990 --> 00:57:41,890 the prediction d of V1 and V2. 728 00:57:53,230 --> 00:57:56,620 So this is each of them, each of these two factors, 729 00:57:56,620 --> 00:58:01,420 there's a probability that little v1 lies in A1, 730 00:58:01,420 --> 00:58:03,790 little v2 lies in A2, and then you 731 00:58:03,790 --> 00:58:09,030 tack on this extra constant, namely, this constant here. 732 00:58:09,030 --> 00:58:11,580 So we're trying to show that this difference is small. 733 00:58:11,580 --> 00:58:14,430 And the claim is that this difference is, indeed, 734 00:58:14,430 --> 00:58:15,900 always small. 735 00:58:15,900 --> 00:58:22,800 There's always, at most, epsilon for every A1 in V1 and A2 736 00:58:22,800 --> 00:58:23,400 in V2. 737 00:58:29,970 --> 00:58:33,950 And here, in particular, so this statement 738 00:58:33,950 --> 00:58:35,990 looks somewhat like the definition of epsilon 739 00:58:35,990 --> 00:58:38,150 regularity, but there's no restrictions 740 00:58:38,150 --> 00:58:42,710 on the sizes of A1 and A2, and they don't have to be big. 741 00:58:46,690 --> 00:58:48,670 And as you can imagine, this statement, 742 00:58:48,670 --> 00:58:51,250 we're not really using all that much. 743 00:58:51,250 --> 00:58:56,170 All we're assuming is the epsilon regularity 744 00:58:56,170 --> 00:58:58,150 between B1 and B2. 745 00:58:58,150 --> 00:59:01,000 So we will deduce this inequality 746 00:59:01,000 --> 00:59:07,535 from the hypotheses of epsilon regularity between B1 and B2. 747 00:59:07,535 --> 00:59:08,160 So let's check. 748 00:59:11,730 --> 00:59:19,950 So we know that B1 and B2 is epsilon regular by hypotheses. 749 00:59:19,950 --> 00:59:30,770 So if either A1 or A2 is too small, 750 00:59:30,770 --> 00:59:40,950 so if A1 is too small or A2 is too small, 751 00:59:40,950 --> 00:59:50,920 then we see that both of these terms 752 00:59:50,920 --> 00:59:52,780 here are, at most, epsilon. 753 01:00:00,040 --> 01:00:02,770 So if the A's are too small, then neither of these terms 754 01:00:02,770 --> 01:00:03,730 can be too large. 755 01:00:03,730 --> 01:00:05,920 Here it's bounded by-- 756 01:00:05,920 --> 01:00:08,640 if you took the product of A1 and 757 01:00:08,640 --> 01:00:11,743 A2 and likewise over there. 758 01:00:11,743 --> 01:00:13,660 So in this case, their difference is, at most, 759 01:00:13,660 --> 01:00:14,910 epsilon, and we're good to go. 760 01:00:17,570 --> 01:00:24,750 Otherwise, if A1 and A2 are both at least an epsilon fraction 761 01:00:24,750 --> 01:00:34,220 of their [INAUDIBLE] sets, then we find that-- 762 01:00:37,200 --> 01:00:38,330 so what happens? 763 01:00:38,330 --> 01:00:45,480 So here, so by the hypothesis of epsilon regularity, 764 01:00:45,480 --> 01:00:52,990 we find that d of V1 and V2 differs 765 01:00:52,990 --> 01:00:56,500 from the number of edges between A1 and A2, 766 01:00:56,500 --> 01:00:59,860 divided by the product of their sizes. 767 01:00:59,860 --> 01:01:03,040 So that's just d of A1 and A2. 768 01:01:03,040 --> 01:01:16,490 So this difference is, at most, epsilon, which then implies 769 01:01:16,490 --> 01:01:17,890 the inequality up there. 770 01:01:21,640 --> 01:01:24,160 So here we're using that the size of A 771 01:01:24,160 --> 01:01:35,670 is, at most, the size of V. 772 01:01:35,670 --> 01:01:38,130 So we have this claim. 773 01:01:38,130 --> 01:01:42,720 And that claim proves this inequality in star. 774 01:01:42,720 --> 01:01:44,850 And basically, what we've done is 775 01:01:44,850 --> 01:01:48,630 we showed that if you took out a single edge, 776 01:01:48,630 --> 01:01:53,190 you change the desired quantity by, at most, an epsilon, 777 01:01:53,190 --> 01:01:54,540 essentially. 778 01:01:54,540 --> 01:02:00,030 So now you do this for every edge of H. Alternatively, 779 01:02:00,030 --> 01:02:06,730 you can do induction on the number of edges of H. 780 01:02:06,730 --> 01:02:09,360 So to complete a proof of the counting 781 01:02:09,360 --> 01:02:24,980 lemma, so we do induction on the number of edges of H. 782 01:02:24,980 --> 01:02:30,520 And when H has exactly one edge, well, that's pretty easy. 783 01:02:30,520 --> 01:02:34,210 But now if you have more edges, well, you 784 01:02:34,210 --> 01:02:37,480 apply induction hypothesis to the graph, 785 01:02:37,480 --> 01:02:41,410 which is H minus the edge 1, 2. 786 01:02:44,330 --> 01:02:58,880 And you find that this quantity here 787 01:02:58,880 --> 01:03:14,850 differs from the predicted quantity 788 01:03:14,850 --> 01:03:20,550 by the number of edges of H minus 1 times epsilon. 789 01:03:24,760 --> 01:03:27,310 In other words, you run this prove that we just 790 01:03:27,310 --> 01:03:28,850 did one edge at a time. 791 01:03:28,850 --> 01:03:30,610 So each time you take out an edge, 792 01:03:30,610 --> 01:03:32,350 you use epsilon regularity to show 793 01:03:32,350 --> 01:03:34,660 that the effect of taking that edge out from H 794 01:03:34,660 --> 01:03:38,890 does not have too big of an effect on the actual number 795 01:03:38,890 --> 01:03:40,570 of embeddings. 796 01:03:40,570 --> 01:03:42,940 Do this for one edge at a time, and eventually you 797 01:03:42,940 --> 01:03:44,745 prove the graph counting lemma. 798 01:03:49,110 --> 01:03:52,320 So this is one of those proofs which 799 01:03:52,320 --> 01:03:57,840 may be less intuitive compared to the one I showed earlier, 800 01:03:57,840 --> 01:04:01,120 in the sense that there's not as nice of a story you can tell 801 01:04:01,120 --> 01:04:04,180 about putting in one vertex at a time. 802 01:04:04,180 --> 01:04:07,560 But on the other hand, if you were to carry out this proof 803 01:04:07,560 --> 01:04:11,010 to bound each time how big the sets have to be, 804 01:04:11,010 --> 01:04:12,900 it gets much hairier over here. 805 01:04:12,900 --> 01:04:17,070 And here, the execution is much cleaner, but maybe less 806 01:04:17,070 --> 01:04:19,650 intuitive, unless you're willing to be 807 01:04:19,650 --> 01:04:21,760 comfortable with these calculations. 808 01:04:21,760 --> 01:04:23,010 And it's really not so bad. 809 01:04:25,810 --> 01:04:29,700 And the strength of these two results are somewhat different. 810 01:04:29,700 --> 01:04:33,500 So again, it's not so much the exact statements that matter, 811 01:04:33,500 --> 01:04:36,830 but the spirit of these statements, which 812 01:04:36,830 --> 01:04:40,550 is that if you have a bunch of epsilon regular pairs, 813 01:04:40,550 --> 01:04:45,800 then you can embed and kind of pretending that everything 814 01:04:45,800 --> 01:04:47,290 behaved roughly like random. 815 01:04:50,890 --> 01:04:52,266 Any questions? 816 01:04:59,410 --> 01:05:02,740 So now we have Szemeredi's graph regularity lemma, 817 01:05:02,740 --> 01:05:06,730 we have the graph counting lemma, embedding lemmas, 818 01:05:06,730 --> 01:05:11,620 we can use it to derive some additional applications that 819 01:05:11,620 --> 01:05:13,540 don't just involve triangles. 820 01:05:13,540 --> 01:05:15,940 So when we only had a triangle counting lemma, 821 01:05:15,940 --> 01:05:18,700 we can only do the triangle removal lemma, 822 01:05:18,700 --> 01:05:22,330 but now we can do other removal lemmas. 823 01:05:22,330 --> 01:05:25,930 So in particular, there's the graph removal lemma, 824 01:05:25,930 --> 01:05:28,490 which generalizes the triangle removal lemma. 825 01:05:38,690 --> 01:05:40,870 So in the graph removal lemma, the statement 826 01:05:40,870 --> 01:05:46,040 is that for every H and epsilon, there 827 01:05:46,040 --> 01:05:55,040 exists a delta, such that every N vertex 828 01:05:55,040 --> 01:06:05,450 graph with fewer than delta N to the vertex 829 01:06:05,450 --> 01:06:10,980 of H number of copies of H-- 830 01:06:10,980 --> 01:06:15,190 so it has very few copies of H-- 831 01:06:15,190 --> 01:06:29,920 such graph can be made H-free by removing 832 01:06:29,920 --> 01:06:33,350 a fairly small number of edges. 833 01:06:33,350 --> 01:06:37,190 All right, so same statement as the triangle removal lemma, 834 01:06:37,190 --> 01:06:41,250 except now for general graph H. 835 01:06:41,250 --> 01:06:44,460 And as you expect, the proof is more or less the same 836 01:06:44,460 --> 01:06:48,140 as that of a triangle removal lemma, 837 01:06:48,140 --> 01:06:51,470 once we have the H counting lemma. 838 01:06:51,470 --> 01:06:53,880 So let me remind you how this goes. 839 01:06:53,880 --> 01:07:01,410 So it's really the same proof as triangle removal, 840 01:07:01,410 --> 01:07:05,130 where there was this recipe for applying the regularity 841 01:07:05,130 --> 01:07:07,020 lemma from last time. 842 01:07:07,020 --> 01:07:08,220 So what is it? 843 01:07:08,220 --> 01:07:14,980 You first-- so what's the first step when you do regularity? 844 01:07:14,980 --> 01:07:15,950 You partition. 845 01:07:15,950 --> 01:07:17,271 So let's do partition. 846 01:07:20,240 --> 01:07:20,740 OK. 847 01:07:20,740 --> 01:07:24,020 So apply the regularity lemma to do partitioning. 848 01:07:24,020 --> 01:07:26,780 And what's the second step? 849 01:07:26,780 --> 01:07:28,470 So we clean this graph. 850 01:07:28,470 --> 01:07:30,870 And you do the same cleaning procedure 851 01:07:30,870 --> 01:07:33,720 as in the triangle removal lemma, 852 01:07:33,720 --> 01:07:36,210 except maybe you have to adjust the parameters somewhat, so 853 01:07:36,210 --> 01:07:44,610 remove edges, remove low density pairs, 854 01:07:44,610 --> 01:07:50,200 irregular pairs, and small vertex sets. 855 01:07:54,810 --> 01:07:58,070 And the last step-- 856 01:07:58,070 --> 01:07:59,810 OK, so what do we do now? 857 01:07:59,810 --> 01:08:00,310 OK. 858 01:08:00,310 --> 01:08:03,020 So you can count. 859 01:08:03,020 --> 01:08:11,350 So if there were any H left, then the counting lemma 860 01:08:11,350 --> 01:08:14,030 shows you that you must have lots of copies of H left. 861 01:08:26,600 --> 01:08:28,705 So now let me show you how to use the strategy. 862 01:08:31,600 --> 01:08:35,104 Now that we have this general graph counting lemma, 863 01:08:35,104 --> 01:08:38,050 we'll prove the Erdos-Stone-Simonovits theorem, 864 01:08:38,050 --> 01:08:40,500 which we omitted, the proof that we 865 01:08:40,500 --> 01:08:42,250 omitted from the first part of the course. 866 01:08:46,020 --> 01:08:54,330 So remind you, the Erdos-Stone-Simonovits theorem 867 01:08:54,330 --> 01:09:01,020 says that if you have a graph H, then the extremal number of H 868 01:09:01,020 --> 01:09:06,910 is equal to this quantity which depends only 869 01:09:06,910 --> 01:09:16,569 on the chromatic number of H. 870 01:09:16,569 --> 01:09:19,840 The lower bound comes from taking the Turan graph. 871 01:09:25,390 --> 01:09:28,140 If you take the Turan graph, you get this lower bound, 872 01:09:28,140 --> 01:09:32,149 so it's really the upper bound that we need to think about. 873 01:09:34,930 --> 01:09:35,913 All right. 874 01:09:35,913 --> 01:09:37,080 So what's the strategy here? 875 01:09:40,760 --> 01:09:42,580 The statement really is that if you 876 01:09:42,580 --> 01:09:50,229 have a graph G that's N vertex whose number of edges 877 01:09:50,229 --> 01:09:58,230 is at least that much-- 878 01:09:58,230 --> 01:10:02,720 OK, so I fixed an epsilon bigger than zero, 879 01:10:02,720 --> 01:10:04,370 fixed a positive epsilon. 880 01:10:04,370 --> 01:10:06,760 So the claim, what we're trying to show with 881 01:10:06,760 --> 01:10:09,500 Erdos-Stone-Simonovits, is that if you have a graph G with too 882 01:10:09,500 --> 01:10:12,880 many edges-- too many meaning this many edges-- 883 01:10:12,880 --> 01:10:19,750 then G contains a copy of H if N is sufficiently large. 884 01:10:27,400 --> 01:10:27,900 OK. 885 01:10:27,900 --> 01:10:30,240 So let's use the regularity method, 886 01:10:30,240 --> 01:10:33,000 so applying this three-step recipe. 887 01:10:33,000 --> 01:10:34,702 First, we partition. 888 01:10:40,840 --> 01:10:50,370 So partition the vertex set of G into m pieces, 889 01:10:50,370 --> 01:10:54,870 and in such a way that it is eta regular-- 890 01:10:54,870 --> 01:10:57,840 and for some parameter eta that we'll decide later. 891 01:11:14,560 --> 01:11:16,360 The second step is cleaning. 892 01:11:23,510 --> 01:11:25,920 The cleaning step, again, it's the same kind 893 01:11:25,920 --> 01:11:27,820 of cleaning as we've done before. 894 01:11:27,820 --> 01:11:37,910 So let's remove an edge from Vi cross Vj, 895 01:11:37,910 --> 01:11:43,180 if any of the following hold-- 896 01:11:43,180 --> 01:11:52,630 if Vi Vj is not A to regular, if the density is 897 01:11:52,630 --> 01:12:05,000 too small, or either the two sets is too small. 898 01:12:09,880 --> 01:12:12,900 So same cleaning as before. 899 01:12:12,900 --> 01:12:20,500 And we can check that the number of edges removed 900 01:12:20,500 --> 01:12:23,980 is not too small. 901 01:12:23,980 --> 01:12:25,540 So in the first case, so again, it's 902 01:12:25,540 --> 01:12:27,360 the same calculation as last time. 903 01:12:27,360 --> 01:12:29,440 In the first case, the number of edges removed 904 01:12:29,440 --> 01:12:32,250 is, at most, eta N squared. 905 01:12:32,250 --> 01:12:41,620 And we'll choose eta to be less than epsilon over 8, 906 01:12:41,620 --> 01:12:43,753 although it will be actually significantly smaller, 907 01:12:43,753 --> 01:12:44,920 as you will see in a second. 908 01:12:47,490 --> 01:12:51,820 In the second step, same as what happened in the triangle 909 01:12:51,820 --> 01:12:56,830 removal stage, the number of edges 910 01:12:56,830 --> 01:13:01,700 removed in the second type is, at most, that amount, 911 01:13:01,700 --> 01:13:04,010 still very small number. 912 01:13:04,010 --> 01:13:11,440 And the third one here is also a very small number. 913 01:13:11,440 --> 01:13:15,620 So the third type, they start with one of these sets, 914 01:13:15,620 --> 01:13:18,580 the m possible-- 915 01:13:18,580 --> 01:13:20,980 so it's a very small number. 916 01:13:20,980 --> 01:13:28,675 And so the total is, at most, an epsilon over 2 N squared. 917 01:13:31,960 --> 01:13:33,465 So maybe I want epsilon over-- 918 01:13:36,140 --> 01:13:42,650 so let's say epsilon over 2 N squared number of edges. 919 01:13:42,650 --> 01:13:46,690 And I would like that to be strictly bigger than. 920 01:13:50,500 --> 01:14:04,640 So now after removing these edges from G, 921 01:14:04,640 --> 01:14:11,340 we have this G prime, which has strictly more than 1 minus 1 922 01:14:11,340 --> 01:14:13,330 over r. 923 01:14:13,330 --> 01:14:18,950 Yeah, 1 minus 1 over r times N squared over 2 edges. 924 01:14:23,785 --> 01:14:24,660 So now what do we do? 925 01:14:28,170 --> 01:14:31,980 So we knew from Turan's theorem that if your graph has strictly 926 01:14:31,980 --> 01:14:38,270 more than this number of edges, you must have a K sub r plus 1. 927 01:14:38,270 --> 01:14:41,700 So even after deleting all these edges, 928 01:14:41,700 --> 01:14:45,000 G still has lots of edges left, in particular, 929 01:14:45,000 --> 01:14:51,560 Turan's theorem implies that G prime contains 930 01:14:51,560 --> 01:14:56,270 a clique on r plus 1 vertices. 931 01:14:56,270 --> 01:15:04,063 So here I should say that r is chromatic number of H minus 1. 932 01:15:08,010 --> 01:15:11,993 So I find one copy of this clique, 933 01:15:11,993 --> 01:15:13,410 but what does that copy look like? 934 01:15:17,080 --> 01:15:19,120 I find this one copy of a clique. 935 01:15:23,310 --> 01:15:26,650 Let's say r equals to 4. 936 01:15:26,650 --> 01:15:30,400 And the point, now, is that the counting lemma will allow 937 01:15:30,400 --> 01:15:44,450 me to amplify that clique into H. 938 01:15:44,450 --> 01:15:50,190 So it will allow me to amplify this clique into a copy of H. 939 01:15:50,190 --> 01:16:03,230 So, for example, if H were this graph over here, 940 01:16:03,230 --> 01:16:07,590 so then you would find a copy of H in G, which is what we want. 941 01:16:07,590 --> 01:16:09,090 So why does the counting lemma allow 942 01:16:09,090 --> 01:16:12,000 you to do this amplification? 943 01:16:12,000 --> 01:16:15,500 So it's this point, the ideas are all there, 944 01:16:15,500 --> 01:16:18,360 but there's a slight wrinkle in the calculations. 945 01:16:18,360 --> 01:16:20,220 I mean, there's, in the executions, 946 01:16:20,220 --> 01:16:23,070 I just want to point out, just in case 947 01:16:23,070 --> 01:16:26,610 some of the vertices of H end up in the same vertex in G. 948 01:16:26,610 --> 01:16:28,780 But that turns out not to be an issue. 949 01:16:28,780 --> 01:16:38,020 So by counting lemma, the number of homomorphisms 950 01:16:38,020 --> 01:16:43,480 from H to G prime, where I'm really only considering 951 01:16:43,480 --> 01:16:47,200 homomorphisms that map each vertex of H 952 01:16:47,200 --> 01:16:51,380 to its assigned part. 953 01:16:51,380 --> 01:16:57,920 It's at least this quantity where 954 01:16:57,920 --> 01:17:04,130 I'm looking at the predicted density of such homomorphisms, 955 01:17:04,130 --> 01:17:08,480 and all of these edge densities are at least epsilon over 8. 956 01:17:08,480 --> 01:17:13,820 So it's at least that amount minus a small error that 957 01:17:13,820 --> 01:17:15,110 comes from the counting lemma. 958 01:17:19,720 --> 01:17:23,720 And, well, all of the vertex parts 959 01:17:23,720 --> 01:17:27,875 are quite large, so all of the vertex parts 960 01:17:27,875 --> 01:17:35,000 are of size like that. 961 01:17:38,055 --> 01:17:40,180 So that's the result of the counting lemma combined 962 01:17:40,180 --> 01:17:43,030 with information about the densities of the parts 963 01:17:43,030 --> 01:17:48,210 and the sizes of the parts that came out of cleaning. 964 01:17:48,210 --> 01:18:00,240 So setting eta to be an appropriate value, 965 01:18:00,240 --> 01:18:08,660 we see that for sufficiently large N, 966 01:18:08,660 --> 01:18:14,400 this quantity here, is on the order of N 967 01:18:14,400 --> 01:18:19,590 to the number of vertices of H. But I'm only 968 01:18:19,590 --> 01:18:21,780 counting homomorphisms, and so it 969 01:18:21,780 --> 01:18:24,720 could be that some of the vertices of H 970 01:18:24,720 --> 01:18:29,130 end up in the same vertex of G. And those would not 971 01:18:29,130 --> 01:18:31,110 be genuine subgraphs, so I shouldn't 972 01:18:31,110 --> 01:18:32,820 consider those as subgraphs. 973 01:18:32,820 --> 01:18:35,110 Because otherwise, if you were to allow those, 974 01:18:35,110 --> 01:18:38,550 then if you found this K4, then you 975 01:18:38,550 --> 01:18:41,040 found all four chromatic graphs. 976 01:18:41,040 --> 01:18:44,400 So you shouldn't consider copies that are degenerate, 977 01:18:44,400 --> 01:18:50,080 but that's OK because the number of maps from the vertex set 978 01:18:50,080 --> 01:18:55,290 of H to the vertex set of G that are non-injective 979 01:18:55,290 --> 01:19:00,420 is of a lower order. 980 01:19:00,420 --> 01:19:02,833 The number of non-injective maps from the vertex set, 981 01:19:02,833 --> 01:19:04,500 well, you have to pick two vertices of H 982 01:19:04,500 --> 01:19:07,050 to map to the same vertex, and then the number of choices, 983 01:19:07,050 --> 01:19:10,750 you have one order less. 984 01:19:10,750 --> 01:19:14,307 So there are negligible fraction of these homomorphisms. 985 01:19:16,990 --> 01:19:19,310 And the conclusion, then, is that G prime 986 01:19:19,310 --> 01:19:27,050 contains a copy of H, which is what we're looking for. 987 01:19:27,050 --> 01:19:29,850 If G prime contains copy of H, then G contains a copy of H, 988 01:19:29,850 --> 01:19:32,460 and that proves the Erdos-Stone-Simonovits theorem. 989 01:19:35,340 --> 01:19:37,950 You get a bit more out of this proof. 990 01:19:37,950 --> 01:19:44,780 So you see that not only does G contain one copy of H, 991 01:19:44,780 --> 01:19:46,700 but the counting lemma actually shows you 992 01:19:46,700 --> 01:19:50,120 it contains many copies of H. 993 01:19:50,120 --> 01:19:53,360 And this is a phenomenon known as supersaturation, which 994 01:19:53,360 --> 01:19:55,340 you already saw in the first problem set, 995 01:19:55,340 --> 01:19:59,360 that often when you are beyond a certain threshold, 996 01:19:59,360 --> 01:20:04,160 an extremal threshold, you don't just gain one extra copy, 997 01:20:04,160 --> 01:20:07,570 but you often gain many copies. 998 01:20:07,570 --> 01:20:11,790 And you see this in this proof here. 999 01:20:11,790 --> 01:20:15,990 So to summarize, we've seen this proof 1000 01:20:15,990 --> 01:20:19,230 of Erdos-Stone-Simonovits, which comes 1001 01:20:19,230 --> 01:20:22,470 from applying regularity and then finding 1002 01:20:22,470 --> 01:20:25,980 a single copy of a clique from Turan's theorem, 1003 01:20:25,980 --> 01:20:27,840 and then using counting lemma to boost 1004 01:20:27,840 --> 01:20:35,030 that copy from Turan's theorem into an actual copy of H. 1005 01:20:35,030 --> 01:20:38,220 So in the second homework, one of the problems 1006 01:20:38,220 --> 01:20:41,790 is to come up with a different proof of Erdos-Stone-Simonovits 1007 01:20:41,790 --> 01:20:46,920 that is more similar to the proof of Kovari-Sos-Turan, 1008 01:20:46,920 --> 01:20:49,400 more through double-counting like arguments. 1009 01:20:49,400 --> 01:20:51,150 And that is closer in spirit, although not 1010 01:20:51,150 --> 01:20:56,610 exactly the same as the original proof in Erdos-Stone. 1011 01:20:56,610 --> 01:21:00,290 So this regularity proof, I think it's more conceptual. 1012 01:21:00,290 --> 01:21:02,540 You get to see how to do this boosting, 1013 01:21:02,540 --> 01:21:04,250 but it gets a terrible bound. 1014 01:21:04,250 --> 01:21:06,350 And the other proof that you see in the homework 1015 01:21:06,350 --> 01:21:08,270 gives you a much more reasonable bound 1016 01:21:08,270 --> 01:21:13,070 on the dependence between how N grows 1017 01:21:13,070 --> 01:21:17,710 versus how quickly this little o has to go to zero.